# Möbius transformation on Upper half plane

If I have the function $\phi (x)= \frac{-z+i} {-iz+1} z \in \mathbb{C}$

from D to H where $D = { z \in \mathbb{C} | |z| <1}$ and H is the upper half plane

It's not that hard to see that $\phi^{-1} (x)= \frac{z-i} {iz-1}$

g is given as the standard metric of the upper half plane

how can i show that: $(\phi^* g)[id](z) = \frac 4 {(1-|z|^2)^2} id_{[\mathbb{R}^2]}$

• Sorry, I'm confused about your notation. Can you unpack the last line a little so I can try to work out an answer? – Nick Jan 7 '14 at 17:31
• I think it should be the pullback of $\phi$ on g. where g the standart metric is. – lukas Jan 7 '14 at 17:34
• Yeah, on second thought I'm not sure why that wasn't apparent to me to begin with. I'm scratching out some notes right now, hopefully I can come up with something. – Nick Jan 7 '14 at 17:38

## 1 Answer

The metric on the upper half-plane is given by

$$g_{|w} = \frac{\left|dw\right|^2}{\mathrm{Im}(w)^2}$$

To find the expression of the the pull-back of this metric by $\phi$ at some point $z$, just do $w = \phi(z)$ in the previous expression. You find $$(\phi^*g)_{|z} = \frac{\left|d\phi(z)\right|^2}{\mathrm{Im}(\phi(z))^2}$$ with $$d\phi(z) = \phi'(z) dz = {2 dz \over (z+i)^2}$$ and $$\mathrm{Im}(\phi(z)) = {1 - \left|z\right|^2 \over \left|z+i\right|^2}$$ you indeed get $$(\phi^*g)_{|z} = \frac{4 \left|dz\right|^2}{( 1 - \left|z\right|^2)^2}$$