Unit 5: The Second Fundamental Form & The Weingarten Map

1. The Spherical Gauss Map

While the First Fundamental Form captures the intrinsic geometry of distances and angles within a surface, the way a surface bends and curves in the surrounding Euclidean space $\mathbb{R}^3$ is governed by how its unit normal vector changes from point to point.

Definition 5.1 (The Gauss Map): Let $S \subset \mathbb{R}^3$ be an oriented regular surface with unit normal field $\mathbf{n}: S \to \mathbb{R}^3$, where $\|\mathbf{n}(p)\| = 1$ for all $p \in S$. The Gauss map is the smooth mapping:

which assigns to each point $p \in S$ the point on the unit sphere $S^2$ corresponding to the unit normal vector at $p$. The image $\mathbf{n}(S) \subseteq S^2$ is called the spherical image of the surface.

2. The Differential of the Gauss Map and the Shape Operator

For each point $p \in S$, consider the differential of the Gauss map $d\mathbf{n}_p: T_p S \to T_{\mathbf{n}(p)} S^2$. Because $\|\mathbf{n}(p)\|^2 = 1$, any variation of $\mathbf{n}$ must be orthogonal to $\mathbf{n}(p)$:

Since $T_p S$ is precisely the plane perpendicular to $\mathbf{n}(p)$, the tangent space to the unit sphere at $\mathbf{n}(p)$ is canonically identical to $T_p S$:

Therefore, $d\mathbf{n}_p$ can be viewed as a linear operator from $T_p S$ to itself!

Definition 5.2 (Shape Operator / Weingarten Map): The Shape Operator (or Weingarten Map) $S_p: T_p S \to T_p S$ is defined as the negative differential of the Gauss map:

The conventional minus sign ensures that for an outwardly convex surface (like a sphere with outward normal), the shape operator has positive eigenvalues.

3. Self-Adjointness of the Shape Operator

Theorem 5.1 (Symmetry / Self-Adjointness): The Shape Operator $S_p: T_p S \to T_p S$ is a self-adjoint (symmetric) linear operator with respect to the metric inner product of $T_p S$:

Proof: It suffices to verify the identity on the basis vectors $\{\mathbf{r}_u, \mathbf{r}_v\}$ of $T_p S$. Note that $S_p(\mathbf{r}_u) = -\mathbf{n}_u$ and $S_p(\mathbf{r}_v) = -\mathbf{n}_v$. We must prove:

Since $\mathbf{n}$ is perpendicular to the tangent vectors everywhere on $S$:

Differentiating $\mathbf{n} \cdot \mathbf{r}_u = 0$ with respect to $v$:

Similarly, differentiating $\mathbf{n} \cdot \mathbf{r}_v = 0$ with respect to $u$:

By Clairaut's theorem for $C^2$ surfaces, mixed partial derivatives commute: $\mathbf{r}_{uv} = \mathbf{r}_{vu}$. Therefore:

Hence $\langle S_p(\mathbf{r}_u), \mathbf{r}_v \rangle = \langle \mathbf{r}_u, S_p(\mathbf{r}_v) \rangle$, proving that $S_p$ is self-adjoint. $\blacksquare$

1. Definition of the Second Fundamental Form

The self-adjoint shape operator $S_p$ naturally induces a symmetric bilinear form and quadratic form on the tangent space $T_p S$.

Definition 5.3 (The Second Fundamental Form): The Second Fundamental Form of a surface $S$ at $p$, denoted by $II$ or $II_p$, is the quadratic form:

For an infinitesimal displacement $d\mathbf{r} = \mathbf{r}_u \, du + \mathbf{r}_v \, dv$, the quadratic form is:

Expanding bilinearly:

(also conventionally written as $e \, du^2 + 2f \, du \, dv + g \, dv^2$ or $b_{11} du^2 + 2b_{12} dudv + b_{22} dv^2$).

2. Formulas for the Coefficients $L, M, N$

Theorem 5.2 (Computation of Coefficients $L, M, N$): The coefficients of the Second Fundamental Form can be evaluated using either first derivatives of the normal vector or second derivatives of the position vector:

Proof: We proved in Section 5.1 that differentiating $\mathbf{n} \cdot \mathbf{r}_u = 0$ with respect to $u$ yields:

Similarly, differentiating $\mathbf{n} \cdot \mathbf{r}_v = 0$ with respect to $v$ gives:

And differentiating $\mathbf{n} \cdot \mathbf{r}_u = 0$ with respect to $v$ gives:

Since $\mathbf{n} = \frac{\mathbf{r}_u \times \mathbf{r}_v}{\|\mathbf{r}_u \times \mathbf{r}_v\|} = \frac{\mathbf{r}_u \times \mathbf{r}_v}{\sqrt{EG - F^2}}$, the dot product $\mathbf{r}_{uu} \cdot \mathbf{n}$ is:

The formulas for $M$ and $N$ follow identically. $\blacksquare$

1. Resolving Normal Derivatives in the Tangent Basis

Since $\mathbf{n}(u, v)$ is a unit vector, its partial derivatives $\mathbf{n}_u$ and $\mathbf{n}_v$ are perpendicular to $\mathbf{n}$, and hence must lie entirely in the tangent plane $T_p S = \operatorname{span}\{\mathbf{r}_u, \mathbf{r}_v\}$. Therefore, there exist scalar coefficients $a_{11}, a_{12}, a_{21}, a_{22}$ such that:

The matrix $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ represents the Shape Operator $S_p$ with respect to the coordinate basis $\{\mathbf{r}_u, \mathbf{r}_v\}$.

2. Derivation of the Weingarten Equations

Taking the dot product of $-\mathbf{n}_u = a_{11} \mathbf{r}_u + a_{21} \mathbf{r}_v$ with $\mathbf{r}_u$ and $\mathbf{r}_v$:

Similarly, for $-\mathbf{n}_v = a_{12} \mathbf{r}_u + a_{22} \mathbf{r}_v$:

In matrix form:

Multiplying by the inverse matrix $g^{-1} = \frac{1}{EG - F^2} \begin{pmatrix} G & -F \\ -F & E \end{pmatrix}$:

Carrying out the matrix multiplication:

Theorem 5.3 (The Weingarten Equations): The partial derivatives of the surface normal $\mathbf{n}$ are expressed in terms of the tangent vectors by:

1. Definition of the Third Fundamental Form

Just as the First Fundamental Form measures the length of tangent vectors on $S$ ($I(d\mathbf{r}) = \|d\mathbf{r}\|^2$), we can measure the length of their spherical images on $S^2$ under the Gauss map.

Definition 5.4 (The Third Fundamental Form): The Third Fundamental Form of a surface $S$, denoted $III$, is the quadratic form defined by the Euclidean inner product of the differentials of the normal vector:

Expanding:

where:

In terms of the Shape Operator $S_p$:

2. Characteristic Polynomial of the Shape Operator

Since $S_p: T_p S \to T_p S$ is a linear operator on a 2D vector space, its characteristic polynomial $P(\lambda)$ is:

The two fundamental scalar invariants of $S_p$ are:

• Mean Curvature: $H = \frac{1}{2} \operatorname{tr}(S_p) = \frac{1}{2}(a_{11} + a_{22}) = \frac{EN - 2FM + GL}{2(EG - F^2)}$
• Gaussian Curvature: $K = \det(S_p) = \det(A) = \frac{LN - M^2}{EG - F^2}$

Thus:

3. The Fundamental Invariant Identity

Theorem 5.4 (Operator Identity and Fundamental Form Relation):

1. By the Cayley-Hamilton Theorem, the Shape Operator satisfies its own characteristic equation:

1. Consequently, the three fundamental forms satisfy the universal linear relation:

Proof: For any tangent vector $\mathbf{w} \in T_p S$, apply the Cayley-Hamilton operator equation:

Taking the inner product with $\mathbf{w}$:

Recognizing the definitions:

Substituting these yields:

1. Curvature of a Surface Curve and Normal Curvature

Let $C: \mathbf{r}(s)$ be a regular curve parametrized by arc-length $s$ lying on a regular surface $S$. The unit tangent vector is $\mathbf{T}(s) = \mathbf{r}'(s) \in T_p S$. By the Serret-Frenet formulas (Unit 2), the derivative of the tangent vector is:

where $\kappa$ is the space curvature of $C$ and $\mathbf{N}$ is the curve's principal normal vector.

Since $\mathbf{n}(s)$ is the surface normal and $\mathbf{T}(s)$ lies in the tangent plane, the acceleration vector $\mathbf{r}''(s)$ can be decomposed into two orthogonal components:

where:

• $\mathbf{k}_n = \kappa_n \mathbf{n}$ is the normal curvature vector, directed along the surface normal $\mathbf{n}$.
• $\mathbf{k}_g$ is the geodesic curvature vector, lying in the tangent plane $T_p S$.

2. Formula for Normal Curvature

Taking the dot product of $\mathbf{r}''(s)$ with the surface unit normal $\mathbf{n}$:

Since $\mathbf{r}'(s) \cdot \mathbf{n}(s) = 0$ along the curve, differentiating with respect to $s$ gives:

Writing $\mathbf{r}'(s) = \mathbf{r}_u u' + \mathbf{r}_v v'$:

For an arbitrary parameter $t$ (not necessarily arc-length):

Theorem 5.5 (Normal Curvature Formula): The normal curvature of a surface along a tangent direction $d\mathbf{r} = (du, dv)$ is the ratio of the Second and First Fundamental Forms:

Notice that $\kappa_n$ depends only on the direction of the tangent vector $(du : dv)$ and the surface at $p$, and is completely independent of the curve chosen passing through $p$ with that tangent!

3. Meusnier's Theorem

Theorem 5.6 (Meusnier's Theorem, 1776): Let $C$ be a regular curve on a surface $S$ passing through $p$ with curvature $\kappa > 0$ and principal normal $\mathbf{N}$. Let $\theta$ be the angle between the principal normal to the curve $\mathbf{N}$ and the surface normal $\mathbf{n}$ ($\cos \theta = \mathbf{N} \cdot \mathbf{n}$). Then:

Equivalently, the radius of curvature $\rho_n = 1/|\kappa_n|$ of the normal section is related to the radius of curvature $\rho = 1/\kappa$ of the curve by:

Proof: Starting from the definition:

By Serret-Frenet, $\mathbf{r}''(s) = \kappa \mathbf{N}$. Therefore:

Geometric Interpretation: All curves lying on a surface passing through $p$ and sharing the same tangent line $\mathbf{T}$ have the same osculating circle projection onto the normal plane. If a plane section of the surface is tilted by an angle $\theta$ from the normal plane, its radius of curvature shrinks by a factor of $\cos \theta$!