Quantum Gravity Seminar

Week 1, Track 2

Toby Bartels

October 2, 2000

The Wizard gazed over the group that sat before him. Most of the people remaining belonged to the Quantum Gravity Crew, but there were a few unfamiliar faces. Not scared of Lie algebras and differential forms, eh? Well, he'd just have to plunge in and see how they'd do.

"Just to make things easy," he began "we'll work with Lie groups of matrices. Typical groups will be the real groups SO(n) and the complex groups SU(n). Here, "n" means n x n matrices, "O" means orthogonal, "U" means unitary, and "S" means determinant 1. Then the Lie algebras will be like so(n) and su(n). Here, "n" means n x n matrices again, "o" means antisymmetric, "u" means anti-Hermitian, and "s" means trace 0. We'll assume that our manifolds are infinitely differentiable (smooth), paracompact and Hausdorff to rule out really icky things, oriented so we can integrate, and usually compact. We will, however, often allow boundaries on the manifolds.

I'll talk about some things we can do with principal G-bundles. If G is a Lie group and M is a manifold, the trivial principal G-bundle over M is the manifold M x G with the projection M x G ->> M. A principal G-bundle is a manifold P with a projection pi: P ->> M which looks *locally* like the trivial principal G-bundle.

The main concept we need is a connection on the bundle. Physicists also call this "gauge field" or "vector potential". Locally, at least, this is a g-valued 1-form A on M. In other words, it's a smooth map taking a point in M to a g-valued covector, a linear beast with the power to turn a vector at that point into an element of the Lie algebra g.

Then there's a "gauge transformation", or at least one of several things which go by that name, which is really just a function g from M to G. g acts on A, turning A into g A g-1 + g dg-1. To understand this formula, just remember that the values of g and A at a given point on M are just matrices (or matrix valued covectors), so it's easy to know how to multiply them together. When you multiply elements of g on either side by elements of G, you're guaranteed always to get back another element of g, and the derivative of a G-valued function is a g-valued function, so g A g-1 + g dg-1 is really a g-valued 1-form.

Then there's the curvature F of the connection A, which physicists might call "field strength". This is defined to be dA + A ^ A. Since A is a g-valued 1-form, dA is a g-valued 2-form. Since A is a 1-form, A ^ A is also a 2-form. Since matrix multiplication doesn't commute, there's no reason to assume that A ^ A = 0, even though that's true for an R-valued 1-form A. In fact, we can express A ^ A as (1/2) [A, A], where "[A, A]" means we use the wedge product to multiply forms and the matrix commutator to multiply matrices. Strictly speaking, we have to be able to express it this way, since otherwise we wouldn't know that A ^ A is a g-valued 2-form, rather than a 2-form taking other values. Ultimately, then, F is, locally, a g-valued 2-form.

How does the gauge transformation g act on the curvature F? I claim that it transforms F to g F g-1. It's really a straightforward calculation:

F = dA + A ^ A |-> d(g A g-1 + g dg-1) + (g A g-1 + g dg-1) ^ (g A g-1 + g dg-1)
                = ... er, this is going to be messier than I thought ...
                = g (dA + A ^ A) g-1
                = g F g-1
I'll leave the messy middle steps as an exercise. Trust me, it's a straightforward calculation.".

Looking a bit less majestic and wizardly now, the Wizard decided it would be best if he changed the subject.

"Now, in physics, we like to express theories in Lagrangian form. If M is an n-dimensional manifold, the Lagrangian L is an n-form on M. Then we can integrate L over M to get the action S. We can then apply variational principles to get classical solutions or Feynman's rules to get quantum transition probabilities or transform this set-up over to a Hamiltonian framework.

Anyway, let's see what kind of Lagrangians we can cook up from A. Remember, the Lagrangian has to be a bona fide n-form, not g-valued. Any guesses?"

Miguel the Acolyte suggested "Well, you could wedge A together with itself until you got an n-form and then take the --".

"No, no," said the Wizard "dumber than that.".

"Er, how about the trace of A?".

"Yes!" said the Wizard "That's wonderfully dumb. Since A is a g-valued 1-form, tr A is an R-valued 1-form, at least assuming that the Lie group G is real. So, this might be a Lagrangian on a 1-dimensional manifold, which is what we use to study mechanics. But why is this so dumb?".

"Well," said Toby the Acolyte "with our examples of Lie groups, SO(n) and SU(n), the trace would automatically be 0.".

"Yeah, but it's dumb anyway." said the Wizard "See, I guess I didn't tell you this before, but we want our Lagrangian to be gauge invariant. That is, the gauge transformation g should act trivially on it. But it doesn't. See here:

tr A |-> tr (g A g-1 + g dg-1)
      = tr (g A g-1) + tr (g dg-1)
      = tr A + tr (g dg-1)
Now, for the first term, I used the cyclic property of the trace. This says that, for any matrices m and n, tr mn = tr nm. So, tr (g A g-1) = tr (g-1 g A) = tr (1 A) = tr A. Since there's no reason to assume tr (g dg-1) = 0 -- any nonconstant gauge transformation is a counterexample -- tr A isn't gauge invariant.

OK, someone guess again."

Jay the Acolyte then suggested "If tr A doesn't work, then tr F should.".

"Let's see." said the Wizard.

tr F |-> tr (g F g-1) 
       = tr (g-1 g F) 
       = tr (1 F) 
       = tr F
"You're right, it works! Mathematicians call tr F "1st Chern form", so you could call this "1st Chern form theory". If the Lie group G is SO(2), this is (2+0)-dimensional gravity. That is, the classical solutions of this theory on a 2-dimensional manifold are just the solutions to Einstein's field equations for general relativity on a (2+0)-dimensional manifold without matter. A bit duller than the real world, but worth studying.

Well, can we get a theory on a 4-dimensional manifold?".

"tr (F ^ F)!" shouted the whole class together.

"Right! The cyclic property of the trace comes in again:

tr (F ^ F) |-> tr (g F g-1 ^ g F g-1)
             = tr (F ^ g-1 g F)
             = tr (F ^ F)
You could call this "2nd Chern form theory". Unfortunately, this doesn't give you general relativity.".

"If you take G = SO(3,1)," asked Toby the Acolyte "it's not general relativity, but is it anything interesting?".

"It's called "topological gravity"." said the Wizard "This is less interesting than actual general relativity, because it has no local degrees of freedom -- like 2-dimensional gravity, it's a topological field theory, and the only degrees of freedom are of a global, topological nature. Still, it's worth studying, if for no other reason than to gain insight.

Well, it's past time to leave, so I'll see you all next week."

Then Miguel, Jay, and Toby the Acolytes went out to dinner with the Wizard and John, the Acolyte the Wizard Didn't Even Know about, to discuss the eigenstates of the observable conjugate to whatever observable spin networks are eigenstates of. But this is too secret for you.


© 2000 Toby Bartels