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 Dec 27 '17 at 12:57
  • $\begingroup$ @Dap how is a symmetric open set works? $\endgroup$ – user99914 Dec 27 '17 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 Dec 27 '17 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 Dec 27 '17 at 13:04
  • 2
    $\begingroup$ I think a sphere minus two antipodal points works though $\endgroup$ – Dap Dec 27 '17 at 13:21

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.