# Diameter estimate of distance sphere of positive curved manifold

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?

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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 $S(r)\ne\emptyset$.

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@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