# Spin connection equations of motion from Einstein-Palatini action

I am working through "Supergravity" from Freedman and Van Proeyen. In exercise 8.11 one is tasked to vary the Einstein-Palatini action $$S = \frac{1}{2\kappa^2}\int d^Dx\ e e^\mu{}_a e^\nu{}_b R_{\mu\nu}{}^{ab}(\omega)$$ wrt. to the spin connection $$\omega_{\mu ab}$$ (here greek indices represent "curved" indices, while latin represent "flat" ones).
Ultimately we are to show that this variation is proportional to $$D_{[\mu}e^a{}_{\nu]}$$, where $$D$$ denotes the covariant derivative wrt. the spin connection $$\omega$$, letting us conclude that the solution of the equations of motion is exactly the torsion free connection.
As a hint, we are reminded that we showed earlier $$\delta R_{\mu\nu ab} = 2 D_{[\mu}\delta\omega_{\nu]ab}.$$ It is also clear, since $$D_\mu e = \partial_\mu e = \partial_\mu \sqrt{-g} = \frac{1}{2}\sqrt{-g}g^{\alpha \beta} \partial_\mu g_{\alpha\beta} = \frac{1}{4}\sqrt{-g}\partial_\mu(g^{\alpha \beta} g_{\alpha\beta})= 0$$, that we can integrate by parts without worrying about $$e$$.
Edit 1: As was pointed out in the comments, this is actually wrong. However, below I corrected the calculation so that it does not use this argument.

However I only managed the following (setting $$\kappa=1$$) $$\delta S = \frac{1}{2}\int d^Dx\ e e^\mu{}_{a}e^\nu{}_b \delta R_{\mu\nu}{}^{ab} = \int d^Dx\ e e^\mu{}_{a}e^\nu{}_bD_{[\mu}\delta\omega_{\nu]}{}^{ab} = \int d^Dx\ e e^{[\mu}{}_{a}e^{\nu]}{}_bD_{\mu}\delta\omega_{\nu}{}^{ab}.$$ We can now rewrite the covariant derivative: $$e e^{[\mu}{}_{a}e^{\nu]}{}_bD_{\mu}\delta\omega_{\nu}{}^{ab} = ee^{[\mu}{}_{a}e^{\nu]}{}_b(\partial_\mu \delta\omega_{\nu}{}^{ab} + \omega_{\mu}{}^a{}_c \delta\omega_{\nu}{}^{cb} +\omega_{\mu}{}^b{}_c \delta\omega_{\nu}{}^{ac})\\ = \partial_\mu(e e^{[\mu}{}_{a}e^{\nu]}{}_b \delta\omega_{\nu}{}^{ab})-\partial_\mu(e e^{[\mu}{}_{a}e^{\nu]}{}_b) \delta\omega_{\nu}{}^{ab} + \omega_{\mu}{}^c{}_{a}(e e^{[\mu}{}_{a}e^{\nu]}{}_b) \delta\omega_{\nu}{}^{ab} + \omega_{\mu}{}^c{}_b(e e^{[\mu}{}_{a}e^{\nu]}{}_b)\delta\omega_{\nu}{}^{ab} \\ = \partial_\mu(e e^{[\mu}{}_{a}e^{\nu]}{}_b \delta\omega_{\nu}{}^{ab})- [\partial_\mu(e e^{[\mu}{}_{a}e^{\nu]}{}_b) + \omega_{\mu a}{}^{c}(e e^{[\mu}{}_{c}e^{\nu]}{}_b) + \omega_{\mu b}{}^c(e e^{[\mu}{}_{a}e^{\nu]}{}_c)]\delta\omega_{\nu}{}^{ab} \\ =\partial_\mu(e e^{[\mu}{}_{a}e^{\nu]}{}_b \delta\omega_{\nu}{}^{ab})- D_\mu(e e^{[\mu}{}_{a}e^{\nu]}{}_b)\delta\omega_{\nu}{}^{ab},$$ where the first term vanishes in the integral and we conclude that $$D_{\mu}(e \ e^{[\mu}{}_{a}e^{\nu]}{}_b) =0.$$
however, I do not see how that brings us the proposed solution.
Any hints and help to reach the conclusion would be much appreciated.

Edit 2: I found a source, which skips almost all actual steps of the derivation, but gave one good hint and an intermediate result.
From the definition of the determinant $$e$$, we can conclude up to factors that: $$e \ e^{[\mu}{}_{a}e^{\nu]}{}_b \sim \varepsilon_{abcd}\varepsilon^{\mu\nu\rho\sigma} e^{c}{}_\rho e^d{}_\sigma,$$ which then would give $$D_{\mu}(\varepsilon_{abcd}\varepsilon^{\mu\nu\rho\sigma} e^{c}{}_\rho e^d{}_\sigma) = 0.$$
If now the covariant derivative of the $$\varepsilon$$s would be zero, we would arrive at
$$\varepsilon_{abcd}\varepsilon^{\mu\nu\rho\sigma} D_{\mu} (e^{c}{}_\rho e^d{}_\sigma) = 0,$$ which by anti symmetry immediately gives $$\varepsilon_{abcd}\varepsilon^{\mu\nu\rho\sigma} D_{\mu} (e^{c}{}_\rho) e^d{}_\sigma = 0,$$ which according to the source, eq. (2.14) is not only the case, but also implies the wanted equations of motion!
However, now I am stuck at proving that actually the covariant derivatives vanish and even if I just assume that, I do not see how then the claimed equations of motion follow.

• The argument that $\partial_\mu e = 0$ doesn't work as $\partial_\mu g^{\alpha \beta}$ is not the same as $\partial_\mu g_{\alpha \beta}$. You can find counterexamples easily enough, e.g. polar coordinates. For the rest, you'll probably need to multiply by the inverse-vielbein one-forms and contract the free indices, rewriting derivatives of the vielbein in terms of the inverse-vielbein. Jan 10, 2021 at 2:20
• You are absolutely correct! I rushed over it, since in my mind i made a similar argument as for EOM in Minkowski space where a contracted term usually can be just written as two times one of the terms. However, I still think that integration by parts should still be valid, but I have to figure out why. Your ideas seem sound enough and I already tried some combinations, but maybe I have missed something. Thank you so far! Jan 10, 2021 at 17:14

## 1 Answer

So I think I finally pieced together everything to solve my question:

Indeed, let's start from $$D_\mu(e\ e ^{[\mu}{}_a e^{\nu]}{}_b) = 0.$$ We multiply the definition of the $$\varepsilon$$ in curved coordinates $$\varepsilon_{a_1...a_D}e^{a_1}{}_{\mu_1} \cdots e^{a_D}{}_{\mu_D} = e\ \varepsilon_{\mu_1...\mu_D}$$ with $$e^{\mu_{D-1}}{}_k e^{\mu_D}{}_l$$ and with $$\varepsilon^{\mu_1...\mu_{D-2}\mu\nu}$$ and use eq. (3.9) from the book, namely $$\varepsilon_{\mu_1...\mu_n\nu_1...\nu_p} \varepsilon^{\mu_1...\mu_n\rho_1...\rho_p} = - p!n!\delta^{\rho_1...\rho_p}_{\nu_1...\nu_p},$$ where $$p = D-n$$ and $$\delta^{\rho_1...\rho_p}_{\nu_1...\nu_p} := \delta^{[\rho_1}_{\nu_1}\delta^{\rho_2}_{\nu_2} \cdots \delta^{\rho_p]}_{\nu_p}$$ and thus get $$\varepsilon_{a_1...a_{D-2}kl}\varepsilon^{\mu_1...\mu_{D-2}\mu\nu}e^{a_1}{}_{\mu_1} \cdots e^{a_{D-2}}{}_{\mu_{D_2}} = - e\ 2!(D-2)! e^{[\mu}{}_ke^{\nu]}{}_l.$$ This is up to factors exactly what we have in our starting equation! So we now make sure, that both epsilons are covariantly constant.
We start with the flat indices, remembering that the symbol has constant coefficients $$\pm 1$$ in any frame: $$D_\mu \varepsilon_{a_1...a_D} = \partial_\mu \varepsilon_{a_1...a_D} + \omega_{\mu a_1}{}^k \varepsilon_{ka_2...a_D} + \omega_{\mu a_2}{}^k \varepsilon_{a_1 k ... a_D} + ... + \omega_{\mu a_D}{}^k \varepsilon_{a_1...a_{D-1} k}.$$ The first summand vanishes and we remember, that any top form only has one independent component, so plug in numbers for the indices: $$D_\mu \varepsilon_{01...D-1} = \omega_{\mu 0}{}^k \varepsilon_{k12...D-1} + \omega_{\mu 1}{}^k\varepsilon_{0k2...D-1} + ... + \omega_{\mu D-1}{}^k\varepsilon_{01...k}.$$ Examining the first term, we see that the epsilon only takes a non-zero value for $$k = 0$$ but since $$\omega_{\mu a}{}^b = - \omega_{\mu}{}^b{}_a$$, we have $$\omega_{\mu 0}{}^0 = 0$$. The same argument works for all other terms and hence we conclude that $$D_\mu \varepsilon_{a_1...a_D} = 0.$$ The curved epsilon seems a little bit more complicated: $$D_\mu \varepsilon^{\mu_1...\mu_D} = D_\mu(e\ e^{\mu_1}{}_{a_1} \cdots e^{\mu_D}{}_{a_D} \varepsilon^{a_1...a_D} )=D_\mu(e\ e^{\mu_1}{}_{a_1} \cdots e^{\mu_D}{}_{a_D}) \varepsilon^{a_1...a_D} \\ = \partial_\mu(e\ e^{\mu_1}{}_{a_1} \cdots e^{\mu_D}{}_{a_D} )\varepsilon^{a_1...a_D} + e(\omega_{\mu a_1}{}^k e^{\mu_1}{}_k e^{\mu_2}{}_{a_2}\cdots e^{\mu_D}{}_{a_D} \varepsilon^{a_1...a_D} \\ + \omega_{\mu a_2}{}^ke^{\mu_1}{}_{a_1} e^{\mu_2}{}_k \cdots e^{\mu_D}{}_{a_D} \varepsilon^{a_1...a_D} + ...).$$ If we now rename the indices $$(k, a_j)$$ in each term, use the antisymmetry of the connection and that $$\varepsilon^{a_1...a_D}$$ is again constant, we can write this (pulling out the vielbeins) as: $$\partial_\mu(e e^{\mu_1}{}_{a_1} \cdots e^{\mu_D}{}_{a_D} \varepsilon^{a_1...a_D}) - e\ e^{\mu_1}{}_{a_1} \cdots e^{\mu_D}{}_{a_D} (\omega_{\mu}{}^{a_1}{}_k \varepsilon^{ka_2...a_D} + \omega_\mu{}^{a_2}{}_k \varepsilon^{a_1 k... a_D} + ...) \\ = \partial_\mu(\varepsilon^{\mu_1...\mu_D}) - e\ e^{\mu_1}{}_{a_1} \cdots e^{\mu_D}{}_{a_D} D_\mu \varepsilon^{a_1...a_D}.$$ But we know, the flat epsilon is covariantly constant, so the second term vanishes and we defined the curved epsilon s.t. it has values $$\pm 1$$ in any coordinate system, so the first term vanishes as well and we have shown that $$D_\mu \varepsilon^{\mu_1...\mu_D} = 0.$$

This now means, that we can rewrite our starting equation via the epsilons and can pull them out directly and we have $$0 = D_\mu(e\ e ^{[\mu}{}_a e^{\nu]}{}_b) \sim \varepsilon_{a_1...a_{D-2}kl}\varepsilon^{\mu_1...\mu_{D-2}\mu\nu}D_\mu (e^{a_1}{}_{\mu_1}\cdots e^{a_{D-2}}{}_{\mu_{D-2}} ) \\\sim \varepsilon_{a_1...a_{D-2}kl}\varepsilon^{\mu_1...\mu_{D-2}\mu\nu}D_\mu (e^{a_1}{}_{\mu_1}) e^{a_2}{}_{\mu_2}\cdots e^{a_{D-2}}{}_{\mu_{D-2}},$$ the last line follows by the product rule and renaming of indices s.t. the terms match.
Now we can do the summation on the vielbein indices, rearrange indices and again use eq. (3.9) to get $$0 = \varepsilon_{a_2...a_{D-2}kl}\varepsilon^{a_2...a_{D-2} \mu_1 \mu \nu}(D_\mu e^{a_1}{}_{\mu_1}) = - 3!(D-3)!e\ \delta^{\mu_1 \mu \nu}_{a_1 k l} (D_\mu e^{a_1}{}_{\mu_1}).$$ From here on, I will be a little bit less verbose, since the equations get quite long, but I will give the directions to get the correct result.
First note that $$\delta^{\mu}_k = e^\mu{}_a\delta^a_k = e^\mu{}_k$$, so we can write the above equation as some antisymmetrization of vielbeins, which can be switched to the covariant derivative and in total reads $$0 = e^\nu{}_l e^{\mu_1}{}_{a_1} e^\mu{}_k D_{[\mu}e^{a_1}{}_{\mu_1]} + e^\nu{}_k e^\mu{}_{a_1} e^{\mu_1}{}_k D_{[\mu}e^{a_1}{}_{\mu_1]} + e^\nu{}_{a_1} e^{\mu_1}{}_k e^{\mu}{}_l D_{[\mu}e^{a_1}{}_{\mu_1]} \tag{1}.$$ Note that we have $$(\nu, k, l)$$ as free indices and if we contract the above with $$e^{\nu}{}_k$$, we arrive at the equation $$0 = (D-2) e^{\mu}{}_{a_1} e^{\mu_1}{}_l D_{[\mu}e^{a_1}{}_{\mu_1]},$$ which for $$D \neq 2$$ means $$e^{\mu}{}_{a_1} e^{\mu_1}{}_l D_{[\mu}e^{a_1}{}_{\mu_1]} = 0$$ and by multiplication with a vielbein also means $$e^{\mu}{}_{a_1} D_{[\mu}e^{a_1}{}_{\nu]} = 0.$$ But two terms in eq. (1) are of that form, so in total this equation simplifies to $$0 = e^{\nu}{}_{a_1} e^{\mu_1}{}_k e^{\mu}{}_l D_{[\mu}e^{a_1}{}_{\mu_1] }$$ and since $$(\nu, k, l)$$ appear as free indices, multiplying by appropriate vielbeins gives the desired equations of motion $$0 = 2 D_{[\mu}e^a{}_{\nu]} = T^a{}_{\mu\nu} .$$