Independence of rho and equivalent quotient representative

Put $a(q)={\rho }_2(x)\mathrm{/}{\rho }_1(x)$ for $q=xH$. The rho covariance makes this ratio independent of representative and positive continuous. If $\varphi =T_{H}f$, the two Weil formulas give ${\int }_X\varphi d{\mu }_{{\rho }_2}={\int }_{G}f{\rho }_{2}dx={\int }_{G}f{\rho }_{1}adx={\int }_X\varphi (q)a(q)d{\mu }_{{\rho }_1}(q)\mathrm{.}$ The measure $a{\mu }_{{\rho }_1}$ is Radon because $a$ is positive continuous and bounded on compact sets. Surjectivity of $T_H$ gives equality of its $C_c$ integrals with those of ${\mu }_{{\rho }_2}$, and [F6] identifies $d{\mu }_{{\rho }_2}=ad{\mu }_{{\rho }_1}$.

Define $U(F)=a^{-1\mathrm{/}2}F=[{\rho }_1\mathrm{/}{\rho }_2{]}^{1\mathrm{/}2}F$. The ratio is $H$-invariant, so covariance is preserved, and step 1.1 gives $\mathrm{\parallel }UF{\mathrm{\parallel }}_{{\rho }_2}^2={\int }_X{a}^{-1}\mathrm{\parallel }F{\mathrm{\parallel }}^{2}d{\mu }_{{\rho }_2}=\mathrm{\parallel }F{\mathrm{\parallel }}_{{\rho }_1}^2\mathrm{.}$ The inverse multiplier is $a^{1\mathrm{/}2}$, hence $U$ extends onto the Hilbert completions.

For the action, both sides of $U{\mathrm{\Pi }}_1(g)F={\mathrm{\Pi }}_2(g)UF$ multiply $F(g^{-1}x)$ by the same scalar: $a(xH{)}^{-1\mathrm{/}2}{D}_g^1(xH{)}^{1\mathrm{/}2}=D_g^2(xH{)}^{1\mathrm{/}2}a(g^{-1}xH{)}^{-1\mathrm{/}2},$ which follows by substituting $a={\rho }_2\mathrm{/}{\rho }_1$ into [F4]. Thus the rho choices give equivalent representations.

For an equivalent quasi-invariant $\nu$, [F3] supplies local densities $w=d\nu \mathrm{/}d{\mu }_{\rho }>0$ on each open sigma-compact component. Multiplication by $w^{-1\mathrm{/}2}$ is a unitary from the ${\mu }_{\rho }$ section space to the $\nu$ section space, since $d\nu =wd{\mu }_{\rho }$. On each open $\sigma$-compact target component, $g^{-1}$ maps it to an open $\sigma$-compact set meeting only countably many components, so the componentwise densities are measurable there. Pushing $w{\mu }_{\rho }$ forward under $q\mapsto gq$ gives the local Radon--Nikodym derivative $D_g^{\nu }(q)=\frac{w(g^{-1}q)}{w(q)}{D}_g^{\rho }(q)$ almost everywhere on that component. The componentwise multiplication maps assemble on the Hilbert direct sum, and substitution in the action formula gives $U{\mathrm{\Pi }}_{\rho }(g)={\mathrm{\Pi }}_{\nu }(g)U$ almost everywhere. Thus the general measure representative gives the same unitary representation. ∎