Let $S$ be a compact surface and let $h: S \longrightarrow \Bbb{R}$ be the height function with respect to a given plane. By compactness, there is a minimum point $p$, at which the hessian $(d^2h)_p$ is semi-definite positive. On the other hand, we know that
$$
(d^2h)_p(v) = \sigma_p(v, v), \quad v \in T_pS
$$
where $\sigma_p$ is the second fundamental form at $p$. Hence $\sigma_p$ is semi-definite positive and hence its determinant is non-negative. Hence the Gauss curvature at $p$ is non-negative.
My question is:
Why can we conclude that $\sigma_p$ is semi-definite positive?
The motivation is Exercise 3.40 in Curves and Surfaces, second edition, by Montiel and Ros, and whose solution is presented above:
Show that there are no compact surfaces that have negative Gauss curvature everywhere.
Best Answer
The most direct answer to your question is obtained by taking $v$ to be the eigenvectors of $\sigma_p$. The determinant of $\sigma_p$ is the product of the eigenvalues, which are nonnegative.