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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]} $

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    $\begingroup$ Sorry, I'm confused about your notation. Can you unpack the last line a little so I can try to work out an answer? $\endgroup$ – Nick Jan 7 '14 at 17:31
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    $\begingroup$ I think it should be the pullback of $\phi$ on g. where g the standart metric is. $\endgroup$ – lukas Jan 7 '14 at 17:34
  • $\begingroup$ 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. $\endgroup$ – Nick Jan 7 '14 at 17:38
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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 $$(\phi^*g)_{|z} = \frac{4 \left|dz\right|^2}{( 1 - \left|z\right|^2)^2}$$

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