# Spin connection in terms of the vielbein/tetrad and their derivatives

Mimicking the process for finding the Christoffel symbol in terms of the metric (and its derivatives), see box 17.4 on page 205 of Moore's GR workbook, we can use the torsion-free (gauge local translations curvature set to zero) condition and some non-trivial index gymnastics to solve for the spin connection in terms of the vielbein (and its derivatives).

The following thesis outlines this. I include it here

I cannot get beyond the following with the first/algebraic Bianchi identity-like equation 2.54 of the thesis

$$0 = R_{\mu\nu}{}^a e_{\rho a} + R_{\rho\mu}{}^a e_{\nu a} -R_{\nu\rho}{}^a e_{\mu a}$$

$$R_{\nu\rho}{}^a e_{\mu a} = R_{\mu\nu}{}^a e_{\rho a} + R_{\rho\mu}{}^a e_{\nu a}$$

$$(\partial_{[\nu} e_{\rho]}{}^a - \omega_{[\nu}{}^{ab} e_{\rho]}{}_b) e_{\mu a} =(\partial_{[\mu} e_{\nu]}{}^a - \omega_{[\mu}{}^{ab} e_{\nu]}{}_b) e_{\rho a}+(\partial_{[\rho} e_{\mu]}{}^a - \omega_{[\rho}{}^{ab} e_{\mu]}{}_b) e_{\nu a}$$

$$\partial_{[\nu} e_{\rho]}{}^a e_{\mu a} - \omega_{[\nu}{}^{ab} e_{\rho]}{}_b e_{\mu a} = \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} - \omega_{[\mu}{}^{ab} e_{\nu]}{}_b e_{\rho a} + \partial_{[\rho} e_{\mu]}{}^a e_{\nu a} - \omega_{[\rho}{}^{ab} e_{\mu]}{}_b e_{\nu a}$$

$$\omega_{[\mu}{}^{ab} e_{\nu]}{}{}_b e_{\rho a} + \omega_{[\rho}{}^{ab} e_{\mu]}{}_b e_{\nu a} - \omega_{[\nu}{}^{ab} e_{\rho]}{}_b e_{\mu a} = \partial_{[\mu} e_{\nu]}{}^a e_{\rho a} + \partial_{[\rho} e_{\mu]}{}^a e_{\nu a} - \partial_{[\nu} e_{\rho]}{}^a e_{\mu a}$$

Any suggestions?

You can just start by noting that $$\psi\gamma^\mu \psi$$ is a vector and that the covariant derivative is linear and obeys the product rule. By noting this and using the format of the covariant derivative of a spinor as $$\nabla_\mu \psi = \partial_\mu \psi + \frac{i}{2} \omega_\mu^{ab}\sigma_{ab}\psi$$ you might arive at the expression for the spin connection in terms of the tetrad. If you want to check, it's the equation (2.32) in Shapiro's paper, and if you set null torsion then K = 0 in (2.31). Technically it is in terms of the Christofell symbol, but you might be able to express it in terms of the vierbein.