**12**

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 ...

**5**

votes

**3**answers

479 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**

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 ...

**3**

votes

**1**answer

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**

votes

**0**answers

115 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**

votes

**0**answers

114 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 ...

**8**

votes

**3**answers

312 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 ...

**20**

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

270 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 ...

**4**

votes

**0**answers

209 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

**0**answers

173 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 ...

**3**

votes

**1**answer

389 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

196 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} = ...

**12**

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 ...

**16**

votes

**0**answers

735 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 ...

**14**

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 ...

**11**

votes

**1**answer

591 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

**4**answers

1k views

### 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**

votes

**4**answers

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

**1**answer

352 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 ...

**7**

votes

**1**answer

754 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

**0**answers

294 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 < ...

**6**

votes

**1**answer

599 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**

votes

**1**answer

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**

votes

**4**answers

784 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

467 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

495 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 ...

**6**

votes

**2**answers

585 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

**1**answer

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 ...

**6**

votes

**0**answers

227 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

130 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) ...

**4**

votes

**2**answers

488 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 ...

**3**

votes

**3**answers

467 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

**2**answers

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

**1**answer

662 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

**2**answers

585 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

**2**answers

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**

votes

**2**answers

292 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**

votes

**0**answers

69 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 ...

**1**

vote

**0**answers

65 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 ...