# Is there an infinite depth irreducible finite index maximal subfactor (other than Temperley Lieb) ?

A subfactor $N \subset M$ is maximal if it admits no non-trivial intermediate subfactors $N \subset P \subset M$.

Is there an infinite depth irreducible finite index maximal subfactor (other than Temperley-Lieb $A_{\infty}$) ?

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## 1 Answer

The infinite depth subfactor coming from SU(3) at any index above 9 gives an example. Here the Q-system is $V_{(1,0)} \otimes V_{(0,1)} \cong V_{(1,1)} \oplus V_{(0,0)}$ so the only possible sub-objects are the whole thing or the trivial, so it's certainly maximal.

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I see, thank you. Maybe there are such examples for every simple Lie groups... –  Sébastien Palcoux Jun 5 '13 at 22:37
In general it'd probably be a bit tricky to work out whether the subfactor is maximal. But for SU(n) the same argument should work. –  Noah Snyder Jun 5 '13 at 22:47
In my opinion, it would be a nice work to exhibit every infinite depth irreducible finite index maximal subfactors coming from the simple Lie groups. –  Sébastien Palcoux Jun 12 '13 at 19:56