MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Actually, is it just the sheafification of the pullback of our sheaf on $E$ to a presheaf on $U$? – David Corwin Jul 5 '10 at 11:18
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. – BCnrd Jul 5 '10 at 14:44
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. – David Corwin May 2 '13 at 20:50
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.