📂Geometry

Lanczos Theorem Proof

Theorem ¹

$\kappa \ne 0$ for a given unit speed curve $\alpha$ being a helix is equivalent to it being a certain constant $c \in \mathbb{R}$ for $\tau = c \kappa$.

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• $\tau, \kappa$ is the torsion, curvature.

Proof

Definition of a Helix: A regular curve $\alpha$ is called a helix if for some fixed unit vector $\mathbf{u}$, $\left< T, \mathbf{u} \right>$ is a constant, and $\mathbf{u}$ is called the axis.

Auxiliary Lemma: In a $n$-dimensional inner product space $V$, if $E = \left\{ e_{1} , \cdots , e_{n} \right\}$ is an orthogonal set, then $E$ is a basis of $V$, and for all $v \in V$, $$v = \sum_{k=1}^{n} \left< v , e_{k} \right> e_{k}$$

Differentiation of Inner Products: $$\left< f, g \right>^{\prime} = \left< f^{\prime}, g \right> + \left< f, g^{\prime} \right>$$

Frenet-Serret Formulas: If $\alpha$ is a $\kappa (s) \ne 0$ unit speed curve, 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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$(\implies)$

Since $\alpha$ is a helix, for some fixed axis $\mathbf{u}$, $\left< T , \mathbf{u} \right>$ is constant. Let’s express this specifically for a fixed angle $\theta$ as follows. $$\left< T , \mathbf{u} \right> = \cos \theta$$ If for $n \in \mathbb{Z}$, it holds that $\theta = n \pi$, then $$\left< T , \mathbf{u} \right> = \pm 1 \implies T = \pm \mathbf{u}$$ Thus, since $T$ is constant, $\alpha$ is a straight line, and since $\kappa = 0$, it contradicts our assumption, hence $\theta \ne n \pi$ must be true.

$$\pm 1 \ne \cos \theta = \left< T, \mathbf{u} \right>$$ We will use the differentiation of inner products in the above equation. Since $\mathbf{u}$ is constant, $\mathbf{u}^{\prime} = 0$, and according to the Frenet-Serret formulas, $$\begin{align*} 0 =& \left< T, \mathbf{u} \right>^{\prime} \\ =& \left< T^{\prime}, \mathbf{u} \right> + \left< T, \mathbf{u}^{\prime} \right> \\ =& \left< \kappa N, \mathbf{u} \right> + \left< T, 0 \right> \\ =& \kappa \left< N, \mathbf{u} \right> \end{align*}$$ Assuming $\kappa \ne 0$, it must be that $\left< N, \mathbf{u} \right> = 0$. According to the auxiliary lemma, $$\begin{align} \mathbf{u} =& \left< \mathbf{u} , T \right> T + \left< \mathbf{u} , N \right> N + \left< \mathbf{u} , B \right> B \\ =& \cos \theta T + \left< \mathbf{u} , B \right> B \end{align}$$ Squaring both sides gives $|T|^{2} = \left| \mathbf{u} \right| = 1$ and hence $T \perp B \implies \left< T, B \right> =0$, therefore $$\begin{align*} 1 =& \left| \mathbf{u} \right|^{2} \\ =& \cos^{2} \theta |T|^{2} + 2 \cos \theta \left< \mathbf{u} , B \right> \left< T, B \right> + \left< T, B \right>^{2} \\ =& \cos^{2} \theta + \left< \mathbf{u} , B \right>^{2} \end{align*}$$ Since $\sin^{2} + \cos^{2} = 1$, then $$\left< \mathbf{u} , B \right>^{2} = \sin^{2} \theta$$ Going back to $(2)$, we eventually obtain the following. $$\mathbf{u} = \cos \theta T +\sin \theta B$$ Differentiating both sides according to the Frenet-Serret formulas, $$\begin{align*} 0 =& \mathbf{u}^{\prime} \\ =& \cos \theta T^{\prime} + \sin \theta B^{\prime} \\ =& \cos \theta \kappa N + \sin \theta (-\tau N) \\ =& \left( \kappa \cos \theta - \tau \sin \theta \right) N \end{align*}$$ Since $N$ is not $0$, $\kappa \cos \theta - \tau \sin \theta$ must be $0$, and since $\theta \ne n \pi$, it cannot be $\sin \theta \ne 0$. Therefore, $$\kappa {{ \cos \theta } \over { \sin \theta }} = \tau$$ Since $\theta$ is a fixed value, it has been shown that torsion is represented as a constant multiple of curvature like $\tau = \kappa \cot \theta$.

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$(\impliedby)$

Let’s assume $c$ for some constant. For some $0 < \theta < \pi$, let $c := \cot \theta$. $$\tau = \cot \theta \kappa$$ Define the vector $\mathbf{u}$ as follows. $$\mathbf{u} := \cos \theta T + \sin \theta B$$ Differentiating the left side according to the Frenet-Serret formulas, $$\begin{align*} \mathbf{u}^{\prime} =& \cos \theta T^{\prime} + \sin \theta B^{\prime} \\ =& \cos \theta \kappa N + \sin \theta ( - \tau N) \\ =& \left( \cos \theta \kappa - \sin \theta \tau \right) N \\ =& \left( \cos \theta \kappa - \sin \theta {{ \cos \theta } \over { \sin \theta }} \kappa \right) N \\ =& 0 \end{align*}$$ Therefore, $\mathbf{u}$ is constant, and since $\alpha$ is a unit speed curve, $$\left< T, \mathbf{u} \right> = \left< T, \cos \theta T + \sin \theta B \right> = \cos \theta \cdot 1 + 0$$ Therefore, $\left< T, \mathbf{u} \right>$ is constant and $\alpha$ is a helix.

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