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The answer is yes for separable Hilbert spaces. If the Hilbert space is separable with basis $\lbrace e_n \mid n \in \mathbb N\rbrace$, you only have to fix countably many inner products and define $\ …

It was proved by Karl H. Hofmann and Karl-H.Neeb here (see here for the preprint), that epimorphisms in the category of $C^{\star}$-algebras are surjective.

There is important work by Alain Connes and Dimitri Shlyakhtenko (see $L^2$-homology for von Neumann algebras (MSN)). They come up with a definition of $\ell^2$-homology for finite von Neumann algebra …

Consider $G=\mathbb Z$ and $\chi = \delta_{-1} + \delta_1 \in \mathbb C[\mathbb Z]$. Then $\chi=\chi^*$. One can check that $$(\chi^{2n})_k= \binom{2n}{n+k}.$$ Hence, with your definition (and the rem …

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 …

Theorem Let $A$ be a unital ring and $I_1,\dots,I_n \subset A$ be 2-sided commutative ideals such that $A=I_1+\dots + I_n$. Then, $A$ is commutative. Proof: If $A=I_1+\dots+I_n$, then $1 = x_1+\dots+ …

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 …

It is well-known that $K_1(C^*_{\rm red}(F_2))={\mathbb Z^2}$ with generators given by $[u]$ and $[v]$, where $F_2=\langle u,v\rangle$. Now, the automorphism of order two associated with the even-odd …

There is an easy example, namely $SL_3(F)$ where $F$ is the algebraic closure of some finite field. This group does not admit non-trivial characters (a result of Kirillov) and is locally finite, hence …

The reduced $C^*$-algebra of a non-abelian free group $G$ has a unique trace. Hence, there is no chance to separate conjugacy classes of group elements using traces on $C^\star_{red} G$. On the other …

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 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 …

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 …

If Kazhdan's property (T) is reflected in the structure of the Cayley graph, then not in a very geometric way. Steve Gersten (that is what I read in the book by B. Bekka, P. de la Harpe and A. Valet …

This is mainly a comment. My first guess would be that if it is true, then a random subset of density $1/2$ works. A result in this direction is Lemma 3.2 in M. Rudelson, Contact points of convex bo …