Can the minimal index of a subfactor take all values in {4cos^2(pi/n);n=3,4,5,...} u [4,infinity]? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:36:13Z http://mathoverflow.net/feeds/question/39968 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39968/can-the-minimal-index-of-a-subfactor-take-all-values-in-4cos2pi-nn3-4-5 Can the minimal index of a subfactor take all values in {4cos^2(pi/n);n=3,4,5,...} u [4,infinity]? André Henriques 2010-09-25T19:12:52Z 2010-09-26T12:30:15Z <p>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 {$4\cdot \cos^2(\pi/n);n=3,4,5,...$} $\cup$ $[4,\infty]$.</p> <p>The minimal conditional expectation is the one that minimizes the value of $Ind(E)$. The <em>minimal index</em> of the subfactor is then defined to be the index of its minimal conditional expectation.</p> <p>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?</p> <hr> <p><b>Remark:</b> 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 <i>Jones index</i>. 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.<br> 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). </p> http://mathoverflow.net/questions/39968/can-the-minimal-index-of-a-subfactor-take-all-values-in-4cos2pi-nn3-4-5/39982#39982 Answer by Dave Penneys for Can the minimal index of a subfactor take all values in {4cos^2(pi/n);n=3,4,5,...} u [4,infinity]? Dave Penneys 2010-09-25T22:52:44Z 2010-09-25T22:52:44Z <p>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?</p>