MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M$ be an $n$-dimensional Riemannian manifold with sectional curvature lower bound 1. Fix a point say $O\in M$, let $S(r)$ denote the distance sphere centered at $O$ with radius $r$. The classical Hessian comparison theorem says that the principle curvatures of $S(r)$ is less than that of standard sphere ${S}^n(1)$. And Toponogov triangle comparison implies that given any two point in $S(r)$ there distance in $M$ is less than or equal to the correspond distance in round sphere with the same openning angle at $O$.

So is there any way to see how the intrinsic diameter (i.e. the length metric induced from ambient metric) upper bound?

How about the Ricci curvature case?

share|cite|improve this question
up vote 3 down vote accepted

I guess you want to ask is it true that $$\mathop{\rm IntrinsicDiameter}[S(r)]\le\mathop{\rm IntrinsicDiameter}[\tilde S(r)],$$ where $\tilde S(r)$ denotes the sphere of radius $r$ in the standard sphere.

  • This is true if $r\ge \tfrac\pi2$; it follows since $S(r)$ has bigger curvature than $\tilde S(r)$ in the sense of Alexandrov.

  • Note that if $r<\tfrac\pi2$ then $S(r)$ might be not connected; in this case $$\mathop{\rm IntrinsicDiameter}[S(r)]=\infty.$$ If sectional curvature $\ge 1$, I do not see other counterexamples. It reminds me some questions related to the conjecture that boundary of Alexandrov space is an Alexandrov space.
    If you find a way to prove it then likely you will get some nontrivial corollaries of this conjecture say if $\Sigma$ is an Alexandrov space with curvature $\ge 1$ then $\mathop{\rm diam}\partial\Sigma\le \pi$ or perimeter of any triangle in $\partial\Sigma$ is at most $2{\cdot}\pi$.
    If $r\le\tfrac\pi2$, it is possible to construct a short map $h_r\colon \tilde S(r)\to M$ so that its image covers $S(r)$. In particular $$\text{area}[S(r)]\le\text{area}[\tilde S(r)]$$ (which is obvious anyway). In general the image of $h_r$ contains creases which stick inside $S(r)$ which in principle might be used as a shortcut.

  • For Ricci curvature the statement does not hold even if $S(r)$ is connected. You may take a small disc in hyperbolic plane and take a warp product with the sphere to make the Ricci curvature of obtained manifold to be colose to $+\infty$. The sphere $S(r)$ will have intrinsic diameter bigger than $\tilde S(r)$ as far as $S(r)\ne\emptyset$.

share|cite|improve this answer
@Anton, Is there any background material for this open problem you mentioned? – J. GE Mar 22 '13 at 10:24
@Sergei, if $r\ge\tfrac\pi2$ then $S(r)$ is the boundary of convex set, so it has to be connected. – Anton Petrunin Mar 23 '13 at 17:35
What is an example where $S(r)$ is not connected? – horse with no name Mar 23 '13 at 23:53
@horse, Here is the example, take a small circle $C$ and consider the spherical suspension over it. it is an Alexandrov space with two singular non-smooth conic point at the north and south poles. Then any point close to the poles will have disconnected distance sphere $S(r)$ when $r$ is small enough, as there won't be geodesic passing through the poles. Smooth the metric in arbitary small neighborhood of poles would give you a smooth Riemannian manifold. – J. GE Mar 25 '13 at 10:49
@horse, the example looks like the surface of a cigar. Its length can be arbitrary close to $\pi$; so if the center is near the middle and say $r=\tfrac\pi3$ then $S_r$ is formed by two circles near surrounding the ends of the cigar. – Anton Petrunin Mar 25 '13 at 20:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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