All Questions
Tagged with oa.operator-algebras von-neumann-algebras
72 questions
37
votes
5
answers
4k
views
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 ...
- 37.8k
17
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
697
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 ...
6
votes
2
answers
919
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?
- 275
8
votes
1
answer
716
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 $\...
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 (...
4
votes
1
answer
710
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 ...
1
vote
0
answers
150
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 ...
- 896
26
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 ...
- 357
23
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 ...
- 9,486
22
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 $H$ ...
14
votes
4
answers
1k
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 ...
- 29.2k
10
votes
1
answer
492
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 ...
- 2,754
9
votes
3
answers
409
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 ...
8
votes
2
answers
961
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 $S^m$ ...
- 619
8
votes
0
answers
306
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
0
answers
502
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{...
6
votes
1
answer
680
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 ...
5
votes
0
answers
254
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 $\...
- 7,047
4
votes
0
answers
151
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
200
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_{\...
2
votes
1
answer
233
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{...
34
votes
2
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 ...
- 37.8k
18
votes
0
answers
756
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 $...
- 10.1k
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 ...
- 43.2k
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 ...
- 7,637
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 ...
- 1,373
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 ...
- 5,425
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 ...
- 9,330
13
votes
1
answer
451
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 ...
- 771
12
votes
1
answer
2k
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 ...
- 6,853
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 ...
- 1,607
12
votes
1
answer
901
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 ...
11
votes
4
answers
2k
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 ...
- 3,937
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 ...
- 37.8k
8
votes
1
answer
426
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 ...
- 7,047
8
votes
0
answers
345
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 < ...
- 7,800
8
votes
0
answers
269
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 $\...
7
votes
1
answer
429
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 $\...
- 619
7
votes
1
answer
732
views
To what extent can a von Neumann algebra be determined by its projection lattice structure?
Let $ M, N $ be von Neumann algebras, $ P $ (resp. $Q$) the projection lattice of $M$ (resp. $N$). Any isomorphism $ \varphi : M \to N $ on the level of involutive algebras induces an isomorphism $ \...
- 1,586
7
votes
1
answer
592
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......
- 43.2k
7
votes
1
answer
545
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
votes
1
answer
701
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 ...
user36116
7
votes
1
answer
990
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 ...
- 1,411
6
votes
0
answers
168
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}(\...
- 1,090
6
votes
1
answer
494
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 $...
- 7,047
6
votes
2
answers
964
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}...
- 585
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 ...
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 ...
- 85
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 ...
- 9,330