§ Second fundamental form

• Let $z = f(x, y)$ be a (local) parametrization of the surface. Taylor expand $f$. we get:
• $f(x + dx, y + dy) = f(x, y) + dx^T a + dy^T b + dx^T L dx + 2 dx^T M dy + dy^T N dy$.
• We must get such a taylor expansion since our output is 1D (a real number), inputs are $dx, dy$ which are 3D vectors, and the infinitesimals must be linear/tensorial. These are the only possible contractions we can make.
• So, the second degree part can be written as:

• the matrix in the middle, or the quadratic form $II \equiv dx^T L dx + 2 dx^T M dy + dy^T N dy$ is the second fundamental form.

§ Classical geometry

• Let $z = f(x, y)$ be a (local) parametrization of the surface.
• At each point $p ≡ (u, v)$ on the surface within the local parametrization, we get tangent vectors $r_u(p) ≡ (\partial_x f(x, y)_p, r_v(p) ≡ (\partial_y f(x, y))_p$ which span the tangent space at $p$
• These define a unique normal vector $n(p) ≡ r_u(p) × r_v(p)$ at each point on the surface. This gives us a normal field.
• The coefficient of the second fundamental form project the second derivative of the function $f$ onto the normals. So they tell us how much the function is escaping the surface (ie, is moving along the normal to the surface) in second order.
• Recall that this is pointless to do for first order, since on a circle, tangent is perpendicular to normal, so any dot product of first order information with normal will be zero.
• Alternatively, first order information lies on tangent plane, and the normal is explicitly constructed as perpendicular to tangent plane, so any dot product of first order info with normal is zero.
• We can only really get meaningful info by dotting with normal at second order.
• So we get that $L(p) = (\partial_x \partial_x f(x, y))(p) \cdot N(p)$, $M(p) = (\partial_x \partial_y f(x, y))(p)$, and $N(p) = (\partial_y \partial_y f(x, y))(p)$, where we define $L, M, N$ via second fundamental form

§ Proof of equivalence between 2nd fundamental form and geometry

§ 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:
• $\partial_i \mathbf N = -S_{ji} \mathbf X_b$