Questions tagged [subfactors]
The subfactors tag has no usage guidance.
170
questions
2
votes
0
answers
104
views
A process generating series of new subfactors
Consider the following process:
Take a maximal finite depth-index irreducible subfactor planar algebra $P^{(1)} = P(A^{(0)} \subset A^{(1)})$.
Choose a composition with itself such that there is no ...
0
votes
1
answer
121
views
Is there an irreducible subfactor with an infinite homogeneous single chain lattice?
We know that we can build an irreducible subfactor realizing a finite single chain lattice containing any finite index irreducible maximal subfactors, by using the free composition (see here).
Now ...
1
vote
1
answer
259
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 single ...
1
vote
0
answers
37
views
Skein theory: How axiomatizing a 2-box space?
Let $(A,+,\times, *)$ with an adjoint operation compatible with $+$, $\times$ and $*$, such that $(A,+,\times)$ and $(A,+,*)$ are finite dimensional ${\rm C}^{*}$-algebras.
What are the axioms on $...
1
vote
0
answers
107
views
Is every irreducible subfactor planar algebra a quotient of the planar algebra of tangles?
Let $\mathcal{T}_{n,\pm}$ be the vector space generated by the planar tangles (up to isomorphism) having $2n$ intervals on their "ouput'' disk and a white (or black) shaded marked interval. Then the ...
1
vote
0
answers
174
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 $\...
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$...
3
votes
1
answer
142
views
Is there a tangle encoding the fusion rules?
Let $(N \subset M)$ be an irreducible finite index depth $n$ subfactor. Let $P = P(N \subset M)$ its planar algebra.
Let $(B_i)$ be the finite sequence of $N$-$N$-bimodules appearing in the principal ...
1
vote
1
answer
131
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 = \mathcal{F}(\mathcal{F}^{-...
1
vote
1
answer
180
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
132
views
Is there a Frobenius reciprocity for the coproduct?
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 = \mathcal{F}(\mathcal{F}^{-...
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 $\...
4
votes
1
answer
399
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 \...
9
votes
1
answer
2k
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
0
answers
306
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 ...
4
votes
2
answers
538
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
299
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 $B$.
...
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 ...
2
votes
0
answers
132
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
1
answer
188
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
1
answer
228
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} = \mathbb{...
3
votes
2
answers
353
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 $\mathcal{A}'$, $...
5
votes
0
answers
305
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 ...
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 $(R^{G_{1}} \subset R^{H_{1}}...
8
votes
0
answers
341
views
Is there a finite-index finite-depth II$_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 < ...
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 ...
2
votes
0
answers
197
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 ...
20
votes
1
answer
813
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 $n=1$), ...
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 ...
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 $II_1$ ...
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$.
...
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_{...
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(...
2
votes
1
answer
178
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 :
"...
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 $...
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 ...
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 ...
6
votes
0
answers
237
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}...
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 $\mathcal{R}_{1}$ ...
11
votes
1
answer
388
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 $\...
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 ...
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 &...
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 ...
4
votes
0
answers
149
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
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 $SU(2)$....
5
votes
1
answer
174
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 ...
13
votes
3
answers
678
views
Does every Frobenius algebra in a monoidal *-category give a Q-system?
Suppose that C is a fusion C*-cateogry and that A is an irreducible Frobenius algebra object in C, is there always a Frobenius algebra A' isomorphic to A such that A' is a Longo Q-system (that is the ...
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 $(I-P)AP \neq 0$ ...
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 ...
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, ...