Suppose we are given a Riemannian manifold $M$ (it has dimension $n$, is metrically complete and connected) with the following property: for every point $p\in M$ and for every radius $0\leq R$ there is another point $q(p,R)\neq p$ and radius $R'(p,R)\in \mathbb{R}$ such that $$S(p,R) = S(q(p,R),R'(p,R))$$ where, provided $Q\geq 0$, $S(a,Q)\subset M$ is the set of points which are a distance $Q$ from $a\in M$ (If $Q<0$, $S(a,Q)$ is agreed to be empty).

Q: Is $M$ isometric to an $n$-dimensional sphere?

(To see why a sphere has the mentioned property, define $q(p,R)$ simply to be the point antipodal to $p$ and define $R'(p,R)=\pi R^*-R$ where $R^*$ is the radius of the sphere)

Be sure to have a look at a, virtually identical, question I asked on mathoverflow

  • $\begingroup$ Maybe you want to add that $M$ is closed and connected. Otherwise you can take a symmetric open subset, or two spheres. $\endgroup$
    – Dap
    Commented Dec 27, 2017 at 12:57
  • $\begingroup$ @Dap how is a symmetric open set works? $\endgroup$
    – user99914
    Commented Dec 27, 2017 at 12:58
  • $\begingroup$ @JohnMa ok, I guess not just any open subset works. I mean e.g. the set of points within ten degrees of the equator. $\endgroup$
    – Dap
    Commented Dec 27, 2017 at 13:01
  • $\begingroup$ you mean convex open subset of the sphere? Sorry what does "ten degree" are supposed to mean here? @Dap $\endgroup$
    – user99914
    Commented Dec 27, 2017 at 13:04
  • 2
    $\begingroup$ I think a sphere minus two antipodal points works though $\endgroup$
    – Dap
    Commented Dec 27, 2017 at 13:21

1 Answer 1


i) For $p$ we have antipodal point $q$ s.t. $$M= B_r(p)\cup B_R(q)$$ and $(S:=)\ \partial B_r(p)=\partial B_R(q)$.

For $x\in S$, there is a minimizing geodesic of unit speed $c$ (resp. $c_2$) from $p$ to $x$ (resp. from $q$ to $x$).

For small $\varepsilon>0$, note that $c'(r)$ is orthogonal to $T_{x}\ \partial B_{\varepsilon } \bigg(c(r-\varepsilon )\bigg)$.

And $c_2 '(R)$ is orthogonal to $T_{x}\ \partial B_{\varepsilon } \bigg(c_2 (R-\varepsilon )\bigg)$.

Here $B_r(p)$ contains $ B_{\varepsilon } \bigg(c(r-\varepsilon )\bigg)$ and $B_R(q)$ contains $ B_{\varepsilon } \bigg(c_{2} (R-\varepsilon )\bigg)$.

Hence interiors of $ B_{\varepsilon } \bigg(c(r-\varepsilon )\bigg)$ and $ B_{\varepsilon } \bigg(c_{qc(r)} (R-\varepsilon )\bigg)$ do not intersect so that union of $c$ and $c_2$ is smooth.

ii) Cut locus ${\rm Cut}\ (p)$ of $p$ is $\{q\}$ : If $x\in {\rm Cut}\ (p)\cap {\rm Int}\ B_R(q)$ s.t. there are at least two minimizing geodesics from $p$ to $x$, then geodesics pass through $S$. Hence $x=q$.

Hence $\exp_p\ tv,\ t\in [0,r+R]$ is minimizing for all $|v|=1$.

If ${\rm diam}\ M=d(a,b)$, then this implies that $d(a,b)\leq r+R$.

So any point ant its antipodal point give a diameter so that it is isometric to a sphere.


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