Vector potentials of a continuous field on an open subset of ${\mathbb{R}}^3$

Let $U\subseteq {\mathbb{R}}^3$ be open and let $B:U\to {\mathbb{R}}^3$ be continuous. Given a map $A:U\to {\mathbb{R}}^3$, we say $A$ is a vector potential for $B$ when $A$ is $C^1$ on $U$ and $\mathrm{curl}A=B$ at every point of $U$, with the curl of Divergence and curl of a $C^1$ vector field and the class $C^1$ of $C^k$ Euclidean maps and diffeomorphisms. A field admitting a vector potential is said to have a vector potential on $U$.

• This is the curl analogue of exactness, not the same notion. A field is exact when it is the gradient of a $C^2$ scalar (Exact and closed C1 vector fields); it has a vector potential when it is the curl of a $C^1$ field. The two conditions constrain a field in different ways: on an open subset of ${\mathbb{R}}^3$ a gradient of a $C^2$ function has vanishing curl and a curl of a $C^2$ field has vanishing divergence.

• Nonuniqueness on a nonempty domain. If $U$ is nonempty and $A$ is a vector potential for $B$, take the $C^2$ coordinate function $\varphi (x)=x_0$. Then $A+\mathrm{\nabla }\varphi =A+e_0$ is distinct from $A$, is $C^1$, and has the same curl by the linearity of curl (Divergence and curl are linear and satisfy the scalar product rules) and $\mathrm{curl}\mathrm{\nabla }\varphi =0$ (The curl of the gradient of a $C^2$ function vanishes). On the empty open set there is only one map to ${\mathbb{R}}^3$, so the nonempty hypothesis is essential to this remark.