I recently found myself curious what explicit formula I would get if I traced through the de Rham cohomology proof that if $\mathbf{F}$ is a vector field defined on all of $\mathbb{R}^3$ which has divergence 0, then $\mathbf{F}$ is the curl of some other vector field. So, tracing through the proof, what I did starting with such a divergence-free vector field was:
- Convert $\mathbf{F}$ to a 2-form.
- Pull back along the contraction homotopy $H : \mathbb{R}^3 \times [0, 1] \to \mathbb{R}^3$, $(x, y, z, t) \mapsto (tx, ty, tz)$.
- Apply the integral operator $\mathfrak{I}$ which intuitively sends $dt \wedge \omega \mapsto \int_{t=0}^{t=1} \omega \, dt$ and for other $\omega$ without a $dt$, only with $dx,dy,dz$, $\omega \mapsto 0$.
- Convert the resulting 1-form back to a vector field.
After working through this, the formula I eventually got could be summarized as: $$(\operatorname{anticurl} \mathbf{F})(\mathbf{x}) := \left[ \int_0^1 t \mathbf{F}(t \mathbf{x})\,dt \right] \times \mathbf{x}.$$ This has the expected property that $\operatorname{div} \mathbf{F} = 0 \implies \mathbf{F} = \operatorname{curl} (\operatorname{anticurl} \mathbf{F})$. (And trying it out on for example $\mathbf{F}(x, y, z) = (y^a z^b, 0, 0)$, I get $(\operatorname{anticurl} \mathbf{F})(x, y, z) = \frac{1}{a+b+2} (0, -y^a z^{b+1}, y^{a+1} z^b)$ which does have the expected curl.)
It is also interesting that this $\operatorname{anticurl}$ is $\mathbb{R}$-linear and in addition it respects rotations about axes through the origin.
So now, what I was curious about was: this explicit partial inverse of curl seems like something that very likely would have shown up before. If there's a standard name for it, that would be interesting to know, or otherwise it would be good to see an example usage of this formula in a textbook or other reference more on the introductory than research level.
