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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 …

Hillel Furstenberg conjectured that the only $2$-$3$-invariant probability measure on the circle without atoms is the Lebesgue measure. More precisely: Question: (Furstenberg) Let $\mu$ be a cont …

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 …

Let $H$ be a Hilbert space and let $u_1 \in U(H)$ be a unitary operator on $H$. Consider the self-map $w: U(H) \to U(H)$ which is given by $$w(v) := v^2 u_1 v^{-1}.$$ Since $U(H)$ is connected, there …

The answer is yes, and much more is true. Any hyperfinite von Neumann algebra (with separable predual) has a unique embedding (up to conjugacy) into the ultra-product of the hyperfinite $II_1$-factor. …

Let $P=\left[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right]$ and $Q(\phi)=\left[\begin{matrix} \cos^2(\phi) & \cos(\phi)\sin(\phi) \\ \cos(\phi)\sin(\phi) & \sin^2(\phi) \end{matrix}\right]$. Then …

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 book Kazhdan’s Property (T) (Appendix A7) by Bekka, de la Harpe and Valette the symmetric Fock space on a Hilbert space is $H$ studied as the analogue of a space of measurable functions on a Hi …

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 …

It is not so clear what you mean. However, every separable $C^\ast$-algebra embeds in $B(\ell^2 \mathbb N)$. Hence, the isomorphism classes of separable $C^\ast$-algebras form a set.

The Gel'fand spectrum of $\ell^1 \mathbb N$ is indeed the closed unit disc. After all, every functional to $\mathbb C$, must be given by sending the generator to some complex number. It is easy to see …

The following theorem is due to Plotkin and Rudin and characterizes $p \neq 2,4,6,\dots.$ Theorem: (Plotkin-Rudin): Let $0< p< \infty$ and $p \neq 2,4,6,\dots$. Let $(\Omega,\mu)$ and $(\Omega',\nu)$ …

You can look at $\mathbb N \subset \mathbb Z$. Then the Beurling densities conincide (and give $1/2$) whereas the invariant measure $$\mu(A) = \lim_{n \to \omega}\frac{|A \cap \{1,\dots,n\}|}n$$ gives …

If $n:=[G:H]$, then $\mathbb C[G] \subset M_n \mathbb C[H]$, where $g \in G$ maps to a permutation matrix decorated with elements from $H$ and the embedding depends essentially only on a choice of a t …