In the past I had the feeling that I understood the mathematics and ideas behind principal connections and connection one-forms rather well. However, while trying to explain these ideas in simple terms to a nonmathematical audience, I noticed that my understanding might not extend flawlessly beyond the pure formulas.

When you have a principal $G$-bundle $\pi:P\rightarrow M$, you can canonically define the vertical subbundle as the kernel of the projection $\pi_*$. A connection is then equivalent to a choice of complement inside the tangent bundle $TP$. Locally, where $P|_U\cong U\times G$, one can identify the vertical spaces with the Lie algebra $\mathfrak{g}$. A complement can then be defined by assigning to every basis $\partial_i$ of a tangent space on $M$ a ''horizontal lift'' $$\widetilde{\partial}_i:=\partial_i+\chi_i,$$ where $\chi_i\in\mathfrak{g}$ (these two terms should be mapped in the right way to a tangent space of $P|_U$ and extended to a local frame).

The associated connection one-form $\omega$ on $P$ is then, as far as I understand, the form that assigns to any vector field on $P$, at any point, the contribution in $\mathfrak{g}$ that does not arise from these $\chi_i$'s: $$X = \sum_i\lambda_i(\partial_i+\chi_i) + \omega(X),$$ for some scalars numbers $\lambda_i$. It measures the change in the fibres that does not arise from a mere change on the base manifold.

When we then, locally, pullback the connection one-form $\omega$ to a one-form on the patch $U$, we get $$A = s^*\omega.$$ By definition of the pullback I would then think that this evaluates to the following formula on any vector field: $A(Y) = \omega(s_*Y)$. By then combining the above statements, I would expect that $A$ assigns to any vector field on $M$ the difference between its pushforward (along a section) to the bundle and its horizontal lift, resulting in a linear combination of the $\chi_i$'s. Is this correct?

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    $\begingroup$ I haven't read the details of your question, but I encourage anyone who is trying to understand how a connection on a principal bundle works to first learn about a connection on a vector bundle. This in turn defines a natural connection on the bundle of frames of the vector bundle, by applying the connection to each section in the frame. In that setting it is, at least for me, a lot easier to derive the formulas for a connection on a principal $GL(k)$-bundle since the action of $G$ on the frame bundle is quite explicit. After that, it's not hard to translate everything into the abstract case. $\endgroup$
    – Deane
    Jan 15, 2022 at 20:05
  • $\begingroup$ I know about the situation on vector bundles and how one obtains the formulas. These are not the problem. Its rather how to obtain a simple interpretation of what the local one-forms do. $\endgroup$
    – NDewolf
    Jan 15, 2022 at 20:35
  • $\begingroup$ I would support Deane's suggestion specifically with what you're saying. Think of the case where $P$ is the frame bundle (say orthonormal frames). Then when you pull back by a section you're looking specifically at how the frame is twisting. If $s(x)=(e_i(x))$, then $s^*\omega = (\omega_{ij}(x))$ and $\nabla e_i = \sum\omega_{ij}e_j$. You can interpret this, of course, in terms of local parallel frame fields. $\endgroup$ Jan 15, 2022 at 21:52
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    $\begingroup$ Right. Twisting by the negative of $(s^*\omega)_p(Y)$ makes the frame $s(p)$ parallel as you move instantaneously in direction $Y$. $\endgroup$ Jan 16, 2022 at 0:16
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    $\begingroup$ I don't think there's any contradiction. I'm saying that if you are driving along turning to the left, then you must compensate by turning to the right in order to go straight. :) $\endgroup$ Jan 16, 2022 at 16:43


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