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

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**1**answer

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### 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 ...

**12**

votes

**1**answer

748 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 & ...

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votes

**3**answers

426 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 ...

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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 ...

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votes

**1**answer

423 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 ...

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votes

**0**answers

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### 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 ...

**2**

votes

**0**answers

243 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 ...

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votes

**6**answers

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### 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 ...

**6**

votes

**2**answers

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### 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 ...

**4**

votes

**2**answers

253 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 ...

**6**

votes

**2**answers

289 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 ...

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vote

**0**answers

70 views

### Are the integer index finite depth irreducible subfactors Kac-coideal?

Is it known whether or not the integer index finite depth irreducible subfactors (planar algebra) are Kac-coideal subfactors: $(R^{\mathbb{A}} \subset R^{\mathbb{I}})$, with $\mathbb{A}$ a finite ...

**26**

votes

**4**answers

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### 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 ...

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votes

**3**answers

924 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 ...

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votes

**3**answers

274 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 ...

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votes

**2**answers

1k 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 ...

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votes

**1**answer

187 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 ...

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votes

**0**answers

169 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}, ..., ...

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votes

**0**answers

154 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 ...

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votes

**1**answer

334 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 ...

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votes

**1**answer

200 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 ...

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votes

**1**answer

162 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} = ...

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votes

**2**answers

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### 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 ...

**32**

votes

**3**answers

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### 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 ...

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votes

**9**answers

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### 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 ...

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votes

**2**answers

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### 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 ...

**11**

votes

**5**answers

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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 ...

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votes

**3**answers

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### 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 $H$ ...

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votes

**1**answer

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### 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 ...

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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 ...

**15**

votes

**1**answer

832 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 ...

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votes

**0**answers

235 views

### Are there only finitely many maximal subfactors of a 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 subfactors of a fixed finite ...

**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 ...

**10**

votes

**3**answers

754 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

239 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 ...

**7**

votes

**0**answers

281 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 < ...

**5**

votes

**1**answer

277 views

### Is the fundamental group of $II_{1}$ factors invariant under a relation?

This issue is in continuation of an answer I gave here about
noncommutative sets.
In order to define the equivalence relation, let's first recall the Tomita-Takesaki modular theory and ...

**15**

votes

**1**answer

293 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 $\mathbb{M}_m(\mathbb{M}_n)$ ...

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votes

**4**answers

692 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 ...

**9**

votes

**1**answer

494 views

### Mono- and epi-morphisms for C*-algebras

This question is motivated by Yemon Choi's answer here: Epimorphisms have dense range in TopHausGrp?
It's well-known that the category of unital commutative C*-algebras and $*$-homomorphisms is dual ...

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votes

**3**answers

548 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 ...

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votes

**2**answers

392 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 ...

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votes

**1**answer

460 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 $S^m$ ...

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votes

**2**answers

366 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 ...

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votes

**1**answer

262 views

### What's the natural equivalence of subfactors in general?

Let $A$ be a factor and $\mathcal{C}_{A}$ be the category of all the subfactors $(M \subset N)$ such that $M$ and $N$ are isomorphic to $A$. The most famous of them is perhaps $\mathcal{C}_{R}$ with ...

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vote

**2**answers

214 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 ...

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votes

**2**answers

420 views

### B(H) as a direct sum of a closed two sided ideal and a subalgebra

Let $B(H)$ is the C*-algebra of all bounded linear operators on
Hilbert space $H$. Are there a closed two-sided ideal $I$ and a
subalgebra $A$ of $B(H)$ such that $B(H)=I \oplus A$ (direct sum I and ...

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votes

**1**answer

295 views

### positive elements in tensor product

Let x be a positive element in the spatial tensor product of two non unital C* algebras
A and B. Is there a single element $a \otimes b \geq x$?
How can we noncommutativize the following proof, in ...

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votes

**0**answers

195 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 III factor.
Question : Is ...

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votes

**1**answer

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### Separability of the C*-algebra in the definition of K-homology

There are (at least) two approaches to K-homoology: one is via the so called dual algebra which is due to Paschke. The second is via the Fredholm modules and is due to Kasparov. In Nigel Higson's book ...