Proof Let $T\colon \mathcal {B}_H\to \mathbb {B}(X)$ be a representation. Fix $t\in G$ and elements $b_1,\ldots ,b_n\in \mathcal {B}_{tH}.$ Take $s_1,\ldots ,s_n\in H$ such that $b_j\in B_{ts_j}$ ( $j=1,\ldots , n$ ). Set $\mathfrak {s}:=(s_1,\ldots , s_n)$ and $t\mathfrak {s}:=(ts_1,\ldots ,ts_n)$ and define the matrix space

$$ \begin{align*} \mathbb{M}_{t\mathfrak{s}}(\mathcal{B}):=\{ (M_{i,j})_{i,j=1}^n\colon M_{i,j}\in B_{(ts_i)^{-1}(ts_j)},\ \forall\ i,j=1,\ldots,n \}. \end{align*} $$

It is of key importance to notice that $\mathbb {M}_{t\mathfrak {s}}(\mathcal {B})\equiv \mathbb {M}_{\mathfrak {s}}(\mathcal {B}_H).$

By [Reference Abadie and Ferraro2, Lemma 2.8], $\mathbb {M}_{t\mathfrak {s}}(\mathcal {B})$ is a C*-algebra with usual matrix multiplication as product and $*$ -transpose as involution. A quick way of proving this is by taking a representation $R\colon \mathcal {B}\to \mathbb {B}(Y)$ with all the restrictions $R|_{B_t}$ being isometric (which exists by [Reference Fell and Doran11, Chapter VIII Section 16.10]) and to identify $\mathbb {M}_{t\mathfrak {s}}(\mathcal {B})$ with the concrete C*-algebra

$$ \begin{align*} \{ (R_{M_{i,j}})_{i,j=1}^n\colon M_{i,j}\in B_{s_i^{-1}s_j},\ \forall\ i,j=1,\ldots,n\}\subset \mathbb{B}(Y^n). \end{align*} $$

The matrix $M:=(b_i^{*}b_j)_{i,j=1}^n$ belongs to $\mathbb {M}_{t\mathfrak {s}}(\mathcal {B})$ and, regarding $\mathcal {B}$ as a $\mathcal {B}-\mathcal {B}$ -equivalence bundle, we can easily adapt the proof of [Reference Abadie and Ferraro2, Lemma 2.8] to show that M is positive in $\mathbb {M}_{t\mathfrak {s}}(\mathcal {B})\equiv \mathbb {M}_{\mathfrak {s}}(\mathcal {B}_H).$ An alternative (direct) proof is as follows.

Notice that for all $M=(M_{i,j})_{i,j=1}^n\in \mathbb {M}_{t\mathfrak {s}}(\mathcal {B})$ and all $i=1,\ldots ,n$ we have $M_{i,i}\in B_e.$ So we may define a trace function $\operatorname {tr}\colon \mathbb {M}_{t\mathfrak {s}}(\mathcal {B})\to B_e$ by $\operatorname {tr}(M):=\sum _{i=1}^n M_{i,i}.$ The operations

$$ \begin{align*} \mathbb{M}_{t\mathfrak{s}}(\mathcal{B})\times & B_e\to B_e & (M,b)&\mapsto (M_{i,j}b)_{i,j=1}^n\\ \langle\ ,\ \rangle_{B_e}\colon \mathbb{M}_{t\mathfrak{s}}(\mathcal{B})\times & \mathbb{M}_{t\mathfrak{s}}(\mathcal{B})\to B_e & (M,N)&\mapsto \langle M,N\rangle_{B_e}:=\operatorname{tr}(M^{*}N) \end{align*} $$

make of $\mathbb {M}_{t\mathfrak {s}}(\mathcal {B})$ a full right $B_e$ -Hilbert module which we denote by $\mathbb {M}_{t\mathfrak {s}}(\mathcal {B})_{B_e}.$

For every $M\in \mathbb {M}_{t\mathfrak {s}}(\mathcal {B}),$ the operator $\phi _M\colon \mathbb {M}_{t\mathfrak {s}}(\mathcal {B})_{B_e}\to \mathbb {M}_{t\mathfrak {s}}(\mathcal {B})_{B_e}, N\mapsto MN,$ is adjointable because for all $N,P\in \mathbb {M}_{t\mathfrak {s}}(\mathcal {B})_{B_e},$

$$ \begin{align*} \langle \phi_M N,P\rangle_{B_e}= \operatorname{tr}((MN)^{*}P)=\operatorname{tr}(N^{*}(M^{*}P))=\langle N,\phi_{M^{*}}P\rangle_{B_e}. \end{align*} $$

Moreover, the natural representation $\phi \colon \mathbb {M}_{t\mathfrak {s}}(\mathcal {B})\to \mathbb {B}(\mathbb {M}_{t\mathfrak {s}}(\mathcal {B})_{B_e})$ , $M\mapsto \phi _M,$ is a *-homomorphism.

To prove that $\phi $ is faithful, we take an approximate unit $\{e_i\}_{i\in I}$ of $\mathbb {M}_{t\mathfrak {s}}(\mathcal {B}).$ For all $M\in \mathbb {M}_{t\mathfrak {s}}(\mathcal {B}),$ we have $\sum _{i,j=1}^n (M_{i,j})^{*}M_{i,j} = \operatorname {tr}(M^{*}M)=\lim _i\operatorname {tr}((Me_i)^{*}M)=\lim _i \langle \phi _Me_i,M\rangle _{B_e}.$ Thus, $\phi _M=0$ implies $\sum _{i,j=1}^n (M_{i,j})^{*}M_{i,j}=0$ and this yields $M=0.$

Now that we know $\phi $ is faithful, to prove that $M:=(b_i^{*}b_j)_{i,j=1}^n \geq 0 $ (in $\mathbb {M}_{t\mathfrak {s}}(\mathcal {B})\equiv \mathbb {M}_{\mathfrak {s}}(\mathcal {B}_H)$ ) it suffices to show that $\phi _M\geq 0.$ This is the case because for all $N\in \mathbb {M}_{t\mathfrak {s}}(\mathcal {B})_{B_e}$ we have

$$ \begin{align*} \langle \phi_M N,N\rangle_{B_e} = \operatorname{tr}((\phi_M N)^{*}N)=\operatorname{tr}(N^{*}M^{*}N)=\sum_{i=1}^n\left(\sum_{j=1}^n b_jN_{j,i}\right)^{*}\left(\sum_{k=1}^n b_kN_{k,i}\right)\geq 0; \end{align*} $$

where the last inequality follows from the fact that the definition of Fell bundle requires $c^{*}c$ to be positive in $B_e$ for all $c\in B.$

If $N:= M^{1/2}\in \mathbb {M}_{\mathfrak {ts}}(\mathcal {B})\equiv \mathbb {M}_{\mathfrak {s}}(\mathcal {B}_H),$ then all the entries $N_{i,j}$ of N belong to $\mathcal {B}_H$ and for all $\xi _1,\ldots , \xi _n\in X$ we have

$$ \begin{align*} \sum_{i,j=1}^n\langle T_{b_j^{*}b_i}\xi_i,\xi_j\rangle = \sum_{i,j,k=1}^n \langle T_{{N_{k,j}}^{*}N_{k,i}} \xi_i,\xi_j\rangle = \sum_{k=1}^n \langle \sum_{i=1}^n T_{N_{k,i}}\xi_i,\sum_{j=1}^nT_{N_{k,j}}\xi_j\rangle\geq 0; \end{align*} $$

proving that T is $\mathcal {B}$ -positive.

Proof Let $J\subset C^{*}(\mathcal {B})$ be the intersection of all the kernels of integrated forms of representations of $\mathcal {B}$ induced from representations of $\mathcal {B}_H.$ We obviously have $J\subset \ker (\tilde { {\mathop {\mathrm {Ind}}} } _H^{\mathcal {B}}(T)),$ so there exists a unique function $\rho \colon C^{*}(\mathcal {B})/J\to \mathbb {B}(X^p_T)$ such that $\rho \circ q^{H\uparrow \mathcal {B}} = \tilde { {\mathop {\mathrm {Ind}}} } _H^{\mathcal {B}}(T)$ (implying $\rho $ is a representation).

Assume $\tilde { T } $ is faithful and let S be any other representation of $\mathcal {B}_H.$ Both S and its nondegenerate part induce the same representation of $\mathcal {B},$ so we may assume both S and T are nondegenerate. We have $\tilde { S } \preceq \tilde { T } $ because $\tilde { T } $ is faithful. By the definition of weak containment for bundles, $S\preceq T$ and the continuity of the induction process with respect the regional topology [Reference Fell and Doran11, Chapter XI Section 12.4] implies $\tilde { {\mathop {\mathrm {Ind}}} } _H^{\mathcal {B}}(S)\preceq \tilde { {\mathop {\mathrm {Ind}}} } _H^{\mathcal {B}}(T).$ Hence, $\ker (\tilde { {\mathop {\mathrm {Ind}}} } _H^{\mathcal {B}}(T))\subset \ker (\tilde { {\mathop {\mathrm {Ind}}} } _H^{\mathcal {B}}(S))$ and it follows that $J=\ker (\tilde { {\mathop {\mathrm {Ind}}} } _H^{\mathcal {B}}(T))$ or, in other words, that $\rho $ is faithful.

Proof By definition, we may think of $C^{*}_{H\uparrow G}(\mathcal {B})$ as the image of the *-homomorphism $\delta \colon C^{*}(\mathcal {B})\to M(C^{*}(\mathcal {B})\otimes Q)$ associated with the quotient map $q\colon C^{*}(G)\to Q:=C^{*}(G)/I;$ with $I\subset C^{*}(G)$ the intersection of all the kernels of integrated form of *-representations induced from $H.$ Hence, $\tilde { {\mathop {\mathrm {Ind}}} } _H^G(U)$ factors through a representation $\pi \colon Q\to \mathbb {B}(Y)$ and we get a representation $\tilde { T } \otimes \pi \colon C^{*}(\mathcal {B})\otimes Q\to \mathbb {B}(X\otimes Y)$ we may extend to the multiplier algebra to get a representation $\overline {\tilde { T } \otimes \pi }.$

When we described $\delta \colon C^{*}(\mathcal {B})\to M(C^{*}(\mathcal {B})\otimes Q)$ in the introduction we made use of canonical unitary representation $\iota \colon G\to M(Q)$ and the action of $\mathcal {B}$ on $C_c(\mathcal {B})\subset C^{*}(\mathcal {B}).$ In fact, $\delta _f (g\otimes z) =\int _G f(t)g\otimes \iota (t)z\, dt$ for all $f,g\in C_c(\mathcal {B})$ and $z\in Q.$ This yields $ \overline {\tilde { T } \otimes \pi }\circ \delta = T\tilde { \otimes } {\mathop {\mathrm {Ind}}}_H^G(U).$ Since we are identifying $C^{*}_{H\uparrow G}(\mathcal {B})=\delta (C^{*}(\mathcal {B})),$ the quotient $q^{H\uparrow G}_{\mathcal {B}}\colon C^{*}(\mathcal {B})\to C^{*}_{H\uparrow G}(\mathcal {B})$ becomes $f\mapsto \delta _f$ and we may therefore define $\rho $ as the restriction of $\overline {\tilde { T } \otimes \pi }$ to $C^{*}_{H\uparrow G}(\mathcal {B}).$ Say we have some other representation $\rho '\colon C^{*}_{H\uparrow G}(\mathcal {B})\to \mathbb {B}(X\otimes Y)$ such that $\rho '\circ q^{H\uparrow G}_{\mathcal {B}}=T\tilde { \otimes } {\mathop {\mathrm {Ind}}}_H^G(U).$ Then $\rho '\circ q^{H\uparrow G}_{\mathcal {B}} =\rho \circ q^{H\uparrow G}_{\mathcal {B}}$ and this yields $\rho =\rho '$ because $q^{H\uparrow G}_{\mathcal {B}}$ is surjective.

Assume the integrated forms of T and U are faithful. Then, $\pi $ is faithful because given any other representation V of H we have $\tilde { V } \preceq \tilde { U } \Rightarrow V\preceq U\Rightarrow {\mathop {\mathrm {Ind}}}_H^G(V)\preceq {\mathop {\mathrm {Ind}}}_H^G(U)\Rightarrow \tilde { {\mathop {\mathrm {Ind}}} } _H^G(V)\preceq \tilde { {\mathop {\mathrm {Ind}}} } _H^G(U)$ and this implies I is the kernel of $\tilde { {\mathop {\mathrm {Ind}}} } _H^G(U)$ or, in other words, $\pi $ is faithful and so must be $\overline {\tilde { T } \otimes \pi };$ which clearly implies $\rho $ is faithful.

Proof Let $T\colon \mathcal {B}_H\to \mathbb {B}(X)$ be a representation. By Proposition 2.5, there exists a *-homomorphism $\theta \colon \mathbb {B}(L^2_H(\mathcal {B}))\to \mathbb {B}(L^2_H(\mathcal {B})\otimes _{\tilde { T } }X)$ such that $\theta (M)=M\otimes _{\tilde { T } }1$ and $\theta \circ \Lambda ^{H\mathcal {B}} = \tilde { {\mathop {\mathrm {Ind}}} } _H^{\mathcal {B}}(T).$ The restriction $\theta |_{\Lambda ^{H\mathcal {B}}(C^{*}(\mathcal {B}))}\colon \Lambda ^{H\mathcal {B}}(C^{*}(\mathcal {B}))\to \mathbb {B}(X^p_T)$ is then the unique representation $\rho $ of $\Lambda ^{H\mathcal {B}}(C^{*}(\mathcal {B}))$ such that $\rho \circ \Lambda ^{H\mathcal {B}}= \tilde { {\mathop {\mathrm {Ind}}} } _H^{\mathcal {B}}(T).$ If $\tilde { T } $ is faithful then $\theta $ is also, which clearly implies $\rho $ is faithful.

Proof For convenience, we introduce the following notation:

$$ \begin{align*} {}_tH&\equiv tH{t^{-1}} & {}_{t|}T & \equiv T|_{\mathcal{B}_{tH{t^{-1}}}} & V&\equiv {\mathop{\mathrm{Ind}}}_H^G(U) & p_t&:=p_{tH{t^{-1}}}. \end{align*} $$

The Hilbert space induced by U is $Y^p_U$ and we regard $X\otimes ^G Y^p_U:=\ell ^2(G)\otimes X\otimes Y^p_U$ as an $\ell ^2$ -direct sum of $\# G$ copies of $X\otimes Y^p_U$ . Similarly, the direct sum of $\#G$ copies of $T\otimes V$ is

$$ \begin{align*} T\otimes^G V &\colon \mathcal{B}\to \mathbb{B}(X\otimes^G Y^p_U) & b&\mapsto 1_{\ell^2(G)}\otimes [T\otimes V]_b, \end{align*} $$

which is the composition of $T\otimes V$ with the faithful *-homomorphism

$$ \begin{align*} \Theta\colon \mathbb{B}(X\otimes Y^p_U)\to \mathbb{B}(X\otimes^G Y^p_U),\ \Theta(R)=1_{\ell^2(G)}\otimes R. \end{align*} $$

If we denote by $T\tilde { \otimes } V$ and $T\tilde { \otimes } ^G V$ the integrated forms of $T\otimes V$ and $T\otimes ^G V,$ respectively, then $\Theta \circ (T\tilde { \otimes } V)=T\tilde { \otimes } ^GV.$ Indeed, for all $f\in C_c(\mathcal {B})$ and every elementary tensor $g\otimes \xi \otimes \eta \in \ell ^2(G)\otimes X\otimes Y^p_U,$ we have

$$ \begin{align*} [T\tilde{ \otimes } ^GV]_f(g\otimes \xi\otimes \eta) & = \int_G g\otimes [T\otimes V]_{f(t)}(\xi\otimes \eta)\, dt =g\otimes \left([T\tilde{ \otimes } V]_f (\xi\otimes\eta)\right)\\ & = \Theta\circ [T\tilde{ \otimes } V]_f(g\otimes \xi\otimes \eta). \end{align*} $$

Since $\Theta $ is an isometry and both $T\otimes V$ and all the members of $\{{\mathop {\mathrm {Ind}}}_{{}_t H}^{\mathcal {B}}({}_{t|}T\otimes {}_tU)\}_{t\in G}$ are nondegenerate, we can use [Reference Fell and Doran11, Chapter XI Section 8.20] to the deduce it suffices to show that

(4.1) $$ \begin{align} \|[T\tilde{ \otimes } ^G V]_f \|= \sup\{ \|\tilde{ {\mathop{\mathrm{Ind}}} } _{{}_t H}^{\mathcal{B}}({}_{t|}T\otimes {}_tU)_f\|\colon t\in G \}\qquad \forall\ f\in C_c(\mathcal{B}). \end{align} $$

The $\ell ^2$ -direct sum

(4.2) $$ \begin{align} S:=\bigoplus_{t\in G} {\mathop{\mathrm{Ind}}}_{{}_tH}^{\mathcal{B}}({}_{t|}T\otimes {}_tU) \end{align} $$

is a nondegenerate representation whose integrated form is $\tilde { S } = \oplus _{t\in G} \tilde { {\mathop {\mathrm {Ind}}} } _{{}_tH}^{\mathcal {B}}({}_{t|}T\otimes {}_tU)$ and $\|\tilde { S } _f\|=\sup \{ \|\tilde { {\mathop {\mathrm {Ind}}} } _{{}_t H}^{\mathcal {B}}({}_{t|}T\otimes {}_tU)_f\|\colon t\in G \}.$ Thus, S is weakly equivalent to $T\otimes ^G V$ if and only if (4.1) holds.

To compare $T\otimes ^G V$ and $S,$ we will construct a linear isometry between their domains,

(4.3) $$ \begin{align} I\colon \bigoplus_{t\in G} (X\otimes Y)^{p_t}_{{}_{t|}T\otimes {}_tU} \to X\otimes^G Y^p_U, \end{align} $$

that intertwines S and $T\otimes ^GV.$ To state the properties defining $I,$ we need to prove the existence of a (unique) linear function

(4.4) $$ \begin{align} L\colon C_c(G,X\otimes^G Y )\to X\otimes^G Y^p_U \end{align} $$

which is continuous in the inductive limit topology and satisfies

(4.5) $$ \begin{align} L( f\odot [\delta_t\otimes \xi\otimes \eta ]) =\delta_t\otimes ( \xi \otimes [f\otimes_{U} \eta]) \end{align} $$

for all $f\in C_c(G)$ , $\xi \in X$ , $\eta \in Y$ and $t\in G,$ where we regard the algebraic tensor product $C_c(G)\odot (X\otimes ^G Y)$ as a subspace of $C_c(G,X\otimes ^G Y)$ in the usual way ( $f\odot u$ is the function $t\mapsto f(t)u$ ).

Uniqueness of L follows from the fact that the functions of the form $f\odot (\delta _t\otimes \xi \otimes \eta )$ span a subset of $C_c(G,X\otimes ^G Y)$ which is dense in the inductive limit topology [Reference Fell and Doran10, Chapter II Section 14.6]. To prove the existence of $L,$ we take $u,v\in C_c(G,X\otimes ^G Y)$ such that $u = f\odot (\delta _r\otimes \xi \otimes \eta )$ and $v=g\odot (\delta _s\otimes \zeta \otimes \kappa )$ for some $f,g\in C_c(G)$ , $r,s\in G$ , $\xi ,\zeta \in X$ and $\eta ,\kappa \in Y.$ The inner product of the candidates for $L(u)$ and $L(v)$ is

(4.6) $$ \begin{align} \langle & \delta_r\otimes \xi \otimes [f\otimes_{U} \eta],\delta_s\otimes \zeta \otimes [g \otimes_{U} \kappa]\rangle \nonumber\\ & = \langle \delta_r,\delta_s\rangle\langle \xi,\zeta\rangle \langle f\otimes_{U} \eta,g \otimes_{U} \kappa\rangle \nonumber \\ & = \langle \delta_r,\delta_s\rangle \langle \xi,\zeta\rangle \int_{H} \int_G \Delta_H^G(t)^{1/2} \overline{f(z)} g(zt) \langle \eta, U_t \kappa\rangle \,dzdt \nonumber \\ & = \int_{H} \int_G \Delta_H^G(t)^{1/2} \overline{f(z)} g(zt) \langle \delta_r \otimes \xi\otimes\eta, (1\otimes 1\otimes U_t)(\delta_s\otimes \zeta\otimes \kappa )\rangle \,dzdt\nonumber \\ &= \int_{H} \int_G \Delta_H^G(t)^{1/2} \langle u(z), (1\otimes 1\otimes U_t)v(zt)\rangle \,dzdt. \end{align} $$

Fix a compact $D\subset G$ and denote by $C_D(G,X\otimes ^G Y)$ the set formed by those $f\in C_c(G,X\otimes ^G Y)$ with ${\mathop {\mathrm { supp}}}(f)\subset D.$ This vector space is a Banach space with the norm $\|\ \|_\infty .$ Let $C_D^\odot $ be the subspace of $C_D(G,X\otimes ^G Y)$ spanned by the functions of the form $f\odot (\delta _t\otimes \xi \otimes \eta )$ with $f\in C_D(G)$ , $t\in G$ , $\xi \in X$ and $\eta \in Y.$ We clearly have $C_c(G)C_D^\odot \subset C_D^\odot .$ If for each $t\in G,$ we define $C_D^\odot (t)$ as the closure of $\{u(t)\colon u\in C_D^\odot \},$ then $C_D^\odot (t)=X\otimes ^G Y$ for every t in the interior of D and $\{0\}$ otherwise. By [Reference Abadie1, Lemma 5.1], the closure of $C_D^\odot $ in $C_c(G,X\otimes ^G Y)$ with respect to the inductive limit topology is $\{f\in C_c(G,X\otimes ^G Y)\colon f(t)\in C_D^\odot (t)\, \forall \, t\in G \}=C_D(G,X\otimes ^G Y).$

Take any $u,v\in C_D(G,X\otimes ^G Y).$ By the preceding paragraph, there are sequences $\{u_n\}_{n\in \mathbb {N}}$ and $\{v_n\}_{n\in \mathbb {N}}$ in $C_D^\odot $ converging uniformly to u and $v,$ respectively. Then for all $n\in \mathbb {N}$ there exists, a positive integer $m_n$ and (for each $j=1,\ldots ,m_n$ ) elements $f_{n,j},g_{n,j}\in C_D(G)$ , $r_{n,j},s_{n,j}\in G$ , $\xi _{n,j},\zeta _{n,j}\in X$ and $\zeta _{n,j},\kappa _{n,j}\in Y$ such that

$$ \begin{align*} u_n &=\sum_{j=1}^{m_n} f_{n,j}\odot (\delta_{r_{n,j}}\otimes \xi_{n,j}\otimes \eta_{n,j} ) & v_n&=\sum_{j=1}^{m_n}g_{n,j}\odot (\delta_{s_{n,j}}\otimes \zeta_{n,j}\otimes \kappa_{n,j}). \end{align*} $$

By (4.6), for all $a,b\in \mathbb {N}$ we have

$$ \begin{align*} &\left\|\sum_{j=1}^{m_a} \delta_{r_{a,j}} \otimes (\xi_{a,j}\otimes [f_{a,j}\otimes_{U} \eta_{a,j}] )-\sum_{k=1}^{m_b}\delta_{s_{b,k}}\otimes ( \zeta_{b,k}\otimes [g_{b,k}\otimes_{U} \kappa_{b,k}] )\right\|^2 \\ &\quad = \left| \int_{H} \int_G \Delta_H^G(t)^{1/2} \langle (u_a-v_b)(z), (1\otimes 1\otimes U_t)(u_a-v_b)(zt)\rangle \,dzdt \right|. \end{align*} $$

For the inner product inside the integral above to be non zero, we must have $z,tz\in D,$ which implies $t= tz z^{-1}\in H\cap (DD^{-1}).$ If $\alpha _D$ and $\beta _D$ are the measures of D and $H\cap (D^{-1}D)$ with respect to the Haar measures of G and $H,$ respectively, and $\gamma _D:=\sup \{ \Delta _H^G(t)^{1/2} \colon t\in H\cap (D^{-1}D)\},$ then

$$ \begin{align*} \left\|\sum_{j=1}^{m_a} \delta_{r_{a,j}} \otimes (\xi_{a,j}\otimes [f_{a,j}\otimes_{U} \eta_{a,j}] )-\sum_{k=1}^{m_b}\delta_{s_{b,k}}\otimes ( \zeta_{b,k}\otimes [g_{b,k}\otimes_{U} \kappa_{b,k}] )\right\|^2 \end{align*} $$

is not greater than $\|u_a-v_b\|_\infty ^2 \alpha _D\beta _D\gamma _D.$

Several conclusion arise from that bound:

1. (1) If we take $u=v$ and $v_b=u_b,$ it follows that $\{\sum _{j=1}^{m_a} \delta _{r_{a,j}} \otimes (\xi _{a,j}\otimes [f_{a,j}\otimes _{U} \eta _{a,j}]) \}_{a\in \mathbb {N}}$ is a Cauchy sequence in $X\otimes ^G Y^p_U;$ the limit of which we denote by $L_D(\{u_n\}_{n\in \mathbb {N}}).$

2. (2) If $u=v$ and we take limit in a and $b,$ we get $L_D(\{u_n\}_{n\in \mathbb {N}})= L_D(\{v_n\}_{n\in \mathbb {N}}).$ Thus, we may define a function $L_D\colon C_D(G,X\otimes ^G Y)\to X\otimes ^G Y^p_U,\ u\mapsto L_D(u):=L_D(\{u_n\}_{n\in \mathbb {N}}).$

3. (3) Taking limit in a and b we obtain $\| L_D(u)-L_D(v) \|\leq \|u-v\|_{\infty } \sqrt {\alpha _D\beta _D\gamma _D},$ so $L_D$ is continuous.

4. (4) $L_D(f\odot (\delta _t\otimes \xi \otimes \eta ))=\delta _t\otimes \xi \otimes (f\otimes _U\eta )$ and $L_D$ is linear when restricted to $C_D^\odot .$ Thus $L_D$ is linear.

5. (5) By (4.6) and the continuity of $L_D,$

   (4.7) $$ \begin{align} \langle L_D(u),L_D(v)\rangle = \int_G \int_H \Delta_H^G(t)^{1/2}\langle u(s), (1\otimes 1 \otimes U_t)v(st)\rangle \,dt ds. \end{align} $$

It follows immediately that $L_E$ is an extension of $L_D$ whenever $E\subset G$ is a compact set containing $D.$ Then, there exists a unique function $L\colon C_c(G,X\otimes ^G Y)\to X\otimes ^G Y^p_U$ extending all the $L_D$ ’s. This extension is linear and continuous in the inductive limit topology by [Reference Fell and Doran10, Chapter II Section 14.3]. Note also that L satisfies (4.5) and, consequently, it has dense range.

Now that we know L exists, we are one step closer of being able to specify the properties defining the map I of (4.3). Given $r\in G$ , $f\in C_c(\mathcal {B})$ , $\xi \in X$ and $\eta \in Y$ we define $[r,f,\xi ,\eta ]\in C_c(G,X\otimes ^G Y)$ by

(4.8) $$ \begin{align} [r,f,\xi,\eta](s)= \Delta_G(r)^{-1/2} \delta_r\otimes T_{f(s{r^{-1}})}\xi\otimes \eta. \end{align} $$

We claim there exists a unique linear and continuous map I with the domains and ranges specified in (4.3) and such that for all $t\in G$ , $f\in C_c(\mathcal {B})$ , $\xi \in X$ and $\eta \in Y,$

(4.9) $$ \begin{align} I( f\otimes_{{}_{t|}T\otimes {}_tU} (\xi\otimes \eta) ) = L([t,f,\xi,\eta]). \end{align} $$

To prove this claim, we start by taking elementary tensors $f\otimes _{{}_{r|}T\otimes {}_r U}(\xi \otimes \eta )$ and $g\otimes _{{}_{t|}T\otimes {}_tU}(\zeta \otimes \kappa )$ on (possibly equal) direct summands of $\bigoplus _{s\in G} (X\otimes Y)^{p_s}_{{}_{s|}T\otimes {}_sU}.$ Notice the r and the t in the subindexes indicate the direct summands the tensors belongs to. By (4.7), we have

$$ \begin{align*} \langle & L([r,f,\xi,\eta]),L([t,g,\zeta,\kappa])\rangle= \\ & = \int_G\int_H \Delta_H^G(w)^{1/2}\Delta_G(rt)^{-1/2} \langle \delta_r\otimes T_{f(z{r^{-1}} )}\xi \otimes \eta ,\delta_t\otimes T_{g(zw{t^{-1}})}\zeta\otimes U_w\kappa\rangle \,dwdz \\ & = \int_H\int_G \Delta_H^G(w)^{1/2} \underbrace{\Delta_G(rt)^{-1/2}\langle \delta_r,\delta_t\rangle}_{=\Delta_G({r^{-1}})\langle \delta_r,\delta_t\rangle} \langle T_{f(z{r^{-1}} )}\xi \otimes \eta ,T_{g(zw{r^{-1}})}\zeta\otimes U_w\kappa\rangle \,dzdw \\ & = \int_H\int_G \Delta_H^G(w)^{1/2} \langle \delta_r,\delta_t\rangle \langle T_{f(z)}\xi \otimes \eta ,T_{g(zr w{r^{-1}})}\zeta\otimes U_w\kappa\rangle \,dzdw\\ &= \int_H\int_G \Delta_{{}_rH}^G(rw{r^{-1}})^{1/2} \langle \delta_r,\delta_t\rangle \langle \xi \otimes \eta ,T_{f(z)^{*}g(zr w{r^{-1}} )}\zeta\otimes {}_r U_{rw{r^{-1}}}\kappa\rangle \,dzdw \\ &= \int_{{}_rH}\int_G \Delta_{{}_rH}^G(w)^{1/2} \langle \delta_r,\delta_t\rangle \langle \xi \otimes \eta ,T_{f(z)^{*}g(zw)}\zeta\otimes {}_r U_{w}\kappa\rangle \,dzdw \\ &= \langle \delta_r,\delta_t\rangle \langle \xi \otimes \eta ,[{}_{r|}T\tilde{ \otimes } {}_r U]_{p_r(f^{*}*g)}(\zeta\otimes \kappa)\rangle \\ &= \langle f\otimes_{{}_{r|}T\otimes {}_r U}(\xi\otimes \eta),g\otimes_{{}_{t|}T\otimes {}_tU}(\zeta\otimes\kappa)\rangle. \end{align*} $$

The existence of I then follows immediately from Remark 4.1 if one considers the functions

$$ \begin{align*} b&\colon G\times C_c(\mathcal{B})\times X\times Y\to \bigoplus_{t\in G}(X\otimes Y)^{p_t}_{{}_{t|}T\otimes{}_tU} & b(r,f,\xi,\eta)&=f\otimes_{{}_{r|}T\otimes {}_r U}(\xi\otimes \eta)\\ c& \colon G\times C_c(\mathcal{B})\times X\times Y\to X\otimes^G Y^p_U & c(r,f,\xi,\eta)&=L([r,f,\xi,\eta]). \end{align*} $$

We are not sure if I is surjective, but we can use the ideas of [Reference Exel and Ng8] to “move” the image of I to “fill” $X\otimes ^G Y.$ The movement is via the map

$$ \begin{align*} \rho & \colon G\to \mathbb{B}(X\otimes^G Y^p_U) & \rho_t &:= \mathtt{lt}_t\otimes 1_{X}\otimes 1_{Y^p_U}, \end{align*} $$

where $\mathtt {lt}\colon G\to \mathbb {B}(\ell ^2(G))$ is the left regular representation of the discrete version of G ( $\mathtt {lt}_t(\delta _s)=\delta _{ts}$ ). Note $\rho $ and $\Theta $ have commuting ranges, so the range of $\rho $ commutes with that of $T\tilde { \otimes } ^GV=\Theta \circ (T\tilde { \otimes } G).$ We remark that the continuity of $\rho $ (which may fail) plays no rôle in the proof.

Let K be the image of $I.$ We claim that $ G\cdot K:= {\mathop {\mathrm {span}}}\{\rho _t K\colon t\in G\}$ is dense in $X\otimes ^G Y^p_U.$ To prove this we define, for each $t\in G,$ the function

$$ \begin{align*} \mu_t \colon & C_c(G,X\otimes^G Y)\to C_c(G,X\otimes^G Y) & (\mu_t f)(z) &=(\mathtt{lt}_t\otimes 1_{X}\otimes 1_{Y})f(z). \end{align*} $$

In particular, $\mu _t (f\odot (\delta _r\otimes \xi \otimes \eta ))=f\odot (\delta _{tr}\otimes \xi \otimes \eta ).$ Hence,

$$ \begin{align*} L\circ \mu_t (f\odot (\delta_r\otimes \xi\otimes \eta)) & = L(f \odot (\delta_{tr}\otimes \xi\otimes \eta)) =\delta_{tr}\otimes \xi \otimes [f\otimes_{U} \eta] \\ & =\rho_t (\delta_{r}\otimes \xi \otimes [f\otimes_{U} \eta]) =\rho_t L(f\odot( \delta_r\otimes \xi\otimes \eta )) \end{align*} $$

and we get $L\circ \mu _t =\rho _t \circ L$ because both $L\circ \mu _t $ and $\rho _t \circ L$ are linear and continuous in the inductive limit topology and agree on a dense set. Thus $\overline {G\cdot K}$ contains the image through L of

$$ \begin{align*} K_0:= {\mathop{\mathrm{span}}}\{ \mu_t [r,f,\xi,\eta]\colon r,t\in G,\, f\in C_c(\mathcal{B}),\, \xi\in X,\, \eta\in Y \}\subset C_c(G,X\otimes^G Y). \end{align*} $$

Note $C(G)K_0\subset K_0.$ Besides,

(4.10) $$ \begin{align} \mu_r [t,f,\xi,\eta](z) =\Delta_G(t)^{-1/2}\delta_{rt}\otimes T_{f(z{t^{-1}})} \xi\otimes \eta. \end{align} $$

Fixing $z\in G$ and varying $r,t\in G$ , $\xi \in X$ , $\eta \in Y$ and $f\in C_c(\mathcal {B}),$ the elements we obtain on the right-hand side of (4.10) are all those of the form $\delta _{s}\otimes T_b \xi \otimes \eta ,$ for arbitrary $s\in G$ , $b\in \mathcal {B}$ , $\xi \in X$ and $\eta \in Y.$ The closed linear span of this vectors is $X\otimes ^G Y$ because T is nondegenerate, and we conclude (using [Reference Fell and Doran10, Chapter II Section 14.3]) that $K_0$ is dense in $C_c(G,X\otimes ^G Y)$ in the inductive limit topology. Hence, $\overline {G\cdot K}$ contains the dense set $L(K_0)$ and it follows that $\overline {G\cdot K}=X\otimes ^G Y^p_U.$

Our next goal is to show that I intertwines the S of (4.2) and $T\otimes ^G V.$ To prove this claim, we fix $r,s,t\in G$ , $b\in B_r$ , $f\in C_c(\mathcal {B})$ , $g\in C_c(G)$ , $\xi ,\zeta \in X$ and $\eta , \kappa \in Y.$ For convenience, we denote by u and v the tensors $f\otimes _{{}_{s|}T\otimes {}_{s}U} (\xi \otimes \eta ) $ and $g\odot (\delta _t\otimes \zeta \otimes \kappa ),$ respectively. Recalling (4.7) we get

$$ \begin{align*} \langle & [T\otimes^G V]_b \circ I (u ),L(v ) \rangle = \langle L([s,f,\xi,\eta]),[T\otimes^G V]_{b^{*}}(\delta_t\otimes \zeta\otimes (g\otimes_{U}\kappa)) \rangle \nonumber\\ &= \langle L([s,f,\xi,\eta]),\delta_t\otimes T_{b^{*}}\zeta\otimes ( {r^{-1}} g\otimes_{U}\kappa ) \rangle \nonumber\\ &= \langle L([s,f,\xi,\eta]),L({r^{-1}} g\odot( \delta_t \otimes T_{b^{*}}\zeta\otimes \kappa))\rangle\\ & =\int_H\int_G \Delta_H^G(w)^{1/2} \langle [s,f,\xi,\eta](z),\delta_t \otimes T_{b^{*}}\zeta\otimes U_w\kappa \rangle g(r zw)\,dzdw\\ & =\int_H\int_G \Delta_H^G(w)^{1/2} \Delta_G(s)^{-1/2} \langle \delta_s\otimes T_{f(zs^{-1})}\xi\otimes \eta,\delta_t \otimes T_{b^{*}}\zeta\otimes U_w\kappa \rangle g(r zw)\,dzdw\\ & =\int_H\int_G \Delta_H^G(w)^{1/2} \Delta_G(s)^{-1/2}\langle \delta_s,\delta_t\rangle \langle T_{bf({r^{-1}} zs^{-1})}\xi\otimes \eta,\zeta\otimes U_w\kappa \rangle g(zw)\,dzdw\\ & =\int_H\int_G \Delta_H^G(w)^{1/2} \Delta_G(s)^{-1/2}\langle \delta_s,\delta_t\rangle \langle T_{(bf)(zs^{-1})}\xi\otimes \eta,\zeta\otimes U_w\kappa \rangle g(zw)\,dzdw\\ & =\langle L([s,bf,\xi,\eta]),L(g\odot (\delta_t\otimes \zeta\otimes \kappa))\rangle =\langle I(bf\otimes_{{}_{|s}T\otimes {}_s U}(\xi\otimes \eta)),L(v)\rangle\\ &= \langle I\circ S_b(u),L(v)\rangle. \end{align*} $$

Since u and v are arbitrary basic tensors and L has dense range, by linearity and continuity we get that $[T\otimes ^G V]_b \circ I = I\circ S_b;$ which implies that for all $f\in C_c(\mathcal {B})$

(4.11) $$ \begin{align} [T\tilde{ \otimes } ^G V]_f \circ I = I\circ \tilde{ S } _f. \end{align} $$

The same identity holds for all $f\in C^{*}(\mathcal {B})$ because $C_c(\mathcal {B})$ is dense in $C^{*}(\mathcal {B}).$

Recall that $G\cdot K$ spans a dense subset of $\ell ^2(G)\otimes X\otimes Y^p_U, $ thus for all $g\in C^{*}(\mathcal {B}), $ we have

$$ \begin{align*} [T\tilde{ \otimes } ^G V]_g=0 & \Leftrightarrow \ [T\tilde{ \otimes } ^G V]_g\rho_t \circ I=0\ \forall \ t\in G \Leftrightarrow \ \rho_t [T\tilde{ \otimes } ^G V]_g\circ I=0\ \forall \ t\in G\\ & \Leftrightarrow \ [T\tilde{ \otimes } ^G V]_g\circ I=0 \Leftrightarrow \ I\circ \tilde{ S } _g = 0 \Leftrightarrow \tilde{ S } _g = 0, \end{align*} $$

where the last equivalence follows from the fact that I is an isometry. We may then define a *-homomorphism $\Omega \colon (T\tilde { \otimes } ^G V)(C^{*}(\mathcal {B}))\to \tilde { S } (C^{*}(\mathcal {B}))$ by $\Omega \left ([T\tilde { \otimes } ^G V]_g\right )=\tilde { S } _g.$

The thesis of the theorem was shown to be the equivalent to (4.1) which, in turn, is equivalent to say $\Omega $ is an isometry. But $\Omega $ is indeed an isometry because it is an injective *-homomorphism between two C*-algebras.

Proof Let $\tilde { T } \colon \mathcal {B}\to \mathbb {B}(X)$ be a faithful nondegenerate representation and $\kappa \colon H\to \mathbb {C}$ the trivial representation. By Propositions 3.1 and 3.2, Lemma 3.5, and Theorem 4.1, for all $f\in C^{*}(\mathcal {B}),$ we have

$$ \begin{align*} \|q^{H\uparrow\mathcal{B}}(f)\|=\| {\mathop{\mathrm{Ind}}}_H^{\mathcal{B}}(T|_{\mathcal{B}_H}\otimes \kappa)_f \|\preceq \|[T\otimes {\mathop{\mathrm{ Ind}}}_H^G(\kappa)]_f\|\leq \|q^{H\uparrow G}_{\mathcal{B}}(f)\|; \end{align*} $$

which implies the existence of $\Theta ^H_{\mathcal {B}}.$ Uniqueness is a consequence of the fact that $q^{H\uparrow G}_{\mathcal {B}}$ is surjective. In case $\mathcal {B}$ is saturated, $\Theta ^H_{\mathcal {B}}$ is the inverse of the $\pi ^H_{\mathcal {B}}$ of (1.3).

To prove the claim about the intersection of the kernels, we take a faithful nondegenerate representation $\tilde { U } $ of $H.$ The integrated form of all the restrictions $T|_{\mathcal {B}_{rH{r^{-1}}}}$ and conjugations ${}_r U$ have faithful integrated forms, so for all $f\in C^{*}(\mathcal {B}),$ we have

$$ \begin{align*} q^{H\uparrow G}_{\mathcal{B}}(f)\in \bigcap_{r\in G} \ker(\Theta^{rH{r^{-1}}}_{\mathcal{B}}) & \Leftrightarrow q^{rH{r^{-1}} \uparrow \mathcal{B}}_{\mathcal{B}}(f)\ \forall\ r\in G \nonumber \\ &\Leftrightarrow \| {\mathop{\mathrm{Ind}}}_{rH{r^{-1}}}^{\mathcal{B}}(T|_{\mathcal{B}_{rH{r^{-1}}}}\otimes {}_r U)_f \|=0\ \forall\ r\in G\nonumber \\ & \Leftrightarrow \|[T\otimes {\mathop{\mathrm{Ind}}}_H^G(U)]_f\|=0 \nonumber\\ & \Leftrightarrow q^{H\uparrow G}_{\mathcal{B}}(f)=0, \end{align*} $$

which completes the proof.

Proof Let $\mathcal {C}$ be the bundle C*-completion of $\mathcal {B}$ and let $\rho \colon \mathcal {B}\to \mathcal {C}$ be the canonical quotient map of [Reference Fell and Doran11, Chapter VIII Section 16.7]. The construction of $\mathcal {C}$ implies the existence of a unique representation $S\colon \mathcal {C}\to \mathbb {B}(X)$ such that $S\circ \rho = T.$

Theorem 2.1 implies $S|_{\mathcal {C}_{tH{t^{-1}}}}\otimes {}_t U $ is $\mathcal {C}$ -positive for every $t\in G.$ Hence, by [Reference Fell and Doran11, Chapter XI Section 12.6], the composition $(S|_{\mathcal {C}_{tH{t^{-1}}}}\otimes {}_t U)\circ (\rho |_{\mathcal {B}_{tH{t^{-1}}}})\equiv T|_{\mathcal {B}_{tH{t^{-1}}}}\otimes {}_tU$ is $\mathcal {B}$ -positive for all $t\in G.$

We know that $S\otimes {\mathop {\mathrm {Ind}}}_H^G(U)\approx \{{\mathop {\mathrm {Ind}}}_{tH{t^{-1}} }^{\mathcal {C}}(S|_{\mathcal {C}_{tH{t^{-1}}}}\otimes {}_tU) \}_{t\in G},$ which implies

(4.13) $$ \begin{align} (S\otimes {\mathop{\mathrm{Ind}}}_H^G(U))\circ \rho \approx \{{\mathop{\mathrm{Ind}}}_{tH{t^{-1}} }^{\mathcal{C}}(S|_{\mathcal{C}_{tH{t^{-1}}}}\otimes {}_tU)\circ \rho \}_{t\in G}. \end{align} $$

It is clear that $(S\otimes {\mathop {\mathrm {Ind}}}_H^G(U))\circ \rho = (S\circ \rho )\otimes {\mathop {\mathrm {Ind}}}_H^G(U) = T\otimes {\mathop {\mathrm { Ind}}}_H^G(U).$ On the other hand, by [Reference Fell and Doran11, Chapter XI Section 12.6], for all $t\in G$ we have

$$ \begin{align*} {\mathop{\mathrm{Ind}}}_{tH{t^{-1}} }^{\mathcal{C}}(S|_{\mathcal{C}_{tH{t^{-1}}}}\otimes {}_tU)\circ \rho & = {\mathop{\mathrm{Ind}}}_{tH{t^{-1}} }^{\mathcal{B}}\left( (S|_{\mathcal{C}_{tH{t^{-1}}}}\otimes {}_tU)\circ (\rho|_{\mathcal{B}_{tH{t^{-1}}}})\right) \\ & ={\mathop{\mathrm{Ind}}}_{tH{t^{-1}} }^{\mathcal{B}}( T|_{\mathcal{B}_{tH{t^{-1}}}}\otimes {}_tU). \end{align*} $$

Thus (4.13) implies (4.12).

Proof The existence of $\mu ^{HK}_{\mathcal {B}}$ is equivalent to the validity of $\| q^{K\uparrow \mathcal {B}}(f) \|\leq \| q^{H\uparrow \mathcal {B}}(f) \|$ for all $f\in C^{*}(\mathcal {B}).$ To prove this, we take a faithful nondegenerate representation $\tilde { T } \colon C^{*}(\mathcal {B}_K)\to \mathbb {B}(X).$ By Proposition 3.1 and induction in stages [Reference Fell and Doran11, Chapter XI Section 12.15],

(5.1) $$ \begin{align} \| q^{K\uparrow \mathcal{B}}(f) \|=\|\tilde{ {\mathop{\mathrm{Ind}}} } _H^{\mathcal{B}}(T)_f\|=\|\tilde{ {\mathop{\mathrm{Ind}}} } _K^{\mathcal{B}}({\mathop{\mathrm{Ind}}}_H^{\mathcal{B}_K}(T))_f\|\leq \|q^{H\uparrow \mathcal{B}}(f)\|. \end{align} $$

The proof of the existence of $\nu ^{HK}_{\mathcal {B}}$ is similar. It combines induction in stages with Proposition 3.2, the details are left to the reader with the suggestion to consult (5.2).

Proof We have $ \mu ^{H\{e\}}_{\mathcal {B}} \circ q^{H\uparrow \mathcal {B}} = q^{\{e\}\uparrow \mathcal {B}} = \lambda _{\mathcal {B}}=q^{\{e\}\uparrow G}_{\mathcal {B}} = \nu ^{H\{e\}}_{\mathcal {B}} \circ q^{H\uparrow G}_{\mathcal {B}},$ so it suffices to consider the case $H=\{e\}.$ In [Reference Fell and Doran11, Chapter VIII Section 16.4] Fell proves that the direct sum of the integrated forms of the generalized regular representations of $\mathcal {B}$ is faithful. This is equivalent to say that $\tilde { {\mathop {\mathrm {Ind}}} } _{\{e\}}^{\mathcal {B}}(\pi )|_{L^1(\mathcal {B})}=\lambda _{\mathcal {B}}\otimes _\pi 1|_{L^1(\mathcal {B})}$ is faithful, with $\pi $ the universal representation of $B_e.$ Thus, $\lambda _{\mathcal {B}}|_{L^1(\mathcal {B})}$ is faithful.

Proof The first claim follows easily after one notices $\mu ^{G\{e\}}_{\mathcal {B}} =\mu ^{H\{e\}}_{\mathcal {B}} \circ \mu ^{GH}_{\mathcal {B}}$ ; $\nu ^{G\{e\}}_{\mathcal {B}} =\nu ^{H\{e\}}_{\mathcal {B}} \circ \nu ^{GH}_{\mathcal {B}}$ and identifies $\mu ^{G\{e\}}_{\mathcal {B}}=\lambda _{\mathcal {B}}=\nu ^{G\{e\}}_{\mathcal {B}}.$ For the second claim, we go back to (5.1). The hypothesis implies the integrated form of ${\mathop {\mathrm {Ind}}}_H^{\mathcal {B}_K}(T)$ is faithful, so the inequality of (5.1) becomes an equality.

The proof of the third claim is similar. Take faithful nondegenerate representations $\tilde { T } \colon C^{*}(\mathcal {B})\to \mathbb {B}(X)$ and $\tilde { U } \colon C^{*}(K)\to \mathbb {B}(Y).$ Then, $\tilde { {\mathop {\mathrm {Ind}}} } _K^H(U)$ is faithful and Proposition 3.2 (together with induction in stages) implies that for all $f\in C^{*}(\mathcal {B})$

(5.2) $$ \begin{align} \| q^{K\uparrow \mathcal{B}}(f) \| = \|[T\tilde{ \otimes } {\mathop{\mathrm{Ind}}}_K^G(U)]_f\| = \|[T\tilde{ \otimes } {\mathop{\mathrm{Ind}}}_H^G({\mathop{\mathrm{Ind}}}_K^H(U))]_f\|=\| q^{H\uparrow \mathcal{B}}(f) \|, \end{align} $$

which clearly implies $\nu ^{HK}_{\mathcal {B}}$ is isometric.

If H is amenable and we set $K=\{e\},$ then $C^{*}_{K\uparrow \mathcal {B}}(\mathcal {B})=C^{*}_{K\uparrow G}(\mathcal {B})=C^{*}_{\mathop {\mathrm { r}}}(\mathcal {B})$ ; $C^{*}(H)=C^{*}_{\mathop {\mathrm {r}}}(H)=Q^H_K$ and $C^{*}(\mathcal {B}_H)= C^{*}_{\mathop {\mathrm {r}}}(\mathcal {B}_H)$ (see [Reference Exel and Ng8]). Thus the last claim is a consequence of the preceding ones.

Proof Assume H is open in G and decompose $L^2(G)$ into a direct sum with respect to the right cosets $H\setminus G=\{Ht\colon t\in G\},$ that is $L^2(G)=\bigoplus _{Z\in H\setminus G} L^2(Z).$ With this decomposition $\lambda ^G|_H$ becomes a direct sum of representations unitary equivalent to $\lambda ^H,$ so $\lambda ^G|_H\approx \lambda ^H.$

We now assume that H is normal in G and that the left invariant Haar measures of $G,H$ and $G/H$ have been normalized so that $ \int _G f(t)\, dt = \int _{G/H} \int _H f(ts)\, dsd(tH)$ for all $f\in C_c(G)$ (exactly as in [Reference Fell and Doran10, Chapter III Section 13.17]). Besides, we let $\Gamma \colon G\to \mathbb {R}^+$ be the continuous homomorphism of [Reference Fell and Doran10, Chapter III Section 13.20], i.e., the unique function such that $\int _H f(sts^{-1})\, dt =\Gamma (s)\int _H f(t)\, dt $ for all $f\in C_c(H)$ and $s\in G.$

For any $f\in C_c(H)$ and $\xi \in C_c(G)\subset L^2(G),$ we have

(5.3) $$ \begin{align} \| \tilde{ (\lambda^G|_H) } _f \xi \|^2 & =\int_G\int_H f^{*}*f(t)\xi({t^{-1}} s)\overline{\xi(s)}dtds\nonumber \\ & = \int_{G/H} \int_H\int_H f^{*}*f(t)\xi({t^{-1}} sr)\overline{\xi(sr)}\, dtdrd(sH) \nonumber \\ & = \int_{G/H} \Gamma(s)\int_H\int_H f^{*}*f(t)\xi({t^{-1}} rs)\overline{\xi(rs)}dtdrd(sH). \end{align} $$

To bound the inner double integral in the last term above we define, for each $s\in G$ , $\xi _s\in C_c(H)\subset L^2(H)$ by $\xi _s(z)=\xi (zs).$ Then

$$ \begin{align*} \int_H\int_H f^{*}*f(t)\xi({t^{-1}} rs)\overline{\xi(rs)}dtdr =\int_H f^{*}*f(t)\langle \lambda^H_t \xi_s,\xi_s\rangle \, dt \leq \|\tilde{ \lambda^H } _f\|^2\|\xi_s\|^2 \end{align*} $$

and we can continue (5.3) to get

$$ \begin{align*} \| \tilde{ (\lambda^G|_H) } _f \xi \|^2 \leq \|\tilde{ \lambda^H } _f\|^2 \int_{G/H} \int_H \xi(sr)\overline{\xi(sr)}\, drd(sH) = \|\tilde{ \lambda^H } _f\|^2\|\xi\|^2. \end{align*} $$

Thus $\| \tilde { (\lambda ^G|_H) } _f\|\leq \|\tilde { \lambda ^H } _f\|$ for all $f\in C_c(H)$ and this implies $\lambda ^G|_H\preceq \lambda ^H.$ Lemma 3.5 gives $\lambda ^H\preceq \lambda ^G|_H,$ so $\lambda ^H\approx \lambda ^G|_H.$

For the general case, we define N as the normalizer of H in G and use the arguments of the proof of Lemma 3.5 to get $ \lambda ^G|_H =\lambda ^G|_N|_H\approx \lambda ^N|_H\approx \lambda ^H,$ which implies $\lambda ^G|_H\approx \lambda ^H.$

Proof We have $\mu ^{H\{e\}}_{\mathcal {B}}\circ \Theta ^H_{\mathcal {B}}=\nu ^{H\{e\}}_{\mathcal {B}}.$ Hence, $C^{*}_{H\uparrow G}(\mathcal {B})=C^{*}_{\mathop {\mathrm {r}}}(\mathcal {B})$ (i.e., $\nu ^{H\{e\}}_{\mathcal {B}}$ an isomorphism) if and only if $C^{*}_{H\uparrow G}(\mathcal {B})=C^{*}_{H\uparrow \mathcal {B}}(\mathcal {B})=C^{*}_{\mathop {\mathrm {r}}}(\mathcal {B})$ (both $\mu ^{H\{e\}}_{\mathcal {B}}$ and $\Theta ^H_{\mathcal {B}}$ are isomorphisms). By Corollary 5.3, $C^{*}(\mathcal {B}_H)=C^{*}_{\mathop {\mathrm {r}}}(\mathcal {B}_H)$ implies $C^{*}_{H\uparrow \mathcal {B} }(\mathcal {B})=C^{*}_{\mathop {\mathrm {r}}}(\mathcal {B}).$

Suppose $C^{*}_{H\uparrow \mathcal {B}}(\mathcal {B})=C^{*}_{\mathop {\mathrm {r}}}(\mathcal {B})$ and take a faithful nondegenerate representation $\tilde { T } $ of $\mathcal {B}.$ By Proposition 3.1, ${\mathop {\mathrm {Ind}}}_H^{\mathcal {B}} (T|_{\mathcal {B}_H})\approx {\mathop {\mathrm { Ind}}}_{\{e\}}^{\mathcal {B}} (T|_{B_e})$ and this, together with Lemma 3.5, yields $T|_{\mathcal {B}_H}\preceq {\mathop {\mathrm {Ind}}}_H^{\mathcal {B}} (T|_{\mathcal {B}_H})|_{\mathcal {B}_H}\approx {\mathop {\mathrm {Ind}}}_{\{e\}}^{\mathcal {B}} (T|_{B_e})|_{\mathcal {B}_H}$ $\Rightarrow $ $T|_{\mathcal {B}_H}\preceq {\mathop {\mathrm {Ind}}}_{\{e\}}^{\mathcal {B}} (T|_{B_e})|_{\mathcal {B}_H}.$ By Exel–Ng’s absorption principle, ${\mathop {\mathrm {Ind}}}_{\{e\}}^{\mathcal {B}} (T|_{B_e})|_{\mathcal {B}_H}\approx T|_{\mathcal {B}_H}\otimes \lambda ^G|_H$ and from Lemma 5.4 we get ${\mathop {\mathrm {Ind}}}_{\{e\}}^{\mathcal {B}} (T|_{B_e})|_{\mathcal {B}_H}\approx T|_{\mathcal {B}_H}\otimes \lambda ^H$ and we conclude $T|_{\mathcal {B}_H}\preceq T|_{\mathcal {B}_H}\otimes \lambda ^H.$ Lemma 3.5 implies the integrated for of $T|_{\mathcal {B}_H}$ is faithful, thus $\mathcal {B}_H$ has the WCP.

Proof This is a straightforward consequence of Corollary 5.3 and Theorem 5.5.