A sum involving irreducible characters of the symmetric group

For any irreducible character $\chi^\lambda$ of $S_{2n}$, the value $\chi^\lambda([2^n])$ can be computed as follows. If $\lambda$ has a nonempty 2-core, then $\chi^\lambda([2^n])=0$. Otherwise $\lambda$ has a 2-quotient $(\mu,\nu)$, where $\mu$ and $\nu$ are partitions satisfying $|\mu|+|\nu|=n$. Say $\mu$ is a partition of $k$, so $\nu$ is a partition of $n-k$. Then $\chi^\lambda([2^n])=\pm{n\choose k}f^\mu f^\nu$, where $f^\rho$ denotes the dimension of the irrep of $S_n$ indexed by $\rho$ (given explicitly by the hook-length formula). The sign is $(-1)^{m/2}$, where $\lambda$ has $m$ odd parts. This result follow from the theory of cores and quotients, e.g., Section 2.7 of James and Kerber, The Representation Theory of the Symmetric Group.

Hi Thomas,

This is very similar to things I have been working on or thinking about. I would be very interested to know the source of this problem. My own source for similar questions was in Random Matrix Theory. What do you want to know about those sums?

First a remark. In your definition of H(n,L), you could see this as a sum over all partitions of $2n$. The term $ \chi^{Y_{i,j,w}}(\tau)$ ensures only the diagrams you describe give non-zero contribution.

You might know this, but the $-p+q$ is the content of the cell $(p,q)$.

You can compute $\chi^{Y_{i,j,w}}([2^n])$ via the Murnaghan-Nakayama rule indeed: it is counting all the different ways to express $Y_{i,j,w}$ as an incremental union of 2-ribbons (with signs!). Another way uses the so-called rim hook lattice. If the 2-core $Y_{i,j,w}$ is non-empty, this character value is 0. If it is empty, take the two 2-quotients of $Y_{i,j,w}$ and compute the dimension of each. Combining this with a binomial sum, you can obtain the character value you are looking for. [*]

Finally, I would send you to Stanley's recent paper (look at the version on his homepage http://math.mit.edu/~rstan/papers.html#hooks ) called ''Some combinatorial properties of hook lengths, contents, and parts of partitions'' There he has sums over partitions of functions of the contents (like you do), but weighted by the Plancherel measure. I think you can make that measure aappear out of your sum. His result would typically take the form of a polynomiality in $n$ of the $L^i$ coefficient of $H(n,L)$.

[Edit], concerning part [*]: This is essentially what's done in the accepted answer to this question: Statistics of irreps of S_n that can be read off the Young diagram, and consequences of Kerov-Vershik (I was describing the Fomin-Lulov method, while that answer to the MO question describes their result).