Let $S$ be a compact oriented surface, the Gauss map of $S$ is a diffeo local iff its Gaus curvature is positive everywhere.
My try:
Since $S$ is oriented we can define the Gaus map $N:S\rightarrow S^2$, my ideia was to use the diferential $dN_{p}: T_{p}S\rightarrow T_{p}S$, $T_{p}S$ is the tangent plane of $S$ at $p$, and show that its injective, so i proved the first part, but i dont know how to prove this because the null vector must not be in $T_{p}S$ to make sense my try.
Besides, i dont know how to prove the second part either, or use the hyphothesis that $S$ is compact, maybe i should define the height function $f(p)=<p,N(p)>$ or something like that, anyone can give me a north to continue?
Best Answer
Since $S$ and $S^2$ are of the same dimension, the Gauss map $N : S \to S^2$ is a local diffeomorphism if and only if $dN$ is invertible everywhere. From the equation $K = \det dN$ we see that this is true exactly if $K$ is nonzero everywhere.
To show that $N$ being a local diffeomorphism implies $K > 0$ (rather than just $K\ne 0$), we need to use the compactness assumption. Since $S$ is compact, the function $f(p) = |p|^2$ attains a maximum $R^2 = \sup_S f$ somewhere on $S,$ which means geometrically that $S$ is contained within the closed ball $\overline{B(0,R)}$ but touches its boundary. At this point of contact, $S$ must have both principal curvatures $\ge 1/R$ in order to stay within the sphere, and thus the Gauss curvature is positive there. But the curvature of a smooth surface is a continuous function, so since we know $K\ne 0$ everywhere and $K > 0$ somewhere we conclude $K>0$ everywhere.