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Nov 19, 2013 at 16:58 history edited Sebastien Palcoux CC BY-SA 3.0
Minor edit : I've removed the tag "functional analysis"
Sep 17, 2013 at 10:23 vote accept Sebastien Palcoux
Sep 17, 2013 at 10:23
Aug 11, 2013 at 20:23 review Close votes
Aug 12, 2013 at 1:53
Jul 13, 2013 at 17:29 comment added Sebastien Palcoux I edited all the details and ambiguities.
Jul 12, 2013 at 15:24 history edited Sebastien Palcoux CC BY-SA 3.0
I add lots of explanations.
Jul 12, 2013 at 8:49 comment added André Henriques What invariant does one use to distinguish those uncountably non-isomorphic index 6 subfactors? Using that invariant, you'll be able to sharpen you question and ask for an equivalence relation on subfactors that preserves that invariant.
Jul 12, 2013 at 7:55 comment added Sebastien Palcoux Ok @MTS I will post some precisions.
Jul 11, 2013 at 20:57 comment added André Henriques Sebasitien: are you looking for a different equivalence relation than "have the same planar algebra"?
Jul 11, 2013 at 20:49 comment added MTS Sebastian, I didn't vote your question down, but it could use some more explanation. How exactly is your class of subfactors defined? How do you "easily" define your equivalence $\tilde_1$? What properties do you want from your equivalence? Do you have examples? I feel that this question is too open-ended to get a good answer, as phrased.
Jul 11, 2013 at 19:47 comment added Sebastien Palcoux Once again, a seemingly non-justified downvote, without comment...
Jul 11, 2013 at 18:48 history edited Sebastien Palcoux CC BY-SA 3.0
I replace "isomorphism" by "equivalence".
Jul 11, 2013 at 16:20 answer added Sebastien Palcoux timeline score: -1
Jul 11, 2013 at 16:19 history edited Sebastien Palcoux CC BY-SA 3.0
I migrate the final part to an attempt of answer
Jul 11, 2013 at 14:39 history asked Sebastien Palcoux CC BY-SA 3.0