Problems and Solutions

Diamond and Shurman:

A First Course in Modular Forms

Author:Alex Youcis

Last Revised:March 12, 2013

1

1.1

Exercise 1.1.1. Prove that SL2(Z) is generated by A =

[1 10 1

]and B =

[0 −11 0

].

Solution This amounts to the Euclidean algorithm. Namely, let’s begin with an element

[a bc d

]∈ SL2(Z). We note

then that

Bm[a bc d

]=

[a+ cm b+ dm

c d

](1)

for every m ∈ Z and

A

[a bc d

]=

[−c −da b

](2)

So, now, by the Euclidean algorithm we can write a = cq + r and so we see that if we let m = −q we see by (1) that

Bm[a bc d

]=

[a− cq b− dqc d

]=

[r b− dqc d

]So, we see then

SBm[a bc d

]=

[−c −dr b− dq

]So, by the Euclidean algorithm we see that there exists r1 and q1 so that −c = rq1 + r1 and so taking m1 = −q1 in (1)we see that

Bm1SBm[a bc d

]=

[−c− rq1 −d− rq1

r d− rq

]=

[r1 −d− rq1

r b− dq

]Continuing in this process, and using the fact that ad− bc = 1 so that (a, c) = 1, we see that eventually we will arriveat a matrix of the form [

±1 ∗0 ∗

]or

[0 ∗±1 ∗

]Since S moves between matrices of these two forms, we may assume without loss of generality that we have a matrixof the first form. But, now since this matrix must have determinant 1 it’s easy to see that the bottom left entry must

be ±1. Thus, we see that, in fact, we may continue the process of applying powers of B and S to

[a bc d

]to get a

matrix of the form [±1 x0 ±1

]Now, since x is an unknown integer, we may up to relabeling, multiply this matrix by −I = S2 to get the matrix[

1 x0 1

]with x ∈ Z. But, since this matrix is precisely Bx we see it lands in the subgroup generated by A and B. Thus, putting

this all together, we see that we can apply combinations of powers of A and B to the left side of

[a bc d

]to get into

the subgroup generated by A and B, and so

[a bc d

]is in the generated subgroup. Since

[a bc d

]was arbitrary the fact

that the subgroup generated by A and B is SL2(Z) follows.

Exercise 1.1.2. Let γ =

(a bc d

)∈ SL2(Z).

1

i) Prove that Im(γ(z)) =Im(z)

|cz + d|2

ii) Show that if γ′ ∈ SL2(Z) then (γγ′)(z) = γ(γ′(z)) for all z ∈ H.

iii) Show thatdγ

dz=

1

(cz + d)2.

Solution i) We merely make the following set of calculations

γ(z) =az + b

cz + d

=(az + b)(cz + d)

|cz + d|2

=ac|z|2 + adz + bcz + bd

|cz + d|2

= (ad− bc) Im(z)i

|cz + d|2+

(ad+ bc)Re(z) + ac|z|2 + bd

|cz + d|2

=Im(z)i

|cz + d|2+

(ad+ bc)Re(z) + ac|z|2 + bd

|cz + d|2

Since the second summand of the last line is real, it follows that Im(γ(z)) =Im(z)

|cz + d|2as desired.

ii) Suppose that γ′ =

(a′ b′

c′ d′

). Then,

(γγ′)(z) =

(aa′ + bc′ ab′ + bd′

ca′ + dc′ cb′ + dd′

)(z)

=(aa′ + bc′)z + (ab′ + bd′)

(ca′ + dc′) + (cb′ + dd′)

and

γ(γ′(z) = γ

(a′z + b′

c′z + d′

)

=

a

(a′z + b′

c′z + d′

)+ b

c

(a′z + b′

c′z + d′

)+ d

=a(a′z + b′) + b(c′z + d′)

c(a′z + b′) + d(c′z + d′)

=(aa′ + bc′)z + (ab′ + bd′)

(ca′ + dc′)z + (cb′ + dd′)

Since this was true for all z ∈ H it follows that γγ′ = γ γ′ as maps, as desired.

iii) This is merely differentiating z 7→ az + b

cz + dwhich gives the desired result by applying the quotient rule.

Exercise 1.1.3. i) Show that the set Mk(SL2(Z)) is a vector space over C.

ii) If f and g are modular forms weight k and ` respectively then fg is a modular form of weight k + ` respectively.

iii) Show that Sk(SL2(Z)) is a vector subspace of Mk(SL2(Z)) and that S(SL2(Z)) is an ideal of M(SL2(Z)).

2

Solution i) Since a linear combination of holomorphic functions on H is holomorphic it suffices to prove that alinear combination of weakly modular functions of weight k are weakly modular of weight k, and that a linearcombination of bounded at i∞ functions are bounded at i∞. The first is clear for if f and g are weakly modular

of weight k, and if γ =

(a bc d

)∈ SL2(Z) then

(λ1f + λ2g)(γ(z)) = λ1f(γ(z)) + λ2g(γ(z))

= λ1(cz + d)kf(z) + λ2(cz + d)kg(z)

= (cz + d)k(λ1f(z) + λ2g(z))

= (cz + d)k(λ1f + λ2g)(z)

To show the second claim let f and g be bounded at i∞ which is equivalent to limIm(z)→∞

f(z) and limIm(z)→∞

g(z) exist-

ing. But, it’s easy to see that, in fact, limIm(z)→∞

(λ1f+λ2g)(z) exists, and in particular, is equal to λ1 limIm(z)→∞

f(z)+

λ2 limIm(z)→∞

g(z). Thus, with all of these properties verified it follows that Mk(SL2(Z)) is a C-vector space as fol-

lows.

ii) Since the product of holomorphic functions is holomorphic, it suffices to prove that fg is bounded at i∞ andthat fg is weakly modular of weight k + `. To see this first fact it suffices to note that, as is easy to verify,

limIm(z)→∞

(fg)(z) =

(lim

Im(z)→∞f(z)

)(lim

Im(z)→∞g(z)

), and in particular, that lim

Im(z)→∞(fg)(z) exists. The second

claim follows since if γ =

(a bc d

)∈ SL2(Z) then

(fg)(γ(z) = f(γ(z))g(γ(z))

= (cz + d)kf(z)(cz + d)`g(z)

= (cz + d)k+`f(z)g(z)

= (cz + d)k+`(fg)(z)

for every z ∈ H from which the conclusion follows.

iii) Consider the map F = limIm(z)→∞

:Mk(SL2(Z)) → C, defined in the obvious way. Now, noting that F |Mk(SL2(Z))

is a C-linear map and that Sk(SL2(Z)) is the kernel of this map we see that Sk(SL2(Z)) is a C-subspace ofMk(SL2(Z)). Now, since S(SL2(Z)) is the kernel of F , and F is easily seen to be a C-algebra map, it followsthat S(SL2(Z)) is an ideal in M(SL2(Z)). Moreover, since F is obviously surjective (being C-linear and nonzero)it follows from the first isomorphism theorem that M(SL2(Z))/S(SL2(Z)) ∼= C and so, in fact, S(SL2(Z)) is amaximal ideal of M(SL2(Z)).

Exercise 1.1.4. Let k > 3 be an integer and let L′ = Z2 − (0, 0).

i) Show that the series∑

(c,d)∈L

(max|c|, |d|)−k converges by considering partial sums over expanding squares.

ii) Fix positive numbers A and B and let

Ω = z ∈ H : |Re(z)| 6 A, Im(z) > B

Prove that there is a constant C > 0 such that |z + δ| > C max1, |δ| for all z ∈ Ω and δ ∈ R.

iii) Use parts i) and ii) to prove that the series defining Gk(z) converges absolutely and uniformly for z ∈ ∆. Concludethat Gk is holomorphic on H.

iv) Show that for γ ∈ SL2(Z), right multiplication by γ defines a bijection L′ → L′.

3

v) Use the calculation from iii) to show that Gk is bounded on Ω. From the text and part iv), Gk is weakly modularso in particular Gk(z + 1) = Gk(z). Show that therefore Gk is bounded as Im(z)→∞.

Solution i) As intuition shows us by considering expanding squares about the origin defined by (x, y) ∈ R2 :max|x|, |y| = m we see that

∑(c,d)∈L′

(max|c|, |d|)−k =

∞∑m=1

Ammk

where Am = #(x, y) ∈ Z2 : max|x|, |y| = m. That said, it’s easy to see that Am 6 8(m + 1) (by consideringthe 8 half edges of the square which each contain m+ 1 points ) so that

∞∑m=1

Amm−k

6 8

∞∑m=1

(m+ 1)

mk6 8

∞∑m=1

(m+ 1)

mk<∞

and so the conclusion follows.

ii) As the book suggests, we break this into four cases. Let z = x+ iy

Case 1) Suppose that |δ| < 1. Then,

|z + δ| =√B2 + (δ −A)2 > B = Bmax1, |δ|.

Case 2) Suppose next that 1 6 |δ| 6 3A and y > A. Note then that

|z + δ| =√

((x− δ)2 + y2 > A >|δ|3

=1

3max1, |delta|.

Case 3) Now, if B > A then Im(z) > B > A for all z ∈ Ω and so we may assume without loss of generalitythat B 6 A. So, assume in particular that 1 6 |δ 6 3A and B 6 Im(z) 6 A. Then, since the region

[1, 3A] × z ∈ Ω : B 6 Im(z) 6 A is compact the mapping (z, δ) 7→ |z + δ

δassumes a minimum, call it

M . Then, by definition

|z + δ| >M |δ| = M max1, |δ|

Case 4) Lastly, if |δ > 3A then

|z + δ| > |δ| −A >2

3|δ =

2

3max1, |δ|

where we used the reverse triangle inequality.

So, if we take C = 12 min 1

3 ,M,B the problem follows from these four cases.

iii) Note that on Ω if we define f(c,d)(z) = (cz + d)−k then we have that

|f(c,d)(z)| 6Ck

max|c|, |d|k

where C is the constant found in i). Then, by applying a double series version of the Weierstrass M -test wemay conclude that Gk(z) converges uniformly and absolutely on Ω. The fact that Gk(z) then converges to aholomorphic function on Ω is simple complex analysis. Indeed, take any z0 ∈ H and choose a ball Bδ(z0) aroundthat point. Since Bδ(z0) is contained in some region Ω we know that Gk(z) converges uniformly and absolutelyon Bδ(z0), and moreover since Gk(z) is extendable to Bδ(z0) which is compact we know that Gk is bounded onBδ(z0). Then, to see that Gk(z) is holomorphic on Bδ(z0) it suffices to note that if T is any triangle in Bδ(z0)then ∮

T

Gk(z) =∑

(c,d)∈L′

∮T

1

(cz + d)k=

∑(c,d)∈L′

0 = 0

where we could interchange the sum and series by uniform convergence, and the integral over the inside is zeroby Cauchy’s theorem (note that the roots of cz + d = 0 aren’t in H). Since T was arbitrary Morera’s theoremimplies that Gk(z) is holomorphic in Bδ(z0) and so, in particular, holomorphic at z0. Since z0 was arbitrary theconclusion follows.

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iv) We merely note that since γ ∈ SL2(Z) ⊆ GL2(Z) that γ is an invertible linear map Z2 → Z2 and so, in particular,is a bijection L′ → L′.

v) We know that if M denotes the constant Ck∑

(c,d)∈L′

1

max|c|, |d|kfound in iii) then we know from iii) that

|Gk(z)| 6 M for all z ∈ Ω. Now, since Gk(z) is 1-periodic that Gk(z) being bounded as Im(z) → ∞ with z ∈ Ω,with Ω being with respect to A = B = 1, implies that Gk(z) is boundeded as Im(z)→∞ with z ∈ H.

Exercise 1.1.5. Establish the following formulae:

(1) π cot(πz) =1

z+

∞∑n=1

(1

z − n+

1

z + n

)(2) π cot(πz) = πi− 2πi

∞∑n=0

e2πinz

Solution To prove (1) start with the well-known formula

sin(πz) = πz

∞∏n=1

(1− zn

n2

)Then, we see that taking the logarithmic derivative gives

πcos(πz)

sin(πz)=

1

z+

∞∑n=1

2z

z2 − n2

Now, by definition the left hand side of the above equation is π cot(πz) and since2z

z2 − n2=

1

z − n+

1

z + nthe equation

(1) follows.

To establish (2) we make the following series of computations

π cot(πz) = πcos(πz)

sin(πz)

= π

eiπz + e−iπz

2eiπz − e−iπz

2i

= πieiπz + e−iπz

eiπz − e−iπz

= πie2πiz + 1

e2πiz − 1

= πi

(1 +

2

e2πiz − 1

)= πi

(1− 2

e2πiz − 1

)= πi− 2πi

1

e2πiz − 1

= πi− 2πi

∞∑n=0

e2πinz

The last step where we expanded1

1− e2πizas a series was valid for if z ∈ h then∣∣e2πiz∣∣ = eRe(2πiz) = e−2πi Im(z) < 1

since Im(z) > 0.

5

Exercise 1.1.6. This exercise obtains formula (1.2) without using the cotangent. Let f(z) =∑d∈Z

1

(z + d)kfor k > 2

and z ∈ H. Since f is holomorphic (by the method of exercise 1.1.4) and Z-periodic and since limIm(z)→∞

f(z) = 0 there

is a Fourier expansion f(z) =

∞∑m=1

amqm = g(q) as in the section where q = e2πiz and

am =1

2πit

∫γ

g(q)

qm+1dq

is a path integral once counterclockwise over a circle about 0 in the punctured disc D′.

i) Show that

am =

∫ 1+iy

z=0+iy

f(z)e−2πimz dz =

∫ ∞+iy

z=−∞+iy

z−ke−2πimz dz

for all y > 0.

ii) Let gm(z) = z−ke−2πimz, a meromorphic function on C with a singularity only at the origin. Show that

−2πi Res(gm(z), 0) =(−2πi)k

(k − 1)!mk−1

iii) Establish (1.2) by integrating gm(z) clockwise about a large rectangular path and applying the Residue Theorem.Argue that the integral along the top side goes to am and the integral along the other three sides go to 0.

iv) Let h : R → C be a function such that the integral

∫ ∞−∞|h(x)|dx is finite and the sum

∑d∈Z

h(z + d) converges

absolutely and uniformly on compact subsets and is infinitely differentiable. Then the Poisson summation formulasays that ∑

d∈Zh(x+ d) =

∑m∈Z

h(m)e2πimx

where h is the Fourier transform of h,

h(x) =

∫ ∞−∞

h(t)e−2πixt dt

We will not prove this, but the idea is that the left side sum symmetries h to a function not period 1 and the right

side sum is the Fourier series of the left side since the mth Fourier coefficient is

∫ 1

0

∑d∈Z

h(t+ d)e−2πimt = h(m).

Letting h(x) =1

(x+ iy)kwhere y > 0 , show that h meets the requirements for Poisson summation. Show that

h(m) = e−2πmyam with am from above for m > 0 and that h(m) = 0 for m 6 0. Establish formula (1.2) again,this time as a special case of Poisson summation.

Solution i) We begin by noting by the homotopy version of Cauchy’s theorem that am is equal no matter whatradius the circle γ is. So, parameterize the circle as γ(t) = re2πit for some r ∈ (0, 1). Note then that by makingthis parameterization we have that

am =1

2πi

∫γ

g(z)

zm+1dz

=1

2πi

∫ 1

0

g(re2πit

)(re2πit)

m+1

(2πire2πit

)dt

=

∫ 1

0

g

(exp

(2πi

(log(r)

2πi+ t

)))r−me−2πimt dt

= r−m∫ 1

0

f

(t− i log(r)

2π

)e−2πimt dt

6

Now, note that this makes sense in terms of domains since Im

(t− i log(r)

2π

)is − log(r)

2π> 0 since r < 1. Since

this was independent of r we have that

am = limr→1−

(r−m

∫ 1

0

f

(t− i log(r)

2π

)e−2πimt dt

)= limr→1−

∫ 1

0

f

(t− i log(r)

2π

)e−2πimt dt

Now, noting that

t− i log(r)

2π: (t, r) ∈ [0, 1]× [ 1

2 , 1)

is contained in a bounded region of H we have by the

continuity of f on H that f is uniformly bounded on this region. Thus, we may apply the Bounded ConvergenceTheorem to conclude that

am = limr→1−

∫ 1

0

f

(t− i log(r)

2π

)e−2πimt dt

=

∫ 1

0

limr→1−

f

(t− i log(r)

2π

)e−2πimt dt

=

∫ 1

0

f

(limr→1−

(t− i log(r)

2π

))e−2πimt dt

=

∫ 1

0

f(t)e−2πimt dt

Now, making the substitution t 7→ t+ iy for any y > 0 we see that

am =

∫ 1

0

f (t+ iy) e−2πim(t+iy) dt =

∫ 1+iy

iy

f(z)e−2πimz dz

where we used the fact that t 7→ t+ iy was a parameterization of a path from iy to 1 + iy (the linear path). Thisproves the first equality. The second is achieved by making the following calculations

am =

∫ 1+iy

iy

f(z)e−2πimz dz

=

∫ 1+iy

iy

∑d∈Z

e−2πimz

(z + d)kdz

=∑d∈Z

∫ 1+iy

iy

e−2πimz

(z + d)kdz

where the interchanging of the limits was valid because the convergence of the series was uniform and f is boundedas z → i∞. Now, making the change of variables z 7→ z + d and using the 2πi periodicity of exp gives us that

am =∑d∈Z

∫ 1+iy

iy

e−2πimz

(z + d)kdz

=∑d∈ Z

∫ d+1+iy

d+iy

e−2πimz

zkdz

=

∫ ∞+iy

−∞+iy

z−ke−2πimz dz

as desired.

ii) To find the residue at 0 we merely find the Laurent series of gm(z) at 0. Indeed,

gm(z) = z−ke−2πimz

= z−k∞∑n=0

(−2πim)nzn

n!

=

∞∑n=0

(−2πim)nzn−k

n!

7

Thus, the residue of gm(z) at 0 is the coefficient of z−1 which occurs when n− k = −1 or n = k− 1. Thus, we see

that the residue is(−2πim)k−1

(k − 1)!. Thus, −2πiRes(gm(z), 0) is

(−2πi)k

(k − 1)!mk−1 as desired.

iii) Let ΓR,T be the symmetric rectangle of width 2R and height 2T centered at the origin. Consider then that by theResidue Theorem and ii) we have that

(−2πi)k

(k − 1)!mk−1 =

∫ΓR,T

z−ke−2πimz dz

= f(R, T ) + g1(R, T ) + g2(R, T ) + g3(R, T )

(1)

where

f(R, T ) =

∫ R+iT

−R+iT

z−ke−2πiz dz

is the integral over the top portion of ΓR,T ,

g1(R, T ) = −ie−2πimR

∫ T

−T(R+ it)−ke2πmt dt

is the integration over the rightmost vertical part of ΓR,T (after making the standard parameterization),

g2(R, T ) = ie2πimR

∫ T

−T(it−R)−ke2πmt dt

is the integration over the leftmost vertical part of ΓR,T (after making the standard parameterization),and

g3(R, T ) = e−2πmT

∫ R

−R(t− iT )−ke−2πimt dt

is the integration over the bottom portion of ΓR,T (after making the standard parameterization). Now, we makethe claim that lim

R→∞gi(R, T ) = 0 for i = 1, 2 and lim

R→∞g3(R, T ) is well-defined (non-infinite) function of T . To see

that the first of these assertions is true we merely note that

|g1(R, T )| 6∫ T

−T|R+ it|−ke2πmt dt

61

Rk

∫ T

−Te2πimt dt

And since, for fixed T , this last term approaches 0 as R → ∞ it follows that limR→∞

g1(R, T ) = 0 for each fixed

T as desired. A nearly identical calculation shows that limR→∞

g2(R, T ) = 0 for each fixed T . Now, the fact that

limR→∞

g3(R, T ) is a well-defined function of T follows by the mere, and obvious fact, that

∫ ∞−∞

(t− iT )−ke−2πimt

is convergent (because the exponential term is modulus 1). We last make the observation that by i) we have that,independent of T , we have that lim

R→∞f(R, T ) = am. Thus, putting this all together and letting R tend to infinity

inf (1) gives

(−2πi)k

(k − 1)!mk−1 = am + e−2πmT

∫ ∞−∞

(t− iT )−ke−2πimt (2)

Now, since

8

∣∣∣∣e−2πmT

∫ ∞−∞

(t− iT )−ke−2πimt

∣∣∣∣ 6 e−2πmT

∫ ∞−∞|t− iT |−k dt

= e−2πmT

∫ ∞−∞

1

(t2 + T 2)k2

dt

6 e−2πmT

∫ ∞−∞

dt

(t2 + 1)k2

and because k is large enough the integral in the last term is convergent, and since e−2πmT → 0 as T →∞ we seein particular that

limT→∞

(e−2πmT

∫ ∞−∞

(t− iT )−ke−2πmt dt

)= 0

So, letting T tend to infinity in (2) gives us

(−2πi)k

(k − 1)!mk−1 = am

as desired.

iv) Since |h(x)| =1

(x2 + y2)k2

and so |h(x)| ∼ 1

|x|k(asymptotically equivalent to) which converges because k > 2.

Moreover, the fact that∑d∈Z

h(z+ d) converges uniformly and absolutely on compact subsets is clear from previous

exercise. Lastly,∑d∈Z

h(z + d) is infinitely differentiable since it is holomorphic which is true precisely because it

converges uniformly on compact subsets of C. Thus, Poisson summation applies. Noting that

h(m) =

∫ ∞−∞

h(x+ iy)e−2πimx dx

= e−2πmy

∫ ∞−∞

h(x+ iy)e−2πim(x+iy) dx

= e−2πmy

∫ ∞+iy

−∞+iy

h(z)e−2πimz dz

= e−2πmyam

for m > 0 where the last step follows from i). Since∑d∈Z

h(z + d) = f(z) we see that

f(z) =∑d∈Z

ame−2πim(x+iy) =

∑d∈Z

amqm

Exercise 1.1.7. The Bernoulli numbers Bk are defined by the formal power series expansion

t

et − 1=

∞∑k=0

Bktk

k!

Thus they are calculable in succession by matching coefficients in the power series identity

t = (et − 1)

∞∑k=0

Bktk

k!=

∞∑n=1

(n−1∑k=0

(n

k

)Bk

)tn

n!

(i.e. the nth parenthesized sum is 1 if n = 1 and 0 otherwise) and they are rational. Since the expression

t

et − 1+t

2=t

2· e

t + 1

et − 1

9

is even it follows that B1 =−1

2and Bk = 0 for all other k. The Bernoulli numbers will be motivated, discussed, and

generalized in Chapter 4.

i) Show that B2 =1

6, B4 =

−1

30, and B6 =

1

42.

ii) Use the expression for π cot(πz) from the section to show

1− 2

∞∑k=1

ζ(2k)z2k = π cot(πz) = πiz +

∞∑k=0

(2πiz)k

k!.

Use these to show that for k > 2 even, the Riemann zeta function satisfies

2ζ(k) = − (2πi)k

k!Bk

so in particular ζ(2) =π2

6, ζ(4) =

π4

90, and ζ(6) =

π6

945. Also, this shows that the normalized Eisenstein series of

weight K

Ek(z) =Gk(z)

2ζ(k)= 1− 2k

Bk

∞∑n=1

σk−1(n)qn

has rational coefficients with a common denominator.

iii) Equation coefficients in the relation E8(z) = E4(z)2 to establish formula (1.3).

iv) Show that a0 = 0 and a1 = (2π)12 in the Fourier expansion of the discriminant function ∆ from the text.

Solution i) We know that B0 = 1, B1 = −12 , and B3 = 0. Then, since

0 =

2∑k=0

(3

k

)Bk = B0 + 3B1 + 3B2

we get that B2 =1

6. The other values of Bk are found using the exact same recursion.

ii)

Exercise 1.1.8. Recall that µ3 denotes the complex cube root of unity e2πi3 . Show that

(0 −11 0

)(µ3) = µ3 + 1

so that by periodicity g2

((0 −11 0

)(µ3)

)= g2(µ3). Show that by modularity g2

((0 −11 0

)(µ3)

)= µ4

3g2(µ3) and

therefore g2(µ3) = 0. Conclude that g3(µ3) 6= 0 and j(µ3) = 0. Argue similarly to show that g3(i) = 0, g2(i) 6= 0, andj(i) = 1728.

1.2

Exercise 1.2.1. f

Exercise 1.2.2. f

Exercise 1.2.3. i) Let p be a prime and let e be a positive integer. Show that |SL2(Z/peZ)| = p3e − p3e−2.

ii) Show that |SL2(Z/NZ)| = N3∏p|N

(1− 1

p2

), so this is also the index [SL2(Z) : Γ(N)].

10

iii) Show that the map Γ1(N)→ Z/NZ given by

(a bc d

)7→ b mod N surjects and has kernel Γ(N).

iv) Show that the map Γ0(N)→ (Z/NZ)× given by

(a bc d

)7→ d mod N surjects and has kernel Γ1(N).

v) Show that [SL2(Z) : Γ0(N)] = N∏p|N

(1 +

1

p

).

Solution i) We induct on e. So, to begin we find |SL2(Z/pZ)| by brute force. Indeed, we are looking for

(a bc d

)∈

Mat2(Z/pZ) such that ad− bc = 1. Suppose first that d = 0, then we are looking for (a, b, c) ∈ (Z/pZ)3 such thatbc = 1, clearly there are p(p− 1) such pairs since a can be arbitrary, c is determined by b and b can be any unit.Now, if d 6= 0 then d is a unit and so there are p− 1 choices for d. Moreover, once we have found b and c we knowthat a is determined. Moreover b and c can be anything for then a is just d−1(bc+ 1). Thus, there are (p− 1)p2

choices. Thus, overall there are (p− 1)p2 + p(p− 1) = p3 − p choices as desired.

Now, suppose that we have proven that |SL2(Z/peZ)| = p3e−p3e−2 and consider SL2(Z/pe+1Z). There is a naturalgroup map SL2(Z/pe+1Z) → SL2(Z/peZ) given by reducing entries modulo pe. Now, something in the kernel of

this map is of the form

(mpe + 1 `pe

kpe npe + 1

). Moreover, we know that the determinant of this matrix must be 1

modulo pe but the determinant modulo pe is just pe(m+ n) + 1. Thus, we need that p | m+ n and that k and `can be arbitrary. Now, since we are working modulo pe+1 the pe in front of m,n, ` and k means that we are reallyonly looking for m,n, ` and k modulo p. So, the fact that k and ` can be arbitrary gives us p2 choices (p for each).To find the number of m and n is equivalent to finding the number of m and n in Z/pZ that sum to 0 in Z/pZ.To find this note that we have the obvious group map (Z/pZ)2 → Z/pZ → Z/pZ : (m,n) 7→ m + n for which weare trying to find the cardinality of the kernel. But, this map is obviously surjective and so the kernel has sizep2

p= p. Thus, we see that there are p choices for (m,n). Thus, overall there are p2 · p = p3 choices. Thus, the

kernel of the map SL2(Z/pe+1Z) → SL2(Z/peZ) has cardinality p3 and so the first isomorphism theorem impliesthat

|SL2(Z/pe+1Z)| = p3|SL2(Z/peZ)| = p3(p3e − p3e−2

)= p3(e+1) − p3(e+1)−2

and so the induction is complete.

ii) Let’s first recall that the formation of SL2(R) for a ring R is functorial, so that if R ∼= S as rings then SL2(R) ∼=SL2(S). Next, note that (

(a1, a2) (b1, b2)(c1, c2) (d1, d2)

)7→((

a1 b1c1 d1

),

(a2 b2c2 d2

))is a group isomorphism SL2(R × S) → SL2(R) × SL2(S). Combining these results and the Chinese RemainderTheorem we know that if N = pe11 · · · penn then

SL2(Z/NZ) ∼=n∏i=1

SL2(Z/peii Z)

Taking orders of both sides gives

|SL2(Z/NZ)| =n∏i=1

p3eii

(1− 1

p2i

)= (pe11 · · · penn )3

∏p|N

(1− 1

p2

)= N3

∏p|N

(1− 1

p2

)

as desired.

iii) To see that this is a group map we merely note that if f is the described map and

(a bc d

),

(a′ b′

c′ d′

)∈ Γ1(N)

then

11

f

((a bc d

)(a′ b′

c′ d′

))= ab′ + bc′ mod N ≡ b+ b′ mod N

since c ≡ c′ ≡ 1 mod N . But, it’s obvious that

b+ b′ mod N = f

((a bc d

))+ f

((a′ b′

c′ d′

))

and so f really is a homomorphism. It’s surjective for if b ∈ 0, · · · , N − 1 then

(1 b0 1

)is in Γ(n) and the image

is precisely b. Finally, the kernel of the map is(a bc d

)∈ Γ1(N) : b ≡ 0 mod N

But, by definition, since

(a bc d

)∈ Γ1(N) we know that a ≡ d ≡ 1 mod N and c ≡ 0 mod N . Thus, with

b ≡ 0 mod N we have that (a bc d

)≡(

1 00 1

)mod N

and so

(a bc d

)∈ Γ(N) and the reverse inclusion is just as trivial.

iv) To see that the map Γ0(N)→ (Z/NZ)×, call this map g, is a group map we merely note that if

(a bc d

),

(a′ b′

c′ d′

)∈

Γ0(N) then

g

((a bc d

)(a′ b′

c′ d′

))= cb′ + dd mod N = dd′

since c ≡ 0 mod N by the definition that

(a bc d

)∈ Γ0(N). But, clearly dd′ is precisely

g

((a bc d

))g

((a′ b′

c′ d′

))and so it follows that g really is a group map. This map surjects since if d ∈ (Z/NZ)× then (d,N) = 1 and so

there exists ba such that ad+ bN = 1. We see then that γ =

(a bN d

)∈ Γ0(N) and g (γ) = d as desired.

v) We have the following chain of subgroups Γ(N) ⊆ Γ1(N) ⊆ Γ0(N) ⊆ SL2(Z). Now, from Lagrange’s theorem weknow that

[SL2(Z) : Γ0(N)] =[SL2(Z) : Γ(N)]

[Γ0(N) : Γ1(N)][Γ1(N) : Γ(N)]

Now, from iii) we know that [Γ1(N) : Γ(N)] = N , and from iv) we know that if N = pe11 · · · penn

= ϕ(N)

= Nϕ(N)

N

= N

N∏i=1

pei−1(p− 1)

n∏i=1

peii

= N

n∏i=1

(1− 1

pi

)

12

Thus,

[ SL2(Z) : Γ0(N)] =[SL2(Z) : Γ(N)]

[Γ0(N) : Γ1(N)][Γ1(N) : Γ(N)]

=

N3n∏i=1

(1− 1

p2i

)N2

n∏i=1

(1− 1

pi

)

= N

n∏i=1

(1 +

1

pi

)= N

∏p|N

(1 +

1

p

)

as desired.

Exercise 1.2.4.

Exercise 1.2.5. If Γ 6 SL2(Z) is a congruence subgroup and γ ∈ SL2(Z), then γΓγ−1 is a congruence subgroup ofSL2(Z).

Solution We merely note that since Γ is a congruence subgroup there exists some N ∈ Z such that Γ(N) 6 Γ.Then, γΓ(N)γ−1 6 γΓγ−1, but Γ(N) E SL2(Z) and so γΓ(N)γ−1 = Γ(N). Thus, Γ(N) 6 γΓγ−1 and so γΓγ−1 is acongruence subgroup as desired.

Exercise 1.2.6. This exercise proves Proposition 1.2.4. Let Γ be a congruence subgroup of SL2(Z), thus containingΓ(N) for some N , and suppose that the function f : H → C is holomorphic and weight-k invariant under Γ. Suppose

also that in the Fourier expansion f(z) =

∞∑n=0

anqnN the coefficients for n > 0 satisfy |an| 6 Cnr for some positive

constants C and r.

i) Show that for any z = x+ iy ∈ H,

|f(z)| 6 |a0|+ C

∞∑n=1

nre−2πnyN (1)

Changing n to a nonnegative real variable t, show that the continuous version g(t) = tre−2πtyN of the summand

increases monotonically on the interval

[0,rN

2πy

]and then decreases monotonically on

[rN

2πy,∞)

. Using this,

represent all but two terms of the sum in (1) as unit-wide boxes under the graph of g and consider the missingterms individually to establish the estimate (where in this exercise C0 and C can denote different constants indifferent places)

|f(z)| 6 C0 + C

(∫ ∞0

g(t) dt+1

yr

)

After a change of variables the integral takes the formC

yr+1

∫ ∞0

tre−t dt. This last integral is a gamma function

integral, to be introduced formally in Section 4.4, but at any rate it converges at both ends and is independent ofy. In sum,

13

|f(z)| 6 C0 +C

yr

as y →∞.

ii) For every α ∈ SL2(Z), the transformed function (f [α]k)(z) is holomorphic and weight-k invariant under α−1Γαand therefore has a Laurent expansion

(f [α]k)(z) =∑n∈Z

a′nqnN

To show that the Laurent series truncates from the left to a power series it suffices to show that

limqN→0

(f [α]k)(z)qN = 0

If α fixes ∞ then this is immediate from the Fourier series of f itself. Otherwise show that the transformedfunction (f [α]k)(z) = (cz + d)−kf(α(z)) satisfies

limqN→0

|(f [α]k])(z)qN | 6 C limqN→0

(yr−k|qN |)

Recalling that qN = e2πi(x+y)

N , show that y = −C log(|qN |), and use the fact that polynomials dominate logarithmsto complete the proof.

Solution i) The equation (1) follows immediately by applying the triangle inequality, the fact that |an| 6 Cnr, andthat

∣∣exp(

2πinzN

)∣∣ = exp(Re(

2πinzN

))= exp

(−2πnyN

).

To establish the claim about g(t) we differentiate to find that g′(t) = −2πytr

N e−2πtyN + rtr−1e

−2πtyN , or g′(t) =

e−2πtyN tr−1

(−2πytN + r

). Since the multiplier is always positive, this is negative precisely when r 6 2πyt

N or Nr2πy 6 t.

Thus, we see that g is decreasing precisely when t ∈[Nr

2πy,∞)

, and thus g must be increasing on

[0,Nr

2πy

]as

desired.

Using the fact that the sign change occurs atNr

2πywe may use Riemann sums to estimate that

∞∑n=1

g(n) =

∞∑n=0

g(n) 6∫ ∞

0

g(t) dt+Nr

(2πye)r1

yr(2)

(the first equality being true because g(0) = 0). In particular, choosing the partition [n−1, n] : n ∈ N and using

Riemann left hand Riemann sums for intervals lying in[0, Nr2πy

], using right hand Riemann sums on intervals lying

inside

[Nr

2πy,∞)

, and bounding the sum on the partition containingNr

2πyby g

(Nr

2πy

)=

Nr

(2πye)r1

yr(since this is

the maximum g takes, and the interval is measure 1) we get (2). Thus, we see that

∞∑n=1

g(n) 6 C ′(∫ ∞

0

g(t) +1

yr

)

where C ′ = max

1,

Nr

(2πye)r

. Thus, we see that

|f(z)| 6 |a0|+ C

∞∑n=1

g(n) 6 |a0|+ CC ′(∫ ∞

0

g(t) dt+1

yr

)

Thus, taking C0 = |a0| and C = CC ′ (where, as was stated, the C means different things on the right and lefthand sides) we get our desired inequality.

14

ii) We have the estimation

|f [α]k(z)qN | = |j(α, z)|−k|f(α(z))qN 6 |j(α, z)|−k(C0 +

C

yr|j(α, z)2r|

)qN

where we used the previous part and made use of the fact that Im(α(z)) =Im(z)

|j(α, z)|2. Now, letting qN → 0 is

equivalent to letting y →∞. Now, from this it’s easy to see that |j(α, z)| is asymptotically C ′y for some constantC ′ (really it’s the c entry of α). Thus, we see that

limqN→0

|f [α]k(z)qN | 6 |j(α, z)|−k(C0 +

C

yr|j(α, z)|2r

)qN

= limqN→0

C

yr|j(α, z)|2r−kqN

= limqN→0

Dyr−kqN

where the first equality is via the (obvious) fact that limqN→0

|j(α, z)|−kC0qN = 0, and the second fact follows

from the asymptotic for |j(α, z)| with D = CC ′. Now, by definition we have that qN = exp(

2πi(x+iy)N

)and so

|qN | = e−2πiyN and so |y| = A log |qN | where A is a constant. Thus,

| limqN→0

Dyr−kqN | = limqN→0

|y|r−k|qN | = A limqN→0

log |qN |r−k|qN | = 0

because |qN dominates log |qN |r−k.

15