# sum of squares of Schur polynomials indexed over partition valued functions on a set

Fix a finite set $X$ and two natural numbers $d$ and $n$.

For a partition $\lambda$ and a number $d$ denote by $s_\lambda^d(x_1,\dots,x_d)$ the Schur polynomial in $d$-many variables $x_1,\dots,x_d$. Denote by $\mathcal P_n(X)$ the set of partition-valued functions on $X$ of total size $n$, i.e. the set of functions $\lambda\colon X\to\{\text{partitions}\}$ such that $\|\lambda\|:=\sum_{x\in X}|\lambda(x)|=n$.

Let $a\colon X\to \mathbb{N}\backslash\{0\}$ be a function on $X$. EDIT: I am mostly interested in the case where $a$ is the constant function with value $1$.

I am trying to evaluate the expression $$F_X(d,a):=\sum_{\lambda\in \mathcal P_n(X)}\left(\prod_{x\in X}s^d_{\lambda(x)}(a(x),\dots,a(x))\right)^2.$$

EDIT:

1. I have now good reason to believe that if we take for $a$ the constant function with value $1$ then we have $$F_X(d,a)=\left(\binom{d^2|X|}{n}\right)$$ Here $\left(\binom{m}{n}\right)$ denotes the multiset number

2. More generally I am hoping that if $G$ is a finite group, $X$ is the set of irreducible characters (over $\mathbb C$) of $G$ and $a$ maps each character to its dimension then we have $$F_X(d,a)=\left(\binom{d^2|G^{\mathrm{ab}}|}{n}\right)$$ Here $G^\mathrm{ab}$ is the abelianisation of $G$. This reduces to case 1 if $G$ is taken to be an abelian group.

EDIT2: my conjecture 2 above is totally wrong (as can be seen by hand in small examples like $G=S_3$) and I have no good substitute hypothesis. I am now mostly interested in a proof of point 1 (where $a$ is the constant function $1$)

The special case where $X=\{\star\}$ is a one-element set can be done using the identity $$\prod_{i,j}(1-x_iy_j)^{-1}=\sum_n\sum_{|\lambda|=n} s_\lambda(x_1,\dots)s_\lambda(y_1,\dots)$$ by putting $x_i=y_j=t$ for $i,j=1,\dots,d$ (and zero otherwise) and picking out the coefficient of $t^{2n}$ obtaining $$c^d_nt^{2n}=\sum_{|\lambda|=n}s^d_\lambda(t,\dots,t)^2,$$ where $c^d_n$ is the coefficient of $t^n$ in $(1-t)^{-d^2}$ and can be calculated to be the multiset number $\left(\binom{d^2}{n}\right)$. All in all we obtain $F_{\{\star\}}(d,a)=\left(\binom{d^2}{n}\right)\cdot a(\star)^{2n}$

• You rrally want the sum of the lengths to be $n$, not the sum of the sizes? – darij grinberg Feb 25 '16 at 15:55
• you are right, of course. It should be the sizes – Tashi Walde Feb 25 '16 at 16:34

assume $a$ is the constant function with value $1$.
as in the case of $X=\{\star\}$ we have the equality $$(1-t^2)^{-d^2|X|}=\left(\sum_{\lambda\in \{\mathrm{partitions}\}}s^d_\lambda(t,\dots,t)^2\right)^{|X|}=\sum_{\lambda\colon X\to\{\mathrm{partitions}\}}\prod_{x\in X}s_{\lambda(x)}^d(t,\dots,t)^2$$ and picking out the monomial $t^2$ gives $$\left(\binom{d^2|X|}{n}\right)t^{2n}=\sum_{\lambda\in P_n(X)}\prod_{x\in X}s_{\lambda(x)}^d(t,\dots,t)^2$$ The result follows.
Note that $s_\lambda(1^d)$ can be computed explicitly via Weyls formula on wikipedia in terms of the parts of $\lambda$.