The hyperbolic lattice point problem in conjugacy classes

arXiv:1504.01307v1 [math.NT] 6 Apr 2015

THE HYPERBOLIC LATTICE POINT PROBLEM IN CONJUGACY CLASSES DIMITRIOS CHATZAKOS AND YIANNIS N. PETRIDIS Abstract. For Γ a cocompact or cofinite Fuchsian group, we study the hyperbolic lattice point problem in conjugacy classes, which is a modification of the classical hyperbolic lattice point problem. We use large sieve inequalities for the Riemann surfaces Γ\H to obtain average results for the error term, which are conjecturally optimal. We give a new proof of the error bound O(X 2/3 ), due to A. Good. For SL2 (Z) we interpret our results in terms of indefinite quadratic forms.

1. Introduction Let H be the hyperbolic plane, z, w be two fixed points in H and ρ(z, w) denote the hyperbolic distance. Let also Γ ⊂ PSL2 (R) be a cocompact or cofinite Fuchsian group. The classical hyperbolic lattice point problem asks to estimate the number of points in the orbit Γz that belong in a disk of radius R and center w, i.e. to give an asymptotic formula for #{γ ∈ Γ : ρ(γz, w) ≤ R}. Let cosh ρ(z, w) = 2u(z, w) + 1, where u(z, w) is the standard point-pair invariant function u(z, w) =

|z − w|2 , 4ℑ(z)ℑ(w)

and define H(X; z, w) = #{γ ∈ Γ : 4u(z, γw) + 2 ≤ X}.

Selberg [22] proved that (1.1)

H(X; z, w) =

X

π 1/2

1/2 0}. Huber’s interpretation shows that N (H, X; z) actually counts γ in Γ/hgi such that cos v ≥ X −1 , where v is the angle defined by the ray from 0 to γz and the geodesic {yi, y > 0}. For Γ cocompact or cofinite, Good [11] proved a general sum formula that covers many cases of decompositions of the group G = SL2 (R). One of these cases corresponds to Huber’s hyperbolic lattice point problem in conjugacy classes. In Good’s notation the hyperbolic lattice point problem in conjugacy classes corresponds to the η Gζ case, whereas the classical one corresponds to the ζ Gζ case [11, p. 20, Eq. (3.12)]. His method is based on defining certain Poincar´e series as sums over cosets of a hyperbolic subgroup of Γ. After a matching of notation Good’s general formula [11, Theorem 4, p. 116] implies X µ 2 X + 2λH λz Nz (t) = aj (H, z)X sj + Ez (t), vol(Γ\H) ν 1 2 0, we define I(A) and J(A) by Z A
0 0 Ps−1 (ix) + Ps−1 (−ix) (A − x)dx, I(A) = 0


−2 −2 J(A) = (A2 + 1) Ps−1 (iA) + Ps−1 (−iA) . Then, it is easy to see that    √ −1 s  I(V ) − I(U ) s+1 + Γ 1− · . d(f , t) = (2 π) Γ 2 2 V −U

HYPERBOLIC LATTICE-POINT COUNTING

7

Lemma 2.2. The functions I(A) and J(A) satisfy the relation −2 I(A) = J(A) − 2Ps−1 (0).

Proof. Using integration by parts, the formula [9, p. 968, eq. 8.752.3], and the fact −1 that the function (z 2 − 1)1/2 Ps−1 (z) is single-valued in the disk with center (1, 0) and radius 2, we get Z A
−1 −1 (−x2 − 1)1/2 Ps−1 (ix) − Ps−1 I(A) = −i (−ix) dx. 0

Using again twice [9, p. 968, eq. 8.752.3] for m = 1, 2 we get   A −2 −2 I(A) = (x2 + 1) Ps−1 (ix) + Ps−1 (−ix) . 0

The result is immediate.



2.3. Proof of Theorem 1.1 for the cocompact case. Lemma 2.2 implies that    √ s+1 s  J(V ) − J(U ) Γ 1− · . (2.8) d(f + , t) = (2 π)−1 Γ 2 2 V −U Relation [9, p. 971, eq. 8.776.1] implies

(2.9)

d(f + , t)


(V 2 + 1)V s−1 − (U 2 + 1)U s−1 · 1 + O(U −2 ) V −U
(V 2 + 1)V −s − (U 2 + 1)U −s · 1 + O(U −2 ) , + D(s) · V −U

= B(s) ·

where (2.10)

(2.11)



Γ 1 − 2s Γ s − 21 B(s) = 2 , πΓ(s + 2)  π  Γ s+1
Γ 1 − s
Γ 1 − s
π 2 2 2 D(s) = ei 2 (−s) + e−i 2 (−s) . πΓ(3 − s)2s+1 s−2

 π Γ π ei 2 (s−1) + e−i 2 (s−1)

s+1 2




Proposition 2.3. a) For any s = 1/2 + it we have       1 3 1 3 + 1/2+it d(f , t) = B + it + it X +D + it − it X 1/2−it 2 2 2 2       1 1 2 −1/2+it 2 −1/2−it + it |t| X Y +D + it |t| X Y + O B 2 2

b) Let t ∈ R (i.e ℜ(s) = 1/2) and t 6= 0. Then, d(f + , t) can be written in the form d(f + , t) = a(t, Y /X)X 1/2+it + b(t, Y /X)X 1/2−it ,

where the coefficients a(t, Y /X) and b(t, Y /X) satisfy the bound
a(t, Y /X), b(t, Y /X) = O t−2 min{t, XY −1 } . Hence

  d(f + , t) = O t−2 min{t, XY −1 }X 1/2 .

c) Let t ∈ / R, i.e s ∈ (1/2, 1]. Then d(f + , t) =

B(s)(s + 1)X s + D(s)(2 − s)X 1−s

+O(Γ(s − 1/2)Y + Γ(1/2 − s)X 1/2 ).

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DIMITRIOS CHATZAKOS AND YIANNIS N. PETRIDIS

d) For t = 0 we get d(f + , 0) = O(X 1/2 log X). Proof. a) First, apply the mean value theorem to the function f (x) = xs+1 + xs−1 to get (V 2 + 1)V s−1 − (U 2 + 1)U s−1 = (s + 1)X s + O(s(s + 1)X s−1 Y + (s − 2)X −1 ). V −U Applying it again to the function g(x) = x2−s + x−s , we have (V 2 + 1)V −s − (U 2 + 1)U −s = (2 − s)X 1−s + O((2 − s)(1 − s)X −s Y + (−s)X −3/2 ). V −U Plugging s = 1/2 + it in (2.9) and using that O(U −2 ) = O(X −2 ) and the above estimates, we get the result. b) First, consider the function f (x) as above. We know from part a) that the terms containing X 1/2+it come from the terms contaning f (x). The mean value theorem imples (V 2 + 1)V s−1 − (U 2 + 1)U s−1 ≪ |t| · X 1/2 , V −U whereas, trivial estimates imply

Hence, if we set

(V 2 + 1)V s−1 − (U 2 + 1)U s−1 ≪ X 3/2 Y −1 . V −U


(V 2 + 1)V s−1 − (U 2 + 1)U s−1 −(1/2+it) a(t, Y /X) = 1 + O(U −2 ) B(s) · X , V −U and use the Stirling’s formula for the Γ function, we get the bound
a(t, Y /X) = O t−2 min{t, XY −1 } .

Doing the same for g(x) as above and the coefficient b(t, Y /X) defined as
(V 2 + 1)V −s − (U 2 + 1)U −s −(1/2−it) b(t, Y /X) = 1 + O(U −2 ) D(s) · X , V −U we get b). c) It follows from a), after estimating three of the Γ-factors in B(s), D(s) and keeping the ones that appear in the error term. d) Putting t = 0 in (2.8) we get √ H(V ) − H(U ) d(f + , 0) = (2 π)−1 Γ2 (3/4) V −U   −2 −2 where H(z) = (z 2 + 1) P−1/2 (iz) + P−1/2 (−iz) . Thus, applying once again the mean value theorem, there exists a ξ ∈ [U, V ] such that √ d(f + , 0) = (2 π)−1 Γ2 (3/4)H ′ (ξ).

For H ′ (z) we have    d  −2 −2 −2 −2 H ′ (z) = 2z P−1/2 (iz) + P−1/2 (−iz) + (z 2 + 1) P−1/2 (iz) + P−1/2 (−iz) . dz Formula [9, p. 964, eq. (8.731.1)] implies  5i   3z  −2 −2 −2 −2 (2.12) H ′ (z) = P−1/2 (iz) − P−1/2 (−iz) − P1/2 (iz) − P1/2 (−iz) . 2 2

HYPERBOLIC LATTICE-POINT COUNTING

9

Consider the first bracket. Using formula [9, p. 961, eq. (8.713.2)] we get Z ∞
−5/4 −2 −2 2 cosh2 t + ξ 2 dt P−1/2 (iξ) − P−1/2 (−iξ) ≪ (ξ + 1) 0

≪ ξ Setting x = cosh t/ξ we get !−5/4 2 Z ∞  cosh t +1 dt ξ 0

∞

Z

−1/2



0

=

Z

=

Z

∞

1

x2 + 1

1/ξ

+

2


−5/4

ξ dx (ξ 2 x2 − 1)1/2

x2 + 1

1/ξ

∞

Z

!−5/4

cosh t ξ


−5/4

x2 + 1

1

+1

dt.

ξ dx (ξ 2 x2 − 1)1/2


−5/4

ξ dx. − 1)1/2

(ξ 2 x2

Since U, V → ∞, we can assume that ξ ≥ 2. We see that Z ∞ Z ∞
−5/4
−5/4 ξ dx = O(1) x2 + 1 dx ≪ x2 + 1 2 2 1/2 (ξ x − 1) 1 1 and, after setting u = xξ, Z 1
−5/4 ξ dx = x2 + 1 2 2 (ξ x − 1)1/2 1/ξ ≤ Combining these estimates we get

Z

ξ

1

Z

1

ξ



ξ2 u2 + ξ 2

5/4

(u2

du ξ 1/2 ξ − 1)

1 √ du ≪ log ξ. 2 u −1

−2 −2 P−1/2 (iξ) + P−1/2 (−iξ) ≪ ξ −1/2 log ξ.

For the second bracket, using once again [9, p. 961, eq. (8.713.2)], we get Z ∞
−5/4 −2 −2 2 cosh t cosh2 t + ξ 2 dt P1/2 (iξ) − P1/2 (−iξ) ≪ (ξ + 1) 0

≪ ξ

1/2

Z

0

∞

cosh t ξ



cosh t ξ

2

+1

!−5/4

dt.

As above, set x = cosh t/ξ and split the integral into two integrals: Z 1 Z ∞
−5/4
−5/4 ξx ξx x2 + 1 x2 + 1 dx + dx. 2 2 1/2 2 x2 − 1)1/2 (ξ x − 1) (ξ 1/ξ 1

As above, assuming ξ ≥ 2, the second integral is easily seen to converge, whereas Rξ the first one, setting u = xξ is again bound by 1 (u2 −1)−1/2 du. Finally, combining all the above estimates, we get d(f + , 0) ≪ H ′ (ξ) ≪ ξ 1/2 log ξ ≪ V 1/2 log V,

which implies the desired bound, since X ∼ U and V < 2U .



In order to bound E(H, X; z), we will also need the following bound (weak type of Weyl’s law) for the period integrals of uˆj ’s defined in Lemma 2.1:

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DIMITRIOS CHATZAKOS AND YIANNIS N. PETRIDIS

Lemma 2.4 (Huber, [15, eq. (63), p. 24] ). For the sequence of the period integrals {ˆ uj }∞ j=0 , the following estimate holds: X |ˆ uj |2 ≪ T. tj ≤T

The exact asymptotic behavior was first proved by Good [11, Theorem 2, p. 108], see also [17] and in bigger generality by Tsuzuki [23, Theorem 1, p. 2]. We can now prove Theorem 1.1 when Γ is cocompact: Theorem 2.5. Let Γ be a cocompact Fuchsian group, and H a hyperbolic conjugacy class of Γ. Then the error term E(H, X; z) satisfies the bound E(H, X; z) = O(X 2/3 ). Proof. We begin with the spectral expansion of A(f + ): X X 2d(f + , tj )ˆ uj uj (z) c(f + , tj )uj (z) = A(f + )(z) = j

j

Using Proposition 2.3, we write it in the form X 2B(sj )(sj + 1)ˆ uj uj (z)X sj + A(f + )(z) = 1/2 X 1/2 , then

(4.1)

R 1 X |E(H, Xm ; z)|2 ≪ X log2 X. R m=1

Letting R go to infinity, we get Z 1 2X (4.2) |E(H, x; z)|2 dx ≪ X log2 X. X X

For the spatial average, we use Theorem 4.2 to prove the following.

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DIMITRIOS CHATZAKOS AND YIANNIS N. PETRIDIS

Proposition 4.5. Let X > 2 and z1 , z2 , ..., zR be points in Γ\H away from the cusps, satisfying the condition ρ(zi , zj ) > δ for some δ > 0, when i 6= j. Then, we have R X (4.3) |E(H, X; zm )|2 ≪ δ −2 X + R1/3 X 4/3 log2 X, m=1

and

(4.4)

R X

m=1

|E(H, X; zm )|4 ≪ δ −2 X 2 log4 X + R1/3 X 8/3 log3 X,

where the ‘≪’ constants depend on Γ, H and z.

Theorem 4.6. If Rδ 2 ≫ 1 and R > X 1/2 , then, for n = 1, 2 R 1 X |E(H, X; zm )|2n ≪ X n log2n X, R m=1

Letting R go to infinity, if Γ is cocompact, we get Z |E(H, X; z)|2n dµ(z) ≪ X n log2n X. Γ\H

Before giving the proof of the above results, we need to fix the following notation. For a function f ∈ C0∗ [1, ∞), denote by Ef (H, X; z) the difference X 2d(f, tj )ˆ uj uj (z). Ef (H, X; z) = A(f )(z) − 1/2≤sj ≤1

In the proofs of Theorems 2.5 in section 2 and 3.4 in section 3 we proved that for Γ cocompact or cofinite we have (4.5)

Ef + (H, X; z) = O(XY −1/2 + X 1/2 ),

(4.6)

Ef − (H, X; z) = O(XY −1/2 + X 1/2 ),

(4.7)

Ef − (H, X; z) < E(H, X; z) + O(Y + X 1/2 log X) < Ef + (H, X; z).

We begin with the proof of Proposition 4.3. Proof. (of Proposition 4.3) We choose Y such that X 1/2 log X ≪ Y ≪ X. We get Ef − (H, X; z) < E(H, X; z) + O(Y ) < Ef + (H, X; z).

We choose f to be f + or f − as in (2.6) and (2.7) with X = Xm and U given by (2.5). We have R X

2

m=1

|E(H, Xm ; z)| ≪

R X

m=1

|Ef (H, Xm ; z)|2 + RY 2 .

The estimates below are true for f = f + or f − . We write X d(f, tj )ˆ uj uj (z) S(X, z, T ) = 2 T X 2 Y −2 },

= {t : |t| > X 2 Y −2 }.

Using the notation Z X 1X d(f, tj )ˆ uj uj (z) + Si (z) := 2 d(f, t)Eˆa (1/2 + it)Ea (z, 1/2 + it)dt, π a Bi tj ∈Ai

Ef (H, X; z) can be written as Ef (H, X; z) = S1 (z) + S2 (z) + S3 (z). We first estimate S3 (z). Using the estimates for tj ∈ R, we get the bound X X |tj |−2 min{tj , X/Y }X 1/2 u ˆj uj (z) 2d(f, tj )ˆ uj uj (z) ≪ tj ∈A3

|tj |>X 2 Y −2

≪

X

3/2 −1 t−2 Y u ˆj uj (z). j X

tj >X 2 Y −2

Using dyadic decomposition, this is bounded by  ∞ X X  X 3/2 Y −1 n=0

2n X 2 Y −2