8

Background: See Noah and Emily's posts on subfactors and planar algebras on the Secret Blogging Seminar.

There are plenty of examples of $3$-super-transitive (3-ST) subfactors; Haagerup, $S_4 < S_5$, and others. There's exactly one known example of a $5$-ST subfactor, the Haagerup-Asaeda subfactor, and one $7$-ST subfactor, the extended Haagerup subfactor.

Below index $4$ there are the $A_n$ and $D_n$ families, which are arbitrarily super-transitive. Ignore those; I'm just interested above index $4$.

Is there anything that's even more super-transitive?

Sebastien Palcoux
  • 25,922
Scott Morrison
  • 7,540
  • 4
    (Aside: I'm not asking this in the expectation that anyone will have an immediate answer --- if you have one, you should answer here and then hurry off and submit to a journal!)

    I'm mostly asking here to record a bet with Emily Peters: if such exists, I owe her a bottle of champagne. If you can prove that none exist, she owes me a bottle of champagne. I win by default on my 60th birthday.

     – Scott Morrison Oct 07 '09 at 20:39
  • Are the group-subgroup subfactors known to be less than $7$-supertransitive? Else by exploring the maximal inclusions of groups $(H \subset G)$ (and so the primitive permutation groups, with GAP), perhaps we can find such a subfactor more than $7$-supertransitive. – Sebastien Palcoux Mar 02 '14 at 18:25
  • 1
    A subgroup subfactor can be at most 3-supertransitive. The 4-box space is at least 15 dimensional. This can be seen by looking at the partition planar algebra (MR2972458). – Dave Penneys Mar 02 '14 at 22:47
  • I see, thanks Dave. In analogy with the Mathieu groups, we could expected the finiteness of finite depth, index $>4$ and (at least) $4$-supertransitive subfactors. – Sebastien Palcoux Mar 03 '14 at 01:32
  • The partition planar algebras are the $(S_{n-1} \subset S_n)$ subfactor planar algebras. Perhaps the $3$-$ST$ finite group-subgroups subfactors are (up to equiv. and dual) limited to the infinite families $(S_{n-1} \subset S_{n})$ and $(A_{n-1} \subset A_{n})$ plus finitely many others possibilities, perhaps coming from the Mathieu groups. In general, perhaps the finite depth finite index irreducible $3$-$ST$ subfactors are limited to some well-known infinite families plus finitely many sporadic examples. Idem for $2$-$ST$ (much harder) and for maximal $1$-$ST$ (much much harder). – Sebastien Palcoux Mar 03 '14 at 12:00
  • 1
    @DavePenneys: thanks to the paper you cite (here p13-14) if I'm not mistaken, for $k=1,2$ or $3$, $(R^G \subset R^H)$ is $k$-supertransitive iff the action of $G$ on $X=G/H$ is $k$-transitive, and so iff $G/H_G$ is a $k$-transitive group (of course it's false for $k > 3$). So the $3$-$ST$ group-subgroup subfactors are given by the $3$-$T$ groups (classification?). Also, if $G/H_G$ is a $2$-$T$ group then $(R^G \subset R^H)$ is $2$-$ST$, so maximal, so is $(H \subset G)$ and then $G/H_G$ is a primitive group. – Sebastien Palcoux Mar 03 '14 at 18:15
  • @DavePenneys: warning $(R^{S_3} \subset R^{S_2})$ is $4$-$ST$ (but of course, what you've written is true if $[G:H] \ge 4$). – Sebastien Palcoux Mar 03 '14 at 18:49
  • See the write-up Classification of triply transitive finite groups and List of 2-transitive permutation groups of Jack Schmidt. – Sebastien Palcoux Mar 04 '14 at 11:38

0 Answers0