# One-form symmetry action in 4d Maxwell - closed or not?

Note: I had an overlong post which I was advised to break into pieces - here is my first, most basic question. Next question is here.

I'm trying to understand how the electric one-form symmetry acts in 4d Maxwell theory on a manifold without boundary, and then what it means to try to gauge it. SE answers I've already read include 1, 3, 4 and 5, but I wasn't able to get a complete picture.

What is the precise action of the global one-form symmetry on the operators, vs a possibly gauged version?

Here is my current understanding.

The Maxwell action is

$$S_0 = \int \frac{1}{2g^2} F\wedge \star F$$ with $$F = dA$$ (locally).

We can see that this action is totally invariant under $$A \rightarrow A + \lambda$$, if $$d \lambda = 0$$. $$\lambda = d\omega$$ is of course the 0-form gauge symmetry.

While $$S_0$$ is invariant, the shift in $$A$$ will be detected by Wilson lines $$W_n[C] = e^{i n \oint A} \rightarrow e^{in \oint \lambda} W_n[C].$$

This SE answer shows that the phase picked up is topological in the sense that it depends only on the homology class of $$C$$, since if $$C \sqcup (-C') = \partial \Sigma$$ then $$\int_C \lambda - \int_{C'} \lambda = \int_{\Sigma} d\lambda = 0.$$ Just as global 0-form symmetries involve shifting a field by a constant, i.e a 0-dimensional operator which doesn't change when you move around the connected manifold, so does a 1-form global symmetry involve shifting a field by a 1-dimensional operator that doesn't depend on how you smoothly deform it. Crucially this requires $$d\lambda=0$$.

On the other hand, consider shifting by some $$\lambda$$ such that $$d\lambda \neq 0$$. Then $$\Delta S_0 = \frac{1}{g^2} \int \star F \wedge d \lambda = \frac{1}{g^2} \int d\star F \wedge \lambda = 0$$ by virtue of the equation of motion for $$A$$. (This is up to some $$d\lambda \wedge \star d\lambda$$ term we are allowed to absorb.).

Elsewhere $$\lambda$$ is referred to as a flat connection, but a field is still said to shift by $$d\lambda$$.

$$\color{orange}{\textbf{THE QUESTION}}$$

Does the global symmetry action have $$d\lambda = 0$$ (as is held to be the case in this answer) or not? If not, what equations does $$\lambda$$ have to obey, and what changes about these equations when we consider trying to gauge the symmetry?

Compare this to the 0-form case where we start with a shift by a constant $$\alpha$$, then, when we gauge, consider the more general $$\alpha(x)$$.

The answer is essentially what @ACuriousMind pointed out in a comment (which I will quote here, because comments are intended to be temporary)

if you use the equations of motion you're not demonstrating a symmetry. $$δS=0$$ under all possible infinitesimal variations is the definition of a solution of the equations of motion, so if you have to use the equation of motions to arrive at $$δS=0$$, you're not showing anything interesting - symmetries have to be off-shell, see also this.

Another way to see this, which is inherently a QFT statement is through the path integral. $$\newcommand{\d}{\mathrm{d}}\newcommand{\D}{\mathrm{D}}$$Namely, the gauge fields are dummy integration variables, i.e. $$\int\D a \exp(-S[a]) = \int\D a' \exp(-S[a']).\tag{1}$$

But now, take $$S[a]$$ to be the Maxwell action$$1$$ $$S[a]=\frac{1}{2g^2}\Vert f\Vert^2,$$ and take $$a' = a+\lambda$$, where $$\lambda$$ is allowed to have some curvature. You can most easily see that $$\D a' = \D a$$ and $$S[a'] = \frac{1}{2g^2}\left(\Vert f\Vert^2 + 2\left + \Vert\d \lambda\Vert^2\right)$$ Then (1) says that $$\int\D a \exp(-S[a]) = \int\D a \exp\!\left(-S[a]+\frac{1}{g^2}\left + \frac{1}{2g^2}\Vert\d \lambda\Vert^2\right),$$ which is true if and only if $$\d\lambda = 0$$.

Therefore only shifts by flat connections give a symmetry of Maxwell theory.

1 where $$\Vert\bullet\Vert^2$$ comes from the usual inner-product $$\left<\bullet,\circ\right>:=\int \bullet\wedge\star\circ$$

• Thanks for this response, that's very clear. Followup questions are 1. If we gauge the symmetry, are we considering non-flat $\lambda$? 2. Are there subtleties involved in the fact that the one-form electric current is only conserved when $A$ is on-shell? Commented Dec 1, 2022 at 14:01
• @quixot for 1. I believe that's one of the main topics in your other question, which I'll answer when/if I have the time, unless someone else answers it sufficiently before I manage to do so. Short answer is yes. I don't understand what you mean by 2. Do you mean when $\lambda$ is not flat? This just tells you that there isn't a global symmetry for non-flat $\lambda$. Commented Dec 1, 2022 at 14:30
• I just mean that $dJ_e=d\star F = 0$ is the equation of motion for $A$. Commented Dec 1, 2022 at 14:37
• Would it be fair to say the following: symmetries must have $\delta S = 0$ off-shell, but their associated current is only conserved on-shell? Commented Dec 1, 2022 at 16:01
• @quixot, yes, that is correct Commented Dec 1, 2022 at 21:33