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Extended Haagerup Exists! March 25, 2009

Posted by Scott Morrison in conferences, planar algebras, small examples, subfactors, talks.
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Following on from Noah’s post about the great Modular Categories conference last weekend in Bloomington, I’ll say a little about the talk I gave: Extended Haagerup exists!

The classification of low index, finite-depth subfactor planar algebras seems to be a difficult problem. Below index 4, there’s a wonderful ADE classification. The type A planar algebras are just Temperley-Lieb at various roots of unity (and so the same as U_q(sl_2), as long as you change the pivotal structure). The type D planar algebras (with principal graphs the Dynkin diagrams D_{2n}) were the subject of Noah’s talk at the conference, and the E_6 and E_8 planar algebras are nicely described in Stephen Bigelow’s recent paper.

But what happens as we go above index 4? In 1994 Haagerup gave a partial classification up to index 3+\sqrt{3} \equiv 4.73205. He showed that the only possible principal graphs come in two infinite families

haagerup-green eh eeh-red

and

hexagon-red ehexagon-red

(in both cases here the initial arm increases in steps of length 4) and another possibility

ha-green

This result really opened a can of worms. Which of these graphs are actually realised? (Hint, they’re nicely colour-coded :-) What about higher index? What does it all mean? Are these graphs part of some quantum analogue of the classification of finite simple groups? Read one for the answer to the first question, at least.

A few years after this classification was announced (it’s actually never been proved in print; Haagerup’s paper only goes up to index 3+\sqrt{2} \equiv 4.41421), Haagerup and Asaeda constructed the first graph in the first family (where “constructing a graph” means “constructing a subfactor or subfactor planar algebra whose principal graph is the graph”, of course), and the last graph. These graphs are now commonly referred to as the Haagerup graph, and the Asaeda-Haagerup graph. Bisch ruled out all the hexagon graphs, and Asaeda and Yasuda recently ruled out all the graphs in the first family except the first two!

This leaves just one, mysterious graph,

eh

called the “extended Haagerup” graph, and no one has known whether it really comes from a subfactor planar algebra or not. Happily, the mystery is now solved, as we (Stephen Bigelow, Emily Peters, Noah Snyder and me) recently found a construction for it. There’s no scary subfactor nonsense, lots of pretty pictures, and even some jellyfish.

We give a “generators mod relations” presentation of the planar algebra, so it’s essentially pure skein theory, and very accessible. We need a few tricks however. Proving that a set of skein theory relations are consistent (that is, they don’t collapse the planar algebra to zero) is in general very hard, and we get around this problem by embedding our planar algebra inside the “graph planar algebra”. This is exactly analogous to proving that a group described by generators and relations isn’t trivial by embedding it into a symmetric group. Proving that you can evaluate every closed diagram in our planar algebra requires some nifty skein theory, shown in the last few slides.

Our construction relies heavily on Jones’ techniques for analysing the representation theory of the annular Temperley-Lieb category, and follows almost identically Emily Peters’ recent re-construction of the Haagerup subfactor. If you want to understand more detail beyond what’s in the slides, you should definitely read her paper. The differences between the two cases are that to construct extended Haagerup, you need to solve a huge family of overdetermined quadratics in a 148475-dimensional vector space, while for Haagerup the corresponding space was only 375-dimensional, and that the skein theory requires a bit more work.

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Comments

1. Emily Peters - March 25, 2009

Nice summary, Scott, just a couple notes I’d like to add:

First, subfactors/planar algebras actually have a *pair* of graphs as their invariants: a principal and dual principal graph. The list above has one from every possible pair (the more symmetric one), but it’s worth noting that, for instance, the graph
o-o-o-o-o-o
| |
o o
is the principal graph of a subfactor, whose dual principal graph is
o-o-o-o<8=8=8 (the Haagerup principal graph).

Second, regarding the “mystery now solved” of whether extended Haagerup exists, most people who work in subfactors have believed for a while that this does exist, based on numerical evidence provided by Ikeda in “Numerical evidence for flatness of Haagerup’s connections.”

Third, regarding the skein theory for extended Haagerup: In general to prove that something is a subfactor planar algebra, you need some relations, and some way to guarantee that the relations you have are sufficient. While the relations we get for extended Haagerup are pretty similar in nature to, and epsilon harder to prove than, my relations for Haagerup — the guarantee that the relations are sufficient in extended Haagerup, based on Stephen’s jellyfish argument, is stronger, more general, and basically just awesomer than the way I did it for Haagerup.

2. Emily Peters - March 25, 2009

Ascii art fails above. The graph I wanted to draw was

o
|
o
|
o
|
o-o
|
o-o
|
o

3. Victor Ostrik - March 25, 2009

What are the fusion rules for extended Haagerup (and its dual)?

4. Scott Morrison - March 25, 2009

I don’t think we’ve calculated these yet. We have nice programs to do so, however, and you can use them too.

Download the FusionAtlas package, by typing

svn checkout http://tqft.net/svn/fusionatlas/code/package/

Start Mathematica, add the “package” directory just created to your $Path, then load the package with

<<FusionAtlas`

Now you can compute stuff! Try

FindFusionAlgebras[EvenPart[haagerupFamilyBigraph[7]]]

to find the fusion algebra for the even part! There are many many more, entirely undocumented functions. :-)

More seriously, if someone calculates them at some point, we’ll tell you, and if you tell us a good reason to know them now, we could do it faster.

5. Victor Ostrik - March 25, 2009

My reason is pure curiosity: I remember that usual Haagerup
is related with abelian group Z/3Z (invertible objects in one of
the corresponding categories form this group) and I wonder is there
something similar for extended Haagerup.

6. Scott Morrison - March 25, 2009

Oh, no, sadly nothing like this happens for extended Haagerup. In the Haagerup graph (the first one in the post above), there’s a 3-fold symmetry of the graph, so you can see that the three univalent vertices must have dimension 1. Indeed, they form a Z/3Z tensor category in the even part.

In extended Haagerup, on the other hand, the only object with dimension one is the trivial object.

7. Scott Morrison - March 25, 2009

The tensor product table is …. (blog comments are not the right medium for telling people matrices, but oh well!)

{{1, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0,
0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 0, 0}, {0, 0,
0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0,
1}}, {{0, 1, 0, 0, 0, 0, 0, 0}, {1, 1, 1, 0, 0, 0, 0, 0}, {0, 1, 1,
1, 0, 0, 0, 0}, {0, 0, 1, 1, 1, 1, 0, 0}, {0, 0, 0, 1, 1, 1, 0,
1}, {0, 0, 0, 1, 1, 1, 1, 0}, {0, 0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0,
1, 0, 0, 0}}, {{0, 0, 1, 0, 0, 0, 0, 0}, {0, 1, 1, 1, 0, 0, 0,
0}, {1, 1, 1, 1, 1, 1, 0, 0}, {0, 1, 1, 2, 2, 2, 1, 1}, {0, 0, 1, 2,
2, 2, 1, 0}, {0, 0, 1, 2, 2, 2, 0, 1}, {0, 0, 0, 1, 1, 0, 0,
0}, {0, 0, 0, 1, 0, 1, 0, 0}}, {{0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 1,
1, 1, 1, 0, 0}, {0, 1, 1, 2, 2, 2, 1, 1}, {1, 1, 2, 4, 4, 4, 1,
1}, {0, 1, 2, 4, 3, 4, 1, 1}, {0, 1, 2, 4, 4, 3, 1, 1}, {0, 0, 1, 1,
1, 1, 0, 1}, {0, 0, 1, 1, 1, 1, 1, 0}}, {{0, 0, 0, 0, 1, 0, 0,
0}, {0, 0, 0, 1, 1, 1, 1, 0}, {0, 0, 1, 2, 2, 2, 0, 1}, {0, 1, 2, 4,
3, 4, 1, 1}, {1, 1, 2, 3, 4, 3, 1, 1}, {0, 1, 2, 4, 3, 3, 1,
1}, {0, 1, 0, 1, 1, 1, 1, 0}, {0, 0, 1, 1, 1, 1, 0, 0}}, {{0, 0, 0,
0, 0, 1, 0, 0}, {0, 0, 0, 1, 1, 1, 0, 1}, {0, 0, 1, 2, 2, 2, 1,
0}, {0, 1, 2, 4, 4, 3, 1, 1}, {0, 1, 2, 4, 3, 3, 1, 1}, {1, 1, 2, 3,
3, 4, 1, 1}, {0, 0, 1, 1, 1, 1, 0, 0}, {0, 1, 0, 1, 1, 1, 0,
1}}, {{0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0,
1, 0, 1, 0, 0}, {0, 0, 1, 1, 1, 1, 0, 1}, {0, 0, 1, 1, 1, 1, 0,
0}, {0, 1, 0, 1, 1, 1, 1, 0}, {0, 0, 0, 1, 0, 0, 0, 0}, {1, 0, 0, 0,
1, 0, 0, 0}}, {{0, 0, 0, 0, 0, 0, 0, 1}, {0, 0, 0, 0, 0, 1, 0,
0}, {0, 0, 0, 1, 1, 0, 0, 0}, {0, 0, 1, 1, 1, 1, 1, 0}, {0, 1, 0, 1,
1, 1, 0, 1}, {0, 0, 1, 1, 1, 1, 0, 0}, {1, 0, 0, 0, 0, 1, 0,
0}, {0, 0, 0, 1, 0, 0, 0, 0}}

8. Noah Snyder - March 25, 2009

What Scott told you was the tensor product table for the even part of the principal graph. There’s also the even part of the dual principal graph which I’ve put below. For the full tensor product rules for the entire subfactor you’d also need to know how these act on the module categories of A-B and B-A bimodules. We don’t currently have that implemented.

Here’s the even part of the dual.

{{{1, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0}, {0, 0,
0, 1, 0, 0}, {0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 1}}, {{0, 1, 0, 0,
0, 0}, {1, 1, 1, 0, 0, 0}, {0, 1, 1, 1, 0, 0}, {0, 0, 1, 1, 1,
1}, {0, 0, 0, 1, 2, 1}, {0, 0, 0, 1, 1, 0}}, {{0, 0, 1, 0, 0,
0}, {0, 1, 1, 1, 0, 0}, {1, 1, 1, 1, 1, 1}, {0, 1, 1, 2, 3, 1}, {0,
0, 1, 3, 3, 2}, {0, 0, 1, 1, 2, 1}}, {{0, 0, 0, 1, 0, 0}, {0, 0,
1, 1, 1, 1}, {0, 1, 1, 2, 3, 1}, {1, 1, 2, 4, 5, 3}, {0, 1, 3, 5,
6, 3}, {0, 1, 1, 3, 3, 2}}, {{0, 0, 0, 0, 1, 0}, {0, 0, 0, 1, 2,
1}, {0, 0, 1, 3, 3, 2}, {0, 1, 3, 5, 6, 3}, {1, 2, 3, 6, 7, 4}, {0,
1, 2, 3, 4, 2}}, {{0, 0, 0, 0, 0, 1}, {0, 0, 0, 1, 1, 0}, {0, 0,
1, 1, 2, 1}, {0, 1, 1, 3, 3, 2}, {0, 1, 2, 3, 4, 2}, {1, 0, 1, 2,
2, 1}}}

9. Noah Snyder - March 25, 2009

The way you’re meant to read these matrices is that if you want to find dim Hom(V_i (x) V_j, V_k) then you look at the ith matrix and its (j,k) entry. The numbering is to read the graph from left-to-right, and within each depth from bottom-to-top.

10. Victor Ostrik - March 25, 2009

Scott and Noah, thanks for the matrices! Actually I am not insisting
on invertible objects, any proper subcategory would be fine.
But I don’t know how extract any from your matrices..

11. Scott Morrison - March 25, 2009

At least the the principal graph, the tensor square of every even vertex includes V_2 (the first interesting even vertex), except the three univalent vertices. However the tensor square of each of the distant univalent vertices is just the last even vertex before the branch. (eg, look up the last row of the last matrix in my comment.)

Therefore every even object generates the entire category (at least assuming by proper subcategory you mean including subobjects).

Not sure about the dual graph, but presumably something similar works.

12. Chris Schommer-Pries - March 26, 2009

You said how Bisch ruled out all the hexagon graphs. Are there any (finite depth, finite index) subfactors whose principal graphs have loops?

13. Noah Snyder - March 26, 2009

Yup Chris there are plenty with loops. The simplest example is any subfactor coming from the standard rep of U_q(su(3)). The principal graph comes from looking at V(x)V*(x)V etc. and is made up of a grid of hexagons.

Another simple example with a hexagon is the group subgroup subfactor S_4<S_5 (whose principal graph has even vertices which are young diagrams of 5 boxes, odd vertices diagrams of 4 boxes, and edges given by adding or removing boxes).

14. The Jellyfish Algorithm « Secret Blogging Seminar - September 24, 2009

[...] up on the arxiv today about the extended Haagerup subfactor and its planar algebra. Scott already explained nicely the story that this paper fits into. Today I wanted to tell you about one of the key elements of [...]

15. Scott Morrison - September 24, 2009

Our paper is now out, available at http://arxiv.org/abs/0909.4099 or http://tqft.net/EH. Noah also has a blog post up (see the comment above) about the ‘jellyfish algorithm’, the key tool in showing we found enough identities to describe the extended Haagerup planar algebra by generators and relations.

16. Noah Snyder - September 24, 2009

Also the answer to Victor Ostrik’s question is nicely formatted in the appendix.


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