Intro to Geometric Algebra

Geometric Algebra is a line of Mathematics which treats vectors as first-class citizens, rather than arrows defined by coordinates on a coordinate system.

For a flat Euclidean space, Geometric Algebra can be derived from a single rule that the square of the magnitude of any vector \(\mathbf{a}\) is equal to the square of the vector itself.

With this one rule we can introduce addition and multiplication of two-dimensional vectors, as well as a simple rotation.

One rule

The square of the magnitude of an arbitrary vector \(\mathbf{a}\) is equal to the square of the vector itself:

\[ |\mathbf{a}|^2 = \mathbf{a}^2 \]

This is similar to the more familiar dot-product of a vector with itself defining the Euclidean magnitude-squared of a vector, but as we will see, much more powerful.

Normal basis vectors

Starting with a single dimension, we define a normal basis vector \(\mathbf{x}\) (bold will indicate a vector) as having a magnitude of one:

\[ \mathbf{x}^2 = 1\]

Any other vector \(\mathbf{a}\) within this single dimension can be represented as some multiple of \(\mathbf{x}\), such as \(5\mathbf{x}\) or \(-3.141579\mathbf{x}\) or, more generally,

\[\mathbf{a} = x_a\mathbf{x}\]

Evaluating the magnitude-squared of \(\mathbf{a}\) results in:

\[ \mathbf{a}^2 = (x_a\mathbf{x})^2 = x_a\mathbf{x}x_a\mathbf{x} = x_a^2\mathbf{x}^2 = x_a^2 \]

as we would expect.

Orthogonal basis vectors

To describe vectors in two dimensions we need to add a second basis vector \(y\) with the similar property of unit-magnitude, so when multiplied by itself,

\[ \mathbf{y}^2 = 1 \]

We also need to ensure that this second vector \(\mathbf{y}\) is orthogonal to \(\mathbf{x}\), though we don't yet know how to represent this orthogonality. Nonetheless, by defining a general two-dimensional vector \(\mathbf{a}\) as

\[ \mathbf{a} = x_a\mathbf{x} + y_a\mathbf{y} \]

we can derive orthogonality by using the rule that the magnitude-squared of any vector is equal to the vector squared, to evaluate that:

\[ \begin{aligned} \mathbf{a}^2 &= (x_a\mathbf{x} + y_a\mathbf{y})(x_a\mathbf{x} + y_a\mathbf{y}) \\ &= x_a\mathbf{x}x_a\mathbf{x} + x_a\mathbf{x}y_a\mathbf{y} + y_a\mathbf{y}x_a\mathbf{x} + y_a\mathbf{y}y_a\mathbf{y} \\ &= x_a^2 + y_a^2 + x_a y_a(\mathbf{xy} + \mathbf{yx}) \end{aligned} \]

So we see that the square of the magnitude of the vector \(\mathbf{a}\) - a scalar number - will only equal the square of the vector if

\[ \mathbf{xy} = -\mathbf{yx} \]

This then is the definition of orthogonal vectors - they are anti-commutative.

With this definition of orthogonality, we have

\[ \mathbf{a}^2 = x^2 + y^2 \]

which is exactly what we are used to - Pythagoras' theorem.

Adding two vectors

The algebraic addition of two arbitrary two-dimensional vectors is straight forward,

\[ \begin{aligned} \mathbf{a} + \mathbf{b} &= (a_x\mathbf{x} + a_y\mathbf{y}) + (b_x\mathbf{x} + b_y\mathbf{y}) \\ &= (a_x + b_x)\mathbf{x} + (a_y + b_y)\mathbf{y} \\ &= \mathbf{b} + \mathbf{a} \end{aligned} \]

The order in which the vectors are added is not relevant which means that vectors are associative under addition (and subtraction), as \(\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}\).

Activity: See if you can calculated the magnitude-squared of \(\mathbf{a} + \mathbf{b}\)

Solution

Multiplying two vectors

The geometric product of two vectors may be a little less familiar:

\[ \begin{aligned} \mathbf{a} \mathbf{b} &= (a_x\mathbf{x} + a_y\mathbf{y})(b_x\mathbf{x} + b_y\mathbf{y}) \\ &= a_x\mathbf{x}b_x\mathbf{x} + a_x\mathbf{x}b_y\mathbf{y} + a_y\mathbf{y}b_x\mathbf{x} + a_y\mathbf{y}b_y\mathbf{y} \\ &= a_x b_x + a_y b_y + a_x b_y \mathbf{xy} + a_y b_x \mathbf{yx} \\ &= a_x b_x + a_y b_y + \mathbf{xy}(a_x b_y - a_y b_x) \\ \mathbf{ba} &= a_x b_x + a_y b_y - \mathbf{xy}(a_x b_y - a_y b_x) \end{aligned} \]

The vector product is not commutative, as \(\mathbf{ab} \neq \mathbf{ba}\) nor is it anti-commutative, as \(\mathbf{ab} \neq -\mathbf{ba}\). If you are familiar with the traditional vector dot and cross products, you may recognise \(\mathbf{ab}\) as some combination of both, but what is relevant is that it is a combination of a scalar (a pure number) and a pseudo-scalar (the \(\mathbf{xy}\) term, which is not scalar, nor is it a vector).

Based on the above, we can add or subtract the two products to end up with either a scalar or pseudo-scalar respectively:

\[ \begin{aligned} \mathbf{ab} + \mathbf{ba} &= 2(a_x b_x + a_y b_y) \\ \mathbf{ab} - \mathbf{ba} &= 2\mathbf{xy}(a_x b_y - a_y b_x) \end{aligned} \]

Of particular interest is that the pseudo-scalar \(\mathbf{xy}\) has the property that:

\[ (\mathbf{xy})^2 = \mathbf{xyxy} = -\mathbf{xyyx} = -1 \]

Rotating vectors in two-dimensions

It is worth noting briefly that the product of two vectors, \(\mathbf{ab}\) appears to define a scale and rotation. We will look at this in more detail later, but we can already see that if \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular and have a magnitude of 1:

\[ \begin{aligned} \mathbf{a} &= \mathbf{x} \\ \mathbf{b} &= \mathbf{y} \end{aligned} \]

so that

\[ \mathbf{ab} = \mathbf{xy}\]

then this product will rotate any other vector \(\mathbf{c}\):

\[ \mathbf{c} = c_x\mathbf{x} + c_y\mathbf{y} \]

by a right-angle each time it is applied:

\[ \begin{aligned} \mathbf{abc} &= \mathbf{xy}(c_x\mathbf{x} + c_y\mathbf{y}) \\ &= c_y\mathbf{x} - c_x\mathbf{y} \\ \mathbf{ababc} &= -c_x\mathbf{x} - c_y\mathbf{y} = -\mathbf{c} \\ \mathbf{abababc} &= -c_y\mathbf{x} + c_x\mathbf{y} \\ \mathbf{ababababc} &= \mathbf{c} \end{aligned} \]

But we will first need to understand Euler's formula to appreciate this rotation in its general form.