Let $X$ be a CW-complex with contractible universal cover $\tilde{X}$ and fundamental group $\pi = \pi_1X$. Twisted (co)homology is found by lifting the cell structure on $X$ to a $\pi$-invariant structure on $\tilde{X}$, regarding the cell complex as a chain complex of $\mathbb{Z}\pi$ modules, and computing its (co)homology.

I am interested in learning whether there is an analogous construction for Cech cohomology, and if so, where I can read about it. Let $X$ now be a manifold and replace the cells in the previous construction by a suitably fine open cover (say, one with all intersections contractible). Similarly to the case of twisted cohomology, the cover lifts to a $\pi$-invariant cover on $\tilde{X}$. Let $F = \mathcal{O}(E)$ be the sheaf of (say) germs of sections of some smooth vector bundle over $X$; this lifts to a sheaf over $\tilde{X}$.

I would like to know if this cover-lifting is used to define a "twisting" of Cech cohomology on $\tilde{X}$ with coefficients in $F$. If so, where can I study it?

As background: my interest arises from studying deformations of geometric structures (in the sense of Thurston). Deformations are often naturally expressed by first cohomology with coefficients in an appropriate sheaf. I have run into a situation where I would like to perform some sort of twisting as described above (although over the model space, not the universal cover - hence thinking in terms of open covers) and if it exists I would like to learn the theory rather than trying to invent it from scratch.

  • $\begingroup$ If you don't receive an answer here, this would probably be appropriate for MathOverflow. (I'm sure you know this, and I'm only reminding you because I want to see an answer too!) $\endgroup$
    – Potato
    Commented Feb 8, 2013 at 0:41
  • $\begingroup$ Do you want specifically this way of twisting? Because you can just immediately use a local-coefficients system. $\endgroup$ Commented Feb 8, 2013 at 0:45
  • $\begingroup$ @ChrisGerig If using a local-coefficients system is essentially what I'm describing, I'd probably accept that as an answer. However, I want to eventually compute (co)homology for a $(G,X)$-structure on the model space $X$, which is not in general the same as the universal cover of the given $(G,X)$-space. $\endgroup$
    – Neal
    Commented Feb 11, 2013 at 21:43

1 Answer 1


Local coefficient homology is a particular case of sheaf cohomology (cohomology of a locally constant sheaf). So, even if I'm not sure I understood precisely your wishes, I think it is possible that what you're trying to do (twisting sheaf cohomology) really amounts to consider sheaf cohomology for something like a tensor product $\mathcal F \otimes \mathcal L$, where $\mathcal L$ is the locally constant sheaf corresponding to your local coefficients. If that's true, you may be happy with the genuine Čech cohomology of that particular sheaf.

Two classical references for sheaf cohomology for topologists are Iversen's Cohomology of Sheaves and Dimca's Sheaves in Topology. The latter is considerably terser than the former, but easier to find. In particular, those books deal with a general expression of Poincaré duality in sheaf cohomology (Poincaré-Verdier duality) which requires twisting (by the orientation local coefficient system $\mathcal L_{\textrm{or}}$) so they will spend some time explaining things quite close to the things you seem to dream about.

A quite ancient reference from the Séminaire Cartan by Frenkel alludes to a construction which seems related, but I must say it seems quite opaque to me.

(Even if it's only distantly related to your question, I'd like to take this opportunity to quote two books which break the omertà on local coefficients: G.W. Whitehead's Elements of homotopy theory and Davis & Kirk's Lectures on Algebraic Topology. Neither of their account on this topic is exhaustive or perfect, but at least, they do not content themselves with a two-line remark.)

  • 1
    $\begingroup$ Thanks! Let me digest your answer. I've been reading and re-reading chapter 5 of Davis-Kirk (in fact, it's open in front of me right now). $\endgroup$
    – Neal
    Commented Feb 14, 2013 at 5:01
  • $\begingroup$ Okay, I think this gives enough direction to warrant awarding the bounty (with 5 minutes left!). Thank you, especially for the references! $\endgroup$
    – Neal
    Commented Feb 14, 2013 at 14:25

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