Subtag of [tag:oa.operator-algebras] for questions about von Neumann algebras, that is, weak operator topology closed, unital, *-subalgebras of bounded operators on a Hilbert space.

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13
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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 (Izumi-Longo-...
5
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3answers
481 views

Jordan-Hölder theorem for subfactors?

All the subfactors $(N\subset M)$ are irreducible and finite index inclusions of II$_1$ factors. First recall that in this paper, D. Bisch characterizes the Jones projections $e_K$ of the ...
28
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4answers
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Reference for the Gelfand-Neumark theorem for commutative von Neumann algebras

The Gelfand-Neumark theorem for commutative von Neumann algebras states that the following three categories are equivalent: (1) The opposite category of the category of commutative von Neumann ...
3
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1answer
548 views

Abelian subfactors, a relevant concept?

Through the questions below, this post asks whether the concept of abelian subfactor is relevant. Remark : here abelian qualifies an inclusion of II$_1$ factors $(N \subset M)$, $N$ is not an abelian ...
4
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118 views

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 ...
5
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117 views

Are the integer index finite depth irreducible subfactors Kac-coideal?

Is every integer index finite depth irreducible subfactors planar algebra, the intermediate of an irreducible finite index depth $2$ subfactors planar algebra? In other words, of the following form (...
20
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3answers
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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 $H$ ...
8
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0answers
272 views

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 ...
8
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3answers
330 views

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 ...
4
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0answers
213 views

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 ...
6
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0answers
233 views

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 $\mathcal{...
3
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0answers
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What are the first non-maximal non-group-subgroup simple irreducible subfactors?

Definition: For an irreducible (finite index) subfactor $(\mathcal{N} \subset \mathcal{M})$, an intermediate $(\mathcal{N} \subset \mathcal{P} \subset \mathcal{M})$ is normal if the biprojections $e_{...
3
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1answer
397 views

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 ...
2
votes
1answer
198 views

${\rm II}_1$-factors with finite commutant: $\mathcal{A} \cap \mathcal{B} = \mathbb{C} \Rightarrow \mathcal{A}' \cap \mathcal{B}'$ hyperfinite?

Let $\mathcal{A} , \mathcal{B} \subset B(H)$ be ${\rm II}_1$-factors such that $\mathcal{A}', \mathcal{B}' $ are also a ${\rm II}_1$-factors. Question: $\mathcal{A} \cap \mathcal{B} = \mathbb{...
12
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5answers
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Non-commutative geometry from von Neumann algebras?

The Gelfand transform gives an equivalence of categories from the category of unital, commutative $C^*$-algebras with unital $*$-homomorphisms to the category of compact Hausdorff spaces with ...
16
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0answers
742 views

Which complete Boolean algebras arise as the algebras of projections of commutative von Neumann algebras?

Projections in an arbitrary commutative von Neumann algebra form a complete Boolean algebra. Moreover, a morphism of commutative von Neumann algebras induces a continuous morphism of the corresponding ...
16
votes
3answers
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Is the group von Neumann algebra construction functorial?

Let $G$ be a group and $CG$ the complex group algebra over the field $C$ of complex number. The group von Neumann algebra $NG$ is the completion of $CG$ wrt weak operator norm in $B(l^2(G))$, the set ...
11
votes
1answer
597 views

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 ...
15
votes
4answers
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Are almost commuting hermitian matrices close to commuting matrices (in the 2-norm)?

I consider on $M_n(\mathbb C)$ the normalized $2$-norm, i.e. the norm given by $\|A\|_2 = \sqrt{Tr(A^* A)/n}$. My question is whether a $k$-uple of hermitian matrices that are almost commuting (with ...
13
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4answers
1k views

Monoidal structures on von Neumann algebras

My question is based on the following vague belief, shared by many people: It should be possible to use von Neumann algebras in order to define the cohomology theory TMF (topological modular forms) in ...
7
votes
1answer
761 views

Can we characterize the spatial tensor product of von Neumann algebras categorically?

The tensor product of commutative algebras is exactly their coproduct in the category of commutative algebras. In other words, if A and B are two commutative algebras, then the covariant functor that ...
7
votes
1answer
358 views

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

In order to define the equivalence relation, let's first recall the Tomita-Takesaki modular theory and conditional expectation for von Neumann algebras. Let $H$ be a separable Hilbert space and $B(H)$...
7
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0answers
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Is there a finite-index finite-depth II$_1$ subfactor which is more than $7$-super-transitive?

Background: See Noah and Emily's posts on subfactors and planar algebras on the Secret Blogging Seminar. There are plenty of examples of $3$-super-transitive (3-ST) subfactors; Haagerup, $S_4 < ...
6
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1answer
605 views

Subfactor theory and Hilbert von Neumann Algebras

There seem to be intimate connections between the different definitions of von Neumann module. The two that I'm aware of are Hilbert von Neumann modules and correspondences (in the sense of Connes). I ...
4
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1answer
338 views

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 \...
12
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4answers
796 views

What is the right definition of “real von Neumann algebra”?

Recall that a real C*-algebra is a Banach $\ast$-algebra $A$ over $\mathbb{R}$ which satisfies the standard C* identity and which also has the property that $1 + a^{\ast}a$ is invertible in the ...
8
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3answers
484 views

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 ...
6
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2answers
507 views

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 P}...
6
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2answers
605 views

The monotone closure of a $C^*$-algebra

Related to Jon's question, I have two questions. Let $\mathcal{A}$ be a concrete $C^*$-algebra on a Hilbert space $\mathcal{H}$. For any selfadjoint subset $S$ of $\mathbb{B}(\mathcal{H})$, let $S^m$ ...
3
votes
1answer
279 views

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 $...
4
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2answers
489 views

The category of subfactors extending the category of groups?

This post was inspired by this answer of Dave Penneys. In the category of (irreducible hyperfinite II$_1$) subfactors, the morphisms of $(N \subset M)$ to $(N' \subset M')$ are usually defined as ...
4
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2answers
132 views

Collection of projection operators in finite dimension and algebraic techinques

Consider a set of linearly independent vectors $\{x_1,\dots,x_n\}$ in some finite-dimensional Hilbert space $H$. For any subset $S \subset [n]$, let $P_S$ be the (orthogonal) projection (operator) ...
3
votes
3answers
472 views

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 ? ...
2
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2answers
251 views

${\rm II}_1$-factors with finite commutant and trivial intersection generate $B(H)$?

Let $H$ be an $\infty$-dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators. Let $\mathcal{A}$, $\mathcal{B} \subset B(H)$ be ${\rm II}_1$-factors such that $\mathcal{A}'$, $...
2
votes
1answer
663 views

Centralizers in C*-algebra

Let $a,b\in A$ be self-adjoint elements in $C^*$-algebra $A$ with equal centralizers, $\{x\in A; [a,x]=0\}=\{x\in A; [b,x]=0\}$. Can we say anything about the correspondence between $a$ and $b$? For ...
1
vote
2answers
591 views

Decomposition of an abelian von Neumann algebra

Hi, I came across the statement below and I couldn't figure out why it is true. I was hoping someone could explain it or give me a good reference. Thank you in advance. "Let $\pi$ be a non-degenerate ...
0
votes
2answers
201 views

Isomorphism theorem for subfactors?

It's about the existence of a generalization of the first isomorphism theorem for groups, for subfactors : Let $(N \subset M)$ and $(N' \subset M')$ be irreducible inclusions of hyperfinite $II_1$ ...
3
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2answers
306 views

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?
2
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0answers
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Are there infinitely many amenable Hadamard-Petrescu subfactors?

The complex Hadamard matrices of dimension $n$ are used to build index $n$ subfactors through the commuting square construction. For more details, see the paper Subfactors and Hadamard Matrices by W....
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0answers
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Why does one only consider one-parameter groups in Borchers-Arveson theorem?

(question from math.stackexchange) The theorem (Operator algebras and Quantum statistical mechanics vol. 1, Bratteli, Robinson, Thm. 3.2.46 p.261) roughly says that if one has a one parameter ...