\classheader{2012-11-05}

A partition of an integer $n\geq 1$ is an integer sequence $\lambda=(\lambda_1\geq \lambda_2\geq\cdots\geq 0)$ such that $\sum \lambda_i=n$. We define $|\lambda|:=\sum \lambda_i$. Let $\eucal{P}_n$ be the set of partitions of $n$.

Let $R_n^d=\Z^d[x_1,\ldots,x_n]^{S_n}$ denote the homogeneous integer polynomials of degree $d$ in $n$ variables. We use mutli-index notation, so that for $\alpha=(\alpha_1,\ldots,\alpha_n)\in\Z_{\geq 0}^n$, the symbol $x^\alpha$ denotes $x_1^{\alpha_1}\cdots x_n^{\alpha_n}$. We claim that $R_n^d$ is a free $\Z$-module with basis
\[m_\lambda=\sum_{\substack{\alpha\text{ which are}\\\text{a permutation}\\ \text{ of }\lambda\,\in\,\eucal{P}_d}}\hspace{-0.1in}x^\alpha\]
as $\lambda$ ranges over the elements of $\eucal{P}_d$ with no more than $n$ non-zero elements. 
These $m_\lambda$ are called the monomial symmetric functions. In our discussion we will be assuming
that $n\ge d$; otherwise, there will be $\lambda\in\eucal{P}_d$ with more than $n$ non-zero elements, so that no $\alpha\in\Z_{\geq 0}^n$ is a permutation of them. Because we would like to talk about $R_n^d$ for every $d\geq 0$, it might seem that working in $\Z[x_1,x_2,\ldots]$ would be preferable, but we can't because symmetric functions in infinitely many variables would require infinite sums, which are nonsensical.  Thus, for this class we will let the value of $n$ be in limbo; some statements will technically only be true up to degree $n$, but since we can choose the value of $n$ to be arbitrarily high, we will omit this issue.

Note that $R=\bigoplus_{d\geq 0}R^d$ is a ring, and that it has the structure of a graded ring. We see that
\[\sum_{d\geq 0}\rank(R^d)t^d=\sum_{d\geq 0}(\#\eucal{P}_d)t^d=\prod_{k\geq 0}\frac{1}{1-t^k},\]
by a theorem of Euler.

For each $r>0$, we let $e_r$ denote the $r$th elementary symmetric function (again, we are not specifying how many variables these symmetric functions are in; it is just arbitrarily large). Then their generating function is
\[E(t)=\sum_r e_rt^r=\prod_{i\geq 1}(1+x_it).\]
(This product is only true up to degree $n$ if we are working with $n$ variables.)

{\small 
%jw
Reader suggestion: to avoid the confusion of $n$ in limbo, perhaps we can consider $R_\infty =
\varprojlim R_n$ where the transition maps $R_{n+1} \twoheadrightarrow R_n$ evaluates $x_{n+1}$ to
zero. Then $R_\infty \twoheadrightarrow R_n$ for any $n$. This agrees
with (and is motivated by) the use of generating functions. Note that
$R_\infty$ is still a graded ring, and $R^d_\infty = R^d_n$ for $n \ge d$.
}

Now define $h_r$ by
\[H(t)=\sum_{r\geq 0}h_rt^r=\prod_{i\geq 1}\frac{1}{1-x_it}.\]
These $h_i$ are called the complete symmetric functions. We can see that in general
\[h_r=\sum_{\lambda\in\eucal{P}_r}m_\lambda,\]
with $h_0=1$ and $h_1=e_1$.

It is easy to see that $H(t)\cdot E(-t)=1$. Therefore, looking at coefficients, %\vspace{-0.05in}
\[\sum_{r=0}^n(-1)^re_rh_{n-r}=0.\]
Because this gives a way of expressing $h_n$ in terms of the $e_i$'s and the $h_k$ for $k< n$, the above result has as a corollary that $R=\Z[h_1,h_2,\ldots]$ (we already knew $R=\Z[e_1,e_2,\ldots]$). In fact, the $h_i$'s freely generate $R$ as a $\Z$-algebra, because if they satisfied any non-trivial relations between each other, we could use the result to turn that into a non-trivial relation between the $e_i$'s, which we know is impossible.

In summary: the $m_\lambda$ freely generate $R$ as a $\Z$-module, the $e_i$'s freely generate $R$ as a $\Z$-algebra, and the $h_i$'s freely generate $R$ as a $\Z$-algebra.

For each $r\geq 1$, we define the power sum $p_r=\sum x_i^r$. Their generating function is
\[P(t)=\sum_{r\geq 1}p_rt^{r-1}.\]
It is a standard computation that
\[P(t)=\sum_{r\geq 1}p_rt^{r-1}=\sum_{i\geq 1}\sum_{r\geq 1}x_i^rt^{r-1}=\sum_{i\geq 1}\frac{x_i}{1-x_it}=\]
\[\sum_{i\geq 1}\frac{d}{dt}\bigg(\log\bigg(\frac{1}{1-x_it}\bigg)\bigg)=\frac{d}{dt}\bigg(\log\bigg(\prod_{i\geq 1}\frac{1}{1-x_it}\bigg)\bigg)=\frac{d\,(\log(H(t)))}{dt}=\frac{H'(t)}{H(t)}.\]
We can also then see that
\[P(-t)=\frac{E'(t)}{E(t)}.\]
In coefficents, these observations tell us that
\[n\cdot h_n=\sum_{r=1}^np_rh_{n-r},\qquad n\cdot e_n=\sum_{r=1}^n(-1)^{r-1}p_re_{n-r}.\]
These are called Newton's identities. As a corollary, if we allow symmetric functions with coefficients in $\Q$ so that we can get rid of the $n$'s, we see that
\[\Q\otimes_\Z R=\Q[p_1,p_2,\ldots].\]
\begin{remark}
%jw
Let $x=\operatorname{diag}(x_1,\ldots,x_n)$, and let $x\lcirclearrowright V=k^n$. Then
$x\lcirclearrowright \underbrace{V\otimes\cdots\otimes V}_{r\text{ times}}$, via $x \otimes \dotsb
\otimes x$. 
%\begin{align*}
% &(x\otimes\id\otimes\cdots\otimes \id)\\
%+&(\id \mathbin{\otimes} x\otimes \cdots\otimes\id)\\
%+ & \cdots\\
%+&(\id\otimes\id\otimes\cdots\otimes x).
%\end{align*}
%\[(x\otimes\id\otimes\cdots\otimes \id)+(\id\otimes x\otimes\id\otimes\cdots\otimes\id)+\cdots\]
Taking the trace of this action,
\[\tr(x|_{\Lambda^rV})=e_r(x_1,\ldots,x_n)\]
\[\tr(x|_{\operatorname{Sym}^rV})=h_r(x_1,\ldots,x_n)\]
You might hope that there is some construction on $V$ such that $x$ has trace $p_r(x_1,\ldots,x_n)$ on it. There is no such thing, but Adams pretended there was such a thing, and his construction is important in topology.
\[\tr(x|_{\operatorname{Adams}(V)})=p_r(x_1,\ldots,x_n).\]
\end{remark}

Given a partition $\lambda=(\lambda_1\geq\lambda_2\geq\cdots)$, we define
\[e_\lambda=e_{\lambda_1}e_{\lambda_2}\cdots,\quad h_\lambda=h_{\lambda_1}h_{\lambda_2}\cdots,\quad p_\lambda=p_{\lambda_1}p_{\lambda_2}\cdots.\]
The $e_\lambda$ form a $\Z$-basis for $R$, the $h_\lambda$ also form a $\Z$-basis for $R$, and the $p_\lambda$ form a $\Q$-basis for $\Q\otimes_\Z R$.

Given a partition $\lambda=(\lambda_1\geq\lambda_2\geq\cdots)$, we define $d_i(\lambda)=\#\{s\mid \lambda_s=i\}$. and then define
\[z_\lambda=\prod_{i\geq 1}i^{d_i(\lambda)}\cdot(d_i(\lambda)!).\]
Now note that
\[H(t)=\exp\bigg(\sum_{r\geq 1}p_r\frac{t^r}{r}\bigg)=\prod_{r\geq 1}e^{p_r\frac{t^r}{r}}=\prod_{r\geq 1}\bigg(\sum_{d_r=0}^\infty(p_rt^r)^{d_r}\frac{1}{r^{d_r}\cdot d_r!}\bigg)=\sum_{\lambda}\frac{1}{z_\lambda}p_\lambda t^{|\lambda|}.\]
Thus,
\[\prod_{i\geq 1}\frac{1}{1-x_it}=\sum_\lambda\frac{1}{z_\lambda}p_\lambda t^{|\lambda|}.\]

\subsection*{Schur functions}

Given $f\in\Z[x_1,\ldots,x_n]$, we define the alternation of $f$ to be
\[a(f)=\sum_{\sigma\in S_n}\sign(\sigma)\,\sigma(f)\in\Z[x_1,\ldots,x_n]^{\sign}.\]
By last time, we know $D_n\mid a(f)$. Thus, for any $\alpha=(\alpha_1,\ldots,a_n)\in\Z_{\geq 0}^n$, we have that $a(x^\alpha)=0$ whenever there are any $i,j$ such that $\alpha_i=\alpha_j$. Also note that letting $\rho=(n-1,n-2,\ldots,0)$, we have $a(x^\rho)=D_n$.

Up to sign, we can rearrange a non-zero $a(x^\alpha)$ so that $\alpha_1>\cdots>\alpha_n$. For any $\alpha$, we can write $\alpha=\lambda+\rho$ for $\lambda=(\lambda_1\geq \lambda_2\geq\cdots)$. Define the Schur polynomial for $\lambda$ to be \[s_\lambda(x)=\frac{a(x^{\lambda+\rho})}{a(x^\rho)}=\frac{\det(x_i^{\lambda_j+n-j})_{i,j}}{\det(x_i^{n-j})_{i,j}}\]
(note that the numerator resembles the Vandermonde determinant, except that we added $\lambda_j$'s to the powers).

The set $\{a(x^\alpha)\mid \alpha=(\alpha_1>\cdots>\alpha_n)\}$ forms a $\Z$-basis of $\Z[x_1,\ldots,x_n]^{\sign}$, which (as we proved last class) is a rank 1 free $\Z[x_1,\ldots,x_n]^{S_n}$-module with $\Z[x_1,\ldots,x_n]^{S_n}$-basis $\{D_n\}$. Because $s_\lambda$ is just defined by dividing by $a(x^\rho)=D_n$, we have that $\{s_\lambda\}$ is a $\Z$-basis of $R$.

The Cauchy identities involve the expression
\[XY=\frac{1}{\textstyle\prod\limits_{i,j=1}^n(1-x_iy_j)}\]
The first Cauchy identity states that
\[XY=\sum_\lambda\frac{p_\lambda(x)p_\lambda(y)}{z_\lambda}\]
This follows from the identity
\[\prod\frac{1}{1-x_it}=\sum_\lambda \frac{1}{z_\lambda}p_\lambda t^{|\lambda|},\]
where instead of $x_1,\ldots,x_n$, we introduce $n^2$ variables so that the left side becomes $XY$ and the right side is what we want.