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Let $K\subset\mathbb{R}^d$ be compact such that $K^\circ\neq\emptyset$. Let $U\subset\mathbb{R}^d$ bounded and open ($U$ could e.g. be chosen to be the open unit ball). Let $Φ: K^\circ\to U$ be a diffeomorphism.

Can I find a set $V \supset K$ open and a diffeomorphism $\Psi: V \to \Psi(V)$ such that $\Psi|_{K^\circ} = \Phi$? The range of $\Psi$ is of no great importance to me, I only care that it's a proper superset of $U$.


A similar question was asked at Extending a diffeomorphism outside a compact set where a counterexample was given for non-contractable sets. However, the question asked for an extension onto the whole space $\mathbb{R}^d$, whereas I'm only interested to extend $\Phi$ a little bit outside of $K$.

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The answer is, in full generality, no. I will use Riemann's uniformisation Theorem:

If $U$ and $V$ are non-empty simply connected open subsets of $\mathbb{C}$, different than $\mathbb{C}$, then there exists a biholomorphism $\varphi\colon U \to V$. In particular, such a biholomorphism is a diffeomorphism.

Consider $\Delta = \left\{ z \in \mathbb{C} \mid |z|<1 \right\}$ and $U$ the interior of a Von Koch curve. Then $\Delta$ and $U$ are non-empty simply connected open subsets of $\mathbb{C}$: there exists a diffeomorphism $\varphi\colon \Delta \to U$.

Suppose by contradiction that $\varphi$ extends to $\psi \colon V \to \psi(V)$ as a diffeomorphism with $\overline\Delta \subset V$. Then $\psi(\partial\Delta) = \partial \psi(\Delta) = \partial U$ and $\psi$ maps the unit circle to the Von Koch curve. But a diffeomorphism sends smooth curves to smooth curves, which is a contradiction. It follows that $\psi$ does not exist.

Comment: there is no reason for a smooth / conformal / holomorphic map to extend to the boundary of a domain, as shows the above example. We indeed could have chosen a square, heptagone or any non-smooth curve instead of the Von Koch curve, but this example shows how pathological it can be (and in fact, it can be even worse).

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    $\begingroup$ In your definition of $\Delta$, did you intend a strict inequality $<$ instead of equality? $\endgroup$
    – peek-a-boo
    Commented Aug 25, 2021 at 9:00
  • $\begingroup$ @peek-a-boo Yes of course. Thank you for pointing that out! $\endgroup$
    – Didier
    Commented Aug 25, 2021 at 9:02
  • $\begingroup$ Thanks for the counter example! Would the result hold, if we assumed more regularity of $K$? Maybe $C^1$ boundary or just "$K$ such that $\Phi$ extends to $\partial K$ in a suitable sense". $\endgroup$
    – CallMeStag
    Commented Aug 25, 2021 at 9:12
  • $\begingroup$ @CallMeStag To be fair, I don't know. Extending conformal maps to the boundary is a whole field of research in complex geometry / CR geometry, and what I remember from a course in geometry is that "we don't know much things about extending maps to boundaries". $\endgroup$
    – Didier
    Commented Aug 25, 2021 at 9:15
  • $\begingroup$ All right, thanks anyway! $\endgroup$
    – CallMeStag
    Commented Aug 25, 2021 at 9:16

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