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 \(\v{a}\) 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 \(\v{a}\) is equal to the square of the vector itself:

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

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 \(\v{x}\) (bold will indicate a vector) as having a magnitude of one:

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

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

\[\v{a} = \vcOne{a}\]

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

\[ \v{a}^2 = (\vcOne{a})^2 = \vcOne{a}\vcOne{a} = a_x^2\v{x}^2 = a_x^2 \]

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,

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

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

\[ \v{a} = \vcTwo{a} \]

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

\[ \begin{aligned} \v{a}^2 &= (\vcTwo{a})(\vcTwo{a}) \\ &= a_x\v{x}a_x\v{x} + a_x\v{x}a_y\v{y} + a_y\v{y}a_x\v{x} + a_y\v{y}a_y\v{y} \\ &= a_x^2 + a_y^2 + a_x a_y(\v{xy} + \v{yx}) \end{aligned} \]

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

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

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

With this definition of orthogonality, we have

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

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,

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

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

Magnitude squared of two vectors

See if you can calculate the magnitude-squared of

\[\v{a} + \v{b}\]

Solution

\[ \begin{aligned} (\v{a} + \v{b})^2 &= (\v{a} + \v{b})(\v{a} + \v{b}) \\ &= \v{a}^2 + \v{ab} + \v{ba} + \v{b}^2 \end{aligned} \]

We can't yet go any further as we haven't yet defined what a multiple of two general vectors means. But soon we'll see that our solution here is actually the law of cosines.

Multiplying two vectors

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

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

While the reverse is,

\[ \v{ba} = a_x b_x + a_y b_y - \v{xy}(a_x b_y - a_y b_x) \]

The general vector product is not commutative, as \(\v{ab} \neq \v{ba}\) nor is it anti-commutative, as \(\v{ab} \neq -\v{ba}\). If you are familiar with the traditional vector dot and cross products, you may recognise \(\v{ab}\) 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 \(\v{xy}\) 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:

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

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

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

Rotating vectors in two-dimensions

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

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

so that

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

then this vector product will rotate any other vector \(\v{c}\). To simplify things, let's multiply the unit vector \(\v{x}\):

\[ \begin{aligned} \v{abx} &= \v{xyx} \\ &= -\v{xxy} \\ &= -\v{y} \\ \v{xy}(-\v{y}) &= -\v{x} \\ \v{xy}(-\v{x}) &= -\v{xyx} \\ &= \v{xxy} \\ &= \v{y} \\ \v{xy(y)} &= \v{x} \\ \end{aligned} \]

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