If for any two points p,q in a regular, compact surface $S\subseteq R^3$, there exists an isometry $f:S\rightarrow S$ s.t. f(p)=q. How to prove that $S$ is locally isometric to the sphere?
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$\begingroup$ Your question is an exercise for students, this place is instead for research problems. Try at the web site math.stackexchange.com as a better option, or ask a classmate or teacher. $\endgroup$– Daniele ZuddasCommented May 17, 2013 at 13:43
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1$\begingroup$ Your question is also ambiguous, as there are several notions of "isometry" one can use. In the easiest case, an isometry is the restriction of an isometry of $R^3$. Then it is an undergraduate-level exercise. If by isometry you mean an isometry $f: S\to S$ as a Riemannian manifold then the answer is still positive, but, now, it is a graduate-level exercise which one can assign in a Riemannian geometry class. $\endgroup$– MishaCommented May 17, 2013 at 15:42
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4$\begingroup$ The "Riemmanian" version of the question is not that trivial IMHO. A possible proof: the gauss curvature of the surface is constant because it is preserved by the isometries. It is positive since every compact surface must have a point where the curvature is positive. Then, the surface is diffeomorph to the sphere or to the projective plane. The latter can not be imbedded in $R^3$ (and the proof of this fact is much easier under the assumption that the curvature is positive) so the surface is the sphere. Now, by the Alexandrov theorem the sphere is standartly imbedded. $\endgroup$– Vladimir S MatveevCommented May 19, 2013 at 7:49
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