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Let $\alpha, \bar{\alpha}: I \mapsto \mathbb{R^3}$ be two regular unit speed curves with non vanishing curvature and torsion. Prove that if the binormal vectors of the curves coincide, i.e $B(s) = \bar{B}(s)$, they are congruent.

I know there is a unique isometry that takes the orthonormal frenet frame of $\alpha$ to that of $\bar{\alpha}$, but I don't know how to prove what the exercise is asking me. I tried a proof by contradiction but that led nowhere. I'd be grateful for any help.

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If $B(s) = \overline{B}(s)$, differentiation gives $\tau(s)N(s) = \overline{\tau}(s)\overline{N}(s)$. Since torsions are positive and normals are unit vectors, applying $|\cdot|$ says that $|\tau(s)| = |\overline{\tau}(s)|$, and hence $N(s) = \pm\overline{N}(s)$. Reparametrizing one of the curves we can assume that $N(s) = \overline{N}(s)$. Differentiating again says $$-\kappa(s) T(s) + \tau(s)B(s) = -\overline{\kappa}(s)\overline{T}(s) + \overline{\tau}(s)\overline{B}(s),$$and by previous work we have that $\kappa(s)T(s) = \overline{\kappa}(s)\overline{T}(s)$. Similarly to what we have done to the torsion, we have $\kappa(s) = \overline{\kappa}(s)$. Since $\kappa = \overline{\kappa}$ and $\tau = \overline{\tau}$, we're done.

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  • $\begingroup$ Perfect. Thanks! Thought of almost everything only to have the obvious elude me: just differentiate (as usual when studying differential geometry...) . $\endgroup$ Feb 10 '18 at 1:45
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    $\begingroup$ The more you know... (but yes, if you get stuck in anything in geometry of curves and surfaces, just go trigger happy differentiating everything, a solution will pop out) $\endgroup$
    – Ivo Terek
    Feb 10 '18 at 1:47

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