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Let $N \subset M$ be an irreducible finite depth and finite index subfactor.

$M$ is a completely reducible (algebraic) $N$-$N$ bimodule, it decomposes into irreducibles as follows : $$M=\bigoplus_{i \in I} V_i \otimes M^{(i)}$$ with $(M^{(i)})_{i \in I}$ the irreducible bimodules, $M^{(0)}=N$, and $(V_i)_{i \in I}$ the multiplicity spaces.

Definition : such a subfactor has property $C$ if : $\exists i \in I$, $ \exists v \in V_i$ such that $\langle N, v \otimes M^{(i)} \rangle=M$

Question : how translate this property into the planar algebra framework ?

Remarks : $M^{(i)}$ is an irreducible $N$-$N$ bimodule, so $\forall x \in M^{(i)}$, $\langle N, v \otimes x \rangle=\langle N, v \otimes M^{(i)} \rangle$.
I suspect that property $C$ characterizes what I call the cyclic subfactors.

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    $\begingroup$ So if I understand you correctly, your property means that there is an irreducible submodule ${}_NH_N$ of ${}_NL^2(M)_N$ such that every other such submodule is contained in ${}_NH_N\boxtimes_N \cdots \boxtimes_N {}_NH_N$ ? $\endgroup$ Dec 11, 2013 at 17:31
  • $\begingroup$ @MarcelBischoff : thank you for your comment, it's a nice idea for trying to reformulate this property. Do you suspect something like : if $S \subset M$ is a sub-$N$-$N$-bimodule of $M$ then $S.N.S$ is the bimodule $S \boxtimes_N S$ ? If not, how do you prove this reformulation ? I'm troubled because the multiplicity spaces $V_i$ have finite dimension, and $\forall n (S.N)^n.S \subset M$, but the dimension of the multiplicity spaces of $S^{\boxtimes_N n}$ are in general not bounded in $n$, so I don't understand. $\endgroup$ Dec 11, 2013 at 20:00
  • $\begingroup$ You see $M$ as an algebraic bimodule I don't know if this spoil the argument. I was actually thinking again in the type III case. $M=\langle N,\Psi_1,\ldots,\Psi_n\rangle$, where $\Psi_i$ are isometries corresponding to the irreducible submodules. Then your statement looks like $M=\langle N,\Psi_i\rangle$ but that means that all other $\Psi_j$ can be obtained by $\Psi_i$ and $N$, this should correspond to the above criterion, though I did not check it. In particular this means trivially there is no proper intermediate $N\subset \langle N, \Psi_i\rangle \subset R \subset M$. $\endgroup$ Dec 11, 2013 at 20:31
  • $\begingroup$ @MarcelBischoff: for a group subfactor, your condition is necessary for the group has a simple representation category (i.e. no non-trivial subcategory). A non-abelian simple group $G$ checks this condition. $G$ doesn't admit non-trivial normal subgroups, but it admits non-trivial subgroups, so the subfactor admits non-trivial intermediate. About my condition, an intermediate subfactor $P$ gives a sub-$N$-$N$-bimodule, but for the non-abelian simple group subfactor, a non-trivial irreducible sub-$N$-$N$-bimodule of $P$ generates all the other irreducibles (by simplicity), but $P\subsetneq M$. $\endgroup$ Dec 11, 2013 at 20:52
  • $\begingroup$ So I was also just thinking my first criterion can just be necessary but don't need to be sufficient, as you just showed, I guess. $\endgroup$ Dec 11, 2013 at 21:20

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