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GEORGIAN MATHEMATICAL JOURNAL: Vol. 3, No. 5, 1996, 485-500
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
AND 29 FOR THE CONGRUENCE GROUP Γ0 (4N )
7
2
G. LOMADZE
Abstract. Entire modular forms of weights 27 and 29 for the congruence group 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 7 and 9
variables.
1.
In this paper N, a, k, q, r, t denote positive integers; u, s are odd positive
integers; H, c, g, h, j, m, n, α, β, γ, δ, ξ are integers; A, B, C, D, G are complex
numbers; z, τ (Im τ > 0) are complex variables,and
H = {τ ∈ C| Im τ > 0}.
Further, ( uh ) is the generalized Jacobi symbol; nt is a binomial coefficient;
ϕ(k) is Euler’s function; e(z) = exp 2πiz; η(γ) = 1 if γ ≥ 0 and η(γ) = −1
if γ < 0.
Let
X
ϑgh (z|τ ; c, N ) =
(−1)h(m−c)/N ×
m≡c (mod N )
×e
hence
1
g
g 2
τ e m+ z ;
m+
2N
2
2
(1.1)
X
∂n
n
;
c,
N
)
=
(πi)
(−1)h(m−c)/N (2m + g)n ×
ϑ
(z|τ
gh
∂z n
m≡c (mod N )
1
g 2
g
×e
m+
τ e m + z (n ≥ 0).
(1.2)
2N
2
2
1991 Mathematics Subject Classification. 11F37, 11F27, 11F11.
Key words and phrases. Entire modular form, congruence subgroup, multiplier system, theta-function, half integral weight.
485
c 1997 Plenum Publishing Corporation
1072-947X/96/0900-0485$12.50/0
486
G. LOMADZE
Put
(n)
ϑgh (τ ; c, N ) =
∂n
ϑgh (z|τ ; c, N )z=0 (n ≥ 0).
∂z n
(1.3)
It is known (see, for e.g., [1], formulas (11) and (12)) that
(n)
(n)
ϑg,h+2j (τ ; c, N ) = ϑgh (τ ; c, N ) (n ≥ 0),
β g
2
(n)
(n)
ϑg,h+βg+βN (τ ; c, N ) (n ≥ 0),
c+
ϑgh (τ + β; c, N ) = e
2N
2
β
g 2
(n)
ϑgh (τ − β; c, N ) = (−1)n e −
c+
×
2N
2
(n)
× ϑ−g,−h+βg−βN (τ ; −c, N ) (n ≥ 0).
(1.4)
(1.5)
From (1.1) and (1.2), according to notation (1.3) it follows, in particular,
that
(n)
ϑgh (τ ; 0, N ) = (πi)n
∞
X
(−1)hm (2N m + g)n ×
m=−∞
1
g 2
τ (n ≥ 0).
Nm +
×e
2N
2
(1.6)
For ξ2 6= 0, ξ1 , g, h, N with ξ1 g + ξ2 h + ξ1 ξ2 N ≡ 0 (mod 2), put
Sgh
ξ1
; c, N =
ξ2
X
mmodN |ξ2 |
m≡c (mod N )
ξ
g 2
1
(−1)h(m−c)/N e
m+
.
2N ξ2
2
Finally, let
n ατ + β
γτ + δ
n ατ + β
Γ0 (4N ) =
γτ + δ
Γ=
For τ ∈ H put
o
αδ − βγ = 1 ,
o
∈ Γ γ ≡ 0 (mod 4N ) .
s
(γτ + δ)s/2 = (γτ + δ)1/2 ,
−
π
π
< arg(γτ + δ)1/2 ≤ .
2
2
Definition. Let M be a matrix of an arbitrary substitution from Γ0 (4N ),
and let v(M ) be a multiplier system on Γ0 (4N ) and of weight 2s . We shall
say that a function F defined on H is an entire modular form of weight 2s
and of multiplier system v(M ) on Γ0 (4N ), if
(1) F is regular on H;
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
487
α β
of substitutions from Γ0 (4N ) and all
(2) for all matrices M =
γ δ
τ ∈ H,
ατ + β
F
= v(M )(γτ + δ)s/2 F (τ );
γτ + δ
(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),
(γτ + δ)s/2 F (τ ) =
∞
X
m ατ + β
.
A′m e
4N γτ + δ
m=0
Lemma 1 ([1], p. 58, Lemma 3). If g is even, then for n ≥ 0 and all
substitutions from Γ0 (4N ) we have
(n) ατ + β
ϑgh
; 0, 2N =
γτ + δ
2βN sgn δ
(γτ + δ)(2n+1)/2 ×
= (sgn δ)n i(2n+1)η(γ)(sgn δ−1)/2 i(1−|δ|)/2
|δ|
αγδ 2 h2 βδg 2 δ 2ϕ(2N )−2
(n)
e
ϑαg,h (τ ; 0, 2N ).
×e −
16N
4
4N
Lemma 2 ([1], p. 61, Lemma 4). If γ 6= 0, then for n ≥ 0
(n)
(γτ + δ)(2n+1)/2 ϑgh (τ ; 0, 2N ) =
−1/2
n
= e (2n + 1) sgn γ/8 2N |γ|
− i sgn γ ×
n
ατ + β
X
(n)
; H, 2N +
×
ϕg′ gh (0, H; 2N ) ϑg′ h′
γτ + δ
H mod2N
t
n X
o
X
Atk
n
(n−t) ατ + β
; H, 2N ,
· ϑg′ h′
+
t
k! z=0
γτ + δ
t=1
k=1
where
g ′ = δg − γh − 2γδN, h′ = −βg + αh − 2αβN,
αβ
g ′ 2 βg
g ′
e −
×
H+
H+
ϕg′ gh (0, H; 2N ) = e
4N
2
4N
2
δ
; −αH, 2N ,
× Sg−αg′ ,h−βg′
−γ
(1.7)
488
G. LOMADZE
Atk z=0 =
(
(2k)!(−2N γπi(γτ + δ))k
0
if t = 2k
if t =
6 2k
(1.8)
(t = 1, 2, . . . , n; k = 1, 2, . . . , t).
From (1.8) it follows that
t
X
Atk
=
k! z=0
k=1
(A
t,t/2
(t/2)! z=0
0
if 2|t,
if 2 ∤ t.
(1.9)
2.
Lemma 3. For given N let
where
Ω(τ ; gl , hl , cl , Nl ) =
= Ω(τ ; g1 , g2 , g3 ; h1 , h2 , h3 ; c1 , c2 , c3 ; N1 , N2 , N3 ) =
n 1
=
ϑ′′ (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
N1 g1 h1
o
1 ′′
−
ϑg2 h2 (τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 ) ϑg3 h3 (τ ; c3 , 2N3 ),
N2
3
X
hk
.
2|gk , Nk |N (k = 1, 2, 3), 4|N
Nk
(2.1)
(2.2)
k=1
Then for all substitutions from Γ in the neighborhood of each rational point
τ = − γδ (γ =
6 0, (γ, δ) = 1), we have
(γτ + δ)7/2 Ω(τ ; gl , hl , 0, Nl ) =
∞
X
n ατ + β
Gn e
.
4N γτ + δ
n=0
(2.3)
Proof. 1. From Lemma 2 for n = 2 and n = 0 (respectively with g1 , h1 , N1 ,
g1′ , h1′ , H1 and g2 , h2 , N2 , g2′ , h′2 , H2 instead of g, h, N, g ′ , h′ , H), according
to (1.8)–(1.9), it follows that
1
(γτ + δ)3 ϑ′′g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) =
N1
3
= −e sgn γ (4N1 N2 γ 2 )−1/2 ×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
×
H1 mod2N1
H2 mod2N2
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
×
ατ + β
n 1
ατ + β
ϑ′′g′ h′
; H1 , 2N1 ϑg2′ h2′
; H2 , 2N2 +
N1 1 1 γτ + δ
γτ + δ
ατ + β
−4γπi(γτ + δ)ϑg1′ h′1
; H1 , 2N1 ×
γτ + δ
o
ατ + β
; H2 , 2N2 .
×ϑg2′ h′2
γτ + δ
489
(2.4)
Replacing N1 , g1 , h1 , H1 , g1′ , h1′ by N2 , g2 , h2 , H2 , g2′ , h′2 in (2.4), and vice
versa, we obtain
1
(γτ + δ)3 ϑ′′g2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 ) =
N2
3
= −e sgn γ (4N1 N2 γ 2 )−1/2 ×
4
X
′
ϕg2 g2 h2 (0, H2 ; 2N2 )ϕg1′ g1 h1 (0, H1 ; 2N1 ) ×
×
H1 mod2N1
H2 mod2N2
n 1
ατ + β
ατ + β
×
ϑ′′g′ h′
; H2 , 2N2 ϑg1′ h1′
; H1 , 2N1 +
N2 2 2 γτ + δ
γτ + δ
ατ + β
−4γπi(γτ + δ)ϑg2′ h′2
; H2 , 2N2 ×
γτ + δ
o
ατ + β
; H1 , 2N1 .
×ϑg1′ h′1
γτ + δ
(2.5)
Subtracting (2.5) from (2.4), we obtain
n 1
(γτ + δ)3
ϑ′′ (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) +
N1 g1 h1
o
1
− ϑ′′g2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 ) =
N2
3
= −e sgn γ (4N1 N2 γ 2 )−1/2 ×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
×
H1 mod2N1
H2 mod2N2
n 1
ατ + β
ατ + β
×
ϑ′′g′ h′
; H1 , 2N1 ϑg2′ h2′
; H2 , 2N2 +
N1 1 1 γτ + δ
γτ + δ
o
1 ′′
ατ + β
ατ + β
ϑg ′ h ′
−
; H2 , 2N2 ϑg1′ h′1
; H1 , 2N1 .
N2 2 2 γτ + δ
γτ + δ
From Lemma 2 for n = 0, we have
(γτ + δ)1/2 ϑg3 h3 (τ ; 0, 2N3 ) = e sgn γ/8 (2N3 |γ|)−1/2 ×
(2.6)
490
G. LOMADZE
X
×
ϕg3′ g3 h3 (0, H3 ; 2N3 )ϑg3′ h3′
H3 mod2N3
ατ + β
; H3 , 2N3 .
γτ + δ
(2.7)
Multiplying (2.6) by (2.7), according to (2.1), we obtain
(γτ + δ)7/2 Ω(τ ; gl , hl , 0, Nl ) =
= −e(7 sgn γ/8)(8N1 N2 N3 |γ|3 )−1/2 ×
3
Y
X
×
ϕgk′ gk hk (0, Hk ; 2Nk )Ω
H1 mod2N1 k=1
H2 mod2N2
H3 mod2N3
ατ + β
; gl′ , h′l , Hl , Nl .
γτ + δ
(2.8)
2. In (2.8) let γ be even. Then, by (1.7), gk′ (k = 1, 2, 3) are also
even. Now in (1.2) and (1.3) instead of m let us introduce new letters of
summations mk defined by the equalities m − Hk = 2Nk mk (k = r, t, 3),
respectively, and for brevity we put
g ′ 2
Tk = Hk + 2Nk mk + k
(k = r, t, 3).
2
Then, by (2.1), in (2.8) for r = 1, t = 2 and r = 2, t = 1, we get
ϑg′′r′ hr′
ατ + β
ατ + β
Y
; Hr , 2Nr
; Hk , 2Nk =
ϑgk′ hk′
γτ + δ
γτ + δ
k=t,3
= (πi)2
∞
X
mr
2 N/Nr
′
ατ + β
(−1)hr mr 2(Hr + 2Nr mr ) + gr′ e
×
Tr
4N
γτ + δ
=−∞
×
∞
X
mt
×
N/N
′
ατ + β
t
(−1)ht mt e
×
Tt
4N
γτ + δ
=−∞
∞
X
m3 =−∞
=
×
N/N
4N
3
T3
ατ + β
=
γτ + δ
∞
n ατ + β X
n ατ + β
r
t
Bnt e
×
Bn r e
δ
4N
γτ
+
4N
γτ + δ
n =0
=0
∞
X
nr
∞
X
n3
′
(−1)h3 m3 e
t
∞
n ατ + β
n ατ + β X
3
Bn 3 e
=
,
Cn(r,t) e
4N γτ + δ
4N γτ + δ
=0
n=0
(2.9)
since by (2.2) nk = NNk Tk are non-negative integers. Thus, for even γ, (2.3)
follows from (2.1),(2.8), and (2.9).
In (2.8) let now γ be odd. If h1 , h2 , h3 are even, then by (1.7), g1′ , g2′ ,
′
g3 are also even, and we obtain the same result. But if hr is odd, then by
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
491
(1.7), hr′ and gr′ will also be odd, and in (1.2) we shall have
2
2
m + gr′ /2 = m + (gr′ − 1)/2 + m + (gr′ − 1)/2 + 1/4.
Hence
ατ + β
h′ ατ + β
r
×
; Hr , 2Nr = (πi)2 e
γτ + δ
16Nr γτ + δ
∞
X
2 N/Nr
′
ατ + β
Wr
(r = 1, 2).
(−1)hr mr 2(Hr + 2Nr mr )+gr′ e
×
4N
γτ + δ
m =−∞
ϑ′′gr′ h′r
r
Here by (1.4) we can imply that h′r = 1, since by (1.7), h′r and hr have the
same parity, and
2
Wr = Hr + 2Nr mr + (gr′ − 1)/2 +
+ Hr + 2Nr mr + (gr′ − 1)/2 (r = 1, 2).
(2.10)
Analogously,
h′ ατ + β
ατ + β
t
; Ht , 2Nt = e
×
γτ + δ
16Nt γτ + δ
∞
N/N
X
′
ατ + β
t
(−1)ht mt e
(t = 2, 1).
Wt
×
4N
γτ + δ
m =−∞
ϑgt′ h′t
t
The same formula is valid for t = 3 with the same remarks concerning h3′
as for ht′ , t = 2, 1. Further, for k = t, 3 we have
Tk
if 2|gk′ ,
2
′
(2.11)
Wk = (Hk + 2Nk mk + (gk − 1)/2) +
′
′
if 2 ∤ gk .
(Hk + 2Nk mk + (gk − 1)/2)
Thus, if among h1′ , h′2 , h′3 at least one is odd, then, as in (2.9), for r = 1,
t = 2 and r = 2, t = 1 we have
ατ + β
Y
ατ + β
ϑgk′ hk′
; Hr , 2Nr
; Hk , 2Nk =
ϑ′′gr′ hr′
γτ + δ
γτ + δ
k=t,3
3
∞
1 1 X
hk′ ατ + β X ′ nr ατ + β
N
=e
×
B e
4N 4
Nk γτ + δ n =0 nr 4N γτ + δ
k=1
∞
X
r
∞
n ατ + β
n ατ + β X
t
3
×
Bn′ 3 e
Bn′ t e
=
4N
γτ
+
4N
γτ + δ
δ
n =0
n =0
t
3
3
∞
1 1 X
h′k ατ + β X (r,t) n′ ατ + β
Dn′ e
=e
N
=
4N 4
Nk γτ + δ
4N γτ + δ
′
k=1
n =0
492
G. LOMADZE
=
∞
X
Dn(r,t) e
n=0
n ατ + β
,
4N γτ + δ
since, by (2.2), (2.10), and (2.11), nk =
N
Nk Wk
and n = 14 N
(2.12)
3
P
k=1
h′k
Nk
+ n′ are
non-negative integers. Thus, for odd γ, (2.3) follows from (2.1), (2.8), and
(2.12).
Theorem 1. For given N the function Ω(τ ; gl , hl , 0, Nl ) is an entire
modular form of weight 7/2 and of the multiplier system
β∆ sgn δ
2
(2.13)
v(M ) = i3η(γ)(sgn δ−1)/2 i3(|δ|−1) /4
|δ|
α β
is a matrix of the substitution from Γ0 (4N ) and ∆ is the
(M =
γ δ
determinant of an arbitrary positive quadratic form with integer coefficients
in 7 variables) on Γ0 (4N ) if the following conditions hold:
(1)
(2)
2|gk , Nk |N (k = 1, 2, 3),
4|N
3
X
k=1
h2k
Nk
4|
s
X
k=1
(2.14)
gk2
,
4Nk
(2.15)
3) for all α and δ with αδ ≡ 1 (mod 4N )
∆
N N N
1 2 3
Ω(τ ; αgl , hl , 0, Nl ) =
Ω(τ ; gl , hl , 0, Nl ).
|δ|
|δ|
(2.16)
Proof. 1. It is well known that theta-series (1.1)–(1.2) are regular on H;
hence the function Ω(τ ; gl , hl , 0, Nl ) satisfies condition (1) of the Definition.
2. It is easily verified that (2.15) implies
4|N δ 2
3
X
k=1
h2k /Nk , 4|
3
X
δ 2ϕ(2Nk )−2 gk2 /4Nk ,
(2.17)
k=1
since 2 ∤ δ, because αδ ≡ 1 (mod 4N ).
From (2.14) it follows that
Γ0 (4N ) ⊂ Γ0 (4Nk ) (k = 1, 2, 3).
(2.18)
By Lemma 1 for n = 2 and n = 0, according to (2.17) and (2.18), for all
substitutions from Γ0 (4N ), we have
ατ + β
ϑg′′r hr
; 0, 2Nr = i5η(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
γτ + δ
2βN sgn δ
r
(2.19)
(γτ + δ)5/2 ϑ′′αgr ,hr (τ ; 0, 2Nr ),
×
|δ|
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
ατ + β
; 0, 2Nt = iη(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
γτ + δ
2βN sgn δ
t
(γτ + δ)1/2 ϑαgt ,ht (τ ; 0, 2Nt ),
×
|δ|
ατ + β
ϑg3 h3
; 0, 2N3 = iη(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
γτ + δ
2βN sgn δ
3
(γτ + δ)1/2 ϑαg3 ,h3 (τ ; 0, 2N3 ).
×
|δ|
493
ϑgt ht
(2.20)
(2.21)
Thus for all substitutions from Γ0 (4N ), if r = 1, t = 2 and r = 2, t = 1,
we have
ατ + β
Y
ατ + β
ϑgk hk
; 0, 2Nr
; 0, 2Nk =
ϑg′′r hr
γτ + δ
γτ + δ
k=t,3
2β sgn δ N N N
r t 3
= i7η(γ)(sgn δ−1)/2 i3(1−|δ|)/2
(γτ + δ)7/2 ×
|δ|
|δ|
ατ + β
Y
ατ + β
′′
;
0,
2N
;
0,
2N
ϑ
. (2.22)
×ϑαg
r
k
αg
,h
,h
k
k
r
r
γτ + δ
γτ + δ
k=t,3
It is not difficult to verify that
i3(1−|δ|)/2
2
2
= i3(|δ|−1) /4 .
|δ|
(2.23)
Hence, by (2.1), (2.22), (2.23), (2.13), and (2.16) we get
ατ + β
2
; gl , hl , 0, Nl = i3η(γ)(sgn δ−1)/2 i3(|δ|−1) /4 ×
Ω
γτ + δ
β sgn δ N N N
1 2 3
×
(γτ + δ)7/2 Ω(τ ; αgl , hl , 0, Nl ) =
|δ|
|δ|
= v(M )(γτ + δ)7/2 Ω(τ ; gl , hl , 0, Nl ).
Thus, the function Ω(τ ; gl , hl , 0, Nl ) satisfies condition (2) of the Definition.
3. From (1.6) it follows for r = 1, t = 2 and r = 2, t = 1 that
ϑ′′gr hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt )ϑg3 h3 (τ ; 0, 2N3 ) =
∞
X
= (πi)2
(−1)hr mr +ht mt +h3 m3 (4Nr mr + gr )2 ×
mr ,mt ,m3 =−∞
X
×e
k=r,t,3
∞
X
1
(2Nk mk + gk /2)2 τ =
Bn(r,t) e(nτ ),
4Nk
n=0
(2.24)
494
G. LOMADZE
where
n=
3
X
1
(2Nk mk + gk /2)2 =
4Nk
k=1
=
3
X
3
(Nk m2k + mk gk /2) +
k=1
1X 2
gk /4Nk ,
4
k=1
by (2.14) and (2.15), is a non-negative integer. Thus, by (2.1) and (2.24),
the function Ω(τ ; gl , hl , 0, Nl ) satisfies condition (3) of the Definition.
4. By Lemma 3, the function Ω(τ ; gl , hl , 0, Nl ) satisfies condition (4) of
the Definition.
3.
Lemma 4. For given N let
Ψ1 (τ ; gl , hl , cl , Nl ) = Ψ1 (τ ; g1 , g2 , g3 ; h1 , h2 , h3 ; c1 , c2 , c3 ; N1 , N2 , N3 ) =
n 1
=
ϑ′′ (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
N1 g 1 h 1
o
1 ′′
(3.1)
ϑg2 h2 (τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 ) ϑ′g3 h3 (τ ; c3 , 2N3 )
−
N2
and
Ψ2 (τ ; gl , hl , cl , Nl ) = Ψ2 (τ ; g1 , g2 , g3 ; h1 , h2 , h3 ; c1 , c2 , c3 ; N1 , N2 , N3 ) =
n 1
ϑ′′′ (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
=
N1 g 1 h 1
o
3 ′′
(3.2)
ϑg2 h2 (τ ; c2 , 2N2 )ϑ′g1 h1 (τ ; c1 , 2N1 ) ϑg3 h3 (τ ; c3 , 2N3 ),
−
N2
where
2|gk , Nk |N (k = 1, 2, 3), 4|N
3
X
hk
.
Nk
(3.3)
k=1
Then for all substitutions from Γ, in the neighborhood of each rational point
6 0, (γ, δ) = 1), we have
τ = − γδ (γ =
(γτ + δ)9/2 Ψj (τ ; gl , hl , 0, Nl ) =
∞
X
n ατ + β
(j = 1, 2).
G(j)
n e
4N γτ + δ
n=0
(3.4)
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
495
Proof. 1. Formula (2.6) is obtained in Lemma 3. From Lemma 2, for n = 1,
we have
(γτ + δ)3/2 ϑ′g3 h3 (τ ; 0, 2N3 ) = −e(3 sgn γ/8)(2N3 |γ|)−1/2 (i sgn γ) ×
ατ + β
X
(3.5)
; H3 , 2N3 .
ϕg3′ g3 h3 (0, H3 ; 2N3 )ϑg′ ′ h′
×
3 3
γτ + δ
H3 mod2N3
Multiplying (2.6) by (3.5), according to (3.1), we obtain
(γτ + δ)9/2 Ψ1 (τ ; gl , hl , 0, Nl ) =
= e(9 sgn γ/8)(8N1 N2 N3 |γ|3 )−1/2 (i sgn γ) ×
X
×
3
Y
ϕgk′ gk hk (0, Hk ; 2Nk )Ψ1
H1 mod2N1 k=1
H2 mod2N2
H3 mod2N3
ατ + β
; gl′ , hl′ , Hl , Nl . (3.6)
γτ + δ
Reasoning
further just as in Lemma
3.2, but taking
everywhere
ατ +β
ατ +β
instead of ϑg3 h3 γτ +δ ; H3 , 2N3 , from (3.1) and
γτ +δ ; H3 , 2N3
ϑg′ 3 h3
(3.6) we obtain (3.4) if j = 1.
2. From Lemma 2 for n = 3, n = 0, and respectively for n = 2, n = 1, it
follows that
×
1
(γτ + δ)4 ϑ′′′
g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) =
N1
1
e(sgn γ)(4N1 N2 γ 2 )−1/2 (i sgn γ) ×
=
N1
n
ατ + β
X
ϕg1′ g1 h1 (0, H1 ; 2N1 ) ϑg′′′′ h′
; H1 , 2N1 +
1 1
γτ + δ
H1 mod2N1
3 X
t
o
X
Atk
3
(3−t) ατ + β
; H1 , 2N1
×
ϑg ′ h ′
t
k! z=0 1 1 γτ + δ
t=1
k=1
ατ + β
X
; H2 , 2N2 =
×
ϕg2′ g2 h2 (0, H2 ; 2N2 )ϑg2′ h′2
γτ + δ
H2 mod2N2
X
= e(sgn γ)(i sgn γ)(4N1 N2 γ 2 )−1/2
ϕg1′ g1 h1 (0, H1 ; 2N1 ) ×
+
H1 mod2N1
H2 mod2N2
ατ + β
n 1
ϑg′′′′ h′
; H1 , 2N1 ϑg2′ h2′ (τ ; 0, 2N2 ) +
N1 1 1 γτ + δ
o
ατ + β
(3.7)
−12γπi(γτ + δ)ϑg′ ′ h′
; H1 , 2N1 ϑg2′ h′2 (τ ; 0, 2N2 )
1 1
γτ + δ
×ϕg2′ g2 h2 (0, H2 ; 2N2 )
496
G. LOMADZE
and
3
(γτ + δ)4 ϑ′′g2 h2 (τ ; 0, 2N2 )ϑ′g1 h1 (τ ; 0, 2N1 ) =
N2
X
= e(sgn γ)(i sgn γ)(4N1 N2 γ 2 )−1/2
ϕg1′ g1 h1 (0, H1 ; 2N1 ) ×
H1 mod2N1
H2 mod2N2
n 3
ατ + β
ϑ′′g′ h′
; H2 , 2N2 +
×ϕg2′ g2 h2 (0, H2 ; 2N2 )
N2 2 2 γτ + δ
o
ατ + β
; H2 , 2N2 ϑ′g′ h′ (τ ; 0, 2N1 ).
−12γπi(γτ + δ)ϑ′g′ h′
1 1
2 2
γτ + δ
(3.8)
Subtracting (3.8) from (3.7) and multiplying the obtained result by (2.7),
according to (3.2), we get
(γτ + δ)9/2 Ψ2 (τ ; gl , hl , 0, Nl ) =
= e(9 sgn γ/8)(8N1 N2 N3 |γ|3 )−1/2 (i sgn γ) ×
3
ατ + β
Y
X
ϕgk′ gk hk (0, Hk ; 2Nk )Ψ2
; gl′ , hl′ , Hl , Nl . (3.9)
γτ + δ
×
H1 mod2N1 k=1
H2 mod2N2
H3 mod2N3
In (3.9) let γ be even. Then, reasoning as in Lemma 3.2, by (3.2), from
(3.9) we obtain
ατ + β
ατ + β
n 1
ϑg′′′′ h′
; H1 , 2N1 ϑg2′ h′2
; H2 , 2N2 +
N1 1 1 γτ + δ
γτ + δ
ατ + β
o
3 ′′ ατ + β
− ϑg ′ h ′
; H2 , 2N2 ϑ′g′ h′
; H1 , 2N1
×
1 1
N2 2 2 γτ + δ
γτ + δ
n (πi)3
ατ + β
; H3 , 2N3 =
×
×ϑg3′ h3′
γτ + δ
N1
∞
X
3 N/N1
′
ατ + β
×
(−1)h1 m1 2(H1 + 2N1 m1 ) + g1′ e
×
T1
4N
γτ + δ
m =−∞
1
×
∞
X
m2
−
3(πi)
N1
×
3
∞
X
m2
∞
X
m1
N/N
′
ατ + β
2
(−1)h2 m2 e
+
T2
4N
γτ + δ
=−∞
2 N/N2
′
ατ + β
×
(−1)h2 m2 2(H2 + 2N2 m2 ) + g2′ e
T2
4N
γτ + δ
=−∞
N/N1
′
ατ + β o
(−1)h1 m1 2(H1 + 2N1 m1 ) + g1′ e
×
T1
4N
γτ + δ
=−∞
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
×
∞
X
′
(−1)h3 m3 e
m3 =−∞
=
n
−
×
T3
ατ + β
=
γτ + δ
2
∞
n ατ + β o
n ατ + β X
2
1
Bn′ 1 e
×
Bn′ 2 e
+
δ
4N
γτ
4N
γτ + δ
=0
n =0
∞
X
n2
4N
3
∞
n ατ + β
n ατ + β X
1
2
Bn2 e
+
Bn1 e
+
4N
γτ
δ
4N
γτ + δ
=0
n =0
∞
X
n1
N/N
497
1
∞
n ατ + β
n ατ + β X
3
Cn e
=
.
Bn 3 e
4N γτ + δ
4N γτ + δ
n=0
=0
∞
X
n3
(3.10)
Here Tk and nk have the same meaning as in Lemma 3.2.
Thus, for even γ, (3.4) follows from (3.2), (3.9) and (3.10) if j = 2.
Further, reasoning as in Lemma 3.2, we obtain (3.4) for odd γ if j = 2.
Theorem 2. For given N the functions Ψ1 (τ ; gl , hl , 0, Nl ) and Ψ2 (τ ; gl ,
hl , 0, Nl ) are entire modular forms of weight 9/2 and of the multiplier system
β∆ sgn δ
2
v(M ) = iη(γ)(sgn δ−1)/2 i(|δ|−1) /4
(3.11)
|δ|
α β
is the matrix of the substitution from Γ0 (4N ) and ∆ is the
(M =
γ δ
determinant of an arbitrary positive quadratic form with integer coefficients
in 9 variables) on Γ0 (4N ) if the following conditions hold:
(1)
2|gk , Nk |N (k = 1, 2, 3),
(2)
4|N
3
X
hk2
Nk
k=1
4|
s
X
gk2
,
4Nk
(3.12)
(3.13)
k=1
(3) for all α and δ with αδ ≡ 1 (mod 4N )
N N N
1 2 3
sgn δ
Ψj (τ ; αgl , hl , 0, Nl ) =
|δ|
−∆
=
Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2).
|δ|
(3.14)
Proof. 1. As in the case of Theorem 1, the functions Ψ1 (τ ; gl , hl , 0, Nl ) and
Ψ2 (τ ; gl , hl , 0, Nl ) satisfy condition (1) and, by Lemma 4, also condition (4)
of the Definition.
2. By Lemma 1 for n = 3 and n = 1, according to (2.17) and (2.18), for
all substitutions from Γ0 (4N ), we have
ατ + β
ϑ′′′
; 0, 2N1 = sgn δi7η(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
g1 h1
γτ + δ
498
G. LOMADZE
2βN sgn δ
1
(γτ + δ)7/2 ϑ′′′
αg1 ,h1 (τ ; 0, 2N1 ),
|δ|
ατ + β
ϑg′ 1 h1
; 0, 2N1 = sgn δi3η(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
γτ + δ
2βN sgn δ
1
×
(γτ + δ)3/2 ϑ′αg1 ,h1 (τ ; 0, 2N1 ),
|δ|
ατ + β
; 0, 2N3 = sgn δi3η(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
ϑg′ 3 h3
γτ + δ
2βN sgn δ
3
′
(τ ; 0, 2N3 ).
(γτ + δ)3/2 ϑαg
×
3 ,h3
|δ|
×
(3.15)
(3.16)
(3.17)
Thus, for all substitutions from Γ0 (4N ) we respectively get:
(1) by (2.19), (2.20), and (3.17)
ατ + β
ατ + β
ατ + β
; 0, 2Nr ϑgt ht
; 0, 2Nt ϑg′ 3 h3
; 0, 2N3 =
ϑ′′gr hr
γτ + δ
γτ + δ
γτ + δ
2β sgn δ N N N
1 2 3
= sgn δiη(γ)(sgn δ−1)/2 i3(1−|δ|)/2
×
|δ|
|δ|
×(γτ + δ)9/2 ϑ′′αgr ,hr (τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt )ϑ′αg3 ,h3 (τ ; 0, 2N3 )
(3.18)
for r = 1, t = 2 and r = 2, t = 1;
(2) by (3.15), (2.20) for t = 2 and (2.21)
ατ + β
ατ + β
ατ + β
ϑg′′′1 h1
; 0, 2N1 ϑg2 h2
; 0, 2N2 ϑg3 h3
; 0, 2N3 =
γτ + δ
γτ + δ
γτ + δ
2β sgn δ N N N
1 2 3
×
= sgn δiη(γ)(sgn δ−1)/2 i3(1−|δ|)/2
|δ|
|δ|
′′′
(τ ; 0, 2N1 )ϑαg2 ,h2 (τ ; 0, 2N2 )ϑαg3 ,h3 (τ ; 0, 2N3 ); (3.19)
×(γτ + δ)9/2 ϑαg
1 ,h1
(3) by (2.19) for r = 2 , (3.16), and (2.21)
ατ + β
ατ + β
ατ + β
; 0, 2N2 ϑ′g1 h1
; 0, 2N1 ϑg3 h3
; 0, 2N3 =
ϑ′′g2 h2
γτ + δ
γτ + δ
γτ + δ
N N N
1 2 3
η(γ)(sgn δ−1)/2 3(1−|δ|)/2 2β sgn δ
= sgn δi
i
×
|δ|
|δ|
′′
(τ ; 0, 2N2 )ϑ′αg1 ,h1 (τ ; 0, 2N1 )ϑαg3 ,h3 (τ ; 0, 2N3 ). (3.20)
×(γτ + δ)9/2 ϑαg
2 ,h2
It is not difficult to verify that
−2
2
i3(1−|δ|)/2
= i(|δ|−1) /4 .
|δ|
(3.21)
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
499
Hence, by (3.1), (3.18), (3.21), (3.11), (3.14) (if j = 1) and (3.2), (3.19)–
(3.21), (3.11), (3.14) (if j = 2) we get
ατ + β
2
Ψj
; gl , hl , 0, Nl = sgn δiη(γ)(sgn δ−1)/2 i(|δ|−1) /4 ×
γτ + δ
−β sgn δ N N N
1 2 3
(γτ + δ)9/2 Ψj (τ ; αgl , hl , 0, Nl ) =
×
|δ|
|δ|
= v(M )(γτ + δ)9/2 Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2).
Thus the functions Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) satisfy condition (2) of
the Definition.
From (1.6) it follows that
ϑ′′gr hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt )ϑ′g3 h3 (τ ; 0, 2N3 ) =
P
∞
X
h m
(−1) k=r,t,3 k k (4Nr mr + gr )2 (4N3 m3 + g3 ) ×
= (πi)3
mr ,mt ,m3 =−∞
X
×e
k=r,t,3
∞
X
1
Cn(r,t) e(nτ )
(2Nk mk + gk /2)2 τ =
4Nk
n=0
for r = 1, t = 2 and r = 2, t = 1,
ϑ′′′
g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 )ϑg3 h3 (τ ; 0, 2N3 ) =
∞
P3
X
= (πi)3
(−1) k=1 hk mk (4N1 m1 + g1 )3 ×
m1 ,m2 ,m3 =−∞
3
∞
X
X
1
Cn e(nτ )
×e
(2Nk mk + gk /2)2 τ =
4Nk
n=0
k=1
and
= (πi)3
ϑ′′g2 h2 (τ ; 0, 2N2 )ϑ′g1 h1 (τ ; 0, 2N1 )ϑg3 h3 (τ ; 0, 2N3 ) =
∞
P3
X
(−1) k=1 hk mk (4N2 m2 + g2 )2 (4N1 m1 + g1 ) ×
m1 ,m2 ,m3 =−∞
×e
∞
3
X
X
1
(2Nk mk + gk /2)2 τ =
Dn e(nτ ),
4Nk
n=0
k=1
since in all these expansions
n=
3
X
1
(2Nk mk + gk /2)2 ,
4Nk
k=1
500
G. LOMADZE
as has been shown by Theorem 1.3, is a non-negative integer. Therefore,
in view of (3.1) and (3.2), the functions Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) satisfy
condition (3) of the Definition.
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 24.11.1994)
Author’s address:
Faculty of Mechanics and Mathematics
I. Javakhishvili Tbilisi State University
2, University St., Tbilisi 380043
Republic of Georgia
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
AND 29 FOR THE CONGRUENCE GROUP Γ0 (4N )
7
2
G. LOMADZE
Abstract. Entire modular forms of weights 27 and 29 for the congruence group 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 7 and 9
variables.
1.
In this paper N, a, k, q, r, t denote positive integers; u, s are odd positive
integers; H, c, g, h, j, m, n, α, β, γ, δ, ξ are integers; A, B, C, D, G are complex
numbers; z, τ (Im τ > 0) are complex variables,and
H = {τ ∈ C| Im τ > 0}.
Further, ( uh ) is the generalized Jacobi symbol; nt is a binomial coefficient;
ϕ(k) is Euler’s function; e(z) = exp 2πiz; η(γ) = 1 if γ ≥ 0 and η(γ) = −1
if γ < 0.
Let
X
ϑgh (z|τ ; c, N ) =
(−1)h(m−c)/N ×
m≡c (mod N )
×e
hence
1
g
g 2
τ e m+ z ;
m+
2N
2
2
(1.1)
X
∂n
n
;
c,
N
)
=
(πi)
(−1)h(m−c)/N (2m + g)n ×
ϑ
(z|τ
gh
∂z n
m≡c (mod N )
1
g 2
g
×e
m+
τ e m + z (n ≥ 0).
(1.2)
2N
2
2
1991 Mathematics Subject Classification. 11F37, 11F27, 11F11.
Key words and phrases. Entire modular form, congruence subgroup, multiplier system, theta-function, half integral weight.
485
c 1997 Plenum Publishing Corporation
1072-947X/96/0900-0485$12.50/0
486
G. LOMADZE
Put
(n)
ϑgh (τ ; c, N ) =
∂n
ϑgh (z|τ ; c, N )z=0 (n ≥ 0).
∂z n
(1.3)
It is known (see, for e.g., [1], formulas (11) and (12)) that
(n)
(n)
ϑg,h+2j (τ ; c, N ) = ϑgh (τ ; c, N ) (n ≥ 0),
β g
2
(n)
(n)
ϑg,h+βg+βN (τ ; c, N ) (n ≥ 0),
c+
ϑgh (τ + β; c, N ) = e
2N
2
β
g 2
(n)
ϑgh (τ − β; c, N ) = (−1)n e −
c+
×
2N
2
(n)
× ϑ−g,−h+βg−βN (τ ; −c, N ) (n ≥ 0).
(1.4)
(1.5)
From (1.1) and (1.2), according to notation (1.3) it follows, in particular,
that
(n)
ϑgh (τ ; 0, N ) = (πi)n
∞
X
(−1)hm (2N m + g)n ×
m=−∞
1
g 2
τ (n ≥ 0).
Nm +
×e
2N
2
(1.6)
For ξ2 6= 0, ξ1 , g, h, N with ξ1 g + ξ2 h + ξ1 ξ2 N ≡ 0 (mod 2), put
Sgh
ξ1
; c, N =
ξ2
X
mmodN |ξ2 |
m≡c (mod N )
ξ
g 2
1
(−1)h(m−c)/N e
m+
.
2N ξ2
2
Finally, let
n ατ + β
γτ + δ
n ατ + β
Γ0 (4N ) =
γτ + δ
Γ=
For τ ∈ H put
o
αδ − βγ = 1 ,
o
∈ Γ γ ≡ 0 (mod 4N ) .
s
(γτ + δ)s/2 = (γτ + δ)1/2 ,
−
π
π
< arg(γτ + δ)1/2 ≤ .
2
2
Definition. Let M be a matrix of an arbitrary substitution from Γ0 (4N ),
and let v(M ) be a multiplier system on Γ0 (4N ) and of weight 2s . We shall
say that a function F defined on H is an entire modular form of weight 2s
and of multiplier system v(M ) on Γ0 (4N ), if
(1) F is regular on H;
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
487
α β
of substitutions from Γ0 (4N ) and all
(2) for all matrices M =
γ δ
τ ∈ H,
ατ + β
F
= v(M )(γτ + δ)s/2 F (τ );
γτ + δ
(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),
(γτ + δ)s/2 F (τ ) =
∞
X
m ατ + β
.
A′m e
4N γτ + δ
m=0
Lemma 1 ([1], p. 58, Lemma 3). If g is even, then for n ≥ 0 and all
substitutions from Γ0 (4N ) we have
(n) ατ + β
ϑgh
; 0, 2N =
γτ + δ
2βN sgn δ
(γτ + δ)(2n+1)/2 ×
= (sgn δ)n i(2n+1)η(γ)(sgn δ−1)/2 i(1−|δ|)/2
|δ|
αγδ 2 h2 βδg 2 δ 2ϕ(2N )−2
(n)
e
ϑαg,h (τ ; 0, 2N ).
×e −
16N
4
4N
Lemma 2 ([1], p. 61, Lemma 4). If γ 6= 0, then for n ≥ 0
(n)
(γτ + δ)(2n+1)/2 ϑgh (τ ; 0, 2N ) =
−1/2
n
= e (2n + 1) sgn γ/8 2N |γ|
− i sgn γ ×
n
ατ + β
X
(n)
; H, 2N +
×
ϕg′ gh (0, H; 2N ) ϑg′ h′
γτ + δ
H mod2N
t
n X
o
X
Atk
n
(n−t) ατ + β
; H, 2N ,
· ϑg′ h′
+
t
k! z=0
γτ + δ
t=1
k=1
where
g ′ = δg − γh − 2γδN, h′ = −βg + αh − 2αβN,
αβ
g ′ 2 βg
g ′
e −
×
H+
H+
ϕg′ gh (0, H; 2N ) = e
4N
2
4N
2
δ
; −αH, 2N ,
× Sg−αg′ ,h−βg′
−γ
(1.7)
488
G. LOMADZE
Atk z=0 =
(
(2k)!(−2N γπi(γτ + δ))k
0
if t = 2k
if t =
6 2k
(1.8)
(t = 1, 2, . . . , n; k = 1, 2, . . . , t).
From (1.8) it follows that
t
X
Atk
=
k! z=0
k=1
(A
t,t/2
(t/2)! z=0
0
if 2|t,
if 2 ∤ t.
(1.9)
2.
Lemma 3. For given N let
where
Ω(τ ; gl , hl , cl , Nl ) =
= Ω(τ ; g1 , g2 , g3 ; h1 , h2 , h3 ; c1 , c2 , c3 ; N1 , N2 , N3 ) =
n 1
=
ϑ′′ (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
N1 g1 h1
o
1 ′′
−
ϑg2 h2 (τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 ) ϑg3 h3 (τ ; c3 , 2N3 ),
N2
3
X
hk
.
2|gk , Nk |N (k = 1, 2, 3), 4|N
Nk
(2.1)
(2.2)
k=1
Then for all substitutions from Γ in the neighborhood of each rational point
τ = − γδ (γ =
6 0, (γ, δ) = 1), we have
(γτ + δ)7/2 Ω(τ ; gl , hl , 0, Nl ) =
∞
X
n ατ + β
Gn e
.
4N γτ + δ
n=0
(2.3)
Proof. 1. From Lemma 2 for n = 2 and n = 0 (respectively with g1 , h1 , N1 ,
g1′ , h1′ , H1 and g2 , h2 , N2 , g2′ , h′2 , H2 instead of g, h, N, g ′ , h′ , H), according
to (1.8)–(1.9), it follows that
1
(γτ + δ)3 ϑ′′g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) =
N1
3
= −e sgn γ (4N1 N2 γ 2 )−1/2 ×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
×
H1 mod2N1
H2 mod2N2
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
×
ατ + β
n 1
ατ + β
ϑ′′g′ h′
; H1 , 2N1 ϑg2′ h2′
; H2 , 2N2 +
N1 1 1 γτ + δ
γτ + δ
ατ + β
−4γπi(γτ + δ)ϑg1′ h′1
; H1 , 2N1 ×
γτ + δ
o
ατ + β
; H2 , 2N2 .
×ϑg2′ h′2
γτ + δ
489
(2.4)
Replacing N1 , g1 , h1 , H1 , g1′ , h1′ by N2 , g2 , h2 , H2 , g2′ , h′2 in (2.4), and vice
versa, we obtain
1
(γτ + δ)3 ϑ′′g2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 ) =
N2
3
= −e sgn γ (4N1 N2 γ 2 )−1/2 ×
4
X
′
ϕg2 g2 h2 (0, H2 ; 2N2 )ϕg1′ g1 h1 (0, H1 ; 2N1 ) ×
×
H1 mod2N1
H2 mod2N2
n 1
ατ + β
ατ + β
×
ϑ′′g′ h′
; H2 , 2N2 ϑg1′ h1′
; H1 , 2N1 +
N2 2 2 γτ + δ
γτ + δ
ατ + β
−4γπi(γτ + δ)ϑg2′ h′2
; H2 , 2N2 ×
γτ + δ
o
ατ + β
; H1 , 2N1 .
×ϑg1′ h′1
γτ + δ
(2.5)
Subtracting (2.5) from (2.4), we obtain
n 1
(γτ + δ)3
ϑ′′ (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) +
N1 g1 h1
o
1
− ϑ′′g2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 ) =
N2
3
= −e sgn γ (4N1 N2 γ 2 )−1/2 ×
4
X
ϕg1′ g1 h1 (0, H1 ; 2N1 )ϕg2′ g2 h2 (0, H2 ; 2N2 ) ×
×
H1 mod2N1
H2 mod2N2
n 1
ατ + β
ατ + β
×
ϑ′′g′ h′
; H1 , 2N1 ϑg2′ h2′
; H2 , 2N2 +
N1 1 1 γτ + δ
γτ + δ
o
1 ′′
ατ + β
ατ + β
ϑg ′ h ′
−
; H2 , 2N2 ϑg1′ h′1
; H1 , 2N1 .
N2 2 2 γτ + δ
γτ + δ
From Lemma 2 for n = 0, we have
(γτ + δ)1/2 ϑg3 h3 (τ ; 0, 2N3 ) = e sgn γ/8 (2N3 |γ|)−1/2 ×
(2.6)
490
G. LOMADZE
X
×
ϕg3′ g3 h3 (0, H3 ; 2N3 )ϑg3′ h3′
H3 mod2N3
ατ + β
; H3 , 2N3 .
γτ + δ
(2.7)
Multiplying (2.6) by (2.7), according to (2.1), we obtain
(γτ + δ)7/2 Ω(τ ; gl , hl , 0, Nl ) =
= −e(7 sgn γ/8)(8N1 N2 N3 |γ|3 )−1/2 ×
3
Y
X
×
ϕgk′ gk hk (0, Hk ; 2Nk )Ω
H1 mod2N1 k=1
H2 mod2N2
H3 mod2N3
ατ + β
; gl′ , h′l , Hl , Nl .
γτ + δ
(2.8)
2. In (2.8) let γ be even. Then, by (1.7), gk′ (k = 1, 2, 3) are also
even. Now in (1.2) and (1.3) instead of m let us introduce new letters of
summations mk defined by the equalities m − Hk = 2Nk mk (k = r, t, 3),
respectively, and for brevity we put
g ′ 2
Tk = Hk + 2Nk mk + k
(k = r, t, 3).
2
Then, by (2.1), in (2.8) for r = 1, t = 2 and r = 2, t = 1, we get
ϑg′′r′ hr′
ατ + β
ατ + β
Y
; Hr , 2Nr
; Hk , 2Nk =
ϑgk′ hk′
γτ + δ
γτ + δ
k=t,3
= (πi)2
∞
X
mr
2 N/Nr
′
ατ + β
(−1)hr mr 2(Hr + 2Nr mr ) + gr′ e
×
Tr
4N
γτ + δ
=−∞
×
∞
X
mt
×
N/N
′
ατ + β
t
(−1)ht mt e
×
Tt
4N
γτ + δ
=−∞
∞
X
m3 =−∞
=
×
N/N
4N
3
T3
ατ + β
=
γτ + δ
∞
n ατ + β X
n ατ + β
r
t
Bnt e
×
Bn r e
δ
4N
γτ
+
4N
γτ + δ
n =0
=0
∞
X
nr
∞
X
n3
′
(−1)h3 m3 e
t
∞
n ατ + β
n ατ + β X
3
Bn 3 e
=
,
Cn(r,t) e
4N γτ + δ
4N γτ + δ
=0
n=0
(2.9)
since by (2.2) nk = NNk Tk are non-negative integers. Thus, for even γ, (2.3)
follows from (2.1),(2.8), and (2.9).
In (2.8) let now γ be odd. If h1 , h2 , h3 are even, then by (1.7), g1′ , g2′ ,
′
g3 are also even, and we obtain the same result. But if hr is odd, then by
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
491
(1.7), hr′ and gr′ will also be odd, and in (1.2) we shall have
2
2
m + gr′ /2 = m + (gr′ − 1)/2 + m + (gr′ − 1)/2 + 1/4.
Hence
ατ + β
h′ ατ + β
r
×
; Hr , 2Nr = (πi)2 e
γτ + δ
16Nr γτ + δ
∞
X
2 N/Nr
′
ατ + β
Wr
(r = 1, 2).
(−1)hr mr 2(Hr + 2Nr mr )+gr′ e
×
4N
γτ + δ
m =−∞
ϑ′′gr′ h′r
r
Here by (1.4) we can imply that h′r = 1, since by (1.7), h′r and hr have the
same parity, and
2
Wr = Hr + 2Nr mr + (gr′ − 1)/2 +
+ Hr + 2Nr mr + (gr′ − 1)/2 (r = 1, 2).
(2.10)
Analogously,
h′ ατ + β
ατ + β
t
; Ht , 2Nt = e
×
γτ + δ
16Nt γτ + δ
∞
N/N
X
′
ατ + β
t
(−1)ht mt e
(t = 2, 1).
Wt
×
4N
γτ + δ
m =−∞
ϑgt′ h′t
t
The same formula is valid for t = 3 with the same remarks concerning h3′
as for ht′ , t = 2, 1. Further, for k = t, 3 we have
Tk
if 2|gk′ ,
2
′
(2.11)
Wk = (Hk + 2Nk mk + (gk − 1)/2) +
′
′
if 2 ∤ gk .
(Hk + 2Nk mk + (gk − 1)/2)
Thus, if among h1′ , h′2 , h′3 at least one is odd, then, as in (2.9), for r = 1,
t = 2 and r = 2, t = 1 we have
ατ + β
Y
ατ + β
ϑgk′ hk′
; Hr , 2Nr
; Hk , 2Nk =
ϑ′′gr′ hr′
γτ + δ
γτ + δ
k=t,3
3
∞
1 1 X
hk′ ατ + β X ′ nr ατ + β
N
=e
×
B e
4N 4
Nk γτ + δ n =0 nr 4N γτ + δ
k=1
∞
X
r
∞
n ατ + β
n ατ + β X
t
3
×
Bn′ 3 e
Bn′ t e
=
4N
γτ
+
4N
γτ + δ
δ
n =0
n =0
t
3
3
∞
1 1 X
h′k ατ + β X (r,t) n′ ατ + β
Dn′ e
=e
N
=
4N 4
Nk γτ + δ
4N γτ + δ
′
k=1
n =0
492
G. LOMADZE
=
∞
X
Dn(r,t) e
n=0
n ατ + β
,
4N γτ + δ
since, by (2.2), (2.10), and (2.11), nk =
N
Nk Wk
and n = 14 N
(2.12)
3
P
k=1
h′k
Nk
+ n′ are
non-negative integers. Thus, for odd γ, (2.3) follows from (2.1), (2.8), and
(2.12).
Theorem 1. For given N the function Ω(τ ; gl , hl , 0, Nl ) is an entire
modular form of weight 7/2 and of the multiplier system
β∆ sgn δ
2
(2.13)
v(M ) = i3η(γ)(sgn δ−1)/2 i3(|δ|−1) /4
|δ|
α β
is a matrix of the substitution from Γ0 (4N ) and ∆ is the
(M =
γ δ
determinant of an arbitrary positive quadratic form with integer coefficients
in 7 variables) on Γ0 (4N ) if the following conditions hold:
(1)
(2)
2|gk , Nk |N (k = 1, 2, 3),
4|N
3
X
k=1
h2k
Nk
4|
s
X
k=1
(2.14)
gk2
,
4Nk
(2.15)
3) for all α and δ with αδ ≡ 1 (mod 4N )
∆
N N N
1 2 3
Ω(τ ; αgl , hl , 0, Nl ) =
Ω(τ ; gl , hl , 0, Nl ).
|δ|
|δ|
(2.16)
Proof. 1. It is well known that theta-series (1.1)–(1.2) are regular on H;
hence the function Ω(τ ; gl , hl , 0, Nl ) satisfies condition (1) of the Definition.
2. It is easily verified that (2.15) implies
4|N δ 2
3
X
k=1
h2k /Nk , 4|
3
X
δ 2ϕ(2Nk )−2 gk2 /4Nk ,
(2.17)
k=1
since 2 ∤ δ, because αδ ≡ 1 (mod 4N ).
From (2.14) it follows that
Γ0 (4N ) ⊂ Γ0 (4Nk ) (k = 1, 2, 3).
(2.18)
By Lemma 1 for n = 2 and n = 0, according to (2.17) and (2.18), for all
substitutions from Γ0 (4N ), we have
ατ + β
ϑg′′r hr
; 0, 2Nr = i5η(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
γτ + δ
2βN sgn δ
r
(2.19)
(γτ + δ)5/2 ϑ′′αgr ,hr (τ ; 0, 2Nr ),
×
|δ|
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
ατ + β
; 0, 2Nt = iη(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
γτ + δ
2βN sgn δ
t
(γτ + δ)1/2 ϑαgt ,ht (τ ; 0, 2Nt ),
×
|δ|
ατ + β
ϑg3 h3
; 0, 2N3 = iη(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
γτ + δ
2βN sgn δ
3
(γτ + δ)1/2 ϑαg3 ,h3 (τ ; 0, 2N3 ).
×
|δ|
493
ϑgt ht
(2.20)
(2.21)
Thus for all substitutions from Γ0 (4N ), if r = 1, t = 2 and r = 2, t = 1,
we have
ατ + β
Y
ατ + β
ϑgk hk
; 0, 2Nr
; 0, 2Nk =
ϑg′′r hr
γτ + δ
γτ + δ
k=t,3
2β sgn δ N N N
r t 3
= i7η(γ)(sgn δ−1)/2 i3(1−|δ|)/2
(γτ + δ)7/2 ×
|δ|
|δ|
ατ + β
Y
ατ + β
′′
;
0,
2N
;
0,
2N
ϑ
. (2.22)
×ϑαg
r
k
αg
,h
,h
k
k
r
r
γτ + δ
γτ + δ
k=t,3
It is not difficult to verify that
i3(1−|δ|)/2
2
2
= i3(|δ|−1) /4 .
|δ|
(2.23)
Hence, by (2.1), (2.22), (2.23), (2.13), and (2.16) we get
ατ + β
2
; gl , hl , 0, Nl = i3η(γ)(sgn δ−1)/2 i3(|δ|−1) /4 ×
Ω
γτ + δ
β sgn δ N N N
1 2 3
×
(γτ + δ)7/2 Ω(τ ; αgl , hl , 0, Nl ) =
|δ|
|δ|
= v(M )(γτ + δ)7/2 Ω(τ ; gl , hl , 0, Nl ).
Thus, the function Ω(τ ; gl , hl , 0, Nl ) satisfies condition (2) of the Definition.
3. From (1.6) it follows for r = 1, t = 2 and r = 2, t = 1 that
ϑ′′gr hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt )ϑg3 h3 (τ ; 0, 2N3 ) =
∞
X
= (πi)2
(−1)hr mr +ht mt +h3 m3 (4Nr mr + gr )2 ×
mr ,mt ,m3 =−∞
X
×e
k=r,t,3
∞
X
1
(2Nk mk + gk /2)2 τ =
Bn(r,t) e(nτ ),
4Nk
n=0
(2.24)
494
G. LOMADZE
where
n=
3
X
1
(2Nk mk + gk /2)2 =
4Nk
k=1
=
3
X
3
(Nk m2k + mk gk /2) +
k=1
1X 2
gk /4Nk ,
4
k=1
by (2.14) and (2.15), is a non-negative integer. Thus, by (2.1) and (2.24),
the function Ω(τ ; gl , hl , 0, Nl ) satisfies condition (3) of the Definition.
4. By Lemma 3, the function Ω(τ ; gl , hl , 0, Nl ) satisfies condition (4) of
the Definition.
3.
Lemma 4. For given N let
Ψ1 (τ ; gl , hl , cl , Nl ) = Ψ1 (τ ; g1 , g2 , g3 ; h1 , h2 , h3 ; c1 , c2 , c3 ; N1 , N2 , N3 ) =
n 1
=
ϑ′′ (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
N1 g 1 h 1
o
1 ′′
(3.1)
ϑg2 h2 (τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 ) ϑ′g3 h3 (τ ; c3 , 2N3 )
−
N2
and
Ψ2 (τ ; gl , hl , cl , Nl ) = Ψ2 (τ ; g1 , g2 , g3 ; h1 , h2 , h3 ; c1 , c2 , c3 ; N1 , N2 , N3 ) =
n 1
ϑ′′′ (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
=
N1 g 1 h 1
o
3 ′′
(3.2)
ϑg2 h2 (τ ; c2 , 2N2 )ϑ′g1 h1 (τ ; c1 , 2N1 ) ϑg3 h3 (τ ; c3 , 2N3 ),
−
N2
where
2|gk , Nk |N (k = 1, 2, 3), 4|N
3
X
hk
.
Nk
(3.3)
k=1
Then for all substitutions from Γ, in the neighborhood of each rational point
6 0, (γ, δ) = 1), we have
τ = − γδ (γ =
(γτ + δ)9/2 Ψj (τ ; gl , hl , 0, Nl ) =
∞
X
n ατ + β
(j = 1, 2).
G(j)
n e
4N γτ + δ
n=0
(3.4)
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
495
Proof. 1. Formula (2.6) is obtained in Lemma 3. From Lemma 2, for n = 1,
we have
(γτ + δ)3/2 ϑ′g3 h3 (τ ; 0, 2N3 ) = −e(3 sgn γ/8)(2N3 |γ|)−1/2 (i sgn γ) ×
ατ + β
X
(3.5)
; H3 , 2N3 .
ϕg3′ g3 h3 (0, H3 ; 2N3 )ϑg′ ′ h′
×
3 3
γτ + δ
H3 mod2N3
Multiplying (2.6) by (3.5), according to (3.1), we obtain
(γτ + δ)9/2 Ψ1 (τ ; gl , hl , 0, Nl ) =
= e(9 sgn γ/8)(8N1 N2 N3 |γ|3 )−1/2 (i sgn γ) ×
X
×
3
Y
ϕgk′ gk hk (0, Hk ; 2Nk )Ψ1
H1 mod2N1 k=1
H2 mod2N2
H3 mod2N3
ατ + β
; gl′ , hl′ , Hl , Nl . (3.6)
γτ + δ
Reasoning
further just as in Lemma
3.2, but taking
everywhere
ατ +β
ατ +β
instead of ϑg3 h3 γτ +δ ; H3 , 2N3 , from (3.1) and
γτ +δ ; H3 , 2N3
ϑg′ 3 h3
(3.6) we obtain (3.4) if j = 1.
2. From Lemma 2 for n = 3, n = 0, and respectively for n = 2, n = 1, it
follows that
×
1
(γτ + δ)4 ϑ′′′
g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) =
N1
1
e(sgn γ)(4N1 N2 γ 2 )−1/2 (i sgn γ) ×
=
N1
n
ατ + β
X
ϕg1′ g1 h1 (0, H1 ; 2N1 ) ϑg′′′′ h′
; H1 , 2N1 +
1 1
γτ + δ
H1 mod2N1
3 X
t
o
X
Atk
3
(3−t) ατ + β
; H1 , 2N1
×
ϑg ′ h ′
t
k! z=0 1 1 γτ + δ
t=1
k=1
ατ + β
X
; H2 , 2N2 =
×
ϕg2′ g2 h2 (0, H2 ; 2N2 )ϑg2′ h′2
γτ + δ
H2 mod2N2
X
= e(sgn γ)(i sgn γ)(4N1 N2 γ 2 )−1/2
ϕg1′ g1 h1 (0, H1 ; 2N1 ) ×
+
H1 mod2N1
H2 mod2N2
ατ + β
n 1
ϑg′′′′ h′
; H1 , 2N1 ϑg2′ h2′ (τ ; 0, 2N2 ) +
N1 1 1 γτ + δ
o
ατ + β
(3.7)
−12γπi(γτ + δ)ϑg′ ′ h′
; H1 , 2N1 ϑg2′ h′2 (τ ; 0, 2N2 )
1 1
γτ + δ
×ϕg2′ g2 h2 (0, H2 ; 2N2 )
496
G. LOMADZE
and
3
(γτ + δ)4 ϑ′′g2 h2 (τ ; 0, 2N2 )ϑ′g1 h1 (τ ; 0, 2N1 ) =
N2
X
= e(sgn γ)(i sgn γ)(4N1 N2 γ 2 )−1/2
ϕg1′ g1 h1 (0, H1 ; 2N1 ) ×
H1 mod2N1
H2 mod2N2
n 3
ατ + β
ϑ′′g′ h′
; H2 , 2N2 +
×ϕg2′ g2 h2 (0, H2 ; 2N2 )
N2 2 2 γτ + δ
o
ατ + β
; H2 , 2N2 ϑ′g′ h′ (τ ; 0, 2N1 ).
−12γπi(γτ + δ)ϑ′g′ h′
1 1
2 2
γτ + δ
(3.8)
Subtracting (3.8) from (3.7) and multiplying the obtained result by (2.7),
according to (3.2), we get
(γτ + δ)9/2 Ψ2 (τ ; gl , hl , 0, Nl ) =
= e(9 sgn γ/8)(8N1 N2 N3 |γ|3 )−1/2 (i sgn γ) ×
3
ατ + β
Y
X
ϕgk′ gk hk (0, Hk ; 2Nk )Ψ2
; gl′ , hl′ , Hl , Nl . (3.9)
γτ + δ
×
H1 mod2N1 k=1
H2 mod2N2
H3 mod2N3
In (3.9) let γ be even. Then, reasoning as in Lemma 3.2, by (3.2), from
(3.9) we obtain
ατ + β
ατ + β
n 1
ϑg′′′′ h′
; H1 , 2N1 ϑg2′ h′2
; H2 , 2N2 +
N1 1 1 γτ + δ
γτ + δ
ατ + β
o
3 ′′ ατ + β
− ϑg ′ h ′
; H2 , 2N2 ϑ′g′ h′
; H1 , 2N1
×
1 1
N2 2 2 γτ + δ
γτ + δ
n (πi)3
ατ + β
; H3 , 2N3 =
×
×ϑg3′ h3′
γτ + δ
N1
∞
X
3 N/N1
′
ατ + β
×
(−1)h1 m1 2(H1 + 2N1 m1 ) + g1′ e
×
T1
4N
γτ + δ
m =−∞
1
×
∞
X
m2
−
3(πi)
N1
×
3
∞
X
m2
∞
X
m1
N/N
′
ατ + β
2
(−1)h2 m2 e
+
T2
4N
γτ + δ
=−∞
2 N/N2
′
ατ + β
×
(−1)h2 m2 2(H2 + 2N2 m2 ) + g2′ e
T2
4N
γτ + δ
=−∞
N/N1
′
ατ + β o
(−1)h1 m1 2(H1 + 2N1 m1 ) + g1′ e
×
T1
4N
γτ + δ
=−∞
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
×
∞
X
′
(−1)h3 m3 e
m3 =−∞
=
n
−
×
T3
ατ + β
=
γτ + δ
2
∞
n ατ + β o
n ατ + β X
2
1
Bn′ 1 e
×
Bn′ 2 e
+
δ
4N
γτ
4N
γτ + δ
=0
n =0
∞
X
n2
4N
3
∞
n ατ + β
n ατ + β X
1
2
Bn2 e
+
Bn1 e
+
4N
γτ
δ
4N
γτ + δ
=0
n =0
∞
X
n1
N/N
497
1
∞
n ατ + β
n ατ + β X
3
Cn e
=
.
Bn 3 e
4N γτ + δ
4N γτ + δ
n=0
=0
∞
X
n3
(3.10)
Here Tk and nk have the same meaning as in Lemma 3.2.
Thus, for even γ, (3.4) follows from (3.2), (3.9) and (3.10) if j = 2.
Further, reasoning as in Lemma 3.2, we obtain (3.4) for odd γ if j = 2.
Theorem 2. For given N the functions Ψ1 (τ ; gl , hl , 0, Nl ) and Ψ2 (τ ; gl ,
hl , 0, Nl ) are entire modular forms of weight 9/2 and of the multiplier system
β∆ sgn δ
2
v(M ) = iη(γ)(sgn δ−1)/2 i(|δ|−1) /4
(3.11)
|δ|
α β
is the matrix of the substitution from Γ0 (4N ) and ∆ is the
(M =
γ δ
determinant of an arbitrary positive quadratic form with integer coefficients
in 9 variables) on Γ0 (4N ) if the following conditions hold:
(1)
2|gk , Nk |N (k = 1, 2, 3),
(2)
4|N
3
X
hk2
Nk
k=1
4|
s
X
gk2
,
4Nk
(3.12)
(3.13)
k=1
(3) for all α and δ with αδ ≡ 1 (mod 4N )
N N N
1 2 3
sgn δ
Ψj (τ ; αgl , hl , 0, Nl ) =
|δ|
−∆
=
Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2).
|δ|
(3.14)
Proof. 1. As in the case of Theorem 1, the functions Ψ1 (τ ; gl , hl , 0, Nl ) and
Ψ2 (τ ; gl , hl , 0, Nl ) satisfy condition (1) and, by Lemma 4, also condition (4)
of the Definition.
2. By Lemma 1 for n = 3 and n = 1, according to (2.17) and (2.18), for
all substitutions from Γ0 (4N ), we have
ατ + β
ϑ′′′
; 0, 2N1 = sgn δi7η(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
g1 h1
γτ + δ
498
G. LOMADZE
2βN sgn δ
1
(γτ + δ)7/2 ϑ′′′
αg1 ,h1 (τ ; 0, 2N1 ),
|δ|
ατ + β
ϑg′ 1 h1
; 0, 2N1 = sgn δi3η(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
γτ + δ
2βN sgn δ
1
×
(γτ + δ)3/2 ϑ′αg1 ,h1 (τ ; 0, 2N1 ),
|δ|
ατ + β
; 0, 2N3 = sgn δi3η(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
ϑg′ 3 h3
γτ + δ
2βN sgn δ
3
′
(τ ; 0, 2N3 ).
(γτ + δ)3/2 ϑαg
×
3 ,h3
|δ|
×
(3.15)
(3.16)
(3.17)
Thus, for all substitutions from Γ0 (4N ) we respectively get:
(1) by (2.19), (2.20), and (3.17)
ατ + β
ατ + β
ατ + β
; 0, 2Nr ϑgt ht
; 0, 2Nt ϑg′ 3 h3
; 0, 2N3 =
ϑ′′gr hr
γτ + δ
γτ + δ
γτ + δ
2β sgn δ N N N
1 2 3
= sgn δiη(γ)(sgn δ−1)/2 i3(1−|δ|)/2
×
|δ|
|δ|
×(γτ + δ)9/2 ϑ′′αgr ,hr (τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt )ϑ′αg3 ,h3 (τ ; 0, 2N3 )
(3.18)
for r = 1, t = 2 and r = 2, t = 1;
(2) by (3.15), (2.20) for t = 2 and (2.21)
ατ + β
ατ + β
ατ + β
ϑg′′′1 h1
; 0, 2N1 ϑg2 h2
; 0, 2N2 ϑg3 h3
; 0, 2N3 =
γτ + δ
γτ + δ
γτ + δ
2β sgn δ N N N
1 2 3
×
= sgn δiη(γ)(sgn δ−1)/2 i3(1−|δ|)/2
|δ|
|δ|
′′′
(τ ; 0, 2N1 )ϑαg2 ,h2 (τ ; 0, 2N2 )ϑαg3 ,h3 (τ ; 0, 2N3 ); (3.19)
×(γτ + δ)9/2 ϑαg
1 ,h1
(3) by (2.19) for r = 2 , (3.16), and (2.21)
ατ + β
ατ + β
ατ + β
; 0, 2N2 ϑ′g1 h1
; 0, 2N1 ϑg3 h3
; 0, 2N3 =
ϑ′′g2 h2
γτ + δ
γτ + δ
γτ + δ
N N N
1 2 3
η(γ)(sgn δ−1)/2 3(1−|δ|)/2 2β sgn δ
= sgn δi
i
×
|δ|
|δ|
′′
(τ ; 0, 2N2 )ϑ′αg1 ,h1 (τ ; 0, 2N1 )ϑαg3 ,h3 (τ ; 0, 2N3 ). (3.20)
×(γτ + δ)9/2 ϑαg
2 ,h2
It is not difficult to verify that
−2
2
i3(1−|δ|)/2
= i(|δ|−1) /4 .
|δ|
(3.21)
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
499
Hence, by (3.1), (3.18), (3.21), (3.11), (3.14) (if j = 1) and (3.2), (3.19)–
(3.21), (3.11), (3.14) (if j = 2) we get
ατ + β
2
Ψj
; gl , hl , 0, Nl = sgn δiη(γ)(sgn δ−1)/2 i(|δ|−1) /4 ×
γτ + δ
−β sgn δ N N N
1 2 3
(γτ + δ)9/2 Ψj (τ ; αgl , hl , 0, Nl ) =
×
|δ|
|δ|
= v(M )(γτ + δ)9/2 Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2).
Thus the functions Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) satisfy condition (2) of
the Definition.
From (1.6) it follows that
ϑ′′gr hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt )ϑ′g3 h3 (τ ; 0, 2N3 ) =
P
∞
X
h m
(−1) k=r,t,3 k k (4Nr mr + gr )2 (4N3 m3 + g3 ) ×
= (πi)3
mr ,mt ,m3 =−∞
X
×e
k=r,t,3
∞
X
1
Cn(r,t) e(nτ )
(2Nk mk + gk /2)2 τ =
4Nk
n=0
for r = 1, t = 2 and r = 2, t = 1,
ϑ′′′
g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 )ϑg3 h3 (τ ; 0, 2N3 ) =
∞
P3
X
= (πi)3
(−1) k=1 hk mk (4N1 m1 + g1 )3 ×
m1 ,m2 ,m3 =−∞
3
∞
X
X
1
Cn e(nτ )
×e
(2Nk mk + gk /2)2 τ =
4Nk
n=0
k=1
and
= (πi)3
ϑ′′g2 h2 (τ ; 0, 2N2 )ϑ′g1 h1 (τ ; 0, 2N1 )ϑg3 h3 (τ ; 0, 2N3 ) =
∞
P3
X
(−1) k=1 hk mk (4N2 m2 + g2 )2 (4N1 m1 + g1 ) ×
m1 ,m2 ,m3 =−∞
×e
∞
3
X
X
1
(2Nk mk + gk /2)2 τ =
Dn e(nτ ),
4Nk
n=0
k=1
since in all these expansions
n=
3
X
1
(2Nk mk + gk /2)2 ,
4Nk
k=1
500
G. LOMADZE
as has been shown by Theorem 1.3, is a non-negative integer. Therefore,
in view of (3.1) and (3.2), the functions Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) satisfy
condition (3) of the Definition.
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 24.11.1994)
Author’s address:
Faculty of Mechanics and Mathematics
I. Javakhishvili Tbilisi State University
2, University St., Tbilisi 380043
Republic of Georgia