Proof of the formula for the exterior derivative in terms of the covariant derivative.

Question

SETUP

Everything is smooth.

Consider the ring $\Omega^{p}(M)$ of sections of the exterior powers of the cotangent bundle $\Lambda^{p}(T^{*}M)$ over a manifold $M$.

Purely in terms of the smooth structure, we define the exterior derivative as the map $\operatorname{d}:\Omega^{p}(M)\to\Omega^{p+1}(M)$ whose action on $\phi\in\Omega^{p}(M)$ is defined by $$\operatorname{d}\!\phi\bigg(\bigotimes_{k=0}^{p}X_{k}\bigg) = \sum_{i=0}^{p}(-1)^{i}X_{i}\left(\phi\Big(\bigotimes_{k\neq i}X_{k}\Big)\right) + \sum_{i<j}(-1)^{i+j}\phi\Big(\left[X_{i},X_{j}\right],\bigotimes_{k\neq i,j}X_{k}\Big)$$ where the $X_{k}$ are $p+1$ sections of the tangent bundle $TM\to M$, and the tensor products are taken in order.

On the other hand, if we have a connection on the frame bundle we may inherit a covariant derivative $\nabla$ in $TM$ and then in every tensor bundle $T^{n}_{m}M$, including the subbundles $\Lambda^{p}(T^{*}M)$. If the connection is torsion free, then we may express the exterior derivative in terms of the covariant derivative by (using abstract index notation)

$$(\operatorname{d}\phi)_{a_0\dots a_p} = (p+1)\nabla_{[a_0}\phi_{a_1\dots a_p]} \in\Omega^{p+1}(M)$$

for every $\phi\in\Omega^{p}(M)$. This construction is well defined for vector valued forms (elements of $\Omega^{p}(M,E)$), however at the moment I'm just interested in the following...

...QUESTION

Is there a coordinate-free proof (say, using abstract index notation) to see that the second definition is equivalent to the first, when $\phi\in\Omega^{p}(M)$?

I've just finished a proof using Penrose graphical notation, however it is rather lengthy. Maybe I was not ingenuous enough. Does anybody here know of a good one?

Answer 1

Althought there are better ways to do this, I wanted to close the question posting my own answer. This is a refined version of the proof I was refering to when I posted the question.

We start with $$(\operatorname{d}\phi)\left(\bigotimes_{k=0}^{p}X_{k}\right) = \sum_{i=0}^{p}(-1)^{i}X_{i}\left(\phi\left(\bigotimes_{k\neq i}X_{k}\right)\right) + \sum_{i<j}(-1)^{i+j}\phi\left(\left[X_{i},X_{j}\right],\bigotimes_{k\neq i,j}X_{k}\right)$$

Let's denote the two terms in the RHS by $(A)$ and $(B)$.

Now we expand $(A)$ as follows

So at the end we recognize that

$$(A) = (p+1)\nabla_{[a_0}\phi_{a_1\dots a_p]}\left(\bigotimes_{k=0}^{p}X_{k}\right)^{a_0 \dots a_p} - (B)$$

Hence

$$(\operatorname{d}\phi)\left(\bigotimes_{k=0}^{p}X_{k}\right) = (A) + (B) = (p+1)\nabla_{[a_0}\phi_{a_1\dots a_p]}\left(\bigotimes_{k=0}^{p}X_{k}\right)^{a_0 \dots a_p}$$

Since the $X_i$ are arbitrary, we finally have

$$(\operatorname{d}\phi)_{a_0 \dots a_p} = (p+1)\nabla_{[a_0}\phi_{a_1\dots a_p]} \hspace{2cm} \square$$