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Given two $C^*$-algebras $A$ and $B$ and a *-homomorphism $\pi:A\odot B\to B(H)$ (note: $A\odot B$ is the *-algebra tensor product).

For every $C^*$-norm $\gamma:A\odot B\to \mathbb{R}_{\ge0}$, is there always a continuous, unique extension $\hat{\pi}:A \otimes_{\gamma} B\to B(H)$ of $\pi$? Here $A \otimes_{\gamma} B$ denotes the $\gamma$-norm closure of $A\odot B$ (i.e. the tensor product $C^*$-algebra with respect to $\gamma$).

I know that $\pi$ can be extended on the maximal tensor product, i.e. if $\gamma =\|\enspace \|_{\max}$. However, I guess that the answer is no in general, but I can't justify it.

Do you have an idea or do you know a suitable example?

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Let $A=B=C^*_r (\mathbb F_2) $, so that we know that the max and min norms are different. Let $\pi $ be the inclusion map into $A\otimes_\max B $. If you now take $\gamma=\min $, then $\pi $ cannot have an extension to $A\otimes_\min B $ (the fact that $*$-homomorphisms are contractive would imply that $\max=\min $).

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  • $\begingroup$ very nice argument, thank you ! $\endgroup$ – user197416 Oct 15 '16 at 12:54

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