"Sure I remember." said Miguel the Acolyte "It's the most work I've ever gone through to derive 0 = 0.".

"Right." said the Wizard "Well, today I'm going to introduce a short cut that will make deriving 0 = 0 much slicker.

First of all, I defined what DS is when S is the action. Now let me define the variation of an arbitrary function f of the connection. If DA is the variation in the connection, then Df is (d/de) f(A + eDA) evaluated at e = 0. For example, if F is the curvature F = dA + A ^ A, then our earlier manipulations with D led to DF = dDA + [A, DA]. Now we can check this rigorously:

F(A + eDA) = d(A + eDA) + (A + eDA) ^ (A + eDA) = dA + edDA + A ^ A + e(A ^ DA) + e(DA ^ A) + eThe derivative of this with respect to e is dDA + [A, DA] + 2e(DA ^ DA); evaluate this at e = 0 to get DF = dDA + [A, DA], just like before.^{2}(DA ^ DA) = dA + edDA + A ^ A + e[A, DA] + e^2(DA ^ DA).

Now that we have some rigor behind us,
I'm going to introduce some neat notation:
The exterior covariant derivative d_{A}.
If C is any Ad(P)-valued n-form, let d_{A} C be dC + [A, C].
Then the formula above for DF is simply d_{A} DA.
(Remember that A is not an Ad(P)-valued 1-form, but DA is!)

OK, now that we've done all that slick stuff, here's the calculation of DS for 1st Chern theory. Ready? If S = \int tr F, then DS is

D \int tr F = \int D tr F = \int tr DF = \int tr dwhere the last step uses Stokes' theorem and the fact that DA is 0 on the boundary of spacetime._{A}DA = \int d tr DA = 0,

Now, there's only one tricky step above,
and that's \int tr d_{A} DA = \int d tr DA.
I need to prove that, in general, tr d_{A} w = d tr w,
for w any Ad(P)-valued n-form. That's easy:

tr dwhere I've used the fact that the trace of a bracket is 0._{A}w = tr (dw + [A, w]) = tr dw + tr [A, w] = d tr w + 0 = d tr w,

Now that we can prove the triviality of 1st Chern theory in a single line, let's try 2nd Chern theory. Will it be more interesting? Recall that

S = \int tr (F ^ F)in this theory. Then

DS = \int tr D(F ^ F) = \int tr (DF ^ F + F ^ DF) = 2 \int tr (DF ^ F) = 2 \int tr (dNow, there is one tricky thing so far, and that was when I changed F ^ DF to DF ^ F. The Ad(P)-valued 4-forms F ^ DF and DF ^ F aren't the same! However, they have the same_{A}DA ^ F).

Now, since DA is an odd form, when we integrate by parts, we get

\int tr (dwhere the 0 again comes from Stokes' theorem and an integral over the boundary. Thus,_{A}DA ^ F) - \int tr (DA ^ d_{A}F) = 0,

DS = 2 \int tr (DA ^ dTherefore, DS = 0 for all possible variations DA if and only if_{A}F).

d_{A}F = 0.

Thank goodness, an equation of motion that isn't 0 = 0, right? On the other hand, something looks strange about that expression.".

"That's just the Bianchi identity!" said one of the Acolytes.

"Goodness," said the Wizard "you're right!
d_{A} F = 0 is an identity obeyed by *all* connections.
Just write it out in full:

dNow, ddA = 0 is a property of the exterior derivative d, [A, [A, A]] = 0 is a consequence of the Jacobi identity, and the remainder is 0 since [dA, A] = -[A, dA]._{A}F = dF + [A, F] = d(dA + [A, A]/2) + [A, dA + [A, A]/2] = ddA + [dA, A]/2 - [A, dA]/2 + [A, dA] + [A, [A, A]]/2.

Darn! Perhaps *next* week we'll get some nonvacuous equations of motion."

toby@math.ucr.edu