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Let me ask this question without too much formalization:

Suppose a smooth surface $M$ has the property that for all spheres $S(p,R)$ (i.e. the set of all points which lie a distance $R\geq 0$ from $p \in M$, with distance as measured inside the surface), there is always a different point $q(p,R) \in M$ and a distance $D(p,R)\geq 0$ so that $$S(p,R)=S(q(p,R),D(p,R))$$ In words: both $p$ and $q(p,R) \neq p$ are a center of the sphere $S(p,R)$.

Q: Is $M$ (a part of) a sphere? More formally: is $M$ isometric to (a subset of) a sphere?

Remark: Spheres clearly possess the mentioned property, with every sub-sphere having one point $p$ ánd its antipodal $\pi(p)$ at its center. This property is maintained when we remove a finite number of pairs of antipodal points from such a sphere or when we take a few disconnected spheres.

This question has a counterpart on stackexchange.

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Blaschke's conjecture might be relevant. In particular, if the surface is $S^2$ with a $C^3$ metric, then it follows from Blaschke's conjecture, proved by Green. This states that a metric on the sphere in which every point has a unique conjugate point must be the round metric on the 2-sphere. In your condition, one may observe that $q(p,R)$ does not depend on $R$, essentially by the Gauss lemma. The point is that the sphere of radius $r$ about $B(p,R)$ (the ball of radius $R$ about $p$) is $S(p,r+R)$. Hence $q(p,R)=q(p,r+R)=q(p)$. Then $q(p)$ is the unique point conjugate to $p$.

For the general case, it might follow by taking the completion of the metric, and showing that it must be a (topological) sphere.

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  • $\begingroup$ Have you had alook at the answer on the similar question on stackexchange? math.stackexchange.com/questions/2581644/… $\endgroup$ – Thibaut Demaerel Jan 2 '18 at 18:59
  • $\begingroup$ While your answer certainly seems pleasing to me, I want to take some more time to analyse Lee's answer and see what's of it. May take a few days, unfortunately. $\endgroup$ – Thibaut Demaerel Jan 2 '18 at 19:02

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