Sheaf cohomology question - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:22:05Z http://mathoverflow.net/feeds/question/29380 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29380/sheaf-cohomology-question Sheaf cohomology question Victor L. 2010-06-24T14:09:23Z 2010-06-25T12:24:58Z <p>For a topological space $X$ and a sheaf of abelian groups $F$ on it sheaf cohomology $H^n(X,F)$ is defined.</p> <p>Singular cohomology of $X$ can be expressed as sheaf cohomology if $X$ is locally contractible and $F$ is the sheaf of locally constant functions.</p> <p>I have two related questions.</p> <p>For an algebraic scheme $X$ one uses the sheaf cohomology with the structure sheaf $F=\mathcal{O}_X$. What happens if $X$ is a topological manifold and $F$ is the sheaf of continuous functions to the real numbers? Or differentiable manifolds? Has this cohomology special name under which I can search for literature?</p> <p>Doesn't one get many interesting cohomology theories beside singular cohomology for a topological space $X$ from sheaf cohomology? I mean for a classical topological space, not a scheme, which is Hausdorff and so forth. Is there a list or an overview in the literature? Thank you.</p> <p>I like to restate the second question. Does any reasonable cohomology theory of topological spaces come from sheaf cohomology?</p> http://mathoverflow.net/questions/29380/sheaf-cohomology-question/29383#29383 Answer by Donu Arapura for Sheaf cohomology question Donu Arapura 2010-06-24T14:26:13Z 2010-06-25T12:24:58Z <p>For the first question: The higher cohomology of a paracompact Hausdorff space (e.g. manifold) in the sheaf of continuous real valued functions is zero, because the sheaf is <a href="http://en.wikipedia.org/wiki/Injective_sheaf#Fine_sheaves" rel="nofollow">fine</a>.</p> <p>If I understand the spirit of your second question correctly, then you can use locally constant sheaves or more generally constructible sheaves to get interesting "topological" cohomologies.</p> <p>A quick follow up: (1) The argument in the first paragraph also applies to sheaves of modules over the ring of continuous functions. (2) I don't know if every generalized cohomology theory (in the sense of algebraic topology) can or should be regarded as sheaf cohomology, but I leave to an expert to give a precise answer. On the flip side, I should point out that sheaf cohomology with <em>arbitrary</em> coefficients is generally not homotopy invariant, so it really is a different sort of beast.</p> http://mathoverflow.net/questions/29380/sheaf-cohomology-question/29388#29388 Answer by coudy for Sheaf cohomology question coudy 2010-06-24T15:01:17Z 2010-06-24T15:01:17Z <p>Sheaves of continuous or smooth functions with values in a vector bundle have zero cohomology groups because of the existence of <em>partitions of unity</em>. </p> <p>Still they provide interesting resolutions for the locally constant sheaf, e.g. by considering the bundle built from exterior algebras on the tangent space, and that's a way to prove that De Rham cohomology coincides with singular or Cech cohomology for paracompact manifolds.</p> <p>Of course, we can look at the De Rham complex obtained by tensoring the exterior forms by an arbitrary bundle on the manifold. In which case we obtain an interesting cohomology with "values" in a vector bundle instead of a coefficient ring. This is similar to singular cohomology with twisted coefficient. And I think that's the kind of examples that gave rise to the concept of sheaf.</p>