§ Shape operator [TODO ]

§ Principal curvature

• take a point $p$. Consider the normal to the surface at the point, $N(p)$.
• Take any normal plane: a plane $Q_p$ which contains $N(p)$. This plane (which is normal to the surface, since it contains the normal) intersecs the surface $S$ at a curve (intuitively, since a plane in 3D is defined by 1 eqn, intersection with plane imposes 1 equation on the surface, cutting it down to 1D).
• The curvature of this curve (normal plane $Q_p$ intersection surface $S$) at point $p$ is the normal curvature of the normal plane $Q_p$.
• The maximum and minimum such normal curvatures at a point (max, min taken across all possible normal planes $Q_p$) are the principal curvatures.

§ Shape operator has principal curvatures as eigenvalues

• https://math.stackexchange.com/questions/36517/shape-operator-and-principal-curvature
• https://math.stackexchange.com/questions/3665865/why-are-the-eigenvalues-of-the-shape-operator-the-principle-curvatures

§ Shape operator in index notation

• Let $X$ be tangent vectors at point $p$, $N$ be normal to surface at point $P$. The shape operator $S_{ij}$is determined by the equation:
• $\partial_i \mathbf N = -S_{ji} \mathbf X_b$