# Hurwitz numbers and $t$-cores

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:

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

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 $$S_d$$)

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

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

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

whose logarithm has a "genus" expansion

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

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:

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

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.

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

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

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

Furthermore

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

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

$$$$\lambda \stackrel{\phi}{\Longleftrightarrow} \vec{n}$$$$

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

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

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

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

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

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

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

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

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

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

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 $$H_3(x \, ; q)$$

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

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

• We have $\nu_T(\lambda)=\sum {\lambda_i\choose 2}-\sum {\lambda'_i\choose 2}$, though I don't know if this is useful. Commented Mar 29, 2021 at 1:31
• 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. Commented Mar 29, 2021 at 2:45
• 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/… Commented Mar 29, 2021 at 19:40
• 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? Commented Mar 29, 2021 at 21:42
• 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 $S_d$.
– JSE
Commented Apr 1, 2021 at 2:33

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

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

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

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

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

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

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

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

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:

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

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:

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

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

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

and thus, when taken altogether, we get

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

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