"O, we know," said the Acolytes "we just thought we'd get up and stretch -- get a pop -- call out for pizza -- see what's on TV -- do some homework --".

"All right, that's enough!" said the Wizard "You can go downstairs and visit the vending machines, but be back here in 5 minutes, or don't come back at all.".

"Yay!" said the Acolytes.

5 minutes later, the Wizard began again:

"Let's recall that we get 2D GR using the connection alone,
while the EF, or BF, trick gives 3D GR
and a special trick in 4 dimensions gives 4D GR.
At least, that's what I've told you.
But you don't really know, because you're mathematicians --
well, most of you -- and you don't know how to *use* Lagrangians.
So, I'll tell you.

Let's start with nonrelativistic particle mechanics.
If we have n particles moving in 3D space,
then a physical evolution of these n particles
is given by a function q: **R** -> **R**^{3n}
which traces out a path in 3nD space through time.
Let's see how this works:

The Lagrangian is L = (m/2)|dq/dt|^2 - V(q),
for V some function from **R**^{3n} to **R**;
V is the potential for a conservative force F = -Del V.
So, the action is S = \int L dt.
We're going to vary the function q
and see what the variation in S will be.
The rule is, when the variation in S is 0,
no matter what the variation in q was,
then the path q which we started from is a solution.

Here's how that whole variation thing works: If q is a path and Dq is a function which vanishes at the endpoints, then we can say that Dq is a variation in the path q which changes q to q + Dq. So, what's DS? Well, we could say DS = S(q + Dq) - S(q), but we're trying to make D some sort of differential thing, so instead say DS = (d/de) S(q + eDq) evaluated at e = 0.".

"Excuse me," said Jay the Acolyte "but, as the only official physicist in this group, I've happened to learn a thing or two about Lagrangians. I'll just take your word on that technical definition of DS -- but shouldn't the "D" be a Delta? And lowercase?".

"Gosh, you're right!" said the Wizard "I hadn't noticed.
And the "e" should be an Epsilon too!
(And I don't want to *think* about what happened to that integral sign.)
I know what's going on, though;
I was posting to USENET earlier today,
and I had my mouth set on ASCII,
and I must have forgotten to turn it off!
Oh, well, I won't be able to switch back
until I get back to my computer upstairs,
so we'll just have to learn to live with it.
You're right, it should be a Delta, and lowercase.
But it's really a good thing that it's uppercase now,
because we're going to be using "d" for something else later --
actually, I guess we already are.

OK, let's return to our example. S = \int ((m/2)|dq/dt|^2 - V(q)) dt, so DS = \int (m(dq/dt).(dDq/dt) - (Del V)(q).Dq) dt. Now, we'd like to get a bunch of stuff times Dq under the integral, because then we can say that bunch of stuff is 0, since DS has to be 0 no matter what Dq is. Of course we do this using integration by parts. What we've really got here is DS = \int m(dq/dt).dDq - \int (Del V)(q).Dq dt. So, if we integrate the first part by parts, remembering that Dq = 0 at the endpoints of the integral, then we get DS = \int Dq.dm(dq/dt) - \int (Del V)(q).Dq dt = \int (-m(d^2q/dt^2) - (Del V)(q)).Dq dt. Therefore, for the reasons I just explained, we know a solution to the physical problem is when -m(d^2q/dt^2) - (Del V)(q) = 0, or, in more prosaic terms, when F = ma.

So, we're up to the 17th century.

OK, let's try this on 1st Chern form theory, where the Lagrangian is L = tr F. Actually, this L is not so much the analogue of the previous L as it is the analogue of the 1-form L dt -- since that 1-form is what we integrated to get the action. But we still call L the "Lagrangian" anyway.

Now, let's vary the connection A. DA is not itself a connection -- it's actually an Ad(P)-valued 1-form. You can't add connections together, after all, but you can add an Ad(P)-valued 1-form to a connection to get another connection. So, that's what DA is. So, recalling that S is \int tr (dA + A ^ A), we see that DS is \int tr (dDA + A ^ DA + DA ^ A) = \int tr (dDA + [A,DA]). You can calculate this using the formal definition of DS easily enough, or you can just think of bringing the D inside everything, treating it as a differential operator when it acts on the product A ^ A. Again, we want to have a bunch of stuff multiplied by DA. Last time, we integrated by parts to get this; now, we're a bit fancier, so we'll use Stokes' theorem.".

"Actually," said Miguel the Acolyte "when you think about it,
integration by parts *is* Stokes' theorem.".

"Well, now you *are* getting fancy." said the Wizard
"Anyway, to apply Stokes' theorem, we need to get a d in front.
There's a trace in the way, but that's not really a problem.
See, the trace acts on the matrix part and the d on the form part,
so they commute. Thus, DS = \int d(tr DA) + \int tr [A,DA].
The integral of d(tr DA) is the integral over the boundary of tr DA.
Just as we asked Dq in the mechanical case to be 0 at the endpoints,
so we also ask DA to be 0 on the boundary -- or maybe just say M is compact.
In any case, the integral over the boundary of tr DA is just 0.
So, DS reduces to \int tr [A,DA] -- now what do we do?"

"That's easy" said Miguel the Acolyte "just use the cyclic property of the trace. The trace of a bracket is always 0.".

"Right." said the Wizard "Now, some of you may not be familiar with the cyclic property of the trace. That says that, if X and Y are any matrices, then tr XY = tr YX. The proof of this is simple enough in component notation -- ah, the heck with that, let's use our diagrams from track 1! See here, this is the trace of XY (remember XY is Y first, then X):

_______ / \ / \ | v V | | | | | / \ | | Y | | \_/ | | | | V ^ v V | | | | | / \ | | X | | \_/ | | | | | v V \ / \_______/Now, just move the X around a bit:

_______ / \ / \ V ^ v V | | | | / \ / \ | X*| | Y | \_/ \_/ | | | | V ^ v V \ / \_______/(Notice I

_______ / \ / \ | v V | | | | | / \ | | X | | \_/ | | | | V ^ v V | | | | | / \ | | Y | | \_/ | | | | | v V \ / \_______/Voila! The trace of YX! Same thing!

So, what does this mean for 1st Chern class theory? It means that our variation DS is always 0, no matter what. In other words, every connection A is a valid solution!".

"So," asked Toby the Acolyte "is this an example of a topological field theory?".

"Yes," said the Wizard "that's one way that can happen. There may be global restrictions, caused by the shape of the manifold, on the possible connections, but that's all the content there is. It's pretty boring -- at least classically. When you quantise, you can find that these global topological effects can get a bit more life into them -- but I won't go into that now.

And I think we'll stop there for today -- for good this time.".

toby@math.ucr.edu