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Prove that curve with zero torsion is planar

How can we use the Frenet-Serret formulas to prove if $\alpha(s)$ is a unit speed curve with $\kappa \neq 0$ and $\tau = 0$, then $\alpha(s)$ lies in a plane.

I am thinking maybe to show that $B$ is contant, but not sure how this helps.

Thanks

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Hint for visualization: If $\tau = 0$, then a moving frame does not twist around the velocity vector. How does this constrain the motion of a particle on the curve? Hint for computation: Using the Frenet-Serret formulas, if $\tau = 0$, can you find a vector that is perpendicular to all velocity vectors of the curve? –  Neal Oct 11 '12 at 20:45
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marked as duplicate by Qiaochu Yuan Oct 12 '12 at 2:33

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By Frenet-Serret formula, we have $B'(s)=\tau(s) N(s)$, where $B(s)$ is binormal and $N(s)$ is normal. Since $\tau(s)\equiv 0$, we have $B'(s)=0$ which implies that $B(s)$ must be a constant vector.

Now choose any point $\alpha(s_0)$ on the curve $\alpha$, consider the function given by $f(s)=(\alpha(s)-\alpha(s_0))\cdot B(s)$. Then $$f'(s)=\alpha'(s)\cdot B(s)+(\alpha(s)-\alpha(s_0))\cdot B'(s)=T(s)\cdot B(s)+(\alpha(s)-\alpha(s_0))\cdot 0=0$$ where $T(s)$ is the tangent and is orthogonal to $B(s)$. Hence, $f(s)$ must be a constant function, i.e. $$(\alpha(s)-\alpha(s_0))\cdot B(s)= c\mbox{ for all }s$$ for some constant $c$. On the other hand, when $s=s_0$, the left hand side is zero. Therefore, $c=0$, i.e. $$(\alpha(s)-\alpha(s_0))\cdot B(s)= 0\mbox{ for all }s.$$ Recall that $B(s)$ is a constant vector, i.e. $B(s)=B$ for some constant vector $B$. Hence, we have $$(\alpha(s)-\alpha(s_0))\cdot B= 0\mbox{ for all }s,$$ which implies that $\alpha(s)$ lies on the plane with $B$ as the unit normal to the plane.

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