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Let $N \subsetneq K_i \subsetneq M$, $i=1,2$, be a square of irreducible finite index unital inclusion of hyperfinite ${\rm II}_1$ factors, such that there is no extra intermediate, with $K_1 \not \subseteq K_2$ and $K_2 \not \subseteq K_1$ (i.e. not flat square).
In other words, the lattice $\mathcal{L}(N \subset M)$ is boolean of rank $2$.

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Question: Is it true that $|M:K_1| = 2$ implies $|K_2 : N| = 2$ ?

Proposition: It is true for the finite group-subgroup subfactors $(R \rtimes H \subset R \rtimes G)$.
Proof by Galois correspondence: Let $H_1$, $H_2$ be two intermediate subgroups $H \subsetneq H_i \subsetneq G$ with $H = H_1 \cap H_2$. By the product formula: $|H_1H_2| \cdot |H_1 \cap H_2| = |H_1| \cdot |H_2|$.
It follows that $\frac{|H_1H_2|}{|H_2|} = \frac{|H_1|}{|H|} $. So $x=|H_1:H| = \frac{|H_1H_2|}{|H_2|} \le |G:H_2|=y$.
If $y=2$ then $2 \le x \le y = 2$, so $x=2$. $\square$

Remark: For $(R^G \subset R^H)$, I have no general proof, but it is true in each of the following cases:

  • $|G|<128$ (checked by GAP)
  • $G$ simple and $|G|<10^6$ (GAP)
  • $|G:H| = 2p$ with $p$ prime (trivial proof)
  • $|G:H|<32$ (GAP)
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1 Answer 1

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Yes.

If $|M:K_1| = 2$, then there is a faithful outer action of $\mathbb{Z}/2 = \langle \tau\rangle$ on $M$ such that $K_1$ is exactly the fixed-point subfactor $M^{\mathbb{Z}/2}$.

Claim: $\tau(K_2) = K_2$.
proof: $N \subsetneq K_2 \subsetneq M$, $\tau(N) = N$ and $\tau(M) = M$, so $N \subsetneq \tau(K_2) \subsetneq M$. If $\tau(K_2) \neq K_2$ then $\tau(K_2) = K_1 = M^{\mathbb{Z}/2}$, but $\tau(M^{\mathbb{Z}/2}) = M^{\mathbb{Z}/2}$, so $K_1 = \tau(K_1) = \tau(\tau(K_2)) = K_2$, contradiction with $K_1 \neq K_2$. $\square$

Now $K_2 \subset M$ but $K_2 \not \subseteq K_1 = M^{\mathbb{Z}/2}$, so under this action $K_2^{\mathbb{Z}/2} \subsetneq K_2$. Because of the no-extra intermediate assumption, $K_2^{\mathbb{Z}/2} = N$, so $|K_2:N| = 2$.

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