Can a gauge transformation eliminate singularity of gauge potential? Suppose I have a gauge potential $A$ which goes to infinity at some point $x_0$. Can I use a gauge transformation
\begin{equation}
A'=U^{-1}AU+U^{-1}dU,~~~U=\exp\{-i\alpha^a(x)T^a\}
\end{equation}
to kill the singularity, i.e, $A'$ is non-singular? I have seen somewhere that for the physical gauge transformation, the functions in $U$ should be non-singular and vanish at infinity. Is this right? If two gauge potentials differ by a singular $U$(i.e., $\alpha^a$'s are singular somewhere), are they "gauge-eqivalent"?
 A: We can identify two cases here.
Case 1: The gauge potential has a singularity which produces a singularity in the physical field too.
Given that the physical field is invariant under gauge transformations, it follows that you won't be able to cure this by making a gauge transformation. We can make a trivial example of this by considering a $U(1)$ theory in two dimensions. Take the gauge field to be $A=(x^{-1},0)$. The physical field is $F_{01}=(\mathrm d A)_{01}= x^{-2}$. These are both singular at the origin. I can try to remove the singularity in the $0$-component of $A$ by performing a gauge transformation $\Lambda=-tx^{-1}$. The new gauge field is $A+\mathrm d \Lambda=(0,tx^{-2})$. So I have only managed to move the singularity to a different component of the gauge field.
Case 2: The gauge potential has a singularity but the physical field is regular.
In this case there is no singular behavior in the physical field, so it should be possible to remove this behavior from the gauge field. Consider the gauge field $A=(x^{-1},-tx^{-2})$. This is singular at the origin, but the physical field is $F_{01}=(\mathrm d A)_{01}=0$. Under the same gauge transformation as above, we can obtain $A=0$.
Commentary: In general, singularities in physical fields are undesirable and indicate a deficiency in the theory, like for example the singular field of a point charge in classical EM or the singularity in a black hole in GR. The point charge in classical EM has both a singular electric field and a singular scalar potential, and there is nothing that can be done about this within the context of the classical theory. However, as long as we are only interested in phenomena sufficiently far away from the charge, there's no reason to worry about it. This is the situation in Case 1 above. In Case 2, the singular potential is artificial: this is really a vacuum solution. So really we should never have written a singular potential at all. Indeed, the potential I gave is not even really defined at the origin, but yet the physical field at the origin is perfectly normal and in fact just zero. But if for some reason someone gave you a singular potential for a non-singular field, you could undo it with a singular gauge transformation. I would view this as just "undoing" the improper step of writing down a singular potential to begin with.
