Sheaf and Cech cohomology $H^*(X,\mathcal{F})$ (which give the same result when applied to good enough topological spaces) are a useful generalisation of the concepts of de Rham and Dolbeault cohomology, just by putting $\mathcal{F}=\mathbb{R}$ or the sheaf of holomorphic functions $\Omega^p(M)$. But each generalisation involves losing intuition about the measured geometric properties.

My question is: is it possible to understand the geometric intuition behind sheaf and Cech cohomology in the same way one can understand the geometry behind de Rham cohomology?

See also this related question.


closed as unclear what you're asking by abx, Stefan Kohl, José Figueroa-O'Farrill, Peter Crooks, Vidit Nanda Feb 18 '15 at 15:28

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    $\begingroup$ If you put $G=\mathbb{C}$ you do not get Dolbeault cohomology. You just get ordinary cohomology with coefficients in $\mathbb{C}$. $\endgroup$ – Steven Gubkin Feb 18 '15 at 12:38
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    $\begingroup$ If you work just a little bit with sheaf cohomology, you'll get very soon the intuition of what it is. $\endgroup$ – abx Feb 18 '15 at 13:19
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    $\begingroup$ Here is a computation which helped my intuition. Take a combinatorially triangulated manifold $X$. For each vertex $v$ of the triangulation, let $U(v)$ be the union of the interiors of all simplices (of all dimensions) containing $v$. So $U(v)$ is an open ball. Let $\underline{\mathbb{Z}}$ be the sheaf of locally constant $\mathbb{Z}$-valued functions on $X$. Compute $H^{\ast}(X, \underline{\mathbb{Z}})$ with respect to the Cech cover $U(v)$. You should get the cochain complex of the triangulation. $\endgroup$ – David E Speyer Feb 18 '15 at 15:05
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    $\begingroup$ Dear Jjm, I have cast the final vote to close since your question is more suited for math stackexchange, where I hope it will receive the comprehensive answer that it deserves. I also suggest that you modify it there by removing the erroneous reference to Dolbeault cohomology. $\endgroup$ – Vidit Nanda Feb 18 '15 at 15:29
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    $\begingroup$ @,Jjm Nothing! We can just leave it here and the stackexchange automatons will deal with it :) $\endgroup$ – Vidit Nanda Feb 18 '15 at 15:58