1.5. Applications of Modular Forms 9
Algorithm 1.19 (Width of Cusp). Given a congruence subgroup Γ of level
N and a cusp α for Γ, this algorithm computes the width h of α. We assume
that Γ is given by congruence conditions, e.g., Γ = Γ0(N) or Γ1(N).
(1) [Find γ] Use the extended Euclidean algorithm to find γ SL2(Z)
such that γ(∞) = α, as follows. If α = ∞, set γ = 1; otherwise,)(
write α = a/b, find c, d such that ad bc = 1, and set γ = a b
c d
.
(2) [Compute Conjugate Matrix] Compute the following element of
Mat2(Z[x]):
δ(x) = γ
1 x
0 1
γ−1.
Note that the entries of δ(x) are constant or linear in x.
(3) [Solve] The congruence conditions that define Γ give rise to four
linear congruence conditions on x. Use techniques from elementary
number theory (or enumeration) to find the smallest simultaneous
positive solution h to these four equations.
Example 1.20. (1) Suppose α = 0 and Γ = Γ0(N) or Γ1(N). Then
γ =
(
0 −1
1 0
)
has the property that γ(∞) = α. Next, the congruence
condition is
δ(x) = γ
1 x
0 1
γ−1
=
1 0
x 1

1
0 1
(mod N).
Thus the smallest positive solution is h = N, so the width of 0
is N.
(2) Suppose N = pq where p, q are distinct primes, and let α = 1/p.
Then γ =
(
1 0
p 1
)
sends to α. The congruence condition for Γ0(pq)
is
δ(x) = γ
1 x
0 1
γ−1
=
1 px x
−p2x
px + 1


0
(mod pq).
Since
p2x
0 (mod pq), we see that x = q is the smallest solution.
Thus 1/p has width q, and symmetrically 1/q has width p.
Remark 1.21. For Γ0(N), once we enforce that the bottom left entry is 0
(mod N) and use that the determinant is 1, the coprimality from the other
two congruences is automatic. So there is one congruence to solve in the
Γ0(N) case. There are two congruences in the Γ1(N) case.
1.5. Applications of Modular Forms
The above definition of modular forms might leave the impression that mod-
ular forms occupy an obscure corner of complex analysis. This is not the
case! Modular forms are highly geometric, arithmetic, and topological ob-
jects that are of extreme interest all over mathematics:
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