**12**

votes

**1**answer

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

**1**answer

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

**3**answers

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

**3**answers

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

**4**answers

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

**2**answers

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

**1**answer

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

**0**answers

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

**0**answers

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

**6**answers

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

**3**answers

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

**2**answers

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

**0**answers

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

**2**answers

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

**3**answers

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

**7**answers

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

**3**answers

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

**0**answers

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

**3**answers

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

**3**answers

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

**3**answers

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

**2**answers

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

**3**answers

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

**0**answers

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

**1**answer

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

**1**answer

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

**0**answers

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

**1**answer

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

**1**answer

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

**3**answers

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

**2**answers

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

**1**answer

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

**0**answers

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

**2**answers

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

**3**answers

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

**2**answers

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

**9**answers

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

**3**answers

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

**5**answers

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

**2**answers

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

**2**answers

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

**1**answer

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

**1**answer

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

**4**answers

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

**4**answers

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

**3**answers

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

**1**answer

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

**1**answer

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

**3**answers

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

**1**answer

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 ...