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 $4k$.

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 $4k+2$).

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.

Thanks in advance.

  • 1
    $\begingroup$ In this paper arxiv.org/abs/0908.4121 Misha Verbitsky considers the hyperkaehler case. $\endgroup$ – algori Aug 11 '10 at 13:48
  • $\begingroup$ In dimension 4 one requires stabilization M-> M#S^2XS^2 to guarantee the existence of a diffeomorphism realising a given automorphism, for a homeomorphism, however, stabilization is unnnecessary. These results are respectively due to Wall and Freedman. As to whether this is true in higher dimensions, I'd put money on you not needing stabilization for a diffeomorphism- this would be analogous to the situation with h-cobordisms. $\endgroup$ – Tom Boardman Aug 11 '10 at 13:51
  • $\begingroup$ ... but the action of the mcg is on the second cohomology group (and the form that is preserved by the action is not the cup product, it's the Beauville-Bogomolov-Fujiki form). Woops... $\endgroup$ – algori Aug 11 '10 at 13:56
  • $\begingroup$ Similarly whoops. My comment is missing the condition of simple connectedness. $\endgroup$ – Tom Boardman Aug 11 '10 at 14:31

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.

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.

  • $\begingroup$ Whoops! You are indeed correct Greg! Schoolboy error.... :( $\endgroup$ – Tom Boardman Aug 11 '10 at 14:26
  • $\begingroup$ Thanks for the counterexample. Actually, I don't really want to put restrictions on the manifold. I would be much more interested in a way of computing the subgroup of the integral orthogonal or symplectic group which is realized by diffeomorphisms. I have to think more about this. Thanks again! $\endgroup$ – Samuel Monnier Aug 11 '10 at 15:43
  • $\begingroup$ A smooth manifold $M$ has a group of simple homotopy equivalences, which is a complicated object but one that can be analyzed algebraically. Then, if $M$ is high-dimensional, there is a "surgery structure set" which tells you whether a simple homotopy equivalence lifts to a diffeomorphism. If $M$ is 4-dimensional, then gauge theory is known to give you some obstructions, but the totality of the obstructions is surely a very open problem. en.wikipedia.org/wiki/Surgery_structure_set $\endgroup$ – Greg Kuperberg Aug 11 '10 at 19:46
  • $\begingroup$ For some reason I missed the comments on this thread... Sorry about this. Thanks for the extra info. After some reflexion, I would be satisfied with an example of a 4k + 2 manifold whose diffeomorphism group surjects on the symplectic group acting on the integral cohomology of degree 2k + 1. A naive way to construct candidates might be to mimick the construction of Riemann surfaces as connected sums of tori by considering connected sums of products of 2k+1-spheres. But I fear I do not know enough topology to prove anything. $\endgroup$ – Samuel Monnier Sep 29 '10 at 9:49

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 $x$.

[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.]

  • $\begingroup$ Thanks! Actually I would be more interested in computing the subgroup realized by diffeomorphism than putting restriction on the manifold. $\endgroup$ – Samuel Monnier Aug 11 '10 at 15:45
  • $\begingroup$ Do you really want a general answer, or is it that you have certain manifolds in mind? Either way, the problem is traditionally divided into two parts. In the simply connected case: (1) Which automorphisms of cohomology are realised by homotopy equivalences? (2) Which homotopy equivalences are homotopic to a diffeomorphism? Surgery theory says that for (2) it's sufficient in high dimensions to understand two obstructions. The first concerns the relation between the tangent bundles, and the second is the obstruction to surgering an optimally-chosen cobordism to an h-cobordism. $\endgroup$ – Tim Perutz Aug 11 '10 at 16:14
  • $\begingroup$ I missed the comments on this thread, very sorry about this. Thanks for the extra info. After reflexion, I would only need an example, if there is one, of a 4k + 2 manifold whose diffeomorphism group surjects on the symplectic group acting on the integral cohomology of degree 2k + 1. See also the comment on Greg Kuperberg's answer... $\endgroup$ – Samuel Monnier Sep 29 '10 at 9:52
  • $\begingroup$ Oh, but that's much easier. Recall that the MCG of a closed genus $g$ surface is generated by Dehn twists along circles (and passing to homology, all of $Sp(2g,\mathbb{Z})$ arises this way). We can mimic this on $M$, the connected sum of $g$ copies of $S^3\times S^3$. By Hurewicz, $\pi_3=H_3$. By Whitney, maps $S^3\to M$ can be approximated by immersions, and by the Whitney trick they are homotopic to embeddings. An embedded 3-sphere $S$ has trivial normal bundle (because $\pi_2 SO(3)=0$). Choose a framing; thereby identify a neighbourhood with $TS^3$. $\endgroup$ – Tim Perutz Sep 29 '10 at 19:54
  • $\begingroup$ But there is a comapctly supported diffeo of $TS^3$ which acts as the antipodal map on the zero-section, namely, a generalised Dehn twist (Picard-Lefschetz transformation). Read about these, e.g., in Voisin's book on Hodge theory and complex geometry (vol. 2 I think; not sure). Transplant this into your manifold. Its action on homology is given by the Picard-Lefschetz formula, just like for Dehn twists on surfaces. $\endgroup$ – Tim Perutz Sep 29 '10 at 19:56

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