Derivative of the metric components - Christoffel symbols

Question

I'm beginning to learn Riemannian geometry and differential forms and got stuck mixing the two.

Let $(M,g)$ be a Riemannian manifold and $\{E_i\}$ a local frame. If $\nabla$ is a connection on $M$ compatible with $g$, then $dg_{ij}$ should be somehow related to the Christoffel symbols of $\nabla$ over $\{E_i\}$ right?

I'm trying to find an explicit formula, but I can't figure out how to apply the exterior derivative $d$ through the metric on $\langle E_i,E_j\rangle=g_{ij}$ in a way to obtain such formula. We know the covariant derivative is well behaved because the connection is compatible, i.e. $$\nabla_Z\langle X,Y\rangle=\langle\nabla_ZX,Y\rangle+\langle X,\nabla_ZY\rangle$$

but how can we use this to work with $d$ and obtain a formula?

Answer 1

So you're avoiding the moving frames with an orthogonal coframe so far ...

So, let's see, the connection $1$-forms $\omega_i^j$ are defined by $\nabla E_i = \sum \omega_i^j\otimes E_j$. So $$dg_{ij} = \nabla(\langle E_i,E_j\rangle) = \langle \nabla E_i,E_j\rangle + \langle E_i,\nabla E_j\rangle = \sum g_{kj}\omega_i^k + \sum g_{ik}\omega^k_j.$$