# Variation of the Spin Connection

I am trying to get an explicit expression for the variation $\delta \omega_{\mu}^{\ ab} / \delta e_\mu^a$, 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 vielbein. 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.

Edit: If someone knows a good Mathematica package to take variational derivatives of the vielbein 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 vielbein \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)

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 vielbein? \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}

• Are you comfortable with the mathematicians' coordinate-independent formalism? I suspect Lawson & Michelsohn's book on Spin geometry has some relevant results, but they will not be written in physicists' notation. – Danu Mar 2 '17 at 14:58
• Could you maybe refer to a specific chapter or page you were thinking of? I have been looking through the book today, but besides a pile of awe-inspiring sections on Index Theorems and K-Theory I cannot seem to find what I am looking for. I would like to point out 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), however I cannot find it anywhere. – JgL Mar 3 '17 at 14:44
• It was just a shot in the dark---I'm sorry to have wasted your time. Indeed, you are probably better off looking through the SUGRA literature, but I wouldn't know anything about that either... – Danu Mar 3 '17 at 15:52
• It's also a shot in the dark, but I don't think your premise is correct, the equation you start with is an almost triviality (it says the covariant derivative of the identity tensor is zero). Instead maybe try starting with $\omega_{\mu\ \ \ b}^{\ a}=\frac{1}{2}(c_{bdc}+c_{cdb}-c_{dcb})e^c_\mu$, where $c^a_{\ bc}e_a=[e_b,e_c]$. – Bence Racskó Mar 5 '17 at 0:59
• The equation I start with follows from the requirement that the spin connection should be torsion free and compatible with the metric. You can derive the same relation from the requirement that that $\nabla_\kappa g_{\mu\nu} =0$. I believe the relation you wrote down for $\omega_{\mu\ b}^{\ a}$ is essentially the same as the relation I eventually obtained this way for $\omega_{\mu}^{\ ab}$ (though I agree, written your way the expression for the spin connection does look somewhat more manageable). I tried calculating the variation in both cases, but without success. – JgL Mar 6 '17 at 4:12