Let $\nabla$ be a symmetric, linear connection on a smooth manifold $X$.

If $p \in X$ is any point, on a normal chart for $\nabla$ around $p$ it holds: $$ \Gamma_{ij}^k (p) = 0 \ , $$ where $\Gamma_{ij}^k$ denote the Christoffel symbols on those coordinates.

I am wondering whether a similar statement holds for principal connections.

To be precise, let $G$ be a Lie group and let $P_0 := G \times X \to X$ be the trivial principal $G$-bundle. Consider a principal connection on it, defined by a 1-form $\alpha$ on $P_0$ with values on the Lie algebra $\mathfrak{g}$ of $G$.

If $(g,x)\in P_0$ is any point, is it possible to find a connection $\bar{\alpha}$ isomorphic to $\alpha$, whose value $\bar{\alpha}_{(g,x)} $ at that point is zero?

  • $\begingroup$ A principal connection form $\alpha$ maps isomorphically every fibre of the vertical tangent bundle $VP_{0} \cong P_{0}\times \mathfrak{g}$ to the Lie algebra $\mathfrak{g}$, so $\alpha$ cannot vanish at any point. $\endgroup$ Jan 27, 2014 at 12:01

1 Answer 1


Yes, consider a chart centered at $p$ and the rays emitting from $p$ with respect to this chart. Now, take a frame at $p$ and consider the parallel transport along the rays. This gives you a local section of $P$ (or a new trivialisation if you prefer), and the connection 1-form with respect to this section vanishes at $p$.

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    $\begingroup$ I guess by $p$ you mean a point on the base $X$ and you want to say that the pullback of the connection form $\alpha$ along your local section vanishes at $p$? $\endgroup$ Jan 27, 2014 at 11:57
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    $\begingroup$ To find such a section, you don't need to do anything with parallel transport (which involves solving differential equations). It suffices to take any section $\sigma:X\to P_0$ such that the tangent space of the image $\sigma(X)\subset P_0$ at $(g,x)$ is equal to the kernel of $\alpha_{(g,x)}:T_{(g,x)}\to{\frak{g}}$, for then $\sigma^*(\alpha)$ will vanish at $x$, which is what you really want (instead of $\alpha_{(g,x)}$ vanishing, which, as Oldfich noted in his comment above, never happens). $\endgroup$ Jan 27, 2014 at 12:24
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    $\begingroup$ You both are right, I was not clear enough with my phrasing "connection 1-form with respect to this section vanishes at p," which should mean pullback of the connection form by the section. And Robert's remark shows us how nicely differential calculus can be approximated linearly, I just oversaw it. $\endgroup$
    – Sebastian
    Jan 27, 2014 at 13:02
  • $\begingroup$ You are all right: I'm not used to this language yet and, by trying to simplify the question, I got confused. What I was trying to argue was the existence of a section $\sigma \colon X \to P_0$ such that $\alpha_{|\sigma}$ vanishes at $p$, and any of your reasonings is enough. $\endgroup$ Jan 27, 2014 at 16:25

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