Eq. (27) in http://arxiv.org/abs/1110.2662 says I can construct the Weyl spinor according to

$$\Psi_{ABCD} = \frac 14 C{}_{\mu\nu\lambda\rho} \left( \sigma^\mu \right){}_A{}^{\dot A} \left( \sigma^\nu \right){}_{B\dot A}\left( \sigma^\lambda \right){}_C{}^{\dot C}\left( \sigma^\rho \right){}_{D\dot C}\tag{27}$$

I understand that the partial contraction in the two index pairs $\mu\leftrightarrow\nu$ and $\lambda\leftrightarrow\rho$ induces a unique way to extract either two left-handed or two right-handed spinor indices. Moreover, the above spinor is completely symmetrical, $\Psi{}_{ABCD} = \Psi{}_{(ABCD)}$ as it must to encompass all ten components of the Weyl tensor.

But why does the above work?


1 Answer 1


I figured it out. Multiplying

$C{}_{ABCD\dot A\dot B\dot C\dot D} = C{}_{\mu\nu\lambda\rho}\left(\sigma{}^\mu\right){}_{A\dot A} \left(\sigma{}^\nu\right){}_{B\dot B}\left(\sigma{}^\lambda\right){}_{C\dot C}\left(\sigma{}^\rho\right){}_{D\dot D} = \Psi{}_{ABCD}\epsilon{}_{\dot A\dot B} \epsilon{}_{\dot C \dot D} + \Psi{}_{\dot A\dot B\dot C\dot D}\epsilon{}_{AB}\epsilon{}_{CD}$

on both sides with $\epsilon{}^{\dot A\dot B}\epsilon{}^{\dot C\dot D}$ and using that $\Psi{}_{\dot A\dot B\dot C\dot D} = \Psi{}_{(\dot A\dot B\dot C\dot D)}$ as well as $\epsilon{}_{\dot A\dot B}\epsilon{}^{\dot A\dot B} = 2$ gives the above result.


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