The Prime Geodesic Theorem

Matt Tyler

Throughout these notes, we write Γ for SL2(Z).

Recall that a quadratic form Q(x, y) = [a, b, c] = ax2+bxy+cy2 is primitive if gcd(a, b, c) =1. The discriminant of [a, b, c] is d = b2−4ac. We will be concerned here with the indefinitequadratic forms, which are those forms with non-square discriminant d > 0. The group Γacts on the set of quadratic forms via the action((

α βγ δ

)Q

)(x, y) = Q(αx+ βy, γx+ δy). (0.1)

This action preserves both the discriminant and the property of being primitive. We saythat two quadratic forms in the same orbit are equivalent.

The stabilizer of the quadratic form [a, b, c] in Γ is the group

Γ[a,b,c] =

{±(x−by

2 −cyay x+by

2

)| x2 − dy2

}∼= Z× Z/2Z. (0.2)

If x0, y0 > 0 are the fundamental solutions to the Pell equation x2 − dy2, then we have

Γ[a,b,c] =⟨±M[a,b,c]

⟩where M[a,b,c] =

(x0−by02 −cy0

ay0x0+by0

2

). (0.3)

The two important quantities depending on the discriminant d are

εd =x0 +

√dy0

2and hd =

∣∣∣{equivalence classes of primitiveforms of discriminant d

}∣∣∣ , (0.4)

which correspond to the fundamental unit and narrow class number for Q(√d), respectively.

Let D be the set of positive discriminants {d > 0 | d ≡ 0, 1 (mod 4), d non-square}, and letD(x) = {d ∈ D | εd ≤ x}. The purpose of these notes is to prove the following asymptoticexpression due to Sarnak for the average value of hd ordered by εd.

Theorem 0.1.1

|D(x)|∑

d∈D(x)

hd =16

35

li(x2)

x+O

(x2/3+ε

)where li(u) is the logarithmic integral

∫ u2

dtlog t , as in the ordinary prime number theorem.

1

Summing by parts, we immediately obtain the following corollary.

Corollary 0.2.1

|D(x)|∑

d∈D(x)

hd log εd =8

35x+O

(x2/3+ε

).

By way of comparison, the following asymptotic expression for the average value of hd log εdordered by discriminant was noticed by Gauss and confirmed by Siegel.

Theorem 0.3.

1

|{d ∈ D | d ≤ x}|∑d∈Dd≤x

hd log εd =π2

9ζ(3)

√x+O(x log x).

In §1, we will prove Theorem 0.1 (assuming the prime geodesic theorem for Γ and an asymp-totic expression for |D(x)|) by exploiting a correspondence between equivalence classes ofprimitive binary quadratic forms, primitive hyperbolic conjugacy classes in Γ, and closedgeodesics on Γ\H. The main ingredient in the proof of the prime geodesic theorem is theSelberg trace formula, which we will explain in §2. We will then prove the prime geodesictheorem in §3, and discuss an improvement due to Soundararajan and Young in §4. Inappendix A, we prove the asymptotic formula for |D(x)|. The material in §1 and appendixA comes from Sarnak [5], the material in §2 and §3 comes primarily from the books ofBergeron [1] and Iwaniec [3], and the material in §4 comes from Soundararajan and Young[6].

Contents

1 Closed Geodesics on SL2(Z)\H 3

2 The Selberg Trace Formula 42.1 The spectral trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Parabolic classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 The identity class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 Hyperbolic classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.5 Elliptic classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.6 The full trace formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 The Prime Geodesic Theorem 9

4 Improvements to the Prime Geodesic Theorem 12

A Asymptotics for |D(x)| 17

2

1 Closed Geodesics on SL2(Z)\HLet H be the Poincare model of the hyperbolic plane, which is the upper half-plane {x+iy |y > 0} with the Riemannian metric. The group SL2(R) acts on H via fractional lineartransformations

(a bc d

)z = az+b

cz+d , which are isometries of H.

Γ\H

i

− 12

12

Our goal in this section is to explain the correspondence betweenequivalence classes of primitive binary quadratic forms, primitivehyperbolic conjugacy classes in Γ, and closed geodesics on Γ\H.We begin with the correspondence between the latter two col-lections (which actually holds for any discrete subgroup Γ′ ofSL2(R)).

We say that g ∈ SL2(R) is hyperbolic if its fixed points as a linearfractional transformation are in R = R ∪ {∞} and distinct. Ifg is hyperbolic, then it is conjugate (in SL2(R)) to a homothety(p1/2 0

0 p−1/2

)for some p > 1 (with p determined by the formula

tr g = p1/2 + p−1/2), and we say the norm Ng of g is p.

We say that a hyperbolic element γ ∈ Γ is primitive if it is nota non-trivial power in Γ, so that every hyperbolic element of Γis a power of a unique primitive hyperbolic in Γ. Note that con-jugates of hyperbolic elements are hyperbolic, and similarly forprimitive hyperbolics, so we may speak of primitive hyperbolicconjugacy classes in Γ.

The geodesics on H are the semicircles perpendicular to the real line and the vertical lines,so there is a unique geodesic connecting any two points in R. Each hyperbolic elementg ∈ SL2(R) maps the geodesic ag (of length logNg) connecting the two fixed points of gback to itself. For γ ∈ Γ hyperbolic, aγ is a closed geodesic on Γ\H, and any Γ-conjugateof γ gives rise to the same geodesic. This gives a map{

primitive hyperbolicconjugacy classes [γ]

}∼−→{

closed geodesics onΓ\H of length logNγ

}, (1.1)

and it is easy to see that this map is bijective.

Returning to binary quadratic forms, recall the matrix M[a,b,c] for a quadratic form [a, b, c]

(defined in (0.3)), which is a primitive hyperbolic with trace x0 = εd+ ε−1d and hence norm

ε2d. The map [a, b, c] 7→M[a,b,c] is Γ-invariant and in fact gives a bijection{equivalence classes of primitive

binary quadratic forms

}∼−→{

primitive hyperbolicconjugacy classes

}(1.2)

which sends a form of discriminant d to conjugacy class of norm ε2d.

3

Combining the correspondences (1.1) and (1.2), we arrive at the following proposition.

Proposition 1.1. The lengths of the closed geodesics on Γ\H are the numbers 2 log εd withmultiplicity hd.

In light of this reformulation, Theorem 0.3 follows from the asymptotic expression

|D(x)| = 35

16x+O

(x2/3+ε

)(1.3)

proved in the appendix, as well as the following result.

Theorem 1.2 (Prime geodesic theorem).

π(x) = li(x) +O

(x3/4

log x

)

where π(x) is the number of closed geodesics on Γ\H of length at most log x.

Proof of Theorem 0.1. By the prime geodesic theorem and the proposition above,

∑d∈D(x)

hd = π(x2) = li(x2) +O

(x3/2

log x

). (1.4)

Dividing by (1.3), we find

1

|D(x)|∑

d∈D(x)

hd =16

35

li(x2)

x+O

(x2/3+ε

), (1.5)

as claimed.

2 The Selberg Trace Formula

Given a function k : H×H→ C, consider the integral operator L with kernel k defined by

(Lf)(z) =

∫Hk(z, w)f(w)dµw. (2.1)

This operator is SL2(R)-invariant (i.e. it commutes with precomposition with g for allg ∈ SL2(R)) if and only if

k(gz, gw) = k(z, w) for all g ∈ SL2(R), (2.2)

4

in which case k depends only on the hyperbolic distance ρ(z, w) between z and w. Thishyperbolic distance satisfies

cosh ρ(z, w) = 1 + 2u(z, w) where u(z, w) =|z − w|2

4 Im z Imw, (2.3)

so we may write k(z, w) as k(u(z, w)) where k(u) is a function in one variable u ≥ 0.

These invariant integral operators have the following important property. Whenever f :

H→ C is an eigenfunction of the Laplacian ∆ = y2(∂2

∂x2+ ∂2

∂y2

), f is also an eigenfunction

of L. In particular, if (∆ + λ)f = 0, then∫Hk(z, w)f(w)dµw = h(t)f(z) (2.4)

where λ = 14 + t2 and h(t) is the Selberg/Harish-Chandra transformation of k defined by

q(v) =

∫ ∞v

k(u)√u− v

du,

h(r) = 2q((sinh r/2)2),

h(t) =

∫ ∞−∞

eirth(r)dr.

(2.5)

If we restrict the domain of L to automorphic functions (i.e. functions invariant under Γ),then we may write

(Lf)(z) =

∫Γ\H

K(z, w)f(w)dµw (2.6)

where K(z, w) =∑

γ∈Γ k(z, γw). The Selberg trace formula comes from evaluating the

trace TrK =∫

Γ\HK(z, z)dµz of K in two different ways, one using the spectrum of the

Laplacian and one using the geometry of the Riemann surface Γ\H. It will be moreconvenient to write the formula in terms of the function h and its Fourier transform h(r) =1

2π

∫∞−∞ e

irth(t)dt. In order for all of our sums and integrals to converge, we impose thefollowing conditions on h:

h(t) is even,

h(t) is holomorphic in the strip | Im t| ≤ 1

2+ ε,

h(t)� (|t|+ 1)−(2+ε) in this strip.

(2.7)

To begin, we may write TrK =∑

γ∈Γ

∫Γ\H k(z, γz)dµz. Let us partiton Γ into its conjugacy

classes and evaluate the sum over each conjugacy class [γ] = {τ−1γτ | τ ∈ Γ} in turn. Wehave ∑

γ′∈[γ]

∫Γ\H

dµz =∑

τ∈Z(γ)\Γ

∫Γ\H

k(z, τ−1γτz)dµz =

∫Z(γ)\H

k(z, γz)dµz, (2.8)

5

where Z(γ) is the centralizer of γ in Γ, so we have

TrK =∑[γ]

∫Z(γ)\H

k(z, γz)dµz. (2.9)

Aside from ±I, the conjugacy classes in SL2(R) (and hence also Γ) break up into threetypes depending on their fixed points:

• Parabolic: One fixed point in R, conjugate over SL2(R) to a translation ( 1 v0 1 ) with

fixed point ∞.

• Hyperbolic: Two distinct fixed points in R, conjugate over SL2(R) to a homothety(p1/2 0

0 p−1/2

)with fixed points 0 and ∞ (discussed in §1).

• Elliptic: One fixed point in H (and the conjugate fixed point in H), conjugate to arotation

(cos θ − sin θsin θ cos θ

)with fixed point i (and −i).

We may therefore break up (2.9) as

TrK =∑γ=±I

+∑

[γ] parabolic

+∑

[γ] hyperbolic

+∑

[γ] elliptic

(2.10)

We will evaluate each of the five pieces of this formula in terms of h in turn. The diagramsbelow give the regions over which we must integrate for the final three terms above (afterconjugating suitably in SL2(R)). The calculations in each case are quite long, so we willskip them and record only the final results.

Parabolic

Z (( 1 10 1 )) \H

i

− 12

12

Hyperbolic

Z((

p1/2 0

0 p−1/2

))\H

i

− 12

12

pi

Elliptic

Z((

cos θ − sin θsin θ cos θ

))\H

i

− 12

12

2θ

6

Γ\H(Y )

i

− 12

12

Y iThere is a small technical point that we will need to take care of,which is that both sides of (2.10) are actually infinite. Therefore,instead of integrating over all of Γ\H, we will instead integrateover Γ\H(Y ) and let Y → ∞, where H(Y ) =

⋃γ∈ΓF(Y ) and

F(Y ) = {z = x + iy | |x| < 12 , 0 < y < Y } is the portion of

the fundamental domain F = {z = x + iy | |x| < 12 , 0 < y}

avoiding the cusp at ∞. See the diagram to the right. Thetrouble comes from the left-hand side of (2.10) and the paraboliccomponent of the right-hand side, which grow like A1 log Y +B1 + o(1) and A2 log Y +B2 + o(1) for constants A1, B1, A2, B2.The equality A1 = A2 will be trivial, so we may subtract thoseterms to find B1 = B2 + B3 with B3 the remaining right-handside contribution, and it is this equation that we call the traceformula.

2.1 The spectral trace

First, let us consider the spectral trace, which is another way of writing the left-hand sideof (2.9). The function K has the spectral expansion

K(z, w) =∑j

h(tj)uj(z)uj(w) +1

4π

∫ ∞−∞

h(t)E(z,1

2+ it)E(w,

1

2+ it)dt (2.11)

where λj = 14 +t2j are the discrete eigenvalues of the Laplacian, uj(z) are the corresponding

eigenfunctions, and E(z, s) is the Eisenstein series 12

∑Γ∞\Γ(Im γz)s. Letting z = w and

integrating (2.11) over Γ\H(Y ), we find

TrK∣∣Γ\H(Y )

=

∫Γ\H(Y )

K(z, z)dµz

=∑j

h(tj) + h(0) log Y − h(0) log π +1

4h(0)

+1

2π

∫ ∞−∞

h(t)Γ′

Γ(1

2+ it)dt− 2

∞∑n=1

Λ(n)

nh(2 log n) + o(1).

(2.12)

7

2.2 Parabolic classes

Every parabolic conjugacy class in SL2(Z) is a power of [γ0] where γ0 = ( 1 10 1 ). Therefore,

we have

∑[γ] parabolic

∫Z(γ)\H(Y )

k(z, γz)dµz =

∞∑m=1

∫Z(γ0)\H(Y )

k(z, z +m)dµz

= h(0) log Y − h(0) log 2 +1

4h(0)

− 1

2π

∫ ∞−∞

h(t)Γ′

Γ(1 + it)dt+ o(1).

(2.13)

At this point, note that the leading terms h(0) log Y in (2.12) and (2.13) are the same.Subtracting h(0) log Y and letting Y → ∞, we obtain constants in both cases, and we nolonger need concern ourselves with Y and integrating over Γ\H(Y ).

2.3 The identity class

In this case, we simply find∫Γ\H

k(z, z)dµz = |Γ\H| k(0) =1

12

∫ ∞−∞

th(t) tanh(πt)dt. (2.14)

2.4 Hyperbolic classes

We have

∑[γ] hyperbolic

∫Z(γ)\H

k(z, γz)dµz =∑P=[γ0]

∞∑m=1

∫Z(γ0)\H

k(z, γm0 z)dµz

=∑P

logNP∞∑m=1

h(m logNP )

NPm/2 −NP−m/2,

(2.15)

where the latter two outer sums are either over primitive hyperbolic conjugacy classes Pin Γ or (in light of the correspondence (1.1)) closed geodesics in Γ\H. Recall that for aprimitive hyperbolic conjugacy class P , trP = NP 1/2 + NP−1/2, and the length of thecorresponding geodesic is logNP .

2.5 Elliptic classes

The elliptic conjugacy elements of Γ are all conjugate to some power of either(

0 −11 0

),

which has order 2 as a function on H, or(

0 −11 1

), which has order 3 as a function on H.

8

The elliptic contribution to the trace amounts to

∑[γ] elliptic

∫Z(γ)\H

k(z, γz)dµz =1

4

∫ ∞−∞

h(t)dt

cosh(πt)+

2√

3

9

∫ ∞−∞

h(t)cosh(πt/3)

cosh(πt)dt. (2.16)

2.6 The full trace formula

Putting together (2.12), (2.13), (2.14), (2.15), (2.16), we obtain the full Selberg traceformula.

Theorem 2.1. For any function h satisfying the conditions (2.7),

∑j

h(tj) = h(0) logπ

2+ 2

∞∑n=1

Λ(n)

nh(2 log n)

− 1

2π

∫ ∞−∞

h(t)

(Γ′

Γ(1

2+ it) +

Γ′

Γ(1 + it)

)dt

+1

12

∫ ∞−∞

th(t) tanh(πt)dt +1

4

∫ ∞−∞

h(t)dt

cosh(πt)+

2√

3

9

∫ ∞−∞

h(t)cosh(πt/3)

cosh(πt)dt

+∑P

logNP∞∑m=1

h(m logNP )

NPm/2 −NP−m/2,

where the first sum is over the discrete eigenvalues λj = 14 + t2j of ∆, and the last sum is

over the closed geodesics P on Γ\H.

3 The Prime Geodesic Theorem

In this section, we will apply the Selberg trace formula to a carefully chosen family offunctions hx,ε in order to prove the prime geodesic theorem.

Let ϕ be an even non-negative smooth bump function supported in [−1, 1] and such that∫∞−∞ ϕ(r)dr = 1. Given ε > 0, let ϕε(r) = ε−1ϕ(r/ε), which is supported in [−ε, ε] and also

satisfies∫∞−∞ ϕε(r)dr = 1.

Given a real number x > 0, define hx to be the function with Fourier transform

hx(r) = 2 cosh(r/2)χ[− log x,log x](r) (3.1)

where χA is the characteristic function of A ⊆ R. We would like to apply the Selberg traceformula to hx, but hx does not satisfy (2.7), so we will instead use the functions hx,ε, which

9

are defined to have Fourier transform hx,ε = hx ∗ ϕε. Note that

hx(t) =

∫ log x

− log x(esr + e(1−s)r)dr =

xs − x−s

s+x1−s − x−(1−s)

1− swhere s =

1

2+ it,

hx,ε(t) = h(t)ϕε(t) = h(t)ϕ(εt).

(3.2)

The Laplacian is a symmetric and non-negative operator, so its eigenvalues λ = 14 + t2 =

s(1 − s) are real and non-negative. Hence, we have either 12 < s ≤ 1 or Re s = 1

2 . Letus estimate hx(t) and hx,ε(t) in each of these regions. We have ϕ(t) = 1 + O(t), so for12 < s ≤ 1,

hx(t) =xs

s+O

(x1/2

)and hx.ε(t) =

xs

s+O

(x1/2 + εx

). (3.3)

Similarly, |ϕ(t)| � 1(1+|t|)2 since ϕ is smooth, so for Re s = 1

2 (i.e. t ∈ R),

|hx(t)| � x1/2

1 + |t|and |hx,ε(t)| �

x1/2

(1 + |t|)(1 + ε|t|)2. (3.4)

Note that this last estimate implies∫ ∞0

t|hx,ε(t)|dt� ε−1x1/2. (3.5)

We will need the following two results due to Selberg regarding the discrete eigenvaluesλj = 1

4 + t2j = sj(1− sj) of the Laplacian:

λj ≥3

16if λj 6= 0,

N(t)� t2 where N(t) = |{j | tj ∈ [−t, t]}|.(3.6)

(In fact, it is possible to show that λj ≥ 14 and N(t) = 1

12 t2 + O(t log t), but we will not

make use of these facts here.) As a consequence, using (3.3) and (3.5), we obtain thefollowing asymptotic expression for the part of the Selberg trace formula coming from thediscrete spectrum.∑

j

hx,ε(tj) =∑

12<sj≤1

h(tj) +

∫ ∞0

hx,ε(t)dN(t)

= x+∑

12<sj≤ 3

4

xsj

sj+O

(x1/2 + εx

)+O

(∫ ∞0

t|hx,ε(t)|dt)

= x+O(εx+ ε−1x1/2

)(3.7)

10

Aside from the hyperbolic contribution, all the other terms in the Selberg trace formula areO(ε−1x1/2

)by (3.5) and therefore contribute only to the error term. Hence, the Selberg

trace formula implies

Hε(x) = x+O(εx+ ε−1x1/2

)where Hε(x) =

∑P

logNP∞∑m=1

hx,ε(m logNP )

NPm/2 −NP−m/2.

(3.8)

We will see that this asymptotic expression implies the prime geodesic theorem. Let

H(x) =∑P

logNP

∞∑m=1

hx(m logNP )

NPm/2 −NP−m/2=

∑NPm≤x

logNP1 +NP−m

1−NP−m. (3.9)

We would like to show H(x) = x+O(x3/4

). To that end, note that

hxe−ε,ε(r) ≤ hx(r + ε) ≤ eε/2hx(r) whenever r ≥ 0,

hxeε,ε(r) ≥ hx(r − ε) ≥ e−ε/2hx(r) whenever r ≥ ε.(3.10)

Therefore, for ε sufficiently small,

e−ε/2Hε(xe−ε) ≤ H(x) ≤ eε/2Hε(xe

ε), (3.11)

so letting ε = x−1/4 and applying (3.8), we obtain H(x) = x+O(x3/4

), as claimed.

In analogy to the usual prime number theorem, we now define

ψ(x) =∑

NPm≤xlogNP and ϑ(x) =

∑NP≤x

logNP. (3.12)

Note that

H(x) =∑

NPm≤xlogNP +O

∑NPm≤x

logNP

NPm

= ψ(x) + o(ψ(x)) (3.13)

since only finitely many of the values NPm are less than a fixed constant and H(x)→∞.Therefore, ψ(x) ∼ x, which means∑

NPm≤x

logNP

NPm=

∫ x

0

1

ydψ(y) = O(log x) (3.14)

and hence ψ(x) = x+O(x3/4

).

11

Now,ψ(x) = ϑ(x) + ϑ(x1/2) + · · ·+ ϑ(x1/k) (3.15)

where the number k of terms in this sum is at most log xδ = O(log x), with δ the length of

the shortest closed geodesic on Γ\H, so ϑ(x) = x+O(x3/4) as well.

Considering

π(x) =∑NP≤x

1 =

∫ x

eδ/2

1

log ydϑ(y), (3.16)

the prime geodesic theorem

π(x) = li(x) +O

(x3/4

log x

)(3.17)

follows.

As with the classical prime number theorem, we can refine this analysis into an explicitformula for ψ(x) as in the following proposition. This will be used in the subsequentsection.

Proposition 3.1. For 1 ≤ T ≤√x

(log x)2,

ψ(x) = x+∑|tj |≤T

xsj

sj+O

( xT

log log x).

4 Improvements to the Prime Geodesic Theorem

In this section, we will discuss improvements to the asymptotic expression ψ(x) = x +O(x3/4

)obtained in §3. Following Soundararajan and Young, we will show ψ(x) = x +

O(x25/36+ε

), and assuming the generalized Lindelof hypothesis for Dirichlet L-functions,

ψ(x) = x + O(x2/3+ε

). As before, whenever we have an asymptotic of the form ψ(x) =

x+O (xα) (for α > 12), the asymptotic π(x) = li(x) +O (xα) follows.

The method we will use is to relate ψ(x) to special values of Dirichlet L-functions, andthen use known asymptotics for Dirichlet L-functions to understand how ψ(x) behaves inshort intervals, and hence asymptotically.

The connection to Dirichlet L-functions is as follows. Let X = x1/2 + x−1/2, so that thecondition NP ≤ x is equivalent to trP ≤ X. By the description of the primitive hyperbolicconjugacy classes in Γ in §1, we have

ψ(x) =∑m≤X

∑m2−dn2=4

hd log ε2d. (4.1)

12

By the class number formula hd log εd =√dL(1, χd), where χd is the real character

(d·)

associated to the discriminant d, we find

ψ(x) = 2∑m≤X

∑m2−dn2=4

√dL(1, χd). (4.2)

Define the Dirichlet seriesL(s, δ) =

∑dn2=δ

n1−2sL(s, χd) (4.3)

for any discriminant δ, so that

ψ(x) =∑m≤X

√m2 − 4L(1,m2 − 4). (4.4)

From the definition of L(s, δ), it follows that if δ = D`2 with D a fundamental discriminant,then

L(s, δ) = L(s, χD)∑

`1`2`3=`

µ(`1)χD(`1)`−s1 `1−2s3 . (4.5)

Moreover, if L(s, δ) =∑∞

q=1λq(δ)qs , then we may evaluate λq(m

2 − 4) explicitly to find

λq(m2 − 4) =

∑q21q2=q

1

q2

∑k (mod q2)

e2πi km

q2 S(k2, 1; q2)

where S(u, v; c) =∑

ad≡1 (mod c)

e2πi du+avc .

(4.6)

By the Weil bound |S(u, v; c)| � c1/2+ε√

gcd(u, v, c), it follows that if q = a2b with bsquare-free, then ∑

m≤Xλq(m

2 − 4) = Xµ(b)

b+O

(q1/2+ε

). (4.7)

Proposition 4.1. Let D range over the fundamental discriminants. If we have the bound

L(1

2+ it, χD)� (1 + |t|)A|D|θ+ε for some constants A > 0, θ ≥ 0,

thenψ(x+ u)− ψ(x)− u = O

(ux−1 + u1/2x1/4+θ/2+ε

).

The Burgess bound allows us to use θ = 316 , but the best known value of θ is 1

6 , due toConrey and Iwaniec [2]. The generalized Lindelof hypothesis permits θ = 0.

13

Proof. Let X ′ = (x+ u)1/2 + (x+ u)−1/2, so that

ψ(x+ u)− ψ(x) = 2∑

X<m≤X′

√m2 − 4L(1,m2 − 4)

= (2 +O(x−1

))

∑X<m≤X′

mL(1,m2 − 4).(4.8)

If we define, for a parameter V to be chosen later,

ΛV (δ) =∞∑q=1

λq(δ)

qe−q/V , (4.9)

then by using the relation 12πi

∫(1) α

−sΓ(s)ds = e−α and shifting the contour from Re s = 1

to Re s = −12 , we find

ΛV (δ) =1

2πi

∫(1)L(1 + s, δ)V sΓ(s)ds

= L(1, δ) +1

2πi

∫(− 1

2)L(1 + s, δ)V sΓ(s)ds.

(4.10)

If δ = D`2 with D a fundamental discriminant, then L(12 + it, δ) � `εL(1

2 + it, χD) by(4.5), so using the assumption L(1

2 + it, χD) � (1 + |t|)A|D|θ+ε, we find that the integralabove is

� V 1/2`ε∫ ∞−∞

∣∣∣∣∣Γ(12 + it)

−12 + it

∣∣∣∣∣ |L(1

2+ it, χd)|dt

� V 1/2`ε|D|θ+ε

� V 1/2δθ+ε.

(4.11)

Therefore, we conclude

ψ(x+ u)− ψ(x) = (2 +O(x−1

))

∑X<m≤X′

mΛV (m2 − 4) +O(uxθ+εV −1/2

). (4.12)

Summing (4.7) by parts, we find that if q = a2b with b square-free, then∑X<m≤X′

mλq(m2 − 4) =

(X ′2 −X2

) µ(b)

2b+O

(X ′q1/2+ε

)=(u+O

(x−1

)) µ(b)

2b+O

(x1/2q1/2+ε

),

(4.13)

and hence∑X<m≤X′

mΛV (m2 − 4) =(u+O

(x−1

)) 1

2

∞∑a,b=1

µ(b)

a2b2e−a

2b/V +O(x1/2V 1/2+ε

). (4.14)

14

Shifting contours as before, we find

∞∑a,b=1

µ(b)

a2b2e−a

2b/V =1

2πi

∫(1)

∞∑a,b=1

µ(b)

a2+2sb2+sV sΓ(s)ds

=1

2πi

∫(1)

ζ(2 + 2s)

ζ(2 + s)V sΓ(s)ds

= 1 +1

2πi

∫(− 1

2)

ζ(2 + 2s)

ζ(2 + s)V sΓ(s)ds

= 1 +O(V −1/2

).

(4.15)

Together with (4.12) and (4.14), this implies

ψ(x+ u)− ψ(x) = u+O(ux−1 + x1/2V 1/2+ε + uxθ+εV −1/2

). (4.16)

The optimal value of V is uxθ−1/2, from which we obtain the result.

We will now see how to use Proposition 4.1 to obtain asymptotics for ψ(x). The idea is toconsider a smoothed averaging of ψ. In particular, let k be a smooth real-valued functionsupported on [0, Y ] where Y is a parameter satisfying x1/2+ε ≤ Y ≤ x/ log x to be chosen

later, and suppose∫∞−∞ k(u)du =

∫ Y0 k(u)du = 1 and

∫∞−∞

∣∣k(j)(u)∣∣ du�j Y

−j for all j ≥ 0.

Define ψ(x; k) =∫ Y

0 ψ(x+ u)k(u)du, and note that

ψ(x) = ψ(x)

∫ Y

0k(u)du

=

(ψ(x; k)−

∫ Y

0uk(u)du

)−(∫ Y

0(ψ(x+ u)− ψ(x)− u)k(u)du

).

(4.17)

The second paranthetical expression above is controlled by Proposition 4.1, so it remainsto control the first.

Proposition 4.2. If we have the bound∑|tj |≤T

xitj � xaT b for some constants a ≥ 0, b ≥ 1,

then

ψ(x; k)−∫ Y

0uk(u)du = x+O

(xa+b− 1

2+εY 1−b

).

15

Since there are O(T 2) eigenvalues tj with |tj | ≤ T , we may certainly let (a, b) = (0, 2) above.We can improve this by invoking (58) of Luo and Sarnak [4], which gives (a, b) = (1

8 ,54 + ε),

and therefore

ψ(x) =

(ψ(x; k)−

∫ Y

0uk(u)du

)−(∫ Y

0(ψ(x+ u)− ψ(x)− u)k(u)du

)=(x+O

(x7/8+εY −1/4

))−(∫ Y

0

(ux−1 + u1/2x1/4+θ/2+ε

)k(u)du

)= x+O

(x7/8+εY −1/4 + x1/4+θ/2+εY 1/2

).

(4.18)

Choosing Y = x5/6−2θ/3, we find

ψ(x) = x+O(x2/3+θ/6+ε

), (4.19)

so ψ(x) = x + O(x25/36+ε

)unconditionally and ψ(x) = x + O

(x2/3+ε

)assuming the

generalize Lindelof hypothesis.

If we instead were to assume that we have square-root cancellation (i.e. (a, b) = (ε, 1 + ε)),this would improve to ψ(x) = x+O

(x1/2+ε

)for any value of θ.

Proof of Proposition 4.2. Let

E(x; k) = ψ(x; k)− x−∫ Y

0uk(u)du =

∫ Y

0(ψ(x+ u)− (x+ u)) k(u)du. (4.20)

From Proposition 3.1 with T =√x

(log x)3, we find

E(x; k) =∑|tj |≤T

1

sj

∫ Y

0(x+ u)sjk(u)du+O

(x1/2+ε

). (4.21)

We will bound this sum separately in the ranges |tj | ≤ x1+ε/Y and x1+ε/Y ≤ |tj | ≤ T .

For the latter range, we integrate by parts ` times and use the fact that∫ Y

0 k(`)(u)du� Y −`

to find∫ Y

0(x+ u)sjk(u)du = (−1)`

∫ Y

0

(x+ u)sj

(sj + 1) . . . (sj + `)k(`)(u)du�`

x1/2+`

|sj |`Y `. (4.22)

Since there are O(T 2) eigenvalues tj with |tj | ≤ T , the contribution of the range x1+ε/Y ≤|tj | ≤ T to (4.21) is

�`

∑x1+ε/Y≤|tj |≤T

1

|sj |x1/2+`

|sj |`Y `�`

x1/2+`

Y `

∫ T

x1+ε/Y

t

(1 + t)l+1dt�` x

3/2+ε(1−`), (4.23)

16

which is absorbed by the error term O(x1/2+ε

)for ` sufficiently large.

It remains only to bound

∑|tj |≤x1+ε/Y

1

sj

∫ Y

0(x+ u)sjk(u)du� x1/2

∑|tj |≤x1+ε/Y

xitj

1 + |tj |(4.24)

Using the assumption∑|tj |≤T x

itj � xaT b and summing by parts we find that this sum is

� xa+b−1/2+εY 1−b, which means

E(x; k)� xa+b−1/2+εY 1−b + x1/2+ε

� xa+b−1/2+εY 1−b,(4.25)

as claimed.

A Asymptotics for |D(x)|We prove here the following result due to Sarnak [5] needed in §1. Recall that D = {d >0 | d ≡ 0, 1 (mod 4), d non-square} and D(x) = {d ∈ D | εd ≤ x}.

Proposition A.1.

|D(x)| = 35

16x+O

(x2/3+ε

).

We begin by reducing the problem to counting the number of solutions to a diophantineinequality. Let

S(x) =∣∣{(m,n, d) | m2 − dn2 = 4, d > 0, d ≡ 0, 1 (mod 4), 0 < m,n ≤ x}

∣∣ . (A.1)

Lemma A.2.|D(x)| = S(x) +O

(x1/2

).

Proof. Given d ∈ D and k ≥ 1, let εkd =xd,k+

√dyd,k

2 , so that xd,k, yd,k ∈ Z satisfy x2d,k −

dy2d,k = 4. The map (d, k) 7→ (xd,k, yd,k, d) induces a bijection

{(d, k) | d ∈ D, k ≥ 1} ∼−→{

(m,n, d) | m2 − dn2 = 4, d ∈ D, m, n > 0}, (A.2)

so we find

S(x) =∣∣{(m,n, d) | m2 − dn2 = 4, d ∈ D, m, n > 0, m ≤ x

}∣∣= |{(d, k) | xd,k ≤ x}| .

(A.3)

17

We will see over the course of the proof of Proposition A.1 that S(x) = O(x), so since

εkd =xd,k +

√x2d,k − 4

2= xd,k +O

(x−1d,k

), (A.4)

we findS(x) = ψ(x) +O(1) where ψ(x) =

∣∣∣{(d, k) | εkd ≤ x}∣∣∣ . (A.5)

Now, let ϑ(x) = |D(x)|, so that

ψ(x) = ϑ(x) + ϑ(x1/2) + · · ·+ ϑ(x1/k) (A.6)

where the number k of terms in this sum is at most log xlog 2 . Again, since ψ(x) = O(x), we

find ϑ(x) = ψ(x) +O(x1/2

), and we are done.

We now break up S(x) according to the value of n as

S(x) =∑

0<n≤xSn(x) =

∑0<n≤x1/2

Sn(x) +∑

x1/2<n≤x

Sn(x)

where Sn(x) =∣∣{(m, d) | m2 − dn2 = 4, d > 0, d ≡ 0, 1 (mod 4), 0 < m ≤ x}

∣∣ . (A.7)

As the following lemma shows, the second of these sums contributes only to the error term,so we may concentrate on the first.

Lemma A.3. ∑x1/2<n≤x

Sn(x) = O(x2/3+ε

).

Proof. LetS∗n(x) =

∣∣{(m, k) | m2 − kn2 = 4, k > 0, 0 < m ≤ x}∣∣ , (A.8)

and note that Sn(x) ≤ S∗n(x).

For n > x1/2, S∗n(x) is bounded from above by the number of solutions to m2 ≡ 4 (mod n2)in Z/n2Z, which is O(nε) for any ε > 0. Therefore,∑

x1/2<n≤x

Sn(x) ≤∑

x1/2<n≤x2/3S∗n(x) +

∑x2/3<n≤x

S∗n(x)

= O(x2/3+ε

)+

∑x2/3<n≤x

S∗n(x).(A.9)

This latter term is the number of solutions to

kn2 = (m− 2)(m+ 2) with k > 0, 0 < m ≤ x, x2/3 < n ≤ x, (A.10)

18

so we may write n in the form

n = 2ηuv with η ∈ {0, 1}, u2 | (m− 2), v2 | (m+ 2). (A.11)

Either u or v is at least 12n

1/2 > 12x

1/3, so the choices of m solving (A.10) are among thosesatisfying

m ≡ ±2 (mod w2) for some1

2x1/3 < w ≤ x1/2 (A.12)

Given such an m, there are at most d2(m2 − 4) pairs (n, k) solving (A.10), where dn(`) isthe number of square divisors of `, which satisfies d2(`) � λε. Therefore, the number ofsolutions to (A.10) is

� xε∑

12x1/3<w≤x1/2

∣∣{0 < m ≤ x | m ≡ ±2 (mod w2)}∣∣

� xε∑

12x1/3<w≤x1/2

x

w2

� x2/3+ε,

(A.13)

which completes the proof of the lemma.

What remains is to estimate∑

0<n≤x1/2 Sn(x). Let

S′n(x) =∣∣{(m, k) | m2 − 4kn2 = 4, k > 0, 0 < m ≤ x}

∣∣ ,S′′n(x) =

∣∣{(m, k) | m2 − (4k + 1)n2 = 4, k > 0, 0 < m ≤ x}∣∣ , (A.14)

so that Sn(x) = S′n(x) + S′′n(x). Let

T ′n =∣∣{m ∈ Z/4n2Z | m2 ≡ 4

}∣∣ ,T ′′n =

∣∣{m ∈ Z/4n2Z | m2 ≡ n2 + 4}∣∣ , (A.15)

and note thatS′n(x) =

x

4n2T ′n +O

(T ′n)

=x

4n2T ′n +O (nε) ,

S′′n(x) =x

4n2T ′′n +O

(T ′′n)

=x

4n2T ′′n +O (nε) .

(A.16)

Therefore, ∑0<n≤x1/2

Sn(x) =∑

0<n≤x1/2

( x

4n2(T ′n + T ′′n ) +O (nε)

)

=x

4

∞∑n=1

(T ′nn2

+T ′′nn2

)+O

(x1/2+ε

).

(A.17)

In order to prove Proposition A.1, it remains only to evaluate∑∞

n=1T ′nn2 and

∑∞n=1

T ′′nn2 ,

which is the content of the following two lemmas.

19

Lemma A.4.∞∑n=1

T ′nn2

=11

2.

Proof. Note thatT ′n = 2

∣∣{m ∈ Z/n2Z | m2 ≡ 1}∣∣ , (A.18)

so T ′n = 2U(n) where U(n) is the multiplicative function defined by

U(2j) =

{2 if j = 1,

4 if j ≥ 2,

U(pj) = 2 for p ≥ 3 prime, j ≥ 1.

(A.19)

Therefore, we have

∞∑n=1

T ′nn2

= 2∞∑n=1

U(n)

n2

= 2∏p

(1 +

U(p)

p2+U(p2)

p4+ . . .

)= 2

(1 +

2

22+

4

24+ . . .

)∏p≥3

(1 +

2

p2+

2

p4+ . . .

)

= 2

(1 +

1

2+

2−2

1− 2−2

)∏p≥3

(1 +

2p−2

1− p−2

)

=

(11

3

)(1− 2−2

1 + 2−2

)∏p

1 + p−2

1− p−2

=11

5

ζ(2)2

ζ(4)

=11

2.

(A.20)

Lemma A.5.∞∑n=1

T ′′nn2

=13

4.

20

Proof. Let n = 2βpe11 · · · pe`` with pi ≥ 3 and ei > 0 be the prime factorization of n, and

note that by the Chinese remainder theorem,

T ′′n =∣∣∣{m ∈ Z/22β+2Z | m2 ≡ n2 + 4

}∣∣∣ ∏i=1

∣∣∣{m ∈ Z/p2eii Z | m2 ≡ n2 + 4

}∣∣∣=∣∣∣{m ∈ Z/22β+2Z | m2 ≡ 22β + 4

}∣∣∣ ∏i=1

∣∣∣{m ∈ Z/p2eii Z | m2 ≡ 4

}∣∣∣ . (A.21)

Hence, T ′′n = 2V (n) where V (n) is the multiplicative function defined by

V (2β) =

{0 if β = 1, 2,

4 if β ≥ 3,

V (pj) = 2 for p ≥ 3 prime, j ≥ 1.

(A.22)

Therefore, we have

∞∑n=1

T ′′nn2

= 2∞∑n=1

V (n)

n2

= 2∏p

(1 +

V (p)

p2+V (p2)

p4+ . . .

)= 2

(1 +

4

26+

4

28+ . . .

)∏p≥3

(1 +

2

p2+

2

p4+ . . .

)

= 2

(1 +

2−4

1− 2−2

)∏p≥3

(1 +

2p−2

1− p−2

)

=

(13

6

)(1− 2−2

1 + 2−2

)∏p

1 + p−2

1− p−2

=13

10

ζ(2)2

ζ(4)

=13

4.

(A.23)

21

Altogether, we have now shown

|D(x)| = S(x) +O(x1/2

)=

∑0<n≤x1/2

Sn(x) +O(x2/3+ε

)=x

4

(11

2+

13

4

)+O

(x2/3+ε

)=

35

16x+O

(x2/3+ε

),

(A.24)

which is Proposition A.1.

22

References

[1] Nicolas Bergeron. The spectrum of hyperbolic surfaces. Springer International Publish-ing, 2016.

[2] J. Brian Conrey and Henryk Iwaniec. The cubic moment of central values of automor-phic l-functions. Annals of Mathematics, 151:1175–1216, 1998.

[3] Henryk Iwaniec. Spectral methods of automorphic forms. American MathematicalSociety, 2003.

[4] Wenzhi Luo and Peter Sarnak. Quantum ergodicity of eigenfunctions on psl 2(Z)/H2.Publications Mathematiques de l’IHES, 81:207–237, 1995.

[5] Peter Sarnak. Class numbers of indefinite binary quadratic forms. Journal of NumberTheory, 15(2):229 – 247, 1982.

[6] Kannan Soundararajan and Matthew P. Young. The prime geodesic theorem. Journalfur die reine und angewandte Mathematik (Crelles Journal), 2013(676), Jan 2013.

23