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No; consider $\mathbb C \oplus \mathbb C \oplus \mathbb C \subset \mathbb C \oplus M_2(\mathbb C)$ (in the obvious way) with the state $\varphi(x_1,x_2,x_3)= \frac12(x_1 + x_2)$. Then, the extension o …

There has been a lot of work on the spectrum of the operator $u+v + (u+v)^{\star}$. For example by Choi-Elliott-Yui (see here). Similar operators have also been studied from the Mathematical Physics c …

Over the last week, I was discussing this question with Narutaka Ozawa. I could come up with a proof that quasi-diagonal, torsion generated Kazhdan groups must have a finite quotient. Here, a group is …

There is an exact sequence $$ 0 \to \oplus_n M_n(\mathbb C) \to A \to \mathcal K \to 0.$$ Thus, $A$ is nuclear as an extension of nuclear $C^*$-algebras, see vor example $IV.3.1.3$ in [Bruce Blackadar …

The spectral measures for self-adjoint elements in $\mathbb C F_2$ are very special. In particular, it is known that non of the elements in $\mathbb C F_2$ has a kernel when acting via the left-regula …

A priori, it does not make sense to talk about complete boundedness, since there are no specified operator space structure on $A_1$ and $A_2$. In general, an infinite-dimensional Banach space can car …

Suppose that there was such a category. Then, all objects would isomorphic to $R$ anyway and the question is how sum and tensor product act on the morphisms. The natural choice of morphisms is the set …

Let $M= L \mathbb F_2$ and $H = \ell^2 \mathbb F_2$, where $\mathbb F_2 = \langle a,b \rangle$ is the free group on two generators and $B(H)$ is a bimodule via the left and right multiplication with t …

The answer is no, in general there is no lifting. A lifting exists if the $C^{\ast}$-algebra has the so-called lifting property (LP), and local liftings exist if it has the local lifting property (LLP …

There is an alternative argument for the free group; not using that free groups are residually finite-dimensional. Let $\pi$ be a faithful representation of $C^{\ast}(F)$ on a Hilbert space $H$. Then …

The group $G:=SL_{\infty}(\mathbb Q) = \cup_n SL_n(\mathbb Q)$ is a concrete example. It is obviously simple and non-amenable. Let $g_n \in SL_{\infty}(\mathbb Q)$ be the matrix which is $$g_n:= 1_n …

The answer is yes, this is true (assuming that the Hilbert space is complex). If $\langle \xi,A\xi \rangle = \sigma$ for some $\sigma \in \mathbb C$ and all $\xi$, then $B:=A - \bar \sigma 1_H$ has t …

Let $A$ be a $C^*$-algebra or some norm-closed algebra of operators on a Hilbert space. In the old paper Hille, E. On Roots and Logarithms of Elements of a Complex Banach Algebra, Math. Annalen, B …

In the late 70s, Cuntz and Behncke had a paper H. Behncke and J. Cuntz, Local Completeness of Operator Algebras, Proceedings of the American Mathematical Society, Vol. 62, No. 1 (Jan., 1977), pp. 95 …

In the theory of subfactors, it frequently happens that a subfactor of finite index $N \subset M$ satisfies $N' \cap M = {\mathbb C}$. Those subfactors are called irreducible. In this case the relativ …