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2

Let $\phi$ be a state. Suppose that $\phi(a_1)=0$. Then, for some fixed $j$, $$ 0\leq\phi(a_ja_j^*)\leq\phi(a_1)=0, $$ so $\phi(a_ja_j^*)=0$. Now, since $\phi$ is completely positive, it satisfies the Kadison-Schwarz inequality; thus $$\tag{1} 0\leq\phi(a_j)\phi(a_j)^*\leq\phi(a_ja_j^*)=0, $$ so $\phi(a_j)\phi(a_j)^*=0$, which implies $\phi(a_j)=0$. Since ...


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Yes. Let $G$ be a finitely-generated discrete group with finite generating set $S$. Then $G$ is amenable if and only if the Cayley graph of $G$ with respect to $S$ has Cheeger constant 0. This is known as Følner's criterion for amenability. For example, $\mathbb{Z}\times\mathbb{Z}$ is amenable because the infinite square grid has Cheeger constant zero, ...


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Yes (Chapter 14 of "Harmonic Analysis", edited by Eymard and Pier)


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Let $\|\cdot\|$ be some C$^*$-norm on $C_c(X)$. Write $A=(C_0 (X),\|\cdot\|_\infty) $, $B=\overline {C_c(X),\|\cdot\|) }$. Fix $f_0\in C_c(X)$. Put $A_0=C^*(f_0)\subset A$, and $\pi:A_0\to B$ the identity map (this works because $A_0\subset C_c (X)\subset B $; since we use the supremum norm to generate it, every element has support inside that of $f_0$). ...


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It's false if $A$ has more than one complex homomorphism. Say $\phi$ and $\psi$ are distinct complex homomorphisms. Show that $\phi$ and $\psi$ are linearly independent. Now choose $x_0\in A$ with $\psi(x_0)\ne0$ and define $T:A\to A$ by $$Tx=\hat x(\phi)x_0.$$So $$\widehat{Tx}(\psi)=\hat x(\phi)\psi(x_0)$$ If $T$ is given by the multiplier $m$ then ...


1

I wouldn't expect a straightforward answer to your question. Suppose that $A$ is finitely generated, i.e. $A=F_0''$ for some finite $F_0$. Then, for any sequence $\{A_n\}$, for any $B\subseteq A$, for any $n$, you can take $F=F_0$ and then $B\subseteq A=F_0''=(A_n\cup F_0)''$. The problem is that it is not known if every von Neumann algebra is finitely ...


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For every positive element $a$ one has $a\leq\Vert a\Vert I$


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The Hilbert space for the GNS construction for that particular state is certainly not $L^2[0,1]$. What you are expected to do is to go through the proof of the GNS theorem and calculate the Hilbert space and the inner product for that concrete C$^*$-algebra and that particular state. It is not hard.


3

The category of $C^*$-algebras is complete. The limit of a diagram $(A_i,\lVert - \rVert)_{i \in I}$ of $C^*$-algebras has as underlying $*$-algebra the $*$-subalgebra of the $*$-algebra $\prod_{i \in I} A_i$ whose elements $x=(x_i)_{i \in I}$ are subject to two conditions: First, the usual matching condition: For edges $i \to j$ the map $A_i \to A_j$ should ...


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You should study Morita equivalence of C*-algebras (in the sense of Rieffel): the fact is that $C(X,M_2)$ is Morita equivalent to $C(X)$. And it is a general fact that Morita equivalent C*-algebras have same closed (two-sided) ideals. It follows that the closed ideals of $C(X,M_2)$ are all of the form $C_0(U,M_2)$ (functions vanishing outside $U$) for ...


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Assume $k$ field and $A$ a finite dimensional $k$ algebra. Then the spectrum of any element is finite. Step 1. Every element $a$ of $A$ satisfies a polynomial equation of degree at most $n= \dim_{k} A$. Indeed, the elements $1$, $a$, $\ldots$, $a^n$ are linearly dependent. Step 2. Let $a$ satisfying a polynomial equation $P(a) = 0$, where $P \in k[X]$ , ...


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Consider abelian C*-algebra $C^*(a)$ which is infinite dimensional ( because $\sigma(a)$ is infinite). Also $C^*(a) \subset A$ which implies that $A$ is infinite dimensional.


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You are right, my previous attempt was wrong: here are examples that answer your question: take $A=C(X)$ to be the Banach algebra of all continuous function on some compact space (say $X=[0,1]$, or take $X$ to be a point so that $A=\mathbb{C}$) and define a new norm on $A$ by $\|f\|_r:= r \sup_{x\in X} |f(x)|$, where $r>1$ is some (fixed) number. Then ...


2

No. Let $$ S=\begin{bmatrix}0&1\\0&0\end{bmatrix},\ \ T=\begin{bmatrix}0&0\\1&0\end{bmatrix}. $$ Then $$ \text{Tr}(TS)=1,\ \ \text{ and } \|T\|\,|\text{Tr}(S)|=0. $$


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The specific unitaries appearing in the OP are a representation of the relations defining the rotation algebra $A_\theta$ that is faithful when $\theta$ is irrational (for otherwise $V^q=I$ for some integer $q$). The more general definition of the rotation algebra $A_\theta$ is as the universal C*-algebra generated by two unitaries $u,v$ that satisfy to the ...


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The key observation is that for any $\alpha,\beta\in\mathbb C$, the operator $\alpha I+\beta A^*A$ is normal. This guarantees the first equality below: \begin{align} \|\alpha I+\beta A^*A\|&=\max\,\{|\lambda|:\ \lambda\in\sigma(\alpha+\beta A^*A)\}=\max \,\{|\lambda|:\ \lambda\in\alpha+\beta \sigma(A^*A)\}\\ &=\max \,\{|\alpha+\beta\lambda|:\ ...


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Of course it is not true in general. As you mentioned, $\|V\|^2=\|\Phi\|$, so if $V^*V\leq\text{Id}_{L(H)}$, this implies that $\|\Phi\|\leq1$. The equality $\Phi(a)=P_H\,\pi(a)|_H$ is not really an equality; there is a unitary hanging around, but it is not worth writing. What happens is this: if $\Phi$ is unital, then $V$ is an isometry. So $H_0=VH$ is a ...


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The trick I know for this uses the multiplicative domain. The thing is to show that $$\tag{1} L=\{x\in M:\ \phi(yx)=0\,\ \forall y\in M\}. $$ The multiplicative domain of $\phi$ is the set where equality holds in the Kadison-Schwarz inequality, i.e. $$ R_\phi=\{x:\in M:\ \phi(x)^*\phi(x)=\phi(x^*x)\}. $$ It can be shown (there are proofs in Paulsen's book ...


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You are almost there. The ideals of $C (Y) $ are given by its closed subsets, and the essential ideals are those that correspond to closed nowhere dense subsets. In other words, the essential ideals of $C (Y) $ are precisely $C _0 (T) $, where $T\subset Y $ is open and dense.


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Since the value of $B(\infty)$ is not a multiple of the identity matrix, the function $B$ does not seem to belong to the unitization of $S^2\mathbb M_2$ as claimed, but rather to $\mathbb M_2$ of the unitization of $S^2\mathbb C$.



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