For $i=1,2$, let $\mathcal{A}_i$ be an abstract $C^*$-algebra and $\pi_i : \mathcal{A}_i \rightarrow \mathcal{B}(\mathcal{H}_i)$ a $C^*$-representation. Let $\alpha: \mathcal{A}_1 \rightarrow \mathcal{A}_2$ a *-homomorphism.
I'm looking for a theorem that may assert that there exists a unique $*$-homomorphism $\tilde{\alpha} : \pi_1 (\mathcal{A}_1)'' \rightarrow \pi_2 (\mathcal{A}_2)''$ of von Neumann algebras such $ \tilde{\alpha}(\pi_1(a))= \pi_2(\alpha(a))$ for all $a \in \mathcal{A}_1$. ¿In which topology $\tilde{\alpha}$ is continuous?
Please, I need references about such a theorem.
Thanks,
D
Such extension does not usually exist. For instance take $\mathcal A_1=\mathcal A_2=UHF(2^\infty)$. Let $\pi_1:\mathcal A_1\to \mathcal B(\mathcal H_1)$ be a Powers representation such that $\pi_1(\mathcal A_1)''$ is a type III factor $M$. And let $\pi_2:\mathcal A_2\to \mathcal B(\mathcal H_2)$ be GNS from the trace, so $\pi_2(\mathcal A_2)''$ is the hyperfinite II$_1$-factor $R$.
Take $\alpha(x)=x$, the identity isomorphism.
Now you want a $*$-homomorphism $\bar\alpha: M\to R$. We may assume $\bar\alpha(1)=1$ (otherwise, replace $R$ with $\bar\alpha(1) R\bar\alpha(1)$, which is still a hyperfinite II$_1$). Given any projection $p\in M$, we have $p\simeq 1$, so $\bar\alpha(p)\simeq 1$; as this last equivalence happens in a II$_1$-factor, we have $\bar\alpha(p)=1$ for all projections $p\in M$. But then $1=\bar\alpha(1)=\bar\alpha(p+1-p)=\bar\alpha(p)+\bar\alpha(1-p)=2$. So $\bar\alpha$ does not exist.