Let $M$ be a manifold with boundary. Consider the following groups:

(1) $\pi_0(\operatorname{Diff}(M,\partial M))$.

(2) $\pi_0(\operatorname{Homeo}(M,\partial M))$.

(3) $\pi_0(\operatorname{HomEq}(M,\partial M))$.

That is, isotopy (resp. isotopy and homotopy) classes of diffeomorphisms (resp. homeomorphisms and homotopy equivalences) relative to $\partial M$.

I would like to find conditions on $M$ which guarantee that one of the groups (1)/(2)/(3) has no element of finite order.

My motivation for this is to generalize from the case $\dim M=2$, when as long as $\partial M\ne\varnothing$, none of the groups (1)/(2)/(3) has an element of finite order (This is left as an exercise for the reader. Hint: an element of finite order in $\operatorname{MCG}(\Sigma_g)$ fixes some hyperbolic structure).

Of course, if $\dim M=2$, then (1)=(2)=(3). This question is really about finding a proper generalization of the result for $\dim M=2$ to higher dimensions, so I'm leaving it open as to which of (1)/(2)/(3) this question is really about. We remark that (1) seems unlikely to be the right group to consider; for instance when $(M,\partial M)=(D^n,S^{n-1})$ and $n>4$, then it is the group of exotic spheres in dimension $n+1$.

  • $\begingroup$ I don't think your question is going to get an actual answer, in that there's no conditions of any real generality known that will ensure an answer to your question. There are extreme special cases -- for example if $M$ is hyperbolic and you're interested in case (3). But I think it's a difficult question to construct high-dimensional manifolds where $\pi_0 Diff(M)$ has no elements of finite order. $\endgroup$ – Ryan Budney Oct 11 '11 at 21:46
  • $\begingroup$ I would be satisfied with an answer that addresses just one of the cases (1)/(2)/(3). $\endgroup$ – John Pardon Oct 11 '11 at 22:05

In the case of simply connected 4-manifolds, a famous theorem of Michael Friedman asserts that $\pi_0 Homeo(M)$ is isomorphic to the group of automorphisms $Aut(Q)$ of the intersection quadratic form $Q$ on the middle homology. This is an arithmetic group, and hence it contains a finite index subgroup that is torsion free. However, $Aut(Q)$ will almost always have torsion, except possibly in some low-rank cases.

One way of thinking about this is: For surfaces, the map from MCG to $Aut(Q)=Sp_{2g}(\mathbb{Z})$ (sending a homeomorphism to the induced automorphism of homology) has a huge kernel, namely the Torelli group. But in dimension 4 under the simply connected hypothesis this map is an isomorphism.


Here's one case where you can answer (3).

Let $M$ be a compact hyperbolic manifold of dimension $ \geq 3$. By Mostow rigidity, $HomEq(M)$ has the same homotopy-type as its isometry group, a finite group. By design, $Isom(M) \simeq Out(\pi_1 M)$. This latter isomorphism is because $M$ is a $K(\pi,1)$ space.

So $\pi_0 HomEq(M)$ contains elements of finite order if and only if $Out(\pi_1 M)$ does, if and only if this hyperbolic manifold has symmetries.


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