How many boundary values determine an automorphism of unit disk?

Question

How many boundary values are needed to uniquely determine an automorphism of unit disk?

More precisely, find the $n \in \mathbb{N} $ such that for any two sets $ \{ x_1, x_2, ..., x_n \} $ and $ \{ y_1, y_2, ..., y_n \} $ of $n$ distinct points on $ \mathbb{S}^1 $, there exists a unique automorphism $ f: \mathbb{D} \rightarrow \mathbb{D} $ with $ f(x_i) = y_i $ for all $i$.

Clearly, $n=2$ is not enough: there are many automorphisms $ \mathbb{D} \rightarrow \mathbb{D} $ fixing the two points $1$ and $-1$ on $ \mathbb{S}^1 $.

My guess is $n=3$, since any automorphism $ \mathbb{D} \rightarrow \mathbb{D} $ is of the form $ f_{e^{i\theta},a} (z) = e^{i\theta} \frac{z-a}{1-\bar{a}z} $, where $ (e^{i\theta},a) \in \mathbb{S}^1 \times \mathbb{D} $. The dimension of the parameter space $ \mathbb{S}^1 \times \mathbb{D} $ suggests that we need three boundary values.

However, when I plug in $ f_{e^{i\theta},a} (x_i) = y_i, $ $ i=1,2,3 $, I can't solve $ (e^{i\theta},a) $ explicitly in terms of $x_i$ and $y_i$. Any suggestion?

Answer 1

Your guess based on dimension is correct.

To come up with a proof, it helps to recognize that your formula for an automorphism of the disc generalizes to the concept of a fractional linear transformation $$T(z) = \frac{az+b}{cz+d} $$ where $a,b,c,d \in \mathbb{C}$ and $ad-bc \ne 0$. Furthermore, after a normalization you may assume $ad-bc=1$. Normalized fractional linear transformations form a group under composition; this group is isomorphic to $SL(2,\mathbb{C})$.

Just to put this in context, each normalized fractional linear transformation is an automorphism of the entire Riemann sphere $\mathbb{C} \cup \{\infty\}$, and these are exactly all of the automorphisms of $\mathbb{C} \cup \{\infty\}$, so every automorphism of $\mathbb{D}$ extends to the whole of $\mathbb{C} \cup \{\infty\}$. Thus, if you know something about automorphisms of $\mathbb{C} \cup \{\infty\}$ then perhaps you can apply it to gain knowledge about automorphisms of $\mathbb{D}$.

The uniqueness property you want then follows from a lemma which says that for any two triples of distinct points $z_1,z_2,z_3$ and $w_1,w_2,w_3$ in $\mathbb{C} \cup \{\infty\}$ (assuming the three $z$'s are all different, as are the three $w$'s) there exists a unique normalized fractional linear transformation $T$ such that $w_1 = T(z_1)$, $w_2 = T(z_2)$, $w_3 = T(z_3)$. It is easiest to prove this lemma first in a special case $w_1=0$, $w_2=1$, $w_3=\infty$, and then use that to prove the general case.

Answer 2

I realize I am perhaps a little late to answer this question, but I was wondering about the same thing and it took me quite some time to figure it out. So I hope other people might find this solution in the future.

I find the question easier to answer in the upper half plane model $\mathbb{H}$. Using for example

$$ \phi\colon \mathbb{D} \rightarrow \mathbb{H}, z \mapsto \frac{z-i}{1-iz} $$

and extending it to $\partial \mathbb{D}$, we see that $\phi(\partial \mathbb{D}) = \mathbb{R} \cup \{\infty\}$ holds.

Furthermore you can check that the automorphism group of $\mathbb{H}$ is given by $\operatorname{PSL}(2,\mathbb{R})$.

What we want to prove next is, that given any two triplebs of boundary points of $\mathbb{H}$ (i.e. $\mathbb{R} \cup \{\infty\}$), we find a unique element in the automorphism group of $\mathbb{H}$ converting the one into the other.

To see that we consider the complex cross-ratio given by

$$ (z; z_1, z_2, z_3) \mapsto \frac{(z - z_2)(z_1 -z_3)}{(z_1 - z_2)(z - z_3)}, $$

where we kill a term, if one of the variables is $\infty$ and the $z_i$ are distinct and fixed. Then we get a Möbius transformation which we denote by $\phi$. We see the following:

$$ \phi(z_1) = 1, \quad \phi(z_2) = 0 \text{ and } \phi(z_3) = \infty. $$

From that we get the following observation: if the $z_i$ are in $\mathbb{R} \cup \infty$, then $\phi$ is represented by a real matrix and we can find a representative in $\operatorname{PSL}(2, \mathbb{R})$. So we can map any $\mathbb{H}$ boundary triple to the triple $(1,0, \infty)$, which is also a $\mathbb{H}$ boundary triple. Concatenation with an inverse of another cross-ratio yields our assertion.

The uniqueness of the tranfsormation is given by the fact, that any Möbius transformation is uniquely determined by three points (which is actually a consequence of the cross-ratio, we used).

Now, going back to $\mathbb{D}$, we get that any automorphism of it, is uniquely determined by three boundary values.