= Solution
Differentiate the product with the first factor close to the identity. A small displacement of its $x'$ coordinate adds to $x$; a small $y'$ displacement adds to $y$; and a small $z'$ displacement changes $(x,y,z)$ by $(-x,y,1)$ times that displacement. The infinitesimal left actions therefore give the <right-invariant vector fields>
$$
R_1=\partial_x,\qquad R_2=\partial_y,\qquad R_3=\partial_z-x\partial_x+y\partial_y.
$$
For right multiplication, regard $(x',y',z')$ as the fixed point and the second factor as infinitesimal. Its displacements are $(e^{-z'}\delta x,e^{z'}\delta y,\delta z)$; relabeling the coordinates of the fixed point gives the <left-invariant vector fields>
$$
L_1=e^{-z}\partial_x,\qquad L_2=e^z\partial_y,\qquad L_3=\partial_z.
$$
The naming distinction matters: generators of left multiplication are right invariant, whereas generators of right multiplication are left invariant.
All mixed <Lie brackets of vector fields> vanish. For example, $R_3(e^{-z})=-e^{-z}$ cancels $L_1(-x)=-e^{-z}$ in $[L_1,R_3]$, and $R_3(e^z)=e^z$ cancels $L_2(y)=e^z$ in $[L_2,R_3]$. The brackets with $R_1,R_2$ are zero because the coefficients of the $L_i$ have no $x$ or $y$ dependence; $[L_3,R_3]=0$ because the coefficients of $R_3$ have no $z$ dependence. Hence
$$
\boxed{[L_i,R_j]=0\quad\text{for every }i,j.}
$$
Conceptually, the corresponding flows commute because $(ag)b=a(gb)$, by associativity of the <Lie group> multiplication.
The dual <coframe> is
$$
\theta^1=e^zdx,\qquad\theta^2=e^{-z}dy,\qquad\theta^3=dz,
$$
which satisfies $\theta^i(L_j)=\delta^i_j$. Under left multiplication by $(x_0,y_0,z_0)$, the transformed coordinates are $X=x_0+e^{-z_0}x$, $Y=y_0+e^{z_0}y$, $Z=z_0+z$. Therefore $e^Z dX=e^zdx$, $e^{-Z}dY=e^{-z}dy$, and $dZ=dz$. Each one-form is a <left-invariant differential form>, so their sum of squares is the <left-invariant metric>
$$
\boxed{g=(\theta^1)^2+(\theta^2)^2+(\theta^3)^2=e^{2z}dx^2+e^{-2z}dy^2+dz^2.}
$$
This is the standard metric of the <Sol Lie group>. As a sign check, $[L_3,L_1]=-L_1$, $[L_3,L_2]=L_2$, while $d\theta^1=\theta^3\wedge\theta^1$ and $d\theta^2=-\theta^3\wedge\theta^2$, in agreement with the <Maurer-Cartan equation>.
Back to article page
