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{$  $}
\vskip5truecm
\centerline{\tit  SPECTRAL PROPERTIES OF ONE-DIMENSIONAL}
\vskip.2truecm
\centerline{\tit SCHR\"ODINGER OPERATORS WITH POTENTIALS}
\vskip.2truecm
\centerline{\tit GENERATED BY SUBSTITUTIONS}
\vskip2truecm
%\centerline{\tit }
%\vskip2.5cm
\centerline{\aut\ Anton Bovier${}^{1,2}$\footnote{}{${}^{1}$\ftn  
Institut f\"ur Angewandte Analysis und Stochastik, Hausvogteiplatz 5-7,
 O-1086 
Berlin, Germany}
\footnote{}{\ftn e-mail: BOVIER@IAAS-BERLIN.DBP.DE}
 and Jean-Michel Ghez${}^{2}$\footnote{}{${}^{2}$\ftn and PHYMAT, 
D\'epartement de Math\'ematiques, Universit\'e de Toulon et du Var,}
\footnote{}{\ftn B.P. 132 - F-83957 La Garde Cedex, France}}
\footnote{}{\ftn e-mail: GHEZ@CPTVAX.IN2P3.FR}

\vskip2truecm\rm
\noindent {\bf Abstract:}  We investigate one-dimensional discrete Schr\"odinger
operators whose potentials are invariant under a substitution rule. The 
spectral properties of these operators can be obtained from the analysis
of a dynamical system, called the trace map. We give a careful derivation
of these maps in the general case and exhibit some specific properties.
Under an additional, easily verifiable hypothesis concerning the 
structure of the trace map we present an analysis of their dynamical 
properties that allows us to prove that the spectrum of the underlying
Schr\"odinger operator is singular and supported on a set of zero 
Lebesgue measure. A condition allowing to exclude point spectrum is also 
given.  The application of our theorems is explained on a series of
examples.



\vfill
\line{\rm May, 1992\hfill}
\line{\rm CPT-92/P. 2705\hfill}

\vfill\eject
\count0=1

\chap{I. Introduction}

In this article we present general results on the spectral
properties of a class of one-dimensional discrete Schr\"odinger
operators of the form
$$
H_v= -\D +V\text{on}l^2(\Z)\tag {1.1}
$$
where $\D$ is the discrete Laplacian and $V$ is a diagonal
operator whose diagonal elements $V_n$ are obtained from a
{\it substitution sequence} [1]. By a substitution sequence we
mean the following. Let $\A$ be a finite set, called an {\it
alphabet}. Let $\A^k$ be the set of {\it words} of length $k$ in
the alphabet, $\A^*\equiv \cup_{k\in \N}\A^k$ the set of all
words of finite length, and  $\A^{\N}$ the set of one-sided
infinite sequences of letters. A map $\xi:\A\rightarrow\A^*$ is
called a {\it substitution}. A substitution $\xi$ naturally
induces maps from $\A^*\rightarrow\A^*$ and
$\A^{\N}\rightarrow\A^{\N}$, which we will denote by the same name
and which are obtained simply be applying $\xi$ to each letter in
the respective words or sequences (e.g.
$\xi(abc)=\xi(a)\xi(b)\xi(c)$). A substitution may possess
fix-points in $\A^{\N}$, and such fix-points, $u$, will be called
substitution sequences. There are two natural conditions that
guarantee the existence of at least one fix-point, namely $\xi^\infty 0$, 
and that we
will assume to be satisfied for all substitutions we discuss [1]:
\item{(C1)} There exists a letter, called $0$, in $\A$, such that
        the word $\xi(0)$ begins with $0$.
\item{(C2)} The length of the words $\xi^n(0)$ tends to infinity,
as $n\uparrow\infty$.

\noindent A class of substitutions we will in general deal with are
the so-called {\it primitive } substitutions [1]. 
They are characterized by the 
fact that there exists an integer, $k$, such that for any two letters
$\a_i$, $\a_j$ in $\A$ the word $\xi^k\a_i$ contains the letter $\a_j$. 

Given a fix-point $u=(\a_0\a_1\a_2\dots )$ of a substitution
$\xi$, the associated sequence of potentials is now obtained as
follows. Consider a map $v:\A\rightarrow\R$ (which we will always
assume to be non-constant), we set, for $n\geq 0$,
$V_n=v(\a_n)$. This sequence is then completed to the negative
side by setting, say, $V_{-n-1}=V_n$.

Schr\"odinger operators with potentials of this type have
attracted considerable attention over the last years in
connection with the discovery of quasi-crystals [2,3]. For,
indeed, the prototypical one-dimensional quasi-crystal is
associated to the Fibonacci-sequences, which are substitution
sequences associated to the substitutions $\xi$ on the alphabet
$\A=\left\{a,b\right\}$, where
$$
\eqalign{
\xi(a)&=ab^n\cr
\xi(b)&=a
}
\tag {1.2}
$$
(The most studied example (also called the Kohmoto model)
corresponds to the case $n=1$ and the Fibonacci sequence
associated to the golden number). There is a host of numerical and
analytical work been done for these models [4], with 
amongst the
most notable mathematical results those by Casdagli [5], S\"ut\"o
[6] and Bellissard et al [7], in which it was shown that the
spectrum of these operators is always singular continuous and
supported on a Cantor set of zero Lebesgue measure. All these
results relied heavily on the very fact that the Fibonacci
sequences are substitution sequences (in more technical terms,
they employed the so called {\it trace map}, whose existence is a
direct consequence of the substitution, as we will discuss in
detail below), and this observation stimulated the investigation
of other examples of substitution sequences.  The first and most
heavily studied [8] example was the {\it Thue-Morse sequence} [9],
defined by the substitution
$$
\eqalign{
\xi(a)&=ab     \cr
\xi(b)&=ba
}
\tag {1.3}
$$
which offers an additional interesting feature in that it is not
quasi-periodic. Again it was proven that the spectrum of the
corresponding
Hamiltonian is purely singular continuous [10,11] and, moreover, a
complete description of the gap-structure of the spectrum,
including the dependence of the gap-width on the potential
strength could be given [10]. A further example, where the same
type of results could be proven [11], is provided by the {\it
period-doubling sequence}, with substitution
$$
\eqalign{
\xi(a)&=ab        \cr
\xi(b)&=aa
}
\tag {1.4}
$$
These results required, in each example, a rather detailed
analysis of some dynamical system associated to the so called
trace map. Unfortunately, for more complicated substitutions
(e.g. on more than two letters), these become prohibitively
complicated. Nonetheless, one would expect that certain
qualitative properties of the spectra of such Hamiltonians should
not depend on the details, but only on some general features of
the substitution.

There are, indeed, two promising approaches attempting to obtain
more general results. One is the perturbative method of Luck [12]
that establishes, on a heuristic level, a connection between the
Fourier spectrum of the sequences themselves and the gap
structure of the spectrum of the Hamiltonians and that allows
even to compute the behaviour of the gap-widths. A shortcoming of
this approach is, besides the difficulties to give mathematically
rigorous justifications of some of the steps involved,  that it
fails to make clear predictions in situations where the Fourier
spectrum of the underlying sequence is not of the pure-point
type. Unfortunately, singular continous and even absolutely
continous Fourier spectra are not at all uncommon for
substitution sequences. Nonetheless we emphasize that this perturbation
method is so far the most powerful tool to get fast quantitative 
predictions.

Another attempt to obtain general information on these systems
is based on the K-theory of $C^*$-algebras. It was realized [13,14] 
that general gap-labelling theorems  [15,16] can be applied
particularly well in these cases as substitution sequences allow
for an easy computation of the corresponding $K_0$-groups. This
allows then to predict all possible spectral gaps from a simple
computation of a Perron-Frobenius eigenvector of a (not too
large) matrix. The shortfall of this approach is, so far, that it
cannot predict whether the allowed gaps will actually be open
for given values of the potentials, and in the known examples,
closed gaps do occasionally occur. In particular, the $K$-theory
makes no predictions on the type of spectrum one may expect.


In this article we attempt to obtain general results on the nature of the 
spectrum from a careful analysis of the trace maps. Indeed, it is natural 
to conjecture that the existence of an exact renormalization group structure,
as is presented by the trace map, is responsible for the particular
spectral properties observed in the examples. In particular, one may be led to 
believe that due to the existence of the trace map the singular spectral type
should be the rule rather than the exception. We will prove here that
this is true in some sense: namely,  that under some conditions
that can be verified fairly easily (there is a simple algorithmic procedure
to verify them) and that appear to hold in most examples (the 
Rudin-Shapiro sequence [17] being a notable exception), the spectrum
of our operators is always singular and supported on a set of zero 
Lebesgue measure. This result is based on the analysis of some general
properties of the trace maps and of the ensuing characteristics of large time
behaviour of the associated dynamical systems. These will allow to identify
the spectrum with the set of energies for which the Lyapunov exponent 
vanishes. A general theorem proven already in [11] which is based on 
a profound lemma of Kotani [18] will then yield the result. 



A more subtle question relates to the existence of
point spectrum: there is a simple supplementary
condition under which the existence of 
eigenvalues can be excluded, but this condition is not  satisfied
in all examples where the singular continuous nature of the 
spectrum was proven.

The remainder of this article is organized as follows. In chapter II we 
review the derivation of the trace maps and exhibit some of their properties.
We will define a new substitution rule on an extended alphabet that encodes
the principal part of the trace map and formulate the assumptions 
entering in our theorem in terms of this substitution. In chapter III
we formulate our main theorem and present  its proof. We also 
discuss the problem of eigenvalues. In chapter IV we elucidate our results
with some examples.

\noindent{\bf Acknowledgements:} We are grateful to Jean Bellissard  for
previous collaborations on this subject and for inspiring discussions.
We also thank Monique Combescure for having brought ref. [20] to our attention.
A.B. thanks the Centre de Physique Th\'eorique, Marseille, for its 
warm hospitality and the Universit\'e de Toulon et du Var for financial
support. 

\vfill\eject
\chap{II. The trace map}

In this section we give a careful review of the derivation of the
so-called {\it trace map} and establish some crucial properties
of these maps. The trace map was originally introduced by Allouche
and Peyri\`ere [19], but we also refer to the paper [20] by Kol\'ar and
Nori in which a more general and systematic construction is given.

As usual for one-dimensional discrete Schr\"odinger operators
like (1.1), the analysis of their spectra is based on the
discussion of the associated Schr\"odinger equation, written in
vector form as
$$
\Psi_E(n+1)=\left(\matrix{E-V_n&-1\cr
1&0}\right)\Psi_E(n)
\tag{2.1}
$$
where $\Psi_E(n)\equiv \left(\matrix{\psi_E(n)\cr \psi_E(n-1)}\right)$
with $\psi_E$ the solution of the usual Schr\"odinger equation
$H_v\psi_E=E\psi_E$.
Iterating equation (2.1) we get, of course, the solution of the
initial value problem in the form of a product of matrices as
$$
\Psi_E(n+1)=\prod_{k=n}^0\left(\matrix{E-V_k&-1\cr
1&0}\right)\Psi_E(1)
\tag{2.2}
$$
In the case of substitution sequences we are naturally led to
define the maps
$T_E:\A\rightarrow SL(2,\R)$ via
$$
T_E(\a)=   \left(\matrix{E-v(\a)&-1\cr
1&0}\right)\tag {2.3}
$$
Again, by some abuse of notation we denote by the same symbol the
maps $T_E:\A^*\rightarrow SL(2,\R)$ where for
$\o=(\a_0\dots\a_{n-1})\in\A^n$,
$$
T_E(\o)\equiv T_E(\a_{n-1})\dots T_E(\a_0) \tag {2.4}
$$
The map $T_E$ allows us to introduce the induced action of $\xi$
on $Im(T_E)$ via
$$
\xi T_E(\o)\equiv T^{(1)}_E(\o)\equiv T_E(\xi \o)\tag {2.5}
$$
and we will also use the notation
$$
\xi^n T_E(\o)\equiv T^{(n)}_E(\o)=T_E(\xi^n\o) \tag {2.6}
$$
It is obvious from (2.4) that the action of $\xi$ defines a
dynamical system on $SL(2,\R)^{|\A|}$, since $T_E^{(n)}(\a)$, $\a\in\A$,
can be expressed as a product of matrices $T^{(n-1)}_E(\a)$,
$\a\in\A$. The analysis of this dynamical system could in principle yield
all desired information on the spectrum of (1.1). In practice, however,
it turns out to be difficult to work with this system directly and it is
advantageous to pass to a new dynamical system based on the traces of the
matrices $T^{(n)}_E(\o)$.

Let us define, for $\o\in \A^*$, $x_E(\o)\equiv \tr T_E(\o)$. Of
course we may write also $x^{(n)}_E(\o)\equiv \tr T_E^{(n)}(\o)$
and obviously we may extend the action of $\xi$ to write
$\xi x_E^{(n-1)}(\o)=x_E^{(n)}(\o)$, however this time there is no immediate
expression of $\xi x_E^{(n-1)}(\a)$ as a function of the
$x_E^{(n-1)}(\a)$, i.e. a realization of this action as a dynamical
system on $\R^{|\A|}$ and in general such a realization will not exist. However,
it is always possible to find a finite subset, $\B\subset\A^*$
such that for all $\o\in\B$,
$x^{(n)}_E(\o)$ can be expressed as a function of the
$x^{(n-1)}_E(\o)$, with $\o\in\B$, that is a realization
of the action of $\xi$ as a dynamical system on $\R^{|\B|}$.
Such a dynamical system is called a trace map. Note that in the sequel we will
use the names $\b$ or $\b_i$ for the elements of $\B$ to distinguish them
from generic words $\o$.
Following [20], such a trace map can be constructed for any substitution
in the following way.

Notice first that for unimodular $2\times 2$-matrices $A,B$, the
Cayley-Hamilton theorem yields
$$
\tr (AB)=\tr A \tr B -\tr (BA^{-1}) \tag {2.7}
$$
It is easy to deduce from this relation  (see [20]) that for
three such matrices $A,B, C$, one has
 $$
\tr (ABAC)=\tr (AB) \tr (AC) +\tr (BC)-\tr B\tr C \tag {2.8}
 $$
Let us label the letters in $\A$ by $\a_1,\dots,\a_K$, with $K\equiv|\A|$.
 Starting with $\a_1$ we write
 $$
 x^{(n+1)}_E(\a_1) = \tr \prod_{\a\in\xi\a_1} T_E^{(n)}(\a)
 \tag {2.9}
 $$
Now there are two possibilities: if $\xi\a_1$ contains no letter
of $\A$ twice, then we set $\b_{K+1}\equiv \xi\a_1$. The word $\b_{K+1}$
will then be considered as a `letter' in the new alphabet $\B$ (which also
contains all the letters $\a_i$ from $\A$) that we will construct.
 More precisely, due to the invariance of the trace under cyclic permutations
it is natural to identify words $\xi\a_1$ that differ only by a cyclic
permutation of their letters, so that the elements of $\B$ will really be
equivalence classes of  words in $\A^*$.

If $\xi\a_1$ contains a letter, say $\a$, in $\A$ twice, then an element
in its equivalence class may be written in the form $\a\o_1\a\o_2$, and thus
by (2.8),
$$
x^{(n+1)}_E(\a_1)=\tr T_E^{(n)}(\a\o_1)\tr T_E^{(n)}(\a\o_2)
+\tr T_E^{(n)}(\o_1\o_2)-\tr T_E^{(n)}(\o_1)\tr T_E^{(n)}(\o_2)
\tag {2.10}
$$
We now proceed with each of the traces appearing in (2.10) just as before,
that is if a corresponding word (say $\a\o_1$) contains no letter twice
it is included into $\B$, whereas for words that still contain a letter
twice, (2.8) is again applied. The important point is that with each
application of (2.8) the words that may appear become strictly shorter
so that this process necessarily terminates after a finite number of steps,
leaving us with $x^{(n+1)}_E(\a_1)$ expressed as a polynomial in
the variables $x_E^{(N)}(\b_i)$, with $\b_i$ elements of some finite
set $\B$. The same procedure is now applied on the remaining letters
$\a_i$ in $\A$, and finally on the new letters $\b_i\in\B$ that have been
introduced in the process. But since the elements of $\B$ are equivalence
classes of words in $\A^*$ that contain no letter twice, the length of these
words is a priori bounded by $K$, and the cardinality of $\B$ by [20]
 $$
 |\B|\leq \sum_{l=1}^K\frac{K!}{l(K-l)!} \tag {2.11}
 $$
so that the algorithm described above will terminate after a
finite number of steps. In the end we have, for each $\b_i\in\B$,
an expression
$$
x^{(n+1)}_E(\b_i)=f_{i}\left(x^{(n)}_E(\b_1),\dots,
x^{(n)}_E(\b_{|\B|})\right)
\tag {2.12}
$$
where each $f_i$ is a polynomial map from  $\R^{|\B|}$ to
$\R^{|\B|}$, and where we have fixed,
for notational convenience,  the numbering of the
letters $\b_i\in \B$ once and for all.

An important further characterization of these maps can be given through the
following notion of a {\it `degree'}, $d$, defined as follows:
Put, for $x\in \R^{|\B|}$,
$$
d(x_i)\equiv |\b_i|
\tag {2.13}
$$
and let for any two polynomials, $p$ and $q$ in the variables
$\{x_i\}_{i=1}^{|\B|}$
$$
d(pq)\equiv d(p)+
d(q)
\tag {2.14}
$$
and
$$
 d(p+q)\equiv\max\left(
d(p),d(q)\right)
\tag {2.15}
$$
Obviously, these three relations allow to compute the `degree',
$d(p)$,
of any such polynomial.


 We collect the properties of the trace map in the following

\proposition {2.1} {\it Let $\xi$ be a substitution on an
alphabet $\A$ of cardinality $K$. Then there exists an alphabet
$\B$ whose elements are words modulo cyclic permutations
in $\cup_{l=1}^K \A^l$, such that
$\A\subset\B$ and $|\B|\leq \sum_{l=1}^K\frac {K!}{l(K-l)!}$,
and a polynomial map $f:\R^{|\B|}\rightarrow\R^{|\B|}$, such that
if
$f^{(n)}(x)$ is the $n$-th iterate of $f$ on the initial vector
$x$, then,
$$
x^{(n)}_E=f^{(n)}(x_E^{(0)})
\tag {2.16}
$$
(2.12) holds for each $\b_i\in\B$. Moreover
$f$ satisfies
$$
d(f_{i})=|\xi\b_i|_{\A}
\tag{2.17}
$$
where $|\o|_{\A}$ denotes the number of letters of $\o$
considered as a word in $\A^*$.
Finally, there exists a unique monomial of highest `degree'
(whose coefficient is one)
  in $f_{i}$
  which we shall denote by $\tilde
  f_{i}$.
}

\proof Most of the proposition is evident from the construction
given above and has already been noticed earlier [20]. The
statement (2.17) on the degree is also evident from the fact that
only (2.8) is used in the construction of the trace map and that
there is exactly one term on the right hand side of (2.8) that
has the same degree as the term on the left. \endproof

\remark The reader may notice that the construction of the trace map
(and not even the alphabet $\B$) is
not unique, and that in general several trace maps can be obtained for the
same substitution. They will all, however, enjoy the properties
stated in proposition 2.1. For practical purposes, one may try
to minimize the size of $\B$ and consider the trace map on
invariant submanifolds. For our general considerations here this will
be of no importance.


The  map $\tilde f$, introduced in proposition 2.1, will be called the
{\it reduced trace map} and is of central importance for our analysis.
We find it useful --  and natural --  to
associate with $\tilde f$ a substitution, $\phi:\B\rightarrow\B^*$,
in the following way: Let us first define the map
$X:\B^*\rightarrow \R$ such that for any
$\o=(\b_{i_1}\dots\b_{i_k})\in \B^*$,
$$
X(\o)\equiv x_{i_1}\dots x_{i_k}
\tag {2.18}
$$
Then $\phi$ is a  substitution such that for any $\b_i\in \B$,
$$
X(\phi\b_i)\equiv \tilde f_{i}\left( x_{1},
\dots,x_{|\B|})\right) \tag {2.19}
$$

Properties of the substitution $\phi$ will be crucial for our analysis.
The substitutions $\phi$ associated to trace maps will typically not be
primitive, but have a structure that we will call {\it semi-primitive}:

\noindent{\thbf Definition 2.1:} {\it
A substitution $\phi$ on an alphabet $\B$ is called semi-primitive, if
\item{(i)} There exists a subset $\C\subset\B$ such that $\phi$ maps
 $\C$ into $\C^*$ and the restriction of $\phi$ to $\C$ is a primitive
substitution on the alphabet $\C$.
\item{(ii)} There exists a positive integer $k$ such that for each
letter $\b\in \B$, $\phi^k\b$ contains at least one letter from $\C$.

}

Note that although (2.19) does not uniquely define the substitution $\phi$,
(since $X(\o)$ does not depend on the order in which the
letters appear in $\o$ but only on their multiplicity)
either all or none of the substitutions satisfying (2.19) for a given
$\tilde f$ are  semi-primitive.
In most examples of trace maps associated to primitive substitutions $\xi$
we have analyzed (see section IV),
the associated substitutions $\phi$ turned out to be
semi-primitive, the Rudin-Shapiro sequence being the only counter-example.

To conclude this section, we note that semi-primitive
substitutions arising from primitive substitutions $\xi$ in the
above described way have the following additional property:

\lemma {2.1} {\it Let $\xi $ be a primitive substitution
satisfying conditions C1 and C2. Let $\phi$ be a substitution on
$\B$ associated to its reduced trace map. Let $\C$ be a subset of $\B$ such
that the restriction of $\phi$ to $\C$ is primitive. Then there
exists a letter $\g_0\in \C$, such that $\g_0$ as a word in
$\A^*$ contains the letter $0$.}

\proof  To prove the lemma, just notice that for any letter
$\b_i\in \B$, the word $\phi\b_i\in \B^*$ considered as a word
in $\A^*$ is made of the same letters as the word $\xi\b_i\in
\A^*$. The same holds true for the $k$-th iterates of $\phi$ and
$\xi$, respectively. Now if $\xi$ is primitive, then there exists
$k$ such that for any letter $\a$, and a fortiori for any word
$\o\in \A^*$, $\xi^k\o$ contains the letter $0$. Therefore, for
any letter $\g\in \C$, $\phi^k\g$ must contain a letter from $\B$
containing $0$. But by assumption, $\phi^k\g$ is a word in
$\C^*$, and thus $\C$ must contain at least one letter which,
considered as a word in $\A^*$, contains the letter $0$, which
was to be proven. \endproof

\vfill\eject
\chap{III. Trace map and spectrum}

In this section we review the determination of the spectrum of
$H$ through the dynamical spectrum of the trace map and some of
its consequences. In particular, we will prove the main result of
this article, that is

\theo 1 {\it Let $\xi$ be a non-constant
primitive substitution defined on a finite
alphabet $\A$. Let $v$ a be non-constant map from $\A$ to $\R$ and $H_v$
the Schr\"odinger operator defined in (1.1). Suppose there exists
a trace map whose associated substitution $\phi$, defined on an alphabet
$\B$, as described in section II, is semi-primitive. Assume further that
there exists $k<\infty$ such that $\xi^k0$ contains the word $\b\b$ for
some $\b\in\B$. Then the spectrum of $H_v$ is singular and supported on
a set of zero Lebesgue measure.}

The strategy of the proof follows the one used in [11] to prove that
the spectrum of $H_v$ is singular continuous in the particular case of
the period doubling sequence.

Let us begin by defining the so-called unstable set $\U$.

\noindent{\thbf Definition 3.1:} {\it Let $\g_0$ denote a letter
in $\C$ that contains the letter $0\in \A$. Let $\U_n$ denote the set
$$
\U_n\equiv \left\{x\in\R^{|\B|}\big|
\forall_{m\geq n} |f_{i(\g_0)}^{(m)}(x)|>2\right\}
\tag {3.1}
$$
where $i(\g_0)$ is defined such that $\g_0=\b_{i(\g_0)}$.
Then
$$
\U\equiv \bigcup_{n=0}^\infty \hbox{int}\,U_n
\tag {3.2}
$$
}

\remark  Notice that in general $\C$ may contain several letters
containing the letter $0$, and thus the set $\U$ depends a priori
on which of these letters was chosen. However, as will become
clear later, $\U$ is really independent of this choice. Note also
that we may choose the labelling of the letters in $\B$ in such a
way that our chosen $\g_0$ is $\b_1$ (i.e. $i(\g_0)=1$),
which will simplify further notation.

Since in the sequel we will want to speak, for fixed $v$, of the set of energies
such that  $x_E^{(0)}$ belongs to $\U$ or in fact other sets we will define
later, it will be convenient to define, for any set  $Y\subset\R^{|B|}$,
 $\E(Y)\subset\R$, by
$$
\E(Y)\equiv\left\{E\bigl|x^{(0)}_E \in Y\right\}
\tag {3.3}
$$
Notice that $\E(Y^c)=\E(Y)^c$, where the superscript $c$ indicates the
complement of a set.
The definition of $\U$ (notice that it differs from the one given
e.g. in S\"ut\H o [6]) implies immediately

\lemma {3.1} {\it For given $v$,  $\E(\U)\subset
 \s\left(H_v\right)^c$.}

The proof of this lemma in this form has been given by
Bellissard [10]. A similar result was also proven by S\"ut\H o
[6]. Essentially it is contained in Theorem VIII.24a of
Reed-Simon, Vol. 1 [21].

In principle we would like to prove also the converse of lemma 3.1
which would allow to compute the spectrum of $H_v$ from the trace map.
In [11] we have seen that if $\U$ is such that the Lyapunov exponent
vanishes for $E\in\E(\U^c)$,
then not only the converse of
lemma 3.1 holds, but also, applying some general results of Kotani [18],
the spectrum has zero Lebesgue measure. However, while
the definition of $\U$ is convenient to prove lemma 3.1, it is
inconvenient to describe $\U$ in more detail since in order to
decide whether $x^{(0)}_E$ is in $\U$ we need to control $x^{(n)}_E$
for all $n$. In [6],[10] and  [11] a simpler characterization was
found in the cases of the Fibonacci, Thue-Morse and period-doubling
sequences which required  information on $x^{(n)}_E$ only for
{\it some} $n$. This implied in particular that the sets $\U_n$
were open. We will give such a characterization in the
general case. In fact, we will define a set $\tilde\U$ that a
priori is contained in $\U$ but that is big enough such that for
energies $E\in\E(\tilde \U)^c$, the Lyapunov exponent
vanishes.




 To define this set, let us introduce
the maps $\r^{(n)}:\B\rightarrow \R$ by
$$
\r^{(n)}(\b_i) \equiv |f_i^{(n)}(x)|^{\frac 1{|\xi^n\b_i|}}
\tag{3.4}
$$
and let
$$
\r_{max}^{(n)}\equiv
\max_{\b_i\in\B}\r^{(n)}(\b_i)
\tag{3.5}
$$
Note that for notational conveniance we have dropped the
explicit mentioning of the
dependence of $\r^{(n)}$ on the initial condition $x$.

>From now on we will always  consider a trace map whose associated
substitution $\phi$ is
semi-primitive. Recall that this means that $\phi$ is primitive
on an alphabet $\C\subset\B$.

\noindent{\thbf Definition 3.2:} {\it Let $\tilde \U_{\e,c,n}$ be
the subset of $\R^{|\B|}$ such that
$x\in\tilde \U_{\e,c,n}$ implies
\item{(i)}
$$
\min_{\g\in\C}\r^{(n)}(\g)>\left[\r_{max}^{(n)}\right]^{1-\e}\tag
{3.6}
$$
\item{(ii)}
$$
\left[\min_{\g\in\C}\r^{(n)}(\g)\right]^{min_{\a\in\A}|\xi^n\a|}> c
\tag {3.7}
$$
}
We have the following

\lemma {3.2} {\it There exists $n_0<\infty$ such that 
for all  $\e>0$ small enough and $\e'>\e$,
there exists $1<c<\infty$ such
that  for all $n\geq n_0$ and for  all $n'>n$,
 $\tilde \U_{\e,c,n}\subset \tilde \U_{\e',c,n'}$.
}

\proof   Note first that a priori $\tilde\U_{\e,c,n}\subset\tilde\U_{\e',c',n}$
if $\e'\geq\e$ and $c'\leq c$. We will now show that
$\tilde\U_{\e,c,n}\subset\tilde\U_{\e',c',n+1}$, for all
$\e'\geq \e+2c^{-\d\tilde\th}$ and
$c'\leq c^{\tilde\th(1-\e')}$, where $\d>0$ is some constant that depends only
on the substitutions $\xi$ and $\phi$, and $\tilde\th>1$ depends only on the
substitution $\xi$ (in fact, as $n\uparrow\infty$, $\tilde \th\uparrow\th$ 
where $\th$ is the largest eigenvalue of the
`substitution matrix', i.e. the matrix whose entries  $M_{ij}$ are the
number of times the letter $\a_i$ appears in the word $\xi\a_j$ [1]).
Iterating this result one sees that $\tilde\U_{\e,c,n}\subset\tilde
\U_{\e_k,c_k,n+k}$, where $c_k$   grows like
$c^{[\tilde\th(1-\tilde\e)]^k}$ and $\e_k$ needs only to satisfy
$\e_k\geq \e+\sum_{l=1}^kc_k^{-\d}$. But since
$\e+\sum_{l=1}^kc_k^{-\d}\leq\e+\sum_{l=1}^\infty c_k^{-\d}\equiv\tilde\e$
 it suffices to choose $\e_k \geq \tilde\e$.
Moreover, if $c$ is chosen sufficiently
large, $\tilde\e$ will be as   close to $\e$ as desired.
This obviously will imply the lemma.

The crucial idea of  the proof is the observation that for $c$
sufficiently large $x\in \tilde \U_{\e,c,n}$ implies that
$f\sim\tilde f$. Indeed, for any $i$,
$$
\eqalign{
\big |\tilde f_i(f^{(n)}(x))\big |&
\leq\sup_{\{n_i\}\,:\sum_{i=1}^{|\B|}n_i |\xi^n\b_i|=|\xi^{n+1}\b|}
\prod_{i=1}^{|\B|}\big |f^{(n)}_i(x)\big |^{n_i}\cr
&=\sup_{\{n_i\}\,:\sum_{i=1}^{|\B|}n_i |\xi^n\b_i|=|\xi^{n+1}\b|}
\prod_{i=1}^{|\B|}\left[\r^{(n)}(\b_i)\right]^{n_i|\xi^n\b_i|}\cr
&\leq\sup_{\{m_i\}\,:\sum_{i=1}^{|\B|}m_i =|\xi^{n+1}\b|}
\prod_{i=1}^{|\B|}\left[\r^{(n)}(\b_i)\right]^{m_i}\cr
&\leq\left[\r_{max}^{(n)}\right]^{|\xi^{n+1}\b|}
}
\tag {3.8}
$$
Using the fact that by assumption for any $\g\in\C$, $\phi(\g)$ contains only
letters in $\C$, in a similar way we obtain for any $\g\in\C$
$$
\eqalign{
\big |\tilde f_{i(\g)}(f^{(n)}(x))\big |&\geq
\inf_{\{n_i\}\,:\sum_{i=1}^{|\C|}n_i |\xi^n\g_i|=|\xi^{n+1}\g|}
\prod_{j=1}^{|\C|}\big |f^{(n)}_{i(\g_j)}(x)\big |^{n_{i(\g_j)}}\cr
&\geq\left[\min_{\g\in\C}\r^{(n)}(\g)\right]^{|\xi^{n+1}\g|}\cr
&\geq\left[\r_{max}^{(n)}\right]^{|\xi^{n+1}\g|(1-\e)}
}
\tag {3.9}
$$
On the other hand
$$
\eqalign{
\big |\tilde f_i(f^{(n)}(x))-f_i(f^{(n)}(x))\big |&\leq const.
\sup_{\{n_i\}\,:\sum_{i=1}^{|\B|}n_i |\xi^n\b_i|<|\xi^{n+1}\b|}
\prod_{i=1}^{|\B|}\big |f_i^{(n)}(x)\big |^{n_i}\cr
&\leq
const.\left[\r_{max}^{(n)}\right]^{|\xi^{n+1}\b|-
\inf_{\b_i\in\B}|\xi^n\b_i|}\cr
&\leq
\left[\r_{max}^{(n)}\right]^{|\xi^{n+1}\b|(1-\k)}
}
\tag {3.10}
$$
where $\k$ is some strictly positive constant.
Here we have used that
$$
|\xi^{n+1}\b|=\sum_{\a\in\b}|\xi^{n+1}\a|\leq
K\max_{\a\in\A}|\xi^{n+1}\a|
\tag {3.11}
$$
and
$$
\inf_{\b_i\in\B}|\xi^n\b_i|=\inf_{\a\in\A}|\xi^n\a|
\tag{3.12}
$$
Moreover, for primitive substitutions (see e.g. [1]), there
exists a finite constant $b$ such that
$$
-r_n+\th\leq\frac{|\xi^{n+1}\a| }{\inf_{\a\in\A}|\xi^n\a|}\leq
b\th+r_n,\text{uniformly in } \a\in\A
\tag{3.13}
$$
where $r_n$ converges to zero exponentially fast as
$n\uparrow\infty$. In particular, (3.13) is uniformly bounded from above by
some constant $1/\k$, and for $n$ sufficiently large, it is bounded from below
by a constant $\tilde\th>1$. The uniform upper bound then
implies the last inequality in (3.10). For $c$ sufficiently large,
the constant in (3.10) can be bounded by an arbitrarily small power of
$\left[\r_{max}^{(n)}\right]^{|\xi^{n+1}\b|}$
and thus it can be absorbed in $\kappa$.

Putting together (3.8) and (3.10), we get for all $\b_i\in\B$ the upper bound
$$
\eqalign{
\big|f_{i}(f^{(n)}(x))\big| &\leq \big|\tilde f_i(f^{(n)}(x)\big|+\big|
f_i(f^{(n)}(x))-\tilde f_i(f^{(n)}(x))\big|\cr
&\leq \left[\r_{max}^{(n)}\right]^{|\xi^{n+1}\b|}+
\left[\r_{max}^{(n)}\right]^{|\xi^{n+1}\b|(1-\k)}\cr
&=\left[\r_{max}^{(n)}\right]^{|\xi^{n+1}\b|}\left(1+
\left[\r_{max}^{(n)}\right]^{-\k|\xi^{n+1}\b|}\right)
}
\tag{3.14}
$$
Thus
$$
\eqalign{
\r^{(n+1)}(\b)&\leq\r_{max}^{(n)}
\left[1+\left[\r_{max}^{(n)}\right]^{
-\k|\xi^{n+1}\b|
}
\right]^{
\frac  1{|\xi^{n+1}\b|}
}
\cr
&\leq \r_{max}^{(n)}
\exp\left\{
\frac
{\left[\r_{max}^{(n)}\right]^{
-\k|\xi^{n+1}\b|
}
}
{|\xi^{n+1}\b|}
\right\}
\cr
&\leq\left[\r_{max}^{(n)}\right]^{
1+
\left(
\ln\left[\r_{max}^{(n)}\right]^{
|\xi^{n+1}\b|
}
\left[\r_{max}^{(n)}\right]^{
\k|\xi^{n+1}\b|
}
\right)^{-1}
}
\cr
&\leq\left[\r_{max}^{(n)}\right]^{1+c^{-\tilde\th\k}}
}
\tag{3.15}
$$
where we have used the lower bound on $\r_{max}^{(n)}$ implied by
(3.8), and the uniform lower bound on (3.13) given by $\th$. We have also 
readjusted the constant $\k$ in the last line 
to absorb the $\ln\left[\r_{max}^{(n)}\right]$ in the exponent.
 Since the bound in (3.15) is uniform in $i$, the
last line in (3.15) is an upper bound for $\r_{max}^{(n+1)}$. In much the
same way we obtain a lower bound on $\r^{(n+1)}(\g)$ for $\g\in\C$, namely
$$
\eqalign{
\r^{(n+1)}(\g)&\geq\left[\r_{max}^{(n)}\right]^{1-\e}
\left[1-\left[\r_{max}^{(n)}\right]^{
-(\k-\e)|\xi^{n+1}\g|
}
\right]^{
\frac  1{|\xi^{n+1}\g|}
}
\cr
&\geq\left[\r_{max}^{(n)}\right]^{
1-\e-
\left(
\ln\left[\r_{max}^{(n)}\right]^{
|\xi^{n+1}\g|
}
\left[\r_{max}^{(n)}\right]^{
(\k-\e)|\xi^{n+1}\g|
}
\right)^{-1}(1+\t)
}
\cr
&\geq\left[\r_{max}^{(n)}\right]^{1-\e-c^{-\tilde\th(\k-\e)}(1+\t)}
}
 \tag {3.16}
$$
 Here we assumed that $\e$ is smaller that $\k$
(since $\k$ is some absolute constant that depends only on the trace map,
we may always choose  $\e$, for instance, smaller than $\k/2$),
and
$\t$ may be chosen as $c^{-(\k-\e)\tilde\th}/2$ (note that the second
inequality in (3.16) uses that for $a>0$, $e^{-a}\leq 1-a+\frac {a^2}2\leq
1-a(1-\t)$, if $\t\geq \frac a2$). 
Putting (3.15) and (3.16) together, we get that
$$
\min_{\g\in\C}\r^{(n+1)}(\g)\geq \left[\r_{max}^{(n+1)}\right]^
{1-\e-c^{-\tilde\th\k}-c^{\tilde\th(\k-\e)}(1+\t)}
\tag {3.17}
$$
and
$$
\left[\min_{\g\in\C}\r^{(n+1)}(\g)\right]^{\min_{\a\in\A}|\xi^{n+1}\a|}
\geq c^{\tilde\th(1-\e-c^{-\tilde\th(\k-\e)}(1+\t))}
\tag {3.18}
$$
as claimed above and the proof of lemma 3.2 is completed. \endproof

\noindent {\thbf Corollary 3.1:} {\it  Let $\e$ and $c>2$ and $n_0$
 be chosen
such that the conclusion of Lemma 3.2 holds. Then, for all $n\geq
0$
$$
\tilde\U_{\e,c,n}\subset \hbox{int}\, \U_n \tag {3.19}
$$
Moreover, defining (for a given choice of $\e$, $c$ and $n_0$)
$$
\tilde \U\equiv \bigcup_{n=n_0}^\infty \tilde\U_{\e,c,n}\tag {3.20}
$$
we have that
$$
\tilde\U\subset\U\tag {3.21}
$$
}

\proof By lemma 3.2 we have that if $x\in\tilde\U_{\e,c,n}$, then
for all $m\geq n$,
$$
\eqalign{
|f^{(m)}_{i(\g_0)}(x)|&\geq \min_{\g\in \C}\left(
\left[\r^{(m)}(\g)\right]^{|\xi^m\g|}\right)
=\min_{\g\in \C}\left(
\left[\r^{(m)}(\g)\right]^{
\min_{\a\in\A}|\xi^{m}\a|\frac{|\xi^m\g|}
{\min_{\a\in\A}|\xi^{m}\a|}
}\right)\cr
&\geq c_{m-n}^{\frac{|\xi^m\g|}{
\min_{\a\in\A}|\xi^{m}\a|}}\geq c>2
}
\tag {3.22}
$$
which by definition of $\U_n$ proves that
$\tilde\U_{\e,c,n}\subset \U_n $. But since
the sets $\tilde\U_{\e,c,n}$ are manifestly open, they are also
contained in the interiors of the sets $\U_n$. The final
conclusion (3.21) is then obvious from the definition (3.20), which proves the
corollary.\endproof



\proposition {3.1} {\it Suppose $\phi$ is semi-primitive. Then
$x\in\tilde
\U^c$ implies that for all $\b_i\in\B$,
$$
\limsup_{n\uparrow\infty}\frac
1{|\xi^n\b_i|}\ln|f_i^{(n)}(x)|\leq 0\tag{3.23}
$$
}

\proof Proving the proposition is equivalent to proving that
$\limsup_{n\uparrow\infty}\maxn \leq 1$.
Notice first that if $\maxn \leq D^{\th^{-n}}$ for some
$n$-independent constant $D$ that may be chosen as large as desired,
then a trivial estimate similar to the one used in obtaining the
upper bound in the proof of lemma 3.2 shows that there exists, for
any $m$, a constant $D_1$, depending only on $D$,$m$ and $f$,
such that
$$
\r^{(n+m)}_{max}\leq D_1^{\th^{-n}}
\tag {3.24}
$$

On the other hand, if $\maxn > D^{\th^{-n}}$, with $D$ chosen
sufficiently large, we will prove that there exist $k$, $k'$ and
$\d>0$, (independent of $n$), such that
$$
\r_{max}^{(n+k+k')}\leq \left[\maxn\right]^{1-\d}
\tag {3.25}
$$
Before proving (3.25), let us show how (3.24) and (3.25) imply the proposition.
Let us fix $D$ and $m=k+k'$.
Obviously, to prove (3.23), it is enough to show that for any $1\leq n_0\leq
m$, we have that
$$
\limsup_{i\uparrow\infty}\rho_{max}^{(n_0+im)}\leq 1
\tag {3.26}
$$
Now by (3.24) and (3.25) we get that for any $n_0$,
$$
\rho_{max}^{(n_0+m)}\leq\max\left\{D_1^{\th^{-n_0}},
\left[\rho_{max}^{(n_0)}\right]^{1-\d}\right\}
\tag{3.27}
$$
and iterating this
$$
\eqalign{
&\rho_{max}^{(n_0+2m)}\leq\max\left\{D_1^{\th^{-n_0-m}},
D_1^{\th^{-n_0}(1-\d)},
\left[\rho_{max}^{(n_0)}\right]^{(1-\d)^2}
\right\},\dots,\cr
&\rho_{max}^{(n_0+im)}\leq\max\left\{
D_1^{\th^{-n_0-(i-1)m}},D_1^{\th^{-n_0-(i-2)m}(1-\d)},\dots,
D_1^{\th^{-n_0}(1-\d)^{i-1}},\dots,
\left[\rho_{max}^{(n_0)}\right]^{(1-\d)^i}
\right\}
}
\tag{3.28}
$$
As a matter of fact, depending on whether $\th^m$ is smaller or greater than
$(1-\d)$ the last line in (3.28) is bounded by either
$\max\left\{
D_1^{\th^{-n_0-im}},
\left[\rho_{max}^{(n_0)}\right]^{(1-\d)^i}
\right\}$ or $\max\left\{
D_1^{\th^{-n_0}(1-\d)^{({i-1})}},
\left[\rho_{max}^{(n_0)}\right]^{(1-\d)^i}
\right\}$
Obviously, whatever $D_1$ or $\rho_{max}^{(n_0)}$, this bound converges to one
as $i\uparrow\infty$, proving (3.26) and hence (3.23).



To prove (3.25) we proceed in two steps:
 First, we use the fact that $\phi$ is
primitive on $\C$ to show that there exist  $k$ and  $\d'>0$ such that
for  all $n$,
$$
\max_{\g\in\C}\r^{(n+k)}(\g)\leq
\left[\r_{max}^{(n)}\right]^{1-\d'}
\tag {3.29}
$$
Then we use this inequality together with the second condition from
the definition of semi-pri\-miti\-vi\-ty
to show that there exist $k'$ and $\d>0$ such that
(3.25) holds.

Let us now  prove (3.29).
For each $\b_i\in\B$,
$$
f^{(n+k)}_i(x)=f^{(k)}_i(f^{(n)}(x))
\tag{3.30}
$$
where $f^{(k)}_i$ is a polynomial s.t.
$$
d\left(f^{(k)}_i\right)=\big| \xi^k\b_i\big|
\tag{3.31}
$$
and
$$
d\left(f^{(k)}_i-\tilde f^{(k)}_i
\right)\leq \big|\xi^k\b_i\big|-1
\tag{3.32}
$$
so that as in the proof of lemma 3.2,
$$
\eqalign{
\Big|f^{(k)}_i(f^{(n)}(x))-\tilde f^{(k)}_i(f^{(n)}(x))\Big|&\leq
const_k\left[\maxn\right]^{|\xi^{n+k}\b_i|-\min_{\a\in\A}|\xi^n\a|}\cr
&\leq\left[\maxn\right]^{|\xi^{n+k}\b_i|(1-\tilde\k)}
}
 \tag{3.33}
 $$
where $1>\tilde\k>0$ depends only on $k$, provided
$ \left[\maxn\right]^{|\xi^{n+k}\b_i|}$ is sufficiently large,
which is guaranteed by the condition $\maxn > D^{\th^{-n}}$. 
In particular, 
$\tilde\k$ can be chosen such that (3.33) holds for all $0\leq k\leq m$.


Now, to prove (3.29), notice that, since $x\in\U^c$,
there exists $\tilde\g\in \C$, such that either
$$
\big|f^{(n)}_{i(\tilde\g)}(x)\big|\leq c_1
\tag {3.34}
$$
with $c_1\leq c^b$ (recall (3.13))
or
$$
\big|f^{(n)}_{i(\tilde\g)}(x)\big|^{\frac
1{|\xi^n\tilde\g|}}\leq\left[\maxn\right]^{1-\e}
\tag {3.35}
$$
Now choose $k$ such that $\phi^k\g$ contains all letters in $\C$ (and in
particular $\tilde\g$) so that
$$
\eqalign{
\left|\tilde f^{(k)}_{i(\g)}(f^{(n)}(x))\right|&\leq
\sup_{\{n_i\}\,:\,\sum_{i=1}^{|\C|}n_i|\xi^n\g_i|=
|\xi^{n+k}\g|-|\xi^n\tilde\g|}
\prod_{j=1}^{|\C|}\big|f^{(n)}_{i(\g_j)}(x)\big|^{n_j}\times
\big|f^{(n)}_{i(\tilde\g)}(x)\big|\cr
&\leq\left[\maxn\right]^{|\xi^{n+k}\g|-|\xi^n\tilde\g|}
\left[\maxn\right]^{(1-\e)|\xi^n\tilde\g |}\cr
&=\left[\maxn\right]^{|\xi^{n+k}\g|-\e|\xi^n\tilde\g|}
}
\tag{3.36}
$$
Since $|\xi^n\tilde\g|\geq\kappa |\xi^{n+k}\g|$, uniformly in $\g$,
$\tilde \g\in\C$, this yields
$$
\left|\tilde f^{(k)}_{i(\g)}(f^{(n)}(x))\right|\leq
\left[\maxn\right]^{|\xi^{n+k}\g|(1-\kappa\e)}
\tag{3.37}
$$
which together with (3.33) gives (3.26).

Finally, we choose $k'$ such that for all $\b_i\in \B$,
$\phi^{k'}\b_i$ contains
a letter, say $\tilde\g$, from $\C$. Then, for all $\b_j \in \B$
$$
\eqalign{
\left|\tilde f^{(k'+k)}_j(f^{(n)}(x))\right|&\leq
\sup_{\{n_i\}\,:\,\sum_{i=1}^{|\B|}n_i|\xi^{n+k}\b_i|=
|\xi^{n+k+k'}\b_j|-|\xi^{n+k}\tilde\g|}
\prod_{i=1}^{|\B|}\big|\tilde f^{(k)}_i(f^{(n)}(x))\big|^{n_i}\cr
&\times
\big|\tilde f^{(k')}_{i(\tilde\g)}(f^{(n)}(x))\big|\cr
&\leq\left[\r_{max}^{(n)}\right]^{|\xi^{n+k+k'}\b_j|-|\xi^{n+k}\tilde\g|}
\left[\maxn\right]^{(1-\d)|\xi^{n+k}\tilde\g |}\cr
&\leq\left[\r_{max}^{(n)}\right]^{|\xi^{n+k+k'}\b_j|-\d|\xi^{n+k}\tilde\g|}
}
\tag {3.38}
$$
from which (3.25) follows as before. This concludes the proof of
the proposition.
 \endproof



\remark Proposition 3.1 provides us with a nice dichotomy: for
substitutions with semi-primitive reduced trace maps, for any initial
condition $x$, {\it either} all components of $f^{(n)}(x)$
diverge in absolute value exponentially fast with the same rate,
{\it or} no component grows exponentially fast.
To prove this it was crucial that for primitive substitutions the
lengths of the words $|\xi^n\a|$ grow with $n$ exponentially fast
with the same rate, i.e. $|\xi^n\a|\sim\th^n$, where $\th$ is the
largest eigenvalue of the substitution matrix (see e.g. [1,14]).



Our next task will be to show that - under some extra
conditions - the Lyapunov exponent will be zero if
$E\in\E(\tilde\U)^c$. This is the contents of

\proposition {3.2} {\it Suppose $\tilde f$ satisfies the
assumptions of proposition 3.1. Assume further that there exists
$k<\infty$ such that $\xi^k 0$ contains the word $\b\b$, for some
$\b\in\B$. Then
$E\in\E(\tilde
\U)^c$ implies that
$$
\g(E,v)\equiv \lim_{n\uparrow\infty}\frac
1{|n|}\ln\big\|T_E(u^{(n)})\big\|=0
\tag{3.39}
$$
(Here $u^{(n)}$ denotes the word consisting of the first $n$ letters of
the substitution sequence $u\equiv\xi^\infty 0$)
}

\proof  We show first that
$$
\lim_{n\uparrow\infty}\frac
1{|\xi^n 0|}\ln\big\|T_E(\xi^n 0)\big\|=0
\tag{3.40}
$$
Now let us denote
$$
\tmax \equiv
\max_{\a\in\A}\big\|T^{(n)}_E(\a)\big\|^{\frac
1{|\xi^n\a|}}
\tag {3.41}
$$
Using the Schwarz inequality, one finds that
$$
\big\|T^{(n+1)}_E(\a)\big\|\leq \left[\tmax\right]^{|\xi^{n+1}\a|}
\tag {3.42}
$$
Now choose $k$ such that $\xi^k 0$ contains $\b\b$, for some
$\b\in\B$, and use that
$$
\left(T_E^{(n)}(\b)\right)^2= x_E^{(n)}(\b)T_E^{(n)}(\b)-\id
\tag {3.43}
$$
where by proposition 3.1
$\limsup_{n\uparrow\infty}\big|x_E^{(n)}(\b)\big|^{1/|\xi^n\b|}\leq 1$.
Thus
$$
\eqalign{
\big\|T_E^{(n+k)}(\a)\big\|&\leq
\big|x_E^{(n)}(\b)\big| \sup_{\{n_i\}\,:\,\sum n_i
|\xi^n\a_i|=|\xi^{n+k} \a|-|\xi^n\b|}
\prod_{i=1}^{|\A|}
\big\|T_E^{(n)}(\a_i)\big\|^{n_i}
\cr
&+ \sup_{\{n_i\}\,:\,\sum n_i
|\xi^n\a_i|=|\xi^{n+k} \a|-2|\xi^n\b|}
\prod_{i=1}^{|\A|}
\big\|T_E^{(n)}(\a_i)\big\|^{n_i}
\cr
&\leq\left[\tmax\right]^{|\xi^{n+k}\a|( 1-\t_k)}
}
\tag {3.44}
$$
from which (3.40) follows as the analogous statement in
proposition 3.1.

>From (3.40) one obtains (3.39) just as in [11]. \endproof

\remark Note that the condition in proposition 3.2 that $\b\in \B$
is not very restrictive. For, if some other word, say $\o$, appears as $\o\o$
in $u$, one may always extend the alphabet $\B$ to include $\o$ and study the
corresponding trace map.

Proposition 3.2 provides in fact two pieces of information: First it shows
 that the Lyapunov exponent vanishes on
$\E(\tilde\U)^c$. However, this also implies that if $E\in \s(H_v)^c$, then
$x_E^{(0)}\in \tilde U$. This is implied by the general fact that for
Schr\"odinger operators the Lyapunov exponent is strictly positive
if $E$ is outside the spectrum (see, e.g. [22]). This allows us to prove

\proposition {3.3} {\it Suppose $H_v$ permits a trace map satisfying the
assumptions of proposition 3.2. Then  $E\in\s(H_v)$ if and only if
$\g(E,v)=0$.}

\proof  To prove the proposition,  set
$$
\OO\equiv \left\{E\bigl|\g(E,v)=0\right\}
\tag {3.45}
$$
We have just seen that $\E(\tilde U)^c\subset \OO$
while in general $\OO\subset\s(H_v)$. On the other hand, lemma 3.1
shows that $\s(H_v)\subset(\E(\U))^c$, while by
corollary 3.1, $\E(\U)^c\subset
\E(\tilde\U)^c$, so that finally we
have the chain of inclusions
$$
\E(\tilde\U)^c\subset\OO\subset\s(H_v)\subset\E(\U)^c\subset\E(\tilde\U)^c
\tag {3.46}
$$
which clearly implies the equality of all these sets and proves the
proposition. \endproof

Theorem 1 is now a direct consequence of the following general theorem
that was proven in [11]:

\theo 2 [11] {\it Let $H_v$ be an operator of the form (1.1)
where $V$ is a potential that takes only finitely many values.
Let $(\O, T)$ denote the topological dynamical system where
$\O$ is the closure of the set of translates of the sequence $V_n$ and $T$ the
shift operator. Assume that $V_n$ is aperiodic and $(\O,T)$
permits a unique ergodic
$T$-invariant probability measure $\mu$. Then, if $\s(H_v)=
\left\{E|\g(E,v)=0\right\}$, $\s(H_v)$ is supported on a set of zero Lebesgue
measure. In particular, $\s(H_v)$ has no absolutely continuous component.}

This theorem is in fact a consequence of a lemma of Kotani [18] which
states that for aperiodic potentials that take only a finite number
of values, the set of energies for which the mean Lyapunov exponent
(where the mean is taken over the hull $\O$ with respect to the $T$-invariant
measure $\mu$) vanishes is of Lebesgue measure zero. Using a result of
Herman [23] one can then show, along the lines of a proof of
Avron and Simon [24] in the case of almost periodic potentials, that
under the assumption of unique ergodicity the sets on which
the Lyapunov exponents for different elements in the hull vanish
may differ only by sets of zero Lebesgue measure.
The detailed proof of this theorem can be found in [11] and will not be
reproduced here.  The assumption of unique ergodicity is satisfied for
substitution sequences based on primitive substitutions.
The proof of this result
is rather elaborate and may be found in the book by Qu\'effelec [1].
Therefore, theorem 1 is proven. \endproof\endproof

Theorem 1 shows that for substitution sequences satisfying our
hypothesis, the spectrum is manifestly different from both periodic
(absolutely continuous spectrum) and random (dense pure point spectrum)
potentials. However, in the examples more precise results were proven
in that also the existence of eigenvalues could be excluded. In our general
setup we can only exclude this possibility under a simple
supplementary hypothesis:

\theo 3 {\it Suppose the hypothesis of theorem 1 are satisfied. If in addition
there exists $n_0<\infty$ s.t. $\xi^{n_0}0=\xi^m(\g_0)\xi^m(\g_0)\o$,
where $\g_0\in\C$ and
$\o\in\A^*$ and $m$ are arbitrary,
then the spectrum of $H_v$ is purely singular continuous and supported on a
Cantor set of zero Lebesgue measure.}

\proof The basic idea of the proof was used already in S\"ut\H o  [6]
to obtain
the same result for the Fibonacci sequence. Namely, note that under
our assumption for all $n\geq n_0$,
$$
T^{(n)}_E(0)=T^{(n-n_0)}_E(\o)T^{(n-n_0+m)}_E(\g_0)T^{(n-n_0+m)}_E(\g_0)
\tag {3.47}
$$
and therefore $T^{(n-n_0+m)}_E(\g_0)$ and ${T^{(n-n_0+m)}_E(\g_0)}^2$ are
transfer matrices over $|\xi^{(n-n_0+m)}\g_0|$ and $2|\xi^{(n-n_0+m)}\g_o|$
sites, respectively.
Now, (2.7) implies (see e.g. [6]) that for any normalized
vector $\Psi\in\R^2$,
$$
\frac 12\leq\max\left\{\|{T^{(n-n_0+m)}_E(\g_0)}^2\Psi\|,|\tr
{T^{(n-n_0+m)}_E(\g_0)}|\,\|{T^{(n-n_0+m)}_E(\g_0)}\Psi\|\right\}
\tag {3.48}
$$

To use (3.48), we only need the following

\lemma {3.4} {\it Let $\g_0\in \C$ be any word such that the
word $\g_0\in\A^* $ contains the letter $0$.
  Then, for all $E\in \s(H_v)$ there exists a sequence of integers
$n_i$ tending to infinity such that for all $i$,
$|x_E^{(n_i)}(\b)|\leq 2$.}

\proof  As we have stated at the beginning of this section, any
letter $\g_0$ satisfying the assumptions of lemma 3.4 may be
used to define the set $\U$, and in each case, $\E(\U)^c =\s(H_v)$,
and thus does not depend on the chosen letter. The conclusion of
the lemma is then obvious from the definition of $\U$ together
with the fact that the sets $\E(\U_n)$ are open.
\endproof

Let now $E\in \s(H_v)$ and let $\Psi_E$ be a solution of (2.1), i.e.
a solution of the Schr\"odinger equation. Let $n_i$ be the sequence
given by lemma 3.4. Assume that $\Psi_E(1)\not=0$ (otherwise, of necessity,
$\Psi_E(0)$ will be nonzero, and the discussion below  can be repeated with
$n_i$ replaced by $-n_i$). Then
$$
\|\Psi_E\|_2^2\geq\sum_{i=1}^\infty \max\left\{
\|\Psi_E(|\xi^{n_i}\b|)\|^2,
\|\Psi_E(|2\xi^{n_i}\b|)\|^2\right\}\geq \sum_{i=1}^\infty \frac 14
\|\Psi_E(1)\|^2=\infty
\tag {3.49}
$$
which proves that $\psi_E$ is not in $l^2(\Z)$ and thus that $E$ is not an
eigenvalue. But since this holds for all energies in the spectrum, the theorem
is proven. \endproof\endproof

\remark The proof of theorem 3 implies the stronger result that for all energies
in the spectrum, no solution of the Schr\"odinger equation tends to zero
at both plus and minus infinity.

\remark  The hypothesis in theorem 3 is clearly not {\it necessary}.
The period doubling sequence provides an example where the
hypothesis does not hold but
the spectrum is still singular continuous. This is also true for the
Thue-Morse sequence [10,11], where, however, an additional symmetry allows to
use essentially the same argument. We feel that in all cases where the
hypothesis of theorem 1 hold, the spectrum should be singular continuous.

\vfill\eject
\line{\ch IV. Examples\hfil}

In this final section  we consider some specific examples, in fact the same
ones as  in [14]: the Fibonacci sequence, the Thue-Morse sequence, the period-doubling
sequence, the circle sequence,  the `binary' and
`ternary' `non-Pisot' sequences and finally, rather as a `counter-example',
the Rudin-Shapiro sequence.

In all examples (except Rudin-Shapiro) the alphabets $\A$ will consist of
at most  three letters that we denote by $a,b,c$.
For the corresponding traces (that we identify with the elements of $\B$)
we will use the simplified notations
$$
\eqalign{
x&\equiv \tr T_E(a)\cr
y&\equiv \tr T_E(b)\cr
z&\equiv \tr T_E(c)\cr
u&\equiv \tr T_E(a)T_E(b)\cr
v&\equiv \tr T_E(b)T_E(c)\cr
w&\equiv \tr T_E(a)T_E(c)\cr
r&\equiv \tr T_E(c)T_E(b)T_E(a)
}
\tag {4.1}
$$

\vskip 12pt \line{\bf 1. The Fibonacci sequence\hfill}

The Fibonacci sequence is the fixpoint of the substitution
$\xi$ on two letters, $a$ and $b$, defined by
$$
\eqalign{
a&\rightarrow \xi(a)=ab\cr
b&\rightarrow \xi(b)=a}
\tag {4.2}
$$

 The substitution  $\xi$ is primitive, since $\xi^2(a)=aba$ and
$\xi^2(b)=ab$ both  contain all the letters of the alphabet.
Using (2.7) the reader verifies easily that  a trace map $f$ is found as
$$
\eqalign{
x&\rightarrow u\cr
y&\rightarrow x\cr
u&\rightarrow xu-y}
\tag {4.3}
$$

Thus the reduced trace map $\tilde f$ is then
$$
\eqalign{
x&\rightarrow u\cr
y&\rightarrow x\cr
u&\rightarrow xu}
\tag {4.4}
$$
Obviously, (4.4) may be viewed directly as the substitution $\phi$, defined in
 section 2, acting on the letters $x,y,u$.
This substitution is semi-primitive, since, with $\C\equiv (x,u)$
\item {(i)} $\phi$ maps $\C$ into $\C^*$ and $\phi^2(x)=xu$ and $\phi^2(u)=uxu$
both contain all the letters of $\C$.
\item {(ii)} $\phi^2(x)$ contains $x$ and $u$, $\phi^2(y)$ contains $u$ and
$\phi^2(u)$ contains $x$ and $u$.

Moreover, since $0$ is given by $a$, $\xi^30=abaab$ and thus contains the
square of the word $a$.

Therefore all the hypothesis of theorem 1 are satified and then the spectrum of
$H_v$ is singular and supported on a set of zero Lebesgue measure.

Moreover, $\xi^40=abaababa$ begins with the square of the word $aba$. Now,
$aba$ is not a word in $\B$, however, following the remark after
proposition 3.2 we may enlarge $\B$ by including the letter
$t\equiv aba$. A simple
calculation shows then that $\phi(t)= xu^2$ and  this extended trace map
is still semi-primitive. Thus,
theorem 3 implies that the spectrum of $H_v$ is purely singular continuous.

This of course  recovers here a result already proven in [6] and [7].


\vskip 12pt \line{\bf 2. The Thue-Morse sequence\hfil}

The substitution this time is defined by [9]
$$
\eqalign{
a&\rightarrow \xi(a)=ab\cr
b&\rightarrow \xi(b)=ba}
\tag {4.5}
$$

Obviously, the substitution is primitive.
Notice that both the letters $a$ and $b$ can be taken as ``$0$'' and that
there are therefore two fixpoints $\xi^\infty(a)$ and $\xi^\infty(b)$.

Using again (2.7) with $A=B$, we can find the following trace map $f$:
$$
\eqalign{
x&\rightarrow u\cr
y&\rightarrow u\cr
u&\rightarrow xyu-x^2-y^2+2}
\tag {4.6}
$$
and the corresponding reduced trace map $\tilde f$ and the substitution $\phi$
are
$$
\eqalign{
x&\rightarrow u\cr
y&\rightarrow u\cr
u&\rightarrow xyu}
\tag {4.7}
$$
This time, the substitution $\phi$ is even primitive since
$\phi^2(x)=\phi^2(y)=xyu$ and $\phi^2(u)=u^2xyu$ contain all the
letters of $\B$.

Finally, chosing $a$ as $0$, $\xi^2a=abba$, which contains the square of the
word $b$.
Therefore theorem 1 holds and thus the spectrum of $H_v$ is singular and
supported on a set of zero Lebesgue measure.

As we noticed in the last remark of chapter 3, although we cannot apply
theorem 3, the spectrum of $H_v$ is purely singular continuous, as was
proven in [10] and [11].


\vskip 12pt \line{\bf 3. The period-doubling sequence\hfil}

It is defined as the fixpoint of the primitive substitution
$$
\eqalign{
a&\rightarrow \xi(a)=ab\cr
b&\rightarrow \xi(b)=aa}
\tag {4.8}
$$
The trace map here is
$$
\eqalign{
x&\rightarrow u\cr
y&\rightarrow x^2-2\cr
u&\rightarrow x^2u-xy-u}
\tag {4.9}
$$
$\phi$ is given by
$$
\eqalign{
x&\rightarrow u\cr
y&\rightarrow x^2\cr
u&\rightarrow x^2u}
\tag {4.10}
$$
With $\C\equiv (x,u)$ one checks that it is  semi-primitive, since
\item {(i)} $\phi$ maps $\C$ into $\C^*$ and $\phi^2(x)=x^2u$ and
$\phi^2(u)=u^2x^2u$
\item {(ii)} $\phi^2(x)$ contains $x$ and $u$, $\phi^2(y)$ contains $u$ and
$\phi^2(u)$ contains $x$ and $u$.

Finally, $\xi^20=abaa$ contains the square of the word $a$ and thus  theorem 1
applies. However, the hypothesis of theorem 3 are not verified, although it
was proven (through a rather cumbersome calculation)
in [11] that the spectrum is singular continuous. Note however that the
`inverted' sequence (obtained by setting $\xi(a)=ba$) satisfies the hypothesis
of theorem 3.

\vskip 12pt \line{\bf 4. The circle sequence\hfil}

The circle sequence is associated to the substitution $\xi$ on three letters
$$
\eqalign{
a&\rightarrow \xi(a)=cac\cr
b&\rightarrow \xi(b)=accac\cr
c&\rightarrow \xi(c)=abcac
}\tag {4.11}
$$
This substitution has no fixpoint, since it does not posses a letter ``$0$'',
but it has  a cycle of length two and the twice iterated substitution has two
fixpoints.


Using then the identities (2.7) and (2.8), we find an alphabet
$\B\equiv(a,b,c,ab,bc,ca,abc)$, identified with $\B\equiv(x,y,z,u,v,w,r)$,
and the following trace map $f$
$$
\eqalign{
x&\rightarrow zw-x\cr
y&\rightarrow zw^2-xw-z\cr
z&\rightarrow wr-y\cr
u&\rightarrow (zw-x)(zw^2-xw-z)-w\cr
v&\rightarrow (zw^2-xw-z)(wr-y)-yz+v\cr
w&\rightarrow (wr-y)(zw-x)-u\cr
r&\rightarrow (wr-y)(zw-x)((zw^2-xw-z)-w)+w^2r-r-yw-u(zw^2-xw-z)}
\tag {4.12}
$$
The reduced trace map $\tilde f$ is
$$
\eqalign{
x&\rightarrow zw\cr
y&\rightarrow zw^2\cr
z&\rightarrow wr\cr
u&\rightarrow z^2w^3\cr
v&\rightarrow zw^3r\cr
w&\rightarrow zw^2r\cr
r&\rightarrow z^2w^4r}
\tag {4.13}
$$
The associated substitution $\phi$ is again semi-primitive with
$\C\equiv (z,w,r)$, since

\item {(i)} $\phi$ maps $\C$ into $\C^*$ and $\phi^2(z)=zw^2rz^2w^4r$,
$\phi^2(w)=wr(zw^2r)^2z^2w^4r$\hfill\break and $\phi^2(r)=(wr)^2(zw^2r)^4z^2w^4r$
\item {(ii)} For any $\b\in \B$, $\phi^2(\b)$ contains $z$, $w$ and $r$.

Moreover $\xi^2c$ begin with the square of the word $ca\in\B$ so that
both theorem 1 and 3 apply and show that the spectrum is singular continuous
in this case, too.

\vskip 12pt \line{\bf 5. Binary non-Pisot sequence\hfil}

This sequence corresponds to the substitution
$$
\eqalign{
a&\rightarrow \xi(a)=ab\cr
b&\rightarrow \xi(b)=aaa}
\tag {4.14}
$$
and the trace map
$$
\eqalign{
x&\rightarrow u\cr
y&\rightarrow x^3-3x\cr
u&\rightarrow x^3u-x^2y-2xu+y}
\tag {4.15}
$$
with reduced trace map
$$
\eqalign{
x&\rightarrow u\cr
y&\rightarrow x^3\cr
u&\rightarrow x^3u}
\tag {4.16}
$$
Here $\C\equiv (x,u)$ and since $\phi^2(x)=x^3u$ and $\phi^2(u)=x^4u^3$, we
see that the substitution $\phi$ is semi-primitive. Moreover,
$\xi^20=abaaa$ contains the square of the word $a$, so theorem 1 applies.

Theorem 3, however, does not apply in this case (although again, as in the
case of the period doubling sequence, the inverted sequence satisfies the
hypothesis of this theorem) and we do not know for sure whether the eigenvalues
are present in this example.

\vskip 12pt \line{\bf 6. Ternary non-Pisot sequence\hfil}

This sequence corresponds to the substitution
$$
\eqalign{
a&\rightarrow \xi(a)=c\cr
b&\rightarrow \xi(b)=a\cr
c&\rightarrow \xi(c)=bab
}\tag {4.17}
$$
As in the case of the circle sequence, this substitution does not posses a
fixpoint, but a cycle of length three, whose three elements can be considered
as substitution sequences.
With the alphabet $\B\equiv(x,y,z,u,v,w)$, we can find the trace map $f$
$$
\eqalign{
x&\rightarrow z\cr
y&\rightarrow x\cr
z&\rightarrow yu+x\cr
u&\rightarrow w\cr
v&\rightarrow u^2-2\cr
w&\rightarrow uv+w-xz}
\tag {4.18}
$$
and the reduced trace map
$$
\eqalign{
x&\rightarrow z\cr
y&\rightarrow x\cr
z&\rightarrow yu\cr
u&\rightarrow w\cr
v&\rightarrow u^2\cr
w&\rightarrow uv}
\tag {4.19}
$$

The substitution $\phi$ is semi-primitive with  $\C\equiv (u,v,w)$
since $\phi^5(u)=wu^2uvuv$,
$\phi^5(v)=uvw^2uvw^2$ and $\phi^5(w)=wvw^3u^2wu^2$
and for any $\b\in \B$, $\phi^3(\b)$ contains $u$, $v$ and $w$.

Moreover, $\xi^5a=\xi^6b=\xi^4c=babacabab$ begins with the square of the word
$ba$.
Therefore, by theorems 1 and 3, the spectrum of $H_v$ is purely singular
continuous.


\vskip 12pt \line{\bf 7. The Rudin-Shapiro sequence\hfil}

The Rudin-Shapiro sequence [17] is defined on an alphabet of four letters. The
substitution rule is
$$
\eqalign{
a&\rightarrow \xi(a)=ac\cr
b&\rightarrow \xi(b)=dc\cr
c&\rightarrow \xi(c)=ab\cr
d&\rightarrow \xi(d)=db
}\tag {4.20}
$$
This final example serves to illustrate that even the hypothesis of theorem
1 are not always satisfied. It has been remarked in different contexts (see
[12]) that the Rudin-Shapiro sequence has quite exceptional properties and
that the analysis of the spectrum of the associated operators eludes
perturbative and even numerical methods.

One may note from the start that no  square of any word may
ever appear in an
iterate of any of the letters $a$, $b$, $c$ or $d$. This already shows that
 we will not be able to  apply
theorem 1.

Moreover, using the trace map computed by [20], we obtain a reduced trace map
$\tilde f$ on an alphabet $\B\equiv(x,y,z,w,s,t,q,r)$ (this trace map was
obtained in [20] in a clever way in order to stay with as few traces as
possible. A straightforward derivation would give a map on twelve letters
which would share the same properties)
$$
\eqalign{
x&\rightarrow s\cr
y&\rightarrow t\cr
z&\rightarrow t\cr
w&\rightarrow s\cr
s&\rightarrow r\cr
t&\rightarrow q\cr
q&\rightarrow xwr\cr
r&\rightarrow yzq}
\tag {4.21}
$$
It is easy to notice that the two alphabets $\C_1\equiv(x,w,t,r)$ and
$\C_2\equiv(y,z,s,q)$ are mutually exchanged by the substitution $\phi$
associated to $\tilde f$.
This implies that $\phi$ is not semi-primitive. Now,  $\C_1$ and $\C_2$
are left invariant under $\phi^2$ and one might hope to
simply study the dynamics of the trace maps on the two sub-alphabets separately.
 However, the subdominant terms in the
trace map (which we have not written, but see [20]) do not respect
this invariance which makes it impossible to even adapt the proof of
propositions 3.1 and 3.2 to this situation. So once again, the Rudin-Shapiro
sequence retains its mistery.
\parskip1pt plus 3pt

\vfill\eject
\parskip=1pt plus 2pt
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