Chapter 2 • Theory & Derivations

Unit 2: The Frenet-Serret Apparatus: Curvature, Torsion & The Moving Trihedron

Comprehensive theory of the Frenet-Serret apparatus: the orthonormal moving trihedron {T, N, B}, the three fundamental planes (osculating, normal, rectifying), geometric definitions of curvature kappa and torsion tau, formulas for arbitrary parameters, the complete line-by-line derivation of the Frenet-Serret equations, the Darboux rotation vector, and the Fundamental Theorem of Space Curves.

§2.1 The Principal Normal, Binormal & The Frenet Moving Trihedron

1. Construction of the Moving Orthonormal Frame

Let $\mathbf{r}: I \to \mathbb{R}^3$ be a regular $C^3$ space curve parametrized by arc-length $s$. By definition, the unit tangent vector is:

\mathbf{T}(s) = \mathbf{r}'(s) = \frac{d\mathbf{r}}{ds}

Since $\mathbf{T}(s)$ has constant unit length for all $s$:

\langle \mathbf{T}(s), \mathbf{T}(s) \rangle = \|\mathbf{T}(s)\|^2 = 1

Differentiating both sides with respect to arc-length $s$ using the product rule:

\frac{d}{ds} \langle \mathbf{T}(s), \mathbf{T}(s) \rangle = 2 \left\langle \frac{d\mathbf{T}}{ds}, \mathbf{T}(s) \right\rangle = 0 \implies \mathbf{T}'(s) \perp \mathbf{T}(s)

The derivative vector $\mathbf{T}'(s)$ is strictly orthogonal to $\mathbf{T}(s)$ at every point where it is non-zero!

2. The Principal Normal Vector $\mathbf{N}$

Definition 2.1 (Curvature and Principal Normal): Let $s$ be a point where $\mathbf{T}'(s) \ne \mathbf{0}$.

The curvature $\kappa(s)$ of the curve is the magnitude of the rate of change of the unit tangent vector:

\kappa(s) = \|\mathbf{T}'(s)\| = \left\| \frac{d^2\mathbf{r}}{ds^2} \right\| > 0

The reciprocal $\rho(s) = \frac{1}{\kappa(s)}$ is the radius of curvature.

The principal normal vector $\mathbf{N}(s)$ is the unit vector pointing in the direction of $\mathbf{T}'(s)$:

\mathbf{N}(s) = \frac{\mathbf{T}'(s)}{\|\mathbf{T}'(s)\|} = \frac{1}{\kappa(s)} \mathbf{T}'(s) \iff \mathbf{T}'(s) = \kappa(s) \mathbf{N}(s)

3. The Binormal Vector $\mathbf{B}$ and the Frenet Trihedron

Definition 2.2 (Binormal Vector $\mathbf{B}$): The binormal vector $\mathbf{B}(s)$ is defined as the cross product of the unit tangent and principal normal vectors:

\mathbf{B}(s) = \mathbf{T}(s) \times \mathbf{N}(s)

Theorem 2.1 (The Frenet Moving Trihedron): The ordered set of vectors $\{\mathbf{T}(s), \mathbf{N}(s), \mathbf{B}(s)\}$ forms a right-handed orthonormal basis of $\mathbb{R}^3$ at each point of the curve where $\kappa(s) > 0$:

\|\mathbf{T}\| = \|\mathbf{N}\| = \|\mathbf{B}\| = 1

\mathbf{T} \cdot \mathbf{N} = \mathbf{N} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{T} = 0

\mathbf{T} \times \mathbf{N} = \mathbf{B}, \quad \mathbf{N} \times \mathbf{B} = \mathbf{T}, \quad \mathbf{B} \times \mathbf{T} = \mathbf{N}

This moving orthonormal coordinate frame is called the Frenet-Serret Moving Trihedron.

§2.2 The Three Fundamental Planes of Curve Theory

1. Geometric Definition of the Fundamental Planes

At every point $\mathbf{r}(s)$ of a regular curve with $\kappa(s) > 0$, the Frenet trihedron $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$ defines three mutually perpendicular coordinate planes.

Definition 2.3 (The Three Fundamental Planes): Let $\mathbf{X} = (X, Y, Z)$ denote an arbitrary point in space.

Osculating Plane (Plane of Curvature):

The plane spanned by the tangent $\mathbf{T}$ and principal normal $\mathbf{N}$. Its normal vector is the binormal $\mathbf{B}$.

\mathbf{B}(s) \cdot (\mathbf{X} - \mathbf{r}(s)) = 0

Normal Plane:

The plane spanned by the principal normal $\mathbf{N}$ and binormal $\mathbf{B}$. Its normal vector is the unit tangent $\mathbf{T}$.

\mathbf{T}(s) \cdot (\mathbf{X} - \mathbf{r}(s)) = 0

Rectifying Plane:

The plane spanned by the unit tangent $\mathbf{T}$ and binormal $\mathbf{B}$. Its normal vector is the principal normal $\mathbf{N}$.

\mathbf{N}(s) \cdot (\mathbf{X} - \mathbf{r}(s)) = 0

2. Physical and Geometric Roles

Osculating Plane: Contains the instantaneous velocity and acceleration vectors ($\mathbf{r}' = \mathbf{T}$, $\mathbf{r}'' = \kappa \mathbf{N}$). Any planar curve lies entirely within its osculating plane.
Normal Plane: Contains all lines passing through $\mathbf{r}(s)$ perpendicular to the curve's direction of motion. The circle of curvature (osculating circle) intersects this plane perpendicularly.
Rectifying Plane: The plane along which the curve can be "unrolled" or rectified. If a curve is a geodesic on a developable surface, the surface is the envelope of the curve's rectifying planes.

§2.3 Curvature, Torsion & Arbitrary Parametrization Formulas

1. Geometric Definition and Interpretation of Torsion

Just as curvature $\kappa$ measures the rate at which the curve turns away from its tangent line, torsion $\tau$ measures the rate at which the curve twists out of its osculating plane.

Definition 2.4 (Torsion): Since $\mathbf{B}(s)$ is a unit vector, $\mathbf{B}'(s) \perp \mathbf{B}(s)$. Furthermore, since $\mathbf{B} = \mathbf{T} \times \mathbf{N}$, differentiating gives:

\mathbf{B}' = \mathbf{T}' \times \mathbf{N} + \mathbf{T} \times \mathbf{N}' = (\kappa \mathbf{N}) \times \mathbf{N} + \mathbf{T} \times \mathbf{N}' = \mathbf{0} + \mathbf{T} \times \mathbf{N}'

This shows $\mathbf{B}'(s) \perp \mathbf{T}(s)$. Since $\mathbf{B}'$ is perpendicular to both $\mathbf{B}$ and $\mathbf{T}$, it must be collinear with $\mathbf{N}$! The torsion $\tau(s)$ is defined by:

\frac{d\mathbf{B}}{ds} = -\tau(s) \mathbf{N}(s) \iff \tau(s) = -\mathbf{N}(s) \cdot \mathbf{B}'(s)

The radius of torsion is $\sigma(s) = \frac{1}{\tau(s)}$.

2. Arbitrary Parametrization Formulas

In practical applications, curves are rarely parametrized by arc-length. We require formulas for $\kappa$ and $\tau$ expressed directly in terms of an arbitrary parameter $t$.

Theorem 2.2 (General Parameter Curvature & Torsion Formulas): Let $\mathbf{r}(t)$ be a regular $C^3$ curve with arbitrary parameter $t$. Then:

Curvature:

\kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}

Torsion:

\tau(t) = \frac{[\mathbf{r}'(t), \mathbf{r}''(t), \mathbf{r}'''(t)]}{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|^2} = \frac{(\mathbf{r}'(t) \times \mathbf{r}''(t)) \cdot \mathbf{r}'''(t)}{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|^2}

The Frenet Vectors:

\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}, \quad \mathbf{B}(t) = \frac{\mathbf{r}'(t) \times \mathbf{r}''(t)}{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}, \quad \mathbf{N}(t) = \mathbf{B}(t) \times \mathbf{T}(t)

Complete Line-by-Line Proof:

Let $s$ be arc-length, and let $v = \frac{ds}{dt} = \|\mathbf{r}'(t)\|$. By the Chain Rule:

\mathbf{r}'(t) = \frac{d\mathbf{r}}{ds} \frac{ds}{dt} = v \mathbf{T}

Differentiating with respect to $t$:

\mathbf{r}''(t) = v' \mathbf{T} + v \frac{d\mathbf{T}}{dt} = v' \mathbf{T} + v \left( \frac{d\mathbf{T}}{ds} \frac{ds}{dt} \right) = v' \mathbf{T} + v^2 \kappa \mathbf{N}

Computing the vector cross product $\mathbf{r}'(t) \times \mathbf{r}''(t)$:

\mathbf{r}'(t) \times \mathbf{r}''(t) = (v \mathbf{T}) \times (v' \mathbf{T} + v^2 \kappa \mathbf{N}) = v v' (\mathbf{T} \times \mathbf{T}) + v^3 \kappa (\mathbf{T} \times \mathbf{N}) = v^3 \kappa \mathbf{B}

Taking the Euclidean norm of both sides (since $\|\mathbf{B}\| = 1$ and $\kappa > 0, v > 0$):

\|\mathbf{r}'(t) \times \mathbf{r}''(t)\| = v^3 \kappa = \|\mathbf{r}'(t)\|^3 \kappa \implies \kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}

Next, differentiating $\mathbf{r}''(t)$ to find $\mathbf{r}'''(t)$:

\mathbf{r}'''(t) = \frac{d}{dt}[v' \mathbf{T} + v^2 \kappa \mathbf{N}] = v'' \mathbf{T} + v' (v \kappa \mathbf{N}) + (v^2 \kappa)' \mathbf{N} + v^2 \kappa (v \mathbf{N}') = \dots + v^3 \kappa \tau \mathbf{B}

Now take the dot product with $\mathbf{r}'(t) \times \mathbf{r}''(t) = v^3 \kappa \mathbf{B}$:

(\mathbf{r}'(t) \times \mathbf{r}''(t)) \cdot \mathbf{r}'''(t) = (v^3 \kappa \mathbf{B}) \cdot (\dots + v^3 \kappa \tau \mathbf{B}) = (v^3 \kappa)^2 \tau = \|\mathbf{r}'(t) \times \mathbf{r}''(t)\|^2 \tau

Solving for $\tau$ yields:

\tau(t) = \frac{[\mathbf{r}'(t), \mathbf{r}''(t), \mathbf{r}'''(t)]}{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|^2} \quad \blacksquare

§2.4 The Frenet-Serret Formulas & The Darboux Vector

1. The Frenet-Serret Formulas

The fundamental differential equations governing space curves were discovered independently by Jean Frédéric Frenet (1847) and Joseph Alfred Serret (1851).

Theorem 2.3 (The Frenet-Serret Equations): Let $\mathbf{r}(s)$ be an arc-length parametrized $C^3$ curve with curvature $\kappa(s) > 0$ and torsion $\tau(s)$. The derivatives of the moving orthonormal frame $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$ with respect to arc-length satisfy:

\begin{aligned} > \frac{d\mathbf{T}}{ds} &= \kappa \mathbf{N} \\ > \frac{d\mathbf{N}}{ds} &= -\kappa \mathbf{T} + \tau \mathbf{B} \\ > \frac{d\mathbf{B}}{ds} &= -\tau \mathbf{N} > \end{aligned}

In matrix notation:

\frac{d}{ds} \begin{pmatrix} \mathbf{T} \\ \mathbf{N} \\ \mathbf{B} \end{pmatrix} = \begin{pmatrix} > 0 & \kappa & 0 \\ > -\kappa & 0 & \tau \\ > 0 & -\tau & 0 > \end{pmatrix} \begin{pmatrix} \mathbf{T} \\ \mathbf{N} \\ \mathbf{B} \end{pmatrix}

Notice that the coefficient matrix is skew-symmetric ($A^T = -A$), reflecting the fact that the frame remains orthonormal at all times.

Complete Line-by-Line Proof:

1. First equation $\mathbf{T}' = \kappa \mathbf{N}$:

This holds by Definition 2.1 of curvature $\kappa$ and principal normal $\mathbf{N}$.

2. Third equation $\mathbf{B}' = -\tau \mathbf{N}$:

This holds by Definition 2.4 of torsion $\tau$.

3. Second equation $\mathbf{N}' = -\kappa \mathbf{T} + \tau \mathbf{B}$:

Since $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$ is an orthonormal basis, we can express $\mathbf{N}'$ as a linear combination:

\mathbf{N}' = c_1 \mathbf{T} + c_2 \mathbf{N} + c_3 \mathbf{B}

where $c_1 = \mathbf{N}' \cdot \mathbf{T}$, $c_2 = \mathbf{N}' \cdot \mathbf{N}$, and $c_3 = \mathbf{N}' \cdot \mathbf{B}$.

Differentiating $\mathbf{N} \cdot \mathbf{N} = 1$:

2 (\mathbf{N}' \cdot \mathbf{N}) = 0 \implies c_2 = 0

Differentiating $\mathbf{N} \cdot \mathbf{T} = 0$:

\mathbf{N}' \cdot \mathbf{T} + \mathbf{N} \cdot \mathbf{T}' = 0 \implies c_1 = -\mathbf{N} \cdot \mathbf{T}' = -\mathbf{N} \cdot (\kappa \mathbf{N}) = -\kappa \|\mathbf{N}\|^2 = -\kappa

Differentiating $\mathbf{N} \cdot \mathbf{B} = 0$:

\mathbf{N}' \cdot \mathbf{B} + \mathbf{N} \cdot \mathbf{B}' = 0 \implies c_3 = -\mathbf{N} \cdot \mathbf{B}' = -\mathbf{N} \cdot (-\tau \mathbf{N}) = \tau \|\mathbf{N}\|^2 = \tau

Substituting $c_1, c_2, c_3$ gives:

\mathbf{N}' = -\kappa \mathbf{T} + \tau \mathbf{B} \quad \blacksquare

2. The Darboux Rotation Vector

Jean Gaston Darboux observed that the Frenet-Serret equations can be unified into a single kinematic angular velocity equation:

Definition 2.5 (Darboux Vector): The Darboux vector (or angular velocity vector of the frame) is:

\boldsymbol{\omega}(s) = \tau(s) \mathbf{T}(s) + \kappa(s) \mathbf{B}(s)

Theorem 2.4 (Darboux Kinematic Law): For each vector $\mathbf{F} \in \{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$, the rate of change is given by:

\frac{d\mathbf{F}}{ds} = \boldsymbol{\omega} \times \mathbf{F}

Proof:

$\boldsymbol{\omega} \times \mathbf{T} = (\tau \mathbf{T} + \kappa \mathbf{B}) \times \mathbf{T} = \tau(\mathbf{T} \times \mathbf{T}) + \kappa(\mathbf{B} \times \mathbf{T}) = \mathbf{0} + \kappa \mathbf{N} = \frac{d\mathbf{T}}{ds}$.

$\boldsymbol{\omega} \times \mathbf{N} = (\tau \mathbf{T} + \kappa \mathbf{B}) \times \mathbf{N} = \tau(\mathbf{T} \times \mathbf{N}) + \kappa(\mathbf{B} \times \mathbf{N}) = \tau \mathbf{B} - \kappa \mathbf{T} = \frac{d\mathbf{N}}{ds}$.

$\boldsymbol{\omega} \times \mathbf{B} = (\tau \mathbf{T} + \kappa \mathbf{B}) \times \mathbf{B} = \tau(\mathbf{T} \times \mathbf{B}) + \kappa(\mathbf{B} \times \mathbf{B}) = -\tau \mathbf{N} + \mathbf{0} = \frac{d\mathbf{B}}{ds}$. $\blacksquare$

§2.5 The Fundamental Theorem of Space Curves

1. The Natural / Intrinsic Equations of a Curve

A remarkable consequence of the Frenet apparatus is that curvature $\kappa(s)$ and torsion $\tau(s)$ contain complete geometric information about the curve. The equations $\kappa = \kappa(s)$ and $\tau = \tau(s)$ are called the natural equations (or intrinsic equations) of the curve.

2. Statement of the Fundamental Theorem

Theorem 2.5 (Fundamental Theorem of Space Curves / Bonnet's Theorem): Let $I \subseteq \mathbb{R}$ be an interval containing $s_0$. Let $\kappa: I \to \mathbb{R}$ and $\tau: I \to \mathbb{R}$ be continuous functions such that $\kappa(s) > 0$ for all $s \in I$.

Existence: There exists a $C^3$ curve $\mathbf{r}: I \to \mathbb{R}^3$ parametrized by arc-length $s$ whose curvature is $\kappa(s)$ and whose torsion is $\tau(s)$.

Uniqueness: If $\tilde{\mathbf{r}}: I \to \mathbb{R}^3$ is another curve with the same curvature $\kappa(s)$ and torsion $\tau(s)$, then $\tilde{\mathbf{r}}$ differs from $\mathbf{r}$ by at most a rigid motion of Euclidean space (a translation and a rotation in $\mathrm{SO}(3)$).

Proof Outline (Linear ODE Systems):

1. Solve for the frame: The Frenet-Serret system $\frac{d\mathbf{F}}{ds} = A(s)\mathbf{F}$ is a linear homogeneous system of ODEs with skew-symmetric coefficient matrix $A(s)$. By the Picard-Lindelöf theorem, given an initial orthonormal frame $\{\mathbf{T}_0, \mathbf{N}_0, \mathbf{B}_0\}$ at $s_0$, there exists a unique solution $\{\mathbf{T}(s), \mathbf{N}(s), \mathbf{B}(s)\}$ on $I$. Because $A(s)$ is skew-symmetric, the frame remains orthonormal for all $s$.

2. Integrate for the curve: Define $\mathbf{r}(s) = \mathbf{r}_0 + \int_{s_0}^s \mathbf{T}(u) \, du$.

Then $\mathbf{r}'(s) = \mathbf{T}(s)$, $\|\mathbf{r}'(s)\| = 1$, and its curvature and torsion match $\kappa(s)$ and $\tau(s)$.

3. Uniqueness: If two curves have identical $\kappa(s)$ and $\tau(s)$, apply a rotation to align their initial frames at $s_0$, and a translation to align their initial positions. By uniqueness of solutions to linear ODEs, the two curves must coincide everywhere on $I$. $\blacksquare$

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 / Problem 2.1 Example 2.1: Complete Frenet Apparatus Computation for the Circular Helix

Consider the standard circular helix with radius $a > 0$ and pitch parameter $b > 0$:

\mathbf{r}(t) = \begin{pmatrix} a \cos t \\ a \sin t \\ b t \end{pmatrix}, \quad t \in \mathbb{R}

Find the speed $v = \|\mathbf{r}'(t)\|$ and the arc-length function $s(t)$ measured from $t_0 = 0$.
Compute the unit tangent vector $\mathbf{T}$, principal normal vector $\mathbf{N}$, and binormal vector $\mathbf{B}$ as functions of $t$.
Compute the curvature $\kappa(t)$ and torsion $\tau(t)$. Show that both are constants and find their ratio $\tau / \kappa$.

1. Speed and Arc-Length Computation

Differentiating $\mathbf{r}(t)$:

\mathbf{r}'(t) = \begin{pmatrix} -a \sin t \\ a \cos t \\ b \end{pmatrix}

The speed is:

v = \|\mathbf{r}'(t)\| = \sqrt{(-a \sin t)^2 + (a \cos t)^2 + b^2} = \sqrt{a^2(\sin^2 t + \cos^2 t) + b^2} = \sqrt{a^2 + b^2}

Let $c = \sqrt{a^2 + b^2} > 0$. The speed is constant $v = c$. The arc-length function from $t_0 = 0$ is:

s(t) = \int_0^t c \, du = c t \iff t = \frac{s}{c}

2. Frenet Moving Trihedron $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$

Unit Tangent Vector $\mathbf{T}$:

\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{1}{c} \begin{pmatrix} -a \sin t \\ a \cos t \\ b \end{pmatrix}

Principal Normal Vector $\mathbf{N}$:

Differentiating $\mathbf{T}$ with respect to arc-length $s$:

\frac{d\mathbf{T}}{ds} = \frac{d\mathbf{T}}{dt} \frac{dt}{ds} = \frac{1}{c} \frac{d}{dt} \left[ \frac{1}{c} \begin{pmatrix} -a \sin t \\ a \cos t \\ b \end{pmatrix} \right] = \frac{1}{c^2} \begin{pmatrix} -a \cos t \\ -a \sin t \\ 0 \end{pmatrix} = -\frac{a}{c^2} \begin{pmatrix} \cos t \\ \sin t \\ 0 \end{pmatrix}

The curvature is the norm:

\kappa = \left\| \frac{d\mathbf{T}}{ds} \right\| = \frac{a}{c^2} \sqrt{\cos^2 t + \sin^2 t} = \frac{a}{c^2} = \frac{a}{a^2 + b^2}

The principal normal vector is:

\mathbf{N} = \frac{1}{\kappa} \frac{d\mathbf{T}}{ds} = \begin{pmatrix} -\cos t \\ -\sin t \\ 0 \end{pmatrix}

Binormal Vector $\mathbf{B}$:

\mathbf{B} = \mathbf{T} \times \mathbf{N} = \frac{1}{c} \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -a \sin t & a \cos t & b \\ -\cos t & -\sin t & 0 \end{vmatrix} = \frac{1}{c} \begin{pmatrix} b \sin t \\ -b \cos t \\ a \sin^2 t + a \cos^2 t \end{pmatrix} = \frac{1}{c} \begin{pmatrix} b \sin t \\ -b \cos t \\ a \end{pmatrix}

3. Torsion and Ratio $\tau / \kappa$

Differentiating $\mathbf{B}$ with respect to $s$:

\frac{d\mathbf{B}}{ds} = \frac{1}{c} \frac{d\mathbf{B}}{dt} = \frac{1}{c} \cdot \frac{1}{c} \begin{pmatrix} b \cos t \\ b \sin t \\ 0 \end{pmatrix} = \frac{b}{c^2} \begin{pmatrix} \cos t \\ \sin t \\ 0 \end{pmatrix} = -\frac{b}{c^2} \mathbf{N}

Comparing with the Frenet formula $\frac{d\mathbf{B}}{ds} = -\tau \mathbf{N}$:

\tau = \frac{b}{c^2} = \frac{b}{a^2 + b^2}

Both $\kappa$ and $\tau$ are strictly constant. Their ratio is:

\frac{\tau}{\kappa} = \frac{b / (a^2 + b^2)}{a / (a^2 + b^2)} = \frac{b}{a} = \text{constant} \quad \blacksquare

Advanced / Problem 2.2 Example 2.2: Complete Rigorous Proof of the Planar Curve Criterion (tau = 0)

Let $\mathbf{r}: I \to \mathbb{R}^3$ be a regular $C^3$ curve parametrized by arc-length $s$ with $\kappa(s) > 0$ for all $s \in I$.

Prove that if $\mathbf{r}(I)$ lies in a fixed plane $\Pi \subset \mathbb{R}^3$, then its torsion $\tau(s) \equiv 0$ everywhere on $I$.
Conversely, prove that if $\tau(s) \equiv 0$ for all $s \in I$, then the curve lies entirely in a fixed plane.
Explicitly construct the equation of this plane in terms of the initial point $\mathbf{r}(s_0)$ and binormal $\mathbf{B}(s_0)$.

1. Forward Direction: Curve Lies in a Plane $\implies \tau(s) \equiv 0$

Assume $\mathbf{r}(I)$ lies in a fixed plane $\Pi$. The equation of $\Pi$ is:

\mathbf{n}_0 \cdot (\mathbf{r}(s) - \mathbf{r}_0) = 0 \quad \forall s \in I

where $\mathbf{n}_0$ is a fixed constant unit normal vector ($\|\mathbf{n}_0\| = 1$), and $\mathbf{r}_0 \in \Pi$.

Differentiating once with respect to $s$:

\mathbf{n}_0 \cdot \mathbf{r}'(s) = 0 \iff \mathbf{n}_0 \cdot \mathbf{T}(s) = 0

Thus, $\mathbf{n}_0$ is perpendicular to $\mathbf{T}(s)$ for all $s$.

Differentiating a second time with respect to $s$:

\mathbf{n}_0 \cdot \mathbf{T}'(s) = 0 \iff \mathbf{n}_0 \cdot (\kappa(s) \mathbf{N}(s)) = 0

Since $\kappa(s) > 0$, dividing by $\kappa(s)$ gives:

\mathbf{n}_0 \cdot \mathbf{N}(s) = 0

Since $\mathbf{n}_0$ is a unit vector orthogonal to both $\mathbf{T}(s)$ and $\mathbf{N}(s)$, and $\{\mathbf{T}, \mathbf{N}, \mathbf{B}\}$ is an orthonormal basis, $\mathbf{n}_0$ must be parallel to $\mathbf{B}(s)$:

\mathbf{B}(s) = \pm \mathbf{n}_0 \quad \text{for all } s \in I

Since $\mathbf{n}_0$ is constant, $\mathbf{B}(s)$ is a constant vector:

\frac{d\mathbf{B}}{ds} = \mathbf{0}

By the third Frenet-Serret formula:

\frac{d\mathbf{B}}{ds} = -\tau(s) \mathbf{N}(s) = \mathbf{0}

Since $\|\mathbf{N}(s)\| = 1 \ne 0$, we must have:

\tau(s) \equiv 0 \quad \forall s \in I \quad \blacksquare

2. Reverse Direction: $\tau(s) \equiv 0 \implies$ Curve Lies in a Plane

Assume $\tau(s) \equiv 0$ for all $s \in I$. By the third Frenet formula:

\frac{d\mathbf{B}}{ds} = -\tau(s) \mathbf{N}(s) = -0 \cdot \mathbf{N}(s) = \mathbf{0}

Since $\frac{d\mathbf{B}}{ds} = \mathbf{0}$ on the connected interval $I$, the binormal vector is constant:

\mathbf{B}(s) = \mathbf{B}_0 \quad \forall s \in I

Now fix a point $s_0 \in I$ and consider the scalar function:

f(s) = \mathbf{B}_0 \cdot (\mathbf{r}(s) - \mathbf{r}(s_0))

Notice that:

At $s = s_0$: $f(s_0) = \mathbf{B}_0 \cdot \mathbf{0} = 0$.
Differentiating with respect to $s$:

f'(s) = \mathbf{B}_0 \cdot \mathbf{r}'(s) = \mathbf{B}(s) \cdot \mathbf{T}(s) = 0

since the binormal and tangent vectors of the Frenet frame are strictly orthogonal. Since $f'(s) = 0$ everywhere on $I$ and $f(s_0) = 0$, $f(s)$ is identically zero for all $s \in I$:

f(s) = \mathbf{B}_0 \cdot (\mathbf{r}(s) - \mathbf{r}(s_0)) \equiv 0 \quad \forall s \in I

This is precisely the equation of a plane passing through $\mathbf{r}(s_0)$ with normal vector $\mathbf{B}_0$. Therefore, the curve lies entirely within this fixed plane! $\blacksquare$

Honors / Problem 2.3 Example 2.3: Picard-Lindelöf Proof of the Fundamental Theorem of Space Curves

Provide a complete, rigorous proof of the Fundamental Theorem of Space Curves (Theorem 2.5):

Formulate the Frenet-Serret equations as a first-order system of linear ordinary differential equations $\frac{d\Phi}{ds} = A(s)\Phi(s)$ for the $3 \times 3$ matrix $\Phi(s) = [\mathbf{T}(s), \mathbf{N}(s), \mathbf{B}(s)]^T$.
Prove that the matrix $A(s)$ is skew-symmetric, and deduce that any solution with orthonormal initial conditions $\Phi(s_0) \in \mathrm{SO}(3)$ remains in $\mathrm{SO}(3)$ for all $s \in I$.
Prove that two curves with identical curvature $\kappa(s)$ and torsion $\tau(s)$ differ by at most a rigid Euclidean transformation $\mathbf{r}_2(s) = R \mathbf{r}_1(s) + \mathbf{r}_0$ where $R \in \mathrm{SO}(3)$.

1. Matrix Formulation of the Frenet System

Let $\mathbf{T}, \mathbf{N}, \mathbf{B}$ be arranged as the rows of a $3 \times 3$ matrix:

\Phi(s) = \begin{pmatrix} \mathbf{T}(s)^T \\ \mathbf{N}(s)^T \\ \mathbf{B}(s)^T \end{pmatrix}

The Frenet-Serret equations can be written compactly as:

\frac{d\Phi}{ds} = A(s) \Phi(s)

where the $3 \times 3$ coefficient matrix $A(s)$ is:

A(s) = \begin{pmatrix} 0 & \kappa(s) & 0 \\ -\kappa(s) & 0 & \tau(s) \\ 0 & -\tau(s) & 0 \end{pmatrix}

Since $\kappa(s)$ and $\tau(s)$ are continuous on $I$, the matrix-valued function $A(s)$ is continuous on $I$. By the Picard-Lindelöf Theorem for linear differential equations, given any initial condition $\Phi(s_0) = \Phi_0$, there exists a unique $C^1$ matrix solution $\Phi(s)$ defined on the entire interval $I$.

2. Preservation of Orthonormality ($\Phi(s) \in \mathrm{SO}(3)$)

Notice that the matrix $A(s)$ is skew-symmetric:

A(s)^T = \begin{pmatrix} 0 & -\kappa(s) & 0 \\ \kappa(s) & 0 & -\tau(s) \\ 0 & \tau(s) & 0 \end{pmatrix} = -A(s)

We examine the Grammian matrix $G(s) = \Phi(s) \Phi(s)^T$. Differentiating with respect to $s$ using the product rule:

\frac{d}{ds} \left( \Phi(s) \Phi(s)^T \right) = \frac{d\Phi}{ds} \Phi(s)^T + \Phi(s) \left( \frac{d\Phi}{ds} \right)^T

Substituting $\frac{d\Phi}{ds} = A(s) \Phi(s)$:

\begin{aligned} \frac{d}{ds} \left( \Phi(s) \Phi(s)^T \right) &= (A(s) \Phi(s)) \Phi(s)^T + \Phi(s) (A(s) \Phi(s))^T \\ &= A(s) (\Phi(s) \Phi(s)^T) + \Phi(s) (\Phi(s)^T A(s)^T) \\ &= A(s) G(s) + G(s) A(s)^T \end{aligned}

Suppose we choose an initial frame $\Phi(s_0) = \Phi_0 \in \mathrm{SO}(3)$, so $G(s_0) = \Phi_0 \Phi_0^T = I_3$ (the $3 \times 3$ identity matrix). Notice that the constant function $\tilde{G}(s) \equiv I_3$ satisfies the differential equation:

$$A(s) I_3 + I_3 A(s)^T = A(s) + A(s)^T = A(s) - A(s) = 0$$

By the uniqueness theorem for linear ODEs, the unique solution satisfying $G(s_0) = I_3$ must be:

G(s) = \Phi(s) \Phi(s)^T = I_3 \quad \forall s \in I

Furthermore, since $\det(\Phi(s_0)) = 1$ and $\det(\Phi(s)) = \pm 1$ continuously, $\det(\Phi(s)) \equiv 1$ for all $s \in I$. Therefore, $\Phi(s) \in \mathrm{SO}(3)$ for all $s \in I$. This guarantees that $\{\mathbf{T}(s), \mathbf{N}(s), \mathbf{B}(s)\}$ remains a valid right-handed orthonormal frame throughout $I$!

3. Proof of Uniqueness up to Euclidean Rigid Motion

Let $\mathbf{r}_1(s)$ and $\mathbf{r}_2(s)$ be two unit-speed curves with identical curvature $\kappa(s)$ and torsion $\tau(s)$. Let their Frenet frames be $\Phi_1(s)$ and $\Phi_2(s)$. At the initial point $s_0$, $\Phi_1(s_0), \Phi_2(s_0) \in \mathrm{SO}(3)$. Define the rotation matrix:

R = \Phi_2(s_0)^T \Phi_1(s_0) \in \mathrm{SO}(3)

which maps the frame of curve 1 to the frame of curve 2 at $s_0$: $\Phi_2(s_0) = \Phi_1(s_0) R^T$. Now define the rotated curve:

\tilde{\mathbf{r}}_1(s) = R \mathbf{r}_1(s) + (\mathbf{r}_2(s_0) - R \mathbf{r}_1(s_0))

The rotated curve $\tilde{\mathbf{r}}_1(s)$ satisfies:

$\tilde{\mathbf{r}}_1(s_0) = \mathbf{r}_2(s_0)$.
Its Frenet frame is $\tilde{\Phi}_1(s) = \Phi_1(s) R^T$.
At $s_0$, $\tilde{\Phi}_1(s_0) = \Phi_1(s_0) R^T = \Phi_2(s_0)$.

Both $\tilde{\Phi}_1(s)$ and $\Phi_2(s)$ satisfy the identical linear ODE system $\frac{d\Phi}{ds} = A(s) \Phi(s)$ with identical initial values at $s_0$. By the uniqueness theorem for ODEs:

\tilde{\Phi}_1(s) = \Phi_2(s) \quad \forall s \in I \implies \tilde{\mathbf{T}}_1(s) = \mathbf{T}_2(s) \quad \forall s \in I

Integrating the velocity vectors:

\mathbf{r}_2(s) - \mathbf{r}_2(s_0) = \int_{s_0}^s \mathbf{T}_2(u) \, du = \int_{s_0}^s \tilde{\mathbf{T}}_1(u) \, du = \tilde{\mathbf{r}}_1(s) - \tilde{\mathbf{r}}_1(s_0)

Since $\tilde{\mathbf{r}}_1(s_0) = \mathbf{r}_2(s_0)$, it follows that:

\mathbf{r}_2(s) = \tilde{\mathbf{r}}_1(s) = R \mathbf{r}_1(s) + \mathbf{r}_0 \quad \forall s \in I

where $R \in \mathrm{SO}(3)$ is a rotation and $\mathbf{r}_0 \in \mathbb{R}^3$ is a translation. This completes the proof of the Fundamental Theorem. $\blacksquare$