The subfactors tag has no usage guidance.

**9**

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 ...

**13**

votes

**1**answer

832 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 & ...

**8**

votes

**2**answers

944 views

### Generalization of a theorem of Øystein Ore in group theory

Theorem (Øystein Ore, 1938): A finite group $G$ is cyclic iff its lattice of subgroups $\mathcal{L}(G)$ is distributive.
Proof: see below.
Let $(H \subset G)$ be an inclusion of finite groups and ...

**4**

votes

**3**answers

465 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 ...

**0**

votes

**2**answers

149 views

### Products of maximal inclusions of finite groups with a non-obvious intermediate

Let $(H_1 \subset G_1)$ and $(H_2 \subset G_2)$ be core-free maximal inclusions of finite groups.
Their product, the inclusion $(H_1 \times H_2 \subset G_1 \times G_2)$, admits four obvious ...

**2**

votes

**1**answer

515 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

109 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 ...

**3**

votes

**0**answers

101 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 ...

**6**

votes

**2**answers

302 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 ...

**4**

votes

**1**answer

243 views

### Normal intermediate subgroup and normal core

Let $G$ be a finite group and $H$ a subgroup.
The normal core of $H$ in $G$ is $core_G(H) := \bigcap_{g \in G}g^{-1}Hg$
Definition: $K$ is a normal intermediate subgroup of the inclusion $(H ...

**3**

votes

**0**answers

268 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 ...

**2**

votes

**1**answer

370 views

### A second isomorphism theorem for the inclusions of groups

The usual second isomorphism theorem for groups is: let $G$ be a group, $S$ and $N$ subgroups with $N$ normal, then $SN$ is a subgroup of $G$, $S\cap N$ is a normal subgroup of $S$ and $SN/N \simeq ...

**8**

votes

**0**answers

265 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 ...

**7**

votes

**3**answers

296 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 ...

**6**

votes

**1**answer

638 views

### Are there workable algebraic geometry approaches for the pentagon equation?

A pentagon equation is a system of polynomial equations of degree $3$ with several variables and integer coefficients, given by a fusion ring.
A fusion ring is given by a finite set of integer ...

**1**

vote

**3**answers

376 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 ...

**5**

votes

**1**answer

200 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 ...

**4**

votes

**0**answers

178 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}, ..., ...

**3**

votes

**0**answers

164 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

209 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 ...

**2**

votes

**1**answer

191 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

187 views

### $\mathcal{L}(H_i \subset G_i)$ distributive $\Rightarrow$ $\mathcal{L}(H_1 \times H_2 \subset G_1 \times G_2)$ modular?

Let $\mathcal{L}( G)$ be the lattice of subgroups of $G$ and $\mathcal{L}(H \subset G)$ the lattice of intermediate subgroups.
Definitions: A lattice $(L, \wedge, \vee)$ is distributive if, ...

**10**

votes

**3**answers

807 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

**0**answers

292 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 < ...

**3**

votes

**1**answer

327 views

### Are all the R-R-bimodules completely reducible?

Let $R$ be the hyperfinite $II_1$ factor and let $X$ be any $R$-$R$-bimodule.
Question: Is $X$ completely reducible (i.e. a direct integral of irreducible $R$-$R$-bimodules)?
Example: If $(N ...

**4**

votes

**0**answers

180 views

### An embedding theorem for a fusion ring planar algebra?

We first recall the embedding theorem for finite depth subfactor planar algebras:
The planar algebra generated by a (finite depth) subfactor, is embeddable into the planar algebra generated by its ...

**3**

votes

**1**answer

272 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 ...

**2**

votes

**0**answers

140 views

### Are there workable numerical approaches for the pentagon equation?

Warning: this post is the "numerical" analog of
Are there workable algebraic geometry approaches for the pentagon equation?
I've replaced "algebraic geometry" by "numerical" in its content,
...

**4**

votes

**2**answers

473 views

### The category of subfactors extending the category of groups?

This post was inspired by this answer of Dave Penneys.
In the category of (irreducible hyperfinite II$_1$) subfactors, the morphisms of $(N \subset M)$ to $(N' \subset M')$ are usually defined as ...

**3**

votes

**0**answers

264 views

### What's the ratio of inclusions of finite groups with a distributive lattice?

Definition: Two inclusions of finite groups are equivalent, $(A \subset B) \sim (C \subset D)$, if: $(A/A_B \subset B/A_B) \simeq (C/C_D \subset D/C_D)$ with $A_B$ the normal core of $A$ in $B$.
...

**2**

votes

**2**answers

247 views

### ${\rm II}_1$-factors with finite commutant and trivial intersection generate $B(H)$?

Let $H$ be an $\infty$-dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Let $\mathcal{A}$, $\mathcal{B} \subset B(H)$ be ${\rm II}_1$-factors such that $\mathcal{A}'$, ...

**1**

vote

**1**answer

152 views

### The coproduct on the 2-boxes space of the group-subgroup subfactor planar algebras

Let $(H \subset G)$ be an inclusion of finite groups.
Let the subfactor $(\mathcal{R} \rtimes H \subset \mathcal{R} \rtimes G)$ with $\mathcal{R}$ the hyperfinite ${\rm II}_1$ factor, and its planar ...

**1**

vote

**1**answer

226 views

### Existence of homogeneous single chain compositions of a given maximal subfactor?

All the subfactors here are irreducible inclusion of hyperfinite II$_1$ factors.
A subfactor $(N \subset M)$ is Homogeneous Single Chain ($HSC$) if its lattice of intermediate subfactors is a ...

**0**

votes

**2**answers

198 views

### Isomorphism theorem for subfactors?

It's about the existence of a generalization of the first isomorphism theorem for groups, for subfactors :
Let $(N \subset M)$ and $(N' \subset M')$ be irreducible inclusions of hyperfinite $II_1$ ...

**2**

votes

**1**answer

470 views

### Image, kernel, quotient and first isomorphism theorem, in a category of monoid objects

Let $\mathcal{C}$ be a monoidal category and Mon$_{\mathcal{C}}$ the category of monoids (also called algebra objects) on $\mathcal{C}$.
Questions: are there definitions of image and kernel for a ...

**1**

vote

**1**answer

45 views

### Are the two-side TLJ subfactors maximal?

Let $(N \subset M)$ be an inclusion of ${\rm II}_1$ factors, with the following principal graph (called two-side TLJ)
Question: Is $(N \subset M)$ a maximal subfactor?

**1**

vote

**0**answers

55 views

### Are there infinitely many amenable Hadamard-Petrescu subfactors?

The complex Hadamard matrices of dimension $n$ are used to build index $n$ subfactors through the commuting square construction. For more details, see the paper Subfactors and Hadamard Matrices by ...

**1**

vote

**0**answers

131 views

### The planar algebra generated by the biprojections

Let $(N_1 \subset M_1)$ and $(N_2 \subset M_2)$ be two irreducible finite index subfactors.
Let $\mathcal{B}_i$ be the set of all the biprojections of $\mathcal{P}_{2+}(N_i \subset M_i)$.
Let ...

**1**

vote

**1**answer

112 views

### Is the coproduct of central operators, also central?

Let $\mathcal{P}$ be an irreducible finite index-depth subfactor planar algebra. The $2$-boxes space $\mathcal{P}_{2,+}$ is equipped with the coproduct $(a,b) \mapsto a*b = ...