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).

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    $\begingroup$ It is no longer a conjecture that every discrete group is exact. Being exact implies being coarsely embeddable into a Hilbert space. Using random groups, Gromov has constructed a countable discrete group that does not embedd into a Hilbert space, and thus is not exact. $\endgroup$ – Owen Sizemore Jul 30 '13 at 22:43
  • $\begingroup$ I just read Martin's answer to your question regarding nuclearity, you might be able to do a similar thing for exactness using a non exact group. $\endgroup$ – Owen Sizemore Jul 30 '13 at 22:56
  • $\begingroup$ Yes @OwenSizemore, but what about the existence of a single generator, and a faithful irreducible representation ? $\endgroup$ – Sebastien Palcoux Jul 30 '13 at 22:59
  • $\begingroup$ Yes, that's where the "might" comes in, unfortunately Gromov's examples are quite complicated, in fact to the point that I have talked to many other operator algebraist and none that I have talked to have studied them sufficiently to be able to answer these two questions. If you are feeling particularly masochistic the paper is "Random walks in Random Groups" $\endgroup$ – Owen Sizemore Jul 30 '13 at 23:03

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.

  • $\begingroup$ Thank you @CalebEckhardt. There are some things not clear for me. In your first paragraph, there are simple and non-exact $C^{*}$-algebras. In your second paragraph, there are singly generated and non-exact $C^{*}$-algebras. In order to have a counter-example, is there in your examples, a simple, non-exact and singly-generated $C^{*}$-algebra ? $\endgroup$ – Sebastien Palcoux Jul 31 '13 at 8:33
  • $\begingroup$ About $\mathcal{A} = C^{*}(\mathbb{F}_{2})$ : I read here (theorem 5 p 16) that $C^{*}(\Gamma)$ is exact iff $\Gamma$ is amenable (Wassermann Kirchberg), so $\mathcal{A}$ non-exact. $\endgroup$ – Sebastien Palcoux Jul 31 '13 at 10:37
  • $\begingroup$ Next, Choi (1980) proved in this paper, that $\mathcal{A}$ is primitive (i.e. admits a faithful irreducible representation), and also that $\mathcal{A} ≃ C^{*}(U,V)$, with $(U,V)$ a universal pair of unitaries operators on H. But $C^{*}(\mathbb{F}_{2})$ is not singly generated by Masaru Nagisa (see here example 5 page 4). $\endgroup$ – Sebastien Palcoux Jul 31 '13 at 12:34
  • $\begingroup$ Yes, take any simple, separable non-exact C*-algebra $A$ (which Dadarlat has constructed for example) and tensor it with a UHF algebra $U$, then $A\otimes U$ will be simple (classic result of Takesaki that the minimal tensor product of simple C*-algebras is simple) and non-exact (since it contains the non-exact subalgebra $A$) and singly-generated by the above-mentioned result of Olsen and Zame. $\endgroup$ – user85775 Jul 31 '13 at 13:10

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