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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.

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    $\begingroup$ Just a guess: it should be possible to move all nontriviality into the $r$-disk, so your group might have $\operatorname{Diff}(D^r,S^r)$ as a deformation retract. This group is well studied, see concordances and pseudoisotopies, but I am no expert here. $\endgroup$
    – Sebastian Goette
    Commented Mar 27, 2017 at 14:14
  • $\begingroup$ I will check the details, but I believe you are right. We can deform the diffeomorphism to be constant outside a disk. I will look at the literature on the spaces $\operatorname{Diff}(D^r,S^r)$. Thanks for the suggestion. $\endgroup$
    – Martins Bruveris
    Commented Mar 27, 2017 at 14:39
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    $\begingroup$ Another name for the space Sebastian mentions is the Pseudo-isotopy diffeomorphism group of a sphere ($S^{d-1}$). It was studied by Cerf in the 70's and it is known to be connected provided $d \geq 8$. It is not known to be connected in general. For example, $d=5$ is an open problem as far as I know. $\endgroup$
    – Ryan Budney
    Commented Mar 27, 2017 at 19:39
  • $\begingroup$ In your definition can you replace the limit conditions with a uniform condition, for example, $|f|<1/2$ for $|x|>2$ and $|Df|<1/2$ for $|x|>2$? If so I believe these two spaces have the same homotopy-type. $\endgroup$
    – Ryan Budney
    Commented Mar 27, 2017 at 19:41
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    $\begingroup$ No, the space of diffeomorphisms supported in the unit $r$-disk is not the same as the space of pseudo-isotopies. The latter is connected for large values of $r$, as Ryan says. The former is very often not connected; for large values of $r$ its components correspond to types of exotic $(r+1)$-spheres. $\endgroup$
    – Tom Goodwillie
    Commented Apr 2, 2017 at 14:06

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