J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
F. M. D
EKKING
Z.-Y. W
EN
Boundedness of oriented walks generated by substitutions
Journal de Théorie des Nombres de Bordeaux, tome
8, n
o2 (1996),
p. 377-386
<http://www.numdam.org/item?id=JTNB_1996__8_2_377_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Boundedness of oriented walks
generated by
substitutions
par F.M. DEKKING ET Z.-Y. WEN
RÉSUMÉ. Soit x = x0x1 ... un
point
fixe de la substitution surl’al-phabet
{a,b},
et soit Ua =$$(
-1 0 -1 1
)
et Ub =1 0 1 1)
. On donneune classification
complète
des substitutions 03C3 :{a, b}*
~{a,
b}*
selon que la suite de matrices(Ux0Ux1 ... Uxn)~n=0
est bornée ou non. Celacor-respond
au fait que les chemins orientésengendrés
par les substitutionssont bornés ou non.
ABSTRACT. Let x = x0x1 ... be a fixed
point
of a substitution on thealphabet
{a,b},
and let Ua =(-10 -11)
andUb = (10 11)
. Wegive
acomplete
classification of the substitutions 03C3 :{a, b}*
~{a, b}*
according
to whether the sequence of matrices(Ux0 Ux1... Uxn)~n=0 is
bounded or unbounded. This
corresponds
to the boundedness orun-boundedness of the oriented walks
generated by
the substitutions.1. Introduction
Let A be the
alphabet
la, bl,
and let x = xoxl ... be an infinitesequence
over A.
Any
such sequencegenerates
an oriented walk(SN) _
(SN, f(x))
on theintegers by
thefollowing
rules:, , - -
-In other words: we move one
step
in the same direction if =b,
and onestep
in the reversed direction if a. Another way to describe is
by
introducing
the matricesthen
Mots-cles : Substitutions, self-similarity, walks. Manuscrit reçu le 3 juin 1995.
In the
probability
literature(SN(x))
is also known aspersistent
random walkor correlated random walk if x is obtained
according
to aproduct
measure onAN.
Here we shall consider the case where the sequence x is a fixedpoint
of aprimitive
substitution o, on{a,
b}.
Non-oriented walks
(SN, f(x))
on R are definedby
So(x)
= 0 andwhere
f :
A -~R,
and A is now anarbitrary
finite set.(It
is convenient to extendf homomorphically
to A* = i.e.wk)
= + ... +f(wk)
forNon-oriented walks with x a fixed
point
of a substitution have been studied in[2], [3], [5], [6], [7], [10].
Here[10]
contains a rathercomplete analysis
of thebehaviour of
(SN, f (x ) )
for two letteralphabets
A ={a, b}.
It follows from a
general
result in[4]
thatby
enlarging
thealphabet
A orientedwalks
generated
by
substitutions may be viewed as non-oriented walksgenerated
by
substitutions. Hence itmight
look as if the main result of[5],[6] -
which admitsalphabets
ofarbitrary
sizes - would answer allquestions
on the twosymbol
oriented walk. Their result is that as N -; o0
where 0 is the Perron-Frobenius
eigenvalue
of the matrixM,
of the substitutionor
(with
entries m9t = number of occurrences of thesymbol s
in theword
u(t)),
82
the secondlargest
(in
absolutevalue)
eigenvalue
ofM,
(which
isrequired
to beunique
andlarger
than1)
and v is the vectorsatisfying
Mav
= 8vand ¿sEA Vs
=1. Furthermore a + 1 is the order of92
in the minimalpolynomial
ofM,
and F :[1, oo)
-~ R is a bounded continuous function which satisfies aself-similarity
property:
However,
even such asimple
property
as boundedness of can oftennot be resolved with
(2).
Let us take forexample
A ={a, b, c}, f (a)
=f (b)
=+1,
f (c)
= -1 and a such that the matrix of crequals
We
quickly
see thatM,
haseigenvalues
0,82
= 2 and 0 =8,
and that the PerronBut
f (Qa)
= =/(7c)
=0,
so = 0 for allN,
and(SN, f(x))
isbounded,
inspite
of the behavioursuggested by
(3). (Of
course(3)
and theself-similarity
property
of Fimply
that F =0.)
The
goal
of thisstudy
is to determine for any substitution a on{a, b}
whetherthe oriented walk in
(1)
will be bounded or not.Although
notexplicitly formulated,
theanalysis
of the oriented twosymbol
case in
[10]
heavily
relies on the fact that a substitution a on a twosymbol
alphabet
with(v, f )
= 0automatically
admits arepresentation
with the samef
in R
(terminology
from[1]),
i.e.,
there exists A E R such thatf (a(s))
=Af (s)
for s =
a, b.
(In
[9]
such a are calledgeornetric,
see also(8~).
However this is nolonger
true forlarger alphabets,
and this is the main reason that our solution tothe boundedness
problem
is rather delicate.2. Four
types
of substitutionsLet a be a substitution on
~a, b}
such that the first letter ofa(a)
is a, and let u be the fixedpoint
of a with uo = a. Letbe the matrix of a.
Here,
asusual,
Ivlw
denotes the number of occurrences of a word w in a word v. It appears that thequestion
of boundedness of(SN(u))
depends
crucially
on the entries ofMa
reduced modulo two. LetMo
be this matrix. Then there are21
= 16 of these matricespossible. However,
since a,~’
and ol
allgenerate
the same fixedpoint
u, weonly
have to consider fourtypes,
namely
Here the dots indicate that the
corresponding
entries are either 0 or 1. Thefour cases cover
respectively
6,1,8
and 1 of the 16possibilities.
Forexample
the Fibonacci substitution a -ab,
b - abelongs
toType
III since(1 1
~) 3
m0 1 0). 1
3.
Type
I substitutionsHere
M, =
(° °) .
The essential feature of this case is that the number of a’sin both
a(a)
andbeing
even, the orientation at thebeginning
is the same asat the end of these words. Hence if we consider
a(a)
andQ(b)
as newsymbols
we can obtain a non-oriented walk which behaves very much as theoriginal
oriented walk.Formally,
define thehomomorphism
(w.r.t. concatenation)
by
p(a(a) )
= a,cp((~(b)) _
{3.
Then define a substitution Q onI a, 0} *
by
but for a
change
ofalphabet!),
and definef :
--~ Rby
where fa
=la(a)1 I (the
length
of(a)),
and
=Let u be the fixed
point
of & withico
= a. Then the non-oriented walkvisits a
subsequence
of theoriginal
oriented walk(SN(u)),
the time instants notbeing
furtherapart
thanmax(ta, ib).
Hence boundedness of(SN(u))
isequivalent
to boundedness of
(9g~~ j (f) ) .
The latter can beeasily
resolved with Theorem 1.27of
[10].
’
Example.
Let Q be the Prouhet-Thue-Morse substitutionQ(a)
=baab. Then
f (a) =
-2,/(/3)
= 2 and&(a)
= It is easy to see that E{-2, 0, 2},
so theoriginal
oriented walk is also bounded(actually
it is confined to the set{20133, 20132, 20131,0,1,2}).
4. An
equivalence
relationWe call words v, w E A* = of
length n
andlength
mequivalent,
anddenote this
by v - w
ifi.e.,
the associated oriented walks end at the sameinteger
with the sameorienta-tion. In terms of the matrices
Ua
andUb
introduced in Section 1 we have v N wiff
(1 0)
Uv
=(10) Uw (here UW1W2...Wk
=UWl UW2
... ifw E Ak ~ .
Note thatconcatenation preserves
equivalence.
We denote theempty
wordby E.
Typical
examples
areSince the orientation
changes
iff an a occurs we haveLEMMA 1.
If v N
w, thenIvla 1WIa
modulo 2.The
following
lemma isimportant
in theanalysis
ofType
II and IV.LEMMA 2. For alt w E A* there exist
r, i
E{0,1}
and n E N such that w-albnar.
.Proof.
Apply a2 ’"
c until w Nalbnlabn2a...
abnk ar for some k. Thenapply
bab N a
min(nk-l, nk)
times on The result is that w isequivalent
to aword of the form above with k one smaller. The lemma then follows
by
induction.0
Proof.
By
Lemma2,
andby
Lemma1,
r + l is odd. Sow2 ~
al.bnabnar.
But since a, we obtainW2 ’"
0Warning:
note that ingeneral u N v
does notimply
a(u) -
u(v).
5.
Type
II substitutionsHere
Ma =
(o o) .
Note that also%£ =
(o o) .
Hence theSquaring
Lemmaimplies
PROPOSITION 1.
If a
isof
Type
II,
then(SN(U))
is boundediff
E.Proof. Note first that
a2 (ab) N
e iff Since if forexample
E,then
«)
Now =u2(u) = QL (uo )Q~ (ul ) ... ,
where each E.This
obviously implies
that(SN(u))
is bounded.~)
We will show thatimplies
that(SN(u))
is unbounded. Because of(4)
any wordun (w)
isequivalent
to one of thefollowing
words for some k > 0Now we take for w the word
o,(ab).
Since the numbers of a’s and b’s inQ(ab)
are both even,
only
the first twopossibilities
aboveremain,
andmoreover, k
iseven. Let us consider the first
possibility, i.e.,
[an(ab)]’.
Then alsoo-n(ab)
= henceContinuing
in this fashion we obtainSince we assume that e, and since
[o,(ab)]",
we have k >0,
so k > 2. Since
ab,
and hence has to occur(5)
implies
that(SN(u))
isunbounded,
becauseQ2 (ab)
contains an even number of a’s whichimplies
thatthe walk
corresponding
to does notchange
orientation. In casethe same
argument
applies
with a and binterchanged. D
Example.
We consider 2 substitutions with matrix(~
~).
A. Let =aabab ,
a(b)
= bba.Then
u2(ab) rv
~Q(ba)~2 ~
b4,
so the walk is unbounded.B. Let
Q(a)
=abbaa ,
Q(b)
= abb.(In
case B alsoa(ab) -
c. The substitutiongiven by
a(a)
=abb,
a(b)
= aprovides
anexample
whereu2(ab) rv f,
butE).
6.
Type
III substitutionsHere
Ma
=(10) .
Now the orientation has notchanged
after occurrence ofo,(b).
The idea is then tokeep
track of theparity
of the number of a’s that haveoccurred until occurrence of un
in u,
and obtain(SN(u))
as non-oriented walkTo this end we consider a four
symbol alphabet A
=~a+, aw, b+, b-~
with a substitution 6r with fixed
point
û.E.g.
fn
=a+
will mean thatu~ = a
and that an even number of a’s have occurred in uo ... un-,. The substitution
if is defined
by exponentiating
thesymbols
a and b ofa(a)
anda(b),
by
+’sand -’s
according
to the rules:(i)
the firstsymbol
obtains a +,(ii)
if asymbol
follows an a the
exponent
is reversed if it follows a b it remainsequal
to thatof its
predecessor.
Theú(a-)
andú(b-)
are obtainedby reversing
thesigns
in6’(a~),
respectively
ú(b+).
Now wedefine f :
A -~ Rby
Then it
maybe
verified(this
is aspecial
case of the construction in[4])
that forN = -1,0,1,...
- I I - 1-1
where ic is the fixed
point
of & withico
= a+.Example.
Let 0- begiven
by
o(a)
=aabab, a(b)
= ababb. This induces asubsti-tution å on the
alphabet
~a+, a-, b+, b-}
by
By
definitionsign changes
(a+
followedby
a- orb-,
etc.)
occur andonly
occur in thedirectly
following
asymbol
a+ or a-. SinceM, =
(1 °) ,
thisimplies
that in7(~)
there isexactly
one morea+,
say y+ 1,
than a -(note
that
ä(a+)
always
starts witha+),
and inä(b+)
there areequal
numbers of a+and a-, say x.
Moreover,
one obtainsu(a-)
fromj(a+)
by
reversing
signs,
andsimilarly
forâ(b-).
Combining
theseconstraints,
we see that the matrixMü
ofwhere all entries are
non-negative.
It turns out that the crucialparameters
are a and(3
definedby
PROPOSITION 2.
If a
if of
Type
III,
then(SN(u))
is boundediff /3
=0,
orif
thereexist even numbers may and mb such that
Proof. For real numbers
K,
L letThen
Therefore one has for 1
Note that
We shall rather concentrate on the occurrences of b+ in 7~. These will
certainly
take
place,
and thusQ"(b+)
will also occur for each n. But from thebeginning
to the end of such an occurrence the walk will travel
steps
(by (7)).
Hence if1,81
>1,
then(SN(u))
is unbounded.There remain 3
possibilities: ,8
= 0 or,0
= ~1.In
case /3
=0,
f (Q(b+))
=f (~(b-))
= 0. Butalso i(ü(a+a-»
= 0. Sincesymbols
a+ and a- alternate in
f,
thesequence f
has adecomposition
in words from theset V =
{(b+)k, a+(b-)ka- :
k >0}.
Moreover,
we may assume this set to bea’s in u is bounded. Each word v in the set V has = 0. This
clearly
implies
that(SN,f(u)),
and hence(SN(u))
is bounded.Now the case
{3
= ~1. We seethat fl
is aneigenvalue
ofMQ
(with
right
eigen-vector
Passing
from a toa2we
may therefore assume(again
by
idempotency
ofV,)
that{3
= +1. In that caseSo if
a ~
0,
then the walk is unbounded. We are left with the case a =0"Q
= 1.Then we see from
(6) that wl,l is
a lefteigenvector
ofMa.
Because of(8)
this is anecessary and sufficient condition for 6, to admit a
representation
by j
in R.(Cf.
the remarks in the last
paragraphs
of theintroduction).
Thisimplies
that the walk is renormalizable in thefollowing
sense: if the 4steps
corresponding
to thesymbols
a+, a-, b+
and b- arereplaced by
the sequences ofsteps
corresponding
to~(a+), 7(a"), o’(~),
respectively Q(b-),
then this new walk isequal
to theoriginal
walk. From this one deduces that if theSN,;(å(a+)),
1 Nlå(a+)1
crosseslevel
1,
thenSN, f (Q’ (a+) ) ,
1 N ~~z (d+) ~
[
will cross level 2. Moregenerally
(Srr,f(~2’~(a+)))
will cross level n and the walk will be unbounded. So supposeremains between the levels -1 and +1. Then
b2
cannot occur inu, as this would lead to three
+’s
or three -’s in u. Also sincej(û(a+))
=1,
and N,f N-Oequals
N,f JJN=O mirrored around zero, ,a2
cannot occur, unless
(SN,j.((a+)))
stays
between 0 and1,
which wouldonly
bepossible
ifa+
and a- alternate inQ(a+),
what contradicts theprimitivity
ofM&.
We have shown that
a(a)
containsneither a2 nor
b 2,
but thencr(a) =
where m is even because a is of
type
III. Since the samearguments
apply
toand
a (b)
has to appear in someC1n(a),C1(b)
will also neither containa2
norb2.
. Sinceo,(ab)
andQ(ba)
will occur, it follows likewise that the first and the lastletter of
Q(b)
areequal
to b. Hencea(b)
has the claimed form. D.Example.
Let r be the Fibonacci substitution definedby
T(a)
=ab,T(b)
= a.Then
r3 (a)
=abaab,
T3 (b)
=aba,
and a =r 3if
ofType
III. We have6,(a+) =
= a+b-a-. hence a = -2 and/3
=-1,
so(SN(u))
isun-bounded.
(The
substitution agiven
by
a(a)
=abb,
o,(b)
= ababgives
a(nonperi-odic)
example
of a boundedwalk.)
7.
Type
IV SubstitutionsHere
Ma
=’ 1).
Let us write v =C1(a),p,
=a (b).
Then,
because a is ofSo if we consider v
and /-i
assymbols,
we have that the two groupsare
isomorphic.
Here w denotes theequivalence
class of a word w under theequivalence
relation introduced in Section 4.But the matrix
(~ ~)
of a overla, b}
transforms to the matrix(i o)
of b overfit, v},
where the substitution & onv}
is defined as in Section 3. We then usethe
analysis
ofType III,
where at an occurrenceof p respectively v
we move K :=respectively
L :=steps
= =lu(b)l).
BecauseM&
=(~
~),
there are an odd number of v’s inû(JL).
But then theparameter
a = s - t of the associated 3 matrix has to beodd, i.e.,
the renormalizable case a =0, (3
= 1 of theType
IIIanalysis
can not occur for these matrices.Furthermore,
using
the vector wK,L instead of wl,l in(9),
we see from(7)
thatthe walk is bounded iff
,Q
=0,
which occurs iffo~(v+)
has anequal
number of v+and v-. Since
3(v+)
already
has anequal
number ofp+
and it- wefinally
obtainPROPOSITION 3.
If a if of
Type IV,
then(SN(u)) is
boundediff r(b) rv f.,
where T is the substitutions obta2nedfrom a by interchanging
a and b.Example.
Let Q be definedby
7(a) =
aabb,
u(b)
= ab. Then T isgiven by
T(a)
=ba, T(b)
= bbaa. Since-r(b) AE,
agenerates
an unbounded oriented walk.Acknowledgement
We thank the referee for hisremarks,
which have lead toimprovements
of theexposition
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