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While working with Schur polynomials I found what seems like a nice identity, and I wonder if it has a simple proof.

Notation: Suppose $d,n\in\mathbb{N}$, and $\lambda =(\lambda_1,\dots,\lambda_n)$ is an ordered partition of $d$, that is $\lambda_1 \ge \dots \ge \lambda_n \ge 0$, and $\lambda_1 + \dots + \lambda_n = d$.

Let $\Lambda(d,n)$ be the set of all such partitions. For each $\lambda \in \Lambda(d,n)$ define: $$ N(\lambda) = \prod_{1\le i < j \le n} \frac{\lambda_i - \lambda_j + j - i}{j - i}, $$ and $$ W(\lambda) = \prod_{1 \le i \le n} (\lambda_i + n - i)!. $$ Remark: $N(\lambda)=s_\lambda(1,\dots,1)$, where $s_\lambda$ is the Schur polynomial associated with the partition $\lambda$. Now, define $$ A(d,n) = \sum_{\lambda \in \Lambda(d,n)} \frac{N(\lambda)^2}{W(\lambda)}. $$ It seems the following identity holds: $$ A(d,n) = \left( \prod_{k=0}^{n-1} k! \right)^{-1} \frac{n^d}{d!}.$$ Is there a simple proof\explanation?

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  • $\begingroup$ It seems that the identity follows from a "Cauchy like identity" (that I guessed): $$\sum_\lambda \frac{s_\lambda(x) s_\lambda(y)}{W(\lambda)} = \prod_{j\ge 1} \frac{e^{x_j y_j}}{(j-1)!}.$$ I don't know if this "identity" (in infinitely many variables) is actually correct\known, or can be made rigorous. $\endgroup$ – sometempname Jul 25 '16 at 22:05
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We can use the fact that $N(\lambda)=\left|\text{SSYT}(\lambda)\right|$, the number of semistandard Young tableaux of shape $\lambda$, and that $d!\cdot\left(\frac{\prod_{1\le i < j\le n}(\lambda_i - \lambda_j + j - i)}{\prod_{1 \le i \le n} (\lambda_i + n - i)!}\right)=\left|\text{SYT}(\lambda)\right|$, the number of standard Young tableaux of shape $\lambda$. By some simple rearranging your identity becomes $$\sum_{|\lambda|=d}\left|\text{SSYT}(\lambda)\right|\left|\text{SYT}(\lambda)\right|=n^d.$$ Now, if you track the left hand side through the Robinson-Schensted-Knuth correspondence, you'll find that it corresponds to $d\times n$ matrices with row sums all $1$. So there are exactly $n^d$ of these.

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  • $\begingroup$ Where do I miss the $d!$ factor in this argument? $\endgroup$ – sometempname Jul 25 '16 at 23:03
  • $\begingroup$ I had a typo in the hook length formula :) $\endgroup$ – Gjergji Zaimi Jul 25 '16 at 23:04
  • $\begingroup$ Nice proof. Do you expect there's a corresponding "weighted Cauchy identity" for Schur polynomials? (I just guessed the form) $\endgroup$ – sometempname Jul 25 '16 at 23:13
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    $\begingroup$ If you like representation theory better than RSK, the displayed equation takes the dimension of both sides of Schur-Weyl duality: $\sum_{|\lambda|=d} Sp_{\lambda} \boxtimes S_{\lambda}(V) \cong V^{\otimes d}$ where $V$ is a vector space of dimension $n$, $Sp_{\lambda}$ is the Specht module and $S_{\lambda}$ is the Schur functor. The isomorphism is of representations of $S_d \times GL(V)$. $\endgroup$ – David E Speyer Jul 26 '16 at 13:45
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I found a proof which I don't really like, but I'll share it.

For two (real) diagonal matrices $A,B$, the Harish-Chandra-Itzykson-Zuber (HCIZ) integral is $$ I(A,B) = \int_{U(n)} e^{\rm{tr}(U^* A U B)} \, \rm{d} U = c_n \frac{\det\left([e^{a_j b_k}]_{j,k=1}^n\right)}{\Delta(a)\Delta(b)}, $$ where $\Delta(a) = \prod_{j<k} (a_k - a_k)$ is the Vandermonde determinant, and $c_n = \Delta([1,\dots,n]) = \prod_{k=1}^{n-1} k!$ (the integration is with respect to Haar measure on the unitary group).

Using an infinite version of the Cauchy-Binet formula, we can write $$ \det\left(\{e^{a_j b_k}\}_{j,k=1}^n\right) = \sum_\lambda \det\left(\left[\frac{a_j^{\lambda_k+n-k}}{\sqrt{(\lambda_k+n-k)!}}\right]_{j,k=1}^n\right) \det\left(\left[\frac{b_k^{\lambda_j+n-j}}{\sqrt{(\lambda_j+n-j)!}}\right]_{j,k=1}^n\right), $$ where the sum is over all partitions $\lambda$ of size $n$. Using the fact $$ s_\lambda(a) = \frac{\det\left(a_j^{\lambda_k+n-k}\right)}{\Delta(a)},$$ we find $$ I(A,B) = c_n \sum_\lambda \frac{s_\lambda(a)s_\lambda(b)}{\prod_{j=1}^n (\lambda_j+n-j)!} = c_n \sum_\lambda \frac{s_\lambda(a)s_\lambda(b)}{W(\lambda)}. $$ If we substitute $A = B = t \cdot \rm{Id}_n$, then we find $$ e^{t^2 n} = c_n \sum_{d\ge 0} t^{2d} \sum_{\lambda\in\Lambda(d,n)} \frac{\left(s_\lambda(1,\dots,1)\right)^2}{W(\lambda)}. $$ We get the identity by comparing coefficients of $t$.

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  • $\begingroup$ What are the vectors $a$ and $b$ in your formulae? I suppose the eigenvalues of $A$ and $B$? $\endgroup$ – Wolfgang Jul 26 '16 at 8:15
  • $\begingroup$ Fixed, thanks. Real diagonal matrices are sufficient for the argument. $\endgroup$ – sometempname Jul 26 '16 at 14:14

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