Physical significance of metric compatibility When we try to construct a covariant derivative, we impose several conditions on it so that the resulting derivative is unique. However, I can't make sense of the condition of metric compatibility. I realize how we use it to get the connection from the metric or write it in terms of the metric. But I don't see any physical significance for it on its own. It seems like an unduly assumption that just makes the math easier, but doesn't have any manifestation in physical reality. At least not one that's obvious to me.
 A: There is a formulation of GR where metric compatibility follows from the equations of motion. If you think of the Einstein-Hilbert action as being a functional of both the connection and the metric (this is called a first-order formulation), and the connection will be determined using the equations of motion. In the absence of matter, or for "standard" matter fields like a scalar field or electromagnetic field with minimal coupling, and if you assume that the connection is torsion-free (symmetric on the lower indices), then the equations of motion for the connection end up being the metric compatibility condition, and the theory is equivalent to general relativity. (See https://en.wikipedia.org/wiki/Palatini_variation). The assumption that the connection is torsion-free is needed for this to work; it is consistent (but more complicated and not necessary to explain observations) to assume there is some non-dynamical torsion.
I think I can detect in your question an undercurrent of dissatisfaction with making a "purely mathematical" assumption without physical reasoning in the usual way GR is presented. I share that feeling, and to that end I'd also like to note that there are other ways to arrive at GR, starting from a more physical set of principles. In particular, you can think of GR as the unique low energy theory of a massless spin-2 particle, with local and Lorentz invariant interactions. See, eg, [1, 2, 3]. Relatedly, Weinberg's textbook on GR takes the point of view that GR is a consequence of the equivalence principle, and does not start from differential geometry.
[1] https://www.amazon.com/Feynman-Lectures-Gravitation-Frontiers-Physics/dp/0813340381
[2] https://inspirehep.net/literature/1461
[3] https://arxiv.org/abs/gr-qc/0411023
A: Metric compatibility follows from the Einstein equivalence principle, which asserts that the laws of special relativity hold for local non-gravitational phenomena in a freely falling laboratory. Here, "local" means up to tidal effects that are second-order in the size of the laboratory. A freely falling laboratory is a reference frame in which $\partial_\lambda g_{\mu\nu} = 0$ at a point, so that the metric locally takes the flat-spacetime Minkowski form as in special relativity. Likewise, we must have $\Gamma^\lambda_{\mu\nu} = 0$ in such a frame so that equations involving the covariant derivative reduce to those of special relativity. The requirement that $\Gamma^\lambda_{\mu\nu} = 0$ when $\partial_\lambda g_{\mu\nu} = 0$ is sufficient for metric compatibility.
Note, one can mathematically define many different connections on a given spacetime. For the question about "the" connection to be meaningful, I take it as referring to the connection that matter couples to, i.e., the one that governs physical parallel transport and that appears in covariant equations such as $\nabla_\mu T^{\mu\nu} = 0$.
Define $\Gamma' = \Gamma - \Gamma_{\mathrm{LC}}$ as the difference between an arbitrary connection $\Gamma$ and the Levi-Civita connection $\Gamma_{\mathrm{LC}}$. Since $\Gamma'$ is a tensor, it cannot be made to vanish by a choice of reference frame, and it could be directly measured if matter coupled to it. This would violate the Einstein equivalence principle because it would affect local non-gravitational phenomena in a freely falling laboratory.
Even with metric compatibility, $\Gamma'$ could contain torsion, which would likewise violate the Einstein equivalence principle if coupled to matter, as in Einstein-Cartan theory. On the other hand, teleparallel gravity defines a connection with torsion but, in a special case of the theory that is equivalent to general relativity, $\Gamma'$ drops out of physical predictions and is unobservable.
