Divergence for differential $n$-form

Question

I'm given that $$L_{\mathbb{X}} \omega=(\mathrm{div}_{\omega}\mathbb{X})\omega$$ and I need to show that for $\omega=dx^1 \wedge dx^2 \wedge \dots \wedge dx^n$, $$(\mathrm{div}_{\omega}\mathbb{X})\omega=\frac{\partial \mathbb{X}^i}{\partial x^i}\omega.$$

These are my initial thoughts:

\begin{align} (\mathrm{div}_{\omega}\mathbb{X})\omega & = L_{\mathbb{X}} \omega \\ & = L_{\mathbb{X}}(dx^1 \wedge dx^2 \wedge \dots \wedge dx^n) \\ & = L_{\mathbb{X}}dx^1 \wedge \dots\wedge L_{\mathbb{X}}dx^n \\ & = dL_{\mathbb{X}}x^1 \wedge \dots \wedge dL_{\mathbb{X}}x^n \\ & = d\mathbb{X}^1 \wedge \dots \wedge d\mathbb{X}^n \end{align}

I'm stuck at this point though. What do I need to do from here?

Answer 1

Write $X=(X_1,...,X_n)$, $L_X\omega=di_X\omega+i_Xd\omega=di_X\omega$.

$i_X\omega=\sum_{i=1}^{i=n}(-1)^{i+1}X_idx_1\wedge..\hat{dx_i}..\wedge dx_n$, this implies that:

$d(i_X\omega)=\sum_{i=1}^{i=n}(-1)^{i+1}(\sum_{j=1}^{j=n}{{\partial X_i}\over{\partial x_j}}dx_j)dx_1\wedge..\hat{dx_i}..\wedge dx_n$

$=\sum_{i=1}^{i=n}{{\partial X_i}\over{\partial x_i}}dx_1\wedge..\wedge dx_n=div(X)\omega$.