# Zhenghe's Blog

## January 30, 2011

### Notes 3: Kotani Theory (II)-Fibred Rotation Number (IDS)

Filed under: Schrodinger Cocycles — Zhenghe @ 10:00 pm
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Up to now, all my posts started from Jan.22.2011 are based on Artur‘s course here in Fields institute, Toronto. Part of the contents of this course are even from Artur’s unpublished work.

This time let me prove the following theorem

Theorem 1: Let $(f, A^{(E-v)})$ as last time, then for a.e. $E,$ $L(E)=o$ implies that $L(E+i\epsilon)=O(\epsilon),$ for small $\epsilon>0$.

As I mentioned last, this Theorem together with the Lemma of last post imply the main theorem of kotani theory.

The proof this theorem make use of the harmonicity of $L(E)$ in upperhalf plane. In particular there is a harmonic function $\rho(E)$ such that $L+i\rho:\mathbb H\rightarrow \mathbb C$ is holomophic. We are going to show this $\rho$ is in fact the fibred rotation number of the correspoding $Schr\ddot odinger$ cocyles and which is also basically the IDS (integrated density of states) of the $Schr\ddot odinger$ operators. This will be the key object of this post. We will prove the following facts about $\rho$:

1. $\rho(E)$ well-defined for all $E\in \overline{\mathbb H}$ and is continuous up to $\partial{\mathbb H}=\mathbb R$.
2. $(f, A^{E-v})$ is monotonic in $E\in\mathbb R$ in some sense which implies the monotonicity of $\rho$ in $E\in\mathbb R$.
3. Thus $\rho(E)$ is differentiable for a.e. $E\in R$ and Cauchy-Riemann equation will imply the conclusion of our theorem.

Let me carry out all the details.

I have to say, for me the fibred rotation number is always a subtle concept. This time I am going to expore as detailed information about it as I can.

From last post we know there exist invariant section $u(E,x)$ for projective dynamics $(f, A^{(E-v)}), \Im E>0$. Thus

$A^{(E-v)}(x)\binom{u(E,x)}{1}=u(E,x)\binom{u(E,f(x))}{1},$

from which it’s easy to see that another way to calculate Lyapunov Exponents via invariant section is

$L(E)=\int_X\ln |u(x)|d\mu,$ thus

$(L+i\rho)(E)=\int_X\ln u(E,x)d\mu:\mathbb H\rightarrow \mathbb C$ is holomorphic.

From which we see that $\rho(E)=\int_X\arg u(E,x)d\mu$ (Here $\arg u(E,x)$ is well-defined. Because by last post, more concretely proof of Lemma 2, it’s easy to see that $u:\mathbb H\times X\rightarrow\mathbb H$. Thus there is no nontrivial loop around origin). For obvious reason, it’s convenient to instead consider $\rho(E)=\frac{1}{2\pi}\int_X\arg u(E,x)d\mu$ (so $L+2\pi \rho i$ is holomorphic functon). By Birkhoff Ergodic Theorem, we have for a.e. x

$\lim\limits_{n\rightarrow\infty}\frac{1}{2\pi n}\sum^{n-1}_{j=0}(\arg u(E,f^{j+1}(x))\cdots u(E,x)-\arg u(E,f^j(x))$ $\cdots u(E,x))$
$=\frac{1}{2\pi n}\sum^{n-1}_{j=0}\arg u(E,f^{j+1}(x))$
$=\rho(E),$

which implies that $\rho$ is some sort of averaged rotation, i.e. a rotation number.

Before proving the next Lemma, I need to do some preparation. To consider rotation number in more general setting, we need go from the $Poincar\acute e$ upperhalf plane $\mathbb H$ to the $Poincar\acute e$ disk $\mathbb D$ via the following matrix

$Q=\frac{-1}{1+i}\begin{pmatrix}1& -i\\1& i\end{pmatrix}\in \mathbb U(2).$

It’s easy to see that $Q\cdot\mathbb H=\mathbb D.$ And $QSL(2,\mathbb R)Q^*=SU(1,1)$, where $SU(1, 1)$ is the subgroup of $SL(2,\mathbb C)$ preserving the unit disk in $\mathbb C\mathbb P^1=\mathbb C\cup\{\infty\}$ under Mobius transformation. For

$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL(2, \mathbb C),$ let

$\hat A=QAQ^*=\begin{pmatrix}\frac{1}{2}[(a+d)+(b-c)i],&\frac{1}{2}[(a-d)-(b+c)i]\\\frac{1}{2}[(a-d)+(b+c)i],&\frac{1}{2}[(a+d)-(b-c)i]\end{pmatrix}=\begin{pmatrix}\hat u&\hat v\\v&u\end{pmatrix}.$

Then it’s easy to see

for $A\in SL(2,\mathbb R),$ i.e. $A\cdot\mathbb H=\mathbb H,$ then $\hat A\cdot\mathbb D=\mathbb D,$
$\hat u=\bar u,\hat v=\bar v,$ and $|u|^2-|v|^2=1;$
for $\overline{A\cdot\mathbb H}\subset\mathbb H,$ then $\overline{\hat A\cdot\mathbb D}\subset\mathbb D,$
$|u|^2-|v|^2>1.$ Let’s denote this class by $\mathfrak{B}$
for $\overline{A\cdot\mathbb H_-}\subset\mathbb H_-,$ then $\overline{\hat A\cdot\overline{\mathbb D}^c}\subset\overline{\mathbb D}^c,$
$|u|^2-|v|^2<1.$

And all these sets of $A$, or equivalently of $\hat A$ are multiplicative.

In the following Lemma, I always consider the equivalent dynamics $(f, \hat A^{(E-v)}).$ The Lemma is

Lemma 2: $\rho(E)$ is well-defined for all $E\in\overline{\mathbb H}$ and is continuous on $\overline{\mathbb H}$.

Proof: First let’s show that, as long as the cocycle map $\hat A^{(E-v)}(x)\in\mathfrak B,$ or equivalently, $\Im E >0$, we can define $\rho(E)$ via any continuous section $m:X\rightarrow \mathbb D$ (not necessary invariant).

Let’s define $m_n, \tau_n(E,x,m)$ be that

$\hat A^{(E-v)}_n (x)\binom{m}{1}=\tau_n(E,x,m)\binom{m_n}{1}.$

Then obviously $\tilde u(E,x)=Q\cdot u(E,x)$ is the unstable invariant section of $(f, \hat A^{(E-v)})$, thus $\rho(E)$ can be defined as

$\rho(E)=\frac{1}{2\pi}\int_X\arg \tau_1(E,x,\tilde u(E,x))d\mu$
$=\lim\limits_{n\rightarrow\infty}\frac{1}{2\pi n}\sum^{n-1}_{j=1}\arg \tau_1(E,f^{j+1}(x),\tilde u(E,f^{j+1}(x)))$

for a.e. $x.$

Then we show that $\tilde u$ can be replaced by any continuous section $m$ and the convergence is independent of the choice of such $m$. Indeed, we always have that for any $m, m'\in \mathbb D,$

$|\arg\tau_n(E,x,m)-\arg\tau_n(E,x,m')|<\pi$

(hence $|\arg\tau_n(E,x,m)-\arg\tau_n(E,x,m')|\leq\pi,$ for all $E\in\overline{\mathbb H}$).

In fact, by our choice of cocyle map, if we denote $\hat A^{(E-v)}_n(x)=\begin{pmatrix}\hat u_n&\hat v_n\\v_n&u_n\end{pmatrix},$  then

$\tau_n(E,x,\mathbb D)=v_n\mathbb D+u_n,$

which is a disk stay away from $0$ with distance at least $1.$ Thus the above estimate follows easily (Note this is not true for Lyapunov exponents, i.e. $\left |\ln|\tau_n(E,x,m)|-\ln|\tau_n(E,x,m')|\right|$ cannot necessary be bounded). Now we may fix constant section $m(x)\equiv m\in\mathbb D$ to do the remaing computation.

It’s easy to see that $\tau_{n+l}(E,x,m)=\tau_{n}(E,x,m)\tau_{l}(E,f^n(x),m_n)$, so

$\int_X\arg\tau_{n+l}(E,x,m)d\mu=\int_X\arg\tau_{n}(E,x,m)d\mu+\int_X\arg\tau_{l}(E,f^n(x),m_n)d\mu.$

For simplicity let $a_n(E,m)=\frac{1}{2\pi}\int_X\arg\tau_{n}(E,x,m)d\mu.$ Then the above formula implies that

$a_{n+l}(E,m)=a_n(E,m)+a_l(E,m_n),$ and obviously $\lim\limits_{n\rightarrow\infty}\frac{a_n(E,m)}{n}=\rho(E).$

We then have

$|\frac{a_n(E,m)}{n}-\frac{a_l(E,m)}{l}|$
$\leq\frac{1}{nl}\sum_{j=1}^{l-1}|a_n(E,m)-a_n(E,m_{jn})|+\frac{1}{nl}\sum_{p=1}^{n-1}|a_l(E,m)-a_n(E,m_{pl})|$
$\leq \pi(\frac{1}{n}+\frac{1}{l})\rightarrow 0,$ as $n, l\rightarrow\infty.$

Hence, the convergence is uniform.  In the similar way, we can show that $\rho(E)$ is uniform contious in $\mathbb H$. Indeed, it’s easy to see for any fixed $n,$ $a_n(E,m)$ is unform continuous on $\overline{\mathbb H}.$ So we can choose $n_0$ such that $a_{l}(E,m), l=0,\cdots, n_0$ are equi-uniform continuous. Now for any $\epsilon>0,$ we can choose sufficiently small $\delta>0$ such that for $|E-E'|<\delta, E, E'\in\overline{\mathbb H},$

$|a_l(E,m)-a_l(E',m)|<\epsilon,$ for all $l=0,\cdots,n_0.$

Now for arbitrary $n\geq0,$ we have $n=kn_0+l, 0\leq l\leq n_0-1.$ Thus
$|\frac{1}{n}(a_n(E,m)-a_n(E',m))|$
$\leq\frac{1}{kn_0}\sum_{j=0}^{k-1}(|a_{n_0}(E,m_{l+(j-1)n_0})-a_{n_0}(E',m_{l+(j-1)n_0})|+|a_l(E,m)-a_l(E',m)|)$
$\leq\frac{\pi}{k}+\frac{1}{kn_0}|a_l(E,m)-a_l(E',m)|+\frac{1}{n_0}|a_{n_0}(E,m)-a_{n_0}(E',m)|<\epsilon,$ for $k$ large.  Since $n$ is arbitrary, we see $|\rho(E)-\rho(E')|\leq\epsilon,$ for $E, E'\in\mathbb H$.

Thus we can extend $\rho(E)$ to $\overline{\mathbb H}$ which is continous up to $\partial{\mathbb H}=\mathbb R.$ Again we denote it by $\rho(E).$  The above computation actually shows that for $E\in\mathbb R,$

$\rho(E)=\lim\limits_{n\rightarrow\infty}\frac{1}{n}a_n(E,m).$

Indeed, if not we may without loss of generality assume

$\lim\limits_{n\rightarrow\infty}\frac{1}{n}{a_n(E,m)}=a\neq \rho(E).$

Then we can choose $E'\in\mathbb H$ sufficient close to $E$ and $n$ sufficiently large such that all the following terms are less than $\frac{1}{4}|a_0-\rho(E)|$:

$|a_0-\frac{1}{n}a_n(E,m)|, |\frac{1}{n}(a_n(E,m)-a_n(E',m))|,$
$|\frac{1}{n}a_n(E',m)-\rho(E')|, |\rho(E')-\rho(E)|,$

which is obvious a contradiction. $\square$

Our next lemma is

Lemma 3: $\rho(E), E\in\mathbb R$ is nonincreasing.

Proof: This in fact follows from the monotonicity of the following function.  Fix arbitrary $u\in R, x\in\mathbb R/\mathbb Z$ consider the function in $E\in\mathbb R$

$g(E)=A^{(E-v)}(x)\cdot u=E-v(x)-\frac{1}{u}.$

To make everything clear, let’s introduce another way to study the fibred rotation number. Fix $m\in \partial{\mathbb D}$, consider

$N(E)=\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum_{j=1}^{n}(\arg m_j(E)-\arg m_{j-1}(E)).$

A direct computation shows that $\arg m_j(E)-\arg m_{j-1}(E)=-2\arg\tau_1(E,f^{j}(x),m_{j-1}).$ Indeed,

$m_j=\begin{pmatrix}u(f^jx)&v(f^jx)\\\bar v(f^jx)&\bar u(f^jx)\end{pmatrix}\cdot m_{j-1}=\frac{u(f^jx)m_{j-1}+v(f^jx)}{\bar v(f^jx)m_{j-1}+\bar u(f^jx)}$. Thus

$\frac{u(f^jx)m_{j-1}+v(f^jx)}{\bar v(f^jx)m_{j-1}+\bar u(f^jx)}/m_{j-1}=\frac{v(f^jx)\bar m_{j-1}+u(f^jx)}{\bar v(f^jx)m_{j-1}+\bar u(f^jx)}(m_j\in\partial{\mathbb D},\forall j\geq 0)$. So

$\arg\frac{v(f^jx)\bar m_{j-1}+u(f^jx)}{\bar v(f^jx)m_{j-1}+\bar u(f^jx)}=-2(\bar v(f^jx)m_{j-1}+\bar u(f^jx))=-2\arg\tau_1(E,f^{j}(x),m_{j-1})$ and $N(E)=-2\rho(E).$

Consider $u=Q^*\cdot m\in\mathbb R.$ Then the relation between $u$ and $m$ are $\cot\theta$ and $e^{-2i\theta}.$ Thus it’s not difficult to see that

$E'>E\Rightarrow g(E')>g(E)\Rightarrow \arg m_1(E')>\arg m_1(E).$

Now since we start with the same $m,$ we obviously have $\arg m_1(E')>\arg m_1(E)$ and

$\arg m_2(E')=\arg (\hat A^{(E'-v)}(x)\cdot m_1(E'))$
$>\arg (\hat A^{(E'-v)}(x)\cdot m_1(E))$
$>\arg (\hat A^{(E-v)}(x)\cdot m_1(E))=\arg m_2(E),$

where the first inequality follows from the fact that $\hat A$ preserves order and the second one follows from monotonicity. So by induction we have

$\arg m_n(E)>\arg m_n(E'),$ for all $n>0.$

Thus  $N(E')\geq N(E)$ and  $\rho(E')\leq\rho(E),$ which completes the proof. $\square$

Now we are ready to prove the theorem of this post.

Proof of Theorem 1: By standard harmonic theorem it’s easy to see in our case, $\rho(E+i\epsilon)$ is Poisson integral of $\rho(E), E\in\mathbb R$. Obviously, $\partial_E{\rho}$ is again harmonic. Since $\rho(E), E\in\mathbb R$ is monotonic, $\partial_E{\rho}$ is in fact the Poisson integral of $d\rho(E), E\in\mathbb R.$ Then Fatou’s Theorem tells us for Leb a.e. $E,$ $\lim\limits_{\epsilon\rightarrow 0}\partial_E\rho(E+i\epsilon)=\frac{\partial\rho}{\partial E}(E).$

Now by Cauchy-Riemann equation we have that

$L(E+i\epsilon)=L(E+i0^+)+\int^{\epsilon}_{0}\partial_s(L(E+is))ds$
$=L(E+i0^+)-\int^{\epsilon}_{0}\partial_E(\rho(E+is))ds.$

Now since Lyapunov exponents is a nonnegative upper semicontinuous  function, it’s continuous at $E$, where $L(E)=0.$  Thus the above discussion shows that for a.e. E with $L(E)=0,$ we have

$\lim\limits_{\epsilon\rightarrow 0}\frac{L(E+i\epsilon)}{\epsilon}=-\partial_E\rho(E),$

which completes the proof the Theorem. $\square$

Now I’ve already finished the proof, but probabily I will show that $\Sigma_{ac}=\overline{\mathcal Z}^{ess}$ in the future. As I said in the last post, $\Sigma_{ac}\subset\overline{\mathcal Z}^{ess}$ is relatively easy. It lies in the fact that the generalized eigenfunctions of absolutely continuous spectrum grow at most polynomially fast, which obviously contradicts with positive Lyapunov exponents. And the other part due to Kotani theory has already been contained in these two posts. Let me go back to this in the future.

Next post I will give some application of Kotani theory on deterministic potential and problems concerning density of positive Lyapunov exponents.

## January 26, 2011

### Notes 2: Kotani theory (I)-Zero Lyapunov Exponents and L^2 rotation conjugacy

This and next posts will be about Kotani theory, which is part of my syllabus. Kotani theory is first found by Shinichi Kotani when he studies spectrum theory of $Schr\ddot odinger$ operator with ergodic potentials. It gives a complete decription of  the absolutely continuous part of spectrum of  $Schr\ddot odinger$ operators in terms of Lyapunov exponents of the  corresponding of $Schr\ddot odinger$ cocycles. Thus it builds a deep and beautiful relation between operator theory and dynamical systems.

Later Artur Avila and Raphael Krikorian generalize Kotani theory to more general coycle dynamics setting and it becomes a powerful tool to study Schrodinger operators. What I am going to introduce here are main results and some preliminary stuff. Let’s start with one dimensional (1D) discrete $Schr\ddot odinger$ operators and $Schr\ddot odinger$ cocyles. From now on I set $L(E)=L(f,A^{(E-v)})$.

For simplicity, I will continue to use the triple $(X,f,\mu)$ as in last post. Assume $v\in C(X,R)$ be  a contiuous function. Then we can define a cocycle map via

$A^{(v)}(x)=\begin{pmatrix}v(x)&-1\\1&0\end{pmatrix}.$

The correspoding cocycle dynamics $(f, A^{(v)})$ is called $Schr\ddot odinger$ cocyles. It arises from the following $Schr\ddot odinger$ operator $H_{f,v,x}$ on $l^2(\mathbb Z)$

$(H_{f,v,x}u)_n=u_{n+1}+u_{n-1}+v(f^n(x))u_n,$

where $u=(u_n)_{n\in\mathbb Z}\in l^2(\mathbb Z).$ Then $u\in\mathbb C^{\mathbb Z}$ solving the eigenfunction equation $H_{f,v,x}u=Eu, E\in\mathbb R$ if and only if

$A^{(E-v)}(f^n(x))\binom{u_n}{u_{n-1}}=\binom{u_{n+1}}{u_n}.$

Thus the growth rate of $|u_n|$ with respect to $n$ characterizes both the spectrum type of the energy $E$ and the dynamics of cocycle $(f,A^{(E-v)})$, which allows one to go back and forth between spectrum theory and dynamical systems. Let $\Sigma_x$ be the spectrum of the bounded linear selfadjoint operator $H_{f,v,x}$, then the first basic fact relate operator and cocycle is for a.e. $x,$

$\Sigma_x=\{E: (f, A^{(E-v)})\notin \mathcal U\mathcal H\}.$

If furthermore $f$ is minimal, then the above relation is in fact true for all $x.$ Thus  for a.e. $x,$ the spectrum $\Sigma_x$ is independent of $x\in X,$ let’s denote it as $\Sigma$. For each $x,$ we can further decompose $\Sigma_x$ as $\Sigma_x=\Sigma_{x,pp}\cup\Sigma_{x,ac}\cup\Sigma_{x,sc}$, which correspond to pure point, absolutely continuous and singular continuous part of the spectrum of the operator $H_{f,v,x}$. These are defined by the spectral measure of the operator, which one can find in any standard functional analysis book.  It turns out that in our case, there exists sets $\Sigma_{\bullet}, \bullet\in\{pp, ac, sc\}$ such that $\Sigma_{x,\bullet}=\Sigma_{\bullet}$ for a.e. $x$ and $\bullet\in\{pp, sc,ac\}.$ (If in addition $f$ is minimal, then in fact $\Sigma_{x,ac}=\Sigma_{ac},$ for all $x,$ which is in general not true for pp and sc.)

Now we denote $\mathcal Z=\{E: L(E)=0\}.$ And for any set $\mathcal S\subset\mathbb R,$ the essential support of the set $\mathcal S$ is given by

$\overline{\mathcal S}^{ess}=\{E\in\mathbb R: Leb(\mathcal S\cap (E-\epsilon, E+\epsilon))>0$ for every $\epsilon>0\}.$

Then the next deep relation between dynamics and spectrum is the following

$\Sigma_{ac}=\overline{\mathcal Z}^{ess}.$

The relation $\Sigma_{ac}\subset\overline{\mathcal Z}^{ess}$ is relatively easy since positive Lyapunov exponents give exponential growth of the solution $u$ to the eigenfunction equation which in some sense contradicts with the absolute continuity of spectrum. The part $\overline{\mathcal Z}^{ess}\subset\Sigma_{ac}$ is a rather deep result called Kotani theory. It in fact  is a theory about that, under the assumption that Lyapunov exponents are zero, when can one in some sense conjugate the $SL(2,\mathbb R)-$valued cocycles to $SO(2,\mathbb R)-$valued cocycles. Obviously, if the cocycle map $A$ takes value in $SO(2,\mathbb R)$, then all orbits $\{A_n(x)w\}_n$ are bounded. While zero Lyapunv exponent in general just means that $\|A_n(x)w\|_n$ grows subexponentially.  What Kotani theory tells us is in fact that the following theorem

Theorem 1: For almost every $E$, if $L(E)=0,$ then there exists a map $B:X\rightarrow SL(2,\mathbb R)$ such that$B(f(x))A^{(E-v)}(x)B(x)^{-1}\in SO(2,\mathbb R)$ and $\int_X\|B(x)\|^2d\mu<\infty.$

Thus the generalized eigenfunctions for absolutely continous spectrum of the operators in question oscillate in some $L^2$ sense. Which are kind of wave like solutions and far from eigenfunctions of real eigenvalues, which decay in $l^2$ sense. As I said this is obviously stronger then zero Lyapunov exponents. Indeed, in this case we have

$\frac{1}{n}\int_X\ln\|A_n(x)\|d\mu\leq\frac{1}{n}\int_X\|A_n(x)\|d\mu\leq\frac{1}{n}\int_X\|B(f^n(x))^{-1}\|\|B(x)\|d\mu$
$\leq\frac{1}{n}\int_X\|B(x)\|^2d\mu\rightarrow 0,$ as $n\rightarrow\infty$

Now let me explain how can one construct the above $B:X\rightarrow SL(2,\mathbb R).$  We first prove the following key lemma

Lemma 2: Assume $E\in\mathbb R$ satisfying $L(E+i\epsilon)=O(\epsilon),$ for $\epsilon>0$ small. Then $(f,A^{(E-v)})$ is $L^2-$conjugate to $SO(2,\mathbb R)-$valued cocycles.
Proof: First we note by the same reason that $(f,R_{i\theta}A)\in\mathcal U\mathcal H$ as in last blog, we have $(f,A^{(E+i\epsilon-v)})\in\mathcal U\mathcal H$. Thus there is invariant section $u^{\epsilon}:X\rightarrow \mathcal H$ which is the unstable direction. There are different ways to calculate Lyapunov exponents of cocylce dynamics via invaiant section of corresponding projective dynamics. We use the following: $-\frac{1}{2}$ of the contraction rate measured in $Poincar\acute e$ metric of mobius transformation at $u^{\epsilon}$. In our case we need to consider the following composition of map. Let $\mathbb H_{\epsilon}=\{E:\Im E>\epsilon\}$ with standard $Poincar\acute e$ metric, then the composition is

$\mathbb H\overset{A^{(E+i\epsilon-v)}}{\longrightarrow}\mathbb H_{\epsilon}\overset{i_{\epsilon}}{\longrightarrow}\mathbb H,$

where the first map is a isometry and the second one (which is the inclusion map) is a contraction. Thus we consider the contraction of the second map at invariant section. Then the Lyapunov exponents is given by

$L(E+i\epsilon)=-\frac{1}{2}\int_X\ln(1-\frac{\epsilon}{\Im u^{\epsilon}})d\mu.$

For simplicity, let me first assume that:

$\lim\limits_{\epsilon\rightarrow 0}u^{\epsilon}(x)$ exists for $a.e. x$ and we denote it by $u(x)$.
(then obviously, $u(x)$ is invariant for a.e. $x$, i.e. $A^{(E-v)}(x)\cdot u(x)=u(f(x))$ for a.e.$x$.)

Assuming this, we have the following straightforward estimate via Fatou’s lemma and our assumption in Lemma :

$\frac{1}{2}\int_X\frac{1}{\Im u}d\mu\leq\liminf\limits_{\epsilon\rightarrow 0}-\frac{1}{2\epsilon}\int_X\ln(1-\frac{\epsilon}{\Im u^{\epsilon}})d\mu=O(1)<\infty.$

Since $u(f(x))=E-v(x)-\frac{1}{u(x)}$, we have

$\frac{1}{2}\int_X\frac{1}{\Im u}d\mu=\frac{1}{2}\int_X\frac{1}{\Im u(f(x)}d\mu=\frac{1}{2}\int_X\frac{|u|^2}{\Im u}d\mu<\infty$.

Hence

$\int_X\frac{1+|u|^2}{\Im u}d\mu<\infty.$

Let $\phi(u)=\frac{1+|u|^2}{\Im u}$. Now we define $B:X\rightarrow SL(2,\mathbb R)$ such that $B(x)\cdot u=i,$ thus $B(f(x))A(x)B(x)^{-1}\cdot i=i$ which implies $B(f(x))A(x)B(x)^{-1}\in SO(2,\mathbb R)$. On the other hand, it’s easy to see for quite general reason $\|B(x)\|_{HS}^2=\phi(u).$ Thus $\int_X\|B(x)\|_{HS}^2d\mu<\infty.$
Here for $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL(2,\mathbb R),$ $\|A\|_{HS}^2$ is the Hilbert-Schmit norm of $A$ which is just $a^2+b^2+c^2+d^2.$ It’s easy to see that if $A\cdot z=i,$ then $\|A\|_{HS}^2=\phi(z)$. Since all norms of $SL(2,R)$ are equivalent, we complete the proof of the Lemma up to the assumption.    $\square$

To get around the assumption, we need conformal barycenter, which is a Borelian function $\mathcal B:\mathcal M\rightarrow\mathbb H,$ where $\mathcal M$ is the space of probability measures on $\mathbb H.$ This function is equivariant with respect to $SL(2,\mathbb R)$ change of coordinates. i. e. for

$A\in SL(2,\mathbb R), \nu\in\mathcal M,$ we have $\mathcal B(A_*\nu)=A\cdot\mathcal B(\nu).$

Now let’s consider the probability measures $\nu_{\epsilon}=\mu\otimes\delta_{u^{\epsilon}}$ on $X\times\overline{\mathbb H}$. Then there is a subsequence converging to some $\nu=\mu\otimes\nu_x$ such that $\int_{X\times\overline{\mathbb H}}\phi d\nu<\infty$. Then apply conformal barycenter to $\nu_x$, we get a point $u(x)=\mathcal B(\nu_x)\in\mathbb H$ such that $u$ is an invariant section and

$\int_X\phi(u(x))d\mu=\int_{X\times\overline{\mathbb H}}\phi d\delta_{u(x)}d\mu\leq\int_{X\times\overline{\mathbb H}}\phi d\nu<\infty.$

Now we can construct the map $B$ as  in the proof of lemma.

Although we introduce conformal barycenter to complete the proof of Lemma, to prove the Theorem we actually only need to apply Fubini and Fatou theorem, to pass the existence of limits from For every $x$, converges for  a.e. $E$  to  For a.e. $E$, converges for a.e. $x$. Then we get the conclusion in assumption for a.e. E.

In the next post, I will prove how can we get the condition in Lemma under the condtion in Theorem, which will complete the proof of theorem.

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