= Solution
Take the frame dual to the <left-invariant coframe of the translation-dilation group>. It is
$$
\boxed{L_0=\rho\partial_\rho,\qquad L_1=\rho\partial_{x^1},\qquad L_2=\rho\partial_{x^2}.}
$$
Indeed $\lambda^i(L_j)=\delta^i_j$. These are <left-invariant vector fields>, either because they are dual to invariant forms or because $L_j(g)=(dL_g)_eX_j$. For instance, the curve $g\exp(tX_0)$ differentiates to $\rho\partial_\rho$, while $g\exp(tX_1)$ differentiates to $\rho\partial_{x^1}$.
The coordinate <Lie brackets> satisfy
$$
[L_0,L_1]=\rho\partial_{x^1}=L_1,\qquad [L_0,L_2]=L_2,\qquad [L_1,L_2]=0.
$$
For the first identity only the derivative of the coefficient $\rho$ contributes. \b[The linear map $\boxed{X_j\longmapsto L_j}$ is a <Lie algebra isomorphism>]: it preserves all basis brackets, is injective by independence of the frame, and is onto the three-dimensional space of <left-invariant vector fields>.
Back to article page
