"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) + e2(DA ^ DA) = dA + edDA + A ^ A + e[A, DA] + e^2(DA ^ DA).The 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.
Now that we have some rigor behind us, I'm going to introduce some neat notation: The exterior covariant derivative dA. If C is any Ad(P)-valued n-form, let dA C be dC + [A, C]. Then the formula above for DF is simply dA 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 dA DA = \int d tr DA = 0,where the last step uses Stokes' theorem and the fact that DA is 0 on the boundary of spacetime.
Now, there's only one tricky step above, and that's \int tr dA DA = \int d tr DA. I need to prove that, in general, tr dA w = d tr w, for w any Ad(P)-valued n-form. That's easy:
tr dA w = tr (dw + [A, w]) = tr dw + tr [A, w] = d tr w + 0 = d tr w,where I've used the fact that the trace of a bracket is 0.
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 (dA DA ^ F).Now, 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 trace. That's because the form parts of F ^ DF and DF ^ F can be switched, and the matrix parts can be switched because of the trace.
Now, since DA is an odd form, when we integrate by parts, we get
\int tr (dA DA ^ F) - \int tr (DA ^ dA F) = 0,where the 0 again comes from Stokes' theorem and an integral over the boundary. Thus,
DS = 2 \int tr (DA ^ dA F).Therefore, DS = 0 for all possible variations DA if and only if
dA 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! dA F = 0 is an identity obeyed by all connections. Just write it out in full:
dA 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.Now, 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].
Darn! Perhaps next week we'll get some nonvacuous equations of motion."
© 2000 Toby Bartels