Weyl Tensor in Geometric Algebra

Question

In Gravity, Gauge theories and Geometric Algebra, p.39, they derive the Weyl Tensor in the following manner:

Six of the degrees of freedom in $\mathcal{R}(B)$ can be removed by arbitrary gauge rotations. It follows that $\mathcal{R}(B)$ can contain only 14 physical degrees of freedom. To see how these are encoded in $\mathcal{R}(B)$ we decompose it into Weyl and 'matter' terms. Since the contraction of $\mathcal{R}(a \wedge b)$ results in the Ricci tensor $\mathcal{R}(a)$, we expect that $\mathcal{R}(a \wedge b)$ will contain a term in $\mathcal{R}(a) \wedge b$. This must be matched with a term in $a \wedge \mathcal{R}(b)$, since it is only the sum of these that is a function of $a \wedge b$. Contracting this sum we obtain $$ \partial_a \cdot(\mathcal{R}(a) \wedge b+a \wedge \mathcal{R}(b)) =b \mathcal{R}-\mathcal{R}(b)+4 \mathcal{R}(b)-\mathcal{R}(b) \\ =2 \mathcal{R}(b)+b \mathcal{R}, $$ and it follows that $$ \partial_a \cdot\left(\frac{1}{2}(\mathcal{R}(a) \wedge b+a \wedge \mathcal{R}(b))-\frac{1}{6} a \wedge b \mathcal{R}\right)=\mathcal{R}(b) . $$ We can therefore write $$ \mathcal{R}(a \wedge b)=\mathcal{W}(a \wedge b)+\frac{1}{2}(\mathcal{R}(a) \wedge b+a \wedge \mathcal{R}(b))-\frac{1}{6} a \wedge b \mathcal{R}, $$ Some questions arise for me:

• Why $R (a \wedge b)$ must contain a term $R(a) \wedge b$?
• Why only a sum of it plus $a \wedge R(b)$ is a function of $a \wedge b$?
• Can someone explain the steps in the first equation?

Answer 1

$ \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\MVects[1]{\mathop{\textstyle\bigwedge^{#1}}} $The exterior power $\MVects 2V$ of a vector space $V$ is characterized by the following "universal property":

• Let $W$ be any other vector space. Every alternating multilinear map $g : V\times V \to W$ extends uniquely to a linear map $g' : \MVects 2V \to W$ such that $g'(a\wedge b) = g(a, b)$ for all $a, b \in V$.

What this means is that a linear function $f$ on $\MVects 2V$ is well-defined so long as $(a, b) \mapsto f(a\wedge b)$ is alternating on vectors $a, b$ (equivalently antisymmetric in the case of e.g. $\mathbb R$ or $\mathbb C$).

The authors aren't so much saying that $\mathcal R(a\wedge b)$ must contain this turn, more so that it ought to and including it will show us what the traceless part of $\mathcal R(a\wedge b)$ is. For any linear function $f : V \to V$ the following identity is true: $$ \partial_a\cdot(f(a)\wedge b) = \partial_a\cdot f(a)\,b - b\cdot\partial_a\,f(a) = (\mathop{\mathrm{Tr}}f)b - f(b). $$ In the particular case of $\mathcal R$ $$ \partial_a\cdot(\mathcal R(a)\wedge b) = \partial_a\cdot\mathcal R(a)\,b - b\cdot\partial_a\,\mathcal R(a) = \mathcal Rb - \mathcal R(b) $$ where $\mathcal R$ without an argument is the Ricci scalar. So we can get $\mathcal R(b)$ from this sort of term, and we know that $\mathcal R(b) = \partial_a\cdot\mathcal R(a\wedge b)$ so including this sort of term in $\mathcal R(a\wedge b)$ is natural. The universal property of $\MVects 2V$ tells us we can form a valid function of $a\wedge b$ by antisymmetrizing, so we want to focus on the form $$ \mathcal R(a)\wedge b - \mathcal R(b)\wedge a $$ where $X$ is arbitrary. We can easily rewrite this as $$ \mathcal R(a)\wedge b + a\wedge\mathcal R(b). $$ Applying $\partial_a\cdot$ in the same way as we did previously yields $$ \partial_a\cdot(\mathcal R(a)\wedge b + a\wedge\mathcal R(b)) = 2\mathcal R(b) + b\mathcal R. $$ Now noting that similarly $$ \partial_a\cdot(a\wedge b) = 4b - b = 3b $$ we see $$ \partial_a\cdot\left[\frac12(\mathcal R(a)\wedge b + a\wedge\mathcal R(b)) - \frac16\mathcal Ra\wedge b\right] = \mathcal R(b) = \partial_a\cdot\mathcal R(a\wedge b). $$ Thus we put $\mathcal R(a\wedge b)$ in the form $$ \mathcal R(a\wedge b) = \mathcal W(a\wedge b) + \frac12(\mathcal R(a)\wedge b + a\wedge\mathcal R(b)) - \frac16\mathcal Ra\wedge b $$ where $\mathcal W(a\wedge b)$ is an arbitrary function. This form is valid because the RHS is antisymmetric in $a, b$ and $\mathcal W$ is arbitrary and does whatever it needs to to make the equality happen. We now define the Weyl tensor as the necessary $\mathcal W$: $$ \mathcal W(a\wedge b) = \mathcal R(a\wedge b) - \frac12(\mathcal R(a)\wedge b + a\wedge\mathcal R(b)) + \frac16\mathcal Ra\wedge b. $$ Note that by construction $\partial_a\cdot\mathcal W(a\wedge b) = 0$, i.e. $\mathcal W$ is traceless.