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This is one of Do Carmo's excersices and I got it as homework. Part (a) is easy and I include it here for the sake of completness. But I am entirely lost on part (b).

A function $g:\mathbb{R} \rightarrow \mathbb{R}$ given by $g(t)=yt+x$, $t,x,y \in \mathbb{R}, y>0$ is called an affine proper function. The subset of all such functions with respect to the usual composition law forms a Lie group $G$.As a differentiable manifold $G$ is just the half upper plane with the differentiable structure induced from $\mathbb{R}^{2}$. 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 $g_{11}=g_{22}=\frac{1}{y^{2}}$, $g_{12}=0$.

(b) Putting $(x,y)= z=x + iy$, $i=\sqrt{-1}$, the transformation $z\rightarrow z'= \frac{az+b}{cz+d}$, $a,b,c,d \in \mathbb{R}$, $ad-bc=1$ is an isometry of $G$.

Hint: Observe that the first fundamental form can be written as:

$ds^{2}= \frac{dx^{2} + dy^{2}}{y^{2}} = \frac{4dzd\overline{z}}{(z- \overline{z})^{2}}$.

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show that the metric is preserved by the moebius transformation (using the complex expression should be easier, which is why it's included). it's just a computational exercise. – yoyo Mar 16 '11 at 16:16
See here:… – t.b. Mar 16 '11 at 16:17
Thank you. I think my problem was in the complex notation. I will post the full solution this afternoon after I get back froms school. – Sak Mar 16 '11 at 16:39
up vote 1 down vote accepted

Let be $p\in \mathbb{H}$ and, $u=(u_1,u_2),v=(v_1,v_2)\in T_p\mathbb{H}$

Put, $dz(u)=u_1+iu_2$ and $d\overline{z}(u)=u_1-iu_2,$ $dx(u)=u_1, dy(u)=u_2.$

Then is easy to see that,

$$ds^2=\dfrac{dx^2+dy^2}{y^2}= \frac{4dzd\overline{z}}{(z- \overline{z})^{2}}.$$

Now consider the following maps,

  1. $T_1 :z\mapsto z+c$

  2. $T_2: z\mapsto cz$

  3. $T_3: z\mapsto \dfrac{1}{z}$

Any Möbius application is a composition of maps above. Then is sufficient to see that the above maps are isometries.

Consider for example $T_2(z)=\dfrac{az+b}{cz+d}$, then $T_2'(z)=c$.

By definition

$$\langle T'(z)u, T'(z)v\rangle_{T(z)}:=\langle dL_{T(z)^{-1}}(T(z))T'(z)u, dL_{T(z)^{-1}}(T(z))T'(z)v\rangle_e=\dfrac{4T'(z)dz(u)T'(z)d\overline{z}(v)}{(T(z)-\overline{T(z)})^2}=\dfrac{4dz(u)d\overline{z}(v)}{(z-\overline{z})^2}=:\langle u,v\rangle_z$$

Similarly you do this for the other Möbius transformations.

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How is $T_2: z\to cz$ isometry? – dmm Sep 25 '12 at 5:28

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