# The space of diffeomorphisms on a manifold

It is well known that given a compact connected smooth manifold without boundary $M$, the set of diffeomorphisms $Diff^{r}(M)$ of $M$ for $r≥1$, is open in $C^{r}(M)$, the set of continuous functions (for example, Thm. 1.7 - Chap 2, in "Differential Topology", M. Hirsch).

My question is: does it follow that $Diff^{r+\alpha}(M) \subset C^{r+\alpha}(M)$ is open w.r.t. the $C^{r + \alpha}$ topology for $0< \alpha \leq 1$?

If it is so, I guess the same can be said about the subset of expanding maps, right? Anything about the subset of hyperbolic systems? I guess here things are strongly dependent on the manifold, along the lines of the work of Crovisier, Bonatti, etc.

Can someone give me some references? Thanks.

• Is the notation $C^{r+a}$ completely standard? What does it mean for you? – Dylan Thurston May 14 '14 at 12:39
• I mean $C^{r + Hölder}$. – user50774 May 14 '14 at 12:41
• JFYI: Many people would write instead of $C^{r+\alpha}$, the notation $C^{r,\alpha}$ for Holder spaces. – Willie Wong May 14 '14 at 14:13
• True. I think in Analysis the $C^{r, \alpha}$ is the standard notation, whereas in Dynamical Systems the $C^{r + \alpha}$ one. – user50774 May 14 '14 at 14:19

Yes, $Diff^{r,\alpha}(M)$ is open in $C^{r,\alpha}(M,M)$. You need a Riemannian metric on $M$ to even say what a $C^{r,\alpha}$-map $M\to M$ is. This is because the $C^1$-topology is coarser than the $C^{r,\alpha}$-topology: so the embedding from $C^{r,\alpha}(M,M)$ with the $C^{r,\alpha}$-topology to $C^{1}(M,M)$ with the $C^1$-topology is continuous, and the inverse image of the open set $Diff^1(M)$ is then the open set $Diff^{r,\alpha}(M)$. See