When does zero divergence imply a vector potential exists?

From electrodynamics we know that $$\boldsymbol{\nabla}\mathbf{B}=\mathbf 0$$ hence we can introduce a vector potential such that $$\mathbf{B}=[\boldsymbol \nabla\times \mathbf{A}]$$.

What is the general mathematical proof that a vector field with zero divergence can be expressed as a curl of another vector field (that this field would necessarily exist)? What are the restrictions on the fields?

I suspect this question was asked before, I just didn't manage to find a clean answer.

• The answer is topological and basically given by de Rham’s theorem: zero divergence on a particular domain gives a vector potential on that domain if and only if the second homology of the domain is trivial. See also math.stackexchange.com/questions/3865242/… Oct 9, 2021 at 19:17
• Is ther an accessible proof I can read? I don't know much about topology, my background is in physics with standard calculus, vector analysis etc Oct 9, 2021 at 19:19
• The proof of the “if and only if” is a deep result and quite difficult. Poincaré’s lemma gives a sufficient condition and is easier to prove but is not as strong. You might wish to glance at the American Mathematical Monthly paper referenced in the answer to the question I linked to above. Oct 9, 2021 at 19:20
• Another way to put it is to identify sufficient conditions for$$\vec{\nabla}\cdot\vec{B}=0,\,\vec{A}\left(t,\,\vec{r}\right):=\frac{1}{4\pi}\int\frac{\vec{\nabla}\times\vec{B}\left(\vec{r}^\prime\right)}{\left|\vec{r}-\vec{r}^\prime\right|}d^3\vec{r}^\prime\implies\vec{\nabla}\times\vec{A}=\vec{B}.$$
– J.G.
Oct 9, 2021 at 20:08
• Let me also be more precise: my comment above elides some quantifier subtleties. On a domain $D$, consider the property $P$ that “every divergence-free field on $D$ admits a vector potential on $D$.” There is a theorem (consequence of de Rham) that says $D$ has property $P$ if and only if it has trivial second homology. In the case that the second homology isn’t trivial for a particular $D$, this only means that some divergence-free fields won’t have vector potentials; others will. So for a particular field on a nontrivial $D$, it could still go either way. Oct 10, 2021 at 7:48

One of them being: a divergence-less $$[\nabla.\vec{X}=0]$$ vector field should wind upon itself, or simply be solenoidal $$[\vec{X} \text{ is curl of some other field}\implies \vec{X}=\nabla\times \vec{Y}]$$ since $$\forall \vec{Y}\,\nabla . (\nabla \times \vec{Y})=0$$. (Unlike many other identities in vector analysis, this one's quite easy to visualize.)
And since EM deals with vector fields representing E and B in space, the divergence-lessness of $$\vec{B}$$ gives its solenoidal character, being $$\vec{B}=\nabla \times \vec{A}$$