📂Geometry

Conditions for a Curve to Lie in a Plane in Three-Dimensional Euclidean Space

Theorem ¹

Given a unit speed curve $\kappa \ne 0$, the following three are equivalent:

• (a): $\alpha$ lies in a plane.
• (b): $B$ is a constant.
• (c): $\tau = 0$ holds.

Explanation

This is a corollary of the Frenet-Serret formulas, allowing us to understand why torsion is defined in such a bizarre way as $\tau := \left< B^{\prime}, N \right>$.

Proof

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

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Part 1. $(b) \implies (c)$

Since $B$ is a constant, $B^{\prime} = \mathbf{0}$, and by the Frenet-Serret formulas, $\tau = - \left< B^{\prime} , N \right> = \mathbf{0}$.

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Part 2. $(c) \implies (b)$

If $\tau = \mathbf{0}$, then by the Frenet-Serret formulas, $B^{\prime} = -\tau N = \mathbf{0}$.

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Part 3. $(a) \implies (b)$

For convenience, assume $\alpha$ is on the $xy$-plane and appears as $\alpha (s) = \left( x(s) , y(s) , 0 \right)$. $$T = \alpha^{\prime} (s) = \left( x^{\prime} , y^{\prime}, 0 \right) \\ N = {{ T^{\prime}(s) } \over { \kappa (s) }} = {{ 1 } \over { \kappa }} \left( x^{\prime \prime} , y^{\prime \prime} , 0 \right)$$ Thus, $$B = T \times N = {{ 1 } \over { \kappa }} \begin{bmatrix} e_{1} & e_{2} & e_{3} \\ x^{\prime} & y^{\prime} & 0 \\ x^{\prime \prime} & y^{\prime \prime} & 0 \end{bmatrix} = {{ 1 } \over { \kappa }} ( 0, 0, \kappa ) = (0 , 0, \pm 1)$$ i.e., $B$ is a constant. Though this is shown in the $xy$-plane, it can be generalized to all planes by merely changing the basis. Here, $\kappa = \left| x^{\prime} y^{\prime \prime} - x^{\prime \prime} y^{\prime} \right|$ is obtained through calculation².

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Part 4. $(b) \implies (a)$

It suffices to show that for some $v \in \mathbb{R}^{3}$, the following holds: $$\left< \alpha (s) - \alpha \left( s_{0} \right) , v \right> = 0$$ For $v = B$, $$\left< \alpha (s) - \alpha \left( s_{0} \right) , B \right>^{\prime} = \left< \alpha^{\prime}(s) , B \right> + \left< \alpha (s) - \alpha \left( s_{0} \right) , B^{\prime} \right> \\ = \left< T , B \right> + \left< \alpha (s) - \alpha \left( s_{0} \right) , 0 \right> \\ = 0 + 0$$ Therefore, $\left< \alpha (s) - \alpha \left( s_{0} \right) , B \right>$ remains unchanged for all $s \in I$, and specifically, if we set $s = s_{0}$, then $$\left< \alpha (s) - \alpha \left( s_{0} \right) , B \right> = 0$$

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