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4

No. For any such sequence $\{p_n\}$ with each $p_n$ of finite rank, take $A=B(\mathcal H)$ and $$ B=\overline{ \{b\in B(\mathcal H):\ \exists n,\ b=p_nbp_n\}}.$$ Then $p_nAp_n=p_nBp_n$ for all $n$, but $B$ is separable while $A$ is non-separable.


3

What you need is the Whitehead Lemma: If $v$ is a unitary in $A$, then $$ \begin{pmatrix} v & 0 \\ 0 & v^{\ast} \end{pmatrix} \sim_h \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \text{ in } U(M_2(A)) $$ which itself follows from the fact that $$ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \sim_h \begin{pmatrix} 1 & 0 \\ 0 & 1 ...


2

These results follow from some quite general results about surjective homomorphisms between $C^{\ast}$-algebras which I will state without proof. In the following, $A$ and $B$ are unital $C^{\ast}$-algebras and $\varphi :A\to B$ is a unital surjective $\ast$-homomorphism. [Proposition 4.3.14 in Higson & Roe's Analytic K-homology ] If $f: [0,1] \to ...


2

Somehow, after I post here a question the solution comes to my mind... So, I'll use the following argument:almost unitaries are close to a unitary element Now, it's enough to show that: If $A$ is a unital $C^*$ algebra and $A=\overline{\bigcup_{k\in \mathbb{N}} A_k}$ ,where each $A_k$ is a unital (same unit of $A$) $C^*$ subalgebra, then for any unitary $...


2

Ignoring your original question: Note that $A = \lim_k A_k$ so $K_1(A) = \lim_k K_1(A_k)$. But $K_1(A_k) = 0$ since $K_1(M_n(\mathbb C)) = 0$ for each natural number $n$. This is a general argument that AF algebras have trivial $K_1$-group.


2

Yes. What you do is show that if $A$ is not simple, then there is a non-faithful irrep. If $J\subset A$ is a non-trivial ideal, then consider an irreducible representation of $A/J$ into $B(H_J)$; then $A\to A/J\to B(H_J)$ is an irreducible representation of $A$ with kernel that at least contains $J$, so it is not faithful.


2

If $\delta\leq1$, then $x^*x$ and $xx^*$ are invertible. In particular, we can do the polar decomposition $x=u(x^*x)^{1/2}$ and we will have $u\in A$. Also, $u$ is a unitary because $$u^*u=(x(x^*x)^{-1/2})^*(x(x^*x)^{1/2} =(x^*x)^{-1/2}x^*x(x^*x)^{-1/2}=1, $$ $$ uu^*=x(x^*x)^{-1/2}(x^*x)^{-1/2}x^*=x(x^*x)^{-1}x^*=1. $$ This last equality is not obvious, but ...


2

Let $A$ be the algebra of all $2\times2$ matrices over $\mathbb{R}$ (or $\mathbb{C}$) of the form $$\begin{bmatrix} a & b \\ 0 & c \end{bmatrix} $$ Then $A$ is a Banach algebra, which is noncommutative, and is not a $C^*$-algebra.


1

If all you assume about your involution is that it's an involution. that is, $(x+y)^*=x^*+y^*$, $(xy)^*=x^*y^*$, $x^{**}=x$ and $(cx)^*=\overline cx^*$, then most of what you expect doesn't follow. In particular you assume above that $\phi(x^*)=\overline{\phi(x)}$, and that doesn't follow: Consider $C([-1,1])$. Define $$f^*(t)=\overline{f(-t).}$$ That's an ...


1

Another similar approach (inspired by Martin Argerami) is the following: If $x \in A$ and $x^*x, xx^*$ are invertible, then $x$ is invertible. To see this, check that $(x^*x)^{-1}x^* = x^*(xx^*)^{-1}$. This is then the inverse of $x$. Now $x = u \sqrt{x^*x}$ for $u$ unitary since $x$ is invertible. Then the last argument of Martin applies, i.e. that $\...



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