An unconventional definition of the $ C^{*} $-algebraic reduced crossed product

Question

Let $ (A,G,\alpha) $ be a $ C^{*} $-dynamical system, i.e., $ A $ is a $ C^{*} $-algebra, $ G $ is a locally compact Hausdorff group and $ \alpha $ is a strongly continuous action of $ G $ on $ A $ by $ * $-automorphisms. Equip $ {C_{c}}(G,A) $, the linear space of continuous $ A $-valued functions on $ G $ with compact support, with an associative multiplication $ \star_{\alpha} $ and an involution $ ^{*_{\alpha}} $ by \begin{align} \forall f,g \in {C_{c}}(G,A), ~ \forall x \in G: \quad (f \star_{\alpha} g)(x) & \stackrel{\text{df}}{=} \int_{G} f(y) ~ {\alpha_{y}}(g(y^{-1} x)) ~ \mathrm{d}{y}, \\ {f^{*_{\alpha}}}(x) & \stackrel{\text{df}}{=} {\alpha_{x}}(f(x^{-1})^{*}) \cdot \Delta(x^{-1}), \end{align} where $ \Delta $ denotes the modular function of $ G $.

Let $ \pi $ be a faithful $ * $-representation of $ A $ on the Hilbert space $ \mathcal{H} $. From this, fashion a $ * $-representation $ \tilde{\pi} $ of $ A $ on $ {L^{2}}(G,\mathcal{H}) \cong {L^{2}}(G) \otimes \mathcal{H} $ by $$ \forall a \in A, ~ \forall \xi \in {L^{2}}(G,\mathcal{H}), ~ \forall x \in G: \quad [[\tilde{\pi}(a)](\xi)](x) \stackrel{\text{df}}{=} [\pi({\alpha_{x^{-1}}}(a))](\xi(x)). $$ Define also a unitary representation $ \lambda $ of $ G $ on $ {L^{2}}(G,\mathcal{H}) $ by $$ \forall x,y \in G, ~ \forall \xi \in {L^{2}}(G,\mathcal{H}): \quad (\lambda_{x} \xi)(y) \stackrel{\text{df}}{=} \xi(x^{-1} y). $$ Then the integrated form $ \tilde{\pi} \rtimes_{\alpha} \lambda $ defines a $ * $-representation of $ ({C_{c}}(G,A),\star_{\alpha},^{*_{\alpha}}) $ on $ {L^{2}}(G,\mathcal{H}) $: $$ \forall f \in {C_{c}}(G,A), ~ \forall \xi \in {L^{2}}(G,\mathcal{H}), ~ \forall x \in G: \\ [[(\tilde{\pi} \rtimes_{\alpha} \lambda)(f)](\xi)](x) \stackrel{\text{df}}{=} \int_{G} [[\tilde{\pi}(f(y))](\lambda_{y} \xi)](x) ~ \mathrm{d}{y} = \int_{G} [\pi({\alpha_{x^{-1}}}(f(y)))](\xi(y^{-1} x)) ~ \mathrm{d}{y}. $$ Finally, the $ C^{*} $-algebraic reduced crossed product $ A \rtimes_{\alpha,\text{r}} G $ is taken to be the completion of $ {C_{c}}(G,A) $ under the $ C^{*} $-norm $ \| \cdot \|_{*} $ defined by $$ \forall f \in {C_{c}}(G,A): \quad \| f \|_{*} \stackrel{\text{df}}{=} \| (\tilde{\pi} \rtimes_{\alpha} \lambda)(f) \|_{\mathscr{B}({L^{2}}(G,\mathcal{H}))}. $$

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I have also seen an unconventional definition of $ A \rtimes_{\alpha,\text{r}} G $. For each $ f \in {C_{c}}(G,A) $, define a function $ f^{\Delta} \in {C_{c}}(G,A) $ by $ f^{\Delta} \stackrel{\text{df}}{=} f \sqrt{\Delta} $. Define a $ * $-representation $ \rho $ of $ ({C_{c}}(G,A),\star_{\alpha},^{*_{\alpha}}) $ on $ {L^{2}}(G,\mathcal{H}) $ by $$ \forall f \in {C_{c}}(G,A), ~ \forall \xi \in {L^{2}}(G,\mathcal{H}), ~ \forall x \in G: \\ [[\rho(f)](\xi)](x) \stackrel{\text{df}}{=} \int_{G} [\pi({\alpha_{x}}({f^{\Delta}}(x^{-1} y)))](\xi(y)) ~ \mathrm{d}{y}. $$ Then the $ C^{*} $-algebraic reduced crossed product $ A \rtimes_{\alpha,\text{r}} G $ is taken to be the completion of $ {C_{c}}(G,A) $ under the $ C^{*} $-norm $ \| \cdot \|_{**} $ defined by $$ \forall f \in {C_{c}}(G,A): \quad \| f \|_{**} \stackrel{\text{df}}{=} \| \rho(f) \|_{\mathscr{B}({L^{2}}(G,\mathcal{H}))}. $$

Question. What is the exact equation that relates $ \tilde{\pi} \rtimes_{\alpha} \lambda $ to $ \rho $?

Thanks for your help!

Answer 1

It seems that I have answered my own question. For the benefit of anyone who might have an interest in this sort of thing, I have decided to post my answer.

My idea is to find a unitary mapping $$ U: {L^{2}}(G,\mathcal{H}) \to {L^{2}}(G,\mathcal{H}) $$ that intertwines $ (\tilde{\pi} \rtimes_{\alpha} \lambda)(f) $ and $ \rho(f) $, i.e., $ U $ satisfies the following commutative diagram: \begin{equation} \require{AMScd} \begin{CD} {L^{2}}(G,\mathcal{H}) @>{(\tilde{\pi} \rtimes_{\alpha} \lambda)(f)}>> {L^{2}}(G,\mathcal{H}) \\ @V{U}VV @VV{U}V \\ {L^{2}}(G,\mathcal{H}) @>>{\rho(f)}> {L^{2}}(G,\mathcal{H}). \end{CD} \end{equation} Naïvely, one can try to define $ U $ by $$ \forall \xi \in {L^{2}}(G,\mathcal{H}), ~ \forall x \in G: \quad (U \xi)(x) \stackrel{\text{df}}{=} \xi(x^{-1}), $$ but this is incorrect because then $ U $ is not isometric in the case that $ G $ is not unimodular. Therefore, one has to modify this flawed definition using the modular function $ \Delta $ of $ G $ so that $ U $ is indeed isometric. The theory of integration on locally compact Hausdorff groups then yields the following correct definition of $ U $: $$ \forall \xi \in {L^{2}}(G,\mathcal{H}), ~ \forall x \in G: \quad (U \xi)(x) \stackrel{\text{df}}{=} \sqrt{\Delta(x^{-1})} \cdot \xi(x^{-1}). $$ Straightforward computations show that $ U = U^{-1} = U^{*} $. Hence, $$ \rho(f) = U \circ (\tilde{\pi} \rtimes_{\alpha} \lambda)(f) \circ U, $$ i.e., $ \rho(f) $ and $ (\tilde{\pi} \rtimes_{\alpha} \lambda)(f) $ are unitarily equivalent.