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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)

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?

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I think the usual definition of finite depth, when dealing with infinite index, is to say that all isomorphism classes are generated by a finite iteration of the basic construction. With this definition, the outer action of a compact group on a factor would provide a positive answer to Question 1 (a depth 2 irr. infinite index subfactor). But you seem to be interested in something different, something like a fusion category where objects are allowed to have infinite dimension, is that right? – pasquale zito Oct 7 '10 at 9:56
@pasquale: Sorry. My use of the term "finite depth" was misguided (what term should I use then?). But you're perfectly right: I care about something like a fusion category where objects are allowed to have infinite dimension. – André Henriques Oct 7 '10 at 15:35
I think "finite rank" is the best term for what you're looking for. The "depth" of a subfactor tells you how high a power of the generating object you need to look in to find everything, while the "rank" is the number of simple objects. Even for finite index finite depth subfactors the depth and the rank are typically different, but in the finite index case you know that each depth only has finitely many new objects. – Noah Snyder Oct 7 '10 at 16:12
Thanks Noah. I've replaced "depth" by "rank". – André Henriques Oct 7 '10 at 19:50
Sorry, Andre. I assumed your notion of depth was the standard one, and posted what knew quickly in order to help. I should have more carefully considered your definition! Answer deleted. – Jon Bannon Oct 9 '10 at 0:24

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.

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Thank you Makoto. So you're saying that this is finite rank. How many isomorphism types of M-N-bimodules will I then find in $K\boxtimes\ldots\boxtimes H$? – André Henriques Oct 11 '10 at 9:48
For a fixed length, $\End((K\boxtimes H)^{\boxtimes n})$ is finite dimensional. At most $n\cdot n!$ irreducible bimodules appear. But in total there is an infinite number of irreducible bimodules (growing at least with $n$). – Steven Deprez Oct 11 '10 at 12:06
@Steven: By "in total there is an infinite number of irreducible bimodules", are you referring to the number of isomorphism classes? I care about a situation in which there is an absolute bound (independent of $n$) on the number of iso classes of bimodules. – André Henriques Oct 11 '10 at 15:54
This is exactely what I mean: for every $n$, $(K\boxtimes H)^{\boxtimes n}$ contains at least $n$ non-isomorphic irreducibles subbimodules. – Steven Deprez Oct 11 '10 at 18:02
@André: It isn't, if you meant 1) $\exist$ finite set $F$ of irred. bimod. classes ( $\forall k$, any irred. submod. of $(K\boxtimes H)^{\boxtimes k}$ is in $F$). It instead satisfies 2) $\forall k$ ($\exist$ finite set $F$ of irred. bimod. classes, s.t. any irred. submod. of $(K\boxtimes H)^{\boxtimes k}$ is in $F$). It is hard to come up with an example of inclusion $N \subset M$ of II$_1$ factors satisfying 1). At least, if it has depth 2 and has infinite index, then there are infinitely many classes of irred. submods. at the second step by Enock-Nest. – Makoto Yamashita Oct 13 '10 at 2:22

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