Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 23H (Differential Geometry)

The Gauss map sends each point of an oriented surface to its chosen unit normal $N(p)$. Both $T_{p}S$ and $T_{N(p)}{S}^2$ are the plane orthogonal to $N(p)$, so the derivative is an endomorphism of that plane. In local coordinates $r(u,v)$, differentiating $⟨N,r_j⟩=0$ gives $⟨N_i,r_j⟩=-⟨N,r_{ij}⟩$. Since $r_{ij}=r_{ji}$, this equals $⟨N_j,r_i⟩$. Bilinearity then gives $⟨dN(v),w⟩=⟨v,dN(w)⟩$ for all tangent vectors. The derivative of the Gauss map is self-adjoint.

If the Gauss map is a diffeomorphism onto $S^2$, the surface is compact and $dN$ is invertible everywhere. Its Gaussian curvature $K=\det (dN)$ is continuous and nowhere zero, so has constant sign on the connected surface. To determine that sign, maximize a height function $r\mapsto ⟨r,a⟩$. At a maximum, $a$ is parallel to the normal and the Hessian of the height function is semidefinite. Thus the second fundamental form is semidefinite, up to a sign. Its determinant is nonnegative and cannot be zero because $dN$ is invertible. Hence $K>0$ at that point and therefore everywhere.

The converse is false under the stated hypotheses. An open hemisphere of the unit sphere is a connected oriented surface with $K=1$, but its outward Gauss map has only that hemisphere as its image, and is not a diffeomorphism onto all of $S^2$. The missing compactness hypothesis matters.