Large gauge transformations and global symmetries

Question

In gauge theories, states in Hilbert space that are related by local gauge transformations are identified as the same physical state. This is necessary because otherwise the Hilbert space contains states of negative norm. My understanding is that getting rid of these unphysical states involves only "small" gauge transformations that are continuously connected with the identity transformation. This leaves "large gauge transformations" (those that are not connected to identity), which can be interpreted as actual symmetry transformations unlike the gauge redundancies, essentially because there is no reason to assume otherwise. My main source here is the book "Quantization of Gauge System" by M. Henneaux and C. Teitelboim, although I admittedly do not understand many details of their discussion. I am aware that large gauge transformations involve non-trivial topology (as described in the answer here: https://physics.stackexchange.com/a/317273) and are important for Yang-Mills instantons.

Therefore, the usual jargon seems to be that the "physical content" of a gauge symmetry is contained in the large transformations, while the small ones are just redundancies. On the other hand, it is often stated that a local gauge invariance implies a global symmetry (same transformation at every $x$) and that this global symmetry gives rise to physically conserved Noether currents, like in electromagnetism. I am having hard time understanding how these two concepts fit together. My questions are:

1. If all gauge transformations that are connected with the identity are mere redundancies, then what gives rise to charge conservation in abelian gauge theory? Please elaborate if the answer is different for different theories (QED, abelian Higgs etc).

Surely $U = e^{i \alpha}$ with $\alpha = \text{constant}$ is in the identity component of $U(1)$?

2. In Yang-Mills theory ($SU(N)$ for concreteness), is there a global $SU(N)$ symmetry with physical consequences associated with the gauge redundancy?

Based on a lot of literature I think the answer is no, there is only a center symmetry associated with the large gauge transformations. But I don't see why Yang-Mills would drastically differ from the abelian case in this regard.

Edit: Question 3. It is obviously true that in an abelian theory, say QED, the gauge field itself is unchanged under a global $U(1)$ transformation, and it looks like there is global symmetry $\psi \rightarrow e^{i\alpha} \psi$ for matter fields. But $\psi$ itself is not a gauge invariant object, so why is this any different from a special case of the gauge redundancy?

Specifically, in Yang-Mills the remnant center symmetry is clearly a true symmetry of the system because there exists a gauge-invariant observable that transforms non-trivially under operation of matrices in the group center (Polyakov loop). Following this logic, if QED has a true U(1) symmetry then shouldn't there exist a gauge-invariant object that is sensitive to the global U(1) symmetry?

Thank you in advance.

Answer 1

Here's what I think is a straightforward way to understand the residual global symmetry in a gauge theory. The story won't actually involve large gauge transformations but I think it will answer your numbered questions.

The part of the Lagrangian involving the fermions is like $$\mathcal{L_\psi} = \bar{\psi}i\gamma\partial\psi + \bar{\psi}\gamma A\psi$$ where although I am not writing the indices, $\psi$ transforms in the fundamental representation of a non-Abelian Lie group and $A$ is a member of the Lie algebra.

We may transform only the fermions and not $A$ under a global transformation $\psi\rightarrow U\psi$. This is distinct from a gauge transformation since it only acts on the matter fields. But due to the coupling term it is only a symmetry of the action if $U^{-1}AU =A$, so only elements of the center of the Lie group should end up as actual global symmetries.

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Let's proceed with the standard Noether's theorem argument anyway and see what happens. Consider now the local transformation on the fermion fields generated by a member of the basis of the Lie algebra $T_a$ $$\delta\psi=i\epsilon_a(x)T_a\psi,\qquad \delta\bar{\psi}=-i\epsilon_a(x)\bar{\psi}T_a.$$ After integration by parts $$\delta\mathcal{L}=\epsilon_a(x)\left[\partial\left(\bar{\psi}i\gamma T_a\psi\right)+\bar{\psi}i\gamma[A,T_a]\psi\right]=\epsilon_a(x)\left(\delta_{ac}\partial+f_{abc}A_b\right)\bar{\psi}i\gamma T_c\psi$$ So when the equations of motion are satisfied the covariant divergence of the current vanishes $$\boxed{\mathcal{D}_\mu\left(\bar{\psi}i\gamma^\mu T_a\psi\right)=0}$$ Due to the non-homogeneous part in the covariant derivative, we can't use this to define a conserved charge. This is similar to how there is trouble with conserved energy in general relativity even though the covariant divergence of the energy-momentum tensor vanishes. But this is still a valid operator equation.