Suddenly, he glanced up. The Wizard was looking at him. "Yes, Sir." said Oz, without knowing way, and sat down quietly.
"Today," said the Wizard "we've had enough of complex vector spaces. It's time to do something a bit more exciting -- vector spaces with orthogonal and symplectic structures!
An orthogonal structure is like an inner product. However, it's not like the complex inner product, it's like the ordinary real inner product. So, it's best not to call it "inner product" at all but something like "metric" instead. Of course, you can call it "orthogonal structure", but that's a tad wordy -- plus, nobody knows what it means.
A metric is a way to combine 2 vectors into a single scalar. So, we know how to draw it in a diagram:
| | | | V v v V \ / \_______/A lot like the cup, except that there are no dual spaces in sight.
Now, a metric has some requirements on it. First, it must be symmetric. In other words, we consider this twisted version of the cup:
| | | | V v v V \ / \ / / / \ / \ V v v V \ / \_/and say that this has to equal the untwisted cup. So, you see that when you put 2 vectors into the metric, it doesn't matter what order you place them in.
The other requirement on the metric is that it is nondegenerate. What does this mean? Well, consider this creature:
| | | V v _______ | / \ | / \ | V v | \ / | \_______/ | V ^ | | |This has the metric down below on the left side and the usual cap up above on the right side. Overall, it's a map from V to V*, called "#", or "sharp". We can abbreviate # like this:
| | V v | | V ^ | |which you get just by pulling the definition of # straight.
When I say that the metric is nondegenerate, what I mean is that # is an *isomorphism*. What does it mean for the map # to be an isomorphism? Well, since all are maps are linear, all there is to it is that # has an inverse; the inverse of # is called "b", or -- Hang on, wait a minute, let me try that again. The inverse of # is called "b" -- I mean, "b" -- Darn it!" exclaimed the Wizard, shaking his wand vigorously, "I can't seem to get that stupid flat to come out right; it just keeps looking like a lowercase letter "B"! I must have forgotten to install all the fonts on my wand. Anyway, the inverse of sharp is called "flat".
b -- I mean, flat -- is drawn like this:
| | V ^ | | V v | |So, to say that # and b are inverses is to say that
| | | | V ^ | | | | | V v = ^ V | | | | V ^ | | | | |and
| | | | V v | | | | | V ^ = v V | | | | V v | | | | |In the first case, you can see that we've got # on top and b down below; in the second case, you can see that we've got b on top and # down below. So, # and b are inverses. It's all pretty natural.".
The class groaned.
"Let's look at what these do to some basis vectors. If ei and ej are basis vectors in V, then we know what the metric does if we know what it does to the basis element ei(x) ej of V (x) V. The metric should map ei(x) ejto a complex number. We'll call this complex number "gij". That's the definition of the symbol "gij".
Now, what does # do to a vector in V? We'll know if we can tell what it does to a basis element ei. So, look at the diagram that defines #. First, we have ei going through the pipe. Now, what does that cap do on the right side? Who remembers?".
Oz was about to speak, but then he quickly huddled down in his chair.
"Oz???" said the Wizard.
"Well," said Oz, "I think that's something you wrote down in terms of basis vectors of V and V*. I mean, if you look at the cap only, then you have V on the left and V* on the right. So, if (ei) is a bunch of basis vector thingies for V, then I guess you get a bunch of basis vector thingies for V*, say (fj). Then you want to tensor these together, so you get ei (x) fj. Something like that.".
The Wizard glared at him. "ei (x) fj" for which i and j?" he said.
"Well, all of them, I guess." said Oz "You just sum them, right? Isn't that the Heisenberg summation convention or something?".
"It's Einstein summation convention -- Einstein's greatest contribution to mathematics, some say -- and you're doing it all wrong. The summation convention applies when you have 2 copies of the *same* index: i and i, not i and j. So, you need ei(x) fi. Also, the summation convention only applies when one of the indices is up and the other is down. Since the basis vector eihas the index down, then we must have ei (x) fi instead. Now the Einstein summation convention is making sense.
But we can't use just any old basis (fi) of V*! We have to use the particular basis which is dual to (ei). We use the same letter "e" for this basis, to indicate that it's the dual. The dual, remember, is the basis which satisfies ei(ej) = deltaij. So, the cap on the right side actually becomes ei (x) ei. Finally, we can't use the letter "i" at all, because it's being used over here on the left side for the basis vector that we put into # in the first place. So, we really have ej (x) ej. And that's the answer, with the Einstein summation convention telling us to sum over j.".
While the Wizard was explaining this, Oz was rapt with attention until something strange slowly dawned on him. "Erm, that's very interesting, Mr. Wizard, Sir, but ... why didn't the fireball hurt me?".
"Why, that's because of the fireball protection I gave you 3 weeks ago." said the Wizard "Don't you remember?".
"But you said that that wore off after, what was it, 6 1/2 days.".
The Wizard shrugged. "So I lied.". Then he gave Oz a cold, hard stare. "But now it really *is* gone, so be good!".
"Eep!" said Oz.
"Now," said the Wizard "where was I? Oh, yes! I put ei into #. So far, we've got ei (x) ej (x) ej. Then ei (x) ejgets tossed into the metric, and I just said that gijis what comes out. So, we find that #(ei) = gijej, where again the Einstein summation convention tells us to sum over j.
Now, what about flat? b(ei) should be gijejfor some matrix (gij). How does this matrix gijrelate to the matrix gij? Well, let's look at the equations that # and b are supposed to satisfy.
One says that if you do # and then b, you get back where you started. So, do that to ei. You get b(#(ei)) = b(gijej). Now, b is linear, so we can pull the complex numbers gijoutside, as well as that implicit sum that we get from the Einstein convention. This gives b(#(ei)) = gijb(ej) = gijgjk ek. But we know that b(#(ei)) is supposed to be just ei! How can we write ei in terms of ek? Any ideas?".
Then Jay the Acolyte, who, as a physicist, knows all about index juggling, said "You use the Kronecker delta: ei = deltak iek.".
"Right! The Kronecker delta - deltaki! It's 1 when i equals k, and zero otherwise. Some say it was Kronecker's greatest contribution to physics." said the Wizard.
"So, we get deltaik ek = gijgjk ek. And, if this is going to be true for any i, it must be that gijgjk = deltaik. Similarly, if you look at #(b(ei)), you'll calculate that gijgjk= deltaik. If you think about matrix multiplication a bit, you'll find that what these 2 equations really say together is that the matrices (gij) and (gij) are inverses of each other. That's good; now we can calculate one from the other.
There is one other construction that we need to know about. Just as the metric is a cup, so it also gives us a cap. That cap looks like this:
_______ / \ / \ V v V v | | | |How do you suppose we define it?".
"Start with the regular cap between V and V*, and then stick a flat on one of the ends of it." said Toby the Acolyte.
"Hmm," said the Wizard "that's not how I'd have put it, but it probably works out like that. Let's see:
_______ / \ / \ V v V ^ | | | |There's the regular cap. Now, the right end points in the wrong direction, so I guess you want me to attach b there. First, we'd better make sure that it fits! b turns V* into V and that is a V* there, so, yes, I guess b *does* fit there. Now we have:
_______ / \ / \ V v V ^ | | | | | V v | | | |And that goes from C to V (x) V, just like we wanted! So, I guess you were right, Toby; that's the metric cap.
Great, we understand orthogonal structures, or metrics. Now, we're going to do it all over again, only for symplectic structures. But don't get discouraged! Now that we've done it once, the next time is easy.
A symplectic structure is like a metric structure, in that we have a cup:
| | | | V v V v \ / \_______/Only this is a symplectic cup, not a metric cup. What makes it symplectic? Well, we have this:
| | | | | | | | V v v V | | \ / | | \ / = - V v v V / | | / \ | | / \ | | V v v V \ / \ / \ / \_/ \_____/Notice that little minus sign! We didn't have it before. That little minus sign is the difference between orthogonal structures (or metrics) and symplectic structures.
Everything else is the same! The definition of # is the same. We demand that # be invertible, defining b. And the defintion of the symplectic cap in terms of b is the same. There's only one difference I'd like to point out, and that's that instead of "g", we like to use the Greek letter "w" -- I mean, "w" -- not Latin letter, dammit, Greek!" The Wizard sighed. "Anyway, a lowercase Greek Omega, which I guess today I'll just write "w".
Now, before we go, I'd like to look at one last diagram:
_______ / \ / \ V v v V \ / \_______/Earlier, we saw a diagram very much like that, only with V* on the left instead of V. That diagram had to be defined using the braiding, but this diagram is direct out of the metric or symplectic cup and cap. The diagram with V* on the left was the dimension of V as a vector space, so we might think of this diagram as the dimension of V as an orthogonal or symplectic vector space, not as just a plain old vector space. What's the difference? What does this circle evaluate to? That's your homework for this week.".
© 2000 Toby Bartels