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Distribution of Energy Levels of a Quantum Free Particle on a Surface of Revolution


IASSNS-HEP-93/35 May 1993 Distribution of Energy Levels of a Quantum Free Particle on a

arXiv:alg-geom/9306008v1 18 Jun 1993

Surface of Revolution Pavel M. Bleher School

of Natural Science Institute for Advanced Study Princeton, NJ 08540

Abstract. We prove that the error term Λ(R) in the Weyl asymptotic formula #{En ≤ R2 } = Vol M 2 R + Λ(R), 4π

for the Laplace operator on a surface of revolution M satisfying a twist hypothesis, has the form Λ(R) = R1/2 F (R) where F (R) is an almost periodic function of the Besicovitch class B 2 , and the Fourier series of F (R) in B 2 is
γ

A(γ) cos(|γ|R ? φ) where the sum goes over all closed geodesics

on M , and A(γ) is computed through simple geometric characteristics of γ. We extend this result to surfaces of revolution, which violate the twist hypothesis and satisfy a more general Diophantine hypothesis. In this case we prove that Λ(R) = R2/3 Φ(R) + R1/2 F (R), where Φ(R) is a ?nite sum of periodic functions and F (R) is an almost periodic function of the Besicovitch class B 2 . The Fourier series of Φ(R) and F (R) are computed.

1

1. Introduction

Let M be a two–dimensional smooth compact manifold which is homeomorphic to a sphere, and which is a surface of revolution in R3 , with an axis A and poles N and S (see Fig.1). The geodesic ?ow on M is a classical integrable system due to the Clairaut integral, r sin α = const . (1.1)

In the present work we are interested in high energy levels of the corresponding quantum system, ??un = En un . Let s be the normal coordinate (the length of geodesic) along meridian and r = f (s), 0 ≤ s ≤ L, (1.3) (1.2)

be the equation of M , where r is the radial coordinate. Then ? = f (s)?1 where ? is the angular coordinate. We will assume that f (s) has a simple structure, so that f ′ (s) = 0, where f (smax ) = max f (s) ≡ fmax .
0≤s≤L

? ?s

f (s)

? ?s

+ f (s)?2

?2 , ??2

(1.4)

s = smax ;

f ′′ (smax ) = 0,

(1.5)

For normalization we put fmax = 1. Another assumption on M is the following twist hypothesis. Consider the equator on M , γE = {s = smax , 0 ≤ ? ≤ 2π}, (we do not assume that M is symmetric with respect to γE , but we still call γE the equator keeping a visual interpretation of objects on M ), and a geodesic γ which starts at x0 = (s = smax , ? = 0) ∈ γE at some angle ?π/2 < α0 < π/2 to the direction to the north. The Clairaut integral on γ is I = sin α0 , and we can parametrize γ by ?1 < I < 1 : 2 γ = γ(I). It follows from the Clairaut

integral that γ(I) oscillates between two parallels, s = s+ and s = s? , where f (s? ) = f (s+ ) = I, so γ(I) intersects γE in?nitely many times. Let xn be the n–th intersection of γ with γE , n ∈ Z. De?ne τ (I) = |γ[x0 , x2 ]|, the length of γ between x0 and x2 , and ω(I) = (2π)?1 (?(x2 ) ? ?(x0 )), the phase of γ between x0 and x2 (see Fig. 1). Observe that ω(I) is de?ned and for I ≥ 0, ω(I) = π ?1 mod 1. To de?ne

ω(I) uniquely, we choose a continuous branch of ω(I) starting at ω(0) = 0. Then ω(?I) = ?ω(I)
s+ s?

d? ds ? 1. ds

(1.6)

De?ne τ (1) = limI→1?0 τ (1) and ω(1) = limI→1?0 ω(I). It is easy to see that a ?nite geodesic γ with the Clairaut integral 0 < I < 1 is closed i? ω(I) is rational. More precisely, let n(γ) denote the number of revolutions of a closed geodesic γ around the axis A and m(γ) denote the number of oscillations of γ along meridian. Then ω(I) = (n(γ)/m(γ)) ? 1. To facilitate formulation of subsequent results we take the convention that a ?nite geodesic γ with I = 1, which goes along the equator, is closed i? both n(γ) and m(γ) ≡ are integers. Twist Hypothesis (TH). ω ′ (I) = 0, ? I ∈ [0, 1]. To illustrate TH consider an ellipsoid of revolution, x2 y2 z2 + 2 + 2 = 1. a2 a b If a < b (oblong ellipsoid), then ω ′ (I) > 0, if a > b (oblate ellipsoid), then ω ′ (I) < 0; so TH holds in both cases. Curves (a) and (b) on Fig.2 are the graphs of ω(I) found with the help of computer for the ellipsoids of revolution with a = 1, b = 2 and a = 2, b = 1, respectively. The cross–sections of the ellipsoids are shown in the lower part of Fig.2. TH is violated for a sphere (a = b), when 3 n(γ) ω(1) + 1

ω ′ (I) ≡ 0. TH can be also violated for a bell–like shape of M shown on Fig.2 (see the cross–section (c) and the graph (c) of ω(I) on this ?gure) and in some other cases. Let N (R) = #{En ≤ R2 } be the counting function of En . Then the Weyl law says that N (R) = Vol M 2 R + Λ(R) 4π (1.7)

where Λ(R) = o(R2 ), R → ∞. A general estimate of H¨rmander [H¨r1] gives o o Λ(R) = O(R). This estimate is sharp for S 2 and some other degenerate surfaces for which closed geodesics cover a set of positive Liouville measure in the phase space. If the Liouville measure of the union of all closed geodesics in the phase space is 0, then as was shown by Duistermaat and Guillemin [DG], Λ(R) = o(R). For surfaces of negative curvature Selberg and B?rard [B?r] proved a better estimate: e e Λ(R) = O(R/ log R), and it is a very di?cult open problem to show that Λ(R) = O(R1?ε ) for some ε > 0, even in the case of constant negative curvature (see recent works [Sar], [LS] and [HR] where statistics of eigenvalues and eigenfunctions of the Laplace operator on surfaces of constant negative curvature is discussed). For a ?at torus Λ(R) reduces to the error term of the classical circle problem, and the best estimate here is due to Huxley [Hux]: Λ(R) = O(R46/73 (log R)315/146 ). A well–known conjecture of Hardy [Har1] Λ(R) = O(R(1/2)+ε ), ?ε > 0,

is probably also a very di?cult open problem. On the other hand Hardy proved [Har2] that lim sup R?1/2 |Λ(R)| = ∞,
R→∞

4

so (1/2) + ε is the best possible exponent. Colin de Verdi`re [CdV1,CdV2] proved that for a generic surface of revolution of simple struce ture Λ(R) = O(R2/3 ). We prove in the present paper the following result: Theorem 1.1. Assume that M is a surface of revolution of simple structure, and M satis?es TH. Then N (R) = Vol M 2 R + R1/2 F (R), 4π (1.9) (1.8)

where F (R) is an almost periodic function of the Besicovitch class B 2 , and the Fourier series of F (R) in B 2 is F (R) =
closed geodesics γ

A(γ) cos(|γ|R ? φ),

(1.10)

where summation goes over all closed (in general, multiple) oriented geodesics γ = 0 on M , φ = (π/2) + (π/4) sgn ω ′ (I), and A(γ) = π ?1 (?1)m(γ) |ω ′ (I)|?1/2 m(γ)?3/2 = π ?1 (?1)m(γ) |ω ′ (I)|?1/2 τ (I)3/2 |γ|?3/2 , (1.11) I = I(γ).

In Theorem 1.2 we extend Theorem 1.1 to the case when TH is violated. In this case we introduce Diophantine Hypothesis (DH). Assume that ω(I) has at most ?nitely many critical points 0 < I1 < · · · < IK < 1 (so that I = 0, 1 are not critical) with ω ′ (Ik ) = 0, and ω ′′ (Ik ) = 0, k = 1, . . . , K. Assume, in addition, that for every k = 1, . . . , K, ω(Ik ) is either rational or Diophantine in the sense that ? 1 > ζ > 0 and C > 0 such that ω(Ik ) ? C p ≥ 2+ζ , q q ? p ∈ Q. q (1.12)

Theorem 1.2. Assume that M is a surface of revolution of simple structure and DH holds. Then N (R) = Vol M 2 R + R2/3 4π Φk (R) + R1/2 F (R),
k : ω(Ik )∈Q

(1.13)

5

where Φk (R) are bounded periodic functions, and F (R) is an almost periodic function of the Besicovitch class B 2 . The Fourier series of Φk (R) is Φk (R) = (1/2)3?2/3 Γ(2/3)π ?4/3 τ (Ik )4/3
γ : I(γ)=Ik

(?1)m(γ) |γ|?4/3 sin(|γ|R),

(1.14)

and the Fourier series of F (R) is F (R) =
γ : I(γ)=I1 ,...,IK ′

A(γ) cos(|γ|R ? φ(γ)),

(1.15)

where φ(γ) = (π/2) + (π/4) sgn ω (I), I = I(γ), and A(γ) is given in (1.11). In the works [H-B], [BCDL], [Ble1], [Ble2] and [BL] some general results were proved on the existence and properties of a limit distribution of any almost periodic function of the Besicovitch class B 2 (see especially Theorems 4.1–4.3 in [Ble1] and Theorems 3.1, 3.3 in [Ble2]). Theorem 1.2 combined with these results lead us to the following Corollary. Assume that M is a surface of revolution of simple structure and M satis?es TH or, more generally, DH. Then the normalized error function F (R) in the formulas (1.9) and (1.13) has a limit distribution ν(dt), i.e., for every bounded continuous function g(t) on the line,
∞ 1 T g(F (R)) dR = g(t) ν(dt). T →∞ T 0 ?∞ If, in addition, the lengths of all primitive closed geodesics on M with I ≥ 0 are linearly independent

lim

over Z, then ν(dt) is absolutely continuous and the density function p(t) = ν(dt)/dt is an entire function of t which satis?es on the real axis the estimates 0 ≤ p(t) ≤ C exp(?λt4 ), C, λ > 0, C ′ , λ′ > 0;
t

P (?t), 1 ? P (t) ≥ C ′ exp(?λ′ t4 ), t ≥ 0;

P (t) =
?∞

p(t′ ) dt′ .

Observe that Theorem 1.1 is a particular case of Theorem 1.2, so we need to prove only Theorem 1.2. The plan of the remainder of the paper is as follows. In Section 2 we present a theorem of Colin de Verdi`re and show how with the help of this theorem to reduce Theorem 1.2 to a lattice–point e problem for the Bohr–Sommerfeld quasi–classical approximation. Section 3 is auxiliary: here we investigate the asymptotics at in?nity of the Fourier transform of the characteristic function of a plane domain with non–degenerate in?ection points and angular points. In Section 4 we prove the lattice–point version of Theorem 1.2 for the number of lattice points inside a dilated plane oval with ?nitely many points of in?ection. Section 5 is technical, and here we prove some lemmas used in Section 4. In Section 6 we prove Theorem 1.2. Finally in an Appendix we prove a 2/3–estimate for ovals with semicubic singularity. This estimate is used in the main part of the paper. 6

2. Quasi–Classical Approximation

Colin de Verdi`re proved in [CdV2] the following result: e Theorem CdV. If M is a surface of revolution of simple structure then Spectrum (??) = {Ekl = Z(k + (1/2), l); k, l ∈ Z, |l| ≤ k} with Z(p) = Z(p1 , p2 ) ∈ C ∞ (R2 ) such that Z(p) = Z2 (p) + Z0 (p) + O(|p|?1 ), where Z2 (p), Z0 (p) ∈ C ∞ (R2 \ {0}); and Zj (λp) = λj Zj (p), ? λ > 0, p ∈ R2 , j = 0, 2. Z2 (p) > 0, p = 0, |p| → ∞, (2.2) (2.1)

In addition, in the sector {p1 ≥ |p2 |}, Z2 (p) satis?es the equation π ?1
a b

Z2 (p) ? p2 f ?2 (s)ds = p1 ? |p2 |, 2

(2.4)

where a, b are the turning points, i.e., Z2 (p) ? p2 f ?2 (s) = 0 2 for s = a, b. (2.5)

It is to be noted that the Bohr–Sommerfeld quantization rule is π ?1
a b

Ekl ? l2 f ?2 (s)ds = k + (1/2) ? |l|,

k ≥ |l|,

(2.6)

which is equivalent to the approximation Ekl = Z2 (k + (1/2), l). (2.1) implies that N (R) = #{Ekl ≤ R2 } = #{(k, l) : Z(k + (1/2), l) ≤ R2 , |l| ≤ k}. 7 (2.8) (2.7)

De?ne NBS (R) = #{(k, l) : Z2 (k + (1/2), l) ≤ R2 , |l| ≤ k} (BS stands for Bohr–Sommerfeld). Theorem 2.1. 1 T →∞ T lim Proof. De?ne M = {n = (k + (1/2), l) : k, l ∈ Z; |l| ≤ k}. Then N (R) ? NBS (R) =
n∈M 0

(2.9)

T

|N (R) ? NBS (R)|2 R?1 dR = 0.

(2.10)

(2.11)

(χ(n; R) ? χBS (n; R)),

(2.12)

where χ(p; R) and χBS (p; R) are the characteristic functions of the domains {Z(p) ≤ R2 } and {Z2 (p) ≤ R2 }, respectively. Hence 1 T with I(n, n′ ) = Now, ? C > 0 such that χ(n; R) ? χBS (n; R) = 0 if where ΓR = {p ∈ R2 : Z2 (p) = R2 } = RΓ, Indeed,
p∈R2 T 0

|N (R) ? NBS (R)|2 R?1 dR =
n,n′ ∈M

I(n, n′ )

(2.13)

1 T

T 0

(χ(n; R) ? χBS (n; R))(χ(n′ ; R) ? χBS (n′ ; R))R?1 dR.

(2.14)

dist {n, ΓR } > C|n|?1 ,

(2.15)

Γ = {p ∈ R2 : Z2 (p) = 1}.

(2.16)

sup |Z(p) ? Z2 (p)| = C0 < ∞,

(2.17)

so if Z(p) = R2 then for |ε| < 1/2, |Z2 (p(1 + ε)) ? R2 | ≥ |Z2 (p(1 + ε)) ? Z2 (p)| ? |Z2 (p) ? Z(p)| ≥ C1 |p|2 ε ? C0 .
?1 Hence if ε = C|p|?2 with C = 2C0 C1 , then

±(Z2 (p(1 ± ε)) ? R2 ) ≥ CC1 ? C0 = C0 > 0, 8

and ? |ε0 | < ε such that Z2 (p(1 + ε0 )) = R2 , |pε0 | ≤ C|p|?1 , which implies (2.15). De?ne X(p) = (Z2 (p))1/2 . Then X(p) > 0 and X(λp) = λX(p) ? λ > 0, p ∈ R2 . (2.15) implies that I(n, n′ ) = 0 In addition, (2.14), (2.15) imply that I(n, n′ ) = 0 if and ? n, n′ , |I(n, n′ )| ≤ CT ?1 |n|?2 . From (2.18)–(2.20) I(n, n′ ) ≤ C
n,n′ ∈M n∈M : X(n)≤2T

if |X(n) ? X(n′ )| > C0 |n|?1 .

(2.18)

X(n) ≥ 2T,

(2.19)

(2.20)

T ?1 |n|?2
n′ ∈M : |X(n)?X(n′ )|≤C0 |n|?1

1.

Due to the 2/3–estimate of Sierpinski–Landau–Randol–Colin de Verdi`re [Sie], [Lan], [Ran], [CdV1], e 1 ≤ C|n|2/3 ,
n′ ∈M : |X(n)?X(n′ )|≤C0 |n|?1

hence I(n, n′ ) ≤ CT ?1
n,n′ ∈M n∈M : X(n)≤2T

|n|?4/3 ≤ C0 T ?1/3 .

(2.21)

Since (2.13) and (2.21) imply (2.10), Theorem 2.1 is proved. By (2.9) NBS (R) is the number of lattice points (k + (1/2), l) of a shifted square lattice in the sectorial domain ?(R) = {p ∈ R2 : Z2 (p) ≤ R2 , |p2 | ≤ p1 }. Observe that ?(R) is ? = ?(1) dilated with the coe?cient R, so the problem of ?nding the asymptotics of NBS (R) as R → ∞ reduces to a lattice–point problem about the asymptotics of the number of lattice points inside R?, where ? is a sectorial domain between diagonals p2 = ±p1 which is bounded by the curve Γ = {Z2 (p) = 1}. By (2.4) Γ is the graph of the function p1 = |p2 | + π ?1
a b

1 ? p2 f ?2 (s)ds. 2

Fig.3 shows the curve Γ found with the help of computer for the surfaces of revolution presented on Fig.2 above. Sections 3–5 below are devoted to the lattice–point problem for arbitrary sectorial domain between diagonals which is bounded by a “generic” curve Γ. 9

3. Asymptotics of Fourier Transform of Nonconvex Domains

In this section we consider the following auxiliary problem. Let ? be a sectorial domain on a plane, which is bounded by two segments [0, z0 ], [0, z1 ] and by a smooth curve Γ which goes from z0 to z1 (see Fig.4). We will assume that Γ has at most ?nitely many points of in?ection where the dσ = 0 at these curvature σ(p) vanishes, and all the points of in?ection are non–degenerate, i.e., ds points. We are interested in the asymptotics of the Fourier transform χ(ξ) = ?
?

exp(ipξ) dp,

as |ξ| → ∞. With the help of a partition of unity this problem can be reduced to a series of local problems of the following type. Let Γ be a smooth curve near some point p0 ∈ Γ and χ(p) = 1 from one side of Γ and χ(p) = 0
∞ from the other side of Γ (locally). Let ?(p) ∈ C0 be a C ∞ –function with compact support near

p0 . Assume in addition that ?(p) = 1 in the vicinity of p0 . Then the local problem is: what is the asymptotics of χ(ξ) = ?
R2

?(p)χ(p) exp(ipξ) dp,

as |ξ| → ∞? We will be also interested in the case when Γ has an angular point at p0 , and we will consider in a sequence the following cases: (a) Γ is either convex or concave near p0 ; (b) Γ has an in?ection point at p0 ; (c) Γ has an angular point at p0 with one side which is either convex or concave and with the other side which is a straight ray; (d) Γ is an angle between two straight rays; and (e) Γ is a straight line (see Fig.5). In the case (a) the answer is the following well–known lemma. Let Γ0 ? Γ be an open arc on Γ such that p0 ∈ Γ0 ? {p ∈ Γ : ?(p) = 1}, and V = {ξ ∈ R2 \ {0} : ? p = p(ξ) ∈ Γ0 such that nΓ (p) = |ξ|?1 ξ}, where nΓ (p) is the vector of normal to Γ at p ∈ Γ which looks in the direction where χ(p) = 0. De?ne Y (ξ) = ξ · p(ξ), 10 ξ ∈ R2 \ {0}. (3.1)

For the sake of brevity we will denote σ(p(ξ)) by σ(ξ). We will assign a sign to the curvature σ(p), so that σ(p) > 0 if the region {χ(p) = 1} is convex near p ∈ Γ, and σ(p) < 0 if this region is concave near p. If p2 = f (p1 ) is the equation of Γ near some p ∈ Γ in the coordinate system with ? an orthonormal basis e1 , e2 such that e2 = nΓ (?), then p σ(?) = ?f ′′ (?1 ). p p (3.2)

Lemma 3.1 (see [Hla]). If σ(p) = 0 on Γ0 then χ(ξ) = (2π)?1/2 |ξ|?3/2 |σ(ξ)|?1/2 cos(i(Y (ξ) ? φ)) + O(|ξ|?5/2 ), ? where φ = (π/2) + (π/4) sgn σ(ξ). (3.4) if ξ ∈ V, (3.3)

dσ 0 (p ) = 0. ds 0 In this case p is the turning point for nΓ (p), so that for small deviations of the direction of ξ from Assume now that p0 is a non–degenerate point of in?ection, i.e., σ(p0 ) = 0 and nΓ (p0 ) we have either two points p = p± (ξ) with nΓ (p) = |ξ|?1 ξ or no such point at all. De?ne for ξ ∈V, θ(ξ) = 1 if ? p ∈ Γ0 with nΓ (p) = |ξ|?1 ξ, 0 otherwise. To be de?nite in the choice of p± (ξ) we will assume that σ(p+ (ξ)) > 0 and σ(p? (ξ)) < 0. Let σ± (ξ) = σ(p± (ξ)) and Y± (ξ) = ξ · p± (ξ). Let α(ξ) be the angular coordinate of ξ. Let ?nally


(3.5)

Ai (t) =
?∞

exp(itτ + iτ 3 /3)dτ

be the Airy function. Recall that Ai (t) ∈ C ∞ (R1 ) and when t → ∞, Ai (?t) = π ?1/2 t?1/4 cos(ζ ? (π/4)) + O(t?7/4 ), Ai ′ (?t) = π ?1/2 t1/4 sin(ζ ? (π/4)) + O(t?5/4 ), |Ai (t)|, |Ai ′ (t)| ≤ C exp(?t), which implies that for t = 0, |Ai (t) ? θ(?t)π ?1/2 |t|?1/4 cos(ζ ? (π/4))| ≤ C|t|?7/4 , |Ai ′ (t) ? θ(?t)π ?1/2 |t|1/4 sin(ζ ? (π/4))| ≤ C|t|?5/4 , 11 (3.6) ζ = (2/3)t3/2 ,

where θ(t) = 1 for t ≥ 0 and θ(t) = 0 for t < 0. dσ 0 (p ) = 0. Then there exist real valued functions ds a(α), b(α) near α0 = α(nΓ (p0 )) such that a(α0 ) = 0, a′ (α0 ) = 0, b(α0 ) = Y (nΓ (p0 )) and Lemma 3.2. Assume that σ(p0 ) = 0 and χ(ξ) = ?i exp i|ξ|b(α(ξ)) ? |ξ|?4/3 Ai |ξ|2/3 a(α(ξ)) u(α(ξ)) (3.7)

+ |ξ|?5/3 Ai ′ |ξ|2/3 a(α(ξ)) v(α(ξ)) + O(|ξ|?7/3 ), if ξ ∈ V, where u(α), v(α) are C ∞ functions and u(α0 ) = 1 dσ 0 (p ) 2 ds
1/3

.

(3.8)

Corollary. If α(ξ) = α0 then |χ(ξ) ? θ(ξ)(χ+ (ξ) + χ? (ξ))| ≤ C|ξ|?5/2 |α(ξ) ? α0 |?7/4 where χ± (ξ) = |ξ|?3/2 |2πσ± (ξ)|?1/2 exp(i(Y± (ξ) ? φ± )), φ± = (π/2) ± (π/4). Proof of Corollary. From (3.6) and (3.7) we obtain that for α(ξ) = α = α0 , χm (ξ) = ?i exp i|ξ|b(α) |ξ|?3/2 θ(ξ) u0 (α) cos |ξ|a0 (α) ? (π/4) + v0 (α) sin |ξ|a0 (α) ? (π/4) + O(|ξ|?5/2 |α ? α0 |?7/4 ) (3.11) with some a0 (α), u0 (α) and v0 (α). On the other hand, Lemma 3.1 gives us that for a ?xed α(ξ) = α = α0 , χm (ξ) = θ(ξ)(χ+ (ξ) + χ? (ξ)) + O(|ξ|?5/2 ), m m |ξ| → ∞ (3.12) (3.10) (3.9)

Comparing (3.11) with (3.12) we obtain that the main terms in these two asymptotics coincide, hence χm (ξ) = θ(ξ)(χ+ (ξ) + χ? (ξ)) + O(|ξ|?5/2 |α ? α0 |?7/4 ), m m which proves Corollary. Proof of Lemma 3.2. Consider an orthonormal basis e1 , e2 on the plane with e2 = nΓ (p0 ). Let p = p0 + p1 e1 + p2 e2 , ξ = ξ1 e1 + ξ2 e2 and p2 = f (p1 ) be the equation of Γ near p0 . Observe that 12

f (0) = f ′ (0) = f ′′ (0) = 0 and we can choose the direction of e1 in such a way that f ′′′ (0) = Integrating by parts in p2 we obtain that χ(ξ) = ?
R2 ?1 ?(p)χ(p) exp(ipξ)dp = ?iξ2 exp(ip0 ξ) ∞ ?∞ ∞

dσ 0 (p ). ds

?(p1 , f (p1 )) exp(ip1 ξ1 + if (p1 )ξ2 )dp1
?∞

?1 + O(|ξ|?N ) = ?iξ2 exp(ip0 ξ)

?(p1 , f (p1 )) exp[i|ξ|Φ(p1 , α(ξ))]dp1 + O(|ξ|?N ),

where Φ(p1 , α) = p1 sin(α ? α0 ) + f (p1 ) cos(α ? α0 ) is a C ∞ real valued function with Φ= ?2Φ ?Φ = = 0, ?p1 ?p2 1 ?3Φ dσ 0 3 = ds (p ), ?p1 and ?2Φ = 0, ?p1 ?α

at p1 = 0, α = α0 . (3.7) follows now from Theorem 7.7.18 in [H¨r2]. Lemma 3.2 is proved. o Let us turn now to angular points. So, assume that Γ is a smooth curve near p0 ∈ Γ and L is a straight line which intersects Γ at p0 transversally. Then near p0 , Γ and L divide the plane into four parts. Let χ(p) = 1 in one of these parts and χ(p) = 0 in the remainder. Again we are interested in asymptotics of χ(ξ) as |ξ| → ∞. ? De?ne the auxiliary functions P± (y) = (2π)?1/2 exp(±πi/4) which are C ∞ bounded functions on R1 such that P± (?∞) = 0, P+ (y) = P? (y), P± (0) = 1/2, P± (∞) = 1; (3.14)
y ?∞

exp(?it2 /2)dt,

(3.13)

P± (y) + P± (?y) = 1;

|P± (y) ? θ(y)| ≤ C(1 + |y|)?1 Lemma 3.3. Assume σ(p) = 0 in vicinity of p0 ∈ Γ and L intersects Γ at p0 transversally. Then there exists C ∞ real valued functions a(α) and b(α) near α0 = α(nΓ (p0 )) such that a(α0 ) = 0, a′ (α0 ) = 0, b(α0 ) = Y (nΓ (p0 )) and χ(ξ) = |ξ|?3/2 exp[i|ξ|b(α(ξ)) ? iφ]Pζ |ξ|1/2 a(α(ξ)) u(α(ξ)) + O(|ξ|?2 ), if ξ ∈ V, ? (3.15)

where φ = (π/2) + (π/4) sgn σ(p0 ), ζ = sgn σ(p0 ) and u(α) is a C ∞ function near α0 with u(α0 ) = |2πσ(p0 )|?1/2 . Corollary. If ξ ∈ V and α(ξ) = α0 = α(nΓ (p0 )), then χ(ξ) ? |ξ|?3/2 |2πσ(ξ)|?1/2 θ(ξ) exp[i(Y (ξ) ? φ)] ≤ C|ξ|?2 |α(ξ) ? α0 |?1 , ? 13 (3.16)

where θ(ξ) = 1 if ? p = p(ξ) ∈ Γ with nΓ (p) = |ξ|?1 |ξ|, and θ(ξ) = 0 otherwise. Proof of Corollary. From (3.14), (3.15), we obtain that for α(ξ) = α = α0 χ(ξ) = |ξ|?3/2 exp[i|ξ|b(α) ? iφ]θ(ξ)u(α) + O(|ξ|?2 |α ? α0 |?1 ), ? On the other hand, when α(ξ) = α = α0 is ?xed, Lemma 3.1 proves that χ(ξ) = |ξ|?3/2 |2πσ(ξ)|?1 θ(ξ) exp[i(Y (ξ) ? φ)] + O(|ξ|?2 ), |ξ| → ∞, ξ ∈ V. ? Comparing these two asymptotics we conclude that the main terms in them coincide, and therefore (3.16) holds. Proof of Lemma 3.3. Consider a basis e1 , e2 on the plane, where e1 is a unit tangent vector to Γ at p0 and e2 is a unit tangent vector to L at p0 . Let e⊥ , e⊥ be a dual basis, e⊥ · ej = δij , and 2 1 i p = p0 + p1 e1 + p2 e2 , ξ = ξ1 e⊥ + ξ2 e⊥ . Then 2 χ(ξ) = ?
R2

ξ∈V

?(p)χ(p) exp(ipξ)dp
0 ?1 ξ)iξ2 ∞ 0

(3.17) ?(p1 , f (p1 )) exp(ip1 ξ1 + if (p1 )ξ2 )dp1 + O(|ξ|
?N

= J exp(ip

)

where p2 = f (p1 ) is the equation of Γ and J is the Jacobian. Observe that p1 ξ1 + f (p1 )ξ2 = |ξ|Φ(p1 ; α(ξ)) where Φ(p1 , α) is a C ∞ real valued function with Φ= at p1 = 0, α = α0 . By theorem 7.5.13 in [H¨r2] there exists a C ∞ change of variable t = t(p1 , α) with t(0, α0 ) = 0, o ?t (0, α0 ) = 0 such that Φ(p1 , α) = ?ζ(t2 /2) + b(α), b(α0 ) = 0, where ζ = sgn σ(p0 ). Hence ?p1


?Φ = 0, ?p1

?2Φ = 0, ?p2 1

?2Φ = 0, ?p1 ?α

I(ξ) ≡
0

?(p1 , f (p1 )) exp(i|ξ|Φ(p1 , α))dp1


= exp(i|ξ|b(α))
t(0,α)

ψ(t; α) exp(?i|ξ|ζt2 /2)

dp1 dt, dt

where ψ(t, α) = ?(p1 , f (p1 )), p1 = p1 (t, α). Observe that ψ(t, α) = 1 near (0, α0 ). Let dx1 = u0 (α) + tu1 (α, t) dt Then I(ξ) = exp(i|ξ|b(α))u0 (α)|ξ|?1/2 Pζ (?t(0, α)|ξ|1/2 ) + O(|ξ|?1 ) 14 (3.18)

From (3.17), (3.18) we obtain (3.15). Lemma 3.3 is proved. Lemma 3.4. Let χ(p) = 1 between two rays L1 = {p0 + λe1 , λ ≥ 0} and L2 = {p0 + λe2 , λ ≥ 0}, where e1 , e2 are linearly independent vectors, and χ(p) = 0 otherwise. Then |χ(ξ)| ≤ C|ξ|?1 (1 + |ξ · e1 |)?1 + (1 + |ξ · e2 |)?1 , ? and for λ → ±∞, χ(λe⊥ ) = Cj λ?1 + O(|λ|?2 ), ? j j = 1, 2; e⊥ · ej = δij . i (3.21) (3.20)

Proof. Let p = p0 + p1 e1 + p2 e2 , ξ = ξ1 e⊥ + ξ2 e⊥ . Without loss of generality we may assume 2 1 that |ξ2 | ≥ |ξ1 |. Then integrating by parts in p2 we obtain that
?1 χ(ξ) = iξ2 exp(ip0 ξ) ? ∞ 0

?(p1 , 0) exp(ip1 ξ1 )dp1 + O(|ξ|?2 ).

Integrating now by parts in p1 we obtain (3.20), while setting ξ1 = 0 we obtain (3.21). Lemma 3.4 is proved. Lemma 3.5. Let χ(p) = 1 from one side of the line L = {p0 + λe, λ ∈ R} and χ(p) = 0 from the other side of L. Then |χ(ξ)| ≤ C|ξ|?1 (1 + |ξ · e|)?1 , ? χ(λe⊥ ) = Cλ?1 exp(iλp0 · e⊥ ) + O(|λ|?2 ), ? |λ| → ∞.

The proof of Lemma 3.5 is similar to the proof of Lemma 3.4.

15

4. Lattice–Point Problem

Let Z(p) ∈ C ∞ (R2 \ {0}) be a C ∞ positive function homogeneous of degree 2. De?ne N (R) = #{n = (n1 , n2 ) ∈ Z2 : Z(n1 + (1/2), n2 ) ≤ R2 , |n2 | ≤ n1 }, ?(R) = {p ∈ R2 : Z(p) ≤ R2 , |p2 | ≤ p1 }, A(R) = AR2 = Area ?(R). The lattice–point problem we are interested in is to evaluate N (R) ? A(R) as R → ∞. Let Γ = {p ∈ R2 : Z(p) = 1, |p2 | ≤ p1 }, and z0 , z1 ∈ Γ be the endpoints of Γ with z01 = ?z02 > 0 and z11 = z12 > 0. For p ∈ Γ denote by nΓ (p) the vector of outer normal to Γ at p. Observe that nΓ (p) = | grad Z(p)|?1 grad Z(p), and p · grad Z(p) = 2Z(p) > 0, hence p · nΓ (p) = 2| grad Z(p)|?1 grad Z(p) > 0. (4.3) (4.2) (4.1)

Denote by σ(p) the curvature of Γ at p ∈ Γ with a sign, so that σ(p) > 0 if ? is convex near p, σ(p) < 0 if ? is concave near p. In what follows we assume the following Hypothesis D. (i) σ(p) = 0 everywhere on Γ except, maybe, a ?nite set W = {w1 , . . . , wK }, z0 , z1 ∈ W , and dσ (wk ) = 0, ds k = 1, . . . , K, (4.5)

where s is the natural coordinate on Γ. (ii) ? wk ∈ W the vector νk = nΓ (wk ) is either rational, i.e., n · νk = 0 for some n ∈ Z2 , n = 0, or Diophantine in the sense that ? 1 > ζ > 0 and C > 0 such that |n · νk | > C , |n|1+ζ ? n ∈ Z2 , n = 0. (4.6)

Without loss of generality we may assume that 0 = s(z0 ) < s(w1 ) < · · · < s(wK ) < s(z1 ), 16 (4.7)

where s(p) is the natural coordinate of p ∈ Γ. We call ξ ∈ R2 \ {0} rational if n · ξ = 0 for some n ∈ Z2 \ {0}. It is to be noted that ξ is rational i? the set L(ξ) = (Z2 \ {0}) ∩ {λξ, λ ∈ R} is non–empty. Let Γrat = {p ∈ Γ : nΓ (p) For p ∈ Γ de?ne Y (p) = p · nΓ (p). By (4.3) Y (p) > 0. Theorem 4.1. Assume Hypothesis D holds. Then N (R) = AR2 + R2/3
k : νk is rational

is rational}.

(4.8)

Φk (R) + R1/2 F (R),

(4.9)

where Φk (R) are continuous periodic functions, Φk (R) = (1/2)3?2/3 Γ(2/3)π ?4/3 dσ (wk ) ds
?1/3

(?1)n1 |n|?4/3 sin(2πY (wk )|n|R)
n∈L(νk )

(4.10)

and F (R) ∈ B 2 . The Fourier series of F (R) is F (R) = π ?1
p∈Γrat \W

θ(p)|σ(p)|?1/2
n∈L(nΓ (p))

(?1)n1 |n|?3/2 cos(2πY (p)|n|R ? φ(p)),

(4.11)

where 1 θ(p) = (1/2) and φ(p) = (π/2) + (π/4) sgn σ(p). We need Theorem 4.1 to prove our main Theorem 1.2. For the sake of completeness we want to formulate another theorem which is not needed for the proof of Theorem 1.2, but which is of interest by itself. Let α ∈ R2 be a ?xed point on the plane. De?ne N (R; α) = #{n ∈ Z2 : Z(n + α) ≤ R2 }. 17 if p = z0 , z1 , if p = z0 , z1 ,

Theorem 4.2. Assume Hypothesis D holds. Then N (R; α) = AR2 + R2/3
k : νk is rational

Φk (R; α) + R1/2 F (R; α),

where Φk (R; α) are continuous periodic functions of R, Φk (R; α) = (1/2)3?2/3 Γ(2/3)π ?4/3 dσ (wk ) ds
?1/3

|n|?4/3 sin(2π(Y (wk )|n|R ? n · α))
n∈L(νk )

and F (R; α) ∈ B 2 in R. The Fourier series of F (R; α) is F (R; α) = π ?1
p∈Γrat \W

|σ(p)|?1/2
n∈L(nΓ (p))

|n|?3/2 cos(2πY (p)|n|R ? φ(p, n, α)),

(4.12)

where φ(p, n, α) = (π/2) + (π/4) sgn σ(p) + 2πn · α. Theorem 4.2 is a generalization of Theorem 1.1 in [Ble1] to non–convex domains. Proof of Theorem 4.1. N (R) can be written as N (R) =
n∈Z2

χ(n + α; R),

α = (1/2, 0),

(4.13)

where χ(p; R) is the characteristic function of ?(R). De?ne for δ > 0, Nδ (R) =
n∈Z2

χδ (n + α; R),

(4.14)

where χδ (p; R) =
?(R)

δ?2 ?(δ?1 (p ? p′ )) dp′ = χ(·; R) ? (δ?2 ?(δ?1 ·))(p),

(4.15)

and
∞ ?(p) ∈ C0 (R2 ); ?(p) = ?0 (|p|); ?(p) ≥ 0;

?(p)dp = 1; ?(p) = 0
R2

when |p| ≥ 1.

Lemma 4.3. For T > 1, 1 T
T 1

|Nδ (R) ? N (R)|2 R?1 dR ≤ CδT (T ?1/3 + δ).

(4.16)

Proof of this and all subsequent lemmas is given in the next section. For what follows we put δ = T ?1 . 18 (4.17)

In this case (4.16) reduces to 1 T
T 1

|Nδ (R) ? N (R)|2 R?1 dR ≤ CT ?1/3 .

(4.18)

By the Poisson summation formula Nδ (R) ? AR2 = R2
n∈Z2 \{0}

?(2πnδ)χ(2πnR)(?1)n1 , ? ?

(4.19)

where χ(ξ) = ?
?

exp(ipξ)dp,

? = {p ∈ R2 : Z(p) ≤ 1, |p2 | ≤ p1 }.

Let us consider a partition of unity on the projective line RP1 ,
L

ψl (ξ) = 1,
l=1

0 ≤ ψl (ξ) ≤ 1,

ψl (ξ) ∈ C ∞ (RP1 ),

and lift it to R2 \ {0} putting ψ(λξ) = ψ(ξ) Let Γ(l) = {p ∈ Γ : nΓ (p) ∈ supp ψl (ξ)}, and Γj , j = 1, . . . , Jl , be connected components of Γ(l) . Without loss of generality we may assume that each Γj contains at most one singular point z0 , z1 , w1 , . . . , wK . Then, for a given l consider a partition of unity in the p–plane,
M (l) (l)

? λ ∈ R1 \ {0}.

χm (p) = 1,
m=1 (l)

0 ≤ χm (p) ≤ 1,

χm (p) ∈ C ∞ (?),
(l)

such that for each Γj , 1 ≤ j ≤ Jl , ? a unique χm (p) which is ≡ 0 on Γj . Without loss of generality we may also assume that for each m, χm (p) contains at most one singular point. De?ne Elm (R) = R2
n∈Z2 \{0}

ψl (n)?(2πnδ)χm (2πnR)(?1)n1 , ? ?

(4.20)

where χm (ξ) = ?
?

χm (p) exp(ipξ)dp. 19

Then by (4.19) Nδ (R) ? AR2 =
l,m

Elm (R).

Now we evaluate Elm (R). Let Γm = Γ ∩ supp χm . Lemma 4.4 (χm out of Γ(l) ). If Γm ∩ Γ(l) = ? then sup |Elm (R)| ≤ C log2 T.

1≤R≤T

Next we consider the case when Γj ? Γm and Γm contains no singular point. In this case ? ξ ∈ supp ψl ? a unique p(ξ) ∈ Γj such that nΓ (p(ξ)) = |ξ|?1 ξ. Denote by Y0 (ξ) = ξ · nΓ (p(ξ)), σ0 (ξ) = σ(p(ξ)).
(l) (l)

(l)

Lemma 4.5 (regular χm on Γ). Assume that z0 , z1 , w1 , . . . , wK ∈ Γm and Γj ? Γm . Then Elm (R) = R1/2 Flm (R), where Flm (R) ∈ B 2 and Flm (R) = π ?1
n∈Z2 \{0}

ψl (n)|n|?3/2 |σ0 (n)|?1/2 cos(2πY0 (n)R ? φ),

(4.21)

where φ = (π/2) + (π/4) sgn σ(p), p ∈ Γm . Assume now that some in?ection point wk lies inside Γj
(l)

and Γj

(l)

? Γm . Then for every
(l)

ξ ∈ supp ψl \ {λνk , λ ∈ R} two possibilities exist: either ? two points p± (ξ) ∈ Γj

such that

nΓ (p± (ξ)) = |ξ|?1 ξ, or there is no such point at all. De?ne the function θ(ξ) which is equal to 1 in the ?rst case, and which is equal to 0 in the second case. For the sake of de?niteness we will assume that ±σ(p± (ξ)) > 0. Denote by Y± (ξ) = ξ · nΓ (p± (ξ)); σ± (ξ) = σ(p± (ξ)).
(l)

Lemma 4.6. (χm on in?ection point). If νk = nΓ (wk ) ∈ Γj ? Γm and νk is rational then Elm (R) = R2/3 Φlm (R) + R1/2 Flm (R), (4.23)

where Φlm (R) is a periodic continuous function, which is given in (4.10), and Flm (R) ∈ B 2 . The Fourier series of Flm (R) is Flm (R) = π ?1
± n∈Z2 \{λνk , λ∈R}

ψl (n)θ(n)|n|?3/2 |σ± (n)|?1/2 cos(2πY± (n)R ? φ± ), 20

(4.25)

where φ± = (π/2) ± (π/4). If νk is Diophantine then Elm (R) = R1/2 Flm (R) where Flm (R) ∈ B 2 , and the Fourier series of Flm (R) coincides with (4.25). Lemma 4.7 (angular χm ). If nΓ (z0 ) ∈ Γj Flm (R) ∈ B 2 , and the Fourier series of Flm (R) is Flm (R) = π ?1
n∈Z2 \{0} (l)

? Γm then Elm (R) = R1/2 Flm (R), where

ψl (n)θ(n)|n|?3/2 |σ0 (n)|?1/2 cos(2πY0 (n)R ? φ),

(4.26)

where θ(n) = (1/2) if n = nΓ (z0 ), 0 otherwise, and φ = (π/2) + (π/4) sgn (z0 ). Proof of Lemmas 4.3–4.7 is given in the next section.

1 if n = nΓ (p) for some p ∈ Γj \ {z0 }, (4.27)

(l)

End of the proof of Theorem 4.1. Let us ?x l. By Lemma 4.4, 1 T →∞ T lim
T 1

|R?1/2 Elm (R)|2 dR = 0,

m : Γm ∩Γ(l) =?

and by Lemmas 4.5–4.7, Elm (R) = R2/3
m : Γm ∩Γ(l) =? m∈I1 (l)

Φlm (R) + R1/2
m∈I2 (l)

Flm (R),

where I1 (l) = {m : ? k, j such that wk ∈ Γj ? Γm and νk is rational}, Making a summation over l we obtain that if we de?ne
L L (l)

I2 (l) = {m : Γm ∩ Γ(l) = ?}.

Φ(R) =
l=1 m∈I1 (l)

Φlm (R),

F (R) =
l=1 m∈I2 (l)

Flm (R),

then 1 T →∞ T lim
1

T

|Nδ (R) ? AR2 ? R2/3 Φ(R) ? R1/2 F (R)|2 R?1 dR = 0.

Lemma 4.3 implies that the same is true for N (R): 1 T →∞ T lim
T 1

|N (R) ? AR2 ? R2/3 Φ(R) ? R1/2 F (R)|2 R?1 dR = 0. 21

Since Flm (R) ∈ B 2 , F (R) ∈ B 2 as well (see [Bes]). The Fourier series for Φ(R) and F (R) are the sum of the Fourier series for Φlm (R) and Flm (R), respectively. This proves the formulas (4.11), (4.12). Theorem 4.1 is proved. Theorem 4.2 is proved in the same way.

5. Proof of Lemmas

Proof of Lemma 4.3. We have: Nδ (R) ? N (R) =
n∈Z2

(χδ (n + α; R) ? χ(n + α; R)).

The support of χδ (p; R) ? χ(p; R) is concentrated in the δ–neighborhood of ??(R). Observe that ??(R) consists of the curve RΓ and two segments [0, Rz0 ] and [0, Rz1 ] oriented along the diagonals p1 ± p2 = 0. If δ is small, then the δ–neghborhood of [0, Rz0 ] ∪ [0, Rz1 ] does not contain points n + α, n ∈ Z2 , α = (1/2, 0), and it does not contribute to Nδ (R) ? N (R). Now, I≡ where Iδ (n, n′ ) = 1 T
T 0 T 0

1 T

|Nδ (R) ? N (R)|2 R?1 dR =
n,n′

Iδ (n, n′ ),

(χδ (n + α; R) ? χ(n + α; R))(χδ (n′ + α; R) ? χ(n′ + α; R))R?1 dR.

Observe that Iδ (n, n′ ) = 0 unless both n + α and n′ + α lie in δ–neighborhood of RΓ for some R, hence ? C > 0 such that Iδ (n, n′ ) = 0, if either |X(n + α) ? X(n′ + α)| ≥ Cδ or X(n + α) ≥ CT,

where X(p) = (Z(p))1/2 . In addition, for all n, n′ , Iδ (n, n′ ) ≤ CT ?1 δX(n)?1 ≤ C0 T ?1 δ|n|?1 . Therefore, I≤
n : X(n+α)≤CT

C0 T ?1 δ|n|?1
n′ : |X(n+α)?X(n′ +α)|≤Cδ

1

By 2/3–estimate (see, e.g., [CdV1]) 1 ≤ C(n2/3 + δ|n|),
n′ : |X(n+α)?X(n′ +α)|≤Cδ

22

hence I ≤ C1 T ?1 δ
n : X(n+α)≤CT

|n|?1 (|n|2/3 + δ|n|) ≤ C2 δT (T ?1/3 + δ).

Lemma 4.3 is proved. Proof of Lemma 4.4. Let us consider two cases: when 0 ∈ supp χm (p) and when z0 ∈ Γm (or z1 ∈ Γm ) and Γm ∩ Γ(l) = ?; all the other cases are simpler. Due to (4.20), Elm =
n∈Z2 \{0}

K(n)

with K(n) = ?(2πnδ)ψl (n)R2 χm (2πRn)(?1)n1 . ? ? By Lemma 3.4, if 0 ∈ supp χm (p) then R2 |χm (Rξ)| ≤ C|ξ|?1 max{R?1 + |ξ1 + ξ2 |?1 , R?1 + |ξ1 ? ξ2 |?1 }, ? so if n ∈ Λ, where Λ = {n ∈ Z2 : |n1 + n2 | ≥ |n1 ? n2 | > 0}, then R2 |χm (2πRn)| ≤ C|n|?1 |n1 ? n2 |?1 . ? Therefore I0 ≡
n∈Λ

(5.1)

K(n) is estimated as I0 ≤ C
n∈Λ

|?(2πnδ)| · |n|?1 |n1 ? n2 |?1 . ?

∞ Since ?(p) ∈ C0 (R2 ),

|?(ξ)| ≤ C(1 + |ξ|)?5 , ? and |?(2πnδ)| ≤ C0 , ?
|n|>T 2

(5.2)

δ = T ?1 .

On the other hand |n|?1 |n1 ? n2 |?1 ≤ C log2 T,
n∈Λ, |n|≤T 2

(5.3)

Hence I0 ≤ C log2 T . The sum over {|n1 ? n2 | ≥ |n1 + n2 | > 0} is estimated similarly hence we obtain K(n) ≤ C log2 T.
n : n1 ?n2 =0, n1 +n2 =0

23

By Lemma 3.4 K(n) = C1 R
n : n1 =n2 =0 n : n1 =n2 =0

?(2πnδ)ψl (n)n?1 (?1)n1 + O(1). ? 1

Since ?(ξ) and ψl (ξ) are even functions, ? ?(2πnδ)ψl (n)n?1 (?1)n1 = 0, ? 1
n : n1 =n2 =0

and thus K(n) = O(1).
n : n1 =n2 =0

Similarly, K(n) = O(1).
n : n1 =?n2 =0

As a result, |Elm (R)| =
n∈Z2 \{0}

K(n) ≤ C log2 T,

which was stated. Assume now that z0 ∈ Γm and Γm ∩ Γ(l) = ?. Let ?0 be the angle with the vertex at z0 and the sides which go along [z0 , 0] and the tangent vector to Γ at z0 , so that ?0 is a linear approximation to ? near z0 . Assume for the sake of de?niteness that σ(z0 ) > 0. Then ?0 ? ? near z0 . Let χ(0) (ξ) = ?m Then χ(0) (ξ) ? χm (ξ) = ?m ? χm (p) exp(ipξ)dp.
?0 \?

χm (p) exp(ipξ)dp.
?0

is estimated as follows. Integrating in the direction orthogonal to ξ we obtain that χ(0) (ξ) ? χm (ξ) = ?m ?
t1 t0

χm (t) exp(it|ξ|)dt,

where χm (t) is equal to zero in vicinity of t1 and χm (t) = a1 (t ? t0 )2 + a2 (t ? t0 )3 + . . . in vicinity of t0 . This implies that |χm (ξ) ? χm (ξ)| ≤ C(1 + |ξ|)?3 , hence ? ? (K(n) ? K (0) (n)) = O(R?1 ),
n∈Z2 \{0} (0)

24

where K (0) (n) = ?(2πnδ)ψl (n)R2 χ(0) (2πRn)(?1)n1 . ? ?m Now, ?0 is an angular domain. Using the same arguments as in the case 0 ∈ supp χm , we obtain that K (0) (n) = O(log2 T ).
n∈Z2 \{0}

This proves that
1≤R≤T

sup |Elm (R)| ≤ C log2 T.

Lemma 4.4 is proved. We omit proof of Lemma 4.5 and pass now to more complicated Lemmas 4.6, 4.7. Lemma 4.5 is proved similarly, with some simpli?cations (see also the proof of Theorem 3.1 in [Ble1]). Proof of Lemma 4.6. Assume νk ∈ Γj ? Γm . Let us split Elm (R) into two parts: Elm (R) =
n∈Z2 ∩L
γ , n=0

(l)

(1)

K(n),

Elm (R) =
n∈Z2 \L
γ

(2)

K(n),

(5.4)

where K(n) is de?ned in (5.1) and Lγ = {ξ ∈ R2 : dist (ξ, L) ≤ γ}, L = {ξ = λνk ,
(1)

λ ∈ R1 }.

with some γ > 0 which will be chosen later. Let us evaluate Elm (R). Assume ?rst that νk is rational. Then we can choose γ > 0 such that Lγ \L contains no integer points. In this case Elm (R) =
n∈L(νk ) (1)

?(2πnδ)R2 χ(2πnR)(?1)n1 . ? ?

By Lemma 3.5, if ξ ∈ L then χ(ξ) = ?i exp(iY (ξ))|ξ|?4/3 Ai (0)|(1/2)σ ′ (wk )|?1/3 + O(|ξ|?5/3 ), ? hence Elm (R) = R2/3 Φlm (R; δ) + O(R1/3 ), with Φlm (R; δ) =
n?L(νk ) (1)

Ai (0) = 3?2/3 Γ?1 (2/3),

?(2πnδ)(?i) sin(2πY (n)R) × |2πn|?4/3 Ai (0)|(1/2)σ ′ (wk )|?1/3 (?1)n1 . ? 25

Since



|1 ? ?(2πqrδ)|q ?4/3 ≤ Cδ1/3 = CT ?1/3 , ?
q=1

we obtain that Elm (R) = R2/3 Φlm (R) + O(R1/3 + R2/3 T ?1/3 ), where Φlm (R) = (1/2)3?2/3 π ?4/3 Γ?1 (2/3)|σ ′ (wk )|?1/3 is a periodic function of R. Assume now that νk is Diophantine, |n · νk | ≥ C|n|?1?ζ , In this case we put γ = 1. De?ne Elm (R; N ) =
n∈Z2 ∩L1 , 0<|n|≤N (1) (1)

(5.5)

(?1)n1 |n|?4/3 sin(2πY (n)R)
n∈L(νk )

(5.6)

0 < ζ < 1.

(5.7)

K(n).

Let us prove that
1≤R<∞

sup R?1/2 |Elm (R) ? Elm (R; N )| ≤ CN (ζ?1)/4 .

(1)

(1)

(5.8)

Indeed, |Ai (y)| ≤ C|y|?1/4 and |ak (α)| ≥ C0 |α ? α(νk )|, hence (3.7) implies that |χm (ξ)| ≤ C|ξ|?3/2 |α(ξ) ? α(νk )|?1/4 ? and |K(n)| ≤ CR1/2 |n|?3/2 |α(n) ? α(νk )|?1/4 . Therefore R?1/2 |Elm (R) ? Elm (R; N )| = R?1/2
n∈Z2 ∩L1 , |n|>N (1) (1)

K(n) (5.9)
?1/4

≤C
n∈Z2 ∩L1 , |n|>N

|n|

?3/2

|α(n) ? α(νk )|

.

Due to the Diophantine condition (5.7), |α(n) ? α(νk )| ≥ C|n|?(2+ζ) . 26

Let J ≥ N . Order all n ∈ Z2 ∩ L1 with 1 ≤ |n| ≤ 2J and n · νk > 0 in the increasing order of |n · νk | = |n2 νk1 ? n1 νk2 |: |n(1) · νk | ≤ |n(2) · νk | ≤ . . . Then (5.7) implies that |n(1) · νk | ≥ C0 J ?(1+ζ) and |n(j+1) · νk | ≥ |n(j) · νk | + C0 J ?(1+ζ) , so that |n(j) · νk | ≥ C0 jJ ?(1+ζ) , hence |α(n(j) ) ? α(νk )| ≥ C1 jJ ?(2+ζ) and
C0 J

j ≥ 1,

|n|
n∈Z2 ∩L1 , J≤|n|≤2J

?3/2

|α(n) ? α(νk )|

?1/4

≤ CJ

?3/2 (2+ζ)/4

J

j ?1/4 ≤ C1 J (ζ?1)/4 .
j=1

This implies |n|?3/2 |α(n) ? α(νk )|?1/4 ≤ CN (ζ?1)/4 ,
n∈Z2 ∩L1 , N >|n|

and hence (5.8) follows from (5.9). Let us evaluate now Elm (R). From (3.9)
? + |Elm (R) ? Elm (R) ? Elm (R)| (2)

(2)

≤ CR2
n∈Z2 \Lγ

ψl (n)|?(2πnδ)| · |Rn|?5/2 |α(n) ? α(νk )|?7/4 ?

(5.11)

where
± Elm (R) = n∈Z2 \Lγ

K ± (n)

and K ± (n) = R1/2 π ?1 ψl (n)?(2πnδ)θ(n)|n|?3/2 |σ± (n)|?1/2 (?1)n1 cos(2πRY± (n) ? φ± ). ? 27 (5.12)

The RHS in (5.11) is estimated as follows. Let d(ξ) = dist (ξ, L). Then RHS ≤ CR?1/2
n∈Z2 \Lγ

ψl (n)|?(2πnδ)| · |n|?5/2 d(n)?7/4 |n|7/4 ? ψl (n)|?(2πnδ)| · |n|?3/4 d(n)?7/4 . ?
n∈Z2 \Lγ

= CR?1/2 Since for p = 1, 2, . . . ,

ψl (n)d(n)?7/4 ≤ C,
n∈Z2 \Lγ , p≤|n|≤p+1

and sup
p≤|n|≤p+1

|?(2πnδ)| ≤ C(1 + pδ)?5 , ?

we obtain


RHS ≤ CR?1/2
p=1

(1 + pδ)?5 p?3/4 ≤ C0 R?1/2 δ?1/4 = C0 R?1/2 T 1/4 ,

so that
+ ? |Elm (R) ? Elm (R) ? Elm (R)| ≤ CR?1/2 T 1/4 . (2)

(5.13)

De?ne

± ± Flm (R) = R?1/2 Elm (R) = R?1/2 n∈Z2 \Lγ ± Flm (R; N, δ) = R?1/2 n∈Z2 \Lγ , |n|≤N

K ± (n), (5.14)

K ± (n)

The central point in our proof is Lemma 5.1. For all N, T ≥ 1, 1 T
T 0 ± ± |Flm (R) ? Flm (R; N, δ)|2 dR ≤ C(N ?1/3 + T ?1/4 ),

δ = T ?1 .

We will give the proof of Lemma 5.1 below, in the end of this section, and now let us derive Lemma 4.6 from Lemma 5.1. Assume that νk is rational. De?ne Flm (R) = R?1/2 (Elm (R) ? R2/3 Φlm (R)) with Φlm (R) given in (5.6). Then by (5.5) Flm (R) = R?1/2 Elm (R) + O(R?1/6 ), 28
(2)

1 ≤ R ≤ T,

and by (5.13)
+ ? R?1/2 Elm (R) = Flm (R) + Flm (R) + O(R?1 T 1/4 ), (2)

1 ≤ R ≤ T,

so that
+ ? Flm (R) = Flm (R) + Flm (R) + O(R?1/6 + R?1 T 1/4 ),

1 ≤ R ≤ T,

which implies that 1 T →∞ T lim De?ne
0

T

? + |Flm (R) ? Flm (R) ? Flm (R)|2 dR = 0.

(5.15)

± Flm (R; N ) = n∈Z2 \Lγ , |n|≤N

K± (n)

where K± (n) = π ?1 ψl (n)θ(n)|n|?3/2 |σ± (n)|?1/2 (?1)n1 cos(2πRY± (n) ? φ± ).
± Observe that Flm (R; N ) is a ?nite trigonometric sum, hence ± ± |Flm (R; N, δ) ? Flm (R; N )| ≤ C(N )δ = C(N )T ?1 .

(5.16)

Therefore Lemma 5.1 implies that lim sup
T →∞

1 T

T 0

± ± |Flm (R) ? Flm (R; N )|2 dR ≤ CN ?1/3 ,

hence from (5.15) we deduce that lim sup
T →∞

1 T

T 0

? + |Flm (R) ? Flm (R; N ) ? Flm (R; N )|2 dR ≤ CN ?1/3 .

(5.17)

This implies that Flm (R) ∈ B 2 and (4.19) is the Fourier expansion of Flm (R). For rational νk Lemma 4.6 is proved. In the case of Diophantine νk we de?ne Flm (R) = R?1/2 Elm (R). Then Flm (R) = Flm (R) + Flm (R), and similarly to (5.17) we have that lim sup
T →∞ (1) (2)

Flm (R) = R?1/2 Elm (R),

(j)

(j)

j = 1, 2,

1 T

T 0

? + |Flm (R) ? Flm (R; N ) ? Flm (R; N )|2 dR ≤ CN ?1/3

(2)

(5.18)

± with Flm (R; N ) de?ned in (5.16).

29

Then, (5.8) implies that lim sup
T →∞

1 T

T 0

|Flm (R) ? Flm (R; N )|dR ≤ CN (ζ?1)/4 ,

(1)

(1)

(5.19)

where Flm (R; N ) = R3/2
n∈Lγ , 0<|n|≤N (1)

ψl (n)?(2πnδ)χ(2πnR)(?1)n1 . ? ?

This is a ?nite sum and from (3.9) we obtain that
(1) ?+ ?? |Flm (R; N ) ? Flm (R; N ) ? Flm (R; N )| ≤ C(N )(R?1 + T ?1 )

(5.20)

with ?± Flm (R) =
n∈Z2 ∩Lγ , 0<|n|≤N

K± (n).

It follows from (5.18)–(5.20) that lim sup
T →∞

1 T ?

+ ?+ |Flm (R) ? Flm (R; N ) ? Flm (R; N ) 0 ? ?? Flm (R; N ) ? Flm (R; N )|2 dR ≤ CN (ζ?1)/4 ,

T

which proves Lemma 4.6 for Diophantine νk . Proof of Lemma 5.1. To simplify notations we will omit ± in sub– and superscripts. From (5.12) and (5.14) 1 T
T 0

|Flm (R) ? Flm (R; N, δ)|2 dR = π ?2
n,n′ ∈Z2 \Lγ ; |n|,|n′ |>N

ψl (n)ψl (n′ )

(5.21)

?(2πnδ)?(2πn′ δ)θ(n)θ(n′ )|n|?3/2 |n′ |?3/2 |σ(n)|1/2 |σ(n′ )|1/2 I(n, n′ ), ? ? where I(n, n′ ) = Observe that |I(n, n′ )| ≤ min{1, T ?1 |Y (n) ? Y (n′ )|?1 }. In addition, |?(2πnδ)| ≤ C(1 + Y (n)δ)?5 , ? C|n| ≤ Y (n) ≤ C ′ |n|, (5.22) 1 T
0 T

cos(2πY (n)R ? φ(n)) cos(2πY (n′ )R ? φ(n′ ))dR.

and the RHS of (5.21) is symmetric in n, n′ . This implies that 1 T
T

|Flm (R) ? Flm (R; N, δ)|2 dR ≤ C0
0 n,n′ ∈Z2 \Lγ ; Y (n)≥Y (n′ )>βN

H(n, n′ )

(5.23)

30

with

H(n, n′ ) = ψl (n)ψl (n′ )(1 + Y (n)δ)?5 θ(n)θ(n′ )Y (n)?3/2 Y (n′ )?3/2 |σ(n)|?1/2 |σ(n′ )|?1/2 min{1, T ?1 (Y (n) ? Y (n′ ))?1 }

and β > 0. First we estimate

(n,n′ )∈S(N )

H(n, n′ ) with (5.24)

S(N ) = {n, n′ ∈ Z2 \ Lγ ; Y (n) ≥ Y (n′ ) > βN ; Y (n) ? Y (n′ ) ≥ 1}. Let us ?x some n with Y (n) ≥ βN and de?ne the layers S(N, n, j) = {n′ : n′ ∈ Z2 \ Lγ ; Y (n′ ) ≥ βN ; j + 1 ≥ Y (n) ? Y (n′ ) ≥ j}, 1 ≤ j ≤ Y (n) ? βN. Observe that |σ(n)| ≤ C|α(n) ? α(νk )|?1/2 , hence ψl (n′ )θ(n′ )|σ(n′ )|1/2 ≤ C Area S(N, n, j) ≤ C0 (Y (n) ? j)
n′ ∈S(N,n,j)

and therefore ψl (n′ )θ(n′ )|σ(n′ )|1/2 Y (n′ )?3/2 T ?1 (Y (n) ? Y (n′ ))?1
n′ ∈S(N,n) Y (n)?1

≤C
j=1

(Y (n) ? j)?1/2 T ?1 j ?1 ≤ C0 T ?1 Y (n)?1/2 log Y (n),

where S(N, n) = ∪j S(N, n, j). Making a summation in n we obtain now that H(n, n′ ) ≤ CT ?1
(n,n′ )∈S(N ) n∈Z2 \Lγ ; Y (n)>βN

ψl (n)(1 + Y (n)δ)?5

× Y (n)?2 log Y (n)|σ(n)|?1/2 . De?ne the layers Sj = {n ∈ Z2 \ Lγ , j ≤ Y (n) ≤ j + 1}, Then ψl (n)θ(n)|σ(n)|?1/2 ≤ C Area Sj ≤ C0 j,
n∈Sj

j > βN.

hence H(n, n′ ) ≤ CT ?1
(n,n′ )∈S(N )



(1 + jδ)?5 j ?1 log j
j=βN

(5.25)

≤ C0 T ?1 | log δ|2 = C0 T ?1 log2 T. 31

It remains to estimate

(n,n′ )∈S0 (N )

H(n, n′ ) where

S0 (N ) = {n, n′ ∈ Z2 \ Lγ ; Y (n) ≥ Y (n′ ) > βN ; 1 ≥ Y (n) ? Y (n′ ) ≥ 0}. Let us ?x some n with T ≥ Y (n) > βN . De?ne the layers S0 (N, n, j) = {n′ ∈ Z2 \ Lγ : Y (n′ ) > βN ; (j + 1)T ?1 ≥ Y (n) ? Y (n′ ) ≥ jT ?1 }, j = 0, 1, . . . , T. To estimate the sum over S0 (N, n, j) we use the following Lemma 5.2 (2/3–estimate). Let Y (ξ) = |ξ|f (|α(ξ) ? α0 |1/2 ), (5.26)

where f (t) ∈ C ∞ ([0, ε]), ε > 0, f (t) > 0, f ′ (0) = 0, f ′′′ (0) = 0. Then if ψ(α) ∈ C ∞ ([α ? 0, α0 + ε]) with supp ψ(α) near α0 then ψ(α(n))|α(n) ? α0 |?1/2 ≤ CR2/3 ,
n∈Π(R), dist (n,L)>1

(5.27)

where Π(R) = {n ∈ Z2 : α0 ≤ α(n) ≤ α0 + ε; R ≤ Y (n) ≤ R + R?1/3 } and L = {ξ : α(ξ) = α0 }. Remark. If we step away from α0 putting α0 + ε0 ≤ α(n) ≤ α0 + ε, ε0 > 0, in Π(R) (instead of α0 ≤ α(n) ≤ α0 + ε), then (5.27) reduces to a well–known 2/3–estimate (see, e.g., [CdV1]). Proof of Lemma 5.2 is given in Appendix to the paper. With the help of Lemma 5.2 we obtain that if T 3 ≥ Y (n) then ψl (n′ )θ(n′ )|σ(n′ )|?1/2 ≤ CY (n)2/3
n′ ∈S0 (N,n,j)

(observe that |σ(n′ )|?1/2 ≤ C|α(n′ ) ? α(νk )|?1/4 ≤ C|α(n′ ) ? α(νk )|?1/2 ), and ψl (n′ )θ(n′ )|σ(n′ )|?1/2 Y (n′ )?3/2 min{1, T ?1 (Y (n) ? Y (n′ ))?1 }
n′ ∈S0 (N,n,j)

≤ CY (n)?3/2 Y (n)2/3 T ?1 j ?1 Y (n) = CY (n)1/6 T ?1 j ?1 , 32

if

T ≥ j ≥ 2,

and ≤ CY (n)?5/6 if j = 0, 1. Hence ψl (n′ )θ(n′ )|σ(n′ )|?1/2 Y (n′ )?3/2 min{1, T ?1 (Y (n) ? Y (n′ ))?1 }
n′ ∈S0 (N,n)

≤ C(Y (n)1/6 T ?1 log T + Y (n)?5/6 ), where S0 (N, n) = {n′ ∈ Z2 \ Lγ : Y (n′ ) > βN ; 1 ≥ Y (n) ? Y (n′ ) ≥ 0}. Making a summation over n, we obtain H(n, n′ ) ≤ C
(n,n′ )∈S0 (N ); Y (n)≤T 3 n∈Z2 \Lγ ; T 3 ≥Y (n)≥βN

ψl (n)(1 + Y (n)δ)?5

θ(n)Y (n)?3/2 |σ(n)|?1/2 (Y (n)1/6 T ?1 log Y (n) + Y (n)?5/6 ) ≡ CI0 . Consider the layers Sj = {n ∈ Z2 \ Lγ , j + 1 ≥ Y (n) ≥ j}, T 3 ≥ j ≥ βN.

The width of Sj is of order of 1, and a simple argument shows that ψl (n)θ(n)|σ(n)|?1/2 ≤ Cj.
n∈Sj

This implies
T3

I0 ≤ C
j=βN

(1 + jδ)?5 j ?3/2 j(j 1/6 T ?1 log j + j ?5/6 )

≤ C0 (δ?2/3 | log δ|T ?1 + N ?1/3 ) = C0 (T ?1/3 log T + N ?1/3 ). Thus H(n, n′ ) ≤ C(T ?1/3 log T + N ?1/3 )
(n,n′ )∈S0 (N ); Y (n)≤T 3

(5.28)

The ?nal step is to estimate H(n, n′ )
(n,n′ )∈S0 (N ); Y (n)≥T 3

and this is quite simple. Since for (n, n′ ) ∈ S0 (N ), H(n, n′ ) ≤ C(1 + Y (n)δ)?5 Y (n)?3 |σ(n)|?1/2 |σ(n′ )|?1/2 and |σ(n′ )|?1/2 ≤ CY (n),
n′ : Y (n)≥Y (n′ )≥Y (n)?1; n′ ∈Lγ

33

we obtain H(n, n′ )
(n,n′ )∈S0 (N ); Y (n)≥T 3

≤C
(n,n′ )∈S0 (N ); Y (n)≥T 3 ∞

(1 + Y (n)δ)?5 Y (n)?2 |σ(n)|?1/2 (1 + jδ)?5 j ≤ C1 T ?2 .
j=T 3

(5.29)

≤ C0

From (5.23), (5.25), (5.28) and (5.29) Lemma 5.1 follows. Proof of Lemma 4.7. The proof of Lemma 4.7 is similar in main steps to the proof of Lemma 4.6. Assume ν0 = nΓ (z0 ) ∈ Γj ? Γm . By (4.15) Flm (R) =
n∈Z2 \{0} (l)

K(n).

(5.30)

with K(n) = R3/2 ?(2πnδ)ψl (n)χm (2πRn)(?1)n1 ? ? De?ne L = {λν0 , Flm (R) =
n∈L1 , n=0 (1) (2) (1)

λ ∈ R},

L1 = {n ∈ Z2 : dist (n, L) ≤ 1}, Flm (R) =
n∈L1 (2)

K(n),

K(n),

so that Flm (R) = Flm (R) + Flm (R). By (3.15) |K(n)| ≤ C|n|?3/2 , hence |K(n)| ≤ CN ?1/2 .
n∈L1 , |n|≥N

(5.31)

De?ne K0 (n) = π ?1 |n|?3/2 ?(2πnδ)|σ0 (n)|?1/2 θ(n) cos(2πRY0 (n) ? φ). ? By (3.18) |K(n) ? K0 (n)|
n∈L1 , |n|≥N

(5.32)

≤ CR?1/2
n∈L1 , |n|≥N

|?(2πnδ)| · |n|?2 |α(n) ? α0 |?1 ≤ R?1/2 log2 T. ?

(5.33)

The following lemma holds: 34

Lemma 5.3. For all N, T ≥ 1, 1 T
T 1 n∈L1 , |n|≥N 2

K0 (n) dR ≤ C(N ?1/3 + T ?1/4 ).

The omit the proof of this lemma since it basically the same as the proof of Lemma 3.3 in [Ble1] (see also the proof of Lemma 5.1 above, where a similar statement was proved in a more complicated situation). Lemma 3.3 implies that for a ?xed n ∈ Z2 \ {0}, K(n) converges to K0 (n) as R → ∞, hence 1 lim T →∞ T and by Lemma 5.3, 1 lim T →∞ T This proves Lemma 4.7.
T T 1 0<|n|<N 2

(K(n) ? K0 (n)) dR = 0,

2

Flm (R) ?
1 0<|n|<N

K0 (n) dR = 0.

6. Energy Levels and Closed Geodesics

In this section we prove Theorem 1.2. Let N (R) = #{En ≤ R2 }, and Γ = {p ∈ R2 : Z2 (p) = 1, |p2 | ≤ p1 }. Recall that DH is de?ned by (1.12) and Hypothesis D by (4.5), (4.6). Lemma 6.1. DH implies Hypothesis D for Γ. Proof. By (2.4) Γ is the graph of the function p1 = g(p2 ) ≡ |p2 | + π ?1
a b

NBS (R) = #{n ∈ Z2 : Z2 (n1 + (1/2), n2 ) ≤ R2 , |n2 | ≤ n1 }

(6.1)

(6.2)

(1 ? p2 f ?2 (s))1/2 ds, 2

f (a) = f (b) = |p2 |,

|p2 | ≤ fmax .

(6.3)

g(p2 ) is an even C ∞ function, so we will assume p2 ≥ 0. The function dp1 = 1 ? π ?1 dp2
b a

p2 f ?2 (s)(1 ? p2 f ?2 (s))?1/2 ds 2 35

(6.4)

has a nice geometric interpretation. Proposition 6.2. dp1 dp2 Proof. An equation of γ(I) is d? = If ?2 (s), dl where l is the normal coordinate on γ. Hence d? = If ?2 (s)(1 ? I 2 f ?2 (s))?1/2 ds and ω(I) = π ?1
a b

= ?ω(I),
p2 =I

I ≥ 0.

(6.5)

ds = (1 ? I 2 f ?2 (s))1/2 , dl

(6.6)

d? ds ? 1 = π ?1 ds

b a

If ?2 (s)(1 ? I 2 f ?2 (s))?1/2 ds ? 1.

Comparing this with (6.4) we obtain (6.5). Proposition 6.2 is proved. d2 p1 = 0. By (6.5) this is equivalent dp2 2 to ω ′ (I) = 0. Similarly, the nondegeneracy of the in?ection point is characterized by ω ′′ (I) = 0. In Observe that an in?ection point on Γ is characterized by addition, the Diophantine condition (4.6) is equivalent to (1.12), hence DH implies Hypothesis D. Lemma 6.1 is proved. Lemma 6.1 implies that Theorem 4.1 holds for NBS (R). Our goal now is to ?nd geometric interpretation of frequencies and amplitudes in formulas (4.11) (4.12). Consider a ?nite geodesic γ which starts at x0 = (smax , 0) at some angle ?π/2 ≤ α ≤ π/2 to the direction to the north. γ is uniquely determined by I = sin α and l = |γ|, γ = γ(I, l). Let G be the set of all γ(I, l), ?1 ≤ I ≤ 1, l > 0. Assume that Γ is de?ned as in (6.3). De?ne two maps: p : G → Γ, where p : γ = γ(I, l) → p(γ) = (g(I), I), ξ : γ = γ(I, l) → ξ(γ) = (lτ ?1 (I), ω(I)lτ ?1 (I)). (6.7) ξ : G → R2 \ {0}, (6.6)

Proposition 6.3. p(γ) and ξ(γ) satisfy nΓ (p(γ)) = |ξ(γ)|?1 ξ(γ) 36 (6.8)

and p(γ) · ξ(γ) = (2π)?1 |γ|. (6.9)

Proof. (6.5) implies that nΓ (p(γ)) is collinear with the vector (1, ω(I)) as well as ξ(γ), hence (6.8) follows. To prove (6.9) observe that the both sides of (6.9) depend linearly on |γ|, so it is su?cient to prove (6.9) in the particular case when |γ| = τ (I). In this case (6.9) reduces to g(I) + Iτ (I) = (2π)?1 τ (I). Since g(I) = I + π ?1
a b

(6.10)

(1 ? I 2 f ?2 (s))1/2 ds,

ω(I) = π ?1
a

b

If ?2 (s)(1 ? I 2 f ?2 (s))?1/2 ds ? 1,

and by (6.6)
b

τ (I) = 2
a

dl ds = 2 ds

b a

(1 ? I 2 f ?2 (s))?1/2 ds,

(6.10) follows. Proposition 6.3 is proved. (6.8) implies that ξ(γ) ∈ L+ (nΓ (p(γ)) = {λnΓ (p(γ)), λ > 0}. Hence we can de?ne the map π : G → N+ Γ, where N+ Γ = ∪p∈Γ L+ (nΓ (p)), as π : γ → (p(γ), ξ(γ)). Observe that π is one–to–one. Proposition 6.4. γ ∈ G is a closed geodesic i? ξ(γ) ∈ Z2 . In this case Y (p(γ))|ξ(γ)| = (2π)?1 |γ|, |σ(p(γ))|?1/2 |ξ(γ)|?3/2 = |ω ′ (I)|?1/2 τ (I)3/2 |γ|?3/2 , sgn σ(p(γ)) = sgn ω ′ (I), γ = γ(I, l). (6.11) (6.12) (6.13)

Proof. Observe that γ(I, l), I = 0, is a closed geodesic with n1 revolutions around the axis and n2 oscillations along the meridian i? l = |γ| = n2 τ (I) and ω(I) = (n1 /n2 ) ? 1, so that 37

ξ(γ) = (n2 , n1 ? n2 ) ∈ Z2 . Similarly, γ = γ(0, l) is a closed geodesic i? l = |γ| = n2 τ (I), so that ξ(γ) = (n2 , 0). This proves the ?rst part of Proposition 6.4. To prove (6.11) let us notice that Y (p(γ))|ξ(γ)| = p(γ) · ξ(γ), hence (6.11) follows from (6.9). Let us prove (6.12). We have: σ(I) = ? and since g′ (I) = ?ω(I), σ(p(γ)) = On the other hand, by (6.7) |ξ(γ)| = |γ|τ (I)?1 (1 + ω(I)2 )1/2 , hence |σ(p(γ))|?1/2 |ξ(γ)|?3/2 = |ω ′ (I)|?1/2 (1 + ω(I)2 )3/4 |γ|?3/2 τ (I)3/2 (1 + ω(I)2 )?3/4 = |ω ′ (I)|?1/2 τ (I)3/2 |γ|?3/2 . (6.12) is proved. (6.13) follows from (6.14). Proposition 6.4 is proved. Proof of Theorem 1.2. From Lemma 6.1 and Theorem 4.1 we obtain that NBS (R) = AR2 + R2/3
k : nΓ (wk ) is rational

g′′ (I) , (1 + (g′ (I))2 )3/2 ω ′ (I) . (1 + ω(I)2 )3/2

(6.14)

Φk (R) + R1/2 F (R),

where Φk (R) are periodic continuous functions and F (R) ∈ B 2 . In addition, the Fourier series of Φk (R) and F (R) are given in formulas (4.11) and (4.12), respectively. From Theorem 2.1 we obtain now that N (R) = AR2 + R2/3
k : nΓ (wk ) is rational

? Φk (R) + R1/2 F (R),

where 1 T →∞ T lim
0

T

? |F (R) ? F (R)|2 dR = 0.

? ? This implies that F (R) ∈ B 2 as well and the Fourier series of F (R) and F (R) coincide. If we substitute formulas of Lemma 6.4 into (4.11), (4.12) we obtain (1.14), (1.15). Theorem 1.2 is proved.

38

Appendix. Proof of Lemma 5.2

For the sake of de?niteness we will assume that f ′′′ (0) > 0. Let Γ? = {Y (ξ) = 1} ∩ {α0 ≤ α(ξ) ≤ α0 + ε} and ξ 0 = Γ? ∩ {α(ξ) = α0 }. Consider a basis e1 , e2 on the plane such that e1 = ξ 0 and e2 is parallel to the tangent vector to Γ? at ξ0 (see Fig.6). In the basis e1 , e2 the equation of Γ? has the form ξ1 = h(ξ2 ), where h(t) ∈ C ∞ , h(0) = 1, h′ (0) = h′′ (0) = 0, h′′′ (0) > 0. Let us choose the length of e2 in such a way that h′′′ (0) = 2, so that h(t) = 1 + t3 /3 + . . . for small t. Let e⊥ , e⊥ be a dual basis to e1 , e2 . Let 0 ≤ λ(t) ≤ 1 be a C ∞ function on a line which is equal to 0 1 2 near 0, and which is equal to 1 when t > 1. We will show that I(R) ≡
Z2 ∩Π(R) ⊥ where n⊥ = n · e⊥ , j = 1, 2. Observe that |α(ξ) ? α0 |?1/2 ≤ C0 |ξ2 |?1/2 R1/2 when ξ ∈ Π(R), hence j j 1/2

λ(n⊥ )ψ(n⊥ )(n⊥ )?1/2 ≤ CR1/6 , 2 2 2

(A.1)

from (A.1) Lemma 5.2 follows.
⊥ ⊥ ⊥ Let Q(ξ; R) = λ(ξ2 )ψ(R?1 ξ2 )|ξ2 |?1/2 , and Q0 (ξ; R) = Q(ξ; R) if ξ ∈ Π(R) and Q0 (ξ; R) = 0

otherwise, so that I(R) =
n∈Z2

Q0 (n; R).

Let Qδ (ξ; R) =
Π(R) ∞ where ?(ξ) ∈ C0 (R2 ), ?(ξ) ≥ 0, ?(ξ) = 0 if |ξ| > 1 and

Q(η; R)δ?2 ?(δ?1 (ξ ? η))dη = Q0 (·; R) ? (δ?r2 ?(δ?1 ·))(ξ), ?(ξ)dξ = 1, and

δ > 0,

R2

Iδ (R) =
n∈Z2

Qδ (n; R).

Put δ = R?1/3 . Observe that Q0 (ξ; R) ≤ CQδ (ξ; R), so I(R) ≤ CIδ (R) and to prove (A.1) it is su?cient to show that Iδ (R) ≤ CR1/6 . 39 (A.3) (A.2)

By the Poisson summation formula Iδ (R) =
n∈Z2

? ?(2πnδ)Q(2πn; R). ?

(A.4).

The term n = 0 is ? Q(0; R) =
Π(R)

Q(ξ; R)dξ ≤ CR?1/3
0

εR

⊥ ⊥ |ξ2 |?1/2 dξ2 ≤ C0 R1/6 ,

hence we may consider only n = 0. Let p1 = p · e1 , p2 = p · e2 . If p2 ≥ γ|p1 |, γ > 0, then ? Q(p; R) =
Π(R)

Q(ξ; R) exp(ipξ)dξ

(A.5)

can be estimated in the following way. First we integrate in (A.5) by the lines |p|?1 p · ξ = c and then in c, so that ? Q(p; R) = where S(c; R) =
Π(R)∩{|p|?1 p|ξ|=c} ⊥ ⊥ ⊥ λ(ξ2 )ψ(R?1 ξ2 )|ξ2 |?1/2 dξ. ∞

exp(i|p|c)S(c; R)dc,
?∞

If ε ? γ, then the lines |p|?1 p · ξ = c cross Π(R) transversally, which implies that S(c; R) = R?1/3 λ0 (c ? c0 )ψ0 (R?1 (c ? c0 ); R)|c ? c0 |?1/2 , where c0 = R|p|?1 p · ξ 0 , λ0 (t) ∈ C ∞ , λ0 (t) = 0 in vicinity of 0, and λ0 (t) = 1 in vicinity of ∞ and
∞ ∞ ψ0 (t; R) ∈ C0 ([0, ∞)) has a limit in C0 –topology as R → ∞. Therefore

? |Q(p; R)| ≤ CR?1/3 |p|?5 and ? Q(2πn; R) ≤ CR?1/3 .
n : |n·e2 |≥γ|n·e1 |, n=0

(A.6)

? The main di?culty is to estimate Q(p; R) when |p2 | < γp1 , γ > 0. We have: ? Q(p; R) = J t=
0 ⊥ ξ2 , ∞

dt exp(ip2 t)λ(t)ψ(R?1 t)t?1/2
a ⊥ ξ1 ,

b

ds exp(ip1 s),
?1/3

(A.7)

s=

a = h(t 40

1/2

; R), b = h(t; R + R

),

where J is Jacobian and h(t; R) = Rh(R?1 t). Let ? Q0 (p; R) = J
0 ∞

dt exp(ip2 t)ψ(R?1 t)t?1/2
a

b

ds exp(ip1 s),

(A.8)

? ? ? The di?erence Q1 (p; R) = Q0 (p; R) ? Q(p; R) can be estimated as follows. Observe that
∞ 0

dt exp(ip2 t)(1 ? λ(t))ψ(R?1 t)t?1/2 ≤ C(1 + p2 )?1/2 ,

hence ? |Q1 (p : R)| ≤ C(1 + p2 )?1/2 R?1/3 and ? Q1 (2πn; R) ≤ CR?1/3
n : |n·e2 |<γ|n·e1 |, n=0 n∈Z2

|?(2πnδ)|(1 + |n · e2 |)?1/2 ?

≤ C0 R?1/3 δ?3/2 = C0 R1/6 . ? Thus it remains to estimate a similar sum with Q0 (2πn; R). Let us integrate in s in (A.8) and make the change of variable u = (R?1 t)1/2 . This gives ? Q0 (p; R) = 2J(ip1 )?1 R1/2 W (p; R) ? U (p; R) , where U (p; R) =
0



du exp(iRp2 u2 )ψ(u2 ) exp(iRp1 h(u)),


(A.9) du exp(iRp2 u )ψ(u ) exp[iR(1 + R
2 2 ?4/3

W (p; R) =
0

)p1 h((1 + R

?4/3

)u)].

Let us evaluate ?rst


U (p; R) =
0

du exp[iRp1 (yu2 + h(u))]ψ(u2 ),

y = p2 /p1 .

(A.10)

There exists a C ∞ change of variable T = T (u, y) such that T (0) = 0, yu2 + h(u) = b(y) + a(y)T + T 3 /3, where a(y), b(y) ∈ C ∞ , a(0) = b(0) = 0 (see [H¨r2]). In addition, o a(y) + T 2 (0, y) = 0, 41

?T ?T (0) = (0) = 1 and ?u ?y (A.11)

which follows if we di?erentiate both sides of (A.11) at u = 0. After this change of variable U (p; R) reduces to


U (p; R) = exp(iRp1 b(y))
c(y)

exp(iRp1 (?c2 (y)T + T 3 /3))ψ0 (T, y)

?u (T, y)dT, ?T

(A.12)

with c(y) = T (0, y) and ψ0 (T, y) = ψ(t2 (T, y)). Following [H¨r2] let us divide (?u/?T )(T, y) by o ?c2 (y) + T 2 with a remainder: ?u (T, y) = r(T, y)(?c2 (y) + T 2 ) + r0 (y) + r1 (y)T. ?T Then the ?rst term after substitution into (A.12) allows integration by parts, which gives an extra (Rp1 )?1 , and the other two terms give the main contribution to U (p; R): U (p; R) = exp(iRp1 b(y)) (Rp1 )?1/3 V (c(y)(Rp1 )1/3 )r0 (y) + (Rp1 )?2/3 V0 (c(y)(Rp1 )1/3 )r1 (y) + O((Rp1 )?1 ), (A.13) where
∞ ∞

V (x) =
x

exp[i(?x2 T + T 3 /3)]dT,

V0 (x) =
x

T exp[i(?x2 T + T 3 /3)]dT.

The method of stationary phase gives the asymptotics of V (x), V0 (x) when |x| → ∞, and from this asymptotics we obtain |V (x)| ≤ C(1 + |x|)?1/2 and |V0 (x)| ≤ C(1 + |x|)1/2 . Therefore |U (p; R)| ≤ C|Rp1 |?1/3 min{1, (|p2 /p1 | |Rp1 |1/3 )?1/2 } = C min{R?1/2 |p2 |?1/2 , R?1/3 |p1 |?1/3 }. A similar estimate holds for W (p; R) and ?nally we obtain ? |Q0 (p; R)| ≤ C|p1 |?1 min{|p2 |?1/2 , R1/6 |p1 |?1/3 }. Hence ? ?(2πnδ)Q0 (2πn : R) ≤ CR1/6 ?
|n·e2 |≤1 |n·e2 |≤1

|n · e1 |?4/3 ≤ C0 R1/6 ,

and ? ?(2πnδ)Q0 (2πn : R) ≤ C ?
1<|n·e2 |≤γ|n·e1 | 1<|n·e2 |≤γ|n·e1 |

|?(2πnδ)||n · e1 |?1 |n · e2 |?1/2 ?

≤ C0 δ?1/2 = C0 R1/6 . Lemma 5.2 is proved. Acknowledgements. The author thanks Freeman Dyson, Dennis Hejhal and Peter Sarnak for useful discussions of the paper. This work was done at the Institute for Advanced Study, Princeton, and the author is grateful to IAS for ?nancial support during his stay at the Institute. The work was also supported by the Ambrose Monell Foundation. 42

References

[B?r] P. H. B?rard, On the wave equation on a compact Riemannian manifold without conjugate e e points, Math. Z., 155, 249–276 (1977). [Bes] A. S. Besicovitch, Almost periodic functions, Dover Publications, New York, 1958. [Ble1] P. M. Bleher, On the distribution of the number of lattice points inside a family of convex ovals, Duke Math. Journ. 67, 3, 461–481 (1992). [Ble2] P. M. Bleher, Distribution of the error term in the Weyl asymptotics for the Laplace operator on a two–dimensional torus and related lattice problems, Preprint, Institute for Advanced Study, IASSNS-HEP-92/80, 1992 (to appear in Duke Math. Journ.). [BCDL] P. M. Bleher, Zh. Cheng, F. J. Dyson and J. L. Lebowitz, Distribution of the error term for the number of lattice points inside a shifted circle, Preprint, Inst. Adv. Study, IASSNS-HEP92/10, 1992 (to appear in Commun. Math. Phys.). [BL] P. M. Bleher and J. L. Lebowitz, Energy–level statistics of model quantum systems: universality and scaling in a lattice–point problem, Preprint, Inst. Adv. Study, IASSNS-HEP-93/14, 1993. [CdV1] Y. Colin de Verdi`re, Nombre de points entiers dans une famille homoth`tique de domaines de e e ? Rn , Ann. Scient. Ec. Norm. Sup., 4e s?rie, 10, 559–576 (1977). e [CdV2] Y. Colin de Verdi`re, Spectre conjoint d’op?rateurs pseudo–di??rentiels qui commutent. II. Le e e e cas int?grable, Math. Z., 171, 51–73 (1980). e [DG] J. J. Duistermaat and V. W. Guillemin, The spectrum of positive elliptic operators and periodic bicharacteristics, Inventiones math., 29, 39–79 (1975). [Har1] G. H. Hardy, The average order of the arithmetical functions P (x) and ?(x), Proc. London Math. Soc., 15, 192–213 (1916). [Har2] G. H. Hardy, On Dirichlet’s divisor problem, Proc. London Math. Soc., 15, 1–25 (1916). [HR] D. A. Hejhal and B. Rackner, On the topography of Maass waveforms for PSL(2,Z): experiments and heuristics, Preprint, University of Minnesota, UMSI 92/162, 1992. ¨ [Hla] E. Hlawka, Uber Integrale auf konvexen K¨rpern. I, Monatsh. Math., 54, 1–36 (1950). o 43

[H¨r1] L. H¨rmander, The spectral function of an elliptic operator, Acta Math., 121, 193–218 (1968). o o [H¨r2] L. H¨rmander, The analysis of linear partial di?erential operators. I. Distribution theory and o o Fourier analysis. Springer–Verlag, Berlin e.a., 1983. [Hux] M. N. Huxley, Exponential sums and lattice points. II. Preprint, University of Wales College of Cardi?, Cardi?, 1992. [Lan] E. Landau, Vorlesungen uber Zahlentheorie, Chelsea, New York, 1969. ¨ [LS] W. Luo and P. Sarnak, Number variance for arithmetic hyperbolic surfaces, Preprint, Princeton University, Princeton, 1993. [Ran] B. Randol, A lattice–point problem, Trans. Amer. Math. Soc. 121, 257–268 (1966). [Sar] P. Sarnak, Arithmetic quantum chaos, Schur Lectures, Tel Aviv, 1992. [Sie] W. Sierpinski, Oeuvres Choisies, Vol. I, P.W.N., Warsaw, 1974, 73–108.

44

Figures Captions

Fig.1. Geodesic on a surface of revolution. Fig.2. The phase function ω(I) for di?erent surfaces of revolution. Cross–sections of the surfaces of revolution are shown in the lower part of the ?gure. Fig.3. Curve Γ for the surfaces of revolution shown on Fig.2. Fig.4. Sectorial domain with points of in?ection on the boundary. Fig.5. Local structures of Γ. Fig.6. Basis e1 , e2 .

45


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