Let $\pi: P \rightarrow M$ be a $G$-principal bundle. Let $\Phi \in \Omega^1(P, \mathfrak{g})$ be a principal connection on $P$. Let $I$ be an open interval with $0 \in I$. Consider two curves $\gamma_1$, $\gamma_2: I \rightarrow M$ with $\gamma_1(0)=\gamma_2(0)=x$. Fix $u \in P_x$. Denote by $\gamma_i^*:I \rightarrow P$ the uniquely determined horizontal lift of $\gamma_i$ with $\gamma_i(0)=u$ for $i=1,2$.

$\gamma_1$, $\gamma_2$ have contact of order $r$ in zero if and only if $\gamma_1^*$ and $\gamma_2^*$ have contact of order $r$ in zero.

Remember that two curves are said to have contact of order $r$ in a point, if there exist (equivalently: for all) local coordinates around that point, such that all derivatives up to the $r$-th derivative of the coordinate representations of the two curves coincide.

Note also, that $\gamma_1$, $\gamma_2$ have contact of order $r$ if and only if for any smooth function $\varphi: M \rightarrow \mathbb{R}$ we have $\varphi \circ \gamma_1 - \varphi \circ \gamma_2 = o(t^r)$.

The direction "$\Leftarrow$" is trivial, when using the second characterization.

I have problems showing the direction "$\Rightarrow$". I would like to use local coordinates $(x_1, \dots, x_n, y_1, \dots, y_m)$ around $u$, such that $\frac{\partial}{\partial x_i}$ are horizontal and $\frac{\partial}{\partial y_j} \in Ker (d \pi)$. Then the claim would follow, because $(x_1 \circ \pi ^{-1}, \dots, x_n \circ \pi^{-1})$ would be a chart of $M$ around $x$. (It needs some extra explanation, why $x_i \circ \pi^{-1}$ is well-defined, i.e. independent of the choice of the preimage under $\pi$) And we could check the first criterion for contact for the curves $\gamma_1^*$, $\gamma_2^*$: The last $m$ coordinates of all derivatives are zero for both curves. The first $n$ coordinates of all derivatives coincide with the derivative of the original curves $\gamma_1$, $\gamma_2$. (In the previously described coordinates respectively)

But why do such coordinates exist? I am not even sure they do exist...


The idea described above (choosing suitable coordinates) is not very fruitful. Such coordinates exist if and only if the principal connection has vanishing curvature, which is not generally the case.

Here is a proof:

Let $\sigma : M \supset U \rightarrow P$ be a local section. This defines a local trivialization of $P$, namely $\theta : (x, g) \mapsto \sigma (x) \cdot g$. Now write $\theta^{-1} \circ \gamma_i^* =(\gamma_i,g_i)$ for some $g_i : I \rightarrow G$. Then $0=\Phi ((\gamma_i^*)'(t))=\Phi ( d \theta (\gamma_i'(t),0))+\Phi(d\theta(0,g_i'(t)))$, i.e. $$\Phi ( d \theta (\gamma_i'(t),0))=-\Phi(d\theta(0,g_i'(t))). \tag{1}$$ We have $$\Phi ( d \theta (\gamma_i'(t),0))=\Phi(d \sigma (\gamma_i'(t)))\tag{2}$$ and $$\Phi(d\theta(0,g_i'(t)))=dL_{g_i^{-1}(t)} (g_i'(t)).\tag{3}$$ By assumption $\gamma_i'$ have contact of order $(r-1)$ at zero. Hence equation (2) gives that the left hand side of equation (1) has $(r-1)$-th order contact. This shows that also the right hand side of (3) has $(r-1)$-th order contact. This means however that the $g_i$ have contact of order $r$. (Check this by writing the curves $g_i$ in the map $\exp$)

So the $\gamma_i$ and $g_i$ have $r$-th order contact, hence $\theta (\gamma_i,g_i)$ also have $r$-th order contact.


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