Mathematics / Differential Geometry Curves, Surfaces, Fundamental Forms & Curvatures 100% Free Open Access
Chapter 7 โ€ข Theory & Derivations

Unit 7: Lines of Curvature, Asymptotic Curves & The Dupin Indicatrix

Directional geometry on surfaces: Rodrigues' formula for lines of curvature, Euler's theorem on normal curvature, the Dupin indicatrix conic sections, asymptotic directions and curves ($II = 0$), the Beltrami-Enneper theorem on the torsion of asymptotic curves ($ au = \pm \sqrt{-K}$), conjugate directions, and triply orthogonal systems.

ยง7.1 Rodrigues' Formula & Differential Equations of Lines of Curvature

1. Definition of Lines of Curvature

Definition 7.1 (Line of Curvature): A regular curve $C$ on a surface $S$ is called a line of curvature (or curvature line) if its tangent vector at every point points along a principal direction of the surface.


2. Rodrigues' Formula

Theorem 7.1 (Rodrigues' Formula, 1815): A regular curve $\mathbf{r}(t)$ on a surface $S$ is a line of curvature if and only if there exists a scalar function $\kappa(t)$ (the principal curvature) such that:

$$d\mathbf{n} + \kappa \, d\mathbf{r} = \mathbf{0} \iff \mathbf{n}'(t) = -\kappa(t) \mathbf{r}'(t)$$

Proof: By definition, $\mathbf{r}'(t)$ is an eigenvector of the Shape Operator $S_p$ if and only if:

$$S_p(\mathbf{r}'(t)) = \kappa(t) \mathbf{r}'(t)$$

Recall that the Shape Operator is defined by $S_p(\mathbf{w}) = -d\mathbf{n}_p(\mathbf{w})$. Applying this to $\mathbf{w} = \mathbf{r}'(t)$:

$$-d\mathbf{n}(\mathbf{r}'(t)) = -\mathbf{n}'(t) = \kappa(t) \mathbf{r}'(t) \iff \mathbf{n}'(t) + \kappa(t) \mathbf{r}'(t) = \mathbf{0}$$

This establishes Rodrigues' formula directly. $\blacksquare$


3. Differential Equation of Lines of Curvature

Let $d\mathbf{r} = \mathbf{r}_u \, du + \mathbf{r}_v \, dv$ and $d\mathbf{n} = \mathbf{n}_u \, du + \mathbf{n}_v \, dv$. By Rodrigues' formula, $d\mathbf{n}$ is collinear with $d\mathbf{r}$. Hence, their cross product must vanish in $\mathbb{R}^3$:

$$d\mathbf{n} \times d\mathbf{r} = \mathbf{0}$$

Since both $d\mathbf{n}$ and $d\mathbf{r}$ lie in the tangent plane $T_p S$, the vector $d\mathbf{n} \times d\mathbf{r}$ is parallel to the normal vector $\mathbf{n}$. Therefore:

$$(d\mathbf{n} \times d\mathbf{r}) \cdot \mathbf{n} = 0 \iff \det(d\mathbf{r}, d\mathbf{n}, \mathbf{n}) = 0$$

Using the Weingarten equations to express $d\mathbf{n}$ in terms of $E, F, G$ and $L, M, N$, this determinant expands into the classical Monge-Darby determinant:

Theorem 7.2 (Differential Equation of Lines of Curvature): The lines of curvature on a surface patch satisfy the second-order quadratic ODE:

$$\det \begin{pmatrix} dv^2 & -du \, dv & du^2 \\ E & F & G \\ L & M & N \end{pmatrix} = 0$$

Expanding this $3 \times 3$ determinant:

$$(EM - FL) \, du^2 + (EN - GL) \, du \, dv + (FN - GM) \, dv^2 = 0$$

Corollary 7.1 (Coordinate Curves as Lines of Curvature): The coordinate curves $u = \text{const}$ and $v = \text{const}$ form a family of lines of curvature if and only if:

$$F = 0 \quad \text{and} \quad M = 0$$

In this case, the principal curvatures are simply $\kappa_1 = L/E$ and $\kappa_2 = N/G$.

ยง7.2 Euler's Theorem on Normal Curvature & Mean Curvature Invariance

1. Statement and Proof of Euler's Formula

Let $p \in S$ be a non-umbilical point ($\kappa_1 \ne \kappa_2$). Choose an orthonormal basis $\{\mathbf{e}_1, \mathbf{e}_2\}$ in $T_p S$ along the principal directions:

$$S_p(\mathbf{e}_1) = \kappa_1 \mathbf{e}_1, \qquad S_p(\mathbf{e}_2) = \kappa_2 \mathbf{e}_2$$

Any unit tangent vector $\mathbf{u} \in T_p S$ can be represented as:

$$\mathbf{u} = \cos\theta \, \mathbf{e}_1 + \sin\theta \, \mathbf{e}_2$$

where $\theta \in [0, 2\pi)$ is the angle made by $\mathbf{u}$ with the first principal direction $\mathbf{e}_1$.

Theorem 7.3 (Euler's Theorem, 1760): The normal curvature $\kappa_n(\theta)$ along the direction making an angle $\theta$ with the first principal direction is given by:

$$\kappa_n(\theta) = \kappa_1 \cos^2\theta + \kappa_2 \sin^2\theta$$

Proof: Recall from Section 5.2 that the normal curvature along a unit tangent vector $\mathbf{u}$ is:

$$\kappa_n = II(\mathbf{u}) = \langle S_p(\mathbf{u}), \mathbf{u} \rangle$$

Substituting $\mathbf{u} = \cos\theta \, \mathbf{e}_1 + \sin\theta \, \mathbf{e}_2$:

$$S_p(\mathbf{u}) = \cos\theta \, S_p(\mathbf{e}_1) + \sin\theta \, S_p(\mathbf{e}_2) = \kappa_1 \cos\theta \, \mathbf{e}_1 + \kappa_2 \sin\theta \, \mathbf{e}_2$$

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

$$\langle S_p(\mathbf{u}), \mathbf{u} \rangle = (\kappa_1 \cos\theta \, \mathbf{e}_1 + \kappa_2 \sin\theta \, \mathbf{e}_2) \cdot (\cos\theta \, \mathbf{e}_1 + \sin\theta \, \mathbf{e}_2)$$

Since $\mathbf{e}_1 \cdot \mathbf{e}_1 = 1$, $\mathbf{e}_2 \cdot \mathbf{e}_2 = 1$, and $\mathbf{e}_1 \cdot \mathbf{e}_2 = 0$:

$$\kappa_n(\theta) = \kappa_1 \cos^2\theta + \kappa_2 \sin^2\theta \quad \blacksquare$$

2. Orthogonal Pairs and the Invariance of Mean Curvature

Consider two mutually perpendicular tangent directions: $\theta$ and $\theta + \pi/2$. The normal curvature in the perpendicular direction is:

$$\kappa_n\left(\theta + \frac{\pi}{2}\right) = \kappa_1 \cos^2\left(\theta + \frac{\pi}{2}\right) + \kappa_2 \sin^2\left(\theta + \frac{\pi}{2}\right) = \kappa_1 \sin^2\theta + \kappa_2 \cos^2\theta$$

Adding the two normal curvatures together:

$$\kappa_n(\theta) + \kappa_n\left(\theta + \frac{\pi}{2}\right) = \kappa_1(\cos^2\theta + \sin^2\theta) + \kappa_2(\sin^2\theta + \cos^2\theta) = \kappa_1 + \kappa_2 = 2H$$

Corollary 7.2 (Invariance of Sum of Orthogonal Curvatures): For any pair of orthogonal unit tangent vectors on a surface, the sum of their normal curvatures is constant and equals twice the Mean Curvature:

$$\kappa_n(\theta) + \kappa_n\left(\theta + \frac{\pi}{2}\right) = 2H$$

ยง7.3 The Dupin Indicatrix & Osculating Quadrics

1. Construction of the Dupin Indicatrix

The Dupin indicatrix is a geometric construction in the tangent plane $T_p S$ that visually characterizes the local second-order shape of the surface. Let Cartesian coordinates $(\xi, \eta)$ be chosen in $T_p S$ along the principal directions $\mathbf{e}_1, \mathbf{e}_2$. For each direction defined by angle $\theta$ ($\xi = r \cos\theta, \eta = r \sin\theta$), plot a segment of length:

$$r = \frac{1}{\sqrt{|\kappa_n(\theta)|}}$$

Squaring and multiplying by $|\kappa_n(\theta)|$:

$$r^2 |\kappa_n(\theta)| = 1 \iff r^2 |\kappa_1 \cos^2\theta + \kappa_2 \sin^2\theta| = 1$$

Substituting $\xi = r \cos\theta$ and $\eta = r \sin\theta$:

$$|\kappa_1 \xi^2 + \kappa_2 \eta^2| = 1$$

Definition 7.2 (The Dupin Indicatrix): The Dupin indicatrix at a point $p \in S$ is the quadratic curve in $T_p S$ given by:

$$\kappa_1 \xi^2 + \kappa_2 \eta^2 = \pm 1$$

2. Geometric Shape Depending on Point Type

1. At an Elliptic Point ($K > 0$):

$\kappa_1$ and $\kappa_2$ have the same sign. The equation is:

$$\kappa_1 \xi^2 + \kappa_2 \eta^2 = 1 \quad (\text{if } \kappa_1, \kappa_2 > 0)$$

This is an ellipse with semi-axes $a = 1/\sqrt{\kappa_1}$ and $b = 1/\sqrt{\kappa_2}$. If the point is an umbilic ($\kappa_1 = \kappa_2$), the ellipse becomes a circle.

2. At a Hyperbolic Point ($K < 0$):

$\kappa_1$ and $\kappa_2$ have opposite signs (say $\kappa_1 > 0, \kappa_2 < 0$). The Dupin indicatrix consists of a pair of conjugate hyperbolas:

$$\kappa_1 \xi^2 - |\kappa_2| \eta^2 = 1 \quad \text{and} \quad \kappa_1 \xi^2 - |\kappa_2| \eta^2 = -1$$

The asymptotes of these hyperbolas are given by $\kappa_1 \xi^2 + \kappa_2 \eta^2 = 0$, which correspond to the asymptotic directions of the surface!

3. At a Parabolic Point ($K = 0, \kappa_1 \ne 0, \kappa_2 = 0$):

The equation reduces to:

$$\kappa_1 \xi^2 = \pm 1 \implies \xi = \pm \frac{1}{\sqrt{|\kappa_1|}}$$

This is a pair of parallel straight lines parallel to the direction of zero curvature.

ยง7.4 Asymptotic Curves & The Beltrami-Enneper Torsion Theorem

1. Asymptotic Directions and Curves

Definition 7.3 (Asymptotic Direction and Curve):

  1. A tangent direction $(du, dv)$ at $p \in S$ is called an asymptotic direction if the normal curvature along that direction is zero:
$$\kappa_n = 0 \iff II(du, dv) = L \, du^2 + 2M \, du \, dv + N \, dv^2 = 0$$
  1. A regular curve on $S$ whose tangent vector at every point points along an asymptotic direction is called an asymptotic curve (or asymptotic line).

From Euler's theorem, $\kappa_n(\theta) = \kappa_1 \cos^2\theta + \kappa_2 \sin^2\theta = 0$:

$$\tan^2\theta = -\frac{\kappa_1}{\kappa_2}$$
  • At an elliptic point ($K > 0$): $\kappa_1 / \kappa_2 > 0$, so $\tan^2\theta < 0$, which has no real solutions. There are no asymptotic directions at elliptic points!
  • At a parabolic point ($K = 0$): there is one unique asymptotic direction (along the direction of $\kappa_2 = 0$).
  • At a hyperbolic point ($K < 0$): $\tan\theta = \pm \sqrt{-\kappa_1 / \kappa_2}$, yielding two distinct real asymptotic directions symmetric about the principal directions!

2. The Beltrami-Enneper Theorem

Along an asymptotic curve, $\kappa_n = \kappa \cos \theta = 0$. Assuming the space curvature $\kappa > 0$, Meusnier's theorem dictates that $\cos \theta = 0 \implies \mathbf{N} \cdot \mathbf{n} = 0$. Thus, the principal normal to the curve $\mathbf{N}$ is perpendicular to the surface normal $\mathbf{n}$! Consequently, the binormal vector $\mathbf{B} = \mathbf{T} \times \mathbf{N}$ is parallel to $\mathbf{n}$:

$$\mathbf{B} = \pm \mathbf{n}$$

The osculating plane of an asymptotic curve coincides with the tangent plane to the surface!

Theorem 7.4 (Beltrami-Enneper Theorem, 1870): Let $C$ be an asymptotic curve with non-vanishing space curvature $\kappa > 0$ on a surface with negative Gaussian curvature $K < 0$. Then the torsion $\tau$ of the asymptotic curve satisfies:

$$\tau^2 = -K \iff \tau = \pm \sqrt{-K}$$

Proof: Along the asymptotic curve, the binormal satisfies $\mathbf{B} = \mathbf{n}$ (up to a sign $\pm 1$). By the Serret-Frenet formulas:

$$\mathbf{B}'(s) = -\tau(s) \mathbf{N}(s)$$

Taking the norm squared:

$$\|\mathbf{B}'(s)\|^2 = \tau^2$$

Since $\mathbf{B} = \mathbf{n}$, $\mathbf{B}'(s) = d\mathbf{n}(\mathbf{T}) = -S_p(\mathbf{T})$. Thus:

$$\tau^2 = \|S_p(\mathbf{T})\|^2 = \langle S_p^2(\mathbf{T}), \mathbf{T} \rangle = III(\mathbf{T})$$

By the fundamental form identity (Theorem 5.4):

$$III(\mathbf{T}) = 2H II(\mathbf{T}) - K I(\mathbf{T})$$

Since the curve is asymptotic, $II(\mathbf{T}) = 0$. Since $\mathbf{T}$ is a unit vector, $I(\mathbf{T}) = 1$. Substituting these values:

$$\tau^2 = 2H(0) - K(1) = -K$$

Since $K < 0$, $-K > 0$, taking the square root gives:

$$\tau = \pm \sqrt{-K} \quad \blacksquare$$

ยง7.5 Conjugate Directions, Koenigs Nets & Triply Orthogonal Systems

1. Conjugate Directions

Definition 7.4 (Conjugate Directions): Two tangent directions $\mathbf{w}_1 = (du, dv)$ and $\mathbf{w}_2 = (\delta u, \delta v)$ at $p \in S$ are called conjugate directions if:

$$\langle S_p(\mathbf{w}_1), \mathbf{w}_2 \rangle = 0 \iff -d\mathbf{n}(\mathbf{w}_1) \cdot \mathbf{w}_2 = 0$$

In coordinates, the conjugacy condition is:

$$L \, du \, \delta u + M(du \, \delta v + dv \, \delta u) + N \, dv \, \delta v = 0$$

Theorem 7.5 (Geometric Properties of Conjugate Directions):

  1. The principal directions are the only mutually orthogonal conjugate directions.
  2. An asymptotic direction is self-conjugate ($II(du, dv) = 0$).
  3. Two directions $\theta_1, \theta_2$ relative to the principal frame are conjugate if and only if:
$$\kappa_1 \cos\theta_1 \cos\theta_2 + \kappa_2 \sin\theta_1 \sin\theta_2 = 0 \iff \tan\theta_1 \tan\theta_2 = -\frac{\kappa_1}{\kappa_2}$$

2. Triply Orthogonal Systems & Dupin's Theorem

Definition 7.5 (Triply Orthogonal System): A system of three families of surfaces in $\mathbb{R}^3$ is called a triply orthogonal system if through each point there passes exactly one surface from each family, and the three surfaces intersect each other pairwise orthogonally.

Theorem 7.6 (Dupin's Theorem, 1813): The intersection curves of the surfaces of any triply orthogonal system are lines of curvature on all three intersecting surfaces!

Example: Confocal quadrics (ellipsoids, hyperboloids of one sheet, and hyperboloids of two sheets sharing the same focal conics) form a triply orthogonal system. By Dupin's Theorem, their curves of intersection are automatically the lines of curvature on each quadric surface!

Tiered Solved Examination Problems & Rigorous Derivations

Step-by-step rigorous derivations with unskipped proofs, categorized into Foundational Concepts, Advanced Structural Analysis, and Honors / Proof Challenge tiers.

Foundational Example 7.1: Verification of Euler's Formula & Directional Average of Normal Curvature

Let $p \in S$ be a regular point with principal curvatures $\kappa_1$ and $\kappa_2$.

  1. By Euler's formula, the normal curvature in direction $\theta$ is $\kappa_n(\theta) = \kappa_1 \cos^2\theta + \kappa_2 \sin^2\theta$. Prove that the maximum and minimum values of $\kappa_n(\theta)$ over $\theta \in [0, 2\pi)$ are indeed $\kappa_1$ and $\kappa_2$.
  2. Compute the angular average of the normal curvature:
$$\bar{\kappa}_n = \frac{1}{2\pi} \int_0^{2\pi} \kappa_n(\theta) \, d\theta$$

and show that this average is identically equal to the Mean Curvature $H$.

1. Extrema of Euler's Formula

Euler's formula gives:

$$\kappa_n(\theta) = \kappa_1 \cos^2\theta + \kappa_2 \sin^2\theta$$

Without loss of generality, assume $\kappa_1 \ge \kappa_2$. Rewriting $\sin^2\theta = 1 - \cos^2\theta$:

$$\kappa_n(\theta) = \kappa_1 \cos^2\theta + \kappa_2 (1 - \cos^2\theta) = \kappa_2 + (\kappa_1 - \kappa_2)\cos^2\theta$$

Since $0 \le \cos^2\theta \le 1$:

  • When $\cos^2\theta = 1$ ($\theta = 0$ or $\theta = \pi$), $\kappa_n(\theta) = \kappa_2 + (\kappa_1 - \kappa_2)(1) = \kappa_1$ (maximum).
  • When $\cos^2\theta = 0$ ($\theta = \pi/2$ or $\theta = 3\pi/2$), $\kappa_n(\theta) = \kappa_2 + (\kappa_1 - \kappa_2)(0) = \kappa_2$ (minimum).

Thus, the principal curvatures $\kappa_1$ and $\kappa_2$ are the absolute maximum and minimum of normal curvature! $\blacksquare$


2. Angular Average of Normal Curvature

Using the half-angle identities:

$$\cos^2\theta = \frac{1 + \cos(2\theta)}{2}, \qquad \sin^2\theta = \frac{1 - \cos(2\theta)}{2}$$

Substitute these into Euler's formula:

$$\kappa_n(\theta) = \kappa_1 \left(\frac{1 + \cos(2\theta)}{2}\right) + \kappa_2 \left(\frac{1 - \cos(2\theta)}{2}\right) = \frac{\kappa_1 + \kappa_2}{2} + \frac{\kappa_1 - \kappa_2}{2} \cos(2\theta)$$

Now integrate over $\theta \in [0, 2\pi)$:

$$\int_0^{2\pi} \kappa_n(\theta) \, d\theta = \int_0^{2\pi} \left[ \frac{\kappa_1 + \kappa_2}{2} + \frac{\kappa_1 - \kappa_2}{2} \cos(2\theta) \right] d\theta$$

Notice that:

$$\int_0^{2\pi} \cos(2\theta) \, d\theta = \left[ \frac{\sin(2\theta)}{2} \right]_0^{2\pi} = \frac{\sin(4\pi) - \sin(0)}{2} = 0$$

Therefore:

$$\int_0^{2\pi} \kappa_n(\theta) \, d\theta = \int_0^{2\pi} \frac{\kappa_1 + \kappa_2}{2} \, d\theta = \left(\frac{\kappa_1 + \kappa_2}{2}\right)(2\pi)$$

Dividing by $2\pi$:

$$\bar{\kappa}_n = \frac{1}{2\pi} \int_0^{2\pi} \kappa_n(\theta) \, d\theta = \frac{\kappa_1 + \kappa_2}{2} = H \quad \blacksquare$$

This provides a beautiful physical and geometric interpretation: the Mean Curvature $H$ is the exact uniform average of normal curvatures over all possible directions!

Advanced Example 7.2: Asymptotic Curves of the Catenoid & Orthogonality

Consider the standard Catenoid parametrized by:

$$\mathbf{r}(u, v) = \begin{pmatrix} \cosh u \cos v \\ \cosh u \sin v \\ u \end{pmatrix}, \quad u \in \mathbb{R}, \; v \in [0, 2\pi)$$

Recall that $E = \cosh^2 u$, $F = 0$, $G = \cosh^2 u$, and $L = -1$, $M = 0$, $N = 1$.

  1. Set up the differential equation of asymptotic curves $II = 0$.
  2. Solve this differential equation explicitly to find the two families of asymptotic curves.
  3. Show that these two families of curves intersect at right angles everywhere on the catenoid.

1. Differential Equation of Asymptotic Curves

The condition for an asymptotic direction is:

$$II(du, dv) = L \, du^2 + 2M \, du \, dv + N \, dv^2 = 0$$

For the catenoid, $L = -1$, $M = 0$, and $N = 1$. Thus:

$$-du^2 + 0 + dv^2 = 0 \iff dv^2 - du^2 = 0 \quad \blacksquare$$

2. Solving for the Families of Curves

Factor the difference of squares:

$$(dv - du)(dv + du) = 0$$

This yields two first-order ODEs:

  1. Family 1: $dv - du = 0 \implies \frac{dv}{du} = 1 \implies v - u = c_1$
  2. Family 2: $dv + du = 0 \implies \frac{dv}{du} = -1 \implies v + u = c_2$

where $c_1, c_2$ are arbitrary integration constants. These are straight diagonal lines in the $uv$-parameter plane! $\blacksquare$


3. Orthogonality of the Asymptotic Net

Let direction 1 be $(du_1, dv_1) = (1, 1) \, dt$ and direction 2 be $(du_2, dv_2) = (1, -1) \, ds$. The inner product of these tangent directions with respect to the First Fundamental Form is:

$$\langle d\mathbf{r}_1, d\mathbf{r}_2 \rangle = E \, du_1 \, du_2 + F(du_1 \, dv_2 + dv_1 \, du_2) + G \, dv_1 \, dv_2$$

Recall that $E = G = \cosh^2 u$ and $F = 0$:

$$\langle d\mathbf{r}_1, d\mathbf{r}_2 \rangle = \cosh^2 u (1)(1) + 0 + \cosh^2 u (1)(-1) = \cosh^2 u - \cosh^2 u = 0$$

Because the inner product vanishes everywhere on the surface, the two families of asymptotic curves form an orthogonal net!

Remark: On any minimal surface ($H = 0$), $\kappa_1 = -\kappa_2$. By Euler's formula, $\kappa_n(\theta) = \kappa_1(\cos^2\theta - \sin^2\theta) = \kappa_1 \cos(2\theta) = 0 \implies 2\theta = \pm \pi/2 \implies \theta = \pm \pi/4$. Hence, on any minimal surface, the asymptotic directions bisect the principal directions and are always mutually orthogonal! $\blacksquare$

Honors / Proof Challenge Example 7.3: Rigorous Proof of the Beltrami-Enneper Torsion Theorem

Let $S \subset \mathbb{R}^3$ be a $C^3$ regular surface with negative Gaussian curvature $K < 0$, and let $C: \mathbf{r}(s)$ be an asymptotic curve on $S$ parametrized by arc-length $s$ with space curvature $\kappa(s) > 0$.

  1. Prove that the binormal vector $\mathbf{B}(s)$ to the curve is collinear with the surface normal $\mathbf{n}(s)$, and determine the sign relationship.
  2. Differentiate $\mathbf{B}(s) = \pm \mathbf{n}(s)$ along the curve and express $\mathbf{n}'(s)$ using the Shape Operator.
  3. Compute the torsion $\tau(s)$ of the curve and prove that $\tau^2 = -K(s)$, so that $\tau(s) = \pm \sqrt{-K(s)}$.

1. Collinearity of Binormal and Surface Normal

Let $C: \mathbf{r}(s)$ be parametrized by arc-length. Its unit tangent is $\mathbf{T}(s) = \mathbf{r}'(s) \in T_p S$, so $\mathbf{T} \cdot \mathbf{n} = 0$. By Serret-Frenet, $\mathbf{r}''(s) = \kappa \mathbf{N}$. The normal curvature of the curve on the surface is:

$$\kappa_n = \mathbf{r}''(s) \cdot \mathbf{n} = \kappa (\mathbf{N} \cdot \mathbf{n})$$

Because $C$ is an asymptotic curve, $\kappa_n = 0$ by definition. Since $\kappa > 0$ by assumption:

$$\mathbf{N} \cdot \mathbf{n} = 0$$

Thus, both $\mathbf{T}$ and $\mathbf{N}$ are orthogonal to the surface normal $\mathbf{n}$. In Euclidean 3-space, the orthogonal complement of the plane spanned by $\{\mathbf{T}, \mathbf{N}\}$ is one-dimensional, spanned by the binormal $\mathbf{B} = \mathbf{T} \times \mathbf{N}$. Since $\mathbf{n}$ is orthogonal to both $\mathbf{T}$ and $\mathbf{N}$, and $\|\mathbf{n}\| = \|\mathbf{B}\| = 1$, we must have:

$$\mathbf{B}(s) = \epsilon \, \mathbf{n}(s), \quad \text{where } \epsilon = \pm 1 \quad \blacksquare$$

2. Derivative of the Normal and Binormal

Differentiating $\mathbf{B}(s) = \epsilon \, \mathbf{n}(s)$ with respect to arc-length $s$:

$$\mathbf{B}'(s) = \epsilon \, \mathbf{n}'(s)$$

By the third Serret-Frenet formula (Unit 2):

$$\mathbf{B}'(s) = -\tau(s) \mathbf{N}(s)$$

On the other hand, applying the chain rule to the Gauss map:

$$\mathbf{n}'(s) = d\mathbf{n}(\mathbf{r}'(s)) = d\mathbf{n}(\mathbf{T}) = -S_p(\mathbf{T})$$

where $S_p$ is the Shape Operator. Equating the two expressions for $\mathbf{B}'(s)$:

$$-\tau(s) \mathbf{N}(s) = -\epsilon S_p(\mathbf{T}) \iff \tau(s) \mathbf{N}(s) = \epsilon S_p(\mathbf{T})$$

3. Evaluating the Torsion $\tau^2 = -K$

Taking the norm squared of both sides:

$$\tau(s)^2 \|\mathbf{N}(s)\|^2 = \epsilon^2 \|S_p(\mathbf{T})\|^2$$

Since $\|\mathbf{N}(s)\| = 1$ and $\epsilon^2 = 1$:

$$\tau^2 = \|S_p(\mathbf{T})\|^2 = \langle S_p(\mathbf{T}), S_p(\mathbf{T}) \rangle = \langle S_p^2(\mathbf{T}), \mathbf{T} \rangle = III(\mathbf{T})$$

Recall the fundamental form operator identity (Theorem 5.4):

$$III(\mathbf{T}) = 2H II(\mathbf{T}) - K I(\mathbf{T})$$

Because $\mathbf{T}$ points along an asymptotic direction, $II(\mathbf{T}) = 0$. Because $\mathbf{T}$ is a unit tangent vector, $I(\mathbf{T}) = \|\mathbf{T}\|^2 = 1$. Substituting these values:

$$\tau^2 = 2H(0) - K(1) = -K$$

Because $S$ has negative Gaussian curvature, $K < 0$, which ensures that $-K > 0$. Taking the square root:

$$\tau(s) = \pm \sqrt{-K(s)} \quad \blacksquare$$

This remarkable theorem proves that the torsion of an asymptotic curve depends only on the Gaussian curvature of the surface at that point!