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I'm trying to calculate the parallel frame $\{T, U, V\}$ of a space curve $\alpha : I \mapsto \mathbb{R}^3$. It's similar to Frenet frame, except we have instead the projection of torsion $\tau$ on $U$: $\tau_U = 0$

It's described by the three equations, $$T' = \kappa_U U + \kappa_V V \quad \quad \quad \quad \ \ (1)$$ $$U' = -\kappa_U T = -\langle \kappa,U \rangle \ T \quad \quad (2)$$ $$V' = -\kappa_V T = -\langle \kappa,V \rangle \ T \quad \quad (3)$$

where $\kappa = T'$ is the curvature vector and $\kappa_U, \kappa_V$ are its components. I know $T$, $\kappa$ and the initial conditions $\{T(0), U(0), V(0)\}$ at $\alpha(0)$ but I don't know how to solve the ODE $(2)$ and $(3)$ in order to get $U$ and $V$.

Any suggestion or a note on what to search for would be great!

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  • $\begingroup$ When the torsion is null, the curve is planar. $\endgroup$ – Yves Daoust May 16 at 15:56
  • $\begingroup$ I meant $\tau_U = 0$, the projection of $\tau$ on $U$. Edited it now. $\endgroup$ – mike May 16 at 16:09
  • $\begingroup$ In the first line, I'm guessing its $T'=\kappa_U U + \kappa_V V$? $\endgroup$ – Calvin Khor May 16 at 17:34
  • $\begingroup$ Yes :D edited it now too! $\endgroup$ – Sptmp May 16 at 18:00
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To find $U$ and $V$ first consider the Frenet frame $\{T,N,B\}$ along a unit speed curve (or where it exists). We know this frame explicitly since $N=T'/\|T'\|$ and $B=T\times N$. Also note that there exists locally a smooth function $\theta\colon I \to \mathbb{R}$ such that $$U = \cos \theta N + \sin \theta B.$$ Deriving $U$ gives $$ U' = -\kappa \cos\theta T + (\theta'+\tau)(-\sin\theta N + \cos\theta B). $$ We see that $U$ satisfies ODE $(2)$ iff $\theta' = -\tau$. Also the initial value $\theta(0)$ is determined by $U(0)$, so we find $\theta$ from this: $$ \theta = -\int \tau\,dt + \mathrm{const.} $$ The vector field $V$ is determined similarly. Note that the inner product $U\cdot V$ is constant along the curve. So if $U(0)$ and $V(0)$ are orthogonal, we can just take $V=\pm T \times U$.

Remark: You can do the same calculation starting with an arbitrary adapted frame $\{T,N_1,N_2\}$. You will find the condition $\theta' = -N_1' \cdot N_2$.

You will probably find this article interesting: There is more than one way to frame a curve. by R.L. Bishop (The American Mathematical Monthly, 1975).

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  • $\begingroup$ Thank you for your answer and the article, I will work through it! $\endgroup$ – mike May 17 at 21:28

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