4
$\begingroup$

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?

$\endgroup$
5
  • 2
    $\begingroup$ There are certainly different principal bundles with the same base manifold and isomorphic fiber: one example I know is $S^1\to S^3 \to S^2$ where $S^1\curvearrowright S^3$ by $\lambda\cdot z= z \lambda^{\pm 1}$, using multiplication in $\Bbb C^2$. These two are not isomorphic principal bundles---one of them is the standard Hopf bundle. $\endgroup$
    – Danu
    Sep 15, 2016 at 15:04
  • 1
    $\begingroup$ If you already know of the existence of a nontrivial principal $G$-bundle, there is at least one more, the trivial bundle! Generally, the "set" of principal $G$-bundles over a space $X$ is in bijection with the set of homotopy classes of maps $X\to BG$ (where $BG$ is a special space associated to $G$, called its "classifying space"). You have to "do physics" first in order to see what "framework" you're really in. That means finding the transition functions by seeing how the physics in each local coordinate systems relates to each other. (Alternatively you guess and check with experiment). $\endgroup$ Sep 15, 2016 at 23:10
  • $\begingroup$ @ChrisGerig Thanks, that makes a lot of sense. I was hoping that "doing physics" just meant going outside once to figure out what $G$ should be, but it makes sense that in physics you need to work locally, experimentally to find the transition functions. But in some path integral where you're integrating over all connections modulo gauge transformations $\mathcal{A}/\mathcal{G}$, does this refer to all connections on a fixed principal bundle, or all connections on all principal bundles? $\endgroup$
    – Benighted
    Sep 16, 2016 at 17:02
  • $\begingroup$ You can't measure it locally, as I tried to explain below. Transition functions only determine topological properties, which by their very nature cannot be observed locally. For that same reason, it doesn't matter whether you take the path integral as integrating over all bundles, or a bundle with a particular set of transition functions: if the integration range is the set of all bundles, that is just the disjoint union of integrating over particular bundles, and quantum trajectories over disjoint spaces cannot influence one another. $\endgroup$ Sep 16, 2016 at 19:32
  • $\begingroup$ (To further underline my point that transition functions cannot be observed locally: note that if your underlying space is topologically trivial (i.e. contractible), then all different choices of transitions functions are in fact equivalent/indistinguishable. Hence to 'measure' transition functions, you need to do something very global, such that whatever you do is sensitive to the topology of your underlying space.) $\endgroup$ Sep 16, 2016 at 19:35

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Right, I understand that the gauge fields are dynamic, but these gauge fields are connections on a fixed underlying principal bundle. So that underlying bundle has to be chosen before you talk about connections on it. Unless the connection canonically determines a bundle, via something I don't understand. $\endgroup$
    – Benighted
    Sep 15, 2016 at 19:21
  • 2
    $\begingroup$ Well, the connection determines the geometry of the bundle, the transitions functions determine the topology. The topology is usually fixed by initial conditions. Indeed: physics enters through the action principle, which only cares about local extrema, which cannot be influenced by topological considerations. $\endgroup$ Sep 15, 2016 at 21:17
  • 1
    $\begingroup$ In case you want more mathematical details: for a given gauge group $G$, it doesn't pick out a single bundle, but rather the topologically distinct bundles are labeled by something called Cech cohomology, or relatedly by the topologically distinct mappings from the base manifold $M$ into the so-called "classifying space" associated to $G$. This set of inequivalent maps is denoted $[M,BG]$. For example, if our base manifold is $S^2$ and $G = U(1)$, then $BG = \mathbb CP^\infty$ and opening a math textbook we see $[S^2, \mathbb CP^\infty] = \pi_2( \mathbb CP^\infty) = \mathbb Z$. $\endgroup$ Sep 15, 2016 at 21:28
  • $\begingroup$ Hence there are an integer number of topologically distinct $U(1)$-bundles on the $2$-sphere. An action principle cannot favor one of these over the others, and so it is usually initial conditions that determines which one 'we are in'. Once we have specified the topology of our bundle, it is then the dynamics/physics which determines the geometry of our bundle. $\endgroup$ Sep 15, 2016 at 21:31
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.