dextercioby said:
This is something that just came to me: what is "i" doing there? Just because it exists, one obtains U(1) by exponentiation, but actually, without the "i", it is simply \mathbb{R}. I know that the "i" is requested by passage to quantum mechanics or QFT, but, from a purely mathematical standpoint, and also from the point of view of classical physics, the gauge symmetry should be simply \mathbb{R}. Actually, mathematicians do not put the "i" in the exponential which links an element in the vecinity of the identity of a Lie group to the element of the Lie algebra, only physicists do, as per the Stone's theorem which is fundamental in the implementation of symmetry groups in quantum physics.
1) I am
free to write \delta A_{\mu} = -i \partial_{\mu}(i\Lambda).
2) If I write e^{isG} , \ s \in \mathbb{R}, then it should be clear to you that G \in \mbox{T}_{e}\left(U(1)\right) \cong \mathfrak{u}(1), i.e., the infinitesimal generator of the group of unitary transformation U(1). Now, If I can define/find such generator G in the (free) Maxwell theory, then my job is done. Well, if you know that Maxwell theory is a
constraint system, then you should also know that G exists and it is called
Gauss’ law, the generator of gauge transformation. Here is how one can find it. Construct the following
conserved current J_{\Lambda}^{\mu} = - F^{\mu\nu}\partial_{\nu} \Lambda \ , and consider its integrated
charge G_{\Lambda} = \int_{\mathbb{R}^{3}} d^{3}x \ J_{\Lambda}^{0} = \int_{\mathbb{R}^{3}} d^{3}x \ \pi^{\nu}(x) \partial_{\nu}\Lambda (x) \ , where \pi^{\nu} (x) \equiv \frac{\delta}{\delta (\partial_{0}A_{\mu})} \left( - \frac{1}{4} \int_{\mathbb{R}^{3}} d^{3}x \ F^{2} \right) = - F^{0 \nu}(x) , and \pi^{0}(x) \approx 0 is the
primary constraint. Now, we first observe that G_{\Lambda} generates an
Abelian Lie algebra (via
Poisson bracket) \{ G_{\Lambda} , G_{\Omega} \} = 0 \ . The second, since the current J_{\Lambda}^{\mu} is conserved, its charge G_{\Lambda} must generate infinitesimal
symmetry transformations on the coordinate pair (A_{\mu} , \pi^{\mu}). Indeed, it is an easy exercise to show that G_{\Lambda} does generate the correct gauge transformations: \{ G_{\Lambda} , \pi^{\mu}(x) \} = 0 = \delta \pi^{\mu} \ , \{ G_{\Lambda} , A_{\mu}(x) \} = \frac{\delta G_{\Lambda}}{\delta \pi^{\mu}(x)} = \partial_{\mu}\Lambda = \delta A_{\mu}\ . The last equation, can be rewritten as \delta A_{\mu} = -i \frac{d}{ds} \{ e^{isG_{\Lambda}} , A_{\mu}(x) \} |_{s = 0} , \ \ e^{isG_{\Lambda}} \in U(1) \ .
3) Since \Lambda (x) is a Lie algebra
element taking values in \mathbb{R}^{(1,3)}, your “\mathbb{R}” must be the 1-dimensional
Lie algebra \mathbb{R}. Now, you are stuck because, as a
Lie algebra, \mathbb{R} has
many locally
isomorphic Lie groups. Indeed, it is the Lie algebra of the following four (among many more) Lie groups: U(1) = \{ z \in \mathbb{C} : \ |z| = 1 \}, the positive reals ( \mathbb{R}^{+} , \times ), the multiplicative group of \mathbb{R} with two connected components, also known as, the non-zero reals (\mathbb{R}^{\times} , \times ) and the additive group ( \mathbb{R} , +). So, tell me, which one of those four (locally isomorphic) groups is “your” gauge group and why?
4) In the presence of a point source, if you take the gauge group to be \mathbb{R}, you end up violating the Dirac charge quantization e \ g = 2 \pi n. Indeed, in this case one can compute the de Rham cohomology groups to be H^{(p \ = \ 0 , 2)}( \mathbb{R}^{4} - \mathbb{R}_{\tau}) = \mathbb{R} \ , where \mathbb{R}_{\tau} is the world line of the point source. The Dirac relation e \ g = 2 \pi n is satisfied
if and only if U(1) is the gauge group.
5) Lastly, and more importantly, we now understand that the
full Maxwell theory emerges naturally from gauging the
global U(1) symmetry of an
arbitrary matter field action, i.e., by sticking an
independent U(1) group to each space-time point (associating a U(1) fibre to each point x^{\mu}).
