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The exterior derivative of a scalar function is

$d f(x,y,z) = ( \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz )$

Am I correct in assuming then that

$d\left( F_x(x,y,z) e_x + F_y(x,y,z) e_y + F_z(x,y,z) e_z \right)$

would be

$\left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) dx \wedge dy + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) dz \wedge dx + \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) dy \wedge dz$

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2 Answers 2

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We only talk about exterior derivatives of differential $k$-forms, not vector fields. However, what we can do is the following: given a vector field $F: \Bbb{R}^3 \to \Bbb{R}^3$, $F = (F_x, F_y, F_z)$, we can consider the following one-form: \begin{align} \omega &= F_x \, dx + F_y \, dy + F_z \, dz \end{align} And yes, the exterior derivative of the one-form $\omega$ is indeed the thing you wrote down: \begin{align} d\omega &= \left(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y}\right) dx \wedge dy + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) dz \wedge dx + \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) dy \wedge dz \end{align}


Just some fun extra tidbits: if you know some vector calculus, the above expression probably looks pretty familiar, almost like the curl of $F$, though not quite. If you want to somehow get the curl of $F$ from here, you need to look at the "Hodge star" operator, which assigns to the above $2$-form $d\omega$ a certain $1$-form $\alpha$, namely \begin{align} \alpha &= \left(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y}\right) dz + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) dy + \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) dx \end{align} then from here, you can get a vector field, $G$, (pretty much by replacing $dx$ with $e_x$, $dy$ with $e_y$ and $dz$ with $e_z$), \begin{align} G:= \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) e_x + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) e_y + \left(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y}\right) e_z, \end{align} and this is precisely the curl of $F$

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  • $\begingroup$ Thank you for taking the time to answer me. Very helpful. $\endgroup$
    – R. Emery
    Jun 19, 2020 at 1:56
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$$d(F_xdx + F_ydy + F_zdz) = dF_x \wedge dx + dF_y \wedge dy + dF_z \wedge dz$$ and use that $$dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy + \frac{\partial F}{\partial z}dz$$ according to your first claim.

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  • $\begingroup$ Seems so obvious now. Thank you $\endgroup$
    – R. Emery
    Jun 19, 2020 at 2:18

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