# Can you tell if a subfactor is finite depth by the growth rate of the standard invariant?

Let $N\subset M$ be a finite index inclusion of $II_1$ factors. To the inclusion we associate the tower of higher relative commutants

$\begin{array}{ccccccc} \mathbb{C} = N'\cap N & \subset & N'\cap M & \subset & N'\cap M_1 & \subset & \cdots \\ & & \cup & & \cup & & \\ & &\mathbb{C} = M'\cap M & \subset & M'\cap M_1 & \subset & \cdots \end{array}$

Where $M_1$ denotes the basic construction associated to $N\subset M$. We say that this inclusion is $\textit{finite depth}$ if the Bratteli diagram associated the the inclusion of the higher relative commutants has finite width (ie. The number of simple factors in the commutants is bounded).

My question is:

Is it possible to determine if a subfactor is finite depth by the growth rate of the dimension of $N'\cap M_k$? I know that this is bounded above by the index. If it attains this bound does that imply finite depth?

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If it attains this bound, the index is a rational number. – Sebastien Palcoux Jul 10 '13 at 21:08
Interesting, do you have a reference? – Owen Sizemore Jul 10 '13 at 21:19
What do you mean by "it attains" ? Do you mean "it converges" or do you mean $\exists k$ such that $dim(N'\cap M_{k+1})/dim(N'\cap M_{k})=[M:N]$. In the second case the index is rational. – Sebastien Palcoux Jul 10 '13 at 21:28
I meant it attains as in there exists a $k$. So this implies rational index? Does it always converge? – Owen Sizemore Jul 10 '13 at 21:40
As you see, the index is rational by definitionn because $dim(N' \cap M_{k+1} )/dim(N' \cap M_{k} )$ is a rational number. Next, in general it converges to the square of the norm of the principal graph, which is the index if the graph is finite, otherwise it's $\le$ the index. For example, there is a $A_{\infty}$-subfactor for every index beyond $4$, but $\Vert A_{\infty} \Vert^{2} = 4$ – Sebastien Palcoux Jul 10 '13 at 22:07