# Manifolds whose diffeomorphism group has the homotopy type of a manifold itself

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?

• 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. Jul 4, 2016 at 18:27
• @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? Jul 4, 2016 at 18:58
• 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 Jul 4, 2016 at 19:10

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.
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.
• 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. Jul 4, 2016 at 18:51
• As I understand it, Milnor found that $Diff(S^6)$ has 28 components, so in particular isn't homotopic to $SO(7)$. Jul 24, 2016 at 21:43