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The $\tau$-function $H^\circ \big(t ;\vec{x} \big)$ associated with counting simple Hurwitz numbers is the formal power series

\begin{equation} (\dagger) \quad H^\circ \big(t ;\vec{x} \big) \, = \, \sum_{\lambda} {\mathrm{dim}(\lambda) \over {|\lambda|!}} s_\lambda\big(\vec{x}\big) \, \mathrm{exp} \big(\mathrm{\bf c}(\lambda) \, t\big) \end{equation}

where the sum is taken over all integer partitions $\lambda$, where $s_\lambda\big(\vec{x}\big)$ is the Schur function associated to $\lambda$ in the variables $\vec{x}=(x_1, x_2, x_3, \dots)$, where $\dim(\lambda)$ is the number of standard tableaux of shape $\lambda$, and $\mathrm{\bf c}(\lambda):= \sum_{\Box \, \in \, \lambda} c(\Box)$ is the sum of contents of $\lambda$'s Young diagram.

Applying the Pieri rule when $t = 0$ we get

\begin{equation} (\dagger\dagger) \quad \sum_{\lambda} {\mathrm{dim}(\lambda) \over {|\lambda|!}} s_\lambda\big(\vec{x}\big) \ = \ \mathrm{exp} \big\{ s_{\Box}\big(\vec{x} \big) \big\} \end{equation}

where $s_{\Box} \big(\vec{x} \big) = \sum_{i \geq 1} x_i$ is the Schur function associated with the partition $\Box = (1)$.

Of course the exponential factor in $H^\circ \big(t ;\vec{x} \big)$ drops out for those partitions $\lambda$ with total content $\mathrm{\bf c}(\lambda) = 0$, namely the self-conjugate partitions.

Question: Is the a closed formula for

\begin{equation} \sum_{\lambda \, = \, \lambda'} {\mathrm{dim}(\lambda) \over {|\lambda|!}} s_\lambda\big(\vec{x}\big) \end{equation}

where the sum is taken over all self-conjugate partitions?

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  • $\begingroup$ Please could you add a reference for $(\dagger)$? Do you know if there is a closed form under any non-trivial specializations, for example, $x_1 = 1, x_2 = q,\ldots, x_m=q^{m-1}, x_{m+1} = \ldots = 0$? $\endgroup$
    – Mark Wildon
    Commented Nov 25, 2022 at 9:05
  • $\begingroup$ Dear Mark, I don't know about specialization(s), and I would also be very happy to find out if nice closed form expression(s) exist in such case(s). Formula $(\dagger)$ can be found on page 13 in the following paper of M. Kazarian and S. Lando, see arxiv.org/abs/1512.07172 $\endgroup$
    – Jeanne Scott
    Commented Nov 25, 2022 at 17:25
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    $\begingroup$ $(\dagger\dagger)$ follows from $\sum_{|\lambda| = n} \mathrm{dim}(\lambda) s_\lambda(\vec{x}) = s_{\Box}(\vec{x})^n$. The corresponding terms when grouping by $|\lambda|$ for self-conjugate partitions don't look straightforward in any of Sage's standard bases for symmetric functions. $\endgroup$
    – Peter Taylor
    Commented Nov 26, 2022 at 19:16
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    $\begingroup$ I don't know of any symmetric function identities that would shed light on your problem. For $\sum_{\lambda=\lambda'} (-1)^{\frac 12(|\lambda|+\mathrm{rank}(\lambda))}s_\lambda$, see Exercise 7.29(a) of Enumerative Combinatorics, volume 2. For $\sum_{\lambda}s_\lambda$, where $\lambda$ ranges over certain partitions "close to self-conjugate," see Exercise 7.29(b). $\endgroup$
    – Richard Stanley
    Commented Nov 27, 2022 at 23:01

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