# Palatini action: variation of spin connection: show that torsion vanishes

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

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?

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. $$D (\varepsilon_{IJKL} A^{IJKL}) = \mathrm{d} (\varepsilon_{IJKL} A^{IJKL})$$

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.