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.

Thanks in advance for help!

  • $\begingroup$ Is $\omega(v_1,\dots,v_{n-1}):= det[v_1|\dots|v_{n-1}|X]$ ? and is it rather $d\omega=(-1)^{n-1} (divX)dv_1 \wedge\dots\wedge dv_n$ ? $\endgroup$ – PilouPili Oct 20 '18 at 13:07
  • $\begingroup$ @NPE No, I've written whatever is in the exercises. $\endgroup$ – reflexive Oct 20 '18 at 13:10
  • $\begingroup$ It must have something to do with $dw_p(v_1,...,v_{n-1})=(-1)^{n-1}\sum_{j=1}^n d(X_p)_j det[v_k, k\neq j] = (-1)^{n-1}\sum_{j=1}^n \sum_{i=1}^n \frac{\partial (X_p)_j}{\partial x_i} dx_i det[v_k, k\neq j]$ $\endgroup$ – PilouPili Oct 20 '18 at 13:43
  • $\begingroup$ @NPE would you please consider writing an answer? $\endgroup$ – reflexive Oct 20 '18 at 13:44
  • 1
    $\begingroup$ 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 $\endgroup$ – 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, $$ d\omega =\frac{\partial }{\partial x_i} X_i (-1)^{n-1} dx_1\wedge \cdots \wedge dx_n $$


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