Selberg's zeta functions for congruence subgroups of modular groups in SL(2,R) and SL(2,C)

0.17 MB·English

by  Yasufumi Hashimoto

Selberg’s zeta functions for congruence subgroups of
modular groups in SL (R) and SL (C)
9 2 2
0
0
2
Yasufumi Hashimoto
∗
n
a
J

9
2

]
T
Abstract
N
.
h It is known that the Selberg zeta function for the modular group has an ex-
t
a pression in terms of the class numbers and the fundamental units of the indefinite
m
binary quadratic forms. In the present paper, we generalize such a expression to
[ any congruence subgroup of the modular groups in SL (R) and SL (C).
2 2

2
v
9
1 Introduction
2
8
4
Let H be the upper half plane with the hyperbolic metric and Γ a discrete subgroup
.
6
of SL (R) such that the volume of Γ H is finite. We denote by Prim(Γ) the set of
0 2
\
8 primitive hyperbolic conjugacy classes of Γ and N(γ) the square of the larger eigenvalue
0
of γ Prim(Γ). The Selberg zeta functions for Γ are defined as follows.
:
v ∈
i
X
∞
r ZΓ(s) := (1 N(γ)−s−n) Res > 1.
a −
γ Prim(Γ)n=0
∈Y Y
It is known that Z (s) is analytically continued to the whole complex plane as a mero-
Γ
morphic function of order 2. By virtue of the analytic properties of Z (s), we can obtain
Γ
the following asymptotic formula called by the prime geodesic theorem (see, e.g., [He]).
π (x) := # γ Prim(Γ) N(γ) < x = li(x)+O(xδ) as x ,
Γ
{ ∈ | } → ∞
where li(x) := x(logt) 1dt and the constant δ (0 < δ < 1) depends on Γ.
2 −
When Γ = SL (Z), since there is a one-to-one correspondence between elements of
R 2
Prim(Γ) and equivalence classes of primitive indefinite binary quadratic forms, and N(γ)
for γ Prim(Γ) coincides the square of the fundamental unit of the quadratic form
∈
corresponding to γ (see e.g. [G]), we can express the Selberg zeta function and the prime
∗
Partially supported by JSPS Grant-in-Aid for Young Scientists (B) no. 20740027.
MSC: primary: 11M36; secondary:11E41
1
2 Y. Hashimoto
geodesic theorem for Γ = SL (Z) as follows (see [Sa1]).
2
∞
Z (s) = (1 ǫ(D) 2(s+n))h(D) Res > 1, (1.1)
SL2(Z) − −
D Dn=0
Y∈ Y
π (x) = h(D) li(x2) as x , (1.2)
SL2(Z) ∼ → ∞
D D
ǫ(XD∈)<x
where D is the set of integers D > 0 such that D is not square and D 1,0 mod 4, h(D)
≡
and ǫ(D) are respectively the class number and the fundamental unit of the discriminant
D in the narrow sense.
For a three dimensional hyperbolic manifolds whose fundamental group is SL ( )
2
O
where is the integer ring of the imaginary quadratic field, similar expressions of the
O
Selberg zeta function and the prime geodesic theorem were obtained in [Sa2].
Such arithmetic expressions of the Selberg zeta functions essentially give those of
the length spectra on the corresponding manifolds. Though it is not easy in general to
study the distributions of the length spectra and its multiplicities in detail, by virtue of
the expressions by h(D) and ǫ(D), the distributions of them for the cases of the modular
group and the congruence subgroups could be studied by the classical tools in the analytic
number theory (see [BLS], [Pe] and [H2]). Moreover, by writing down the prime geodesic
theorem for the modular group with h(D) and ǫ(D), one can get the density formula of
the sum of the class numbers with a different form to that conjectured by Gauss and
proven by Siegel [Si] (see [Sa1], [H1] and [H3]).
The aimof thepresent paper isto generalize such arithmetic expressions of theSelberg
zeta functions for congruence subgroups of SL (Z) and SL ( ). For the groups Γ (N),
2 2 0
O
Γ (N) and Γ(N) of SL (Z), the author [H1] already gave explicit expressions of the
1 2
logarithmic derivatives of the Selberg zeta functions in terms of the class numbers and
the fundamental units. In the present paper, we show that the Selberg zeta functions
for any congruence subgroup of SL (Z) and SL ( ) are expressed in terms of the class
2 2 K
O
numbers and the fundamental units such like (1.1).
2 Quadratic forms and modular groups
In this section, we give notations of the quadratic forms, explain the relations between
the quadratic forms and elements in the modular group and study the difference between
the conjugation in SL ( ) and that in SL ( /n) for an integer ring of an algebraic
2 2
O O O
number fields and an ideal n in .
O
Let K be an algebraic number field over Q such that [K : Q] < and the ring of
∞ O
integers of K. We denote by
Q(x,y) = [a,b,c] := ax2 +bxy +cy2
a binary quadratic form over where a,b,c . We write Q Q if there exists
′
O ∈ O ∼
g SL ( ) such that Q(x,y) = Q g.(x,y) . Let D := b2 4ac be the discriminant of
2 ′
∈ O −
(cid:0) (cid:1)
Selberg’s zeta functions for congruence subgroups 3
[a,b,c] and P(D) the set of solutions (t,u) 2 of the Pell equation t2 Du2 = 4. The
∈ O −
set P(D) is an abelian group with the identity (2,0) and the product
t t +u u D t u +t u
1 2 1 2 1 2 2 1
(t ,u ) (t ,u ) = , .
1 1 2 2
∗ 2 2
(cid:16) (cid:17)
It is known that P(d) (Z/lZ) Z where l is the number of roots of 1 in , r is
≃ × OK(√D) 1
the number of embeddings K(√D) into R and 2r is the number of embeddings K(√D)
2
into C with Im(K(√D)) R. Note that r +2r = [K(√D) : Q]. For a quadratic form
1 2
6⊂
Q = [a,b,c] and a solution (t,u) P(D), we put
∈
t+bu
cu
γ Q,(t,u) := 2 − SL ( ). (2.1)
 t bu 2
∈ O
au −
(cid:0) (cid:1) 2
 
 
Conversely, for γ = (γ ) SL ( ), we put
i,j 1 i,j 2 2
≤ ≤ ∈ O
t := γ +γ , u := gcd(γ ,γ γ , γ ),
γ 11 22 γ 21 11 22 12
− −
a := γ /u , b := (γ γ )/u , c := γ /u ,
γ 21 γ γ 11 22 γ γ 12 γ
− −
t2 4
Q := [a ,b ,c ], D := γ − = b2 4a c .
γ γ γ γ γ u2 γ − γ γ
γ
We note the following elementary facts without proofs.
Fact 2.1. Suppose that γ,γ ,γ SL ( ) are not in the center of SL ( ). Then we have
1 2 2 2
∈ O O
1. γ(Q ,(t ,u )) = g for any g SL ( ).
g g g 2
∈ O
2. D(Q) = D for any (t,u) P(D) and D = D(Q ).
γ(Q,(t,u)) γ γ
∈
3. (t ,u ) = (t,u) and (t ,u ) P(d ).
γ(Q,(t,u)) γ(Q,(t,u)) γ γ γ
∈
4. If Q = Q then Q = Q = Q and (t ,u ) = (t ,u ) (t ,u ).
γ1 γ2 γ1γ2 γ1 γ2 γ1γ2 γ1γ2 γ1 γ1 ∗ γ2 γ2
5. Qg−1γg(x,y) = Qγ g.(x,y) for any g SL2( ).
∈ O
For a fixed discrimin(cid:0)ant D (cid:1) , let I(D) be the set of equivalence classes of quadratic
∈ O
forms [a,b,c] with b2 4ac = D. It is known that h(D) := #I(D) is finite. According to
−
Fact 2.1, we see that
h(d) =# γ Conj(SL ( )) D = D /P(D)
2 γ
{ ∈ O | }
=# γ Conj(SL ( )) t = t,u = u , (2.2)
2 γ γ
{ ∈ O | }
where (t,u) is a non-trivial solution of t2 Du2 = 4.
−
For quadratic forms [a ,b ,c ],[a ,b ,c ] I(d), we give the following product.
1 1 1 2 2 2
∈
a a a a b b +D
1 2 1 2 1 2
[a ,b ,c ] [a ,b ,c ] = ,v b +v b +w ,C , (2.3)
1 1 1 ∗ 2 2 2 β2 1 β 2 2 β 1 2β
h i
4 Y. Hashimoto
where C satisfies that the discriminant of the quadratic form of the right hand side is
∈ O
d,β := gcd(a ,a ,(b +b )/2)andv ,v ,w arechosenasv a +v a2+w(b +b )/2 = β.
1 2 1 2 1 2 1 1 2 1 2
∈ O
It is easy to see that I(D) is a finite abelian group with the identity [1,δ,(D δ2)/4] and
−
the product above, where δ is taken as 4 d δ2.
∈ O | −
For an ideal a , let
⊂ O
(2) := α α2 1 mod a ,
Oa { ∈ O | ≡ }
Γˆ(a) :=Ker SL ( ) proj. PSL ( /a) ,
2 2
O → O
= γ SL ( ) γ αI mod a,α (2) .
{ ∈(cid:0) 2 O | ≡ (cid:1) ∈ Oa }
Note that if the factorization ofa in is a = pr and # /p = q (a power of a rational
O p OK
prime) then
Q
q3r 2(q2 1)
Γˆ(a) = Γˆ(pr), [SL ( ) : Γˆ(a)] = #PSL ( /a) = p − − .
\p 2 O 2 O Q #Oa(2)
We now study the conjugation in PSL ( /a).
2
O
Lemma 2.1. Let a be an ideal in . For a,b,c /a, d = b2 4ac and t,u /a
O ∈ O − ∈ O
satisfying t2 Du2 = 4 without (t,u) = ( 2,0), we denote by
− ±
1 t+δu D δ2
(t+bu) cu − ν 1u
−
γ := 2 − , γ := 2 4 ,
 1  ν  t δu 
au (t bu) νu −
2 − 2
   
   
where ν ( /a) and δ is taken as 2 b+δ when gcd(a,2) = 1 and is taken by 0 when
∗
∈ O | 6
gcd(a,2) = 1. Then we have γ γ in PSL ( /a) for some ν (2).
ν 2 a
∼ O ∈ O
Proof. When gcd(a,a) = 1, it is easy to see that there exists g PSL ( /a) such that
2
6 ∈ O
gcd (g 1γg) /u,a = 1. Then it is sufficiently to consider onlythe case forgcd(a,a) = 1.
− 21
When a = να2 for ν (2) and α /a, we have g 1γg = γ where
(cid:0) (cid:1) a ′− ′ ν
∈ O ∈ O
α 1 (να) 1(b+δ)/2
− −
g := .
′
0 α
(cid:18) (cid:19)
This completes the proof.
Lemma 2.2. Let a be an ideal in . For (t,u) 2 ( 2,0) , we denote by (t,u) :=
O ∈ O −{ ± } I
γ Conj(SL ( )) t = t,u = u and (t,u) := γ (t,u) γ γ in PSL ( /a) .
2 γ γ ν ν 2
{ ∈ O | } J { ∈ I | ∼ O }
(2)
For ν,ν′ a , define the equivalence ν ν′ such that γν is conjugate to γν′ in
PSL ( /a∈). OPut (t,u) := ν (2)/ ∼J (t,u) = φ and µ (t,u) := # (t,u).
2 a a ν a a
O L { ∈ O ∼ | 6 } L
Number the elements of (t,u) by ν (= 1),ν ,...,ν . For simplicity, rewrite by .
La 1 2 µ Jνi Ji
Then we have # = h(D)/µ for any 1 i µ.
i
J ≤ ≤
Selberg’s zeta functions for congruence subgroups 5
Proof. For γ ,γ (t,u), denote by γ γ := γ(Q Q ,(t,u)), where the product
1 2 ∈ I 1 ∗ 2 γ1 ∗ γ2 ∗
in the right hand side is given by (2.3).
Fix γ . It is easy to see that γ γ for any γ . When γ ,γ are
′ i ′ i 1 1 2 i
∈ J ∗ ∈ J ∈ J ∈ J
not conjugate in SL ( ), it is trivial that γ γ γ γ in PSL ( ). Then we have
2 1 ′ 2 ′ 2
O ∗ 6∼ ∗ O
γ . Conversely, γˆ γ for any γˆ . Then we can get = γ
i ′ 1 ′ 1 i 1 ′ i
J ⊃ ∗ J ∗ ∈ J ∈ J J ∗ J
similarly. This yields that # = # for any 1 i µ. Since µ # = h(d), we
J1 Ji ≤ ≤ i=1 Ji
conclude that # = h(d)/µ for any 1 i µ.
i
J ≤ ≤ P
Consider the case of = Z. Note that, when = Z.
O O
1 , p = 2,r = 1,
{ }
1,3 , p = 2,r = 2,
Z(2) = { }
pr  1,3,5,7 , p = 2,r 3,
{ } ≥
1,η , p 3,
{ } ≥




where η (Z/pZ) is a non-quadratic residue of p.
×
∈
Lemma 2.3. Let p 5 be a prime number. If p ∤ α then there exists l Z/pZ such that
≥ ∈
1+αl2 is a non-quadratic residue.
Proof. When (α/p) = 1, it is easy to see that αl2 l (Z/pZ) coincides β
∗
{ | ∈ } { ∈
(Z/pZ) (β/p) = 1 . Since # β (Z/pZ) (β/p) = 1 = 1/2#(Z/pZ) , the claim
∗ ∗ ∗
| } { ∈ | }
of the lemma is trivial.
When (α/p) = 1, then αl2 l (Z/pZ) = β (Z/pZ) (β/p) = 1 . Assume
∗ ∗
− { | ∈ } { ∈ | − }
that1+αl2 isquadraticresidueforanyl (Z/pZ) ,namely 1+αl2 l (Z/pZ) = β
∗ ∗
∈ { | ∈ } { ∈
(Z/pZ) (β/p) = 1 . Since # β (Z/pZ) (β/p) = 1 = # β (Z/pZ) (β/p) =
∗ ∗ ∗
| } { ∈ | } { ∈ |
1 , it should hold that (1/p) = (3/p) = = 1, (2/p) = (4/p) = = 1 or
− } ··· ··· −
(1/p) = (3/p) = = 1, (2/p) = (4/p) = = 1. However (1/p) = (4/p) = 1. This
··· − ···
contradicts to the assumption. Therefore the claim holds.
Lemma 2.4. (i) When p = 2 we have
if d 1 mod 4 then γ γ in PSL (Z/2rZ) for ν = 3,5,7,
1 ν 2
≡ ∼
if d 3 mod 4 then γ γ , γ γ , γ3 γ and γ3 γ in PSL (Z/2rZ),
≡ 1 ∼ 5 3 ∼ 7 1 ∼ 3 3 ∼ 1 2
if d 2 mod 8 then γ γ , γ γ , γ3 γ and γ3 γ in PSL (Z/2rZ),
≡ 1 ∼ 7 3 ∼ 5 1 ∼ 3 3 ∼ 1 2
if d 6 mod 8 then γ γ , γ γ , γ5 γ and γ5 γ in PSL (Z/2rZ),
≡ 1 ∼ 3 5 ∼ 7 1 ∼ 5 5 ∼ 1 2
if d 4 mod 8 then γ γ , γ γ , γ3 γ and γ3 γ in PSL (Z/2rZ),
≡ 1 ∼ 5 3 ∼ 7 1 ∼ 3 3 ∼ 1 2
if d 0 mod 8 then γν γ and γν γ in PSL (Z/2rZ) for ν = 3,5,7,
≡ 1 ∼ ν ν ∼ 1 2
where d = D if the square free factor of D is equivalent to 1 modulo 4 and d = D/4
otherwise.
(ii) When p 3 we have
≥
if p ∤ D then γ γ in PSL (Z/prZ),
1 η 2
∼
if p D then γη γ and γµ γ (ηµ 1 mod pr) in PSL (Z/prZ).
| 1 ∼ η η ∼ 1 ≡ 2
6 Y. Hashimoto
Proof. When p = 2, let
1 0
g := .
i i 1
(cid:18) (cid:19)
We have
6 d 2, 3 d 1,
≡ ≡
(g 1γ g ) /u 5 d 4, (g 1γ g ) /u 5 d 3,
1− 1 1 21 ≡ ≡ 2− 1 2 21 ≡ ≡
3 d 6, 7 d 5,
≡ ≡
3 d 5, 5 d 5,
(g 1γ g ) /u  ≡ (g 1γ g ) /u  ≡
4− 1 4 21 ≡(5 d 1, 6− 1 6 21 ≡(7 d 1.
≡ ≡
Then the relations γ γ in (i) of the lemma are true.
1 ν
∼
When p = 3 and 3 ∤ D , we have (g 1γ g) 2u mod 3r, where
γ − 1 21 γ
≡
0 ( D/8) 1/2
−
− − , D 1 mod 3,
 ( 8/D)1/2 0 ! γ ≡
g := −

 (2+D/4)1/2 (2+D/4)1/2 1
 − , D 2 mod 3.
γ
1 1 ! ≡


Then we see that γ1∼ γ2 in PSL2(Z/3rZ) if 3 ∤ D.
When p 5, according to Lemma 2.3, we see that if p ∤ α the equation x2 αy2 η
≥ − ≡
mod pr has a solution such that p ∤ x or p ∤ y. For p ∤ D , let (l,m) (Z/prZ)2 be the
γ
∈
solution of x2 4 1Dy2 η and
−
− ≡
l
g := ∗ .
m
(cid:18) ∗(cid:19)
Then we have (g 1γ g) /u η mod pr. This means that γ is conjugate to γ in
− 1 21 1 η
≡
PSL (Z/prZ).
2
By virtue of 4. of Fact 2.1, we have
(γν) u (t+u√D)/2 ν (t u√D)/2 ν
1 21 = γν = − −
u u 2u√d
(cid:0) (cid:1) (cid:0) (cid:1)
[(ν 1)/2]
− ν t ν 1 Du2 l
= − .
2l+1 2 Du2 +4
Xl=0 (cid:18) (cid:19)(cid:16) (cid:17) (cid:16) (cid:17)
When ν = 3, we have
(γν) t 2 Du2
1 21 = ν + 3 mod 8
u 2 Du2 +4 ≡
(cid:16) (cid:17) (cid:16) (cid:17)
for d 0,2,3,4,7 mod 8. Then in these cases ν3 ν . Similarly we can see that ν5 ν
≡ 1 ∼ 3 1 ∼ 5
for d 6 mod 8. When 8 d or p d, we have
≡ | |
[(ν 1)/2]
(γν) t ν 1 − ν t ν 1 Du2 l
1 21 =ν − + − ν mod 8 or p.
u 2 2l+1 2 Du2 +4 ≡
(cid:16) (cid:17) Xl=1 (cid:18) (cid:19)(cid:16) (cid:17) (cid:16) (cid:17)
Then γν γ holds for p = 2, ν = 3,5,7 and p 3, ν = η.
1 ∼ ν ≥
Selberg’s zeta functions for congruence subgroups 7
3 Generalized Venkov-Zograf’s formula
In this section, we give a generalization of Venkov-Zograf’s formula [VZ] used in the proof
of the main theorem.
Let G be a connected non-compact semi-simple Lie group with finite center, and K
a maximal compact subgroup of G. We put d = dim(G/K). We denote by g, k the Lie
algebras of G, K respectively and g = k+p a Cartan decomposition with respect to the
Cartan involution θ. Let a be a maximal abelian subspace of p. Throughout this paper
p
we assume that rank(G/K) = 1, namely dima = 1. We extend a to a θ-stable maximal
p p
abelian subalgebra a of g, so that a = a +a , where a = a p and a = a k. We put
p k p k
∩ ∩
A = expa, A = expa and A = expa .
p p k k
We denote by gC, aC the complexification of g, a respectively. Let Φ be the set of
roots of (gC,aC), Φ+ the set of positive roots in Φ, P = α Φ+ α 0 on a , and
+ p
{ ∈ | 6≡ }
P = Φ+ P . We put ρ = 1/2 α. For h A and linear form λ on a, we denote
by−ξ the−char+acter of a given by ξ α(∈hP)+= expλ(lo∈gh). Let Σ be the set of restrictions to
λ λ
P
a of the elements of P . Then the set Σ is either of the form β or β,2β with some
p +
{ } { }
element β Σ. We fix an element H a such that β(H ) = 1, and put ρ = ρ(H ).
0 p 0 0 0
∈ ∈
Note that
m 1, m, (G = SO(m,1),m 2),
− ≥
2m, 2m, (G = SU(m,1),m 2),
2ρ =  d =  ≥
0 4m+2, 4m, (G = SP(m,1),m 1),
  ≥
22, 16, (G = F ).
4
 
 
 
 
Let Γ be a discrete subgroup of G such that the volume of X = Γ G/K is finite. We
Γ
\
denote by Prim(Γ) the set of primitive hyperbolic conjugacy classes of Γ, and Z(Γ) the
center of Γ. For γ Prim(Γ), we denote by h(γ) an element of A which is conjugate to
∈
γ, and h (γ), h (γ) the elements of A , A respectively such that h(γ) = h (γ)h (γ). Let
p k p k p k
N(γ) be the norm of γ given by N(γ) = exp(β(log(h (γ)))).
p
For a finite dimensional representation χ of Γ, we define the Selberg zeta function as
follows.
ZΓ(s,χ) := det(Id χ(γ)ξλ(h(γ))−1N(γ)−s)mλ Res > 2ρ0,
−
γ Prim(Γ)λ L
∈Y Y∈
where L is the semi-lattice of linear forms on a given by L := l m α α P ,m
{ i=1 i i | i ∈ + i ∈
Z and m is the number of distinct (m , ,m ) (Z )l such that λ = l m α
≥0} λ 1 ··· l ∈ ≥0 P i=1 i i ∈
L. We note that the logarithmic derivative of the above is written as follows.
P
Z (s,χ)
Γ′ = trχ(γk)logN(γ)D(γk) 1N(γ) ks Res > 2ρ , (3.1)
− − 0
Z (s,χ)
Γ
γ Prim(Γ),k 1
∈ X ≥
where D(γ) := 1 ξ (h(γ)) 1 .
α P+| − α − |
We nowgeneral∈ize the Venkov-Zograf formula [VZ]for two dimensional case asfollows.
Q
8 Y. Hashimoto
Lemma 3.1. Let Γ be a discrete subgroup of G such that the volume of Γ G/K is finite
\
and Γ a subgroup of Γ with the finite index. Then, for a finite dimensional representation
′
χ of Γ, we have
′
ZΓ′(s,χ) = ZΓ(s,IndΓΓ′χ),
where IndΓ χ is a representation of Γ induced by χ.
Γ′
Proof. It is easy to see that, for γ Γ, there exist positive integers m , ,m and a
1 k
complete system of representatives ∈a(i) of Γ/Γ such that m m··,· k m =
{ j } ′ 1 ≥ ··· ≥ k i=1 i
[Γ : Γ] and (a(i)) 1γa(i) for m 2, 1 j m 1 and (a(i)) 1γa(i) are in Γ. Then by
′ j − j+1 i ≥ ≤ ≤ i − mi − 1 P′
the definition of the induced representation, we see that
S 0
m1 ···
(IndΓΓ′χ)(γ) ∼  ... ... ... , (3.2)
0 S
 ··· mk 
 
where S is given by
mi
χ(a(i)−1γa(i)) (m = 1),
1 1 i
 0 χ(a(i)−1γa(i)) 0
1 2 ···
Smi =  ... ... ... ...  (m 2).
 0 0 χ(a(i) −1γa(i))  i ≥
 ··· mi 1 mi 
 χ(am(i)i−1γa1(i)) 0 ··· −0 
 

Then we have
ZΓ(s,IndΓΓ′χ) = det(Id−IndΓΓ′χ(γ)ξλ(h(γ))−1N(γ)−s)mλ
γ Prim(Γ)λ L
∈Y Y∈
k
= det(Id−χ(a(1i)−1γmia(1i))ξλ(h(γmi))−1N(γmi)−s)mλ.
γ Prim(Γ)λ Li=1
∈Y Y∈ Y
According to Lemma 2.3 of [HW], we see that (IndΓ χ)(γ) is expressed as (3.2) if and only
Γ′
if a(i)−1γmia(i) (1 i k) are primitive and are not conjugate to each other in Γ. This
1 1 ≤ ≤ ′
means that
a(i)−1γmia(i) γ Prim(Γ),1 i k = Prim(Γ).
{ 1 1 | ∈ ≤ ≤ } ′
Then we have
ZΓ(s,IndΓΓ′χ) = det(Id−χ(γ′)ξλ(h(γ′))−1N(γ′)−s)mλ = ZΓ′(s,χ).
γ′ Prim(Γ′)λ L
∈Y Y∈
This completes the proof.
Selberg’s zeta functions for congruence subgroups 9
We furthermore prepare the following two lemmas.
Lemma 3.2. (see [HW]) Let Γ,Γ ,Γ be discrete subgroups of G such that vol(Γ G/K) <
1 2
\
, Γ ,Γ Γ, Γ Γ = Γ and [Γ : Γ Γ ] < . Then, for any γ Γ, we have
1 2 1 2 1 2
∞ ⊂ ∩ ∞ ∈
IndΓ 1 (γ) = IndΓ 1 (γ) IndΓ 1 (γ).
Γ1 Γ2 Γ1 ⊗ Γ2
∩
Lemma 3.3. Let Γ b(cid:0)e a congr(cid:1)uence s(cid:0)ubgroup(cid:1) of lev(cid:0)el N =(cid:1) pr. Then we have
pN
Γ = Γ for some Γ Γˆ(pr). |
∩p|N pr pr ⊃ Q
Proof. Put Γ = ΓΓˆ(pr). It is easy to check that Γ = Γ.
pr pN pr
∩ |
4 Main theorem and its proof
In the case of G = SO(2,1) SL (R) and G = SO(3,1) SL (C), the Selberg zeta
2 2
≃ ≃
function is written as follows.
∞
Z (s) = (1 N(γ) s n), Res > 1, G = SL (R),
Γ − − 2
−
γ Prim(Γ)n=0
∈Y Y
2n
Z (s) = (1 a(γ) 2ma(γ)− N(γ) s), Res > 2, G = SL (C),
Γ − − 2
−
γ Prim(Γ) m,n 0
∈Y m nmYo≥dm(γ)
≡
where a(γ) is the eigenvalue of γ Prim(Γ) such that a(γ) > 1 and m(γ) is the order
∈ | |
of the torsion of the centralizer of γ. Note that a(γ) 2 = N(γ).
| |
It is known that Γ = SL (Z) and SL ( ) where is the ring of integers of Q(√ M)
2 2
O O −
and M > 0 is not square are discrete subgroups of G = SL (R) and SL (C) respectively
2 2
such that the volume of Γ G/K is finite.
\
Let
D Z D is not a square and D 0,1 mod 4 K = Q,
>0
D = D := { ∈ | ≡ }
K ( D D is not a square and d α2 mod 4 K = Q(√ M).
{ ∈ O | ≡ ∃ } −
For D D , we see that P(D) (Z/lZ) Z. We call that ǫ(D) := (t +u √D)/2 by
K 1 1
∈ ≃ ×
the fundamental unit of D. Then the primitive element γ of SL ( ) is written as (2.1)
2
O
for (t,u) = (t ,u ). For such a γ, we see that a(γ) = ǫ(D) and N(γ) = ǫ(D) 2. Since
1 1
| |
m(γ) = l, by using (t,u) P(D) with 0 argǫ(D) π/2, we see that the Selberg zeta
∈ ≤ ≤
function for Γ = SL ( ) as follows (see [Sa1] and [Sa2]).
2
O
∞
Z (s) = (1 ǫ(D) 2(s+n))h(D) Res > 1,
SL2(Z) − −
d Dn=0
Y∈ Y
∞ 2n
Z (s) = (1 ǫ(D) 2mǫ(D)− ǫ(D) 2s)h(D) Res > 2.
SL2(O) − − | |−
D Dm,n=0
Y∈ Y
Let a be an ideal in . We call that Γ SL ( ) is a congruence subgroup of SL ( )
2 2
O ⊂ O O
with the level a if Γ Γˆ(a) and Γ Γˆ(a) for any a a. The main result of the present
′ ′
⊃ 6⊃ |
paper is as follows.
10 Y. Hashimoto
Theorem 4.1. Let be the integer ring of Q or its imaginary quadratic extension, a an
O
ideal in and Γ a congruence subgroup of SL ( ) of level a. Then the following formulas
2
O O
hold:
(i) The case of = Z,
O
∞
Z (s) = det(I χ (γ )ǫ(D) 2(s+n))h(D),
Γ Γ 1 −
−
D Dn=0
Y∈ Y
Z (s) 2logǫ(D)
Γ′ = trχ (γ )h(D) ǫ(D) 2js.
Z (s) Γ 1 1 ǫ(D) 2j −
Γ −
D Dj 1 −
X∈ X≥
(ii) The case of Q(√ M),
O ⊂ −
∞ 2n
Z (s) = det(I χ (γ )ǫ(D) 2mǫ(D)− ǫ(D) 2s)h(D)/µa(D),
Γ Γ ν − −
− | |
DY∈Dn,Ym=0ν∈La(t1Y(D),u1(D))
Z (s) h(D) 2logǫ(D)
Γ′ = trχ (γ ) ǫ(D) 2js.
Z (s) Γ ν µ (D) 1 ǫ(D) 2j 2| |−
Γ a −
DX∈DXj≥1 ν∈La(t1X(D),u1(D)) | − |
where γ and (t,u) are respectively defined in Lemma 2.1 and 2.2.
ν a
L
Proof. Since the expressions of the logarithmic derivatives of Z (s) can be obtained easily
Γ
from those of Z (s), we treat only Z (s).
Γ Γ
According to Lemma 3.1, we have
Z (s) =Z (s,χ )
Γ SL2( ) Γ
O
= det(I χΓ(γ)ξλ(h(γ))−1N(γ)−s)mλ
−
γ∈PrimY(SL2(O))λY∈L
= det(I χΓ(γ)ξλ(h(γ))−1N(γ)−s)mλ
−
DY∈DλY∈Lγ∈I(t1(YD),u1(D))
µa(D)
= det(I χΓ(γ)ξλ(h(γ))−1N(γ)−s)mλ.
−
dY∈DλY∈L Yi=1 γ∈Ji(t1Y(D),u1(D))
where χΓ := IndΓSL2(Z)1. Since γ ∼ γνi in PSL2(O/a) and χΓ ∼ IndΓSL/Γˆ2((Oa)/a)1, we see that
χ (γ) χ (γ ). Then, from Lemma 2.2, we have
Γ ∼ Γ νi
Z (s)
Γ
µa(D)
∞
det(I χ (γ )ǫ(D) 2(s+n))h(D)/µa(D), = Z,
 − Γ νi − O
D Dn=0 i=1
= Y∈ Y Y



 µa(D)
∞ n
det(I χ (γ )ǫ(D) mǫ(D)− ǫ(D) 2(s+n))h(D)/µa(D), Q(√ M).
− Γ νi − | |− O ⊂ −

DY∈DmY,n=0 Yi=1



