*3.3. Connection Form in a G-Principal Bundle*

We now focus on the case where the smooth fiber bundle is a *G*-principal bundle with smooth action P.

Here, we need a group *G*, that we generally take to be a matrix Lie group. We then have the corresponding algebra g, a matrix vector space in the present case.

The action <sup>P</sup> defines a map *<sup>σ</sup>* : <sup>g</sup> <sup>→</sup> <sup>Γ</sup>(*VE*) called the *fundamental map*10, where at *<sup>p</sup>* <sup>∈</sup> *<sup>P</sup>*, for an element *ξ* ∈ g, it is given via the exponential map *Exp* : g → *G*.

$$
\sigma\_p(\xi) = \frac{\mathbf{d}}{\mathbf{d}t} \mathfrak{P}\_{\mathfrak{e}^{\mathfrak{d}\xi}}(p)\big|\_{t=0}.\tag{12}
$$

The map is vertical because

$$\left. \pi\_\* \sigma\_p(\xi) = \frac{\mathbf{d}}{\mathbf{d}t} \pi(\mathfrak{P}\_{\mathfrak{e}^{\mathfrak{f}}}(p)) \right|\_{t=0} = \frac{\mathbf{d}}{\mathbf{d}t} \pi(p) = 0. \tag{13}$$

Thus, the vector *σp*(*ξ*) is vertical and it is called the *fundamental vector*. Before proceeding, we need some Lie group theory.

**Recall of Lie machinery:** Let *G* be a Lie group (a differentiable manifold) with g as its Lie algebra and ∀*g*, *h* ∈ *G*. We define:


$$\begin{split} \mathbf{Ad}\_{\mathcal{S}} \mathfrak{f} &= \frac{\mathbf{d}}{\mathbf{d}t} \Big( (L\_{\mathcal{S}} \circ R\_{\mathcal{S}^{-1}}) (e^{t\xi}) \big)|\_{t=0} = \frac{\mathbf{d}}{\mathbf{d}t}|\_{t=0} \\ &= \mathfrak{g} \mathfrak{f} \mathfrak{g}^{-1} \in \mathfrak{g}. \end{split} \tag{14}$$

where the last two equalities hold in the present case of matrix Lie groups. This is not to be confused with the adjoint action ad : g × g → g, which is generated by the derivative of the adjoint map with *g* = *e <sup>t</sup><sup>χ</sup>* and *<sup>χ</sup>* <sup>∈</sup> <sup>g</sup>, such that ad*χ<sup>ξ</sup>* = [*χ*, *<sup>ξ</sup>*];


$$\theta\_{\mathcal{S}} := L\_{\mathcal{S}^{-1} \*} \, : \, T\_{\mathcal{S}} \mathcal{G} \to T\_{\mathcal{C}} \mathcal{G} \cong \mathfrak{g}. \tag{15}$$

For any left invariant vector field *v*, it holds ∀*g* ∈ *G* that *θg*(*v*(*g*)) = *v*(*e*). Therefore, left invariant vector fields are identified by their values over the identity thanks to the Maurer–Cartan form *θ*. So we can state ([8]) that this identification *v*(*e*) 7→ *v* defines an isomorphism between the space of left

<sup>10</sup> It turns out that it is an isomorphism, since P is regular.

<sup>11</sup> We stress that the exponential map is not an isomorphism for all Lie groups; thus, the elements generated by the exponential map belong, in general, to a connected subgroup of the total group, which is usually homeomorphic to its simply connected double cover. More in general, the isomorphism is between a subset of the algebra containing 0 and a subset of the group containing the identity. Moreover, for a compact, connected, and simply connected Lie group, the algebra always generates the whole group via the exponential map.

invariant vector fields on *G* and the space of vectors in *TeG*, thus, the Lie algebra g. For matrix Lie groups, it holds that *θ<sup>g</sup>* = *g* <sup>−</sup>1*dg*.

By definition, the action of an element of the group on *P* is P*<sup>g</sup>* : *P* → *P*, and therefore, it defines a tangent map P*g*∗: *TP* → *TP*, for which the following Lemma holds:

**Lemma 1.**

$$\mathfrak{P}\_{\mathcal{S}^\*} \circ \sigma(\mathfrak{F}) = \sigma(Ad\_{\mathcal{S}^{-1}}\mathfrak{F}).\tag{16}$$

**Proof.** At *p* ∈ *P*

$$\left. \mathfrak{P}\_{\mathcal{S}^\approx} \sigma\_p(\xi) = \frac{\mathbf{d}}{\mathbf{d}t} \left( (\mathfrak{P}\_{\mathcal{S}} \circ \mathfrak{P}\_{\mathcal{C}^\lessdot})(p) \right) \right|\_{t=0} = \frac{\mathbf{d}}{\mathbf{d}t} \left( (\mathfrak{P}\_{\mathcal{S}} \circ \mathfrak{P}\_{\mathcal{C}^\lessdot} \circ \mathfrak{P}\_{\mathcal{S}^{-1}} \circ \mathfrak{P}\_{\mathcal{S}})(p) \right) \big|\_{t=0'} \tag{17}$$

we then use the fact that P*<sup>g</sup>* ◦ P*<sup>e</sup> <sup>t</sup><sup>ξ</sup>* ◦ P*g*−<sup>1</sup> = P*g*−1*<sup>e</sup> <sup>t</sup><sup>ξ</sup> <sup>g</sup>* <sup>=</sup> <sup>P</sup>Ad*g*−<sup>1</sup> *e <sup>t</sup><sup>ξ</sup>* and the identity for matrix groups Ad*ge <sup>t</sup><sup>ξ</sup>* = *e t*Ad*gξ* to get the following:

$$\mathfrak{P}\_{\mathfrak{F}^s} \sigma\_p(\mathfrak{f}) = \frac{\mathbf{d}}{\mathbf{d}t} \langle \mathfrak{P}\_{\mathfrak{e}^{t(\mathrm{Ad}\_{\mathfrak{g}^{-1}} \mathbb{S}^1}} (\mathfrak{P}\_{\mathfrak{g}}(p)) \rangle \big|\_{t=0} = \sigma\_{\mathfrak{P}\_{\mathfrak{k}}(p)} (\mathrm{Ad}\_{\mathfrak{g}^{-1}} \mathfrak{f}). \tag{18}$$

It is time to define what we were aiming to define at the beginning of the section: the *connection form*.

**Definition 12.** *Let P be a smooth G-principal bundle and HE* ⊂ *TP be an Ehresmann connection. We call the* <sup>g</sup>*-valued* <sup>1</sup>*-form <sup>ω</sup>* <sup>∈</sup> <sup>Ω</sup><sup>1</sup> (*P*, g)*, satisfying*

$$
\omega(v) = \begin{cases}
\mathfrak{F} & \text{if } \ v = \sigma(\mathfrak{F}), \ \mathfrak{F} \in \mathcal{C}^{\infty}(P, \mathfrak{g}) \\
0 & \text{if } \ v \quad \text{horizontal},
\end{cases}
\tag{19}
$$

*the connection* 1*-form.*

**Proposition 1.**

$$
\mathfrak{P}\_{\mathcal{S}}^{\*}\omega = A d\_{\mathcal{S}^{-1}} \circ \omega. \tag{20}
$$

**Proof.** Suppose *v* = *σ*(*ξ*), since the other case left is trivial.

We can carry out some calculations on the left-hand side, and following from the definition of pull-back and Lemma 1, we have

$$\omega\left(\mathfrak{P}\_{\mathcal{S}}^{\*}\omega\right)\left(\sigma(\mathfrak{f})\right) = \omega\left(\mathfrak{P}\_{\mathcal{S}^{\*}}\circ\sigma(\mathfrak{f})\right) = \omega\left(\sigma(\operatorname{Ad}\_{\mathcal{S}^{-1}}(\mathfrak{f})) = \operatorname{Ad}\_{\mathcal{S}^{-1}}(\mathfrak{f}).\tag{21}$$

Then, we only need to manipulate the right-hand side as

$$\operatorname{Ad}\_{\mathcal{S}^{-1}}\left(\omega\left(\sigma(\mathfrak{f})\right)\right) = \operatorname{Ad}\_{\mathcal{S}^{-1}}(\mathfrak{f}).\tag{22}$$

Both times, we used just the given definition of connection 1-form (Equation (19)).

**Remark 1.** *This last Proposition is called G-equivariance. It can be imposed instead of by assuming that HE is an Ehresmann connection, and then HE can be shown to be such an Ehresmann connection.*

Another fundamental concept is given in the following:

**Definition 13** (Tensorial form)**.** *Let <sup>ρ</sup>* : *<sup>G</sup>* <sup>→</sup> Aut(*V*) *be a representation over a vector space <sup>V</sup> and <sup>α</sup>* <sup>∈</sup> <sup>Ω</sup>*<sup>k</sup>* (*P*, *V*) *be a vector valued differential form.*

*We call α a tensorial form if it is the following:*


*We define horizontal and equivariant forms as maps belonging to* Ω*<sup>k</sup> G* (*P*, *V*)*.*

**Observation 5:** The connection form *ω* is not, in general, horizontal; thus, it is not a tensorial form, *<sup>ω</sup>* <sup>∈</sup>/ <sup>Ω</sup><sup>1</sup> *G* (*P*, g). This will be clear when taking into account how the gauge field transforms under a change of trivialization in Section 4.
