G Teyssedre

C Laurent

L Boudou

S Le Roy

High Electric Field and Energetic Processes in Poly(ethylene naphthalate)

Poly(ethylene naphthalate) is a semi-crystalline polymeric material having high field withstanding and strong luminescence properties owing to its aromatic structure. Because of these two features it is the insulating polymer having strongest electroluminescence -EL yield investigated so far. EL is detected in an electric field regime where strong non-linear conduction properties have been identified. The light excitation mechanism, though not completely elucidated could be charge recombination with a contribution from impact excitation. It is believed that copious space charges effects are at play in the field regime when EL is detected. The EL spectrum shape evolves with field and some of the bands were ascribed to electrical ageing of the material. All of these features make of PEN a fascinating model material for investigating fundamental processes at play under high fields in insulating polymers. Fig. 1. Molecular structure of the (a) poly(ethylene terephthalate) -PET and (b) poly(ethylene naphthalate) -PEN [5].

I. INTRODUCTION

Poly(ethylene 2,6-naphthalate) -PEN is an unsaturated polyester structurally similar to the most commercially important polyester PET -Poly(ethylene terephthalate) [START_REF] Tonelli | PET versus PEN: what difference can a ring make[END_REF]. PEN is competitive with PET in certain performance-driven markets based on its superior strength, heat stability and barrier properties to gases [START_REF] Morse | PEN: The new polyester[END_REF].

Whereas PET is widely and even massively produced for many applications ranging from textile, to packaging and to capacitors, PEN is relatively newer, more expensive such that its superior properties are not necessarily exploited. Replacing a phenyl ring by a naphtyl ring, cf. Fig. 1, has some consequences on the mechanical properties, making the polymer chain stiffer [START_REF] Tonelli | PET versus PEN: what difference can a ring make[END_REF] and leading in general to a shift to higher temperature of characteristic transitions. It also has a great impact on the photo-physical properties of the material, PEN being a very efficiently luminescent material. Many recent studies focus on the application of PEN to scintillators for radiation detection [START_REF] Nakamura | Evidence of deep-blue photon emission at high efficiency by common plastic[END_REF][START_REF] Garankin | Identification of particles of ionizing radiation by the analysis of fluorescence pulse form of the thin PEN film scintillator[END_REF]. Detailed investigations of the optical properties of PET and PEN were reported by spectroscopic ellipsometry, in terms of their optical, electronic, and vibrational response and with consideration of anisotropic response in bioriented films [START_REF] Laskarakis | Study of the electronic and vibrational properties of poly(ethylene terephthalate) and poly(ethylene naphthalate) films[END_REF].

From the dielectric properties point of view, Ohki et al. [START_REF] Yang | Dielectric properties of poly(ethylene terephthalate) and poly(ethylene 2,6-naphthalate)[END_REF] made recently a comparative study on the (low field) dielectric properties of PET and PEN: due to similar chemical structures, PET and PEN have many parallel dielectric properties mainly controlled by an increase by about 50°C (from ≈70 to ≈120°C) [START_REF] Cañadas | Relaxational study of poly(ethylene-2,6naphthalene dicarboxylate) by t.s.d.c., d.e.a. and d.m.a[END_REF] of the glass transition temperature. Both are polar materials, producing relatively high dielectric losses in a given frequency range [START_REF] Yang | Dielectric properties of poly(ethylene terephthalate) and poly(ethylene 2,6-naphthalate)[END_REF][START_REF] Cañadas | Relaxational study of poly(ethylene-2,6naphthalene dicarboxylate) by t.s.d.c., d.e.a. and d.m.a[END_REF].

We have investigated the high field behavior of PEN over the last 20 years, considering especially its electroluminescence properties [START_REF] Mary | Understanding optical emissions from electrically stressed insulating polymers: electroluminescence in poly(ethylene terephthalate) and poly(ethylene 2, 6naphthalate) films[END_REF][START_REF] Teyssedre | Optical emission due to space charge recombination in insulating polymers: An insight into electrical ageing[END_REF][START_REF] Teyssedre | Electroluminescence and photoluminescence of UV-aged poly(ethylene naphthalate) films[END_REF][START_REF] Teyssedre | Evidence of hot electron-induced chemical degradation in electroluminescence spectra of insulating polymers[END_REF]. Indeed, owing to its high luminescence yield, strong non-linear behavior at high field and high field withstanding, it has been considered as a model material for electroluminescence analysis. Our purpose in this communication is to review the behavior of PEN under high electric field in relation to its photo-physical properties.

II. STRUCTURE AND OPTICAL PROPERTIES OF PEN

A. Microstructure and dynamics

PEN is a semicrystalline material that can be easily obtained in a broad range of crystalline degree, ranging from purely amorphous to ≈ 60% crystallinity by thermal annealing. The glass transition temperature Tg is at about 125°C, and the melting point at about 270°C. Upon heating of amorphous material, cold crystallization occurs in the range 160-180°C [START_REF] Cañadas | Comparative study of amorphous and partially crystalline poly(ethylene-2,6-naphthalene dicarboxylate) by TSDC, DEA, DMA and DSC[END_REF].

For PEN, four relaxation peaks (labelled β, β * , α and ρ) have been reported by using different experimental techniques (Dielectric Electrical Analysis, Dynamic Mechanical Analysis or Thermally Stimulated Depolarization Current). Fig. 2 shows an example of relaxation map obtained by impedance spectroscopy [START_REF] Hardy | Dielectric relaxation behaviour of poly(ethylene naphthalene 2,6 dicarboxylate) (PEN)[END_REF].

The β relaxation, which is observed at the lower temperature, is assigned to the local motion of ester groups (O-C=O) of the polymer chain [START_REF] Hardy | Dielectric relaxation behaviour of poly(ethylene naphthalene 2,6 dicarboxylate) (PEN)[END_REF][START_REF] Schartel | Dielectric investigations on secondary relaxation of polyarylates: comparison of low molecular models and polymeric compounds[END_REF][START_REF] Coburn | Dielectric relaxation in poly(ethylene terephthalate)[END_REF]. For Hakme et al. [START_REF] Hakme | Uniaxially stretched poly(ethylene naphthalene 2,6dicarboxylate) films studied by broadband dielectric spectroscopy[END_REF], the asymmetric shape of the β relaxation is due to the overlapping of two symmetric relaxations β1 and β2, which amplitudes strongly depend on the humidity content, the drawing ratio and the stretched temperature. The β * relaxation, observed at a temperature between the β-peak temperature and Tg, is related to the relative motion of the two parts of the naphtyl ring or is thought to be due to the presence of COOgroups adjacent to the naphtyl ring in different configurations [START_REF] Bravard | Dielectric relaxation in amorphous poly(ethylene terephthalate) and poly(ethylene 2,6-naphthalene dicarboxylate) and their copolymers[END_REF][START_REF] Ezquerra | Dielectric relaxation of amorphous random copolymers of poly(ethylene terephthalate) and poly(ethylene-2, 6-naphthalene dicarboxylate)[END_REF]. The α relaxation is clearly related to the glass transition of PEN, involving long-range cooperative segments motion. Finally the interpretation of the ρ peak, observed at a temperature higher than Tg, is more difficult. For Martinez-Vega et al. [START_REF] Martinez-Vega | Molecular mobility in amorphous and partially crystalline PEN after isothermal crystallization[END_REF] the ρ-peak is linked to the cold crystallization process. For Canãdas et al. the ρ-peak is directly linked to the rearrangement of the amorphous-crystalline interphases, which can act as traps for charges [START_REF] Cañadas | Comparative study of amorphous and partially crystalline poly(ethylene-2,6-naphthalene dicarboxylate) by TSDC, DEA, DMA and DSC[END_REF]. Hoang et al. [START_REF] Hoang | Dissociating space charge processes from orientation polarization in Poly(ethylene naphthalate) films[END_REF], by coupling TSDC and space charge measurements, concluded that the ρ-peak is predominantly associated with the release of the negative charges trapped in the material, associated with a structural change due to an annealing phenomenon.

B. Optical and electronic properties

The existence of a naphtyl group in PEN, instead of the phenyl ring in PET, leads to a significant shift of the characteristic absorption bands to lower energies, as well as a characteristic split in all of them [START_REF] Laskarakis | Study of the electronic and vibrational properties of poly(ethylene terephthalate) and poly(ethylene naphthalate) films[END_REF]. The main absorption peaks of PEN in the UV-vis region are found at ≈3.4 eV (365 nm), 3.6 eV (345 nm), 4.3eV (290 nm) and 5.1 eV (240 nm). The 2 lowest energy peaks are attributed to the electronic transition of the non-bonded electron of the carbonyl O atom from the n state to the π* unoccupied valence state orbital (n→ π* transition). The other two peaks are splitted in three components each and correspond to the electronic transition due to the π → π* excitation of the π-electron structures. Because of these features, the PEN films are strongly absorbing: for a 25 µm thick film, all the light is absorbed under 380 nm, see Fig. 3 [START_REF] Teyssedre | Electroluminescence and photoluminescence of UV-aged poly(ethylene naphthalate) films[END_REF].

The photoluminescence spectrum of PEN has been reported in different works [START_REF] Mary | Understanding optical emissions from electrically stressed insulating polymers: electroluminescence in poly(ethylene terephthalate) and poly(ethylene 2, 6naphthalate) films[END_REF][START_REF] Teyssedre | Optical emission due to space charge recombination in insulating polymers: An insight into electrical ageing[END_REF][START_REF] Teyssedre | Electroluminescence and photoluminescence of UV-aged poly(ethylene naphthalate) films[END_REF]. At room temperature, the fluorescence emission shows a peak at 430 nm, irrespective of the excitation wavelength: the spectrum remains strictly unchanged when exciting in the range 220-360 nm [START_REF] Teyssedre | Electroluminescence and photoluminescence of UV-aged poly(ethylene naphthalate) films[END_REF]. Fig. 3 shows an example of spectrum obtained at low temperature (about -110°C) [START_REF] Teyssedre | Optical emission due to space charge recombination in insulating polymers: An insight into electrical ageing[END_REF]. The fluorescence exhibits a structured emission with bands at 390, 412 and 430 nm. It has been associated to a 1 (π, π*) fluorescence transition of an excimer state formed between two adjacent naphthalene-dicarboxylate units [START_REF] Phillips | Luminescence from aromatic polymers, monomers, and dimers under high energy electron excitation[END_REF][START_REF] Allen | Photochemical reactions in commercial poly(ethylene 2,6-naphthalate)[END_REF][START_REF] Hashimoto | Fluorescence study of thermotropic liquid-crystalline polyesters[END_REF]. A weak phosphorescence emission, not detected at room temperature, is visible in Fig. 3 with peaks at 578 and 638 nm. The emission has been assigned to a 3 (π, π*) phosphorescence transition originating from the monomeric units [START_REF] Allen | Photochemical reactions in commercial poly(ethylene 2,6-naphthalate)[END_REF][START_REF] Cheug | Synthesis, photodegradation, and energy transfer in a series of poly(ethylene terephthalate-co-2,6-naphthalenedicarboxylate) copolymers[END_REF]. The low yield of PEN phosphorescence is due to a low probability of intersystem crossing between 1 (π, π*) and 3 (π, π*) states as detailed in [START_REF] Cheug | Synthesis, photodegradation, and energy transfer in a series of poly(ethylene terephthalate-co-2,6-naphthalenedicarboxylate) copolymers[END_REF].

The various transitions and energy band diagram for PEN are represented in Fig. 4 [START_REF] Mary | Understanding optical emissions from electrically stressed insulating polymers: electroluminescence in poly(ethylene terephthalate) and poly(ethylene 2, 6naphthalate) films[END_REF][START_REF] Teyssedre | Optical emission due to space charge recombination in insulating polymers: An insight into electrical ageing[END_REF]. The relative positions of the energy levels are from quantities given above, considering particularly data from Cheung et al. [START_REF] Cheug | Synthesis, photodegradation, and energy transfer in a series of poly(ethylene terephthalate-co-2,6-naphthalenedicarboxylate) copolymers[END_REF].

S 3 (π, π * ) (π, π * ) (π, π * ) S 2 S 1 (n, π * ) (n, π * ) (π, π * )

II. ELECTRICAL PROPERTIES

A. Current-voltage characteristics

PEN is a polar material, as manifested by the large amplitude of relaxation peaks shown in Fig. 2. For this reason, reaching conduction current is not a straightforward task as the transient current is dominated by polarization contributions (see below). Long charging times are needed to avoid In red: Photoluminescence spectrum of PEN measured at -110°C. The phosphorescence region is magnified for clarity. From [START_REF] Teyssedre | Optical emission due to space charge recombination in insulating polymers: An insight into electrical ageing[END_REF].

orientation current, but charge build up can be at play. Indeed, in the high field range, 300 kV/mm or so, evidence of massive space charge build was reported with a continuous decrease of the charging current over days [START_REF] Teyssedre | Dependence of electroluminescence intensity and spectral distribution on ageing time in polyethylene naphthalate as modelled by space charge -modified internal field[END_REF]. Chavez et al. investigated the field dependence of conductivity for PEN with various degrees of crystallinity as obtained with 'intermediate' charging times of 2 h at each voltage level, cf. Fig. 5 [START_REF] Chavez | Influence of morphology on electric conductance in poly(ethylene naphthalene-2,6-dicarboxylate)[END_REF]. A ohmic regime is evidenced up to about 80 kV/mm. As shown below, a second change in conduction regime occurs at about 140 kV/mm [START_REF] Augé | Combined electroluminescence and charge profile measurements in Poly(ethylene-2,6-naphthalate) under a DC field[END_REF]. Surprisingly the crystallinity does not influence much the resistivity with a variation by a factor of less than 2 between the two extreme values in the ohmic region (2.9 vs. 5.7 10 16 Ω.m).

B. Polarization and space charges

Space charges were measured on PEN in very different conditions and using different methods. Because PEN is polar, and owing to the slow release of dipoles in the glassy state, the space charge measured in volt-off contains a response due to orientation polarization [START_REF] Hoang | Dissociating space charge processes from orientation polarization in Poly(ethylene naphthalate) films[END_REF][START_REF] Augé | Combined electroluminescence and charge profile measurements in Poly(ethylene-2,6-naphthalate) under a DC field[END_REF]. By laser-induced pressure pulse -LIPP measurements on relatively thin samples (50 µm), homocharges were detected near the both electrodes, in similar amount, for fields in the range 100-225 kV/mm. By lightinduced modulation method -LIMM similar space charge profiles with homocharges at both electrodes were detected in 25µm thick films under 200 kV/mm, cf. Fig. 6 [START_REF] Anh | Effect of trapped space charge on mechanical deformation induced by electric field[END_REF][START_REF] Teyssedre | Some aspects of coupled electrical-mechanical effects in dielectric materials[END_REF]. Fukuma et al. [START_REF] Fukuma | Millisecondtime range analysis of space charge distribution and electroluminescence in insulating polymers under transient electric stress[END_REF] investigated transient charge processes by pulsed electro-acoustic -PEA method in 75 µm thick PEN films under ramped/square signal with various rising rate, at fields up to 200 kV/mm. Here negative charges were dominant, in high density -of the order of 50 C/m 3 , and occupied the all thickness, cf. Fig. 7. Most of the charges appeared to be released quickly but transient processes in EL and charge measurements during voltage rising revealed the presence of stored positive charges near the electrodes. All the above experiments were carried out at room temperature. More recently Hoang et al. [START_REF] Hoang | Dissociating space charge processes from orientation polarization in Poly(ethylene naphthalate) films[END_REF] carried out space charge measurements by PEA after previously charging the film (188 µm thick) at 170°C under a field of 10 kV/mm. After cooling, the space charge is frozen. The charge profile appears dominated by negative charges occupying the volume. For a polarization temperature of 130°C no charges were detected.

III. ELECTROLUMINESCENCE PROPERTIES

A. Electroluminescence under DC stress

The EL-field and current-field characteristics can be recorded simultaneously under DC stress using a home-made set-up [START_REF] Mary | Understanding optical emissions from electrically stressed insulating polymers: electroluminescence in poly(ethylene terephthalate) and poly(ethylene 2, 6naphthalate) films[END_REF]. Example of such results is given in Fig. 8. Here a step-up voltage was applied, keeping relatively short time (5min) at each step in an objective to limit charge build up into the material. Once the maximum field of 300 kV/mm was reached the field was stepped down with 2.5 min duration at each step. During stepping up the stress, a clear threshold for EL detection is found at about 140 kV/mm. The current rises sharply beyond the threshold for EL detection. Beyond the field threshold, the EL appears proportional to the applied field and it is such over 5 decades in signal [START_REF] Teyssedre | Optical emission due to space charge recombination in insulating polymers: An insight into electrical ageing[END_REF]. When stepping down the voltage, the characteristic is clearly not symmetrical: even though EL is still proportional to current for fields beyond 180 kV/mm, the drop is faster than the rise. More importantly, a change of sign of the current is observed when the applied field is less than about 140 kV/mm. Still with Fig. 7. Space charge profiles obtained on 75 µm thick PEN film by PEA measurements under fast varying stress. Profiles correspond to different times during the 5 ms rising time from 0 to 200 kV/mm followed by a plateau. The period of signal is 50 ms. From [START_REF] Fukuma | Millisecondtime range analysis of space charge distribution and electroluminescence in insulating polymers under transient electric stress[END_REF].

decreasing the applied field, the EL tends to rise and reaches a steady value well above the noise level even at zero volts. These features are the sign that there is a slow release of previously generated charges, in large amount for inducing substantial field redistribution. The EL can be due to bipolar charge recombination if one resorts to some of the space charge patterns featuring homocharge build-up at each electrode, or to impact excitation if the field at the electrode is the driving process for current and EL. The monitoring of EL, current and EL spectrum for several days under high field indicated that strong field distribution occurs under DC stress [START_REF] Teyssedre | Dependence of electroluminescence intensity and spectral distribution on ageing time in polyethylene naphthalate as modelled by space charge -modified internal field[END_REF]. The analysis of light decay following DC charging of the material indicated that charge recombination seems to be a major mechanism in EL excitation [START_REF] Teyssedre | Analysis of the luminescence decay following excitation of polyethylene naphthalate films by an electric field[END_REF]. The memory effect of previous DC stressing on field-strain curves [START_REF] Anh | Effect of trapped space charge on mechanical deformation induced by electric field[END_REF][START_REF] Teyssedre | Some aspects of coupled electrical-mechanical effects in dielectric materials[END_REF] is a further mark of charge build up at high field in PEN.

B. Electroluminescence under AC stress

In view of the large space charge effects likely to occur in PEN at high fields, it is interesting to consider the case of AC stress. The effect of frequency on the EL phase patterns has been investigated along with the EL-field characteristics. A feature of EL under ac stress, not specific to PEN, is the occurrence of surface plasmons emission when using gold electrodes [START_REF] Cisse | Influence of voltage frequency, the electrode material and a superimposed dc bias on ac electroluminescence of polymer films[END_REF]. Recent characterization achieved using ITO electrodes allows getting rid of these effects [START_REF] Qiao | AC electroluminescence spectra of polyethylene naphthalate: impact of the nature of electrodes[END_REF]. Typically EL-field characteristics obtained on PEN are shown in Fig. 9. Compared to DC stress, the EL threshold is found at at a much lower field (about 30 kV/mm), and the increase of the EL with field is much milder. The EL is in advance in respect to the peak voltage; the advance of phase tends to decrease with increasing the field (from 18° to 9° when going from 160 to 270 kV/mm). The EL under AC is controlled by the recombination of injected carriers and trapped carriers with opposite charges. Injected charges either are trapped or recombine with trapped carriers of opposite polarity from last half field cycle, generating advance of EL peak. Compared to DC, AC-EL is presumably localized next to the electrode upon injection/extraction of charges. It is more sensitive to electrode nature and interface properties than DC-EL.

C. Electroluminescence spectra

We finally consider the spectral distribution of EL in PEN.

A typical EL spectrum obtained with 25 µm thick film provided with semitransparent silver electrodes is shown in Fig. 10 for different DC fields. The spectrum has been interpreted as constituted by 3 main contributions [START_REF] Teyssedre | Electroluminescence and photoluminescence of UV-aged poly(ethylene naphthalate) films[END_REF][START_REF] Teyssedre | Dependence of electroluminescence intensity and spectral distribution on ageing time in polyethylene naphthalate as modelled by space charge -modified internal field[END_REF]: -Bands at 578 and 638 nm: They correspond to the phosphorescence of the material, cf. Fig. 3. The EL spectrum obtained at low temperature is clearly dominated by these two bands. The 638 nm band is probably overlapped by other contributions in EL spectra recorded at room temperature.

-Bands at 618 and 675 nm (620-680 nm in Fig. 9): Both strengthen, with respect to the phosphorescence of PEN, as the field increases. We detected an identical evolution of these two bands as a function of field, which means that they are probably paired. Note the shift by about 40 nm with respect to phosphorescence of PEN. They do not appear in the EL spectrum at low temperature. These bands most probably originate from the degradation: chemical groups derived from the polymer's structure would be formed. -Band at 500 nm: It has been initially attributed to a tail of the fluorescence spectrum [START_REF] Mary | Understanding optical emissions from electrically stressed insulating polymers: electroluminescence in poly(ethylene terephthalate) and poly(ethylene 2, 6naphthalate) films[END_REF]. Its relative lowering for increasing field would be explained by quenching of excited states by the field: the excitons formed in the singlet state would be dissociated rather than relax. A problem is that its spectrum does not really match with the fluorescence of PEN. So an alternative would be that degradation products at the surface of PEN, as e.g. oxidized groups would be responsible for the emission. It would explain why this band tends to be larger under ac stress when the EL is more related to interface effects.

PL measurements on UV-aged PEN films showed the appearance of a new band in the region of 500 nm which corresponds to the EL spectrum at low field and so there is a possible common origin [START_REF] Teyssedre | Electroluminescence and photoluminescence of UV-aged poly(ethylene naphthalate) films[END_REF]. Its relative decrease as the field increases remains to be explained. Work is in progress for clarifying this behavior.

The evolution of the EL spectrum as a function of stressing time under 340 kV/mm has been exploited to infer the maximum field in the material [START_REF] Teyssedre | Dependence of electroluminescence intensity and spectral distribution on ageing time in polyethylene naphthalate as modelled by space charge -modified internal field[END_REF]: the spectrum evolved as if the field was increasing. With considering the total EL variation, current as a function of time and EL spectrum shape evolution, we showed that the maximum field could exceed 400 kV/mm and that the charge density could be of the order of 300 C/m 3 over 70% of the thickness after stressing continuously for several days. Such amounts are perfectly compatible with order of magnitudes of charge density given above.

The relation between EL spectrum shape and field magnitude is relatively well understood. However, the origin of the emission, particularly the longer wavelength doublet is not yet identified. The best reproduction of the EL spectrum has been obtained using electron beam irradiation, so-called cathodoluminescence -CL spectrum, cf. Fig. 11 [START_REF] Teyssedre | Evidence of hot electron-induced chemical degradation in electroluminescence spectra of insulating polymers[END_REF]. Intense luminescence is obtained comparatively to EL; however the spectrum evolves quickly due presumably to degradation under impinging electrons. The evolution in time of the spectrum is similar to that obtained under applied field. The main difference is that fluorescence is clearly detected at the beginning of the irradiation. The consistency of EL and CL spectra points to EL excitation by the impact of hot electrons.

IV. CONCLUSION

We have reviewed works carried out by our group over the last 20 years about the behavior of Poly(ethylene naphthalate), particularly as regards the high field behavior. PEN is a good model material for its high field withstanding, for the coexistence of polarization and transport properties and for its relatively intense luminescence activity. We have given some evidence that at high fields space charges are settled into the material and impart to the material memory effects of electrical pre-stressing, through electroluminescence on one hand and through it electro-mechanical behavior on the other hand. A more complete characterization from the space charge standpoint at high field would be useful to discriminate between bipolar or unipolar space charge build-up.

Regarding ageing effects, we have touched the possibility that part of the signatures found in EL is related to ongoing chemical modification of the materials during excitation by an electric field. Identification of the nature of the emitting species would constitute a direct proof of electrical ageing process with a fingerprint of processes and by-products. at room temperature. From [START_REF] Teyssedre | Evidence of hot electron-induced chemical degradation in electroluminescence spectra of insulating polymers[END_REF].

Fig. 4 .

 4 (a) The electron transitions for dimethyl 2,6-naphthalene dicarboxylate in ethanol solution (taken from[START_REF] Cheug | Synthesis, photodegradation, and energy transfer in a series of poly(ethylene terephthalate-co-2,6-naphthalenedicarboxylate) copolymers[END_REF]). (b): Tentative energy band diagram for dimethyl 2, 6-naphthalene dicarboxylate. The order of magnitude of the energy gap has been inferred on the basis of the transition σ─>σ* of C atoms.

Fig. 2 .Fig. 3 .

 23 Fig. 2. Dielectric loss factor map of biaxially oriented PEN heat-treated at 250°C showing the 3 relaxation processes: β, β* ad α. From [13].

Fig. 5 .Fig. 6 .

 56 Fig. 5. Current-voltage characteristics for 70µm thich PEN films in amorphous state (diamond) and semi-crystalline state(40% crystallinity) (square, triangle) [26].

Fig. 8 .Fig. 9 .Fig. 10 .

 8910 Fig. 8. EL and current vs. field characteristics obtained for a 25 µm thick PEN at room temperature. A step-wise increase/decrease of the field was applied, with steps of 10kV/mm. The steps duration were 300 s in step-up and 150 s in step down. Reported current and EL data are averaged over the last 200 s in step-up and 100 s in step-down. The current is obtained for an electrode area of 20 cm². A change of sign of the current occurred during step-down as indicated by vertical line.

Fig. 11 .

 11 Fig. 11. Cathodoluminescence spectra (open symbols) under 10 keV electron beam for 3 different irradiation times, EL spectrum (solid circles) under 250 kV/mm DC stress and photoluminescence spectrum (solid line)at room temperature. From[START_REF] Teyssedre | Evidence of hot electron-induced chemical degradation in electroluminescence spectra of insulating polymers[END_REF].


F Durand

A Messaoudi

BOUNDARY OF THE RAUZY FRACTAL SETS IN R × C GENERATED BY P (x) = x 4 -x 3 -x 2 -x -1

We study the boundary of the 3-dimensional Rauzy fractal E ⊂ R × C generated by the polynomial P (x) = x 4 -x 3 -x 2 -x -1. The finite automaton characterizing the boundary of E is given explicitly. As a consequence we prove that the set E has 18 neighborhoods where 6 of them intersect the central tile E in a point. Our construction shows that the boundary is generated by an iterated function system starting with 2 compact sets.

Introduction

Consider A = {1, 2, 3} as an alphabet. Let A * be the set of finite words on A and σ : A → A * be the map (called Tribonacci substitution) defined by σ(1) = 12, σ(2) = 13, σ(3) = 1.

We extend σ to A N by concatenation : σ(a 0 • • • a n . . .) = σ(a 0 ) • • • σ(a n ) . . .. It is clear that σ has a unique fixed point u : σ(u) = u ∈ A N . The dynamical system associated to σ is the couple (Ω, S) where S : A N → A N is the shift map (S((x n ) n∈N ) = (x n+1 ) n∈N ) and Ω is the S-orbit closure of u : Ω = {S n u|n ∈ N}. It is well known that (Ω, S) is minimal, uniquely ergodic and of zero entropy (see [Q87, F02] for more details). In 1982, G. Rauzy [R82] studied the Tribonacci substitution σ. He proved that the dynamical system generated by σ is measure theoretically conjugate to an exchange of domains X 1 , X 2 , X 3 in a compact tile X = X 1 ∪ X 2 ∪ X 3 . The set X is the classical two-dimensional Rauzy fractal. It has been extensively studied and is related to many topics : numeration systems [M00, M06, M05], geometrical representation of symbolic dynamical systems [START_REF] Arnoux | Pisot substitutions and Rauzy fractals[END_REF][START_REF] Arnoux | Higher dimensional extensions of substitutions and their dual maps[END_REF][START_REF] Canterini | Geometric representation of substitutions of Pisot type[END_REF][START_REF] Holton | Geometric realizations of susbstitutions[END_REF][START_REF] Messaoudi | Propriétés arithmétiques et dynamiques du fractal de Rauzy[END_REF][START_REF] Thuswaldner | Unimodular Pisot substitutions and their associated tiles[END_REF][START_REF] Sirvent | Relationships between the dynamical systems associated to the Rauzy substitutions[END_REF], multidimensional continued fractions and simultaneous approximations [START_REF] Arnoux | Discrete planes, Z 2 -actions, Jacobi-Perron algorithm and substitutions[END_REF][START_REF] Chekhova | Propriétés combinatoires, ergodiques et arithmétiques de la substitution de Tribonacci[END_REF][START_REF] Chevallier | Best simultaneous diophantine approximations of some cubic algebraic number[END_REF][START_REF] Hubert | Best simultaneous diophantine approximations of Pisot numbers and Rauzy fractals[END_REF], self-similar tilings [A99, A00, AI01, P99] and Markov partitions of Hyperbolic automorphisms of the torus [START_REF] Kenyon | Arithmetic Construction of sofic partitions of hyperbolic toral automorphisms[END_REF][START_REF] Messaoudi | Propriétés arithmétiques et dynamiques du fractal de Rauzy[END_REF][START_REF] Praggastis | Numeration systems and Markov partitions from self-similar tilings[END_REF]. Among the main properties of the set X, let us recall it is compact, connected, its interior is simply connected, its boundary is fractal and it induces a periodic tiling of R 2 ( [R82]). It is possible to associate such a fractal set to a large class of substitutions over an alphabet with d letters (called unimodular Pisot substitutions). Let us call them Rauzy fractals. P. Arnoux and S. Ito [START_REF] Arnoux | Pisot substitutions and Rauzy fractals[END_REF] (see also [START_REF] Canterini | Geometric representation of substitutions of Pisot type[END_REF]) proved that the dynamical system associated to such a substitution σ is measure theoretically conjugate to an exchange of domains X 1 , . . . , X d in the Rauzy fractal X σ = X 1 ∪ . . . ∪ X d ⊂ R d-1 provided that a certain combinatorial condition is true. All these sets X σ are compact and generate a periodic tiling of R d-1 .

There are different ways to define the Rauzy fractal associated to a given substitution σ over an alphabet of d + 1 letters. One is through numeration systems. Let d ≥ 2 and a 1 , a 2 , . . . , a d be integers such that a 1 ≥ a 2 ≥ . . . ≥ a d ≥ 1. Consider A = {1, 2, . . . , d + 1} as an alphabet. Let σ d be the substitution defined by σ d (i) = 11 . . . 1 ai (i + 1) if i ≤ d and σ d (d + 1) = 1.

We define the Rauzy fractal associated to σ d as follows. Consider the sequence (F n ) n≥0 defined by

F n+d+1 = a 1 F n+d + a 2 F n+d-1 + • • • + a d F n+1 + F n , ∀n ≥ 0,
with initial conditions (called Parry conditions)

F 0 = 1, F n = a 1 F n-1 + • • • + a n F 0 + 1, ∀ 1 ≤ n ≤ d.
It is well-known (using the greedy algorithm) that for every integer n we have n = N i=0 c i F i with (c i ) 0≤i≤N ∈ D a1,...,a d , where D a1,...,a d is the set of sequences (ε i ) l≤i≤k , l, k ∈ Z, such that for all l ≤ i ≤ k :

(1)

ε i ∈ {0, 1, . . . , a 1 } ; (2) ε i ε i-1 . . . ε i-d < lex a 1 a 2 . . . a d 1 when i ≥ l + d . ( 3
) ε i ε i-1 . . . ε l 0 d-i+l < lex a 1 a 2 . . . a d 1 when l ≤ i ≤ l + d, where < lex is the usual lexicographic ordering. We set D ∞ a1,...,a d = {(ε i ) i≥l ; l ∈ Z, (ε i ) l≤i≤n ∈ D a1,.
..,a d , ∀n ≥ l} . Now, consider the following polynomial

P a1,...a d (x) = x d+1 -a 1 x d -a 2 x d-1 -• • • -a d x -1.
It can be checked that P has a root β = β 1 ∈]1, +∞[ and d roots with modulus less than 1. Let β 1 , β 2 , β 3 , . . . , β r be the roots of P belonging to R and β r+1 , . . . , β r+s , β r+1 , . . . , β r+s its complex roots. For all i ∈ Z, we set

α i = (β i 2 , . . . , β i r , β i r+1 , .
. . , β i r+s ). We also put α 0 = 1 = (1, . . . , 1). Then, the Rauzy fractal associated to σ is the set

E a1,...a d ⊂ R r-1 × C s ≈ R d defined by E a1,...a d = +∞ i=d+1 ε i α i ; (ε i ) i≥d+1 ∈ D ∞ a1,...,a d . The set E 1,1 = X is the classical two-dimensional Rauzy fractal.
The structure of the boundary of Rauzy fractals has been first investigated by Ito and M. Kimura in [START_REF] Ito | On the Rauzy Fractal[END_REF]. They showed that the boundary of E 1,1 is a Jordan curve generated by the Dekking method [D82] and they calculated its Hausdorff dimension. Relating the boundary of E a1,1 to the complex numbers having at least two expansions in base α, A. Messaoudi [M00, M05] constructed a finite automaton characterizing and generating this boundary. As a consequence it permitted to parameterize the boundary of E a1,1 , to compute its the Hausdorff dimension and to show it is a quasi circle. In [T06], J. M. Thuswaldner studied the set E a1,a2 . In particular, he gave an explicit formula for the fractal dimension of the boundary of this set.

In this paper we propose to study the boundary of the Rauzy fractal set E 1,1,1 ⊂ R× C. We construct a finite automaton characterizing its boundary. As a consequence we prove that E 1,1,1 has 18 neighborhoods where 6 of them intersect the central tile E 1,1,1 in a point. We also prove that the boundary can be generated by two subregions. More precisely, the boundary of E 1,1,1 is 18 i=1 X i where X i , i = 1, . . . , 6 are singletons, and for all i ∈ [7, 18], there exist affine functions f ij , j = 1, . . . , m i and g ij , j = 1, . . . , n i from R × C to itself such that

X i = mi j=1 f ij (X 7 ) ni j=1 g ij (X 8 ).
2. Notations, definitions and background

2.1. β-expansions. Let β > 1 be a real number. A β-representation of a non- negative real number x is an infinite sequence (x i ) i≤k , x i ∈ Z + = [0, +∞[, such that x = x k β k + x k-1 β k-1 + • • • + x 1 β + x 0 + x -1 β -1 + x -2 β -2 + • • • .
where k is an integer. It is denoted by

x = x k x k-1 . . . x 1 x 0 .x -1 x -2 . . . .
A particular β-representation, called the β-expansion, is computed by the "greedy algorithm" (see [P60]): denote by ⌊y⌋ and {y} respectively the integer part and the fractional part of a number y. There exists k ∈ Z such that β k ≤ x < β k+1 . Let x k = ⌊x/β k ⌋ and r k = {x/β k }. Then for i < k, put x i = ⌊βr i+1 ⌋ and r i = {βr i+1 }. We get

x = x k β k + x k-1 β k-1 + • • • If k < 0 (x < 1), we put x 0 = x -1 = • • • = x k+1 = 0.
If an expansion ends by infinitely many zeros, it is said to be finite, and the ending zeros are omitted.

The digits x i belong to the set

A = {0, • • • , β -1} if β is an integer, or to the set A = {0, • • • , ⌊β⌋} if β is not an integer.
The β-expansion of every positive real number x is the lexicographically greatest among all β-representations of x.

We denote by Fin(β) the set of numbers which have finite greedy β-expansion.

Let N ∈ Z, we denote by Fin N (β) the set of numbers x such that in their βexpansion (x i ) i≤k , x i = 0 for all i < N . We will sometimes denote a β-expansion

x n • • • x k , n ≥ k by (x i ) k≤i≤n .
We put

E β = {(x i ) i≥k ; k ∈ Z, ∀n ≥ k, (x i ) k≤i≤n is a finite β-expansion}.
In the case where β is the dominant root of the polynomial P a1,...a d , it is known (see [START_REF] Frougny | Finite β-expansions[END_REF]) that E β = D ∞ a1,...,a d . We will need the two following classical lemmas. Lemma 1 ([P60]). Let x n • • • x 0 and y m • • • y 0 be two β-expansions. Then, the following are equivalent

• n i=0 x i β i < m i=0 y i β i , • x n • • • x 0 < lex y m • • • y 0 , where < lex is the lexicographical order. Lemma 2 ([FS92]). If β = β 1 , then Z[β] ∩ [0, +∞[⊂ Fin(β).
2.2. Boundary of E a1,...,a d . The coordinates of α have modulus strictly less than 1. Moreover, 0 belongs to the interior of E a1,...,a d ( [A00], see also [R82] for E 1,1 ). Hence, for all z ∈ R r-1 × C s there exists k ∈ N such that α k z ∈ E a1,...,a d . Then, all z ∈ R r-1 × C s can be written as follows z = ∞ i=l ε i α i , where l ∈ Z and (ε i ) i≥l ∈ D ∞ a1,...,a d . The sequence (ε i ) i≥l is called α-expansion of z. We should remark that these α-expansions are not unique : some z can have many different α-expansions. In [M05] it is proven that the points belonging to the boundary of E a1,...,a d have at least two different α-expansions. These points are characterized by the following theorem which is a straightforward consequence of a result due to W. Thurston [T90] (see also [M05]).

Theorem 3. There exists a finite automaton B such that for all distinct elements of D ∞ a1,...,a d , (b i ) i≥l and (c i ) i≥l , the following are equivalent :

• ∞ i=l b i α i = ∞ i=l c i α i • ((b i , c i ))
i≥l is recognizable by B (i.e an infinite path in B beginning in the initial state).

The proof of this result does not give explicitly the states of the automaton. In [M98] is given an algorithm that gives these states for E 1,1 . In [M06], they were given for E a1,1 where a 1 ≥ 2.

Characterization of the boundary of E 1,1,1

In the sequel we suppose d = 3 and a 1 = a 2 = a 3 = 1, and

P (x) = P 1,1,1 (x) = x 4 -x 3 -x 2 -x -1 = (x -β 1 )(x -β 2 )(x -β 3 )(x -β 3 )
where β 1 , β 2 , β 3 are defined in Section 1. Approximations of these numbers are β = β 1 = 1.9275 . . . , β 2 = -0.7748 . . . and β 3 = -0.0763 • • • + i0.8147 . . . . We recall that we defined for all i ∈ Z, α i = (β i 2 , β i 3 ). In this situation

D = D 1,1,1 = {(ε i ) l≤i≤n ; l, n ∈ Z, ε i ∈ {0, 1}, ε i ε i-1 ε i-2 ε i-3 = 1111, l ≤ i ≤ n} , D ∞ = D ∞ 1,1,1 = {(ε i ) i≥l ; l ∈ Z, (ε i ) l≤i≤n ∈ D 1,1,1 , n ≥ l} and E = E 1,1,1 = +∞ i=4 ε i α i ; (ε i ) i≥4 ∈ D ∞ = +∞ i=4 ε i α i ; ε i ∈ {0, 1}, ε i ε i-1 ε i-2 ε i-3 = 1111, i ≥ 4 .
An important and known result is:

Proposition 4. The set E is compact, connected and generates a periodic tiling of R × C with group periods G = Zα 0 + Zα + Zα 2 : R × C = p∈G (E + p),
and the Lebesgue measure of (E + p) ∩ (E + q) is zero whenever p = q, p, q ∈ G.

Proof. The proof can easily be deduced from [R82] or [START_REF] Arnoux | Pisot substitutions and Rauzy fractals[END_REF].

3.1. Definition of the automaton recognizing the points with at least two expansions. In the sequel we proceed to the construction of the automaton A that characterize the boundary of E. This characterization will be proven in Section 3.2. The set of states of the automaton A is

S = ± 3 i=0 c i α i ; c 0 c 1 c 2 c 3 = 1111, c i ∈ {0, 1}, 0 ≤ i ≤ 3 ±(α -1 + 1 + α 2 ), ±(α -2 + α -1 + α), ±(α -3 + α -2 + 1 + α 3 ) .
Let s and t be two states. The set of edges is the set of (s, (a, b), t)

∈ S × {0, 1} 2 × S satisfying t = s α + (a -b)α 3 . The set of initial states is {0} and the set of states is S. A path (resp. infinite path) of A is a sequence (a n , b n ) k≤n≤l (resp. (a n , b n ) n≥k ) such that there exists a sequence (e n ) k≤n≤l+1 (resp. (e n ) n≥k ) of elements of S for which (e n , (a n , b n ), e n+1 ) belongs to S for all n ∈ {k, k + 1, . . . , l + 1} (resp. n ≥ k).
We say it starts in the initial state when e k = 0. The automaton is explicitly defined in the Annexe at the end of this paper. Let us explain the behavior of this automaton.

Let ε = (ε i ) i≥l and ε ′ = (ε ′ i ) i≥l belonging to D ∞ , x = ∞ i=l ε i α i and y = ∞ i=l ε ′ i α i . For all k ≥ l we set A k (ε, ε ′ ) = α -k+3 k i=l (ε i -ε ′ i )α i (1)
In Subsection 3.2 we will prove that x = y if and only if all the A k , k ≥ l, belong to S. But as, for all k ≥ l, we have

A k+1 (ε, ε ′ ) = A k (ε, ε ′ ) α + (ε k+1 -ε ′ k+1 )α 3 , (2) this means that x = y if and only if (0, (ε l , ε ′ l ) , A l (ε, ε ′ )) (A k (ε, ε ′ ), (ε k+1 , ε ′ k+1 ), A k+1 (ε, ε ′ ))
k≥l is an infinite sequence of edges of S starting in the initial state. And, this is equivalent to say that (ε i , ε ′ i ) i≥l is an infinite path of A starting in the initial state. Let us give an example on how we can use this automaton to obtain information about the digits of x and y. Let s be the smallest integer such that

ε s = ε ′ s . Hence A i (ε, ε ′ ) = 0 for i ∈ {l, • • • , s -1}. Suppose ε s > ε ′ s , that is ε s = 1 and ε ′ s = 0. Then, A s = α 3 . From (2) we deduce A s+1 (ε, ε ′ ) = α 2 + (ε s+1 -ε ′ s+1 )α 3 which should belong to S. Hence A s+1 (ε, ε ′ ) = α 2 ∈ S if ε s+1 = ε ′ s+1 , and, A s+1 (ε, ε ′ ) = α 2 + α 3 ∈ S if (ε s+1 , ε ′ s+1 ) = (1, 0)
. Hence, (α 3 , (1, 0), α 2 + α 3 ), (α 3 , (0, 0), α 2 ) and (α 3 , (1, 1), α 2 ) are edges coming from the state α 3 . Let us explain why (α 3 , (0, 1), α 2 -α 3 ) is not an edge, and hence why we cannot have (ε s+1 , ε ′ s+1 ) = (0, 1). We should have that α 2 -α 3 = -α -1 -1 -α belongs to S. Then β should satisfies the same equality. Hence

β -1 + 1 + β should belong to 3 i=0 c i β i ; c 0 c 1 c 2 c 3 = 1111, c i ∈ {0, 1}, 0 ≤ i ≤ 3 (β -1 + 1 + β 2 ), (β -2 + β -1 + β), (β -3 + β -2 + 1 + β 3 ) ,
which is not possible by Lemma 1.

3.2. Characterization of the points with at least two expansions.

Lemma 5. Let (ε i ) i≥0 , (ε ′ i ) i≥0 ∈ D ∞ . Then, +∞ i=0 (ε i -ε ′ i )β i 2 ≤ 1 1 + β 2 , +∞ i=0 (ε i -ε ′ i )β i 3 ≤ C 1 -|β 3 | 6 . where C = max 5 i=0 (c i -d i )β i 3 ; (c i ) 0≤i≤5 ∈ D, (d i ) 0≤i≤5 ∈ D .
Proof. The second inequality is easy to establish. For the first inequality, as -1 < β 2 < 0, all sequences (c i ) i≥0 which terms are 0 or 1 satisfy the following inequality :

β 2 1 -β 2 2 = +∞ i=0 β 2i+1 2 ≤ +∞ i=0 c i β i 2 ≤ +∞ i=0 β 2i 2 = 1 1 -β 2 2 .
This achieves the proof.

For all ε = (ε i ) i≥l and ε ′ = (ε ′ i ) i≥l belonging to D ∞ , we set S(ε, ε ′ ) = {A k (ε, ε ′ ); k ≥ l} = α -k+3 k i=l (ε i -ε ′ i )α i ; k ≥ l . Proposition 6. Let x = ∞ i=l ε i α i , y = ∞ i=l ε ′ i α i , where ε = (ε i ) i≥l and ε ′ = (ε ′ i ) i≥l belong to D ∞ . Then, x = y if and only if the set S(ε, ε ′ ) is finite. Moreover S(ε, ε ′ ) ⊂ S = ± 3 i=0 c i α i ; (c i ) 0≤i≤3 ∈ D ±(α -1 + 1 + α 2 ), ±(α -2 + α -1 + α), ±(α -3 + α -2 + 1 + α 3 ) and S = (ε,ε ′ )∈∆ S(ε, ε ′ ), where ∆ = ((ε i ) i≥l , (ε ′ i ) i≥l ) ∈ D ∞ × D ∞ ; ∞ i=l ε i α i = ∞ i=l ε ′ i α i . Proof. It is easy to establish that if S(ε, ε ′ ) is finite then x = y. Let us prove the reciprocal. Let x = ∞ i=l ε i α i = ∞ i=l ε ′ i α i = y with ε = (ε i ) i≥l and ε ′ = (ε ′ i ) i≥l belonging to D ∞ . Let us prove that A k = A k (ε, ε ′ ) belongs to S for all k ≥ l. As x = y, for all k ≥ l, we have A k = ∞ i=k+1 (ε ′ i -ε i )α i-k+3 = ∞ i=4 (ε ′ i+k-3 -ε i+k-3 )α i . (3)
Let us fix k ≥ l and assume A k = 0. From (1), we deduce there exist n, p, q, r ∈ Z such that

A k = nα 3 + pα 2 + qα + r. (4) But nβ 3 + pβ 2 + qβ + r or -(nβ 3 + pβ 2 + qβ + r) belongs to Z[β] ∩ R + ,
which is contained in Fin(β) (see Lemma 2). We deduce there exists (c i ) s≤i≤m ∈ D such that c m = 1 and

nβ 3 + pβ 2 + qβ + r = ± m i=s c i β i . (5)
We suppose it is equal to m i=s c i β i . The other case can be treated in the same way. As β, β 2 and β 3 are algebraically conjugate, from (1), ( 4) and ( 5) we have

β -k+3 k i=l ε i β i = β -k+3 k i=l ε ′ i β i + m i=s c i β i . (6) From Lemma 1, β -k+3 k i=l ε i β i < β 4 , consequently m ≤ 3. Setting c i = 0 for i > m, we have A k = 3 i=s c i α i . (7)
Remark that if s ≥ 0 then A k belongs to S. Hence we suppose s ≤ -1. Suppose s = -1 and c -1 = 1. Let us show that A k is equal to α -1 + 1 + α 2 and consequently belongs to S. In order to do so, we show that the other cases are not possible. Using Lemma 5 and (3), the first entry of (

A k ), (A k ) 1 , should satisfy |(A k ) 1 | ≤ β 4 2 (1 + β 2 ) -1
which is less than a = 1.6004. This excludes the following points : α -1 + α+ α 2 + α 3 , α -1 + α+ α 3 , α -1 + α and α -1 + α 3 because the absolute value of their first entries is greater than the value below it in the following array :

β -1 2 + β 2 + β 2 2 + β 3 2 β -1 2 + β 2 + β 3 2 β -1 2 + β 2 β -1 2 + β 3 2
1.9 2.5 2.0 1.7

In the same way we should have

|(A k ) 2 | ≤ C|β 3 | 4 (1 -|β 3 | 6
) -1 which is less than b = 1, 8120. This excludes the following points : α -1 + 1 + α 3 , α -1 + α 2 + α 3 and α -1 + 1 + α 2 + α 3 , because the absolute value of their second entries is greater than the value below it in the following array :

β -1 3 + 1 + β 3 3 β -1 3 + β 2 3 + β 3 3 β -1 3 + 1 + β 2 3 + β 3 3 2.0 1.9 1.9
In order to exclude the other cases, except 1 α + 1 + α 2 , we used (2) to compute A k+i , i ≥ 1. Let us explain the strategy. Suppose neither (A k ) 1 nor (A k ) 2 is greater than respectively a and b. Then, we compute A k+1 using (2). We have three possible values :

A k α , A k α + α 3 and A k α -α 3 .
To check that A k does not belong to S, it suffices to show that for all these values, either the first entry or the second is respectively greater than a or b. If it is not the case, for each value that does not satisfy this (both entries are less than, respectively, a and b) we apply again this strategy. Applying this just once we show that 1 α + 1 + α + α 3 does not belong to S. The values of the relevant entries are in the following array and should be read in the following way : The value (1.9 for example) below a relevant entry of A k+1 (resp. 1

β 2 2 + 1 β2 + 1 + β 2 2
) is greater than the absolute value of the relevant entry :

| 1 β 2 2 + 1 β2 + 1 + β 2 2 | > 1.9. A k 1 α + 1 + α + α 3 A k+1 1 β 2 2 + 1 β2 + 1 + β 2 2 1 β 2 3 + 1 β3 + 1 + β 2 3 + β 3 3 1 β 2 2 + 1 β2 + 1 + β 2 2 -β 3 2
1.9 1.9 2.4

For the following case, 1 α + 1, we need to apply the strategy twice because for

A k+1 = 1 β 2 3 + 1
β3 -β 3 3 both entries are respectively less than a and b.

A

k 1 α + 1 A k+1 1 β 2 3 + 1 β3 1 β 2 3 + 1 β3 + β 3 3 1.83 2.0 A k+2 1 β 3 3 + 1 β 2 3 -β 2 3 1 β 3 3 + 1 β 2 3 -β 2 3 + β 3 3 1 β 3 3 + 1 β 2 3 -β 2 3 -β 3 3 2.1 1.63 2.7
For the case A k = 1 α + 1 + α we need two steps because at the first one both

1 β 2 2 + 1 β2 + 1 and 1 β 2 2 + 1 β2 + 1 + β 3
3 have entries less than, respectively, a and b.

A k

1 α + 1 + α A k+1 1 β 2 2 + 1 β2 + 1 -β 3 2 1.84 A k+2 1 β 3 2 + 1 β 2 2 + 1 β2 1 β 3 2 + 1 β 2 2 + 1 β2 + β 3 2 1 β 3 3 + 1 β 2 3 + 1 β3 -β 3 3
1.77 2.23 1.818

1 β 3 3 + 1 β 2 3 + 1 β3 + β 2 3 1 β 3 2 + 1 β 2 2 + 1 β2 + β 2 2 + β 3 2 1 β 3 3 + 1 β 2 3 + 1 β3 + β 2 3 -β 3 3 1.86 1.63 2.24
For the three following cases, 1 α + α 2 , 1 α and 1 α + α + α 2 , we need three steps.

A k 1 α + α 2 A k+1 1 β 2 3 + β 3 1 β 2 3 + β 3 -β 3 3 1.89 2.34 A k+2 1 β 3 3 + 1 + β 2 3 1 β 3 3 + 1 + β 2 3 -β 3 3 1.83 2.26 A k+3 1 β 4 3 + 1 β3 + β 3 + β 2 3 1 β 4 3 + 1 β3 + β 3 + β 2 3 + β 3 3 1 β 4 2 + 1 β2 + β 2 + β 2 2 -β 3 2 1.818 2.32 1.77 A k 1 α A k+1 1 β 2 2 1 β 2 2 -β 3 2 1.66 2.13 A k+2 1 β 3 2 + β 2 2 + β 3 2 1 β 3 3 + β 2 3 -β 3 3 2.01 2.17 A k+3 1 β 4 3 + β 3 1 β 4 3 + β 3 + β 3 3 1 β 4 3 + β 3 -β 3 3 2.00 2.21 1.92 A k 1 α + α + α 2 A k+1 1 β 2 2 + 1 + β 2 1 β 2 2 + 1 + β 2 -β 3 2 1.89 2.35 A k+2 1 β 3 2 + 1 β2 + 1 + β 2 2 1 β 3 2 + 1 β2 + 1 + β 2 2 + β 3 2 1.84 2.30 A k+3 1 β 4 2 + 1 β 2 2 + 1 β2 + β 2 1 β 4 3 + 1 β 2 3 + 1 β3 + β 3 1 β 4 2 + 1 β 2 2 + 1 β2 + β 2 -β 2 2 -β 2 3 + β 3 3 -β 2 3 -β 3 3 1.77 1.818 2.23
Hence the only possible A k (with c -1 = 1) is 1 α + 1 + α 2 . Suppose now s ≤ -2 and c s = 1. It is useful for the sequel to remark that U = (u i ) i≥s = (c s , c s+1 , . . . , c 2 , c 3 , ε k+1 , ε k+2 , ε k+3 , . . . ) belongs to D ∞ . Indeed, if c 3 = 0, it is clear. If c 3 = 1 and c 2 = 0 then by ( 6), ε k = 1. Hence ε k+1 ε k+2 ε k+3 = 111 and U belongs to D ∞ . The other cases can be treated in the same way. Using ( 3) and ( 7) we obtain

u = 3 i=s c i α i + ∞ i=4 ε i+k-3 α i = ∞ i=4 ε ′ i+k-3 α i = v. We set V = (ε ′ i+k-3 ) i≥s where ε ′ i+k-3 = 0 when s ≤ i ≤ 3. Then (U, V ) belongs to ∆, A s+4 (U, V ) = c s α -1 +c s+1 +c s+2 α+c s+3 α 2 +c s+4 α 3 and A 3 (U, V ) = 3 i=s c i α i
. Doing what we did for x and y to u and v we obtain that A s+4 (U, V ) = α -1 +1+α 2 . Let us show that for all n ≥ s + 5, A n (U, V ) belongs to

C = ±(α -2 + α -1 + α), ±(α -3 + α -2 + 1 + α 3 ) .
This will imply that A k belongs to S for all k ≥ l. We have that A s+5 (U, V ) belongs to

1 α 2 + 1 α + α, 1 α 2 + 1 α + α + α 3 , 1 α 2 + 1 α + α -α 3 .
The third one can be excluded because 1

β 2 3 + 1 β3 + β 3 -β 3 3 ≥ 1.85.
We proceed as before to exclude the second element :

A k 1 α 2 + 1 α + α + α 3 A k+1 1 β 3 3 + 1 β 2 3 + 1 + β 2 3 1 β 3 3 + 1 β 2 3 + 1 + β 2 3 -β 3 3 2.00 2.55 A k+2 1 β 4 3 + 1 β 3 3 + 1 β3 + β 3 1 β 4 3 + 1 β 3 3 + 1 β3 + β 3 + β 3 3 1 β 4 3 + 1 β 3 3 + 1 β3 + β 3 -β 3 3 2.45 2.54 2.47 Consequently, A s+5 (U, V ) = 1 α 2 + 1 α + α. We deduce A s+6 (U, V ) = 1 α 3 + 1 α 2 + 1 + α 3 because 1 β 3 3 + 1 β 2 3 + 1 > 2.03 and 1 β 3 3 + 1 β 2 3 + 1 -β 3 3 > 2.56. Once again, A s+7 (U, V ) = 1 α 4 + 1 α 3 + 1 α + α 2 -α 3 because 1 β 4 3 + 1 β 3 3 + 1 β3 + β 2 3 > 1.
85 and

1 β 4 3 + 1 β 3 3 + 1 β3 + β 2 3 + β 3 3 > 2.16. But an easy computation leads to 1 α 4 + 1 α 3 + 1 α + α 2 -α 3 = -( 1 α 2 + 1 α + α) = -A s+5 (U, V ).
Then continuing in the same way we can check

A n (U, V ) = -A n+2 (U, V ) and A n ∈ C for n ≥ s + 5. As 3 ≥ s + 5, we obtain that A k (ε, ε ′ ) = A 3 (U, V ) belongs to C. Thus S(ε, ε ′ ) is included in S.
To complete the proof we should show that each element of S belongs to Γ = ∪ (ε,ε ′ )∈∆ S(ε, ε ′ ). Remark that if A k belongs to Γ then -A k also belongs to Γ. Consequently it is sufficient to consider the cases where

A k = 3 i=0 c i α i with (c i ) 0≤i≤3 ∈ D or A k = α -1 + 1 + α 2 , α -2 + α -1 + α or α -3 + α -2 + 1 + α 3 . Notice that we have -α 3 = +∞ i=1 (α 4i + α 4i+1 + α 4i+2 ) = 1 + α + α 2 + +∞ i=1 (α 4i+1 + α 4i+2 + α 4i+3 ) = α + α 2 + α 4 + +∞ i=1 (α 4i+2 + α 4i+3 + α 4i+4 ) = α 2 + α 4 + α 5 + +∞ i=1 (α 4i+3 + α 4i+4 + α 4i+5 ).
Hence, 1+α+α 2 , α+α 2 and α 2 belong to Γ. Multiplying by α we deduce α+α 2 +α 3 , α 2 + α 3 and α 3 belong to Γ. Now subtracting -α 2 we obtain 1 + α and α belong to Γ. We have that 1 belongs to Γ because

+∞ i=1 α 4i = 1 + +∞ i=1 α 4i+1 . Now, 1 + α 2 belongs to Γ because ∞ i=2 α 2i = 1 + α 2 + ∞ i=2 α 2i+1 .
Multiplying by α we deduce α + α 3 belongs to Γ. Because

α -3 + α -2 + 1 + α 3 + ∞ i=1 (α 4i+2 + α 4i+3 ) = ∞ i=1 (α 4i + α 4i+1 ), ( 8
)
we obtain that α -3 + α -2 + 1 + α 3 belongs to Γ. Multiplying (8) by, respectively, α and α 2 we obtain, respectively, that α -2 + α -1 + α and α -1 + 1 + α 2 belong to Γ. From

α 4 + ∞ i=1 α 4i+3 = 1 + α 3 + ∞ i=1 α 4i+1 = 1 + α + α 3 + ∞ i=1 α 4i+2
it is clear 1 + α 3 and 1 + α + α 3 belong to Γ. The equality

α 4 + ∞ i=1 α 4i+2 = 1 + α 2 + α 3 + ∞ i=1 α 4i+1
implies that 1 + α 2 + α 3 belongs to Γ and achieves the proof.

Corollary 7. Let A be the automaton defined in Subsection 3.1. Then, for all (ε i ) i≥l and (ε ′ i ) i≥l belonging to D ∞ the following assertions are equivalent :

• i≥l ε i α i = i≥l ε ′ i α i ; • (ε i , ε ′ i ) i≥l is an infinite path in A beginning in the initial state. i=12 h i (X), where h 0 (z) = α 4 + α 4 z, h 1 (z) = α 4 + α 6 + α 4 z h 2 (z) = α 4 + α 5 + α 4 z, h 3 (z) = α 4 + α 5 + α 6 + α 4 z, h 4 (z) = 1 + α + α 2 + α 7 + α 5 z, h 5 (z) = h 2 (z) h 6 (z) = α 4 + α 7 + α 4 z, h 7 (z) = α 4 + α 8 + α 9 + α 7 z, h 8 (z) = h 0 (z), h 9 (z) = α 4 + α 5 + α 7 + α 4 z, h 10 (z) = α 4 + α 5 + α 8 + α 9 + α 7 z, h 11 (z) = 1 + α + α 6 + α 7 + α 5 z h 12 (z) = α 4 + α 8 + α 7 z, h 13 (z) = h 7 (z), h 14 (z) = α 4 + α 5 + α 8 + α 7 z, h 15 (z) = h 10 (z) h 16 (z) = 1 + α + α 6 + α 5 z, h 17 (z) = h 11 (z),
Proof. a) The set 1 + αX is clearly included in 1 + αE. Moreover it is easy to check that 1 + αX is a subset of α 4 + αE which is included in E. Hence 1 + αX ⊂ E(1).

On the other hand, let z ∈ E(1). Then, there exist

(ε i ) i≥4 and (ε ′ i ) i≥4 in D ∞ such that z = 1+ i≥4 ε i α i = i≥4 ε ′ i α i . From Corollary 7, (1, 0)(0, 0)(0, 0)(0, 0)(ε 4 , ε ′ 4 ) (ε 5 , ε ′ 5
) is a finite path in the automaton A starting at the initial state. Following this path in the automaton we deduce (ε 4 , ε ′ 4 ) = (0, 1) and (ε 5 , ε

′ 5 ) = (1, 0). It gives z = 1 + α 5 + α 2 w = α 4 + α 2 w ′ where w, w ′ ∈ E. Consequently E(1) ⊂ (1 + αE) ∩ (α 4 + αE) = 1 + α(E ∩ (1 + α + α 2 + E)) = 1 + αX. b) We have 1+α+α 2 +αE(α 2 ) = (α 4 +αE)∩(1+α+α 2 +αE) ⊂ E ∩(1+α+α 2 +E) = X. Hence E(α 2 ) ⊂ -1 α -1 -α + X α .
To prove the other inclusion, let z ∈ X. Then by the automaton we deduce that z

= 1 + α + α 2 + αw = α 4 + αw ′ , w, w ′ ∈ E. Hence -1 α -1 -α + z α = w = α 2 + w ′ and -1 α -1 -α + X α ⊂ E(α 2 ). c) Let z ∈ E(1 + α 2 ) : z = i≥4 ε i α i = 1 + α 2 + i≥4 ε ′ i α i , (ε i ) i≥4 , (ε ′ i ) i≥4 ∈ D ∞ .
Corollary 7 and the automaton show that (0, 1)(0, 0)(0, 1)(0, 0)(ε 4 , ε ′ 4 ) . . . is a infinite path starting in the initial state and (ε i , ε ′ i ) i≥4 is equal to uuu . . . where u = (1, 0)(0, 1). Then,

z = α 4 + α 6 + α 8 + • • • = 1 + α 2 + α 5 + α 7 + α 9 + . . . and E(1 + α 2 ) = α 4 (1 -α 2 ) -1 . d) Let z ∈ E(α -2 + α -1 + α).
From Corollary 7 and using the automaton we deduce that z

= α -2 + α -1 + α + α 4 + ∞ i=2 (α 4i-1 + α 4i ) = ∞ i=1 (α 4i+1 + α 4i+2 ). Hence E(α -2 + α -1 + α) = (α 5 + α 6 )(1 -α 4 ) -1 .
e) Corollary 7 and the automaton give the result. f) Let z ∈ E(α). Then, there exist (ε i ) i≥4 and (ε

′ i ) i≥4 in D ∞ such that z = i≥4 ε i α i = α + i≥4 ε ′ i α i . From Corollary 7, (0, 0)(0, 1)(0, 0)(0, 0)(ε 4 , ε ′ 4 ) . . . is a path in the automaton starting in the initial state. Hence, (ε 4 , ε ′ 4 )(ε 5 , ε ′ 5 )(ε 6 , ε ′ 6 ) belongs to {(0, 0), (1, 1), (0, 1)}(1, 0)(0, 1). Consequently, z belongs to the union of (α 5 +α 2 E)∩(α+α 2 E), (α 4 +α 5 +α 2 E)∩(α+α 4 +α 2 E) and (α 5 +α 2 E)∩(α+α 4 +α 2 E) which is equal to f 0 (X) ∪ f 1 (X) ∪ f 1 (Y ). Hence E(α) = f 0 (X) ∪ f 1 (X) ∪ f 1 (Y ). g) Let z ∈ E(α + α 2 ). Then, there exist (ε i ) i≥4 and (ε ′ i ) i≥4 in D ∞ such that z = i≥4 ε i α i = α + α 2 + i≥4 ε ′ i α i .
From Corollary 7, (0, 0)(0, 1)(0, 1)(0, 0)(ε 4 , ε ′ 4 ) . . . is a path in the automaton starting in the initial state. Hence, we either have (1) ((ε i , ε ′ i )) 4≤i≤7 ∈ (0, 1)(1, 0){(0, 0), (1, 1), (1, 0)}(0, 1), (2) (ε 4 , ε ′ 4 ) ∈ {(0, 0), (1, 1)}, or (3) ((ε i , ε ′ i )) i≥4 ∈ (0, 1){(0, 0)(0, 0), (0, 0)(1, 1), (1, 1)(0, 0)}ww . . . , where w = (0, 1)(1, 0)(1, 0)(0, 1). This means that z belongs to

(α 5 + α 4 E) ∩ (α + α 2 + α 4 + α 7 + α 4 E) ∪ (α 5 + α 6 + α 4 E) ∩ (α + α 2 + α 4 + α 6 + α 7 + α 4 E) ∪ (α 5 + α 6 + α 4 E) ∩ (α + α 2 + α 4 + α 7 + α 4 E) ∪ (αE ∩ (α + α 2 + αE) ∪ (α 4 + αE) ∩ (α + α 2 + α 4 + αE) ∪ {z 1 , z 2 , z 3 } = g 0 (X) ∪ g 1 (X) ∪ g 1 (Y ) ∪ g 2 (Y ) ∪ g 3 (Y ) ∪ {z 1 , z 2 , z 3 } where z 1 = +∞ i=2 (α 4i + α 4i+1 ) = α + α 2 + α 4 + α 7 + +∞ i=2 (α 4i+2 + α 4i+3
), z 2 = α 6 +z 1 and z 3 = α 5 +z 1 . We can also check that (1, 0)(1, 0)(0, 0)(0, 0)uuu . . . , where u = (0, 1)(1, 1)(1, 0)(1, 0), is an infinite path of the automaton starting in the initial state. Consequently, z 1 = α 4 + α 5 + +∞ i=2 (α 4i+1 + α 4i+2 ) and

z 1 ∈ (α 4 + αE) ∩ (α + α 2 + α 4 + αE) = g 3 (Y ).
Moreover, it shows that z 2 belongs to g 3 (Y ). In the same way,

z 1 = α 5 + α 6 + α 8 + +∞ i=2 (α 4i+3 + α 4i+4 ). Thus, z 3 = 2α 5 + α 6 + α 8 + +∞ i=2 (α 4i+3 + α 4i+4 ) = α 5 + +∞ i=2 (α 4i + α 4i+1 ). But 2α 5 + α 6 = α + α 2 + α 7 , consequently z 3 belongs to αE) ∩ (α + α 2 + αE) = g 2 (Y ). h) Let z ∈ X = E(1 + α + α 2 ). Then, there exist (ε i ) i≥4 and (ε ′ i ) i≥4 in D ∞ such that z = i≥4 ε i α i = 1 + α + α 2 + i≥4 ε ′ i α i .
From Corollary 7, we necessarily have (ε 4 , ε ′ 4 ) = (1, 0) and one of the following situations :

(1) ((ε i , ε ′ i )) i≥5 ∈ (1, 0){(0, 0), (1, 1)}ww . . . where w = (0, 1)(1, 0); (2) ((ε i , ε ′ i )) i≥5 ∈ (0, 1){(0, 0), (1, 1)}ww . . . where w = (0, 1)(1, 0)(1, 0)(0, 1); (3) (ε i , ε ′ i ) 5≤i≤8 ∈ {(0, 0), (1, 1)} 2 (0, 1)(1, 0); (4) (ε i , ε ′ i ) 5≤i≤9 = (1, 0)(1, 0)(0, 1)(1, 0)(0, 1); (5) (ε i , ε ′ i ) 5≤i≤8 ∈ {(0, 0), (1, 1)}(1, 0)(0, 1)(1, 0). This means z belongs to 4 i=0 h i (X) ∪ h 1 (Y ) ∪ h 3 (Y ) ∪ {x 1 , x 2 , x 3 , x 4 } where x 1 =α 4 + α 5 + +∞ i=4 α 2i = 1 + α + α 2 + +∞ i=3 α 2i+1 , x 2 =x 1 + α 6 , x 3 =α 4 + +∞ i=2 (α 4i + α 4i+1 ) = 1 + α + α 2 + α 5 + α 7 + +∞ i=2 (α 4i+2 + α 4i+3 ), x 4 =x 3 + α 6 , h 0 (X) =(α 4 + α 4 E) ∩ (1 + α + α 2 + α 7 + α 4 E), h 1 (X) =(α 4 + α 6 + α 4 E) ∩ (1 + α + α 2 + α 6 + α 7 + α 4 E), h 2 (X) =(α 4 + α 5 + α 4 E) ∩ (1 + α + α 2 + α 5 + α 7 + α 4 E), h 3 (X) =(α 4 + α 5 + α 6 + α 4 E) ∩ (1 + α + α 2 + α 5 + α 6 + α 7 + α 4 E), h 4 (X) =(α 4 + α 5 + α 6 + α 8 + α 5 E) ∩ (1 + α + α 2 + α 7 + α 5 E), h 1 (Y ) =(α 4 + α 6 + α 4 E) ∩ (1 + α + α 2 + α 7 + α 4 E) and h 3 (Y ) =(α 4 + α 5 + α 6 + α 4 E) ∩ (1 + α + α 2 + α 5 + α 7 + α 4 E).
We easily can check (using Corollary 7 and the automaton) that

x 1 = x 3 = α 4 + +∞ i=2 (α 4i + α 4i+1 ),
and thus x 1 ∈ h 0 (X), x 2 ∈ h 1 (X), and x 2 = x 4 , which concludes the proof of h). i) Let z ∈ X = E(1 + α). Then, there exist (ε i ) i≥4 and (ε

′ i ) i≥4 in D ∞ such that z = i≥4 ε i α i = 1 + α + i≥4 ε ′ i α i .
From Corollary 7, we necessarily have (ε 4 , ε ′ 4 ) = (1, 0) and one of the following situations :

(1) ((ε i , ε ′ i )) i≥5 ∈ {(0, 0), (1, 1)}(0, 1) 2 {(0, 0), (1, 1)} 2 ww . . . ; (2) ((ε i , ε ′ i )) i≥5 ∈ (1, 0){(0, 0), (1, 1)} 2 (1, 0)(0, 1)ww . . . ; (3) ((ε i , ε ′ i )) 5≤i≤7 ∈ {(0, 0), (1, 1)}(0, 1){(0, 0), (1, 1)}; (4) ((ε i , ε ′ i )) 5≤i≤8 ∈ {(0, 0), (1, 1)}(0, 1) 2 (1, 0){(0, 0), (1, 1), (1, 0)}(0, 1)(1, 0); (5) ((ε i , ε ′ i )) 5≤i≥8 ∈
(1, 0)(0, 1){(0, 0), (1, 1), (0, 1)}(1, 0)(0, 1). where w = (0, 1)(1, 0)(1, 0)(0, 1). Hence z belongs to

11 i=5 h i (Y ) ∪ 17 i=12 h i (X) ∪ {y i ; 1 ≤ i ≤ 8},
where

y 1 =α 4 + +∞ i=2 (α 4i+3 + α 4i+4 ) = 1 + α + α 6 + α 7 + α 10 + +∞ i=3 (α 4i+1 + α 4i+2 ), y 2 =y 1 + α 9 , y 3 = y 1 + α 8 , y 4 = y 1 + α 5 , y 5 = y 1 + α 5 + α 9 . y 6 =y 1 + α 5 + α 8 = 1 + α + +∞ i=2 (α 4i+1 + α 4i+2 ), y 7 = y 6 + α 7 , y 8 = y 6 + α 6 , h 5 (Y ) =(α 4 + α 5 + α 4 E) ∩ (1 + α + α 5 + α 6 + α 4 E), h 6 (Y ) =(α 4 + α 7 + α 4 E) ∩ (1 + α + α 6 + α 7 + α 4 E), h 7 (Y ) =(α 4 + α 8 + α 9 + α 7 E) ∩ (1 + α + α 6 + α 7 + α 10 + α 7 E), h 8 (Y ) =(α 4 + α 4 E) ∩ (1 + α + α 6 + α 4 E), h 9 (Y ) =(α 4 + α 5 + α 7 + α 4 E) ∩ (1 + α + α 5 + α 6 + α 7 + α 4 E), h 10 (Y ) =(α 4 + α 5 + α 8 + α 9 + α 7 E) ∩ (1 + α + α 5 + α 6 + α 7 + α 10 + α 7 E), h 11 (Y ) =(α 4 + α 5 + α 8 + α 5 E) ∩ (1 + α + α 6 + α 7 + α 5 E), h 12 (Y ) =(α 4 + α 8 + α 7 E) ∩ (1 + α + α 6 + α 7 + α 10 + α 7 E), h 13 (X) =(α 4 + α 8 + α 9 + α 7 E) ∩ (1 + α + α 6 + α 7 + α 9 + α 10 + α 7 E), h 14 (X) =(α 4 + α 5 + α 8 + α 7 E) ∩ (1 + α + α 5 + α 6 + α 7 + α 10 + α 7 E), h 15 (X) =(α 4 + α 5 + α 8 + α 9 + α 7 E) ∩ (1 + α + α 5 + α 6 + α 7 + α 9 + α 10 + α 7 E), h 16 (X) =(α 4 + α 5 + α 8 + α 5 E) ∩ (1 + α + α 6 + α 5 E), h 17 (X) =(α 4 + α 5 + α 7 + α 8 + α 5 E) ∩ (1 + α + α 6 + α 7 + α 5 E),
Let us prove that for each integer i ∈ {1, . . . , 8}, there exists j ∈ {5, . . . , 11} or k ∈ {12, . . . , 17} such that y i belongs to h j (X) or to h k (Y ). Indeed, since y 1 = α 4 + α 3 z 1 (see case g)), then y 1 ∈ (α 4 + α 7 + α 4 E) ∩ (2α 4 + α 5 + α 7 + α 4 E) = h 6 (Y ). We deduce that y 2 and y 3 belong also to h 6 (Y ), and, y 4 and y 5 belong to h 9 (Y ). Using the automaton we can verify that y 6 = 1+α+α 5 +α 6 + +∞ i=2 (α 4i+2 +α 4i+3 ). Hence, y 6 belongs to 1 + α + α 5 + α 6 + α 4 E. But it also belongs to α 4 + α 5 + α 4 E. Thus y 6 ∈ h 5 (Y ) and y 7 ∈ h 9 (Y ). We have y 8 = y 6 + α 6 ∈ (1 + α + α 6 + α 5 E). On the other hand we can check using the automaton that y 8 = α 4 + α 5 + α 8 + +∞ i=5 α 2i , hence y 8 ∈ (α 4 + α 5 + α 8 + α 5 E) and y 8 belongs to h 16 (X).

Remarks and comments. There are points which has at least 6 expansions in base α. For example:

α + +∞ i=2 α 2i = +∞ i=1 (α 4i + α 4i+1 ) = 1 + α + α 2 + ∞ i=2 α 2i+1 = 1 + α + ∞ i=1 (α 4i+1 + α 4i+2 )) = α + α 2 + α 4 + ∞ i=1 (α 4i+3 + α 4i+4 ) = α -3 + α -2 + 1 + α 3 + ∞ i=1 (α 4i+2 + α 4i+3 ).
We address the two following questions :

(1) Can you parameterize the boundary of E 1,1,1 ?

(2) Does this boundary be homeomorphic to the sphere ? The technics used in this work can be used to study E a1,a2,...,a d with the assumption that a 1 ≥ a 2 ≥ • • • ≥ a d ≥ 1.

acknowledgements. Both authors would like to thank the Brazil-France agreement for cooperation in Mathematics and the CNRS that permitted the authors to visit each other. This strongly contributed to this work. The first author would like to thank the hospitality of the Mathematic department of the State University of São Paolo in São José do Rio Preto (Brazil). The second author thanks the Laboratoire Amiénois de Mathématiques Fondamentales et Appliquées, CNRS-UMR 6140, from the University of Picardie Jules Verne (France) for his hospitality. The second author was supported by a CNPq grant Proc. 305043/2006-4.

Annexe.

0 -α -α 2 -α 3 α + α 2 + α 3 -α 3 α 3 -1 -α 1 + α -α 2 -α 3 -α -α 2 α 2 + α 3 α + α 2 -1 -α 2 1 + α 2 1 + α + α 2 -α 2 -α -α 3 α + α 3 α 2 -1 -α -α 2 1 + α + α 3 -1 -α -α 3 -α α -1 1 1 α + 1 + α 2 -1 α -1 -α 2 1 α 2 + 1 α + α -1 α 2 -1 α -α -1 -α 2 -α 3 1 + α 2 + α 3 1 α 3 + 1 α 2 + 1 + α 3 -1 α 3 -1 α 2 -1 -α 3 -1 -α 3 1 + α 3
(0, 0), (1, 1) (0, 1) (1, 0) (0, 0), (1, 1) (0, 0), (1, 1)

(1, 0) (0, 1) (0, 0), (1, 1) (0, 1) (0, 0), (1, 1)

(1, 0) (0, 1) (0, 0)

(1, 1) (0, 1)

(1, 1) (0, 0) (0, 1)

(1, 1), (0, 0) (0, 1)

(1, 1), (0, 0)

(1, 0) (0, 1)

(1, 0) (0, 1) (0, 1) (0, 0)

(1, 1)

(1, 0)

(1, 0) (0, 0)

(1, 1) (0, 1) (0, 1) (0, 0) (1, 1)

(1, 0) (0, 0) (1, 1) (0, 1)

(1, 0) (0, 0) (1, 1) (0, 0)

(1, 1) (0, 1) (1, 0) (0, 0)

(1, 1) (0, 0)

(1, 1) (0, 0) (1, 1) (0, 0) (1, 1)

(1, 0) (0, 1) (0, 1) (1, 0)

(1, 0) (0, 1)

(1, 0) (0, 1)

(1, 0) (0, 1)

Laboratoire Amiénois de Mathématiques Fondamentales et Appliquées, CNRS-UMR 6140, Université de Picardie Jules Verne, 33 rue Saint Leu, 80039 Amiens Cedex, France. E-mail address: fabien.durand@u-picardie.fr Departamento de Matemática, Unesp-Universidade Estadual Paulista, Rua Cristovão Colombo, 2265, Jardim Nazareth, 15054-000, São José do Rio Preto, SP, Brasil. E-mail address: messaoud@ibilce.unesp.br

Neighbors of E.

Here we prove that the set E has 18 neighbors where 6 of them have an intersection with E reduced to a singleton, and that the boundary can be generated by just 2 subregions.

Lemma 8. Let (ε i ) i≥4 and (ε ′ i ) i≥l be two elements of D ∞ such that

, where l < 4 and ε ′ l = 1, then ε ′ l α l + ε ′ l+1 α l+1 • • • + ε ′ 3 α 3 belongs to S. In particular l ≥ -3 and

In particular, for i = l, we obtain the result.

Lemma 9. Let u ∈ S. Then, there exists

Proof. This comes from Proposition 6 and the identity (3).

In our context, Lemma 2 in [M05] can be formulated in the following way.

Lemma 10. Let x ∈ R × C, then x belongs to the boundary of E if and only if there exists l ≤ 3 such that x = +∞ i=4 ǫ i α i = +∞ i=l ǫ ′ i α i , where (ǫ i ) i≥4 and (ǫ ′ i ) i≥l belong to D ∞ , and, ǫ ′ l = 0. Proposition 11. The boundary of E is the union of the 18 non empty regions E(u), u ∈ {a, -a; a ∈ A}, whose pairwise intersections have measure zero, where

Proof. Let u be an element of A, then u is a state of the automaton A. From Lemma 9, there exist

is not empty and with measure zero. It will be useful to check that

Consequently, Proposition 4 implies u∈A E(u)∪E(-u) is contained in the boundary of E. Now, let z be an element of the boundary of E, then by Lemma 10 there exist two elements of D ∞ , (ε i ) i≥4 and (ε

We can suppose ε l = 1. Let us consider the following four cases. Suppose l = -3. From Lemma 8, we deduce that z ∈ E(α -3 + α -2 + 1 + α 3 ). Suppose l = -2. From Lemma 8, we deduce that z ∈ E(α -2 + α -1 + α).

Corollary 7 implies that t = (0, 1)(0, 1)(0, 0)(0, 1)(0, 0)(ε 4 , ε ′ 4 ) . . . is an infinite path of the automaton A starting at the initial state. Using the automaton, we see that t = (0, 1)(0, 1), (0, 0)(0, 1)(0, 0)awww . . . , where a = (1, 1) or (0, 0) and w = (0, 1)(1, 0)(1, 0)(0, 1). Consequently,

. is an infinite path in the automaton beginning in the initial state. This implies that t = (0, 1)(0, 1)(0, 0)(0, 1)(1, 1)(0, 0)ww . . . where w = (0, 1)(1, 0)(1, 0)(0, 1). Hence z = 1+α+α

, there is no infinite path in the automaton starting in the initial state and beginning with (0, ε ′ 0 )(0, ε ′ 1 )(0, 0)(0, 1)(ε 4 , 1)(ε 5 , 1). • Hence it remains to consider the case : ε ′ 0 + ε ′ 1 = 2. But it is easy to check that this implies ε ′ 6 = 1 which is not possible. This ends the proof.

Using the automaton given in the Annexe, we deduce the following result.

g) E(α + α 2 ) = g 0 (X) ∪ g 1 (X) ∪ g 1 (Y ) ∪ g 2 (Y ) ∪ g 3 (Y ), where f 0 (z) = α + α 2 z, f 1 (z) = α + α 4 + α 2 z, g 0 (z) = α 5 + α 4 z, g 1 (z) = α 5 + α 6 + α 4 z, g 2 (z) = αz, g 3 (z) = α 4 + αz, h) X =