III Modular Forms and L-functions - Modular forms for subgroups of SL<sub>2</sub>(ℤ)

8Modular forms for subgroups of SL₂(ℤ)

III Modular Forms and L-functions

8.3 Examples of modular forms

We now look at some examples of (non-trivial) modular forms for different

subgroups. And the end we will completely characterize

(Γ

(4)). This seems

like a rather peculiar thing to completely characterize, but it turns out this

understanding M

(Γ

(4)) can allow us to prove a rather remarkable result!

Eisenstein series

At the beginning of the course, the first example of a modular form we had was

Eisenstein series. It turns out a generalized version of Eisenstein series will give

us more modular forms.

Definition (G

r,k

). Let k ≥ 3. Pick any vector r = (r

, r

) ∈ Q

. We define

r,k

(z) =

m∈Z

((m

+ r

)z + m

+ r

)

where

means we omit any m such that m + r = 0.

For future purposes, we will consider r as a row vector.

As before, we can check that this converges absolutely for

k ≥

3 and

z ∈ H

This obviously depends only on r mod Z

, and

0,k

= G

Theorem.

(i) If γ ∈ Γ(1), then

r,k

γ = G

rγ,k

(ii) If Nr ∈ Z

, then G

r,k

∈ M

(Γ(N)).

Proof.

(i) If g ∈ GL

(R)

and u ∈ R

, then

z + u

)

g =

(det g)

k/2

((au

+ cu

)z + (bu

+ du

))

(det g)

k/2

z + v

)

where v = n · g. So

r,k

γ =

(((m + r)

γ)z + ((m + r)γ)

)

((m

+ r

)z + m

+ r

)

= G

rγ,k

(z),

where m

= mγ and r

= rγ.

(ii)

By absolute convergence,

r,k

is holomorphic on the upper half plane.

Now if

Nr ∈ Z

and

γ ∈

Γ(

), then

Nrγ ≡ Nr

(

mod N

). So

rγ ≡ r

(mod Z

). So we have

r,k

γ = G

rγ,k

= G

r,k

So we get invariance under Γ(

). So it is enough to prove

r,k

is holo-

morphic at cusps, i.e.

r,k

is holomorphic at

∞

for all

γ ∈

Γ(1). So it is

enough to prove that for all r, G

r,k

is holomorphic at ∞.

We can write

r,k

((m

+ r

)z + m

+ r

)

The first sum is

>−r

∈Z

([(m

+ r

)z + r

] + m

)

We know that (

)

∈ H

. So we can write this as a Fourier series

>−r

d≥1

(−2π)

(k −1)!

k−1

2πr

We now see that all powers of q are positive. So this is holomorphic.

The sum over m

+ r

= 0 is just a constant. So it is fine.

For the last term, we have

<−r

∈Z

(−1)

((−m

− r

)z −r

− m

)

which is again a series in positive powers of q

−m

−r

ϑ functions

Our next example of modular forms is going to come from

functions. We

previously defined a ϑ function, and now we are going to call it ϑ

(z) = ϑ(z) =

n∈Z

πin

= 1 + 2

n≥1

We proved a functional equation for this, which was rather useful.

Unfortunately, this is not a modular form. Applying elements of Γ(1) to ϑ

will give us some new functions, which we shall call ϑ

and ϑ

Definition (ϑ

and ϑ

(z) =

n∈Z

πi(n+1/2)

= q

1/8

n∈Z

n(n+1)/2

= 2q

1/8

n≥0

n(n+1)/2

(z) =

n∈Z

(−1)

πin

= 1 + 2

n≥1

(−1)

Theorem.

(i) ϑ

(z) = ϑ

(z ±1) and θ

(z + 1) = e

πi/4

(z).

(ii)



−







1/2

(z)



−







1/2

(z)

Proof.

(i) Immediate from definition, e.g. from the fact that e

πi

= 1.

(ii)

The first part we’ve seen already. To do the last part, we use the Poisson

summation formula. Let

(x) = e

−πt(x+1/2)

= g



x +



where

(x) = e

−πtx

We previously saw

ˆg

(y) = t

−1/2

−πy

We also have

(y) =

−2πixy



x +



−2πi(x−1/2)y

(x) dx

= e

πiy

ˆg

(y).

So by the Poisson summation formula,

(it) =

n∈Z

(n) =

n∈Z

(n) =

n∈Z

(−1)

−1/2

−πn

= t

−1/2





There is also a

, but we have

= 0. Of course, we did not just invent

a funny name for the zero function for no purpose. In general, we can define

functions

(

u, z

), and the

functions we defined are what we get when we set

u = 0. It happens that ϑ

has the property that

(u, z) = −ϑ

(−u, z),

which implies ϑ

(z) = 0.

We now see that the action of

(

) send us between

and

, up to

simple factors.

Corollary.

(i) Let

F =









Then

F (z + 1) =





−1 0 0

0 0 1

0 1 0





F, z

−2



−







0 0 −1

0 −1 0

−1 0 0





(ii) ϑ

∈ M

(Γ) for a subgroup Γ

≤

Γ(1) of finite index. In particular,

holomorphic at ∞ for any γ ∈ GL

(Q)

Proof.

(i) Immediate from the theorem.

(ii)

We know

Γ(1)

hS, T i

, where

±(

1 1

0 1

)

and



0 −1

1 0



. So by (i),

there is a homomorphism ρ : Γ(1) → GL

(Z) and ρ(−I) = I with

F |

γ = ρ(γ)F,

where

(

) is a signed permutation. In particular, the image of

is finite,

so the kernel Γ = ker ρ has finite index, and this is the Γ we want.

It remains to check holomorphicity. But each

is holomorphic at

∞

Since

F |

(

)

, we know

is a sum of things holomorphic at

∞

and is hence holomorphic at ∞.

It would be nice to be more specific about exactly what subgroup

invariant under. Of course, whenever

γ ∈

Γ, then we have

. But in

fact ϑ

is invariant under a larger group.

To do this, it suffices to do it for

, and the remaining follow by

conjugation.

We introduce a bit of notation

Notation. We write



0 −1

N 0



Note that in general, W

does not have determinant 1.

Theorem.

Let

(

) =

)

. Then

(

)

∈ M

(Γ

(4)), and moreover,

f |

−f.

To prove this, we first note the following lemma:

Lemma. Γ

(4) is generated by

−I, T =



1 1

0 1



, U =



1 0

4 1



= W



1 −1

0 1



−1

Four is special. This is not a general phenomenon.

Proof. It suffices to prove that Γ

(4) is generated by T and U = ±(

1 0

4 1

Let

γ = ±



a b

c d



∈ Γ

(4).

We let

s(γ) = a

+ b

is even, we know

a ≡

1 (

mod

2). So

(

)

≥

1, and moreover

(

) = 1 iff

b = 0, a = ±1, iff γ = T

for some n.

We now claim that if

(

)

= 1, then there exists

δ ∈ {T

±1

, U

±1

}

such that

s(γδ) < s(γ). If this is true, then we are immediately done.

To prove the claim, if s(γ) 6= 1, then note that |a| 6= |2b| as a is odd.

–

|a| < |

, then

min{|b±a|} < |b|

. This means

(

γT

±1

) =

b±a

)

s(γ).

–

|a| > |

, then

min{|a ±

b|} < |a|

, so

(

γU

±1

) = (

a ±

)

s(γ).

Proof of theorem. It is enough to prove that

f |

T = f |

U = f.

This is now easy to prove, as it is just a computation. Since

(

+ 2) =

(

), we

know

f |

T = f(z + 1) = f(z).

We also know that

f |

= 4(4z)

−2



−1





−



= −f(z),



−







1/2

ϑ(z).

So we have

f |

U = f |

−1

= (−1)(−1)f = f.

We look at Γ

(2) and Γ

(4) more closely. We have

(2) =



a b

2c d





γ ∈ Γ(1) : γ =



1 ∗

0 1



mod 2



We know

|SL

(

)

= 6, and so (Γ(1) : Γ

(2)) = 3. We have coset representa-

tives



0 −1

0 1





1 0

1 1



We also have a map

(2)  Z/2Z



a b

2c d



7→ c,

which one can directly check to be a homomorphism. This has kernel Γ

(4). So

we know (Γ

(2) : Γ

(4)) = 2, and has coset representatives



1 0

2 1



(2) =



T, ±



1 0

2 1



= W



1 1

0 1



−1



We can draw fundamental domains for Γ

(2):

and Γ

(4):

We are actually interested in Γ

(2) and Γ

(4) instead, and their fundamental

domains look “dual”.

Consider

g(z) = E

(

2 0

0 1

) − E

= 2E

(2z) − E

(z).

Recall that we had

(z) = 1 − 24

n≥1

(n)q

= z

−2



−



−

2πiz

Proposition. We have g ∈ M

(Γ

(2)), and g|

= −g.

Proof. We compute

(2z)



−





−



−

(2z)



−1



= E

(z) +

2πiz

− 2



(2z) +

2πi · 2z



= −g(z).

We also have

T = g(z + 1) = g(z),

and so

(

1 0

2 1

) = g|

−1

= g.

Moreover,

is holomorphic at

∞

, and hence so is

−g

. So

is also

holomorphic at 0 =

(

∞

). As

∞

has width 1 and 0 has width 2, we see

that these are all the cusps, and so

is holomorphic at the cusps. So

g ∈

(Γ

(2)).

Now we can do this again. We let

h = g(2z) =

(

2 0

0 1

) = 2E

(4z) − E

(2z).

Since g ∈ M

(Γ

(2)), this implies h ∈ M

(Γ

(4)) ⊇ M

(Γ

(2)).

The functions

and

are obviously linearly independent. Recall that we

have

dim M

(Γ

(4)) ≤ 1 +

k(Γ(1) : Γ

(4))

= 2.

So the inequality is actually an equality. We have therefore shown that

Theorem.

(Γ

(4)) = Cg ⊕Ch.

Recall we also found an

(

) =

)

∈ M

(Γ

(4)). So we know we must

have

f = ag + bh

for some constants a, b ∈ C.

It is easy to find what

and

are. We just write down the

-expansions. We

have

f = ϑ(2z)

= (1 + 2q + 2q

+ ···)

= 1 + 8q + 24q

+ 32q

+ ···

g = 2E

(2z) − E

(z)

= 1 + 24

n≥1

(n)(q

− 2q

)

= 1 + 24q + 24q

+ 96q

+ ···

h = g(2z)

= 1 + 24q

+ ···

By looking at the first two terms, we find that we must have

f =

g +

h =

(4E

(4z) − E

(z)) = 1 + 8

k≥1



(n) − 4σ





where σ





= 0 if

6∈ Z.

But recall that

f =

n∈Z

a,b,c,d∈Z

n∈N

(n)q

where

(

) is the number of ways to write

as a sum of 4 squares (where order

matters). Therefore,

Theorem (Lagrange’s 4-square theorem). For all n ≥ 1, we have

(n) = 8



(n) − 4σ





= 8

d|n4-d

In particular, r

(n) > 0.

We could imagine ourselves looking at other sums of squares. For example,

we can look instead at

)

, which turns out to be an element of

(Γ

(4)),

one can get a similar formula for the number of ways of writing

as a sum of 2

squares.

We can also consider higher powers, and then get approximate formulae for

(

), because the appropriate Eisenstein series no longer generate

. There

may be a cusp form as well, which gives an error term.

In general, if we have

γ =



a b

Nc d



∈ Γ

(N),

then we find

γW

−1

∈



d −c

−Nb a



∈ Γ

(N).

normalizes the group Γ

(

). Then if

f ∈ M

(Γ

(

)), then

f |

∈

(Γ

(N)), and this also preserves cusp forms.

Moreover, we have

f |

= f |



−N 0

0 −N



= f,

as −I ∈ Γ

(N). So

(Γ

(N)) = M

(Γ

(N))

⊕ M

(Γ

(N))

−

where we split into the (

1)-eigenspaces for

, and the cusp forms decompose

similarly. This

is the Atkin-Lehner involution. This is the “substitute” for

the the operator S =



0 −1

1 0



in Γ(1).