1

$SL(2, R)$ acts on $H^2$ by Möbius transformations $$ g\cdot z=\frac{az+b}{cz+d}, \quad g=\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in SL(2,R), \quad z\in H^2,$$ where $H^2=\{z\in C \mid \operatorname{Im} (z)>0\}$, i.e., the complex upper half plane.

My question is how to view the Jacobian of $M_g $ as a linear transformation on $R^2$, and how to compute its determinant. Here, $M_g$ denote the associated Möbius transformations.

The answer is $|cz+d|^{-4}$, but I cannot get that. please help me , thanks in advance.

dfeuer
  • 9,069
user63788
  • 206
  • 2
    The Jacobian of a holomorphic function is $\lvert f'(z)\rvert^2$. The derivative of the Möbius transformation is $$\frac{ad-bc}{(cz+d)^2}.$$ – Daniel Fischer Dec 20 '13 at 04:56
  • @DanielFischer The OP has already normalized out the determinant, so your numerator is $1.$ – Igor Rivin Dec 20 '13 at 06:26
  • That step was intentionally left out, @IgorRivin. Replacing it with $1$ in the normalised case should not be a problem, but as written, it's also correct in the non-normalised case. – Daniel Fischer Dec 20 '13 at 06:30

1 Answers1

1

Answered by Daniel Fischer:

The Jacobian of a holomorphic function is $\lvert f'(z)\rvert^2$. The derivative of the Möbius transformation is $$\frac{ad-bc}{(cz+d)^2}.$$

With the $SL(2,\mathbb R)$ normalization, the numerator is $1$. Thus, the Jacobian is $|cz+d|^{-4}$.

Something like this is true in higher dimensions (in $\mathbb R^n$) too. Namely, the Jacobian of the inversion $x\mapsto x/|x|^2$ is $1/|x|^{2n}$. A general fractional linear transformation is the composition of linear map with inversion, followed by another linear map.

Community
  • 1
user127096
  • 9,683