Caleb Eckhardt
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I think Nik Weaver is right that the ideal mentioned is the unique maximal ideal. This simultaneously answers both questions (since the quotient is clearly infinite dimensional). Let $\tau$ be the ...

I am not sure if it is in Brown and Ozawa, but it is in Pisier's recent book "Tensor Products of C*-algebras and Operator Spaces" as Corollary 10.16. It may also be in his earlier Operator ...

No. This will almost never be true (subalgebras of the compacts are the only cases I can think of where it could work). The easiest example is probably $C[0,1].$ It has an injective homomorphism to $... View answer Accepted answer 10 votes I think the answer is: It's always continuous if$A$is subhomogeneous and never continuous otherwise(min or max). First notice that if the product map is continuous for min, then it's continuous for ... View answer 9 votes This is a partial answer that shows an obstruction to certain eigenvalue sequences. First, I claim that if$M$is rank one then it isn't similar to something of the stated form. Take a matrix$M$... View answer Accepted answer 9 votes I don't think so. If A is a noncommutative C*-algebra with the min operator space structure then it's double dual$A^{**}$will also have the min operator space structure (this fact can be found in ... View answer 8 votes (Unless I messed up the arithmetic, here's a counterexample.) Let$B=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$and$A=\begin{bmatrix} 1 & 0 \\ 0 & R \end{bmatrix}$where$R$is a ... View answer Accepted answer 7 votes It is not Type I in general. Probably it is not Type I whenever$G$is infinite. Here is an argument when$G=\mathbb{Z}.$Consider the projections in$\ell^\infty(\mathbb{Z})$defined by ... View answer Accepted answer 7 votes No. Here's an argument that shows the answer is no for non-abelian, torsion free, finitely generated two-step nilpotent groups (this argument could easily be pushed to nilpotent groups of steps ... View answer 7 votes This is not an answer, but too long for a comment and just an idea for an approach. Kirchberg algebras (unital, purely infinite, separable, nuclear) in the UCT class are classified by a triple$(G_0, ...

Your algebra is Type I and residually finite dimensional (RFD) Therefore it's unitization is also Type I and RFD. Let's call $B$ it's unitization. Since $B$ is Type I it satisfies the UCT and ...
I think the transpose map on the compacts gives a counterexample. Let $K(H)$ be the compacts on a separable infinite dimensional Hilbert space with orthonormal basis $\{ e_n \}.$ Let $T:K(H)\... View answer Accepted answer 6 votes This is true in the separable case (and more generally) and a consequence of Larry Brown's stable isomorphism theorem (1977 Pacific Journal of Math). A special case of his theorem states: If$A$is ... View answer Accepted answer 6 votes No. I use facts about Schur multipliers which can be found, for example in Paulsen's monograph on completely bounded maps. Basically you are searching for a positive semidefinite matrix with positive ... View answer Accepted answer 6 votes Yes. Since$C^{**}\cong A^{**}\oplus B^{**}$, uniform bounds on the dimension of the irreducible representations of$A$and$B$impose a uniform bound on the dimension of the irreducible ... View answer 6 votes An explicit example to question 1 is given by$A=B=C^*_r(G)$where$G$is the free group on two generators. Takesaki produced the first systematic study of nuclearity (then called Property (T)) in ... View answer Accepted answer 4 votes In practice, there are several non-obvious corollaries of Voiculescu's theorem that are collectively called "by Voiculescu's Theorem." In this case it is a consequence of Voiculescu's theorem (the one ... View answer 4 votes It will be rare to find a C*-algebra which has one folium for all states. Most likely you need to require that it's dual space is norm separable (so you can do it for say subalgebras of compact ... View answer 4 votes Ext is a group for the full case. Kirchberg proved in Lemma 3.3 of [Commutants of unitaries in UHF algebras and functorial properties of exactness. J. Reine Angew. Math. 452 (1994), 39–77] that the ... View answer 3 votes The conditional expectation it is not unique. Basically the different positive definite functions on the quotient group$\mathbb{Z}/k\mathbb{Z}$will produce other conditional expectations. Here is ... View answer Accepted answer 3 votes There are "cheap" examples. If a C*-algebra$A$has stable rank 1 then it is stably finite. On the other hand every simple, unital purely infinite C*-algebra has real rank 0. So every simple, ... View answer 3 votes No. To obtain a counterexample, you just need a C*-algebra$A$with finitely generated K-theory and a quotient$A/I$of$A$which does not have finitely generated K-theory and the quotient ... View answer Accepted answer 3 votes Using your notation, we have diag$(x',...,x')\in M_n\otimes M_m(A).$Then$M_n$is isomorphic to the algebra (let's call it$B$) generated by$e_{ij}\otimes x'$where$e_{ij}$are matrix units for$...