The Beltrami Equation and Measurable Riemann Mapping: Examples and Counterexamples

Put $s:=\mathrm{\mid }\nu \mathrm{\mid }<1$. The inverse formula follows from $w-\nu \bar{w}=(1-\mathrm{\mid }\nu {\mathrm{\mid }}^2)z$. By [F3], $A_z=1$ and $A_{\bar{z}}=\nu$, so $J_A=1-\mathrm{\mid }\nu {\mathrm{\mid }}^2>0$; the inverse makes $A$ a homeomorphism. By [F4], it is analytically $K_A$-quasiconformal with coefficient $\nu$, where $K_A=(1+s)\mathrm{/}(1-s)$; [F5] gives orientation preservation.

If $g(z)=az+b\bar{z}+c$, then $g_z=a$ and $g_{\bar{z}}=b$. Thus the equation is equivalent to $b=\nu a$. Its Jacobian is $\mathrm{\mid }a{\mathrm{\mid }}^2-\mathrm{\mid }b{\mathrm{\mid }}^2=(1-\mathrm{\mid }\nu {\mathrm{\mid }}^2)\mathrm{\mid }a{\mathrm{\mid }}^2$, so an orientation-preserving solution has $a\mathrm{\ne }0$; conversely any such $a$ gives the stated real-affine solution.

Since $\mathrm{\mid }\nu \mathrm{\mid }<1$, $1+\nu \mathrm{\ne }0$ and $f^{\nu }=(1+\nu {)}^{-1}A$ is invertible on $\mathbb{C}$. Its lower bound $\mathrm{\mid }{f}^{\nu }(z)\mathrm{\mid }\ge (1-\mathrm{\mid }\nu \mathrm{\mid })\mathrm{\mid }z\mathrm{\mid }\mathrm{/}\mathrm{\mid }1+\nu \mathrm{\mid }$ shows it extends continuously by $f^{\nu }(\mathrm{\infty })=\mathrm{\infty }$, and its inverse extends likewise. In the source and target infinity coordinates $w=1\mathrm{/}z$ and $\eta =1\mathrm{/}{f}^{\nu }(z)$, the expression is $\eta (w)=\frac{(1+\nu )w\bar{w}}{\bar{w}+\nu w},\eta (0)=0.$ For $w\mathrm{\ne }0$, $\mathrm{\mid }\bar{w}+\nu w\mathrm{\mid }\ge (1-\mathrm{\mid }\nu \mathrm{\mid })\mathrm{\mid }w\mathrm{\mid }$; the expression is homogeneous of degree one and smooth on the punctured disk, so its derivative is bounded there by its bound on the unit circle, while $\mathrm{\mid }\eta (w)\mathrm{\mid }=O(\mathrm{\mid }w\mathrm{\mid })$. The bounded derivative gives a Lipschitz bound along segments avoiding $0$, and continuity extends that bound across $0$. Its coordinate-line restrictions are therefore absolutely continuous: the sum of their increments is bounded by the Lipschitz constant times the total interval length. Their derivatives are bounded off $0$, hence locally square-integrable, so the ACL characterization in [F7] gives $W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$ at $0$. In the finite chart, $f_{\bar{z}}^{\nu }=\nu f_z^{\nu }$; the coefficient pullback in [F7] gives the same weak equation in the infinity chart, with the point $w=0$ immaterial. The local orientation sign remains positive by [F6]. Hence $f^{\nu }$ is a sphere weak solution and orientation-preserving quasiconformal homeomorphism. Direct substitution gives $f^{\nu }(0)=0$ and $f^{\nu }(1)=1$.

The Axiom of Choice permits use of [F8]. Step 2.1 proves that $f^{\nu }$ is a normalized orientation-preserving quasiconformal homeomorphic solution, so uniqueness in [F8] identifies it with the normalized MRMT solution. Every other orientation-preserving quasiconformal homeomorphic sphere solution is its Möbius postcomposition by [F8], and every such postcomposition is a solution.

If $\nu =0$, then $A$ and $f^0$ are the identity and the ellipse ratio is $1$. Otherwise write $\nu =s{e}^{i\theta }$ and set $z=e^{i\theta \mathrm{/}2}(x+iy)$. Then $A(z)=e^{i\theta \mathrm{/}2}((1+s)x+i(1-s)y)\mathrm{.}$ Thus $A(S^1)$ has semiaxes $1+s$ and $1-s$, so its ratio is $(1+s)\mathrm{/}(1-s)=K_A$, which also equals the coefficient ellipse ratio by [F9]. Multiplication by $(1+\nu {)}^{-1}$ scales and rotates both axes equally.

Choose a measurable representative ${\mu }_0$ of the given a.e. class and extend it by zero off $\mathrm{\Omega }$, obtaining $\tilde{\mu }$. Since $\mathrm{\Omega }$ is open and hence Borel, [F2] makes the extension measurable. It satisfies $\mathrm{\parallel }\tilde{\mu }{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }\mu {\mathrm{\parallel }}_{\mathrm{\infty }}\le k$. For every ball $B$, [F4] gives ${\mathbf{1}}_B\in L^1$ with norm ${\lambda }_2(B)<\mathrm{\infty }$, and [F5] gives ${\int }_B\mathrm{\mid }\tilde{\mu }\mathrm{\mid }\le k{\lambda }_2(B)$. Thus $\tilde{\mu }\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^2)$ by [F10].

Write $h_n=2^{-n}$. Each $Q\in {\mathcal{Q}}_n$ has ${\lambda }_2(Q)=h_n^2>0$ by [F6], so its average $c_Q$ is defined. A.e. changes of ${\mu }_0$ do not change any $c_Q$ by [F7]. The half-open squares form a countable measurable partition of ${\mathbb{R}}^2$, so on $\mathrm{\Omega }$ the function ${\mu }_n$ is measurable and constant on each $\mathrm{\Omega }\cap Q$. Moreover [F5] gives $\mathrm{\mid }{c}_Q\mathrm{\mid }=\frac{1}{h_n^2}\mid {\int }_Q\tilde{\mu }\mid \le \frac{\mathrm{\parallel }\tilde{\mu }{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{\mathbf{1}}_Q{\mathrm{\parallel }}_1}{h_n^2}\le k,$ hence $\mathrm{\parallel }{\mu }_n{\mathrm{\parallel }}_{\mathrm{\infty }}\le k$.

Let $x\in \mathrm{\Omega }$ be a Lebesgue point of $\tilde{\mu }$, and let $Q_n(x)$ be its unique half-open dyadic square. Every point of $Q_n(x)$ is within distance $\sqrt{2}{h}_n$ of $x$, so $Q_n(x)\subset B(x,\sqrt{2}{h}_n)$. The containing square of side $2\sqrt{2}{h}_n$ has area $8h_n^2$ by [F6], whence ${\lambda }_2(B(x,\sqrt{2}{h}_n))\mathrm{/}{h}_n^2\le 8$. Therefore, writing $g_x(y)=\mathrm{\mid }\tilde{\mu }(y)-\tilde{\mu }(x)\mathrm{\mid }$, $\mathrm{\mid }{\mu }_n(x)-\tilde{\mu }(x)\mathrm{\mid }\le \frac{1}{h_n^2}{\int }_{Q_n(x)}{g}_x\le 8A_{\sqrt{2}{h}_n}{g}_x(x)⟶0$ by [F8]. The Lebesgue points of $\tilde{\mu }$ have full measure by [F9], proving (b) on $\mathrm{\Omega }$.

For a countable locally finite triangulation ${\mathcal{T}}_j$ with mesh ${\delta }_j\to 0$, fix an enumeration of its triangles and assign shared faces to the first incident cell, giving a Borel partition. For a triangle $T$, write $d_T=\mathrm{diam}T>0$, let $c_T$ be its barycenter, and assign the constant $a_T:=\frac{1}{{\lambda }_2(B(c_T,d_T))}{\int }_{B(c_T,d_T)}\tilde{\mu }$ to its cell in $\mathrm{\Omega }$. The resulting function is measurable by [F2]. If $x$ belongs to that cell, then $\mathrm{\mid }x-c_T\mathrm{\mid }\le d_T$, so $B(c_T,d_T)\subset B(x,2d_T)$. The inner ball contains a square of side $\sqrt{2}{d}_T$ and the outer ball lies in a square of side $4d_T$; [F4] and [F6] therefore give $\frac{{\lambda }_2(B(x,2d_T))}{{\lambda }_2(B(c_T,d_T))}\le \frac{16d_T^2}{2d_T^2}=8.$ Thus at every Lebesgue point $x$ the same estimate as in step 3.1 gives $\mathrm{\mid }{a}_T-\tilde{\mu }(x)\mathrm{\mid }\le 8A_{2d_T}{g}_x(x)$, which tends to zero uniformly as $d_T\le {\delta }_j\to 0$. The averages remain bounded by $k$ by [F5]. If averages over $T$ itself are used and ${\lambda }_2(T)\ge c(\mathrm{diam}T{)}^2$ uniformly, then $T\subset B(x,2d_T)$ and the outer-to-cell measure ratio is at most $16\mathrm{/}c$, giving the analogous estimate; this is the uniform shape-regularity condition stated in (c).

Steps 2.1 and 3.1 prove (a) and (b), and step 4.1 proves the shape-independent triangulation version of (c).

Direct calculation gives $A_z=1$, $A_{\bar{z}}=\nu$, $J_A=1-\mathrm{\mid }\nu {\mathrm{\mid }}^2>0$, and $A^{-1}(w)=(w-\nu \bar{w})\mathrm{/}(1-\mathrm{\mid }\nu {\mathrm{\mid }}^2)$. Thus $A$ is a real-linear homeomorphism of $\mathbb{C}$; since $\mathrm{\mid }A(z)\mathrm{\mid }\ge (1-\mathrm{\mid }\nu \mathrm{\mid })\mathrm{\mid }z\mathrm{\mid }$, it extends by $A(\mathrm{\infty })=\mathrm{\infty }$ to a sphere homeomorphism. Its affine coordinate-line restrictions are absolutely continuous with locally square-integrable derivatives. In source and target infinity charts its expression is $a(w)=w\bar{w}\mathrm{/}(\bar{w}+\nu w)$ for $w\mathrm{\ne }0$, with $a(0)=0$. The denominator has modulus at least $(1-\mathrm{\mid }\nu \mathrm{\mid })\mathrm{\mid }w\mathrm{\mid }$, so $a$ is continuous at $0$; its degree-one homogeneity and smoothness off $0$ give bounded derivatives there. Integrating along segments, splitting at $0$ if needed, makes $a$ Lipschitz. Thus its coordinate-line restrictions are absolutely continuous with locally square-integrable derivatives, and [F5] gives local $W^{1,2}$ in both charts. The pullback law [F2] transports the equation off $0$, a null point. Hence $A$ is analytically quasiconformal with coefficient $\nu$ and $K=(1+\mathrm{\mid }\nu \mathrm{\mid })\mathrm{/}(1-\mathrm{\mid }\nu \mathrm{\mid })$; [F6] gives orientation preservation. The values $A(0)=0$ and $A(1)=1+\nu$ show it is already normalized exactly when $\nu =0$.

Since $\mathrm{\mid }\nu \mathrm{\mid }<1$, $1+\nu \mathrm{\ne }0$. By [F3], there is a unique Möbius map $M$ sending $(0,1+\nu ,\mathrm{\infty })$ to $(0,1,\mathrm{\infty })$. Write $M(w)=(aw+b)\mathrm{/}(cw+d)$ with $ad-bc\mathrm{\ne }0$. The conditions $M(0)=0$ and $M(\mathrm{\infty })=\mathrm{\infty }$ force $b=c=0$, so $M(w)=\lambda w$ with $\lambda =a\mathrm{/}d\mathrm{\ne }0$; then $M(1+\nu )=1$ gives $\lambda =(1+\nu {)}^{-1}$. Thus $M(w)=w\mathrm{/}(1+\nu )$, proving (b).

Put $f^{\nu }=M\circ A$. In the finite chart, $f_z^{\nu }=1\mathrm{/}(1+\nu )$ and $f_{\bar{z}}^{\nu }=\nu \mathrm{/}(1+\nu )$, so $f_{\bar{z}}^{\nu }=\nu f_z^{\nu }$ and ${\mu }_{f^{\nu }}=\nu$. It is a sphere homeomorphism fixing $\mathrm{\infty }$; in the infinity coordinates $w=1\mathrm{/}z$ and $\eta =1\mathrm{/}{f}^{\nu }(z)$ its expression is $\eta (w)=\frac{(1+\nu )w\bar{w}}{\bar{w}+\nu w},\eta (0)=0.$ This is $(1+\nu )a(w)$, with $a$ from step 1.1, so it has the same local $W^{1,2}$ regularity by linearity of weak derivatives. The pullback law in [F2] transports the weak equation to the infinity chart, where the coefficient still has modulus $\mathrm{\mid }\nu \mathrm{\mid }$; [F5] gives the analytic quasiconformal inequality there. Thus $f^{\nu }$ is a quasiconformal sphere homeomorphism and weak solution; its three values are $0,1,\mathrm{\infty }$. By [F7] it is the unique normalized solution, proving (c). The local orientation sign remains positive by [F6].

Let $f$ be any quasiconformal homeomorphic solution. Its homeomorphism property makes $f(0),f(1),f(\mathrm{\infty })$ distinct. By [F3], the unique Möbius map carrying this triple to $(0,1,\mathrm{\infty })$ exists. By [F7], all solutions are Möbius postcompositions of $f^{\nu }$ and the normalized solution is unique; therefore the normalizing map is the inverse of the unique Möbius map in the family, and the full solution family has exactly the stated form.

Since $\lambda \mathrm{\ne }0$, $\psi (\zeta )=\lambda \zeta$ has holomorphic inverse $\zeta \mapsto \zeta \mathrm{/}\lambda$. The pullback formula [F1] gives ${\psi }^{\ast }\mu =(\mu \circ \psi )\bar{\lambda }\mathrm{/}\lambda$. Writing $\lambda =r{e}^{i\varphi }$ yields $\bar{\lambda }\mathrm{/}\lambda =e^{-2i\varphi }$, so a constant $\nu$ keeps modulus $\mathrm{\mid }\nu \mathrm{\mid }$ and, when $\nu \mathrm{\ne }0$, its phase changes by $-2\varphi$. The direction formula in [F2] therefore gives the pulled-back major-axis line at angle $\frac{1}{2}\arg \nu -\varphi (\mathrm{m}\mathrm{o}\mathrm{d}\pi )$; when $\nu =0$, [F2] says the ellipse is a circle and no direction is defined.

For ${\rho }_{\theta }(\zeta )=e^{i\theta }\zeta$, [F1] reads ${\rho }_{\theta }^{\ast }\mu =(\mu \circ {\rho }_{\theta })e^{-2i\theta }$. Equality as almost-everywhere coefficient classes is equivalent, after multiplication by the nonzero constant $e^{2i\theta }$, to $\mu (e^{i\theta }\zeta )=e^{2i\theta }\mu (\zeta )$ almost everywhere. This proves both directions of the stated equivalence.

The map $j(w)=1\mathrm{/}w$ on ${\mathbb{C}}^{\times }$ is its own holomorphic inverse. Its derivative is $j^{\mathrm{\prime }}(w)=-w^{-2}$, so $\overline{j^{\mathrm{\prime }}(w)}\mathrm{/}{j}^{\mathrm{\prime }}(w)=w^2\mathrm{/}{\bar{w}}^2$. Substitution in [F1] gives ${\mu }_0(1\mathrm{/}w)w^2\mathrm{/}{\bar{w}}^2$ on the chart overlap; [F9] makes the value at $w=0$ immaterial to the chartwise coefficient.

On each relatively compact coordinate patch, use [F4] and the weak equation to obtain $(f\circ \psi {)}_{\bar{\zeta }}=(\mu \circ \psi )(f_z\circ \psi )\overline{{\psi }^{\mathrm{\prime }}}$. The other formula in [F4] gives $(f\circ \psi {)}_{\zeta }=(f_z\circ \psi ){\psi }^{\mathrm{\prime }}$, so multiplying it by ${\psi }^{\ast }\mu =(\mu \circ \psi )\overline{{\psi }^{\mathrm{\prime }}}\mathrm{/}{\psi }^{\mathrm{\prime }}$ from [F1] gives the same expression. The $W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$ membership is also part of [F4], proving the pulled-back weak-solution claim.

By [F7] the model is measurable, and by [F8] it satisfies $\mathrm{\parallel }\mu {\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\mid }c\mathrm{\mid }<1$, so it is a Beltrami coefficient. For $\zeta \mathrm{\ne }0$, $\mu (e^{i\theta }\zeta )=c{e}^{2i\theta }\zeta \mathrm{/}\bar{\zeta }=e^{2i\theta }\mu (\zeta )$; at $\zeta =0$ both sides are $0$. Thus the covariance holds everywhere and step 1.2 gives rotational invariance.

The affine map $A(z)=z+\nu \bar{z}$ is $C^1$, and [F5] gives $A_z=1$ and $A_{\bar{z}}=\nu$, so it is a weak solution for the constant coefficient $\nu$. Directly, $(A\circ {\rho }_{\theta })(\zeta )=e^{i\theta }\zeta +\nu e^{-i\theta }\bar{\zeta }$, whose Wirtinger derivatives are $e^{i\theta }$ and $\nu e^{-i\theta }$; their ratio is $\nu e^{-2i\theta }$, as in step 1.1.

The pullback and norm formulas [F1, F3] give $\mathrm{\parallel }{\psi }^{\ast }\mu {\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }\mu {\mathrm{\parallel }}_{\mathrm{\infty }}$; since $K(\mu )=(1+\mathrm{\parallel }\mu {\mathrm{\parallel }}_{\mathrm{\infty }})\mathrm{/}(1-\mathrm{\parallel }\mu {\mathrm{\parallel }}_{\mathrm{\infty }})$, this implies $K({\psi }^{\ast }\mu )=K(\mu )$. The pointwise modulus identity also preserves each ellipse's eccentricity under coordinate pullback.

By [F2], $\mu \equiv 0$ is an admissible sphere coefficient. By [F3], $f_1(z)=z$ and $f_2(z)=1\mathrm{/}z$ are biholomorphic sphere maps. By [F4], both are $1$-quasiconformal and have Beltrami coefficient $0$; [F5] makes each a weak solution of $f_{\bar{z}}=0$.

The two maps are distinct, since $f_1(2)=2$ while $f_2(2)=1\mathrm{/}2$. Thus the refuted uniqueness statement fails for $\mu =0$. The general Möbius ambiguity and three-point uniqueness stated above are exactly the conclusions of [F6].