Let $\Sigma$ be a surface (let's say oriented and of finite type). We can consider the configuration space $F(\Sigma,n)$ of $n$ ordered distinct points on $\Sigma$, i.e. $\Sigma^n\setminus \Delta$ where $\Delta$ is the "big diagonal". The cohomology of $F(\Sigma,n)$ can be computed in a very explicit way, as explained in Burt Totaro's paper "Configuration spaces of algebraic varieties": the Leray spectral sequence for $F(\Sigma,n) \hookrightarrow \Sigma^n$ degenerates after the first differential and can be written down in a concrete way, which gives us a completely explicit differential graded algebra whose cohomology is $H^\bullet(F(\Sigma,n))$.

Now fix a finite group $G$ and consider the space $F(\Sigma,G,n)$ which parametrizes $n$ points on $\Sigma$ and a principal $G$-bundle over the complement of the $n$ points on $\Sigma$. So we have a finite sheeted covering $F(\Sigma,G,n) \to F(\Sigma,n)$ such that the fiber over the point $(s_1,\ldots,s_n) \in \Sigma^n$ is the set $\mathrm{Hom}(\Pi_1(\Sigma \setminus \{s_1,\ldots,s_n\}),G)/G$, where $G$ acts on the set of maps by conjugation.

Q1. Is there a good way to describe or compute the cohomology of $F(\Sigma,G,n)$?

Q2. (A vaguer question.) Let's say $\Sigma$ is compact for simplicity and let $FM(\Sigma,n)$ be the Fulton-MacPherson compactification of $F(\Sigma,n)$. I think there is a natural compactification $$F(\Sigma,G,n) \hookrightarrow FM(\Sigma,G,n)$$ covering $F(\Sigma,n) \hookrightarrow FM(\Sigma,n)$ and such that the covering $FM(\Sigma,G,n) \to FM(\Sigma,n)$ ramifies along the boundary; $FM(\Sigma,G,n)$ should parametrize principal $G$-bundles which are allowed to ramify over the nodes. Is there a natural smaller compactification of $F(\Sigma,G,n)$ which is still smooth, analogous to the compactification $$F(\Sigma,n) \hookrightarrow \Sigma^n$$ (which of course is smaller than the Fulton-MacPherson)?

Craig Westerland suggests an alternative description of the cohomology of $F(\Sigma,G,n)$. The covering $p \colon F(\Sigma,G,n) \to F(\Sigma,n)$ satisfies $R^ip_\ast\mathbf Z =0$ for $i>0$, so $$ H^\bullet(F(\Sigma,G,n),\mathbf Z) = H^\bullet(F(\Sigma,n),p_\ast\mathbf Z). $$ Now $F(\Sigma,n)$ is a $K(\pi,1)$ space where $\pi$ is by definition the pure braid group on $n$ strands of the surface $\Sigma$, $P_n(\Sigma)$. Hence this cohomology is given by $$ H^\bullet(P_n(\Sigma), \mathbf Z [ \mathrm{hom}(\pi_1(\Sigma \setminus \{s_1,\ldots,s_n\},G)/G]).$$ Also the action of $P_n(\Sigma)$ is the restriction of an action of the $n$-strand surface braid group $B_n(\Sigma)$ (i.e. where the points are unordered).

The simplest example should be $\Sigma = \mathbf R^2$, where we get the usual pure braid group. Since the fundamental group of the punctured plane is free we find that $\mathrm{hom}(\pi_1(\Sigma \setminus \{s_1,\ldots,s_n\},G)/G = G^n/G$, where $G$ acts by elementwise conjugation on $G^n$. The action of the braid group $B_n$ can be written down in terms of Artin's generators $\sigma_i$: the element $\sigma_i$ acts by $$ (g_1,\ldots,g_i,g_{i+1},\ldots,g_n) \mapsto (g_1,\ldots,g_ig_{i+1}g_i^{-1},g_i,\ldots,g_n)$$ (which is well defined on equivalence classes modulo conjugation). Then I guess even these cohomology groups are hard to compute in general?

  • $\begingroup$ Do you even know the connected components of $F(\Sigma,G,n)$, i.e., $H_0$? I am not a geometric topologist, but my understanding is that this is already a fairly delicate question when $n$ is "small" compared to the genus of $\Sigma$ and $G$, cf. Deligne, Kluitman, thesis of Richard Hamilton, Gabai-Kazez, etc. $\endgroup$ Jan 14, 2013 at 13:57
  • $\begingroup$ Let me say a little more. There is a decomposition $F=\sqcup F_a$ where $a$ ranges over $G$-conjugacy classes of data $(H,g_1,\dots,g_n)$: $H$, the image subgroup of $\pi_1$, and $g_i$ the images of small loops about the punctures $s_1,\dots,s_n$ (considered up to the obvious action of $\mathfrak{S}_n$). However, even if you take $G$ to be the symmetric group $\mathfrak{S}_d$, and even if you take $a$ to the <I>most generic</I> datum, namely $H=\mathfrak{S}_d$ and $g_i$ are transpositions, it is still tricky to prove $F_a$ is connected. It is true if $n \gg_{g,d} 0$. $\endgroup$ Jan 14, 2013 at 14:10
  • $\begingroup$ Thanks for the comments! You're right, in general I have no clue even about $H_0$, so maybe the question is too ambitious. $\endgroup$ Jan 15, 2013 at 10:27

1 Answer 1


As Jason Starr mentions, the set of components of this space is somewhat difficult to pin down. I will work over the Riemann sphere $\Sigma =S^2$ for simplicity (it's already complicated enough). If you restrict your focus to the subspace $CF(\Sigma, G,n)$ consisting of branched covers $G$-covers which are connected, then for large $n$, there is a description of the set of components in terms of a certain quotient of the $H_2(G)$ by commutators. When $G = S_n$ (and monodromy is constrained to be given by transpositions), this case was studied by Clebsch and Hurwitz, with the result being that the space is connected. For more general $G$, this is an unpublished result of Conway-Parker, which was expanded upon by Fried-Völklein in the appendix to their paper "The inverse Galois problem and rational points on moduli spaces."

Computing the cohomology of this space is quite difficult. It follows from your description above that the cohomology is the same as the cohomology of the mapping class group of the surface being fibred over, with (nontrivial twisted) coefficients in the free abelian group generated by $Hom(\pi_1(\Sigma \setminus \{ s_1, \dots, s_n\}), G)$; i.e.,

$$H^\ast(F(\Sigma, G, n)) \cong H^\ast(mcg(\Sigma); \mathbb{Z}Hom(\pi_1(\Sigma \setminus \{ s_1, \dots, s_n\}), G)).$$

Note that you can identify the cohomology of individual components as the group cohomology with coefficients in the subrepresentation generated by a given orbit of homomorphisms.

Abstractly, this is nice, but it doesn't actually tell you what the groups are. One thing that you could hope for is a form of homological stability (i.e., that the homology of a single component stabilizes as $n \to \infty$), as it does for the configuration space (here I'm using the unordered configuration space, not the ordered one). Then, if you figure out the limiting homology, you can at least compute some of the homology groups of $F(\Sigma, G, n)$ (those to which your stability theorem applies).

To advertise some recent work of Ellenberg, Venkatesh, and myself, in "Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields, I and II," we have proven a rational homological stability result for certain choices of $G$ (also, we often restrict the sort of monodromy around branch points that is allowed to lie in a given collection of conjugacy classes), as well as giving a method of determining the limiting homology.

It's far from an optimal answer: we can give you partial, rational information only under certain assumptions on $G$. Also, if you're really after the ordered moduli space (we treat the unordered), our results don't quite answer your question (though I think that you can bootstrap them to get some information at least). Any improvements in the computations of the cohomology of these spaces would be very interesting indeed.

  • $\begingroup$ Also, compactifications of these spaces do exist; see, for instance, the work of Abramovich-Corti-Vistoli, amongst others. $\endgroup$ Jan 14, 2013 at 19:18
  • $\begingroup$ Thanks for your comments, they are useful! But I guess you mean the surface braid group of $\Sigma$, not its mapping class group? I tried to work out the details to see if I understood things, but it was too long for a comment so I added it to my question. $\endgroup$ Jan 15, 2013 at 10:23
  • $\begingroup$ Yeah, sorry, that's right. The setup in my response is appropriate for computing the homology of the moduli space of these branched covers; your setup (fibering over the configuration space) will only use the braid group. And indeed, computing the homology of these spaces even when $\Sigma = \mathbb{R}^2$ is really quite difficult for non-abelian $G$. $\endgroup$ Jan 15, 2013 at 11:23

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