Magnetic monopole and vector potential Does anyone know how to prove (in a simple way if possible) that it is impossible to define a single-valued globally defined magnetic vector potential $\vec{A}$ on the manifold $M=\mathbb{R}^3\backslash\{0\}$ for a static magnetic monopole placed at the origin?
Hypothesis:  $\vec{B} = g \vec{r}/(4\pi r^3)$
 A: You seek a 1-form $A$ on $\mathbb{R} - \{0\}$ such that $\mathrm{d}A = B$. On all of $\mathbb{R} - \{0\}$, $\mathrm{d}B = 0$, so this could exist. But, since you have magnetic flux, you require that the integral of $B$ over any 2-sphere around the origin should be $g$. Therefore, by Stokes' theorem,
$$ g = \int_{S^2}B = \int_{S^2} \mathrm{d}A = \int_{\partial S^2} A = \int_{\emptyset} A = 0$$
which is a contradiction for $g \neq 0$. Therefore, such an $A$ cannot exist.
This is a more formal rephrasing of Holographer's answer.
A: The magnetic flux through any closed surface enclosing the origin is just $g$ (the magnetic charge enclosed). If the magnetic field comes from a vector potential $\vec{B}=\nabla\times\vec{A}$, this surface integral by Stokes' theorem is an integral of $\vec{A}$ around the boundary of the surface. But the surface is closed, so has no boundary, so the answer must vanish. This is a contradiction if $g$ is nonzero, so no such $\vec{A}$ exists.
(This has a more sophisticated interpretation in the language of differential forms and de Rham cohomology, but that's not really necessary here!)
A: user23873 answered my question in the comments. I quote: 
"
Try reading the book 'Geometry, Topology and Gauge Fields: Foundations', the author (Naber) have this discussion right on the introductory chapter and points how the impossibility to define a proper vector potential on ℝ3−0 is linked with it's topology (the second homotopy group is non-trivial), and also how classical dirac monopoles rise from this. Obs: Your hypothesis have non-zero flux integral, just like the equivalent solution for the electric point charge field.
"
