# Existence of a third intermediate if there are two intermediate subfactors of index 2

Let $$(N \subset M)$$ be an irreducible finite index unital inclusion of hyperfinite $${\rm II}_1$$ factors.
Let $$K_1$$ and $$K_2$$ be two distinct intermediate subfactors $$N \subset K_i \subset M$$, such that $$|M:K_i| = 2$$.

Question: Is there (at least) a third intermediate subfactor strictly between $$K_1 \cap K_2$$ and $$M$$?

Remark: It is true for the finite group-subgroup subfactors $$(R \rtimes H \subset R \rtimes G)$$ and $$(R^G \subset R^H)$$.
Proof using Galois correspondence: Let $$H_1$$, $$H_2$$ be two intermediate subgroups $$H \subset H_i \subset G$$.

• First, if $$|G:H_i| = 2$$ then $$H_i$$ is a normal subgroup of $$G$$. It follows that $$L=H_1 \cap H_2$$ is also normal and $$[L,G] \simeq [1,G/L]$$. If there is not a third intermediate, then $$[1,G/L]$$ is boolean, so by Ore's theorem $$G/L$$ is cyclic, but with two subgroups of index $$2$$, contradiction.
• Next if $$|H_i:H| = 2$$ then $$H$$ is a normal subgroup of $$H_i$$. It follows that $$H$$ is also a normal subgroup of $$T=\langle H_1 , H_2 \rangle$$. If there is no third, then by the same argument $$T/H$$ is cyclic with two subgroups of order $$2$$, contradiction. $$\square$$
• See Sano-Watatani, Angles between two subfactors, Remark and Corollary 6.1 p230. Commented Feb 12, 2017 at 21:15

The answer is yes. Moreover, $K_1\cap K_2 \subset M$ is a dihedral group subfactor, so the lattice of intermediate subfactors between $K_1 \cap K_2$ and $M$ is clear.
Proof: Let us look at the dual lattice. Suppose $\hat{K_1},\hat{K_2}$ are index two intermediate subfactors of $M\subset \hat{N}$. Let the $e+p_1, e+p_2$ be the corresponding biprojections. The biprojection $q$ generated by the two projections is determined by the coproduct of $p_1$ and $p_2$, see Theorem 4.8 in this paper. Note that $p_1, p_2$ have trace 1, so their coproduct is the same as their fusion rule. We know that the quotient of $Z_2*Z_2$ is a dihedral group. That means $(p_1\otimes p_2)^{\otimes n} \sim 1$ for some smallest $n$. So the biprojection $q$ gives a dihedral group subfactor with index $2n$. In particular, $p_1*p_2$ generates a third biprojection with index $n$.
Suppose $P_1,P_2$ are intermediate subfactors of a finite index irreducible subfactor $N \subset M$, and $P$ is the intermediate subfactor generated by $P_1$ and $P_2$. If $N\subset P_1, N\subset P_2$ are group crossed product, then $N\subset P$ is also a group crossed product. If $N\subset P_1, N\subset P_2$ are depth 2, i. e. Kac algebra crossed product, then $N\subset P$ is also depth 2, see Theorem 4.9 in this paper.
• Using your result that $(N \subset P_i)$ is group crossed product implies that $(N \subset P_1 \vee P_2)$ is also a group crossed product, then we can directly see the existence of a third intermediate as follows: suppose there is no third, then by Galois correspondence, the group has a distributive subgroup lattice (in fact boolean of rank $2$), so by Ore's theorem, the group is cyclic, but with two subgroups of order $2$, contradiction. Of course, your characterization of the dihedral group is more precise. Commented Sep 14, 2016 at 0:18