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Consider the tetrad-Palatini action:

$$S[e,\omega] = \int e \wedge e \wedge F[\omega]^\star,$$

where $\star$ denotes the Hodge dual, i.e. $F_{IJ}^\star = \frac{1}{2} \varepsilon_{IJKL} F^{KL}$. The curvature 2-form is

${F^I}_J = {d\omega^I}_J + {\omega^I}_K \wedge {\omega^K}_J$

Using this (and the fact that $a \wedge b = - b \wedge a$), I should be able to rewrite the action as

$ S[e,\omega] = \frac{1}{2} \int e^I \wedge e^J \wedge F^{KL} \varepsilon_{IJKL} = \frac{1}{2} \int \left( F^{KL} \wedge e^I \wedge e^J \right) \varepsilon_{IJKL}$

According to my textbook, a variation of this action w.r.t. the connection should yield

$de^I + {\omega^I}_{J} \wedge e^J = 0$ ,

namely that the torsion vanishes. I have been trying to show this but to no avail. If I consider a variation w.r.t. the connection, I get:

$\delta F^{KL} = \delta (d \omega^{KL}) + \delta({\omega^K}_A \wedge \omega^{AL}) = \delta (d \omega^{KL}) + \delta {\omega^K}_A \wedge \omega^{AL} + {\omega^K}_A \wedge \delta \omega^{AL}$

Thus for the variation of the action:

$\delta S[e,\omega] = \frac{1}{2} \int \left( \left( \delta (d \omega^{KL}) + \delta {\omega^K}_A \wedge \omega^{AL} + {\omega^K}_A \wedge \delta \omega^{AL} \right) \wedge e^I \wedge e^J \right) \varepsilon_{IJKL}$

How do I get from here to the vanishing of torsion?

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The key point to understandinging how this works is to realize that you can partially integrate covariant derivatives.

First, note that the variation of the curvature 2-form can be written as a covariant derivative, i.e.

$$ \delta F^{IJ} = D \delta \omega^{IJ} = \mathrm{d} \delta \omega^{IJ} + \omega^I{}_K \delta \omega^{LJ} + \omega^J{}_K \delta \omega^{KJ} $$

Variation of the action

$$S[e,\omega] = \int e^I \wedge e^J \wedge F^{KL} \varepsilon_{IJKL}$$

with respect to $\omega^{IJ}$ then is

$$ \begin{align} \delta S[e,\omega] &= \int e^I \wedge e^J \wedge D \delta \omega^{KL} \varepsilon_{IJKL} \\ &= - \int D ( e^I \wedge e^J )\wedge \delta \omega^{KL} \varepsilon_{IJKL} \\ &= - 2 \int D e^I \wedge e^J \wedge \delta \omega^{KL} \varepsilon_{IJKL} \end{align} $$

This gives the desired equation $D e^I = 0$.

To see why you can partially integrate note that you want the covariant derivative to satisfy a (graded) Leibniz rule $$ D ( a^I \wedge b^J ) = D a^I \wedge b^J + (-1)^{|a|} a^I \wedge D b^J $$ and furthermore agree with the exterior derivative on scalars, e.g. $$ D (\varepsilon_{IJKL} A^{IJKL}) = \mathrm{d} (\varepsilon_{IJKL} A^{IJKL}) $$

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Torsion doesn't just ''vanish,'' there are no true reasons why the torsion should vanish. Some have taken a gauge invariant approach, meaning you treat General relativity in terms of curvature alone - so when curvature is zero, torsion is not, and vice versa. But this doesn't really make any sense, when would we have a torsion but no curvature in reality?

There are no underlying assumptions which exist that explains why torsion ''must'' vanish. The vanishing of torsion is unwarranted, but simpler.

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