Assuming that $g$ is a metric on the n-dimensional sphere $S^n(n\geq 3)$ satisfying sectional curvature bounded: $|K(g)|\leq 1$, is the diameter of $g$ at least $\pi$? How about if we only assume that $K(g)\leq 1$?
1
-
3$\begingroup$ I don't have time for a full answer now, but there is a paper by Buser and Gromoll (with title something about almost non-negative metrics on $S^3$) which shows that under the assumption $K(g)\leq 1$, the diameter may be as small as you wish. $\endgroup$– Jason DeVito - on hiatusCommented Mar 2, 2017 at 19:52
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
If $0<h\leq K \leq H$ everywhere, then it is a result of Mei-Chin Ku (PAMS, 1976) that
$$D\geq \frac{2\pi}{3\sqrt{H}}.$$
answered Mar 2, 2017 at 20:15
-
1$\begingroup$ In my understanding, his proof is wrong because he used Klingenberg's estimate of injectivity radius without the assumption of even dimension or pinching of positive curvature. $\endgroup$ Commented Mar 4, 2017 at 4:19
-
$\begingroup$ @FengWang Interesting! Is this well-known (that this is wrong), or did you just notice it? $\endgroup$ Commented Mar 4, 2017 at 17:21
-
$\begingroup$ I don't hear of the paper before. I just read and thought it's wrong. $\endgroup$ Commented Mar 4, 2017 at 18:45