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Let $(S,\Delta)$ be a compact quantum group and $U\in M(K(H)\otimes S)$ be a unitary corepresentation of $S$ on $H$. Let $\phi $ be a state of $S$. Let $A$ be a sub-$C^*$-algebra of $B(H)$. If $a\in A$, we let $Ad_U(a)=U(a\otimes 1)U^*$.

Why the element $(Id\otimes \phi)\circ Ad_U(a)$ is well-defined?

I don't understand the action of $Id\otimes \phi$ on $Ad_U(a)$.

Remark: I try to understand Definition 2.1 (page 3) of

http://arxiv.org/PS_cache/arxiv/pdf/1002/1002.2551v2.pdf

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Well, $U(a\otimes 1)U^* \in M(K(H)\otimes S)$. So you could verify that $Id\otimes\phi:K(H)\otimes S\rightarrow\mc K(H)$ is a strictly continuous map (it is!) and so it extends to the multiplier algebra. Lance's book on Hilbert modules is a good place to start trying to understand multiplier algebra (in my opinion). –  Matthew Daws Jan 14 '12 at 13:33
    
Now I understand very well this definition. Thank you very much! –  Zouba Jan 14 '12 at 15:58

1 Answer 1

Well, $U(a\otimes 1)U^*\in M(K(H)\otimes S)$. So you could verify that $Id\otimes \varphi$ is a strictly continuous map (it is!) and so it extends to the multiplier algebra. Lance's book on Hilbert modules is a good place to start trying to understand multiplier algebra (in my opinion).

Credit goes to Matthew Daws

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