Consider an inverse- square vector field
$$ \vec{F} = \frac{x}{r^3}\hat{x} + \frac{y}{r^3}\hat{y} + \frac{z}{r^3}\hat{z} = \frac{\hat{r}}{r^2}$$
where $r = \sqrt{x^2 + y^2 + z^2}$. The curl $\nabla \times \vec{F} = \vec{0}$, therefore we might go looking for a potential $V$. I find that $V = -1/r$ works and therefore one can say that $\vec{F} = \nabla V$ is derivable from a potential function $V$. I'll point out right now that $\vec{F}$ is undefined at the origin.
The divergence $\nabla \cdot \vec{F} = 0$. Therefore, we might go looking for a vector potential $\vec{A}$ such that $\vec{F} = \nabla \times \vec{A}$. One would say that $\vec{F}$ is derivable from a vector potential $\vec{A}$. But I'm having trouble seeing that an inverse-square vector field is derivable from both a vector potential and a scalar potential. So I know we have a trouble point at the origin. Yet this trouble point doesn't really seem to affect the "conservativeness" or path-independence of the vector field. But this trouble point does seem to affect the surface-independence of the vector field. As long as the surface doesn't wrap around the origin, I'd expect the inverse-square vector field to be surface-independent for a given boundary curve.
Can an inverse-square vector field be derivable from both a scalar potential and a separate a vector potential? (Helmholtz theorem comes to mind. But the question I'm asking involves two separate equations. One $\nabla V$ gives $\vec{F}$ and another $\nabla \times \vec{A}$ gives $\vec{F}$ as well).
Best Answer
If all you want is an inverse-square vector field in a topologically trivial region, then the Dirac string vector potential does the trick: $$ \vec{A} = \frac{1 - \cos \theta}{r \sin \theta} \hat{\phi} = \frac{\tan (\theta/2)}{r} \hat{\phi}. $$ It is not too hard to show that the curl of this vector field is $$ \vec{\nabla} \times \vec{A} = \frac{1}{r^2} \hat{r}. $$ However, $\vec{A}$ is ill-defined when $\theta = \pi$, which means that it can't be extended to all of space. As noted in my previous answer, there are topological obstructions to extending any such vector field over all of space.