3
$\begingroup$

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? $$ \begin{align} \delta \Gamma^\mu_{\nu\rho} =& \Gamma_{\lambda\nu\rho} \delta g^{\mu\lambda} + g^{\mu\lambda} \delta \Gamma_{\lambda\nu\rho} \\ =& \frac{1}{2}g^{\mu\lambda} \left( \nabla_\lambda \delta g_{\nu\rho} + \nabla_\nu \delta g_{\rho\lambda} - \nabla_\rho \delta g_{\lambda\nu} \right) \end{align} $$

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – Danu
    Commented Mar 2, 2017 at 14:58
  • $\begingroup$ 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. $\endgroup$
    – Jeroen
    Commented Mar 3, 2017 at 14:44
  • $\begingroup$ 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... $\endgroup$
    – Danu
    Commented Mar 3, 2017 at 15:52
  • $\begingroup$ 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 $c^a_{\ bc}e_a=[e_b,e_c]$. $\endgroup$ Commented Mar 5, 2017 at 0:59
  • 1
    $\begingroup$ You can get the variation of torsion free spin-connection from eq. (7.96) of the book 'Supergravity' by Freedman and Proeyen. Also look at the equation right before sec. 3.1.2 and eq. (49) in arxiv.org/abs/2212.10044. A Mathematica package to derive this result yourself can be 'Spinors' arxiv.org/abs/1110.2662. $\endgroup$
    – vyali
    Commented Aug 29, 2023 at 7:04

1 Answer 1

2
$\begingroup$

Long time no answer! I only just saw the question and maybe someone will find this useful. We want the change in the spin connection as a result of varying the tetrad ${\bf e}_a$ while preserving the torsion-free condition.

The calculation is indeed a bit tedious, and when I needed it it took me several days. Here is what I get:

Define $$ \delta e_{ij} = {\bf e}_i\cdot \delta {\bf e}_j= \eta_{ib}[e^{*b}_\alpha \delta e_j^\alpha], $$ then $$ (\delta \omega_{ij\mu}) e^\mu_k =-\frac 12\left\{(\nabla_j \delta e_{ik} - \nabla_k \delta e_{ij}) +( \nabla_k \delta e_{ji}- \nabla_i \delta e_{jk}) -( \nabla_i \delta e_{kj}- \nabla_j \delta e_{ki})\right\}.\nonumber $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.