Does a gauge group $G$ determine the Principal $G$-bundle?

Question

I'm trying to understand the mathematical underpinnings of gauge theories in the language of principal $G$-bundles and associated vector bundles. Not long ago, I had assumed that the physical choice of a symmetry group $G$ (compact Lie group) immediately and uniquely determined an infinite dimensional group $\mathcal{G}$ of gauge transformations. Then, I thought $\mathcal{G}$ was simply providing the transition functions of a principal $G$-bundle. I know in other contexts a bundle can be uniquely recovered from its transition functions. Thus, I thought that the physically motivated choice of $G$, immediately determines the principal bundle.

I now understand that the gauge transformations are different from the transition functions. There is a lovely discussion (Global vs. local gauge group in mathematical sense - physics examples?)

So it now appears to me that choosing the transition functions of the principal bundle taking values in $G$ is actually extra data that needs to be provided in addition to $G$. As opposed to being uniquely determined by $G$ and the physical gauge transformations.

Is this correct? If so, how do physicists decide which principal $G$-bundle they need?

Answer 1

Maybe it helps to think of the example of GR: knowing the local symmetries of spacetime doesn't fix its metric (or topology). It only fixes that it locally 'looks like' $\mathbb R^{1,3}$. In that case we know how this extra information is fixed: by initial conditions, boundary conditions and dynamics. Similarly for the non-Abelian gauge theories/bundles encountered in the standard model, it is only fixed that locally it looks like $\mathcal M \times G$. And similarly, its `geometry' is in principle free and has to be obtained through the triad of initial conditions, boundary conditions and dynamics. This is exactly why gauge fields are dynamic, with the geometry being captured by the field strength $F = dA$. You can think of me sending a packet of light your way as me sending a ripple through the $U(1)$ line bundle.

Answer 2

In General Relativity the classical solutions are spacetimes $(M,g)$ which are Lorentzian manifolds. I urge you to observe that the topology of the underlying manifold is part of the classical solution. So the unknown is not just the metric tensor!

In gauge theory things are quite similar. Given a compact and semi-simple Lie group $G$ we can construct several principal $G$-bundles over the same base manifold $M$. One of them is the trivial bundle $\pi_1:M\times G\to M$, where $\pi_1$ is the projection onto the first factor, and where the right $G$-action is $$(x,g)\cdot h=(x,gh)\tag{1}.$$

But obviously this isn't everything. We have non-trivial bundles which do not take this simple product form with this simple $G$-action (1). They are topologically different from the trivial bundle.

Now, the gauge field is in fact a connection on a principal $G$-bundle, and what is the specific $G$-bundle is part of the specification of the solution inasmuch as the spacetime topology is part of the specification of the classical GR solution!

It turns out that every principal $G$-bundle is, by definition, locally isomorphic to the trivial bundle. This correspondence is specified by choosing a local section $\sigma : U\subset M\to \pi^{-1}(U)$ and defining $h:U\times M\to \pi^{-1}(U)$ to be $h(x,g)=\sigma(x)\cdot g$. On such a locally trivial open set the connection is codified in a Lie-algebra valued one-form $A : U\to T^\ast U\otimes \mathfrak{g}$. This is the object we are used to in Yang-Mills theory.

But beware! When the principal bundle is not the trivial one then $A$ is not globally defined in the whole spacetime. In that case of non-trivial topology you cannot represent the connection by a single $A$. Rather you must cover the underlying base manifold by open sets $\{U_i\}$ over which the bundle can be trivialized. In each of the $U_i$ then you do have one $A_i$ and in order that they give a well-defined connection in the principal bundle they must obey certain compatibility conditions in the overlaps.

Now compare again to GR. The metric $g$ in each coordinate domain is specified by the components $g_{\mu\nu}$. Often a single chart won't cover the whole manifold and you'll have several in which you have the quantities $g_{\mu\nu}$, which obey compatibility conditions in the overlaps in order for them to give rise to a well-define intrinsic object $g$.

So in summary, the answer to "Does a gauge group G determine the Principal G-bundle?" is that no, there are several topologically inequivalent principal $G$-bundles over the same base manifold and this data is part of the specification of the gauge theory gauge field configuration.