# 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 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 $G \simeq H$.

• 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 '14 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. – Sebastien Palcoux Feb 16 '14 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. – Sebastien Palcoux Feb 16 '14 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 $2$. – Sebastien Palcoux Feb 16 '14 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 '14 at 22:43

This answer came after a discussion with Feng Xu. The following more general result is true:

Theorem: Let $(N_i \subset M_i)$, $i=1,2$, be irreducible finite index subfactors. Then $$\mathcal{L}(N_1 \subset M_1) \times \mathcal{L}(N_2 \subset M_2) \subsetneq \mathcal{L}(N_1 \otimes N_2 \subset M_1 \otimes M_2)$$ if and only if they are intermediate subfactors $N_i \subseteq P_i \subset Q_i \subseteq M_i$, $i=1,2$, such that $(P_i \subset Q_i)$ is depth $2$ and isomorphic to $(R^{\mathbb{A}_i} \subset R)$, with $\mathbb{A}_2 \simeq \mathbb{A}_1^{cop}$ which is the Kac algebra $\mathbb{A}_1$ with the opposite coproduct.

Proof: This theorem was proved in the $2$-supertransitive case by Y. Watatani. The general case was conjectured by the OP, and proved by a discussion with Feng Xu as follows:
Let the intermediate subfactors $$N_1 \otimes N_2 \subseteq P_1 \otimes P_2 \subset R \subset Q_1 \otimes Q_2 \subseteq M_1 \otimes M_2$$ with $R$ not of tensor product form, $P_1 \otimes P_2$ and $Q_1 \otimes Q_2$ the closest (below and above resp.) to $R$ among them of tensor product form. Now using Proposition 3.5 (2) of [xu], $(P_i \subseteq Q_i)$, $i=1,2$, are depth $2$, there corresponding Kac algebras, $\mathbb{A}_i$, $i=1,2$, are very simple and $\mathbb{A}_2 \simeq \mathbb{A}_1^{cop}$ (see Definition 3.6 and Proposition 3.10 of [xu]). The converse is given by Theorem 3.14 (2) of [xu].

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 $(R^{G_1} \otimes R\rtimes{H_2} \subset R^{H_1} \otimes R \rtimes {G_2})$ ?

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 $2$.

• "isomorphic" should be replaced by "cop-isomorphic" (see Theorem 3.14 of Feng Xu's paper). – Sebastien Palcoux Jun 5 '16 at 9:08