When is a spherical polygon convex suppose we have a polygon on the sphere $\mathbb{S}^2$, that is, a domain with piecewise geodesic boundary. Is it true that this domain is convex if and only if all angles of the polygon are in $(0,\pi)$?
A subset $U\subset\mathbb{S}^2$ is called convex if for any pair of points $p,q\in U$ there exist a geodesic from p to q which is contained in $U$. (For example a hemisphere is convex.)
For polygons in the euclidean plane this is known to be true. But I do not know the answer for the sphere. 
Best wishes


Is it true that this domain is convex if and only if all angles of the polygon are in $(0,π)$?

No, according to your definitions, this is not true. Take three quarter slices of a sphere. That gives you an inner angle of $\frac34\cdot2\pi>\pi$, but for every pair of points the plane spanned by these two points and the center of the sphere will intersect the sphere in a circle with some continuous portion missing. There is still a continouous portion which will contain a connecting geodesic, though.
