Let $E=\mathbb{C}/L$ be an elliptic curve. Then $\mathbb{C}$ is contractible, and $L$ is the fundamental group of $E$. What's interesting is that we can find the cohomology of $E$, which is the same as the sheaf cohomology $H^i(E,\mathbb{Z})$ for the constant sheaf $\mathbb{Z}$, by taking the group cohomology of $L$ with trivial action on $\mathbb{Z}$. What's more interesting is that if $\mathcal{O}^\times$ is the sheaf of invertible holomorphic functions on $E$, then we can find the sheaf cohomology of this by considering the group cohomology of the action of $L$ on the group of invertible holomorphic functions on $\mathbb{C}$ defined by $(\omega f)(z)=f(z+\omega)$ for $\omega \in L$. In the case of $H^0$, this is obvious, since the holomorphic functions which are fixed by $L$ ($H^0$ in group cohomology), which is the same as invertible holomorphic functions on $E$, which is the same as the global sections of the sheaf on $E$ (and this is true for many other sheaves). In the case of $H^1$, one can see that both are the group of invertible line bundles on $E$.

My question is: why is the group ($L$-module) of holomorphic functions on $\mathbb{C}$ the natural one to consider? In this case, it's clearer, since we know what holomorphic functions are. If in general we have a covering map $U \to X$ for some space $X$, with $U$ a contractible universal cover, then given a sheaf on $X$, how do we know what sheaf on $U$ to consider? Specifically, which sheaf on $U$ gives the sheaf cohomology when we take the group cohomology of its global sections?

  • $\begingroup$ Actually, is it just the sheafification of the pullback of our sheaf on $E$ to a presheaf on $U$? $\endgroup$ Jul 5, 2010 at 11:18
  • 2
    $\begingroup$ There is a nice appendix on all of this (with some minor typos) at the end of chapter 1 of Mumford's book on abelian varieties. $\endgroup$
    – BCnrd
    Jul 5, 2010 at 14:44
  • $\begingroup$ To expand on Brian's comment: The pullback of the sheaf $G_m$ to the universal cover has vanishing higher cohomology, and the results in Mumford then imply that group cohomology agrees with sheaf cohomology. $\endgroup$ May 2, 2013 at 20:50

2 Answers 2


I doubt that in general one can construct a reasonable sheaf on $U$ with the required properties. To see what kind of bad things can happen, let us try to understand why this works for $X$ an elliptic curve and the sheaf $\cal{O}^{\times}$ on it.

We have the derived global sections functor from the $D^b$ of sheaves on $X$ to the $D^b$ of sheaves on a point, i.e. graded vector spaces. But given a sheaf $F$ on $X$ we can compute its global sections in a roundabout way: we can first take the pullback to $U$, then take global sections and then take the $G$-invariants where $G=\pi_1(X)$. Passing to the derived categories we get $$R\Gamma (F)=R(R\Gamma f^{-1}(F))^G$$ where $F\in D^b(X)$, $f:U\to X$ is the projection, $R\Gamma f^{-1}$ is the right derived functor of the left exact functor $\Gamma f^{-1}$ and $R(\cdot)^G$ is the right derived functor of the functor of $G$-invariants (this functor goes from the $D^b$ of $G$-modules to graded vector spaces).

We have the Grothendieck spectral sequence that converges to $H^\ast(X,F)$ with the $E_2$ sheet given by $$E_2^{p,q}=H^p(G,H^q(U,f^{-1}(F)).$$

Now if $X$ is an elliptic curve, $F=\cal{O}^{\times}$ and $U=\mathbf{C}$, then it follows from the exponential exact sequence that $H^q(U,f^{-1}(F))=0$ for $q\neq 0$, so the above spectral sequence collapses and we get the required isomorphism $H^\ast (X,F)=H^\ast(G,H^0(U,f^{-1}(F))$. This also happens when say $F$ is locally constant and $U$ is contractible. But in general there seems no reason to expect the spectral sequence to collapse, let alone to be concentrated in one row only.


Short answer: I believe the answer is you can always do this, as long as you do it in the right way (which isn't always by taking global sections, so this complements the previous answer). I will focus only on a certain class of sheaves, however (though it contains most of the usual sheaves).

The observation you are making is essentially about Morita invariance of Lie groupoid cohomology. In this example, the Lie groupoids are $\mathbb{C}\rtimes L\rightrightarrows \mathbb{C}$ and $E\rightrightarrows E\,.$ These Lie groupids are Morita equivalent for the following reason: whenever you have a principal $G$-bundle $\pi:P\to X\,,$ the Lie groupoids $P\rtimes G\rightrightarrows P$ and $X\rightrightarrows X$ are Morita equivalent, via the map $(p,g)\mapsto \pi(p)\,.$

Now the cohomology of Lie groupoids with values in a module is invariant under Morita equivalence, and the topology of $X$ is irrelevant. To compute the cohomology of a Lie groupoid $G\rightrightarrows G^0\,,$ you need to form the nerve $B^{\bullet}G$, take the induced sheaf on each $B^iG\,,$ take a resolution of the sheaf on each $B^iG$ and then compute the cohomology of the total complex on $B^{\bullet}G\,.$

A module is like a Lie groupoid representation, except that the fibers over the base don't have to be vector spaces, they can be any abelian Lie group. The simplest examples of modules for a Lie groupoid $G\rightrightarrows G^0$ are ones of the form $G^0\times A\,,$ where $A$ is an abelian Lie group. A module comes with a compatible action of $G\,,$ and in this case there is a canonical one: if $g$ is an arrow between $x$ and $y\,,$ then $g$ sends $a$ over $x$ to $a$ over $y\,.$ Note that we get a sheaf by taking the sheaf of sections of the module.

Now, in the example you gave a major simplification occurs: since $H^i(\mathbb{C},\mathcal{O}^*)$ is trivial for $i>0\,,$ and since $L$ is discrete (meaning that the cohomology of $L$ with respect to the sheaf $\mathcal{O}^*$ is automatically trivial in positive degree) we don't have to take a resolution of the sheaf, ie. we can compute the sheaf cohomology using global sections, ie. the usual groupoid cocycles, the same as how it's done for smooth representations.

Now to get to your question about how do we know which sheaf to use: if you have a Morita equivalence of Lie groupoids $H\rightrightarrows H^0\,,$ $G\rightrightarrows G^0$ given by a homomorphism $f:H\to G\,,$ and if you have a module $M\to H^0\,,$ one can pullback the module to get a module over $G^0\,,$ ie. $f^*M\to G^0\,.$ This is the sheaf that you use. For the example where the module is $M=G^0\times A$ we simply have that $f^*M=H^0\times A\,.$

To connect this with your example: on $E$ we are taking cohomology in $\mathcal{O}^*\,,$ which is the sheaf of sections of $M=E\times \mathbb{C}^*\,.$ From the above then, we know that the module we need to use on $\mathbb{C}\rtimes L\rightrightarrows \mathbb{C}$ is just $\mathbb{C}\times \mathbb{C}^*\,,$ whose sheaf of sections is just $\mathcal{O}^*\,.$ Therefore, together with the fact about $\mathbb{C}$ having no higher cohomology in $\mathcal{O}^*\,,$ we get that the cohomology can be computed as the group cohomology of $L\rightrightarrows *$ in the naive way (ie. where no resolution is taken), and where we are taking cohomology of the module $H^0(\mathbb{C}\,,\mathcal{O}^*)$ (this is a module over $*$) with the action given by the one you wrote.

There's a lot more to say, but hopefully what I said is somewhat comprehensible. I wrote about this in my thesis, if you're interested in more details (also I know this is question is super old): https://arxiv.org/abs/2205.02109v2


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