[proofplan] We identify a vertical principal-bundle automorphism with the unique smooth function $\gamma: P \to G$ measuring how far the automorphism moves each point along its fibre. Principal equivariance of the automorphism is exactly the conjugation equivariance condition $\gamma(pg)=g^{-1}\gamma(p)g$, which is precisely the condition needed for $p \mapsto [p,\gamma(p)]$ to descend to a section of the associated bundle $P \times_G G$. Conversely, a section of the adjoint bundle can be represented uniquely in the form $[p,\gamma(p)]$ over $\pi(p)$, and the resulting function $\gamma$ defines a smooth gauge transformation $p \mapsto p\gamma(p)$. Finally, fibrewise multiplication of sections matches composition of gauge transformations by a direct computation using conjugation equivariance. [/proofplan] [step:Extract the equivariant multiplier from a gauge transformation] Let $\pi: P \to M$ denote the smooth principal-bundle projection, and let $\Phi \in \operatorname{Gau}(P)$. For each $p \in P$, the equality $\pi(\Phi(p))=\pi(p)$ implies that $p$ and $\Phi(p)$ lie in the same principal $G$-orbit. Since the right $G$-action on each fibre is free and transitive, there is a unique element $\gamma_\Phi(p) \in G$ such that \begin{align*} \Phi(p)=p\gamma_\Phi(p). \end{align*} This defines a smooth map $\gamma_\Phi: P \to G$ after the local smoothness verification below. The map $\gamma_\Phi$ is smooth because it is obtained locally from a smooth trivialization. Indeed, let $U \subset M$ be an [open set](/page/Open%20Set) and let $\tau: U \to P$ be a smooth local section of $\pi$. Every $p \in \pi^{-1}(U)$ has a unique expression \begin{align*} p=\tau(x)g \end{align*} with $x=\pi(p)$ and $g \in G$. Since $\Phi(\tau(x))$ lies in $P_x$, there is a unique smooth map $a_\Phi: U \to G$ defined by \begin{align*} \Phi(\tau(x))=\tau(x)a_\Phi(x). \end{align*} Using $G$-equivariance of $\Phi$, we obtain \begin{align*} \Phi(\tau(x)g)=\Phi(\tau(x))g=\tau(x)a_\Phi(x)g. \end{align*} Since $\tau(x)g$ multiplied on the right by $g^{-1}a_\Phi(x)g$ is $\tau(x)a_\Phi(x)g$, uniqueness in the fibre gives \begin{align*} \gamma_\Phi(\tau(x)g)=g^{-1}a_\Phi(x)g. \end{align*} This expression is smooth in the local coordinates $(x,g)$ on $\pi^{-1}(U)$, so $\gamma_\Phi$ is smooth. For $p \in P$ and $g \in G$, we compute using $G$-equivariance of $\Phi$: \begin{align*} p g \gamma_\Phi(pg)=\Phi(pg)=\Phi(p)g=p\gamma_\Phi(p)g. \end{align*} Freeness of the right action gives \begin{align*} \gamma_\Phi(pg)=g^{-1}\gamma_\Phi(p)g. \end{align*} [guided] The first point is that a gauge transformation cannot move a point to a different fibre: by definition, $\Phi$ covers $\operatorname{id}_M$, so $\pi \circ \Phi=\pi$. Hence the points $p$ and $\Phi(p)$ both lie in the fibre $P_{\pi(p)}$. A principal fibre is, by the defining property of a principal $G$-bundle, a free and transitive right $G$-space, so there is exactly one group element $\gamma_\Phi(p) \in G$ satisfying \begin{align*} \Phi(p)=p\gamma_\Phi(p). \end{align*} This defines a map $\gamma_\Phi: P \to G$. We must check that this map is smooth, not merely set-theoretic. Let $U \subset M$ be an open set over which $P$ admits a smooth local section $\tau: U \to P$. Every point of $\pi^{-1}(U)$ is uniquely of the form $\tau(x)g$, with $x \in U$ and $g \in G$. Since $\Phi$ preserves fibres, $\Phi(\tau(x))$ lies in $P_x$, so there is a unique element $a_\Phi(x) \in G$ such that \begin{align*} \Phi(\tau(x))=\tau(x)a_\Phi(x). \end{align*} This gives a map $a_\Phi: U \to G$. In a smooth trivialization, the coordinate expression of $\Phi$ is smooth, so $a_\Phi$ is smooth. Now use equivariance of $\Phi$: \begin{align*} \Phi(\tau(x)g)=\Phi(\tau(x))g=\tau(x)a_\Phi(x)g. \end{align*} On the other hand, by definition of $\gamma_\Phi$, \begin{align*} \Phi(\tau(x)g)=\tau(x)g\gamma_\Phi(\tau(x)g). \end{align*} Freeness of the right action forces \begin{align*} g\gamma_\Phi(\tau(x)g)=a_\Phi(x)g, \end{align*} hence \begin{align*} \gamma_\Phi(\tau(x)g)=g^{-1}a_\Phi(x)g. \end{align*} This local formula is smooth in $(x,g)$, so $\gamma_\Phi$ is smooth. Finally, the same calculation gives the precise equivariance law. For arbitrary $p \in P$ and $g \in G$, the definition of $\gamma_\Phi$ gives \begin{align*} \Phi(pg)=pg\gamma_\Phi(pg). \end{align*} Equivariance of $\Phi$ gives \begin{align*} \Phi(pg)=\Phi(p)g=p\gamma_\Phi(p)g. \end{align*} Therefore \begin{align*} pg\gamma_\Phi(pg)=p\gamma_\Phi(p)g. \end{align*} Cancelling $p$ by freeness of the right action gives \begin{align*} \gamma_\Phi(pg)=g^{-1}\gamma_\Phi(p)g. \end{align*} This is exactly the conjugation transformation rule needed to descend from $P$ to the associated bundle $P \times_G G$. [/guided] [/step] [step:Descend the multiplier to a smooth section of the adjoint bundle] We use the convention \begin{align*} \operatorname{Ad}(P):=P \times_G G, \end{align*} where $G$ acts on itself by conjugation, so the associated-bundle relation is $(pg,h) \sim (p,ghg^{-1})$. Define $s_\Phi: M \to \operatorname{Ad}(P)$ by choosing any $p \in P_x$ and setting \begin{align*} s_\Phi(x)=[p,\gamma_\Phi(p)]. \end{align*} This is independent of the chosen point $p \in P_x$. Indeed, if $p' = pg$ for some $g \in G$, then the equivariance law from the previous step gives \begin{align*} \gamma_\Phi(p')=\gamma_\Phi(pg)=g^{-1}\gamma_\Phi(p)g. \end{align*} The [equivalence relation](/page/Equivalence%20Relation) defining $P \times_G G$ for the conjugation action identifies \begin{align*} [pg,g^{-1}\gamma_\Phi(p)g]=[p,\gamma_\Phi(p)]. \end{align*} Thus $s_\Phi$ is well-defined. Smoothness is local. In the local section $\tau: U \to P$ used above, \begin{align*} s_\Phi(x)=[\tau(x),a_\Phi(x)] \end{align*} for $x \in U$, and $a_\Phi: U \to G$ is smooth. Hence $s_\Phi$ is a smooth section of $\operatorname{Ad}(P)$. [/step] [step:Lift a smooth adjoint section to an equivariant multiplier] Let $s: M \to \operatorname{Ad}(P)$ be a smooth section. We continue to use the convention that $\operatorname{Ad}(P)=P \times_G G$ is formed using the conjugation action $(g,h) \mapsto g^{-1}hg$ in the local-coordinate formula below, equivalently $(pg,h) \sim (p,ghg^{-1})$. For each $p \in P$, set $x=\pi(p)$. To prove existence of a representative with first entry $p$, choose any representative $[q,h]$ of $s(x)$ with $q \in P_x$ and $h \in G$. Freeness and transitivity of the principal right action on $P_x$ give a unique $k \in G$ such that $q=pk$. The associated-bundle equivalence relation gives \begin{align*} [q,h]=[pk,h]=[p,khk^{-1}]. \end{align*} Hence there exists an element $\gamma_s(p) \in G$ such that \begin{align*} s(x)=[p,\gamma_s(p)]. \end{align*} Uniqueness holds because if $[p,h_1]=[p,h_2]$, then the defining equivalence relation and freeness of the right action give $h_1=h_2$. This defines a map $\gamma_s: P \to G$. It is smooth by local trivialization. If $\tau: U \to P$ is a smooth local section, then $\tau$ induces the associated-bundle trivialization \begin{align*} \operatorname{Ad}(P)|_U \to U \times G, \end{align*} which sends $[\tau(x)g,h]$ to $(x,ghg^{-1})$. Since $s$ is smooth, its second coordinate in this trivialization is a smooth map $a_s: U \to G$, characterized by \begin{align*} s(x)=[\tau(x),a_s(x)]. \end{align*} Then for $p=\tau(x)g$ we have \begin{align*} s(x)=[\tau(x),a_s(x)]=[\tau(x)g,g^{-1}a_s(x)g]. \end{align*} Therefore \begin{align*} \gamma_s(\tau(x)g)=g^{-1}a_s(x)g, \end{align*} which is smooth in local coordinates. The same formula also gives the conjugation equivariance law \begin{align*} \gamma_s(pg)=g^{-1}\gamma_s(p)g \end{align*} for all $p \in P$ and $g \in G$. [guided] The section $s$ assigns to each $x \in M$ an element of the fibre $\operatorname{Ad}(P)_x$. The convention is that $\operatorname{Ad}(P)=P \times_G G$ is formed using conjugation, with $(pg,h) \sim (p,ghg^{-1})$. To recover an equivariant function on $P$, fix $p \in P$ and write $x=\pi(p)$. Because every element of $\operatorname{Ad}(P)_x$ has a representative whose first component is this chosen point $p$, there is a unique element $\gamma_s(p) \in G$ such that \begin{align*} s(x)=[p,\gamma_s(p)]. \end{align*} Uniqueness follows from the associated-bundle equivalence relation: if $[p,h_1]=[p,h_2]$, then the only group element carrying the first representative $p$ to itself is the identity, by freeness of the principal right action, and hence $h_1=h_2$. This defines a map $\gamma_s: P \to G$. We now verify smoothness. Let $\tau: U \to P$ be a smooth local section over an open set $U \subset M$. The section $\tau$ induces the local trivialization $\operatorname{Ad}(P)|_U \to U \times G$ sending $[\tau(x)g,h]$ to $(x,ghg^{-1})$. Since $s$ is smooth, its second coordinate in this trivialization is a smooth map $a_s: U \to G$, so \begin{align*} s(x)=[\tau(x),a_s(x)]. \end{align*} If $p=\tau(x)g$, then the associated-bundle relation for the conjugation action gives \begin{align*} [\tau(x),a_s(x)]=[\tau(x)g,g^{-1}a_s(x)g]. \end{align*} By the defining uniqueness of $\gamma_s$ with first representative $\tau(x)g$, we obtain \begin{align*} \gamma_s(\tau(x)g)=g^{-1}a_s(x)g. \end{align*} The right-hand side is smooth in the local coordinates $(x,g)$, because multiplication, inversion, and conjugation in the Lie group $G$ are smooth and $a_s$ is smooth. The same formula gives the equivariance law. Replacing $p$ by $pg$ in the defining formula and using the associated-bundle relation gives \begin{align*} \gamma_s(pg)=g^{-1}\gamma_s(p)g \end{align*} for all $p \in P$ and $g \in G$. [/guided] [/step] [step:Build the gauge transformation from the lifted multiplier] Define $\Phi_s: P \to P$ by \begin{align*} \Phi_s(p)=p\gamma_s(p). \end{align*} The map $\Phi_s$ is smooth because the principal right action $P \times G \to P$ and the map $\gamma_s: P \to G$ are smooth. It preserves fibres since \begin{align*} \pi(\Phi_s(p))=\pi(p\gamma_s(p))=\pi(p). \end{align*} It is $G$-equivariant. For $p \in P$ and $g \in G$, using the equivariance law for $\gamma_s$ gives \begin{align*} \Phi_s(pg)=pg\gamma_s(pg)=pgg^{-1}\gamma_s(p)g=p\gamma_s(p)g=\Phi_s(p)g. \end{align*} It is bijective, with inverse $\Psi_s: P \to P$ defined by \begin{align*} \Psi_s(p)=p\gamma_s(p)^{-1}. \end{align*} The map $\Psi_s$ is smooth because inversion in the Lie group $G$ and the principal right action $P \times G \to P$ are smooth. It preserves fibres because right multiplication stays inside a principal fibre. It is $G$-equivariant: for $p \in P$ and $g \in G$, the equivariance law for $\gamma_s$ gives \begin{align*} \gamma_s(pg)^{-1}=g^{-1}\gamma_s(p)^{-1}g, \end{align*} so \begin{align*} \Psi_s(pg)=pg\gamma_s(pg)^{-1}=pgg^{-1}\gamma_s(p)^{-1}g=p\gamma_s(p)^{-1}g=\Psi_s(p)g. \end{align*} We now verify that $\Psi_s$ is the two-sided inverse of $\Phi_s$. First, using the equivariance law with $g=\gamma_s(p)$, \begin{align*} \gamma_s(p\gamma_s(p))=\gamma_s(p)^{-1}\gamma_s(p)\gamma_s(p)=\gamma_s(p). \end{align*} Therefore \begin{align*} \Psi_s(\Phi_s(p))=\Psi_s(p\gamma_s(p))=p\gamma_s(p)\gamma_s(p\gamma_s(p))^{-1}=p\gamma_s(p)\gamma_s(p)^{-1}=p. \end{align*} Second, using the equivariance law with $g=\gamma_s(p)^{-1}$, \begin{align*} \gamma_s(p\gamma_s(p)^{-1})=\gamma_s(p)\gamma_s(p)\gamma_s(p)^{-1}=\gamma_s(p). \end{align*} Therefore \begin{align*} \Phi_s(\Psi_s(p))=\Phi_s(p\gamma_s(p)^{-1})=p\gamma_s(p)^{-1}\gamma_s(p\gamma_s(p)^{-1})=p\gamma_s(p)^{-1}\gamma_s(p)=p. \end{align*} Thus $\Phi_s \in \operatorname{Gau}(P)$. [/step] [step:Verify that the two constructions are inverse maps] Starting with $\Phi \in \operatorname{Gau}(P)$, the constructed section satisfies \begin{align*} s_\Phi(\pi(p))=[p,\gamma_\Phi(p)]. \end{align*} When this section is lifted back using the rule defining $\gamma_{s_\Phi}$, uniqueness of the representative with first entry $p$ gives \begin{align*} \gamma_{s_\Phi}(p)=\gamma_\Phi(p). \end{align*} Therefore $\Phi_{s_\Phi}(p)=p\gamma_\Phi(p)=\Phi(p)$ for all $p \in P$. Conversely, starting with $s \in \Gamma^\infty(\operatorname{Ad}(P))$, the lifted multiplier satisfies \begin{align*} s(\pi(p))=[p,\gamma_s(p)]. \end{align*} The section associated to $\Phi_s$ is therefore \begin{align*} s_{\Phi_s}(\pi(p))=[p,\gamma_s(p)]=s(\pi(p)). \end{align*} Since every $x \in M$ has some $p \in P_x$, this proves $s_{\Phi_s}=s$. Hence the constructions are mutually inverse bijections. [/step] [step:Match fibrewise multiplication with composition of gauge transformations] Let $s,t \in \Gamma^\infty(\operatorname{Ad}(P))$, and let $\gamma_s,\gamma_t: P \to G$ be their associated equivariant multipliers. The fibrewise product section $st: M \to \operatorname{Ad}(P)$ is defined by \begin{align*} (st)(x)=[p,\gamma_s(p)\gamma_t(p)] \end{align*} for any $p \in P_x$. This is well-defined because both factors transform by conjugation under the associated-bundle equivalence relation for $P \times_G G$. For $p \in P$, compute the composition of the corresponding gauge transformations: \begin{align*} (\Phi_s \circ \Phi_t)(p)=\Phi_s(p\gamma_t(p)). \end{align*} Using the definition of $\Phi_s$ and the equivariance law for $\gamma_s$, we get \begin{align*} \Phi_s(p\gamma_t(p))=p\gamma_t(p)\gamma_s(p\gamma_t(p)). \end{align*} Since \begin{align*} \gamma_s(p\gamma_t(p))=\gamma_t(p)^{-1}\gamma_s(p)\gamma_t(p), \end{align*} we obtain \begin{align*} (\Phi_s \circ \Phi_t)(p)=p\gamma_s(p)\gamma_t(p). \end{align*} Thus \begin{align*} \Phi_s \circ \Phi_t=\Phi_{st}. \end{align*} Therefore the bijection $\Theta: \Gamma^\infty(\operatorname{Ad}(P)) \to \operatorname{Gau}(P)$ defined by $\Theta(s)=\Phi_s$ for every $s \in \Gamma^\infty(\operatorname{Ad}(P))$ is a [group homomorphism](/page/Group%20Homomorphism). Equivalently, its inverse gives the stated natural group isomorphism \begin{align*} \operatorname{Gau}(P)\cong \Gamma^\infty(\operatorname{Ad}(P)). \end{align*} More precisely, for an isomorphism of principal $G$-bundles $F: P \to P'$ covering a diffeomorphism $f: M \to M'$, conjugation of gauge transformations by $F$ corresponds to transporting adjoint-bundle sections by the induced bundle isomorphism $\operatorname{Ad}(F): \operatorname{Ad}(P) \to \operatorname{Ad}(P')$, $[p,h] \mapsto [F(p),h]$. Indeed, if $\Phi(p)=p\gamma(p)$, then \begin{align*} (F\circ \Phi\circ F^{-1})(F(p))=F(p\gamma(p))=F(p)\gamma(p), \end{align*} so the corresponding adjoint section is carried from $[p,\gamma(p)]$ to $[F(p),\gamma(p)]$. Thus the isomorphism is natural under principal-bundle isomorphisms. [guided] The group law must be checked because composition of gauge transformations reverses the order in which the maps are applied, while the adjoint-bundle multiplication is fibrewise multiplication in $G$. Let $s,t \in \Gamma^\infty(\operatorname{Ad}(P))$, and let $\gamma_s,\gamma_t: P \to G$ be their equivariant multipliers. The product section $st: M \to \operatorname{Ad}(P)$ is defined fibrewise by \begin{align*} (st)(x)=[p,\gamma_s(p)\gamma_t(p)] \end{align*} for any $p \in P_x$. This definition is independent of $p$: replacing $p$ by $pg$ changes both factors by conjugation, so their product changes to \begin{align*} g^{-1}\gamma_s(p)g g^{-1}\gamma_t(p)g=g^{-1}\gamma_s(p)\gamma_t(p)g, \end{align*} which represents the same element of the associated bundle. Now compute the composition at a point $p \in P$: \begin{align*} (\Phi_s \circ \Phi_t)(p)=\Phi_s(p\gamma_t(p)). \end{align*} By definition of $\Phi_s$, \begin{align*} \Phi_s(p\gamma_t(p))=p\gamma_t(p)\gamma_s(p\gamma_t(p)). \end{align*} Using the equivariance law for $\gamma_s$ with $g=\gamma_t(p)$ gives \begin{align*} \gamma_s(p\gamma_t(p))=\gamma_t(p)^{-1}\gamma_s(p)\gamma_t(p). \end{align*} Substituting this into the previous display yields \begin{align*} (\Phi_s \circ \Phi_t)(p)=p\gamma_s(p)\gamma_t(p)=\Phi_{st}(p). \end{align*} Thus $\Theta(st)=\Theta(s)\circ\Theta(t)$ for the map $\Theta(s)=\Phi_s$, so $\Theta: \Gamma^\infty(\operatorname{Ad}(P)) \to \operatorname{Gau}(P)$ is a group homomorphism. The naturality statement says that this construction does not depend on arbitrary choices. If $F: P \to P'$ is a principal $G$-bundle isomorphism covering a diffeomorphism $f: M \to M'$, then it induces $\operatorname{Ad}(F): \operatorname{Ad}(P) \to \operatorname{Ad}(P')$ by $[p,h] \mapsto [F(p),h]$. If $\Phi(p)=p\gamma(p)$, then \begin{align*} (F\circ \Phi\circ F^{-1})(F(p))=F(p\gamma(p))=F(p)\gamma(p). \end{align*} Therefore the transported adjoint section sends $[p,\gamma(p)]$ to $[F(p),\gamma(p)]$, exactly matching conjugation of gauge transformations by $F$. This proves the stated natural group isomorphism. [/guided] [/step]