Interpolating symmetric functions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 21 July 2012

Interpolating symmetric functions

Define $q_r(X_n;q,t)=q_r(x_1,...,x_n;q,t)$ by the generating function $$\prod _{i=1}^n\frac{1-x_{i}tz}{1-x_{i}qz}=(q-t)\sum _{r\ge 0}{q}_r(x_1,...,x_n;q,t)z^r.$$ Note that with this definition $q_0=\frac{1}{(q-t)}.$ Define the elementary symmetric functions, the complete symmetric functions and the power symmetric functions by the formulas $$\begin{array}{ll}\begin{array}{rcll}{(-t)}^{r-1}{e}_r\left(X_n\right)& =& q_r(X_n;0,t),& \\ q^{r-1}{h}_r\left(X_n\right)& =& q_r(X_n;q,0),& and\\ q^{r-1}{p}_r\left(X_n\right)& =& q_r(X_n;q,q),& respectively.\end{array}& (ISF\; 1)\end{array}$$ The elementary symmetric functions have special importance because of the following ways in which they appear naturally.

1. If $f\left(t\right)$ is a polynomial in $t$ with roots ${\gamma }_1,...,{\gamma }_n$ then $$\begin{array}{ll}the\; coefficient\; of\; t^r \; in\; f\left(t\right) \; is\; {(-1)}^{n-r}{e}_r({\gamma }_1,...,{\gamma }_n).& (ISF\; 2)\end{array}$$
2. If $A$ is an $n\times n$ matrix with entries in $𝔽$ with eigenvalues ${\gamma }_1,...,{\gamma }_n$ then the trace of the action of $A$ on the $r^{\mathrm{th}}$ exterior power of the vector space ${𝔽}^n$ is $$\begin{array}{ll}\begin{array}{rcll}\mathrm{tr}(A,{\bigwedge }^r{𝔽}^n)& =& e_r({\gamma }_1,...,{\gamma }_n),& so\; that\\ \mathrm{Tr}\left(A\right)& =& e_1({\gamma }_1,...,{\gamma }_n),& and\\ \det \left(A\right)& =& e_n({\gamma }_1,...,{\gamma }_n),& \end{array}& (ISF\; 3)\end{array}$$ and the characteristic polynomial of $A$ is $$\begin{array}{ll}{\mathrm{char}}_t\left(A\right)=\sum _{r=0}^n{(-1)}^{n-r}{e}_{n-r}({\gamma }_1,...,{\gamma }_n)t^r.& (ISF\; 4)\end{array}$$

Expanding $\frac{1-x_{i}tz}{1-x_{i}qz}=1+(q-t)\sum _{l>0}{q}^{l-1}{x}_i^l{z}^l$ and multiplying out $$\prod _{i=1}^n\frac{1-x_{i}tz}{1-x_{i}qz}=\frac{1-x_{1}tz}{1-x_{1}qz}\cdots \frac{1-x_{n}tz}{1-x_{n}qz}$$ gives $$\begin{array}{ll}{q}_r=\sum _{1\le i_1\le \cdots \le i_r\le n}{(q-t)}^{\mathrm{Card}\left(\left\{j\right|i_j<i_{j+1}\}\right)}{q}^{\mathrm{Card}\left(\left\{j\right|i_j=i_{j+1}\}\right)}{x}_{i_1}{x}_{i_2}\cdots x_{i_r},& (ISF\; 5)\end{array}$$ from which it follows that $$q_r=\sum _{\lambda ⊢r}{(q-t)}^{l\left(\lambda \right)-1}{q}^{r-l\left(\lambda \right)}{m}_{\lambda }(x_1,...,x_n).$$ For an $n\times n$ matrix $a=\left(a_{ij}\right)$ with entries from ${\mathbb{Z}}_{\ge 0}$ let $$x^a=\prod _{i=1}^n{\left(x_i\right)}^{a_{ij}},and\mathrm{wt}\left(a\right)=q^{\left|\lambda \right|-l\left(a\right)}{(q-t)}^{l\left(a\right)-l\left(\lambda \right)},$$ where $l\left(a\right)$ is the number of nonzero entries in $a,$ $l\left(\lambda \right)$ is the number of nonzero entries in $\lambda ,$ and $\left|\lambda \right|$ is the sum of the entries of $\lambda .$ Define $$\begin{array}{rcl}\mathrm{rs}\left(a\right)& =& ({\rho }_1,...,{\rho }_n),\\ \mathrm{cs}\left(a\right)& =& ({\gamma }_1,...,{\gamma }_n),\end{array}where{\rho }_i=\sum _{j=1}^l{a}_{ij}and{\gamma }_j=\sum _{i=1}^n{a}_{ij},$$ so that $\mathrm{rs}\left(a\right)$ and $\mathrm{cs}\left(a\right)$ are the sequences of row sums and column sums of $a,$ respectively.

For a sequence of nonnegative integers $\lambda =({\lambda }_1,...,{\lambda }_l)$ define $$q_{\lambda }(X_n;q,t)=q_{{\lambda }_1}(X_n;q,t)\cdot q_{{\lambda }_2}(X_n;q,t)\cdot \cdots \cdot q_{{\lambda }_l}(X_n;q,t).$$ Then $$q_{\lambda }=\sum _{\mu }{a}_{\lambda \mu }(q,t)m_{\mu },where{a}_{\lambda \mu }(q,t)=\sum _{a\in A_{\lambda \mu }}\mathrm{wt}\left(a\right),$$ and the sum is over partitions $\mu$ such that $\left|\mu \right|=\left|\lambda \right|.$

		Proof.
	If $\lambda =({\lambda }_1,...,{\lambda }_l)$ then $$q_{\lambda }=\prod _{j=1}^l{q}_{{\lambda }_j}=\sum _{\mathrm{rs}\left(a\right)=\lambda }\mathrm{wt}\left(a\right)x^a=\sum _{\gamma \in {\mathbb{Z}}_{\ge 0}^n}\sum _{\genfrac{}{}{0ex}{}{\mathrm{rs}\left(a\right)=\lambda }{\mathrm{cs}\left(a\right)=\gamma }}\mathrm{wt}\left(a\right)x^{\gamma }=\sum _{\mu }{a}_{\lambda \mu }(q,t)m_{\mu }.$$ $\square$

Multiplying out $$\prod _{i=1}^n\frac{1-x_{i}tz}{1-x_{i}qz}=\frac{1}{1-x_{1}qz}\cdot \frac{1}{1-x_{2}qz}\cdot \cdots \cdot \frac{1}{1-x_{n}qz}(1-x_{n}tz)(1-x_{n-1}tz)\cdot \cdots \cdot (1-x_{1}tz)$$ gives $$\begin{array}{ll}{q}_r=\sum _{i_1\le i_2\le \cdots \le i_k>i_{k+1}>\cdots >i_r}{q}^{k-1}{(-t)}^{r-k}{x}_{i_1}\cdots x_{i_k}{x}_{i_{k+1}}\cdots x_{i_r}.& (ISF\; 6)\end{array}$$ The bijection ???? between sequences $i_1\le i_2\le \cdots \le i_k>i_{k+1}>\cdots >i_r$ and column strict tableaux of shape $\left(k{1}^{r-k}\right)$ yields $$\begin{array}{ll}{q}_r=\sum _{k=1}^r{(-t)}^{r-k}{q}^{k-1}{s}_{\left(k{1}^{r-k}\right)}\left(X_n\right).& (ISF\; 7)\end{array}$$

For each positive integer $k$ define $${\left[k\right]}_{q,t}=\frac{q^k-t^k}{q-t}=q^{k-1}+t{q}^{k-2}+\cdots +t^{k-2}q+t^{k-1}.$$

Comparing coefficients of $z^r$ on each side of $$\left(\prod _{i=1}^n\frac{1-x_{i}sz}{1-x_{i}tz}\right)\left(\prod _{i=1}^n\frac{1-x_{i}tz}{1-x_{i}qz}\right)=\prod _{i=1}^n\frac{1-x_{i}sz}{1-x_{i}qz}$$ gives $$\begin{array}{ll}\begin{array}{rcl}(t-s)q_r(X_n;t,s)& +& (q-t)(t-s)\left(\sum _{j=1}^{r-1}{q}_j(X_n;q,t)q_{r-j}(X_n;t,s)\right)\\ & +& (q-t)q_r(X_n;q,t)=(q-s)q_r(X_n;qs).\end{array}& (ISF\; 8)\end{array}$$

Example. Putting $s=0, q=0$ and $t=0$ in (???) yield, respectively, $$\begin{array}{rcl}{q}_r(X_n;q,t)+\left(\sum _{j=1}^{r-1}{h}_j\left(X_n\right)t^j{q}_{r-j}(X_n;q,t)\right)-h_r\left(X_n\right){\left[r\right]}_{q,t}& =& 0\\ q_r(X_n;q,t)+\left(\sum _{j=1}^{r-1}{e}_j\left(X_n\right){(-q)}^j{q}_{r-j}(X_n;q,t)\right)+{(-1)}^r{e}_r\left(X_n\right){\left[r\right]}_{q,t}& =& 0\\ \sum _{j=0}^r{(-t)}^{r-j}{\left[j\right]}_{q,t}{h}_j\left(X_n\right)e_{r-j}\left(X_n\right)& =& (q-t)q_r(X_n;q,t).\end{array}$$ Then putting ????? (???) gives $$r{q}_r(X_n;q,t)-\left(\sum _{j=1}^{r-1}{p}_j\left(X_n\right)(q^j-t^j)q_{r-j}(X_n;q,t)\right)-p_r\left(X_n\right){\left[r\right]}_{q,t}=0,$$ and the Newton identities $$k{h}_k=\sum _{i=1}^k{p}_i{h}_{k-i}andk{e}_k=\sum _{i=1}^k{(-1)}^{i-1}{p}_i{e}_{k-i},$$ are obtained by putting ??? in (???).

Let $\lambda =({\lambda }_1,...,{\lambda }_n)$ be a partition. Then

1. $e_{\lambda \prime }=\sum _{\mu \le \lambda }{a}_{\lambda \prime \mu }{m}_{\mu },$ where $a_{\lambda \mu }$ is the number of matrices with entries from $\{0,1\}$ with row sums $\lambda \prime$ and column sums $\mu .$
2. $a_{\lambda \prime \lambda }=1$ and $a_{\lambda \prime \mu }=0$ unless $\mu \le \lambda .$
3. $\left\{e_{\lambda }\right|l(\lambda \prime )\le n\}$ is a $\mathbb{Z}-$basis of $\mathbb{Z}{\left[X_n\right]}^{S_n}.$
4. $\mathbb{Z}{[x_1,...,x_n]}^{S_n}=\mathbb{Z}[e_1,...,e_n]=\mathbb{Z}[h_1,...,h_n]$ and $\mathbb{Q}{[x_1,...,x_n]}^{S_n}=\mathbb{Q}[p_1,...,p_n].$
5. The set $\left\{h_{\lambda }\right|l(\lambda \prime )\le n\}$ is a basis of $\mathbb{Z}{\left[X_n\right]}^{S_n}.$

		Proof.
	Follows by putting $q=0$ in (???). Since there is a unique matrix $A$ with $\mathrm{rs}\left(A\right)=\lambda \prime$ and $\mathrm{cs}\left(A\right)=\lambda ,$ $a_{\lambda \prime \lambda }=1.$ If $A$ is a 0, 1 matrix with $\mathrm{rs}\left(A\right)=\lambda \prime$ and $\mathrm{cs}\left(A\right)=\mu$ then ${\mu }_1+\cdots +{\mu }_i\le {\lambda }_1+\cdots +{\lambda }_i$ since there are at most ${\lambda }_1+\cdots +{\lambda }_i$ nonzero entries in the first $i$ columns of $A.$ Thus $a_{\lambda \prime \mu }=0$ unless $\mu \le \lambda .$ This is a consequence of (b) and the fact that $\left\{m_{\lambda }\right|l\left(\lambda \right)\le n\}$ is a basis of $\mathbb{Z}{\left[X_n\right]}^{S_n}.$ The first equality is an immediate consequence of (c). The second equality follows from the identity (???), which allows one to, inductively, expand $h_r$ in terms of $e_r,e_{r-1},...,e_1.$ Similarly, the third equality follows from the Newton identity (???) which allows one to, inductively, expand $p_r$ in terms of $e_r,e_{r-1},...,e_1$ (with coefficients in $\mathbb{Q}$). $\square$

1. There is an involutive automorphism $\omega$ of $\mathbb{Z}{\left[X_n\right]}^{S_n}$ defined by $$\begin{array}{rrcl}\omega :& \mathbb{Z}{\left[X_n\right]}^{S_n}& \to & \mathbb{Z}{\left[X_n\right]}^{S_n}\\ & e_k& \mapsto & h_k\end{array}.$$
2. $\omega \left(q_r(X_n;q,t)\right)=q_r(X_n;-t,-q)$ and $\omega \left(p_k\right)={(-1)}^{k-1}{p}_k.$

		Proof.
	The map $\omega$ is a well defined ring homomorphism since $\mathbb{Z}{\left[X_n\right]}^{S_n}=\mathbb{Z}[e_1,...,e_n]$ is a polynomial ring. Comparing coefficients of $z^k$ on each side of $$1=\left(\prod _{i=1}^n(1-x_{i}z)\right)\left(\prod _{i=1}^n\frac{1}{1-x_{i}z}\right)yields0=\sum _{r=1}^k{(-1)}^r{e}_r{h}_{n-r}.$$ Thus $e_1=h_1,$ and $$\begin{array}{ll}{h}_k=\sum _{i=1}^k(-1)e_i{h}_{k-i}and{e}_k={(-1)}^{-k}\sum _{i=0}^k(-1)e_i{h}_{k-i}=\sum _{j=1}^k{(-1)}^{-j-1}{e}_{k-j}{h}_j.& (ISF\; 9)\end{array}$$ From the first of these relations, by induction on $k,$ $$\omega \left(h_k\right)=\sum _{i=1}^k{(-1)}^{i+1}{h}_i{e}_{k-i},$$ and, by comparing this identity with the second relation in (???) shows that $\omega \left(h_k\right)=e_k.$ Hence ${\omega }^2=\mathrm{id}.$ ???? $\square$

For a partition $\lambda =(1^{m_1}{2}^{m_2}\cdots )$ of $k$ define $$\begin{array}{ll}{z}_{\lambda }=1^{m_1}{m}_1!2^{m_2}{m}_2!\cdots so\; that\frac{n!}{z_{\lambda }}=\mathrm{Card}\left(\{w\in S_k|whas\; cycle\; type\lambda \}\right)& (ISF\; 10)\end{array}$$ is the size of the conjugacy class indexed by $\lambda$ in the symmetric group $S_k.$ Recalling that $$\ln (1-x_i{y}_j)=\sum _{k\ge 1}\frac{x_i^k{y}_j^k}{k}since\ln (1-t)=\int \frac{1}{1-t}\mathrm{dt}=\int (1+t+t^2+\cdots )\mathrm{dt},$$ we have $$\begin{array}{rcl}\prod _{i,j}\frac{1}{1-x_i{y}_j}& =& \exp \ln \left(\prod _{i,j}\frac{1}{1-x_i{y}_j}\right)\\ & =& \exp \left(\sum _{i,j}\ln (1-x_i{y}_j)\right)\\ & =& \exp \left(\sum _k\sum _{i,j}\frac{x_i^k{y}_j^k}{k}\right)\\ & =& \exp \left(\sum _k\frac{p_k\left(x\right)p_k\left(y\right)}{k}\right)\\ & =& \prod _k\exp \left(\frac{p_k\left(x\right)p_k\left(y\right)}{k}\right)\\ & =& \prod _k\sum _{m_k\ge 0}\left(\frac{p_k^{m_k}\left(x\right)p_k^{m_k}\left(y\right)}{k^{m_k}{m}_k!}\right)\\ & =& \sum _{m_1,m_2,...}\left(\frac{p_1^{m_1}\left(x\right)p_2^{m_2}\left(x\right)\cdots p_1^{m_1}\left(y\right)p_2^{m_2}\left(y\right)\cdots }{1^{m_1}{m}_1!2^{m_2}{m}_2!\cdots }\right)\\ & =& \sum _{\lambda }\frac{p_{\lambda }\left(x\right)p_{\lambda }\left(y\right)}{z_{\lambda }}.\end{array}$$

Notes and References

[Mac] I.G. Macdonald, Symmetric functions and Hall polynomials, Second edition, Oxford University Press, 1995.

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