Does classical electrodynamics have $U(1)$ symmetry? If yes, how? Quantum electrodynamics (QED) is based on $U(1)$ symmetry. What happens to this symmetry in classical electrodynamics?
Addendum The books on classical electrodynamics such as J. D. Jackson, does not mention about $U(1)$ symmetry in the context of gauge invariance (as far as I know). Gauge invariance is simply understood, in classical electrodynamics books, as the invariance of Maxwell's equations under $A_\mu\to A_\mu+\partial_\mu\chi(x)$. There is no sign of U(1) invariance that I can discover here. On the other hand, when something like Dirac equation or Dirac field is brought into the scene then the implementation U(1) transformation is clear. But that is always discussed in quantum field theory books. It appears that it is essential to have a Dirac field to understand U(1) symmetry. So the question is whether it is possible to understand the existence of U(1) symmetry in classical electrodynamics without bringing in the Dirac field into the picture?
 A: *

*A free "$\mathrm{U}(1)$" gauge theory can never tell whether the gauge group is $\mathrm{U}(1)$ or $\mathbb{R}$ because the only field in the theory, the gauge potential $A$, transforms as
$$ A\mapsto A + \partial_\mu \chi,$$ 
where $\chi$ is just a real-valued function, and the real numbers are the Lie algebra of both $\mathrm{U}(1)$ and $\mathbb{R}$. This is not a classical or quantum property, you simply cannot tell the difference. So, in a sense, asking whether this theory has $\mathrm{U}(1)$ symmetry or not is meaningless - it has $\mathfrak{u}(1)$ symmetry, and there is no meaningful notion of the symmetry group.

*Electromagnetism coupled to an external conserved current still cannot tell what the gauge group is, since the current is gauge-invariant.

*Electromagnetism coupled to other fields can tell what the gauge group is, since part of coupling it to other fields is specifying how these fields transform under gauge transformations. There we have a choice between (infinitesimally) $\psi \mapsto \psi + \chi \psi$ and $\psi\mapsto \psi + \mathrm{i}\chi \psi$, which lead to finite transformations $\psi\mapsto \mathrm{e}^{\chi}\psi$ and $\psi\mapsto \mathrm{e}^{\mathrm{i}\chi}\psi$, respectively. The former corresponds to a gauge group $\mathbb{R}$, the latter to $\mathrm{U}(1)$. Again, none of this is classical or quantum.
The reason you likely think that the $\mathrm{U}(1)$ is a quantum feature is that it much more natural in quantum field theory than in classical field theory to have complex-valued fields, but in fact we can consider e.g. classical electromagnetism coupled to a classical complex scalar field and then we are likewise forced to specify the gauge group.
A: Using compact notation of differential forms, the differential operator $D=d+A$ is a covariant operator with the gauge potential $1$-form $A$. The field $2$-form is $F~=~D\wedge D$ $=~(d~+~A)\wedge(d~+~A)$. This acting on a unit with $d\wedge d~=~0$ (boundary of boundary is 0) gives 
$$
F~=~dA~+~A\wedge A,
$$
The $U(1)$ group simply means that $A\wedge A~=~0$. We of course know that a $1$-form in an exterior product with itself is zero. However, if there are internal color or charge indices you can have a number of these. In $SU(2)$ there are $3$ gauge field and in $SU(3)$ there are $8$. With $U(1)$ there is only one gauge potential with no charge index.
The field $2$-form has components $F_{\mu\nu}$ and with some effort you can show that
$$
F_{0i}~=~\frac{\partial A_i}{\partial t}~-~\frac{\partial A_0}{\partial x^i}~=~E_i
$$
$$
F_{ij}~=~\frac{\partial A_i}{\partial x_j}~-~\frac{\partial A_j}{\partial x^i}~\rightarrow~B_i~=~-(\nabla\times A)_i.
$$
These are standard electromagnetic calculations. This would all be very different if there were the nonlinear terms $A\wedge A$ included. So classical electromagnetism is abelian.
