Cohomology of fibrations over the circle - MathOverflow

Cohomology of fibrations over the circle

2009-11-06T10:48:20Z

Are there any general results on the (integral) cohomology of manifolds that are fibrations over the circle? Any literature references much appreciated.

Answer by Carsten Schultz for Cohomology of fibrations over the circle

2009-11-06T11:48:53Z

Not very exciting, but the Euler characteristic is zero. I assume that you are aware of the Serre spectral sequence?

Answer by Richard Kent for Cohomology of fibrations over the circle

2009-11-06T11:57:16Z

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).

Answer by yeshengkui for Cohomology of fibrations over the circle

2009-11-06T13:56:08Z

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.

Answer by Ryan Budney for Cohomology of fibrations over the circle

2009-11-06T18:37:16Z

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 $R \times_f F, f: F --> F$ 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.

Answer by Andy Putman for Cohomology of fibrations over the circle

2009-11-06T20:16:04Z

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.

Answer by Andy Putman for Cohomology of fibrations over the circle

2009-11-07T01:28:05Z

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.