6
$\begingroup$

I'm trying to understand some Yang-Mills and Chern-Simons theory but I'm getting tripped up by some of the mathematics.

I'm confused about the exterior covariant derivative of Lie algebra valued k-forms on a principal bundle P. In particular, I understand the derivation done in section 2.2.2 of https://empg.maths.ed.ac.uk/Activities/GT/Lect2.pdf, but I'm failing to generalize to k-forms (Exercise 2.5 of the same section) and understand the case where V is the Lie algebra of G where the wedge is replaced by a commutator.

Also, there seem to be different conventions(?) for the coefficient in front of the second term of the exterior covariant derivative. I sometimes see a factor of 1/2 (for example in section 3.2.2 of https://empg.maths.ed.ac.uk/Activities/GT/Lect3.pdf) and other places without it. I believe my understanding has to do with commutators and wedge products of Lie algebra valued forms, but I could be mistaken. Could someone explain and perhaps identify my misunderstandings?

(I've looked in Nakahara, but his section on connections on a principal bundle didn't quite address my question.)

EDIT: I think I've managed to narrow down my confusion. In the first set of notes (Lecture 2), they derive an equation for the exterior covariant derivative of the connection form (proposition 2.1) as well as the exterior covariant derivative of a vector valued form (exercise 2.5). My understanding is that the result of proposition 2.1 should be a special case of exercise 2.5, with the Lie algebra action being that of conjugation (commutator), but there's a factor of 1/2 on the former. Can somebody explain this point to me?

$\endgroup$
1
$\begingroup$

There are two reasonable ways of defining the wedge product of forms. One is to say that $$\alpha\wedge\beta(X,Y) = \frac{1}{2}\left(\alpha(X)\beta(Y) - \alpha(Y)\beta(X)\right)$$ while the other doesn't include this factor of $\frac{1}{2}$ $$\alpha\wedge\beta(X,Y) = \alpha(X)\beta(Y) - \alpha(Y)\beta(X).$$ Here, $\alpha$ and $\beta$ are 1-forms and $X$ and $Y$ are vector fields. It seems like these notes are using the former convention when the "$\wedge$" is explicitly written (so that $$ (\rho(\omega)\wedge\omega)(X,Y) = \frac{1}{2}\left([\omega(X),\omega(Y)] - [\omega(Y),\omega(X)]\right) = [\omega(X),\omega(Y)] $$ in exercise 2.5) while putting the $\frac{1}{2}$ in manually in the notation of prop 2.1 $$ \frac{1}{2}[\omega,\omega](X,Y) = \frac{1}{2}\left([\omega(X),\omega(Y)] - [\omega(Y),\omega(X)]\right) = [\omega(X),\omega(Y)]. $$ One possible motivation for this convention is that, due to the anti-symmetry of both the wedge product and the Lie bracket, the combination of the two is actually symmetric: $$[\alpha,\beta] = [\beta,\alpha]$$ where $\alpha$ and $\beta$ are both Lie algebra valued forms on $P$. Many times when we plug the same argument into both slots of a symmetric operation like this it is convenient to have the factor of $\frac{1}{2}$ for various reasons. Possibly the notation is there to emphasize this sort of behavior of the "bracket and wedge" operation.

One last thing: it's not strictly true that 2.1 is a special case of 2.5 since the covariant derivative as defined is only meant to be applied to basic forms (which are, in particular, horizontal) whereas $\omega$ is as far from being horizontal as possible ($h^*\omega = 0$). When treating everything as forms on $P$, these details don't really matter, but if you want to associate the basic forms with certain forms on the manifold (see 2.2.1) then the two equations you mention are quite different things. It is important to the story that the connection one form $\omega$ cannot be thought of as any type of global form on $M$. Instead, it transforms as a connection should (this is mentioned in lecture 1 from this series).

$\endgroup$
  • $\begingroup$ Thanks so much for your answer! I'm not sure about their convention for wedge products, but your last paragraph helped clear things up for me. I didn't realize that it was intended to apply for horizontal and invariant forms. I saw that the pull-back of the connection form satisfied the invariant property and jumped to conclusions. I'll go through the derivations again to double check the convention and post an update when I do. $\endgroup$ – qm-arv Mar 1 '17 at 5:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.