subfactor of finite rank but infinite index: is this possible?

Question

A subfactor $N\subset M$ is essentially the same thing as an $N$-$M$-bimodule. I'll recall the basic definitions in the language of bimodules, and I hope that subfactor people will excuse me.

Explanation: To go from a subfactor $N\subset M$ to a bimodule, consider the actions of $N$ amd $M$ on the standard form $L^2(M)$. To go the other way around, given a bimodule ${_N}H_M$, you get an inclusion $N\hookrightarrow M'$.

A subfactor $N\subset M$ is said to have finite index if the corresponding bimodule ${_N}H_M$ is dualizable. This means that there is a dual bimodule ${_M}K_N$, a unit map ${}_NL^2(N)_N\to {_N}H\ \boxtimes_M K_N$ and a counit map ${_M}K\ \boxtimes_N H_M \to {}_ML^2(M)_M$ that satisfy the usual zigzag identities $(H\to H\boxtimes K\boxtimes H \to H) = 1_H$ and $(K\to K\boxtimes H\boxtimes K \to K) = 1_K$.

A subfactor is said to have finite rank if the irreducible summands of
$H\boxtimes K\boxtimes H\boxtimes K\boxtimes H\ldots \boxtimes K$,
the irreducible direct summands of
$H\boxtimes K\boxtimes H\boxtimes K\boxtimes H\ldots \boxtimes H$,                      ($*$)
and the irreducible direct summands of
$K\boxtimes H\boxtimes K\boxtimes H\boxtimes K\ldots \boxtimes K$ lie in finitely many isomorphism classes.
(see my last comment to Makoto's answer for a disambiguation)

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Question 1: Does there exist a subfactor that is of finite rank but of infinite index?

Question 2: If I furthermore assume that all the branching multiplicites are finite, (i.e. that every ($*$) splits as a finite direct sum of irreducible bimodules), is it still possible?

Answer 1

If $G$ is the group of finite permutations of $\mathbb{N}$ and $H$ is the stabilizer subgroup of $1 \in \mathbb{N}$, the inclusion $N = LH \subset LG = M$ of left regular von Neumann algebras gives an affirmative answer to Question 2. The index of $H$ in $G$ is infinite, hence the index $[M : N]$ is also infinite.

The algebra ${\rm End}(K \boxtimes H \boxtimes \cdots \boxtimes H)$ ($k$ times $H$) is a direct sum of ${\rm End}(\ell^2 \mathbb{N}^j)$ for $j \le k$ with finite multiplicities. This is because the Jones tower associated to $N \subset M$ consists of the algebras $B(\ell^2 \mathbb{N}^k) \rtimes G$ and $(B(\ell^2 \mathbb{N}^k) \otimes \ell^\infty \mathbb{N}) \rtimes G$. Then the relative commutants are (contained in one of) $G' \cap B(\ell^2 \mathbb{N}^k) \rtimes G$. If you ignore the difficulty coming from the type II$_\infty$ situation, it is just ${\rm End}_G(\ell^2 \mathbb{N}^k)$ because $G$ is ICC. These algebras are finite dimensional by Lieberman's work [1].

[1] A. Lieberman. The structure of certain unitary representations of infinite symmetric groups. Trans. Amer. Math. Soc., 164:189–198, 1972.

Answer 2

Here is a trivial example for Question 1. Take $R\otimes 1\subset R\otimes \mathrm{B}(\ell^2(\mathbb N))$, then the only irreducible bimodules in the summands are the respective trivial ones, thus the rank is finite. The index is infinite, since the relative commutant is infinite dimensional. Similarly, for a type III factor $M$ you can take the subfactor $\rho(M)\subset M$, where $\rho =\sum_i^\infty \mathrm{Ad} v_i$, where $\sum_i^\infty v_iv_i^\ast=1$ and $v^\ast_iv_j=\delta_{ij}$.

More general, one can take a reducible subfactor with infinite relative commutant, such that the local subfactors generate the same finite set of bimodules (or a subset of it). Examples are easy to construct, start with a finite index finite depth subfactor $\rho(M)\subset M$ of type III and take $\rho_\oplus(M)\subset M$ with $\rho_\oplus=\bigoplus_{i=1}^\infty \rho$. The same construction can be adapted to a type II${}_1$ factor and yields an inclusion into a type II${}_\infty$ subfactor.