# de Rham Cohomology of Non-Flat Bundle

Let $E$ be a smooth vector bundle on a smooth manifold $M$. If $E$ is flat, there is a connection $\nabla$ which is a differential which we can use to define the de Rham cohomology of $E$.

If $E$ is a non-flat bundle on $M$, is there any meaningful way to construct a de Rham cohomology of $E$?

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I suppose you want a cohomology theory defined as closed forms modulo exact ones. Your comment on flat bundles says this cannot happen for bundles that do not admit flat connections; there is no way to extend the $d$ operator to sections of the bundle in such a way that the square of the extension will be zero. The closest analogue is perhaps in complex geometry, where the $\bar\partial$ operator extends to sections of any holomorphic bundle, giving rise to Dolbeault cohomology groups $H^q(X,E)$. If there's a fancy version of this in general Riemannian geometry, then I haven't heard of it. –  Gunnar Magnusson Sep 24 '12 at 14:49
This is not to say that there aren't cohomology theories of non-flat bundles over a Riemannian (or smooth) manifold. Cech cohomology gives a perfectly fine cohomology theory on sheaves over a topological space, and I'm sure there are modern approaches that work even better. The only difficulty lies in wanting these theories to rise via differential forms. That they don't in general perhaps goes some way to explaining the success of Bochner-Kodaira-Nakano methods in complex geometry. But these are just idle thoughts, better suited for a blog post than anything else. –  Gunnar Magnusson Sep 24 '12 at 14:54
The relation between Cech and deRham cohomology is that the deRham cohomology of $(E, \nabla)$ is the Cech cohomology of the sheaf $\mathcal{E}$ where $\mathcal{E}(U)$ is $\nabla$-flat sections of $E$ over $U$. If $\nabla$ is not flat, I suppose you could still consider this sheaf, but it would be likely to just be the zero sheaf and thus have no interesting cohomology. –  David Speyer Sep 24 '12 at 19:09
@GunnarMagnusson: I was thinking of a cohomology theory defined using forms. I know that for a non-flat bundle you can't use a connection to define a complex, but I was wondering whether there is a way to twist a connection that would allow for such an approach. Alternatively, I imagine that you could use derived categories to define a cohomology theory for non-flat bundles, I'm just not sure how it would manifest itself geometrically. –  Michael Albanese Oct 2 '12 at 2:18