## TQFT – elementary examples and consequences

In TQFT Axioms – 1, I shared the axioms for a topological quantum field theory as detailed in Kock.  Any functor meeting these axioms count.  We’re going to explore some consequences.  There are some surprising consequences (they’re bound to be better as I understand them more).

Let’s consider a compact oriented n-manifold $\Sigma$ and the oppositely oriented version $\bar{\Sigma}$.  Let’s suppose our TQFT is called F and that $F(\Sigma)=V$ and $F(\bar{\Sigma})=W$.  Just as with 2Cob we have a ‘macaroni’ cobordism $\Sigma \sqcup \bar{\Sigma}\rightarrow M \leftarrow \{\}$,

This will be sent, under F, to a linear transformation $\beta : V\otimes W\to \mathbb{K}$.  The co-macaroni cobordism,

From $\{\}\to \Sigma \sqcup \bar{\Sigma}$ will be sent, under F, to a linear transformation $\gamma : \mathbb{K} \to V\otimes W$.  These two linear transformations will have important consequences.

NOTE: If you’re questioning the orientation of the above co-macaroni cobordism I don’t blame you.  The orientation is Kock’s choice versus the following

I think it’s mostly for convenience in the below picture-calculations but I’m not sure.  It boils down to the difference between $A\sqcup B$ and $B\sqcup A$ which are isomorphic.

Next we look at the cylinder $\Sigma\times I$ which is sent, under F, to the identity on V, $id_V$.  We can decompose this cobordism into a composition of other cobordisms and see what their image is under F.  Then we can re-compose them in Vect and see what happens.  The following decomposition is also covered in Kock and is called the snake decomposition.

We start with the cylinder and consider embedding another copy of $\Sigma$ but oriented opposite of the ends.

This wouldn’t be a valid cobordism, because (I think) when we treat it as the composition of two cobordisms it doesn’t work. The illustration below will be helpful.

Trying to treat this decomposition of the cylinder as two cobordisms that can be composed into the cylinder results in a cobordism that isn’t well defined.  The domain looks like the empty set and….what?  Less than nothing?  Somehow, during the ‘middle’ of the cobordism, a manifold leaps into existence as input for the lower macaroni cobordism but at the top another manifold vanishes!

However we can obtain a valid decomposition with a couple more embeddings.  The following is the called the Snake Decomposition.

As Kock says, “interpret it correctly.”  The decomposition results in four cobordisms, each of them is something we’ve seen before.

Kock says this is going to be really important later on, why I don’t know.  For now it does demonstrate a couple ideas.  We can look at this as the composition of two cobordisms, each of which are disjoint unions of cobordisms.    This gives us two ways of making new cobordisms.

What happens to this cobordism under the TQFT functor?  Look at step 2 of the above illustration.  There’s a left and right piece of the composition.  Let’s call them L and R respectively.  Each of these will be sent to a linear transformation between vector spaces.  But each cobordism, L and R, are themselves disjoint unions and so will be sent to tensor products of maps.  We can see that L is the disjoint union of a macaroni cobordism and a cylinder.  This disjoint will be sent to $id_V\otimes\beta : V\otimes\mathbb{K}\to \mathbb{K}\otimes V$ .  I’m guessing that this process of chopping up the identity cobordism like this is important because we know that it’s sent to the identity map in Vect even though the decomposition of the cylinder results in lots of component cobordisms.

There are a couple of similar examples we can try that I didn’t see in Kock.  What if we compose a macaroni cobordism with a co-macaroni cobordism.  There are two ways to do this.  One results in a map $\{ \}\to \{ \}$ which will be sent to $\mathbb{K}\to\mathbb{K}$ which is the composition $\beta\circ\gamma$.

The other results in a map $\Sigma\sqcup \bar{\Sigma}\to \Sigma\sqcup \bar{\Sigma}$

These do not necessarily get mapped to identity transformations.  We’ll see more details later when we work a problem from Kock’s book.

Earlier I mentioned that the maps $\beta, \gamma$ that arise from the macaroni and co-macaroni cobordisms would be important.  It’s time to look more closely that those.  The snake cobordism gives us $V\otimes \mathbb{K}\to V\otimes W\otimes V \to \mathbb{K}\otimes V$ the first map to act is $id_V \otimes \gamma$ and the second is $\beta \otimes id_V$.

I’ve been trying to fill this part of the post in but am embarrassed to say I haven’t.  Kock claims that the pairing $\beta$ is nondegenerate.  I’ve been working on this a bit but keep getting turned around.  Turns out Kock has a definition for this in chapter 2, I suspect it’ll end up being consistent with the usual definition but it’s not clear to me right now.

If this is the case then we can  construct an isomorphism between V* and W.   Let $w\in W$ and restrict $\beta$ to the subspace $V\otimes \{w \}$.  This is a linear map sending V to $\mathbb{K}$ and so is an element of V*.  Now let $\nu \in V^*$ be a linear functional, we need to construct a $w\in W$ that this corresponds to.  The linear functional $\nu$ is completely determined by its values on a basis of V.  Let $e_k$ be such a basis, then the numbers $\nu_k$ completely determine $\nu$.  If $f_j$ is a basis for W then we can specify the action of $\beta$ on $V\otimes W$ by its evaluation on the basis $e_k\otimes f_j$.  Let $\beta ( e_k\otimes f_j)=\beta_{kj}$.  Let $w\in W$ then

$\beta(e_k\otimes w)=w^j \beta(e_k\otimes f_j)=w^j \beta_{kj}$

Set this equal to the $\nu_k$,

$\nu_k = w^j \beta_{kj}$

This is where the nondegeneracy of $\beta$ comes in, so I think.  The values $w^j$ determine the vector $w\in W$ that corresponds to $\nu \in V^*$.  We’ve constructed an isomorphism between W and V*.

I can’t say that I appreciate the full utility of this result but it feels right.  The macaroni cobordism will always result in a map like $\beta$ and so what’s happening is that $\Sigma \to V$ and $\bar{\Sigma}\to V^*$  that’s neat.  Spaces of kets and bras resulting from oriented and oppositely oriented manifolds?

The natural rise of dual spaces though is going to allow for things like differential forms to manifest naturally as well, and while I don’t understand all of these nearly well enough, I know that there are some workarounds when you don’t have a metric tensor but do have differential forms.

Kock does all of these very briefly, I’ve had to think about this for quite a while to work all the details out.  One punchline he gives us is that the dimension of the vector spaces under the TQFT must be finite dimensional.

Let's become better.
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### 3 Responses to TQFT – elementary examples and consequences

1. sheabrowne says:

Thanks for these two posts, it’s nice to see some details on TQFT. I’m looking forward to seeing where this connects up with gravity and QFT.

2. sheabrowne says:

Ok, I’ve just read the introduction of Koch’s book online, and I see that he is not attempting here to connect up with gravity or QFT, only in the sense that the structure of TQFT has several properties that one would want for a quantum theory of gravity… I’ll need to look elsewhere for the detailed connection. Love his writing style by the way.

• True, but I think Kock is the most accessible introduction. You can check out Witten’s original article. Those types of articles focus on the construction of the functor which involves an insane amount of both high level physics and mathematics. My hope is that by working on the mathematical theory of TQFT i can break into a very dense area of research.

By and by, I’m reading an interesting book called “On Space and Time” edited by a fellow named Shahn Majid who’s a pure mathematician who does work in quantum gravity. Other contributors are Connes and Penrose. So far, Majid’s article has been fantastic. He takes a long hard look at the nature of a scientific theory of quantum gravity and makes some bold claims that dovetail with others, who are increasingly looking to Category Theory for a big picture of physics. I’ll try to give a review of Majid’s contribution to the book soon.

Let me know what you find elsewhere.

Cheers,

Kevin