Cohomology of fibrations over the circle - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T10:33:06Z http://mathoverflow.net/feeds/question/4361 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle Cohomology of fibrations over the circle Dan 2009-11-06T10:48:20Z 2010-01-15T01:02:14Z <p>Are there any general results on the (integral) cohomology of manifolds that are fibrations over the circle? Any literature references much appreciated.</p> http://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle/4366#4366 Answer by Carsten Schultz for Cohomology of fibrations over the circle Carsten Schultz 2009-11-06T11:48:53Z 2009-11-06T11:48:53Z <p>Not very exciting, but the Euler characteristic is zero. I assume that you are aware of the <a href="http://en.wikipedia.org/wiki/Serre%5Fspectral%5Fsequence" rel="nofollow">Serre spectral sequence</a>?</p> http://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle/4369#4369 Answer by Richard Kent for Cohomology of fibrations over the circle Richard Kent 2009-11-06T11:57:16Z 2009-11-06T11:57:16Z <p>For 3-manifolds, the cohomology could be almost anything surjecting $\mathbb{Z}$.</p> <p>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).</p> http://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle/4380#4380 Answer by yeshengkui for Cohomology of fibrations over the circle yeshengkui 2009-11-06T13:56:08Z 2009-11-06T13:56:08Z <p>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.</p> http://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle/4412#4412 Answer by Ryan Budney for Cohomology of fibrations over the circle Ryan Budney 2009-11-06T18:37:16Z 2010-01-15T01:02:14Z <p>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:</p> <p>$$0 \to coker( f_n - I ) \to H_n(M) \to ker( f_{n-1} - I ) \to 0$$</p> <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle/4428#4428 Answer by Andy Putman for Cohomology of fibrations over the circle Andy Putman 2009-11-06T20:16:04Z 2009-11-06T20:16:04Z <p>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).</p> <p>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.</p> <p>A reference for this material is McCleary's "User's guide to spectral sequences", page 145.</p> http://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle/4479#4479 Answer by Andy Putman for Cohomology of fibrations over the circle Andy Putman 2009-11-07T01:28:05Z 2009-11-07T01:28:05Z <p>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 $\Sigma_g$. </p> <p>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 $f$.</p> <p>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.</p> <p>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).</p> <p>The story does not end here. A topological space has a "higher-order" intersection theory given by the so-called <a href="http://en.wikipedia.org/wiki/Massey_product" rel="nofollow">Massey products</a>. 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 $\pi_1(\Sigma_g)$. </p> <p>For the details of this plus references to Johnson's papers, see the following paper:</p> <p>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. </p>