\documentstyle[]{article}
\title{Singular Continuous Spectrum for Palindromic Schr\"odinger Operators}
\author{A. Hof\thanks{Department of Mathematics and Statistics,
McMaster University, Hamilton, Ontario, Canada L8S 4K1.
Work partially supported by NSERC.} \ ,
O. Knill\thanks{Division of Physics, Mathematics and Astronomy,
California Institute of Technology, 253-37,
Pasadena, CA, 91125 USA. This material is based upon work
supported by the National Science Foundation under Grant
No. DMS-9101715.
The Government has certain rights in this material.}
\ , and
B. ${\rm Simon}^{\dagger}$}
\date{}
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\begin{abstract}
We give new examples of discrete Schr\"odinger operators
with potentials taking finitely many values that have purely
singular continuous spectrum.
If the hull $X$ of the potential is strictly ergodic,
then the existence of just one
potential $x$ in $X$ for which the operator has no
eigenvalues implies that there is a generic set in $X$ for which
the operator has purely singular continuous spectrum.
A sufficient condition for the existence of
such an $x$ is that there is a $z\in X$
that contains arbitrarily long palindromes.
Thus we can define a large class of primitive substitutions for
which the operators are purely singularly continuous for a generic
subset in $X$.
The class includes well-known substitutions like Fibonacci, Thue-Morse,
Period Doubling, binary non-Pisot and ternary non-Pisot.
We also show that the operator has no absolutely continuous spectrum for
all $x\in X$ if $X$ derives from a primitive substitution.
For potentials defined by circle maps, $x_n = 1_J (\theta_0+ n\alpha)$,
we show that the operator has purely singular continuous spectrum
for a generic subset in $X$ for all irrational $\alpha$ and every
half-open interval $J$.
\end{abstract}
\section{Introduction}
Discrete Schr\"odinger operators with
potentials taking values in a finite set $A \subset \RR$
have interesting spectral properties. Topological spaces
of such operators are obtained by
choosing a compact shift-invariant subset $X$ of
the compact metric space $A^{\ZZ}$.
If $T$ denotes the left shift on $X$, the dynamical system $(X,T)$ is called
a subshift.
Every point $x\in X$ defines an operator on $l^2(\ZZ)$ by
$$ (L(x) u)_n=u_{n+1} + u_{n-1} + x_n u_n. $$
In $X = A^{\ZZ}$, any spectral type can occur: the
dense set of periodic operators in $A^{\ZZ}$ have
purely absolutely continuous spectrum; for almost
all $x$ with respect to any non-trivial
product measure on $X=A^{\ZZ}$ the spectrum
is pure point \cite{CaKlMa87}. Hence, by the wonderland theorem \cite{Sim94},
there exists a generic set in $A^{\ZZ}$ for which $L(x)$ has purely
singular continuous spectrum.
The shift on $A^{\ZZ}$ has many invariant measures and there are
many orbits which are not dense.
It is therefore convenient to consider the case of a compact shift-invariant
$X \subset A^{\ZZ}$ that is minimal (i.e., the set of translates of
every $ x\in X$ is dense in $X$) and uniquely ergodic (i.e., there exists
only one $T$-invariant measure).
A system $(X,T)$ that is both minimal and uniquely ergodic is
called strictly ergodic.
Two common methods for generating strictly ergodic subshifts are
the following: \medskip
$\bullet$ A {\em primitive substitution\/}, which is a map $S$
from the alphabet $A$ to the
set of finite words $A^*$, defines a fixed point $z^+ \in A^{\NN}$. Taking
any $z \in A^{\ZZ}$ satisfying $z_n=z^+_{n}$ for $n \in \NN$ and defining
$X_S$ as the set of accumulation points of $\{T^nz \; | \; n \in \NN\}$
gives a strictly ergodic system. \medskip
$\bullet$ A {\em circle map} $\theta \mapsto \theta + \alpha$ with irrational
$\alpha$ together with a countable union $J$ of half-open intervals and an
initial point $\theta_0 \in \TT^1$
defines a sequence $z_n=1_J(\theta_0 + n \alpha)$. The orbit closure
$X_J=X(J,\theta_0,\alpha)$ is independent of $\theta_0$ and is a
strictly ergodic dynamical system (see \cite{Hof93} Proposition A.1.)
\medskip
We do not know of any strictly ergodic aperiodic $X \subset A^{\ZZ}$
for which some $L(x)$ has a spectral type different from
purely singular continuous.
The aim of this paper is to extend the set of examples
that have purely singular continuous spectrum.
We therefore need to exclude absolutely continuous spectrum
and eigenvalues.
The belief that singular continuous spectrum is the rule for such
aperiodic operators is supported
by Kotani's theorem \cite{Kot89} which says
that almost all $L(x)$ have no absolutely continuous spectrum.
Although the spectrum of $L(x)$ does not depend on $x \in X$ if $X$ is
strictly ergodic \cite{Hof93}, it does not follow that there is no absolutely
continuous spectrum for all $x \in X$.
If, however, the Lyapunov exponent exists for all $x \in X$ and is
independent of $x$, then indeed no $L(x)$ has absolutely continuous spectrum
as we will see below. In \cite{Hof93} it is shown that the
Lyapunov exponent exists uniformly if $X$ is defined by a primitive
substitution.
Thus we exclude absolutely continuous spectrum for all operators $L(x)$
generated by substitutions.
In order to get singular continuous spectrum one has also to exclude
eigenvalues.
This was done by a Gordon-type criterion for circle maps \cite{DePe86}
and by using the so-called trace map in
\cite{Cas86,Suto87,Suto89,Bel+89,Bel90,DePe91,Bel+91a,BoGh93}
for circle maps and substitutions.
In this paper we note that the criterion of Jitomirskaya and
Simon \cite{JS94}
can be used for sequences that we call {\em strongly palindromic}.
A strongly palindromic sequence contains palindromes $w_i$
of length $l_i$ centered at $m_i \rightarrow \infty$
such that $l_i$ grows exponentially fast with respect to $m_i$.
We show that if a sequence $x \in X$ contains arbitrary large
palindromes (we call this palindromic), then there is an uncountable
set in $X$ which is strongly palindromic.
Many kinds of sequences defined by circle maps and substitutions
are palindromic.
In addition, we show that the existence of just one $x\in X$
for which $L(x)$ has no eigenvalues implies that there is a generic
set in $X$ for which $L(x)$ has purely singular continuous spectrum.
Our work gives new examples of operators with purely singular
continuous spectrum for both substitutions and circle maps.
For circle maps, Delyon and Petritis \cite{DePe86} have shown absence of
eigenvalues for almost all $\alpha$, all intervals $[0,\beta)$
and almost every $\theta_0$.
They do not exclude eigenvalues for all irrational $\alpha$.
Eigenvalues have been excluded for all irrational $\alpha$
in the case $\beta=\alpha$ by Bellissard et al.\ \cite{Bel+89},
but only for $\theta_0=0$.
This was proved independently by S\"ut\H{o} \cite{Suto89}
in the ``golden case'' $\alpha=(\sqrt 5 -1)/2$.
We prove here that for all irrational $\alpha$ and all intervals
$J=[0,\beta)$, there is a generic set in $X_J$ for which $L(x)$
has purely singular continuous spectrum.
All papers on Schr\"odinger operators with potentials defined
by primitive substitutions \cite{Bel90,Bel+91a,DePe91,BoGh93}
exclude eigenvalues for just one $x$; \cite{Bel+91a,BoGh93}
claim that the spectrum is a Cantor set of Lebesgue measure 0.
We get purely singular continuous spectrum for a generic set in $X_S$
for a large class of substitutions.
Only Bovier and Ghez \cite{BoGh93} claim to exclude eigenvalues for a class
of substitutions.
Our class of substitutions contains all but one of the examples
(the ``circle sequence'') given in \cite{BoGh93} and the Bovier-Ghez
class does not contain the ``period doubling'' and the ``Binary-Non-Pisot''
substitutions, which are contained in ours.
(It should be noted, however, that the results in \cite{Bel+91a,BoGh93}
do not apply to the sequences stated in those papers, but to different
ones (see the erratum \cite{BoGh94}) that are not necessarily in $X_S$.)
\section{Palindromic sequences}
Let $A$ be a finite set called {\em alphabet\/}
and $A^*$ the set of (finite) {\em words\/} $w_1w_2\cdots w_n$
with $w_i \in A$.
A word $w \in A^*$ is called a {\em palindrome} if it is the same
when read backwards.
The empty word is considered a palindrome.
An element $x \in A^{\ZZ}$ is called {\em palindromic} if $x$
contains arbitrarily long palindromes.
We say that a word $w$ is {\em centered} at $(n+m)/2$ in $x$ if
$w=x_{n} \cdots x_m$.
The {\em shift\/} $T$ on $A^\ZZ$ is defined by $T(z)_n=z_{n+1}$.
The {\em orbit\/} $\Orb (z)$ of $z\in A^\ZZ$ is the set $\{T^n z\}_{n\in \ZZ}$.
The closure of $\Orb(z)$ in the product topology is called
the {\em orbit closure\/} of $z$ and is denoted by $X_z$.
A compact shift-invariant subset $X$ of $A^\ZZ$ is called {\em minimal\/} if
$X_z=X$ for all $z\in X$.
If $X$ is minimal then every word occuring in some $y \in X$ occurs in
all $x \in X$.
Thus it makes sense to speak of {\em palindromic sets\/}
$X \subset A^{\ZZ}$, compact shift-invariant minimal sets containing
one (and therefore only) palindromic sequences.
Similarly, $X$ is called aperiodic if the elements in $X$ are
aperiodic.
Define $x \in A^{\ZZ}$ to be {\em strongly palindromic}, if there exists
for every $B>0$ a sequence $w_i$ of palindromes of length $l_i$
centered at $m_i \rightarrow \infty$ such that
$ e^{B m_i}/ l_i \to 0$.
If $x$ is strongly palindromic then it is clearly palindromic.
\begin{propo}
\label{strongly palindromic}
If an aperiodic minimal set $X \subset A^{\ZZ}$ is palindromic, then it
contains uncountably many strongly palindromic sequences.
\end{propo}
\begin{proof}
Let $\{w^i\}_{i=1}^\infty$ be a sequence of palindromes of length
$l_i\to\infty$ that occur in the sequences in $X$.
We will construct sequences $y_n\in A^{\ZZ}$ such
that \\
1) $y_n$ is equal to one of the $w^i$ on an interval $I_n$ containing 0;\\
2) $I_n\subset I_{n+1}$ and every $m \in \ZZ$ is eventually
in these intervals; \\
3) $y_{n+1}$ is equal to $y_n$ on $I_n$; \\
4) for every $y_n$, there are two possibilities for $y_{n+1}$ that
are different on the intersection of the intervals where they are
equal to one of the $w^i$. \\
Clearly, every such sequence $y_n$ converges to a different $y\in X$.
Since there are infinitely many choices to be made, there are uncountably
many different limit points.
Conditions will be imposed to make sure that the limit points are
strongly palindromic.
Choose $B$ and a sequence of integers $m_i$ such that $m_i\to\infty$ and
$e^{B{m_i}}/l_i \to 0$ as $i\to\infty$.
For every word $w$ there is a $d(w) \in \NN$ such that in every $x\in X$
there occurs a copy of $w$ in every interval of length $d(w)$
(see, e.g., \ \cite{Queffelec}, p.~71).
Let $y_n$ be equal to $w^{j_n}$ on an interval $I_n$ of
length $l_{j_n}$ containing 0.
We show how two choices arise for $y_{n+1}$.
There exists an integer $i_{n+1}$ such that $d(w^{j_n}) < m_{i_{n+1}}/2$.
This means there is a copy of $w^{j_n}$ in each of the first two
subwords of length $m_{i_{n+1}}/2$ to the left of the center of
each $w^j$ with $j\geq i_{n+1}$.
There exists a $j_{n+1} \geq i_{n+1}$ such that $w^{j_{n+1}}$ is not
periodic with any period less than or equal to $m_{i_{n+1}}$ over
an interval of length $l_{j_{n+1}} - m_{i_{n+1}}$, for otherwise
$X$ would contain a periodic sequence by minimality.
The two occurrences of $w^{j_n}$ in those first two subwords
to the left of the center of $w^{j_{n+1}}$ give the two possibilities
for $y_{n+1}$.
Denote the two possibilities by $z$ and $z'$.
Let $I$ and $I'$ denote the intervals on which $z$ and $z'$ are equal
to $w^{j_{n+1}}$, respectively.
We are left to show that $z$ and $z'$ differ on $I\cap I'$.
Note that the length of $I\cap I'$ is at least $l_{j_{n+1}}-m_{i_{n+1}}$
and that $z'$ on $I\cap I'$ is equal to $z$ translated by at most
$m_{i_{n+1}}$.
Now suppose $z$ and $z'$ are equal on $I\cap I'$.
Then $w^{j_{n+1}}$ would be periodic with period at most $m_{i_{n+1}}$
over a length of at least $l_{j_{n+1}}-m_{i_{n+1}}$.
But $j_{n+1}$ was chosen such that this does not happen.
Note that each limit point of the $y_n$ must be strongly palindromic
because $l_{j_n} \geq l_{i_n}$ and because the center of $w^{j_n}$
is within distance $m_{j_n}$ of 0.
\end{proof}
\section{Substitution sequences}
A {\em substitution} is a map $S:A \mapsto A^*$; it will be extended
to a map $A^*\mapsto A^*$ and $A^\NN\mapsto A^\NN$ by concatenation,
that is, $S(a_1 a_2\cdots a_n) = S(a_1) S(a_2) \cdots S(a_n)$.
A substitution $S$ is called {\em primitive} if there exists a $k$ such that
for all $a \in A$ the word $S^k(a)$ contains at least one copy of every
symbol.
Without loss of generality, one can assume that there exists $a \in A$
such that $S(a)$ starts with $a$ (see \cite{Hof93}).
Iterating $S$ on $a$ then gives a fixed point $z^+ \in A^\NN$.
Let $z$ be any element in $A^{\ZZ}$ satisfying $z_n=z^+_n$ for $n \in \NN$.
The substitution dynamical system $(X_S,T)$ is now defined by
\[
X_S =\{x\in A^\ZZ \bigm| x=\lim_{j\to\infty} T^{n_j} z
\quad \mbox{\rm and } \quad n_j \to\infty\} \; .
\]
It is possible to generate the sequences in $X_S$ directly from the
substitution (see, e.g., \cite{Dek78,Hof92}).
The system $(X_S,T)$ is strictly ergodic (see, e.g., \ \cite{Queffelec}).
A substitution $S$ is called palindromic if $X_S$ is palindromic.
We say that a primitive substitution $S$ is in the {\em class P}
if there exists a palindrome $p$
and for each $b \in A$ a palindrome $q_b$ such that $S(b)=p q_b$
for all $b \in A$.
We allow $p$ to be the empty word; if $p$ is not the empty word
then $q_b$ is allowed to be the empty word.
\begin{lemma}
\label{class P}
Class $P$ substitutions are palindromic.
\end{lemma}
\begin{proof}
We show first by induction that if $w$ is a palindrome, then $S(w)=p u$
with a palindrome $u$. We can write $w=bvb$ with some $b \in A$ and some
palindrome $v$.
We have $S(w)=S(b) S(v) S(b)= p q_b S(v) p q_b$. By induction, $S(v)$ is of
the form $p u$ with some palindrome $u$. We get
$S(w)= p q_b p u p q_b$ which is of the form $p \tilde{u}$ with
palindrome $\tilde{u}$.
All words $S^k(a)$ occur in all $x\in X_S$, so the sequences contain
arbitrarily long palindromes.
\end{proof}
\begin{coro}
\label{uncountably}
If $S$ is class $P$, then there are uncountably many
elements in $X_S$ that are strongly palindromic.
\end{coro}
\begin{proof} Apply \ref{strongly palindromic} and \ref{class P}.
\end{proof}
{\bf Remarks}.
1) It is sufficient that some power of $S$ is of class $P$ in order that
$S$ is palindromic. \\
2) A subclass of class $P$ are substitutions for which
$S(b)$ is a palindrome for all $b \in A$.
Just take for $p$ the empty palindrome. \\
3) Clearly, we could include into class $P$ substitutions of the form
$S(b)=q_b p$.
We do not know whether all palindromic $X_S$ arise from
substitutions that are in this extended class $P$.\\
4) Not all substitutions are palindromic, as the following example
shows: \\
$$ S: a \mapsto abbaaabbba, b \mapsto ab \; . $$
The word $bbaaabbb$ occurs in the sequences but the word
$bbbaaabb$ does not occur. \\
5) The set of strongly palindromic sequences is invariant and therefore
has either measure 0 or 1; but which we do not know.
We cannot even exclude that all $x$ are strongly palindromic. \smallskip
Here are some examples of substitutions in class $P$ that have appeared
in the literature on Schr\"odinger operators:\smallskip
{\bf Thue-Morse}: $S^2: a \mapsto abba, b \mapsto baab$
\cite{Luc89,Bel90,DePe91}. \\
{\bf Fibonacci}: $S: a \mapsto ab, b \mapsto a$
\cite{Koh+83,Ost+83,Cas86,Suto87,Suto89,Bel+89,Bel+92a,BoGh93}.\\
{\bf Period doubling}: $S: a \mapsto ab, b \mapsto aa$
\cite{Luc89,Bel+91a,Bel+92a,BoGh93}.\\
{\bf Binary non-Pisot}: $S: a \mapsto ab, b \mapsto aaa$
\cite{Bel+92a,BoGh93}. \\
{\bf Ternary non-Pisot}: $S: a \mapsto c, b \mapsto a, c \mapsto bab$
\cite{Bel+92a,BoGh93}. \smallskip
{\bf Remarks}.
6)
The papers \cite{Bel+91a,Bel+92a,BoGh93} implicitly consider the set $X_z$
where $z$ is the symmetric extension of $z^+$: for $n\in\NN$, $z_{-n-1}=z^+_n$
and $z_n = z^+_n$.
This is not natural.
The system $(X_z,T)$ is, in general, neither uniquely ergodic nor
minimal. \\
7) We do not know if the Rudin-Shapiro sequence
($r_n=(-1)^{f_n}$, where $f_n$ is the number of pairs $11$ in the
binary expansion of $n$) is palindromic.
Computer experiments suggest it is.
A Schr\"odinger operator with potential based on the Rudin-Shapiro
sequence has appeared in \cite{Luc89}.
\section{Sequences defined by circle maps}
Consider the irrational rotation $\theta \mapsto \theta+\alpha$ on
the circle $\TT^1=\RR/\ZZ$. For a half-open interval
$J=[0,\beta) \subset \TT^1$ and
a point $\theta_0$ in $\TT^1$, we consider the sequence $z=z(\theta_0)$
given by
$$ z_n = 1_J (n \alpha + \theta_0) \in \{0,1\}^{\ZZ} \; . $$
We denote by $X_J$ the compact subset of $A^{\ZZ}$ obtained by taking the
closure of the orbit of $z(\theta_0)$. It is independent of $\theta_0$
and the system $(X_J,T)$ is strictly ergodic. This remains true if
the half-open interval $J$ is replaced by a countable union of half-open
intervals \cite{Hof93}.
\begin{lemma}
\label{palindromic circle}
For all irrational $\alpha$ and every half-open interval $J$
there exists $\theta_0 \in \TT^1$
such that $z(\theta_0)$ is palindromic.
\end{lemma}
\begin{proof}
If $\beta \neq 2 k \alpha$, then the orbit of $\theta_0=\beta/2$ is disjoint
from $\{0,\beta\}$. If $\beta=2k \alpha \; ({\rm mod} \; 1)$, the orbit of
$\theta_0=(\beta+1)/2$ is disjoint from $\{0,\beta\}$.
By symmetry, these orbits are even and therefore palindromic
if they are disjoint from $\{0,\beta\}$.
\end{proof}
{\bf Remarks}.
1) The lemma can be generalized.
Suppose $J$ is a symmetric countable union of half-open intervals
with a point of symmetry $\theta_0$.
Then if one of the orbits of the points $\theta_0+ (i + j\alpha)/2$
($i,j\in\{0,1\})$ does not hit the boundary of $J$, then that orbit is even,
and $X_J$ is palindromic. \\
2) Another generalization of the
lemma can be obtained by taking an ergodic
group translation $\theta \mapsto \theta + \alpha$ on $\TT^{\nu}$ and taking
$J=[0,\beta_1) \times \dots \times [0,\beta_{\nu})$. The sequence
$z_n=1_J(\theta+n \alpha)$ generates a strictly ergodic
set $X$ independent of $\theta$ as in the case $\nu=1$.
This set $X$ is palindromic: it can be shown (by induction in $\nu$) that
one of the orbits starting at
$\theta_0=(\beta+v)/2$, where $v_i \in \{0,1\}$, does
not hit the boundary of $J$. \\
3) There are (in cardinality) more
sequences generated by circle maps than sequences
generated by substitutions: the length of the interval $[0,\beta)$
measures the average frequency of occurrences of $1$
in the sequence $z_n$. In this way one obtains uncountably many different
sequences for different $\beta$.
On the other hand, there are only countably many substitutions. \\
4) The circle $\TT^1$ is embedded in $X_J$ by
$z(\theta) = 1_J (n \alpha + \theta)$ but
not every sequence $z \in X_J$ is of this form.
\section{The point spectrum}
In the preceding three sections $A$ was a finite set of symbols.
In the remainder of the paper we map the elements to $A$ to real
numbers and again denote the image by $A$.
So from now on sequences $x\in A^\ZZ$ take values in $\RR$.
Every $x \in A^{\ZZ}$ defines a discrete Schr\"odinger operator
on $l^2(\ZZ)$ by
$$ (L(x)u)_n= u_{n+1} + u_{n-1} + x_n u_n \; . $$
\begin{thm}[Jitomirskaya-Simon]\label{JS}
If $x$ is strongly palindromic, then $L(x)$ has no eigenvalues.
\end{thm}
\begin{proof} The proof is essentially \cite{JS94}. We give a proof in
the appendix.
\end{proof}
{\bf Remarks}.
1) Most (with respect to the product measure) sequences in $A^{\ZZ}$
are palindromic since almost all sequences contain all possible finite
words. But most sequences in $A^{\ZZ}$ are not strongly palindromic
since most sequences give Schr\"odinger operators with Anderson localization
\cite{CaKlMa87}
and strongly palindromic sequences give operators
without eigenvalues. \\
2) In the golden case treated in \cite{Suto89,Bel+89} the sequence
$x$ is symmetric around 0.
The sequence in \cite{DePe91} is antisymmetric around 0.
We do not know whether these sequences are strongly palindromic, so
we can not exclude eigenvalues for those particular sequences,
although in both cases $X$ is palindromic.
\section{The absolutely continuous spectrum}
\begin{thm}
\label{absolutely continuous}
If the dynamical system $(X, T)$ has the property that
the Lyapunov exponent $\lambda_x(E)=\lambda(E)$
of $L(x)$ exists for all $x \in X$ and is independent of $x \in X$,
then $L(x)$ has no absolutely continuous spectrum for all $x \in X$.
\end{thm}
\begin{proof} For all $E \notin \sigma(L)$ or
$E \in \sigma(L(x))$ with $\lambda(E)>0$, there exist two
solutions $u^{\pm} \in R^{\ZZ}$ of $Lu=Eu$ which
are in $l^2(\pm \NN)$.
By Kotani's result \cite{Kot89} and the Lyapunov assumption,
this is the case for a set $Y \subset \RR$ of full measure
that is independent of $x$.
The vectors $\Phi^{\pm}(n)=(u^{\pm}(n+1),u^{\pm}(n))$
satisfy $A_E(n) \Phi^{\pm}(n)=\Phi^{\pm}(n+1)$.
If $\Phi^{+} \neq \Phi^-$, the Titchmarsh-Weyl functions
$m^{\pm}_E(n)=u^{\pm}(n+1)/u^{\pm}(n)$
are different and real (but they may be infinite).
If $\Phi^+=\Phi^-$, $E$ is an eigenvalue.
The resolvent $(L-E)^{-1}_{nn}=G_E(n,n)$ of $L$ satisfies
$G_E(n,n)=1/(m_E^+(n)-m_E^-(n))$ and is the Borel transform of
the spectral measure $dk_{e_n}$. The absolutely continuous part of
$dk_{e_n}$ is by Fatou's theorem given by
${\rm Im}(G_{E+i 0}(n,n)) = {\rm Im}(1/(m_{E+i0}^+(n)-m_{E+i0}^-(n)))$,
which is zero for all $E \in Y$ which are not eigenvalues.
\end{proof}
{\bf Remark}. A different proof can be obtained by using a result of Berezanskii
(see \cite{Sim82}):
if $\mu_{\phi}$ is any spectral measure of any discrete Schr\"odinger
operator $L$ then for $\mu_{\phi}$-almost every $E \in \RR$
there exist polynomially bounded solutions of $Lu=Eu$.
If there is an $x\in X$ such that $L(x)$ has absolutely continuous
spectrum then it has an
absolutely continuous spectral measure $\mu_\phi$ supported
on a set $A$ of positive Lebesgue measure.
For $\mu_\phi$-a.e. $E\in A$ --- which is equivalent to Lebesgue a.e.\ $E$ ---
the equation $L(x) u = E u$ has polynomially bounded solutions.
So $\lambda_x(E)=0$ for those $E$.
But we assumed that $\lambda_x(E)$ does not depend on $x$ and it is
strictly positive Lebesgue almost everywhere by Kotani \cite{Kot89}.
Thus the assumption that $L(x)$ has absolutely continuous spectrum
leads to a contradiction.
\begin{coro}
If $(X_S,T)$ is a substitution dynamical system, then $L(x)$
has no absolutely continuous spectrum for all $x \in X_S$.
\end{coro}
\begin{proof}
>From \cite{Hof93} we know that substitution sequences give operators
with Lyapunov exponents independent of $x \in X_S$.
\end{proof}
{\bf Remarks}.
1) Another way to exclude absolutely continuous spectrum for all $x$
is to show that the spectrum has Lebesgue measure zero and is independent
of $x$.
This was done in \cite{Bel+89} for circle maps with $\alpha=\beta$.
Circle maps give spectrum independent of $x$ for all values of $\alpha$
and $\beta$ \cite{Hof93}.
Kotani's theorem \cite{Kot89} does not imply that the spectrum has
Lebesgue measure 0; a singular spectrum can have positive Lebesgue
measure.
We do not know whether the Lyapunov exponent is independent of $x$
for sequences generated by circle maps. \\
2) One might guess that if $(X,T)$ is a strictly ergodic dynamical
system and $V: X \rightarrow \RR$ is continuous, then
the absolutely continuous spectrum of the operator
$(L(x)u)_n=u_{n+1}+u_{n-1}+ V(T^nx)u_n$ is independent of $x \in X$.
We have shown that this is the case
for dynamical systems $(X_S,T)$ obtained by substitutions if
$V(T^n x) = x_n$.
It is also true for substitution dynamical systems if $V(x)$
only depends on $x_{-l} \cdots x_l$, because then
the Lyapunov exponent remains independent of $V(x)$ by Proposition~5.2
in \cite{Hof93}.
It was already known for circle maps with $\alpha=\beta$ \cite{Bel+89}
and $V(T^nx)=x_n$. \\
3) Assume that $(X,T)$ is strictly ergodic, that $\lambda_x(E)$
exists for all $x \in X$ and is independent of $x$ and that
there exists $x \in X$ such that $L(x)$ has no eigenvalues.
Then it follows for example from the theorem of de la Vall\'ee Poussin that the
spectrum of $L(x)$ is the closure of $\{ E \; | \; \lambda_x(E) = 0 \}$.
We expect (but can't prove)
that the spectrum of $L(x)$ is equal to $\{ E \; | \; \lambda_x(E) = 0 \}$
from which it would follow that it is a Cantor set of zero Lebesgue measure.
The spectrum has been shown to be equal to the set $\{E \;|\;
\int d\mu(x) \lambda_x(E)=0 \}$, where the integration is with respect to
the invariant measure on $X$,
for circle maps with $\alpha=\beta$ in \cite{Bel+89}.
It has been shown that
$\mbox{{\rm spec}(L(v))}=\{E\;|\;\lambda_v(E)=0\}$
for a sequence $v$ defined by the Period
Doubling substitution in \cite{Bel+91a} and this result was extended
to a class of substitutions
including Period Doubling in \cite{BoGh93}.
\section{The singular continuous spectrum}
\begin{thm}[Simon]
\label{generic}
Let $(X,d)$ be a complete metric space of selfadjoint operators
on a Hilbert space $H$ such that the metric $d$ is stronger
than the strong resolvent convergence. Then both of the sets
$ \{ A \in X \; | \; \sigma_{pp} = \emptyset \}, \;
\{ A \in X \; | \; \sigma_{ac} = \emptyset \} $
are $G_{\delta}$'s, countable intersections of open sets.
\end{thm}
See \cite{Sim94}.
\begin{coro}
\label{singular}
Let $X \subset A^{\ZZ}$ be strictly ergodic.
If there is one $x\in X$ such that $L(x)$ has no eigenvalues,
then there exists a generic subset in $X$ on which $L$ is
purely singular continuous.
\end{coro}
\begin{proof}
By Kotani \cite{Kot89} there is a set of full measure
with no absolutely continuous spectrum.
By strict ergodicity, this set is dense and by Theorem~\ref{generic}
it is a dense $G_\delta$.
Since $\Orb(x)$ is dense by minimality, there is dense set on
which $L$ has no eigenvalues.
Again, this set is a dense $G_\delta$ by Theorem~\ref{generic}.
Since $X$ is a complete metric space, the intersection of two
dense $G_\delta$'s is a dense $G_\delta$ by the Baire category theorem.
\end{proof}
By \ref{JS}, \ref{strongly palindromic}, \ref{class P} and
\ref{palindromic circle} we get:
\begin{coro}
If $X$ is strictly ergodic and palindromic, then there is a generic
$Y\subset X$ such that for $x\in Y$ the operator $L(x)$ has purely singular
continuous spectrum.
\end{coro}
\begin{coro}[Case of substitutions]
For substitutions of class $P$, there exists a generic set $Y$ in $X_S$
such that for $x \in Y$ the operator $L(x)$ has purely singular
continuous spectrum.
\end{coro}
\begin{coro}[Case of circle maps]
For all irrational $\alpha$ and every half-open interval $J$, there exists
a generic set $Y$ in $X_J$ such that for $x \in Y$ the operator
$L(x)$ has purely singular continuous spectrum.
\end{coro}
{\bf Remark}. Strictly ergodic palindromic systems other then those
defined by circle maps and class P substitutions can be constructed
from so-called Toeplitz sequences \cite{JaKe69,Nev69}.
\section{Appendix: The criterion of Jitomirskaya and Simon}
%
For the convenience of the reader we repeat the proof of the theorem
of Jitomirskaya and Simon \cite{JS94} for the case of potentials taking
finitely many values.
\begin{thm} [Jitomirskaya-Simon]
Let $x \in A^{Z}$, where $A \subset \RR$ is finite.
Consider the operator on $l^2(Z)$ given by
$(Lu)(n)=u(n+1)+u(n-1) + x_n u(n)$.
Suppose there is a sequence of intervals $I_i$ of length $|I_i|$
centered at $m_i$ such that $x$ is a palindrome on $I_i$, $0\in I_i$
and $|I_i|\to\infty$ as $i \to \infty$.
Then there exists a constant $C$ depending only on $A$ such that
$L$ has no eigenvalues if $C^{m_i}/|I_i| \rightarrow 0$ as $i\to\infty$.
\end{thm}
\begin{proof}
We can assume that $|I_i|$ is either odd or even for all $i$.
We prove the theorem for the case that $|I_i|$ is odd
and then indicate what changes if it is even.
So first let $I_i =\{ m_i-l_i, m_i + l_i\}$.
Assume $Lu=Eu$ with $||u||=1$. Define $u_i(n)=u(2m_i-n)$, the
vector reflected at $m_i$. Let $W(u,u_i)(n)=u(n+1) u_i(n)-u_i(n+1) u(n)$
be the Wronskian of $u$ and $u_i$. Let $\Phi(n)=(u(n+1),u(n))$ and
$\Phi_i(n)=(u_i(n+1),u_i(n))$ and $u^\pm_i = u \pm u_i$
and $\Phi_i^{\pm}=\Phi \pm \Phi_i$.
Step 1. $W(n):=W(u,u_i)(n)$ is constant for
$n \in I_i$. \\
{\it Proof}. Compute using $u(n+1)=-u(n-1)+(E-x_n)u(n)$ and
$u_i(n+1)=-u_i(n-1)+(E-x_n)u_i(n)$
$$ W(n)-W(n-1)= ( x_{2m_i-n}-x_n ) u(2m_i-n) u(n)=0 \; .$$
Step 2. $|W(m_i)| \leq 2/|I_i|=2/(2l_i+1)$. \\
{\it Proof}. $\sum_n |W(n)| \leq 2$ by Schwarz inequality and $||u||=||u_i||=1$.
Use Step 1.
Step 3. Either (i) $\; |u_i^+(m_i)| \leq 2 |I_i|^{-1/2}$ or
(ii) $ \; |u_i^-(m_i+1)| \leq 2 |I_i|^{-1/2}$. \\
{\it Proof}. Since $u_i^-(m_i)=0$, one has
$$u_i^+(m_i) u_i^-(m_i+1)=W(u_i^-,u_i^+)(m_i)=2 \cdot W(u,u_i)(m_i) =
2 \cdot W(m_i)$$
Use Step 2.
Step 4. There exists a constant $C_1$ such that $\| \Phi_i^{\pm}(m_i)\|
\leq C_1 |I_i|^{-1/2}$ holds for all $i$, with either the sign $+$ or
$-$. \\
{\it Proof}. In case (ii), we have $||\Phi_i^{-}(m_i)||=||(u_i^-(m_i+1),0)||
\leq 2 |I_i|^{-1/2}$. Assume now case (i). We have $u_i(m_i)=u(m_i)$ and
$u_i(m_i-1)=u(m_i+1)$, rsp. $u_i(m_i+1)=u(m_i-1)$ so that
$u_i^+(m_i-1)=u_i^+(m_i+1)$.
Therefore $u_i^+(m_i+1)=\frac{1}{2}(E-x_{m_i}) u_i^+(m_i)$
so that $\|\Phi^+ (m_i)\| \leq C_1' |u^+(m_i)| \leq C_1 \cdot |I_i|^{-1/2}$. \\
%\medskip
Step 5. For some $C>0$,
$\|\Phi_i^{\pm}(0)\| \leq C^{m_i} C_1 \cdot |2l_1+1|^{-1/2}$, for
either sign $+$ or sign $-$. \\
{\it Proof}.
There exists $C$ such that $||A(n)|| \leq C$ for all transfer
matrices $A(n)$, for all $E$ in the spectrum of $L$.
Denote by $A_i$ the transfer matrix that maps vectors at $m_i$
to vectors at $0$. In particular,
because $x$ is exactly symmetric on $I_i$,
$$ A_i \Phi(m_i) = \Phi(0), \qquad A_i \Phi_i(m_i) = \Phi_i(0) \;.$$
Then also
$$ \Phi_i^{\pm}(0) = A_i \Phi_i^{\pm}(m_i) \; $$
so that
$$ ||\Phi_i^{\pm}(0)|| \leq C^{m_i} C_1 \cdot |2l_1+1|^{-1/2} \; . $$
Step 6. Proof of the theorem. \\
By Step 5,
$$\bigl| \|\Phi(0)\|-\|\Phi(2m_i)\| \bigr|\leq \|\Phi^\pm(0)\|\to 0$$
for suitable choices of $+$ and $-$ as $i\to\infty$.
Since $\|\Phi(m_i)\|\to 0$ as $m_i\to\infty$ because $u\in l^2$,
it follows that $\Phi(0)=0$.
But then $u=0$, contradicting $\|u\|=1$.
If $I_i$ is even for all $i$ we can put
$I_i:=\{m_i-l_i,\ldots, m_i, m_i+1, \ldots, m_i+l_i+1\}$.
Define $u_i(n)=u(2m_i-n+1)$, the at $m_i+1/2$ reflected vector.
We have $x_{2m_i-n+1}=x_n$ on $I_i$.
Step 1 and Step 2 do not change. \\
In Step 3, the claim stays the same. Note that $u_i^-(m_i) = 0$
is no longer true in general. However, the claim follows from
$$ W(u_i^-,u_i^+)=2 u_i^+(m_i) u_i^-(m_i+1) $$
(even with a factor 2 deleted), which is a consequence of
$u_i^+(m_i) u_i^-(m_i+1) = - u_i^+(m_i+1) u_i^-(m_i)$. \\
Step 4. In case (ii) we take
$$ \|\Phi_i^+(m_i)\|=||(u_i^+(m_i+1),u_i^+(m_i))||
\leq \sqrt{2} \cdot |u_i^+(m_i)| $$
since $u_i^+(m_i)=u_i^+(m_i+1)$.
Case (i) with a different $C_1$ follows from
$u_i^-(m_i)=-u_i^-(m_i+1)$
Step 5. Define $A_i$ by $A_i \Phi(m_i) = \Phi(0)$.
Because $x$ is symmetric on $I_i$, we have
$$ A_i A(m_i) \Phi(m_i) = A_i \Phi(m_i+1)= \Phi(2m_i+1)=\Phi_i(0) \; . $$
Together
$$ \|\Phi_i^{\pm}(0)\|
= \|\Phi(0) \pm \Phi_i(0)\| \leq 2 C^{m_i+1} \| \Phi^{\pm}(m_i) \| $$
for either the sign $+$ or the sign $-$.
Step 6 is the same (replace $\Phi(2m_i)$ by $\Phi(2m_1+1)$).
\end{proof}
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\end{document}