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Timeline for Two definitions of Čech cohomology

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May 20, 2011 at 15:09 comment added José Figueroa-O'Farrill @André: I stand corrected! I had in fact not seen the unordered Čech complex before.
May 20, 2011 at 13:49 comment added David E Speyer This question was asked before mathoverflow.net/questions/10056/…
May 20, 2011 at 12:31 comment added Neil Strickland A key point is that the normalised and unnormalised chain complexes associated to any simplicial abelian group are homotopy equivalent. This is explained in section 22 of May's "Simplicial objects in algebraic topology" (goo.gl/HhBlj) for example, and probably many other places.
May 20, 2011 at 12:18 comment added Philipp Hartwig This is Proposition 2 in Section 20 in [Serre, Faisceaux algébriques cohérents], available here: mat.uniroma1.it/people/arbarello/FAC.pdf.
May 20, 2011 at 11:54 comment added André Henriques @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.
May 20, 2011 at 11:52 comment added André Henriques 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?
May 20, 2011 at 11:52 comment added José Figueroa-O'Farrill 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.
May 20, 2011 at 11:41 history asked Rafael Mrden CC BY-SA 3.0