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

  • $\begingroup$ 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 $\endgroup$ Commented Aug 31, 2022 at 13:15
  • $\begingroup$ @JustDroppendIn Thanks for the answer! Could you explain why it's 'yes' for the minimal one? $\endgroup$
    – Lau
    Commented Aug 31, 2022 at 17:21

1 Answer 1


This is not a complete answer.

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".

For the general question, I don't know, I don't see an obvious reason for this to be true. I don't even see a reason for the property to pass to the maximal tensor product.


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