# extension of a representation on $C^*$-algebra tensor products

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?

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 $\max=\min$).