3
$\begingroup$

I have a very stupid question. Let $M$ be a closed smooth manifold. In particular cases the homotopy type of the diffeomorphism group $Diff(M)$ can be very pathological. For example, in the case of $M=T^{n-2}\times S^2$ we have that the group $\pi_i(Diff(M))$ has infinite rank for any $n\geq 5$ and $0\leq i\leq n-3$. But my question is about an opposite situation. Does someone know when the diffeomorphism group $Diff(M)$ has the homotopy type of another closed smooth manifold? Or a more simple question, when $Diff(M)$ is at least a Poincare space?

$\endgroup$
3
  • 1
    $\begingroup$ There appear to be very few such cases, and they are only in low dimensions. ThiKu's response covers many of them, but of course most geometric 3-manifolds satisfy your condition. For higher-dimensional manifolds the answer could very well be never, as far as I know. $\endgroup$ Jul 4, 2016 at 18:27
  • $\begingroup$ @RyanBudney Yes, I am more interesting in higher dimensional manifolds. And, please, could you say what kind of reason for failing $Diff(M)$ to be a manifold in this case? Is there a computation of the cohomology ring $H^*(Diff(M),\mathbb{Q})$? Or is there a more elementary argument? $\endgroup$ Jul 4, 2016 at 18:58
  • $\begingroup$ Yes, there are some results of that kind. They go back to the work of Antonelli, Burghelea and Kahn. I believe they show that cohomology is infinitely-generated, but it's been a while since I've looked at their paper. P.L. Antonelli, D. Burghelea, P.J. Kahn, "The non-finite homotopy type of some diffeomorphism groups" Topology , 11 : 1 (1972) pp. 1–49 $\endgroup$ Jul 4, 2016 at 19:10

1 Answer 1

6
$\begingroup$

If $M$ is a surface of genus $g\ge 2$, then the Earle-Eells Theorem asserts that the connected components of Diff(M) are contractible, so Diff(M) has the homotopy type of a non-compact 0-dimensional manifold.

The same is true for hyperbolic manifolds of dimension 3, except that there are only finitely many components then. (Gabai)

Moreover, it is a Theorem of Smale that $Diff(S^2)$ deformation retracts to $SO(3)$. The analogous statement that $SO(4)$ is a deformation retract of $Diff(S^3)$ was known as the Smale conjecture and was proved by Hatcher.

$\endgroup$
2
  • $\begingroup$ O, thanks. It's nice examples. And of course the same statement is true in the case of the torus $T^2$. The connected component $Diff_0(T^2)$ is homotopy equivalent to $T^2$ itself. $\endgroup$ Jul 4, 2016 at 18:51
  • $\begingroup$ As I understand it, Milnor found that $Diff(S^6)$ has 28 components, so in particular isn't homotopic to $SO(7)$. $\endgroup$ Jul 24, 2016 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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