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\title[Hopf algebras in combinatorics]{Hopf algebras in combinatorics}

%\dedicatory{To ...}

\author{Victor Reiner}
\email{reiner@math.umn.edu}
\address{School of Mathematics\\
University of  Minnesota\\
Minneapolis, MN 55455\\
USA}

% \thanks{NSF ??

% \subjclass{05A15, 05A10}

\date{\today}
\keywords{Hopf algebra, combinatorics}


\begin{document}


\begin{abstract}
Certain Hopf algebras arise in combinatorics because they have bases 
naturally parametrized by combinatorial objects (partitions, compositions, 
permutations, tableaux, graphs, trees, posets, polytopes, etc). 
The rigidity in the structure of 
a Hopf algebra can lead to enlightening proofs, and many 
interesting invariants of combinatorial objects turn out to be evaluations 
of Hopf morphisms.

These are lecture notes for Fall 2012 Math 8680 Topics in Combinatorics
at the University of Minnesota.  The course is an attempt to focus on 
examples that I find interesting, but which are hard to find 
fully explained currently in books or in one paper.  
Be warned that these notes are highly idiosyncratic in choice of topics, 
and they steal heavily from the sources in the bibliography.
\end{abstract}

\maketitle

\tableofcontents


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{What is a Hopf algebra?}
\label{Hopf-intro-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The standard references are Abe \cite{Abe} and Sweedler \cite{Sweedler}, 
and some other good ones are \cite{ChariPressley, DascalescuNastasescuRaianu, Kassel, Montgomery}.  
A reference which we discovered late, having a great deal of overlap with these notes
is Hazewinkel, Gubareni, and Kirichenko \cite{HazewinkelGubareniKirichenko}.

Let's build up the definition of Hopf algebra 
structure bit-by-bit, starting with the more familiar definition of algebras.


\vskip.2in
\noindent
{\bf Warnings:} Unless otherwise specified ...
\begin{enumerate}
\item[$\bullet$]
$\kk$ here usually denotes a field, but sometimes 
we'll want to take $\kk=\ZZ$,  
\item[$\bullet$]
all maps are $\kk$-linear,
\item[$\bullet$]
all tensor products are over $\kk$, and
\item[$\bullet$]
$\one$ will denote the multiplicative identity in some ring like $\kk$ or in some $\kk$-algebra, but also denote as the identity map on various spaces.
\end{enumerate}

Hopefully context will resolve some of the ambiguities.

\subsection{Algebras}

\begin{definition}
An {\it associative $k$-algebra} $A$ is a $\kk$-vector space with
an associative operation $A \otimes A \overset{m}{\rightarrow} A$,
and a {\it unit} $\kk \overset{u}{\rightarrow} A$ 
sending $\one$ in $\kk$ to the two-sided multiplicative identity 
element $\one$ in $A$.  One can rephrase this by saying that these 
diagrams commute:

\begin{equation}
\label{associativity-diagram}
\xymatrix{
& A  \otimes A \otimes A \ar[dl]_{m \otimes \one} \ar[dr]^{\one \otimes m} &  \\
A \otimes A \ar[dr]^m & &  A \otimes A \ar[dl]_m \\
& A &
}
\end{equation}


\begin{equation}
\label{unit-diagram}
\xymatrix{
A \otimes \kk \ar[d]_{\one \otimes u} & A \ar[l] \ar[d]_{\one} \ar[r] & \kk \otimes A \ar[d]^{u \otimes \one} \\
A \otimes A \ar[r]^m & A & A \otimes A \ar[l]_m
}
\end{equation}
where the maps $A \rightarrow A \otimes \kk$ and
$A \rightarrow \kk \otimes A$ are the isomorphisms
sending $a \mapsto  a \otimes \one$ and
$a \mapsto \one \otimes a$.
\end{definition}

Well-known examples of $\kk$-algebras are
{\it tensor} and {\it symmetric algebras}, which we can think of 
as algebras of {\it words} and {\it multisets}, respectively.  

\begin{example}
The {\it tensor algebra} $T(V)=\oplus_{n \geq 0} V^{\otimes n}$
on a $\kk$-vector space $V$, say with $\kk$-basis $\{x_i\}_{i \in I}$, is an
associative $\kk$-algebra with a $\kk$-basis of decomposable tensors
$x_{i_1} \cdots x_{i_k}:=x_{i_1} \otimes \cdots \otimes x_{i_k}$ indexed by {\it words} $(i_1,\ldots,i_k)$ in the alphabet $I$,
and multiplication defined $\kk$-linearly by concatenation of words:
$$
m( x_{i_1} \cdots x_{i_k} \otimes  x_{j_1} \cdots x_{j_\ell}):=
x_{i_1} \cdots x_{i_k} x_{j_1} \cdots x_{j_\ell}.
$$
\end{example}

Recall that in an algebra $A$, when one 
has a {\it two-sided ideal} $J \subset A$, meaning a
$\kk$-linear subspace with $m( J \otimes A), m(A \otimes J) \subset J$,
then one can form a {\it quotient algebra} $A/J$.  
%We further assume that $1$ is not in $J$ 
%so that $A/J$ does not collapse to the trivial ring in which $1 = 0$.

\begin{example}
The {\it symmetric algebra} $\Sym(V)=\oplus_{n \geq 0} \Sym^n(V)$ is the
quotient of $T(V)$ by the two-sided ideal generated by
all elements $x y - y x$ with $x,y$ in $V$.  It can be identified
with a (commutative) polynomial algebra $\kk[x_i]_{i \in I}$,
having a $\kk$-basis of (commutative) monomials 
$x_{i_1} \cdots x_{i_k}$ as
$(i_1,\ldots,i_k)$ runs through all finite multisubsets of $I$, and
with multiplication defined $\kk$-linearly via
multiset union.
\end{example}

Topology and group theory give more examples.

\begin{example}
The {\it cohomology algebra} $H^*(X;\kk)=\oplus_{i \geq 0}H^i(X;\kk)$ 
with coefficients in $\kk$ for 
a topological space $X$ has an associative {\it cup product}.  Its unit
$\kk =H^*(\pt;\kk) \overset{u}{\rightarrow} H^*(X;\kk)$ is induced
from the unique (continuous) map $X \rightarrow \pt$, where $\pt$ is a one-point space.
\end{example}

\begin{example}
For a group $G$, the {\it group algebra} $\kk G$ has $\kk$-basis $\{t_g\}_{g \in G}$ and multiplication defined $\kk$-linearly by $t_g t_h = t_{gh}$, and
unit defined by $u(1) = t_e$, where $e$ is the identity element of $G$.
\end{example}



\subsection{Coalgebras}

If we are to think of the multiplication $A \otimes A  \rightarrow A$
in an algebra as {\it putting together} two basis elements of $A$ to get a
sum of basis elements of $A$, then coalgebra structure should be thought of 
as {\it taking basis elements apart}.

\begin{definition}
A {\it co-associative coalgebra} $C$ is 
a $\kk$-vector space $C$ with
a {\it comultiplication}, that is, a $\kk$-linear map
$C \overset{\Delta}{\rightarrow} C \otimes C$,
and a {\it counit} $C \overset{\epsilon}{\rightarrow} \kk$ 
making commutative the diagrams as in \eqref{associativity-diagram},
\eqref{unit-diagram} but with {\it all arrows reversed}:

\begin{equation}
\label{coassociativity-diagram}
\xymatrix{
& C  \otimes C \otimes C &  \\
C \otimes C \ar[ur]^{\Delta \otimes \one} & &  C \otimes C \ar[ul]_{\one \otimes \Delta} \\
& C \ar[ur]^\Delta \ar[ul]_\Delta&
}
\end{equation}

\begin{equation}
\label{counit-diagram}
\xymatrix{
C \otimes \kk \ar[r] & C & \kk \otimes C \ar[l] \\
C \otimes C \ar[u]^{\one \otimes \epsilon} & C \ar[l]^{\Delta} \ar[u]^\one \ar[r]_{\Delta} & C \otimes C \ar[u]^{\epsilon \otimes \one}
}
\end{equation}
Here the maps $C \otimes \kk \rightarrow C$ and
$\kk \otimes C \rightarrow C$ are the isomorphisms
sending $c \otimes \one \mapsto c$ and
$\one \otimes c \mapsto c$.
\end{definition}

One often uses the {\it Sweedler notation}
$$
\Delta(c) = \sum_{(c)} c_1 \otimes c_2 = \sum c_1 \otimes c_2
$$
to abbreviate formulas involving $\Delta$.  For example, commutativity of the
left square in \eqref{counit-diagram} asserts that
$\sum_{(c)} c_1 \epsilon(c_2) = c$.

\begin{example}
The {\it homology} $H_*(X;\kk)=\oplus_{i \geq 0} H_*(X;\kk)$ 
for a topological space $X$ is naturally a coalgebra: the
(continuous) {\it diagonal embedding}
$X \rightarrow X \times X$ sending $x \mapsto (x,x)$ induces
a coassociative map 
$$
H_*(X; \kk) \rightarrow H_*(X \times X;\kk) \cong H_*(X;\kk) \otimes H_*(X;\kk)
$$
in which the last isomorphism comes from the {\it K\"unneth theorem}
with field coefficients $\kk$.
As before, the unique (continuous) map $X \rightarrow \pt$ induces the
counit $H_*(X;\kk) \overset{\epsilon}{\rightarrow} H_*(\pt;\kk) \cong \kk$.
\end{example}



\subsection{Morphisms, tensor products, and bialgebras}

\begin{definition}
A {\it morphism} of algebras $A \overset{\varphi}{\rightarrow} B$
makes these diagrams commute:

\begin{equation}
\qquad
\xymatrix{
A  \ar[r]^{\varphi} & B  \\
A \otimes A \ar[u]^{m_A} \ar[r]^{\varphi \otimes \varphi} & B \otimes B \ar[u]_{m_B}
}
\qquad
\xymatrix{
A \ar[rr]^{\varphi}& & B  \\
&\kk \ar[ul]^{u_A} \ar[ur]_{u_B}&
}
\end{equation}
Here the subscripts on $m_A,m_B,u_A,u_B$ indicate for which algebra they are 
part of the structure-- we will occasionally use such conventions from now on.

Similarly a {\it morphism of coalgebras}
is a $\kk$-linear map $C \overset{\varphi}{\rightarrow} D$
making the reverse diagrams commute:

\begin{equation}
\label{coalgebra-morphism-diagrams}
\xymatrix{
C \ar[d]_{\Delta_C} \ar[r]^{\varphi} & D \ar[d]^{\Delta_D} \\
C \otimes C \ar[r]^{\varphi \otimes \varphi} & D \otimes D
}
\qquad
\xymatrix{
C \ar[dr]_{\epsilon_C} \ar[rr]^{\varphi}& & D \ar[dl]^{\epsilon_D}  \\
&\kk&
}
\end{equation}
\end{definition}

\begin{example}
Continuous maps $X \overset{f}{\rightarrow} Y$ of topological spaces induce algebra morphisms $H^*(Y;\kk) \rightarrow H^*(X;\kk)$, and coalgebra morphisms
$H_*(X;\kk) \rightarrow H_*(Y;\kk)$.
\end{example}

\begin{definition}
Given two $\kk$-algebras $A, B$, their tensor product $A \otimes B$ also
becomes a $\kk$-algebra defining the multiplication bilinearly via
$$
m((a \otimes b) \otimes (a' \otimes b') := aa' \otimes bb'
$$
or in other words $m_{A \otimes B}$ is the composite map
$$
A \otimes B \otimes A \otimes B 
\overset{\one \otimes T \otimes \one}{\longrightarrow}
A \otimes A \otimes B \otimes B \overset{m_A \otimes m_B}{\longrightarrow}
A \otimes B
$$
where $T$ is the {\it twist map} $B \otimes A \rightarrow A \otimes B$
that sends $b \otimes a \mapsto a \otimes b$.

Here we are omitting the topologist's sign 
in the twist map which should be present for graded algebras 
and coalgebras that come from cohomology and homology: 
for homogeneous elements $a$ and $b$ the topologist's twist map sends
\begin{equation}
\label{topologist-twist}
b \otimes a \longmapsto  (-1)^{\deg(a)\deg(b)} a \otimes b.
\end{equation}
This means that most of our examples which we later call {\it graded} 
should actually be considered to live in only {\it even} degrees, e.g. 
by artificially doubling their grading.  We will ignore this issue, and hope that it causes no confusion later!

The unit element of $A \otimes B$ is $\one_A \otimes \one_B$, 
meaning that the
unit map $\kk \overset{u_{A \otimes B}}{\rightarrow} A \otimes B$ is the composite
$$
\kk \longrightarrow \kk \otimes \kk \overset{u_A \otimes u_B}{\rightarrow} A \otimes B.
$$

Similarly, given two coalgebras $C, D$, one can make $C \otimes D$ a coalgebra in which the comultiplication and counit maps are the composites of
$$
C \otimes D \overset{\Delta_C \otimes \Delta_D}{\longrightarrow}
C \otimes C \otimes D \otimes D
\overset{\one \otimes T \otimes \one}{\longrightarrow}
C \otimes D \otimes C \otimes D
$$
and
$$
C \otimes D \overset{\epsilon_C \otimes \epsilon_D}{\longrightarrow}
\kk \otimes \kk \longrightarrow \kk.
$$
\end{definition}

\noindent
One of the first signs that these definitions interact nicely is the following straightforward proposition.

\begin{proposition}
\label{bialgebra-prop}
When $A$ is both a $\kk$-algebra and a
$\kk$-coalgebra, the following are equivalent:
\begin{itemize}
\item $(\Delta,\epsilon)$ are morphisms for the algebra structure $(m,u)$.
\item $(m,u)$ are morphisms for the coalgebra structure $(\Delta,\epsilon)$.
\item These four diagrams commute:
\begin{equation}
\label{bialgebra-diagrams}
\begin{array}{ccc}
& \xymatrix{
 & A \otimes A \ar[dl]_{\Delta \otimes \Delta} \ar[ddr]_m & \\
A \otimes A \otimes A \otimes A \ar[dd]_{\one \otimes T \otimes \one} &  & \\
 & & A \ar[ddl]_\Delta\\
A \otimes A \otimes A \otimes A \ar[dr]_{m \otimes m} &  & \\
 & A \otimes A & 
} & \\
\xymatrix{
A \otimes A \ar[r]^{\epsilon \otimes \epsilon} \ar[d]_{m} &\kk \otimes \kk \ar[d]^m \\
A \ar[r]_{\epsilon}  &\kk
} & &  
\xymatrix{
\kk \ar[r]^{u} \ar[d]_{\Delta} &A \ar[d]^\Delta \\
\kk \otimes \kk \ar[r]_{u \otimes u}  &A \otimes A
}\\
& \xymatrix{
\kk \ar[dr]_{u} \ar[rr]^{\one} & & \kk \\
 &A \ar[ur]_{\epsilon} &
} &
\end{array}
\end{equation}
\end{itemize}
\end{proposition}

\begin{definition}
Call the $\kk$-vector space $A$ a {\it $\kk$-bialgebra} if it is
a $\kk$-algebra and $\kk$-coalgebra satisfying the three
equivalent conditions in Proposition~\ref{bialgebra-prop}.
\end{definition}

\begin{example}
For a group $G$, one can make the group algebra $\kk G$ a coalgebra with
counit $\kk G \overset{\epsilon}{\rightarrow} \kk$ mapping
$t_g \mapsto 1$ for all $g$ in $G$, and with comultiplication
$\kk G \overset{\Delta}{\rightarrow} \kk G \otimes \kk G$ given by
$\Delta(t_g) := t_g \otimes t_g$.  Checking the various diagrams in
\eqref{bialgebra-diagrams} commute is easy.  For example, one
can check the pentagonal diagram on each basis element $t_g \otimes t_h$:
$$
\xymatrix{
 & t_g \otimes t_h \ar[dl]_{\Delta \otimes \Delta} \ar[ddr]_m & \\
t_g \otimes t_g \otimes t_h \otimes t_h \ar[dd]_{\one \otimes T \otimes \one} &  & \\
 & & t_{gh} \ar[ddl]_\Delta\\
t_g \otimes t_h \otimes t_g \otimes t_h \ar[dr]_{m \otimes m} &  & \\
 & t_{gh} \otimes t_{gh} & 
}
$$
\end{example}

\begin{remark}
\label{comultiplication-as-tensor-action}
In fact, one can think of adding a bialgebra structure to a $\kk$-algebra $A$
as a way of making $A$-modules $M,N$ have an $A$-module structure on their
tensor product $M \otimes N$:  the algebra $A \otimes A$ already acts naturally
on $M \otimes N$, so one can let $a$ in $A$ act via $\Delta(a)$ in $A \otimes A$.  In the theory of group representations over $\kk$, that is, $\kk G$-modules $M$, this is how one defines the {\it diagonal action} of $G$ on $M \otimes N$, namely $t_g$ acts as $t_g \otimes t_g$.  
\end{remark}

\begin{definition}
An element $x$ in a bialgebra for which $\Delta(x) = x \otimes x$ and
$\epsilon(x)=1$ is called {\it group-like}.

An element $x$ in a bialgebra for which $\Delta(x) = 1 \otimes x+x\otimes 1$ 
is called {\it primitive}.
\end{definition}

\begin{example}
The {\it tensor algebra} $T(V)=\oplus_{n \geq 0} V^{\otimes n}$ is
a coalgebra, with counit $\epsilon$ equal to 
the identity on $V^{\otimes 0} = \kk$ and the zero map on
$V^{\otimes n}$ for $n > 0$, and with comultiplication defined
to make the elements $x$ in $V^{\otimes 1}=V$ all primitive:
$$
\Delta(x) := \one \otimes x + x \otimes \one \text{ for }x \in V^{\otimes 1}.
$$  
Since the elements of $V$ 
generate $T(V)$ as a $\kk$-algebra, and since $T(V) \otimes T(V)$
is also an associative $\kk$-algebra, the universal property of $T(V)$ as
the free associative $\kk$-algebra on the generators $V$ 
allows one to define $T(V) \overset{\Delta}{\rightarrow} T(V) \otimes T(V)$ arbitrarily on $V$, and extend it as
an algebra morphism.  

It may not be obvious that this $\Delta$ is coassociative,
but one can note that 
$$
\left(
(\one \otimes \Delta) \circ \Delta
\right)(x) =
x \otimes \one \otimes \one +
\one \otimes x \otimes \one +
\one \otimes \one \otimes x
=\left( (\Delta \otimes \one) \circ \Delta \right)(x)
$$
for every $x$ in $V$.  Hence the two 
maps $(\one \otimes \Delta) \circ \Delta$ and
$(\Delta \otimes \one) \circ \Delta$, considered
as algebra morphisms 
$T(V) \rightarrow T(V) \otimes T(V) \otimes T(V)$,
must coincide on every element of $T(V)$ since they coincide on $V$.
We leave it as an exercise to check the map $\epsilon$ defined as
above satisfies the counit axioms \eqref{counit-diagram}.

Here is a sample calculation in $T(V)$ when $V$ has basis $\{x,y,z\}$:
$$
\begin{aligned}
\Delta(xyz) &= \Delta(x) \Delta(y) \Delta(z) \\
&=(1 \otimes x+x\otimes 1)(1 \otimes y+y\otimes 1)(1 \otimes z+z\otimes 1) \\
&=(1 \otimes xy + x \otimes y+ y \otimes x + xy \otimes 1)(1 \otimes z+z\otimes 1) \\
&=1 \otimes xyz 
+ x \otimes yz + y \otimes xz + z \otimes xy\\
&\qquad + xy \otimes z + xz \otimes y + yz \otimes x
+xyz \otimes 1
\end{aligned}
$$
This illustrates the idea that comultiplication ``takes basis elements apart''.
Here for any $v_{1},v_{2},...,v_{n}$ in $V$ one has
$$
\Delta\left(  v_{1}v_{2}...v_{n}\right) =
\sum
v_{j_{1}}\cdots v_{j_{r}}\otimes v_{k_{1}}\cdots
v_{k_{n-r}}
$$
where the sum is over ordered pairs 
$\left(  j_{1},j_{2},...,j_{r}\right)  ,\left(  k_{1},k_{2},...,k_{n-r}\right)$
of complementary subwords of the word $\left(  1,2,...,n\right)$.
\end{example}

Recall one can quotient a $\kk$-algebra $A$ by a two-sided ideal $J$ to
obtain a quotient algebra $A/J$.

\begin{definition}
In a coalgebra $C$, a {\it two-sided coideal} is a $\kk$-linear subspace 
$J \subset C$ for which 
$$
\begin{aligned}
\Delta(J) &\subset J \otimes C + C \otimes J \\
\epsilon(J) &= 0
\end{aligned}
$$
The quotient space $C/J$ then inherits a coalgebra structure.
Similarly, in a bialgebra $A$, a subset $J \subset A$ which is both
a two-sided ideal and two-sided coideal gives rise to a quotient
bialgebra $A/J$.
\end{definition}

\begin{example}
\label{symmetric-algebra-as-bialgebra-example}
The {\it symmetric algebra} $\Sym(V)$ was the 
quotient of $T(V)$ by the two-sided ideal $J$ generated by
all {\it commutators} $[x,y]=x y - y x$ for $x,y$ in $V$. 
Note that $x,y$ are primitive elements in $T(V)$,
and the following very reusable calculation 
shows that {\it the commutator of two primitives is primitive}:
\begin{equation}
\label{commutator-of-primitives}
\begin{aligned}
\Delta [x,y] &= \Delta( xy-yx ) \\
&=(\one \otimes x+x\otimes \one)(\one \otimes y+y\otimes \one)
-(\one \otimes y+y\otimes \one)(\one \otimes x+x\otimes \one)\\
&=\one \otimes xy - \one \otimes yx +
  xy \otimes \one - yx \otimes \one \\
&\qquad +
  x \otimes y + y \otimes x
 - x \otimes y - y \otimes x \\
&=\one \otimes (xy-yx) + (xy-yx) \otimes \one\\
&=\one \otimes [x,y] + [x,y] \otimes \one.
\end{aligned}
\end{equation}
In particular, the commutators $[x,y]$ have $\Delta[x,y]$ in
$J \otimes T(V) + T(V) \otimes J$.  They also satisfy
$\epsilon([x,y])=0$.  Since they are generators for
$J$ as a two-sided ideal, it is not hard to see this implies
$\Delta (J) \subset J \otimes T(V) + T(V) \otimes J$,
and $\epsilon(J) = 0$.  Thus $J$ is
also a two-sided coideal, and 
$\Sym(V)=T(V)/J$ inherits a bialgebra structure.
\end{example}

In fact we will see in Section~\ref{PSH-implies-polynomial-section} 
that symmetric algebras are the universal example of bialgebras which are {\it graded, connected, commutative, cocommutative}.  But first we should define some of these concepts.


\begin{definition}
A {\it graded $\kk$-vector space} $V$ is one with a $\kk$-vector space
direct sum decomposition $V=\oplus_{n \geq 0} V_n$.  Elements $x$ in $V_n$
are called {\it homogeneous} of degree $n$, or $\deg(x)=n$.  
When we are working with $\kk=\ZZ$, we will always assume
that {\it graded} $\ZZ$-modules $V=\oplus_{n \geq 0} V_n$ have
each $V_n$ a {\it free} $\ZZ$-module.

One endows tensor products $V \otimes W$ of graded vector spaces $V,W$
with graded vector space structure in which $(V \otimes W)_n:=\oplus_{i+j=n} V_i \otimes W_j$. 

A $\kk$-linear map $V \overset{\varphi}{\rightarrow} W$ between two graded $\kk$-vector spaces is called
{\it graded} if $\varphi(V_n) \subset W_n$ for all $n$.
Say that a $\kk$-algebra (coalgebra, bialgebra) is {\it graded} if
it is a graded $\kk$-vector space and 
all of the relevant structure maps ($u,\epsilon,m,\Delta)$ are graded.

Say that a graded vector space $V$ is {\it connected} if $V_0 \cong \kk$.
\end{definition}

\begin{example}
A path-connected space $X$ has its homology and cohomology
$$
\begin{aligned}
H_*(X;\kk)&=\bigoplus_{i \geq 0} H_i(X;\kk)\\
H^*(X;\kk)&=\bigoplus_{i \geq 0} H^i(X;\kk) 
\end{aligned}
$$
carrying the structure of graded connected coalgebras and algebras, respectively.  If in addition, $X$ is a topological group,
or even less strongly, an {\it $H$-space} (e.g. the {\it loop space} $\Omega Y$ on some other space $Y$), 
the continuous multiplication map
$
X \times X \rightarrow X
$
induces an algebra structure on $H_*(X;\kk)$ and a coalgebra structure on
$H^*(X;\kk)$, so that each become bialgebras
(and these bialgebras are dual to each other in a sense soon to be discussed).
This was Hopf's motivation:  the (co-)homology of a compact Lie group carries bialgebra structure that explains 
why it takes a certain form;  see Cartier \cite[\S 2]{Cartier}.
\end{example}

\begin{example}
Tensor algebras $T(V)$ and symmetric algebras $\Sym(V)$ are graded, once one picks a graded vector space structure for $V$;  then 
$$
\deg(x_{i_1} \cdots x_{i_k}) =\deg(x_{i_1}) + \cdots + \deg(x_{i_k}).
$$ 
Assuming that $V_0=0$, the algebras $T(V)$ and $\Sym(V)$ are connected.
For example, we will often say that all elements of $V$ are homogeneous of degree $1$, but at other times, it will make sense to have $V$ live in different (positive) degrees.
\end{example}

\begin{exercise}
\label{graded-connected-exercise}
Check that for a graded connected $\kk$-bialgebra $A$, the 
gradedness of the unit $u$ and counit $\epsilon$ maps, along with
commutativity of diagrams \eqref{unit-diagram}, \eqref{counit-diagram},
and \eqref{bialgebra-diagrams} imply
\begin{itemize}
\item[(a)] $\kk$ lies in $A_0$,
\item[(b)] $u$ is an isomorphism $\kk \overset{u}{\rightarrow} A_0$, while
\item[(c)] the two-sided ideal 
$
\ker \epsilon 
$
is the space of positive degree elements 
$I = \bigoplus_{n > 0} A_n$.
\item[(d)] $\epsilon$ restricted to $A_0$ is the inverse isomorphism
$A_0 \overset{\epsilon}{\rightarrow} \kk$ to $u$, and
\item[(e)] every $x$ in $I$ has comultiplication of the form
$$
\Delta(x) = 1 \otimes x + x \otimes 1 + \Delta_+(x)
$$
where $\Delta_+(x)$ lies in $I \otimes I$.
\end{itemize}
\end{exercise}


\subsection{Antipodes and Hopf algebras}

There is one more piece of structure needed to make a bialgebra
a Hopf algebra, although it will come for free in the graded connected case.

\begin{definition}
For any coalgebra $C$ and algebra $A$, one can endow the $\kk$-linear
maps $\Hom(C,A)$ with an associative algebra structure called
the {\it convolution algebra}:  send $f,g$ in $\Hom(C,A)$ to
$f \star g$ defined by
$(f \star g)(c) = \sum f(c_1) g(c_2)$, using
the {\it Sweedler notation} $\Delta(c) = \sum c_1 \otimes c_2$.
Equivalently, $f \star g$ is the composite
$$
C \overset{\Delta}{\longrightarrow} C \otimes C
\overset{f \otimes g}{\longrightarrow} A \otimes A
\overset{m}{\longrightarrow} A.
$$
One sees that $u \circ \epsilon$ is a two-sided
identity element for $\star$, meaning that
$$
\sum f(c_1) \epsilon(c_2) = f(c) = \sum \epsilon(c_1) f(c_2)
$$
by adding a top row to \eqref{counit-diagram}:


\begin{equation}
\label{added-to-counit-diagram}
\xymatrix{
A \otimes \kk \ar[r] & A & \kk \otimes A \ar[l] \\
C \otimes \kk \ar[u]^{f \otimes \one} \ar[r] & C \ar[u]^{f} & \kk \otimes C \ar[u]^{\one \otimes f} \ar[l] \\
C \otimes C \ar[u]^{\one \otimes \epsilon} & C \ar[l]^{\Delta} \ar[u]^\one \ar[r]_{\Delta} & C \otimes C \ar[u]^{\epsilon \otimes \one}
}
\end{equation}
\end{definition}

In particular, when one has a bialgebra $A$, the convolution product
$\star$ gives an associative algebra structure on $\End(A):=\Hom(A,A)$.

\begin{definition}
A bialgebra $A$ is called a {\it Hopf algebra} if there is an
element $S$ (called an {\it antipode} for $A$) in $\End(A)$ 
which is a $2$-sided inverse under $\star$ for the identity map $1_A$.
In other words, this diagram commutes:
\begin{equation}
\label{antipode-diagram}
\xymatrix{
&A \otimes A \ar[rr]^{S \otimes \one_A}& &A \otimes A \ar[dr]^m& \\
A \ar[ur]^\Delta \ar[rr]^{\epsilon} \ar[dr]_\Delta& &\kk \ar[rr]^{u} & & A\\
&A \otimes A \ar[rr]_{\one_A \otimes S}& &A \otimes A \ar[ur]_m& \\
}
\end{equation}
Or equivalently, if $\Delta(a) = \sum a_1 \otimes a_2$,
then
\begin{equation}
\label{antipode-defining-relation}
\sum_{(a)} S(a_1) a_2 = u(\epsilon(a)) = \sum_{(a)} a_1 S(a_2).
\end{equation}
\end{definition}

\begin{example}
For a group algebra $\kk G$, one can define an antipode $\kk$-linearly
via $S(t_g)=t_{g^{-1}}$.  The top pentagon in the above diagram commutes because
$$
(S \star \one)(t_g) = m ((S \otimes \one)(t_g \otimes t_g)) =
S(t_g) t_g = t_{g^{-1}} t_g = t_e = (u \circ \epsilon)(t_g).
$$
\end{example}


Note that when it exists, the antipode $S$ is unique, as with all 2-sided inverses in associative algebras: if $S, S'$ are both 2-sided $\star$-inverses to $\one_A$ then
$$
S'=(u\circ \epsilon) \star S'
=(S \star \one_A) \star S'
=S \star (\one_A \star S')
=S \star (u \circ \epsilon)= S.
$$

On the other hand, the next property is not quite as obvious, but is useful when one wants to check that a certain map {\it is} the antipode in a particular Hopf algebra, by checking it on an algebra generating set.

\begin{proposition}
\label{antipodes-are-antiendomorphisms}
The antipode $S$ in a Hopf algebra $A$ is an algebra anti-endomorphism:
$S(\one)=\one$, and $S(ab)=S(b)S(a)$ for all $a,b$ in $A$.
\end{proposition}
\begin{proof}
(see \cite[Chap. 4]{Sweedler})
Since $\Delta$ is an algebra map, one has
$\Delta(1)=1\otimes 1$, and therefore 
$
\one= u \epsilon (\one) = S(\one) \cdot \one = S(\one).
$

To show $S(ab)=S(b)S(a)$,
consider $A \otimes A$ as a coalgebra and $A$ as an algebra.
Then $\Hom(A \otimes A,A)$ is an associative algebra with a convolution
product $\starstar$ (to be distinguished from the convolution $\star$
on $\End(A)$), having two-sided identity element $u_A \epsilon_{A \otimes A}$.
We will show below that these three elements of $\Hom(A \otimes A,A)$
$$
\begin{aligned}
f(a \otimes b) &= ab \\
g(a \otimes b) &=S(b) S(a) \\
h(a \otimes b) &=S(ab)
\end{aligned}
$$
have the property that 
\begin{equation}
\label{three-antipode-calculations}
h \starstar f 
= u_A \epsilon_{A \otimes A}
= f \starstar g
\end{equation}
which would then show the desired equality $h=g$ via associativity:
$$
h=h \starstar (u_A \epsilon_{A \otimes A})
=h \starstar (f \starstar g)
=(h \starstar f) \starstar g
=(u_A \epsilon_{A \otimes A}) \starstar g=g.
$$
So we evaluate the three elements in \eqref{three-antipode-calculations}
on $a \otimes b$, assuming $\Delta(a)=\sum_{(a)} a_1 \otimes a_2$ and
$\Delta(b) = \sum_{(b)} b_1 \otimes b_2$, and hence $\Delta(ab)= \sum_{(a),(b)} a_1 b_1 \otimes a_2 b_2$.
One has 
$$
\begin{aligned}
(u_A \epsilon_{A \otimes A})(a \otimes b)
&=u_A( \epsilon_A(a) \epsilon_A(b))= u_A(\epsilon_A (ab)).\\
& \\
(h \starstar f)(a \otimes b)
&=\sum_{(a),(b)} h(a_1 \otimes b_1) f(a_2 \otimes b_2) \\
&=\sum_{(a),(b)} S(a_1 b_1) a_2 b_2 \\
&= (S \star \one_A) (ab) 
= u_A (\epsilon_A (ab)).\\
& \\
(f \starstar g)(a \otimes b)
&=\sum_{(a),(b)} f(a_1 \otimes b_1) g(a_2 \otimes b_2) \\
&=\sum_{(a),(b)} a_1 b_1 S(b_2) S(a_2) \\
&=\sum_{(a)} a_1 \cdot (\one_A \star S)(b) \cdot S(a_2) \\
&= u_A(\epsilon_A(b)) \sum_{(a)} a_1 S(a_2)
= u_A (\epsilon_A (b)) u_A (\epsilon_A (a))
= u_A (\epsilon_A (ab))
\end{aligned}
$$
\end{proof}

\begin{remark}
\label{antipode-as-dual-action}
Recall from Remark~\ref{comultiplication-as-tensor-action}
that the comultiplication on a bialgebra $A$ allows one
to define an $A$-module structure on the tensor product $M \otimes N$
of two $A$-modules $M,N$.  Similarly, the anti-endomorphism $S$
in a Hopf algebra allows one to turn {\it left} $A$-modules into
{\it right} $A$-modules, or vice-versa.  E.g., left $A$-modules
$M$ naturally have a right $A$-module structure on the
dual space $M^*:=\Hom(M,\kk)$, defined via $(fa)(m):=f(am)$
for $f$ in $M^*$ and $a$ in $A$.  The antipode $S$
can be used to turn this back 
into a left $A$-module $M^*$, via $(af)(m) = f(S(a)m)$.

For groups $G$ and left $\kk G$-modules 
(group representations) $M$, this is 
how one defines the {\it contragredient action} of $G$ on $M^*$,
namely $t_g$ acts as $(t_g f)(m) = f( t_{g^{-1}} m)$.  

Along the same lines, we are supposed to think of the counit
$A \overset{\epsilon}{\rightarrow} \kk$ as giving a way to
make $\kk$ into a {\it trivial} $A$-module.
\end{remark}

\begin{corollary}
\label{commutative-implies-involutive-antipode-cor}
Commutativity of $A$ implies the antipode is an involution: $S^2=\one_A$.
\end{corollary}
\begin{proof}
One checks that $S^2=S \circ S$ is a right $\star$-inverse to $S$, and hence $S^2=\one_A$:
$$
\begin{aligned}
(S \star S^2)(a) 
  &= \sum_{(a)} S(a_1) S^2(a_2) \\
  &= S \left( \sum_{(a)} S(a_2) a_1 \right)  
    \text{ by Proposition~\ref{antipodes-are-antiendomorphisms}} \\
  &= S \left( u (\epsilon(a)) \right)  \\
  & = u (\epsilon(a)) \text{ since }S(\one)=\one \text{ by Proposition~\ref{antipodes-are-antiendomorphisms}}.
\end{aligned}
$$
\end{proof}

\begin{remark}
We won't need it, but it is easy to adapt the above proof to show that
$S^2=\one_A$ also holds for {\it cocommutative} Hopf algebras; 
see \cite[Corollary 1.5.12]{Montgomery} or \cite[Chapter 4]{Sweedler}.
For a general Hopf algebra which is not finite-dimensional over $\kk$, the antipode
$S$ may not even have finite order, even in the graded connected setting.
E.g., Aguiar and Sottile \cite{AguiarSottile}
show that the Malvenuto-Reutenauer Hopf algebra of permutations has antipode of infinite
order.  In general, antipodes need not even be invertible \cite{Takeuchi}.
\end{remark}

In our frequent setting of graded connected bialgebras,
antipodes come for free.

\begin{proposition}
\label{graded-connected-bialgebras-have-antipodes}
A graded connected bialgebra $A$ has a unique antipode $S$,
which is a graded map $A \overset{S}{\longrightarrow} A$, endowing it with a Hopf structure.  
\end{proposition}
\begin{proof}
Let us try to define a ($\kk$-linear) left $\star$-inverse $S$ to $\one_A$
on each homogeneous component $A_n$, via induction on $n$.

In the base case $n=0$, Proposition~\ref{antipodes-are-antiendomorphisms} and its proof show that one
must define $S(\one)=\one$ so $S$ is the identity on $A_0=\kk$.

In the inductive step, recall from 
Exercise~\ref{graded-connected-exercise} that
a homogeneous element $a$ of degree $n > 0$
has $\Delta(a) = a \otimes \one  + \sum a_1 \otimes a_2$,
with each $\deg(a_1) < n$.
Hence in order to have $S \star \one_A = u \epsilon$,
one must define $S(a)$ so that
$
S(a) \cdot \one + \sum S(a_1) a_2 = u \epsilon (a) = 0
$
and hence
$
S(a) := - \sum S(a_1) a_2,
$
where $S(a_1)$ have already been uniquely defined
by induction.  This does indeed define such a left $\star$-inverse $S$ to
$\one_A$, by induction.  It is also a graded map by induction.

The same argument shows how to define a right $\star$-inverse 
$S'$ to $\one_A$.  Then $S=S'$ is a 
two-sided $\star$-inverse to $\one_A$ by the associativity of $\star$.
\end{proof}

Here is another consequence of the fact that $S(\one)=\one$.

\begin{proposition}
\label{antipode-on-primitives}
In bialgebras, primitive elements $x$ have $\epsilon(x)=0$,
and in Hopf algebras, they have $S(x)=-x$.
\end{proposition}
\begin{proof}
In a bialgebra, $\epsilon(1)=1$.  Hence
$\Delta(x) = \one \otimes x + x \otimes \one$ implies
via \eqref{counit-diagram} that
$1 \cdot \epsilon(x) + \epsilon(1) x = x$, so $\epsilon(x) = 0$.
It also implies 
$S(x) \one + S(\one) x=u \epsilon(x)=u(0)=0$, so $S(x)=-x$.  
\end{proof}

\noindent
Thus whenever $A$ is a Hopf algebra generated as an algebra by its 
primitive elements, $S$ is the unique anti-endomorphism that negates
all primitive elements.

\begin{example}
The tensor and symmetric algebras 
$T(V)$ and $\Sym(V)$ are each generated by $V$, which contains
only primitive elements in either case.  Hence one
has in $T(V)$ that
\begin{equation}
\label{antipode-in-tensor-algebra}
S(x_{i_1} x_{i_2} \cdots x_{i_k})=(-x_{i_k}) \cdots (-x_{i_2})(-x_{i_1}) =
(-1)^k x_{i_k} \cdots x_{i_2}x_{i_1}
\end{equation}
for each word $(i_1,\ldots,i_k)$ in the alphabet $I$.
The same holds in $\Sym(V)$ for each multiset $(i_1,\ldots,i_k)$,
recalling that the monomials are now commutative.  In other words,
for a commutative polynomial $f(\xx)$ in $\Sym(V)$, the antipode
$S$ sends $f$ to $f(-\xx)$, negating all the variables.
\end{example}





The antipode for a graded connected Hopf algebra
has an interesting formula due to
Takeuchi \cite{Takeuchi}, reminiscent of P. Hall's formula for the M\"obius function of a
poset\footnote{In fact, for incidence Hopf algebras, Takeuchi's formula generalizes Hall's formula-- see Corollary~\ref{P.Hall-formula}.}.  For the sake of stating this, define the $k$-fold {\it tensor power}
$
A^{\otimes k}=A \otimes \cdots \otimes A
$
and define
{\it iterated multiplication
and comultiplication} maps
$$
\begin{array}{rcl}
A^{\otimes k} & \overset{m^{(k-1)}}{\longrightarrow} & A \\
A & \overset{\Delta^{(k-1)}}{\longrightarrow} & A^{\otimes k} \\
\end{array}
$$ 
by $m^{(-1)}=u$, $\Delta^{(-1)}=\epsilon$, $m^{(0)}=\Delta^{(0)}=\one_A$,
and 
$$
\begin{array}{rcl}
m^{(k)}&=m \circ (\one_A \otimes m^{(k-1)}) 
    &=m \circ (m^{(k-1)} \otimes \one_A)\\
\Delta^{(k)}&=(\one_A \otimes \Delta^{(k-1)}) \circ \Delta
    &=(\Delta^{(k-1)} \otimes \one_A) \circ \Delta
\end{array}
$$
using associativity, coassociativity to show that these
are well-defined.  They are how one gives meaning
to the right sides of these equations:
$$
\begin{aligned}
m^{(k)}(a^{(1)} \otimes \cdots \otimes a^{(k+1)}) 
&= a^{(1)} \cdots a^{(k+1)}\\
\Delta^{(k)}(b) 
&= \sum b_1 \otimes \cdots \otimes b_{k+1} \text{ in Sweedler notation.}
\end{aligned}
$$





\begin{proposition}
\label{Takeuchi-prop}
In a graded connected Hopf algebra $A$, the
antipode has formula
\begin{equation}
\label{Takeuchi-formula}
\begin{aligned}
S&=\sum_{k \geq 0} (-1)^k m^{(k-1)} f^{\otimes k} \Delta^{(k-1)} \\
 &=u\epsilon - f + m \circ f^{\otimes 2} \circ \Delta 
    - m^{(2)} \circ f^{\otimes 3} \circ \Delta^{(2)} + \cdots
\end{aligned}
\end{equation}
where $f:=1_A - u\epsilon$ in $\End(A)$.
\end{proposition}
\begin{proof}
We argue as in \cite[Lemma 14]{Takeuchi} or \cite[\S 5]{AguiarSottile}.  
For any $f$ in $\End(A)$ one has this explicit formula for
its $k$-fold convolution power $f^{\star k}:=f \star \cdots \star f$
in terms of its tensor powers $f^{\otimes k}:=f \otimes \cdots \otimes f$:
$$
f^{\star k} = m^{(k-1)} \circ f^{\otimes k} \circ \Delta^{(k-1)}.
$$
Therefore any $f$ annihilating $A_0$ will be
{\it locally $\star$-nilpotent} on $A$, meaning that for each $n$ one
has that $A_n$ is annihilated by $f^{\star m}$ with $m > n$:  homogeneity forces that
for $a$ in $A_n$, every summand of $\Delta^{(m-1)}(a)$ must contain among its 
$m$ tensor factors at least one factor lying in $A_0$,
so each summand is annihilated by $f^{\otimes m}$, and
$f^{\star m}(a)=0$.

In particular such $f$ have the property that 
$u \epsilon + f$ has as two-sided $*$-inverse
$$
\begin{aligned}
(u\epsilon + f)^{\star (-1)} &= u\epsilon - f 
    + f \star f - f \star f \star f + \cdots \\
& = \sum_{k \geq 0} (-1)^k f^{\star k}
= \sum_{k \geq 0} (-1)^k m^{(k-1)} \circ f^{\otimes k} \circ \Delta^{(k-1)}.
\end{aligned}
$$
The proposition follows upon taking 
$f:=\one_A - u\epsilon$, which annihilates $A_0$.
\end{proof}

\begin{remark}
\label{non-graded-Takeuchi-remark}
In fact, one can see that Takeuchi's formula applies more generally to
define an antipode $A \overset{S}{\longrightarrow} A$
in any (not necessarily graded)
bialgebra $A$ where the map $\one_A - u\epsilon$ is locally 
$\star$-nilpotent.

It is also worth noting that the proof of Proposition~\ref{Takeuchi-prop}
gives an alternate proof of 
Proposition~\ref{graded-connected-bialgebras-have-antipodes}
\end{remark}


To finish our discussion of antipodes, we mention some properties
(taken from \cite[Chap. 4]{Sweedler}) relating antipodes
to convolutional inverses.  It also shows that
a bialgebra morphism between Hopf algebras automatically respects the 
antipodes. 

\begin{proposition}
\label{antipodes-give-convolution-inverses}
Let $H$ be a Hopf algebra with antipode $S$.
\begin{enumerate}
\item[(a)] For any algebra $A$ and algebra morphism $H \overset{\alpha}{\rightarrow} A$,
one has $\alpha \circ S = \alpha^{\star -1}$, the convolutional inverse to $\alpha$ in $\Hom(H,A)$.
\item[(b)] For any coalgebra $C$ and coalgebra morphism $C \overset{\gamma}{\rightarrow} H$,
one has $S \circ \gamma = \gamma^{\star -1}$, the convolutional inverse to $\gamma$ in $\Hom(C,H)$.
\item[(c)] If $H_1,H_2$ are Hopf algebras with antipodes $S_1,S_2$, then any bialgebra morphism $H_1 \overset{\beta}{\rightarrow} H_2$ is a Hopf morphism, that is, it commutes with the antipodes, since 
$
\beta \circ S_1 \overset{(a)}{=} \beta^{\star -1}\overset{(b)}{=} S_2 \circ \beta.
$
\end{enumerate}
\end{proposition}
\begin{proof}
We prove (a); the proof of (b) is similar, and (c) follows immediately from (a),(b) as indicated in its statement.  Begin with the following:
given an algebra morphism
$A \overset{\alpha}{\rightarrow} A'$ and
coalgebra morphism
$C \overset{\gamma}{\rightarrow} C'$,
the pre- and post-composition
$f \longmapsto \alpha \circ f \circ \gamma$
is a convolution algebra morphism
$
\Hom(C',A) \overset{\varphi}{\longrightarrow}\Hom(C,A'),
$
as one can check that
$
\varphi(u_A \circ \epsilon_{C'})
=\alpha \circ u_A \circ \epsilon_{C'} \circ \gamma 
=u_{A'} \circ \epsilon_{C}, 
$
and
\begin{equation}
\label{pre-and-post-composition-in-convolution}
\begin{aligned}
\varphi(f \star g)
&=\alpha \circ (f \star g) \circ \gamma \\
&=\alpha \circ m \circ (f \otimes g) \circ \Delta \circ \gamma \\
&=m \circ (\alpha \otimes \alpha) \circ (f \otimes g) \circ 
     (\gamma \otimes \gamma) \circ \Delta \\
&=m \circ \left( (\alpha \circ f \circ \gamma) \otimes 
          (\alpha \circ g \circ \gamma) \right) \circ \Delta \\
&=\varphi(f) \star \varphi(g).
\end{aligned}
\end{equation}
For assertion (a), note that a special case of the above
observation shows $f \longmapsto \alpha \circ f$ gives
a convolution-algebra morphism $\Hom(H,H) \overset{\varphi}{\rightarrow} \Hom(H,A)$, and hence
$$
\alpha \circ S = \varphi(S) 
               = \varphi\left((1_H)^{\star -1}\right) 
               = \left(\varphi(1_H)\right)^{\star -1} 
               = \left(\alpha \circ 1_H\right)^{\star -1} 
               = \alpha^{\star -1}.
$$
\end{proof}





\subsection{Commutativity, cocommutativity}

\begin{definition}
Say that the $\kk$-algebra $A$ is {\it commutative} if $ab=ba$, that is,
this diagram commutes:
\begin{equation}
\label{commutativity-diagram}
\xymatrix{
A \otimes A \ar[rr]^T \ar[dr]^m& &A \otimes A \ar[dl]_m\\
 &A&
}
\end{equation}
Say that the $\kk$-coalgebra $C$ is {\it cocommutative} if this diagram
commutes:
\begin{equation}
\label{cocommutativity-diagram}
\xymatrix{
C \otimes C \ar[rr]^T & &C \otimes C \\
 &C \ar[ul]^\Delta \ar[ur]_\Delta&
}
\end{equation}
\end{definition}

\begin{example}
Group algebras $\kk G$ are always cocommutative, but commutative if and only if $G$ is abelian.

Tensor algebras $T(V)$ are always cocommutative, 
but commutative if and only if $\dim_\kk V \leq 1$.

Symmetric algebras $\Sym(V)$ are always cocommutative and commutative.

Homology and cohomology of $H$-spaces are always cocommutative and commutative
{\it in the topologists sense} where one reinterprets that 
twist map $A \otimes A \overset{T}{\rightarrow} A \otimes A$ to have
the extra sign as in \eqref{topologist-twist}.
\end{example}

Note how the cocommutative Hopf algebras $T(V), \Sym(V)$
have much of their structure controlled by their subspace $V$ of
primitive elements.  This is not far from the truth in general,
and closely related to Lie algebras.

\begin{exercise}
Recall that a {\it Lie algebra} over $\kk$ is a $\kk$-vector space
$\Lieg$ with a $\kk$-bilinear map $[\cdot,\cdot]$ that
satisfies $[x,x]=0$ for $x$ in $\Lieg$, and the {\it Jacobi identity}
$$
\begin{aligned}[ ]
[x,[y,z]]&=[[x,y],z] + [y,[x,z]], \text{ or equivalently }\\
[x,[y,z]]&+[z,[x,y]]+[y,[z,x]]=0. 
\end{aligned}
$$
\noindent
(a) Check that any associative algebra $A$ gives rise to a Lie algebra by means of the commutator operation $[a,b]:=ab-ba$.
\newline
(b) If $A$ is also a bialgebra, show that the $\kk$-subspace
of primitive elements $\Liep \subset A$ is closed under the Lie
bracket, that is, $[\Liep, \Liep] \subset \Liep$, and hence forms
a Lie subalgebra.

Conversely, given a Lie algebra $\Liep$,
one constructs the {\it universal enveloping algebra}
$\UUU(\Liep):=T(\Liep)/J$ as the quotient of the tensor algebra
$T(\Liep)$ by the two-sided ideal $J$ generated by all elements
$xy-yx-[x,y]$ for $x,y$ in $\Liep$.
\newline
(c) Show that $J$ is also a two-sided coideal in $T(\Liep)$ for
its usual coalgebra structure, and hence the quotient
$\UUU(\Liep)$ inherits the structure of a cocommutative bialgebra.
\newline
(d) Show that the antipode $S$ on $T(\Liep)$ preserves $J$, meaning
that $S(J) \subset J$, and hence $\UUU(\Liep)$ inherits the
structure of a (cocommutative) Hopf algebra.
\end{exercise}

There are theorems,
discussed in \cite[\S3.8]{Cartier}, \cite[Chap. 5]{Montgomery},
giving various mild hypotheses in addition to  cocommutativity
which imply that the inclusion of the space $\Liep$ of primitives in
a Hopf algebra $A$ extends to a Hopf isomorphism 
$\UUU(\Liep) \cong A$.


\subsection{Duals}

Recall that for finite dimensional $\kk$-vector spaces $V$, taking the dual
space $V^* :=\Hom(V,\kk)$ reverses $\kk$-linear maps.  That is, 
$V \overset{\varphi}{\rightarrow} W$ induces
$W^* \overset{\varphi^*}{\rightarrow} V^*$ defined uniquely by
$$
( f, \varphi(v) ) = ( \varphi^*(f), v )
$$
in which $(f,v)$ is the bilinear pairing $V^* \times V \rightarrow \kk$
sending $(f,v) \mapsto f(v)$.  When $\varphi$ is
expressed in terms of a basis $\{v_i\}_{i \in I}$ for $V$,
the map $\varphi^*$ is expressed by the transpose matrix in terms
of the dual basis $\{f_i\}_{i \in I}$ for $V^*$ that satisfies 
$(f_i,v_j)=\delta_{i,j}$, where $\delta_{i,j}$ is
the {\it Kronecker delta}: $\delta_{i,j}= 1$ if $i=j$ and $0$ else.

When discussing graded vector spaces $V=\oplus_{n \geq 0} V_n$ {\it of finite type}, meaning that each $V_n$ is finite-dimensional, note that $V^*$ 
can contain functionals $f$ supported on infinitely many $V_n$.  
Instead we will consider the subspace
$V^o:=\oplus_{n \geq 0} (V_n)^* \subset V^*$, 
sometimes called the {\it restricted dual},
consisting of the functions $f$ that vanish on all but finitely many $V_n$,
which is again a graded vector space of finite type.

Reversing the diagrams should then make it clear that, in the finite-dimensional or finite-type situation, duals of algebras are coalgebras, and vice-versa,
and duals of bialgebras or Hopf algebras are bialgebras or Hopf algebras.
For example, the product in a Hopf algebra $A$ uniquely defines 
the coproduct of $A^o$ via adjointness:
$$
\left(\Delta_{A^o}(f) , a \otimes b\right)_{A \otimes A}
 = (f , a b)_{A}.
$$
Thus if $A$ has a basis $\{a_i\}_{i \in I}$ with {\it product structure
constants} $\{c^i_{j,k}\}$, meaning
$$
a_j a_k = \sum_{i \in I} c^i_{j,k} a_i, 
$$
then the dual basis $\{f_i\}_{i \in I}$ has the same  $\{c^i_{j,k}\}$ as its
{\it coproduct structure constants}:
$$
\Delta_{A^o}(f_i) = \sum_{(j,k) \in I \times I} c^i_{j,k} f_j \otimes f_k.
$$

\begin{example}
When $A=T(V)$ is the tensor algebra for a finite-dimensional $\kk$-vector space
$V$, having $\kk$-basis 
$\{x_i\}_{i \in I}$, its restricted dual $A^o$ is another Hopf
algebra whose basis $\{y_{(i_1,\ldots,i_\ell)} \}$ is indexed by
words in the alphabet $I$, called the {\it shuffle algebra} of $V^*$.  
Duality shows that the {\it cut} coproduct in $A^o$ is defined by
$$
\Delta y_{(i_1,\ldots,i_\ell)} =
\sum_{j=0}^{\ell} y_{(i_1,\ldots,i_j)} \otimes y_{(i_{j+1},i_{j+2},\ldots,i_\ell)}.
$$
For example,
$$
\Delta y_{abcb}  = y_{\varnothing} \otimes y_{abcb}
+y_{a} \otimes y_{bcb}
+y_{ab} \otimes y_{cb}
+y_{abc} \otimes y_{b}
+y_{abcb} \otimes y_{\varnothing} 
$$
Duality also shows that the
{\it shuffle} product in $A^o$ will be given by
$$
y_{(i_1,\ldots,i_\ell)} y_{(j_1,\ldots,j_m)} =
\sum_{\kk=(k_1,\ldots,k_{\ell+m}) 
        \in \shuf{\ii}{\jj} } y_{(k_1,\ldots,k_{\ell+m})}
$$
where $\shuf{\ii}{\jj}$ denotes the multiset of $\binom{\ell+m}{\ell}$
words obtained as {\it shuffles} of the two
words $\ii=(i_1,\ldots,i_\ell)$ and $\jj=(j_1,\ldots,j_m)$. 
For example,
$$
\begin{aligned}
y_{ab} y_{cb} &= y_{abcb} + y_{acbb} + y_{cabb} + y_{cabb}+ y_{acbb} + y_{cbab}\\
&= y_{abcb} + 2 y_{acbb} + 2 y_{cabb} +  y_{cbab}
\end{aligned}
$$
More formally, one has
$$
y_{(i_1,i_2,\ldots,i_\ell)} y_{(i_{\ell+1},i_{\ell+2},\ldots,i_{\ell+m})} =
\sum_{\substack{w \in \Symm_{\ell+m}:\\w(1)<\cdots<w(\ell),\\
       w(\ell+1) < \cdots < w(\ell+m)}}
       y_{(i_{w^{-1}(1)},i_{w^{-1}(2)},\ldots,i_{w^{-1}(\ell+m)})}.
$$
Lastly, the antipode $S$ of $A^o$ is the adjoint of the antipode of $A=T(V)$ described in \eqref{antipode-in-tensor-algebra}:
$$
S y_{(i_1,i_2,\ldots,i_\ell)}
=(-1)^{\ell} y_{(i_\ell,\ldots,i_2,i_1)}.
$$

\end{example}




\begin{exercise}
Let $V$ be a $1$-dimensional vector space with basis element $x$,
so $\Sym(V) \cong \kk[x]$, with
$\kk$-basis $\{1=x^0,x^1,x^2,\ldots\}$.
\newline
\noindent
(a) Check that the powers $x^i$ satisfy
$$
\begin{aligned}
x^i \cdot x^j &= x^{i+j} \\
\Delta(x^n) &= \sum_{i+j=n} \binom{n}{i} x^i \otimes x^j \\
S(x^n) &= (-1)^n x^n
\end{aligned}
$$
\noindent
(b) Check that the dual basis elements 
$\{f^{(0)},f^{(1)},f^{(2)},\ldots\}$ for $\Sym(V)^o$,
defined by $f^{(i)}(x^j)=\delta_{i,j}$, satisfy 
$$
\begin{aligned}
f^{(i)} f^{(j)} &= \binom{i+j}{i} f^{(i+j)} \\
\Delta(f^{(n)}) &= \sum_{i+j=n} f^{(i)} \otimes f^{(j)} \\
S(f^{(n)}) &= (-1)^n f^{(n)}
\end{aligned}
$$
\noindent
(c) Show that if $\kk$ has characteristic zero, then the
map $\Sym(V)^{o} \rightarrow \Sym(V)$ sending
$f^{(n)} \mapsto \frac{x^n}{n!}$ is a graded Hopf isomorphism.

For this reason, the Hopf structure on $\Sym(V)^o$ is 
called a {\it divided power algebra}.
\newline
\noindent
(d) Show that when $\kk$ has characteristic $p >0$, 
one has $(f^{(1)})^p = 0$, and hence why there can
be no Hopf isomorphism $\Sym(V)^{o} \rightarrow \Sym(V)$.
\end{exercise}


\begin{exercise}
Let $V$ have $\kk$-basis $\{x_1,\ldots,x_n\}$,  and let $V \oplus V$ have 
$\kk$-basis $\{x_1,\ldots,x_n, y_1,\ldots,y_n\}$, so that one has
isomorphisms
$$
\Sym(V \oplus V) \cong \kk[\xx,\yy] \cong \kk[\xx] \otimes \kk[\yy] \cong
\Sym(V) \otimes \Sym(V).
$$
\newline
\noindent
(a) Show that our usual coproduct on $\Sym(V)$ can be re-expressed
as follows:
$$
\begin{array}{ccc}
\Sym(V) & &  \Sym(V) \otimes \Sym(V) \\
\Vert & & \Vert \\
\kk[\xx] & \overset{\Delta}{\longrightarrow} & \kk[\xx,\yy] \\
 f(x_1,\ldots,x_n) & \longmapsto & f(x_1+y_1,\ldots,x_n+y_n)\\
\end{array}
$$
In other words, it is induced from the diagonal map
\begin{equation}
\label{vector-space-diagonal-map}
\begin{array}{rcl}
V & \longrightarrow & V \oplus V\\
x_i & \longmapsto & x_i+y_i\\
\end{array}
\end{equation}
\newline
\noindent
(b) One can similarly define a coproduct
on the {\it exterior algebra} $\wedge V$, which is the quotient $T(V)/J$
where $J$ is the two-sided ideal generated by the elements
$\{x^2(=x \otimes x)\}_{x \in V}$ in $T^2(V)$.
This becomes a graded commutative algebra
$$
\wedge V
 = \bigoplus_{d=0}^n \wedge^d V \left( = \bigoplus_{d=0}^\infty \wedge^d V \right),
$$
if one views the elements of $V=\wedge^1 V$ as having {\it odd} degree, and 
uses the topologist's sign conventions (as in
\eqref{topologist-twist}).  One again has
$
\wedge(V \oplus V) = \wedge V \otimes \wedge V
$
as graded algebras.  Show that one can again let the diagonal map 
\eqref{vector-space-diagonal-map} induce a map
$$
\begin{array}{ccc}
\wedge(V) & \overset{\Delta}{\longrightarrow} 
 & \wedge V \otimes \wedge V \\
  f(x_1,\ldots,x_n) & \longmapsto & f(x_1+y_1,\ldots,x_n+y_n) \\
 \Vert & & \Vert \\
 \sum c_{i_1,\ldots,i_d} \cdot x_{i_1} \wedge \cdots \wedge x_{i_d} &  
  & \sum c_{i_1,\ldots,i_d} \cdot (x_{i_1}+y_{i_1} )\wedge \cdots \wedge (x_{i_d}+y_{i_d}) \\
\end{array}
$$
which makes $\wedge V$ into a graded connected Hopf algebra.
\newline
\noindent
(c) Show that in the tensor algebra $T(V)$, if 
one views the elements of $V=V^{\otimes 1}$ as having odd degree,
then for any $x$ in $V$ one has
$
\Delta(x^2) = 1 \otimes x^2 + x^2 \otimes 1.
$
\newline
(Hint: Make sure you use the convention \eqref{topologist-twist} in
the twist map!)
\newline
\noindent
(d) Use part (d) to show that the two-sided ideal $J \subset T(V)$
generated by $\{x^2\}_{x \in V}$ is also a two-sided coideal,
and hence the quotient $\wedge V = T(V)/J$ inherits the structure
of a bialgebra.  Check that the coproduct on $\wedge V$ 
inherited from $T(V)$
is the same as the one defined in part (b).
\end{exercise}

\newpage


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Review of symmetric functions $\Lambda$ as Hopf algebra}
\label{Sym-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Here we review the ring of symmetric functions, borrowing heavily from
standard treatments, such as Macdonald \cite[Chap. I]{Macdonald}, 
Sagan \cite[Chap. 4]{Sagan}, and Stanley \cite[Chap. 7]{Stanley}, 
but emphasizing the Hopf structure early on.

\subsection{Definition of $\Lambda$}

As before, $\kk$ here could either be a field or the integers $\ZZ$.
Given an infinite variable set $\xx=(x_1,x_2,\ldots)$, a monomial
$\xx^\alpha:=x_1^{\alpha_1} x_2^{\alpha_2} \cdots$ is indexed by an element
$\alpha=(\alpha_1,\alpha_2,\ldots)$ in $\NN^\infty$ having finite support;
such $\alpha$ are called {\it weak compositions}.  We will 
consider the ring $R(\xx)$ of formal power series $f(\xx)=\sum_{\alpha} c_\alpha \xx^\alpha$ 
with $c_\alpha$ in $\kk$ {\it of bounded degree}, that is, where there exists some
bound $d=d(f)$ for which $\deg(\xx^{\alpha}):= \sum_i \alpha_i > d$ implies
$c_\alpha = 0$.  It is easy to see that the product of two such power
series is well-defined, and also has bounded degree.

The symmetric group $\Symm_n$ permuting the first $n$ variables
$x_1,\ldots,x_n$ acts a group of automorphisms on $R(\xx)$, as does
the union $\Symm_{(\infty)}=\cup_{n \geq 1} \Symm_n$
of the infinite ascending chain
$
\Symm_1 \subset \Symm_2 \subset \cdots
$
of symmetric groups.


\begin{definition}
The {\it ring of symmetric functions in $\xx$ with coefficients in $\kk$}, denoted
$
\Lambda = \Lambda_\kk = \Lambda(\xx) = \Lambda_\kk(\xx),
$
is the $\Symm_{(\infty)}$-invariant subalgebra $R(\xx)^{\Symm_{(\infty)}}$
of $R(\xx)$:
$$
\Lambda :=\left\{ f=\sum_\alpha c_\alpha \xx^\alpha \in R(\xx):
c_\alpha=c_\beta\text{ if }\alpha, \beta\text{ lie in the same }
\Symm_{(\infty)}\text{-orbit} \right\}.
$$
\end{definition}

Note that $\Lambda$ is a graded $\kk$-algebra, since
$\Lambda = \oplus_{n \geq 0} \Lambda_n$ where $\Lambda_n$ are
the symmetric functions $f=\sum_\alpha c_\alpha \xx^\alpha$ 
which are {\it homogeneous of degree $n$},
meaning $\deg(\xx^\alpha)=n$ for all $c_\alpha \neq 0$.  

\begin{definition}
A {\it partition} $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_\ell,0,0,\ldots)$
is a weak composition whose parts weakly decrease: $\lambda_1 \geq \cdots \geq \lambda_\ell > 0$.  One sometimes omits trailing zeroes from a partition.

Every weak composition $\alpha$ lies in the  $\Symm_{(\infty)}$-orbit
of a unique {\it partition} 
$\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_\ell,0,0,\ldots)$
with $\lambda_1 \geq \cdots \geq \lambda_\ell > 0$.  
Define the {\it monomial symmetric function} 
\begin{equation}
\label{monomial-symmetric-function-definition}
m_\lambda:=\sum_{\alpha \in \Symm_{(\infty)}\lambda}  \xx^\alpha.
\end{equation}
Letting $\lambda$ run through the set $\Par$ of all partitions, 
this gives the {\it monomial $\kk$-basis} $\{m_{\lambda}\}$ of $\Lambda$.
Letting $\lambda$ run only through the set $\Par_n$ of partitions of $n$
gives the monomial $\kk$-basis for $\Lambda_n$.
\end{definition}

\begin{example}
For $n=3$, one has
$$
\begin{aligned}
m_{(3)} &= x_1^3+x_2^3+x_3^3+ \cdots \\
m_{(2,1)} &= x_1^2 x_2 + x_1 x_2^2 + x_1^2 x_3 +x_1 x_3^2 + \cdots \\
m_{(1,1,1)} &= x_1 x_2 x_3 +x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4 + x_1 x_2 x_5 + \cdots 
\end{aligned}
$$
\end{example}

\begin{remark}
\label{finite-variable-set-remark}
It is sometimes convenient to work with finite variable sets $x_1,\ldots,x_n$,
which one justifies as follows.  Note that the algebra homomorphism 
$$
R(\xx) \rightarrow R(x_1,\ldots,x_n)=\kk[x_1,\ldots,x_n]
$$
which sends $x_{n+1},x_{n+2},\ldots$ to $0$ restricts to
an algebra homomorphism 
$$
\Lambda_\kk(\xx) \rightarrow \Lambda_\kk(x_1,\ldots,x_n)=\kk[x_1,\ldots,x_n]^{\Symm_n}.
$$
Furthermore, this last homomorphism is a $\kk$-linear isomorphism when restricted
to $\Lambda_i$ for $0 \leq i \leq n$, since it sends the monomial basis
elements $m_\lambda(\xx)$ to the monomial basis elements 
$m_\lambda(x_1,\ldots,x_n)$.
Thus when one proves identities in $\Lambda_\kk(x_1,\ldots,x_n)$ for all $n$,
they are valid in $\Lambda$, that is, $\Lambda$ is the inverse limit of 
the $\Lambda(x_1,\ldots,x_n)$ in the category of graded $\kk$-algebras.
\end{remark}

One can also define a comultiplication on $\Lambda$ as follows.
Note that when one decomposes the variables into two sets
$(\xx,\yy)=(x_1,x_2,\ldots,y_1,y_2,\ldots)$, one has a ring homomorphism
$$
\begin{array}{rcl}
R(\xx) \otimes R(\xx) & \longrightarrow& R(\xx,\yy)  \\
f(\xx) \otimes g(\xx) & \longmapsto& f(\xx)g(\yy).
\end{array}
$$
This restricts to an isomorphism
\begin{equation}
\label{symmetric-functions-in-two-sets-iso}
\Lambda \otimes \Lambda 
=R(\xx)^{\Symm_{(\infty)}} \otimes R(\xx)^{\Symm_{(\infty)}} 
\longrightarrow R(\xx,\yy)^{\Symm_{(\infty)} \times \Symm_{(\infty)} }
\end{equation}
where $\Symm_{(\infty)} \times \Symm_{(\infty)}$ denotes
permutations of (finite subsets of) the
$\xx$  and separate permutations of (finite subsets of)
the $\yy$, because $R(\xx,\yy)^{\Symm_{(\infty)} \times \Symm_{(\infty)}}$ has $\ZZ$-basis $\{ m_\lambda(\xx) m_\mu(\yy) \}_{\lambda, \mu \in \Par}$.  
As $\Symm_{(\infty)} \times \Symm_{(\infty)}$ is a subgroup of the group $\Symm_{(\infty)}$ acting on
all of $(\xx,\yy)$, one gets an inclusion of rings
$$
\Lambda(\xx,\yy) =  R(\xx,\yy)^{\Symm_{(\infty)}} \hookrightarrow 
     R(\xx,\yy)^{\Symm_{(\infty)} \times \Symm_{(\infty)} } \cong \Lambda \otimes \Lambda
$$
where the last isomorphism is the inverse of the one in \eqref{symmetric-functions-in-two-sets-iso}.  This gives a comultiplication
$$
\begin{array}{rcl}
\Lambda=\Lambda(\xx) & \overset{\Delta}{\longrightarrow} 
   & \Lambda(\xx,\yy) \hookrightarrow \Lambda \otimes \Lambda\\
f(\xx)=f(x_1,x_2,\ldots) & \longmapsto & f(\xx,\yy)=f(x_1,x_2,\ldots,y_1,y_2,\ldots).
\end{array}
$$

\begin{example}
One has 
$$
\begin{aligned}
\Delta m_{(2,1)} &= m_{(2,1)}(x_1,x_2,\ldots,y_1,y_2,\ldots) \\
&= x_1^2 x_2 + x_1 x_2^2 + \cdots \\
 &\quad + x_1^2 y_1 + x_1^2 y_2 + \cdots \\
 &\quad + x_1 y_1^2 + x_1 y_2^2 + \cdots \\
 &\quad + y_1^2 y_2 + y_1 y_2^2 + \cdots\\
&= m_{(2,1)}(\xx) + m_{(2)}(\xx) m_{(1)}(\yy) +
 m_{(1)}(\xx) m_{(2)}(\yy) + m_{(2,1)}(\yy) \\
&= m_{(2,1)} \otimes \one + m_{(2)} \otimes m_{(1)} +
m_{(1)}\otimes m_{(2)} + \one \otimes m_{(2,1)}.
\end{aligned}
$$
This example generalizes easily to the following formula
\begin{equation}
\label{comultiplication-of-monomial-symmetric-function}
\Delta m_\lambda = 
  \sum_{\substack{ (\mu, \nu):\\ \mu \sqcup \nu = \lambda} } m_\mu \otimes m_\nu.
\end{equation}
in which $\mu \sqcup \nu$ is the partition obtained by taking the multiset union  of the parts of $\mu$ and $\nu$, and then reordering them to make them weakly decreasing.

\end{example}

Checking that $\Delta$ is coassociative amounts to checking that
$$
(\Delta \otimes \one) \circ \Delta f =
f(\xx,\yy,\zz)=
(\one \otimes \Delta) \circ \Delta f
$$
inside $\Lambda(\xx,\yy,\zz)$ as a subring of
$\Lambda \otimes \Lambda \otimes \Lambda$.
The counit $\Lambda \overset{\epsilon}{\rightarrow} \kk$
is defined in the usual fashion for graded connected coalgebras,
namely $\epsilon$ annihilates $I=\oplus_{n > 0} \Lambda_n$, and
$\epsilon$ is the identity on $\Lambda_0=\kk$;  alternatively $\epsilon$
sends a symmetric function $f(\xx)$ to its constant term $f(0,0,\ldots)$.

Note that $\Delta$ is an algebra morphism 
$\Lambda \rightarrow \Lambda \otimes \Lambda$
because it is a composition of maps which are all algebra morphisms.
As the unit and counit axioms are easily checked, 
$\Lambda$ becomes a graded connected $\kk$-bialgebra of finite type,
and hence also a Hopf algebra by 
Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
We will identify its antipode more explicitly in Section~\ref{symm-antipode-section} below.



\subsection{Other Bases}

We introduce the usual other bases of $\Lambda$,
and explain their significance later.
 
\begin{definition}
Define the families of {\it power sum symmetric functions} $p_n$,
{\it elementary symmetric functions} $e_n$,
and {\it complete homogeneous symmetric functions} $h_n$,
for $n=1,2,3,\ldots$ by
$$
\begin{array}{rlll}
p_n&:=x_1^n +x_2^n + \cdots &=m_{(n)} \\
e_n&:=\sum_{i_1 < \cdots < i_n}x_{i_1} \cdots x_{i_n}&=m_{(1^n)} \\
h_n&:=\sum_{i_1 \leq \cdots \leq i_n}x_{i_1} \cdots x_{i_n}&=\sum_{\lambda \in \Par_n} m_\lambda
\end{array}
$$
where $(1^n)=(1,1,\ldots,1)$, using a multiplicative notation
$\lambda=(1^{m_1}2^{m_2}\cdots)$ if the multiplicity of the part $i$ in $\lambda$ is $m_i$.  By convention, also define $h_0=e_0=1$, and $h_n=e_n=0$ if $n < 0$.  Extend these multiplicatively to 
partitions $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_\ell)$ 
with $\lambda_1 \geq \cdots \geq \lambda_\ell > 0$: 
$$
\begin{aligned}
p_\lambda&:=p_{\lambda_1} p_{\lambda_2} \cdots p_{\lambda_\ell}\\
e_\lambda&:=e_{\lambda_1} e_{\lambda_2} \cdots e_{\lambda_\ell} \\
h_\lambda&:=h_{\lambda_1} h_{\lambda_2} \cdots h_{\lambda_\ell}\\
\end{aligned}
$$
Also define the {\it Schur function} 
\begin{equation}
\label{Schur-function-definition}
s_\lambda:=\sum_T \xx^{\cont(T)}
\end{equation}
where $T$ runs through all {\it column-strict tableaux} of
shape $\lambda$, that is, $T$ is an assignment of entries in $\{1,2,3,\ldots\}$
to the cells of the {\it Ferrers diagram} for $\lambda$, weakly
increasing left-to-right in rows, and strictly increasing top-to-bottom
in columns.  Here $\xx^{\cont(T)}=\prod_i x_i^{|T^{-1}(i)|}$.  For example,
$$
T=\begin{matrix}
1 & 1 & 1 & 4 & 7\\
2 & 3 & 3 &   & \\
4 & 4 & 6 &   & \\
6 & 7
\end{matrix}
$$
is a column-strict tableau of shape $\lambda=(5,3,3,2)$
with $\xx^{\cont(T)}=x_1^3 x_2^1 x_3^2 x_4^3 x_5^0 x_6^2 x_7^2$.
\end{definition}

\begin{example}
One has
$$
\begin{aligned}
m_{(1)}&=p_{(1)}=e_{(1)}=h_{(1)}=s_{(1)}=x_1+x_2+x_3+\cdots \\
s_{(n)}&=h_n\\
s_{(1^n)}&=e_n
\end{aligned}
$$
\end{example}

\begin{example}
\label{transition-matrices-example}
One has for $\lambda=(2,1)$ that
$$
\begin{array}{rl}
p_{(2,1)}&=p_2 p_1 = (x_1^2+x_2^2+\cdots)(x_1+x_2+\cdots) \\
         &          = m_{(2,1)}+m_{(3)}\\
& \\
e_{(2,1)}&=e_2 e_1 = (x_1 x_2+x_1 x_3 + \cdots)(x_1+x_2+\cdots)\\
         &          = m_{(2,1)}+3m_{(1,1,1)}\\
& \\
h_{(2,1)}&=h_2 h_1 = (x_1^2 +x_2^2+\cdots + x_1x_2+x_1 x_3 + \cdots)(x_1+x_2+\cdots)\\
         &          = m_{(3)}+2m_{(2,1)}+3m_{(1,1,1)} \\

\end{array}
$$
and
$$
\begin{aligned}
&\begin{matrix}
s_{(2,1)}&=x_1^2x_2 &+x_1^2x_3&+x_1x_2^2&+x_1x_3^2&+x_1x_2x_3&+x_1x_2x_3&+x_1x_2x_4&+\cdots\\
&11            &11       &12            &13       &12        &13        &12 & \\
&2\phantom{1}        &3\phantom{1}   &2\phantom{1}        &3\phantom{1}   &3\phantom{1}    &2\phantom{1}    &4\phantom{1}  & \\
\end{matrix} \\
& \qquad \quad = m_{(2,1)}+2m_{(1,1,1)} 
\end{aligned}
$$
In fact, one has these transition matrices for $n=3$ expressing elements in terms of the monomial basis $m_\lambda$:
$$
\bordermatrix{
~          & p_{(3)} & p_{(2,1)} & p_{(1,1,1)} \cr
m_{(3)}    & 1       & 1         & 1  \cr
m_{(2,1)}  & 0       & 1         & 3\cr
m_{(1,1,1)}& 0       & 0         & 6
}
\qquad
\bordermatrix{
~          & e_{(3)} & e_{(2,1)} & e_{(1,1,1)} \cr
m_{(3)}    & 0       & 0         & 1  \cr
m_{(2,1)}  & 0       & 1         & 3\cr
m_{(1,1,1)}& 1       & 3         & 6
}
$$
\vskip.1in
$$
\bordermatrix{
~          & h_{(3)} & h_{(2,1)} & h_{(1,1,1)} \cr
m_{(3)}    & 1       & 1         & 1  \cr
m_{(2,1)}  & 1       & 2         & 3\cr
m_{(1,1,1)}& 1       & 3         & 6
}
\qquad
\bordermatrix{
~          & s_{(3)} & s_{(2,1)} & s_{(1,1,1)} \cr
m_{(3)}    & 1       & 0         & 0  \cr
m_{(2,1)}  & 1       & 1         & 0\cr
m_{(1,1,1)}& 1       & 2         & 1
}
$$
\end{example}

Our next goal is to show that $p_\lambda, e_\lambda, s_\lambda, h_\lambda$
all give bases for $\Lambda$.  However at the moment it is not yet
even clear that $s_\lambda$ are symmetric!

\begin{proposition}
\label{Schur-functions-are-symmetric-prop}
Schur functions $s_\lambda$ are symmetric, that is, they lie
in $\Lambda$.
\end{proposition}
\begin{proof}
It suffices to show $s_\lambda$ is symmetric under swapping the variables
$x_i, x_{i+1}$, by providing an involution $\iota$ on the set
of all column-strict tableaux $T$ of shape $\lambda$ which switches
the $\cont(T)$ for $(i,i+1) \cont(T)$.  
Restrict attention to the entries $i,i+1$
in $T$, which must look something like this:
$$
\begin{matrix}
   &   & & & & &   &   &i  &i  &i  &i&i+1&i+1 \\
   &i  &i&i&i&i&i+1&i+1&i+1&i+1&i+1& &   & \\
i+1&i+1&i+1& & &   &   &   &   &   & &   &
\end{matrix}
$$
One finds several vertically aligned pairs 
$\begin{matrix} i\\ i+1 \end{matrix}$.  If one were to remove all such pairs, the remaining entries would be a sequence of rows, each looking like this:
\begin{equation}
\label{i-i+1-row-to-be-swapped}
\underbrace{i,i,\ldots,i}_{r\text{ occurrences}},
\underbrace{i+1,i+1,\ldots,i+1}_{s\text{ occurrences}}
\end{equation}
An involution due to Bender and Knuth tells us to
leave fixed all the vertically aligned pairs 
$\begin{matrix} i\\ i+1 \end{matrix}$,
but change each sequence as in \eqref{i-i+1-row-to-be-swapped} to this:
$$
\underbrace{i,i,\ldots,i}_{s\text{ occurrences}},
\underbrace{i+1,i+1,\ldots,i+1}_{r\text{ occurrences}}
$$
For example, the above configuration in $T$ would change
to
$$
\begin{matrix}
   &   & & & &   &   &   &i  &i  &i  &i  &i  &i+1 \\
   &i  &i&i&i&i+1&i+1&i+1&i+1&i+1&i+1&   &   & \\
i  &i+1&i+1& &   &   &   &   &   &   &   &   &
\end{matrix}
$$
It is easily checked that this map is an involution,
and that it has the effect of swapping $(i,i+1)$ in $\cont(T)$.
\end{proof}

\begin{remark}
\label{other-linear-orders-remark}
The symmetry of Schur functions allows one to
reformulate them via column-strict tableaux defined with respect to
{\it any} total ordering $\LLL$ on the positive integers, rather than the usual
$1< 2 < 3<\cdots$.  For example, one can use 
the {\it reverse order}\footnote{This reverse order is what one uses when one defines a Schur function as a generating function for {\it reverse semistandard tableaux} or {\it column-strict plane partitions}; see Stanley \cite[Proposition 7.10.4]{Stanley}.} 
$
\cdots <3<2<1,
$
or even more exotic orders, such as
$$
1<3<5<7<\cdots<2<4<6<8<\cdots.
$$
Say that an assignment $T$ of entries in $\{1,2,3,\ldots\}$ to the cells of
the Ferrers diagram of $\lambda$ is an {\it $\LLL$-column-strict tableau} if
it is weakly $\LLL$-increasing left-to-right in rows, and strictly
$\LLL$-increasing top-to-bottom in columns.

\begin{proposition}
For any total order  $\LLL$ on the positive integers,
\begin{equation}
\label{different-total-order-Schur-function}
s_\lambda = \sum_T \xx^{\cont(T)}
\end{equation}
as $T$ runs through all $\LLL$-column-strict tableaux of shape $\lambda$.
\end{proposition}
\begin{proof}
Given a weak composition $\alpha=(\alpha_1,\alpha_2,\ldots)$  with
$\alpha_{n+1}=\alpha_{n+2}=\cdots=0$, 
assume that the integers $1,2,\ldots,n$ are totally ordered by $\LLL$ as
$w(1) <_{\LLL} \cdots <_{\LLL} w(n)$ for some $w$ in $\Symm_n$.
Then the coefficient of 
$\xx^{\alpha}=x_1^{\alpha_1} \cdots x_{n}^{\alpha_n}$ 
on the right side of \eqref{different-total-order-Schur-function} 
is the same as the coefficient of $\xx^{w^{-1}(\alpha)}$
on the right side of \eqref{Schur-function-definition}
defining $s_\lambda$,
which by symmetry of $s_\lambda$ is the same
as the coefficient of $\xx^{\alpha}$ on the right side of \eqref{Schur-function-definition}.
\end{proof}
\end{remark}

It is now not hard to show that
$p_\lambda, e_\lambda, s_\lambda$ give bases by a triangularity
argument.  For this purpose, let us introduce a useful partial
order on partitions.

\begin{definition}
The {\it dominance} or {\it majorization} partial order on $\Par_n$,
written $\lambda \triangleright \mu$, is defined by
$$
\lambda_1+\lambda_2+ \cdots+\lambda_k \geq \mu_1 + \mu_2 + \cdots+\mu_k
\quad \text{ for } k=1,2,\ldots,n.
$$
\end{definition}

\begin{definition}
For a partition $\lambda$, its {\it conjugate}
or {\it transpose} partition $\lambda^t$, is the one whose 
Ferrers diagram is obtained from that of $\lambda$ 
by exchanging rows for columns.  Alternatively, one has this formula
for its $i^{th}$ part:
$$
(\lambda^t)_i:=|\{ j: \lambda_j \geq i \}|.
$$
\end{definition}

It is an interesting (and useful) exercise to check that
$\lambda \triangleright \mu$ if and only if
$\mu^t \triangleright \lambda^t$.

\begin{proposition}
\label{many-symmetric-function-bases-prop}
The sets $\{e_\lambda\}, \{s_\lambda\}$ as $\lambda$ runs through
all partitions give $\kk$-bases for $\Lambda_\kk$ for any field $\kk$
or $\kk=\ZZ$.  The same holds for $\{p_\lambda\}$ when $\kk$ is a field
of characteristic zero.
\end{proposition}
\begin{proof}
One can restrict attention to each homogeneous component $\Lambda_n$
and partitions $\lambda$ of $n$.  We check that in each case, 
the proposed basis expands triangularly in the $\{m_\lambda\}$
with some choice of orderings on $\Par_n$ indexing the rows and columns,
as illustrated in Example~\ref{transition-matrices-example}.


One has $s_\lambda=\sum_\mu K_{\lambda,\mu} m_\mu$
where the coefficient $K_{\lambda,\mu}$ is the {\it Kostka number} counting
the column-strict tableaux $T$ of shape $\lambda$ having $\cont(T)=\mu$;
this follows because both sides are symmetric functions, and
$K_{\lambda,\mu}$ is the coefficient of $\xx^\mu$ on both sides.
Since for each positive integer $k$, 
the entries $1,2,\ldots,k$ in $T$ must all lie within the
first $k$ rows of $\lambda$, one has that
$\lambda_1+\lambda_2+ \cdots+\lambda_k \geq \mu_1 + \mu_2 + \cdots+\mu_k$,
that is, $\lambda \triangleright \mu$, 
so $s_\lambda=\sum_{\mu: \lambda \triangleright \mu} K_{\lambda,\mu} m_\mu$.
One can also check that $K_{\lambda,\lambda}=1$, so this expansion is
unitriangular with appropriate ordering of rows and columns.


One has $e_\lambda=\sum_\mu a_{\lambda,\mu} m_\mu$
where $a_{\lambda,\mu}$ counts matrices with $\{0,1\}$ entries having
row sum $\lambda$ and column-sum $\mu$: when one
expands $e_{\lambda_1} e_{\lambda_2} \cdots $, 
choosing the monomial $x_{j_1} \ldots x_{j_{\lambda_i}}$ in the
$e_{\lambda_i}$ factor corresponds to putting 
$1$'s in the $i^{th}$ row and columns $j_1,\ldots,j_{\lambda_i}$ of the 
$\{0,1\}$-matrix.
It is not hard to check\footnote{This is the easy implication in the {\it Gale-Ryser Theorem}.} that $a_{\lambda,\mu}$ vanishes unless
$\lambda^t \triangleright \mu$.
One can also check that $a_{\lambda,\lambda^t}=1$, so this expansion is
again unitriangular with appropriate ordering of rows and columns.



One has $p_\lambda=\sum_\lambda b_{\lambda,\mu} m_\mu$
where $b_{\lambda,\mu}$ counts the ways to partition the nonzero parts
$\lambda_1,\ldots,\lambda_\ell$ into blocks so that the sums of the blocks
give $\mu$; more formally, $b_{\lambda,\mu}$ is the number of maps
$\varphi: \{1,2,\ldots,\ell\} \rightarrow \{1,2,3,\ldots\}$ having
$\mu_j = \sum_{i: \varphi(i)=j} \lambda_i$ for $j=1,2,\ldots$.  
Again it is not hard to check that $b_{\lambda,\mu}$ vanishes
unless $\lambda \triangleleft \mu$,
and hence this expansion is triangular, for
appropriate ordering of rows and columns (but not unitriangular,
as $b_{\lambda,\lambda} \neq 1$ in general).
\end{proof}



\subsection{Comultiplications}

Thinking about comultiplication $\Lambda \overset{\Delta}{\rightarrow} \Lambda \otimes \Lambda$
on Schur functions forces us to immediately confront the following.

\begin{definition}
\label{skew-Schur-function-definition}
For partitions $\mu, \lambda$ say that $\mu \subseteq \lambda$ if
$\mu_i \leq \lambda_i$ for $i=1,2,\ldots$, so the Ferrers diagram for
$\mu$ is a subset of the cells for the Ferrers diagram of $\lambda$.
In this case, define the {\it skew (Ferrers) diagram} $\lambda/\mu$ to
be their set difference.  

Then define the {\it skew Schur function} $s_{\lambda/\mu}(\xx)$
to be the sum $s_{\lambda/\mu}:=\sum_{T} \xx^{\cont(T)}$, where the sum ranges
over all {\it column-strict tableaux} $T$ of shape $\lambda/\mu$, that is,
assignments of a value in $\{1,2,3,\ldots\}$ to each cell of $\lambda/\mu$,
weakly increasing left-to-right in rows, and strictly increasing top-to-bottom
in columns.  
\end{definition}
\begin{example}
\label{skew-tableau-example}
$$
T=\begin{matrix}
\cdot & \cdot & \cdot & 2 & 5\\
\cdot & 1     & 1     &   & \\
2     & 2     & 4     &   & \\
4     & 5
\end{matrix}
$$
is a column-strict tableau of shape $\lambda/\mu=(5,3,3,2)/(3,1,0,0)$
and it has $\xx^{\cont(T)}=x_1^2 x_2^3 x_3^0 x_4^2 x_5^2$.
\end{example}

\begin{proposition}
\label{symm-comultiplication-formulas-prop}
The comultiplication $\Lambda \overset{\Delta}{\rightarrow} \Lambda \otimes \Lambda$
has the following effect on the symmetric functions discussed so far\footnote{
The abbreviated summation indexing $\sum_{i+j=n} t_{i,j}$ used here is intended to mean
$$
\sum_{ \substack{ (i,j):\\ 0 \leq i,j \leq n, \\ i+j=n} } t_{i,j}.
$$}:
\begin{enumerate}
\item[ (i) ] $\Delta p_n = \one \otimes p_n + p_n \otimes \one$, that is, the
power sums $p_n$ are primitive.
\item[ (ii) ] $\Delta e_n = \sum_{i+j=n} e_i \otimes e_j$
\item[ (iii) ] $\Delta h_n = \sum_{i+j=n} h_i \otimes h_j$
\item[ (iv) ] $\Delta s_{\lambda} = \sum_{\mu \subseteq \lambda} s_{\mu} \otimes s_{\lambda/\mu}$
\end{enumerate}

\end{proposition}
\begin{proof}
Recall that $\Delta$ sends $f(\xx) \mapsto f(\xx,\yy)$, and one can easily check that
\begin{enumerate}
\item[ (i) ] $p_n(\xx,\yy) = \sum_i x_i^n + \sum_i y_i^n 
= p_n(\xx) \cdot 1 + 1 \cdot p_n(\yy)$
\item[ (ii) ] $e_n(\xx,\yy) = \sum_{i+j=n} e_i(\xx)e_j(\yy)$
\item[ (iii) ] $h_n(\xx,\yy) = \sum_{i+j=n} h_i(\xx)h_j(\yy)$
\end{enumerate}
For assertion (iv), note that by Remark~\ref{different-total-order-Schur-function}, one has
$$
s_\lambda(\xx,\yy)=\sum_T (\xx,\yy)^{\cont(T)}
$$
where the sum is over column-strict tableaux $T$ of shape $\lambda$ having entries
in the linearly ordered alphabet
%\footnote{
%There is a subtle point here:  
%the linear ordering \eqref{two-variable-set-ordering} is not equivalent under
%permutations in $\Symm_{(infty)}$ to some other orderings, such as,
%$x_1 < y_1 < x_2 < y_2 < \cdots$.   Given $\lambda$ in $\Par_n$,
%it would be better to first assume one has
%{\it finite} variable sets $\xx=(x_1,\ldots,x_n),
%\yy=(y_1,\ldots,y_n)$ in proving the identity
%$s_\lambda(\xx,\yy)=\sum_{\mu \subseteq \lambda} s_\mu(\xx) s_{\lambda/\mu}(\yy%)$
%as above;  the result then follows for infinite variable sets by 
%the usual considerations.}
\begin{equation}
\label{two-variable-set-ordering}
x_1 < x_2 < \cdots < y_1 < y_2 < \cdots .
\end{equation}
For example, 
$$
T=\begin{matrix}
x_1   & x_1 & x_1 & y_2 & y_5\\
x_2   & y_1 & y_1 &   & \\
y_2   & y_2 & y_4 &   & \\
y_4   & y_5
\end{matrix}
$$
is such a tableau of shape $\lambda=(5,3,3,2)$.  Note that the restriction of
$T$ to the alphabet $\xx$ gives a column-strict tableau $T_\xx$ of some shape 
$\mu \subseteq \lambda$, and the restriction of $T$ to the alphabet $\yy$ gives
a column-strict tableau $T_\yy$ of shape $\lambda/\mu$ (e.g. for $T$ in the example above,
the tableau $T_\yy$ appeared in Example~\ref{skew-tableau-example}).
Consequently, one has 
$$
\begin{aligned}
s_{\lambda}(\xx,\yy) &= \sum_T \xx^{\cont(T_\xx)} \cdot \yy^{\cont(T_\yy)} \\
       &= \sum_{\mu \subseteq \lambda} \left( \sum_{T_\xx} \xx^{\cont(T_\xx)} \right)
                                      \left( \sum_{T_\yy} \yy^{\cont(T_\yy)} \right)
       = \sum_{\mu \subseteq \lambda} s_{\mu}(\xx)  s_{\lambda/\mu}(\yy).
\end{aligned}
$$
\end{proof}



\subsection{The antipode, the involution $\omega$, and algebra generators}
\label{symm-antipode-section}

Since $\Lambda$ is a graded connected $\kk$-bialgebra, it will have an
antipode $\Lambda \overset{S}{\rightarrow} \Lambda$ making it a Hopf algebra
by Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
However, several issues will be resolved by identifying $S$ more explicitly now.


\begin{proposition}
\label{symm-antipode-and-h-basis}
Each of $\{e_n\}_{n=1,2,\ldots}, \{h_n\}_{n=1,2,\ldots}$
are algebraically independent, and generate $\Lambda_\kk$ as a polynomial
algebra for any field $\kk$ or $\kk=\ZZ$.  The same holds 
for $\{p_n\}_{n=1,2,\ldots}$ when $\kk$ is a field of characteristic zero.

Furthermore, the antipode $S$ acts as follows:
\begin{enumerate}
\item[ (i) ] $S(p_n) = -p_n$
\item[ (ii) ] $S(e_n) = (-1)^n h_n$
\item[ (iii) ] $S(h_n) = (-1)^n e_n$
\end{enumerate}
\end{proposition}

\begin{proof}
The assertions that $\{e_n\},\{p_n\}$ are algebraically independent and generate
$\Lambda$ are equivalent to Proposition~\ref{many-symmetric-function-bases-prop}
asserting $\{e_\lambda\}, \{p_\lambda\}$ give bases for $\Lambda$.
The assertion $S(p_n) = -p_n$ follows from Proposition~\ref{antipode-on-primitives}
since $p_n$ is primitive by Proposition~\ref{symm-comultiplication-formulas-prop}(i).

For the remaining assertions, start with the easy generating function identities
\begin{align}
\label{h-generating-function}
H(t)&:=\prod_{i=1}^\infty (1-x_it)^{-1} 
     = 1+h_1(\xx)t+h_2(\xx)t^2+\cdots = \sum_{n \geq 0} h_n(\xx) t^n \\
\label{e-generating-function}
E(t)&:=\prod_{i=1}^\infty (1+x_it) 
     = 1+e_1(\xx)t+e_2(\xx)t^2+\cdots = \sum_{n \geq 0} e_n(\xx) t^n
\end{align}
which shows that 
\begin{equation}
\label{E(t)-H(t)-relation}
1=E(-t)H(t)=\left( \sum_{n \geq 0} e_n(\xx) (-t)^n \right)
            \left( \sum_{n \geq 0} h_n(\xx) t^n \right)
\end{equation}
and hence, equating coefficients of powers of $t$, that for $n=0,1,2,\ldots$ one has
\begin{equation}
\label{e-h-relation}
\sum_{i+j=n} (-1)^i e_i h_j = \delta_{0,n}.
\end{equation}
This lets one recursively express the $e_n$ in terms of $h_n$ and vice-versa:
\begin{equation}
\label{e-h-recursions}
\begin{aligned}
e_0&:=1=:h_0\\
e_n&= e_{n-1} h_1 - e_{n-2} h_2 + e_{n-3} h_3 - \cdots\\
h_n&= h_{n-1} e_1 - h_{n-2} e_2 + h_{n-3} e_3- \cdots
\end{aligned}
\end{equation}
for $n =1,2,\ldots$  
Thus if one defines an algebra homomorphism
$\Lambda \overset{\omega}{\rightarrow} \Lambda$
on the algebraically independent polynomial generators $\{e_n\}$ by
sending $e_n \mapsto h_n$,
the identical form of the two recursions in \eqref{e-h-recursions}
shows that this map $\omega$ also sends $h_n \mapsto e_n$. 
Therefore $\omega$ is an involutive automorphism of $\Lambda$, 
and the $\{h_n\}$ are another algebraically independent generating set 
for $\Lambda$.

For the assertion about the antipode $S$ applied to $e_n$ or $h_n$,
note that the coproduct formulas for $e_n, h_n$ in
Proposition~\ref{symm-comultiplication-formulas-prop}(ii),(iii)
show that the defining relations for their
antipodes \eqref{antipode-defining-relation} 
will in this case be
$$
\begin{aligned}
\sum_{i+j=n} S(e_i) e_j &= \delta_{0,n} = \sum_{i+j=n} e_i S(e_j) \\
\sum_{i+j=n} S(h_i) h_j &= \delta_{0,n} = \sum_{i+j=n} h_i S(h_j)
\end{aligned}
$$
because $u \epsilon(e_n)=u \epsilon(h_n)=\delta_{0,n}$.
Comparing these to \eqref{e-h-relation}, one concludes via induction on $n$
that $S(e_n)=(-1)^n h_n$ and $S(h_n)=(-1)^n e_n$.
\end{proof}

Proposition~\ref{symm-antipode-and-h-basis} shows that the 
antipode $S$ on $\Lambda$ is, up to sign,
the same as the fundamental involution $\omega$:  one has 
\begin{equation}
\label{antipode-versus-omega-in-Sym}
S(f)=(-1)^n \omega(f) \text{ for }f \in \Lambda_n
\end{equation}
since this formula holds for all elements
of the generating set $\{e_n\}$ (or $\{h_n\}$).

\begin{remark}
Up to now we have not yet derived how the involution $\omega$ and the
antipode $S$ act on (skew) Schur functions, which is quite beautiful:
\begin{equation}
\label{skew-Schur-antipodes}
\begin{aligned}
\omega(s_{\lambda/\mu}) &= s_{\lambda^t/\mu^t}  \\
S(s_{\lambda/\mu})      &= (-1)^{|\lambda/\mu|} s_{\lambda^t/\mu^t} \\
\end{aligned}
\end{equation}
where recall that $\lambda^t$ is the transpose or conjugate partition
to $\lambda$, and $|\lambda/\mu|$ is the number of squares in
the skew diagram $\lambda/\mu$, that is, $|\lambda/\mu|=n-k$ 
if $\lambda, \mu$ lie in
$\Par_n, \Par_k$ respectively.


We will deduce this later in two ways, once as an exercise using
skewing operators in Exercise~\eqref{Darij's-skewing-exercise},
and for the second time from the action of
the antipode in $\Qsym$ on $P$-partition enumerators
in Corollary~\ref{antipode-on-skew-Schur-corollary}.
However, one could also deduce it immediately from our knowledge of the action of
$\omega$ and $S$ on $e_n, h_n$, if we were to prove the following 
famous {\it Jacobi-Trudi} and {\it dual Jacobi-Trudi} formulas.

\begin{theorem}
\label{Jacobi-Trudi-formulae}
Skew Schur functions are the following polynomials in $\{h_n\}, \{e_n\}$:
$$
\begin{aligned}
s_{\lambda/\mu} &= 
\det( h_{\lambda_i - \mu_j - i + j} )_{i,j=1,2,\ldots,\ell} \\
s_{\lambda^t/\mu^t} &= 
\det( e_{\lambda_i - \mu_j - i + j} )_{i,j=1,2,\ldots,\ell}.
\end{aligned}
$$
if $\lambda$ has at most $\ell$ nonzero parts.
\end{theorem}

\noindent
Since we appear not to need these formulas in the sequel, we will omit the proofs.  
Various proofs are well-explained in \cite[\S I.5]{Macdonald}, 
\cite[\S 4.5]{Sagan}, \cite[\S 7.16]{Stanley}.
An elegant treatment of Schur polynomials taking the
Jacobi-Trudi formula as the {\it definition} of $s_\lambda$ 
is given by Tamvakis \cite{Tamvakis}.
\end{remark}





\subsection{Cauchy product, Hall inner product, self-duality}

The Schur functions, although a bit unmotivated right now, have 
special properties with regard to the Hopf structure.  One property is
intimately connected with the following {\it Cauchy identity}.

\begin{theorem}
One has the following Schur function expansion for the
{\it Cauchy product/Cauchy kernel}
\begin{equation}
\label{Cauchy-identity}
\prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
=\sum_{\lambda \in \Par} s_{\lambda}(\xx) s_\lambda(\yy).
\end{equation}
\end{theorem}

\begin{remark}
Some readers may be bothered by the ambient ring in which this expansion takes place, which is a certain completion of $R(\xx) \otimes R(\yy)$.  One simple way to understand it is to replace each $x_i$ by $x_i t$, and write the equivalent identity in the 
power series ring $R(\xx) \otimes R(\yy)[[t]]$ 
\begin{equation}
\label{Cauchy-identity-with-t}
\prod_{i,j=1}^\infty (1-tx_i y_j)^{-1}
=\sum_{\lambda \in \Par} t^{|\lambda|} s_{\lambda}(\xx) s_\lambda(\yy).
\end{equation}
where $|\lambda|=\lambda_1 + \lambda_2 + \cdots +\lambda_\ell$.
\end{remark}

\begin{proof}
We follow the standard combinatorial proof 
(see \cite[\S 4.8]{Sagan},\cite[\S 7.11,7.12]{Stanley}), which rewrites the 
left and right sides of \eqref{Cauchy-identity-with-t}, and then compares them with the
{\it Robinson-Schensted-Knuth} (RSK) bijection.
On the left side, expanding out each geometric series
$$
(1-tx_i y_j)^{-1}=1+tx_iy_j+(tx_iy_j)^2 +(tx_iy_j)^3+\cdots
$$
and thinking of $(x_i y_j)^m$ as $m$ occurrences of a {\it biletter} $\binom{i}{j}$,
one can think of the left side as a sum over multisets of biletters
$\binom{i_1}{j_1}, \ldots, \binom{i_\ell}{j_\ell}$.  
Order the biletters in such a multiset in a
lexicographic order $\leq_{lex}$ that first checks if $i_1 \leq i_2$ and 
then if $i_1 = i_2$ checks if $j_1 \leq j_2$.  
Defining a {\it biword} to be an array 
$\binom{\ii}{\jj} = \binom{i_1 \cdots i_\ell}{j_1 \cdots j_\ell}$
in which the biletters are ordered
$\binom{i_1}{j_1} \leq_{lex} \cdots \leq_{lex} \binom{i_\ell}{j_\ell}$,
then the left side of \eqref{Cauchy-identity-with-t}
is the sum $\sum t^{\ell} \xx^{\cont(\ii)} \yy^{\cont(\jj)}$ over
all biwords $\binom{\ii}{\jj}$, with $\ell$ its number of biletters.
On the right side, expanding out the Schur functions as sums of tableaux
gives $\sum_{(P,Q)} t^{\ell} \xx^{\cont(Q)} \yy^{\cont(P)}$ in which the sum is over all ordered
pairs $(P,Q)$ of column-strict tableaux {\it having the same shape},
with $\ell$ cells.

The Robinson-Schensted-Knuth insertion algorithm gives us a bijection between
the biwords $\binom{\ii}{\jj}$ and the tableau pairs $(P,Q)$, with the property that 
$$
\begin{aligned}
\cont(\ii) &= \cont(Q),\\
\cont(\jj) &= \cont(P).
\end{aligned}
$$  
Starting with the pair $(P_0,Q_0)=(\varnothing,\varnothing)$ and 
$m=0$, it inserts one at a time the next biletter
$\binom{i_{m+1}}{j_{m+1}}$ of the biword into the pair of 
tableaux $(P_{m},Q_{m})$ already built; see Example~\ref{RSK-example} below.  
The bottom letter $j_{m+1}$ tries to insert itself
into the first row of $P_m$ by either bumping out the leftmost letter in the first row
strictly larger than $j_{m+1}$, or else placing itself at the right end of the row if
no such larger letter exists.  If a letter was bumped from the first row, it follows
the same rules to insert itself into the second row, and so on.  At the end of the bumping,
the tableau $P_{m+1}$ created has an extra corner cell not present in $P_m$, and one creates
$Q_{m+1}$ from $Q_m$ by adding the top letter $i_{m+1}$ of $\binom{i_{m+1}}{j_{m+1}}$ to $Q_m$
in this extra corner cell location.  After all of the biletters have been inserted,
the result is $(P_\ell,Q_\ell)=:(P,Q)$.



\begin{example}
\label{RSK-example}
The term in the expansion of the left side of \eqref{Cauchy-identity} corresponding to
$$
(x_1 y_2)^1(x_1 y_4)^1(x_2 y_1)^1(x_4 y_1)^1(x_4 y_3)^2 (x_5 y_2)^1
$$ 
is the biword
$\binom{\ii}{\jj}=\binom{1124445}{2411332}$, whose RSK insertion goes
as follows:
$$
\begin{array}{rclrcl}
P_0&= &\varnothing & Q_0 &=& \varnothing \\
 & & & & &\\
P_1&= &\begin{matrix} 2 \end{matrix}
 & Q_1 &= &\begin{matrix} 1 \end{matrix} \\
 & & & & &\\
P_2&= &\begin{matrix} 2&4 \end{matrix}
 & Q_2 &= &\begin{matrix} 1&1  \end{matrix} \\
 & & & & &\\
P_3&= &\begin{matrix} 1&4\\ 2 \end{matrix}
 & Q_3 &= &\begin{matrix} 1&1\\ 2 \end{matrix} \\
 & & & & &\\
P_4&= &\begin{matrix} 1&1\\ 2&4 \end{matrix}
 & Q_4 &= &\begin{matrix} 1&1\\ 2&4 \end{matrix} \\
 & & & & &\\
P_5&= &\begin{matrix} 1&1&3\\ 2&4& \end{matrix}
 & Q_5 &= &\begin{matrix} 1&1&4\\ 2&4& \end{matrix} \\
 & & & & &\\
P_6&= &\begin{matrix} 1&1&3&3\\ 2&4& & \end{matrix}
 & Q_6 &= &\begin{matrix} 1&1&4&4\\ 2&4& & \end{matrix} \\
 & & & & &\\
P:=P_7&= &\begin{matrix} 1&1&2&3\\ 2&3& & \\ 4& & & \end{matrix}
 & Q:=Q_7 &= &\begin{matrix} 1&1&4&4\\ 2&4& & \\5& & & \end{matrix} \\
\end{array}
$$
\end{example}

It requires some thought, but is not too hard, to see
that the bumping rule maintains the property 
that $P_m$ is a column-strict tableau of some Ferrers shape throughout.
It should be clear that $(P_m,Q_m)$ have the same shape at each stage.
Also, the construction of $Q_m$ shows that it is at least weakly 
increasing in rows and weakly increasing in columns throughout.  
What is perhaps least clear is
that $Q_m$ remains strictly increasing down columns.   
That is, when one has a string of equal
letters on top $i_m=i_{m+1}=\cdots=i_{m+r}$,
so that on bottom one bumps in $j_m \leq j_{m+1} \leq \cdots \leq j_{m+r}$, 
one needs to know that the new cells form a {\it horizontal strip}, that is,
no two of them lie in the same column.  This follows once one observes that
when one bumps in two letters $j \leq j'$, with $j$ bumped in first, 
the {\it bumping path} for $j'$ (the cells into which letters bump) 
stays strictly to the right, within each row, of the bumping path for $j$.  As an example, when $j_{m+1}=1$ is inserted into the tableau $P_m$ shown below, the result $P_{m+1}$ is shown with bumping path entries underlined:
$$
P_m=\begin{matrix} 
1&1&2&2&3\\ 
2&2&4&4&\\ 
3&4&5& &\\
4&6&6& & 
\end{matrix}
\qquad
\underset{j_{m+1}=1}{\overset{\text{insert}}{\longmapsto}}
\qquad
P_{m+1}=\begin{matrix} 
1&1&\underline{1}&2&3\\ 
2&2&\underline{2}&4&\\ 
3&4&\underline{4}& &\\
4&\underline{5}&6& & \\
\underline{6}& & & &
\end{matrix}
$$

To see that the map is a bijection, we show how to 
recover $\binom{\ii}{\jj}$ from $(P,Q)$.  This is done by {\it reverse bumping} from $(P_{m+1},Q_{m+1})$ to recover
both the biletter $\binom{i_{m+1}}{j_{m+1}}$ and the tableaux $(P_m,Q_m)$, as follows.
Firstly, $i_{m+1}$ is the maximum entry of $Q_{m+1}$, and $Q_m$ is obtained by removing the rightmost occurrence of this letter $i_{m+1}$ from $Q_{m+1}$.
To produce $P_m$ and $j_{m+1}$, find the position of the 
rightmost occurrence of $i_{m+1}$ in $Q_{m+1}$,
and start {\it reverse bumping} in $P_{m+1}$ from the entry in
this same position, where reverse bumping an 
entry means inserting it into one row higher by having it bump
out the rightmost entry which is strictly smaller.  The entry bumped out of the
first row is $j_{m+1}$, and the resulting tableau is $P_{m}$.

Finally, to see that the RSK map is surjective, one needs to show that the reverse bumping procedure can be applied to any pair $(P,Q)$ of column-strict tableaux of the same shape, and will result in a (lexicographically ordered)
biword $\binom{\ii}{\jj}$.  We leave this verification to the reader.
\end{proof}






\begin{corollary}
\label{self-duality-corollary}
In the Schur function basis $\{s_\lambda\}$ for $\Lambda$, the
structure constants for multiplication and comultiplication are the same, that is,
if one defines $c^\lambda_{\mu,\nu}, \hat{c}^\lambda_{\mu,\nu}$ via the unique expansions
\begin{equation}
\label{symm-structure-constants}
\begin{aligned}
s_\mu s_\nu &= \sum_{\lambda} c^\lambda_{\mu,\nu} s_\lambda\\
\Delta(s_\lambda) &= \sum_{\mu, \nu} \hat{c}^\lambda_{\mu,\nu} s_\mu \otimes s_\nu
\end{aligned}
\end{equation}
then $c^\lambda_{\mu,\nu}=\hat{c}^\lambda_{\mu,\nu}$.
\end{corollary}
\begin{proof}
The identity \eqref{Cauchy-identity}
lets one interpret both $c^\lambda_{\mu,\nu}, \hat{c}^\lambda_{\mu,\nu}$ 
as the coefficient of $s_\mu(\xx) s_\nu(\yy) s_\lambda(\zz)$ in
the product 
$$
\begin{aligned}
\prod_{i,j=1}^\infty (1-x_i z_j)^{-1}
\prod_{i,j=1}^\infty (1-y_i z_j)^{-1}
&=\left( \sum_\mu s_\mu(\xx) s_\mu(\zz) \right)
\left( \sum_\nu s_\nu(\yy) s_\nu(\zz) \right) \\
&=\sum_{\mu,\nu} s_\mu(\xx) s_\nu(\yy) \cdot s_\mu(\zz) s_\nu(\zz) \\
&=
\sum_{\mu,\nu} s_\mu(\xx) s_\nu(\yy) \left( \sum_\lambda c^\lambda_{\mu,\nu} s_\lambda(\zz) \right)
\end{aligned}
$$
since, regarding $x_1,x_2,\ldots,y_1,y_2,\ldots$ as lying in a single variable set
$(\xx,\yy)$, separate from the variables $\zz$, the Cauchy identity \eqref{Cauchy-identity}
expands the same product as
$$
\begin{aligned}
\prod_{i,j=1}^\infty (1-x_i z_j)^{-1}
\prod_{i,j=1}^\infty (1-y_i z_j)^{-1}
&=\sum_{\lambda} s_\lambda(\xx,\yy) s_\lambda(\zz)\\
&= \sum_\lambda \left( \sum_{\mu,\nu} \hat{c}^\lambda_{\mu,\nu}  
     s_\mu(\xx) s_\nu(\yy) \right) s_\lambda(\zz).
\end{aligned}
$$
\end{proof}

\begin{definition}
The coefficients $c^\lambda_{\mu,\nu}=\hat{c}^\lambda_{\mu,\nu}$ appearing in 
the expansions \eqref{symm-structure-constants}
are called {\it Littlewood-Richardson coefficients}.
\end{definition}

\begin{remark}
\label{third-L-R-interpretation-remark}
Noting on one hand the expansion
$$
s_\lambda(\xx,\yy) = \Delta(s_\lambda) 
   = \sum_{\mu,\nu} c^\lambda_{\mu,\nu} s_\mu(\xx) s_\nu(\yy)
$$
and on the other hand
$$
s_\lambda(\xx,\yy) = \sum_{\mu \subseteq \lambda} s_\mu(\xx) s_{\lambda/\mu}(\yy)
$$
one concludes another standard interpretation for $c^\lambda_{\mu,\nu}$:
$$
s_{\lambda/\mu} = \sum_{\nu} c^\lambda_{\mu,\nu} s_\nu.
$$
In particular, $c^\lambda_{\mu,\nu}$ vanishes unless
$\mu, \nu \subseteq \lambda$, and $|\mu|+|\nu|=|\lambda|$.
Note also that $c_{\mu,\nu}^\lambda = c^\lambda_{\nu,\mu}$.
 We will interpret $c^\lambda_{\mu,\nu}$ 
combinatorially in Section~\ref{bialternant-section}.
\end{remark}


\begin{definition}
Define the {\it Hall inner product} on $\Lambda$ to be the $\kk$-bilinear
form $(\cdot,\cdot)$ which makes $\{s_\lambda\}$ an orthonormal basis,
that is, $(s_\lambda, s_\nu)=\delta_{\lambda,\nu}$.
\end{definition}

\begin{corollary}
The isomorphism $\Lambda^o \cong \Lambda$ induced by the
Hall inner product is an isomorphism of Hopf algebras.
\end{corollary}

\begin{proof}
We have seen that the orthonormal basis  $\{s_\lambda\}$ of Schur functions is 
{\bf self-dual}, in the sense that its
multiplication and comultiplication structure constants
are the same.  Thus the isomorphism $\Lambda^o \cong \Lambda$ induced by the
Hall inner product is an isomorphism of bialgebras, and hence also a Hopf algebra isomorphism by Proposition~\ref{antipodes-give-convolution-inverses}(c). 
\end{proof}

As is customary, we identify two other dual pairs of bases, by expanding
the Cauchy product in two other ways.

\begin{proposition}
One can also expand
\begin{equation}
\label{Two-other-Cauchy-expansions}
\prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
=\sum_{\lambda \in \Par} h_{\lambda}(\xx) m_\lambda(\yy)
=\sum_{\lambda \in \Par} z_\lambda^{-1} p_{\lambda}(\xx) p_\lambda(\yy)
\end{equation}
where $z_\lambda:=m_1! \cdot 1^{m_1} \cdot m_2! \cdot 2^{m_2} \cdots$
if $\lambda=(1^{m_1},2^{m_2},\ldots)$ with multiplicity $m_i$ for the part $i$.
\end{proposition}
\begin{remark}
It is relevant later (and explains the notation) that $z_\lambda$ is
the size of the $\Symm_n$-centralizer subgroup for a permutation having
cycle type $\lambda$ with $|\lambda|=n$.
\end{remark}
\begin{proof}
For the first expansion, note that \eqref{h-generating-function} shows
$$
\begin{aligned}
\prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
&=\prod_{j=1}^\infty \sum_{n \geq 0} h_n(\xx) y_j^n\\
&=\sum_{\substack{ \text{weak }\\\text{ compositions } \\ (n_1,n_2,\ldots)}}
     ( h_{n_1}(\xx)h_{n_2}(\xx)\cdots  ) (y_1^{n_1} y_2^{n_2}\cdots )\\
&=\sum_{\lambda \in \Par} h_{\lambda}(\xx) m_\lambda(\yy).
\end{aligned}
$$
For the second expansion (and for later use in
the proof of Theorem~\ref{Sym-to-Hall-isomorphism-thm}) 
note that 
$$
\log H(t) 
= \log \prod_{i=1}^\infty (1-x_i t)^{-1}
=\sum_{i=1}^\infty -\log(1-x_i t) 
=\sum_{i=1}^\infty \sum_{m=1}^\infty \frac{(x_it)^m}{m}
=\sum_{m=1}^\infty \frac{1}{m} p_m(\xx) t^m
$$
so that taking $\frac{d}{dt}$ then shows that
\begin{equation}
\label{E(t)-H(t)-P(t)-relation}
P(t):=\sum_{m \geq 0} p_{m+1}t^m = \frac{H'(t)}{H(t)}= H'(t) E(-t).
\end{equation}
A similar calculation shows that
$$
%\begin{aligned}
\log \prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
%&=\sum_{i,j=1}^\infty -\log(1-x_i y_j) \\
%&=\sum_{i,j=1}^\infty \left( \frac{(x_iy_j)^1}{1}+\frac{(x_iy_j)^2}{2}+\frac{(x_iy_j)^3}{3}+\cdots \right) \\
%&=\sum_{m=1}^\infty \frac{1}{m} \sum_{i,j=1}^\infty (x_iy_j)^m 
=\sum_{m=1}^\infty \frac{1}{m} p_m(\xx) p_m(\yy)
%\end{aligned}
$$
and hence
$$
\begin{aligned}
&\prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
=\exp\left( \sum_{m=1}^\infty \frac{1}{m} p_m(\xx) p_m(\yy) \right) 
=\sum_{k=0}^\infty \frac{1}{k!} 
      \left( \sum_{m=1}^\infty \frac{1}{m} p_m(\xx) p_m(\yy) \right)^k\\
&=\sum_{k=0}^\infty \frac{1}{k!} 
           \sum_{\substack{(m_1,m_2,\ldots) \in \NN^\infty:\\ m_1+m_2+\cdots=k}}
             \binom{k}{m_1,m_2,\ldots} 
               \left( \frac{p_1(\xx) p_1(\yy)}{1} \right)^{m_1}
               \left( \frac{p_2(\xx) p_2(\yy)}{2} \right)^{m_2} \cdots \\
&=\sum_{\substack{ \text{weak }\\\text{ compositions } \\ (m_1,m_2,\ldots)}}
             \frac{ \left( p_1(\xx) p_1(\yy) \right)^{m_1} }{ 1^{m_1} m_1! }
             \cdot \frac{ \left( p_2(\xx) p_2(\yy)\right)^{m_2} }{ 2^{m_2} m_2! }
             \cdots 
=\sum_{\lambda \in \Par} \frac{ p_\lambda(\xx) p_\lambda(\yy) }{z_\lambda}.
\end{aligned}
$$
\end{proof}

\begin{corollary}
With respect to the Hall inner product on $\Lambda$, one also has
dual bases $\{h_\lambda\}$ and $\{m_\lambda\}$, as well as another
orthonormal basis $\{\frac{p_\lambda}{\sqrt{z_\lambda}} \}$.
\end{corollary}
\begin{proof}
Since \eqref{Cauchy-identity} and \eqref{Two-other-Cauchy-expansions}
showed
$$
\prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
=\sum_{\lambda \in \Par} s_{\lambda}(\xx) s_\lambda(\yy)\\
=\sum_{\lambda \in \Par} h_{\lambda}(\xx) m_\lambda(\yy)\\
=\sum_{\lambda \in \Par} \frac{p_{\lambda}(\xx)}{\sqrt{z_\lambda}} 
                       \frac{p_{\lambda}(\yy)}{\sqrt{z_\lambda}} 
$$
it suffices to show that any pair of bases $\{u_\lambda\},\{v_\lambda\}$ having 
$$
\sum_{\lambda \in \Par} s_{\lambda}(\xx) s_\lambda(\yy)=
\sum_{\lambda \in \Par} u_{\lambda}(\xx) v_\lambda(\yy)
$$ 
will be dual with respect to $(\cdot,\cdot)$.  Write transition matrices
$A=(a_{\nu,\lambda}), B=(b_{\nu,\lambda})$ uniquely expressing 
$$
\begin{aligned}
u_\lambda&=\sum_\nu a_{\nu,\lambda} s_\nu, \\
v_\lambda&=\sum_\nu b_{\nu,\lambda} s_\nu. \\
\end{aligned}
$$
Then orthonormality of the $s_\lambda$ gives
$(u_\alpha,v_\beta) = \sum_{\nu} a_{\nu,\alpha} b_{\nu,\beta}$,
and hence we want $A^t B = I$, that is, $B^{-1}=A^t$.  On the other hand, one has
$$
\sum_{\lambda } s_{\lambda}(\xx) s_\lambda(\yy)
=\sum_{\lambda } u_{\lambda}(\xx) v_\lambda(\yy)
=\sum_\lambda \sum_\nu a_{\nu,\lambda} s_\nu(\xx) 
  \sum_\rho b_{\rho,\lambda} s_\rho(\yy).
$$
Comparing coefficients of $s_\nu(\xx) s_\rho(\yy)$ forces
$\sum_\lambda a_{\nu,\lambda} b_{\rho,\lambda} = \delta_{\nu,\rho}$,
or in other words, $AB^t=I$, so $B^tA = I$, and hence $A^tB=I$, as desired.
\end{proof}






\subsection{Bialternants, Littlewood-Richardson:  Stembridge's concise proof}
\label{bialternant-section}

There is a more natural way in which Schur functions arise as a $\kk$-basis
for $\Lambda$, coming from consideration of polynomials in a finite variable set,
and the relation between those which are symmetric and those which are {\it alternating}.

For the remainder of this section, 
fix a positive integer $n$, and let $\xx=(x_1,\ldots,x_n)$ be a finite variable set.
This means that $s_{\lambda/\mu}=s_{\lambda/\mu}(\xx)=\sum_T \xx^{\cont(T)}$ is a 
generating function for column-strict tableaux $T$ as in 
Definition~\ref{skew-Schur-function-definition}, but with the extra condition
that $T$ have entries in $\{1,2,\ldots,n\}$.  We will assume without 
further mention that all partitions appearing in the section
have at most $n$ parts.  Lastly, we also take $\kk=\ZZ$ or a field of 
characteristic not equal to $2$,
to avoid certain annoyances in the discussion of alternating polynomials
in characteristic $2$.


\begin{definition}
Say that a polynomial $f(\xx)=f(x_1,\ldots,x_n)$ is {\it alternating} if
for every permutation $w$ in $\Symm_n$ one has that 
$$
(wf)(\xx) = f(x_{w(1)},\ldots,x_{w(n)})= \sgn(w) f(\xx).
$$
Let $\Lambda^{\sgn} \subset \kk[x_1,\ldots,x_n]$ 
denote the subset of alternating polynomials\footnote{When $\kk$ has characteristic $2$, 
it is probably best to define the alternating polynomials 
$\Lambda^{\sgn}_\kk$ as the subspace
$\Lambda^{\sgn} \otimes_{\ZZ} \kk 
\subset \ZZ[x_1,\ldots,x_n] \otimes_{\ZZ} \kk  \cong \kk[x_1,\ldots,x_n]$}.
\end{definition}

As with $\Lambda$ and its monomial basis $\{m_\lambda\}$, there is 
an obvious $\kk$-basis for $\Lambda^{\sgn}$, coming from the fact that
a polynomial $f=\sum_\alpha c_\alpha \xx^\alpha$
is alternating if and only if $c_{w(\alpha)} = \sgn(w) c_\alpha$ for every $w$ in $\Symm_n$
and every weak composition $\alpha$.  This means that every alternating $f$ 
is a $\kk$-linear combination of the following elements.

\begin{definition}
For $\alpha=(\alpha_1,\ldots,\alpha_n)$ in $\NN^n$, define
the {\it alternant} 
$$
a_{\alpha}:= \sum_{ w \in \Symm_n} \sgn(w) w(\xx^{\alpha}) 
= \det \left[ \begin{matrix}
x_1^{\alpha_1} & \cdots & x_1^{\alpha_n} \\
x_2^{\alpha_1} & \cdots & x_2^{\alpha_n} \\
\vdots & \ddots & \vdots \\
x_n^{\alpha_1} & \cdots & x_n^{\alpha_n} 
\end{matrix}
\right].
$$
\end{definition}

\begin{example}
One has
$$
a_{(1,5,0)} = x_1^1 x_2^5 x_3^0 
               -  x_1^5 x_2^1 x_3^0 -  x_1^1 x_2^0 x_3^5 -  x_1^0 x_2^5 x_3^1
               + x_1^0 x_2^1 x_3^5 + x_1^5 x_2^0 x_3^1 
            = - a_{(5,1,0)}.
$$
Meanwhile, $a_{(5,2,2)} = 0$ since the transposition $t=\binom{123}{132}$ fixes
$(5,2,2)$ and hence 
$$
a_{(5,2,2)} = t(a_{(5,2,2)}) = \sgn(t) a_{(5,2,2)} = -a_{(5,2,2)}.
$$
Alternatively, $a_{(5,2,2)}=0$ as it is a determinant of a matrix with two equal columns.
\end{example}

\noindent
This example illustrates that, for a $\kk$-basis for  $\Lambda^{\sgn}$,
one can restrict attention to alternants $a_\alpha$ in which $\alpha$ is a
{\it strict partition} $\alpha_1 > \alpha_2 > \cdots > \alpha_n$.
One can therefore uniquely express $\alpha = \lambda + \rho$,
where $\lambda$ is a (weak) partition
$\lambda_1 \geq \cdots \geq \lambda_n \geq 0$ and where 
$
\rho:=(n-1,n-2,\ldots,2,1,0)
$
is sometimes called the {\it staircase partition} due to its Ferrers shape.
For example $\alpha=(5,1,0)=(3,0,0) + (2,1,0)=\lambda+\rho$.  

\begin{proposition}
The alternants $\{ a_{\lambda +\rho} \}$ as $\lambda$ runs through the partitions
with at most $n$ parts form a $\kk$-basis for $\Lambda^{\sgn}$.
In addition, the {\bf bialternants} $\{ \frac{ a_{\lambda +\rho} }{ a_\rho } \}$
as $\lambda$ runs through the same set form a $\kk$-basis for 
$\Lambda(x_1,\ldots,x_n)=\kk[x_1,\ldots,x_n]^{\Symm_n}$.
\end{proposition}
\begin{proof}
The first assertion should be clear from our previous discussion:  the
alternants $\{ a_{\lambda +\rho} \}$ span $\Lambda^{\sgn}$ by definition, and
they are $\kk$-linearly independent because they are supported on disjoint
sets of monomials $\xx^\alpha$.

The second assertion follows from the first,
after proving the following {\bf Claim}: $f(\xx)$ lies in
$\Lambda^{\sgn}$ if and only if $f(\xx)=a_\rho \cdot g(\xx)$ where
$g(\xx)$ lies in $\kk[\xx]^{\Symm_n}$ and where
$$
a_{\rho} = \det( x_i^{n-j} )_{i,j=1,2,\ldots,n} =\prod_{1 \leq i < j \leq n} (x_i - x_j)
$$
is the {\it Vandermonde determinant/product}.
In other words 
$$
\Lambda^{\sgn}= a_{\rho}  \cdot \kk[\xx]^{\Symm_n}
$$
is a free $\kk[\xx]^{\Symm_n}$-module of rank one, with $a_\rho$ as its
$\kk[\xx]^{\Symm_n}$-basis element.

To see the Claim, first note the inclusion
$$
\Lambda^{\sgn} \supseteq a_{\rho} \cdot \kk[\xx]^{\Symm_n}
$$
since the product of a symmetric polynomial and an alternating polynomial is
an alternating polynomial.  For the reverse inclusion, note that since an alternating polynomial $f(\xx)$ changes
sign whenever one exchanges two distinct variables $x_i,x_j$, it must vanish upon setting
$x_i = x_j$, and therefore be divisible by $x_i - x_j$, so divisible by
the entire product $\prod_{1 \leq i < j \leq n} (x_i - x_j)=a_\rho$.  But then
the quotient $g(\xx) = \frac{f(\xx)}{a_\rho}$ is symmetric, as 
it is a quotient of two alternating polynomials.
\end{proof}

Our goal is to show that the mysterious bialternant basis 
$\{ \frac{ a_{\lambda +\rho} }{ a_\rho } \}$
are actually the Schur functions $\{ s_\lambda(\xx) \}$.  
Stembridge \cite{Stembridge} noted
that one could give a remarkably concise proof of an even stronger
assertion, which simultaneously gives one of the standard combinatorial
interpretations for the Littlewood-Richardson coefficients $c^{\lambda}_{\mu,\nu}$.  For the purposes of stating it, 
we introduce for a tableau $T$ the notation $T|_{\cols \geq j}$ (resp. $T|_{\cols \leq j}$)
to indicate the subtableau which is the restriction
of $T$ to the union of its columns $j,j+1,j+2,\ldots$ (resp. columns $1,2,\ldots,j$).

\begin{theorem}
\label{Stembridge-theorem}
For partitions $\lambda,\mu,\nu$ with $\mu \subseteq \lambda$, one has
$$
a_{\nu+\rho} s_{\lambda/\mu} = \sum_T a_{\nu+\cont(T)+\rho}
$$
where $T$ runs through all column-strict tableaux with entries in $\{1,2,\ldots,n\}$
of shape $\lambda/\mu$ with
the property that for all $j=1,2,\ldots$ one
has $\nu + \cont(T|_{\cols\geq j})$  a partition.
\end{theorem}

\noindent
Before proving Theorem~\ref{Stembridge-theorem}, let us see some of its consequences.

\begin{corollary}
$$
s_\lambda(\xx) = \frac{ a_{\lambda +\rho} }{ a_\rho }.
$$
\end{corollary}
\begin{proof}
Take $\nu=\mu=\varnothing$ in Theorem~\ref{Stembridge-theorem}.  Note
that for any $\lambda$, there is only one column-strict tableau $T$
of shape $\lambda$ having each $\cont(T|_{\cols\geq j})$ a partition,
namely the one having every entry in row $i$ equal to $i$:
$$
\begin{matrix}
1 & 1 & 1 & 1 & 1\\
2 & 2 & 2 &   & \\
3 & 3 & 3 &   & \\
4 & 4
\end{matrix}
$$
Furthermore, this $T$ has $\cont(T)=\lambda$, so the theorem says $a_{\rho} s_\lambda= a_{\lambda+\rho}$.
\end{proof}

\begin{example}
For $n=2$, so that $\rho=(1,0)$, if we take $\lambda=(4,2)$, then one
has 
$$
\begin{aligned}
\frac{a_{\lambda+\rho}}{a_\rho}
&=\frac{a_{(4,2)+(1,0)}}{a_{(1,0)}} 
=\frac{a_{(5,2)}}{a_{(1,0)}} \\
&=\frac{x_1^5 x_2^2 - x_1^2 x_2^5}{x_1 - x_2}\\
&= x_1^4 x_2^2 + x_1^3 x_2^3 + x_1^2 x_2^4\\
&= \xx^{\cont{
\left(\begin{matrix}
1111\\
22\phantom{1}\phantom{1}
\end{matrix}\right)
}}
+
\xx^{\cont{
\left(\begin{matrix}
1112\\
22\phantom{1}\phantom{1}
\end{matrix}\right)
}}
+
\xx^{\cont{
\left(\begin{matrix}
1122\\
22\phantom{1}\phantom{1}
\end{matrix}\right)
}}\\
&=s_{(4,2)}
=s_\lambda.
\end{aligned}
$$
\end{example}



Next divide through by $a_\rho$ on both sides of the theorem to give the following.
\begin{corollary}
For partitions $\lambda, \mu, \nu$ having at most $n$ parts, one has
\label{finite-L-R-rule}
$$
s_{\nu} s_{\lambda/\mu} = \sum_T s_{\nu+\cont(T)}
$$
where $T$ runs through the same set as in Theorem~\ref{Stembridge-theorem}.
In particular, taking $\nu=\varnothing$, 
$$
s_{\lambda/\mu} = \sum_T s_{\cont(T)}
$$
where in the sum $T$ runs through all column-strict tableaux of shape $\lambda/\mu$
for which each $\cont(T|_{\cols\geq j})$ is a partition.

%Rephrasing, this,  whenever, $\lambda, \mu, \nu$ all have at most $n$ parts
%the Littlewood-Richardson coefficient $c^\lambda_{\mu,\nu}$ is the
%number of such tableaux $T$ having $\cont(T)=\nu$.
\end{corollary}

By taking the number of variables $n$ sufficiently large, one deduces from
this last assertion the following version of the Littlewood-Richardson rule.
\begin{corollary}
\label{L-R-rule}
For partitions $\lambda, \mu, \nu$ (of any lengths),
the Littlewood-Richardson coefficient $c^\lambda_{\mu,\nu}$ counts
column-strict tableaux $T$ of shape $\lambda/\mu$ with $\cont(T)=\nu$ having
the property that each $\cont(T|_{\cols\geq j})$ is a partition.
\end{corollary}



\begin{proof}[Proof of Theorem~\ref{Stembridge-theorem}]
Start by rewriting the left side of the theorem, and using the fact that
$w(s_{\lambda/\mu})=s_{\lambda/\mu}$ for any $w$ in $\Symm_n$:
$$
\begin{aligned}
a_{\nu+\rho} s_{\lambda/\mu} 
&=\sum_{w \in \Symm_n} \sgn(w) \xx^{w(\nu+\rho)} 
     w(s_{\lambda/\mu}) \\
&=\sum_{w \in \Symm_n} \sgn(w) \xx^{w(\nu+\rho)} 
     \sum_{\substack{\text{column-strict }T\\\text{ of shape }\lambda/\mu}} \xx^{w(\cont(T))}  \\
&=\sum_{\substack{\text{column-strict }T\\\text{ of shape }\lambda/\mu}} 
         \qquad \sum_{w \in \Symm_n} \sgn(w) \xx^{w(\nu+\cont(T)+\rho)}  \\
&= \sum_{\substack{\text{column-strict }T\\\text{ of shape }\lambda/\mu}}
      a_{\nu+\cont(T)+\rho}
\end{aligned}
$$
We wish to cancel out all the summands indexed by column-strict tableaux $T$ which
fail any of the conditions that $\nu + \cont(T|_{\cols\geq j})$ be a partition.
Given such a $T$, find the maximal $j$ for which it fails this condition, and
then find the minimal $k$ for which
$$
\nu_k+\cont_k(T|_{\cols\geq j}) <
\nu_{k+1}+\cont_{k+1}(T|_{\cols\geq j}).
$$
Maximality of $j$ forces 
$$
\nu_k+\cont_k(T|_{\cols\geq j+1}) \geq
\nu_{k+1}+\cont_{k+1}(T|_{\cols\geq j+1}).
$$
Since column-strictness implies that column $j$ of $T$ can contain at most one
occurrence of $k$ or of $k+1$ (or neither or both), 
the previous two inequalities imply that
column $j$ must contain an occurrence of $k+1$ and no occurrence of $k$, so
that
$$
\nu_k+\cont_k(T|_{\cols\geq j}) + 1 
= \nu_{k+1}+\cont_{k+1}(T|_{\cols\geq j}).
$$
This implies that the adjacent transposition $t_{k,k+1}$ swapping $k$ and $k+1$
fixes the vector $\nu+\cont(T|_{\cols\geq j})+\rho$.

Now create a new tableau $T^*$ from $T$ by
applying the Bender-Knuth involution (from the proof of
Proposition~\ref{Schur-functions-are-symmetric-prop})
on letters $k,k+1$,
but {\it only to 
columns} $1,2,\ldots,j-1$ of $T$, leaving columns $j,j+1,j+2,\ldots$ unchanged.
One should check that $T^*$ is still column-strict, but this holds 
because column $j$ of $T$ has no occurrences of letter $k$.
Note that   
$$
t_{k,k+1} \cont(T|_{\cols\leq j-1}) =
              \cont(T^*|_{\cols\leq j-1})
$$
and hence
$$
t_{k,k+1} ( \nu+\cont(T)+\rho ) = \nu+\cont(T^*)+\rho
$$
so that $a_{\nu+\cont(T)+\rho} = - a_{\nu+\cont(T^*)+\rho}$.

Because $T, T^*$ have exactly the same columns  $j,j+1,j+2,\ldots$,
the tableau $T^*$ is also a violator of at least one of the conditions that 
$\nu + \cont(T^*|_{\cols\geq j})$ be a partition, and 
has the same choice of maximal $j$ and minimal $k$ as did $T$.
Hence the map $T \mapsto T^*$ is an involution on the violators
that lets one cancel their summands 
$a_{\nu+\cont(T)+\rho}$ and $a_{\nu+\cont(T^*)+\rho}$ in pairs.
\end{proof}





\subsection{The Pieri and Assaf-McNamara skew Pieri rule}

The classical Pieri rule refers to two special cases of the Littlewood-Richardson rule.
To state them, recall that a skew shape with $n$ cells
is called a {\it horizontal (resp. vertical) $n$-strip}
if no two of its cells lie in the same column (resp. row).

\begin{theorem}
\label{Pieri-rules}
$$
\begin{aligned}
s_\lambda h_n &= 
\sum_{\substack{\lambda^+: \lambda^+/\lambda \text{ is a }\\\text{horizontal }n\text{-strip}}}  s_{\lambda^+} \\
s_\lambda e_n &= 
\sum_{\substack{\lambda^+: \lambda^+/\lambda \text{ is a }\\\text{vertical }n\text{-strip}}}  s_{\lambda^+}
\end{aligned}
$$
\end{theorem}

\begin{example}
$$
\begin{aligned}
&\begin{matrix}
s &   &   & \\
  &\sq&\sq&\sq\\
  &\sq&\sq&   \\
  &\sq&\sq&
\end{matrix}
\qquad \bullet \qquad
\begin{matrix}
h &   &   \\
  &\sq&\sq \\
  &   & \\
  &   & \\
\end{matrix}
\\
 & \\
&=\qquad
\begin{matrix}
s &   &   & \\
  &\sq&\sq&\sq\\
  &\sq&\sq&   \\
  &\sq&\sq& \\
  &\bsq&\bsq& 
\end{matrix}
+\begin{matrix}
s &   &   & \\
  &\sq&\sq&\sq\\
  &\sq&\sq&\bsq\\
  &\sq&\sq& \\
  &\bsq&  &
\end{matrix}
+\begin{matrix}
s &   &   &   &\\
  &\sq&\sq&\sq&\bsq\\
  &\sq&\sq&   &\\
  &\sq&\sq&   &\\
  &\bsq&  &   &
\end{matrix} 
\\
& \\
&\qquad \qquad \qquad 
+\begin{matrix}
s &   &   &   &\\
  &\sq&\sq&\sq&\bsq\\
  &\sq&\sq&\bsq &\\
  &\sq&\sq&   &\\
\end{matrix}
+\begin{matrix}
s &   &   &   &    &\\
  &\sq&\sq&\sq&\bsq&\bsq\\
  &\sq&\sq&   &    &\\
  &\sq&\sq&   &    &\\
\end{matrix}
\end{aligned}
$$
\end{example}

\begin{proof}
For the first Pieri formula involving $h_n$, as $h_n=s_{(n)}$ one has
$$
s_\lambda h_n = \sum_{\lambda^+} c^{\lambda^+}_{\lambda,(n)} s_{\lambda^+}.
$$
Corollary~\ref{L-R-rule} says $c^{\lambda^+}_{\lambda,(n)}$ counts
column-strict tableaux $T$ of shape $\lambda^+/\lambda$ 
having $\cont(T)=(n)$ (i.e. all entries of $T$ are $1$'s),
with an extra condition.  Since its entries are all equal,
such a $T$ must certainly have shape being a horizontal strip.  Conversely
for any horizontal strip, there is a unique such filling,
and it will trivially satisfy the extra condition 
that $\cont(T|_{\cols\geq j})$ is a partition for each $j$. 
Hence $c^{\lambda^+}_{\lambda,(n)}$ is $1$ if $\lambda^+/\lambda$ is
a horizontal $n$-strip, and $0$ else.

For the second Pieri formula involving $e_n$, using $e_n=s_{(n)}$ one has
$$
s_\lambda e_n = \sum_{\lambda^+} c^{\lambda^+}_{\lambda,(1^n)} s_{\lambda^+}.
$$
Corollary~\ref{L-R-rule} says $c^{\lambda^+}_{\lambda,(1^n)}$
counts column-strict tableaux $T$ of shape $\lambda^+/\lambda$ 
having $\cont(T)=(1^n)$, so its entries are $1,2,\ldots,n$ each
occurring once, with the extra condition that $1,2,\ldots,n$
appear from right-to-left.  Together with
the tableau condition, this forces at most one entry in each row, that is
$\lambda^+/\lambda$ is a vertical strip, and then there is a unique
way to fill it  maintaining column-strictness.  Thus $c^{\lambda^+}_{\lambda,(1^n)}$ 
is $1$ if $\lambda^+/\lambda$ is a vertical $n$-strip, and $0$ else.
\end{proof}


Assaf and McNamara \cite{AssafMcNamara} recently proved an elegant generalization.

\begin{theorem}
\label{Assaf-McNamara-skew-Pieri}
$$
\begin{aligned}
s_{\lambda/\mu} h_n &= 
\sum_{\substack{\lambda^+,\mu^-:\\ 
                  \lambda^+/\lambda \text{ a horizontal strip}\\
                   \mu/\mu^{-}\text{ a vertical strip}\\
                   |\lambda^+/\lambda|+|\mu/\mu^-|=n}} 
           (-1)^{|\mu/\mu^{-}|} s_{\lambda^+/\mu^-} \\
s_{\lambda/\mu} e_n &= 
\sum_{\substack{\lambda^+,\mu^-: \\
                 \lambda^+/\lambda \text{ a vertical strip}\\
                    \mu/\mu^{-}\text{ a horizontal strip}\\
                    |\lambda^+/\lambda|+|\mu/\mu^-|=n}} 
           (-1)^{|\mu/\mu^{-}|} s_{\lambda^+/\mu^-}
\end{aligned}
$$
\end{theorem}

\begin{example}
$$
\begin{aligned}
&\begin{matrix}
s &   &   & \\
  &   &\sq&\sq\\
  &   &\sq&   \\
  &\sq&\sq&
\end{matrix}
\qquad \bullet \qquad
\begin{matrix}
h &   &   \\
  &\sq&\sq \\
  &   & \\
  &   & \\
\end{matrix}
\\
 & \\
&=\qquad
\begin{matrix}
s &   &   & \\
  &   &\sq&\sq\\
  &   &\sq&   \\
  &\sq&\sq& \\
  &\bsq&\bsq& 
\end{matrix}
\qquad+\qquad
\begin{matrix}
s &   &   & \\
  &   &\sq&\sq\\
  &   &\sq&\bsq\\
  &\sq&\sq& \\
  &\bsq&  &
\end{matrix}
\qquad+\qquad
\begin{matrix}
s &   &   &   &\\
  &   &\sq&\sq&\bsq\\
  &   &\sq&   &\\
  &\sq&\sq&   &\\
  &\bsq&  &   &
\end{matrix} 
\\
& \\
&\qquad \qquad \qquad 
+\qquad
\begin{matrix}
s &   &   &   &\\
  &   &\sq&\sq&\bsq\\
  &   &\sq&\bsq &\\
  &\sq&\sq&   &\\
\end{matrix}
\qquad+\qquad\begin{matrix}
s &   &   &   &    &\\
  &   &\sq&\sq&\bsq&\bsq\\
  &   &\sq&   &    &\\
  &\sq&\sq&   &    &\\
\end{matrix}
\\
 & \\
&\qquad \qquad
-\quad
\begin{matrix}
s &    &   &   \\
  &    &\sq&\sq\\
  &\bsq&\sq&   \\
  &\sq &\sq&   \\
  &\bsq&   &
\end{matrix}
\qquad-\qquad
\begin{matrix}
s &    &   &   \\
  &    &\sq&\sq\\
  &\bsq&\sq&\bsq\\
  &\sq &\sq&   \\
\end{matrix}
\qquad-\qquad
\begin{matrix}
s &    &   &   &\\
  &    &\sq&\sq&\bsq\\
  &\bsq&\sq&   &\\
  &\sq &\sq&   &
\end{matrix}
\\
 & \\
&\qquad\qquad\qquad\qquad
+\qquad
\begin{matrix}
s &    &   &   \\
  &\bsq&\sq&\sq\\
  &\bsq&\sq&   \\
  &\sq &\sq&   
\end{matrix}
\end{aligned} 
$$
\end{example}

\noindent
Theorem~\ref{Assaf-McNamara-skew-Pieri} is proven
in the next section, using an important Hopf algebra tool.


\subsection{Skewing and Lam's proof of the skew Pieri rule}

We codify here the operation $s_\mu^\perp$ of {\it skewing by $s_\mu$},
acting on Schur functions via
$$
s_\mu^\perp(s_\lambda) = s_{\lambda/\mu}
$$
where one defines $s_{\lambda/\mu}=0$ if $\mu \not\subseteq \lambda$.
These operations play a crucial role 
\begin{itemize}
\item
in Lam's proof of the
skew Pieri rule, 
\item in Lam, Lauve, and Sottile's proof \cite{LamLauveSottile} 
of a more general  skew Littlewood-Richardson rule 
that had been conjectured by Assaf and McNamara, and
\item in Zelevinsky's structure theory of PSH algebras to be developed in the next chapter.
\end{itemize}

\begin{definition}
Given a Hopf algebra $A$ of finite type, and its (restricted) dual $A^o$, let 
$(\cdot,\cdot)=(\cdot,\cdot)_A$ be 
the pairing $(f,a):=f(a)$ for $f$ in $A^o$ and $a$ in $A$.
Then define for each
$f$ in $A^o$ an operator 
$A \overset{f^\perp}{\rightarrow} A$ as follows\footnote{This $f^\perp(a)$ is
called $a \leftharpoonup f$ in Montgomery \cite[Example 1.6.5]{Montgomery}}:  for $a$ in $A$
with $\Delta(a) = \sum a_1 \otimes a_2$, let
$$
f^\perp(a) = \sum (f,a_1) a_2.
$$
\end{definition}

\noindent
Note that when one takes $A=\Lambda=A^o$, the element
$a=s_\lambda$ has $\Delta s_\lambda = \sum_\mu s_\mu \otimes s_{\lambda/\mu}$.
Hence if $f=s_\mu$, then one has $f^\perp(a) = s_{\lambda/\mu}=s_\mu^\perp(s_\lambda)$
as desired.

\begin{proposition}
\label{skewing-properties-prop}
The $f^\perp$ operators $A \rightarrow A$ have the following properties.
\begin{enumerate}
\item[(i)] $f^\perp$ is adjoint to left multiplication $A^o \overset{f \cdot}{\rightarrow} A^o$ in the sense that
$$
(g, f^\perp(a) ) = (fg, a).
$$
\item[(ii)] $(fg)^\perp(a) = g^\perp(f^\perp(a))$, that is,
$A$ becomes a right $A^o$-module via the $f^\perp$ action.
\item[(iii)] If $\Delta(f) = \sum f_1 \otimes f_2$, then
$$
f^\perp (a b) = \sum f_1^\perp(a) f_2^\perp(b).
$$
In particular, if $f$ is primitive in $A^o$, 
so that $\Delta(f) = f \otimes \one + \one \otimes f$,
then $f^\perp$ is a {\bf derivation}:
$$
f^\perp (a b) = f^\perp(a) \cdot b + a \cdot f^\perp(b).
$$
\end{enumerate}
\end{proposition}
\begin{proof}
For (i), note that
$$
(g, f^\perp(a) )
=\sum (f,a_1) (g,a_2)
=( f \otimes g , \Delta_A (a) )
=( m_{A^o}(f \otimes g) , a )
= (fg, a).
$$
For (ii), using (i) and considering any $h$ in $A^o$, one has that
$$
(h, (fg)^\perp(a)) 
= (fgh, a)
= (gh, f^\perp(a))
= (h, g^\perp(f^\perp(a)).
$$
For (iii), noting that
$$
\Delta(ab)= \Delta(a)\Delta(b) 
= \left( \sum_{(a)} a_1 \otimes a_2 \right)
\left( \sum_{(b)} b_1 \otimes b_2 \right)
= \sum_{(a),(b)} a_1 b_1 \otimes a_2 b_2,
$$
one has that
$$
\begin{aligned}
f^\perp(ab) 
&= \sum_{(a),(b)} (f,a_1 b_1)_A \,\, a_2 b_2 
= \sum_{(a),(b)} (\Delta(f),a_1 \otimes b_1)_{A \otimes A} \,\, a_2 b_2 \\
&= \sum_{(f),(a),(b)} (f_1,a_1)_A (f_2,b_1)_A \,\, a_2 b_2 \\
&= \sum_{(f)} \left( \sum_{(a)} (f_1,a_1)_A a_2 \right)  \left( \sum_{(b)} (f_2,b_1)_A  b_2 \right)
=\sum_{(f)} f_1^\perp(a) f_2^\perp(b).
\end{aligned}
$$

%considering any $g$ in $A^o$, one has that
%$$
%\begin{aligned}
%(g,f^\perp(ab))
%&= (fg, ab)_A \\
%&= (fg, m_A(a \otimes b))_A \\
%&= ( \Delta_{A^o}(fg), a \otimes b )_{A \otimes A}\\
%&= ( \Delta_{A^o}(f) \Delta_{A^o}(g), a \otimes b )_{A \otimes A}\\
%&= \left( \left( \sum f_1 \otimes f_2 \right) \Delta_{A^o}(g), a \otimes b \right)_{A \otimes A}\\
%&= \left(  \Delta_{A^o}(g), \sum f_1^\perp(a) \otimes f_2^\perp(b) \right)_{A \otimes A}\\
%&= \left(  g, \sum f_1^\perp(a) f_2^\perp(b) \right)_A.\\
%\end{aligned}
%$$
\end{proof}

\noindent
The following interaction between multiplication and $h^\perp$ is the key to
deducing the skew Pieri formula from the usual Pieri formulas.
\begin{lemma}
\label{Lam's-lemma}
For any $f,g$ in $\Lambda$, one has
$$
f \cdot h_n^\perp(g) 
= \sum_{k=0}^n (-1)^k h_{n-k}^\perp(e_k^\perp(f) \cdot g).
$$
\end{lemma}
\begin{proof}
Starting with the right side, first apply 
Proposition~\ref{skewing-properties-prop}(iii):
$$
\begin{aligned}
&\sum_{k=0}^n (-1)^k h_{n-k}^\perp(e_k^\perp(f) \cdot g)\\
&=\sum_{k=0}^n (-1)^{k} \sum_{j=0}^{n-k} 
                       h_j^\perp(e_k^\perp(f)) 
                      \cdot h_{n-k-j}^\perp(g)\\
&=\sum_{i=0}^n (-1)^{n-i} 
       \left( \sum_{j=0}^{n-i} 
                      (-1)^j h_j^\perp(e_{n-i-j}^\perp(f)) \right)
                      \cdot h_i^\perp(g) \quad (\text{ reindexing }i:=n-k-j\,\,) \\
&=\sum_{i=0}^n (-1)^{n-i} 
       \left( \sum_{j=0}^{n-i} (-1)^j e_{n-i-j} h_j\right)^\perp \left( f \right) 
                      \cdot h_i^\perp(g) \quad 
                       (\text{ by Proposition~\ref{skewing-properties-prop}(ii)}\,\,) \\
&=1^\perp(f) \cdot h_n^\perp(g) = f \cdot h_n^\perp(g)
\end{aligned}
$$
where the second-to-last equality used \eqref{e-h-relation}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{Assaf-McNamara-skew-Pieri}]
We prove the first skew Pieri rule; the second is analogous, swapping 
$h_i \leftrightarrow e_i$ and swapping the words ``vertical'' $\leftrightarrow$ ``horizontal''.
Symmetry of $(\cdot,\cdot)_\Lambda$, commutativity of $\Lambda$, and
Proposition~\ref{skewing-properties-prop}(i) imply for $f$ in $\Lambda$,
\begin{equation}
\label{skewing-versus-multiplication-in-Sym}
\left( s_{\lambda/\mu}, f \right) =
\left( s_\mu^\perp(s_\lambda),f \right) =
\left( s_\lambda ,s_\mu f \right)
\end{equation}
Hence for any $g$ in $\Lambda$, one can compute that
\begin{equation}
\label{Lam-proof-calculation}
\begin{aligned}
( h_n s_{\lambda/\mu} \,\, ,\,\, g )
&\underset{~\ref{skewing-properties-prop}(i)}{\overset{Prop.}{=}}
( s_{\lambda/\mu} \,\, ,\,\, h_n^\perp g ) 
\overset{\eqref{skewing-versus-multiplication-in-Sym}}{=} 
 ( s_{\lambda}\,\, , \,\, s_\mu \cdot h_n^\perp g ) \\
&\underset{\ref{Lam's-lemma}}{\overset{Lemma}{=}}
 \sum_{k=0}^n (-1)^k ( s_\lambda \,\, , \,\,
                 h_{n-k}^\perp (e_k^\perp(s_\mu) \cdot g ) ) \\
&\underset{~\ref{skewing-properties-prop}(i)}{\overset{Prop.}{=}}
\sum_{k=0}^n (-1)^k ( h_{n-k} s_\lambda \,\, , e_k^\perp(s_\mu) \cdot g )
\end{aligned}
\end{equation}
The first Pieri rule in Theorem~\ref{Pieri-rules} lets one rewrite
$h_{n-k} s_\lambda = \sum_{\lambda^+} s_{\lambda^+}$, with the sum running through
$\lambda^+$ for which $\lambda^+/\lambda$ is a horizontal $(n-k)$-strip.
The second Pieri rule in Theorem~\ref{Pieri-rules} lets one rewrite
$e_k^\perp s_\mu = \sum_{\mu^-} s_{\mu^-}$, with the sum running through
$\mu^-$ for which $\mu/\mu^-$ is a vertical $k$-strip, since
$(s_{\mu^-},e_k^\perp s_\mu) = (e_k s_{\mu^-}, s_\mu)$.
Thus the last line of \eqref{Lam-proof-calculation} becomes
$$
\sum_{k=0}^n (-1)^k \left( \sum_{\lambda^+}  s_{\lambda^+} \,\, , \,\, 
                         \sum_{\mu^-} s_{\mu^-} \cdot g \right) \\
\overset{\eqref{skewing-versus-multiplication-in-Sym}}{=}
\left( \sum_{k=0}^n  (-1)^k \sum_{(\lambda^+,\mu^-)}  s_{\lambda^+/\mu^-} , g \right)
$$
where the sum is over the pairs $(\lambda^+, \mu^-)$
for which $\lambda^+/\lambda$ is a horizontal $(n-k)$-strip and
$\mu/\mu^-$ is a vertical $k$-strip.
\end{proof}


\begin{exercise}(due to D. Grinberg.)
\label{Darij's-skewing-exercise}
The goal of this exercise is to prove
\eqref{skew-Schur-antipodes} using the skewing operators that
we have developed.
\newline
(a) Show that $\omega\left(  p_{\lambda
}\right)  =\left(  -1\right)  ^{\left\vert \lambda\right\vert -\ell\left(
\lambda\right)  }p_{\lambda}$ for any $\lambda\in\operatorname*{Par}$, where
$\ell\left(  \lambda\right)  $ denotes the length of the partition $\lambda$.
\newline
(b) Recall the involution $\omega:\Lambda\rightarrow\Lambda$ defined in
the proof of Proposition 2.18.  Show that $\omega$ is an isometry.
\newline
(c) Show that this same map $\omega:\Lambda\rightarrow\Lambda$ is a 
Hopf automorphism.
\newline
(d) Prove that $\omega\left(  a^{\perp}b\right)  =\left(
\omega\left(  a\right)  \right)  ^{\perp}\left(  \omega\left(  b\right)
\right)  $ for every $a\in\Lambda$ and $b\in\Lambda$.
\newline
(e) For any partition $\lambda=(\lambda_1,\ldots,\lambda_\ell)$ 
with length $\ell(\lambda)=\ell$, prove that 
$$
e_{\ell}^{\perp
}s_{\lambda}=s_{\left(  \lambda_{1}-1,\lambda_{2}-1,...,\lambda_{\ell
}-1\right)  }.
$$
\noindent
(f) For any partition 
$\lambda=(\lambda_1,\lambda_2,\ldots)$ , prove that 
$$
h_{\lambda_{1}}^{\perp
}s_{\lambda}=s_{\left(  \lambda_{2},\lambda_{3},\lambda_{4},...\right)  }.
$$
\noindent
(g) Prove \eqref{skew-Schur-antipodes}.
\end{exercise}

\begin{comment}
Darij's solution.
I hope that the solution of this exercise is mostly straightforward given the
order of the parts. The solution of \textbf{(a)} only uses the primitivity of
$p_{n}$ (and Proposition 1.31). Then, \textbf{(b)} follows from \textbf{(a)}
and the fact that $\left\{  \dfrac{p_{\lambda}}{\sqrt{z_{\lambda}}}\right\}
_{\lambda\in\operatorname*{Par}}$ is an orthonormal basis of $\Lambda
_{\mathbb{R}}$. For \textbf{(c)}, one only needs to check that $\omega$ is a
coalgebra homomorphism, which is easily checked on the generators $h_{n}$ and
thus holds everywhere. Then, \textbf{(d)} follows from \textbf{(b)} and
\textbf{(c)}. Parts \textbf{(e)} and \textbf{(f)} follow from the fact that
$s_{\mu}^{\perp}\left(  s_{\lambda}\right)  =s_{\lambda\diagup\mu}$, and the
fact that a parallel translation of a skew Ferrers shape doesn't change the
corresponding skew Schur function. The only nontrivial part is \textbf{(g)}.
It should be clear that $S\left(  s_{\lambda\diagup\mu}\right)  =\left(
-1\right)  ^{\left\vert \lambda\diagup\mu\right\vert }s_{\lambda^{t}\diagup
\mu^{t}}$ follows from $\omega\left(  s_{\lambda\diagup\mu}\right)
=s_{\lambda^{t}\diagup\mu^{t}}$, and this, in turn, follows from \textbf{(d)}
if one knows that $\omega\left(  s_{\lambda}\right)  =s_{\lambda^{t}}$ for
every partition $\lambda$.

So it remains to prove that $\omega\left(  s_{\lambda}\right)  =s_{\lambda
^{t}}$ for every partition $\lambda$. To do so, we use induction over
$\left\vert \lambda\right\vert $, and fix $\lambda$. Part \textbf{(b)} yields
$\omega\left(  s_{\lambda}\right)  =\pm s_{\nu}$ for some partition $\nu$ (a
foretaste of Section 3). Hence, $\omega\left(  s_{\nu}\right)  =\pm
s_{\lambda}$. Writing $\lambda$ as $\left(  \lambda_{1},\lambda_{2}%
,...,\lambda_{\ell}\right)  $ with all of $\lambda_{1}$, $\lambda_{2}$, $...$,
$\lambda_{\ell}$ being positive, and writing $\nu$ as $\left(  \nu_{1},\nu
_{2},\nu_{3},...\right)  $, we easily see (using \textbf{(d)} and
\textbf{(f)}) that $\omega\left(  e_{\nu_{1}}^{\perp}s_{\lambda}\right)  =\pm
h_{\nu_{1}}^{\perp}\left(  s_{\nu}\right)  =\pm s_{\left(  \nu_{2},\nu_{3}%
,\nu_{4},...\right)  }\neq0$, so that $e_{\nu_{1}}^{\perp}s_{\lambda}\neq0$,
and therefore $\ell\geq\nu_{1}$. On the other hand, \textbf{(d)} and
\textbf{(e)} yield $\omega\left(  h_{\ell}^{\perp}s_{\nu}\right)  =\pm
e_{\ell}^{\perp}\left(  s_{\lambda}\right)  =\pm s_{\left(  \lambda
_{1}-1,\lambda_{2}-1,...,\lambda_{\ell}-1\right)  }\neq0$, thus $h_{\ell
}^{\perp}s_{\nu}\neq0$, so that $\ell\leq\nu_{1}$. Combined with $\ell\geq
\nu_{1}$, this yields $\ell=\nu_{1}$. Hence, $h_{\ell}^{\perp}s_{\nu}%
=h_{\nu_{1}}^{\perp}s_{\nu_{1}}=s_{\left(  \nu_{2},\nu_{3},\nu_{4},...\right)
}$ (by \textbf{(f)}), so that%
\[
\omega\left(  h_{\ell}^{\perp}s_{\nu}\right)  =\omega\left(  s_{\left(
\nu_{2},\nu_{3},\nu_{4},...\right)  }\right)  =s_{\left(  \nu_{2},\nu_{3}%
,\nu_{4},...\right)  ^{t}}%
\]
by the induction assumption. Compared with%
\[
\omega\left(  h_{\ell}^{\perp}s_{\nu}\right)  =\pm e_{\ell}^{\perp}\left(
s_{\lambda}\right)  =\pm s_{\left(  \lambda_{1}-1,\lambda_{2}-1,...,\lambda
_{\ell}-1\right)  }%
\]
(a consequence of \textbf{(d)} and \textbf{(e)}), this yields $s_{\left(
\nu_{2},\nu_{3},\nu_{4},...\right)  ^{t}}=\pm s_{\left(  \lambda_{1}%
-1,\lambda_{2}-1,...,\lambda_{\ell}-1\right)  }$, from which it directly
follows that $\pm=+$ and $\left(  \nu_{2},\nu_{3},\nu_{4},...\right)
^{t}=\left(  \lambda_{1}-1,\lambda_{2}-1,...,\lambda_{\ell}-1\right)  $. This
quickly yields $\nu=\lambda^{t}$, so that $\omega\left(  s_{\lambda}\right)
=\pm s_{\nu}$ becomes $\omega\left(  s_{\lambda}\right)  =s_{\lambda^{t}}$
(the $\pm$ in $\omega\left(  s_{\lambda}\right)  =\pm s_{\nu}$ is the same as
in $s_{\left(  \nu_{2},\nu_{3},\nu_{4},...\right)  ^{t}}=\pm s_{\left(
\lambda_{1}-1,\lambda_{2}-1,...,\lambda_{\ell}-1\right)  }$), so we are done.
\end{comment}

\newpage



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Zelevinsky's structure theory of positive self-dual Hopf algebras}
\label{PSH-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Section~\ref{Sym-section} showed that, 
as a $\ZZ$-basis for the Hopf algebra $\Lambda=\Lambda_\ZZ$,
the Schur functions $\{s_\lambda\}$ have two special properties:
they have the {\it same} structure constants $ c^{\lambda}_{\mu,\nu}$
for their multiplication as for their comultiplication (Corollary~\ref{self-duality-corollary}), and these structure constants 
are all {\it nonnegative} integers (Corollary~\ref{L-R-rule}).
Zelevinsky \cite[\S 2,3]{Zelevinsky} isolated these two properties as crucial.


\begin{definition}
Say that a graded connected Hopf algebra $A$ over $\kk=\ZZ$ with a 
distinguished $\ZZ$-basis  $\{\sigma_\lambda\}$ consisting of homogeneous
elements is a {\it positive self-dual Hopf algebra} (or PSH)
if it satisfies the two further axioms
\begin{itemize}
\item{\bf (self-duality)} The same structure constants
$a^{\lambda}_{\mu,\nu}$ appear for the product
$\sigma_\mu \sigma_\nu = \sum_{\lambda} a^{\lambda}_{\mu,\nu} \sigma_\lambda$ and
the coproduct 
$\Delta \sigma_\lambda = \sum_{\mu,\nu} a^{\lambda}_{\mu,\nu} \sigma_\mu \otimes \sigma_\nu$.
\item{\bf (positivity)} The $a^\lambda_{\mu,\nu}$ are all nonnegative (integers).
\end{itemize}
Call $\{\sigma_\lambda\}$ the {\it PSH-basis} of $A$.
\end{definition}
He then developed a beautiful structure theory for PSH-algebras, explaining
how they can be uniquely expressed as tensor products of copies of PSH's each
isomorphic to $\Lambda$ after rescaling their grading.
The next few sections explain this, following his exposition closely.

\subsection{Self-duality implies polynomiality}
\label{PSH-implies-polynomial-section}

We begin with a property that forces a Hopf algebra to have
algebra structure which is a {\it polynomial} algebra, specifically the
symmetric algebra $\Sym(\Liep)$, where $\Liep$ is the subspace of primitive elements.

Recall from Exercise~\ref{graded-connected-exercise}
that for a graded connected Hopf algebra $A = \bigoplus_{n=0}^\infty A_n$, 
every $x$ in the two-sided ideal 
$
I:=\ker \epsilon = \bigoplus_{n > 0} A_n
$
has the property that its comultiplication takes the form
$$
\Delta(x) = 1 \otimes x + x \otimes 1 + \Delta_+(x)
$$
where $\Delta_+(x)$ lies in $I \otimes I$.
Recall also that the elements $x$ for which $\Delta_+(x)=0$ are
called the {\it primitives}.  Denote by $\Liep$ the $\ZZ$-submodule
of primitive elements inside $A$.

Given a Hopf algebra $A$ over $\kk=\ZZ$ with a PSH-basis $\{\sigma_\lambda\}$, 
we identify $A^o$ with $A$ via the bilinear  
form $(\cdot,\cdot)_A$ on $A$ that makes this basis orthonormal.
Similarly, the elements $\{\sigma_\lambda \otimes\sigma_\mu\}$ give
an orthonormal basis for a form $(\cdot,\cdot)_{A \otimes A}$ on $A \otimes A$. 
This is an instance of the following notion of self-duality.

\begin{definition}
Say that a bialgebra $A$ is {\it self-dual} with respect to a 
given symmetric bilinear form $(\cdot, \cdot)$ if 
one has $(a, m(b \otimes c))_A = (\Delta(a) , b \otimes c)$
and $(\one_A,a)=\epsilon(a)$ for $a,b,c$ in $A$.
If $A$ is a graded Hopf algebra of finite type
then this is equivalent to the $\kk$-module map $A \rightarrow A^o$
induced by $(\cdot, \cdot)_A$ giving a Hopf algebra isomorphism. 
\end{definition}

\begin{proposition}
\label{primitives-split-off-prop}
Let $A$ be a Hopf algebra over $\kk=\ZZ$ or $\kk=\QQ$ which is graded, connected, and self-dual
with respect to a positive definite graded\footnote{That is,
$(A_i,A_j)=0$ for $i \neq j$.}  bilinear form.  
Then within the ideal $I$,
the subspace of primitives $\Liep$ is the orthogonal complement to the subspace $I^2$.
In particular, $\Liep \cap I^2=0$, and when $\kk=\QQ$, one has $I = \Liep \oplus I^2$.
\end{proposition}
\begin{proof}
Note that $I^2=m(I \otimes I)$.  Hence an element $x$ in $I$ lies in the perpendicular
space to $I^2$ if and only if one has for all $y$ in $I \otimes I$ that
$$
0= ( x , m(y) )_A = (\Delta(x), y)_{A \otimes A} = (\Delta_+(x), y)_{A \otimes A}
$$ 
where the second equality uses self-duality, while the third equality uses
the fact that $y$ lies in $I \otimes I$ and the form $(\cdot,\cdot)_{A \otimes A}$ 
makes distinct homogeneous components orthogonal.  
Since $y$ was arbitrary, this means $x$ is perpendicular
to $I^2$ if and only if $\Delta_+(x)=0$, that is, $x$ lies in $\Liep$.
\end{proof}

\begin{remark}
One might wonder why we didn't just say $I = \Liep \oplus I^2$ even when $\kk=\ZZ$
in Proposition~\ref{primitives-split-off-prop}.
However, this is false\footnote{Thanks to Darij Grinberg for pointing this out to me.},
even for $A=\Lambda_\ZZ$:
the second homogeneous component $(\Liep \oplus I^2)_2$ is the 
index $2$ sublattice of $\Lambda_2$ which is $\ZZ$-spanned by
$\{p_2,e_1^2\}$, containing $2e_2$, but not containing $e_2$ itself.
\end{remark}

Already the fact that $\Liep \cap I^2=0$ has a strong implication.

\begin{lemma}
\label{general-commutativity-lemma}
A graded connected Hopf algebra $A$ over any ring $\kk$ having $\Liep \cap I^2=0$
has algebra structure which is commutative.
\end{lemma}
\begin{proof}
The component $A_0 = \kk$ commutes with all of $A$.  This forms the base case for
an induction on $i+j$ in which one shows that any elements $x$ in $A_i$ and $y$ in $A_j$ 
with $i,j>0$ will have $[x,y]:=xy-yx = 0$.  
Since $[x,y]$ lies in $I^2$, it suffices to show that
$[x,y]$ also lies in $\Liep$:
$$
\begin{aligned}
\Delta[x,y] 
&= [\Delta(x),\Delta(y)]\\
&= [1 \otimes x + x \otimes 1+\Delta_+(x),
    1 \otimes y + y \otimes 1+\Delta_+(y)]\\
&= [1 \otimes x + x \otimes 1,1 \otimes y + y \otimes 1] \\
&\qquad   +[1 \otimes x + x \otimes 1,\Delta_+(y)]
   +[\Delta_+(x), 1 \otimes y + y \otimes 1]
   +[\Delta_+(x),\Delta_+(y)]\\
&=[1 \otimes x + x \otimes 1,1 \otimes y + y \otimes 1] \\
&=1 \otimes [x,y] + [x,y] \otimes 1
\end{aligned}
$$
showing that $[x,y]$ lies in $\Liep$.
Here the second-to-last equality used the inductive hypotheses:
 homogeneity implies that $\Delta_+(x)$ is a sum of 
homogeneous tensors of the form $z_1 \otimes z_2$
satisfying $\deg (z_1 ),\deg (z_2 ) < i$, so that by 
induction they will commute with $\one \otimes y, y \otimes \one$, 
thus proving that $[\Delta_+ (x) , \one \otimes y + y \otimes \one] = 0$; 
a symmetric argument shows $[\one \otimes x+x \otimes \one,\Delta_+ (y)] = 0$,
and, a similar argument shows $[\Delta_+ (x) , \Delta_+ (y)] = 0$.
The last equality is an easy calculation, and was done already in
\eqref{commutator-of-primitives}.
\end{proof}

\begin{remark}
Zelevinsky actually shows \cite[Proof of A.1.3,p. 150]{Zelevinsky} that the 
self-duality assumption (along with hypotheses of unit, counit, graded, connected,
and $\Delta$ being a morphism for multiplication)
already implies the {\it associativity} of the multiplication in $A$ ! 
One shows by induction on $i+j+k$ that any $x,y,z$ in $A_i,A_j,A_k$ 
with $i,j,k>0$ have vanishing {\it associator} 
$\assoc(x,y,z) := x(yz) - (xy)z$.  In the inductive step, one first
notes that $\assoc(x,y,z)$ lies in $I^2$, and then checks 
that $\assoc(x,y,z)$ also lies in $\Liep$, by a calculation very similar to the one above,
repeatedly using the fact that $\assoc(x,y,z)$ is multilinear in its three arguments.
\end{remark}

This leads to a general structure theorem.
\begin{theorem}
\label{general-structure-theorem}
If a graded, connected Hopf algebra $A$ over 
a field $\kk$ of characteristic zero has 
$I=\Liep \oplus I^2$,
then the inclusion $\Liep \hookrightarrow A$ extends to a
Hopf algebra isomorphism from the symmetric algebra 
$\Sym_\kk(\Liep) \rightarrow A$.  In particular, $A$ is
both commutative and cocommutative.
\end{theorem}

\noindent
Note that these hypotheses are valid, using Proposition~\ref{primitives-split-off-prop},
whenever $A$ is obtained from a PSH (over $\ZZ$) by tensoring with $\QQ$. 

\begin{proof}
Since Lemma~\ref{general-commutativity-lemma} implies that $A$ is commutative,
the universal property of $\Sym_\kk(\Liep)$ as a free commutative algebra
on generators $\Liep$ shows that the inclusion 
$\Liep \hookrightarrow A$ at least extends to an algebra morphism 
$\Sym_\kk(\Liep) \overset{\varphi}{\rightarrow} A$.
Since the Hopf structure on $\Sym_\kk(\Liep)$
makes the elements of $\Liep$ primitive (see
Example~\ref{symmetric-algebra-as-bialgebra-example}),
this $\varphi$ is actually a coalgebra morphism 
(since $\Delta \circ \varphi = (\varphi \otimes \varphi) \circ \Delta$ 
and $\epsilon \circ \varphi =\epsilon$ 
need only to be checked on algebra generators), 
hence a bialgebra morphism, 
hence a Hopf algebra morphism 
(by Proposition~\ref{antipodes-give-convolution-inverses} (c)).
It remains to show that $\varphi$ is surjective, and injective.

For the surjectivity of $\varphi$, note that the hypothesis 
$I=\Liep \oplus I^2$ implies that the composite $\Liep \hookrightarrow I \rightarrow I/I^2$ 
gives a $\kk$-vector space isomorphism.
What follows is a standard argument
to deduce that $\Liep$ generates $A$ as a commutative graded $\kk$-algebra.
One shows by induction 
on $n$ that any homogeneous element $a$ in $A_n$ lies in the 
$\kk$-subalgebra generated 
by $\Liep$.  The base case $n=0$ is trivial as $a$ lies in $\kk$.  
In the inductive step where $a$ lies in $I$, write $a \equiv p \bmod{I^2}$ 
for some $p$ in $\Liep$.  Thus $a=p + \sum_i b_i c_i$,
where $b_i,c_i$ lie in $I$ but have strictly smaller degree, 
so that by induction they lie in the 
subalgebra generated by $\Liep$, and hence so does $a$.

Note that the surjectivity argument did not use the assumption that
$\kk$ has characteristic zero, but we will now use it in the injectivity
argument for $\varphi$, to establish the following 
\begin{equation}
\label{primitives-in-symmetric-algebra}
\text{{\bf Claim:} Every primitive element of $\Sym(\Liep)$ lies in $\Liep=\Sym^1(\Liep)$.}
\end{equation}
Note that this claim fails in positive characteristic, e.g. if $\kk$ has characteristic $2$
then $x^2$ lies in $\Sym^2(\Liep)$, however
$$
\Delta (x^2) = 1 \otimes x^2 + 2 x \otimes x + x^2 \otimes 1 
= 1 \otimes x^2  + x^2 \otimes 1.
$$
To see the claim, assume not, so that by gradedness, there must exist some
primitive element $y \neq 0$ lying in some $\Sym^n(\Liep)$ with $n \geq 2$.
This would mean that the composite map $f$ that follows the coproduct with
a component projection
$$
\Sym^n(\Liep) \overset{\Delta}{\longrightarrow} 
 \bigoplus_{i+j=n} \Sym^i(\Liep) \otimes \Sym^j(\Liep)
   \longrightarrow \Sym^1(\Liep) \otimes \Sym^{n-1}(\Liep)
$$
has $f(y)=0$.  However, one can check on a basis
that the multiplication backward 
$\Sym^1(\Liep) \otimes \Sym^{n-1}(\Liep) \overset{m}{\rightarrow} \Sym^n(\Liep)$
has the property that $m \circ f = n \cdot \one_{\Sym^n(\Liep)}$:
$$
(m \circ f)(x_1 \cdots x_n) 
 = m \left(\sum_{j=1}^n x_j \otimes x_1 \cdots \widehat{x_{j}} \cdots x_n \right)
 = n \cdot x_1 \cdots x_n
$$
for $x_1,\ldots,x_n$ in $\Liep$.
Then $n \cdot y = m(f(y)) = m(0) = 0$ leads to the contradiction that $y=0$,
since $\kk$ has characteristic zero.

Now one can argue the injectivity of the (graded) map\footnote{The grading on $\Sym(\Liep)$ is induced from the grading on $\Liep$, a homogeneous subspace of $I \subset A$ as it is the kernel of the graded map $I \overset{\Delta_+}{\longrightarrow} A \otimes A$.} $\varphi$ by assuming that one has a nonzero homogeneous
element $u$ in $\ker(\varphi)$ of minimum degree.
In particular, $\deg(u) \geq 1$.  Also since $\Liep \hookrightarrow A$, one has that $u$ is not in $\Sym^1(\Liep) = \Liep$, and hence $u$ is not primitive
by the previous Claim.  Consequently $\Delta_+(u) \neq 0$, and one can find 
a nonzero component $u^{(i,j)}$ of $\Delta_+(u)$ lying in 
$\Sym(\Liep)_i \otimes \Sym(\Liep)_j$ for some $i,j > 0$.
Since this forces $i,j < \deg(u)$, 
one has that $\varphi$ maps both $\Sym(\Liep)_i,\Sym(\Liep)_j$ injectively into $A_i,A_j$.  
Hence the tensor product map 
$$
\Sym(\Liep)_i \otimes \Sym(\Liep)_j 
\overset{\varphi \otimes \varphi}{\longrightarrow} 
A_i \otimes A_j
$$
is also injective\footnote{One needs to know that for two
injective maps $V_i \overset{\varphi_i}{\rightarrow} W_i$
of $\kk$-vector spaces $V_i,W_i$ with $i=1,2$, the
tensor product $\varphi_1 \otimes \varphi_2$ is also injective.
Factoring it as 
$
\varphi_1 \otimes \varphi_2 =
\left( \one \otimes \varphi_2 \right) \circ 
\left( \varphi_1 \otimes \one \right),
$
one sees that it suffices to show that for an injective
map $V \overset{\varphi}{\hookrightarrow} W$ of free $\kk$-modules,
and any free $\kk$-module $U$, the map 
$V \otimes U \overset{\varphi \otimes \one}{\longrightarrow} W\otimes U$ is
also injective.  Since tensor products commute with direct sums,
and $U \cong \kk^I$ for some $I$, this reduces to the easy-to-check 
case where $U=\kk$.

Note that some kind of freeness or flatness hypothesis on $U$ is needed here since, e.g. the injective $\ZZ$-module maps 
$\ZZ \overset{\varphi_1 = (\cdot \times 2)}{\longrightarrow} \ZZ$ and 
$\ZZ/2\ZZ \overset{\varphi_2= \one}{\longrightarrow} \ZZ/2\ZZ$
have $\varphi_1 \otimes \varphi_2=0$ on
$\ZZ \otimes_\ZZ \ZZ/2\ZZ \cong \ZZ/2\ZZ \neq 0$.}.
This implies $(\varphi \otimes \varphi)(u^{(i,j)}) \neq 0$, giving the contradiction that
$$
0=\Delta^A_+(0) = \Delta^A_+(\varphi(u)) = (\varphi \otimes \varphi)(\Delta^{\Sym(\Liep)}_+(u)) 
$$
contains the nonzero $A_i \otimes A_j$-component 
$(\varphi\otimes \varphi)(u^{(i,j)})$.
\end{proof}

Before closing this section, we mention one nonobvious corollary of
the Claim \eqref{primitives-in-symmetric-algebra}, when applied to
the ring of symmetric functions $\Lambda_\QQ$ with $\QQ$-coefficients,
since Proposition~\ref{symm-antipode-and-h-basis} says that
$\Lambda_\QQ=\QQ[p_1,p_2,\ldots]=\Sym(V)$ where $V=\QQ\{p_1,p_2,\ldots\}$.


\begin{corollary}
\label{primitives-in-Lambda}
The subspace $\Liep$ of primitives in $\Lambda_\QQ$ is
one-dimensional in each degree $n=1,2,\ldots$, and 
spanned by $\{p_1,p_2,\ldots\}$.
\end{corollary}


\subsection{The decomposition theorem}

Our goal here is Zelevinsky's theorem \cite[Theorem 2.2]{Zelevinsky} giving a canonical
decomposition of any PSH as a tensor product into PSH's that each
have only one primitive element in their PSH-basis.  For the sake
of stating it, we introduce some notation.

\begin{definition}
\label{indecomposable-PSH-factors-defn}
Given $A$ a PSH with 
PSH-basis $\Sigma$,
let $\CCC:=\Sigma \cap \Liep$ be the  
primitive elements in  $\Sigma$.
For each $\rho$ in $\CCC$, let
$A(\rho) \subset A$ be the $\ZZ$-span of
$$
\Sigma(\rho):=\{\sigma \in \Sigma: 
              \text{ there exists }n \geq 0\text{ with } 
                (\sigma, \rho^n) \neq 0\}.
$$
\end{definition}

\begin{theorem}
\label{Zelevinsky-decomposition-theorem}
Any PSH $A$ has a canonical tensor product decomposition
$$
A = \bigotimes_{ \rho \in \CCC } A(\rho)
$$
with $A(\rho)$ a PSH, and $\rho$ the only  
primitive element in its PSH-basis $\Sigma(\rho)$.
\end{theorem}
\noindent
Although in all the applications, $\CCC$ will be finite,
when $\CCC$ is infinite one should interpret the tensor product
in the theorem as the inductive limit of tensor products over finite subsets of $\CCC$,
that is, linear combinations of basic tensors $\otimes_\rho {a_\rho}$
in which there are only finitely many factors $a_\rho \neq 1$.

The first step toward the theorem uses a certain unique
factorization property.

\begin{lemma}
\label{unique-factorization-lemma}
Given $\PPP$ a set of pairwise orthogonal primitives in a PSH $A$, 
$$
( \rho_1 \cdots \rho_r , \pi_1 \cdots \pi_s ) = 0
$$
for $\rho_i, \pi_j$ in $\PPP$ unless $r=s$ and
one can reindex so that $\rho_i=\pi_i$.
\end{lemma}
\begin{proof}
Induct on $\min(r,s)$.  One has
$$
\begin{aligned}
(\rho_1 \cdots \rho_r , \pi_1 \cdots \pi_s ) 
&= (\rho_2 \cdots \rho_r , \rho_1^{\perp}(\pi_1 \cdots \pi_s) ) \\
&= (\rho_2 \cdots \rho_r , 
  \sum_{j=1}^s (\pi_1 \cdots \pi_{j-1} \cdot \rho_1^{\perp}(\pi_j) \cdot \pi_{j+1} \cdots \pi_s) )
\end{aligned}
$$
from Proposition~\ref{skewing-properties-prop}(iii) 
because $\rho_1$ is primitive.  On the other hand, since each $\pi_j$ is
primitive, one has 
$
\rho_1^\perp(\pi_j) = (\rho_1,\one) \cdot \pi_j + (\rho_1,\pi_j) \cdot \one
=(\rho_1,\pi_j)
$
which vanishes unless $\rho_1 = \pi_j$.
Hence $(\rho_1 \cdots \rho_r , \pi_1 \cdots \pi_s )=0$ unless $\rho_1 \in
\{\pi_1,\ldots,\pi_s\}$, in which case after reindexing so that
$\pi_1=\rho_1$, it equals 
$$
n \cdot (\rho_1,\rho_1) \cdot 
(\rho_2 \cdots \rho_r , \pi_2 \cdots \pi_s ) 
$$
if there are exactly $n$ occurrences of $\rho_1$ among
$\pi_1,\ldots,\pi_s$.  Now apply induction.
\end{proof}

So far the positivity hypothesis for a PSH has played little role.
Now we use it to introduce a certain partial order on the PSH $A$,
and then a semigroup grading.

\begin{definition}
Let $\NN:=\{0,1,2,\ldots\}$, and for a subset $S$ of an abelian group,
let $\ZZ S$ (resp. $\NN S$) denote the subgroup of $\ZZ$-linear combinations
(resp. subsemigroup of $\NN$-linear combinations) of the elements of $S$.

In a PSH $A$ with PSH-basis $\Sigma$, the subset $\NN \Sigma$ forms
a subsemigroup, and lets one define a partial order on $A$ via $a \leq b$
if $b-a$ lies in $\NN\Sigma$.
\end{definition}

We note a few trivial properties of this partial order:
\begin{itemize}
\item
The positivity hypothesis implies that 
$
\NN \Sigma \cdot \NN \Sigma \subset \NN \Sigma.
$
\item
Hence multiplication by an element $c \geq 0$
(meaning $c$ lies in $\NN \Sigma$) preserves the order: 
$a \leq b$ implies $ac \leq bc$ since
$(b-a)c$ lies in $\NN \Sigma$.  
\item
Thus $0 \leq c \leq d$ and $0 \leq a \leq b$ implies
$
ac \leq bc \leq bd.
$
\end{itemize}

\noindent
This allows one to introduce a semigroup grading on $A$.

\begin{definition}
Let $\NN^{\CCC}_{\fin}$ denote the additive subsemigroup
of $\NN^{\CCC}$ consisting of those $\alpha=(\alpha_\rho)_{\rho \in \CCC}$
with finite support.  

Note that for any $\alpha$ in $\NN^{\CCC}_{\fin}$, one has that the product
$\prod_{ \rho \in \CCC } \rho^{\alpha_\rho} \geq 0$.  Define 
$$
\Sigma(\alpha):=\{ \sigma \in \Sigma: \sigma \leq \prod_{\rho \in \CCC} \rho^{\alpha_\rho} \},
$$
that is, the subset of $\Sigma$ on which $\prod_{ \rho \in \CCC } \rho^{\alpha_\rho}$
has support.  Also define 
$$
A_{(\alpha)} := \ZZ \Sigma(\alpha) \subset A.
$$
\end{definition}

\begin{proposition}
\label{semigroup-grading-prop}
The PSH $A$ has an $\NN^{\CCC}_{\fin}$-semigroup-grading:
one has an orthogonal direct sum decomposition
$$
A = \bigoplus_{ \alpha \in \NN^{\CCC}_{\fin}} A_{(\alpha)}
$$
for which
\begin{align}
\label{product-semigroup-graded}
A_{(\alpha)} A_{(\beta)}  &\subset A_{(\alpha+\beta)}  \\
\label{comultiplication-semigroup-graded}
\Delta A_{(\alpha)} &\subset 
  \bigoplus_{\alpha=\beta+\gamma}A_{(\beta)} \otimes A_{{(\gamma)}}.  
\end{align}
\end{proposition}
\begin{proof}
We will make free use of the fact that a PSH $A$ is commutative,
since it embeds in $A \otimes_\ZZ \QQ$, which is commutative by 
Theorem~\ref{general-structure-theorem}.

Note that the orthogonality $\left( A_{(\alpha)}, A_{(\beta)} \right) = 0$ 
for $\alpha \neq \beta$ is equivalent to the assertion that 
$\left(\prod_{ \rho \in \CCC } \rho^{\alpha_\rho},
\prod_{ \rho \in \CCC } \rho^{\beta_\rho} \right)=0$,
which follows from Lemma~\ref{unique-factorization-lemma}.

Next let us deal with the assertion \eqref{product-semigroup-graded}.
It suffices to check that when $\tau,\omega$ in $\Sigma$ 
lie in $A_{(\alpha)}, A_{(\beta)}$, respectively, then $\tau\omega$
lies in $A_{(\alpha+\beta)}$.  But note that any $\sigma$ in $\Sigma$
having $\sigma \leq \tau \omega$ will then have
$$
\sigma \leq \tau \omega 
\leq 
\prod_{ \rho \in \CCC } \rho^{\alpha_\rho} \cdot
\prod_{ \rho \in \CCC } \rho^{\beta_\rho}
=\prod_{ \rho \in \CCC } \rho^{\alpha_\rho+\beta_\rho}
$$
so that $\sigma$ lies in $A_{(\alpha+\beta)}$.  This means
that $\tau\omega$ lies in $A_{(\alpha+\beta)}$.  

This lets us check that $\bigoplus_{ \alpha \in \NN^{\CCC}_{\fin}} A_{(\alpha)}$ exhaust $A$.  It suffices
to check that any $\sigma$ in $\Sigma$ lies in some $A_{(\alpha)}$.  Proceed
by induction on $\deg(\sigma)$, with the case $\sigma=\one$ being
trivial;  the element $\one$ always lies in $\Sigma$, and hence lies in $A_{(\alpha)}$ for $\alpha=0$.  For $\sigma$ lying in $I$,
by Proposition~\ref{primitives-split-off-prop}, one either has 
$(\sigma,a) \neq 0$ for some $a$ in $I^2$, or
else $\sigma$ lies in $(I^2)^\perp=\Liep$, so that $\sigma$ is in $\CCC$
and we are done.  
If $(\sigma,a) \neq 0$ with $a$ in $I^2$, then 
$\sigma$ appears in the support of some
$\ZZ$-linear combination of elements $\tau\omega$ where $\tau,\omega$
lie in $\Sigma$ and have strictly smaller degree.  There exists at least
one such pair $\tau, \omega$ for which $(\sigma, \tau\omega) \neq 0$,
and therefore $\sigma \leq \tau\omega$.  Then by induction $\tau, \omega$
lie in some $A_{(\alpha)}, A_{(\beta)}$, respectively, so $\tau\omega$ lies in $A_{(\alpha+\beta)}$, and hence $\sigma$ lies
in $A_{(\alpha+\beta)}$ also.

Self-duality shows that 
\eqref{product-semigroup-graded}
implies \eqref{comultiplication-semigroup-graded}:
if $a,b,c$ lie in $A_{(\alpha)}, A_{(\beta)}, A_{(\gamma)}$,
respectively, then
$
(\Delta a, b \otimes c)_{A \otimes A} = (a, bc)_A = 0
$
unless $\alpha=\beta+\gamma$.
\end{proof}


\begin{proposition}
\label{disjoint-support-prop}
For $\alpha, \beta$ in $\NN^\CCC_{\fin}$ with disjoint support, one has a bijection
$$
\begin{array}{rcl}
\Sigma(\alpha) \times \Sigma(\beta) & \longrightarrow & \Sigma(\alpha+\beta) \\
(\sigma, \tau) & \longmapsto & \sigma \tau.
\end{array}
$$
so that the multiplication map 
$
A_{(\alpha)} \otimes A_{(\beta)} \rightarrow A_{(\alpha+\beta)}$
is an isomorphism.
\end{proposition}
\begin{proof}
We first check that for $\sigma_1,\sigma_2$ in  $\Sigma(\alpha)$
and $\tau_1,\tau_2$ in $\Sigma(\beta)$, one has  
\begin{equation}
\label{disjoint-support-orthonormality}
(\sigma_1 \tau_1, \sigma_2 \tau_2) = \delta_{(\sigma_1,\tau_1),(\sigma_2,\tau_2)}.
\end{equation}
Note that this is equivalent to showing both 
\begin{enumerate}
\item[$\bullet$]
that $\sigma \tau$ lie in $\Sigma(\alpha+\beta)$
so that the map is well-defined, since it shows
$(\sigma \tau, \sigma \tau)=1$, and
\item[$\bullet$]
that the map is injective.
\end{enumerate}  
One calculates
$$
\begin{aligned}
(\sigma_1 \tau_1, \sigma_2 \tau_2)_A
&=(\sigma_1 \tau_1, m(\sigma_2 \otimes \tau_2))_A \\
&=(\Delta(\sigma_1 \tau_1), \sigma_2 \otimes \tau_2)_{A \otimes A}\\
&=(\Delta(\sigma_1) \Delta(\tau_1), \sigma_2 \otimes \tau_2)_{A \otimes A}
\end{aligned}
$$
Note that due to \eqref{comultiplication-semigroup-graded}, $\Delta(\sigma_1) \Delta(\tau_1)$ lies in $\bigoplus A_{(\alpha'+\beta')} \otimes A_{(\alpha'' + \beta'')}$ where 
$$
\begin{aligned}
\alpha'+\alpha''&=\alpha\\
\beta'+\beta''&=\beta.
\end{aligned}
$$
Since $\sigma_2 \otimes \tau_2$
lies in $A_{(\alpha)} \otimes A_{(\beta)}$, the only nonvanishing terms in the inner product come
from those with 
$$
\begin{aligned}
\alpha'+\beta'&=\alpha\\
\alpha''+\beta''&=\beta.
\end{aligned}
$$  
As $\alpha,\beta$ have disjoint support, this can only happen if 
$$
\alpha'=\alpha, \,\, \alpha''=0, \,\, \beta'=0,\,\, \beta''=\beta,
$$
that is, the only nonvanishing term comes from
$(\sigma_1 \otimes \one)(\one \otimes \tau_1)=\sigma_1 \otimes \tau_1$.
Hence 
$$
(\sigma_1 \tau_1, \sigma_2 \tau_2)_A
=(\sigma_1 \otimes \tau_1,\sigma_2 \otimes \tau_2)_{A \otimes A}\\
= \delta_{(\sigma_1,\tau_1),(\sigma_2,\tau_2)}.
$$

To see that the map is surjective, express 
$$
\begin{aligned}
\prod_{\rho \in \CCC} \rho^{\alpha_\rho}&=\sum_i \sigma_i \\
\prod_{\rho \in \CCC} \rho^{\beta_\rho}&=\sum_j \tau_j
\end{aligned}
$$
with $\sigma_i \in \Sigma(\alpha)$ and $\tau_j$ in $\Sigma(\beta)$.
Then each product $\sigma_i \tau_j$ is in $\Sigma(\alpha +\beta)$ by
\eqref{disjoint-support-orthonormality}, and 
$$
\prod_{\rho \in \CCC} \rho^{\alpha_\rho+\beta_\rho}=\sum_{i,j} \sigma_i \tau_j
$$
shows that $\{\sigma_i \tau_j\}$ exhausts $\Sigma(\alpha+\beta)$.  This
gives surjectivity.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{Zelevinsky-decomposition-theorem}]
Recall from Definition~\ref{indecomposable-PSH-factors-defn} that
for each $\rho$ in $\CCC$, one defines $A(\rho) \subset A$ to
be the $\ZZ$-span of
$$
\Sigma(\rho):=\{\sigma \in \Sigma: 
              \text{ there exists }n \geq 0\text{ with } 
                (\sigma, \rho^n) \neq 0\}.
$$
In other words, $A(\rho):=\bigoplus_{n \geq 0} A_{(n \cdot e_\rho)}$
where $e_\rho$ in $\NN^{\CCC}_{\fin}$ is the standard basis element indexed by $\rho$.
Proposition~\ref{semigroup-grading-prop} then shows that $A(\rho)$ is a subHopf algebra of $A$.
Since every $\alpha$ in $\NN^{\CCC}_{\fin}$ can be expressed as the
(finite) sum $\sum_{\rho} \alpha_{\rho} e_\rho$, and the $e_\rho$ have disjoint support,
iterating Proposition~\ref{disjoint-support-prop} shows that
$A = \otimes_{ \rho \in \CCC } A(\rho)$.  Lastly, $\Sigma(\rho)$ is clearly a PSH-basis
for $A(\rho)$, and if $\sigma$ is any primitive
element in $\Sigma(\rho)$ then $(\sigma,\rho^n) \neq 0$ lets one conclude
via Lemma~\ref{unique-factorization-lemma} that $\sigma=\rho$ (and $n=1$).
\end{proof}




\subsection{$\Lambda$ is the unique indecomposable PSH}

The goal here is to prove the rest of Zelevinsky's structure theory for PSH's.
Namely, if $A$ has only one primitive element $\rho$
in its PSH-basis $\Sigma$, then $A$ must be isomorphic as a PSH to
the ring of symmetric functions $\Lambda$, after one rescales the grading of $A$.
Note that every $\sigma$ in $\Sigma$ has
$\sigma \leq \rho^n$ for some $n$, and hence has degree divisible by
the degree of $\rho$.  Thus one can divide all degrees by that of $\rho$
and assume $\rho$ has degree $1$.

The idea is to find within $A$ and $\Sigma$ a set of elements that
play the role of 
$$
\{ h_n=s_{(n)} \}_{n=0,1,2,\ldots}, \{ e_n=s_{(1^n)} \}_{n=0,1,2,\ldots}
$$
within $A=\Lambda$ and its PSH-basis of Schur 
functions $\Sigma=\{s_\lambda\}$.  Zelevinsky's argument does this by
isolating some properties that turn out to characterize these elements:

\begin{enumerate}
\item[(a)] $h_0=e_0=1$, and $h_1=e_1=:\rho$ has $\rho^2$ a sum of two elements of $\Sigma$, namely 
$$
\rho^2=h_2+e_2.
$$
\item[(b)] For all $n=0,1,2,\ldots$, there exist unique elements
$h_n, e_n$ in $A_n \cap \Sigma$ that satisfy
$$
\begin{aligned}
h_2^\perp e_n &= 0,\\
e_2^\perp h_n &= 0
\end{aligned}
$$
with $h_2,e_2$ being the two elements of $\Sigma$ introduced in (a).
\item[(c)] For $k=0,1,2,\ldots,n$ one has
$$
\begin{aligned}
h_k^\perp h_n & = h_{n-k} \text{ and } 
  \sigma^\perp h_n = 0 \text{ for } \sigma \in \Sigma \setminus \{h_0,h_1,\ldots,h_n\} \\
e_k^\perp e_n & = e_{n-k} \text{ and } 
  \sigma^\perp e_n = 0 \text{ for } \sigma \in \Sigma \setminus \{e_0,e_1,\ldots,e_n\}. \\
\end{aligned}
$$
In particular, $e_k^\perp h_n = 0=h_k^\perp e_n$ for $k \geq 2$.
\item[(d)] Their coproducts are
$$
\begin{aligned}
\Delta(h_n) &= \sum_{i+j=n} h_i \otimes h_j,\\
\Delta(e_n) &= \sum_{i+j=n} e_i \otimes e_j.\\
\end{aligned}
$$
\end{enumerate}

We will prove Zelevinsky's result \cite[Theorem 3.1]{Zelevinsky} 
as a combination of the following two theorems.  

\begin{theorem}
\label{uniqueness-of-sym-parts-a-b-c-d} 
Let $A$ be a PSH with PSH-basis $\Sigma$ containing only one primitive 
$\rho$, and assume that the grading has been rescaled so that
$\rho$ has degree $1$.  
Then, after renaming $\rho=e_1=h_1$, one can find unique sequences
$
\{ h_n\}_{n=0,1,2,\ldots}, \{ e_n \}_{n=0,1,2,\ldots}
$
of elements of $\Sigma$ having properties (a),(b),(c),(d) listed above.
\end{theorem}

The second theorem uses the following notion.

\begin{definition}
A {\it PSH-morphism } $A \overset{\varphi}{\rightarrow} A'$
between two PSH's $A,A'$ having PSH-bases $\Sigma, \Sigma'$
is a graded Hopf algebra morphism for which $\varphi(\NN \Sigma) \subset \NN \Sigma'$.  If $A=A'$ and $\Sigma=\Sigma'$ it will be called a {\it PSH-endomorphism}. If $\varphi$ is an isomorphism and restricts to a bijection
$\Sigma \rightarrow \Sigma'$, it will be called a {\it PSH-isomorphism};
if it is both an isomorphism and an endomorphism, it is
a {\it PSH-automorphism}.
\end{definition}

\begin{theorem}
\label{uniqueness-of-sym-parts-e-f-g} 
The elements
$
\{ h_n\}_{n=0,1,2,\ldots}, \{ e_n \}_{n=0,1,2,\ldots}
$
in Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d} also satisfy the following.

\begin{enumerate}
\item[(e)] The elements $h_n,e_n$ in $A$ satisfy the same relation \eqref{e-h-relation}
$$
\sum_{i+j=n} (-1)^i e_i h_j = \delta_{0,n}.
$$
as their counterparts in $\Lambda$, along with the
property that 
$$
A=\ZZ[h_1,h_2,\ldots]=\ZZ[e_1,e_2,\ldots].
$$
\item[(f)]  There is exactly one nontrivial automorphism $A \overset{\omega}{\rightarrow} A$ as
a PSH, swapping $h_n \leftrightarrow e_n$.

\item[(g)]  There are exactly two PSH-isomorphisms $A \rightarrow \Lambda$,
\begin{itemize}
\item one sending $h_n$ to the complete homogeneous symmetric functions $h_n(\xx)$, while sending $e_n$ to the elementary symmetric functions $e_n(\xx)$,
\item the second one (obtained by composing the first with $\omega$) sending
$h_n \mapsto e_n(\xx)$ and $e_n \mapsto h_n(\xx)$.
\end{itemize}
\end{enumerate}
\end{theorem}

Before embarking on the proof, we mention one more bit of convenient
terminology:  say that an element $\sigma$ in $\Sigma$ is a {\it constituent}
of $a$ in $\NN \Sigma$ when $\sigma \leq a$, that is, $\sigma$ appears with nonzero coefficient $c_\sigma$ 
in the  unique expansion $a=\sum_{\tau \in \Sigma} c_\tau \tau$.

\begin{proof}[Proof of Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}]
One fact that occurs frequently is this:
\begin{equation}
\label{every-sigma-is-a-constituent-of-powers-of-rho}
\text{Every }\sigma\text{ in }\Sigma \cap A_n\text{ is a constituent of }\rho^n.
\end{equation}
This follows from Theorem~\ref{Zelevinsky-decomposition-theorem},
since $\rho$ is the only primitive element of $\Sigma$:  one has
$A=A(\rho)$ and $\Sigma=\Sigma(\rho)$, so that $\sigma$ is a constituent
of some $\rho^m$, and homogeneity considerations force $m=n$.
\vskip.1in
\noindent
{\sf Assertion (a).} Note that
$$
(\rho^2,\rho^2)=(\rho^\perp(\rho^2),\rho)=(2\rho,\rho)=2
$$
using the fact that $\rho^\perp$ is a derivation since $\rho$ is primitive 
(Proposition~\ref{skewing-properties-prop}{(iii)}).  On the other hand, expressing 
$\rho^2 = \sum_{ \sigma \in \Sigma } c_\sigma \sigma$ with $c_\sigma$ in $\NN$,
one has
$
(\rho^2,\rho^2)=\sum_\sigma c_\sigma^2.
$
Hence exactly two of the $c_\sigma=1$, so $\rho^2$ has exactly two distinct
constituents.   Denote them by $h_2$ and $e_2$.  
One concludes that $\Sigma \cap A_2=\{h_2,e_2\}$ from
\eqref{every-sigma-is-a-constituent-of-powers-of-rho}.

Note also that the same argument shows $\Sigma \cap A_1 = \{\rho\}$,
so that $A_1=\ZZ \rho$.  Since $\rho^\perp h_2$ lies in $A_1=\ZZ\rho$
and $(\rho^\perp h_2,\rho)=(h_2,\rho^2)=1$, we have $\rho^\perp h_2=\rho$.
Similarly $\rho^\perp e_2=\rho$.
\vskip.1in
\noindent
{\sf Assertion (b).} 
We will show via induction on $n$ the following three assertions for $n \geq 1$:
\begin{equation}
\label{h_n-inductive-properties}
\begin{aligned}
& \bullet \text{There exists an 
element }h_n\text{ in }\Sigma \cap A_n \text{ with }e_2^\perp h_n=0.\\
& \bullet \text{This element }h_n \text{ is unique.} \\
& \bullet \text{Furthermore }\rho^\perp h_n = h_{n-1}.\\
\end{aligned}
\end{equation}
In the base cases $n=1,2$, it is not hard to check that
our previously labelled elements, $h_1, h_2$ (namely $h_1:=\rho$, 
and $h_2$ as named in part (a)) really {\it are} 
the unique elements satisfying these hypotheses.

In the inductive step, it turns out that 
we will find $h_n$ as a constituent of $\rho h_{n-1}$.
Thus we again use the derivation property of $\rho^\perp$ to compute
that $\rho h_{n-1}$ has exactly two constituents:
$$
\begin{aligned}
( \rho h_{n-1}, \rho h_{n-1} ) 
&=( \rho^\perp (\rho h_{n-1}), h_{n-1} ) \\
&=( h_{n-1} + \rho \cdot \rho^\perp h_{n-1}, h_{n-1} )\\
&=( h_{n-1} + \rho h_{n-2}, h_{n-1} )\\
&=1+ ( h_{n-2}, \rho^\perp h_{n-1} )\\
&=1+ ( h_{n-2}, h_{n-2} )=1+1=2
\end{aligned}
$$
where the inductive hypothesis $\rho^\perp h_{n-1}=h_{n-2}$ was used twice.
We next show that exactly one of the two constituents of $\rho h_{n-1}$
is annihilated by $e_2^\perp$.  Note that since $e_2$ lies in $A_2$,
and $A_1$ has $\ZZ$-basis element $\rho$, there is a constant $c$ in $\ZZ$
such that
\begin{equation}
\label{coproduct-of-e2}
\Delta(e_2) = e_2 \otimes \one + c \rho \otimes \rho + \one \otimes e_2.
\end{equation}
On the other hand, (a) showed
$$
1=( e_2, \rho^2 )_A = (\Delta (e_2), \rho \otimes \rho )_{A \otimes A}
$$
so one must have $c=1$.
Therefore by Proposition~\ref{skewing-properties-prop}{(iii)} again,
\begin{equation}
\label{e_2-skewing-calculation}
\begin{array}{rcccccl}
e_2^\perp ( \rho h_{n-1} ) 
&=& e_2^\perp(\rho) h_{n-1} &+& \rho^\perp(\rho) \rho^\perp(h_{n-1}) &+& \rho e_2^{\perp}(h_{n-1}) \\
&=& 0&+&h_{n-2}&+&0\\
&=&h_{n-2}& & & &.
\end{array}
\end{equation}
where the first term vanished due to degree considerations and the last term
vanished by the inductive hypothesis.
Bearing in mind that $\rho h_{n-1}$ lies in $\NN \Sigma$, and
in a PSH with PSH-basis $\Sigma$, any skewing operator 
$\sigma^\perp$ for $\sigma$ in $\Sigma$ will 
preserve $\NN \Sigma$, one concludes from
\eqref{e_2-skewing-calculation} that 
\begin{enumerate}
\item[$\bullet$]
one of the two distinct constituents of the element $\rho h_{n-1}$ must be sent by $e_2^\perp$ to $h_{n-2}$, and
\item[$\bullet$]
the other constituent of $\rho h_{n-1}$ must be annihilated by 
$e_2^\perp$; call this second constituent $h_n$.
\end{enumerate}

Lastly, to see that this $h_n$ is unique, it suffices to show that 
any element $\sigma$ of $\Sigma \cap A_n$ which is killed by $e_2^\perp$ must be a constituent of
$\rho h_{n-1}$.  This holds for the following reason.
We know $\sigma \leq \rho^n$ by
\eqref{every-sigma-is-a-constituent-of-powers-of-rho},
and hence $0 \neq (\rho^n,\sigma) = (\rho^{n-1},\rho^\perp \sigma)$,
implying that $\rho^\perp \sigma \neq 0$.
On the other hand, since 
$0= \rho^\perp e_2^\perp \sigma = e_2^\perp \rho^\perp \sigma$,
one has that $\rho^\perp \sigma$ is annihilated by $e_2^\perp$, and
hence $\rho^\perp \sigma$ must be a (positive) multiple of $h_{n-1}$ by part of 
our inductive hypothesis.  Therefore $(\sigma, \rho h_{n-1}) = (\rho^\perp \sigma, h_{n-1})$
is positive, that is, $\sigma$ is a constituent of $\rho h_{n-1}$.

The preceding argument, applied to $\sigma=h_n$, shows that 
$\rho^\perp h_n=ch_{n-1}$ for some $c$ in $\{1,2,\ldots\}$.  Since
$(\rho^\perp h_n,h_{n-1})=(h_n,\rho h_{n-1})=1$, this $c$ must be $1$, so
that $\rho^\perp h_n=h_{n-1}$.  This completes the induction step in
the proof of \eqref{h_n-inductive-properties}.

One can then argue, swapping the roles of $e_n, h_n$ in the
above argument, the existence and uniqueness of 
a sequence $\{e_n\}_{n=0}^\infty$ in $\Sigma$ satisfying 
the properties analogous to
\eqref{h_n-inductive-properties}, with $e_0:=1, e_1:=\rho$.



\vskip.1in
\noindent
{\sf Assertion (c).}
Iterating the property from (b) that $\rho^\perp h_n = h_{n-1}$ shows
that $(\rho^k)^\perp h_n = h_{n-k}$ for $0 \leq k \leq n$.  However one
also has an expansion 
$$
\rho^k = c h_k + \sum_{\substack{\sigma \in \Sigma \cap A_k:\\\sigma \neq h_k}} c_ \sigma \sigma
$$ 
for some integers $c, c_\sigma > 0$, since every $\sigma$ in $\Sigma \cap A_k$
is a constituent of $\rho^k$.  Hence
$$
1 = (h_{n-k}, h_{n-k}) =
((\rho^k)^\perp h_n,(\rho^k)^\perp h_n)
\geq c^2 (h_k^\perp h_n, h_k^\perp h_n)
$$
using Proposition~\ref{skewing-properties-prop}(ii).
Hence if we knew that $h_k^\perp h_n \neq 0$ this would force 
$$
h_k^\perp h_n = (\rho^k)^\perp h_n = h_{n-k}
$$
as well as $\sigma^\perp h_n=0$ for all $\sigma \not\in \{h_0,h_1,\ldots,h_n\}$.  But 
$$
(\rho^{n-k})^\perp h_k^\perp h_n = 
h_k^\perp (\rho^{n-k})^\perp h_n = 
h_k^\perp h_k = 1 \neq 0 
$$
so $h_k^\perp h_n \neq 0$, as desired.
The argument for $e_k^\perp e_n = e_{n-k}$ is symmetric.

The last assertion in (c) follows if one checks that
$e_n \neq h_n$ for each $n \geq 2$, but this holds since
$e_2^\perp(h_n)=0$ but $e_2^\perp(e_n) = e_{n-2}$.

\vskip.1in
\noindent
{\sf Assertion (d).}
Part (c) implies that
$$
(\Delta h_n, \sigma \otimes \tau)_{A \otimes A}
=(h_n , \sigma \tau)_A
=(\sigma^\perp h_n, \tau)_A = 0
$$
unless $\sigma = h_k$ for some $k=0,1,2,\ldots,n$
and $\tau = h_{n-k}$.  Also one can compute
$$
(\Delta h_n, h_k \otimes h_{n-k})
= (h_n, h_k h_{n-k}) 
= (h_k^{\perp} h_n, h_{n-k})
\overset{(c)}{=}(h_{n-k}, h_{n-k}) =1.
$$
This is equivalent to the assertion
for $\Delta h_n$ in (d).  The argument for $\Delta e_n$ is symmetric.
\end{proof}


Before proving Theorem~\ref{uniqueness-of-sym-parts-e-f-g}, we note
some consequences of Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}.
Define for each partition 
$\lambda=(\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_\ell)$
the elements of $A$
$$
\begin{aligned}
h_\lambda=h_{\lambda_1} h_{\lambda_2} \cdots, \\
e_\lambda=e_{\lambda_1} e_{\lambda_2} \cdots.
\end{aligned}
$$
Also, define the {\it lexicographic order} on $\Par_n$
by saying $\lambda <_{\lex} \mu$ if $\lambda \neq \mu$ and
the smallest index $i$ for which $\lambda_i \neq \mu_i$ has
$\lambda_i < \mu_i$.  Recall also that $\lambda^t$ denotes the
{\it conjugate} or {\it transpose} partition to $\lambda$, obtained by
swapping rows and columns in the Ferrers diagram.

The following unitriangularity lemma will play a role in the proof
of Theorem~\ref{uniqueness-of-sym-parts-e-f-g}(e).
\begin{lemma}
\label{triangularity-lemma}
Under the hypotheses of Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d},
for $\lambda, \mu$ in $\Par_n$, one has
$$
e_\mu^\perp h_\lambda =
\begin{cases}
1 & \text{ if }\mu=\lambda^t \\
0 & \text{ if }\mu >_{\lex} \lambda^t.
\end{cases}
$$
Consequently 
$$
\det\left[ (e_{\mu^t}, h_\lambda) \right]_{\lambda,\mu \in \Par_n}=\pm 1.
$$
\end{lemma}
\begin{proof}
Induct on the length of $\mu$.  If $\lambda$ has length $\ell$, so that $\lambda_1^t=\ell$, then 
$$
\begin{aligned}
e_\mu^\perp h_\lambda 
&= e_{(\mu_2,\mu_3,\ldots)}^\perp 
\left(
e_{\mu_1}^\perp (h_{\lambda_1} \cdots h_{\lambda_\ell}) 
\right) \\
&=e_{(\mu_2,\mu_3,\ldots)}^\perp 
\sum_{i_1+\ldots+i_\ell=\mu_1}
e_{i_1}^\perp(h_{\lambda_1}) \cdots
e_{i_\ell}^\perp(h_{\lambda_\ell}) \\
&=\begin{cases}
0 & \text{ if }\mu_1 > \ell=\lambda_1^t \\
e_{(\mu_2,\mu_3,\ldots)}^\perp  h_{(\lambda_1-1, \ldots,\lambda_\ell-1)} & \text{ if }\mu_1 = \ell=\lambda_1^t
\end{cases}
\end{aligned}
$$
where the second-to-last equality 
used Proposition~\ref{skewing-properties-prop}(iii)
along with (the iterates of) the coproduct formula for $\Delta(e_n)$ in 
Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}(d), and
the last equality used 
$$
e_k^\perp(h_n)=\begin{cases} h_{n-1} & \text{ if }k=1,\\ 0 &\text{ if }k \geq 2 \end{cases}
$$ 
Now apply induction, since 
$(\lambda_1-1, \ldots,\lambda_\ell-1)^t = (\lambda_2^t,\lambda_3^t,\ldots)$.

For the last assertion, note 
$(e_{\mu^t}, h_\lambda)=(e_{\mu^t}^\perp(h_\lambda), 1)=e_{\mu^t}^\perp(h_\lambda)$
for $\lambda,\mu$ in $\Par_n$.
\end{proof}

The following proposition will be the crux of the
proof of Theorem~\ref{uniqueness-of-sym-parts-e-f-g}(f) and (g),
and turns out to be closely related to Kerov's {\it asymptotic theory of
characters of the symmetric groups} \cite{Kerov}.

\begin{proposition}
\label{normalized-character-prop}
Given a PSH $A$ with PSH-basis $\Sigma$ containing only one primitive $\rho$,
the two maps $A \rightarrow \ZZ$
defined on $A=\bigoplus_{n \geq 0} A_n$ via 
$$
\begin{aligned}
\delta_h&=\bigoplus_n h_n^\perp,\\
\delta_e&=\bigoplus_n e_n^\perp
\end{aligned}
$$
are characterized as the only two 
$\ZZ$-linear maps $A \overset{\delta}{\rightarrow} \ZZ$ with the three
properties of being
\begin{itemize}
\item {\bf positive}: $\delta(\NN \Sigma) \subset \NN$,
\item {\bf multiplicative}: $\delta(a_1 a_2) = \delta(a_1) \delta(a_2)$, and
\item {\bf normalized}: $\delta(\rho)=1$.
\end{itemize}
\end{proposition}
\begin{proof}
It should be clear from their definitions that $\delta_h, \delta_e$ are $\ZZ$-linear,
positive and normalized.  To see that $\delta_h$ is multiplicative, by $\ZZ$-linearity,
it suffices to check that for $a_1,a_2$ in $A_{n_1}, A_{n_2}$ with $n_1+n_2=n$, one has
$$
\delta_h(a_1 a_2) = h_n^\perp(a_1 a_2) 
= \sum_{i_1 + i_2=n} h_{i_1}^\perp(a_1) h_{i_2}^\perp(a_2)
= h_{n_1}^\perp(a_1) h_{n_2}^\perp(a_2) = \delta_h(a_1) \delta_h(a_2)
$$
in which the second equality used Proposition~\ref{skewing-properties-prop}(iii)
and Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}(d).  The argument for $\delta_e$
is symmetric.

Conversely, given $A \overset{\delta}{\rightarrow} \ZZ$ which is $\ZZ$-linear,
positive, multiplicative, and normalized, note that
$$
\delta(h_2)+\delta(e_2) = \delta(h_2+e_2)
= \delta(\rho^2) 
= \delta(\rho)^2
= 1^2 = 1
$$
and hence positivity implies that either $\delta(h_2)=0$ or $\delta(e_2)=0$.
Assume the latter holds, and we will show that $\delta=\delta_h$.

Given any $\sigma$ in $\Sigma \cap A_n \setminus \{h_n\}$, note that
$e_2^\perp \sigma \neq 0$ by Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}(b), and hence
$
0 \neq (e_2^\perp \sigma, \rho^{n-2}) =( \sigma, e_2 \rho^{n-2}).
$
Thus $\sigma$ is a constituent of $e_2 \rho^{n-2}$, so positivity implies
$$
0 \leq \delta(\sigma) \leq \delta(e_2 \rho^{n-2}) = \delta(e_2) \delta(\rho^{n-2}) = 0.
$$
Thus $\delta(\sigma)=0$ for $\sigma$ in $\Sigma \cap A_n \setminus \{h_n\}$.  Since
$\delta(\rho^n)=\delta(\rho)^n =1^n=1$, this forces $\delta(h_n)=1$, for each $n \geq 0$
(including $n=0$, as 
$1=\delta(\rho)=\delta(\rho \cdot 1)=\delta(\rho)\delta(1)=1 \cdot \delta(1)=\delta(1)$.)
Thus $\delta=\delta_h$.  The argument when $\delta(h_2)=0$ showing $\delta=\delta_e$ is
symmetric.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{uniqueness-of-sym-parts-e-f-g}]
Many of the assertions of parts (e) and (f) will come from constructing
the unique nontrivial PSH-automorphism $\omega$ of $A$ from the antipode $S$:
for homogeneous $a$ in $A_n$, define $\omega(a):=(-1)^n S(a)$.  We now study some
of the properties of $S$ and $\omega$.  

Since $A$ is a PSH, it is commutative by Theorem~\ref{general-structure-theorem}.  This implies both that
$S, \omega$ are actually algebra endomorphisms by 
Proposition~\ref{antipodes-are-antiendomorphisms}, and that
$S^2=\one_A=\omega^2$ by Corollary~\ref{commutative-implies-involutive-antipode-cor}.

Since $A$ is self-dual and the defining diagram \eqref{antipode-diagram} 
satisfied by the antipode $S$ is sent to itself when one replaces $A$ by $A^o$
and all maps by their adjoints, one concludes that $S=S^t$, i.e., $S$ is self-adjoint.
Since $S$ is an algebra endomorphism, and $S=S^t$, in fact $S$ is also a 
coalgebra endomorphism, a bialgebra endomorphism, and a Hopf endomorphism 
(by Proposition~\ref{antipodes-give-convolution-inverses}(c))  The same properties
are shared by $\omega$.

Since $\one_A = S^2 = S S^t$, one concludes that $S$ is an isometry, and hence so
is $\omega$.

Since $\rho$ is primitive, one has $S(\rho)=-\rho$ and $\omega(\rho)=\rho$.
Therefore $\omega(\rho^n)=\rho^n$ for $n=1,2,\ldots$.  Use this as follows
to check that $\omega$ is a PSH-automorphism, which amounts to checking that
every $\sigma$ in $\Sigma$ has $\omega(\sigma)$ in $\Sigma$:
$$
( \omega(\sigma),\omega(\sigma) )= (\sigma,\sigma)=1
$$
so that $\pm \omega(\sigma)$ lies in $\Sigma$, but also if $\sigma$ lies in $A_n$, then
$$
( \omega(\sigma), \rho^n )= (\sigma , \omega(\rho^n) )= (\sigma , \rho^n )>0.
$$
In summary, $\omega$ is a PSH-automorphism of $A$, an isometry, and an involution.

Let us try to determine the action of $\omega$ on the $\{h_n\}$.
By similar reasoning as in \eqref{coproduct-of-e2}, one has
$$
\Delta(h_2) = h_2 \otimes \one + \rho \otimes \rho + \one \otimes h_2.
$$
Thus $0 = S(h_2) + S(\rho)\rho + h_2$, and combining this with $S(\rho)=-\rho$,
one has $S(h_2) = e_2$.  Thus also $\omega(h_2)=(-1)^2 S(h_2)=e_2$.

We claim that this forces $\omega(h_n)=e_n$, because $h_2^\perp \omega(h_n)= 0$ via
the following calculation:  for any $a$ in $A$ one has
$$
\begin{aligned}
(h_2^\perp \omega(h_n), a) 
&=  (\omega(h_n), h_2 a) \\
&=  ( h_n, \omega(h_2 a) ) \\
&=  ( h_n, e_2 \omega(a) ) \\
&=  ( e_2^\perp h_n, \omega(a) ) =  ( 0 , \omega(a) )  =0.
\end{aligned}
$$

Consequently the involution $\omega$ swaps $h_n$ and $e_n$, while the
antipode $S$ has $S(h_n)=(-1)^ne_n$ and $S(e_n)=(-1)^n h_n$.  Thus  
the coproduct formulas in (d) and definition of the  antipode $S$ imply
the relation \eqref{e-h-relation} between $\{h_n\}$ and $\{e_n\}$.

This relation  \eqref{e-h-relation} also lets one recursively express the
$h_n$ as polynomials with integer coefficients in the $\{e_n\}$, 
and vice-versa, so that $\{h_n\}$ and $\{e_n\}$  each generate the 
same $\ZZ$-subalgebra $A'$ of $A$.  We wish to show that $A'$ exhausts $A$.

We argue that Lemma~\ref{triangularity-lemma} implies the {\it Gram matrix} 
$\left[ (h_{\mu},h_{\lambda}) \right]_{\mu,\lambda \in \Par_n}$ has determinant $\pm 1$ as
follows.  Since $\{h_n\}$ and $\{e_n\}$  both generate $A'$, 
there exists a $\ZZ$-matrix $(a_{\mu,\lambda})$ expressing
$e_{\mu^t} = \sum_{\lambda} a_{\mu,\lambda} h_\lambda$, and one has
$$
 \left[ (e_{\mu^t}, h_{\lambda}) \right]
= \left[ a_{\mu,\lambda} \right] \cdot
\left[ (h_{\mu},h_{\lambda}) \right].
$$
Taking determinants of these three $\ZZ$-matrices, and 
using the fact that the determinant on the left is $\pm 1$, 
both determinants on the right must also be $\pm 1$.

Now we will show that every $\sigma \in \Sigma \cap A_n$ lies in $A_n'$.
Uniquely express $\sigma=\sigma'+\sigma''$ in which $\sigma'$ lies
in the $\RR$-span $\RR A_n'$ and  $\sigma''$ lies in the real perpendicular
space  $(\RR A_n')^\perp$ inside $\RR \otimes_\ZZ A_n$.  
One can compute $\RR$-coefficients $(c_\mu)_{\mu \in \Par_n}$
that express $\sigma' = \sum_\mu c_\mu h_\mu$ by solving the system
$$
\left( \sum_\mu c_\mu h_\mu, h_\lambda \right) 
 = (\sigma, h_\lambda) \text{ for }\lambda \in \Par_n.
$$
This linear system is governed by the Gram matrix 
$\left[ (h_{\mu},h_{\lambda}) \right]_{\mu,\lambda \in \Par_n}$
with determinant $\pm 1$, and its right side has $\ZZ$-entries since $\sigma, h_\lambda$
lie in $A$.  Hence the solution $(c_\mu)_{\mu \in \Par_n}$ will have $\ZZ$-entries,
so $\sigma'$ lies in $A'$.  Furthermore, $\sigma''= \sigma - \sigma'$ will lie in $A$,
and hence by the orthogonality of $\sigma',\sigma''$,
$$
1=(\sigma,\sigma) = (\sigma',\sigma')+(\sigma'',\sigma'').
$$
One concludes that either $\sigma''=0$, or $\sigma'=0$.  The latter cannot
occur since it would mean that $\sigma=\sigma''$ is perpendicular to all of $A'$.
But $\rho^n=h_1^n$ lies in $A'$, and $(\sigma,\rho^n) \neq 0$.  Thus $\sigma''=0$,
meaning $\sigma=\sigma'$ lies in $A'$.  This completes the proof of assertion (e).
Note that in the process, having shown 
$\det(h_{\mu},h_{\lambda})_{\lambda,\mu \in \Par_n}=\pm 1$, one also knows 
that $\{h_\lambda\}_{\lambda \in \Par_n}$ are $\ZZ$-linearly
independent, so that $\{h_1,h_2,\ldots\}$ are algebraically independent,
and $A=\ZZ[h_1,h_2,\ldots]$ is the polynomial algebra generated by
$\{h_1,h_2,\ldots\}$.

For assertion (f), we have seen that $\omega$ gives such a PSH-automorphism $A \rightarrow A$, swapping $h_n \leftrightarrow e_n$.  Conversely, given a
PSH-automorphism $A \overset{\varphi}{\rightarrow} A$, consider the
positive, multiplicative, normalized $\ZZ$-linear map 
$\delta:=\delta_h \circ \varphi: A \rightarrow \ZZ$.
Proposition~\ref{normalized-character-prop} shows that either
\begin{itemize}
\item $\delta=\delta_h$, which then forces $\varphi(h_n)=h_n$ for all $n$,
so $\varphi=\one_A$, or 
\item
$\delta=\delta_e$,
which then forces $\varphi(e_n)=h_n$ for all $n$,
so $\varphi=\omega$.
\end{itemize}

For assertion (g), given a PSH $A$ with PSH-basis $\Sigma$ having
exactly one primitive $\rho$, since we have seen $A=\ZZ[h_1,h_2,\ldots]$,
where $h_n$ in $A$ is as defined in 
Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}, 
one can uniquely define an algebra morphism
$A \overset{\varphi}{\rightarrow} \Lambda$ that sends the element
$h_n$ to the
complete homogeneous symmetric function $h_n(\xx)$.  Assertions (d) and (e) 
show that $\varphi$ is a bialgebra isomorphism, and hence it is a Hopf
isomorphism.  To show that it is a PSH-isomorphism, we first note that
it is an isometry because one can iterate 
Proposition~\ref{skewing-properties-prop}(iii) together with assertions (c) and
(d) to compute all inner products
$$
( h_\mu, h_\lambda )_A 
=( \one , h_\mu^\perp h_\lambda )_A 
=(1, h_{\mu_1}^\perp h_{\mu_2}^\perp\cdots \left( h_{\lambda_1} h_{\lambda_2} \cdots \right) )_A
$$
for $\mu, \lambda$ in $\Par_n$.  Hence 
$$
( h_\mu, h_\lambda )_A 
= ( h_\mu(\xx), h_\lambda(\xx) )_\Lambda
= ( \varphi(h_\mu), \varphi(h_\lambda) )_\Lambda
$$
Once one knows $\varphi$ is an isometry, then elements 
$\omega$ in $\Sigma \cap A_n$
are characterized in terms of the form $(\cdot,\cdot)$
by $(\omega,\omega)=1$ and $(\omega, \rho^n) > 0$.  
Hence $\varphi$ sends each $\sigma$ in $\Sigma$ to a 
Schur function $s_\lambda$, and is a PSH-isomorphism.
\end{proof}




\newpage

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Complex representations 
for $\Symm_n$, wreath products, $GL_n(\FF_q)$}
\label{representations-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

After reviewing the basics that we will need from representation and character theory of finite groups, we give Zelevinsky's three main examples of PSH's arising as spaces of virtual characters for three towers of finite groups:  
\begin{enumerate}
\item[$\bullet$]
{\it symmetric} groups, 
\item[$\bullet$]
their {\it wreath products} with any finite group, and 
\item[$\bullet$]
the finite {\it general linear} groups.
\end{enumerate}

\subsection{Review of complex character theory}
\label{character-theory-review-section}

A good source for this material, including the crucial Mackey formula, is Serre \cite[Chaps. 1-7]{Serre}.

\subsubsection{Basic definitions, Maschke, Schur}

For a group $G$, a {\it representation} is a homomorphism $G \overset{\varphi}{\rightarrow} GL(V)$ for some vector space $V$ over a field.  We will take the field to be $\CC$ from now on, and we will also assume that $V$ is finite-dimensional over $\CC$.  Thus a representation of $G$ is the same as a finite-dimensional (left-)$\CC G$-module $V$. 

We also assume that $G$ is finite, so that
Maschke's Theorem\footnote{... which has a beautiful generalization to finite-dimensional Hopf algebras due to Larson and Sweedler; see Montgomery \cite[\S 2.2]{Montgomery}.} says that $\CC G$ is semisimple, meaning that every $\CC G$-module $U \subset V$ has a $\CC G$-module complement $U'$ with $V = U \oplus U'$.  Equivalently, {\it indecomposable} $\CC G$-modules are the same thing as {\it simple} (={\it irreducible}) $\CC G$-modules.

Schur's Lemma implies that for two simple $\CC G$-modules $V_1,V_2$, one has
$$
\Hom_{\CC G}(V_1,V_2) \cong 
\begin{cases}
\CC & \text{ if }V_1 \cong V_2,\\
0 & \text{ if } V_1 \not\cong V_2.
\end{cases}
$$


\subsubsection{Characters and Hom spaces}

A $\CC G$-module $V$ is completely determined up to isomorphism by its
{\it character} 
$$
\begin{array}{rcl}
G & \overset{\chi_V}{\longrightarrow} &\CC \\
g & \longmapsto &\chi_V(g):=\trace(g: V \rightarrow V).
\end{array}
$$
This character $\chi_V$ is a {\it class function}, meaning it is 
constant on $G$-conjugacy classes.  The space $R_\CC(G)$ of 
class functions $G \rightarrow \CC$ has a Hermitian, positive definite
form 
$$
(f_1,f_2)_G:=\frac{1}{|G|} \sum_{g \in G} f_1(g) \overline{f_2(g)}.
$$
For any two $\CC G$-modules $V_1, V_2$,
$$
( \chi_{V_1}, \chi_{V_2} )_G=
\dim_\CC \Hom_{\CC G}(V_1,V_2). 
$$
The set of all {\it irreducible characters}
$$
\Irr(G)=\{ \chi_V : V \text{ is a simple }\CC G\text{-module}\}
$$
forms an orthonormal basis of $R_\CC(G)$
with respect to this form, and spans a $\ZZ$-sublattice 
$$
R(G):=\ZZ \Irr(G) \subset R_\CC(G)
$$
sometimes called the {\it virtual characters} of $G$.  


\subsubsection{Tensor products}

Given two groups $G_1,G_2$ and $\CC G_i$-modules $V_i$ for $i=1,2$, 
their tensor product $V_1 \otimes_\CC V_2$ becomes a $\CC[G_1 \times G_2]$-module
via $(g_1,g_2)(v_1 \otimes v_2)=g_1(v_1) \otimes g_2(v_2)$.  When $V_1, V_2$ are
both simple, then so is $V_1 \otimes V_2$, and all 
simple $\CC[G_1 \times G_2]$-modules arise this way.  
Thus one has identifications and isomorphisms
$$
\begin{aligned}
\Irr(G_1 \times G_2) &= \Irr(G_1) \times \Irr(G_2)\\
R(G_1 \times G_2) &\cong R(G_1) \otimes_{\ZZ} R(G_2).
\end{aligned}
$$


\subsubsection{Induction and restriction}

Given a subgroup $H < G$ and $\CC H$-module $U$, one can use the fact that $\CC G$ is a $(\CC G, \CC H)$-bimodule to form the {\it induced $\CC G$-module} 
$$
\Ind_H^G U : = \CC G \otimes_{\CC H} U.
$$
The fact that $\CC G$ is free as a (right-)$\CC H$-module\footnote{... which also has a beautiful generalization to finite-dimensional Hopf algebras due to Nichols and Zoeller; see \cite[\S 3.1]{Montgomery}.}
on basis elements $\{t_{g}\}_{gH \in G/H}$
makes this tensor product easy to analyze.  For example one can
compute its character
\begin{equation}
\label{induced-character-formula}
\chi_{\Ind_H^G U}(g) = \Ind_H^G(\chi_U)(g)
 :=\frac{1}{|H|} \sum_{\substack{k \in G:\\ k g k^{-1} \in H}} \chi_U(k g k^{-1}).
\end{equation}
One can also recognize when a $\CC G$-module 
$V$ is isomorphic to $\Ind_H^G U$ for some $\CC H$-module $U$:  this happens
if and only if there is an $H$-stable subspace $U \subset V$ having the property that $V = \bigoplus_{gH \in G/H} gU$.

The {\it restriction} operation 
$V \mapsto \Res^G_H V$ restricts a $\CC G$-module $V$ to a $\CC H$-module.
{\it Frobenius reciprocity} asserts the adjointness between 
$\Ind_H^G$ and $\Res^G_H$
\begin{equation}
\label{Frobenius-reciprocity}
\Hom_{\CC G}( \Ind_H^G U, V ) \cong
\Hom_{\CC H}( U, \Res^G_H V ),
\end{equation}
as a special case ($S=A=\CC G, R=\CC H, B=U,C=V$)
of the general {\it adjoint associativity}
$$
\Hom_S(A \otimes_R B,C) \cong \Hom_R (B,\Hom_S(A,C)) 
$$
for $S,R$ two rings,
$A$ an $(S,R)$-bimodule,
$B$ a left $R$-module,
$C$ a left $S$-module.


\subsubsection{Mackey's formula}

Mackey gave an alternate description of a 
module which has been induced and then restricted.  To state it, for a subgroup $H < G$ and $g$ in $G$,
let $H^g:=g^{-1}Hg$ and ${}^gH:=gHg^{-1}$.  
Given a $\CC H$-module $U$, say defined by a homomorphism 
$H \overset{\varphi}{\rightarrow} GL(U)$, 
let $U^{g}$ denote the $\CC[gHg^{-1}]$-module 
on the same $\CC$-vector space $U$ defined by
the composite homomorphism
$$
\begin{array}{rcccl}
{}^gH &\longrightarrow&H &\overset{\varphi}{\longrightarrow}&GL(U).\\
h &\longmapsto &g^{-1}hg& & 
\end{array}
$$

\begin{theorem}(Mackey's formula)
\label{Mackey-formula}
Consider subgroups $H, K < G$,  and any $\CC H$-module $U$.
If $\{g_1,\ldots,g_t\}$ are double coset representatives for
$K\backslash G /H$, then 
$$
\Res^G_K \Ind_H^G U 
\cong \bigoplus_{i=1}^t 
\Ind^K_{{}^{g_i}H \cap K} \left( \Res^H_{H \cap K^{g_i}} U \right)^{g_i}
$$
\end{theorem}
\begin{proof}
In this proof, all tensor product symbols $\otimes$
should be interpreted as $\otimes_{\CC H}$.
Recall $\CC G$ has $\CC$-basis $\{t_g\}_{g \in G}$.
For subsets $S \subset G$, 
let $\CC[S]$ denote the $\CC$-span of $\{t_g\}_{g \in S}$ in $\CC G$.

Note that each double coset $K g H$ gives rise to a sub-$(K,H)$-bimodule
$\CC[K g H]$ within $\CC G$, and one has
a $\CC K$-module direct sum decomposition
$$
\Ind_H^G U = \CC G \otimes U
= \bigoplus_{i=1}^t \CC[K g_i H] \otimes U.
$$
Hence it suffices to check for any element $g$ in $G$ that
$$
\CC[K g H] \otimes U 
\cong
\Ind^K_{{}^{g}H \cap K} \left( \Res^H_{H \cap K^{g}} U \right)^{g}.
$$
Note that ${}^{g}H \cap K$ is the subgroup of $K$ consisting of
the elements $k$ in $K$ for which $kg H =gH$.  Hence by picking $\{k_1,\ldots,k_s\}$ to be coset representatives for
$K/({}^gH \cap K)$,
one disjointly decompose the double coset
$$
KgH = \bigsqcup_{j=1}^s k_j({}^gH \cap K)gH,
$$
giving a $\CC$-vector space direct sum decomposition
$$
\begin{aligned}
\CC[K g H] \otimes U 
&=\bigoplus_{j=1}^s \CC\left[k_j \left( {}^gH \cap K\right) g H\right]  \otimes U \\
&\cong
\Ind_{{}^gH \cap K}^K 
\CC [\left( {}^gH \cap K \right) g H]  \otimes U.
\end{aligned}
$$
So it remains to check that one has a $\CC [{}^gH \cap K]$-module isomorphism
$$
\CC [\left( {}^gH \cap K \right) g H]  \otimes U
\cong \left( \Res^H_{H \cap K^g} U \right)^g.
$$
Bearing in mind that, for each $k$ in ${}^g H \cap K$ and $h$ in $H$, one has
$g^{-1}kg$ in $H$ and hence
$$
t_{kgh}
= t_g \cdot t_{g^{-1}kg \cdot h}  \otimes u 
=t_g \otimes g^{-1}kgh \cdot u 
$$
one sees that this isomorphism can be defined by mapping 
$$
t_{kgh} \otimes u \longmapsto  g^{-1}kgh \cdot u.
$$
\end{proof}


\subsubsection{Inflation and fixed points}


There are two (adjoint) constructions on representations
that apply when one has a normal subgroup $K \triangleleft G$.
Given a $\CC[G/K]$-module $U$, say defined by the 
homomorphism $G/K \overset{\varphi}{\rightarrow} GL(U)$,
the {\it inflation} of $U$ to a 
$\CC G$-module $\Inf_{G/K}^G U$ has the same underlying space $U$,
and is defined by the composite 
homomorphism $G \rightarrow G/K \overset{\varphi}{\rightarrow} GL(U)$.
We will later use the easily-checked
fact that when $H < G$ is any other subgroup, one has
\begin{equation}
\label{restriction-inflation-commutation}
\Res^G_H \Inf^G_{G/K} U = 
\Inf^H_{H/H \cap K} \Res^{G/K}_{H/H \cap K} U.
\end{equation}

Inflation turns out to be adjoint to the {\it $K$-fixed space construction}
sending a $\CC G$-module $V$ to the $\CC[G/K]$-module
$$
V^K:=\{v \in V: k(v)=v \text{ for }k \in K\}
$$
Note that $V^K$ is indeed a $G$-stable subspace:  
for any $v$ in $V^K$ and $g$ in $G$, one
has that $g(v)$ lies in $V^K$ since an element $k$ in $K$ satsifies
$kg(v) = g \cdot g^{-1}kg(v) = g(v)$ as $g^{-1}kg$ lies in $K$.
One has this adjointness
\begin{equation}
\label{inflation-reciprocity}
\Hom_{\CC G}(\Inf_{G/K}^G U, V) = \Hom_{\CC[G/K]}(U, V^K)
\end{equation}
because any $\CC G$-module homomorphism $\varphi$ on the left must
have the property that $k \varphi(u) = \varphi(k(u)) = \varphi(u)$
for all $k$ in $K$, so that $\varphi$ actually lies on the right.
We will also need the following formula for the
character $\chi_{V^K}$ in terms of the character $\chi_V$:
\begin{equation}
\label{deflation-character-formula}
\chi_{V^K}(gK) = \frac{1}{|K|} \sum_{k \in K} \chi_V(gk).
\end{equation}
To see this, note that when one has a $\CC$-linear
endomorphism $\varphi$ on a space $V$ that preserves some
$\CC$-subspace $W \subset V$, if $V \overset{\pi}{\rightarrow} W$
is any idempotent projection onto $W$, then the trace of the restriction
$\varphi|_W$ equals the trace of $\varphi \circ \pi$ on $V$.
Applying this to $W=V^K$ and $\varphi = g$, with 
$\pi=\frac{1}{|K|} \sum_{k \in K} k$, gives
\eqref{deflation-character-formula}.


\subsubsection{Semidirect products}
Recall that a {\it semidirect product} is a group $G \ltimes K$
having two subgroups $G,K$ with 
\begin{enumerate}
\item[$\bullet$]
$K \triangleleft (G \ltimes K)$ is a normal subgroup,
\item[$\bullet$]
$G \ltimes K=GK=KG$, and 
\item[$\bullet$]
$G \cap K=\{e\}$.
\end{enumerate}
In this setting one has two interesting adjoint constructions,
applied in Section~\ref{wreath-product-section}.

\begin{proposition}
\label{semidirect-product-adjoint-functors}
Fix a $\CC[G \ltimes K]$-module $V$.
\begin{enumerate}
\item[(i)] 
For any $\CC G$-module $U$, one has $\CC[G \ltimes K]$-module 
structure 
$$
\Phi(U):=U \otimes V,
$$ 
determined via
$$
\begin{aligned}
k(u \otimes v)&= u \otimes k(v),\\
g(u \otimes v)&= g(u) \otimes g(v).\\
\end{aligned}
$$
\item[(ii)]
For any $\CC[G \ltimes K]$-module $W$, one has $\CC G$-module 
structure 
$$
\Psi(W):=
\Hom_{\CC K}(\Res^{G \ltimes K}_K V, \Res^{G \ltimes K}_K W),
$$ 
determined via
$
g(\varphi)=g \circ \varphi \circ g^{-1}.
$
\item[(iii)]
The maps
$$
\CC G-\mods 
\underset{\Psi}{\overset{\Phi}{\rightleftharpoons}} 
\CC[G \ltimes K]-\mods
$$
are adjoint in the sense that one has an isomorphism 
$$
\begin{array}{ccc}
\Hom_{\CC G}(U,\Psi(W))  
  &\longrightarrow & 
      \Hom_{\CC[G\ltimes K]}(\Phi(U),W) \\
\Vert &  & \Vert\\
\Hom_{\CC G}(U,\Hom_{\CC K}(\Res^{G \ltimes K}_K V, \Res^{G \ltimes K}_K W)) & &
\Hom_{\CC[G\ltimes K]}(U \otimes V,W) \\
& & \\
\varphi & \longmapsto & \overline{\varphi}(u \otimes v):=\varphi(u)(v)
\end{array}
$$
\item[(iv)]
One has a $\CC G$-module isomorphism
$$
(\Psi \circ \Phi)(U) \cong U \otimes \End_{\CC K}(\Res^{G \ltimes K}_K  V).
$$ 
In particular, if $\Res^{G \ltimes K}_K  V$
is a simple $\CC K$-module, then $(\Psi \circ \Phi)(U) \cong U$.

\end{enumerate}
\end{proposition}
\begin{proof}
These are mostly straightforward exercises in the definitions.  
To check assertion (iv), for example, note that $K$ acts only
in the right tensor factor in 
$\Res^{G \ltimes K}_K(U \otimes V)$,
and hence as $\CC G$-modules one has
$$
\begin{aligned}
(\Psi \circ \Phi)(U)
        &= \Hom_{\CC K}(\Res^{G \ltimes K}_K V, \,\, 
             \Res^{G \ltimes K}_K(U \otimes V)) \\
        &= \Hom_{\CC K}(\Res^{G \ltimes K}_K V, \,\, 
             U \otimes \Res^{G \ltimes K}_K V) \\
        &= U \otimes \Hom_{\CC K}(\Res^{G \ltimes K}_K V, \,\,
                                  \Res^{G \ltimes K}_K V) \\
        &= U \otimes \End_{\CC K}(\Res^{G \ltimes K}_K V) 
\end{aligned}
$$ 
\end{proof}





\subsection{Three towers of groups}
\label{three-towers-section}

Here we consider three towers of groups
$$
G_*= (G_1 < G_2 < G_3 < \cdots)
$$
where either
\begin{enumerate}
\item[$\bullet$] $G_n=\Symm_n$, the {\it symmetric group}, or
\item[$\bullet$] $G_n=\Symm_n[\Gamma]$, the {\it wreath product} of the 
symmetric group with some arbitrary finite group $\Gamma$, or
\item[$\bullet$] $G_n=GL_n(\FF_q)$, the finite general linear group.
\end{enumerate}
Here the wreath product $\Symm_n[\Gamma]$ can be thought of informally as the group of {\it monomial} $n \times n$ matrices with entries in $\Gamma$, that is,
$n \times n$ matrices having exactly one nonzero entry in each row and column, and that entry is an element of $\Gamma$.  E.g.
$$
\left[
\begin{matrix}
0 & g_2 & 0 \\
g_1 & 0 & 0 \\
0   & 0 & g_3
\end{matrix}
\right]
\left[
\begin{matrix}
0   & 0   & g_6 \\
0   & g_5 & 0 \\
g_4   & 0 & 0
\end{matrix}
\right]
=
\left[
\begin{matrix}
0        & g_2g_5   & 0 \\
0        & 0        & g_1 g_6 \\
g_3g_4   & 0 & 0
\end{matrix}
\right].
$$
More formally, $\Symm_n[\Gamma]$ is 
the semidirect product $\Symm_n \ltimes \Gamma^n$
in which $\Symm_n$ acts on $\Gamma^n$ via 
$\sigma(\gamma_1,\ldots,\gamma_n)=
(\gamma_{\sigma^{-1}(1)},\ldots,\gamma_{\sigma^{-1}(n)}).
$

For each of the three towers $G_*$, 
there are embeddings $G_i \times G_j \hookrightarrow G_{i+j}$ 
and we introduce maps $\ind^{i+j}_{i,j}$ taking $\CC[G_i \times G_j]$-modules to
$\CC G_{i+j}$-modules, as well as maps $\res^{i+j}_{i,j}$ carrying modules
in the reverse direction which are adjoint:
$$
\Hom_{\CC G_{i+j}}( \ind^{i+j}_{i,j} U , V ) = \Hom_{\CC[G_i \times G_j]}( U, \res^{i+j}_{i,j} V ).
$$

\begin{definition}
\label{ind-res-defn}
For $G_n=\Symm_n$, one embeds $\Symm_i \times \Symm_j$ into $\Symm_{i+j}$
as the permutations that permute $\{1,2,\ldots,i\}$ and $\{i+1,i+2,\ldots,i+j\}$
separately.  Here one defines
$$
\begin{aligned}
\ind_{i,j}^{i+j}&:=\Ind_{\Symm_i \times \Symm_j}^{\Symm_{i+j}},\\
\res_{i,j}^{i+j}&:=\Res_{\Symm_i \times \Symm_j}^{\Symm_{i+j}}.
\end{aligned}
$$
For $G_n=\Symm_n[\Gamma]$, similarly embed 
$\Symm_i[\Gamma] \times \Symm_j[\Gamma]$ into 
$\Symm_{i+j}[\Gamma]$ as block monomial matrices whose two diagonal blocks
have sizes $i,j$ respectively, and define
$$
\begin{aligned}
\ind_{i,j}^{i+j}&:=\Ind_{\Symm_i[\Gamma] \times \Symm_j[\Gamma]}^{\Symm_{i+j}[\Gamma]},\\
\res_{i,j}^{i+j}&:=\Res_{\Symm_i[\Gamma] \times \Symm_j[\Gamma]}^{\Symm_{i+j}[\Gamma]}.
\end{aligned}
$$
For $G_n=GL_n(\FF_q)$, which we will denote just $GL_n$,
similarly embed $GL_i \times GL_j$ into 
$GL_{i+j}$ as block diagonal matrices whose two diagonal blocks
have sizes $i,j$ respectively.  However, one also introduces
as an intermediate the {\it parabolic subgroup} $P_{i,j}$ consisting
of the block upper-triangular matrices of the form
$$
\left[
\begin{matrix} 
g_i & \ell\\
0   & g_j
\end{matrix}
\right]
$$
where $g_i, g_j$ lie in $GL_i,GL_j$, respectively, and
$\ell$ in $\FF_q^{i \times j}$ is arbitrary.  One has 
a quotient map $P_{i,j} \rightarrow GL_i \times GL_j$ whose kernel $K_{i,j}$
is the set of matrices of the form
$$
\left[
\begin{matrix} 
I_i & \ell\\
0   & I_j
\end{matrix}
\right]
$$
with $\ell$ again arbitrary.
Here one defines
$$
\begin{aligned}
\ind_{i,j}^{i+j}&:=\Ind_{P_{i,j}}^{GL_{i+j}} \Inf^{P_{i,j}}_{GL_i \times GL_j},\\
\res_{i,j}^{i+j}&:=\left(\Res^{GL_{i+j}}_{P_{i,j}}(-)\right)^{K_{i,j}}.
\end{aligned}
$$
\end{definition}

The operation $\ind_{i,j}^{i+j}$ is sometimes 
called {\it parabolic induction} or {\it Harish-Chandra induction}.
The operation $\res_{i,j}^{i+j}$ is essentially just the $K_{i,j}$-fixed point construction $V \mapsto V^{K_{i,j}}$.
However writing it as the above two-step composite makes it more obvious,
(via \eqref{Frobenius-reciprocity} and \eqref{inflation-reciprocity})
that $\res_{i,j}^{i+j}$ is again adjoint to $\ind_{i,j}^{i+j}$.

\begin{definition}
For each of the three towers $G_*$, define a graded $\ZZ$-module
$$
A:=A(G_*)=\bigoplus_{n \geq 0} R(G_n)
$$
with a bilinear form $(\cdot,\cdot)_A$ whose restriction to $A_n:=R(G_n)$
is the usual form $(\cdot,\cdot)_{G_n}$, and so that
$\Sigma=\{\chi\}=\sqcup_{n \geq 0} \Irr(G_n)$ gives an orthonormal
$\ZZ$-basis.  Here we adopt the convention that $A_0=\ZZ$ has
its basis element $\one$ equal to the unique irreducible character
of the trivial group $G_0$.  
\end{definition}


Bearing in mind that $A_n=R(G_n)$ and
$$
A_i \otimes A_j =
R(G_i) \otimes R(G_j) \cong
R(G_i \times G_j)
$$
one then has candidates for product and coproducts
defined by
$$
\begin{array}{rrcl}
m:=\ind_{i,j}^{i+j}:&A_i \otimes A_j & \longrightarrow & A_{i+j} \\
\Delta:=\bigoplus_{i+j=n} \res_{i,j}^{i+j}: &A_n & \longrightarrow & \bigoplus_{i+j=n} A_i \otimes A_j. \\
\end{array}
$$
Associativity of $m$ and coassociativity of $\Delta$
are easy consequences of transitivity of the 
constructions of induction, restriction, inflation and fixed points.
Endowing $A=\bigoplus_{n \geq 0} R(G_n)$ with the 
obvious unit and counit maps, it thus becomes a graded, finite-type
$\ZZ$-algebra and $\ZZ$-coalgebra.  

The next section address the issue of why 
they form a bialgebra.  However, assuming this for the moment,
it should be clear that each of these algebras $A$ is a PSH having
$\Sigma=\sqcup_{n \geq 0} \Irr(G_n)$ as its PSH-basis.
$\Sigma$ is self-dual because $m, \Delta$ are defined by adjoint maps,
and it is positive because $m, \Delta$ take irreducible representations
to genuine representations not just virtual ones, and
hence have characters which are nonnegative sums of irreducible characters.


\subsection{Bialgebra and double cosets}
\label{double-coset-section}

To show that the algebra and coalgebras $A=A(G_*)$ are bialgebras,
the central issue is checking the pentagonal diagram in \eqref{bialgebra-diagrams}, that is, as maps $A \otimes A \rightarrow A \otimes A$, one has
\begin{equation}
\label{pentagonal-diagram-written-out}
(\Delta \otimes \Delta) \circ m = (m \otimes m) \circ (\one \otimes T \otimes \one) \circ (\Delta \otimes \Delta).
\end{equation}

In checking this, it is convenient to have a lighter notation for
various subgroups of the groups $G_n$ corresponding to compositions $\alpha$.

\begin{definition}
Given a composition $\alpha=(\alpha_1,\ldots,\alpha_\ell)$ of $n$,
define subgroups
$$
G_\alpha \cong G_{\alpha_1} \times \cdots \times G_{\alpha_\ell}
< G_n
$$
via block-diagonal embeddings with diagonal blocks of
size $(\alpha_1,\ldots,\alpha_\ell)$, 
called {\it Young subgroups} $\Symm_\alpha$ when $G_n =\Symm_n$, 
and {\it Levi subgroups} when $G_n=GL_n$.
In the case where $G_n=GL_n$, also define the {\it parabolic subgroups}
$
P_\alpha 
$
to be the subgroup of $G_n$ consisting of block-upper triangular
matrices whose diagonal blocks have sizes $(\alpha_1,\ldots,\alpha_\ell)$,
and the kernels
$
K_\alpha
$
of the obvious surjections $P_\alpha \rightarrow G_\alpha$ which sends
a block upper-triangular matrix to the product of its diagonal block
matrices whose sizes are $\alpha$.  Also define operators $\ind^n_\alpha,
\res^n_\alpha$ for the three towers analogous to those defined in
Definition~\ref{ind-res-defn} when $\alpha=(i,j)$ has only two parts.
\end{definition}


Using homogeneity, checking the bialgebra condition 
\eqref{pentagonal-diagram-written-out}
in the homogeneous component $(A \otimes A)_n$ 
amounts to the following:
for each pair of representations $U_1,U_2$ of
$G_{r_1}, G_{r_2}$ with $r_1+r_2=n$, and 
for each $(c_1,c_2)$ with $c_1+c_2=n$, one
must verify that
\begin{equation}
\label{bialgebra-as-Mackey}
\begin{aligned}
&\res_{c_1,c_2}^n\left(  \ind_{r_1,r_2}^n U_1 \otimes U_2 \right) \\
&\quad \cong \bigoplus_A 
  \left( \ind_{a_{11},a_{21}}^{c_1} \otimes \ind^{c_2}_{a_{12},a_{22}}\right)
   \left( \left( \res^{r_1}_{a_{11},a_{12}} U_1 \otimes \res^{r_2}_{a_{21},a_{22}} U_2 \right)^{\tau_A} \right)
\end{aligned}
\end{equation}
where the direct sum is over all matrices 
$
A=\left[
\begin{matrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{matrix}
\right]
$
in $\NN^{2 \times 2}$ with row sums $(r_1,r_2)$ and column sums $(c_1,c_2)$,
and where $\tau_A$ is the map that uses
the obvious isomorphism between the subgroups
\begin{equation}
\label{two-by-two-matrix-Mackey-subgroups}
\begin{aligned}
 G_{a_{11},a_{12},a_{21},a_{22}} 
&\quad \left( < G_{r_1, r_2} \right)\\
 G_{a_{11},a_{21},a_{12},a_{22}} 
&\quad \left( < G_{c_1,c_2} \right)\\
\end{aligned}
\end{equation}
to identify modules for the first subgroup with modules
for the second subgroup.

As one might guess, \eqref{bialgebra-as-Mackey} comes from the Mackey formula
(Theorem~\ref{Mackey-formula}), 
once one identifies the appropriate double coset representatives.
This is just as easy to do in a slightly more general setting.


\begin{definition}
Given compositions $\alpha,\beta$ of $n$ having lengths $\ell,m$
and a matrix $A$ in $\NN^{\ell \times m}$ with
row sums $\alpha$ and column sums $\beta$, define a permutation $w_A$
in $\Symm_n$ as follows.  Disjointly decompose
$[n]=\{1,2,\ldots,n\}$ into consecutive intervals of numbers
$$
\begin{aligned}[]
[n]&=I_1 \sqcup \cdots \sqcup I_\ell \\
[n]&=J_1 \sqcup \cdots \sqcup J_m
\end{aligned}
$$
so that $|I_i|=\alpha_i, |J_j|=\beta_j$.
Then let $w$ map a consecutive string of $a_{ij}$ of the letters of 
$J_j$ into a consecutive string of $a_{ij}$ of the letters of $I_i$, 
in an increasing fashion.  
\end{definition}

\begin{example}
Taking 
$n=9$ and $\alpha=(4,5), \beta=(3,4,2)$, one has
$$
\begin{array}{ccc}
I_1=\{1,2,3,4\},&I_2=\{5,6,7,8,9\}& \\
J_1=\{1,2,3\}, &J_2=\{4,5,6,7\},&J_3=\{8,9\}
\end{array}
$$
Then one possible matrix $A$ having row and column sums $\alpha, \beta$ is
$
A=\left[ 
\begin{matrix}
2 & 2 & 0\\
1 & 2 & 2
\end{matrix}
\right],
$
and its associated permutation $w_A$ written in two-line notation is
$$
\left(
\begin{matrix}
1 & 2 & 3 &|& 4 & 5 & 6 & 7&|& 8 & 9 \\
\underline{1} & \underline{2} & \underline{\underline{5}}&| 
&\underline{3} & \underline{4} & \underline{\underline{6}} 
&\underline{\underline{7}} &|
&\underline{\underline{8}}
&\underline{\underline{9}}
\end{matrix}
\right)
$$
with vertical lines dividing the sets $J_j$ on top, and
with elements of $I_i$ underlined $i$ times on the bottom.
\end{example}

\begin{proposition}
\label{double-coset-prop}
The permutations $\{w_A\}$ as $A$ runs over matrices $A$ having row, column sums
$\alpha,\beta$ give a system of double coset representatives for
$$
\begin{aligned}
&\Symm_\alpha \backslash \Symm_n / \Symm_\beta \\
&\Symm_\alpha[\Gamma] \backslash \Symm_n[\Gamma] / \Symm_\beta[\Gamma] \\
&P_\alpha \backslash GL_n / P_\beta 
\end{aligned}
$$
\end{proposition}
\begin{proof}
First note that double coset representatives for
$\Symm_\alpha \backslash \Symm_n / \Symm_\beta$ should also provide
double coset representatives for
$\Symm_\alpha[\Gamma] \backslash \Symm_n[\Gamma] / \Symm_\beta[\Gamma]$,
since 
$$
\Symm_\alpha[\Gamma]=\Symm_\alpha  \Gamma^n=\Gamma^n \Symm_\alpha.
$$
We give an algorithm to show 
that every double coset $\Symm_\alpha w \Symm_\beta$
contains some $w_A$.  Start by altering
$w$ within its coset $w\Symm_\beta$, that is, by permuting the
{\it positions} within each set $J_j$, 
to obtain a representative $w'$ for $w\Symm_\beta$ in 
which each set $w'(J_j)$ appears in increasing order in the second line of the
two-line notation for $w'$.  Then alter $w'$ within its coset
$\Symm_\alpha w'$, that is, by permuting the {\it values} within each set
$I_i$, to obtain a representative
$w_A$ having the elements of each set $I_i$ appearing
in increasing order in the second line;  because the values within
each set $I_i$ are consecutive, this alteration will not ruin the property
that one had each set $w'(J_i)$ appearing in increasing order.  
For example, one might have
$$
\begin{aligned}
w&=\left(
\begin{matrix}
1 & 2 & 3 &|& 4 & 5 & 6 & 7&|& 8 & 9 \\
\underline{4} & \underline{\underline{8}} & \underline{2}&| 
&\underline{\underline{5}} & \underline{3} & \underline{\underline{9}} 
&\underline{1} &|
&\underline{\underline{7}}
&\underline{\underline{6}}
\end{matrix}
\right)\\
w'&=\left(
\begin{matrix}
1 & 2 & 3 &|& 4 & 5 & 6 & 7&|& 8 & 9 \\
\underline{2} & \underline{4} & \underline{\underline{8}}&| 
&\underline{1} & \underline{3} & \underline{\underline{5}} 
&\underline{\underline{9}} &|
&\underline{\underline{6}}
&\underline{\underline{7}}
\end{matrix}
\right) \in w \Symm_\beta\\
w_A&=\left(
\begin{matrix}
1 & 2 & 3 &|& 4 & 5 & 6 & 7&|& 8 & 9 \\
\underline{1} & \underline{2} & \underline{\underline{5}}&| 
&\underline{3} & \underline{4} & \underline{\underline{6}} 
&\underline{\underline{7}} &|
&\underline{\underline{8}}
&\underline{\underline{9}}
\end{matrix}
\right) \in \Symm_\alpha w' \subset \Symm_\alpha w' \Symm_\beta = \Symm_\alpha w \Symm_\beta\\
\end{aligned}
$$

Next note that 
$\Symm_\alpha w_A \Symm_\beta=\Symm_\alpha w_B \Symm_\beta$ implies
$A=B$, since the quantities
$$
a_{i,j}(w):=|w(J_j) \cap I_i|
$$ 
are easily seen to be constant on double cosets
$\Symm_\beta w \Symm_\beta$.  

A similar argument shows that
$P_\alpha w_A P_\beta=P_\alpha w_B P_\beta$ implies
$A=B$:  for $g$ in $GL_n$,
the rank $r_{ij}(g)$ of the matrix obtained by restricting $g$ to
rows $I_i \sqcup I_{i+1} \sqcup \cdots \sqcup I_\ell$
and columns $J_1 \sqcup J_2 \sqcup \cdots \sqcup J_j$ is
constant on double cosets $P_\alpha g P_\beta$, and 
for a permuation matrix $w$ one can recover $a_{i,j}(w)$ from the formula
$$
a_{i,j}(w) = 
r_{i,j}(w)-r_{i,j-1}(w)-r_{i+1,j}(w) + r_{i+1,j-1}(w).
$$ 

Thus it only remains to show that every double coset $P_\alpha g P_\beta$ contains
some $w_A$.  Since $\Symm_\alpha < P_\alpha$, and we have seen already that
every double coset $\Symm_\alpha w \Symm_\beta$ contains some $w_A$,
it suffices to show that every double coset 
$P_\alpha g P_\beta$ contains some permutation $w$.  However, we claim that 
this is  already true for the smaller double cosets $BwB$ where $B=P_{1^n}$ is 
the {\it Borel subgroup} of upper triangular invertible matrices, that is, one has the usual {\it Bruhat decomposition} 
$$
GL_n = \bigsqcup_{w \in \Symm_n} BwB.
$$
To prove this decomposition, we show how to
find a permutation $w$ in each double coset $BgB$.
The freedom to alter $g$ within its coset $gB$ allows one to scale columns and
adding scalar multiples of earlier columns to later columns.  We claim that
using such column operations, one can
always find a representative $g'$ for coset $gB$ in which
\begin{enumerate}
\item[$\bullet$]
the bottommost nonzero entry of each column has been scaled to $1$ (call this a {\it pivot}), 
\item[$\bullet$]
the entries to right of each pivot within its row are all $0$, and
\item[$\bullet$]
there is one pivot in each row and each column, so that they
lie in the positions of some permutation matrix $w$.
\end{enumerate}
In fact, we will see below that $BgB=BwB$ in this case.
The algorithm which produces $g'$ from $g$ is simple:  starting with the 
leftmost column, find its bottommost nonzero entry, and scale the column
to make this entry a $1$, creating the pivot in this column.  Now use this 
pivot to clear out all entries in its row to its right, using
column operations that subtract multiples of this column from later columns.
Having done this, move on to the next column to the right, and repeat, scaling to create a pivot, and using it to eliminate entries to its right.

For example, the typical matrix $g$ lying in the double coset $BwB$ where
$$
w=\left(
\begin{matrix}
1 & 2 & 3 &|& 4 & 5 & 6 & 7&|& 8 & 9 \\
\underline{4} & \underline{\underline{8}} & \underline{2}&| 
&\underline{\underline{5}} & \underline{3} & \underline{\underline{9}} 
&\underline{1} &|
&\underline{\underline{7}}
&\underline{\underline{6}}
\end{matrix}
\right)\\
$$
from before is one that can be altered within its coset $gB$ to look like this:
$$
g'=\left[
\begin{matrix}
* & * & * & * & * & * & 1 & 0 & 0 \\
* & * & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
* & * & 0 & * & 1 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & * & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & * & 0 & 0 & 0 & * & 0 & * & 1 \\
0 & * & 0 & 0 & 0 & * & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
\end{matrix}
\right] \in gB.
$$
Having found this $g'$ in $gB$, a similar algorithm using left multiplication
by $B$ shows that $w$ lies in 
$Bg' \subset Bg'B=BgB$.  This time no scalings are required
to create the pivot entries:  starting with the bottom row, one
uses its pivot to eliminate all the entries above it in the same column
(shown by stars $*$ above)
by adding multiples of the bottom row to higher rows. Then do the same using the pivot in the next-to-bottom row, etc.  The result is the permutation matrix for $w$.
\end{proof}

\begin{corollary}
\label{three-towers-give-bialgebras}
For each of the three towers of groups $G_*$, the product and coproduct
structures on $A=A(G_*)$ endow it with a bialgebra structure, and hence
they form PSH algebras.
\end{corollary}
\begin{proof}
The first two towers $G_n = \Symm_n$ and $G_n=\Symm_n[\Gamma]$ have
product, coproduct defined by induction, restriction along 
embeddings $G_i \times G_j < G_{i+j}$.  Hence the desired bialgebra equality
\eqref{bialgebra-as-Mackey} follows from Mackey's Theorem~\ref{Mackey-formula}
and Proposition~\ref{double-coset-prop},
taking 
$
G=G_n,
H=G_{(r_1,r_2)},
K=G_{(c_1,c_2)}
$
and checking for a given double coset 
$$
HgK=(G_{r_1,r_2})w_A(G_{c_1,c_2})
$$
indexed by a matrix $A$ in $\NN^{2 \times 2}$ with row sums $(r_1,r_2)$ and
column sums $(c_1,c_2)$, that the two subgroups appearing on the left in 
\eqref{two-by-two-matrix-Mackey-subgroups} are exactly
$$
\begin{aligned}
{g}^H \cap K &= {}^w(G_{r_1,r_2}) \cap G_{c_1,c_2},\\
H \cap K^g &= G_{r_1,r_2} \cap (G_{c_1,c_2})^w.
\end{aligned}
$$
One should also check that the isomorphism $\tau_A$ between the two subgroups
in \eqref{two-by-two-matrix-Mackey-subgroups} is the conjugation isomorphism
by $w$.  We leave all of these bookkeeping details to the reader to check.

For the tower with $G_n=GL_n$, there is slightly more work to be done
to check the equality \eqref{bialgebra-as-Mackey}.
Via Mackey's Theorem~\ref{Mackey-formula} and 
Proposition~\ref{double-coset-prop}, the left side is
\begin{equation}
\label{left-side-GL-bialgebra}
\begin{aligned}
&\res_{c_1,c_2}^n\left(  \ind_{r_1,r_2}^n U_1 \otimes U_2 \right) \\
&= \left( 
 \Res^{G_n}_{P_{c_1,c_2}} \Ind^{G_n}_{P_{r_1,r_2}} \Inf^{P_{r_1,r_2}}_{G_{r_1,r_2}} U_1 \otimes U_2
\right)^{K_{c_1,c_2}} \\
&= \bigoplus_A 
\left( 
 \Ind^{P_{c_1,c_2}}_{P_{r_1,r_2} \cap P_{c_1,c_2}^{w_A}} 
   \left( \Res^{P_{r_1,r_2}}_{P_{r_1,r_2} \cap P_{c_1,c_2}^{w_A}} 
      \Inf^{P_{r_1,r_2}}_{G_{r_1,r_2}} U_1 \otimes U_2
   \right)^{\tau_A}
\right)^{K_{c_1,c_2}}
\end{aligned}
\end{equation}
where $A$ runs over the usual $2 \times 2$ matrices.
The right side is a direct sum over the same set of $2 \times 2$ matrices $A$:
\begin{equation}
\label{right-side-GL-bialgebra}
\begin{aligned}
&\bigoplus_A 
  \left( \ind_{a_{11},a_{21}}^{c_1} \otimes \ind^{c_2}_{a_{12},a_{22}}\right)
   \left( \left( \res^{r_1}_{a_{11},a_{12}} U_1 \otimes \res^{r_2}_{a_{21},a_{22}} U_2 \right)^{\tau_A} \right) \\
&= \bigoplus_A 
 \Ind^{G_{c_1,c_2}}_{P_{a_{11},a_{21}} \times P_{a_{12},a_{22}}}  \\
&\qquad \qquad
  \Inf^{P_{a_{11},a_{21}} \times P_{a_{12},a_{22}}}_{G_{a_{11},a_{21},a_{12},a_{22}}}
    \left(
     \left(
      \Res^{G_{r_1,r_2}}_{P_{a_{11},a_{12}} \times P_{a_{21},a_{22}}}  U_1 \otimes U_2
     \right)^{K_{a_{11},a_{21}} \times K_{a_{12},a_{22}}} 
    \right)^{\tau_A}
\end{aligned}
\end{equation}
Thus it suffices to check for each $2 \times 2$ matrix $A$
that any $\CC G_{c_1,c_2}$-module of the
form $V_1 \otimes V_2$ has the same inner product with the $A$-summands of
\eqref{left-side-GL-bialgebra} and \eqref{right-side-GL-bialgebra}.
Abbreviate $w:=w_A$ and $\tau:=\tau_A$.  Computing first the inner product 
with the $A$-summand of
\eqref{left-side-GL-bialgebra}, and using adjointness properties and \eqref{restriction-inflation-commutation}, one gets
$$
\begin{aligned}
&
\left(
\left( \Res^{P_{r_1,r_2}}_{P_{r_1,r_2} \cap P_{c_1,c_2}^{w}} 
      \Inf^{P_{r_1,r_2}}_{G_{r_1,r_2}} U_1 \otimes U_2
   \right)^{\tau} , \right. \\
& \qquad \qquad \qquad \left.
\Res^{P_{c_1,c_2}}_{{}^{w}P_{c_1,c_2} \cap P_{r_1,r_2}} 
      \Inf^{P_{c_1,c_2}}_{G_{c_1,c_2}} V_1 \otimes V_2
\right)_{{}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}} \\
&=\left(
\left( 
  \Inf^{P_{r_1,r_2} \cap P_{c_1,c_2}^{w}}_{G_{r_1,r_2} \cap P_{c_1,c_2}^{w}} 
   \Res^{G_{r_1,r_2}}_{G_{r_1,r_2} \cap P_{c_1,c_2}^{w}} U_1 \otimes U_2
   \right)^{\tau} , \right. \\
& \qquad \qquad \qquad \left.
  \Inf^{{}^wP_{r_1,r_2} \cap P_{c_1,c_2}}_{{}^wP_{r_1,r_2} \cap G_{c_1,c_2}} 
   \Res^{G_{c_1,c_2}}_{{}^wP_{r_1,r_2} \cap G_{c_1,c_2}} V_1 \otimes V_2
\right)_{{}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}} 
\end{aligned}
$$
One can compute this inner product 
by first checking that ${}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}$ is
the group of matrices having the block form
$$
\left[
\begin{matrix}
g_{11} & h    & i & j \\
0     &g_{21} & 0 & k \\
0     & 0    & g_{12} & \ell \\
0     & 0    & 0 & g_{22} 
\end{matrix}
\right]
$$
in which the diagonal blocks $g_{ij}$ for $i,j=1,2$ are invertible of
size $a_{ij} \times a_{ij}$, 
while the blocks $h,i,j,k,\ell$ are all arbitrary matrices
of the appropriate (rectangular) block sizes;  then ${}^{w}P_{r_1,r_2} \cap G_{c_1,c_2}$ is the subgroup where the blocks $i,j,k$ all vanish.  The inner product above then becomes
\begin{equation}
\label{first-big-inner-product-sum}
\begin{aligned}
&\frac{1}{|{}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}|}
\sum_{\substack{(g_{i,j})\\(h,i,j,k,\ell)}}
  \chi_{U_1}\left(\begin{matrix} g_{11}& i \\ 0 & g_{12} \end{matrix}\right)
  \chi_{U_2}\left(\begin{matrix} g_{21}& k \\ 0 & g_{22} \end{matrix}\right) \\
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad 
  \overline{\chi}_{V_1}\left(\begin{matrix} g_{11}& h \\ 0 & g_{21} \end{matrix}\right)
  \overline{\chi}_{V_2}\left(\begin{matrix} g_{12}& \ell \\ 0 & g_{22} \end{matrix}\right).
\end{aligned}
\end{equation}
If one instead computes the inner product of $V_1 \otimes V_2$ with the $A$-summand of
\eqref{left-side-GL-bialgebra}, using adjointness properties and \eqref{deflation-character-formula} one gets
$$
\begin{aligned}
&\left(
\left(
 \left( 
  \Res^{G_{r_1,r_2}}_{P_{a_{11},a_{12}} \times P_{a_{21},a_{22}}} U_1 \otimes U_2
 \right)^{K_{a_{11},a_{12}} \times K_{a_{21},a_{22}}}
\right)^{\tau} , \right.\\
&\qquad\qquad
 \left.
 \left( 
  \Res^{G_{c_1,c_2}}_{P_{a_{11},a_{21}} \times P_{a_{12},a_{22}}} V_1 \otimes V_2
 \right)^{K_{a_{11},a_{21}} \times K_{a_{12},a_{22}}}
\right)_{G_{a_{11},a_{21},a_{12},a_{22}}} \\
&=\frac{1}{|G_{a_{11},a_{21},a_{12},a_{22}}|}
   \sum_{(g_{ij})} 
     \frac{1}{|K_{a_{11},a_{21}}|}
       \sum_{(i,k)}
          \chi_{U_1}\left(\begin{matrix} g_{11}& i \\ 0 & g_{12} \end{matrix}\right)
          \chi_{U_2}\left(\begin{matrix} g_{21}& k \\ 0 & g_{22} \end{matrix}\right)\\
&\qquad \qquad \qquad \qquad \qquad
     \frac{1}{|K_{a_{12},a_{22}}|} \sum_{(h,\ell)}
  \overline{\chi}_{V_1}\left(\begin{matrix} g_{11}& h \\ 0 & g_{21} \end{matrix}\right)
  \overline{\chi}_{V_2}\left(\begin{matrix} g_{12}& \ell \\ 0 & g_{22} \end{matrix}\right).
\end{aligned}
$$
But this last sum can be seen to equal the sum 
\eqref{first-big-inner-product-sum}, after one notes that
$$
|{}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}|
=
|G_{a_{11},a_{21},a_{12},a_{22}}|
\cdot
|K_{a_{11},a_{21}}|
\cdot
|K_{a_{12},a_{22}}| \cdot \#\{j \in \FF_q^{a_{11} \times a_{22}} \}
$$
and that the summands in  \eqref{first-big-inner-product-sum} are independent of the
matrix $j$ in the summation.
\end{proof}



\subsection{Symmetric groups}
\label{symmeric-group-section}

Finally, some payoff.  
Consider the tower of symmetric groups
$G_n=\Symm_n$, and $A=A(G_*)=A(\Symm)$.  Denote by
$\one_{\Symm_n}, \sgn_{\Symm_n}$ the {\it trivial} and {\it sign} characters on
$\Symm_n$.  For a partition $\lambda$ of $n$, denote by 
$\one_{\Symm_\lambda}, \sgn_{\Symm_\lambda}$ the trivial and sign characters 
restricted to the Young subgroup 
$
\Symm_\lambda = \Symm_{\lambda_1} \times \Symm_{\lambda_2 } \times \cdots,
$
and denote by 
$\one_\lambda$ the class function 
which is the characteristic function
for the $\Symm_n$-conjugacy class of permutations of cycle type $\lambda$.



\begin{theorem}
\label{symmetric-group-Frobenius-map-theorem}
Irreducible complex characters $\{ \chi^\lambda \}$ of
$\Symm_n$ are indexed by partitions $\lambda$ in $\Par_n$,
and one has a PSH-isomorphism, the {\bf Frobenius characteristic map},
$$
A=A(\Symm) \overset{\ch}{\longrightarrow} \Lambda
$$
that sends
$$
\begin{array}{rcl}
\one_{\Symm_n}& \longmapsto & h_n \\
\sgn_{\Symm_n}& \longmapsto & e_n \\
\one_{(n)}& \longmapsto &  \frac{p_n}{n} \\
\chi^\lambda & \longmapsto & s_\lambda \\
\Ind_{\Symm_\lambda}^{\Symm_n} \one_{\Symm_\lambda} & \longmapsto &h_\lambda\\
\Ind_{\Symm_\lambda}^{\Symm_n} \sgn_{\Symm_\lambda} & \longmapsto &e_\lambda\\
\one_\lambda & \longmapsto &\frac{p_\lambda}{z_\lambda}
\end{array}
$$
and where the involution $\omega$ on $\Lambda$ corresponds under $\ch^{-1}$
to the map on each virtual character space $R(\Symm_n)$ given by tensoring
with the sign character $\sgn_{\Symm_n}$.
\end{theorem}
\begin{proof}
Corollary \ref{three-towers-give-bialgebras}
implies that the set $\Sigma=\{\chi\}=\sqcup_{n \geq 0} \Irr(\Symm_n)$
gives a PSH-basis for $A$.
Since a character $\chi$ of $\Symm_n$ has
\begin{equation}
\label{character-coproduct-in-symm}
\Delta(\chi) = \bigoplus_{i+j=n} \Res^{\Symm_n}_{\Symm_i \times \Symm_j} \chi,
\end{equation}
such an element $\chi$ is never primitive for $n \geq 2$.
Hence the unique irreducible character $\rho=\one_{\Symm_1}$ of $\Symm_1$
is the only element of $\CCC = \Sigma \cap \Liep$.  

Thus Theorem~\ref{uniqueness-of-sym-parts-e-f-g} 
tells us that there are two isomorphisms
$A \rightarrow \Lambda$, each of which sends $\Sigma$ to the
PSH-basis of Schur functions $\{s_\lambda\}$ for $\Lambda$.
It also tells us that we can pin down one of the two isomorphisms
to call $\ch$, by insisting that it map the two characters 
$\one_{\Symm_2},\sgn_{\Symm_2}$ in $\Irr(\Symm_2)$ to $h_2,e_2$ (and not
$e_2, h_2$).

Bearing in mind the coproduct formula \eqref{character-coproduct-in-symm},
and the fact that $\one_{\Symm_n}, \sgn_{\Symm_n}$ restrict, respectively, 
to trivial and sign characters of $\Symm_i \times \Symm_j$ for
$i+j=n$, one finds that for $n \geq 2$ one has
$\sgn_{\Symm_2}^\perp$ annihilating $\one_{\Symm_n}$,
and $\one_{\Symm_2}^\perp$ annihilating $\sgn_{\Symm_n}$.
Therefore Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}(b) implies
$
\one_{\Symm_n}, \sgn_{\Symm_n}
$
are sent under $\ch$ to $h_n, e_n$.
Then the fact that
$\Ind_{\Symm_\lambda}^{\Symm_n} \one_{\Symm_\lambda}, \Ind_{\Symm_\lambda}^{\Symm_n} \sgn_{\Symm_\lambda}$
are sent to $h_{\lambda}, e_\lambda$ follows via induction products.

For the assertion about $\one_{(n)}$, note that it is primitive in $A$ 
for $n \geq 1$, because as a class function, the indicator function
of $n$-cycles vanishes upon restriction to $\Symm_i \times \Symm_j$
for $i+j=n$ if both $i,j \geq 1$;  these subgroups contain no $n$-cycles.  
Hence Corollary~\ref{primitives-in-Lambda} implies that
$\ch(\one_{(n)})$ is a scalar multiple of $p_n$.  To pin down the scalar,
note $p_n=m_{(n)}$ so 
$
(h_n,p_n)_{\Lambda}=(h_n,m_n)_{\Lambda}=1,
$
while $\ch^{-1}(h_n)=\one_{\Symm_n}$ has
$$
(\one_{\Symm_n},\one_{(n)})=\frac{1}{n!} \cdot (n-1)!=\frac{1}{n}.
$$
Thus $\ch(\one_{(n)})=\frac{p_n}{n}$.  The fact that 
$\ch(\one_{\lambda})=\frac{p_\lambda}{z_\lambda}$ then follows via induction
product calculations, e.g. using \eqref{induced-character-formula}.

The fact that 
$\omega$ corresponds on $A_n=R(\Symm_n)$ to the operation
$\chi \longmapsto \chi \otimes \sgn_{\Symm_n}$, comes from
Theorem~\ref{uniqueness-of-sym-parts-e-f-g}(f), once one notes that the latter
operation induces a (nontrivial) PSH-algebra automorphism of $A$.
\end{proof}


\begin{remark}
The paper of Liulevicius \cite{Liulevicius} gives
a very elegant alternate approach to the Frobenius map
as a Hopf isomorphism $A(\Symm) \overset{\ch}{\longrightarrow} \Lambda$,
inspired by equivariant $K$-theory and vector bundles over spaces which
are finite sets of points!
\end{remark}





\subsection{Wreath products}
\label{wreath-product-section}

Next consider the tower of groups
$G_n=\Symm_n[\Gamma]$ for a finite group $\Gamma$,
and the Hopf algebra $A=A(G_*)=:A(\Symm[\Gamma])$.  Recall from the previous
section that irreducible complex representations $\chi^\lambda$ of $\Symm_n$ are
indexed by partitions $\lambda$ in $\Par_n$.  Index the 
irreducible complex representations of $\Gamma$ as
$
\Irr(\Gamma) = \{ \rho_1,\ldots,\rho_d\}.
$

\begin{definition}
Define for a partition $\lambda$ in $\Par_n$ 
and $\rho$ in $\Irr(\Gamma)$
a representation $\chi^{\lambda,\rho}$
of $\Symm_n[\Gamma]$ in which $\sigma$ in $\Symm_n$ and
$\gamma=(\gamma_1,\ldots,\gamma_n)$ in $\Gamma^n$ act 
on the space 
$\chi^\lambda \otimes \left( \rho^{\otimes n} \right)$
as follows
\begin{equation}
\label{wreath-irreducible-construction}
\begin{aligned}
\sigma( u \otimes (v_1 \otimes \cdots \otimes v_n) )
&= \sigma(u) \otimes (v_{\sigma^{-1}(1)} \otimes \cdots \otimes v_{\sigma^{-1}(n)}) \\
\gamma( u \otimes (v_1 \otimes \cdots \otimes v_n) )
&= u \otimes (\gamma_1 v_1 \otimes \cdots \otimes \gamma_n v_n) \\
\end{aligned}
\end{equation}
\end{definition}

\begin{theorem}
\label{wreath-product-irreducibles}
The irreducible $\CC \Symm_n[\Gamma]$-modules
are the induced characters 
$$
 \chi^{\underline{\lambda}}: =
\Ind_{\Symm_\lambda[\Gamma]}^{\Symm_n[\Gamma]} 
  \left( 
   \chi^{\lambda^{(1)},\rho_1} \otimes \cdots 
   \otimes \chi^{\lambda^{(d)},\rho_d} 
  \right)
$$
as $\underline{\lambda}$ runs through all functions 
$$
\begin{array}{rcl}
\Irr(\Gamma) &\overset{\underline{\lambda}}{\longrightarrow}&   \Par\\
\rho_i &\longmapsto&  \lambda^{(i)}
\end{array}
$$
with the property that $\sum_{i=1}^d |\lambda^{(i)}|=n$.
Furthermore, one has a PSH-isomorphism
$$
\begin{array}{rcl}
A(\Symm[\Gamma]) &\longrightarrow &\Lambda^{\otimes n} \\
\chi^{\underline{\lambda}} & \longmapsto &
 s_{\lambda^{(1)}} \otimes \cdots \otimes s_{\lambda^{(d)}}.
\end{array}
$$
\end{theorem}
\begin{proof}
We know from Corollary~\ref{three-towers-give-bialgebras} that
$A(\Symm[\Gamma])$ is a PSH, with PSH-basis $\Sigma$
given by the union of all irreducible characters of all
groups $\Symm_n[\Gamma]$.  Therefore 
Theorem~\ref{Zelevinsky-decomposition-theorem}
tells us that 
$A(\Symm[\Gamma]) \cong \bigotimes_{\rho \in \CCC} A(\Symm[\Gamma])(\rho)$
where $\CCC$ is the set of irreducible characters which are
also primitive.  Just as in the case of $\Symm_n$,
it is clear from the definition of the coproduct that
an irreducible character $\rho$ of $\Symm_n[\Gamma]$ is
primitive if and only if $n=1$, that $\Symm_n[\Gamma]=\Gamma$,
and $\rho$ lies in $\Irr(\Gamma)=\{\rho_1,\ldots,\rho_d\}$.

The remaining assertions of the theorem will then follow from the definition
of the induction product algebra structure on $A(\Symm[\Gamma])$, once
we have shown that there is a PSH-isomorphism sending
\begin{equation}
\label{desired-wreath-PSH-iso}
\begin{array}{rcl}
A(\Symm) &\longrightarrow &A(\Symm[\Gamma])(\rho) \\
\chi^{\lambda} &\longmapsto &\chi^{\lambda,\rho}.
\end{array}
\end{equation}
Such an isomorphism comes from applying 
Proposition~\ref{semidirect-product-adjoint-functors}
to the semidirect product $\Symm_n(\Gamma) = \Symm_n \ltimes \Gamma^n$,
so that $K=\Gamma^n, G=\Symm_n$, and fixing $V=\rho^{\otimes n}$ as
$\CC \Symm_n[\Gamma]$-module with structure as defined 
in \eqref{wreath-irreducible-construction}.  
One obtains for each $n$, maps 
$$
A(\Symm_n) \underset{\Psi}{\overset{\Phi}{\rightleftharpoons}} 
A(\Symm_n[\Gamma])
$$
where 
$$
\begin{array}{rcl}
\chi &\overset{\Phi}{\longmapsto} &\chi \otimes (\rho^{\otimes n})\\
\alpha&\overset{\Psi}{\longmapsto}& 
  \Hom_{\CC \Gamma^n}(\rho^{\otimes n},\alpha).
\end{array}
$$
Taking the direct sum of these maps for all $n$ gives maps 
$
A(\Symm) \underset{\Psi}{\overset{\Phi}{\rightleftharpoons}} 
A(\Symm[\Gamma]).
$

These maps are coalgebra morphisms because of
their interaction with restriction to $\Symm_i \times \Symm_j$.
Since Proposition~\ref{semidirect-product-adjoint-functors}(iii)
gives the adjointness property that 
$$
(\chi,\Psi(\alpha))_{A(\Symm)} = (\Phi(\chi),\alpha)_{A(\Symm[\Gamma])}.
$$
one concludes from the self-duality of $A(\Symm), A(\Symm[\Gamma])$
that $\Phi,\Psi$ are also algebra morphisms.  Since they take genuine
characters to genuine characters, they are PSH-morphisms.
Since $\rho$ being a simple $\CC \Gamma$-module implies that
$V=\rho^{\otimes n}$ is a simple $\CC \Gamma^n$-module,
Proposition~\ref{semidirect-product-adjoint-functors}(iv) shows that
\begin{equation}
\label{Psi-Phi-composite}
(\Psi \circ \Phi)(\chi) = \chi
\end{equation}
for all $\Symm_n$-characters $\chi$.   Hence $\Phi$ is an injective
PSH-morphism.   Using adjointness, \eqref{Psi-Phi-composite} also shows that
$\Phi$ sends $\CC \Symm_n$-simples $\chi$ 
to $\CC[\Symm_n[\Gamma]]$-simples $\Phi(\chi)$:
$$
(\Phi(\chi),\Phi(\chi))_{A(\Symm[\Gamma])} 
 = ((\Psi \circ \Phi)(\chi),\chi)_{A(\Symm)}
 = (\chi,\chi)_{A(\Symm)} = 1.
$$
Since $\Phi(\chi)= \chi \otimes (\rho^{\otimes n})$
has $V=\rho^{\otimes n}$ as a constituent upon restriction to $\Gamma^n$,
Frobenius Reciprocity shows that the irreducible character 
$\Phi(\chi)$ is a constituent of 
$\Ind_{\Gamma^n}^{\Symm_n[\Gamma]} \rho^{\otimes n}= \rho^n$.  Hence the
entire image of $\Phi$ lies in $A(\Symm[\Gamma])(\rho)$, and so $\Phi$
must restrict to an isomorphism as desired in \eqref{desired-wreath-PSH-iso}.
\end{proof}

One of Zelevinsky's sample applications of the theorem is this branching rule.

\begin{corollary}
\label{wreath-product-branching-rule}
Given $\underline{\lambda}=(\lambda^{(1)},\ldots,\lambda^{(d)})$ with $\sum_{i=1}^d |\lambda^{(i)}|=n$, 
one has
$$
\Res^{\Symm_n[\Gamma]}_{\Symm_{n-1}[\Gamma] \times \Gamma}  \left( \chi^{\underline{\lambda}} \right)
= \sum_{i=1}^d \sum_{\substack{ \lambda^{(i)}_- \subseteq \lambda^{(i)}: \\ |\lambda^{(i)}/\lambda^{(i)}_-|=1}}
\chi^{(\lambda^{(1)}, \ldots,\lambda^{(i)}_-, \ldots,\lambda^{(d)})} \otimes \rho_i.
$$
\end{corollary}

\begin{example}
For $\Gamma$ a two-element group, so $\Irr(\Gamma)=\{\rho_1,\rho_2\}$ and
$d=2$, then
$$
\Res^{\Symm_6[\Gamma]}_{\Symm_{5}[\Gamma] \times \Gamma}  
\left( \chi^{((3,1),(1,1))} \right)
=
\chi^{((3),(1,1))} \otimes \rho_1 
+\chi^{((2,1),(1,1))} \otimes \rho_1 
+\chi^{((3,1),(1))} \otimes \rho_2.
$$
\end{example}

\begin{proof}[Proof of Corollary~\ref{wreath-product-branching-rule}]
By Theorem~\ref{wreath-product-irreducibles}, this is equivalent to
computing in the Hopf algebra $A:=\Lambda^{\otimes d}$ the component of the
coproduct of $s_{\lambda^{(1)}} \otimes \cdots \otimes s_{\lambda^{(d)}}$
that lies in $A_{n-1} \otimes A_1$.  Working within each tensor factor $\Lambda$,
the Pieri formula implies that the $\Lambda_{n-1} \otimes \Lambda_1$-component
of $\Delta(s_\lambda)$ is
$$
\sum_{\substack{\lambda_- \subseteq \lambda: \\|\lambda/\lambda_-|=1}} s_{\lambda_-} \otimes \rho.
$$
One must apply this in each of the $d$ tensor factors of $A=\Lambda^{\otimes d}$, then sum on $i$.
\end{proof}






\subsection{General linear groups}
\label{general-linear-section}

We now consider the tower of finite general linear groups
$G_n=GL_n=GL_n(\FF_q)$ and $A=A(G_*)=:A(GL)$.  
Corollary~\ref{three-towers-give-bialgebras} tells us that
$A(GL)$ is a PSH, with PSH-basis $\Sigma$
given by the union of all irreducible characters of all
groups $GL_n$.  Therefore Theorem~\ref{Zelevinsky-decomposition-theorem}
tells us that 
\begin{equation}
\label{GL-Zelevinsky-decomposition}
A(GL) \cong \bigotimes_{\rho \in \CCC} A(GL)(\rho)
\end{equation}
where $\CCC=\Sigma \cap \Liep$ is the set of primitive irreducible characters.

\begin{definition}
Call an irreducible representation $\rho$ of $GL_n$ {\it cuspidal}
for $n \geq 1$ if it lies in $\CCC$, that is, its restriction to proper parabolic subgroups $P_{i,j}$ with $i+j=n$
and $i,j>0$ contain no nonzero vectors which are $K_{i,j}$-invariant.
Given an irreducible character $\sigma$ of $GL_n$, say that $d(\sigma)=n$,
and let $\CCC_n:=\{ \rho \in \CCC: d(\rho)=n\}$ for $n \geq 1$
denote the subset of cuspidal characters of $GL_n$.
\end{definition}

Just as was the case for $\Symm_1$ and $\Symm_1[\Gamma]=\Gamma$, {\it every}
irreducible character $\rho$ of $GL_1(\FF_q)=\FF_q^\times$ is cuspidal.
However, this does not exhaust the cuspidal characters.  In fact, one
can predict the number of cuspidal characters in $\CCC_n$, using
knowledge of the number of conjugacy classes in $GL_n$.
Let $\FFF$ denote the set of all nonconstant monic irreducible polynomials $f(x) \neq x$ in $\FF_q[x]$.
Let $\FFF_n:=\{ f \in \FFF: \deg(f)=n\}$ for $n \geq 1$.

\begin{proposition}
The number $|\CCC_n|$ of cuspidal characters of $GL_n(\FF_q)$
is the number of $|\FFF_n|$ of irreducible monic degree $n$ polynomials $f(x) \neq x$
in $\FF_q[x]$ with nonzero constant term.
\end{proposition}
\begin{proof}
We show  $|\CCC_n|=|\FFF_n|$ for $n \geq 1$ by induction on $n$.
For the base case $n=1$, just as with the families $G_n=\Symm_n$ and $G_n=\Symm_n[\Gamma]$, when $n=1$ any irreducible character $\chi$ of $G_1=GL_1(\FF_q)$
gives a primitive element of $A=A(GL)$, and hence is cuspidal.
Since $GL_1(\FF_q)=\FF_q^\times$ is abelian, there are $|\FF_q^\times|=q-1$
such cuspidal characters in $\CCC_1$, which agrees with the fact that
there are $q-1$ monic (irreducible) 
linear polynomials $f(x) \neq x$ in $\FF_q[x]$, namely 
$
\FFF_1:=\{ f(x)=x-c: c \in \FF_q^\times\}.
$


In the inductive step, use the fact that
the number $|\Sigma_n|$ of irreducible complex characters $\chi$
of $GL_n(\FF_q)$ equals its number of conjugacy classes.
These conjugacy classes are 
uniquely represented by {\it rational canonical forms}, 
which are parametrized
by functions $\underline{\lambda}: \FFF \rightarrow \Par$ with
the property that $\sum_{f \in \FFF} \deg(f) |\lambda(f)|=n$.
On the other hand, \eqref{GL-Zelevinsky-decomposition} tells us that
$|\Sigma_n|$ is similarly parametrized by the functions 
$\underline{\lambda}: \CCC \rightarrow \Par$ having
the property that $\sum_{\rho \in \CCC} d(\rho) |\lambda(f)|=n$.
Thus we have parallel disjoint decompositions 
$$
\begin{array}{rll}
\CCC &= \bigsqcup_{ n \geq 1} \CCC_n 
         & \text{ where }\CCC_n=\left\{\rho \in \CCC:d(\rho)=n\right\}\\
\FFF &= \bigsqcup_{ n \geq 1} \FFF_n 
         & \text{ where }\FFF_n=\left\{\rho \in \FFF:\deg(f)=n\right\}
\end{array}
$$
and hence an equality for all $n \geq 1$ 
$$
\left|\left\{ \CCC\overset{\underline{\lambda}}{\longrightarrow} \Par:
\quad 
\sum_{\rho \in \CCC} d(\rho) |\lambda(f)| =n\right\}\right| \\
=|\Sigma_n|
=\left|\left\{ \FFF\overset{\underline{\lambda}}{\longrightarrow} \Par:
\quad 
\sum_{f \in \FFF} \deg(f) |\lambda(f)| =n\right\}\right|.
$$
Since there is only one partition $\lambda=(1)$ of having $|\lambda|=1$,
this leads to parallel recursions
$$
\begin{aligned}
|\CCC_n|
&=|\Sigma_n|-
\left|\left\{ \bigsqcup_{i=1}^{n-1} \CCC_i \overset{\underline{\lambda}}{\longrightarrow} \Par: \quad
\sum_{\rho} d(\rho) |\lambda(\rho)| =n\right\}\right| \\
|\FFF_n|
&=|\Sigma_n|-
\left|\left\{ \bigsqcup_{i=1}^{n-1} \FFF_i \overset{\underline{\lambda}}{\longrightarrow} \Par: \quad
\sum_{f} \deg(f) |\lambda(f)| =n\right\}\right| 
\end{aligned}
$$
and induction implies that $|\CCC_n|=|\FFF_n|$.
\end{proof}


\begin{example}
\label{low-degree-cuspidal-labelling-example}
Taking $q=2$, let us list the sets $\FFF_n$ of monic
irreducible polynomials $f(x) \neq x$ in $\FF_2[x]$ of degree $n$ 
for $n \leq 3$, so that we know how many cuspidal characters
of $GL_n(\FF_q)$ in $\CCC_n$ to expect:
$$
\begin{aligned}
\FFF_1&=\{ x+1\}\\
\FFF_2&=\{ x^2+x+1 \}\\
\FFF_3&=\{ x^3+x+1, x^3+x^2+1\}
\end{aligned}
$$
Thus we expect 
\begin{enumerate}
\item[$\bullet$]
one cuspidal character of $GL_1(\FF_2)$, namely
$\rho_1(=\one_{GL_1(\FF_2)})$, 
\item[$\bullet$]
one cuspidal character $\rho_2$ of $GL_2(\FF_2)$, and
\item[$\bullet$]
two cuspidal characters $\rho_3,\rho_3'$ of $GL_3(\FF_2)$.
\end{enumerate}
We will say more about $\rho_2, \rho_3,\rho_3'$ in the next section.
\end{example}


\begin{exercise}
(a) Show that for $n \geq 2$,
\begin{equation}
\label{primitive-necklace-number}
|\CCC_n| (=|\FFF_n|)= 
\frac{1}{n} \sum_{d \text{ dividing }n} 
\mu\left(\frac{n}{d} \right) q^{d}
\end{equation}
where $\mu(m)$ is the {\it number-theoretic M\"obius function} of $m$,
that is $\mu(m)=(-1)^d$ if $m=p_1 \cdots p_d$ for $d$ distinct primes,
and $\mu(m)=0$ if $m$ is not squarefree.
\newline
(b) Show that \eqref{primitive-necklace-number} also counts 
the {\it necklaces} with $n$ beads of $q$ colors 
(=equivalence classes under the $\ZZ/n\ZZ$-action of
cyclic rotation on sequences $(a_1,\ldots,a_n)$ in $\FF_q^n$)
which are {\it primitive} in the sense that no nontrivial rotation
fixes any of the sequences within the equivalence class.
For example, when $q=2$, here are representatives of these
primitive necklaces for $n=2,3,4$:
$$
\begin{aligned}
n=2:& \{(0,1)\}\\
n=3:& \{(0,0,1),(0,1,1)\}\\
n=4:& \{(0,0,0,1),(0,0,1,1),(0,1,1,1)\}.
\end{aligned}
$$
\end{exercise}


\subsection{Steinberg's unipotent characters}

Not surprisingly, the (cuspidal) character $\iota:=\one_{GL_1}$ of $GL_1(\FF_q)$
plays a distinguished role.  The parabolic subgroup $P_{(1^n)}$ is the 
Borel subgroup
$B$ of upper triangular matrices, 
and the subalgebra $A(GL)(\iota)$ of $A(GL)$ is the $\ZZ$-span of
the irreducible characters $\sigma$ that appear as constituents of 
$
\iota^n = \Ind_B^{GL_n} \one_B = \CC[GL_n/B]
$ 
for some $n$.

\begin{definition}
An irreducible character $\sigma$ of $GL_n$ appearing as a constituent
of $\Ind_B^{GL_n} \one_B = \CC[GL_n/B]$ is called a {\it unipotent character}.
Equivalently, by Frobenius reciprocity, $\sigma$ is unipotent if it contains a nonzero
$B$-invariant vector.
\end{definition}

In particular, $\one_{GL_n}$ is a unipotent character of $GL_n$ for each $n$.

\begin{proposition}
\label{Sym-to-Steinberg-isomorphism}
One can choose $\Lambda \cong A(GL)(\iota)$ in 
Theorem~\ref{uniqueness-of-sym-parts-e-f-g}(g) so that
$h_n \longmapsto \one_{GL_n}$.
\end{proposition}
\begin{proof}
Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}(a) 
tells us $\iota^2 = \Ind_B^{GL_2}$ must have exactly two irreducible
constituents, one of which is $\one_{GL_2}$;  call the other one $\St_2$.
Choose the isomorphism so as to send $h_2 \longmapsto \one_{GL_2}$.
Then $h_n \mapsto \one_{GL_n}$ follows from the 
claim that $\St_2^\perp(\one_{GL_n})=0$ for $n \geq 2$:
one has 
$$
\Delta(\one_{GL_n}) 
= \sum_{i+j=n} \Res^{G_n}_{P_{i,j}} (\one_{GL_n})^{K_{i,j}}
=\sum_{i+j=n}\one_{GL_i} \otimes \one_{GL_j}
$$
so that $\St_2 ^\perp(\one_{GL_n})=(\St_2,\one_{GL_2}) \one_{GL_{n-2}} = 0$ since
$\St_2 \neq \one_{GL_2}$.
  
\end{proof}


This subalgebra $A(GL)(\iota)$, and the unipotent characters $\chi^\lambda_q$ corresponding
under this isomorphism to the Schur functions $s_\lambda$, were introduced
by Steinberg \cite{Steinberg}.  He wrote down $\chi^\lambda_q$ as a virtual sum of induced
characters $\Ind_{P_\alpha}^{GL_n} \one_{P_\alpha} (=\one_{G_{\alpha_1}} \cdots \one_{G_{\alpha_\ell}})$, 
modelled on the 
Jacobi-Trudi determinantal expression for $s_\lambda=\det(h_{\lambda_i-i+j})$.
Note that  $\Ind_{P_\alpha}^{GL_n} \one_{P_\alpha}$ is the transitive
permutation representation $\CC[G/P_\alpha]$ for $GL_n$ 
permuting the {\it finite partial flag variety} $G/P_\alpha$, that is, the set
of {\it $\alpha$-flags of subspaces}
$$
\{0\} \subset V_{\alpha_1} \subset V_{\alpha_1+\alpha_2} 
 \subset \cdots \subset  V_{\alpha_1+\alpha_2+\cdots+\alpha_{\ell-1}} 
\subset \FF_q^n
$$
where $\dim_{\FF_q} V_d = d$ in each case.  This character has dimension equal
to $|G/P_\alpha|$, with formula given by the {\it $q$-multinomial coefficient}
(see e.g. Stanley \cite[\S 1.3]{Stanley}):
$$
\qbin{n}{\alpha}{q}=\frac{[n]!_q}{[\alpha_1]!_q \cdots [\alpha_\ell]!_q}
$$
where $[n]!_q:=[n]_q [n-1]_q \cdots [2]_q [1]_q$ and 
$[n]_q:=1+q+\cdots+q^{n-1}=\frac{q^n-1}{q-1}$.


Our terminology $\St_2$ is motivated by the $n=2$ special case of 
the {\it Steinberg character} $\St_n$, which is the
unipotent character corresponding under the isomorphism in
Proposition~\ref{Sym-to-Steinberg-isomorphism} to $e_n = s_{(1^n)}$.
It can be defined by the virtual sum
$$
\St_n:=
\chi^{(1^n)}_q = \sum_{\alpha} (-1)^{n-\ell(\alpha)} \Ind_{P_\alpha}^{GL_n} \one_{P_\alpha}
$$
in which the sum runs through all compositions $\alpha$ of $n$.
This turns out to be the genuine character for
$GL_n(\FF_q)$ acting on the top homology group of its {\it Tits building}: 
the simplicial complex whose vertices are 
nonzero proper subspaces $V$ of $\FF_q^n$, and whose simplices correspond
to flags of nested subspaces.  One needs to know that this Tits building has
only top homology, so that one can deduce the above character formula from
the Hopf trace formula; see Bj\"orner \cite{Bjorner}.



\subsection{Examples: $GL_2(\FF_2)$ and $GL_3(\FF_2)$}

Let's get our hands dirty.


\begin{example}
For $n=2$, there are two unipotent characters, $\chi^{(2)}_q = \one_{GL_2}$
and 
\begin{equation}
\label{Steinberg-on-lines-description}
\St_2:=\chi^{(1,1)}_q 
= \one_{GL_1}^2 - \one_{GL_2}
= \Ind_B^{GL_2} \one_B -   \one_{GL_2} 
\end{equation}
since the Jacobi-Trudi formula gives
$
s_{(1,1)}= \det \left[
\begin{matrix}
h_1 & h_2 \\
1  & h_1 
\end{matrix} \right]
=h_1^2-h_2.
$
The description \eqref{Steinberg-on-lines-description} for
this Steinberg character $\St_2$ shows that it 
has dimension 
$$
|GL_2/B|-1 = (q+1)-1 = q
$$
and that one can think of it as follows:
consider the permutation action of $GL_2$ on the $q+1$ lines
$\{\ell_0,\ell_1,\ldots,\ell_q\}$
in the projective space $\Proj^1_{\FF_q}=GL_2(\FF_q)/B$, and
take the invariant subspace perpendicular to the sum of basis
elements $e_{\ell_0} + \cdots +e_{\ell_q}$.
\end{example}


\begin{example}
Continuing the previous example, but taking $q=2$,
we find that we have constructed two unipotent characters:
$\one_{GL_2}=\chi^{(2)}_{q=2}$ of dimension $1$, and
$\St_2=\chi^{(1,1)}_{q=2}$ of dimension $q=2$.
This lets us identify the unique cuspidal character
$\rho_2$ of $GL_2(\FF_2)$, using knowledge of
the character table of $GL_2(\FF_2) \cong \Symm_3$:

\vskip.1in
\begin{tabular}{|c||c|c|c|c|}\hline
 & & & & \\
&
& $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$
&
$\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right]$,
$\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right]$,
$\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$ 
& 
$\left[ \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right]$,
$\left[ \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right]$\\
 & & & & \\\hline\hline
$\one_{GL_2}=\chi^{(2)}_{q=2}$ &\text{unipotent}& 1 & 1 & 1 \\\hline
$\St_2=\chi^{(1,1)}_{q=2}$ &\text{unipotent}& 2 & 0 & -1 \\\hline
$\rho_2$ &\text{cuspidal}& 1 & -1 & 1 \\\hline
\end{tabular}

\vskip.1in
\noindent
In other words, the cuspidal character $\rho_2$ of $GL_2(\FF_2)$
corresponds under the isomorphism $GL_2(\FF_2) \cong \Symm_3$
to the sign character $\sgn_{\Symm_3}$.
\end{example}

\begin{example}
Continuing the previous example to $q=2$ and $n=3$ lets us analyze
the irreducible characters of $GL_3(\FF_2)$.
Recalling our labelling
$\rho_1,\rho_2,\rho_3,\rho_3'$ from Example~\ref{low-degree-cuspidal-labelling-example} of the cuspidal characters of $GL_n(\FF_2)$ for $n=1,2,3$,
Zelevinsky's Theorem~\ref{Zelevinsky-decomposition-theorem} 
tells us that the $GL_3(\FF_2)$-irreducible
characters should be labelled by function 
$
\{\rho_1,\rho_2,\rho_3,\rho_3'\}
\overset{\underline{\lambda}}{\longrightarrow} \Par
$
for which 
$$
1 \cdot |\lambda(\rho_1)| + 
2 \cdot |\lambda(\rho_2)| + 
3 \cdot |\lambda(\rho_3)| + 
3 \cdot |\lambda(\rho'_3)| = 3
$$
We will label such an irreducible character
$\chi^{\underline{\lambda}}=\chi^{(\lambda(\rho_1),\lambda(\rho_2),\lambda(\rho_3),\lambda(\rho'_3))}$.  

Three of these irreducibles will be the unipotent characters, 
mapping under the isomorphism
from Proposition~\ref{Sym-to-Steinberg-isomorphism} as follows:
\begin{enumerate}
\item[$\bullet$]
$
s_{(3)} =h_3 \longmapsto
\chi^{((3),\varnothing,\varnothing,\varnothing)} = \one_{GL_3} 
$
of dimension $1$.
\item[$\bullet$]
$$
s_{(2,1)}=\det \left[
\begin{matrix}
h_2 & h_3 \\
1  & h_1 
\end{matrix} \right]
=h_2h_1-h_3
\longmapsto \chi^{((2,1),\varnothing,\varnothing,\varnothing)} 
            =\Ind_{P_{2,1}}^{GL_3} \one_{P_{2,1}} -   \one_{GL_3},
$$
of dimension $\qbin{3}{2,1}{q}-\qbin{3}{3}{q}=[3]_q-1=q^2+q 
\overset{q=2}{\rightsquigarrow} 6$.
\item[$\bullet$]
Lastly, 
$$
\begin{aligned}
s_{(1,1,1)}&=\det \left[
\begin{matrix}
h_1 & h_2 & h_3 \\
1  & h_1 & h_2 \\
0  & 1  & h_1
\end{matrix} \right]
=h_1^3-h_2 h_1 - h_1 h_2 + h_3 \\
&\longmapsto \St_3=\chi^{((1,1,1),\varnothing,\varnothing,\varnothing)} 
=\Ind_{B}^{GL_3} \one_{B} -  
\Ind_{P_{1,2}}^{GL_3} \one_{P_{1,2}} -  
\Ind_{P_{2,1}}^{GL_3} \one_{P_{2,1}} +   \one_{GL_3}. 
\end{aligned}
$$
of dimension 
$$
\begin{aligned}
&\qbin{3}{1,1,1}{q}-\qbin{3}{2,1}{q}-
\qbin{3}{1,2}{q}-\qbin{3}{3}{q} \\
&=[3]!_q - [3]_q-[3]_q + 1=q^3
\overset{q=2}{\rightsquigarrow} 8.
\end{aligned}
$$
\end{enumerate}

There should also be one non-unipotent, non-cuspidal character, namely
$$
\chi^{((1),(1),\varnothing,\varnothing)} 
= \rho_1 \rho_2 
= \Ind_{P_{1,2}}^{GL_3} \Inf_{GL_1 \times GL_2}^{P_{1,2}} 
      \one_{GL_1} \otimes \rho_2
$$
having dimension $\qbin{3}{1,2}{q} \cdot 1 \cdot 1=[3]_q
\overset{q=2}{\rightsquigarrow} 7$.

Finally, we expect cuspidal characters 
$\rho_3=\chi^{(\varnothing,\varnothing,(1),\varnothing)},
\rho_3'=\chi^{(\varnothing,\varnothing,\varnothing,(1))}$,
whose dimensions $d_3, d_3'$ can be deduced from the equation 
$$
1^2+6^2+8^2+7^2+d_3^2+(d_3')^2=|GL_3(\FF_2)|
=\left[(q^3-q^0)(q^3-q^1)(q^3-q^2)\right]_{q=2}=168.
$$
This forces $d_3^2+(d_3')^2=18$, 
whose only solution in positive integers
is $d_3=d_3'=3$.  


We can check our predictions of the dimensions 
for the various  $GL_3(\FF_2)$-irreducible
characters since $GL_3(\FF_2)$ is the finite simple group of
order $168$ (also isomorphic to $PSL_2(\FF_7)$), with known
character table (see James and Liebeck \cite[p. 318]{JamesLiebeck}):\\
\begin{tabular}{|c||c||c|c|c|c|c|c|}\hline
&\text{centralizer order} &168&8 &4 &3  &7 &7 \\\hline
 &unipotent?/cuspidal? &   &  &  &   &  &\\\hline\hline
$\one_{GL_3}=\chi^{((3),\varnothing,\varnothing,\varnothing)}$ 
  &\text{unipotent}& 1 & 1 & 1 & 1 & 1 & 1 \\\hline
$\phantom{\St_3=}\chi^{((2,1),\varnothing,\varnothing,\varnothing)}$ 
  &\text{unipotent}& 6 & 2 & 0 & 0 & -1 & -1 \\\hline
$\St_3=\chi^{((1,1,1),\varnothing,\varnothing,\varnothing)}$ 
  &\text{unipotent}& 8 & 0 & 0 & -1 & 1 & 1 \\\hline
$\phantom{\St_3=}\chi^{((1),(1),\varnothing,\varnothing)}$ 
  &\text{ }& 7 & -1 & -1 & 1 & 0 & 0 \\\hline
$\rho_3=\chi^{(\varnothing,\varnothing,\varnothing,(1))}$ 
  &\text{cuspidal}& 3 & -1 & 1 & 0 & $\alpha$ & $\overline{\alpha}$ \\\hline
$\rho_3'=\chi^{(\varnothing,\varnothing,(1),\varnothing)}$ 
  &\text{cuspidal}& 3 & -1 & 1 & 0 & $\overline{\alpha}$ & $\alpha$ \\\hline
\end{tabular}\\
Here $\alpha:=-1/2+i\sqrt{7}/2$.

\begin{remark}
It is known (see e.g. Bump \cite[Cor. 7.4]{Bump}) 
that, for $n \geq 2$, the dimension of any cuspidal irreducible character
$\rho$ of $GL_n(\FF_q)$ is 
$$
(q^{n-1}-1)(q^{n-2}-1)\cdots(q^2-1)(q-1).
$$
Note that when $q=2$, 
\begin{enumerate}
\item[$\bullet$]
for $n=2$ this gives $2^1-1=1$ for the dimension of $\rho_2$, and
\item[$\bullet$]
for $n=3$ it gives $(2^2-1)(2-1)=3$ for the dimensions of $\rho_3,\rho_3'$,
\end{enumerate}
agreeing with our calculations above.
Much more is known about the character table of $GL_n(\FF_q)$; see
Remark~\ref{Green-polynomial-remark} below, 
Zelevinsky \cite[Chap. 11]{Zelevinsky}, and 
Macdonald \cite[Chap. IV]{Macdonald}
\end{remark}


\end{example}




\subsection{The Hall algebra}
\label{Hall-algebra-section}

There is another interesting Hopf subalgebra (and quotient Hopf algebra) of $A(GL)$,
related to unipotent conjugacy classes in $GL_n(\FF_q)$.

\begin{definition}
Say that an element $g$ in $GL_n(\FF_q)$ is {\it unipotent} if its eigenvalues are
all equal to $1$.  Denote by $\HHH_n$ the $\CC$-subspace of $R_\CC(GL_n)$ consisting
of those class functions which are supported only on unipotent conjugacy classes,
and let $\HHH=\oplus_{n \geq 0} \HHH_n$ as a  $\CC$-subspace of 
$A_\CC(GL)=\oplus_{n \geq 0} R_\CC(GL_n)$.  
\end{definition}




\begin{proposition}
\label{Hall-algebra-is-Hopf-sub-quotient}
The subspace $\HHH$ is 
a Hopf-subalgebra of $A_\CC(GL)$, which 
is graded, connected, and of finite type, and self-dual with
respect to the inner product on class functions inherited
from $A_\CC(GL)$.
It is also a quotient Hopf algebra of $A_\CC(GL)$, as the $\CC$-linear surjection 
$
A_\CC(GL) \twoheadrightarrow \HHH
$
restricting class functions to unipotent classes
has kernel $\HHH^\perp$ which is both an ideal and a coideal.
\end{proposition}
\begin{proof}
Given two class functions $\chi_i, \chi_j$ on $GL_i, GL_j$
and $g$ in $GL_{i+j}$, one has
\begin{equation}
\label{Hall-algebra-product}
\left( \chi_i \cdot \chi_j \right)(g) =
\frac{1}{|P_{i,j}|} 
  \sum_{\substack{h \in GL_{i+j}:\\ 
    h^{-1} g h=\left[ \begin{matrix} g_i & * \\0 & g_j\end{matrix} \right] \in P_{i,j}}}
  \chi_i(g_i) \chi_j(g_j).
\end{equation}
Since $g$ is unipotent if and only if $h^{-1}gh$ is unipotent if and only if both 
$g_i,g_j$ are unipotent, the formula \eqref{Hall-algebra-product} shows 
both that $\HHH$ is a subalgebra and that $\HHH^\perp$ is a two-sided ideal:
$\chi_i$ and $\chi_j$ are both supported only on unipotent classes if and
only if the same holds for $\chi_i \cdot \chi_j$.
Similarly, for class functions $\chi$ on $GL_n$ and $(g_i, g_j)$ in $GL_{i,j}=GL_i \times GL_j$, one has
$$
\Delta(\chi)(g_i, g_j) = \frac{1}{q^{ij}} \sum_{k \in \FF_q^{i \times j}} 
\chi\left[ \begin{matrix} g_i & k \\0 & g_j\end{matrix} \right]
$$
using \eqref{deflation-character-formula}.  This shows both that $\HHH$ is a sub-coalgebra 
of $A=A_\CC(GL)$
$$
\Delta \HHH \subset \HHH \otimes \HHH
$$
and that $\HHH^\perp$ is a co-ideal 
$$
\Delta(\HHH^\perp) \subset \HHH^\perp \otimes A + A \otimes \HHH^\perp
$$
since it shows that 
$\chi$ is supported only on unipotent classes if and only if $\Delta(\chi)$ vanishes on
$(g_1,g_2)$ that have either $g_1$ or $g_2$ non-unipotent.
The rest follows.
\end{proof}

The subspace $\HHH$ is called the {\it Hall algebra}. It has
an obvious orthogonal $\CC$-basis, with interesting structure constants.

\begin{definition}
Given a partition $\lambda$ of $n$, let $J_{\lambda}$ denote the $GL_n$-conjugacy class of 
unipotent matrices whose {\it Jordan type} (= Jordan block sizes) is given by $\lambda$, and let $z_\lambda(q)$ denote the size of this conjugacy class $J_{\lambda}$.

The indicator class functions $\{ \one_{J_\lambda} \}_{\lambda \in \Par}$ 
form a $\CC$-basis for $\HHH$ whose multiplicative structure constants
are called the {\it Hall coefficients} $g^{\lambda}_{\mu,\nu}(q)$:
$$
\one_{J_\mu} \one_{J_\nu} = \sum_\lambda g^{\lambda}_{\mu,\nu}(q) \,\, \one_{J_\lambda}.
$$
\end{definition}

\noindent
Because the dual basis to $\{\one_{J_\lambda}\}$ is
$\{ z_\lambda(q)^{-1} \one_{J_\lambda} \}$, 
self-duality of $\HHH$ shows that the Hall coefficients are 
(essentially) also structure constants for the comultiplication:
$$
\Delta \one_{J_\lambda}
= \sum_{\mu,\nu} g^{\lambda}_{\mu,\nu}(q) 
\frac{z_{\mu}(q) z_{\nu}(q)}{z_{\lambda}(q)} \cdot
\one_{J_\mu} \otimes \one_{J_\nu}. 
$$

The Hall coefficient $g^{\lambda}_{\mu,\nu}(q)$ has the following interpretation.

\begin{proposition}
Fix any $g$ in $GL_n(\FF_q)$ 
acting unipotently on $\FF_q^n$ with Jordan type $\lambda$.
Then $g^{\lambda}_{\mu,\nu}(q)$ counts the $g$-stable $\FF_q$-subspaces $V \subset \FF_q^n$ 
for which the restriction 
$g|V$ acts with Jordan type $\mu$, and the induced map $\bar{g}$ 
on the quotient space $\FF_q^n/V$ has Jordan type $\nu$.
\end{proposition}
\begin{proof}
Given $\mu,\nu$ partitions of $i,j$ with $i+j=n$, taking $\chi_i, \chi_j$ equal
to $\one_{J_\mu}, \one_{J_\nu}$ in 
\eqref{Hall-algebra-product}
shows that for any $g$ in $GL_n$, the value of 
$\left( \one_{J_\mu} \cdot \one_{J_{\nu}} \right)(g)$ is given by
\begin{equation}
\label{Hall-coefficient-crux}
\frac{1}{|P_{i,j}|} 
  \left|\left\{ h \in GL_n: h^{-1} g h=\left[ \begin{matrix} g_i & * \\0 & g_j\end{matrix} \right] \text{ with }g_i \in J_{\mu},g_j \in J_{\nu} \right\}\right|
\end{equation}
Let $S$ denote the set appearing in \eqref{Hall-coefficient-crux},
and let $\FF_q^i$ denote the $i$-dimensional
subspace of $\FF_q^n$ spanned by the first $i$ standard basis vectors.
Note that the condition on an element $h$ in $S$ saying that $h^{-1}gh$ is in block upper-triangular form can be re-expressed by saying that the subspace $V:=h(\FF_q^i)$ is $g$-stable.
One then sees that the map $h \overset{\varphi}{\longmapsto} V=h(\FF_q^i)$ surjects
$S$ onto the set of $i$-dimensional $g$-stable subspaces $V$ of $\FF_q^n$ for which
$g|V$ and $\bar{g}$ are unipotent of types $\mu,\nu$, respectively.  Furthermore, for
any particular such $V$, 
its fiber $\varphi^{-1}(V)$ in $S$ is the stabilizer within
$GL_n$ of $V$, which is conjugate to $P_{i,j}$, and hence has cardinality
$|\varphi^{-1}(V)|=|P_{i,j}|$.  This proves the assertion of the proposition.
\end{proof}

The Hall algebra $\HHH$ will turn out to be isomorphic to the ring $\Lambda_\CC$ of symmetric
functions with $\CC$ coefficients, via a composite $\varphi$ of three maps
$$
\Lambda_\CC \longrightarrow 
A(GL)(\iota)_\CC \longrightarrow 
A(GL)_\CC \longrightarrow 
\HHH
$$
in which the first map is the isomorphism from 
Proposition~\ref{Sym-to-Steinberg-isomorphism}, the second is inclusion,
and the third is the quotient map from
Proposition~\ref{Hall-algebra-is-Hopf-sub-quotient}.

\begin{theorem}
\label{Sym-to-Hall-isomorphism-thm}
The above composite $\varphi$ is a Hopf algebra isomorphism, sending
$$
\begin{array}{rcl}
h_n &\longmapsto & \sum_{\lambda \in \Par_n} \one_{J_\lambda}, \\
e_n &\longmapsto & q^{\binom{n}{2}} \one_{J_{(1^n)}}, \\
p_n &\longmapsto & \sum_{\lambda \in \Par_n} (q;q)_{\ell(\lambda)}\one_{J_\lambda}.
\end{array}
$$
where 
$$
(x;q)_\ell:=(1-x)(1-qx)(1-q^2x) \cdots (1-q^{\ell-1}x).
$$
\end{theorem}
\begin{proof}
That $\varphi$ is a Hopf morphism follows because it is a composite
of three such morphisms.  We claim that once one shows the formula for the
(nonzero) image of $\varphi(p_n)$ given above is correct, 
then this will already show $\varphi$ is an isomorphism, 
by the following argument.  Note first that $\Lambda_\CC$ and $\HHH$ 
both have dimension $|\Par_n|$ for their $n^{th}$ homogeneous
components, so it suffices to show that the graded map $\varphi$ is injective.
On the other hand, both $\Lambda_\CC$ and $\HHH$ 
are (graded, connected, finite type) {\it self-dual} Hopf algebras, 
so Theorem~\ref{general-structure-theorem} says that each is the symmetric algebra
on its space of primitive elements.  Thus it suffices to check that
$\varphi$ is injective when restricted to their subspaces of primitives.
For $\Lambda_\CC$, by Corollary~\ref{primitives-in-Lambda} the primitives
are spanned by $\{p_1,p_2,\ldots\}$, with only
one basis element in each degree.  Hence 
$\varphi$ is injective on the subspace of primitives if and only if
it does not annihilate any $p_n$.

Thus it only remains to show the above formulas for the images of
$h_n,e_n,p_n$ under $\varphi$.
This is clear for $h_n$, since
Proposition~\ref{Sym-to-Steinberg-isomorphism} shows
that it maps under the first two composites to the indicator function
$\one_{GL_n}$ which then restricts to the sum of indicators
$\sum_{\lambda \in \Par_n} \one_{J_\lambda}$ in $\HHH$.  
For $e_n, p_n$, we resort to generating functions.
Let $\tilde{h}_n,\tilde{e_n},\tilde{p}_n$ denote the
three putative images in $\HHH$ of $h_n,e_n,p_n$,
appearing on the right side in the theorem, and define
generating functions in $\HHH[[t]]$
$$
\tilde{H}(t):=\sum_{n \geq 0} \tilde{h}_n t^n, \quad
\tilde{E}(t):=\sum_{n \geq 0} \tilde{e}_n t^n, \quad
\tilde{P}(t):=\sum_{n \geq 0} \tilde{p}_{n+1} t^n.
$$
We wish to show that $\varphi$ maps $H(t),E(t),P(t)$ in $\Lambda[[t]]$
to these three generating functions.  Since we have already shown this is
correct for $H(t)$, by \eqref{E(t)-H(t)-relation},
\eqref{E(t)-H(t)-P(t)-relation}, it suffices to check that in $\HHH[[t]]$ one has
$$
\begin{array}{rccl}
\tilde{H}(t) \tilde{E}(-t) = 1,& \text{ or equivalently, }
  &\sum_{k=0}^n (-1)^k \tilde{e}_k \tilde{h}_{n-k} = \delta_{0,n}\\
\tilde{H}'(t) \tilde{E}(-t) = \tilde{P}(t),& \text{  or equivalently, } 
  &\sum_{k=0}^n (-1)^k (n-k) \tilde{e}_k \tilde{h}_{n-k} = \tilde{p}_n.
\end{array}
$$
Thus it would be helpful to evaluate the class function
$\tilde{e}_k \tilde{h}_{n-k}$.  Note that a unipotent $g$ in $GL_n$
having $\ell$ Jordan blocks has an $\ell$-dimensional $1$-eigenspace, so that
$$
(\tilde{e}_k \tilde{h}_{n-k})(g)
= q^{\binom{k}{2}} \cdot (\one_{J_{(1^k)}} \cdot \tilde{h}_{n-k})(g)
=  q^{\binom{k}{2}} \qbin{\ell}{k}{q} 
$$
where 
$$
\qbin{\ell}{k}{q} = \frac{(q;q)_\ell}{(q;q)_k (q;q)_{\ell-k}}
$$
is the {\it $q$-binomial coefficient}
counting $k$-dimensional $\FF_q$-subspaces $V$ of
an $\ell$-dimensional $\FF_q$-vector space; see, e.g., \cite[\S 1.3]{Stanley}.
Thus one needs for $\ell \geq 1$ that
\begin{align}
\sum_{k=0}^\ell (-1)^k 
q^{\binom{k}{2}} \qbin{\ell}{k}{q} 
&= 0,\label{first-q-binomial-identity}\\
\sum_{k=0}^\ell (-1)^k (n-k) 
q^{\binom{k}{2}} \qbin{\ell}{k}{q} 
&= (q;q)_{\ell}.\label{second-q-binomial-identity}
\end{align}
Identity \eqref{first-q-binomial-identity}
comes from setting $x=1$ in the {\it $q$-binomial theorem} \cite[Exer. 3.45]{Stanley}:
$$
\sum_{k=0}^\ell (-1)^k q^{\binom{k}{2}} \qbin{\ell}{k}{q} x^{\ell-k}
=(x-1)(x-q)(x-q^2)\cdots (x-q^{\ell-1}).
$$
Identity \eqref{second-q-binomial-identity} 
comes from taking $\frac{d}{dx}$ in the
$q$-binomial theorem, then setting $x=1$, and finally
adding $(n-\ell)$ times \eqref{first-q-binomial-identity}.
\end{proof}

We next indicate, without proof, how $\HHH$ relates to the
classical Hall algebra.

\begin{definition}
The usual {\it Hall algebra}, or what Schiffmann \cite[\S 2.3]{Schiffmann}
calls {\it Steinitz's classical Hall algebra} 
(see also Macdonald \cite[Chap. II]{Macdonald}),
has $\ZZ$-basis elements 
$\{ u_\lambda \}_{\lambda \in \Par}$, with the multiplicative structure constants
$g^{\lambda}_{\mu,\nu}(p)$ in 
$$
u_\mu u_\nu = \sum_\lambda
g^{\lambda}_{\mu,\nu}(p)\,\, u_\lambda
$$ 
defined as follows: fix a finite abelian $p$-group $L$ of {\it type $\lambda$},
meaning that 
$$
L \cong \bigoplus_{i=1}^{\ell(\lambda)} \ZZ/p^{\lambda_i}\ZZ,
$$
and let $g^{\lambda}_{\mu,\nu}(p)$ be the number of subgroups $M$ of $L$ of
type $\mu$, for which the quotient $N:=L/M$ is of type $\nu$.  In other words,
$g^{\lambda}_{\mu,\nu}(p)$ counts, for a fixed abelian 
$p$-group $L$ of type $\lambda$,  the number of short exact sequences
$
0 \rightarrow M
\rightarrow L
\rightarrow N
\rightarrow 0
$
in which $M,N$ have types $\mu,\nu$, respectively.
\end{definition}

We claim that when one takes the finite field $\FF_q$ of order
$q=p$ a {\it prime}, the map  
\begin{equation}
\label{classical-Hall-to-Zelevinsky-Hall}
u_\lambda \longmapsto \one_{J_\lambda}
\end{equation}
gives an isomorphism from this classical Hall algebra 
to the $\ZZ$-algebra $\HHH_\ZZ \subset \HHH$.
The key point is {\it Hall's Theorem}, a non-obvious
statement for which Macdonald includes two proofs in
\cite[Chap. II]{Macdonald}, one of them due to 
Zelevinsky\footnote{See also \cite[Thm. 2.6, Prop. 2.7]{Schiffmann} 
for quick proofs of part of it, similar to Zelevinsky's.}.
To state it, we first recall some notions about discrete valuation rings.

\begin{definition}
A {\it discrete valuation ring (DVR)} $\ooo$ is
a principal ideal domain having only one maximal ideal $\mmm$, with
quotient $k=\ooo/\mmm$ called its {\it residue field}.  

The structure theorem for finitely generated modules 
over a PID implies that an $\ooo$-module $L$
with finite composition series of composition length $n$  
must have $L \cong \oplus_{i=1}^{\ell(\lambda)} \ooo/\mmm^{\lambda_i}$ 
for some partition $\lambda$ of $n$; say $L$ has {\it type } $\lambda$
in this situation.  
\end{definition}

Here are the two crucial examples for us.

\begin{example}
\label{DVR-example-F[[t]]}
For any field $\FF$, the power series ring $\ooo=\FF[[t]]$ is a DVR
with maximal ideal $\mmm=(t)$ and 
residue field $k=\ooo/\mmm=\FF[[t]]/(t)\cong\FF$.
An $\ooo$-module $L$ of type $\lambda$ is an $\FF$-vector space together
with an $\FF$-linear transformation $T$ that acts on $M$ nilpotently 
(so that $g:=T+1$ acts unipotently) with 
Jordan blocks of sizes given by $\lambda$:  each summand
$\ooo/\mmm^{\lambda_i}=\FF[[t]]/(t^{\lambda_i})$ of $L$ has 
an $\FF$-basis $\{1,t,t^2,\ldots,t^{\lambda_i-1}\}$ on
which the map $T$ that multiplies by $t$ acts as a nilpotent Jordan block of
size $\lambda_i$.
Note also that, in this setting, $\ooo$-submodules are the same as
$T$-stable (or $g$-stable) $\FF$-subspaces.  
\end{example}

\begin{example}
\label{DVR-example-Z_p}
The ring of $p$-adic integers $\ooo=\ZZ_p$ is a DVR with
maximal ideal $\mmm=(p)$ and 
residue field $k=\ooo/\mmm=\ZZ_p/p\ZZ_p \cong \ZZ/p\ZZ$.
An $\ooo$-module $L$ of type $\lambda$ is an abelian $p$-group of
type $\lambda$:  for each summand, 
$\ooo/\mmm^{\lambda_i}=\ZZ_p/p^{\lambda_i}\ZZ_p \cong \ZZ/p^{\lambda_i}\ZZ$.
Note also that, in this setting, $\ooo$-submodules are the same as
subgroups.
\end{example}

\noindent
One last notation: $n(\lambda):=\sum_{i \geq 1} (i-1)\lambda_i$,
for $\lambda$ in $\Par$.  Hall's Theorem is as follows.

\begin{theorem}
\label{Hall's-theorem}
Assume $\ooo$ is a DVR with maximal ideal $\mmm$, and that its residue field
$k=\ooo/\mmm$ is finite of cardinality $q$.  
Fix an $\ooo$-module $L$ of type $\lambda$.
Then the number of $\ooo$-submodules $M$ of
type $\mu$ for which the quotient $N=L/M$ is of type $\nu$ can be
written as the specialization 
$$
[g^{\lambda}_{\mu,\nu}(t)]_{t=q}
$$ 
of a polynomial $g^{\lambda}_{\mu,\nu}(t)$ in $\ZZ[t]$, called the {\bf Hall polynomial}.  

Furthermore, the Hall polynomial $g^{\lambda}_{\mu,\nu}(t)$
has degree at most $n(\lambda)-(n(\mu)+n(\nu))$, and its
coefficient of $t^{n(\lambda)-(n(\mu)+n(\nu))}$ is the Littlewood-Richardson
coefficient $c^{\lambda}_{\mu,\nu}$.
\end{theorem}

\noindent
Comparing what Hall's Theorem says in 
Examples~\ref{DVR-example-F[[t]]}
and \ref{DVR-example-Z_p}, shows that
the map \eqref{classical-Hall-to-Zelevinsky-Hall}
gives the desired isomorphism from the classical Hall algebra
to $\HHH_\ZZ$.

We close this section with some remarks on the vast literature
on Hall algebras that we will {\it not} discuss here.  

\begin{remark}
Macdonald's version of Hall's Theorem \cite[(4.3)]{Macdonald} is stronger
than Theorem~\ref{Hall's-theorem}, and useful for certain applications:
he shows that $g^{\lambda}_{\mu,\nu}(t)$ is the zero polynomial whenever
the Littlewood-Richardson coefficient $c^{\lambda}_{\mu,\nu}$ is zero. 
\end{remark}


\begin{remark}
\label{Green-polynomial-remark}
Zelevinsky in \cite[Chaps 10, 11]{Zelevinsky} uses
the isomorphism $\Lambda_\CC \rightarrow \HHH$ to derive J. Green's formula
for the value of any irreducible character $\chi$ of $GL_n$ on any unipotent
class $J_\lambda$.  The answer involves values of irreducible characters
of $\Symm_n$ along with {\it Green's polynomials}
$Q^\lambda_\mu(q)$ 
(see Macdonald \cite[\S III.7]{Macdonald};  they are 
denoted $Q(\lambda,\mu)$ by Zelevinsky),
which express the images under the isomorphism 
of Theorem~\ref{Sym-to-Hall-isomorphism-thm}
of the symmetric function basis $\{p_\mu\}$ in terms of the 
basis $\{\one_{J_\lambda}\}$.
\end{remark}

\begin{remark}
The Hall polynomials $g^{\lambda}_{\mu,\nu}(t)$ also essentially give the
multiplicative structure constants for $\Lambda(\xx)[t]$ with
respect to its basis of {\it Hall-Littlewood symmetric functions}
$P_\lambda=P_{\lambda}(\xx;t)$:
$$
P_{\mu} P_{\nu} 
= \sum_{\lambda} t^{n(\lambda)-(n(\mu)+n(\nu))} g^{\lambda}_{\mu,\nu}(t^{-1}) P_{\lambda}.
$$
See Macdonald \cite[\S III.3]{Macdonald}.
\end{remark}

\begin{remark}
Schiffmann \cite{Schiffmann} discusses self-dual Hopf algebras which
vastly generalize the classical Hall algebra called {\it Ringel-Hall algebras}, 
associated to abelian categories which are hereditary.  Examples
come from categories of nilpotent representations of quivers;
the quiver having exactly one node and one arc recovers the
classical Hall algebra $\HHH_\ZZ$ discussed above.
\end{remark}

\begin{remark}
Work of Thiem and Vinroot \cite{ThiemVinroot} shows that
the tower $\{G_*\}$ of 
{\it finite unitary groups} $U_n(\FF_{q^2})$
give rise to another positive self-dual Hopf algebra 
$A=\bigoplus_{n \geq 0}R( U_n(\FF_{q^2}) )$,
in which the role of Harish-Chandra induction 
is played by {\it Deligne-Lusztig induction}.  In this theory,
character and degree formulas for $U_n(\FF_{q^2})$ are related
to those of $GL_n(\FF_q)$ by substituting $q \mapsto -q$, along
with appropriate scalings by $\pm 1$, a phenomenon sometimes called
{\it Ennola duality}.  See also \cite[\S 4]{RStantonThiem}.
\end{remark}

\newpage




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Quasisymmetric functions and $P$-partitions}
\label{Qsym-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

We discuss here our next important example of a Hopf algebra arising in combinatorics:  the {\it quasisymmetric functions} of Gessel \cite{Gessel}, 
with roots in work of Stanley \cite{Stanley-thesis} on $P$-partitions.

\subsection{Definitions, and Hopf structure}

The definitions of quasisymmetric functions require a totally ordered variable set.
Usually we will use a variable set denoted $\xx=(x_1,x_2,\ldots)$ with the
usual ordering $x_1 < x_2 < \ldots$.  However, it is good to have some flexibility
in changing the ordering, which is why we make the following definition.

\begin{definition}
Given any totally ordered set $I$, create a totally ordered
variable set $\{x_i\}_{i\in I}$, and then let $R(\{x_i\}_{i\in I})$
denote the power series 
of bounded degree in $\{x_i\}_{i \in I}$
having coefficients in $\kk$.

The {\it quasisymmetric functions} $\Qsym:=\Qsym(\{x_i\}_{i \in I})$
will be the $\kk$-submodule consisting of the elements $f$ in
$R(\{x_i\}_{i\in I})$ that have the same coefficient on 
the monomials $x_{i_1}^{\alpha_1} \cdots x_{i_\ell}^{\alpha_\ell}$
and $x_{j_1}^{\alpha_1} \cdots x_{j_\ell}^{\alpha_\ell}$
whenever both $i_1 < \cdots < i_\ell$ and $j_1 < \cdots < j_\ell$
in the total order on $I$.
\end{definition}

It immediately follows from this definition that
$\Qsym(\{x_i\}_{i \in I})$ forms a free $\kk$-submodule of 
$R(\{x_i\}_{i\in I})$, having as $\kk$-basis elements
the {\it monomial quasisymmetric functions}
$$
M_\alpha(\{x_i\}_{i\in I}):=
\sum_{i_1 < \cdots < i_\ell\text{ in }I}
x_{i_1}^{\alpha_1} \cdots x_{i_\ell}^{\alpha_\ell}
$$
for all $\alpha$ in $\Comp$.
Note that $\Qsym= \oplus_{n \geq 0} \Qsym_n$ is a 
graded $\kk$-vector space of finite type, where $\Qsym_n$ is the subspace
of quasisymmetric functions which are homogeneous of degree $n$.
Letting $\Comp$ denote the set of all compositions $\alpha$, and
$\Comp_n$ the compositions $\alpha$ of $n$, that is, $\alpha_1+\cdots+\alpha_\ell=n$,
the subset $\{ M_\alpha \}_{\alpha \in \Comp_n}$ gives a $\kk$-basis for $\Qsym_n$.


\begin{example}
Taking the variable set $\xx=(x_1 < x_2 < \cdots)$ to define $\Qsym=\Qsym(\xx)$,
for $n=0,1,2,3$, one has these basis elements in $\Qsym_n$:
$$
\begin{array}{rll}
M_{()} =M_\varnothing&= 1 &\\
& & \\
M_{(1)} &= x_1+x_2+x_3+ \cdots &= m_{(1)}=s_{(1)}=e_1=h_1=p_1\\
& & \\
M_{(2)} &= x_1^2+x_2^2+x_3^2+ \cdots &=m_{(2)}=p_2\\
M_{(1,1)} &= x_1x_2+x_1x_3+x_2x_3+ \cdots&=m_{(1,1)}=e_2 \\
& & \\
M_{(3)} &= x_1^3+x_2^3+x_3^3+ \cdots &=m_{(3)}=p_3\\
M_{(2,1)} &= x_1^2 x_2 + x_1^2 x_2 + x_2^2 x_3 + \cdots &\\
M_{(1,2)} &= x_1 x_2^2 + x_1 x_3^2 + x_2 x_3^2 + \cdots &\\
M_{(1,1,1)} &= x_1 x_2 x_3 +x_1 x_2 x_4 + x_1 x_3 x_4 +  \cdots &=m_{(1,1,1)}=e_3
\end{array}
$$
\end{example}

It is not obvious that $\Qsym$ is a subalgebra of $R(\xx)$,
but we will show this momentarily.  For example,
$$
\begin{aligned}
M_{(a)}M_{(b, c)} 
&=(x_1^a + x_2^a + x_3^a+\cdots) 
(x_1^b x_2^c + x_1^b x_3^c + x_2^b x_3^c + \cdots)\\
&=(x_1^{a+b} x_2^c + \cdots +x_1^b x_3^{a+c} + \cdots 
  + x_1^a x_2^b x_3^c + \cdots
  +x_1^b x_2^a x_3^c + \cdots
  + x_1^b x_2^c x_3^a + \cdots)\\
&=M_{(a+b, c)} + M_{(b, a+c)} + M_{(a, b, c)} + M_{(b, a, c)} + M_{(b, c, a)}
\end{aligned}
$$

\begin{proposition}
For any infinite totally ordered set $I$, one has that
$\Qsym=\Qsym(\{x_i\}_{i \in I})$  is $\kk$-subalgebra of 
$R(\{x_i\}_{i\in I})$, with multiplication in the 
$\{M_\alpha\}$-basis as follows:
if $\alpha=(\alpha_1,\ldots,\alpha_\ell), \beta=(\beta_1,\ldots,\beta_m)$
then
\begin{equation}
\label{Qsym-product-on-monomials}
M_\alpha M_\beta = \sum_f M_{\wt f} 
\end{equation}
in which the sum is over all maps $f$ from the disjoint union of two chains to a chain
\begin{equation}
\label{map-from-two-chains-to-one}
(i_1 < \cdots <i_\ell) \sqcup (j_1 < \cdots < j_m) 
\overset{f}{\longrightarrow} 
(k_1 < \cdots <k_p)
\end{equation}
which are both surjective and strictly order-preserving ($x < y$ implies $f(x) < f(y)$),
and where the composition $\wt(f):=(\wt_1(f),\ldots,\wt_p(f))$ is defined by
$
\wt_s(f):= \sum_{a_i \in f^{-1}(k_s)} \alpha_i + \sum_{b_j \in f^{-1}(k_s)} \beta_j.
$

In particular, all such algebras are isomorphic to a single
algebra $\Qsym$, defined as having $\kk$-basis $\{M_\alpha\}_{\alpha \in \Comp}$
and with multiplication defined $\kk$-linearly by 
\eqref{Qsym-product-on-monomials}.  The isomorphism sends
$M_\alpha \longmapsto M_\alpha(\{x_i\}_{i\in I})$.
\end{proposition}
\begin{proof}
Formula \eqref{Qsym-product-on-monomials} comes from
considering a typical product of two monomials in the expansion
of $M_\alpha M_\beta$:
$$
\left( x_{i_1}^{\alpha_1} \cdots x_{i_\ell}^{\alpha_\ell} \right) 
\left( x_{j_1}^{\beta_1} \cdots x_{j_m}^{\beta_m} \right)
=
x_{k_1}^{\gamma_1} \cdots x_{k_p}^{\gamma_p}
$$
for subscript sequences
$i_1 < \cdots <i_\ell$
and  
$j_1 < \cdots < j_m$
and
$k_1 < \cdots <k_p$
with 
$$
\{k_1,\ldots,k_p\}=\{i_1,\ldots,i_\ell\} \cup \{j_1 ,\ldots, j_m\}.
$$
Thinking of $\{i_s\}$ and $\{j_t\}$ as disjoint sets, 
multiplication gives a surjective map $f$ 
as in \eqref{map-from-two-chains-to-one},
with $\gamma_{s}=\wt_s(f)$.

Once one has this multiplication rule \eqref{Qsym-product-on-monomials},
the last assertion follows.  
\end{proof}


The comultiplication $\Qsym$ will extend the one that we defined
for $\Lambda$.  That is, one considers the linear order from \eqref{two-variable-set-ordering} on two sets of variables
$
(\xx,\yy)=(x_1< x_2 < \ldots < y_1 < y_2 < \ldots),
$
and notes that 
$$
\Qsym(\xx,\yy) \subset \Qsym(\xx) \otimes \Qsym(\yy)
$$
so that one can define $\Qsym \overset{\Delta}{\longrightarrow} \Qsym\otimes \Qsym$
%\begin{equation}
%\label{Qsym-coproduct-defintion}
%\begin{array}{rcl}
%\Qsym=\Qsym(\xx) & \overset{\Delta}{\longrightarrow} 
%   & \Qsym(\xx,\yy) \hookrightarrow \Qsym\otimes \Qsym\\
%f(\xx)=f(x_1,x_2,\ldots) & \longmapsto & f(\xx,\yy)=f(x_1,x_2,\ldots,y_1,y_2,\ldots).
%\end{array}
%\end{equation}
as the  composite of the maps in the bottom row here:
\begin{equation}
\label{Qsym-coproduct-defintion}
\begin{array}{rccccl}
      &                 & R(\xx,\yy)     & \cong & R(\xx) \otimes R(\yy) \\
      &                 & \cup           &       & \cup\\

\Qsym & \cong & \Qsym(\xx,\yy) & \hookrightarrow &
\Qsym(\xx) \otimes \Qsym(\yy) & \cong \Qsym \otimes \Qsym\\
f&\longmapsto & f(\xx,\yy)=f(x_1,x_2,\ldots,y_1,y_2,\ldots)& & 
\end{array}
\end{equation}

\begin{example}
For example,
$$
\begin{aligned}
\Delta M_{(a,b,c)} &= M_{(a,b,c)}(x_1,x_2,\ldots,y_1,y_2,\ldots) \\
&= x_1^a x_2^b x_3^c+ x_1^a x_2^b x_4^c + \cdots \\
 &\quad + x_1^a x_2^b \cdot y_1^c+ x_1^a x_2^b \cdot y_2^c + \cdots \\
 &\quad + x_1^a \cdot y_1^b y_2^c + x_1^a \cdot y_1^b y_3^c + \cdots \\
 &\quad + y_1^a y_2^b y_3^c + y_1^a y_2^b y_4^c+ \cdots\\
&= M_{(a,b,c)}(\xx) + M_{(a,b)}(\xx) M_{(c)}(\yy) +  M_{(a)}(\xx) M_{(b,c)}(\yy)+ M_{(a,b,c)}(\yy) \\
&= M_{(a,b,c)} \otimes \one + M_{(a,b)} \otimes M_{(c)} 
+ M_{(a)} \otimes M_{(b,c)} + \one \otimes M_{(a,b,c)}
\end{aligned}
$$
\end{example}
\noindent
Defining the {\it concatenation} $\beta \cdot \gamma$ of 
two composition $\beta=(\beta_1,\ldots,\beta_r),
\gamma=(\gamma_1,\ldots,\gamma_s)$ to be the composition
$(\beta_1,\ldots,\beta_r,\gamma_1,\ldots,\gamma_s)$, one has the following
description of the coproduct in the $\{M_\alpha\}$ basis.

\begin{proposition}
\label{Qsym-coproduct-on-monomials}
For a composition $\alpha=(\alpha_1,\ldots,\alpha_\ell)$,
one has
$$
\Delta M_\alpha = \sum_{k=1}^\ell M_{(\alpha_1,\ldots,\alpha_k)} \otimes M_{(\alpha_{k+1},\ldots,\alpha_\ell)} = \sum_{\substack{(\beta,\gamma):\\ \beta \cdot \gamma=\alpha}} M_\beta \otimes M_\gamma
$$
\end{proposition}
\begin{proof}
This comes from expressing a monomial in $\Delta M_\alpha = M_\alpha(\xx,\yy)$
uniquely in the form 
$x_{i_1}^{\alpha_1} \cdots x_{i_k}^{\alpha_k} \cdot
y_{j_1}^{\alpha_{k+1}} \cdots y_{j_{\ell-k}}^{\alpha_\ell}$
for some subscripts $i_1 < \cdots < i_k$ and $j_1 < \cdots < j_{\ell-k}$.
\end{proof}

\begin{proposition}
The quasisymmetric functions $\Qsym$ form a graded connected Hopf algebra of finite
type, which is commutative, and contains 
the symmetric functions $\Lambda$ as a Hopf subalgebra.  
\end{proposition}
\begin{proof}
Coassociativity of $\Delta$ again comes from checking 
$
(\Delta \otimes \one) \circ \Delta f =
f(\xx,\yy,\zz)=
(\one \otimes \Delta) \circ \Delta f.
$
The counit is as usual for a graded connected coalgebra,
and just as in the case of $\Lambda$,
sends a quasisymmetric function 
$f(\xx)$ to its constant term $f(0,0,\ldots)$.
This is an evaluation, and hence an algebra map.
The coproduct $\Delta$ is an algebra morphism because
it is defined as a composite of algebra morphisms
in the bottom row of \eqref{Qsym-coproduct-defintion}.
Hence $\Qsym$ forms a bialgebra, and as it is
graded, connected, and of finite type, also
a Hopf algebra by Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
\end{proof}

We will identify the antipode in $\Qsym$ shortly, but we first
deal with another slightly subtle issue.
In addition to the counit evaluation $\epsilon(f) = f(0,0,\ldots)$,
starting in Section \ref{ABS-part-I-section}, we will
want to specialize elements in $\Qsym(\xx)$ by making other 
variable substitutions, in which all but a finite list of variables 
are set to zero.  We justify this here.

\begin{proposition}
Fix a totally ordered set $I$, a $\kk$-algebra $A$,
a finite list of variables $x_{i_1}, \ldots,x_{i_m}$, say
with $i_1 < \ldots < i_m$ in $I$, 
and an ordered list of elements $(a_1,\ldots,a_m) \in A^m$.

Then there is a well-defined evaluation homomorphism
$$
\begin{array}{rcl}
\Qsym(\{x_i\}_{i \in I}) &\longrightarrow &A \\
f &\longmapsto& 
\left[ f 
\right]_{\substack{x_{i_1}=a_1,\ldots,x_{i_m}=a_m\\
x_j=0 \text{ for }j \not\in \{i_1,\ldots,i_m\}}}.
\end{array}
$$
Furthermore, this homomorphism
depends only upon the list $(a_1,\ldots,a_m)$, as it coincides
with the following:
$$
\begin{array}{rccl}
\Qsym(\{x_i\}_{i \in I}) &\cong
\Qsym(x_1,x_2,\ldots) &\longrightarrow &A \\
&f(x_1,x_2,\ldots)& \longmapsto& f(a_1,\ldots,a_m,0,0\ldots).
\end{array}
$$
\end{proposition}
\begin{proof}
One already can make sense of evaluating 
$x_{i_1}=a_1,\ldots,x_{i_m}=a_m$ and $x_j=0$ for $j \not \in \{i_1,\ldots,i_m\}$
in the ambient ring $R(\{x_i\}_{i \in I})$ containing $\Qsym(\{x_i\}_{i \in I})$,
since a power series $f$ of bounded degree will have finitely
many monomials that only involve the variables 
$x_{i_1},\ldots,x_{i_m}$.  The last assertion follows from quasisymmetry
of $f$, and is perhaps checked most easily when $f=M_\alpha(\{x_i\}_{i \in I})$
for some $\alpha$.
\end{proof}


The antipode in $\Qsym$ has a reasonably simple expression 
in the $\{M_\alpha\}$ basis, but requiring a definition.

\begin{definition}
For $\alpha,\beta$ in $\Comp_n$, say that $\alpha$ {\it refines} $\beta$
or $\beta$ {\it coarsens} $\alpha$ if, informally, one can obtain $\beta$ from $\alpha$ by combining some of its parts.  Alternatively,
one has a bijection $\Comp_n \rightarrow 2^{[n-1]}$ where $[n-1]:=\{1,2,\ldots,n-1\}$ which sends $\alpha=(\alpha_1,\ldots,\alpha_\ell)$ having
length $\ell(\alpha)=\ell$ to its subset of partial sums
$$
D(\alpha):= (\alpha_1,\alpha_1+\alpha_2,\ldots,\alpha_1+\cdots+\alpha_{\ell-1}),
$$
and this sends the refinement ordering to the inclusion ordering
on the Boolean algebra $2^{[n-1]}$.  There is also a bijection sending
$\alpha$ to its {\it ribbon} diagram:  the skew diagram $\lambda/\mu$
having rows of sizes $(\alpha_1,\ldots,\alpha_\ell)$ read from bottom-to-top
with exactly one column of overlap between adjacent rows.
These bijections and the refinement partial order are illustrated here for $n=4$:
$$
\tiny
\xymatrix{
 & \{1,2,3\} & \\
\{1,2\} \ar@{-}[ur]& \{1,3\} \ar@{-}[u] & \{2,3\} \ar@{-}[ul] \\
\{1\} \ar@{-}[ur]\ar@{-}[u]& \{2\}\ar@{-}[ur]\ar@{-}[ul] & \{3\} \ar@{-}[ul]\ar@{-}[u]\\
&\varnothing \ar@{-}[ur]\ar@{-}[u]\ar@{-}[ul]&
}
\xymatrix{
 & (1,1,1,1)& \\
(1,1,2) \ar@{-}[ur]& (1,2,1) \ar@{-}[u] & (2,1,1) \ar@{-}[ul] \\
(1,3) \ar@{-}[ur]\ar@{-}[u]& (2,2) \ar@{-}[ur]\ar@{-}[ul] & (3,1) \ar@{-}[ul]\ar@{-}[u]\\
&(4) \ar@{-}[ur]\ar@{-}[u]\ar@{-}[ul]&
}
\xymatrix{
 &\txt{$\sq$ \\ $\sq$ \\ $\sq$ \\  $\sq$} & \\
  \txt{$\sq \sq$\\$\sq\phantom{\sq}$ \\$\sq\phantom{\sq}$} \ar@{-}[ur]
 & \txt{$\phantom{\sq}\sq$ \\  $\sq \sq$ \\$\sq\phantom{\sq}$ } \ar@{-}[u] 
 & \txt{$\phantom{\sq}\sq$ \\ $\phantom{\sq}\sq$ \\  $\sq \sq$} \ar@{-}[ul] \\
 \txt{$\sq \sq \sq$\\$\sq\phantom{\sq}\phantom{\sq}$} \ar@{-}[ur]\ar@{-}[u]
 & \txt{$\phantom{\sq}\sq\sq$\\$\sq \sq \phantom{\sq}$} \ar@{-}[ur]\ar@{-}[ul] 
 & \txt{$\phantom{\sq}\phantom{\sq}\sq$\\$\sq \sq \sq$} \ar@{-}[ul]\ar@{-}[u]\\
&\txt{$\sq\sq\sq\sq$} \ar@{-}[ur]\ar@{-}[u]\ar@{-}[ul]&
}
$$
Given $\alpha=(\alpha_1,\ldots,\alpha_\ell)$,
its {\it reverse} composition is
$\rev(\alpha)=(\alpha_{\ell},\alpha_{\ell-1},\ldots,\alpha_2,\alpha_1)$.
Note that $\alpha \mapsto \rev(\alpha)$ is a poset automorphism for the refinement ordering.
\end{definition}


\begin{theorem}
For any composition $\alpha$ in $\Comp$,
\label{Qsym-antipode-on-monomials}
$$
S(M_\alpha)= (-1)^{\ell(\alpha)} \sum_{\substack{\gamma \in \Comp:\\ \gamma \text{ coarsens }\rev(\alpha)}} M_{\gamma} 
$$
\end{theorem}

For example,
$$
S(M_{(a,b,c)})=-\left( M_{(c,b,a)} + M_{(b+c,a)} + M_{(c,a+b)} + M_{(a+b+c)} \right)
$$

\begin{proof}
We give Ehrenborg's proof\footnote{A different proof was 
given by Malvenuto and Reutenauer \cite[Cor. 2.3]{MalvenutoReutenauer}, and is sketched in Remark~\ref{second-antipode-proof} below.} \cite[Prop. 3.4]{Ehrenborg} 
via induction on $\ell=\ell(\alpha)$.
One has easy base cases when
$\ell(\alpha)=0$, where
$
S(M_\varnothing) = S(1)=1=(-1)^0 M_{\rev(\varnothing)},
$
and when 
$\ell(\alpha)=1$, where $M_{(n)}$ is primitive by 
\eqref{Qsym-coproduct-on-monomials}, so Proposition~\ref{antipode-on-primitives} shows
$
S(M_{(n)})=-M_{(n)}=(-1)^1 M_{\rev((n))}.
$

For the inductive step, apply the inductive definition of $S$
from Proposition~\ref{graded-connected-bialgebras-have-antipodes}:
$$
\begin{aligned}
S(M_{\alpha_1,\ldots,\alpha_\ell})
&=-\sum_{i=0}^{\ell-1} S(M_{(\alpha_1,\ldots,\alpha_i)})
M_{(\alpha_{i+1},\ldots,\alpha_\ell)}  \\
&=\sum_{i=0}^{\ell-1}
\sum_{\substack{ \beta\text{ coarsening } \\ (\alpha_i,\alpha_{i-1},\ldots,\alpha_1)}} 
(-1)^{i+1} M_{\beta} M_{(\alpha_{i+1},\ldots,\alpha_\ell)} 
\end{aligned}
$$
The idea will be to cancel terms of opposite sign
that appear in the expansions of the products 
$M_\beta  M_{(\alpha_{i+1},\ldots,\alpha_\ell)}$.  Note that
each composition $\beta$ appearing above has 
first part $\beta_1$ of the form $\alpha_i+\alpha_{i-1}+\cdots+\alpha_h$ for some $h \leq i$, and hence each term $M_\gamma$ in the 
expansion of the product 
$
M_\beta  M_{(\alpha_{i+1},\ldots,\alpha_\ell)}
$
has $\gamma_1$ a sum that can take one of these three forms:
\begin{enumerate}
\item[$\bullet$] 
$\alpha_i+\alpha_{i-1}+\cdots+\alpha_h$,
\item[$\bullet$] 
$\alpha_{i+1}+(\alpha_i+\alpha_{i-1}+\cdots+\alpha_h)$,
\item[$\bullet$] 
$\alpha_{i+1}$.
\end{enumerate}
Say that the {\it type} of $\gamma$
is $i$ in the first case, and $i+1$ in the second two cases;
in other words, the type is the largest subscript
$k$ on a part $\alpha_k$ which was combined in the sum $\gamma_1$.
It is not hard to see that a given $\gamma$ for which the type $k$
is strictly smaller than $\ell$ arises from exactly two
pairs $(\beta,\gamma),(\beta',\gamma)$, having opposite signs $(-1)^k$
and $(-1)^{k+1}$ in the above sum.  
For example, if $\alpha=(\alpha_1,\ldots,\alpha_8)$,
then the composition 
$\gamma=(\alpha_6+\alpha_5+\alpha_4,\alpha_3,\alpha_7,
          \alpha_8+\alpha_2+\alpha_1)$
of type $6$
can arise from either of
$$
\begin{aligned}
\beta&=(\alpha_6+\alpha_5+\alpha_4,\alpha_3,\alpha_2+\alpha_1)
\text{ with }i=6\text{ and sign }(-1)^7\\ 
\beta'&=(\alpha_5+\alpha_4,\alpha_3,\alpha_2+\alpha_1)
\text{ with }i=5 \text{ and sign }(-1)^6.
\end{aligned}
$$
Similarly,
$\gamma=(\alpha_6,\alpha_5+\alpha_4,\alpha_3,\alpha_7,
          \alpha_8+\alpha_2+\alpha_1)$ can arise from either of
$$
\begin{aligned}
\beta&=(\alpha_6,\alpha_5+\alpha_4,\alpha_3,\alpha_2+\alpha_1)
\text{ with }i=6\text{ and sign }(-1)^7\\
\beta'&=(\alpha_5+\alpha_4,\alpha_3,\alpha_2+\alpha_1)
\text{ with }i=5\text{ and sign }(-1)^6.
\end{aligned}
$$
Thus one can cancel almost all the terms, excepting those
with $\gamma$ of type $\ell$ among the terms $M_\gamma$
in the expansion of the last ($i=\ell-1$) summand $M_\beta M_{(\alpha_\ell)}$.
A bit of thought shows that these are the 
$\gamma$ coarsening $\rev(\alpha)$, and all have sign $(-1)^\ell$.
\end{proof}











\subsection{The fundamental basis and $P$-partitions}

There is a second important basis for $\Qsym$ which arose originally
in Stanley's $P$-partition theory \cite{Stanley-thesis}.

\begin{definition}
A {\it labelled poset} will here mean 
a partially ordered set $P$ whose underlying set
is some subset of the integers.  A {\it $P$-partition} is a function
$P \overset{f}{\rightarrow} \{1,2,\ldots\}$ with the property
that 
\begin{enumerate}
\item[$\bullet$]
$i <_P j$ and $i <_\ZZ j$ implies $f(i) \leq f(j)$, and
\item[$\bullet$]
$i <_P j$ and $i >_\ZZ j$ implies $f(i) < f(j)$.
\end{enumerate}
Denote by $\AAA(P)$ the set of all $P$-partitions $f$, and let 
$
F_P(\xx):=\sum_{f \in \AAA(P)} \xx_f
$
where $\xx_f:=\prod_{i \in P} x_{f(i)}$
\end{definition}

\begin{example}
\label{labelled-poset-example}
Depicted is a labelled poset $P$, along with the relations among the
four values $f=(f(1),f(2),f(3),f(4))$ that define its $P$-partitions $f$:
$$
\xymatrix{
 & &2\\
4& &1\ar@{-}[u]\\
 &3\ar@{-}[ur]\ar@{-}[ul]&
} \qquad
\xymatrix{
    & &f(2)\\
f(4)& &f(1)\ar@{-}[u]^{\leq}\\
    &f(3)\ar@{-}[ul]^{\leq} \ar@{-}[ur]^{<}&
}
$$
\end{example}

The following is an important special case.

\begin{proposition}
\label{fundamental-P-partition-prop}
When $P$ is a total or linear order $w=(w_1 < \ldots < w_n)$,
the generating function $F_w(\xx)$ depends only upon the {\it descent set} 
$$
\Des(w):=\{i:w_i >_{\ZZ} w_{i+1}\} \subset \{1,2,\ldots,n-1\}
$$
and its associated composition $\alpha$ in $\Comp_n$ having
partial sums $D(\alpha)=\Des(w)$: one has that
$F_w(\xx)$ equals the {\bf fundamental quasisymmetric function}
\begin{equation}
\label{fundamental-qsym-fn-defn}
L_\alpha := L_\alpha(\xx)
:=\sum_{\substack{(1 \leq )i_1 \leq \cdots \leq i_n:\\ i_j < i_{j+1} \text{ if } j \in D(\alpha)}} 
         x_{i_1} \cdots x_{i_n}
:=\sum_{\substack{\beta \in \Comp_n:\\ 
                               \beta \text{ refines }\alpha}} M_\beta.
\end{equation}
\end{proposition}
\noindent
E.g., total order $w=35142$ has $\Des(w)=\{2,4\}$ and 
composition $\alpha=(2,2,1)$, so 
$$
\begin{aligned}
F_{35142}(\xx) 
&=\sum_{f(3) \leq f(5) < f(1) \leq f(4) < f(2)} 
         x_{f(3)}x_{f(5)}x_{f(1)}x_{f(4)}x_{f(2)}\\
&=\sum_{i_1 \leq i_2 < i_3 \leq i_4 < i_5} 
         x_{i_1}x_{i_2}x_{i_3}x_{i_4}x_{i_5}\\
&= L_{(2,2,1)} = M_{(2,2,1)}+M_{(2,1,1,1)}
                               +M_{(1,1,2,1)}+M_{(1,1,1,1,1)}.
\end{aligned}
$$

\begin{proof}
Write $F_w(\xx)$ as a sum of monomials $x_{f(w_1)} \cdots x_{f(w_n)}$
over all sequences $f(w_1) \leq \cdots \leq f(w_n)$ having
strict inequalities $f(w_i) < f(w_{i+1})$ whenever $i$ is in $\Des(w)$.
The underlying set $\{f(w_i)\}_{i=1}^n$ will equal $\{j_1 < \ldots < j_\ell\}$
with indices in increasing order having 
a multiplicity sequence $\beta=(\beta_1,\ldots,\beta_\ell)$
that gives a composition $\beta$ refining $\alpha$.
\end{proof}

\begin{example}
\label{the-two-extreme-fundamentals-example}
The extreme cases for $\alpha$ in $\Comp_n$ give quasisymmetric functions
$L_\alpha$ which are symmetric:
$$
\begin{aligned}
L_{(1^n)}&=M_{(1^n)}=e_n,\\
L_{(n)}&=\sum_{\alpha \in \Comp_n} M_{\alpha}= h_n\\
\end{aligned}
$$
\end{example}


One sees that the $\{L_\alpha\}_{\alpha \in \Comp}$
are a $\ZZ$-basis for $\Qsym$, as inclusion-exclusion applied 
to \eqref{fundamental-qsym-fn-defn} gives
$$
M_\alpha = \sum_{\substack{\beta \in \Comp_n:\\ 
           \beta \text{ refines }\alpha}} (-1)^{\ell(\beta)-\ell(\alpha)} L_\beta.
$$

The next proposition is Stanley's main 
lemma \cite[Thm. 7.19.4]{Stanley} in $P$-partition theory.  It expands any
$F_P(\xx)$ in the $\{L_\alpha\}$ basis, 
as a sum over the set $\LLL(P)$ of all {\it linear extensions} $w$
of $P$, that is, the set of all extensions of $P$ to a linear order.
E.g., the poset $P$ from Example~\ref{labelled-poset-example} has 
$\LLL(P) = \{3124,3142,3412\}$.


\begin{theorem}
\label{Stanley's-P-partition-theorem}
For any labelled poset $P$,
$$
F_P(\xx) = \sum_{w \in \LLL(P)} F_w(\xx).
$$
\end{theorem}
\begin{proof}
We give Gessel's proof \cite[Thm. 1]{Gessel}, via induction on the
number of pairs $i,j$ which are incomparable in $P$.  When this quantity
is $0$, then $P$ is itself a linear order $w$, so that $\LLL(P)=\{w\}$
and there is nothing to prove.  

In the inductive step, 
let $i,j$ be incomparable elements.  Consider the two 
posets $P_{i<j}$ and $P_{j<i}$ which are obtained
from $P$ by adding in an order relation between $i$ and $j$, and then taking
the transitive closure;  it is not hard to see that these transitive closures
cannot contain a cycle, so that these really do define two posets.
The result then follows by induction applied to  $P_{i<j}, P_{j<i}$, once
one notices that
$
\LLL(P) = \LLL(P_{i<j}) \sqcup  \LLL(P_{j<i})
$
since every linear extension $w$ of $P$ either has $i$ before $j$ or vice-versa, and 
$
\AAA(P) = \AAA(P_{i<j}) \sqcup \AAA(P_{j<i})
$
since, assuming that $i <_\ZZ j$ without loss of generality, every
$f$ in $\AAA(P)$ either satisfies $f(i) \leq f(j)$ or $f(i) > f(j)$.
\end{proof}

\begin{example}
To illustrate the induction in the above proof, consider the poset
$P$ from Example~\ref{labelled-poset-example},
having $\LLL(P)=\{3124,3142,3412\}$.  Then choosing as incomparable pair 
$\{i,j\}=\{1,4\}$, one has
$$
\xymatrix{
        &4& &2\\
P_{i<j}=& &1\ar@{-}[ur]\ar@{-}[ul]& \\
        & &3\ar@{-}[u]&}
\qquad
\xymatrix{
f(4)& &f(2)\\
 &f(1)\ar@{-}[ur]^{\leq} \ar@{-}[ul]^{\leq}& \\
 &f(3)\ar@{-}[u]^{<}&}
\qquad 
\xymatrix{ \\ \LLL(P_{i<j})=\{3124,3142\}\\ }
$$
\vskip.1in
$$
\xymatrix{
 &2\\
P_{j<i}=&1\ar@{-}[u]\\
 &4\ar@{-}[u]\\
 &3\ar@{-}[u]}
\qquad \qquad \qquad \qquad \qquad \qquad
\xymatrix{
  f(2)\\
  f(1)\ar@{-}[u]^{\leq} \\
  f(4)\ar@{-}[u]^{<} \\
  f(3)\ar@{-}[u]^{\leq}}
\qquad
\xymatrix{ \\ \LLL(P_{j<i})=\{3412\}\\ }
$$
\end{example}

We next wish to describe the 
structure maps for the Hopf algebra $\Qsym$ in the basis $\{L_\alpha\}$
of fundamental quasisymmetric functions.  For this purpose, two more
definitions are useful.

\begin{definition}
Given two words $a=(a_1,\ldots,a_m), b=(b_1,\ldots,b_n)$,
the set of 
{\it shuffles} $\shuf{a}{b}$ is the collection of 
$\binom{m}{n}$ words that contain both $a,b$ as subsequences.

Given two nonempty compositions $\alpha=(\alpha_1,\ldots,\alpha_\ell), \beta=(\beta_1,\ldots,\beta_m)$, their {\it near-concatentation} is
$$
\alpha \odot \beta:=
(\alpha_1,\ldots,\alpha_{\ell-1},\alpha_\ell+\beta_1,\beta_2,\ldots,\beta_m)
$$
For example, the figure below depicts for $\alpha=(1,3,3)$ 
(black squares) and $\beta=(4,2)$ (white squares) the concatentations and
near-concatenations as ribbons:
$$
\begin{aligned}
\alpha \cdot \beta&=
\begin{matrix}
   &   &   &   &   &   &   &\sq&\sq\\
   &   &   &   &\sq&\sq&\sq&\sq&   \\
   &   &\bsq&\bsq&\bsq&   &   &   &   \\
\bsq&\bsq&\bsq&   &   &   &   &   &   \\
\bsq&   &   &   &   &   &   &   & 
\end{matrix} \\
\alpha \odot \beta&=
\begin{matrix}
   &   &   &   &   &   &   &   &\sq&\sq\\
   &   &\bsq&\bsq&\bsq&\sq&\sq&\sq&\sq&\\
\bsq&\bsq&\bsq&   &   &   &   &   &   &\\
\bsq&   &   &   &   &   &   &   &   &
\end{matrix}
\end{aligned}
$$

Lastly, given $\alpha$ in $\Comp_n$, let $\omega(\alpha)$ be the unique composition in $\Comp_n$ whose partial sums $D(\alpha)$ 
form the complementary set within $[n-1]$ to the partial sums $D(\rev(\alpha))$;  alternatively, one can check this means that the ribbon for $\omega(\alpha)$ is obtained from that of $\alpha$ 
by conjugation or transposing, that is, if $\alpha =\lambda/\mu$ then
$\omega(\alpha) =\lambda^t/\mu^t$.  E.g. if $\alpha=(4,2,2)$ so that $n=8$,
then $\rev(\alpha)=(2,2,4)$ has $D(\rev(\alpha))=\{2,4\} \subset [7]$, 
complementary to the set $\{1,3,5,6,7\}$ which are the partial sums
for $\omega(\alpha)=(1,2,2,1,1,1)$, and the ribbon diagrams of $\alpha,\omega(\alpha)$ are
$$
\alpha =
\begin{matrix}
   &   &   &   &\sq&\sq\\
   &   &   &\sq&\sq& \\
\sq&\sq&\sq&\sq&   & \\
\end{matrix}
\qquad 
\omega(\alpha) =
\begin{matrix}
   &   &\sq\\
   &   &\sq\\
   &   &\sq\\
   &\sq&\sq\\
\sq&\sq&   \\
\sq&   &   \\
\end{matrix}
$$

\end{definition}

\begin{proposition}
\label{fundamental-basis-structure-maps}
The structure maps for the Hopf algebra $\Qsym$ in the basis $\{L_\alpha\}$
of fundamental quasisymmetric functions are as follows:
\begin{align}
\Delta L_\alpha &= \sum_{\substack{(\beta,\gamma):\\
\beta \cdot \gamma=\alpha \text{ or } 
\beta \odot \gamma=\alpha}} L_\beta \otimes L_\gamma
\label{Qsym-coproduct-on-fundamentals}\\
L_\alpha L_\beta &= \sum_{w \in \shuf{w_\alpha}{w_\beta}} L_{\gamma(w)}
\label{Qsym-product-on-fundamentals}\\
S(L_\alpha)&= (-1)^{|\alpha|} L_{\omega(\alpha)}
\label{Qsym-antipode-on-fundamentals}
\end{align}
where 
\begin{enumerate}
\item[$\bullet$] $w_\alpha$ is any permutation of 
$\{1,2,\ldots,|\alpha|\}$ with $D(w_\alpha)=D(\alpha)$,
\item[$\bullet$] $w_\beta$ is any permutation of 
$\{|\alpha|+1,|\alpha|+2,\ldots,|\alpha|+|\beta|\}$ with 
$D(w_\beta)=D(\beta)$
\end{enumerate}
and where $\gamma(w)$ is the unique composition with $D(\gamma(w))=\Des(w)$.
\end{proposition}

\noindent
At first glance the formula \eqref{Qsym-coproduct-on-fundamentals} for $\Delta L_\alpha$
might seem more complicated than the formula
\eqref{Qsym-coproduct-on-monomials} for $\Delta M_\alpha$.  However, it is equally simple
when viewed in terms of ribbon
diagrams:  it cuts the ribbon diagram $\alpha$ into two smaller ribbons $\beta$ and $\gamma$, in all $|\alpha|+1$
possible ways, via {\it horizontal} cuts ($\beta \cdot \gamma=\alpha$) or {\it vertical} cuts ($\beta \odot \gamma=\alpha$).  For example,
$$
\begin{array}{rcccccl}
\Delta L_{(3,2)}
&=1 \otimes L_{(3,2)}
&+L_{(1)} \otimes L_{(2,2)}
&+L_{(2)} \otimes L_{(1,2)}
&+L_{(3)} \otimes L_{(2)}
&+L_{(3,1)} \otimes L_{(1)}
&+L_{(3,2)} \otimes 1 \\
    &\begin{matrix}  & & \sq & \sq \\ |\sq & \sq & \sq & \end{matrix} 
    &\begin{matrix}  & & \sq & \sq \\ \sq & |\sq & \sq & \end{matrix} 
    &\begin{matrix}  & & \sq & \sq \\ \sq & \sq & |\sq & \end{matrix} 
    &\begin{matrix}  & & \underline{\sq} & \sq \\ \sq & \sq & \sq & \end{matrix} 
    &\begin{matrix}  & & \sq & |\sq \\ \sq & \sq & \sq & \end{matrix} 
    &\begin{matrix}  & & \sq & \sq| \\ \sq & \sq & \sq & \end{matrix} 
\end{array}
$$

\begin{example}  
To multiply $L_{(1,1)} L_{(2)}$, one could pick $w_\alpha = 21$ and $w_\beta=34$,
and then
$$
\begin{array}{rcccccl}
L_{(1,1)} L_{(2)}
=\sum_{w \in \shuf{21}{34}} L_{\gamma(w)}
&= L_{\gamma(2134)}
&+L_{\gamma(2314)}
&+L_{\gamma(3214)}
&+L_{\gamma(2341)}
&+L_{\gamma(3241)}
&+L_{\gamma(3421)}\\
&= L_{(1,3)}
&+L_{(2,2)}
&+L_{(1,1,2)}
&+L_{(3,1)}
&+L_{(1,2,1)}
&+L_{(2,1,1)}.
\end{array}
$$
\end{example}

\begin{proof}[Proof of Proposition~\ref{fundamental-basis-structure-maps}.]
To prove formula \eqref{Qsym-coproduct-on-fundamentals}
for $\alpha$ in $\Comp_n$, note that
$$
\Delta L_\alpha = L_\alpha(\xx,\yy) 
= \sum_{k=0}^n  
\sum_{\substack{
1 \leq i_1 \leq \cdots \leq i_k,\\
1 \leq i_{k+1} \leq \cdots \leq i_n:\\
i_r < i_{r+1} \text{ for }r \in D(\alpha)
}}
 x_{i_1} \cdots x_{i_k} \cdot y_{i_{k+1}} \cdots y_{i_n}.
$$
One then realizes that the inner sums corresponding
to values of $k$ that lie (resp. do not lie) in $D(\alpha)$ 
correspond to the terms of $L_\beta(\xx) L_\gamma(\yy)$ 
for pairs $(\beta,\gamma)$ in which 
$\beta \cdot \gamma =\alpha$ (resp. $\beta \odot \gamma =\alpha$).

%Expanding
%in the monomial basis lets one write both sides as
%sums $\sum_{(\beta,\gamma)} M_\beta \otimes M_\gamma$ 
%over certain pairs $(\beta,\gamma)$.  On the left side,
%one ranges over all such pairs with $\beta \cdot \gamma$ refining $\alpha$.
%On the right side, one ranges over such pairs in which
%$(\beta,\gamma)$ have coarsenings $(\hat{\beta},\hat{\gamma})$
%for which either $\hat{\beta} \cdot \hat{\gamma} =\alpha$
%or $\hat{\beta} \odot \hat{\gamma} =\alpha$.  
%
%One can see that these are the same pairs $(\beta,\gamma)$
%by thinking of them as equivalent to a choice
%of extra dividing lines between 
%some of the squares adjacent in the rows of the ribbon for $\alpha$,
%and one distinguished double-dividing line designating the end of 
%$\beta$ and the beginning of $\gamma$.  If this double-dividing line
%occurs at the end of row, then $\hat{\beta} \cdot \hat{\gamma} =\alpha$;
%if it occurs in the middle of a row,
%then $\hat{\beta} \cdot \hat{\gamma} =\alpha$.  
%For example, if 
%$\alpha=(4,1,2,3)$, here are two
%choices of dividing lines:
%$$
%\begin{aligned}
%(\beta,\gamma)&=((1,3,1),(1,1,2,1))\\
%(\hat\beta,\hat\gamma)&=((4,1),(2,3))
%\end{aligned}
%\begin{matrix}
%   &   &    &                  &\phantom{|}\sq &\sq&|\sq \\
%   &   &    &\sq\phantom{\Vert}&|\sq&   &\\
%   &   &    &\sq\Vert&   &   &\\
%\sq&|\sq&\sq&\sq\phantom{\Vert}&   &   &\\
%\end{matrix}
%$$
%$$
%\begin{aligned}
%(\beta,\gamma)&=((1,3,1,1),(1,2,1))\\
%(\hat\beta,\hat\gamma)&=((4,1,1),(1,3))
%\end{aligned}
%\begin{matrix}
%   &   &    &       &\phantom{\Vert}\sq &\sq&|\sq \\
%   &   &    &\sq    &\Vert\sq&   &\\
%   &   &    &\sq&   &   &\\
%\sq&|\sq&\sq&\sq&   &   &\\
%\end{matrix}
%$$
%
For formula \eqref{Qsym-product-on-fundamentals}, let
$P$ be the labelled poset which is the disjoint union of linear orders $w_\alpha, w_\beta$.
Then 
$$
L_\alpha L_\beta
=F_{w_\alpha}(\xx) F_{w_\beta}(\xx)
=F_P(\xx) = \sum_{w \in \LLL(P)} F_w(\xx)
=\sum_{w \in \shuf{w_\alpha}{w_\beta}} L_{\gamma(w)}
$$
where the first equality used Proposition~\ref{fundamental-P-partition-prop},
the second equality comes from the definition of a $P$-partition,
the third equality from Theorem~\ref{Stanley's-P-partition-theorem},
and the fourth from the equality $\LLL(P)=\shuf{w_\alpha}{w_\beta}$.

To prove formula \eqref{Qsym-antipode-on-fundamentals}, compute
using \eqref{Qsym-antipode-on-monomials} that 
$$
%\begin{aligned}
S(L_\alpha) 
=\sum_{\beta \text{ refining } \alpha} S(M_\beta) 
=\sum_{\substack{(\beta,\gamma):\\
         \beta \text{ refines } \alpha,\\
         \gamma \text{ coarsens }\rev(\beta)}} (-1)^{\ell(\beta)} M_\gamma 
=\sum_{\gamma} M_\gamma 
     \sum_{\beta}  (-1)^{\ell(\beta)}
%&=(-1)^{|\alpha|} 
%    \sum_{\gamma: D(\alpha) \cap D(\rev(\gamma))= \varnothing} M_\gamma \\
%&=(-1)^{|\alpha|} \sum_{\gamma: D(\gamma) \supseteq D(\omega(\alpha))} M_\gamma% \\
%&=(-1)^{|\alpha|} L_{\omega(\alpha)} \\
%\end{aligned}
$$
in which the last inner sum is over $\beta$ for which
$$
D(\beta) \supseteq D(\alpha) \cup D(\rev(\gamma)).
$$
The alternating signs make such inner sums vanish
unless they have only the single
term where $D(\beta)=[n-1]$ (that is, $\beta=(1^n)$).  This happens
exactly when 
$
D(\rev(\gamma)) \cup D(\alpha)=[n-1]
$ 
or equivalently, when $\rev(\gamma)$ contains the complement
of $D(\alpha)$, that is, when $\gamma$ contains the complement of
$D(\rev(\alpha))$, that is, when 
$\gamma$ refines $\omega(\alpha)$.  Thus 
$$
S(L_\alpha)=
\sum_{\substack{\gamma \in \Comp_n:\\\gamma \text{ refines }\omega(\alpha)}} 
M_\gamma  \cdot (-1)^n
=(-1)^{|\alpha|} L_{\omega(\alpha)}.
$$
\end{proof}

The antipode formula \eqref{Qsym-antipode-on-fundamentals}
for $L_\alpha$ leads to a general interpretation for the antipode
of $\Qsym$ acting on $P$-partition enumerators $F_P(\xx)$.

\begin{definition}
Given a labelled poset $P$ on $\{1,2,\ldots,n\}$, let the {\it opposite} or dual labelled poset $P^{\opp}$ have $i<_{P^{\opp}} j$ if and only if $j <_P i$.
\end{definition}
For example,
$$
P=
\xymatrix{
 & &2\\
4& &1\ar@{-}[u]\\
 &3\ar@{-}[ur]\ar@{-}[ul]&
} \qquad \qquad
P^{\opp}=
\xymatrix{
    &3\ar@{-}[dl]\ar@{-}[dr]&\\
  4& & 1\ar@{-}[d]\\
    & &2
} 
$$

The following observation is straightforward.

\begin{proposition}
\label{rotating-ribbons-is-reversing-permutations}
When $P$ is a linear order 
corresponding to some permutation $w=(w_1,\ldots,w_n)$ in $\Symm_n$, 
then $w^{\opp}=w w_0$ where $w_0$ is the longest permutation that swaps $i \leftrightarrow n+1-i$.  Furthermore, in this situation 
one has $F_w(\xx)=L_\alpha$, that is, $\Des(w) = D(\alpha)$ if and only if
$\Des(w^{\opp})= D(\omega(\alpha))$, that is $F_{w^{\opp}}(\xx) = L_{\omega(\alpha)}$,
so that 
$$
S(F_w(\xx)) =(-1)^n F_{w^{\opp}}(\xx).
$$
\end{proposition}

For example, given the compositions considered earlier
$$
\alpha =(4,2,2)=
\begin{matrix}
   &   &   &   &\sq&\sq\\
   &   &   &\sq&\sq& \\
\sq&\sq&\sq&\sq&   & \\
\end{matrix}
\qquad 
\omega(\alpha) =(1,2,2,1,1,1)=
\begin{matrix}
   &   &\sq\\
   &   &\sq\\
   &   &\sq\\
   &\sq&\sq\\
\sq&\sq&   \\
\sq&   &   \\
\end{matrix}
$$
if one picks $w=1235\cdot47\cdot68$ 
(with descent positions marked by dots) having
$\Des(w)=\{4,6\}=D(\alpha)$, then 
$w^{\opp}=ww_0=8\cdot67\cdot45\cdot3\cdot2\cdot1$ 
has $\Des(w^{\opp})=\{1,3,5,6,7\}=D(\omega(\alpha))$.


\begin{corollary}
\label{antipode-in-Qsym-as-reciprocity-corollary}
For any labelled poset $P$ on $\{1,2,\ldots,n\}$, one has
$$
S \left( F_P(\xx) \right)= (-1)^n F_{P^{\opp}}(\xx).
$$
\end{corollary}
\begin{proof}
Since $S$ is linear, one can apply Theorem~\ref{Stanley's-P-partition-theorem}:
and Proposition~\ref{rotating-ribbons-is-reversing-permutations}
$$
S \left( F_P(\xx) \right)=
\sum_{w \in \LLL(P)} S( F_w(\xx) ) 
=\sum_{w \in \LLL(P)} (-1)^{n} F_{w^{\opp}}(\xx)=
(-1)^n F_{P^{\opp}}(\xx).
$$
as $\LLL(P^{\opp}) = \{w^{\opp}: w \in \LLL(P)\}$.
\end{proof}

We remark on a special case of this corollary 
to which we alluded earlier, related to skew Schur functions.

\begin{corollary}
\label{antipode-on-skew-Schur-corollary}
In $\Lambda$, the action of $\omega$ and
the antipode $S$ on skew Schur functions $s_{\lambda/\mu}$
are as follows:
\begin{align}
\omega(s_{\lambda/\mu})&=s_{\lambda^t/\mu^t} \label{omega-on-skew-Schur}\\
S(s_{\lambda/\mu})&=(-1)^{|\lambda/\mu|} s_{\lambda^t/\mu^t} \label{antipode-on-skew-Schur}.
\end{align}
\end{corollary}
\begin{proof}
Given a skew shape $\lambda/\mu$, one can always create a labelled
poset $P$ which is its {\it skew Ferrers poset}, together with one of many
{\it column-strict labellings}, in such a way that 
$F_P(\xx)=s_{\lambda/\mu}(\xx)$.  An example is shown here
for $\lambda/\mu=(4,4,2)/(1,1,0)$:
$$
\lambda/\mu=
\begin{matrix}
   &\sq&\sq&\sq\\
   &\sq&\sq&\sq\\
\sq&\sq&  &  
\end{matrix}
\qquad \qquad
P=\xymatrix@C=8pt@R=8pt{
                      &5&                         &       &       & \\
8\ar@{-}[ur]&       &4\ar@{-}[ul]&       &2& \\
                      &7\ar@{-}[ul]\ar@{-}[ur]&       &3\ar@{-}[ul]\ar@{-}[ur]&       &1\ar@{-}[ul]\\
       &       &6\ar@{-}[ul]\ar@{-}[ur]&       &       &   \\
}
\qquad
\xymatrix@C=8pt@R=8pt{
                      &f(5)&                         &       &       & \\
f(8)\ar@{-}[ur]^{<}&       &f(4)\ar@{-}[ul]^{\leq}&       &f(2)& \\
                      &f(7)\ar@{-}[ul]^{\leq}\ar@{-}[ur]^{<}&       &f(3)\ar@{-}[ul]^{\leq}\ar@{-}[ur]^{<}&       &f(1)\ar@{-}[ul]^{\leq}\\
       &       &f(6)\ar@{-}[ul]^{\leq}\ar@{-}[ur]^{<}&       &       &   \\
}
$$
It is not hard to see that in this situation, $F_{P^{\opp}}(\xx)=\sum_T \xx^{\cont(T)}$ as $T$ ranges over all 
{\it reverse column-strict tableaux} or {\it column-strict plane partitions}  
of $\lambda^t/\mu^t$:
$$
\lambda^t/\mu^t=
\begin{matrix}
   &   &\sq\\
\sq&\sq&\sq&\\
\sq&\sq&  &\\
\sq&\sq&  &
\end{matrix}
\qquad \qquad
P^{\opp}=\xymatrix@C=8pt@R=8pt{
 & &6& & &\\
 &7\ar@{-}[ur]& &3\ar@{-}[ul]& &1\\
8\ar@{-}[ur]& &4\ar@{-}[ur]\ar@{-}[ul]& &2\ar@{-}[ur]\ar@{-}[ul]& \\
 &5\ar@{-}[ur]\ar@{-}[ul]& & & &
}
\qquad
\xymatrix@C=8pt@R=8pt{
 & &f(6)& & &\\
 &f(7)\ar@{-}[ur]^{<}& &f(3)\ar@{-}[ul]^{\leq}& &f(1)\\
f(8)\ar@{-}[ur]^{<}& &f(4)\ar@{-}[ur]^{<}\ar@{-}[ul]^{\leq}& &f(2)\ar@{-}[ur]^{<}\ar@{-}[ul]^{\leq}& \\
 &f(5)\ar@{-}[ur]^{<}\ar@{-}[ul]^{\leq}& & & &
}
$$
But this means that $F_{P^{\opp}}(\xx)=s_{\lambda^t/\mu^t}(\xx)$, since
the fact that skew Schur functions lie in $\Lambda$ implies that they
can be defined either as generating functions for
column-strict tableaux or reverse column-strict tableaux; 
see Remark~\ref{other-linear-orders-remark} above, 
or \cite[Prop. 7.10.4]{Stanley}.

Thus we have 
$$
\begin{aligned}
F_P(\xx)&=s_{\lambda/\mu}(\xx)\\
F_{P^{\opp}}(\xx)&=s_{\lambda^t/\mu^t}(\xx).
\end{aligned}
$$
Proposition~\ref{antipodes-give-convolution-inverses}(c)
tell us that the antipode for $\Qsym$ must specialize
to the antipode for $\Lambda$ (see also Remark~\ref{antipode-comparison-remark} below), so \eqref{antipode-on-skew-Schur}
is a special case of 
Corollary~\ref{antipode-in-Qsym-as-reciprocity-corollary}.
Then \eqref{omega-on-skew-Schur}
follows from the relation \eqref{antipode-versus-omega-in-Sym} that
$S(f)=(-1)^n \omega(f)$ for $f$ in $\Lambda_n$.
\end{proof}



\begin{remark}
Before leaving $P$-partitions temporarily, we mention two
open questions about them.  


The first is a conjecture of Stanley from his thesis \cite{Stanley-thesis}.
As mentioned in the proof of Corollary~\ref{antipode-on-skew-Schur-corollary},
each skew Schur function $s_{\lambda/\mu}(\xx)$ is a 
special instance of $P$-partition enumerator $F_P(\xx)$.

\begin{conjecture}
A labelled poset $P$ has $F_P(\xx)$ symmetric, and not just quasisymmetric,
{\it if and only if} $P$ is a column-strict labelling of some
skew Ferrers poset $\lambda/\mu$.
\end{conjecture}
\noindent
A somewhat weaker result in this direction
was proven Malvenuto in her thesis \cite[Thm. 6.4]{Malvenuto}, showing
that if a labelled poset $P$ has the stronger property that its set of linear extensions
$\LLL(P)$ is a union of {\it plactic} or {\it Knuth equvalence classes}, then
$P$ must be a column-strict labelling of a skew Ferrers poset.

The next question is due to P. McNamara, and is suggested by the
obvious factorizations of $P$-partition enumerators
$F_{P_1 \sqcup P_2}(\xx) = F_{P_1}(\xx) F_{P_2}(\xx)$.

\begin{question}
\label{McNamara's-question}
Does a {\it connected} labelled poset $P$ always have 
$F_P(\xx)$ {\it irreducible} within the ring $\Qsym$?
\end{question}

The phrasing of this question requires further comment.
It is assumed here that $\xx=(x_1,x_2,\ldots)$ is infinite;
for example when $P$ is a $2$-element chain
then $F_P(\xx)=e_2(\xx)$ is irreducible, but its specialization
to two variables $\xx=(x_1,x_2)$ is $e_2(x_1,x_2)=x_1 x_2$,
which is reducible.  If one wishes to work in finitely many variables
$\xx=(x_1,\ldots,x_m)$ one can perhaps assume that $m$ is at least
$|P|+1$.  

When working in $\Qsym=\Qsym(\xx)$ in 
infinitely many variables, it is perhaps not so clear where factorizations occur.
For example, if $f$ lies in $\Qsym$ and factors 
$f=g \cdot h$ with $g, h$ in $R(\xx)$, does this imply that $g, h$ also lie in $\Qsym$? 
The answer is ``Yes'', but this is not obvious, and
was proven by P. Pylyavskyy in \cite[Chap. 11]{Pylyavskyy-thesis}.


One also might wonder whether $\Qsym$ is a unique factorization domain, but
this follows from the result of M. Hazewinkel \cite{Hazewinkel} who
proved a conjecture of Ditters that $\Qsym(:=\Qsym_\ZZ)$ 
is a polynomial algebra;  earlier 
Malvenuto and Reutenauer \cite[Cor. 2.2]{MalvenutoReutenauer} had shown
that $\Qsym_\QQ$ is a polynomial algebra.  In fact, 
one can find polynomial generators $\{P_{\alpha}\}$ for $\Qsym_\QQ$ as a 
subset of the dual basis to the $\QQ$-basis $\{\pi_\alpha\}$ for $\Nsym_\QQ$ 
which comes from taking 
products $\pi_{\alpha}:= \pi_{\alpha_1} \cdots \pi_{\alpha_\ell}$
of the elements $\{\pi_n\}$ defined in Remark~\ref{second-antipode-proof} below.  Specifically,
one takes those $P_\alpha$ for which the composition $\alpha$ is a {\it Lyndon composition}.

An affirmative answer to Question~\ref{McNamara's-question}
is known at least in the special case where $P$ is a connected column-strict labelling
of a skew Ferrers diagram, that is, when $F_P(\xx)=s_{\lambda/\mu}(\xx)$
for some connected skew diagram $\lambda/\mu$;  see \cite{BarekatRWilligenburg}.
\end{remark}




\subsection{The Hopf algebra $\Nsym$ dual to $\Qsym$}

We introduce here the dual Hopf algebra to $\Qsym$.

\begin{definition}
Let $\Nsym:=\Qsym^o$, with dual pairing 
$\Nsym \otimes \Qsym \overset{(\cdot,\cdot)}{\longrightarrow} \ZZ$.
Let $\{H_\alpha\}$ be the $\ZZ$-basis of $\Nsym$ dual to the $\ZZ$-basis 
$\{M_\alpha\}$ of $\Qsym$, so that 
$$
(H_\alpha,M_\beta)=\delta_{\alpha,\beta}.
$$
\end{definition}

\begin{theorem}
Letting $H_n:=H_{(n)}$ for $n=0,1,2,\ldots$, with $H_0=1$,
one has that 
$$
\Nsym \cong \ZZ\langle H_1,H_2,\ldots \rangle,
$$
the free associative (but not commutative) algebra on 
generators $\{H_1,H_2,\ldots\}$ with coproduct determined by
\begin{equation}
\label{Nsym-coproduct-on-H}
\Delta H_n = \sum_{i+j=n} H_i \otimes H_j.
\end{equation}
\end{theorem}
\begin{proof}
Since \eqref{Qsym-coproduct-on-monomials} asserts that
$\Delta M_\alpha 
      = \sum_{(\beta,\gamma): \beta \cdot \gamma=\alpha} 
       M_\beta \otimes M_\gamma$, and since $\{H_\alpha\}$ are
dual to $\{M_\alpha\}$, one concludes that for any compositions $\beta, \gamma$, one has
$$
H_\beta H_\gamma= H_{\beta \cdot \gamma}
$$
Iterating this gives
$$
H_{\alpha}=H_{(\alpha_1,\ldots,\alpha_\ell)}
    =H_{\alpha_1} \cdots H_{\alpha_\ell}.
$$
Since the $H_\alpha$ are a $\ZZ$-basis for $\Nsym$, this shows
$\Nsym \cong \ZZ\langle H_1,H_2,\ldots \rangle$.

Note that $H_n=H_{(n)}$ is dual to $M_{(n)}$, so to
understand $\Delta H_n$, one should understand how  $M_{(n)}$ can appear
as a term in the product $M_{\alpha} M_{\beta}$.
By \eqref{Qsym-product-on-monomials} this occurs only
if $\alpha=(i), \beta=(j)$ where $i+j=n$, where
$$
M_{(i)} M_{(j)}= M_{(i+j)}+M_{(i,j)}+M_{(j,i)}.
$$
By duality, this implies the formula \eqref{Nsym-coproduct-on-H}.
\end{proof}


\begin{corollary}
The algebra homomorphism defined by
$$
\begin{array}{rcl}
\Nsym &\overset{\pi}{\longrightarrow} &\Lambda\\
H_n &\longmapsto &h_n
\end{array}
$$
is a Hopf algebra surjection, and adjoint to the 
inclusion $\Lambda \overset{i}{\hookrightarrow} \Qsym$.
\end{corollary}
\begin{proof}
As an algebra map $\pi$ may be identified with the surjection
$T(V) \rightarrow \Sym(V)$ from the tensor algebra on a graded
free $\ZZ$-module $V$ with basis $\{H_1, H_2,\ldots\}$ to the
symmetric algebra on $V$, since 
$$
\begin{aligned}
\Nsym &\cong \ZZ\langle H_1,H_2,\ldots\rangle\\
\Lambda &\cong \ZZ[h_1,h_2,\ldots]
\end{aligned}
$$
As \eqref{Nsym-coproduct-on-H} and Proposition~\ref{symm-comultiplication-formulas-prop}(iii)
assert that
$$
\begin{aligned}
\Delta H_n &= \sum_{i+j=n} H_i \otimes H_j\\
\Delta h_n &= \sum_{i+j=n} h_i \otimes h_j\\
\end{aligned}
$$
this map $\pi$ is also a bialgebra morphism, and hence a Hopf morphism by 
Proposition~\ref{antipodes-give-convolution-inverses}.

To check $\pi$ is adjoint to $i$, let $\lambda(\alpha)$
denote the partition which is the weakly decreasing rearrangement of the composition $\alpha$, 
and note that the bases $\{H_\alpha\}$
of $\Nsym$ and $\{m_\lambda\}$ of $\Lambda$ satisfy
$$
(\pi(H_\alpha),m_\lambda)
=(h_\alpha,m_\lambda)
=\left\{ \begin{matrix} 
1 & \text{ if }\lambda(\alpha)=\lambda \\
0 & \text{ otherwise }
\end{matrix} \right\}
=\left( H_{\alpha},\sum_{\beta: \lambda(\beta)=\lambda} M_\beta \right)
=(H_\alpha, i(m_\lambda)).
$$
\end{proof}

\begin{remark}
\label{second-antipode-proof}
For those who prefer generating functions to sign-reversing involutions, 
we sketch here  Malvenuto and Reutenauer's elegant proof 
\cite[Cor. 2.3]{MalvenutoReutenauer} of the antipode
formula \eqref{Qsym-antipode-on-monomials}.  One needs to
know that when $A$ is a $\kk$-algebra (possibly noncommutative)
with $\kk$ of characteristic zero,
in the ring of power series $A[[t]]$ where $t$ commutes with
all of $A$, one still has familiar facts, such as
$$
a(t) = \log b(t) \quad \text{ if and only if }\quad b(t) = \exp a(t)
$$
and whenever $a(t), b(t)$ commute in $A[[t]]$, one has
\begin{align}
\exp \left( a(t)+b(t) \right) &= \exp a(t) \exp b(t) \label{exp-of-sum} \\
\log \left( a(t) b(t) \right) &= \log a(t) + \log b(t)\label{log-of-product}
\end{align}
Start by defining in $\Nsym_\QQ:=\Nsym \otimes_\ZZ \QQ$ the elements
$\{\pi_1,\pi_2,\ldots\}$ via generating functions
in $\Nsym_\QQ[[t]]$:
\begin{equation}
\begin{aligned}
\label{relation-definining-pi's}
H(t)&:=\sum_{n \geq 0} H_n t^n,\\
\pi(t)&:=\sum_{n \geq 1} \pi_n t^n=\log H(t)
\end{aligned}
\end{equation}
One first checks that this makes each $\pi_n$ primitive, via
a computation in $(\Nsym_\QQ \otimes \Nsym_\QQ)[[t]]$:
$$
\begin{aligned}
\Delta \pi(t) &= \log \sum_{n \geq 0} \Delta(H_n) t^n
= \log \sum_{n \geq 0} \left( \sum_{i+j=n} H_i \otimes H_j \right) t^n\\
&= \log \left( \left(\sum_{i \geq 0} H_i t^i\right)  \otimes
                \left(\sum_{j \geq 0} H_j t^j\right) \right) 
= \log \left( \left(\sum_{i \geq 0} H_i t^i \otimes \one  \right) 
                \left( \one \otimes \sum_{j \geq 0} H_j t^j\right)  \right)\\
&\overset{\eqref{log-of-product}}{=}\log H(t) \otimes \one + \one \otimes \log H(t) 
= \pi(t) \otimes \one + \one \otimes \pi(t)
\end{aligned}
$$
Thus $S(\pi_n) = -\pi_n$, by Proposition~\ref{antipode-on-primitives}.
This allows one to determine $S(H_n)$ and $S(H_\alpha)$, after
one first inverts the relation \eqref{relation-definining-pi's}
to get that $H(t) = \exp \pi(t)$, and hence 
$$
\begin{aligned}
S(H(t)) &= S(\exp \pi(t)) 
=\exp S(\pi(t)) 
=\exp\left( -\pi(t) \right)
=\exp \pi(t) ^{-1} \\
&\overset{\eqref{exp-of-sum}}{=}
H(t)^{-1}=\left( 1 + H_1 t + H_2 t^2 + \cdots \right)^{-1}.
\end{aligned}
$$
Upon expanding the right side, and comparing coefficients of $t^n$, this gives
$$
S(H_n)= \sum_{\alpha \in \Comp_n} (-1)^{\ell(\beta)} H_\beta
$$
and hence 
$$
S(H_\alpha) = S(H_{\alpha_\ell}) \cdots S(H_{\alpha_2}) S(H_{\alpha_1})
=\sum_{\substack{\gamma: \\ \gamma \text{ refines }\rev(\alpha)}} 
(-1)^{\ell(\gamma)} H_\gamma.
$$
As $S_{\Nsym}, S_{\Qsym}$ are adjoint, and $\{H_\alpha\},
\{M_\alpha\}$ are dual bases, this is equivalent to
formula \eqref{Qsym-antipode-on-monomials}:
$$
S(M_\alpha)= (-1)^{\ell(\alpha)} \sum_{\substack{\gamma:\\ \gamma \text{ coarsens }\rev(\alpha)}} M_{\gamma}.
$$
\end{remark}


We next explore the basis for $\Nsym$ dual to the $\{L_\alpha\}$ in $\Qsym$.
\begin{definition}
Define the {\it noncommutative ribbon functions}
$\{R_\alpha\}$ to be the $\kk$-basis of $\Nsym$ dual to 
the fundamental basis $\{L_\alpha\}$ of $\Qsym$,
so that $(R_\alpha,L_\beta)=\delta_{\alpha,\beta}$.
\end{definition}

\begin{theorem}
\label{Nsym-structure-on-ribbons-theorem}
One has that
\begin{align}
H_\alpha&=\sum_{\beta \text{ coarsens }\alpha} R_\beta 
\label{H-as-sum-of-ribbons}\\
R_\alpha&=\sum_{\beta \text{ coarsens }\alpha} (-1)^{\ell(\beta)-\ell(\alpha)} H_\beta 
\label{ribbon-as-sum-of-Hs}
\end{align}
and the surjection $\Nsym \overset{\pi}{\longrightarrow} \Lambda$
sends $R_\alpha \longmapsto s_\alpha$, the skew Schur function associated
to the ribbon $\alpha$.  Furthermore,
\begin{align}
R_\alpha R_\beta &= R_{\alpha \cdot \beta} + R_{\alpha \odot \beta}  
\label{Nsym-product-in-ribbons}\\
S(R_\alpha) &= (-1)^{|\alpha|} R_{\omega(\alpha)} 
\label{Nsym-antipode-in-ribbons}
\end{align}
\end{theorem}

\begin{proof}
For the first assertion, note that
$$
H_\alpha
= \sum_{\beta} (H_\alpha,L_\beta) R_\beta 
= \sum_{\beta} 
 \left( H_\alpha, \sum_{\substack{\gamma:\\\gamma \text{ refines }\beta}} M_\gamma \right) 
    R_\beta 
=  \sum_{\substack{\beta:\\ \beta \text{ coarsens }\alpha}}     R_\beta.
$$
The second assertion follows from the first
by inclusion-exclusion.

To show that $\pi(R_\alpha)=s_\alpha$, we instead examine $\pi(H_\alpha)$:
$$
\pi(H_\alpha)
=h_\alpha 
=h_{\alpha_1} \cdots h_{\alpha_\ell}
= s_{(\alpha_1)} \cdots s_{(\alpha_\ell)}
= s_{(\alpha_1) \oplus \cdots \oplus (\alpha_\ell)}
$$
where $(\alpha_1) \oplus \cdots \oplus (\alpha_\ell)$ is the
skew shape which is a horizontal strip having rows of lengths $\alpha_1,\ldots,\alpha_\ell$
from bottom to top.  We claim 
$$
s_{(\alpha_1) \oplus \cdots \oplus (\alpha_\ell)}
=\sum_{\substack{\beta:\\\beta\text{ coarsens }\alpha}} s_\beta
$$
because column-strict tableaux $T$ of shape 
$(\alpha_1) \oplus \cdots \oplus (\alpha_\ell)$ biject to column-strict 
tableaux $T'$ of some ribbon $\beta$ coarsening $\alpha$, as follows:  
let $a_i,b_i$ denote the leftmost, rightmost entries of the the $i^{th}$ row from the
bottom in $T$, of length $\alpha_i$, and 
\begin{enumerate}
\item[$\bullet$] if $b_i \leq a_{i-1}$,
merge parts $\alpha_i,\alpha_{i-1}$ in $\beta$,
and concatenate the rows of length $\alpha_i,\alpha_{i-1}$ in $T$, or
\item[$\bullet$] if $b_i > a_{i-1}$,
do not merge parts $\alpha_i,\alpha_{i-1}$ in $\beta$,
and let these two rows overlap in one column in $T'$
\end{enumerate}
E.g., if $\alpha=(3,3,2,3,2)$, this $T$ of shape $(\alpha_1) \oplus \cdots \oplus (\alpha_\ell)$ maps to this $T'$ of shape $\beta=(3,8,2)$:
$$
T=\begin{matrix}
 & & & & & & & & & & &3&4 \\
 & & & & & & & &4&4&5& &\\
 & & & & & &4&4& & & & &\\
 & & &2&2&3& & & & & & &\\
1&1&3& & & & & & & & & &
\end{matrix}
\quad \longmapsto \quad
\begin{matrix}
 & & & & & & & & &3&4 \\
 & &2&2&3&4&4&4&4&5& &\\
1&1&3& & & & & & & & & &
\end{matrix}
$$
The reverse bijection breaks the rows of $T'$ into the rows of $T$ of lengths dictated by
the parts of $\alpha$.  Having shown $\pi(H_\alpha) =\sum_{\beta:\\\beta\text{ coarsens }\alpha} s_\beta$,
the relation \eqref{H-as-sum-of-ribbons} and inclusion-exclusion show $\pi(R_\alpha)=s_\alpha$.
\end{proof}

\begin{remark}
\label{antipode-comparison-remark}
Since the maps 
$$
\xymatrix{
\Nsym \ar@{->>}[dr]_\pi & & \Qsym \\
      &\Lambda \ar@{^{(}->}[ur]_i& 
}
$$
are Hopf morphisms, they must respect the antipodes $S_{\Lambda}, S_{\Qsym}, S_{\Nsym}$,
but it is interesting to compare them explicitly using the fundamental basis for
$\Qsym$ and the ribbon basis for $\Nsym$.  

On one hand \eqref{Qsym-antipode-on-fundamentals} 
shows that $S_{\Qsym}(L_\alpha)= (-1)^{|\alpha|} L_{\omega(\alpha)}$
extends the map $S_{\Lambda}$ since $L_{(1^n)} = e_n$ and $L_{(n)}=h_n$, as observed in
Example~\ref{the-two-extreme-fundamentals-example}, and $\omega((n))=(1^n)$.

On the other hand, \eqref{Nsym-antipode-in-ribbons} shows that
$S_{\Nsym}(R_\alpha)= (-1)^{|\alpha|} R_{\omega(\alpha)}$  lifts the map $S_{\Lambda}$ to $S_{\Nsym}$:
Theorem~\ref{Nsym-structure-on-ribbons-theorem} showed that
$R_\alpha$ lifts the skew Schur function $s_\alpha$, while
\eqref{skew-Schur-antipodes} asserted that
$S(s_{\lambda/\mu}) = (-1)^{|\lambda/\mu|} s_{\lambda^t/\mu^t}$,
and a ribbon $\alpha=\lambda/\mu$ has $\omega(\alpha)=\lambda^t/\mu^t$.
\end{remark}











\subsection{Polynomial generators for $\Qsym$ and Lyndon words}

Perhaps to be filled in later....

\subsection{Application:  Multiple zeta values and
Hoffman's stuffle conjecture}

Perhaps to be filled in later....





\newpage





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Aguiar-Bergeron-Sottile character theory Part I: $\Qsym$ as a terminal object} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

It turns out that the 
universal mapping property of $\Nsym$ as a free associative algebra leads
via duality to a universal property for its dual $\Qsym$, elegantly explaining several
combinatorial invariants that take the form of quasisymmetric or symmetric functions: 

\begin{enumerate}
\item[$\bullet$]  Ehrenborg's quasisymmetric function of a {\it ranked poset} \cite{Ehrenborg},
\item[$\bullet$]  Stanley's {\it chromatic} symmetric function of a {\it graph} \cite{Stanley-chromatic-qsym-fn},
\item[$\bullet$]  the quasisymmetric function of a {\it matroid} considered in \cite{BilleraJiaR}
\end{enumerate}


\subsection{Characters and the universal property}
\label{ABS-part-I-section}


\begin{definition}
Given a Hopf algebra $A$ over $\kk$, a {\it character} is an algebra morphism 
$A \overset{\zeta}{\longrightarrow} \kk$, that is, 
\begin{enumerate}
\item[$\bullet$]
$\zeta(\one_A)=\one_\kk$,
\item[$\bullet$]
$\zeta$ is $\kk$-linear, and
\item[$\bullet$]
$\zeta(a b)=\zeta(a)\zeta(b)$ for $a,b$ in $A$.
\end{enumerate}
\end{definition}

\begin{example}
\label{Qsym-zeta-character-example}
A particularly important character for $A=\Qsym$ is defined as follows:
$$
\begin{array}{rcl}
\Qsym & \overset{\zeta_Q}{\longrightarrow} & \kk \\
f(\xx)& \longmapsto & f(1,0,0,\ldots) = \left[ f(\xx) \right]_{x_1=1,x_2=x_3=\cdots=0}.
\end{array}
$$
so that 
$$
\zeta_Q(M_\alpha)=
\zeta_Q(L_\alpha)=
\begin{cases}
1 & \text{ if } \alpha=(n) \text{ for some }n,\\
0 & \text{ otherwise.}
\end{cases}
$$
In other words, the restriction $\zeta_Q|_{{\Qsym}_n}$ coincides with the
functional $H_n$ in $\Nsym_n=\Hom_\kk(\Qsym_n,\kk)$:  one has for $f$ in $\Qsym_n$ that
$\zeta_Q(f)=(H_n,f)$.

It is worth remarking that there is nothing special about setting $x_1=1$
and $x_2=x_3=\cdots=0$:  for quasisymmetric $f$, could have defined the same character
$\zeta_Q$ by picking any variable, say $x_n$, and sending
$$
f(\xx) \longmapsto \left[ f(\xx) \right]_{\substack{x_n=1,\text{ and}\\x_m=0 \text{ for }m\neq n}}.
$$
\end{example}

This character  $\Qsym \overset{\zeta_Q}{\longrightarrow} \kk$ 
has a certain universal property.

\begin{theorem}
\label{Qsym-as-terminal-object-theorem}
A graded Hopf algebra $A$ of finite type together with
a character $A \overset{\zeta}{\longrightarrow} \kk$ induces a unique
graded Hopf morphism $A \overset{\Psi}{\longrightarrow} \Qsym$ making this diagram commute:
\begin{equation}
\label{ABS-terminal-morphism-diagram}
\xymatrix{
A\ar[rr]^{\Psi} \ar[dr]_{\zeta} & & \Qsym \ar[dl]^{\zeta_Q}\\
  &\kk&
}
\end{equation}
Furthermore, $\Psi$ has this formula on elements of $A_n$:
\begin{equation}
\label{ABS-terminal-morphism-formula}
\Psi(a)=\sum_{\alpha \in \Comp_n} \zeta_\alpha(a) M_\alpha
\end{equation}
where for $\alpha=(\alpha_1,\ldots,\alpha_\ell)$, the map $\zeta_\alpha$ is the composite 
$$
A_n \overset{\Delta^{(\ell-1)}}{\longrightarrow} A^{\otimes \ell} 
    \overset{\pi_\alpha}{\longrightarrow} A_{\alpha_1} \otimes \cdots \otimes A_{\alpha_\ell}
     \overset{\zeta^{\otimes \ell}}{\longrightarrow} \kk
$$
in which 
$
A^{\otimes \ell} 
\overset{\pi_\alpha}{\longrightarrow} 
A_{\alpha_1} \otimes \cdots \otimes A_{\alpha_\ell}
$
is a projection.
\end{theorem}
\begin{proof}
One argues that $\Psi$ is unique, and has formula \eqref{ABS-terminal-morphism-formula},
using only that $\zeta$ is $\kk$-linear and that
$\Psi$ is a graded {\it $\kk$-coalgebra} map making \eqref{ABS-terminal-morphism-diagram} commute.
Equivalently, consider the adjoint {\it $\kk$-algebra} map 
$$
\Nsym = \Qsym^o \overset{\Psi^*}{\longrightarrow} A^o.
$$
Commutativity of \eqref{ABS-terminal-morphism-diagram} implies that for $a$ in $A_n$,
$$
(\Psi^*(H_n),a) = (H_n,\Psi(a)) = \zeta_Q( \Psi(a) ) = \zeta(a)
$$
where the second equality used Example~\ref{Qsym-zeta-character-example}.
In other words, $\Psi^*(H_n)$ is the element of $A^o$ defined as the following functional on $A$:
\begin{equation}
\label{image-of-Hn-as-functional}
\Psi^*(H_n)(a) = 
\begin{cases} 
\zeta(a) & \text{ if } a \in A_n,\\
0 & \text{ if }a \in A_m\text{ for some }m \neq n.
\end{cases}
\end{equation}
By the universal property for $\Nsym \cong \kk\langle H_1,H_2,\ldots \rangle$
as free associative $\kk$-algebra, \eqref{image-of-Hn-as-functional},
any choice of a $\kk$-linear map $\Qsym \overset{\zeta}{\rightarrow} \kk$ 
would uniquely produce such a $\Psi^*$, and hence also $\Psi$.  Formula
\eqref{ABS-terminal-morphism-formula} follows as
$$
\Psi(a) = \sum_{\alpha \in \Comp} (H_\alpha, \Psi(a)) \,\, M_\alpha
$$
and for a composition $\alpha=(\alpha_1,\ldots,\alpha_\ell)$, one has
$$
\begin{aligned}
(H_\alpha, \Psi(a))
=(\Psi^*(H_\alpha),a) 
&=( \Psi^*(H_{\alpha_1}) \cdots \Psi^*(H_{\alpha_\ell}), a )\\
&=( \Psi^*(H_{\alpha_1}) \otimes \cdots \otimes \Psi^*(H_{\alpha_\ell}),( \pi_\alpha \circ \Delta^{(\ell-1)})(a))
=\zeta_{\alpha}(a)
\end{aligned}
$$
using \eqref{image-of-Hn-as-functional} and the definition of $\zeta_\alpha$.

We wish to show that if, in addition, $A$ is a Hopf algebra and $A \overset{\zeta}{\longrightarrow} \kk$ is a character (algebra map), then  $A \overset{\Psi}{\longrightarrow} \Qsym$ will be an algebra map, that is, the two maps $A \otimes A \longrightarrow \Qsym$ 
given by $\Psi \circ m$ and $m \circ (\Psi \otimes \Psi)$ coincide.  To see this, consider
these two diagrams having the two maps in question as the composites of their top rows:
\begin{equation}
\xymatrix{
A \otimes A \ar[r]^{m}\ar[dr]_{\zeta \otimes \zeta} & A\ar[r]^\Psi \ar[d]_{\zeta} &\Qsym \ar[dl]^{\zeta_Q}\\
  &\kk&
}
\qquad
\xymatrix{
A \otimes A \ar[r]^{\Psi \otimes \Psi} \ar[dr]_{\zeta \otimes \zeta} 
  & \Qsym^{\otimes 2} \ar[r]^{m} \ar[d]_{\zeta_Q \otimes \zeta_Q} 
    &\Qsym \ar[dl]^{\zeta_Q}\\
  &\kk&
}
\end{equation}
The fact that $\zeta, \zeta_Q$ are algebra maps makes the above diagrams commute, so that
applying the uniqueness in the first part of the proof to the 
character $A \otimes A \overset{\zeta \otimes \zeta}{\longrightarrow} \kk$ proves 
the desired equality $\Psi \circ m=m \circ (\Psi \otimes \Psi)$. 
\end{proof}

\begin{remark}
\label{Sym-as-terminal-object-remark}
When one assumes in addition that $A$ is cocommutative,
it follows that the image of $\Psi$ will lie
in the subalgebra $\Lambda \subset \Qsym$, e.g. from the explicit 
formula \eqref{ABS-terminal-morphism-formula} and the fact
that one will have $\zeta_\alpha = \zeta_\beta$ whenever $\beta$ is a rearrangement of $\alpha$.  In other words, 
the character
$\Lambda \overset{\zeta_\Lambda}{\longrightarrow} \kk$ defined by
restricting $\zeta_Q$ to $\Lambda$, or by 
$$
\zeta_\Lambda(m_\lambda)=
\begin{cases}
1 &\text{ if }\lambda=(n) \text{ for some }n,\\
0 &\text{ otherwise,}
\end{cases}
$$
has a universal property as terminal object
with respect to characters on cocommutative co- or Hopf algebras.
\end{remark}

We close this section by discussing a well-known polynomiality and reciprocity
phenomenon; see, e.g., Humpert and Martin \cite[Prop. 2.2]{HumpertMartin}, 
Stanley \cite[\S 4]{Stanley-chromatic-qsym-fn}.

\begin{definition}
For a field $\kk$, the {\it binomial Hopf algebra}
is the polynomial algebra $\kk[m]$ in a single variable $m$,
having coalgebra structure as in $\Sym(V)$ where $V\cong \kk^1$,
whose $\kk$-basis is the primitive element $m$,
as in Example~\ref{symmetric-algebra-as-bialgebra-example}.  Thus 
$\Delta m = \one \otimes m + m \otimes \one$ and $S(m) = -m$.
As $S$ is an algebra anti-endomorphism by
Proposition~\ref{antipodes-are-antiendomorphisms} and $\kk[m]$ is commutative,
one has $S(g)(m) = g(-m)$ for all polynomials $g(m)$ in $\kk[m]$.
\end{definition}


\begin{definition}
\label{principal-specialization-at-1-defn}
For an element $f(\xx)$ in $\Qsym$ and a nonnegative integer $m$, 
let $\ps^1(f)(m)$ denote the element of $\kk$
obtained by {\it principal specialization at $q=1$}
$$
\begin{aligned}
\ps^1(f)(m)&=\left[ f(\xx)
\right]_{\substack{x_1=x_2=\cdots =x_m=1,\\x_{m+1}=x_{m+2}= \cdots =0}} \\
&=f(\underbrace{1,1,\ldots,1}_{m\text{ ones}},0,0,\ldots).
\end{aligned}
$$
\end{definition}



\begin{proposition}
\label{zeta-polynomiality-proposition}
The map $\ps^1$ has the following properties.
\begin{enumerate}
\item[(i)]
There is polynomial in $\kk[m]$ which agrees for each nonnegative integer $m$
with $\ps^1(f)(m)$, and which, by abuse of notation, we will also denote
$\ps^1(f)(m)$.  If $f$ lies in $\Qsym_n$, then $\ps^1(f)(m)$ is a polynomial
of degree at most $n$, taking these values on $M_\alpha, L_\alpha$
for $\alpha=(\alpha_1,\ldots,\alpha_\ell)$ in $\Comp_n$:
$$
\begin{aligned}
\ps^1(M_\alpha)(m) &=\binom{m}{\ell}, \\
\ps^1(L_\alpha)(m) &=\binom{m-\ell+n}{n}.
\end{aligned}
$$
\item[(ii)] The map $\Qsym \overset{\ps^1}{\longrightarrow} \kk[m]$
is a Hopf morphism into the binomial Hopf algebra.
\item[(iii)] For all $m$ in $\ZZ$ and $f$ in $\Qsym$ one has
$$
\zeta_Q^{\star m}(f) = \ps^1(f)(m).
$$
In particular, one also has 
$$
\zeta_Q^{\star (-m)}(f) = \ps^1(S(f))(m) =\ps^1(f)(-m).
$$
\item[(iv)]
For a graded Hopf algebra $A$ of finite type with a character
$A \overset{\zeta}{\longrightarrow} \kk$, and any element $a$ in $A_n$,
the polynomial $\ps^1(\Psi(a))(m)$ in $\kk[m]$ has degree at most $m$,
and when specialized to $m$ in $\ZZ$ satisfies
$$
\zeta^{\star m}(a) = \ps^1(\Psi(a))(m).
$$
\end{enumerate}
\end{proposition}
\begin{proof}
To prove assertion (i), note that one has
$$
\begin{aligned}
\ps^1(M_\alpha)(m) 
=M_\alpha(1,1,\dots,1,0,0,\ldots)
&=\sum_{1 \leq i_1 < \cdots < i_\ell < m} 
 \left[ x_{i_1}^{\alpha_1} \cdots x_{i_\ell}^{\alpha_\ell} \right]_{x_j=1} 
= \binom{m}{\ell}\\
\ps^1(L_\alpha)(m) 
=L_\alpha(1,1,\dots,1,0,0,\ldots)
&=\sum_{\substack{1 \leq i_1 \leq \cdots \leq i_n \leq m\\i_k < i_{k+1} \text{ if }k \in D(\alpha)}}  \left[ x_{i_1} \cdots x_{i_n} \right]_{x_j=1}\\
&=|\{1 \leq j_1 \leq j_1 \cdots \leq j_n \leq m-\ell+1\}|
=\binom{m-\ell+n}{n}.
\end{aligned}
$$
As $\{M_\alpha\}_{\alpha \in \Comp_n}$ form a basis for $\Qsym_n$, and 
$\binom{m}{\ell}$ is a polynomial function in $m$ of degree $\ell (\leq n)$, 
one concludes that for $f$ in $\Qsym_n$ one has that $\ps^1(f)(m)$ is a polynomial function in $m$ of
degree at most $n$.

To prove assertion (ii), note that $\ps^1$ is an algebra morphism because it 
as an evaluation homomorphism.  
To check that it is a coalgebra morphism, it suffices to check
$\Delta \circ \ps^1 = (\ps^1 \otimes \ps^1) \circ \Delta$ on each $M_\alpha$ for
$\alpha=(\alpha_1,\ldots,\alpha_\ell)$ in $\Comp_n$. Using the  Vandermonde
summation $\binom{A+B}{\ell} = \sum_{k} \binom{A}{k} \binom{B}{\ell-k}$, one has
$$
(\Delta \circ \ps^1)(M_\alpha)
= \Delta \binom{m}{\ell} 
= \binom{ m \otimes \one + \one \otimes m }{\ell} 
=\sum_{k=0}^\ell \binom{ m \otimes \one }{k}  \binom{ 1 \otimes m }{\ell-k}  
=\sum_{k=0}^\ell \binom{ m }{k} \otimes \binom{  m }{\ell-k}  
$$
while at the same time
$$
\left((\ps^1 \otimes \ps^1) \circ \Delta\right)(M_\alpha)
= \sum_{k=0}^{\ell} \ps^1(M_{\alpha_1,\ldots,\alpha_k})
                     \otimes \ps^1(M_{\alpha_{k+1},\ldots,\alpha_\ell)})
=\sum_{k=0}^{\ell} \binom{m}{k} \otimes \binom{m}{\ell-k}.
$$
Thus $\ps^1$ is a bialgebra map, and hence also a Hopf map, by 
Proposition~\ref{antipodes-give-convolution-inverses}(c).

For assertion (iii), first assume $m$ lies in $\{0,1,2,\ldots\}$.
Since $\zeta_Q(f)=f(1,0,0,\ldots)$, one has
$$
\begin{aligned}
\zeta_Q^{\star m}(f) 
&= \zeta_Q ^{\otimes m}\circ \Delta^{(m-1)} f(\xx) 
=  \zeta_Q ^{\otimes m} \left( f(\xx^{(1)},\xx^{(2)},\ldots,\xx^{(m)}) \right) \\
&=  \left[ f(\xx^{(1)},\xx^{(2)},\ldots,\xx^{(m)}) 
         \right]_{\substack{x^{(1)}_1=x^{(2)}_1=\cdots=x^{(m)}_1=1,
                    \\x^{(j)}_2=x^{(j)}_3=\cdots=0 \text{ for all }j}}\\
&=f(1,0,0,\ldots,1,0,0,\ldots,\cdots,1,0,0,\ldots)
=f(\underbrace{1,1,\ldots,1}_{m\text{ ones}},0,0,\ldots) = \ps^1(f)(m)
\end{aligned}
$$
But then Proposition~\ref{antipodes-give-convolution-inverses}(a) also implies
$$
\begin{aligned}
\zeta_Q^{\star (-m)}(f) 
&=\left( \zeta_Q^{\star (-1)} \right)^{\star m}(f)
= \left( \zeta_Q \circ S \right)^{\star m}(f)
= \zeta_Q^{\star m}(S(f))\\
&= \ps^1(S(f))(m) 
= S(\ps^1(f))(m) 
=\ps^1(f)(-m).
\end{aligned}
$$
For assertion (iv), note that
$$
\zeta^{\star m}(a)
= (\zeta_Q \circ \Psi)^{\star m}(a)
= (\zeta_Q^{\star m})(\Psi(a))
= \ps^1(\Psi(a))(m).
$$
where the three equalities come from
\eqref{ABS-terminal-morphism-diagram}, 
Proposition~\ref{antipodes-give-convolution-inverses}(a),
and assertion (iii) above, respectively.
\end{proof}


\subsection{Example: Ehrenborg's quasisymmetric function of a ranked poset} 

Here we consider incidence algebras, coalgebras and Hopf algebras generally,
and then particularize to the case of graded posets, to recover Ehrenborg's interesting quasisymmetric function invariant 
via Theorem~\ref{Qsym-as-terminal-object-theorem}.





\subsubsection{Incidence algebras, coalgebras, Hopf algebras}


\begin{definition}
Given a family $\PPP$ of finite partially ordered sets $P$,
let $\kk[\PPP]$ denote the free $\kk$-module
whose basis consists of symbols $[P]$ corresponding
to isomorphism classes of posets $P$ in $\PPP$.

We will assume throughout that each $P$ in $\PPP$ is {\it bounded},
that is, it has a unique minimal element $\hat{0}:=\hat{0}_P$ and unique
maximal element $\hat{1}:=\hat{1}_P$.  In particular, $P \neq \varnothing$,
although it is allowed that $|P|=1$, so that $\hat{0}=\hat{1}$;
denote this isomorphism class of posets with one element by $[o]$.

If $\PPP$ is closed under taking intervals 
$$
[x,y]:=[x,y]_P:=\{ z \in P: x \leq_P z \leq_P y \}
$$
then one can easily that the following coproduct and counit
endow $\kk[\PPP]$ with the structure of a coalgebra,
called the {\it (reduced) incidence coalgebra}:
$$
\begin{aligned}
\Delta [P]&:= \sum_{x \in P} [\hat{0},x] \otimes [x,\hat{1}],\\
\epsilon [P]&:= 
\begin{cases} 
1 & \text{ if }|P|=1 \\
0 & \text{ otherwise.}
\end{cases}  
\end{aligned}                       
$$
The dual algebra $\kk[\PPP]^o$ is generally called the
{\it (reduced) incidence algebra} for the family $\PPP$.
It contains the important element 
$\kk[\PPP] \overset{\zeta}{\longrightarrow} \kk$,
called the {\it $\zeta$-function}
that takes the value $\zeta[P]=1$ for all $P$.

If $\PPP$ (is not empty and) satisfies the further property of being {\it hereditary}
in the sense that for every $P_1,P_2$ in $\PPP$, the
{\it Cartesian product poset} $P_1 \times P_2$ with componentwise
partial order is also in $\PPP$, then one can check that 
the following product and unit endow $\kk[\PPP]$ with
the structure of a (commutative) algebra:
$$
\begin{aligned}[]
[P_1] \cdot [P_2]&:=m([P_1] \otimes [P_2]) := [P_1 \times P_2],\\
\one_{\kk[\PPP]}&:=[o].
\end{aligned}
$$
\end{definition}

\begin{proposition}
\label{incidence-coalgebras-are-Hopf}
For any hereditary family $\PPP$ of finite posets,
$\kk[\PPP]$ is a bialgebra, and even a Hopf algebra
with antipode $S$ given as in Theorem~\ref{Takeuchi-formula} (Takeuchi's formula):
$$
S[P]=\sum_{k \geq 0} (-1)^k 
     \sum_{\hat{0}=x_0 < \cdots <x_k=\hat{1}} 
        [x_0,x_1] \cdots [x_{k-1},x_k].
$$
\end{proposition}
\begin{proof}
Checking the commutativity of the pentagonal diagram in
\eqref{bialgebra-diagrams}
amounts to the fact that, for any 
$(x_1,x_2)  <_{P_1 \times P_2} (y_1,y_2)$,
one has a poset isomorphism
$$
\left[ (x_1,x_2) \,\, , \,\, (y_1,y_2) \right]_{P_1 \times P_2}
\cong [x_1,y_1]_{P_1} \times [x_2,y_2]_{P_2}.
$$
Commutativity of the remaining diagrams in 
\eqref{bialgebra-diagrams} is straightforward, and so $\kk[\PPP]$
is a bialgebra.  But then Remark~\ref{non-graded-Takeuchi-remark}
implies that it is a Hopf algebra, with antipode $S$ as in the theorem,
because the map $f:=\one_{\kk[\PPP]} - u\epsilon$ (sending the class $[o]$ to $0$,
and fixing all other $[P]$) is locally $\star$-nilpotent:
$$
f^{\star k}[P]
=\sum_{\hat{0}=x_0 < \cdots <x_{k}=\hat{1}}
[x_0,x_1] \cdots [x_{k-1},x_{k}]
$$
will vanish due to an empty sum whenever $k$ exceeds the maximum length of
a chain in the finite poset $P$.
\end{proof}


It is perhaps worth remarking how this generalizes the M\"obius function
formula of P. Hall.  Note that the zeta function 
$\kk[\PPP] \overset{\zeta}{\longrightarrow} \kk$ is a {\it character}, 
that is, an algebra morphism.  
Proposition~\ref{antipodes-give-convolution-inverses}(a)
then tells us that $\zeta$ should have a convolutional
inverse $\kk[\PPP] \overset{\mu=\zeta^{\star -1}}{\longrightarrow} \kk$,
traditionally called the {\it M\"obius function}, with the formula
$\mu = \zeta^{\star -1} = \zeta \circ S$.  Rewriting this via
the antipode formula for $S$ given 
in Proposition~\ref{incidence-coalgebras-are-Hopf} yields 
P. Hall's formula.

\begin{corollary}
\label{P.Hall-formula}
For a finite bounded poset $P$, one has
$$
\mu[P] = 
\sum_{k \geq 0} (-1)^k 
|\{ \text{chains }\hat{0}=x_0 < \cdots <x_k=\hat{1}\text{ in }P\}|.
$$
\end{corollary}


\subsubsection{The incidence Hopf algebras for ranked posets and Ehrenborg's function}

\begin{definition}
Take $\PPP$ to be the class of bounded {\it ranked} finite
posets $P$, that is, those for which all maximal chains from
$\hat{0}$ to $\hat{1}$ have the same length $r(P)$.  This is a hereditary
class, as it implies that any interval is $[x,y]_P$ is also ranked, 
and the product of two bounded ranked posets is also bounded and ranked.
It also uniquely defines a 
{\it rank function} $P \overset{r}{\longrightarrow} \NN$
in which $r(\hat{0})=0$ and $r(x)$ is the length of any maximal chain
from $\hat{0}$ to $x$.
\end{definition}

\begin{example}
\label{polytope-example}
Consider a pyramid with apex vertex $a$ over a square base with vertices $b,c,d,e$:
$$
\xymatrix{
 & &a\ar@{-}[dll]\ar@{-}[ddl]\ar@{-}[drr]\ar@{-}[ddrrr]&&& & \\
b\ar@{--}[rrrr]\ar@{-}[dr]& & &&e\ar@{-}[dr]& \\
 &c\ar@{-}[rrrr]& & &&d
}
$$
Ordering its faces by inclusion gives a bounded ranked poset $P$,
where the rank of an element is one more than the dimension
of the face it represents:
$$
\xymatrix{
 & & & & & & & & \text{rank:}\\
   &   &   &abcd\ar@{-}[drrrr]\ar@{-}[d]\ar@{-}[dl]\ar@{-}[dll]\ar@{-}[dlll]&  &  &  & & 4\\
abc\ar@{-}[d]\ar@{-}[dr]&acd\ar@{-}[d]\ar@{-}[dr]&ade\ar@{-}[d]\ar@{-}[dr]&abe\ar@{-}[d]\ar@{-}[dlll] &  &  &  &bcde\ar@{-}[d]\ar@{-}[dl]\ar@{-}[dll]\ar@{-}[dlll]& 3\\
ab &ac &ad &ae  &be\ar@{-}[ul]&bc\ar@{-}[ulllll]&cd\ar@{-}[ulllll]&de\ar@{-}[ulllll]& 2\\ 
a\ar@{-}[u]\ar@{-}[ur]\ar@{-}[urr]\ar@{-}[urrr]  &   &   &    &b\ar@{-}[u]\ar@{-}[ur]\ar@{-}[ullll] &c\ar@{-}[u]\ar@{-}[ur]\ar@{-}[ullll] &d\ar@{-}[u]\ar@{-}[ur]\ar@{-}[ullll] &e\ar@{-}[u]\ar@{-}[ulll]\ar@{-}[ullll] & 1\\
   &   &   &    &\varnothing\ar@{-}[ullll]\ar@{-}[u]\ar@{-}[ur]\ar@{-}[urr]\ar@{-}[urrr]& & & & 0\\
}
$$
\end{example}

\begin{definition}
{\it Ehrenborg's quasisymmetric function} $\Psi[P]$ for a bounded ranked poset $P$ 
is the image of $[P]$ under the map 
$\kk[\PPP] \overset{\Psi}{\longrightarrow} \Qsym$
induced by the zeta function 
$\kk[\PPP] \overset{\zeta}{\longrightarrow} \kk$ as
a character, via Theorem~\ref{Qsym-as-terminal-object-theorem}.
\end{definition}

The quasisymmetric function $\Psi[P]$ captures several interesting combinatorial invariants of $P$; see Stanley \cite[Chap. 3]{Stanley} for more background on these notions.

\begin{definition}
Let $P$ be a bounded ranked poset $P$ of rank $r(P):=r(\hat{1})$.
Define its {\it rank-generating function}
$$
RGF(P,q):=\sum_{p \in P} q^{r(p)},
$$
its {\it characteristic polynomial}
$$
\chi(P,q):=\sum_{p \in P} \mu(\hat{0},p)q^{r(p)},
$$
its {\it zeta polynomial} 
\begin{align}
Z(P,m)&=|\{\text{multichains }
\hat{0} \leq  p_1 \leq_P \cdots \leq_P p_{m-1}  \leq_P \hat{1} \} |\\
 &= \sum_{s=0}^{r(P)-1} \binom{m}{s+1} 
          |\{\text{chains }\hat{0} < p_1 < \cdots < p_s < \hat{1}\}| 
\label{zeta-polynomial-as-chain-sum}
\end{align}
and for a subset  $S \subset \{1,2,\ldots,r-1\}$,
its {\it flag number} $f_S$, as a component of its {\it flag $f$-vector} $(f_S)_{S \subset [r-1]}$
$$
f_S=|\{\text{chains } \hat{0} <_P p_1 <_P \cdots <_P p_s  <_P \hat{1}
\text{ with }\{r(p_1),\ldots,r(p_s)\}=S\}|,
$$
as well as the {\it flag $h$-vector} entry $h_T$ given by 
$f_S=\sum_{ T \subset S} h_T$, or by inclusion-exclusion,
$h_S=\sum_{ T \subset S} (-1)^{|S \setminus T|} f_T$.
\end{definition}


\begin{example}
For the poset $P$ in Example~\ref{polytope-example}, one
has $RGF(P,q)=1+5q+8q^2+5q^3+q^4$.  Since $P$ is the poset of
faces of a polytope, the M\"obius function values for its intervals are easily predicted:  $\mu(x,y)=(-1)^{r[x,y]}$, that is, $P$ is an {\it Eulerian ranked poset}; see Stanley \cite[\S 3.14]{Stanley}.  Hence
its characteristic polynomial is trivially related to the
rank generating function, sending $q \mapsto -q$, that is,
$$
\chi(P,q)= RGF(P,-q)=1-5q+8q^2-5q^3+q^4.
$$
Its flag $f$-vector and $h$-vector entries are given in the following
table.
\vskip.1in
\begin{tabular}{|c|c|cl|}\hline
$S$           & $f_S$ &       &$h_S$  \\ \hline\hline
$\varnothing$ & $1$   &       &$1$ \\ \hline
$\{1\}$       & $5$   & $5-1=$&$4$ \\ \hline
$\{2\}$       & $8$   & $8-1=$&$7$ \\ \hline
$\{3\}$       & $5$   & $5-1=$&$4$ \\ \hline
$\{1,2\}$     & $16$  & $16-(5+8)+1=$&$4$ \\ \hline
$\{1,3\}$     & $16$  & $16-(5+5)+1=$&$7$ \\ \hline
$\{2,3\}$     & $16$  & $16-(5+8)+1=$&$4$ \\ \hline
$\{1,2,3\}$   & $32$  & $32-(16+16+16)+(5+8+5)-1=$&$1$ \\ \hline
\end{tabular}
\vskip.1in
and using \eqref{zeta-polynomial-as-chain-sum}, its zeta polynomial is
$$
Z(P,m)=1 \binom{m}{1} + (5+8+5) \binom{m}{2} + (16+16+16) \binom{m}{3} + 
32 \binom{m}{4}=\frac{m^2(2m-1)(2m+1)}{3}.
$$
\end{example}

\begin{theorem}
\label{Ehrenborg's-theorem}
Ehrenborg's quasisymmetric function $\Psi[P]$ for a bounded ranked poset $P$
encodes
\begin{enumerate}
\item[(i)]
the flag $f$-vector entries $f_S$ and flag $h$-vector entries $h_S$
as its $M_\alpha$ and $L_\alpha$ expansion coefficients\footnote{In fact,
Ehrenborg {\it defined} $\Psi[P]$  in \cite[Defn. 4.1]{Ehrenborg} 
via this $M_\alpha$ expansion, and then showed that it gave a Hopf morphism.} :
$$
\Psi[P] = \sum_\alpha f_{D(\alpha)}(P) \,\, M_\alpha
= \sum_\alpha h_{D(\alpha)}(P) \,\, L_\alpha ,
$$
\item[(ii)]
the zeta polynomial as the specialization from 
Definition~\ref{principal-specialization-at-1-defn}
$$
Z(P,m)=\ps^1(\Psi[P])(m)
=\left[ \Psi[P] \right]_{\substack{x_1=x_2=\cdots =x_m=1,\\x_{m+1}=x_{m+2}= \cdots =0}}, and
$$
\item[(iii)]
the rank-generating function as the specialization
$$
RGF(P,q) 
=\left[ \Psi[P] \right]_{\substack{x_1=q,x_2=1,\\x_3=x_4= \cdots =0}} ,
$$
\item[(iv)]
the characteristic polynomial as the convolution
$$
\chi(P,q)= ((\psi_q \circ S) \star \zeta_Q) \circ \Psi[P] 
$$
where $\Qsym \overset{\psi_q}{\longrightarrow} \kk[q]$
maps $f(\xx) \longmapsto f(q,0,0,\ldots)$.
\end{enumerate}
\end{theorem}
\begin{proof}
In assertion (i), the 
expansion $\Psi[P]=\sum_\alpha f_{D(\alpha)}(P) \, M_\alpha$ is 
\eqref{ABS-terminal-morphism-formula}, since
$\zeta_\alpha[P]=f_{D(\alpha)}(P)$.  The $L_\alpha$ expansion
follows by inclusion-exclusion, as 
$L_\alpha=\sum_{\beta: D(\beta) \supset D(\alpha)} M_\beta$
and $f_S(P)=\sum_{T \subset S} h_T$.


Assertion (ii) is immediate from 
Proposition~\ref{zeta-polynomiality-proposition}(iv),
since $Z(P,m)=\zeta^{\star m}[P]$.

Assertion (iii), can be deduced from assertion (i), 
but it is perhaps more fun and in the spirit of things to proceed
as follows. Note that 
$\psi_q(M_\alpha)=q^{n}$ for $\alpha=(n)$, and $\psi_q(M_\alpha)$ vanishes for
all other $\alpha \neq (n)$ in $\Comp_n$.  Hence
for a bounded ranked poset $P$ one has
\begin{equation}
\label{q-rank-functional-equation}
(\psi_q \circ \Psi)[P]= q^{r(P)}.
\end{equation}
Consequently, using \eqref{pre-and-post-composition-in-convolution}
one can compute
$$
\begin{aligned}
RGF(P,q) & = \sum_{p \in P} q^{r(P)} \cdot \zeta[p,\hat{1}]
=\sum_{p \in P} (\psi_q \circ \Psi)[\hat{0},p] \cdot
               (\zeta_Q \circ \Psi)[p,\hat{1}] \\
&=(\psi_q \star \zeta_Q)(\Psi[P]) 
=(\psi_q \otimes \zeta_Q)\left( \Delta \Psi[P])\right) \\
&=\left[ \Psi[P](\xx,\yy) \right]_{\substack{x_1=q,x_2=x_3=\cdots=0\\y_1=1,y_2=y_3=\cdots=0}}
=\left[ \Psi[P](\xx) \right]_{\substack{x_1=q,x_2=1,\\x_3=x_4=\cdots=0}}
\end{aligned}
$$

%Assertion (ii) follows from observing the evaluation for $\alpha$ in $\Comp_n$
%$$
%\left[ M_\alpha\right]_{\substack{x_1=q,x_2=1,\\x_3=x_4= \cdots =0}}
%=\begin{cases}
%1+q^n & \text{ if }\alpha=(n),\\
%q^k & \text{ if } \alpha=(k,n-k), 0 < k < n\\
%0 & \text{ if }\ell(\alpha) \geq 3.
%\end{cases}
%$$
%This implies that 
%$$
%\left[ \Psi[P] \right]_{\substack{x_1=q,x_2=1,\\x_3=x_4= \cdots =0}}
%= 1+q^{r(P)} + \sum_{r=1}^{r(P)-1} f_{(k,n-k)}(P) q^k
%=RGF(P,q).
%$$

Similarly, for assertion (iv) first note  that
Proposition~\ref{incidence-coalgebras-are-Hopf} and
Corollary~\ref{P.Hall-formula} let one calculate that
$$
\begin{aligned}
(\psi_q \circ \Psi \circ S)[P] 
&= \sum_k (-1)^k \sum_{\hat{0}=x_0 < \cdots < x_k=\hat{1}}
            (\psi_q \circ \Psi)[x_0,x_1] \cdot (\psi_q \circ \Psi)([x_{k-1},x_k]) \\
& 
\overset{\eqref{q-rank-functional-equation}}{=} \sum_k (-1)^k \sum_{\hat{0}=x_0 < \cdots < x_k=\hat{1}} q^{r(P)} 
= \mu(\hat{0},\hat{1}) q^{r(P)}.
\end{aligned}
$$
This is used in the penultimate equality here:
$$
\begin{aligned}
((\psi_q \circ S) \star \zeta_Q) \circ \Psi[P]
%&= m \circ ((\psi_q \circ S) \otimes \zeta_Q) \circ \Delta \circ \Psi[P] \\
%&= m \circ ((\psi_q \circ S) \otimes \zeta_Q) \circ (\Psi \otimes \Psi) \circ\Delta [P] \\
&\overset{\eqref{pre-and-post-composition-in-convolution}}{=} 
  ((\psi_q \circ S \circ \Psi) \star (\zeta_Q \circ \Psi)) [P] 
= ((\psi_q \circ \Psi \circ S) \star \zeta) [P] \\
&= \sum_{p \in P}  (\psi_q \circ \Psi \circ S)[\hat{0},p] \cdot \zeta[p,\hat{1}]
=\sum_{p \in P} \mu[\hat{0},p] q^{r(p)} = \chi(P,q).
\end{aligned}
$$
\end{proof}









\subsection{Example: Stanley's chromatic symmetric function of a graph}

We introduce the {\it chromatic Hopf algebra of graphs} and an associated
character $\zeta$ so that the map $\Psi$ from 
Theorem~\ref{Qsym-as-terminal-object-theorem}
sends a graph $G$ to Stanley's {\it chromatic symmetric function} 
of $G$.  Then principal specialization $\ps^1$ sends this
to the {\it chromatic polynomial} of the graph.

\subsubsection{The chromatic Hopf algebra of graphs}


\begin{definition}
The {\it chromatic Hopf algebra} (see Schmitt \cite[\S 3.2]{Schmitt-structures})
$\GGG$ is a free $\kk$-module whose
$\kk$-basis elements $[G]$ are indexed by isomorphism classes of
(finite) simple graphs $G=(V,E)$.
Define for $G_1=(V_1,E_1), G_2=(V_2,E_2)$ the multiplication 
$$
[G_1] \cdot [G_2]:=[G_1 \sqcup G_2]
$$
where $[G_1 \sqcup G_2]$ denote the isomorphism class of the
disjoint union, on vertex set $V=V_1 \sqcup V_2$ which
is a disjoint union of copies of their vertex sets $V_1,V_2$, with
edge set $E=E_1 \sqcup E_2$.  For example,
$$
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right] \cdot
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\\
\bullet\ar@{-}[u]
}
\right]
=
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]& &\bullet\ar@{-}[u]
}
\right] 
$$
  Thus the class $[\varnothing]$
of the empty graph $\varnothing$ having $V=\varnothing, E=\varnothing$
is a unit element. 

Given $V' \subset V$ of $G|_V':=(V',E')$
the {\it subgraph induced on vertex set $V'$} has
edge set $E'=\{e \in E: e={v_1,v_2} \subset V'\}$.
This lets one define a comultiplication 
$$
\Delta[G]:=\sum_{(V_1,V_2): V_1 \sqcup V_2=V} [G|_{V_1}] \otimes [G|_{V_2}].
$$
Define a counit
$$
\epsilon [G]:= 
\begin{cases} 
1 & \text{ if }G=\varnothing \\
0 & \text{ otherwise.}
\end{cases}  
$$
\end{definition}

\begin{proposition}
The above maps endow $\GGG$ with the structure of a graded connected
finite type Hopf algebra over $\kk$, which is both commutative and cocommutative.
\end{proposition}

\begin{example}
Here are some examples of these structure maps:
$$
\begin{aligned}
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right] \cdot
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\\
\bullet\ar@{-}[u]
}
\right]
&=
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]& &\bullet\ar@{-}[u]
}
\right] \\
%%%%%%%%%%%%%%
\Delta \left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right]
&= \one 
\otimes 
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right]
+
2
\left[
\xymatrix@C=4pt@R=4pt{\bullet}
\right]
\otimes
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\\
\bullet\ar@{-}[u]
}
\right]
+
\left[
\xymatrix@C=4pt@R=4pt{\bullet & \bullet}
\right]
\otimes
\left[
\xymatrix@C=4pt@R=4pt{\bullet}
\right]
+
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right]
\otimes \one
\end{aligned}
$$
\end{example}

\begin{proof}
The associativity of the multiplication and comultiplication
should be clear as 
$$
\begin{aligned}
m^{(2)}([G_1] \otimes [G_2] \otimes [G_3])
&=[G_1 \sqcup G_2 \sqcup G_3]\\
\Delta^{(2)}[G]
&=\sum_{\substack{(V_1,V_2,V_3):\\V=V_1 \sqcup V_2 \sqcup V_3}}
       [G|_{V_1}] \otimes [G|_{V_2}] \otimes [G|_{V_3}].
\end{aligned}
$$
Checking the unit and counit conditions are straightforward. 
Commutativity of the pentagonal bialgebra diagram
in \eqref{bialgebra-diagrams} comes down to check that,
given graphs $G_1,G_2$ on disjoint vertex
sets $V_1,V_2$ , when one applies to $[G_1] \otimes [G_2]$
either the composite $\Delta \circ m$ or the composite
$(m \otimes m) \circ (\one \otimes T \otimes \one) \circ (\Delta \otimes \Delta)$,
the result is the same:
$$
\sum_{\substack{(V_{11},V_{12},V_{21},V_{22}):\\
        V_1=V_{11} \sqcup V_{12}\\
        V_2=V_{21} \sqcup V_{22}}}
[G_1|_{V_{11}} \sqcup G_2|_{V_{21}}] \otimes
[G_1|_{V_{12}} \sqcup G_2|_{V_{22}}].
$$
Letting $\GGG_n$ be the $\kk$-span of $[G]$ having $n$ vertices makes $\GGG$
a bialgebra which is graded, connected, and of finite type, and hence also a
Hopf algebra by Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
Cocommutativity should be clear, and commutativity follows from the
graph isomorphism $G_1 \sqcup G_2 \cong G_2 \sqcup G_1$.
\end{proof}

\begin{remark}
\label{Humpert-Martin-antipode-remark}
Humpert and Martin \cite[Theorem 3.1]{HumpertMartin}
gave the following expansion for the antipode in the
chromatic Hopf algebra, containing fewer terms than Takeuchi's general
formula \eqref{Takeuchi-formula}:  given a graph $G=(V,E)$, one has
$$
S[G]=\sum_{F} (-1)^{|V|-\rank(F)} \acyc(G/F) [G_{V,F}].
$$
Here $F$ runs over all subsets of edges that form {\it flats}
in the graphic matroid for $G$, meaning that if $e=\{v,v'\}$ is an
edge in $E$ for which one has a path of edges in $F$ connecting $v$ to $v'$,
then $e$ also lies in $F$.  Here $G/F$ denotes the quotient graph in which all of the
edges of $F$ have been {\it contracted}, while $\acyc(G/F)$ denotes its number
of {\it acyclic orientations}, and $G_{V,F}:=(V,F)$ as a simple graph.
\end{remark}


It turns out that the chromatic Hopf algebra $\GGG$ is self-dual.
In the dual Hopf algebra $\GGG^o$,
let $\{[G]^*\}$ denote the dual basis elements, so that $([H]^*,[G])=\delta_{[H],[G]}$.
To describe the structure maps in $\GGG^o$ explicitly, for graphs $H,H_1,H_2$ one has
\begin{equation}
\label{dual-chromatic-Hopf-structure-maps}
\begin{aligned}
\Delta[H]^* 
&=\sum_{\substack{(V_1,V_2):\\ V=V_1 \sqcup V_2\\H=H|_{V_1} \sqcup H|_{V_2}}}
[H|_{V_1}]^* \otimes [H|_{V_2}]^*\\
[H_1]^* [H_2]^*
 &= \sum_{\substack{H=(V_1 \sqcup V_2,E)\\H|_{V_1}=H_1\\H|_{V_2}=H_2}} [H]^*.
\end{aligned}
\end{equation}


\begin{proposition}
One has a Hopf isomorphism 
$\GGG \overset{\varphi}{\longrightarrow} \GGG^o$
defined by
$$
[G] \longmapsto \sum_{\substack{H=(V,E'):\\ E' \cap E = \varnothing}} [H]^*.
$$
\end{proposition}
For example, this isomorphism maps
$$
\varphi \left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right]
= 
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\ar@{-}[rr]& &\bullet\\
       &\bullet&
}
\right]^*
+
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet&
}
\right]^*.
$$

\begin{proof}
First note that $\varphi$ is a $\kk$-module
isomorphism via triangularity:  one has $H=(V,E')$
with $E' \cap E =\varnothing$ if and only 
if $H$ is an edge subgraph of the {\it complementary graph} $\overline{G}$
to $G$ on the same vertex set $V$.

One can then check that for graphs $G_1,G_2$ on vertex sets $V_1, V_2$,
the fact that $\varphi([G_1][G_2])=
\varphi[G_1]\varphi[G_2]$
amounts, using
\eqref{dual-chromatic-Hopf-structure-maps},
to both being a sum of $[H]^*$
over graphs $H$ on $V_1 \sqcup V_2$ that share no
edges with $G_1 \sqcup G_2$.

Similarly, one can check that the fact
that $\Delta \varphi[G]=(\varphi \otimes \varphi)(\Delta[G])$
amounts, using
\eqref{dual-chromatic-Hopf-structure-maps},
to both being a sum of $[H_1]^* \otimes [H_2]^*$
over triples $(H,H_1,H_2)$ where 
$H$ is a graph on the same vertex
set $V$ as $G$ but sharing no edges with $G$, and
with $H=H_1 \sqcup H_2$.
\end{proof}



\subsubsection{Stanley's chromatic symmetric function of a graph}

\begin{definition}
{\it Stanley's chromatic symmetric function} $\Psi[G]$ for a simple graph $G=(V,E)$
is the image of $[P]$ under the map 
$\GGG \overset{\Psi}{\longrightarrow} \Qsym$
induced via Theorem~\ref{Qsym-as-terminal-object-theorem}
from the {\it edge-free character}
$\GGG \overset{\zeta}{\longrightarrow} \kk$
\begin{equation}
\label{edge-free-character-defn}
\zeta[G]=\begin{cases}
1 & \text{ if }G\text{ has no edges, that is, }G\text{ is an independent/stable set of vertices},\\
0 & \text{ otherwise.}
\end{cases}
\end{equation}
Note that, because $\GGG$ is cocommutative, $\Psi[G]$ is symmetric and not just quasisymmetric;
see Remark \ref{Sym-as-terminal-object-remark}.
\end{definition}

Recall that for a graph $G=(V,E)$,
a (vertex-)coloring $f: V \rightarrow \{1,2,\ldots\}$
is called {\it proper} if no edge $e=\{v,v'\}$ in $E$ has $f(v)=f(v')$.

\begin{proposition}
For a graph $G=(V,E)$, the symmetric function $\Psi[G]$ has the expansion
\footnote{In fact, Stanley {\it defined} $\Psi[G]$  in 
\cite[Defn. 2.1]{Stanley-chromatic-qsym-fn} 
via this expansion.} 
$$
\Psi[G]=\sum_{\substack{\text{proper colorings}\\f:V \rightarrow \{1,2,\ldots\}}}
         \xx_f
$$
where $\xx_f:=\prod_{v \in V} x_{f(v)}$.
In particular, its specialization from Proposition~\ref{principal-specialization-at-1-defn}
gives the chromatic polynomial of $G$:
$$
\ps^1 \Psi[G](m)=\chi_G(m)
=|\{\text{proper colorings }f:V \rightarrow \{1,2,\ldots,m\}|
$$
\end{proposition}
\begin{proof}
The iterated coproduct 
$\GGG \overset{\Delta^{(\ell-1)}}{\longrightarrow} \GGG^{\otimes \ell}$
sends
$$
[G] \longmapsto \sum_{\substack{(V_1,\ldots,V_{\ell}):\\ V=V_1 \sqcup \cdots \sqcup V_\ell}} 
[G|_{V_1}] \otimes \cdots \otimes [G|_{V_\ell}]
$$
and the map $\zeta^{\otimes \ell}$ sends the element on the right to
$1$ or $0$, depending upon whether each $V_i \subset V$ is a stable set or not,
that is, whether the assignment of color $i$ to the vertices in $V_i$ gives
a proper coloring.  Thus formula \eqref{ABS-terminal-morphism-formula}
shows that the coefficient $\zeta_\alpha$ of $x_1^{\alpha_1} \cdots x_\ell^{\alpha_\ell}$ in $\Psi[G]$ counts the
proper colorings $f$ in which $|f^{-1}(i)|=\alpha_i$ for each $i$.
\end{proof}

\begin{example}
For the complete graph $K_n$ on $n$ vertices,
one has 
$$
\begin{aligned}
\Psi[K_n]&=n! e_n\\
\ps^1 (\Psi[K_n])(m)
&=n! e_n(\underbrace{1,1,\ldots,1}_{m\text{ ones}})
=n! \binom{m}{n}\\
& =m(m-1) \cdots (m-(n-1))=\chi(K_n,m).
\end{aligned}
$$
In particular, the single vertex graph $K_1$ has
$\Psi[K_1]=e_1$, and since the Hopf morphism $\Psi$ is in particular
an algebra morphism, a graph $K_1^{\sqcup n}$ having $n$ isolated
vertices and no edges will have $\Psi[K_1^{\sqcup n}]=e_1^n$.

As a slightly more interesting example, the graph $P_3$ which is
a path having three vertices and two edges will have
$$
\Psi[P_3]=m_{(2,1)}+6m_{(1,1,1)}=e_2 e_1 + 3 e_3
$$
\end{example}

One might wonder, based on the previous examples, when $\Psi[G]$ is
{\it $e$-positive}, that is, when does its unique expansion in the $\{e_\lambda\}$
basis for $\Lambda$ have nonnegative coefficients?  This is an even stronger
assertion than {\it $s$-positivity}, that is, having nonnegative coefficients for the expansion in terms
of Schur functions $\{s_\lambda\}$, since each $e_\lambda$ is $s$-positive.
This weaker property fails, starting with the {\it claw graph $K_{3,1}$}, which has
$$
\Psi[K_{3,1}]=s_{(3,1)}-s_{(2,2)}+5s_{(2,1,1)}+8s_{(1,1,1,1)}.
$$
On the other hand, a result of Gasharov \cite{Gasharov} shows that one at
least has $s$-positivity for $\Psi[\inc(P)]$ where
$\inc(P)$ is the {\it incomparability graph} of a poset
which is $(\mathbf{3}+\mathbf{1})$-free;  we refer the
reader to Stanley\cite[\S 5]{Stanley-chromatic-qsym-fn} for a discussion
of the following conjecture, which remains open\footnote{A recent
refinement for incomparability graphs of posets which are both $(\mathbf{3}+\mathbf{1})$- and
$(\mathbf{2}+\mathbf{2})$-free, also known as {\it unit interval orders}
is discussed by Shareshian and Wachs \cite{ShareshianWachs}.}

\begin{conjecture}
For any $(\mathbf{3}+\mathbf{1})$-free poset $P$, the incomparability graph
$\inc(P)$ has $\Psi[\inc(P)]$ an $e$-positive symmetric function.
\end{conjecture}

Here is another question about $\Psi[G]$:  how well does it distinguish nonisomorphic
graphs? Stanley gave this example of two graphs $G_1, G_2$ having $\Psi[G_1]=\Psi[G_2]$:
$$
G_1=
\xymatrix@C=6pt@R=6pt{
\bullet\ar@{-}[rr] &  &\bullet\\
        &\bullet\ar@{-}[ul]\ar@{-}[ur]\ar@{-}[dr]\ar@{-}[dl]& \\
\bullet\ar@{-}[rr] &  &\bullet
}
\qquad
G_2=
\xymatrix{
\bullet\ar@{-}[r] \ar@{-}[dr]\ar@{-}[d]&\bullet\ar@{-}[d]\ar@{-}[dr]  &\\
\bullet\ar@{-}[r] &\bullet &\bullet
}
$$
\noindent
At least $\Psi[G]$ appears to do better at distinguishing {\it trees}, much
better than its specialization, the chromatic polynomial $\chi(G,m)$, which takes
the same value $m(m-1)^{n-1}$ on all trees with $n$ vertices.  
\begin{question}
Does the chromatic symmetric function distinguish trees?
\end{question}
\noindent
It has been checked that the answer is affirmative for trees on $23$ vertices or less.
There are also interesting partial results on this question
by Martin, Morin and Wagner \cite{MartinMorinWagner}.

\vskip.1in
We close this section with a few other properties of $\Psi[G]$ proven by Stanley
which follow easily from the theory we have developed.  
For example, his work makes
no explicit mention of the chromatic Hopf algebra $\GGG$, and the fact that $\Psi$ is
a Hopf morphism (although he certainly notes the trivial algebra morphism property 
$\Psi[G_1 \sqcup G_2])=\Psi[G_1] \Psi[G_2]$).  One property he proves is implicitly related to
$\Psi$ as a coalgebra morphism:  he considers the effect on $\Psi$ of the operator
$
\frac{\partial}{\partial p_1}:
\Lambda \longrightarrow \Lambda
$ 
which acts by first expressing a symmetric function
$f=f(p_1,p_2,\ldots)$ as a polynomial in the powers sums $\{ p_n \}$, and then applies
$\frac{\partial}{\partial p_1}$.  It is not hard to see that  $\frac{\partial}{\partial p_1}$
is the same as the skewing operator $s_1^\perp=p_1^\perp$:  both act as  derivations
on $\Lambda=\QQ[p_1,p_2,\ldots]$, and agree in their effect on each $p_n$, in that  
both send $p_1 \mapsto  1$, and both annihilate $p_2,p_3,\ldots$.


\begin{proposition}(Stanley \cite[Cor. 2.12(a)]{Stanley-chromatic-qsym-fn})
For any graph $G=(V,E)$, one has 
$$
\frac{\partial}{\partial p_1} \Psi[G] = \sum_{v \in V} \Psi[G|_{V\setminus v}].
$$
\end{proposition}
\begin{proof}
One first computes
$$
\Delta \Psi[G] =
 (\Psi \otimes \Psi) \Delta[G]
=\sum_{\substack{(V_1,V_2):\\V=V_1 \sqcup V_2}} \Psi[G|_{V_1}] \otimes \Psi[G|_{V_2}].
$$
Since degree considerations force $(s_1, \Psi[G|_{V_1}]) = 0$ unless $|V_1|=1$,
in which case $\Psi[G|_{V_1}]=s_1$, 
one has
$$
\frac{\partial}{\partial p_1} \Psi[G] = s_1^\perp \Psi[G] =
\sum_{\substack{(V_1,V_2):\\V=V_1 \sqcup V_2}} (s_1,\Psi[G|_{V_1}])  \cdot \Psi[G|_{V_2}]
=\sum_{v \in V} \Psi[G|_{V\setminus v}].
$$
\end{proof}


\begin{definition}
Given a graph $G=(V,E)$, an acyclic orientation $\Omega$ of the edges $E$,
and a vertex-coloring $f: V \rightarrow \{1,2,\ldots\}$,
say that the pair $(\Omega,f)$ are {\it weakly compatible} if
if whenever $\Omega$ orients an edge $\{v,v'\}$ in $E$ as $v \rightarrow v'$,
one has $f(v) \leq f(v')$.   Note that a {\it proper} vertex-coloring $f$ of a graph $G=(V,E)$
is compatible with a unique acyclic orientation $\Omega$.
\end{definition}

\begin{proposition}(Stanley \cite[Prop. 4.1, Thm. 4.2]{Stanley-chromatic-qsym-fn})
\label{Stanley-chromatic-reciprocity}
The involution $\omega$ of $\Lambda$ sends $\Psi[G]$ to
$
\omega \left( \Psi[G] \right) = \sum_{(\Omega,f)} \xx_f
$
in which the sum runs over weakly compatible 
pairs $(\Omega,f)$ of an acyclic orientation $\Omega$ and
vertex-coloring $f$.

Furthermore, the chromatic polynomial $\chi_G(m)$ has the property that
$(-1)^{|V|} \chi(G,-m)$ counts all such weakly compatible pairs $(\Omega,f)$
in which $f: V \rightarrow \{1,2,\ldots,m\}$ is a vertex-$m$-coloring.
\end{proposition}
\begin{proof}
As observed above, a proper coloring $f$ is compatible with a unique
acyclic orientation $\Omega$ of $G$.  Denote by $P_\Omega$ the
poset on $V$ which is the transitive closure of $\Omega$, endowed
with a {\it strict labelling} by integers, that is, $i<_P j$ implies $i >_\ZZ j$.
Then proper colorings $f$ that induce $\Omega$ are the same as $P_{\Omega}$-partitions,
so that
\begin{equation}
\label{chromatic-symm-fn-as-sum-of-posets}
\Psi[G] = \sum_{\Omega} F_{P_{\Omega}}(\xx).
\end{equation}
Applying the antipode $S$ and
using Corollary~\ref{antipode-in-Qsym-as-reciprocity-corollary} gives
$$
\omega \left( \Psi[G] \right)
=(-1)^{|V|} S \left( \Psi[G] \right) 
= \sum_{\Omega}  F_{P_{\Omega}^{\opp}}(\xx)
= \sum_{(\Omega,f)} \xx_{f}
$$
where in the last line one sums over weakly compatible pairs as in the proposition.  The last equality comes from the fact that since each $P_\Omega$ has been given a strict labelling, $P_\Omega^{\opp}$ acquires a {\it weak (or natural) labelling},
that is $i <_{P_\Omega^{\opp}} j$ implies $i <_\ZZ j$.

The last assertion follows from Proposition~\ref{zeta-polynomiality-proposition}(iii).
\end{proof}

\begin{remark}
The interpretation
of $\chi(G,-m)$ in Proposition~\ref{Stanley-chromatic-reciprocity}
is a much older result of Stanley \cite{Stanley-negative-colors}.
The special case interpreting $\chi(G,-1)$ as $(-1)^{|V|}$ times
the number of acyclic orientations of $G$ 
has sometimes been called Stanley's {\it (-1)-color theorem}.
It also follows (via Proposition~\ref{zeta-polynomiality-proposition})
from Humpert and Martin's antipode formula for $\GGG$ discussed in
Remark~\ref{Humpert-Martin-antipode-remark}: taking $\zeta$ to be
the character of $\GGG$ given in \eqref{edge-free-character-defn},
$$
\chi(G,-1)=\zeta^{*(-1)}[G] = \zeta(S[G])
=\sum_{F} (-1)^{|V|-\rank(F)} \acyc(G/F) S[G_{V,F}]
=(-1)^{|V|}\acyc(G)
$$
where the last equality uses the vanishing of $\zeta$ 
on graphs that have edges, so only the $F=\varnothing$ term survives.
\end{remark}



\subsection{Example: The quasisymmetric function of a matroid} 

We introduce the {\it matroid-minor Hopf algebra} of Schmitt \cite{Schmitt}, 
and studied extensively by 
Crapo and Schmitt \cite{CrapoSchmitt1, CrapoSchmitt2, CrapoSchmitt3}.   A very simple character
$\zeta$ on this Hopf algebra will then give rise, via the map $\Psi$ from
Theorem~\ref{Qsym-as-terminal-object-theorem}, to the quasisymmetric function
invariant of matroids from the work of Billera, Jia and the author \cite{BilleraJiaR}.

\subsubsection{The matroid-minor Hopf algebra}

We begin by reviewing some notions from matroid theory; see Oxley \cite{Oxley} for background,
undefined terms and unproven facts.

\begin{definition}
A {\it matroid} $M$ of rank $r$ on a (finite) ground set $E$
is specified by a nonempty collection $\BBB(M)$ of $r$-element subsets of $E$ with the following
{\it exchange property}: 
\begin{quote} 
For any $B, B'$ in $\BBB(M)$ and $b$ in $B$,
there exists $b'$ in $B'$ with $(B \setminus \{b\}) \cup \{b'\}$ in
$\BBB(M)$.
\end{quote}
\end{definition}

\begin{example}
\label{vector-matroid-example}
A matroid $M$ is {\it represented} by a collection of vectors $E=\{e_1,\ldots,e_n\}$
in a vector space if $\BBB(M)$ is the collection of subsets $B=\{e_{i_1},\ldots,e_{i_r}\}$
having the property that $B$ forms a basis for the span of all of the vectors in $E$.
For example, if $E=\{a,b,c,d\}$ are the four vectors $a=(1,0),b=(1,1),c=(0,1)=d$ in $\RR^2$
depicted here
$$
\xymatrix{
c,d & b \\
  \ar[r] \ar@{->>}[u] \ar[ur]     & a
}
$$
then $\BBB(M)=\{\{a,b\},\{a,c\},\{a,d\},\{b,c\},\{b,d\}\}$.
\end{example}

\begin{example}
\label{graphic-matroid-example}
A special case of matroids $M$ represented by vectors are {\it graphic matroids}, coming
from a graph $G=(V,E)$, with parallel edges and self-loops allowed.  One represents
these by vectors in $\RR^V$ with standard basis $\{ \epsilon_v\}_{v \in V}$ by
associating to the edge $e=\{v,v'\}$ the vector $\epsilon_v -\epsilon_{v'}$.  
One can check (or see \cite[\S 1.2]{Oxley}) that the bases $B$ in $\BBB(M)$ correspond to the edge
sets of {\it spanning forests} for $G$, that is, edge sets which are acyclic and contain
one spanning tree for each connected component of $G$.
For example, the graph $G=(V,E)$ shown below has the same matroid $\BBB(M)$
as the one represented by the vectors in Example~\ref{vector-matroid-example}:
$$
\xymatrix{
&\bullet \ar@{-}[dl]_{a}\ar@{-}[dr]^{b}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{c}  \ar@(dr,dl)@{-}[rr]_{d}& &\bullet
}
$$
whose spanning trees are the edge sets $\BBB(M)=\{ \{a,b\},\{a,c\},\{a,d\},\{b,c\},\{b,d\}\}$.
\end{example}


To define the matroid-minor Hopf algebra one needs the basic matroid
operations of {\it deletion} and {\it contraction}.
These  model the operations of deleting or contracting an edge in a graph.
For configuations of vectors they model the deletion of a vector, or the passage to 
images in the quotient space modulo the span of a vector.

\begin{definition}
Given a matroid $M$ of rank $r$ and an element $e$ of its ground set $E$,  say that
$e$ is {\it loop} (resp. {\it coloop}) of $M$ if $e$ lies in no basis (resp. every basis) $B$ in $\BBB(M)$.
If $e$ is not a coloop, the {\it deletion} $M \setminus e$ is a matroid of rank $r$ 
on ground set $E \setminus \{e\}$ having bases 
\begin{equation}
\label{deletion-bases}
\BBB(M \setminus e):=\{ B \in \BBB(M): e \not\in B\}.
\end{equation}
If $e$ is not a loop, the {\it contraction} $M / e$ is a matroid of rank $r$
on ground set $E \setminus \{e\}$ having bases 
\begin{equation}
\label{contraction-bases}
\BBB(M/ e):=\{ B \setminus \{e\}: e \in B \in \BBB(M)\}.
\end{equation}
When $e$ is a loop of $M$, then $M/e$ has rank $r$ instead of $r-1$ and one defines its
bases as in \eqref{deletion-bases} rather than \eqref{contraction-bases}; similarly, if $e$ is a coloop
of $M$ then $M \setminus e$ has rank $r-1$ instead of $r$ and one defines its
bases as in \eqref{contraction-bases} rather than \eqref{deletion-bases}.
\end{definition}

\begin{example}
Starting with the graph $G$ and its graphic matroid $M$ from Example ~\ref{graphic-matroid-example}, the deletion $M\setminus a$ and contraction $M/c$
correspond to the graphs $G \setminus a$ and $G/c$ shown here:
$$
G \setminus a=
\xymatrix{
&\bullet \ar@{-}[dr]^{b}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{c}  \ar@(dr,dl)@{-}[rr]_{d}& &\bullet
}
\qquad
G/c=
\xymatrix{
\bullet \ar@(dl,ul)@{-}[d]^{a}\ar@(dr,ur)@{-}[d]^{b}\\
\bullet\ar@(dr,dl) @{-}_{d}
}
$$
One has
\begin{enumerate}
\item[$\bullet$]
$\BBB(M \setminus a)=\{\{b,c\},\{b,d\}\}$, 
so that $b$ has become a coloop in $M \setminus a$, and
\item[$\bullet$]
$\BBB(M/c)=\{\{a\},\{b\}\}$,
so that $d$ has become a loop in $M/c$.  
\end{enumerate}
\end{example}

\begin{definition}
Deletions and contractions commute with each other, leading to well-defined operations for subsets $A \subset E$ of the
\begin{enumerate}
\item[$\bullet$] 
{\it restriction} $M|_A$ on ground set $A$, obtained by deleting all
$e$ of $E \setminus A$ in any order, and
\item[$\bullet$]
{\it quotient/contraction} $M/A$ on ground set $E \setminus A$, 
obtained by contracting all $e$ in $A$ in any order.
\end{enumerate}

We will also need the {\it direct sum} $M_1 \oplus M_2$,
whose ground set $E=E_1 \sqcup E_2$ is the disjoint union of a copy of the
ground sets $E_1,E_2$ for $M_1,M_2$, and having bases
$$
\BBB(M_1\oplus M_2):=\{B_1 \sqcup B_2: B_i \in \BBB(M_i) \text{ for }i=1,2\}.
$$ 

Lastly, say that two matroids $M_1, M_2$ are {\it isomorphic} if there is a bijection
of their ground sets $E_1 \overset{\varphi}{\longrightarrow} E_2$ having the property
that $\varphi \BBB(M_1) = \BBB(M_2)$.
\end{definition}

Now one can define the matroid-minor Hopf algebra, originally introduced by Schmitt \cite[\S 15]{Schmitt}, and studied further by Crapo and Schmitt \cite{CrapoSchmitt1, CrapoSchmitt2, CrapoSchmitt3}.
\begin{definition}
Let $\MMM$ have $\kk$-basis  elements $[M]$ indexed by isomorphism classes of
matroids. Define the multiplication via 
$$
[M_1] \cdot [M_2]:=[M_1 \oplus M_2]
$$
so that the class $[\varnothing]$ of the {\it empty matroid} $\varnothing$ having
empty ground set gives a unit.  Define the comultiplication for $M$ a matroid on ground set $E$ via
$$
\Delta[M]:=\sum_{A \subset E} [M|_A] \otimes [M/A],
$$
and a counit
$$
\epsilon [M]:= 
\begin{cases} 
1 & \text{ if }M=\varnothing \\
0 & \text{ otherwise.}
\end{cases}  
$$
\end{definition}


\begin{proposition}
The above maps endow $\MMM$ with the structure of a graded connected
finite type Hopf algebra over $\kk$, which is commutative.
\end{proposition}
\begin{proof}
Checking the unit and counit conditions are straightforward. 
Associativity and commutativity of the multiplication follow because the direct sum operation
$\oplus$ for matroids is associative, and commutative up to isomorphism.  Coassociativity
follows because for a matroid $M$ on ground set $E$,
one has this equality between the two candidates for $\Delta^{(2)}[M]$ 
$$
\begin{aligned}
&\sum_{\varnothing \subseteq A_1 \subseteq A_2 \subseteq E}
[M|_{A_1}] \otimes [(M|_{A_2})/A_1] \otimes [M/A_2] \\
&=
\sum_{\varnothing \subseteq A_1 \subseteq A_2 \subseteq E}[M|_{A_1}] \otimes [(M/A_1)|_{A_2 \setminus A_1}] \otimes [M/A_2]
\end{aligned}
$$
due to the matroid isomorphism $(M|_{A_2})/A_1 \cong (M/A_1)|_{A_2 \setminus A_1}$.
Commutativity of the bialgebra diagram in \eqref{bialgebra-diagrams} amounts
to the fact that for a pair of matroids $M_1,M_2$ and subsets $A_1,A_2$ of their
(disjoint) ground sets $E_1,E_2$, one has isomorphisms
$$
\begin{aligned}
M_1|_{A_1} \oplus M_2|_{A_2} 
  & \cong \left( M_1 \oplus M_2\right)|_{A_1 \sqcup A_2}, \\
M_1/A_1 \oplus M_2/A_2 
  & \cong \left( M_1 \oplus M_2\right)/(A_1 \sqcup A_2). \\
\end{aligned}
$$
Letting $\MMM_n$ be the $\kk$-span of $[M]$ for matroids whose ground set $E$
has cardinality $|E|=n$, one can then easily check that $\MMM$
becomes a bialgebra which is 
graded, connected, and of finite type, hence also a
Hopf algebra by Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
\end{proof}











\subsubsection{A quasisymmetric function for matroids}

\begin{definition}
Define  a character $\MMM \overset{\zeta}{\longrightarrow} \kk$ by
$$
\zeta[M]=\begin{cases}
1 & \text{ if }M\text{ has only one basis},\\
0 & \text{ otherwise.}
\end{cases}
$$
It is easily checked that this is a character, that is, an algebra map 
$\MMM \overset{\zeta}{\longrightarrow} \kk$.
Note that if $M$ has only one basis, say $\BBB(M)=\{B\}$, then
$B:=\coloops(M)$ is the set of coloops of $M$, and
$E\setminus B=\loops(M)$ is the set of loops of $M$.  Equivalently,
$M = \bigoplus_{e \in E} M|_{\{e\}}$ is the direct sum of matroids each 
having one element, each a coloop or loop.

Define $\Psi[M]$ for a matroid $M$ to be
the image of $[M]$ under the map 
$\MMM\overset{\Psi}{\longrightarrow} \Qsym$
induced via Theorem~\ref{Qsym-as-terminal-object-theorem}
from the above character $\zeta$.
\end{definition}

It turns out that $\Psi[M]$ is intimately related with greedy algorithms and
finding minimum cost bases. 
A fundamental property of matroids (and one that characterizes them, in fact; see \cite[\S 1.8]{Oxley})
is that no matter how one assigns costs $f: E \rightarrow \RR$ to the elements of $E$,
the following {\it greedy algorithm} (generalizing {\it Kruskal's algorithm} for finding
minimum cost spanning trees) always succeeds
in finding one basis $B$ in $\BBB(M)$ achieving the minimum {\it total cost} $f(B):=\sum_{b \in B} f(b)$:
\begin{quote}
\label{greedy-algorithm}
Start with the empty subset $I_0=\varnothing$ of $E$.
For $j=1,2,\ldots,r$, having already defined the set $I_{j-1}$, let
$e$ be the element of $E \setminus I_{j-1}$ having the lowest cost $f(e)$ among all those
for which $I_{j-1} \cup \{e\}$ is {\it independent}, that is, still a subset of at least one
basis $B$ in $\BBB(M)$.  Then define $I_j:=I_{j-1} \cup \{e\}$.
Repeat this until $j=r$, and $B=I_r$ will be among the bases that achieve the minimum cost.
\end{quote}

\begin{definition}
Say that a cost function $f: E \rightarrow \{1,2,\ldots\}$ is
{\it $M$-generic} if there is a {\it unique} basis $B$ in $\BBB(M)$
achieving the minimum cost $f(B)$.
\end{definition}

\begin{example}
For the graphic matroid $M$ of Example~\ref{graphic-matroid-example},
this cost function $f_1: E \rightarrow \{1,2,\ldots\}$ 
$$
\xymatrix{
&\bullet \ar@{-}[dl]_{f_1(a)=1}\ar@{-}[dr]^{f_1(b)=3}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{f_1(c)={3}}  \ar@(dr,dl)@{-}[rr]_{f_1(d)=2}& &\bullet
}
$$
is $M$-generic, as it minimizes uniquely on the basis $\{a,d\}$,
whereas this cost function $f_2: E \rightarrow \{1,2,\ldots\}$
$$
\xymatrix{
&\bullet \ar@{-}[dl]_{f_2(a)=1}\ar@{-}[dr]^{f_2(b)=3}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{f_2(c)=2}  \ar@(dr,dl)@{-}[rr]_{f_2(d)=2}& &\bullet
}
$$
is {\it not} $M$-generic, as it achieves its minimum value on the two
bases $\{a,c\}, \{a,d\}$.
\end{example}

\begin{proposition}
\label{matroid-qsym-expansion}
For a matroid $M$ on ground set $E$,
one has this expansion\footnote{In fact, 
this expansion was the original definition of $\Psi[M]$  in 
\cite[Defn. 1.1]{BilleraJiaR}.} 
$$
\Psi[M]=\sum_{\substack{M \text{-generic}\\f:E \rightarrow \{1,2,\ldots\}}}
         \xx_f
$$
where $\xx_f:=\prod_{e \in E} x_{f(e)}$.
In particular, for $m \geq 0$, its specialization $ps^1$ 
from Definition~\ref{principal-specialization-at-1-defn}
has this interpretation:
$$
\ps^1 \Psi[M](m)
=|\{M\text{-generic }f:E \rightarrow \{1,2,\ldots,m\}|.
$$
\end{proposition}
\begin{proof}
The iterated coproduct 
$\MMM \overset{\Delta^{(\ell-1)}}{\longrightarrow} \MMM^{\otimes \ell}$
sends
$$
[M] \longmapsto 
\sum
[M|_{A_1}] \otimes [(M|_{A_2})/A_1] \otimes \cdots \otimes [(M|_{A_{\ell}})/A_{\ell-1}]
$$
where the sum is over flags of nested subsets
\begin{equation}
\label{flag-of-nested-subsets}
\varnothing = A_0 \subseteq A_1 \subseteq \cdots 
\subseteq A_{\ell-1} \subseteq A_{\ell}=E.
\end{equation}
The map $\zeta^{\otimes \ell}$ sends each summand to
$1$ or $0$, depending upon whether each $(M|_{A_j})/A_{j-1}$ has
a unique basis or not.  Thus formula \eqref{ABS-terminal-morphism-formula}
shows that the coefficient $\zeta_\alpha$ of 
$x_{i_1}^{\alpha_1} \cdots x_{i_\ell}^{\alpha_\ell}$ in $\Psi[M]$ 
counts the flags of subsets in \eqref{flag-of-nested-subsets} for
which $|A_j \setminus A_{j-1}|=\alpha_j$ and $(M|_{A_j})/A_{j-1}$
has a unique basis, for each $j$.

Given a flag as in  \eqref{flag-of-nested-subsets}, 
associate the cost function $f: E \rightarrow \{1,2,\ldots\}$
whose value on each element of $A_j \setminus A_{j-1}$ is $i_j$;  conversely, given
any cost function, say whose distinct values are $i_1 < \ldots < i_\ell$,
one associates the flag having $A_j \setminus A_{j-1}=f^{-1}(i_j)$ for each $j$.

We will prove below, using induction on $s=0,1,2\ldots,\ell$ 
the following {\bf claim}:
After having completed $\alpha_1+\alpha_2+\cdots+\alpha_s$ steps in the 
greedy algorithm \eqref{greedy-algorithm}, there is {\it a unique choice}
for the independent set produced thus far,
namely 
\begin{equation}
\label{t-step-unique-greedy-basis}
I_{\alpha_1+\alpha_2+\cdots+\alpha_s}=\bigsqcup_{j=1}^{s} \coloops( (M|_{A_j})/A_{j-1} ),
\end{equation}
if and only if each of the matroids $(M|_{A_j})/A_{j-1}$ for $j=1,2,\ldots,s$
{\it has a unique basis}.

The case $s=\ell$ in this claim would show what we want, namely that 
$f$ is $M$-generic, minimizing uniquely on the basis 
shown in \eqref{t-step-unique-greedy-basis} with $s=\ell$,
if and only if each $(M|_{A_j})/A_{j-1}$ has a unique basis.

The assertion of the claim is trivially true for $s=0$.  
In the inductive step, one may assume that 
\begin{enumerate}
\item[$\bullet$]
the independent set $I_{\alpha_1+\alpha_2+\cdots+\alpha_{s-1}}$ takes the form in
\eqref{t-step-unique-greedy-basis}, replacing $s$ by $s-1$,
\item[$\bullet$]
it is the unique $f$-minimizing basis 
for $M|_{A_{s-1}}$, and 
\item[$\bullet$]
$(M|_{A_j})/A_{j-1}$ has a unique basis for $j=1,2,\ldots,s-1$.
\end{enumerate}
Since $A_{s-1}$ exactly consists of all of the elements $e$ of $E$ whose
costs $f(e)$ lie in the range
$\{i_1,i_2,\ldots,i_{s-1}\}$, in the next $\alpha_s$ steps of the algorithm 
will work in the quotient matroid $M/A_{j-1}$ and
attempt to augment $I_{\alpha_1+\alpha_2+\cdots+\alpha_{s-1}}$ 
using the next-cheapest elements, namely the 
elements of $A_{s} \setminus A_{s-1}$, which all have cost $f$ equal to $i_s$.  
Thus the algorithm will have no choices about how to do this augmentation
if and only if $(M|_{A_s})/M_{A_{s-1}}$ has a unique basis, namely its set of
coloops, in which case the algorithm will choose to add all of these coloops,
giving $I_{\alpha_1+\alpha_2+\cdots+\alpha_s}$ as described in \eqref{t-step-unique-greedy-basis}.
This completes the induction.

The last assertion follows from Proposition~\ref{zeta-polynomiality-proposition}.
\end{proof}

\begin{example}
If $M$ has one basis then every function $f:E \rightarrow \{1,2,\ldots\}$ is $M$-generic, and
$$
\Psi[M]=\sum_{f: E \rightarrow \{1,2,\ldots\}} \xx_f = (x_1+x_2+ \cdots)^{|E|} =
M_{(1)}^{|E|}.
$$
\end{example}
\begin{example}
Let $U_{r,n}$ denote the {\it uniform matroid} of rank $r$ on $n$ elements $E$, having
$\BBB(U_{r,n})$ equal to all of the $r$-element subsets of $E$. 

As $U_{1,2}$ has $E=\{1,2\}$ and $\BBB=\{\{1\},\{2\}\}$,
genericity means $f(1) \neq f(2)$, so 
$$
\Psi[U_{1,2}]=\sum_{\substack{(f(1),f(2)):\\f(1) \neq f(2)}} x_{f(1)} x_{f(2)}
=x_1 x_2 + x_2 x_1 + x_1 x_3+x_3 x_1 + \cdots = 2M_{(1,1)}.
$$

Similarly $U_{1,3}$ has $E=\{1,2,3\}$ with $\BBB=\{\{1\},\{2\},\{3\}\}$,
and genericity means either that $f(1),f(2),f(3)$ 
are all distinct, or that 
two of them are the same and the third is smaller.
This shows
$$
\begin{aligned}
\Psi[U_{1,3}]&=3\sum_{i<j} x_i x_j^2+6\sum_{i<j<k} x_i x_j x_k\\
&=3M_{(1,2)}+6M_{(1,1,1)}\\
\ps^1 \Psi[U_{1,3}](m) &= 3 \binom{m}{2} + 6 \binom{m}{3}=\frac{m(m-1)(2m-1)}{2}
\end{aligned}
$$
One can similarly analyze $U_{2,3}$ and check that
$$
\begin{aligned}
\Psi[U_{2,3}]&=3M_{(2,1)}+6M_{(1,1,1)} \\
\ps^1 \Psi[U_{2,3}](m) &= 3 \binom{m}{2} + 6 \binom{m}{3}=\frac{m(m-1)(2m-1)}{2}
\end{aligned}
$$
\end{example}

These last examples illustrate the behavior of $\Psi$ under
the duality operation on matroids.

\begin{definition}
Given a matroid $M$ of rank $r$ on ground set $E$, its
{\it dual} or {\it orthogonal matroid} $M^\perp$ is a matroid of
rank $|E|-r$ on the same ground set $E$, having
$$
\BBB(M^\perp):=\{ E \setminus B \}_{B \in \BBB(M)}.
$$
\end{definition}

Here are a few examples of dual matroids.

\begin{example}
The dual of a uniform matroid is another uniform matroid:
$$
U_{r,n}^\perp = U_{n-r,n}.
$$
\end{example}

\begin{example}
If $M$ is matroid of rank $r$ represented by collection of vectors $E=\{e_1,\ldots,e_n\}$ in a vector space over some field $\kk$, 
one can find a collection of vectors $\{e_1^\perp,\ldots,e_n^\perp\}$ that represent $M^\perp$ in the following way.  Pick a basis for the span of the vectors $\{e_i\}_{i=1}^n$, and create a matrix $A$ in $\kk^{r \times n}$ whose columns express the $e_i$ in terms of this basis.  Then pick any matrix $A^\perp$ whose row space is the null space of $A$, and one finds that the columns $\{e_i^\perp\}_{i=1}^n$ of $A^\perp$ represent $M^\perp$.  See Oxley \cite[\S 2.2]{Oxley}.
\end{example}

\begin{example}
\label{graphical-duality}
Let $G=(V,E)$ be graph embedded in the plane with edge set $E$, giving rise to a graphic matroid $M$ on ground set $E$.  Let $G^\perp$ be a planar dual of $G$, so that, in particular, for each edge $e$ in $E$, the graph
$G^\perp$ has one edge $e^\perp$, crossing $e$ transversely.  
Then the graphic matroid of $G^\perp$ is $M^\perp$.  See Oxley \cite[\S 2.3]{Oxley}.
\end{example}

\begin{proposition}
If $\Psi[M]=\sum_\alpha c_\alpha M_\alpha$ then
$\Psi[M^\perp] = \sum_\alpha c_\alpha M_{\rev(\alpha)}$.

Consequently, $\ps^1 \Psi[M](m) = \ps^1 \Psi[M^\perp](m)$.
\end{proposition}
\begin{proof}
This amounts to showing that
for any composition $\alpha=(\alpha_1,\ldots,\alpha_\ell)$,
the cardinality of the set of $M$-generic $f$ having $\xx_f=x^{\alpha}$
is the same as the cardinality of the set of
$M^\perp$-generic $f^\perp$ having $\xx_{f^\perp}=\xx^{\rev(\alpha)}$.
We claim that the map $f \longmapsto f^\perp$ in which
$f^\perp(e)=\ell+1-f(e)$ gives a bijection between these sets.
To see this, note that
\begin{align}
f(B) + f(E \setminus B) &= \sum_{e\in E}f(e) \label{first-dual-base-relation}\\
f(E \setminus B) + f^\perp(E \setminus B) & = (\ell+1)(|E|-r)
\label{second-dual-base-relation}
\end{align}
Thus $B$ is $f$-minimizing if and only if $E\setminus B$ is $f$-maximizing
(by \eqref{first-dual-base-relation}) if and only $E \setminus B$ is $f^\perp$-minimizing (by \eqref{second-dual-base-relation}).  Consequently $f$ is $M$-generic if and only if $f^\perp$ is $M^\perp$-generic.

The last assertion follows, for example, from the calculation in
Proposition~\ref{zeta-polynomiality-proposition}(i) that
$\ps^1 M_\alpha = \binom{m}{\ell(\alpha)}$ together with the fact that
$\ell(\rev(\alpha))= \ell(\alpha)$.
\end{proof}

Just as \eqref{chromatic-symm-fn-as-sum-of-posets} showed that 
Stanley's chromatic symmetric function of a graph has an expansion as a sum of $P$-partition enumerators for certain strictly labelled posets $P$, the same
holds for $\Psi[M]$.

\begin{definition}
Given a matroid $M$ on ground set $E$, and a basis $B$ in $\BBB(M)$,
define the {\it base-cobase} poset $P_B$ to have $b < b'$ whenever
$b$ lies $B$ and $b'$ lies in $E \setminus B$ and 
$\left(B  \setminus \{b\}\right) \cup \{b'\}$ is in $\BBB(M)$.
\end{definition}

\begin{proposition}
\label{matroid-qsym-fn-as-sum-of-posets-prop}
For any matroid $M$, one has
$\Psi[M]=\sum_{B \in \BBB(M)} F_{(P_B,\strict)}(\xx)$
where $F_{(P_,\strict)}(\xx)$ for a poset $P$ means the $P$-partition
enumerator for any strict labelling of $P$, i.e.
a labelling such that 
the $P$-partitions satisfy $f(i) < f(j)$ whenever $i<_P j$.

In particular, $\Psi[M]$ expands nonnegatively in the
$\{L_\alpha\}$ basis.
\end{proposition}
\begin{proof}
A basic result about matroids, due to Edmonds \cite{Edmonds}, describes 
the {\it edges} in the {\it matroid base polytope} which is
the convex hull of all vectors $\{ \sum_{b \in B} \epsilon_b\}_{B \in \BBB(M)}$
inside $\RR^E$ with standard basis $\{\epsilon_e\}_{e \in E}$.  He shows
that all such edges connect two bases $B, B'$ that differ by a single
{\it basis exchange}, that is, $B'=
\left(B  \setminus \{b\}\right) \cup \{b'\}$ for some $b$ in $B$
and $b'$ in $E \setminus B$.  

Polyhedral theory then says that
a cost function $f$ on $E$ will minimize uniquely at $B$ if and only if
one has a strict increase 
$f(B)<f(B')$ along each such edge $B \rightarrow B'$ emanating from $B$,
that is, if and only if $f(b) < f(b')$ whenever $b<_{P_B} b'$ in the
base-cobase poset $P_B$, that is, $f$ lies in $\AAA(P_B,\strict)$.
\end{proof}

\begin{example}
\label{graphic-matroid-example-continued}
The graphic matroid from Example~\ref{graphic-matroid-example} has
this matroid base polytope, with the bases $B$ in $\BBB(M)$ labelling the
vertices:
$$
\xymatrix{
 & &cd\ar@{-}[dll]\ar@{-}[ddl]\ar@{-}[drr]\ar@{-}[ddrrr]&&& & \\
ac\ar@{--}[rrrr]\ar@{-}[dr]& & &&ad\ar@{-}[dr]& \\
 &bc\ar@{-}[rrrr]& & &&bd
}
$$
The base-cobase posets $P_B$ for its five vertices $B$ are as follows:
$$
\xymatrix@C=6pt@R=6pt{
a\ar@{-}[d]\ar@{-}[dr] & b\ar@{-}[d] \ar@{-}[dl] \\
c & d
}
$$
$$
\xymatrix@C=6pt@R=6pt{
b\ar@{-}[d]& d\ar@{-}[d] \ar@{-}[dl] \\
a & c
}\qquad
\xymatrix@C=6pt@R=6pt{
a\ar@{-}[d]& d\ar@{-}[d] \ar@{-}[dl] \\
b & c
}\qquad
\xymatrix@C=6pt@R=6pt{
a\ar@{-}[d]& c\ar@{-}[d] \ar@{-}[dl] \\
b & d
}\qquad
\xymatrix@C=6pt@R=6pt{
b\ar@{-}[d]& c\ar@{-}[d] \ar@{-}[dl] \\
a & d
}
$$
One can label the first of these five strictly as
$$
\xymatrix@C=6pt@R=6pt{
1\ar@{-}[d]\ar@{-}[dr] & 2\ar@{-}[d] \ar@{-}[dl] \\
3 & 4
}
$$
and compute its strict $P$-partition enumerator from the linear
extensions $\{3412,3421,4312,4321\}$ as
$$
L_{(2,2)}+L_{(2,1,1)}+L_{(1,1,2)}+L_{(1,1,1,1)}
$$
while any of the last four can be labelled strictly as
$$
\xymatrix@C=6pt@R=6pt{
1\ar@{-}[d]& 2\ar@{-}[d] \ar@{-}[dl] \\
3 & 4
}
$$
and they each have an extra linear extension $3142$ giving their
strict $P$-partition enumerators as
$$
L_{(2,2)}+L_{(2,1,1)}+L_{(1,1,2)}+L_{(1,1,1,1)}+L_{(1,2,1)}.
$$
Hence one has
$$
\Psi[M]=5L_{(2,2)}+5L_{(1,1,2)}+4L_{(1,2,1)}+5L_{(2,1,1)}+5L_{(1,1,1,1)}.
$$
As $M$ is a graphic matroid $M$ for a self-dual planar graph, one has a matroid isomorphism $M \cong M^\perp$ (see Example~\ref{graphical-duality}), reflected in the fact that $\Psi[M]$ is invariant under the symmetry
swapping $M_\alpha \leftrightarrow M_{\rev(\alpha)}$ (and simultaneously
swapping  $L_\alpha \leftrightarrow L_{\rev(\alpha)}$).
\end{example}


This $P$-partition expansion for $\Psi[M]$ also allows us to identify
its image under the antipode of $\Qsym$.

\begin{proposition}
\label{matroid-qsym-reciprocity}
For a matroid $M$ on ground set $E$, one has
$$
S(\Psi[M]) = (-1)^{|E|} \sum_{f: E \rightarrow \{1,2,\ldots\}}
                  |\{f\text{-maximizing bases }B\}| \cdot \xx_f
$$
and
$$
\ps^1\Psi[M](-m) = (-1)^{|E|} \sum_{f: E \rightarrow \{1,2,\ldots,m\}}
                  |\{f\text{-maximizing bases }B\}|. 
$$
In particular, the expected number of $f$-maximizing bases among
all cost functions that take at most $m$ values is $(-m)^{-|E|} \ps^1\Psi[M](-m)$.
\end{proposition}
\begin{proof}
Corollary~\ref{antipode-in-Qsym-as-reciprocity-corollary} implies
$$
S(\Psi[M]) 
=\sum_{B \in \BBB(M)} S(F_{(P_B,\strict)}(\xx)) \\
=(-1)^{|E|} \sum_{B \in \BBB(M)} F_{(P_B^{\opp},\weak)}(\xx)
$$
where $F_{(P,\weak)}(\xx)$ is the enumerator
for $P$-partitions in which $P$ has been {\it naturally} labelled,
so that they satisfy $f(i) \leq f(j)$ whenever $i <_P j$.
When $P=P_B^{\opp}$, this is exactly the condition for $f$ to achieve
its maximum value at $f(B)$ (possibly not uniquely), that is,
for $f$ to lie in the {\it closed} normal cone to the vertex indexed
by $B$ in the matroid base polytope;  compare this with the 
discussion in the proof of
Proposition \ref{matroid-qsym-fn-as-sum-of-posets-prop}.  Thus one has
$$
S(\Psi[M])=(-1)^{|E|} \sum_{\substack{(B,f):\\B \in \BBB(M)\\f\text{ maximizing at }B}}
\xx_f
$$
which agrees with the statement of the proposition, after reversing
the order of the summation.  

The rest follows from Proposition~\ref{zeta-polynomiality-proposition}.
\end{proof}

\begin{example}
We saw in Example~\ref{graphic-matroid-example-continued}
that the matroid $M$ from Example~\ref{graphic-matroid-example} has
$$
\Psi[M]=5L_{(2,2)}+5L_{(1,1,2)}+4L_{(1,2,1)}+5L_{(2,1,1)}+5L_{(1,1,1,1)}.
$$
and therefore will have
$$
%\begin{aligned}
\ps^1 \Psi[M](m)= 5 \binom{m-2+4}{4} + (5+4+5) \binom{m-3+4}{4} + 5 \binom{m-4+4}{4} 
            = \frac{m(m-1)(2m^2-2m+1)}{2}.
%\end{aligned}
$$
using $\ps^1(L_\alpha)(m) =\binom{m-\ell+|\alpha|}{|\alpha|}$ from
Proposition~\ref{zeta-polynomiality-proposition} (i).
Let us first do a reality-check on a few of its values with
$m \geq 0$ using Proposition~\ref{matroid-qsym-expansion},
and for negative $m$ using Proposition~\ref{matroid-qsym-reciprocity}:
\vskip.1in
\begin{tabular}{|c||c|c|c|c|}\hline
$m$ & $-1$ & $0$ & $1$ & $2$ \\\hline
$\ps^1 \Psi[M](m)$ & $5$ & $0$ & $0$ & $5$\\ \hline
\end{tabular}
\vskip.1in
When $m=0$, interpreting the set of cost functions $f:E \rightarrow \{1,2,\ldots,m\}$
as being empty explains why the value shown is $0$.
When $m=1$, there is only one function $f:E \rightarrow \{1\}$, and it is
not $M$-generic; any of the $5$ bases in $\BBB(M)$ will minimize $f(B)$, explaining
both why the value for $m=1$ is $0$, but also explaining the
value of $5$ for $m=-1$.  
The value of $5$ for $m=2$ counts these
$M$-generic cost functions $f:E \rightarrow \{1,2\}$:
$$
\xymatrix{
&\bullet \ar@{-}[dl]_{1}\ar@{-}[dr]^{1}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{2}  \ar@(dr,dl)@{-}[rr]_{2}& &\bullet
}\quad
\xymatrix{
&\bullet \ar@{-}[dl]_{1}\ar@{-}[dr]^{2}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{1}  \ar@(dr,dl)@{-}[rr]_{2}& &\bullet
}\quad
\xymatrix{
&\bullet \ar@{-}[dl]_{1}\ar@{-}[dr]^{2}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{2}  \ar@(dr,dl)@{-}[rr]_{1}& &\bullet
}\quad
\xymatrix{
&\bullet \ar@{-}[dl]_{2}\ar@{-}[dr]^{1}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{1}  \ar@(dr,dl)@{-}[rr]_{2}& &\bullet
}\quad
\xymatrix{
&\bullet \ar@{-}[dl]_{2}\ar@{-}[dr]^{1}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{2}  \ar@(dr,dl)@{-}[rr]_{1}& &\bullet
}
$$
\noindent
Lastly, Proposition~\ref{matroid-qsym-reciprocity} predicts
the expected number of $f$-minimizing bases for 
$f: E \rightarrow \{1,2,\ldots,m\}$ as
$$
(-m)^{-|E|}\ps^1 \Psi[M](-m)= (-m)^{-4}\frac{m(m+1)(2m^2+2m+1)}{2}
=\frac{(m+1)(2m^2+2m+1)}{2m^3},
$$
whose limit as $m \rightarrow \infty$ is $1$, consistent with the notion
that ``most'' cost functions should be generic with respect to the bases of
$M$, and maximize/minimize on a unique basis.
\end{example}


\begin{remark}
It is not coincidental that there is a
similarity of results for Stanley's chromatic symmetric
function of a graph $\Psi[G]$ and for the matroid quasisymmetric
function $\Psi[M]$, such as the $P$-partition expansions
\eqref{chromatic-symm-fn-as-sum-of-posets} versus Proposition~\ref{matroid-qsym-fn-as-sum-of-posets-prop}, and the reciprocity results 
Proposition~\ref{Stanley-chromatic-reciprocity} versus
Proposition~\ref{matroid-qsym-reciprocity}.  
It was noted in \cite[\S 9]{BilleraJiaR} that one can associate a similar
quasisymmetric function invariant to any {\it generalized permutohedra}
in the sense of Postnikov \cite{Postnikov}.  Furthermore, recent work of
Ardila and Aguiar \cite{AguiarArdila} has shown that there is a Hopf algebra
of such generalized permutohedra, arising from a {\it Hopf monoid} in the sense of
Aguiar and Mahajan \cite{AguiarMahajan}.  This Hopf algebra
generalizes the chromatic Hopf algebra of
graphs and the matroid-minor Hopf algebra, and its quasisymmetric function
invariant derives as usual from Theorem~\ref{Qsym-as-terminal-object-theorem}.
Their work \cite{AguiarArdila} also provides a generalization of the chromatic Hopf algebra
antipode formula of Humpert and Martin \cite{HumpertMartin} discussed 
in Remark~\ref{Humpert-Martin-antipode-remark} above.
\end{remark}









\newpage




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Malvenuto-Reutenauer Hopf algebra of permutations}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Like so many Hopf algebras we have seen, the {\it Malvenuto-Reutenauer
Hopf algebra} $\FQsym$ can be 
thought of fruitfully in more than one way.  One is that it gives
a natural noncommutative lift of the quasisymmetric 
$P$-partition enumerators and the 
fundamental basis $\{L_\alpha\}$ of $\Qsym$, rendering their
product and coproduct formulas even more natural.


\subsection{Definition and Hopf structure}

\begin{definition}
Define $\FQsym=\bigoplus_{n \geq 0} \FQsym_n$ to be a graded $\kk$-module
in which $\FQsym_n$ has $\kk$-basis $\{F_w\}_{w \in \Symm_n}$ indexed by
the permutations $w=(w_1,\ldots,w_n)$ in $\Symm_n$.
\end{definition}

We first attempt to lift the product and coproduct formulas
\eqref{Qsym-product-on-fundamentals},
\eqref{Qsym-coproduct-on-fundamentals} 
in the $\{L_\alpha\}$ basis of $\Qsym$.
We attempt to define a product for $u \in \Symm_k, v \in \Symm_{\ell}$
\begin{equation}
\label{FQsym-product-on-F's}
F_u F_v:= \sum_{w \in \shuf{u}{v[k]}} F_w
\end{equation}
where for $v=(v_1,\ldots,v_\ell)$ one sets
$v[k]:=(k+v_1,\ldots,k+v_\ell)$.

The coproduct will be defined using the notation of {\it standardization}
$\std(\ii)$ of a word $\ii=(i_1,\ldots,i_n)$ in 
some linearly ordered alphabet, 
which is the permutation in $\Symm_n$ obtained by
replacing all the occurrences of the smallest letter in $\ii$ 
by the numbers $1,2,\ldots,m_1$ from left-to-right, then replacing all
occurrences of the next smallest letter by the numbers $m_1+1,m_1+2,\ldots,m_1+m_2$, from left-to-right, etc.  

\begin{example}
\label{standardization-example}
Considering words in the Roman alphabet $a<b<c< \cdots$
$$
\begin{array}{rccccccccccl}
\std(b&a&c&c &b&a&a&b&a&c &b)\\
=   (5&1&9&10&6&2&3&7&4&11&8).
\end{array}
$$
\end{example}

Using this, define for $w=(w_1,\ldots,w_n)$ in $\Symm_n$
\begin{equation}
\label{FQsym-coproduct-on-F's}
\Delta F_w := \sum_{k=0}^n 
F_{\std(w_1,w_2,\ldots,w_k)} \otimes F_{\std(w_{k+1},w_{k+2},\ldots,w_n)}.
\end{equation}

It is possible to check directly that the maps defined in 
\eqref{FQsym-product-on-F's} and
\eqref{FQsym-coproduct-on-F's}
endow $\FQsym$ with the structure of a graded connected finite type Hopf 
algebra; see Hazewinkel, Gubareni, Kirichenko \cite[Thm. 7.1.8]{HazewinkelGubareniKirichenko}.  
However in justifying this here, we
will follow the approach of Duchamp, Hivert and Thibon 
\cite{NCSF6}, which exhibits $\FQsym$ as a subalgebra of a larger
ring of (noncommutative) power series of bounded degree
in a totally ordered alphabet.

\begin{definition}
Given a totally ordered set $I$, create a totally ordered variable
set $\{X_i\}_{i \in I}$, and the ring $R\langle \{X_i\}_{i\in I} \rangle$ of
noncommutative power series of bounded degree
in this alphabet.  Many times, we will use a variable
set $\XX:=(X_1 < X_2 < \cdots)$, and call the ring $R \langle \XX \rangle$.
\end{definition}

We first identify the algebra structure for $\FQsym$ as the subalgebra 
of finite type within $R\langle \{X_i\}_{i\in I} \rangle$ spanned
by the elements
$$
F_w = F_w( \{X_i\}_{i\in I}):= \sum_{\substack{\ii=(i_1,\ldots,i_n):\\ \std(\ii) = w^{-1}}} X_\ii
$$
where $X_{\ii}:=X_{i_1} \cdots X_{i_n}$,
as $w$ ranges over $\cup_{n \geq 0} \Symm_n$ .

\begin{example}
\label{F's-example}
For the alphabet $\XX=(X_1 < X_2 < \cdots)$, in $R\langle \XX \rangle$ one has
$$
\begin{aligned}
F_{1}&=\sum_{1 \leq i} X_i=X_1 + X_2 + \cdots \\
F_{12}&=\sum_{1 \leq i \leq j} X_i X_j=X_1^2 + X_2^2 + \cdots + X_1 X_2 + X_1 X_3 + X_2 X_3 + X_1 X_4+\cdots \\
F_{21}&=\sum_{1 \leq i < j} X_j X_i =X_2 X_1 + X_3 X_1 + X_3 X_2 + X_4 X_1 + \cdots \\
F_{312}&=\sum_{\ii: \std(\ii)=231} X_\ii 
        =\sum_{1 \leq i < j \leq k} X_j X_k X_i \\
        &=X_2^2 X_1 + X_3^2 X_1 + X_3^2 X_2 + \cdots 
         + X_2 X_3 X_1 + X_2 X_4 X_1 + \cdots
\end{aligned}
$$
\end{example}

\begin{proposition}
For any totally ordered infinite set $I$,
the elements $\{F_w\}$ as $w$ ranges over $\cup_{n \geq 0} \Symm_n$ 
form the $\kk$-basis for a subalgebra $\FQsym( \{X_i\}_{i\in I})$
of $R \langle \XX \rangle$, which is graded connected and of finite type, 
having multiplication defined $\kk$-linearly by
\eqref{FQsym-product-on-F's}.

Consequently all such algebras are isomorphic
to a single algebra $\FQsym$, having basis $\{F_w\}$ and
multiplication given by the rule \eqref{FQsym-product-on-F's},
with the isomorphism mapping $F_w \longmapsto F_w (\{X_i\}_{i\in I})$.
\end{proposition}

For example,
$$
\begin{aligned}
F_1 F_{21} 
&= (X_1 + X_2 + X_3 + \cdots)
    (X_2 X_1 + X_3 X_1 + X_3 X_2 + X_4 X_1 + \cdots) \\
&= X_1 \cdot X_3 X_2 + X_1 \cdot X_4 X_2 + \cdots
 +X_2 \cdot X_3 X_2 + X_2 \cdot X_4 X_2 + \cdots \\
&\qquad   +X_2 \cdot X_3 X_1 + X_2 \cdot X_4 X_1 + \cdots 
   +X_3 \cdot X_3 X_1 + X_3 \cdot X_3 X_2 + \cdots \\
&\qquad   +X_3 \cdot X_2 X_1 + X_4 \cdot X_2 X_1 + \cdots \\
& = \sum_{\ii:\std(\ii)=132} X_\ii
+\sum_{\ii:\std(\ii)=231} X_\ii
+\sum_{\ii:\std(\ii)=321} X_\ii
= F_{132}+F_{312}+F_{321} = \sum_{w \in \shuf{1}{32}} F_w
\end{aligned}
$$

\begin{proof}
The elements $\{F_w(\{X_i\}_{i\in I})\}$ are linearly independent as
they are supported on disjoint monomials, and so
form a $\kk$-basis for their span.  The fact that they multiply
via rule \eqref{FQsym-product-on-F's} is the equivalence of
conditions (i) and (iii) in the following Lemma~\ref{standardization-lemma},
from which all the remaining assertions follow.
\end{proof}


\begin{lemma}
\label{standardization-lemma}
For a triple of permutations
$$
\begin{aligned}
u&=(u_1,\ldots,u_k) \text{ in }\Symm_k,\\
v&=(v_1,\ldots,v_{n-k}) \text{ in }\Symm_{n-k},\\
w&=(w_1,\ldots,w_n) \text{ in }\Symm_n,
\end{aligned}
$$
the following conditions are equivalent:
\begin{enumerate}
\item[(i)]
$w^{-1}$ lies in the set $\shuf{u^{-1}}{v^{-1}[k]}$.
\item[(ii)]
$u=\std(w_1,\ldots,w_k)$ and 
$v=\std(w_{k+1},\ldots,w_n)$,
\item[(iii)]
for some word $\ii=(i_1,\ldots,i_n)$
with $\std(\ii)=w$ one has
$u=\std(i_1,\ldots,i_k)$ and
$v=\std(i_{k+1},\ldots,i_n)$.
\end{enumerate}
\end{lemma}
\begin{proof}[Proof of Lemma.]
The implication (ii) $\Rightarrow$ (iii) is clear since $\std(w)=w$.
The reverse implication (iii) $\Rightarrow$ (ii) is best illustrated
by example, e.g. considering Example~\ref{standardization-example}
as concatentated, with $n=11, k=6, n-k=5$:
$$
\begin{array}{rccccccccccccccl}
w=\std&(b&a&c&c&b&a&|             &a&b&a&c &b)\\
 =   &(5&1&9&10&6&2&|             &3&7&4&11&8)\\
       & & &  & & &              & & & &  &\\

 u=\std&(5&1&9&10 &6&2&)\quad v= \std(&3&7&4&11&8)\\
  =&  (3&1&5&6 &4&2&)\quad =    (&1&3&2&5 &4)\\
=\std&(b&a&c&c &b&a&)\quad  = \std(&a&b&a&c &b)
\end{array}
$$

The equivalence of (i) and (ii) is a fairly standard consequence
of unique parabolic factorization $W=W^J W_J$
where $W=\Symm_n$ and $W_J=\Symm_k \times \Symm_{n-k}$,
so that $W^J$ are the  minimum-length coset representatives for cosets
$xW_J$, satisfying $x_1 < \cdots < x_k, x_{k+1} < \cdots < x_n$. 
One can uniquely express any $w$ in $W$ as
$w=x y$ with $x$ in $W^J$ and $y$ in $W_J$, which here means
that $y=u \cdot v[k]=v[k] \cdot u$ for some $u$ in $\Symm_k$
and $v$ in $\Symm_{n-k}$.  Therefore $w=x u v[k]$,
if and only if $w^{-1}=u^{-1} v^{-1}[k] x^{-1}$, which means that
$w^{-1}$ is the shuffle of the sequences $u^{-1}$ in positions
$\{x_1,\ldots,x_k\}$ and $v^{-1}[k]$ in positions $\{x_{k+1},\ldots,x_n\}$.
\end{proof}

\begin{example}
To illustrate the equivalence of (i) and (ii) and the parabolic
factorization in the preceding proof, let $n=9$ and $k=5$ with
$$
\begin{aligned}
w&=\left(
\begin{matrix}
1 & 2& 3 & 4  & 5 & | & 6 & 7 & 8 & 9 \\
4 & 9 & 6 & 1 & 5 & | & 8 & 2 & 3 & 7
\end{matrix}
\right)\\
&=
\left(
\begin{matrix}
1 & 2& 3 & 4  & 5 & | & 6 & 7 & 8 & 9 \\
1 & 4 & 5 & 6 & 9 & | & 2 & 3 & 7 & 8
\end{matrix}
\right)
\left(\begin{matrix}
1 & 2& 3 & 4  & 5 \\
2 & 5 & 4 & 1 & 3 
\end{matrix}
\right)
\left(\begin{matrix}
6 & 7& 8 & 9 \\
9 & 6 & 7 & 8
\end{matrix}
\right)\\
&=x \cdot u \cdot v[k]\\
 &  \\
w^{-1}&=\left(
\begin{matrix}
1 & 2& 3 & 4  & 5  & 6 & 7 & 8 & 9 \\
4 & 9 & 6 & 1 & 5  & 8 & 2 & 3 & 7
\end{matrix}
\right)\\
&=
\left(\begin{matrix}
1 & 2& 3 & 4  & 5 \\
4 & 1 & 5 & 3 & 2 
\end{matrix}
\right)
\left(\begin{matrix}
6 & 7& 8 & 9 \\
7 & 8 & 9 & 6
\end{matrix}
\right)
\left(
\begin{matrix}
1 & 2& 3 & 4  & 5 & 6 & 7 & 8 & 9 \\
\underline{1} & \underline{\underline{6}} & \underline{\underline{7}} & \underline{2} & \underline{3}  & \underline{4} & \underline{\underline{8}} & \underline{\underline{9}} & \underline{5}
\end{matrix}
\right) \\
&= u^{-1} \cdot v^{-1}[k] \cdot x^{-1}
\end{aligned}
$$
\end{example}


One can now use this to 
define a coalgebra structure on $R\langle \XX \rangle$ as follows.
Given the ordered variable set 
$$
(\XX,\YY) :=(X_1 < X_2 < \cdots < Y_1 < Y_2 < \cdots )
$$
form the ring $R \langle \XX,\YY \rangle$ and its
quotient $R \langle \XX,\YY \rangle/[\XX,\YY]$ by the two-sided ideal
generated by all commutators $[X_i,Y_j] = X_i Y_j - Y_j X_i$, in which one
has forced the $\XX$ variables to commute with $\YY$ variables.
One can check that 
$$
R \langle \XX,\YY \rangle/[\XX,\YY]
\cong R \langle \XX \rangle \otimes R \langle \YY \rangle
$$
giving a ring homomorphism
$\FQsym \overset{\Delta}{\longrightarrow}
R \langle \XX \rangle \otimes R \langle \YY \rangle$
which is the composite of these ring morphisms:
\begin{equation}
\label{coproduct-composite-for-Qsym}
\begin{array}{rcl}
\FQsym &
  \cong &\FQsym(\XX,\YY) 
  \hookrightarrow R \langle \XX,\YY \rangle 
   \longrightarrow R \langle \XX,\YY \rangle/[\XX,\YY] 
  \cong R \langle \XX \rangle \otimes R \langle \YY \rangle \\
f(\XX) & \longmapsto &f(\XX,\YY).
\end{array}
\end{equation}

\begin{example}
Recall from Example~\ref{F's-example} that one has
$$
F_{312}=\sum_{\ii: \std(\ii)=231} X_\ii 
        =\sum_{1 \leq i < j \leq k} X_j X_k X_i 
$$
and therefore its coproduct is
$$
\begin{aligned}
\Delta F_{312}
&=F_{312}(X_1,X_2,\ldots,Y_1,Y_2,\ldots) \\
&=\sum_{i < j \leq k} X_j X_k X_i
+\sum_{\substack{i < j,\\ k}} X_j Y_k X_i 
+\sum_{\substack{i,\\ j \leq k}} Y_j Y_k X_i
+\sum_{i < j \leq k} Y_j Y_k Y_i\\
&= \sum_{i < j \leq k} X_j X_k X_i \cdot 1
+\sum_{\substack{i < j,\\ k}} X_j X_i \cdot Y_k 
+\sum_{\substack{i,\\ j \leq k}} X_i \cdot Y_j Y_k
+\sum_{i < j \leq k} 1 \cdot Y_j Y_k Y_i \\
&=F_{312}(\XX)\cdot 1 
+ F_{21}(\XX) \cdot F_1(\YY) 
+ F_1(\XX) \cdot F_{12}(\YY)
+ 1 \cdot F_{312}(\YY) \\
&=F_{312} \otimes \one
+ F_{21} \otimes F_1
+ F_1 \otimes F_{12}
+ \one \otimes F_{312} 
\end{aligned}
$$
\end{example}


\begin{proposition}
The image of the composite map 
in \eqref{coproduct-composite-for-Qsym}
lies in $\FQsym(\XX) \otimes \FQsym(\YY)$, giving rise to a coproduct
$$
\begin{array}{ccc}
\FQsym(\XX,\YY)  &\overset{\Delta}{\longrightarrow}
 &\FQsym(\XX) \otimes \FQsym(\YY) \\
\Vert & & \Vert \\
\FQsym & & \FQsym \otimes \FQsym
\end{array}
$$
whose action on the $\{F_w\}$ basis is given by
the rule \eqref{FQsym-coproduct-on-F's}.  This endows $\FQsym$ with
the structure of a graded connected finite type Hopf algebra.
\end{proposition}
\begin{proof}
One has
$$
\begin{aligned}
\Delta F_w &= F_w(\XX,\YY) 
 = \sum_{\ii: \std(\ii)=w^{-1} } (\XX,\YY)_{\ii} \\
 &= \text{ ( .... insert some careful explanation here ??? ... ) } \\
 &= \sum_{k=0}^n F_{\std(w_1,\ldots,w_k)}(\XX) F_{\std(w_{k+1},\ldots,w_n)}(\YY) \end{aligned}
$$
relying again on Lemma~\ref{standardization-lemma}.
\end{proof}


\begin{corollary}
\label{FQsym-self-dual}
The Hopf algebra $\FQsym$ is self-dual:
the map sending $F_w \longmapsto F_{w^{-1}}$
gives an isomorphism $\FQsym \longmapsto \FQsym^{o}$.
\end{corollary}
\begin{proof}
One has
$$
F_{u^{-1}} F_{v^{-1}} 
= \sum_{w^{-1} \in \shuf{u^{-1}}{v^{-1}[k]}} F_{w^{-1}}
= \sum_{\substack{w:\\ 
           \std(w_1,\ldots,w_k)=u \\
           \std(w_{k+1},\ldots,w_n)=v }} F_{w^{-1}}
$$
via the equivalence of (i) and (ii) in Lemma~\ref{standardization-lemma}.
On the other hand, in $\FQsym^o$, the dual $\kk$-basis $\{G_w\}$
to the $\kk$-basis $\{F_w\}$ for $\FQsym$ should have product
formula  
$$
G_u G_v = \sum_{\substack{w:\\ 
           \std(w_1,\ldots,w_k)=u \\
           \std(w_{k+1},\ldots,w_n)=v }} G_w 
$$
coming from the coproduct formula \eqref{FQsym-coproduct-on-F's}
for $\FQsym$ in the $\{F_w\}$-basis.
\end{proof}

We can now be a bit more precise about the relations between
the various algebras
$$
\Lambda, \Qsym, \Nsym, \FQsym, R \langle \XX \rangle, R(\xx).
$$
Not only does $\FQsym$ allow one to {\it lift} the Hopf structure
of $\Qsym$, it dually allows one to {\it extend} the Hopf structure
of $\Nsym$.  To set up this duality,
note that Corollary~\ref{FQsym-self-dual} 
motivates the choice of an inner product on $\FQsym$ in which 
$$
(F_u, F_v):=\delta_{u^{-1},v}.
$$
We wish to identify the images of the ribbon basis
$\{R_\alpha\}$ of $\Nsym$ when included in $\FQsym$.


\begin{definition}
Define the {\it free quasi-ribbon function}
$$
\RRR_\alpha:=\sum_{w: \Des(w)=D(\alpha) } F_{w^{-1}}
=\sum_{\substack{(w,\ii): \\ \Des(w)=D(\alpha)\\ \std(\ii)=w}} \XX_\ii
=\sum_{\ii: \Des(\ii)=D(\alpha)} \XX_\ii
$$
where the {\it descent set} of a sequence $\ii=(i_1,\ldots,i_n)$
is defined by 
$$
\Des(\ii):=\{j \in \{1,2,\ldots,n-1\}: i_j > i_{j+1} \}=
\Des(\std(\ii)).
$$
Alternatively, $\RRR_\alpha= \sum_T \XX_T$
in which the sum is over column-strict tableaux of
the ribbon skew shape $\alpha$, and $\XX_T=\XX_{\ii}$ 
in which $\ii$ is the sequence of entries of $T$ read in
order from the southwest toward the northeast.

\end{definition}


\begin{example}
Taking $\alpha=(1,3,2)$, with ribbon shape and column-strict fillings $T$ as 
shown
$$
\begin{matrix}
   &   &\sq&\sq\\
\sq&\sq&\sq& \\
\sq&   &   &
\end{matrix}
\qquad
T=\quad
\begin{matrix}
   &   &    &   &i_5&\leq&i_6\\
   &   &    &   &\wedge&    &   \\
i_2&\leq&i_3&\leq&i_4& \\
\wedge&   &   &     &  & \\
i_1&   &   &     &  &
\end{matrix}
$$
one has that
$$
\RRR_{(1,3,2)}=
\sum_{\substack{ \ii=(i_1,i_2,i_3,i_4,i_5,i_6):\\
                   \Des(\ii)=D(\alpha)=\{1,4\}}} \XX_\ii
=\sum_{i_1 > i_2 \leq i_3 \leq i_4 \geq i_5 \leq i_6}
   X_{i_1}  X_{i_2} X_{i_3}  X_{i_4}  X_{i_5} X_{i_6}
=\sum_T \XX_T
$$
\end{example}




\begin{corollary}
The following surjection and injection are Hopf morphisms 
$$
\begin{array}{rcl}
\FQsym &\overset{\pi}{\twoheadrightarrow}& \Qsym \\
F_w & \longmapsto &L_{\gamma(w)} \\
& & \\
\Nsym &\overset{\iota}{\hookrightarrow}& \FQsym \\
R_\alpha & \longmapsto & \RRR_\alpha
\end{array}
$$
and are also adjoint maps with respect to the above choice
of inner product on $\FQsym$
and the usual dual pairing between $\Nsym$ and $\Qsym$.
Furthermore, the second map $\iota$ lets one factor  
the surjection $\Nsym \twoheadrightarrow \Lambda$ as follows
$$
\begin{array}{rcccl}
\Nsym & \rightarrow & \FQsym \hookrightarrow R\langle \XX \rangle & \rightarrow & R(\xx) \\
 R_\alpha &\longmapsto & \RRR_\alpha & \longmapsto & s_\alpha(\xx)
\end{array}
$$
through the abelianization map 
$R\langle \XX \rangle \twoheadrightarrow R(\xx)$
that sends the noncommutative variable $X_i$ to its commutative image.
\end{corollary}
\begin{proof}
The fact that $\FQsym \overset{\pi}{\twoheadrightarrow} \Qsym$
is a Hopf map comes from checking that it respects the product
(compare \eqref{Qsym-product-on-fundamentals}
and \eqref{FQsym-product-on-F's})
\eqref{FQsym-product-on-F's} and
and the coproducts (compare \eqref{Qsym-coproduct-on-fundamentals}
and \eqref{FQsym-coproduct-on-F's})
then applying Proposition~\ref{antipodes-give-convolution-inverses}(c).
It follows that $\Nsym \overset{\iota}{\hookrightarrow} \FQsym$
is a Hopf map once we check the adjointness assertion, via the
following calculation:
$$
\begin{aligned}
(\iota(R_\alpha), F_w) 
=(\RRR_\alpha,F_w) 
=\sum_{u: \Des(u)=D(\alpha)} (F_{u^{-1}}, F_w)
&=\left\{
\begin{array}{rl}
1 &\text{ if }\Des(w)=D(\alpha) \\
0 &\text{ otherwise }
\end{array}
\right\} \\
&=(R_\alpha, L_{\gamma(w)}) 
=(R_\alpha,\pi(F_w)).
\end{aligned}
$$
The last assertion is clear: the abelianization map
sends the noncommutative tableau monomial $\XX_T$ to the commutative tableau
monomial $\xx_T$.
\end{proof}

We summarize some of this picture as follows:
$$
\xymatrix{
\FQsym\ar@{<->}[rr] &         & \FQsym \ar@{->>}[d]\\
\Nsym \ar@{->>}[dr] \ar@{^{(}->}[u] &         & \Qsym \\
       &\Lambda\ar@{^{(}->}[ur]& 
}
$$





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{$0$-Hecke algebras}
\label{0-Hecke-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\subsection{Review of representation theory of finite-dimensional algebras}

Review the notions of indecomposables, simples, projectives, along with the theorems of Krull-Remak-Schmidt, of Jordan-H\"older, 
and the two kinds of Grothendieck groups dual to each other.

\subsection{$0$-Hecke algebra representation theory}
Describe  the simples and projectives, following
Denton, Hivert, Schilling, Thiery on $\JJJ$-trivial monoids.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Nsym and Qsym as Grothendieck groups}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Give Krob and Thibon's intepretation of 
\begin{enumerate}
\item[$\bullet$]
$\Qsym$ and the Grothendieck group of 
composition series, and
\item[$\bullet$]
$\Nsym$ and the Grothendieck group of projectives.
\end{enumerate}

\begin{remark}
Mention P. McNamara's interpretation,in the case of {\it supersolvable lattices}, of the Ehrenborg quasisymmetric function as the composition series enumerator for an $H_n(0)$-action on the maximal chains
\end{remark}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Aguiar-Bergeron-Sottile character theory Part II: Odd and even
characters, subalgebras} 
\label{ABS-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%







%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Face enumeration, Eulerian posets, and cd-indices}
\label{cd-index-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Borrowing from Billera's ICM notes.

\subsection{f-vectors, h-vectors}

\subsection{flag f-vectors, flag h-vectors}

\subsection{ab-indices and cd-indices}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Further topics}
\label{further-topics-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Some of these we may touch on in class, others are appropriate for student talks.

\begin{itemize}
\item Loday-Ronco Hopf algebra of planar binary trees
\item Poirier-Reutenauer Hopf algebra of tableaux
\item Reading Hopf algebra of Baxter permutations
\item Hopf monoids, e.g. of Hopf algebra of generalized permutohedra, of matroids, of graphs, Stanley chromatic symmetric functions and Tutte polynomials
\item Lam-Pylyavskyy Hopf algebra of set-valued tableaux
\item Connes-Kreimer Hopf algebra and renormalization
\item Noncommutative symmetric functions and $\Omega \Sigma \CC P^\infty$
\item Maschke's theorem and ``integrals'' for Hopf algebras
\item Nichols-Zoeller structure theorem and group-like elements
\item Cartier-Milnor-Moore structure theorem and primitive elements
\item Quasi-triangular Hopf algebras and quantum groups
\item The Steenrod algebra, its dual, and tree Hopf algebras
\item Ringel-Hall algebras of quivers
\item Ellis-Khovanov odd symmetric function Hopf algebras
(see also Lauda-Russell)
\end{itemize}

Student talks given in class were:
\begin{enumerate}
\item Al Garver, on Maschke's theorem for finite-dimensional Hopf algebras
\item Jonathan Hahn, on the paper by Humpert and Martin.
\item Emily Gunawan, on the paper by Lam, Lauve and Sottile.
\item Jonas Karlsson, on the paper by Connes and Kreimer
\item Thomas McConville, on Butcher's group and generalized Runge-Kutta methods.
\item Cihan Bahran, on universal enveloping algebras and the Poincar\'e-Birkhoff-Witt theorem.
\item Theodosios Douvropolis, on the Cartier-Milnor-Moore theorem.
\item Alex Csar, on the Loday-Ronco Hopf algebra of binary trees
\item Kevin Dilks, on Reading's Hopf algebra of (twisted) Baxter permutations
\item Becky Patrias, on the paper by Lam and Pylyavskyy
\item Meng Wu, on multiple zeta values and Hoffman's homomorphism from $\Qsym$
\end{enumerate}




\section{Some open problems and conjectures}
\begin{itemize}
\item Is there a proof of the Assaf-McNamara skew Pieri rule that
gives a resolution of Specht or Schur/Weyl modules whose character
corresponds to $s_{\lambda/\mu} h_n$, whose terms model their alternating sum?

\item Explicit antipodes in the Lam-Pylyavskyy Hopf algebras

\item P. McNamara's question:  are $P$-partition enumerators irreducible for connected posets $P$?

\item Stanley's question:  are the only $P$-partition enumerators 
which are symmetric (not just quasisymmetric) those for which $P$ is
a skew-shape with a column-strict labelling?

\item Does Stanley's chromatic symmetric function distinguish trees?

\item Hoffman's stuffle conjecture

\item Billera-Brenti's nonnegativity conjecture for 
the total $cd$-index of  Bruhat intervals

\end{itemize}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Acknowledgements}
The author thanks the following for helpful comments and/or teaching 
him about Hopf algebras: 
Marcelo Aguiar, Federico Ardila, Lou Billera, Richard Ehrenborg, Darij Grinberg, Mark Haiman, Florent Hivert,
Christophe Hohlweg, Jia Huang, Jang Soo Kim, Aaron Lauve, John Palmieri, Margie Readdy, Nathan Reading, Muge Taskin, Jean-Yves Thibon.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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\end{document}
