Space curve defined by it's normal vector field

Question

Let $N: (a,b) \to \mathbb{R}^3$ be a vector field such that $||N(s)|| =1$ for all $s\in (a,b)$. Prove that there is unique (up to isometry) curve $\alpha: (a,b) \to \mathbb{R}^3$ such that $N$ is the normal vector field of the curve $\alpha$.

My idea is to express tangent and binormal vector fields in terms of $N$ and then I can easily calculate curvature and torsion. Finally you can use fundamental theorem of curves to get the result. My problem is how to get $T$ in terms of $N$?

Edit

There is a counterexample for any planar curve. Can we prove the theorem assuming that the torsion is not zero, i.e. the curve is not planar.

Answer 1

I'm not sure where you found this question, but I suggest you look for counterexamples.

For example: Start with the unit circle as one curve. Then, parametrizing it by arclength, we'll have $\alpha(s)=(\cos s,\sin s,0)$, say, and $N(s)=(-\cos s,\sin s,0)$. Now, let's take any $a,b>0$ with $a^2+b^2=1$ and consider the circular helix $\beta(s)=(a\cos s,a\sin s, bs)$. The reader can now verify that $\beta$ is likewise arclength parametrized and that $N_\beta(s)=(-\cos s,-\sin s,0)$, as well.

Answer 2

I think the statement is wrong. Let us consider the biregular curve which describes a circle, $\alpha(s) = (\cos(s), \sin(s),0)$ and the biregular curve which describes a helix, $\beta(s)= (\sqrt{2}^{-1}\cos(s),\sqrt{2}^{-1}\sin(s),\sqrt{2}^{-1}s)$.

The normal vector for a biregular curve is given by the normalized vector of $x''(s)$, which in both cases is:$$N(s) = (-\cos(s),-\sin(s),0).$$

In this case, the two curves are not isometric as they have torsion $0$ and $1/\sqrt{2}.$

Edit

Yes, we can still find counterexamples. As the other person posted, we can find infinitely many helices with different torsion, but with the same normal vector $$N(s)=(-\cos{s},-\sin{s},0).$$

The difference between torsions comes from $a^2+b^2=1$. If I remember well, the torsion is given by $\frac{b}{a^2+b^2}$. I may be wrong about the formula, but as we can find infinitely many $a$ and $b$, we can also find infinitely many torsion values.