# As a physicist, why are associated bundles important?

I have a good grasp on principal bundles as providing a lie group on some fibers of our field. So for example, the wavefunction tells us the phase of a particle in space and time, and this can be a section of a $$U(1)$$ bundle. I guess the gauge symmetry is related to rotational symmetry on the circle, since phase is $$\theta = 2\pi \cdot n \cdot \theta$$, so we use a covariant derivative in terms of the lie algebra connection $$\mathfrak{u}(1)$$. What I don't understand, is: why do we care about some left multiplication or some rep of G on a different fiber? What is the associated bundle here, and what does it tell us physically?

• Keep in mind that most (95%+?) of physicists don’t care, so (notwithstanding that your question is entirely legitimate given this proportion) maybe a slight sharpening of the phraseology of the question might be in order... – ZeroTheHero Apr 10 at 16:25
• A slight variation on @ZeroTheHero's comment: Maybe we should say that most physicists don't know they care. Associated bundles are everywhere, even if we don't usually recognize them as such. – Chiral Anomaly Apr 10 at 16:59
• @ChiralAnomaly yeah that's probably better in fact. – ZeroTheHero Apr 10 at 18:03
• @ZeroTheHero this is a good point :-) – rage_man Apr 10 at 20:12

Classically speaking, gauge fields are connections on principle $$G$$ bundles. Matter fields, on the other hand, are sections of associated bundles. To construct an associated bundle, you must choose a representation of $$G$$. For instance, if $$G = U(1)$$, then this amounts to choosing an integer. When you construct the associated bundle, this integer is the electric charge of the matter field.
• OK, maybe I am overcomplicating it. Would you say its as simple as the group can be represented as acting on a thing, and when it is a principal bundle/gauge field the representation means it acts on a matter field? The specific representation of $G=U(1)$ is the charge of this field? If I'm with you so far, then that's good, but I'm still confused about wavefunctions. That's a separate question, I guess, so I just want to confirm I have fundamental understanding for now. – rage_man Apr 10 at 23:23
• It is difficult for me to judge your understanding based on what you have written. If your bundle is trivial, then a gauge transformation is as simple as a map from $M \to G$, where $M$ is the base spacetime manifold. We can call this map $g(x)$. If your matter field, written in local coordinates, is the vector field $v(x)$, then it is true that the gauge transformations acts on the matter field as $v(x) \mapsto \rho(g(x)) v(x)$, where $\rho$ is the representation of $G$ in question. – user1379857 Apr 11 at 2:05