UNIVERSITY OF DUBLIN
XMA342J
TRINITY COLLEGE
Faculty of Engineering, Mathematics
and Science
school of mathematics
JS & SS Mathematics
Trinity Term 2012
MA342J — Introduction to Modular Forms
??, May ??
?? Exam Hall
9.30 — 11.30
Dr. M. Vlasenko
All 3 questions have equal weight.
It is sufficient to solve any 3 of the 4 parts a)-d) to get the full mark for questions 2 and 3.
There is a page with some notations and formulas in attachment.
Non-programmable calculators are permitted for this examination,—please indicate the make
and model of your calculator on each answer book used.
Page 2 of 7
XMA342J
1. The Fourier expansion of a modular form f of weight 16 on the full modular group
SL(2, Z) starts as
f = 3 + 1152 q + . . .
Find the Fourier coefficient of this form near q 2 .
2. Jacobi’s theta function
θ(z) =
X
2
q n = 1 + 2q + 2q 4 + 2q 9 + . . .
q = e2πiz
n∈Z
is a modular form of weight 21 for the modular group Γ0 (4). In particular, θ4 ∈ M2 (Γ0 (4))
is a modular form of weight 2.
a) Recall that for the Eisenstein series
E2 (z) = 1 − 24
X X d qn
n≥1
d|n
3
the (nonholomorphic) function E2∗ (z) = E2 (z)− πIm(z)
is invariant under SL(2, Z)
in weight 2. Show that f1 , f2 ∈ M2 (Γ0 (4)) where
f1 = E2 (z) − 2E2 (2z) , f2 = E2 (2z) − 2E2 (4z) .
b) Check that modular forms f1 and f2 are linearly independent.
c) Dimension formula shows that dim M2 (Γ0 (4)) = 2, hence f1 and f2 is a basis for
this space and θ4 must be a linear combination of these two functions. Find the
coefficients of this linear combination.
d) Compute the number of ways to represent 2012 as a sum of four squares of integer
numbers.
3. Consider the one-parametric family of cubic curves given by
Et : y 2 = x3 + x2 + −3t +
4
x+t
9
(t ∈ C) .
a) Determine singular values of t for this family, i.e. those t0 for which Et0 is not an
elliptic curve
Page 3 of 7
XMA342J
b) Write j-invariant for this family j(Et ) as a rational function of t.
c) Show that for nonsingular t1 6= t2 the curves Et1 and Et2 are nonisomorphic.
Show that in this family one can find a representative of every isomorphism class
of elliptic curves except two of them. Which two isomorphism classes are missing?
d) It follows that t can be written as a meromorphic SL(2, Z)-invariant function on
the upper half-plane t(z), z ∈ H. This t(z) then must have a Laurent series
expansion in the variable q = e2πiz . Write the beginning terms of this expansion.
Page 4 of 7
Γ = SL(2, Z)/{±I} ,
Some notation

n a
Γ0 (N ) = 
c
For even k ≥ 0
dim Mk (Γ) =
12
Ek (z) =
1
2ζ(k)
and formulas

b
 ∈ SL(2, Z) c ≡ 0
d
h i

k



12




h i



 k
X
(m,n)6=(0,0)
XMA342J
when k ≡ 2
+1
mod 12
otherwise
∞
1
2k X
σk−1 (n)q n
=
1
−
(mz + n)k
Bk n=1
E4 = 1 + 240q + 2160q 2 + . . .
E6 = 1 − 504q − 16632q 2 + . . .
j = 1728
E43
1
= + 744 + . . .
3
2
E4 − E6
q
A cubic curve given in the Weierstrass form
y 2 = 4x3 − ax − b
E :
o
mod N .
is an elliptic curve if its discriminant
∆(E) = a3 − 27 b2
is nonzero. The j-invariant of E is then given by
a3
a3
j(E) = 1728
= 1728 3
.
∆(E)
a − 27 b2
Page 5 of 7
XMA342J
MA342J: Introduction to Modular Forms
Solutions
1. According to the dimension formula dim(M16 ) =
h i
16
12
+ 1 = 2. Since the modular
forms
E44 =
E4 E62
1 + 240q + 2160q 2 + . . .
4
= 1 + 960q + 354240q 2 + . . .
2
= 1 + 240q + 2160q 2 + . . . 1 − 504q − 16632q 2 + . . .
= 1 − 768q − 19008q 2 + . . .
of weight 16 are obviously not proportional they could be taken as a basis in this space.
Then
αE44 + βE4 E62 = (α + β) + (960α − 768β)q + . . . = 3 + 1152 q + . . .
for α = 2 and β = 1. Hence
f = 2E44 + E4 E62 = 3 + 1152 q + 689472 q 2 + . . .
and the second Fourier coefficient equals 689472.
2.
a) We checked in class that for any modular form f (z) ∈ Mk (Γ) one has f (N z) ∈
Mk (Γ0 (N )). The function E2∗ (z) is SL(2, Z)-equivariant in weight 2 and bounded
at ∞, i.e. satisfies all properties of a modular form of weight 2 apart from being
holomorphic. Therefore E2∗ (N z) is also equivariant w.r.t. Γ0 (N ) in weight 2 and
bounded at all cusps of this group. It remains to notice that
f1 (z) = E2 (z) − 2E2 (2z) = E2∗ (z) − 2E2∗ (2z)
(the non-holomorphic correction terms cancel) and therefore f1 ∈ M2 (Γ0 (2)).
Since Γ0 (4) ⊂ Γ0 (2) we have M2 (Γ0 (2)) ⊂ M2 (Γ0 (4)). Analogously
f2 (z) = E2 (2z) − 2E2 (4z) = E2∗ (2z) − 2E2∗ (4z) ∈ M2 (Γ0 (4)) .
b) We see this from the beginning two terms of their Fourier expansion:
2
2
f1 = 1 − 24q − 72q + . . . − 2 1 − 24q + . . . = −1 − 24q − 24q 2 + . . .
2
4
f2 = 1 − 24q + . . . − 2 1 − 24q + . . . = −1 − 24q 2 + . . .
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XMA342J
c) Since
θ4 = (1 + 2q + . . . )4 = 1 + 8 q + . . .
we find that
2
1
1
4E2 (4z) − E2 (z)
θ 4 = − f1 − f2 =
3
3
3
d) Continuing the last computation we get
θ4 =
X X 1
4E2 (4z) − E2 (z) = 1 + 8
d qn .
3
n≥1
d|n,4-d
On the other hand
X
4
θ =
q
n21 +n22 +n23 +n24
= 1+
n1 ,n2 ,n3 ,n4 ∈Z
∞
X
r4 (n)q n
n=1
where r4 (n) is the number of ways to represent n as a sum of four squares. Hence
X
r4 (2012) = 8
d = 8(1 + 2 + 503 + 1006) = 12096.
d|2012,4-d
3. A simple change of variables x = X − 31 , y =
Y
2
brings the equation of Et into the form
Et : Y 2 = 4X 3 − a(t)X − b(t)
1
a(t) = 12 t −
27
1
b(t) = −8 t −
27
which is convenient for our computations.
a) Et is an elliptic curve iff its discriminant is nonzero:
1 2 28 ∆(Et ) = a(t) − 27b(t) = 1728 t −
t−
6 0.
=
27
27
3
2
Therefore the singular values of the parameter are
t0 =
1
27
and
t0 =
28
27
b)
j(Et ) = 1728
t−
a(t)3
= 1728
∆(Et )
t−
1
27
28
27
.
Page 7 of 7
XMA342J
c) We can invert the above formula and determine t as a function of j, namely
t =
1
j
+
.
27
j − 1728
Since elliptic curves are isomorphic iff their j-invariants are equal we see that curves
Et with different t must be nonisomorphic. Also we see that singular parameters
t0 =
1 28
,
27 27
and ∞ correspond to j = 0, ∞ and 1728, therefore isomorphism classes
of curves with j = 0 and j = 1728 are not represented by any fibers of this family.
d) Substituting the q-expansion
j =
1
+ 744 + . . .
q
into the formula for t we get
t =
28
+ 1728q + . . .
27
c UNIVERSITY OF DUBLIN 2012