1
$\begingroup$

Let $(N \subset M)$ be a unital inclusion of ${\rm II}_1$ factors, with the following principal graph (called two-side TLJ) enter image description here
Question: Is $(N \subset M)$ a maximal subfactor?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

This depends on your definition of maximal. There's an intermediate von Neumann algebra which is not a factor, but no intermediate subfactors.

I'll translate this question into a question about tensor categories. The even part of the subfactor is the category of graded vector spaces $\mathrm{Vec}(\mathbb{Z})$ where I'll denote the 1-dimensional vector space in grade $n$ by $V_n$. The algebra object is just the matrix algebra $\mathrm{End}(V_0 \oplus V_1)$ whose underlying object is $(V_0 \oplus V_1)(V_0 \oplus V_1)^* = V_0^{\oplus 2} \oplus V_1 \oplus V_{-1}$. Your question is about whether this algebra has any subalgebras. Since subalgebras are always self-dual as objects, if it contains $V_1$ it would also have to contain $V_{-1}$ and from that it's pretty easy to see it's the whole thing. So the only way to get a nontrivial proper subalgebra is $V_0^{\oplus 2}$. This is obviously a subalgebra because $V_0 \otimes V_0 = V_0$ so the product of any two things in the $V_0$'s has to land back in one of the $V_0$'s. But this algebra has a nontrivial center. So it corresponds to an intermediate von Neumann algebra but not an intermediate subfactor.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ What you should have in mind here as an example of a similar flavor is $\mathbb{C} \subset M_2(\mathbb{C})$ sitting as the scalar matrices. The only intermediate $C^*$ algebra is the diagonal matrices, which has nontrivial center. This example looks exactly the same, except with gradings around that don't actually do anything important. $\endgroup$
    – Noah Snyder
    Commented May 7, 2015 at 17:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .