Closely related: part (a)
Consider the Lie group $G$ of the functions $g: \mathbb R \to \mathbb R$ such that $g(t) = yt + x$, with $y > 0$ with the operation of composition. The metric of the Lobatchevski Geometry in this group is $g_{11} = g_{22} = \frac{1}{y^2}, g_{12} = g_{21} = 0$. The statement of the problem reads
(b) Let $(x, y) = z = x + iy$. Prove that the map $\varphi: z \mapsto z' = \frac{az + b}{cz + d}$ with $a, b, c, d \in \mathbb R$ such that $ad – bc = 1$ is an isometry of $G$.
My attempt of solution:
First we show that $\varphi$ is indeed a map from $G$ to $G$. We have:
$$
\frac{az + b}{cz + d} = \frac{(az + b)(c \overline z + d)}{|cz + d|^2} = \frac{ac |z|^2 + bd + x + iy}{|cz + d|^2}
$$
which is indeed in $G$.
My next step was trying to show that for $u, v \in T_eG$, where $e = (0, 1)$ is the identity element, it holds that
$$
\langle u, v \rangle_e = \langle d \varphi_e(u), d \varphi_e(v) \rangle_{\varphi(e)}.
$$
I computed the derivative on $z$ by choosing a curve $\alpha$ such that $\alpha(0) = z, \alpha'(0) = u$ so
$$
d \varphi_z(u) = \left. \frac{d}{dt}\right|_{t = 0} \varphi(\alpha(t)) = \frac{1}{(cz+d)^2} u.
$$
Now, since the metric in $e$ coincides with the euclidean metric, we have that if $u = u_1 e_1 + u_2 e_2$ and $v = v_1 e_1 + v_2 e_2$ then
$$
\langle u, v \rangle_e = u_1v_1 + u_2v_2.
$$
On the other hand, making the computations I end up with
$$
\langle d \varphi_e(u), d \varphi_e(v) \rangle_{\varphi(e)} = \frac{1}{(d + ic)^4} \langle u, v \rangle_{\frac{ac + bd +i }{(d + ci)(d – ci)}} = \frac{(d – ci)^2}{(d + ic)^2} (u_1v_1 + u_2v_2).
$$
Where is my mistake? Is the line of thought correct?
Thanks in advance and kind regards.
Best Answer
Consider first $df_z(u)$: \begin{align*} df_z(u) = \frac{ad - bc}{(cz+d)^2} u = \frac{1}{(cz+d)^2}u \end{align*} Notice that $g_{11} = g_{22}$ can be rewritten as: \begin{align*} g_{11} = g_{22} = \frac{1}{y^2} = - \frac{4}{(z-\overline{z})^2} \end{align*} Also pre-compute the value: \begin{align*} f(z) - f(\overline{z}) = \frac{(ad - bc)(z-\overline{z})}{(cz+d)(c\overline{z}+d)} = \frac{z-\overline{z}}{(cz+d)(c\overline{z}+d)} \end{align*} Now look at the inner products before and after the map: \begin{align*} \langle v, u \rangle_{z} & = - \overline{v} \frac{4}{(z-\overline{z})^2} u \\ \langle df_z(v), df_z(u) \rangle_{f(z)} &= - \frac{4}{(f(z) - f(\overline{z}))^2} \overline{df_z(u)} df_z(u) \\ &= - \frac{4((cz+d)(c\overline{z}+d))^2}{(z-\overline{z})^2} \frac{1}{(cz+d)^2}u \frac{1}{(c\overline{z}+d)^2}\overline{v} \\ &= - \overline{v} \frac{4}{(z-\overline{z})^2} u = \langle v, u \rangle_{z} \end{align*} So we have proved isometry.