17
$\begingroup$

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?

$\endgroup$
11
  • $\begingroup$ 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? $\endgroup$ Commented Nov 21, 2010 at 5:08
  • $\begingroup$ (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.) $\endgroup$ Commented Nov 21, 2010 at 5:10
  • 2
    $\begingroup$ @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. $\endgroup$ Commented Nov 21, 2010 at 5:42
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Nov 21, 2010 at 6:01
  • 1
    $\begingroup$ 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! $\endgroup$ Commented Nov 21, 2010 at 7:13

2 Answers 2

9
$\begingroup$

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!

$\endgroup$
14
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Nov 21, 2010 at 5:13
  • 4
    $\begingroup$ 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. $\endgroup$
    – T_P
    Commented Nov 21, 2010 at 5:18
  • 1
    $\begingroup$ @Mariano: historically the answer to the question is "tautologically yes". Serre was interested in the Leray SS in the special case of a fibration. $\endgroup$ Commented Nov 21, 2010 at 5:44
  • 1
    $\begingroup$ @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). $\endgroup$ Commented Nov 21, 2010 at 5:57
  • 1
    $\begingroup$ @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. $\endgroup$
    – T_P
    Commented Nov 21, 2010 at 6:04
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .