Let $\mathbb{G}$ be a compact (quantum group) with function algebra $(C(\mathbb{G}), \Delta)$ and Haar state $\varphi_{\mathbb{G}}$. Consider the associated GNS-representation $\pi_{\mathbb{G}}: C(\mathbb{G})\to B(L^2(\mathbb{G}))$ with GNS-vector $\xi_\mathbb{G}$. Let $V \in M(B_0(L^2(\mathbb{G}))\otimes C(\mathbb{G}))$ the the right regular representation, which can be defined as follows: Consider a faithful-non degenerate representation $\pi: C(\mathbb{G})\subseteq B(K)$ and define the unitary
$$V: L^2(\mathbb{G})\otimes K\to L^2(\mathbb{G})\otimes K$$
by
$$V(\pi_\mathbb{G}(a)\xi_\mathbb{G}\otimes \pi(b)\eta) = (\pi_{\mathbb{G}}\otimes \pi)(\Delta(a)(1\otimes b))(\xi_{\mathbb{G}}\otimes \eta).$$ I am trying to show that it is false that
$$V(B(H)\otimes 1)V^*\subseteq B(H)\otimes C(\mathbb{G}).$$
However, I am not sure how to produce an explicit example of $x\in B(L^2(\mathbb{G}))$ with $V(x\otimes 1)V^*\notin B(L^2(\mathbb{G}))\otimes C(\mathbb{G})$. I'm also interested in examples where $C(\mathbb{G})$ is replaced by another $C^*$-algebra, $L^2(\mathbb{G})$ by another Hilbert space and $V$ a unitary of choice.
1 Answer
Take for $\mathbb{G}$ the compact group $S^1$ viewed as the unit circle in the plane. Denote by $(\lambda_y)_{y \in S^1}$ the regular representation on $L^2(S^1)$. Your question then becomes if the map $y \mapsto \lambda_y T \lambda_y^*$ is continuous from $S^1$ to $B(L^2(S^1))$ equipped with the operator norm. This is typically not the case. For instance, take $T = 1_A$, the multiplication operator with the indicator function of a half circle $A$. Then $\|\lambda_y T \lambda_y^* - T\|$ equals $1$ if $y \neq 1$ and equals $0$ if $y=1$.
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$\begingroup$ @Andromeda Yes, of course. I have edited the answer and replaced $2$ by $1$. $\endgroup$ Commented May 29, 2023 at 14:54