0
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $N$.)

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user3342072 May 14 '18 at 9:22
  • 1
    $\begingroup$ 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). $\endgroup$ – Chappers May 14 '18 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.