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I am trying to imagine normal balls which are not convex. Are there any nice exmaples?

Obviously, in the Euclidean space all normal balls are convex and it seems to me this is also true for the sphere.

Perhaps a space of negative curvature? (Is there a necessary condition on the curvature for the existence of non-convex normal balls?)

Reminders:

Let $M$ be a Riemannian manifold. A normal ball around $p$, is a set of the form of $\exp_p(B_r(0))$ where $\bar B_r(0)$ (the closed ball in $T_pM$ with radius $r$) is contained in an open set $V \subseteq T_pM$ such that $\exp_p:V \to \exp_p(V)$ is a diffeomorphism.

A subset $A \subseteq M$ is convex, if every two points in $A$ can be joined by a minimizing geodesic (Some people call this weak convexity I think, since I do not require uniqueness).

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    $\begingroup$ In the round unit sphere, a geodesic ball of radius greater than $\pi/2$, which compactly contains a closed hemisphere, is not geodesically convex. $\endgroup$ – Andrew D. Hwang Dec 22 '16 at 16:52
  • $\begingroup$ True! Ok, this is nice, but this only works for "large" balls (I was implicitly thinking on small balls). Is it true (in general) that every sufficiently small ball is convex? $\endgroup$ – Asaf Shachar Dec 22 '16 at 16:56
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    $\begingroup$ Sufficiently small geodesic balls in a Riemannian manifold are convex, see e.g., Are small $\varepsilon$-balls convex in geodesic metric spaces? at MathOverflow. $\endgroup$ – Andrew D. Hwang Dec 22 '16 at 17:03
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    $\begingroup$ Thanks! I looked for something like this but did not googled the right words it seems... I will probably delete the question then. $\endgroup$ – Asaf Shachar Dec 22 '16 at 17:07
  • $\begingroup$ (That upvote signifies "glad you found an answer". :) $\endgroup$ – Andrew D. Hwang Dec 22 '16 at 17:18
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(The question turned out to be rather easy, I am posting an answer based on Andrew D. Hwang's comments.

In the round unit sphere, a geodesic ball of radius greater than $\frac{\pi}{2}$ , which compactly contains a closed hemisphere, is not geodesically convex.

Also, as stated here, sufficiently small geodesic balls in a Riemannian manifold are convex.

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  • $\begingroup$ Have you a reference for the fact stated in the link you gave? $\endgroup$ – Cronus Jun 5 '19 at 9:47

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