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Disclaimer: This question is probably based on a misconception or flaw in the understanding.

Can we use the term $U(1)$ gauge invariance for the free electromagnetic field? Let me explain why I ask this question.

Response to ACuriousMind's query As far as I know, gauge invariance is another name for local invariance, and free electromagnetic field is not a local gauge theory but QED is (I may be wrong!). In QED, where there is a fermion field $\psi(x)$, we demand local gauge invariance as $\psi(x)\to e^{i\theta(x)}\psi(x)$ where $e^{i\theta(x)}\in U(1)$. In case of a free electromagnetic field, I do not see any trace of the group $U(1)$. All I know is that the free Lagrangian is invariant under $A_\mu\to A_\mu+\partial_\mu\Lambda(x)$ but I fail to see any $U(1)$ group transformation property or any U(1) group element associated with this.

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    $\begingroup$ "free electromagnetic field is not a local gauge theory" - why do you say this? It is precisely the free Yang-Mills theory for the case of $\mathrm{U}(1)$ as the gauge group. $\endgroup$ – ACuriousMind Aug 8 '17 at 12:10
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OP has a point. The field $A^\mu$ is a connection, and therefore it lives in the algebra of the gauge group, not in the group itself. In this case, $\mathfrak u(1)=\mathbb R$. At first sight, this is all we may conclude from $A\to A+\mathrm d\Lambda$. The group $\mathrm U(1)$ is, apparently, not here yet.

The correct statement is that the theory described by $A^\mu$ has a $\mathfrak{u}(1)$ gauge symmetry. By exponentiation, we may get either $\mathrm U(1)$ or its universal cover, $\mathbb R$. Which of these groups is the "correct" gauge group depends on the global properties of $A$, which are not fixed by the algebra. Instead, these are fixed by the system under consideration: some $\mathfrak u(1)$ theories exponentiate to $\mathrm U(1)$ and some others to $\mathbb R$. And which one of these is the correct group can only be discerned from the physics of the problem under consideration.

In the case of YM+matter, the correct option is $\mathrm U(1)$ (because we demand $\psi$ to be single-valued). In some other systems (such as the theory of the fractional Hall effect), the algebra $\mathfrak u(1)$ actually exponentiates to $\mathbb R$. In general terms, there is not a single option: both are in principle valid. In this sense, it is better to say that free electromagnetism is the theory of a $\mathfrak u(1)$ gauge symmetry (which does not necessarily correspond to $\mathrm U(1)$, but it may correspond to $\mathbb R$ instead).

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  • $\begingroup$ Is the reason that the correct option is $\mathrm{U}(1)$ really that we want $\psi$ to be single-valued, or that we want charge to be quantized as observed in nature? Please correct me if I'm wrong, but my understanding is that in the absence of magnetic monopoles, QED with a non-compact gauge group $\mathbb{R}$ has single-valued wavefunctions but with arbitrary possible values for the electric charge. $\endgroup$ – tparker Feb 1 '18 at 0:03
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    $\begingroup$ @tparker I actually had that same question not-so-long ago: Are U(1) charges quantised?. It turns out that $\mathrm U(1)$ does not, by itself, lead to quantisation of charge. You need either to embed your theory into a larger (and simple) gauge group, or have a sort of Kaluza-Klein mechanism. $\endgroup$ – AccidentalFourierTransform Feb 1 '18 at 15:03
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You can see the $U(1)$ transformation for the free electromagnetic field in the Wilson loops (holonomies): With $A'_\mu = A_\mu+\partial_\mu\Lambda(x)$

$$e^{i \int_{x_1}^{x_2} A'_\mu dx^\mu } = e^{i \int_{x_1}^{x_2} A_\mu dx^\mu } e^{i\Lambda(x_2)} e^{-i\Lambda(x_1)}$$

The holonomies of a principal bundle, in general, reflect the structure group of the bundle. When the path is closed, the above formula shows zero holonomy.

However, when a magnetic flux flows inside the loop, then $d\Lambda$ will not be exact but closed there will be a net phase after a full rotation. When the space time manifold has nontrivial toplogy (non-vanihing fundamental group). These holonomy factors can be measured and they belong to the group $U(1)$.

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Yes we can; the theory still enjoys local invariance.

In covariant EM you have the field strength tensor $F^{\mu\nu}$ defined in terms of the potential $A^\mu$ as $$ F^{\mu\nu} = \partial^\mu A^{\nu} - \partial^{\nu}A^{\mu} $$ that is invariant under the transformation $A^{\mu} \rightarrow A^{\mu} + \partial^\mu \Lambda$ with $\Lambda(x)$ any local function.

This implies the invariance of the lagrangian $\mathcal{L} = -\frac{1}{4} F^{\mu\nu}F_{\mu\nu}$.

Addendum:

We saw that the invariance is encoded in a function "with no internal indices" (and no spacetime indices as well, but this is unimportant). If we make the transformation $\Lambda(x)$ a rigid transformation, $\Lambda$ is just a (real) constant. So, rigid transformations are parametrized by real numbers, and $\mathbb{R}$ is the Lie algebra of the group $U(1)$.

We can see see the connection with $U(1)$ in a different way. For reasons one wants the gauge symmetry group to be compact, so the symmetry group must be (isomorphic to) $\mathbb{R}/Z$ where $Z$ is some discrete subgroup of $\mathbb{R}$; all these quotients are isomorphic to $U(1)$.

Then, one can study how matter fields can possibly transform under gauge transformation by studying the representations of $U(1)$; it turns out that indeed they are all of the form $\exp(i n \Lambda)$ with integer $n$ and real $\Lambda \in \text{Lie } U(1) \simeq \mathbb{R}$.

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We have the same notion of gauge invariance in both theories, i.e EM and QED. In both, gauge transformation is given by a real function ($\theta (x)$ in QED and $\Lambda (x)$ in EM). In QED, We talk about U(1) symmetry because this real function appears as an arbitrary phase for the wave function $\psi (x) $ and so the geometry of the symmetry group is $S^1$. However, in EM the geometry of gauge symmetry is a real line. In principle, topologically there is no difference between $S^1$ and $R^1$.

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