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4

Yes, because entrywise convergence in $M_n(A)$ implies convergence. Indeed, if $a^{(t)}=(a^{(t)}_{kj})$, $a=(a_{kj})$, and $a_{kj}^{(t)}\to a_{kj}$ for each pair $k,j$, then $a^{(t)}\to a$. Below is an argument. The norm in $M_n(A)$ is given by the operator norm of the action of $M_n(A)$ on $H^n$, when we represent $A\subset B(H)$. Then, for any $b\in M_n(...


3

Since the properties involved are about boundedness and convergence, they survive $*$-homomorphisms, which means that they pass to quotients. And any separable C$^*$-algebra is a quotient of $C^*(\mathbb F_\infty)$. In particular, $C[0,1]$ is, so $C^*( \mathbb F_\infty)$ does not have a boundedly complete Schauder basis by the article you linked to.


3

You have $$C_0(\mathbb R)=\{f\in C(\beta\mathbb R):\ f|_{\beta\mathbb R\setminus\mathbb R}=0\}.$$


2

If $B$ abelian, then $\varphi:A\to B$ is a unital completely isometry if and only if it is a unital isometry. Indeed, if $\varphi$ is a unital isometry, then $\varphi^{-1}$ is unital and contractive on $\varphi(A)\subset B$. Let $f\in S(A)$ (that is, a state on $A$). Then $f\circ\varphi^{-1}$ is a unital and contractive functional on the operator system ...


2

(the argument below is extracted from Lemmas 7.2.13 and 7.2.14 of Kadison-Ringrose; the relevant more general theorems are Theorem 7.2.15 and Corollary 7.2.16) If $x,y\in A'$ are selfadjoint, then we can find $\{a_n\}, \{b_n\}\subset A$, selfadjoint, with $a_n\xi\to x\xi$ and $b_n\xi\to y\xi$. Indeed, since $\xi$ is cyclic we can get $c_n$ in $A$ with $c_n\...


2

As noted in the comments, this is false. Intuitively, on the left-hand side in 1) you add one point to both spaces and then take the cartesian product whereas on the right-hand side you first take the product and then add one point. For example let $A=B=C_0((0,1))$. Then $\widetilde{A\otimes B}=\widetilde{C_0((0,1)\times (0,1))}=C(S^2)$ and $\tilde A\otimes ...


2

You have, since $0\leq q\leq I$ and $0\leq p\leq I$, $$ -I\leq -q\leq p-q\leq I-q\leq I. $$ So, as you mentioned, it follows that $\sigma(p-q)\subset[-1,1]$. Note also that the argument does not use that $p,q$ are projections, only that they are positive elements of the unit ball.


2

for $2\implies 3$: you have, since $a\geq0$, $$ \tau(uau^*)=\tau(ua^{1/2}a^{1/2}u^*)=\tau((a^{1/2}u^*)^*a^{1/2}u^*) =\tau(a^{1/2}u^*(a^{1/2}u^*)^*)=\tau(a^{1/2}u^*ua^{1/2})=\tau(a^{1/2}a^{1/2}) =\tau(a) $$ For $3\implies 1$: since the condition holds for positive $a$ and every operator is a linear combination of positives, the equality holds for arbitrary $...


1

Your first inequality is usually known as the Kadison-Schwarz inequality. It only requires $2$-positivity. Claim. $A=\begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\geq0$ if and only if $a^*a\leq b$. Proof. If $A\geq0$, then for any $\xi\in H$, $$ \langle (b-a^*a)\xi,\xi\rangle=\left\langle \begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\,\begin{...


1

The GNS construction doesn't have a canonical way of choosing "the right amount" of states. For an arbitrary algebra one uses all states (see for instance Corollary I.9.11 in Davidson's C$^*$-Algebras by Example). But this is often way too much. Let's think of the easiest concrete example: let $A=M_2(\mathbb C)$. If we follow the abstract recipe, we need ...


1

I claim that if $A$ is an infinite-dimensional C*-algebra then it does not have a boundedly complete basis. Proof. Assume that $A$ has such a basis. A space with a basis is separable, so $A$ is separable. Moreover, $A$ being infinite-dimensional, contains an infinite-dimensional abelian sub-C*-algebra. That abelian subalgebra will contain a sequence of ...


1

The only solution I can see uses Tomita-Takesaki theory: Let $M=A''$ be the von Neumann algebra generated by $A$, so that $M$ is also abelian, i.e., $M\subseteq M'$. Note that $M'=A'''=A'$, so we are done if we show that $M'=M$ (in fact, this implies that $M$ is maximal abelian). We already have one inclusion $M\subseteq M'$. Since $\xi$ is cyclic for $A$, ...


1

The only thing that is wrong is that the inclusion $\bigoplus_{i\in\mathbb{N}}A_i \to \bigoplus_{i\in\mathbb{N}}A_i^1 $ is given by $$(a_n)_n\mapsto \left(a_n\right)_n.$$


1

You have proper inclusions of finite-dimensional C $^*$-algebras $$\tag{1}A_1\subset A_2\subset\cdots $$ embedded in a unital way. Each of these is a direct sum of full matrix algebras, and the embeddings are tracked by the Bratelli diagram. By looking at a single path in the diagram, we get an inclusion $$\tag{2} M_{n_1}(\mathbb C)\hookrightarrow M_{n_2}(...


1

Each $B_n$ can be seen as $M_n(\mathbb C)$, with the embeddings $$\gamma_{n,n+1}:X\longmapsto \begin{bmatrix}B&0\\0&1\end{bmatrix}.$$ Let $\{e_{kj}^{(n)}\}$ denote the matrix units of $M_n(\mathbb C)$, and define $x_n=e_{nn}$. Then, for all $n$, $$ \|x_1+\cdots+x_n\|=1, $$ while $$\|(x_1+\cdots+x_m)-(x_1+\cdots+x_n)\|=1$$ if $m\ne n$.


1

This is exactly the uniqueness in the polar decomposition. You have, since $v $ is a partial isometry, $$\tag {2}{\text {ran}\,v^*v}= {\text {ran}\,v^*}=(\ker v)^\perp=(\ker y)^\perp=\overline {\text {ran}\,y}. $$ Suppose that $w,z $ gives another such decomposition of $x $. Let $p=v^*v=w^*w $. Then, since $py=y$, we have $$ y^2=y^*y=y^*py=y^*v^*vy=x^*x=|x|...


1

Here $$g=\text{diag}\,(1,-1)=\begin{bmatrix}1&0\\0&-1\end{bmatrix}.$$ You can easily check that $$ M_2(A)^{(0)}=\{a:\ gag=a\},\ \ \ \ M_2(A)^{(1)}=\{a:\ gag=-a\}. $$ The standard even grading on $A\otimes\mathbb K$ is obtained by doing the above on $M_2(A\otimes\mathbb K)$ (diagonal matrices and matrices with diagonal zero). How it looks like ...


1

Note that in your definition of $I_n$ you have some confusing use of $n$, which is fixed but also appears in $\cup_n\{x_n\}$, which is probably better written as $\overline{\{x_k:\ k\in\mathbb N\}}$. I'm not particularly comfortable with the way you argue that $I\ne A$. It is enough to show that $I$ cannot contain any function that is nonzero on $\overline{...


1

Your question is unclear to me but I will comment on it anyway. What generating relations do you impose on your generators? Note that it is an open problem whether the reduced group algebra of the free group on two generators has a Schauder basis. Before Enflo showed that there exist Banach spaces without the AP, people had looked at this object as a natural ...


1

Your proof is fine. What you are missing to work at points other than zero is the following lemma: Lemma. Let $f:X\to\mathbb C$ with $X$ a compact subset of $\mathbb R$, $\varepsilon>0$ and $R>0$. Then there exists $\delta=\delta(\varepsilon,R)>0$ such that if $a,b\in A^+$, with $\|a\|+\|b\|<R$, with $\sigma(a)\cup\sigma(b)\subset X$, and such ...


1

$\def\c{\complement} \def\cl{\mathrm{cl}\,}$ Taking this definition: $\partial A:=\cl A\cap \cl A^\c$, the spectrum $A:=\sigma(a)$ is closed, and its complement is dense as misses at most parts of the real line. This proves $\sigma(a)=\partial\sigma(a)$.


1

Invertible in $qAq$ is not the same as invertible in $A$. For instance, $q$ is invertible in $qAq$ (it is the unit there). That $a$ is invertible in $qAq$ means that there exists $y\in qAq$ such that $ay=ay=q$. So, you have $x=va$; then $$ v=vq=vay=xy\in A. $$ About the spectrum, in general you cannot say anything: if $A=M_2(\mathbb C)$ and $$ p=\begin{...


1

Martin gave a beautiful answer to your question but let me take this opportunity to advertise a solution to a more general question of which C*-algebras may be written as tensor products of two infinite-dimensional C*-algebras. In particular, it is proved that $B(\ell_2)$ as well as the Calkin algebra cannot be decomposed like that no matter which tensor ...



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