$\pi_T(B)|_{H_k}=0$. Fix $\gamma,\delta\in F$. Then $$T^*_{\gamma}\pi_T(B)T_{\delta}=\pi_T(w^*_{\gamma})\pi_T(B)\pi_T(w_{\delta})=\pi_T(D_{\gamma,\delta})$$ Since $T_{\delta}$ is an injection from $H_{e}$ to $H_k$ and $\pi_T(B)|_{H_k}=0$, it follows that $\pi_T(B)T_{\delta}|_{H_e}=0$. Thus $\pi_T(D_{\gamma,\delta})|_{H_e}=0$.

But the restriction $$(\pi_T|_{C^*\big(\mu^{-1}(e),\mu^{-1}(e)\cap P\big)})|_{H_e}$$ is the Toeplitz representation of $\big(\mu^{-1}(e),\mu^{-1}(e)\cap P\big)$, and hence is faithful by amenability. Thus $D_{\gamma,\delta}=0$. Repeating the argument finitely many times shows that all the $D_{\sigma,\tau}=0$ and hence that $B=0$. Thus $\pi_T|_{B_{k,F}}$ is faithful on $H_k$, and therefore is isometric.

\[faithful on Bk\] Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups, and suppose that $\mu:G\to K$ is a controlled map. Let $B_k=\overline{{\operatorname{span}}}\{w_pw^*_q:\mu(p)=\mu(q)=k\}$. Let ${\mathcal{F}}$ be the set of all finite sets $F\subseteq \Sigma_k$. Then $B_k=\overline{\cup_{F\in{\mathcal{F}}}B_{k,F}}$. If $\big(\mu^{-1}(e),\mu^{-1}(e)\cap P\big)$ is amenable, then $\pi_T|_{B_{k}}$ is isometric on $H_k$.

Observe that ${\mathcal{F}}$ is a directed set partially ordered by inclusion with $E,F\in {\mathcal{F}}$ majorismajorizeded by $E\cup F$. If $E\subseteq F$, then $B_{k,E}\subseteq B_{k,F}$. Thus $\{B_{k,F}:F\in {\mathcal{F}}\}$ is an inductive system with limit $\overline{\cup_{F\in{\mathcal{F}}}B_{k,F}}$.

For each $F\in{\mathcal{F}}$ we have $B_{k,F}\subseteq B_k$, and $B_k$ is closed. Therefore $\overline{\cup_{F\in{\mathcal{F}}}B_{k,F}}\subseteq B_k$. To prove the reverse inclusion it suffices to show that the spanning elements of $B_k$ are in $B_{k,F}$ for some $F$. Fix $p,q\in P$ such that $\mu(p)=\mu(q)=k$ and consider $w_pw^*_q$. By Definition \[expanded controlled map definition\], the set 