The Möbius ambiguity in conformal welding

The Möbius extension $H$ carries $\mathbb{T}$ onto itself, so it permutes the two complementary components $\mathbb{D}$ and ${\mathbb{D}}^{\ast }$. If $H(\mathbb{D})={\mathbb{D}}^{\ast }$, its orientation-preserving sphere map carries the counterclockwise boundary orientation of $\mathbb{D}$ to the clockwise boundary orientation induced by ${\mathbb{D}}^{\ast }$; then $h=H{\mathrm{\mid }}_{\mathbb{T}}$ would reverse the circle orientation. Since $h$ is orientation-preserving, $H(\mathbb{D})=\mathbb{D}$ and $H({\mathbb{D}}^{\ast })={\mathbb{D}}^{\ast }$.

For any Möbius map $M$, postcomposition gives $(M\circ f{)}^{-1}\circ (M\circ g)=f^{-1}\circ M^{-1}\circ M\circ g=f^{-1}\circ g$; the welding map is unchanged while the curve becomes $M(\mathrm{\Gamma })$. Closure and inversion in [F7] make this composition calculation valid in the Möbius group.

Let $(\mathbb{T},f_1,g_1)$ and $(\mathbb{T},f_2,g_2)$ weld the same $h$. The first curve is globally removable by [F4], so [F5] gives one Möbius map $M$ with $f_2=M\circ f_1$ and $g_2=M\circ g_1$. Since both curves are $\mathbb{T}$, $M(\mathbb{T})=\mathbb{T}$. This proves the stated ambiguity for weldings with the round curve fixed.

The maps $f={\mathrm{i}\mathrm{d}}_{\mathbb{D}}$ and $g=H{\mathrm{\mid }}_{{\mathbb{D}}^{\ast }}$ are conformal bijections of the two complementary components and extend continuously to $\mathbb{T}$. Therefore [F1] gives $f^{-1}\circ g{\mathrm{\mid }}_{\mathbb{T}}=H{\mathrm{\mid }}_{\mathbb{T}}=h$. In particular, for the displayed Blaschke map, its disk automorphism extension restricted to ${\mathbb{D}}^{\ast }$ is the required exterior parameter map.

Take any welding $({\mathrm{\Gamma }}^{\mathrm{\prime }},f^{\mathrm{\prime }},g^{\mathrm{\prime }})$ of $h$. Compare it by [F5] with the round welding $(\mathbb{T},{\mathrm{i}\mathrm{d}}_{\mathbb{D}},H{\mathrm{\mid }}_{{\mathbb{D}}^{\ast }})$ from step 2.1, using $\mathbb{T}$ first. Then ${\mathrm{\Gamma }}^{\mathrm{\prime }}=M(\mathbb{T})$ and both parameter maps are postcomposed by $M$. If two such weldings have the same images of three distinct boundary points under their first parameter map, their relative Möbius map fixes those three distinct points; [F6] forces it to be the identity. Hence the normalized maps and curve are unique. The inherited AC premise is recorded in [F8].

Write $M(z)=(az+b)\mathrm{/}(cz+d)$ with $ad-bc\mathrm{\ne }0$. For $w=M(z)$, the condition $\mathrm{\mid }z\mathrm{\mid }=1$ becomes $\mathrm{\mid }dw-b\mathrm{\mid }=\mathrm{\mid }a-cw\mathrm{\mid }$, which is a nondegenerate circle equation or line equation in the finite plane; the line case includes $\mathrm{\infty }$ on the sphere. Therefore every curve ${\mathrm{\Gamma }}^{\mathrm{\prime }}=M(\mathbb{T})$ is a generalized round circle, proving the Statement.