# SF&PA – the Temperley-Lieb algebra

Hi all,

First, I’d like to thank the organizers for inviting me to post on their blog, and apologize for the low tech pictures in what follows.

As Noah mentioned, my name is Emily, I study subfactors and planar algebras, and that’s the back of my head at the top of this page (still). While Noah is taking you through the delights of subfactors sans analysis, I’ll say a few words about planar algebras to set the stage for their later appearance in subfactorland. For now, let’s leave definitions to a future post, and say a little bit about my favorite planar algebra: the Temperley-Lieb algebra.

To get a Temperley-Lieb picture, arrange $n$ points at the bottom of your page, and $n$ points at the top, and connect the points up among themselves in a non-crossing way: We only consider such pictures up to isotopy — then the number of such pictures is exactly the $n^{th}$ Catalan number (since you can, for instance, read matching parenthesizations as directions for connecting up the $2n$ points). Now, form a vector space $TL_n$ whose basis is Temperley-Lieb pictures on $2n$ points. For instance, We turn this vector space into an algebra by defining multiplication: The product of two boxes is the picture you get by stacking them: But what about that loop in the middle? It’s not part of the data of a Temperley-Lieb picture, so we have to throw it out — but let’s remember it was there by multiplying the resulting picture by $\delta$ (If there had been $k$ circles, we’d have multiplied the picture by $\delta^k$).

If you enjoy multiplying Temperley-Lieb pictures, try this fun exercise: show that Temperley-Lieb is multiplicatively generated by elements $e_i$, which consist of $n-2$ through strings and a cup and a cap starting at the $i^{th}$ string: and satisfy the relations $e_i^2 = \delta e_i$, $e_i e_j = e_j e_i$ if $|i-j|>1$ and $e_i e_{i \pm 1} e_i = e_i$ (hmm, don’t those last two relations sort of remind you of the braid group?)

One of the reasons we subfactoralists (subfactorers?) like Temperley-Lieb is that it has a lot of structure to it. For instance, we can define an involution $^*$ on $TL_n$ by horizontal reflection: So, for example: and we can also define a trace by connecting the top points to the bottom points — the result is some number of loops in a $TL_0$ diagram, ie a power of $\delta$: We call this a trace because it doesn’t care about the order of multiplication (just slide the bottom picture along the strings until it ends up on top).

This combination of a trace and an involution is pretty powerful, as it lets us define a bilinear form $\left< x, y \right> := \text{tr}(y^* x)$ on $TL_n$. Here’s a hard one for you: For which values of $\delta$ is this form positive definite?

Maybe that’s a good place to stop for now. Coming soon: why is Temperley-Lieb a planar algebra, instead of a just plain algebra?

# SF&PA: An example

Alright, let’s try to build a bi-oidal category with a good theory of duals. The dumbest possible way to do this is to start with a good monoidal category and put in meaningless labels by hand to make it bi-oidal. To be concrete, let’s consider the group S_3 and its category of unitary representations. The irreps of S_3 are the trivial irrep $V^1$, the sign irrep $V^{-1}$, and the standard rep $V^2$.

Let ${}_A X_B = V^2$ where A and B are meaningless labels. Note that ${}_B X^*_A = V^2$ since the standard rep is self-dual. Let’s see what bi-oidal category they tensor generate. Well, ${}_A X_B \otimes_B {}_B X^*_A$ should mean take $V^2 \otimes V^2$ and think of that as an A-A object (again the labels here are totally meaningless). Note that $V^2 \otimes V^2 \cong V^1 \oplus V^{-1} \oplus V^2.$ Continuing on in this fashion we get a bi-oidal category whose A-A, A-B, B-A, and B-B sectors are all just a copy of the representation theory of S_3, and all tensor products are just the tensor products as S_3 representations.

Now let’s try doing this same process with the dihedral group $D_{2 \cdot 4}$ again starting with X being the two dimensional irrep. An easy computation shows that you end up with a slightly different picture. The A-A and B-B sectors consist of direct sums of the 1-dimensional representations, while the A-B and B-A sectors consist only of direct sums of the 2-dimensional representation. This is also a subfactor category.

You can go through the same process with your favorite group and your favorite representation and get a new subfactor category.

(Warning! This construction is not what’s called a group subfactor, which I’ll get to soon.)

# SF&PA: What is a Subfactor?

By a subfactor I will mean a pair of rings A < B, such that (and I apologize in advance if I mess this up, as I said I don’t know any analysis):

• A and B are both Von Neumann algebras
• A and B have trivial centers (that is they are factors)
• A and B are both II_1-factors (that is they have a unique normalized trace)
• A<B has finite index (perhaps I’ll define this later)
• A<B is irreducible (that is the centralizer of A in B is trivial)

Whew! What on earth does all of this mean? And why would a representation theorist care about these conditions?

Well, as we saw before in the Galois theory example, given a pair A<B we get a bi-oidal category C(A<B) which is tensor generated by ${}_A B_B$ and ${}_B B_A$. It turns out that you can use the above conditions to prove a lot of nice algebraic properties about C(A<B). Furthermore, due to a converse theorem of Ocneanu Popa’s given a suitable category C(A<B) it must come from a subfactor. With this (hard!) theorem in hand we can happily ignore the subfactor setup and instead just think about categories that look like C(A<B).

Furthermore, if category theory scares you, tomorrow we’ll be doing the first thing you should do when someone tries to talk to you about categories: translate everything into pictures! At that point you’ll be able to forget both the Subfactor and the category and just start drawing pictures.

# SF&PA: How complicated are groups?

One more warmup post before I get to actual Subfactors. If you were asked to rank finite groups in order of how complicated they were, what measurement would you use?

There are three candidates that come to mind quickly, and a fourth which is a little subtler.

# Subfactors and Planar Algebras: Galois Theory

Suppose you were to run across two fields K < L. In a blatant attempt at foreshadowing instead of a field extension, we’ll refer to this pair as a “subfield.” Further let’s suppose that this “subfield” has “finite index”, namely that [L:K] is finite. As a representation theorist how might you go about studying this “subfield”?

# Subfactors and Planar Algebras: The Series

This is the first post in a many part series of posts on Subfactors and Planar Algebras. The study of Subfactors is a topic in the theory of Von Neumann algebras which at first might seem very far from our blog’s usual topics. However, it turns out to be intimately related to quantum algebra and higher category theory. In fact, at Berkeley almost everyone who studies quantum groups or related topics also learns at least the rudiments of Subfactors (due largely to Vaughan Jones’s excellent Subfactor seminar). Since algebraists outside of Berkeley don’t seem to get as much exposure to this beautiful topic, hopefully all you out there in internet land will enjoy learning a little bit about it. I’ll be taking a very indeosyncratic representation theory approach to the topic, and so it should be accessible to people who (like me) know very little analysis.

I’m also very happy to welcome our friend and colleague Emily Peters who will be guest blogging on this topic. She’s a 5th year graduate student at Berkeley working with Vaughan Jones. Emily’s main research is on the Haagerup subfactor, which makes her far more expert on subfactors than I. She’s collaborating with Scott Morrison and me on a knot theory/subfactor topic which we may have more to say about in the future. Perhaps most importantly the back of Emily’s head features prominently on an certain illustrious math blog.

I hope to put up the a few posts with mathematical content on this topic up in the next few days.