# Spin covariant derivative of gamma matrices?

Where can I find a general expression (on curved manifolds) in local coordinates, for the following: $$\nabla^S_{\mu}\gamma^{\nu} = ?$$ $\nabla^S_{\mu} = \partial_{\mu} + \omega^S_{\mu}$ is the spin covariant derivative associated to the spin connection $\omega^S_{\mu}$ and the gamma matrices satisfy: $$\{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}.$$

• I suppose $\nu$ is a space time (coordinate) index? Please define what $\gamma^{\nu}$ is then. The usual definition of gamma matrices have an internal Lorentz index, but it is also just convenient (and confusing) notation. Gamma matrices are just pre-defined constants, they transform trivially under diffeos and local Lorentz. The reason we write an uppercase Lorentz index is that $\bar{\psi} \gamma^I \psi$ actually transforms like a Lorentz vector, but not because of gamma matrices: it is $\psi$ that’s transformation law is responsible for this. – Prof. Legolasov Feb 9 '18 at 2:46
• @SolenodonParadoxus they would be constants in the flat case, hence there is no question of their derivatives there. However, in the case of curved manifold, I'm not sure why that should be the case. – phydev Feb 12 '18 at 11:35
• I am telling you that it is still the case in the curved spacetime. The coordinate-dependent version is related to constant matrices by $\gamma^{\mu} = e^{\mu}_I \gamma^I$ where $e$ is the tetrad field. – Prof. Legolasov Feb 12 '18 at 11:56
• @SolenodonParadoxus Yes, indeed, that's right. Then, the derivative would basically entail the derivative of the tetrad field, i.e. $\nabla^S_{\mu}\gamma^{\mu} = (\nabla^S_{\mu}e^{\mu}_I)\gamma^I$? – phydev Feb 14 '18 at 11:30
• Exactly. And the compatibility condition says that the tetras is covariantly constant, so your derivative is still zero. – Prof. Legolasov Feb 14 '18 at 13:07