For $I$ an open interval in $\mathbb{R}$, let $\gamma: I \to \mathbb{R^3}$ be a smooth unit speed curve whose curvature vanish nowhere.

Define the normal surface of $\gamma$ as the image set of the map

$g:I \times\mathbb{R} \to \mathbb{R^3},g(t,u)= \gamma(t)+u N(t); N= \frac{\ddot{\gamma}}{\Vert \ddot{\gamma} \Vert}$

I figured out when $u= \frac{1}{\kappa},\tau=0$, $g(t,u)$is not a regular point, since the differential here is not monomorphic.

I am confused because I need to prove $g$ is actually a 2-dimensional submanifold of $\mathbb{R^3}$ (regular surface), but here we have a contradiction. Can anyone explain it?

Any help would be appreciated.

  • $\begingroup$ @Matematleta Sorry for my bad notation, $N= \frac{\ddot{\gamma}}{\Vert \ddot{\gamma} \Vert}$, that is the second derivative of curve. $\endgroup$
    – Flashhh
    Oct 8, 2019 at 2:04

1 Answer 1


Carefully check the definition of your problem. If the domain of $g$ is actually $I \times \mathbb{R}$ the claim is false because the image of $g$ can be self-intersecting in $\mathbb{R}^3$. My guess would be that the domain of $g$ is $I \times (-\varepsilon, \varepsilon)$ for some sufficiently small $\varepsilon$. In that case the image of $g$ is a submanifold of $\mathbb{R}^3$.


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