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

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 $$\forall f \in {C_{c}}(G,A): \quad \| f \|_{*} \stackrel{\text{df}}{=} \| (\tilde{\pi} \rtimes_{\alpha} \lambda)(f) \|_{\mathscr{B}({L^{2}}(G,\mathcal{H}))}.$$

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 $\rho$?

• These constructions should, I think, be identical---$\tilde{\pi} \ltimes_\alpha \lambda$ uses translations on the left, $\rho$ uses translations on the right, but the appearance of the modular function $\Delta$ in the construction of $\rho$ should guarantee that they actually yield the same representation. Sep 13 '14 at 12:36
• @BranimirĆaćić: Thanks! I’ve managed to show that $\rho(f)$ and $\tilde{\pi} \rtimes_{\alpha} \lambda$ are unitarily equivalent $*$-representations, just as you’ve said. Sep 13 '14 at 19:11
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: $$\require{AMScd} \begin{CD} {L^{2}}(G,\mathcal{H}) @>{(\tilde{\pi} \rtimes_{\alpha} \lambda)(f)}>> {L^{2}}(G,\mathcal{H}) \\ @V{U}VV @VV{U}V \\ {L^{2}}(G,\mathcal{H}) @>>{\rho(f)}> {L^{2}}(G,\mathcal{H}). \end{CD}$$ 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.