# 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.

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If the torsion is identically zero, the curve must be planar. Do you want a condition for torsion to be zero at a particular point (but not necessarily elsewhere)? – Rahul Oct 18 '10 at 11:57
No, it should be zero through out. Any ideas or it is just not possible? – vener Oct 18 '10 at 12:47
I know for certain that it's not possible for any regular curve. – WWright Oct 18 '10 at 13:21
Notice that torsion is usually only defined for curves at points where curvature does not vanish, so to make sense of this some assumption in that direction is needed. – Mariano Suárez-Alvarez Oct 18 '10 at 21:45

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 $$\frac{d}{ds}\left(\frac{1}{\kappa}\frac{d^2\alpha}{ds^2}\right) + \kappa\frac{d\alpha}{ds} = 0$$

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As Rahul points out, this is assuming that the curve is regular. Otherwise the 1/\kappa is a problem. – yasmar Oct 18 '10 at 14:12
@vener: Remember that intuitively, torsion measures how much your space curve departs from planarity. It stands to reason that zero torsion gets you a flat curve, as removing the appropriate terms in the Frenet-Serret formulae show. – J. M. Oct 18 '10 at 14:15

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.

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For those having trouble imagining the beastie: it's a curve with two flat sections joined at right angles at the origin. – J. M. Oct 18 '10 at 21:41
Torsion is usually only defined for curves at points where curvature does not vanish. – Mariano Suárez-Alvarez Oct 18 '10 at 21:44
@Mariano: Oh, good point. I had completely overlooked that. So I guess my example is, at best, a curve with torsion that's identically zero where it's defined, which is everywhere but the origin. – Jack Lee Oct 18 '10 at 21:54
In his book 'Differential Geometry of Curves and Surfaces' do Carmo gives a similar example and then asks: "Show that $\tau$ can be defined so that $\tau \equiv 0$, even though $\alpha$ is not a plane curve." Can someone explain what does he means? Is it possible to define torsion when the curvature vanish? – Nuno Oct 18 '10 at 23:07
@Nuno: At all points where the curvature does not vanish, the torsion is zero. I guess do Carmo means that one can extend the torsion from that set to the whole curve continuously. – Mariano Suárez-Alvarez Oct 19 '10 at 0:11

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.

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It seems that the Wikipedia statement you referred to is false. The usual definition of a regular curve (confirmed at en.wikipedia.org/wiki/Curve) is a differentiable curve whose velocity never vanishes. With this definition of regular, it's not true that a regular curve with zero torsion must be a plane curve. (See my answer to this question.) The correct theorem statement should be "If the torsion of a regular curve with nonvanishing curvature is zero, then this curve belongs to a fixed plane." – Jack Lee Oct 18 '10 at 21:35
@Jack Lee: Thanks. Seeing Mariano's and your comments, I've edited the Wikipedia page. (Ah, the perils of citing Wikipedia.) – Rahul Oct 18 '10 at 23:38
@Jack Lee @Rahul I was assuming regular was as Rahul defined here, i.e., the curvature doesn't vanish. It sounds like this may not be a universal definition? It does agree with the definition given on this wikipedia page: en.wikipedia.org/wiki/Differential_geometry_of_curves I don't have DoCarmo's book on curves and surfaces here, but I would take that as a good authority on what the 'standard' definition should be. – yasmar Oct 19 '10 at 2:21