Generalisations of the Harer–Zagier recursion for 1-point functions

Abstract

Harer and Zagier proved a recursion to enumerate gluings of a 2d-gon that result in an orientable genus g surface, in their work on Euler characteristics of moduli spaces of curves. Analogous results have been discovered for other enumerative problems, so it is natural to pose the following question: how large is the family of problems for which these so-called 1-point recursions exist? In this paper, we prove the existence of 1-point recursions for a class of enumerative problems that have Schur function expansions. In particular, we recover the Harer–Zagier recursion, but our methodology also applies to the enumeration of dessins d’enfant, to Bousquet-Mélou–Schaeffer numbers, to monotone Hurwitz numbers, and more. On the other hand, we prove that there is no 1-point recursion that governs single Hurwitz numbers. Our results are effective in the sense that one can explicitly compute particular instances of 1-point recursions, and we provide several examples. We conclude the paper with a brief discussion and a conjecture relating 1-point recursions to the theory of topological recursion.

Example Maple code for 1-point recursions

In Example 3, we asserted that a 1-point recursion for monotone Hurwitz numbers could be derived from several lines of code, using the gfun package for Maple [58]. We reproduce such code below, which may be adapted for other enumerative problems.

We next provide some explanatory notes to indicate how the code above produces the desired 1-point recursion. Recall that monotone Hurwitz numbers satisfy the relation

$$\begin{aligned} m(d) = \sum _{g=0}^\infty m_g(d) \,\hbar ^{2g-1} = \frac{1}{d} \sum _{k=1}^d u_k \, v_{d-k}, \end{aligned}$$

where $\frac{u_{k+1}}{u_k} = \frac{G(k\hbar )}{k\hbar }$ and $\frac{v_{k+1}}{v_k} = -\frac{G(-(k+1)\hbar )}{(k+1) \hbar }$.

• Line 1 loads the gfun package into Maple.

• Line 2 defines the weight generating function G(z) that produces monotone Hurwitz numbers.

• Line 3 expresses the recursion satisfied by the sequence $u_0, u_1, u_2, \ldots$ above.

• Line 4 expresses the recursion satisfied by the sequence $v_0, v_1, v_2, \ldots$ above.

• Line 5 expresses the recursion satisfied by the sequence $\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \ldots$.

• Line 6 determines a recursion for the Cauchy product of the sequences $u_0, u_1, u_2, \ldots$ and $v_0, v_1, v_2, \ldots$.

• Line 7 determines a recursion for the Hadamard product of the Cauchy product from the previous line and the sequence $\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \ldots$.

• The output asserts that

  $$\begin{aligned} (-2\hbar +4d\hbar ) \, m(d) + (-d-1+\hbar ^2 d^3 + \hbar ^2 d^2) \, m(d+1) = 0. \end{aligned}$$

  By collecting the coefficient of $h^{2g-1}$ and shifting the index, we obtain the 1-point recursion

  $$\begin{aligned} d \, m_g(d) = 2(2d-3) \, m_g(d-1) + d(d-1)^2 \, m_{g-1}(d). \end{aligned}$$