3
$\begingroup$

Let $M$ be a Riemannian manifold with $\text{Sec}\ge 0$. From Topogonov Theorem follows that for every $p \in M$ the quantity $$ \frac{\text{Diam}(B_r(p))}{r} $$ is non-increasing in $(0,\infty)$. Does this hold also if we only assume $\text{Ric}\ge 0$? I suspect that this is no longer true, however I am not able to build a counterexaple.

Clarification: $\text{Diam}(B_r(p)):=\sup_{x,y\in B_r(p)}d(x,y).$

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ It is a theorem due to Abresch-Gromoll that diameter growth is linear. Any complete Riemannian manifold with $Ric \geq 0$ has diameter growth of order $O(r)$ with respect to any point. $\endgroup$
    – C.F.G
    Sep 11, 2020 at 13:06
  • 1
    $\begingroup$ The notion of diameter growth in that paper is not the one I am interested in, here by $\text{Diam}(B_r)$ I mean the supremum of $d(x,y)$ among all $x,y \in B_r$, which clearly implies $\text{Diam}(B_r)\le 2r.$ $\endgroup$
    – mrprottolo
    Sep 11, 2020 at 13:19
  • $\begingroup$ I would try to modify the Perelman's example library.msri.org/books/Book30/files/perex.pdf $\endgroup$
    – Anton Petrunin
    Dec 2, 2022 at 20:21

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.