## TQFTs via Planar Algebras (Part 2) December 6, 2008

Posted by Chris Schommer-Pries in low-dimensional topology, Pictorial Algebra, planar algebras, QFT, subfactors, talks, tqft.

In my last post I explained a strategy for using n-dimensional algebraic objects to construct (n+1)-dimensional TQFTs, and I went through the n=1 case: Showing how a semi-simple symmetric Frobenius algebra gives rise to a 2-dimensional TQFT. But then I had to disappear and go give my talk. I didn’t make it to the punchline, which is how planar algebras can give rise to 3D TQFTs!

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 $\sigma_i$ for each hom spaces $Hom( X \otimes Y, D)$. Note that each hom space $Hom(A, A)$ 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 $\sigma^*_i$ of the dual space, $Hom(D, X \otimes Y)$.

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 $\sigma_5$. 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!

1. Noah Snyder - December 7, 2008

“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.”

Under the normalization I’m used to using cutting this open at different edges is going to give you results that differ by factors corresponding to the dimension of the edge that you cut. Is this the “other problem you glossed over”?

2. Chris Schommer-Pries - December 7, 2008

@ Noah,

Yes, that’s right. In fact the two identities that I kept using force the normalization that you are probably used to.

This is what I want to start my next post talking about.

3. Scott Carnahan - December 7, 2008

This is pretty wild. What do you use to draw those diagrams?

4. Chris Schommer-Pries - December 8, 2008

@ Scott

I use tikz/pgf.

This is a latex package and so I end up coding them. These pictures are actually a result of my first attempt at tikz. The whole file which is on my website took me about a day to create.

I needed to learn a flexible way to get graphics into latex and I asked Anton, who suggested tikz as an alternative to pstricks and others.

5. Bruce Bartlett - December 10, 2008