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Random Unitary Operators, Density of States, Thouless Formula, Lyapunov Exponents
0303191201104
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\begin{document}
\title{Density of States and Thouless Formula for Random Unitary Band Matrices}
\author{Alain Joye \\ Institut Fourier \\
Universit\'e de Grenoble 1,
BP 74 \\
38402 SaintMartin d'H\`eres Cedex, France}
\date{ }
\maketitle
\abstract{We study the density of states measure for some class of random
unitary band matrices and prove a Thouless formula
relating it to the associated Lyapunov exponent.
This class of random matrices appears in the study of the
dynamical stability of certain quantum systems and can also be considered
as a unitary version of the Anderson model. We further determine
the support of the density of states measure and
provide a condition ensuring it possesses an analytic density.}
%*********************
\setcounter{equation}{0}
\section{Introduction}
The stability of quantum dynamical systems generated by time periodic hamiltonians
is sometimes characterized by means of the spectral properties of the corresponding unitary
evolution operator over a period, also called monodromy operator, see \cite{be, h1, c2}.
Unfortunately, even for this relatively simple timedependence, except for certain specific
models, e.g. \cite{c1,df, bo}, it is rarely the case that one has enough information
about the actual monodromy operator so that a complete
spectral analysis can be performed. Therefore, one resorts to different approximation
techniques in some specific regimes to say something about the spectrum. For example,
KAM inspired techniques, see e.g. \cite{be, c0, ds1, ade, dlsv, gy}, or adiabatic related
approaches, see e.g. \cite{h2, h3, h4, n1, j, n2}, have been used to tackle this problem.
In case the complexity of the monodromy operator is important enough to forbid
of a complete description of it, one may resort to a statistical modelization.
It is the case in particular in the study of
the quantum dynamics of electrons confined to a ring threaded by a time dependent
magnetic flux, see e.g. the paper \cite{bb} and references therein. A modelization
of this dynamics by means of an effective random monodromy operator taking into account
the details of the metallic structure of the ring is considered and tested numerically in
\cite{bb}. We refer the reader to this paper and \cite{bhj} for a more detailed account of
the construction of the monodromy operator.
Motivated by this approach, the spectral analysis of a class of random and deterministic unitary
operators, which contains the above monodromy operator, is performed in \cite{bhj}.
The main characteristics of these unitaries is that, when expressed as matrices in some basis,
they display a band structure: more precisely they are fivediagonal.
While the construction of the set of unitaries studied in \cite{bhj} is patterned after
the above mentionned physical model, we believe it may be useful for a wider class of problems.
Moreover, in the regime we consider here, this set of unitaries bears strong resemblances
with the Jacobi matrices related to selfadjoint discrete onedimensional Schr\"odinger
operators.
Another motivation in that direction stems from the recent paper \cite{cmv} where certain unitary
infinite matrices associated to the construction of orthonormal polynomials on the unit circle
are shown to display the same fivediagonal structure as our set of monodromy operators.
These matrices are shown in \cite{cmv} to be unitarily equivalent to unitary operators
introduced almost ten years ago in \cite{gt} for the study of a related trigonometric
moment problem. Moreover, in the latter paper, some effects of randomness in the
coefficients of these operators were investigated.\\
The goal of the present paper is to pursue the analysis of such random unitaries
in the setting considered in the paper \cite{bhj}.
The {\it phases} of the matrix elements of the
fivediagonal operators are random variables and the deterministic
modulus depend on one parameter only: if the phases are all set to zero, what
we will call the "free case", the unitary operator depends on a "reflexion"
coefficient $r\in ]0,1[$, see below. However, while the analysis of \cite{bhj}
focused on spectral issues, i.e. proving singularity of the almost sure spectrum by means
of a unitary version of the IshiiPastur theorem and the positivity of the Lyapunov
exponent obtained via Furstenberg's Theorem, the main object of the present
study of the density of states measure and its links with the corresponding
Lyapunov exponent.
More precisely, expressing the density of states as the density of eigenvalues of
a series of unitary operators restricted to "boxes",
we are able to state this relation as what is known as a Thouless formula. This
formula allows to compute the Lyapunov exponent by means of the density of states
and to recover the a.c. component of the density of
states measure by means of a derivative of the Lyapunov exponent.
A consequence
of our version of Thouless formula is the extension of some results of \cite{bhj}
providing, in particular, an explicit value of the
Lyapunov exponent in these cases. We also prove the validity of the Thouless
formula for the deterministic free case, by explicit computations of the relevant
quantities.
Taking advantage of the analogy of our unitary matrices with the one
dimensional discrete Schr\"odinger operator, we characterize the support of the
density of states in term of that of the distribution of the i.i.d. random phases.
Finally, we provide an effective criterion ensuring analyticity of the integrated
density of states in terms of the exponential decay rate of the Fourier coefficients
of the the distribution of the phases. This result relies on some kind of
propagation estimates for the free evolution.
We note here also that a Thouless formula is proven for the unitary random operator
studied by Geronimo and
Teplyaev in \cite{gt}. The corresponding random matrix $V_{\omega}$ is
defined in the canonical basis of $l^2(\Z)$ as well but displays a different
structure: for all $k\in \Z$, the vector $V_{\omega}\ffi_k$ has non zero
coefficients along $\ffi_j$, for $j=\infty,\cdots, k+1$ only. Such matrices
are also called Hessenberg matrices.
However, the operator under consideration here presents different characteristics
from the one of \cite{gt}, or \cite{cmv}, in particular regarding the way randomness
appears in the phases of the matrix elements.
The plan of the paper is as follows. Section 2 is devoted to the definition of
the model and its basic properties. The density of states is introduced in the next
section and Thouless formula is proven in Section 4. The statements about the
support of the density of state and ist analyticity properties are made in Section
5, whereas an Appendix contains some technical items.
%***********************************************
\section{The Model}
We present here the unitary matrices we will be concerned
with and recall some of its basic properties to be used later.
The unitary operator we consider has the following explicit form
in the canonical basis $\{\ffi_k\}_{k\in\Z}$ of $l^2(\Z)$
\bea\label{matel}
U_{\omega}\ffi_{2k}&=&irt e^{i\eta_{2k}^{\omega}}\ffi_{2k1}
+
r^2 e^{i\eta_{2k}^{\omega}}\ffi_{2k}\nonumber\\
&+&irt e^{i\eta_{2k+1}^{\omega}}\ffi_{2k+1}

t^2 e^{i\eta_{2k+1}^{\omega}}\ffi_{2k+2}\nonumber\\
& & \nonumber\\
U_{\omega}\ffi_{2k+1}&=&t^2 e^{i\eta_{2k}^{\omega}}\ffi_{2k1}
+
itr e^{i\eta_{2k}^{\omega}}\ffi_{2k}\nonumber\\
&+&r^2 e^{i\eta_{2k+1}^{\omega}}\ffi_{2k+1}
+
irt e^{i\eta_{2k+1}^{\omega}}\ffi_{2k+2},
\eea
for any $k\in\Z$. According to \cite{bhj}, the random phases
$\{\eta_k^{\omega}\}_{k\in\Z}$ are functions of some physically relevant
i.i.d. random variables $\{(\theta_k^{\omega}, \alpha_k^{\omega})\}_{k\in\Z}$
on the torus given by
\be\label{defeta}
\eta_k^{\omega}=\theta_k^{\omega}+\theta_{k1}^{\omega}+
\alpha_k^{\omega}\alpha_{k1}^{\omega},
\ee
for all $k\in\Z$ and the coefficients $r, t\in]0,1[$ are interpreted as
reflexion and transition coefficients linked by $r^2+t^2$.
We will identify the operator and its matrix representation (\ref{matel}). Let
us recall that these parameters are assumed to be different from their extreme
values $0$ and $1$, because in case $r=1\Longleftrightarrow t=0$ the operator
$U_{\omega}$ is diagonal and if $r=0\Longleftrightarrow t=1$, it is unitarily
equivalent to the direct sum of two shifts. Let us finally mention that
$U_{\omega}$ is constructed in section 2 of \cite{bhj} as a product of two unitaries
given by infinite direct sums of $2\times 2$ unitary blocks.
\subsection{Ergodic properties}
More precisely, let us introduce a probabilistic space
$(\Omega, {\cal F}, \P)$, where $\Omega$ is identified with $\{{\T}^{\Z} \}$,
$\T$ being the torus,
and $\P=\otimes_{k\in\Z}\P_k$, where $\P_{2k}=\P_0$ and $\P_{2k+1}=\P_1$ for any $k\in\Z$
are probability distributions on $\T$ and ${\cal F}$ the $\sigma$algebra generated by the
cylinders. We introduce
the set of random vectors on
$(\Omega, {\cal F}, \P)$ given by
\bea\label{beta}
& &\beta_k=(\theta_k, \alpha_k): \Omega \rightarrow \T^2,
\,\,\, k\in \Z,\nonumber\\
& &\theta_k^{\omega}=\omega_{2k}, \,\,\,\, \alpha_k^{\omega}=\omega_{2k+1}.
\eea
The random vectors $\{\beta_k\}_{k\in\Z}$ are thus i.i.d on $\T^2$.
We denote by $U_{\omega}$ the random unitary operator corresponding to the random
infinite matrix (\ref{matel}). In analogy with Jacobi matrices describing
the discrete Schr\"odinger equation, we will also denote the vector $\ffi_k$
by the site $k$, $k\in\Z$.
Introducing the shift operator $S$ on $\Omega$ by
\be\label{shift}
S(\omega)_k=\omega_{k+2}, k\in\Z,
\ee
we get an ergodic set $\{S^j\}_{j\in\Z}$ of translations.
With the unitary operator $V_j$ defined on the canonical basis of
$l^2(\Z)$ by
\be
V_j\ffi_k=\ffi_{k2j}, \forall k\in\Z,
\ee
we observe that for any $j\in\Z$
\be\label{ero}
U_{S^j\omega}=V_jU_{\omega}V_j^*.
\ee
Therefore, our random operator $U_{\omega}$ is a an ergodic unitary operator.
Now, general arguments on the properties of the spectral resolution
of ergodic operators $E_{\omega}(\Delta)$, where $\Delta$ is a Borel set of the torus
$\T$, ensure that this projector is weakly measurable, as well as
$E^{x}_{\omega}(\Delta)=P^x_{\omega}E_{\omega}(\Delta)$,
where $x=p.p., \;\;a.c.$ and $s.c.$, denote the pure point, absolutely continuous
and singular continuous components, see \cite{cl}, chapter V.
The analysis performed in \cite{bhj} for the case where
$\{(\theta_k^{\omega}, \alpha_k^{\omega})\}_{k\in\Z}$ are uniformly distributed on
the torus shows that the a.c. component of the spectrum of $U_{\omega}$ is
almost surely empty.
\subsection{Lyapunov Exponent}
Let us proceed by recalling some facts concerning the Lyapunov exponent.
It is shown in \cite{bb} and \cite{bhj} that generalized eigenvectors defined by
\bea\label{eveq}
&&U_{\omega}\psi=e^{i\lambda}\psi, \nonumber \\
&&\psi=\sum_{k\in\Z}c_k\ffi_k, \,\, c_k\in \C, \,\, \lambda\in\C
\eea
in our unitary setting can be computed by means of $2\times 2$ transfer matrices
due to the structure of the matrix $U_{\omega}$. They are such that for all
$k\in\Z$, (\cite{bhj})
\be\label{trans}
\pmatrix{c_{2k} \cr c_{2k+1}}=T(k)\pmatrix{c_{2k2} \cr c_{2k1}}
\ee
where the randomness lies in the phases $\eta_k(\lambda)\equiv \eta_k^{\omega}(\lambda)$
defined by
\be
\eta_k(\lambda)=\eta_k+\lambda,
\ee
and
\bea\label{tren}
T(k)_{11}&=& e^{i\eta_{2k1}(\lambda)} \\
T(k)_{12}&=&i\frac{r}{t}\left(e^{i\eta_{2k1}(\lambda)}
1\right)\nonumber\\
T(k)_{21}&=&i\frac{r}{t}\left(e^{i(\eta_{2k}(\lambda)\eta_{2k1}(\lambda))}
e^{i\eta_{2k1}(\lambda)}\right)\nonumber\\
T(k)_{22}&=&\frac{1}{t^2}\,e^{i\eta_{2k}(\lambda)}
+\frac{r^2}{t^2}\left(e^{i(\eta_{2k}(\lambda)\eta_{2k1}(\lambda))}
+1e^{i\eta_{2k1}(\lambda)}\right).\nonumber
\eea
Note the properties
\be\label{mapt}
T(k) \equiv T(\eta_{2k}(\lambda), \eta_{2k1}(\lambda))
\ee
whereas $ \det T(k)=e^{i(\eta_{2k}\eta_{2k1})}$ is independent of $\lambda$.
Therefore, knowing {\em e.g.} the coefficients $(c_0, c_1)$, we compute
for any $k\in\N$,
\bea\label{cocycle}
\pmatrix{c_{2k} \cr c_{2k+1}}&=&T(k)\cdots T(2)T(1)\pmatrix{c_{0} \cr c_{1}}
\equiv \Phi(k)\pmatrix{c_{0} \cr c_{1}}\nonumber \\
\pmatrix{c_{2k} \cr c_{2k+1}}&=&T(k+1)^{1}\cdots
T(1)^{1}T(0)^{1}\pmatrix{c_{0} \cr c_{1}}
\equiv \Phi(k)\pmatrix{c_{0} \cr c_{1}}.
\eea
The dynamical system at hand being ergodic and the determinant of the transfer
matrices being of modulus one, we get the existence of a deterministic Lyapunov exponent
$\gamma(e^{i\lambda})$, for any $\lambda\in\C$, such that
\be\label{lyapu}
\lim_{k\ra\pm\infty}\frac{1}{k}\ln\\Phi(k)\=\gamma(e^{i\lambda})\;\;\; \mbox{a.s.}.
\ee
Writing $e^{i\lambda}=z\in\C\setminus\{0\}$, we also know from classical arguments,
see {\it e.g.} \cite{cfks}, that $\gamma$ is a subharmonic function of $z$.
\setcounter{equation}{0}
\section{Density of States}
Following the standard approach in the selfadjoint case, we start by a
definition of the density of states by averaging over the phases and invoking
the RieszMarkov theorem. Then we relate the density of state with alternative
definitions in terms of the density of eigenvalues of truncations of
the original operator to $l^2([M,N])$, as $NM\ra \infty$. \\
{\bf Definition:} {\em The density of states is the (nonrandom)
measure $dk$ on $\T$ defined by
\be\label{dos}
\int_{\T}f(e^{i\lambda})dk(\lambda):=\E[\bra\ffi_0f(U_{\omega})\ffi_0\ket+
\bra\ffi_1f(U_{\omega})\ffi_1\ket
]/2,
\ee
for any continuous function $f: S^1\ra\C$.}\\
The average over the $\ffi_0$ and $\ffi_1$ matrix elements is motivated by
the forms of the matrix (\ref{matel}) and shift (\ref{shift}). Note also that
this definition makes $dk$ a probability measure.
Now we turn to the definition of appropriate finite size unitary matrices
constructed from (\ref{matel}). There are several possible constructions
suited to our purpose. Those we use below result from
considering $U_{\omega}$ provided with boundary conditions at certain sites
forbidding transitions through these sites.
Although such an interpretation is not needed in the sequel, let us mention
it can be seen in \cite{bhj}. There, a more general unitary matrix
than (\ref{matel}) is considered, whose reflection and transition coefficients
$(r_k,t_k)$ may depend on the index $k$, whereas (\ref{matel}) is a special case
with $r_k=r$ and $t_k=t$. Imposing $t_N=0$ there, one gets that the matrix takes
a block structure which decouples the sites with indices smaller than $N$ from
those with indices larger than $N$.
Let us drop temporarily the sub and superscripts $\omega$ in the notation.
Fix $N\in\Z$ and consider the unitary operator $U^{2N}$ on $l^2(\Z)$
obtained from the original operator $U$ by imposing the following boundary
conditions at the sites $2N$.
Let $U^{2N}$ be defined by (\ref{matel}) for $k\notin \{2N, 2N+1\}$ where
\be
\eta_{2N1}=\eta_{2N}=\eta_{2N+1}=\eta_{2N+2}=0
\ee
and, for $k\in \{2N, 2N+1\}$
\bea\label{even}
&& U^{2N}\ffi_{2N}=it\ffi_{2N1}+r\ffi_{2N}\nonumber\\
&& U^{2N}\ffi_{2N+1}=r\ffi_{2N+1}+it\ffi_{2N+2}.
\eea
Similarly, a boundary condition imposed at site $2N+1$ defines $U^{2N+1}$
by (\ref{matel}) for $k\notin \{2N, 2N+1, 2N+2, 2N+3\}$ where
\be
\eta_{2N+1}=\eta_{2N+2}=0
\ee
and, for $k\in \{2N, 2N+1, 2N+2, 2N+3\}$
\bea\label{odd}
&& U^{2N+1}\ffi_{2N}=irte^{i\eta_{2N}}\ffi_{2N1}+r^2e^{i\eta_{2N}}
\ffi_{2N}+it\ffi_{2N+1}\nonumber\\
&& U^{2N+1}\ffi_{2N+1}=t^2e^{i\eta_{2N}}\ffi_{2N1}+
irte^{i\eta_{2N}}\ffi_{2N}+r\ffi_{2N+1}\nonumber\\
&& U^{2N+1}\ffi_{2N+2}=r\ffi_{2N+2}+irte^{i\eta_{2N+3}}\ffi_{2N+3}
t^2e^{i\eta_{2N+3}}\ffi_{2N+4}\nonumber\\
&& U^{2N+1}\ffi_{2N+3}=+it\ffi_{2N+2}+r^2e^{i\eta_{2N+3}}\ffi_{2N+3}+
irte^{i\eta_{2N+3}}\ffi_{2N+4}.
\eea
For any $M\in\Z$, the corresponding operator $U^M$ has a the block structure
mentionned above and it is unitary. Then, given $(M,N)\in\Z^2$ such that $M+40$
is given, there exists $R(\eps)<\infty$ so that
\be
\sup_{\theta\in\T}\leftf(\theta)P_{R(\eps)}(\theta)\right\leq \eps.
\ee
Hence we get using (\ref{rela}),
\bea
&&\mbox{ tr }(f(V^{M,N})\chi_{M,N}f(U)\chi_{M,N})=
\mbox{ tr }(\chi_{M,N}(f(U^{M,N})f(U))\chi_{M,N})=\nonumber\\
&&\mbox{ tr }(\chi_{M,N}(P_{R(\eps)}(U^{M,N})P_{R(\eps)}(U))\chi_{M,N})+\nonumber\\
&&\mbox{ tr }(\chi_{M,N}((fP_{R(\eps)})(U^{M,N})(fP_{R(\eps)})(U))\chi_{M,N}),
\eea
where the trace norm of the last term is bounded by
$2\eps (NM)$, so that it becomes negligeable when divided by $(NM)$.
We are thus to consider $z^s$ and $\bar{z}^s$, with $s\in\N$.
We can write for any $s\geq 1$
\be
U^s(U^{N,M})^s=\sum_{j=0}^{s1}U^j(UU^{N,M})(U^{N,M})^{sj1},
\ee
so that
\be
\chi_{M,N}(U^s(U^{N,M})^s)\chi_{M,N}=\sum_{j=0}^{s1}\chi_{M,N}U^j
(UU^{N,M})\chi_{M,N}(U^{N,M})^{sj1}.
\ee
Therefore,
\be
\frac{\mbox{ tr }(\chi_{M,N}(U^s(U^{N,M})^s)\chi_{M,N})}{NM}\leq
\frac{s\(UU^{N,M})\chi^{M,N}\_1}{NM}.
\ee
The same result is true if $s<0$, with all unitaries replaced by their adjoints.
Thus, $R(\eps)\leq s\leq R(\eps)$ and the hypothesis on the trace norm
of $(UU^{N,M})\chi^{M,N}$ yield the result.\hfill \ep
Then, restoring the dependence on $\omega$ in the notation, we get by
the same arguments as in the self adjoint case,
that the density of states is almost surely the limit in the vague sense of
the measures $dk_{M,N}$ and $\tilde{dk}_{M,N}$ as $NM\ra\infty$. A proof is
provided in Appendix for completeness.
\begin{prop} \label{p2}For any continuous function $f: S^1\ra\C$,
\be\label{eqdef}
\lim_{NM\ra\infty}\int_{\T}f(e^{i\lambda})\tilde{dk}_{M,N}^{\omega}(\lambda)=
\int_{\T}f(e^{i\lambda})dk(\lambda)\;\;\;\mbox{ a.s. },
\ee
and the support of the density of states $dk$ coincides with $\Sigma$, the a.s.
spectrum of $U_{\omega}$.
\end{prop}
\setcounter{equation}{0}
\section{Thouless Formula}
The link between the density of states and the Lyapunov exponent is
provided by an analysis of the spectrum of the finite unitary matrices
$V^{M,N}$. It reads
\begin{thm}\label{ttf} [Thouless Formula ]
For any $z\in\C\setminus\{0\}$
\be\label{tf}
\gamma(z)=2\int_{\T}\lnze^{i\lambda'}dk(\lambda')+\ln(1/t^2)\ln z.
\ee
\end{thm}
{\bf Remarks:}\\
0) The identity $\gamma(1/\bar{z})=\gamma(z)$ holds.
\\
i) It follows from the above formula, as in Theorem 4.6 in \cite{gt}, that the integrated
density of states is continuous and satisfies
\be
N(\lambda_1 ) N( \lambda_2)\leq
\frac{\ln(2/t^2)}{\lne^{i\lambda_1} e^{i\lambda_2}}\, ,\,\,\mbox{ where }\, \,
N(\lambda)=\int_{\pi}^{\lambda}dk(\lambda'),
\ee
by an argument of Craig and Simon \cite{cs}.\\
ii) In case $z=e^{i\lambda}\in S^1$, the formula can be cast into the form
\be
\gamma(e^{i\lambda})= \int_{\T}\ln(\sin^2((\lambda\lambda')/2))dk(\lambda')+
\ln(4/t^2),
\ee
from which we recover the estimate
$0\leq \gamma(e^{i\lambda})\leq \ln(4/t^2)$
that follows from the form of the transfer matrices (\ref{tren}).
The proof of this version of Thouless formula is given at the end of the section and
we proceed with a Corollary and an application of this formula. The Corollary
essentially expresses the radial derivative of the Lyapunov exponent as the
Poisson integral of the density of states measure $dk$, which allows to recover
the a.c. component of $dk$ by a limiting procedure.
\begin{cor} For any $\eps >0$ and any $\lambda'\in\T$,
\bea
&& \lim_{\eps\ra 0^+}\gamma(e^{i\lambda'}e^{\eps})=\gamma(e^{i\lambda'}),\\
&& \frac{\partial}{\partial \eps} \gamma(e^{i\lambda'}e^{\pm \eps})=\mp
\int_{T}\frac{1e^{i\lambda'} e^{\pm \eps}^2}{e^{i\lambda}
e^{i\lambda'} e^{\pm \eps}^2}dk(\lambda)\equiv \mp
P[dk](e^{i\lambda'} e^{\pm \eps}).
\eea
Therefore, if $n(\lambda)d\lambda/2\pi$ denotes the a.c. component
of $dk(\lambda)$,
\be\label{limd}
\lim_{\eps\ra 0^+} \frac{\partial}{\partial \eps}
\gamma(e^{i\lambda'}e^{\eps})=n(\lambda')= \frac{\partial}{\partial \eps}
\gamma(e^{i\lambda'}),
\ee
where the limit and the derivative exist for Lebesgue almost all $\lambda'\in\T$.
\end{cor}
{\bf Remark:}\\ As in \cite{cs}, it follows also from the subharmonicity
of $\gamma(z)$, that if $\gamma(e^{i\lambda_0})=0$, then $\gamma: S^1 \ra \R^+$
is continuous at $e^{i\lambda_0}$.\\
{\bf Proof:}\\ Let us first consider the second statement with lower
indices only. We compute
\be\label{eeps}
\gamma(e^{i\lambda'}e^{\eps})=\eps+\ln(1/t^2)+\int_{\T}
\ln(1+e^{2\eps}e^{\eps}2\cos(\lambda\lambda'))dk(\lambda),
\ee
which we can differentiate under the integral sign as long as $\eps>0$ to get
\bea
\frac{\partial}{\partial \eps} \gamma(e^{i\lambda'}e^{\eps})&=&1
+\int_{\T}\frac{2e^{2\eps}+e^{\eps}2\cos(\lambda\lambda')}{1+e^{2\eps}
e^{\eps}2\cos(\lambda\lambda')}dk(\lambda)\nonumber\\
&=&\int_{\T}\frac{1e^{2\eps}}{1+e^{2\eps}e^{\eps}2\cos(\lambda\lambda')}
dk(\lambda)=P[dk](e^{i\lambda'}e^{\eps}).
\eea
The existence for almost all $\lambda'\in\T$ of the limit and the
first equality in (\ref{limd}) is a direct consequence of the above equality.
The existence and equality with the derivative at zero
for such $\lambda'$ follows from the mean value Theorem.
To get the first statement, notice that
$1+e^{2\eps}e^{\eps}2\cos(x)>2e^{\eps}(1\cos(x))$ in formula (\ref{eeps}) above
yields
\bea
&&0\leq \ln((1+e^{2\eps}e^{\eps}2\cos(\lambda\lambda'))/4)<
\ln(2e^{\eps}(1\cos(\lambda\lambda'))/4)=\nonumber\\
&&\eps \ln((1\cos(\lambda\lambda'))/2),
\eea
where the last function is in $L^1(\T, dk)$ by Thouless formula.
Therefore, an application of the dominated convergence
Theorem shows we can take the limit
$\eps\ra 0$ inside the integral to get the result.
\hfill \ep\\
We consider now the properties of
$U_{\omega}$ characterized by i.i.d. phases
$\theta_k^{\omega}$ and $\alpha_k^{\omega}$ in the definition (\ref{defeta}),
assuming one set of phases is uniformly distributed on $\T$. In that situation,
not only can we can prove the transfer matrices have a (positive) Lyapunov behaviour, but
we can also exactly compute the Lyapunov exponent $\gamma(e^{i\lambda})$. This
shows that in this situation, the spectrum of $U_{\omega}$ is almost surely singular,
in view of the unitary version of the IshiiPastur Theorem proven in \cite{bhj}.
This strengthens the corresponding results of \cite{bhj}, Theorem 4.1 and
Propositions 5.4. There Furstenberg's Theorem is applied to prove positivity
of the Lyapunov exponent, so that no value for $\gamma(e^{i\lambda})$ is provided.
\begin{thm}\label{co}
Let $(\theta_k^{\omega})_{k\in\Z}$ and $(\alpha_k^{\omega})_{k\in\Z}$ be i.i.d.
on $\T$ and assume the distribution of either the $\theta_k^{\omega}$'s or the
$\alpha_k^{\omega}$'s is uniform on $\T$. Then, for any $\lambda\in \T$,
\be
dk(\lambda)=d\lambda/2\pi,\,\,\,\mbox{ and }\,\,\,
\gamma(e^{i\lambda})=\ln(1/t^2)>0,
\ee
therefore,
\be
\sigma(U_{\omega})_{a.c}=\emptyset\,\,\, \mbox{ and }
\,\,\,\sigma(U_{\omega})_{sing.}=S^1 \,\,\,\,\mbox{almost surely.}
\ee
\end{thm}
{\bf Remark:}\\
The assumption on the distribution of the phases actually implies that the $\eta_k$'s are
i.i.d. and uniform on $T$, see Lemma \ref{uni} below. This explains why the a.s.
spectrum coincides with $S^1$ and why the density of states is flat.\\
{\bf Proof of Theorem \ref{co}:}\\
We first use the following lemma of purely probabilistic nature proven in Appendix.
\begin{lem}\label{uni}
Under the hypotheses of Theorem \ref{co}, the $\eta_k^{\omega}$'s are i.i.d.
and uniform on $T$.
\end{lem}
Then we show the density of states is uniform for uniformly distributed phases.
Expanding (\ref{defeta}) of the $\eta_k(\omega)$'s
we can write for any $n\neq 0$,
\bea \label{exploit}
&&\bra\ffi_j U^n_{\omega}\ffi_j\ket=\sum_{\vec{k}=k_1,k_2,\cdots,k_{n1}}(U_{\omega})_{j,k_1}
(U_{\omega})_{k_1,k_2}\cdots (U_{\omega})_{k_{n1},j}=\nonumber\\
&&\sum_{\vec{k}}\exp\left(i\sum_{l\in {\cal L}}p_l\eta_l(\omega)\right)(U_{0})_{j,k_1}
(U_{0})_{k_1,k_2}\cdots (U_{0})_{k_{n1},j},
\eea
where $U_0$ corresponds to $U_{\omega}$ when all phases $\eta_k=0$ and where
${\cal L}$ is a finite set of indices depending on $j,\vec{k},n$ and $p_l$ are integers.
Observing that the variables $\eta_k(\omega)$'s all appear with the same sign
in (\ref{matel}), no compensation can take place between contributions of
different matrix elements above and one at least among the integers $p_l$, for
$l\in{\cal L}$ is stricly positive when $n\neq 0$. Using independence and the
characterization $\E(e^{im\eta_k(\omega)})=\delta_{m,0}$ of the uniform distribution,
we get
\be
\E(\bra\ffi_j U^n_{\omega}\ffi_j\ket)=\delta_{n,0}\,\,\Longrightarrow
\int_{\T}e^{in\lambda}dk(\lambda)=\delta_{n,0}
\ee
and the first statement follows.
The second equality is a consequence of Thouless formula together with the
identity
\be
\int_0^{2\pi}\ln1e^{i\lambda}d\lambda=0.
\ee
The singular nature of the almost sure spectrum of $U_{\omega}$ comes from the
unitary version of IshiiPastur Theorem proven as Theorem 5.3 in
\cite{bhj}, which is independent of the properties of the common distributions of the
$\alpha_k$'s and $\theta_k$'s and only requires ergodicity. Finally, Proposition \ref{p2}
yields the result about the support of the a.s. singular spectrum.
\hfill\ep
We compute here, for the sake of completeness, the density of
states and Lyapunov exponent for the deterministic free operator $U_0$
corresponding to $U_{\omega}$ in case $\eta_k=0, \forall k\in\Z$. In this
case, equation (\ref{eqdef}) of Proposition \ref{p2} becomes a definition
of the free density of states $dk_0$, provided the limit exists.
That the limit exists, is the content of the next
\begin{lem}\label{deflim} The free density of states $dk_0$ exists when defined
for any $f\in C(S^1)$ by
\be
\int_{T}f(e^{i\lambda})dk_0(\lambda)=\lim_{NM\ra \infty}\int_{\T}f(e^{i\lambda})
d\tilde{k}_{M,N}(\lambda).
\ee
\end{lem}
As we know essentially everything about the purely a.c. operator $U_0$, we can also
use a direct approach to perform these computations. In particular, the integrated
density of states of $U_0$ can be defined as the distribution function on $\T$ of
the band functions yielding the spectrum $\Sigma_0$ of $U_0$. This direct approach
of the density of states coincides with the above definition,
see the proofs of
Proposition \ref{freep} and Lemma \ref{deflim} in Appendix.
We note here that the spectrum of $U_0$ consists in the set
\be
\Sigma_0=\{e^{\pm i(\arccos(r^2 t^2\cos(y)))}, y\in\T\}.
\ee
We get in particular that $\Sigma_0$ is the
support of the density of states whereas $\Sigma_0^c$ is that of the Lyapunov exponent:
\begin{prop}\label{freep} If $N_0$, $dk_0$ and $\gamma_0$ denote the integrated density of
states, the density
of states and Lyapunov
exponents of $U_0$, respectively. We have for $\lambda\in\T\simeq ]\pi,\pi]$,
\bea
dk_0(\lambda)&=&\left\{\matrix{\frac{\sin(\lambda)}{2\pi\sqrt{t^4(r^2\cos(\lambda))^2}}d\lambda &
\mbox{\em if }\lambda <\arccos(r^2t^2)\cr 0 & \mbox{\em otherwise }}\right.\\
\label{nzero}
N_0(\lambda)&=&\left\{\matrix{\frac{1}{2\pi}\arccos\left(\frac{r^2\cos(\lambda)}{t^2}\right) &
\mbox{\em if }\lambda\in [\arccos(r^2t^2),0]\cr 1\frac{1}{2\pi}
\arccos\left(\frac{r^2\cos(\lambda)}{t^2}\right) &\mbox{\em if } \lambda\in [0,\arccos(r^2t^2)]}\right.\\
\gamma_0(e^{i\lambda})&=&\left\{\matrix{0 &
\mbox{\em if } \lambda\leq \arccos(r^2t^2)\cr
\cosh^{1}\left(\frac{r^2\cos(\lambda)}{t^2}\right)
& \mbox{\em otherwise. }}\right.
\eea
Finally, Thouless formula (\ref{tf}) holds true for these quantities with $z=e^{i\lambda}$,
$\lambda\in\T$.
\end{prop}
{\bf Remarks:}\\ Note that the density of $dk_0(\lambda)$ diverges as
$1/\sqrt{\lambda \arccos(r^2t^2)}$ at the band edges
and behaves as $1/2\pi t$ as $\lambda\ra 0$. \\
The integrated density of states $N_0(\lambda)$ tends to its values $0$ and $1$ as
$\sqrt{\lambda \arccos(r^2t^2)}$ at the band edges.\\
Also, in keeping with the fact that $U_0$
becomes a shift if $t=1$ and the identity as $r=1$, $N_0(\lambda)$ becomes linear in $\lambda$ as
$t\ra 1$ and a step function as $r\ra 1$.\\
The Lyapunov exponent, where non zero, is equivalently given by
\be
\gamma_0(e^{i\lambda})=\ln \left( \frac{r^2\cos(\lambda)}{t^2}+
\sqrt{\left(\frac{r^2\cos(\lambda)}{t^2}\right)^21}\right).
\ee
It is an even $C^{\infty}$ function of $\lambda$ on $\{ \lambda> \arccos(r^2t^2) \}$,
strictly increasing
on $[\arccos(r^2t^2),\pi]$. And $d \gamma_0(e^{i\lambda})/d\lambda$ behaves as
$1/\sqrt{\lambda \arccos(r^2t^2)}$ as $\lambda\ra \arccos(r^2t^2)^+$.\\
Given Lemma \ref{deflim} above, it is clear that Thouless formula holds
for the above quantities. A direct proof of this fact is nevertheless given in Appendix.
\subsection{Proof of Thouless Formula}
We now turn to the proof of Theorem \ref{ttf}.
Writing down explicitely the effect of the boundary conditions
at $N>M$ on the coefficients of the eigenvector (\ref{eveq}) we obtain
the following relations, which depend on the parity of $N$ and $M$.
Let $\psi^{M,N}=\chi_{M,N}\psi$ and consider
\be\label{redeq}
V^{M,N}\psi^{M,N}=e^{i\lambda}\psi^{M,N} \,\,\,\, \mbox{ in } l^{2}[M+1,N].
\ee
We get by inspection,
\begin{lem} Assume (\ref{redeq}) is satisfied. Then, if $M$ is even
\be
\pmatrix{c_{M+2}\cr c_{M+3}}=c_{M+1}b_1(e^{i\lambda})\equiv
c_{M+1} \frac{1}{t^2}\pmatrix{it(re^{i\lambda})
\cr (re^{i\lambda})+r(re^{i\lambda})}.
\ee
If $M$ is odd,
\be
\pmatrix{c_{M+1}\cr c_{M+2}}=c_{M+1}b_2(e^{i\lambda})\equiv
c_{M+1}\frac{1}{it}\pmatrix{it \cr e^{i\lambda}r}.
\ee
Similarly, if $N$ is even,
\be
\pmatrix{c_{N2}\cr c_{N1}}=c_{N}b_3(e^{i\lambda})\equiv
c_{N} \frac{1}{t^2}\pmatrix{
(re^{i\lambda})+r(re^{i\lambda})\cr it(re^{i\lambda})}.
\ee
If $N$ is odd,
\be
\pmatrix{c_{N1}\cr c_{N}}=c_{N1}b_4(e^{i\lambda})\equiv
c_{N1}\frac{1}{it}\pmatrix{e^{i\lambda}r \cr it}.
\ee
\end{lem}
These relations together with the formulas (\ref{cocycle}) allow
to describe the spectrum of $V^{M,N}$ in a convenient manner.
\begin{cor}\label{corol}
Let $M0$. The last property can be checked also by means of the
transfer matrices (\ref{tren})
Let $\alpha\in \mbox{supp} \mu$. Then, for all $\eps >0$, there exists a set
$I_{\eps}\ni \alpha$ such that $I_{\eps}\leq \eps$, and $\mu( I_{\eps})>0$.
With the notation $\omega(k)=\eta_k(\omega)$, $k\in\Z$, we define for all $n\in\N$ and
$k\in\Z$,
\be
A_n(k)=\{\omega(kn)\in I_{\eps},\omega(kn+1)\in I_{\eps}, \cdots ,
\omega(kn+ n1)\in I_{\eps} \}.
\ee
Due to the assumed independence, we have for any $k$,
$ \P(A_n(k))=\mu( I_{\eps})^n >0$
so that for any $n>0$, by BorelCantelli,
$\P(\cup_{k\in \Z}A_n(k))=1$.
Let $\Delta_n(k)=\{kn, kn+1, \cdots, kn+n1\}$ denote the set of indices appearing in
$A_n(k)$ and consider now
\be
\psi_{n,k}(\lambda)=\sum_{j\in\Delta_n(k)}c_j(\lambda)\ffi_j=
\chi(\Delta_n(k))\psi(\lambda),
\ee where $\chi(\Delta_n(k))$ is the projector on the span of $\{\ffi_j\}_{j\in \Delta_n(k)}$
Because of (\ref{propgen}),
\be
U_0\psi_{n,k}(\lambda)=e^{i\lambda}\psi_{n,k}(\lambda) + R^_{kn}(\lambda)+R^+_{k(n+1)},
\ee
where the vectors $R^{\pm}_{j}$ have at most four components close to the index $j$ and
\be
\R^{\pm}_{j}\\leq R, \mbox{ where } R \mbox{ is uniform in } j.
\ee
Also, by construction of $A_n(k)$, $U_0$ and $U_{\omega}$, we have
\bea
\ U_{\omega}\psi_{n,k}(\lambda)e^{i\alpha}U_0\psi_{n,k}(\lambda)\
&\leq &\ (U_{\omega}e^{i\alpha}U_0)
\chi(\Delta_n(k)) \\psi_{n,k}(\lambda)\\nonumber\\
& = &O(\eps) \\psi_{n,k}(\lambda)\,
\eea
where the estimate $O(\eps)$ is uniform in $n$ and $ k$. Therefore, for all $\eps>0$
and all $n>0$, there exists, with probability one, a $k$ such that $A_n(k)$ and the
corresponding $\psi_{n,k}(\lambda)$ have the above properties so that
\bea
&&\U_{\omega}\psi_{n,k}(\lambda)e^{i(\alpha+\lambda)}\psi_{n,k}(\lambda)\/
\\psi_{n,k}(\lambda)\=\nonumber\\
&&(\(U_{\omega}e^{i\alpha}U_0)\psi_{n,k}(\lambda)+
e^{i\alpha}(U_0e^{i\lambda})\psi_{n,k}(\lambda)\)/\\psi_{n,k}(\lambda)\\leq \nonumber\\
&& O(\eps)+2R/\\psi_{n,k}(\lambda)\=O(\eps + 1/n).
\eea
It remains to chose $n= [1/\eps]$ to conclude that $e^{i(\alpha+\lambda)}\in
\sigma(U_{\omega})$ almost surely.
Let us now show that $S^1\setminus \Sigma$ belongs to the resolvent set of
$U_{\omega}$. In order to do so we use Lemma \ref{anderson}
Therefore, we can consider as well the spectrum of the product $D_{\omega}S_0$
to which the perturbation theory recalled in Chap.1, \S 11 of \cite{yaf} for example,
applies. In particular, dropping the $\omega$ in the notation as randomness plays
no role here, if we know that for all $j\in\Z$, $\eta_j\in[\alpha,\beta]\subset \T$, then
$\sigma(D)\subseteq (\delta_1,\delta_2)$ where $(\delta_1,\delta_2)$ denotes the corresponding
arc on the unit circle swept in the positive direction from $\delta_1\in S^1$ to
$\delta_2\in S^1$. We denote by $(\delta_1,\delta_2)$ the length on the torus of this
arc. Since $\sigma(S_0)=\Sigma_0$ corresponds to the symmetric arc $(e^{i\arccos(r^2t^2)},
e^{i\arccos(r^2t^2)})$, perturbation theory tells us that after (multiplicative)
perturbation by $S_0$, the spectrum of $U\simeq DS_0$ is a subset of an arc of wider aperture than
$(\delta_1,\delta_2)$. Quantitatively, Theorem 8, p.65 in \cite{yaf} tells us that the arc
$(e^{i\arccos(r^2t^2)}\delta_2,e^{i\arccos(r^2t^2)}\delta_1)$ belongs to the
resolvent set of $U$, provided $(\delta_1,\delta_2)<(e^{i\arccos(r^2t^2)},
e^{i\arccos(r^2t^2)})$. This condition simply insures that the subset of the resolvent
set we are talking about is not reduced to the empty set. This is enough to get the result
in case the support of $\mu$ is such that $\Sigma$ is connected. In case this set is not
connected, as $\Sigma_0>0$, it consists of a finite set of connected components, each
of which can be associated with the convex hull of sufficiently far apart subsets of
the support of $\mu$. Denoting these subsets by $m_j$, $j=1,\cdots,N$ and the associated
arcs on $S^1$ by $(M_1(j),M_2(j))$, we have that the spectrum of $D$ is the disjoint union
of subsets $\sigma_j$ satisfying $\sigma_j\subseteq (M_1(j),M_2(j))$. The same argument as above
says that the spectrum of $DS_0$ is confined to the finite union of arcs
$((e^{i\arccos(r^2t^2)}M_1(j),(e^{i\arccos(r^2t^2)}M_2(j))$, which ends the
proof of the Theorem.
\hfill \ep
\subsection{Analyticity of the density of states}
At the price of some combinatorics, we can further exploit the relation
(\ref{exploit}) in order to
obtain a condition on the common distribution of the $\eta_k$'s ensuring the analyticity
of the density of states. Recall that a function $f$ on $\T$ is analytic, if and only if
its Fourier coefficients $\hat{f}$ satisfy an estimate of the form
\be\label{four}
\hat{f}(n)\leq A e^{Bn}, \,\,\, \forall n\in\Z,
\ee
for some positive constants $A,B$. We have
\begin{thm}\label{52} Assume the $\eta_k$'s are distributed according to a law that
has an
analytic density $f$ characterized by the estimate (\ref{four}) with $A,B>0$.
Then, if
\be
B>\ln(1+2rt)+\ln A,
\ee
the density of states $dk$ admits an analytic density, so that the integrated
density of states $N$ is analytic as well.
\end{thm}
{\bf Remarks:}\\ As $\hat{f}(0)=\int_{\T}f(\eta)d\eta =1$, $A\geq 1$.\\
When the Theorem applies, it prevents the Lyapunov exponent from being zero
on a set of positive measure.\\
This result has to be compared with the Proposition VI. 3.1. of \cite{cl} stating
a similar result for the $d$dimensional Anderson model.\\
As an immediate consequence, using $r^2+t^2=1$, we get the following
\begin{cor}
If the $\eta_k$'s have an analytic density $f$, characterized by
(\ref{four}) with $B>\ln A$, then there exist $r^+(f)$ and $r^(f)$ in
$]0,1[$ such that the density of states is analytic provided the reflexion coefficient
$r$ satisfies $1>r>r^+(f)$ or $0\ln(2A)$, The density of state
is analytic $\forall r\in[0,1]$.
\end{cor}
{\bf Remark:}\\ It is easy to check that in both the extreme cases $r=1$ and $r=0$,
the density of states is analytic. Indeed, if $r=1$, $dk(\lambda)=f(\lambda)d\lambda$, where
$f$ is the density of the $\eta_k$'s, whereas if $r=0$, $dk(\lambda)=d\lambda /(2\pi)$.\\
{\bf Proof of Theorem \ref{52}:}\\
By hypothesis, for any $n\in\Z$,
\be
\Phi_{\eta}(n)=\left\int_{\T}e^{i\eta n}f(\eta)d\eta\right\leq A e^{Bn}.
\ee
Then, in (\ref{exploit}) above, $\sum_{l\in {\cal L}}p_l=n$, so that using independence
\be\label{rhss}
\E\bra\ffi_j U^n_{\omega}\ffi_j\ket \leq A^ne^{Bn}
\left.\sum_{k_1,k_2,\cdots,k_{n1}}(U_{0})_{j,k_1}
(U_{0})_{k_1,k_2}\cdots (U_{0})_{k_{n1},j}\right.
\ee
Here the sum carries over a set of indices that form paths of length $n+1$
from index $j$ to index $j$. The allowed paths are those giving rise to non zero
matrix elements $(U_0)_{l,m}$ in the sum above. In order
to compute this last sum, we proceed as follows. Let us
introduce more general $j$dependent subsets ${\cal C}_{n1}(j)$ of indices of $\Z^{n1}$
that appear in the computation of the matrix element $\bra\ffi_0U_{\omega}^n\ffi_j\ket$.
This set consists of paths of the form
$\{k_0=0, k_1, k_2, \cdots, k_{n1}, k_n=j\}$ of length $n+1$ in $\Z$ from $0$ to $j$
with the condition that
\bea
&& k_{m+1}k_m\in\{0,+1,1,+2\} \quad\mbox{ if } \quad k_{m} \mbox{ is odd}\nonumber\\
&& k_{m+1}k_m\in\{0,+1,1,2\} \quad\mbox{ if } \quad k_{m} \mbox{ is even},
\eea
for all $m=0,1,\cdot, n1$. Let us define
\be
S_{n1}(j):=\sum_{{\cal C}_{n1}(0)}(U_{0})_{0,k_1}
(U_{0})_{k_1,k_2}\cdots (U_{0})_{k_{n1},j},
\ee
where the matrix elements $(U_{0})_{l,m}$ are given by $r^2, rt$ and $t^2$ respectively,
when $lm$ equals $0$, $1$ and $2$ respectively.
This quantity actually gives a crude upper bound on the probability to go from
site $0$ to $j$ in $n$ time steps, under the free evolution. It is crude in the sense
that it does not take the phases into account during that free evolution.
We are actually interested in the computation of $S_{n1}(0)$ and of the similar quantity
appearing in the computation of
$\bra\ffi_1U_{\omega}^n\ffi_1\ket$, which correspond the the sum in the right hand side
of (\ref{rhss}), in the asymptotic regime $n\ra\infty$. The case of the matrix element
$\bra\ffi_1U_{\omega}^n\ffi_1\ket$ being similar, we only consider $S_{n1}(0)$.
The plan is to use a transfer matrix formalism to evaluate the generating
function associated with $S_{n1}(j)$ and then to compute the asymptotics of
$S_{n1}(0)$.
In view of (\ref{rhss}), the following proposition implies the Theorem.
\begin{prop} \label{comb} For some constant $c>0$,
\begin{eqnarray}
S_{n1}(0)= \frac{c(r+t)^{2n}}{\sqrt{n}}(1+o(1)) \,\,\,\mbox{ as }\, n\ra\infty .
\end{eqnarray}
\end{prop}
{\bf Proof of Proposition \ref{comb}:}\\
Let
\be
P_n(x)=\sum_{2n\leq j\leq 2n}{S}_{n1}(j)x^j
\ee
be this generating function which we split into two parts
$P_n(x)=P_n^+(x)+P_n^(x)$ where
\be
P_n^{\pm}(x)=\sum_{\scriptsize \matrix{2n\leq j \leq 2n \cr
j \matrix{ \mbox{ even}\cr\mbox{ odd}} }}{S}_{n1}(j)x^j.
\ee
Clearly we have for $n=0, 1$,
\be
P_0^+(x)=r^2, P_0^(x)=0, P_1^+(x)=r^2+t^2x^{2}, P_1^(x)=rt(x+x^{1}).
\ee
It is readily shown by induction that a transfer matrix allows
to compute $P_n(x)$ for any $n$:
\begin{lem}\label{tramat}
For any $n\geq0$,
$$
\pmatrix{P_{n+1}^+(x)\cr P_{n+1}^(x)}=\pmatrix{r^2+t^2x^{2} & rt(x+x^{1}) \cr
rt(x+x^{1}) & r^2+t^2x^{2}}\pmatrix{P_n^+(x)\cr P_n^(x)},
$$
with $P_0^+(x)=r^2, P_0^(x)=0$.
\end{lem}
Denoting by $T(x)$ the transfer matrix defined in this Lemma, and introducing the
parameter
\be
\tau=t/r \in ]0,\infty[,
\ee
we rewrite it as
\be
T(x)=r^2\pmatrix{1+\tau^2x^{2} & \tau(x+x^{1}) \cr
\tau(x+x^{1}) & 1+\tau^2x^{2}}.
\ee
We will consider first the case $t\neq r \Longleftrightarrow \tau\neq 1$. The
case $\tau=1$, for which more can be said about $S_{n1}(j)$, see Proposition
\ref{better}, is dealt with below.
\subsubsection{Case $\tau\neq 1$}
The eigenvalues of $T(x)$ are given by $r^2$ times $\lambda_{\pm}(x)$, where
\be
\lambda_{\pm}(x)=\left\{1+\tau(x^2+x^{2})/2\pm
\sqrt{(1+\tau(x^2+x^{2})/2)^2(1\tau^2)^2}\right\},
\ee
so that
\be
T^n(x)=r^{2n}A(x)\pmatrix{\lambda_{+}^n(x)& 0 \cr 0 & \lambda_{}(x)^{n}}A(x)^{1}
\ee
with
\be
A(x)=\pmatrix{\lambda_+(x)(1+\tau^2 x^{2}) & \lambda_(x)(1+\tau^2 x^{2})\cr
\tau(x+x^{1}) & \tau(x+x^{1})}.
\ee
For the moment, $x$ is just book keeping parameter, so that we ignore the potential
problems of the definition of $A(x)$ in case the eigenvalues are degenerate
and we further compute
\bea
&&\pmatrix{P_n^+(x)\cr P_n^(x)}=T^n(x)\pmatrix{r^2\cr 0}=\\
&&\frac{r^{2n}\tau(x+x^{1})}
{2\sqrt{(1+\tau(x^2+x^{2})/2)^2(1\tau^2)^2}}\times\nonumber\\
&&\quad\quad \quad\quad\quad\quad\pmatrix{\lambda_+(x)^{n+1}\lambda_(x)^{n+1}(\lambda_+(x)^n\lambda_(x)^n)
(1+\tau^2x^2)\cr \tau(x+x^{1})(\lambda_+(x)^n\lambda_(x)^n)}.\nonumber
\eea
We note at this point that one checks, using the binomial Theorem, that
despite the presence of square roots in the expressions for $P_n^{\pm}(x)$,
these quantities actually are given by finite Laurent expansions in $x$, as they
should. Focusing on $P_n^+(x)$ we can rewrite with the shorthand $\sqrt{\cdot}$
for the square root of the denominator above
\bea
&& P_n^+(x)=\\
&& \quad \frac{r^{2n}\tau(x+x^{1})}
{2\sqrt{\cdot}}\left((\lambda_+(x)^n\lambda_(x)^n)
\frac{\tau^2}{2}(x^{2}+x^2)+\frac{\sqrt{\cdot}}{2}
(\lambda_+(x)^n+\lambda_(x)^n)\right).\nonumber
\eea
The quantity of interest to us is $S_{n1}(0)$, the coefficient of
$x^0$ in the expansion of $ P_n^+(x)$. Substituting $e^{i\theta}$ for
$x$ in $P_n^+$, we get a trigonometric polynomial whose zero'th Fourier
coefficient is obtained by integration
\be\label{fourier}
S_{n1}(0)=\int_{\T}P_n^+(e^{i\theta})d\theta/(2\pi).
\ee
It remains to perform the asymptotic analysis of the above integral as
$n\ra\infty$. It is a matter of routine
to verify the following propereties: The eigenvalues, as functions of
$\theta\in \T\simeq ]\pi, \pi]$, are
continuous. If $\tau<1$, they are real valued, with discontinuity
of the derivative at $\theta=\pm\pi/2$, where they cross and are given by
$1\tau^2$. At all other values of $\theta$, they are $C^{\infty}$ and
they satisfy
\be\label{satis}
\lambda_+(e^{i\theta})>\lambda_(e^{i\theta}), \mbox{ with }
\lambda_+(e^{i\theta})>1\tau^2.
\ee
If $\tau>1$, the eigenvalues become complex conjugate. Let
$\theta_c=\arccos(\frac{\tau^22}{\tau^2})/2$ be the critical value where
the square root becomes zero. If
$\theta \in [\theta_c,\pi\theta_c]\cup [\pi+\theta_c, \theta_c]$, the
eigenvalues are complex conjugate, of modulus $1\tau^2$. Otherwise they
are real valued, and satisfy (\ref{satis}) as well. Therefore, the asymptotics
as $n\ra\infty$ of (\ref{fourier}) is determined by $\lambda_+$ only.
Moreover, in both cases, $\ln(\lambda_+(e^{i\theta}))$ admits
non degenerate maxima at $\theta=0$ and $\pi$, where $\lambda_+$ reaches its
maximum value $(1+\tau^2)$.
Therefore, Laplace's method yields the asymptotics of the Proposition. \hfill\ep
\subsubsection{case $\tau=1$}
The course of the proof being the same, it is presented in Appendix.
However, instead of computing
$S_{n1}(0)$ as $n\ra\infty$, we can get exact forms for all $S_{n1}(j)$'s. The
Proposition we actually show is
\begin{prop}\label{better}
\begin{eqnarray}
{ S}_{n1}(j)&=&\frac{1}{2^n}\pmatrix{2n1 \cr j/2+n},
\quad \quad 2n\leq j\leq 2(n1), \quad \quad j \mbox{ \em even}\nonumber\\
{S}_{n1}(j)&=&\frac{1}{2^n}\pmatrix{2n1 \cr (j1)/2+n},
\quad \quad 2n+1\leq j\leq 2n1, \quad \quad j \mbox{ \em odd}
\end{eqnarray}
{\bf Remark:}\\ Of course, Stirling's formula for $n$ large yields
proposition \ref{comb} with $r=t=1/\sqrt{2}$:
\be
S_{n1}(0)=\frac{1}{2^{n}}\pmatrix{2n1 \cr n}\simeq \frac{2^{n}}{\sqrt{\pi n}}.
\ee
\end{prop}
\setcounter{equation}{0}
\section{Appendix}
{\bf Proof of Proposition \ref{p2}:}\\
We have by definition,
\be
\int_{\T}f(e^{i\lambda})\tilde{dk}_{M,N}^{\omega}(\lambda)=\frac{1}{NM}\sum_{j=M+1}^N\bra
\ffi_j f(U_{\omega})\ffi_j\ket,
\ee
where, depending on the parity of $M$ and $N$ and due to the fact that $f$ is uniformly bounded,
the right hand side can be rewritten as
\bea
&& \frac{1}{NM}\left(\sum_{k=(M+1)/2}^{N/2}
\bra \ffi_{2k} f(U_{\omega})\ffi_{2k}\ket+\bra \ffi_{2k+1} f(U_{\omega})\ffi_{2k+1}\ket\right)
+O_f(\frac{1}{NM})=\nonumber\\
&&\frac{1}{NM}\left(\sum_{k=(M+1)/2}^{N/2}
\bra \ffi_{0} f(U_{S^k(\omega)})\ffi_{0}\ket+\bra \ffi_{1} f(U_{S^k(\omega)})\ffi_{1}\ket\right)
+O_f(\frac{1}{NM}).
\eea
Now, by Birkhoff ergodic theorem, there exists $\Omega_f$ of measure one such that for all
$\omega\in\Omega_f$,
\be
\lim_{NM\ra \infty}\frac{1}{NM}\sum_{k=(M+1)/2}^{N/2}\bra \ffi_{j} f(U_{S^k(\omega)})\ffi_{j}\ket=
\frac{1}{2}\E(\bra \ffi_{j} f(U_{\omega})\ffi_{j}\ket) , \forall j\in\Z,
\ee
therefore,
\be
\frac{1}{NM}\mbox{ tr } (\chi^{M,N}f(U_{\omega}))\ra \frac{1}{2}
\left(\E(\bra \ffi_{0} f(U_{\omega})\ffi_{0}\ket+\bra \ffi_{0} f(U_{\omega})\ffi_{0}\ket)\right).
\ee
Then, $C(S^1)$ being separable, we have the existence of a countable set of $\{f_j\}_{j\in\N}$, dense
in $C(S^1)$, for which
the above is true, on a set of probability one, which proves the almost sure convergence stated
in the proposition.
Now assume $e^{i\lambda_0}\not\in \Sigma$ and take a continuous non negative $f$ such that
$f(e^{i\lambda_0})=1$ and $f_{\Sigma}=0$. Then $f(U_{\omega})=0$ a.s. so that
$\int f(e^{i\lambda}) dk(\lambda)=0$ and $e^{i\lambda_0}\not\in \mbox{ supp } k$. Conversely,
if $e^{i\lambda_0}\not\in \mbox{ supp } k$, there exists a non negative continuous $f$
with $f(e^{i\lambda_0})=1$ and $\int f(e^{i\lambda}) dk(\lambda)=0$. Hence, a.s.,
$\bra \ffi_{0} f(U_{\omega})\ffi_{0}\ket+\bra \ffi_{1} f(U_{\omega})\ffi_{1}\ket=0$,
therefore, by ergodicity, $\bra \ffi_{j} f(U_{\omega})\ffi_{j}\ket=0$ a.s. for any $j$ and
$f(U_{\omega})=0$. As $f$ is continuous and equals one at $e^{i\lambda_0}$, we get that
$e^{i\lambda_0}\not\in \Sigma$.
\hfill \ep
{\bf Proof of Lemma \ref{uni}:}\\
We only deal with the case where the $\theta_k^{\omega}$'s are i.i.d.
and uniform, the other case beeing similar. Let $\Phi_{\eta}(n)=\E(e^{in\eta_k^{\omega}})$
be the characteristic function of the random variable $\eta_k^{\omega}$, and similarly
for $\alpha_k^{\omega}$, and $\Phi_{\theta}(n)=\delta_{n,0}$. Then, using independence,
\be
\Phi_{\eta}(n)=\Phi_{\theta}(n)^2\Phi_{\alpha}(n)\Phi_{\alpha}(n)=\delta_{n,0}
\Phi_{\alpha}(n)^2=\delta_{n,0},
\ee
so that the $\eta_k$'s are uniformly distributed. Consider now
\be
\Phi_{\eta_{k_0},\eta_{ k_1},\cdots,\eta_{ k_j}}(n_0, n_1,\cdots, n_j)=\E(e^{i\sum_{l=0}^jk_l\eta_l}).
\ee
We can assume the $k_j$'s are ordered and we observe that $\eta_k$ and $\eta_{k+j}$ are
independent as soon as $j\geq 2$, see (\ref{defeta}). Therefore, we can consider
consecutive indices $k_l$ and deal with
\bea
&& \Phi_{\eta_{k},\eta_{ k+1},\cdots,\eta_{ k+j}}(n_1, n_2,\cdots, n_j)=\\
&&\E(e^{in_0\theta_{k1}+i(n_0+n_1)\theta_k+\cdots+i(n_{j1}+n_j)\theta_{k+j1}+n_j\theta_j)}
\E(f(\alpha, \vec{n})),\nonumber
\eea
where the second expectation contains $\alpha_k$'s only. Then
\bea
&&\Phi_{\eta_{k},\eta_{ k+1},\cdots,\eta_{ k+j}}(n_1, n_2,\cdots, n_j)=\nonumber\\
&&\Phi_{\theta}(n_0)\Phi_{\theta}(n_0+n_1)\cdots\Phi_{\theta}(n_{j1}+n_{j})
\Phi_{\theta}(n_j)\E(f(\alpha))=\nonumber\\
&&\delta_{n_0,0}\delta_{n_1,0}\cdots \delta_{n_j,0}\E(f(\alpha, \vec{n}))=
\delta_{\vec{n},\vec{0}}\E(f(\alpha, \vec{0}))=\delta_{\vec{n},\vec{0}},
\eea
whith the obvious notation, which yields the result.
\hfill\ep\\
{\bf Proof of Proposition \ref{freep}:}\\
We first prove this Proposition with the definition of the density
of states as the distribution function of the "band functions" of $U_0$,
to be defined below. Then we'll see in the course of the proof
of Lemma \ref{deflim} below the equivalence with the definiton as an average
counting measure.
The proof of Proposition 6.2 in \cite{bhj} shows that $U_0$ on
$l^2(\Z)$ is unitarily equivalent to the operator multiplication by the matrix
\be\label{demul}
V(x)=\pmatrix{r^2t^2e^{2ix}& 2itr\cos(x)\cr 2itr\cos(x)&r^2t^2e^{2ix}} \,\,\mbox{ on }
\,L^2(\T)\simeq L^2_+(\T)\oplus L^2_(\T),
\ee
by the unitary mapping that sends $\ffi_k\mapsto e^{ikx}/\sqrt{2\pi}$, and where
$ L^2_{\pm}(\T)$ is the susbspace generated by even/odd harmonics $\{e^{ikx}\}_{k\in\Z}$.
The eigenvalues of $V(x)$ are
\be\label{alpha}
\lambda_{\pm}(x)=e^{\pm i\alpha(x)}, \,\,\mbox{ where } \,\, \alpha(x)=
\arccos(r^2t^2\cos(2x)).
\ee
We note that $\lambda_{\pm}(x)=\lambda_{\pm}(x)$ and
\be
V(x)=JV(x)J\, \,\,\mbox{ where } \,\, J=\pmatrix{0&1\cr 1&0}.
\ee
Hence, the corresponding eigenvectors $\chi_{\pm}(x)$ satisfy
\be\label{symv}
V(x)\chi_{\pm}(x)=\lambda_{\pm}(x)\chi_{\pm}(x) \, \,\,\mbox{ and } \, \,
V(x)J\chi_{\pm}(x)=\lambda_{\pm}(x)J\chi_{\pm}(x),
\ee
so that $\chi_{\pm}(x)$ and $J\chi_{\pm}(x)$ are linearly dependent. This is in keeping with
the fact that the subspace of generalized eigenvectors is of dimension 2, see (\ref{trans}).
Also, one checks that
for any phase $\beta \in ]\arccos(r^2t^2),0[\cup ]0,\arccos(r^2t^2)[$,
\be
\alpha^{1}(\beta)=\{x_1,x_2,x_2x_1\}\subset ]\pi, \pi[ .
\ee
Therefore, due to (\ref{symv}), only half these points contribute for the computation
of the density of states. We can now compute the integrated density
of states $N_0(\beta)$
%\be
% :=\int_{\pi}^{\beta} dk(\lambda)\,\, , \,\,\,\,\, \forall \beta\in]\pi,\pi]
%\ee
as follows: Taking into account the normalisation by a factor $1/2\pi$
in the definition (\ref{dos}), the fact that $\mbox{ supp }k\subset[\arccos(r^2t^2),
\arccos(r^2t^2)] $ and the symmetries, we have for any $\beta\in [\arccos(r^2t^2),0]$
\bea
N_0(\beta)&=&\frac{1}{4\pi}\int_{\T}d\lambda\chi_{\{\alpha(\lambda )<\beta\leq 0\}}=
\frac{1}{2\pi}\int_{\pi/2}^{\pi/2}d\lambda\chi_{\{\cos(2\lambda)>(r^2\cos(\beta))/t^2\}}\\
&=&\frac{1}{2\pi}\int_0^{\arccos((r^2\cos(\beta))/t^2)}=
\frac{1}{2\pi}\arccos\left(\frac{r^2\cos(\beta)}{t^2}\right).
\eea
A similar computation for $\beta\in [0,\arccos(r^2t^2))$ yields (\ref{nzero}).
Therefore, $dk_0$ is absolutely continuous w.r.t. Lebesgue and, for any $\lambda<\arccos(r^2t^2)$,
$dk_0(\lambda)=N'(\lambda)d\lambda$,
from which the result on the density of states follows.
In order to obtain the Lyapunov exponent, it is enough to observe that the transfer matrices (\ref{trans})
$T$, now independent of $k$, are of determinant one and trace equal to $2(r^2\cos(\lambda))/t^2$.
Therefore, it is readily checked that when the eigenvalues $\tau_{\pm}(\lambda)$ of $T$
\be
\tau_{\pm}(\lambda)=(r^2\cos(\lambda)\pm\sqrt{(r^2\cos(\lambda))^2t^4})/t^2
\ee
are complex conjugates, i.e. when $\lambda<\arccos(r^2t^2)$, they are of modulus one, whereas
\be
\max\{\tau_+,\tau_\}=(r^2\cos(\lambda)\pm\sqrt{(r^2\cos(\lambda))^2t^4})/t^2 ,
\ee
if $\lambda\geq \arccos(r^2t^2)$. It remains to use definition (\ref{lyapu}) to get
$\gamma_0(e^{i\lambda})$. In order to prove the last statement, we first rewrite the right
hand side of Thouless formula with $dk_0(\lambda')$ above as
\be\label{rew}
\frac{1}{2\pi}\int_{1}^1 \frac{\ln((xy)^2)}{\sqrt{1x^2}}dx+\ln 2
\ee
by means elementary manipulations, changing variables to $x=(r^2\cos(\lambda'))/t^2$ and introducing
$y=(r^2\cos(\lambda))/t^2\in [1, (r^2+1)/t^2]$. Hence we are to show that (\ref{rew}) above equals
$0$ if $y\leq 1$ and $\ln(y+\sqrt{y^21})$ if $y >1$. We first deal with the case $y>1$. We can
differentiate (\ref{rew}) with respect to $y$ under the integral sign to get
\be\label{cint}
\frac{1}{\pi}\int_{1}^1 \frac{dx}{\sqrt{1x^2}(yx)}=\frac{1}{2\pi}\int_{C}
\frac{dz}{\sqrt{1z^2}(yz)},
\ee
where $C$ is a contour in the complex plane surrounding the segment $[1,1]$ in the positive
direction which does not contain $y$ in its interior. By deforming the contour to a circle centered
at the origin and of radius $R>0$ large enough, we pick a residue at $y$. As the integral
on the circle
is of order $1/R$, we eventually get in the limit $R\ra\infty$
\be
\frac{d}{dy}\left\{\frac{1}{2\pi}\int_{1}^1 \frac{\ln((xy)^2)}{\sqrt{1x^2}}dx+\ln 2\right \}=
\frac{1}{\sqrt{y^21}},
\ee
as expected. The limit as $y\ra 1^+$ of the Lyapunov exponent fixes the constant to $0$. Now, if
$y\in ]1,1[$, we first convert (\ref{rew}) to a contour integral along a path similar to
the one above with te following difference. As the $\ln$ is multivalued, with a cut from $y$
along the real axis towards $\infty$, the contour is attached to the point $1$. By assumption,
$y$ does not belong to the contour of integration, so that we can now differentiate with respect
to $y$ under the integral sign and thus get the same contour integral (\ref{cint}) as above
to consider. However, by expanding the contour to infinity, we get to residue this time, so that
(\ref{rew}) is constant for $y\in ]1,1[$. As it is known (\cite{gr}, \# 4.224, p.526) that,
\be
\frac{1}{2\pi}\int_{1}^1 \frac{\ln(x^2)}{\sqrt{1x^2}}dx=\frac{2}{\pi}\int_{0}^1
\frac{\ln(x)}{\sqrt{1x^2}}dx=\frac{2}{\pi}\int_0^{\pi/2}\ln(\sin(t))dt=\ln 2,
\ee
we have, by continuity, that the integral is equal to zero on $[1,1]$.
\hfill\ep
{\bf Proof of Lemma \ref{deflim}:}\\
We use freely the notations above. Let us introduce the eigenprojectors
$P_{\pm}(x)$ associated with $\lambda_{\pm}(x)$ such that
\be
V(x)=P_+(x)\lambda_+(x)+P_(x)\lambda_(x).
\ee
These quantities are analytic in $x$, in a strip including the real axis.
Let $f\in C(S^1)$ and let us compute by means of (\ref{demul}) and the definition
of $L^2_{\pm}(\T)$
\bea
&&\mbox{ tr }\bra \chi_{M,N} f(U_0)\chi_{M,N}\ket =\sum_{M