Here, differentiable means $C^\infty$.

Let $I\subset \Bbb{R}$ be an open interval and $\alpha:I\to \Bbb{R}^3$ be a regular curve parameterized by arc length ($|\alpha'(s)|=1,\forall s\in I$) with curvature $|\alpha''(s)|=k(s)>0,\forall s\in I$. Define $n(s)$ as the normal vector, by $n(s)=\dfrac{1}{k(s)}\alpha''(s)$, which we also know to be differentiable, i.e. $n:I\to \Bbb{R}^3$ is differentiable. We also denote $b$ the binormal vector, $\tau$ the torsion, etc.

Suppose then that all the normal lines to the curve meet together at the origin. This implies that, for each $s\in S$, there exists a unique $\lambda(s)\in \Bbb{R}$ such that $\alpha(s)=\lambda(s)n(s)$.

So, I can define a function $\lambda:I\to \Bbb{R}$ such that each $\lambda(s)$ is as above.

Question: Is $\lambda:I\to \Bbb{R}$ differentiable? Why?

I know that $\alpha$ and $n$ are differentiable, but does this $\alpha=\lambda n$ alone suffice to know that $\lambda$ is differentiable?

I need this in order to show that $\alpha$ is in fact a part of a circle. At Manfredo's Differentiable Geometry of Curves and Surfaces this is the exercise 4, of §1.5 and he suggests simply to take the derivative of $\alpha=\lambda n$, (which furthermore he claims to be constant!) to get $$(1-\lambda k)t+\lambda'n-\lambda \tau b=0,$$ but I don't know if it is allowed to take $\lambda'$...

Edit: to say that $\alpha(s)=\lambda(s)n(s)=const.$ seems wrong, since I'm trying to prove that $\alpha(s)$ lies on a circle. Maybe the author did mean that $\lambda(s)=const.$ (?)


1 Answer 1


If $\alpha=\lambda n$ then we have that

$$\langle \alpha ,n\rangle=\langle \lambda n,n\rangle =\lambda.$$ Thus $\lambda$ is differentiable since $\alpha$ and $n$ are differentiable.


If $\alpha=\lambda n$ is constant then $\alpha$ is a point. In this case we can't define the tangent vector nor the normal vector. So the author means that $\lambda$ is constant.

Note that

$$\lambda'=\langle \alpha',n\rangle+\langle \alpha,n'\rangle=\langle T,n\rangle+\langle \alpha,n'\rangle=0+\langle \alpha,n'\rangle=\langle \alpha,n'\rangle.$$ So

$$\lambda'=\langle \alpha,n'\rangle=\langle \lambda n,n'\rangle=\lambda\langle n,n'\rangle=0,$$ from where we get that $\lambda$ is constant.

  • $\begingroup$ Thank you! It was so simple... $\endgroup$
    – Ders
    Dec 16, 2016 at 19:34

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