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Let $\mathcal{O}$ be a finite-dimensional, paracompact, Hausdorff, smooth (and compact, if it helps) orbifold. Is there an isomorphism between the real Čech cohomology and singular cohomology of the underlying space $|\mathcal{O}|$?

(NB. The answers of Simon Rose and David C refer to an earlier version of the question and relate the Čech cohomology of the orbifold $\mathcal{O}$ to the singular cohomology of it's underlying space $|\mathcal{O}|$)

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  • $\begingroup$ Aren't Čech and singular cohomology isomorphic for paracompact Hausdorff spaces? (and you are only considering the cohomology of an underlying space, right?) cf mathoverflow.net/a/4229/4177 $\endgroup$
    – David Roberts ♦
    Commented Nov 10, 2016 at 7:12
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    $\begingroup$ @DavidRoberts, not even for metric compact spaces, far from it. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Nov 10, 2016 at 7:39
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    $\begingroup$ Can you clarify: are you trying to prove that the Cech cohomology of $\vert\mathcal O \vert$ is isomorphic to the singular cohomology of $\vert\mathcal O \vert$, or are you trying to compare the cohomologies of $\mathcal O$ and $\vert\mathcal O \vert$? In the latter case, the answers of David and Simon are on point: the cohomologies are isomorphic with coefficients in a field of characteristic zero, but not in general. In the former case, the statement is true with arbitrary coefficients: the underlying space of an orbifold is locally contractible, so Cech and singular cohomology always agree. $\endgroup$
    – Dan Petersen
    Commented Nov 10, 2016 at 10:24
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    $\begingroup$ @DanPetersen, DavidRoberts: Sorry it was unclear, I edited the question. I did not know that the cohomologies aways agree for locally contractible spaces. It makes the question a little silly, I gess. Anyway, I learned from that and thats the point so thanks! $\endgroup$
    – Caramello
    Commented Nov 10, 2016 at 18:07
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    $\begingroup$ @DanPetersen why not put the relevant part of your comment as an answer? Then Caramello can accept it. $\endgroup$
    – David Roberts ♦
    Commented Nov 11, 2016 at 0:58

3 Answers 3

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It's a classical result that if $X$ is paracompact and locally contractible, then singular cohomology and Cech cohomology of $X$ coincide, with coefficients in any abelian group. A reference is Spanier's textbook. This applies in particular to the underlying space of an orbifold, in which case a small neighborhood of any point is the cone on the link of the point.

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  • $\begingroup$ It is not the main question but can you please add where exactly in Spanier's book I can find that result? I tried seeing roughly but could not find it in that book.. $\endgroup$
    – Praphulla Koushik
    Commented Sep 10, 2018 at 20:23
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As a reference I recommend this paper by K. Behrend:

http://users.ictp.it/~pub_off/lectures/lns019/Behrend/Behrend.pdf

Cech cohomology and De Rham cohomology can be defined for differentiable stacks using a double complex associated to the underlying Lie groupoid. These two cohomologies are isomorphic (see remark 10 p. 262).

Singular cohomology is also defined in this paper (see p. 280), and for real coefficients we have a De Rham isomorphism (see p. 289).

Proposition 36 gives an isomorphism, for any Deligne-Mumford stack $\mathfrak{X}$, between singular cohomology $H^*(\mathfrak{X};\mathbb{Q})$ and the singular cohomology of its coarse moduli space $H^*(\overline{\mathfrak{X}};\mathbb{Q})$.

Example: consider the stack $BG$ where $G$ is a finite group, its coarse moduli space is just a pointy and you recover that rational coholology of $BG$ which is the cohomology of group of $G$ with rational coefficients is trivial.

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David C's answer gets this nicely, but another way to intuitively think of it is the following.

Choose a fine enough open cover $\{U_i\}_i$ of $\mathcal{O}$ such that each $U_i$ is isomorphic to $\mathbb{R}^N$ and such that the local groups are all sitting in $O(N)$. Then each $U_i$ and $U_i/G_i$ are contractible, and so all of the terms in your Čech complex are the same.

The maps may vary a little, but over $\mathbb{Q}$ you will get the same result up to isomorphism.

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  • $\begingroup$ What about the intersections $\ U_i\cap U_j\ $ etc. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Nov 10, 2016 at 8:14
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    $\begingroup$ Can't you just choose a fine enough refinement such that all intersections are contractible? $\endgroup$
    – Simon Rose
    Commented Nov 10, 2016 at 8:51
  • $\begingroup$ @SimonRose that is a special case of a project I'm vainly wishing to attract a masters student with (the fact my academic position is as tenuous as it can get isn't helping...) cf my question mathoverflow.net/questions/156142/…, which should cover orbifolds by taking almost free actions of compact Lie groups. $\endgroup$
    – David Roberts ♦
    Commented Nov 10, 2016 at 10:56

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