= Solution
Acting on the column $(x,1)^T$ gives the faithful <Matrix Lie group> representation
$$
\begin{pmatrix}e^\alpha&\beta\\0&1\end{pmatrix}
\begin{pmatrix}x\\1\end{pmatrix}
=\begin{pmatrix}e^\alpha x+\beta\\1\end{pmatrix}.
$$
Multiplication and inversion are
$$
(\alpha,\beta)(a,b)=(\alpha+a,\beta+e^\alpha b),\qquad
(\alpha,\beta)^{-1}=(-\alpha,-e^{-\alpha}\beta).
$$
Thus $G$ is the <real affine group>. Differentiating the matrices at the identity gives its <Lie algebra>
$$
\mathfrak g=\left\{\begin{pmatrix}a&b\\0&0\end{pmatrix}:a,b\in\mathbb R\right\}.
$$
With
$$
D=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad
T=\begin{pmatrix}0&1\\0&0\end{pmatrix},
$$
the matrix <commutator> is \b[$[D,T]=T$], and $[D,D]=[T,T]=0$.
The left <Maurer-Cartan form> and its right analogue are
$$
\theta_L=g^{-1}dg
=\begin{pmatrix}d\alpha&e^{-\alpha}d\beta\\0&0\end{pmatrix}
=D\,d\alpha+T\,e^{-\alpha}d\beta,
$$
$$
\theta_R=dg\,g^{-1}
=\begin{pmatrix}d\alpha&d\beta-\beta\,d\alpha\\0&0\end{pmatrix}
=D\,d\alpha+T(d\beta-\beta\,d\alpha).
$$
For a constant $h$, left translation leaves $\theta_L$ unchanged, and right translation leaves $\theta_R$ unchanged. Taking their dual <vector fields> gives
$$
\boxed{D_L=\partial_\alpha,\qquad T_L=e^\alpha\partial_\beta},
\qquad
\boxed{D_R=\partial_\alpha+\beta\partial_\beta,\qquad T_R=\partial_\beta}.
$$
The 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> yields
$$
[D_L,T_L]=T_L,\qquad [D_R,T_R]=-T_R.
$$
All 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 $\theta_L=D\ell^D+T\ell^T$, then
$$
d\ell^D=0,\qquad d\ell^T=-\ell^D\wedge\ell^T.
$$
For $\theta_R=Dr^D+Tr^T$, instead
$$
dr^D=0,\qquad dr^T=r^D\wedge r^T.
$$
Thus 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 $\exp(\epsilon T)$ and $\exp(\epsilon D)$. Their left action on a general group element is
$$
g_1(\epsilon)g(\alpha,\beta)=g(\alpha,\beta+\epsilon),\qquad
g_2(\epsilon)g(\alpha,\beta)=g(\alpha+\epsilon,e^\epsilon\beta).
$$
Differentiating at $\epsilon=0$ gives $T_R$ and $D_R$ respectively. Hence \b[<Left translations on a Lie group> are generated by <right-invariant vector fields>]. More generally, the velocity of $\exp(t\xi)g$ is $(dR_g)_e\xi$, 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>.
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