# 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 – ClassicStyle Sep 20 '14 at 18:48
• See Prove that curve with zero torsion is planar. Or is there a specific prove you don't understand? – Florian Sep 21 '14 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. – Kainui Sep 21 '14 at 8:13

## 1 Answer

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 deomonstration. 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

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

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 $\int_{s_0}^s \kappa(u)du$ 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

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

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

$\Theta(s) = e^{(\int_{s_0}^s \kappa(u) du)J} \Theta(s_0) \tag{28}$

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

$\dot {\Theta}(s) = \dfrac{d}{ds}(\int_{s_0}^s \kappa(u)du)Je^{(\int_{s_0}^s \kappa(u) du)J} = \kappa(s)J \Theta(s_0) = \kappa(s) J \Theta(s), \tag{29}$

showing that (28) satisfies (26); furthermore (28) is consistent with the initial condition at $s = s_0$;

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

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

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

viz.

$\dfrac{d}{ds}e^{(\int_{s_0}^s \kappa(u) du)J} = \dfrac{d(\int_{s_0}^s \kappa(u) du)J}{ds}e^{(\int_{s_0}^s \kappa(u) du)J} = \kappa(s)Je^{(\int_{s_0}^s \kappa(u) du)J}. \tag{32}$

(32) applies by virtue of the fact that $\int_{s_0}^s \kappa(u) du)J$ 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 $e^{(\int_{s_0}^s \kappa(u) du)J}$ 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 $e^{(\int_{s_0}^s \kappa(u) du)J}$ as a power series

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

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? – nwsteg Jun 14 '19 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! – Robert Lewis Jun 14 '19 at 16:04