I'm trying to understand Proposition 4.3 (page 562) in S. Morita's article Characteristic Classes of Surface Bundles, which can be found on Andy Putman's website here. I don't think that my question is terribly particular to the situation at hand, but I'll try and give some context in case particular details turn out to be important.

We have an oriented fiber bundle $\pi:M \to X$ with fiber a closed oriented surface of genus $g$, which I'll call $\Sigma_g$. The base $X$ need not be simply-connected. We are interested in the cohomology with coefficients in $\mathbb Z / m$ (but I don't think this particular point is essential). Part of our construction of $M \to X$ ensures that the coefficient system is trivial, and so we have the Serre spectral sequence with $E_2$-page $$ E_2^{p,q} = H^p(X; H^q(\Sigma_g; \mathbb Z/m)). $$

Partway through his argument, Morita makes the following claim: $$ E_\infty^{2,0} = \operatorname{Im}(H^2(X;\mathbb Z / m) \to H^2(M; \mathbb Z)). $$ Why is this?

There is a filtration (I'll suppress coefficients here for simplicity) $$ 0 \subset F^2_2 \subset F^2_1 \subset F^2_0 = H^2(M) $$ with $E_\infty^{2-i,i} = F^2_i / F^2_{i+1}$. The filtration comes by taking (at least following Allen Hatcher's construction in his spectral sequences book) $$ F^2_i = \operatorname{Ker}(H^2(M) \to H^2(M_{i})), $$ where $M_i$ denotes the fiber of the $i$-skeleton of $X$. This should mean that $$ E_\infty^{2,0} = F^2_2 = \operatorname{Ker}(H^2(M) \to H^2(M_{1})). $$ Why is this also realizable as the image of the pullback of the projection map? Is there some exact sequence lurking somewhere?

  • $\begingroup$ I think your first bit about $E_2$ is wrong, it should be $H^*(X; H^*(\Sigma_g; \mathbb{Z}/m))$ $\endgroup$ – Sean Tilson Sep 19 '12 at 2:11
  • $\begingroup$ Yes, you are right. This is also written up in his book, where he uses $M$ for the base space... $\endgroup$ – NKS Sep 19 '12 at 2:12

One way to see this is that the Serre spectral sequence is natural (we only need this naturality in fiber bundles over $X$). There is a map from the bundle $\Sigma_g \to M \to X$ to the trivial bundle $* \to X \to X$, and so you get an associated map of spectral sequences: $$ H^p(X; H^q(*; \mathbb{Z}/m)) \to H^p(X; H^q(\Sigma_g; \mathbb{Z}/m)) $$ For the right-hand spectral sequence, $E_\infty^{p,0}$ is a subgroup of $H^p(M)$ coming from the filtration you list. On the left, however, the spectral sequence is computing $H^*(X)$ and degenerates at the $E^2$-term to $H^p(X; \mathbb{Z}/m)$, concentrated on the $q=0$ line.

This induces a map of $E_\infty$ terms, which is a filtration of the map $H^*(X) \to H^*(M)$. For $q=0$ it is a surjection from $H^p(X;\mathbb{Z}/m)$ to some subobject of $H^*(M)$. This exhibits the $q=0$ line in the $E_\infty$ term as the image of $H^*(X)$.


At each stage (or page, if you prefer) of the spectral sequence there are differentials $d_r: E_r^{p,q} \to E_r^{p+r, q - r + 1}.$ The cohomology of these on $E_r^{p,q}$ gives $E_{r+1}^{p,q}$. "Passing to the limit" over all $r$ gives $E_{\infty}^{p,q}$. Your spectral sequence (which is a case of what I would call the Leray spectral sequence) is a first quadrant s.s., and so there are no convergence issues: if $r$ is large enough (with respect to any fixed $p,q$) the source and target of $d_r$ are zero (draw the picture!) and so the $E_r^{p,q}$ stablize.

Now $E_r^{2,0}$ sits along the $p$-axis, and so $d_r$ maps out of the first quadrant (since $r \geq 2$) and is necessarily zero. Thus in this case every element of $E_r^{2,0}$ is a cocycle for $d_r$, and so $E_{r+1}^{2,0}$ is the quotient of $E_r^{2,0}$ by the image of $d_r$. Thus $E_{\infty}^{2,0}$ is a quotient of $E_2^{2,0} = H^2(X)$ which is also a subobject of $H^2(M)$. Now a consideration of the construction of the s.s. (which may or may not be easy, depending on how you think of it as being constructed) shows that in fact it is the image of the natural map $H^2(X) \to H^2(M)$. (And what else could it be?!)

Thinking in terms of skeleta, as you do, is normally not helpful. There are lots of ways to build this s.s.; using skeleta and a simplicial/singular approach is one way; using sheaves is another. Delving into any particular construction normally will lead to madness. (This is not a definitive rule, of course, but I have found it to be a useful general principle.) It is normally better to use the "internal logic" of the spectral sequence itself, i.e. the knowledge that there are various pages $E_r$ linked by the differentials $d_r$ and their cohomology. (As a vague justification for this philosophy, the whole point of a s.s. is to wrap up the complicated details of some situation into a nice package and delving too much into the construction just unwraps that package, which defeats the purpose to some extent.)

In general, the $d_r$ are hard to compute in cases where you can't identify them for some easy reason, but in the case of $E_r^{p,0}$ and $E_r^{0,q}$, where the $d_r$ mapping out or in necessarily vanishes, one is usually in better shape, and the resulting map of $E_2^{p,0}$ to the limit of the s.s., or of the limit to $E_2^{q,0}$ can often be described. (These maps, which describe the two extreme "ends" of the filtration on the limit of the s.s., are called edge maps, because they come terms on the two edges of the quadrant.)

So in this case, the edge map $E_2^{p,0} \to H^p(M)$ is just the pull-back $H^p(X) \to H^p(M),$ while the edge map $H^q(M) \to E_2^{0,q}$ is the natural map $H^q(M) \to H^q(\Sigma_g)$ given by restricting a cohomology class to a fibre.

When you are trying to understand a spectral sequence, working out the edge maps is a good first step. It is reasonable to expect them to be given by certain natural maps in the given situation. The other differentials are usually harder, and one shouldn't typically expect to compute them (unless you can show that they are zero, say, by showing that their source or target vanishes).

  • $\begingroup$ Thanks for your reply. While I appreciate the intuition that you give ($E^{2,0}_\infty$ is a subobject of $H^2(M)$ that is also a quotient of $E^{2,0}_2 = H^2(X)$) and understand how the obvious map that we've been talking about is the natural candidate, I'd like to be able to understand this in more detail. I thought about this before ("unwrapping the package" as you said) but ended up finding myself slogging through the constructions and couldn't make sense of things. Do you know of a reasonably painless way to understand this? You hinted that this might be easier in other constructions. $\endgroup$ – NKS Sep 19 '12 at 3:07
  • $\begingroup$ Dear Matt, thanks for posting this. I'd like to comment, though, that many of the spectral sequences of interest in homotopy theory seem to arise in this manner. For instance, the Adams spectral sequence, the spectral sequence of a filtered complex, the Atiyah-Hirzebruch spectral sequence are all instances of trying to get at the homotopy (or homology) groups of a complicated object by "filtering" (in some way) it by simpler pieces (e.g., $H \mathbb{Z}/2$-modules in the case of the Adams spectral sequence, skeleta in the Atiyah-Hirzebruch case). $\endgroup$ – Akhil Mathew Sep 20 '12 at 2:52
  • $\begingroup$ ...at least, I just attended two topology classes today that expounded at length on spectral sequences as arising from filtered objects! (I am certainly no expert, but insofar as I've managed to acquire any understanding of spectral sequences, it's primarily from this point of view.) $\endgroup$ – Akhil Mathew Sep 20 '12 at 2:53
  • $\begingroup$ @AkhilMathew: Dear Akhil, I know that spectral sequences are supposed to arise from filtrations in this way, but personally I've only ever understood them in terms of Cartan--Eilenberg-type double complexes/resolutions, where the filtration is just by one of the indices in the double complex. This has sufficed to understand and analyze all the spectral sequences I've ever had to care about (Leray, Hochschild--Serre, Grothendieck, ...), but I realize that as an algebraic/arithmetic geometer and number theorist, my level of communion with spectral sequences pales in comparison to ... $\endgroup$ – Matt E Sep 20 '12 at 2:57
  • $\begingroup$ ... an algebraic topologists! Cheers, $\endgroup$ – Matt E Sep 20 '12 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.