Are closed ball convex in a translation surface?

Question

Let $(X,\omega)$ be a translation surface and $x$ any point (smooth or not) in it. Let $r\in \mathbb{R}^+$ be such that it is smaller than the diameter of $(X,\omega)$. Is the closed ball $B_r(x)$ always convex?

My guess is no.

I tried to figure it out using a simple translation surfaces: the regular octahedron with sides identified (it has one point of conical singularity of total angle $6\pi$). Then if I'm not wrong I can find a smooth point $x$ and an $r>0$ such that the closed ball $B_r(x)$ "overlaps" about the singular point giving non convexity. In the figure below I've drawn the situation I mean: the ball $B_r(x)$ is the dark part of the octahedron and I drew two segments not entirely contained in it.

Are my guess and my construction right?

Thank you

Answer 1

You are correct that the answer is no. Your example seems correct as well.

Perhaps the simplest example is an infinite circular cylinder of radius $r$ (and circumference $2\pi r$). If $p$ is any point on this cylinder, the ball of radius $\pi r$ centered at $p$ is "tangent" to itself on the back side, and is thus clearly not convex. The following picture shows this ball:

Indeed, balls on this cylinder are convex if and only if their radius is less than $\pi r/2$.

Of course, this example is non-compact, but basically the same geometry works on a flat torus of sufficient size.