Directory UMM :Journals:Journal_of_mathematics:GMJ:
Georgian Mathematical Journal
1(1994), No. 1, 53-76
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS 5
AND 6 FOR THE CONGRUENCE GROUP Γ0 (4N )
G. LOMADZE
Abstract. Two entire modular forms of weight 5 and two of weight
6 for the congruence subgroup Γ0 (4N ) are constructed, which will
be useful for revealing the arithmetical sense of additional terms in
formulas for the number of representations of positive integers by
quadratic forms in 10 and 12 variables.
In this paper N, a, k, n, r, s, t denote positive integers; u are odd positive integers; H, c, g, h, j, m, α, β, γ, δ, ξ, η are integers; A, B, C, D
are complex numbers, and z, τ (Im τ > 0) are complex variables. Further,
n
( hu ) is the generalized Jacobi symbol,
a binomial coefficient, ϕ(k) Eut
ler’s function, and e(z) = exp 2πiz, η(γ) = 1 if γ ≥ 0 and η(γ) = −1 if
γ < 0.
Let
o
n ατ + β
Γ=
αδ − βγ = 1 ,
γτ + δ
n ατ + β
o
Γ0 (4N ) =
∈ Γγ ≡ 0 (mod 4N ) .
γτ + δ
Definition. We shall say that a function F defined on H = {τ ∈
C | Im τ > 0} is an entire modular form of weight r and character χ(δ)
for the congruence subgroup Γ0 (4N ) if
1) F is regular on H,
2) for all substitutions from Γ0 (4N ) and all τ ∈ H
F
ατ + β
= χ(δ)(γτ + δ)r F (τ ),
γτ + δ
1991 Mathematics Subject Classification. 11F11,11F03,11F27.
53
54
G. LOMADZE
3) in the neighborhood of the point τ = i∞
F (τ ) =
∞
X
Am e(mτ ),
m=0
4) for all substitutions from Γ, in the neighborhood of each rational point
τ = − γδ (γ 6= 0, (γ, δ) = 1)
r
(γτ + δ) F (τ ) =
∞
X
n ατ + β
A′m e
.
4N γτ + δ
m=0
For η 6= 0, ξ, g, h, N with ξg + ηh + ξηN ≡ 0 (mod 2), put
ξ
X
g 2
ξ
m+
.
Sgh
; c, N =
(−1)h(m−c)/N e
η
2N η
2
mmodN |η|
m≡c (mod N )
It is known ([2] p. 323, formula (2.4)–(2.6)) that
ξ
ξ
Sg+2j,h
; c, N = Sgh
; c + j, N ,
η
η
ξ
ξ
; c, N = Sgh
; c, N ,
Sg,h+2j
η
η
ξ
ξ
; c, N .
; c + Nj , N = (−1)hj Sgh
Sgh
η
η
(1)
Let
=
X
m≡c (mod N )
ϑgh (z|τ ; c, N ) =
1
g 2
g
(−1)h(m−c)/N e
m+
τ e m+ z ,
2N
2
2
(2)
hence
X
∂n
(−1)h(m−c)/N (2m + g)n ×
ϑgh (z|τ ; c, N ) = (πi)n
n
∂z
m≡c (mod N )
1
g 2
g
τ e m+ z .
m+
×e
2N
2
2
(3)
Put
∂n
ϑ
(z|τ
;
c,
N
)
,
gh
∂z n
z=0
(0)
ϑgh (τ ; c, N ) = ϑgh (τ ; c, N ) = ϑgh (0|τ ; c, N ).
(n)
ϑgh (τ ; c, N ) =
(4)
ON SOME ENTIRE MODULAR FORMS
55
It is known ([2], p.318, formula (1.3); p. 321, formula (1.12); p.327,
formulas (3.9), (3.5), (3.3), (3.7); p.324, formula (2.16); p.327, formulas
(3.10), (3.11)) that
(5)
ϑg,h+2j (z|τ ; c, N ) = ϑgh (z|τ ; c, N );
2
iτ 1/2 N z
z 1
e
×
ϑgh − ; c, N = −
τ
τ
N
2τ
X
g
h
1
(6)
×
c+
H+
ϑhg (z|τ ; H, N );
e −
N
2
2
H modN
−i(γτ + δ) sgn γ 21 N γz 2
z ατ + β
ϑgh
; c, N =
e
×
γτ + δ γτ + δ
N |γ|
2(γτ + δ)
X
×
ϕg′ gh (c, H; N )ϑg′ h′ (z|τ ; H, N ) (γ =
6 0),
(7)
H modN
where
g ′ = αg + γh + αγN, h′ = βg + δh + βδN,
(8)
βδ
′
′
g 2
β
g
g
c+
H+
e
×
H+
ϕg′ gh (c, H; N ) = e −
2N
2
N
2
2
α
(9)
×Sg−δg′ ,h+βg′
; c − δH, N ;
γ
β
g 2
c+
ϑgh (z|τ + β; c, N ) = e
ϑg,h+βg+βN (z|τ ; c, N ),
2N
2
ϑgh (−z|τ − β; c, N ) =
(10)
g 2
β
=e −
ϑ−g,−h+βg−βN (z|τ ; −c, N ).
c+
2N
2
From (5) and (10), according to the notations (4), it follows that
ϑg,h+2j (τ ; c, N ) = ϑgh (τ ; c, N ),
(n)
(n)
ϑg,h+2j (τ ; c, N ) = ϑgh (τ ; c, N );
β
g 2 (n)
(n)
ϑg,h+βg+βN (τ ; c, N ),
c+
ϑgh (τ + β; c, N ) = e
2N
2
(n)
ϑgh (τ − β; c, N ) =
g 2 (n)
β
c+
ϑ−g,−h+βg−βN (τ ; −c, N ).
= (−1)n e −
2N
2
(11)
(12)
From (2) and (3), according to the notations (4), in particular, it follows
56
G. LOMADZE
that
ϑgh (τ ; 0, N ) =
∞
X
1
g 2
Nm +
τ ,
(−1)hm e
2N
2
m=−∞
(n)
(13)
ϑgh (τ ; 0, N ) =
= (πi)n
∞
X
1
g 2
τ .
Nm +
(−1)hm (2N m + g)n e
2N
2
m=−∞
1.
Lemma 1. For n ≥ 0
iτ (2n+1)/2
1
(n)
×
ϑgh − ; c, N = (N i)n −
τ
N
X
g
h
1
e −
×
c+
H+
×
N
2
2
H modN
n
n
t
o
X
n X Atk
(n−t)
(n)
× ϑhg (τ ; H, N ) +
· ϑhg (τ ; H, N ) ,
t
k! z=0
t=1
k=1
where
Atk
z=0
=
(
k
(2k)! Nτπi
if t = 2k
0
if t 6= 2k
(t = 1, 2, . . . , n; k = 1, 2, . . . , t).
(1.1)
Proof. From (6), by Leibniz’s formula, we obtain
−iτ 1/2 ∂ n n N z 2
z 1
∂n
ϑgh − ; c, N ) = τ n
×
e
n
∂z
τ
τ
N
∂z n
2τ
o
X
1
g
h
c+
H+
ϑhg (z|τ ; H, N ) =
e −
×
N
2
2
H modN
= (N i)n
×
X
H modN
−iτ (2n+1)/2
×
N
o
1
g
h n N z 2 ∂ n
e
ϑ
(z|τ
;
H,
N
+
e −
c+
H+
)
hg
N
2
2
2τ ∂z n
+
n
o
X
n ∂ n N z 2 ∂ n−t
e
ϑ
(z|τ
H,
N
)
;
.
hg
t ∂z n
2τ ∂z n−t
t=1
(1.2)
ON SOME ENTIRE MODULAR FORMS
57
According to formulae (a) and (b) of [1], p. 371
N z 2 At2 N z 2
Att N z 2
∂t N z2
e
e
e
=
A
+
+
·
·
·
+
e
t1
∂z t
2τ
2τ
2!
2τ
t!
2τ
(t = 1, 2, . . . , n),
where
N πiz 2 ∂ t N πiz 2k−1
∂ t N πiz 2 k
−
+
k
∂z t
τ
τ
∂z t
τ
k(k − 1) N πiz 22 ∂ t N πiz 2k−2
+
+
2!
τ
∂z t
τ
N πiz 2 k−1 ∂ t N πiz 2
+ · · · + (−1)k−1 k
(k = 1, 2, . . . , t),
τ
∂z t
τ
Atk =
hence
t
X
∂ t N z 2
Atk
e
=
∂z t
2τ z=0
k! z=0
(t = 1, 2, . . . , n)
(1.3)
k=1
and
∂ t N πiz 2 k
Atk z=0 = t
∂z
τ
z=0
(k = 1, 2, . . . , t).
(1.4)
Thus, in view of notations (4), the lemma follows from (1.2)–(1.4).
6 0, then for n ≥ 0
Lemma 2. If γ =
(n)
ϑgh
ατ + β
sgn γ (2n+1)/2
×
; c, N = (N |γ|i sgn γ)n − i(γτ + δ)
γτ + δ
N |γ|
n
X
(n)
ϕg′ gh (c, H; N ) ϑg′ h′ (τ ; H, N ) +
×
+
H modN
n
X
t=1
n
t
t
o
X
Atk
(n−t)
· ϑg′ h′ (τ ; H, N ) ,
k! z=0
k=1
where g ′ h′ and ϕg′ gh (c, H; N ) are defined by the formulas (8) and (9),
(
γπi k
(2k)! N
if t = 2k
γτ +δ
Atk
=
z=0
if t 6= 2k
0
(t = 1, 2, . . . , n; k = 1, 2, . . . , t).
1 Page
75 in the Russian version of [1] published in 1933.
58
G. LOMADZE
Proof. From (7), according to Leibnitz’s formula, we obtain
z ατ + β
sgn γ 1/2
∂n
c,
N
ϑ
;
×
= (γτ + δ)n − i(γτ + δ)
gh
∂z n
γτ + δ γτ + δ
N |γ|
o
∂ n n N γz 2 X
ϕg′ gh (c, H; N )ϑg′ h′ (z|τ ; H, N ) =
× n e
∂z
2(γτ + δ)
H modN
sgn γ (2n+1)/2
×
= (N |γ|i sgn γ)n − i(γτ + δ)
N |γ|
n N γz 2 ∂ n
X
ϕg′ gh (c, H; N ) e
×
ϑg′ h′ (z|τ ; H, N ) +
2(γτ + δ) ∂z n
H modN
n
X
+
t=1
o
N γz 2 ∂ n−t
n ∂t
′ h′ (z|τ ; H, N ) .
ϑ
e
g
t ∂z t 2(γτ + δ) ∂z n−t
As in Lemma 1, but with e
N γz 2
2(γτ +δ)
(1.5)
2
instead of e N2τz , we have
t
X
Atk
∂ t N γz 2
(t = 1, 2, . . . , n)
e
=
∂z t
2(γτ + δ) z=0
k! z=0
(1.6)
k=1
and
Atk
z=0
=
∂ t N γπiz 2k
(k = 1, 2, . . . , t).
∂z t γτ + δ
z=0
(1.7)
Thus, according to the notations (4), the lemma follows from (1.5)(1.7).
Lemma 3. If g is even, then for n ≥ 0 and all substitutions from Γ0 (4N )
we have
(n) ατ
ϑgh
+β
; 0, 2N =
γτ + δ
= ( sgn δ)n i(2n+1)η(γ)( sgn δ−1)/2 i(1−|δ|)/2
2βN sgn δ
×
|δ|
αγδ 2 h2 βδg 2 δ 2ϕ(2N )−2
(n)
×(γτ + δ)(2n+1)/2 e −
e
ϑαg,h (τ ; 0, 2N ).
16N
4
4N
Proof. 1) Let γ 6= 0. In [4] (p.18, formula (5.1)) it is shown that
β
; 0, 2N =
Sg0
δ
βδg 2 δ 2ϕ(2n)−2
2βN sgn δ
|δ|1/2 .
i(1−|δ|)/2
=e
4
4N
|δ|
(1.8)
ON SOME ENTIRE MODULAR FORMS
59
Replacing α, β, γ, δ, τ, c, N by β, −α, δ, −γ, τ ′ , 0, 2N in Lemma 2,
we obtain
(n) βτ
ϑgh
′
δτ ′
sgn δ (2n+1)/2
−α
; 0, 2N = (2N |δ|i sgn δ)n − i(δτ ′ − γ)
×
−γ
2N |δ|
n
X
(n)
×
ϕh′ gh (0, H; 2N ) ϑh′ ,αg (τ ′ ; H, 2N ) +
H mod2N
n
X
t
o
n X Atk
(n−t)
· ϑh′ ,αg (τ ′ ; H, 2N ) ,
t
k! z=0
+
t=1
(1.9)
k=1
where by (8),(9),(11) and (1)
h′ = βg + δh + 2βδN,
(1.10)
ϕh′ gh (0, H; 2N ) =
αγh
h′ β
αg
e −
H+
Sg0
; 0, 2N ,
=e −
δ
16N
4N
2
(
k
δπi
if t = 2k
(2k)! 2N
δτ ′ −γ
Atk
=
z=0
0
if t 6= 2k
′2
(k = 1, 2, . . . , t).
(1.11)
(1.12)
Writing − τ1 instead of τ ′ in (1.9) and (1.12), according to (1.11), we
obtain
−i(− δ − γ) sgn δ (2n+1)/2
+β
τ
×
; 0, 2N = (2N i)n (|δ| sgn δ)n
γτ + δ
2N |δ|
′
αγh 2 βδg 2 δ 2ϕ(2N )−2 (1−|δ|)/2 2βN sgn δ 1/2
e
i
|δ| ×
×e −
16N
4
4N
|δ|
αg
X
h′ n (n) 1
e −
ϑh′ ,αg − ; H, 2N +
×
H+
4N
2
τ
(n) ατ
ϑgh
H mod2N
n
X
+
t=1
where
Atk
z=0
=
(
n
t
t
X
o
Atk
1
(n−t)
· ϑh′ ,αg − ; H, 2N ,
k! z=0
τ
k=1
δπiτ k
(2k)! 2N
if t = 2k
γτ +δ
0
if t 6= 2k
(k = 1, 2, . . . , t).
(1.13)
60
G. LOMADZE
Writing αg, h′ , − τ1 , 0, 2N instead of g, h, τ , c, N in Lemma 1, we
obtain
(n)
ϑαg,h′ (τ ; 0, 2N ) = (2N i)n
i (2n+1)/2
2N τ
X
H mod2N
αg
h′
e −
H+
×
4N
2
n
1
(n)
× ϑh′ ,αg − ; H, 2N +
τ
n
t
X
o
1
n X Atk
(n−t)
· ϑh′ ,αg − ; H, 2N ,
+
t
k! z=0
τ
t=1
(1.14)
k=1
where
Atk
z=0
=
(
(2k)!(−2N πiτ )k if t = 2k
0
if t 6= 2k
(k = 1, 2, . . . , t).
From (1.14) it follows that
X
H mod2N
n
X
+
t=1
αg
h′ n (n) 1
e −
H+
ϑh′ ,αg − ; H, 2N +
4N
2
τ
n
t
t
X
o
Atk
1
(n−t)
=
· ϑh′ ,αg − ; H, 2N
k! z=0
τ
= (2πi)
k=1
−n
(n)
(−2N iτ )(2n+1)/2 ϑαg,h′ (τ ; 0, 2N ).
(1.15)
From (1.13) and (1.15) we obtain
(n) ατ
+β
; 0, 2N = (−2N iτ )(2n+1)/2 (|δ| sgn δ)n ×
γτ + δ
′
−i(− δ − γ) sgn δ (2n+1)/2
αγh 2
(1−|δ|)/2
τ
×
e −
i
×
2N |δ|
16N
βδg 2 δ 2ϕ(2N )−2 2βN sgn δ
(n)
×e
|δ|1/2 ϑαg,h′ (τ ; 0, 2N ).
4
4N
|δ|
ϑgh
(1.16)
In [3] (p.85, formula (6.16) with 2N instead of N ) it is shown that
−i(− δ − γ) sgn δ 1/2
τ
(−2N iτ )1/2 =
2N |δ|
γτ + δ 1/2
.
= i sgn γ( sgn δ−1)/2
|δ|
(1.17)
ON SOME ENTIRE MODULAR FORMS
61
From (1.16) and (1.17) it follows that
ατ + β
; 0, 2N =
γτ + δ
γτ + δ (2n+1)/2 (1−|δ|)/2
= (|δ| sgn δ)n i sgn γ( sgn δ−1)
i
×
|δ|
′
αγh 2 βδg 2 δ 2ϕ(2N )−2 2βN sgn δ 1/2 (n)
e
|δ| ϑαg,h′ (τ ; 0, 2N ).
×e −
16N
4
4N
|δ|
(n)
ϑgh
Thus, in view of (1.10) and (1.11), the lemma is proved for γ 6= 0.
2) Let γ = 0. Then α = δ = 1 or α = δ = −1 in (12). Putting c = 0 in
(12) and writing 2N instead of N , by (11), we obtain
βg 2 (n)
(n)
ϑgh (τ + β; 0, 2N ) = e
ϑ (τ ; 0, 2N ),
16N gh
βg 2 (n)
(n)
ϑ
(τ ; 0, 2N ),
ϑgh (τ − β; 0, 2N ) = (−1)n e −
16N −g,h
i.e., the lemma is also proved for γ = 0.
Lemma 4. If γ =
6 0, then for n ≥ 0
(n)
(γτ + δ)(2n+1)/2 ϑgh (τ ; 0, 2N ) =
= e (2n + 1) sgn γ/8 (2N |γ|)−1/2 (−i sgn γ)n ×
ατ + β
n
X
(n)
×
ϕg′ gh (0, H; 2N ) ϑg′ h′
; H, 2N +
γτ + δ
H mod2N
n
X
+
t=1
n
t
where
Atk
z=0
=
t
X
o
Atk
(n−t) ατ + β
· ϑg ′ h ′
; H, 2N ,
k! z=0
γτ + δ
k=1
(
(2k)!(−2N γπi(γτ + δ))k if t = 2k
if t 6= 2k
0
(t = 1, 2, . . . ; k = 1, 2, . . . , t).
Remark. From (1.18) it follows that
( A
t
t,t/2
X
Atk
(t/2)! z=0 if 2|t
=
k! z=0
0 if 2†t.
k=1
(1.18)
(1.19)
62
G. LOMADZE
Proof. Replacing α, β, γ, δ, τ, c, N by δ, −β, −γ, α, τ ′ , 0, 2N in
Lemma 2, we obtain
δτ ′ − β
(n)
2N
ϑgh
;
0,
=
−γτ ′ + α
sgn γ (2n+1)/2
= (−2N |γ|i sgn γ)n i(−γτ ′ + α)
×
2N |γ|
n
X
(n)
×
ϕg′ gh (0, H; 2N ) ϑg′ h′ (τ ′ ; H, 2N ) +
+
H mod2N
n
X
t=1
where
t
o
n X Atk
(n−t)
· ϑg′ h′ (τ ′ ; H, 2N ) ,
t
k! z=0
(1.20)
k=1
g ′ = δg − γh − 2γδN, h′ = −βg + αh − 2αβN,
(1.21)
αβ
′
′
βg
g 2
g
e −
×
H+
H+
ϕg′ gh (0, H; 2N ) = e −
4N
2
4N
2
δ
(1.22)
; −αH, 2N ,
×Sg−αg′ ,h−βg′
−γ
(
γπi k
(2k)! 2N
if t = 2k,
γτ ′ −α
=
(1.23)
Atk
z=0
0 if t 6= 2k.
In [3, p.87, formula (6.23)] (with 2N instead of N ) it is shown that
1/2
e( sgn γ/8)
i sgn γ
.
=
2N |γ|(γτ + δ)
(2N |γ|)1/2 (γτ + δ)1/2
(1.24)
+β
′
Taking ατ
γτ +δ instead of τ in (1.20) and (1.23) and using (1.24), we complete
the proof of the lemma.
2.
Lemma 5. For a given N let
Ψ1 (τ ) = Ψ1 (τ ; g1 , g2 , h1 , h2 , c1 , c2 , N1 , N2 ) =
1 ′′′
(τ ; c1 , 2N1 )ϑg′ 2 h2 (τ ; c2 , 2N2 ) −
ϑ
=
N1 g1 h1
1
− ϑg′′′2 h2 (τ ; c2 , 2N2 )ϑ′g1 h1 (τ ; c1 , 2N1 )
N2
and
Ψ2 (τ ) = Ψ2 (τ ; g1 , g2 , h1 , h2 , c1 , c2 , N1 , N2 ) =
(2.1)
ON SOME ENTIRE MODULAR FORMS
1 (4)
ϑ
(τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
N12 g1 h1
1 (4)
+ 2 ϑg2 h2 (τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 ) −
N2
6
ϑ′′ (τ ; c1 , 2N1 )ϑ′′g2 h2 (τ ; c2 , 2N2 ),
−
N1 N2 g1 h1
63
=
(2.2)
where
2|g1 , 2|g2 , N1 |N, N2 |N, 4|N
h1
h2
.
+
N1
N2
(2.3)
For all substitutions from Γ in the neighborhood of each rational point τ =
− γδ (γ 6= 0, (γ, δ) = 1), we then have
(γτ + δ)5 Ψj (τ ; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 ) =
=
∞
X
n ατ + β
Cn(j) e
4N γτ + δ
n=0
(j = 1, 2).
(2.4)
Proof. I. From Lemma 4 for n = 3 (with g1 , h1 , N1 , g1′ , h1′ , H1 instead
of g, h, N , g ′ , h′ , H) and n = 1 (with g2 , h2 , N2 , g2′ , h′2 , H2 instead of g,h,
N , g ′ , h′ , H), according to (1.19), it follows that
1
′
(γτ + δ)5 ϑ′′′
g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) =
N1
1 5
e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
=
N1 4
n
X
ατ + β
; H1 , 2N1 +
×
ϕg1′ g1 h1 (0, H1 ; 2N1 ) ϑ′′′
g1′ h′1
γτ + δ
H1 mod2N1
ατ + β
o
3·2
+
· ϑg′ ′ h′
A21
; H1 , 2N1 ×
1 1
2!
γτ + δ
z=0
X
ατ + β
×
; H2 , 2N2 =
ϕg2′ g2 h2 (0, H2 ; 2N2 )ϑ′g′ h′
2 2
γτ + δ
H2 mod2N2
5
−1
sgn γ 2|γ|(N1 N2 )1/2
×
4
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 )×
=e
×
X
H1 mod2N1
H2 mod2N2
×
n 1
ατ + β
ατ + β
ϑ′′′′ ′
; H1 , 2N1 ϑ′g′ h′
; H2 , 2N2 −
2 2
N1 g1 h1 γτ + δ
γτ + δ
ατ + β
; H1 , 2N1 ×
−12γπi(γτ + δ)ϑg′ ′ h′
1 1
γτ + δ
64
G. LOMADZE
ατ + β
o
; H2 , 2N2 .
2
γτ + δ
×ϑg′ ′ h′
2
(2.5)
Replacing N1 , g1 , h1 , H1 , g1′ , h′1 by N2 , g2 , h2 , H2 , g2′ , h2′ in (2.5) and
vice versa, we obtain
1
′
(γτ + δ)5 ϑ′′′
g2′ h′2 (τ ; 0, 2N2 )ϑg1′ h′1 (τ ; 0, 2N1 ) =
N2
5
= e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
4
X
′
ϕg2 g2 h2 (0, H2 ; 2N2 )ϕg1′ g1 h1 (τ ; 0, H1 ; 2N1 ) ×
×
H2 mod2N2
H1 mod2N1
n 1
ατ + β
ατ + β
×
ϑ′′′
; H2 , 2N2 ϑ′g′ h′
; H1 , 2N1 −
g2′ h′2
1
1
N2
γτ + δ
γτ + δ
+
β
ατ
; H2 , 2N2 ×
−12γπi(γτ + δ)ϑ′g′ h′
2 2
γτ + δ
o
ατ + β
′
; H1 , 2N1 .
×ϑg′ h′
1 1
γτ + δ
(2.6)
Subtracting (2.6) from (2.5), according to (2.1), we obtain
×
(γτ + δ)5 Ψ1 (τ ; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 ) =
5
= e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
H1 mod2N1
H2 mod2N2
×Ψ1
ατ + β ′ ′ ′ ′
; g1 , g2 , h1 , h2 , H1 , H2 , N1 , N2 .
γτ + δ
(2.7)
In (2.7) let γ be even. Then, by (1.21) and (2.3), g1′ and g2′ are also even.
Therefore, according to (3) and the notations (4), we have
ατ + β
ατ + β
; Hr , 2Nr ϑg′ s′ h′s
; Hs , 2Ns =
γτ + δ
γτ + δ
∞
∞
X
N n1 /Nr ατ + β X
N n2 /Ns ατ + β
=
Bn 2 e
Bn1 e
=
4N
γτ + δ n =0
4N
γτ + δ
n =0
ϑg′′′r′ hr′
2
1
=
∞
X
n ατ + β
Bn e
4N γτ + δ
n=0
for r = 1, s = 2 and r = 2, s = 1.
(2.8)
ON SOME ENTIRE MODULAR FORMS
65
Hence, for even γ, (2.4) follows from (2.1), (2.7), and (2.8) if j = 1.
Now let γ in (2.7) be odd. If h1 and h2 are both even, then by (1.21)
and (2.3), g1′ and g2′ are also even, and we obtain the same result. But if hr
is odd, then by (1.21) gr′ will also be odd, and in (3) we have
hence
m+
2
1
1
gr′ 2
1
= m + (gr′ − 1) + m + (gr′ − 1) + ,
2
2
2
4
ϑg′′′r′ hr′
×
ατ + β
1 ατ + β
; Hr , 2Nr = −π 3 ie
×
γτ + δ
16Nr γτ + δ
X
′
(−1)hr (m−Hr )/2Nr (2m + gr′ )3 ×
m≡Hr (mod 2Nr )
n 1
ατ + β o
1
1
=
(m + (gr′ − 1))2 + (m + (gr′ − 1))
×e
4Nr
2
2
γτ + δ
∞
hr ατ + β X
n1 ατ + β
=e
B′ e
,
16Nr γτ + δ n =0 n1 4Nr γτ + δ
1
since (5) implies that hr = 1. Analogously,
ϑ′gs′ h′s
∞
n2 ατ + β
ατ + β
hs ατ + β X
Bn′ 2 e
,
; Hs , 2Ns = e
γτ + δ
16Ns γτ + δ n =0
4Ns γτ + δ
2
which implies that hs = 1 if hs is odd and hs = 0 if hs is even. Thus, if
among h1 and h2 at least one is odd, then we have, for r = 1, s = 2 and
r = 2, s = 1,
ατ + β
ατ + β
; Hr , 2Nr ϑg′ s′ h′s
; Hs , 2Ns =
γτ + δ
γτ + δ
∞
N/4(hr /Nr + hs /Ns ) ατ + β X
n1 ατ + β
=e
×
Bn′ 1 e
4N
γτ + δ n =0
4Nr γτ + δ
ϑ′′′
gr′ h′r
1
×
∞
X
n2
∞
n ατ + β
n2 ατ + β X
Bn′ e
Bn′ 2 e
=
,
4Ns γτ + δ
4N γτ + δ
n=0
=0
(2.9)
hr
hs
+N
) is an integer. Hence for odd γ (2.4) follows
since, by (2.3), N4 ( N
r
s
from (2.1),(2.7), and (2.9) if j = 1.
II. From Lemma 4 for n = 4 and n = 0, as in I, it follows that
1
(4)
(γτ + δ)5 ϑg1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) =
N22
1 5
= 2 e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
N1 4
66
G. LOMADZE
n
(4) ατ + β
ϕg1′ g1 h1 (0, H1 ; 2N1 ) ϑg′ h′
; H1 , 2N1 +
1 1
γτ + δ
H1 mod2N1
ατ + β
4·3
· ϑ′′g′ h′
; H1 , 2N1 +
A21
+
1 1
2!
γτ + δ
z=0
o
ατ + β
A42
+
· ϑg1′ h′1
; H1 , 2N1 ×
2! z=0
γτ + δ
X
ατ + β
ϕg2′ g2 h2 (0, H2 ; 2N2 )ϑg2′ h2′
; H2 , 2N2 =
×
γτ + δ
X
×
H2 mod2N2
×
5
−1
= e sgn γ 2|γ|(N1 N2 )1/2
×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
H1 mod2N1
H2 mod2N2
n 1
(4) ατ + β
ϑg′ h′
; H1 , 2N1 ×
2
N1 1 1 γτ + δ
ατ + β
ατ + β
24γπi
×ϑg2′ h′2
(γτ + δ)ϑ′′g′ h′
; H2 , 2N2 −
; H1 , 2N1 ×
1
1
γτ + δ
N1
γτ + δ
ατ + β
+β
ατ
; H2 , 2N2 − 48γ 2 π 2 (γτ + δ)2 ϑg1′ h′1
; H1 , 2N1 ×
ϑg2′ h′2
γτ + δ
γτ + δ
ατ + β
o
×ϑg2′ h2′
; H2 , 2N2 .
(2.10)
γτ + δ
×
Replacing N1 , g1 , h1 , H1 , g1′ , h′1 by N2 , g2 , H2 , g2′ , h′2 in (2.10) and vice
versa, we obtain
1
(4)
(γτ + δ)5 ϑg2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 ) =
N12
5
= e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
4
X
ϕg2′ g2 h2 (0, H2 ; 2N2 )ϕg1′ g1 h1 (0, H1 ; 2N1 )×
×
H2 mod2N2
H1 mod2N1
n 1
(4) ατ + β
×
ϑg′ h′
; H2 , 2N2 ×
2
2
2
N2
γτ + δ
ατ + β
ατ + β
24γπi
(γτ + δ)ϑ′′g′ h′
×ϑg1′ h′1
; H1 , 2N1 −
; H2 , 2N2 ×
2 2
γτ + δ
N2
γτ + δ
ατ + β
ατ + β
; H1 , 2N1 − 48γ 2 π 2 (γτ + δ)2 ϑg2′ h2′
; H2 , 2N2 ×
×ϑg1′ h1′
γτ + δ
γτ + δ
ON SOME ENTIRE MODULAR FORMS
×ϑg1′ h′1
o
ατ + β
; H1 , 2N1 .
γτ + δ
67
(2.11)
Analogously, from Lemma 4 for n = 2 it follows that
6
(γτ + δ)5 ϑg′′1 h1 (τ ; 0, 2N1 )ϑ′′g2 h2 (τ ; 0, 2N2 ) =
N1 N2
5
= e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
×
H1 mod2N1
H2 mod2N2
ατ + β
ατ + β
6
; H1 , 2N1 ϑ′′g′ h′
; H2 , 2N2 −
ϑ′′g′ h′
2 2
N1 N2 1 1 γτ + δ
γτ + δ
ατ + β
ατ + β
24γπi
−
(γτ + δ)ϑg1′ h′1
; H1 , 2N1 ϑg2′ h′2
; H2 , 2N2 −
N2
γτ + δ
γτ + δ
+
β
+β
ατ
ατ
24γπi
(γτ + δ)ϑ′′g′ h′
; H1 , 2N1 ϑ′′g′ h′
; H2 , 2N2 −
−
1 1
2 2
N1
γτ + δ
γτ + δ
ατ + β
−96γ 2 π 2 (γτ + δ)2 ϑg1′ h′1
; H1 , 2N1 ×
γτ + δ
o
ατ + β
; H2 , 2N2 .
(2.12)
×ϑg2′ h′2
γτ + δ
×
n
Subtracting (2.12) from the sum of (2.10) and (2.11), according to (2.2),
we obtain
×
(γτ + δ)5 Ψ2 (τ ; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 ) =
5
−1
= e sgn γ 2|γ|(N1 N2 )1/2
×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
H1 mod2N1
H2 mod2N2
×Ψ2
ατ + β ′ ′ ′ ′
; g , g , h , h , H 1 , H2 , N 1 , N 2 .
γτ + δ 1 2 1 2
(2.13)
Further, reasoning as in I, from (2.2) and (2.13) we obtain (2.4) if
j = 2.
Theorem 1. For a given N the functions Ψ1 (τ ) and Ψ2 (τ ) with c1 =
c2 = 0 are entire modular forms of weight 10 and character χ(δ) = sgn δ( −∆
|δ| )
(∆ is the determinant of an arbitrary positive quadratic form in 5 variables)
68
G. LOMADZE
for the congruence group Γ0 (4N ) if the following conditions hold:
1)
2|g1 , 2|g2 , N1 |N, N2 |N,
h2
g2
h2 g 2
2) 4N 1 + 2 , 4 1 + 2 ,
N1
N2
4N1
4N2
3) for all α and δ with α ≡ δ ≡ 1 (mod 4N )
N 1 N2
Ψj (τ ; αg1 , αg2 , h1 , h2 , 0, 0, N1 , N2 ) =
|δ|
∆
=
Ψj (τ ; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 ) (j = 1, 2).
|δ|
(2.14)
(2.15)
(2.16)
Proof.
I. It is wellknown that the thetaseries (2)-(3) are regular on H, hence the
functions Ψ1 (τ ) and Ψ2 (τ ) satisfy condition 1 of the definition.
II. From (2.15), since δ 2 ≡ 1 (mod 4), it follows that
h2
h2 g 2
g2
4N δ 2 1 + 2 , 4 1 δ 2ϕ(2N1 )−2 + 2 δ 2ϕ(2N2 −2) .
N1
N2
4N1
4N2
(2.17)
By Lemma 3 for n = 3 and n = 1 (with gr , hr , Nr and gs , hs , Ns instead of
g, h, N ), according to (2.14) and (2.17), for each substitution from Γ0 (4N ),
we have
ατ + β
ατ + β
ϑ′′′
; 0, 2Nr ϑ′gs hs
; 0, 2Ns = iη(γ)(sgnδ−1) i1−|δ| ×
gr hr
γτ + δ
γτ + δ
N
N
r
s
′
ϑ′′′
×(γτ + δ)5
αgr ,hr (τ ; 0, 2Nr )ϑαgs ,hs (τ ; 0, 2Ns ) =
|δ|
−Nr Ns
(γτ + δ)5 ×
= sgn δ
|δ|
′
×ϑ′′′
(2.18)
αgr ,hr (τ ; 0, 2Nr )ϑαgs ,hs (τ ; 0, 2Ns )
for r = 1, s = 2 and r = 2, s = 1.
Analogously, by Lemma 3, we have:
1) for n = 4 and n = 0
(4)
ϑgr hr
ατ + β
ατ + β
; 0, 2Nr ϑgs hs
; 0, 2Ns = iη(γ)(sgnδ−1) i1−|δ| ×
γτ + δ
γτ + δ
Nr Ns (4)
ϑαgr ,hr (τ ; 0, 2Nr )ϑαgs ,hs (τ ; 0, 2Ns ) =
×(γτ + δ)5
|δ|
−Nr Ns
= sgn δ
(γτ + δ)5 ×
|δ|
(4)
′
(τ ; 0, 2Ns )
×ϑαgr ,hr (τ ; 0, 2Nr )ϑαg
s ,hs
if r = 1, s = 2 and r = 2, s = 1;
(2.19)
ON SOME ENTIRE MODULAR FORMS
69
2) for n = 2
ϑ′′g1 h1
ατ + β
ατ + β
; 0, 2N1 ϑg′′2 h2
; 0, 2N2 =
γτ + δ
γτ + δ
−N1 N2
(γτ + δ)5 ×
= sgn δ
|δ|
′′
×ϑ′′αg1 ,h1 (τ ; 0, 2N1 )ϑαg
(τ ; 0, 2N2 ).
2 ,h2
(2.20)
Hence, according to (2.16), for all α and δ with αδ ≡ 1 (mod 4N ), we
have
ατ + β
Ψj
; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 =
γτ + δ
−N1 N2
(γτ + δ)5 Ψj (τ ; αg1 , αg2 , h1 , h2 , 0, 0, N1 , N2 ) =
= sgn δ
|δ|
−∆
(γτ + δ)5 Ψj (τ ; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 )
= sgn δ
|δ|
for j = 1, by (2.1) and (2.16), and for j = 2, by (2.2),(2.19) and (2.20).
Thus the functions Ψ1 (τ ) and Ψ2 (τ ) with c1 = c2 = 0 satisfy condition 2 of
the definition.
III. From (13) it follows that, for r = 1, s = 2 and r = 2, s = 1,
′
4
ϑ′′′
gr hr (τ ; 0, 2Nr )ϑgs hs (τ ; 0, 2Ns ) = π
∞
X
(−1)hr mr +hs ms ×
mr ,ms =−∞
3
×(4Nr mr + gr ) (4Ns ms + gs )e(Λτ ),
(4)
ϑgr hr (τ ; 0, 2Nr )ϑgs hs (τ ; 0, 2Ns ) = π 4
∞
X
(−1)hr mr +hs ms ×
mr ,ms =−∞
×(4Nr mr + gr )4 e(Λτ )
(2.21)
and also
ϑ′′g1 h1 (τ ; 0, 2N1 )ϑ′′g2 h2 (τ ; 0, 2N2 ) = π 4
∞
X
(−1)h1 m1 +h2 m2 ×
m1 ,m2 =−∞
×(4N1 m1 + g1 ) (4N2 m2 + g2 )2 e(Λτ ),
2
where
Λ=
2
2
2
X
X
1X 2
1
(2Nk mk + gk /2)2 =
(Nk m2k + mk gk /2) +
gk /4Nk ,
4Nk
4
k=1
k=1
k=1
by (2.14) and (2.15), is an integer. Thus, the functions Ψ1 (τ ) and Ψ2 (τ )
with c1 = c2 = 0 satisfy condition 3 of the definition.
70
G. LOMADZE
IV. By Lemma 5 the functions Ψ1 (τ ) and Ψ2 (τ ) with c1 = c2 = 0 also
satisfy condition 4 of the definition.
3.
Lemma 6. For a given N let
Φ1 (τ ) = Φ1 (τ ; g1 , . . . , g4 , h1 , . . . , h4 , c1 , . . . , c4 , N1 , . . . , N4 ) =
4
=
Y
1 ′′′
ϑgk hk (τ ; ck , 2Nk ) −
ϑg1 h1 (τ ; c1 , 2N1 )ϑ′g2 h2 (τ ; c2 , 2N2 )
N1
k=3
4
Y
1
′
− ϑ′′′
ϑgk hk (τ ; ck , 2Nk )
g2 h2 (τ ; c2 , 2N2 ))ϑg1 h1 (τ ; c1 , 2N1 )
N2
(3.1)
k=3
and
Φ2 (τ ) = Φ2 (τ ; g1 , . . . , g4 , h1 , . . . , h4 , c1 , . . . , c4 , N1 , . . . , N4 ) =
4
=
Y
1 (4)
ϑgk hk (τ ; ck , 2Nk ) +
ϑg1 h1 (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 )
2
N1
k=3
4
+
Y
1 (4)
ϑgk hk (τ ; ck , 2Nk ) −
ϑg2 h2 (τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 )
2
N2
k=3
4
−
Y
6
ϑgk hk (τ ; ck , 2Nk ), (3.2)
ϑ′′g1 h1 (τ ; c1 , 2N1 )ϑ′′g2 h2 (τ ; c2 , 2N2 )
N1 N2
k=3
where
4
X
hk
2gk , Nk N (k = 1, 2, 3, 4), 4N
.
Nk
(3.3)
k=1
For all substitutions from Γ in the neighborhood of each rational point τ =
− γδ (γ 6= 0, (γ, δ) = 1), we then have
(γτ + δ)6 Φj (τ ; g1 , . . . , g4 , h1 , . . . , h4 , 0, . . . , 0, N1 , . . . , N4 ) =
∞
n ατ + β
X
Dn(j) e
=
(j = 1, 2).
4N γτ + δ
n=0
(3.4)
Proof. From Lemma 4 for n = 3, n = 1, and n = 0, as in the proof of
Lemma 5, it follows that
4
Y
1
′
ϑgk hk (τ ; 0, 2Nk ) =
(γτ + δ)6 ϑ′′′
g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 )
N1
k=3
ON SOME ENTIRE MODULAR FORMS
4
2 Y
3
1/2 −1
= e sgn γ 4γ
Nk
2
k=1
X
4
Y
71
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
Hk mod2Nk k=1
(k=1,2,3,4)
n 1
ατ + β
ϑ′′′
; H1 , 2N1 −
g1′ h1′
N1
γτ + δ
ατ + β
; H1 , 2N1 ×
−12γπi(γτ + δ)ϑ′g′ h′
1 1
γτ + δ
4
Y
ατ + β
o
ατ + β
ϑgk′ h′k
×ϑg′ ′ h′
; H2 , 2N2
; Hk , 2Nk
2 2
γτ + δ
γτ + δ
×
(3.5)
k=3
and
4
Y
1
′
(γτ + δ)6 ϑ′′′
ϑgk hk (τ ; 0, 2Nk ) =
g2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 )
N2
k=3
=e
3
sgn γ
2
4γ 2
4
Y
Nk
k=1
1/2
−1
X
4
Y
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
Hk mod2Nk k=1
(k=1,2,3,4)
n 1
ατ + β
ϑ′′′′ ′
; H2 , 2N2 −
N2 g2 h2 γτ + δ
ατ + β
; H2 , 2N2 ×
−12γπi(γτ + δ)ϑ′g′ h′
2 2
γτ + δ
4
Y
o
ατ + β
ατ + β
×ϑ′g′ h′
ϑgk′ h′k
; H1 , 2N1
; Hk , 2Nk .
1 1
γτ + δ
γτ + δ
×
(3.6)
k=3
As in the proof of Lemma 5, from Lemma 4 it follows that
1) for n = 4 and n = 0
4
Y
1
(4)
(γτ + δ)6 ϑg1 h1 (τ ; 0, 2N1 )
ϑgk hk (τ ; 0, 2Nk ) =
2
N1
k=2
3
= e sgn γ
2
4γ 2
4
Y
k=1
Nk
1/2
−1
X
4
Y
Hk mod2Nk k=1
(k=1,2,3,4)
ϕgk′ gk hk 0, Hk ; 2Nk ×
n 1
(4) ατ + β
ϑg′ h′
; H1 , 2N1 −
2
1
1
N1
γτ + δ
ατ + β
24γπi
(γτ + δ)ϑ′′g′ h′
; H1 , 2N1 −
−
1 1
N1
γτ + δ
ατ + β
; H1 , 2N1 ×
−48γ 2 π 2 (γτ + δ)2 ϑg1′ h′1
γτ + δ
×
72
G. LOMADZE
×
4
Y
ϑgk′ h′k
k=2
and
o
ατ + β
; Hk , 2Nk
γτ + δ
(3.7)
4
Y
1
(4)
ϑgk hk (τ ; 0, 2Nk ) =
(γτ + δ)6 ϑg2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 )
2
N2
k=3
4
1/2 −1
3
Y
Nk
= e sgn γ 4γ 2
2
k=1
X
4
Y
Hk mod2Nk k=1
(k=1,2,3,4)
ϕgk′ gk hk 0, Hk ; 2Nk ×
n 1
(4) ατ + β
; H2 , 2N2 −
ϑg′ h′
2
N2 2 2 γτ + δ
ατ + β
24γπi
(γτ + δ)ϑ′′g′ h′
; H2 , 2N2 ×
−
2 2
N2
γτ + δ
4
Y
ατ + β
ατ + β
; H1 , 2N1
; Hk , 2Nk −
×ϑg1′ h1′
ϑgk′ h′k
γτ + δ
γτ + δ
×
k=3
−48γ 2 π 2 (γτ + δ)2
4
Y
ϑgk′ hk′
k=1
o
ατ + β
; Hk , 2Nk ,
γτ + δ
(3.8)
2) for n = 2 and n = 0
4
Y
6
ϑgk hk (τ ; 0, 2Nk ) =
(γτ + δ)6 ϑg′′1 h1 (τ ; 0, 2N1 )ϑ′′g2 h2 (τ ; 0, 2N2 )
N1 N2
k=3
4
Y
1/2 −1
3
Nk
= e sgn γ (4γ 2
2
k=1
X
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
Hk mod2Nk
(k=1,2,3,4)
ατ + β
6
ϑg′′′ h′
; H1 , 2N1 −
1
1
N1 N 2
γτ + δ
ατ + β
24γπi
; H1 , 2N1 ×
−
(γτ + δ)ϑg1′ h1′
N2
γτ + δ
4
Y
ατ + β
ατ + β
ϑgk′ hk′
×ϑ′′g′ h′
; H2 , 2N2
; Hk , 2Nk −
2 2
γτ + δ
γτ + δ
k=3
24γπi
ατ + β
−
(γτ + δ)ϑ′′g1 h1
; H1 , 2N1 +
N1
γτ + δ
ατ + β
; H1 , 2N1 ×
+96γ 2 π 2 (γτ + δ)2 ϑg1 h1
γτ + δ
×
n
ON SOME ENTIRE MODULAR FORMS
×
4
Y
ϑgk′ h′k
k=2
73
o
ατ + β
; Hk , 2Nk .
γτ + δ
(3.9)
Subtracting (3.6) from (3.5), and (3.9) from the sum of (3.7) and (3.8),
according to (3.1) and (3.2) respectively, we obtain
(γτ + δ)6 Φj (τ ; g1 , . . . , g4 , h1 , . . . , h4 , 0, . . . , N1 , . . . , N4 ) =
=e
4
3
Y
1/2 −1
sgn γ 4γ 2
Nk
2
k=1
×Φj
X
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
Hk mod2Nk
(k=1,2,3,4)
ατ + β
; g1 , . . . , g4 , h1 , . . . , h4 , H1 , . . . , H4 , N1 , . . . , N4
γτ + δ
(j = 1, 2).
(3.10)
Further, reasoning as in Lemma 5, from (3.10) we obtain (3.4).
Theorem 2. For a given N the functions Φ1 (τ ) and Φ2 (τ ) with c1 =
c2 = c3 = c4 = 0 are entire modular forms of weight 12 and character
∆
) (∆ is the determinant of an arbitrary positive quadratic form
χ(δ) = ( |δ|
in 6 variables) for the congruence group Γ0 (4N ) if the following conditions
hold:
1)
2)
3)
2|gk , Nk |N
4N
4
X
k=1
(
(k = 1, 2, 3, 4),
h2k
,4
Nk
4
X
k=1
(3.11)
gk2
,
4Nk
(3.12)
for all α and δ with αδ ≡ 1 (mod 4N )
Q4 N
k=1 k
Φj (τ ; αg1 , . . . , αg4 , h1 , . . . , h4 , 0, . . . , 0, N1 , . . . , N4 ) =
|δ|
∆
Φj (τ ; g1 , . . . , g4 , h1 , . . . , h4 , 0, . . . , 0, N1 , . . . , N4 )
=
|δ|
(j = 1, 2).
(3.13)
Proof.
I. As in the case of Theorem 1, the functions Φ1 (τ ) and Φ2 (τ ) with
c1 = c2 = c3 = c4 = 0 satisfy condition 1 and, by Lemma 6, also condition
4 of the definition.
II. From (3.12), since δ 2 ≡ 1 (mod 4), it follows that
4
4
X
h2k X gk2 2ϕ(2Nk )−2
,4
δ
.
4 N δ 2
Nk
4Nk
k=1
k=1
(3.14)
74
G. LOMADZE
By Lemma 3, for n = 3, n = 1 and n = 0, according to (3.11) and (3.14),
for each substitution from Γ0 (4N ), we have
ατ + β
ατ + β
; 0, 2Nr ϑ′gs hs
; 0, 2Ns ×
γτ + δ
γτ + δ
4
Y
ατ + β
; 0, 2Nk =
ϑgk hk
×
γτ + δ
k=3
Q
4 N
′′′
k=1 k
= (γτ + δ)6
ϑαg
(τ ; 0, 2Nr )ϑ′αgs ,hs (τ ; 0, 2Ns ) ×
r ,hr
|δ|
ϑ′′′
gr hr
×
4
Y
ϑαgk ,hk (τ ; 0, 2Nk )
(3.15)
k=3
for r = 1, s = 2 and r = 2, s = 1.
Analogously, by Lemma 3, we have:
1) for n = 4 and n = 0
ατ + β
ατ + β
; 0, 2Nr ϑgs hs
; 0, 2Ns ×
γτ + δ
γτ + δ
4
Y
ατ + β
ϑgk hk
×
; 0, 2Nk =
γτ + δ
k=3
Q4 N
(4)
k=1 k
ϑαgr ,hr (τ ; 0, 2Nr )ϑαgs ,hs (τ ; 0, 2Ns ) ×
= (γτ + δ)6
|δ|
(4)
ϑgr hr
×
4
Y
ϑαgk ,hk (τ ; 0, 2Nk )
(3.16)
k=3
for r = 1, s = 2 and r = 2, s = 1,
2) for n = 2 and n = 0
2
Y
4
ατ + β
ατ + β
Y
ϑgk hk
; 0, 2Nk
; 0, 2Nk =
γτ + δ
γτ + δ
k=3
k=1
4
2
Q4 N Y
Y
′′
6
k=1 k
ϑαgk ,hk (τ ; 0, 2Nk )
ϑαgk ,hk (τ ; 0, 2Nk ).
= (γτ + δ)
|δ|
ϑ′′gk hk
k=1
k=3
Hence, according to (3.13), for all α and δ with αδ ≡ 1 (mod 4N ), we
have
ατ + β
Φj
; g1 , . . . , g4 , h1 , . . . , h4 , 0, . . . , N1 , . . . , N4 =
γτ + δ
Q4 N
k=1 k
(γτ + δ)6 ×
=
|δ|
ON SOME ENTIRE MODULAR FORMS
75
×Φj (τ ; αg 1 , . . . , αg 4 , h1 , . . . , h4 , 0, . . . , 0, N1 , . . . , N4 ) =
∆
(γτ + δ)6 Φj (τ ; g1 , . . . , g4 , h1 , . . . , h4 , 0, . . . , N1 , . . . , N4 )
=
|δ|
for j = 1, by (3.1) and (3.15), and for j = 2, by (3.2), (3.16), and (3.17).
Thus the functions Φ1 (τ ) and Φ2 (τ ) satisfy condition 2 of the definition.
III. From (13) it follows that, for r = 1, s = 2 and r = 2, s = 1,
′
ϑ′′′
gr hr (τ ; 0, 2Nr )ϑgs hs (τ ; 0, 2Ns )
4
Y
ϑgk hk (τ ; 0, 2Nk ) =
k=3
∞
X
= π4
(−1)mr +ms +m3 +m4 (4Nr mr + gr )3 (4Ns ms + gs )e(Λτ ),
mr ,ms ,m3 ,m4 =−∞
(4)
ϑgr hr (τ ; 0, 2Nr )ϑgs hs (τ ; 0, 2Ns )
4
Y
ϑgk hk (τ ; 0, 2Nk ) =
k=3
∞
X
= π4
(−1)mr +ms +m3 +m4 (4Nr mr + gr )4 e(Λτ ),
mr ,ms ,m3 ,m4 =−∞
and also
ϑ′′g1 h1 (τ ; 0, 2N1 )ϑ′′g2 h2 (τ ; 0, 2N2 )
4
Y
ϑgk hk (τ ; 0, 2Nk ) =
k=3
=π
∞
X
4
m1 ,... ,m4 =−∞
P4
(−1) k=1 hk mk (4N1 m1 + g1 )2 (4N2 m2 + g2 )2 e(Λτ ),
where
Λ=
4
4
4
X
1X
gk2
1
gk 2 X
1
,
2Nk mk +
=
Nk m2k + mk gk +
4Nk
2
2
4
4Nk
k=1
k=1
k=1
by (3.11) and (3.12), is an integer. Thus the functions Φ1 (τ ) and Φ2 (τ )
with c1 = c2 = c3 = c4 = 0 satisfy condition 3 of the definition.
References
´ Goursat, Cours d’analyse mathematique. T. I. Gauthier-Villars,
1. E.
Paris, 1902.
2. H.D. Kloosterman, The behaviour of general theta functions under the
modular group and the characters of binary modular congruence groups. I.
Ann. Math. 47(1946), No.3, 317-375.
76
G. LOMADZE
3. G.A.Lomadze, On the representation of numbers by sums of generalized polygonal numbers. I. (Russian) Trudy Tbiliss. Mat. Inst. Razmadze
22(1956), 77-102.
4. —–, On the representation of numbers by certain quadratic forms
in six variables. I. (Russian) Trudy Tbilis. Univ. Mat. Mech. Astron.
117(1966), 7-43.
(Received 17.11.1992)
Author’s address:
Faculty of Mechanics and Mathematics
I. Javakhishvili Tbilisi State University
2 University St., Tbilisi 380043
Republic of Georgia
1(1994), No. 1, 53-76
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS 5
AND 6 FOR THE CONGRUENCE GROUP Γ0 (4N )
G. LOMADZE
Abstract. Two entire modular forms of weight 5 and two of weight
6 for the congruence subgroup Γ0 (4N ) are constructed, which will
be useful for revealing the arithmetical sense of additional terms in
formulas for the number of representations of positive integers by
quadratic forms in 10 and 12 variables.
In this paper N, a, k, n, r, s, t denote positive integers; u are odd positive integers; H, c, g, h, j, m, α, β, γ, δ, ξ, η are integers; A, B, C, D
are complex numbers, and z, τ (Im τ > 0) are complex variables. Further,
n
( hu ) is the generalized Jacobi symbol,
a binomial coefficient, ϕ(k) Eut
ler’s function, and e(z) = exp 2πiz, η(γ) = 1 if γ ≥ 0 and η(γ) = −1 if
γ < 0.
Let
o
n ατ + β
Γ=
αδ − βγ = 1 ,
γτ + δ
n ατ + β
o
Γ0 (4N ) =
∈ Γγ ≡ 0 (mod 4N ) .
γτ + δ
Definition. We shall say that a function F defined on H = {τ ∈
C | Im τ > 0} is an entire modular form of weight r and character χ(δ)
for the congruence subgroup Γ0 (4N ) if
1) F is regular on H,
2) for all substitutions from Γ0 (4N ) and all τ ∈ H
F
ατ + β
= χ(δ)(γτ + δ)r F (τ ),
γτ + δ
1991 Mathematics Subject Classification. 11F11,11F03,11F27.
53
54
G. LOMADZE
3) in the neighborhood of the point τ = i∞
F (τ ) =
∞
X
Am e(mτ ),
m=0
4) for all substitutions from Γ, in the neighborhood of each rational point
τ = − γδ (γ 6= 0, (γ, δ) = 1)
r
(γτ + δ) F (τ ) =
∞
X
n ατ + β
A′m e
.
4N γτ + δ
m=0
For η 6= 0, ξ, g, h, N with ξg + ηh + ξηN ≡ 0 (mod 2), put
ξ
X
g 2
ξ
m+
.
Sgh
; c, N =
(−1)h(m−c)/N e
η
2N η
2
mmodN |η|
m≡c (mod N )
It is known ([2] p. 323, formula (2.4)–(2.6)) that
ξ
ξ
Sg+2j,h
; c, N = Sgh
; c + j, N ,
η
η
ξ
ξ
; c, N = Sgh
; c, N ,
Sg,h+2j
η
η
ξ
ξ
; c, N .
; c + Nj , N = (−1)hj Sgh
Sgh
η
η
(1)
Let
=
X
m≡c (mod N )
ϑgh (z|τ ; c, N ) =
1
g 2
g
(−1)h(m−c)/N e
m+
τ e m+ z ,
2N
2
2
(2)
hence
X
∂n
(−1)h(m−c)/N (2m + g)n ×
ϑgh (z|τ ; c, N ) = (πi)n
n
∂z
m≡c (mod N )
1
g 2
g
τ e m+ z .
m+
×e
2N
2
2
(3)
Put
∂n
ϑ
(z|τ
;
c,
N
)
,
gh
∂z n
z=0
(0)
ϑgh (τ ; c, N ) = ϑgh (τ ; c, N ) = ϑgh (0|τ ; c, N ).
(n)
ϑgh (τ ; c, N ) =
(4)
ON SOME ENTIRE MODULAR FORMS
55
It is known ([2], p.318, formula (1.3); p. 321, formula (1.12); p.327,
formulas (3.9), (3.5), (3.3), (3.7); p.324, formula (2.16); p.327, formulas
(3.10), (3.11)) that
(5)
ϑg,h+2j (z|τ ; c, N ) = ϑgh (z|τ ; c, N );
2
iτ 1/2 N z
z 1
e
×
ϑgh − ; c, N = −
τ
τ
N
2τ
X
g
h
1
(6)
×
c+
H+
ϑhg (z|τ ; H, N );
e −
N
2
2
H modN
−i(γτ + δ) sgn γ 21 N γz 2
z ατ + β
ϑgh
; c, N =
e
×
γτ + δ γτ + δ
N |γ|
2(γτ + δ)
X
×
ϕg′ gh (c, H; N )ϑg′ h′ (z|τ ; H, N ) (γ =
6 0),
(7)
H modN
where
g ′ = αg + γh + αγN, h′ = βg + δh + βδN,
(8)
βδ
′
′
g 2
β
g
g
c+
H+
e
×
H+
ϕg′ gh (c, H; N ) = e −
2N
2
N
2
2
α
(9)
×Sg−δg′ ,h+βg′
; c − δH, N ;
γ
β
g 2
c+
ϑgh (z|τ + β; c, N ) = e
ϑg,h+βg+βN (z|τ ; c, N ),
2N
2
ϑgh (−z|τ − β; c, N ) =
(10)
g 2
β
=e −
ϑ−g,−h+βg−βN (z|τ ; −c, N ).
c+
2N
2
From (5) and (10), according to the notations (4), it follows that
ϑg,h+2j (τ ; c, N ) = ϑgh (τ ; c, N ),
(n)
(n)
ϑg,h+2j (τ ; c, N ) = ϑgh (τ ; c, N );
β
g 2 (n)
(n)
ϑg,h+βg+βN (τ ; c, N ),
c+
ϑgh (τ + β; c, N ) = e
2N
2
(n)
ϑgh (τ − β; c, N ) =
g 2 (n)
β
c+
ϑ−g,−h+βg−βN (τ ; −c, N ).
= (−1)n e −
2N
2
(11)
(12)
From (2) and (3), according to the notations (4), in particular, it follows
56
G. LOMADZE
that
ϑgh (τ ; 0, N ) =
∞
X
1
g 2
Nm +
τ ,
(−1)hm e
2N
2
m=−∞
(n)
(13)
ϑgh (τ ; 0, N ) =
= (πi)n
∞
X
1
g 2
τ .
Nm +
(−1)hm (2N m + g)n e
2N
2
m=−∞
1.
Lemma 1. For n ≥ 0
iτ (2n+1)/2
1
(n)
×
ϑgh − ; c, N = (N i)n −
τ
N
X
g
h
1
e −
×
c+
H+
×
N
2
2
H modN
n
n
t
o
X
n X Atk
(n−t)
(n)
× ϑhg (τ ; H, N ) +
· ϑhg (τ ; H, N ) ,
t
k! z=0
t=1
k=1
where
Atk
z=0
=
(
k
(2k)! Nτπi
if t = 2k
0
if t 6= 2k
(t = 1, 2, . . . , n; k = 1, 2, . . . , t).
(1.1)
Proof. From (6), by Leibniz’s formula, we obtain
−iτ 1/2 ∂ n n N z 2
z 1
∂n
ϑgh − ; c, N ) = τ n
×
e
n
∂z
τ
τ
N
∂z n
2τ
o
X
1
g
h
c+
H+
ϑhg (z|τ ; H, N ) =
e −
×
N
2
2
H modN
= (N i)n
×
X
H modN
−iτ (2n+1)/2
×
N
o
1
g
h n N z 2 ∂ n
e
ϑ
(z|τ
;
H,
N
+
e −
c+
H+
)
hg
N
2
2
2τ ∂z n
+
n
o
X
n ∂ n N z 2 ∂ n−t
e
ϑ
(z|τ
H,
N
)
;
.
hg
t ∂z n
2τ ∂z n−t
t=1
(1.2)
ON SOME ENTIRE MODULAR FORMS
57
According to formulae (a) and (b) of [1], p. 371
N z 2 At2 N z 2
Att N z 2
∂t N z2
e
e
e
=
A
+
+
·
·
·
+
e
t1
∂z t
2τ
2τ
2!
2τ
t!
2τ
(t = 1, 2, . . . , n),
where
N πiz 2 ∂ t N πiz 2k−1
∂ t N πiz 2 k
−
+
k
∂z t
τ
τ
∂z t
τ
k(k − 1) N πiz 22 ∂ t N πiz 2k−2
+
+
2!
τ
∂z t
τ
N πiz 2 k−1 ∂ t N πiz 2
+ · · · + (−1)k−1 k
(k = 1, 2, . . . , t),
τ
∂z t
τ
Atk =
hence
t
X
∂ t N z 2
Atk
e
=
∂z t
2τ z=0
k! z=0
(t = 1, 2, . . . , n)
(1.3)
k=1
and
∂ t N πiz 2 k
Atk z=0 = t
∂z
τ
z=0
(k = 1, 2, . . . , t).
(1.4)
Thus, in view of notations (4), the lemma follows from (1.2)–(1.4).
6 0, then for n ≥ 0
Lemma 2. If γ =
(n)
ϑgh
ατ + β
sgn γ (2n+1)/2
×
; c, N = (N |γ|i sgn γ)n − i(γτ + δ)
γτ + δ
N |γ|
n
X
(n)
ϕg′ gh (c, H; N ) ϑg′ h′ (τ ; H, N ) +
×
+
H modN
n
X
t=1
n
t
t
o
X
Atk
(n−t)
· ϑg′ h′ (τ ; H, N ) ,
k! z=0
k=1
where g ′ h′ and ϕg′ gh (c, H; N ) are defined by the formulas (8) and (9),
(
γπi k
(2k)! N
if t = 2k
γτ +δ
Atk
=
z=0
if t 6= 2k
0
(t = 1, 2, . . . , n; k = 1, 2, . . . , t).
1 Page
75 in the Russian version of [1] published in 1933.
58
G. LOMADZE
Proof. From (7), according to Leibnitz’s formula, we obtain
z ατ + β
sgn γ 1/2
∂n
c,
N
ϑ
;
×
= (γτ + δ)n − i(γτ + δ)
gh
∂z n
γτ + δ γτ + δ
N |γ|
o
∂ n n N γz 2 X
ϕg′ gh (c, H; N )ϑg′ h′ (z|τ ; H, N ) =
× n e
∂z
2(γτ + δ)
H modN
sgn γ (2n+1)/2
×
= (N |γ|i sgn γ)n − i(γτ + δ)
N |γ|
n N γz 2 ∂ n
X
ϕg′ gh (c, H; N ) e
×
ϑg′ h′ (z|τ ; H, N ) +
2(γτ + δ) ∂z n
H modN
n
X
+
t=1
o
N γz 2 ∂ n−t
n ∂t
′ h′ (z|τ ; H, N ) .
ϑ
e
g
t ∂z t 2(γτ + δ) ∂z n−t
As in Lemma 1, but with e
N γz 2
2(γτ +δ)
(1.5)
2
instead of e N2τz , we have
t
X
Atk
∂ t N γz 2
(t = 1, 2, . . . , n)
e
=
∂z t
2(γτ + δ) z=0
k! z=0
(1.6)
k=1
and
Atk
z=0
=
∂ t N γπiz 2k
(k = 1, 2, . . . , t).
∂z t γτ + δ
z=0
(1.7)
Thus, according to the notations (4), the lemma follows from (1.5)(1.7).
Lemma 3. If g is even, then for n ≥ 0 and all substitutions from Γ0 (4N )
we have
(n) ατ
ϑgh
+β
; 0, 2N =
γτ + δ
= ( sgn δ)n i(2n+1)η(γ)( sgn δ−1)/2 i(1−|δ|)/2
2βN sgn δ
×
|δ|
αγδ 2 h2 βδg 2 δ 2ϕ(2N )−2
(n)
×(γτ + δ)(2n+1)/2 e −
e
ϑαg,h (τ ; 0, 2N ).
16N
4
4N
Proof. 1) Let γ 6= 0. In [4] (p.18, formula (5.1)) it is shown that
β
; 0, 2N =
Sg0
δ
βδg 2 δ 2ϕ(2n)−2
2βN sgn δ
|δ|1/2 .
i(1−|δ|)/2
=e
4
4N
|δ|
(1.8)
ON SOME ENTIRE MODULAR FORMS
59
Replacing α, β, γ, δ, τ, c, N by β, −α, δ, −γ, τ ′ , 0, 2N in Lemma 2,
we obtain
(n) βτ
ϑgh
′
δτ ′
sgn δ (2n+1)/2
−α
; 0, 2N = (2N |δ|i sgn δ)n − i(δτ ′ − γ)
×
−γ
2N |δ|
n
X
(n)
×
ϕh′ gh (0, H; 2N ) ϑh′ ,αg (τ ′ ; H, 2N ) +
H mod2N
n
X
t
o
n X Atk
(n−t)
· ϑh′ ,αg (τ ′ ; H, 2N ) ,
t
k! z=0
+
t=1
(1.9)
k=1
where by (8),(9),(11) and (1)
h′ = βg + δh + 2βδN,
(1.10)
ϕh′ gh (0, H; 2N ) =
αγh
h′ β
αg
e −
H+
Sg0
; 0, 2N ,
=e −
δ
16N
4N
2
(
k
δπi
if t = 2k
(2k)! 2N
δτ ′ −γ
Atk
=
z=0
0
if t 6= 2k
′2
(k = 1, 2, . . . , t).
(1.11)
(1.12)
Writing − τ1 instead of τ ′ in (1.9) and (1.12), according to (1.11), we
obtain
−i(− δ − γ) sgn δ (2n+1)/2
+β
τ
×
; 0, 2N = (2N i)n (|δ| sgn δ)n
γτ + δ
2N |δ|
′
αγh 2 βδg 2 δ 2ϕ(2N )−2 (1−|δ|)/2 2βN sgn δ 1/2
e
i
|δ| ×
×e −
16N
4
4N
|δ|
αg
X
h′ n (n) 1
e −
ϑh′ ,αg − ; H, 2N +
×
H+
4N
2
τ
(n) ατ
ϑgh
H mod2N
n
X
+
t=1
where
Atk
z=0
=
(
n
t
t
X
o
Atk
1
(n−t)
· ϑh′ ,αg − ; H, 2N ,
k! z=0
τ
k=1
δπiτ k
(2k)! 2N
if t = 2k
γτ +δ
0
if t 6= 2k
(k = 1, 2, . . . , t).
(1.13)
60
G. LOMADZE
Writing αg, h′ , − τ1 , 0, 2N instead of g, h, τ , c, N in Lemma 1, we
obtain
(n)
ϑαg,h′ (τ ; 0, 2N ) = (2N i)n
i (2n+1)/2
2N τ
X
H mod2N
αg
h′
e −
H+
×
4N
2
n
1
(n)
× ϑh′ ,αg − ; H, 2N +
τ
n
t
X
o
1
n X Atk
(n−t)
· ϑh′ ,αg − ; H, 2N ,
+
t
k! z=0
τ
t=1
(1.14)
k=1
where
Atk
z=0
=
(
(2k)!(−2N πiτ )k if t = 2k
0
if t 6= 2k
(k = 1, 2, . . . , t).
From (1.14) it follows that
X
H mod2N
n
X
+
t=1
αg
h′ n (n) 1
e −
H+
ϑh′ ,αg − ; H, 2N +
4N
2
τ
n
t
t
X
o
Atk
1
(n−t)
=
· ϑh′ ,αg − ; H, 2N
k! z=0
τ
= (2πi)
k=1
−n
(n)
(−2N iτ )(2n+1)/2 ϑαg,h′ (τ ; 0, 2N ).
(1.15)
From (1.13) and (1.15) we obtain
(n) ατ
+β
; 0, 2N = (−2N iτ )(2n+1)/2 (|δ| sgn δ)n ×
γτ + δ
′
−i(− δ − γ) sgn δ (2n+1)/2
αγh 2
(1−|δ|)/2
τ
×
e −
i
×
2N |δ|
16N
βδg 2 δ 2ϕ(2N )−2 2βN sgn δ
(n)
×e
|δ|1/2 ϑαg,h′ (τ ; 0, 2N ).
4
4N
|δ|
ϑgh
(1.16)
In [3] (p.85, formula (6.16) with 2N instead of N ) it is shown that
−i(− δ − γ) sgn δ 1/2
τ
(−2N iτ )1/2 =
2N |δ|
γτ + δ 1/2
.
= i sgn γ( sgn δ−1)/2
|δ|
(1.17)
ON SOME ENTIRE MODULAR FORMS
61
From (1.16) and (1.17) it follows that
ατ + β
; 0, 2N =
γτ + δ
γτ + δ (2n+1)/2 (1−|δ|)/2
= (|δ| sgn δ)n i sgn γ( sgn δ−1)
i
×
|δ|
′
αγh 2 βδg 2 δ 2ϕ(2N )−2 2βN sgn δ 1/2 (n)
e
|δ| ϑαg,h′ (τ ; 0, 2N ).
×e −
16N
4
4N
|δ|
(n)
ϑgh
Thus, in view of (1.10) and (1.11), the lemma is proved for γ 6= 0.
2) Let γ = 0. Then α = δ = 1 or α = δ = −1 in (12). Putting c = 0 in
(12) and writing 2N instead of N , by (11), we obtain
βg 2 (n)
(n)
ϑgh (τ + β; 0, 2N ) = e
ϑ (τ ; 0, 2N ),
16N gh
βg 2 (n)
(n)
ϑ
(τ ; 0, 2N ),
ϑgh (τ − β; 0, 2N ) = (−1)n e −
16N −g,h
i.e., the lemma is also proved for γ = 0.
Lemma 4. If γ =
6 0, then for n ≥ 0
(n)
(γτ + δ)(2n+1)/2 ϑgh (τ ; 0, 2N ) =
= e (2n + 1) sgn γ/8 (2N |γ|)−1/2 (−i sgn γ)n ×
ατ + β
n
X
(n)
×
ϕg′ gh (0, H; 2N ) ϑg′ h′
; H, 2N +
γτ + δ
H mod2N
n
X
+
t=1
n
t
where
Atk
z=0
=
t
X
o
Atk
(n−t) ατ + β
· ϑg ′ h ′
; H, 2N ,
k! z=0
γτ + δ
k=1
(
(2k)!(−2N γπi(γτ + δ))k if t = 2k
if t 6= 2k
0
(t = 1, 2, . . . ; k = 1, 2, . . . , t).
Remark. From (1.18) it follows that
( A
t
t,t/2
X
Atk
(t/2)! z=0 if 2|t
=
k! z=0
0 if 2†t.
k=1
(1.18)
(1.19)
62
G. LOMADZE
Proof. Replacing α, β, γ, δ, τ, c, N by δ, −β, −γ, α, τ ′ , 0, 2N in
Lemma 2, we obtain
δτ ′ − β
(n)
2N
ϑgh
;
0,
=
−γτ ′ + α
sgn γ (2n+1)/2
= (−2N |γ|i sgn γ)n i(−γτ ′ + α)
×
2N |γ|
n
X
(n)
×
ϕg′ gh (0, H; 2N ) ϑg′ h′ (τ ′ ; H, 2N ) +
+
H mod2N
n
X
t=1
where
t
o
n X Atk
(n−t)
· ϑg′ h′ (τ ′ ; H, 2N ) ,
t
k! z=0
(1.20)
k=1
g ′ = δg − γh − 2γδN, h′ = −βg + αh − 2αβN,
(1.21)
αβ
′
′
βg
g 2
g
e −
×
H+
H+
ϕg′ gh (0, H; 2N ) = e −
4N
2
4N
2
δ
(1.22)
; −αH, 2N ,
×Sg−αg′ ,h−βg′
−γ
(
γπi k
(2k)! 2N
if t = 2k,
γτ ′ −α
=
(1.23)
Atk
z=0
0 if t 6= 2k.
In [3, p.87, formula (6.23)] (with 2N instead of N ) it is shown that
1/2
e( sgn γ/8)
i sgn γ
.
=
2N |γ|(γτ + δ)
(2N |γ|)1/2 (γτ + δ)1/2
(1.24)
+β
′
Taking ατ
γτ +δ instead of τ in (1.20) and (1.23) and using (1.24), we complete
the proof of the lemma.
2.
Lemma 5. For a given N let
Ψ1 (τ ) = Ψ1 (τ ; g1 , g2 , h1 , h2 , c1 , c2 , N1 , N2 ) =
1 ′′′
(τ ; c1 , 2N1 )ϑg′ 2 h2 (τ ; c2 , 2N2 ) −
ϑ
=
N1 g1 h1
1
− ϑg′′′2 h2 (τ ; c2 , 2N2 )ϑ′g1 h1 (τ ; c1 , 2N1 )
N2
and
Ψ2 (τ ) = Ψ2 (τ ; g1 , g2 , h1 , h2 , c1 , c2 , N1 , N2 ) =
(2.1)
ON SOME ENTIRE MODULAR FORMS
1 (4)
ϑ
(τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
N12 g1 h1
1 (4)
+ 2 ϑg2 h2 (τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 ) −
N2
6
ϑ′′ (τ ; c1 , 2N1 )ϑ′′g2 h2 (τ ; c2 , 2N2 ),
−
N1 N2 g1 h1
63
=
(2.2)
where
2|g1 , 2|g2 , N1 |N, N2 |N, 4|N
h1
h2
.
+
N1
N2
(2.3)
For all substitutions from Γ in the neighborhood of each rational point τ =
− γδ (γ 6= 0, (γ, δ) = 1), we then have
(γτ + δ)5 Ψj (τ ; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 ) =
=
∞
X
n ατ + β
Cn(j) e
4N γτ + δ
n=0
(j = 1, 2).
(2.4)
Proof. I. From Lemma 4 for n = 3 (with g1 , h1 , N1 , g1′ , h1′ , H1 instead
of g, h, N , g ′ , h′ , H) and n = 1 (with g2 , h2 , N2 , g2′ , h′2 , H2 instead of g,h,
N , g ′ , h′ , H), according to (1.19), it follows that
1
′
(γτ + δ)5 ϑ′′′
g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) =
N1
1 5
e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
=
N1 4
n
X
ατ + β
; H1 , 2N1 +
×
ϕg1′ g1 h1 (0, H1 ; 2N1 ) ϑ′′′
g1′ h′1
γτ + δ
H1 mod2N1
ατ + β
o
3·2
+
· ϑg′ ′ h′
A21
; H1 , 2N1 ×
1 1
2!
γτ + δ
z=0
X
ατ + β
×
; H2 , 2N2 =
ϕg2′ g2 h2 (0, H2 ; 2N2 )ϑ′g′ h′
2 2
γτ + δ
H2 mod2N2
5
−1
sgn γ 2|γ|(N1 N2 )1/2
×
4
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 )×
=e
×
X
H1 mod2N1
H2 mod2N2
×
n 1
ατ + β
ατ + β
ϑ′′′′ ′
; H1 , 2N1 ϑ′g′ h′
; H2 , 2N2 −
2 2
N1 g1 h1 γτ + δ
γτ + δ
ατ + β
; H1 , 2N1 ×
−12γπi(γτ + δ)ϑg′ ′ h′
1 1
γτ + δ
64
G. LOMADZE
ατ + β
o
; H2 , 2N2 .
2
γτ + δ
×ϑg′ ′ h′
2
(2.5)
Replacing N1 , g1 , h1 , H1 , g1′ , h′1 by N2 , g2 , h2 , H2 , g2′ , h2′ in (2.5) and
vice versa, we obtain
1
′
(γτ + δ)5 ϑ′′′
g2′ h′2 (τ ; 0, 2N2 )ϑg1′ h′1 (τ ; 0, 2N1 ) =
N2
5
= e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
4
X
′
ϕg2 g2 h2 (0, H2 ; 2N2 )ϕg1′ g1 h1 (τ ; 0, H1 ; 2N1 ) ×
×
H2 mod2N2
H1 mod2N1
n 1
ατ + β
ατ + β
×
ϑ′′′
; H2 , 2N2 ϑ′g′ h′
; H1 , 2N1 −
g2′ h′2
1
1
N2
γτ + δ
γτ + δ
+
β
ατ
; H2 , 2N2 ×
−12γπi(γτ + δ)ϑ′g′ h′
2 2
γτ + δ
o
ατ + β
′
; H1 , 2N1 .
×ϑg′ h′
1 1
γτ + δ
(2.6)
Subtracting (2.6) from (2.5), according to (2.1), we obtain
×
(γτ + δ)5 Ψ1 (τ ; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 ) =
5
= e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
H1 mod2N1
H2 mod2N2
×Ψ1
ατ + β ′ ′ ′ ′
; g1 , g2 , h1 , h2 , H1 , H2 , N1 , N2 .
γτ + δ
(2.7)
In (2.7) let γ be even. Then, by (1.21) and (2.3), g1′ and g2′ are also even.
Therefore, according to (3) and the notations (4), we have
ατ + β
ατ + β
; Hr , 2Nr ϑg′ s′ h′s
; Hs , 2Ns =
γτ + δ
γτ + δ
∞
∞
X
N n1 /Nr ατ + β X
N n2 /Ns ατ + β
=
Bn 2 e
Bn1 e
=
4N
γτ + δ n =0
4N
γτ + δ
n =0
ϑg′′′r′ hr′
2
1
=
∞
X
n ατ + β
Bn e
4N γτ + δ
n=0
for r = 1, s = 2 and r = 2, s = 1.
(2.8)
ON SOME ENTIRE MODULAR FORMS
65
Hence, for even γ, (2.4) follows from (2.1), (2.7), and (2.8) if j = 1.
Now let γ in (2.7) be odd. If h1 and h2 are both even, then by (1.21)
and (2.3), g1′ and g2′ are also even, and we obtain the same result. But if hr
is odd, then by (1.21) gr′ will also be odd, and in (3) we have
hence
m+
2
1
1
gr′ 2
1
= m + (gr′ − 1) + m + (gr′ − 1) + ,
2
2
2
4
ϑg′′′r′ hr′
×
ατ + β
1 ατ + β
; Hr , 2Nr = −π 3 ie
×
γτ + δ
16Nr γτ + δ
X
′
(−1)hr (m−Hr )/2Nr (2m + gr′ )3 ×
m≡Hr (mod 2Nr )
n 1
ατ + β o
1
1
=
(m + (gr′ − 1))2 + (m + (gr′ − 1))
×e
4Nr
2
2
γτ + δ
∞
hr ατ + β X
n1 ατ + β
=e
B′ e
,
16Nr γτ + δ n =0 n1 4Nr γτ + δ
1
since (5) implies that hr = 1. Analogously,
ϑ′gs′ h′s
∞
n2 ατ + β
ατ + β
hs ατ + β X
Bn′ 2 e
,
; Hs , 2Ns = e
γτ + δ
16Ns γτ + δ n =0
4Ns γτ + δ
2
which implies that hs = 1 if hs is odd and hs = 0 if hs is even. Thus, if
among h1 and h2 at least one is odd, then we have, for r = 1, s = 2 and
r = 2, s = 1,
ατ + β
ατ + β
; Hr , 2Nr ϑg′ s′ h′s
; Hs , 2Ns =
γτ + δ
γτ + δ
∞
N/4(hr /Nr + hs /Ns ) ατ + β X
n1 ατ + β
=e
×
Bn′ 1 e
4N
γτ + δ n =0
4Nr γτ + δ
ϑ′′′
gr′ h′r
1
×
∞
X
n2
∞
n ατ + β
n2 ατ + β X
Bn′ e
Bn′ 2 e
=
,
4Ns γτ + δ
4N γτ + δ
n=0
=0
(2.9)
hr
hs
+N
) is an integer. Hence for odd γ (2.4) follows
since, by (2.3), N4 ( N
r
s
from (2.1),(2.7), and (2.9) if j = 1.
II. From Lemma 4 for n = 4 and n = 0, as in I, it follows that
1
(4)
(γτ + δ)5 ϑg1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) =
N22
1 5
= 2 e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
N1 4
66
G. LOMADZE
n
(4) ατ + β
ϕg1′ g1 h1 (0, H1 ; 2N1 ) ϑg′ h′
; H1 , 2N1 +
1 1
γτ + δ
H1 mod2N1
ατ + β
4·3
· ϑ′′g′ h′
; H1 , 2N1 +
A21
+
1 1
2!
γτ + δ
z=0
o
ατ + β
A42
+
· ϑg1′ h′1
; H1 , 2N1 ×
2! z=0
γτ + δ
X
ατ + β
ϕg2′ g2 h2 (0, H2 ; 2N2 )ϑg2′ h2′
; H2 , 2N2 =
×
γτ + δ
X
×
H2 mod2N2
×
5
−1
= e sgn γ 2|γ|(N1 N2 )1/2
×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
H1 mod2N1
H2 mod2N2
n 1
(4) ατ + β
ϑg′ h′
; H1 , 2N1 ×
2
N1 1 1 γτ + δ
ατ + β
ατ + β
24γπi
×ϑg2′ h′2
(γτ + δ)ϑ′′g′ h′
; H2 , 2N2 −
; H1 , 2N1 ×
1
1
γτ + δ
N1
γτ + δ
ατ + β
+β
ατ
; H2 , 2N2 − 48γ 2 π 2 (γτ + δ)2 ϑg1′ h′1
; H1 , 2N1 ×
ϑg2′ h′2
γτ + δ
γτ + δ
ατ + β
o
×ϑg2′ h2′
; H2 , 2N2 .
(2.10)
γτ + δ
×
Replacing N1 , g1 , h1 , H1 , g1′ , h′1 by N2 , g2 , H2 , g2′ , h′2 in (2.10) and vice
versa, we obtain
1
(4)
(γτ + δ)5 ϑg2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 ) =
N12
5
= e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
4
X
ϕg2′ g2 h2 (0, H2 ; 2N2 )ϕg1′ g1 h1 (0, H1 ; 2N1 )×
×
H2 mod2N2
H1 mod2N1
n 1
(4) ατ + β
×
ϑg′ h′
; H2 , 2N2 ×
2
2
2
N2
γτ + δ
ατ + β
ατ + β
24γπi
(γτ + δ)ϑ′′g′ h′
×ϑg1′ h′1
; H1 , 2N1 −
; H2 , 2N2 ×
2 2
γτ + δ
N2
γτ + δ
ατ + β
ατ + β
; H1 , 2N1 − 48γ 2 π 2 (γτ + δ)2 ϑg2′ h2′
; H2 , 2N2 ×
×ϑg1′ h1′
γτ + δ
γτ + δ
ON SOME ENTIRE MODULAR FORMS
×ϑg1′ h′1
o
ατ + β
; H1 , 2N1 .
γτ + δ
67
(2.11)
Analogously, from Lemma 4 for n = 2 it follows that
6
(γτ + δ)5 ϑg′′1 h1 (τ ; 0, 2N1 )ϑ′′g2 h2 (τ ; 0, 2N2 ) =
N1 N2
5
= e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
×
H1 mod2N1
H2 mod2N2
ατ + β
ατ + β
6
; H1 , 2N1 ϑ′′g′ h′
; H2 , 2N2 −
ϑ′′g′ h′
2 2
N1 N2 1 1 γτ + δ
γτ + δ
ατ + β
ατ + β
24γπi
−
(γτ + δ)ϑg1′ h′1
; H1 , 2N1 ϑg2′ h′2
; H2 , 2N2 −
N2
γτ + δ
γτ + δ
+
β
+β
ατ
ατ
24γπi
(γτ + δ)ϑ′′g′ h′
; H1 , 2N1 ϑ′′g′ h′
; H2 , 2N2 −
−
1 1
2 2
N1
γτ + δ
γτ + δ
ατ + β
−96γ 2 π 2 (γτ + δ)2 ϑg1′ h′1
; H1 , 2N1 ×
γτ + δ
o
ατ + β
; H2 , 2N2 .
(2.12)
×ϑg2′ h′2
γτ + δ
×
n
Subtracting (2.12) from the sum of (2.10) and (2.11), according to (2.2),
we obtain
×
(γτ + δ)5 Ψ2 (τ ; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 ) =
5
−1
= e sgn γ 2|γ|(N1 N2 )1/2
×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
H1 mod2N1
H2 mod2N2
×Ψ2
ατ + β ′ ′ ′ ′
; g , g , h , h , H 1 , H2 , N 1 , N 2 .
γτ + δ 1 2 1 2
(2.13)
Further, reasoning as in I, from (2.2) and (2.13) we obtain (2.4) if
j = 2.
Theorem 1. For a given N the functions Ψ1 (τ ) and Ψ2 (τ ) with c1 =
c2 = 0 are entire modular forms of weight 10 and character χ(δ) = sgn δ( −∆
|δ| )
(∆ is the determinant of an arbitrary positive quadratic form in 5 variables)
68
G. LOMADZE
for the congruence group Γ0 (4N ) if the following conditions hold:
1)
2|g1 , 2|g2 , N1 |N, N2 |N,
h2
g2
h2 g 2
2) 4N 1 + 2 , 4 1 + 2 ,
N1
N2
4N1
4N2
3) for all α and δ with α ≡ δ ≡ 1 (mod 4N )
N 1 N2
Ψj (τ ; αg1 , αg2 , h1 , h2 , 0, 0, N1 , N2 ) =
|δ|
∆
=
Ψj (τ ; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 ) (j = 1, 2).
|δ|
(2.14)
(2.15)
(2.16)
Proof.
I. It is wellknown that the thetaseries (2)-(3) are regular on H, hence the
functions Ψ1 (τ ) and Ψ2 (τ ) satisfy condition 1 of the definition.
II. From (2.15), since δ 2 ≡ 1 (mod 4), it follows that
h2
h2 g 2
g2
4N δ 2 1 + 2 , 4 1 δ 2ϕ(2N1 )−2 + 2 δ 2ϕ(2N2 −2) .
N1
N2
4N1
4N2
(2.17)
By Lemma 3 for n = 3 and n = 1 (with gr , hr , Nr and gs , hs , Ns instead of
g, h, N ), according to (2.14) and (2.17), for each substitution from Γ0 (4N ),
we have
ατ + β
ατ + β
ϑ′′′
; 0, 2Nr ϑ′gs hs
; 0, 2Ns = iη(γ)(sgnδ−1) i1−|δ| ×
gr hr
γτ + δ
γτ + δ
N
N
r
s
′
ϑ′′′
×(γτ + δ)5
αgr ,hr (τ ; 0, 2Nr )ϑαgs ,hs (τ ; 0, 2Ns ) =
|δ|
−Nr Ns
(γτ + δ)5 ×
= sgn δ
|δ|
′
×ϑ′′′
(2.18)
αgr ,hr (τ ; 0, 2Nr )ϑαgs ,hs (τ ; 0, 2Ns )
for r = 1, s = 2 and r = 2, s = 1.
Analogously, by Lemma 3, we have:
1) for n = 4 and n = 0
(4)
ϑgr hr
ατ + β
ατ + β
; 0, 2Nr ϑgs hs
; 0, 2Ns = iη(γ)(sgnδ−1) i1−|δ| ×
γτ + δ
γτ + δ
Nr Ns (4)
ϑαgr ,hr (τ ; 0, 2Nr )ϑαgs ,hs (τ ; 0, 2Ns ) =
×(γτ + δ)5
|δ|
−Nr Ns
= sgn δ
(γτ + δ)5 ×
|δ|
(4)
′
(τ ; 0, 2Ns )
×ϑαgr ,hr (τ ; 0, 2Nr )ϑαg
s ,hs
if r = 1, s = 2 and r = 2, s = 1;
(2.19)
ON SOME ENTIRE MODULAR FORMS
69
2) for n = 2
ϑ′′g1 h1
ατ + β
ατ + β
; 0, 2N1 ϑg′′2 h2
; 0, 2N2 =
γτ + δ
γτ + δ
−N1 N2
(γτ + δ)5 ×
= sgn δ
|δ|
′′
×ϑ′′αg1 ,h1 (τ ; 0, 2N1 )ϑαg
(τ ; 0, 2N2 ).
2 ,h2
(2.20)
Hence, according to (2.16), for all α and δ with αδ ≡ 1 (mod 4N ), we
have
ατ + β
Ψj
; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 =
γτ + δ
−N1 N2
(γτ + δ)5 Ψj (τ ; αg1 , αg2 , h1 , h2 , 0, 0, N1 , N2 ) =
= sgn δ
|δ|
−∆
(γτ + δ)5 Ψj (τ ; g1 , g2 , h1 , h2 , 0, 0, N1 , N2 )
= sgn δ
|δ|
for j = 1, by (2.1) and (2.16), and for j = 2, by (2.2),(2.19) and (2.20).
Thus the functions Ψ1 (τ ) and Ψ2 (τ ) with c1 = c2 = 0 satisfy condition 2 of
the definition.
III. From (13) it follows that, for r = 1, s = 2 and r = 2, s = 1,
′
4
ϑ′′′
gr hr (τ ; 0, 2Nr )ϑgs hs (τ ; 0, 2Ns ) = π
∞
X
(−1)hr mr +hs ms ×
mr ,ms =−∞
3
×(4Nr mr + gr ) (4Ns ms + gs )e(Λτ ),
(4)
ϑgr hr (τ ; 0, 2Nr )ϑgs hs (τ ; 0, 2Ns ) = π 4
∞
X
(−1)hr mr +hs ms ×
mr ,ms =−∞
×(4Nr mr + gr )4 e(Λτ )
(2.21)
and also
ϑ′′g1 h1 (τ ; 0, 2N1 )ϑ′′g2 h2 (τ ; 0, 2N2 ) = π 4
∞
X
(−1)h1 m1 +h2 m2 ×
m1 ,m2 =−∞
×(4N1 m1 + g1 ) (4N2 m2 + g2 )2 e(Λτ ),
2
where
Λ=
2
2
2
X
X
1X 2
1
(2Nk mk + gk /2)2 =
(Nk m2k + mk gk /2) +
gk /4Nk ,
4Nk
4
k=1
k=1
k=1
by (2.14) and (2.15), is an integer. Thus, the functions Ψ1 (τ ) and Ψ2 (τ )
with c1 = c2 = 0 satisfy condition 3 of the definition.
70
G. LOMADZE
IV. By Lemma 5 the functions Ψ1 (τ ) and Ψ2 (τ ) with c1 = c2 = 0 also
satisfy condition 4 of the definition.
3.
Lemma 6. For a given N let
Φ1 (τ ) = Φ1 (τ ; g1 , . . . , g4 , h1 , . . . , h4 , c1 , . . . , c4 , N1 , . . . , N4 ) =
4
=
Y
1 ′′′
ϑgk hk (τ ; ck , 2Nk ) −
ϑg1 h1 (τ ; c1 , 2N1 )ϑ′g2 h2 (τ ; c2 , 2N2 )
N1
k=3
4
Y
1
′
− ϑ′′′
ϑgk hk (τ ; ck , 2Nk )
g2 h2 (τ ; c2 , 2N2 ))ϑg1 h1 (τ ; c1 , 2N1 )
N2
(3.1)
k=3
and
Φ2 (τ ) = Φ2 (τ ; g1 , . . . , g4 , h1 , . . . , h4 , c1 , . . . , c4 , N1 , . . . , N4 ) =
4
=
Y
1 (4)
ϑgk hk (τ ; ck , 2Nk ) +
ϑg1 h1 (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 )
2
N1
k=3
4
+
Y
1 (4)
ϑgk hk (τ ; ck , 2Nk ) −
ϑg2 h2 (τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 )
2
N2
k=3
4
−
Y
6
ϑgk hk (τ ; ck , 2Nk ), (3.2)
ϑ′′g1 h1 (τ ; c1 , 2N1 )ϑ′′g2 h2 (τ ; c2 , 2N2 )
N1 N2
k=3
where
4
X
hk
2gk , Nk N (k = 1, 2, 3, 4), 4N
.
Nk
(3.3)
k=1
For all substitutions from Γ in the neighborhood of each rational point τ =
− γδ (γ 6= 0, (γ, δ) = 1), we then have
(γτ + δ)6 Φj (τ ; g1 , . . . , g4 , h1 , . . . , h4 , 0, . . . , 0, N1 , . . . , N4 ) =
∞
n ατ + β
X
Dn(j) e
=
(j = 1, 2).
4N γτ + δ
n=0
(3.4)
Proof. From Lemma 4 for n = 3, n = 1, and n = 0, as in the proof of
Lemma 5, it follows that
4
Y
1
′
ϑgk hk (τ ; 0, 2Nk ) =
(γτ + δ)6 ϑ′′′
g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 )
N1
k=3
ON SOME ENTIRE MODULAR FORMS
4
2 Y
3
1/2 −1
= e sgn γ 4γ
Nk
2
k=1
X
4
Y
71
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
Hk mod2Nk k=1
(k=1,2,3,4)
n 1
ατ + β
ϑ′′′
; H1 , 2N1 −
g1′ h1′
N1
γτ + δ
ατ + β
; H1 , 2N1 ×
−12γπi(γτ + δ)ϑ′g′ h′
1 1
γτ + δ
4
Y
ατ + β
o
ατ + β
ϑgk′ h′k
×ϑg′ ′ h′
; H2 , 2N2
; Hk , 2Nk
2 2
γτ + δ
γτ + δ
×
(3.5)
k=3
and
4
Y
1
′
(γτ + δ)6 ϑ′′′
ϑgk hk (τ ; 0, 2Nk ) =
g2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 )
N2
k=3
=e
3
sgn γ
2
4γ 2
4
Y
Nk
k=1
1/2
−1
X
4
Y
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
Hk mod2Nk k=1
(k=1,2,3,4)
n 1
ατ + β
ϑ′′′′ ′
; H2 , 2N2 −
N2 g2 h2 γτ + δ
ατ + β
; H2 , 2N2 ×
−12γπi(γτ + δ)ϑ′g′ h′
2 2
γτ + δ
4
Y
o
ατ + β
ατ + β
×ϑ′g′ h′
ϑgk′ h′k
; H1 , 2N1
; Hk , 2Nk .
1 1
γτ + δ
γτ + δ
×
(3.6)
k=3
As in the proof of Lemma 5, from Lemma 4 it follows that
1) for n = 4 and n = 0
4
Y
1
(4)
(γτ + δ)6 ϑg1 h1 (τ ; 0, 2N1 )
ϑgk hk (τ ; 0, 2Nk ) =
2
N1
k=2
3
= e sgn γ
2
4γ 2
4
Y
k=1
Nk
1/2
−1
X
4
Y
Hk mod2Nk k=1
(k=1,2,3,4)
ϕgk′ gk hk 0, Hk ; 2Nk ×
n 1
(4) ατ + β
ϑg′ h′
; H1 , 2N1 −
2
1
1
N1
γτ + δ
ατ + β
24γπi
(γτ + δ)ϑ′′g′ h′
; H1 , 2N1 −
−
1 1
N1
γτ + δ
ατ + β
; H1 , 2N1 ×
−48γ 2 π 2 (γτ + δ)2 ϑg1′ h′1
γτ + δ
×
72
G. LOMADZE
×
4
Y
ϑgk′ h′k
k=2
and
o
ατ + β
; Hk , 2Nk
γτ + δ
(3.7)
4
Y
1
(4)
ϑgk hk (τ ; 0, 2Nk ) =
(γτ + δ)6 ϑg2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 )
2
N2
k=3
4
1/2 −1
3
Y
Nk
= e sgn γ 4γ 2
2
k=1
X
4
Y
Hk mod2Nk k=1
(k=1,2,3,4)
ϕgk′ gk hk 0, Hk ; 2Nk ×
n 1
(4) ατ + β
; H2 , 2N2 −
ϑg′ h′
2
N2 2 2 γτ + δ
ατ + β
24γπi
(γτ + δ)ϑ′′g′ h′
; H2 , 2N2 ×
−
2 2
N2
γτ + δ
4
Y
ατ + β
ατ + β
; H1 , 2N1
; Hk , 2Nk −
×ϑg1′ h1′
ϑgk′ h′k
γτ + δ
γτ + δ
×
k=3
−48γ 2 π 2 (γτ + δ)2
4
Y
ϑgk′ hk′
k=1
o
ατ + β
; Hk , 2Nk ,
γτ + δ
(3.8)
2) for n = 2 and n = 0
4
Y
6
ϑgk hk (τ ; 0, 2Nk ) =
(γτ + δ)6 ϑg′′1 h1 (τ ; 0, 2N1 )ϑ′′g2 h2 (τ ; 0, 2N2 )
N1 N2
k=3
4
Y
1/2 −1
3
Nk
= e sgn γ (4γ 2
2
k=1
X
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
Hk mod2Nk
(k=1,2,3,4)
ατ + β
6
ϑg′′′ h′
; H1 , 2N1 −
1
1
N1 N 2
γτ + δ
ατ + β
24γπi
; H1 , 2N1 ×
−
(γτ + δ)ϑg1′ h1′
N2
γτ + δ
4
Y
ατ + β
ατ + β
ϑgk′ hk′
×ϑ′′g′ h′
; H2 , 2N2
; Hk , 2Nk −
2 2
γτ + δ
γτ + δ
k=3
24γπi
ατ + β
−
(γτ + δ)ϑ′′g1 h1
; H1 , 2N1 +
N1
γτ + δ
ατ + β
; H1 , 2N1 ×
+96γ 2 π 2 (γτ + δ)2 ϑg1 h1
γτ + δ
×
n
ON SOME ENTIRE MODULAR FORMS
×
4
Y
ϑgk′ h′k
k=2
73
o
ατ + β
; Hk , 2Nk .
γτ + δ
(3.9)
Subtracting (3.6) from (3.5), and (3.9) from the sum of (3.7) and (3.8),
according to (3.1) and (3.2) respectively, we obtain
(γτ + δ)6 Φj (τ ; g1 , . . . , g4 , h1 , . . . , h4 , 0, . . . , N1 , . . . , N4 ) =
=e
4
3
Y
1/2 −1
sgn γ 4γ 2
Nk
2
k=1
×Φj
X
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
Hk mod2Nk
(k=1,2,3,4)
ατ + β
; g1 , . . . , g4 , h1 , . . . , h4 , H1 , . . . , H4 , N1 , . . . , N4
γτ + δ
(j = 1, 2).
(3.10)
Further, reasoning as in Lemma 5, from (3.10) we obtain (3.4).
Theorem 2. For a given N the functions Φ1 (τ ) and Φ2 (τ ) with c1 =
c2 = c3 = c4 = 0 are entire modular forms of weight 12 and character
∆
) (∆ is the determinant of an arbitrary positive quadratic form
χ(δ) = ( |δ|
in 6 variables) for the congruence group Γ0 (4N ) if the following conditions
hold:
1)
2)
3)
2|gk , Nk |N
4N
4
X
k=1
(
(k = 1, 2, 3, 4),
h2k
,4
Nk
4
X
k=1
(3.11)
gk2
,
4Nk
(3.12)
for all α and δ with αδ ≡ 1 (mod 4N )
Q4 N
k=1 k
Φj (τ ; αg1 , . . . , αg4 , h1 , . . . , h4 , 0, . . . , 0, N1 , . . . , N4 ) =
|δ|
∆
Φj (τ ; g1 , . . . , g4 , h1 , . . . , h4 , 0, . . . , 0, N1 , . . . , N4 )
=
|δ|
(j = 1, 2).
(3.13)
Proof.
I. As in the case of Theorem 1, the functions Φ1 (τ ) and Φ2 (τ ) with
c1 = c2 = c3 = c4 = 0 satisfy condition 1 and, by Lemma 6, also condition
4 of the definition.
II. From (3.12), since δ 2 ≡ 1 (mod 4), it follows that
4
4
X
h2k X gk2 2ϕ(2Nk )−2
,4
δ
.
4 N δ 2
Nk
4Nk
k=1
k=1
(3.14)
74
G. LOMADZE
By Lemma 3, for n = 3, n = 1 and n = 0, according to (3.11) and (3.14),
for each substitution from Γ0 (4N ), we have
ατ + β
ατ + β
; 0, 2Nr ϑ′gs hs
; 0, 2Ns ×
γτ + δ
γτ + δ
4
Y
ατ + β
; 0, 2Nk =
ϑgk hk
×
γτ + δ
k=3
Q
4 N
′′′
k=1 k
= (γτ + δ)6
ϑαg
(τ ; 0, 2Nr )ϑ′αgs ,hs (τ ; 0, 2Ns ) ×
r ,hr
|δ|
ϑ′′′
gr hr
×
4
Y
ϑαgk ,hk (τ ; 0, 2Nk )
(3.15)
k=3
for r = 1, s = 2 and r = 2, s = 1.
Analogously, by Lemma 3, we have:
1) for n = 4 and n = 0
ατ + β
ατ + β
; 0, 2Nr ϑgs hs
; 0, 2Ns ×
γτ + δ
γτ + δ
4
Y
ατ + β
ϑgk hk
×
; 0, 2Nk =
γτ + δ
k=3
Q4 N
(4)
k=1 k
ϑαgr ,hr (τ ; 0, 2Nr )ϑαgs ,hs (τ ; 0, 2Ns ) ×
= (γτ + δ)6
|δ|
(4)
ϑgr hr
×
4
Y
ϑαgk ,hk (τ ; 0, 2Nk )
(3.16)
k=3
for r = 1, s = 2 and r = 2, s = 1,
2) for n = 2 and n = 0
2
Y
4
ατ + β
ατ + β
Y
ϑgk hk
; 0, 2Nk
; 0, 2Nk =
γτ + δ
γτ + δ
k=3
k=1
4
2
Q4 N Y
Y
′′
6
k=1 k
ϑαgk ,hk (τ ; 0, 2Nk )
ϑαgk ,hk (τ ; 0, 2Nk ).
= (γτ + δ)
|δ|
ϑ′′gk hk
k=1
k=3
Hence, according to (3.13), for all α and δ with αδ ≡ 1 (mod 4N ), we
have
ατ + β
Φj
; g1 , . . . , g4 , h1 , . . . , h4 , 0, . . . , N1 , . . . , N4 =
γτ + δ
Q4 N
k=1 k
(γτ + δ)6 ×
=
|δ|
ON SOME ENTIRE MODULAR FORMS
75
×Φj (τ ; αg 1 , . . . , αg 4 , h1 , . . . , h4 , 0, . . . , 0, N1 , . . . , N4 ) =
∆
(γτ + δ)6 Φj (τ ; g1 , . . . , g4 , h1 , . . . , h4 , 0, . . . , N1 , . . . , N4 )
=
|δ|
for j = 1, by (3.1) and (3.15), and for j = 2, by (3.2), (3.16), and (3.17).
Thus the functions Φ1 (τ ) and Φ2 (τ ) satisfy condition 2 of the definition.
III. From (13) it follows that, for r = 1, s = 2 and r = 2, s = 1,
′
ϑ′′′
gr hr (τ ; 0, 2Nr )ϑgs hs (τ ; 0, 2Ns )
4
Y
ϑgk hk (τ ; 0, 2Nk ) =
k=3
∞
X
= π4
(−1)mr +ms +m3 +m4 (4Nr mr + gr )3 (4Ns ms + gs )e(Λτ ),
mr ,ms ,m3 ,m4 =−∞
(4)
ϑgr hr (τ ; 0, 2Nr )ϑgs hs (τ ; 0, 2Ns )
4
Y
ϑgk hk (τ ; 0, 2Nk ) =
k=3
∞
X
= π4
(−1)mr +ms +m3 +m4 (4Nr mr + gr )4 e(Λτ ),
mr ,ms ,m3 ,m4 =−∞
and also
ϑ′′g1 h1 (τ ; 0, 2N1 )ϑ′′g2 h2 (τ ; 0, 2N2 )
4
Y
ϑgk hk (τ ; 0, 2Nk ) =
k=3
=π
∞
X
4
m1 ,... ,m4 =−∞
P4
(−1) k=1 hk mk (4N1 m1 + g1 )2 (4N2 m2 + g2 )2 e(Λτ ),
where
Λ=
4
4
4
X
1X
gk2
1
gk 2 X
1
,
2Nk mk +
=
Nk m2k + mk gk +
4Nk
2
2
4
4Nk
k=1
k=1
k=1
by (3.11) and (3.12), is an integer. Thus the functions Φ1 (τ ) and Φ2 (τ )
with c1 = c2 = c3 = c4 = 0 satisfy condition 3 of the definition.
References
´ Goursat, Cours d’analyse mathematique. T. I. Gauthier-Villars,
1. E.
Paris, 1902.
2. H.D. Kloosterman, The behaviour of general theta functions under the
modular group and the characters of binary modular congruence groups. I.
Ann. Math. 47(1946), No.3, 317-375.
76
G. LOMADZE
3. G.A.Lomadze, On the representation of numbers by sums of generalized polygonal numbers. I. (Russian) Trudy Tbiliss. Mat. Inst. Razmadze
22(1956), 77-102.
4. —–, On the representation of numbers by certain quadratic forms
in six variables. I. (Russian) Trudy Tbilis. Univ. Mat. Mech. Astron.
117(1966), 7-43.
(Received 17.11.1992)
Author’s address:
Faculty of Mechanics and Mathematics
I. Javakhishvili Tbilisi State University
2 University St., Tbilisi 380043
Republic of Georgia