Single-valuedness of the monopole harmonics

Question

It is known that it is impossible to have a non-singular vector potential for the monopole magnetic field.

How about the monopole harmonics?

Could the wave functions be made single-valued and finite on the sphere?

Answer 1

The monopole harmonics are proportional to the Rotation matrices $$ D^J_{mm'}(\theta, \phi, \psi) $$ where $\theta,\phi, \psi$ are the Euler angles $0<\theta<\pi$, $0<\phi<2\pi$, $0<\psi<4\pi$ describing the rotation via $$ D^J_{mm'}(\theta, \phi, \psi)= \langle J,m| \exp\{-i\phi J_3\} \exp\{-i\theta J_2\} \exp\{-i\psi J_3\} |J,m'\rangle $$ When $m'=0$ we have $$ Y^{L}_m(\theta, \phi) \propto [D^L_{m0}(\theta, \phi, \psi)]^* $$ which does not depend on the third Euler angle. These are the usual spherical harmonics.

If the monopole harmonics are to be single valued on the two sphere parameterised by the $\theta$ and $\phi$, we must make a smooth choice a value of $\psi$ for each pair $\theta,\phi$. (This is our choice of gauge.) However we cannot do this because the map $$ {\rm Hopf}: (\theta,\phi,\psi)\mapsto (\theta, \phi), $$ the Hopf map from the three sphere to the two sphere is topologically non trivial. There will always be a point on the two sphere where one is forced to have $\psi$ wind through $4 \pi$ as you circle close to the the point.