Bounded turning, quasiconformal images of the circle, and quasiconformal reflections

The two-point and four-point formulations in (ii) are the Gehring equivalence in [F1]. For the forward constant, order the four points so $\mathrm{\mid }{z}_1-z_3\mathrm{\mid }\le \mathrm{\mid }{z}_2-z_4\mathrm{\mid }$ and label the two arcs from $z_1$ to $z_3$ so the arc through $z_2$ has smaller diameter. Then $\mathrm{\mid }{z}_1-z_2\mathrm{\mid },\mathrm{\mid }{z}_2-z_3\mathrm{\mid }\le M\mathrm{\mid }{z}_1-z_3\mathrm{\mid }$, while the triangle inequality gives $\mathrm{\mid }{z}_3-z_4\mathrm{\mid },\mathrm{\mid }{z}_4-z_1\mathrm{\mid }\le (M+1)\mathrm{\mid }{z}_2-z_4\mathrm{\mid }$; adding the two products gives $b=2M(M+1)$. Conversely, if both arcs from $z_1$ to $z_3$ had diameter greater than $2b\mathrm{\mid }{z}_1-z_3\mathrm{\mid }$, choose $z_2,z_4$ on the two arcs with $\mathrm{\mid }{z}_1-z_2\mathrm{\mid },\mathrm{\mid }{z}_1-z_4\mathrm{\mid }>b\mathrm{\mid }{z}_1-z_3\mathrm{\mid }$. The two products on the left would sum to more than $b\mathrm{\mid }{z}_1-z_3\mathrm{\mid }(\mathrm{\mid }{z}_2-z_3\mathrm{\mid }+\mathrm{\mid }{z}_3-z_4\mathrm{\mid })$, at least the right side by the triangle inequality.

Suppose (i), and take a $K$-quasiconformal sphere map $W$ with $W({\mathbb{S}}^1)=\mathrm{\Gamma }$. Then $\sigma =W\circ \tau \circ W^{-1}$ is an orientation-reversing quasiconformal involution with fixed set exactly $\mathrm{\Gamma }$, interchanging the two components and with dilatation at most $K^2$ by [F6]. To prove (ii), put $W_0=\chi \circ W$ and $q=W_0^{-1}(\mathrm{\infty })\mathrm{\notin }{\mathbb{S}}^1$. There is a circle-preserving Möbius map $A$ taking $\mathrm{\infty }$ to $q$: if $q\in \mathbb{D}$, compose $z\mapsto 1\mathrm{/}z$ with a disc automorphism taking $0$ to $q$; if $q$ is in the exterior, conjugate the analogous disc map by $z\mapsto 1\mathrm{/}z$; for $q=\mathrm{\infty }$ use the identity. Thus $V=W_0\circ A$ fixes infinity and is a $K$-quasiconformal plane homeomorphism. Given $x,y\in {\mathbb{S}}^1$, every point $u$ of their shorter circle arc satisfies $\mathrm{\mid }u-x\mathrm{\mid }\le \mathrm{\mid }y-x\mathrm{\mid }$. By [F2], $\mathrm{\mid }V(u)-V(x)\mathrm{\mid }\le {\eta }_K(1)\mathrm{\mid }V(y)-V(x)\mathrm{\mid }$. The image arc consequently has diameter at most $2{\eta }_K(1)\mathrm{\mid }V(y)-V(x)\mathrm{\mid }$. This gives (ii) with a bound depending only on $K$, in the stipulated finite chart.

Assume (ii). In the four-point inequality each product transforms under a Möbius map by the same factor, as follows from $\mathrm{\mid }T(z)-T(w)\mathrm{\mid }=\mathrm{\mid }\det T\mathrm{\mid }\mathrm{\mid }z-w\mathrm{\mid }\mathrm{/}(\mathrm{\mid }cz+d\mathrm{\mid }\mathrm{\mid }cw+d\mathrm{\mid })$; hence it is invariant, including poles by limits. Send one curve point to infinity and write $\mathrm{\Lambda }=T(\chi (\mathrm{\Gamma }))$. Let $C=b$. Letting the fourth point tend to infinity gives, for consecutive finite points $P_1,P_2,P_3$ on this generalized line, $\mathrm{\mid }{P}_1-P_2\mathrm{\mid }+\mathrm{\mid }{P}_2-P_3\mathrm{\mid }\le C\mathrm{\mid }{P}_1-P_3\mathrm{\mid }$. Choose boundary-extended conformal maps $f:\mathbb{H}\to {\mathrm{\Omega }}_{+}$ and $g:{\mathbb{H}}^{-}\to {\mathrm{\Omega }}_{-}$ with $f(\mathrm{\infty })=g(\mathrm{\infty })=\mathrm{\infty }$. The boundary correspondence $h=g^{-1}\circ f$ fixes infinity and is increasing because the source and target boundary orientations on the two sides are both opposite. Countable Choice required by the extremal-length interfaces follows from AC by [F8]. We establish the adjacent-interval bound by the following length-area calculation. For an ordered triple on $\mathrm{\Lambda }$, put $\alpha =P_2{P}_3$, ${\alpha }^{\mathrm{\prime }}=P_1{P}_2$, $\beta =P_1\mathrm{\infty }$ on the ray avoiding $P_3$, and ${\beta }^{\mathrm{\prime }}=P_3\mathrm{\infty }$ on the other ray. By [F9], their joining extremal distances $D,D^{\mathrm{\prime }}$ in ${\mathrm{\Omega }}_{+}$ satisfy $D{D}^{\mathrm{\prime }}=1$, and likewise $D_{\ast }{D}_{\ast }^{\mathrm{\prime }}=1$ in ${\mathrm{\Omega }}_{-}$. For $P_j=f(x_j)$ with $x_2-x_1=x_3-x_2$, $D=D^{\mathrm{\prime }}=1$: after affine normalization the upper-half-plane quadruple is $(0,1,2,\mathrm{\infty })$, and the upper-half-plane automorphism $z\mapsto (2z-2)\mathrm{/}z$ interchanges the complementary marked pairs. Reciprocity then forces their equal positive values to be one.

Put $a=\mathrm{\mid }{P}_1-P_2\mathrm{\mid }$, $d=\mathrm{\mid }{P}_2-P_3\mathrm{\mid }$ for the triple in step 2.1. The ordered-triple bound puts $\alpha$ inside the disk about $P_2$ of radius $Cd$ and keeps $\beta$ outside the disk of radius $a\mathrm{/}C$. If $a>C^{2}d$, every joining path crosses the intervening round annulus. Its radial density $1\mathrm{/}\mathrm{\mid }z-P_2\mathrm{\mid }$ gives $D\ge (2\pi {)}^{-1}\log (a\mathrm{/}(C^{2}d))$; restriction to either side only lowers its density area. Since $D=1$, $a\mathrm{/}d\le C^2{e}^{2\pi }$. Applying the same argument to ${\alpha }^{\mathrm{\prime }},{\beta }^{\mathrm{\prime }}$ and $D^{\mathrm{\prime }}=1$ gives $d\mathrm{/}a\le C^2{e}^{2\pi }$. For $Q_1\in \alpha$, $Q_2\in \beta$, repeated ordered-triple bounds give $\mathrm{\mid }{Q}_1-Q_2\mathrm{\mid }\ge C^{-1}\mathrm{\mid }{Q}_1-P_1\mathrm{\mid }\ge C^{-2}a\ge C^{-4}{e}^{-2\pi }d$. Set $\delta =C^{-4}{e}^{-2\pi }d$ and $R=Cd$. Use density one on the disk $B(P_2,R+\delta )$ in ${\mathrm{\Omega }}_{-}$. Every path from $\alpha$ to $\beta$ has density length at least $\delta$: if it stays in the disk this follows from endpoint separation, and if it leaves, the initial portion from $\alpha \subseteq \bar{B}(P_2,R)$ has that length already. Thus $D_{\ast }\ge c(C):={\delta }^2\mathrm{/}(\pi (R+\delta {)}^2)>0$, a constant independent of the triple and scale. The same estimate for the complementary pair gives $D_{\ast }^{\mathrm{\prime }}\ge c(C)$, and reciprocity yields $D_{\ast }\le 1\mathrm{/}c(C)$.

Write $y_j=h(x_j)$ and $r=(y_3-y_2)\mathrm{/}(y_2-y_1)>0$. Affinely normalize this lower-half-plane quadruple to $(0,1,1+r,\mathrm{\infty })$. If $r<1$, a joining path from $[1,1+r]$ to the ray ending at $0$ crosses the annulus about $1$ of radii $r,1$; the same radial-density estimate gives $D_{\ast }\ge (2\pi {)}^{-1}\log (1\mathrm{/}r)$. If $r>1$, the complementary joining paths cross the annulus about $1$ of radii $1,r$, giving $D_{\ast }^{\mathrm{\prime }}\ge (2\pi {)}^{-1}\log r$. The two upper bounds $D_{\ast },D_{\ast }^{\mathrm{\prime }}\le 1\mathrm{/}c(C)$ therefore imply $e^{-2\pi \mathrm{/}c(C)}\le r\le e^{2\pi \mathrm{/}c(C)}$. This is exactly the two-order adjacent-interval quasisymmetry condition for $h$, with constant depending only on $b$, hence only on $M$.

Extend $h$ by [F4] to a quasiconformal sphere map $H$ preserving both half-planes. Define $W=f$ on $\bar{\mathbb{H}}$ and $W=g\circ H$ on $\overline{{\mathbb{H}}^{-}}$. On the common boundary, $g(H(t))=g(h(t))=f(t)$; the boundary extensions in [F3] make the pasted map a sphere homeomorphism. It is quasiconformal off $\mathbb{R}\cup \{\mathrm{\infty }\}$, hence globally quasiconformal by [F5], and maps that generalized line onto $\mathrm{\Lambda }$. If $C$ is a Möbius map from ${\mathbb{S}}^1$ onto $\mathbb{R}\cup \{\mathrm{\infty }\}$, then ${\chi }^{-1}\circ T^{-1}\circ W\circ C$ is a quasiconformal sphere homeomorphism carrying ${\mathbb{S}}^1$ onto $\mathrm{\Gamma }$. This proves (i) from (ii).

Suppose (iii). Let $\mathrm{\Omega }$ be either complementary component and take a conformal map $f:\mathbb{D}\to \mathrm{\Omega }$ with its homeomorphic boundary extension from [F3]. Define $W=f$ on $\bar{\mathbb{D}}$ and $W=\sigma \circ f\circ \tau$ on the closed exterior disc. The second formula maps the exterior disc onto the other component, is quasiconformal there, and agrees with $f$ on ${\mathbb{S}}^1$ because $\sigma$ fixes $\mathrm{\Gamma }$ pointwise. Hence $W$ is a sphere homeomorphism; [F5] makes it quasiconformal globally and $W({\mathbb{S}}^1)=\mathrm{\Gamma }$. This proves (i) from (iii), completing the equivalence.