# 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

• Piecewise smooth or piecewise straight? And what does convex mean in this context? In a Euclidean space a set is convex if for any pair of points all points on the connecting segment are included, too. But on the sphere there are two such segments. Do you always pick the shorter one? What if both have equal length? Is the hemisphere convex?
– MvG
Jan 8, 2017 at 23:18
• Sorry for being not clear enough. I have meant piecewise geodesic (straight) boundary. I have edited the question such that the other questions should be clarified now Jan 9, 2017 at 6:06

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.
• Thank you for your help. Your example has angles greater than $\pi$. Jan 9, 2017 at 8:09
• @MoisheCohen: Originally I had written an angle $<\pi$ by mistake. I assume Thorsten just wanted to point that out.