 12:23 AM
@ACuriousMind I think you understand this stuff better than me: Can we get a Noether current without knowing the trafo of $A$ explicitly? @Danu Since all that appears in the action is $F$, it should suffice know the trafo of that
Just treat $F$ as the "fundamental field" for the purpose of the derivation
Hm, wait @ACuriousMind Well check out the Noether current
It doesn't care about the trafo of $F$ That could go wrong because the equation of motion is for the variation of $A$, not $F$ Right, that could be one way to phrase the cause of trouble. @Danu Yeah, right.
If your equation of motion is for $A$, then you need to have the transformation behaviour of $A$, yes 12:26 AM
So, wat do?
Given that that's not so easy (is it even uniquely determined?) So unless you can phrase the duality transformation on $A$, I don't think you can use Noether's theorem here Alternatively, how do I current without Noether Without Noether, you don't even know there is a conserved current! @ACuriousMind Okay, but from the paper we know there is one. @Danu Yeah, so there is some symmetry transformation that generates it, meaning either one can phrase the duality transformation in terms of $A$, or this current is not actually associated to the duality transformation 12:30 AM
@ACuriousMind Have you given it any thought, this current?
Did you have a look at the paper by Anco? @Danu Oh god, jets Yeah, it's kind of stupid terminology IMO but the idea is clear. @ACuriousMind Oh, so the jet thing is in fact over-horrified? I mean... the $n$-jet of an object is just [all the information contained in up to the $n$-th derivative] AFAIK.
The terminology is kinda uncalled for. @Danu Yes...but I don't even understand why that thing they call a conserved current is...a conserved current! Look at that weird "total derivative operator", how does that give us a "normal" conserved current? 12:43 AM
@ACuriousMind Their "total derivative operator" just takes a derivative.
It just has a special action on the $F$'s
and even that special action is rather trivial
it takes the $q$-jet and gives back the $q+1$-jet Oh
Right It's stupid terminology, I agree. That's written in a terrbily convoluted way, yeah Yeah, so it's mostly just accounting, right?
They're trying to layer the conservation laws into orders
So they introduce the jets so they can assign an order to a given current Yeah, but you can just do that in more simple terminology without any problem, of course :P 12:47 AM
@Danu Probably, yeah In any case, on the expressions (2.13), (2.14) it just acts as $\partial_\mu$ so let's forget about it for now.
So, how do we reproduce 2.14 @Danu The "proper" way to get the symmetry corresponding to a conserved quantity is to take the Noether charge and calculate the canonical transformation it generates in the Hamiltonian formulation, but I'm not sure that actually works in a gauge theory
It might be that there is no symmetry corresponding to this conserved current. 1:04 AM
@ACuriousMind So how does one derive it? @ACuriousMind That would be really weird.
I thought Noether did have a converse.
In what situations does it fail to have one? @EmilioPisanty It has for non-degenerate Lagrangian systems, i.e. those that have a unconstrained Hamiltonian formulation, because then the Noether charge just generates the symmetry. But for a gauge theory, you can't pass to the Hamiltonian formulation in a straightforward manner, in particular, you can't express the velocities in terms of the momenta
I'm not aware of an inverse Noether theorem for such degenerate systems - there might be one, but it is not obvious to me how one would prove it
@Danu By that method of characteristics and "adjoint symmetries" in section 3 of Anco, I guess
Note that they seem to really only work on the level of the equations
Symmetries of the equations of motion can fail to lift to symmetries of the action, I think @ACuriousMind Oh, ok, that's a big bummer there. But that makes sense. But there's a Qmechanic answer on that somewhere, I think, searching...
Yeah, the duality transformation does not lift to a symmetry of the action, not even to a quasi-symmetry, see here
So I think that if their derivation is really based on symmetries of the e.o.m, trying to get that from Noether's theorem is bound to fail 1:19 AM  