$I_{n}:=\int\widetilde{\zeta}_{n}(\gamma_{n})d\gamma_{n}=\sum_{s\in\mathrm{Sing}^{\prime}}\operatorname{res}(s)\int\frac{d\gamma_{n}}{\gamma_{n}-s}.$

Then,

$I_{n}=\sum_{s\in\mathrm{Sing}^{\prime}}\operatorname{res}(s)\Psi_{n}(s)=\langle\tilde{m}_{n}^{\prime},\Psi_{n}\rangle$

where $\Psi_{n}(s)=\int\frac{d\gamma_{n}}{\gamma_{n}-s}$, and is a
determination of $\log\left(\frac{\gamma_{n}(1)-s}{\gamma_{n}(0)-s}\right)$
and where $\tilde{m}^{\prime}_{n}$ is the sum of Dirac masses at points
$s\in\mathrm{Sing}^{\prime}$ ponderated by $\operatorname{res}s$; note that
$\tilde{m}^{\prime}_{n}$ is also equal to $(f_{n})_{*}m_{n}$ with at most two
residudes removed.

We have for $n$ big enough

$|{\operatorname{Im}\,}\Psi_{n}|\leq C_{0}$

where $C_{0}$ is independent from $\varepsilon$ and $n$.

###### Proof.

There exists a real $m>0$ such that for all $n$,
$\min|\gamma^{\prime}_{n}|\geq m$. There exists $\eta>0$ such that for all $n$
big enough, $|t-s|\leq\eta$ $\implies$
$|\gamma^{\prime}_{n}(t)-\gamma^{\prime}_{n}(s)|\leq m/10$ (by uniform
convergence of $\gamma^{\prime}_{n}$ to a continuous function). For any
$z\in{\mathbb{C}}$, on an interval $[t_{0},t_{1}]$ of values of $t$ of size
$\eta$ (or less), the argument of $\gamma^{\prime}_{n}(t)$ deviates less than
$\operatorname{asin}(1/10)$ from its value at $t=t_{0}$ and it follows that
the variation of argument of $\gamma_{n}(t)-s$ is at most
$\pi+\operatorname{asin}(1/10)$ on $[t_{0},t_{1}]$. Slicing $[0,1]$ into at
most $\left\lceil\frac{1}{\eta}\right\rceil$ intervals of length at most
$\eta$, we get
$|{\operatorname{Im}\,}\Psi_{n}|\leq\left\lceil\frac{1}{\eta}\right\rceil(\pi+\operatorname{asin}(1/10))$.
∎

Let $\varepsilon>0$ and consider the set $G_{n}=V_{\varepsilon}(\gamma_{n})$
defined as the $\varepsilon$-neighbourhood $\gamma_{n}$. By Proposition 57,
the sets $f_{n}^{-1}(G_{n})$ are all contained in some common closed ball
$K=\overline{B}(0,R)$. By increasing $R$, we also assume that $K$ contains for
all $n$ the corners of the litte squares on which the average of $\mu$ is not
$0$. In particular, $f_{n}(K)$ contains $\operatorname{Sing}$ and $G_{n}$.

Let us write

$I_{n}=J_{n}+J^{\prime}_{n}$

where

$J^{\prime}_{n}=\sum_{s\in
G_{n}\cap\mathrm{Sing}^{\prime}}\operatorname{res}(s)\Psi_{n}(s)$
$J_{n}=\sum_{s\in T_{n}}\operatorname{res}(s)\Psi_{n}(s)$

where $T_{n}=\mathrm{Sing}^{\prime}-G_{n}$.

Note: In the rest of this proof, inequalities involving constants $C_{k}$ for
some $k\in{\mathbb{N}}$ will appear. They will be valid for all $\varepsilon$
and $n$ subjected to the conditions $\varepsilon<\varepsilon_{k}$ and
$n>n_{k}$, where $\varepsilon_{k}$ and $n_{k}$ are constants that will be
implied.

Bound on $J^{\prime}_{n}$:

We have

$\operatorname{Leb}G_{n}\leq C_{1}\varepsilon$

for some $C_{1}$, and for all $s\in\mathrm{Sing}$,

$|\operatorname{res}s|\leq C_{2}/n^{2}$

for some $C_{2}$.

Because of Lemma 58, the disks of radius $C_{3}/2n$ around the actuals
singularities are disjoint, because their preimage by $f_{n}$ lie in $K$. So
this is the case for the $s\in G_{n}\cap\mathrm{Sing}^{\prime}$. Moreover, as
soon as $C_{3}/2n<\varepsilon$ then these disks are contained in
$G_{n}(2\varepsilon)$, so the sum of the areas of these disks is at most
$2C_{1}\varepsilon$. It follows that

$\\#(G_{n}\cap\mathrm{Sing}^{\prime})\leq\frac{2C_{1}\varepsilon}{\pi(C_{3}/2n)^{2}}=C_{4}\varepsilon
n^{2}$

for some $C_{4}$.

Case 1: $\gamma_{n}(0)=\gamma_{n}(1)$. Then $\Psi_{n}$ is purely imaginary and
hence is bounded, according to one of the remarks above. Hence

$|J^{\prime}_{n}|\leq\sum_{s\in\mathrm{Sing^{\prime}}\cap
G_{n}}|\operatorname{res}(s)\Psi_{n}(s)|\leq\\#(\mathrm{Sing^{\prime}}\cap
G_{n})\times\max|\operatorname{res}s|\times\max|\Psi_{n}(s)|$ $\leq
C_{4}\varepsilon n^{2}\times\frac{C_{2}}{n^{2}}\times
C_{0}=C_{4}C_{2}C_{0}\varepsilon.$

Case 2: $\gamma_{n}(0)\neq\gamma_{n}(1)$.

We have

$|\Psi_{n}(s)|<\log\frac{1}{|\gamma_{n}(0)-s|}+\log\frac{1}{|\gamma_{n}(1)-s|}+C_{5}$

for some $C_{5}$.

###### Proof.

We have already seen that $|{\operatorname{Im}\,}\Psi_{n}(s)|\leq C_{0}$. The
real part is easier to deal with, since

${\operatorname{Re}\,}\Psi_{n}(s)=\log\frac{|\gamma_{n}(1)-s|}{|\gamma_{n}(0)-s|}=\log\frac{1}{|\gamma_{n}(0)-s|}-\log\frac{1}{|\gamma_{n}(1)-s|}.$

For any $t$,
$\left|\log\frac{1}{|\gamma_{n}(t)-s|}\right|\leq\log\frac{1}{|\gamma_{n}(t)-s|}+C$
for $C=\max(0,2\log\operatorname{diam}B)$ where $B$ is a compact set
containing all the curves $\gamma_{n}$ and the singularities of non-vanishing
residue of $\zeta_{n}$ for all $n$. The constant $C$ is independent of
$\varepsilon$ and $n$. ∎

For any singularity $s$ at distance $\geq\varepsilon$ from $\gamma_{n}(0)$ we
have

$\log\frac{1}{|\gamma_{n}(0)-s|}\leq\log\frac{1}{\varepsilon}.$

For the singularities $s\in\mathrm{Sing}^{\prime}$ at distance $<\varepsilon$
from $\gamma_{n}(0)$ we have $r_{n}\leq|s-\gamma_{n}(0)|<\varepsilon$ with
$r_{n}=\exp(-n)$. Recall that the disks $B(s,C_{3}/2n)$ are disjoint. There is
hence at most one $s_{0}\in\mathrm{Sing}^{\prime}$ with
$|s_{0}-\gamma_{n}(0)|<C_{3}/2n$ and since $|s_{0}-\gamma_{n}(0)|>e^{-n}$ we
have

$\log\frac{1}{|\gamma_{n}(0)-s_{0}|}\leq n.$

For the other ones, let $k$ be the smallest integer such that
$2^{k}\frac{C_{3}}{2n}>\varepsilon$. Recall that we assume that $n$ is big
enough so that $C_{3}/2n<\varepsilon$, hence $k>0$. We have

$2^{k}\leq\frac{4n\varepsilon}{C_{3}}.$

Let $j\in{\mathbb{N}}\cap[0,k)$. The number of $s\in\mathrm{Sing}^{\prime}$
such that $|\gamma_{n}(0)-s|$ belongs to $[2^{j}C_{3}/2n,2^{j+1}C_{3}/2n)$ is
at most $\pi\big{(}(2^{j+1}+1)^{2}-(2^{j}-1)^{2}\big{)}=\pi(3\times
4^{j}+6\times 2^{j}\leq 9\pi\times 4^{j}$, because the union of the disjoint
disks $B(s,C_{3}/2n)$ is contained in the round annulus of $z\in{\mathbb{C}}$
such that $(2^{j+1}+1)C_{3}/2n<|z-s|<(2^{j}-1)C_{3}/2n$. For such an $s$, we
have

$\log\frac{1}{|\gamma_{n}(0)-s|}\leq\log\frac{2n}{2^{j}C_{3}}.$

Hence

$\sum_{s\in\mathrm{Sing^{\prime}}\cap
B(\gamma_{n}(0),\varepsilon)}\log\frac{1}{|\gamma_{n}(0)-s|}\leq
n+\sum_{j=0}^{k-1}9\pi\times 4^{j}\times\log\frac{2n}{2^{j}C_{3}}$

Now

$\sum_{j=0}^{k-1}4^{j}\times\log\frac{2n}{2^{j}C_{3}}=4^{k}\sum_{j=0}^{k-1}\frac{1}{4^{k-j}}\log
2^{k-j}\frac{2n}{2^{k}C_{3}}=4^{k}\sum_{p=1}^{k}\frac{1}{4^{p}}\left(p\log
2+\log\frac{2n}{2^{k}C_{3}}\right)$ $\leq
4^{k}\sum_{p=1}^{k}\frac{1}{4^{p}}\left(p\log
2+\log\frac{1}{\varepsilon}\right)\leq 4^{k}\left(\frac{4}{9}\log
2+\frac{1}{3}\log\frac{1}{\varepsilon}\right)$
$\leq\left(\frac{4n\varepsilon}{C_{3}}\right)^{2}\times\left(\frac{4}{9}\log
2+\frac{1}{3}\log\frac{1}{\varepsilon}\right)$

Hence

$\sum_{s\in\mathrm{Sing^{\prime}}\cap
B(\gamma_{n}(0),\varepsilon)}\log\frac{1}{|\gamma_{n}(0)-s|}\leq
n+C_{6}n^{2}\varepsilon^{2}+C_{7}n^{2}\varepsilon^{2}\log\frac{1}{\varepsilon}.$

Let us put it all together (assuming $\varepsilon\leq 1$): using
$\max|\operatorname{res}s|\leq C_{2}/n^{2}$ we have

$|J^{\prime}_{n}|\leq\frac{C_{2}}{n^{2}}\left(\sum\log\frac{1}{|\gamma_{n}(0)-s|}+\sum\log\frac{1}{|\gamma_{n}(1)-s|}+\sum
C_{5}\right);$

separating the sum of $\log\frac{1}{|\gamma_{n}(0)-s|}$ according to whether
or not $|\gamma_{n}(0)-s|\geq\varepsilon$, and then regrouping the terms and
using that there are at most $C_{4}\varepsilon n^{2}$ values of $s$ as we saw
earlier:

$|J^{\prime}_{n}|\leq\frac{C_{2}}{n^{2}}\left(C_{4}\varepsilon
n^{2}\times\Big{(}2\log\frac{1}{\varepsilon}+C_{5}\Big{)}+n+C_{6}n^{2}\varepsilon^{2}+C_{7}n^{2}\varepsilon^{2}\log\frac{1}{\varepsilon}\right)$
$=C_{2}\times\left(C_{4}\varepsilon\times\Big{(}2\log\frac{1}{\varepsilon}+C_{5}\Big{)}+\frac{1}{n}+C_{6}\varepsilon^{2}+C_{7}\varepsilon^{2}\log\frac{1}{\varepsilon}\right).$

Hence

$\limsup|J^{\prime}_{n}|\leq C_{8}\varepsilon\log\frac{1}{\varepsilon}$

for some $C_{8}$. The important fact is that this bound tends to $0$ as
$\varepsilon\longrightarrow 0$.

Bound on $J_{n}$:

Similarly to the definition of $G_{n}$ as the $\varepsilon$-neighbourhood of
$\gamma_{n}$, we let $G$ be the $\varepsilon$-neighbourhood of $\gamma$. For
$s\notin\gamma([0,1])$, let $\Psi(s)=\int\frac{d\gamma}{\gamma-s}$, which is
also the limit of the functions $\Psi_{n}$. Similarly as for $G_{n}$ we have
that $|{\operatorname{Im}\,}\Psi|\leq C_{0}$ and

${\operatorname{Re}\,}\Psi(s)=\log\frac{|\gamma(1)-s|}{|\gamma(0)-s|}=\log\frac{1}{|\gamma(0)-s|}-\log\frac{1}{|\gamma(1)-s|}.$

Note that $|\Psi(z)|$ and $|\Psi_{n}(z)|$ tend to $0$ as
$|z|\longrightarrow+\infty$. There exists a sequence $\tilde{\Psi}_{n}$ of
continuous extensions of the restriction of $\Psi_{n}$ to the closed set
${\mathbb{C}}-G_{n}$ and a continuous extension $\tilde{\Psi}$ of the
restriction of $\Psi$ to the closed set ${\mathbb{C}}-G$, such that moreover
$\tilde{\Psi}_{n}$ tends to $\tilde{\Psi}$ uniformly on compact subsets of
${\mathbb{C}}$ and such that
$\|\tilde{\Psi}\|_{\infty}=\|\Psi|_{{\mathbb{C}}-G}\|_{\infty}$. This follows
from the Tietze extension theorem (Theorem 35.1 page 219 of [Mun00]). Apply it
first to extend $\Psi|_{{\mathbb{C}}-G}$ into $\tilde{\Psi}$. Then apply it
again to the metric space
$X=(\\{0\\}\cup\big{\\{}{1/n}\,\big{|}\,{n\in{\mathbb{N}}^{*}}\big{\\}})\times{\mathbb{C}}$,
the continuous function that is the disjoint union of $\tilde{\Psi}$ and the
restrictions of the $\Psi_{n}$ (the disjoint union of the domains of these
maps is indeed a closed subset of $X$).

Now

$J_{n}-\int\tilde{m}^{\prime}_{n}\tilde{\Psi}_{n}=\sum_{s\in
G_{n}\cap\mathrm{Sing}^{\prime}}\operatorname{res}(s)\tilde{\Psi}_{n}(s)$

Hence

$\left|J_{n}-\int\tilde{m}^{\prime}_{n}\tilde{\Psi}_{n}\right|\leq\max_{s\in\mathrm{Sing}^{\prime}}|\operatorname{res}(s)|\times\\#(G_{n}\cap\mathrm{Sing}^{\prime})\times\max_{G_{n}}\big{|}\tilde{\Psi}_{n}\big{|}$
$\leq\frac{C_{2}}{n^{2}}\times
C_{4}n^{2}\varepsilon^{2}\times\max_{n}\max_{G_{n}}|\tilde{\Psi}_{n}|\leq
C_{7}\varepsilon^{2}$

for some $C_{7}$. Also,

$\left|\int\tilde{m}^{\prime}_{n}\tilde{\Psi}_{n}-\int\tilde{m}^{\prime}_{n}\tilde{\Psi}\right|\leq\|\tilde{m}^{\prime}_{n}\|\|\tilde{\Psi}_{n}-\tilde{\Psi}\|_{\infty}$

and as $n\longrightarrow+\infty$, the left factor stays bounded and the right
one tends to $0$. Let $m^{\prime}=f_{*}m$. By Corollary 52 (removing two
singularities does not change the convergence, since the total mass removed is
$\cal O(1/n^{2})$) we have

$\int\tilde{m}^{\prime}_{n}\tilde{\Psi}\longrightarrow\int
m^{\prime}\tilde{\Psi}$

when $n\longrightarrow+\infty$. Recall that
$\|\tilde{\Psi}\|_{\infty}=\|\Psi|_{{\mathbb{C}}-G}\|_{\infty}$. Since $\Psi$
is a branch of $\log\left(\frac{\gamma(1)-s}{\gamma(0)-s}\right)$, and since
on ${\mathbb{C}}-G$ the distance to $\gamma$ is at least $\varepsilon$, we get
for all $s\in{\mathbb{C}}-G$: $|{\operatorname{Re}\,}\Psi(s)|\leq
C_{8}+\log\frac{1}{\varepsilon}$ for some $C_{8}>0$. Since, as we saw,
$|{\operatorname{Im}\,}\Psi|\leq C_{0}$ we get

$\|\tilde{\Psi}\|_{\infty}\leq C_{0}+C_{8}+\log\frac{1}{\varepsilon}.$

Now $m=h\operatorname{Leb}$ where $h$ is a continuous function so
$m^{\prime}=f_{*}m=g\operatorname{Leb}$ with $g(f(z))=h(z)/\det Df$ is
continuous too. In particular $g$ is bounded, say by $C_{9}$. We have
$\operatorname{Leb}G\leq C_{1}\varepsilon$, like for $G_{n}$. It follows that

$\left|\int_{G}m^{\prime}\tilde{\Psi}\right|\leq
C_{9}\varepsilon\log\frac{1}{\varepsilon}$

for some $C_{9}$, and this quantity tends to $0$ as
$\varepsilon\longrightarrow 0$. Concerning the original function $\Psi$, from
the expression of ${\operatorname{Re}\,}\Psi$ and the bound on
${\operatorname{Im}\,}\Psi$, we get

$\left|\int_{G}m^{\prime}\Psi\right|\leq
C_{10}\varepsilon+C_{9}\varepsilon^{2}\log\frac{1}{\varepsilon}.$

If we collect all the results, we know that for all $\eta>0$, by taking
$\varepsilon$ small enough, then for all $n$ big enough, the quantity $I_{n}$
to be evaluated, differs from $\int_{\mathbb{C}}m^{\prime}\Psi$ by at most
$\eta$, i.e.

$\left|I_{n}-\int_{\mathbb{C}}m^{\prime}\Psi\right|<\eta.$

This integral $\int_{\mathbb{C}}m^{\prime}\Psi$ is equal to
$\int\zeta(\gamma(t))\gamma^{\prime}(t)dt$: indeed, recall that
$\zeta=m^{\prime}*\frac{1}{z}$, whence

$\int\zeta(\gamma(t))\gamma^{\prime}(t)dt=\int_{t\in[0,1]}\left(\int_{s\in{\mathbb{C}}}\frac{1}{\gamma(t)-s}dm^{\prime}(s)\right)\gamma^{\prime}(t)dt$
$=\int_{s\in{\mathbb{C}}}\left(\int_{t\in[0,1]}\frac{d\gamma(t)}{\gamma(t)-s}\right)dm^{\prime}(s)=\int_{\mathbb{C}}m^{\prime}\Psi.$

The permutation in the integral is justified because the integral of the
absolute value

$\int_{s\in{\mathbb{C}}}\frac{1}{|z-s|}dm^{\prime}(s)=\int_{s\in{\mathbb{C}}}\frac{g(s)}{|z-s|}d\operatorname{Leb}(s)$

is convergent ($g$ is bounded with compact support, $z\mapsto 1/|z|$ is
$L^{1}$ on any compact set) and bounded for $z\in\gamma([0,1])$, so we are in
the conditions of the Fubini Tonelli theorem. ∎

######  Remark.

In the statement of Theorem 5, we decided to restrict to pathes avoiding the
singularities and to remove the 0, 1 or 2 singularities that are too close to
the extremities of the path. We could have proceeded differently. For instance
we could have allowed for every paths by either:

  * $-$

removing the 0, 1 or 2 aforementioned singularities, and for every other
singularity $s$ crossed by $\gamma_{n}$, give a meaning to
$\operatorname{res}(s)\int\frac{d\gamma_{n}}{\gamma_{n}-s}$ if
$\gamma_{n}^{\prime}$ is continuous at all $t$ for which $\gamma_{n}(t)=s$, or
if the jump of $\arg\gamma_{n}^{\prime}$ at $t$ is smaller than, say, $\pi/2$;

  * $-$

chopping off $\zeta_{n}$ close to some/all poles, for instance by setting
$\zeta_{n}(z)=0$ when $\exists s$ singularity such that
$|z-s|<r^{\prime}_{n}$, for an appropriate sequence $r^{\prime}_{n}$;

  * $-$

there are several choice in $r^{\prime}_{n}$ and the way $\zeta_{n}$ is
chopped off; for instance one can arrange so that $\zeta_{n}$ remains
continuous;

  * $-$

one could instead subtract all the poles of $\zeta$ that are at distance
$<r^{\prime}_{n}$ to $\gamma$, for an appropriate sequence $r^{\prime}_{n}$.

Other approaches are possible.

Interpreting the limit

Let us sum up what we have done up to now. Recall the similarity surfaces
$\cal S_{n}$ associated to our approximating Beltrami forms $\mu_{n}$, and
recall their completions $\cal S^{\prime}_{n}$ into a Riemann surface
isomorphic to $\hat{\mathbb{C}}$. To this global chart $\hat{\mathbb{C}}$ we
associated a singular connection, of symbol $\zeta_{n}$ that is a meromorphic
function444444… A rational map in fact, since for each $n$ there are only
finitely many non-erasable poles. all whose poles are all simple, but get more
and more densely packed as $n$ tends to infinity. We proved that $\zeta_{n}$
has a weak limit454545In the previous section, Theorem 5, we improved the weak
limit statement into a statement in terms of limit of associated parallel
transport. and that this weak limit is the function $\zeta$ introduced in
Proposition 53 via a convolution product.

######  Theorem 6.

Let $(e_{1},e_{2})$ be the canonical basis of ${\mathbb{R}}^{2}$. Let $A$ be
the push-forward by $f$ of the constant vector field $e_{1}$ of
${\mathbb{R}}^{2}$, and $B$ the push-forward of the constant vector field
$e_{2}$. Note that $A$ and $B$ commute. Consider the conformal symmetric
connection of symbol

$\tilde{\zeta}(z)=-\frac{\partial_{A}B}{AB}$

defined on $U$. Then $\tilde{\zeta}$ is equal to the symbol $\zeta$ introduced
in Proposition 53:

$\tilde{\zeta}=\zeta.$

The connection of symbol $\tilde{\zeta}$ expresses as

$\nabla_{X}Y=\partial_{X}Y-\partial_{B}A\frac{XY}{AB}.$

Before we prove the theorem, let us introduce an easy lemma.

######  Lemma 59.

Let $A$ and $B$ be two vector fields in an open subset
$U\subset{\mathbb{R}}^{2}\equiv{\mathbb{C}}$. Assume that they vanish nowhere.
Then there is a unique conformal symmetric connection $\nabla$ such that the
parallel transport under integral lines of $A$ leaves $B$ invariant. The
symbol of $\nabla$ in the chart $U$ is

$-\frac{\partial_{A}B}{AB}.$

###### Proof.

A conformal symmetric connection $\nabla$ takes the form

$\nabla_{X}Y=\partial_{X}Y+\zeta XY$

where $\zeta$ is a continuous function from $U$ to ${\mathbb{C}}$. The
condition expresses as the pair the equation

$\nabla_{A}B=0,$

i.e.

$\partial_{A}B+\zeta AB=0.$

∎

Incidentally, if $A$ and $B$ commute, then

$-\frac{\partial_{A}B}{AB}=-\frac{\partial_{B}A}{AB},$

i.e. the unique conformal symmetric connection such that the parallel
transport under integral lines of $A$ leaves $B$ invariant coincides with the
unique conformal symmetric connection such that the parallel transport under
integral lines of $B$ leaves $A$ invariant.

We will also need the following, harder to get, result. Let, as earlier,

$r_{n}=\exp(-n)$

and recall that $f$ is $C^{1}$ (Lemma 54).

######  Proposition 60.

$\sup_{\operatorname{dist}(z,\frac{1}{n}{\mathbb{Z}})>r_{n}}\|D_{z}f_{n}-D_{z}f\|\underset{n\to\infty}{\longrightarrow}0$

where the norm is any operator norm on the set of ${\mathbb{R}}$-linear self-
maps of ${\mathbb{C}}$.

###### Proof.

In Section 8. ∎

###### Proof of Theorem 6.

By Lemma 59, to prove the claim, it is enough to prove that for $\zeta$, the
associated parallel transport along integral lines of $A$ leave $B$ invariant,
where $A=f_{*}e_{1}$ and $B=f_{*}e_{2}$.

We will use the fact that this parallel transport is the limit of the parallel
transport associated to $\zeta_{n}$ (Theorem 5). Let
$\gamma:[0,1]\to{\mathbb{C}}$ defined by

$\gamma(z)=f(x_{0}+(x_{1}-x_{0})t+iy_{0})$

for some $\ell>0$ and let for $n$ big enough

$\gamma_{n}(t)=f_{n}(x_{0}+(x_{1}-x_{0})t+iy_{n})$

where $y_{n}$ is chosen so that $y_{n}\longrightarrow y_{0}$ and
$d(y_{n},\frac{1}{n}{\mathbb{Z}})>r_{n}$: for instance $y_{n}=(0.5+\lfloor
ny_{0}\rfloor)/n$.

The path $\gamma$ is $C^{1}$ and the path $\gamma_{1}$ is $C^{1}$ by parts. By
Section 5.4, $\gamma_{n}\longrightarrow\gamma$ uniformly, and by Proposition
60 $\gamma_{n}^{\prime}\longrightarrow\gamma^{\prime}$ uniformly.

$f_{n}$ Figure 16. Illustration of an argument in the proof of Proposition 60.

Now, recall that the parallel transport for $\zeta_{n}$ is also the parallel
transport associated to the similarity surface $\cal S_{n}$ built from gluing
together the images of the small squares by maps of the form $az+b\bar{z}$
where $a$ and $b$ are any complex constants such that $b/a\in{\mathbb{D}}$ is
the value of $\mu_{n}$ on this square. The image of such a square is a
parallelogram The horizontal line $x_{0}+(x_{1}-x_{0})t+iy_{n}$ runs through a
row of those squares. Consider the corresponding parallelograms. We can place
the initial parallelogram so that the image of the vertical edge of the square
is a vertical edge with the same orientation and same size $1/n$. Then its
opposite side is parallel and has the same size. If we scale rotate and place
the other parallelograms in this row so as to match this side, all the images
of the vertical sides remain vertical with the same length $1/n$. See Figure
16. One deduces that if we define the vector $v_{n}$ as the image by $f_{n}$
of the vertical unit vector $e_{2}$ attached to $x_{0}+iy_{n}$, then the
parallel transport by $\nabla_{\zeta_{n}}$ of $v_{n}$ along $\gamma_{n}$
remains for all $t$ the image $v_{n}(t)$ by $f_{n}$ of the vertical vector
$e_{2}$ attached to $x_{0}+(x_{1}-x_{0})t+iy_{n}$. Now by Proposition 60 and
continuity of $Df$, for each fixed $t$, the vector $v_{n}(t)$ tends as
$n\longrightarrow+\infty$ to the push-forward $v(t)$ by $f$ of $e_{2}$
attached to $x_{0}+(x_{1}-x_{0})t+iy_{0}$. Note that $v(t)=B(\gamma(t))$.

We can now invoke Theorem 5: at the limit, the parallel transport along
$\gamma$ of $v(0)=B(\gamma(0))$ is the vector $v(t)=B(\gamma(t))$. ∎

Unlike the case of $\tilde{\zeta}$, we have not been able to give an
interesting expression of the curvature form
$\omega=\bar{\partial}\tilde{\zeta}$ in terms of $A$ and $B$: a direct
application of $\bar{\partial}$ to the expression $(\partial_{B}A)/AB$ seems
to give rise to a complicated expression.

##  8\. On averaging the Beltrami form

In this article, up to now we used the average of the complex-valued function
$\mu(z)$, assumed $L^{\infty}$ in LABEL:part:beltrami or $C^{2}$ in
LABEL:sec:sslim, over the Lebesgue measure on squares. But the Beltrami
differential is just one way to represent ellipse fields. We could have taken
another quantity $\nu=\nu(\mu)$, like $(1+\mu)/(1-\mu)$, or even functions of
$\mu$ that are not holomorphic like $\frac{1+|\mu|}{1-|\mu|}e^{i\arg\mu}$.

So, what if we averaged $\nu$ instead of $\mu$? In other words, given $n$, we
take $\mu_{n}$ on each small square $S$ in the construction to be constant
equal to $\nu^{-1}\left(\frac{\int\nu\circ\mu}{\int 1}\right)$, where the
integrals are still against the Lebesgue measure on $S$. Note that the
quantity $\frac{\int\nu\circ\mu}{\int 1}$ belongs to the convex hull of the
image of $\nu$.

We assume that $\mu\mapsto\nu(\mu)$ is a $C^{1}$-diffeomorphism from the unit
disk to a convex open subset of ${\mathbb{R}}^{2}$ and ask two questions:

  1. (1)

Does the sequence $f_{n}$ of normalized straightenings of $\mu_{n}$ still
converge to the straightening of $\mu$?

  2. (2)

Assuming that $\nu$ is $C^{2}$, in the setting of LABEL:sec:sslim, are the
limit complex measure $m$, curvature form $\omega$ and Christoffel symbol
$\zeta$ the same, and does the parallel transport still converge?

The answer to both questions is: yes.

For point (1), the proof of convergence (Section 5.4) relied on uniqueness of
the straightening of $\mu$ (Theorem 3) and on the sole facts that
$\|\mu_{n}\|_{\infty}$ is bounded away from $1$ and that locally, the $L^{1}$
norm of $\mu_{n}-\mu$ tends to $0$ (Lemma 40), which still hold: indeed
$\nu\circ\mu_{n}$ is the average of the bounded measurable function
$\nu\circ\mu$ on the little squares of the $n$-the generation; using the same
proof as in Lemma 40 we get that for all compact subset $S$ of ${\mathbb{C}}$,
$\|\nu\circ\mu_{n}-\nu\circ\mu\|_{L^{1}(S)}\longrightarrow 0$ as
$n\longrightarrow+\infty$. The set $\nu\circ\mu({\mathbb{C}})$ in
${\mathbb{C}}$ is compactly contained in the image of $\nu$ hence its convex
$C$ hull too. It follows that $\|\mu_{n}\|_{\infty}$ is bounded away from $1$.
It also follows that the map $\nu^{-1}$ is uniformly Lipschitz on $C$, hence
$\|\mu_{n}-\mu\|_{L^{1}(S)}=\|\nu^{-1}\circ\nu\circ\mu_{n}-\nu^{-1}\circ\nu\circ\mu\|_{L^{1}(S)}\leq
C\|\nu\circ\mu_{n}-\nu\circ\mu\|_{L^{1}(S)}$ for some constant $C>0$
independent of $n$.

For point (2), recall that $\mu$ is in this case also $C^{2}$, hence
continuous. So in four squares around a corner, $\mu$ takes values that remain
close to the value $\mu_{0}$ of $\mu$ at the corner. Below, all the $o(\cdot)$
are uniform in $|m|$ small enough, $\mu_{0}$ subject to $|\mu_{0}|<\kappa<1$
and $z_{0}\in{\mathbb{C}}$. Let

$\nu(\mu_{0}+m)=\nu_{0}+Q_{1}(m)+Q_{2}(m)+o(|m|^{2})$

be an expansion where $Q_{k}$ is a degree-$k$ ${\mathbb{R}}$-homogeneous
polynomial in $({\operatorname{Re}\,}m,{\operatorname{Im}\,}m)$ taking values
in ${\mathbb{C}}$ seen as a dimension 2 ${\mathbb{R}}$-vector space. Let

$\mu(z_{0}+z)=\mu_{0}+P_{1}(z)+P_{2}(z)+o(|z|^{2})$

with a similar convention. Note that $P_{1}$ and $Q_{1}$ are
${\mathbb{R}}$-linear endomorphisms of ${\mathbb{C}}$ and that $Q_{1}$ is
invertible by hypothesis that $\nu$ is a diffeomorphism. Then

$\nu^{-1}(\nu_{0}+v)=\mu_{0}+Q_{1}^{-1}(v)-Q_{1}^{-1}\circ Q_{2}\circ
Q_{1}^{-1}(v)+o(|v|^{2}),$

and

$\nu\circ\mu(z_{0}+z)=\nu_{0}+Q_{1}\circ P_{1}(z)+Q_{1}\circ
P_{2}(z)+Q_{2}\circ P_{1}(z)+o(|z|)^{2}.$

Let $M_{j}$ be the average of $\mu$ on the squares $C_{j}$ of side length
$\varepsilon$ appearing in the proof of LABEL:prop:lim-mes. We have

$M_{j}=\mu_{0}+P_{1}\big{(}\operatorname{\underset{C_{j}}{\operatorname{avg}}}z\big{)}+\operatorname{\underset{C_{j}}{\operatorname{avg}}}\big{(}P_{2}(z)\big{)}+o(\varepsilon^{2})$

Denote $N_{j}$ the average of $\nu\circ\mu$ on $C_{j}$. We get

$N_{j}=\nu_{0}+Q_{1}\circ
P_{1}\big{(}\operatorname{\underset{C_{j}}{\operatorname{avg}}}z\big{)}+Q_{1}\Big{(}\operatorname{\underset{C_{j}}{\operatorname{avg}}}P_{2}(z)\Big{)}+\operatorname{\underset{C_{j}}{\operatorname{avg}}}\big{(}Q_{2}\circ
P_{1}(z)\big{)}+o(\varepsilon^{2})$

Let $M_{j}^{\prime}=\nu^{-1}(N_{j})$. Composing the previous formula with the
expansion of $\nu^{-1}$ and comparing to $M_{j}$ we get

$M^{\prime}_{j}=M_{j}+Q_{1}^{-1}\Big{(}\operatorname{\underset{C_{j}}{\operatorname{avg}}}\big{(}Q_{2}\circ
P_{1}(z)\big{)}-Q_{2}\circ
P_{1}\big{(}\operatorname{\underset{C_{j}}{\operatorname{avg}}}z\big{)}\Big{)}+o(\varepsilon^{2})$

We have $Q_{2}\circ P_{1}(x+iy)=a_{xx}x^{2}+2a_{xy}xy+a_{yy}y^{2}$ for some
coefficients $a_{\ldots}\in{\mathbb{C}}={\mathbb{R}}+i{\mathbb{R}}$. Then

$\operatorname{\underset{C_{j}}{\operatorname{avg}}}\big{(}Q_{2}\circ
P_{1}(z)\big{)}-Q_{2}\circ
P_{1}\big{(}\operatorname{\underset{C_{j}}{\operatorname{avg}}}z\big{)}=(a_{xx}+a_{yy})\left(\frac{1}{3}-\frac{1}{4}\right)\varepsilon^{2}=B\varepsilon^{2}$

where $B\in{\mathbb{C}}$ depends on $z_{0}$. So

$M^{\prime}_{j}=M_{j}+Q_{1}^{-1}(B)\varepsilon^{2}+o(\varepsilon^{2}).$

Note that the term $Q_{1}^{-1}(B)\varepsilon^{2}$ depends on $z_{0}$ but is
independent of $j$. Then the monodromy factor of the singularity for the new
scheme is

$\Lambda^{\prime}=\frac{L^{\prime}_{1}\times
L^{\prime}_{3}}{L^{\prime}_{0}\times L^{\prime}_{2}}$

where $L^{\prime}_{j}=(1+M^{\prime}_{j})(1-M^{\prime}_{j})$, whereas for the
original one it is

$\Lambda=\frac{L_{1}\times L_{3}}{L_{0}\times L_{2}}$

where $L_{j}=(1+M_{j})(1-M_{j})$. Recall that $1+M_{j}$ is close to
$1+\mu_{0}$ when $\varepsilon$ is close to $0$. A computation gives

$\frac{L^{\prime}_{j}}{L_{j}}=1+Q_{1}^{-1}(B)\left(\frac{1}{1+\mu_{0}}+\frac{1}{1-\mu_{0}}\right)\varepsilon^{2}+o(\varepsilon^{2})=1+B^{\prime}\varepsilon^{2}+o(\varepsilon^{2})$

Where $B^{\prime}$ depends on $z_{0}$ but not on $j$. As a consequence,

$\frac{\Lambda^{\prime}}{\Lambda}=1+o(\varepsilon^{2}).$

Hence the limit of $\log_{p}\Lambda^{\prime}/\varepsilon^{2}$ is the same as
the limit of $\log_{p}\Lambda/\varepsilon^{2}$. One can check that the
convergence is still uniform. The proof of Proposition 53 adapts verbatim to
the new situation so we still have

$\zeta_{n}\xrightharpoonup{\phantom{m}}\zeta:=(f_{*}m)*\frac{1}{z}$

for the new $\zeta_{n}$, where $\zeta$ and $m$ stay the same. In Lemma 58 one
argument is that $\sup_{\mathbb{C}}|\hat{\mu}_{n}|\leq M/n$ for some $M>0$. It
is based on the fact that $\|\mu_{n}-\mu\|_{\infty}=\cal O(1/n)$, which is
still true here ($\nu\circ\mu$ is $C^{2}$, hence $C^{1}$, with compact
support, so its variation on a square of side $1/n$ is $\cal O(1/n)$; hence
its average will differ from the central value by at most $\cal O(1/n)$ (in
fact we have better than that); the preimage of this average by $\nu$ also
differs by the same order since everything takes place in a compact subset of
the domain of $\nu^{-1}$ which is at least $C^{1}$). Several of the arguments
of Theorem 5, are based on the fact that that the residues are $\cal
O(1/n^{2})$, and this is still the case here. Also, we have seen that $f_{n}$
still tends to $f$. So Point (2) holds.

Non-holomorphy

On the other hand, in the case of holomorphic families of Beltrami forms, if
$\mu\mapsto\nu(\mu)$ is not assumed holomorphic, we lose the holomorphic
dependence, with respect to the parameter, of $\mu_{n}$. For instance let
$\nu(x+iy)=x+x^{3}+iy$ and $\mu(x+iy)=tx$ for $|x+iy|<0.1$ where $t=u+iv$ is a
complex parameter in ${\mathbb{D}}$ (choose any $C^{2}$ extension for
$|x+iy|>0.1$). Then when $n>100$, near $0$ we have on the square
${\operatorname{Re}\,}z\in[0,1/n]$, ${\operatorname{Im}\,}z\in[0,1/n]$ that
$\mu_{n}=P^{-1}(\frac{u}{2n}+\frac{u^{3}}{4n^{3}})+\frac{v}{2n}i$ where
$P(x)=x+x^{3}:{\mathbb{R}}\to{\mathbb{R}}$. This quantity does not depend
holomorphically on $u+iv$.

This has the consequence that we also loose holomorphic dependence in the
parameter of the normalized straightening $f_{n}$ of $\mu_{n}$. Indeed, if
$f_{n}$ depends holomorphically on $t$, then we prove below that
$\mu_{n}=\bar{\partial}f_{n}/\partial f_{n}$ also depends holomorphically on
$t$, leading to a contradiction.

In fact we will prove a slightly more general lemma, probably already known:

######  Lemma 61.

Let $t\in B(0,\varepsilon)\to f_{t}$ be a family of normalized
$K$-quasiconformal homemomorphisms of ${\mathbb{C}}$ for some $K>1$. Assume
that for all $z$, the map $t\mapsto f_{t}(z)$ is holomorphic. Then the
Beltrami differential $\mu_{t}$ of $f_{t}$, as an element of the Banach space
$\cal B=L^{\infty}({\mathbb{C}})$, depends analytically on $t$ in the sense of
Proposition 47.

###### Proof.

It is enough to prove holomorphic dependence near $t=0$. We fix some ball $B$
in ${\mathbb{C}}$ that we will allow to grow near the end of the proof. For
all test function $\tau$ we have $\langle f_{t},\tau\rangle$ that depends
holomorphically too ($f_{t}$ satisfies local bounds that are uniform w.r.t.
$t$ varying in a compact set). Applying this to a test function that is a
partial derivative of another test function, we get that criterion (5) of
Proposition 46 is satisfied for the Banach space $L^{1}(B)$ (a bound on
$Df_{t}$ in $L^{1}(B)$ is needed: it follows for instancee from the $L^{2}$
bound of Lemma 43). It follows that the distribution derivatives of $f_{t}$,
seen as $L^{1}$ functions on $B$, are holomorphic w.r.t. $t$ in the sense of
criterion (1) of Proposition 46, and so are $\partial f_{t}$ and
$\bar{\partial}f_{t}$: there are $a_{n}$, $b_{n}$ in $L^{1}(B)$ and some $r>0$
such that for all $t$ with $|t|<r$, $\partial f_{t}=\sum a_{n}t^{n}$ and
$\bar{\partial}f_{t}=\sum b_{n}t^{n}$ in the sense of distributions, with
$\sum\left(\int_{B}|a_{n}|\right)r^{n}<+\infty$ and
$\sum\left(\int_{B}|b_{n}|\right)r^{n}<+\infty$. Consider representatives
$A_{n}$ and $B_{n}$ of $a_{n}$ and $b_{n}$. There is a subset $E$ of $B$ of
Lebesgue measure $0$ for which, for all $z\in B-E$ the sums
$\sum|A_{n}(z)|r^{n}$ and $\sum|A_{n}(z)|r^{n}$ converge. For those $z$, the
functions $t\mapsto A(z,t)=\sum A_{n}(z)t^{n}$ and $B(z,t)=\sum B_{n}(z)r^{n}$
are holomorphic. We define $A(z,t)=0$ and $B(z,t)=0$ when $z\in E$. The
functions $A$ and $B$ are measurable. The partial sums of the series $\sum
a_{n}t^{n}$ tend to function $z\mapsto A(z,t)$ in $L^{1}(B)$ so the latter is
a representative of $\sum a_{n}t^{n}$. A similar statement holds for $\sum
b_{n}t^{n}$. Then for all $t$, $z\mapsto A(z,t)$ is a representative of
$\partial f_{t}$ and $z\mapsto B(z,t)$ is a representative of
$\bar{\partial}f_{t}$. Moreover we know by properties of quasiconformal maps
(see LABEL:sub:prelim) that for all $t$, $\partial f_{t}$ is non-zero on the
complement of a set (that depends a priori on $t$) of measure $0$. So the set
of $z$ such that the function $t\mapsto A(z,t)$ is identically equal to $0$
can only have measure $0$. So the function $B/A$ is meromorphic in $t$ for
almost every $z$. For all $t$ with $|t|<r$, the function $z\mapsto
B(z,t)/A(z,t)$ is the quotient of representatives of respectively
$\bar{\partial}f_{t}$ and $\partial f_{t}$, so it is almost everywhere equal
to $\mu_{t}$, which is bounded by some $\kappa<1$. By measurability,
$|B/A|>\kappa$ defines a measurable set, hence for almost every $z$, the
inequality $|B/A|\leq\kappa$ holds for almost every $t$. In particular the
meromorphic function $t\mapsto B(z,t)/A(z,t)$ has all its poles removable. Let
us extend this function into a holomorphic function of $t\in B(0,r)$. Then the
extended function is measurable in $(z,t)$ and can serve as the function
$\mu_{t}(z)$ in criterion (3) of Proposition 47. ∎

Appendix K Proof of Proposition 60

Outside a uniform ball $\overline{B}(0,R^{\prime})$, the maps $f$ and $f_{n}$
are all holomorphic, injective, and tend to infinity at infinity. In
particular their derivatives $f_{n}^{\prime}(z)$ and $f^{\prime}(z)$ all have
a limit as $z\to\infty$. Moreover $f_{n}$ tends to $f$ uniformly on compact
subsets of ${\mathbb{C}}$, according to Section 5.4. It follows, by properties
of analytic maps, that $f_{n}^{\prime}$ tends to $f_{n}$ uniformly on
${\mathbb{C}}-\overline{B}(0,R^{\prime}+1)$. So the claim is proved near
infinity and there remains to prove it on $\overline{B}(0,R^{\prime}+1)$.

Consider as in Lemma 58

$\widehat{f}_{n}=f_{n}\circ f^{-1}.$

Since $f$ is a $C^{1}$ diffeomorphism, it is enough to prove the following
claim: for all $\varepsilon>0$, and all compact subset $K$ of ${\mathbb{C}}$,

$\sup_{\scriptstyle z\in
K,\,\operatorname{dist}(z,\operatorname{Sing})>\varepsilon
r_{n}}\|D_{z}\widehat{f}_{n}-\operatorname{Id}\|\underset{n\to\infty}{\longrightarrow}0,$

where we recall that $\operatorname{Sing}$ is the set of actual singularities
of $\zeta_{n}$ (with $\infty$ omitted). We have the following stronger version
of Lemma 58:

######  Lemma 62.

For all compact subset $K$ of ${\mathbb{C}}$ and any sequence
$\varepsilon_{n}$ satisfying

$\log 1/\varepsilon_{n}=o(n)$

there exists $b_{n}\longrightarrow 0$ such that for all $n$ big enough, for
any two points $u$, $v$ in $K$, satisfying $|u-v|\geq\varepsilon_{n}$ we have

$\left|\frac{\widehat{f}_{n}(v)-\widehat{f}_{n}(u)}{v-u}-1\right|\leq b_{n}.$

###### Proof.

Using the notation of Lemma 58 consider again the function $\Phi:\lambda\in
B(0,R_{n}/2)\mapsto\widehat{f}_{n}(\lambda,v)-\widehat{f}_{n}(\lambda,u)\in
B^{\prime}=B(0,2R)-\\{0\\}$ and recall that $R_{n}\geq n/M$. We still have
that the hyperbolic distance in $B(0,2R)-\\{0\\}$ from $\Phi(0)=v-u$ to
$\Phi(1)=v_{n}-u_{n}$ is at most the hyperbolic distance in $B(0,R_{n}/2)$
from $0$ to $1$, which is at most $\operatorname{cst}/n$. Let
$\Phi(0)=re^{i\theta}$ and $\tau=\log(2R/r)>0$. We now use the fact that the
map $z\in{\mathbb{H}}_{+}\mapsto 2R\exp(i\theta-\tau z)\in B^{\prime}$
preserves the respective infinitesimal hyperbolic metrics and sends $1$ to
$\Phi(0)$. Consider the lift $\delta$ of the path $t\in[0,1]\mapsto\Phi(t)$
starting from $\delta(0)=1$. We have

$\Phi(1)/\Phi(0)=\exp(\tau\delta(0)-\tau\delta(1)).$

The point $\delta(1)$ is, in ${\mathbb{H}}_{+}$, at a hyperbolic distance from
$\delta(0)=1$ that is at most $\operatorname{cst}/n$. So
$|\delta(1)-\delta(0)|\leq\frac{\operatorname{cst}}{n}$ for another constant,
provided $n$ is big enough (independently of $u$ and $v$). Multiplying by
$\tau$ (which depends on $u$ and $v$), we get

$\left|\log_{p}\frac{\Phi(1)}{\Phi(0)}\right|\leq|\tau\delta(1)-\tau\delta(0)|\leq\frac{\operatorname{cst}}{n}\tau.$

Now

$\frac{\tau}{n}=\frac{1}{n}\log\left|\frac{2R}{\Phi(0)}\right|\leq\frac{\operatorname{cst}+\log
1/\varepsilon_{n}}{n}\underset{n\to\infty}{\longrightarrow}0$

since $|\Phi(0)|\geq\varepsilon_{n}$ and $\log 1/\varepsilon_{n}=o(n)$. The
claim follows. ∎

We assume that $n$ is big enough so that $K$ is contained in the array formed
by the small squares, which we recall is the big square given by
$|{\operatorname{Re}\,}z|\leq n$ and $|{\operatorname{Im}\,}z|\leq n$. To get
some margin we in fact take $n$ be big enough so that $K$ is contained in
$|{\operatorname{Re}\,}z|\leq n-1/n$ and $|{\operatorname{Im}\,}z|\leq n-1/n$.

Consider now a set of four small squares around a square corner $c$. The
Beltrami form $\mu_{n}$ takes four values on these squares. Consider the four
quadrants defined by $\arg(z-c)$ belonging to the intervals
$(k\pi/2,(k+1)\pi/2)$, $k\in\\{0,1,2,3\\}$. Consider the Beltrami form
$\check{\mu}$ that is constant on each of these quadrants and takes the same
value $\check{\mu}_{k}$ as the one of the four small squares with corner $c$
that the quadrant contains (we omit $n$ in these notations). We know an
explicit straightening of $\check{\mu}$, obtained as follows: first apply

$z\mapsto z-c,$

then on each quadrant based on $0$, indexed by $k\in\\{0,1,2,3\\}$, consider
the map

$z\mapsto z+\check{\mu}_{k}\bar{z}.$

Glue the four sectors thus obtained together along three of the four boundary
pairs, using ${\mathbb{C}}$-affine maps. For instance use the identity for
$k=0$, then

$z\mapsto\frac{i+\check{\mu}_{0}\bar{\imath}}{i+\check{\mu}_{1}\bar{\imath}}\,z$

for $k=1$, then

$z\mapsto\frac{-1-\check{\mu}_{1}}{-1-\check{\mu}_{2}}\cdot\frac{i+\check{\mu}_{0}\bar{\imath}}{i+\check{\mu}_{1}\bar{\imath}}\,z$

for $k=2$ and

$z\mapsto\frac{-i-\check{\mu}_{2}\bar{\imath}}{-i-\check{\mu}_{3}\bar{\imath}}\cdot\frac{-1-\check{\mu}_{1}}{-1-\check{\mu}_{2}}\cdot\frac{i+\check{\mu}_{0}\bar{\imath}}{i+\check{\mu}_{1}\bar{\imath}}\,z$

for $k=3$. Note that each of these maps is close to the identity because the
$\check{\mu}_{k}$ are close to each other, as averages of the continuous
function $\mu$ on nearby small squares, and their modulus is never close to
$1$ since $\|\mu\|_{\infty}<1$. We obtain a sector of angle close to $2\pi$
(it may be bigger than $2\pi$). The sector is then closed appropriately as
explained in Section 2 by applying

$z\mapsto z^{\alpha}=\exp(\alpha\log z)$

for the branch of $\log z$ whose imaginary part belongs to $[0,2\pi)$, where

$\alpha=\frac{2\pi i}{2\pi i+\log\tau}$

and
$\tau=\frac{1+\check{\mu}_{3}}{1+\check{\mu}_{0}}\cdot\frac{-i-\check{\mu}_{2}\bar{\imath}}{-i-\check{\mu}_{3}\bar{\imath}}\cdot\frac{-1-\check{\mu}_{1}}{-1-\check{\mu}_{2}}\cdot\frac{i+\check{\mu}_{0}\bar{\imath}}{i+\check{\mu}_{1}\bar{\imath}}$,
i.e.

$\tau=\frac{1+\check{\mu}_{3}}{1-\check{\mu}_{3}}\cdot\frac{1-\check{\mu}_{2}}{1+\check{\mu}_{2}}\cdot\frac{1+\check{\mu}_{1}}{1-\check{\mu}_{1}}\cdot\frac{1-\check{\mu}_{0}}{1+\check{\mu}_{0}}$

which is close to $1$, and where the determination of $\log\tau$ is the
principal one. In the proof of LABEL:prop:lim-mes we have in fact evaluated
$\tau$ (it corresponds to $\Lambda$ in that proposition) and found that
$\log\tau=\cal O(1/n^{2})$, uniformly. As a consequence, $\alpha$ is close to
$1$ and

$|\alpha-1|=\cal O(1/n^{2}),$

uniformly. Last we apply

$z\mapsto az+b$

where $a$ and $b$ will be chosen later. Denote $\Phi_{n}$ this composition of
maps:

$\Phi_{n}:F\to{\mathbb{C}}$

where
$F=\big{\\{}{z\in{\mathbb{C}}}\,\big{|}\,{|{\operatorname{Re}\,}(z-c)<1/n,\
|{\operatorname{Im}\,}(z-c)|<1/n}\big{\\}}$ is the interior of the union of
the four square. The purpose of $\Phi_{n}$ is to serve as a model of the map
$f_{n}$ near $c$. Note that $\Phi_{n}$ is quasiconformal, and is
${\mathbb{R}}$-differentiable inside each square.

######  Lemma 63.

There exists $\eta_{n}\longrightarrow 0$ such that for all $n$ big enough, for
all $c$ as above, the differential of the corresponding $\Phi_{n}$ on
$\big{\\{}{z\in F}\,\big{|}\,{|z-c|>r^{\prime}_{n}}\big{\\}}$ varies less than
$\eta_{n}$, in the sense that for any two points $z$, $z^{\prime}$ in this
set, the corresponding differentials $L_{z}$ and $L_{z^{\prime}}$
satisfy464646Where $\|\cdot\|$ is any operator norm on chosen in advance on
the set of ${\mathbb{R}}$-linear endomorphisms of ${\mathbb{C}}$.
$\|L_{z^{\prime}}\circ L_{z}^{-1}-\operatorname{Id}\|<\eta_{n}$.

###### Proof.

Of course the differential of $z\mapsto z-c$ is the identity. Each $z\mapsto
z+\check{\mu}_{k}\bar{z}$ is ${\mathbb{R}}$-linear and close to $z\mapsto
z+\mu(c)\bar{z}$, because $|\check{\mu}_{k}-\mu(c)|<\operatorname{cst}/n$
where $\operatorname{cst}$ is independent of $c$ and $n$. The differential of
$z\mapsto z^{\alpha}$ is $\alpha z^{\alpha-1}dz$. Recall that $\alpha$ is
close to $1$ and $z^{\alpha-1}=\exp((\alpha-1)\log z)$ is close to $1$
provided $(\alpha-1)\log|z|$ is close to $0$. Since $(\alpha-1)=\cal
O(1/n^{2})$ uniformly, it is enough that $|z|>r_{n}=\exp(-n)$ (we see that we
could even have taken a much smaller $r_{n}$). ∎

Though we do not formally need it, it helps, for a clearer mental picture of
the situation, to realize that, as we can see from the proof, $\Phi_{n}$ is
close to the similitude $z\mapsto a(z-c)+b$ in some sense that we do not need
to make explicit.

$f$$\widehat{f}_{n}$$f_{n}$$\Phi_{n}$$\check{f}_{n}$$F$$c$$z^{*}$$F^{\prime}$
Figure 17. This commutative diagram shows the objects involved in the end of
the proof of Proposition 60. The model map $\Phi_{n}$ is (voluntarily) only
defined on the four small squares depicted in gray. Circled are the places
where the derivative of $\check{f}_{n}$ is not controlled, and the image of
the places where the derivative of $f_{n}$ and $\widehat{f}_{n}$ is not
controlled. These places are of size of order $r_{n}=\exp(-n)$, so in fact
much smaller than the grid step $1/n$, so one should imagine the circles
nearly invisible. Dashed is the image by $\Phi_{n}$ of the gray square scaled
about its centre by a factor of $0.75$.

It follows from Lemma 63 that the differential of $\Phi_{n}^{-1}$ also varies
less than $\eta_{n}$ on the image by $\Phi_{n}$ of $\big{\\{}{z\in
F}\,\big{|}\,{|z-c|>r^{\prime}_{n}}\big{\\}}$. The image $\Phi_{n}(F)$ of
$\Phi_{n}$ is close to be a parallelogram and the part that must be removed is
contained in a very small disk near its centre, of size of order $r_{n}$ times
$\|L_{z}\|$, which is of the order of $|a|$ where $a$ is the constant yet to
be chosen in the definition of $\Phi_{n}$, whereas the parallelogram has a
diameter of order $|a|/n$, and bounded geometry (recall that
$\|\mu\|_{\infty}<1$). In the sequel we denote

$F^{\prime}=\Phi_{n}(\big{\\{}{z\in F}\,\big{|}\,{|z-c|>r_{n}}\big{\\}}).$

Let $\check{f}_{n}=f_{n}\circ\Phi_{n}^{-1}$ so that
$f_{n}=\check{f}_{n}\circ\Phi_{n}$. Then $\check{f}_{n}$ is an injective
holomorphic map because it is a composition of quasiconformal maps that sends
the circle ellipse field to itself. Moreover,

$\check{f}_{n}=\widehat{f}_{n}\circ(f\circ\Phi_{n}^{-1}).$

In the definition of $\Phi_{n}$ there was to choose constants $a$ and $b$. We
choose $b=f(c)$, so that $\Phi_{n}(c)=f(c)$. Consider the point
$z^{*}=c+\frac{1+i}{2n}$, which is the centre of one of the four squares. We
choose $a$ so that $\check{f}_{n}^{\prime}(\Phi_{n}(z^{*}))=1$.

Since $f$ is $C^{1}$ (actually $C^{2}$) and $F$ has small diameter, the
differential of $f$ is nearly constant on $F$. It follows that the map
$f\circ\Phi_{n}^{-1}$ has a nearly constant differential on $F^{\prime}$. In
particular,474747The set $F^{\prime}$ is not convex but it is nearly a
parallelogram with a removed hole that is nearly an ellipse. pairs of points
$u$, $v$ in $F^{\prime}$ have images $u^{\prime}$, $v^{\prime}$ by
$f\circ\Phi_{n}^{-1}$ such that $(u^{\prime}-v^{\prime})/L(u-v)$ is close to
$1$, where $L$ is some ${\mathbb{R}}$-linear map that depends on $n$ and $c$
but not on $u$ and $v$.

We have also seen that $\widehat{f}_{n}$ maps pairs of points $u,v\in K$ at
distance $>r_{n}$ to pairs of points $u^{\prime}$, $v^{\prime}$ whose
associated vector is nearly the same: $(u^{\prime}-v^{\prime})/(u-v)$ is close
to $1$, uniformly.

It follows that pairs of points $u$, $v$ in $F^{\prime}$ with distance at
least $|a|r_{n}/M^{\prime}$ for some $M^{\prime}>0$ independent of $n$ and
$c$, have images $u^{\prime}$, $v^{\prime}$ by $\check{f}_{n}$ such that
$(u^{\prime}-v^{\prime})/L(u-v)$ is close to $1$. But recall that
$\check{f}_{n}$ is holomorphic and that its derivative at $\Phi_{n}(z^{*})$ is
equal to $1$. It follow by properties of univalent maps that $\check{f}_{n}$
is close to the identity on the compact subset of $\Phi_{n}(F^{\prime\prime})$
of $\Phi_{n}(F)$ where $F^{\prime\prime}$ is defined by the equations
$|{\operatorname{Re}\,}z-c|<0.75/n$ and $|{\operatorname{Im}\,}z-c|<0.75/n$,
and that $\check{f}_{n}^{\prime}$ is close to $1$ on that set. It also follows
that $L$ is actually close to the identity.

Recall that $f\circ\Phi_{n}^{-1}$ has a differential nearly constant and close
to $L$ on $F^{\prime}$. Hence $f\circ\Phi_{n}^{-1}$ has actually a
differential close to the identity on $F^{\prime}$.

Since, $\widehat{f}_{n}=\check{f}_{n}\circ(f\circ\Phi_{n}^{-1})^{-1}$ we get
Proposition 60 by the chain rule.

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