# Tag Info

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 \$\...

Only top voted, non community-wiki answers of a minimum length are eligible