Projective GA

VGA gives us the tools to represent vectors, planes, etc. that intersect the origin, and rotations/reflections around the origin. PGA (Projective Geometric Algebra) adds a new basis vector \(w\), and uses it to represent points, lines, and planes anywhere in Euclidean space¹, and any isometry of Euclidean space.

What is an isometry?

An isometry is a distance-preserving transformation. (The word "isometry" literally means "same measure.") In Euclidean space, this is some combination of translations, rotations, and reflections.

Is \(w\) the fourth dimension ?

No, we're not talking about 4D space yet. \(w\) is not an ordinary spatial dimension, it's just an algebraic tool. Everything discussed here can be generalized to any number of dimensions.

Projective plane

In projective GA, we keep \(x^2 = y^2 = z^2 = 1\). But our new basis vector \(w\) is a null vector, which means it squares to zero: \(w^2 = 0\). Let's stick to 2D for now, so that we only have three basis vectors: \(x\), \(y\), and \(w\).

Points, lines, and planes

PGA gives us the ability to represent a point as a vector. A point at coordinates \(\langle a, b \rangle\) is represented with the vector \(ax + by + w\). The origin \(\langle 0, 0 \rangle\) is represented with the vector \(w\). These vectors can be scaled arbitrarily and still represent the same point. For example, these blades all represent \(\langle 2, -1 \rangle\):

• \(2x-y+w\)
• \(-2x+y-w\)
• \(10x-5y+5w\)

This gives the geometry of the projective plane (or in general, a projective space), where we have all the finite points \(\langle x, y \rangle\) plus a line at infinity that can be reached by traveling infinitely far in any direction. (In 3D+, I like to think of it as half of a skybox.) Points on the line at infinity are represented using vectors with zero \(w\) component, such as \(-3x+2y\).

Projective geometry

The projective plane has some nice properties:

• Every pair of distinct points has exactly one line passing through both.
• Every pair of lines intersects at exactly one point, even if they are parallel.

Here's some examples of how this works:

• Given two points \(p\) and \(q\) represented using vectors, \(p \wedge q\) is the line containing \(p\) and \(q\). Iff² \(p\) and \(q\) are the same point, then \(p \wedge q = 0\).
• Given a line \(l\) represented using a bivector and a point \(p\) represented using a vector, \(l \wedge p\) is the plane containing \(l\) and \(p\). Iff \(p\) is already on \(l\), then \(l \wedge p = 0\).

In general, any blade \(C_r\) intersects the projective plane \(w=1\) at an \((r-1)\)-dimensional subspace \(S\), so we say that \(C_r\) represents \(S\). If \(C_r\) has no \(w\) component, we say it represents a subspace at infinity. (In 2D, that's the line at infinity or a point at infinity.)

The sign of a blade indicates the orientation of the subspace. So \(-C_r\) represents the same \(r\)-dimensional subspace as \(C_r\), but with the opposite orientation. Orientation will be very important later on.

Outer product

The outer product works here too: The outer product \(C_r \wedge D_s\) gives the representation of the unique \((r+s-1)\)-dimensional subspace containing \(C_r\) and \(D_s\). If \(C_r\) and \(D_s\) are in the same \((r+s-2)\)-dimensional subspace, then there isn't a unique \((r+s-1)\)-dimensional subspace containing them, so \(C_r \wedge D_s = 0\).

Which side?

We can use this to check if a point \(p\) is contained in a subspace \(C_r\): just check whether \(C_r \wedge p\) is zero. In \(N\)-dimensional space, we can check which side of a hyperplane \(H\) (represented as an \(N\)-blade) contains \(p\) by computing \(C_r \wedge p\), which gives a single pseudoscalar component whose sign is positive if \(p\) is on one side of \(C_r\) and negative if it's on the other side. To get a scalar, we take the dual: \((C_r \wedge p) \rfloor I\). Note that negating either \(p\) or \(C_r\) negates the sign of the result.

We can generalize this to work in subspaces as well: If \(C_r\) is a subspace of \(D_{r+1}\), then the sign of the scalar \((C_r \wedge p) \rfloor D_{r+1}\) indicates which side of \(C_r\) contains \(p\). Note that negating either \(p\), \(C_r\), or \(D_{r+1}\) negates the sign of the result.

Exercise

Think about smooth transformations of a point \(p\), a line \(C_1\), and a plane \(D_2\) such that \(p\) and \(C_1\) stay in \(D_2\) but \(p\) ends up on the other side of \(C_1\). By the intermediate value theorem, the sign of \((C_r \wedge p) \rfloor D_s\) can only change when \(p\) touches \(C_1\), so the sign is the same. But \(p\) is on the same side of \(C_1\). How did this happen?

Answer

In order to get \(p\) to the other side of \(C_r\), \(D_s\) got flipped over and changed sign.

This pattern of replacing the pseudoscalar with another blade to simulate lower-dimensional environments will show up a lot.

Motors and flectors

Remember rotors, our friends from the even subalgebra of VGA? They've gotten an upgrade in PGA: they're now called motors and can represent arbitrary translations in addition to rotations! And of course we have flectors, which use the odd subalgebra and represent any combination of translations, rotations, and an odd number of reflections. You can use them exactly the same way you use rotors.

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1. Actually, anywhere in projective space, which is a superset of Euclidean space! ↩

2. "iff" is short for if and only if. ↩