# TQFTs via Planar Algebras (Part 3)

This is the third and final post in my series about using planar algebras to construct TQFTs. In the first post we looked at the 2D case and came up with a master strategy for constructing TQFTs. In the last post we began carrying out that strategy in the 3-dimensional setting, but ran into some difficulties. In this post we will overcome those difficulties and build a TQFT.

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 $\sigma$ represents a morphism from $X \otimes Y$ 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 $C^*$-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 $Hom(X,X) = \mathbb C \cdot id_X$, 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 $Hom(X \otimes Y, Z)$ a Hermitian inner product given by sending $\sigma$ and $\mu$ to $\sigma \circ \mu^* \in \mathbb{C} = Hom(Z, Z)$. 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 $\overline{\text{X}}$, which is a left and right dual. This means that there are morphisms $r_X \in Hom( 1, \overline{X} \otimes X)$ and $\overline{r}_X \in Hom( 1, X \otimes \overline{X})$,  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 $d(X) \in \{ 2 \cos \frac{\pi}{n}, \; n = 3, 4, \dots \} \cup [2, \infty)$

and they satisfy $d(X \otimes Y) = d(X) d(Y)$ and $d(X \oplus Y) = d(X) + d(Y)$.

#### 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 $Hom( X\otimes Y, Z)$ 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 $\sigma \in Hom( X \otimes Y, Z)$ of norm one. Then we can get a new element of $Hom(Z \otimes \overline{Y}, X)$ by the planar diagram: In fact, applying this is an invertible process and so applying it to a basis  of $Hom(X \otimes Y, Z)$  will give us a basis for $Hom(Z \otimes \overline{Y}, X)$. 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 $\sigma$ 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 $d(Z) d(X)^{-1}$. This is the same as and Finally we simplify to get: So we need to rescale this element by $\sqrt{d(Z)^{-1} d(X)}$ 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 $\overline{\text{A}}$ and the $\sigma_i$, 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 $\sqrt{ d(B) d(C)}$, 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 $(\sqrt{ d(B) d(C)} \sqrt { d(H) d(Y) })^{-1}$

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: $(\sqrt{ d(B) d(G)} \sqrt { d(H) d(C) } \sqrt { d(G) d(Y) })^{-1}$

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 $d(G)^{-1}$.

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 $d(E)$. 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 $(\sqrt{ d(B) d(C)} )^{-1} d(A) d(B) d(C) d(D) d(X) d(Y)$

while has a symmetry factor of $\left( \sqrt{d(A) d(G) d(H) d(A) d(H) d(C) d(B) d(G)} \right)^{-1}$

and an edge factor of $d(A) d(B) d(C) d(D) d(E) d(F) d(G) d(H) d(X) d(Y)$

Thus the difference between the symmetry and edge factors of the two diagrams becomes: $d(A)^{-1} d(F) d(E)$

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 $\sigma_5$: 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: $w = \sum_E d(E)^2$

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 $w^{-1}$. 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: $Z(M, T) = w^{-v } \cdot\sum_{\text{labelings of } T} (\text{ edge factors})$ $\cdot ( \text{Tetra. Sym. factors}) \cdot (\text{ Original guess})$

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, $\oplus_{\text{labelings of } \partial_\text{in}} \otimes H_{X, Y}^Z \to \oplus_{\text{labelings of } \partial_\text{out}} \otimes H_{X, Y}^Z$

where the sum over labelings means labeling the edges of the boundary triangulations with our choice of simples and the vector spaces are $H_{X, Y}^Z = Hom( X \otimes Y, Z)$. 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 $S \times [0,1]$ is supposed to be the identity. Here, however, the linear map associated to $(S, T) \times [0,1]$, where (S,T) is a surfce with a triangulation, is not the identity but only a projection.

We can glue this $(S,T) \times [0,1]$ 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.

## One thought on “TQFTs via Planar Algebras (Part 3)”

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