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.

  • 2
    $\begingroup$ "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
    Commented Aug 8, 2017 at 12:10

4 Answers 4


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).

  • 1
    $\begingroup$ 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
    Commented Feb 1, 2018 at 0:03
  • 3
    $\begingroup$ @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. $\endgroup$ Commented Feb 1, 2018 at 15:03

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 $U(1)$.


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 $\mathcal{L} = -\frac{1}{4} F^{\mu\nu}F_{\mu\nu}$.


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 $\Lambda \in \text{Lie } U(1) \simeq \mathbb{R}$.


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 $R^1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.