Frenet-Serret and Vector Fields

Question

The well-known Frenet-Serret equations,

$\dot T(s) = \kappa N(s), \tag 1$

$\dot N(s) = -\kappa(s) T(s) + \tau(s) B(s), \tag 2$

$\dot B(s) = -\tau(s) N(s), \tag 3$

where

$T = \dot \alpha(s), \tag 4$

$\alpha(s)$ being a unit speed curve in $\Bbb R^3$ with arc-length $s$, are most often applied to discover and describe properties of such space curves.

Given an open set

$U \subset \Bbb R^3, \tag 5$

and a vector field

$X \in C^\infty(U, \Bbb R^3) \tag 6$

on $U$, we may of course consider the flow $\phi_X(x, t)$ of $X$; the reader will recall it is defined as, more or less, the entire family of integral curves of the vector field $X$ in the sense that

$\phi_X(x, 0) = x, \; \forall x \in U, \tag 7$

and

$\dfrac{d}{dt}\phi(x, t) = X(\phi(x, t)). \tag 8$

A somewhat natural area of inquiry based upon these two concepts, the Frenet-Serret apparatus and vector fields and their flows, is the relationship 'twixt the Frenet-Serret formulas and the integral curves of $X$; that is, finding the expressions for $T(s)$, $N(s)$, $B(s)$, $\kappa(s)$ and $\tau(s)$ in terms of $X$ and related quantities such as its magnitude $\vert X \vert = \langle X, X \rangle^{1/2}$ and its derivatives $\nabla X$ etc.

The Question then becomes:

Given a (sufficiently smooth) vector field $X$ on an open set $U \subset \Bbb R^3$, find the vector fields $T(s)$, $N(s)$, and $B(s)$ and the scalar quantities $\kappa(s)$ and $\tau(s)$ associated with the integral curves of $X$, expressed in terms of $X$ and it's associated quantities such as $\vert X \vert$ and so forth.

A Few Observations: Given such an open set $U$ and vector field $X$, of course it is true that the flow $\phi_X(x, t)$ may not exist for all values of $t$, but this is of no consequence here since all calculations are local in nature. Indeed, for all $x \in U$ the flow $\phi_X(x, t)$ is defined for sufficiently small values of $t$, and this is sufficient for the present purposes.

A Useful Starting Point may be the observation that $X/\vert X \vert$ is a unit vector field, and that in fact

$T(s) = \dfrac{X(\alpha(s))}{\vert X(\alpha(s))\vert} \tag 9$

along the arc-length parametrized integral curve $\alpha(s)$ of $X/ \vert X \vert$. Of course, we may also adopt and employ the given parametrization of the integral curves of $X$ by $t$, as in (7), (8); in fact we have

$\dfrac{ds}{dt} = \vert X(\alpha(t)) \vert, \; \dfrac{dt}{ds} = \vert X(\alpha(s)) \vert^{-1}, \tag{10}$

which allow the conversion 'twixt $t$ and $s$ via integration:

$s - s_0 = \displaystyle \int_{t_0}^t \vert X(\alpha(u)) \vert \; du, \; t - t_0 = \displaystyle \int_{s_0}^s \vert X(\alpha(u)) \vert^{-1} \; du. \tag{11}$

We can also express the unit tangent vector $T$ in terms of the parameter $t$:

$T(t) = \dfrac{X(\alpha(t))}{\vert X(\alpha(t))\vert}. \tag{12}$

In these formulas the reader will recognize that $\alpha(t)$ and $\alpha(s)$ represent the same curves in the geometrical sense, that is, he same paths in $\Bbb R^3$, though they are differently parametrized.

Answer 1

I will start with the simpler case where the vector field $\mathbf{X}$ is a $C^{\infty}$ vector field on $\mathbb{R}^{2}$. For simplicity, let us assume that $\mathbf{X}$ is non-singular on all of $\mathbb{R}^{2}$ and that it satisfies $\mathbf{X}\cdot\mathbf{X} = 1$, i.e., the vector field $\mathbf{X}$ has been normalized so that the integral curves of $\mathbf{X}$ are parametrized by arc-length. (Here, `$\cdot$' represents the ordinary Euclidean dot product. I will make every effort to keep the response in terms of ordinary Euclidean geometry, although some tensor analysis could possibly simplify the exposition.)

Write $\mathbf{X}$ in standard Cartesian coordinates as $$ \mathbf{X}(x, y) = f(x, y)\frac{\partial}{\partial x} + g(x, y) \frac{\partial}{\partial y} = \begin{pmatrix}f(x, y)\\ g(x, y)\end{pmatrix}. $$ (I will employ the convention of vector coordinates relative to a basis being column vectors.)

Now consider a point $P = P(x_{0}, y_{0})$ in the plane and denote the integral curve of $\mathbf{X}$ passing through $P$ at time zero by $\alpha(s) = (x(s), y(s))$. By definition we then have that $\alpha(0) = P$ and $$\frac{d\alpha}{ds} = \mathbf{X}\circ\alpha(s) = \begin{pmatrix}f\left(\alpha\left(s\right)\right)\\ g\left(\alpha\left(s\right)\right)\end{pmatrix}.$$

Note that our assumptions on the vector field $\mathbf{X}$ imply that $\frac{d\alpha}{ds}\cdot \frac{d\alpha}{ds} = 1$. It follows that the unit tangent vector field along our curve $\alpha$ is $$ \mathbf{T}(s) = \frac{d\alpha}{ds} = \mathbf{X}\circ\alpha(s) = \begin{pmatrix}f\left(\alpha\left(s\right)\right)\\ g\left(\alpha\left(s\right)\right) \end{pmatrix}, $$ while the (oriented) unit normal vector field $\mathbf{N}$ along $\alpha$ is obtained by a positive counterclockwise rotation of $\pi/2$ radians and is given in coordinates by $$ \mathbf{N}(s) = \begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix} \begin{pmatrix}f\left(\alpha\left(s\right)\right)\\ g\left(\alpha\left(s\right)\right)\end{pmatrix} = \begin{pmatrix}-g\left(\alpha\left(s\right)\right)\\ f\left(\alpha\left(s\right)\right)\end{pmatrix} $$

Standard properties of differentiation and the dot product give that the vector field $\frac{d\mathbf{T}}{ds}$ along $\alpha$ is perpendicular to $\mathbf{T}$ and a scalar multiple of $\mathbf{N}$. The (oriented) curvature function $\kappa$ along $\alpha$ is then found by differentiating the vector field $\mathbf{T}$ and the relation $$ \frac{d\mathbf{T}}{ds} = \frac{d^{2}\alpha}{ds^2} = \kappa(s)\mathbf{N}(s). $$

Since $\mathbf{T}(s) = \frac{d\alpha}{ds} = \mathbf{X}\circ \alpha(s)$ we have the following: (To avoid a proliferation of parentheses, all functions and vector fields defined on $\mathbb{R}^{2}$ are assumed to be evaluated along the integral curve $\alpha(s)$.)

\begin{align*} \frac{d\mathbf{T}}{ds} &= \frac{d}{ds}\left(\mathbf{X}\circ\alpha(s)\right) = \begin{pmatrix} \nabla f \cdot \mathbf{X} \\ \nabla g \cdot \mathbf{X} \end{pmatrix} = \nabla \mathbf{X}\bullet \mathbf{X}\\ \end{align*} On the last expression in the string of equalities above, $\displaystyle \nabla \mathbf{X} = \begin{pmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\\ \end{pmatrix}$ is the Jacobian of the vector field $\mathbf{X}$ and $\nabla \mathbf{X}\bullet \mathbf{X}$ understood to be matrix multiplication. (Again, for emphasis, all functions/vector fields on $\mathbb{R}^{2}$ are understood to be evaluated along the integral curve $\alpha$.)

At this point, one can solve the equation $$ \frac{d\mathbf{T}}{ds} = \begin{pmatrix} \nabla f \cdot \mathbf{X} \\ \nabla g \cdot \mathbf{X} \end{pmatrix} = \kappa(s) \begin{pmatrix} -g\\ f\\ \end{pmatrix} $$ for the (oriented) curvature function $\kappa$, or use a well-known identity for unit-speed curves in the plane such as $$ \kappa(s) = \det \left(\frac{d\alpha}{ds}, \frac{d^2\alpha}{ds^2}\right) = \det\begin{pmatrix} f & \nabla f\cdot \mathbf{X}\\ g & \nabla g \cdot \mathbf{X}\end{pmatrix}. $$ Either way, one finds $$ \kappa(s) = \left(f \nabla g - g \nabla f\right)\cdot \mathbf{X}, $$ where the right hand side is evaluated along the integral curve $\alpha(s)$.

---

We now turn our attention to the case where $\mathbf{X}$ is a vector field on the three-dimensional Euclidean space $\mathbb{R}^{3}$. As before, we will assume that the vector field $\mathbf{X}$ satisfies $\mathbf{X}\cdot\mathbf{X} = 1$. We will express $\mathbf{X}$ in the standard coordinate frame as $$ \mathbf{X}\left(x, y, z\right) = f(x, y, z) \frac{\partial }{\partial x} + g(x, y, z) \frac{\partial }{\partial y} + h(x, y, z) \frac{\partial }{\partial z} = \begin{pmatrix} f(x, y, z)\\ g(x, y, z)\\ h(x, y, z)\\ \end{pmatrix}. $$

Let $\alpha : \mathbb{R} \to \mathbb{R}^{3}$ be the integral curve of $\mathbf{X}$ that passes through $\displaystyle P = P\left(x_{0}, y_{0}, z_{0}\right)$ at time $s = 0$. As before it follows that $$ \frac{d\alpha}{ds} = \mathbf{T}(s) = \mathbf{X}\circ \alpha(s) $$ is the unit tangent vector field along $\alpha$.

Standard considerations give that $\mathbf{T}$ is perpendicular to $\mathbf{T}^\prime = \frac{d\mathbf{T}}{ds}$. Assuming that $\mathbf{T}^{\prime}(s)$ is non-zero (which is at least the case on an appropriate open interval about 0 when $\mathbf{T}'(0)$ is not the zero vector and the integral curve is not a line/geodesic), then the unit normal vector field $\mathbf{N}$ along $\alpha$ is defined by $$ \mathbf{N} = \frac{\mathbf{T}^{\prime}}{\lvert\lvert \mathbf{T}^{\prime}\rvert\rvert}, $$ with the curvature $\displaystyle \kappa(s) = \lvert\lvert \mathbf{T}^{\prime}\rvert\rvert$.

A similar calculation to the case when the vector field $\mathbf{X}$ is a vector field on the plane gives \begin{equation} \frac{d\mathbf{T}}{ds} = \frac{d}{ds}\left(\mathbf{X}\circ\alpha(s)\right) = \begin{pmatrix} \nabla f \cdot \mathbf{X} \\ \nabla g \cdot \mathbf{X}\\ \nabla h \cdot \mathbf{X} \end{pmatrix} = \nabla \mathbf{X}\bullet \mathbf{X},\\ \end{equation} where $\displaystyle \nabla \mathbf{X} = \begin{pmatrix} \nabla f & \nabla g &\nabla h \end{pmatrix}^{t}$ is the Jacobian of $\mathbf{X}$ and $\nabla \mathbf{X}\bullet \mathbf{X}$ is matrix multiplication. (Again, everything defined on all of $\mathbb{R}^{3}$ is assumed to be evaluated along the integral curve $\alpha$. )

The curvature function $\kappa$ of the integral curve $\alpha$ is then given by $$ \kappa(s) = \sqrt{\left(\nabla \mathbf{X}\bullet \mathbf{X}\right) \cdot \left(\nabla \mathbf{X}\bullet \mathbf{X}\right)}. $$

The unit binormal vector field $\mathbf{B}$ along $\alpha$ is defined by $\mathbf{B} = \mathbf{T} \times \mathbf{N}$. Identifying $$ \mathbf{N} = \frac{\mathbf{T}^\prime}{\lvert\lvert \mathbf{T}^{\prime}\rvert\rvert} = \frac{1}{\kappa} \begin{pmatrix} \nabla f \cdot \mathbf{X} \\ \nabla g \cdot \mathbf{X}\\ \nabla h \cdot \mathbf{X} \end{pmatrix},$$ a cross-product calculation gives \begin{align*} \mathbf{B} &= \frac{1}{\kappa} \begin{pmatrix} g \nabla h \cdot \mathbf{X} - h\nabla g \cdot\mathbf{X}\\ h\nabla f\cdot \mathbf{X} - f \nabla h \cdot \mathbf{X}\\ f \nabla g\cdot\mathbf{X} - g \nabla f \cdot \mathbf{X} \end{pmatrix}\\ &= \frac{1}{\kappa} \begin{pmatrix} \left(g \nabla h - h\nabla g\right) \cdot\mathbf{X}\\ \left(h\nabla f- f \nabla h\right) \cdot \mathbf{X}\\ \left(f \nabla g - g \nabla f\right) \cdot \mathbf{X} \end{pmatrix}\\ &= \frac{1}{\kappa} \begin{pmatrix} \left(g \nabla h - h\nabla g\right)^{t} \\ \left(h\nabla f- f \nabla h\right)^{t} \\ \left(f \nabla g - g \nabla f\right)^{t} \end{pmatrix} \bullet \mathbf{X} \end{align*}

At this point, it is not clear to me the most efficient way to calculate the torsion function $\tau$ of the integral curve $\alpha$, although one option is as follows. For a unit-speed curve $\alpha$ with non-zero curvature in three-dimensional Euclidean space, we have that the torsion $\tau$ is \begin{align*} \tau &= \frac{1}{\kappa^2} \left(\mathbf{T} \times \mathbf{T}^{\prime}\right)\cdot \mathbf{T}^{\prime\prime}\\ &\\ &=\frac{1}{\kappa}\mathbf{B}\cdot\mathbf{T}^{\prime\prime}. \end{align*}

Additionally, we have everything except $\mathbf{T}^{\prime\prime}$ expressed in terms of the components s the vector field $\mathbf{X}$. Differentiating (1) with respect to $s$ yields $$ \frac{d}{ds}\left(\frac{d\mathbf{T}}{ds}\right) = \mathbf{T}^{\prime\prime} = \begin{pmatrix} \left(f\nabla f_{x} + f_{x} \nabla f + g \nabla f_{y} + f_{y} \nabla g + h \nabla f_{z} + f_{z}\nabla h \right)\cdot \mathbf{X}\\ \left(f\nabla g_{x} + g_{x} \nabla f + g \nabla g_{y} + g_{y} \nabla g + h \nabla g_{z} + g_{z}\nabla h \right)\cdot \mathbf{X}\\ \left(f\nabla h_{x} + h_{x} \nabla f + g \nabla h_{y} + h_{y} \nabla g + h \nabla h_{z} + h_{z}\nabla h \right)\cdot \mathbf{X} \end{pmatrix} $$ As before, all expressions above are evaluated along the integral curve $\alpha = \alpha(s)$. (It would be nice if the expression for $\mathbf{T}^{\prime\prime}$ could be succinctly summarized using some vector identities, but I do not see it at the moment.)

One can then use the indicated expressions to calculate the torsion of the integral curve $\alpha$ entirely in terms of the vector field $\mathbf{X}$.

Answer 2

The first thing we need to do is construct the unit tangent vector field $T$ to the integral curves of $X$ from $X$ itself; now the magnitude of $X$ is

$\vert X \vert = \langle X, X \rangle^{1/2}; \tag 1$

thus normalizing $X$ yields

$T = \dfrac{X}{\vert X \vert} = \dfrac{X}{\langle X, X \rangle^{1/2}} = \langle X, X \rangle^{-1/2}X, \tag 2$

where $T$ is the unit tangent field to the integral curves of $X$;

$\nabla_{X/\vert X \vert} \left(\dfrac{X}{\vert X \vert} \right) = \nabla_{\langle X, X \rangle^{-1/2}X} \left( \langle X, X \rangle^{-1/2}X \right) = \langle X, X \rangle^{-1/2}\nabla_X \left( \langle X, X \rangle^{-1/2}X \right); \tag 3$

$\nabla_X \left( \langle X, X \rangle^{-1/2}X \right) = \left ( \nabla_X\langle X, X \rangle^{-1/2} \right ) X + \langle X, X \rangle^{-1/2} \nabla_X X; \tag 4$

$ \nabla_X\langle X, X \rangle^{-1/2} = -\dfrac{1}{2}\langle X, X \rangle^{-3/2} \nabla_X \langle X, X \rangle = -\dfrac{1}{2}\langle X, X \rangle^{-3/2} \left ( \langle \nabla_X X, X \rangle + \langle X, \nabla_X X \rangle \right )$ $= -\dfrac{1}{2}\langle X, X \rangle^{-3/2} \left (2 \langle X, \nabla_X X \rangle \right ) = -\langle X, X \rangle^{-3/2} \langle X, \nabla_X X \rangle; \tag 5$

$\nabla_X \left( \langle X, X \rangle^{-1/2}X \right) = -\langle X, X \rangle^{-3/2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1/2} \nabla_X X; \tag 6$

$\nabla_{X/\vert X \vert} \left(\dfrac{X}{\vert X \vert} \right) = \langle X, X \rangle^{-1/2}\nabla_X \left( \langle X, X \rangle^{-1/2}X \right)$ $= -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X; \tag 7$

we confirm that this vector is normal to $X$:

$\left \langle \nabla_{X/\vert X \vert} \left(\dfrac{X}{\vert X \vert} \right), X \right \rangle = \langle -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X, X \rangle$ $= \langle -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X, X \rangle + \langle \langle X, X \rangle^{-1} \nabla_X X, X \rangle$ $= -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle \langle X, X \rangle + \langle X, X \rangle^{-1} \langle \nabla_X X, X \rangle$ $= -\langle X, X \rangle^{-1} \langle X, \nabla_X X \rangle + \langle X, X \rangle^{-1} \langle \nabla_X X, X \rangle = 0; \tag 8$

(7) is thus normal to $X/\vert X \vert$ as well; thus we have $N$, the unit normal to our integral curves, expressible as

$N = \dfrac{ -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X }{\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X \vert }; \tag 9$

now

$\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X \vert^2$ $= \langle -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X, -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X \rangle$ $+ 2 \langle -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X, \langle X, X \rangle^{-1} \nabla_X X \rangle + \langle \langle X, X \rangle^{-1} \nabla_X X, \langle X, X \rangle^{-1} \nabla_X X \rangle$ $= \langle X, X \rangle^{-4} \langle X, \nabla_X X \rangle^2 \langle X, X \rangle - 2 \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2 + \langle X, X \rangle^{-2} \langle \nabla_X X, \nabla_X X \rangle$ $= \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2 - 2 \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2 + \langle X, X \rangle^{-2} \langle \nabla_X X, \nabla_X X \rangle$ $= \langle X, X \rangle^{-2} \langle \nabla_X X, \nabla_X X \rangle - \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2; \tag{10}$

thus,

$\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X \vert$ $= \sqrt{ \langle X, X \rangle^{-2} \langle \nabla_X X, \nabla_X X \rangle - \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2}$ $= (\langle X, X \rangle^{-2} \langle \nabla_X X, \nabla_X X \rangle - \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2)^{1/2}; \tag{11}$

this quantity (11), being the magnitude of $\nabla_{X / \vert X \vert} (X / \vert X \vert)$, is in fact the curvature of the integral curves of $X$, so we write

$\kappa = (\langle X, X \rangle^{-2} \langle \nabla_X X, \nabla_X X \rangle - \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2)^{1/2}, \tag{12}$

which leads to

$\kappa^2 = \langle X, X \rangle^{-2} \langle \nabla_X X, \nabla_X X \rangle - \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2$ $= \langle X, X \rangle^{-3} \langle \nabla_X X, \nabla_X X \rangle\langle X, X \rangle - \langle X, X \rangle^{-3} \langle X, \nabla_X X \rangle^2$ $= \dfrac{ \langle \nabla_X X, \nabla_X X \rangle\langle X, X \rangle - \langle X, \nabla_X X \rangle^2}{\langle X, X \rangle^3} = \dfrac{\vert \nabla_X X \vert^2 \vert X \vert^2 - \langle X, \nabla_X X \rangle^2}{\vert X \vert^6}, \tag{13}$

and returning once again to

$\kappa = \dfrac{(\vert \nabla_X X \vert^2 \vert X \vert^2 - \langle X, \nabla_X X \rangle^2)^{1/2}}{\vert X \vert^3}, \tag{14}$

if $\alpha(t)$ is an integral curve of $X$, that is, if $\alpha(t)$ satisfies

$\dot \alpha(t) = X(\alpha(t)), \tag{15}$

then

$\ddot \alpha(t) = \nabla_X X, \tag{16}$

whence

$\kappa = \dfrac{\vert \ddot \alpha(t) \vert^2 \vert \dot \alpha(t) \vert^2 - \langle \dot \alpha(t), \ddot \alpha(t) \rangle^2}{\vert \dot \alpha(t) \vert^3}, \tag{17}$

which is the usual formula for $\kappa$ in terms of the derivatives of a (not necessarily arc-length) parametrized curve $\alpha(t)$. (14) expresses this quantity in terms of the vector field $X$, and hence applies to all the integral curves of this vector field. We also observe that (13) ensures

$\kappa^2 > 0, \tag{18}$

so that $\kappa$ is well-defined (i.e., real) from (14), (18); this follows from the Cauchy-Schwarz inequality applied to the numerator of (13), that is, to

$\vert \nabla_X X \vert^2 \vert X \vert^2 - \langle X, \nabla_X X \rangle^2, \tag{19}$

for indeed we have

$\langle X, \nabla_X X \rangle^2 \le \vert \nabla_X X \vert^2 \vert X \vert^2. \tag{20}$

We have now derived expressions for $T$, $N$ and $\kappa$ in terms of the vector field $X$; we pause to note that $T$ and $N$ are in this case vector fields themselves, and $\kappa$ a scalar field, on any open set $U$ where $X$ is defined; new fields from old.

Having built $\kappa$, $T$ and $N$ from $X$, we proceed to calculate

$B = T \times N; \tag{21}$

by means of (2) and (9) we write

$T \times N = \dfrac{X}{\vert X \vert} \times \dfrac{ -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X }{\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X \vert }$ $= \dfrac{X \times (-\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X)}{\vert X \vert\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X \vert}; \tag{22}$

we evalutate the numerator:

$X \times (-\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X)$ $= -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X \times X + \langle X, X \rangle^{-1} X \times \nabla_X X = \langle X, X \rangle^{-1} X \times \nabla_X X, \tag{23}$

since $X \times X = 0$; thus (22) becomes

$T \times N = \dfrac{\langle X, X \rangle^{-1} X \times \nabla_X X}{\vert X \vert\vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X \vert}; \tag{24}$

we multiply the numerator and denominator through by $\langle X, X \rangle > 0$ and obtain

$B = T \times N = \dfrac{ X \times \nabla_X X}{\vert X \vert \langle X, X \rangle \vert -\langle X, X \rangle^{-2} \langle X, \nabla_X X \rangle X + \langle X, X \rangle^{-1} \nabla_X X \vert}$ $= \dfrac{ X \times \nabla_X X}{\vert X \vert \vert - \langle X, X \rangle^{-1}\langle X, \nabla_X X \rangle X + \nabla_X X \vert} = \dfrac{ X \times \nabla_X X}{\vert X \vert \vert \nabla_X X - \langle X, X \rangle^{-1}\langle X, \nabla_X X \rangle X \vert}; \tag{25}$

Nota Bene: Whoops! Looks like I hit that old "Post" button a tad prematurely! So stay tuned, there is a bit more to come! End of Note.