Content uploaded by Silviu Radu
Author content
All content in this area was uploaded by Silviu Radu on Feb 18, 2016
Content may be subject to copyright.
AN ALGORITHMIC APPROACH TO RAMANUJAN
CONGRUENCES
SILVIU RADU
July 3, 2008
Abstract. In this paper we present an algorithm that takes as input a gener-
ating function of the form
Q
δ|M
Q
∞
n=1
(1 − q
δn
)
r
δ
=
P
∞
n=0
a(n)q
n
and three
positive integers m, t, p, and which returns true if a(mn+t) ≡ 0 (mod p), n ≥ 0,
or false otherwise. Our method builds on work by Rademacher [12], Kolberg
[6], Sturm [17], Eichhorn and Ono [3].
Keywords: partition congruences, number theoretic algorithm, modular
forms
Introduction
Throughout this article M denotes a positive integer, and r = (r
δ
) denotes a se-
quence of integers r
δ
indexed by all positive integer divisors δ of M.
In this paper we present an algorithm that takes as input a generating function of
the form
Q
δ |M
Q
∞
n=1
(1 − q
δn
)
r
δ
=
P
∞
n=0
a(n)q
n
and three positive integers m, t, p,
and which returns true if a(mn + t) ≡ 0 (mod p), n ≥ 0, or false otherwise. A
similar algorithm for generating functions of the form
Q
∞
n=1
(1 − q
n
)
r
1
(i.e. the case
M = 1) has already been given in [3]. Our original plan was to implement that
algorithm in order to prove some congruences from [1]. The algorithm we present
here and the one in [3] both have in common that at the end one has to check that
the congruence is true for the first coefficients up to a bound ν that the algorithm
returns, and then to use the theorem of Sturm [17] to conclude that it is true for
all coefficients. However we noticed that for our purpose the bound ν given in [3]
was extremely high for some inputs. Encouraged by comments of Peter Paule we
examined the problem in more detail. Finally our study resulted in a significant
improvement of estimating the bound ν a priori. Our main tools to derive a better
bound ν are the ones used by Rademacher [12], Newman [10]; Kolberg [6] was
another major source of inspiration.
The organization of this paper is as follows: In section 1 we present the basic
terminology. In section 2 we prepare some results needed to apply the theorem
of Sturm. The main result, Theorem 2.13, can be viewed as a generalization of a
theorem of R. Lewis [8]. In section 3 we estimate functions at different points; this
is needed in order to prove they are indeed modular forms. In section 4 we show
how to apply the theorem of Sturm in order to prove our desired congruence. In
section 5 we conclude by giving some examples.
sradu@risc.uni-linz.ac.at, supported by the FWF grant SFB F1305.
1
2 SILVIU RADU
1. Basic Terminology and Formulas
We use the notation X ≡
v
Y if X and Y are congruent modulo v.
For integers m and n we let throughout gcd(m, n) denote the greatest common
divisor of m and n which is always normalized to return positive values.
Let a be an integer relatively prime to 6, i.e. gcd(a, 6) = 1. For such a one can
easily show that a
2
− 1 ≡
24
0. Similarly if gcd(a, 3) = 1 then a
2
− 1 ≡
3
0, and
finally, if gcd(a, 2) = 1 then a
2
− 1 ≡
8
0. This facts will be used throughout the
text.
For a positive integer N we define the following matrix groups:
M
2
(Z)
∗
:=
a b
c d
| a, b, c, d ∈ Z, ad − bc > 0
,
Γ :=
a b
c d
∈ M
2
(Z)
∗
| ad − bc = 1
,
Γ
∞
:=
a b
c d
∈ Γ | c = 0
,
Γ
0
(N) :=
a b
c d
∈ Γ | c ≡
N
0
.
There is an explicit formula for the index (e.g. [15]):
(1) [Γ : Γ
0
(N)] = N
Y
p|N
(1 + p
−1
).
Throughout we use the following conventions:
• N
∗
denotes the positive integers.
• q := e
2π iτ
.
• η(τ) denotes the Dedekind eta function for which
(2) η(τ) := q
1
24
∞
Y
n=1
(1 − q
n
).
• H := {x ∈ C | Im(x) > 0}.
• H
∗
:= H ∪ Q ∪ {∞}.
• γ =
a b
c d
∈ M
2
(Z)
∗
acts on elements τ ∈ H
∗
as γτ :=
aτ+b
cτ+d
because of
the formula Im(γτ) = (ad − bc)
Im(τ)
|cτ+d|
2
(e.g.[15]). Let f(τ) be a function of
τ. We will later use that f(γ
1
(γ
2
τ)) = f ((γ
1
γ
2
)τ) where γ
1
, γ
2
∈ M
2
(Z)
∗
.
• [x]
m
denotes an element of Z
m
. (Note: [x]
m
= [y]
m
iff x ≡
m
y.)
Definition 1.1. Let k ∈ Z. A modular form of weight k for a subgroup G of Γ is
a function f (τ) defined on H
∗
such that :
(1) f(τ ) is holomorphic in H;
(2) (cτ + d)
−k
f(γτ) = f(τ ) for all τ ∈ H
∗
and all γ ∈ G;
(3) for all γ ∈ Γ the function (cτ + d)
−k
f(γτ) has a Taylor series expansion in
powers of q
1
n
, n a positive integer, which converges in a nontrivial neigh-
borhood of 0.
AN ALGORITHMIC APPROACH TO RAMANUJAN CONGRUENCES 3
Definition 1.2. Let a ∈ Z. For an odd integer n > 0 we define:
• If n = 1 then:
(3)
a
n
=
a
1
:= 1.
• If n is a prime p then:
(4)
a
n
=
a
p
:=
0 if p | a
1 if a is a square modulo p
−1 otherwise
.
• If p
α
1
1
· . . . · p
α
k
k
is the prime factorization of n then:
(5)
a
n
:=
a
p
1
α
1
· . . . ·
a
p
k
α
k
.
The symbol
a
n
is called the Legendre-Jacobi symbol.
Lemma 1.3. Let n > 0 be an odd integer, then the following relations hold:
• If a and b are integers then
(6)
a
n
b
n
=
ab
n
.
•
(7)
2
n
= (−1)
n
2
−1
8
.
• If m is an odd integer then
(8)
m
n
=
n
m
(−1)
m−1
2
n−1
2
.
Proof. See [13], page 71.
Definition 1.4. We define ǫ : {(a, b, c, d) |
a b
c d
∈ Γ} 7→ C to be the unique
mapping that satisfies
(9) η(γτ ) = (−i(cτ + d))
1
2
ǫ(a, b, c, d)η(τ),
for all γ =
a b
c d
∈ Γ and τ ∈ H
∗
.
Remark 1.5. This definition is meaningful because for all γ =
a b
c d
∈ Γ and
τ ∈ H
∗
we have η
24
(γτ ) = (cτ + b)
12
η
24
(τ) (e.g. [14]). This also implies that
ǫ
24
(a, b, c, d) = 1 for all
a b
c d
∈ Γ.
For γ ∈ Γ with gcd(a, 6) = 1, a > 0 and c > 0 Newman [10] determined ǫ as
(10) ǫ(a, b, c, d) =
c
a
e
−
aπi
12
(c−b−3)
.
4 SILVIU RADU
Lemma 1.6 (Newman [10]). Let N ∈ N
∗
, k ∈ Z and f : H
∗
7→ C a function
such that for all γ =
a b
c d
∈ Γ
0
(N) with gcd(a, 6) = 1, a > 0, c > 0 we
have f(γτ) = (cτ + d)
2k
f(τ ). Then for all γ =
a b
c d
∈ Γ
0
(N) we have
f(γτ) = (cτ + d)
2k
f(τ ).
Definition 1.7. Given a positive integer m let ϕ(t, τ) : [0, m − 1] × H
∗
7→ H
∗
be
a function with expansion ϕ(t, τ) = q
−t
P
∞
n=0
a(n)q
n
. Let S
m
be a complete set
of non-equivalent representatives of the residue classes modulo m. For κ ∈ N with
gcd(m, κ) = 1 we define:
(11) M
m,κ
(ϕ(t, τ)) :=
X
λ∈S
m
ϕ
t,
τ + κλ
m
.
In this paper we are always choosing
(12) κ := gcd(1 − m
2
, 24).
With this choice clearly gcd(κ, m) = 1.
Another property needed later is as follows:
Lemma 1.8. Let κ be as defined in (12), then 6|κm.
Proof. One can proceed by case distinction. For instance, if 2 ∤ m and 3|m, then
m
2
− 1 ≡
8
0 because of gcd(m, 2) = 1. Hence by (12) we have 8|κ, thus 6|κm. The
other cases are similar.
Lemma 1.9. Given positive integers m and κ, let ϕ(t, τ) be as in Definition 1.7.
Then we have:
(13) M
m,κ
(ϕ(t, τ)) = m
∞
X
n=0
a(mn + t)q
n
.
Proof. By Definition 1.7 we have
M
m,κ
(ϕ(t, τ)) =
X
λ∈S
m
e
−2π it
τ+κλ
m
∞
X
n=0
a(n)e
2π in
τ+κλ
m
=
∞
X
n=0
a(n)e
−
2πiτ t
m
e
2πinτ
m
X
λ∈S
m
e
2π iλ
−κt+κn
m
=
X
n≥0
n≡
m
t
ma(n)e
−
2πiτ t
m
e
2πinτ
m
=m
∞
X
n=0
a(mn + t)q
−t
m
q
t
m
q
n
=m
∞
X
n=0
a(mn + t)q
n
.
Note that the sum
P
λ∈S
m
e
2π iλ
−κt+κn
m
equals m if −κt + κn ≡
m
0. This is exactly
the case when n ≡
m
t. For n 6≡
m
t the sum is 0.
AN ALGORITHMIC APPROACH TO RAMANUJAN CONGRUENCES 5
Definition 1.10. Let M ∈ N
∗
. By R(M) we denote the set of all integer sequences
(r
δ
) indexed by all positive divisors δ of M .
Definition 1.11. For m, M ∈ N
∗
, t ∈ N such that 0 ≤ t ≤ m − 1 and r = (r
δ
) ∈
R(M), we define:
(14) f(τ, r) :=
Y
δ |M
∞
Y
n=0
(1 − q
δn
)
r
δ
=
∞
X
n=0
a(n)q
n
,
and
(15) g
m,t
(τ, r) := q
24t+
P
δ|M
δr
δ
24m
∞
X
n=0
a(mn + t)q
n
.
Lemma 1.12. For m, M ∈ N
∗
, t ∈ N such that 0 ≤ t ≤ m−1 and r = (r
δ
) ∈ R(M )
we obtain the following representation:
(16) g
m,t
(τ, r) =
1
m
m−1
X
λ=0
e
2πiκλ(−24t−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κλ)
m
.
Proof. Using (2) we see that f(τ, r) = q
−
P
δ|M
δr
δ
24
Q
δ |M
η
r
δ
(δτ). Next applying
M
m,κ
to ϕ(t, τ) := q
−t
f(τ, r), by Definition 1.7 we see that:
M
m,κ
(ϕ(t, τ)) =
m−1
X
λ=0
e
2π i(
τ+κλ
m
)(−t−
P
δ|M
δr
δ
24
)
Y
δ |M
η
r
δ
δ(τ + κλ)
m
= q
−24t−
P
δ|M
δr
δ
24m
m−1
X
λ=0
e
2πiκλ(−24t−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κλ)
m
.
Alternatively by Lemma 1.9 we obtain:
M
m,κ
(ϕ(t, τ)) = m
∞
X
n=0
a(mn + t)q
n
= mq
−
24t+
P
δ|M
δr
δ
24m
g
m,t
(τ, r).
Comparing the two expressions for M
m,κ
(ϕ(t, τ)) we obtain our assertion.
The following lemma will be used at several occasions:
Lemma 1.13. Given a real number k and maps f : H
∗
7→ C and g : Γ × H
∗
7→ C.
Suppose for all γ =
a b
c d
∈ Γ and for all τ ∈ H
∗
:
(cτ + d)
−k
f(γτ) = g(γ, τ).
Then for all ξ =
A B
C D
∈ M
2
(Z)
∗
and for all τ ∈ H
∗
:
gcd(A, C)
AD − BC
(Cτ + D)
−k
f(ξτ)
= g
A
gcd(A,C)
−y
C
gcd(A,C)
x
!
,
gcd(A, C)τ + Bx + Dy
AD−BC
gcd(A,C)
!
where the integers x and y are chosen such that Ax + Cy = gcd(A, C).
6 SILVIU RADU
Proof. Define
γ :=
A
gcd(A,C)
−y
C
gcd(A,C)
x
!
and γ
′
:=
gcd(A, C) Bx + Dy
0
AD−BC
gcd(A,C)
.
Then the statement follows from the relation ξ = γγ
′
and by
f(ξτ) = f (γ(γ
′
τ)) =
C
gcd(A, C)
(γ
′
τ) + x
k
g(γ, γ
′
τ).
2. The function g
m,t
(τ, r) under modular substitutions
Throughout this section we will assume that gcd(a, 6) = 1, a > 0 and c > 0 so
that (10) will always apply and a
2
≡
24
1. For this reason it will be convenient to
introduce the following notation:
(17) Γ
0
(N)
∗
:= {γ ∈ Γ
0
(N)|a > 0, c > 0, gcd(a, 6) = 1}.
Because M and r = (r
δ
) are assumed as fixed we will write g
m,t
(τ) := g
m,t
(τ, r)
and f(τ ) := f(τ, r) throughout.
We are interested in deriving a formula for g
m,t
(γτ ) with γ ∈ Γ
0
(N)
∗
where N is
an integer such that for every prime p with p|m we have also p|N, i.e.,
(18) p|m implies p|N,
and such that for every δ|M with r
δ
6= 0 we have δ|mN, i.e.,
(19) δ|M implies δ|mN;
and some additional properties which we will specify later. For our purpose it is
convenient to define the following set:
Definition 2.1. We define
∆ :=
(m, M, N, (r
δ
)) ∈ (N
∗
)
3
× R(M) |
m, M, N and (r
δ
) satisfy
the conditions (18) and (19).
.
Lemma 2.2. Let (m, M, N, (r
δ
)) ∈ ∆, γ =
a b
c d
∈ Γ
0
(N)
∗
and λ a nonnega-
tive integer. Then:
(i) There exist integers x and y such that
(20) (a + κλc)x + mcy = 1
and where y := y
0
(mκc)
3
for some integer y
0
.
(ii) There exists an integer a
′
satisfying a
′
a ≡
24c
1.
Let x, y and a
′
be as in (i) and (ii). Then for
(21) µ := λdx +
bx − ba
′
m
2
κ
the following statements hold:
AN ALGORITHMIC APPROACH TO RAMANUJAN CONGRUENCES 7
(iii) For ǫ as in Definition 1.4, τ ∈ H
∗
and δ|M with r
δ
6= 0 we have
η
δ(γτ + κλ)
m
= (−i(cτ + d))
1
2
ǫ(a + κλc, −δy,
mc
δ
, x)η
δ(τ + κµ)
m
e
2πiabmδ
24
,
(22)
and
(23) ǫ(a + κλc, −δy,
mc
δ
, x) =
mcδ
a + κλc
e
−
(a+κλc)πi
12
(mc/δ−3)
.
(iv) The value µ is an integer, and if λ runs through a complete set of represen-
tatives of residue classes modulo m then so does µ; i.e., λ 7→ µ is a bijection
of Z
m
.
(v)
(24) λ ≡
c
µa
2
− ab
1 − m
2
κ
.
Proof. We prove each part of Lemma 2.2 separately.
(i). We know that the equation
(25) (a + κλc)x + mcy
0
(mκc)
3
= 1
has integer solutions x and y
0
iff
(26) gcd(a + κλc, mc(mκc)
3
) = 1.
To prove (26) it suffices to prove gcd(a + κλc, m) = 1 and gcd(a + κλc, κc) = 1. We
have that
gcd(a + κλc, κc) = gcd(a, κc).
But gcd(a, c) = 1 because of ad − bc = 1, and gcd(a, κ) = 1 because of gcd(a, 6) = 1
by assumption and κ being a divisor of 24. Next we see that gcd(a + κλc, c) =
1 implies gcd(a + κλc, N) = 1 because N|c. But gcd(a + κλc, N) = 1 implies
gcd(a + κλc, m) = 1 by (18). This proves (26).
Note: Because of y = y
0
(mκc)
3
Lemma 1.8 gives
(27) y ≡
24
0.
(ii). The assumptions gcd(a, 6) = 1 and gcd(a, c) = 1 imply that gcd(a, 24c) = 1,
which is equivalent to the existence of an integer a
′
such that a
′
a ≡
24c
1.
(22). For ǫ as in Definition 1.4 let
(28) C := (−i(cτ + d))
1
2
ǫ(a + κλc, −δy,
mc
δ
, x).
For this part of the proof we exploit the relation:
(29) η
δ((a + κλc)τ + b + κ λd )
mcτ + md
= Cη
δτ + δ(b + κλd)x + mdδ y
m
which is valid under the assumption that (a + κλc)x + δy
mc
δ
= 1.
Note that
mc
δ
is a positive integer (N|c and, by (19), δ|mN), and that (a + κλc)x +
δy
mc
δ
= 1 because of (20).
8 SILVIU RADU
Relation (29) is proven by applying Lemma 1.13 with f(τ) = η(τ ), k = 1/2,
g(γ, τ) = (−i)
1
2
ǫ(a
0
, b
0
, c
0
, d
0
)η(τ), γ =
a
0
b
0
c
0
d
0
∈ Γ
and ξ =
δ(a + κλc) δ(b + κλd)
mc md
.
We will also need that for all integers j we have as a trivial consequence of (2):
(30) η(τ + j) = η(τ)e
2πij
24
.
Consequently,
η
δ(γτ + κλ)
m
=η
δ((a + κλc)τ + b + κ λd )
mcτ + md
(by substituting for γ)
=Cη
δτ + δ(b + κλd)x + mdδ y
m
(by (29))
=Cη
δτ + δ(b + κλd)x
m
(by (30) and (27))
=Cη
δτ + δ(b + κλd)x − δ ba
′
m
2
m
+ δba
′
m
=Cη
δτ + δ(b + κλd)x − δ ba
′
m
2
m
e
2πiδa
′
bm
24
(by (30))
=Cη
δ(τ + κµ)
m
e
2πiδa
′
bm
24
(by (21))
=Cη
δ(τ + κµ)
m
e
2πiδabm
24
(because of a
′
≡
24
a).
In the last line we used fact (ii), namely aa
′
≡
24c
1. This together with a
2
≡
24
1
implies that a ≡
24
a
′
because of uniqueness of the inverse.
(23). First note that
(31) gcd(a + κλc, 6) = 1,
because of κc ≡
6
0 by Lemma 1.8 and (18) together with N|c.
We have that
AN ALGORITHMIC APPROACH TO RAMANUJAN CONGRUENCES 9
ǫ(a + κλc, −δy,
mc
δ
, x)
=
mc/δ
a + κλc
e
−
(a+κλc)πi
12
(mc/δ+δy−3)
(by (10) and (31))
=
mc/δ
a + κλc
e
−
(a+κλc)πi
12
(mc/δ−3)
(by (27))
=
mc/δ
a + κλc
δ
2
a + κλc
e
−
(a+κλc)πi
12
(mc/δ−3)
(see below)
=
mcδ
a + κλc
e
−
(a+κλc)πi
12
(mc/δ−3)
(by (6)).
The third equality is shown as follows. If gcd(a + κλc, δ) = 1 then Definition
1.2 implies that
δ
2
a+κλc
= 1. To prove relative primeness we see by (18) and
(19) that each prime p dividing δ also divides N and consequently also c. So
gcd(a + κλc, p) = gcd(a, p). But since p|c and gcd(a, c) = 1 by ad − bc = 1, we
conclude that gcd(a + κλc, δ) = 1.
(iv). In order to prove that µ is an integer we need to show that bx − ba
′
m
2
≡
κ
0.
By (25) we obtain ax ≡
κ
1. We also know by (ii) that aa
′
≡
24c
1. Because of κ|24
by (12), we have that aa
′
≡
κ
1. From this follows that x ≡
κ
(a
′
a)x ≡
κ
a
′
(ax) ≡
κ
a
′
.
Consequently,
bx − ba
′
m
2
≡
κ
bx − bxm
2
≡
κ
bx(1 − m
2
) ≡
κ
0,
using κ|(1 − m
2
) from (12).
Next we show that the mapping λ 7→ µ is a bijection of Z
m
by providing an inverse
using the observation that:
µ −
bx − ba
′
m
2
κ
≡
m
λdx implies λ ≡
m
(xd)
−1
(µ −
bx − ba
′
m
2
κ
).
The only non-trivial step is to show that d and x are indeed invertible modulo m.
First of all, x is invertible modulo m because of (25). Because of ad − bc = 1 we
have that gcd(c, d) = 1, and since N|c we have that gcd(N, d) = 1. By (18) we get
that gcd(m, d) = 1 which shows that also d is invertible modulo m.
(v). By (25) we have that ax ≡
κc
1. From aa
′
≡
24c
1 and κ|24 we conclude that
aa
′
≡
κc
1 which implies x ≡
κc
a
′
by uniqueness of the inverse.
Because of the relation ad − bc = 1 we have that ad ≡
c
1. From ax ≡
κc
1 it follows
that ax ≡
c
1 which implies d ≡
c
x by uniqueness of the inverse.
Next we will show the validity of
(32) µ ≡
c
λd
2
+ bd
1 − m
2
κ
by the following chain of arguments starting with the Definition (21):
κµ ≡
κc
κλdx + bx − ba
′
m
2
≡
κc
κλdx + bx − bxm
2
≡
κc
κ(λdx + bx
1 − m
2
κ
)
which implies that
µ ≡
c
λdx + bx
1 − m
2
κ
≡
c
λd
2
+ bd
1 − m
2
κ
.
10 SILVIU RADU
We thus have proven (32). By multiplying the last congruence with a
2
, we obtain:
µa
2
− ba
1 − m
2
κ
≡
c
λ.
We have again used that the inverse of d is a modulo c.
In order to arrive at our main result, Theorem 2.13, we need to introduce some
additional assertions, Lemmas 2.3 to 2.10.
Lemma 2.3. Let l, j be integers and C, a, s non-negative integers such that:
(1) the relation p|l implies p|C for any prime p;
(2) gcd(a, l) = 1;
(3) l = 2
s
j where j is odd;
(4) a is odd and C is even.
Then for any non-negative integer λ:
(33)
l
a + λC
=
l
a
(−1)
λC(j−1)
4
(−1)
s
2aλC+λ
2
C
2
8
.
Proof. By a similar reasoning as in the proof of (23) we see that gcd(a + λC, l) = 1
for all integers λ.
Next we can write j = j
1
j
2
where j
1
is squarefree and j
2
is a square. Clearly j
1
|C
by assumption. Then:
j
a + λC
=
j
1
a + λC
j
2
a + λC
(by (6))
=
j
1
a + λC
(because of gcd(a + λC, j) = 1)
=(−1)
a+λC−1
2
j
1
−1
2
a + λC
j
1
(by (8))
=(−1)
a+λC−1
2
j
1
j
2
−1
2
a + λC
j
1
(because of j
2
≡
4
1)
=(−1)
λC(j−1)
4
(−1)
a−1
2
j
1
−1
2
a
j
1
(because of a + λC ≡
j
1
a)
=(−1)
λC(j−1)
4
j
1
a
(by (8))
=(−1)
λC(j−1)
4
j
a
(by (6) and because of
j
2
a
= 1).
Summarizing, we have proven:
(34)
j
a + λC
= (−1)
λC(j−1)
4
j
a
.
Next,
(35)
2
a + λC
= (−1)
2aλC+λ
2
C
2
8
2
a
AN ALGORITHMIC APPROACH TO RAMANUJAN CONGRUENCES 11
is easily seen by
2
a + λC
=(−1)
(a+λC)
2
−1
8
(by (7))
=(−1)
2aλC+λ
2
C
2
8
2
a
(by (7)).
The following derivation concludes the proof:
l
a + λC
=
2
a + λC
s
j
a + λC
(by (6))
=
2
a
s
j
a
(−1)
λC(j−1)
4
(−1)
s
2aλC+λ
2
C
2
8
(by (34) and (35))
=
l
a
(−1)
λC(j−1)
4
(−1)
s
2aλC+λ
2
C
2
8
(by (6)).
In order to make the next lemmas more readable we need to introduce some helpful
definitions:
Definition 2.4. A tuple (m, M, N, (r
δ
)) ∈ ∆ is said to be κ-proper, if
(36) κN
X
δ |M
r
δ
mN
δ
≡
24
0,
and
(37) κN
X
δ |M
r
δ
≡
8
0,
for κ = gcd(1 − m
2
, 24).
Definition 2.5. For (m, M, N, (r
δ
)) ∈ ∆, γ =
a b
c d
∈ Γ
0
(N)
∗
and λ a non-
negative integer we define:
(38) β(γ, λ) := e
P
δ|M
2πir
δ
δamb
24
Y
δ |M
mcδ
a + κλc
|r
δ
|
e
−
(a+κλc)πi
12
P
δ|M
r
δ
(mc/δ−3)
.
Definition 2.6. For M a positive integer and (r
δ
) ∈ R(M ) let π(M, (r
δ
)) := (s, j)
where s is a non-negative integer and j an odd integer uniquely determined by
Q
δ |M
δ
|r
δ
|
= 2
s
j.
Lemma 2.7. Let (m, M, N, (r
δ
)) ∈ ∆ be κ-proper, γ =
a b
c d
∈ Γ
0
(N)
∗
,
(s, j) := π(M, (r
δ
)). Then for λ a non-negative integer the following relations hold:
(39) β(γ, λ) =
Y
δ |M
mcδ
a + κλc
|r
δ
|
e
−
πia
12
(
P
δ|M
mc
δ
r
δ
−
P
δ|N
r
δ
δm b−3
P
δ|M
r
δ
)
,
and
(40) β(γ, λ) =
(
β(γ, 0) , if κc ≡
8
0
β(γ, 0)(−1)
κλc(j−1)
4
(−1)
s
2aκλc+κ
2
λ
2
c
2
8
, if
P
δ |M
r
δ
≡
2
0
.
12 SILVIU RADU
Proof. (39): From its definition β(γ, λ) can be rewritten as
=
Y
δ |M
mcδ
a + κλc
|r
δ
|
e
−
πia
12
(
P
δ|M
r
δ
mc
δ
−
P
δ|M
mbδ r
δ
−3
P
δ|M
r
δ
)
· e
−
πiκλc
12
(
P
δ|M
r
δ
mc
δ
−3
P
δ|M
r
δ
)
.
Because of N|c, (36) and (37) we can conclude that
P
δ |M
r
δ
cκ
mc
δ
≡
24
0 and
κc
P
δ |M
r
δ
≡
8
0. Hence
(41) e
−
πiκλc
12
(
P
δ|M
r
δ
mc
δ
−3
P
δ|M
r
δ
)
= 1.
(40): Condition (37) implies that either
P
δ |M
r
δ
≡
2
0 or κN ≡
8
0. From (39) by
Lemma 2.3 we see that if κc ≡
8
0 then β(γ, λ) = β(γ, 0), λ ≥ 0.
If
P
δ |M
r
δ
≡
2
0 we have
Y
δ |M
δmc
a + κλc
|r
δ
|
=
Q
δ |M
δ
|r
δ
|
a + κλc
!
(by (6))
=
Q
δ |M
δ
|r
δ
|
a
!
(−1)
κλc(j−1)
4
(−1)
s
2aκλc+κ
2
λ
2
c
2
8
(by Lemma 2.3)
=
Y
δ |M
δmc
a
|r
δ
|
(−1)
κλc(j−1)
4
(−1)
s
2aκλc+κ
2
λ
2
c
2
8
(by (6)).
In view of (39) this implies that
(42) β(γ, λ) = β(γ, 0)(−1)
κλc(j−1)
4
(−1)
s
2aκλc+κ
2
λ
2
c
2
8
.
Note that in order to apply Lemma 2.3 above we need to verify that p|
Q
δ |M
δ
|r
δ
|
implies p|κc and that gcd(a,
Q
δ |M
δ
|r
δ
|
) = 1. This follows from (18) and (19)
together with gcd(a, c) = 1 because of ad − bc = 1.
Lemma 2.8. Let (m, M, N, (r
δ
)) ∈ ∆ be κ-proper, γ =
a b
c d
∈ Γ
0
(N)
∗
and t
an integer with 0 ≤ t ≤ m − 1 such that the relation
(43)
24m
gcd(κ(−24t −
P
δ |M
δr
δ
), 24m)
| N,
holds, then for τ ∈ H
∗
we have that
g
m,t
(γτ ) = (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
·
1
m
m−1
X
λ=0
β(γ, λ)e
2πiκµa
2
(−24t−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κµ)
m
,
(44)
where µ is defined as in (21).
Proof. Given two integers λ, λ
′
such that λ ≡
c
λ
′
, relation (43) implies
λ ≡
24m
gcd(24m,−24t−
P
δ|M
δr
δ
)
λ
′
,
consequently
e
2πiλκ(−24t−
P
δ|M
δr
δ
)
24m
= e
2πiλ
′
κ(−24t−
P
δ|M
δr
δ
)
24m
.
AN ALGORITHMIC APPROACH TO RAMANUJAN CONGRUENCES 13
Therefore by (v) in Lemma 2.2 we conclude that:
(45) e
2πiλκ(−24t−
P
δ|M
δr
δ
)
24m
= e
2πiκ(µa
2
−
ab(1−m
2
)
κ
)(−24t−
P
δ|M
δr
δ
)
24m
.
Hence,
g
m,t
(γτ ) =
1
m
m−1
X
λ=0
e
2πiκλ(−24t−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(γτ + κλ)
m
(by (16))
= (−i(cτ + d))
P
δ|M
r
δ
2
·
1
m
m−1
X
λ=0
β(γ, λ)e
2πiκλ(−24t−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κµ)
m
(by (22), (23) and (38))
= (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
·
1
m
m−1
X
λ=0
β(γ, λ)e
2πiκµa
2
(−24t−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κµ)
m
(by (45)).
Lemma 2.9. Let (m, M, N, (r
δ
)) ∈ ∆ be κ-proper, γ =
a b
c d
∈ Γ
0
(N)
∗
, and
t an integer with 0 ≤ t ≤ m − 1 such that (43) holds. Let t
′
be the unique integer
satisfying 0 ≤ t
′
≤ m − 1 and t
′
≡
m
ta
2
+
a
2
−1
24
P
δ |M
δr
δ
. Assume that κN ≡
8
0,
then for τ ∈ H
∗
we have that
(46) g
m,t
(γτ ) = β(γ, 0) (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
g
m,t
′
(τ).
Proof.
g
m,t
(γτ ) = β(γ, 0) (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
·
1
m
m−1
X
λ=0
e
2πiκµ(−24t
′
−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κµ)
m
(by (44) and because β(γ, 0) = β(γ, λ), λ ∈ Z by (40))
=β(γ, 0) (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
g
m,t
′
(τ)
(by (16) and (iv) in L emma 2.2).
Lemma 2.10. Let (m, M, N, (r
δ
)) ∈ ∆ be κ-proper , γ =
a b
c d
∈ Γ
0
(N)
∗
,
(s, j) := π(M, (r
δ
)) and t an integer with 0 ≤ t ≤ m − 1 such that (43) holds.
Assume further that
P
δ |M
r
δ
≡
2
0 and 2|m.
14 SILVIU RADU
(i) If s ≡
2
0 let t
′
be the unique integer satisfying t
′
≡
m
ta
2
+
a
2
−1
24
P
δ |M
δr
δ
−
3mca
2
(j−1)
24
and 0 ≤ t
′
≤ m − 1. Then for τ ∈ H
∗
we have that
g
m,t
(γτ ) = (−1)
abc(1−m
2
)(j−1)
4
β(γ, 0)
· (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
g
m,t
′
(τ).
(47)
(ii) If κc ≡
4
0 let t
′
be the unique integer satisfying t
′
≡
m
−
3mcsa
2
24
+ ta
2
+
a
2
−1
24
P
δ |M
δr
δ
and 0 ≤ t
′
≤ m − 1. Then for τ ∈ H
∗
we have that
g
m,t
(γτ ) = (−1)
sa
2
bc(1−m
2
)
4
β(γ, 0)
· (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
g
m,t
′
(τ).
(48)
Proof. (i):
g
m,t
(γτ ) = (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
β(γ, 0)
·
1
m
m−1
X
λ=0
(−1)
κλc(j−1)
4
e
2πiκµa
2
(−24t−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κµ)
m
(by (44) and (42), together with 2|m which implies 2|c because of (18))
=(−1)
abc(1−m
2
)(j−1)
4
β(γ, 0) (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
·
1
m
m−1
X
λ=0
(−1)
κµa
2
c(j−1)
4
e
2πiκµa
2
(−24t−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κµ)
m
(by (24) and c ≡
2
0)
=(−1)
abc(1−m
2
)(j−1)
4
β(γ, 0) (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
·
1
m
m−1
X
λ=0
e
2πiκµa
2
(3mc(j−1)−24t−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κµ)
m
=(−1)
abc(1−m
2
)(j−1)
4
β(γ, 0) (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
·
1
m
m−1
X
λ=0
e
2πiκµ(−24t
′
−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κµ)
m
(by substituting for t
′
)
=(−1)
abc(1−m
2
)(j−1)
4
β(γ, 0) (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
g
m,t
′
(τ)
(by (16) and (iv) in L emma 2.2).
AN ALGORITHMIC APPROACH TO RAMANUJAN CONGRUENCES 15
(ii):
g
m,t
(γτ ) = (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
β(γ, 0)
·
1
m
m−1
X
λ=0
(−1)
saκλc
4
e
2πiκµa
2
(−24t−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κµ)
m
(by (44) and (42))
=(−1)
sa
2
bc(1−m
2
)
4
β(γ, 0) (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
·
1
m
m−1
X
λ=0
(−1)
κµa
3
cs
4
e
2πiκµa
2
(−24t−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κµ)
m
(by (24) and c ≡
2
0))
=(−1)
sa
2
bc(1−m
2
)
4
β(γ, 0) (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
·
1
m
m−1
X
λ=0
e
2πiκµ(−24t
′
−
P
δ|M
δr
δ
)
24m
Y
δ |M
η
r
δ
δ(τ + κµ)
m
(by substituting for t
′
)
=(−1)
sa
2
bc(1−m
2
)
4
β(γ, 0) (−i(cτ + d))
P
δ|M
r
δ
2
e
2πiab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
g
m,t
′
(τ)
(by (16) and (iv) in L emma 2.2).
Note that if 2 ∤ m then κN ≡
8
0 and Lemma 2.9 applies. If 2|m and κN 6≡
8
0 then
the Lemma 2.10 applies.
Let (m, M, N, (r
δ
)) ∈ ∆ and s, j integers such that π(M, (r
δ
)) = (s, j). In the next
theorem we will also assume that:
(49) κN ≡
4
0 or s ≡
2
0,
and
(50) 2 ∤ N or 8|N (1 − j) or 8|N s.
Definition 2.11. We define
∆
∗
:= { all tuples (m, N, N, t, (r
δ
)) with properties as listed in (51)} :
(m, M, N, (r
δ
)) ∈ ∆ is κ-proper, t ∈ N, 0 ≤ t ≤ m − 1;
in addition (43), (49) and (50) hold.
(51)
16 SILVIU RADU
Definition 2.12. Let m, M, N ∈ N
∗
and (r
δ
) ∈ R(M ). Define the operation ⊙ :
Γ
0
(N)
∗
× {0, . . . , m − 1} 7→ {0, . . . , m − 1}, (γ, t) 7→ γ ⊙ t, where for γ =
a b
c d
the image γ ⊙ t is uniquely defined by the relation
(52) γ ⊙ t ≡
m
ta
2
+
a
2
− 1
24
X
δ |M
δr
δ
.
Finally we arrive at the main theorem of this section which can be viewed as a
generalization of a theorem of R. Lewis; see Remark 2.14 below.
Theorem 2.13. Let (m, M, N, t, (r
δ
) = r) ∈ ∆
∗
, g
m,t
(τ, r) be as in Definition 1.11,
γ =
a b
c d
∈ Γ
0
(N)
∗
, and β as in Definition 2.5. Then for all τ ∈ H
∗
we have
that
(53)
g
m,t
(γτ, r) = β(γ, 0) (−i(cτ + d))
P
δ|M
r
δ
2
e
2π i
ab(1−m
2
)(24t+
P
δ|M
δr
δ
)
24m
· g
m,γ⊙t
(τ, r).
Proof. If κN ≡
8
0 then (53) follows from Lemma 2.9. If κN 6≡
8
0 then 2|m, and
because of (49), either (i) or (ii) in L emma 2.10 apply. When (i) applies we have
3mca
2
(j−1)
24
≡
m
0 because of (50), and consequently t
′
= γ ⊙ t. Also because of (50)
we have that (−1)
abc(1−m
2
)(j−1)
4
in (47) is equal to 1 for all
a b
c d
∈ Γ
0
(N)
∗
and
(53) holds. When (ii) applies, the proof is analogous.
Remark 2.14. Theorem 2.13 extends Theorem 1 in [8] which covers products of the
form
Q
∞
n=1
(1 − q
n
)
r
1
where r
1
is a fixed integer.
3. Formulas for g
m,t
(γτ ) when γ ∈ Γ
Usually g
m,t
(τ) = g
m,t
(τ, r) as defined in Definition 1.11 is not a modular form.
But if we choose a sequence (a
δ
) ∈ R(N) properly, we can always make sure that
Q
t
′
∈P (t)
g
m,t
′
(τ)
Q
δ |N
η
a
δ
(δτ)
(with P (t) as in (66)) is a modular form. To
prove this we need some formulas for
Q
δ |N
η
a
δ
(δ(γτ )) and for g
m,t
(γτ ) that are
valid for all γ in Γ, in order to check condition (3) in Definition 1.1 of a modular
form. This is done in the Lemmas 3.1 to 3.6 below.
Recall that κ = gcd(1 − m
2
, 24).
Lemma 3.1. Let (m, M, N, (r
δ
)) ∈ ∆ and γ =
a b
c d
∈ Γ. For δ|M with
δ > 0 and λ an integer let x(δ, λ) and y(δ, λ) be any fixed solutions to the equation
δ(a + κλc) · x(δ, λ) + mc · y(δ, λ) = gcd(δ(a + κλc), mc). Further define
(54) w(δ, λ, γ) :=
gcd(δ(a + κλc), mc)τ + δ(b + κλd)x(δ, λ) + mdy(δ, λ)
δm
gcd(δ(a+κλc),mc)
.
Then there exists a map C : Γ 7→ C such that for all γ ∈ Γ and τ ∈ H
∗
the following
relation holds:
(55)
Y
δ |M
η
r
δ
δ(γτ + κλ)
m
= C(γ)(cτ + d)
1
2
P
δ|M
r
δ
Y
δ |M
η
r
δ
(w (δ, λ, γ)) .
AN ALGORITHMIC APPROACH TO RAMANUJAN CONGRUENCES 17
In addition, there exist mappings C
′
: Γ 7→ C and µ : Z 7→ Z such that for all
γ ∈ Γ
0
(N) and τ ∈ H
∗
the following relation holds:
(56)
Y
δ |M
η
r
δ
δ(γτ + κλ)
m
= C
′
(γ)(cτ + d)
1
2
P
δ|M
r
δ
Y
δ |M
η
r
δ
δ(τ + κµ(λ))
m
,
where µ is chosen such that [λ]
m
7→ [µ(λ)]
m
is a bijection of Z
m
.
Proof. (55): Let γ =
a b
c d
∈ Γ. Apply Lemma 1.13 with k set to
1
2
, f(τ)
to η(τ ), g(γ, τ) to (−i)
1
2
ǫ(a, b, c, d)η(τ) and ξ to
δ(a + κλc) δ(b + κλd)
mc md
; then
for all δ|M with δ > 0 the following relation holds:
gcd(δ(a + κλc), mc)
δm
m(cτ + d)
−
1
2
η
δ((a + κλc)τ + b + κ λd )
m(cτ + d)
=(−i)
1
2
ǫ(
δ(a + κλc)
gcd(δ(a + κλc), mc)
, −y(δ, λ),
mc
gcd(δ(a + κλc), mc)
, x(δ, λ)))η (w(δ, λ, γ)).
Taking the product over δ|M on both sides and using that
η
δ(γτ + κλ)
m
= η
δ((a + κλc)τ + b + κ λd )
m(cτ + d)
proves (55).
(56): In order to prove (56) we first will prove that gcd(δ(a + κλc), mc) = δ if N|c.
By (19) we see that δ|mc hence gcd(δ(a + κλc), mc) = δ gcd(a + κλc,
mc
δ
). Also
since gcd(a + κλc, c) = 1 because of ad − bc = 1, and gcd(a + κλc, m) = 1 because
of (18), we can conclude that gcd(a + κλc,
mc
δ
) = 1. Next, for λ ∈ Z let x
0
(λ) and
y
0
(λ) be any solutions to the equation (a + κλc)x
0
(λ) + mcy
0
(λ) = 1. Then we can
define x(δ, λ) := x
0
(λ) and y(δ, λ) := δy
0
(λ) because of gcd(δ(a + κλc), mc) = δ.
Consequently,
(57) η(w(δ, λ, γ)) = η
δτ + δ(b + κλd )x
0
(λ)
m
+ δdy
0
(λ)
.
Next, let X and Y be integers such that κX + mY = 1. Such integers clearly exist
by (12). Define µ(λ) := (b + κλd)Xx
0
(λ). Then
η
δ(τ + κµ(λ))
m
= η
δ(τ + κ(b + κλd)Xx
0
(λ))
m
=η
δ(τ + (b + κλd)x
0
(λ))
m
− Y (b + κλd)x
0
(λ)
.
(58)
This shows that
η(w(δ, λ, γ)) = ǫη
δ(τ + κµ(λ))
m
for some 24-th root of unity ǫ because of (30) and by (57) and (58). It only remains
to show that µ is a bijection of Z
m
. Note that x
0
(λ) is invertible modulo m because
of (a + κλc)x
0
(λ) + mcy
0
(λ) = 1 implying (µ(λ)X
−1
x
0
(λ)
−1
− b)κ
−1
d
−1
≡
m
λ.
Note that d is invertible modulo m because of gcd(c, d) = 1 which by (18) implies
gcd(m, d) = 1.
Remark 3.2. Note that (56) is very similar to (22) in Lemma 2.2 but here we lifted
the restriction gcd(a, 6) = 1, a > 0, c > 0.
18 SILVIU RADU
Lemma 3.3. Let γ
0
∈ Γ, (m, M, N, (r
δ
)) ∈ ∆, t ∈ Z with 0 ≤ t ≤ m − 1, and
define the mappings p : Γ × [0, . . . , m − 1] 7→ Q and p : Γ 7→ Q by
(59) p(γ, λ) :=
1
24
X
δ |M
r
δ
gcd
2
(δ(a + κλc), mc)
δm
,
and
(60) p(γ) := min
λ∈{0,...,m−1}
p(γ, λ).
Then for all γ =
a b
c d
∈ Γ
0
(N)γ
0
Γ
∞
there exists a positive integer k and a
Taylor series h(q) in powers of q
1
k
such that for τ ∈ H
∗
we have
(61) (cτ + d)
−
1
2
P
δ|M
r
δ
g
m,t
(γτ ) = h(q)q
p(γ
0
)
.
Proof. We write γ =
a b
c d
= γ
N
γ
0
γ
∞
where γ
N
=
a
N
b
N
c
N
d
N
∈ Γ
0
(N), γ
∞
=
1 b
∞
0 1
∈ Γ
∞
and γ
0
=
a
0
b
0
c
0
d
0
∈ Γ. Then
g
m,t
(γτ ) =
1
m
m−1
X
λ=0
C
1
(λ)
Y
δ |M
η
r
δ
δ(γτ + κλ)
m
(by (16)) with suitably chosen C
1
: {0, . . . , m − 1} 7→ C)
=(c
N
(γ
0
γ
∞
τ) + d
N
)
1
2
P
δ|M
r
δ
·
1
m
m−1
X
µ=0
C
2
(µ(λ))
Y
δ |M
η
r
δ
δ(γ
0
γ
∞
τ + κµ(λ))
m
(by (56) with suitably chosen C
2
: {0, . . . , m − 1} 7→ C)
=((c
N
(γ
0
γ
∞
τ) + d
N
)(c
0
(γ
∞
τ) + d
0
))
1
2
P
δ|M
r
δ
·
1
m
m−1
X
µ=0
C
3
(µ(λ))
Y
δ |M
η
r
δ
gcd
2
(δ(a
0
+ κµ(λ)c
0
), mc)τ + C
4
(µ(λ))
δm
(by (55) with suitably chosen C
3
: {0, . . . , m−1} 7→ C and C
4
: {0, . . . , m−1} 7→ C)
=(cτ + d)
1
2
P
δ|M
r
δ
·
1
m
m−1
X
µ=0
C
3
(µ(λ))
Y
δ |M
η
r
δ
gcd
2
(δ(a
0
+ κµ(λ)c
0
), mc)τ + C
4
(µ(λ))
δm
(because of
a
N
b
N
c
N
d
N
a
0
b
0
c
0
d
0
1 b
∞
0 1
=
a b
c d
)
=(cτ + d)
1
2
P
δ|M
r
δ
m−1
X
µ(λ)=0
C
3
(µ(λ))q
p(γ
0
,µ(λ))
h(µ(λ), q)
AN ALGORITHMIC APPROACH TO RAMANUJAN CONGRUENCES 19
(where for each µ(λ), h(µ(λ), q ) is a Taylor series in powers of q
24p(γ
0
,µ(λ))
by (2))
=(cτ + d)
1
2
P
δ|M
r
δ
q
p(γ
0
)
h(q)
(with h(q) := q
p(γ
0
)
P
m−1
µ=0
C
3
(µ(λ))q
p(γ
0
,µ(λ))−p(γ
0
)
h(µ(λ), q)).
Lemma 3.4. Let N ∈ N
∗
, (a
δ
) ∈ R(N), f(τ) :=
Q
δ |N
η
a
δ
(δτ), and define the
mapping p
∗
: Γ 7→ C by p
∗
a
0
b
0
c
0
d
0
:=
1
24
P
δ |N
a
δ
gcd
2
(δ,c
0
)
δ
. Then for all
γ =
a b
c d
∈ Γ there exists an integer k and a Taylor series h
∗
(q) in powers of
q
1
k
such that
(62) (cτ + d)
−
1
2
P
δ|N
a
δ
f(γτ) = h
∗
(q)q
p
∗
(γ)
.
Furthermore, for γ
1
∈ Γ and γ
2
∈ Γ
0
(N)γ
1
Γ
∞
we have p
∗
(γ
1
) = p
∗
(γ
2
).
Proof. Let w
δ
:= gcd(δa, c)
gcd(δa,c)τ+δbx
δ
+dy
δ
δ
where x
δ
, y
δ
∈ Z such that aδx
δ
+
cy
δ
= gcd(aδ, c) for any fixed δ|N with δ > 0. Then
(cτ + d)
−
1
2
P
δ|N
a
δ
Y
δ |N
η
a
δ
(δγτ ) =(cτ + d)
−
1
2
P
δ|N
a
δ
Y
δ |N
η
a
δ
δa
gcd(δ a,c )
w
δ
− y
δ
c
gcd(δ a,c )
w
δ
+ x
δ
!
=C
Y
δ |N
η
a
δ
(w
δ
)
(by (9)) with suitably chosen C ∈ C)
=Cq
p
∗
(γ)
Y
δ |N
h
∗
(δ, q)
(by (2) for some Taylor series h
∗
(δ, q) where δ|N (with constant term 1)). This
proves (62).
To prove the remaining part of Lemma 3.4 let γ
1
=
a b
c d
and γ
2
=
A B
C D
.
Because of γ
2
∈ Γ
0
(N)γ
1
Γ
∞
we have that γ
2
= γ
N
γ
1
γ
∞
for some γ
N
=
a
′
b
′
c
′
N d
′
∈
Γ
0
(N) and γ
∞
=
1 b
∞
0 1
∈ Γ
∞
. This shows that C = ac
′
N + d
′
c and
clearly gcd(d
′
, c
′
N) = 1 because of a
′
d
′
− c
′
Nd
′
= 1. For δ|N this implies that
gcd(δ, C) = gcd(δ, ac
′
δ
N
δ
+ d
′
c) = gcd(δ, d
′
c) = gcd(δ, c). By this we have shown
that the sums p
∗
(γ
1
) and p
∗
(γ
2
) have the same summands which proves that they
are identical.
20 SILVIU RADU
Theorem 3.5. Let (m, M, N, (r
δ
)) ∈ ∆, t ∈ Z with 0 ≤ t ≤ m−1, p be as in Lemma
3.3, (a
δ
) and p
∗
be as in Lemma 3.4, and γ
0
∈ Γ. Then for all γ =
a b
c d
∈
Γ
0
(N)γ
0
Γ
∞
the expression
(63) q
−(p(γ
0
)+p
∗
(γ
0
))
(cτ + d)
−
1
2
P
δ|M
r
δ
−
1
2
P
δ|N
a
δ
g
m,t
(γτ )
Y
δ |N
η
a
δ
(δ(γτ ))
finds a representation as a Taylor series in powers of q
1
k
for some positive integer
k.
Proof. By Lemmas 3.3 and 3.4, for each γ =
a b
c d
∈ Γ
0
(N)γ
0
Γ
∞
there exists
a positive integer k, and Taylor series h(q) and h
∗
(q) in powers of q
1
k
such that
(64) (cτ + d)
−
1
2
(
P
δ|M
r
δ
+
P
δ|N
a
δ
)
g
m,t
(γτ )
Y
δ |N
η
a
δ
(δ(γτ )) = h(q)h
∗
(q)q
p(γ
0
)+p
∗
(γ
0
)
.
Lemma 3.6. Let F : H
∗
7→ C be a mapping, k an integer, and l a positive integer.
Assume that for all γ =
a b
c d
∈ Γ there exists a positive integer n and a Taylor
series h(γ, q) in powers of q
1
n
such that for all τ ∈ H
∗
the relation (cτ +d)
−k
F (γτ ) =
h(γ, q) holds. Then for all γ =
a b
c d
∈ Γ there exists a positive integer n
′
and
a Taylor series h
∗
(γ, q) in powers of q
1
n
′
such that for all τ ∈ H
∗
the relation
(cτ + d)
−k
F (l(γτ)) = h
∗
(γ, q) holds.
Proof. We apply Lemma 1.13 with f(τ) = F (τ), g(γ, τ ) = h(γ, q), ξ =
al bl
c d
,
g := gcd(al, c), and x, y some integers such that alx + cy = g. As a consequence we
have that
(65)
g
l
(cτ + d)
−k
f(l(γτ )) = h
∗
al
g
−y
c
g
x
!
, q
g
2
l
e
2πig
l
(blx+dy)
!
.
Choosing n
′
=
g
2
l
n and