A von Neumann algebra is a unital *-subalgebra of the algebra of bounded operators on a Hilbert space, closed in the weak operator topology. Also called a $W^*$-algebra, and may be regarded as a non-commutative generalization of $L^\infty$ space. These algebras are extensively used in knot theory, ...

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Some doubts concerning spectral theory.

Probably I'm saying something wrong (that's why the conclusions are strange) so please correct me! There is the continuous functional calculus for a normal element $N$ of a C*-Algebra. This means ...
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21 views

If $N$ is normal, $W^\ast(N)=\{N\}''$

I want to prove that if $N$ is normal, then $W^\ast(N)=\{N\}''$, where $W^\ast(N)$ is the Von Neumann algebra generated by $N$ and $\{N\}''$ is the bicommutant of $N$. for the inclusion $W^\ast(N) ...
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29 views

norm of a matrix that its entries are operators in B(H).

Let S is a subset of B(H). Define $M_2(S)=\{T= \left( \begin{array}{ccc} A & B \\ C & D \\ \end{array} \right) : A,B,C,D \in S\}$. what is the relationship between $||T||$ and ...
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25 views

polar decomposition of multiplicative operator on L^2 induced by identity function.

We know that every operator in B(H) has a polar decomposition. $T=VP$ that $P=|T|$ and V is a partial isometry with initial space closure of ImP and final space ImT. How can i obtain polar ...
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32 views

Tensor Product of Factors of type $II_{1}$

Let $\mathcal{R}$ be a type $II_{1}$ factor. Is it true that $\mathcal{R}\otimes\mathcal{R}$ and $M_{2}(\mathcal{R})$ are still type $II_{1}$ factors ? If so, any reference ? (I am asking this ...
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Closed unit ball of B(H) is not compact in strong operator topology of B(H).

In operator theory we prove that closed unit ball of B(H) is compact in weak operator topology and is closed in strong operator topology. But a book of operator theory states that closed unit ball of ...
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39 views

The map $T\longmapsto \|T\|$ is not continuous in the strong operator topology of $\mathscr B(H)$

In the context of Strong and Weak operator topologies on $\mathscr B(H)$ there is an statement that says: the map on $\mathscr B(H)$ that $T\longmapsto \|T\|$ is not continuous in the strong operator ...
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34 views

A detail in Rădulescu's Theorem proof

I've been following one Rădulescu's Theorem proof ($G$ is hyperlinear if and only if $\mathcal{L}_{G}$ can be embedded in a ultrapower of the hyperfinite type-$II_{1}$ factor $\mathcal{R}$, where ...
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23 views

Why is this statement true for two equivalent projections in $B(H)$?

In a book of operator theory it is stated that two projections $P$ and $Q$ in a von Neumann algebra $A$ are equivalent if there exist $V$ in $A$ that $V^*V=P$ and $VV^*=Q$. After this definition, it ...
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55 views

Why is the weak operator closure of a commutative $\boldsymbol{C^*\!\!\!\!-}$algebra also commutative?

In a book on Operator Theory there is the following statement: If $\mathscr A$ is a commutative $C^*$-subalgebra of $\mathscr B(\mathcal H)$, where $\mathcal H$ is a Hilbert space, then the weak ...
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33 views

Strong closure of a C*-algebra of operators.

In Arveson's book, the Kaplansky density theorem is proved in order to have this corollary: "Let $A$ be a self-adjoint algebra of operators on a separable Hilbert space $H$. Then for every operator ...
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Dual subfactor and commutant

Let $(N \subset M)$ be a subfactor and $N \subset M \subset M_1$ the basic construction. Question: Is $(M \subset M_1) \simeq (M' \subset N')$? Else in which generic case it's true? What's the ...
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31 views

Question about a passage in the Bicommutant Theorem's proof.

In the Averson's book, in the proof of the Von Neumann's Bicommutant theorem there is this passage: ($A $ is a self-adjoint algebra of operators in $L(H)$) "Let $\xi_1$ be an element of the Hilbert ...
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32 views

Is $vN(M_1,M_2) \cap M_3= vN(M_1,M3) \cap vN(M_2,M3)$?

Let $M_1,M_2,M_3$ be von Neumann algebras (i.e. weakly closed subalgebras of $B(H)$ where $H$ is a Hilbert space). Let $vN(M_1,M_2)$ denote the von Neumann algebra generated by $M_1$ and $M_2$ inside ...
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38 views

Abelian projection in the von-Neumann algebras

As we know, a minimal projection must be Abelian, An Abelian projection must be finite. A minimal projection correspond to a rank one operator, a finite projection correspond to a finite rank ...
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36 views

Existence of proper I.C.C. subgroup

A countable discrete group $G$ is called I.C.C.(infinite conjugacy class) if for any $e\neq g\in G$, $\#\{sgs^{-1}\mid s\in G\}=\infty$. My question is: Is it possible for a group $G$ to be ...
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32 views

Is the double commutant of a unital $\ast$ algebra equal to its double dual?

The Sherman Takeda theorem says that the double dual $A^{\ast \ast}$ of a $C^{\ast}$ algebra $A$ can be identified as a Banach space with $A''$, the enveloping von Neumann algebra of $A$, i.e. the ...
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38 views

Do two II$_1$-factors with trivial intersection generate $B(H)$?

Let $H$ be an infinite dim. separable Hilbert space and $B(H)$ the algebra of bounded operators. Let $A$, $B \subset B(H)$ be II$_1$-factors such that $A \cap B = \mathbb{C}I$. Examples: (1) Take ...
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81 views

Question on amenable direct summand

Given a finite Von Neumann algebra $(N,\tau)$, and Von Neumann subalgebras $A\subset B$ with the same identity, I came across the fact saying that $B$ has an amenable direct summand implies $A$ has an ...
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31 views

find element in relative commutant of a matrix subalgebra

Let $M=A*P$ to be a free product von Neumann algebra, and $A$ is a finite dimensional subalgebra, for simplicity, we may assume $A=M_2(\mathbb{C})$, and $P\neq \mathbb{C}$. A standard fact is that ...
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Understanding minimal projections

This might be very easy but it is not quite clear for me. Detailed explanation appreciated! I went through the commutative case but beyond that I lack intuition. Let $A$ be a C*-algebra and let ...
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49 views

von Neumann Algebras and measures

I read that any abelian von Neumann algebra is isomorphic to $L^\infty(X,\mu)$ for some $X$ and $\mu$. This seems to be reasons, to consider any von Neumann Algebra as non-commutative measurable ...
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80 views

No trace on $B(H)$ if $H$ is infinite dimensional

Let $H$ be an infinite dimensional Hilbert space and $B(H)$ the bounded linear operators on $H$. Then thre is no ultra weakly continous non-zero positve trace $tr:B(H)\rightarrow \mathbb{C}$. I ...
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Ultra weakly continuous trace on a von Neumann Algebra

Let $M$ be a infinite dimensional von Neumann Algebra with a positive, faithful, ultra weakly continuous trace $tr:M\rightarrow \mathbb{C}$. Is it possible to show that $tr$ is strongly continuous?
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A question on a lemma about the product map

Here is a Lemma in the book “C*-algebras and Finite-Dimensional Approximations”: Lemma 3.8.4. Let $A$ be a C*-algebra, $M\subset B(H)$ be a con Neumann algebra and $\phi: A\rightarrow M$ be a ...
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24 views

Unique trace on a type $II_1$ von Neumann Algebra

Let $M \subseteq B(H)$ be a type $II_1$ von Neumann Algebra. Then any two non-zero ultraweakly continious normalised traces $Tr,tr : \rightarrow \mathbb{C}$ are equal. I'm trying to understand this ...
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27 views

Minimal projections and Type II von Neumann Algebras.

Let $M \subseteq B(H)$ be a type $II_1$ factor. Can it contain a minimal projection? If it can't, what would go wrong? I assume something about the trace being faithful?
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Ultra weakly closed *-subalgebra of B(H)

I'm currently working on a text about von Neumann algebras and the author used without further clarifying that any ultra weakly closed *-subalgebra of $B(H)$ contains a largest projection. Could ...
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36 views

Polar decomposition in a von Neumann algebra

Let $M \subseteq B(H)$ be a von Neumann algebra and $T \in M$. If $T=U|T|$ is the polar decomposition of T, why is $U \in M$? I'm thinking it's because $M$ is SOT-closed, but I'm not entirely sure.
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69 views

Kaplansky's Density Theorem for Unitary Operators

Let $M \subseteq B(H)$ be a *-subalgebra containing the identity on H. If there is a unitary T in the unit ball of the SOT-closure of $M$, is there a net of unitary oprators in the unit ball of $M$ ...
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17 views

type I algebras definition

I'm studying von Neumann algebras type decomposition and I already noticed two "different" definitions of, for instance, type I : (i) a vN algebra is said to be of type I, if it has an abelian ...
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63 views

Isn't the center of a von Neumann algebra on a separable Hilbert space a hyperfinite von Neumann subalgebra?

this is a very quick, probably dumb, question, I was reading this chapter from "Hochschild cohomology of von Neumann algebras" by Allan Sinclair and Roger M. Smith and I came across this theorem on ...
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1answer
19 views

projection generated by intersection of two projection

let $H$ be a Hilbert space and $P,Q$ be projections on $H$. suppose $P,Q$ do not commute. $P\wedge Q$ is a projection on $PH\cap QH$. I want to calculate $P\wedge Q$ but I can not. Please help me. ...
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26 views

About what happens to eigenspace under functional calculus for Unbounded Operator

Let $T$ be an unbounded self adjoint positive operator on a Hilbert Space $\mathcal{H}$. Let $x \in \mathcal{H}$ be a vector such that $Tx = x$. Is it true that $T^{\frac{1}{2}} x = x$. For what $f$ ...
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30 views

$E(a)=0\Longrightarrow E(a^{n})=0$?

Let $(M; \tau)$ be the hyperfinite $II_{1}$-Factor and consider a W${}^{\ast}$-subalgebra, $N$. Is there a (trace-preserving) conditional expectation, $E:M\to N$? Considering, now, a more general ...
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A symbol of commuting ranges in tensor product

Here is a proposition of tensor product: ($A,~B,~C$ are C*-algebras) Proposition 3.1.17 Given two *-homomorphisms $\pi_{A}: A\rightarrow C$ and $\pi: B\rightarrow C$ with commuting ranges (i.e., ...
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38 views

The multiplication of tensor product

Proposition 3.1.15 (Multiplication). Let $A$, $B$ be C*-algebra, the tensor product $A\odot B$ (denotes the algebraic tensor product) has a multiplication defined by $$(\sum\limits_{i}a_{i}\otimes ...
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The involution of tensor product

Proposition 3.1.8 (Linear independence). If $\{x_{1},...,x_{n}\}\subset X$ are linearly independent, $\{y_{1},...,y_{n}\}\subset Y$ are arbitrary and $$0=\sum\limits_{i=1}^{n}x_{i}\otimes y_{i}\in ...
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Continuity of linear functionals

We know from Operator Theory the following Theorem: A linear functional $f:A\rightarrow\mathbb{C}$ on a van Neumann algebra $A$ on an Hilbert space $H$ is ultra-weak continuous iff there is an trace ...
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35 views

Zhou operator theory book, Kaplanskys formula

In Zhou's operator theory book, Kaplanskys formula has stated that if $P$ and $Q$ are projection in a von neumann algebra $A$ acting on $H$, then $P\vee Q-Q\sim P-P\wedge Q$. In the proof, it says ...
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42 views

A representation of von Neumann algebra of type I

I am reading a book "C*-algebras and Finite-Dimensional Approximations". There is a quotation below: For infinite-dimensional Hilbert space $H$ and a abelian von Neumann algebra $A$, we can represent ...
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23 views

A inequality about pointwise absolute value vectors

Let $\Gamma$ be a discrete group and $\xi\in l^{2}(\Gamma)$ be a unit vector. If $|\xi|$ be the pointwise absolute value of $\xi$, then how to verify: ($S$ is a linear bounded operator on ...
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66 views

Representations of C*-algebras and projections

I had come across two links between (sub)representations (of von Neumann algebra actually in one case) and projections and I just realized that they are not the same. Let $(\mathcal{H},\pi)$ be a ...
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33 views

The integral about probability measures

Definition For a discrete group $\Gamma$, we let Prob$(\Gamma)$ be the space of all probability measures on $\Gamma$: $$Prob(\Gamma)=\{\mu\in l^{1}(\Gamma): ...
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1answer
49 views

Universal property about discrete group in C*-algebra

Universal property: Let $u:\Gamma \rightarrow B(H)$ be any unitary representation of $\Gamma$. Then, there is a unique $*-$homomorphism $\pi:C^{*}(\Gamma) \rightarrow B(H)$ such that ...
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Amenable group in C*-algebra

Definition 2.6.1. A group $\Gamma$ is amenable if there exists a state $\mu$ on $l^{\infty}(\Gamma)$ which is invariant under the left translation action: for all $s\in \Gamma$ and $f\in ...
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47 views

Positive definite function on discrete group in C*-algebra

Recall A function $\phi: \Gamma\rightarrow\mathbb{C}$ is said to be positive definite if the matrix $$[\phi(s^{-1}t)]_{s,t\in F}\in M_{F}(\mathbb{C})$$ is positive for every finite set $F\subset ...
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30 views

A simple question about 1-norm

Let $\Gamma$ be a discrete group, if $\mu \in l^{1}(\Gamma)$, then what is the 1-norm of $\mu$, I mean $||\mu||_{1}=?$. As we know, $l^{1}(\Gamma)=\{(\alpha_{x})_{x\in\Gamma}: ...
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73 views

A question about reduced C*-algebra of discrete group

There is a quotation below: Let $\Gamma$ be a discrete group and $\Lambda\subset \Gamma$ be a subgroup. The right cosets give a direct sum decomposition $$l^{2}(\Gamma)\cong\bigoplus l^{2}(\Lambda ...
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An easy (I guess) question about vector state in C*-algebra

I meet with some problems when I read a book about C*-algebra. Definition 2.5.10. Let $\phi:\Gamma \rightarrow \mathbb{C}$ be a function ($\Gamma$ is a discrete group here). We define a corresponding ...