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The well known Pogorelov’s rigidity theorem says that if two convex closed surfaces in Euclidean 3-space are isometric to each other being equipped with their intrinsic metrics, then they are congruent, namely there is an isometry of the Euclidean space which maps one surface onto the other.

Are there counter examples to Pogorelov’s theorem in higher dimensions?

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  • $\begingroup$ False in 2-space? Two simple closed (convex) curves of the same length are isometric in their intrinsic metrics? But of course I can have ellipses of the same length but different eccentricity. $\endgroup$ Commented Sep 14, 2019 at 11:57
  • $\begingroup$ @GeraldEdgar : You are right: in dimension 2 this is false. I suspect it should be false in any dimension other than 3, but do not have counter example. $\endgroup$
    – asv
    Commented Sep 14, 2019 at 12:15
  • $\begingroup$ How about if we take $A \times B$ where $A$ comes from a counterexample in $2$-space, and $B$ is a sphere? But I suppose that gives us a counterexample of codimension $2$ not $1$. $\endgroup$ Commented Sep 14, 2019 at 13:26

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There is definitely no counterexample for convex polyhedra and no counterexample for bodies with smooth boundary with non-degenerate second fundamental form. (In the latter case even small open subsets of the boundary cannot be deformed, the argument can be found e.g. in Spivak volume 5.)

I do not know if the question was asked somewhere in full generality, although it sounds very natural.

If one could prove a sort of stability estimate: if the intrinsic metrics of surfaces are $\epsilon$-close, then the bodies are $\delta$-close, where $\delta$ depends on $\epsilon$ and some "rough" geometric characteristics of the bodies like their diameter, then one would get the uniqueness in the general case by approximation. Such a stability estimate is claimed to be proved in Volkov's article reproduced as appendix to the new edition of Alexandrov's "Convex polyhedra", but the proof is flawed.

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  • $\begingroup$ Thank you. You mean Volkov claims the result in any dimension? In particular, if one believes Volkov then the answer to my question is positive in full generality? $\endgroup$
    – asv
    Commented Sep 14, 2019 at 13:46
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    $\begingroup$ No, Volkov claims this for dimension 3 only. This does not imply higher dimensional result directly. One could try to do induction on the dimension (considering the vertex links, then one needs stability for spherical polytopes as well), but this is not straightforward, because the number of vertices may be arbitrarily large. On the other hand, if one finds a correct proof in dimension 3, maybe it easily generalizes to higher dimensions... $\endgroup$ Commented Sep 14, 2019 at 13:52

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