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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 ...
Dmitri Pavlov's user avatar
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 ...
Dmitri Pavlov's user avatar
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)...
André Henriques's user avatar
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 $B \subset A$. ...
Andreas Thom's user avatar
  • 25.5k
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 ...
Oliver's user avatar
  • 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 ...
Mikael de la Salle's user avatar
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 ...
Jesse Peterson's user avatar
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 $H$ ...
Sebastien Palcoux's user avatar
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 $\...
Andre's user avatar
  • 1,199
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 $...
Julien's user avatar
  • 660
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
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 $...
Narutaka OZAWA's user avatar
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 ...
Martin Argerami's user avatar
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 ...
Super-Measurable Analyst's user avatar
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 ...
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
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
  • 5,425
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 ...
Toby Bartels's user avatar
  • 2,754
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 ...
Jon Bannon's user avatar
  • 7,057
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 ...
Hannes Thiel's user avatar
  • 3,497
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 $\...
André Henriques's user avatar
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 ...
André Henriques's user avatar
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 ...
Jonas Meyer's user avatar
  • 7,329
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 ...
Dave Penneys's user avatar
  • 5,425
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 ...
Paul Siegel's user avatar
  • 29.2k
14 votes
1 answer
1k views

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 ...
Jon Bannon's user avatar
  • 7,057
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 ...
André Henriques's user avatar
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; ...
Nik Weaver's user avatar
  • 42.8k
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
  • 9,330
13 votes
1 answer
1k views

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 ...
Jon Bannon's user avatar
  • 7,057
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 $...
Dave Penneys's user avatar
  • 5,425
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 &...
Dave Penneys's user avatar
  • 5,425
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 $...
Jon Bannon's user avatar
  • 7,057
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 ...
Isaac's user avatar
  • 771
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 $Ind(E)$ ...
André Henriques's user avatar
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 ...
Dave Penneys's user avatar
  • 5,425
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 ...
Yemon Choi's user avatar
  • 25.8k
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 ...
André Henriques's user avatar
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 ...
Martin Argerami'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
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 ...
Sebastien Palcoux's user avatar
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 ...
dohmatob's user avatar
  • 6,853
12 votes
2 answers
1k views

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 ...
Jon Bannon's user avatar
  • 7,057
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 ...
Chris Heunen's user avatar
  • 3,937
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 ...
André Henriques's user avatar
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(\...
Yemon Choi's user avatar
  • 25.8k
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 ...
Andreas Thom's user avatar
  • 25.5k
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 ...
MaoWao's user avatar
  • 1,027
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 &...
Andromeda's user avatar
  • 175

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