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A subfactor $N \subset M $ is irreducible if $N' \cap M = \mathbb{C} $.
It's maximal if it admits no non-trivial intermediate subfactors $N \subset P \subset M $.
It's cyclic if its lattice of intermediate subfactors is distributive (see here).
For "index" and "depth", see Jones-Sunder (1997) pages 29 and 91.

$\begin{array}{c|c|c|c|c|c} \text{cyclic}^{1} & \text{maximal} & \text{finite}& \text{finite}&\text{irreducible}& \text{examples}^{23} \newline & & \text{depth} & \text{index} & & \newline \hline & \times & & & \times& L(S_{\mathbb{N}^{*}}) \subset L(S_{\mathbb{N}}) \newline \hline & \times & & \times & \times & A_{\infty}\text{-subfactors} \newline \hline & \times & \times & \times& \times& \mathbb{Z}/p\mathbb{Z} , \ S_{n} \subset S_{n+1} , \text{Haagerup}\newline \hline \times & & \times & \times& \times& \mathbb{Z}/m\mathbb{Z}, A_{n}\otimes A_{n+1}\text{-subfactor} \newline \hline \times & & \times & & \times& \mathbb{Z} \newline \hline & \times & \times & & \times& \textbf{?} \end{array}$
$^{1}$ cyclic and non-maximal (obviously, maximal $\Rightarrow$ cyclic)
$^{2}$ $p \in \mathbb{P}$ , $n \ge 2$ and $ m\notin \mathbb{P}$
$^{3}$ the groups $G$ or $G_{1} \subset G_{2}$ correspond to the subfactors $R^{G} \subset R$ or $R^{G_{2}} \subset R^{G_{1}}$

Question: is there a maximal finite depth infinite index irreducible subfactor ?

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  • $\begingroup$ Perhaps an Izumi subfactor for $\mathbb{Z}$, if it exists... $\endgroup$ – Sebastien Palcoux Jun 13 '13 at 15:59

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