*3.2. Ehresmann Connection and Horizontal Lift*

We recall the case of the linear connection ∇; it was uniquely determined by a parallel transport procedure. In the case of a principal connection, we have an analogous.

**Definition 11** (Lift)**.** *Let π* : *E* → *M be a fiber bundle, M be a differentiable manifold, x* ∈ *M and e* ∈ *E such that π*(*e*) = *x.*

*Given a smooth curve <sup>γ</sup>* : <sup>R</sup> <sup>→</sup> *M such that <sup>γ</sup>*(0) = *x, we define a lift of <sup>γ</sup> through e as the curve <sup>γ</sup>*˜*, satisfying*

$$
\tau \dot{\gamma}(0) = e \quad \text{and} \quad \pi(\dot{\gamma}(t)) = \gamma(t) \quad \forall t. \tag{10}
$$

*If E is smooth, then a lift is horizontal if every tangent to γ*˜ *lies in a fiber of HE, namely*

$$
\dot{\gamma}(t) \in H E\_{\tilde{\gamma}(t)} \,\,\forall t. \tag{11}
$$

It can be shown that an Ehresmann connection uniquely determines a horizontal lift. Here, it is the analogy with parallel transport.

<sup>9</sup> Think of (*U*(1), *Aµ*) for electromagnetism.
