In three dimensions, a vector potential for a vector field $\mathbf F$ is a vector field $\mathbf A$ such that

$$\mathbf F=\nabla\times\mathbf A.$$

Vector potentials play for divergence-free fields a role analogous to scalar potentials for curl-free fields, but the global existence conditions are different.

Every curl is divergence-free

Let

$$\mathbf A=(A_x,A_y,A_z).$$

Then

$$\nabla\times\mathbf A

\left( \frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}, \frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}, \frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y} \right).$$

Taking the divergence gives pairs of mixed second derivatives with opposite signs. If the components have continuous second partial derivatives, the mixed derivatives agree and cancel, so

$$\boxed{\nabla\cdot(\nabla\times\mathbf A)=0}.$$

Therefore a field that has a smooth vector potential must be divergence-free.

When a divergence-free field has a vector potential

The converse requires conditions on the domain. On a suitable contractible region—one that can be continuously shrunk to a point without leaving the region—a sufficiently smooth divergence-free field admits a vector potential:

$$\nabla\cdot\mathbf F=0 \quad\Longrightarrow\quad \mathbf F=\nabla\times\mathbf A.$$

The potential $\mathbf A$ is not unique. Later applications can exploit this freedom to choose a particularly convenient representation.

As with scalar potentials, the domain condition is not cosmetic. Local differential information does not always determine global behavior when the domain contains a topological obstruction.

A divergence-free field with no global vector potential

Consider the field

$$\mathbf F(\mathbf r)=\frac{\mathbf r}{\lVert\mathbf r\rVert^3}$$

on $\mathbb R^3\setminus{\mathbf0}$. Away from the missing origin,

$$\nabla\cdot\mathbf F=0.$$

Take a sphere of radius $R$ centered at the origin. On the sphere,

$$\mathbf F=\frac{1}{R^2}\mathbf n,$$

where $\mathbf n$ is the outward unit normal. Its outward flux is therefore

$$\iint_S\mathbf F\cdot\mathbf n,dS

\frac{1}{R^2}\iint_S dS

\frac{1}{R^2}(4\pi R^2) =4\pi.$$

Suppose a single smooth vector potential $\mathbf A$ existed on the whole punctured domain with

$$\mathbf F=\nabla\times\mathbf A.$$

Applying Stokes' theorem to the closed sphere would give

$$\iint_S(\nabla\times\mathbf A)\cdot\mathbf n,dS

\oint_{\partial S}\mathbf A\cdot d\mathbf r =0,$$

because a closed sphere has no boundary. But the same flux is $4\pi$, a contradiction.

So the field is divergence-free everywhere in its domain but has no single globally defined smooth vector potential there. The missing point creates the obstruction.

Local versus global vector potentials

A smooth divergence-free field can still admit vector potentials on sufficiently small uncomplicated neighborhoods even when a global one fails to exist. The distinction is therefore the same broad lesson seen with scalar potentials: differential equations such as zero divergence are local, while the existence of one potential over an entire region can depend on topology.