Amélie Piveteau

Alastair A Abbott

Sadiq Muhammad

Mohamed Bourennane

Armin Tavakoli

Armin Tavakoli Weak

Weak entanglement improves quantum communication using only product measurements

Introduction.-Entanglement is well-known to be the crucial resource for a wide variety of quantum information applications. A major domain of application is in quantum communication where it can, e.g., increase the classical capacity of a quantum channel [START_REF] Bennett | Entanglement-assisted classical capacity of noisy quantum channels[END_REF] or reduce classical communication complexity [START_REF] Buhrman | Nonlocality and communication complexity[END_REF]. However, not all forms of entanglement are evidently useful. For instance, while entanglement that is strong enough to generate nonlocality has been found to improve noiseless classical communication beyond its conventional limitations (see, e.g., [START_REF] Cleve | Substituting quantum entanglement for communication[END_REF][START_REF] Brukner | Bell's inequalities and quantum communication complexity[END_REF][START_REF] Cubitt | Improving zero-error classical communication with entanglement[END_REF][START_REF] Brunner | Connection between Bell nonlocality and Bayesian game theory[END_REF][START_REF] Tavakoli | Higher-dimensional communication complexity problems: Classical protocols versus quantum ones based on Bell's theorem or prepare-transmit-measure schemes[END_REF][START_REF] Tavakoli | Does violation of a Bell inequality always imply quantum advantage in a communication complexity problem?[END_REF]), weaker entangled states that cannot violate any Bell inequality do not have that ability [START_REF] Pauwels | Adaptive advantage in entanglement-assisted communications[END_REF]. However, if the system communicated is itself quantum, e.g. a qubit instead of a bit, then some weaker forms of entanglement also become useful. This can be seen in the celebrated dense coding protocol [START_REF] Bennett | Communication via one-and two-particle operators on Einstein-Podolsky-Rosen states[END_REF] where a maximally entangled state, |φ

+ = |00 +|11 √ 2
, is exploited to allow a qubit message to transmit two bits instead of one bit. Then, if the maximally entangled state is replaced with an isotropic state

ρ v = v|φ + φ + | + 1 -v 4 1 1, (1)
for some visibility v ∈ [0, 1], then an advantage over an unassisted qubit message exists whenever v > 1 3 [START_REF] Tavakoli | Semi-device-independent characterisation of multipartite entangled states and measurements[END_REF][START_REF] Moreno | Semi-device-independent certification of entanglement in superdense coding[END_REF]. This coincides with the visibility at which the state becomes separable, v sep = 1 3 , and is considerably lower than the critical visibility for Einstein-Podolsky-Rosen steering under general projective measurements, v unsteer = 1 2 [START_REF] Wiseman | Steering, entanglement, nonlocality, and the Einstein-Podolsky-Rosen paradox[END_REF]. Moreover, it is even further below the largest known visibility at which the isotropic state (1) cannot violate any Bell inequality, v local ≈ 0.6875 [START_REF] Designolle | Improved local models and new Bell inequalities via Frank-Wolfe algorithms[END_REF].

However, to harness the dense coding advantage one must measure in the Bell basis {Φ + , Φ -, Ψ + , Ψ -}, where Φ ± = |φ ± φ ± | and Ψ ± = |ψ ± ψ ± | are the projectors onto the states |φ ± = |00 ±|11 √ 2

and |ψ ± = |01 ±|10 √ 2

. In optical systems, which are the most relevant for quantum communication, it is impossible to implement a linear optics Bell basis measurement on separate photons without the use of auxiliary photons [START_REF] Lütkenhaus | Bell mea-surements for teleportation[END_REF]. While dense coding experiments have been reported [START_REF] Mattle | Dense coding in experimental quantum communication[END_REF][START_REF] Li | Quantum dense coding exploiting a bright Einstein-Podolsky-Rosen beam[END_REF][START_REF] Barreiro | Beating the channel capacity limit for linear photonic superdense coding[END_REF][START_REF] Williams | Superdense coding over optical fiber links with complete Bell-state measurements[END_REF][START_REF] Hu | Beating the channel capacity limit for superdense coding with entangled ququarts[END_REF], implementation of the Bell basis is not expected to be scalable in optical systems in the near future. Nevertheless, it has recently been found that entanglement can yield advantages in one-shot quantum communication scenarios by using much more limited, yet much less experimentally demanding, optical measurements that are compatible with passive linear optics [START_REF] Piveteau | Entanglement-assisted quantum communication with simple measurements[END_REF]. However, the schemes considered thus far come with greater requirements on the quality of entanglement, in particular demanding states that can be used to violate a Bell inequality.

Here, we show that when restricting to simple optical measurements, compatible with ancilla-free linear optics, weak forms of entanglement still constitute a resource for enhancing communication. To this end, we introduce a communication task which can be interpreted as a stochastic version of the cryptographic primitive known as secret sharing. In secret sharing a secret is distributed between two parties in such a way that they must cooperate to reconstruct it [START_REF] Blakley | Safeguarding cryptographic keys[END_REF][START_REF] Shamir | How to share a secret[END_REF]. This task is of considerable interest for quantum cryptography, and has consequently received much attention (see e.g. [START_REF] Hillery | Quantum secret sharing[END_REF][START_REF] Karlsson | Quantum entanglement for secret sharing and secret splitting[END_REF][START_REF] Guo | Quantum secret sharing without entanglement[END_REF][START_REF] Tavakoli | Secret sharing with a single d-level quantum system[END_REF][START_REF] Karimipour | Quantum secret sharing and random hopping: Using single states instead of entanglement[END_REF]). We prove that for our task there exist entangled but unsteerable isotropic states which enable an advantage over unassisted qubits using only conventional optical partial Bell state analysers [START_REF] Weinfurter | Experimental Bell-state analysis[END_REF][START_REF] Braunstein | Measurement of the Bell operator and quantum teleportation[END_REF]. Using polarisation qubits generated via spontaneous parametric down-conversion, we use both a maximally entangled state to experimentally demonstrate an optimal stochastic secret sharing protocol, and an unsteerable isotropic state to outperform the best possible entanglementunassisted qubit protocol for the task. We also consider whether an advantage could be obtained using the simplest type of measurements, namely product measurements of separate photons that require no two-photon interference in the decoding procedure. Even with such measurements, we find that some isotropic states that are steerable but cannot violate any Bell inequality still enable an advantage over unassisted qubit communication.

Stochastic secret sharing.-Suppose that Alice wishes to generate a random secret bit a ∈ {0, 1} that is shared with Bob and Charlie in such a way that they individually have no knowledge of a but can learn its value if they cooperate. To this end, we consider the scenario illustrated in Fig 1 .  In this scenario, Bob and Charlie each privately select two uniformly random bits x ≡ (x 0 , x 1 ) ∈ {0, 1} 2 and y ≡ (y 0 , y 1 ) ∈ {0, 1} to Alice. Alice privately selects a binary input z ∈ {0, 1} and accordingly decodes the two incoming messages. The decoding yields one of three possible values a ∈ {0, 1, ⊥}. The values a ∈ {0, 1} are interpreted as Alice's secret bit, whereas the value a =⊥ is associated to rejecting and discarding the round of the secret sharing protocol (after which it can be re-initiated). Specifically, the conditions for successfully completing the task are as follows. When Alice selects input z, the binary value of x z ⊕ y z determines whether the round contributes to the secret sharing or is a "discarding round". If it is a discarding round (x z ⊕y z = 0), then the round is deemed successful if a =⊥. Otherwise, if it is a secret sharing round (x z ⊕ y z = 1), then it is successful if Bob and Charlie can reconstruct Alice's output through the relation a = x z ⊕ y z , where z = z ⊕ 1. Thus, by announcing z, Alice informs Bob and Charlie which of their bits hold the shared secret. The average success probabilities in the discarding rounds and the secret sharing rounds, respectively, becomes

S discard = 1 16 z x,y: xz⊕yz=0 p(a =⊥ |x, y, z), S secret = 1 16 z x,y: xz⊕yz=1 p(a = x z ⊕ y z |x, y, z).
(2) Note that this task could be equally well formulated using only the secret sharing rounds, i.e., by removing the outcome a =⊥. However, including S discard in the analysis turns out to allow advantages to be obtained from even more weakly entangled states, while still admitting a simple interpretation. One may think of S discard as a control parameter which certifies the nonclassical nature of the secret sharing correlations. While we may consider the pair (S discard , S secret ), it is both simpler and sufficient for our purposes to instead consider the single success metric obtained by averaging the two,

S = 1 2 (S discard + S secret ) . (3)
Naturally, in order to draw meaningful conclusions, the parties must have some physical limitations. On the one hand, we are interested in the situation in which Bob and Charlie share no prior entanglement and simply send messages encoded in qubit states β x and γ y , respectively, while Alice can decode using a general quantum measurement {M a|z } a . The observed correlations are then given by

p qubit (a|x, y, z) = tr (β x ⊗ γ y )M a|z , (4)
where {M a|z } a are positive operator-valued measures for each z. On the other hand, we also consider the situation illustrated in Fig. 1, where Bob and Charlie may additionally share a twoqubit entangled state ρ and then encode their qubit messages through local quantum channels Λ B x and Λ C y , respectively. In this entanglement assisted case, the correlations are given by

p EAqubit (a|x, y, z) = tr Λ B x ⊗ Λ C y (ρ) M a|z . (5)
We now show that there exists an entanglement-assisted quantum protocol that simultaneously achieves a perfect discarding rate and a perfect secret sharing rate, while requiring no more sophisticated measurements than those compatible with standard linear optics.

Ideal entanglement-assisted and unassisted protocols.-Consider a maximally entangled state ρ = |φ + φ + | and let Bob's and Charlie's local channels correspond to implementing the four unitaries U B x = σ x0 X σ x1 Z and U C y = σ y0 X σ y1 Z respectively, where σ X and σ Z are the Pauli bit-flip and phase-flip operators. This means that the two-qubit state arriving to Alice is one of the four Bell states. When z = 0, Alice performs a three-outcome measurement corresponding to a projection onto {Ψ + , Ψ -, Φ + + Φ -}, i.e., she discriminates the states Ψ ± . When z = 1, she instead discriminates the states Φ -and Ψ -by projecting the two qubits onto {Φ -, Ψ -, Φ + + Ψ + }. A direct calculation then shows that S discard = S secret = 1 and hence S EAqubit = 1. Importantly, the employed measurements may be seen as partial Bell state analysers and it is known that passive linear optics can discriminate no more than two of the four Bell states [START_REF] Calsamiglia | Maximum efficiency of a linear-optical Bell-state analyzer[END_REF]. This is indeed the case in our quantum protocol.

In contrast, in the scenario when entanglement is absent, it is no longer possible to succeed deterministically. Indeed, we were able to determine the maximum value of S achievable with entanglement-unassisted qubits and general projective measurements for Alice. To this end, we used a straightforward modification of a hierarchy of semidefinite programs for bounding dimensionally-restricted quantum correlation [START_REF] Navascués | Bounding the set of finite di-mensional quantum correlations[END_REF]. In order to obtain sufficiently tight upper bounds on S by solving numerically the semidefinite programs, we needed to consider terms appears in the first three levels of this hierarchy (the full third level being too computationally difficult to implement) [START_REF]Specifically, we used what would[END_REF]. To render these intensive calculations more readily tractable on a standard desktop computer, we employed recently developed symmetrisation techniques [START_REF] Tavakoli | Enabling computation of correlation bounds for finite-dimensional quantum systems via symmetrization[END_REF]. The symmetrisation is based on the observation that the objective function S remains invariant under i) simultaneous bit-flip of x 0 and y 0 , ii) simultaneous bit-flips of x 1 and y 1 and iii) simultaneous swaps x 0 ↔ x 1 and y 0 ↔ y 1 . Up to solver precision, we determine the upper bound S qubit ≤ 5 8 . This bound in fact holds even if all three parties share some pre-agreed classical randomness.

Although the approach described provides only an upper bound on S qubit , we demonstrate its tightness by exhibiting a strategy-in fact, an entirely classical one-which saturates it, thereby showing that there exists no quantum-over-classical advantage without using entanglement. To this end, consider the strategy in which Bob and Charlie relay the bits x 0 and y 0 , respectively, to Alice. Alice then outputs a =⊥ unless z = 0 and she receives (0, 1) or (1, 0), in which case she simply outputs a = 0. A direct calculation gives S discard = 1 2 and S discard = 1 8 , giving an overall winning probability of S classical = 5 8 . We conclude that any value in the range 5 8 < S ≤ 1 implies an advantage over both the best classical and the best entanglement-unassisted quantum model, and hence is powered by the consumption of entanglement.

Advantage from unsteerable states.-If, in the above ideal entanglement-assisted protocol, we substitute the maximally entangled state for the isotropic state (1), we find that S = 3+5v 8 Thus, we observe an advantage ascribed to entanglement whenever S > S qubit , which occurs whenever v > 2 5 . Consequently, when v ∈ 2 5 , 1 2 , the isotropic state is both unsteerable and a communication resource with the employed decoding resources.

Bell-local state and product measurements.-While the partial Bell state measurements used above are compatible with passive linear optics, they still require Alice to perform precise two-photon interferences. This can become costly to scale to many qubits. It motivates the question of whether the simplest conceivable joint measurements can also reveal entanglementassisted advantages [START_REF] Piveteau | Entanglement-assisted quantum communication with simple measurements[END_REF]. This entails separately measuring each of the two qubits and then post-processing the two separate outcomes into the final output a. As we now show, such simple measurements can nonetheless still lead to an advantage from a relatively weak form of entanglement, namely that which is too weak to violate any Bell inequality but still strong enough to manifest Einstein-Podolsky-Rosen steering.

To this end, consider once more the isotropic state (1) and the unitaries U which outperforms the unassisted qubit limit whenever v > 2 3 . Hence, when v ∈ 2 3 , 0.6875 , the isotropic state is both Bell-local and a communication resource even when using only such minimal decoding resources.

Experimental realisation.-We report here on an experimental implementation of the entanglement-assisted commu- The unitaries of Bob and Charlie are implemented using a combination of quarter wave plate (QWP), half wave plates (HWP) and phase shifters (PS). Alice's partial Bell state measurements are implemented using HWP, polarising beam splitters (PBS), beam splitters (BS) and then detected by single photon detectors (DET). See main text for further details.

nication advantage in stochastic secret sharing developed in this letter. We give a proof-of-principle demonstration of both the ideal quantum protocol using maximally entangled states encoded in the polarisation of photos, and an advantage from a weakly entangled, unsteerable, state.

To generate the entangled states, ultraviolet light centered at a wavelength of 390 nm is focused onto two 2 mm thick β barium borate (BBO) nonlinear crystals placed in an interferometric configuration to produce photon pairs emitted into two spatial modes through the second order degenerate type-I spontaneous parametric down-conversion process (SPDC). The spectral, and temporal distinguishability between the down-converted photons is carefully removed by passing through narrow-bandwidth interference filters and quartz wedges respectively (see Fig. 2).

To prepare the desired states for the secret sharing protocol, we first prepared the polarisation entangled pairs of photons in the state |φ

+ = |HH +|V V √ 2
, where H and V are, respectively, the horizontal and vertical photonic polarisation modes. By taking the standard encoding of |0 := |H and |1 := |V , we recover the desired maximally entangled state. In order to prepare an isotropic state, ρ v , |φ + must be mixed with white noise. We achieve this by, with probability 1 -v, randomly transforming |φ + into one of the four Bell states |φ ± , |ψ ± . These transformations were experimentally realised by a motorised rotation of two quarter wave plates (QWP) placed in each of the modes (a) and (b) (see Fig. 2 and Supplemental Material for the QWP settings).

The polarisation measurements are performed using half wave plates (HWP) and polarising beam splitters (PBS), beam splitters (BS) placed at the two output modes of the PBS, and then by single photon detectors (actively quenched Siavalanche photodiodes). The partial Bell analyser is implemented through two-photon interference, using PBS and HWPs set at 22.5 • . The two-photon Hong-Ou-Mandel dip visibility is 99% ± 4, where the substantial statistical error is due to the low two-photon coincidence rate used in the experiment (one per second) and a measurement time of 2400 seconds per point (see Supplemental Material). To switch from a Bell measurement discriminating between {Ψ + , Ψ -, Φ + +Φ -} to {Φ -, Ψ -, Φ + +Ψ + }, Alice uses three HWPs placed before the first PBS (see Supplemental Material the HWP settings). All single-detection events were registered using a VHDL-programmed multichannel coincidence logic unit, with a time coincidence window of 1.7 ns.

Bob's and Charlie's transformations U B x and U C y , respectively, are performed using two quarter wave plates (QWP), two half wave plates (HWP) and a phase shifter (PS) to change the relative phase between the two modes (a) and (b). Since they have a total of 16 settings but only four qualitatively different global operations on the states we consider, we have considered a simplified setting in which only the latter cases are realised (see Supplemental Material for details and HWP and PS settings).

To perform state tomography of isotropic states ρ v for different v, we performed tomography of the four Bell states generated in the randomisation procedure. Measurements were made at a rate of one two-coincidence per second over 1400 seconds for each of the nine settings needed to perform the state tomography for each Bell state. These results were then recombined a posteriori at different ratios to establish the density matrices ρ v . We considered the reconstructed states ρ v for v = 0.4 to v = 0.5, with a step-size of 0.01, corresponding to the resourceful but unsteerable range. Naturally, the reconstructed density matrices are not exactly isotropic states. To ensure the unsteerability of the experimentally realised states, we used the linear programming method of Ref. [START_REF] Nguyen | Geometry of Einstein-Podolsky-Rosen correlations[END_REF] which allows one to obtain a certificate of unsteerability for an arbitrary two-qubit state. We chose to proceed during the experiment with v = 0.47, as this represented a good balance between being below the steering threshold of v unsteer = 1 2 while allowing for good enough statistics to show a significant quantum advantage. The fidelity of the reconstructed state with the target isotropic state for v = 0.47 is 0.9983 ± 0.0004. Detailed tomography results for the density matrices, the state fidelities, and the certificates of unsteerability are presented in the Supplemental Material.

The protocol was then carried out with the isotropic state at v = 0.47 with the noise added by randomly changing the Bell state between each two-photon coincidence while maintaining the necessary ratio between these states. To obtain at most one event per change of Bell state and thus ensure the randomness and therefore unpredictability of each event, we chose to work at a rate one two-photon detection coincidence per second. The effective measurement time per setting was 2.8 hours. We obtained a success probability of S = 0.655 ± 0.003 which significantly goes beyond the theoretical entanglementunassisted limit of S qubit ≤ 5 8 , hence showing an advantage from unsteerable states.

The same experiment was also performed for a maximally entangled state to realise the ideal protocol and show that Bob's and Charlie's unit rotations coupled with Alice's measurements FIG. 3. Nonlocal properties (red) and quantum communication avantages (blue) of the two-qubit isotropic state [START_REF] Bennett | Entanglement-assisted classical capacity of noisy quantum channels[END_REF]. The ranges for separability and unsteerability are tight, whereas the limit for Bell-local models is a lower bound [START_REF] Designolle | Improved local models and new Bell inequalities via Frank-Wolfe algorithms[END_REF]. The range for quantum communication advantages from general entangled measurements follows from dense coding-like protocols [START_REF] Tavakoli | Semi-device-independent characterisation of multipartite entangled states and measurements[END_REF][START_REF] Moreno | Semi-device-independent certification of entanglement in superdense coding[END_REF] and coincides with the range in which the state is entangled. The ranges shown for quantum communication advantages with passive linear optics measurements and product measurements are established in this work and are (potentially suboptimal) upper bounds on the critical visibilities.

give the expected results. This amounts to effectively setting v = 1 and thus no randomisation over Bell states was required. This allowed the experiment to be performed at an average rate of 800 two-photon detection coincidences per second and a measurement time per setting of 2 hours. The fidelity of the prepared state and the maximally entangled state |φ + was measured to be 0.9947 ± 0.0009. We observed a success probability of S = 0.9748 ± 0.0001 in the secret sharing protocol, close to the ideal maximum value of S = 1 and showing a large advantage due to entanglement.

Discussion.

-In this letter we demonstrated theoretically, and confirmed experimentally, that one can obtain quantum communication advantages using weakly entangled unsteerable states in a quantum secret sharing task. The advantage is rendered experimentally accessible by its achievability with passive linear optics, notably exploiting a Bell state analyser. Figure 3 summarises the ranges of noise v for which quantum communication advantages are achievable with two-qubit isotropic states and the relation to the nonlocal properties of the isotropic states. By focusing on easily realisable measurements, we showed that advantages can be obtained using partial Bell state measurements when v ≥ 2/5 and hence for some unsteerable isotropic states, and even using the simplest product measurements when v ≥ 2/3, a range that still includes some Bell-local states. It remains an open question whether advantages can be obtained in both these cases for smaller visibilities. In the former case, our bound is tight for the task at hand so an advantage would require a different task, whereas in the latter case it remains open whether v = 2/3 is tight for product measurements. Of particular relevance is to investigate whether product measurements also can generate advantages from highly noisy multi-qubit states, as this would pave the way for experiments that go beyond proof-of-principle demonstrations.

We finish by noting that, while we focused on the communication advantage itself, obtaining a sufficiently high success probability in the task can also be interpreted as a semi-deviceindependent certification of the entanglement in the shared resource state in the spirit of [START_REF] Tavakoli | Semi-device-independent characterisation of multipartite entangled states and measurements[END_REF]. Moreover, if there is a strict separation between the critical visibilities for product and more general measurements in this task, a sufficiently high success probability in the task would also certify the succesful implementation of an entangled measurement in a semi-device-independent manner. generation of Werner states, to set the orientation of the wave-plates with a repeatably high precision of 0.02 • . The use of different settings (x, y) induces a systematic error, which we estimate using Monte Carlo simulation. We assume that the wave plates' setting errors are normally distributed with a standard deviation of 0.02 • . Alice's wave plates are not motorised. In order to reduce systematic errors in the angle settings, all settings for z = 0 are measured, then all settings for z = 1, so the results are initially presented separately for each case. We estimate the accuracy of the angles of these WPs to be half a degree. This, together with the Poissonian error in photon counting statistics, comprises the final error reported here. Due to inefficiency in the single photon detectors, the photons are detected randomly and their counting is Poissonian. To decrease Poissonian counting error, we have chosen for v = 1 a measurement time of two hours for every setting and collected approximately 93 million events.

To guarantee that both parties receive single qubits, we worked at a low rate (≈ 800 coincidence per sec) to suppress higher order coincidences to almost 0.9 per sec. For v = 0.47, we have collected around one hundred thousand two-photon coincidences and the low rate renders the higher order coincidences completely negligible.

SUPPLEMENTARY NOTE 6: TWO-FOLD HONG-OU-MANDEL DIP VISIBILITY

Bell state measurements are implemented through two-photon interference, using PBS and HWP plates set at 22.5°. The photons are detected by Si avalanche photodiodes and the coincidences are registered with an eight channel multifold coincidence counting unit. This Bell analyser consists of coherent interference at a polarising beam splitter (PBS). To achieve the necessary indistinguishability of the photons, due to their arrival times, we adjusted the path length of one of the photons using a delay line [START_REF] Hong | Measurement of subpicosecond time intervals between two photons by interference[END_REF]. In Figure 4, the coincidences between the detectors versus the delay path length is shown. The zero delay corresponds to a maximal overlap (maximum indistinguishability). The interfering photons bunch (i.e., they both exit in the same output arm of the PBS) causing the coincidences to vanish. The measured visibility of the two-fold Hong-Ou-Mandel dip is 99% ± 4, with a coincidence rate outside the dip around 0.9224 coincidences per second, identical to the rate at which we performed the experiment for v = 0.47.

SUPPLEMENTARY NOTE 7: DENSITY MATRIX TOMOGRAPHY

To ensure that the prepared state is the desired one, a tomography measurement is performed for a perfect Φ + state and for isotropic states ρ v with a visibility between v = 0.4 and v = 0.5. The reconstructed density matrices obtained for the cases of v = 1 (i.e., the ideal case of ρ 1 = Φ + ) and v = 0.47, which are the two states for which the experimental demonstration of the protocol was carried out, are presented in Figures 5 and6, respectively.

The fidelity of each state obtained during the tomography with the ideal isotropic state ρ v was found to be between 0.9947 and 0.9983. Table VIII gives the fidelity for each of these states.

  B x and U C y for Bob and Charlie. Now, let Alice perform product measurements σ Z ⊗ σ Z for z = 0 and σ X ⊗ σ X for z = 1. The individual qubit measurements yield binary outcomes b B , b C ∈ {0, 1}. Alice then decides her final outcome by a classical post-processing where she assigns (b B , b C ) ∈ {(0, 0), (1, 1)} to a =⊥ and (b B , b C ) ∈ {(0, 1), (1, 0)} to a = 1. Evaluating this strategy gives S prod = 3(1+v) 8

FIG. 2 .

 2 FIG. 2. Experimental setup. Entangled photon pairs in spatial modes (a) and (b) are generated through the SPDC process. Isotropic states are prepared by randomly transforming the maximally entangled |φ + state into one of the other Bell states using quarter wave plates (QWP).The unitaries of Bob and Charlie are implemented using a combination of quarter wave plate (QWP), half wave plates (HWP) and phase shifters (PS). Alice's partial Bell state measurements are implemented using HWP, polarising beam splitters (PBS), beam splitters (BS) and then detected by single photon detectors (DET). See main text for further details.

FIG. 4 .

 4 FIG. 4. Two-fold Hong-Ou-Mandel dip. The plot displays the two-fold photon counting coincidences versus the delay (the path difference between the two arms). The error bars indicate the Poissonian photon counting error statistics. The data is fitted with a Gaussian curve.

Adv. entangled measurements Adv. linear optics measurements Adv. product measurements

	Separable
	Unsteerable
	Bell-local
							𝒗
	0	1/3	2/5	1/2	2/3	0.6875	1

TABLE VII .

 VII Theoretical and experimental success probabilities for the isotropic state ρv with v = 0.47, conditioned on z and then averaged to give the overall success probability S.

	Parameter	Theoretical	Experimental	error
	S discard [z=0] S [z=0] secret	0.735 0.6025	0.743 0.591	0.005 0.006
	S S	[z=1] discard [z=1] secret	0.735 0.6025	0.712 0.576	0.005 0.006
	S		0.66875	0.655	0.003

ACKNOWLEDGMENTS

A.A.A. and A.T. thank Anthony Martin and Alek Lagarrigue for discussions on an early iteration of the task presented here. This work supported by the Swedish research council, the Wenner-Gren Foundation and by the Knut and Alice Wallenberg Foundation through the Wallenberg Center for Quantum Technology (WACQT).

SUPPLEMENTARY NOTE 1: UNSTEERABILITY OF EXPERIMENTAL ISOTROPIC STATES

Recall that the isotropic state ρ v is defined as

where

It is well known that ρ v is unsteerable for v ≤ 1 2 and entangled for v > 1 3 [START_REF] Wiseman | Steering, entanglement, nonlocality, and the Einstein-Podolsky-Rosen paradox[END_REF]. The states prepared experimentally, however, are necessarily not exactly the isotropic states. To exhibit an advantage in the stochastic secret sharing task from entangled but unsteerable states, it is necessary to verify that the experimentally prepared states, as reconstructed via state tomography, is indeed unsteerable.

While two-qubit entanglement can easily be verified by calculating its negativity, verifying unsteerability is more challenging. Here, we use a recently developed method [START_REF] Nguyen | Geometry of Einstein-Podolsky-Rosen correlations[END_REF] that allows one to obtain certificates of unsteerability via linear programming.

In particular, the approach of Ref. [START_REF] Nguyen | Geometry of Einstein-Podolsky-Rosen correlations[END_REF] computes, for an arbitrary two-qubit state and a given direction of steering (say, from Bob to Charlie or Charlie to Bob), both upper and lower bounds, r upper crit and r lower crit , respectively, on a "critical radius" r crit satisfying r lower crit < r crit < r upper crit . They show that a two-qubit state ρ is steerable (in a given direction) if and only if r crit < 1. Thus, by finding r lower crit ≥ 1 for both possible directions of steering, we are guaranteed a state is unsteerable. The precision of the upper and lower bounds can be increased at the cost of computational time and memory. On a standard desktop computer, using the CPLEX packaged provided by Ref. [START_REF] Nguyen | Geometry of Einstein-Podolsky-Rosen correlations[END_REF], we were typically able to obtain, for the experimental states approximating ρ v , a gap r upper crit -r lower crit ≈ 0.02 with a few minutes of computation. This can be reduced by an order of magnitude by allocating somewhat more computational time.

For the tomographically reconstructed density matrices ρv , for v = 0.4 to v = 0.5 (with a step size of 0.1), we computed r lower crit and found the states to be unsteerable for all values except v = 0.5. However, to ensure that this result is robust (recalling, of course, the experimental error in the description of ρv ) and ensure we choose a state providing a large enough quantum advantage (which diminishes as v decreases), we chose to perform the experiment with v = 0.47.

For this state, we performed a finer analysis of the unsteerability of ρ0.47 , taking into account the errors in each term of the density matrix. Adopting a conservative approach, we applied the worst case errors to all subsets of density matrix elements (up to Hermiticity) before calculating r lower crit for the (renormalised) perturbed density matrix. We found that, in all cases, r lower crit 1.031 (while for the reconstructed state ρ0.47 we found r lower crit ≈ 1.055), showing that the prepared state is clearly unsteerable, even when errors are taken into account.

SUPPLEMENTARY NOTE 2: STATE PREPARATION, TRANSFORMATIONS AND MEASUREMENTS 1. State implementation

The setup used generates an entangled Φ + state with a visibility close to 100%. In order to generate an isotropic state ρ v [START_REF] Brunner | Connection between Bell nonlocality and Bayesian game theory[END_REF], this state needs to be mixed with probability (1 -v) with the maximally mixed state 1 4 . We implement the maximally mixed state by randomly switching between the four Bell states, Φ ± , Ψ ± , exploiting the fact that 1

We can thereby generate ρ v for any v by choosing correctly the weights with which we randomly change Φ + to one of the other Bell states.

The switching between these different states is done with the help of four quarter-wave plates and a phase plate (which is not necessary in theory, but which we use to compensate for the phase that rotating QWPs can bring); see Fig. 2 of the main text.

These plates are distributed so that two are placed on the path going to Bob (mode (a)), 4 . These plates are motorised and switched randomly (with the desired probability) from one state to another every second. The number of coincidences being less than one per second, the result is to switch from one Bell state to another randomly and independently between each pair of successive coincidences. The values of the angles allowing one to pass from one state to another are given in Table I.

Bob's and Charlie's rotations

As described in the main text, on inputs x = (x 0 , x 1 ) ∈ {0, 1} 2 and y = (y 0 , y 1 ) ∈ {0, 1} 2 , Bob and Charlie apply the unitaries U B x = σ y0 X σ y1 Z and U C y = σ y0 X σ y1 Z , respectively. Table II summarises explicitly the operations that Bob and Charlie must perform in inputs x and y, as well as the Bell state expected to be received by Alice in the ideal case (assuming Bob and Charlie share a Φ + state). The settings to be applied to achieve each of the four different possible rotations are given in Table III Experimentally, the phase of a single photon cannot be determined, and only the relative phase between two photons can be measured. Moreover, as a QWP at 90 • is equivalent to a QWP at 0 • plus a phase, we have chosen to keep them at 0 • in order to limit the number of rotations and therefore the systematic error and to compensate with the phase plate. In addition, to limit motor problems due to fast setting changes (once per second), we decided to limit rotations as much as possible and therefore to apply the equivalent global operations. The rotations actually performed by the two HWPs, the two QWPs and a phase shifter, as described in the main text (see Fig. 2 therein), are given in Table IV. TABLE IV. Experimental rotations: For each initial Bell state that was prepared, the table gives the actual equivalent rotation that was (locally) performed so as to implement the global operations equivalent to the transformations specified in Table II.

State prepared

Equivalent rotation

Alice's measurements

Depending of her setting z ∈ {0, 1}, Alice performs one of the three-outcome partial Bell measurement {Ψ + , Ψ -, Φ + + Φ -} or {Φ -, Ψ -, Φ + + Ψ + } and obtains an outcome c ∈ {0, 1, ⊥}. To make these measurements, she uses three HWPs λ A1 2 , λ A2 2 and λ A3 2 , with the first two (A1 and A2) placed before one of the input ports of the PBS, and last one (A3) placed before the other input port (see Fig. 2 of the main text). The values of the angles of these HWPs are given in Table V. Tables VI and VII list our experimental results alongside the theoretical success probabilities S with the related errors for each combination of inputs (x, y, z). To calculate these errors, following [START_REF] Anwer | Experimental characterization of unsharp qubit observables and sequential measurement incompatibility via quantum random access codes[END_REF] we consider errors originating from the measurement side only. To reduce experimental errors in the measurements, we used computer controlled high precision motorised rotation stages, for Bob, Charlie and the


Julien Trevisan

Limit laws in the lattice problem. V. The case of analytic and stricly convex sets

établissements d'enseignement et de recherche français ou étrangers, des laboratoires

Introduction

The lattice point problem is an open problem in the Geometry of Numbers, at least since Carl Friedrich Gauss took interest in which became the famous Gauss circle problem. The general problem states as followed. Let d be an integer greater than 1. We recall the following definition :

Definition 1. A subset L of R d is a lattice if it is a subgroup of R d such that L is discrete and span(L) = R d .
Let P be a measurable subset of R d of non-zero finite Lebesgue measure. We want to evaluate the following cardinal number when t → ∞ :

N (tP + X, L) = |(tP + X) ∩ L|
where X ∈ R d , L is a lattice of R d and tP +X denotes the set P dilated by a factor t relatively to 0 and then translated by the vector X. Under mild regularity conditions on the set P , one can show that :

N (tP + X, L) = t d Vol(P ) Covol(L) + o(t d )
where o(f (t)) denotes a quantity such that, when divided by f (t), it goes to 0 when t → ∞ and where Covol(L) is defined in the following definition :

Definition 2. The covolume of a lattice L of R d , covol(L), is the Lebesgue measure of a measurable fundamental set of L. Furthermore, a lattice is said to be unimodular if its covolume is equal to 1. When d = 2, instead of using the term covolume, we use the term coarea.

We are interested in the error term R(tP + X, L) = N (tP + X, L) -t d Vol(P ) Covol(L) .

In the case where d = 2 and where P is the unit disk D 2 , Hardy's conjecture in [START_REF] Hardy | The average order of the arithmetical functions p (x) and δ (x)[END_REF] stipulates that we should have for all ǫ > 0,

R(tD 2 , Z 2 ) = O(t 1 2 +ǫ ).
One of the result in this direction has been established by Iwaniec and Mozzochi in [START_REF] Iwaniec | On the divisor and circle problems[END_REF]. They have proven that for all ǫ > 0, R(tD 2 , Z 2 ) = O(t 7 11 +ǫ ). This result has been recently improved by Huxley in [START_REF] Martin | Exponential sums and lattice points iii[END_REF]. Indeed, he has proven that : R(tD 2 , Z 2 ) = O(t K log(t) Λ ) where K = 131 208 and Λ = 18627 8320 . In dimension 3, Heath-Brown has proven in [START_REF] Heath-Brown | Lattice points in the sphere[END_REF] that : R(tD 3 , Z 3 ) = O(t 21 16 +ǫ ).

These last two results are all based on estimating what are called exponential sums. Another approach was followed first by Heath-Brown in [START_REF] Heath-Brown | The distribution and moments of the error term in the dirichlet divisor problem[END_REF] and then by Bleher, Cheng, Dyson and Lebowitz in [START_REF] Pavel | Distribution of the error term for the number of lattice points inside a shifted circle[END_REF]. They took interest in the case where the dilatation parameter t is random. More precisely, they assumed that t was being distributed according to the measure ρ( t T )dt (that is absolutely continuous relatively to Lebesgue measure) and where ρ is a probability density on [0, 1] and T is parameter that goes to infinity. In that case, Bleher, Cheng, Dyson and Lebowitz showed the following result (which generalizes the result of Heath-Brown) :

Theorem ([3]). Let α ∈ [0, 1[ 2 . There exists a probability density p α on R such that for every piecewise continuous and bounded function g : R -→ R,

lim T →∞ 1 T T 0 g( R(tD 2 + α, Z 2 ) √ t )ρ( t T )dt = R g(x)p α (x)dx.
Furthermore p α can be extended as an analytic function over C and verifies that for every ǫ > 0, p α (x) = O(e -|x| 4-ǫ ) when x ∈ R and when |x| → ∞.

In our case, we keep t deterministic as in the original Gauss problem but we let the lattice L be a random unimodular lattice and we study R. This approach was first initiated by Kesten in [START_REF] Kesten | Uniform distribution mod 1[END_REF] and in [START_REF] Kesten | Uniform distribution mod 1 (ii)[END_REF]. It should be noted that several counting problems have followed this approach : we can cite, for example, [START_REF] Björklund | Counting in generic lattices and higher rank actions[END_REF], [START_REF] Dolgopyat | Deviations of ergodic sums for toral translations i. convex bodies[END_REF], [START_REF] Marklof | The n-point correlations between values of a linear form[END_REF], [?], [?] and [START_REF] Vinogradov | Limiting distribution of visits of sereval rotations to shrinking intervals[END_REF]. We denote by S 2 the space of unimodular lattices and it can be seen as the quotient space SL 2 (R)/SL 2 (Z). We denote by µ 2 the unique Haar probability measure on it. Let us set :

Π = {(k 1 , k 2 ) ∈ Z 2 | k 1 ∧ k 2 = 1, k 1 0} (1)
where we agree that if k 1 = 0, k 1 ∧ k 2 = 1 means that k 2 = 1. We denote by • the usual euclidean norm over R 2 . We need to define some additional objects.

Definition 3.

For every i ∈ {1, 2}, we call

L i = min{r > 0 | B f (0, r) contains i vectors of L lineary independent}
where B f (0, r) is the closed centred ball on 0 for the norm • of radius r.

In fact, for almost all L ∈ S 2 , L 2 > L 1 and there exists only one couple of vectors (e 1 (L), e 2 (L)) such that (e 1 (L)) 1 > 0, e 1 (L) = L 1 , (e 2 (L)) 1 > 0 and e 2 (L) = L 2 .

In the rest of the article, for the sake of simplicity, for L a lattice, we will use the notation L instead of L 1 .

For a lattice L ∈ S 2 , we also say that a vector of L is prime if it is not a non-trivial integer multiple of another vector of L.

In fact, for every M ∈ SL 2 (R), a vector l ∈ L is prime if, and only if, M l ∈ M L is prime and a vector (k 1 , k 2 ) ∈ Z 2 is prime if, and only if, k 1 ∧ k 2 = 1. With these notations, one has that, for a generic lattice L ∈ S 2 , e ∈ L is prime if, and only if, e can be written as e = k 1 e 1 (L) + k 2 e 2 (L) with

k 1 ∧ k 2 = 1.
Finally, for a generic lattice L ∈ S 2 , we call P + (L) the set of vectors e of L such that e = k 1 e 1 (L) + k 2 e 2 (L) with (k 1 , k 2 ) ∈ Π. All the vectors of P + (L) are prime vectors according to the previous remark.

We recall also the fact that we say that a real random variable Z is symmetrical if P Z = P -Z where, for every random variable X, P X stands for the law of the random variable X. Let μ2 be a probability measure that has a smooth bounded density σ with respect to µ 2 . There are two different cases that are addressed in our main result, which is Theorem 1. The first one is when μ2 is compactly supported, id est when there exists α > 0 such that :

μ2 ({L ∈ S 2 | L 1 < α}) = 0. (2)
The second one is when μ2 is non-compactly supported under the following condition : there exists m > 0, there exists α > 0 such that for all L that belongs to the event (

L < α), σ(L) m. ( 3
)
An example of such a measure μ2 is given by the normalized Haar measure µ 2 .

Let E be an ellipse centred around 0. Let's call M a matrix that transforms E into a disk and that belongs to SL 2 (R). M is unique modulo the natural action of SL 2 (Z). Let's set : T ∞ = (T 1 ) Π where T 1 = R/Z and let's call λ ∞ the normalized Lebesgue measure product over T ∞ . The main result of the previous article [START_REF] Trevisan | Limit laws in the lattice problem. ii. the case of ovals[END_REF] is the following theorem :

Theorem 1. For every real numbers a < b,

lim t→∞ μ2 L ∈ S 2 | R(tE, L) √ t ∈ [a, b] = (λ ∞ ×(M -1 ) * μ2 ) (θ, L) ∈ T ∞ × S 2 | S(θ, L ⊥ ) ∈ [a, b]
where (M -1 ) * μ2 is the push-forward of μ2 by M -1 and where θ = (θ e ) ∈ T ∞ ,

S(θ, L) = 2 π e∈P + (L) φ(θ e ) e 3 2 with φ(θ e ) = m 1 cos(2πmθ e -3π 4 ) m 3 2 . ( 4
)
Furthermore, S(θ, L) (and S(θ, L ⊥ ) :

• converges almost surely

• is symmetrical and its expectation is equal to 0

• admits a moment of order 1 + κ for any 0 κ < 1 3

• S(θ, L) admits moments of all order 1 p < ∞ when μ2 is compactly supported

• does not admit a moment of order 4 3 when μ2 is non-compactly supported under the condition [START_REF] Pavel | Distribution of the error term for the number of lattice points inside a shifted circle[END_REF].

In this article, we want to extend this last result. More precisely, let's suppose that E = Ω γ where γ is analytic curve that is simple, closed and strictly convex and where 0 ∈ Ω γ . Let's also call x γ (ξ) the point on γ where the outer normal to γ coincides with γ γ and ρ γ (ξ) the curvature radius of γ at x γ (ξ). Let's set finally T ∞,2 = (T 1 ) Π × (T 1 ) Π and let's call λ ∞,2 the normalized Lebesgue measure product over T ∞,2 . Then, we want to prove the following theorem : Theorem 2. There exists a distribution function D γ (z) such that for every real z we have :

lim t→∞ μ2 R(tΩ γ , L) √ t ∈] -∞, z] = D γ (z).
In the case where

Ω γ is symmetric, D γ (z) is the distribution function of S γ (θ, L) = 2 π e∈P + (L) ρ γ (e)φ(θ e ) e 3 2
with θ = (θ e ) ∈ T ∞ being distributed according to λ ∞ and L being distributed according to μ2 . In the non symmetric case, D γ (z) is the distribution function of

S γ (θ, L) = 1 π e∈P + (L) φ γ,2 (θ e , e) e 3 2
where θ = (θ 1,e , θ 2,e ) ∈ T ∞,2 being distributed according to λ ∞,2 and L being distributed according to μ2 and with

φ γ,2 (θ e , e) = m 1 ρ γ (e) cos(2πmθ

.

(5)

Furthermore, in both cases, S γ admits the same properties of S(θ, L) listed in Theorem 1.

In fact, this theorem can be generalized as followed :

Theorem 3. For every α ∈ R 2 , there exists a distribution function D γ,α (z) such that for every real z we have :

lim t→∞ μ2 R(tΩ γ + α, L) √ t ∈] -∞, z] = D γ,α (z).
In the case where Ω γ is symmetric, D γ (z) is the distribution function of

S γ (θ, L) = 2 π e∈P + (L) ρ γ (e)φ α (θ e , e) e 3 2
with θ = (θ e ) ∈ T ∞ being distributed according to λ ∞ and L being distributed according to μ2 and where

φ α (θ e , e) = m 1 cos(2πmθ e + 2πm < α, e > -3π 4 ) m 3 2 .
In the non symmetric case, D γ,α (z) is the distribution function of

S γ (θ, L) = 1 π e∈P + (L) φ α,γ,2 (θ e , e) e 3 2
where θ = (θ 1,e , θ 2,e ) ∈ T ∞,2 being distributed according to λ ∞,2 and L being distributed according to μ2 and with

φ α,γ,2 (θ e , e) = m 1 ρ γ (e) cos(2πmθ 1,e + 2πm < α, e > -3π 4 ) + ρ γ (-e) cos(2πmθ 2,e -2πm < α, e > -3π 4 ) m 3 2
. Furthermore, in both cases, S γ admits the same properties of S(θ, L) listed in Theorem 1. This last theorem is a generalization of Theorem 1 on two planes. The first one is the shape of the sets : it treats the more general case of analytic curves, not only the case of the ellipses centred on 0. The second one is the presence of a translation parameter α whereas Theorem 1 can be deduced by assuming that α = 0 (no translation of tE). In the next section, we give a brief heuristic explanation of the approach that we will follow. It is basically the same as in [START_REF] Trevisan | Limit laws in the lattice problem. ii. the case of ovals[END_REF]. At the end of the section, we will give the plan of the paper.

Heuristic explanation and plan of the proof

First, let's explain the different steps of the proof of Theorem 2, Theorem 3 being a simple generalization of Theorem 2. First step. By regularizing the problem and using the Poisson summation formula, we are going to show that the quantity S A,prime (L, t), for A > 0 a fixed parameter that is taken large enough, when t → ∞, is close, in probability, to

R(tΩγ ,L) √ t
where S A,prime (L, t) is defined by :

S A,prime (L, t) = 1 π l∈L ⊥ prime 0< l A 1 l 3 2 m∈N-{0} ρ γ (l) cos(2πtmY γ (l) -3π 4 ) + ρ γ (-l) cos(2πtmY γ (-l) -3π 4 ) m 3 2 (6)
where Y γ (ξ), for all ξ ∈ R 2 -{0}, is defined by

Y γ (ξ) =< ξ, x γ (ξ) >
(it is a positive homogeneous function of order 1). Let's say a few remarks. First, the vectors l prime considered are such that l A. We have to limit the norm of the considered vectors because of a convergence problem. Second, we take into account a phenomenon of multiplicity (if l appears in the sum, 2l is also going to appear). In [START_REF] Pavel | Distribution of the error term for the number of lattice points inside a shifted circle[END_REF] such a phenomenon was also taken into account. This was done in order to get independence at infinity, as it was also done in [START_REF] Dolgopyat | Deviations of ergodic sums for toral translations ii[END_REF], and we do that for the same goal. By using the remark that is in Definition 3 and by replacing L by L ⊥ (which is done only for a matter of convenience), one has that :

S A,prime (L ⊥ , t) = 1 π k 1 ∧k 2 =1 k 1 0 k 1 e 1 (L)+k 2 e 2 (L) A 1 k 1 e 1 (L) + k 2 e 2 (L) 3 2 (7) φ γ,2 ((tY γ (k 1 e 1 (L) + k 2 e 2 (L)), tY γ (-(k 1 e 1 (L) + k 2 e 2 (L)))), k 1 e 1 (L) + k 2 e 2 (L))
where the function φ γ,2 was defined by the Equation [START_REF] Dolgopyat | Deviations of ergodic sums for toral translations i. convex bodies[END_REF]. We have done that so from this stage onwards we consider vectors of P + (L) with a fixed indexation (that does not depend on L). Furthermore, this indexation will be very useful for the second step (for more details, see Section 4).

Second step. In the non symmetric case, we will show that the family of variables (tY γ (k 1 e 1 (L)+ k 2 e 2 (L)), tY γ (-(k 1 e 1 (L) + k 2 e 2 (L)))), whose values are in (R/Z) 2 , become, when t → ∞, independent from one another and indeed converge towards independent and identically distributed random variables whose common distribution is given by the normalized Haar measure on (R/Z) 2 . The idea here is basically the same as in [START_REF] Pavel | Distribution of the error term for the number of lattice points inside a shifted circle[END_REF] and in [START_REF] Heath-Brown | The distribution and moments of the error term in the dirichlet divisor problem[END_REF] where the respective authors used the fact that the square roots of square free integers are Z-free. It is a generalization of what was done in [START_REF] Trevisan | Limit laws in the lattice problem. ii. the case of ovals[END_REF]. In our case, to prove the result, we will decompose the space of unimodular lattices into small geodesic segments, calculate the Taylor series of Y γ (k 1 e 1 (L) + k 2 e 2 (L)) and of Y γ (-(k 1 e 1 (L) + k 2 e 2 (L))) at order 1 on such a segment and show that the coefficients of order 1 are Z-free. We will also prove that these variables become independent, when t → ∞, from the variable L due to the presence of the factor t. In the symmetric case, instead of considering (tY γ (k 1 e 1 (L)+k 2 e 2 (L)), tY γ (-(k 1 e 1 (L)+k 2 e 2 (L)))), we consider the family of variables ((tY γ (k 1 e 1 (L) + k 2 e 2 (L))), whose values are in (R/Z), and show the same results.

Third step. Thanks to the first and second step, we will see that the asymptotic distribution of

R(tΩγ ,L) √ t
is the distribution of S γ (θ, L) (see Theorem 2) under the assumption that the quantity S(θ, L) is well-defined. This last fact will be quasi immediate because what was done in the last section of [START_REF] Trevisan | Limit laws in the lattice problem. ii. the case of ovals[END_REF] can be generalized directly in our case. Furthermore, all the listed properties of S γ (θ, L) are also going to be obtained immediately. After doing all of that, we will finally get the validity of Theorem 2.

Plan of the paper. The next section will be dedicated to deal with the first step of the proof, namely it will show that

R(tΩγ ,L) √ t
is close in probability with S A,prime (L ⊥ , t) when A is a fixed parameter taken large enough and t goes to infinity (see Proposition 1). We have to "cut" the sum because of the problem of convergence of the Fourier series of X -→ which is due to the lack of regularity of the indicator function 1 tΩγ . To prove this, we are going to proceed by regularization which means here that we are going to smooth the indicator function 1 tΩγ via a Gaussian kernel. In Section 4, we tackle the second step of the proof, that is the fact, in the non symmetric case, that the (tY γ (k 1 e 1 (L)+ k 2 e 2 (L), tY γ (-(k 1 e 1 (L)+ k 2 e 2 (L))) become independent when t → ∞.

We also show that they converge towards random variables that are identically distributed according to the normalized Haar measure over (R/Z) 2 and that they become independent, when t → ∞, from L and so that

R(tΩγ ,L) √ t
has the same distribution of S γ (θ, L). We also deal with the symmetric case, which is simpler. In Section 5 we are going to tackle the third step of the proof, namely study the convergence of SA (ω, L) when A → ∞ and the existence of moments of its limit. In Section 6, which is the last section, we give the approach, based on the approach to prove Theorem 2, to prove Theorem 3. In the rest of the article, all the calculus of expectation E, of variance Var and of probability P will be made according to the measure μ2 . Furthermore, the expression typical is going to signify μ2almost surely. In fact, like we have said in Section 1, we are going to suppose that μ2 = µ 2 in Section 3 and in Section 4 because all the results extend to the general case.

Reduction to the study of the Siegel transform

The main object of this section is to show the following proposition : Proposition 1. For every α > 0, for every A > 0 large enough, for every t large enough, one has that :

P(∆ A,prime (L, t) α) α where ∆ A,prime (L, t) = | R(tΩ γ , L) √ t -S A,prime (L, t)| . ( 8
)
This proposition basically says that we can reduce the asymptotical study of R(tΩγ ,L)

√ t
to the study of its Fourier transform, taking into account a phenomenon of multiplicity. In fact, due to the triangle inequality, we only have to prove the following two lemmas : Lemma 1. For every α > 0, for every A > 0 large enough, for every t large enough, one has that :

P(∆ A (L, t) α) α where ∆ A (L, t) = | R(tΩ γ , L) √ t -H A (L, t)| with (9) H A (L, t) = 1 2π l∈L ⊥ 0< l A 1 l 3 2 ρ γ (l) cos(2πtY γ (l) - 3π 4 ) + ρ γ (-l) cos(2πtY γ (-l) - 3π 4
) [START_REF] Iwaniec | On the divisor and circle problems[END_REF] which becomes when Ω γ is symmetric

H A (L, t) = 1 π l∈L ⊥ 0< l A ρ γ (l) cos(2πtY γ (l) -3π 4 )) l 3 2 . ( 11
)
Lemma 2. For every α > 0, for every A > 0 large enough, for every t large enough, one has that :

P(|S A,prime (L, t) -H A (L, t)| α) α.
Proof of Proposition 1. One has that :

∆ A,prime (L, t) ∆ A (L, t) + |S A,prime (L, t) -H A (L, t)|. ( 12
)
The Lemma 1 and the Lemma 2 imply then the wanted result.

Let's say a few words about Lemmas 1 and 2 before following with their respective proofs.

The Lemma 1 says that the study of

R(tΩγ ,L) √ t
can be reduced to the study of its Fourier transform. The Lemma 2 says that the phenomenon of multiplicity (the fact that for a prime vector l, 2l, 3l etc. appear in the sum H A (L, t) when A → ∞) is not so important. We only have to gather all the multiples of a prime vector (which corresponds to the infinite sum over m, see equation ( 6)), so that we focus on prime vectors.

Proof of Lemma 1

First, we are going to prove the Lemma 1. To do so, we are following closely the approach of [START_REF] Bleher | On the distribution of the number of lattice points inside a family of convex ovals[END_REF], yet with some differences because in our case it is not the radius of dilatation that is random but the lattice (or, equivalently and in a certain sense, the oval). For x ∈ R 2 and t > 0, let's define

λ(x; t) = t 2 4π e -t 2 4π x 2 and, for M ∈ SL 2 (R), λ M (x; t) = λ(M x; t). ( 13
)
We recall that :

R 2 λ M (x; t)dx = 1 ( 14
)
and that the Fourier transform of λ M (•; t) can be expressed as

λ M (ξ; t) = e -(M -1 ) T ξ 2 t 2 . ( 15
)
We introduce the following function :

χ γ,M (x; t) = (1 tΩ M -1 γ * λ M (•; t))(x) = R 2 1 tΩ M -1 γ (y)λ M (x -y; t)dy ( 16
)
(it is a regularization of the function

1 tΩ M -1 γ ).
Let us also set :

N reg (tΩ γ , M ) = n∈Z 2 χ γ,M (n; t) and (17)
(the index "reg" stands for regularized)

F (M, t) = N reg (tΩ γ , M ) -Area(tΩ γ ) √ t . ( 18
)
Let L be a unimodular lattice such that e 1 (L) and e 2 (L) are well-defined and let

M = [e 1 (L), e 2 (L)] if det([e 1 (L), e 2 (L)]) > 0 (19) and M = [e 2 (L), e 1 (L)] if det([e 2 (L), e 1 (L)]) > 0. ( 20
)
Then M is a matrix that represents L and one has immediately that :

R(tΩ γ , L) = R(tΩ M -1 γ , Z 2 ).
Now, let's call :

∆ 1 (L, t) = | R(tΩ γ , L) √ t -F (M, t)| and (∆ 2 ) A (L, t) = |F (M, t) -H A (L, t)| (21)
so one has that :

∆ A (L, t) ∆ 1 (L, t) + (∆ 2 ) A (L, t). ( 22
)
The proof of Lemma 1 lies on the two following lemmas :

Lemma 3. The quantity ∆ 1 (L, t) converges to 0 when t → ∞.
Lemma 4. For all α > 0, for all A large enough, for all t large enough,

P((∆ 2 ) A (L, t) α) α.
Proof of Lemma 1. It is the direct consequence of Equation ( 22) and of Lemma 3 and Lemma 4.

Lemma 3 basically tells us that the study of R(tΩγ ,L)

√ t
can be reduced to the study of one of its regularized Fourier series, whereas Lemma 4 means that the asymptotical study of this regularized Fourier series can be brought back to the study of the non-regularized Fourier series. The next subsubsection is dedicated to the proof of Lemma 3 and the subsubsection after it is dedicated to the proof of Lemma 4.

Proof of Lemma 3

The proof of Lemma 3 is based on two sublemmas. The first one is the following :

Sublemma 1. For all x ∈ R 2 , for all t > 0, |χ γ,M (x; t) -1 tΩ M -1 γ (x)| e -t 2 4 dist(M x,tγ) 2 where for all z ∈ R 2 , dist(z, tγ) = inf y∈tγ |z -y|.
Proof. One has that :

|χ γ,M (x; t) -1 tΩ M -1 γ (x)| = | y / ∈tΩγ t 2 4π e -M x-y 2 4 dy| if M x ∈ tΩ γ and |χ γ,M (x; t) -1 tΩ M -1 γ (x)| = | y∈tΩγ t 2 4π e -M x-y 2 4 dy| if M x / ∈ tΩ γ
because of Equation ( 14) and by making the change of variable y = M u. The proof of Lemma 3.2 from [START_REF] Bleher | On the distribution of the number of lattice points inside a family of convex ovals[END_REF] gives the wanted result.

The second sublemma gives an estimate of dist(M n, tγ). To state it, we need some notations.

Like in [START_REF] Bleher | On the distribution of the number of lattice points inside a family of convex ovals[END_REF], let the curve γ be defined in the polar coordinates (r, ϕ) by the equation

r = Γ(ϕ). (23)
Let's define :

r γ (x) = x Γ(ϕ(x)) ( 24
)
where ϕ(x) is the angular coordinate of x. Then, one has that there exists C > 0 small enough so that for every

x ∈ R 2 , dist(x, tγ) C|r γ (x) -t|. ( 25
)
We deduce the following sublemma : Sublemma 2. For all L ∈ S 2 , for all t > 0, for all n ∈ Z 2 , we have that :

dist(M n, tγ) C|r γ (M n) -t|.
Now we can prove Lemma 3.

Proof of Lemma 3. By using the Equation ( 21), we have that :

∆ 1 (L, t) 1 √ t n∈Z 2 |χ γ,M (n; t) -1 tΩ M -1 γ (n)| because, also, R(tΩ γ , L) = R(tΩ M -1 γ , Z 2
). So, Sublemma 1 and Sublemma 2 imply that :

∆ 1 (L, t) 1 √ t n∈Z 2 e -C 2 t 2 4 |rγ(M n)-t| 2 .
The essential part of the right-hand side are the terms such that n ∈ Z 2 verify that

|r γ (M n) -t| 1 t 3/4 . ( 26
)
Yet the number of n ∈ Z 2 that belong to such an annulus is of order t 1 4 . So, one has finally :

∆ 1 (L, t) = O( 1 t 1 4
).

Proof of Lemma 4

To prove Lemma 4, we first need to give another expression of F (M, t), obtained via the Poisson formula. It is the object of the following lemma :

Lemma 5. F (M, t) = 1 2π n∈Z 2 -{0} λ M (2πn; t) (M -1 ) T n 3 2 (27) ρ γ ((M -1 ) T n) cos(2πtY γ ((M -1 ) T n) - 3π 4 ) + ρ γ (-(M -1 ) T n) cos(2πtY γ (-(M -1 ) T n) - 3π 4 ) + O M (t -1 )
which becomes in the symmetric case

F (M, t) = 1 π n∈Z 2 -{0} λ M (2πn; t) (M -1 ) T n 3 2 ρ γ ((M -1 ) T n) cos(2πtY γ ((M -1 ) T n) - 3π 4 ) + O M (t -1 )
where the M in index of O M is to signal that it depends on M (or, equivalently, on the lattice L).

To prove it, we first need a calculatory sublemma :

Sublemma 3. Let γ be a simple, closed, analytic, strictly convex curve such that 0 ∈ Ω γ . Let D ∈ SL 2 (R). Let γ = Dγ.
Then one has for every ξ ∈ R2 -{0} :

x γ (ξ) = Dx γ (D T ξ), ρ γ (ξ) = ξ 3 D T ξ 3 ρ γ (D T ξ) and Y γ (ξ) = Y γ (D T ξ)
where D T is the transpose of the matrix D.

Proof. Let's call T the unit tangent vector at x γ (D T ξ) such that (D T ξ, T ) is a orthogonal and direct basis of R 2 . Then, DT is a unit tangent vector at Dx γ (D T ξ).

Let's call k the unit normal exterior vector of γ at Dx γ (D T ξ). Then, one knows that k is orthogonal to DT and (k, DT ) is direct because det(D) = 1. By property of the adjoint operator, one knows that D T k is orthogonal to T and thus there exists α ∈ R -{0} such that

D T k = αD T ξ. ( 28
)
So, one gets that :

k = αξ. ( 29
)
Yet, one has also that (k, DT ) and (D T ξ, T ) are direct and orthogonal basis of R 2 . So, one must have α > 0 and it gives us the first wanted result. Now, concerning the third equality, one has that, by definition :

Y γ (ξ) =< ξ, x γ (ξ) > (30)
So, the first equality of Sublemma 3 gives us that :

Y γ (ξ) =< ξ, Dx γ (D T ξ) > . ( 31
)
By using the adjoint property, one finds the wanted result :

Y γ (ξ) =< D T ξ, x γ (D T ξ) >= Y γ (D T ξ). ( 32
)
So, one gets the third equality. Concerning the second equality, one knows that t -→ γ(t) is a parametrization of the curve γ and that t -→ Dγ(t) is a parametrization of the curve Dγ. So, one can use these parametrizations to compute ρ γ and ρ γ . By using the fact that D ∈ SL 2 (R), one has that :

ρ γ (Dγ(t)) = ( Dγ ′ (t) γ ′ (t) )
Let's call t 0 the instant such that

γ(t 0 ) = x γ (D T ξ)
and so, according to the first result of the Sublemma 3, one has that

Dγ(t 0 ) = x γ (ξ). (34)
Let's set γ(t) = (x(t), y(t)) and Dγ(t) = (φ 1 (t), φ 2 (t)).

Then one has at the instant t = t 0 :

αRD T ξ = (x ′ (t 0 ), y ′ (t 0 )) (35) and βRξ = (φ ′ 1 (t 0 ), φ ′ 2 (t 0 )) (36) with α, β > 0 and R = 0 -1 1 0 ∈ SO 2 (R).
By using the fact that DRD T = R, one finds that :

α = β. ( 37
)
By using Equation (34), Equation (35), Equation (36), Equation (37) and by using Equation (33) at the instant t = t 0 , one has the third wanted equality.

We can now tackle the proof of Lemma 5.

Proof of Lemma 5. According to the Equation ( 18), the Poisson summation formula and because of the fact that

1 tΩ M -1 γ (0) = Area(tΩ M -1 γ
) one has that :

F (M, t) = 1 √ t n∈Z 2 -{0} 1 tΩ M -1 γ (2πn) λ M (2πn; t). ( 38
)
Yet, according to Lemma 2.1 from [START_REF] Bleher | On the distribution of the number of lattice points inside a family of convex ovals[END_REF], one has that :

1 tΩ M -1 γ (ξ) = √ t ξ -3 2 ± 2πρ M -1 γ (±ξ) exp(±i(tY M -1 γ (±ξ)- 3π 4
))+O M (t -1 2 ξ -5 2 ) (39)

By using Sublemma 3 with M -1 = D, we get with Equation (38) and by grouping the n and -n terms in the Fourier series, we get the wanted result.

Let's set ν(l, t) = ρ γ (l) cos(2πtY γ (l) - 3π 4 ). ( 40
)
Using the Equation ( 15), the Equation ( 21), the fact that if M represents a lattice L, (M -1 ) T represents the dual lattice L ⊥ , and the previous lemma, that is Lemma 5, one gets that :

(∆ 2 ) A (L, t) ∆ 2,1 (L, t) + (∆ 2,2 ) A (L, t) + (∆ 2,3 ) A (L, t) (41) where ∆ 2,1 (L, t) = O M (t -1 ) (42) (∆ 2,2 ) A (L, t) = 1 2π | l∈L ⊥ 0< l A 1 l 3 2 (ν(l, t) + ν(-l, t))(1 -e -(2π) 2 l 2 t 2 )| (43) (∆ 2,3 ) A (L, t) = 1 2π | l∈L ⊥ A< l 1 l 3 2 (ν(l, t) + ν(-l, t))e -(2π) 2 l 2 t 2 |. (44)
So, if we prove the following lemmas, we will get Lemma 4 and, in fine, get Lemma 1 : Lemma 6. ∆ 2,1 (L, t) converges almost surely to 0 when t → ∞.

Let's remark, by the way, that this last lemma is immediate according to equation (42).

Lemma 7.

For all A > 0, (∆ 2,2 ) A (L, t) converges to 0 when t → ∞.

Lemma 8. For all α > 0, for all A large enough, for all t large enough,

P((∆ 2,3 ) A (L, t) α) α.
Proof of Lemma 4. Let α > 0. Let's take A large enough so that for all t large enough,

P((∆ 2,3 ) A (L, t) α) α.
It is possible according to Lemma 8. According to Lemma 7, according to Lemma 6 and because the almost-sure convergence imply the convergence in probability, even if it means taking t larger, one can suppose that :

P((∆ 2,1 )(L, t) α) α and P((∆ 2,2 ) A (L, t) α) α.
By using equation 41, one gets the wanted result.

Before following with the proof of Lemma 1, let's say a few words about Lemma 7 and Lemma 8. The first tells us that the non-regularized Fourier series is "close" enough to the regularized Fourier series whereas the second one tells us that the large terms of the regularized Fourier series do not matter, in a certain sense, for our study.

It remains only to prove Lemma 7 and Lemma 8. Because the density of μ2 is bounded, we only need to prove these lemmas for mu 2 = µ 2 and we will make this assumption for the rest of the section. We are now going to prove Lemma 7 and Lemma 8.

Proof of Lemma 7

Proof of Lemma 7. Let l ∈ L ⊥ . Then one has :

|1 -e -(2π) 2 l 2 t 2 | (2π) 2 l 2 t 2 . ( 45
)
With this equation and with Equation (43), one gets that :

(∆ 2,2 ) A (L, t) l∈L ⊥ 0< l A M t 2 l 1 2 (46)
with M > 0 because ν is a bounded function. It follows that there exists C(L) > 0 such that : Lemma 9 ([14], [START_REF] Vinogradov | Limiting distribution of visits of sereval rotations to shrinking intervals[END_REF], [START_REF] Kelmer | The second moment of the siegel transform in the space of symplectic lattices[END_REF]). For f a piecewise smooth function with compact support on R 2 , one has :

(∆ 2,2 ) A (L, t) C(L) A 5 2 t 2 . ( 47
• S 2 S(f )dµ 2 = c 1 R 2 f dλ • When f is even, (b) S 2 S(f ) 2 dµ 2 C R 2 f 2 dλ + c 2 ( R 2 f dλ) 2
where C > 0.

With this lemma, we are going to prove two lemmas that will enable us to prove Lemma 8 by using Chebyshev's inequality : the first one is intended to estimate the expectation of (∆ 2,3 ) A (L, t) to see that it goes to 0 when t → ∞ (uniformly in A), the second one is intended to estimate its variance to see that it can be as uniformly small in t as one wants if A is chosen large enough. Until the end of this section, we are going to suppose A > 1.

Lemma 10. E((∆ 2,3 ) A (L, t)) = O( 1 t ).
Proof. One has :

(∆ 2,3 ) A (L, t) = l∈L ⊥ A< l f (l) (48)
where

f (l) = 1 2π 1 l 3 2 (ν(l, t) + ν(-l, t))e -(2π) 2 l 2 t 2 . ( 49
)
The Lemma 9 gives us then that :

|E((∆ 2,3 ) A (L, t))| = C| R 2 f (x)1 x >A dx|. ( 50
)
By passing into polar coordinates (r, θ), one gets that :

|E((∆ 2,3 ) A (L, t))| = 2C| 2π θ=0 r A cos(2πtrh(θ) -3π 4 )e -(2π) 2 r t 2 r 1 2 drdθ| (51)
by setting h(θ) = Y γ ((cos(θ), sin(θ)) and by using the fact that Y γ is positively homogeneous. Furthermore, an integration by part gives us that :

r A cos(2πtrh(θ) -3π 4 )e -(2π) 2 r t 2 r 1 2 dr (52) = - e -(2π) 2 r 2 t 2 sin(2πtAh(θ) -3π 4 ) 2πth(θ)A 1 2 + 1 4πth(θ) r A sin(2πtrh(θ) - 3π 4 ) e -(2π) 2 r 2 t 2 r 3 2 dr + 4π t 3 h(θ) r A sin(2πtrh(θ) - 3π 4 )e -(2π) 2 r 2 t 2 r 1 2 dr.
By using that r

1 2
r (because A > 1) and the fact that h(θ) admits a positive lower bound (because 0 ∈ Ω γ ), by estimating the three terms of the right member, one gets that :

| r A cos(2πtrh(θ) -3π 4 )e -(2π) 2 r t 2 r 1 2 dr| C t ( 53
)
with C > 0 that does not depend on θ. By using Equation (51) and Equation ( 53), one gets that :

E((∆ 2,3 ) A (L, t)) = O( 1 t ). Lemma 11. Var((∆ 2,3 ) A (L, t)) = O( 1 A )
where the O can be chosen independent from t.

Proof. By using the same notation as before, by using again the Lemma 9 and by using the Lemma 10, one gets that :

Var((∆ 2,3 ) A (L, t)) C R 2 f 2 (x)1 x >A dx. ( 54
)
So, by passing into polar coordinates and by using the fact that ν is bounded, one gets that :

Var((∆ 2,3 ) A (L, t)) C ∞ r=A 1 r 2 dr. ( 55
)
By integrating, we get the wanted result.

We can now prove the Lemma 8.

Proof of Lemma 8. The Chebyshev's inequality gives the wanted result if, first, we choose A large enough and, second, we choose t large enough so that E((∆ 2,3 ) A (L, t)) and Var((∆ 2,3 ) A (L, t)) are small enough. These choices are possible according to Lemmas 10 and 11.

So, now the proof of Lemma 1 is complete and we will conclude this section by proving the Lemma 2 so that the proof of Proposition 1 will be complete.

Proof of Lemma 2

To prove the Lemma 2, we are going to take the same kind of approach as before : estimate the expectation and the variance of the quantity S A -S A,prime and get the result via the Chebyshev's inequality. We have that :

H A (L, t) -S A,prime (L, t) = l∈L ⊥ prime f (l) (56)
where

f (l) = 1 2π l 3 2 1 0< l A k ⌊ A l ⌋+1 (ν(kl, t) + ν(-kl, t)) k 3 2 . ( 57
)
With this expression, we see that we are going to have a little problem of integrability at 0 if we use the Lemma 9. That's why, we have to exclude 0 and we will suppose that L is chosen so that L ⊥ 1 ǫ where 0 < ǫ < 1. Only a small number of lattices is excluded according to this lemma :

Lemma 12. For every 0 < ǫ < 1, one has that

P( L 1 < ǫ) = O(ǫ 2 ).
Proof. It is a consequence of Lemma 9 by taking

S(f )(L) = l∈L 1 B f (0,ǫ) (l)
where 1 B f (0,ǫ) (l) is the indicator function of the closed ball for the norm • centred on 0 of radius ǫ.

Thus, for the chosen lattices, we have :

H A (L, t) -S A,prime (L, t) = l∈L ⊥ prime f (l)1 l ǫ = ∆ 3,ǫ,A,t (L) (58)
(this equation defines ∆ 3,ǫ,A,t (L)).

Lemma 13. E(∆ 3,ǫ,A,t (L)) = O ǫ,A ( 1 t ).
Proof. By using the Lemma 9, one gets that :

E(∆ 3,ǫ,A,t (L)) C A r=ǫ 2π θ=0 1 r 1 2 k A r ν(kr(cos(θ), sin(θ)), t) k 3 2 dr. ( 59
)
Lebesgue's dominated convergence theorem gives us that :

A ǫ 1 r 1 2 k A r ν(kr(cos(θ), sin(θ)), t) k 3 2 dr = ∞ k=1 1 k 3 2 A max( A k ,ǫ) ν(kr(cos(θ), sin(θ)), t) r 1 2 dr. ( 60
)
An integration by part (on the variable r) as in the proof of Lemma 10 and the Equation (59) give us finally that :

E(∆ 3,ǫ,A,t (L)) = O ǫ,A ( 1 t ). ( 61
)
Lemma 14. There exists K > 0 such that :

Var(∆ 3,ǫ,A,t (L)) K(- log(ǫ) A + log(A) A ).
Proof. Lemma 9 gives us that :

Var(∆ 3,ǫ,A,t (L)) C A r=ǫ 2π θ=0 1 r 2 ( k A r ν(kr(cos(θ), sin(θ)), t) k 3 2 ) 2 dr. ( 62
)
Yet one also has that for all x > 0 :

k x 1 k 3 2 D x 1 2 (63)
where D > 0. Thus, Equation (62), Equation ( 63) and the fact that ν is bounded imply that :

Var(∆ 3,ǫ,A,t (L)) 2πCD A A ǫ 1 r dr. ( 64
)
We can now give the proof of Lemma 2.

Proof of Lemma 2. First we take 1 > ǫ > 0 small enough so that the measure of the neglected lattices, id est the lattices such that L 1 < ǫ, is small enough. It is possible according to Lemma 12. Then we take A large enough so that Var(∆ 3,ǫ,A,t (L)) is small enough. It is possible according to Lemma 14. Finally, we take t large enough so that E(∆ 3,ǫ,A,t (L)) is small enough, which is possible according to Lemma 13, and conclude by using Chebyshev's inequality.

So, we are now brought back to the study of S A,prime (L, t) when t → ∞ and the next section is dedicated to it. We are going to replace L ⊥ by L (it changes nothing because we are studying the asymptotic convergence in law with L ∈ S 2 distributed according to μ2 ).

4 Study of S A,prime (L, t) when t → ∞

Reductions for the study of S A,prime (L, t) and proof of Theorem 1

Before entering in the main object of this section, we need to do a small rewriting of S A,prime (L, t). We recall that a vector l ∈ L is prime if, and only if, Kl ∈ KL is prime where K ∈ SL 2 (R). Furthermore, a vector (l 1 , l 2 ) ∈ Z 2 is prime if, and only if, l 1 ∧ l 2 = 1. By using the symmetry l -→ -l, we deduce that S A,prime (L, t) can be rewritten as followed :

S A,prime (L, t) = 1 π k∈Π A (L) Z k (L, t) W k (L) (65)
where, for k

= (k 1 , k 2 ) ∈ Z 2 , W k (L) = k 1 e 1 (L) + k 2 e 2 (L) 3 2 , ( 66
) Z k (L, t) = m∈N-{0} ν(m(k 1 e 1 (L) + k 2 e 2 (L)), t) + ν(-m(k 1 e 1 (L) + k 2 e 2 (L)), t) m 3 2 , ( 67
)
and where

Π A (L) = {(k 1 , k 2 ) ∈ Z 2 | k 1 ∧ k 2 = 1, k 1 0 k 1 e 1 (L) + k 2 e 2 (L) A} ( 68
)
and here we agree that if (k 1 , k 2 ) ∈ Π A (L) then k 1 = 0 implies that k 2 = 1 (for the definition of e 1 (L) and e 2 (L) see Definition 3).

Let's recall that :

Π = {(k 1 , k 2 ) ∈ Z 2 | k 1 ∧ k 2 = 1, k 1 0}. ( 69
)
Our goal now is to prove the following proposition :

Proposition 2. In the symmetric case, {Z k (L, t)} k∈Π converge, when t → ∞, in distribution towards {ρ γ (k 1 e 1 (L) + k 2 e 2 (L)) Zk (ω)} k∈Π where Zk (ω), with ω ∈ Ω, are independent identically distributed real random variables that have a compact support, are symmetrical and are non-zero.

In the non symmetric case, {Z k (L, t)} k∈Π converge, when t → ∞, in distribution towards

{ρ γ (k 1 e 1 (L) + k 2 e 2 (L)) Zk (ω 1 ) + ρ γ (-(k 1 e 1 (L) + k 2 e 2 (L))) Zk (ω 2 )} k∈Π where (ω 1 , ω 2 ) ∈ Ω × Ω,
where Zk (ω) are independent identically distributed real random variables that have a compact support, are symmetrical and are non-zero.

In the next section we are going to consider, in the symmetric case, the sums of the type

SA (ω, L) = k∈Π A (L) ρ γ (k 1 e 1 (L) + k 2 e 2 (L)) Zk (ω) k 1 e 1 (L) + k 2 e 2 (L) 3 2
and the sums of the type, in the non symmetric case,

SA (ω 1 , ω 2 , L) = k∈Π A (L) ρ γ (k 1 e 1 (L) + k 2 e 2 (L)) Zk (ω 1 ) + ρ γ (-(k 1 e 1 (L) + k 2 e 2 (L))) Zk (ω 2 ) k 1 e 1 (L) + k 2 e 2 (L) 3 2
where Z k are non-zero real independent identically distributed random variables from Ω ∋ ω that are symmetrical and have a compact support and where (ω 1 , ω 2 ) ∈ Ω × Ω. Proposition 3 tells us that :

Proposition 3. The sums of these types :

• converge almost surely

• their respective limits are symmetrical and their expectations are equal to 0

• their respective limits admit moment of order 1 + κ for any 0 κ < 1 3

• their respective limits do not admit a moment of order 4 3 when σ(L) m where m > 0 and where L belongs to an event of the form ( L < α) with α > 0.

• when there exists α > 0 such that μ2 ({L ∈ S 2 | L 1 < α}) = 0 then their respective limits admit moments of all order 1 p < ∞.

We are going to see now that it is enough to prove Proposition 2 and Proposition 3 to establish Theorem 1, with the exception of the exact form of the limiting law (yet it is given by Proposition 4).

Proof of Theorem 1. Let ψ ∈ C ∞ c (R). Let ǫ > 0. According to Proposition 1, we can take A as large as we want and then t as large as we want so that :

|E ψ( R(tΩ γ , L) √ t ) -E (ψ(S A,prime (L, t))) | ǫ. ( 70
)
Thanks to Proposition 2, one has also that :

|E (ψ(S A,prime (L, t))) -E ψ( SA (ω, L)) | ǫ (71)
where the Z k (ω) in SA (ω, L) are given by Proposition 2. Furthermore, Proposition 3 gives us that :

|E ψ( SA (ω, L)) -E ψ( lim A→∞ SA (ω, L)) | ǫ (72)
with lim A→∞ SA (ω, L) that verify all the listed properties. So, Equation (70), Equation (71) and Equation ( 72) give the wanted result.

The main reason why the Z k are going to be independent from L is the presence of the factor t.

The main reasons why the rest of Proposition 2 will be true are the presence of the factor t in Z k and the fact that the coefficients of order 1 of the Taylor series of (Y (k 1 e 1 (L) + k 2 e 2 (L))) k∈Π A (L) on a small geodesic segment are Z-free and, in the non symmetric case, (Y (k 1 e 1 (L) + k 2 e 2 (L)), Y (-(k 1 e 1 (L) + k 2 e 2 (L))) k∈Π A (L) on a small geodesic segment are Zfree.

In order to prove Proposition 2, it is actually enough to prove the following proposition by using the definition of ν (see Equation ( 40)) :

Proposition 4. For k = (k 1 , k 2 ), let θ k (L, t) = tY γ (k 1 e 1 (L) + k 2 e 2 (L))) mod 1. ( 73
)
Then, we have that {θ k (L, t)} k∈Π converge, when t → ∞, towards random variable that are independent identically distributed, are distributed according to the Lebesgue measure λ over R/Z and are independent from L.

In the non symmetric case, we have that {θ k (L, t)} k∈Π and {θ -k (L, t)} k∈Π converge, when t → ∞, towards random variable that are independent identically distributed, are distributed according to the Lebesgue measure λ over R/Z and are independent from L.

Thanks to this proposition, we now understand why the limit law of R(tD 2 ,L) √ t is given by S(θ, L). To prove this last proposition, it is sufficient to prove the following proposition where e(θ) stands for exp(i2πθ) :

Proposition 5. For every l ∈ N -{0}, for every ψ ∈ C ∞ c (S 2 ), for every (p 1 , • • • , p l ) ∈ Z l -{0}, one has : E ψ(L)e( l h=1 p h θ k h ) → t→∞ 0 ( 74
)
where the k h ∈ Π are all distinct.

In the non symmetric case, one has that for every l ∈ N -{0}, for every

ψ ∈ C ∞ c (S 2 ), for every (p -l , • • • , p -1 , p 1 , • • • , p l ) ∈ Z 2l -{0}, one has : E ψ(L)e( l h=1 p h θ k h + l h=1 p -h θ -k h ) → t→∞ 0 (75)
where the k h ∈ Π are all distinct.

Before passing to the proof of Proposition 5, let's give some heuristic about it in the symmetric case, the non symmetric case being similar here. Basically, by working with a foliation of the space S 2 given by small enough geodesic segments, we are first going to have :

E ψ(L)e( l h=1 p h θ k h ) ≈ E(ψ)E e( l h=1 p h θ k h )
due to the presence of the factor t in θ.

The right member will go to 0 when t goes to infinity because a Riemann-Lebesgue lemma will apply because quantities "close" to the variables θ k are typically Z-free (see the heuristic explanation of the second step). The rest of this section is now dedicated to the proof of the Proposition 5.

Foliation and local estimates

We recall that a foliation of the space S 2 is given by the orbits of the group δ where

δ(λ) = λ 0 0 1 λ .
To prove Proposition 5, we are going to look at what it is happening on a small "segment" of the form

J ǫ (L) = {δ(λ)L | λ ∈ [ 1 1 + ǫ , 1 + ǫ]} (76)
where L ∈ S 2 and ǫ > 0 can be taken as small as possible. More precisely, we are going to show, when t → ∞, the independence of the (θ k ) and of L over smalls segments of the form J ǫ (L), as well as the fact that the (θ k ) are identically distributed and distributed according to the normalized Haar measure over R/Z.

Let's call s : (x, y) ∈ R 2 -→ (x, -y) and let's set for k = (k 1 , k 2 ) ∈ Π or k ∈ -Π : W k (L) = dY γ (k 1 e 1 (L) + k 2 e 2 (L)) (s(k 1 e 1 (L) + k 2 e 2 (L))) (77)
where dY γ (k 1 e 1 (L)+k 2 e 2 (L))(•) stands for the differential of Y γ at the point k 1 e 1 (L)+k 2 e 2 (L). On a segment of the form J ǫ (L), the following lemma basically tells us how we can estimate the quantities Y γ (k 1 e 1 (L) + k 2 e 2 (L)) :

Lemma 15. For a typical L ∈ S 2 , there exists ǫ > 0 small enough such that for every

λ ∈ [ 1 1+ǫ , 1 + ǫ], e 1 (δ(λ)L) = δ(λ)e 1 (L) and e 2 (δ(λ)L) = δ(λ)e 2 (L).
Furthermore, for such a lattice L, for such λ,

for k = (k 1 , k 2 ) ∈ Π or for k ∈ -Π, we have for h = λ -1, Y γ (k 1 e 1 (δ(λ)L) + k 2 e 2 (δ(λ)L)) = Y γ (k 1 e 1 (L) + k 2 e 2 (L)) (78)
+ hW k (L) + O k 1 ,k 2 ,L (h 2 ).
Proof. The first fact was proven in [START_REF] Trevisan | Limit laws in the lattice problem. ii. the case of ovals[END_REF] (see Lemma 15). Let's note that, as γ is analytical, Y γ is regular. As a consequence, the second fact is obtained from the first fact of Lemma 15 and by a simple calculus of Taylor series.

To prove Proposition 5 we see, in light of Lemma 15, that it would be convenient to prove the following proposition : Proposition 6. For a typical L ∈ S 2 , for every m ∈ N -{0}, for every family

(p 1 , • • • , p m ) ∈ Z m , for every k 1 , • • • , k m ∈ Π all distinct if m i=1 p i W k i (L) = 0 (79) then p 1 = • • • = p m = 0.
In other words, for a typical L ∈ S 2 ,

(W k (L)) k∈Π is a Z-free family.
In the non symmetric case, for a typical L ∈ S 2 , for every m ∈ N -{0}, for every family

(p -m , • • • , p -1 , p 1 , • • • , p m ) ∈ Z 2m , for every k 1 , • • • , k m ∈ Π all distinct if m i=1 p i W k i (L) + m i=1 p -i W -k i (L) = 0 (80) then p -m = • • • = p -1 = p 1 = • • • = p m = 0.
In other words, in the non symmetric case, for a typical L ∈ S 2 ,

(W k (L)) k∈-Π∪ • Π is a Z-free family.
The next subsection is dedicated to prove this proposition.

Proof of Proposition 6

To prove Proposition 6, we are following closely what was done in the Section 5 of [START_REF] Dolgopyat | Deviations of ergodic sums for toral translations i. convex bodies[END_REF] and we need four preliminary lemmas.

To state these lemmas we need to put in place some notations. Let's call

P : X ∈ R 2 -→ dY γ (X)(s(X)). (81)
Let's also set for every

k = (k 1 , k 2 ) ∈ -Π ∪ • Π, f k : (A = (X 1 , Y 1 ), B = (X 2 , Y 2 )) ∈ R 2 × R 2 -→ P (k 1 X + k 2 Y ) (82)
where

X = (X 1 , X 2 ) and Y = (Y 1 , Y 2 ) ∈ R 2 .
Let's set for every L ∈ GL 2 (R),

g L : δ ∈ R -→ P (L(1, δ)). ( 83
)
Let's also set for every L ∈ GL 2 (R),

gL : δ ∈ R -→ P (L(-1, -δ)). (84)
Lemma 16. If γ is analytic, we have that for any L ∈ GL 2 (R), g L is (real) analytic and not equal to a polynomial.

Proof. First, let's note that we X ∈ R 2 -→ dY γ (X)(s(X)) is positively homogeneous because X -→ Y γ (X) also is positively homogeneous. So, one has that for every δ ∈ R,

g L (δ) = 1 + δ 2 P (L( 1 √ 1 + δ 2 , δ √ 1 + δ 2 )). ( 85
)
We know that g L is analytic because γ is analytic. So, let's suppose that f L is equal to a polynomial. Let's observe that P (L( 1

√ 1+δ 2 , δ √ 1+δ 2
)) is bounded so that g L can only be of degree at most one. From this fact, one has that, for every λ, δ ∈ R, (P L)(λ, δ) can be written as :

(P • L)(λ, δ) = a 0 λ + a 1 δ ( 86
)
where a 0 , a 1 ∈ R.

By making the change of variable u = L(λ, δ), one gets finally that there exists b 0 , b 1 ∈ R such that for all λ, δ ∈ R :

P (λ, δ) = b 0 λ + b 1 δ. ( 87
)
Yet, by using Equation (81), one gets that :

x∂ x Y γ -y∂ y Y γ = b 0 x + b 1 y. ( 88
)
Yet, Y γ is positively homogeneous and so for x > 0, one has Y γ (x, y) = xY γ (1, y x ). Using differentiation, one gets that :

∂ y Y γ (x, y) = ∂ y Y γ (1, y x ). (89)
Using Equation (88) and Equation (89), by diving by x and making x goes to infinity one gets that :

∂ x Y γ (x, y) → b 0 (90) when x → ∞.
So, one gets that (because Y γ is analytic) :

∂ x Y γ (x, y) = K(y) (91)
where K(y) is a regular function. By interchanging the role of x and y, one gets that :

Y γ (x, y) = A 0 xy + A 1 ( 92
)
where A 0 and A 1 belong to R. Yet by setting x = 1 and y = δ and by using Lemma 5.2 from [START_REF] Dolgopyat | Deviations of ergodic sums for toral translations i. convex bodies[END_REF], we obtain that it is impossible.

In the non symmetric case, we will need the following lemma.

Lemma 17. The following alternative holds. Let k 2, k ∈ 2N. Either (i) There exists L ∈ GL 2 (R) and δ, δ ′ ∈ R such that

g (k) L (δ) g (k) L (δ ′ ) = g(k) L (δ) g(k) L (δ ′ ) (93)
or (ii) Ω γ has a center of symmetry.

Proof. Let's suppose that (i) does not hold. Let L = Id. We have that g (k)

Id (•) = cg (k)
Id (•) for some constant c. In other words

( ∂ ∂δ ) k P (1, δ) = c( ∂ ∂δ ) k P (-1, -δ). (94)
Since for x > 0 we have P (x, y) = xP (1, y x ), it follows that

∂ k y P (x, y) = c∂ k y P (-x, -y). ( 95
)
Since γ is analytic, this equality in fact holds identically. In particular :

∂ k y P (-x, -y) = c∂ k y P (x, y). ( 96
)
As a consequence, and because of Lemma 16, one must have c = ±1. Furthermore, from Equation ( 95), one has necessarily that :

P (x, y) -cP (-x, -y) = k-1 l=0 a l (x)y l . ( 97
)
Let's set now H(x, y) = Y γ (x, y) -cY γ (-x, -y). Equation (97) and Equation (97) give us that :

x∂ x H(x, y) -y∂ y H(x, y) = k-1 l=0 a l (x)y l . ( 98
)
Furthermore, one has, by positive homogeneity of H(x, y), that ∂ x H(x, y) = ∂ x H(x/y, 1). Equation (98) gives us then that the growth of ∂ y H(x, y), at x fixed, is of order y d-1 . As H(x, y) is analytic, we must have :

∂ y H(x, y) = k-1 l=0 bl (x)y l (99)
and so

H(x, y) = k l=0 b l (x)y l (100)
where the b l (x) are analytic functions. Yet, H(x, y) = yH( x y , 1). So, one must have : H(x, y) = b 0 (x) + b 1 (x)y. So, we see that we have reached the same point of the demonstration of Lemma 5.3 of [START_REF] Dolgopyat | Deviations of ergodic sums for toral translations i. convex bodies[END_REF]. So, by reasoning the same way, one gets finally that :

Y γ (X) -Y γ (-X) =< X, v > (101)
with v being a vector of R 2 . By shifting the origin to x 0 , Y γ (X) is replaced by Y γ (X)+ < X, x 0 > and Y γ (-X) by Y γ (-X)-< X, x 0 >. So, after shifting the origin to v 2 , we get Y γ (X) = Y γ (-X) so that Ω γ is symmetric.

Let's assume WLOG that Equation (93) holds for L = Id.

Lemma 18. For every m ∈ N-{0}, for every family

(p 1 , • • • , p m ) ∈ Z m , for every k 1 , • • • , k m ∈
Π all distinct, one has the following implication : if m i=1 p i f k i is a polynomial function then all the p i must be equal to 0. In the non symmetric case, for every m ∈ N-{0}, for every family

(p -m , • • • , p -1 , p 1 , • • • , p m ) ∈ Z 2m , for every k 1 , • • • , k m ∈ Π all distinct if m i=1 p i f k i + m i=1 p -i f -k i is
a polynomial function then all the p i must be equal to 0.

Proof. Let's suppose that m i=1 p i f k i is a polynomial. Let's give us j ∈ [1, m]. We are going to show that p j = 0. Let β ∈ R 2 such that < k j , β > = 0. Let α ∈ R 2 and let X = α and Y = δα + θβ. Then, one has :

f k (A, B) = | < k, α > |g Id (δ + θ < k, β > | < k, α > | ) if < k, α >> 0 or f k (A, B) = | < k, α > |g Id (δ + θ < k, β > | < k, α > | ) if < k, α >< 0. As m i=1 p i f k i is a polynomial then m i=1 p i f k (A, B
) is a polynomial in β whose degree is bounded by a number that does not depend on α (nor in β). So there exists K 2 such that the K-th derivative of m i=1 f k (A, B) relatively to θ is equal to 0. It means that the terms in front of θ K is equal to 0. Hence the following equation :

m i=1 h i < k i , β > K | < k i , α > | K-1 = 0 ( 102
)
where

h i = p i g (K) Id (δ) if < k i , α >> 0 and h i = p i g (K)
Id (δ) if < k i , α >< 0. Now, since the k i belong to Π and are all distinct, it is possible to choose α so that < k j , α >> 0 is arbitrary small while < k i , α > remain bounded away from zero for every i = j. Thus, we must have h j = 0. Yet, g is not a polynomial so there exists δ ∈ R such that g (K)

Id (δ) = 0 and it gives us that p j = 0. In the non symmetric case, Equation (102) becomes

m i=1 h i < k i , β > K | < k i , α > | K-1 = 0. ( 103
)
where

h i = p i g (K) Id (δ) + p -i g(K) Id (δ) if < k i , α >> 0 and h i = p i g(K) Id (δ) + p -i g (K)
Id (δ) if < k i , α >< 0. Let's consider for example the case where the first alternative holds. As before, we must have p j g (K)

Id (δ) + p -j g(K) Id (δ) = h j = 0 for any choice of δ. Since we assume that Equation (93), this implies that p j = p -j = 0.

Lemma 18 enables us to prove the following lemma.

Lemma 19. For a typical L ∈ S 2 , for every m ∈ N-{0}, for every family

(p 1 , • • • , p m ) ∈ Z m , for every k 1 , • • • , k m ∈ Π all distinct if m i=1 p i W k i (L) = 0 (104) then m i=1 p i f k i = 0.
In the non symmetric case, for a typical L ∈ S 2 , for every m ∈ N -{0}, for every family

(p -m , • • • , p -1 , p 1 , • • • , p m ) ∈ Z 2m , for every k 1 , • • • , k m ∈ Π all distinct if m i=1 p i W k i (L) + m i=1 p -i W -k i (L) = 0 (105) then m i=1 p i f k i + m i=1 p -i f -k i = 0.
Proof. First, we note that, for every k We can now give the proof of Proposition 6.

∈ -Π∪ • Π, W k (L) = f k ((e 1 (L))
Proof of Proposition 6. It is a direct consequence of Lemma 18 and Lemma 19.

Proof of Proposition 5

As we have at our disposal Proposition 6, the proof of Proposition 5 is the same as the proof of Proposition 5 in [START_REF] Trevisan | Limit laws in the lattice problem. ii. the case of ovals[END_REF]. We are going to present it again for completeness. Before starting the proof of Proposition 5, we only need a simple lemma : Lemma 20. For every m ∈ N -{0}, for every family

(p 1 , • • • , p m ) ∈ Z m -{0}, for every k 1 , • • • , k m ∈ Π,
all distinct, for every 0 < ǫ < 1, there exists a > 0 and K ǫ a measurable set of S 2 such that μ2 (S 2 -K ǫ ) ǫ and such that for all

L ∈ K ǫ , | m i=1 p i W k i (L)| a .
In the non symmetric case, for every m ∈ N-{0}, for every family

(p -m , • • • , p -1 , p 1 , • • • , p m ) ∈ Z 2m -{0}, for every k 1 , • • • , k m ∈ Π,
all distinct, for every 0 < ǫ < 1, there exists a > 0 and K ǫ a measurable set of S 2 such that μ2 (S 2 -K ǫ ) ǫ and such that for all L ∈ K ǫ ,

| m i=1 p i W k i (L) + m i=1 p -i W -k i (L)| a .
Proof. It is a direct consequence of Proposition 6. Now, by using the foliation given by δ(λ) and previous results, we can now prove Proposition 5.

Proof of Proposition 5. The proof in all its generality can be made as in the case where μ2 = µ 2 . So, we will suppose for simplicity that μ2 = µ 2 . We consider the case where Ω γ is symmetric. The non symmetric case is similar.

Let l 1. Let ψ ∈ C ∞ c (S 2 ). Let (p 1 , • • • , p l ) ∈ Z l -{0}
. For all ǫ > 0, we call F ǫ the tribe on S 2 generated by the J ǫ (L). Let 1 > ǫ 1 ǫ 2 > 0. According to Lemma 15 and Lemma 20, there exists a measurable part K ǫ 1 such that µ 2 (K ǫ 1 ) 1 -ǫ 1 , a real M > 0 and a real a > 0, such that

• for every L ∈ K ǫ 1 , for every λ ∈ [ 1 1+ǫ 1 , 1 + ǫ 1 ] : |ψ(δ(h)L) -ψ(L)| M |h| (106)
where

h = λ -1 • for every L ∈ K ǫ 1 , for every λ ∈ [ 1 1+ǫ 1 , 1 + ǫ 1 ], Equation (78) is verified. • for every L ∈ K ǫ 1 , | l i=1 p i W k i (L)| a . ( 107
)
Furthermore, we are going to suppose, even if it means making ǫ 2 goes to 0, that for every

L ∈ K ǫ 1 , J ǫ 2 (L) ⊂ K ǫ 1 . ( 108
)
Claim. With these notations, we have, for all L ∈ K ǫ 1 , that :

E ψe( l i=1 p i θ k i )|F ǫ 2 (L) = O(ǫ 2 ) + O( 1 atǫ 2 ) + 1 a O(ǫ 2 ). ( 109
)
To this end, let's set, for all ǫ > 0,

δ(ǫ) = 1 + ǫ - 1 1 + ǫ .
Then one has for every L ∈ K ǫ 1 according to Lemma 15 :

E(ψe( l j=1 p j θ k j )|F ǫ 2 )(L) = 1 δ(ǫ 2 )
p j θ k j )1 Kǫ 1   | ψ ∞ ǫ 1 + O(ǫ 2 ) + O( 1 atǫ 2 ) + 1 a O(ǫ 2 ). ( 115
)
By first choosing ǫ 1 > 0 small enough (note that a depends on ǫ 1 ), then choosing ǫ 2 > 0 and finally choosing t large enough, we obtain the wanted result.

We are now brought back to the study of the convergence, in the symmetric case, of the sums of the type SA (ω, L) = where Z k are non-zero real independent identically distributed random variables from Ω ∋ ω are symmetrical and have a compact support and where (ω 1 , ω 2 ) ∈ Ω × Ω. In the next section, we are going to study the sums of this type and prove Proposition 3, which will conclude the proof of Theorem 1.

5 Asymptotic study of SA (ω, L) and of SA (ω 1 , ω 2 , L)

The study is in fact essentially done in Section 5 of [START_REF] Trevisan | Limit laws in the lattice problem. ii. the case of ovals[END_REF]. We only need to notice the following fact to apply the same method :

Lemma 21. There exists m, M > 0 such that for every ξ ∈ R 2 -{0},

M ρ γ (ξ) m.
Proof. As γ is strictly convex, from the definition of ρ γ , ρ γ > 0. Furthermore, for every ξ ∈ R 2 -{0},

ρ γ (ξ) = ρ γ ( ξ ξ ).
S 1 being compact, we obtain immediately the wanted result.

Proof of Proposition 3. We use the same method of the last section of [START_REF] Trevisan | Limit laws in the lattice problem. ii. the case of ovals[END_REF] and use Lemma 21.

Elements of proof of Theorem 3

To prove Theorem 3 we follow the same approach that was followed to prove Theorem 2. Namely, as a first step, we prove the equivalent of Proposition ρ γ (l) cos(2πtmY γ (l) + 2πm < α, l > -3π 4 ) + ρ γ (-l) cos(2πtmY γ (-l) -2πm < α, l > -3π 4 ) m Then, one can prove, as Proposition 1 was proven : Proposition 7. For every β > 0, for every A > 0 large enough, for every t large enough, one has that : P(∆ A,prime (L, α, t) β) β.

Second, as Proposition 4 was already proven, the limit distribution of S A,prime (L, α, t) when t → ∞ and with α ∈ R 2 fixed and with L being distributed according to μ2 is, in the symmetric case, is the distribution of the almost-sure limit of In the non symmetric case, the limit distribution of S A,prime (L, α, t) when t → ∞ and with α ∈ R 2 fixed and with L being distributed according to μ2 is the distribution of the almost-sure limit of Finally, we conclude the proof by proving the equivalent of Proposition 3. To do so, we follow the exact same approach that was used to prove Proposition 3.

) 3 . 1 . 4

 314 Proof of Lemma 8 To prove Lemma 8 we need to use what are called Siegel and Rogers formulas. Theses formulas will also be useful later in this paper. By setting c k = ζ(2) -k for k an integer larger than 1 and where ζ denotes the ζ function of Riemann, one has the following formulas :

1 ep 1 e

 11 itD 1 (L)h+itD 2 (L,h) dh + O(ǫ 2 ) j θ k j (δ(λ)L) -l j=1 p j θ k j (L) -D 1 (L)h such that (112) D 2 (L, h) = O(h 2 ) and (113) D 2 (L, •) is smooth around 0.Thus, by integrating by part and by using Equation (107), one gets that for allL ∈ K ǫ 1 itD 1 (L)h+itD 2 (L,h) dh = 1 δ(ǫ 2 )   e itD 1 (L)h+itD 2 (L,h) itD 1 (L)

1 (

 1 D 2 (L, •)) ′ (h)e itD 1 (L)h+itD 2 (L,h) dh (110) and Equation (114) give the wanted claim.Thanks to Equation (109), the fact that µ 2 (K ǫ 1 ) 1 -ǫ 1 and because of Equation (108), we have that :

2

 2 k∈Π A (L) ρ γ (k 1 e 1 (L) + k 2 e 2 (L)) Zk (ω) k 1 e 1 (L) + k 2 e 2 (L) 3and of the sums of the type, in the non symmetric case,SA (ω 1 , ω 2 , L) = k∈Π A (L) ρ γ (k 1 e 1 (L) + k 2 e 2 (L)) Zk (ω 1 ) + ρ γ (-(k 1 e 1 (L) + k 2 e 2 (L))) Zk (ω 2 ) k 1 e 1 (L) + k 2 e 2 (L) 3 2

S

  γ,A (θ, L) = 2 π (k 1 ,k 2 )∈Π e(k 1 ,k 2 ,L) A ρ γ (e(k 1 , k 2 , L))φ α (θ (k 1 ,k 2 ) , e(k 1 , k 2 , L)) e(k 1 , k 2 , L)when A → ∞ and where θ = (θ (k 1 ,k 2 ) ) ∈ T ∞ being distributed according to λ ∞ and L being distributed according to μ2 and where e(k 1 , k 2 , L) = k 1 e 1 (L) + k 2 e 2 (L).

S

  γ,A (θ, L) = 1 π (k 1 ,k 2 )∈Π e(k 1 ,k 2 ,L) A φ α,γ,2 (θ (k 1 ,k 2 ) , e(k 1 , k 2 , L)) e(k 1 , k 2 , L) when A → ∞ and where θ = (θ (k 1 ,k 2 ) ) = (θ 1,(k 1 ,k 2 ) , θ 2,(k 1 ,k 2 ) ) ∈ T ∞,2 being distributed according to λ ∞,2 and L being distributed according to μ2 .

  [START_REF] Björklund | Counting in generic lattices and higher rank actions[END_REF] , (e 2 (L)) 1 , (e 1 (L)) 2 , (e 2 (L)) 2 ). Thanks to this remark, we see that Lemma 19 is a direct consequence of the facts that S 2 = SL 2 (R)/SL 2 (Z), that det is a polynomial function whereas m i=1 p i f k i and m i=1 p i f k i + m i=1 p -i f -k i are polynomial functions, for a typical L, only in the trivial case according to Lemma 18.

  1. Namely, let's set ∆ A,prime (L, α, t) = | R(tΩ γ + α, L)

					√	t	-S A,prime (L, α, t)|	(116)
	with
	S A,prime (L, α, t) =	1 π	l∈L ⊥ prime	1 l	2 3		(117)
			0< l A
	m∈N-{0}

R(tΩγ +X,L)√ t

ρ γ (γ(t)). (33)