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?

  • $\begingroup$ 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. $\endgroup$ Jul 25, 2016 at 22:05

2 Answers 2


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.

  • $\begingroup$ Where do I miss the $d!$ factor in this argument? $\endgroup$ Jul 25, 2016 at 23:03
  • $\begingroup$ I had a typo in the hook length formula :) $\endgroup$ Jul 25, 2016 at 23:04
  • $\begingroup$ Nice proof. Do you expect there's a corresponding "weighted Cauchy identity" for Schur polynomials? (I just guessed the form) $\endgroup$ Jul 25, 2016 at 23:13
  • 3
    $\begingroup$ 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 $S_d \times GL(V)$. $\endgroup$ Jul 26, 2016 at 13:45

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 $t$.

  • $\begingroup$ What are the vectors $a$ and $b$ in your formulae? I suppose the eigenvalues of $A$ and $B$? $\endgroup$
    – Wolfgang
    Jul 26, 2016 at 8:15
  • $\begingroup$ Fixed, thanks. Real diagonal matrices are sufficient for the argument. $\endgroup$ Jul 26, 2016 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.