Let us assume that we have constructed a $G$-principal bundle $P$ over the manifold $M$ (for a curved space-time this is a $GL$-bundle, for a gauge theory I take $U(1)$ = electrodynamics) and the corresponding associated bundle $P_F$ with a typical fibre $F$ being a vector space.

1) At first, I am confused about the meaning of a local section $\sigma: U \rightarrow P$ on the principal bundle, where $U \subset M$. I understand it as assigning some "point of reference" with respect to $G$ to the corresponding point in $M$. This can be seen by the induced local trivialization which sets $\sigma(x) = (x, e)$ so that the section always corresponds to a neutral element of $G$. The associated bundle $P_F$ is constructed as the set of equivalence classes of $P \times F$ with respect to the equivalence relation $(p, f) \sim (pg, g^{-1}f)$, which means that the simultaneous transformation of the basis and the components does not change the vector. Then the section $\sigma$ fixes a representative in each equivalence class in $P \times F$, and this is interpreted as fixing the frame/gauge, is this correct?

2) If so, how does a section on the associated bundle $\psi: U \rightarrow P_F$, which is some matter field for the gauge $U(1)$-bundle $P$, look like? If I assume that $\sigma$ picks up different elements of $U(1)$, does it mean that $\psi (x)$ has different phases when I go through $U$ so that I have something like $\psi(x) e^{i \theta(x)}$? For me this sounds like a mistake because this is already a gauge transformation since the phase $\theta$ depends on the point on the manifold.

3) The connection form $\omega$ acts on the tangent to $G$ components of a tangent vector $X_p$ on the principle bundle. What is the intuition behind tangent vectors in $P$?

4) Furthermore, the connection form gives a separation of tangent spaces in $P$ into the vertical and horizontal spaces, which I see intuitively as "parallel" to $G$ and $M$, correspondingly. Since I lack intuition for this choice and its relation to the local sections on $P$ I would like to consider the following example. Let us consider a flat 2-dimensional manifold with a unique chart with polar coordinates which correspond (?) to some section $\sigma$ in $P = LM$. Will it be correct to say that taking horizontal spaces $H_p P$ in such a way that tangent vectors to $\sigma(x)$ always lie in $H_p P$ means that the parallel transport of the vectors will consist in projecting the vectors to the coordinate lines during the transport, so that the corresponding connection coefficients are zero?


2 Answers 2


There are too many questions to answer in one go. I'll address the questions of sections of associated bundles and gauge choices.

Let's work in a local patch in $P$ with coordinates $(x,g)$ such that $\pi:P\to M$ is $\pi(x,g)=x$.

We want a section of an associated bundle $P_V= P\times _G V$ where $V$ is a representation of $G$ under which a vector with components $\varphi_i$ transforms as $g: \varphi_i \mapsto D_{ij}(g) \varphi_j$.

I like to think of such a section as a $V$ valued function $\varphi_i(x,g)$ on the total space $P$ that obeys $$ \varphi_i(x,gh)= D_{ij}(h^{-1})\varphi_j(x,g). $$ This defines the section in all gauges at once. A gauge choice is given by selecting a $g(x)$ for each point $x$ in the base space so that $$ \varphi_i(x,g(x)\equiv \varphi_i(x) $$ is the usual (gauge chosen) matter field.

Note that we have to have an $h^{-1}$ for consistency $$ \varphi_i(x,gh_1h_2)=D_{ij}(h_2^{-1})\varphi_j(x,gh_1)\\ =D_{ij}(h_2^{-1})D_{jk}(h_1^{-1})\varphi_k(x,g)\\ = D_{ik}(h_2^{-1}h_1^{-1})\varphi_k(x,g)\\ = D_{ik}((h_1h_2)^{-1})\varphi_k(x,g). $$

Covariant derivatives $\nabla_\mu$ are now directional derivatives on the total space $P$ that lie in the horizontal subspace at each $(x,g)$ and project down to $\partial_\mu$ on the basespace $M$.

  • $\begingroup$ Ok, I see the point; in the case of $LM$ your $V$-valued function says that I can change the frame (right action of $G$) or, equivalently, the vector components in the reversed direction (left action of $G$). Therefore, I can choose by this $V$-valued function a vector on $V$ without fixing any frame (this is what I mean by equivalence class with respect to transformations from $G$). In turn, the section $g(x)$ on $P$ fixes a concrete frame and the components of the vector on $V$, so a definite representative from the equivalence class, correct? $\endgroup$ Apr 29, 2020 at 20:27
  • 1
    $\begingroup$ yes. An exaple is that the "monopole harmonics" solution of the wavefunction for a particle on a sphere around a charge $\Lambda$ monopole are given by the ${\rm SU}(2)$ rotation matrices as ${\mathcal Y}^J_{m,\Lambda}(\theta,\phi)= [D^{J}_{m,\Lambda/2}(\theta,\phi,\psi)]^*$ where the third Euler angle $\psi(\theta,\phi)$ is the gauge choice. $\endgroup$
    – mike stone
    Apr 29, 2020 at 20:39
  • $\begingroup$ Just to check that I understand correctly the idea: if I have a $U(1)$-bundle and fix the gauge by some section $g(x)$, do I say that the chosen points on the principal bundle represent zero phases for the corresponding matter field, i.e. section on the corresponding associated $\mathcal{C}$-bundle? $\endgroup$ Apr 29, 2020 at 20:49

Then the section $\sigma$ fixes a representative in each equivalence class in $P\times F$, and this is interpreted as fixing the frame/gauge, is this correct?

Yes. You can also think of $P$ as a collection of all (abstract) reference frames of a given type over all spacetime points. A local section of $P$ is a choice of reference frames over the region it is defined on.

Here reference frame is meant in the abstract sense, since 1) for any associated vector bundle it serves as the vector bundle's own local frame field, 2) it provides a "local frame" for nonlinear associated bundles as well, yet the usual concept of local frame fields is meaningful only for vector bundles.

If so, how does a section on the associated bundle look like?

The thing is, it does not look like at all. A local section $\psi:U\rightarrow P_F$ associates to an $x\in M$ an element of $P_F=P\times_GF$ at the fibre above $x$.

The fibre $(P_F)_x$ consists of equivalence classes $[(p_x,v)]$, where $p_x\in P_x$ is a "frame" at $x$, and $v\in F$ is a vector in the typical fibre, with pairs $(p_x,v)$ and $(p^\prime_x,v^\prime)$ being equivalent if $p^\prime_x=p_xg$ and $v^\prime=g^{-1}v$.

So the value $\psi(x)$ at $x$ not really an object in the usual sense, but an entire equivalence class.

If you have a local section $\sigma$ defined in a neighborhood of $U$, then $\sigma$ identifies $P_x$ with $G$ by writing $p_x=\sigma(x)g$, and then $p_x\sim g$, and it identifies the fibre $(P_F)_x$ with $F$ by mapping the class $[(p_x,v)]=[(\sigma(x)g,v)]=[(\sigma(x),gv)]$ to the element $gv\in F$.

This allows us to construct a map $\bar\psi:U\rightarrow F$ by mapping $x$ to $gv$, which is the local representative of $\psi$, one that takes values in "concrete" objects, rather than abstract equivalence classes.

There is a more intuitive way to look at this however. It is known (see Kobayashi/Nomizu for more details, by the notation I assume OP is reading Bleecker, and he doesn't go into this a lot) that the set of all sections of $P_F$ is in one-to-one correspondance with maps $P\rightarrow F$ that are equivariant, i.e. $f(pg)=g^{-1}f(p)$.

Suppose that $f_\psi$ is this map corresponding to your section $\psi$. The correspondance is that (above $x\in M$ for simplicity) $\psi(x)=[(p_x,f_\psi(p_x))]$. Then if $\sigma$ is a local section of $P$, then simply $$ \bar\psi(x)=f_\psi(\sigma(x)). $$

What is the intuition behind tangent vectors in $P$?

This question is rather ambigous, but here is what can be said. We have for any $p\in P$ the vertical space $V_pP<T_pP$. An important property of principal bundles is that $V_pP\cong \mathfrak g$ for all $p\in P$. The connection form is a vertical projection, i.e. it projects a tangent vector on $P$ at $p\in P$ into $V_pP$, but then it also identifies it with an element of $\mathfrak g$ via this isomorphism.

Will it be correct to say that...

If I understand what is written correctly, then yes. If a local section $\sigma:U\rightarrow P$ exists so that $T_x\sigma(T_xM)\le H_{\sigma(x)}P$, then the pullback of the conenction form along $\sigma$ (which is the local representative of the connection form on the base space) vanishes, and the connection is flat.

The curvature can be seen as the obstruction to the Frobenius integrability of the horizontal tangent distribution, and these "horizontal sections" $\sigma$ essentially exist locally if and only if the horizontal distribution is Frobenius integrable and thus the curvature vanishes.

  • $\begingroup$ Thank you very much for your detailed answer! Maybe I'll try to reformulate the question concerning the tangent vector in the principal bundle. Why the connection coefficients are equal to the components of the corresponding element of the Lie algebra? For example, if I take $LM$, how is it connected to the "undergraduate" definition of connection coefficients as covariante derivatives of the basis vectors in direction of basis vectors, $\nabla_\mu e_\nu = \Gamma^\lambda_{\nu\mu} e_\lambda$? $\endgroup$ Apr 29, 2020 at 21:17

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