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4

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 = ...

3

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 ...

0

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 ...

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