My understanding of strictly convexity of the compact set in Euclidean space is that if we take any straight line joining any two boundary points then the line must be in a compact set with out intersecting any other boundary point.
But for compact Riemannian manifold $(M,g)$, the definition of strict convexity is that the second fundamental form of boundary must be positive definite. Which depends upon metric.
To better understand this concept in the Riemannian setup, I was trying to find an example of metric on Sphere such that in that case, it becomes non strictly convex. But I could not able to find it.
Can anyone please help to relate the positive definiteness of the second fundamental form with strictly convexity in the Riemannian manifold?
Any Help or hint will be greatly appreciated.
Just take any non-convex closed hypersurface of $\mathbb{R}^³$ that is not -strictly- convex but is diffeomorphic to the sphere. Take the pullback metric from the ambiant space .
Comment Convexity is an extrinsic notion. A Riemannian manifold $(M,g)$ itself cannot be convex. But given a Riemannian embedding $(M,g) \to (N,g')$, one can study the extrinsic geometry of that embedding via its second fundamental form (or Weingarten map, or Shape operator). This closely depends on the choosen embedding, so that the usal round metric on the sphere can be convex, strictly convex or not, depending on the Riemannian embedding.