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Let $A$ be a C$^*$-algebra, let $\Delta:A\rightarrow A \otimes_{\min} A$ be a $*$-homomorphism, and let $\phi$ be a state on $A$. Let $(H,\pi,\xi_0)$ be the GNS construction for $\phi$; let $B=\pi(A) \subseteq \mathcal B(H)$. I do not assume that $\pi:A\rightarrow B$ is injective.

Suppose that $(\phi\otimes\phi)\circ\Delta = \phi$. Is this enough to ensure that there is a $*$-homomorphism $\Delta_B:B\rightarrow B\otimes_{\min} B$ with $\Delta_B\circ\pi = (\pi\otimes\pi)\circ\Delta$?

(Motivation comes from compact quantum groups, where we know more. But I want to know if the result is true in this rather general setting.)

Actually, this motivates a more general question:

For $i=1,2$, let $A_i$ be a C$^*$-algebra with a state $\phi_i$, and suppose that $\theta:A_1\rightarrow A_2$ is a $*$-homomorphism with $\phi_2\circ\theta = \phi_1$. Let $(H_i,\pi_i,\xi_i)$ be the GNS constructions. Does there exist a $*$-homorphism $\varphi:\pi_1(A_1) \rightarrow \pi_2(A_2)$ with $\varphi\circ\pi_1 = \pi_2\circ\theta$?

We'd need that $\pi_1(a)=0 \implies \pi_2(\theta(a))=0$. Now, $\pi_1(a)=0$ if and only if $\pi_1(ab)\xi_1=0$ for all $b$, if and only if $\phi_1(b^*a^*ab)=0$ for all $b\in A_1$. However, $\pi_2(\theta(a))=0$ if and only if $\phi_2(c^* \theta(a^*a) c)=0$ for all $c\in A_2$. I don't see why this need to true in general (I think it would be true if $\phi_1$ were KMS for example (so in particular if $A_1$ is abelian)).

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I assume everything in sight (algebras, homomorphisms) is unital? –  user16299 Apr 2 '12 at 22:53
    
@Yemon: Ah, yes! Good point. Indeed, I assume everything to be unital... –  Matthew Daws Apr 4 '12 at 17:11

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