[proofplan]
The curvature form is defined by applying $d\omega$ only to the horizontal components of tangent vectors, so it vanishes as soon as one input is vertical. For equivariance, we use the structure equation $\Omega=d\omega+\frac12[\omega\wedge\omega]$ together with the connection-form transformation law $(R_g)^*\omega=\operatorname{Ad}_{g^{-1}}\omega$. Naturality of exterior differentiation gives the transformation of $d\omega$, and the fact that $\operatorname{Ad}_{g^{-1}}:\mathfrak{g}\to\mathfrak{g}$ is a Lie algebra automorphism gives the transformation of the bracket term.
[/proofplan]
[step:Use the horizontal projection definition to prove horizontality]
For each $p\in P$, let $H_pP:=\ker \omega_p\subset T_pP$ be the horizontal subspace and let $V_pP:=\ker d\pi_p\subset T_pP$ be the vertical subspace. The connection form gives a direct sum decomposition
\begin{align*}
T_pP=H_pP\oplus V_pP.
\end{align*}
Let
\begin{align*}
h_p:T_pP\to H_pP
\end{align*}
denote the projection onto $H_pP$ along $V_pP$. Let
\begin{align*}
\exp:\mathfrak{g}\to G
\end{align*}
denote the Lie group exponential map of $G$.
We use the standard convention for the curvature of a principal connection: the horizontal-projection formula and the Cartan structure equation used below are equivalent descriptions of the same $\mathfrak{g}$-valued $2$-form. By the horizontal-projection description, for every $p\in P$ and every $X,Y\in T_pP$,
\begin{align*}
\Omega_p(X,Y)=d\omega_p(h_pX,h_pY).
\end{align*}
For $\xi\in\mathfrak{g}$, the fundamental vector $\xi_P(p)$ is vertical because its curve $t\mapsto p\cdot \exp(t\xi)$ lies entirely in the fibre $\pi^{-1}(\pi(p))$. Hence
\begin{align*}
h_p(\xi_P(p))=0.
\end{align*}
Therefore, for every $X\in T_pP$,
\begin{align*}
\Omega_p(\xi_P(p),X)=d\omega_p(0,h_pX)=0.
\end{align*}
Since $\Omega$ is a $2$-form, it is alternating, so
\begin{align*}
\Omega_p(X,\xi_P(p))=-\Omega_p(\xi_P(p),X)=0.
\end{align*}
[guided]
The point of horizontality is that curvature ignores vertical directions. We make that precise using the connection splitting. For each $p\in P$, the connection form determines the horizontal subspace
\begin{align*}
H_pP:=\ker \omega_p\subset T_pP
\end{align*}
and the vertical subspace
\begin{align*}
V_pP:=\ker d\pi_p\subset T_pP.
\end{align*}
These satisfy
\begin{align*}
T_pP=H_pP\oplus V_pP.
\end{align*}
Let
\begin{align*}
h_p:T_pP\to H_pP
\end{align*}
be the projection onto the horizontal summand along the vertical summand. Let
\begin{align*}
\exp:\mathfrak{g}\to G
\end{align*}
denote the Lie group exponential map of $G$.
We are using the standard convention for curvature of a principal connection. Under this convention, the horizontal-projection formula in this step and the Cartan structure equation used in the equivariance argument are equivalent descriptions of one and the same $\mathfrak{g}$-valued $2$-form. The horizontal-projection description says that the curvature form is obtained from the connection by taking $d\omega$ only after projecting both inputs horizontally:
\begin{align*}
\Omega_p(X,Y)=d\omega_p(h_pX,h_pY)
\end{align*}
for all $X,Y\in T_pP$. Now fix $\xi\in\mathfrak{g}$. The fundamental vector $\xi_P(p)$ is tangent to the curve
\begin{align*}
t\mapsto p\cdot \exp(t\xi).
\end{align*}
This curve stays inside the fibre over $\pi(p)$ because $\pi(p\cdot \exp(t\xi))=\pi(p)$. Therefore $\xi_P(p)\in V_pP$, and the horizontal projection of a vertical vector is zero:
\begin{align*}
h_p(\xi_P(p))=0.
\end{align*}
Substituting this into the definition of curvature gives
\begin{align*}
\Omega_p(\xi_P(p),X)=d\omega_p(0,h_pX)=0.
\end{align*}
The second displayed horizontality identity follows from alternation of the $2$-form $\Omega$:
\begin{align*}
\Omega_p(X,\xi_P(p))=-\Omega_p(\xi_P(p),X)=0.
\end{align*}
[/guided]
[/step]
[step:Transform the exterior derivative term under the principal right action]
Fix $g\in G$. The connection form satisfies the principal equivariance identity
\begin{align*}
(R_g)^*\omega=\operatorname{Ad}_{g^{-1}}\omega.
\end{align*}
Applying exterior differentiation to both sides and using naturality of $d$ under pullback gives
\begin{align*}
(R_g)^*d\omega=d((R_g)^*\omega)=d(\operatorname{Ad}_{g^{-1}}\omega).
\end{align*}
Since $\operatorname{Ad}_{g^{-1}}:\mathfrak{g}\to\mathfrak{g}$ is a fixed [linear map](/page/Linear%20Map), it commutes with exterior differentiation of $\mathfrak{g}$-valued forms. Thus
\begin{align*}
(R_g)^*d\omega=\operatorname{Ad}_{g^{-1}}d\omega.
\end{align*}
[/step]
[step:Transform the Lie bracket term under the adjoint action]
Define the bracket product of $\mathfrak{g}$-valued $1$-forms by
\begin{align*}
[\omega\wedge\omega]_p(X,Y):=[\omega_p(X),\omega_p(Y)]-[\omega_p(Y),\omega_p(X)]
\end{align*}
for $p\in P$ and $X,Y\in T_pP$. Because the Lie bracket on $\mathfrak{g}$ is alternating, this is equivalently
\begin{align*}
[\omega\wedge\omega]_p(X,Y)=2[\omega_p(X),\omega_p(Y)].
\end{align*}
Let $p\in P$ and let $X,Y\in T_pP$. Using the definition of pullback and the connection-form equivariance identity,
\begin{align*}
((R_g)^*[\omega\wedge\omega])_p(X,Y)=[\omega_{p\cdot g}(d(R_g)_pX),\omega_{p\cdot g}(d(R_g)_pY)]-[\omega_{p\cdot g}(d(R_g)_pY),\omega_{p\cdot g}(d(R_g)_pX)].
\end{align*}
The identity $(R_g)^*\omega=\operatorname{Ad}_{g^{-1}}\omega$ gives
\begin{align*}
\omega_{p\cdot g}(d(R_g)_pX)=\operatorname{Ad}_{g^{-1}}(\omega_p(X))
\end{align*}
and
\begin{align*}
\omega_{p\cdot g}(d(R_g)_pY)=\operatorname{Ad}_{g^{-1}}(\omega_p(Y)).
\end{align*}
Since $\operatorname{Ad}_{g^{-1}}$ is a Lie algebra automorphism,
\begin{align*}
[\operatorname{Ad}_{g^{-1}}A,\operatorname{Ad}_{g^{-1}}B]=\operatorname{Ad}_{g^{-1}}[A,B]
\end{align*}
for all $A,B\in\mathfrak{g}$. Substituting $A=\omega_p(X)$ and $B=\omega_p(Y)$ gives
\begin{align*}
((R_g)^*[\omega\wedge\omega])_p(X,Y)=\operatorname{Ad}_{g^{-1}}([\omega\wedge\omega]_p(X,Y)).
\end{align*}
Therefore
\begin{align*}
(R_g)^*[\omega\wedge\omega]=\operatorname{Ad}_{g^{-1}}[\omega\wedge\omega].
\end{align*}
[/step]
[step:Combine the transformed terms in the structure equation]
Under the same standard convention for the curvature of a principal connection, the curvature form satisfies the Cartan structure equation
\begin{align*}
\Omega=d\omega+\frac12[\omega\wedge\omega].
\end{align*}
Using the two transformation identities already proved,
\begin{align*}
(R_g)^*\Omega=(R_g)^*d\omega+\frac12(R_g)^*[\omega\wedge\omega].
\end{align*}
Hence
\begin{align*}
(R_g)^*\Omega=\operatorname{Ad}_{g^{-1}}d\omega+\frac12\operatorname{Ad}_{g^{-1}}[\omega\wedge\omega].
\end{align*}
By linearity of $\operatorname{Ad}_{g^{-1}}:\mathfrak{g}\to\mathfrak{g}$,
\begin{align*}
(R_g)^*\Omega=\operatorname{Ad}_{g^{-1}}\left(d\omega+\frac12[\omega\wedge\omega]\right).
\end{align*}
Using the structure equation again,
\begin{align*}
(R_g)^*\Omega=\operatorname{Ad}_{g^{-1}}\Omega.
\end{align*}
Together with horizontality, this proves that the curvature form is horizontal and $\operatorname{Ad}$-equivariant.
[/step]
