I have the following question. In physics, when one talks about (Yang-Mills) gauge theories, one often states that it is enough to specify the following data:

  1. The gauge group $G$, which is usually a compact Lie group.
  2. The field, their potentails as well as the representation under which the fields transform.

As an example, the standard model is a Yang-Mills gauge theory with gauge group

$$G:=\mathrm{SU}(3)\times\mathrm{SU}(2)\times U(1)$$

with three left-handed Weyl fields, which transform in the representation $$(1,\textbf{2},-1/2)\oplus (1,1,1)\oplus (\textbf{3,2},1/6)\oplus (\overline{\textbf{3}},1,-2/3)\oplus(\overline{\textbf{3}},1,1/3)$$ as well as a single complex scalar field transforming in the representation $(1,\textbf{2},-1/2)$.

Now, I would like to ask the following: When when talks about gauge theories in mathematics, one does usually specify more, in particular, one also has to specify the principal bundle. For example, in Yang-Mills theory, we have to specify a principal $G$-bundle $P$. The action is then defined by

$$\mathcal{S}_{\mathrm{YM}}[A]:=\int_{\mathcal{M}}\mathrm{tr}(F^{A}\wedge\ast F^{A}),$$ where $F^{A}\in\Omega^{2}(\mathcal{M},\mathrm{Ad}(P))$ denotes the curvature, which is a $2$-form on the adjoint bundle $\mathrm{Ad}:=P\times_{\mathrm{Ad}}\mathfrak{g}$.

My question:

Why does one never specify the choice of principal bundle? A very similar question arises when discussing Dirac fields, in which, from the mathematical point of view, we have to a priori choose a spin bundle. Why does one nevery talk about these choices? As an example, when discussing the standard model as specifies above, which principal $G$-bundle do we choose? Different choices should lead to different models, right?

EDIT: Of course, when discussing Minkowski space, these choices might not be relevant as every principal bundle is trivial on a contractible space, but it should definitely matter when discussing the standard model on some curved background Lorentzian manifolds $(\mathcal{M},g)$.

  • 2
    $\begingroup$ The quantum gauge theory sums over all possible configurations of gauge fields in the (Euclidean) path integral, which basically means summing over all bundles. $\endgroup$
    – Meng Cheng
    Jul 1, 2022 at 17:28
  • 1
    $\begingroup$ @MengCheng In QED, one can choose to insert a Dirac monopole. The path integral is performed under such boundary conditions. This means that one does not sum over all bundles. Bundles are classified by its Chern classes. aka integers. $\endgroup$
    – Valac
    Jul 1, 2022 at 17:42
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    $\begingroup$ @Libertarian You can of course choose to have any background field you like when the gauge field is classical, but when the gauge field is fully dynamical, all bundles must be summed over. $\endgroup$
    – Meng Cheng
    Jul 1, 2022 at 17:58

1 Answer 1

  1. You have correctly diagnosed at the end of your question why many texts never bother with bundles: As long as we're only doing physics on $\mathbb{R}^4$ or are only interested in local phenomena that happen in a coordinate patch, all bundles are trivial and there's no point in mentioning them.

  2. The idea that physics never talks about bundles is wrong, but it often does so in a disguised way, namely in terms of the "gauge at infinity", "instantons", the "Chern-Simons current" and a bunch of other constructions that are - apart from perhaps issues with mathematical rigor that physics often does not particularly care for - equivalent to specifying a principal bundle. See, for instance, this and this answer of mine.

    Since you mention the Standard Model: In quantum theories there isn't a single bundle choice, but we sum over all possible bundles in the form of summing over all instantons with different Chern numbers. Again, many physics texts will not frame this in terms of "bundles", but that doesn't mean it's not there.


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