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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.

Matheus Andrade
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2 Answers2

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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.

Ivo Terek
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  • Perfect. Thanks! Thought of almost everything only to have the obvious elude me: just differentiate (as usual when studying differential geometry...) . – Matheus Andrade Feb 10 '18 at 01:45
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    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) – Ivo Terek Feb 10 '18 at 01:47
  • Liked your answer very much... as at first it seemed counter intuitive to me. Next, how does congruence change by including non zero geodesic curvature $\kappa_g$ in the TNB frame? Are there some examples? – Narasimham Oct 27 '21 at 00:55
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We have $$B'(s) = \tau (s) N(s) \qquad \text{and} \qquad \overline{B}'(s) = \overline{\tau} (s) \overline{N}(s)$$ since $$B(s) = \overline{B}(s), \quad B'(s) = \overline{B}'(s)$$ then $$\tau(s) N(s) = \overline{\tau}(s) \overline{N}(s).$$ We have that $\tau(s) \neq 0$ so $$N(s) = \frac{\overline{\tau}(s)}{\tau(s)} \cdot \overline{N}(s).$$

Taking the norm $$\|N(s)\| = \left| \frac{\overline{\tau}(s)}{\tau(s)}\right| \cdot \|\overline{N}(s)\| \Rightarrow \left| \frac{\overline{\tau}(s)}{\tau(s)}\right| =1 \Rightarrow |\overline{\tau}(s)| = |\tau(s)|$$

then, $N(s) = \pm \overline{N}(s)$. So $$\frac{\alpha''(s)}{\kappa(s)} = \pm \frac{\alpha''(s)}{\overline{\kappa(s)}} \Rightarrow \frac{\|\alpha''(s)\|}{\kappa(s)} = \frac{\|\alpha''(s)\|}{\overline{\kappa}(s)} = 1$$ $$\Rightarrow \frac{\overline{\kappa(s)}}{\kappa(s)}=1 \Rightarrow \overline{\kappa}(s) = \kappa(s).$$

As the curves are equal and the torsions coincide at less than sign, so by the fundamental theorem of the curves, there is an isometry $F: \mathbb{R}^3 \to \mathbb{R}^3$ such that $\overline{\alpha} = \alpha \circ F$. Therefore, $\alpha$ and $\overline{\alpha}$ are congruents.

user908109
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