Georgian Mathematical Journal 1(1994), No. 1, 53-76

ON SOME ENTIRE MODULAR FORMS OF WEIGHTS 5 AND 6 FOR THE CONGRUENCE GROUP Γ_{0}(4N)

G. LOMADZE

Abstract. Two entire modular forms of weight 5 and two of weight 6 for the congruence subgroup Γ_{0}(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,

(\frac{h}{u}) is the generalized Jacobi symbol, ( \begin{matrix}n \\ t \end{matrix} ) a binomial coefficient, φ(k) Eu-

ler’s function, and e(z) = exp 2πiz, η(γ) = 1 if γ ≥ 0 and η(γ) = −1 if γ < 0.

Let

Γ = \{\frac{ατ + β}{γτ + δ} |αδ − βγ = 1\},

Γ_{0}(4N) = \{\frac{ατ + β}{γτ + δ} ∈ Γ|γ ≡ 0(mod 4N)\}.

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 Γ_{0}(4N) if

1) F is regular on H,

2) for all substitutions from Γ_{0}(4N) and all τ ∈ H

F (\frac{ατ + β}{γτ + δ} ) = χ(δ)(γτ + δ)^{r}F (τ),

1991 Mathematics Subject Classification. 11F11,11F03,11F27.

54 G. LOMADZE

3) in the neighborhood of the point τ = i∞

F (τ) = ∑_{m=0}^{∞} A_{m}e(mτ),

4) for all substitutions from Γ, in the neighborhood of each rational point τ = −\frac{δ}{γ} (γ ≠ 0, (γ, δ) = 1)

(γτ + δ)^{r}F (τ) = ∑_{m=0}^{∞} A′_{m}e(\frac{n}{4N} \frac{ατ + β}{γτ + δ} ).

For η ≠ 0, ξ, g, h, N with ξg + ηh + ξηN ≡ 0 (mod 2), put

S_{gh} ( \begin{matrix}ξ \\ η \end{matrix} ; c, N) = ∑_{\begin{matrix}mmodN|η| \\ m≡c (mod N) \end{matrix}}(−1)^{h(m−c)/N}e(\frac{ξ}{2Nη} (m + \frac{g}{2})^{2}).

It is known ([2] p. 323, formula (2.4)–(2.6)) that

\begin{matrix}S_{g+2j,h} ( \begin{matrix}ξ \\ η \end{matrix} ; c, N) = S_{gh} ( \begin{matrix}ξ \\ η \end{matrix} ; c + j, N) , \\ S_{g,h+2j} ( \begin{matrix}ξ \\ η \end{matrix} ; c, N) = S_{gh} ( \begin{matrix}ξ \\ η \end{matrix} ; c, N) , \\ S_{gh} ( \begin{matrix}ξ \\ η \end{matrix} ; c + N_{j}, N) = (−1)^{hj}S_{gh} ( \begin{matrix}ξ \\ η \end{matrix} ; c, N) . \end{matrix}  (1)

Let

ϑ_{gh}(z|τ; c, N) =
= ∑_{m≡c (mod N)}(−1)^{h(m−c)/N}e(\frac{1}{2N} (m + \frac{g}{2})^{2}τ)e((m + \frac{g}{2})z),  (2)

hence

\frac{∂^{n}}{∂z^{n}} ϑ_{gh}(z|τ; c, N) = (πi)^{n} ∑_{m≡c (mod N)}(−1)^{h(m−c)/N}(2m + g)^{n} ×

×e(\frac{1}{2N} (m + \frac{g}{2})^{2}τ)e((m + \frac{g}{2})z).  (3)

Put

\begin{matrix}ϑ_{gh}^{(n)}(τ; c, N) = \frac{∂^{n}}{∂z^{n}} ϑ_{gh}(z|τ; c, N)|_{z=0}, \\ ϑ_{gh}^{(0)}(τ; c, N) = ϑ_{gh}(τ; c, N) = ϑ_{gh}(0|τ; c, N). \end{matrix}  (4)

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

ϑ_{g,h+2j}(z|τ; c, N) = ϑ_{gh}(z|τ; c, N);  (5)

ϑ_{gh}(\frac{z}{τ} | − \frac{1}{τ}; c, N) = ( − \frac{iτ}{N} )^{1/2}e(\frac{Nz^{2}}{2τ} ) ×

× ∑_{H modN} e( − \frac{1}{N} (c + \frac{g}{2})(H + \frac{h}{2}))ϑ_{hg}(z|τ; H, N);  (6)

ϑ_{gh}(\frac{z}{γτ + δ} |\frac{ατ + β}{γτ + δ}; c, N) = (\frac{−i(γτ + δ) sgn γ}{N|γ|} )^{\frac{1}{2}} e(\frac{Nγz^{2}}{2(γτ + δ)}) ×

× ∑_{H modN} φ_{g′gh}(c, H; N)ϑ_{g′h′}(z|τ; H, N) (γ ≠ 0),  (7)

where

g′ = αg + γh + αγN, h′ = βg + δh + βδN,  (8)

φ_{g′gh}(c, H; N) = e( − \frac{βδ}{2N} (H + \frac{g′}{2} )^{2})e(\frac{β}{N} (c + \frac{g}{2})(H + \frac{g′}{2} )) ×

×S_{g−δg′,h+βg′}( \begin{matrix}α \\ γ \end{matrix} ; c − δH, N);  (9)

\begin{matrix}ϑ_{gh}(z|τ + β; c, N) = e(\frac{β}{2N} (c + \frac{g}{2})^{2})ϑ_{g,h+βg+βN}(z|τ; c, N), \\ ϑ_{gh}(−z|τ − β; c, N) = \\ = e( − \frac{β}{2N} (c + \frac{g}{2})^{2})ϑ_{−g,−h+βg−βN}(z|τ; −c, N). \end{matrix} (10)

From (5) and (10), according to the notations (4), it follows that

\begin{matrix}ϑ_{g,h+2j}(τ; c, N) = ϑ_{gh}(τ; c, N), \\ ϑ_{g,h+2j}^{(n)}(τ; c, N) = ϑ_{gh}^{(n)}(τ; c, N); \end{matrix}  (11)

\begin{matrix}ϑ_{gh}^{(n)}(τ + β; c, N) = e(\frac{β}{2N} (c + \frac{g}{2})^{2})ϑ_{g,h+βg+βN}^{(n)}(τ; c, N), \\ ϑ_{gh}^{(n)}(τ − β; c, N) = \\ = (−1)^{n}e( − \frac{β}{2N} (c + \frac{g}{2})^{2})ϑ_{−g,−h+βg−βN}^{(n)}(τ; −c, N). \end{matrix}  (12)

From (2) and (3), according to the notations (4), in particular, it follows that

\begin{matrix}ϑ_{gh}(τ; 0, N) = ∑_{m=−∞}^{∞}(−1)^{hm}e(\frac{1}{2N} (Nm + \frac{g}{2})^{2}τ), \\ ϑ_{gh}^{(n)}(τ; 0, N) = \\ = (πi)^{n} ∑_{m=−∞}^{∞}(−1)^{hm}(2Nm + g)^{n}e(\frac{1}{2N} (Nm + \frac{g}{2})^{2}τ). \end{matrix}  (13)

1.

Lemma 1. For n ≥ 0

ϑ_{gh}^{(n)}( − \frac{1}{τ}; c, N) = (Ni)^{n}( − \frac{iτ}{N} )^{(2n+1)/2} ×

× ∑_{H modN} e( − \frac{1}{N} (c + \frac{g}{2})(H + \frac{h}{2})) ×

×\{ϑ_{hg}^{(n)}(τ; H, N) + ∑_{t=1}^{n} ( \begin{matrix}n \\ t \end{matrix} ) ∑_{k=1}^{t} \frac{A_{tk}}{k!} |_{z=0} · ϑ_{hg}^{(n−t)}(τ; H, N)\},

where

A_{tk}|_{z=0} = \{ \begin{matrix}(2k)!(\frac{Nπi}{τ})^{k} if t = 2k \\ 0 if t ≠ 2k \end{matrix}

(t = 1, 2, …, n; k = 1, 2, …, t).  (1.1)

Proof. From (6), by Leibniz’s formula, we obtain

\frac{∂^{n}}{∂z^{n}} ϑ_{gh}(\frac{z}{τ} | − \frac{1}{τ}; c, N) = τ^{n}(\frac{−iτ}{N} )^{1/2} \frac{∂^{n}}{∂z^{n}} \{e(\frac{Nz^{2}}{2τ} ) ×

× ∑_{H modN} e( − \frac{1}{N} (c + \frac{g}{2})(H + \frac{h}{2}))ϑ_{hg}(z|τ; H, N)\} =

= (Ni)^{n}(\frac{−iτ}{N} )^{(2n+1)/2}×

× ∑_{H modN} e( − \frac{1}{N} (c + \frac{g}{2})(H + \frac{h}{2}))\{e(\frac{Nz^{2}}{2τ} )\frac{∂^{n}}{∂z^{n}} ϑ_{hg}(z|τ; H, N)\} +

+ ∑_{t=1}^{n} ( \begin{matrix}n \\ t \end{matrix} )\frac{∂^{n}}{∂z^{n}} e(\frac{Nz^{2}}{2τ} )\frac{∂^{n−t}}{∂z^{n−t}} ϑ_{hg}(z|τ; H, N)\}.  (1.2)

ON SOME ENTIRE MODULAR FORMS 57

According to formulae (a) and (b) of [1], p. 37^{1}

\frac{∂^{t}}{∂z^{t}} e(\frac{Nz^{2}}{2τ} ) = A_{t1}e(\frac{Nz^{2}}{2τ} ) + \frac{A_{t2}}{2!} e(\frac{Nz^{2}}{2τ} ) + ⋯ + \frac{A_{tt}}{t!} e(\frac{Nz^{2}}{2τ} )

(t = 1, 2, …, n),

where

A_{tk} = \frac{∂^{t}}{∂z^{t}} (\frac{Nπiz^{2}}{τ} )^{k} − k(\frac{Nπiz^{2}}{τ})\frac{∂^{t}}{∂z^{t}} (\frac{Nπiz^{2}}{τ})^{k−1} +

+\frac{k(k − 1)}{2!}(\frac{Nπiz^{2}}{τ})^{2} \frac{∂^{t}}{∂z^{t}} (\frac{Nπiz^{2}}{τ})^{k−2} +

+ ⋯ + (−1)^{k−1}k(\frac{Nπiz^{2}}{τ} )^{k−1} \frac{∂^{t}}{∂z^{t}} (\frac{Nπiz^{2}}{τ} ) (k = 1, 2, …, t),

hence

\frac{∂^{t}}{∂z^{t}} e(\frac{Nz^{2}}{2τ} )|_{z=0} = ∑_{k=1}^{t} \frac{A_{tk}}{k!} |_{z=0} (t = 1, 2, …, n)  (1.3)

and

A_{tk}|_{z=0} = \frac{∂^{t}}{∂z^{t}} (\frac{Nπiz^{2}}{τ} )^{k}|_{z=0} (k = 1, 2, …, t).  (1.4)

Thus, in view of notations (4), the lemma follows from (1.2)–(1.4).

Lemma 2. If γ ≠ 0, then for n ≥ 0

ϑ_{gh}^{(n)}(\frac{ατ + β}{γτ + δ}; c, N) = (N|γ|i sgn γ)^{n}( − i(γτ + δ) \frac{sgn γ}{N|γ|} )^{(2n+1)/2} ×

× ∑_{H modN} φ_{g′gh}(c, H; N){ϑ_{g′h′}^{(n)}(τ; H, N) +

+ ∑_{t=1}^{n} ( \begin{matrix}n \\ t \end{matrix} ) ∑_{k=1}^{t} \frac{A_{tk}}{k!} |_{z=0} · ϑ_{g′h′}^{(n−t)}(τ; H, N)\},

where g′ h′ and φ_{g′gh}(c, H; N) are defined by the formulas (8) and (9),

A_{tk}|_{z=0} = \{ \begin{matrix}(2k)!(\frac{Nγπi}{γτ+δ})^{k} if t = 2k \\ 0 if t ≠ 2k \end{matrix}

(t = 1, 2, …, n; k = 1, 2, …, t).

1Page 75 in the Russian version of [1] published in 1933.

Proof. From (7), according to Leibnitz’s formula, we obtain

\frac{∂^{n}}{∂z^{n}} ϑ_{gh}(\frac{z}{γτ + δ} |\frac{ατ + β}{γτ + δ}; c, N) = (γτ + δ)^{n}( − i(γτ + δ) \frac{sgn γ}{N|γ|} )^{1/2} ×

×\frac{∂^{n}}{∂z^{n}} \{e(\frac{Nγz^{2}}{2(γτ + δ)}) ∑_{H modN} φ_{g′gh}(c, H; N)ϑ_{g′h′}(z|τ; H, N)\} =

= (N|γ|i sgn γ)^{n}( − i(γτ + δ) \frac{sgn γ}{N|γ|} )^{(2n+1)/2} ×

× ∑_{H modN} φ_{g′gh}(c, H; N)\{e(\frac{Nγz^{2}}{2(γτ + δ)})\frac{∂^{n}}{∂z^{n}} ϑ_{g′h′}(z|τ; H, N) +

+ ∑_{t=1}^{n} ( \begin{matrix}n \\ t \end{matrix} )\frac{∂^{t}}{∂z^{t}} e(\frac{Nγz^{2}}{2(γτ + δ)})\frac{∂^{n−t}}{∂z^{n−t}} ϑ_{g′h′}(z|τ; H, N)\}.  (1.5)

As in Lemma 1, but with e(\frac{Nγz^{2}}{2(γτ+δ)}) instead of e(\frac{Nz^{2}}{2τ} ), we have

\frac{∂^{t}}{∂z^{t}} e(\frac{Nγz^{2}}{2(γτ + δ)})|_{z=0} = ∑_{k=1}^{t} \frac{A_{tk}}{k!} |_{z=0} (t = 1, 2, …, n)  (1.6)

and

A_{tk}|_{z=0} = \frac{∂^{t}}{∂z^{t}} (\frac{Nγπiz}{γτ + δ} )^{2k}|_{z=0} (k = 1, 2, …, t).  (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 Γ_{0}(4N) we have

ϑ_{gh}^{(n)}(\frac{ατ + β}{γτ + δ}; 0, 2N) =

= ( sgn δ)^{n} i^{(2n+1)η(γ)( sgn δ−1)/2} i^{(1−|δ|)/2}(\frac{2βN sgn δ}{|δ|} ) ×

×(γτ + δ)^{(2n+1)/2} e( − \frac{αγδ^{2}h^{2}}{16N} ) e(\frac{βδg^{2}}{4} \frac{δ^{2φ(2N)−2}}{4N} )ϑ_{αg,h}^{(n)}(τ; 0, 2N).

Proof. 1) Let γ ≠ 0. In [4] (p.18, formula (5.1)) it is shown that

S_{g0}( \begin{matrix}β \\ δ \end{matrix} ; 0, 2N) =

= e(\frac{βδg^{2}}{4} \frac{δ^{2φ(2n)−2}}{4N} )i^{(1−|δ|)/2}(\frac{2βN sgn δ}{|δ|} )|δ|^{1/2}.  (1.8)

Replacing α, β, γ, δ, τ, c, N by β, −α, δ, −γ, τ′, 0, 2N in Lemma 2, we obtain

ϑ_{gh}^{(n)}(\frac{βτ′ − α}{δτ′ − γ}; 0, 2N) = (2N|δ|i sgn δ)^{n}( − i(δτ′ − γ)\frac{sgn δ}{2N|δ|})^{(2n+1)/2} ×

× ∑_{H mod2N} φ_{h′gh}(0, H; 2N){ϑ_{h′,αg}^{(n)}(τ′; H, 2N) +

+ ∑_{t=1}^{n} ( \begin{matrix}n \\ t \end{matrix} ) ∑_{k=1}^{t} \frac{A_{tk}}{k!} |_{z=0} · ϑ_{h′,αg}^{(n−t)}(τ′; H, 2N)\},  (1.9)

where by (8),(9),(11) and (1)

h′ = βg + δh + 2βδN,  (1.10)
φ_{h′gh}(0, H; 2N) =
= e( − \frac{αγh′^{2}}{16N} )e( − \frac{αg}{4N} (H + \frac{h′}{2} ))S_{g0}( \begin{matrix}β \\ δ \end{matrix} ; 0, 2N),  (1.11)

A_{tk}|_{z=0} = \{ \begin{matrix}(2k)!(\frac{2Nδπi}{δτ′−γ})^{k} if t = 2k \\ 0 if t ≠ 2k \end{matrix}

(k = 1, 2, …, t).  (1.12)

Writing −\frac{1}{τ} instead of τ′ in (1.9) and (1.12), according to (1.11), we obtain

ϑ_{gh}^{(n)}(\frac{ατ + β}{γτ + δ}; 0, 2N) = (2Ni)^{n}(|δ| sgn δ)^{n}(\frac{−i(−\frac{δ}{τ} − γ) sgn δ}{2N|δ|} )^{(2n+1)/2}×

×e( − \frac{αγh^{′2}}{16N} )e(\frac{βδg^{2}}{4} \frac{δ^{2φ(2N)−2}}{4N} )i^{(1−|δ|)/2}(\frac{2βN sgn δ}{|δ|} )|δ|^{1/2}×

× ∑_{H mod2N} e( − \frac{αg}{4N} (H + \frac{h′}{2} ))\{ϑ_{h′,αg}^{(n)}( − \frac{1}{τ}; H, 2N) +

+ ∑_{t=1}^{n} ( \begin{matrix}n \\ t \end{matrix} ) ∑_{k=1}^{t} \frac{A_{tk}}{k!} |_{z=0} · ϑ_{h′,αg}^{(n−t)}( − \frac{1}{τ}; H, 2N)\},  (1.13)

where

A_{tk}|_{z=0} = \{ \begin{matrix}(2k)!(\frac{2Nδπiτ}{γτ+δ} )^{k} if t = 2k \\ 0 if t ≠ 2k \end{matrix} (k = 1, 2, …, t).

60 G. LOMADZE

Writing αg, h′, −\frac{1}{τ} , 0, 2N instead of g, h, τ, c, N in Lemma 1, we obtain

ϑ_{αg,h′}^{(n)}(τ; 0, 2N) = (2Ni)^{n}(\frac{i}{2Nτ} )^{(2n+1)/2} ∑_{H mod2N} e( − \frac{αg}{4N} (H + \frac{h′}{2} ))×

×\{ϑ_{h′,αg}^{(n)}( − \frac{1}{τ}; H, 2N) +

+ ∑_{t=1}^{n} ( \begin{matrix}n \\ t \end{matrix} ) ∑_{k=1}^{t} \frac{A_{tk}}{k!} |_{z=0} · ϑ_{h′,αg}^{(n−t)}( − \frac{1}{τ}; H, 2N)\},  (1.14)

where

A_{tk}|_{z=0} = \{ \begin{matrix}(2k)!(−2Nπiτ)^{k} if t = 2k \\ 0 if t ≠ 2k \end{matrix} (k = 1, 2, …, t).

From (1.14) it follows that

∑_{H mod2N} e( − \frac{αg}{4N} (H + \frac{h′}{2} ))\{ϑ_{h′,αg}^{(n)}( − \frac{1}{τ}; H, 2N) +

+ ∑_{t=1}^{n} ( \begin{matrix}n \\ t \end{matrix} ) ∑_{k=1}^{t} \frac{A_{tk}}{k!} |_{z=0} · ϑ_{h′,αg}^{(n−t)}( − \frac{1}{τ}; H, 2N)\} =

= (2πi)^{−n}(−2Niτ)^{(2n+1)/2}ϑ_{αg,h′}^{(n)}(τ; 0, 2N).  (1.15)

From (1.13) and (1.15) we obtain

ϑ_{gh}^{(n)}(\frac{ατ + β}{γτ + δ}; 0, 2N) = (−2Niτ)^{(2n+1)/2}(|δ| sgn δ)^{n} ×

×(\frac{−i(−\frac{δ}{τ} − γ) sgn δ}{2N|δ|} )^{(2n+1)/2} i^{(1−|δ|)/2}e( − \frac{αγh^{′2}}{16N} ) ×

×e(\frac{βδg^{2}}{4} \frac{δ^{2φ(2N)−2}}{4N} )(\frac{2βN sgn δ}{|δ|} )|δ|^{1/2} ϑ_{αg,h′}^{(n)}(τ; 0, 2N). (1.16)

In [3] (p.85, formula (6.16) with 2N instead of N) it is shown that

(\frac{−i(−\frac{δ}{τ} − γ) sgn δ}{2N|δ|} )^{1/2}(−2Niτ)^{1/2} =

= i^{sgn γ( sgn δ−1)/2}(\frac{γτ + δ}{|δ|} )^{1/2}.  (1.17)

ON SOME ENTIRE MODULAR FORMS 61

From (1.16) and (1.17) it follows that

ϑ_{gh}^{(n)}(\frac{ατ + β}{γτ + δ}; 0, 2N) =

= (|δ| sgn δ)^{n}(i^{sgn γ( sgn δ−1)} \frac{γτ + δ}{|δ|} )^{(2n+1)/2}i^{(1−|δ|)/2} ×

×e( − \frac{αγh^{′2}}{16N} )e(\frac{βδg^{2}}{4} \frac{δ^{2φ(2N)−2}}{4N} )(\frac{2βN sgn δ}{|δ|} )|δ|^{1/2}ϑ_{αg,h′}^{(n)}(τ; 0, 2N).

Thus, in view of (1.10) and (1.11), the lemma is proved for γ ≠ 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

ϑ_{gh}^{(n)}(τ + β; 0, 2N) = e(\frac{βg^{2}}{16N} )ϑ_{gh}^{(n)}(τ; 0, 2N),

ϑ_{gh}^{(n)}(τ − β; 0, 2N) = (−1)^{n}e( − \frac{βg^{2}}{16N} )ϑ_{−g,h}^{(n)}(τ; 0, 2N),

i.e., the lemma is also proved for γ = 0.

Lemma 4. If γ ≠ 0, then for n ≥ 0

(γτ + δ)^{(2n+1)/2}ϑ_{gh}^{(n)}(τ; 0, 2N) =

= e((2n + 1) sgn γ/8)(2N|γ|)^{−1/2}(−i sgn γ)^{n} ×

× ∑_{H mod2N} φ_{g′gh}(0, H; 2N)\{ϑ_{g′h′}^{(n)}(\frac{ατ + β}{γτ + δ}; H, 2N) +

+ ∑_{t=1}^{n} ( \begin{matrix}n \\ t \end{matrix} ) ∑_{k=1}^{t} \frac{A_{tk}}{k!} |_{z=0} · ϑ_{g′h′}^{(n−t)}(\frac{ατ + β}{γτ + δ}; H, 2N)\},

where

A_{tk}|_{z=0} = \{ \begin{matrix}(2k)!(−2Nγπi(γτ + δ))^{k} if t = 2k \\ 0 if t ≠ 2k \end{matrix}

(t = 1, 2, …; k = 1, 2, …, t).  (1.18)

Remark. From (1.18) it follows that

∑_{k=1}^{t} \frac{A_{tk}}{k!} |_{z=0} = \{ \begin{matrix}\frac{A_{t,t/2}}{(t/2)!} |_{z=0} if 2|t \\ 0 if 2†t. \end{matrix}  (1.19)

Proof. Replacing α, β, γ, δ, τ, c, N by δ, −β, −γ, α, τ′, 0, 2N in Lemma 2, we obtain

ϑ_{gh}^{(n)}(\frac{δτ′ − β}{−γτ′ + α}; 0, 2N) =

= (−2N|γ|i sgn γ)^{n}(i(−γτ′ + α)\frac{sgn γ}{2N|γ|})^{(2n+1)/2} ×

× ∑_{H mod2N} φ_{g′gh}(0, H; 2N){ϑ_{g′h′}^{(n)}(τ′; H, 2N) +

+ ∑_{t=1}^{n} ( \begin{matrix}n \\ t \end{matrix} ) ∑_{k=1}^{t} \frac{A_{tk}}{k!} |_{z=0} · ϑ_{g′h′}^{(n−t)}(τ′; H, 2N)\},  (1.20)

where

g′ = δg − γh − 2γδN, h′ = −βg + αh − 2αβN,  (1.21)

φ_{g′gh}(0, H; 2N) = e( − \frac{αβ}{4N} (H + \frac{g′}{2} )^{2})e( − \frac{βg}{4N} (H + \frac{g′}{2} )) ×

×S_{g−αg′,h−βg′}( \begin{matrix}δ \\ −γ \end{matrix} ; −αH, 2N),  (1.22)

A_{tk}|_{z=0} = \{ \begin{matrix}(2k)!(\frac{2Nγπi}{γτ′−α})^{k} if t = 2k, \\ 0 if t ≠ 2k. \end{matrix}  (1.23)

In [3, p.87, formula (6.23)] (with 2N instead of N) it is shown that

(\frac{i sgn γ}{2N|γ|(γτ + δ)})^{1/2} = \frac{e( sgn γ/8)}{(2N|γ|)^{1/2}(γτ + δ)^{1/2}} .  (1.24)

Taking \frac{ατ+β}{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

Ψ_{1}(τ) = Ψ_{1}(τ; g_{1}, g_{2}, h_{1}, h_{2}, c_{1}, c_{2}, N_{1}, N_{2}) =

= \frac{1}{N_{1}} ϑ′′′_{g_{1}h_{1}}(τ; c_{1}, 2N_{1})ϑ′_{g_{2}h_{2}}(τ; c_{2}, 2N_{2}) −

−\frac{1}{N_{2}} ϑ′′′_{g_{2}h_{2}}(τ; c_{2}, 2N_{2})ϑ′_{g_{1}h_{1}}(τ; c_{1}, 2N_{1})  (2.1)

and

Ψ_{2}(τ) = Ψ_{2}(τ; g_{1}, g_{2}, h_{1}, h_{2}, c_{1}, c_{2}, N_{1}, N_{2}) =

= \frac{1}{N_{1}^{2}} ϑ_{g_{1}h_{1}}^{(4)}(τ; c_{1}, 2N_{1})ϑ_{g_{2}h_{2}}(τ; c_{2}, 2N_{2}) +

+\frac{1}{N_{2}^{2}} ϑ_{g_{2}h_{2}}^{(4)}(τ; c_{2}, 2N_{2})ϑ_{g_{1}h_{1}}(τ; c_{1}, 2N_{1}) −

−\frac{6}{N_{1}N_{2}} ϑ′′_{g_{1}h_{1}}(τ; c_{1}, 2N_{1})ϑ′′_{g_{2}h_{2}}(τ; c_{2}, 2N_{2}),  (2.2)

where

2|g_{1}, 2|g_{2}, N_{1}|N, N_{2}|N, 4|N(\frac{h_{1}}{N_{1}} + \frac{h_{2}}{N_{2}} ).  (2.3)

For all substitutions from Γ in the neighborhood of each rational point τ = −\frac{δ}{γ} (γ ≠ 0, (γ, δ) = 1), we then have

(γτ + δ)^{5}Ψ_{j}(τ; g_{1}, g_{2}, h_{1}, h_{2}, 0, 0, N_{1}, N_{2}) =
= ∑_{n=0}^{∞} C_{n}^{(j)}e(\frac{n}{4N} \frac{ατ + β}{γτ + δ} ) (j = 1, 2).  (2.4)

Proof. I. From Lemma 4 for n = 3 (with g_{1}, h_{1}, N_{1}, g′_{1}, h′_{1}, H_{1} instead of g, h, N, g′, h′, H) and n = 1 (with g_{2}, h_{2}, N_{2}, g′_{2}, h′_{2}, H_{2} instead of g,h, N, g′, h′, H), according to (1.19), it follows that

\frac{1}{N_{1}} (γτ + δ)^{5}ϑ′′′_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ′_{g_{2}h_{2}}(τ; 0, 2N_{2}) =

= \frac{1}{N_{1}} e(\frac{5}{4} sgn γ)(2|γ|(N_{1}N_{2})^{1/2})^{−1} ×

× ∑_{H_{1} mod2N_{1}} φ_{g′_{1}g_{1}h_{1}}(0, H_{1}; 2N_{1})\{ϑ′′′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) +

+\frac{3 · 2}{2!}A_{21}|_{z=0} · ϑ′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})\} ×

× ∑_{H_{2} mod2N_{2}} φ_{g′_{2}g_{2}h_{2}}(0, H_{2}; 2N_{2})ϑ′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) =

= e(\frac{5}{4} sgn γ)(2|γ|(N_{1}N_{2})^{1/2})^{−1} ×

× ∑_{\begin{matrix}H_{1} mod2N_{1} \\ H_{2} mod2N_{2} \end{matrix}} φ_{g′_{1}g_{1}h_{1}}(0, H_{1}; 2N_{1})φ_{g′_{2}g_{2}h_{2}}(0, H_{2}; 2N_{2})×

×\{\frac{1}{N_{1}} ϑ′′′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})ϑ′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) −

−12γπi(γτ + δ)ϑ′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) ×

×ϑ′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2})\}.  (2.5)

Replacing N_{1}, g_{1}, h_{1}, H_{1}, g′_{1}, h′_{1} by N_{2}, g_{2}, h_{2}, H_{2}, g′_{2}, h′_{2} in (2.5) and vice versa, we obtain

\frac{1}{N_{2}} (γτ + δ)^{5}ϑ′′′_{g′_{2}h′_{2}}(τ; 0, 2N_{2})ϑ′_{g′_{1}h′_{1}}(τ; 0, 2N_{1}) =

= e(\frac{5}{4} sgn γ)(2|γ|(N_{1}N_{2})^{1/2})^{−1} ×

× ∑_{\begin{matrix}H_{2} mod2N_{2} \\ H_{1} mod2N_{1} \end{matrix}} φ_{g′_{2}g_{2}h_{2}}(0, H_{2}; 2N_{2})φ_{g′_{1}g_{1}h_{1}}(τ; 0, H_{1}; 2N_{1}) ×

×\{\frac{1}{N_{2}} ϑ′′′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) ϑ′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) −

−12γπi(γτ + δ)ϑ′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) ×

×ϑ′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})\}.  (2.6)

Subtracting (2.6) from (2.5), according to (2.1), we obtain

(γτ + δ)^{5}Ψ_{1}(τ; g_{1}, g_{2}, h_{1}, h_{2}, 0, 0, N_{1}, N_{2}) =

= e(\frac{5}{4} sgn γ)(2|γ|(N_{1}N_{2})^{1/2})^{−1} ×

× ∑_{\begin{matrix}H_{1} mod2N_{1} \\ H_{2} mod2N_{2} \end{matrix}} φ_{g′_{1}g_{1}h_{1}}(0, H_{1}; 2N_{1})φ_{g′_{2}g_{2}h_{2}}(0, H_{2}; 2N_{2}) ×

×Ψ_{1}(\frac{ατ + β}{γτ + δ}; g′_{1}, g′_{2}, h′_{1}, h′_{2}, H_{1}, H_{2}, N_{1}, N_{2}).  (2.7)

In (2.7) let γ be even. Then, by (1.21) and (2.3), g′_{1} and g′_{2} are also even. Therefore, according to (3) and the notations (4), we have

ϑ′′′_{g′_{r}h′_{r}}(\frac{ατ + β}{γτ + δ}; H_{r}, 2N_{r})ϑ′_{g′_{s}h′_{s}}(\frac{ατ + β}{γτ + δ}; H_{s}, 2N_{s}) =

= ∑_{n_{1}=0}^{∞} B_{n_{1}}e(\frac{Nn_{1}/N_{r}}{4N} \frac{ατ + β}{γτ + δ} ) ∑_{n_{2}=0}^{∞} B_{n_{2}}e(\frac{Nn_{2}/N_{s}}{4N} \frac{ατ + β}{γτ + δ} ) =

= ∑_{n=0}^{∞} B_{n}e(\frac{n}{4N} \frac{ατ + β}{γτ + δ} )

for r = 1, s = 2 and r = 2, s = 1.  (2.8)

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 h_{1} and h_{2} are both even, then by (1.21) and (2.3), g′_{1} and g′_{2} are also even, and we obtain the same result. But if h_{r} is odd, then by (1.21) g′_{r} will also be odd, and in (3) we have

(m + \frac{g′_{r}}{2} )^{2} = (m + \frac{1}{2}(g′_{r} − 1))^{2} + (m + \frac{1}{2}(g′_{r} − 1)) + \frac{1}{4},

hence

ϑ′′′_{g′_{r}h′_{r}}(\frac{ατ + β}{γτ + δ}; H_{r}, 2N_{r}) = −π^{3}ie(\frac{1}{16N_{r}} \frac{ατ + β}{γτ + δ} ) ×

× ∑_{m≡H_{r} (mod 2N_{r})}(−1)^{h′_{r}(m−H_{r})/2N_{r}}(2m + g′_{r})^{3} ×

×e\{\frac{1}{4N_{r}} ((m + \frac{1}{2}(g′_{r} − 1))^{2} + (m + \frac{1}{2}(g′_{r} − 1)))\frac{ατ + β}{γτ + δ} \} =

= e(\frac{h_{r}}{16N_{r}} \frac{ατ + β}{γτ + δ} ) ∑_{n_{1}=0}^{∞} B′_{n_{1}}e(\frac{n_{1}}{4N_{r}} \frac{ατ + β}{γτ + δ} ),

since (5) implies that h_{r} = 1. Analogously,

ϑ′_{g′_{s}h′_{s}}(\frac{ατ + β}{γτ + δ}; H_{s}, 2N_{s}) = e(\frac{h_{s}}{16N_{s}} \frac{ατ + β}{γτ + δ} ) ∑_{n_{2}=0}^{∞} B′_{n_{2}}e(\frac{n_{2}}{4N_{s}} \frac{ατ + β}{γτ + δ} ),

which implies that h_{s} = 1 if h_{s} is odd and h_{s} = 0 if h_{s} is even. Thus, if among h_{1} and h_{2} at least one is odd, then we have, for r = 1, s = 2 and r = 2, s = 1,

ϑ′′′_{g′_{r}h′_{r}}(\frac{ατ + β}{γτ + δ}; H_{r}, 2N_{r})ϑ′_{g′_{s}h′_{s}}(\frac{ατ + β}{γτ + δ}; H_{s}, 2N_{s}) =

= e(\frac{N/4(h_{r}/N_{r} + h_{s}/N_{s})}{4N} \frac{ατ + β}{γτ + δ} ) ∑_{n_{1}=0}^{∞} B′_{n_{1}}e(\frac{n_{1}}{4N_{r}} \frac{ατ + β}{γτ + δ} ) ×

× ∑_{n_{2}=0}^{∞} B′_{n_{2}}e(\frac{n_{2}}{4N_{s}} \frac{ατ + β}{γτ + δ} ) = ∑_{n=0}^{∞} B′_{n}e(\frac{n}{4N} \frac{ατ + β}{γτ + δ} ),  (2.9)

since, by (2.3), \frac{N}{4} (\frac{h_{r}}{N_{r}} + \frac{h_{s}}{N_{s}}) is an integer. Hence for odd γ (2.4) follows 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

\frac{1}{N_{2}^{2}} (γτ + δ)^{5}ϑ_{g_{1}h_{1}}^{(4)}(τ; 0, 2N_{1})ϑ_{g_{2}h_{2}}(τ; 0, 2N_{2}) =

= \frac{1}{N_{1}^{2}} e(\frac{5}{4} sgn γ)(2|γ|(N_{1}N_{2})^{1/2})^{−1} ×

66 G. LOMADZE

× ∑_{H_{1} mod2N_{1}} φ_{g′_{1}g_{1}h_{1}}(0, H_{1}; 2N_{1})\{ϑ_{g′_{1}h′_{1}}^{(4)}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) +

+\frac{4 · 3}{2!}A_{21}|_{z=0} · ϑ′′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) +

+\frac{A_{42}}{2!} |_{z=0} · ϑ_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})\} ×

× ∑_{H_{2} mod2N_{2}} φ_{g′_{2}g_{2}h_{2}}(0, H_{2}; 2N_{2})ϑ_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) =

= e(\frac{5}{4} sgn γ)(2|γ|(N_{1}N_{2})^{1/2})^{−1} ×

× ∑_{\begin{matrix}H_{1} mod2N_{1} \\ H_{2} mod2N_{2} \end{matrix}} φ_{g′_{1}g_{1}h_{1}}(0, H_{1}; 2N_{1})φ_{g′_{2}g_{2}h_{2}}(0, H_{2}; 2N_{2}) ×

×\{\frac{1}{N_{1}^{2}} ϑ_{g′_{1}h′_{1}}^{(4)}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) ×

×ϑ_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) − \frac{24γπi}{N_{1}}(γτ + δ)ϑ′′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) ×

ϑ_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) − 48γ^{2}π^{2}(γτ + δ)^{2}ϑ_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) ×

×ϑ_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2})\}.  (2.10)

Replacing N_{1}, g_{1}, h_{1}, H_{1}, g′_{1}, h′_{1} by N_{2}, g_{2}, H_{2}, g′_{2}, h′_{2} in (2.10) and vice versa, we obtain

\frac{1}{N_{1}^{2}} (γτ + δ)^{5}ϑ_{g_{2}h_{2}}^{(4)}(τ; 0, 2N_{2})ϑ_{g_{1}h_{1}}(τ; 0, 2N_{1}) =

= e(\frac{5}{4} sgn γ)(2|γ|(N_{1}N_{2})^{1/2})^{−1}×

× ∑_{\begin{matrix}H_{2} mod2N_{2} \\ H_{1} mod2N_{1} \end{matrix}} φ_{g′_{2}g_{2}h_{2}}(0, H_{2}; 2N_{2})φ_{g′_{1}g_{1}h_{1}}(0, H_{1}; 2N_{1})×

×\{\frac{1}{N_{2}^{2}} ϑ_{g′_{2}h′_{2}}^{(4)}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2})×

×ϑ_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) − \frac{24γπi}{N_{2}}(γτ + δ)ϑ′′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) ×

×ϑ_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) − 48γ^{2}π^{2}(γτ + δ)^{2}ϑ_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2})×

×ϑ_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})\}.  (2.11)

Analogously, from Lemma 4 for n = 2 it follows that

\frac{6}{N_{1}N_{2}} (γτ + δ)^{5}ϑ′′_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ′′_{g_{2}h_{2}}(τ; 0, 2N_{2}) =

= e(\frac{5}{4} sgn γ)(2|γ|(N_{1}N_{2})^{1/2})^{−1} ×

× ∑_{\begin{matrix}H_{1} mod2N_{1} \\ H_{2} mod2N_{2} \end{matrix}} φ_{g′_{1}g_{1}h_{1}}(0, H_{1}; 2N_{1})φ_{g′_{2}g_{2}h_{2}}(0, H_{2}; 2N_{2}) ×

×\{\frac{6}{N_{1}N_{2}} ϑ′′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})ϑ′′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) −

−\frac{24γπi}{N_{2}}(γτ + δ)ϑ_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})ϑ_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) −

−\frac{24γπi}{N_{1}}(γτ + δ)ϑ′′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})ϑ′′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) −

−96γ^{2}π^{2}(γτ + δ)^{2}ϑ_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) ×

×ϑ_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2})\}.  (2.12)

Subtracting (2.12) from the sum of (2.10) and (2.11), according to (2.2), we obtain

(γτ + δ)^{5}Ψ_{2}(τ; g_{1}, g_{2}, h_{1}, h_{2}, 0, 0, N_{1}, N_{2}) =

= e(\frac{5}{4} sgn γ)(2|γ|(N_{1}N_{2})^{1/2})^{−1} ×

× ∑_{\begin{matrix}H_{1} mod2N_{1} \\ H_{2} mod2N_{2} \end{matrix}} φ_{g′_{1}g_{1}h_{1}}(0, H_{1}; 2N_{1})φ_{g′_{2}g_{2}h_{2}}(0, H_{2}; 2N_{2}) ×

×Ψ_{2}(\frac{ατ + β}{γτ + δ}; g′_{1}, g′_{2}, h′_{1}, h′_{2}, H_{1}, H_{2}, N_{1}, N_{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 Ψ_{1}(τ) and Ψ_{2}(τ) with c_{1} = c_{2} = 0 are entire modular forms of weight 10 and character χ(δ) = sgn δ(\frac{−Δ}{|δ|}) (Δ is the determinant of an arbitrary positive quadratic form in 5 variables) for the congruence group Γ_{0}(4N) if the following conditions hold:

1) 2|g_{1}, 2|g_{2}, N_{1}|N, N_{2}|N,  (2.14)
2) 4|N(\frac{h_{1}^{2}}{N_{1}} + \frac{h_{2}^{2}}{N_{2}} ), 4|\frac{g_{1}^{2}}{4N_{1}} + \frac{g_{2}^{2}}{4N_{2}} ,  (2.15)

3) for all α and δ with α ≡ δ ≡ 1 (mod 4N)

(\frac{N_{1}N_{2}}{|δ|} )Ψ_{j}(τ; αg_{1}, αg_{2}, h_{1}, h_{2}, 0, 0, N_{1}, N_{2}) =

=(\frac{Δ}{|δ|})Ψ_{j}(τ; g_{1}, g_{2}, h_{1}, h_{2}, 0, 0, N_{1}, N_{2}) (j = 1, 2).  (2.16)

Proof. I. It is wellknown that the thetaseries (2)-(3) are regular on H, hence the

functions Ψ_{1}(τ) and Ψ_{2}(τ) satisfy condition 1 of the definition.

II. From (2.15), since δ^{2} ≡ 1 (mod 4), it follows that

4|Nδ^{2}(\frac{h_{1}^{2}}{N_{1}} + \frac{h_{2}^{2}}{N_{2}} ), 4|\frac{g_{1}^{2}}{4N_{1}} δ^{2φ(2N_{1})−2} + \frac{g_{2}^{2}}{4N_{2}} δ^{2φ(2N_{2}−2)}. (2.17)

By Lemma 3 for n = 3 and n = 1 (with g_{r}, h_{r}, N_{r} and g_{s}, h_{s}, N_{s} instead of g, h, N), according to (2.14) and (2.17), for each substitution from Γ_{0}(4N), we have

ϑ′′′_{g_{r}h_{r}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{r})ϑ′_{g_{s}h_{s}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{s}) = i^{η(γ)(sgnδ−1)}i^{1−|δ|} ×

×(γτ + δ)^{5}(\frac{N_{r}N_{s}}{|δ|} )ϑ′′′_{αg_{r},h_{r}}(τ; 0, 2N_{r})ϑ′_{αg_{s},h_{s}}(τ; 0, 2N_{s}) =

= sgn δ(\frac{−N_{r}N_{s}}{|δ|} )(γτ + δ)^{5} ×

×ϑ′′′_{αg_{r},h_{r}}(τ; 0, 2N_{r})ϑ′_{αg_{s},h_{s}}(τ; 0, 2N_{s})  (2.18)

for r = 1, s = 2 and r = 2, s = 1. Analogously, by Lemma 3, we have:

1) for n = 4 and n = 0

ϑ_{g_{r}h_{r}}^{(4)}(\frac{ατ + β}{γτ + δ}; 0, 2N_{r})ϑ_{g_{s}h_{s}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{s}) = i^{η(γ)(sgnδ−1)}i^{1−|δ|} ×

×(γτ + δ)^{5}(\frac{N_{r}N_{s}}{|δ|} )ϑ_{αg_{r},h_{r}}^{(4)}(τ; 0, 2N_{r})ϑ_{αg_{s},h_{s}}(τ; 0, 2N_{s}) =

= sgn δ(\frac{−N_{r}N_{s}}{|δ|} )(γτ + δ)^{5} ×

×ϑ_{αg_{r},h_{r}}^{(4)}(τ; 0, 2N_{r})ϑ′_{αg_{s},h_{s}}(τ; 0, 2N_{s})  (2.19)

if r = 1, s = 2 and r = 2, s = 1; 2) for n = 2

ϑ′′_{g_{1}h_{1}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{1})ϑ′′_{g_{2}h_{2}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{2}) =

= sgn δ(\frac{−N_{1}N_{2}}{|δ|} )(γτ + δ)^{5} ×

×ϑ′′_{αg_{1},h_{1}}(τ; 0, 2N_{1})ϑ′′_{αg_{2},h_{2}}(τ; 0, 2N_{2}).  (2.20)

Hence, according to (2.16), for all α and δ with αδ ≡ 1 (mod 4N), we have

Ψ_{j}(\frac{ατ + β}{γτ + δ}; g_{1}, g_{2}, h_{1}, h_{2}, 0, 0, N_{1}, N_{2}) =

= sgn δ(\frac{−N_{1}N_{2}}{|δ|} )(γτ + δ)^{5}Ψ_{j}(τ; αg_{1}, αg_{2}, h_{1}, h_{2}, 0, 0, N_{1}, N_{2}) =

= sgn δ(\frac{−Δ}{|δ|})(γτ + δ)^{5}Ψ_{j}(τ; g_{1}, g_{2}, h_{1}, h_{2}, 0, 0, N_{1}, N_{2})

for j = 1, by (2.1) and (2.16), and for j = 2, by (2.2),(2.19) and (2.20). Thus the functions Ψ_{1}(τ) and Ψ_{2}(τ) with c_{1} = c_{2} = 0 satisfy condition 2 of the definition.

III. From (13) it follows that, for r = 1, s = 2 and r = 2, s = 1,

ϑ′′′_{g_{r}h_{r}}(τ; 0, 2N_{r})ϑ′_{g_{s}h_{s}}(τ; 0, 2N_{s}) = π^{4} ∑_{m_{r},m_{s}=−∞}^{∞}(−1)^{h_{r}m_{r}+h_{s}m_{s}} ×

×(4N_{r}m_{r} + g_{r})^{3}(4N_{s}m_{s} + g_{s})e(Λτ),

ϑ_{g_{r}h_{r}}^{(4)}(τ; 0, 2N_{r})ϑ_{g_{s}h_{s}}(τ; 0, 2N_{s}) = π^{4} ∑_{m_{r},m_{s}=−∞}^{∞}(−1)^{h_{r}m_{r}+h_{s}m_{s}} ×

×(4N_{r}m_{r} + g_{r})^{4}e(Λτ)  (2.21)

and also

ϑ′′_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ′′_{g_{2}h_{2}}(τ; 0, 2N_{2}) = π^{4} ∑_{m_{1},m_{2}=−∞}^{∞}(−1)^{h_{1}m_{1}+h_{2}m_{2}} ×

×(4N_{1}m_{1} + g_{1})^{2}(4N_{2}m_{2} + g_{2})^{2}e(Λτ),

where

Λ = ∑_{k=1}^{2} \frac{1}{4N_{k}} (2N_{k}m_{k} + g_{k}/2)^{2} = ∑_{k=1}^{2}(N_{k}m_{k}^{2} + m_{k}g_{k}/2) + \frac{1}{4} ∑_{k=1}^{2} g_{k}^{2}/4N_{k},

by (2.14) and (2.15), is an integer. Thus, the functions Ψ_{1}(τ) and Ψ_{2}(τ) with c_{1} = c_{2} = 0 satisfy condition 3 of the definition.

IV. By Lemma 5 the functions Ψ_{1}(τ) and Ψ_{2}(τ) with c_{1} = c_{2} = 0 also satisfy condition 4 of the definition.

3.

Lemma 6. For a given N let

Φ_{1}(τ) = Φ_{1}(τ; g_{1}, …, g_{4}, h_{1}, …, h_{4}, c_{1}, …, c_{4}, N_{1}, …, N_{4}) =

= \frac{1}{N_{1}} ϑ′′′_{g_{1}h_{1}}(τ; c_{1}, 2N_{1})ϑ′_{g_{2}h_{2}}(τ; c_{2}, 2N_{2}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; c_{k}, 2N_{k}) −

−\frac{1}{N_{2}} ϑ′′′_{g_{2}h_{2}}(τ; c_{2}, 2N_{2}))ϑ′_{g_{1}h_{1}}(τ; c_{1}, 2N_{1}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; c_{k}, 2N_{k}) (3.1)

and

Φ_{2}(τ) = Φ_{2}(τ; g_{1}, …, g_{4}, h_{1}, …, h_{4}, c_{1}, …, c_{4}, N_{1}, …, N_{4}) =

= \frac{1}{N_{1}^{2}} ϑ_{g_{1}h_{1}}^{(4)}(τ; c_{1}, 2N_{1})ϑ_{g_{2}h_{2}}(τ; c_{2}, 2N_{2}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; c_{k}, 2N_{k}) +

+\frac{1}{N_{2}^{2}} ϑ_{g_{2}h_{2}}^{(4)}(τ; c_{2}, 2N_{2})ϑ_{g_{1}h_{1}}(τ; c_{1}, 2N_{1}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; c_{k}, 2N_{k}) −

−\frac{6}{N_{1}N_{2}} ϑ′′_{g_{1}h_{1}}(τ; c_{1}, 2N_{1})ϑ′′_{g_{2}h_{2}}(τ; c_{2}, 2N_{2}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; c_{k}, 2N_{k}), (3.2)

where

2|g_{k}, N_{k}|N (k = 1, 2, 3, 4), 4|N ∑_{k=1}^{4} \frac{h_{k}}{N_{k}} .  (3.3)

For all substitutions from Γ in the neighborhood of each rational point τ = −\frac{δ}{γ} (γ ≠ 0, (γ, δ) = 1), we then have

(γτ + δ)^{6}Φ_{j}(τ; g_{1}, …, g_{4}, h_{1}, …, h_{4}, 0, …, 0, N_{1}, …, N_{4}) =
= ∑_{n=0}^{∞} D_{n}^{(j)}e(\frac{n}{4N} \frac{ατ + β}{γτ + δ} ) (j = 1, 2).  (3.4)

Proof. From Lemma 4 for n = 3, n = 1, and n = 0, as in the proof of Lemma 5, it follows that

\frac{1}{N_{1}} (γτ + δ)^{6}ϑ′′′_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ′_{g_{2}h_{2}}(τ; 0, 2N_{2}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; 0, 2N_{k}) =

ON SOME ENTIRE MODULAR FORMS 71

= e(\frac{3}{2} sgn γ)(4γ^{2}( ∏_{k=1}^{4} N_{k})^{1/2})^{−1} ∑_{\begin{matrix}H_{k} mod2N_{k} \\ (k=1,2,3,4) \end{matrix}} ∏_{k=1}^{4} φ_{g′_{k}g_{k}h_{k}}(0, H_{k}; 2N_{k}) ×

×\{(\frac{1}{N_{1}} ϑ′′′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) −

−12γπi(γτ + δ)ϑ′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})) ×

×ϑ′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) ∏_{k=3}^{4} ϑ_{g′_{k}h′_{k}}(\frac{ατ + β}{γτ + δ}; H_{k}, 2N_{k})\}  (3.5)

and

\frac{1}{N_{2}} (γτ + δ)^{6}ϑ′′′_{g_{2}h_{2}}(τ; 0, 2N_{2})ϑ′_{g_{1}h_{1}}(τ; 0, 2N_{1}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; 0, 2N_{k}) =

= e(\frac{3}{2} sgn γ)(4γ^{2}( ∏_{k=1}^{4} N_{k})^{1/2})^{−1} ∑_{\begin{matrix}H_{k} mod2N_{k} \\ (k=1,2,3,4) \end{matrix}} ∏_{k=1}^{4} φ_{g′_{k}g_{k}h_{k}}(0, H_{k}; 2N_{k}) ×

×\{(\frac{1}{N_{2}} ϑ′′′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) −

−12γπi(γτ + δ)ϑ′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2})) ×

×ϑ′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) ∏_{k=3}^{4} ϑ_{g′_{k}h′_{k}}(\frac{ατ + β}{γτ + δ}; H_{k}, 2N_{k})\}.  (3.6)

As in the proof of Lemma 5, from Lemma 4 it follows that

1) for n = 4 and n = 0

\frac{1}{N_{1}^{2}} (γτ + δ)^{6}ϑ_{g_{1}h_{1}}^{(4)}(τ; 0, 2N_{1}) ∏_{k=2}^{4} ϑ_{g_{k}h_{k}}(τ; 0, 2N_{k}) =

= e(\frac{3}{2} sgn γ)(4γ^{2}( ∏_{k=1}^{4} N_{k})^{1/2})^{−1} ∑_{\begin{matrix}H_{k} mod2N_{k} \\ (k=1,2,3,4) \end{matrix}} ∏_{k=1}^{4} φ_{g′_{k}g_{k}h_{k}}(0, H_{k}; 2N_{k}) ×

×\{(\frac{1}{N_{1}^{2}} ϑ_{g′_{1}h′_{1}}^{(4)}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) −

−\frac{24γπi}{N_{1}}(γτ + δ)ϑ′′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) −

−48γ^{2}π^{2}(γτ + δ)^{2}ϑ_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})) ×

72 G. LOMADZE

× ∏_{k=2}^{4} ϑ_{g′_{k}h′_{k}}(\frac{ατ + β}{γτ + δ}; H_{k}, 2N_{k})\}  (3.7)

and

\frac{1}{N_{2}^{2}} (γτ + δ)^{6}ϑ_{g_{2}h_{2}}^{(4)}(τ; 0, 2N_{2})ϑ_{g_{1}h_{1}}(τ; 0, 2N_{1}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; 0, 2N_{k}) =

= e(\frac{3}{2} sgn γ)(4γ^{2}( ∏_{k=1}^{4} N_{k})^{1/2})^{−1} ∑_{\begin{matrix}H_{k} mod2N_{k} \\ (k=1,2,3,4) \end{matrix}} ∏_{k=1}^{4} φ_{g′_{k}g_{k}h_{k}}(0, H_{k}; 2N_{k}) ×

×\{(\frac{1}{N_{2}^{2}} ϑ_{g′_{2}h′_{2}}^{(4)}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) −

−\frac{24γπi}{N_{2}}(γτ + δ)ϑ′′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2})) ×

×ϑ_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) ∏_{k=3}^{4} ϑ_{g′_{k}h′_{k}}(\frac{ατ + β}{γτ + δ}; H_{k}, 2N_{k}) −

−48γ^{2}π^{2}(γτ + δ)^{2} ∏_{k=1}^{4} ϑ_{g′_{k}h′_{k}}(\frac{ατ + β}{γτ + δ}; H_{k}, 2N_{k})\},  (3.8)

2) for n = 2 and n = 0

\frac{6}{N_{1}N_{2}} (γτ + δ)^{6}ϑ′′_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ′′_{g_{2}h_{2}}(τ; 0, 2N_{2}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; 0, 2N_{k}) =

= e(\frac{3}{2} sgn γ)(4γ^{2}( ∏_{k=1}^{4} N_{k})^{1/2})^{−1} ∑_{\begin{matrix}H_{k} mod2N_{k} \\ (k=1,2,3,4) \end{matrix}} φ_{g′_{k}g_{k}h_{k}}(0, H_{k}; 2N_{k}) ×

×\{(\frac{6}{N_{1}N_{2}} ϑ′′_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) −

−\frac{24γπi}{N_{2}}(γτ + δ)ϑ_{g′_{1}h′_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})) ×

×ϑ′′_{g′_{2}h′_{2}}(\frac{ατ + β}{γτ + δ}; H_{2}, 2N_{2}) ∏_{k=3}^{4} ϑ_{g′_{k}h′_{k}}(\frac{ατ + β}{γτ + δ}; H_{k}, 2N_{k}) −

−(\frac{24γπi}{N_{1}}(γτ + δ)ϑ′′_{g_{1}h_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1}) +

+96γ^{2}π^{2}(γτ + δ)^{2}ϑ_{g_{1}h_{1}}(\frac{ατ + β}{γτ + δ}; H_{1}, 2N_{1})) ×

× ∏_{k=2}^{4} ϑ_{g′_{k}h′_{k}}(\frac{ατ + β}{γτ + δ}; H_{k}, 2N_{k})\}.  (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}Φ_{j}(τ; g_{1}, …, g_{4}, h_{1}, …, h_{4}, 0, …, N_{1}, …, N_{4}) =

= e(\frac{3}{2} sgn γ)(4γ^{2}( ∏_{k=1}^{4} N_{k})^{1/2})^{−1} ∑_{\begin{matrix}H_{k} mod2N_{k} \\ (k=1,2,3,4) \end{matrix}} φ_{g′_{k}g_{k}h_{k}}(0, H_{k}; 2N_{k}) ×

×Φ_{j}(\frac{ατ + β}{γτ + δ}; g_{1}, …, g_{4}, h_{1}, …, h_{4}, H_{1}, …, H_{4}, N_{1}, …, N_{4})

(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 Φ_{1}(τ) and Φ_{2}(τ) with c_{1} = c_{2} = c_{3} = c_{4} = 0 are entire modular forms of weight 12 and character χ(δ) = (\frac{Δ}{|δ|}) (Δ is the determinant of an arbitrary positive quadratic form in 6 variables) for the congruence group Γ_{0}(4N) if the following conditions hold:

1) 2|g_{k}, N_{k}|N  (k = 1, 2, 3, 4),  (3.11)
2) 4|N ∑_{k=1}^{4}(\frac{h_{k}^{2}}{N_{k}} , 4| ∑_{k=1}^{4} \frac{g_{k}^{2}}{4N_{k}} ,  (3.12)

3) for all α and δ with αδ ≡ 1 (mod 4N)

(\frac{∏_{k=1}^{4} N_{k}}{|δ|} )Φ_{j}(τ; αg_{1}, …, αg_{4}, h_{1}, …, h_{4}, 0, …, 0, N_{1}, …, N_{4}) =

= (\frac{Δ}{|δ|})Φ_{j}(τ; g_{1}, …, g_{4}, h_{1}, …, h_{4}, 0, …, 0, N_{1}, …, N_{4})

(j = 1, 2).  (3.13)

Proof.

I. As in the case of Theorem 1, the functions Φ_{1}(τ) and Φ_{2}(τ) with c_{1} = c_{2} = c_{3} = c_{4} = 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|Nδ^{2} ∑_{k=1}^{4} \frac{h_{k}^{2}}{N_{k}} , 4| ∑_{k=1}^{4} \frac{g_{k}^{2}}{4N_{k}} δ^{2φ(2N_{k})−2}.  (3.14)

By Lemma 3, for n = 3, n = 1 and n = 0, according to (3.11) and (3.14), for each substitution from Γ_{0}(4N), we have

ϑ′′′_{g_{r}h_{r}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{r})ϑ′_{g_{s}h_{s}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{s}) ×

× ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{k}) =

= (γτ + δ)^{6}(\frac{∏_{k=1}^{4} N_{k}}{|δ|} )ϑ′′′_{αg_{r},h_{r}}(τ; 0, 2N_{r})ϑ′_{αg_{s},h_{s}}(τ; 0, 2N_{s}) ×

× ∏_{k=3}^{4} ϑ_{αg_{k},h_{k}}(τ; 0, 2N_{k})  (3.15)

for r = 1, s = 2 and r = 2, s = 1. Analogously, by Lemma 3, we have:

1) for n = 4 and n = 0

ϑ_{g_{r}h_{r}}^{(4)}(\frac{ατ + β}{γτ + δ}; 0, 2N_{r})ϑ_{g_{s}h_{s}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{s}) ×

× ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{k}) =

= (γτ + δ)^{6}(\frac{∏_{k=1}^{4} N_{k}}{|δ|} )ϑ_{αg_{r},h_{r}}^{(4)}(τ; 0, 2N_{r})ϑ_{αg_{s},h_{s}}(τ; 0, 2N_{s}) ×

× ∏_{k=3}^{4} ϑ_{αg_{k},h_{k}}(τ; 0, 2N_{k})  (3.16)

for r = 1, s = 2 and r = 2, s = 1, 2) for n = 2 and n = 0

∏_{k=1}^{2} ϑ′′_{g_{k}h_{k}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{k}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{k}) =

= (γτ + δ)^{6}(\frac{∏_{k=1}^{4} N_{k}}{|δ|} ) ∏_{k=1}^{2} ϑ′′_{αg_{k},h_{k}}(τ; 0, 2N_{k}) ∏_{k=3}^{4} ϑ_{αg_{k},h_{k}}(τ; 0, 2N_{k}).

Hence, according to (3.13), for all α and δ with αδ ≡ 1 (mod 4N), we have

Φ_{j}(\frac{ατ + β}{γτ + δ}; g_{1}, …, g_{4}, h_{1}, …, h_{4}, 0, …, N_{1}, …, N_{4}) =

= (\frac{∏_{k=1}^{4} N_{k}}{|δ|} )(γτ + δ)^{6} ×

×Φ_{j}(τ; αg_{1}, …, αg_{4}, h_{1}, …, h_{4}, 0, …, 0, N_{1}, …, N_{4}) =

= (\frac{Δ}{|δ|})(γτ + δ)^{6}Φ_{j}(τ; g_{1}, …, g_{4}, h_{1}, …, h_{4}, 0, …, N_{1}, …, N_{4})

for j = 1, by (3.1) and (3.15), and for j = 2, by (3.2), (3.16), and (3.17). Thus the functions Φ_{1}(τ) and Φ_{2}(τ) satisfy condition 2 of the definition.

III. From (13) it follows that, for r = 1, s = 2 and r = 2, s = 1,

ϑ′′′_{g_{r}h_{r}}(τ; 0, 2N_{r})ϑ′_{g_{s}h_{s}}(τ; 0, 2N_{s}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; 0, 2N_{k}) =

= π^{4} ∑_{m_{r},m_{s},m_{3},m_{4}=−∞}^{∞} (−1)^{m_{r}+m_{s}+m_{3}+m_{4}}(4N_{r}m_{r} + g_{r})^{3}(4N_{s}m_{s} + g_{s})e(Λτ),

ϑ_{g_{r}h_{r}}^{(4)}(τ; 0, 2N_{r})ϑ_{g_{s}h_{s}}(τ; 0, 2N_{s}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; 0, 2N_{k}) =

= π^{4} ∑_{m_{r},m_{s},m_{3},m_{4}=−∞}^{∞}(−1)^{m_{r}+m_{s}+m_{3}+m_{4}}(4N_{r}m_{r} + g_{r})^{4}e(Λτ),

and also

ϑ′′_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ′′_{g_{2}h_{2}}(τ; 0, 2N_{2}) ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; 0, 2N_{k}) =

= π^{4} ∑_{m_{1},... ,m_{4}=−∞}^{∞}(−1)^{∑_{k=1}^{4} h_{k}m_{k}}(4N_{1}m_{1} + g_{1})^{2}(4N_{2}m_{2} + g_{2})^{2}e(Λτ),

where

Λ = ∑_{k=1}^{4} \frac{1}{4N_{k}} (2N_{k}m_{k} + \frac{g_{k}}{2} )^{2} = ∑_{k=1}^{4} (N_{k}m_{k}^{2} + \frac{1}{2}m_{k}g_{k}) + \frac{1}{4} ∑_{k=1}^{4} \frac{g_{k}^{2}}{4N_{k}} ,

by (3.11) and (3.12), is an integer. Thus the functions Φ_{1}(τ) and Φ_{2}(τ) with c_{1} = c_{2} = c_{3} = c_{4} = 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.

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