Are there any general results on the (integral) cohomology of manifolds that are fibrations over the circle? Any literature references much appreciated.
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1$\begingroup$ There's all kinds of general results as the problem is pretty general and fits into a standard formalism where many of the details are already worked out. Is there a more specific issue you're interested in? $\endgroup$– Ryan BudneyCommented Nov 6, 2009 at 18:50
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6 Answers
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The above Mayer-Vietoris argument gives the cohomology of a fiber bundle over a circle in a concrete fashion. For sake of "mathematical culture", I thought I'd mention what happens for fiber bundles over a higher dimensional sphere (this is also a good excuse for me to test drive the new latex support).
For fiber bundles $F \hookrightarrow E \rightarrow S^n$ with $F$ connected and $n \geq 2$, the Serre spectral sequence degenerates in a very simple fashion into what is known as the Wang exact sequence. Namely, we have a long exact sequence of the form $$\cdots \rightarrow H^k(E) \rightarrow H^k(F) \rightarrow H^{k-n+1}(F) \rightarrow H^{k+1}(E) \rightarrow \cdots$$ The proof of this is completely analogous to the proof of the better known Gysin exact sequence, which tells you what happens for fiber bundles whose fibers are spheres.
A reference for this material is McCleary's "User's guide to spectral sequences", page 145.
answered Nov 6, 2009 at 20:16
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2$\begingroup$ This is also a Mayer-Vietoris argument, decomposing E as the pull-back of the decomposition of the sphere into two discs. But you also need the computation of the homology of F x S^{n-1} so it's a smidgen more complicated. This is also why the map H^k(F) --> H^{k-n+1} does not preserve dimension, the the homology of S^{n-1} introduces a degree shift via the Kunneth Theorem. $\endgroup$ Commented Nov 6, 2009 at 20:54
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$\begingroup$ An excellent point. One thing to remark about the Wang exact sequence, though, is that it also works for simply connected homology spheres. $\endgroup$ Commented Nov 6, 2009 at 21:13
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7$\begingroup$ The meta-point of bringing up the Wang/Gysin exact sequence is that they are illustrations of the fact that spectral sequences with only two nonzero rows/cols degenerate nicely into long exact sequences. This bit of algebra shows up a lot, so it is useful to internalize it. $\endgroup$ Commented Nov 6, 2009 at 21:22
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1$\begingroup$ The Mayer-Vietoris argument also works for simply-connected homology spheres. You decompose the homology sphere into the union of a disc and a homology ball, and then the remainder follows analogously. Basically, this boils down to the observation that the Serre spectral sequence is just Mayer-Vietoris + Kunneth applied to a cell decomposition of the base space. $\endgroup$ Commented Nov 6, 2009 at 21:39
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3$\begingroup$ Ah, okay. But I like cheating. :) $\endgroup$ Commented Nov 7, 2009 at 1:31
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This began as a comment, but it is interesting enough that I decided to make it an answer instead. Let's consider bundles over the circle whose fibers are closed genus $g$ surfaces $\Sigma_g$.
The diffeomorphism type of the total space of our bundle only depends on the isotopy class of the monodromy map. Denote by $M_g$ the mapping class group, ie the group of isotopy classes of orientation-preserving diffeomorphisms of $\Sigma_g$. For $f \in M_g$, denote by $B_f$ the surface bundle over the circle determined by $f$.
In a comment, Tom Church observed that the homology of $B_f$ will be the same as the homology of the trivial bundle if and only if $f$ acts trivially on $H_1(\Sigma_g)$. The group of mapping classes that act trivially on $H_1(\Sigma_g)$ is known as the Torelli group.
One could demand more. Namely, we could require that the cup-product structure on $H^{\ast}(B_f)$ be the same as the cup product structure on the trivial bundle. As Tom observed, a beautiful theorem of Dennis Johnson gives a precise characterization of the subgroup of $I_g$ consisting of monodromies with this property. One easy way of describing it is that it is the kernel of the (outer) action of $M_g$ on the second nilpotent truncation of $\pi_1(\Sigma_g)$ (the group $H_1(\Sigma_g)$ is the first nilpotent, ie abelian, trunctation).
The story does not end here. A topological space has a "higher-order" intersection theory given by the so-called Massey products. They are sort of like generalized cup products. Anyway, Kitano generalized Johnson's work and gave a precise and beautiful description of the monodromies of surface bundles over the circle in which these higher intersection products (up to a certain level) are trivial. Namely, all the degree at most $k$ Massey products of $B_f$ will be trivial if and only if $f$ acts trivially on the (k+1)st nilpotent truncation of $\pi_1(\Sigma_g)$.
For the details of this plus references to Johnson's papers, see the following paper:
MR1381688 (97f:57014) Kitano, Teruaki(J-TOKYTE) Johnson's homomorphisms of subgroups of the mapping class group, the Magnus expansion and Massey higher products of mapping tori. (English summary) Topology Appl. 69 (1996), no. 2, 165--172.
answered Nov 7, 2009 at 1:28
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Given a bundle $F \to M \to S^1$, the Mayer-Vietoris sequence corresponding to the decomposition of $M$ coming from writing $S^1$ as the union of two intervals tells you there's a short exact sequence:
$$0 \to coker( f_n - I ) \to H_n(M) \to ker( f_{n-1} - I ) \to 0$$
Here $f_n : H_n F \to H_n F$ is the induced map from the monodromy of the bundle, ie: you think of the bundle as $R \times_f F, f: F \to F$ a homeomorphism / diffeomorphism / whatever. And $I$ is the identity map on $H_n(M)$ and $H_{n-1} M$ respectively.
There's a similar decomposition for cohomology, and this is what the Serre spectral sequence gives you, too. The short-exact sequence basically encodes the extension problem from the spectral sequence.
answered Nov 6, 2009 at 18:37
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$\begingroup$ @RyanBundey : Could you write the decomposition for cohomology? I wonder whether this will give some relations between Stiefel-Whitney classes on oriented mapping torus? See mathoverflow.net/questions/165712 $\endgroup$ Commented May 10, 2014 at 12:20
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For 3-manifolds, the cohomology could be almost anything surjecting $\mathbb{Z}$.
In this case, the fiber is a surface, and since $\mathrm{Mod}(S) \to \mathrm{Sp}(2g, \mathbb{Z})$ is surjective, you can make the action of the monodromy on the homology of the fiber anything you like (as long as it preserves the intersection form).
answered Nov 6, 2009 at 11:57
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2$\begingroup$ If the 3-manifold is the complement of a fibered knot K in S^3, so the fiber is a punctured surface of genus g(K), then H^1(S^3-K) is exactly Z and the characteristic polynomial of that monodromy has to be the Alexander polynomial of K. $\endgroup$ Commented Nov 6, 2009 at 13:42
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$\begingroup$ Thanks for these interesting examples. Is there anything like the Kunneth formula in these cases? $\endgroup$– DanCommented Nov 6, 2009 at 14:33
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$\begingroup$ In general you have the Leray-Hirsch condition; from this you can show that for a closed 3-manifold fibered over the circle, H*(M) = H*(S) tensor H*(S^1) as groups iff the monodromy action on the homology of the fiber is trivial. Work of Johnson implies that for this to be an isomorphism of rings (i.e. cup product behaves like in Kunneth) you need the monodromy to lie in the Johnson kernel (a certain subgroup of the mapping class group). This should tell you that Kunneth-like behavior in higher dimensions will be very difficult to guarantee. $\endgroup$ Commented Nov 6, 2009 at 23:57
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If the manifold happens to be aspherial manifold(that means the universal covering spacec is contractible), then the cohomology groups is just cohomology of the fundamental groups, which can be detecked by the fundamental group of the fiber and S^1.
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Not very exciting, but the Euler characteristic is zero. I assume that you are aware of the Serre spectral sequence?
