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 + ⠌
Œ
Γ=
Œαδ − βγ = 1 ,
γτ + δ
Œ
o
n ατ + β
Œ
∈ Ìγ ≡ 0 (mod 4N ) .
Γ0 (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 + ⠁
= 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
m=0
A0m e
 n ατ + β ‘
.
4N γτ + δ
For η 6= 0, ξ, g, h, N with ξg + ηh + ξηN ≡ 0 (mod 2), put
’
“
 ξ €
X
g 2 ‘
ξ
Sgh
; c, N =
m+
.
(−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 .
×e
m+
2N
2
2
(3)
Put
Œ
∂n
Œ
ϑ
(z|τ
;
c,
N
)
Π,
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 
€zŒ 1
ϑgh Œ − ; c, N = −
e
×
τ
τ
N
2τ
‘

X
g €
h
1€
×
(6)
e −
c+
H+
ϑhg (z|τ ; H, N );
N
2
2
H modN
‘  −i(γτ + δ) sgn γ ‘ 21  N γz 2 ‘
 z Œ ατ + β
Œ
ϑgh
; c, N =
e
×
Œ
γτ + δ γτ + δ
N |γ|
2(γτ + δ)
X
×
ϕg0 gh (c, H; N )ϑg0 h0 (z|τ ; H, N ) (γ =
6 0),
(7)
H modN
where
g 0 = αg + γh + αγN, h0 = βg + δh + βδN,
(8)
 βδ €
‘
0 ‘ 
0
g 2
β€
g €
g 
H+
e
c+
H+
×
ϕg0 gh (c, H; N ) = e −
2N
2
N
2
2
 α
‘
×Sg−δg0 ,h+βg0
; c − δH, N ;
(9)
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 −
c+
ϑ−g,−h+βg−βN (z|τ ; −c, N ).
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 ) =

β €
g 2 ‘ (n)
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)hm e
m=−∞
(n)
 1 €
g 2 ‘
Nm +
τ ,
2N
2
(13)
ϑgh (τ ; 0, N ) =
∞
X
= (πi)n
(−1)hm (2N m + g)n e
m=−∞
 1 €
g 2 ‘
τ .
Nm +
2N
2
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 
Œ
‘ t
n
o
X
n X Atk Œ
(n−t)
(n)
× ϑhg (τ ; H, N ) +
· ϑhg (τ ; H, N ) ,
Œ
t
k! z=0
t=1
k=1
where
Œ
Œ
Atk Œ
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 Œ 1
∂n
ϑgh Œ − ; 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
=
A
e
+
e
+
·
·
·
+
e
t1
∂z t
2τ
2τ
2!
2τ
t!
2τ
(t = 1, 2, . . . , n),
where
N πiz 2‘ ∂ t N πiz 2‘k−1
∂ t  N πiz 2 ‘k
−
+
k
∂z t
τ
τ
∂z t
τ
k(k − 1) N πiz 2‘2 ∂ t N πiz 2‘k−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 ŒŒ
Atk ŒŒ
e
=
Œ
Œ
∂z t
2τ z=0
k! z=0
(t = 1, 2, . . . , n)
(1.3)
k=1
and
Œ
∂ t € N πiz 2 k ŒŒ
Atk Œz=0 = t
Œ
∂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 㠁(2n+1)/2
×
; c, N = (N |γ|i sgn γ)n − i(γτ + δ)
γτ + δ
N |γ|
n
X
(n)
ϕg0 gh (c, H; N ) ϑg0 h0 (τ ; H, N ) +
×
+
H modN
n 
X
t=1
t
Œ
o
n ‘ X Atk Œ
(n−t)
· ϑg0 h0 (τ ; H, N ) ,
Œ
t
k! z=0
k=1
where g 0 h0 and ϕg0 gh (c, H; N ) are defined by the formulas (8) and (9),
(
€ γπi k
Œ
(2k)! N
if t = 2k
Œ
γτ +δ
Atk Œ
=
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 Œ ατ + β
‘
€
sgn 㠁1/2
∂n
Œ
c,
N
ϑ
;
= (γτ + δ)n − i(γτ + δ)
×
Œ
gh
∂z n
γτ + δ γτ + δ
N |γ|
o
∂ n n  N γz 2 ‘ X
ϕg0 gh (c, H; N )ϑg0 h0 (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
ϕg0 gh (c, H; N ) e
×
ϑg0 h0 (z|τ ; H, N ) +
2(γτ + δ) ∂z n
H modN
n 
X
+
t=1
o
 N γz 2 ‘ ∂ n−t
n ‘ ∂t
0 h0 (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 ŒŒ
∂ t  N γz 2 ‘ŒŒ
e
(t = 1, 2, . . . , n)
=
Œ
Œ
∂z t
2(γτ + δ) z=0
k! z=0
(1.6)
k=1
and
Œ
Œ
Atk Œ
z=0
=
∂ t € N γπiz 2k ŒŒ
(k = 1, 2, . . . , t).
Œ
∂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
‘
Sg0
; 0, 2N =
δ
 2βN sgn δ ‘
 βδg 2 δ 2ϕ(2n)−2 ‘
i(1−|δ|)/2
|δ|1/2 .
=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 , 0, 2N in Lemma 2,
we obtain
(n) € βτ
ϑgh
0
δτ 0

€
sgn δ (2n+1)/2
−α
; 0, 2N = (2N |δ|i sgn δ)n − i(δτ 0 − γ)
×
−γ
2N |δ|
n
X
(n)
×
ϕh0 gh (0, H; 2N ) ϑh0 ,αg (τ 0 ; H, 2N ) +
H mod2N
n 
X
t
Œ
o
n ‘ X Atk Œ
(n−t)
· ϑh0 ,αg (τ 0 ; H, 2N ) ,
Œ
t
k! z=0
+
t=1
(1.9)
k=1
where by (8),(9),(11) and (1)
h0 = βg + δh + 2βδN,
ϕh0 gh (0, H; 2N ) =
2

‘
0
€ αγh 
αg €
h0  ‘  β
e −
H+
Sg0
; 0, 2N ,
=e −
δ
16N
4N
2
(

€
k
Œ
δπi
if t = 2k
(2k)! 2N
Œ
δτ 0 −γ
Atk Œ
=
z=0
0
if t 6= 2k
(k = 1, 2, . . . , t).
(1.10)
(1.11)
(1.12)
Writing − τ1 instead of τ 0 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|
0
€ 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

h0 ‘n (n) € 1
e −
ϑh0 ,αg − ; H, 2N +
×
H+
4N
2
τ
(n) € ατ
ϑgh
H mod2N
n 
X
+
t=1
where
Œ
Œ
Atk Œ
z=0
=
(
n
t
t
‘X
o
Atk ŒŒ
1
(n−t) €
· ϑh0 ,αg − ; H, 2N ,
Œ
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, h0 , − τ1 , 0, 2N instead of g, h, τ , c, N in Lemma 1, we
obtain
(n)
ϑαg,h0 (τ ; 0, 2N ) = (2N i)n
€ i (2n+1)/2
2N τ
X
H mod2N
 αg €
h0 ‘
e −
H+
×
4N
2
n

1
(n) €
× ϑh0 ,αg − ; H, 2N +
τ
n 
Œ
‘ t
X
o
1
n X Atk Œ
(n−t) €
· ϑh0 ,αg − ; H, 2N ,
+
Œ
t
k! z=0
τ
t=1
(1.14)
k=1
where
Œ
Œ
Atk Œ
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 €

h0 ‘n (n) € 1
e −
H+
ϑh0 ,αg − ; H, 2N +
4N
2
τ
n
t
t
‘X
o
Atk ŒŒ
1
(n−t) €
=
· ϑh0 ,αg − ; H, 2N
Œ
k! z=0
τ
= (2πi)
k=1
−n
(n)
(−2N iτ )(2n+1)/2 ϑαg,h0 (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
0
 −i(− δ − γ) sgn δ ‘(2n+1)/2
€ αγh 2 
(1−|δ|)/2
τ
i
e −
×
×
2N |δ|
16N
 βδg 2 δ 2ϕ(2N )−2 ‘ 2βN sgn δ ‘
(n)
×e
|δ|1/2 ϑαg,h0 (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|
0
€ αγh 2   βδg 2 δ 2ϕ(2N )−2 ‘ 2βN sgn δ ‘ 1/2 (n)
e
×e −
|δ| ϑαg,h0 (τ ; 0, 2N ).
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)
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
(n)
ϑgh (τ + β; 0, 2N ) = e
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)
×
ϕg0 gh (0, H; 2N ) ϑg0 h0
; H, 2N +
γτ + δ
H mod2N
n 
X
+
t=1
n
t
where
Œ
Œ
Atk Œ
z=0
=
t
‘X
o
Atk ŒŒ
(n−t) € ατ + β
; H, 2N ,
· ϑg0 h0
Œ
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
Œ
t
t,t/2 Œ
X
Atk ŒŒ
(t/2)! z=0 if 2|t
=
Œ
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 , 0, 2N in
Lemma 2, we obtain
 δτ 0 − β
‘
(n)
ϑgh
2N
;
0,
=
−γτ 0 + α
€
sgn 㠁(2n+1)/2
= (−2N |γ|i sgn γ)n i(−γτ 0 + α)
×
2N |γ|
n
X
(n)
×
ϕg0 gh (0, H; 2N ) ϑg0 h0 (τ 0 ; H, 2N ) +
+
H mod2N
n 
X
t=1
where
t
Œ
o
n ‘ X Atk Œ
(n−t)
· ϑg0 h0 (τ 0 ; H, 2N ) ,
Œ
t
k! z=0
(1.20)
k=1
g 0 = δg − γh − 2γδN, h0 = −βg + αh − 2αβN,
(1.21)
‘
 αβ €
0 ‘ 
0
βg €
g 2
g 
e −
×
ϕg0 gh (0, H; 2N ) = e −
H+
H+
4N
2
4N
2

‘
δ
(1.22)
; −αH, 2N ,
×Sg−αg0 ,h−βg0
−γ
(
€ γπi k
Œ
if t = 2k,
(2k)! 2N
Œ
γτ 0 −α
=
(1.23)
Atk Œ
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
i sgn γ
e( sgn γ/8)
.
=
2N |γ|(γτ + δ)
(2N |γ|)1/2 (γτ + δ)1/2
(1.24)
+β
0
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 ) =
1 000
ϑ
(τ ; c1 , 2N1 )ϑg0 2 h2 (τ ; c2 , 2N2 ) −
=
N1 g1 h1
1
− ϑg0002 h2 (τ ; c2 , 2N2 )ϑ0g1 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
ϑ00 (τ ; c1 , 2N1 )ϑ00g2 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 + ⠁
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 , g10 , h10 , H1 instead
of g, h, N , g 0 , h0 , H) and n = 1 (with g2 , h2 , N2 , g20 , h02 , H2 instead of g,h,
N , g 0 , h0 , H), according to (1.19), it follows that
1
0
(γτ + δ)5 ϑ000
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

×
ϕg10 g1 h1 (0, H1 ; 2N1 ) ϑ000
; H1 , 2N1 +
g10 h01
γτ + δ
H1 mod2N1
Œ
€ ατ + β
o
3·2
Œ
+
· ϑg0 0 h0
A21 Œ
; H1 , 2N1 ×
1 1
2!
γτ + δ
z=0
X

€ ατ + β
×
; H2 , 2N2 =
ϕg20 g2 h2 (0, H2 ; 2N2 )ϑ0g0 h0
2 2
γτ + δ
H2 mod2N2
€
€5
−1
sgn γ 2|γ|(N1 N2 )1/2
×
4
ϕg10 g1 h1 (0, H1 ; 2N1 )ϕg20 g2 h2 (0, H2 ; 2N2 )×
=e
×
×
X
H1 mod2N1
H2 mod2N2
n 1
€ ατ + β
€ ατ + β


ϑ0000 0
; H1 , 2N1 ϑ0g0 h0
; H2 , 2N2 −
2 2
N1 g1 h1 γτ + δ
γτ + δ

€ ατ + β
; H1 , 2N1 ×
−12γπi(γτ + δ)ϑg0 0 h0
1 1
γτ + δ
64
G. LOMADZE
€ ατ + β
o
; H2 , 2N2 .
2
γτ + δ
×ϑg0 0 h0
2
(2.5)
Replacing N1 , g1 , h1 , H1 , g10 , h01 by N2 , g2 , h2 , H2 , g20 , h20 in (2.5) and
vice versa, we obtain
×
1
0
(γτ + δ)5 ϑ000
g20 h02 (τ ; 0, 2N2 )ϑg10 h01 (τ ; 0, 2N1 ) =
N2
€5

= e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
4
X
0
ϕg2 g2 h2 (0, H2 ; 2N2 )ϕg10 g1 h1 (τ ; 0, H1 ; 2N1 ) ×
H2 mod2N2
H1 mod2N1
n 1
€ ατ + β
€ ατ + β


×
ϑ000
; H2 , 2N2 ϑ0g0 h0
; H1 , 2N1 −
g20 h02
1
1
N2
γτ + δ
γτ + δ
€

+
β
ατ
; H2 , 2N2 ×
−12γπi(γτ + δ)ϑ0g0 h0
2 2
γτ + δ
o
€ ατ + β
0
; H1 , 2N1 .
×ϑg0 h0
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
ϕg10 g1 h1 (0, H1 ; 2N1 )ϕg20 g2 h2 (0, H2 ; 2N2 ) ×
H1 mod2N1
H2 mod2N2
×Ψ1

€ ατ + β 0 0 0 0
; 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), g10 and g20 are also even.
Therefore, according to (3) and the notations (4), we have
€ ατ + β
€ ατ + β


; Hr , 2Nr ϑg0 s0 h0s
; Hs , 2Ns =
γτ + δ
γτ + δ
∞
∞
X
€ N n1 /Nr ατ + ⠁ X
€ N n2 /Ns ατ + ⠁
=
Bn 2 e
=
Bn1 e
4N
γτ + δ n =0
4N
γτ + δ
n =0
ϑg000r0 hr0
2
1
=
∞
X
n=0
Bn e
€ n ατ + ⠁
4N γτ + δ
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), g10 and g20 are also even, and we obtain the same result. But if hr
is odd, then by (1.21) gr0 will also be odd, and in (3) we have
€
hence
m+
2 €
 1
1
gr0 2 €
1
= m + (gr0 − 1) + m + (gr0 − 1) + ,
2
2
2
4
ϑg000r0 hr0
×
€ ατ + β

€ 1 ατ + ⠁
; Hr , 2Nr = −π 3 ie
×
γτ + δ
16Nr γτ + δ
X
0
(−1)hr (m−Hr )/2Nr (2m + gr0 )3 ×
m≡Hr (mod 2Nr )
n 1 €
 ατ + β o
1
1
=
(m + (gr0 − 1))2 + (m + (gr0 − 1))
×e
4Nr
2
2
γτ + δ
∞
€ hr ατ + ⠁ X
€ n1 ατ + ⠁
=e
B0 e
,
16Nr γτ + δ n =0 n1 4Nr γτ + δ
1
since (5) implies that hr = 1. Analogously,
ϑ0gs0 h0s
∞
€ ατ + β
€ n2 ατ + ⠁

€ hs ατ + ⠁ X
Bn0 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 ϑg0 s0 h0s
; Hs , 2Ns =
γτ + δ
γτ + δ
∞
€ N/4(hr /Nr + hs /Ns ) ατ + ⠁ X
€ n1 ατ + ⠁
=e
×
Bn0 1 e
4N
γτ + δ n =0
4Nr γτ + δ
ϑ000
gr0 h0r
1
×
∞
X
n2
∞
€ n2 ατ + ⠁ X
€ n ατ + ⠁
Bn0 2 e
=
Bn0 e
,
4Ns γτ + δ
4N γτ + δ
n=0
=0
(2.9)
hr
hs
since, by (2.3), N4 ( N
+N
) is an integer. Hence for odd γ (2.4) follows
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
ϕg10 g1 h1 (0, H1 ; 2N1 ) ϑg0 h0
; H1 , 2N1 +
1 1
γτ + δ
H1 mod2N1
Œ
€ ατ + β

4·3
Œ
+
A21 Œ
· ϑ00g0 h0
; H1 , 2N1 +
1 1
2!
γτ + δ
z=0
o
€ ατ + β
A42 ŒŒ
+
· ϑg10 h01
; H1 , 2N1 ×
Œ
2! z=0
γτ + δ
X

€ ατ + β
ϕg20 g2 h2 (0, H2 ; 2N2 )ϑg20 h20
×
; H2 , 2N2 =
γτ + δ
X
×
H2 mod2N2
€5
€
−1
sgn γ 2|γ|(N1 N2 )1/2
×
4
ϕg10 g1 h1 (0, H1 ; 2N1 )ϕg20 g2 h2 (0, H2 ; 2N2 ) ×
=e
×
X
H1 mod2N1
H2 mod2N2
n 1

(4) € ατ + β
ϑg0 h0
; H1 , 2N1 ×
2
N1 1 1 γτ + δ
€ ατ + β
€ ατ + β
 24γπi

×ϑg20 h02
; H2 , 2N2 −
(γτ + δ)ϑ00g0 h0
; H1 , 2N1 ×
1
1
γτ + δ
N1
γτ + δ
€
€ ατ + β


+β
ατ
ϑg20 h02
; H2 , 2N2 − 48γ 2 π 2 (γτ + δ)2 ϑg10 h01
; H1 , 2N1 ×
γτ + δ
γτ + δ
o
€ ατ + β
; H2 , 2N2 .
×ϑg20 h20
(2.10)
γτ + δ
×
Replacing N1 , g1 , h1 , H1 , g10 , h01 by N2 , g2 , H2 , g20 , h02 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
ϕg20 g2 h2 (0, H2 ; 2N2 )ϕg10 g1 h1 (0, H1 ; 2N1 )×
×
H2 mod2N2
H1 mod2N1
n 1

(4) € ατ + β
×
ϑg0 h0
; H2 , 2N2 ×
2
2
2
N2
γτ + δ
€ ατ + β
€ ατ + β
 24γπi

×ϑg10 h01
(γτ + δ)ϑ00g0 h0
; H1 , 2N1 −
; H2 , 2N2 ×
2 2
γτ + δ
N2
γτ + δ
€ ατ + β

€ ατ + β

; H1 , 2N1 − 48γ 2 π 2 (γτ + δ)2 ϑg20 h20
; H2 , 2N2 ×
×ϑg10 h10
γτ + δ
γτ + δ
ON SOME ENTIRE MODULAR FORMS
×ϑg10 h01
o
€ ατ + β
; H1 , 2N1 .
γτ + δ
67
(2.11)
Analogously, from Lemma 4 for n = 2 it follows that
6
(γτ + δ)5 ϑg001 h1 (τ ; 0, 2N1 )ϑ00g2 h2 (τ ; 0, 2N2 ) =
N1 N2
€5

= e sgn γ (2|γ|(N1 N2 )1/2 )−1 ×
4
X
×
ϕg10 g1 h1 (0, H1 ; 2N1 )ϕg20 g2 h2 (0, H2 ; 2N2 ) ×
H1 mod2N1
H2 mod2N2
n
€ ατ + β
€ ατ + β


6
ϑ00g0 h0
; H1 , 2N1 ϑ00g0 h0
; H2 , 2N2 −
2 2
N1 N2 1 1 γτ + δ
γτ + δ


€ ατ + β
€ ατ + β
24γπi
−
(γτ + δ)ϑg10 h01
; H1 , 2N1 ϑg20 h02
; H2 , 2N2 −
N2
γτ + δ
γτ + δ

€

€
+β
+
β
ατ
ατ
24γπi
(γτ + δ)ϑ00g0 h0
; H1 , 2N1 ϑ00g0 h0
; H2 , 2N2 −
−
1 1
2 2
N1
γτ + δ
γτ + δ
€ ατ + β

−96γ 2 π 2 (γτ + δ)2 ϑg10 h01
; H1 , 2N1 ×
γτ + δ
o
€ ατ + β
; H2 , 2N2 .
(2.12)
×ϑg20 h02
γτ + δ
×
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
ϕg10 g1 h1 (0, H1 ; 2N1 )ϕg20 g2 h2 (0, H2 ; 2N2 ) ×
H1 mod2N1
H2 mod2N2
×Ψ2

€ ατ + β 0 0 0 0
; 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,
Œ € h2
g2
h2  Œ g 2
2) 4ŒN 1 + 2 , 4Œ 1 + 2 ,
N1
N2
4N1
4N2
3) for all α and δ with α ≡ δ ≡ 1 (mod 4N )
€ N1 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
Œ
€ h2
h2  Œ g 2
g2
4ŒN δ 2 1 + 2 , 4Œ 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
ϑ000
; 0, 2Nr ϑ0gs hs
; 0, 2Ns = iη(γ)(sgnδ−1) i1−|δ| ×
gr h r
γτ + δ
γτ + δ

€
N
N
r
s
0
ϑ000
×(γτ + δ)5
αgr ,hr (τ ; 0, 2Nr )ϑαgs ,hs (τ ; 0, 2Ns ) =
|δ|
€ −Nr Ns 
(γτ + δ)5 ×
= sgn δ
|δ|
0
×ϑ000
(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)
0
×ϑαgr ,hr (τ ; 0, 2Nr )ϑαg
(τ ; 0, 2Ns )
s ,hs
if r = 1, s = 2 and r = 2, s = 1;
(2.19)
ON SOME ENTIRE MODULAR FORMS
69
2) for n = 2
ϑ00g1 h1
€ ατ + β


€ ατ + β
; 0, 2N1 ϑg002 h2
; 0, 2N2 =
γτ + δ
γτ + δ
€ −N1 N2 
(γτ + δ)5 ×
= sgn δ
|δ|
00
×ϑ00αg1 ,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
0
ϑ000
gr hr (τ ; 0, 2Nr )ϑgs hs (τ ; 0, 2Ns ) = π
∞
X
(−1)hr mr +hs ms ×
∞
X
(−1)hr mr +hs ms ×
∞
X
(−1)h1 m1 +h2 m2 ×
mr ,ms =−∞
3
×(4Nr mr + gr ) (4Ns ms + gs )e(Λτ ),
(4)
ϑgr hr (τ ; 0, 2Nr )ϑgs hs (τ ; 0, 2Ns ) = π 4
mr ,ms =−∞
×(4Nr mr + gr )4 e(Λτ )
(2.21)
and also
ϑ00g1 h1 (τ ; 0, 2N1 )ϑ00g2 h2 (τ ; 0, 2N2 ) = π 4
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
1 000
ϑgk hk (τ ; ck , 2Nk ) −
ϑg1 h1 (τ ; c1 , 2N1 )ϑ0g2 h2 (τ ; c2 , 2N2 )
N1
k=3
4
Y
1
0
− ϑ000
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)
ϑg2 h2 (τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 )
ϑgk hk (τ ; ck , 2Nk ) −
2
N2
k=3
4
−
Y
6
ϑgk hk (τ ; ck , 2Nk ), (3.2)
ϑ00g1 h1 (τ ; c1 , 2N1 )ϑ00g2 h2 (τ ; c2 , 2N2 )
N1 N2
k=3
where
4
Œ
ΠX
Œ
hk
2Œgk , Nk ŒN (k = 1, 2, 3, 4), 4ŒN
.
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
0
ϑgk hk (τ ; 0, 2Nk ) =
(γτ + δ)6 ϑ000
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
Nk
= e sgn γ 4γ
2
k=1
X
4
Y
Hk mod2Nk k=1
(k=1,2,3,4)
71
ϕgk0 gk hk (0, Hk ; 2Nk ) ×
n 1
€ ατ + β

ϑ000
; H1 , 2N1 −
g10 h10
N1
γτ + δ
‘
€ ατ + β
; H1 , 2N1 ×
−12γπi(γτ + δ)ϑ0g0 h0
1 1
γτ + δ
4
Y
€ ατ + β

o
€ ατ + β
ϑgk0 h0k
×ϑg0 0 h0
; H2 , 2N2
; Hk , 2Nk
2 2
γτ + δ
γτ + δ
×
(3.5)
k=3
and
4
Y
1
0
(γτ + δ)6 ϑ000
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
Hk mod2Nk k=1
(k=1,2,3,4)
ϕgk0 gk hk (0, Hk ; 2Nk ) ×
n 1
€ ατ + β

ϑ0000 0
; H2 , 2N2 −
N2 g2 h2 γτ + δ
‘
€ ατ + β
; H2 , 2N2 ×
−12γπi(γτ + δ)ϑ0g0 h0
2 2
γτ + δ
4
€ ατ + β
Y
o
€ ατ + β
×ϑ0g0 h0
; H1 , 2N1
ϑgk0 h0k
; 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
=e
€3
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)

€
ϕgk0 gk hk 0, Hk ; 2Nk ×
n 1

(4) € ατ + β
ϑg0 h0
; H1 , 2N1 −
2
1
1
N1
γτ + δ
€ ατ + β

24γπi
(γτ + δ)ϑ00g0 h0
−
; H1 , 2N1 −
1 1
N1
γτ + δ
‘
€ ατ + β
; H1 , 2N1 ×
−48γ 2 π 2 (γτ + δ)2 ϑg10 h01
γτ + δ
×
72
G. LOMADZE
×
4
Y
ϑgk0 h0k
k=2
and
€ ατ + β
o
; 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)
€

ϕgk0 gk hk 0, Hk ; 2Nk ×
n 1

(4) € ατ + β
; H2 , 2N2 −
ϑg0 h0
2
N2 2 2 γτ + δ
‘
€ ατ + β
24γπi
−
(γτ + δ)ϑ00g0 h0
; H2 , 2N2 ×
2 2
N2
γτ + δ
4
Y
€ ατ + β
€ ατ + β


; H1 , 2N1
; Hk , 2Nk −
×ϑg10 h10
ϑgk0 h0k
γτ + δ
γτ + δ
×
k=3
−48γ 2 π 2 (γτ + δ)2
4
Y
ϑgk0 hk0
k=1
o
€ ατ + β
; Hk , 2Nk ,
γτ + δ
(3.8)
2) for n = 2 and n = 0
4
Y
6
(γτ + δ)6 ϑg001 h1 (τ ; 0, 2N1 )ϑ00g2 h2 (τ ; 0, 2N2 )
ϑgk hk (τ ; 0, 2Nk ) =
N1 N2
k=3
=e
4

€Y
1/2 −1
€3
sgn γ (4γ 2
Nk
2
k=1
n
X
Hk mod2Nk
(k=1,2,3,4)
ϕgk0 gk hk (0, Hk ; 2Nk ) ×
€ ατ + β

6
ϑg000 h0
; H1 , 2N1 −
1
1
N1 N2
γτ + δ
‘
€ ατ + β
24γπi
; H1 , 2N1 ×
−
(γτ + δ)ϑg10 h10
N2
γτ + δ
4
Y

€ ατ + β
€ ατ + β
×ϑ00g0 h0
; H2 , 2N2
ϑgk0 hk0
; Hk , 2Nk −
2 2
γτ + δ
γτ + δ
k=3
 24γπi
€ ατ + β

−
(γτ + δ)ϑ00g1 h1
; H1 , 2N1 +
N1
γτ + δ
‘
€ ατ + β
; H1 , 2N1 ×
+96γ 2 π 2 (γτ + δ)2 ϑg1 h1
γτ + δ
×
ON SOME ENTIRE MODULAR FORMS
×
4
Y
ϑgk0 h0k
k=2
73
€ ατ + β
o
; 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
Hk mod2Nk
(k=1,2,3,4)
ϕgk0 gk hk (0, Hk ; 2Nk ) ×

€ ατ + β
; g 1 , . . . , g 4 , h 1 , . . . , h 4 , H1 , . . . , H 4 , N 1 , . . . , N 4
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
Œ
4ŒN
4
X
k=1
(
(k = 1, 2, 3, 4),
h2k ŒŒ
,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
Œ
h2k ŒŒ X gk2 2ϕ(2Nk )−2
,4
δ
.
4ŒN 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 ϑ0gs hs
; 0, 2Ns ×
γτ + δ
γτ + δ
4
Y

€ ατ + β
ϑgk hk
; 0, 2Nk =
×
γτ + δ
k=3
Q
 4 N ‘
000
k=1 k
= (γτ + δ)6
ϑαg
(τ ; 0, 2Nr )ϑ0αgs ,hs (τ ; 0, 2Ns ) ×
r ,hr
|δ|
ϑ000
gr h r
×
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
Y
€ ατ + β

; 0, 2Nk
ϑgk hk
; 0, 2Nk =
γτ + δ
γτ + δ
k=3
k=1
4
2
 Q4 N ‘ Y
Y
6
00
k=1 k
= (γτ + δ)
ϑαgk ,hk (τ ; 0, 2Nk )
ϑαgk ,hk (τ ; 0, 2Nk ).
|δ|
ϑ00gk 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,
0
ϑ000
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
ϑ00g1 h1 (τ ; 0, 2N1 )ϑ00g2 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
1. É. Goursat, Cours d’analyse mathematique. T. I. Gauthier-Villars,
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