Solutions to practice exam for Modular Forms
by Raymond van Bommel
1. (a) First we define
Γ(N ) =a b c d ∈ SL2(Z) : a b c d ≡1 0 0 1 mod N , Γ1(N ) = a b c d ∈ SL2(Z) : a b c d ≡1 b 0 1 mod N , Γ0(N ) = a b c d ∈ SL2(Z) : c ≡ 0 mod N .
A congruence subgroup of SL2(Z) is a subgroup Γ ⊂ SL2(Z) containing
Γ(N ) for some positive integer N . The smallest N such that Γ(N ) ⊂ Γ is called the level of Γ.
(b) We can calculate α−1a b c d α = a bp c/p d , for any a b c d ∈ Γ1(N ).
This is an element of Γ1(N ) if and only if cp is an integer. Hence, we get
Γ0 =a b c d ∈ Γ1(N ) : b ≡ 0 mod p .
Now it is immediate that Γ(N p) ⊂ Γ0. Hence, Γ0 is a congruence sub-group. Moreover if d < N p is strictly smaller than N p, then
1 d
d d2+ 1
is an example of a matrix that is in Γ(d), but not in Γ0, as either d 6≡ 0 mod N or d 6≡ 0 mod p.
(c) The action is defined by SL2(Z) × P1(Q) → P1(Q) : a b c d , (p : q) 7→ (ap + bq : cp + dq). For (p : q) ∈ P1(Q) with gcd(p, q) = 1, let r, s ∈ Z be such that pr + qs = 1. Then ρ · (p : q) = (0 : 1) = ∞, for ρ =q −p r s ∈ SL2(Z). Then γ = ρ−1 satisfies γ∞ = ρ−1ρ · (p : q) = (p : q).
(d) The set of cusps is
Cusps(Γ) = Γ\P1(Q).
From part (c) we deduce that the action of SL2(Z) on P1(Q) is transitive.
In other words, Cusps(SL2(Z)) consists of one element.
(e) Let r be the rank of the finitely generated abelian group E(Q). Then the (weak) Birch and Swinnerton-Dyer conjecture states that the order of vanishing of L(E, s) at s = 1 equals r.
2. (a) Recall the valence formula ord∞(f ) + 1 2ordi(f ) + 1 3ordρ(f ) + X w∈W ordw(f ) = k 12.
Let f ∈ S12(SL2(Z)) be a normalised form. Then f − ∆ has a double
zero at ∞. Hence, the valence formula for f − ∆ cannot be satisfied and hence f − ∆ = 0. Therefore, there is only one normalised form in S12(SL2(Z)). As S12(SL2(Z)) is known to have a basis of normalised
eigenfoms by Atkin-Lehner, we see that ∆ must be such a normalised eigenform.
(b) For e = 1, 2, 3, 6, consider the maps i1,6
e : S12(Γ1(1)) → S12(Γ1(6)). The
exercise statement tells us that we may assume that these maps are also defined as maps S12(Γ0(1)) → S12(Γ0(6)).
We know that ∆(z) is a cusp form of weight 12 for SL2(Z) = Γ0(1),
having a simple zero in ∞. Now we have that i1,6e (∆(z)) = ∆(ez) are cusp forms of weight 12 for Γ0(6). Moreover, the q-expansion of ∆(ez)
at ∞ in q∞ = e2πiz is given by an(∆(ez)) = aen(∆(z)). Therefore, ∆(ez)
has a zero of order e at ∞. In particular, we see that ∆(z), ∆(2z), ∆(3z) and ∆(6z) are four linearly independent elements of S12(Γ0(6)).
(c) The Fricke operator ωN is defined as
ωN(f ) = Tα = X γ∈α−1Γ 0(N )α∩Γ0(N ) f |kαγ, where α = 0 −1 N 0 .
Using the calculation
α−1a b c d α = d −c/N −N b a ,
(d) Analogous to what we saw in part (b) we see that ∆(z) and ∆(5z) are cusp forms of weight 12 for Γ0(5). Hence, their product F is a cusp form
of weight 24 for Γ0(5).
First we compute the order of F at the cusp ∞. It is immediate that the period at ∞ is ehΓ0(5)(∞) = 1 (which is the same for SL2(Z)). We already
know that ∆(z) has a simple zero at ∞. Hence, ∆(5z) has a zero of order 5 at ∞, as the q-expansion was given by an(∆(5z)) = a5n(∆(z)). Then
the product F has a zero of order 6 at ∞.
To calculate the order of F at the cusp 0, we can do two things. We could use the valence formula or we could do an explicit calculation. We choose for the latter. We calculate
G(z) := F |24 0 −1 1 0 (z) = z−24∆ −1 z ∆ −5 z = 512∆(z)∆ (z/5) , using the modularity of ∆ for SL2(Z). Using the calculation
0 1 −1 0 a b c d 0 −1 1 0 = d −c −b a ,
we see that the period at the cusp 0 is ehΓ0(5)(0) = 5. Then, the order of
the zero at ∞ for G(z) equals 6 again, as the q-expansion of ∆(z/5) and ∆(z) in q0 = e2πiz/5 has the same coefficients as the q-expansion of ∆(z)
and ∆(5z), respectively, in q∞ = e2πiz. Hence, ord0(F ) = 6.
(e) This is just a matter of calculation. F |24 0 −1 5 0 (z) = 5 24 (5z)24F − 1 5z = 512· 1 (5z)12∆ − 1 5z · 1 z12F −1 z = 512∆(z)∆(5z) = 512F (z) Here, we use the modularity for ∆ again, i.e. that
∆(w) = 1 w12∆ −1 w , both for w = z and w = 5z.
3. (a) For every M | N and e | N/M such that M 6= N we define the map iM,Ne : Sk(Γ1(M )) → Sk(Γ1(N )) : F (z) 7→ F (ez).
Then the space Sk(Γ1(N ))old is defined as the sum of the images of all
these maps. The space Sk(Γ1(N ))new is the orthogonal complement of
this subspace with respect to the Petersson inproduct. The Petersson inproduct is defined as follows
Sk(Γ1(N )) × Sk(Γ1(N )) → R
(f, g) 7→ Z
DΓ1(N)
F (z)G(z)yk−2dxdy,
where DΓ1(N ) is a fundamental domain for Γ\H.
(b) The proof is using the following steps. By explicit calculations you can prove that Tm, for m coprime to N , and hdi all commute with each other
and with their adjoints. Hence, there is a basis of eigenvectors for these operators.
The old- and newspaces are respected by all operators Tmand hdi. Hence,
Sk(Γ1(N ))newhas a basis of eigenvectors for Tm, for m coprime to N , and
hdi. Let f be an element of such basis. We will prove that f is also an eigenvector for Tm with m not coprime to N .
For each m, consider fm := Tmf −am(f )f . On the one hand, this is an
el-ement of Sk(Γ1(N ))new (as this space was respected by the operator Tm),
on the other hand, by explicit calculation, we find a1(fm) = 0. As Tm
commutes with Tdand hdi for d coprime to N , we find that fmis an
eigen-vector for the latter operators. But then we find that ad(fm) = a1(Tdfm)
is some multiple of a1(fm) = 0. Therefore, ad(fm) = 0 for all d coprime
to m.
Then there is a lemma which tells us that under these conditions fm is
an oldform. As fm is also in the newspace, we find that fm = 0 must
hold. Therefore, f is also an eigenvector for Tm for m not coprime to N
and we are done.
4. (a) The completed L-function is defined as Λ(f, s) = Ns/2Γ(s)
(2π)sL(f, s),
where L(f, s) is the L-functionP
n>1an(f )n−sof f and Γ is the
(b) First we do a calculation for the first term. ∞ X n=1 an(2πn)−s Z ∞ 2πn√ N e−ttsdt t = ∞ X n=1 an(2πn)−s Z ∞ 1 √ N e−2πnt(2πnt)sd(2πnt) 2πnt = ∞ X n=1 Z ∞ 1 √ N ane−2πntts dt t = Z ∞ 1 √ N ∞ X n=1 ane2πni·(it)ts dt t = Z ∞ 1 √ N f (it)tsdt t .
Here we used that we can swap the integral and the infinite summation sign, as the whole is absolutely convergent, as
∞ X n=1 Z ∞ 1 √ N |an| · e−2πnt· ts dt t 6 ∞ X n=1 Cnd Z ∞ 1 √ N e−2πntts−1dt 6 ∞ X n=1 Cnde(−2πn+1) √ N Z ∞ 1 √ N e−tts−1dt < ∞,
as both the sum and the integral converge.
For the other term, the calculation is the same. The only extra thing we need to use is that an(f∗) = an(f ). This can be proved as follows. Let
f (z) =P∞
n=0an(f )e2πiz be the q-expansion for f . Then we see that
f∗(z) = f (−z) = ∞ X n=0 an(f )e2πiz, as we have
e−2πiz = e−2πb· (cos (−a)) + i · sin (−a)
= e−2πb· (cos (a) − i · (− sin (a))) = e2πiz,
where z = a + bi with a, b ∈ R. We find the desired result by the uniqueness of the q-expansion.
(c) We plug in s = 1 and k = 2 into the formula of part (b) to get Λ(f, 1) =√N ∞ X n=1 an 2πnΓ 1,2πn√ N − √ηf N ∞ X n=1 an 2πnsΓ 1,√2πn N .
Taking these sums together (with finite sums, this is always allowed), we get the desired equality, using
Γ 1,√2πn N = Z ∞ 2πN√ N e−tdt = [−e−t]∞t=2πn√ N = e−2πn√N.
