A space curve is planar if and only if its torsion is everywhere 0

Can someone please explain this proof to me. I know that a circle is planar and has nonzero constant curvature, so this must be an exception, but I am a little lost on the proof. Thanks!

• You are getting torsion and curvature confused I think. en.m.wikipedia.org/wiki/Torsion_of_a_curve Sep 20, 2014 at 18:48
• See Prove that curve with zero torsion is planar. Or is there a specific prove you don't understand? Sep 21, 2014 at 6:20
• There are no exceptions. Torsion is a movement out of the plane of curve. A helix is like drawing a circle, except instead of staying in the plane, it has torsion that brings it out of the plane spiraling outwards. Sep 21, 2014 at 8:13
• Note that you need to add “with nonvanishing curvature” for this to be true. Otherwise consider, say, a circular arc in one plane followed by a straight line segment followed by a circular arc in another plane. Aug 24, 2021 at 3:44

A circle is indeed planar, and has constant nonzero curvature, but the torsion of a circle is zero; it's not an exception.

Having said this, let $$\gamma(s)$$ be a regular planar curve in $$\Bbb R^3$$ parametrized by arc length, say $$\gamma:I \to \Bbb R^3$$, where $$I \subset \Bbb R$$ is an open interval. $$\gamma(s)$$ regular means

$$\dot{\gamma}(s) = \dfrac{d\gamma(s)}{ds} \ne 0 \tag{1}$$

for any $$s \in I$$. If $$P \subset \Bbb R^3$$ is any plane passing through the point $$\vec p_0 \in \Bbb R^3$$, then the points $$\vec r = (x, y, z) \in P$$ satisfy an equation of the form

$$(\vec r - \vec p_0) \cdot \vec n = 0, \tag{2}$$

where $$\vec n$$ is the unit normal vector to $$P$$; that $$P$$ may be so described is well-known, and will be taken so here without further demonstration. If we insert $$\gamma(s) = (\gamma_x(s), \gamma_y(s), \gamma_z(s))$$ into (2), we see that

$$(\gamma(s) - \vec p_0) \cdot \vec n = 0, \tag{3}$$

and differentiating (3) we obtain

$$\dot \gamma(s) \cdot \vec n = 0. \tag{4}$$

We next recall that

$$\dot \gamma(s) = \vec T(s), \tag{5}$$

where $$\vec T(s)$$ is the unit tangent vector to $$\gamma(s)$$, $$\vec T(s) = \dot \gamma(s)$$ (since $$s$$ is arc-length); then by (4)

$$\vec T(s) \cdot \vec n = 0; \tag{6}$$

furthermore, by the Frenet-Serret equation

$$\dot {\vec T}(s) = \kappa \vec N(s), \tag{7}$$

applied to (6) we have

$$\kappa(s) \vec N(s) \cdot \vec n = \dot {\vec T}(s) \cdot \vec n = 0; \tag{8}$$

from (8) we see that, as long as $$\kappa(s) \ne 0$$, that is, as long as $$\vec N(s)$$ may be defined, we have

$$\vec N(s) \cdot \vec n = 0 \tag{9}$$

holding as well as (6); thus both $$\vec T(s)$$ and $$\vec N(s)$$ are normal to $$\vec n$$ as long as they are defined. Now $$T(s)$$ and $$N(s)$$ form an orthonormal system; that is $$\Vert \vec T(s) \Vert = \Vert \vec N(s) \Vert = 1$$ and $$\vec T(s) \cdot \vec N(s) = 0$$, and since the unit binormal vector along $$\gamma(s)$$, $$\vec B(s) = \vec T(s) \times \vec N(s)$$ also satisifies

$$\vec B(s) \cdot \vec T(s) = (\vec T(s) \times \vec N(s)) \cdot \vec T(s) = 0, \tag{10}$$

$$\vec B(s) \cdot \vec N(s) = (\vec T(s) \times \vec N(s)) \cdot \vec N(s) = 0, \tag{11}$$

$$\vec B(s) \cdot \vec B(s) = 1, \tag{12}$$

we may conclude from the continuity of

$$\vec B(s)$$ that $$\vec B(s) = \pm \vec n, \tag{13}$$

a constant; thus from

$$\dot {\vec B}(s) = -\tau(s) \vec N(s), \tag{14}$$

the Frenet-Serret equation for $$\dot {\vec B}(s)$$, we may infer that

$$\tau(s) \vec N(s) = 0 \Rightarrow \tau(s) = 0, \tag{15}$$

since $$\vec N(s) \ne 0$$ wherever it is defined; we have shown that the torsion $$\tau(s)$$ of any plane curve $$\gamma(s)$$ vanishes.

Of course, there are a couple of caveats in the above argument, most notably the assumptions of regularity (so that $$\vec T(s)$$ exists), and non-vanishing curvature (so that $$\vec N(s)$$ exists); but I think these can be covered pretty easily; I'll defer the discussion until after we have handled the "if" direction of the assertion's logic.

So suppose $$\tau(s) = 0$$; that is, that $$\gamma(s)$$ is a regular curve in $$\Bbb R^3$$ with vanishing torsion. Then $$\vec B(s)$$ must be constant along $$\gamma(s)$$, by (14); choosing $$s_0 \in I$$ we have $$\vec B(s_0) = \vec B(s)$$ for all $$s \in I$$; thus by (10)-(11) $$\vec T(s)$$ and $$\vec N(s)$$ both belong to the subspace $$V \subset \Bbb R^3$$ with $$\vec B(s_0) \bot V$$; indeed, we may take this subspace to be spanned by $$\vec T(s_0)$$, $$\vec N(s_0)$$, since they form an orthonormal pair in $$V$$; $$V = \text{span} \{ \vec T(s_0), \vec N(s_0) \}$$. This being the case, we may write

$$\dot {\gamma}(s) = \vec T(s) = \langle \vec T(s), \vec T(s_0) \rangle \vec T(s_0) + \langle \vec T(s), \vec N(s_0) \rangle \vec N(s_0); \tag{16}$$

upon integrating (16) we find

$$\gamma(s) - \gamma(s_0) = \displaystyle \int_{s_0}^s \dot {\gamma}(u) du = \int_{s_0}^s \vec T(u) du$$ $$= \left (\displaystyle \int_{s_0}^s \langle \vec T(u), \vec T(s_0) \rangle du \right ) \vec T(s_0) + \left (\displaystyle \int_{s_0}^s \langle \vec T(u), \vec N(s_0) \rangle du \right ) \vec N(s_0), \tag{17}$$

which implies that

$$(\gamma(s) - \gamma(s_0)) \cdot \vec B(s_0)$$ $$=(\int_{s_0}^s \langle \vec T(u), \vec T(s_0) \rangle du) \langle \vec T(s_0), \vec B(s_0) \rangle$$ $$+ (\int_{s_0}^s \langle \vec T(u), \vec N(s_0) \rangle du) \langle \vec N(s_0), \vec B(s_0) \rangle = 0 \tag{18}$$

for all $$s \in I$$; but the equation of the plane normal to $$\vec B(s_0)$$ passing through the point $$\gamma(s_0)$$ is in fact

$$(\vec r - \gamma(s_0)) \cdot \vec B(s_0) = 0, \tag{19}$$

where $$\vec r = (x, y, z)$$. Thus $$\gamma(s)$$ lies in this plane. QED.

We can actually take things a step further and present concise formulas for $$\vec T(s)$$ and $$\vec N(s)$$ in terms of $$\displaystyle \int_{s_0}^s \kappa(u)du$$ as follows: When $$\tau(s) = 0$$, the Frenet-Serret equations become

$$\dot{\vec T}(s) = \kappa(s) \vec N(s), \tag{20}$$

$$\dot{\vec N}(s) = -\kappa(s) \vec T(s), \tag{21}$$

and

$$\dot {\vec B}(s) = 0. \tag{22}$$

(22) implies $$B(s)$$ is constant; inspecting (20)-(21) reveals they may be written in combined form by introducing the six-dimensional column vector $$\vec \Theta(s)$$:

$$\vec \Theta(s) = (\vec T(s), \vec N(s))^T, \tag{23}$$

so that

$$\dot {\vec \Theta}(s) = (\dot {\vec T}(s), \dot {\vec N}(s))^T; \tag{24}$$

with this convention, (20)-(21) may be written

$$\dot {\vec {\Theta}}(s) = \begin{bmatrix} 0 & \kappa(s)I_3 \\ -\kappa(s)I_3 & 0 \end{bmatrix} \vec {\Theta}(s) = \kappa(s) J \vec{\Theta}(s), \tag{25}$$

where $$I_3$$ is the $$3 \times 3$$ identity matrix and

$$J = \begin{bmatrix} 0 & I_3 \\ -I_3 & 0 \end{bmatrix}; \tag{26}$$

here it is understood that $$J$$ is presented in the from of $$3 \times 3$$ blocks. It is easy to see that

$$J^2 = \begin{bmatrix} 0 & I_3 \\ -I_3 & 0 \end{bmatrix}\begin{bmatrix} 0 & I_3 \\ -I_3 & 0 \end{bmatrix} = \begin{bmatrix} -I_3 & 0 \\ 0 & -I_3 \end{bmatrix} = -I_6, \tag{27}$$

$$I_6$$ being the $$6 \times 6$$ identity matrix. Careful scrutiny of (25) suggests that

$$\vec \Theta(s) = \exp \left (\left ( \displaystyle \int_{s_0}^s \kappa(u) du \right ) J \right ) \vec \Theta(s_0) \tag{28}$$

might be its unique solution taking the value $$\vec \Theta(s_0)$$ at $$s = s_0$$; indeed, we may differentiate (28) with respect to $$s$$ to obtain

$$\dot {\vec \Theta}(s) = \dfrac{d}{ds}\left (\displaystyle \int_{s_0}^s \kappa(u)du \right )J\exp \left (\left ( \displaystyle \int_{s_0}^s \kappa(u) du \right ) J \right ) \vec \Theta(s_0) = \kappa(s) J \vec \Theta(s), \tag{29}$$

showing that (28) satisfies (25); furthermore (28) is consistent with the initial condition at $$s = s_0$$;

$$\vec \Theta (s_0) = \exp \left (\left ( \displaystyle \int_{s_0}^{s_0} \kappa(u) du \right ) J \right ) \vec \Theta(s_0) = e^{0J} \vec \Theta(s_0) = \vec \Theta(s_0). \tag{30}$$

It is worth pointing out that the reason (28) works as a solution is basically that the $$s$$-derivative of the matrix $$\exp \left (\left ( \displaystyle \int_{s_0}^s \kappa(u) du \right ) J \right )$$ follows the scalar pattern

$$\dfrac{d}{ds}e^{u(s)} = \dfrac{du(s)}{ds}e^{u(s)}, \tag{31}$$

viz.

$$\dfrac{d}{ds}\exp \left (\left ( \displaystyle \int_{s_0}^s \kappa(u) du \right ) J \right ) = \dfrac{d}{ds}\left (\displaystyle \int_{s_0}^s \kappa(u) du \right)J\exp \left (\left ( \displaystyle \int_{s_0}^s \kappa(u) du \right ) J \right )$$ $$= \kappa(s)J\exp \left (\left ( \displaystyle \int_{s_0}^s \kappa(u) du \right ) J \right ). \tag{32}$$

(32) applies by virtue of the fact that $$\left ( \displaystyle\int_{s_0}^s \kappa(u) du \right)J$$ and its derivative $$\kappa(s) J$$ commute with one another, being scalar function multiples of the same matrix $$J$$; for general matrix functions $$A(s)$$, it is not true that $$A'(s)A(s) = A(s)A'(s)$$, and the evaluation of $$(d/ds)A(s)$$ becomes much more complicated; we do not in general have

$$\dfrac{d}{ds}e^{A(s)} = \dfrac{A(s)}{ds}e^{A(s)} \tag{33}$$

in parallel with the scalar formula (31); the interested reader may consult my answer to this question (especially the material surrounding equations (15)-(20)) for a more detailed discussion. However, under the special circumstances that $$A(s) = f(s)B$$ for a constant matrix $$B$$, then $$A'(s) = f'(s)B$$ and $$A(s)A'(s) = f(s)f'(s)B^2 = A'(s)A(s)$$; $$A(s)$$ and its derivative always commute in this special case, which is what we have here. (32) applies and thus we have that (28) solves (25).

We examine the matrix $$\exp \left (\left ( \displaystyle \int_{s_0}^s \kappa(u) du \right ) J \right )$$ occurring in (28) with an eye to determining its structure, and the structure of the solutions to (25). That $$J^2 = - I_6$$ has been noted. Thus we have

$$J^2 = -I_6; \; \; J^3 = J^2J = -J; \;\;$$ $$J^4 = J^3J= -J^2 = I_6; \; \; J^5 = (J^4)J = I_6J = J, \tag{34}$$

and in general,

$$J^{4n + p} = J^{4n}J^p = (J^4)^nJ^p = (I_6)^n J^p = J^p, \tag{35}$$

which shows that all cases of $$J^m$$, $$m \in \Bbb Z$$, are in fact covered by (34), i.e. for $$0 \le p \le 3$$. If we expand the matrix $$\exp \left (\left ( \displaystyle \int_{s_0}^s \kappa(u) du \right ) J \right )$$ as a power series

$$\exp \left (\left ( \displaystyle \int_{s_0}^s \kappa(u) du \right ) J \right )$$ $$= \displaystyle \sum_0^\infty \dfrac{\left ( \left (\displaystyle \int_{s_0}^s \kappa(u) du) \right )J \right )^n}{n!} = \sum_0^\infty \dfrac{\left (\displaystyle \int_{s_0}^s \kappa(u) du \right )^nJ^n}{n!}, \tag{36}$$

we may decompose the right-hand sum in accord with the periodicity relations of the powers of $$J$$, (34)-(35), as follows: we first observe that the pattern of powers of $$J$$ is essentially the same as that of the powers of the ordinary complex number $$i$$, viz.,

$$i^2 = -1; \; \; i^3 = i^2i = -i; \;\;$$ $$i^4 = i^3i = -i^2 = 1; \; \; i^5 = (i^4)i = 1i = i, \tag{37}$$

$$i^{4n + p} = i^{4n}i^p = (i^4)^ni^p = 1^n i^p = i^p, \tag{38}$$

which give rise to the well-known Euler formula

$$e^{ix} = \cos x + i\sin x \tag{39}$$

for

$$e^{ix} = \displaystyle \sum_0^\infty \dfrac{(ix)^n}{n!} \tag{40}$$

when this sum is split into its real and imaginary parts; that is,

$$\Re[e^{ix}] = \cos x = \displaystyle \sum_0^\infty (-1)^n \dfrac{x^{2n}}{(2n)!} \tag{41}$$

and

$$\Im[e^{ix}] = \sin x = \displaystyle \sum_0^\infty (-1)^n \dfrac{x^{2n + 1}}{(2n + 1)!}, \tag{42}$$

To be continued/completed; stay tuned!!?

• The restriction to a regular plane curve rules out cases where, for example, the curve spirals in to a smaller and smaller radius, correct? Jun 14, 2019 at 15:16
• No, consider the clothoid in which $\kappa \sim s$, where $s$ is arc-length from a point on the curve. See math.stackexchange.com/questions/3097067/…. Also known as Euler spirals. Check it out! Cheers! Jun 14, 2019 at 16:04
• My man! unbelievable that you are writing such big answers in this age! Hats off Aug 21, 2021 at 8:09
• @RobertLewis You are an inspiration for me :) Aug 21, 2021 at 8:18
• 7 years and still not completed...... :( Aug 21, 2021 at 8:19