The students stopped chatting and prepared to take notes, but before the Wiz could say anything, there was a huge explosion:

The noise was deafening, the flash was blinding, and it took everyone a while to recover. When they did, they saw a strange sight: another copy of the Wizard standing beside the original one! They were almost indistinguishable. The only difference was that the original Wiz was blackened by smoke, while the copy, WizKABLOOWIE!!!

"What are you doing here?!" asked the original Wiz in an angry but somehow not terribly surprised tone of voice.

"You can guess: I'm changing the topic of today's lecture," replied Wiz_{2}.

"But I'm talking about something important! I was just going to tell them --"

"Yes," said Wiz_{2}, "But next quarter you'll regret doing that.
It's more important to explain the Biedenharn-Elliot identity."

The Wiz pondered this news. "Hmm... maybe you're right."

"Of course I am! I've been there!"

"Okay. The trouble is, I'm not prepared to talk about that subject."

"I'll do it," said Wiz_{2}. "Here, just sit down and take notes;
that way when the next quarter rolls around, you -- I mean I --
will be prepared."

The Wiz rolled his eyes. "The old time travel cliches, eh? All this work on quantum gravity, and for what -- just another cheap plot device?"

Wiz_{2} became annoyed. "Shut up and take notes!
If you'd planned ahead for next quarter's lectures,
we wouldn't be in this jam."

The original Wiz apologized to the class, went over to a desk in the front row, sat down, and opened his notebook. "Okay, go ahead."

Wiz_{2} went to the front of class and began to lecture.

"We defined the 6j symbols as follows:

a\ b/ c/ a\ b\ c/ \ / / \ \ / \/ / \ \/ \ / \ / j\ / = sum {a b i} \ /i \ / i {c k j} \ / | | | | k| k|and we saw how they're related to the tetrahedron:

_________ / \ /\j | _____ _____ / \ | / b \ / a \ / /\ | / \ / \ a/ b/ \c |k /_________\ /_________\ \ / / | \ c / \ i / \/ / | \ / \ / \ / | \_____/ \_____/ i\/ | i k \_________/ {a b i} = __________________________________________________ {c k j} _____ / \ / \i \ / \_____/We can prove various identities for the 6j symbols using our picture tricks. These deepest of these is the Biedenharn-Elliot identity. To prove this, we express the operator

a\ b/ c/ d/ \ / / / \/ / / \ / / e\ / / \/ / \ / f\ / \ / | | g|as a linear combination of these:

a\ b\ c\ d/ \ \ \ / \ \ \/ \ \ / \ \ /j \ \/ \ / \ /i \ / | | g|in two different ways, and equate the coefficients.

The first way goes like this:

a\ b/ c/ d/ a\ b\ c/ d/ \ / / / \ \ / / \/ / / \ \/ / \ / / \ / / e\ / / \ /h / \/ / = sum {a b h} \/ / \ / h {c f e} \ / f\ / f\ / \ / \ / | | | | g| g| a\ b\ c/ d/ \ \ / / \ \/ / \ \ / = sum {a b h} {a h i} \ h\ / h,i {c f e} {d g f} \ \/ \ / \ /i \ / | | g| a\ b\ c\ d/ \ \ \ / \ \ \/ \ \ / = sum {a b h} {a h i} {b c j} \ \ /j h,i,j {c f e} {d g f} {d i h} \ \/ \ / \ /i \ / | | | g|That took 3 steps. The second way goes like this:

a\ b/ c/ d/ a\ b/ c\ d/ \ / / / \ / \ / \/ / / \/ \/ \ / / \ / e\ / / e\ /j \/ / = sum {e c j} \ / \ / j {d g f} \ / f\ / \ / \ / | | | | | g| g| a\ b\ c\ d/ \ \ \ / \ \ \/ \ \ / \ \ /j = sum {e c j} {a b i} \ \/ i,j {d g f} {j g e} \ / \ /i \ / | | | g|That took just 2 steps! Equating the coefficients, we get the Biedenharn-Elliot identity:

sum {a b h} {a h i} {b c j} = {e c j} {a b i} h {c f e} {d g f} {d i h} {d g f} {j g e}You'd have to be insane to remember this formula. What you should remember instead is that there are two ways to go from this tree:

\ / / / \/ / / \ / / \/ / \ / \ / | |to this one:

\ \ \ / \ \ \/ \ \ / \ \/ \ / \ / | |using this basic move:

\ / / \ \ / \/ / \ \/ \ / ===> \ / \ / \ / | | | |One way takes 2 steps, the other takes 3, but both give the same result. In other words, we have a commutative diagram like this!" He drew an elaborate pentagonal diagram in the air with his staff:

\ / \ / \/ \/ \ / \ / \ / \ / / / | \ \ \ / \/ / / =======> | ========> \ \ \/ \ / / | \ \ / \/ / \ \/ \ / \ / \ / \ / | \ \ / / \ \ / / | | ===> \ \/ / \ \/ / ===> | | \ / / ===> \ \ / | \/ / \ \/ \ / \ / \ / \ / | | | |It glowed blue-white as it hovered delicately in the air, and the class gasped with awe, except for the original Wiz, who looked a bit jealous.

"The vertices of this diagram consist of all 5 binary trees with 4 leaves. You've probably heard that wizards use pentagrams for all sorts of magical purposes. But now you see why! It's no coincidence that the Biedenharn-Eliott identity is also called the pentagram equation."

"You mean *pentagon* equation," interrupted the original Wiz.

"Right, sorry. Now, as I hinted last week, the Biedenharn-Elliot identity is the key to seeing why the partition function for 3d quantum gravity doesn't depend on how you chop spacetime into tetrahedra. To understand this, you need to know Pachner's Theorem.

Any compact 3-manifold can be triangulated -- i.e., chopped into tetrahedra. But there are usually lots of ways to do this. Pachner's Theorem says you can go between any two of these triangulations using a finite sequence of moves called "Pachner moves".

The "1-4 Pachner move" lets you take 1 tetrahedron

/|\ / | \ / | \ / | \ / | \ /____ | ____\ \ | / \ | / \ | / \ | / \ | / \|/and chop it into 4 tetrahedra by adding a new vertex in the middle, and connecting it with new edges to all 4 vertices of tetrahedron you started with. The result looks like this:

**
ERROR - ASCII GRAPHICS CAPABILITIES EXCEEDED
**

Hey! What's that?" Instead of the desired picture,
a big ugly error message hovered in front of Wiz_{2}.
He tried again to draw the picture, waving his staff in the air as before,
but only got another error message.

"You forgot," said the original Wizard, "that we're in an ASCII environment here."

"Fire and damnation!" cried Wiz_{2}.
"How can I teach this stuff without decent graphics?"
He muttered a spell of banishment and the error messages disappeared.
Turning sadly to the class -- and to you, the reader -- he said,
"Well, my Acolytes, I'm afraid you'll just have to visualize this yourself:
1 tetrahedron chopped into 4."

"We can also turn 4 back into 1, right?" asked Toby.

"Yes, the Pachner moves are all reversible. Besides the 1-4 move, there's another: the "2-3 Pachner move". This lets you take 2 tetrahedra glued together along a triangle:

/|\ / | \ / | \ / | \ / | \ /.....|.....\ \. | ./ \ . | . / \ .|. / \ | / \ | / \|/and chop it into 3 tetrahedra by drawing a new edge from the top vertex to the bottom, like this:

**
ERROR - ASCII GRAPHICS CAPABILITIES EXCEEDED
**

Curses -- I did it again!" In frustration and rage, he swung his staff and hurled a thunderbolt at the error message, blowing it to smithereens.

"Ahem. Pachner's Theorem says that the 1-4 and 2-3 moves are all we need to go between any two triangulations of a compact 3-dimensional manifold. More generally, his theorem describes moves like this in any dimension! But we'll talk about that later. For now, let's see how the 2-3 move is related to the Biedenharn-Elliot identity.

The idea is simple: apart from some fudge factors, the 6j symbol

{a b i} {c k j}is basically just the evaluation of this tetrahedral spin network, or "tet net" for short:

/|\ / | \ / | \j / k| \ / | \ a/___________\ \ b / \ | / \ | /c i\ | / \ | / \|/The Biedenharn-Elliot identity has a product of two 6j symbols on one side, and three on the other. If we write it out using tet nets, we get the 2-3 move! Of course, we also get some fudge factors. But these follow a simple pattern: we get a loop

____ / \ / \j \ / \____/for each edge labelled by some spin j, and the inverse of a theta net

1 __________________ _____ / j1 \ / \ /_________\ \ j2 / \ / \_____/ j3for each triangle with edges labelled by spins j1, j2, j3."

Oz was getting very confused. "Umm, I don't follow this at all."
Both the original Wiz and Wiz_{2} turned towards Oz
with identical sinister smiles on their face.
He began shaking in his boots. One Wizard was bad enough!
What would two do?

He was actually relieved when they said, in unison, "Okay, then, this will be homework: use our spin network formula for the 6j symbols to write out the Biedenharn-Elliot identity using diagrams. You should get the 2-3 Pachner move together with some extra fudge factors. Show these fudge factors satisfy the above rule."

The class groaned. More homework, thanks to Oz!

"No, you really *must* do this to understand 3d quantum gravity,"
insisted the Wiz. "You must see for yourself how, just as
the 6j symbols, associator, and tetrahedron
are three faces of the same concept,
the Biedenharn-Elliot identity, pentagon equation, and 2-3 Pachner move
are also secretly the same."

And with a wave of his arm, he conjured up the following chart:

SU(2) REPRESENTATION THEORY TOPOLOGY CATEGORY THEORY Clebsch-Gordon coefficients triangle tensor product 6j symbols tetrahedron associator Biedenharn-Elliot identity 2-3 Pachner move pentagon equationWiz

"Of course, this is just the tip of the iceberg. And by the way, those fudge factors aren't so bad: just remember the Euler characteristic! To compute the Euler characteristic of a triangulated manifold, we count the number of vertices, minus the number of edges, plus the number of triangles, minus the number of tetrahedra, and so - an alternating sum. The result is independent of the triangulation.

The partition function in 3d quantum gravity is similar, but subtler. We triangulate our 3-manifold, label the edges with spins, and compute a tet net for each tetrahedron, a theta net for each triangle, and a loop for each vertex. Then we multiply all the tet nets, divide by all the theta nets, and multiply by all the loops. Finally we sum over labellings. The Biedenharn-Elliot identity says the result is preserved by the 2-3 move."

"And it's also invariant under the 1-4 move?" asked Toby.

Wiz_{2} smiled craftily. "No, you'll notice I never said that!
In fact, the sum over labellings usually *diverges*.
To see this, just write down the 1-4 move. If you stare at it,
you'll see the sum over spins labelling the 4 new edges
has infinitely many terms, and no good reason to converge.
This problem doesn't afflict the 2-3 move,
for reasons the wise among you can easily discover.
(Hint: admissible triples must satisfy the triangle inequality.)

To fix this problem, we need to replace SU(2) by the corresponding quantum group. This has only finitely many relevant irreducible representations, corresponding to spins j = 0, 1/2, 1, ... on up to k/2 for some integer k. Then the sum over ways of labelling edges becomes a finite sum, and everything works fine. So: our partition function won't really converge, and the 1-4 move won't work, until we replace SU(2) by a quantum group."

Jay looked unhappy. "But what does it mean *physically*
to replace SU(2) by a quantum group? Is it just a trick?"

Both the Wizard and Wiz_{2} glared.
"It's not just a trick!
It comes from putting a nonzero cosmological constant into our theory.
If we do this with the right sign,
it makes the action in the classical theory grow in proportion to
the volume of spacetime.
In the path integral for the quantum theory,
this damps out the contribution of large-volume spacetimes.
In other words, it imposes a kind of "infrared cutoff",
which gets rid of infinities in the path integral.
When we sum over labellings of the edges of a triangulated 3-manifold by spins,
we're doing a discretized version of this path integral.
The spins labelling the edges correspond to their *lengths*,
so we are really summing over all possible geometries of spacetime.
When we do quantum gravity with cosmological constant, there's
a cutoff k on the spins we need to sum over -- a maximum length!
That's how the infrared cutoff shows up in this discretized theory."

Miguel had a question too: "You say the Euler characteristic is an alternating sum, while the partition function for 3d quantum gravity is an alternating product.

To get from a sum to a product, you take the exponential. So: is there some TQFT whose partition function is the exponential of the Euler characteristic?"

"Yes!" cried Wiz_{2}.
"It's perhaps the most basic TQFT of all: the Euler theory.
We'll talk about that more later.
But now, I have to leave. Class dismissed! See you next quarter!"

And with an enormous implosion, he disappeared:

leaving the original Wizard and the class to collect their wits and eventually have dinner.BANG!!!

baez@math.ucr.edu © 2001 John Baez