Let $G$ be a finite group, $M$ a 2-dimensional manifold. There are

$$ \#\{ (a_1,b_1,a_2,b_2,\cdots,a_g,b_g) \in G^{2g}\,|\, \Pi_i[a_i,b_i]=1\} $$

many $G$ bundles over $M$ up to isomorphism, where $g$ is the genus of $M$.

There is only one bundle if $g=0$, and $|\mbox{Conj}(G)|$ many bundles if $g=1$. However, when $g$ is larger than $1$, things get trickier.

Even the easiest case $g=2$ and $G=S_3$ is hard. Are there ways to cleverly compute the number? If not in general, can I find the answer somewhere for some $g$ and some usual groups $S_n$ (symmetric), $A_n$ (alternative), $D_n$ (dihedral)..?


2 Answers 2


G-bundles are classified by $BG$. One can calculate that $[X,BG]_*=\operatorname{Hom}(\pi_1(X),G)$. Unpointed maps then correspond to $\operatorname{Hom}(\pi_1(X),G)/\sim$ where $\sim$ is the action of conjugation (in particular, you only have half the bundles you want in your genus 2 case). This agrees with your description in the surface case.

  • $\begingroup$ I appreciate your help! I should have been more clear but what I'm asking for is a concrete number: e.g. @Moishe's answer below. $\endgroup$
    – Student
    Apr 23, 2020 at 13:56

Computations of the size of $Hom(\pi_1(M), G)$ when $M$ is a surface and $G$ is finite were done in

M.Mulase, J.Yu, A generating function of the number of homomorphisms from a surface group into a finite group, math/0209008v1.

In particular, they reprove a pretty formula originally due to Mednykh:

$$\frac{|\text{Hom}(\pi_1(M), G)|}{|G|} = \sum_\lambda \left( \frac{\dim V_\lambda}{|G|} \right)^{\chi(M)}$$ where the sum is taken over all isomorphism classes of irreducible complex representations of $G$ ($V_\lambda$ is a complex vector space equipped with an irreducible representation $\lambda$ of $G$) and $M$ is closed, connected and orientable. They also give a formula in the case of non-orientable surfaces. In particular, given $G$, once you have the table of irreducible representations of $G$, you can compute the size of the set of homomorphisms $\pi_1(M)\to G$. Nice thing is that the dependence on the surface $M$ is rather weak: The hard part of the formula is entirely encoded in the representation theory of $G$ which you can look up in the given special case. For instance, in the case of dihedral groups $G=D_n$, see this question. If $n$ is odd and $M$ has genus $g$, we get $$ \frac{|\text{Hom}(\pi_1(M), G)|}{2n} = 2 \left( 2n \right)^{2g-2} + \frac{n+3}{2} n^{2g-2} $$

  • $\begingroup$ This is a beautiful formula! $\endgroup$
    – Stephen
    Apr 23, 2020 at 14:34
  • $\begingroup$ @Stephen: Yes, it is! By the way, the author of the formula (Alexander Mednykh) is still alive and you can email him to express your appreciation of his work. $\endgroup$ Apr 23, 2020 at 20:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .