GEORGIAN MATHEMATICAL JOURNAL: Vol. 2, No. 2, 1995, 189-199

CONSTRUCTION OF ENTIRE MODULAR FORMS OF WEIGHTS 5 AND 6 FOR THE CONGRUENCE GROUP

Γ_{0}(4N)

G. LOMADZE

Abstract. Two classes of entire modular forms of weight 5 and two of weight 6 are constructed for the congruence subgroup Γ_{0}(4N). The constructed modular forms as well as the modular forms from [1] will be helpful in the theory of representation of numbers by the quadratic forms in 10 and 12 variables.

The present paper is a direct continuation of [1] whose notation will be preserved here.

1.

Lemma 1. For a given N let

Ψ_{3}(τ) = Ψ_{3}(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}) −

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

×ϑ′_{g_{3}h_{3}}(τ; c_{3}, 2N_{3})ϑ_{g_{4}h_{4}}(τ; c_{4}, 2N_{4})  (1.1)

and

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

= ∏_{k=1}^{3} ϑ′_{g_{k}h_{k}}(τ; c_{k}, 2N_{k})ϑ_{g_{4}h_{4}}(τ; c_{4}, 2N_{4}),  (1.2)

189 where

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

Key words and phrases. Quadratic form, entire modular form, congruence subgroup, theta-series, generalized Gauss’ sum.

1072-947X/95/0300-0189$07.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1995 Plenum Publishing Corporation

2|g_{k}, N_{k}|N (k = 1, 2, 3, 4), 4|N ∑_{k=1}^{4} \frac{h_{k}}{N_{k}} .  (1.3)

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_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =
= ∑_{n=0}^{∞} C_{n}^{(j)}e(\frac{n}{4N} \frac{ατ + β}{γτ + δ} ) (j = 3, 4).  (1.4)

Proof. I. Taking into account (1.19) from [1], by Lemma 4 from [1], it follows for n = 2 (with g_{1}, h_{1}, g′_{1}, h′_{1}, N_{1}, H_{1} instead of g, h, g′, h′, N, H) and n = 0 (with g_{2}, h_{2}, g′_{2}, h′_{2}, N_{2}, H_{2} instead of g, h, g′, h′, N, H) that

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

= −\frac{1}{N_{1}} e(\frac{3}{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}) + 2A_{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{3}{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}) − 4γπi(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})\}.  (1.5)

If in (1.5) N_{1}, g_{1}, h_{1}, H_{1}, g′_{1}, h′_{1} are replaced by N_{2}, g_{2}, h_{2}, H_{2}, g′_{2}, h′_{2}, and vice versa, then we have

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

= −e(\frac{3}{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}) −

−4γπi(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})\}.  (1.6)

Subtracting (1.6) from (1.5), we obtain

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

−\frac{1}{N_{2}} ϑ_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ′′_{g_{2}h_{2}}(τ; 0, 2N_{2})\} = −e(\frac{3}{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}) −

−\frac{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})\}.  (1.7)

Analogously, by Lemma 4 from [1], for n − 1 and n = 0 we obtain

(γτ + δ)^{2}ϑ′_{g_{3}h_{3}}(τ; 0, 2N_{3})ϑ_{g_{4}h_{4}}(τ; 0, 2N_{4}) =

= −e(\frac{1}{2} sgn γ)(2|γ|(N_{3}N_{4})^{1/2})^{−1}i sgn γ ×

× ∑_{\begin{matrix}H_{3} mod2N_{3} \\ H_{4} mod2N_{4} \end{matrix}} φ_{g′_{3}g_{3}h_{3}}(0, H_{3}; 2N_{3})φ_{g′_{4}g_{4}h_{4}}(0, H_{4}; 2N_{4}) ×

×ϑ′_{g′_{3}h′_{3}}(\frac{ατ + β}{γτ + δ}; H_{3}, 2N_{3})ϑ_{g′_{4}h′_{4}}(\frac{ατ + β}{γτ + δ}; H_{4}, 2N_{4}).  (1.8)

Multiplying (1.7) by (1.8), on account of (1.1), we obtain

(γτ + δ)^{5}Ψ_{3}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =

= e(\frac{5}{4} sgn γ)(4γ^{2}( ∏_{k=1}^{4} N_{k})^{1/2})^{−1}i sgn γ ×

× ∑_{\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}) ×

×Ψ_{3}(\frac{ατ + β}{γτ + δ}; g′_{1}, …, g′_{4}; h′_{1}, …, h′_{4}; H_{1}, …, H_{4}; N_{1}, …, N_{4}). (1.9)

Further, applying the same reasoning as in [1, Lemma 5, pp. 62-63], we obtain (1.4) if j = 3.

II. As in Subsection I, by Lemma 4 from [1], for n = 1 (with g_{k}, h_{k}, N_{k}, g′_{k}, h′_{k}, H_{k} for all 1 ≤ k ≤ 4 instead of g, h, N, g′, h′, H) and n = 0 (with g_{4}, h_{4}, N_{4}, g′_{4}, h′_{4}, H_{4} instead of g, h, N, g′, h′, H), we obtain

(γτ + δ)^{5} ∏_{k=1}^{3} ϑ′_{g_{k}h_{k}}(τ; 0, 2N_{k})ϑ_{g_{4}h_{4}}(τ; 0, 2N_{4}) =

= e(\frac{5}{4} sgn γ)(4γ^{2}( ∏_{k=1}^{4} N_{k})^{1/2})^{−1}i sgn γ ×

× ∑_{\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}) ×

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

Hence, according to (1.2), it follows that

(γτ + δ)^{5}Ψ_{4}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =

= e(\frac{5}{4} sgn γ)(4γ^{2}( ∏_{k=1}^{4} N_{k})^{1/2})^{−1}i sgn γ ×

× ∑_{\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}) ×

×Ψ_{4}(\frac{ατ + β}{γτ + δ}; g′_{1}, …, g′_{4}; h′_{1}, …, h′_{4}; H_{1}, …, H_{4}; N_{1}, …, N_{4}). (1.10)

Further, applying the same reasoning as in [1, Lemma 5], we obtain (1.4) if j = 4.

Theorem 1. For a given N the functions Ψ_{3}(τ) and Ψ_{4}(τ) with c_{1} = c_{2} = c_{3} = c_{4} = 0 are entire modular forms of weight 5 and character χ(δ) = sgn γ(\frac{−Δ}{|δ|}) (Δ is the determinant of an arbitrary positive quadratic form in 10 variables) for the group Γ_{0}(4N) if the following conditions hold:

1) 2|g_{k}, N_{k}|N  (k = 1, 2, 3, 4),  (1.11)

2) 4|N ∑_{k=1}^{4} \frac{h_{k}^{2}}{N_{k}} , 4| ∑_{k=1}^{4} \frac{g_{k}^{2}}{4N_{k}} ,  (1.12)

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

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

=sgn δ(\frac{−Δ}{|δ|})Ψ_{j}(τ; g_{1}, ..., g_{4}; h_{1}, ..., h_{4}; 0, ..., 0;N_{1}, ..., N_{4})  (1.13)
(j = 3, 4).

Proof. I. As in the case of Theorem 1 from [1], the functions Ψ_{3}(τ) and Ψ_{4}(τ) with c_{1} = c_{2} = c_{3} = c_{4} = 0 satisfy the condition 1) and, by Lemma 1, also the condition 4) of the definition from [1, p. 53–54].

II. From (1.12), since 2 ∤ δ, it follows that

4|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}.  (1.14)

Taking into account (1.11), by Lemma 3 from [1], for n = 2 and n = 0 (with g_{r}, h_{r}, N_{r} and g_{s}, h_{s}, N_{s} instead of g, h, N), we obtain for each substitution from Γ_{0}(4N)

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^{3η(γ)(sgnδ−1)}i^{1−|δ|} ×

×(\frac{N_{r}N_{s}}{|δ|} )(γτ + δ)^{3}e(\frac{βδ}{4} (\frac{g_{r}^{2}}{4N_{r}} δ^{2φ(2N_{r})−2} + \frac{g_{s}^{2}}{4N_{s}} δ^{2φ(2N_{s})−2})) ×

×e( − \frac{αγδ^{2}}{4} (\frac{h_{r}^{2}}{4N_{r}} + \frac{h_{s}^{2}}{4N_{s}} )) ×

×ϑ′′_{αg_{r},h_{r}}(τ; 0, 2N_{r})ϑ′_{αg_{s},h_{s}}(τ; 0, 2N_{s})  (1.15)

for r = 1, s = 2 and r = 2, s = 1. Analogously, by Lemma 3 from [1], for n = 1 and n = 0, we have

ϑ′_{g_{3}h_{3}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{3})ϑ_{g_{4}h_{4}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{4}) = sgn δ i^{2η(γ)(sgnδ−1)} ×

×i^{1−|δ|}(\frac{N_{3}N_{4}}{|δ|} )(γτ + δ)^{2}e(\frac{βδ}{4} (\frac{g_{3}^{2}}{4N_{3}} δ^{2φ(2N_{3})−2} + \frac{g_{4}^{2}}{4N_{4}} δ^{2φ(2N_{4})−2})) ×

×e( − \frac{αγδ^{2}}{4} (\frac{h_{3}^{2}}{4N_{3}} + \frac{h_{4}^{2}}{4N_{4}} )) ×

×ϑ′_{αg_{3},h_{3}}(τ; 0, 2N_{3})ϑ_{αg_{4},h_{4}}(τ; 0, 2N_{s}).  (1.16)

Hence, by (1.1), (1.15), (1.16), and (1.14), we obtain

Ψ_{3}(\frac{ατ + β}{γτ + δ}; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =

= sgn δ i^{η(γ)(sgn δ−1)}(−1)^{1−|δ|}(\frac{∏_{k=1}^{4} N_{k}}{|δ|} )(γτ + δ)^{5} ×

×Ψ_{3}(τ; αg_{1}, …, αg_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}),

from which according to (1.13) it follows for all α and δ with αδ ≡ 1 (mod 4N) that

Ψ_{3}(\frac{ατ + β}{γτ + δ}; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =

= sgn δ (\frac{−Δ}{|δ|})(γτ + δ)^{5}Ψ_{3}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}).

Analogously, by the just mentioned Lemma 3, for n = 1 and n = 0 (with g_{k}, h_{k}, N_{k} (k = 1, 2, 3, 4) instead of g, h, N), according to (1.11) and (1.14), we find for all substitutions from Γ_{0}(4N) that

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

= sgn δ i^{η(γ)(sgn δ−1)}(−1)^{1−|δ|}(\frac{∏_{k=1}^{4} N_{k}}{|δ|} )(γτ + δ)^{5} ×

× ∏_{k=1}^{3} ϑ′_{αg_{k},h_{k}}(τ; 0, 2N_{k})ϑ′_{αg_{4},h_{4}}(τ; 0, 2N_{4}).  (1.17)

Hence, by (1.2), (1.17), and (1.13), for all α and δ with αδ ≡ 1 (mod 4N) we have

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

= sgn δ (\frac{−Δ}{|δ|})(γτ + δ)^{5}Ψ_{4}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}).

Thus the functions Ψ_{3}(τ) and Ψ_{4}(τ) with c_{1} = c_{2} = c_{3} = c_{4} = 0 satisfy the condition 2) of the definition from [1].

III. According to (13) from [1] we have

1) ϑ′′_{g_{r}h_{r}}(τ; 0, 2N_{r})ϑ_{g_{s}h_{s}}(τ; 0, 2N_{s}) =
=−π^{2} ∑_{m_{r},m_{s}=−∞}^{∞} (−1)^{h_{r}m_{r}+h_{s}m_{s}}(4N_{r}m_{r}+g_{r})^{2}e(Λ_{1}τ)  (1.18)

for r = 1, s = 2 and r = 2, s = 1, where

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

= ∑_{k=1}^{2} (N_{k}m_{k}^{2} + \frac{1}{2}m_{k}g_{k}) + \frac{1}{4} ∑_{k=1}^{2} g_{k}^{2}/4N_{k};  (1.19)

2) ϑ′_{g_{3}h_{3}}(τ; 0, 2N_{3})ϑ_{g_{4}h_{4}}(τ; 0, 2N_{4}) =
=πi ∑_{m_{3},m_{4}=−∞}^{∞}(−1)^{h_{3}m_{3}+h_{4}m_{4}}(4N_{3}m_{3} + g_{3})e(Λ_{2}τ),  (1.20)

where

Λ_{2} = ∑_{k=3}^{4} \frac{1}{4N_{k}} (2N_{k}m_{k} + g_{k}/2)^{2} =

= ∑_{k=3}^{4} (N_{k}m_{k}^{2} + \frac{1}{2}m_{k}g_{k}) + \frac{1}{4} ∑_{k=3}^{4} g_{k}^{2}/4N_{k};  (1.21)

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

=−π^{3}i ∑_{m_{1},...,m_{4}=−∞}^{∞} (−1)^{∑_{k=1}^{4} h_{k}m_{k}}∏_{k=1}^{3}(4N_{k}m_{k} + g_{k})e(Λτ),

where

Λ = ∑_{k=1}^{4} \frac{1}{4N_{k}} (2N_{k}m_{k} + 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} g_{k}^{2}/4N_{k}.  (1.22)

Λ_{1} + Λ_{2} and Λ are integers by virtue of (1.11) and (1.12). Therefore the functions Ψ_{3}(τ) and Ψ_{4}(τ) satisfy the condition 3) of the definition from [1].

2.

Lemma 2. For a given N let

Φ_{3}(τ) = Φ_{3}(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}) −

−\frac{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}) (2.1)

and

Φ_{4}(τ) = Φ_{4}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; c_{1}, …, c_{4}; N_{1}, …, N_{4}) =
= ∏_{k=1}^{4} ϑ′_{g_{k}h_{k}}(τ; c_{k}, 2N_{k}),  (2.2)

where

2|g_{k}, N_{k}|N (k = 1, 2, 3, 4), 4|N ∑_{k=1}^{4} \frac{h_{k}}{N_{k}} .  (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)^{6}Φ_{j}(τ) = ∑_{n=0}^{∞} D_{n}^{(j)}e(\frac{n}{4N} \frac{ατ + β}{γτ + δ} ) (j = 3, 4).  (2.4)

Proof. I. As in the proof of Lemma 1, by Lemma 4 from [1], it follows for n = 1 that

(γτ + δ)^{3}ϑ′_{g_{3}h_{3}}(τ; 0, 2N_{3})ϑ′_{g_{4}h_{4}}(τ; 0, 2N_{4}) =

= −e(\frac{3}{4} sgn γ)(2|γ|(N_{3}N_{4})^{1/2})^{−1} ×

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

×ϑ′_{g′_{3}h′_{3}}(\frac{ατ + β}{γτ + δ}; H_{3}, 2N_{3})ϑ′_{g′_{4}h′_{4}}(\frac{ατ + β}{γτ + δ}; H_{4}, 2N_{4}).  (2.5)

Multiplying (1.7) by (2.5), by virtue of (2.1), we obtain

(γτ + δ)^{6}Φ_{3}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 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_{3}(\frac{ατ + β}{γτ + δ}; g′_{1}, …, g′_{4}; h′_{1}, …, h′_{4}; H_{1}, …, H_{4}; N_{1}, …, N_{4}). (2.6)

Further, applying the same reasoning as in [1, Lemma 5], we obtain (2.4) if j = 3.

II. By Lemma 4 from [1], for n = 1 we have

(γτ + δ)^{6} ∏_{k=1}^{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}) ∏_{k=1}^{4} ϑ′_{g′_{k}h′_{k}}(\frac{ατ + β}{γτ + δ}; H_{k}, 2N_{k}).

Hence, according to (2.2), it follows that

(γτ + δ)^{6}Φ_{4}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 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}} ∏_{k=1}^{4} φ_{g′_{k}g_{k}h_{k}}(0, H_{k}; 2N_{k}) ×

×Φ_{4}(\frac{ατ + β}{γτ + δ}; g′_{1}, …, g′_{4}; h′_{1}, …, h′_{4}; H_{1}, …, H_{4}; N_{1}, …, N_{4}).

Further, applying the same reasoning as in [1, Lemma 5], we obtain (2.4) if j = 4.

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

1) 2|g_{k}, N_{k}|N  (k = 1, 2, 3, 4),  (2.7)
2) 4|N ∑_{k=1}^{4} \frac{h_{k}^{2}}{N_{k}} , 4| ∑_{k=1}^{4} \frac{g_{k}^{2}}{4N_{k}} ,  (2.8)

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})  (2.9)
(j = 3, 4).

Proof. I. As in the case of Theorem 1, the functions Φ_{3}(τ) and Φ_{4}(τ) with c_{1} = c_{2} = c_{3} = c_{4} = 0 satisfy the condition 1) and, on account of Lemma 2, also the condition 4) of the definition from [1].

II. Since 2 ∤ δ, from (2.8) it follows that

4|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}.  (2.10)

Taking into account (2.7), by Lemma 3 from [1], for n = 1 we find that for all substitutions from Γ_{0}(4N) we have

∏_{k=3}^{4} ϑ′_{g_{k}h_{k}}(\frac{ατ + β}{γτ + δ}; 0, 2N_{k}) = i^{3η(γ)(sgnδ−1)}i^{1−|δ|}(\frac{N_{3}N_{4}}{|δ|} ) ×

×(γτ + δ)^{3}e(\frac{βδ}{4} (\frac{g_{3}^{2}}{4N_{3}} δ^{2φ(2N_{3})−2} + \frac{g_{4}^{2}}{4N_{4}} δ^{2φ(2N_{4})−2})) ×

×e( − \frac{αγδ^{2}}{4} (\frac{h_{3}^{2}}{4N_{3}} + \frac{h_{4}^{2}}{4N_{4}} )) ∏_{k=3}^{4} ϑ′_{αg_{k},h_{k}}(τ; 0, 2N_{k}).  (2.11)

Taking into account (2.10) and (2.9), by (2.1), (1.15), and (2.11), for all α and δ with αδ ≡ 1 (mod 4N) we obtain

Φ_{3}(\frac{ατ + β}{γτ + δ}; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =

= i^{2η(γ)(sgn δ−1)}(−1)^{1−|δ|}(\frac{∏_{k=1}^{4} N_{k}}{|δ|} )(γτ + δ)^{6} ×

×Φ_{3}(τ; αg_{1}, …, αg_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =

= (\frac{Δ}{|δ|})(γτ + δ)^{6}Φ_{3}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}).

Analogously, taking into account (2.7) and (2.10), by Lemma 3 from [1], for n = 1, we have

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

= (\frac{∏_{k=1}^{4} N_{k}}{|δ|} )(γτ + δ)^{6} ∏_{k=1}^{4} ϑ_{αg_{k},h_{k}}(τ; 0, 2N_{k})  (2.12)

for all substitutions from Γ_{0}(4N).

Hence, by (2.2), (2.12), and (2.9), for all α and δ with αδ ≡ 1 (mod 4N), we have

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

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

Thus the functions Φ_{3}(τ) and Φ_{4}(τ) with c_{1} = c_{2} = c_{3} = c_{4} = 0 satisfy the condition 2) of the definition from [1].

III. According to (13) from [1], we have

1) (1.18) with Λ_{1} defined by (1.19); 2)

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

= −π^{2} ∑_{m_{3},m_{4}=−∞}^{∞}(−1)^{h_{3}m_{3}+h_{4}m_{4}} ∏_{k=3}^{4}(4N_{k}m_{k} + g_{k})e(Λ_{2}τ),

where Λ_{2} is defined by (1.20);

3)

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

= π^{4} ∑_{m_{1},...,m_{4}=−∞}^{∞}(−1)^{∑_{k=1}^{4} h_{k}m_{k}} ∏_{k=3}^{4}(4N_{k}m_{k} + g_{k})e(Λτ),

where Λ_{2} is defined by (1.21).

Thus the functions Φ_{3}(τ) and Φ_{4}(τ) with c_{1} = c_{2} = c_{3} = c_{4} = 0 satisfy the condition 3) of the definition from [1].

References

1. G. Lomadze, On some entire modular forms of weights 5 and 6 for the congruence group Γ_{0}(4N). Georgian Math. J. 1(1994), No. 1, 53-76.

(Received 28.06.1993)

Author’s address: Faculty of Mechanics and Mathematics I. Javakhishvili Tbilisi State University 2, University St., Tbilisi 380043 Republic of Georgia