"All right," said the Wizard "remember where we were last time. We had a connection A on a G-bundle P over a manifold M, and we were trying to think of gauge invariant Lagrangians. Now, we know the curvature F is a 2-form, and this gets us some Lagrangians in certain dimensions -- but there's a trick that works in any dimension!
Assume that the manifold M has a metric on it, and assume that the metric gets along with the G-bundle structure, in the sense that the metric transforms trivially under g. Since we agreed that our manifolds are oriented, we can define a Hodge star operator *.
So, if F is a 2-form, then *F must be an (n-2)-form. If we wedge F with *F, we get the n-form F ^ *F, and such an n-form is just what we want in a Lagrangian! Since the metric gets along with the G-action, g maps *F to *(g F g-1) = g (*F) g-1, so tr (F ^ *F) really is a gauge independent Lagrangian. If M is spacetime, this is called "Yang-Mills" theory.".
Then Toby the Acolyte spoke up "Since we're really interested in the action S, which is the integral over M of the Lagrangian L, then the orientation of L cancels out, appearing once in * and once in the integral. Thus, we only need the manifold to be orientable; we don't need an orientation to define the action.".
"Right," said the Wizard "and this is good, because the most famous example of Yang-Mills theory, electromagnetism, doesn't care about the orientation of spacetime.
Now, in general, we can pull this trick with any (n-2)-form in the place of *F, as long as it acts the right way under action by g. In fancy language, an Ad(P)-valued p-form is a locally g-valued p-form which g acts on by conjugation. For example, F is an Ad(P)-valued 2-form, because it's a locally g-valued p-form which g maps to g F g-1.
So, with or without a metric, if we have an Ad(P)-valued (n-2)-form E, then we get a gauge invariant Lagrangian with L = tr (E ^ F). This is called "EF theory". Well, it *should* be called "EF theory". It's actually called "BF theory", because people say "B" instead of "E". But we will see that E is really analogous to the electric field and not to the magnetic field, so "E" is better than "B".
If n = 3 and G is, respectively, SO(3) or SO(2,1), then EF theory is, respectively, Riemannian or Lorentzian 3-dimensional general relativity.
Now, we can pull some tricks here occasionally. For example, in 4 dimensions, F ^ F, E ^ F, and even E ^ E are all 4-forms, so the Lagrangian can be built from linear combinations of all these.
In 3 dimensions, we can make linear combinations of E ^ F and E ^ E ^ E. Now, this should be something similar to 3-dimensional gravity, but with an added twist because of the E ^ E ^ E term. It's actually a very old idea! Can anyone guess what it is?".
Guessing wildly at an old idea related to general relativity, Toby the Acolyte blurted out "Kaluza-Klein theory?".
Thinking more reasonably (or perhaps having looked at the book), Jay the Acolyte suggested "How about a cosmological constant?".
"Right!" said the Wizard "It's the Kaluzma-logical -- cosmo-Lugical -- damn it, Toby, look what you made me do with your bone-headed idea!".
Foom!
"Gee," thought Toby the Acolyte "when Oz gets hit with a fireball, his hair gets singed. But I have a beard, and that gets singed too!".
"Yes, 3-dimensional general relativity with a cosmological constant Lambda is just EF theory with L = tr (E ^ F + Lambda E ^ E ^ E) (unless I'm off by a factor of 2 or something). Interestingly, when you quantize this, you want to use the cosmological constant to tweak your group SO(3) into a quantum group. That's why I think quantum groups are really more about cosmological constants than quantization.
Let's continue this into 2 dimensions. There, E is just a simple g-valued function. So, you can make linear combinations of F, E ^ F, E ^ E ^ F, and so on.".
"If f is a function defined by a power series," said Miguel the Acolyte "or at least a polynomial, can't you just say f(E) ^ F?".
"Yes," said the Wizard "I suppose you could.".
"Then it reduces to E ^ F with f(E) in the place of E.".
"No, it doesn't, because f(E) won't be g-valued. Remember that g is a Lie algebra, closed under commutators. E ^ E is found by *multiplying* matrices, not commuting them. Indeed, in an abstract setting where your Lie groups aren't matrix groups, you have to show that an expression like E ^ E actually makes sense -- it does, but it will never be simply a g-valued function itself.".
Then some more stuff was said, and I forget how the conversation went, but Miguel the Acolyte suggested using an (n-4)form E and tr (E ^ F ^ F), and the Wizard joked that people who study SO(2,2) are simply "dirty, two-timing rats". Then came the good stuff.
"Now," said the Wizard "to define the Lie algebra so(4), you have to put a metric on R4. And, if you do this, then you see that so(4) is naturally isomorphic to A2 R4, the 2nd alternating (or antisymmetric) tensor power of R4. In other words, if you wedge 2 forms that take values in R4, you get a form that takes values in A2 R4 = so(4). So, this is how we do 4-dimensional Riemannian gravity. We say that e is an R4-valued 1-form -- not g-valued at all! e can be called "the cotetrad field", "the soldering form", "the coframe field", or "the vierbein" if you're in a German mood. Then the Lagrangian is L = tr (e ^ e ^ F). If you want a cosmological constant Lambda, add Lambda tr (e ^ e ^ e ^ e).
Notice that e is not vector-valued; it doesn't take values in the tangent space. After all, there's no metric on the tangent space -- not in *this* formulation of general relativity, at least not until *after* you solve the equations. It's an abstract R4 which does have a metric.".
"If you want to do Lorentzian general relativity, then," said Toby the Acolyte "you use R3,1 instead of R4, because A2 R3,1 is naturally isomorphic to so(3,1). And R3,1 is just Minkowski spacetime, from special relativity. So, we see that e isn't a vector valued 1form -- it's a Minkowski vector valued 1form instead!".
"Yes," said the Wizard "you can look at it that way.".
But this last bit was said at the dinner after the seminar, so I've already told you too much.