*5.1. Make It Clear*

**Definition 17** (Gauge field)**.** *Let P* → *M be a G-principal bundle, G be a Lie group with* g *as the respective Lie algebra,* {*Uβ*} *be a cover of M, and s<sup>β</sup>* : *U<sup>β</sup>* → *P be a section.*

*We define the gauge field as the pull-back of the connection form <sup>ω</sup>* <sup>∈</sup> <sup>Ω</sup><sup>1</sup> (*P*, g) *as*

$$A\_{\beta} = s\_{\beta}^{\*}\omega \; \; \in \, \Omega^{1}(\mathcal{U}\_{\beta}, \mathfrak{g}). \tag{29}$$

We notice that, under a change of trivialization, such a gauge field changes via the action of the adjoint map.

In fact, we have the following:

**Lemma 2.** *The restriction of ω to π* −1 (*Uβ*) *agrees with*

$$
\omega\_{\beta} = A d\_{\mathcal{g}\_{\beta}^{-1}} \circ \pi^\* A\_{\beta} + \mathcal{g}\_{\beta}^\* \theta\_{\prime} \tag{30}
$$

*where g<sup>β</sup>* : *π* −1 (*Uβ*) → *G is the map induced by the inverse of the trivialization map ϕ<sup>β</sup> defined in Equation* (2)*, and with Ad g* −1 *β , we intend for the adjoint map at the group element given by gβ*(*p*) −1 *at a point p* ∈ *π* −1 (*Uβ*)*.*

The proof comes from the observation that Equations (19) and (30) coincide in *π* −1 (*Uβ*) for both a horizontal (for which they are zero) and a vertical vector field.

Thanks to this, we easily have the following:

**Proposition 2.** *Let G be a matrix Lie group. Then it holds the following transformation for a gauge field:*

$$A\_{\beta} = \mathcal{g}\_{\beta\gamma} A\_{\gamma} \mathcal{g}\_{\beta\gamma}^{-1} - d \mathcal{g}\_{\beta\gamma} \mathcal{g}\_{\beta\gamma}^{-1}. \tag{31}$$

**Proof.** Using Equations (29) and (30) for all *x* ∈ *U<sup>β</sup>* ∩ *Uγ*,

$$\begin{split} \mathcal{A}\_{\beta} &= s\_{\beta}^{\*}\omega \\ &= s\_{\beta}^{\*}\omega\_{\beta} = s\_{\beta}^{\*}\omega\_{\gamma} \\ &= s\_{\beta}^{\*} \{ \operatorname{Ad}\_{\operatorname{g}\_{\gamma}^{-1}} \circ \pi^{\*} A\_{\gamma} + \operatorname{g}\_{\gamma}^{\*}\theta \} \\ &= \operatorname{Ad}\_{\operatorname{\mathcal{S}}\_{\beta\gamma}^{-1}} \circ A\_{\gamma} + \operatorname{g}\_{\gamma\beta}^{\*}\theta \qquad \qquad \text{(using } \operatorname{g}\_{\gamma} \circ s\_{\beta} := \operatorname{g}\_{\beta\gamma} : \operatorname{U}\_{\beta} \cap \operatorname{U}\_{\gamma} \to \operatorname{G)} \\ &= \operatorname{Ad}\_{\operatorname{\mathcal{S}}\_{\beta\gamma}} \circ \left( A\_{\gamma} - \operatorname{g}\_{\beta\gamma}^{\*} \theta \right) \qquad \qquad \text{(\operatorname{Ad}\_{\operatorname{\mathcal{S}}\_{\beta\gamma}} \circ \operatorname{g}\_{\beta\gamma}^{\*} \theta = -\operatorname{g}\_{\gamma\beta}^{\*} \theta). \end{split} \tag{32}$$

which reduces to the assert for matrix Lie groups.
