Suppose $H$ is the upper half complex plane with Riemann metric $ds^2=\frac{dxdx+dydy}{y^2} $,then the lie group $GL_2(2,\mathbb{R} )$ can give an action on $H$.
For $A=\left( \begin{matrix} a& b\\ c& d\\ \end{matrix} \right) \in GL_2(2,\mathbb{R} )$ ,$A$ acts on $H$ by $f_A$, which $f_A:z\rightarrow \frac{az+b}{cz+d} $.
Then the Riemann metric is invariant under the action. I confused that how to use the function $f_A$ to calculate $dx,dy$ the to verify the Riemann metric is invariant,we can find $f_A$ is difficult,and even we calculated the $df_A/dx$ or something else,how can I get $dx$?.Or $dx,dy$ has other meaning can help us compute them?
Thankyou for sharing your mind.
Consider $z=x+iy$, so $$dz=dx+idy,\,d\overline z=dx-idy\implies dzd\overline{z}=dx^2+dy^2.$$ Since $z-\overline{z}=-2iy$, we have that $y^2=-4(z-\overline{z})^2$. With this change the metric transforms into $$ds^2=-\frac{4dzd\overline{z}}{(z-\overline{z})^2}.$$ Now let's consider the transformation $z'=\frac{az+b}{cz+d}$. For it we have that $$dz'=\frac{-dz}{(cz+d)^2},\,d\overline{z}'=\frac{-d\overline{z}}{(c\overline{z}+d)^2}.$$ Also note that $$z'-\overline{z}'=-\frac{z-\overline{z}}{(cz+d)(c\overline{z}+d)}.$$ Now just susbtituting in the equation of the metric we get that $$\frac{dz'd\overline{z}'}{(z'-\overline{z}')^2}=\frac{-\frac{-4dzd\overline{z}}{(cz+d)^2(c\overline{z}+d)^2}}{\frac{(z-\overline{z})^2}{(cz+d)^2(c\overline{z}+d)^2}}=-\frac{dzd\overline{z}}{(z-\overline{z})^2}=ds^2.$$ As we wanted to show.