• The mathematical framework of a gauge theory is that of principal $G$-bundles, where $G$ is a Lie group (representing some physical symmetry); let ${\pi}\colon{P}\to{M}$ be such a bundle.
  • A connection 1-form on $P$ corresponds to a gauge field, and the curvature 2-form associated to it corresponds to the field strength tensor.
  • A (local) section of $\pi$ serves as a (local) reference frame (alias, gauge) for an observer on $M$. In particular, if $s\colon{U}\to{P}$ is such a section, where $U$ is an open subset of $M$, then an observer on $U$ describes elements of $\Omega^{k}\left({\pi}^{-1}(U),\mathfrak{g}\right)$ by pulling them back by $s$.
  • Specifically, let $A\in\Omega_{con}^{1}\left(P,\mathfrak{g}\right)$ be a connection 1-form and ${F^A}\in\Omega_{hor}^{2}\left(P,\mathfrak{g}\right)^{\operatorname{Ad}}$ the associated curvature 2-form. Then, our observer on $U$ interprets ${A_s}:={s^*}A$ and ${F_s^A}:={s^*}{F^A}$ as the gauge field and the field strength tensor, respectively, in that specific gauge, viz. $s$.

Conclusion: In a certain gauge, any connection 1-form and any curvature 2-form define, respectively, a unique element of $\Omega^{1}\left(M,\mathfrak{g}\right)$ and $\Omega^{2}\left(M,\mathfrak{g}\right)$.

Now, a physicist's approach on gauge theory is to define the space of gauge fields as $\Omega^{1}\left(M,\mathfrak{g}\right)$, when a particular gauge has (implicitly) been chosen. So my question is the following:

"does the whole $\Omega^{1}\left(M,\mathfrak{g}\right)$ qualify as the set of gauge fields, in the sence that there is one-to-one correspondence between $\Omega_{con}^{1}\left(P,\mathfrak{g}\right)$ and $\Omega^{1}\left(M,\mathfrak{g}\right)$, when a gauge is chosen? Or are there elements in $\Omega^{1}\left(M,\mathfrak{g}\right)$ that do not qualify as gauge fields, in the sence that they are not pullbacks of connection 1-forms by the chosen gauge?"

Probably, the question is of mathematical nature and can be answered by rigorous calculations. If one's mathematical background is such that they can provide a rigorous proof, then they are kindly requested to do so (or at least provide a sketch of that proof). However, if there is an intuitive way to approach the answer, I would be glad to know about that.

  • $\begingroup$ $\pi : P \to M$ is the projection; the more common notation to refer to the bundle is $P \to^{\pi} M$ (in better LaTeX normally) and in short to refer to the bundle through the total space, $P$, rather than talking of a section of $\pi$. $\endgroup$
    – JamalS
    Mar 18, 2017 at 19:30
  • $\begingroup$ unessential formalities... Also you may want to check here: en.wikipedia.org/wiki/Principal_bundle $\endgroup$ Mar 18, 2017 at 19:39
  • $\begingroup$ I'm aware it's wrong on Wikipedia :) Probably some guy too lazy to use the proper notation as it's not so easy to typeset. $\endgroup$
    – JamalS
    Mar 18, 2017 at 20:20
  • 4
    $\begingroup$ Anyway, unessential formalities. Both terms are used interchangeably, say it an "abuse of notation" if you like. By the way, are you Nicolas Bourbaki? $\endgroup$ Mar 18, 2017 at 20:40

1 Answer 1


It is not true in general that a connection form $A$ on the bundle descends to a well-defined form on all of $M$. In fact, it does so if and only if the bundle is trivial - since the existence of a global section of a principal bundle by which we can pull back $A$ necessarily implies that $P$ is trivial.

Instead each connection form $A$ descends to local $A_i\in\Omega^1(U_i,\mathfrak{g})$ where the $U_i$ are a cover of $M$ that trivializes $P$, i.e. there are sections $s_i : U_i \to \pi^{-1}(U_i)\cong U_i\times G$ with $s_i(x) = (x,1)$ and we define $A_i := s_i^\ast A$. With regards to the transition functions $t_{ij} : U_i \cap U_j \to G$, these fulfill $$ A_i = \mathrm{ad}_{t_{ij}}(A_j - t_{ij}^\ast \theta),$$ where $\theta$ is the Cartan-Maurer form, or, in a more familiar notation, $$ A_i = t^{-1}_{ij}A_j t_{ij} - t_{ij}^{-1}\mathrm{d}t_{ij}.\tag{1}$$ Conversely, every system of $A_i$ that fulfills these relations defines a connection form on $P$. When the bundle is trivial, this relation is vacuous since we only need on $U_i = M$, and we recover the claimed bijection between connection forms on $P$ and $A\in\Omega^1(M,\mathfrak{g})$ for this special case.

Now, a gauge-transformation is a fiber-preserving automorphism $g : P\to P$ that descends to local functions $g_i : U_i\to G$ with $g_i = g_j t_{ij}$, or, in the trivial case, a function $g : M\to G$, which act on the local connection forms much as the transition functions do (which is the reason the physics literature has trouble distinguishing these two conpcets at times). Such a transformation amounts to choosing another point in every fiber $\pi^{-1}(x) \cong G$ as the identity, which we may do, since the fiber carries a group action but is not naturally a group itself, and so has no distinguished identity element.

If you now note that the sections $s_i$ include such a choice of identity in their definition, then you can see that we may view such a gauge transformation as changing the sections by which we defined the connection form. So the physicist, who usually does not care about the connection form on $P$ but about its local description on the $U_i$, declares those $A_i$ which are related by such transformations to be equivalent, and quotienting them out of the space of $\Omega^1(U_i,\mathfrak{g})$ with the correct transformation property (or equivalently quotienting the space of connection forms on $P$ by gauge transformation $P\to P$) yields the space of gauge equivalence classes.

So, in summary, we answer the question like this: A gauge field is a collection of local gauge potentials $U_i \to \mathfrak{g}$ with the compatibility condition (1), which are in bijection to principal connections on $P$, but physically those that are related by a gauge transformation are declared equivalent and essentially indistinguishable.

  • $\begingroup$ Of course! In the principal bundle level, a connection is a fixed geometrical object, whereas in the base manifold level it forms an equivalence class of gauge fields, where the elements of that class are in 1-1 correspondence with those of $\Gamma(P)$. However, does the set of those classes form a partition of ${\Omega^1}(M,\mathfrak{g})$? Equivalently, is there a bijection between ${\Omega^1}(M,\mathfrak{g})$ and $\Omega_{con}^{1}(P,\mathfrak{g})\times\Gamma(P)$? $\endgroup$ Mar 18, 2017 at 20:34
  • $\begingroup$ And something more, do the defining properties of a connection 1-form on $P$ manifest as certain conditions on 1-forms on M? If yes, how? $\endgroup$ Mar 18, 2017 at 20:37
  • $\begingroup$ ...If they do so and are trivially satisfied, then the anwser would be affirmative. $\endgroup$ Mar 18, 2017 at 20:45
  • $\begingroup$ @user3257624 I clearly stated that there's a bijection between connection forms on $P$ and local forms $\Omega^1(U_i,\mathfrak{g})$, and that this reduces to a bijection between connection forms and $\Omega^1(M,\mathfrak{g})$ in the case of a trivial bundle. I'm not sure what exactly you want more than that. The properties of a connection 1-form are essentially what leads to the compatibility conditions (1). $\endgroup$
    – ACuriousMind
    Mar 18, 2017 at 20:49

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