How many $G$ bundles are there over a surface?

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)..?

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.

• 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. 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 $$\frac{|\text{Hom}(\pi_1(M), G)|}{2n} = 2 \left( 2n \right)^{2g-2} + \frac{n+3}{2} n^{2g-2}$$

• This is a beautiful formula! Apr 23, 2020 at 14:34
• @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. Apr 23, 2020 at 20:54