GEORGIAN MATHEMATICAL JOURNAL: Vol. 2, No. 2, 1995, 189-199
CONSTRUCTION OF ENTIRE MODULAR FORMS OF
WEIGHTS 5 AND 6 FOR THE CONGRUENCE GROUP
Γ0 (4N )
G. LOMADZE
Abstract. Two classes of entire modular forms of weight 5 and two
of weight 6 are constructed for the congruence subgroup Γ0 (4N ). The
constructed modular forms as well as the modular forms from [1] will
be helpful in the theory of representation of numbers by the quadratic
forms in 10 and 12 variables.
The present paper is a direct continuation of [1] whose notation will be
preserved here.
1.
Lemma 1. For a given N let
Ψ3 (τ ) = Ψ3 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; c1 , . . . , c4 ; N1 , . . . , N4 ) =
n 1
ϑ00 (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) −
=
N1 g1 h1
o
1
− ϑg1 h1 (τ ; c1 , 2N1 )ϑg002 h2 (τ ; c2 , 2N2 ) ×
N2
×ϑg0 3 h3 (τ ; c3 , 2N3 )ϑg4 h4 (τ ; c4 , 2N4 )
(1.1)
and
Ψ4 (τ ) = Ψ4 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; c1 , . . . , c4 ; N1 , . . . , N4 ) =
=
3
Y
ϑ0gk hk (τ ; ck , 2Nk )ϑg4 h4 (τ ; c4 , 2N4 ),
(1.2)
k=1
1991 Mathematics Subject Classification. 11F11,11F03,11F27.
Key words and phrases. Quadratic form, entire modular form, congruence subgroup,
theta-series, generalized Gauss’ sum.
189
c 1995 Plenum Publishing Corporation
1072-947X/95/0300-0189$07.50/0 
190
G. LOMADZE
where
2|gk , Nk |N (k = 1, 2, 3, 4),
4
ΠX
hk
Œ
.
4ŒN
Nk
(1.3)
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)5 Ψj (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ) =
∞
 n ατ + β ‘
X
=
Cn(j) e
(j = 3, 4).
4N γτ + δ
n=0
(1.4)
Proof. I. Taking into account (1.19) from [1], by Lemma 4 from [1], it
follows for n = 2 (with g1 , h1 , g10 , h10 , N1 , H1 instead of g, h, g 0 , h0 , N , H)
and n = 0 (with g2 , h2 , g20 , h02 , N2 , H2 instead of g, h, g 0 , h0 , N , H) that
1
(γτ + δ)3 ϑ00g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) =
N1
X

1 €3
= − e sgn γ (2|γ|(N1 N2 )1/2 )−1
ϕg10 g1 h1 (0, H1 ; 2N1 ) ×
N1 4
H1 mod2N1
Œ
 ατ + β
‘
‘o
n
 ατ + β
Œ
; H1 , 2N1 + 2A21 Œ
; H1 , 2N1
×
· ϑg10 h01
× ϑg000 h0
1 1
γτ + δ
γτ + δ
z=0
‘
 ατ + β
X
ϕg20 g2 h2 (0, H2 ; 2N2 )ϑg20 h02
; H2 , 2N2 =
×
γτ + δ
H2 mod2N2
X
−1
€3
€
ϕg10 g1 h1 (0, H1 ; 2N1 ) ×
= −e sgn γ 2|γ|(N1 N2 )1/2
4
H1 mod2N1
H2 mod2N2
n 1
‘
 ατ + β
; H1 , 2N1 ×
ϑ00g0 h0
N1 1 1 γτ + δ
‘
 ατ + β
 ατ + β

; H2 , 2N2 − 4γπi(γτ + δ)ϑg10 h01
; H1 , 2N1 ×
×ϑg20 h02
γτ + δ
γτ + δ
 ατ + β
‘o
×ϑg20 h02
(1.5)
; H2 , 2N2 .
γτ + δ
×ϕg20 g2 h2 (0, H2 ; 2N2 )
If in (1.5) N1 , g1 , h1 , H1 , g10 , h10 are replaced by N2 , g2 , h2 , H2 , g20 , h20 , and
vice versa, then we have
1
(γτ + δ)3 ϑ00g2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 ) =
N2
X
€3
€
−1
ϕg20 g2 h2 (0, H2 ; 2N2 ) ×
= −e sgn γ 2|γ|(N1 N2 )1/2
4
H2 mod2N2
H1 mod2N1
CONSTRUCTION OF ENTIRE MODULAR FORMS
×ϕg10 g1 h1 (τ ; 0, H1 ; 2N1 ) ×
‘
‘
n 1
 ατ + β
 ατ + β
×
; H2 , 2N2 ϑg10 h01
; H1 , 2N1 −
ϑ00g0 h0
N2 2 2 γτ + δ
γτ + δ
 ατ + β
‘
−4γπi(γτ + δ)ϑg20 h02
; H2 , 2N2 ×
γτ + δ
‘o
 ατ + β
×ϑg10 h01
; H1 , 2N1 .
γτ + δ
191
(1.6)
Subtracting (1.6) from (1.5), we obtain
n 1
ϑ00 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) −
N1 g1 h1
o

€3
1
− ϑg1 h1 (τ ; 0, 2N1 )ϑ00g2 h2 (τ ; 0, 2N2 ) = −e sgn γ ×
N2
4
X
€

1/2 −1
ϕg10 g1 h1 (0, H1 ; 2N1 )ϕg20 g2 h2 (0, H2 ; 2N2 ) ×
× 2|γ|(N1 N2 )
(γτ + δ)3
H1 mod2N1
H2 mod2N2
n 1
‘
 ατ + β
‘
 ατ + β
ϑ00g0 h0
; H1 , 2N1 ϑg20 h02
; H2 , 2N2 −
N1 1 1 γτ + δ
γτ + δ

‘
‘o

1
ατ + β
ατ + β
; H1 , 2N1 ϑg000 h0
; H2 , 2N2 .
− ϑg10 h01
2 2
N2
γτ + δ
γτ + δ
×
(1.7)
Analogously, by Lemma 4 from [1], for n − 1 and n = 0 we obtain
×
(γτ + δ)2 ϑg0 3 h3 (τ ; 0, 2N3 )ϑg4 h4 (τ ; 0, 2N4 ) =
€
€1
−1
= −e sgn γ 2|γ|(N3 N4 )1/2
i sgn γ ×
2
X
ϕg30 g3 h3 (0, H3 ; 2N3 )ϕg40 g4 h4 (0, H4 ; 2N4 ) ×
H3 mod2N3
H4 mod2N4
 ατ + β
‘
‘
 ατ + β
; H3 , 2N3 ϑg40 h04
; H4 , 2N4 .
3
γτ + δ
γτ + δ
×ϑ0g0 h0
3
Multiplying (1.7) by (1.8), on account of (1.1), we obtain
(γτ + δ)5 Ψ3 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ) =
4
Y
‘1/2 ‘−1
€5

= e sgn γ 4γ 2
Nk
i sgn γ ×
4
k=1
×
X
4
Y
Hk mod2Nk k=1
(k=1,2,3,4)
ϕgk0 gk hk (0, Hk ; 2Nk ) ×
(1.8)
192
G. LOMADZE
×Ψ3
 ατ + β
‘
; g10 , . . . , g40 ; h01 , . . . , h04 ; H1 , . . . , H4 ; N1 , . . . , N4 .
γτ + δ
(1.9)
Further, applying the same reasoning as in [1, Lemma 5, pp. 62-63], we
obtain (1.4) if j = 3.
II. As in Subsection I, by Lemma 4 from [1], for n = 1 (with gk , hk , Nk ,
gk0 , h0k , Hk for all 1 ≤ k ≤ 4 instead of g, h, N , g 0 , h0 , H) and n = 0 (with
g4 , h4 , N4 , g40 , h04 , H4 instead of g, h, N , g 0 , h0 , H), we obtain
(γτ + δ)
5
3
Y
ϑ0gk hk (τ ; 0, 2Nk )ϑg4 h4 (τ ; 0, 2N4 ) =
k=1
4
‘1/2 ‘−1
Y

€5
Nk
i sgn γ ×
= e sgn γ 4γ 2
4
k=1
×
×
3
Y
X
4
Y
Hk mod2Nk k=1
(k=1,2,3,4)
ϕgk0 gk hk (0, Hk ; 2Nk ) ×
‘
‘
 ατ + β
 ατ + β
; Hk , 2Nk ϑg40 h04
; H4 , 2N4 .
k
γτ + δ
γτ + δ
ϑ0g0 h0
k
k=1
Hence, according to (1.2), it follows that
(γτ + δ)5 Ψ4 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ) =
=e
4
Y
‘1/2 ‘−1

€5
sgn γ 4γ 2
Nk
i sgn γ ×
4
k=1
×
×Ψ4
X
4
Y
Hk mod2Nk k=1
(k=1,2,3,4)
ϕgk0 gk hk (0, Hk ; 2Nk ) ×
‘
 ατ + β
; g10 , . . . , g40 ; h01 , . . . , h04 ; H1 , . . . , H4 ; N1 , . . . , N4 . (1.10)
γτ + δ
Further, applying the same reasoning as in [1, Lemma 5], we obtain (1.4)
if j = 4.
Theorem 1. For a given N the functions Ψ3 (τ ) and Ψ4 (τ ) with c1 =
c2 = c3 = c4 = 0 are entire modular forms of weight 5 and character
χ(δ) = sgn γ( −∆
|δ| ) (∆ is the determinant of an arbitrary positive quadratic
form in 10 variables) for the group Γ0 (4N ) if the following conditions hold:
1)
2|gk , Nk |N
(k = 1, 2, 3, 4),
(1.11)
CONSTRUCTION OF ENTIRE MODULAR FORMS
2)
4
4
ΠX
h2k ŒŒ X gk2
Œ
,4
4N
,
Nk
4Nk
k=1
3)
193
(1.12)
k=1
for all α and δ with αδ ≡ 1 (mod 4N )
 Q4 N ‘
k=1 k
Ψj (τ ; αg1 , . . . , αg4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ) =
|δ|
 −∆ ‘
(1.13)
= sgn δ
Ψj (τ ; g1 , ..., g4 ; h1 , ..., h4 ; 0, ..., 0;N1 , ..., N4 )
|δ|
(j = 3, 4).
Proof. I. As in the case of Theorem 1 from [1], the functions Ψ3 (τ ) and
Ψ4 (τ ) with c1 = c2 = c3 = c4 = 0 satisfy the condition 1) and, by Lemma
1, also the condition 4) of the definition from [1, p. 53–54].
II. From (1.12), since 2 - δ, it follows that
4
4
Œ
ŒX
X
h2k
gk2 2ϕ(2Nk )−2
Œ
Œ
δ
4ŒN δ 2
, 4Œ
.
Nk
4Nk
k=1
(1.14)
k=1
Taking into account (1.11), by Lemma 3 from [1], for n = 2 and n = 0
(with gr , hr , Nr and gs , hs , Ns instead of g, h, N ), we obtain for each
substitution from Γ0 (4N )
‘
 ατ + β
‘
 ατ + β
; 0, 2Nr ϑgs hs
; 0, 2Ns = i3η(γ)(sgnδ−1) i1−|δ| ×
γτ + δ
γτ + δ
N N ‘
 βδ  g 2
‘‘
g2
r s
r
×
(γτ + δ)3 e
δ 2ϕ(2Nr )−2 + s δ 2ϕ(2Ns )−2 ×
|δ|
4 4Nr
4Ns
 αγδ 2  h2
2 ‘‘
h
r
+ s
×
×e −
4
4Nr
4Ns
00
(τ ; 0, 2Nr )ϑ0αgs ,hs (τ ; 0, 2Ns )
×ϑαg
(1.15)
r ,hr
ϑg00r hr
for r = 1, s = 2 and r = 2, s = 1.
Analogously, by Lemma 3 from [1], for n = 1 and n = 0, we have
‘
 ατ + β
‘
 ατ + β
; 0, 2N3 ϑg4 h4
; 0, 2N4 = sgn δ i2η(γ)(sgnδ−1) ×
γτ + δ
γτ + δ
 βδ  g 2
N N ‘
‘‘
g2
3 4
3
×i1−|δ|
(γτ + δ)2 e
δ 2ϕ(2N3 )−2 + 4 δ 2ϕ(2N4 )−2 ×
|δ|
4 4N3
4N4
 αγδ 2  h2
2 ‘‘
h
3
×
×e −
+ 4
4
4N3
4N4
(1.16)
×ϑ0αg3 ,h3 (τ ; 0, 2N3 )ϑαg4 ,h4 (τ ; 0, 2Ns ).
ϑ0g3 h3
194
G. LOMADZE
Hence, by (1.1), (1.15), (1.16), and (1.14), we obtain
 ατ + β
‘
; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 =
γτ + δ
 Q4 N ‘
k=1 k
(γτ + δ)5 ×
= sgn δ iη(γ)(sgn δ−1) (−1)1−|δ|
|δ|
×Ψ3 (τ ; αg 1 , . . . , αg 4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ),
Ψ3
from which according to (1.13) it follows for all α and δ with αδ ≡ 1
(mod 4N ) that
‘
 ατ + β
; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 =
γτ + δ
 −∆ ‘
= sgn δ
(γτ + δ)5 Ψ3 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ).
|δ|
Ψ3
Analogously, by the just mentioned Lemma 3, for n = 1 and n = 0 (with
gk , hk , Nk (k = 1, 2, 3, 4) instead of g, h, N ), according to (1.11) and (1.14),
we find for all substitutions from Γ0 (4N ) that
 ατ + β
 ατ + β
‘
‘
; 0, 2Nk ϑg4 h4
; 0, 2N4 =
γτ + δ
γτ + δ
k=1
 Q4 N ‘
k=1 k
= sgn δ iη(γ)(sgn δ−1) (−1)1−|δ|
(γτ + δ)5 ×
|δ|
3
Y
ϑ0gk hk
×
3
Y
ϑ0αgk ,hk (τ ; 0, 2Nk )ϑ0αg4 ,h4 (τ ; 0, 2N4 ).
(1.17)
k=1
Hence, by (1.2), (1.17), and (1.13), for all α and δ with αδ ≡ 1 (mod 4N )
we have
 ατ + β
‘
Ψ4
; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 =
γτ + δ
 −∆ ‘
(γτ + δ)5 Ψ4 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ).
= sgn δ
|δ|
Thus the functions Ψ3 (τ ) and Ψ4 (τ ) with c1 = c2 = c3 = c4 = 0 satisfy
the condition 2) of the definition from [1].
III. According to (13) from [1] we have
1) ϑ00gr hr (τ ; 0, 2Nr )ϑgs hs (τ ; 0, 2Ns ) =
= −π 2
∞
X
(−1)hr mr+hs ms (4Nr mr +gr )2 e(Λ1 τ )
mr ,ms=−∞
(1.18)
CONSTRUCTION OF ENTIRE MODULAR FORMS
195
for r = 1, s = 2 and r = 2, s = 1, where
Λ1 =
2
X
1
(2Nk mk + gk /2)2 =
4Nk
k=1
=
2
X
€
k=1
2
 1X 2
1
Nk m2k + mk gk +
gk /4Nk ;
2
4
(1.19)
k=1
2) ϑ0g3 h3 (τ ; 0, 2N3 )ϑg4 h4 (τ ; 0, 2N4 ) =
∞
X
= πi
(−1)h3 m3 +h4 m4 (4N3 m3 + g3 )e(Λ2 τ ),
(1.20)
m3 ,m4 =−∞
where
Λ2 =
4
X
1
(2Nk mk + gk /2)2 =
4Nk
k=3
=
4
X
€
k=3
3)
3
Y
4
 1X 2
1
Nk m2k + mk gk +
gk /4Nk ;
2
4
(1.21)
k=3
ϑg0 k hk (τ ; 0, 2Nk )ϑg4 h4 (τ ; 0, 2N4 ) =
k=1
= −π 3 i
∞
X
3
P4
Y
(−1) k=1 hk mk (4Nk mk + gk )e(Λτ ),
m1 ,...,m4 =−∞
k=1
where
Λ=
=
4
X
1
(2Nk mk + gk /2)2 =
4Nk
k=1
4
X
k=1
€
4
 1X 2
1
Nk m2k + mk gk +
gk /4Nk .
2
4
(1.22)
k=1
Λ1 + Λ2 and Λ are integers by virtue of (1.11) and (1.12). Therefore the
functions Ψ3 (τ ) and Ψ4 (τ ) satisfy the condition 3) of the definition from
[1].
196
G. LOMADZE
2.
Lemma 2. For a given N let
Φ3 (τ ) = Φ3 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; c1 , . . . , c4 ; N1 , . . . , N4 ) =
n 1
ϑ00 (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) −
=
N1 g1 h1
4
oY
1
ϑ0gk hk (τ ; ck , 2Nk ) (2.1)
− ϑg1 h1 (τ ; c1 , 2N1 ))ϑ00g2 h2 (τ ; c2 , 2N2 )
N2
k=3
and
Φ4 (τ ) = Φ4 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; c1 , . . . , c4 ; N1 , . . . , N4 ) =
=
4
Y
ϑ0gk hk (τ ; ck , 2Nk ),
(2.2)
k=1
where
4
Œ
Œ
ΠX
hk
.
2Œgk , Nk ŒN (k = 1, 2, 3, 4), 4ŒN
Nk
(2.3)
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 ) =
∞
X
n=0
Dn(j) e
 n ατ + β ‘
4N γτ + δ
(j = 3, 4).
(2.4)
Proof. I. As in the proof of Lemma 1, by Lemma 4 from [1], it follows for
n = 1 that
×
(γτ + δ)3 ϑg0 3 h3 (τ ; 0, 2N3 )ϑ0g4 h4 (τ ; 0, 2N4 ) =
€3
€
1/2 −1
= −e sgn γ 2|γ|(N3 N4
) ×
4
4
X
Y
ϕg30 g3 h3 (0, H3 ; 2N3 )ϕg40 g4 h4 (τ ; 0, H4 ; 2N4 ) ×
H3 mod2N3 k=1
H4 mod2N4
‘
 ατ + β
‘
 ατ + β
; H3 , 2N3 ϑ0g0 h0
; H4 , 2N4 .
3
4 4
γτ + δ
γτ + δ
×ϑ0g0 h0
3
Multiplying (1.7) by (2.5), by virtue of (2.1), we obtain
(γτ + δ)6 Φ3 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ) =
(2.5)
CONSTRUCTION OF ENTIRE MODULAR FORMS
4
‘1/2 ‘−1
 2  Y
€3
Nk
= e sgn γ 4γ
2
k=1
×Φ3
X
Hk mod2Nk
(k=1,2,3,4)
197
ϕgk0 gk hk (0, Hk ; 2Nk ) ×
 ατ + β
‘
; g 0 1 , . . . , g 0 4 ; h0 1 , . . . , h0 4 ; H1 , . . . , H4 ; N1 , . . . , N4 . (2.6)
γτ + δ
Further, applying the same reasoning as in [1, Lemma 5], we obtain (2.4)
if j = 3.
II. By Lemma 4 from [1], for n = 1 we have
(γτ + δ)6
4
Y
k=1
×
X
4
‘1/2 ‘−1
Y
€3

ϑ0gk hk (τ ; 0, 2Nk ) = e sgn γ 4γ 2
Nk
×
2
k=1
4
Y
ϕgk0 gk hk (0, Hk ; 2Nk )
4
Y
k=1
Hk mod2Nk k=1
(k=1,2,3,4)
 ατ + β
‘
ϑ0g0 h0
; Hk , 2Nk .
k k
γτ + δ
Hence, according to (2.2), it follows that
(γτ + δ)6 Φ4 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ) =
=e
4
Y
‘1/2 ‘−1
€3

Nk
sgn γ 4γ 2
2
k=1
×Φ4
X
4
Y
Hk mod2Nk k=1
(k=1,2,3,4)
ϕgk0 gk hk (0, Hk ; 2Nk ) ×
‘
 ατ + β
; g 0 1 , . . . , g 0 4 ; h0 1 , . . . , h0 4 ; H1 , . . . , H4 ; N1 , . . . , N4 .
γτ + δ
Further, applying the same reasoning as in [1, Lemma 5], we obtain (2.4) if
j = 4.
Theorem 2. For a given N the functions Φ3 (τ ) and Φ4 (τ ) with c1 =
c2 = c3 = c4 = 0 are entire modular forms of weight 6 and character
∆
) (∆ is the determinant of an arbitrary positive quadratic form
χ(δ) = ( |δ|
in 12 variables) for the group Γ0 (4N ) provided that the following conditions
hold:
1)
2)
2|gk , Nk |N
4
4
ΠX
h2k ŒŒ X gk2
4 ŒN
,
,4
Nk
4Nk
k=1
3)
(k = 1, 2, 3, 4),
(2.7)
(2.8)
k=1
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|
198
G. LOMADZE
=
∆‘
|δ|
Φj (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 )
(2.9)
(j = 3, 4).
Proof. I. As in the case of Theorem 1, the functions Φ3 (τ ) and Φ4 (τ )
with c1 = c2 = c3 = c4 = 0 satisfy the condition 1) and, on account of
Lemma 2, also the condition 4) of the definition from [1].
II. Since 2 - δ, from (2.8) it follows that
4
Œ
Π4
Œ 2 X hk2 Œ X gk2 2ϕ(2Nk )−2
4ŒN δ
.
, 4Œ
δ
Nk
4Nk
k=1
(2.10)
k=1
Taking into account (2.7), by Lemma 3 from [1], for n = 1 we find that
for all substitutions from Γ0 (4N ) we have
 ατ + β
‘
N N ‘
3 4
; 0, 2Nk = i3η(γ)(sgnδ−1) i1−|δ|
×
γτ + δ
|δ|
k=3
 βδ  g 2
‘‘
g2
3
×(γτ + δ)3 e
δ 2ϕ(2N3 )−2 + 4 δ 2ϕ(2N4 )−2 ×
4 4N3
4N4
4
 αγδ 2  h2
2 ‘‘ Y
h
3
0
+ 4
ϑαg
(τ ; 0, 2Nk ).
(2.11)
×e −
k ,hk
4
4N3
4N4
4
Y
ϑ0gk hk
k=3
Taking into account (2.10) and (2.9), by (2.1), (1.15), and (2.11), for all α
and δ with αδ ≡ 1 (mod 4N ) we obtain
‘
 ατ + β
Φ3
; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 =
γτ + δ
 Q4 N ‘
1−|δ|
2η(γ)(sgn δ−1)
k=1 k
(γτ + δ)6 ×
=i
(−1)
|δ|
×Φ3 (τ ; αg 1 , . . . , αg 4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ) =
∆‘
=
(γτ + δ)6 Φ3 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ).
|δ|
Analogously, taking into account (2.7) and (2.10), by Lemma 3 from [1],
for n = 1, we have
=
 Q4
4
Y
ϑ0gk hk
k=1
k=1
|δ|
Nk ‘
 ατ + β
‘
; 0, 2Nk =
γτ + δ
6
(γτ + δ)
for all substitutions from Γ0 (4N ).
4
Y
k=1
ϑαgk ,hk (τ ; 0, 2Nk )
(2.12)
CONSTRUCTION OF ENTIRE MODULAR FORMS
199
Hence, by (2.2), (2.12), and (2.9), for all α and δ with αδ ≡ 1 (mod 4N ),
we have
 ατ + β
‘
Φ4
; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 =
γτ + δ
∆‘
(γτ + δ)6 Φ4 (τ ; g1 , . . . , g4 ; h1 , . . . , h4 ; 0, . . . , 0; N1 , . . . , N4 ).
=
|δ|
Thus the functions Φ3 (τ ) and Φ4 (τ ) with c1 = c2 = c3 = c4 = 0 satisfy
the condition 2) of the definition from [1].
III. According to (13) from [1], we have
1) (1.18) with Λ1 defined by (1.19);
2)
4
Y
ϑ0gk hk (τ ; 0, 2Nk ) =
k=3
= −π 2
∞
X
(−1)h3 m3 +h4 m4
m3 ,m4 =−∞
4
Y
(4Nk mk + gk )e(Λ2 τ ),
k=3
where Λ2 is defined by (1.20);
3)
4
Y
ϑ0gk hk (τ ; 0, 2Nk ) =
k=1
= π4
∞
X
m1 ,...,m4 =−∞
4
P4
Y
(−1) k=1 hk mk
(4Nk mk + gk )e(Λτ ),
k=3
where Λ2 is defined by (1.21).
Thus the functions Φ3 (τ ) and Φ4 (τ ) with c1 = c2 = c3 = c4 = 0 satisfy
the condition 3) of the definition from [1].
References
1. G. Lomadze, On some entire modular forms of weights 5 and 6 for the
congruence group Γ0 (4N ). Georgian Math. J. 1(1994), No. 1, 53-76.
(Received 28.06.1993)
Author’s address:
Faculty of Mechanics and Mathematics
I. Javakhishvili Tbilisi State University
2, University St., Tbilisi 380043
Republic of Georgia