Algebras of operators on Hilbert space, $C^*-$algebras, von Neumann algebras, non-commutative geometry

learn more… | top users | synonyms (1)

12
votes
1answer
1k 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 ...
15
votes
1answer
901 views

Non-“weakly group theoretical” integral fusion categories?

Is there an integral fusion category of global dimension $210$, such that the simple objects have dimensions $\{1,5,5,5,6,7,7\}$ and the following fusion matrices? $\small{\begin{smallmatrix} 1 & ...
5
votes
3answers
478 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 ...
11
votes
3answers
1k views

Is there a purely group-theoretic reformulation of an equivalence of subgroups?

There is an equivalence relation between inclusion of finite groups coming from the world of subfactors: Definition: $(H_{1} \subset G_{1}) \sim(H_{2} \subset G_{2})$ if $(R^{G_{1}} \subset ...
28
votes
4answers
2k views

Reference for the Gelfand-Neumark theorem for commutative von Neumann algebras

The Gelfand-Neumark 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 ...
8
votes
2answers
1k views

Riemann zeta function at positive integers and an Appell sequence of polynomials related to fractional calculus

I was exploring some raising and lowering operators related to an infinitesimal generator for fractional integro-derivatives and found an Appell sequence of polynomials, i.e., an infinite sequence of ...
3
votes
1answer
534 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 ...
4
votes
0answers
113 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 ...
5
votes
0answers
111 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 ...
29
votes
6answers
4k views

A remark of Connes

In an interview (at http://www.alainconnes.org/docs/Inteng.pdf) Connes remarks that I had been working on non-standard analysis, but after a while I had found a catch in the theory.... The point ...
7
votes
3answers
307 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 ...
5
votes
2answers
318 views

Topological K-theory for commutative C*-algebras

It is in some sense folklore that given two arbitrary abelian groups $G,H$ one can find a $C^*$ algebra $A$ such that $K_0(A)=G$ and $K_1(A)=H$. My question is the following: what is known in the case ...
4
votes
0answers
273 views

Are the homogeneous single chain subfactors, Dedekind?

Background: See here and there. Recall that a subfactor is Dedekind if all its intermediate subfactors are normal. A subfactor $(N \subset M)$ is Homogeneous Single Chain (HSC) if its lattice ...
6
votes
2answers
308 views

Are subfactor planar algebras hard to classify at index 6?

Given a finite index inclusion, $N\subset M$, of $II_1$ factors we can construct two towers of finite dimensional algebras known as the $\textit{standard invariant}$. For low index, this has allowed ...
17
votes
3answers
2k views

Noncommutative smooth manifolds

Connes defined a noncommutative analog of a closed oriented Riemannian spin^c manifold using spectral triples. Using his definition it is unclear how to separate the smooth structure from the metric. ...
12
votes
7answers
1k views

Positive but not completely positive?

The only example I know of a positive map which is not completely positive is the transpose map on $M_n(\mathbb{C})$. Of course, one can come up with minor perturbations of this (compose it with, or ...
18
votes
3answers
1k 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 $H$ ...
8
votes
0answers
269 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 ...
11
votes
3answers
846 views

Universal $C^*$-algebra with generators and relations

We say that the $C^*$-algebra $A$ generated by $a_1,...,a_n$ is universal subject to relations $R_1,...,R_m$ if for every $C^*$-algebra $B$ with elements $b_1,...,b_n$ satisfying relations ...
9
votes
3answers
1k views

positive elements in tensor products

Let $A \otimes B$ be the algebraic tensor of two $C^{\ast}$ -algebras, and an element x in $A\otimes B$ is positive if $x=yy^{\ast}$. Then is it always possible to write x in the form $x=\sum ...
8
votes
3answers
725 views

subgroup of SU(N) with maximal manifold dimension

Given the group SU(N) of NxN unitary matrices, does there exist a subgroup S with a manifold dimension larger than the SU(N-1) manifold dimension and smaller than the SU(N) one? S should not ...
5
votes
2answers
569 views

von neumann algebras and measurable spaces

I've read some pages on links between von neumann (VN) algebras and measurable spaces (Spectra of $C^*$ algebras and Non-commutative geometry from von Neumann algebras?), but I can't get the ...
8
votes
3answers
2k views

Zero divisor conjecture and idempotent conjecture

Let $G$ be a torsion-free group and $C$ the ring of complex numbers. The zero divisor (idempotent, resp.) conjecture is that there is no nontrivial zerodivisor (idempotent, resp.) in $CG$. The wiki ...
4
votes
0answers
185 views

Existence of a Kac algebra for a given fusion ring in a particular class

A $n$-dimensional Kac algebra (i.e., a Hopf C*-algebra), admits finitely many irreducible representations, whose cardinal $r$ is called its rank, the increasing sequence $(d_{1},d_{2},d_{3}, ..., ...
6
votes
1answer
482 views

An equivalence relation on the space of polynomials in one complex variable

Let $P(z)$ be a polynomial with complex variable $z$. We consider the following distribution for the roots of $P(z)=0$: the distribution is a triple $(n_{1},n_{2},n_{3})$ where these integers are ...
5
votes
1answer
202 views

Jordan-Hölder theorem for planar algebras?

First recall the Jordan-Hölder theorem for groups: Theorem (Jordan-Hölder): Let $G$ be a group, and let $$ G=G_1 \supset G_2 \supset \dots \supset G_r = \{ e \} $$ be a normal tower such that ...
3
votes
0answers
171 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 ...
2
votes
1answer
372 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 ...
2
votes
1answer
210 views

An upper bound for the maximal subgroups at fixed index?

Let us call a subgroup an injective homomorphism between groups. I warn the reader that a subgroup designates here an inclusion $(H \subset G)$, not $H$ alone. A subgroup $H \subset G$ is ...
1
vote
3answers
405 views

What's the relation between fusion and coproduct?

For an irreducible finite depth finite index subfactor $(N \subset M)$, there is a structure of fusion category given by the even part of its principal graph. The simple objects $(X_i)_{i \in I}$ of ...
16
votes
2answers
1k views

The letters of the word “ART”

Edit: According to the Gelfand duality between topological spaces and commutative $C^{*}$algebras, I add some new tags. So the question is that what is the structure of $ Ext (A,A)$ where $A$ is ...
2
votes
1answer
194 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} = ...
1
vote
0answers
130 views

Determining the primitive ideal space of C-star algebras

Is there a general way of finding a primitive ideal space of $C^*$-algebra? For example, if $C^*$-algebra is given by the universal $C^*$-algebra generated by two self-adjoint unitary elements, how ...
64
votes
2answers
4k views

Norms of Commutators

If an $n$ by $n$ complex matrix $A$ has trace zero, then it is a commutator, which means that there are $n$ by $n$ matrices $B$ and $C$ so that $A= BC-CB$. What is the order of the best constant ...
35
votes
3answers
4k views

Quantum mechanics formalism and C*-algebras

Many authors (e.g Landsman, Gleason) have stated that in quantum mechanics, the observables of a system can be taken to be the self-adjoint elements of an appropriate C*-algebra. However, many ...
20
votes
2answers
1k views

Does left-invertible imply invertible in full group C*-algebras (discrete case)?

The following question/problem has been bugging me on and off for some time now: so I thought it might be worth broaching here on MO, as a case of "ask the experts". Let $G$ be a discrete group. ...
13
votes
9answers
1k views

Examples of noncommutative analogs outside operator algebras?

Theo's question made me wonder if there are other "noncommutative analogs" outside of operator algebras. Some noncommutative analogs from operator algebras include: A $C^\ast$-algebra is a ...
18
votes
3answers
1k views

Reference request for translating from Top to C*-alg

Some recent questions on MO (for example, Do subalgebras of C(X) admit a description in terms of the compact Hausdorff space X?) have been about Gelfand duality-- namely, that the categories of ...
11
votes
5answers
2k 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 ...
19
votes
2answers
5k views

Is the Invariant Subspace Problem interesting?

There's an amusing comment in Peter Lax's Functional Analysis book. After a brief description of the Invariant Subspace Problem, he says (paraphrasing) "...this question is still open. It is also an ...
14
votes
2answers
853 views

Discrete groups G whose full C*-algebra C*(G) is not quasidiagonal?

Is there a known example of a countable discrete group G whose full group C*-algebra C*(G) is not quasidiagonal? Let us recall that a separable C*-algebra A is quasidiagonal if it admits a faithful ...
11
votes
1answer
575 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 ...
15
votes
1answer
885 views

Convolution algebras for double groupoids?

There is a lot of work of course on convolution algebras of measured groupoids, and this gives "Noncommutative geometry". However there is a lot of interest in algebraically structured groupoids, for ...
15
votes
4answers
998 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 $\|A\|_2 = \sqrt{Tr(A^* A)/n}$. My question is whether a $k$-uple of hermitian matrices that are almost commuting (with ...
12
votes
4answers
1k 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 ...
10
votes
3answers
827 views

Subfactor summer reading list

Many people I talk to lament the nonexistence of a coherent source for learning the theory of subfactors. Could someone suggest a nice (ordered) list of books/papers to work through to obtain a ...
7
votes
1answer
733 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 ...
17
votes
1answer
368 views

A possible extension of a determinant inequality

It is well known that if $A, B$ are positive semidefinite matrices, then $$\det (A+B)\ge \det A+\det B.$$ I am considering a possible extension of this result. Let $\mathbb{M}_m(\mathbb{M}_n)$ ...
13
votes
3answers
1k 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 ...
7
votes
1answer
291 views

Infinite tensor product of states

Tensor products of finite number of different objects are always well described in the literature. However, the situation of infinite tensor products seems to be much tougher. Even in the simplest ...