do Carmo's Riemannian Geometry exercise 1.4(a) - Metric of Lobatchevski Geometry Consider the following problem:

A function $g:\mathbb R\to\mathbb R$ given by $g(t)=yt+x$, $t$,$x$,$y\in\mathbb R$, $y>0$, is called a proper affine function. The subset of all such functions with respect to the usual composition law is a Lie group $G$. As a differentiable manifold $G$ is simply the upper half-plane, that is, $\{(x,y)\in\mathbb R^2|y>0\}$ with the usual differentiable structure. Prove that:
(a) The left-invariant Riemannian metric of $G$ which at the neutral element $e=(0,1)$ coincides with the Euclidean metric ($g_{11}=g_{22}=1,g_{12}=0$) is given by $g_{11}=g_{22}=\frac{1}{y^2},g_{12}=0$.

This question already has an answer here, but I did not understand the argument. Also, I took another path and I would like to know if what I did is reasonable and how to finnish the question following this line of thought.
Here is my attempt:
Note that the tangent space of $G$ at the point $g \in G$ is nothing but $\mathbb R^2$. The parametrization $x$ is just the identity of $\mathbb R^2$. Then
$$
d x_g(1, 0) = (1, 0), \quad d x_g(0, 1) = (0, 1)
$$
for all $g \in G$ and therefore
$$
g_{ij}(0, 1) = \delta_{ij},
$$
since $\langle \cdot, \cdot \rangle_e$ coincides with the euclidean inner product.
Now, for $g \in G$ we have
\begin{align*}
g_{11}( x^{-1}(g)) = & \langle d x_g(1, 0), d x_g(1, 0) \rangle_g \\
 = & \langle d (L_{g^{-1}})_g(1, 0), d (L_{g^{-1}})_g(1, 0) \rangle_e
\end{align*}
I am stuck on how to compute the derivative $d(L_{g^{-1}})_g$. Any hints will be the most appreciated.
Thanks in advance and kind regards.
 A: There's no point in computing $g_{ij}(0,1) = \delta_{ij}$. That was given to you and the point of the exercise is to spread that to the whole half-plane. Let $h(t) = a+bt$ with $b>0$. Then for $g(t) = x+yt$, with $y>0$ we have that $$(L_gh)(t) = g(h(t)) = x+yh(t) = x+y(bt+a) = ya+x+ybt.$$This means that under the identification of the manifold of proper affine functions with the upper half-plane, the group operation is $$L_gh = L_{(x,y)}(a,b) = (x,y)(a,b) = (ya+x, yb).$$Since $L_{(x,y)}$ differs from a linear map (namely, $y\,{\rm Id}$) by the constant $(x,0)$, the derivative is $${\rm d}(L_{(x,y)})_{(a,b)} = \begin{pmatrix} y & 0 \\ 0 & y \end{pmatrix}\implies{\rm d}(L_{(x,y)^{-1}})_{(x,y)} = {\rm d}(L_{(x,y)})_{(0,1)}^{-1} =  \begin{pmatrix} 1/y & 0 \\ 0 & 1/y \end{pmatrix} $$Since all left translations are isometries, then $g_{(x,y)} = (L_{(x,y)^{-1}})^*(g_{(0,1)})$. In matrix form this means that $$(g_{ij}(x,y))_{i,j=1}^2  =\begin{pmatrix} 1/y & 0 \\ 0 & 1/y \end{pmatrix}^\top \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1/y & 0 \\ 0 & 1/y \end{pmatrix} = \frac{1}{y^2}\begin{pmatrix} 1 & 0  \\ 0 & 1\end{pmatrix}.  $$
A: Ivo's answer is very good. Here is another way to calculate $d(L_{g^{-1}})\partial_i$ which is more in line with do Carmo's presentation in my opinion. We will use proposition 2.7 from chapter zero. The obvious choice of curves are $\alpha(s) = (x+s,y),\beta(s) = (x,y+s)$. Then at $g=(x,y) = yt +x$ we have $\alpha(0) = \beta(0) = (x,y) = g$ and $\alpha'(0) = (1,0)=\partial_1$, $\beta'(0) = (0,1) = \partial_2$ (check this by converting from coordinate form to function form and that these coincide). Then we can calculate:
\begin{align*}
(dL_{g^{-1}})_g\partial_1 &= (dL_{g^{-1}})_g\alpha'(0)\\
&= (L_{g^{-1}}\circ \alpha)'(0)\\
&= \frac{\partial}{\partial s}\bigg|_{s=0} L_{g^{-1}}(x+s,y)\\
&= \frac{\partial}{\partial s}\bigg|_{s=0} \left(-\frac{x}{y},\frac{1}{y}\right)\circ (x+s,y)\\
&= \frac{\partial}{\partial s}\bigg|_{s=0} \left(\frac{1}{y}(x+s)-\frac{x}{y},\frac{1}{y}y\right)\\
&= \left(\frac{1}{y},0\right)
\end{align*}
A similar computation holds for $(dL_{g^{-1}})_g\partial_2$ and you should get the same metric as in Ivo's answer. You should also verify that $g^{-1} = (x,y)^{-1} = \left(-\frac{x}{y},\frac{1}{y}\right)$.
