Consider the field in $\mathbb R^3$ given by $X = -ye_1 + xe_2 + e_3$. And consider the form given by $(w_X)_{p} (v) = \langle X(p),v\rangle$. I want to compute the exterior derivative $dw_X$.
My attempt:
Using the definition we have that
$$ (w_X)_{p} (v) = \langle X(p),v\rangle = -y\langle e_1,v\rangle + x \langle e_2,v\rangle + \langle e_3,v\rangle$$
As $\langle e_j,v\rangle = v_j$, for $v = (v_1,v_2,v_3)$, we can write
$$ (w_X)_{p} = -y \,dx + x\, dy + dz $$
And so, $$ d w_X = -dy \wedge dx + dx \wedge dy + 1 \wedge dz $$
And then,should it be $d w_X = 2 dx \wedge dy$, since $dx \wedge dy = - dy \wedge dx$?
My questions are: Is this correct? And if it's why $1 \wedge dy = 0$ and not $dz$?
Thank you.
By duality we have $w_X = -y\,{\rm d}x + x\,{\rm d}y + {\rm d}z$, ok. Then $${\rm d}w_X = -{\rm d}y \wedge {\rm d}x + {\rm d}x\wedge {\rm d}y,$$and there's nothing with ${\rm d}z$, because ${\rm d}({\rm d}z) = 0$. You can even think that $${\rm d}z = 1\,{\rm d}z \implies {\rm d}({\rm d}z) = {\rm d}(1)\wedge {\rm d}z = 0 \wedge {\rm d}z = 0,$$ since $1$ is just a constant. Since we're dealing with one-forms, it results that ${\rm d}w_X = 2\,{\rm d}x\wedge {\rm d}y$. Recall that in general, if $\omega = \sum_{i=1}^3 f_i\,{\rm d}x^i$, then $${\rm d}\omega = \sum_{i=1}^3{\rm d}f_i \wedge {\rm d}x^i,$$where you have that ${\rm d}f_i$ is $${\rm d}f_i = \sum_{j=1}^3 \frac{\partial f_i}{\partial x^j}\,{\rm d}x^j,$$just like you learnt in early calculus courses.