Commensurator, unitary characters and monomial induced representations in the transversal model

Define $K\sim L$ when $K\cap L$ has finite index in both subgroups. Reflexivity and symmetry are immediate. If $K\sim L$ and $L\sim M$, then $K\cap L\cap M$ has finite index in $K\cap L$ because $L\cap M$ has finite index in $L$; it therefore has finite index in $K$. The same argument, starting with $M\cap L$, shows it has finite index in $M$. Thus $\sim$ is transitive. Conjugation preserves finite indices and intersections. For $g,h\in {\mathrm{Comm}}_G(H)$, conjugating $H\sim g^{-1}Hg$ by $h^{-1}$ gives $h^{-1}Hh\sim h^{-1}{g}^{-1}Hgh$, while $H\sim h^{-1}Hh$; hence $H\sim (gh{)}^{-1}H(gh)$ and $gh\in {\mathrm{Comm}}_G(H)$. Conjugating $H\sim g^{-1}Hg$ by $g$ gives $gH{g}^{-1}\sim H$, so $g^{-1}\in {\mathrm{Comm}}_G(H)$. Every $h\in H$ satisfies $h^{-1}Hh=H$. Therefore the commensurator is a subgroup containing $H$.

Uniqueness of $tg=\alpha (t,g)(t\cdot g)$ gives $(t\cdot g_1)\cdot g_2=t\cdot (g_1{g}_2)$ and $\alpha (t,g_1{g}_2)=\alpha (t,g_1)\alpha (t\cdot g_1,g_2)$. Substitution into the formula for $\pi$ yields $\pi (g_1)\pi (g_2)=\pi (g_1{g}_2)$. Right multiplication permutes $T$, and every multiplier $\chi (\alpha (t,g))$ has modulus $1$, so each $\pi (g)$ is unitary.

Suppose now $G$ is locally compact. The open subgroup $H$ is also closed. Its right-coset space $G\mathrm{/}H$ is discrete. Restrict a left Haar measure on $G$ to $H$; this is a left Haar measure on $H$, and partitioning $G$ into the cosets $t^{-1}H$ shows that the Weil quotient formula with constant $\rho =1$ gives counting measure on $G\mathrm{/}H$. For the covariant model in [F2], define $UF(t)=F(t^{-1})$. Every finitely supported function on $T$ arises this way: on each open coset $t^{-1}H$ set $F(t^{-1}h)=\chi (h{)}^{-1}f(t)$, and set it to zero on cosets outside the finite support. This is continuous and covariant, so $U$ extends to a unitary from the completed model to ${\mathrm{\ell }}^2(T)$. If $tg=\alpha (t,g)(t\cdot g)$, then $g^{-1}{t}^{-1}=(t\cdot g{)}^{-1}\alpha (t,g{)}^{-1}$ and covariance gives $F(g^{-1}{t}^{-1})=\chi (\alpha (t,g))F((t\cdot g{)}^{-1})$. Hence $U$ intertwines the covariant left action with $\pi$. By [F3], this is standard unitary induction.

For each $t\in T$, the subgroup $t^{-1}Ht$ is an open neighborhood of $e$. On it $t\cdot g=t$ and $\alpha (t,g)=tg{t}^{-1}$, so $\pi (g){\delta }_t=\chi (tg{t}^{-1}){\delta }_t\to {\delta }_t$ as $g\to e$. Continuity follows on finite-support vectors by linearity. For arbitrary $f\in {\mathrm{\ell }}^2(T)$, approximate by a finite-support $f_0$ and use $\mathrm{\parallel }\pi (g)f-f\mathrm{\parallel }\le 2\mathrm{\parallel }f-f_0\mathrm{\parallel }+\mathrm{\parallel }\pi (g)f_0-f_0\mathrm{\parallel }$; thus continuity holds at $e$ on all vectors, and the representation law gives it at every $g$. Since $\pi (t^{-1}){\delta }_e={\delta }_t$ for every $t\in T$, ${\delta }_e$ is cyclic.

If $G$ is second-countable, let $(B_n{)}_{n\in \mathbb{N}}$ be a countable base. Each left coset $C=Ht$ is nonempty and open, so let $n(C)$ be the least $n$ with $\mathrm{\varnothing }\mathrm{\ne }{B}_n\subseteq C$. Disjoint cosets have distinct such basis elements; thus $H\mathrm{\backslash }G$ is countable. Under AC choose a representative from each coset, so $T$ is countable. Finite-support functions with rational real and imaginary parts form a countable dense subset of ${\mathrm{\ell }}^2(T)$, proving separability. Once a transversal is given, all constructions and calculations above use no further choice.