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Let $(N \subset M)$ be an irreducible finite index 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. $\square$

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  • $\begingroup$ See Sano-Watatani, Angles between two subfactors, Remark and Corollary 6.1 p230. $\endgroup$ – Sebastien Palcoux Feb 12 '17 at 21:15
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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 $n$.

In general, we have the following result:

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.

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  • $\begingroup$ 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. $\endgroup$ – Sebastien Palcoux Sep 14 '16 at 0:18

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