What's the meaning of this path integral measure?

Question

I don't understand the meaning of following path integral measure $$ \frac{[df]}{U(1)} $$ What is the difference between $[df]$ and $[df]/U(1)$? A naive idea is the latter measure is more physical since it removes some gauge degrees of freedom from $U(1)$? The symbols are from equation (1) of the supplementary material from the PRL paper Isometric Evolution in de Sitter Quantum Gravity (2023). However, a detailed definition is not provided in the paper.

Answer 1

In the context you are asking it, that is, the Schwarzian theory and its $\mathrm{U}(1)$ cousins, the measure you are after is $$\mathrm{d}{\mu(f)} = \left.\frac{[\mathrm{d}f]}{\mathrm{U}(1)}\right|_{\text{Cotler & Jensen}}\equiv \left.\frac{\mathrm{D}f}{\operatorname{vol}(\mathrm{U}(1))}\right|_{\text{clearer notation}}.\tag{1}\label{1}$$ The reason I prefer the latter rewriting is that it makes clear what we are dividing by: the volume of a $\mathrm{U}(1)$ group, i.e. the length of a circle with a prescribed radius (remember, $\mathrm{U}(1)\cong\mathbb{S}^1$). The radius of the circle will correspond to a coupling constant, but I will come to that later. (The choice of notation $\mathrm{D}f$ instead of $[\mathrm{d}f]$ is merely aesthetic.) In the case that the $\mathrm{U}(1)$ symmetry is compact, i.e. the radius of the circle is finite, dividing or not is (relatively) immaterial. The final path integral differs by the desired one by a finite constant. You can get to the correct answer by demanding a physicality condition (cf. this Phys.SE answer). Trouble starts when the radius of the circle is infinite and there you get a strictly infinite result if you don't divide by that volume.

$\newcommand{\cF}{\mathcal{F}}$Let's see where all this comes from. Let's say you have a theory with a bunch of dynamical fields, $\phi\in\cF$ ($\cF$ is some functional space), and a non-compact global symmetry group, $G$. As explained in arxiv:2112.03793, the partition function of this theory will be, formally, divergent, due to the volume of the group. To get a meaningful expression you path-integrate not over $\cF$, but over $\cF/G$ instead. This has the effect of dividing by the volume of $G$: $$\int_{\cF/G} \mathrm{d}{\mu(\phi)}=\int_{\cF}\frac{\mathrm{d}\mu(\phi)}{\operatorname{vol}(G)},\tag{2}\label{2}$$ away from fixed points of the $G$-action on $\cF$. In order to compare, you can do the same thing for compact symmetry groups, even though you don't need to. It is useful when you want to compare answers though. This is why people have been doing it in the compact $\mathrm{U}(1)$'s.

In the context of the Schwarzian that is common, because there is a continuous family of theories with a compact symmetry ($\mathrm{U}(1)$), that gets enhanced to a larger, non-compact, symmetry ($\mathrm{SL}(2,\mathbb{R})$) at a specific point. Moreover, in this context, $\cF=\mathrm{diff}\!\left(\mathbb{S}^1\right)$, the space of diffeomorphisms of the circle, and it is known, see e.g. arxiv:1703.04612 that the spaces $$\mathrm{diff}\!\left(\mathbb{S}^1\right)\!\big/\mathrm{U}(1) \qquad\text{and}\qquad \mathrm{diff}\!\left(\mathbb{S}^1\right)\!\big/\mathrm{SL}(2,\mathbb{R})$$ are symplectic manifolds, so \eqref{2} holds everywhere.

Finally, let me just clear up some confusion. The above discussion has nothing to do with gauge symmetries. It applies to global (usually non-compact) symmetries. In systems with gauge symmetries, however, a very similar story pans out. There you also get divergent answers, if you do nothing, even with compact structure groups, but not because you have a group volume in the denominator but because you have, roughly, a group volume per point in spacetime in the denominator. And you have densely infinitely many points in spacetime. So this is, in some sense, a more dangerous divergence. So you do the same, instead of integrating over $\cF$ you integrate over $\cF/\mathcal{G}$. This time, $\mathcal{G}$ is the gauge group (cf. this Phys.SE answer for the difference between structure and gauge group). Annoyingly, in most places in the literature, people don't bother differentiating between the structure group and the gauge group when writing down path integrals of gauge theories, so you end up with formulas identical to \eqref{1}, but different meaning, and lead to confusions like yours (when you tried giving a naive explanation).

Hope this helps.