This week, Oz didn't show up to class. The Wiz waited and waited, but even after Toby walked in, Oz was still missing.
"Has anyone seen that rascal lately?" asked the Wiz. Jay squirmed in his seat but said nothing. "Hmm..." said the Wiz. "That's most unlike him."
He looked at the clock, sighed, and proceeded:
"Okay. Today we'll see how to get TQFTs from topological lattice field theories. We'll only do the 2-dimensional case, but the basic strategy works in any dimension. Next quarter we'll use it to get 3d TQFTs from spin networks.
So, remember what the basic data is for a 2d topological lattice field theory: a semisimple algebra! For starters, this is a vector space V equipped with a product:
\ / V (x) V \ / \ / | \ / | | | | v | | Vand a metric, or "cup":
| | V (x) V | | | | | \ / | \______/ v \CSince the metric is nondegenerate, it defines a "cap" in the usual way:
\C ______ / \ | / \ | | | v | | | | V (x) VUsing these maps, our picture tricks let us interpret any trivalent graph as a linear operator from some tensor power of V to some other tensor power of V."
Still feeling a bit upset by Oz's disappearance, he only drew a rather small example:
\ \ / V (x) V (x) V \ \ / \ \ / | \ / v \ / \ / V (x) V | | | | v | | VMiguel raised an eyebrow, and the Wizard coughed and admitted:
"Yes, I'm using "trivalent graph" in a funny way here. For starters, I'm allowing loose ends at the top and the bottom of the picture - physicists who like Feynman diagrams would call them "external legs". Also, to get an unambiguous operator, I need to pick a cyclic ordering of the 3 edges that meet at any vertex. We talked about that last time, so I won't make a big deal out of it now... though it's important.
Now, we automatically get a trivalent graph like this from any triangulated cobordism between triangulated 1-manifolds...."
"How?" asked John.
"Simple! Just wave the magic wand of Poincare duality over it: each triangle gives a vertex of our graph, and each edge gives an edge of our graph. The orientation of our cobordism gives the cyclic ordering of the 3 edges that meet at a vertex."
"Umm, could you show me an example?"
The Wizard growled with impatience. "Okay, here!" And he drew a cobordism in the air:
______ ______ / \ / \ | | | | v ^ v ^ | | | | |\______/| |\______/| | | | | | | | | \ \ / / \ \_/ / \ / \ / \ / \ / | ....... | |' `| | | | | v ^ \_______/Next he chopped it up into triangles. Finally, with a flourish, he pulled out his wand and carefully waved it over the entire surface, repeatedly intoning the name of Henri Poincare. As he did so, all the little triangles:
o / \ / \ / \ / \ / \ / \ / \ / \ / \ o-------------------ofaded away, leaving trivalent vertices in their place:
. \ . . / \ . . / \. ./ .\ /. . \ / . . \ / . . | . . | . . | . ..........|.......... |so that the whole thing became a trivalent graph with loose ends coming in from the top and going out the bottom.
"My gosh, it looks like an enormous Feynman diagram!" cried Jay.
"That's exactly right," said the Wiz. "Like I keep saying, our ultimate goal is to build spacetime from quantum processes... or in other words, to eliminate the distinction between spacetime and the processes that go on "in" spacetime. Right now we're just playing with a toy model of how one might go about this. But it's very cute.
So: a triangulated cobordism determines a linear operator. We're getting something a bit like a TQFT, but not quite - not yet, anyway - so let's call it a "preliminary TQFT".
We can formalize it like this. To each triangulated 1-manifold S, our preliminary TQFT assigns a vector space Y(S). A triangulated 1-manifold is just a bunch of circles chopped up into edges, and if S has n edges, Y(S) is the n-fold tensor power of the vector space V. To each triangulated cobordism between 1-manifolds, say
M: S -> S',our preliminary TQFT assigns a linear operator
Y(M): Y(S) -> Y(S')following the recipe just described.
This "preliminary TQFT" has some but not all of the properties of an actual TQFT. We can compose triangulated cobordisms in an obvious way, and it's easy to see that Y gets along with this kind of composition:
Y(MM') = Y(M) Y(M').Most of the other TQFT axioms work as well... only the axioms involving identity morphisms that break down. The reason is that there are no triangulated cobordisms that serve as identity morphisms."
"Why not?" asked Jay.
"Well, what would be the identity 1_S of a triangulated 1-manifold S? You can take S x [0,1] and triangulate it somehow, compatible with the triangulation of S on each end... but if you compose this guy with some other triangulated cobordism, you'll get a new triangulated cobordism with more triangles... so it doesn't act like an identity morphism.
So: there's not really a symmetric monoidal category with triangulated manifolds as objects and triangulated cobordisms as morphisms, and Y is not really a symmetric monoidal functor. However, the only problem is the stuff involving identity morphisms.
We can fix this and get an honest TQFT. Here's how. We've seen that the definition of a semisimple algebra ensures that Y(M) doesn't change when we apply any of the Pachner moves to the triangulated cobordism M. In fact, we can even take this as the rest of the definition of a semisimple algebra! And it's a theorem that we can go between any two triangulations of a cobordism using the Pachner moves, as long as the triangulation of the boundary stays the same. It follows that whenever S and S' are triangulated 1-manifolds and
M, M': S -> S'are two triangulated cobordisms with the same topology, we have
Y(M) = Y(M').In short, we actually have quite a bit of triangulation-independence in our preliminary TQFT... but Y(S) really does depend on the triangulation of S.
So this is what we do. For any triangulated 1-manifold S, pick a triangulation of S x [0,1] compatible with the triangulation of S at both ends. Think of this as a triangulated cobordism IS: S -> S. From this cobordism we get an operator
Y(IS): Y(S) -> Y(S).This operator doesn't depend on our choice of IS , thanks to the fact mentioned above. But the really cool fact is that
Y(IS)2 = Y(IS).In other words, Y(IS) is a projection operator!"
"Wait a minute," said Jay. "Why is that?"
"Simple: if we compose two copies of IS, we get a triangulated cobordism IS2: S -> S. Since Y gets along with composition, we have
Y(IS)2 = Y(IS2).Using the Pachner moves, we can simplify the triangulation of IS2 until it becomes IS. From what I've said, this implies
Y(IS2) = Y(IS).Get it?"
"Aha," said Jay.
"So: Y(IS) is a projection operator. Now, define Z(S) to be the range of this operator; this is some subspace of Y(S). Do this for every triangulated 1-manifold S.
Next, given any triangulated cobordism
M: S -> S'define Z(M) to be the operator Y(M) restricted to the subspace Z(S). This gives an operator
Z(M): Z(S) -> Y(S')but in fact this maps Z(S) into Z(S'), so we get
Z(M): Z(S) -> Z(S')Now let's show that Z is a TQFT...."
"Wait a minute!" cried Miguel. First let's prove that Z(M) maps Z(S) to Z(S')! I don't see how that works."
"Okay. It's not hard. If you take M and compose it with the triangulated cylinder IS', you can then use Pachner moves to simplify the result until it looks like M again, so we have
Z(M) = Z(IS' M) = Z(IS') Z(M).This means that the range of Z(M) is contained in the range of the projection Z(IS'). But that's just Z(S')! So we have
Z(M): Z(S) -> Z(S')as desired."
"Aha," said Miguel. "These projections coming from triangulated cylinders are very nice."
"Yes, they work wonders. For example, our preliminary TQFT had trouble with identity morphisms, but any projection operator acts as the identity on its range, so we get
Z(IS) = 1Z(S)just as one would like in a TQFT. This trick comes up a lot in physics - even in more customary approaches where we start with a Lagrangian, and use it to cook up a classical theory and then a quantum theory. In those games, the preliminary Hilbert space Y(S) consists of "kinematical states", while the subspace Z(S) consists of "physical states" - states satisfying all the constraint equations in our theory. The classic example of the kind of constraint equation I'm talking about is the "Hamiltonian constraint", also known as the "Wheeler-DeWitt equation". This says that time evolution acts as the identity operator on physical states in quantum gravity. And this is really just what the equation up there really says!"
All of a sudden the Wiz noticed that the eyes of the Track 1 students were beginning to glaze over from an overdose of physics jargon.
"But never mind," he hastily continued. "Let's prove that this Z fellow is a TQFT. It starts out being defined on triangulated 1-manifolds and triangulated cobordisms, but it really doesn't depend on the triangulations. To see this, first suppose S and S' are two triangulations of the same 1-manifold; let's show that Z(S) is the same as Z(S').
To do this, pick any triangulated cylinder M: S -> S'. This gives an operator
Z(M): Z(S) -> Z(S')To show this is an isomorphism, just flip M over and get a cobordism M*: S' -> S. This gives an operator
Z(M*): Z(S') -> Z(S)Using the Pachner moves, we can retriangulate the composite M* M until we get IS, so
Z(M*) Z(M) = Z(M*M) = Z(IS) = 1Z(S)The same sort of argument shows that
Z(M) Z(M*) = 1Z(S')so Z(M) and Z(M*) are inverses. This shows that Z(M) is an isomorphism. But it's even better than that: it's a canonical isomorphism, since it doesn't really depend on our choice of M. Why? Well, If we'd picked some other choice, say M', we could use the Pachner moves to change the triangulation until we got back M, so we'd have Z(M') = Z(M).
In short, Z(S) only depends on the 1-manifold S, not on its triangulation. The same is true for the operators Z(M). I'll leave the proof of this to you; it's not hard.
So we really do have a map
Z: 1Cob -> Vectand the question is just whether this is a TQFT - that is, a symmetric monoidal functor.
Now, we've already seen that
Y(MM') = Y(M) Y(M'),and the operators Z(M) are restrictions of the operators Y(M), so we certainly have
Z(MM') = Z(M) Z(M').We've also seen that
Z(IS) = 1Z(S).It follows that Z is a functor. It's a little tiresome to check all the details to show it's a symmetric monoidal functor, so I'll leave that to you. It's not hard, just time-consuming.
So, we get a 2d TQFT from any 2d topological lattice field theory!
We can summarize our results so far as follows:
2d TLFTs -----------------------> 2d TQFTs ^ ^ | | | | | | v v semisimple commutative Frobenius algebras algebrasThe vertical arrows are 1-1 correspondences, while the horizontal one, which we constructed today, is just a map.
Now, what does this diagram make you want to do?"
"Fill in bottom edge!" said Toby.
"Right!" All of a sudden the Wiz was relieved that Oz was gone - he would have answered the question very differently.
"What happens if we take a semisimple algebra, use our tricks to get a 2d topological lattice field theory, then get a 2d TQFT from that, and finally get a commutative Frobenius algebra from that? We'll figure that out next time, in our final class of the quarter."
As the class filed out to get snacks during the break, Toby whispered to the Wiz, "You said those vertical arrows were 1-1 correspondences, but that's a bit of a lie. All the arrows in your diagram are really functors, and the vertical ones are equivalences."
"You're right," whispered the Wiz, "but don't tell anyone. Of course there's a 1-1 correspondence between isomorphism classes of semisimple algebras and isomorphism classes of 2d TLFTs, and similarly for commutative Frobenius algebras and 2d TQFTs. But you're right, we've really got something much better: an equivalence of categories in each case.
"What?" said Miguel, who had been eavesdropping. "You mean there's really a category of TQFTs, and a category of TLFTs, and so on?"
"Of course," said the Wizard. "There's a category of pretty much anything. TQFTs are symmetric monoidal functors, so symmetric monoidal natural transformations between these will be morphisms between TQFTs. This gives us a category of all n-dimensional TQFTs"
"Whoa!" said Miguel, "Let me stop and think about that!"
But the Wiz plunged on heedlessly: "In fact, it's not just a category! We can define direct sums and tensor products of n-dimensional TQFTs, so we get a "symmetric 2-rig" of n-dimensional TQFTs. A symmetric 2-rig is just the categorified version of a commutative rig, i.e. a "ring without negatives". A simpler example is Vect, of course, with its usual direct sum and tensor product."
Miguel started making choking noises at around this point, but the Wiz continued, "We can also take direct sums and tensor products of commutative Frobenius algebras, making them into a symmetric 2-rig... and this symmetric 2-rig is equivalent to the symmetric 2-rig of 2d TQFTs. Similarly, the symmetric 2-rig of 2d TLFTs is equivalent to the symmetric 2-rig of semisimple algebras. The horizontal arrow on the top of that diagram is also a morphism of symmetric 2-rigs, and so is the one on the bottom, which we haven't filled in yet. When we're done, the diagram will commute up to a suitable sort of natural isomorphism - a 2-isomorphism in the 2-category of symmetric 2-rigs, I guess."
Miguel plugged his ears and ran out of the room screaming. "Now look what you've done!" said the Wiz to Toby.