Identity involving dimensions and contents of partitions

Question

Let $d_\lambda$ with $\lambda\vdash n$ be the dimensions of irreducible representions of the permutation group $S_n$. Let $C^{\nu}_{\lambda,\mu}$ be the standard Littlewood-Richardson coefficients.

I have been led to define the quantity

$$ F_n=\frac{1}{n!(n+1)!}\sum_{\lambda\vdash n}\sum_{\nu\vdash n+1}C^{\nu}_{\lambda,(1)} d_\lambda d_\nu M_\nu^2,$$

where $M_\nu$ is the product of all non-vanishing contents in the partition $\nu$, i.e.

$$ M_\nu=\prod_{(i,j)\in\nu, i\neq j}(j-i).$$

By comparing two different ways of computing the same result, I was expecting that $F_n$ should be equal to 1 for all $n$. Instead, when I actually calculated the values, I got $F_0=F_1=F_2=1$, fine, but $F_3=\frac{37}{36}$, which is curious.

Does anyone know how to evaluate $F_n$ in general, or something very similar to it (maybe by some mistake I ended up with a slightly wrong definition of $F_n$) that could shed some light into this situation?

Answer 1

This is a partial observation: The Littlewood-Richardson coefficient in your case is simply 1, if and only if $\nu/\lambda$ is a skew shape, and 0 otherwise. Hence, (by using more standard notation for dimensions) $$ F_n = \frac{1}{n!(n+1)!}\sum_{\lambda \vdash n} \sum_{\substack{\nu\vdash n+1 \\ \nu \supset \lambda}} (f^\lambda) (f^\nu) (M_\nu)^2. $$ Moreover, you also have that for $\nu \vdash n+1$, $$ f^\nu = \sum_{\substack{\mu\vdash n \\ \nu \supset \mu}} f^\mu, $$ since every SYT of shape $\nu$ can be constructed by adding a box to a SYT of shape $\mu$.

Rearranging a bit, $$ F_n = \frac{1}{n!(n+1)!}\sum_{\lambda\vdash n} f^\lambda \sum_{\substack{\nu\vdash n+1 \\ \nu \supset \lambda}} \left( \sum_{\substack{\mu\vdash n \\ \nu \supset \mu}} f^\mu\right) (M_\nu)^2. $$ Further rearranging, $$ F_n = \frac{1}{n!(n+1)!}\sum_{\lambda,\mu\vdash n} f^\lambda f^\mu \sum_{\substack{\nu\vdash n+1 \\ \nu \supset \lambda,\\ \nu \supset \mu}} (M_\nu)^2. $$

Now it might be a good idea to consider the separate cases $\lambda=\mu$, and $\lambda \neq \mu$. This might help a bit.

Answer 2

I realize now that, in the original quantity $$ F_n(\mu)=\frac{1}{n!(n+m)!}\sum_{\lambda\vdash n}\sum_{\nu\vdash n+m}C^{\nu}_{\lambda,\mu} d_\lambda d_\nu M_\nu^2 \delta_{D(\nu),D(\mu)},$$ I should have imposed the condition that the Durfee square of $\nu$ is equal to the Durfee square of $\mu$.

In the particular case $\mu=(1)$, mentioned in the question, this implies that $\nu$ is a hook, which of course simplifies things enormously. Once this condition is imposed, we indeed get $F_n((1))=1$, as I had expected.