\end{aligned}$$ Hence $(\delta_\mu\otimes{{\operatorname{id}}})\circ\delta_\mu=({{\operatorname{id}}}\otimes\delta_K)\circ \delta_\mu$. Thus $\delta_\mu$ is a coaction.

To show that $\delta_\mu$ is injective, let $\pi:C^*(G,P)\to B(H)$ be a faithful representation. We will show that $\pi$ can be written as a composition of $\delta_\mu$ and another representation. Let $\epsilon:C^*(K)\to{\mathbb{C}}$ be the trivial representation on ${\mathbb{C}}$ such that $\epsilon(u_k)=1$ for all $k\in K$. By the properties of the minimal tensor product (see [@RaeburnWilliams1998 Proposition B.13]) there exists a homomorphism $$\pi\otimes \epsilon:C^*(G,P)\otimes_{\min} C^*(G)\to B(H)\otimes{\mathbb{C}}=B(H).$$ Since $$(\pi\otimes \epsilon)\circ \delta_\mu(w_p)=(\pi\otimes \epsilon)(w_p\otimes u_{\mu(p)})=\pi(w_p),$$ we have $\pi=(\pi\otimes \epsilon)\circ \delta_\mu$. Now suppose that $\delta_\mu(a)=0$ for some $a\in C^*(G,P)$. Then $0=(\pi\otimes \epsilon)\circ \delta_\mu(a)=\pi(a)$. Since $\pi$ is faithful, $a=0$. Hence $\delta_\mu$ is injective.

To prove that $\delta_\mu$ is a nondegenerate coaction we must show that $$\delta_\mu(C^*(G,P))(1\otimes C^*(K))=C^*(G,P)\otimes_{\min} C^*(K).$$ It suffices to show that we can get the spanning elements $w_pw^*_q\otimes u_k$, and this is easy: $$\delta_\mu(w_pw^*_q)(1\otimes u_{\mu(qp^{-1}) k})=w_pw^*_q\otimes u_{\mu(pq^{-1})}(1\otimes u_{\mu(qp^{-1}) k})=w_pw^*_q\otimes u_k.$$ Thus $\delta_\mu$ is nondegenerate.

Let $\lambda$ be the left-regular representation of a discrete group $K$. There is a trace $\tau$ on $C^*(K)$ characterised by $$ \tau(u_k)= (\lambda_k\epsilon_e\mid \epsilon_e)=\begin{cases}1 &\text{if $k=e$}\\ 0&\text{otherwise}.  \end{cases}$$ It is well-known that if $K$ is an amenable group, then $\tau$ is faithful.

\[conditional expectation construction\] Let $(G,P)$ be a quasi-lattice ordered group. 