The Wiz charged into class, excited and out of breath. "Okay! We're starting to ascend the dimensional ladder. Everyone hang on! Last time we classified all 1d TQFTs. Now let's do 2d!
Suppose we have a 2d TQFT:
Z: 2Cob -> Hilb.What is it like? Let's see.
We'll start by figuring out what it does to objects. What are the objects of 2Cob?"
Looking through his notes, Oz made a guess: "Compact oriented 1-dimensional manifolds?"
"Yeah, yeah," said the Wiz impatiently. "But what are they like? What are the options?"
"Well, there's the circle," said Miguel. "And..."
"Stop right there!" interrupted the Wiz. "Yes, we've got the circle -- but it's an oriented circle. Let's use the standard orientation. This was counterclockwise when I last checked. Have you ever been to Paris, where they keep all the standards? Right next to the standard kilogram, they've got a copy of the unit circle with its standard orientation. It's a wonderful sight: pure platinum, one meter in radius... if you ever visit Paris, make sure to see it."
And in the air he swung out an enormous glowing circle with his staff:
_____
/ \
| |
v ^
| |
\_____/
Okay: what does our TQFT do to this guy?"
"It assigns it a Hilbert space," said Oz.
"Right!" The Wiz drew this as follows:
_____
/ \
| | Z
v ^ |-------------------> Z(S1)
| |
\_____/
"Okay, now what other 1-dimensional manifolds do we have?"
"Well," said Miguel, "as I was about to say, any finite disjoint union of circles."
"Right!" said the Wiz. "And disjoint union gets carried to tensor product, so we have, for example:
_____ _____ / \ / \ | | | | Z v ^ v ^ |-------------------> Z(S1) (x) Z(S1) | | | | \_____/ \_____/and so on for any number of circles."
"Including no circles," said Toby.
"Indeed -- never forget the empty set! As usual, our TQFT assigns to this the Hilbert space C:
Z
|-------------------> C.
Great. But what about this?" And in the air he swung out another
enormous circle with his staff, but clockwise this time:
_____
/ \
| |
^ v
| |
\_____/
"Last time", said Toby, "we saw that if we reverse the
orientation of a manifold representing space, we have to take the dual
of its Hilbert space."
"Good!" said the Wiz. "So we have
_____
/ \
| | Z
^ v |-------------------> Z(S1)*
| |
\_____/
So if we call this orientation-reversed circle S1*, we have
Z(S1*) = Z(S1)*.
Okay. Now we know what our TQFT does to any finite disjoint union of copies of S1 and S1*, so we know what it does to all the objects in 2Cob. So let's see what our TQFT does to morphisms in 2Cob -- or in other words, cobordisms between 1-manifolds."
But before he could continue, Miguel got up and walked over to the copy of S1. He went around behind it inspected it carefully for a minute. Then he let out a yell: "Hey! You were lying. This isn't oriented counterclockwise: it's oriented clockwise!"
The Wiz was embarrassed. "Whoops! Well, err... umm... it only looks like that from behind. On the front," he continued, motioning to it like a salesman, "it's equipped with a fully functioning counterclockwise orientation. Honest."
Angrily, Miguel grabbed the circle and turned it around. "Not any more!"
The Wiz stared at it, shocked. "My gosh, you're right! It must be defective." He repeatedly looked at it from both sides.
"Maybe you should draw a new one," suggested Oz.
Then Miguel had a brilliant idea. "Wait a minute! It's not defective: all we're seeing is that the counterclockwise oriented circle is the clockwise oriented one. In other words, S1 = S1*."
"You're saying they're EQUAL?" said the Wiz in horror.
At this point Toby stepped in. "No, they're not EQUAL, they're just ISOMORPHIC. And the process of turning them around IS THE ISOMORPHISM."
The Wiz slapped his forehead. "Holy sakes alive! You're right! So what we actually have here is an interesting morphism in 2Cob: an isomorphism from S1 to S1*. Let's see if we can visualize that."
He grabbed the counterclockwise oriented circle, lifted it high, and then flipped it around while swinging it downwards. All of a sudden, by a strange trick of magic, time slowed down... and the circle left an iridescent trail, so the process could be seen as a surface:
_____
/ \
| |
v ^
| |
|\_____/|
| | ______
| | / \
| | |\ .\
\ \ | \ . \
\ \ | \ . \
\ \ | / . /
\ \__/ . /
\ . /
\ . /
\ . /|
\__________/ |
| |
| ..... |
|' `|
| |
^ v
| |
\_____/
Then time resumed its normal pace, but the surface continued to hang
glowing in the air,
"Whoa!" said Oz. "That was disorienting."
"Don't worry, you'll get over it. It's just a cobordism from S1 to S1*. It's an isomorphism because it has an inverse, namely this:
_____
/ \
| |
^ v
| |
|\_____/|
| | ______
| | / \
| | |\ .\
\ \ | \ . \
\ \ | \ . \
\ \ | / . /
\ \__/ . /
\ . /
\ . /
\ . /|
\__________/ |
| |
| ..... |
|' `|
| |
v ^
| |
\_____/
Compose these in either order, and you can straighten out the result
to get a cylinder -- i.e., an identity morphism.
Okay, so we're starting to become acquainted with some of the morphisms in 2Cob. What are some others?"
"There's the pair of trousers," said Miguel.
"Right! One circle splits and becomes two, like this," said the Wiz. He drew a counterclockwise circle with his staff, and then he pulled it downwards with his bare hands, pulling it in two as he did so. It split with a popping sound, and traced out a surface which hung glowing in the air:
_______
/ \
| |
v ^
| |
|\_______/|
| |
| |
| |
| |
/ \
/ \
/ \
/ \
/ _ \
/ / \ \
/..... / \ .....\
/ ` / \ ' \
/ `/ \' \
| | | |
v ^ v ^
\ / \ /
\______/ \______/
"People call this the pair of trousers, or pair of pants. The also
call it the sphere with three disks removed -- or the
"trinion", if they're feeling more lofty. What else is
there?"
"The upside-down pair of trousers, obviously," said Miguel.
"Right:
______ ______
/ \ / \
| | | |
v ^ v ^
| | | |
|\______/| |\______/|
| | | |
| | | |
\ \ / /
\ \ / /
\ \_/ /
\ /
\ /
\ /
\ /
| |
| |
| ....... |
|' `|
| |
| |
v ^
\_______/
I like this upside-down version better than the original pair of
trousers, and I call it
m: S1 (x) S1 -> S1.Don't forget, the tensor product here stands for disjoint union. The original pair of trousers is then
m*: S1 -> S1 (x) S1.Remember, applying * to a morphism is our way of "time-reversing" it; we can always do this, since nCob is a *-category.
Now here's a puzzle: do you know why I'm using the letter "m" for this cobordism from S1 (x) S1 to S1? What does "m" stand for?"
"Multiplication!" said Toby. "You gave us a big fat clue last time: you said that in any dimension, there's a cobordism where two copies of the sphere fuse together to form one, and this gives the Hilbert for the sphere an associative product."
"Exactly! Applying our TQFT to the upside-down trousers, we get:
______ ______
/ \ / \
| | | |
v ^ v ^ Z(S1) (x) Z(S1)
| | | |
|\______/| |\______/| |
| | | | |
| | | | |
\ \ / / |
\ \_/ / Z |
\ / |-------------------> | Z(m)
\ / |
\ / |
\ / |
| ....... | |
|' `| |
| | |
| | |
v ^ v
\_______/
Z(S1)
so Z(m) is a product on the Hilbert space Z(S1)... and it's easy to
see using pictures that this product is associative. It's a lot like
what I did in Week 3.
Okay, what are some other morphisms in 2Cob?"
"Well," said Toby, "since Z(S1) is an algebra, there's also the cobordism that gives the unit in this algebra. It looks like this:
_____
/ \
/ \
| |
| |
| |
| |
| ....... |
|' `|
| |
| |
v ^
\_______/
It's a cobordism from the empty set to S1."
"Right," said the Wiz. "Sometimes people call this "the birth of a circle". For short, I call it i, since when we hit it with the functor Z, it gives the unit for our algebra Z(S1):
_____ C
/ \
/ \ |
| | |
| | |
| | |
| | |-------------------> | Z(i)
| ....... | |
|' `| |
| | |
| | v
v ^
\_______/ Z(S1)
Great! Any other simple examples of morphisms in 2Cob?"
"There's the upside-down version of i!" said Miguel.
"Otherwise known as i*, or "the death of a circle"," agreed the Wiz. "It looks like this:
_______
/ \
| |
v ^
| |
|\_______/|
| |
| |
| |
| |
\ /
\_____/
This gives us a linear functional on Z(S1):
_______ Z(S1)
/ \
| | |
v ^ |
| | |
|\_______/| |
| | |
| | |-------------------> | Z(i)
| | |
| | |
\ / v
\_____/
C
And now, my friends, we have all the cobordisms we really need.
i, m, i* and m*, together with those funky orientation-reversing
cobordisms... every cobordism can be built from these using the
machinery present in a braided monoidal category! In other words:
if we compose these, tensor them, and use braidings and identity
morphisms to our heart's content, we can build up every morphism
in 2Cob. Take my word for it. Or better yet, prove it yourself!
So now let's see if we can say what a 2d TQFT really amounts to. Once we know the Hilbert space Z(S1), we know what our TQFT does to all objects in 2Cob. But to say what it does to all the morphisms, we need to know a bit more. First of all, we need to know
Z(m): Z(S1) (x) Z(S1) -> Z(S1)and
Z(i): C -> Z(S1).These make Z(S1) into an algebra. And then...."
Miguel said, "... we need to know
Z(m*): Z(S1) -> Z(S1) (x) Z(S1)and
Z(i*): Z(S1) -> CThese make Z(S1) into a coalgebra!"
"Ah, good," said the Wiz. "So you know how to cothink! It's just like thinking, only backwards. The definition of a coalgebra is just like the definition of an algebra, but with all the arrows turned around. So when we have a 2d TQFT, the Hilbert space of the circle is both an algebra and a coalgebra. But there's a bit more to it than that: these two structures fit together in a nice way."
"Do we get a bialgebra?" asked Miguel. "I've heard that a bialgebra is both an algebra and a coalgebra, but with some nice compatibility conditions relating the multiplication and comultiplication...."
"NO!" said the Wiz, and chucked a small fireball at Miguel. It exploded harmlessly with a small bang. "We'll learn about bialgebras later in this class, but we don't need to know about them here. In fact, we don't even need to know about coalgebras!"
"Whew," thought Oz. "That's a relief -- I didn't understand that stuff about coalgebras. `Just like an algebra, but with all the arrows turned around' -- now what the heck does that mean?"
"Here's the deal," said the the Wizard. "Whenever we have a 2d TQFT, Z(S1) becomes an algebra. But it's better than that: it's a commutative algebra, since switching two elements and multiplying is the same as just multiplying them. Here's the proof:
__ __ __ __
/ \ / \ / \ / \
v ^ v ^ v ^ v ^
|\__/| |\__/| |\__/| |\__/|
| | | | | | | |
\ \/ / | | | |
\ /` / | | | |
\ / ` / \ \ / /
/ /\ \ \ / /
/ ` / \ \ \_/ /
/ `/ \ \ /
/ /\ \ \ /
| / \ | \ /
| \__/ | = | |
\ / | |
\ / | |
\ / | |
| | | |
| | | |
| ... | | ... |
|' `| |' `|
v ^ v ^
\___/ \___/
We couldn't do this trick back when we were studying 1-dimensional TQFTs
last week: Z(S0) was an algebra, but not commutative. The
extra dimension gives us commutativity!
Okay, great, so Z(S1) is a commutative algebra. But it's even better than that: it's an algebra with a nondegenerate trace!"
"Huh?" said Oz.
"Oh, right, I need to define that. A "trace" on an algebra A is a linear functional
tr: A -> Cwith the property that
tr(ab) = tr(ba).This property is automatic when A is commutative. But of course, the classic example is when A is a matrix algebra and tr is the usual trace.
Whenever we have a trace on an algebra, we can define a bilinear form
g: A (x) A -> Cby
g(a,b) = tr(ab).Of course, this is always symmetric:
g(a,b) = g(a,b)We say the trace is "nondegenerate" if g is nondegenerate:
g(a,b) = 0 for all b ==> a = 0You may recall that a nondegenerate symmetric bilinear form on a vector space is also called a "metric". We studied these using our picture tricks in Week 5... and now this will really start to pay off!"
"But wait," said Oz. "You haven't said why Z(S1) has a nondegenerate trace!"
"Oh, right," admitted the Wiz. He threw a little fireball at himself. "Ouch! Okay: we already saw that Z(S1) has a linear functional on it, like this:
_______ Z(S1)
/ \
| | |
v ^ |
| | |
|\_______/| |
| | |
| | |-------------------> | Z(i)
| | |
| | |
\ / v
\_____/
C
Since Z(S1) is commutative, this is automatically a trace.
So the only question is: why is it nondegenerate? Well, in Week 5 we
saw that that a symmetric bilinear form
| |
| |
V v v V
\ /
\_______/
g
is nondegenerate iff this map
|
|
| iV
V v _______
| / \
| / \
| V v |
\ / |
\_______/ |
g V ^
|
|
|
is an isomorphism. We called this map
#: V -> V*since we can use it raise indices. Now, if you take the trace on Z(S1) that we just defined, and follow my instructions to construct
g: Z(S1) (x) Z(S1) -> Cand then use this to construct
#: Z(S1) -> Z(S1)*what do you get?"
The class tried to work through it in their heads, but the Wizard was too impatient. "You get something we've already seen... this!
_____
/ \ Z(S1)
| |
v ^ |
| | |
|\_____/| |
| | ______ |
| | / \ |
| | |\ .\ |
\ \ | \ . \ |
\ \ | \ . \ |
\ \ | / . / |
\ \__/ . / |
\ . / | #
\ . / |
\ . /| |
\__________/ | |
| | |
| ..... | |
|' `| |
| | |
^ v |
| | v
\_____/
Z(S1*) = Z(S1)*
The class gasped. "Think about it," urged the Wiz.
"We've already seen this is an isomorphism... so the trace on
Z(S1) is nondegenerate."
"Now here's the payoff. An algebra equipped with a nondegenerate trace is called a Frobenius algebra. We've just seen that given any TQFT, the Hilbert space Z(S1) is a commutative Frobenius algebra with multiplication given by
Z(m): Z(S1) (x) Z(S1) -> Z(S1),unit given by
Z(i): C -> Z(S1),and trace given by
Z(i*): Z(S1) -> C.But the cool part is the converse: for any Hilbert space with the structure of a commutative Frobenius algebra, there exists a unique 2d TQFT. Actually, uniqueness isn't hard. The only real work is to figure out a formula for Z(m*) in terms of the 3 maps just listed. You can do this using pictures. Hmm, I guess I'll make this a homework exercise."
Oz gulped, and some of the other Acolytes started looking nervous.
"Of course, you may also need to use
#: Z(S1) -> Z((S1)*)and its inverse
b: Z((S1)*) -> Z(S1)but that's okay, since we've just seen what these must be."
"That proves uniqueness. Existence is more work, and let's not worry about it just now. Let me just summarize the theorem as follows:
Theorem: A 2d TQFT is a Hilbert space with the structure of a commutative Frobenius algebra.
Nice, eh? But actually, this isn't the best way to state this result. Until we get around to studying unitary TQFTs, we never really use the inner product on our Hilbert spaces, so they might as well just be vector spaces, so we might as well redefine a TQFT to be symmetric monoidal functor
Z: nCob -> Vect.If we use this definition, we get these theorems:
Theorem: A 1d TQFT is a vector space.
and
Theorem: A 2d TQFT is a commutative Frobenius algebra.
You'll have to admit that's prettier."
"What about the unitary TQFTs?" said Miguel. "In physics we certainly want unitarity."
"True," said the Wiz. "But first let's finish classifying the 2d TQFTs. We've reduced it to an algebra problem; now let's polish it off! One nice example of a Frobenius algebra is the algebra of n x n complex matrices, Mn(C), with its usual trace. Do you see why this trace is nondegenerate? Anyway, we can also modify the trace by multiplying it by any number c, or any nonzero number if we want it to remain nondegenerate. Let's call this modified trace
trc(x) = tr(cx).What other Frobenius algebras are there? Well, the direct sum of Frobenius algebras becomes a Frobenius algebra if we add the traces in the obvious way. And that's all there is!
Theorem: Every Frobenius algebra is a finite direct sum of Frobenius algebras of the form (Mn(C), trc).
I won't prove this, but for the algebraists among you, let me just mention that it follows from a famous old result called Wedderburn's theorem. Every Frobenius algebra is semisimple, since the orthogonal complement of an ideal is an ideal. Wedderburn's theorem says that every semisimple algebra is a finite direct sum of matrix algebras! So the only work is to classify the nondegenerate traces on this sort of algebra."
"Zounds!" cried Oz, his eyes bulging in amazement. "That went way over my head."
"Let's see if this does, too!" said the Wiz, hurling a fireball at Oz. FOOM! Alas, it did not.
"Anyway, if you don't know what that stuff means, don't worry. The point is that now we can completely classify 2d TQFTs. 2d TQFTs correspond to commutative Frobenius algebras, and n x n matrix algebras are only commutative when n = 1. Of course, M1(C) is just C. So the possibilities are very limited: Z(S1) must be a direct sum of copies of (C, trc) for various nonzero values of c."
"Hey," said Miguel, "Why did you bother to classify all Frobenius algebras, when we only needed the commutative ones?"
The Wiz waggled his eyebrows mysteriously. "Time will tell."
Then he glanced at the clock. "Ulp! I've only got a negative amount of time left here, so I'd better hurry. I'll just tell you when our TQFTs are unitary! In dimension 1 they always are; in dimension 2 they are iff all those numbers c are positive, in which case the inner product on Z(S1) is related to the metric as follows: