0
votes
0answers
150 views

Is a finite depth-index irreducible subfactor, intermediate of a depth ≤ 3 one?

Let $(N \subset M)$ be a finite depth-index irreducible subfactor. Main question: Is $(N \subset M)$ the intermediate of a finite index depth $\le 3$ irreducible subfactor? (In others words, is ...
1
vote
0answers
68 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 ...
3
votes
0answers
146 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
0answers
91 views

Uniqueness of the tensor product decomposition of subfactors

A subfactor $(N \subset M)$ is indecomposable if (for $N_i \subset M_i)$: $$(N \subset M) = (N_1 \otimes N_2 \subset M_1 \otimes M_2) \Rightarrow \exists i \ N_i = M_i$$ Then, a subfactor $(N ...
1
vote
1answer
137 views

Infinite amenable group subfactors

Let amenable groups $\Gamma$ and $\Gamma'$. They act outerly of only one manner on the hyperfinite ${\rm II}_1$-factor $\mathcal{R}$. Question: $(\mathcal{R} \subset \mathcal{R} \rtimes \Gamma) ...
2
votes
1answer
159 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} = ...
2
votes
2answers
225 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 $\mathcal{A}'$, ...
2
votes
0answers
113 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 ...
2
votes
1answer
421 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
3answers
417 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 ...
3
votes
1answer
390 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 ...
0
votes
2answers
188 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
0answers
127 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 : ...
2
votes
1answer
263 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 : Let $N ...
4
votes
0answers
162 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
0answers
170 views

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

The fundamental group $\mathcal{F}(N \subset M)$ of an inclusion of $II_{1}$ factors $N \subset M$ is defined as : $\mathcal{F}(N \subset M) =\{t >0 \ | \ (N \subset M)^{t} \simeq (N \subset M) ...
3
votes
1answer
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 ...
7
votes
3answers
272 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 ...
4
votes
0answers
104 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 ...
2
votes
0answers
118 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 ...
7
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 ...
5
votes
1answer
104 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 ...
8
votes
0answers
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 ...
6
votes
0answers
124 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 ...
5
votes
1answer
205 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, ...
6
votes
1answer
278 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 $a \in A$, $$ ...
14
votes
1answer
346 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 ...
5
votes
1answer
285 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 ...
5
votes
1answer
700 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 ...
5
votes
1answer
377 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 ...
13
votes
1answer
973 views

Endomorphism of type III factor: can it satisfy $\phi\circ\phi = \phi\oplus\phi$?

I'm still trying to get some feeling about this question... Given Jesse Peterson's answer to this question (he showed that $\phi\circ\phi\sim\phi$ is impossible), I suspect that the following is also ...
5
votes
1answer
524 views

Subfactor theory and Hilbert von Neumann Algebras

There seem to be intimate connections between the different definitions of von Neumann module. The two that I'm aware of are Hilbert von Neumann modules and correspondences (in the sense of Connes). I ...
7
votes
2answers
408 views

endomorphism of factor: can it be idempotent up to congugacy?

Let $M$ be a factor, and let $\phi:M\to M$ be an irreducible endomorphism ("irreducible" means that the relative commutant of $\phi(M)$ in $M$ is trivial). Let's also assume that $\phi$ is not ...
6
votes
1answer
611 views

subfactor of finite rank but infinite index: is this possible?

A subfactor $N\subset M$ is essentially the same thing as an $N$-$M$-bimodule. I'll recall the basic definitions in the language of bimodules, and I hope that subfactor people will excuse me. ...
12
votes
1answer
401 views

Can the minimal index of a subfactor take all values in {4cos^2(pi/n);n=3,4,5,…} u [4,infinity]?

Given a subfactor $N\to M$ and a conditional expectation $E:M\to N$, there is a numerical invariant $Ind(E)$ associated to to this situation, called the index of $E$. The possible values of $Ind(E)$ ...
9
votes
1answer
325 views

Do subgroups have “two sided bases”?

Let $H\leq G$ be an inclusion of finite groups. Define a map $E\colon \mathbb{C}[G]\to \mathbb{C}[H]$ to be the $\mathbb{C}$-linear extension of $$ E(g)=\begin{cases} g &\text{if } g\in H\\\ 0 ...
13
votes
3answers
715 views

Categorifying the Reals via von Neumann Algebras?

So one way to categorify the natural numbers is to replace them with vector spaces. Then the dimension of the vector space reproduces the natural number. More generally you can categorify integers to ...
4
votes
2answers
628 views

Operator Valued Weights

One of the basic tools in subfactors is the conditional expectation. If $N\subset M$ is a $II_1$-subfactor (or an inclusion of finite factors), then there is a unique trace-preserving conditional ...
8
votes
0answers
491 views

Pimsner-Popa Bases

Let $N\subset M$ be a finite index $II_1$-subfactor. Let $B=\{b_i\}$ be a finite orthonormal (Pimsner-Popa) basis for $M$ over $N$. Let $d=[M\colon N]^{1/2}$. It is well known that $B_1=\{d b_{i_1} ...
7
votes
0answers
280 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 < ...