# maximal tensor product of simple $C^*$algebras is non-simple

Let $A$ and $B$ simple $C^*$-algebras. One can prove that the minimal tensor product $A\otimes _{min}B$ is simple. This is wrong for the maximal tensor product $A\otimes_{max}B$ .

1.Do you know an explicit example?

One idea is to consider (non-nuclear) $C^*$-algebras $A$ and $B$, which are simple and then consider the canonical map $i:A\otimes_{max} B\to A\otimes_{min}B$. One can take $A$ and $B$ such that the map has non-trivial kernel to obtain $\ker(i)\subseteq A\otimes_{max} B$ as a non-trivial closed ideal. Also the example should satisfy: $\ker(i)\neq A\otimes_{max} B$.

Therefore if you take this idea, for example, $A=K(H)=B$ for a separable Hilbert space $H$ doesn't work because these $C^*$-algebras are nuclear, simple and the minimal tensor product is simple.

Then I tried to take (non-nuclear) group $C^*$-algebras. But I'm not sure for which groups $\Gamma$ the reduced group $C^*$-algebra $C_r^*(\Gamma)$ is non-simple, my knowledge is still very limited. For example, there is a result in a paper that if $\Gamma$ is a nonabelian free group, the $C_r^*(\Gamma)$ is simple (R.T. Powers, "Simplicity of the $C^*$-algebra associated with the free group on two generators.").

Therefore, my 2. question is: Do you know a (non-abelian) group $\Gamma$, such that $C_r^*(\Gamma)$ is not amenable and non-simple?

• If a group has a nontrivial amenable normal subgroup than its reduced C*-algebra is non-simple. You have many non-amenable groups like that. Moreover, quite recently an example was given (by Le Boudec, following a criterion by Kalantar-Kennedey) of a group which has no nontrivial normal amenable subgroup and yet, its reduced C*-algebra is nonsimple. But wait: what exactly is the question here? – Uri Bader Aug 13 '16 at 12:40
• There seem to be two separate questions here. Your first question asks about non-simplicit of the maximal tensor product – Yemon Choi Aug 13 '16 at 12:48
• Then, in the question near the end, you seem to ask for a non-amenable discrete group $\Gamma$ whose C*-algebra is non-simple. Just take $F_2\times ({\bf Z}/2{\bf Z})$, surely? – Yemon Choi Aug 13 '16 at 12:50
• @Uri Bader thank you. I didn't know that (my knowledge about $C^*$-algebras is still very limited). Maybe this example by Boudec will be useful for me, I will have a look. Yes, but the questions relate to each other I think. Maybe I should edit my question to clearify what I want to know. – Sabrina Gemsa Aug 13 '16 at 12:51
• I would also like to point out that just because you know $A$ and $B$ are non-nuclear, this does not allow you to deduce that $A\otimes_{\rm max} B \to A\otimes_{\rm min} B$ has a non-trivial kernel. Think about the quantifiers in the definition of nuclearity... – Yemon Choi Aug 13 '16 at 12:52

## 2 Answers

Here is an answer to the second question: yes, there exist such groups. Yemon Choi already gave you one in a comment: take $F_2\times \mathbb{Z}/2$, a product of the free group on 2 generators and a nontrivial finite group. More generally, take any nonameanble group which has a nontrivial amenable normal subgroup. For a very interesting related discussion take a look at the papers by Kalantar-Kennedey and Breuillard-Kalantar-Kennedy-Ozawa (well defined by lists of authors).

Moreover, following the papers above Adrien Le Boudec gave a first example of a group which has no non-trivial normal amenable subgroup which reduced C*-algebra is not simple.

An explicit example to question 1 is given by $A=B=C^*_r(G)$ where $G$ is the free group on two generators.

Takesaki produced the first systematic study of nuclearity (then called Property (T)) in his 1964 paper "On the cross-norm of the direct product of C*-algebras." In Theorem 6 (actually look at the discussion preceding Theorem 6) he shows that $A\otimes_{max} B\neq A\otimes_{min}B.$ And, as you mentioned, Powers showed that $A$ is simple.

• Great, thanks! First I wasn't sure if this example works. – Sabrina Gemsa Aug 14 '16 at 12:48