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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 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 …

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 …

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 answer is that $G=SU(\infty)$ (with the direct limit topology of the usual Hilbert-Schmidt topologies) is extremely amenable. This means (by definition) that every continuous action of $G$ on a co …

Alex Lubotzky and Yehuda Shalom have shown in Finite representations in the unitary dual and Ramanujan groups., (Discrete geometric analysis, 173–189, Contemp. Math., 347, Amer. Math. Soc., Providence …

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 …