Euler's Formula and Geometric Algebra

While working to understand and derive Euler's formula we introduced an imaginary unit \(i\) with the property that \(i^2 = -1\). But we've already seen that the product of the two basis vectors has this same property in that \( (\mathbf{xy})^2 = -1 \).

In this section we will investigate the properties of various rotations using Euler's formula and Geometric Algebra in two dimensional space.

Properties of spatial rotations

Rather than introducing the imaginary unit \(i\) we could have evaluated the exponent function using the value \(\mathbf{xy}\theta\) instead. Doing so leads to a similarly periodic result:

\[ e^{\mathbf{xy}\theta} = \cos \theta + \mathbf{xy}\sin \theta \]

Rather than representing a complex rotation, in this form it represents a rotation in the \(\mathbf{xy}\) plane of our two dimensional space. To see that this is the case, consider multiplying an arbitrary vector \(\mathbf{a}\) by this rotation:

\[ \begin{aligned} \mathbf{a}e^{\mathbf{xy}\theta} &= (a_x\mathbf{x} + a_y\mathbf{y})(\cos \theta + \mathbf{xy}\sin \theta) \\ &= a_x \cos\theta \mathbf{x} + a_x\mathbf{xxy}\sin\theta + a_y\cos\theta\mathbf{y} + a_y\mathbf{yxy}\sin\theta \\ &= (a_x \cos\theta - a_y\sin\theta)\mathbf{x} + (a_x\sin\theta + a_y\cos\theta)\mathbf{y} \end{aligned} \]

which matches the normal definition for a two-dimensional rotation of a vector with components \((a_x, a_y)\).

In the following section, 2d rotations in space, we will use this rotation to simplify the way we represent arbitrary vectors, but for now it will help to note a few properties of this rotation \(e^{\mathbf{xy}\theta}\).

First, the exponent differs in sign depending on whether you left or right multiply your vector, so:

\[ \mathbf{x}e^{\mathbf{xy}\theta} = e^{-\mathbf{xy}\theta} \mathbf{x} \\ \mathbf{y}e^{\mathbf{xy}\theta} = e^{-\mathbf{xy}\theta} \mathbf{y} \]

and therefore more generally

\[ \mathbf{a}e^{\mathbf{xy}\theta} = e^{-\mathbf{xy}\theta}\mathbf{a} \]

This can be seen by expanding to the trigonometric values as follows:

\[ \begin{aligned} \mathbf{x}e^{\mathbf{xy}\theta} &= \mathbf{x}\cos \theta + \mathbf{xxy}\sin \theta \\ &= \mathbf{x} \cos \theta -\mathbf{xyx}\sin \theta \\ &= (\cos \theta -\mathbf{xy}\sin \theta)\mathbf{x}\\ &= e^{-\mathbf{xy}\theta} \mathbf{x} \end{aligned} \]

Next, the exponent does not differ in sign when left or right multiplying by the pseudo-scalar \(\mathbf{xy}\):

\[ \mathbf{xy}e^{\mathbf{xy}\theta} = e^{\mathbf{xy}\theta} \mathbf{xy} \]

which can be seen again by expanding as per the previous example.

Finally, it's worth noting that, based on the above definition of \(e^{\mathbf{xy}\theta}\), both \(\cos\theta\) and \(\sin\theta\) can be represented as a combination of exponentials as follows:

\[ \cos\theta = \frac{e^{\mathbf{xy}\theta} + e^{-\mathbf{xy}\theta}}{2} \]

\[ \sin\theta = \frac{e^{\mathbf{xy}\theta} - e^{-\mathbf{xy}\theta}}{2\mathbf{xy}} \]

Properties of hyperbolic rotations

We can also investigate using another exponent, such as \(\mathbf{x}\theta\), which results in something less familiar, but just as significant when looking at the geometry of relativity.

In this case, we need to return to the full definition of our exponential function which we derived with Euler's formula, to see how it behaves:

\[ \begin{aligned} e^{\mathbf{x}\theta} &= \sum_{k=0}^{\infty} \frac{(\mathbf{x}\theta)^k}{k!} \\ &= \frac{(\mathbf{x}\theta)^0}{0!} + \frac{(\mathbf{x}\theta)^1}{1!} + \frac{(\mathbf{x}\theta)^2}{2!} + \frac{(\mathbf{x}\theta)^3}{3!} + ... \\ &= \sum_{k=0}^{\infty} \frac{\theta^{2k}}{2k!} + \mathbf{x}\sum_{k=0}^{\infty} \frac{\theta^{2k + 1}}{(2k +1)!} \end{aligned} \]

The two summations in the last line are one way to define the hyperbolic functions \(\cosh\theta\) and \(\sinh\theta\) respectively, so that

\[ e^{\mathbf{x}\theta} = \cosh\theta + \mathbf{x}\sinh\theta \]

We will see later how this hyperbolic rotation becomes important for two-dimensional rotations in space-time but for now, a few analogous but different properties to note, all of which can be seen by expanding \(e^{\mathbf{x}\theta}\) to the equivalent trigonometric values:

\[\mathbf{x}e^{\mathbf{x}\theta} = e^{\mathbf{x}\theta}\mathbf{x}\]

\[\mathbf{y}e^{\mathbf{x}\theta} = e^{-\mathbf{x}\theta}\mathbf{y}\]

\[\mathbf{x}e^{\mathbf{y}\theta} = e^{-\mathbf{y}\theta}\mathbf{x}\]

\[\mathbf{y}e^{\mathbf{y}\theta} = e^{\mathbf{y}\theta}\mathbf{y}\]

\[\mathbf{xy}e^{\mathbf{x}\theta} = e^{-\mathbf{x}\theta}\mathbf{xy}\]

\[\mathbf{xy}e^{\mathbf{y}\theta} = e^{-\mathbf{y}\theta}\mathbf{xy}\]

\[ \cosh\theta = \frac{e^{\mathbf{x}\theta} + e^{-\mathbf{x}\theta}}{2} \]

\[ \sinh\theta = \frac{e^{\mathbf{x}\theta} - e^{-\mathbf{x}\theta}}{2\mathbf{x}} \]