Covariant derivative of tetrad/vielbein

Question

I am learning about the tetrad basis for manifolds from this lecture notes. On pg 52, the spin connections ${{w_\mu}^a}_b$ are defined as $${{w_\mu}^a}_b=e^a_\nu e^\lambda_b\Gamma^\nu_{\mu\lambda}-e^\lambda_b\partial_\mu e^a_\lambda,$$

where the coordinate basis vectors $\hat{e}_\mu$ and tetrad basis vectors $\hat{e}_a$ are related by $$\hat{e}_\mu=e^a_\mu\hat{e}_a,$$ $$\hat{e}_a=e^\mu_a\hat{e}_\mu.$$

The author claimed that the first equation is equivalent to $$\nabla_\mu e^a_\nu=0.$$ How can this be shown to be true?

Answer 1

Simply calculate $\nabla_{\mu} e^{a}_{\nu}$ as you usually would (Eq. (3.178) in the notes you're using): $$\begin{align} \nabla_{\mu} e^{a}_{\nu} &= \partial_{\mu} e^{a}_{\nu} - \Gamma_{\mu \nu}^{\eta}e_{\eta}^{a}+ w_{\mu}{}^{a}{}_{c}e^{c}_{\nu} \\ &=\partial_{\mu} e^{a}_{\nu} - \Gamma_{\mu \nu}^{\eta}e_{\eta}^{a} + e^a_\eta e^\lambda_c\Gamma^\eta_{\mu\lambda} e^c_{\nu} -e^\lambda_c\partial_\mu (e^a_\lambda) e^{c}_{\nu} \\ &= \partial_{\mu} e^{a}_{\nu} - \Gamma_{\mu \nu}^{\eta}e_{\eta}^{a} + e^a_{\eta}\Gamma^{\eta}_{\mu \nu} - \partial_{\mu} e^{a}_{\nu} =0 \ , \end{align} $$ where we've just used the standard definitions of the covariant derivative, spin connection and the tetrad with $e^{a}_{\mu} e_{a}^{\nu} = \delta^{\nu}_{\mu}$ and $e^{a}_{\mu} e_{b}^{\mu} = \delta^{a}_{b}$.