C* tensor product admitting a faithful separable representation

Let $$A$$ and $$B$$ be $$C^*$$ algebras which admit faithful representations on separable Hilbert spaces. Is this property passed on to $$C^*$$ tensor products $$A \otimes_\beta B$$?

• Well the answer is obviously yes for the minimal tensor product, thus obviously yes if $A$ or $B$ is nuclear and it is also obviously yes if $A$ and $B$ are both separable, but I'm not sure about the general case Commented Aug 31, 2022 at 13:15
• @JustDroppendIn Thanks for the answer! Could you explain why it's 'yes' for the minimal one?
– Lau
Commented Aug 31, 2022 at 17:21

For the minimal tensor product this is direct: If $$\pi:A\to B(H)$$ and $$\rho:B\to B(K)$$ are representations of $$A,B$$, then one gets a $$*$$-homomorphism $$\pi\odot\rho:A\odot B\to B(H\otimes K)$$ that satisfies on elementary tensors $$\pi\odot\rho(a\otimes b)=\pi(a)\otimes\rho(b)$$. If $$\pi,\rho$$ are faithful, then $$\pi\odot\rho$$ is injective. So, in general one defines a seminorm on $$A\odot B$$ by setting $$\|x\|:=\|\pi\odot\rho(x)\|_{B(H\otimes K)}$$ for all $$x\in A\odot B$$. When $$\pi,\rho$$ are faithful, this is actually a norm. It can be proved (see e.g. Brown-Ozawa) that this norm is independent of the choice of $$\pi,\rho$$, i.e., if $$\pi':A\to B(H')$$ and $$\rho':B\to B(K')$$ are other faithful representations of $$A,B$$, then $$\|\pi\odot\rho(x)\|_{B(H\otimes K)}=\|\pi'\odot\rho'(x)\|_{B(H'\otimes K')}$$ for all $$x\in A\odot B$$.
Now $$A\otimes_{\min}B$$ is defined as the completion of $$A\odot B$$ with respect to this norm that is obtained by any choice of faithful representations. In particular, if $$A,B$$ admit faithful representations on separable Hilbert spaces $$H$$ and $$K$$ respectively, then $$A\otimes B$$ is faithfully represented on the Hilbert space $$H\otimes K$$, which is separable.
As a corollary, if $$A$$ or $$B$$ are nuclear, the answer to your question is "yes".
Also (exercise) it is evident that any separable $$C^*$$-algebra admits a faithful representation on a separable Hilbert space, the reason being that every separable $$C^*$$-algebra has a faithful state; the GNS Hilbert space of a state of a separable $$C^*$$-algebra is separable. As a corollary, if $$A,B$$ are separable, then so is $$A\otimes_\beta B$$ for any $$C^*$$-tensor product, so again in this case the answer to your question is "yes".