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There is the following rigidity theorem of Cohn-Vossen as stated on p. 86 of these lecture notes: http://www.math.brown.edu/~deigen/chern.pdf

Any isometry between two closed smooth convex surfaces (equipped with the induced path metrics) in the Euclidean space $\mathbb{R}^3$ is established by an isometry of $\mathbb{R}^3$.

In his book "Extrinsic geometry of covex sufaces" Pogorelov generalizes (partly?) this result to a pair of convex surfaces without any extra assumptions on regularity in the following form:

If two closed convex surfaces in $\mathbb{R}^3$ are isometric with respect to the induced inner (path) metric then they are congruent, i.e. there is an isometry of $\mathbb{R}^3$ which maps one surface to another one.

It seems to me that the second statement, being specialized to smooth surfaces, is weaker than the first one.

ADDED: Consider say the special case when the two surfaces are smooth and coincide with each other. Then the Pogorelov theorem is trivial in this case, while the Cohn-Vossen theorem gives non-trivial information on isometries of the surface (all of them extend to isometries of $\mathbb{R}^3$).

Thus I am wondering whether any isometry between two closed (not necessarily smooth) convex surfaces in the Euclidean space $\mathbb{R}^3$ is established by an isometry of $\mathbb{R}^3$. In particular, is it true that any self-isometry of a closed convex surface is established by an isometry of $\mathbb{R}^3$?

A reference would be very helpful.

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  • $\begingroup$ Any distance-preserving map of smooth Riemanian manifolds is a smooth isometry in the Riemannian sense (for a long history of this problem and minimal regularity results see shttps://arxiv.org/abs/1605.03850). Thus Pogorelov's theorem is stronger than Cohn-Vossen's. Incidentally, a presumably easier proof of Pogorelov's theorem was given in the 1950s by Volkov who actually proved a much stronger stability result. In your question it is unclear which metric on a convex surface you consider. The induced path metric? The restriction of the distance function? $\endgroup$ Aug 9, 2019 at 13:01
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    $\begingroup$ I do not understand what the issue is. Cohn-Vossen's and Pogorelov’s have the same conclusion. In the smooth case they also have the same assumption. $\endgroup$ Aug 9, 2019 at 13:39
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    $\begingroup$ Oh, I see the issue. This is explained in the 4th paragraph on p.5 of arxiv.org/abs/1705.01223. $\endgroup$ Aug 9, 2019 at 14:12
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    $\begingroup$ @DeaneYang: There is a survey "Geometry of surfaces in Euclidean spaces" (1989) by Yu. Burago (in Russian). He claims that there are three approaches to prove Pogorelov's theorem. Two of them are due to Pogorelov, and the third one due to Yu.A. Volkov. I have not studied any of them. $\endgroup$
    – asv
    Aug 9, 2019 at 19:08
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    $\begingroup$ Together with several colleagues we tried to read Volkov's paper (both the Russian original and English translation) and could not understand some of his arguments. His approach is very interesting, but many things must be clarified, if not reproved. $\endgroup$ Aug 29, 2019 at 15:47

1 Answer 1

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Below is the answer from the comments. First some terminology.

A convex surface is the boundary of a compact convex body in $\mathbb R^3$. Each convex surface comes with two metrics: the path-metric and the metric obtained by restricting the distance function on $\mathbb R^3$, which we call intrinsic and extrinsic, respectively. A homeomorphism $f:(A, d_A)\to (B, d_B)$ of metric spaces is a $\delta$-isometry if $$|d_B(f(x), f(y))-d_A(x, y)| < \delta$$ for any $x, y\in A$. Of course, $f$ is an isometry if it a $\delta$ isometry for every $\delta$.

A stability theorem of Volkov (translated to English as an appendix in [A. D. Alexandrov selected works. Part II, Chapman & Hall/CRC, Boca Raton, FL, 2006, Intrinsic geometry of convex surfaces] says that if $S_1, S_2$ are convex surfaces with intrinsic metrics $\rho_1$, $\rho_2$, and extrinsic metrics $d_1, d_2$ , and if $f:(S_1 , \rho_1) \to (S_2, \rho_2)$ is an $\epsilon$-isometry, then $f:(S_1 , d_1) \to (S_2, d_2)$ is an $C_1 \epsilon^\beta$-isometry where $C$ depends onto on diameters of $\rho_1$, $\rho_2$ and $\beta$ is a positive universal constant.

By Theorem 2 in [Alestalo, P.; Trotsenko, D. A.; Väisälä, J. Isometric approximation. Israel J. Math. 125 (2001), 61–82] any $\delta$-isometry between compacta in $\mathbb R^n$ can be approximated by the restriction of an isometry of $\mathbb R^n$ with the additive error at most $C_2\sqrt{\delta}$, where $C_2$ depends only on $n$ and the diameters of the compacta. Here the metric on a compactum is the restriction of the distance on $\mathbb R^n$.

Thus every path-isometry of convex surfaces extends to an isometry of $\mathbb R^3$.

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  • $\begingroup$ Let me try to summarize my understanding of your answer. 1) By Volkov's theorem any intrinsic isometry between two convex surfaces is an extrinsic isometry. 2) Any extrinsic isometry between two compact subsets of $\mathbb{R}^n$ extends to an isometry of $\mathbb{R}^n$. ( The second step seems to be elemetary and I think I can prove it.) Please correct me if I am wrong. $\endgroup$
    – asv
    Aug 10, 2019 at 8:06
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    $\begingroup$ @MKO: this is correct. $\endgroup$ Aug 10, 2019 at 11:03
  • $\begingroup$ Thanks very much. In that case I am wondering if one needs the full power of Volkov's theorem to prove step 1). But this is a different question... $\endgroup$
    – asv
    Aug 10, 2019 at 12:10
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    $\begingroup$ Like I said initially I think everything should already follow from (a more technical version of ) Pogorelov's uniqueness theorem. I appealed to Volkov because he gives a stronger more satisfying result with allegedly easier proof. $\endgroup$ Aug 10, 2019 at 12:15

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