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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 subfactors : $N_1 \otimes N_2$, $M_1 \otimes M_2$, $N_1 \otimes M_2$ and $M_1 \otimes N_2$, but what about the non-obvious ?

For example, the tensor product of $(R^{\mathbb{Z}/p\mathbb{Z}} \subset R)$ and $(R^{\mathbb{Z}/q\mathbb{Z}} \subset R)$, with $p$ and $q$ prime numbers, admits (at least) one non-obvious intermediate subfactor if and only if $p=q$.

So in general we could speculate that there is (at least) one non-obvious intermediate subfactor if and only if the initial subfactors are isomorphic, but it's false if they are 2-supertransitive and of index > 2, because in this case, even if they are isomorphic, there is no room for a single one non-obvious.
(see Watatani (1996) prop5.1 p329)

What are the possible cases about non-obvious for the tensor product of two maximal subfactors ?


Remark : the intermediate subfactor lattices $\mathcal{L}(R^{\mathbb{Z}/p\mathbb{Z}} \otimes R^{\mathbb{Z}/p\mathbb{Z}} \subset R \otimes R)$ are different for each $p$.
Theorem (Lukacs-Palfy 1986): Let $G$ be a finite abelian group, $H$ a group. If the subgroups lattice of $G \times G$ and $H \times H$ are isomorphic then $G \simeq H$.

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I think this question is quite interesting. I thought about it for an hour or two, but wasn't able to prove any good results. Certainly the "generic" situation is that there's no other intermediates, but there might be a lot of interesting non-generic cases. –  Noah Snyder Feb 16 at 20:51
    
@NoahSnyder: For the time being, my conjectural answer would be that there is no other intermediates iff the maximal subfactors are not isomorphic, or, are isomorphic and $2$-supertransitive. I want to test this conjecture on the group-subgroup subfactor of $(\mathbb{Z}_2 \subset D_{10})$, which is maximal and not $2$-supertransitive. I have to check if its tensor product with itself admits an other intermediate. –  Sébastien Palcoux Feb 16 at 21:36
    
@NoahSnyder : there is no other intermediate, because: let $(g_1,g_2) \in D_{10} \times D_{10}$ such that $(g_1,g_2) \not\in \mathbb{Z}_2 \times D_{10}$ or $D_{10} \times \mathbb{Z}_2$, and let $K= \langle \mathbb{Z}_2 \times \mathbb{Z}_2 , (g_1,g_2) \rangle$, then $\exists (h_1,h_2) \in K$ such that $\langle h_i \rangle = \mathbb{Z}_5$. But through the action of $\mathbb{Z}_2$, $(h^{-1}_1,h_2), (h_1,h^{-1}_2) \in K$, so $(h_1^2,e),(e,h_2^2) \in K$, but $\langle h^2_i \rangle = \mathbb{Z}_5$, so $K=D_{10} \times D_{10}$. Conclusion, my conjecture is false. –  Sébastien Palcoux Feb 16 at 22:19
    
An update of my conjectural answer would be that there is an other intermediate iff the maximal subfactors are isomorphic and depth $2$. –  Sébastien Palcoux Feb 16 at 22:41
    
I would be shocked if you could get an iff statement along those lines. Generically there's no more intermediates, but there might be many very different special ways to get intermediates. Can you even prove that when the two subfactors are not isomorphic there can't be other intermediates? It's not even clear to me that the indices need to be the same! –  Noah Snyder Feb 16 at 22:43
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3 Answers 3

up vote 2 down vote accepted

I report here an answer of Feng Xu :

Conjecture : The tensor product of two irreducible maximal subfactors, admits a non-obvious intermediate subfactor $\Leftrightarrow$ the subfactors are isomorphic and depth $2$.

Answer of Feng Xu: << The conjecture is true and it is a consequence from my results in
On a problem about tensor products of subfactors.
Briefly here is an arugment: Let $P$ be such non-trivial intermediate subfactor, then the roof of $P$ as in (2) of Prop. 3.5 is necessarily of depth $2$ and by maximality is the original subfator, and by Prop. 3.10 these two Kac algebra and hence the corresponding subfactors are isomorphic. In fact the Kac algebras are very simple (def. 3.6). >>

Remark : this nice answer of Feng Xu proves $(\Rightarrow)$.
$(\Leftarrow)$ is clear : let $\mathbb{A}$ be a maximal Kac algebra and $x \not\in \mathbb{C}1$.
Let $\langle x \rangle$ be the left coideal generated by $x$, then by maximalilty $\langle x \rangle = \mathbb{A}$.
So $\mathbb{A} \simeq \langle x \otimes x \rangle \not\in \{ \mathbb{C}\otimes \mathbb{C}, \mathbb{C}\otimes \mathbb{A} , \mathbb{A}\otimes \mathbb{C} , \mathbb{A}\otimes \mathbb{A} \} $, i.e. a non-obvious left coideal of $\mathbb{A}\otimes \mathbb{A}$.

A next step could be to prove whether or not non-trivial maximal Kac algebras exist, and if yes, to understand how compute the left coideal lattice of their tensor square.

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Partial answer : for the group-subgroup subfactors $(R^G \subset R^H)$

Theorem: Let $(H_i \subset G_i)$ be core-free maximal inclusions of groups, then $(H_1 \times H_2 \subset G_1 \times G_2)$ admits a non-obvious intermediate subgroup iff $G_1 \simeq G_2 \simeq \mathbb{Z}_p$.
Proof : see this answer. $\square$

Corollary: The tensor product of two group-subgroup maximal subfactors admits a non-obvious intermediate iff the subfactors are isomorphic and depth $2$.

Proof : if $K \subset H$ is a normal subgroup of $G$ then $(R^G \subset R^H) \simeq (R^{G/K} \subset R^{H/K})$.
So we can restrict to the group-subgroup subfactors $(R^G \subset R^H)$ with $H$ a core-free subgroup of $G$.
But $(R^{G_1} \otimes R^{G_2} \subset R^{H_1} \otimes R^{H_2}) \simeq (R^{G_1 \times G_2} \subset R^{H_1 \times H_2})$, and $(R^G \subset R^H)$ is maximal iff $(H \subset G)$ is maximal, by the Galois correspondence, which proves the result by the theorem. $\square$


Problem : Is the corollary true for all the (irreducible) maximal subfactors ?

Remark : It's also true for the $2$-supertransitive subfactors thanks to the result of Watatani cited above.
Now what's about if at least one of them is not $2$-supertransitive ?
And what's about $(R^{G_1} \otimes R\rtimes{H_2} \subset R^{H_1} \otimes R \rtimes {G_2}) $ ?

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Direct proof (to be completed) generalizing this argument of groups theory :

Let $(N_i \subset M_i)$ be irreducible maximal subfactors.
Let $P$ an intermediate subfactor: $N_1 \otimes N_2 \subset P \subset M_1 \otimes M_2$.

Let $P^1 = \{ x_1 \in M_1 \text{ such that } \exists x_2 \in M_2 \text{ with } x_1 \otimes x_2 \in P \}''$,
and $P_1 = \{ x_1 \in M_1 \text{ such that } x_1 \otimes N_2 \subset P \}''$
Idem, we define $P^2$ and $P_2$.

($\star$) To be proved: $(P_i \subset P^i)$ is depth $2$.

Then $(P_i \subset P^i) \simeq (R \subset R \rtimes \mathbb{A}_i)$ with $ \mathbb{A}_i$ a Kac algebra.
Let a $W^*$-isomorphism $\psi_i : R \rtimes \mathbb{A}_i \to P^i $ with $\psi_i(R) = P_i $.
Let $\phi : \mathbb{A}_1 \to \mathbb{A}_2$ with $\phi(a_1) = a_2$ such that $\psi_1(a_1) \otimes \psi_2(a_2) \in P$

($\star$) To be proved: $\phi$ is a well-defined isomorphism of Kac algebras.
Then $(P_1 \subset P^1) \simeq (P_2 \subset P^2)$.

Now $N_i \subset P_i \subset P^i \subset M_i$, so by maximality: $P_i, P^i \in \{N_i , M_i \}$.

If $(N_1 \subset M_1)$ is depth $>2$ then $P_1=P^1=N_1$ or $M_1$, because $(P_1 \subset P^1)$ is depth $2$,
and $P_2=P^2=N_2$ or $M_2$ because $(P_1 \subset P^1) \simeq (P_2 \subset P^2)$.

($\star$) To be proved: $P = \{ x_1 \otimes x_2 \in P \text{ such that } x_1 \in M_1 \text{ and } x_2 \in M_2 \}''$
But $x_1 \otimes x_2 \in P$ implies $x_1 \otimes N_2$, $N_1 \otimes x_2 \subset P$, because $P_i=P^i$.
So $P = P_1 \otimes P_2 \in \{N_1 \otimes N_2 , N_1 \otimes M_2 , M_1 \otimes N_2 , M_1 \otimes M_2 \} $

If $(N_2 \subset M_2)$ is depth $>2$, idem...

If $(N_1 \subset M_1)$ and $ (N_2 \subset M_2)$ are depth $2$, but are not isomorphic, idem...

Else $(N_1 \subset M_1) \simeq (N_2 \subset M_2) \simeq (R \subset R \rtimes \mathbb{A})$ with $\mathbb{A}$ a maximal Kac algebra.
Let $a \not\in \mathbb{C}1$ and $\langle a \rangle$ the left coideal generated by $a$, then by maximalilty $\langle a \rangle = \mathbb{A}$.
So $\mathbb{A} \simeq \langle a \otimes a \rangle \not\in \{ \mathbb{C}\otimes \mathbb{C}, \mathbb{C}\otimes \mathbb{A} , \mathbb{A}\otimes \mathbb{C} , \mathbb{A}\otimes \mathbb{A} \} $, i.e. a non-obvious left coideal of $\mathbb{A}\otimes \mathbb{A}$.
The result follows by Galois correspondence. $\square$


Corollary of the proof : the lattice of intermediate subfactors of the tensor product of finitely many irreducible subfactors $(N_i \subset M_i)_{i}$ is the direct product of the lattices $\mathcal{L}(N_i \subset M_i)$ of each subfactors if and only if they are pairwise without isomorphic maximal intermediate depth $2$ inclusions.
Proof : the result follows by using $(P_i \subset P^i)$ as above and induction.

Corollary : the tensor product of finitely many irreducible cyclic subfactors is a cyclic subfactor if and only if they are pairwise without isomorphic maximal intermediate depth $2$ inclusions.
Proof : the direct product of distributive lattices is also distributive.

Remark: In particular, the tensor product of finitely many irreducible maximal subfactors is a cyclic subfactor if and only if they don't contain two isomorphic subfactors of depth $2$.

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