Uniform lattice quotient and quasi-regular action

Since $\mathrm{\Gamma }$ is discrete, [F1] gives ${\mathrm{\Delta }}_{\mathrm{\Gamma }}=1$. The function $\rho (x)={\mathrm{\Delta }}_G(x{)}^{-1}$ obeys $\rho (x\gamma )={\mathrm{\Delta }}_{\mathrm{\Gamma }}(\gamma ){\mathrm{\Delta }}_G(\gamma {)}^{-1}\rho (x)$, so it is a rho-function by [F5]. Its cocycle is constant: $D_g(x\mathrm{\Gamma })=\rho (g^{-1}x)\mathrm{/}\rho (x)={\mathrm{\Delta }}_G(g)$.

Let ${\mu }_{\rho }$ be its quotient measure. Since $G\mathrm{/}\mathrm{\Gamma }$ is compact, $0<{\mu }_{\rho }(G\mathrm{/}\mathrm{\Gamma })<\mathrm{\infty }$: finiteness is Radon compact-finiteness and positivity follows from full support. The pushforward $g_{\ast }{\mu }_{\rho }$ has the same total mass as ${\mu }_{\rho }$, while [F2] gives $g_{\ast }{\mu }_{\rho }={\mathrm{\Delta }}_G(g){\mu }_{\rho }$. Therefore ${\mathrm{\Delta }}_G(g)=1$ for every $g$, and $G$ is unimodular.

Now ${\mathrm{\Delta }}_G{\mathrm{\mid }}_{\mathrm{\Gamma }}={\mathrm{\Delta }}_{\mathrm{\Gamma }}=1$, so [F3] gives a nonzero invariant Radon measure on $G\mathrm{/}\mathrm{\Gamma }$; compactness makes it finite. For $\rho =1$ its cocycle is identically one. Scalar covariance says $F(x\gamma )=F(x)$, so sections are exactly functions on $G\mathrm{/}\mathrm{\Gamma }$, and the action is $F(q)\mapsto F(g^{-1}q)$ with the quotient $L^2$ norm. This is the quasi-regular representation, as claimed by [F4]. ∎