So far we have considered operations on vectors only - linear combinations of the two basis vectors, \(\mathbf{x}\) and \(\mathbf{y}\). Here we will consider a more general combination of a vector together with an \(\mathbf{xy}\) component to form a "multivector". Furthermore, we will see that this extra component behaves geometrically as we expect time to behave in the space-time of Special Relativity and, by extension, in our every day experience. We will then see that we can rotate these space-time multivectors in space without affecting the \(\mathbf{xy}\) component, corresponding to changing your frame of reference within a single inertial frame.

At this point we are deviating significantly from the normal understanding of a space-time algebra using Geometric Algebra, where a separate basis vector is normally defined for the time dimension. You have been warned, you may be entering crack-pot territory

To begin with, let's define an arbitrary two-dimensional space-time multivector as a vector plus a component of \(\mathbf{xy}\):

\[ \mathbf{A} = \mathbf{a} + t_a\mathbf{xy} \]

Multiplying this multivector by itself, we see:

\[ \begin{aligned} \mathbf{A}^2 &= (\mathbf{a} + t_a\mathbf{xy})(\mathbf{a} + t_a\mathbf{xy}) \\ &= \mathbf{a}^2 + t_a\mathbf{axy} + t_a\mathbf{xya} + t_a^2\mathbf{xyxy} \\ &= \mathbf{a}^2 + t_a\mathbf{axy} - t_a\mathbf{axy} - t_a^2 \\ &= \mathbf{a}^2 - t_a^2 \\ &= r_a^2 - t_a^2 \end{aligned} \]

If you are familiar with Special Relativity, you will recognise this as the Minkowski space-time metric. Note, we have used above some of the earlier properties of rotations to equate \(\mathbf{xya} = -\mathbf{axy}\), but in full for clarity:

\[ \begin{aligned} \mathbf{xya} &= \mathbf{xy}r_a \mathbf{x} e^{\mathbf{xy}\theta_a} \\ &= - r_a \mathbf{xxy} e^{\mathbf{xy}\theta_a} \\ &= -r_a \mathbf{x}e^{\mathbf{xy}\theta_a}\mathbf{xy} \\ &= -\mathbf{axy} \end{aligned} \]

So the square of a space-time multivector results in a scalar value, a value which matches the Minkowski space-time metric.

\[ |\mathbf{A}|^2 = \mathbf{A}^2 = r_a^2 - t_a^2 \]

An arbitrary space-time multivector \(\mathbf{A}\) is normal if

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

Using the trigonometric identity that

\[\cosh^2\phi - \sinh^2\phi = 1\]

then \(\mathbf{A}\) will be a normalised multivector if \(\mathbf{a}^2 = 1\) and:

\[ \mathbf{A} = \mathbf{a}\cosh \phi + \mathbf{xy}\sinh \phi \]

We can verify this by expanding:

\[ \begin{aligned} \mathbf{A}^2 &= (\mathbf{a}\cosh \phi + \mathbf{xy}\sinh \phi)(\mathbf{a}\cosh \phi + \mathbf{xy}\sinh \phi) \\ &= \mathbf{a}^2\cosh^2\phi + \mathbf{axy}\cosh\phi\sinh\phi + \mathbf{xya}\cosh\phi\sinh\phi + \mathbf{xyxy}\sinh^2\phi \\ &= \cosh^2\phi + \mathbf{axy}\cosh\phi\sinh\phi - \mathbf{axy}\cosh\phi\sinh\phi - \sinh^2\phi \\ &= 1 \end{aligned} \]

We can simplify our definition of a normal space-time multivector by looking back to Euler's formula and geometric algebra where we saw that if \( \mathbf{a}^2 = 1 \), then

\[ e^{\mathbf{a}\phi} = \cosh\phi + \mathbf{a}\sinh\phi \]

This allows us to simplify \(\mathbf{A}\) as follows:

\[ \begin{aligned} \mathbf{A} &= \mathbf{a}\cosh\phi + \mathbf{xy}\sinh\phi \\ &= \mathbf{a}(\cosh\phi + \mathbf{axy}\sinh\phi) & \text{, since }\mathbf{a}^2 = 1\\ &= \mathbf{a}e^{\mathbf{axy}\phi} \end{aligned} \]

since \(\mathbf{axy}\) is just the vector \(\mathbf{a}\) rotated by 90 degrees. With this simpler definition of a normal space-time multivector, seeing that it is indeed normal is trivial:

\[ \begin{aligned} \mathbf{A}^2 &= \mathbf{a}e^{\mathbf{axy}\phi}\mathbf{a}e^{\mathbf{axy}\phi}\\ &= \mathbf{a}e^{\mathbf{axy}\phi}e^{-\mathbf{axy}\phi}\mathbf{a} \\ &= 1 \end{aligned} \]

If we define a spatial rotation using two normalised vectors, \(\mathbf{b}\) and \(\mathbf{c}\), where the difference in orientation between the two is \(\theta\), so that:

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

we find that it no longer defines a rotation when simply applied to an arbitrary multivector, because the magnitude-squared after the multiplication is not the same as before:

\[ \begin{aligned} (\mathbf{Abc})^2 &= \mathbf{AbcAbc} \\ &= (\mathbf{a} + t_a\mathbf{xy})\mathbf{bc}(\mathbf{a} + t_a\mathbf{xy})\mathbf{bc} \\ &= (\mathbf{a} + t_a\mathbf{xy})(\mathbf{a}\mathbf{cb} + t_a\mathbf{xy}\mathbf{bc})\mathbf{bc} \\ &= (\mathbf{a} + t_a\mathbf{xy})(\mathbf{a}\mathbf{cbbc} + t_a\mathbf{xy}\mathbf{bcbc}) \\ &= (\mathbf{a} + t_a\mathbf{xy})(\mathbf{a} + t_a\mathbf{xy}e^{\mathbf{xy}2\theta}) \\ &\neq \mathbf{A}^2 \end{aligned} \]

So instead we halve the angle of rotation, defining:

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

and apply it twice - once from the left and once from the right: \(\mathbf{cbAbc}\).

The fact that \(\mathbf{cbAbc}\) is a rotation can be verified simply by confirming the magnitude-squared is now unchanged:

\[ \begin{aligned} (\mathbf{cbAbc})^2 &= \mathbf{cbAbccbAbc} &\\ &= \mathbf{cb}\mathbf{A}^2\mathbf{bc} \\ &= \mathbf{A}^2\mathbf{cbbc} &\text{, as }\mathbf{A}^2\text{is scalar}\\ &= \mathbf{A}^2 \end{aligned} \]

Now, evaluating the actual rotation, we find:

\[ \begin{aligned} \mathbf{cbAbc} &= \mathbf{cb}(\mathbf{a} + t_a\mathbf{xy})\mathbf{bc} \\ &= \mathbf{cbabc} + t_a\mathbf{cbxybc} \\ &= \mathbf{abcbc} + t_a\mathbf{xycbbc} \\ &= \mathbf{a}e^{\mathbf{xy}\theta} + t_a\mathbf{xy} \end{aligned} \]

That is, when applying the spatial rotation in this way, the vector component of \(\mathbf{A}\) is rotated by \(\theta\), while the time component of the multi-vector is not affected at all. We can change position and orientation within the one inertial frame without affecting the measured time. Note, we have again used above some of the earlier properties of rotations to equate \(\mathbf{cba} = \mathbf{abc}\) and \(\mathbf{cbxy} = \mathbf{xycb}\).

But what if, rather than defining a rotation between two spatial vectors, we instead define a rotation between a spatial vector and the pseudo-scalar component \(\mathbf{xy}\) - a two-dimensional rotation in space-time?