# Show that $d\omega=(-1)^{n-1} (divX)dx_1 \wedge\dots\wedge dx_n$

Let, $$X = (X_1,\dots,X_n): D \subset \Bbb R^n \to \Bbb R^n$$ be a $$C^1$$ vectr field on a domain $$D$$. Define $$divX := \sum_{i=1}^n {\frac{\partial X_i}{\partial x_i}}$$. Define an $$(n-1)$$ form $$\omega$$ on $$D$$ by,$$\omega_p(v_1,\dots,v_{n-1}):= det[v_1|\dots|v_{n-1}|X_p] , \forall p \in D, v_1,\dots v_{n-1} \in \Bbb R^n$$ Show that $$d\omega=(-1)^{n-1} (divX)dx_1 \wedge\dots\wedge dx_n$$

My attempt:

I know how to compute the exterior derivative of a form when it is of the form :$$\omega = \sum_{I}{a_I dx_I}$$ i.e. to find,$$d\omega=\sum_{I}{da_I \wedge dx_I}$$. But I am unable to recognize the given form in that form.

• Is $\omega(v_1,\dots,v_{n-1}):= det[v_1|\dots|v_{n-1}|X]$ ? and is it rather $d\omega=(-1)^{n-1} (divX)dv_1 \wedge\dots\wedge dv_n$ ? – PilouPili Oct 20 '18 at 13:07
• It must have something to do with $dw_p(v_1,...,v_{n-1})=(-1)^{n-1}\sum_{j=1}^n d(X_p)_j det[v_k, k\neq j] = (-1)^{n-1}\sum_{j=1}^n \sum_{i=1}^n \frac{\partial (X_p)_j}{\partial x_i} dx_i det[v_k, k\neq j]$ – PilouPili Oct 20 '18 at 13:43
• Sorry the link between $dx_1\wedge…\wedge dx_n$ and $\sum_j dx_i det[v_k k\neq j]$ is too far fetch. Only trying to help, not answer – PilouPili Oct 20 '18 at 13:46
$$\omega(\frac{\partial }{\partial x_1},\cdots, \frac{\partial }{\partial x_{n-1}}, X)= X_n$$ so that $$\omega = (-1)^{n-i} X_i dx_1\wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n$$
That is, $$d\omega =\frac{\partial }{\partial x_i} X_i (-1)^{n-1} dx_1\wedge \cdots \wedge dx_n$$