Conditions that torsion is zero in a space curve What are the conditions for torsion to be zero other than having a plane curve?
The only thing I can thing of is an equation that have the torsion that cancels out each other.
 A: Here's an example of an infinitely differentiable regular curve with identically zero torsion but not contained in any plane.  (By regular, I mean that the velocity never vanishes.)
$$
\alpha(t) = 
\left\{
\begin{aligned}
&(t,e^{-1/t},0), & t>0,\\
&(0,0,0), &t=0,\\
&(t,0,e^{1/t}), &t<0.
\end{aligned}
\right.
$$
EDIT: As Mariano Suárez-Alvarez points out in his comment, the torsion is only defined at points where the curvature is nonzero.  Since the curvature of this curve is zero at the origin, the torsion is not defined there.  
Thus the argument given by yasmar shows that if the curve is regular, its curvature is nowhere zero, and its torsion is everywhere zero, then it's a plane curve.
A: As has been said, the curve is planar iff the torsion is zero. This is clear if you look at the Frenet formulas. For fun, you can get from those a formula that the curve $\alpha: \mathbb{R} \rightarrow \mathbb{R}^3$ must satisfy. We need
$$
\frac{d\mathbf{N}}{ds} + \kappa\mathbf{T} = 0.
$$
Assuming $\alpha$ is parameterized by arc length, this could be written
$$
\frac{d}{ds}\left(\frac{1}{\kappa}\frac{d^2\alpha}{ds^2}\right) + \kappa\frac{d\alpha}{ds} = 0
$$
A: Wikipedia states that "if the torsion of a regular curve is identically zero then this curve belongs to a fixed plane." By "regular curve" I expect they mean that the curve's first and second derivatives are never zero. I imagine that a planar curve connected via a linear segment to another planar curve lying in a different plane would still have zero torsion everywhere.
This is the limit of my knowledge; I'm posting it as an answer so others can see it and point out if there are any errors.
