Action of the mapping class group on middle-dimensional cohomology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:55:12Z http://mathoverflow.net/feeds/question/35224 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35224/action-of-the-mapping-class-group-on-middle-dimensional-cohomology Action of the mapping class group on middle-dimensional cohomology Samuel Monnier 2010-08-11T13:32:21Z 2010-08-11T14:20:49Z <p>Given an even dimensional manifold, the mapping class group acts on middle dimensional cohomology (or homology) and this action preserves the intersection form. For manifold of dimension $4k+2$, the action symplectic, while it is orthogonal for manifold of dimension $4k$. </p> <p>In dimension 2, it is well-known that any integral symplectic transformation on the cohomology of degree 1 can be realized by some diffeomorphism. I would like to know if this is still true in higher dimension. I am interested mostly in the symplectic case (dimension $4k+2$).</p> <p>More generally, does anyone know a good reference about mapping class groups of manifolds of dimension higher than 2? All the references I found treat exclusively the case of surfaces. </p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/35224/action-of-the-mapping-class-group-on-middle-dimensional-cohomology/35230#35230 Answer by Greg Kuperberg for Action of the mapping class group on middle-dimensional cohomology Greg Kuperberg 2010-08-11T14:17:13Z 2010-08-11T14:17:13Z <p>Without other assumptions, the answer is an easy no. For instance, if $N^3$ is a homology 3-sphere with infinite fundamental group, then $M^6 = N^3 \times S^3$ does not have very many lifts of automorphisms of its middle homology, because there exists a degree one map $S^3 \to S^3$ but no map $S^3 \to N^3$ with non-zero degree. There are also obstructions from other cup products besides the middle one, and from algebraic operations on cohomology other than cup products.</p> <p>So the question is much more reasonable if $M$ is simply connected (unless it is 2-dimensional) and has no homology other than the middle homology and at the ends. In this case, Tom Boardman says in the comment that Wall and Freedman showed that the answer is yes for homeomorphisms, although they surely assume that $M$ is simply connected. In higher dimensions, I don't know the answer to this restricted question, but I imagine that it could be yes using surgery theory.</p> http://mathoverflow.net/questions/35224/action-of-the-mapping-class-group-on-middle-dimensional-cohomology/35231#35231 Answer by Tim Perutz for Action of the mapping class group on middle-dimensional cohomology Tim Perutz 2010-08-11T14:20:49Z 2010-08-11T14:20:49Z <p>Here's an answer for simply connected (closed) 4-manifolds $X$. Freedman showed that every automorphism of the intersection form is realised by a unique (up to homotopy) orientation-preserving homeomorphism. But the Seiberg-Witten invariants, which can be formulated as a finitely supported function $SW: H^2(X;\mathbb{Z})\to \mathbb{Z}$, are invariant under orientation-preserving diffeomorphisms. For instance, if $X$ admits an integrable complex structure making it a general type surface with first Chern class $c$ then every diffeomorphism preserves $c$ up to sign, because $SW(\pm c)=1$ and $SW(x)=0$ for all other $x$. </p> <p>[Also some general remarks about how to frame the question in higher dimensions, echoing Greg's. The group of homeomorphisms acts on the fundamental group and on the graded cohomology ring, and (as Greg says) it respects all cohomology operations. The subgroup of diffeomorphisms fixes the characteristic classes of the tangent bundle; which of these are also preserved by homeomorphisms is a subtle question. To get a tractable question about the action on middle-dimensional cohomology, it's therefore sensible to consider $(n-1)$-connected closed $2n$-manifolds. There is a classification theorem for such manifolds due to C.T.C. Wall.]</p>