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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 $\ …

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 …

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 …

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 …

This is not an answer but too long for a comment. It was shown in Popa, S. and Takesaki,M., The Topological Structure of the Unitary and Automorphism Groups of a Factor, Commun. Math. Phys. 155, 93- …

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 …

A bit more exotic, a finitely generated subalgebra (no matter $*$-subalgebra or not) corresponds to a continuous map to an affine variety over $\mathbb C$ (continuous in the euclidean topology), such …

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 $A$ be a (say finitely generated, unital) commutative complex $\star$-subalgebra of $B(H)$. Then, the self-adjoint elements form a real subalgebra $B:=A_h \subset A$, such that $B[i] = A$. Moroeve …

A unital $*$-ring $A$ (commutative or not) is a subring of $B(H)$ if and only if for each $a \in A$, there exists a linear functional $\varphi \colon A \to \mathbb R$, such that 1) $\varphi(1)=1$ and …

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. …

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 …

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 no. Consider the Toeplitz algebra $\mathcal T$ with its canonical representation on $\ell^2 \mathbb N$, which is generated as a $C^\star$-algebra by the shift $S(e_n)=e_{n+1}$. It is we …