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Hello,

I have found different definitions of Čech complex for sheaf $F$ od abelian groups on topological space $X$ with respect to the cover $\mathcal U$. One in Gelfand-Manin says to take product of $F(U_{i_0} \cap \ldots \cap U_{i_n})$ for all $n$-tuples of indices $(i_0, \ldots ,i_n)$, and one in Hartshorne says that one considers only stricly increasing indices $i_0 < \ldots < i_n$.

How can I see that cohomologies of these "two different" Čech complexes coincide?

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In the Gelfand-Manin defintion it is assumed that a cochain $f_{i_0i_1\cdots i_n}$ is totally skewsymmetric (for the relevant definition of skewsymmetry) and there is a bijection between those and the cochains $f_{i_0i_1\cdots i_n}$ where the $i_j$ form a strictly increasing sequence. –  José Figueroa-O'Farrill May 20 '11 at 11:52
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These two chain complexes are different, so you can't show that they coincide: they don't. However, there is an inclusion of the second into the first. That inclusion induces a map at the level of cohomology, which turns out to be an isomorphism. Your question is then: why is it an isomorphism? –  André Henriques May 20 '11 at 11:52
    
@Jose: If you assume total antisymmetry, then the two complexes are indeed isomoprhic. But there is another question to be asked: what about if you don't ask for antisymmetry? Then the first complex is bigger... but still has the same cohomology. –  André Henriques May 20 '11 at 11:54
    
This is Proposition 2 in Section 20 in [Serre, Faisceaux algébriques cohérents], available here: mat.uniroma1.it/people/arbarello/FAC.pdf. –  Philipp Hartwig May 20 '11 at 12:18
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This question was asked before mathoverflow.net/questions/10056/… –  David Speyer May 20 '11 at 13:49

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