$\Sigma_k$ of minimal elements is complete, and there exists $\sigma,\tau\in\Sigma_k$ such that $\sigma\leq p$ and $\tau \leq q$. Hence there exists $x,y\in  P$ such that $p=\sigma x$ and $q=\tau y$. Thus $w_pw^*_q=w_{\sigma x}w^*_{\tau y}=w_{\sigma}(w_xw^*_y)w^*_{\tau}$ and $w_xw^*_y\in  C^*\big(\mu^{-1}(e),\mu^{-1}(e)\cap P\big)$. Since $\{\sigma,\tau\}\in{\mathcal{F}}$ we have $w_pw^*_q\in B_{k,\{\sigma,\tau\}}$. Thus $B_k\subseteq \overline{\cup_{F\in{\mathcal{F}}}B_{k,F}}$, and equality follows.

Finally, suppose that $\big(\mu^{-1}(e),\mu^{-1}(e)\cap P\big)$ is amenable. Then $\pi_T|_{B_{k,F}}$ is isometric on $H_k$ for all $F\in{\mathcal{F}}$ by Lemma \[isometric on H\_k\]. Since $\pi_T$ is isometric on every $B_{k,F}$, its extension to the closure is also isometric.

Let ${\mathcal{I}}$ be the set of all finite sets $I\subset Q$ that are closed under $\vee$ in the sense that $s,t\in I$ and $s\vee t<\infty$ implies that $s\vee t\in I$.

\[core is limit of CI\] Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups, and suppose that $\mu:G\to K$ is a controlled map. For each $I\in{\mathcal{I}}$ let $$C_I=\overline{\operatorname{span}}\{w_pw^*_q:\mu(p)=\mu(q)\in I\}.$$ Then $C_I$ is a $C^*$-subalgebra of ${\overline{{\operatorname{span}}}}\{w_pw_q^*:\mu(p)=\mu(q)\}$, $C_I={\operatorname{span}}\{B_k:k\in I\}$ and ${\overline{{\operatorname{span}}}}\{w_pw_q^*:\mu(p)=\mu(q)\}=\overline{\cup_{I\in{\mathcal{I}}}C_I}$.

Fix $I\in{\mathcal{I}}$. To see that $C_I$ is a $C^*$-subalgebra, it suffices to show that ${\operatorname{span}}\{w_pw^*_q:\mu(p)=\mu(q)\in I\}$ is a $*$-subalgebra. It’s clearly closed under taking adjoints. Let $p,q,r,s\in P$ such that $\mu(p)=\mu(q)\in I$ and $\mu(r)=\mu(s)\in I$. Then $$w_pw^*_qw_rw^*_s=\begin{cases}w_{pq^{-1}(q\vee r)}w^*_{sr^{-1}(q\vee r)}&\text{if $q\vee r<\infty$}\\0&\text{otherwise.}\end{cases}$$ If $w_pw^*_qw_rw^*_s=0$ we are 