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
Best Answer
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.