# Section of adjoint bundle in gauge theory

I have a couple of questions on gauge theory. Considering a principal $$G$$-bundle $$P\xrightarrow{\pi} M$$, we have a connection, a local Lie algebra-valued 1-form $$A$$. This is the photon field, correct?

Moreover, for a representation $$\rho : G\to GL(V)$$, we can construct the associated vector bundle $$E=P\times_{\rho} V\to M$$. Then a section of such bundle amounts to a particle field $$\phi:M\to V$$. What happens if we take $$V=\mathfrak{g}$$, i.e. the adjoint representation of $$G$$? What is the relation between this particle field $$\phi$$ and the one defined by $$A$$? Can one find a vector field $$v$$ on $$U$$ such that $$A(v)=\phi$$ on $$U$$? In the case of $$G=U(1)$$, since $$\mathfrak{u}(1)=\mathbb{R}$$, the adjoint representation is trivial. What does this imply for the fields $$\phi$$?

• To your first question: If $G=U(1)$, then yes, the connection one-form represents the photon. Mar 25, 2021 at 19:06
• @NDewolf, my question arises from the fact that I have seen particle fields be defined as in the second paragraph, that they should be sections of such a bundle. In the case of the Lie algebra-valued 1 form, I guess we could look at it as a section $A:U\to T^*U\otimes \mathfrak{g}$ for it to match this definition. However in this case $G$ is supposed to have an action on $T^*U\otimes \mathfrak{g}$. I guess the one on $\mathfrak{g}$ is the adjoint action but I don't think there is a canonical action on $T^*M$? Mar 25, 2021 at 19:12
• Or is it that force carrier fields are sections of the gauge bundle, while matter fields are sections of the associated bundles? If so why is that the case? Mar 25, 2021 at 19:19
• A connection is not the same as a section, if that's what you're wondering about. This can be seen immediately by noting that a connection does not transform as a section under a gauge transformation. Section 2 of these lectures may help straighten some things out. Mar 25, 2021 at 19:29
• @RichardMyers, isn't a globally-defined Lie algebra-valued 1-form just a section of the bundle $T^*M \otimes (M\times \mathfrak{g}) \to M$? Mar 25, 2021 at 19:35

There is no relation at all between the particle fields $$\phi$$ that are sections of associated bundles and the gauge field $$A$$. $$A$$ is the field of the gauge bosons, $$\phi$$ is a field of a particle charged under the interaction with that boson. The Standard Model does not contain particles charged in the adjoint of some gauge group, so there is no analogue to your $$\phi$$ in the Standard Model - unless you're willing to admit $$\mathrm{U}(1)$$ as the gauge group, since then the adjoint is the trivial representation and all uncharged fields would be examples. Outside of the Standard Model, such adjoint matter is fairly common, however.

Your equation $$A(v) = \phi$$ is ill-defined, since the two sides have different behaviour under gauge transformations - $$A$$ transforms as $$A\mapsto gAg^{-1} + g^{-1}\mathrm{d}g$$, while $$\phi$$ transforms as $$\phi\mapsto g\phi g^{-1}$$. The gauge field does not transform as the section of an associated bundle - if you want to see it as a section of some bundle, you can express a connection as a section of the first jet bundle of the principal bundle.

It seems you are essentially asking two questions which aren't really related. One of the two can be phrased as: given a Lie algebra-valued 1-form $$A$$ (whether it's a connection or not is irrelevant to this question), does there exist a spacetime vector field $$v$$ such that on some patch $$U$$ of spacetime we may obtain $$A(v)=\phi$$ for an arbitrary Lie algebra-valued scalar field $$\phi$$. To be honest, I'm not certain this problem would even have a good, gauge invariant, answer. But fixing a gauge certainly we can answer this question.

This question may be written locally in components as $$A^a_\mu v^\mu T^a = \phi^a T^a$$ where I have taken $$T^a$$ to be some basis of the Lie algebra. Since the $$T^a$$ are, by definition, linearly independent, it follows that we must have $$A^a_\mu v^\mu = \phi^a.$$ This is a linear system of equations for the components $$v^\mu$$. It may have a solution, no solution, or infinitely many solutions. What the case may be will generically depend on the dimension of the Lie group, the dimension of spacetime, and the values of the components of $$A$$. At least for the case of a 1-dimensional Lie group, like $$U(1)$$, we are guaranteed at least one solution exists so long as at least one component of $$A$$ is non-zero, which we can always make happen as the result of a gauge transformation.

Now, that was really a question about mathematics. You have also asked about how fields are defined and whether there is a relationship between the connection and matter fields (everything that isn't a connection field). The short answer to the latter question is yes, but almost certainly not what you're hoping to find.

So first of all, yes, fields are generally sections of an associated bundle to some principle bundle with connection $$A$$. These fields can transform under any representation of the gauge group you like, including the trivial representation, in which case you'll often hear such fields referred to as "uncharged" under the gauge group.

These fields are completely independent unless you define them not to be...in which case you are not defining a new field but a functional of fields you've already introduced, potentially including the gauge connection. The important point is that such a thing would be defined by you, by hand, by fiat.

So what is the connection between the gauge connection and the matter fields? Only that they all transform at the same time under a gauge transformation. Each field will transform under its respective representation, whatever that might be, including the gauge connection.

It occurs to me that you might find the description in terms of frame fields to be somewhat more satisfying. I'd rather not go into how this is set up here, but you can find some information about it in Nair's QFT book.