6
$\begingroup$

Suppose $(M, g)$ is an open complete nonnegatively curved Riemannian manifold with $d$ its distance.

A totally convex set $C\subset M$ has the property that for any two point $x, y \in C$ any geodesic (not only the minimal ones) joining them must lie in $C$.

If $C$ is totally convex, the fattened set $C^a=\{x \in M \mid d(x,C)\leq a\}$ is still totally convex? Is it at least for small $a$?

This question is somehow related to question "Examples on small cut radius of totally convex set in non-negatively curved manifold", Examples on small cut radius of totally convex set in non-negatively curved manifold

$\endgroup$

1 Answer 1

6
$\begingroup$

No. Consider a rotation-symmetric metric on $\mathbb R^2$ resembling a small spherical cap extented by a flat cone. A sufficiently short geodesic segment at the origin is totally convex in your sense. But a small neighborhood of such a segment is not convex due to positive curvature.

$\endgroup$
0

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.