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What is known about the homeomorphism (or homotopy) type of the group of $C^\infty$ diffeomorphisms of $S^2$ equipped with $C^0$ topology?

The group is the image of the inclusion $\mathrm{Diff}(S^2)\to \mathrm{Homeo}(S^2)$, where $\mathrm{Diff}(S^2)$ has $C^\infty$ topology. The inclusion is (I think) a homotopy equivalence, yet it is not a homeomorphism onto its image, so it does not seem to help answer the question.

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  • $\begingroup$ I'm not sure exactly what you're fishing for. The isomorphism type of $Diff(S^2)$ is enough to recover the topology, so weakening the topology doesn't lose any information about the smooth structure. $\endgroup$
    – Ian Agol
    Commented Sep 10, 2013 at 22:35
  • $\begingroup$ Ian: the homeomorphism type of the group I ask about is directly related to the homeomorphism type of the space of smooth nonnegatively curved metrcs on $S^2$ equipped with $C^0$ topology, cf. my paper arxiv.org/abs/1307.3167 where I deal with $\mathbb R^2$ instead of $S^2$. $\endgroup$
    – Igor Belegradek
    Commented Sep 10, 2013 at 22:53
  • $\begingroup$ For example, is it true that the group I am asking about is a Hilbert manifold, i.e. is every point has a neighborhood homeomorphic to the separable Hilbert space? $\endgroup$
    – Igor Belegradek
    Commented Sep 11, 2013 at 1:21

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The homotopy type seems to be independent of considering this group with $C^0$ or $C^{\infty}$ topology. As far as I know, the group of oriented diffeomorphisms of $S^2$ is homotopy equivalent to the group $PSL(2, C)$ (or $SO(3, R))$. I do not know direct reference, but maybe I can give a sketch of the proof here:

We are going to prove that the stabilizer of the triple of points $(0, 1, \infty)$ is homotopy equaivalent to a point. First of all, we need to establish the isomorphism between this stabilizer and the space of all smooth almost complex (<=> complex, in 2 dimensions these notions are the same) structures on the sphere $S^2$. In order to do it we use the Riemann's existence theorem, which states, that any complex curve of genus $0$ is isomorphic to $CP_1$:

Step 1: We fix the standard complex structure $I$ on the sphere $S^2$.

Step 2: We map the diffeomorphism $f$ to the structure $f^*I$

Step 3: For $f$ and $g$ if $f^*I = g^*I$ then $g^{-1}f$ preserves $I$ and also $(0, 1, \infty)$ and hence it's identity. So $f = g$. Our mapping is injective.

Step 4: By the Riemann's existence theorem our mapping is surjective.

The space of all almost complex structures on a 2-dimensional sphere is contractable, because it is just a quotient space of the space of all riemannian metrics modulo conformal factors. Only thing we need to check is that nearby almost complex tensors will correspond to the nearby diffeomorphisms, and it can be done using smth like infinite-dimensional inverse function theorem...

And then, weakening the topology to the $C^0$ obviously won't make this space uncontractable.

And on a homeomorphism type... Maybe it makes sense to ask what is the corresponding to the $C^0$ topology on the space of complex structures. I suspect that it's topology given by the norms of quasiconformal mappings, but actually, I dunno.

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    $\begingroup$ I am aware of all this (and the reference is Earle-Eeels, 1969 JDG), but I do not see why the argument works in $C^0$ topology, in particular I do not understand your "obviously won't make this space uncontractable". $\endgroup$
    – Igor Belegradek
    Commented Sep 11, 2013 at 2:52
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    $\begingroup$ What I wanted to say is that the space of riemmanian metrics is contractable in any topology, $C^0$ or $C^\infty$. Maybe, the problem with my argument is in the check that inverse function theorem can be applied (i.e. the morphism from diffeomorphisms to the almost-complex structures is differentiable as the morphism of Banach spaces with $C^0$ topology?) $\endgroup$
    – Lev Soukhanov
    Commented Sep 23, 2013 at 18:18
  • $\begingroup$ I agree with what you say about contractibility in $C^0$ topology (thanks for this). The inverse function theorem seems problematic but looks like it is not needed: every metric on $S^2$ is written uniquely as $\phi^*(e^u g_1)$ where $\phi$ is a self-diffeomorphism of $S^2$ fixing $0,1, \infty$, $u$ is a smooth function on $S^2$ and $g_1$ is the curvature $1$ metric, and the map $\phi^*(e^u g_1)\to \phi$ is probably a fiber bundle with contractible total space and fibers. If so, the base is contractible. $\endgroup$
    – Igor Belegradek
    Commented Sep 26, 2013 at 15:05
  • $\begingroup$ Actually, I cannot convince myself why the $\phi$ depends continously on the metric - I wanted to use inverse function theorem for it... $\endgroup$
    – Lev Soukhanov
    Commented Sep 26, 2013 at 18:29
  • $\begingroup$ The diffeomorphism $\phi$ solves Beltrami equation in which the dilatation $\mu$ continuously depends on the metric. Ahlfors (or maybe Ahlfors-Bers?) proved that the solution of Beltrami equation depends continuously on $\mu$. $\endgroup$
    – Igor Belegradek
    Commented Sep 27, 2013 at 2:41

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