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I was considering the following question for 3D curves: Does zero curvature imply zero torsion?

I think it's reasonable, because zero curvature implies the curve is a straight line, which lies in a plane, making the torsion zero.

However, as I checked the definitions, the Frenet-Serret frame is not even defined if $\kappa=0$ (even a single point with zero curvature seems problematic). What is the procedure if the curvature vanishes at a point? Does it mean trouble? What happens to the FS frame?

Thank you

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up vote 3 down vote accepted

If you have zero curvature for an open interval of the parameter, you just have a straight line, and you can assign constant $T,N,B$ if convenient, although the $N$ is a choice.

For isolated zero curvature, there is a problem. Here is a $C^\infty$ example: For $t=0,$ let $\gamma(t) = (0,0,0).$ For $t>0,$ let $\gamma(t) = (t,e^{-1/t},0).$ For $t<0,$ let $\gamma(t) = (t,0, e^{1/t}).$ If you prefer, you can just use $-1/t^2$ for both positive and negative $t$ exponents. Here, the field $N$ changes discontinuously at $t=0,$ not much to be done about it.

For $C^\omega$ I think you can work something out using the first nonvanishing derivatives, not sure.

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More simply, $\gamma(t) = (t,t^3,0)$ has $N \to (0,1,0)$ and $B = (0,0,1)$ as $t \to 0+$ while $N \to (0,-1,0)$ and $B = (0,0,-1)$ as $t \to 0-$. – Robert Israel Mar 24 '13 at 20:46
So if I have a straight line, does $\tau$ even exists? – user1337 Mar 26 '13 at 18:14
@MichaelTouitou, yes, then you have $\tau = 0.$ You choose a constant $N,$ then $B = T \times N$ is also constant. To get the derivative formulas you get $\kappa = 0$ and $\tau = 0.$ – Will Jagy Mar 26 '13 at 18:19
So is this fixable if the sign of $\kappa$ is allowed to swap sign after such a point? At least for @RobertIsrael's example one can directly see that this would work, since $\kappa\to-\kappa$ would imply $N\to -N$ and $B = T\times N\to T\times -N = -B$ – Tobias Kienzler Jun 17 '13 at 14:21
@TobiasKienzler, if you want to break up a curve into pieces with nonzero curvature in the middle and zero at the endpoints, that is fine. You just can no longer write some things as single formulas any more. My example lacks the symmetry of Robert's example: $N$ changes discontinuously by a right angle, that is $90^\circ.$ In Robert's example that change is by $180 ^\circ,$ so there is the appearance that negating $\kappa$ would somehow fix things. – Will Jagy Jun 17 '13 at 18:06

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