[proofplan] Let $\omega\in\Omega^1(P;\mathfrak g)$ be the principal connection form whose local representative in the section $s$ is the local connection form $A=s^*\omega$. We compute the pullback of $\omega$ along the transformed section $s'=s\cdot h$. The derivative of $s'$ has two contributions: one from moving the point $s(x)$ in $P$ and then translating by $h(x)$, and one vertical contribution from differentiating the gauge function $h$. The connection equivariance axiom gives the adjoint term from the first contribution, while the defining normalization of a connection form on fundamental vertical vectors identifies the second contribution with the pullback of the left Maurer-Cartan form. [/proofplan]

[step:Write the transformed section through the bundle action] Let $\pi:P\to M$ denote the principal $G$-bundle projection, and let $U\subset M$ be the [open set](/page/Open%20Set) on which the local sections $s:U\to P$ and $s':U\to P$ are defined. Let $\omega\in\Omega^1(P;\mathfrak g)$ denote the principal connection form, so the local connection forms are $A=s^*\omega$ and $A'=(s')^*\omega$. Let $\rho: P \times G \to P$ be the smooth right action map, so $\rho(p,g)=p\cdot g$. Let $F: U \to P \times G$ be the smooth map defined by $F(x)=(s(x),h(x))$. Then \begin{align*} s'=\rho\circ F. \end{align*} Thus, for $x \in U$ and $v \in T_xU$, \begin{align*} A'_x(v)=\omega_{s'(x)}\bigl(d\rho_{(s(x),h(x))}(ds_x(v),dh_x(v))\bigr). \end{align*} [/step]

[step:Split the derivative into horizontal motion and gauge motion]Fix $x \in U$ and $v \in T_xU$. Define $p=s(x)\in P$, $g=h(x)\in G$, $\xi=ds_x(v)\in T_pP$, and $\eta=dh_x(v)\in T_gG$. Since $\rho$ is smooth, the differential of $\rho$ at $(p,g)$ is linear in the two tangent variables, so \begin{align*} d\rho_{(p,g)}(\xi,\eta)=d(R_g)_p(\xi)+d(\rho_p)_g(\eta), \end{align*} where $\rho_p:G\to P$ is the orbit map defined by $\rho_p(a)=p\cdot a$. Therefore \begin{align*} A'_x(v)=\omega_{p\cdot g}\bigl(d(R_g)_p(\xi)\bigr)+\omega_{p\cdot g}\bigl(d(\rho_p)_g(\eta)\bigr). \end{align*}[/step]

[guided]The transformed section changes for two independent reasons. First, $s(x)$ changes inside $P$ as $x$ moves in $U$. Second, the group element $h(x)$ also changes. We separate these two effects at the level of tangent vectors. For the fixed point $x \in U$ and tangent vector $v \in T_xU$, set \begin{align*} p=s(x), \qquad g=h(x), \qquad \xi=ds_x(v), \qquad \eta=dh_x(v). \end{align*} The action map is $\rho:P\times G\to P$, with $\rho(p,g)=p\cdot g$. Holding $g$ fixed gives the right translation map $R_g:P\to P$, defined by $R_g(q)=q\cdot g$, and holding $p$ fixed gives the orbit map $\rho_p:G\to P$, defined by $\rho_p(a)=p\cdot a$. The canonical tangent-space identification $T_{(p,g)}(P\times G)=T_pP\times T_gG$ lets us regard $(\xi,\eta)$ as the sum of $(\xi,0)$ and $(0,\eta)$. By linearity of $d\rho_{(p,g)}$, its value on $(\xi,\eta)$ is therefore the sum of the two partial contributions: \begin{align*} d\rho_{(p,g)}(\xi,\eta)=d(R_g)_p(\xi)+d(\rho_p)_g(\eta). \end{align*} Since $s'=\rho\circ(s,h)$, we obtain \begin{align*} A'_x(v)=\omega_{p\cdot g}\bigl(d(R_g)_p(\xi)\bigr)+\omega_{p\cdot g}\bigl(d(\rho_p)_g(\eta)\bigr). \end{align*} This is the basic decomposition: the first term is the transported old connection form, and the second term is the purely vertical correction produced by the gauge function.[/guided]

[step:Use equivariance of the connection form for the translated section term] By the equivariance axiom in the definition of a principal connection form, the connection form satisfies \begin{align*} (R_g)^*\omega=\operatorname{Ad}_{g^{-1}}\omega \end{align*} for every $g \in G$. Evaluating this identity at $p=s(x)$ on the tangent vector $\xi=ds_x(v)$ gives \begin{align*} \omega_{p\cdot g}\bigl(d(R_g)_p(\xi)\bigr)=\operatorname{Ad}_{g^{-1}}\bigl(\omega_p(\xi)\bigr). \end{align*} Since $A=s^*\omega$, we have $\omega_p(\xi)=A_x(v)$. Hence \begin{align*} \omega_{p\cdot g}\bigl(d(R_g)_p(\xi)\bigr)=\operatorname{Ad}_{h(x)^{-1}}\bigl(A_x(v)\bigr). \end{align*} [/step]

[step:Identify the gauge derivative term with the Maurer-Cartan form] For each $a \in G$, let $L_a:G\to G$ denote left multiplication, defined by $L_a(b)=ab$. Define \begin{align*} \zeta=\theta_g(\eta)\in\mathfrak g. \end{align*} By the definition of the left Maurer-Cartan form, $\zeta=d(L_{g^{-1}})_g(\eta)$. Equivalently, \begin{align*} \eta=d(L_g)_e(\zeta). \end{align*} Let $\exp:\mathfrak g\to G$ denote the Lie group exponential map. Choose $\varepsilon>0$ and define the smooth curve $c:(-\varepsilon,\varepsilon)\to G$ by \begin{align*} c(t)=g\exp(t\zeta). \end{align*} Then $c(0)=g$ and $c'(0)=d(L_g)_e(\zeta)=\eta$. Therefore \begin{align*} d(\rho_p)_g(\eta)=\frac{d}{dt}\Big|_{t=0} p\cdot g\exp(t\zeta). \end{align*} With the standard right-action convention for principal bundles, the fundamental vertical vector generated by $\zeta\in\mathfrak g$ at $q\in P$ is \begin{align*} \zeta_P(q)=\frac{d}{dt}\Big|_{t=0}q\cdot\exp(t\zeta). \end{align*} Thus $d(\rho_p)_g(\eta)=\zeta_P(p\cdot g)$. The defining normalization of a principal connection form on fundamental vertical vectors is $\omega_q(\zeta_P(q))=\zeta$, so \begin{align*} \omega_{p\cdot g}\bigl(d(\rho_p)_g(\eta)\bigr)=\zeta. \end{align*} Using the definition of $\zeta$, this becomes \begin{align*} \omega_{p\cdot g}\bigl(d(\rho_p)_g(\eta)\bigr)=\theta_{h(x)}(dh_x(v)). \end{align*} [/step]

[step:Combine the two contributions] Substituting the two computed terms into the decomposition of $A'_x(v)$ gives \begin{align*} A'_x(v)=\operatorname{Ad}_{h(x)^{-1}}\bigl(A_x(v)\bigr)+\theta_{h(x)}(dh_x(v)). \end{align*} The second term is precisely $(h^*\theta)_x(v)$, so \begin{align*} A'_x(v)=\bigl(\operatorname{Ad}_{h^{-1}}A\bigr)_x(v)+(h^*\theta)_x(v). \end{align*} Since this holds for every $x \in U$ and every $v \in T_xU$, the equality of $\mathfrak g$-valued $1$-forms follows: \begin{align*} A'=\operatorname{Ad}_{h^{-1}}A+h^*\theta. \end{align*} [/step]