# Identity involving Schur polynomials, binomial coefficients and contents of partition

Let $$C_{\lambda,\mu}$$ be the coefficients defined as $$s_\lambda\left(\frac{x_1}{1-x_1},...,\frac{x_N}{1-x_N}\right)=\sum_{\mu\supset \lambda}C_{\lambda\mu}s_\mu(x_1,...,x_N),$$ where $$s$$ are the Schur polynomials.

These coefficients have an explicit form in terms of a determinant of binomial coefficients, $$C_{\lambda\mu}=\det\left[\binom{\mu_j-j}{\lambda_i-i}\right].$$

Given a partition $$\lambda$$, the set $$\{j-i,1\le j\le \lambda_i,1\le i\le \ell(\lambda)\}$$ is its content set. Let $$t_\lambda$$ be the product of all non-zero contents of $$\lambda$$.

I have met the following identity which I believe to be true: $$\sum_{\mu\supset\lambda}C_{\lambda\mu}\frac{t_\mu^2}{s_\mu(1^N)}=\frac{t_\lambda^2}{s_{\lambda'}(1^N)},$$ where $$\lambda'$$ is the partition conjugated to $$\lambda$$.

Notice the infinite amount of cancellation: Since $$s_\mu(1^N)=0$$ for $$N<\ell(\mu)$$, the left hand side might in principle be singular for all integer $$N$$, but the right hand side is actually only singular for $$N<\ell(\lambda')$$.

Has anyone seen this identity before?

Edit. Richard Stanley has mentioned that $$C_{\lambda\mu}=\frac{|\lambda|!}{|\mu|!}f^{\mu/\lambda}T_{\mu/\lambda}$$, where $$T_{\mu/\lambda}$$ is the product of all contents of the skew partition $$\mu/\lambda$$. Then the identity can also be written as $$\sum_\mu\frac{f^{\mu/\lambda}}{|\mu|!}\frac{T_{\mu/\lambda}^3}{s_\mu(1^N)}=\frac{1}{|\lambda|!s_{\lambda'}(1^N)}$$ (here $$f^{\mu/\lambda}$$ is the number of standard Young tableaux of that skew shape).

Edit2. Let me give an example that will also help clarify the meaning of the equality. I have in mind a large $$N$$ expansion. Take the case $$\lambda=(1,1)$$. If I add the terms corresponding to $$\mu\in\{(1,1),(2,1),(1,1,1)\}$$ I get $$\frac{N^2-3N-8}{N(N-2)(N^2-1)}$$ for the left hand side, which is $$\frac{1}{N^2}- \frac{1}{N^3}+O\left(\frac{1}{N^4}\right)$$ for large $$N$$. If I include further values of $$\mu$$, more terms in the large $$N$$ expansion of the left hand side agree with $$\frac{1}{N^2}-\frac{1}{N^3}+\frac{1}{N^4}-\frac{1}{N^5}+\cdots,$$ which is the large $$N$$ series of $$\frac{1}{N(N+1)}$$, the right hand side.

• A natural approach would be to expand the function $\mu\to t_\mu^2/s_\mu(1^N)$ as a mixture of valuations of $s_\mu$. Are you aware of such representation? Jan 18 at 20:19
• Surely you are aware of this, but there is a formula for the evaluation $s_{\mu}(1^N)$: namely, Stanley's hook-content formula. Jan 18 at 20:41
• The determinant $C_{\lambda\mu}$ can be simplified. See problem 87 of klein.mit.edu/~rstan/ec/ch7suppsol.pdf. Put $n=0$ in the definition of $d_{\lambda\mu}$. Jan 18 at 21:31
• @RichardStanley I see that $C_{\lambda\mu}$ can be written in terms of $f^{\mu/\lambda}$, which is the number of standard Young tableaux of shape $\mu/\lambda$. This seems more like an interpretation than a simplification... unless there is an independent way of computing $f^{\mu/\lambda}$ Jan 18 at 22:24
• @Marcel: there are many known identities involving $f^{\mu/\lambda}$. Perhaps one of them is relevant to your question. Jan 19 at 15:47