Questions tagged [von-neumann-algebras]

Subtag of the [oa.operator-algebras] tag 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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Reference for the Gelfand duality theorem for commutative von Neumann algebras

The Gelfand duality 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 ...
Dmitri Pavlov's user avatar
16 votes
1 answer
2k 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 (Izumi-Longo-Popa ...
7 votes
3 answers
689 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 ...
Sebastien Palcoux's user avatar
6 votes
2 answers
858 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?
moppio89's user avatar
  • 275
21 votes
1 answer
1k 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 ...
Dmitri Pavlov's user avatar
8 votes
1 answer
684 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 ${\rm III}$ factor. Question : Is $\...
Sebastien Palcoux's user avatar
5 votes
0 answers
161 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 (...
Sebastien Palcoux's user avatar
4 votes
1 answer
697 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 ...
Sebastien Palcoux's user avatar
1 vote
0 answers
122 views

Intersection of finitely many type-I von-Neumann algebra factors

If $\mathcal M,\mathcal N$ are type-I von-Neumann algebra factors (over the same Hilbert space $\mathcal H$), is $\mathcal M\cap\mathcal N$ a type-I von-Neumann algebra factor? Notes: An elementary ...
Dominique Unruh's user avatar
23 votes
4 answers
6k 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 $H$ ...
Sebastien Palcoux's user avatar
17 votes
4 answers
1k 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 ...
Simon Henry's user avatar
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14 votes
4 answers
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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 ...
Paul Siegel's user avatar
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10 votes
1 answer
460 views

Which W*-algebras are the duals of C*-coalgebras?

A Banach algebra (assumed associative and unital) is precisely a monoid object in the monoidal category of Banach spaces, short linear maps, and the projective tensor product. A Banach coalgebra is ...
Toby Bartels's user avatar
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9 votes
3 answers
406 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 ...
Sebastien Palcoux's user avatar
8 votes
2 answers
897 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 $S^m$ ...
Masayoshi Kaneda's user avatar
8 votes
0 answers
304 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 ...
Sebastien Palcoux's user avatar
8 votes
2 answers
437 views

Which complete orthomodular lattices arise from von Neumann algebras?

Let $A$ be a von Neumann algebra. Then a classic observation is that the set of projections $\Pi(A)$ is naturally a complete orthomodular lattice. Question 1: Is the construction $A \mapsto \Pi(A)$ a ...
Tim Campion's user avatar
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7 votes
0 answers
469 views

Abstract characterization of group von Neumann algebra (II1 factor)

The group von Neumann algebra $L\Gamma$ is a factor if and only if the group $\Gamma$ is ICC (i.e. infinite conjugacy class property). Moreover if $\Gamma$ is nontrivial then $L\Gamma$ is a $\mathrm{...
Sebastien Palcoux's user avatar
6 votes
1 answer
662 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 ...
Sebastien Palcoux's user avatar
5 votes
0 answers
250 views

Examples of non-proper conditional expectations onto von Neumann subalgebras of $II_{1}$ factors

Denote by $M$ be a type $II_{1}$ factor with trace $\tau$, and $1\in N\subseteq M$ a von Neumann subalgebra. What are some examples of such inclusions for which the normal conditional expectation $\...
Jon Bannon's user avatar
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4 votes
0 answers
249 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 ...
Sebastien Palcoux's user avatar
4 votes
2 answers
202 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) ...
passerby51's user avatar
  • 1,639
4 votes
0 answers
148 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 ...
Sebastien Palcoux's user avatar
3 votes
0 answers
199 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 $e_{\...
Sebastien Palcoux's user avatar
2 votes
1 answer
228 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{...
Sebastien Palcoux's user avatar
34 votes
3 answers
3k views

Can we recover a von Neumann algebra from its predual?

By definition, a von Neumann algebra is a C*‑algebra A that admits a predual, i.e., a Banach space Z such that Z* is isomorphic to the underlying Banach space of A. (We require that isomorphisms in ...
Dmitri Pavlov's user avatar
25 votes
3 answers
2k views

About the category of von neumann algebras

I am looking for one (or more) reference about properties of the category of von Neumann algebra. More precisely, in an answer of a previous question, Dmitri Pavlov mentions that the $W^*$ category ...
Oliver's user avatar
  • 347
21 votes
4 answers
2k 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 $\|A\|_2 = \sqrt{\mathrm{Tr}(A^* A)/n}$. My question is whether a $k$-uple of hermitian matrices that are almost ...
Mikael de la Salle's user avatar
21 votes
1 answer
1k views

Is Dependent Choice all we really need?

http://en.wikipedia.org/wiki/Axiom_of_dependent_choice Is DC sufficient for the understanding of objects that are countable in some suitable sense? For example, is DC sufficient for the full ...
Andre's user avatar
  • 1,199
18 votes
0 answers
689 views

An "exercise" on von Neumann algebra tensor product

The following problem appears to be an easy exercise on von Neumann algebra tensor products, but since I've been failing to find a rigorous proof, I'd like to make sure it's not that trivial. Suppose $...
Narutaka OZAWA's user avatar
18 votes
4 answers
2k 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 ...
André Henriques's user avatar
16 votes
5 answers
3k 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 ...
Dave Penneys's user avatar
  • 5,327
16 votes
4 answers
1k views

Von Neumann algebra associated to the infinite Cuntz algebra

The Cuntz algebra $\mathcal{O}_{\infty}$ is the universal $C^*$-algebra generated by countably infinitely many isometries $s_i$ satisfying the relations $s_i^*s_j = \delta_{ij}$ (there is no condition ...
Ulrich Pennig's user avatar
16 votes
3 answers
2k 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 ...
yeshengkui's user avatar
  • 1,373
15 votes
2 answers
1k views

Categorifying the Reals via von Neumann Algebras?

So one way to categorify the natural numbers is to replace them with vector spaces. Then the dimension of the vector space reproduces the natural number. More generally you can categorify integers to ...
Chris Schommer-Pries's user avatar
13 votes
1 answer
417 views

Factor states on C*-algebras

Which C$^*$-algebras admit factor states for which the von Neumann algebra it generates in the corresponding GNS representation is a type III$_1$ factor? For example, do all purely infinite algebras ...
Isaac's user avatar
  • 771
13 votes
3 answers
1k views

Separable von Neumann algebra

What is the simplest argument which shows that each infinite dimensional von Neumann algebra is not separable (in the norm topology)? It seems that this is a kind of folklore: at least I never saw the ...
truebaran's user avatar
  • 9,140
12 votes
1 answer
1k views

Making sense of "every non-commutative algebra has its own internal time evolution (aka a one-parameter group)"?

I've listened to many interviews and lectures of Alain Connes, in which he says something which goes roughly as follows "Every non-commutative algebra has its own time (evolution of), by which I ...
dohmatob's user avatar
  • 6,706
12 votes
1 answer
876 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 ...
Sebastien Palcoux's user avatar
12 votes
2 answers
3k views

Mysterious quotes (at least for me)

I heard two quotes, one from Alain Connes and an other one from Orlov. Alain Connes was talking about noncommutative geometry and he said the following: " a noncommutative algebra creates its own ...
Max's user avatar
  • 1,607
11 votes
4 answers
1k views

What kind of completion is this?

Let $X$ be a compact Hausdorff space, and $C(X)$ the unital commutative C*-algebra of continuous functions on it. The double Banach dual $C(X)^{**}$ is a commutative von Neumann algebra and hence has ...
Chris Heunen's user avatar
  • 3,919
10 votes
1 answer
1k 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 ...
Dmitri Pavlov's user avatar
8 votes
0 answers
257 views

Shift on trivalent directed tree, operator and von Neumann algebra

Let $\mathcal{T}$ be the trivalent directed tree, with two parents and one child for each vertex (see below). Let $\mathcal{V}$ be the set of vertices of $\mathcal{T}$ and $H$ be the Hilbert space $\...
Sebastien Palcoux's user avatar
8 votes
1 answer
422 views

Is $L(\mathbb{Z}*\mathbb{Z}_{2})$ a free group factor?

This is a reference request for something that is likely to be well-known to operator algebraists. I will not, therefore, include the technical definition of free product of finite von Neumann ...
Jon Bannon's user avatar
  • 6,977
8 votes
0 answers
341 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 < ...
Scott Morrison's user avatar
7 votes
1 answer
579 views

topologies on U(H)

There are many topologies on the algebra $B(H)$ of bounded operators on Hilbert space: the weak, strong, ultraweak (also called σ-weak), ultrastrong (also called σ-strong), and some more......
André Henriques's user avatar
7 votes
1 answer
660 views

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 ...
user avatar
7 votes
1 answer
420 views

Open projections and Murray-von Neumann equivalence

Let $\mathcal{A}$ be a $C^*$-algebra and $p\in\mathcal{A}^{**}$ be an open projection, that is, $p=p^*=p^2$ and $p\in\overline{(p\mathcal{A}^{**}p\cap\hat{\mathcal{A}})}^{\operatorname{w}^*}$, where $\...
Masayoshi Kaneda's user avatar
7 votes
1 answer
949 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 ...
Ollie's user avatar
  • 1,391
7 votes
1 answer
527 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 $B(H)$...
Sebastien Palcoux's user avatar