J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
B
RUNO
P
ARVAIX
Substitution invariant sturmian bisequences
Journal de Théorie des Nombres de Bordeaux, tome
11, n
o1 (1999),
p. 201-210
<http://www.numdam.org/item?id=JTNB_1999__11_1_201_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Substitution
invariant Sturmian
bisequences
par BRUNO PARVAIX
RÉSUMÉ. Les suites sturmiennes indexées sur
Z,
de pente 03B1 etd’intercept
p, sont laissées fixes par une substitution non trivialesi et seulement si 03B1 est un nombre de Sturm et p appartient à
Q(03B1).
On remarque aussi que les suites deBeatty
permettent de définir des partitions de l’ensemble des entiers relatifs.ABSTRACT. We prove that a Sturmian
bisequence,
withslope
03B1and intercept p, is fixed
by
some non-trivial substitution if andonly
if 03B1 is a Sturm number and pbelongs
toQ(03B1).
We also detail acomplementary
system ofintegers
connected withBeatty
bisequences.
1. INTRODUCTION
Beatty
sequences( ( ncY
+ and([na
+ have been studiedextensively.
Many
papers deal with the case p =0,
see[1,
9,
10, 14, 15, 28,
29].
Theinhomogeneous
case is also discussed from severalpoints
of view[6,
7, 16, 20, 21,
22~.
By
the way, this Noteprovides
a new contribution aboutcomplementary
systems
ofintegers.
Thisproblem
arose, in variousforms,
in the works of A. S. Fraenkel[13],
R. L. Graham[17]
and R.Tijdeman
[30,
31].
A natural way to examine
Beatty
sequences is to consider the class of Sturmian words definedby
G. A. Hedlund and M. Morse in the contextof
topological dynamics,
see[25,
26].
For furtherdetails,
both[3]
and[8]
contain extensive lists of references. Here we are
especially
interested insubstitution invariant Sturmian words. In
[27]
we elicitedproperties
aboutsome
right-sided
infinite Sturmian words theintercept
of which is aparticu-lar
homography
of theslope.
We therefore obtained apartial
generalization
of
Crisp
et al.’s main Theoremconcerning cutting
sequences~12~.
The aimof this Note is the full characterization of Sturmian
bisequences
which are+
b~ ~
k
E withfl
irrational and 6 real. As usuallxJ
is theinteger
part and theceiling
of any real number x. Let andr’,j
bethe
generating bisequences
of andZ’,6:
we setfor each n E Z. We say that two subsets of Z are a
complementary
system
if
they
form apartition
of Z.Let ,A* be the free monoid
generated by
the two-letteralphabet
,A. =~0,1~.
The set ofright-sided
infinite words is denotedby
AW and is the set of left-sided infinite words. Abisequence
is adoubly
infinite wordand is the set of
bisequences
over A. We say that thebisequences
... ... and ... ... are
equal
if there exists aninteger k
E Z suchthat vi
=v’ i+k
for each i E Z. In this event, we note...
... cti
... ...
... V-2V-lVOVlV2 --- - ---V-2V- 0 1 2
Let a be irrational and p be real. Consider the
bisequences
and z’ alp
defined
by
’
and
for each n E Z. A
bisequence
x is said to be Sturmian if x - zo.,p or x -z’ a,p
for a suitable choice of a and p. It is clear that
za,P(n)
=za+i,p(n)
and = foreach n E 7G,
so without loss ofgenerality,
wemay take 0 cx 1.
Finally,
aright-sided
infinite word y is Sturmian if thereexist a Sturmian
bisequence x
and a left-sided infinite wordy’
such that x -y’y.
Noting
that Sturmian words areintimately
related tostraight
lines in the
plane,
the number a is theslope
and p theintercept.
A substitution
f
is a map from A* into itself such thatf
(uu’)
=f
(u) f (u’)
for all finite words u and u’. Let w =wowlw2... be a
right-sided
infi-nite word. Let Inv be the
operator
definedby
Inv(w) =
... W2Wlwo andInv(Inv(w))
= w. Asusual,
we setf (w)
= andThe
image
of ...un-der
f
is ...f (v_2) f (v_1) f (vo) f (vl) f (v2)
... A one-sided infiniteword y
isfixed
by
f
iff (y)
=y, and a
bisequence x
is fixedby f
iff (x) -
x.Moreover a substitution
f
is Sturmian iff (w)
is aright-sided
infiniteSturmian word whenever w is. F.
Mignosi
and P. Seeboldproved
that asubstitutions in any order
and number
[24].
A substitutionf
islocally
Sturmian if there exists aright-sided
infinite Sturmian word w such thatf (w)
is Sturmian. J. Berstel and P. S66bold stated that anylocally
Sturmian substitution isactually
Sturmian
[4, 5].
Furthermore a substitution is non-trivial if it differs from the identical
transformation over ,A. In
[27]
weproved
that if aright-sided
infiniteStur-mian word is fixed
by
some non-trivial substitution then itsslope
a, with0 a
1,
is a Sturrnnumber,
thatis,
there exists aninteger n >
2 suchthat:
or
where the
partial quotients k2, ... ,
belong
to I~1* . Remark that thesenumbers were
introduced,
in aslightly
different way,by Crisp
et al.[12].
3. RESULTS
As
usual,
for anyquadratic
irrational a, letQ(a) =
I
be the
splitting
field of a overQ.
The main result of this Note is the fullcharacterization of Sturmian
bisequences
which are invariant under somenon-trivial substitution:
Theorem 1. Let x be a Sturmian
bisequence
withslope
0 a 1. Theword x is
fixed by
sorne non-trivial substitutionif
andonly if a is
a Sturmnumber
and p belongs
toQ(a) .
In
~27],
wecomputed
theslope
and theintercept
off (x)
for any Sturmiansubstitution
f
and anyright-sided
infinite Sturmian word x. Lemma 2 is atranslation of these formulas for Sturmian
bisequences:
Lemma 2. Let a be irrational with 0 a 1 and
let p
be real. ThenMoreover
The
proof
of theseproperties requires
a carefulstudy
ofgenerating
bise-quences of
Beatty
bisequences:
.As an immediate
corollary,
we can characterize the occurrences of a letterin any Sturmian
bisequence.
Moreprecisely
we remark thatfor each ~y irrational with 0 1 1 and v real. This result is a
general-ization of earlier work of A. S.
Fraenkel,
M. Mushkin and U. Tassadealing
with the
homogeneous
case[15].
From Lemma 3 we also obtain aproperty
about
complementary
systems
ofintegers:
Proposition
4. Let/3
> 1 be irrational and 6 be real. Then and_a , as well as and
Z 3
-=L+1’ arecomplementary
systems
~_i ~a_i
of
integers.
4. PROOFS
First of
all,
we examine thegenerating bisequences
ofBeatty bisequences:
Proof of
Lemma 3. Let n E Z. If we state thatthus Next comes
I
andConversely,
if = 1 there exists aninteger
k E Z such that[k,3
+b~
= n. We therefore observe thatIt follows that
The truth of
the first statement
is
nowclear,
and we turn to the secondpart
of theproof.
Let n E Z. If z 1-8 (n)
=1 thenhence we have
In order to describe the
complementary
system
ofintegers,
connected witha
Beatty
bisequence,
we need to introduce thefollowing
Lemma:Lemma 5. Let 0 C cx 1 be irrationnal and p be real. Then
. - - , . ,
Proof.
Weonly
detail theproof
concerning
the first result. Let n E Z. Since the relation[a]
= -~-a~
holds for each real number a, weverify
Proof of
Proposition l~.
Let n E Z. From Lemma3,
we remark thatThen Lemma 5
implies
thatIn other
words,
weget
Furthermore,
since{3
> 1 we can affirm that anyinteger
occurs at most onetime in
.~,~,~ .
Clearly
thisproperty
also holds for,~’ ~
5. In
short,
theB
sets and
.’ p a _ 1
are acomplementary
system
ofintegers.
The0-1, 0-1
part
ofproof concerning
.’,
, and Z p - +1 is similar in allrespects.
DFrom now on we
study properties
of substitution invariant Sturmianbise-quences.
Proof
of
Lemma 2. Assume first that 0 p 1. Wesplit
thebisequence
za, p into the wordsLet =
yoyl ... with yj E A
for j
=0,1, ...
We observe thatyo = 0 =
Let nq+1 be the
(q
+l)-th
occurrence of the letter 0 in the wordW(w)
for eachq >
1. Weeasily
check:For each n > 1 we state that:
From Lemma
5,
we prove that yn = 0 if andonly
ishort we obtain cp rIb
compute
W(w’),
we remarkthat
Indeed,
for each n E N* it is clear thathence
If we write w’ = ... ar,.t ... a1aO over we
get
because
cp(0)
= 01 andcp(1)
= 0. We can deduce thatMuch as
above,
weverify
Moreover we observe that =
11-ao
andcp(d)
=oll-a
for each a E{0,1}.
Next comesW (u)
=0§3(u)
for any u EAW,
andconsequently
). Bearing
in mind thatw’w,
andfor
arbitrarily fl
irrational and 5real,
wedirectly
claim:The
computation
of becomes trivial because we havecp(v)
for each v E
Finally,
thepart
ofproof
concerning z"
is similar in allrespects.
0’
For each Sturmian substitution
f
it is therefore clear thatf (x)
is a Sturmianbisequence
whenever x is. Now we turn to theproof
of Theorem 1. Somepreliminaries
arerequired.
Let x and y be two Sturmianbisequences.
Letf
be a substitution such thatf (x) ^_~
y. There exist a word x’ E LV A and aright-sided
infinite Sturmian word x" such that z ri x’x". Since wef (x’ ) f (x" ) ,
the wordf (x")
is aright-sided
infinite Sturmianword. Thus
f
islocally
Sturmian andconsequently f belongs
to the monoidE
Let us recall some basic
properties
about Sturmianbisequences.
For any irrational a we set Z + Za ={a
+ baI
(a, b) E
7~2~ .
Let A be the set ofcouples
(,3, 6)
with 0/3
1 irrational and 6 real. We also set U ={({3,8)
e A Let e A and(cx’~ P’)
Wehave
ZO/,p’ if
andonly
if a = a’ and p -p’
E Z +Za,
see[26].
Asimilar result can be stated from the relation
2~ ,.
Furthermore,
if
za,,p,
thenbelongs
to U and Inshort,
if twoSturmian
bisequences
areequal
thenthey
have the sameslope
in~0,1 ~.
Bearing
these remarks inmind,
we therefore obtain:Lemma 6. Let x be a Sturmian
bisequence
withslope
0 a 1.If
x isinvariant under some non-trivial substitutions then a is a Sturm number.
Proof
(Sketch).
Assume that there exists a non-trivial substitutionf
suchthat
f (x) ^_J
x. Thenf belongs
to Let/3
E]0, 1[
be theslope
off (x)
which is obtainedby
Lemma 2.Clearly
thiscomputation
can be doneregardless
ofintercepts,
and there exists anhomography
h,
withinteger
coefficients,
such that13
=h(a).
Therefore itonly
remains to solve theequation
a =h(a).
In this context, we haveyet
observed that a is a Sturmnumber: for a full characterization of the
homographies
connected withSturmian
substitutions,
see theproof
of Theorem 1 in[27].
DIn order to prove our main
result,
we add here a new necessary conditionof invariance:
Lemma 7. Let x be a Sturmian
bisequence,
withslope
cx andintercept
p.Proof.
Assume,
without loss ofgenerality,
that 0 a 1. Letf
be anon-trivial substitution such that
f (x) -
x. Lemma 6implies
that a is aSturm number. Since a is a
quadratic irrational,
theimage
of a under anyhomography,
withinteger
coefficients,
belongs
to~(a).
Using
Lemma2,
wecompute
theimage
of x underf .
Let{3
be theslope
and 6 be theintercept
we obtain. It is clear that
13
EQ(oa)
and 0fl
1. We also remark that 6 EQ(a)+pQ(a).
Sincef (x) -
x, we must checkfl
= a and6 - p
E Z + Za.Next comes p E
Q(a) .
DCombining
Lemmas 6 and7,
we establish the"only
ifpart"
of Theorem 1.Now we turn to the
proof
of the "ifpart":
the idea is to use someproperties
that we
reported
in[27].
First ofall,
a technical resultconcerning
Sturmiancontinuations is
required
[26].
Definition 8
(cf.
[26]).
Let y be aright-sided
infinite Sturmian word. ASturmian continuation
of y
is a left-sided infinite wordy’
such thaty’y
is aSturmian
bisequence.
Lemma 9
(cf.
[26]).
Let a be irrational with 0 cx 1 and p be real. Eachright-sided infinite
Sturmian word y, withslope
a andintercept
p, admitsat least one and at most two Sturmian continuations. In the case
where y
admits
different
StuTmian continuations there exist twointegers kl
E Z andk2
E N* such that p =kl
+k2a.
Definition 10
(cf.
(27~).
For each m >1,
we setA
right-sided
infinite Sturmian word y is said to bepermitted
if there existan irrational a with 0 a
1,
aninteger
m > 1 and acouple
ofintegers
(a, b)
EC’(m)
such that y = ory =
m m a
Proposition
11(cf.
[27]).
Let cx be a Sturm number. Eachpermitted
wordy, with
slope
ce, is invariant under some non-trivial substitution.Proof
of
Theorem 1. Let a be a Sturm number and p EQ(()).
Let x be aSturmian
bisequence
such that x -Clearly
there exists(a,
b,
n)
EZ3
with n > - 1 such that p =
Moreover,
since za,j+l - za.an ",
for each
real 6,
weactually
(inod n)--(b (mod n))a . Asusual,
the~ n
residue i
(mod n)
is theinteger j,
with 0j
n, such that there existsan
integer
k E Zsatisfying j
= i + kn. For each real 6 we setis a
right-sided
infinitepermitted
word. FromProposition
11,
it followsthat there exists a non-trivial Sturmian substitution
f
such thatf (y) _
y.Noting
thatwe have
Hence the word y admits.
as Sturmian continuations. If the
relation
is not valid then Lemma 9
implies
that there exists(kl, k2) E
7Gz
withk2
> 1 such thatIn this
event,
since a is irrational we observe thatk2 = 0,
which leads to acontradiction. We therefore obtain
f (x) ^-,
x.If n-~-1 a
(mod n)+b (mod n)
we state that(a (mod n), (b (mod n))-n)
belongs
toC’ (n) .
(mod n)-t-((b(mod
weeasily verify
thatthere exists a non-trivial
substitution g
such that x.There are no other
possibilities
and the truth of the claim is now clear forthe word za,p. The
proof
concerning z"
is similar in allrespects.
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Bruno PARVAIX
LACO, Faculte des Sciences,
123 avenue Albert Thomas
F-87060 LIMOGES