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New answers tagged operator-algebras

1

Well, I think I have an answer to myself... The natural maps $\phi^n:K_0(A_n)\to K_0(A)$ are compatible with $T^{-n}:\Bbb{Z}^2 \to \Bbb{Z}^2$, which are isomorphisms. Thus, $\Bbb{Z}^2 \cong K_0(A)=\cup \phi^n (K_0(A_n)) \cong \cup T^{-n}(\Bbb{Z}^2)$. From continuity of $K_0(A)_+$ we get: $K_0(A)_+=\varinjlim K_0(A_n)_+ =\cup \phi^n(K_0(A_n)_+)\cong \cup T^{-... 1 I just want to add a couple of observations to Jose Brox's answer. Firstly your question had the additional assumption that$A$has an anti-involution$*$fixing each of the$x_i$, but this doesn't make a difference. Indeed consider unital$\mathbb C$-algebra with presentation $$F=\mathbb C\langle x_1,\ldots,x_n \mid x_i^2=x_i,\;x_1+\ldots+x_n=1\rangle. ... 0 Let A=M_2(\mathcal O_n). Fix a nontrivial automorphism \alpha of \mathcal O_n, and let$$ A_0=\left\{\begin{bmatrix}a&0\\0&\alpha(a)\end{bmatrix}:\ a\in\mathcal O_n\right\}. $$Then A_0\simeq\mathcal O_n. Now let$$ A_t=u_t\,A_0\,u_t^*, $$where$$ u_t=\begin{bmatrix}\cos t&\sin t\\ -\sin t&\cos t\end{bmatrix}. $$The continuity of ... 0 Let R be a ring free of all torsion, with unit and idempotents e_1,\ldots,e_n such that e_1+\ldots+e_n=1. If n\leq 3 then necessarily e_ie_j=0 for i\neq j, but if n\geq 4 the claim is false. This second part is proved in Nonorthogonal idempotents whose sums is idempotent (Mauldon, 1964): Nonorthogonal idempotents whose sums is idempotent Let ... 3 False. The closed span of e_k for k \ge n is invariant. 0 It has nothing to do with pure, not even with positivity. States extend uniquely to the unitization of A, so we may assume that A is unital . Let x\in A. Then x-f(x)I\in \ker f=\ker f'. Then$$ 0=f'(x-f(x)I)=f'(x)-f(x). $$So all we are using is linearity and that f(I)=f'(I)=1. For two arbitrary functional we the same kernel, we would obtain f'... 1 I don't know enough to give you a very authoritative answer. But, as far as I can tell, there is no "general theory" of non-selfadjoint subalgebras of B(H) the way that there is a theory for c^*-algebras or von Neumann algebras. There is a rather complete classification of nest algebras, and some generalizations. The original source for non-selfadjoint ... 1 Your first inequality is usually known as the Kadison-Schwarz inequality. It only requires 2-positivity. Claim. A=\begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\geq0 if and only if a^*a\leq b. Proof. If A\geq0, then for any \xi\in H,$$ \langle (b-a^*a)\xi,\xi\rangle=\left\langle \begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\,\begin{... 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

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.

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 $A$, ...

0

Suppose that $a$ is positive, that $\|a\|=1$ and that $a-1$ is invertible. This means that $0\not\in\sigma(a-1)$, so $1\not\in\sigma(a)$. As the spectrum is closed, there exists $\varepsilon>0$ such that $(1-\varepsilon,1+\varepsilon)\cap\sigma(a)=\varnothing$. Consider the function $$f(t)=\begin{cases}t,&\ 0\leq t\leq 1-\varepsilon,\\ 1-\varepsilon,... 0 I tend to think of \beta G differently. Say B(G) is the space of bounded (continuous) complex-valued functions on G. Then B(G) is a Banach algebra, and \beta G is the maximal ideal space of B(G), which is to say the space of complex homomorphisms of B(G). Say \phi\in\beta G, which in our current formulation says \phi:B(G)\to\Bbb C is a ... 2 Let g \in G. Then the translation x \to gx defines an action from G onto itself. This action extends to a continuous action from \beta G to \beta G. 2 The following formula for \|T^{-1}\| is relevant for the question posted. Let (\mathcal E, \|\cdot\|_{\mathcal E}) and (\mathcal F, \|\cdot\|_{\mathcal F}) be Banach spaces and let \mathcal L(\mathcal E,\mathcal F) be the space of all bounded operators from \mathcal E into \mathcal F. Let T \in \mathcal L(\mathcal E,\mathcal F). The following ... 1 This is exactly the uniqueness in the polar decomposition. You have, since v  is a partial isometry,$$\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

Not true as stated. Let $y$ be the operator on $\ell_2$ that maps each sequence $(t_j)_{j\in\mathbb{N}}$ to $(2^{-j}t_j)_{j\in\mathbb{N}}$. This is a positive operator with zero kernel. Let $c$ be the orthogonal projection onto the orthogonal complement of the vector $z = (1,1/2,1/3,\dots)$. Since $z\notin \operatorname{ran} y$, it follows that $cy$ also has ...

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 ...

2

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 $\overline{... 1 Your proof is fine. What you are missing to work at points other than zero is the following lemma: Lemma. Let$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$, and such ... 0 Take$b $to be a nontrivial projection,$u $a unitary, and$a=ub $. Then$\ker a\ne\ker u=\{0\} $. For Q2, take$a=u=b $. 1 Invertible in$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{...

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