subfactor of finite rank but infinite index: is this possible? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T19:53:52Z http://mathoverflow.net/feeds/question/41327 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41327/subfactor-of-finite-rank-but-infinite-index-is-this-possible subfactor of finite rank but infinite index: is this possible? André Henriques 2010-10-06T20:42:15Z 2010-11-29T07:43:58Z <p>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. </p> <blockquote> <p><b>Explanation:</b> 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 $N\hookrightarrow M'$.</p> </blockquote> <p>A subfactor $N\subset M$ is said to have <b><i>finite index</i></b> 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 $(K\to K\boxtimes H\boxtimes K \to K) = 1_K$.</p> <p>A subfactor is said to have <b><i>finite rank</i></b> if the irreducible summands of<br> $H\boxtimes K\boxtimes H\boxtimes K\boxtimes H\ldots \boxtimes K$,<br> the irreducible direct summands of<br> $H\boxtimes K\boxtimes H\boxtimes K\boxtimes H\ldots \boxtimes H$,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;($*$)<br> and the irreducible direct summands of<br> $K\boxtimes H\boxtimes K\boxtimes H\boxtimes K\ldots \boxtimes K$ lie in finitely many isomorphism classes.<br> (see my last comment to Makoto's answer for a disambiguation)<br></p> <hr> <p><b>Question 1:</b> Does there exist a subfactor that is of finite rank but of infinite index?</p> <p><b>Question 2:</b> If I furthermore assume that all the branching multiplicites are finite, (i.e. that every ($*$) splits as a <em>finite</em> direct sum of irreducible bimodules), is it still possible?</p> http://mathoverflow.net/questions/41327/subfactor-of-finite-rank-but-infinite-index-is-this-possible/41764#41764 Answer by Makoto Yamashita for subfactor of finite rank but infinite index: is this possible? Makoto Yamashita 2010-10-11T08:57:50Z 2010-10-11T08:57:50Z <p>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.</p> <p>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].</p> <p>[1] A. Lieberman. The structure of certain unitary representations of infinite symmetric groups. Trans. Amer. Math. Soc., 164:189–198, 1972.</p>