# $Z_n$ gauge theory from $U(1)$

In Appendix A of the paper, "Generalized Global Symmetries" by Gaiotto et al., they have considered an action, which for the purpose of the question, can be taken to be

$$S=\frac{n}{2\pi}\int B dA$$

where $$B$$ is a periodic scalar and $$A$$ is a $$d-1$$ form. Then they went on to construct the operators

$$U=e^{i\oint A} \ \ \ and \ \ \ V=e^{iB}$$

They claim that from the equation of motion, it follows that $$U^n=1$$ and $$V^n=1$$. I do not quite understand why this is so. The equations of motions are $$dA=0$$ and $$dB=0$$, which says that $$B$$ is a constant and $$A$$ is locally a pure gauge. Why should $$B$$ equal to $$2\pi/n$$? Why should,

$$\oint A \ \ \ be \ equal \ to \ \ \ 2\pi/n \ ?$$

Naively, it seems that $$B$$ can be some arbitrary constant, and for example, if we take the space to be two dimensional, $$\oint A$$ should be equal to $$2\pi Z$$ for non-contractible loops. It seems that we need some additional considerations, other than just EOM to obtain the result they quote.

You need to think about what "$$B$$ is a periodic scalar" means: When $$B$$ is $$2\pi$$-periodic, then since $$V$$ acts on $$B$$ as $$B\mapsto B + \frac{2\pi}{n}$$, you necessarily have $$V^n = 1$$ since $$B+2\pi$$ is identified with $$B$$ so $$V^n$$ is a "do nothing" operator, i.e. the identity.
Similarly, when Gaiotto et. al. say that the transformations act on $$A$$ as $$A\mapsto A + \frac{1}{n}\zeta$$ with $$\zeta$$ a "properly normalized flat gauge field", the normalization they're talking about is such that $$S[B = B_0,A = \zeta] = 1$$, i.e. the transformation $$A\mapsto A+\zeta$$ is a do-nothing transformation in terms of $$\mathrm{e}^{2\pi\mathrm{i}S[B,A]}$$ or, in another diction, the shift in $$A$$ is "absorbed" by the periodicity of $$B$$.
• I was wondering, in what sense one talks about $U$ acting on $B$ without canonical quantization. This might be just nonsense, but is it inferred from calculating the correlation function $\langle e^{iB(x_0)} e^{\oint_C A} \rangle$? If we calculate this correlation function, we get $e^{\frac{2\pi i}{n}}$ for all $x_0$ is inside $C$. Maybe this can be interpreted as $e^{i\oint_C A}$ introducing a $2\pi/n$ discontinuity in $B$ across the surface defined by $C$. Commented Sep 20, 2022 at 16:35
• @TuhinSubhraMukherjee The talk about operators is always after quantization. There is no sense to something like $\mathrm{e}^{\mathrm{i}B}$ in the classical theory. Commented Sep 20, 2022 at 16:37