In both cases what happened was that I had:

• A question to which I didn’t know the answer
• An answer to which I didn’t know the question

When talking about planar algebras one often runs into the unfortunate fact that sometimes you want to think about shaded planar algebras (which correspond to 2-categories with two objects) and sometimes you want to think about unshaded planar algebras (which correspond to monoidal categories, or 2-categories with one object). A natural question is given a shaded planar algebra can you turn it into an unshaded one, and if so in how many ways? I ran across this question after we wrote this paper (with Emily and Scott) where there’s two different unshaded D_2n planar algebras but only one shaded one (basically because -1 has two different square roots). A very natural question I really wanted to know the answer to was “what happens if you look at one of the simplest examples: group subfactors.” On the other hand, for a long time I’d wondered what a Tambara-Yamagami category was (since they come up as examples all over the case in the fusion category literature).

I went to a conference in Waco (run by Eric Rowell and Deepak Naidu). Friend of the blog David Jordan was speaking about his joint work with (former mathcamper!) Eric Larson, and to make a long story short, during the talk it became clear to me that TY categories are exactly the ones that come from unshaded versions of group subfactors.

The other example concerned GHJ (Goodman, de la Harpe, Jones) subfactors (which I didn’t understand the construction of), and the question (which I didn’t know the answer to) was “what subfactors do you get from the ADE-type module categories over quantum su(2)?” The tricky thing here is that the answer to the latter question is *not* the ADE subfactors! Instead in discussions after a talk of Wenzl‘s at the 2010 Shanks workshop at Vanderbilt, it became clear to me that again these two confusions anihilated. This story is a little harder to tell in a blog post, but Dave Penneys wrote up some notes of a talk I gave about this on a retreat.

The outline of the post is as follows. First I’ll explain a standard reduction of the 4-color theorem to a question about 3-coloring edges of trivalent graphs. Second I’ll explain why 3-colorings of edges is a question about finding a positive evaluation algorithm for a certain planar algebra. Third, I’ll discuss “Euler characteristic” evaluation algorithms. Fourth I’ll explain how this technique almost answers the 4-color theorem.

Suppose you have a planar graph and you want to 4-color the faces (note, I said faces, not vertices). First notice that you can reduce to the case of 3-valent planar graphs by “adding a new country at every higher valency vertex” (see Week 22 in Mathematical Physics for a delightful ASCII drawing). This observation is from the early days of the 4-color problem and due to Cayley and Kempe. The next reduction also comes from the 19th century, and is due to Tait. There is a bijection between “4-colorings of faces” and “4-coloring one distinguished face and then 3-coloring the edges.” The way to understand this bijection is to label the edges with “rules for changing face colorings.” Namely there are 3 ways to assign a pairing of the four colors. If you’ve chosen such a rule for every edge then you can propogate your single face coloring to the whole graph. It’s a simple exercise to check that this gives a bijection between valid colorings.

The invariant of planar graphs “number of 3-colorings of edges” extends to a planar algebra/TQFT/tensor category by the recipe similar to that Ben outlined here. Any planar graph with boundary can be thought of as a functional on the space Span{labelings of the boundary}. Extending by linearity any linear combination of planar graphs with boundary can also be thought of as such a functional. Mod out by the equivalence setting two diagrams equal to each other if they give the same functional. The key fact is that these relations play well with gluing together diagrams and hence gives a planar algebra (or tensor category together with a chosen generating object).

What explicitly do these relations look like? You can remove a bigon for a multiplicative cost of 2, a triangle for a multiplicative cost of 1, and a circle with no trivalent vertices for a multiplicative cost of 3. Furthermore there is a version of the I = H relation (I strongly encourage you to check yourself that this relation between number of 3-colorings of edges holds for any coloring of the boundary):

So what do we need to do to prove the 4-color theorem? We need to give a manifestly positive algorithm for evaluating closed diagrams in this planar algebra! That’s it.

One common technique for evaluating closed planar diagrams is to concentrate on the faces. Notice above we know that we can remove all bigons and triangles. What about bigger faces? Well the beautiful thing is that by Euler characteristic arguments all you need to do is prove that you can remove squares and pentagons because any planar diagram has a pentagonal or smaller face. In fact this technique and small variations on it have been remarkably successful at answer planar algebraic questions (once for quantum groups and once for an exotic subfactor).

So let’s try that. Here’s the square (finding this formula using the above relations is kinda fun):

Now there’s just one case left, the pentagon. Just find a positive formula there and you’d be done!

What is an evaluation algorithm? Well suppose you have a bunch of diagrams (think formal sums of links) constructed out of a few generators (think crossings) modulo some relations (think the HOMFLY relations). There are two obvious questions. Are these rules enough to reduce every link to a constant times the empty diagram? Does any way of reducing a given diagram give the same answer? Usually the former question is substantially easier than the latter. Often the first question is answered by giving an explicit algorithm for simplifying diagrams.

One simple example is the Kauffman bracket description of the Jones polynomial. Here the evaluation algorithm is very easy: find a crossing and resolve it, wash, rinse, repeat. This clearly terminates because at every step you get fewer crossings.

A slightly trickier example is HOMFLY. Here you can’t just reduce the number of crossings straight away. However, every knot can be unknotted by switching a certain number of crossings (the minimal such number is called the “crossing number”). So first you find some crossing such that switching it decreases crossing number. Using the HOMFLY relation on that crossing gives a sum of two terms one of which has fewer crossings, and the other of which has the same number of crossings but smaller crossing number. Continue until your knot is unknotted.

For extended Haagerup the situation turns out to be much much trickier. Here’s the basic setup. We have a generator with 16 strands coming into it (for comparison a crossing has 4 strands coming into it). Instead of the HOMFLY relation and the Reidemeister moves we instead have much more complicated relations. Rather than giving the exact form of the relations (which you can find on pages 13 and 14 of the preprint) I’ll sketch the general form of them.

• If a generator connects back to itself then the diagram is zero.
• If you rotate a generator you get a multiple of the generator you started with
• If two generators are connected by half their strands then you can replace that with a sum of diagrams involving no generators
• If you have a generator next to some strands then you can replace it with a sum of a bunch of diagrams (possibly with lots of generators) but where all the generators are on the “far side of the strands.”

Let me elaborate a little on the last relation. In our particular case we have two “pulling through strands” relations, because of the shading. One pulls through a single strand and the other pulls through two. I’ll show you one of them.

In this image the rectangles on the bottom are “Jones-Wenzl” idempotents. These are given by a recurrence relation and correspond to the irreducible representations of quantum su_2. All you need to know about them for this post is that they’re a complicated linear combination of diagrams with no copies of the generator. What happens when you expand all of those out? Well exactly one term (the LHS when you look at all vertical strands) is blocked from the top by a strand. All the other terms are zero or have all generators at the top of the screen.

Ok, now that we have relations of this form how do we evaluate a closed diagram? It proceeds in two totally different phases.

1. Pull all the generators to the outside of the diagram
2. Once they’re all on the outside of the diagram find two of them that are connected by half their strands and reduce them

(Why does the second phase work? It’s a simple lemma about planar graphs and I wouldn’t want to spoil your fun.)

Notice that this is quite different from the algorithms described above. The first phase wildly increases the number of generators. Only in the second phase do we make any progress on removing generators. Here’s a beautiful picture Scott made illustrating the first stage.

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 , as long as you change the pivotal structure). The type D planar algebras (with principal graphs the Dynkin diagrams ) were the subject of Noah’s talk at the conference, and the and 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 . He showed that the only possible principal graphs come in two infinite families

and

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

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 ), 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,

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.

But first we need to talk a little more about planar algebras.

As Noah pointed out, in the last post I was being a little sloppy about normalizations, so I want to spend some time discussing this and duals in categories.

This will pay off, because we will then be able to fix the problem we ran into last time and also another problem I didn’t even mention.

Quantum Dimensions

If you look at what happened in the last post, you will realize that we never made much use of the planar aspect of the planar algebra. Sure, I drew some pictures in the  plane, but really I was just stacking picture. This is an essentially linear operation. I didn’t use or say anything about duals and it is the duals which really make a planar algebra planar.

Let’s recall what a planar algebra consists of. For us, in this post, a planar algebra is a way of encoding a monoidal category with duals. So we only have one shading for our planar algebra. In general a planar algebra could have multiple shadings. For example subfactor planar algebras have two shadings. If this were the case, we would no longer be in the context of a monoidal category with duals, but rather a bicategory with duals. So today there is just one shading.

Now a planar algebra also has objects/labeled strands and morphisms/boxes. We can draw these in the plane in pictures such as:

here represents a morphism from to D.

We are going to assume that the underlying category is nice. Nice is going to mean a couple things. First of all, our categories are all going to be -categories, with finite dimensional hom-spaces. Thus for each morphism, there is an adjoint morphism which goes the other way:

Notice that nothing about the objects has changed, just that the direction of the morphism has changed. We also assume our category has compatible direct sums. An object X will be called simple if , and we will assume that our category is semi-simple in the sense that every object is a direct sum of simple objects. We will also assume our category is finite depth, which means that there are a finite number of isomorphism classes of simple objects. I will also need that the monoidal unit, 1, is a simple object.

Let Z be a simple object. Then the above structures give the spaces a Hermitian inner product given by sending and to . We will assume that our hom spaces are Hilbert spaces with respect to this inner product.

The adjoint morphisms can be thought of as a kind of linear duality for morphisms. There is another duality that planar algebras have, a duality for objects. For each object X there is another object , which is a left and right dual. This means that there are morphisms and ,

These satisfy the following and similar “s”-relations:

Thus for any object we can define it’s left and right quantum dimensions:

We will assume that our planar algebra is spherical, i.e. that these numbers are the same, the quantum dimension d(X). For complete detail see for example the paper by Müger.

All of these properties will be satisfied by the even sector of a planar algebra coming from a finite depth, finite index subfactor.

These quantum dimensions are quantized. We always have

and they satisfy and .

Back to TQFTs

Let’s remember what we were doing to construct a TQFT. We were not worrying about the whole TQFT, but only the closed manifold invariant. Our strategy was to choose a triangulation of the manifold, choose an ordering of the vertices (which orients everything in sight) and to assign a number to this triangulation using the planar algebra. Then we want to show this doesn’t depend on the triangulation and so is really an invariant of the manifold.

To do this we need to show two things:

1. We need to show the number doesn’t depend on the ordering of the vertices.
2. We need to show that the number is invariant under the 2-3 and 1-4 Pachner moves.

Last time we started with a guess.  We got our number by first labeling all the edges and faces of the triangulation by elements of our planar algebra. Then we unwrapped each tetrahedron into a planar diagram, which gave us a number for the tetrahedron. Multiplying these numbers and summing over labelings gave us a number which only depended on the ordered triangulation.

Then we showed that this number was invariant under the 2-3 Pachner move. However we ran into trouble with the 1-4 move and we completely ignored the first issue about reordering the vertices.

To show that the number we assign to the triangulation is independent of the ordering, it is enough to show that the number assigned to each individual tetrahedron is invariant under reordering. It turns out our original guess fails to be invariant under these re-orderings. It doesn’t satisfy tetrahedral symmetry.

To see this we need to look closer at how we labeled our triangulation. Remember that we chose a complete set of simple objects and we used these to label each of the edges of our triangulation. Then we also chose an orthonormal basis for the hom spaces for each set of simples.

However, because we have duals, we can relate these different hom spaces. For example, suppose we are given an element of norm one. Then we can get a new element of by the planar diagram:

In fact, applying this is an invertible process and so applying it to a basis  of   will give us a basis for . When we choose orthonormal bases for all these inner product spaces, we can try to require that they are compatible in this way. However, there is a problem. Even if has norm one, the above morphism might not have norm one! In fact we can calculate the norm. The inner product is:

Which is the same as the next diagram. Notice that breaking and reforming edges has a cost. We must multiply the diagram be the factor .

This is the same as

and

Finally we simplify to get:

So we need to rescale this element by to get a norm one element.

Tetrahedral Symmetry

Now let’s see if our invariant is actually invariant under reordering the vertices. If we are given a labeled tetrahedron we turn it into the following planar diagram:

Let’s compare this to the value of the same tetrahedron, but where we have reordered the vertices. The planar diagram has exactly the same value as the following planar diagram:

If we switch the order of the 1 and 2 vertices (ordered as 0,1,2,3) we get a new ordered tetrahedron.  The diagram of the reodered tetrahedron is:

These are very close, but not the same. Even if we take into account that we are summing over both A and and the , these give different results. The problem is exactly the fact that

is not normalized. However we can fix this! Consider the standard tetrahedron and it’s planar diagram:

If we multiply each standard tetrahedral diagram by the factor , then the difference between the two different orderings exactly cancels and we will have restored tetrahedral symmetry.

So now we can try to alter our prescription for a manifold invariant. We label the triangulation, just as before, and we associate numbers to tetrahedra, but now we scale the number associated to each tetrahedron by exactly this factor. Multiplying and summing gives us a number associated to the ordered triangulation which is manifestly independent of the ordering of the triangulation.

2-3 Pachner Move

Now let’s see if this number is invariant under the 2-3 Pachner move. Our calculation from before is still valid, we just need to add in the new tetrahedral factors. The first half of the 2-3 move had two tetrahedra like this:

Thus, adding the tetrahedral symmetry factor, we have to multiply the old diagram by a total extra factor of

While the second half had three tetrahedra which gave three planar diagrams as in the left of,

adding in the symmetry factors means we have to multiply the old diagram by a total factor of:

This means that we have broken the invariance under the 2-3 move! Now we are at step 4 of the master strategy, we must return to step 2. All is not lost however. The difference between the two sides of the Pachner move, geometrically, is the addition of the edge “G”. And the difference of the two sides of the Pachner move, numerically, is the quantum dimension .

This means we can exactly cancel this discrepency by multiplying our number assoicated to the triangulation by an “edge factor”. For every edge, labeled by say E, we multiply by a factor . This doesn’t break our tetrahedral symmetry, but it fixes the 2-3 Pachner move.

1-4 Pachner Move

Now lets look at the 1-4 Pachner move. Again, our calculations from last time are valid, but we need to add in the tetrahedral symmetry factors and the edge factors.

gives a symmetry and edge factor of

while

has a symmetry factor of

and an edge factor of

Thus the difference between the symmetry and edge factors of the two diagrams becomes:

Remember that the old diagrams already differed. One was the right-hand-side of:

While the other was the following, summed over F, G and :

All is not lost (again)! We can fix this discrepancy, but first we need to do a planar calculation. In fact these planar algebra calculations are so fun, I wanted to leave one for you to try at home:

Exercise: Show the following planar identity:

Thus there is still a difference between the two sides of the Pachner 1-4 move, but it is the manageable quatity:

where the sum is over a complete set of irreducibles. So even with both the tetrahedral symmetry and edge factors we still don’t have invariance under the 1-4 Pachner move. However you might guess a theme here. The geometric difference between the initial phase of the 1-4 Pachner move and the final phase, besides the edges which we have already dealt with, is the addition of a vertex, and the numerical difference is this factor w.

Thus we can again alter our prescription by including a “vertex factor”. Every time we see a vertex we are going to multiply by . This doesn’t affect the tetrahedral symmetry, nor does it change invariance under the 2-3 Pachner move, but it does fix the 1-4 Pachner move.

So our final invariant looks like:

where v is the number of vertices. What we have shown is that this number is independent of the triangulation T.

How to get the rest of the TQFT

So now that we’ve managed to produce a 3-manifold invariant, how do we get the rest of the TQFT?

Imagine that we have a 3-manifold with incoming and outgoing surface boundaries. Triangulate the everything. Essentially the same formula for the 3-manifold invariant gives us a linear map,

where the sum over labelings means labeling the edges of the boundary triangulations with our choice of simples and the vector spaces are . What our above calculations show is that this map doesn’t depend on the triangulation of what’s in the middle of the 3-manifold, away from the boundary.

So what we have is something which is note quite a TQFT, but is similar. It associates a vector space to a surface equipped with a triangulation. We want to eliminate the dependence on the triangulation. How can we do that?

Here is a trick. In any 3D TQFT, if S is a surface, then the linear map associated to is supposed to be the identity. Here, however, the linear map associated to , where (S,T) is a surfce with a triangulation, is not the identity but only a projection.

We can glue this to any bordism with (S,T) a boundary component, without changing the bordism. This means that the kernel of the linear operators associated to 3-manifolds contain the kernel of this projection. In short the linear operators are defined on the cokernel of the projection, and we can make a new asignment which associates to a surface with triangulation the cokernel of this projection.

I’ll leave it as an exercise for you that this doesn’t depend on the triangulation and now gives a completely honest TQFT.

In this post I will start explaining the 3D part of the talk. I won’t be able to finish before I run out of steam; that will have to wait for another post. But I will promise to use lots of pretty pictures!

These TQFTs are known as state-sum TQFTs, or in the 3D case as Turaev-Viro or Turaev-Viro-Ocneaunu TQFTs. In the 2D case we know that not all TQFTs arise this way and to my knowledge the 3D case is even less well understood.

I should also mention that this talk is somewhat a product of laziness. The material is basically recycled from a talk I gave at Oberwolfach almost two years ago. It was for the CFT Arbeitsgemeinschaft. These Arbeitsgemeinschaft are neat. The idea is to learn by giving a lecture about some new results which have been found recently by other researchers working on some specific topic. I was supposed to explain the how subfactors give rise to the Turaev-Viro-Ocneanu TQFTs.

When I looked up the constructions of these TQFTs, however, I found a hopelessly complicated series of calculations. How was I supposed to explain this to other people in a talk? Fortunately, I was coming from Berkeley, where I had been steeping in subfactor planar algebra yoga. I found the only way I could really understand the calculations was to translate them into planar algebra diagrams, and -voilà!- I had a talk. Also whenever you can replace a nasty page-long equation by a beautiful picture aren’t you morally obligated to do so?

How to get a 3D TQFT from a Planar Algebra

So let’s recall where we were. We are going to be using a 2-dimensional algebraic gadget to construct a 3-dimensional TQFT. Now a 3D TQFT has lots of data, but in particular assigns a number to each closed 3-manifold. Rather then construct the whole TQFT, I’ll just construct this number, but the rest of the TQFT is there, lurking in the shadows.

The key fact we are going to exploit is that every 3-manifold admits a triangulation and that any two triangulations are related by a series of Pachner moves. In dimension 3 there are again just two such moves: the 2-3 move and the 1-4 move.

Ignore the labeling on these diagrams for right now. Maybe the second one, the 1-4 move, is the easiest to describe. It is just like the 1-3 move in the 2D case. We take a tetrahedron and add a single vertex in the center, creating 4 smaller tetrahedra. The 2-3 move is similar. We start with two tetrahedra glued together by a common face. Now if we add an edge down the middle, you can see that this new arrangement has 3 tetrahedra. Take a minute and stare at these diagrams until you see what is going on.

Recall our master strategy for getting 3-manifold invariants:

1. Given a manifold, choose a triangulation.
2. Use a planar algebra to get a number associated to the triangulation.
3. Show that this number is invariant under the Pachner moves.
4. If step 3 fails, go back to step 2 and try again.
5. Repeat.

So now we are ready to begin step 2. Given a triangulation how are we going to use a planar algebra to get a number? Well let’s remember what we did in the 2D case, using a symmetric Frobenius algebra. First we labeled all the edges of our triangulation by elements of our chosen basis. We associated a number to each individual triangle, multiplied the numbers, and summed over the labelings.

How did we get the number for each triangle? We ran our finger around the outside of the triangle getting a circular, 1-dimensional configuration of elements of our algebra. Then we multiplied and took the trace. Since we started with a symmetric Frobenius algebra, only the cyclic ordering really mattered. Another way to think about this is that we take the 1-dimension boundary of our triangles and unwrap them to get a linear, 1-dimensional configuration, which we can manipulate using our 1-dimensional algebraic gadget.

The 3D case is similar. We are going to choose some basis-like data for our planar algebra, and label the triangulation with this data. Then we are going to unwrap the boundary of each 3-dimensional tetrahedron to get a planar, 2-dimensional configuration. We will get a number out of this configuration. We can then multiply the numbers, and sum over labeling, and so on and so forth.

The real power of this planar approach will be when we want to check invariance under the Pachner moves. We will want to show that two different numbers agree, but since these numbers arise out of planar algebra configurations, we can use planar algebraic techniques to manipulate these configurations. This makes our life much easier.

Planar Algebras Revisited

So we’ve seen several posts on planar algebras here on this blog before, especially in the context of subfactor planar algebras. But today we are going to use planar algebras which are both more general and less general.

Basically we are using Planar algebras as a way to encode all the structure of a nice pivotal category C. So our planar algebra has:

• Shading: There is only one shading here. A subfactor planar algebra usually has two shadings. We can get the planar algebra I want by restricting to just one shaded sector, i.e. to the M-M bimodule sector.
• Labeled 1-dimensional strings. A subfactor planar algebra usually has only one black-white strand and one white-black strand, but if you lay a bunch of strands next to each other you can consider that a new label. These labeled strands correspond to the objects of the pivotal category. We will also call these labels objects.
• vector spaces for boxes with labeled strands comming in. These correspond to the Hom spaces in the pivotal category. We will also call these morphisms.

One of the good things about subfactor planar algebras is that they have a good star-structure which can be realized by flipping the planar diagram up-side down. The Hom vector spaces are actually inner-product spaces and the inner product is compatible with this flipping.We are also going to assume our underlying category is semi-simple and generally nice.

The data in the 3D case that plays the role of the basis in the 2D case is the following: A complete choice of irreducible objects, which we denote A, B, C, D, etc. and a choice of basis for each hom spaces . Note that each hom space can be identified with our ground field, which I will assume is the complex numbers.

In terms of the planar algebra we get diagrams like:

We will also choose a compatible basis of the dual space, .

These elements satisfy some identities which are analogous to the Frobenius normalization identities I used last time. Since we will need these identities I will record them here. The following two planar diagrams are equal:

Also the following two planar diagrams are equal:

Take a minute and convince yourself that these identities make sense and are actually true.

Getting Numbers from Tetrahedra

Now we are ready. Given a triangulated 3-manifold, with ordered vertices, we will use our choice of basis data to label the triangulation. We will label every edge of the triangulation by an object/strand of our planar algebra and we will label each face by compatible basis elements. Then we will unwrap the outside of each tetrahedron, creating a planar diagram.

If you look at this carefully you will see that we have taken the outside of the tetrahedron, which is a triangulation of the 2D sphere, and turned into the dual triangulation. Then we unwrapped the sphere onto the plane.

Notice that this whole diagram represents a morphism from D to itself, but since D is irreducible this is just a number. This number is the number will associate to the tetrahedron. At least it is our first guess.

Now if our planar algebra is spherical then it doesn’t really matter where we cut open the boundary of the tetrahedron, we get the same number. This is the analog of having a symmetric Frobenius algebra. We will assume we have a spherical planar algebra.

So we now have a way to take a labeled triangulation and associate a number to each of the tetrahedra. Multiplying these numbers gives an overall number for each labeled triangulation and summing over labels gives us a number which is associated with the triangulation. We want this not to depend on the triangulation, so we need to show invariance under the Pachner moves.

2-3 Invariance

Let’s begin with the 2-3 Pachner move. This is a move that relates two triangulations. One side of the move has two tetrahedra, while the other has three. Let’s look at the side with two tetrahedra. We can label these tetrahedra as shown:

and turn turn the labeled tetrahedra into planar diagrams…

Here the circled morphism in green corresponds to the common face that the two tetrahedra share. It is enough to fix the rest and sum over these circled elements.

The number of the whole diagram is then the product of these two numbers, summed over this common element. Note that both planar diagrams represent automorphisms of D, so multiplying the numbers is the same as composing/stacking the diagrams. So the following sum represents the total number associated to this diagram if we fix the outside labelings:

Look at this carefully. I’ve put a red box around the part of the diagram I want to highlight. We are exactly in the situation of the first identity. That identity says we can delete the sum and get the following diagram for the value of this triangulation:

Now we can move on to the other half of the 2-3 Pachner move. This triangulation has three tetrahedra, which share faces pair-wise, and also share the new edge. Keeping the outside labeled as before, we can sum over the common inside labelings, which I’ve circled in green. “G” is the common edge, which we are also summing over.

Then we turn these into three planar diagrams, each which represents a number, and also an automorphism of an object. Since composition is bilinear, instead of multiplying we can insert one diagram into another one, to get a bigger planar diagram representing the same number. This is done on the right of the diagram below…

Now take a careful look at this diagram and the same one below, where I’ve highlighted some things:

We can again apply the first of the identities, but now we get to use it twice. After applying the identity we have:

Remember that we are also summing over the common edge, labeled “G”. This is exactly the context of the second identity. Above, I’ve circled the relevant part of the diagram. After applying the identity we get exactly the diagram we wanted:

So the number we associate to a triangulation is in fact invariant under the 2-3 Pachner move. This is one reason it is a good initial guess.

1-4 Invariance (or not)

The 1-4 Pachner move is more problematic. Just like the 2D case, our first attempt at a manifold invariant will fail. Lets look at both sides of the Pachner move. The side with one tetrahedron is easy, we simply unwrap it and get the following diagram:

The other side has four tetrahedra, which share faces pair-wise. We also need to keep track of the four new edges. We also need to be careful about orientations. I forgot to say this earlier. If the orientation of a tetrahedron disagrees with the orientation of the 3-manifold we associate the flipped/stared planar diagram. Bellow shows the four oriented and labeled tetrahedra with the things we are summing over labeled in green. The two with “*” by them have the wrong orientation.

Then we turn these into planar diagrams…

and combine the diagrams into a single planar diagram. Remember that we are summing over the new edges “E” “F” “G” and “H” as well.

I’ve circled where we want to apply the first identity. We do it three times to get…

I’ve circle where we can now apply the second identity. At the end of the day we get the following diagram:

This is almost what we wanted, but it has this annoying sum over E, F and . This is very much like that extra “d” factor from the 2D case.

So there is a problem with the 1-4 move. If you’re astute, you might also catch another problem that I glossed over. In the next post I will explain how we can fix these two problems and get a TQFT.

I think the thing that really amazes me about these TQFTs is that they are actually interesting! I don’t find it supprising that by labeling and summing and averaging we can get a number for each manifold and that these glue nicely and give a TQFT. What amazes me is that we don’t always get trivial results like the constant TQFT, or the zero TQFT.

In fact these TQFTs are vary interesting. Many can distinguish between different orientations on the same manifold. The Haagerup TQFT, associated to the Haagerup subfactor, can distinguish between the 3-sphere and the Poincaré homology sphere!

This talk is about how you can use Planar algebras planar techniques to construct 3D topological quantum field theories (TQFTs) and is supposed to be introductory. We’ve discussed planar algebras on this blog here and here.

So the first order of buisness: What is a TQFT?

According to the Atyiah-Segal axioms a TQFT is a functor.

Specifically a TQFT is a functor from a bordism category of manifolds to the category of vector spaces.  This bordism category has a dimension D associated to it, and so we get D-dimensional TQFTs. Sometimes these are called “N+1 TQFTs” where N+1 = D. Thus a 1+1 TQFT is really 2-dimensional and a 2+1 TQFT is 3-dimensional. I swear this is just a notation and not because we can’t add.

There are a couple variations on the bordism category that people consider depending on what bells and whistles they want to add to their manifolds. Our bells and whistles will be orientations. For us, the bordism category of dimension D will be the following:

• Objects: Closed oriented (D-1)-manifolds.
• Morphisms: A morphism from to is an Equivalence class of Triples , where W is an oriented compact D-manifold with boundary, and are inclusions, such that is a decomposition of into disjoint submanifolds.

Two triples are equivalent if there is an orientation preserving diffeomorphism making everything commute. Composition is given by gluing the bordisms W together, which invloves choosing collar neighborhoods and is not associative at the level of bordisms. However when we pass to equivalence classes, composition becomes associative and the choices we made go away.

The case we are most interested in today are D= 2 and 3.

So a TQFT is a functor from this bordism category to the category of vector spaces. Thus it associates a vector space to each (D-1)-manifold and a linear map to each D-manifold.

Both the bordism category and the category of vector spaces have additional structure and a TQFT is required to respect this structure. Precisely, the disjoint union for manifolds and the tensor prodct for vector spaces make both categories into symmetric monoidal categories and a TQFT is required to be a symmetric monoidal functor. Thus a TQFT sends the disjoint union of manifolds to the tensor product of vector spaces.

Why do people care about TQFTs?

We’ll I like TQFTs because, as we’ll see, they relate algebra and topology in many interesting ways. But other people like TQFTs for other reasons. For 3D TQFTs, one of the most important parts of a TQFT is that they give invariants of 3-manifolds. In fact any TQFT can be thought of as giving invariants for any bordism, but lets look at some simple manifolds and bordisms.

Suppose we have a TQFT called Z. Thus we get vector spaces Z(M) for each (D-1)-manifold M. One of my favorite (D-1)-manifolds is the empty manifold . It is the unit for disjoint union. Since the TQFT is a monoidal functor, it must send the unit to the unit, hence must be the trivial 1-dimensional vector space K.

What are the endomorphisms of in the bordism category? These are compact oriented D-manifolds W, but since the incoming and outgoing boundaries are both empty, W is actually closed. Well actually the morphisms are equivalence classes of these, thus the morphisms from to itself are exactly closed D-manifolds up to diffeomorphism.

So what does the TQFT do to these? Well a TQFT will give us a map from K to K, which is also known as a “number”. So, for example, part of a 3D TQFT associates to each closed 3-manifold a number, and this number only depends on the isomorphism class of the 3-manifold. Thus for any 3D TQFT we get 3-manifold invariants. In fact, as we discussed before, it might be that these invariants distinguish all 3-manifolds. Exciting stuff.

In the rest of this talk due to time constraints, we will be focusing on this number, but the rest of the TQFT is there too, lurking in the background.

Okay, so how does any of this relate to planar algebras?

One way to relate them is that we can use planar algebras (i.e. 2-dimensional algebras)  to construct 3D TQFTs. We’ll see how in a bit, but let’s start with the easier case of how “linear algebras” (i.e. 1-dimensional algberas) allow us to build 2D TQFTs.

It all rests on the following key facts:

1. For 2- and 3-manifolds the classes of TOP, PL, and DIFF are all the same. This means that any 3-manifold has a triangulation and that these give the same PL structure.
2. Pachner’s theorem: Two closed triangulated n-manifolds are PL-homeomorphic if and only if it is possible to move between their trinagulations using a sequence of Pachner moves.

In each dimension, the Pachner moves are a finite list of local moves which relate triangulations. For example in dimension 2 we have:

These two moves are called the 2-2 move and the 3-1 move, respectively. This suggests the following strategy for constructing TQFTs/manifold invariants:

1. Given a manifold, choose a triangulation.
2. Use some algebraic gadget, such as an algebra or a planar algebra, to associate a number to the triangulation.
3. Show that this number is invariant under the Pachner moves.
4. If step 3 fails, go back to step 2 and try again.
5. Repeat.

This really just constructs the closed manifold sector, but with a little extra work we can get the rest of the TQFT too.

Now we make the following loose analogy:

2D Case:

1. Algebras.
2. *-algebras.
3. Frobenius Algberas
4. symmetric Frobenius algebras

3D Case:

1. Monoidal Categories.
2. Monoidal Categories with duals.
3. Pivotal Categories/Planar algebras.
4. sphereical categories/planar algebras.

[I might explain a little about this analogy if time permits. I'm going to assume the audience knows what a planar algebra is. A  Frobenius algebra is finite dimensional,  has a trace which gives a non-degenerate inner product, and hence we can choose a basis and a compatible dual basis for A.]

How to build a 2D TQFT out of a symmetric Frobenius algebra:

We start be taking our 2-manifold and triangulating it. We also order the vertices. This gives each edge and face an orientation. The orientation of the face may agree or disagree with orientation of the surface.

Next we choose a basis for the Frobenius algebra. The idea is that we are going to label the edges of the triangulation by elements of this basis. Given this labeling we are going to get a number for each triangle and we are going to multiply these numbers to get an overall number for the labeled triangulation. Summing over all possible labelings gives us a number for the triangulation, completing step 2 of our overall strategy.

So how are we going to get this number? Well a labeled simplex looks like this:

where a,b, and c are basis elements of our algebra. If the orientation agrees with our surface, what we can do is multiply them and take the trace:

Here denotes taking the -dual of the basis element. If the orientation disagrees, then we do the reversed thing:

As a first remark, we can see a problem immediately. Suppose we reorder the vertices? This changes the orientations of some edges and some faces and gives us an a priori different number for each face. One way to help fix this is to assume that we have a symmetric frobenius algebra. Then the same cyclic ordering of the elements gives the same number.

Now we can check if this is invariant under the Pachner moves. Looking first at the 2-2 pachner move, and labeling we find that it is enough to show that

is equal to

.

To get this to work out, we need the trace to satisfy the Frobenius normalization:

for all elements . The sum is over all x which are from our given basis.Using this identity one can check that each equation is just

.

Now the 1-3 Pachner move is more problematic. To get invariance under the 1-3 Pachner move we need

to be equal to

Now it turns out that is equal to for some number d. So if d is not one, these numbers are different. This is where we reach step 4 in our master strategy.

We can fix this if this number d is invertible and we change our scheme for assigning numbers to triangulations. Notice that the key difference between the 2-2 and the 1-3 cases is that in the 1-3 case we change the number of triangles. We can use this to cancel out the error term above.

What we do to get a number associated to a triangulation is we label, just as before, and get the same number, just as before, but we scale the answer. Every time we see a triangle we multiply by

Now we get a number which is invariant under both pachner moves and hence a manifold invariant.

How to build 3D TQFTs out of Planar Algebras:

Well, I’ll have to come back to this since I have to go give my talk now, but the idea is similar.

Wish me luck!

But wait … why are these two stackings the only candidates for multiplication?  Why shouldn’t we multiply by connecting the right side of a box to the left side of the next box?

or by connecting some top points and some bottom points of each box?  

The observation that there are lots of different multiplications on Temperley-Lieb might lead you to wonder about other operators on Temperley-Lieb.  For instance, we can map to by connecting any point of the first box to a point of the second, and the rest of the points to the boundary:

Everything I’ve drawn above is an example of a “planar tangle” – and the trace we used last week is also a planar tangle, which takes to :

In general, a planar tangle is a diagram where the strings of k input boxes and an output box are connected among themselves in non-crossing ways.  Here’s another example – which is a fine planar tangle, although it’s not clear that it should have any particular meaning if we let it act on Temperley-Lieb inputs.

Planar tangles can sometimes be composed with each other:  we can connect the output of one tangle to the input of another tangle, if both have the same number of strings.  Here’s an unnecessarily complicated example:

Notice that in the LHS, we have labels 1, 2 and 3 in the boxes — this is just so we know what order to do the compositions in.  In the MHS, we’ve stuck the tangles in the boxes they go into; and on the RHS, we’ve discarded the information of the old boundaries and isotoped the strings to make a nicer picture.

The set of planar tangles, with the operation of composition, is an operad.  (I’m not going to tell you what an operad is in general, but if you’re curious  http://homepages.ulb.ac.be/~fschlenk/Maths/What/operad.pdf is a nice introduction.)  A planar algebra is, basically, a representation of the planar operad:  a family of vector spaces with an action by planar tangles which is compatible with composition.  

Temperley-Lieb is not just the simplest and most natural example of a planar algebra; it’s also one of the most important ones.  Coming soon:  Some other examples!

Take a finite group G. Let C[G] denote the group algebra and let F(G) denote the algebra of functions on G with pointwise product (that is F(G) has a basis of elements of the form and is 0 unless in which case it is ). Recall that for both C[G] and F(G) there is a notion of tensor product for modules (for the former g acts on a tensor product via , whereas for the latter acts on a tensor product via

We define a bioidal category called the “group Subfactor” as follows

• The A-A objects are C[G]-modules
• The A-B objects are vector spaces (thought of as representations of the trivial group)
• The B-A objects are vector spaces (thought of as representations of the trivial group)
• The B-B objects are F(G)-modules

Now we need to define the four flavors of tensor products. All of these are of the following form: first push things around (using restriction/induction) until they live where the tensor product is supposed to live, then take the tensor product there.

For example if we take an A-B object (i.e. vector space) V and a B-A object (i.e. vector space) W, their tensor product should be a C[G]-module. To do this we first induce V and W up to G getting two C[G]-modules, then we take their tensor product as C[G]-modules. In particular, if V and W are 1-dimensional, then their tensor product is the regular representation. For another example, suppose we want to take the tensor product of an A-A object (that is a representation V) and an A-B object (that is a representation of the trivial). The answer is supposed to be an A-B object, so we first turn V into a vector space by restriction (forgetting the C[G]-module structure) and then we take the tensor product. One last example, we want to take the tensor product of two vector spaces and get a F(G)-module. So first we take their tensor product as Vector spaces, and then we push that up to an F(G)-module by tensoring with the regular.

This whole tensor product process is a bit more confusing than it ought to be. Can anyone figure out what’s really going on here?

In order to make this a Subfactor category, I also need to fix a simple A-B object called X which tensor-generates. That’s easy here since there’s only one such simple: the one-dimensional vector space. It is easy to see that this tensor generates as and arejust the regular representation of the corresponding ring and thus contains all simples.

What is the dimension of X? (I’m going to leave the definition of dimension vague for the moment.) The A-A part of the category is just the representation theory of G, so we know the dimensions of objects there. In particular, the regular representation has dimension #G. Hence, X has dimension .

Since I haven’t told you much about how actual Subfactor people think about Subfactors, let me sketch the construction in this case. Let N be the von Neumann completion of the countable tensor product of 2×2 matrix rings. N is a hyperfinite II_1 factor. Pick your favorite faithful action of G on a set, and use that action to give an outer action of G on N by permuting the tensor factors in N. Now consider the fixed points N^G = M. Because the action is outer you can prove that M is also a factor, and since G is finite the inclusion M<N is finite index. This is the group subfactor.

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 points at the bottom of your page, and 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 Catalan number (since you can, for instance, read matching parenthesizations as directions for connecting up the points). Now, form a vector space whose basis is Temperley-Lieb pictures on 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 (If there had been circles, we’d have multiplied the picture by ).

If you enjoy multiplying Temperley-Lieb pictures, try this fun exercise: show that Temperley-Lieb is multiplicatively generated by elements , which consist of through strings and a cup and a cap starting at the string:

and satisfy the relations , if and (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 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 diagram, ie a power of :

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 on . Here’s a hard one for you: For which values of 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?

Let where A and B are meaningless labels. Note that since the standard rep is self-dual. Let’s see what bi-oidal category they tensor generate. Well, should mean take and think of that as an A-A object (again the labels here are totally meaningless). Note that 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 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.)

• 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 and . 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.

The fact that A and B are both Von Neumann algebras has two important consequences. First, all the Hom-spaces in this category are not only vector spaces over , they also inherit a positive definite Hermitian form. Secondly, the analysis allows one to prove (and this is substantially harder) that C(A<B) is a semisimple category. Morally this isn’t surprising, as positive definite forms are often helpful for proving semi-simplicity, but the proof is far from obvious.

The fact that A and B are factors says that the trivial bimodules and are irreducible. The fact that A<B is irreducible says that the generating modules and are also irreducible.

Finiteness of the index tells you that there is a good dimension theory for C(A<B). The II_1 property together with finiteness of the index (and maybe irreduciblity?) imply that C(A<B) has a good theory of duals.

Let’s unpack this last bit for a moment. When we were talking about Galois theory, C(L/K) had the property that had a unique copy of the trivial bimodule (and this “evaluation map” is just given by multiplication in L). So, behaves like the dual of . However, on the other hand if you took the tensor product in the opposite order it didn’t work out so well. That is, we didn’t have a unique map In the Subfactor world, there is a unique A-A bilinear map that is called the “conditional expectation.”

To summarize, we’ll call a bi-oidal category C (bi-oidal means that every object comes with two labels A-A, A-B, B-A, or B-B, and that we have two associative tensor products and which are defined when the labels match up) together with a choice of object X in the A-B sector a subfactor category if:

• C is enriched over Hilbert spaces (that is every Hom space has a Hilbert space structure)
• C is semi-simple
• C has a really nice theory of duals. (The right condition here is “spherical” in the sense of Barrett and Westbury, modified appropriately for the bi-oidal setting. Included in this condition is the fact that C has a good notion of dimensions.)
• The trivial objects and are irreducible
• X is irreducible, furthermore X is a tensor generator in the sense that every object in C occurs as a sub-object of or .

One thing I want to emphasize here is that the choice of X is part of the data of a Subfactor category.

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

First you could use #G. This has some nice properties, for example there are only finitely many groups of a given size, so you could hope to enumerate all groups of size say less than 100. Representation theoretically we can define #G as the sum of the squares of the dimensions of the irreps. In Subfactor land we will be calling this measurement of complicatedness the “global index.”

However, as a representation theorist this description has some problems. You could have a small group that nonetheless had many representations. So let’s define the rank of a group to be the number of irreps. This also gives a measure of complicatedness of a group. It is a fun excercise (due originally to Landau) to prove that there are only finitely many groups of a given rank (hint: 1 is the sum of #[g]/#G where [g] is the size of a conjugacy class). For example:

• Rank 1: The trivial group
• Rank 2: Z/2
• Rank 3: Z/3 and S_3

In Subfactor land, we will still call the analogous notion the rank.

Finally, suppose you actually wanted to do computations in a group. You’ll quickly discover that it’s much easier to do computations in the symmetric group S_100 than it is in Monster group, even though the former has much larger size and much larger rank. The reason for this is that S_100 has a 99-dimensional faithful representation where you can do calculations, where for the Monster you’re stuck with a 196882-dimensional irreducible representation. So we could instead measure the complicatedness by the dimension of the smallest faithful irrep.

Now the simplest groups are the cyclic groups, the next are the dihedral groups together with the three platonic solid groups, etc. Notice that we don’t have finitely many examples at each level anymore.

In Subfactor land things will be slightly different, the analogy with groups is that a Subfactor is like a group with a fixed choice of faithful representation. The dimension of this irrep is (the square root of) the index. This is the usual measure of complicatedness for subfactors, which was a source of confusion for me for a long time since I was thinking internally about size (or global index).

One final comment about using the size of a faithful irrep as the definition of complicatedness: it works just as well for infinite groups so long as they have faithful finite dimensional representations.

Finally there is a measure for complicatedness in subfactors which is a bit less intuitive than the above three. It is a standard exercise in group representation theory that given a faithful representation, every other irrep appears in a sufficiently high tensor power. Exactly how high a tensor power do you need? That is to say, how hard is it to describe the other irreps in terms of the irrep you already have? In Subfactor land this tensor exponent is called the depth.

For example, if we take S_4 with the standard representation, it has size 24, rank 5, its smallest faithful irrep has dimension 3, and it has depth 3 (since you need to go to the tensor cube of the standard in order to find the sign rep).

These four are the main measurements used in subfactor theory, but for groups I can think of at least one more. Let’s say that a group is least complicated when the group ring is closest to a matrix ring. That is to say it’s simple when the largest irrep is really large (hence that matrix factor takes up most of the group ring). Perhaps surprisingly it’s possible to say some nice things about what groups are “simple” with respect to this measurement (shameless plug).

A representation theorist studies a ring by studying the modules over it. So a representation theorist should study a pair of rings by studying the bimodules you can make from them. However, if we study all L-K bimodules than we’ve forgotten the inclusion structure. In particular if we had another inclusion of K into L we would still have the same category of bimodules. So instead let’s study the bimodules that we can make starting with the L-K bimodule which we get from the inclusion of K into L and the K-L bimodule .

What sort of operations can we perform on this bimodules? Well, we have a tensor product! So let’s consider the “tensor category” we get generated by and . (I put “tensor” in scare quotes here because we have two tensor products, tensor over K and tensor over L. So it isn’t a monoidal category, instead it’s what I like to call a bi-oidal category. More on this later.) What does this category C(L/K) look like?

First let’s decompose . This is isomorphic (using the multiplication map) to (where both actions come from the inclusion K<L). Clearly this breaks up into [L:K] copies of the trivial bimodule . Further tensoring these with anything else won’t do anything interesting.

So we’re halfway there, now let’s try decomposing . This is a bit trickier, let’s try an example. Let K be the real numbers and L the complex numbers. A basis over K for is given by , , , . We want to decompose this as an L-L bimodule.

A little computation yields the two following simple submodules:

• and . Notice that so this is isomorphic to
• and . Notice that so this is not isomorphic to . We denote it

Okay, what can we do with these new modules? Tensoring with will never do anything interesting. The only new computation is that

It is not difficult to see how this generalizes to an arbitrary Galois extension:

• The only K-K bimodule that occurs is . In any tensor product that lands in the K-K part of the category you can easily compute how many times this bimodule occurs by computing the K-dimension.
• The only L-K bimodule that occurs is . Again counting its multiplicity in tensor products is easy.
• The only K-L bimodule that occurs is . Again counting its multiplicity in tensor products is easy.
• The L-L bimodules are interesting. For every element we have the bimodule . Furthemore the tensor product decompositions are given by and

In summary, the Galois group appears naturally as the set of irreducible L-L bimodules that appear in C(L/K)!

Bonus questions: What if L/K is separable but non-Galois? What if L/K is totally inseparable?

(I’ve done examples in both cases, but they’re fun to look at yourself. I’ll answer questions in comments if people want hints or details.)

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.