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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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If $M$ is a complex algebraic variety of dimension $n$, then its real dimension is $2n$, and so its singular cohomology vanishes in degrees $i > 2n$. One can also try to say things about what happens in the range $0 \leq i \leq 2n$.

E.g. if $M$ is smooth, projective, and connected, then thought of as a real manifold, $M$ is compact, orientable, and connected of dimension $2n$, and so $H^{2n}$ is one-dimensional. More generally, Poincare duality relates $H^i$ and $H^{2n - i}$ (say with $\mathbb C$ coefficients).

E.g. if $M$ is affine, then in fact $H^i$ vanishes if $i > n$. This is a result of Mike Artin, which provides a reinterpretation of an earlier result of Lefschetz (what is commonly called weak Lefschetz).


In general, there is a lot of theory about singular cohomology of complex varieties. In the smooth projective case, in addition to the information provided by real manifold theory (i.e. Poincare duality) one has additional input from Hodge theory.

In the general case, one has Deligne's mixed Hodge theory.

Again in the smooth projective case, one also has the hard Lefschetz theorem, which more or less can be interpreted as saying that the interesting cohomology appears in the middle degree, i.e. in $H^n$.

There is also the theory of nearby and vanishing cycles, which provides tools for describing how cohomology changes when a smooth variety degenerates under deformation to a singular variety. (This is a complex analytic analogue of Morse theory.)


To conclude: this is a major area of investigation, and I have only described some of the basic tools and themes.

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    So the vanishing of cycles corresponds to degeneration of (any) cohomology? Would degeneration of cohomology be related to deformation of the variety?2012-06-15
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    @math-visitor: Dear math-visitor, I'm not sure I understand what you are asking. When a variety degenerates under specialization (e.g. a family of elliptic curves degenerates to a nodal curve) the cohomology changes, and the theory of nearby/vanishing cycles provides tools for understanding and organizing the information about this change of cohomology. If you want to know more , perhaps you should ask a separate question. Regards,2012-06-15
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    Thank you for the reply Matt. After rereading your post, and together with the above comment, it definitely clarified a lot.2012-06-15
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I think the singular cohomology does vanish when $i>n$. You may prove it using long exact sequence or other standard tools.

There is an in-depth discussion on your second question by Milne in first section of his notes.

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    @Zhou, could you please let me know your final answer to my quesions? Somehow I don't care much about the proof, but I found that in one paper the author talked about higher cohomology of a variety over $\mathbb{C}$ which I thought didn't make sence.2012-05-12
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    @LiZhan: I do not have a final answer for the second one. My field is not algebraic geometry. You may ask Matt Emerton who is an expert.2012-05-12