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Given an abelian subalgebra $N$ of a von Neumann algebra $M$, such as the center of $M$, one can always write $N=L^\infty (X)$, for some measure space $X$. Furthermore one can decompose $M$ as a "direct integral" $$M=\int_X^\oplus M_x\,dx$$ (sort of a continuous direct sum) of von Neumann algebras indexed by $X$. When $N$ is the center of $M$ ...

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Use the following statement: If $A$ is a unital $C^*$-algebra and $x\in A$ there are unitaries $u_1,...,u_4$ and complex numbers $a_1,...,a_4$ with $x=\sum_i a_i u_i$. So: $$\varphi(x^*x)=\varphi\left(\sum_i(a_iu_i)^*\sum_j(a_ju_j)\right)= \sum_{ij} \overline{a_i}a_j\varphi(u_i^*u_j)$$ now note that $\varphi(u_i^*u_j)= \varphi\left(u_i(u_i^*u_j)u_i^*\right)... 3 Note that: $$\|(E-E_i)h\|^2 = \|Eh\|^2+\|E_ih\|^2 - \langle Eh, E_ih\rangle - \langle E_ih,Eh\rangle$$ As you have seen,$E_i\to E$in WOT implies$\|E_ih\|^2\to \|Eh\|^2$, this is the only step that uses$E_i$and$E$being projections. From the definition WOT convergence it follows that$\langle Eh, E_ih\rangle\to\langle Eh,Eh\rangle = \|Eh\|^2$and that$\...

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So $f$ is a state. Lets first show that it is faithful, meaning that $f(a^*a)>0$ whenever $a\neq0$: Suppose $a\in A$ and $a\neq0$, then there is a state $\varphi$ with $\varphi(a^*a)>0$. Since $f_n$ is weak* dense you have a sub-sequence with $f_{n_k}\to \varphi$ in the weak* topology, but this topology is just so that you can recover $f_{n_k}(a^*a)\... 2 Something that in my opinion is not encouraged enough, is to try basic examples. You can produce a faithful state on$\mathbb C\oplus\mathbb C$by$\tau(a,b)=\frac{a+b}2$. Then you can easily calculate $$\ker\tau=\{(a,-a):\ a\in\mathbb C\},$$ while$(\ker\tau)^+=\{0\}$. Also, note that linear functionals always have big kernels, as$\dim A/\ker\tau=1$. ... 3 You wrote as if$B$is the finite-dimensional one, so I'll stick with that. Let$\gamma$be a C$^*$-norm on$A\otimes M_n(\mathbb C)$(one certainly exists, because we can represent$A\subset B(H)$and then$A\otimes M_n(\mathbb C)$can be represented in$B(H\otimes\mathbb C^n)$). For any$k$, the map$a\longmapsto \gamma(a\otimes E_{kk})$defines a C$^*$-... 1 I know this a pretty old question, but I would like to give an incomplete answer, in the sense, that I will assume that$\mathcal{H}$is finite-dim and that we are dealing with the anti-symmetrized Fock space so that everything is finite-dimensional. The generalization should be relatively straightforward, except that we would have to drown in details such ... 3 Recently, I wrote all this out in detail for myself, so here I share my notes with you. Note that the assumption that$X$is Tychonoff can be ommitted. The construction works for every topological space. The Tychnoff assumption is only there to ensure that the canonical inclusion is injective. Recall that if$A$is a commutative$C^*$-algebra, then we can ... 3 Consider a special set of characters of$C_b(X)$, for each$x\in X$define: $$\delta_x: C_b(X)\to\Bbb C, \quad g\mapsto g(x)$$ Since the (non-zero) characters of$C_b(X)$are the points of$\beta X$this gives you a way of embedding$X$into$\beta X$. Now if$f$is some continuous function on$\beta X$we may identify it also with an element$\tilde f\in ...

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In the algebra of all bounded operators on $\ell^2$, consider the operator $a_n$ defined by $$a_n(\xi ) = \frac 1{\sqrt n}\langle \xi , e_1\rangle e_n, \quad\forall \xi \in \ell^2,$$ where $\{e_n\}_{n\geq 1}$ is the canonical basis. I'll leave it up to you to verify that $$\sum_{n=1}^\infty a_na_n^*$$ converges in norm, but $$\sum_{n=1}^\infty a_n^*... 1 Not all representations of B(X) satisfy the continuity condition mentioned by the OP. To exhibit an example, let us first notice that B(X) is a unital commutative C*-algebra, and hence it is isomorphic to C(K), where K is its spectrum. Given any Borel set E\subseteq X, we have that the characteristic function \chi_E is an idempotent element in ... 1 The answer is negative and here is a counter-example. Let us begin with the following: Lemma. If T and S are positive operators on a Hilbert space, with 0\leq T\leq S\leq 1, and if \xi  is any vector such that T\xi =\xi , then S\xi =\xi , as well. Proof. We have$$ \|\xi \|^2 = \langle \xi , \xi \rangle = \langle T\xi , \xi \rangle \leq \...

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This is not true. Consider $A=M_2(\mathbb C)$, and $$B=\left\{\begin{bmatrix}b&0\\0&0\end{bmatrix}:b\in\mathbb C\right\}.$$ Note that $$\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\begin{bmatrix}b&0\\0&0\end{bmatrix}=\begin{bmatrix}a_{11}b&0\\a_{21}b&0\end{bmatrix},$$ so clearly $AB$ cannot be all of $A$.

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This is done in several places, for instance in chapter 8 in Kadison-Ringrose; it's done in more generality, as they show that any finite von Neumann algebras has a unique central valued trace. The "Murray-von Neumann way" is done in Sunder's An Invitation to von Neumann Algebras. Here is the (very rough) idea. Show that two projections with the ...

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The functional $\varphi_+$ is positive, so (since $s(\varphi_+)\leq e$ and so $f\,s(\varphi_+)=0$), $$\|\varphi_+\|=\varphi_+(1)=\varphi_+(s(\varphi_+))=\varphi(e\,s(\varphi_+))=\varphi(e\,s(\varphi_+)-f\,s(\varphi_+))=\varphi(s(\varphi_+)).$$ You have that $\|\varphi_+\|\leq\|\varphi_1\|$, $\|\varphi_-\|\leq\|\varphi_2\|$, and $\|\varphi_+\|+\|\varphi_-\|=... 0 Here is an example. Consider the closure of graph$\Gamma$of $$\sin\left(\frac{1}{|x|-1}\right), \qquad x\in(-1,1)$$ This is a compact space which looks sort of like this: Basically there are two intervals and that are connected via some line. Chose some homoemorphism of the line in the middle with$\Bbb R$to embed the algebra$C(\overline\Gamma)$into$...

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From $u$ positive and $\|u\|\leq1$, you get $0\leq u\leq 1$. Also, being positive, $u$ has a (unique) positive square root. Then $$u-u^2=u^{1/2}(1-u)u^{1/2}\geq0.$$

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Without loss of generality, assume $A$ is unital. Then $C^*(u,1)\cong C(\sigma(u)$, the continuous functions on the spectrum of $u$, and this isomorphism takes $u$ to the inclusion function, $\sigma(u)\ni\lambda\mapsto\lambda\in\mathbb C$. The conditions that $u$ is positive and $\|u\|\leq1$ then imply that $0\leq \lambda\leq 1$ for all $\lambda\in\sigma$, ...

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