{GAME23}

GAME23 Sneak Peek

Martin Roelfs, PhD

Lattice Quantum Chromodynamics @ KULAK

{Part 1}

An Algebra of Geometry

{Computer Geometry/Physics}

An Algebra for Geometry

Martin Roelfs

// construct lines
line_from_points (p1, p2)
line_from_points_and_dir (p, d)
line_from_plucker (a, b, c, d, e, f)

// construct planes
plane_from_points (p1, p2, p3)
plane_from_point_and_dirs (p, d1, d2)
plane_from_points_and_dir (p1, p2, d)
plane_from_point_and_line (p, l)
plane_from_equation (a, b, c, d)

// Intersections
intersect_line_plane (l, P)
intersect_plane_plane (P1, P2)
intersect_planes (P1, P2, P3)

// Projections
project_point_plane (p, P)
project_line_plane (l, P)
project_point_line (p, l)

// construct transformations
mtx_translate (x, y, z)
mtx_rotate_euler (h, p, b)
mtx_rotate_axis_angle (x, y, z, a)
mtx_look_at (from, too, pole)

// apply transformations
transform_point (M, p)
transform_direction (M, d)
transform_line (M, l)
transform_plane (M, P)

Current state of affairs: linear algebra representation of geometry.

Use multiple bags of real numbers and do all bookkeeping ourselves.
- No inherent geometry in matrices.
Results in a different function for each distinct operation. (Just a subset shown here.)
Tied to a particular number of dimensions

{Computer Geometry}

An Algebra for Geometry

Martin Roelfs

Wouldn't it be better if we had a single bag of numbers, where each element in this bag is an element of geometry?

a \to bab^{-1}

a \to bab^{-1}

(a \cdot b) b^{-1}

(a \cdot b) b^{-1}

Transform $a$ by $b$ :

Project $a$ onto $b$ :

(Projective) Geometric Algebra

Teaser: let $a$ and $b$ be any elements of geometry (points, lines, planes, translation, rotations, etc)

- Mathematical operations such as multiplication and addition are always geometrically meaningful

- Not tied to a particular number of dimensions

The idea of Geometric Algebra in just 30 mins.

Start from a simple example and Vector Algebra solution
Inspect that method and highlight some issues
Introduce the Geometric Algebra approach
Inspect that method and explore some of the consequences

A simple example ..

$d_i = \big ( \frac 1 f - \frac 1 {d_0} \big )^{-1}\qquad h_i=-\frac {d_i} {d_0} h_0$

Martin Roelfs

An Algebra for Geometry

{GA}

The idea of Geometric Algebra in just 30 mins.

Start from a simple example and Vector Algebra solution
Inspect that method and highlight some issues
Introduce the Geometric Algebra approach
Inspect that method and explore some of the consequences

$d_i = \big ( \frac 1 f - \frac 1 {d_0} \big )^{-1}\qquad h_i=-\frac {d_i} {d_0} h_0$

The paraxial approximation of the thin lens equation

Martin Roelfs

An Algebra for Geometry

{GA}

$d_i = \big ( \frac 1 f - \frac 1 {d_0} \big )^{-1}\qquad h_i=-\frac {d_i} {d_0} h_0$

The paraxial approximation of the thin lens equation

$\vec a = \overline {\vec c - \vec f}$

$\vec b= \vec c - \vec p$

$d_0 = \vec a \cdot \vec b$

Martin Roelfs

An Algebra for Geometry

{GA}

$d_i = \big ( \frac 1 f - \frac 1 {d_0} \big )^{-1}\qquad h_i=-\frac {d_i} {d_0} h_0$

The paraxial approximation of the thin lens equation

$(\overline {\vec c - \vec f}) \, \cdot$

$(\vec c - \vec p)$

$d_0 =$

$h_0 = (\overline{\vec c - \vec g} )\,\cdot\,(\vec c - \vec p)$

Martin Roelfs

An Algebra for Geometry

{GA}

$p \vee f$ = line through p and f
$(p \vee f) \wedge L$ = intersection with lens
$(p \vee f) \wedge L \cdot L$ = line orthogonal to L
$p \vee c$ = line through p and c
$(p \vee f) \wedge L \cdot L \wedge (p \vee c)$ = intersect with line through center

$\vee$ = join $\wedge$ = meet $\bullet$ = dot

Martin Roelfs

An Algebra for Geometry

{GA}

$d_i = \big ( \frac 1 f - \frac 1 {d_0} \big )^{-1}\qquad h_i=-\frac {d_i} {d_0} h_0$

$h_0 = (\overline{\vec c - \vec g} )\,\cdot\,(\vec c - \vec p)$

$d_0 = (\overline{\vec c - \vec f} )\,\cdot\,(\vec c - \vec p)$

$i = -d_i\overline{(\vec c - \vec f )} - h_i \overline{(\vec c - \vec g)} + \vec c$

no need for :

coordinates
origin
dimensionality
arrows
'solving'

Martin Roelfs

An Algebra for Geometry

{GA}

{Part 2}

Implementing the Algebra of Geometry

All well and good, but how do I actually compute something?

Key observation: performing the same reflection twice is the same as doing nothing.

$vvXv^{-1}v^{-1} = v^2 X v^{-2}$

I.e., $v^2 = \pm 1$ .

Martin Roelfs

An Algebra for Geometry

{GA}

In a Clifford (Geometric) Algebra, vectors square to scalars. I.e., for $v \in \mathbb{R}_{p,q,r}, v^2 \in \mathbb{R}$
It is well-known that the products of vectors in a Clifford Algebra are elements of $\text{Pin}(p,q,r)$ .

Therefore our Algebra of Geometry can be represented using normalized vectors in a Clifford Algebra!

What on earth is the Clifford Algebra $\mathbb{R}_{p,q,r}$ ?

Martin Roelfs

An Algebra for Geometry

{GA}

i^2 = -1

i^2 = -1

j^2 = 1

j^2 = 1

\epsilon^2 = 0

\epsilon^2 = 0

The Clifford Algebra $\mathbb{R}_{p,q,r}$ has $p$ positive, $q$ negative, and $r$ null basis vectors.

x^2 + 1 = 0 \implies x^2 = -1

x^2 + 1 = 0 \implies x^2 = -1

Consider the 2D example, $\mathbb{R}_{2,0,0}$ . This has basis vectors $e_1, e_2$ , satisfying $e_1^2 = e_2^2 = 1$ . Any vector is now of the form:

$u = a e_1 + b e_2.$

Martin Roelfs

An Algebra for Geometry

{GA}

What is the square of $u$ ?

\begin{aligned} u^2 &= a^2 e_1^2 + b^2 e_2^2 + ab e_1 e_2 + ab e_2 e_1 \\ &= a^2 + b^2 + ab (e_1 e_2 + e_2 e_1) \end{aligned}

\begin{aligned} u^2 &= a^2 e_1^2 + b^2 e_2^2 + ab e_1 e_2 + ab e_2 e_1 \\ &= a^2 + b^2 + ab (e_1 e_2 + e_2 e_1) \end{aligned}

But there is nothing special about $e_1$ and $e_2$ ! In fact, you might have chosen what I called $u$ as one of your basis vectors. Therefore we must have

\begin{aligned} e_1 e_2 + e_2 e_1 = 0 \quad \implies \quad e_1 e_2 = - e_2 e_1 \end{aligned}

\begin{aligned} e_1 e_2 + e_2 e_1 = 0 \quad \implies \quad e_1 e_2 = - e_2 e_1 \end{aligned}

such that $u^2 = a^2 + b^2$ . So orthogonal vectors anti-commute!

So we can now form lines through the origin $e_{12} := e_1 e_2$ as $u = a e_1 + b e_2$ . What is the square of the point $e_{12}$ ?

Martin Roelfs

An Algebra for Geometry

{GA}

\begin{aligned} u v &= (a e_1 + be_2) (c e_1 + d e_2) \\ &= (ac e_1^2 + bd e_2^2) + (a d e_1 e_2 + b c e_2 e_1) \\ &= (ac + bd) + (a d - b c) e_{12} \end{aligned}

\begin{aligned} u v &= (a e_1 + be_2) (c e_1 + d e_2) \\ &= (ac e_1^2 + bd e_2^2) + (a d e_1 e_2 + b c e_2 e_1) \\ &= (ac + bd) + (a d - b c) e_{12} \end{aligned}

We recognize the scalar part as the dot product $u \cdot v = |u||v| \cos\theta$ of the two vectors, whereas the second part is the exterior product $u \wedge v = |u||v| e_{12} \sin\theta$ .

How about the product of two arbitrary lines through the origin?

\begin{aligned} e_{12}^2 = e_1 e_2 e_1 e_2 = - e_1 e_1 e_2 e_2 = - e_1^2 e_2^2 = -1 \end{aligned}

\begin{aligned} e_{12}^2 = e_1 e_2 e_1 e_2 = - e_1 e_1 e_2 e_2 = - e_1^2 e_2^2 = -1 \end{aligned}

It behaves like the imaginary unit $i$ !

\begin{aligned} u v &= |u||v| e^{\theta e_{12}} \end{aligned}

\begin{aligned} u v &= |u||v| e^{\theta e_{12}} \end{aligned}

The product of two lines yields Euler's formula!

Martin Roelfs

An Algebra for Geometry

{GA}

The line at infinity, $e_0$ , is orthogonal to $e_1$ and $e_2$ . Besides finding a new line, we also find two new points: $e_{01}, e_{02}$ . Because these lie at infinity, they are called directions.

Recall that lines are specified by the linear equation

$ax + by + \delta = 0$

We can thus represent arbitrary lines in 2D as

$u = ae_1 + be_2 + \delta e_0,$

where $\delta$ is the distance from the origin. Similarly, arbitrary directions (points on the horizon)

$p_\infty = y e_{01} + x e_{20}$

whereas arbitrary points are given by

$p = e_{12} + p_\infty = e_{12} + y e_{01} + x e_{20}$

What is the square of $e_0$ ? Well, you cannot reflect in the horizon and live to tell about it!

e_0^2 = 0

e_0^2 = 0

\begin{pmatrix}x \\ y \\ 0\end{pmatrix}

\begin{pmatrix}x \\ y \\ 0\end{pmatrix}

\begin{pmatrix}x \\ y \\ 1\end{pmatrix}

\begin{pmatrix}x \\ y \\ 1\end{pmatrix}

\mathbb{R}_{2,0,1}

\mathbb{R}_{2,0,1}

Martin Roelfs

An Algebra for Geometry

{GA}

Question: how would we now rotate around a given point $(x, y)$ by an angle $\theta$ ?

p = e_{12} + y e_{01} + x e_{20}

p = e_{12} + y e_{01} + x e_{20}

Since $e_0^2 = 0$ , all points are created equal:

$p^2 = e_{12}^2 = -1$

Thus all points generate their own local Euler's formula under exponentiation:

$R = e^{\theta p} = \cos\theta + p \sin\theta$

Directions are even easier, because they square to $p_\infty^2 = 0$ :

$R = e^{a p_\infty} = 1 + a p_\infty$

Martin Roelfs

An Algebra for Geometry

{GA}

Towards nD

In 2D we reflect in a line, in 3D in a plane, in nD in a hyperplane. In (pseudo-)Euclidean space $E(p,q)$ , hyperplanes satisfy the linear equation

$ax + by + cz + ... + \delta = 0$

Therefore, hyperplanes can be represented by vectors in $\mathbb{R}_{p,q,1}$ , with basis vectors $\mathbf e_0^2= 0$ and

$\mathbf e_1^2= \ldots = \mathbf e_p^2 = -\mathbf e_{p+1}^2= \ldots = -\mathbf e_{p+q}^2 = 1,$

$v = a \mathbf e_1 + b\mathbf e_2 + c\mathbf e_3 + ... + \delta \mathbf e_0$

To reflect a hyperplane $w$ in a hyperplane $v$ , use conjugation:

$w \to v[w] = v w v^{-1}$

Martin Roelfs

An Algebra for Geometry

{GA}

Towards nD

Noteworthy spaces

Describes	Group	Name
2D Plane	E(2)	"Planar quaternions"
3D Plane	E(3)	Dual quaternions
Spacetime*	SO(1,3)	Lorentz transformations
Spacetime	Poin(3,1)	Poincare transformations

\mathbb{R}_{2,0,1}

\mathbb{R}_{2,0,1}

\mathbb{R}_{3,0,1}

\mathbb{R}_{3,0,1}

\mathbb{R}_{1,3,1}

\mathbb{R}_{1,3,1}

\mathbb{R}_{1,3,0}

\mathbb{R}_{1,3,0}

Question: what is the difference between $\mathbb{R}_{1,3,0}$ and $\mathbb{R}_{1,3,1}$ ?

not just a new way to shuffle coordinates.
not just the quotient of the tensor algebra with some ideal.
not just the extension of vector algebra to multivectors.

scalar

vector

bivector

Martin Roelfs

An Algebra for Geometry

{GA}

Duality

Vectors are always $d-1$ dimensional subspaces (hyperplanes). What if we want to make a true point, which remains a point if we change the dimension?

The dual of a $d-k$ dimensional quantity is always a $k$ dimensional quantity. E.g. in 2D, the dual of a line is a point. But in 3D, the dual of a plane, is also a point.

not just a new way to shuffle coordinates.
not just the quotient of the tensor algebra with some ideal.
not just the extension of vector algebra to multivectors.

scalar

vector

bivector

Martin Roelfs

An Algebra for Geometry

{GA}

In 2D we reflect in a line, in 3D in a plane, in nD in a hyperplane. In (pseudo-)Euclidean space $E(p,q)$ , hyperplanes satisfy the linear equation

$ax + by + cz + ... + \delta = 0$

Therefore, hyperplanes can be represented by vectors in $\mathbb{R}_{p,q,1}$ , with basis vectors $\mathbf e_0^2= 0$ and

$\mathbf e_1^2= \ldots = \mathbf e_p^2 = -\mathbf e_{p+1}^2= \ldots = -\mathbf e_{p+q}^2 = 1,$

$v = a \mathbf e_1 + b\mathbf e_2 + c\mathbf e_3 + ... + \delta \mathbf e_0$

To reflect a hyperplane $w$ in a hyperplane $v$ , use conjugation:

$w \to v[w] = - v w v^{-1}$

Reflections are involutory, and thus demanding

$v[v[w]] = w$

leads to $v^2 = v^{-2} = \pm 1$ .

Importantly, $v[v] = -v$ .

not just a new way to shuffle coordinates.
not just the quotient of the tensor algebra with some ideal.
not just the extension of vector algebra to multivectors.

scalar

vector

bivector

Martin Roelfs

An Algebra for Geometry

{GA}

u_1 \wedge u_2 \wedge \cdots \wedge u_k

u_1 \wedge u_2 \wedge \cdots \wedge u_k

In $d$ dimensions, we start from reflections in $d-1$ dimensional hyperplanes. $k$ of these can then be intersected to form $d - k$ dimensional elements of geometry.

not just a new way to shuffle coordinates.
not just the quotient of the tensor algebra with some ideal.
not just the extension of vector algebra to multivectors.

scalar

vector

bivector

Martin Roelfs

An Algebra for Geometry

{GA}

The composition of two reflections leads to a handedness preserving isometry.

We call such transformations

Bireflections

\begin{aligned} u[v[A]] &= uvA v^{-1}u^{-1} \\ &= (uv) A (uv)^{-1} \end{aligned}

\begin{aligned} u[v[A]] &= uvA v^{-1}u^{-1} \\ &= (uv) A (uv)^{-1} \end{aligned}

Bireflections $R = uv$ preserve handedness, and are continuous transformations.

Why two-sided transformations? If I ask you to write upside down, what is the easiest way to do it?

Comparison to matrices

not just a new way to shuffle coordinates.
not just the quotient of the tensor algebra with some ideal.
not just the extension of vector algebra to multivectors.

scalar

vector

bivector

Martin Roelfs

An Algebra for Geometry

{GA}

\begin{pmatrix} \vec{p}_\infty' \\ p_o' \end{pmatrix} = \begin{pmatrix} R & t \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \vec{p}_\infty \\ p_o \end{pmatrix}, % \mqty[\textcolor{Fuchsia}{\mathbf R} & \textcolor{Fuchsia}{\mathbf{t}} \\ \textcolor{Fuchsia}{\mathbf{0}} & \textcolor{Fuchsia}{1}] \mqty[\cev{t}_\infty \\ \cev{t}_o].

\begin{pmatrix} \vec{p}_\infty' \\ p_o' \end{pmatrix} = \begin{pmatrix} R & t \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \vec{p}_\infty \\ p_o \end{pmatrix}, % \mqty[\textcolor{Fuchsia}{\mathbf R} & \textcolor{Fuchsia}{\mathbf{t}} \\ \textcolor{Fuchsia}{\mathbf{0}} & \textcolor{Fuchsia}{1}] \mqty[\cev{t}_\infty \\ \cev{t}_o].

In order to transform a point by a rotation and a translation in linear algebra, e.g. open3d.geometry.PointCloud.transform, we use the covariant transformation

where $\vec{p}_\infty$ is a 3-vector indicating a point on the horizon, $p_o$ is the origin, $R$ is a 3x3 rotation matrix, and $t$ is a 3-vector encoding the translation.

This 4x4 matrix uses 16 floats, and can only transform points. For planes we have to use the 4x4 contravariant rep., and for lines the 6x6 adjoint rep. By contrast, the PGA equivalent rotor

$R = a + b\mathbf{e}_{23} + c\mathbf{e}_{31} + d\mathbf{e}_{12} + e\mathbf{e}_{01} + f\mathbf{e}_{02} + g\mathbf{e}_{03} + h\mathbf{e}_{0123}$

transforms points, lines and planes identically using only 8 floats.

1 . 1

{GAME23} GAME23 Sneak Peek Martin Roelfs, PhD Lattice Quantum Chromodynamics @ KULAK

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Mathematical Experimentalist

enkimute.github.io

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