I Compute EOM for Spin Connection from Einstein-Palatini Action

[Lactose]

I was trying to compute the equations of motion for the spin $\omega_{\mu ab}$ connection from the Einstein-Palatini action
$$S := \int d^4 x e^\mu{}_ae^\nu{}_b R_{\mu \nu}{}^{ab}.$$
I managed to get the variation wrt. the connection to the form
$$ \varepsilon_{abcd}\varepsilon^{\mu \nu \rho \sigma} (D_\mu e^c{}_{\rho})e^d{}_\sigma = 0,$$
where latin indices denote frame (or flat) components and greek indices denote coordinate (or curved) components and $D_\mu V^a= \partial_\mu V^a + \omega_{\mu}{}^a{}_c V^c$ is the covariant derivative wrt. to the spin connection.
By the source (eq. 2.14) on page 14, the above is actually correct and then implies
$$T^a{}_{\mu\nu}:= 2D_{[\mu}e^a{}_{\nu]} = 0.$$
However, I do not see how this implication should work. I am pretty sure it has something to do with the contraction rules for epsilons and I already tried out:

1. carrying out the summation in $(d, \sigma)$, which gives contracted epsilons - did not work.
2. multiplying the equation by epsilons corresponding to the free indices, giving contracted epsilons as well - did not work either.

Any hints or help to reach the conclusion would be much appreciated.

 

[Lactose]

As it is often the case, once one poses the question, one finds the answer.

For anyone interested in this, I give a quick summary now and try to write the full solution in the next few days.

Indeed my first ansatz was correct, but needed refinement:
So one first carries out the summation in $(d, \sigma)$ which gives the two contracted $\varepsilon$, which then can be written as an antisymmetrisation of terms involving deltas like $\delta^\mu_k \equiv e^\mu{}_k$, which we sort such that we see the antisymmetrization in $[\mu, \nu]$.
Contracting the resulting equation now with a vielbein $e^\nu{}_a$ let's us deduce that
$$ e^\mu{}_k e^\nu{}_l D_{[\mu}e^k{}_{\nu]} = 0,$$
which also gives us
$$e^\mu{}_k D_{[\mu} e^k_{\nu]} = 0.$$
This we can then plug back into the equation which we contracted with $e^\nu{}_a$ and then all terms except one, namely
$$e^\rho{}_a e^\mu{}_k e^\mu{}_l D_{[\mu}e^a{}_{\nu]}$$
dissapear, so this on its own has to be zero.
We note that $(\rho, k, l)$ appear as free indices and hence can be multiplied away with vielbeins, so we are left with the desired equality.

I did my calculations for arbitrary dimension and the result holds for all dimensions unequal to 2.