I originally asked this question on the Physics Stack Exchange, but as it is more a mathematics question and nobody of the physicists there were able to answer it I thought it might move it over here.
I am trying to get an explicit expression for the variation of the spin connection $\omega_{\mu}^{\ ab}$ with respect to the vierbein $e_\mu^a$ (where $g_{\mu\nu} = e^a_\mu e^b_\nu \eta_{ab}$) in terms of covariant derivatives of $e_\mu^a$, but when doing the actual variation I end up with a series of 16 terms that I cannot simplify any further, though it should be possible to rewrite the resulting expression as a sum of covariant derivatives of the vierbein. The resulting identity is relevant for several calculations that involve spinors on curved spacetimes and Supergravity, but I cannot find a similar calculation or result anywhere for reference.
If someone knows a good Mathematica package to take variational derivatives of the vierbein and spin connection, that would also be very helpful.
Requiring the spin connection to be torsion free and compatible with the metric gives us the following constraint $$ \nabla_\mu e_\nu^a = \partial_\mu e_\nu^a + \omega_{\mu\ b}^{\ a} e^b_\nu - \Gamma^{\ \lambda}_{\mu\ \nu} e^a_\lambda = 0 $$ This allows us to write the following expression for the spin connection in terms of the vierbein $$ \begin{align} \omega_{\mu}^{\ ab} =& e^{\nu a} \Gamma^\lambda_{\ \mu\nu} e^b_\lambda - e^{\nu a} \partial_\mu e_\nu ^{\ b} \\ =& \frac{1}{2} e^{\nu a} g^{\lambda\sigma} \left( \partial_\mu g_{\sigma\nu} + \partial_\sigma g_{\nu\mu} - \partial_\nu g_{\mu\sigma} \right) - e^{\nu a} \partial_\mu e_\nu ^{\ b} \\ =& \frac{1}{2}e^{\nu a}(\partial_\mu e_\nu^{\ b}-\partial_\nu e_\mu^{\ b})-\frac{1}{2}e^{\nu b}(\partial_\mu e_\nu^{\ a}-\partial_\nu e_\mu^{\ a})-\frac{1}{2}e^{\rho a}e^{\sigma b}(\partial_\rho e_{\sigma d}-\partial_\sigma e_{\rho d})e_\mu^{\ d} \\ =& \frac{1}{2} e^{\nu [a} \left( \partial_{\mu} e_{\nu}^{b]} - \partial_{\nu} e_{\mu}^{b]} + e^{b]\sigma}e_\mu^d\partial_\sigma e_ {\nu d} \right) \end{align} $$ where I first wrote the Christoffel connection in terms of the metric and then used that $g_{\mu\nu} = e^a_\mu e^b_\nu \eta_{ab}$.
Does someone know a source where there is an expression for the following variational derivative? $$ \frac{\delta \omega_{\mu}^{\ ab}}{\delta e_\mu^a} $$ I tried calculating it myself, but since the above expression has four bilinear terms in $e$ and two quartic terms in $e$, you will end up with $2 \times 4 + 4 \times 2 = 16$ terms after doing the actual variation. I have a lot of trouble rewriting this variation as a neat expression (and cannot find any references that mention it). I would like to point out that my question is not so much a conceptual one, it is a rather simple identity that I am looking for, the kind of which I would expect to be on the back page of most books on Supergravity (like the one by Freedman and Van Proeyen) or other books dealing with field theory on curved backgrounds (as it is relevant in order to do several calculations), however I cannot find it anywhere.
As a comparison, consider the variation of the Christoffel connection with respect to the metric. $$ \Gamma^\mu_{\nu\rho} = g^{\mu\lambda} \Gamma_{\lambda\nu\rho} = \frac{1}{2} g^{\mu\lambda} \left( \partial_\lambda g_{\nu\rho} + \partial_\nu g_{\rho\lambda} - \partial_\rho g_{\lambda\nu} \right) $$ When varying this expression you will also get a lengthy series of terms, but the trick is to see they can be rearanged as covariant derivatives of the Christoffel connection. I assume there must be a similar trick and expression to rewrite the variation of the spin connection in terms of covariant derivatives of the vierbein? $$ \begin{align} \delta \Gamma^\mu_{\nu\rho} =& \Gamma_{\lambda\nu\rho} \delta g^{\mu\lambda} + g^{\mu\lambda} \delta \Gamma_{\lambda\nu\rho} \\ =& \frac{1}{2}g^{\mu\lambda} \left( \nabla_\lambda \delta g_{\nu\rho} + \nabla_\nu \delta g_{\rho\lambda} - \nabla_\rho \delta g_{\lambda\nu} \right) \end{align} $$
I don't know if this will help but here is a suggestion. Consider Cartan's first structure equation:
$$T^{a} \;=\; de^{a} + \omega^{\,a}_{\,\;b} \wedge e^{b}$$
where $\{e^{a}\}$ are a $g-$orthonormal coframe and $\{\omega^{\,a}_{\,\;b}\}$ are the connection 1-forms. For the torsion-free connection you are interested in, consider its variation with respect to the coframe:
$$d\delta e^{a} + \delta \omega^{\,a}_{\,\;b} \wedge e^{b} + \omega^{\,a}_{\,\;b} \wedge \delta e^{b} \;=\; 0$$
where we have used the commutation of the exterior derivative $d$ and variations. Then
$$\delta \omega^{\,a}_{\,\;b} \wedge e^{b} \;=\; - d\delta e^{a} + \delta \omega^{\,a}_{\,\;b}-\omega^{\,a}_{\,\;b} \wedge \delta e^{b} \;=\; -\left(d + \omega^{\,a}_{\,\;b} \,\wedge\,\right)\delta e^{a} \;=\; D\delta e^{a} $$
in terms of the covariant exterior derivative $D$. Unpacking this with respect to the vierbein frame fields should give you what you want.