Covariant derivative along a horizontal lift in an associated vector bundle

[n.evans]

I am trying to familiarize myself with the use of fibre bundles and associated bundles but am having some problems actually making calculations. I would like to show that the covariant derivative along the horizontal lift of a curve in the base space vanishes (which should be just a matter of employing definitions correctly I think):

Consider a principal fibre bundle $(E, \pi, M)$ with a structure group $G$ associated to a vector bundle $(E_F, \pi_F, M)$, where $F$ is a vector space. Let $\alpha(t): [a, b] \rightarrow M$ and $\alpha^{\uparrow}_F(t): M \rightarrow E_F$ be the horizontal lift of $\alpha$ in the associated bundle. Let $\Psi(x): M \rightarrow E_F$, with $x \in M$ be a section of the associated bundle, such that $\Psi(\alpha(t)) = \alpha^{\uparrow}_F(t)$

Show that the covariant derivative $\nabla_{\alpha}\Psi$ evaluated along $\alpha(t)$ vanishes.

I know that the covariant derivative can be written as

$\nabla_{\mu}\Psi(x) = \partial_{\mu}\Psi(x) + A_{\mu}(x)\Psi(x)$

but I cannot work out how to use the relation between $\Psi$ and $\alpha^{\uparrow}_F(t)$ to show that it vanishes (if indeed it should).

 

[pat_connell]

Let \gamma(t) denote the path in E_F given by \alpha^{\uparrow}_F(t). Then we have that\nabla_{\alpha}\Psi(x) = \frac{d}{dt}\Psi(\gamma(t)) = \frac{d}{dt}\alpha^{\uparrow}_F(t)Now, since \alpha^{\uparrow}_F(t) is the horizontal lift of \alpha(t), it follows that \alpha^{\uparrow}_F(t) is constant along \alpha(t). Thus we have\nabla_{\alpha}\Psi(x) = \frac{d}{dt}\alpha^{\uparrow}_F(t) = 0.Hence, the covariant derivative \nabla_{\alpha}\Psi evaluated along \alpha(t) vanishes.