Geometric algebra sweeping surface/change of basis

Question

Someone has asked a similar question before but the answer was not that useful.

I want to make a sweeping surface with PGA, i.e. to transport a curve along a parametric curve.

With linear algebra one computes the tangent, Normal and Binormal of a point in the path curve, then transforms the swept curve from 2D into 3D by applying the induced transformation (i.e. multiplying the 4x4 homogeneous matrix made by the 3 axes and the position vector).

Or alternatively by multiplying the x and y coordinates by the normal and binormal and adding the point.

I want to do the same with PGA, i.e. I want to use the frenet trihedron to move my curve from 2D to 3D.

As requested in the comments. This is how we would achieve this with lienar algebra.

Given 2 parametric curves $P(t), C(t)$ the first defined in 3D the second in 2D/ the xy plane embedded in 3D and the frenet trihedron operator $F(f, x)->(T,N,B)_x$ We can create a parametric surface as:

compute the frenet trihedron along the first curve $(T_u, N_u, B_u) = F(P, u)$

Then a level curve of the parametric surface is, if $(x, y) = C(v)$ then $S(u_t, v) = xN_{u_t} + y B_{u_t} + P(u_t)$

Or alternatively:

$S(u_t, v) = M \cdot [x, y, 0, 1]$

Where $M$ is a $4\times4$ matrix where the first column is $[T,0]$ the second $[N,0]$ the third $[B,0]$ and the last is $[P(u_t), 1]$.

Answer 1

$ \newcommand\plane\mathbf \newcommand\lcontr{\mathbin\rfloor} \newcommand\rcontr{\mathbin\lfloor} $Let our basis of orthonormal planes be $\plane e_1, \plane e_2, \plane e_3, \infty$ with $\infty$ the plane at infinity. A free vector $V$, representing a pure displacement, is an element of the form $$ V = (v_1\plane e_2\plane e_3 + v_2\plane e_3\plane e_1 + v_3\plane e_1\plane e_2)\infty. $$ A choice of origin $\oslash = \plane e_1\plane e_2\plane e_3$ recovers the unique plane passing through $\oslash$ orthogonal to $V$ $$ \plane V = \oslash^{-1}\rcontr\oslash\vee V = v_1\plane e_1 + v_2\plane e_2 + v_3\plane e_3. $$ We take $T, N, B$ to be such free vectors with orthogonal planes $\plane T, \plane N, \plane B$. We take path functions $P, C$ as returning points (i.e. trivectors); if $$ P_\infty(u) = \bigl(p_x(u)\plane e_2\plane e_3 + p_y(u)\plane e_3\plane e_1 + p_z(u)\plane e_1\plane e_2\bigr)\infty $$ is the free position vector from $\oslash$ to $P(u)$ then $P(u) = \oslash + P_\infty(u)$.

The transformation represented by $M$ performs a rotation taking $\plane e_1\infty$ to $N$ and $\plane e_2\infty$ to $B$. We can write such a rotor $R$ as $$ R = \frac{(\plane N - \plane e_1)(\plane B - \plane e_2')}{|\plane N - \plane e_1||\plane B - \plane e_2'|},\quad \plane e_2' = \frac{-(\plane N - \plane e_1)\plane e_2(\plane N - \plane e_1)}{(\plane N - \plane e_1)^2} = \plane e_2 - 2\frac{\plane N\cdot\plane e_2}{(\plane N - \plane e_1)^2}(\plane N - \plane e_1). $$ Note that this particular expression assumes $N$ and $B$ being unit vectors. This is derived by reflecting $\plane e_1$ to $\plane N$, applying this to $\plane e_2$ to get $\plane e_2'$, and then reflecting $\plane e_2'$ to $\plane B$. As constructed, $R$ is a rotation about the origin $\oslash$.

$M$ then performs a translation taking $\oslash$ to $P(u)$; the corresponding displacement is $P_\infty(u)$ so the corresponding translator $D$ is $$\begin{aligned} D &= \exp\left[\frac12\oslash^{-1}\times P_\infty(u)\right] = 1 + \frac12\oslash^{-1}\times P_u(\infty) \\ &= 1 + \frac12\plane P_\infty(u) = 1 + \frac12\bigl(p_x(u)\plane e_1 + p_y(u)\plane e_2 + p_z(u)\plane e_3\bigr)\infty. \end{aligned}$$ $\plane P_\infty(u)$ can be interpreted as the unique free plane orthogonal to $P_\infty(u)$.

All together, our motor representation of $M$ is $$ M(u) = RD = \frac{(\plane N - \plane e_1)(\plane B - \plane e_2')(2 + \plane P_\infty(u))}{2|\plane N - \plane e_1||\plane B - \plane e_2'|},\quad \plane e_2' = \frac{-(\plane N - \plane e_1)\plane e_2(\plane N - \plane e_1)}{(\plane N - \plane e_1)^2}. $$ It is applied to an object $X$ via $$ X \mapsto \widetilde MXM $$ where $\widetilde M$ is the reverse of $M$. Thus if $S(u, v)$ and $C(v) = \oslash + (x(v)\plane e_2\plane e_3 + y(v)\plane e_3\plane e_1)\infty$ are point-valued then $$ S(u, v) = \widetilde M(u)C(v)M(u). $$