"What's wrong?" asked the Wizard. "Is the clock in your office running a bit fast?"
"No," said Toby. "Don't you remember? I'm a fictional character now! I don't have anything to do except attend this course!"
"Oh, right," said the absent-minded Wizard, who had forgotten the events of the last episode. "But that's no excuse for behaving so unrealistically! Unless, of course, I make you. From now on, please wait until the bell rings."
"Okay," said Toby, exchanging a dubious glance with Miguel.
"Good," said the the Wiz. "Now it's time to get good at spin network calculations. Last time we finished Track 1 with a puzzle: show that
| |
1| 1|
| |
/ \ |
/ \ = c |
1\ /1 |
\ / |
| |
1| |
| |
for some constant c, and work out the value of this constant."
"You know," said Miguel, "we actually solved this puzzle last week, in the discussion about vector networks and the four-color theorem. We were using different notation, but if you translate it back, you get
| |
1| 1|
| |
/ \ |
/ \ = - |
1\ /1 |
\ / |
| |
1| |
| |
Remember?"
The Wiz grinned. "You're right! You saw through my trick! But now we'll solve this puzzle a different way, and see something really cool: any spin network with one spin-j edge coming in and one spin-j edge coming out must be a multiple of the identity. In short:
| |
j| j|
__|__ |
| | |
| | |
| S | = c |
| | |
|_____| |
| |
j| |
| |
where the box labelled S stands for a spin network.
Let's see how it goes. Someone hands us any spin network with one spin-j edge coming in and one spin-j edge coming out, say this one:
|
1|
|
/ \
/ \
1\ /1
\ /
|
1|
|
|
How can we simplify it?"
"First go back to the definitions and write it out using caps, cups and symmetrizers," said Jay.
"Okay, let's try that! By definition we have
| 1|
1| |
| -----
/ \ / _ \
/ \ / / \ \
1\ /1 = --- ---
\ / \ \_/ /
| \ /
1| -----
| |
| 1|
where the unlabelled strands are the spin-1/2 representation,
and the horizontal lines are symmetrizers.
We can always do this. Then what?"
A bit hesitantly, Jay said "We can expand out some of the symmetrizers..."
"Right!" said the Wiz. "Which ones? All four?"
"No," said Jay. "Just the two in the middle."
"Okay, great!" said the Wizard, obviously pleased. "In general, we expand out all the symmetrizers except those at the very top and bottom. In our example, we get:
1| 1| 1| 1|
| | | |
----- ----- ----- -----
/ _ \ / _ \ / _ \ / _ \
/ / \ \ / / \ \ / / \ \ / / \ \
| | | | / | | | | | | \ / | | \
| | | | \ / | | | | \ / \ / \ /
(1/4) | | | | + (1/4) / | | + (1/4) | | / + (1/4) / /
| | | | / \ | | | | / \ / \ / \
| | | | \ | | | | | | / \ | | /
\ \_/ / \ \_/ / \ \_/ / \ \_/ /
\ / \ / \ / \ /
----- ----- ----- -----
| | | |
1| 1| 1| 1|
Now what?"
Miguel said, "We can simplify each term using the skein relations."
"Right! Each term is a mess of spaghetti, and our goal is to untangle this spaghetti until we're left with a bunch of straight vertical strands together with some closed loops, called, umm -- what do they call pasta that's shaped like a loop? Rigatoni? Tortellini? No. It's on the tip of my tongue...."
Some genius yelled out the right answer: "Spaghettios!"
The Wizard laughed. "Right! To do this disentangling, remember that our pictures are really living in 4-dimensional spacetime, so we can untie any knot and unlink any link. We just need to remember that a twist gives a minus sign:
| |
____ | |
/ \ | |
/ \ / |
/ \ / |
| / = - |
\ / \ |
\ / \ |
\____/ | |
| |
| |
For example, if we disentangle the four terms above, we get
1| 1| 1| 1|
| | | |
--- --- --- ---
| | | | | | / \
| | | | | | | |
| | | | | | | |
| | _ | | | | \_/
(1/4) | | / \ - (1/4) | | - (1/4) | | + (1/4) _
| | \_/ | | | | / \
| | | | | | | |
| | | | | | | |
| | | | | | \ /
--- --- --- ---
| | | |
1| 1| 1| 1|
Now what?"
Miguel said, "We can always eliminate spaghettios using this formula:
______
/ \
/ \
| | = -2
\ /
\______/
But for the last term..."
"Don't worry about that yet!" interrupted the Wizard. We're outlining a general proof here, so let's be systematic. First we eliminate all the loops, getting factors of -2. In our example we get:
1| 1| 1| 1|
| | | |
--- --- --- ---
| | | | | | / \
| | | | | | | |
| | | | | | | |
| | | | | | \_/
- (1/2) | | - (1/4) | | - (1/4) | | + (1/4) _
| | | | | | / \
| | | | | | | |
| | | | | | | |
| | | | | | \ /
--- --- --- ---
| | | |
1| 1| 1| 1|
We're left with terms having only strands going straight down,
but maybe also terms with strands that double back,
like the last one here.
The terms with strands going straight down are just multiples of the identity operator, since
j|
|
-------
||...||
||...||
||...||
||...||
||...||
-------
|
j|
is just the inclusion of the spin-j representation
into the (2j)th tensor power of the spin-1/2 representation,
followed by the projection back down to the spin-j rep --
or in other words, the identity!
So far, so good. But what about the terms with strands that double back and go out the same way they came in:
1|
|
---
/ \
| |
| |
\_/
_
/ \
| |
| |
\ /
---
|
1|
For example, what's this?"
The class stared at it doubtfully. Then a strange light shone in Toby's eyes, and he shouted "It's zero!"
The Wiz smiled. "Why?"
"Because the spin-1/2 representation is a symplectic vector space, so the cup is skew-symmetric:
| | | |
\ / | |
/ = - | |
/ \ | |
\_/ \_/
On the other hand, the symmetrizer is symmetric,
even when you turn it upside down:
1| 1|
| |
----- -----
| | | |
\ / | |
/ = | |
/ \ | |
/ \ | |
so we have
1| 1| 1|
| | |
--- ----- ---
/ \ | | / \
| | \ / | |
| | = / = - | |
| | / \ | |
\_/ \_/ \_/
and thus
1|
|
-----
/ \
| |
| | = 0
| |
\_/
In fact, the same kind of argument shows we get zero
whenever we use a cup to hook up two spin-1/2 strands
coming out of the same symmetrizer, like this:
3/2|
|
-----
| | |
| | \
\| \ = 0
\ \
|\ \
| \___/
|
If something is symmetric and also skew-symmetric, it's zero!"
"Right!" said the Wiz. "That's a good observation. I'm glad I fictionalized you, Toby: now we'll always have someone who answers my questions correctly when everybody else gets stuck. Have you ever read the Platonic dialogs, or the ones Galileo wrote? That's how it works there, too!"
"But it's so artificial!" said Miguel. "I wish you'd turn him back into a real person."
The Wizard glared at Miguel and continued, "Anyway, we see that terms with spin-1/2 strands that reenter the same symmetrizer they came out of must vanish, so
1| 1| 1| 1| |
| | | | 1|
| --- --- --- |
| | | | | | | |
/ \ | | | | | | |
/ \ | | | | | | |
\ / = - (1/2) | | - (1/4) | | - (1/4) | | = - |
1\ /1 | | | | | | |
| | | | | | | |
| --- --- --- |
| | | | |
1| 1| 1| 1| |
More generally, our tricks suffice to show that
| |
j| j|
__|__ |
| | |
| | |
| S | = c |
| | |
|_____| |
| |
j| |
| |
whenever S has one spin-j strand coming in
and one spin-j strand going out.
But what about this sort of spin network, where the spin coming in is different from the spin going out?
|
j|
__|__
| |
| |
| S |
| |
|_____|
|
k|
|
What does it equal? What could it equal?
It can't be a multiple of the identity."
Toby said: "It's zero!"
"Right! And the proof is easy: we expand out all the symmetrizers except at the top and bottom, get a lot of terms, disentangle all the spaghetti in each term, remove the spaghettios, and get a bunch of terms where 2j strands come in at the top and 2k go out the bottom. If j is not equal to k, each of these terms must have strands that double back and reenter the same symmetrizer! So they're all zero.
In short,
|
j|
__|__
| |
| |
| S | = 0 if j is not equal to k
| |
|_____|
|
k|
|
Now, do you know whose famous Lemma we have secretly been discussing?"
Cyrill the part-time Acolyte knew the answer: "Schur!"
"Right! Schur's Lemma says that any intertwining operator between irreducible group representations is a multiple of the identity if they're the same, but zero if they're inequivalent. Once you prove that the spin-j representation of SL(2,\C) is irreducible (which is not too hard), and that every spin network defines an intertwining operator (which is obvious from stuff we've done), our results become a puny special case of this Lemma. But the fun thing is that to state and prove our results we never needed to talk about groups or representations. We just played around with pictures!
In the jargon of logic, we've given a "syntactic" proof, where we show (for example) that
| |
j| j|
__|__ |
| | |
| | |
| S | = c |
| | |
|_____| |
| |
j| |
| |
by taking any expression of the left-hand sort
and manipulating it according to our rules
until it turns into one of the right-hand sort!
If our proof seemed strange, perhaps it's because
you're used to syntactic proofs where one manipulates 1-dimensional strings,
rather than higher-dimensional patterns of symbols.
But now we're doing "higher-dimensional algebra" --
and therefore, secretly, n-category theory.
But I digress. You're not supposed to know this stuff yet!
The next question is: given that
| |
j| j|
__|__ |
| | |
| | |
| S | = c |
| | |
|_____| |
| |
j| |
| |
how can we find the constant c?
The answer is simple. Close up both sides in a loop like this:
______ _____
/ \ / \
| | | |
j| | j| |
__|__ | | |
| | | | |
| | | | |
| S | | = c | |
| | | | |
|_____| | | |
| | | |
j| | | |
| | | |
\______/ \_____/
to get an equation between numbers, and then solve for c:
______
__/__ \
| | |
| | |j
| S | |
| | |
|_____| |
\______/
c = ______________
_____
/ \
| |j
| |
\_____/
Voila!" The Wizard chortled with pleasure.
Some of the Acolytes looked pleased, but others stared suspiciously at this equation, so the Wiz explained: "Yes, we're stretching our notation a bit more than ever before. We've added open spin networks before, which is okay, because they represent operators. We've also multiplied them by constants. But now we're dividing one closed spin network by another! That's okay too, because they represent numbers. Well, at least it's okay if the denominator is nonzero.... Here
_____
/ \
| |j
| |
\_____/
is the "superdimension" of the spin-j representation:
its ordinary dimension, 2j+1, if j is "bosonic" (j = 0, 1, 2, ...)
but minus its ordinary dimension if j is "fermionic" (j = 1/2, 3/2, 5/2, ...)
Similarly,
______
__/__ \
| | |
| | |j
| S | |
| | |
|_____| |
\______/
is the "supertrace" of the operator S: j -> j, that is,
its ordinary trace if j is bosonic,
but minus its ordinary trace if j is fermionic.
Anyway, here's the end result:
| ______ | | __/__ \ | j| | | | j| | | | | | | | S | |j | __|__ | | | | | | |_____| | | | | \______/ | | S | = ______________ | | | _____ | |_____| / \ | | | |j | | | | | | \_____/ | j| | | |This will be really useful later.
Okay... let's take a short break." The students stood up, and chatting amongst themselves, they wandered down to the soda machine for refreshments.