"Okay," said the Wiz. "Let's see if you really understand all those abstractions I've been throwing at you lately. Let's see what a 1-dimensional TQFT amounts to. It's a symmetric monoidal functor

Z: 1Cob -> Hilbbut let's unpack that definition and see what it amounts to.

First of all, Z assigns a Hilbert space to every 0-dimensional manifold. And remember: in this game, all our Hilbert spaces are finite-dimensional and all our manifolds are compact and oriented! I won't keep saying that.

So, what are all the compact oriented 0-dimensional manifolds?"

"Well, there's the *point*..." said Miguel.

"Yes indeed -- but it's an *oriented* point!
I'll admit, it's a bit
confusing to think about oriented 0-dimensional manifolds. An
orientation is usually specified by giving ordered basis of tangent
vectors at each point of your manifold. In the 0-dimensional case that
basis is the *empty set*, so you might think there was only one way to
orient the point. But take my word for it: it's good to decree that
the point has two orientations: positively oriented, and negatively
oriented.

Let's call the positively oriented point +, and the negatively oriented point -. Our TQFT assigns to these two vector spaces:

Z + |-------------------> Z(+) Z - |-------------------> Z(-)Now, what other oriented 0-dimensional manifolds are there?"

"Any finite disjoint union of these guys," said Toby.

"Right! So, for example, there's this:

+ -What does Hilbert space does our TQFT assign to this?"

"Well," said Toby, "1Cob is a monoidal category with disjoint union as the tensor product, and our TQFT is a monoidal functor, so we get the Hilbert space Z(+) (x) Z(-)."

"Right! Disjoint union of manifolds gets mapped to tensor product of Hilbert spaces. So:

Z + - |-------------------> Z(+) (x) Z(-)It works this way for any finite disjoint union of copies of + and -. For example,

Z - + |-------------------> Z(-) (x) Z(+)

"But - + is just the same as + -!" protested Miguel.

"Hmm," said the Wizard thoughtfully. "They're certainly
*isomorphic*." He grabbed the glowing dots labelled - and +,
one in each hand, and interchanged them by crossing his arms, rapidly
swinging his hands down as he did so. They left behind a glowing trail
like this:

- + \ / \ / \ / / / \ / \ / \ + -"But are they

Anyway, let me do an example that looks a bit different. What about this 0-dimensional manifold?" he asked, and with his staff he drew in the air.... nothing!

Toby smiled and said, "The empty set!"

"Right -- what Hilbert space does our TQFT assign to this 0-dimensional manifold?"

"The complex numbers," said Toby. "We worked this out
last time: the empty set is the unit object in 1Cob, so it gets sent to
the unit object in Hilb, namely **C**:

Z |------------------->And this is nice, since the empty set is the disjoint union of theC

"Exactly!" said the Wiz. "If this freaks you out," he said to the other Acolytes, some of whom were looking nervous, "you must learn to love nothingness."

"Okay," he continued. "Now let's see what our TQFT
does to *morphisms* in 1Cob -- or in other words, cobordisms between
0-manifolds. For example, what linear operator does it assign to *this*
cobordism?

- + | | v ^ | | \ / \__/Here I'm using arrows to draw the orientation of our cobordism. Remember, our cobordisms are always compact oriented manifolds with boundary. Also, while I haven't explained this yet, the orientation on the cobordism must be compatible the orientation of the boundary. For now I'll just let you guess how this works... it'll be easier to explain when we get to 2d TQFTs.

Anyway, what does our TQFT do to this guy? Let me give this guy a
name... say e_{+}, since in Week 2 we used "e" to
stand for the "cup" or "counit" maps in the category
of vector spaces, and this guy looks like that."

"Well," said John, "it must do something like this:

- + Z(-) (x) Z(+) | | ^ v Z | | | |-------------------> | Z(e"Right!" said the Wiz. "And can you guess what I'm about to ask next?"_{+}) \ / V \__/C

"Yeah," said Richard. "You're going to ask what our TQFT does to this guy:

__ / \ / \ | | v ^ | | + -the "cap" or "unit"."

"Right. Following our notation from last quarter, let's call this
guy i_{+}. And the answer is...?"

"It's got to be something like this," said Jay:

__"Right," said the Wiz. "And what does this suggest?"C/ \ / \ Z | | | |-------------------> | Z(i_{+}) v ^ v | | + - Z(+) (x) Z(-)

"It suggests that Z(-) is the dual of the space Z(+)," said Miguel,
and that Z(i_{+}) and Z(i_-) are the unit and counit for this pair of
dual spaces."

"Right!" said the Wiz, "And it's true! Whenever you have a pair of vector spaces together with a cap and cup like this, they're duals when the cap and cup satisfy these equations:

| | | | | | _______ | | / \ v | / \ | | | ^ | = v | \ / | ^ \_______/ | | | | | | | | |and

| | | | | | ^ _______ | | / \ | | / \ = v | v | | \ / | | \_______/ | | ^ | | | | | | |Since equations like this hold over in 1Cob -- where we take the pictures very literally! -- we get similar equations over in Hilb when we apply the functor Z to everything in sight. In other words, we get

Z(+) Z(+) | | | | Z(iand_{+}) | | _______ | | / \ v | / \ | | | ^ | = v | \ / | ^ \_______/ | | | | Z(e_{+}) | | | | | Z(+) Z(+)

Z(-) Z(-) | | | | | Z(iSo Z(-) is the dual of Z(+), and Z(i_{+}) | ^ _______ | | / \ | | / \ = ^ | v | | \ / | | \_______/ | | ^ | Z(e_{+}) | | | | | | Z(-) Z(-)

This are a lot of nice lessons here. First of all, we see that we can't choose Z(-) and Z(+) independently: Z(-) must be the dual of Z(+). In fact, this is true much more generally! Suppose we have an n-dimensional TQFT

Z: nCob -> HilbGiven an object S in nCob, we can always switch the orientation on S and get an object S*. There will always be morphisms

eand_{S}: S* (x) S -> 1

iwhere 1 is the unit object in nCob (the empty set) and (x) is the tensor product in nCob (disjoint union). We get these just by taking the cylinder [0,1] x S and folding it over to look like a cup or cap! And they always satisfy the above equations, for geometrically obvious reasons. So S* really acts like the "dual" of S. If we apply Z to everything in sight, it follows that Z(S*) is the dual of Z(S). In other words:_{S}: 1 -> S (x) S*

Z(S*) = Z(S)* Z(eSo a TQFT automatically gets along with duality for objects._{S}) = e_{Z(S)}Z(i_{S}) = i_{Z(S)}

But there's more to learn from what we've done! Let's go back and see if we can completely classify 1d TQFTs. We're pretty close.

Suppose we have a 1d TQFT

Z: 1Cob -> Hilband we know Z(+). Then we know Z(-) is the dual of Z(+), and from these we can figure out Z on all objects of 1Cob, since they're all built from copies of + and -. But what about the morphisms? Well, a typical morphism in 1Cob looks something like this:

+ - - | | | v ^ ^ | | | \ / | \ / | / ___ | / \ / \ | / \ / | | | / v | | / \ | | | | \ / | | ^ \ / / | | \ / | \___/ \_/ ^ | -As you can see, it's built from caps, cups and braidings by means of composition and tensoring. A TQFT preserves composition, since it's a functor. It preserves tensor products, since it's a monoidal functor. It preserves the braiding, since it's a braided monoidal functor. And we've just seen that it preserves caps and cups! So the value of Z on any morphism in 1Cob is completely determined as soon as we know Z(+).

In short, a 1-dimensional TQFT is determined by its value on the positively oriented point. And if you think about it a while, you'll see this can be any Hilbert space you like. So the quick way to summarize the classification of 1d TQFTs is this:

**Theorem: A 1d TQFT is a Hilbert space.**

Next time we'll get a similarly snappy classification of 2d TQFTs.

But before we quit for now, let's take a Hilbert space H, use it to build a 1d TQFT with Z(+) = H, and calculate some things in this TQFT. For example, what's the partition function of the circle?"

"It's just the dimension of H," said Miguel.

"Right, we actually saw this last quarter in Week 3. It goes like this:

The circle is built by composing a cap and a cup, so our TQFT sends it to the composite of a cap and a cup, which gives a map fromC__ / \ | / \ v | | Z v ^ |-------------------> H (x) H* | | \ / | \__/ vC

Z(SBy the way, does anyone notice the subtlety I hid under the rug?"^{1}) = dim(H)

Everyone stared at the picture a minute.

"Well," said Toby, "the map H (x) H* -> **C** is not the cup for H;
it's the cup for H*. But that's not terribly subtle! We worked this
out last quarter."

"Right," conceded the Wiz. "I just felt like reminding people. If Z(+) = H, we've already seen what our TQFT does to this cup:

- + | | ^ v | | \ / \__/The answer is:

- + H* (x) H | | ^ v Z | | | |-------------------> | eBut we also need to know what our TQFT does to this cup:_{H}\ / V \__/C

+ - | | v ^ | | \ / \__/There are two answers, both correct. One is this:

+ - H** (x) H* | | v ^ Z | | | |-------------------> | eAnd the other is this:_{H*}\ / V \__/C

+ - + - H (x) H* | | | | v ^ v ^ | | | | | | | | \ / Z | BBut these are the same, if we use the usual trick for identifying H with its double dual H**._{H, H*}| | = \ / |--------> | \ / / v \_/ / \ / \ H* (x) H | | \ / | e_{H}\_/ vC

Okay, one final question. What does Z assign to the 0-sphere?"

The class was silent.

"Well, the 0-sphere consists of two points, and the usual way to orient it is to say one of these point is positively oriented, and the other is negatively oriented. For example, this:

+ -What does our TQFT assign to this?"

"Well," said Toby, "if Z(+) = H, this gets assigned H (x) H*, which is just the algebra of endomorphisms of H."

"Endomorphisms!" said the Wiz, pretending to be disgusted. "Yeah, that's what you call them when you have a Ph.D. in math. If you only have an bachelor's degree in math, you call them "linear transformations". And if you're a normal person, like a physicist, you just call 'em "matrices".

But whatever you call them, they form an algebra! And way back in Week 2, we saw that this comes straight from our diagrams. The product comes from this cobordism:

+ - + - | | | | | | | | v ^ v ^ \ \ / / \ \___/ / \ / \ / v ^ | | | | + -while the unit comes from this cobordism:

/\ | | v ^ | | | | + -Associativity and the left and right unit laws for matrix multiplication are a snap to prove using these pictures. And this is the tip of an interesting iceberg: whenever you have an n-dimensional TQFT, you can draw pictures like this to see that the Hilbert space of the (n-1)-sphere is an algebra."

"Is it always a matrix algebra?" asked Miguel.

"We'll see next week, when we do 2d TQFTs."

baez@math.ucr.edu © 2001 John Baez