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

Question

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?

PS: I have also asked the same question in Physics.SE here. I hope this doesn't violate any rules.

Answer 1

The class of holomorphic maps satisfying 1 and 2 is well known and is frequently used. Let us begin with rational functions satisfying 1,2. They are all of the form: $$az+b-\sum_{n}\frac{c_n}{z-z_n}$$ where $a\geq 0$, $b$ is real, $c_j\geq 0$ and $z_j$ are real. And this is a parametric description (every such function satisfies 1, 2). To get rid of the restriction that $f$ is rational, one passes to the limit. The result depends on your exact assumptions. If you want your function to be meromorphic in the plane, you obtain a similar formula, with finite sum replaced by an infinite sum. If you want a function which is only holomorphic in the upper and lower half planes, then you replace the sum by an integral. The most general formula which gives a holomorphic function in the union of upper and lower half-planes, and maps each half-plane into itself is $$az+b+\int_{-\infty}^\infty\left(\frac{1}{t-z}-\frac{t}{1+t^2}\right)d\mu(t),$$ where $\mu$ is any non-decreasing function such that $$\int_{-\infty}^\infty \frac{d\mu}{1+t^2}<\infty.$$ on the real line. (The extra term is added to make the integral convergent). As above, this is a parametric description. As you say, these formulas can be transplanted to describe the corresponding classes in the unit disk. These are called additive representations.

There are also multiplicative representations (which are more convenient to write for the unit disk). For example, a rational function $f$ which satisfies $|f(z)|<1$ in the unit disk and $|f(z)|>1$ outside, is a Blaschke product: $$f(z)=e^{i\theta}\prod_n\frac{|a_n|}{a_n}\frac{z-a_n}{1-\overline{a_n}z},$$ where $\theta$ is real and $|a_n|<1$. This can be generalized to infinite products, and to an integral. The general multiplicative representation is $$f(z)=B(z)\exp\int_0^{2\pi}\left(\frac{z+e^{it}}{z-e^{it}}d\nu(t)\right),$$ where $B$ is Blaschke product, finite or infinite, and $d\nu$ is a (non-negative) measure on the unit circle.

The correspondence between these two parametric representations of essentially the same class is non-trivial.