Does a classical charge naturally have spin in a non-flat FLRW universe? I was recently reading a paper https://arxiv.org/abs/0912.0225 which describes Coulomb's law in a closed universe. I've seen the argument from several sources that the total electric charge in a closed universe must be zero (and the same formula has been derived elsewhere); however, I noticed that given the electric field of a test charge at the “North Pole” of the three-sphere:
$$\overrightarrow{E}=\frac{q}{4\pi\epsilon_{0}R^{2}sin^{2}(\chi)}\hat{\chi}$$
{Where we are using the fairly standard angular coordinates on a three-sphere universe ( $S^{3}$ with Radius R): $\chi,\theta,\phi$.} 
If we Consider someone someone measuring the field of this charge at some  distance $d_{0}$ (in say meters) $d_{0}=R\chi_{0}$
Then the electric field is:
$$\overrightarrow{E}_{R}(d_{0})=\frac{q}{4\pi\epsilon_{0}R^{2}sin^{2}(\frac{d_{0}}{R})}\hat{\chi}$$
Now let us double the radius of the universe, the same observer, the same distance $(d_{0}=2R(\frac{1}{2}\chi_{0}))$ from the same charge now measures the electric field:$$\overrightarrow{E}_{2R}(d_{0})=\frac{q}{4\pi\epsilon_{0}(2R)^{2}sin^{2}(\frac{d_{0}}{2R})}\hat{\chi}$$
Which is already strange in that we might claim (somewhat naively) that the charge appears to be different! But consider now  (letting the scale parameter R vary with time as is the case for our physical universe) that this must also be indicative of an induced magnetic field:
$$\overrightarrow{\Delta}\times\overrightarrow{B}=\mu_{0}\epsilon_{0}\frac{\partial\overrightarrow{E}}{\partial t}=\frac{q\mu_{0}}{4\pi}\frac{-1}{[(R^{2}sin^{2}(\frac{d_{0}}{R})]^{2}}\left[2Rsin^{2}(\frac{d_{0}}{R})-R^{2}2sin(\frac{d_{0}}{R})cos(\frac{d_{0}}{R})\frac{d_{0}}{R^{2}}\right]\frac{\partial R}{\partial t}\hat{\chi}$$
$$=\frac{-q\mu_{0}}{2\pi}\left[\frac{1}{R^{3}sin^{2}(\frac{d_{0}}{R})}-\frac{cos(\frac{d_{0}}{R})}{sin^{3}(\frac{d_{0}}{R})}\frac{d_{0}}{R^{4}}\right]\frac{\partial R}{\partial t}\hat{\chi}$$
Where $\frac{\partial R}{\partial t}$ can directly related to the Hubble factor. This seems to indicate that any charge would have an intrinsic magnetic moment in a closed expanding universe. 
Note there is no unique orientation for the magnetic moment either, if you were to pick a direction arbitrarily, it would satisfy the above equation.
Finally, it is worth noting that the Pauli matrices are a natural choice for the representation of $S^3$ (since $SU(2)$ is diffeomorphic to the three-sphere).
My question: Would these effects still persist in a more rigorous treatment? Is anyone aware of how to proceed? 
I thought it was really curious at the least and was wondering if anyone's seen a more rigorous way to go about this. I found this excellent paper, but wasn't sure how to apply it: http://www.numdam.org/item/AIHPA_1990__53_3_319_0
Edit:
It occured to me that the same type of effect should also happen in a negatively curved spacetime. Where difference in the initial equation would simply be $sin(\chi)\longrightarrow sinh(\chi)$. I have therefore edited the question title appropriately.
 A: No, the B field is zero everywhere. Rotational symmetry implies that the tangential components must be zero, and reflection symmetry implies that the radial component must be zero.
I think the error in your calculation is simply that $\nabla\times B = \partial E/\partial t$ is only valid in inertial coordinates.
It's interesting to look at a toy special-relativistic example. Put a spherically symmetric charge at the origin of Minkowski space and a field-measuring device moving radially away from it. The device will measure a constant zero B field, a time-varying E field, and therefore a nonzero curl of B, implying that B is nonzero nearby.
But put other radially moving devices near the first one and they'll all measure a zero B field, even though they're at a location where other nearby devices say the field should be nonzero.
The resolution of this paradox is that they're measuring the field with respect to different inertial frames. To measure the B field that another device says is there, they would have to accelerate to rest relative to the other device, which breaks the symmetry.
