3
$\begingroup$

Let $(N \subset M)$ be an inclusion of ${\rm II}_1$ factors, the basic construction is $N \subset M \subset M_1 = \langle M , e^M_N \rangle$.

Question: For any intermediate subfactor $N \subset P \subset M_1$, is it true that $N \subset P \subseteq M$ or $M \subset P \subset M_1$ (i.e. no extra intermediate)?

$\endgroup$
1
$\begingroup$

This answer came by a discussion with Keshab Chandra Bakshi.
Consider the subfactor $(R \subset R \rtimes \mathbb{Z}/2)$, then $R \subset R \rtimes \mathbb{Z}/2 \subset M_2(R) = R \otimes M_2(\mathbb{C})$ is the basic construcition, but $(R \subset M_2(R))$ admits continuously many intermediate subfactors which are conjugate to $R \rtimes \mathbb{Z}/2$ by unitaries in $\mathbb{C} \otimes M_2(\mathbb{C})$. So the answer is no.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.