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?

• Have you tried working out the last equation both in terms of the spin and the ordinary Christoffel symbols? Mar 4, 2021 at 11:07
• @NDewolf I am confused about how to do this because I'm not sure what object $e^a_\nu$ is. Is it the components of a rank 2 tensor? Mar 5, 2021 at 2:44

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 $$e^{a}_{\mu} e_{b}^{\mu} = \delta^{a}_{b}$$.
• This seems to be treating $e^a_\nu$ as the components of a rank 2 tensor? I thought it is the components of a rank 1 tensor (vector) since $\hat{e}_\mu=e^a_\mu\hat{e}_a$, which is similar to $\vec{V}=V^a \hat{e}_a$? Mar 5, 2021 at 2:40