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For integers $k \geq 0$ and $d \geq 1$ let $H(k,d)$ be the Hurwitz number which, for the purposes of this posting, will be defined by:

\begin{equation} H(k,d) \, := \ d! \, \sum_{\lambda \, \vdash d} \, \nu_{\scriptscriptstyle T}^k(\lambda) \ \ \text{where} \ \nu_{\scriptscriptstyle T}(\lambda) := \binom{d}{2} \cdot {{\chi^\lambda_{\scriptscriptstyle T}} \over {\dim(\lambda)} } \end{equation}

and where $\chi^\lambda$ is the character value of the irreducible representation $V_\lambda$ of the symmetric group $S_d$ corresponding to the partition $\lambda \vdash d$ evaluated at any representative transposition (taken from the conjugacy class $T$ of all transpositions) and where $\dim(\lambda)$ is the dimension of $V_\lambda$. The Hurwitz number $H(k,d)$ can be interpreted, using the Verlinde formula, as counting the the number of homomorphisms (up to conjugation in $S_d$)

\begin{equation} \rho : \pi_1 \Big( \Bbb{T}^2_k , \, \mathrm{base \, point}\Big) \longrightarrow S_d \end{equation}

where $\Bbb{T}^2_k$ is the 2-torus with $k$ punctures. We can assemble these Hurwitz numbers into the following bivariate generating function

\begin{equation} \begin{array}{ll} H(x;q) &\displaystyle = \ 1 \ + \ \sum_{d \geq 1} \, \sum_{k \geq 0} \, H(k,d) \, {x^k \over {k!}} \, q^d \\ &\displaystyle = \ 1 \ + \ \sum_{\lambda \ne \emptyset} \, q^{|\lambda|} \, \exp \big\{ x \, \nu_{\scriptscriptstyle T}(\lambda) \big\} \end{array} \end{equation}

whose logarithm has a "genus" expansion

\begin{equation} \log H(x;\tau) = F_1(\tau) \ + \ \sum_{g \geq 2} \, F_g(\tau) \, {x^{2g-2} \over {(2g-2)!}} \end{equation}

where we set $q = e^{2\pi i \tau}$ and each $\tau$-series $F_g(\tau)$ is known to be a quasi-modular form.

Now let $t \geq 2$ be an integer and let us consider the following $t$-core analogues:

\begin{equation} \begin{array}{l} \displaystyle H_t(x;q) \, := \ 1 \ + \ \sum_{\stackrel{\scriptstyle \text{$t$-cores}}{\lambda \,\ne \, \emptyset}} \, q^{|\lambda|} \, \exp \ \big\{ x \, \nu_{\scriptscriptstyle T}(\lambda) \big\} \\ \displaystyle F_{g; \, t}(\tau) \, := \ \text{ the coefficient of} \ {x^{2g-2} \over {(2g-2)!}} \ \text{in} \ \log H_t(x;\tau) \end{array} \end{equation}

Question 1: Does the generating function $H_t(x;q)$ have a nice closed expression, e.g. some sort of product formula?

Question 2: Does the $\tau$-series $F_{g; \, t}(\tau)$ have any kind of modular property?

thanks, ines.

Post Script: As a kind of stupid example, consider the case of $2$-cores, which are precisely the stair-case partitions. The Murnaghan-Nakayama rule tells us that $\chi^\lambda_{\scriptscriptstyle T}$ can be evaluated recursively as the (signed) sum of dimensions $\dim(\mu)$ of partitions $\mu \vdash |\lambda| -2$ obtained by removing skew-hooks of size $2$ from the border of $\lambda$, i.e.

\begin{equation} \chi^\lambda_{\scriptscriptstyle T} \ = \ \sum_{\stackrel{\scriptstyle \lambda \, = \, \mu + \sigma}{\sigma \, \vdash \, 2}} \, \big(-1 \big)^{\#(\sigma)-1} \, \dim(\mu) \end{equation}

where $\#(\sigma)$ is the number of parts of $\sigma$. Of course there are no skew-hooks of size $2$ which can be excised from a stair-case partition, so $\chi^\lambda_{\scriptscriptstyle T} = 0$ for any $2$-core partition $\lambda$ and consequently

\begin{equation} \begin{array}{ll} H_2(x \, ;q) &\displaystyle = \ 1 \ + \ \sum_{\stackrel{\scriptstyle \text{$2$-cores}}{\lambda \, \ne \, \emptyset}} \, q^{|\lambda|} \\ &\displaystyle = \ 1 \ + \ \sum_{d \geq 1} \, q^{{1 \over 2}d(d+1)} \\ &\displaystyle = \ \prod_{d \geq 1} \big(1 - q^{2d}\big) \cdot \big(1 + q^d \big) \end{array} \end{equation}

Furthermore

\begin{equation} \begin{array}{ll} \log H_2(x \, ; q) &\displaystyle = \ \sum_{d \geq 1} \, \log(1 - q^{2d}) \, + \, \log(1 + q^d) \\ &\displaystyle = \ F_{1 ; 2}(\tau) \end{array} \end{equation}

The interesting computation begins with $3$-cores.

Post-Post Script: One possible approach to the problem may be to take advantage of the Garvan-Kim-Stanton correspondence (GKS for short) which is a bijection

\begin{equation} \lambda \stackrel{\phi}{\Longleftrightarrow} \vec{n} \end{equation}

between $t$-cores $\lambda$ and integer vectors $\vec{n} = \big(n_0, n_1, \dots, n_{t-1} \big)$ with zero coordinate sum $n_0 + \cdots + n_{t-1} = 0$ such that

\begin{equation} |\lambda| \ = \ {t \over 2} \| \vec{n} \|^2 \, + \, \vec{b} \cdot \vec{n} \end{equation}

where $\vec{b} = \big(0 ,1 , \dots, t-1 \big)$. The trick might be to express the quantity $\nu_{\scriptscriptstyle T}(\lambda)$ in terms of the coordinates of the corresponding GKS-vector $\vec{n}$.

Consider the case of $3$-cores: If my understanding of O. Brunat and R. Nath's pointed abacus construction is correct (see https://arxiv.org/pdf/2101.01512.pdf) a $3$-core partition $\lambda$ with GKS-vector $\vec{n}= \big(n_0, n_1, n_2 \big)$ has an arm of length $3p + r$ with residue $0 \leq r \leq 2$ if and only if $n_r$ is positive and $0 \leq p \leq n_r - 1$. Likewise $\lambda$ will have a leg of length $3p + 2 - r$ with $0 \leq r \leq 2$ if and only if $n_r$ is negative and $0 \leq p \leq | n_r | - 1 $. As mentioned in the comments

\begin{equation} \begin{array}{ll} \nu_{\scriptscriptstyle T}(\lambda) &\displaystyle = \ {1 \over 2} \, \sum_{j=1}^k \, \Big(a_j + {1 \over 2} \Big)^2 - \Big(b_j + {1 \over 2} \Big)^2 \\ &\displaystyle = \ \sum_{j=1}^k \, a_j + {1 \over 2} a_j^2 \ - \ \sum_{j=1}^k b_j + {1 \over 2} b_j^2 \end{array} \end{equation}

where $a_j$ and $b_j$ are the respective $j$-th arm and length lengths of the partition $\lambda$. So it should be possible to write $H_3(x \, ; q)$ as a piecewise polynomial function of the GKS-coordinates $n_0$, $n_1$, $n_2$. As illustration consider the situation where $n_0 < 0$ and $n_1 \geq -n_0$ and $n_2 = -n_0 - n_1 \leq 0$ which is one of the of six possible sign configurations of the three GKS-coordinates $n_0$, $n_1$, and $n_2$. By the Brunat-Nath recipe only $n_1$ will contribute arm lengths while $n_0$ and $n_2$ will contribute leg lengths. The arm contribution to $\nu_{\scriptscriptstyle T}(\lambda)$ will be

\begin{equation} \begin{array}{ll} \displaystyle \sum_{j=1}^k a_j + {1 \over 2}a_j^2 &\displaystyle = \ \sum_{p=0}^{n_1 - 1} \, (3p+1) + {1 \over 2}(3p+1)^2 \\ &\displaystyle = \ {1 \over 2} 3n_1 \, + \, 3n_1(n_1-1) \, + \, {3 \over 4}n_1(n_1-1)(2n_1-1) \end{array} \end{equation}

while the leg contribution to $\nu_{\scriptscriptstyle T}(\lambda)$ will be

\begin{equation} \begin{array}{l} \displaystyle \sum_{j=1}^k b_j + {1 \over 2}b_j^2 \ \displaystyle = \ \left\{ \begin{array}{c} \displaystyle \sum_{p=0}^{|n_0| -1} \, (3p + 2) + {1 \over 2}(3p+ 2)^2 \\ + \\ \displaystyle \sum_{p=0}^{|n_2| -1} \, (3p) + {1 \over 2}(3p)^2 \end{array} \right. \\ = \, \left\{ \begin{array}{c} \displaystyle -4n_0 \, + \, {9 \over 2}n_0(n_0+1) \, - \, {3 \over 4}n_0(n_0+1)(2n_0+1) \\ + \\ \displaystyle {3 \over 2}n_2(n_2+1) \, - \, {3 \over 4}n_2(n_2+1)(2n_2+1) \end{array} \right. \\ = \, \left\{ \begin{array}{c} \displaystyle -4n_0 \, + \, {9 \over 2}n_0(n_0+1) \, - \, {3 \over 4}n_0(n_0+1)(2n_0+1) \\ + \\ \displaystyle -{3 \over 2}(n_0+n_1)(1-n_0-n_1) \, + \, {3 \over 4}(n_0+n_1)(1-n_0 -n_1)(1-2n_0 - 2n_1) \end{array} \right. \\ \end{array} \end{equation}

Taking the difference of the arm and leg contributions we get the value of $\nu_{\scriptscriptstyle T}(\lambda)$ namely

\begin{equation} \nu_{\scriptscriptstyle T}(\lambda) \ = \ {1 \over 2}\Big(3n_1^2 - 3n_0^2\big(1 + 3n_1\big) + n_0\big(2 + 3n_1 - 9n_1^2 \big) \Big) \end{equation}

So the lattice points of the cone in $\Bbb{Z}^2$ cut out by the inequalities $n_0 < 0$ and $n_1 \geq -n_0$ and $n_2 = -n_0 - n_1 \leq 0$ make the following contribution to $H_3(x \, ; q)$

\begin{equation} \displaystyle \sum_{n_0 < 0} \sum_{n_1 \geq -n_0} \, q^{3n_0^2 + 3n_1^2 + 3n_0n_1 - 2n_0 - n_1} \exp \Big\{ {x \over 2}\Big(3n_1^2 - 3n_0^2\big(1 + 3n_1\big) + n_0\big(2 + 3n_1 - 9n_1^2 \big) \Big) \Big\} \end{equation}

A similiar calculation can be undertaken for the remaining five cones in $\Bbb{Z}^2$. Does anyone recognize this kind of sum?

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    $\begingroup$ We have $\nu_T(\lambda)=\sum {\lambda_i\choose 2}-\sum {\lambda'_i\choose 2}$, though I don't know if this is useful. $\endgroup$ – Richard Stanley Mar 29 at 1:31
  • $\begingroup$ There's also $\nu_{\scriptscriptstyle T} (\lambda) = {1 \over 2} \sum_{i=1}^r \tilde{a}_i^2 - \tilde{b}_i^2$ where $\lambda = (a_1, \dots, a_r \| b_1, \dots, b_r)$ are the half-integer Frobenius coordinates but this probably is just a reformulation of what you wrote. $\endgroup$ – Ines Institoris Mar 29 at 2:45
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    $\begingroup$ The Murnaghan-Nakayama rule is cancellation-free whenever all strips have equal size, and then there is a hook-formula for computing the actual value: www2.math.upenn.edu/~peal/polynomials/… $\endgroup$ – Per Alexandersson Mar 29 at 19:40
  • $\begingroup$ Sorry Per, I don't follow. The Murnaghan-Nakayama rule only involves strips of equal size to begin with. Can you elaborate a bit more? $\endgroup$ – Ines Institoris Mar 29 at 21:42
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    $\begingroup$ This is a very naive question, but is your homomorphism $\rho$ written in the wrong direction? I would have thought it was the fundamental group of the punctured torus mapping to $S_d$. $\endgroup$ – JSE Apr 1 at 2:33
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This posting shows how to handle the case of $3$-cores and compute $H_3(x \, ; q)$.

Let's begin with the straight forward observation that $\lambda$ is a $t$-core if and only if its conjugate partition $\lambda'$ is a $t$-core. Furthermore, taking advantage of R. Stanley's comment

\begin{equation} \begin{array}{ll} \nu_{\scriptscriptstyle T}(\lambda) &\displaystyle = \ \eta( \lambda) \ - \ \eta(\lambda') \\ &\displaystyle = \ \sum_{i \geq 1} \, \binom{\lambda_i}{2} \ - \ \sum_{i \geq 1} \, \binom{\lambda_i'}{2} \end{array} \end{equation}

we see that $\nu_{\scriptscriptstyle T}(\lambda') = -\nu_{\scriptscriptstyle T}(\lambda)$. In particular $\nu_{\scriptscriptstyle T}(\lambda)$ vanishes when $\lambda$ is a self-conjugate partition. This means that

\begin{equation} H_t( x \, ; q) \ = \ 1 + \displaystyle \sum_{\ \lambda \, = \, \lambda'} \, q^{|\lambda|} \ + \displaystyle \sum_{\nu_{\scriptscriptstyle T}(\lambda) \, > \, 0} 2 \cosh \big\{ \nu_{\scriptscriptstyle T}(\lambda) \, x \big\} \, q^{|\lambda|} \end{equation}

where the sums are taken over non-empty $t$-core partitions. The self-conjugate $3$-cores are precisely those $3$-cores whose GKS-vectors are of the form $(a,0,-a)$ with $a \in \Bbb{Z}$. For GKS-vectors $(a,0,-a)$ with $a > 0$ the corresponding $3$-core partition $\lambda$ will have size $|\lambda| = a(3a-2)$ and first part $\lambda_1 = 3a-2$. For GKS-vectors $(-a,0,a)$ with $a \geq 0$ the corresponding $3$-core partition $\lambda$ will have size $|\lambda| = a(3a+2)$ and first part $\lambda_1 = 3a$. These calculation are made using the Brunat-Nath set-up mentioned in second post-script of my original post.

The set of $3$-cores can be arranged into the following triangular hierarchy as depicted on page 142 of these notes (https://qcpages.qc.cuny.edu/~chanusa/courses/636/14/notes/636fa14ch50.pdf) by Christopher Hanusa. After staring at the alcove pattern a bit, I'll guess that (1) a $3$-core partition is self-conjugate if and only if $\nu_{\scriptscriptstyle T}(\lambda) = 0$ and (2) any $3$-core partition with with positive $\nu_{\scriptscriptstyle T}$-value can be uniquely expressed as $\rho^k \cdot \lambda$ where $\lambda = \big(\lambda_1, \dots, \lambda_\ell \big)$ is a self-conjugate $3$-core partition, $k \geq 1$ is an integer, and $\rho \cdot \lambda := \big(\lambda_1 + 2, \lambda_1, \dots, \lambda_\ell \big)$. Here $\rho^k \cdot \lambda$ denotes the $k$-fold iteration of $\rho$.

Note that for a general partition $\lambda$ we have

\begin{equation} \begin{array}{rl} \displaystyle \big| \rho \cdot \lambda \big| &\displaystyle = \ |\lambda| \, + \, \lambda_1 +2 \\ \displaystyle \nu_{\scriptscriptstyle T}\big( \rho \cdot \lambda \big) &\displaystyle = \ \nu_{\scriptscriptstyle T}(\lambda) \, - \, |\lambda| \, + \, \binom{\lambda_1 +2}{2} \end{array} \end{equation}

and using the Faulhaber formulae it's not too hard to see that

\begin{equation} \begin{array}{rl} \big| \rho^k \cdot \lambda \big| &\displaystyle = \ |\lambda| \ + \ k\lambda_1 \ + \ k(k+1) \\ \displaystyle \nu_{\scriptscriptstyle T}\big(\rho^k \cdot \lambda \big) &\displaystyle = \ \left\{ \begin{array}{l} \displaystyle \ \ \ \, \nu_{\scriptscriptstyle T}(\lambda) \, - \, k|\lambda| \\ \displaystyle + \ {1 \over 2} \, k\lambda_1^2 \, + \, {1 \over 2} \, k(k+2)\lambda_1 \\ \displaystyle + \ {1 \over 6} \, k(k+1)(2k+1) \end{array} \right. \end{array} \end{equation}

If we iteratively apply $\rho$ to a self-conjugate $3$-core $\lambda$ with GKS-vector $(a,0,-a)$ with $a > 0$ and tally the total contribution to $H_3(x \, ; q)$ made by $\rho^k \cdot \lambda$ and its conjugate partition as $k \geq 1$ varies we get:

\begin{equation} \begin{array}{ll} G^+_a(x \, ; q) &\displaystyle := \ \sum_{k \geq 1} 2 \cosh \big\{ U_+(a,k) \, x \big\} \, q^{D_+(a,k)} \ \ \text{where} \\ U_+(a,k) &\displaystyle := \ \left\{ \begin{array}{l} \displaystyle \ \ \ \, {1 \over 2} \, (a-2)(3a-2)k \\ \displaystyle + \ {1 \over 2} \, (3a-2)k(k+2) \\ \displaystyle + \ {1 \over 6} \, k(k+1)(2k+1) \end{array} \right. \\ D_+(a,k) &\displaystyle = \ a(3a-2) \ + \ (3a-1)k \ + \ k^2 \end{array} \end{equation}

Similarly the total contribution to $H_3(x \, ; q)$ made by $\rho^k \cdot \lambda$ and its conjugate partition as $k \geq 1$ varies and where $\lambda$ is a self-conjugate $3$-core with GKS-vector $(-a,0,a)$ and $a \geq 0$ is:

\begin{equation} \begin{array}{ll} G^{-}_a(x \, ; q) &\displaystyle := \ \sum_{k \geq 1} 2 \cosh \big\{ U_{-}(a,k) \, x \big\} \, q^{D_{-}(a,k)} \ \ \text{where} \\ U_{-}(a,k) &\displaystyle := \ \left\{ \begin{array}{l} \displaystyle \ \ \ \, {1 \over 2} \, a(3a-4)k \\ \displaystyle + \ {3 \over 2} \, ak(k+2) \\ \displaystyle + \ {1 \over 6} \, k(k+1)(2k+1) \end{array} \right. \\ D_{-}(a,k) &\displaystyle = \ a(3a+2) \ + \ (3a+1)k \ + \ k^2 \end{array} \end{equation}

The self-conjugate $3$-cores on the own make a contribution of

\begin{equation} G(q) \ = \ 1 + \ \sum_{a \, > \, 0} \ q^{a(3a+2)} \ + \ q^{a(3a-2)} \end{equation}

and thus, when taken altogether, we get

\begin{equation} H_3(x \, ; q) \ = \ G(q) \ + \ \sum_{a > 0} G^+_a(x \, ; q ) \ + \sum_{a \geq 0} G^{-}_a(x \, ; q) \end{equation}

Perhaps the right-hand sum will be familiar to the readership of this posting.

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