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Let $(M,g)$ be a $k$ dimensional compact Riemannan manifold which is isometrically embeded in $\mathbb{R}^{k+1}$. The distance arising from the Riemannian metric is denoted by $d_g$.The Euclidian distance of two points $x,y$ in $\mathbb{R}^{k+1}$ is denoted by $|x-y|$. Assume that for every $x,y,z,w\in M$ $d_g(x,y)=d_g(z,w) \iff |x-y|=|z-w|$

Does this imply that $M$ is a round sphere?

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    $\begingroup$ @EricCanton it is assumed that $M$ is compact $\endgroup$
    – S. carmeli
    Dec 13, 2019 at 16:17
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    $\begingroup$ I missed that. I'll leave my comment for other uncareful readers. $\endgroup$
    – Eric Canton
    Dec 13, 2019 at 16:19
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    $\begingroup$ I think it is true. Let $m\in M$. Then the distance of a geodesic of length $\epsilon$ for $\epsilon$ small enough in a direction of principal curvature can be approximated to second order in terms of the eigen-values of the first (or second? im terrible with names) fundamental form. In particular, if it is the same in both principal directions, then the principal curvatures agree up to sign. But it is known that if the second fundamental form of a surface is constant then it is a round sphere. $\endgroup$
    – S. carmeli
    Dec 13, 2019 at 17:06
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    $\begingroup$ Note that this would be false for a $1$-dimensional manifold embedded in $R^4$, as with $\{(5\sin t, 5\cos t, \sin 2t, \cos 2t)\}$, which has the property but is not a circle. $\endgroup$
    – user44143
    Dec 13, 2019 at 20:33
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    $\begingroup$ @S.carmeli I guess you can turn your argument into a proof. Note that since $M$ is compact, there is a point of maximal distance from the origin, where necessarily all principal curvatures are nonzero and of the same sign. And yes, it is the second fundamental form you want. $\endgroup$
    – Sebastian Goette
    Dec 14, 2019 at 15:20

1 Answer 1

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Suppose $\gamma$ is a geodesic in $M$. Then there is a function $\ell$ such that $|\gamma(t_0)-\gamma(t_1)|=\ell(|t_0-t_1|)$, assuming that $|t_0-t_1|$ is small. It follows that $\gamma$ has constant curvature.

Further, note that any two geodesics in $M$ are congruent. In particular, $M$ has constant principle curvatures; so $M$ is a round sphere.

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  • $\begingroup$ Thank you very much and my +1 for your answer $\endgroup$
    – Ali Taghavi
    Dec 5, 2022 at 12:06

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