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

**8**

votes

**1**answer

1k views

### 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 ...

**4**

votes

**3**answers

459 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 ...

**2**

votes

**1**answer

510 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**

votes

**0**answers

108 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 ...

**3**

votes

**0**answers

97 views

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

Is it known whether or not the integer index finite depth irreducible subfactors (planar algebra) are Kac-coideal subfactors: $(R^{\mathbb{A}} \subset R^{\mathbb{I}})$, with $\mathbb{A}$ a finite ...

**27**

votes

**4**answers

2k views

### 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 ...

**17**

votes

**3**answers

1k views

### 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**

votes

**0**answers

265 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 ...

**7**

votes

**3**answers

295 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 ...

**3**

votes

**0**answers

162 views

### 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 ...

**2**

votes

**1**answer

359 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

**1**answer

190 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} = ...

**11**

votes

**5**answers

2k views

### 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 ...

**11**

votes

**1**answer

539 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 ...

**13**

votes

**3**answers

1k views

### 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 ...

**7**

votes

**0**answers

289 views

### 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 < ...

**5**

votes

**1**answer

290 views

### 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 ...

**11**

votes

**4**answers

746 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**

votes

**3**answers

391 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**

votes

**2**answers

437 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 ...

**3**

votes

**1**answer

327 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 ...

**6**

votes

**2**answers

539 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 $S^m$ ...

**4**

votes

**0**answers

180 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 ...

**3**

votes

**1**answer

269 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 ...

**6**

votes

**0**answers

203 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 ...

**4**

votes

**2**answers

472 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 ...

**2**

votes

**2**answers

247 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

**3**answers

379 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**

votes

**1**answer

657 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 ...

**0**

votes

**2**answers

196 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**

votes

**2**answers

261 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?

**1**

vote

**0**answers

58 views

### 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 ...

**1**

vote

**0**answers

54 views

### 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 ...