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Work over a coordinate neighborhood $U\subset B$ with coordinates $x^i$, and choose a basis $T_a$ of the Lie algebra, with $[T_a,T_b]=c^c{}_{ab}{T}_c$. Put $A=A_i^a{T}_{a}d{x}^i$. For each $T_a$ let $R_a$ be its right-invariant vector field on the fiber group. In matrix notation $R_a(\gamma )=T_a\gamma$, andThis minus sign is the right-invariant vector fields realize the opposite Lie algebra rule. It is essential here. Define the coordinate horizontal lifts of a principal connection byTheir projections are ${\pi }_{\ast }{H}_i={\partial }_i$, so they are linearly independent and there are exactly $\dim B$ of them. Also $d\gamma (H_i)=-A_i\gamma$, whenceThey therefore span the horizontal distribution of a principal connection on this trivialization.

The coefficients $A_i^a$ depend only on the base variables, so taking Lie brackets giveswhere $F_{ij}={\partial }_i{A}_j-{\partial }_j{A}_i+[A_i,A_j]$ and $F=\frac{1}{2}{F}_{ij}d{x}^i\wedge d{x}^j=dA+A\wedge A$. Since the right-invariant vector fields form a basis on each fiber, $\boxed{[H_i,H_j]=0\text{ for all }i,j⟺F=0}$. This is the local coordinate version of vanishing curvature of a principal connection and, by the Frobenius theorem, integrability of the horizontal distribution.

The coordinate qualification is necessary: horizontal lifts of arbitrary noncommuting base fields need not commute even for a flat principal connection. Nor does flatness guarantee a globally defined coordinate frame or a global horizontal section; holonomy of a connection can obstruct the latter. If the requested frame were read globally, the trivial flat bundle $S^2\times G$ would already be a counterexample: restricting a global horizontal frame to the identity-fiber section would trivialize $T{S}^2$, contrary to the Hairy ball theorem. The construction on each coordinate trivialization supplies the intended result without that global claim.

Acting on the column $(x,1{)}^T$ gives the faithful Matrix Lie group representationMultiplication and inversion areThus $G$ is the real affine group. Differentiating the matrices at the identity gives its Lie algebraWiththe matrix commutator is $[D,T]=T$, and $[D,D]=[T,T]=0$.

The left Maurer-Cartan form and its right analogue areFor a constant $h$, left translation leaves ${\theta }_L$ unchanged, and right translation leaves ${\theta }_R$ unchanged. Taking their dual vector fields givesThe first pair consists of left-invariant vector fields; the second consists of right-invariant vector fields. These are the invariant frames of the real affine group. Computing their Lie brackets of vector fields yieldsAll brackets of a field with itself vanish. The sign difference is necessary: evaluation at the identity identifies left-invariant vector fields with the matrix Lie algebra as a homomorphism, whereas right-invariant vector fields realize the opposite Lie algebra. Equivalently, $\xi \mapsto -{\xi }_R$ is a homomorphism for the same matrix commutator. There is no convention in which these particular coordinate fields both have the positive structure constant while retaining the usual Lie bracket of vector fields.

The Maurer-Cartan equation gives the same check. If ${\theta }_L=D{\ell }^D+T{\ell }^T$, thenFor ${\theta }_R=D{r}^D+T{r}^T$, insteadThus the left equation is $d{\theta }_L+{\theta }_L\wedge {\theta }_L=0$ and the right equation is $d{\theta }_R-{\theta }_R\wedge {\theta }_R=0$, explaining the opposite bracket signs directly.

Finally, the two given matrix curves are $\exp (ϵT)$ and $\exp (ϵD)$. Their left action on a general group element isDifferentiating at $ϵ=0$ gives $T_R$ and $D_R$ respectively. Hence Left translations on a Lie group are generated by right-invariant vector fields. More generally, the velocity of $\exp (t\xi )g$ is $(d{R}_g{)}_e\xi$, which is right-invariant because left and right multiplication commute. In the other direction, the flows $g\exp (t\xi )$ generate left-invariant vector fields and are right translations on a Lie group.