# Holomorphic maps from upper half plane to itself (or equivalently Poincare disc to itself)

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Suppose I parametrize complex plane by coordinates,$$z = x+i y,\ \bar z=x-i y$$
then the upper half plane, $\mathbb H_+$ is given by $y>0$. I am looking for chiral coordinate transformations, $f(z)$, such that

1. they map the boundary ($y=0$) to itself,$$f(z)=\bar f(\bar z) \ \text{whenever } z=\bar z~.$$
2. $z+\bar z >0 \Leftrightarrow f(z)+\bar f(\bar z)>0~.$

Since there is a conformal map between $\mathbb H_+$ and unit disc (Poincare disc), $\mathbb D = \{w:|w|<1\}$, where the above conditions become:

1. the coordinate transformations map the boundary of $\mathbb D$ ($|w|=1$) to itself,$$g(w) \bar g(\bar w)=1 \ \text{whenever } w\bar w=1~.$$
2. $w \bar w<1 \Leftrightarrow g(w)\bar g(\bar w)<1~.$

The Schwarz-Pick Lemma seems to suggest that a general holomorphic transformation brings the boundary of the disc closer than $1$ in the Poincaré metric (I am interested in AdS$_2$ so I can equivalently say that the *new* boundary after the coordinate transformation is at a finite distance from any interior point),$$\frac{g'(w)\bar g'(\bar w)}{\left(1-g(w)\bar g(\bar w)\right)^2}dw d\bar w \le \frac1{(1-w\bar w)^2}dw d\bar w$$ and the equality folds only for Mobius transformations (which can be seen as isometries of AdS$_2$).

- Is it correct to deduce that there are no (non-trivial, of course not the Mobius transformation) holomorphic transformations that satisfy the conditions 1 and 2 above, or am I interpreting the Schwarz-Pick lemma incorrectly?
- If my interpretation of Schwarz-Pick lemma is correct and there are no *holomorphic* maps with the given constraints, then what are the less restrictive class of functions that obey the above constraints?

asked May 30, 2019

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