# Prove transformation $z \mapsto \frac{az+b}{cz+d}, z = x+ iy, ad-bc = 1$ is isometry of Hyperbolic plane.

Prove transformation $$f: z \mapsto \frac{az+b}{cz+d},\ z = x+ iy,\ ad-bc = 1$$ is isometry of Hyperbolic plane $$M=\{(x,y)\in \Bbb R^2:y>0\} \text{ with Riemannian metric } g= \frac{1}{y^2}(dx \otimes dx + dy \otimes dy).$$

To solve this problem, we can write $$ds^2=-\frac{dzd \bar z}{(z- \bar z)^2}$$, for $$w = f(z) = \frac{az+b}{cz+d}, f^*ds^2=-\frac{dzd \bar z}{(z- \bar z)^2}$$.

Thus $$f$$ is isometry. I'm trying the tensor calculus approach, but I meet some problems...

My effort:

Suppose $$f:(M,\tilde g)\to (M,g)$$, $$f$$ is isometry means $$f^*\tilde g=g.$$

Write $$z = x +iy,\ \bar z= x -iy$$, then

$$g|_z=\frac{1}{y^2}(dx \otimes dx + dy \otimes dy)=\frac{1}{2(\Im z)^2}(dz\otimes d \bar z + d\bar z\otimes dz).$$

For $$f(z)=\frac{az+b}{cz+d},\ \frac{\partial f}{\partial z}=\frac{1}{(cz+d)^2}, \ \frac{\partial f}{\partial \bar z}=0.$$

$$df=\frac{\partial f}{\partial z}dz+\frac{\partial f}{\partial \bar z}d\bar z=\frac{dz}{(cz+d)^2}.\ \Im f(z)=\frac{1}{2i}\frac{z-\bar z}{(cz+d)(c\bar z+d)}.$$

$$g|_z(\frac{\partial}{\partial z},\frac{\partial}{\partial z})= g|_z(\frac{\partial}{\partial \bar z},\frac{\partial}{\partial \bar z})=0,\ g|_z(\frac{\partial}{\partial z},\frac{\partial}{\partial \bar z})=g|_z(\frac{\partial}{\partial \bar z},\frac{\partial}{\partial z})=\frac{1}{2(\Im z)^2}.$$

$$f^*g|_z(\frac{\partial}{\partial z},\frac{\partial}{\partial z})=g|_{f(z)}(df(\frac{\partial}{\partial z}),df(\frac{\partial}{\partial z}))=g|_{f(z)}(f'(z), f'(z))=?$$

How to insert a number $$f'(z)$$ into tensor $$g|_{f(z)}(\cdot,\cdot)$$? And how to proceed?

Thank you for your time and effort.

• Since everything is complex analytic and the metrics are conformal, it is much easier to represent the metric as $|dz|/y$ and go from there. Tensor calculus is really formal overkill here. – Lukas Geyer May 13 at 14:27
• @LukasGeyer Thanks. I know that method, write $ds^2=-\frac{dzd \bar z}{(z- \bar z)^2}$, for $w = f(z) = \frac{az+b}{cz+d}$, we'll finally have $f^*ds^2=-\frac{dzd \bar z}{(z- \bar z)^2}$ so $f$ is isometry. I'm just trying the tensor calculus approach, and I meet some problems... – Andrews May 13 at 14:53