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
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.