Does an irreducible operator generate an exact $C^{*}$-algebra?

Question

Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.

Definition : An operator $T \in B(H)$ is irreducible if $W^{*}(T)=B(H)$.

Definition : A $C^{*}$-algebra is exact if it preserves exact sequences under the minimum tensor product.

Property : A $C^{*}$-algebra is exact if and only if :

• it's nuclearly embeddable into $B(H)$.
• it's isomorphic to a subalgebra of the Cuntz algebra $\mathcal{O}_2$.

Does an irreducible operator generate an exact $C^{*}$-algebra ?

Counter-example : Is there a non-exact singly generated $C^{*}$-algebra ? And does every $C^{*}$-algebra admit an irreducible faithful representation ? (it's ok for the simple $C^{*}$-algebras).

Remark : $C^{∗}$-algebras book (Eds Cuntz Echterhoff) : by definition, a discrete group $Γ$ is exact if $C^{∗}_{r}(\Gamma)$ is exact. If $Γ$ is amenable, then $C^{∗}_{r}(\Gamma)$ is nuclear and so exact. Then, the amenable groups are exact. Next, Adams (1994) proves also that the hyperbolic groups are exact, in particular the free groups are exact. Gromov built non-exact discrete random groups (see @OwenSizemore comment).

Answer 1

For the main question: Not always. Any simple C*-algebra has a faithful irreducible representation. Dadarlat constructs many examples of simple, non-exact C*-algebras in [Nonnuclear subalgebras of AF algebras. Amer. J. Math. 122 (2000), no. 3, 581–597]. It is also known (see Davidson's C*-algebra book) that the full group C*-algebra of free groups have faithful, irreducible representations (and these C*-algebras are not exact).

For the first question labeled ``Counterexample": Yes. By the paper of Olsen and Zame [Some C∗-algebras with a single generator. Trans. Amer. Math. Soc. 215 (1976), 205–217] for any separable C*-algebra $A$, the C*-algebra $A\otimes U$ is singly generated where $U$ is a UHF algebra. So take any separable non-exact C*-algebra $A$, then $A\otimes U$ is separable, non-exact and singly-generated.

For the second question labeled ``Counterexample." No. Take any nontrivial abelian C*-algebra as a counerexample.