14
$\begingroup$

I was wondering if compact surfaces were the only compact manifolds with a "big" or "complicated" mapping class group.

Are there higher dimensional manifolds (which are not in some way reducible to a surface) that have an infinite mapping class group? What kind of group can arise as the mapping class group of a compact manifold?

Thanks for your attention :)

|cite|improve this question|||||
$\endgroup$
  • 6
    $\begingroup$ Is "mapping class group" defined as $\pi_0$ of diffeomorphisms or $\pi_0$ of homotopy equivalences? For surfaces these are nontrivially the same, but I'm never sure which is considered the definition. $\endgroup$ – Tom Goodwillie Feb 4 '17 at 19:01
  • 2
    $\begingroup$ what about the $n$-torus? its MCG is $GL_n(\mathbf{Z})$ as far as I know (whatever the definition). More generally many nilmanifolds have a big MCG. $\endgroup$ – YCor Feb 4 '17 at 19:05
  • 5
    $\begingroup$ $\mathbb{Z}_2^\infty \leq \pi_0(Diff(T^n))$, $n >5$. p. 5: math.cornell.edu/~hatcher/Papers/Diff%28M%292012.pdf $\endgroup$ – Ian Agol Feb 5 '17 at 4:30
  • 1
    $\begingroup$ @TomGoodwillie Indeed, I'm not sure what is the good notion to look at here. I'd say homotopy equivalence not to add subtleties of analytic nature. $\endgroup$ – Selim G Feb 5 '17 at 9:05
13
$\begingroup$

Take $M^d$ to be a connected sum of $n$ copies of $S^1\times S^{d-1}$, where $d\ge 3$. Then $M$ is a closed, orientable manifold of dimension $d$ with $\pi_1(M)=F_n$, the free group of rank $n$. If $n>1$, the mapping class group $\pi_0({\rm Diff}(M))$ surjects onto the outer automorphism group ${\rm Out}(F_n)$, which is infinite, and very complicated.

For an explanation of the relation between mapping class groups and outer automorphism groups, see for instance here. For the surjectivity of the natural map $\pi_0({\rm Diff}(M)) \to {\rm Out}(F_n)$, see here.

|cite|improve this answer|||||
$\endgroup$
  • 2
    $\begingroup$ Why is the mapping class group in that case the outer automorphism group of the fundamental group? $\endgroup$ – Igor Rivin Feb 4 '17 at 17:05
  • 1
    $\begingroup$ Good question. I edited my answer, accordingly. Is this now better? $\endgroup$ – Alex Suciu Feb 4 '17 at 19:39
  • $\begingroup$ While I know that the isomorphism $\pi_0(\text{Diff}(M)) \to \text{Out}(F_n)$ holds for $d=3$, due to Hatcher and based on prime decomposition theory for 3-manifolds, I don't know of a reference for higher dimensions. $\endgroup$ – Lee Mosher Feb 4 '17 at 21:09
12
$\begingroup$

In Infinitesimal computations in topology Sullivan shows in Theorem 13.3 that if $M$ is a simply-connected manifold of dimension $>5$, then $\pi_0(\mathrm{Diff\,} M)$ is commensurable to an arithmetic group (see p.295 for definitions of an arithmetic group).

The group of homotopy classes of homotopy self-equivalences is also commensurable to an arithmetic one (Theorem 10.3)

In Corollary 13.3 Sullivan explains how the forgetful map between the two is largely controlled by the rational Pontryagin class of the manifold (which is preserved by diffeomorphisms by not by homotopy equivalences).

|cite|improve this answer|||||
$\endgroup$
8
$\begingroup$

There are situations in which surfaces are the "unique" examples with big mapping class groups. One such is closed manifolds of negative sectional curvature.

Theorem (Paulin): If $M$ is a closed $n$-dimensional manifold of negative sectional curvature and $n>2$ then $\mathrm{Out}(\pi_1M)=\mathrm{MCG}(M)$ is finite.

Proof: Paulin actually proved that if $\Gamma$ is a torsion-free word-hyperbolic group (such as the fundamental group of a closed manifold of negative curvature) and $\mathrm{Out}(\Gamma)$ is infinite then $\Gamma$ splits over a cyclic subgroup, say as $\Gamma=A*_CB$ for $C$ cyclic. (There is also the HNN-extension case, which is similar.)

Mayer--Vietoris now gives that

$\ldots\to H^{n-1}(C)\to H^n(M)\to H^n(A)\oplus H^n(B) \to H^n(C)\to\ldots$

is exact.

Since $n>2$, $H^{n-1}(C)\cong H^n(C)\cong 0$ and so $H^n(M)\cong H^n(A)\oplus H^n(B)$. But Strebel's theorem implies that $H^n(A)\cong H^n(B)\cong 0$ , a contradiction. QED

(Here, I'm taking $\mathrm{MCG}(M)$ to mean $\pi_0$ of self-homotopy-equivalences. Since negatively curved manifolds are aspherical, this coincides with $\mathrm{Out}(\pi_1M)$.)

This is actually an instance of a much more general phenomenon. In good situations, JSJ theory tells us that $\mathrm{Out}(\Gamma)$ can be 'broken up' into pieces which are essentially mapping class groups of surfaces and tori. This happens in the case of irreducible 3-manifolds, for instance.

|cite|improve this answer|||||
$\endgroup$
  • $\begingroup$ Where does the assumption $n>2$ come into play? $\endgroup$ – Victor Protsak Feb 5 '17 at 1:08
  • 1
    $\begingroup$ @VictorProtsak: The Mayer-Vietoris computation. $\endgroup$ – HJRW Feb 5 '17 at 6:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.