All Questions
Tagged with oa.operator-algebras von-neumann-algebras
504 questions
37
votes
5
answers
4k
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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 ...
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 ...
32
votes
3
answers
2k
views
What does it mean for a category to admit direct integrals?
Given an infinite countable group $G$, the category of unitary representations of $G$ admits direct integrals.
Namely, given a measure space $(X,\mu)$ and a measurable family of unitary $G$-reps $(H_x)...
28
votes
0
answers
2k
views
Finite-dimensional subalgebras of $C^\star$-algebras
Let $A$ be a unital $C^\star$-algebra and let $a_1,\dots,a_n$ be a finite list of normal elements in $A$ which (together with their adjoints) generate a norm-dense $\star$-subalgebra $B \subset A$. ...
26
votes
3
answers
2k
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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 ...
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 ...
23
votes
1
answer
1k
views
Relative commutants of abelian von Neumann algebras
This question arose from the discussion over at the question Centralizers in $C^*$-algebras.
Which von Neumann algebras $N$ satisfy the property that $A' \cap N = B' \cap N \implies A = B$, for all ...
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 $H$ ...
22
votes
1
answer
745
views
The Mackey Topology on a Von Neumann Algebra
Every von Neumann algebra $\mathcal M$ is the dual of a unique Banach space $\mathcal M_* $. The Mackey topology on $\mathcal M$ is the topology of uniform convergence on weakly compact subsets of $\...
21
votes
1
answer
835
views
On complemented von Neumann algebras
Edit: according to Narutaka Ozawa, question 3) is still open in the type $\mathrm{II}_1$ case. In other terms, it is not known whether every topologically complemented type $\mathrm{II}_1$ factor in $...
18
votes
4
answers
2k
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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 ...
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 $...
17
votes
5
answers
2k
views
If two projections are close, then they are unitarily equivalent
Given two projections $p,q\in B(H)$, it is well-known that if $\|p-q\|<1$, then there exists a unitary $u\in B(H)$ with $q=upu^*$.
The proof that immediately occurs to me uses comparison of ...
17
votes
3
answers
3k
views
Which sigma-ideals in a sigma-algebra are ideals of null sets?
My question is motivated, to be somewhat vague, by an attempt to see how much a measure space is defined by the set of null sets. In other words, assume we are not given a concrete measure on a space ...
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 ...
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 ...
16
votes
3
answers
2k
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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 ...
16
votes
5
answers
3k
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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
1
answer
2k
views
W*-completion of a C*-algebra?
tl;dr: Is there such a thing as a W*-completion of a C*-algebra, and if so, where can I read about it?
I'm wondering about the relationship between (abstract) C*-algebras and W*-algebras. On the one ...
15
votes
2
answers
917
views
Can one associate a "nice" topos to a von Neumann algebra?
The question here inspires my present question.
Reyes proves here that the contravariant functor Spec from the category of commutative rings to the category of sets cannot be extended to the category ...
15
votes
2
answers
1k
views
Is a C*-algebra with an isomorphic predual a von Neumann algebra?
It is well-known that a C*-algebra $A$ is a von Neumann algebra if and only if it has an isometric predual, that is, if and only if there exists a Banach space $X$ such that $A$ is isometrically ...
15
votes
1
answer
547
views
Denseness of inner automorphisms inside automorphisms of hyperfinite type III_1 factor
Let $R$ be the hyperfinite type $III_1$ factor,
and let $Aut(R)$ be its group of automorphisms, equipped with the $u$-topology
(topology of pointwise convergence on the predual).
An automorphism $\...
15
votes
1
answer
540
views
On spatial tensor products of von Neumann algebras
Let $H$ be a Hilbert space, and let $A_1,A_2,A_3\subset B(H)$ be three commuting von Neumann algebras.
We write $\odot$ for the algebraic tensor product,
and $\bar\otimes$ for the spatial tensor ...
15
votes
0
answers
790
views
Must we close weakly to apply the spectral theorem?
Let $H$ be an infinite dimensional separable complex Hilbert space. All C*-subalgebras of $B(H)$ are assumed to be non-degenerate.
The spectral projections of a self-adjoint element $T$ of $B(H)$ lie ...
14
votes
2
answers
899
views
Do torsion-free groups give projectionless group ($C^\ast$) algebras?
One of the reasons I study von Neumann algebras is that they always have plenty of projections. There are many projectionless $C^\ast$-algebras ($0$ and possibly $1$ are the only projections), but the ...
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 ...
14
votes
1
answer
1k
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Connes's unpublished manuscript on correspondences, anyone?
There exist unpublished notes on correspondences of von Neumann algebras due to Connes. This is often cited, but I've never seen a copy. It would be nice to have this, say, to maybe look further into ...
14
votes
0
answers
224
views
Unitary group of a von Neumann algebra: is it a retract of $U(H)$?
Let $M\subset B(H)$ be a properly infinite von Neumann algebra (the case I care about is $M=$ hyperfinite $\mathrm{III}_1$).
Consider the unitary groups $U(M)$ and $U(H)$ in their strong operator ...
14
votes
0
answers
647
views
Countably decomposable von Neumann algebras
A von Neumann algebra is countably decomposable if every family of mutually orthogonal nonzero projections is countable. Even a singly-generated von Neumann algebra need not be countably decomposable; ...
13
votes
3
answers
1k
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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 ...
13
votes
1
answer
1k
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Does the hyperfinite II_1 factor admit two irreducible representations that are not unitarily equivalent?
Regarding the hyperfinite $II_{1}$ factor $R$ as $C^{*}$-algebra, is it known whether any two irreducible representations of $R$ are unitarily equivalent? If it is known that there exists a pair of ...
13
votes
2
answers
723
views
Ideals in Factors
One can easily prove that factors have no nontrivial ultraweakly closed 2-sided ideals as these are equivalent to nontrivial central projections. One can also show type $I_n$, type $II_1$, and type $...
13
votes
2
answers
582
views
Do subgroups have "two sided bases"?
Let $H\leq G$ be an inclusion of finite groups. Define a map $E\colon \mathbb{C}[G]\to \mathbb{C}[H]$ to be the $\mathbb{C}$-linear extension of
$$
E(g)=\begin{cases}
g &\text{if } g\in H\\\
0 &...
13
votes
1
answer
1k
views
Do Baumslag-Solitar Group von Neumann algebras have Property $\Gamma$?
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 $...
13
votes
1
answer
451
views
Factor states on C*-algebras
Which C$^*$-algebras admit factor states for which the von Neumann algebra it generates in the corresponding GNS representation is a type III$_1$ factor? For example, do all purely infinite algebras ...
13
votes
1
answer
481
views
Can the minimal index of a subfactor take all values in {4cos^2(pi/n);n=3,4,5,...} u [4,infinity]?
Given a subfactor $N\to M$ and a conditional expectation $E:M\to N$,
there is a numerical invariant $Ind(E)$ associated to to this situation, called the index of $E$.
The possible values of $Ind(E)$ ...
13
votes
2
answers
349
views
Is the domain of an operator valued weight closed under Hahn-Jordan decomposition?
Let $N\subseteq M$ be an inclusion of semi-finite factors with normal faithful semi-finite traces $\operatorname{Tr}_N$ and $\operatorname{Tr}_M$ respectively. Let $T: M^+\to \widehat{N^+}$ be the ...
13
votes
1
answer
706
views
A left inverse for the comultiplication on a Hopf von Neumann algebra
Edit: incorrect claim at end of earlier version; thanks to Matthew Daws for pointing this out in comments.
$\newcommand{\cM}{{\mathcal M}}\newcommand{\stp}{{\overline{\otimes}}}$The following ...
13
votes
1
answer
1k
views
Endomorphism of type III factor: can it satisfy $\phi\circ\phi = \phi\oplus\phi$?
I'm still trying to get some feeling about this question...
Given Jesse Peterson's answer to this question (he showed that $\phi\circ\phi\sim\phi$ is impossible), I suspect that the following is also ...
13
votes
0
answers
3k
views
Connes Embedding Conjecture is false [closed]
This preprint from yesterday claims to prove that Connes Embedding Conjecture fails.
Since the paper is from outside Operator Algebras (Computer Science/Quantum Computing) and they actually work on ...
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 ...
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 ...
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 ...
12
votes
2
answers
1k
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Do Burnside Group Factors have Gamma?
The Free Burnside group $G=B(2,665)=\langle a,b|g^{665} \rangle$ is infinite, by the work of Adyan and Novikov. Furthermore, the centralizer of any nonidentity element in $G$ is finite cyclic, and so ...
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 ...
11
votes
2
answers
490
views
Actions of locally compact groups on the hyperfinite $II_1$ factor
Let $R$ be the hyperfinite $II_1$ factor, and let $G$ be a locally compact group.
(1) Does there always exist a continuous, (faithful) outer action of $G$ on $R$?
(2) If so, how does one ...
11
votes
1
answer
634
views
Embedding the group von Neumann algebra into an injective von Neumann algebra on the same Hilbert space
Let $\Gamma$ be a discrete group, $\newcommand{\VN}{\rm VN}$
and let $\VN(\Gamma)$ denote its von Neumann algebra, regarded as a subalgebra of ${\sf B}(\ell^2(\Gamma))$. It is well known that $\VN(\...
11
votes
1
answer
514
views
Transfinite induction, a theorem of Pedersen, and chains of subalgebras of $B(H)$
This post is closely related to this one. (In fact I copied some of its content.)
Let $H$ be an infinite dimensional separable complex Hilbert space. All $C^{\star}$-subalgebras of $B(H)$ are assumed ...
11
votes
1
answer
381
views
Are groups with the Haagerup property hyperlinear?
In his 2008 paper Hyperlinear and Sofic Groups: A Brief Guide, Pestov asked (Open Question 9.5) whether every group with the Haagerup property is hyperlinear (or sofic). Has this question been ...
10
votes
3
answers
860
views
Takesaki theorem 2.6
I originally posted this question on MSE and didn't get a satisfactory answer, even after putting a bounty on it. Hence, I thought I should ask here:
Consider the following theorem in Takesaki's book &...