Euler's theorem on normal curvatures

Remarks

The proof is one line: $I\!I_p(w) = \ip{S_p w}{w}$ with the shape operator $S_p$, and $S_p e_i = k_i e_i$, so the cross terms vanish. It follows that $k_1$ and $k_2$ are the maximum and the minimum of the normal curvature over all directions. Example: on a cylinder of radius $r$ with the inner normal, $k_1 = 1/r$ around the cylinder and $k_2 = 0$ along it. A direction at angle $\theta$ to the circular sections has normal curvature $\cos^2 \theta / r$. The normal curvature $k_n(w)$ equals the signed curvature of the normal section, the curve cut out of $S$ by the plane through $p$ spanned by $w$ and the normal.

Going deeper

For a student

The theorem is the diagonal form of the quadratic form $I\!I_p$ in an orthonormal eigenbasis of the self-adjoint map $S_p$. It explains why the Gaussian curvature $k_1 k_2$ and the mean curvature $(k_1 + k_2)/2$ capture the whole second-order shape, up to the choice of principal directions. The sign of $k_1 k_2$ decides whether the normal curvature keeps one sign (elliptic point), changes sign (hyperbolic point), or vanishes in one direction (parabolic point). The directions with $k_n(w) = 0$ at a hyperbolic point are the asymptotic directions, at angles $\theta$ with $\tan^2 \theta = -k_1 / k_2$.