# Prove $g(t,u)$ is a regular surface when it may have singular points.

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.

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

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 $$\mathbb{R}^3$$.