I Solving Spinorial Maxwell's Equations with Wald

[ergospherical]

I'm trying to figure out how to do these sorts of calculations but I'm having a lot of trouble figuring out where to start. Take problem 3) of Chapter 13 of Wald, i.e. given that a real antisymmetric tensor $F_{ab}$, corresponding to the spinorial tensor $F_{AA' BB'}$ by the map ${\sigma^a}_{AA'}$, can be written as\begin{align*}
F_{AA' BB'} = \phi_{AB} \bar{\epsilon}_{A'B'} + \bar{\phi}_{A'B'} \epsilon_{AB}
\end{align*}to show that $\partial^a F_{ab} = 0$ and $\partial_{[a} F_{bc]} = 0$ if and only if $\phi^{AB}$ satisfies $\partial_{A_1' A_1} \phi^{A_1, \dots, A_n} = 0$. I've really not much idea where to start; the spinor equivalent of the first Maxwell equation should be\begin{align*}
\partial^{AA'} F_{AA'BB'} = \phi_{AB} \partial^{AA'} \bar{\epsilon}_{A'B'} + \bar{\epsilon}_{A'B'} \partial^{AA'} \phi_{AB} + \bar{\phi}_{A'B'} \partial^{AA'} \epsilon_{AB} + \epsilon_{AB} \partial^{AA'} \bar{\phi}_{A'B'} = 0
\end{align*}In the text it's mentioned that $\partial_{AA'} \epsilon_{BC} = 0$, but no proof is given. Maybe as a starter, how can I show that it follows from the definition $\partial_{\Lambda \Lambda'} \epsilon_{\Sigma \Omega} = \sum_{\mu} {\sigma^{\mu}}_{\Lambda \Lambda'} \dfrac{\partial \epsilon_{\Sigma \Omega}}{\partial x^{\mu}}$? I reckon its simply because $\epsilon^{\Sigma \Omega} = o^\Sigma \iota^\Omega - \iota^\Sigma o^\Omega$ is independent of the spacetime coordinates, with $\{o, \iota\}$ being a fixed basis of $W$...?

 

[martinbn]

Yes, the components are constant, so all the derivatives will be zero. In curved space-time you need the theorem that there is a unique connection with the given properties, one of which is that the $\epsilon$ has zero covariant derivative.