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PGA: Plane-Based Geometric Algebra $\mathbb{ R}^{d,0,1}$

Leo Dorst

l.dorst@uva.nl

Steven De Keninck

enkimute@gmail.com

Contents (ADAPT)

Plane-based: it's all done with mirrors
The algebra of multiple reflections
Specification by invariants
Objects and operators unified

PGA: Why, What's New?

Bullet One
Bullet Two
Bullet Three

It's All Done with Mirrors

Bullet One
Bullet Two
Bullet Three

Plane Reflection Algebra

Define a sensible geometric product such that reflection becomes:

x ~ \mapsto~ -a\,x\,a^{-1}

The mirror plane reflects to its negative:

\[a \mapsto -a\,a\,a^{-1} = -a\]

So planes are oriented, they have sides.

To make the geometric product work in the right way, it needs to be bilinear, associative, not generally commutative, and square to a scalar (which is $a^2 = a \cdot a$ ).

A Plane is a PGA Vector

You may have used the homogeneous coordinates, $\[vec{n},-\delta]$ for a plane.

In PGA we use:

$$p = \mathbf{n} + \delta\,{\sf e}.$$

Here ${\sf e}$ is a basis vector for the extra dimension.

It can be interpreted as the plane at infinity and satisfies \[{\sf e}^2 = 0\]

Commutation and Perpendicularity

A plane $b$ perpendicular to a mirror $a$ reflects to itself: $ -a \,b \, a^{-1} = b$. So we have:

algebra $\rightarrow$ $a \,b = - b \, a \Longleftrightarrow b \, \bot a$ $\leftarrow$ geometry

perpendicular planes anticommute

This holds especially for the Euclidean coordinate planes:

$e_1\,e_2 = -e_2 \, e_1$ etc.

We also demand it for the special plane $\mathsf{e}$: $e_i \, \mathsf{e} = -\mathsf{e}\, e_i$.

Of course, parallel planes commute: $e_i \, e_i = e_i \, e_i $.

Multiple Reflections Give Operator Elements

Perform two reflections in succession:

$x ~~~\mapsto~~-a_2 \, (-a_1\, x \,a_1^{-1})\, a_2^{-1}= (a_2\,a_1)\, x\, (a_2\,a_1)^{-1}$

This shows that elements of the form $(a_2\,a_1)$ act as rotation/translation operators. We can clearly continue this with more reflections.

A geometric product of planes produces a motion, when applied as sandwiching. We call it a motor.

Motors are computational elements of the plane-based geometric algebra PGA: Euclidean motion operators are thus naturally included!

Parallel Reflections: Translation Operators

Double reflection in two parallel planes:

$p_2\,p_1 =$ $(\mathbf{n} + \delta_2 \mathsf{e}) \, (\mathbf{n} + \delta_1 \mathsf{e}) $

$= \mathbf{n}\,\mathbf{n} + \delta_2\, \mathsf{e} \,\mathbf{n} + \delta_1\,\mathbf{n}\,\mathsf{e} + \delta_2\delta_1 \,\mathsf{e}\,\mathbf{n}\,\mathsf{e}\,\mathbf{n} $

$= 1 + (\delta_2-\delta_1) \mathbf{n} \, \mathsf{e} + 0$

$= 1 + \mathbf{t} \mathsf{e}/2$

$= 1- \mathsf{e} \,\mathbf{t}/2 + \frac{1}{2!}(-\mathsf{e} \,\mathbf{t}/2)^2+\cdots$

$= e^{-\mathsf{e} \,\mathbf{t}/2}$

Translations are additive:

$(1+\mathbf{t}_2\mathsf{e}/2)\,(1+\mathbf{t}_1\mathsf{e}/2)$ $ = 1 + (\mathbf{t}_2+\mathbf{t}_1)\mathsf{e}/2 + \mathbf{t}_2\mathsf{e}\,\mathbf{t}_1\mathsf{e}/4$ $= 1 +(\mathbf{t}_2+\mathbf{t}_1)\,\mathsf{e}/2$

You see how ${\sf e}^2 = 0$ is crucial to making this work.

Tilted Plane Reflections: Rotation Operators

Let us choose handy local coordinates so that the two planes are $p_1= e_1$, and $p_2 = e_1 \cos\phi' + e_2\sin\phi'$. Then

$p_2\, p_1 =$ $(e_1 \cos\phi' + e_2\sin\phi')\,e_1$

$= e_1^2 \cos\phi' + e_2e_1\,\sin\phi'$

$= \cos\phi' - L \sin\phi' $

$= e^{-L\phi/2}$

with $e_1e_2 =$$ L$ the common line and $\phi$ the rotation angle.

Note that a line has a negative square:

$L^2 = e_1e_2e_1e_2 = - e_1^2e_2^2 = -1$,

motivating the exponential notation.

The Line Common to Two Planes Is a Bivector

The line common to two planes $a$ and $b$ is the element

$L = a \wedge b = \frac{1}{2} (a \, b - b\, a)$.

It is not normalized, but the norm $|L| = \sin(\angle(a,b))$ is a useful numerical quantity.

Element $a \wedge b$ is called a bivector or 2-vector.

The wedge product $\wedge$ (also known as the meet or outer product) is bilinear, anti-symmetric, and extended to be associative. Note that $a \wedge a = 0$.

(Associativity by defining $x \wedge A = \frac{1}{2}(x \, A + \hat{A}\,x)$, with $\hat{A}$ giving minus sign for odd $A$ ).

Bonus Material:

The Line Is Indeed Invariant Under Rotation

The element $a \wedge b$ commutes with the rotation operator $a \, b$.

$ (a \wedge b)\, (a\,b) = \frac{1}{2} (a \, b \, a \, b - b\,a\,a\,b) $

$= \frac{1}{2} (a \, b \, a \, b - 1) $

$= \frac{1}{2} (a \, b \, a \, b - a\,b\, b\,a)$

$= (a \, b) \, (a \wedge b) $

Therefore, the element $L = a \wedge b$ is invariant under the rotation $R_{L\phi} =a \, b$, for

$ R_{L\phi}\, L \,R^{-1}_{L\phi} = L\, R_{L\phi}\,R^{-1}_{L\phi} = L$.

So it must be the common line. Circular reasoning is OK for rotations!

A Point: a Trivector Common to Three Planes

Standard example:

$ X $ $= (e_1+x_1\mathsf{e}) \wedge (e_2+x_2\mathsf{e}) \wedge (e_3+x_3\mathsf{e})$

$ = e_1e_2e_3 + x_1 \mathsf{e}e_2e_3 + x_2 \mathsf{e} e_3e_1 + x_3 \mathsf{e} e_1e_2$

$ = e_1e_2e_3 - (x_1e_1 + x_2 e_2 + x_3e_3)\,\mathsf{e}e_1e_2e_3$

$= O - \mathbf{x} \, \mathcal{I}$

Just like with homogeneous coordinates, a point $X$ is characterized as $[1, x_1, x_2, x_3]$, but now on the PGA trivector basis $[e_1e_2e_3, -ee_2e_3, -ee_3e_1, -ee_1e_2]$.

The trivector $a \wedge b \wedge c$ is the point

common to three planes $a, b, c$.

Operators and Objects

Only Axis and Angle Matter

REFER TO GAUGE FREEDOM, REPLAY DEMO?

DESIRE TO REPRESENT THAT WAY

TWO ISSUES: LINE REPR, AND EXP

Automatic Equivariance

SHOW PRINCIPLE

CONSEQUENCES:

THEORY - SIMPLE DERIVATIONS

PRACTICE - UNIVERSAL CODE IN TWO WAY: NOT JUST ANYWHERE, BUT ALSO ANYTHING