3
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Lukas Geyer May 13 at 14:27
  • $\begingroup$ @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... $\endgroup$ – Andrews May 13 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.