Questions tagged [subfactors]

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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 ...
Sebastien Palcoux's user avatar
3 votes
0 answers
156 views

A section from subfactors to transitive groups

A finite group-subgroup subfactor is a subfactor $(N \subset M)$ isomorphic to $(R^G \subset R^H)$ with $(H \subset G)$ an inclusion of finite groups acting as outer automorphism on the hyperfinite II$...
Sebastien Palcoux's user avatar
3 votes
1 answer
588 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 ...
Sebastien Palcoux's user avatar
5 votes
1 answer
680 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 \subset ...
Sebastien Palcoux's user avatar
0 votes
2 answers
197 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 ...
Sebastien Palcoux's user avatar
2 votes
0 answers
196 views

Existence of inclusions of finite groups with a particular lattice property

Definition : Let $\sim$ be the equivalence relation on inclusions of finite groups, generated by : $(H \subset G) \sim (\phi(H) \subset \phi(G))$, with $ \phi: G \to L$ a finite group morphism and ...
Sebastien Palcoux's user avatar
4 votes
1 answer
697 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 ...
Sebastien Palcoux's user avatar
7 votes
3 answers
691 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 ...
Sebastien Palcoux's user avatar
5 votes
1 answer
258 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 ...
Sebastien Palcoux's user avatar
4 votes
2 answers
537 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 ...
Sebastien Palcoux's user avatar
0 votes
2 answers
222 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 $II_1$ ...
Sebastien Palcoux's user avatar
2 votes
0 answers
249 views

Fusion categories with permutation "associativity matrices"

Let $\mathcal{C}$ be a fusion category and let $(H_1,...,H_r)$ be its simple objects. $\mathcal{C}$ is non-pointed if at least one of its simple object has Perron-Frobenius dimension $ \neq 1$. ...
Sebastien Palcoux's user avatar
2 votes
0 answers
149 views

Planar algebraic translation of a subfactor property

Let $N \subset M$ be an irreducible finite depth and finite index subfactor. $M$ is a completely reducible (algebraic) $N$-$N$ bimodule, it decomposes into irreducibles as follows : $$M=\bigoplus_{...
Sebastien Palcoux's user avatar
4 votes
0 answers
238 views

Is an integral simple fusion ring, categorifiable?

A fusion ring $\mathcal{F}$ (see here p 28) is integral if the Perron-Frobenius dimension $d(h_i)$ of its basic elements $\{h_1,...,h_r\}$, are integers. Its rank is $r$ and its dimension is $\sum d(...
Sebastien Palcoux's user avatar
4 votes
1 answer
398 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 \...
Sebastien Palcoux's user avatar
2 votes
1 answer
177 views

Are every finitely generated planar algebras, also singly generated?

Let $\mathcal{P}$ be a finitely generated planar algebra. Question : Is it also singly generated ? I ask this question, because, on one hand I've read on this paper of V. Jones and D. Bisch : "...
Sebastien Palcoux's user avatar
8 votes
1 answer
258 views

realizing fusion categories as subfactors of the hyperfinite

Let R be the hyperfinite II_1 or the hyperfinite III_1 factor (pick which ever one you prefer), and let Bim(R) denote the tensor category of R-R-bimodules. This question is inspired by the recent ...
André Henriques's user avatar
4 votes
0 answers
249 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 ...
Sebastien Palcoux's user avatar
6 votes
0 answers
236 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}, ..., d_{r}...
Sebastien Palcoux's user avatar
2 votes
0 answers
165 views

How simplify the pentagonal equation from two fusion rings?

A semi-simple finite dimensional Hopf algebra $\mathbb{A}$, and its dual $\mathbb{A}^{*}$ produce two fusion categories $\mathcal{C}$ and $\mathcal{C}^{*}$ and then two fusion rings $\mathcal{R}_{1}$ ...
Sebastien Palcoux's user avatar
3 votes
0 answers
302 views

Is the fundamental group of a maximal subfactor always $\mathbb{R}_{+}^{*}$?

The fundamental group $\mathcal{F}(N \subset M)$ of a unital inclusion of II$_{1}$ factors $N \subset M$ is defined as : $\mathcal{F}(N \subset M) =\{t >0 \ | \ (N \subset M)^{t} \simeq (N \...
Sebastien Palcoux's user avatar
3 votes
1 answer
330 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 $...
Sebastien Palcoux's user avatar
3 votes
0 answers
132 views

Can you tell if a subfactor is finite depth by the growth rate of the standard invariant?

Let $N\subset M$ be a finite index inclusion of $II_1$ factors. To the inclusion we associate the tower of higher relative commutants $\begin{array}{ccccccc} \mathbb{C} = N'\cap N & \subset &...
Owen Sizemore's user avatar
2 votes
1 answer
294 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 ...
Sebastien Palcoux's user avatar
12 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 R^{H_{1}}...
Sebastien Palcoux's user avatar
7 votes
2 answers
400 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 ...
Owen Sizemore's user avatar
9 votes
3 answers
406 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 ...
Sebastien Palcoux's user avatar
4 votes
0 answers
148 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 ...
Sebastien Palcoux's user avatar
2 votes
0 answers
156 views

About the classification of infinite depth irreducible finite index maximal subfactors

The Temperley Lieb subfactors $A_{\infty}$ are the first examples of infinite depth irreducible finite index maximal subfactors. We can see these subfactors as coming from the simple Lie group $SU(2)$....
Sebastien Palcoux's user avatar
16 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 ...
5 votes
1 answer
173 views

Is there an infinite depth irreducible finite index maximal subfactor (other than Temperley Lieb) ?

A subfactor $N \subset M$ is maximal if it admits no non-trivial intermediate subfactors $N \subset P \subset M$. Is there an infinite depth irreducible finite index maximal subfactor (other than ...
Sebastien Palcoux's user avatar
8 votes
0 answers
304 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 ...
Sebastien Palcoux's user avatar
22 votes
2 answers
2k views

Non weakly-group-theoretical integral fusion category

Is there an integral fusion category of rank $7$, FPdim $210$ and type $(1,5,5,5,6,7,7)$ with the following fusion rules (or the little $\color{purple}{\text{variation}}$ below)? $$\scriptsize{\begin{...
Sebastien Palcoux's user avatar
18 votes
1 answer
1k views

Unitary structures on fusion categories

A unitary fusion category is a fusion category with a $C^*$-tensor structure. Hence, in principle, a fusion category could have more than one unitary structure. Does exist a fusion category with more ...
César Galindo's user avatar
20 votes
1 answer
812 views

Fuss-Catalan algebras and non-commutative Hilbert schemes

Hello, this is a question regarding Reineke's paper "Cohomology of non-commutative Hilbert schemes", http://arxiv.org/abs/math/0306185, and more precisely the formula on page 4 there (at $n=1$), ...
Richard's user avatar
  • 1,363
11 votes
2 answers
392 views

What is known about arbitrary subfactors of integer index?

Let $N\subset M$ be an inclusion of ${\rm II}_1$ factors of finite index, $[M:N]<\infty$. I would be mostly interested in the hyperfinite case, $N\simeq M\simeq R$, but let us just take them ...
Richard's user avatar
  • 1,363
1 vote
0 answers
189 views

Non-invariant subspaces for subfactors.

Let $\mathcal{M}$ be a II_1 factor. If $\mathcal{N}$ is a subfactor of $\mathcal{M}$ ($\mathcal{N} \neq \mathcal{M}$), does there always exist a projection in $\mathcal{M}$ such that $(I-P)AP \neq 0$ ...
heller's user avatar
  • 61
6 votes
0 answers
137 views

Is it true that there are exactly two conjugacy classes of order two elements in Out(R)?

In the title, $R$ stands for the hyperfinite III1 factor. An order two element $\alpha\in Out(M)$ ($M$ any factor) has an invariant $c(\alpha)\in H^3(\mathbb Z/2,S^1)=\mathbb Z/2$. Q: Is $c$ the ...
André Henriques's user avatar
5 votes
1 answer
304 views

How well do we know relative commutants in $L(\mathbb{F}_\infty)$?

Let $H=K_1\oplus K_2$ be infinite dimensional Hilbert spaces. Voiculescu's free Gaussian functor gives us free group factors $L(H)$, $L(K_1)$, $L(K_2)$ acting on the full Fock space $\Gamma(H)$ and, ...
Ollie's user avatar
  • 1,391
6 votes
1 answer
354 views

von Neumann automorphisms: does convergence on a dense algebra imply $u$-convergence?

Let $M$ be a separable von Neumann algebra and let $A$ be a (von Neumann-)dense *-subalgebra. Suppose that $\alpha,\alpha_1,\alpha_2,\dots$ are automorphisms of $M$, such that for every $a \in A$, $$ \...
Aaron Tikuisis's user avatar
15 votes
1 answer
524 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
12 votes
0 answers
306 views

Subfactors of $L(F_{\infty})$

It is a well known result that any subfactor of the hyperfinite $II_{1}$ factor is hyperfinite. I wonder if there is any finite index version of this for free group factors. In particular is it true ...
Mike Hartglass's user avatar
11 votes
1 answer
387 views

A question about maximal subgroups

Let $G$ be a finite group and $H_1,\ldots, H_n$ a set of maximal subgroups of $G$. Let $\delta_{H_i}$ be delta functions with support on $H_i$, and let $A$ be the commutative algebra generated by $\...
Feng Xu's user avatar
  • 111
41 votes
3 answers
3k views

Why is there such a close resemblance between the unitary representation theory of the Virasoro algebra and that of the Temperley-Lieb algebra?

For those who aren't familiar with the Virasoro or Temperley-Lieb algebras, I include some definitions: • The (universal envelopping algebra of the) Virasoro algebra is the $\star$-algebra $...
André Henriques's user avatar
5 votes
1 answer
325 views

representing tensor-C*-categories in BIM

Given a factor M (=von Neumann alg. with center ℂ), let us write BIM for the ⊗-C*-category of M-M-bimodules. Which ⊗-C*-categories can one faithfully embed into BIM? ⓵ Are ...
André Henriques's user avatar
5 votes
1 answer
720 views

Subspaces of a Subfactor

Is the following true? Let $\mathcal N \subset \mathcal M$ be a subfactor. There is a bijective correspondence between the ultraweakly closed subspaces of $\mathcal M$ that are bimodules over $\...
Andre's user avatar
  • 1,199
5 votes
1 answer
560 views

Is there a trivial construction of the trace on the Jones basic construction?

Let $N$ be a type $II_{1}$-factor with trace $\tau$, and $B$ a von Neumann subalgebra. The existence of the semifinite trace on the Jones basic construction $\langle N, e_{B} \rangle$ is reasonably ...
Jon Bannon's user avatar
  • 6,977
6 votes
2 answers
309 views

What is the subfactor planar algebra of type $\tilde{A}_n$, of index 4?

As I understand it, there is a subfactor whose principal graph is the affine Dynkin diagram $\tilde{A}_n$. Since every vertex has two neighbors, does that mean the space of 1-boxes is two dimensional? ...
Stephen Bigelow's user avatar
4 votes
2 answers
353 views

Invertibility of the planar algebra-subfactor correspondence

In Jones's paper "Planar Algebras I", Theorem 4.2.1 establishes that an extremal finite index subfactor admits a spherical C*-planar algebra structure, and Theorem 4.3.1 establishes that spherical C*-...
Jon Bannon's user avatar
  • 6,977
8 votes
4 answers
449 views

Possible values of the index for subfactor inclusions coming from conformal nets

This question is related to Can the minimal index of a subfactor take all values in {4cos^2(pi/n);n=3,4,5,...} u [4,infinity]? I was wondering what one knows for the special case of conformal nets ...
Marcel Bischoff's user avatar