*3.4. Curvature Forms*

Given our connection 1-form *ω*, we can proceed in two ways: the first consists in taking a map called the *horizontal projection* and in defining the curvature as this projection applied on the exterior derivative of *ω*. In this way, we naturally see that curvature measures the displacement of the commutator of two vectors from being horizontal.

We will proceed in a different way though. We will define the curvature through a *structure equation*.

**Definition 14.** *Given <sup>ω</sup>* <sup>∈</sup> <sup>Ω</sup><sup>1</sup> (*P*, <sup>g</sup>)*, a principal connection* <sup>1</sup>*-form, the* <sup>2</sup>*-form* <sup>Ω</sup> <sup>∈</sup> <sup>Ω</sup><sup>2</sup> *G* (*P*, g) *satisfies the following:*

$$
\Omega = d\omega + \frac{1}{2} [\omega \wedge \omega] \tag{23}
$$

*whic is called curvature* 2*-form.*

In Equation (23), [*ω* ∧ *ω*] denotes the bilinear operation on the Lie algebra g called differential Lie bracket. It is defined as follows:

$$[\omega \wedge \eta](\iota, \upsilon) = \frac{1}{2} \{ [\omega(\iota), \eta(\upsilon)] - [\omega(\upsilon), \eta(\iota)] \},\tag{24}$$

where *u* and *v* are vector fields.

It follows straightforwardly that, if we take two general horizontal vector fields *<sup>u</sup>*, *<sup>v</sup>* ∈ <sup>Γ</sup>(*HE*) and we use the ordinary formula<sup>12</sup> for the exterior derivative of a 1-form *<sup>d</sup>ω*(*u*, *<sup>v</sup>*) = *<sup>u</sup>ω*(*v*) <sup>−</sup> *<sup>v</sup>ω*(*u*) <sup>−</sup> *<sup>ω</sup>*([*u*, *<sup>v</sup>*]), since *ω*(*u*) = *ω*(*v*) = 0, we get

$$
\Omega(u, v) = -\omega([u, v]).\tag{25}
$$

We see that Ω measures how the commutator of two horizontal vector fields is far from being horizontal as well.

<sup>12</sup> Here, we regard *<sup>ω</sup>*(*u*) as a function *<sup>ω</sup>*(*u*) : *<sup>P</sup>* <sup>→</sup> <sup>g</sup> belonging to the algebra of smooth functions to <sup>g</sup>, <sup>C</sup> <sup>∞</sup>(*P*, g).
