Why does Gauss's law for magnetism imply existence of a magnetic vector potential?

Question

Gauss's law for magnetism is $\vec{\nabla} \cdot \vec{B} = 0$, and I read that this implies we can write $\vec{B} = \vec{\nabla} \times \vec{A}$, for some vector potential $\vec{A}$.

I understand that having $\vec{B} = \vec{\nabla} \times \vec{A}$ implies the statement of Gauss's law. But I am having trouble proving the converse; that $\vec{\nabla} \cdot \vec{B} = 0$ implies the magnetic field must take the form $\vec{B} = \vec{\nabla} \times \vec{A}$.

My attempt: By the Helmholtz decomposition theorem (https://en.wikipedia.org/wiki/Helmholtz_decomposition), we can write any vector field $$\vec{B} = -\nabla \phi + \vec{\nabla} \times \vec{A}.$$ Then $$\vec{\nabla} \cdot \vec{B} = 0 \Rightarrow -\nabla^2 \phi = 0.$$ However, this does not imply that $\phi \equiv 0$. So why can't magnetic fields be described in terms of some vector potential $\vec{A}$ but also some scalar potential $\phi$ that has $0$ Laplacian?

I don't think this has anything to do with the existence of a magnetic monopole or not, since I'm taking $\vec{\nabla} \cdot \vec{B} = 0$, but I may also be wrong.

Answer 1

1. Use Hodge duality to rewrite the magnetic vector field $\vec{B}$ as a 2-form $B=\frac{1}{2}\epsilon_{ijk}B^i\mathrm{d}x^j\wedge\mathrm{d}x^k$.

2. Gauss's law for magnetism then becomes the fact that the 2-form is closed $\mathrm{d}B=0$.

3. It then follows from Poincare Lemma that the 2-form $B=\mathrm{d}A$ is exact, where $A=A_i\mathrm{d}x^i$ is a 1-form.

4. This translates back into the sought-for formula $\vec{B}=\vec{\nabla}\times \vec{A}$.

5. (Poincare Lemma fails for a magnetic monopole because the punctured 3-space $\mathbb{R}^3\backslash\{0\}$ is not contractible.)