$\def\alt{\textrm{Alt}} \def\d{\mathrm{d}} \def\sgn{\mathrm{sgn}\,}$Let $\nabla$ be a symmetric linear connection in $M$, and $\omega$ be a $k$-form in $M$. I'm trying to find a relation between $\d\omega$ and $\nabla \omega$. I follow Spivak's notation in Calculus on Manifolds and write $$\alt(\nabla \omega)(X_1,\cdots,X_{k+1}) = \frac{1}{(k+1)!}\sum_{\sigma \in S_{k+1}} (\sgn \sigma)\nabla\omega(X_{\sigma(1)},\cdots,X_{\sigma(k+1)}).$$I also assume the formula $$\d\omega(X_1,\cdots,X_{k+1})=\sum_{i=1}^k(-1)^{i+1}X_i(\omega(X_1,\cdots,\widehat{X_i},\cdots,X_{k+1})) + \sum_{i<j}(-1)^{i+j}\omega([X_i,X_j],X_1,\cdots,\widehat{X_i},\cdots,\widehat{X_j},\cdots,X_{k+1}).$$I'd expect something like $\alt(\nabla\omega) = \d \omega$, apart from a multiplicative constant, maybe. I'm stuck. For $k=1$ I got $$\alt(\nabla \omega) = -\frac{1}{2}\d \omega.$$The $1/2$ I can accept, but the minus sign puts me off. I tried brute forcing my way through $k=2$ but I guess I just suck at doing computations like this.
The best I can come up with is
$$\begin{align} (k+1)!\alt(&\nabla\omega)(X_1,\cdots,X_{k+1}) = \sum_{\sigma \in S_{k+1}} (\sgn\sigma) \nabla\omega(X_{\sigma(1)},\cdots,X_{\sigma(k+1)}) \\ &= \sum_{\sigma \in S_{k+1}} (\sgn\sigma) \nabla_{X_{\sigma(k+1)}}\omega(X_{\sigma(1)},\cdots,X_{\sigma(k)}) \\ &= \sum_{\sigma \in S_{k+1}}(\sgn\sigma)\left(X_{\sigma(k+1)}(\omega(X_{\sigma(1)},\cdots,X_{\sigma(k)})) – \sum_{i=1}^n\omega(X_{\sigma(1)},\cdots, \nabla_{X_{\sigma(k+1)}}X_{\sigma(i)},\cdots,X_{\sigma(k)})\right)\\ &= \sum_{\sigma \in S_{k+1}}(\sgn\sigma)X_{\sigma(k+1)}(\omega(X_{\sigma(1)},\cdots,X_{\sigma(k)})) – \sum_{\sigma \in S_{k+1}}\sum_{i=1}^n(\sgn \sigma)\omega(X_{\sigma(1)},\cdots, \nabla_{X_{\sigma(k+1)}}X_{\sigma(i)},\cdots,X_{\sigma(k)}) \\ &\stackrel{\color{red}{(\ast)}}{=} \sum_{\sigma \in S_{k+1}}X_{\sigma(k+1)}(\omega(X_1,\cdots,X_k)) – \sum_{\sigma \in S_{k+1}}\sum_{i=1}^n(\sgn \sigma)\omega(X_{\sigma(1)},\cdots, \nabla_{X_{\sigma(k+1)}}X_{\sigma(i)},\cdots,X_{\sigma(k)})\end{align}$$
and I'm stuck. I'm not sure of the $\color{red}{(\ast)}$ step either. I know that we must use that $\nabla$ is symmetric to get rid of all these $\nabla$ but I don't know how to do this. What is the smart way to do this?
This answer is the most similar thing to what I'm trying to do that I found, but I don't find it easy to see such relation, as it is said there. Also, I'd like to avoid coordinate computations if possible.
I'll ignore the $\color{red}{(\ast)}$ step since I think it is wrong. I don't know how to write it neatly, but doing it for $k=1$ and $k=2$ suggests $$\sum_{\sigma \in S_{k+1}}(\sgn \sigma)X_{\sigma(k+1)}(\omega(X_{\sigma(1)},\cdots,X_{\sigma(k)})) = (-1)^kk!\sum_{i=1}^k(-1)^iX_i(\omega(X_1,\cdots,\widehat{X_i},\cdots,X_{k+1})).$$Also, it seems that $$\sum_{\sigma \in S_{k+1}}\sum_{i=1}^k(\sgn \sigma)\omega(X_{\sigma(1)},\cdots, \nabla_{X_{\sigma(k+1)}}X_{\sigma(i)},\cdots,X_{\sigma(k)})=(-1)^{k+1}k!\sum_{i<j}(-1)^{i+j}\omega([X_i,X_j],X_1,\cdots,\widehat{X_i},\cdots,\widehat{X_j},\cdots,X_{k+1})$$
So: $$(k+1)! {\rm Alt}(\nabla \omega) = (-1)^kk!{\rm d}\omega \implies (-1)^k(k+1)\alt(\nabla \omega) = {\rm d}\omega.$$
Best Answer
Three comments: