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Let $F \to E \to B$ be a fibration with $B$ simply connected (more generally, such that $\pi_1(B)$ acts trivially on the homology of $F$). Then there is a Serre spectral sequence $H_p(B, H_q(F)) \to H_{p+q}(E)$. One can do the same for singular cohomology. However, for reasonable spaces (specifically, locally contractible spaces, e.g. CW complexes), singular cohomology is the same as sheaf cohomology of the constant sheaf $\mathbb{Z}$.

But there is another spectral sequence for sheaf cohomology: the Leray spectral sequence. Given spaces $X, Y$ and $f: X \to Y$, and a sheaf $\mathcal{F}$ on $X$, there is a spectral sequence $H^p(Y, R^q_f(\mathcal{F})) \to H^{p+q}(X, \mathcal{F})$. The Wikipedia article hints that the topological implications of this include in particular the Serre spectral sequence. I would be interested in this, because I like the machinery of the Grothendieck spectral sequence (from which the Leray spectral sequence easily follows), and would be curious if the Serre spectral sequence could be obtained as a corollary.

Is this possible?

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What happens if you try to check the hypotheses of the Grothendieck s.s. theorem in the case of the Serre s.s. for cohomology? –  Mariano Suárez-Alvarez Nov 21 '10 at 5:08
    
(By the way, note that even if using the Grothendieck s.s. you manage to get a s.s. whith $E_2$ term of the same shape as that of the Serre s.s., that is not enough (sadly!) to know that the s.s. you got is the same one as the one Serre got.) –  Mariano Suárez-Alvarez Nov 21 '10 at 5:10
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@Akhil, Serre did precisely as you suspect. He looked at the Leray spectral sequence and applied it to a fibration over a CW complex, and noticed that it cleans up quite a bit in that situation. If you read Serre's papers or Dieudonne's history you'll see this. –  Ryan Budney Nov 21 '10 at 5:42
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Tohoku is from 57, Serre's thesis from 51, and Leray's work predates that. The original arguments, Akhil, did not involve Grothendieck's s.s. –  Mariano Suárez-Alvarez Nov 21 '10 at 6:01
    
@Ryan: Sure, I'll take a look at Homologie des espaces fibres and see how it goes; thanks for the recommendation (the idea hadn't occurred to me). –  Akhil Mathew Nov 21 '10 at 6:13

1 Answer 1

Yes. In fact, the result is basically obvious if you use Czech cohomology on the base.

Serre really had two key insights. First, sheaf cohomology is a pain to compute, but if there is no fundamental group then for fiber bundles the Leray spectral sequence is really just using normal old-fashioned untwisted cohomology. Second, you don't really need to work with fiber bundles -- all you need are Serre fibrations, and those are easy to construct. In particular, you have the standard Serre fibration $\Omega X \rightarrow PX \rightarrow X$, where $\Omega X$ is the loop space of $X$ and $PX$ is the space of paths starting at the basepoint of $X$ and the map $PX \rightarrow X$ is "evaluation at the endpoint". Clearly $PX$ is contractible! An amazing amount of milage can be had from this silly observation!

Serre also really developed many of the key algebraic tricks one needs to work with spectral sequences. For instance, he had the amazing idea that one can work modulo "Serre classes", and thus ignore things like torsion. It's like pretending to localize spaces long before Sullivan and Quillen realized you could do so for real!

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Your first paragraph answers the question by saying the answer is obvious! On the other hand, I don't see the connection of your second paragraph to the question. –  Mariano Suárez-Alvarez Nov 21 '10 at 5:13
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It is obvious if you use Czech cohomology -- I don't really think there is anything more one can say there. Just write down the definition. My guess is that he was confused by trying to do it using singular cohomology (or, worse, by thinking of sheaf cohomology using derived functors or some other such nonsense). The rest of my answer is answering the the meta-question "Why does Serre get all the credit if all he did was construct a special case of something Leray had already done"? But I'm on my 7th beer for the night, so I might be reading too much into the OP's question. –  T_P Nov 21 '10 at 5:18
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@Mariano: historically the answer to the question is "tautologically yes". Serre was interested in the Leray SS in the special case of a fibration. –  Ryan Budney Nov 21 '10 at 5:44
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@Akhil : What you have to prove is that your sheaf is really a constant sheaf in a strong sense. Namely, not only does it give the same thing on all sufficiently small open sets, but it does so in a natural way. At that point, you're just computing Czech cohomology with coefficients. The fiber bundle condition shows that the sheaf gives the same thing on all sufficiently small open sets, and the fundamental group condition shows that there is no monodromy, so your identification of what the sheaf is is natural. –  T_P Nov 21 '10 at 6:04
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@Ryan, the history may be such (is, even!) but Akhil wants a proof not a story :) –  Mariano Suárez-Alvarez Nov 21 '10 at 7:16

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