Complexity of Hartman sequences
par CHRISTIAN STEINEDER et REINHARD WINKLER
RÉSUMÉ. Soit T : x ~ x + g une translation
ergodique
sur ungroupe abélien compact C et soit M une partie de C dont la
frontière est de measure de Haar nulle. La suite binaire infinie
a : Z ~
{0,1}
définie para(k) =
1 siTk(OC)
~ M eta(k) =
0 sinon, est dite de Hartman. NotonsPa(n)
le nombre de motsbinaires de
longueur n
qui apparaissent dans la suite a vue commeun mot bi-infini. Cet article étudie la vitesse de croissance de
Pa (n) .
Celle-ci est toujourssous-exponentielle
et ce résultat estoptimal.
Dans le cas où T est une translationergodique x~x+03B1 (03B1
=(03B11,... 03B1s))
sur Ts et M unparallélotope rectangle
pourlequel la longueur
duj-ème coté 03C1j
n’est pas dans + Z pourtout j = 1, ... , s, on obtient
Pa(n)/ns
= 2s03A0sj=1 03C1s-1j.
ABSTRACT. Let x + g be an
ergodic
translation on the compact group C and M C C a continuity set, i.e. a subset withtopological boundary
of Haar measure 0. An infinitebinary
se-quence
{0,1}
definedby
= 1 if ~ M anda(k) =
0 otherwise, is called a Hartman sequence. This paper studies thegrowth
rate ofPa(n),
wherePa(n)
denotes the num-ber of
binary
words oflength n
~ N occurring in a. Thegrowth
rate is
always subexponential
and this result isoptimal.
If T is anergodic
+ 03B1(03B1 = (03B11, ... , 03B1s))
on Ts and M isa box with side
lengths 03C1j
notequal 03B1j Z
+ Z forall j
= 1, ... , s,we show that limn
Pa(n)/ns =
2s03C1s-1j.
Acknowledgement.
The authors thank the Austrian Science Foundation FWFfor supporting
the researchfor
this paperthrough projects
no. S8312and no. S8302.
1. Motivation and Notation
The notion of a Hartman sequence has
recently
been introduced and studied(cf. ~5~, ~10~, [12]).
It can be seen as ageneralisation
of the notionof a Sturmian sequence. Sturmian sequences
(and
their closerelatives,
theBeatty sequences)
are veryinteresting objects,
as well from the combinato-rial,
the number theoretic and thedynamical point
of view. Let us sketchtwo
approaches.
Consider sequences a =
(ak)
of twosymbols,
say 0 and1,
where k runsthrough
the set Z of allintegers
or N of allpositive integers.
Such sequencesare also called two resp. one sided infinite
binary
words. LetPa(n)
be thenumber of different
binary
words oflength
noccurring
in a. Themapping
n H
Pa(n)
is called thecomplexity
function of a. It iseasily
seen that thecomplexity
function is bounded if andonly
if a is(in
the one sided case:eventually) periodic. Among
allaperiodic
sequences Sturmian sequences have minimalcomplexity, namely P(n)
= n + 1. This is the combinatorialapproach
to characterise Sturmian sequences, which has been introduced in[6]
and[7].
A different characterisation uses the
symbolic coding
of irrational rota- tions. If T =R/Z
denotes the circle group(one
dimensionaltorus)
and Mis a
segment
of T ofangle
27ra with irrational a E then the definitionak = 1 if and
only
if ka E M defines a Sturmian sequence(cf.
for instance[2]).
For
understanding
the definition of a Hartman sequence the secondapproach
is moreappropriate. Replace T by
moregeneral
compact abeliangroups C with normalised Haar measure p _ pc and
replace
aby
anyergodic
group translation. This means that we are interested in the trans-formation
T : C H C, T : m---~ x -~ g, where g
is agenerating
elementof
C,
i.e. the orbit EZ}
is dense in C. Thus C isrequired
tobe monothetic. C can also be
interpreted
as a groupcompactification
ofZ since C is the closure of the
image
of Z under the densehomomorphic embedding t :
Z 2013~C, t(k)
=kg. (Note
that for groupcompactifications
one
usually
does notrequire t
to beinjective.
But to avoid trivial casedistinctions we will demand that is
infinite.)
Thisapproach
isparticu- larly appropriate
for Theorems 1 and 2. For the classicaltheory
ofergodic
group translations we refer for instance to
[11].
A set M C C is called a
(/-Lc- ) continuity
set iffLc( 8M)
= 0 holds forits
topological boundary
8M. For acontinuity
set M consider the inducedbinary
sequence a =(ak)’-.
definedby
ak = 1ifTk(Oc)
E M and ak = 0otherwise. Such sequences are called Hartman sequences. The set H C Z defined
by k
E H if andonly
if ak = 1 isaccordingly
called a Hartman set.Thus, by definition,
a Hartman set H is thepreimage H
=~-1(M)
of acontinuity
set M C C where(C, t)
is a groupcompactification
of Z and wecan write a =
1H
Note that for C = T’and g
=(al,
... ,as),
where thefamily 11, a .. , a, I
islinearly independent
overZ,
Hartman sequencesare
binary coding
sequences of Kronecker sequences.As a consequence of uniform distribution of
ergodic
group translations(for
thetheory
of uniform distribution of sequences we refer to[9]),
Hartmansequences share some nice
properties
with Sturmian sequences(cf. [5], ~10~~.
In
particular
each finite subword occurs with a uniformasymptotic density.
More
precisely:
If the measure of a Hartman set H== ¿ -1 (M)
is definedby p(H)
:=tic(M),
thenholds
uniformly
in 1~ EZ,
cf.[5].
Inparticular,
for agiven
Hartman setH,
this value neitherdepends
on(C, t)
nor on M. In terms of Hartmansequences:
If,
for a =(ak)
=1H, Ak (n)
denotes the number of occur- rences of l ’s in the block ak+n-l oflength
rt, there exists a bound=
o(l),
n ~ oo, such thatfor all k E N. This has been used in
[12]
todevelop
a newapproach
toidentify
theunderlying dynamical system
from itssymbolic coding.
In this paper we start to
investigate
thecomplexity
of Hartman se-quences.
Corresponding
to the zeroentropy
of theunderlying dynamical system,
thegrowth
rate of thecomplexity
function of any Hartman se-quence is
subexponential.
This upper bound turns out to be bestpossible.
In
particular,
thecomplexity
of a Hartman sequencemight
be muchhigher than,
forinstance,
intervalcoding
sequences for which thecomplexity
es-sentially
is linear(cf.
for instance[1]
and[3]).
In some sense this fact is due to the moregeneral
choice of M rather than to the moregeneral
choice ofthe
compact
group C.Nevertheless,
for the case of an M with a verysimple geometric structure, namely
a box in a finite dimensionalcompactification
C = 1r8 the
complexity
growspolynomially
of maximal order s. A moresystematic investigation
of the role of thegeometric properties
of M andthe further
development
of thearguments
used here is to be theobject
offuture research. So the results in this paper are the
following:
~ For any Hartman sequence a and A > 1 we have
Pa(n)
= forn ~ oo
(Theorem 1).
~ For any sequence
P~,
ofsubexponential growth
rate and any compac- tification(C, t)
there is a Hartman sequence acoming
from(C, t)
with
Pn
=o(Pa(n)).
This is even true, ifPa(n)
countsonly binary
words which occur in a with
strictly positive density (Theorem 2).
~ Assume that M is a box of side
lengths pj, j
=1, ... ,
s, in Thenthe
complexity Pa(rz)
of the induced Hartman sequence a has theasymptotic growth
rate cn’ ifaj Z
+ Z forall j
=1, ... ,
s. Themultiplicative
constant c isgiven by
c = 2’pj-l (Theorem 3).
2. A universal upper bound for the
complexity
of Hartmansequences
Corresponding
to the fact thatergodic
group translations haveentropy 0,
oneconjectures
that thecomplexity
of a Hartman sequence issubexpo-
nential. Note that the
topological entropy
of shift spaces andcomplexity
are very
closely
related. Nevertheless we cannotapply immediately perti-
nent theorems in textbooks as
[4]
or[11]
to obtain an upper bound for thecomplexity
of a Hartman sequence in terms of theentropy
of theunderlying
group rotation. Therefore we
give
a directproof
that the aboveconjecture
is true.
Theorem 1. For any
compactification (C, t) of Z
and anycontinuity set
M C C the
complexity Pa(n) of
thecorresponding
Hartman sequence a =1H
with H =~-1 (M) satisfies
Proof. By
Theorem 4 in[12]
we may presume that there is a metric d for thetopology
on C.Let g
= E C denote thegenerating
element of thecompactification.
We write M’ for the
complement C ~
M andM6
for the set of all x E Cwith
d(x, y)
6 for some y E M. Fix F > 0.Using
theregularity
of theHaar measure pc and the
/-tc-continuity
ofM,
we obtainAC(R)
e forR =
(Ms, B M)
U(M~1 ~ M~)
whenever61
> 0 issufficiently
small.By
astandard argument we may assume that R is a
continuity
set. At least oneof the sets M and
M’,
sayM,
hasnonempty
interior. This means that there is some open ball B with center x andpositive
diameter 6bl/2
with B C M. For the sake of
simpler
notation we assume x = 0.Let denote the set of all
binary
words oflength 1
withak = 1 whenever
kg
+ B C M and ak = 0 wheneverkg
-~ B C M’.By
compactness of C there is someNo E
N such thatshowing
that for every y E C there is some n Ef 0, l, ... , No - 1 ~
withy-~ngEB.
Thus any word w of
length No
-~ loccurring
in a lies in some of the setsi
No - 1, consisting
of all wordswith E
Wi
and Sincethis shows
Note that each translate y + B is
totally
contained either in M or in M’whenever y 1:.
R.Thus, by
the uniform distribution of(ng)n
inC,
the subsetT C Z of all k E Z such
that y = kg ¢
R hasdensity R)
> 1 - e.It follows that
22E1,
hencePa(No-~-l)
for lsufficiently large.
Thisyields
and,
for n =Since e > 0 can be chosen
arbitrarily
small this proves the theorem. D 3. A Hartman sequence ofarbitrarily subexponential
complexity
We are
going
to show that the bound deduced in Theorem 1 is bestpossible.
Let
(C, t)
be any groupcompactification
of Zand § :
N )2013~ N.Suppose 0(n)
= e~n n with e~ = 0. We have to show that there existsa
continuity
set M C C such that the Hartman sequence a := with H =~-1(M)
fulfillsPa(n) > 2~~n~.
By
Theorem 4 in[12]
it suffices to prove the assertion for metrisable C andby
Theorem 8.3 in[8]
there is an invariant metric d for thetopology
on C. For c E C we write
I I c I I
=d(c, 0).
For each n E N choose a subsetH (n)
off 0, ... ,
n -11
ofcardinality An >
enn andcontaining
0 such thatthe diameter
dn
of is minimal. We claim thatdn
= 0.Otherwise we had a sequence nl n2 G ... and a 6 > 0 such that
dnk >
2J for all k. There is some r E(0, 6)
such that the open ball B with center 0 E C and radius r is acontinuity
set.By construction,
the lowerdensity
of the set of all n with¿(n)
E B is at most Enk for all k.By
uniformdistribution of
t(n),
the lowerdensity
is adensity
and coincides with the Haar measure, hence~,(B)
Enk = 0. This contradicts the fact thatnonempty
open sets havepositive
measure.Let now
Hn(0), Hn(l~, ... , Hn (2 An - 1)
be an enumeration of all sub- sets ofH(n).
Definerecursively mn(0)
= 0 and +1)
to be theminimal
integer > mn(i)
+ n such that11¿(mn(i
+1))Il dn.
We putHn
= +Obviously Hn
is a finite set ofnonnegative integers
boundedby,
sayhn
E N. Observe furthermore thatby
construction2dn
for all h EHn. Define, again recursively, lo
= 1 and to be the minimalinteger > In + hn
such thatII¿(ln+l)11 dn.
For the union H =in)
thisimplies t(n)
= 0. Thus M =t(H)
isa countable closed subset of C with the
only
accumulationpoint 0,
hencea
continuity
set of measure 0 and H =~-1(M)
is a Hartman set.In the
corresponding
Hartman sequence eachHn
induces at least2 An
different
binary
words oflength
n. Thus thecomplexity
functionP(n)
isbounded below
by
This construction
generates
a zero set M. Hence each word in a con-taining
the letter 1 hasasymptotic density
0. It would be nice toget
apositive frequency
for many words. Let therefore M =fO,
m 2... ~
bean enumeration of M. There are
6n
> 0with 6n
- 0 such that ballsBn,
n E
N,
with center mn andradius 6n
arepairwise disjoint continuity
sets.Replace
Mby
the union of allBn,
which isagain
acontinuity
set. Thisshows:
Theorem 2. Let
(C, c)
be any groupcompactification of Z.
Assume~(n)
n and
0(n) = o(n) for n -
oo. Then there exists acontinuity
set M cC such that its Hartman sequence a :=
fulfills Pa(n)
>2~~n~ .
Furthermore M can be chosen in such a way that
for
each n E l~ at least2’0(n)
wordsof length n
occur in a withstrictly positive density.
4. The case of Kronecker sequences and Boxes
We
finally
restrict our attention to thespecial
case that C = is a fi-nite dimensional
compactification
withgenerating element g
=(aI, ... , as)
modulo
1,
i.e. ~ : ~ ’2013~kg
= where thefamily
is
linearly independent
over Z(such (kg)k
are also called Kronecker se-quences),
and M a box in ~S . To be moreprecise
we use thefollowing
notational convention.
As usual, {r} =r-[r]
and[r]
= 6 Z : ~r}
denote thefractional
respectively integer part
of r E R. Thus Ts = = is considered to be theimage
of the additive group R~ under themapping K =
Ks :
(x 1, ... ,
-Although
has no order structure itis useful to think about intervals in T as
images
of intervals in R underboxes in ~s as
images
of boxes in R~ etc. To avoid too cumbersomenotation we therefore
write,
forinstance,
pj E(0,1)
also for the set M = is natural to call a set M = Tnj +
pj)
9 1rs an s-dimensional box in 1rs with sidelengths pj, j
=1, ... ,
s. We areespecially
interested in Hartman sequences a =1H,
H =
~-1 (M),
for this kind of M and call such a Bohr sequences.Let us fix a box M of side
lengths pj, j
=1, ... ,
s, and assume that nopj is in
ajZ
+ Z. We aregoing
to determine theasymptotic
behaviour ofPb (n)
for the Bohr sequence b =1H, H
=~-1 (M) .
We will use the
following
notation: For a word w = ao ... Ef 0, lln
we introduce the set
and write w =
w(x)
for x EAw.
Notethat, provided 0, Aw
has innerpoints.
Because of thedensity
of the set n EN~,
thecontinuity
of Tand the
special
form of M thisimplies
To
compute
the number of allnonempty A~"
we first consider a half open cubeCo
:= co +[-a/2, Q~2)S
C 1r8 with center co and sidelength
Qpj
for
all j
=1, ... ,
s. We aregoing
to estimate the localcomplexity
functionP(Co, n)
:=IWI
for W =W(Co)
:={w
E10, lln : Aw
f10}.
Notethat for k cubes
Cl, ... , Ck
in 1r8 withdisjoint
closures we havefor
sufficiently large
n. Asabove, Aw
f1implies Co)
> 0.So
P(Co, n)
is the number of different words w= bi
...bi+n-l
oflength
nin b
with ig
ECo.
Defineand
furthermore,
foreach j
=1, ... ,
s,Observe that the sets
(in
contrast to arepairwise disjoint.
For w = ao ... in W note that
and
=~ ~ == 0).
This shows that for w = ao ... an-l and w’ =
aQ
... in W the lettersai and
ai
can differonly
if co-~- ig
e r. Since r =Oi ?
wedefine,
forj
=1,... ,5 and ~ = 0,1,
Due to the
special geometric
situation(Co
and M areboxes),
(xl, ... , xs)
ECo,
w =w(r) =
E Ef 1, ... , s},
thetuple (ai(x)). 1(j) depends only
on Xj,namely
in thefollowing
way. Let2E o
Xj = [yo,
yo +a)
be the interval for thej-th
coordinate ofpoints
inCo.
Then for each i E there is one
point yi (namely
eitherig
or-pj/2 - ig)
suchthat y2 splits
the intervalXj
into two subintervalsxjO)
and
XY)
such thatai(x)
= 0 for Xj ExjO)
andai(x)
= 1 for Xj ESince pj g
+Z,
all Yi, i e7 ,
are distinct. As a consequence themapping rj -
takes at least +l [
differentvalues,
henceE 0
Since the sets
Io ; j
=1
... ; s; arepairwise disjoint
and allcoordinates j
.Since the sets
B
=1...
s, arepairwise disjoint
and allcoordinates j
can be treated
independently,
we concludeFor e > 0 we know
by
uniform distribution of the sequence(ng)n
thatfor n
sufficiently large.
Since~(Q~)
=2 r[~=i 7~(~ " ~’)~’? .7 = ~ " ’ ? ’~
we
get
’
for n
sufhciently large.
Thus we obtainfor all E > 0 and therefore
As a consequence of uniform distribution we know that
x’)
> 6implies w(x) =I- w(~’)
if the words aresufficiently long.
ThusW (Co)
andW (Co)
aredisjoint
whenever two cubesCo
andCo
areseparated by
astrictly positive
distance 6. Fix now k e N and consider the
disjoint
cubesCl, ... , Cks
withcenters ci =
(mi/k),
mi E{0, ... , l~ - 11
and sidelength
Q =1/k - J,
., iJ /
Since this holds for all 5 > 0 we can consider the limit 6 - 0 to
get
For k - oo this
finally
shows the lower boundTo obtain an upper bound for the
complexity
we consider instead ofAj
asdefined above the numbers
Note that the sets
1(j), j
=l, ... , s,
are(in
contrast to the sets notdisjoint.
Thisimplies
thatpossibly depends
on more than onecomponent
of x.Comparison
with theargument
for the lower bound shows that the relevantmapping r -
1 EG’o,
canonly
take one1
additional
value, namely
the zero wordai (x)
= 0 for all i C Thusarguments
similar(in
fact evensimpler)
to those above show thatI Bj
IIij) +
2 andfinally
Since the same
argument applies
if M is not centered at 0 we haveproved:
Theorem 3. Consider an
ergodic
translation T : x H x + g on 7s with g =(al, ... , as).
Assume pj E(0, 1) B
+Z) for all j
=1, ... , s.
For mj E
(0,1), j
=1, ... , s,
let M = +pj)
denote ans-dimensional box
of
sidelengths
pj, and b thecorresponding
Bohr se-quence. Then the
complexity function of
bsatisfies
We add the
following
remarks.Complexity
and volume versus surface: LetV(M)
denote the volumeof a box M in S and
vj (M) =
the(s - 1 )-dimensional
measures(surfaces)
of the facets of M. Then we can express the formula in Theorem 3 in terms ofV(M)
as well as in terms of thevj (M):
Consider first M’ :=
Mo
UMl,
where theMi
aredisjoint
translates of M.The same
arguments
as in theproof
of Theorem 3 show that M’ induces aHartman sequence a’ of
complexity
Comparison
with the value 28 vj(M)
for M indicates that the com-plexity
is related to the surface rather than to the volume.On the other hand we can consider an
automorphism
A of 1r8(i.e.
A E
SL(s, Z) ).
Thetopological generator A(g)
and theparallelepiped
A(M)
induce the same Hartman sequence as g and M. Here the volumeV(M)
=V(A(M))
is invariant while theproduct
of the surface measuresmay
change.
The
interplay
between thegeometry
of moregeneral
sets M and thecomplexity
of thecorresponding
Hartman sequencesmight
be an interest-ing topic
of moresystematic
futureinvestigations.
Dropping
linearindependence:
If theassumption pj ajZ
+ Z failsthen a modification of the
proof
of Theorem 3 based on a careful investi-gation
of theresulting
cancellation effectsyields
acorresponding
formula.Complexity
determines dimension: GivenPb(n)
and the information that M is a box(of
some unknown dimension s and unknown sidelengths
pj),
Theorem 3 tells us how s can be determined.References
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Christian STEINEDER Technische Universitat Wien
Institut fur Diskrete Mathematik und Geometrie Wiedner Hauptstraf3e 8-10
1040 Vienne, Autriche
E-mail : christian.steineder@tuwien.ac.at
Reinhard WINKLER
Technische Universitat Wien
Institut fur Diskrete Mathematik und Geometrie Wiedner Hauptstrafie 8-10
1040 Vienne, Autriche
E-mail : reinhard. winkleretuwien. ac . at