The Wizard walked into class with a sheaf of papers under his arm. "I want you all to take a look at this." he said "I made a copy of everything for each of you. Amanda -- the Track 1 Acolyte who did a problem on the board back in Week 2 -- has done such an excellent job on the homework that I'm going to hand it out to all of you to look at.".
After handing it out, the Wizard took on a sterner look. "On the other hand, Jianjun, the Acolyte that's Never Been Mentioned Yet, hasn't done any of the homework. You may notice he's not here today. Regretfully, I had to turn him into a frog.".
Then the Wizard turned his stern gaze through space and time into your eyes, as you sit reading this. "I hope you have been doing the homework." he said.
"Hic!" said Toby the Acolyte, involuntarily.
"Now let's take a look at how orthogonal and symplectic structures relate to the Reidemeister moves. You may recall --".
"Excuse me, Mr Wizard, Sir," said Oz "but I wonder if you might be persuaded to give a summary of what we've done so far. I must say I'm getting rather tangled up in knots here.".
"Well," said the Wizard "tangled up in knots is an excellent position for studying the Reidemeister moves. But I take it that you were speaking metaphorically -- so I'll try to straighten you out.
Fundamentally, we've been dealing with vector spaces (complex vector spaces of finite dimension, to be precise) and linear maps between the vector spaces. The collection of vector spaces and linear maps, which we can call "Vect" for short, has several structures.
First of all, there's a 1-dimensional structure; you can put linear maps on top of one another vertically to compose them. This makes Vect into what we wizards call a "category". Then you can add another dimension to the structure by placing things side by side using the tensor product; wizards have a name for this too, which is "monoidal category". Then you can add a 3rd dimension by including braiding in front and behind, making this what wizards call a "braided monoidal category". Finally, you can use the 4th dimension by allowing the 2 different kinds of braiding to pass through each other; wizards call this a "symmetric monoidal category".
That's all you can do with the dimensions; 5 dimensions gives you nothing new. Topologically, this is because all knots, or even tangles, can be untied already in 4 dimensions, so the 5th dimension adds nothing. Algebraically, category theory explains this by saying that a symmetric monoidal category is a stably monoidal category -- but I'm not going to explain what that means today.
So, we can conclude that Vect is a 4-dimensional kind of thing. Spacetime is also 4-dimensional -- string theorists notwithstanding. Remember our spacetime diagram from week 1? We'll see how this fits in with symmetric monoidal categories later.
And, oh, yes, there is one other sort of structure in Vect; we have a notion of duals on objects, taking V to V* and correspondingly taking T to T* for T a linear map. But, if we stick an orthogonal or symplectic structure on V, then V and V* can be identified using the isomorphisms # and b.
Feel better, Oz?".
"Much." said Oz "Thank you.".
"No problem." said the Wizard.
"Hic!" said Toby the Acolyte, involuntarily.
"Now then, let's consider the Reidemeister moves. The 2nd and 3rd Reidemeister moves are only about the braiding, so they don't have any problems with orthogonal and symplectic structures. But the 1st Reidemeister move is different. Let's recall it:
|
____ |
/ \ |
/ \ /
/ \ /
| /
\ / \
\ / \
\____/ |
|
|
Before, we constructed this out of eV and iV*.
But now we'll want to construct it out of the cup and cap
that we get from the symplectic or orthogonal structure.
The question is: Will this still equal the identity
when you pull it straight? Any takers?".
John the Acolyte, on the grounds that he hasn't spoken for a while, suggested "For an orthogonal structure, yes, but not for a symplectic structure.".
"Exactly right." said the Wizard "Symplectic structures have those annoying minus signs, and this is one of the places where they show up. This picture is equal to the identity for an orthogonal structure but equal to minus the identity for a symplectic structure.
Logically, this isn't a problem. Nobody ever said that you have to get the same results using the symplectic cup and cap as you got using the natural cup for V and cap for V*. But it's a problem for our geometric intuition. After all, if the vector space V is just a string, then we *should* be able to pull that picture straight, right?
The solution is to say that V is not a string but instead a ribbon If you lay a ribbon out flat to match this picture and pull it straight, then you'll get a 360 degree twist -- not quite the identity. We see that a 360 degree twist is the identity for an orthogonal structure but minus the identity for a symplectic structure. Either way, a 720 degree twist is the identity, which might sound a bit familiar to those of you who know a little about electrons.".
"Does this ribbon picture work even in 5 dimensions?" asked Richard the Acolyte.
The Wizard thought for a moment. "Yes, it does." he said "Anyway, I've been telling you what the 1st Reidemeister move is; let's prove it! First, we start with
|
____ |
/ \ |
/ \ /
/ \ /
| /
\ / \
\ / \
\____/ |
|
|
Then use the 2nd Reidemeister move to turn this into
|
____ |
/ \ |
/ \ /
/ \ /
| /
| ______/ \
\ / \
/ |
/ \ |
| | |
\ / |
\ |
/ \ |
| | |
\_/ |
|
|
Next, use this general fact:
| | | |
| | | |
V v | V v |
| | \ /
| | \ /
/ \ | /
| T | | / \
\_/ | / \
| v X = X v |
\ / | / \
\ / | | T |
/ | \_/
/ \ | |
/ \ | |
| v W | v W
| | | |
| | | |
applied to the case where T is the symplectic or orthogonal cap, to get
|
|
|
/
/ ______
/ / \
| / \
\ / |
\ |
/ \ |
| | |
\_/ |
|
|
Now we know how to untwist this,
getting a minus sign in the symplectic case,
and then we really can pull things straight.
So, the 1st Reidemeister moves works just like I said it did.".
"But what about that "general fact" up above?" asked Oz "You never proved that!".
"Well, you can prove that for yourself." said the Wizard "This will be homework.".
"Gulp!" said Oz.
"Hic!" said Toby the Acolyte, involuntarily.
"Have people taken to drinking in class?" asked the Wizard, looking around to spot the source of these sounds, "No? Good.".
"Now," he continued "let me show you how to solve the puzzle posed by my alter ego in Week 4. This will be the basis of much magic to come, so watch carefully.
First, let me remind you of his puzzle! He wanted you to find operators
| |
| |
V v v V
\ /
\_______/
and
_______
/ \
/ \
V v v V
| |
| |
such that these equations hold:
| | |
| | |
| | |
V v _______ | _______ V v
| / \ = | = / \ |
| / \ | / \ |
| V v | V v | V v |
\ / | | | \ /
\_______/ | | | \______/
V v | V v
| | |
| | |
| | |
_______
/ \
/ \
| |
| |
V v v V = -2
| |
| |
\ /
\_______/
and most importantly of all,
| | | | | |
| | | | | |
V v v V V v v V | |
\ / \ / | |
\ / \_______/ | |
\ / | |
/ = + V v v V
/ \ _______ | |
/ \ / \ | |
/ \ / \ | |
V v v V V v v V | |
| | | | | |
| | | | | |
which is called the "binor identity".
To solve this puzzle, start with the binor identity. Put one of his strangely modified cups on bottom to get this:
| | | | | |
| | | | | |
V v v V V v v V | |
\ / \ / | |
\ / \_______/ | |
\ / | |
/ = + V v v V
/ \ _______ | |
/ \ / \ | |
/ \ / \ | |
V v v V V v v V | |
| | | | | |
| | | | | |
\ / \ / \____/
\___/ \_______/
Now, since he said
_______
/ \
/ \
| |
| |
V v v V = -2
| |
| |
\ /
\_______/
we get:
| | | |
| | | |
V v v V | |
\ / | |
\ / = - V v v V
/ | |
/ \ | |
/ \ | |
V v v V \ /
\ / \ /
\_/ \_____/
And what does this imply?".
"We've got a symplectic structure!" said Jay the Acolyte.
"Not quite." said the Wizard "It says our operator
| |
| |
V v v V
\ /
\_______/
is antisymmetric -- but a symplectic structure must also be nondegenerate.
Luckily, the equation
| | |
| | |
| | |
V v _______ | _______ V v
| / \ | / \ |
| / \ | / \ |
| V v | = V v = | V v |
\ / | | | \ /
\_______/ | | | \______/
V v | V v
| | |
| | |
| | |
comes to the rescue here, and gives us nondegeneracy.
Once know that
| |
| |
V v v V
\ /
\_______/
is a symplectic structure on V, the equation
_______
/ \
/ \
| |
| |
V v v V = -2
| |
| |
\ /
\_______/
says the dimension of V must be 2.
Here I'm using the homework from Week 5.
So the answer is: V is 2-dimensional, the cup
| |
| |
V v v V
\ /
\_______/
is any symplectic structure on V, and
_______
/ \
/ \
V v V v
| |
| |
is the corresponding cap.
Since V is 2-dimensional, we might as well call it C2. Vectors in this space describe states of a spin-1/2 particle, so people often call them "spinors" -- but the wizard Roger Penrose preferred to call them "binors", since there are also other things called spinors.
Next week, I'll tell you why this stuff is so great!"
"Hic!" said Toby the Acolyte involuntarily.