6
$\begingroup$

Let $(M,g)$ be a Riemannian manifold such that for each $C>0$ there is $p\in M$ and $X,Y\in T_pM$ unitary such that $K(X,Y) > C.$ Does this imply that the diameter of $(M,g)$ is infinite?

I just have an intuition about it, for example, by the neck singularity on Ricci flow, or by looking to the Gabriel's Horn: Gabriel.

I searched a lot for a counter-example and possible known theorem's on Petersen's book and other references, but I could conclude nothing, does anyone has a clue?

$\endgroup$
4
  • 11
    $\begingroup$ Since you did not ask for $(M,g)$ to be complete, can't you just take something like $(0,1)^2$ and add a bunch of "dimples" that get very sharp as you approach the boundary? If $(M,g)$ is complete, then the statement is true, simply because a manifold with unbounded sectional curvature must be non-compact and therefore by Hopf-Rinow it has infinite diameter. $\endgroup$ Dec 20, 2019 at 4:21
  • 2
    $\begingroup$ thank you @NateEldredge, I forgot the completeness assumption, I do appreciate your answer, that is what I was looking for. $\endgroup$ Dec 20, 2019 at 4:22
  • 3
    $\begingroup$ I’m voting to close this question because it was answered in the comments. $\endgroup$ Dec 1, 2022 at 1:56
  • $\begingroup$ @NateEldredge could you make an answer from your comment (so the question would disappear from unanswered). $\endgroup$ Dec 18, 2022 at 17:28

1 Answer 1

7
$\begingroup$

If $(M,g)$ is complete, then yes. Since sectional curvature is continuous, if it is unbounded then $(M,g)$ is not compact, and by the contrapositive of the Hopf-Rinow theorem, a complete non-compact Riemannian manifold has infinite diameter.

If $(M,g)$ is not complete, then no. You can take something like $(0,1)^n$ and modify it with little dimples that become smaller but sharper as you approach the boundary, so that their sectional curvature becomes arbitrarily large but the manifold's diameter does not increase significantly.

$\endgroup$

Your Answer

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

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