# The relation between the connection and the equivalence principle?

So I'm reading A relativist's toolkit (Page 21) it states:

$$\Gamma^{\alpha}_{\gamma \beta} = \Gamma^{\alpha}_{ \beta \gamma}$$

And

$$g_{\alpha \beta ; \gamma} = 0$$

In general relativity, these properties come as a consequence of Einstien's equivalence principle.

where $$\Gamma$$ is the connection and $$g$$ is the metric. Using these properties he then derives the connection. It is not obvious to me how the equivalence principle implies this. How does the equivalence principle imply this?

So, the first identity can be derived in the following way: consider a general scalar field $$\Phi$$. It's first covariant derivative is a one form and coincides with the normal derivative. It's second covariant derivative is a $$(0,2)$$-tensor with components $$\Phi_{;\mu;\nu} = \Phi_{,\mu; \nu}$$. For the equivalence principle we can always find a locally minkowskian frame in which the components of the second covariant derivative are just $$\Phi_{,\mu;\nu} = \Phi_{,\mu,\nu} = \frac{\partial}{\partial x^\mu}\frac{\partial}{\partial x^\nu}\Phi$$ which is symmetric. But if a tensor is symmetric in one basis it's symmetric in any basis and so $$\Phi_{,\mu;\nu} = \Phi_{,\mu,\nu}-\Phi_{,\alpha}\Gamma^{\alpha}_{\mu\nu} = \Phi_{,\nu,\mu}-\Phi_{,\alpha}\Gamma^{\alpha}_{\nu\mu} = \Phi_{,\nu;\mu}$$ and since the equality holds for any scalar field, we get that $$\Gamma^{\alpha}_{\mu\nu} = \Gamma^{\alpha}_{\nu\mu}$$
The second identity follows from the same line of reasoning. We know from the equivalence principle that we can always find a locally minkowskian frame in which the metric tensor $$g_{\mu\nu}$$ reduces to $$\eta_{\mu\nu}$$ (the Minkowski metric). The coordinate basis associated with the minkovskian frame coordinates has constant basis vectors, so the affine connections vanish. In this frame $$g_{\mu\nu;\alpha} = \eta_{\mu\nu;\alpha} = \eta_{\mu\nu,\alpha}-\Gamma^{\beta}_{\mu\alpha}\eta_{\beta\nu}-\Gamma^\beta_{\nu\alpha}\eta_{\mu\beta} = 0$$ If all the components of a tensor are zero in a frame will be zero in any coordinate frame, and so $$g_{\mu\nu;\alpha} = 0$$