# Yang-Mills Action for Non-Trivial Bundle

Suppose we have a principal $$G$$ bundle $$(P,M,π)$$ where $$M$$ is a 4-dimenational manifold and $$G$$ a Lie group (and $$\mathfrak{g}$$ its Lie algebra).The Yang Mills action is a functional of the gauge potential $$A$$ (which is nothing more than the connection on the principal bundle), expressed in terms of the curvature $$F=DA=dA +A\wedge A$$ of the connection. I know there is an isomorphism between $$\mathfrak{g}$$-valued tensorial forms of type $$Ad$$ ($$Ad$$ is the adjoint representation) and forms on $$M$$ with values on the Adjoint Bundle $$P \times _{Ad} \mathfrak{g}$$, so my understanding is that in the YM action: $$S_{YM} = \int_M \frac{1}{2} \operatorname{tr}(F \wedge \star F)$$ this $$F$$ is not the curvature, but the image of the curvature 2-form (which is tensorial of type Ad) under the aforementioned isomorphism, is this correct? This is my frist question.

Now in the case where $$P$$ is a trivial bundle, there is a global section $$s:M \to P$$ and the local curvature form $$F_s = s^* F$$ is actually defined globally on $$M$$ and in coordinates it is $$F_s=\frac12 F_{μν} dx^μ \wedge dx^n$$ . So the action takes the more familiar form: $$S_{YM}=\frac{1}{4} \int_{M} \operatorname{tr}\left({F}_{\mu \nu} {F}^{\mu \nu}\right) vol$$My second question is, how is this global local-form $$F_s$$ related to the one we get from the above isomorphism? They are the same up to a gauge transformation?

My third question is, what do we do when the bundle is not trivial. We can still make the local curvature, but we have more than 1. However the isomorphism guarantees there is a global form on $$M$$ as the picture of the curvature (which is what's inside the action integral right?), so how is this related to the local ones? Also, the derivations of the Yang Mills equations I've seen all use coordinates, but how can we express the integral if we need more than one coordinate charts due to non-triviality?

This is my frist question.

It depends on how you define the action. It may be either. You can show that the differential form $$\hat L[F]=\text{Tr}(F\wedge\star F),$$ which is a horizontal $$n$$-form on $$P$$ is actually $$R_g^\ast$$-invariant, so there exists a unique $$n$$-form $$\hat L_{\text{base}}[F]$$ on $$M$$ for which $$\hat L[F]=\pi^\ast(\hat L_{\text{base}}[F])$$. Hence, even if $$F$$ is defined on $$P$$, you can simply integrate this over $$M$$. You can go with either interpretation.

My second question is, how is this global local-form 𝐹𝑠 related to the one we get from the above isomorphism? They are the same up to a gauge transformation?

I am trying to be schematic here somewhat. If you have a global trivialization, you have three kinds of objects. An $$F\in\Omega^2(P,\mathfrak g)$$, which is a $$\mathfrak g$$-valued 2-form on $$P$$, an $$\bar F\in\Omega^2(M,\text{Ad}(P)),$$ which is an $$\text{Ad}(P)$$-valued (vector bundle-valued) 2-form on $$M$$, and an $$F_s\in\Omega^2(M,\mathfrak g),$$ which is a $$\mathfrak g$$-valued 2-form on $$M$$. The latter is what you get via pulling back $$F$$ via the global section.

You already know how $$F$$ and $$F_s$$ are related. To get the relationship between $$F_s$$ and $$\bar F$$, one needs to consider that $$P$$ having a global section $$s$$ also meands there is a global trivialization $$\phi:\text{Ad}(P)\rightarrow M\times\mathfrak g$$ of the adjoint bundle.

We may see $$\bar F$$ as an assignment $$\bar F:(X,Y)\mapsto \bar F(X,Y),$$ where $$\bar F(X,Y)$$ is a section of $$\text Ad(P)$$. Then using the trivialization we have $$\phi\circ\bar F(X,Y)=(\text{Id},F_s(X,Y)),$$ so $$F_s=\text{pr}_2\circ\phi\circ\bar F,$$ where $$\text{pr}_2$$ is the projection into the second factor of $$M\times\mathfrak g$$.

My third question is, what do we do when the bundle is not trivial. We can still make the local curvature, but we have more than 1.

• As explained about the first question, the action integral can be defined globally on the bundlespace, even if the bundle is nontrivial.

• Although the local curvature forms differ by gauge transformations on each overlapping trivialization domain, the "Lagrangian" you build (in whatever interpretation of it) will be an $$\text{Ad}$$-invariant function of the (local) curvature forms. Since gauge transformations happen by Ad, it means that even if the curvature forms do not fit together nicely, the "Lagrangian expressions" will. So you can work with patched together local representatives if you will.

• Although I guess quantum path integrals might spit into the soup, the Euler-Lagrange equations are local differential equations, so you don't even need the action integral to be globally taken.