When people talk about the gauge theory and fiber bundles, mostly what is talked about is simply the group and the connection that is put on the principal bundle. But the principal bundle has a delicate definition and structure and why I need to mention the principal bundle at all from the physics point of view is not obvious to me.

So if I pick a gauge group $G$, one must convince me that this indeed comes from a principal G bundle. On the principal bundle I have the action of the group on the total space and fibers that are isomorphic to G.

  1. It seems natural that the base manifold is space-time but it is not clear to me what the projection map actually is.
  2. Locally I expect the G-bundle to be a Cartesian product of some open set and a fiber with the action of the gauge group on the fiber but I don't know what the fibers are from the physics point of view.
  3. There also should be transition maps which are continuous maps from the intersection of open sets on the base manifold to the structure group which is the group $G$, how do you construct them from the physics point of view?
  4. From the physics I have the gauge group and the Lagrangian with the gauge group acting on the fields. But it seems natural to me once we have the gauge group acting on the fields then it is more natural to talk about an associated bundle. So that I have some representation of the group acting on a vector space. At this point it seems that in order to make the connection between differential geometry and gauge theory precise I at least have to have an associated bundle and a principal bundle.

I suppose my point is I would like to move beyond buzzwords and phrases but I can't seem to put all the pieces together neatly. There is a lot more to principal bundles than the elements of gauge group and the connection one can put on the bundles but that is all people seem to talk about.

Note: It may be unclear by what I mean by physical point of view. So let me give an example where the I can make the picture a bit more clear. Take a classical free particle with spin. It has the configuration space $ \mathbf{R}^3 \times SU(2) $ . The Lagrangian $L =\frac{1}{2}m\dot{x}^2 + i\lambda Tr(s^{-1}\dot{s}) $ where $s\in SU(2) $ and $S_i \sigma_i = s\sigma_3s^{-1}$. We also have the constraint $S_i^2=\lambda^2 $. If we concentrate on the spin degrees of freedom we have the fiber bundle which turns out to be a hopf bundle. The total space is given by the fields in $SU(2)$, the gauge group is $U(1)$ and to get the base manifold we realize the constraint $S_i^2 = \lambda $ gives us a sphere. Here the fibers come from the fields in the Lagrangian, the gauge group is $U(1)$ the base manifold is given by the constraint on the spin degrees of freedom. The local structure is $S^2 \times U(1) $. From this example note how the physics constrains and determines the mathematical structure I get. The gauge group comes from the Lagrangian, the total space comes from the fields in the Lagrangian and the base manifold comes from the constraint on the fields. I am not simply talking about $U(1)$


2 Answers 2


After a bit of thought, I have to come to some sort of understanding that I hope will be useful in making the ideas a bit more precise. In gauge theories physicists begin with a Lagrangian $ \mathcal{L}[\phi,\dot{\phi}] $. The claim is that this $ \mathcal{L} $ is invariant under the action of some group. What is a bit tricky is that to make this statement a bit more precise requires that we have two fiber bundles at once namely the principal-$G$ bundle and its associated vector bundle. I will try to describe things in such a way that the mathematical structure emerges rather than merely providing a dictionary.

1.For each patch of spacetime,$\mathcal{U}_i$ where $\mathcal{M}$ is base manifold we pick a map $S: \mathcal{U}_i \rightarrow G $ $(\textit{this will later be the gauge group})$. Then we pick a certain representation of the group i.e $\rho: G \rightarrow V $ where $V$ is a vector space. We now define what we later be a section $ \psi: \mathcal{U}_i \rightarrow [x, \phi] $ where $x$ is a point on the manifold. Gauge invariance means that $[x,\phi] \sim [x,\rho(g^{-1})\phi] $. We will come back to this construction but I think it is better at this point to concentrate on the map $S$ at this point

2.So rememeber we are at the spacetime patch $\mathcal{U}_i$ with the map $S$ in our hand. With this we construct the cartesian product $ \mathcal{U}_i \times G $. If we happen to find two overlapping open sets $ \mathcal{U}_i \text{ and } \mathcal{U}_j $ then for the sets of points in the intersection we have to make sure things are consistent and so we define functions $ t_{ij}: \mathcal{U}_i \cap \mathcal{U}_j \rightarrow G $ that will act on G i.e $(x, G) \rightarrow (x,t_{ij}(x)G)$. Doing this for the whole manifold $\mathcal{M}$ gives us another manifold $\mathcal{P}$ that is locally $\mathcal{U}_i \times G $. This the principal-$G$ bundle.

3.As we all know, requiring local gauge invariance in step one meets a glitch. The problem is the map $S$ because as we go around on the manifold $\mathcal{M}$ we need a method to go from one fiber to another fiber on the principal-$G$ bundle. To do this requires we introduce a connection $\Omega$ on the principal-$G$ bundle. But physicists always work on the base manifold so we need to pull back $\Omega$ to the base manifold by some section $\sigma $ i.e calculate $\sigma^*\Omega \equiv A $. Now of course these are locally defined sections and so when we are in intersection of two spacetime patches $\mathcal{U}_i\cap \mathcal{U}_j $ we will have two sections $\sigma, \sigma'$. This means we will get $ \sigma^*\Omega = A $ and $\sigma'^*\Omega=A' $. The two sections are related by the map $S$ i.e $\sigma' = R_{S(x)}\sigma = \sigma S(x) = \sigma g $ i.e the group acts by a right action.

4.To calculate $\sigma^*\Omega$ we note that $<\sigma^*\Omega, v>= <\Omega, \sigma_* v >$ where $v$ is a vector on the principal bundle. A tricky calculation shows $\sigma'_*v = R_{g*}(\sigma_{*}v)+\eta_x(p) $ where $\eta_X$ is a fundamental vector field, $X = <S^*\theta,v>$ and $\theta $ is the Maurer-Cartan form and $g$ is the image of $S$. To show this works we do the calculation $(\sigma'^* \omega)(v)= <\Omega,\sigma'v> = <\Omega,R_{g*}(\sigma_*v)>+<\Omega,\eta_X(pg)= <R^*_g\Omega,\sigma_*v> +X = <Ad_{g^-1}\Omega,\sigma_*v>+X=<Ad_{g^{-1}}A + S^*\theta,v> $ This of course is the usual transformation rule for the gauge field on the base manifold

5.We can now state the fact that we have an associated bundle $\mathcal{A}$ which is $\mathcal{P}\times_G V = (\mathcal{P}\times V)/G $ and is locally $\mathcal{U}_i \times V $ with sections as defined in step 1. The sections on this bundle are what physicists call the fields.


My understanding is the projection is the surjective map from the total space to the base manifold.

So locally, the bundle looks like the product space $M \times F,$ where $M$ is the base manifold and $F$ is the fiber. Thus a point in the total space $\left(m,f\right)$ is mapped to $\left(m,0\right)$ by the projection.


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