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?
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2$\begingroup$ 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... $\endgroup$– ZeroTheHeroApr 10, 2021 at 16:25
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2$\begingroup$ 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. $\endgroup$– Chiral AnomalyApr 10, 2021 at 16:59
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2$\begingroup$ @ChiralAnomaly yeah that's probably better in fact. $\endgroup$– ZeroTheHeroApr 10, 2021 at 18:03
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$\begingroup$ @ZeroTheHero this is a good point :-) $\endgroup$– rage_manApr 10, 2021 at 20:12
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1 Answer
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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.
(When you transition from classical field theory to quantum field theory, gauge fields are actually wavefunctionals of connections on a time slice, and matter fields are wavefunctionals of sections of the associated bundle on a time slice, with a lot of extra subtleties due to the redundancy of the gauge symmetry.)
answered Apr 10, 2021 at 16:16
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$\begingroup$ Oh, I see, so we take the associated bundle as a vector field under the representation of G as a completely separate thing. This actually makes sense to me... G acts on something and this relates (for example) the wavefunction to the thing it is describing (a particle). Is this correct? If so, in hindsight, I'm not sure what I was confused about $\endgroup$– rage_manApr 10, 2021 at 20:14
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$\begingroup$ And, as a follow up, if connections are potentials like EM potential and what not, and curvature is field strength, is the distinction between principal and associated bundle anything similar to the distinction between boson and fermion? $\endgroup$– rage_manApr 10, 2021 at 20:15
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$\begingroup$ Some authors say that the sections of the associated bundle represents the wave function of a single quantum particle. I, and many others, think that this is pedagogically a bad way to think about it. In my answer everything involved is a classical field. Yes, you can also consider single particle wave functions, but that is a less "fundamental" construction. Anyway, there's no reason that the matter field needs to be fermionic. It can be either bosonic or fermionic. It just so happens that electrons are fermions, but that is besides the point. Gauge fields must be bosonic though. $\endgroup$ Apr 10, 2021 at 22:00
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$\begingroup$ 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. $\endgroup$– rage_manApr 10, 2021 at 23:23
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1$\begingroup$ 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. $\endgroup$ Apr 11, 2021 at 2:05