# Frenet–Serret formulas in terms of a curve and cross product

Let $\alpha: I\rightarrow \mathbb R^3$ arc length parametric curve with positive curvature. Show $\exists \ \omega: I\rightarrow \mathbb R^3$ a curve such that $$T'=\omega\times T\quad N'=\omega\times N \quad B'=\omega\times T$$ with $\{ T,N,B\}$ Frenet Frame of $\alpha$.

I'm not sure how to approach this problem. Any hints are appreciated!

The Frenet–Serret formulae give you three equations that such a $\omega$ has to satisfy: $$\kappa N = \omega \times T, \\ -\kappa T + \tau B = \omega \times N, \\ -\tau N = \omega \times B.$$ If we write $\omega = aT+bN+cB$, as we may since $\{T,N,B\}$ form a basis, these give three equations for $a,b,c$ using $N \times T = -B$, $B \times T = N$, $T \times N = B$, $B \times N = -T$, $T \times B = -N$, and $N \times B = T$: $$\kappa N = -bB + c N \\ -\kappa T+ \tau B = aB - cT \\ -\tau N = -aN + bT.$$ Equating coefficients then gives $c=\kappa$, $b=0$ and $a=\tau$. We verify $$T' = (\tau T + \kappa B) \times T = \kappa (B \times T) = \kappa N$$ and so on, so $\omega = \tau T + \kappa B$ will work.
(The condition $\kappa \neq 0$ is required for the normal and binormal to be well-defined: otherwise you don't know which direction to take for $N$.)
• Perfect! I was as far as equating coefficients, but what bothers me is existence of $a,b,c$ functions. Is it reasonable to say $a:=\langle\omega,T \rangle$ and so on? – user3342072 May 14 '18 at 9:22
• I'm not really sure what you mean: $\omega$ is the unknown thing, so I don't see how writing the coefficients like that is helpful to you. Existence of $a,b,c$ is simply a result of $\{T,N,B\}$ being a basis, so if there is a solution, it is of the form given. One then checks that this is a solution (i.e. that the conditions are consistent). – Chappers May 14 '18 at 17:20