# Is it possible to use the Faddeev-Popov trick for discrete gauge symmetries?

I was thinking of this previous question of mine, where I was trying to implement a path-integral over the half-line: $$Z=\int_{\varphi\ge0}\ \mathrm e^{iS[\varphi]}\mathrm d\varphi\tag1$$

It seems to me that a possible approach is to integrate over all $\varphi$, but to regard the configurations $\varphi$ and $-\varphi$ as physically equivalent. In other words, we take the $\mathrm d\varphi$ to be unconstrained, but we introduce the gauge equivalence $\varphi\to-\varphi$; the orbits are of the form $\{\pm\varphi\}$, and a representative of each orbit is, for example, $+\varphi$. Integrating over one representative for each orbit brings us back to $(1)$.

Instead of eliminating the gauge redundancy explicitly (which would lead to the integral I don't know how to evaluate, $(1)$), we leave the gauge symmetry as is, but treat the resulting path-integral using the standard methods of gauge-theory. The problem is that I don't know how to implement the Faddeev-Popov trick for $\mathbb Z_2$-valued gauge fields or, more generally, fields over a discrete group.

Has this problem been analysed? (How) Can we implement the Faddeev-Popov trick for system with discrete gauge symmetries?

• Sorry, I am lost, why do you think that Faddeev-Popov eliminates a redundancy, in particular a scalar field? – user178876 Jun 20 '18 at 1:41
• @marmot It doesn't eliminate the scalar field, it eliminates the redundancy in $\varphi\sim-\varphi$. Just like in standard gauge theory, where the redundancy is $\phi(x)\sim \mathrm e^{i\lambda(x)}\phi(x)$ (in the case of $\mathrm U(1)$, with $\phi(x)$ complex). – AccidentalFourierTransform Jun 20 '18 at 1:43
• This is completely new to me. As far as I know, the Faddeev-Popov ghosts exist already in a pure gauge theory, and are used to fix a gauge, i.e. to make the Klein-Gordon operator for gauge fields invertible. Why do you think they eliminate the phase of the scalar field? What if you do not gauge the theory? – user178876 Jun 20 '18 at 1:49
• Well, if you are sure about this, why don't you just see what happens if you get the discrete gauge symmetry in the good old-fashioned way, namely by breaking a U(1) symmetry by giving a scalar $\Phi$ of charge 2 a VEV and look at a downstairs theory with a scalar $\phi$ that had charge 1 under the U(1). It is straightforward to see that the Faddeev-Popov ghosts do nothing of the sort you have in mind (but they also did not in the upstairs theory). – user178876 Jun 20 '18 at 2:02
• Ah, one more thing, when you are talking about discrete gauge symmetries, you are taking them to be those of arXiv:1710.01791, right? – user178876 Jun 20 '18 at 2:04