Pullback of a measurable ellipse field under biholomorphic maps

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.