When is the Gauss Map on a surface bijective?

Question

If $\mathcal{S}$ is a smooth, compact hyper-surface in $\mathbb{R}^n$ with positive definite second fundamental form, can we say that its Gauss map is bijective? If so, why?

Thanks!

Answer 1

Yes, because it's the boundary of strictly convex compact set $K$. Strict convexity implies that every linear function has one point of maximum and one point of minimum. Therefore, each vector $v$ is a normal vector at exactly two points of the hypersurface.

To prove the convexity claim: given two points on the surface, consider a two dimensional plane $P$ containing them. The intersection of $S$ with $P$ is a closed curve of nonvanishing curvature. The rest should be clear.