# Construction of a quasiconformal mapping of the closed disk to itself

My problem relates to the construction of the quasiconformal map $T:\overline{\mathbb{D}} \longrightarrow \overline{\mathbb{D}}$ from beginning of the proof of Lemma 2.2. in Marshall-Rhode.

Question

The text in question writes as follows:

Let $T:\overline{\mathbb{D}} \longrightarrow \overline{\mathbb{D}}$ be a quasiconformal map with $T(i)=\alpha ^+, T(-i)=\alpha ^-, T(1)=1$ and $T^{-1}(0)\in (-1,1)$. For instance, $T$ can be constructed as a composition of a quasiconformal map that sends $\alpha^+$ and $\alpha^-$ to symmetric points, followed by a Möbius transformation.

Here $\alpha ^+$ and $\alpha ^-$ are fixed points on $\partial \mathbb{D}$ such that the oriented arc from $\alpha ^-$ to $\alpha ^+$ (denoted $\langle \alpha ^-,\alpha^+\rangle$ in the text) on $\partial \mathbb{D}$ is seperated by 1 (they are defined above Lemma 2.2, but I don't think that it is relevant how).

My questions are these: Why does the quasiconformal map sending $\alpha^-$ and $\alpha^+$ to symmetric points exist? Is it symmetric in the real axis? And how does it look?

Attempt to solve

To me it would seem as though we need to either

1) Use some sort of "quasiconformal stretching" of the the arcs $\langle \alpha^-, 1\rangle$ and $\langle 1, \alpha^+\rangle$, where we somehow try to preserve the disk and map $\alpha^{-}$ and $\alpha^+$ into $c$ and $\bar{c}$ respectively for some $c\in \partial \mathbb{D}$. Then we could use the Möbius transformation which maps $(\bar{c},c,1)$ into $(-i,i,1)$ and so kinda push'' $0$ around on the real line (satisfying $T^{-1}(0)\in (-1,1)$), or

2) Quasiconformally stretch $\overline{\mathbb{D}}$ to an elliptical disk such that the two points become symmetric and then map the elliptical disk conformally to $\overline{\mathbb{D}}$ with a Möbius transformation.

Any help would be greatly appreciated.

In the unlikely event that anyone, besides me, will need this result, I shall answer for myself. I actually consulted Doctor Marshall and he was most kind to provide me with a more manageable solution to the problem:

There is a typo in the paper: the suggestion for the construction should refer to $T^{-1}$ not $T$, but you probably already understood that because it talks about the image of $\alpha^+$ and $\alpha^-$.

Here's an explicit way to construct $T^{-1}$ which might be easier than the suggestion: first find a Moebius map f with $f(a^-)= -1, f(1)=0$ and $f(a^+)=1$. Then $f$ will map the unit disk onto the upper half plane. Write $f(0)=A+iB$ then set

$g(x+iy)=x+(i-A/B)y = z +(iA/(2B) (z-{\bar z})$

Then $g$ is the identity on the real line and $g(A+iB)=iB$ and $g$ maps the upper half plane onto the upper half plane and the lower half plane onto the lower half plane.

Also $g_{\bar z}= -iA/(2B)$ and $g_z=1+(A/2B)i$

Then $g$ is QC because $|g_{\bar z}/g_z| =c <1$. Now let $h$ be the Moebius map $(i-z)/(i+z)$. The composition $h(g(f(z)))$ is QC and maps the unit disk onto itself with $a^-,1,a^+$ mapped to $-i,1,i$ respectively and $0$ is mapped to some point in $(-1,1)$. Let $T$ be the inverse of this map.

The map $g$ is a flow or shift along each horizontal line, but the amount of the shift depends on the height of the line. This corresponds to a flow along circles through $-1$ which are tangent to the unit circle, by conjugating with the Cayley transform.

• Thank you for sharing this result and your explanations! Great work! – ComplexFlo Sep 20 '16 at 7:38