One-form symmetry and its spontaneously breaking

Question

In the paper "Generalized global symmetry" by Gaiotto, Kapustin, Seiberg and Willett: https://arxiv.org/abs/1412.5148, in section 5 they describe the spontaneously breaking of 1-form symmetry in 4d as following:

We interpret an area law for a charged loop operator as reflecting the fact that the corresponding one-form symmetry is unbroken. Indeed, the expectation value of the loop vanishes as its size is taken to infinity. Perimeter law can be set to zero by redefining the operator by a local geometric counterterm. Then, perimeter law and Coulomb behavior mean that the loop has a nonzero expectation value when it is large and correspondingly the symmetry is spontaneously broken.

This means if the vev of the Wilson line $\langle W(C) \rangle$ along a contour $C$ satisfies the area law $\langle W(C) \rangle \sim e^{- \textrm{Area}(C)}$ then the 1-form symmetry is preserved. On the other hand, if that satisfies the perimeter law $\langle W(C) \rangle \sim e^{-\textrm{Perimeter}(C)}$ then the 1-form symmetry is spontaneously broken.

On the other hand, for the 0-form symmetry acting on a field $\phi$, we know that if $\phi$ develops a background $\langle \phi \rangle \neq 0$, then 0-form symmetry is broken because $\langle \phi \rangle$ is not invariant under the symmetry.

My question is the following. For the 1-form symmetry, when the vev of the Wilson loop $\langle W(C) \rangle \sim e^{- \textrm{Area}(C)}$ satisfies the area law, under the 1-form symmetry this vev will develop a phase and is not invariant. Then why do we say the 1-form symmetry is unbroken in this case? Is it important to take the size to infinity as mentioned above?

Answer 1

It is not important to take the size to infinity. If the symmetry is preserved, any operator that is charged under that symmetry will not develop a vacuum expectation value (vev).

The reason the area law for the unbroken phase does not contradict the above statement is because the operator $W(C)$ is not charged under the global 1-form symmetry, when $C$ is a contractible loop.

Consider (Euclidean) spacetime to be on a $(d+1)$-dimensional torus. For $p$-form symmetries, the symmetry operator lives on a $(d-p)$-dimensional manifold $\Sigma_{d-p}$ that wraps nontrivially around this torus, usually considered to be a time-slice for 0-form symmetries. (Refer Fig. 2 in this review paper). If you have $p=1$ and a contractible loop $C$, it is easy to see that $W(C)$ is uncharged under the symmetry because $C$ and $\Sigma_{d-1}$ intersect an even number of times which cancels out.

Why do we use the expectation value of an uncharged object to characterize spontaneous symmetry breaking (SSB)? This can be viewed in analogy to conventional (0-form) SSB, and the concept of long-range order. One way to diagnose SSB is by creating a particle and an antiparticle at the ends of a string, and ask how the vev of this operator depends on the length of this string. If the symmetry is preserved, it decays exponentially in the length of the string ("area law") whereas if it is broken, there is a constant value with a subleading exponential decay ("perimeter law", perimeter = constant).

We can take this analogy to 1-form symmetries. The particle-antiparticle pair is replaced by an operator $W(C)$ living on the contractible loop $C$. The string connecting the two particles is replaced by the membrane enclosed by $C$. We say the symmetry is preserved when the vev of $W(C)$ decays exponentially in the area, and broken if it decays exponentially in the perimeter.

To drive this point home, one can go back to imagining spacetime on a torus, and smoothly deforming $C$ to wrap around one non-trivial cycle of the torus, such that $C$ becomes two separated non-trivial cycles of the torus with opposite direction, which I will call $C_1$ and $C_2$, with a cylinder shaped membrane between them. Note that $W(C) = W(C_1)W(C_2)$, and $W(C_1), W(C_2)$ are charged under the 1-form global symmetry! Next, I imagine separating $C_1$ and $C_2$ from each other by increasing the height of the cylinder connecting them, let's call this height/separation $h$. The vev decays exponentially in $h$ if we are in area law, but remains constant (with a subleasing decay) if we are in perimeter law. For sufficiently large separation $h$, we can use the usual long-range order argument to say $\langle W(C)\rangle = \langle W(C_1)\rangle \langle W(C_2)\rangle$ and claim that in the case of perimeter law, the charged objects $W(C_1), W(C_2)$ gain an expectation value. Therefore perimeter law implies SSB of 1-form symmetry.