📂Geometry

Proof of the Fundamental Theorem of Curves

Theorem ¹

$a,b$ contains $0$ as an interval. Let’s assume the following holds true:

• (i): $\overline{\kappa}(s) > 0$ is $(a,b)$ at $C^{1}$
• (ii): $\overline{\tau}(s)$ is continuous at $(a,b)$
• (iii): $\mathbf{x}_{0}$ is a fixed point of $\mathbb{R}^{3}$
• (iv): $\left\{ D,E,F \right\}$ is the right-handed orthonormal basis of $\mathbb{R}^{3}$

Then there exists a unique $C^{3}$ regular curve $\alpha : (a,b) \to \mathbb{R}^{3}$ that satisfies the conditions with the parameter being the arc length from $\alpha (0)$: $$\begin{align*} \alpha (0) =& \left( \mathbf{x}_{0} \right) \\ T(0) =& D \\ N(0) =& E \\ B(0) =& F \\ \kappa (s) =& \overline{\kappa} (s) \\ \tau (s) =& \overline{\tau} (s) \end{align*}$$

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• $\left\{ T, N, B , \kappa, \tau \right\}$ is a Frenet-Serret frame.
• $C^{1}$ is a set of continuous differentiable functions.

Explanation

Fundamental Theorem of Curves is a powerful theorem which states that a curve in a $3$-dimensional space can be determined by its curvature and torsion, showing that the following results are not mere coincidence:

• If $\kappa = 0$, then it is a straight line.
• If $\kappa \ne 0 , \tau = 0$, then it is a planar curve.
• If $\kappa / \tau$ is constant, then it is a helix.
• If $\tau = 0$ and $\kappa > 0$ are constant, then it is a circle.
• If $\tau \ne 0$ is constant and $\kappa > 0$ is constant, then it is a circular helix.

The name “fundamental theorem” is well-deserved as it guarantees both uniqueness and existence.

Proof

Picard’s Theorem: For an initial value problem of a system of first-order ordinary differential equations, a unique solution exists.

Frenet-Serret formulas: Given a unit speed curve $\alpha$ satisfying $\kappa (s) \ne 0$, $$\begin{align*} T^{\prime}(s) =& \kappa (s) N(s) \\ N^{\prime}(s) =& - \kappa (s) T(s) + \tau (s) B(s) \\ B^{\prime}(s) =& - \tau (s) N(s) \end{align*}$$

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$$\mathbf{u}_{j}^{\prime} = \sum_{i=1}^{3} a_{ij} (s) u_{i} \\ \left( a_{ij} \right) = \begin{bmatrix} 0 & \overline{\kappa} & 0 \\ -\overline{\kappa} & 0 & \overline{\tau} \\ 0 & \overline{\tau} & 0 \end{bmatrix}$$

Considering the above ODE system and applying Picard’s theorem, a unique solution $\mathbf{u}_{j}(s)$ that satisfies these conditions exists. $$\begin{align*} \mathbf{u}_{1} (0) =& D \\ \mathbf{u}_{2} (0) =& E \\ \mathbf{u}_{3} (0) =& F \end{align*}$$ Next, we need to show that the solution satisfies our required conditions.

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Step 1. $\mathbf{u}_{i}(t)$ are orthonormal.

Let $p_{ij} := \left< \mathbf{u}_{i}, \mathbf{u}_{j} \right>$, $$\begin{align*} p_{ij}^{\prime} =& \left< \mathbf{u}_{i}^{\prime}, \mathbf{u}_{j} \right> + \left< \mathbf{u}_{i}, \mathbf{u}_{j}^{\prime} \right> \\ =& \left< \sum_{k=1}^{3} a_{ki} \mathbf{u}_{k} , \mathbf{u}_{j} \right> + \left< \mathbf{u}_{i}, \sum_{k=1}^{3} a_{kj} \mathbf{u}_{k} \right> \\ =& \sum_{k=1}^{3} a_{ki} p_{kj} + \sum_{k=1}^{3} a_{kj} p_{ik} \end{align*}$$ Thus, $p_{ij}$ is the unique solution to the differential equation with initial values given by Picard’s theorem, $$p_{ij}^{\prime} = \sum_{k=1}^{3} \left( a_{ki} p_{kj} + a_{kj} p_{ik} \right)$$ And $t = 0$ becomes the Kronecker delta function $p_{ij} (0) = \delta_{ij}$. Meanwhile, $$\sum_{k=1}^{3} \left( a_{ki} \delta_{kj} + a_{kj} \delta_{} \right) = a_{ji} + a_{ij} = 0 = \delta_{ij}^{\prime}$$ Thus, $\delta_{ij} = p_{ij}$ itself exists as the unique solution to the given differential equation. Therefore, we have: $$\left< \mathbf{u}_{i} , \mathbf{u}_{j} \right> = \delta_{ij}$$

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Step 2. Regularity of the unit speed curve $\alpha$

$$\alpha (s) := \mathbf{x}_{0} + \int_{0}^{s} \mathbf{u}_{1} (\sigma) d \sigma$$ Regarding $s \in (a,b)$, let $\alpha (s)$ be as previously set. First differentiating once gives, according to the fundamental theorem of calculus, $${{ d \alpha } \over { ds }} = \mathbf{u}_{1} (s)$$ Differentiating once more according to the initial differential equation gives, $${{ d^{2} \alpha } \over { ds^{2} }} = \mathbf{u}_{1}^{\prime} = \overline{\kappa} \mathbf{u}_{2}$$ Since $\overline{\kappa}$ and $\mathbf{u}_{2}$ are differentiable by assumption, differentiating once more results in $${{ d^{3} \alpha } \over { ds^{3} }} = \overline{\kappa}^{\prime} \mathbf{u}_{2} + \overline{\kappa} \mathbf{u}_{2}^{\prime} = \overline{\kappa}^{\prime} \mathbf{u}_{2} + \overline{\kappa} \left( -\overline{\kappa} \mathbf{u}_{1} + \overline{\tau} \mathbf{u}_{3} \right)$$ Since $\overline{\kappa}$ and $\overline{\tau}$ are continuous and $\mathbf{u}_{i}$ are differentiable, they are continuous. Thus ${{ d^{3} \alpha } \over { ds^{3} }}$ is also continuous, thereby $\alpha$ is $C^{3}$. We have already seen in Step 1 that $$\left| {{ d \alpha } \over { ds }} \right| = \left| \mathbf{u}_{1} \right| = 1$$ therefore, $\alpha$ is a unit speed curve.

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Step 3. $\overline{\kappa} = \kappa, \overline{\tau} = \tau, \mathbf{u}_{1} = T, \mathbf{u}_{2} = N, \mathbf{u}_{3} = B$

Since $\alpha^{\prime} = \mathbf{u}_{1}$, it naturally follows that $\mathbf{u}_{1} = T$. According to the Frenet-Serret formulas, $$\kappa N = T^{\prime} = \mathbf{u}_{1}^{\prime} = \overline{\kappa} \mathbf{u}_{2}$$ Since $N$ and $\mathbf{u}_{2}$ are unit vectors and $\overline{\kappa} > 0$, $\overline{\kappa} = \kappa$ must hold, thus $N = \mathbf{u}_{2}$. Because $\left\{ \mathbf{u}_{1} , \mathbf{u}_{2}, \mathbf{u}_{3} \right\}$ is an orthonormal basis of $\mathbb{R}^{3}$, $\left[ \mathbf{u}_{1} , \mathbf{u}_{2}, \mathbf{u}_{3} \right] = \pm 1$ holds and from $s = 0$, $$\left[ D, E, F \right] = \left[ \mathbf{u}_{1} , \mathbf{u}_{2}, \mathbf{u}_{3} \right] = \pm 1$$ the scalar triple product $\left[ \mathbf{u}_{1} , \mathbf{u}_{2}, \mathbf{u}_{3} \right]$ is continuous[^2], thus it must always be $$\left[ \mathbf{u}_{1} , \mathbf{u}_{2}, \mathbf{u}_{3} \right] = 1$$ Therefore, $$B = T \times N = \mathbf{u}_{1} \times \mathbf{u}_{2} = \mathbf{u}_{3}$$ Finally, once more applying the Frenet-Serret formulas, $$-\tau N = B^{\prime} = \mathbf{u}_{3}^{\prime} = - \overline{\tau} \mathbf{u}_{2}$$ yields $N = \mathbf{u}_{2}$.

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