If by a gauge group, I mean the Lie group corresponding to a local continuous symmetry of the Lagrangian of a system, is it true that the Lie group is necessarily infinite dimensional? If so, what is the proof?

By a local symmetry, I mean one that differs from one space-time point to another.

NOTE: This question arises from a study of Noether's Second Theorem which is a statement regarding a infinite dimensional group of transformations.

P.S. Maybe I am confusing between the group of gauge transformations and the Lie group associated with a gauge symmetry (local continuous symmetries). If so, please tell me the difference.

  • $\begingroup$ Lie groups are by definition finite dimensional. Why do you think they are infinite-dimensional? or rather, what do you mean by dimension? $\endgroup$ Mar 4, 2017 at 18:06
  • $\begingroup$ @AccidentalFourierTransform Well then, what infinite dimensional group does Noether talk about? And how is it related to the Lie group, if it is not that already? I hope you see my confusion. $\endgroup$ Mar 4, 2017 at 18:27
  • $\begingroup$ @AccidentalFourierTransform: it's a Schwartz-Lie group, cf dx.doi.org/10.1063/1.526680 $\endgroup$
    – Christoph
    Mar 4, 2017 at 18:31
  • $\begingroup$ I think what causes confusions is that for local gauge transformations, the representing matrices $\Lambda^a_{\ b}$ are functions, so they contain infinite degrees of freedoms. With that said, in a simpler perspective, the gauge group $G$ is a finite dimensional Lie group, and what we have is $U\subset M$ an open region of spacetime and a function $\Lambda:U\rightarrow\rho(G)$, where $\rho$ is a representation. In a more abstract perspective, a gauge transformation is a right action $P\times G\rightarrow P$, where $P$ is a principal fiber bundle whose structure group is $G$. $\endgroup$ Mar 4, 2017 at 18:33
  • $\begingroup$ Nontheless, the group $G$ itself is strictly finite dimensional. $\endgroup$ Mar 4, 2017 at 18:33

1 Answer 1

  1. In Yang-Mills theory the underlying gauge Lie group $G$ is finite-dimensional. E.g. the gauge Lie group $G$ in the standard model is $1+3+8=12$ dimensional, which is a finite number.

  2. However, the corresponding group ${\cal G}= \Gamma(P\times_G G)$ of gauge transformations, i.e. the set of global sections in the associated bundle bundle $P\times_G G$ of the principal $G$-bundle $P$ over spacetime $M$, is necessarily infinite dimensional, if $\dim M >0$.

  3. The latter group ${\cal G}$ (as opposed to $G$) is what is relevant for Noether's second theorem.

  • $\begingroup$ re 1, $G$ is also known as the structure group of the principal bundle; re 2, the group of gauge transformations is not the set of maps $M\to G$ (or equivalently, global sections of the trivial bundle $M\times G$), but global sections of the associated bundle $P\times_G G$ ($P$ being the principal bundle in question) where the $G$-action is given by inner automorphisms $\endgroup$
    – Christoph
    Mar 4, 2017 at 18:59
  • $\begingroup$ @Christoph Isn't the group of gauge transformations defined as the group of automorphisms of the structure group? $\endgroup$ Mar 5, 2017 at 14:12
  • $\begingroup$ @J.Rahman: It's the group of vertical equivariant diffeomorphisms of the principal bundle. This group is isomorphic to the group of equivariant maps from the bundle into the structure group with the action of the group on itself given by conjugation, which in turn is isomorphic to the global sections of the associated bundle mentioned above. $\endgroup$
    – Christoph
    Mar 5, 2017 at 15:03

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