1
$\begingroup$

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 $?

Thanks for your help!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Branimir Ćaćić Sep 13 '14 at 12:36
  • $\begingroup$ @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. $\endgroup$ – Leonard Sep 13 '14 at 19:11
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.