# Homotopy equivalence of diffeomorphism groups

Let $M$ be a closed connected smooth manifold and let ${\rm Diff}^r(M)$ be the group of $C^r$-diffeomorphisms equipped with the compact-open $C^r$-topology. I am looking for a reference to the fact that the natural inclusion ${\rm Diff}^r(M)\to {\rm Diff}^1(M)$ is a homotopy equivalence for $1\leq r\leq \infty$.

I am not aware of a detailed reference but here is a sketch.

1. $\mathrm{Diff}^r(M)$ is a Hilbert manifold (i.e., it is locally homeomorphic to a separable Hilbert space). This can be found e.g., in section 10 of Michor's "Manifolds of differentiable mappings". EDIT: Instead of referring to Michor who indeed focuses on the $C^\infty$ case let me give a direct proof. Let $\exp$ be the exponential map of some smooth Riemannian metric on $M$. Then any self-diffeomorphism $\phi$ of $M$ defines a vector field $X$ given by $\phi(p)=\exp_p X(p)$. Conversely, any vector field on $M$ can be exponentiated to a self-map of $M$ so that the vector field is close to zero if and only if the corresponding diffeomorphism is close to the identity. Also $\phi$ is $C^r$ if and only if $X$ is $C^r$. This defines a homeomorphism of a neighborhood of the identity in $\mathrm{Diff}^r(M)$ onto a neighborhood of the origin in the separable Frechet space of $C^r$ vector fields on $M$. Since $\mathrm{Diff}^r(M)$ is topologically homogeneous we get a chart in every point (not just at the identity). Finally any separable Frechet space is homeomorphic to $\ell^2$, so $\mathrm{Diff}^r(M)$ is a Hilbert manifold.

2. Any Hilbert manifold is an ANR (because $\ell^2$ is an ANR, and any local ANR is an ANR). Also any ANR it is homotopy equivalent to a CW complex. In particular, there is Whitehead's theorem and to show that the inclusion $\mathrm{Diff}^r(M)\to\mathrm{Diff}^1(M)$ is a homotopy equivalence it is enough to prove that it induces an isomorphism on homotopy groups.

3. Build a continuous smoothing operator that takes as an input a $C^r$ self-map of $M$, and instantly makes it $C^\infty$. This is a good elementary exercise in differential topology. (Sketch: $C^\infty$ embed the compatible $C^\infty$ structure on $M$ into some $\mathbb R^n$, use methods of Chapter 2 section 2 of Hirsch's "Differential topology" to continuously approximate $C^r$ maps $M\to\mathbb R^n$ by $C^\infty$ maps, and then use tubular neighborhood projection to push the maps to $M$).

4. The smoothing operator instantly pushes a continuous map from a disk to $\mathrm{Diff}^1(M)$ into $C^\infty(M, M)$, and hence into $\mathrm{Diff}^\infty(M)$ because by the inverse function theorem diffeomorphisms form an open subset in smooth maps. Thus any singular sphere in $\mathrm{Diff}^1(M)$ can be deformed to $\mathrm{Diff}^r(M)$, and if a singular sphere $\mathrm{Diff}^r(M)$ contracts in $\mathrm{Diff}^1(M)$, then the null-homotopy can be pushed into $\mathrm{Diff}^r(M)$.

In summary, the inclusion $\mathrm{Diff}^r(M)\to \mathrm{Diff}^1(M)$ induces an isomorphism on homotopy groups, and hence a homotopy equivalence.

• Dear Igor, thank you for a great answer. The book by Michor you mention in (1) seems to deal with ${\rm Diff}^{\infty}$ only (or maybe I am looking at a wrong place); this group is a Frechet manifold only, see [Hamilton MR0656198]. – Jarek Kędra Jan 7 '18 at 10:16
• Also I don't think ${\rm Diff}^r(M)$ is a Hilbert manifold. You need to extend it to somewhat bigger space of Sobolev maps. And this bigger space should be homotopy equivalent to ${\rm Diff}^r(M)$. – Jarek Kędra Jan 7 '18 at 12:08
• @JarekKędra: Any separable Frechet space is homeomorphic to $\ell^2$ so the notions of a Frechet manifold and Hilbert manifold are the same topologically. I replaced the reference to Michor with a direct proof. The main work is in part 3 where one has to be take care to construct a continuous (!) smoothing. – Igor Belegradek Jan 7 '18 at 13:59
• Regarding item 3 it seems the easiest to replace $M$ with its tubular neighborhood in some $\mathbb R^n$. Any diffeomorphism of $M$ extends to a diffeomorphism of such a neighborhood. Then we just deal with $C^r$ maps of open subsets of $\mathbb R^n$ and one can smooth them by convolution explicitly, and the smoothing would be continuous. The advantage here is that we deal with one chart and won't have to fit together smoothings on different charts (on $M$). – Igor Belegradek Jan 7 '18 at 15:37