\lim_{k \to \infty} (V^{\otimes k})^* (1 \otimes x) V^{\otimes k}.$$ Here the limit makes sense because the sequence is eventually constant for any $x \in ({\mathcal{O}}_\infty)^{\otimes l}$. Namely, for any $k \geq l$ we have $$(V^{\otimes k})^* (1 \otimes x) V^{\otimes k} = (V^{\otimes l})^* (1 \otimes x) V^{\otimes l}.$$

\[thm:model\] Let $G$ be a Hodgkin Lie group. The crossed product ${\mathcal{O}}(G):={\hat{G}}_\alpha \ltimes A$ with the dual action of $G$ is a Kirchberg $G$-algebra with continuous Rokhlin property which is ${\mathrm{KK}}^G$-equivalent to $C(G)$ by the canonical unital inclusion $C(G) = C^*({\hat{G}}) \to {\hat{G}}{\, }_{\alpha}\! \ltimes A$.

Asymptotic representability of $\alpha$, which implies continuous Rokhlin property of $\hat{\alpha}$, follows from connectedness of $G$ since we have a homotopy of diagonal unitary representations $U_t$ connecting $U_S$ and $1$. The desired ${\mathrm{KK}}^G$-equivalence follows from Proposition \[prp:BC\] and the Baaj-Skandalis duality. Separability and nuclearity of ${\mathcal{O}}(G)$ also follow by definition because $G$ is second countable and coamenable. Hence the rest part of the proof is to show that ${\mathcal{O}}(G)$ is purely infinite. For this, we follow the argument appeared in .

For any $\varepsilon > 0$, finite subset $X \subset A$ and $1 \not \in F \subset {\rm Irr}(G)$, we have $x \in A_+$ 