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Is there a convenient theorem about which diffeomorphisms $f: \mathbb R^3\rightarrow\mathbb R^3$ can be extended to diffeomorphisms $\overline{f}: S^3\rightarrow S^3$?

That is, given a diffeomorphism $f:\mathbb R^3\rightarrow\mathbb R^3$, when does there exist an inclusion $i:\mathbb R^3\rightarrow S^3$ inducing a homeomorphism between $R^3$ and $i(\mathbb R^3)$ such that the map $\overline{f}: S^3\rightarrow S^3$ which fixes the point at infinity and is equal to $i\circ f\circ i^{-1}(x)$ for every other $x\in S^3$ is a diffeomorphism?

Kenta S
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George K
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  • Could you elaborate on what you mean by "extension"? Do you have a particular inclusion $\mathbb{R}^3\hookrightarrow S^3$ in mind? – Kajelad Sep 17 '20 at 13:47
  • @Kajelad Yes, sorry about that. Any inclusion $i$ will do, as long as it yields a homeomorphism between $R^3$ and $i(R^3)$. For instance, the stereographic projection would be great. I have now edited the question. – George K Sep 17 '20 at 14:03
  • @LeeMosher Could you please clarify? How does the fact that the one-point compactification of $R^3$ is $S^3$ resolve the question? It certainly means that every homeomorphism of $R^3$ can be extended to a homeomorphism of $S^3$, but why is this also true for diffeomorphisms? – George K Sep 17 '20 at 17:26
  • I see that I missed your emphasis on diffeomorphisms at the end of the second paragraph, so I'll delete my comment. – Lee Mosher Sep 17 '20 at 17:36

1 Answers1

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Here's one approach for stereographic projection, which can be generalized to any inclusion which maps onto the sphere minus one point.

Let $i:\mathbb{R}^3\to S^3$ be the standard stereographic projection, and $p\in S^3$ such that $i(\mathbb{R}^3)=S^3\setminus p$.

The extension $\bar{f}$ is automatically determined on $S^3\setminus p$, and if a smooth extension exists, then the limits at $p$ of $\bar{f}$ and all of its derivatives must be well behaved. This corresponds to a set of conditions "at infinity" for $f$ and its derivatives. A simple way to keep track of these is by choosing a convenient chart around $p$.

For instance, in this case we can use the reversed stereographic projection as a chart. The local representative $g:\mathbb{R}^3\setminus 0\to\mathbb{R}^3\setminus 0$ is given by $$ g(x)=f\left(\frac{x}{\|x\|^2}\right) $$ and $f$ has a smooth extension iff $g$ and all of its partial derivatives have finite limits at zero. Furthermore, this extension is a diffeomorphism if the limit of $dg$ at zero is full rank, i.e. if the limit of the Jacobian determinant of $g$ at zero is nonzero.

Kajelad
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