tom.stoer said:
I do not see why the gauge symmetry must vanish at spatial infinity. You may need some boundary conditions, but gauge theories can be formulated (e.g.) in compact space as well; then periodic boundary conditions will do the job. I don't think this is of any relevance in the present context.
In the context of Yang-Mills theory, the global charges (which vanish on compact manifolds, so I need to have spatial infinity) are like gauge transformations, except that they are constant over space. The algebra of gauge transformations is generated by all g-valued functions, where g is a finite-dimensional Lie algebra with generators T^a. If we consider 3D space and spherical coordinates, a gauge transformation is thus given by a function
f(r, theta, phi) T^a.
In contrast, the global charges are the g generators T^a.
tom.stoer said:
Your most interesting point is the global symmetry. In my opinion the global symmetry is nothing else but a special sub-sector of the local symmetry.
In the Yang-Mills example, the global charge symmetry corresponds to constant functions f = 1. However, there is a crucial difference: local gauge generators annihilate physical states, but global charge does not (for charged states).
tom.stoer said:
Look at QED (QCD): the operator that generates gauge transformations is the abelian (non-abelian) Gauss law.
My argument does not work for QCD, because there are no charged physical states due to confinement. But for QED everything is fine.
Let me make my point specifically for Yang-Mills theory. A useful basis for the space of Gauss law generators consists of
J^a(n,l,m) = r^n Y_lm(theta,phi) T^a,
where Y_lm are the spherical harmonics. However, only the basis elements with n < 0 are good gauge transformations, because the others do not vanish when r = infinity. In particular,
J^a(0,0,0) = T^a
are identified with global charges (more precisely, the Cartan subalgebra of g is), which are obviously nonzero on charged states.
My observation is now that not even the proper gauge transformations J^a(-n,l,m) can annihilate all physical states in the presence of divergent operators, because
[J^a(-n,0,0), J^b(n,0,0)] = i f^abc J^c(0,0,0) = i f^abc T^c
is a linear combinatinos of charge operators. Hence there must be some physical states such that
J^a(-n,0,0) |phys> != 0.
This does not contradict the usual picture, because the divergent operators generate new physical states, which lie outside the original Hilbert space. To have a well-defined action of the full gauge algebra, one must complete the Hilbert space by adding the new states. On the completed Hilbert space, the original gauge symmetry acts in a non-trivial way. It still annihilates the original subspace, of course.
There are different ways to react on this observation. One possibility is to say that divergent operators violate boundary conditions, and that thinking about them is prohibited. However, I fail to see why this is any worse than the situation in CFT; why is it ok that operators diverge when z -> infinity but not when r -> infinity? A more fruitful option is to study the action of the completed gauge algebra in the completed Hilbert space. Who knows, one may even learn something new by doing so...
tom.stoer said:
In non-abelian gauge theories it's rather simple: Once you factor away the gauge degrees of freedom
Evidently, the gauge dofs cannot be factored out anymore in the completed Hilbert space.