13
$\begingroup$

Given a subfactor $N\to M$ and a conditional expectation $E:M\to N$, there is a numerical invariant $Ind(E)$ associated to to this situation, called the index of $E$. The possible values of $Ind(E)$ are restricted to the set {$4\cdot \cos^2(\pi/n);n=3,4,5,...$} $\cup$ $[4,\infty]$.

The minimal conditional expectation is the one that minimizes the value of $Ind(E)$. The minimal index of the subfactor is then defined to be the index of its minimal conditional expectation.

Can the minimal index take all values in {$4\cdot \cos^2(\pi/n);n=3,4,5,...$} $\cup$ $[4,\infty]$? In other words, given a real number in the above set, is there a subfactor whose minimal index is that real number?

Remark: If the factors are of type $II_1$, there is another preferred conditional expectation: the one that is compatible with the traces. The corresponding index is called the Jones index. This is not the index I care about. Jones' index agrees with the minimal index in the case of irreducible subfactors, but not in general.
Jones' index is known to take all the above values. But the subfactors used in the construction are not irreducible (and one can also check that their minimal index is different from their Jones index).

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

There is an irreducible Temperley-Lieb subfactor at every allowed index. For $n\geq 3$, it has index $4\cos^2(\pi/n)$ and principal graph $A_{n-1}$ (in fact all subfactors of index less than $4$ are irreducible), and for every $r\geq 4$, it has index $r$ and principal graph $A_\infty$. Doesn't that do the job by your remark?

Improve this answer
$\endgroup$
3
  • $\begingroup$ Very good. Could you either describe the construction a little bit, or point to some references? Is the construction going to produce a planar algebra first, and claim that it's one coming from a subfactor? $\endgroup$
    – André Henriques
    Commented Sep 26, 2010 at 15:29
  • $\begingroup$ These were first constructed in Popa's "Markov traces on universal Jones algebras and subfactors of finite index" MR1198815. This construction has later been interpreted in terms of planar algebras, see arxiv.org/abs/0807.4146 and the papers referenced therein. The factor M in this case is a free group factor (you might as well take it to be L(F_\infty)). Probably you want a type III construction, I don't know enough about structure theory for factors to figure out if you can replace L(F_\infty) by something type III. $\endgroup$
    – Noah Snyder
    Commented Sep 26, 2010 at 16:34
  • $\begingroup$ Every standard invariant that arises from a finite index type $II_1$ subfactor also arises as the standard invariant of a type $III$ subfactor, and vice versa. See Izumi's paper "On type II and type III principal graphs of subfactors." $\endgroup$
    – Dave Penneys
    Commented Sep 26, 2010 at 20:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .