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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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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
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6 votes
0 answers
169 views

Characterizing fullness of a von Neumann algebra by the topology of its bimodules

Let $\mathcal{M}$ be a $\mathrm{II}_1$ factor. Among other characterizations, it is said to be full iff the adjoint map: $$ \mathrm{Ad}: U(\mathcal{M})/\mathbb{T} \longrightarrow \mathrm{Aut}(\...
Adrián González Pérez's user avatar
6 votes
1 answer
495 views

Property $\Gamma$ in terms of Correspondences

A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset $\{ x_{1}, x_{2},..., x_{n} \} \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $...
Jon Bannon's user avatar
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6 votes
2 answers
965 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}...
Issam Ibnouhsein's user avatar
5 votes
2 answers
988 views

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 ...
Rennela's user avatar
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5 votes
1 answer
177 views

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 ...
Sebastien Palcoux's user avatar
5 votes
1 answer
638 views

Hyperfinite type II_1 factor as the Clifford algebra

In Connes' book Noncommutative geometry, there is a presentation of all hyperfinite factors. He reffers to type $II_1$ as the Clifford algebra of infinite dimensional Euclidean space. This factor can ...
truebaran's user avatar
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4 votes
1 answer
407 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 \...
Sebastien Palcoux's user avatar
4 votes
2 answers
542 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 ...
Sebastien Palcoux's user avatar
4 votes
0 answers
508 views

Good reference for noncommutative $L^p$ spaces

I'm looking for good references to learn about $L^p$ spaces associated with von Neumann algebras. I already know about Uffe Haagerup's paper "$L^p$-spaces associated with an arbitrary von Neumann ...
Screwdriver's user avatar
4 votes
0 answers
115 views

Questions related to a paper of Cowling-Haagerup and uniform lattices of $\mathrm{Sp}(1,n)$

I am reading the following paper of Cowling and Haagerup for my master’s thesis. I am new to this area so I am not very conversant. So I do apologize if the questions are silly. Question 1. In the ...
Y. Paka's user avatar
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3 votes
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138 views

Is there a finite depth irreducible subfactor of prime index and not group-subgroup?

Let $N \subset M$ be a finite depth unital inclusion of II$_1$ factors. By Theorem 3.2 in this paper (Bisch, 1994), if the index $|M:N|$ is integer then for any intermediate subfactor $N \subset P \...
Sebastien Palcoux's user avatar
3 votes
2 answers
361 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 $\mathcal{A}'$, $...
Sebastien Palcoux's user avatar
3 votes
2 answers
273 views

Representing measurable map to compact space as a continuous map

Let $\Omega$ be a measurable space equipped with a $\sigma$-ideal $\mathcal{N}$ (though of as the "null sets"). Define the compact Hausdorff space $$ \tilde{\Omega} := \mathrm{Spec}(L^\infty(...
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3 votes
1 answer
740 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 ...
spelas's user avatar
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3 votes
3 answers
1k 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 ? ...
Sebastien Palcoux's user avatar
3 votes
1 answer
335 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 $...
Sebastien Palcoux's user avatar
3 votes
1 answer
253 views

What is the story behind this Hilbert space in the definition of Hilbert Modules

Here is Deflnition 1.5 of Hilbert module in "L^2-invariants: theory and applications to geometry and K-theory", Springer-Verlag, 2002, by W. Lück: A Hilbert $\mathcal N(G)$-module $V$ is a Hilbert ...
MSMalekan's user avatar
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3 votes
2 answers
397 views

Is the ideal property of $X^{**}$ inheritable to $X$?

Let $X$ be an operator space such that there is a weak$^*$-continuous complete isometry $\phi$ from its second dual $X^{**}$ into a $W^*$-algebra $M$ in which $\phi(X^{**})$ is a (necessarily weak$^*$-...
Masayoshi Kaneda's user avatar
3 votes
2 answers
207 views

Commutative direct summands of C*-algebras

I have a question about commutative direct summands of $C$*-algebras. Let $A$ be a $C$*-algebra (with unit) and suppose that its bidual $A^{**}$ has a commutative direct summand, that is, $A^{**}=B\...
Mark Roelands's user avatar
2 votes
1 answer
471 views

Takesaki II proposition 3.15 proof about modular automorphism groups: mistake in book?

Consider the following fragment from Takesaki's second volume "Theory of operator algebras" in chapter VIII "Modular automorphism groups" p122-123. Why is it possible to choose an ...
Andromeda's user avatar
  • 175
2 votes
2 answers
867 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 ...
Wishiwere Smarter's user avatar
2 votes
1 answer
757 views

Minimal Tensor product of von Neumann algebras

Let $M$ and $N$ be two von Neumann algebras. A natural way to define $M\overline{\otimes}N$ is the von Neumann algebra generated by the algebraic tensor product $M\otimes N$ in $B(H\otimes K)$ where $...
Mathbuff's user avatar
  • 455
2 votes
1 answer
369 views

On diagonal part of tensor product of $C^*$-algebras

Suppose we have a $C^*$-algebra $\mathcal{U}$, Consider the $C^*$-subalgebra generated by elements of the form $a\otimes a$, what is it isomorphic to? Is it isomorphic to $\mathcal{U}$ itself?
user136400's user avatar
2 votes
0 answers
156 views

Extension of a theorem of Bisch to cyclotomic integers of fixed degree

Theorem 3.2 in this paper (Bisch, 1994) states that for a finite depth ${\rm II}_1$ subfactor of integral index, every intermediate subfactor are also of integral index. As an application, every such ...
Sebastien Palcoux's user avatar
2 votes
0 answers
101 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 W....
Sebastien Palcoux's user avatar
2 votes
1 answer
462 views

Direct integral decomposition relative to a given measure space

It is well known that a separable Hilbert space $H$ decomposes as a direct integral in the presence of an abelian von Neumann algebra $\mathscr A\subseteq B(H)$. More precisely, and quoting from ...
Black's user avatar
  • 483
1 vote
1 answer
181 views

Existence of a third intermediate if there are two intermediate subfactors of index 2

Let $(N \subset M)$ be an irreducible finite index unital inclusion of hyperfinite ${\rm II}_1$ factors. Let $K_1$ and $K_2$ be two distinct intermediate subfactors $N \subset K_i \subset M$, such ...
Sebastien Palcoux's user avatar
1 vote
1 answer
683 views

About separability of von Neumann algebras [closed]

Is a von Neumann algebra always separable in the $\sigma$-weak topology? If not, give a counterexample. Under what conditions will it be separable?
mathlover's user avatar
  • 227
1 vote
0 answers
81 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 ...
Noix07's user avatar
  • 189
0 votes
2 answers
225 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 $II_1$ ...
Sebastien Palcoux's user avatar
0 votes
0 answers
104 views

Intersection of type-I von-Neumann algebra factors

Is the intersection of a (possibly infinite) family $\{\mathcal M_i\}$ of type-I von-Neumann algebra factors (over the same Hilbert space $\mathcal H$) again a type-I von-Neumann algebra factor?
Dominique Unruh's user avatar

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