What is the expression for the covariant derivative of a Weyl spinor?
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Let $\mu$ be a spacetime index, $a,b$ be an internal space indices and $\alpha,\beta$ be spinor indices. The covariant derivative has the form: $$ \nabla_\mu \psi_\alpha = \partial_\mu \psi_\alpha + \frac{1}{4}\, \omega_{\mu ab} (\sigma^{ab})_\alpha{}^\beta \psi_\beta $$ where $$ \sigma^{ab} = \frac{1}{4} [ \sigma^a {\bar \sigma}^b - \sigma^b {\bar \sigma}^a ] $$ with $$ \sigma^a = ( I_{2\times2} , \sigma_1 , \sigma_2 , \sigma_3 ) , \qquad {\bar \sigma}^a = ( -I_{2\times2} , \sigma_1 , \sigma_2 , \sigma_3 ) . $$ The signs and factors of $2$ and $4$ are all convention based and I'm not certain I have the right factors in this answer. Your homework is to figure out the right factors.
HINT: With the right factors, we must have $$ [ \nabla_\mu , \nabla_\nu ] \psi_\alpha = \# R_{\mu\nu\rho\sigma} e^\rho_c e^\sigma_d ( \sigma^{cd} )_\alpha{}^\beta \psi_\beta $$
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$\begingroup$ Thanks! I am supposed to follow the convention of Wess and Bagger. Your definition of $\sigma^{ab}$ matches that of Wess & Bagger (eq. A.14). But I'm not sure how to check if the factor of $\frac14$ in your first equation matches with the convention of Wess and Bagger. There is a definition of the covariant derivative below eq. (19.20). But that uses $\omega_{\mu\alpha}{}^{\beta}$. Not sure what factor of 2 has been used in the definition of $\omega_{\mu\alpha}{}^{\beta}$ in terms of $\omega_{\mu a b}(\sigma^{ab})_\alpha{}^\beta$. Any suggestion how to check? $\endgroup$– vyaliCommented Apr 13, 2023 at 3:52
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$\begingroup$ What is # supposed to be in the hint equation? $\endgroup$– vyaliCommented Apr 15, 2023 at 14:14
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1$\begingroup$ that is again dependent on your conventions for $\sigma^{ab}$. If I remember correctly, the proper formula is $[ \nabla_\mu , \nabla_\nu ] \psi_\alpha = R_{\mu\nu\rho\sigma} e^\rho_c e^\sigma_d ( \Sigma^{cd} )_\alpha{}^{\beta} \psi_\beta$ where $\Sigma^{ab}$ is a (spinor) representation of the Lorentz algebra which is normalized to satisfy $[\Sigma_{ab} , \Sigma_{cd} ] = i ( \eta_{ac} \Sigma_{bd} - a \leftrightarrow b ) - c \leftrightarrow d$. Using the properties of $\sigma^{ab}$ you should be able to work out how this is related to the $\Sigma_{ab} = \# \sigma_{ab}$ $\endgroup$– PraharCommented Apr 15, 2023 at 16:47