ON SOME ENTIRE MODULAR FORMS OF WEIGHTS AND N
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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 92 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
m=−∞
(−1)hm (2N m + g)n ×
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
(−1)h(m−c)/N e
mmodN |ξ2 |
m≡c (mod N )
ξ
g 2
1
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
= v(M )(γτ + δ)s/2 F (τ );
F
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
A0m e
m=0
m ατ + β
.
4N γτ + δ
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
= (sgn δ)n i(2n+1)η(γ)(sgn δ−1)/2 i(1−|δ|)/2
(γτ + δ)(2n+1)/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 +
ϕg0 gh (0, H; 2N ) ϑg0 h0
γτ + δ
H mod2N
t
n X
o
X
Atk
n
(n−t) ατ + β
; H, 2N ,
· ϑg0 h0
+
t
k! z=0
γτ + δ
t=1
k=1
where
g 0 = δg − γh − 2γδN, h0 = −βg + αh − 2αβN,
αβ
g 0 2 βg
g 0
e −
×
H+
H+
ϕg0 gh (0, H; 2N ) = e
4N
2
4N
2
δ
× Sg−αg0 ,h−βg0
; −αH, 2N ,
−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
=
ϑ00 (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
N1 g1 h1
o
1 00
−
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=0
Gn e
n ατ + β
.
4N γτ + δ
(2.3)
Proof. 1. From Lemma 2 for n = 2 and n = 0 (respectively with g1 , h1 , N1 ,
g10 , h10 , H1 and g2 , h2 , N2 , g20 , h02 , H2 instead of g, h, N, g 0 , h0 , H), according
to (1.8)–(1.9), it follows that
1
(γτ + δ)3 ϑ00g1 h1 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) =
N1
3
= −e sgn γ (4N1 N2 γ 2 )−1/2 ×
4
X
ϕg10 g1 h1 (0, H1 ; 2N1 )ϕg20 g2 h2 (0, H2 ; 2N2 ) ×
×
H1 mod2N1
H2 mod2N2
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
×
ατ + β
n 1
ατ + β
; H1 , 2N1 ϑg20 h20
; H2 , 2N2 +
ϑ00g0 h0
N1 1 1 γτ + δ
γτ + δ
ατ + β
−4γπi(γτ + δ)ϑg10 h01
; H1 , 2N1 ×
γτ + δ
o
ατ + β
; H2 , 2N2 .
×ϑg20 h02
γτ + δ
489
(2.4)
Replacing N1 , g1 , h1 , H1 , g10 , h10 by N2 , g2 , h2 , H2 , g20 , h02 in (2.4), and vice
versa, we obtain
1
(γτ + δ)3 ϑ00g2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 ) =
N2
3
= −e sgn γ (4N1 N2 γ 2 )−1/2 ×
4
X
0
ϕg2 g2 h2 (0, H2 ; 2N2 )ϕg10 g1 h1 (0, H1 ; 2N1 ) ×
×
H1 mod2N1
H2 mod2N2
n 1
ατ + β
ατ + β
×
ϑ00g0 h0
; H2 , 2N2 ϑg10 h10
; H1 , 2N1 +
N2 2 2 γτ + δ
γτ + δ
ατ + β
−4γπi(γτ + δ)ϑg20 h02
; H2 , 2N2 ×
γτ + δ
o
ατ + β
; H1 , 2N1 .
×ϑg10 h01
γτ + δ
(2.5)
Subtracting (2.5) from (2.4), we obtain
n 1
(γτ + δ)3
ϑ00 (τ ; 0, 2N1 )ϑg2 h2 (τ ; 0, 2N2 ) +
N1 g1 h1
o
1
− ϑ00g2 h2 (τ ; 0, 2N2 )ϑg1 h1 (τ ; 0, 2N1 ) =
N2
3
= −e sgn γ (4N1 N2 γ 2 )−1/2 ×
4
X
ϕg10 g1 h1 (0, H1 ; 2N1 )ϕg20 g2 h2 (0, H2 ; 2N2 ) ×
×
H1 mod2N1
H2 mod2N2
n 1
ατ + β
ατ + β
×
ϑ00g0 h0
; H1 , 2N1 ϑg20 h20
; H2 , 2N2 +
N1 1 1 γτ + δ
γτ + δ
o
1 00
ατ + β
ατ + β
ϑg0 h0
−
; H2 , 2N2 ϑg10 h01
; 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
ϕg30 g3 h3 (0, H3 ; 2N3 )ϑg30 h30
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
ϕgk0 gk hk (0, Hk ; 2Nk )Ω
H1 mod2N1 k=1
H2 mod2N2
H3 mod2N3
ατ + β
; gl0 , h0l , Hl , Nl .
γτ + δ
(2.8)
2. In (2.8) let γ be even. Then, by (1.7), gk0 (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 0 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
ϑg00r0 hr0
ατ + β
ατ + β
Y
; Hr , 2Nr
; Hk , 2Nk =
ϑgk0 hk0
γτ + δ
γτ + δ
k=t,3
= (πi)2
2 N/Nr
0
ατ + β
(−1)hr mr 2(Hr + 2Nr mr ) + gr0 e
Tr
×
4N
γτ + δ
=−∞
∞
X
mr
×
×
=
×
0
(−1)ht mt e
mt =−∞
∞
X
0
(−1)h3 m3 e
m3 =−∞
N/N
t
Tt
3
T3
4N
N/N
4N
ατ + β
×
γτ + δ
ατ + β
=
γτ + δ
∞
n ατ + β X
n ατ + β
r
t
Bn r e
×
Bnt e
δ
4N
γτ
+
4N
γτ + δ
n =0
=0
∞
X
nr
∞
X
n3
∞
X
t
∞
n ατ + β X
n ατ + β
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), g10 , g20 ,
0
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), hr0 and gr0 will also be odd, and in (1.2) we shall have
2
2
m + gr0 /2 = m + (gr0 − 1)/2 + m + (gr0 − 1)/2 + 1/4.
Hence
ατ + β
h0 ατ + β
r
×
; Hr , 2Nr = (πi)2 e
γτ + δ
16Nr γτ + δ
∞
X
2 N/Nr
0
ατ + β
(−1)hr mr 2(Hr + 2Nr mr )+gr0 e
×
Wr
(r = 1, 2).
4N
γτ + δ
m =−∞
ϑ00gr0 h0r
r
Here by (1.4) we can imply that h0r = 1, since by (1.7), h0r and hr have the
same parity, and
2
Wr = Hr + 2Nr mr + (gr0 − 1)/2 +
+ Hr + 2Nr mr + (gr0 − 1)/2 (r = 1, 2).
(2.10)
Analogously,
h0 ατ + β
ατ + β
t
; Ht , 2Nt = e
×
γτ + δ
16Nt γτ + δ
∞
N/N
X
0
ατ + β
t
(−1)ht mt e
×
Wt
(t = 2, 1).
4N
γτ + δ
m =−∞
ϑgt0 h0t
t
The same formula is valid for t = 3 with the same remarks concerning h30
as for ht0 , t = 2, 1. Further, for k = t, 3 we have
Tk
if 2|gk0 ,
0
2
(2.11)
Wk = (Hk + 2Nk mk + (gk − 1)/2) +
0
0
if 2 - gk .
(Hk + 2Nk mk + (gk − 1)/2)
Thus, if among h10 , h02 , h03 at least one is odd, then, as in (2.9), for r = 1,
t = 2 and r = 2, t = 1 we have
Y
ατ + β
ατ + β
; Hr , 2Nr
ϑgk0 hk0
; Hk , 2Nk =
ϑ00gr0 hr0
γτ + δ
γτ + δ
k=t,3
3
∞
1 1 X
hk0 ατ + β X 0 nr ατ + β
=e
N
×
B e
4N 4
Nk γτ + δ n =0 nr 4N γτ + δ
k=1
r
∞
n ατ + β
n ατ + β X
3
t
=
×
Bn0 3 e
Bn0 t e
4N
γτ
+
4N
γτ + δ
δ
n =0
n =0
∞
X
t
3
3
∞
1 1 X
h0k ατ + β X (r,t) n0 ατ + β
=e
N
=
Dn0 e
4N 4
Nk γτ + δ
4N γτ + δ
0
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
h0k
Nk
+ n0 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
v(M ) = i3η(γ)(sgn δ−1)/2 i3(|δ|−1) /4
(2.13)
|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
ϑg00r 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 ϑ00αgr ,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 =
ϑg00r 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
00
;
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
ϑ00gr hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt )ϑg3 h3 (τ ; 0, 2N3 ) =
= (πi)2
∞
X
(−1)hr mr +ht mt +h3 m3 (4Nr mr + gr )2 ×
mr ,mt ,m3 =−∞
×e
X
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
=
ϑ00 (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
N1 g1 h1
o
1 00
ϑg2 h2 (τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 ) ϑ0g3 h3 (τ ; c3 , 2N3 )
(3.1)
−
N2
and
Ψ2 (τ ; gl , hl , cl , Nl ) = Ψ2 (τ ; g1 , g2 , g3 ; h1 , h2 , h3 ; c1 , c2 , c3 ; N1 , N2 , N3 ) =
n 1
ϑ000 (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
=
N1 g1 h1
o
3 00
(3.2)
ϑg2 h2 (τ ; c2 , 2N2 )ϑ0g1 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=0
G(j)
n e
n ατ + β
(j = 1, 2).
4N γτ + δ
(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 ϑ0g3 h3 (τ ; 0, 2N3 ) = −e(3 sgn γ/8)(2N3 |γ|)−1/2 (i sgn γ) ×
ατ + β
X
(3.5)
ϕg30 g3 h3 (0, H3 ; 2N3 )ϑg0 0 h0
; H3 , 2N3 .
×
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 ) =
×
X
= e(9 sgn γ/8)(8N1 N2 N3 |γ|3 )−1/2 (i sgn γ) ×
3
Y
ϕgk0 gk hk (0, Hk ; 2Nk )Ψ1
H1 mod2N1 k=1
H2 mod2N2
H3 mod2N3
ατ + β
; gl0 , hl0 , 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
(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
ϑg0 3 h3
×
1
(γτ + δ)4 ϑ000
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
ϕg10 g1 h1 (0, H1 ; 2N1 ) ϑg0000 h0
; H1 , 2N1 +
1 1
γτ + δ
H1 mod2N1
3 X
t
o
X
Atk
3
(3−t) ατ + β
; H1 , 2N1
×
ϑg0 h0
k! z=0 1 1 γτ + δ
t
t=1
k=1
ατ + β
X
; H2 , 2N2 =
×
ϕg20 g2 h2 (0, H2 ; 2N2 )ϑg20 h02
γτ + δ
H2 mod2N2
X
= e(sgn γ)(i sgn γ)(4N1 N2 γ 2 )−1/2
ϕg10 g1 h1 (0, H1 ; 2N1 ) ×
+
H1 mod2N1
H2 mod2N2
ατ + β
n 1
ϑg0000 h0
; H1 , 2N1 ϑg20 h20 (τ ; 0, 2N2 ) +
N1 1 1 γτ + δ
o
ατ + β
(3.7)
−12γπi(γτ + δ)ϑg0 0 h0
; H1 , 2N1 ϑg20 h02 (τ ; 0, 2N2 )
1 1
γτ + δ
×ϕg20 g2 h2 (0, H2 ; 2N2 )
496
G. LOMADZE
and
3
(γτ + δ)4 ϑ00g2 h2 (τ ; 0, 2N2 )ϑ0g1 h1 (τ ; 0, 2N1 ) =
N2
X
= e(sgn γ)(i sgn γ)(4N1 N2 γ 2 )−1/2
ϕg10 g1 h1 (0, H1 ; 2N1 ) ×
H1 mod2N1
H2 mod2N2
n 3
ατ + β
ϑ00g0 h0
; H2 , 2N2 +
×ϕg20 g2 h2 (0, H2 ; 2N2 )
N2 2 2 γτ + δ
o
ατ + β
; H2 , 2N2 ϑ0g0 h0 (τ ; 0, 2N1 ).
−12γπi(γτ + δ)ϑ0g0 h0
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 ) =
×
X
= e(9 sgn γ/8)(8N1 N2 N3 |γ|3 )−1/2 (i sgn γ) ×
3
Y
ϕgk0 gk hk (0, Hk ; 2Nk )Ψ2
H1 mod2N1 k=1
H2 mod2N2
H3 mod2N3
ατ + β
; gl0 , hl0 , Hl , Nl . (3.9)
γτ + δ
In (3.9) let γ be even. Then, reasoning as in Lemma 3.2, by (3.2), from
(3.9) we obtain
ατ + β
n 1
ατ + β
ϑg0000 h0
; H1 , 2N1 ϑg20 h02
; H2 , 2N2 +
N1 1 1 γτ + δ
γτ + δ
ατ + β
o
3 00 ατ + β
− ϑg0 h0
; H2 , 2N2 ϑ0g0 h0
; H1 , 2N1
×
1 1
N2 2 2 γτ + δ
γτ + δ
n (πi)3
ατ + β
; H3 , 2N3 =
×ϑg30 h30
×
γτ + δ
N1
∞
X
3 N/N1
0
ατ + β
×
×
(−1)h1 m1 2(H1 + 2N1 m1 ) + g10 e
T1
4N
γτ + δ
m =−∞
1
×
−
3(πi)3
N1
×
N/N
4N
2
T2
ατ + β
+
γτ + δ
N/N1
0
ατ + β o
(−1)h1 m1 2(H1 + 2N1 m1 ) + g10 e
×
T1
4N
γτ + δ
=−∞
∞
X
m1
m2 =−∞
0
(−1)h2 m2 e
2 N/N2
0
ατ + β
(−1)h2 m2 2(H2 + 2N2 m2 ) + g20 e
×
T2
4N
γτ + δ
=−∞
∞
X
m2
∞
X
ON SOME ENTIRE MODULAR FORMS OF WEIGHTS
×
=
−
n
×
N/N
4N
3
T3
ατ + β
=
γτ + δ
∞
n ατ + β X
n ατ + β
1
2
Bn2 e
Bn1 e
+
+
4N
γτ
δ
4N
γτ + δ
=0
n =0
2
∞
n ατ + β o
n ατ + β X
1
2
Bn0 1 e
×
Bn0 2 e
+
δ
4N
γτ
4N
γτ + δ
n =0
=0
∞
X
n2
0
(−1)h3 m3 e
m3 =−∞
∞
X
n1
∞
X
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
(M =
is the matrix of the substitution from Γ0 (4N ) and ∆ is the
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
ϑ000
; 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 ϑ000
αg1 ,h1 (τ ; 0, 2N1 ),
|δ|
ατ + β
ϑg0 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 ϑ0αg1 ,h1 (τ ; 0, 2N1 ),
|δ|
ατ + β
; 0, 2N3 = sgn δi3η(γ)(sgn δ−1)/2 i(1−|δ|)/2 ×
ϑg0 3 h3
γτ + δ
2βN sgn δ
3
0
(γτ + δ)3/2 ϑαg
(τ ; 0, 2N3 ).
×
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 ϑg0 3 h3
; 0, 2N3 =
ϑ00gr 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 ϑ00αgr ,hr (τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt )ϑ0αg3 ,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
ϑg0001 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|
000
(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 ϑ0g1 h1
; 0, 2N1 ϑg3 h3
; 0, 2N3 =
ϑ00g2 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|
00
(τ ; 0, 2N2 )ϑ0αg1 ,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
ϑ00gr hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt )ϑ0g3 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,
ϑ000
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
×e
Cn e(nτ )
(2Nk mk + gk /2)2 τ =
4Nk
n=0
k=1
and
= (πi)3
ϑ00g2 h2 (τ ; 0, 2N2 )ϑ0g1 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
