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Let $K$ be a 3-dimensional convex compact subset either in $\mathbb{R}^3$ or in the 3-dimensional real hyperbolic space $\mathbb{H}^3$. Consider its boundary $\partial K$ equipped with the inner (path) metric induced from the ambient space.

Is it true that any isometric map $T\colon \partial K\to \partial K$ is a restriction of an isometry of the ambient space ($\mathbb{R}^3$ or $\mathbb{H}^3$)?

If it matters, I am particularly interested in the case when $T$ is an isometric involution without fixed points.

A reference would be very helpful.

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    $\begingroup$ If the answer is yes, there is some subtle reason, because it's obviously false in dimension 2 (the boundary with its path metric has many self-isometries, but they don't all extend unless $K$ is a disc). $\endgroup$
    – YCor
    Commented Aug 7, 2019 at 15:28
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    $\begingroup$ The answer to your question is yes. It follows from the Cohn-Vossen rigidity theorem. This is stated and proved starting on page 86 of math.brown.edu/~deigen/chern.pdf. $\endgroup$
    – Deane Yang
    Commented Aug 7, 2019 at 16:17
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    $\begingroup$ In higher dimensions the result is easy to prove when the second fundamental form is positive definite. That's because at a point where the second fundamental form is positive definite, the Riemann curvature tensor uniquely determines the second fundamental form. $\endgroup$
    – Deane Yang
    Commented Aug 7, 2019 at 16:21
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    $\begingroup$ If the boundary of $K$ is smooth, then, as Deane Yang points out, this is the Cohn-Vossen rigidity theorem. The same is true without the smoothness assumption, which is the Pogorelov rigidity theorem (with a notoriously complicated proof). $\endgroup$
    – Ivan Izmestiev
    Commented Aug 7, 2019 at 18:10
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    $\begingroup$ I believe that Pogorelov proved it in complete generality. Any two isometric convex surfaces in $\mathbb{R}^3$ differ by a rigid motion. It’s in his book books.google.com/books/about/… $\endgroup$
    – Deane Yang
    Commented Aug 8, 2019 at 0:36

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