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Suppose $M$ is a 1-connected closed manifold with sectional curvature $\ge 1$. So the diameter $D$ of $M$ satisfies $$ D \le \pi $$ When equality holds $M$ is isometric to round sphere. In fact this rigidity holds under the assumption of $Ric \ge n-1$. Hence it is natural to ask what happens to almost extreme case. i.e. when $$D \ge \pi -\epsilon $$Under the Ricci assumption only, the manifold is not necessary sphere by a counter example of M. Andersen. However with extra assumption: sectional curvature is bounded from below, Perelman proved it is homeomorphic to a twist sphere. (Is Perelman the first one prove this?)

By Grove-Shiohama's Diameter Sphere Thereom, under the assumption sectional curvature $\ge 1$ and $D\ge \pi-\epsilon$, $M$ is a twist sphere. (It also follows from Perelman's theorem above)

My question is: Is $M$ diffeomorphic to the standard sphere?

Either Perelman's proof or Grove-Shiohama's Diameter sphere theorem uses the 'soft' approach, i.e. It is not by convergence argument, nor a Lipschitz distance between $M$ and $S^n$ is derived.

Actually, I suspect that it is not Gromov-Hausdorff close to the round sphere, as one might round off two tips of $S^2/\mathbb Z_p$, where $\mathbb Z_p$ acts on $S^2$ by rotation along the $z$-axis.

What if one assume there is also an upper bound $K$ on sectional curvatures, i.e. $$1\le sec(M) \le K$$

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@J Ge: This is an open problem, if memory doesn't fail me. I actually remember talking about it with Grove in one of our group meetings. Also, I believe that the recent work of Curtis Pro might shed some light on that, although, if I remember well, he assumes almost maximal volume. Perhaps you should contact those two people to figure out the state of the art regarding this question. –  Renato G Bettiol May 2 '13 at 2:14
    
@Renato, Thanks for the comment. –  J. GE May 6 '13 at 13:31
    
@J Ge: Your question is essentialy Problem 1.2 in the "Diameter pinching" section of the problem list from the "Manifolds with non-negative sectional curvature" workshop at AIM (aimpl.org/nnsectcurvature/1). The problem is still open, to the best of my knowledge. –  F.G. May 17 '13 at 12:29

1 Answer 1

up vote 3 down vote accepted

I will answer the last question. Namely, let me show that if $1\le \mathrm{sec}\,M \le K$ and $\mathrm{diam}\, M>\pi-\varepsilon$ for sufficiently small $\varepsilon>0$ then it has to be diffeomorphic to $\mathbb S^m$.

Asssume contrary, then there is a sequence of $m$-dimensional manifolds $M_n$ such that $1\le \mathrm{sec}\,M_n \le K$ and $\mathrm{diam}\, M_n\to\pi$ as $n\to \infty$.

Let's pass to a converging sequence $M_n\to A$. Note that $A$ is $m$-dimensional and $1\le \mathrm{sec}\, A \le K$ in the sense of Alexandrov and $\mathrm{diam}\, A=\pi$. It follows that $A$ is isometric to $\mathbb S^m$

Finally note that $M_n$ is diffeomorphic to $A$ for all large $n$; the later was essentally proved by Cheeger in his thesis.

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