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In what follows, a ‘$ * $-representation’ always means a non-degenerate $ * $-representation.


Let $ (\mathscr{A},G,\alpha) $ be a $ C^{*} $-dynamical system, and let $ \pi: \mathscr{A} \to B(\mathcal{H}) $ be a faithful $ * $-representation of $ \mathscr{A} $ on a Hilbert space $ \mathcal{H} $. Define an associated faithful $ * $-representation $ \tilde{\pi}: \mathscr{A} \to B({L^{2}}(G,\mathcal{H})) $ and a unitary group representation $ \lambda: G \to U({L^{2}}(G,\mathcal{H})) $ as follows.

For all $ a \in \mathscr{A} $, $ g \in G $, $ x \in G $ and $ \xi \in {L^{2}}(G,\mathcal{H}) $, \begin{align*} [[\tilde{\pi}(a)](\xi)](x) & \stackrel{\text{def}}{=} [\pi({\alpha_{x^{-1}}}(a))](\xi(x)), \\ [[\lambda(g)](\xi)](x) & \stackrel{\text{def}}{=} \xi(g^{-1} x). \end{align*}

$ \lambda $ is usually called the left regular representation of $ G $.

It is well-known that the pair $ (\tilde{\pi},\lambda) $ — called the left regular representation induced by $ \pi $ — is a covariant representation of $ (\mathscr{A},G,\alpha) $ on $ {L^{2}}(G,\mathcal{H}) $ and that the integrated form $ \tilde{\pi} \rtimes \lambda $ is a faithful $ * $-representation of the twisted convolution $ * $-algebra $ ({C_{c}}(G,\mathscr{A}),\star,^{*}) $ on $ {L^{2}}(G,\mathcal{H}) $.

It is also well-known that $ \tilde{\pi} \rtimes \lambda $ extends uniquely to a faithful $ * $-representation of the reduced crossed product $ C^{*} $-algebra $ \mathscr{A} \rtimes_{\alpha,\text{r}} G $ on $ {L^{2}}(G,\mathcal{H}) $.

My question: Suppose that we have a faithful $ * $-representation $ \rho: \mathscr{A} \to B({L^{2}}(G,\mathcal{H})) $ that is not associated to $ \pi $, i.e. is not of the form $ \tilde{\pi} $, for any faithful $ * $-representation $ \pi: \mathscr{A} \to B(\mathcal{H}) $. Suppose further that the pair $ (\rho,\lambda) $ — which I shall call a non-induced left regular representation — is a covariant representation of $ (\mathscr{A},G,\alpha) $ on $ {L^{2}}(G,\mathcal{H}) $. Then is it necessarily true that the integrated form $ \rho \rtimes \lambda $ gives rise to a faithful $ * $-representation of $ \mathscr{A} \rtimes_{\alpha,\text{r}} G $ on $ {L^{2}}(G,\mathcal{H}) $?

Thank you very much for your help!

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The answer is no in two ways. Let $G$ be a discrete group and consider the trivial action of $G$ on the reduced group $\mathrm{C}^*$-algebra $\mathrm{C}^*_{\mathrm{r}}G$. The pair of the right regular representation $\rho\colon \mathrm{C}^*_{\mathrm{r}}G\to B(\ell_2G)$ and the left regular representation $\lambda\colon G\to U(\ell_2G)$ is covariant. The resulting $*$-homomorphism $\rho\rtimes\lambda$ is continuous on the reduced crossed product (which is nothing but the minimal tensor product $\mathrm{C}^*_{\rho}G \otimes_{\min} \mathrm{C}^*_{\lambda}G$) if and only if $G$ is amenable. It is faithful on the algebraic tensor product $\mathrm{C}^*_{\rho}G \otimes_{\mathrm{alg}} \mathrm{C}^*_{\lambda}G$ if and only if $G$ is ICC.

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