All Questions
13 questions
2
votes
1
answer
159
views
Question on tensor product of von Neumann algebras and subfactors
Let $M_1$ and $M_2$ be von Neumann algebras acting on Hilbert spaces $H_1,H_2$ and consider $M=M_1\overline\otimes M_2$ acting on $H_1\otimes H_2$. Let $K$ be an $M$-invariant subspace (so that $P_K\...
- 759
5
votes
1
answer
239
views
Completely isometric coaction of discrete quantum group is multiplicative?
Let $\mathbb{G}$ be a compact quantum group (in the sense of Woronowicz) with discrete dual $\widehat{\mathbb{G}}$ which we view as a von Neumann algebraic locally compact quantum group (in the sense ...
- 525
3
votes
0
answers
138
views
Is there a finite depth irreducible subfactor of prime index and not group-subgroup?
Let $N \subset M$ be a finite depth unital inclusion of II$_1$ factors. By Theorem 3.2 in this paper (Bisch, 1994), if the index $|M:N|$ is integer then for any intermediate subfactor $N \subset P \...
3
votes
0
answers
111
views
Is there a non-irreducible maximal subfactor other than two-sided TLJ?
A subfactor $N \subseteq M$ is called:
irreducible if $N' \cap M = \mathbb{C}$,
maximal if for any intermediate subfactor $N \subseteq P \subseteq M$ then $P=\{N,M \}$.
The two-sided ...
5
votes
1
answer
203
views
What are the applications of the depth 2 reduction to the subfactors theory?
Let $(N \subset M)$ be an irreducible finite depth ($>2$) finite index unital inclusion of hyperfinite ${\rm II}_1$ factors, then for $n$ sufficiently large the subfactor $(N \subset M_n)$ is depth ...
7
votes
3
answers
698
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 ...
4
votes
2
answers
542
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 ...
9
votes
3
answers
409
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
0
answers
151
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
158
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 $SU(2)$....
17
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
177
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
0
answers
306
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 ...