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The algebraic properties of the pseudoscalar $i$ follows the ordinary rules for imaginary numbers: So its algebraic properties are ~ $i^2 = -1$ the amazing geometric property is that it is an orientated unit 4-volume element for spacetime! Hestene's arrives at the identity:

$i \gamma_{\mu} = -i\gamma_{\mu}$

In which $\gamma_{\mu}$ is the ordinary notation for a 4x4 matrices involving three spatial and one time component. We will concentrate on this, because multiplication of this with $\gamma_0$ yields the pseudovector,

$\gamma_1 \gamma_2 \gamma_3 = i\gamma_0$

You get this from:

$i = \gamma_0\gamma_1 \gamma_2 \gamma_3$

and the matrices contain the same geometric properties as:

$(\gamma_2\gamma_1)^2 = -1$

In such a case, Hestenes correctly notices that geometrically $\gamma_2\gamma_1$ can be interpreted as the unit (directed) area element for the plane containing $\gamma_1$ and $\gamma_2$.

A surprising feature to fall out of this is that it is also the generator of rotations in that plane, since multiplication of those two gamma matrices can be easily proves to rotate a vector in that plane by a quarter term.

(http://geocalc.clas.asu.edu/pdf-preAdobe8/ZBW_I_QM.pdf)

Using these geometric interpretations, we can do something similar to our own imaginary number and would incorporate the goemetric meaning behind We will concentrate on this interpretation within our own opening equation: But first, some more on geometric algebra - the metric is given as:

$g_{\mu \nu} = \gamma_{\mu} \cdot \gamma_{\nu}$

and we used the anticommutator rule

$\gamma_{\mu}\cdot\gamma_{\nu} + \gamma_{\nu} \cdot \gamma_{\mu} = 2 \delta_{\mu \nu} \mathbf{I}$

We can also see in the equation, the relationship

$(\gamma_0\gamma_1 \gamma_2 \gamma_3)^2 = i^2 = -1$

We recognize we have written the metric in a new way using the gamma matrices under Hestene's approach:

$g_{\mu \nu}\mathbf{B} = \gamma_\mu \cdot (\gamma_{\nu} \cdot \mathbf{B}) = (\gamma_\mu \wedge \gamma_{\nu})\cdot \mathbf{B}$

Which is a bi-vector. Normally orthogonality ensures that

$\gamma_{\mu} \cdot \gamma_{\nu} = 0$

A geometric product of two vectors u and v can be interpreted by decomposing it into symmetric and skew-symmetric parts:

$uv = u \cdot v + u \wedge v$

$u \cdot v = \frac{1}{2}(uv + vu)$

$u \wedge v = \frac{1}{2}(uv - vu) = -v \wedge u$

Hestene deducts:

''Thus, the geometric relation of orthogonality is expressed algebraically by ananticommutative geometric product. Similarly, collinearity is expressed by a commutative geometric product.''

The idea is you can decompose any product of vectors into their scalar and vector components as found from:

$uv = u \cdot v + u \wedge v = \frac{1}{2}(uv + vu) + \frac{1}{2}(uv - vu) = \frac{1}{2}(uv + vu) - v \wedge u$

In which we have already identified:

$u \cdot v = \frac{1}{2}(uv + vu)$

$u \wedge v = \frac{1}{2}(uv - vu) = -v \wedge u$

I suspect the same form can be provided into the Einstein equations:

$\mathbf{G}_{\mu \nu} = \mathbf{T}(\gamma_\mu \gamma_{\nu}) = \frac{1}{2}\mathbf{T}(\gamma_{\mu}\cdot \gamma_{\nu} + \gamma_{\mu} \wedge \gamma_{\nu}) = \frac{1}{2}\mathbf{T}[(\gamma_{\mu}\gamma_{\nu} + \gamma_{\nu}\gamma_{\mu}) - \gamma_{\nu} \wedge \gamma_{\mu}]$

Which also breaks the equation up into symmetric and skew symmetric parts. In fact, the product is related to the Pauli pseudovector:

$(\sigma \cdot u)(\sigma \cdot v) = u \cdot v + u \wedge v = u \cdot v + i \sigma \cdot (u \times v)$

It enters an imaginary part (complexification) on the antisymmetric matrix, in which case, using the information I have gathered, it is clear the Einstein equation can also be written as:

$\mathbf{G}_{\mu \nu} = \mathbf{T}(\gamma_\mu \gamma_{\nu}) = \frac{1}{2}\mathbf{T}(\gamma_{\mu}\cdot \gamma_{\nu} + \gamma_{\mu} \wedge \gamma_{\nu}) = \frac{1}{2}\mathbf{T}[\gamma_{\mu} \cdot \gamma_{\nu} + i \sigma \cdot (\gamma_{\mu} \times \gamma_{\nu})]$

Which invites an idea that the Einstein tensor is represented geometrically through a symmetric and antisymmetric part but it also naturally incorporates the Pauli vector involving the spin matrices. Let’s define the square of the Pauli vector:

$\sigma^2_1 = \sigma^2_2 = \sigma^2_3 = -i \sigma_1\sigma_2\sigma_3 = \begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix} = \mathbf{I}$

The Pauli vector is defined by:

$\vec{\sigma} = \sigma_1\hat{x} + \sigma_2 \hat{y} + \sigma_3\hat{z}$

This provides a mapping basis to a Pauli matrix see Pauli matrices -

Wikipedia (https://en.wikipedia.org/wiki/Pauli_matrices)

Recall the three matrices are:

$\sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $

$\sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} $

$\sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

QUESTION

When we see the relationship to the Einstein equations:

$\mathbf{G}_{\mu \nu} = \mathbf{T}(\gamma_\mu \gamma_{\nu}) = \frac{1}{2}\mathbf{T}(\gamma_{\mu}\cdot \gamma_{\nu} + \gamma_{\mu} \wedge \gamma_{\nu}) = \frac{1}{2}\mathbf{T}[\gamma_{\mu} \cdot \gamma_{\nu} + i \sigma \cdot (\gamma_{\mu} \times \gamma_{\nu})]$

my question resides on the cross product because it is generally true that the stress energy tensor is a 4x4 matrix (for four dimensional spacetime) but cross products do not exist in four dimensions as I have come to learn, they only exist in three dimensions or seven. How would in this case, you define or interpret the cross product

$\gamma_{\mu} \times \gamma_{\nu}$

I assume it is a strict no when it comes to interpreting this in four dimensions? How is it possible to take the spatial component and take the cross product with a timelike component without introducing a four dimensional case?

edit

Because it is a bivector theory, I assume we can only deal with one spatial dimension in this case and one time dimension? Wiki states:

''More precisely, a bivector that can be expressed as an exterior product is called simple; in up to three dimensions all bivectors are simple, but in higher dimensions this is not the case.[1] The exterior product of two vectors is anticommutative and alternating, so b ∧ a is the negation of the bivector a ∧ b, producing the opposite orientation, and a ∧ a is the zero bivector.''

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I think I can answer this now.

In the 3D space $R^3$ the outer product of any four vectors $(a,b,c,d)$ is zero, an automatic consequence of the outer product properties.

In three dimensions only three vectors can be independent and therefore a fourth must be expressible as a weighted sum of the other three:

$d = \alpha a + \beta b + \gamma c$

Associativity, distributivity and antisymmetry make the outer product of the four vectors zero:

$ a \wedge b \wedge c \wedge d = a \wedge b \wedge c \wedge (\alpha a + \beta b + \gamma c) = 0$

Where $\gamma$ in this case is not a gamma matrix and $ a \wedge b \wedge c \wedge d$ is known as a quadvector.

The link states so the highest order element that can exist in the subspace algebra of $R^3$ is a trivector.

Apparently this has nothing to do with a limitation of the outer product but rather it satisfies a geometric uselessness of the construction of elements of higher dimension than $R^3$ - it can however be interpreted that the element $0$ as the empty subspace of any dimensionality so this one element $0$ is the zero scalar, the zero bivector and so on. There is no algebraic or geometric reason to distinguish between those, for the empty subspace has no orientation or weight.

https://books.google.co.uk/books?id=-1-zRTeCXwgC&pg=PA32&redir_esc=y#v=onepage&q&f=false

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