On Some Higher Order Counting Functions for PSL(2,R)

DŽENAN GUŠIĆUniversity of Sarajevo

Faculty of Sciences and MathematicsDepartment of Mathematics

Zmaja od Bosne 33-35, 71000 SarajevoBOSNIA AND HERZEGOVINA

dzenang@pmf.unsa.ba

Abstract: This paper is devoted to some counting functions of level one and level three in the case of quotientspace generated by some strictly hyperbolic Fuchsian group and the upper half-plane. Each of the functions isrepresented as a sum of some explicit part plus the error term. The explicit part is indexed over singularities of thecorresponding Selberg zeta function. In particular, the obtained error term is not larger than O

(x

34

). The method

applied in this paper follows traditional approach for achieving the error terms in the case of locally symmetricspaces of real rank one. In order to establish an analogy with the classical case, we consider the counting functionsdivided by x and x3, respectively.

Key–Words: Counting functions, upper half-plane, Fuchsian groups, error terms

Received: February 8, 2020. Revised: March 23, 2020. Accepted: March 27, 2020. Published: March 31, 2020.

1 IntroductionOur notation will be based od [9] and [15].

Let F be a compact Riemann surface of genus g≥ 2. We can therefore represent F as a quotient spaceΓ\H , where Γ is a strictly hyperbolic Fuchsian groupand H is the upper half-plane.

Thus, Γ ⊆ PSL (2,R).If I is the fundamental polygon of F , then ∂I =

α+1 β+1 α−1 β−1 ...α

+g β

+g α−g β−g , where the sides α

−k , α

+k

and β−k , β+k are identified in pairs.

We assume that the sides of I are piecewisesmooth.

As it is known, the upper half-plane H comeswith the Poincare metric ds= |dz|y , whose correspond-ing area element is dµ (z) = y−2dxdy.

Note that the Poincare metric has Gaussian curva-ture K = −1 (see, e.g., [18]).

Suppose that π :H → F is the universal coveringmap.

By projecting the Poincare metric onto F via π,F becomes a compact Rirmannian manifold.

Furthermore, by the Gauss-Bonnet theorem (see,e.g., [17])

A = Area (F ) = µ (I) = 4π (g − 1) .

Note that the automorphic group Γ is determinedup to a conjugation in PSL (2,R). However, the

Poincare metric on F is not affected by such con-jugations because of the invariance properties of thePoincare metric on H (we assume through the rest ofthe paper that F carries the Poincare metric). Further-more, the Gaussian curvature is still K = −1.

In [10], the author concluded that the asymptoticdistribution of the closed geodesics is highly influ-enced by the eigenvalues of the Laplace operator ∆on F , i.e., by the eigenvalues for the problem ∆f +λf = 0 on F .

The Selberg zeta function Sel (s) for the group Γis an entire function of order 2, having a sequence ofzeros at 0, −1, −2,..., with the zero at s = 1 simple,and having additional zeros in the critical strip 0 <Re (s)< 1. The zeros in the critical strip are located atpoints which are solutions of the equations s (s− 1)= λn, where λn ranges through the sequence of eigen-values, omitting λ0 = 0, for the problem ∆f + λf =0 on F .

The multiplicity of such a zero is the same as themultiplicity of the corresponding eigenvalue (see, e.g.,[13], [16]).

Note that the detailed description of the locationsand the orders of the zeros of Sel (s) will be given inthe sequel.

In [15, p. 245, Th. 2] (see also, [11]), the authorderived the following length spectrum.

If Sel (s) has zeros α1, α2,..., αn in(34 , 1), then,

there exist constants c1, c2,..., cn, such that

WSEAS TRANSACTIONS on SYSTEMS and CONTROL DOI: 10.37394/23203.2020.15.9 Dzenan Gusic

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ϕ0 (x) = x+ c1xα1 + ...+ cnx

αn +O(x

34

)

as x → +∞ (the functions ϕn (x), n ∈ N ∪ {0} areintroduced below).

If, however, Sel (s) has no zeros in(34 , 1), then

ϕ0 (x) = x+O(x

34

)as x→ +∞.

The main goal of this research is to derive an anal-ogous result for the function ϕ1(x)x , i.e., a weightedform of the corresponding length spectrum.

In our earlier research [1, p. 466, Th. 1],we derived one such result applying quite complexmathematical apparatus. The obtained remainderO(x

12 log x

), however, was much better than the

classical one O(x

34

).

Motivated by this fact, that a ϕ1 analogue of theclassical length spectrum yields a better result, wegive yet another proof of Theorem 1 [1] based on ap-plication of much simpler mathematical techniques.

2 Preliminaries

We adopt the functions ϕ0 (x) and ϕn (x) from [15,p. 245].

The first one is given as a sum indexed overΓhyp with Norm (γ0) ≤ x, where Γhyp denotes theset of Γ-conjugacy classes of hyperbolic elements inΓ, Norm (γ0) = exp (len (Geoγ0)), len (Geoγ0) isthe length of the prime geodesic Geoγ0 associated tothe conjugacy class γ0 (it is known fact that primegeodesics over F correspond to the conjugacy classesof primitive hyperbolic elements in Γ). More pre-cisely, if γ ∈ Γhyp, then γ is an exponent of someprimitive γ0, with the degree j (γ) ∈ N.

For such γ, Φ (γ) is defined by Φ (γ) =len (Geoγ0).

The functions ϕn (x), n ∈ N are defined induc-tively.

Through the rest of the paper, we shall assumethat l is a number, l ∈N, and that d and T are constantswhich will be fixed later.

For a function f (x), we define Dl−1,+ by

Dl−1,+f (x)=f (x+ (l − 1) d)− (l − 1) f (x+ (l − 2) d)

+(l − 1) (l − 2)

2f (x+ (l − 3) d)− ...

+ (−1)l−1 f (x) .

By [3, p. 315, (22)], Dl−1,+f (x) can be repre-sented as an iterated integral (if f is a differentiablefunction of appropriate order).

Thus, for some x ≤ x̃ ≤ x + (l − 1) d, an ana-logue of the estimate (23) in [3] holds also true.

The number of zeros of Sel (s) on the critical linewill be denoted by n (t).

We shall apply the fact that n (t) can be estimatedas A4π t

2.

3 Main resultWe shall prove the following theorem.

Theorem 1. Let F be a compact Riemann surface ofgenus g ≥ 2. Then,

ϕ1 (x)

x

=1

2x+

∑s0∈S0,R12

The non-trivial zeros of Sel (s+ p) are all con-tained in the union of the interval (0− p, 1− p) withthe vertical line 12 − p + iR, i.e., in (−p, 1− p) ∪(12 − p+ iR

).

As we already noted, the values 0, −1,..., −l arezeros of Sel (s+ p). Hence, these values are poles oforder two of the corresponding function.

The values − (l + 1), − (l + 2),... are then thesimple poles of the same function.

Finally, the value 1 is a simple pole if p = 0.The set of zeros sp of Sel (s+ p) such that sp /∈

Z will be denoted by Sp.In other words, Sp ⊆ (−p, 1− p) ∪(

12 − p+ iR

).

Note that the values sp, where sp ∈ Sp are alsosimple poles of the function in the case at hand.

Now, we determine the residues az,p,l’s for z ∈Ap,l.

Let z be a zero of Sel (s+ p) of multiplicityord (z, p).

Furthermore, let qi (z, p)’s be the correspondingcoefficients in the expansion of the logarithmic deriva-tive of Sel (s+ p) near z.

If z ∈ Ap,l, and z = sp ∈ Sp, then asp,p,l’s arecalculated in the same way as in [3, p. 314, (13)].

Suppose that z ∈ Ap,l, and z = −j ∈{0,−1, ...,−l}.

For the final form of the corresponding residuea−j,p,l in this case, we refer to [3, (14)].

Now, suppose that z ∈ Ap,l, z = −j ∈{− (l + 1) ,− (l + 2) , ...}.

In this case, a−j,p,l is given by

ord (−j, p)l∏

q=0(−j + q)−1 x−j+l.

Furthermore, if z ∈ A0,l, z = 1, then a1,0,l =x1+l

(l+1)! .We shall consider the following subsets of the set

Sp: Sp,R = Sp ∩ R, and Sp, 12−p = Sp \ Sp,R for p ∈

{0, 1}.Thus, Sp,R ⊆ (−p, 1− p) and Sp, 1

2−p ⊆

12 − p +

iR.Assume that z ∈ Ap,l, and z ∈ Sp, 1

2−p.

By the very definition of the operator Dl−1,+, itimmediately follows that d−(l−1)Dl−1,+az,p,l can beestimated by O

(d−(l−1) |z|−l−1 x

12+l)

On the other side, an application of the meanvalue theorem, yields that

|ord (z, p)| |z|−1 |z + 1|−1(x+

l∑i=2

qi

) 12−p+1

dom-

inates∣∣d−(l−1)Dl−1,+az,p,l∣∣ for some 0 ≤ qi ≤ d, i

∈ {2, 3, ..., l}.

Thus, O(|z|−2 x

32

)is the second estimate for

d−(l−1)Dl−1,+az,p,l.Having in mind these two

estimates, one easily concludes that the sum∑z∈S

p, 12−p

d−(l−1)Dl−1,+az,p,l, and hence the

sum1∑p=0

(−1)p∑

z∈Sp, 12−p

d−(l−1)Dl−1,+az,p,l are

O(x

32 log T

)+ O

(d−(l−1)x

12+lT−l+1

).

Furthermore, 12x2 + O (dx) + O

(d2)

dominatesd−(l−1)Dl−1,+a1,0,l.

Suppose that d = xα (log x)β and T =xγ (log x)δ.

Then,

d2 =x2α (log x)2β ,

x32 log T

=γx32 log x+ δx

32 log log x,

d−l+1x12+lT−l+1

=x(−l+1)α+12+l+(−l+1)γ (log x)(−l+1)β+(−l+1)δ .

Since log log x is dominated by log x, we wantthat

2α =3

2= (−l + 1)α+ 1

2+ l + (−l + 1) γ,

2β =1 = (−l + 1)β + (−l + 1) δ.

It is not so hard to conclude that α = 34 , β =12 ,

γ = 14 and δ =1+l

2(1−l) , i.e., d = x34 (log x)

12 and T =

x14 (log x)

1+l2(1−l) .

Since, in this case, O (dx) = O(x

74 (log x)

12

),

it follows that the aforementioned remainders are allO(x

74 (log x)

12

).

Next, we consider

dx =xα+1 (log x)β ,

x32 log T

=γx32 log x+ δx

32 log log x,

d−l+1x12+lT−l+1

=x(−l+1)α+12+l+(−l+1)γ (log x)(−l+1)β+(−l+1)δ .

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Reasoning in the same way as in the previouscase, we are interested in

α+ 1 =3

2= (−l + 1)α+ 1

2+ l + (−l + 1) γ,

β =1 = (−l + 1)β + (−l + 1) δ.

Thus, α = 12 , β = 1, γ =12 and δ =

l1−l , that is,

d = x12 log x and T = x

12 (log x)

l(1−l) .

Also, in this case, the error term O(d2)

is

O(x (log x)2

).

Consequently, the error terms given above are allO(x

32 log x

).

Since the equalityO (dx) =O(d2)

yields at leastthe error termO

(x2), the remaining two cases are not

interesting for our research.The discussion conducted above, yields that

O(x

32 log x

)is the remainder we are looking for, and

is established for d = x12 log x and T =

x12 (log x)

l(1−l) .

The sum over sp ∈ Sp,R, 12 < sp < 1, p ∈ {0, 1},is obviously equal to the sum over s0 ∈ S0,R, 12 < s0< 1.

Thus, for d = x12 log x, the sum is equal to

∑s0∈S0,R12

o (−j, p, l)= (−1)j−2 ord (−j, p) (l − j)! (j − 2)!×

×l∏

q=0q 6=j

(−j + q)−1 ,

it follows that the corresponding sum is estimated byO(x−1

).

Reasoning in the same way as above (applying themain properties of Dl−1,+), we obtain that the sumcorresponding to a−1,p,l, p ∈ {0, 1} resp. the sumcorresponding to a0,p,l, p ∈ {0, 1}, is O (log x) resp.O (x log x).

Note that the sum over sp ∈ Sp,R, −1 < sp < 0, p∈ {0, 1} resp. the sum over sp ∈ Sp,R, 0 < sp ≤ 12 , p∈ {0, 1}, is actually the sum over s1 ∈ S1,R, −1 < s1< 0 resp. the sum over s0 ∈ S0,R, 0 < s0 ≤ 12 .

These two sums, however, areO (x) andO(x

32

),

respectively.Having in mind the subsets of Ap,l we have con-

sidered above, we may write

d−(l−1)Dl−1,+ϕl (x)=d−(l−1)Dl−1,+a1,0,l+

1∑p=0

(−1)p∑

z∈Sp, 12−p

d−(l−1)Dl−1,+az,p,l+

1∑p=0

(−1)p∑

sp∈Sp,R12

T is a constant that will be fixed later (the constant dwill also be fixed later, but, for now, it is assumed thatd is dominated by O (x)).

Now we apply the operator d−l+3Dl−3,+ to theresidue a1,0,l whose position is on the right.

If we use the fact that d−l+3Dl−3,+ applied tosome differentiable function f (x) (of appropriate or-der) is f (l−3) (x̃) for some x ≤ x̃ ≤ x + (l − 3) d,then, we obtain that d−l+3Dl−3,+a1,0,l is given by14!x

4 + O(x3d)

+ O(x2d2

)+ O

(xd3)

+ O(d4).

One should be interested to determine d and Tsuch that the error term O

(x3d), and the error terms

O(x

72T−2

), O(d−l+3 |z|−l−1 T−l+1

)be the same.

As it is usual, denoting d and T by some xα andxβ , respectively, we easily conclude that 3 + α, 72− 2β and −α (l − 3) + 12 + l +

(14 −

12α)

(−l + 1)must be the same.

Solving this system, we obtain thatα= 32 and β =−12 . In other words d is not bounded by O (x), whichmeans that the calculations obtained previously couldnot be valid in this case. Hence, we don’t consider thiscase anymore.

For our needs, it will be sufficient to deter-mine d and T such that O

(d4), and the error terms

O(x

72T−2

), O(d−l+3 |z|−l−1 T−l+1

)be the same.

Proceeding in the same way as in the previouscase, we obtain that 4α, and 72 − 2β, −α (l − 3) +12 + l +

(74 − 2α

)(−l + 1) must be the same. It fol-

low that α = 34 and β =14 . Hence, the condition that

d is dominated by O (x) is provided. Thus, the O−terms: O

(x3d), O

(x2d2

), O

(xd3), O

(d4)

are alldominated by O

(x3d). This error term, however, is

given now by O(x

154

).

We may say now that we are interested in achiev-ing the error term O

(x

154

)in the final form of the

corresponding weighted prime geodesic theorem.Now, we consider the sum of the elements

d−l+3Dl−3,+asp,p,l along p ∈ {0, 1} and sp ∈ Sp,R, 0< sp < 1. It is not so hard to conclude that this sum isgiven by

∑s0∈S0,R0

sp ∈ Sp,R, −1 < sp < 0. Obviously, this sum is theminus sum of the elements d−l+3Dl−3,+as1,1,l overs1 ∈ S1,R, −1 < s1 < 0. The last sum, however, isO(x3).

Combining the estimates derived above, we con-clude that ϕ3 (x) is not larger than

1

4!x4 +

∑s0∈S0,R0

[10] H. Huber, Zur analytischen Theorie hyperbolis-cher Raumformen und Bewegungsgruppen. II,Math. Annalen 142, 1961, pp. 385–398.

[11] H. Huber, Zur analytischen Theorie hyperbolis-cher Raumformen und Bewegungsgruppen. II(Nachtrag zu Math. Annalen 142, 385-398,1961), Math. Annalen 143, 1961, pp. 463–464.

[12] A. E. Ingham, The Distribution of Prime Num-bers, Cambridge Mathematical Library, 1990

[13] H. P. McKean, Selberg’s trace formula as ap-plied to a compact Riemann surface, Comm.Pure Appl. Math. 25, 1972, pp. 225–246.

[14] J. Park, Ruelle zeta function and prime geodesictheorem for hyperbolic manifolds with cusps,in: G. van Dijk, M. Wakayama (eds.), Casimirforce, Casimir operators and Riemann hypothe-sis, de Gruyter, Berlin 2010, pp. 89–104.

[15] B. Randol, On the asymptotic distributon ofclosed geodesics on compact Riemann surfaces,Trans. Amer. Math. Soc. 233, 1977, pp. 241–247.

[16] A. Selberg, Harmonic analysis and discontin-uous groups in weakly symmetric Riemannianspaces with applications to Diriclet series, J. In-dian Math. 20, 1956, pp. 47–87.

[17] C. L. Siegel, Topics in Complex Function The-ory, Wiley–Interscience, 1971

[18] J. J. Stocker, Differential Geometry,Wiley–Interscience, 1969

[19] H. Tang, The Generalized Prime Number The-orem for Automorphic L-Functions, Chin. Ann.Math. 3, 2009, pp. 251–260.

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