[proofplan] We write the transformed connection as the sum of the adjointly transformed old connection and the Maurer-Cartan term. The proof rests on two intrinsic differential identities: differentiating an adjointly transformed form produces a commutator with $u^{-1}du$, and the Maurer-Cartan form satisfies its structure equation. Expanding the curvature of $A'$ then shows that the commutator terms cancel, while the remaining terms combine to $\operatorname{Ad}_{u^{-1}}F_A$. [/proofplan]

[step:Introduce the Maurer-Cartan term and the transformed connection] Let $\theta \in \Omega^1(U;\mathfrak g)$ denote the pullback by $u$ of the left Maurer-Cartan form on $G$, so \begin{align*} \theta = u^{-1}du. \end{align*} Define $\alpha \in \Omega^1(U;\mathfrak g)$ by \begin{align*} \alpha = \operatorname{Ad}_{u^{-1}}A. \end{align*} With this notation the transformed connection form is \begin{align*} A' = \alpha + \theta. \end{align*} The curvature of $A'$ is therefore \begin{align*} F_{A'} = d(\alpha + \theta) + \frac{1}{2}[(\alpha + \theta) \wedge (\alpha + \theta)]. \end{align*} Since the bracket wedge product is bilinear and, for $\mathfrak g$-valued $1$-forms, satisfies $[\alpha \wedge \theta] = [\theta \wedge \alpha]$, this expands to \begin{align*} F_{A'} = d\alpha + d\theta + \frac{1}{2}[\alpha \wedge \alpha] + [\alpha \wedge \theta] + \frac{1}{2}[\theta \wedge \theta]. \end{align*} [/step]

[step:Record the two differential identities needed for the expansion]We use the following two identities for the smooth map $u: U \to G$ and the form $\theta = u^{-1}du$. Since $u$ is smooth, $\theta$ is the smooth pullback of the left Maurer-Cartan form on $G$, so the Maurer-Cartan structure equation applies after pullback: \begin{align*} d\theta + \frac{1}{2}[\theta \wedge \theta] = 0. \end{align*} Also, for the $1$-form $\alpha = \operatorname{Ad}_{u^{-1}}A$, \begin{align*} d\alpha = \operatorname{Ad}_{u^{-1}}dA - [\theta \wedge \alpha]. \end{align*} The first identity is the Maurer-Cartan structure equation for the pullback of the left Maurer-Cartan form. For the second identity, let $X$ and $Y$ be smooth vector fields on $U$. For every smooth map $B: U \to \mathfrak g$, differentiating the pointwise adjoint action gives \begin{align*} X\bigl(\operatorname{Ad}_{u^{-1}}B\bigr) = \operatorname{Ad}_{u^{-1}}(XB) - [\theta(X), \operatorname{Ad}_{u^{-1}}B]. \end{align*} Using the [exterior derivative](/theorems/1525) formula for a $\mathfrak g$-valued $1$-form, we compute \begin{align*} d\alpha(X,Y)=X(\alpha(Y))-Y(\alpha(X))-\alpha([X,Y]). \end{align*} Substituting $\alpha(Z)=\operatorname{Ad}_{u^{-1}}A(Z)$ for each smooth vector field $Z$ and applying the preceding coefficient identity to $B=A(Y)$ and $B=A(X)$ gives \begin{align*} d\alpha(X,Y)=\operatorname{Ad}_{u^{-1}}\bigl(X(A(Y))-Y(A(X))-A([X,Y])\bigr)-[\theta(X),\alpha(Y)]+[\theta(Y),\alpha(X)]. \end{align*} The first term is $\operatorname{Ad}_{u^{-1}}(dA(X,Y))$. By definition of the bracket wedge product on $\mathfrak g$-valued $1$-forms, \begin{align*} [\theta\wedge\alpha](X,Y)=[\theta(X),\alpha(Y)]-[\theta(Y),\alpha(X)]. \end{align*} Therefore \begin{align*} d\alpha(X,Y)=\bigl(\operatorname{Ad}_{u^{-1}}dA-[\theta\wedge\alpha]\bigr)(X,Y). \end{align*} Since $X$ and $Y$ were arbitrary smooth vector fields on $U$, this proves $d\alpha=\operatorname{Ad}_{u^{-1}}dA-[\theta\wedge\alpha]$.[/step]

[guided]The new connection has two parts: the old connection transported by the adjoint action, and the correction term $\theta = u^{-1}du$. To control the curvature, we need to know how these two parts differentiate. First, the form $\theta \in \Omega^1(U;\mathfrak g)$ is the pullback of the left Maurer-Cartan form on $G$. Since $u: U \to G$ is smooth, this pullback is a smooth $\mathfrak g$-valued $1$-form on $U$, and the Maurer-Cartan structure equation applies after pullback. It says that \begin{align*} d\theta + \frac{1}{2}[\theta \wedge \theta] = 0. \end{align*} This identity is exactly the cancellation mechanism for the pure gauge part of the curvature. If $A = 0$, then $A' = \theta$, and the identity says that the curvature of this pure gauge connection is zero. Second, we need to differentiate the adjointly transformed form $\alpha = \operatorname{Ad}_{u^{-1}}A$. The pointwise map $x \mapsto \operatorname{Ad}_{u(x)^{-1}}$ varies with $x$, so $d\alpha$ is not merely $\operatorname{Ad}_{u^{-1}}dA$. The missing term is the commutator with $\theta$. More precisely, for every smooth vector field $X$ on $U$ and every smooth map $B: U \to \mathfrak g$, differentiating the curve $x \mapsto \operatorname{Ad}_{u(x)^{-1}}B(x)$ in the direction $X$ gives \begin{align*} X\bigl(\operatorname{Ad}_{u^{-1}}B\bigr) = \operatorname{Ad}_{u^{-1}}(XB) - [\theta(X), \operatorname{Ad}_{u^{-1}}B]. \end{align*} The term $\operatorname{Ad}_{u^{-1}}(XB)$ differentiates the $\mathfrak g$-valued coefficient $B$, while the commutator term differentiates the moving frame $u^{-1}$. To pass from coefficient functions to the full $1$-form, evaluate on smooth vector fields $X$ and $Y$ on $U$. The exterior derivative formula gives \begin{align*} d\alpha(X,Y)=X(\alpha(Y))-Y(\alpha(X))-\alpha([X,Y]). \end{align*} Because $\alpha(Z)=\operatorname{Ad}_{u^{-1}}A(Z)$ for every smooth vector field $Z$, the coefficient identity applied to $B=A(Y)$ and $B=A(X)$ yields \begin{align*} d\alpha(X,Y)=\operatorname{Ad}_{u^{-1}}\bigl(X(A(Y))-Y(A(X))-A([X,Y])\bigr)-[\theta(X),\alpha(Y)]+[\theta(Y),\alpha(X)]. \end{align*} The expression inside $\operatorname{Ad}_{u^{-1}}$ is $dA(X,Y)$. Also, \begin{align*} [\theta\wedge\alpha](X,Y)=[\theta(X),\alpha(Y)]-[\theta(Y),\alpha(X)]. \end{align*} Hence \begin{align*} d\alpha(X,Y)=\bigl(\operatorname{Ad}_{u^{-1}}dA-[\theta\wedge\alpha]\bigr)(X,Y). \end{align*} Since $X$ and $Y$ were arbitrary, we have proved \begin{align*} d\alpha = \operatorname{Ad}_{u^{-1}}dA - [\theta \wedge \alpha]. \end{align*} This is the exact term that will cancel the mixed bracket term appearing when the curvature of $A' = \alpha + \theta$ is expanded.[/guided]

[step:Use equivariance of the Lie bracket under the adjoint action] The adjoint representation preserves the Lie bracket on $\mathfrak g$: for all $g \in G$ and all $\xi,\eta \in \mathfrak g$, \begin{align*} [\operatorname{Ad}_{g}\xi,\operatorname{Ad}_{g}\eta] = \operatorname{Ad}_{g}[\xi,\eta]. \end{align*} Applying this pointwise with $g = u(x)^{-1}$ gives the identity of $\mathfrak g$-valued $2$-forms \begin{align*} [\alpha \wedge \alpha] = \operatorname{Ad}_{u^{-1}}[A \wedge A]. \end{align*} [/step]

[step:Expand the transformed curvature and cancel the gauge terms]Substitute the identities from the previous steps into the curvature expansion: \begin{align*} F_{A'} = d\alpha + d\theta + \frac{1}{2}[\alpha \wedge \alpha] + [\alpha \wedge \theta] + \frac{1}{2}[\theta \wedge \theta]. \end{align*} Using \begin{align*} d\alpha = \operatorname{Ad}_{u^{-1}}dA - [\theta \wedge \alpha] \end{align*} and \begin{align*} d\theta + \frac{1}{2}[\theta \wedge \theta] = 0, \end{align*} we obtain \begin{align*} F_{A'} = \operatorname{Ad}_{u^{-1}}dA - [\theta \wedge \alpha] + \frac{1}{2}[\alpha \wedge \alpha] + [\alpha \wedge \theta]. \end{align*} Because $\alpha$ and $\theta$ are both $1$-forms, the bracket wedge product is symmetric in these two arguments: \begin{align*} [\theta \wedge \alpha] = [\alpha \wedge \theta]. \end{align*} Thus the mixed terms cancel, and hence \begin{align*} F_{A'} = \operatorname{Ad}_{u^{-1}}dA + \frac{1}{2}[\alpha \wedge \alpha]. \end{align*} Using $[\alpha \wedge \alpha] = \operatorname{Ad}_{u^{-1}}[A \wedge A]$, this becomes \begin{align*} F_{A'} = \operatorname{Ad}_{u^{-1}}dA + \frac{1}{2}\operatorname{Ad}_{u^{-1}}[A \wedge A]. \end{align*} By linearity of $\operatorname{Ad}_{u^{-1}}$ on $\mathfrak g$, \begin{align*} F_{A'} = \operatorname{Ad}_{u^{-1}}\left(dA + \frac{1}{2}[A \wedge A]\right). \end{align*} Since $F_A = dA + \frac{1}{2}[A \wedge A]$, we conclude \begin{align*} F_{A'} = \operatorname{Ad}_{u^{-1}}F_A. \end{align*}[/step]

[guided]We now combine the identities rather than doing any new geometry. The curvature of the transformed connection is \begin{align*} F_{A'} = d\alpha + d\theta + \frac{1}{2}[\alpha \wedge \alpha] + [\alpha \wedge \theta] + \frac{1}{2}[\theta \wedge \theta]. \end{align*} The differential identity for the adjointly transformed part gives \begin{align*} d\alpha = \operatorname{Ad}_{u^{-1}}dA - [\theta \wedge \alpha]. \end{align*} The Maurer-Cartan equation gives \begin{align*} d\theta + \frac{1}{2}[\theta \wedge \theta] = 0. \end{align*} Substituting these two formulas into the curvature expansion gives \begin{align*} F_{A'} = \operatorname{Ad}_{u^{-1}}dA - [\theta \wedge \alpha] + \frac{1}{2}[\alpha \wedge \alpha] + [\alpha \wedge \theta]. \end{align*} For $\mathfrak g$-valued $1$-forms, the bracket wedge product is symmetric in its two $1$-form inputs, so \begin{align*} [\theta \wedge \alpha]=[\alpha \wedge \theta]. \end{align*} Thus the two mixed terms cancel, leaving \begin{align*} F_{A'} = \operatorname{Ad}_{u^{-1}}dA + \frac{1}{2}[\alpha \wedge \alpha]. \end{align*} The adjoint representation preserves the Lie bracket, and $\alpha=\operatorname{Ad}_{u^{-1}}A$, so \begin{align*} [\alpha\wedge\alpha]=\operatorname{Ad}_{u^{-1}}[A\wedge A]. \end{align*} Therefore \begin{align*} F_{A'}=\operatorname{Ad}_{u^{-1}}dA+\frac{1}{2}\operatorname{Ad}_{u^{-1}}[A\wedge A]. \end{align*} Since $\operatorname{Ad}_{u^{-1}}:\mathfrak g\to\mathfrak g$ is linear at each point of $U$, this is \begin{align*} F_{A'}=\operatorname{Ad}_{u^{-1}}\left(dA+\frac{1}{2}[A\wedge A]\right). \end{align*} By the definition $F_A=dA+\frac{1}{2}[A\wedge A]$, we obtain \begin{align*} F_{A'}=\operatorname{Ad}_{u^{-1}}F_A. \end{align*}[/guided]

[step:Identify the adjoint action with conjugation for matrix Lie groups] Assume now that $n$ is a positive integer and that $G$ is a matrix Lie group contained in $GL(n,\mathbb R)$ or $GL(n,\mathbb C)$, with Lie algebra $\mathfrak g$ realised as a matrix Lie algebra. For every $g \in G$ and every $\xi \in \mathfrak g$, the adjoint action is matrix conjugation: \begin{align*} \operatorname{Ad}_{g}\xi = g\xi g^{-1}. \end{align*} Taking $g = u(x)^{-1}$ at each point $x \in U$ gives \begin{align*} \operatorname{Ad}_{u^{-1}}F_A = u^{-1}F_Au. \end{align*} Therefore the general identity $F_{A'} = \operatorname{Ad}_{u^{-1}}F_A$ becomes \begin{align*} F_{A'} = u^{-1}F_Au. \end{align*} This is the stated matrix Lie group formula. [/step]