How to prove $d\omega=(\nabla_\mu\omega)_\nu dx^\mu\wedge dx^\nu$ without using coordinates

Question

This is exercise 7.8 b) of Nakahara's GTaP: Let $\omega\in\Omega^1(M)$ be a 1-form on a Riemannian manifold with Levi-Civita connection $\nabla$. Prove that

$$ \mathrm{d}\omega=(\nabla_\mu\omega)_\nu\, \mathrm dx^\mu\wedge\mathrm dx^\nu $$

I proved it using the fact that $\mathrm dx^\mu\wedge\mathrm dx^\nu=-\mathrm dx^\nu\wedge\mathrm dx^\mu$, so:

\begin{align} (\nabla_\mu\omega)_\nu\,\mathrm dx^\mu\wedge\mathrm dx^\nu & = (\partial_\mu\omega_\nu-{\Gamma^\lambda}_{\mu\nu}\,\omega_\lambda)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \sum_{\mu<\nu}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu+({\Gamma^\lambda}_{\mu\nu}-{\Gamma^\lambda}_{\nu\mu})\omega_\lambda\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \sum_{\mu<\nu}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \mathrm d\omega \end{align}

I hope this is correct and makes sense.

I don't like my solution because starting at the second line, it "quits" Einstein summation convention and needs an explicit summation symbol.

1. Is there a way to prove this without "quitting" Einstein summation convention?
2. Is there maybe even a way to prove it in a coordinate-free way?

Answer 1

Here's an outline of a coordinate-free proof.

(1) For any $1$-form $\omega$ and any vector fields $X,Y$, there is the formula $$d\omega(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y]).$$ (2) For any affine connection $\nabla$, there is the formula $$X (\omega(Y)) = (\nabla_X\omega)(Y) + \omega(\nabla_XY).$$ By switching the roles of $X, Y$ and subtracting, this gives $$X \omega(Y) - Y \omega(X) = (\nabla_X \omega)(Y) - (\nabla_Y \omega)(X) + \omega(\nabla_XY - \nabla_YX).$$ But $\nabla$ is torsion-free, so.... (left to you)

(3) By plugging in $X = \frac{\partial}{\partial x^\mu}$ and $Y = \frac{\partial}{\partial x^\nu}$, we get..... (left to you)

Remarks: Note that the proof works for any torsion-free affine connection $\nabla$, not just the Levi-Civita connection. This formula is an instance of "Cartan's First Structure Equation." Generalizations exist to $k$-forms for any $k \geq 1$.

Answer 2

There are some wrong expressions in your solution. A possible correct solution is as follows.

\begin{align} (\nabla_\mu\omega)_\nu\,\mathrm dx^\mu\wedge\mathrm dx^\nu & = (\partial_\mu\omega_\nu-{\Gamma^\lambda}_{\mu\nu}\omega_\lambda)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \frac{1}{2}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu-({\Gamma^\lambda}_{\mu\nu}-{\Gamma^\lambda}_{\nu\mu})\omega_\lambda\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \frac{1}{2}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \mathrm d\omega \end{align}

Do you agree?