Let $G$ be a finite dimensional connected Lie group and $Diffeo(G)$ be the diffeomorphism group of the underlying manifold. Is it true that $Diffeo(G)$ has the homotopy type of a finite dimensional Lie group? I can't seem to find a counterexample.

  • $\begingroup$ After a bit of searching I have found that if $G$ is a connected Lie group then $G$ is homeomorphic to $K\times\mathbb{R}^n$ where $K$ is a maximal compact subgroup. Moreover, $G$ deformation retracts to $K$. I am not sure if this can be used to say something about $Diffeo(G)$ in terms of $K$ plus some terms due to $\mathbb{R}^n$ (perhaps pertaining to orientations and various smooth structures in dimensions greater than 3). $\endgroup$ – Joseph Zambrano Apr 14 '14 at 17:48
  • $\begingroup$ You say you can't find any counter examples to this, but what are your examples where this is true? It seems plausible that $\mathrm{Diffeo}^+(S^1)$ should have the homotopy type of $S^1$, and I think it is a theorem that $\mathrm{Diffeo}^+(S^3)$ has the homotopy type of $\mathrm{SO}(4)$. But I don't even know what the homotopy type of $\mathrm{Diffeo}^+(\Bbb R^n)$ should be for $n\geq 2$, and wether it somehow relates to $\mathrm{SO}(n)$ or not. So what is your evidence? $\endgroup$ – Olivier Bégassat Apr 14 '14 at 21:43
  • $\begingroup$ Yes, this is Hatcher's theorem. $\endgroup$ – Moishe Kohan Apr 14 '14 at 22:18
  • $\begingroup$ @studiosus Do you know where I can find a statement of this theorem? $\endgroup$ – Joseph Zambrano Apr 14 '14 at 22:21
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    $\begingroup$ Google "Smale conjecture". Incidentally, I think already the group $U(2)$ is a counter example to your question, but I would have to do some computations to prove it. $\endgroup$ – Moishe Kohan Apr 14 '14 at 22:31

Torus of dimension $\ge 25$ is a counter example. See here.

  • $\begingroup$ Thank you for this example as well as the other helpful comments. $\endgroup$ – Joseph Zambrano Apr 16 '14 at 0:41

A lower dimensional example is $S^1\times S^2$.

Hatcher calculated the homotopy type of $Diff(S^1\times S^2)$: It is the one of $O(2)\times O(3)\times \Omega SO(3)$. As the second homotopy group of this space does not vanish since $\pi_2(O(2)\times O(3)\times \Omega SO(3))\cong \mathbb{Z}$ holds, it cannot have the homotopy type of a finite dimensional Lie group.


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