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If $M$ is a manifold of dimension $n$, does singular cohomology $H^i(M, \mathbb{C})$ vanish when $i > n$ ?

If $M$ is an algebraic variety over $\mathbb{C}$, equipped with ordinary topology, can one say something about the vanishing of higer singular cohomology?

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    You could try to carefully look at this question: http://math.stackexchange.com/questions/4201/singular-and-sheaf-cohomology And then use the Vanishing theorem of Grothendieck (Hartshorne, Chapter III, Par. 2) to get some results.2012-05-11
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    @Giovanni: The question is about the metric topology, not the Zariski topology. However, Mariano's answer in the linked question is probably what the OP is looking for.2012-05-11
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    @Zhen, thank you for pointing it out. I'm always confused about what "ordinary topology" means!2012-05-11
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    @Zhen: Isn't there a comparison theorem that says the algebraic cohomology groups are the same as the usual cohomology groups?2012-05-12
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    @Hurkyl: That's GAGA. But there are hypotheses: $M$ has to be a compact smooth algebraic variety and $\mathscr{F}$ has to be a coherent sheaf.2012-05-12
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    @Hurkyl: The comparison theorem assumes the base ring to be finite, also you need etale cohomology, which is difficult to calculate in general.2012-05-12
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    For your first question, this answer is "yes" by e.g. the universal coefficient theorem. (Or you can say that you can induce up from cohomology with $\mathbb{R}$ coefficients, and then it follows from de Rham cohomology.)2012-05-14
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    ... and since it's related, I'll mention that I just read on wikipedia that Grothendieck proved (in '63) that the algebraic de Rham cohomology of a smooth complex variety is isomorphic to its smooth de Rham cohomology (and hence to its singular cohomology with $\mathbb{C}$ coefficients).2012-05-14

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