Palatini action: variation of spin connection: show that torsion vanishes 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?
 A: 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.
$$
D (\varepsilon_{IJKL} A^{IJKL})
=
\mathrm{d} (\varepsilon_{IJKL} A^{IJKL})
$$
A: 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. 
