Quick Intro to Geometric Algebra

Here are some concepts to bear in mind. This is a very brief introduction to geometric algebra; a longer one is here.

Bases and geometric products

  • Every dimension of space comes with a basis vector: an arrow of length 1 unit pointed towards the positive end of the axis.

    • In our 3-dimensional world, there are the basis vectors \(\hat x\), \(\hat y\), and \(\hat z\), which are pointed towards the positive ends of the \(x\), \(y\), and \(z\) axes respectively.

    • The fourth dimensional basis vector is \(\hat w\). In higher dimensions, usually all bases are numbered instead of lettered: the fifth dimensional basis vectors are \(\hat e_1\), \(\hat e_2\), \(\hat e_3\), \(\hat e_4\), and \(\hat e_5\).

  • The geometric product of two basis vectors is their simple multiplication - not the dot or cross product! The geometric product of \(\hat x\) and \(\hat y\) is simply \(\hat x \hat y\).

    • The geometric product of two basis vectors is a basis plane. \(\hat x \hat y\) is the basis plane of the \(xy\) plane. The other basis planes are \(\hat y \hat z\) and \(\hat x \hat z\).

    • The geometric product of three basis vectors is a basis volume. \(\hat x \hat y \hat z\) is the basis volume of 3D space, which only has one basis volume, but also one of the four basis volumes of 4D space.

  • The geometric product of a basis vector with itself is 1. That is, \(\hat x \hat x = \hat x^2 = \hat y \hat y = \hat y^2 = \hat z \hat z = \hat z^2 = 1\).

  • The geometric product of different basis vectors anticommutes: \(\hat x \hat y = - \hat y \hat x\) and \(\hat x \hat y \hat z = - \hat x \hat z \hat y = \hat z \hat x \hat y = - \hat z \hat y \hat x\).

Blades and multivectors

  • A blade is a scaled basis: a scalar (regular real number) multiplied by a basis. For example, \(3 \hat x \hat y\) is a blade. Note that this means all bases are blades scaled by 1.

    • A \(k\)-blade is a blade of grade \(k\): the geometric product of a scalar and \(k\) different basis vectors. \(3 \hat x \hat y\) has grade \(2\); it is a \(2\)-blade.

    • Scalars are \(0\)-blades - blades consisting of no basis vectors.

  • A multivector is a sum of multiple blades. For example, \(1 + 2 \hat x - 3 \hat y \hat z\) is a multivector.

    • The sum of multiple (and only) 1-blades is usually called a simple vector. For example, \(3 \hat x + 2 \hat y\) is a vector.

    • The sum of multiple (and only) 2-blades is a bivector. Basis planes are also known as basis bivectors. For example, \(3 \hat x \hat y\) is a bivector.

  • The rules of linearity, associativity and distributivity in multiplication apply, as long as order of arguments is maintained:

    • \((\hat x)(a \hat y) = a \hat x \hat y\) (linearity, for scalar \(a\))

    • \((\hat x \hat y)(\hat z) = \hat x (\hat y \hat z)\) (associativity)

    • \(\hat x (\hat y + \hat z) = \hat x \hat y + \hat x \hat z\) (distributivity)

    • \((\hat y + \hat z)(a \hat x) = a (\hat y + \hat z)(\hat x)\) (linearity) \(= a (\hat y \hat x + \hat z \hat x)\) (distributivity) \(= a (- \hat x \hat y - \hat x \hat z)\) (anticommutativity) \(= -a (\hat x \hat y + \hat x \hat z)\) (converse of distributivity)

  • However, some things which require commutativity break down, such as the binomial theorem.

The choose operator, inner (dot) and outer (wedge) products

  • \(\langle V \rangle_n\) chooses all \(n\)-blades from the multivector \(V\). For example, if \(V = 1 + 2 \hat x + 3 \hat y + 4 \hat x \hat y + 5 \hat y \hat z\), then \(\langle V \rangle_0 = 1\) and \(\langle V \rangle_1 = 2 \hat x + 3 \hat y\) and \(\langle V \rangle_2 = 4 \hat x \hat y + 5 \hat y \hat z\).

  • \(U \cdot V = \langle UV \rangle_n\) where \(U\) is of grade \(r\), \(V\) is of grade \(s\), and \(n = |r - s|\). This is the inner or dot product.

    • The dot product associates and distributes the same way the geometric product does.

    • From this, for arbitrary vectors \(a \hat x + b \hat y\) and \(c \hat x + d \hat y\), we recover the typical meaning of the dot product:

      \[\begin{split}& (a \hat x + b \hat y) \cdot (c \hat x + d \hat y) \\ &= ac (\hat x \cdot \hat x) + ad (\hat x \cdot \hat y) + bc (\hat y \cdot \hat x) + bd (\hat y \cdot \hat y) \\ &= ac \langle \hat x \hat x \rangle_0 + ad \langle \hat x \hat y \rangle_0 + bc \langle \hat y \hat x \rangle_0 + bd \langle \hat y \hat y \rangle_0 \\ &= ac \langle 1 \rangle_0 + ad (0) + bc (0) + bd \langle 1 \rangle_0 \\ &\text{(because }\hat x \hat y\text{ and }\hat y \hat z\text{ have no part with grade }0\text{)} \\ &= ac + bd\end{split}\]
  • \(U \wedge V = \langle UV \rangle_n\) where \(U\) is of grade \(r\), \(V\) is of grade \(s\), and \(n = r + s\). This is the outer or wedge product.

    • The outer product associates and distributes the same way the geometric product does.

    • From this, for arbitrary vectors \(a \hat x + b \hat y + c \hat z\) and \(d \hat x + e \hat y + f \hat z\), we recover something that looks very much like a cross product:

      \[\begin{split}&(a \hat x + b \hat y + c \hat z) \wedge (d \hat x + e \hat y + f \hat z) \\ &= ad (\hat x \wedge \hat x) + bd (\hat y \wedge \hat x) + cd (\hat z \wedge \hat x) \\ &\quad + ae (\hat x \wedge \hat y) + be (\hat y \wedge \hat y) + ce (\hat z \wedge \hat y) \\ &\quad + af (\hat x \wedge \hat z) + bf (\hat y \wedge \hat z) + cf (\hat z \wedge \hat z) \\ &= ad \langle \hat x \hat x \rangle_2 + be \langle \hat y \hat y \rangle_2 + cf \langle \hat z \hat z \rangle_2 \\ &\quad + (ae - bd) \langle \hat x \hat y \rangle_2 + (af - cd) \langle \hat x \hat z \rangle_2 + (bf - ce) \langle \hat y \hat z \rangle_2 \\ &= 0 + 0 + 0 + \begin{vmatrix} a & b \\ d & e \end{vmatrix} (\hat x \hat y) + \begin{vmatrix} a & c \\ d & f \end{vmatrix} (\hat x \hat z) + \begin{vmatrix} b & c \\ e & f \end{vmatrix} (\hat y \hat z) \\ &\text{(because }\hat x \hat x\text{ etc.} = 1\text{, which has no grade }2\text{ part)} \\ &= \begin{vmatrix} \hat y \hat z & \hat x \hat z & \hat x \hat y \\ a & b & c \\ d & e & f \end{vmatrix}\end{split}\]

Euler’s formula applied to multivectors

  • \(e^{\theta B} = \cos \theta + B \sin \theta\) where \(\theta\) is a scalar in radians and \(B\) is a basis multivector.

    Note

    This formula only works when \(B^2 = -1\) (in the same way as the imaginary unit \(i\)), such as \((\hat x \hat y)^2 = \hat x \hat y \hat x \hat y = - \hat x \hat y \hat y \hat x = - \hat x \hat x = -1\).

    • For purposes of interest, the more general formula for \(e\) raised to a multivector power is the Taylor series:

      \[e^V = \exp(V) = \sum_{n=0}^{\infty} \frac{V^n}{n!}\]
  • For reasons that are beyond my power to explain, the rotation of a multivector \(V\) by \(\theta\) through the plane \(B\) is \(e^{-\frac{\theta B}{2}} V e^{\frac{\theta B}{2}}\).