If the vector field $\mathbf B$ on $\mathbb{R}^3$ is constant, then the vector field $$ \mathbf A = \frac 1 2 \mathbf B \times \mathbf r $$ satisfies $$ \nabla \times \mathbf A = \mathbf B. $$
This can be proven by writing out in components. I would like to phrase this identity and prove it in differential form language (I am not sure if this is possible, but it's worth a try). I have tried the following:
- $\mathbf A$ becomes a 1-form $A$
- $\mathbf B$ becomes a 2-form $\mathcal B$
- $\nabla \times \mathbf A = \mathbf B $ becomes $\mathcal B = dA $
How to continue? I realize I will need to put hodge duals in appropriate places to realize the cross product.
@Ted Shifrin mentioned in the comments that this translation is possible using Geometric Algebra. That's true, but also shows how to further transform this relation to one with differential forms.
With
$$I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3,$$
and the identity $ \mathbf{x} \times \mathbf{y} = I (\mathbf{x} \wedge \mathbf{y}) $, the pair of identities becomes
$$I \mathbf{A} = \frac{1}{{2}} \mathbf{B} \wedge \mathbf{r}$$ $$\boldsymbol{\nabla} \wedge \mathbf{A} = I \mathbf{B}.$$
Introducing a bivector dual $ \boldsymbol{\mathcal{B}} = -I \mathbf{B} $, these can be written
$$\mathbf{A} = \frac{1}{{2}} \boldsymbol{\mathcal{B}} \cdot \mathbf{r}$$ $$\boldsymbol{\nabla} \wedge \mathbf{A} = -\boldsymbol{\mathcal{B}}, \qquad (1)$$
As with the coordinate expansion of the original 3D vector result, this can be verified by direct expansion. Assuming an antisymmetric scalar $ B_{ab} = -B_{ba} $, the bivector coordinate expansion is
$$\boldsymbol{\mathcal{B}} = \frac{1}{{2}} B_{ab} \mathbf{e}_a \wedge \mathbf{e}_b.$$
Expanding out (1) gives
$$\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{A}&=\mathbf{e}_i \partial_i \wedge \left( \frac{1}{{4}} B_{ab} (\mathbf{e}_a \wedge \mathbf{e}_b) \cdot (\mathbf{e}_s x_s)\right) \\ &=\frac{1}{{4}} B_{ab}\mathbf{e}_i \wedge \left( (\mathbf{e}_a \wedge \mathbf{e}_b) \cdot (\mathbf{e}_s \delta_{i s})\right) \\ &=\frac{1}{{4}}B_{ab}\mathbf{e}_i \wedge \left( (\mathbf{e}_a \wedge \mathbf{e}_b) \cdot \mathbf{e}_i\right) \\ &=\frac{1}{{4}}B_{ab}\mathbf{e}_i \wedge \left( \mathbf{e}_a \delta_{b i} - \mathbf{e}_b \delta_{a i}\right) \\ &=\frac{1}{{4}}B_{ab}\left( \mathbf{e}_b \wedge \mathbf{e}_a - \mathbf{e}_a \wedge \mathbf{e}_b\right) \\ &=-\frac{1}{{2}} B_{ab} \mathbf{e}_a \wedge \mathbf{e}_b \\ &= -\boldsymbol{\mathcal{B}}\end{aligned}$$
Mapping the Geometric Algebra form to differential forms should (I believe) just require using differentials as the basis, and using the $ d $ operator in place of the bivector curl $ \boldsymbol{\nabla} \wedge $. With
$$\begin{aligned}\boldsymbol{\mathcal{B}} &= \frac{1}{{2}} B_{ab} dx_a \wedge dx_b \\ \mathbf{r} &= x_a dx_a,\end{aligned}$$
that is
$$\begin{aligned}\mathbf{A} &= \frac{1}{{2}} \boldsymbol{\mathcal{B}} \cdot \mathbf{r} \\ d \mathbf{A} &= -\boldsymbol{\mathcal{B}}.\qquad (2) \end{aligned}$$
A verification by coordinate expansion appears to give the right results. First, explicitly expanding the 1-form $\mathbf{A}$:
$$\begin{aligned}\mathbf{A} &= \frac{1}{{2}} \boldsymbol{\mathcal{B}} \cdot \mathbf{r} \\ &=\frac{1}{{4}} B_{ab} (dx_a \wedge dx_b) \cdot (dx_s x_s) \\ &=\frac{1}{{4}} B_{ab} x_s ( dx_a \delta_{bs} - dx_b \delta_{as} ) \\ &=\frac{1}{{4}} B_{ab} (x_b dx_a - x_a dx_b).\end{aligned}$$
Application of the differential operator gives
$$\begin{aligned}d\mathbf{A} &= \frac{1}{{4}} B_{ab} d(x_b dx_a - x_a dx_b) \\ &=\frac{1}{{4}} B_{ab} (dx_b \wedge dx_a - dx_a \wedge dx_b) \\ &=-\frac{1}{{2}} B_{ab} dx_a \wedge dx_b \\ &=-\boldsymbol{\mathcal{B}}.\end{aligned}$$
While the 2-form $\boldsymbol{\mathcal{B}}$ can be related to a 1-form using the hodge-star operator, that doesn't really appear required for the identity (2) to be valid.