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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?

Akhil Mathew
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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-Álvarez Nov 21 '10 at 05:08
  • (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-Álvarez Nov 21 '10 at 05:10
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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 05:42
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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-Álvarez Nov 21 '10 at 06:01
  • @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 06:13
  • @Mariano: Sure, but I just like to think of the Leray SS as a special case of the Grothendieck SS, even if Leray's came before Grothendieck's. – Akhil Mathew Nov 21 '10 at 06:14
  • What are you asking exactly? If Serre's ss is a corollary of Leray's, or if (both?) follow from Grothendieck's. – Mariano Suárez-Álvarez Nov 21 '10 at 06:18
  • The former. Isn't Leray's a direct corollary of Grothendieck's? (Take the first functor to be the pushforward $f_$ from $Sh(X) \to Sh(Y)$ and the second to be global sections. $f_$ preserves flabbiness, so Grothendieck ss hypotheses apply.) – Akhil Mathew Nov 21 '10 at 06:39
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    Akhil, as I said before, it follows from G's s.s. that a spectral sequence with the same initial term as Leray's exists. But a s.s. is much more than its initial term! – Mariano Suárez-Álvarez Nov 21 '10 at 07:13
  • @Mariano: Good point. All I have ever seen used is the initial term and the existence of a product structure for the cohomology SS (which may be derivable in the Grothendieck case from cup-products). I have so far been able to avoid thinking about the differentials (except in deducing what they are from $E_\infty$, e.g. if the fiber space is contractible), but I have not gotten very far in this subject. – Akhil Mathew Nov 21 '10 at 16:02
  • I really would like to see an answer here which does not just say "it's obvious". After all, the Leray spectral sequence is cohomological. whereas the Serre spectral sequence has a homological (and a cohomological) version. How to deduce it? – Martin Brandenburg Jun 03 '15 at 18:42

2 Answers2

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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-Álvarez Nov 21 '10 at 05:13
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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. –  Nov 21 '10 at 05:18
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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 05:44
  • I can't comment on the OP's question (I dont' have enough rep), so I'll comment here. Mariano asked whether you actually get the same spectral sequence (not just the same E^2 page). Well, here the literature is not in good shape. There are SEVERAL different constructions of the Serre spectral sequence (via the Leray spectral sequence, via cubical cochains, via singular cochains, Dress's bisimplicial constuction, etc). As far as I know, no one has bothered to prove that they give the same differentials, just that they have the same transgression and E^2 page. (continued) –  Nov 21 '10 at 05:52
  • (continued) It would be interesting to try to axiomatize the Serre spectral sequence and then prove some kind of uniqueness statement for it... –  Nov 21 '10 at 05:53
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    @T_P: Thanks! Sorry, but I don't see why it is obvious. (For instance, I don't understand where the condition on the fundamental group pops in---there's nothing about that in the Leray SS a priori.) Also, what's wrong with looking at sheaf cohomology at a derived functor? In pathological cases (e.g. nonseparated schemes), there is no a priori reason to conclude that it will agree with Cech cohomology in dimensions >1. Though I am willing to believe that Cech cohomology is equivalent in this case (since paracompact spaces, e.g. CW complexes, are OK). – Akhil Mathew Nov 21 '10 at 05:57
  • (There is a paper by Matthias Künzer where some useful comparisons are made, iirc.) – Mariano Suárez-Álvarez Nov 21 '10 at 05:58
  • OK, here's what I do see. Cover the base $B$ by contractible nbds $U_\alpha$ and consider the preimages $p^{-1}(U_\alpha) \subset E$; these are fiber homotopy equivalent to $U_\alpha \times E$ by contractibility. So homology is the same. The Cech complex w.r.t. the cover $p^{-1}(U_\alpha)$ is something like the tensor product of the Cech complex of the covering $U_\alpha$ tensored with the Cech complex of $F$ (by the Kunneth formula?). So apply the SS for tensor products or something like that. I'm willing to buy this but will need to check some justifications for myself. – Akhil Mathew Nov 21 '10 at 06:03
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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. –  Nov 21 '10 at 06:04
  • By the way, there is a version of the Serre spectral sequence with no fundamental group assumptions, but there you have to take twisted cohomology groups. For these, I recommend either looking at the chapter in Hatcher or at the original paper (due to Steenrod I think -- Hatcher has a reference to it. I remember it being very readable). –  Nov 21 '10 at 06:06
  • And I have no problem with computing sheaf cohomology with derived functors on weird spaces. I have just seen too many young people learn the subject from Hartschorne and then be unable to work with easy situations like the one at hand, where Czech cohomology works nicely. –  Nov 21 '10 at 06:08
  • I just looked up the paper on cohomology with twisted coefficient systems. It is indeed due to Steenrod, and it is called "Homology with local coefficients". I also remembered that there is a (somewhat less well-written) account of this in McCleary's book on spectral sequences. –  Nov 21 '10 at 06:13
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    @Ryan, the history may be such (is, even!) but Akhil wants a proof not a story :) – Mariano Suárez-Álvarez Nov 21 '10 at 07:16
  • Dear @T_P: thanks! It looks like I'm going to have to learn more about algebraic topology (e.g. I don't know anything about local coefficients) before I can fully understand this answer, so I'll come back here later. – Akhil Mathew Nov 22 '10 at 02:59
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This is not intended to be an answer but rather a long comment.

First, on the question why does Serre take all the credit if his theorem is a particular case of Leray's?. Well, John McCleary says on page 139 of his book User's Guide to Spectral Sequences (2ed):

For the Cêch or Alexander-Spanier cohomology theories, the multiplicative structure is carried along transparently in the construction of the spectral sequence and so we get a spectral sequence of algebras directly with converges to $H^*(E;R)$ as an algebra. The result for singular theory, however, is more difficult --- it is one of the technical triumphs of Serre's celebrated thesis.

And, as the same wiki page that the OP cites say:

Earlier (1948/9) the implications for fiber bundles were extracted in a form formally identical to that of the Serre spectral sequence, which makes no use of sheaves. This treatment, however, applied to Alexander–Spanier cohomology with compact supports (....). Jean-Pierre Serre, who needed a spectral sequence in homology that applied to path space fibrations, whose total spaces are almost never locally compact, thus was unable to use the original Leray spectral sequence (...).

On another topic, and as @T_P points out, Serre's original result does not impose any condition on the homotopy group of the base. That assumption is added so that one can simply the $E_2$-term by getting rid of (co)homology with local coefficients.

Finally, I would like to say a few words about spectral sequences and their initial terms. Apparently, there is a Grotehndieck's spectral sequence whose $E_2$-term is exactly as in Serre's, but the equality of both is questioned, since nothing is said about the differentials. This is completely true. However, there are (not a few) case where sheaf cohomology and singular cohomology provide the same result. Thus, one is found dealing with two spectral sequences which have the same initial term and the same limit. This is still not enough to guarantee that both are the same spectral sequence in general. However, for the particular case when one of them degenerates, so must the other; in this case both become eventually the same. I know it is a very particular situation, but it is also one that becomes very handy sometimes.

Dog_69
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