Extending a $ * $-Representation of $ ({C_{c}}(G,\mathscr{A}),\star,^{*}) $ to a $ * $-Representation of $ \mathscr{A} \rtimes_{\alpha} G $

Question

Let $ (\mathscr{A},G,\alpha) $ be a $ C^{*} $-dynamical system, and consider the twisted convolution $ * $-algebra $ ({L^{1}}(G,\mathscr{A}),\star,^{*}) $ defined by \begin{align*} \forall \phi,\psi \in {L^{1}}(G,\mathscr{A}), ~ \forall g \in G: \quad (\phi \star \psi)(g) & \stackrel{\text{def}}{=} \int_{G} \phi(x) {\alpha_{x}}(\psi(x^{-1} g)) \, d{{\mu_{G}}(x)}, \\ {\phi^{*}}(g) & \stackrel{\text{def}}{=} \Delta(g^{-1}) \cdot {\alpha_{g}}(\phi(g^{-1})^{*}). \end{align*} Note: $ \mu_{G} $ is a Haar measure on $ G $ and $ \Delta: G \to \mathbb{R}_{> 0} $ is the modular function of $ G $.

If $ \pi $ is an algebraic $ * $-representation of $ ({L^{1}}(G,\mathscr{A}),\star,^{*}) $ on some Hilbert space $ \mathcal{H} $, then $ \pi $ is automatically bounded by the $ L^{1} $-norm on $ {L^{1}}(G,\mathscr{A}) $, i.e., $ \| \pi(\phi) \|_{B(\mathcal{H})} \leq \| \phi \|_{L^{1}} $ for all $ \phi \in {L^{1}}(G,\mathscr{A}) $. (This is because an algebraic *-homomorphism from a Banach $ * $-algebra to a $ C^{*} $-algebra is automatically norm-decreasing.) Consequently, we can extend $ \pi $ to a $ * $-representation $ \tilde{\pi} $ of the crossed-product $ C^{*} $-algebra $ \mathscr{A} \rtimes_{\alpha} G $ on $ \mathcal{H} $.

If we replace $ {L^{1}}(G,\mathscr{A}) $ by $ {C_{c}}(G,\mathscr{A}) $, then I suspect that the condition

$ \pi $ is bounded by the $ L^{1} $-norm on $ {C_{c}}(G,\mathscr{A}) $

is not for free and that we have to explicitly assume it if we want to extend $ \pi $ to a $ \tilde{\pi}: \mathscr{A} \rtimes_{\alpha} G \to B(\mathcal{H}) $. My suspicion is substantiated by the following result, which is given in Dana P. William’s book Crossed Products of $ C^{*} $-Algebras together with a ‘clever’ proof due to Iain Raeburn.

Theorem: If $ \pi: ({C_{c}}(G,\mathscr{A}),\star,^{*}) \to B(\mathcal{H}) $ is an algebraic $ * $-representation that is continuous with respect to the inductive limit topology on $ {C_{c}}(G,\mathscr{A}) $, then $ \pi $ is bounded by the universal norm $ \| \cdot \|_{\mathscr{A} \rtimes_{\alpha} G} $ on $ {C_{c}}(G,\mathscr{A}) $ and so is bounded by the $ L^{1} $-norm on $ {C_{c}}(G,\mathscr{A}) $.

Hence,

My question: Is there a ‘simple’ example of an algebraic $ * $-representation $ \pi: ({C_{c}}(G,\mathscr{A}),\star,^{*}) \to B(\mathcal{H}) $ that is not bounded by the $ L^{1} $-norm on $ {C_{c}}(G,\mathscr{A}) $?

To keep things simple, we can assume that $ \mathscr{A} = \mathbb{C} $. Thank you very much for your help!

Answer 1

Here's the discrete group argument, when $A=\mathbb C$. Then $C_c(G,A)$ is just the algebra of finitely supported functions $G\rightarrow\mathbb C$, which has a basis $(\delta_g)_{g\in G}$ say. That $\pi:C_c(G,A)\rightarrow B(H)$ is a $*$-representation means that firstly $\pi(\delta_e)$ is a self-adjoint idempotent; so wlog we may assume $\pi(\delta_e) = 1_H$ (or else just restrict to the invariant subspace of $H$). Then, as $\delta_g^*=\delta_{g^{-1}}$, $$ 1_H = \phi(\delta_g^* \delta_g) = \phi(\delta_g)^* \phi(\delta_g). $$ Similarly $\phi(\delta_g) \phi(\delta_g)^* = 1_H$. So $\phi(\delta_g)$ is a unitary. It now follows that $\phi$ extends to a contraction from $\ell^1(G)$. I think a similar argument works for any $A$.

Edit: For a compact group, it's easier, by virtue of hiding the work in known structure results. Let $Pol(G)$ be the collection of matrix coefficients of finite dimensional (unitary) representations-- this a is a dense $*$-subalgebra of $C(G)$. If we view $C^*(G)$, the universal $C^*$-completion of $L^1(G)$, as a direct sum of full matrix algebras, then $Pol(G)$ exactly corresponds to the algebraic direct sum of full matrix algebras. From this picture, you can see that any $*$-representation of $Pol(G)$ is contractive for the $C^*(G)$ norm.

Answer 2

Here is a complete argument proving automatic continuity for any $ C^{\ast} $-dynamical system $ (G,A,\alpha) $ where $ G $ is discrete.

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Haar’s Theorem essentially states that every Haar measure on $ G $ is a positive scalar multiple of the counting measure on $ G $, so let $ k \in \mathbb{R}_{> 0} $ and equip $ G $ with $ k $ times the counting measure on $ G $.

For each $ a \in A $ and $ g \in G $, let $ a \bullet \delta_{g} $ denote the function in $ {C_{c}}(G,A) $ defined by $$ \forall x \in G: \qquad (a \bullet \delta_{g})(x) \stackrel{\text{df}}{=} \begin{cases} a & \text{if} ~ x = g; \\ 0_{A} & \text{if} ~ x \in G \setminus \{ g \}. \end{cases} $$ We then have the following relations for the convolution $ \star $ and the involution $ ^{\ast} $ on $ {C_{c}}(G,A) $:

• $ (a \bullet \delta_{g}) \star (b \bullet \delta_{h}) = (k \cdot a {\alpha_{g}}(b)) \bullet \delta_{g h} $ for all $ a,b \in A $ and $ g,h \in G $.
• $ (a \bullet \delta_{g})^{\ast} = {\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{g^{- 1}} $ for all $ a \in A $ and $ g \in G $.

Let us suppose that $ \pi $ is an algebraic $ \ast $-homomorphism from $ ({C_{c}}(G,A),\star,^{\ast}) $ to a $ C^{\ast} $-algebra $ C $. We will prove that $ \pi $ is automatically contractive with respect to the $ L^{1} $-norm on $ {C_{c}}(G,A) $ and the norm on $ C $. Pick any $ a \in A $ and $ g \in G $. Then $$ \pi(a \bullet \delta_{g})^{\ast} = \pi((a \bullet \delta_{g})^{\ast}) = \pi({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{g^{- 1}}), $$ which yields \begin{align} \pi(a \bullet \delta_{g})^{\ast} \pi(a \bullet \delta_{g}) & = \pi({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{g^{- 1}}) \pi(a \bullet \delta_{g}) \\ & = \pi( ({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{g^{- 1}}) \star (a \bullet \delta_{g}) ) \\ & = \pi( (k \cdot {\alpha_{g^{- 1}}}(a^{\ast}) {\alpha_{g^{- 1}}}(a)) \bullet \delta_{g^{- 1} g} ) \\ & = \pi( (k \cdot {\alpha_{g^{- 1}}}(a^{\ast}) {\alpha_{g^{- 1}}}(a)) \bullet \delta_{e} ) \\ & = \pi( ( k \cdot {\alpha_{g^{- 1}}}(a^{\ast}) {\alpha_{e}}({\alpha_{g^{- 1}}}(a)) ) \bullet \delta_{e e} ) \\ & = \pi( ({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{e}) \star ({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) ) \\ & = \pi({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{e}) \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \\ & = \pi({\alpha_{g^{- 1}}}(a)^{\ast} \bullet \delta_{e}) \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \\ & = \pi( {\alpha_{e^{- 1}}}({\alpha_{g^{- 1}}}(a)^{\ast}) \bullet \delta_{e^{- 1}} ) \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \\ & = \pi(({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})^{\ast}) \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \\ & = \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})^{\ast} \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}). \end{align} Using the $ C^{\ast} $-norm identity, we thus obtain \begin{align} \| \pi(a \bullet \delta_{g}) \|_{C}^{2} & = \| \pi(a \bullet \delta_{g})^{\ast} \pi(a \bullet \delta_{g}) \|_{C} \\ & = \| \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})^{\ast} \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \|_{C} \\ & = \| \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \|_{C}^{2}, \end{align} so $$ (\spadesuit) \qquad \| \pi(a \bullet \delta_{g}) \|_{C} = \| \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \|_{C}. $$

For each $ g \in G $, define a linear map $ \rho_{g}: A \to C $ by $$ \forall a \in A: \qquad {\rho_{g}}(a) \stackrel{\text{df}}{=} \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}). $$ Then we have for all $ g \in G $ and $ a,b \in A $ that \begin{align} {\rho_{g}}(a b) & = \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a b) \bullet \delta_{e}) \\ & = \frac{1}{k} \cdot \pi(({\alpha_{g^{- 1}}}(a) {\alpha_{g^{- 1}}}(b)) \bullet \delta_{e}) \\ & = \frac{1}{k^{2}} \cdot \left[ k \cdot \pi(({\alpha_{g^{- 1}}}(a) {\alpha_{g^{- 1}}}(b)) \bullet \delta_{e}) \right] \\ & = \frac{1}{k^{2}} \cdot \pi( (k \cdot {\alpha_{g^{- 1}}}(a) {\alpha_{g^{- 1}}}(b)) \bullet \delta_{e} ) \\ & = \frac{1}{k^{2}} \cdot \pi( (k \cdot {\alpha_{g^{- 1}}}(a) {\alpha_{e}}({\alpha_{g^{- 1}}}(b))) \bullet \delta_{e e} ) \\ & = \frac{1}{k^{2}} \cdot \pi( ({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \star ({\alpha_{g^{- 1}}}(b) \bullet \delta_{e}) ) \\ & = \frac{1}{k^{2}} \cdot \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \pi({\alpha_{g^{- 1}}}(b) \bullet \delta_{e}) \\ & = \left[ \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \right] \left[ \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(b) \bullet \delta_{e}) \right] \\ & = {\rho_{g}}(a) {\rho_{g}}(b), \\ {\rho_{g}}(a^{\ast}) & = \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a^{\ast}) \bullet \delta_{e}) \\ & = \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a)^{\ast} \bullet \delta_{e}) \\ & = \frac{1}{k} \cdot \pi( {\alpha_{e^{- 1}}}({\alpha_{g^{- 1}}}(a)^{\ast}) \bullet \delta_{e^{- 1}} ) \\ & = \frac{1}{k} \cdot \pi(({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})^{\ast}) \\ & = \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e})^{\ast} \\ & = \left[ \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \right]^{\ast} \\ & = {\rho_{g}}(a)^{\ast}. \end{align} We thus see for all $ g \in G $ that $ \rho_{g}: A \to C $ is an algebraic $ \ast $-homomorphism, so it is automatically contractive with respect to the norms on $ A $ and $ C $. In other words, $$ \forall g \in G, ~ \forall a \in A: \qquad \left\| \frac{1}{k} \cdot \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \right\|_{C} \leq \| a \|_{A}, $$ or equivalently, $$ (\clubsuit) \qquad \forall g \in G, ~ \forall a \in A: \qquad \| \pi({\alpha_{g^{- 1}}}(a) \bullet \delta_{e}) \|_{C} \leq k \| a \|_{A}. $$

Finally, let $ f \in {C_{c}}(G,A) $, and let $ S $ denote the compact/finite support of $ f $. Then \begin{align} \| \pi(f) \|_{C} & = \left\| \pi \! \left( \sum_{g \in S} f(g) \bullet \delta_{g} \right) \right\|_{C} \\ & = \left\| \sum_{g \in S} \pi(f(g) \bullet \delta_{g}) \right\|_{C} \\ & \leq \sum_{g \in S} \| \pi(f(g) \bullet \delta_{g}) \|_{C} \\ & = \sum_{g \in S} \| \pi({\alpha_{g^{- 1}}}(f(g)) \bullet \delta_{e}) \|_{C} \qquad (\text{From $ (\spadesuit) $.}) \\ & \leq \sum_{g \in S} k \| f(g) \|_{A} \qquad (\text{From $ (\clubsuit) $.}) \\ & = \| f \|_{1}. \end{align} Therefore, $ \pi $ is contractive with respect to the $ L^{1} $-norm on $ {C_{c}}(G,A) $ and the norm on $ C $. $ \quad \blacksquare $