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Matrix-coefficient properties of the transversal model of a monomial representation

Statement

Assume AC (The Axiom of Choice). Let $G$ be a topological group. For $i \in {1, 2}$ , let $H_{i} \leq G$ be open, let $χ_{i} : H_{i} \to T$ be a unitary character, and let $T_{i}$ be a right transversal for the left cosets $H_{i} \ G$ with $e \in T_{i}$ . Write the unique factorization $t g = α_{i} (t, g) (t \cdot_{i} g)$ from Commensurator, unitary characters and monomial induced representations in the transversal model, and let $π_{i}$ be its transversal representation on $ℓ^{2} (T_{i})$ . A bounded intertwiner is a bounded linear map $S : ℓ^{2} (T_{1}) \to ℓ^{2} (T_{2})$ (Hilbert space, A bounded linear operator between normed spaces) satisfying $S π_{1} (g) = π_{2} (g) S$ for every $g \in G$ . Put $f : = S δ_{e} \in ℓ^{2} (T_{2})$ , where $δ_{e}$ is the distinguished basis vector. Then:

$f = 0$ if and only if $S = 0$ .
If $π_{1} = π_{2}$ on the same Hilbert space, then $f$ is a scalar multiple of $δ_{e}$ if and only if $S$ is a scalar operator.
For every $t \in T_{2}$ and $h \in H_{1}$ , $χ_{2} (α_{2} (t, h)) f (t \cdot_{2} h) = χ_{1} (h) f (t)$ .
If $t \in T_{2}$ has an infinite $H_{1}$ -orbit under $t \cdot_{2} h$ , then $f (t) = 0$ .
If $t \in T_{2}$ and $h \in H_{1}$ satisfy $f (t) \neq 0$ and $t \cdot_{2} h = t$ , then $t h t^{- 1} \in H_{2}$ and $χ_{1} (h) = χ_{2} (t h t^{- 1})$ .

Facts & Assumptions

Given: AC, the fixed transversal data $T_{i}$ , their transversal representations $π_{i}$ , and a bounded intertwiner $S$ .

[F1]

AC is inherited from the transversal convention of the preceding definition; the present lemma takes both transversals as data and uses no additional choice (The Axiom of Choice).

[F2]

The transversal action is $(π_{i} (g) u) (t) = χ_{i} (α_{i} (t, g)) u (t \cdot_{i} g)$ , $π_{i} (t^{- 1}) δ_{e} = δ_{t}$ , and $δ_{e}$ is cyclic (Commensurator, unitary characters and monomial induced representations in the transversal model).

[F3]

For any index set $I$ and $u \in ℓ^{2} (I)$ , $∥ u ∥_{2}^{2} = \sup_{F \subseteq I finite} \sum_{s \in F} ∣ u (s) ∣^{2}$ ; hence each finite subsum is at most $∥ u ∥_{2}^{2}$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

[F6]

If $u \in ℓ^{2} (I)$ and $ε > 0$ , there is a finite $F \subseteq I$ such that $\sum_{s \in I ∖ F} ∣ u (s) ∣^{2} < ε$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

[F4]

The real field is Archimedean, so for every real bound $b$ some natural number $n$ satisfies $b < n$ (Every complete ordered field is Archimedean).

[F5]

Each $ℓ^{2} (T_{i})$ is a Hilbert space and a bounded linear map between normed spaces is continuous (Hilbert space, A bounded linear operator between normed spaces).

Proof

technique · direct

Given: AC, $G, H_{i}, χ_{i}, T_{i}, π_{i}$ , and $S$ as in the Statement.

1.1F1F2F3F6given

For every $t \in T_{1}$ , the transversal action gives $π_{1} (t^{- 1}) δ_{e} = δ_{t}$ . To check density from [F3, F6], take $u \in ℓ^{2} (T_{1})$ and $ε > 0$ . Apply [F6] with tolerance $ε^{2}$ to obtain a finite $F \subseteq T_{1}$ with $\sum_{t \in T_{1} ∖ F} ∣ u (t) ∣^{2} < ε^{2}$ . The vector $u_{F}$ equal to $u$ on $F$ and $0$ elsewhere has finite support and $∥ u - u_{F} ∥_{2}^{2} = \sum_{t \in T_{1} ∖ F} ∣ u (t) ∣^{2} < ε^{2}$ by the norm definition [F3], hence $∥ u - u_{F} ∥_{2} < ε$ . Thus finite-support vectors are dense, and since each is a finite linear combination of the vectors $δ_{t} = π_{1} (t^{- 1}) δ_{e}$ , $δ_{e}$ is cyclic. AC is only the inherited transversal convention; the fixed $T_{i}$ are given.

1.2F2F5

For $h \in H_{1}$ , the transversal identities give $α_{1} (e, h) = h$ and $e \cdot_{1} h = e$ , while $s \cdot_{1} h \neq e$ for $s \in T_{1} ∖ {e}$ . Therefore $π_{1} (h) δ_{e} = χ_{1} (h) δ_{e}$ . Intertwining now gives $π_{2} (h) f = S π_{1} (h) δ_{e} = χ_{1} (h) f$ ; evaluating at $t \in T_{2}$ with the formula in [F2] yields $χ_{2} (α_{2} (t, h)) f (t \cdot_{2} h) = χ_{1} (h) f (t)$ .

2.1F5step 1.1

If $f = 0$ , then for every $g \in G$ , $S π_{1} (g) δ_{e} = π_{2} (g) S δ_{e} = 0$ . By cyclicity from step 1.1, $S$ vanishes on a dense subspace; continuity from [F5] gives $S = 0$ . Conversely, $S = 0$ immediately gives $f = 0$ .

2.2F5step 1.1

Suppose $π_{1} = π_{2}$ on the common carrier and $f = λ δ_{e}$ . For every $g \in G$ , $S π_{1} (g) δ_{e} = π_{1} (g) S δ_{e} = λ π_{1} (g) δ_{e}$ . Step 1.1 makes the orbit span dense, so continuity gives $S = λ I$ . Conversely, if $S = λ I$ , then $f = λ δ_{e}$ .

2.3F2F3F4step 1.2

By step 1.2 and $∣ χ_{i} ∣ = 1$ , the modulus $∣ f ∣$ is constant on each $H_{1}$ -orbit in $T_{2}$ . If the orbit $O$ of $t$ is infinite and $c : = ∣ f (t) ∣ > 0$ , choose a natural number $n > ∥ f ∥_{2}^{2} / c^{2}$ by [F4]. There are $n$ distinct points in $O$ ; their finite square sum is $n c^{2}$ , contradicting the finite-subsum bound [F3]. Hence $f (t) = 0$ .

3.1F2step 1.2∎

Suppose $f (t) \neq 0$ and $t \cdot_{2} h = t$ . Step 1.2 then gives $χ_{2} (α_{2} (t, h)) = χ_{1} (h)$ . The factorization $t h = α_{2} (t, h) (t \cdot_{2} h) = α_{2} (t, h) t$ implies $t h t^{- 1} = α_{2} (t, h) \in H_{2}$ , and therefore $χ_{1} (h) = χ_{2} (t h t^{- 1})$ .

Source notes

Bekka–de la Harpe's Lemma 1.F.10, printed pp. 53–54, states exactly the five claims and gives a complete short proof. Each cyclicity, intertwining, coordinate, orbit, and stabilizer calculation is written out above. Blackadar's Part II §10 is only background: pp. 212–213 discuss group C*-algebra functoriality and mention induction while omitting its construction; pp. 219–220 discuss cocycle conjugacy of actions. Those passages do not prove this monomial-transversal lemma and are not used as proof substitutes.

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Sources

Bachir Bekka and Pierre de la Harpe, Unitary Representations of Groups, Duals, and Characters (author-hosted complete book draft, arXiv:1912.07262v1, 16 December 2019) (standard reference, not scraped)
Bruce Blackadar, Operator Algebras: Theory of C*-Algebras and von Neumann Algebras (author-hosted complete text) (standard reference, not scraped)