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Is every integer index finite depth irreducible subfactors planar algebra, the intermediate of an irreducible finite index depth $2$ subfactors planar algebra?
In other words, of the following form (called Kac-coideal): $(R^{\mathbb{J}} \subset R^{\mathbb{I}})$ with $\mathbb{A}$ a finite dimensional Kac algebra and $\mathbb{I} \subset \mathbb{J} \subset \mathbb{A}$ left coideal subalgebras.

Remark: it seems relevant because without the integer index assumption, they are weak-Kac-coideal (see here prop. 9.1.1 p 37) : an irreducible depth $d$ finite index irreducible subfactor $(N \subset M)$ is an intermediate of the depth $2$ (non nec. irreducible) subfactor $(N \subset M_{d-2})$ which is (planar algebra) equivalent to $ (R^{\mathbb{A}} \subset R)$, with $\mathbb{A}$ a finite dim. weak Kac algebra, so that $(N \subset M)$ is (planar algebra) equivalent to $(R^{\mathbb{A}} \subset R^{\mathbb{I}})$ with $\mathbb{I}$ a left coideal subalgebra of $\mathbb{A}$, thanks to the Galois correspondence.

Remark: without the "finite depth" assumption, it's false because you can take a $A_{\infty} * A_{\infty}$ subfactor planar algebra whose first component is non-integer index $\alpha > 4$ and the second $n/\alpha$ with an integer $n > 4 \alpha$, so that the whole subfactor is index $n$, but can't be Kac-coideal (because the indices of the intermediate subfactors of a Kac-coideal subfactor are integer). Moreover, I don't know whether or not every $A_n$ or $A_{\infty}$ subfactor planar algebra of integer index are Kac-coideal.

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