Motivation $\newcommand{\T}{\mathscr{T}}$

I have many times found myself saying some variant of the following. Let $\T_g$ be the Teichmüller space of a surface of genus $g$, and $\Gamma_g$ its mapping class group. The quotient $\T_g/\Gamma_g$ is the moduli space of curves $M_g$. It is a deep fact that $\T_g$ is in fact diffeomorphic a ball, which implies that $M_g$ is a model for the classifying space $B\Gamma_g$ of the mapping class group. In particular, the cohomology of $M_g$ is just the group cohomology of $\Gamma_g$.

...well, almost. Since $\Gamma_g$ does not act freely, $M_g$ is in fact not a $B\Gamma_g$. However, all the stabilizers are finite groups, and this implies via a spectral sequence argument that the rational cohomology of $M_g$ coincides with the rational cohomology of $\Gamma_g$.

Most algebraic geometers seem to ignore these issues by working instead with stacks or orbifolds. Indeed, the stack quotient $[\T_g/\Gamma_g]$ is the moduli stack of curves $\mathcal M_g$, which is in any case the more fundamental object of study.


My question is whether the topological arguments in the first two paragraph can be carried out in a more highbrow way using orbifolds or topological stacks. I am vaguely aware that Noohi's work on topological stacks includes setting up a homotopy theory of topological stacks, but I know almost nothing about any of this. So the question should be interpreted as "does there exist a developed homotopy theory of topological stacks where the following question can be asked and answered".

Question 1. Are any two quotients $[E/G]$ and $[E'/G]$, where $E$ and $E'$ are contractible spaces with a not necessarily free group action by $G$, homotopy equivalent as topological stacks?

My second question is more speculative since I know even less about rational homotopy theory.

Question 2. Let $\mathcal X$ be a topological stack, with coarse moduli space $X$. Suppose that all isotropy groups of $\mathcal X$ are torsion. Is $\mathcal X \to X$ a rational homotopy equivalence?

  • $\begingroup$ This is only tangentially relevant, but Teichmueller space is not in fact a complex ball (as is usually defined). It is a complex manifold diffeomorphic to a real ball. $\endgroup$
    – naf
    Aug 15, 2012 at 12:15
  • $\begingroup$ I don't think a positive answer to Q1 is reasonable since the analogue for spaces is false. $\endgroup$ Aug 15, 2012 at 13:01
  • $\begingroup$ Jeff answer below corrects my intuition. I should have thought twice about Q1. Indeed, stacks are designed to make all actions (virtually) free! $\endgroup$ Aug 15, 2012 at 15:12

1 Answer 1


Here is a simple way to talk about the homotopy type of a stack. Let $\mathfrak{X}$ be a stack and $f: U \to \mathfrak{X}$ a representable surjective submersion (an atlas) from a space $U$ (e.g., the map from the Teichmuller space to the moduli stack.) Now, form the pullback of $f$ along itself: $U\times_{\mathfrak{X}} U$. This comes with two maps to $U$ and there is a diagonal map from $U \to U\times_{\mathfrak{X}} U$. All together, these maps give a topological groupoid. The nerve of this topological groupoid is a simplicial space and the geometric realization of which is a space that one can regard as representing the homotopy type of the stack.

Here are some easy/nice properties of the above notion of homotopy type that are easy to check.

  1. Given a space $X$ with a $G$ action, the homotopy type of $[X/G]$ is the Borel construciton, aka homotopy quotient, $EG \times_G X$. In particular, the answer to your Question 1 is affirmative, and the homotopy type of the moduli stack of curves is exactly $B\Gamma_g$.

  2. One can define singular and de Rham cohomology of a stack and these invariants coincide with the integral and rational cohomology of the homotopy type of the stack. This is in fact almost a tautology since, for example, the de Rham cohomology can be defined by taking a covering by a manifold, forming iterated pullbacks (to produce a simplicial manifold), taking the de Rham algebra of this simplicial manifold to get a cosimplicial dga, and then taking the totalization to get a dga.

  3. It follows from property 1 above that the answer to your question 2 is also affirmative.

  4. [Edit] This notion of homotopy type is well-defined because one can check that any two atlases determine Morita equivalent topological groupoids which then have weakly equivalent nerves.

If I remember correctly, Noohi uses a slightly more sophisticated notion of homotopy type. He defines a universal weak equivalence to be a representable morphism from a space $U$ to a stack $\mathfrak{X}$ such that the pullback along any morphism from a space to $\mathfrak{X}$ is a weak equivalence. $U$ can then be regarded as the homotopy type of the stack. I think this is more of less equivalent to the naive version I explained above, but it has the advantage of being a bit more functorial and there might be some other technical advantages I can't remember. David Carchedi will probably be able to give more details.

  • $\begingroup$ Thanks! This definition of homotopy type does make things clear. $\endgroup$ Aug 15, 2012 at 17:26
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    $\begingroup$ can I ask a naive question? If I start with a (complex) algebraic stack X then I imagine I'd get an associated (analytic/topological) stack X(C) and run the same argument to obtain a homotopy type. Are there any pathologies when the stack X is singular (as an algebraic stack)? $\endgroup$
    – Jacob Bell
    Aug 16, 2012 at 8:19
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    $\begingroup$ Also, are there any good references for the cohomology of simplicial spaces? $\endgroup$
    – Jacob Bell
    Aug 16, 2012 at 8:19
  • $\begingroup$ @Jacob, I should have been a little more careful. I was starting with a stack in the category of smooth manifolds (or topological spaces). You're correct in that if we start with a complex algebraic stack then we should first pass to the associated analytic stack. If it is nonsingular then forgetting the complex analytic structure gives a stack in manifolds. If it is singular then you still get a stack in topological spaces, which still has a homotopy type as I described The only part where you need manifolds is for de Rham forms. $\endgroup$ Aug 16, 2012 at 8:43
  • $\begingroup$ I was going to say, but you clarified in your comment, that the original question never mentions manifolds, as far as I can tell, whereas you do use de Rham forms. But for stacks in the category of manifolds, I totally agree with what you say there. $\endgroup$ Aug 21, 2012 at 17:08

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