The von-neumann-algebras tag has no wiki summary.

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### Type III factor representation

Does there exist any theorem which permits, under suitable hypotheses, to represent a particular complete orthomodular lattice as the projection lattice of a Type III von Neumann factor?

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### two concepts of positivity for elements of $C(X)$ when $X$ is hyper-stonean

Suppose that $X$ is a compact space. Let $M(X)=C(X)^*$ denote the Banach space of regular measures. Is the following statement true:
$F:M(X)\to\mathbb{C}$ is a positive functional if and only if the ...

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### If two projections are close, then they are unitarily equivalent

Given two projections $p,q\in B(H)$, it is well-known that if $\|p-q\|<1$, then there exists a unitary $u\in B(H)$ with $q=upu^*$.
The proof that immediately occurs to me uses comparison of ...

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### Planar algebraic translation of a subfactor property

Let $N \subset M$ be an irreducible finite depth and finite index subfactor.
$M$ is a completely reducible (algebraic) $N$-$N$ bimodule, it decomposes into irreducibles as follows :
...

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### A Question About Pure States, Support Projections and Central Covers

I am trying to study the paper Consistency of a Counterexample to Naimark’s Problem by Charles Akemann and Nik Weaver, and there is a claim in Lemma 1 of the paper that I am stuck at, which is as ...

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### Finding the commutant of a von Neumann algebra

Suppose you have a von Neumann algebra $A$ of operators on $H$ and would like to compute its commutant. You have constructed a collection $B\subset A'$ which you suspect generates it (i.e. you think ...

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### Are all the R-R-bimodules completely reducible?

Let $R$ be the hyperfinite $II_1$ factor and let $X$ be any $R$-$R$-bimodule.
Question: Is $X$ completely reducible (i.e. a direct integral of irreducible $R$-$R$-bimodules)?
Example: If $(N ...

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### Are the finite dimensional von Neumann algebras, singly generated?

Let $\mathcal{M}$ be a finite dimensional von Neumann algebra, then :
$$\mathcal{M} \simeq \bigoplus_i M_{n_i}(\mathbb{C})$$
Question : Is it singly generated (as von Neumann algebra)? how ?
...

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### Are the reduced group Von Neumann algebra/ Group $C^{\ast}$ algebra functorial in the case of LCH groups

Let $G$ be a LCH group and $\mu$ be its left Haar measure. Call $\lambda_G : G \to U(L_2(G,\mu))$ the left regular representation. We can define the reduced $C^{\ast}$ algebra and reduced Von Neumann ...

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### ITPFI factors with restricted growth

You will have to excuse my lack of understanding of von Neumann algebras. I do not know if my question is trivial or nonsensical.
There are ITPFI factors of bounded type, and ITPFI factors of ...

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### The groupoid VN algebra of the transversal to a uniquely ergodic action

I have a uniquely ergodic dynamical system preserving a finite ergodic measure (specifically, I have a nice aperiodic tiling space with an action of $\mathbb{R}^d$). Thus the transformation group von ...

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### Question on structure of von Neumann algebras, clarification in Conway's “A course in operator theory”

I was reading the section on the structure of type I von Neumann algebras in John B. Conway's "A course in operator theory" and a few questions about certain definitions and references arose, I was ...

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### Reference request for a type III action of a group on a manifold

Let an action of a group $\Gamma$ on a manifold $M$ such that $L^{∞}(M)⋊Γ$ is a type $III$ factor.
André Henriques posted here the following comment :
I don't know the literature, so I can't ...

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### is this von Neumann algebra a tensor product?

Let $H_1$ and $H_2$ be Hilbert spaces. Let $A\subset B(H_1)$ be a factor and $A'$ its commutant.
If a von Neumann algebra $M\subset B(H_1\otimes H_2)$ contains $A\otimes 1$ and commutes with ...

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### An embedding theorem for a fusion ring planar algebra?

We first recall the embedding theorem for finite depth subfactor planar algebras:
The planar algebra generated by a (finite depth) subfactor, is embeddable into the planar algebra generated by its ...

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### reference request : constructive measure theory

As the title said, I would like to know if constructive measure theory has been developed somewhere ?
I am more precisely interested in the (constructive) theory of completely continuous valuation on ...

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### A non-hyperfinite type III factor from an action of the free group on the circle

We define below a von Neumann algebra $\mathcal{M}$ from an action of the free group on the circle, and we prove that $\mathcal{M}$ is a non-hyperfinite type III factor.
Question : Is ...

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### Versions of the spectral theorem

Since any $C^*$-algebra can be represented as an algebra of bounded operators $\mathcal{B(H)}$ on a Hilbert space $\mathcal{H}$, the spectral theorem applies to all $C^*$-algebras:
($*$) ...

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### Folium in GNS construction and von Neumann algebras

The GNS construction allows one to represent a $C^*$-algebra as the algebra of bounded operators on a Hilbert space when a state is fixed, this state being represented as a vector on the Hilbert ...

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### C*-algebras and quantum fields

One can represent a quantum system by the Weyl algebra (which is a C*-algebra). For instance, a 1 degree of freedom system can be represented by the algebra generated by $e^{\imath t Q}, e^{\imath s ...

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### Is there a proof that the $C^{*}$-algebras don't see the invariant subspace problem?

This post is an appendix of this one.
Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Invariant subspace problem: Let $T \in B(H)$. Is ...

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### About generator and isomorphism problems for free groups operator algebras

Let $H$ be an infinite dimensional separable Hilbert space.
The $C^{*}$-algebras and von Neumann here unital and subalgebras of $B(H)$.
Definition : Let $\mathcal{A}$ be $C^{*}$-algebra (resp. a ...

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### Is there an operator algebraic reformulation of the invariant subspace problem?

Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Invariant subspace problem: Let $T \in B(H)$. Is there a non-trivial closed $T$-invariant ...

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### Is the fundamental group of $II_{1}$ factors invariant under a relation?

This issue is in continuation of an answer I gave here about
noncommutative sets.
In order to define the equivalence relation, let's first recall the Tomita-Takesaki modular theory and ...

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### What's a noncommutative set?

This issue is for logicians and operator algebraists (but also for anyone who is interested).
Let's start by short reminders on von Neumann algebra (for more details, see [J], [T], [W]):
Let $H$ ...

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### Is there an irreducible, noncompact commuting, nonnormal operator, with spectrum strictly continuous?

Let $H$ be an infinite dimensional separable Hilbert space.
Definition: The commutant $\mathcal{S}'$ of a subset $\mathcal{S} \subset B(H)$ is $ \{A \in B(H) : AB=BA \ , \ \forall B \in \mathcal{S} ...

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### Is the fundamental group of a maximal subfactor always $\mathbb{R}_{+}^{*}$?

The fundamental group $\mathcal{F}(N \subset M)$ of an inclusion of $II_{1}$ factors $N \subset M$ is defined as : $\mathcal{F}(N \subset M) =\{t >0 \ | \ (N \subset M)^{t} \simeq (N \subset M) ...

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### What's the natural equivalence of subfactors in general?

Let $A$ be a factor and $\mathcal{C}_{A}$ be the category of all the subfactors $(M \subset N)$ such that $M$ and $N$ are isomorphic to $A$. The most famous of them is perhaps $\mathcal{C}_{R}$ with ...

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### On complemented von Neumann algebras

Edit: according to Narutaka Ozawa, question 3) is still open in the type $\mathrm{II}_1$ case. In other terms, it is not known whether every topologically complemented type $\mathrm{II}_1$ factor in ...

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### What are the intermediate subfactors of the tensor product of two maximal subfactors?

Let $(N_1 \subset M_1)$ and $(N_2 \subset M_2)$ be two maximal subfactors.
Their tensor product, the subfactor $(N_1 \otimes N_2 \subset M_1 \otimes M_2)$, admits four obvious intermediate ...

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### Is there a maximal finite depth infinite index irreducible subfactor?

A subfactor $N \subset M $ is irreducible if $N' \cap M = \mathbb{C} $.
It's maximal if it admits no non-trivial intermediate subfactors $N \subset P \subset M $.
It's cyclic if its lattice of ...

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### About the classification of infinite depth irreducible finite index maximal subfactors

The Temperley Lieb subfactors $A_{\infty}$ are the first examples of infinite depth irreducible finite index maximal subfactors. We can see these subfactors as coming from the simple Lie group ...

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### The cyclic subfactors theory: a quantum arithmetic?

Context: First recall some results:
- Actions of finite groups on the hyperfinite type $II_{1}$ factor $R$ (Jones 1980).
- A Galois correspondence for depth 2 irreducible subfactors ...

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### Projections in a W*-algebra as a continuous lattice?

A continuous lattice is a complete lattice $L$ in which every element $y$ is equal to $\bigvee${$x \in L \mid x \ll y$} where $x \ll y$ ("x approximates y" or "x is way below y") if for any directed ...

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### Is there an infinite depth irreducible finite index maximal subfactor (other than Temperley Lieb) ?

A subfactor $N \subset M$ is maximal if it admits no non-trivial intermediate subfactors $N \subset P \subset M$.
Is there an infinite depth irreducible finite index maximal subfactor (other than ...

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### Are there only finitely many maximal irreducible amenable subfactors at fixed finite index?

A subfactor $N \subset M $ is maximal if it admits no non-trivial intermediate subfactors $N \subset P \subset M $.
Question: Are there only finitely many maximal irreducible amenable subfactors at ...

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### When does a $W^*$-algebra have a standard Borel spectrum?

EDIT: André Henriques has commented below that the correct separability condition is not weak-* separability as I have written below, but separability of the predual.
This post came out a bit long, ...

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### Weights on Von Neuman factors

Let A be a type $I$ factor on a Hilbert space H. Let $\varphi$ be a semi-finite normal weight on $A^{+}$ is it possible to say that then there exist Hilbert spaces $H_2 \subset H_1$ and an isomorphism ...

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### Find a lower bound for a pre-invariant $Fol(L(F_m), X_m)$

In the paper of Bannon and Ravichandran, A Folner invariant for type $\rm{II}_1$ factors, they defined an invariant $Fol(M)$ for a separable type $\rm{II}_1$ factor $M$, especially for the free group ...

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### Is the image of a vNA under a SOT-continuous morphism still a vNA?

It looks very easy but I must admit I am struggling with this problem.
Okay, let $M$ be a von Neumann algebra acting on a Hilbert space $H$ and let $K$ be another Hilbert space. Suppose $h\colon M\to ...

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### How to classify von Neumann algebra bundles?

If we consider algebra bundles over X where the fiber is an algebra of bounded operators in a separable Hilbert space H over the complex numbers. I learn from "Isomorphism Classification of Operator ...

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### Is this result of Spain correct?

Let us have a look on the proof of Theorem 2 in [P. G. Spain, Boolean algebras of projections, Proceedings of the Edinburgh Mathematical Society (Series 2) 19, 03, March 1975, 287-289]
The author ...

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### Kadison-Singer problem in exotic Hilbert spaces

The Kadison-Singer problem is considered in relation to the separable Hilbert space:
KS: Does every pure state on the diagonal (atomic) masa of $B(\ell_2)$ has a unique extension to $B(\ell_2)$?
...

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### Is the von Neumann algebra associated to a unitary representation of an amenable group always injective?

I should be tarred and feathered for not knowing at least the status of the following question.
Question: Let $\Gamma$ be a discrete amenable group. If $\pi:\Gamma \rightarrow B(\mathcal{H})$ is ...

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### Number of II${}_1$ factors

McDuff proved that there exist continuum many non-isomorphic (separable) II${}_1$ factors. I would like to politely ask whether it is known/open if one can find $2^{\mathfrak{c}}$ (or at least ...

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### Clarifying the link between deformation/rigidity and dual cocycles

Suppose that a type $II_{1}$ factor $M$ decomposes in two ways as a group von Neumann algebra, e.g. as $L\Gamma$ and as $L\Lambda$. The decomposition $L\Gamma$ gives rise to a comultiplication ...

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### Strong convergence of projections in $B(H)$

(I asked this question at math stackexchange 4 months ago, but received no answers)
Let $\{e_{kj}\}$ be the canonical matrix units in $B(H)$, with $H$ separable. Define projections $q_k$ by
$$
...

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### Commutant of a von Neumann algebra as the linear span of unitaries.

I'm reading chapter 4 of Gerard Murphy's C*-algebras book and am confused by a statement in his proof of theorem 4.1.10. In his proof, he says, "$A'$ is the linear span of its unitaries" (where $A'$ ...

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### Limits of von Neumann algebras

Consider (abstract) von Neumann algebras topologised by their weak*-topology arising from the unique predual.
In the theory of topological vector spaces, there is a natural notion of an inductive ...

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### Finite projection in Von Neumann algebra

I had the following question when I am learning von Neumann algebras:
Let p be a finite projection in a finite von Neumann algebra $M$, let $p>p_1>p_2>\cdots$ be a decreasing sequence of ...