# Derivative of the metric components - Christoffel symbols

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?

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 $$dg_{ij} = \nabla(\langle E_i,E_j\rangle) = \langle \nabla E_i,E_j\rangle + \langle E_i,\nabla E_j\rangle = \sum g_{kj}\omega_i^k + \sum g_{ik}\omega^k_j.$$

• Perfect! Thank you so much. – Dave Dmitriyev Sep 29 '17 at 23:27
• Welcome to MSE. In general, when you are satisfied with an answer, you should accept it so that the question will not stay on the "unanswered" list. Of course, often there are follow-up questions in the comments, etc. Have fun with differential geometry. It's great stuff. :) – Ted Shifrin Sep 29 '17 at 23:30
• Done. Thanks again – Dave Dmitriyev Sep 29 '17 at 23:52
• You're most welcome, Dave. I suspect we'll stumble across one another again in the future :) – Ted Shifrin Sep 30 '17 at 0:08