general relativity - Why can we assume torsion is zero in GR?

The first Cartan equation is

$$\mathrm{d}\omega^{a} + \theta^{a}_{b} \wedge \omega^{b} = T^{a}$$

where $\omega^{a}$ is an orthonormal basis, $T^{a}$ is the torsion and $\theta^{a}_{b}$ is the spin connection. In Misner, Thorne and Wheeler, as well as several lectures in general relativity I've seen using the Cartan formalism, they assume $T^{a} = 0$ to then use the first equation to determine the connections.

Why is it valid to assume in general relativity that the torsion is zero?

Answer

There are (at least) two approaches to torsion in the geometric framework behind general relativity:

First, we can use it to encode a new degree of freedom of the theory: The coupling of spin to the gravitational field. This is Einstein-Cartan theory, which is (as far as I'm aware) neither supported nor excluded by observational evidence.

Second, we can use it to encode the existing gravitational degrees of freedom. There's a sort of gauge symmetry between curvature and torsion. Gauge fixing torsion to 0, we end up with general relativity, whereas fixing curvature to 0, we end up with its teleparallel equivalent.