Is the group $$ \operatorname{Diff}^1_0(\mathbb R^d) = \operatorname{Diff}^1(\mathbb R^d) \cap \big(\operatorname{Id}_{\mathbb R^d} + C^1_0(\mathbb R^d,\mathbb R^d)\big) $$ connected? Here $$ C^1_0(\mathbb R^d,\mathbb R^d) = \{ f \in C^1(\mathbb R^d,\mathbb R^d) \,:\, |f(x)|\to 0,\, |Df(x)| \to 0\text{ as } x \to \infty \}\,, $$ and we equip $\operatorname{Diff}^1_0(\mathbb R^d)$ with the uniform $C^1$-topology. Thus $\operatorname{Diff}^1_0(\mathbb R^d)$ consists of diffeomorphisms that decay to the identity towards infinity in the $C^1$-norm; in particular diffeomorphisms in $\operatorname{Diff}^1_0(\mathbb R^d)$ are orientation preserving.

The group $\operatorname{Diff}^1(\mathbb R^d)$ of $C^1$-diffeomorphisms equipped with the weak Whitney topology has two connected components—the orientation preserving and orientation reversing diffeomorphisms. To see this one can use the homotopy $$ \varphi_t(x) = \frac 1t \big(\varphi(tx) - \varphi(0)\big) + t\varphi(0)\,, $$ with $\varphi_0(x) = D\varphi(0)x$. This is a deformation retraction of $\operatorname{Diff}^1(\mathbb R^d)$ unto $GL(\mathbb R^d)$. Since this homotopy works by pushing all nonlinearities out to infinity, it is not continuous in the strong Whitney topology or in the uniform $C^1$-topology used above.

Is there some other construction that can be used to show connectedness of $\operatorname{Diff}^1_0(\mathbb R^d)$? More generally, I am interested in whether the Sobolev diffeomorphism groups
$$ \mathcal D^s(\mathbb R^d) = \operatorname{Diff}^1(\mathbb R^d) \cap \big(\operatorname{Id}_{\mathbb R^d} + H^s(\mathbb R^d,\mathbb R^d)\big)\,,$$
with $s > d/2+1$ are connected. However, a positive answer for $\operatorname{Diff}^1_0(\mathbb R^d)$ together with results from Palais' *Homotopy Theory of Infinite Dimensional Manifols* will imply the result for Sobolev diffeomorphisms.

concordancesandpseudoisotopies, but I am no expert here. $\endgroup$ – Sebastian Goette Mar 27 '17 at 14:14