Problems Problems and Solutions Solutions
Diamond Diamond and Sh Shurm urman: an:
A Fi Firs rst t Cour Course se in Modu Modula lar r Forms orms
Author:
Alex Youcis
Last Revised:
March 12, 2013
1 1.1 Exercise 1.1.1.
Solution
1 1 0 Prove that SL2 (Z) is generated by A = and B = 0 1 1
−1 . 0
This amounts amounts to the Euclidean Euclidean algorithm. algorithm. Namely Namely, let’s begin with an element
then that
B
for every m ∈ Z and
m
a b ∈ SL2 (Z). We note c d
a b a + cm b + dm = c d c d
A
(1)
a b −c −d = c d a b
(2)
So, now, by the Euclidean algorithm we can write a = cq + cq + r and so we see that if we let m = −q we see by (1) that Bm So, we see then
a b a − cq = c d c
SB m
b − dq r = d c
b − dq d
−c −d a b = c d r b − dq
So, by the Euclidean algorithm we see that there exists r1 and q 1 so that −c = rq 1 + r1 and so taking m1 = −q 1 in (1) we see that B
m1
SB
m
a b −c − rq 1 = c d r
r −d − rq 1 = 1 d − rq r
−d − rq 1 b − dq
Continuing in this process, and using the fact that ad − bc = 1 so that (a, ( a, c) = 1, we see that eventually we will arrive at a matrix of the form
±1 ∗ 0 ∗
0 ∗ ±1 ∗
or
Since S moves between matrices of these two forms, we may assume without loss of generality that we have a matrix of the first form. But, now since this matrix must have determinant 1 it’s easy to see that the bottom left entry must a b be ±1. Thus, Thus, we see that, in fact, we may continue continue the process of applying powers powers of B and S to to get a c d matrix of the form
±1 x 0 ±1
Now, since x is an unknown integer, we may up to relabeling, multiply this matrix by −I = S 2 to get the matrix 1 0
x 1
with x ∈ Z. But, since this matrix matrix is precisely B x we see it lands in the subgroup generated by A and B . Thus, putting a b this all together, we see that we can apply combinations of powers of A and B to the left side of to get into c d a b a b the subgroup generated by A and B , and so is in the generated subgroup. Since was arbitrary the fact c d c d that the subgroup generated by A and B is SL2 (Z) follows.
Exercise 1.1.2.
Let γ =
a b c d
∈ SL2 (Z).
i) Prove Prove that Im(γ Im(γ (z )) =
Im(z Im(z ) |cz + d|2
ii) Show that if γ ∈ SL2 (Z) then (γγ )(z )(z ) = γ (γ (z )) for all z ∈ H. iii) Show that that Solution
dγ 1 = . dz (cz + d)2
i) We merely make the following set of calculation calculationss az + b cz + d (az + b)(cz )(cz + d) = |cz + d|2 ac|z |2 + adz + bcz + bd = |cz + d|2 Im(z Im(z )i (ad + bc)Re( bc)Re(zz ) + ac|z |2 + bd = (ad ( ad − bc) bc) + |cz + d|2 |cz + d|2 Im(z Im(z )i (ad + bc)Re( bc)Re(zz ) + ac|z |2 + bd = + |cz + d|2 |cz + d|2
γ (z ) =
Since the second summand of the last line is real, it follows that Im( γ (z )) = ii) Suppose that γ =
a c
Im(z Im(z ) as desired. |cz + d|2
b . Then, d
aa + bc ab + bd (γγ )(z )(z ) = (z ) ca + dc cb + dd (aa + bc )z + (ab (ab + bd ) = (ca + dc ) + (cb (cb + dd )
and
a z + b c z + d a z + b a +b c z + d = a z + b c +d c z + d a(a z + b ) + b(c z + d ) = c(a z + b ) + d(c z + d ) (aa + bc )z + (ab (ab + bd ) = (ca + dc )z + (cb (cb + dd )
γ (γ (z ) = γ
Since this was true for all z ∈ H it follows that γγ = γ ◦ γ as maps, as desired.
iii) This is merely merely differentiatin differentiatingg z →
Exercise 1.1.3.
az + b which gives the desired result by applying the quotient rule. cz + d
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 respectively then f g is a modular form of weight k + respectively. iii) Show that that S k (SL2 (Z)) is a vector subspace of Mk (SL2 (Z)) and that S (SL (SL2 (Z)) is an ideal of M(SL2 (Z)). )).
i) Since a linear combination of holomorphic functions on H is holomorphic it suffices to prove that a linear combination of weakly modular functions of weight k are weakly modular of weight k, and that a linear combination of bounded at i∞ functions are bounded at i∞. The first is clear for if f and g are weakly modular a b of weight k, and if γ = ∈ SL2 (Z) then c d
Solution
(λ1 f + λ2 g)(γ )(γ (z )) = λ1 f ( f (γ (z )) + λ2 g (γ (z )) = λ1 (cz + d)k f (z ) + λ2 (cz + d)k g(z ) = (cz + d)k (λ1 f ( f (z ) + λ2 g (z )) = (cz + d)k (λ1 f + λ2 g)(z )(z ) To show the second claim let f and g be bounded at i∞ which which is equiv equivale alent nt to ing. ing. But, But, it’s it’s easy easy to to see see that, that, in in fact fact,, λ2
lim
Im(z )→∞
lim
lim
Im(z )→∞
f ( f (z ) and
l im
Im(z )→∞
(λ1 f + f + λ2 g)(z )(z ) exists, and in particular, is equal to λ1
Im(z )→∞
g (z ). Thus, with all of these properties verified it follows that Mk (SL2 (Z)) is a
g(z ) exist-
lim
Im(z )→∞
C-vector
f ( f (z )+
space as fol-
lows. ii) Since the product of holomorphi holomorphicc functions is holomorphic, holomorphic, it suffices suffices to prove prove that f g is bounded at i∞ and that f g is weakly modular of weight k + . To see this first fact it suffices suffices to note that, that, as is easy easy to verif verify y, lim
(f g )(z )(z ) =
Im(z )→∞
lim
Im(z )→∞
claim follows since if γ =
f ( f (z )
lim
Im(z )→∞
and in part partic icul ular ar,, tha thatt g(z ) , and
a b ∈ SL2 (Z) then c d
lim
(f g )(z )(z ) exists. exists. The secon second d
Im(z )→∞
(f g )(γ )(γ (z ) = f ( f (γ (z ))g ))g(γ (z )) = (cz ( cz + d)k f (z )(cz )(cz + d) g (z ) = (cz ( cz + d)k+ f ( f (z )g(z ) = (cz ( cz + d)k+ (f g)(z )(z ) for every z ∈ H from which the conclusion follows. iii) Consider Consider the map F =
li m
Im(z )→∞
: Mk (SL2 (Z)) → C, defined defined in the obvious obvious way. way. Now, noting that F |Mk (SL2 (Z))
is a C-linear map and that S k (SL2 (Z)) is the kernel of this map we see that S k (SL2 (Z)) is a C-subspace -subspace of Mk (SL2 (Z)). No Now, w, since since S (SL (SL2 (Z)) is the kernel of F , F , and F is easily seen to be a C-algebra map, it follows that S (SL (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 (SL (SL2 (Z)) ∼ (SL2 (Z)) is a = C and so, in fact, S (SL maximal ideal of M(SL2 (Z)).
Exercise 1.1.4.
Let k
i) Show that the series series
3 be an integer and let L = Z2 − {(0, (0, 0)}.
(max{|c|, |d|})−k converges by considering partial sums over expanding squares.
(c,d)∈L
ii) Fix positive positive numbers numbers A and B and let Ω = {z ∈ H : | Re(z Re(z )| A, Im(z Im(z ) B } Prove that there is a constant C > 0 such that |z + δ | > C max C max{1, |δ |} |} for all z ∈ Ω and δ ∈ R. iii) Use parts parts i) and ii) to prove prove that the series defining G defining Gk (z ) converges absolutely and uniformly for z ∈ ∆. Conclude that Gk is holomorphic on H. iv) Show that that for γ ∈ SL2 (Z), right multiplication by γ defines a bijection L → L .
v) Use the calculatio calculation n from iii) to show that Gk is bounded on Ω. From the text and part iv), Gk is weakly modular so in particular Gk (z + 1) = Gk (z ). Show that therefore Gk is bounded as Im(z Im(z ) → ∞ . i) As intuition shows us by considering expanding squares about the origin defined by {(x, y) ∈ max{|x|, |y|} = m} we see that
Solution
(max{|c|, |d|})
−k
=
Am mk m=1
where Am = #{(x, y) ∈ Z2 : max{|x|, |y |} = m}. That said, it’s easy to see that Am the 8 half edges of the square which each contain m + 1 points ) so that ∞
Am m−k m=1
:
∞
(c,d)∈L
R2
∞
(m + 1) 8 mk m=1
8(m 8(m + 1) (by considering considering
∞
8
(m + 1) <∞ mk m=1
and so the conclusion follows. ii) As the book suggests, suggests, we break this into into four cases. cases. Let z = x + iy Case
1) Suppose that |δ | < 1. Then,
|z + δ | = Case
2) Suppose next that 1
(δ − A)2 B 2 + (δ
|}. B = B max{1, |δ |}
|δ | 3A and y > A A.. Note then that
1 |δ | = max{1, |delta|}. 3 3 3) Now, if B > A then Im(z Im(z ) B > A for all z ∈ Ω and so we may assume without loss of generality that B A. So, assume assume in particul particular ar that 1 |δ 3A and B Im(z Im(z ) A. Then, Then, since the the region region |z + δ [1, [1, 3A] × {z ∈ Ω : B Im(z Im(z ) A} is compact the mapping (z, ( z, δ ) → assumes a minimum, call it δ M . Then, by definition
|z + δ | =
Case
((x ((x − δ )2 + y 2
A>
M max{1, |δ |} |z + δ | M |δ | = M max |} Case
4) Lastly Lastly, if |δ > 3A then 2 2 |z + δ | |δ | − A |δ = max{1, |δ |} |} 3 3 where we used the reverse triangle inequality.
So, if we take C =
1 2
min{ 13 , M , B } the problem follows from these four cases.
iii) Note that that on Ω if we define f (c,d) (z ) = (cz + d)−k then we have that
|f (c,d) (z )|
C k max{|c|, |d|}k
where C is the constan constantt found found in i). Then, Then, by applying applying a double double series series versi version on of the Weie Weierst rstras rasss M -test M -test we may conclude that Gk (z ) conve converge rgess unifor uniformly mly and absolute absolutely ly on Ω. The fact that Gk (z) then converges to a holomorphic function on Ω is simple complex analysis. Indeed, take any z0 ∈ H and choose a ball Bδ (z0 ) around that point. point. Since Bδ (z0 ) is contained in some region Ω we know that Gk (z ) converges uniformly and absolutely on Bδ (z0 ), and moreover since Gk (z) is extendable to Bδ (z0 ) which is compact we know that Gk is bounded on Bδ (z0 ). Then, Then, to see that 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
0= 0
(c,d)∈L
where we could interchange the sum and series by uniform convergence, and the integral over the inside is zero by Cauchy’s theorem (note that the roots of cz + d = 0 aren’t in H). Since Since T was arbitrary Morera’s theorem implies implies that Gk (z) is holomorphic in Bδ (z0 ) and so, in particular, holomorphic at z0 . Since z0 was arbitrary the conclusion follows.
iv) We merely note that since γ ∈ SL2 (Z) ⊆ GL2 (Z) that γ is an invertible linear map is a bijection L → L . v) We know that if M denotes denotes the constant constant C k
(c,d)∈L
Z2
→ Z2 and so, in particular,
1 found in iii) then we know from iii) that max{|c|, |d|}k
Im( z ) → ∞ with z ∈ Ω, |Gk (z )| M for all z ∈ Ω. Now, since Gk (z) is 1-periodic that Gk (z) being bounded as Im(z with Ω being with respect to A = B = 1, implies that Gk (z) is boundeded as Im(z Im( z ) → ∞ with z ∈ H .
Exercise 1.1.5.
Establish the following formulae:
∞
(1) Solution
1 π cot(πz cot(πz)) = + z n=1
1 1 + z−n z+n
∞
(2)
π cot(πz cot(πz)) = πi − 2πi
e2πinz
n=0
To prove ( 1) start with the well-known formula
∞
sin(πz sin(πz)) = πz
1−
n=1
Then, we see that taking the logarithmic derivative gives
zn n2
∞
π
cos(πz cos(πz)) 1 2z = + sin(πz sin(πz)) z n=1 z 2 − n2
Now, by definition the left hand side of the above equation is π cot(πz cot(πz)) and since (1) follows.
z2
To establish (2) we make the following series of computations cos(πz cos(πz)) sin(πz sin(πz)) eiπz + e−iπz = π iπz 2 −iπz e −e 2i iπz e + e−iπz = πi iπz e − e−iπz e2πiz + 1 = πi 2πiz e −1 2 = πi 1 + 2πiz e −1 2 = πi 1 − 2πiz e −1 1 = πi − 2πi 2πiz e −1
π cot(πz cot(πz)) = π
∞
= πi − 2πi
e2πinz
n=0
The last step where we expanded
1 as a series was valid for if z ∈ h then 1 − e2πiz
e2πiz = eRe(2πiz ) = e−2πi Im(z) < 1
since Im(z Im(z ) > 0.
2z 1 1 = + the equation 2 −n z−n z+n
Exercise 1.1.6.
This exercise obtains formula (1. (1.2) without without using the cotangent. cotangent. Let Let f ( f (z ) =
d∈Z
and z ∈ H. Since f is holomorphic (by the method of exercise 1.1.4) and Z-periodic and since
1 for k (z + d)k
lim
Im(z )→∞
2
f ( f (z ) = 0 there
∞
is a Fourier expansion f ( f (z ) =
am q m = g (q ) as in the section where q = e2πiz and
m=1
am
1 = 2πit
g (q ) dq m+1 γ q
is a path integral once counterclockwise over a circle about 0 in the punctured disc D . i) Show that
1+iy
am =
f ( f (z )e
−2πimz
∞+iy
dz =
z =0+iy
z −k e−2πimz dz
z =−∞+ −∞+iy
for all y > 0. ii) Let Let gm (z ) = z −k e−2πimz , a meromorphic function on C with a singularity only at the origin. Show that Res(gm (z ), 0) = −2πi Res(g
(−2πi) πi )k k−1 m (k − 1)!
iii) Establish Establish (1. (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 Let h :
R
→
C
be a function such that the integral
|h(x)|dx is finite and the sum
−∞
h(z + d) converges
d∈Z
absolutely and uniformly on compact subsets and is infinitely differentiable. Then the Poisson summation formula says that
h(x + d) =
d∈Z
h(m)e2πimx
m∈Z
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
1
side sum is the Fourier series of the left side since the mth Fourier coefficient is
0 d∈Z
h(t + d)e−2πimt = h(m).
1 Letting h(x) = where y > 0 , show that h meets meets the requir requirements ements for Poisson Poisson summation. summation. Show that (x + iy) iy )k h(m) = e−2πmy am with am from above for m > 0 and that h(m) = 0 for m 0. Establish Establish formula formula (1. (1.2) again, this time as a special case of Poisson summation.
i) We begin by noting by the homotopy version of Cauchy’s theorem that am is equal no matter what radius the circle γ is. So, parameteri parameterize ze the circle as γ (t) = re2πit for some r ∈ (0, (0, 1). Note then that that by making this parameterization we have that
Solution
am
1 = 2πi =
g (z ) dz m+1 γ z 1
1 2πi
g re2πit
m+1 (re2πit )
0
1
=
g exp 2πi
0
1
= r−m
f t − i
0
2πire2πit dt
log(r log(r) +t 2πi
log(r log(r) 2π
r−m e−2πimt dt
e−2πimt dt
Now, note that this makes sense in terms of domains since Im this was independent of r we have that am = lim r →1
−
r
1
−m
0
log(r log(r) f t − i 2π
log(r log(r) t−i 2π 1
−2πimt
e
dt = lim r→1
−
0
is −
log(r log(r) > 0 since r < 1. Since Since 2π
log(r log(r ) f t − i 2π
e−2πimt dt
log(r log(r) : (t, ( t, r) ∈ [0, [0, 1] × [ 12 , 1) is contained in a bounded region of H we have by the 2π continuity of f on H that f is uniformly bounded on this region. Thus, we may apply the Bounded Convergence Theorem Theorem to conclude that
Now, noting that
t−i
1
am = lim
r →1−
0
log(r log(r) f t − i 2π
1
=
lim f t − i
− 0 r →1
log(r log(r) 2π
1
=
f
r→1−
0
1
=
t−i
lim
e−2πimt dt e−2πimt dt
log(r log(r) 2π
e−2πimt dt
f ( f (t)e−2πimt dt
0
Now, making the substitution t → t + iy for any y > 0 we see that
1
am =
−2πim(t+iy )
f ( f (t + iy) iy ) e
1+iy
dt =
0
f ( f (z )e−2πimz dz
iy
t + iy was a parameterization of a path from iy to 1 + iy (the linear path). This where we used the fact that t → proves the first equality. The second is achieved by making the following calculations
1+iy
am =
f ( f (z )e−2πimz dz
iy
1+iy
=
iy
=
d∈Z 1+iy
iy
d∈Z
e−2πimz dz (z + d)k e−2πimz dz (z + d)k
where the interchanging of the limits was valid because the convergence of the series was uniform and f is bounde b ounded d z + d and using the 2πi as z → i∞. Now, making the change of variables z → 2πi periodicity of exp gives us that
1+iy
am =
iy
d∈Z
e−2πimz dz (z + d)k
d+1+iy
=
d∈ Z d+iy
∞+iy
=
e−2πimz dz zk
z −k e−2πimz dz
−∞+ −∞+iy
as desired. ii) To find the residue at 0 we merely merely find the Laurent series series of gm (z) at 0. Indeed, gm (z ) = z −k e−2πimz ∞
=z
−k
∞
=
(−2πim) πim)n z n n! n=0
(−2πim) πim )n z n−k n! n=0
Thus, the residue of gm (z ) at 0 is the coefficient of z −1 which which occurs when n − k = −1 or n = k − 1. Thus, we see (−2πim) πim )k−1 (−2πi) πi )k k−1 that the residue is . Thus, −2πi Res(g Res(gm (z ), 0) is m as desired. (k − 1)! (k − 1)! iii) Let ΓR,T be the symmetric rectangle of width 2R 2 R and height 2T 2T centered centered at the origin. Consider Consider then that by the Residue Theorem and ii) we have that (−2πi) πi )k k−1 m = (k − 1)!
z −k e−2πimz dz
(1)
ΓR,T
= f ( f (R, T ) T ) + g1 (R, T ) T ) + g2 (R, T ) T ) + g3 (R, T ) T ) where
R+iT
f ( f (R, T ) T ) =
z −k e−2πiz dz
−R+iT
is the integral over the top portion of Γ R,T ,
T
−2πimR
g1 (R, T ) T ) = −ie
(R + it) it)−k e2πmt dt
−T
is the integration over the rightmost vertical part of Γ R,T (after making the standard parameterization),
T
g2 (R, T ) T ) = ie
2πimR
(it − R)−k e2πmt dt
−T
is the integration over the leftmost vertical part of Γ R,T (after making the standard parameterization),and
R
g3 (R, T ) T ) = e
−2πmT
(t − iT ) iT )−k e−2πimt dt
−R
is the integration over the bottom portion of Γ R,T (after making the standard parameterization). parameterization). Now, we make the claim claim that that lim gi (R, T ) T ) = 0 for i = 1, 2 and and lim lim g3 (R, T ) T ) is well-defined (non-infinite) function of T . T . To see R→∞
R→∞
that the first of these assertions is true we merely note that
T
|g1 (R, T ) T )|
|R + it|−k e2πmt dt
−T
1 Rk
T
e2πimt dt
−T
And since, for fixed T , T , this last term approaches 0 as R → ∞ it follo follows ws that that lim g1 (R, T ) T ) = 0 for each fixed R→∞
T as desired. desired. A nearly identi identical cal calculatio calculation n shows that that lim g2 (R, T ) T ) = 0 for each fixed T . T . No Now, w, the fact fact that R→∞
lim g3 (R, T ) T ) is a well-defined function of T follows by the mere, and obvious fact, that
R→∞
∞
(t − iT ) iT )−k e−2πimt
−∞
is convergent (because the exponential term is modulus 1). We last make the observation that by i) we have that, independent of T , T , we have have that that lim f ( f (R, T ) T ) = am . Thus, putting this all together and letting R tend to infinity R→∞
inf (1) gives (−2πi) πi )k k−1 m = am + e−2πmT (k − 1)! Now, since
∞
−∞
(t − iT ) iT )−k e−2πimt
(2)
−2πmT
∞
e
−k −2πimt
(t − iT ) iT )
e
−∞
−2πmT
e
= e−2πmT
e−2πmT
∞
|t − iT |−k dt
−∞ ∞
1 k
−∞ ∞
(t2
T 2 ) 2
−∞
(t2 + 1) 2
+ dt
dt
k
and becaus b ecausee k is large enough the integral in the last term is convergent, and since e−2πmT → 0 as T → ∞ we see in particular that lim
T →∞
∞
−2πmT
(t − iT ) iT )
e
−k −2πmt
e
dt
−∞
=0
So, letting T tend to infinity in ( 2) gives us (−2πi) πi )k k−1 m = am (k − 1)! as desired. iv) Since Since |h(x)| =
1
1 (asymptotically equivalent to) which converges because k |x|k
and so |h(x)| ∼
2. (x2 + y 2 ) 2 Moreover, the fact that h(z + d) converges uniformly and absolutely on compact subsets is clear from previous
exercise. Lastly, Lastly,
k
d∈Z
h(z + d) is infinitely differentiable since it is holomorphic which is true precisely because it
d∈Z
converges uniformly on compact subsets of C. Thus, Poisson summation applies. Noting that
∞
h(m) =
h(x + iy) iy )e−2πimx dx
−∞
∞
−2πmy
=e
h(x + iy) iy )e−2πim(x+iy) dx
−∞ ∞+iy
= e−2πmy
h(z )e−2πimz dz
−∞+ −∞+iy
−2πmy
=e
am
for m > 0 where the last step follows from i). Since
h(z + d) = f ( f (z) we see that
d∈Z
am e−2πim (x+iy) =
f ( f (z ) =
d∈Z
Exercise 1.1.7.
am q m
d∈Z
The Bernoulli numbers Bk are defined by the formal power series expansion t = t e −1
∞
Bk
k=0
tk k!
Thus they are calculable in succession by matching coefficients in the power series identity ∞
t
t = (e − 1)
k=0
∞
tk Bk = k ! n=1
n−1
k=0
n Bk k
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 t et + 1 = · t 2 2 e 1
−1 is even it follows that B1 = and Bk = 0 for all other k . The Bernoulli numbers numbers will be motivated motivated,, discusse discussed, d, and 2 generalized in Chapter 4. 1 1 −1 i) Show that B2 = , B4 = , and B6 = . 6 30 42 ii) Use the expression expression for π cot(πz cot(πz)) from the section to show ∞
1−2
∞
ζ (2k (2k)z
2k
= π cot(πz cot(πz)) = πiz +
k=1
Use these to show that for k
k=0
(2πiz (2πiz))k . k!
2 even, the Riemann zeta function satisfies 2ζ (k ) = −
(2πi (2πi))k Bk k!
π2 π4 π6 so in particular ζ (2) (2) = , ζ (4) (4) = , and ζ (6) (6) = . Also, Also, this shows that the normalized normalized Eisenstein series series of 6 90 945 weight K 2k Gk (z ) E k (z ) = =1 − 2ζ (k ) Bk
∞
σk−1 (n)q n
n=1
has rational coefficients with a common denominator. iii) Equation coefficients coefficients in the relation E 8 (z ) = E 4 (z )2 to establish formula (1. (1.3). 3). iv) Show that that a0 = 0 and a1 = (2π (2 π)12 in the Fourier expansion of the discriminant function ∆ from the text. Solution
i) We know that B0 = 1, B1 =
−1 2 ,
and B3 = 0. Then, since
2
0=
k =0
3 Bk = B0 + 3B 3B1 + 3B 3B2 k
1 we get that B2 = . The other values of Bk are found using the exact same recursion. 6 ii)
Exercise 1.1.8.
Recall Recall that µ3 denotes denotes the complex complex cube root root of unity e
2πi 3
. Show Show tha that t
0 −1 (µ3 ) = µ3 + 1 1 0
0 −1 0 −1 (µ3 ) = g2 (µ3 ). Show Show that that by modula modularit rity y g2 (µ3 ) = µ43 g2 (µ3 ) and 1 0 1 0 therefore g2 (µ3 ) = 0. Conclude that g3 (µ3 ) = Argue similarly similarly to show that g3 (i) = 0, g2 (i) = 0 and j (µ3 ) = 0. Argue 0, and j( j (i) = 1728. 1728. so that by periodicity g2
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/pe Z)| = p3e − p3e−2 . 3
ii) Show that that | SL2 (Z/N Z)| = N
1−
p|N
1 , so this is also the index [SL2 (Z) : Γ(N Γ(N )] )].. p2
iii) Show that the map Γ1 (N ) N ) → Z/N Z given by
a b c d
iv) Show Show that the map Γ0 (N ) N ) → (Z/N Z)× given by v) Show that [SL2 (Z) : Γ0 (N )] N )] = N
1+
p|N
Γ(N )). → b mod N surjects and has kernel Γ(N
a b mod N surjects and has kernel Γ1 (N ) N ). → d mod N c d
1 . p
a b ∈ c d Mat2 (Z/pZ) such that ad − bc = 1. Suppose first that d = 0, then we are looking for (a,b,c ( a,b,c)) ∈ (Z/pZ)3 such that bc = 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 if d = 0 then d is a unit and so there are p − 1 choices for d. Moreover, once we have found b and c we know that a is determined. Moreover Moreover b and c can be anything for then a is just d−1 (bc + 1). Thus, there are ( p ( p − 1) p 1) p2 choices. Thus, overall there are ( p ( p − 1) p2 + p( p( p − 1) = p3 − p choices as desired.
Solution
i) We induct on e. So, to begin we find | SL2 (Z/pZ)| by brute force. force. Indeed, Indeed, we are lo oking for
Now, suppose that we have proven that | SL2 (Z/pe Z)| = p3e − p3e−2 and consider SL2 (Z/pe+1 Z). There is a natural group map SL2 (Z/pe+1 Z) → SL2 (Z/pe Z) given by reducing entries modulo pe . Now, something something in the kernel kernel of e e mp + 1 p this map is of the form . Moreover, we know that the determinant of this matrix must be 1 e kp npe + 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 really only looking for m,n, and k modulo p. So, the fact that k and can be arbitrary gives us p2 choices ( p ( 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) → m + n for which we are trying to find the cardinalit cardinality y of the kernel. kernel. But, this map is obviously obviously surjectiv surjectivee and so the kernel has size p2 = p. Thus, Thus, we see that that there there are p choices for (m, (m, n). Thus, Thus, overall overall there are p2 · p = p3 choices. choices. Thus, Thus, the p kernel of the map SL 2 (Z/pe+1 Z) → SL2 (Z/pe Z) has cardinality p3 and so the first isomorphism theorem implies that
+1)−2 | SL2 (Z/pe+1 Z)| = p3 | SL2 (Z/pe Z)| = p3 p3e − p3e−2 = p3(e+1) − p3(e+1)−
and so the induction is complete.
ii) Let’s first recall recall that the formation formation of SL 2 (R) for a ring R is functorial, so that if R ∼ = S as rings then SL 2 (R) ∼ = SL2 (S ). ). Next, note that
(a1 , a2 ) (b1 , b2 ) (c1 , c2 ) (d1 , d2 )
a1 c1
→
b1 a , 2 d1 c2
b2 d2
is a group isomorphism SL2 (R × S ) → SL2 (R) × SL2 (S ). ). Combining Combining these results results and the Chinese Remainder Remainder e1 en Theorem we know that if N = p1 · · · pn then n
SL2 (Z/N Z) ∼ =
SL2 (Z/pei i Z)
i=1
Taking orders of both sides gives
n
| SL2 (Z/N Z)| =
3ei
pi
i=1
as desired.
1 1− 2 pi
=
e1
( p1 · · · penn )3
1−
p|N
1 p2
= N 3
iii) To see that this is a group group map we merely merely note that that if f is the described map and then
1−
p|N
1 p2
a b a , c d c
b d
N ) ∈ Γ1 (N )
f
a c
a b c d
b d
= ab + bc mod N ≡ b + b mod N
since c ≡ c ≡ 1 mod N . N . But, it’s obvious that
b + b mod N = f
a b c d
+ f
a c
b d
and so f really is a homomorp homomorphism. hism. It’s surjectiv surjectivee for if b ∈ { 0, · · · , N − 1} then is precisely precisely b. Finally, the kernel of the map is
1 b 0 1
is in Γ(n Γ(n) and the image
a b N ) : b ≡ 0 mod N ∈ Γ1 (N ) c d
But, by definition, since
a b c d
∈ Γ1 (N ) N ) we know that a ≡ d ≡ 1 mod N and c ≡ 0 mod N . N . Thus Thus,, with with
b ≡ 0 mod N we have that
a b c d
and so
1 0 0 1
≡
mod N
a b Γ(N )) and the reverse inclusion is just as trivial. ∈ Γ(N c d
iv) To see that the map Γ0 (N ) N ) → (Z/N Z)× , call this map g , is a group map we merely note that if Γ0 (N ) N ) then g
a b a , c d c
b d
∈
a c
a b c d
since c ≡ 0 mod N by the definition that
b d
= cb + dd mod N = dd
a b N ). But, clearly dd is precisely ∈ Γ0 (N ). c d
g
a b c d
a c
g
b d
and so it follows that g really is a group map. This map surjects surjects since since if d ∈ (Z/N Z)× then (d, (d, N ) = 1 and so a b ∈ Γ0 (N ) and g (γ ) = d as desired. there exists ba such that ad + bN = 1. We see then that γ = N d v) We have the following chain chain of subgroups subgroups Γ(N Γ( N )) ⊆ Γ1 (N ) N ) ⊆ Γ0 (N ) N ) ⊆ SL2 (Z). Now, from Lagrange’s Lagrange’s theorem theorem we know that [SL2 (Z) : Γ0 (N )] N )] =
[SL2 (Z) : Γ(N Γ(N )] )] [Γ0 (N ) N ) : Γ1 (N )][Γ N )][Γ1 (N ) N ) : Γ(N Γ(N )] )]
Now, from iii) we know that [Γ1 (N ) N ) : Γ(N Γ(N )] )] = N , N , and from iv) we know that if N = pe11 · · · penn = ϕ(N ) N ) = N
ϕ(N ) N ) N N
pei −1 ( p − 1)
= N i=1
n
pei i
i=1
n
= N
1−
i=1
1 pi
Thus, [ SL2 (Z) : Γ0 (N )] N )] =
[SL2 (Z) : Γ(N Γ(N )] )] [Γ0 (N ) N ) : Γ1 (N )][Γ N )][Γ1 (N ) N ) : Γ(N Γ(N )] )]
n
N
i=1 n
=
N 2
1−
i=1 n
= N
1 pi
1+
1 pi
1+
1 p
i=1
= N
1 1− 2 pi
3
p|N
as desired.
Exercise 1.2.4.
Exercise 1.2.5.
If Γ
SL2 (Z) is a congruence subgroup and γ ∈ SL2 (Z), then γ Γγ −1 is a congruence subgroup of
SL2 (Z). We merely note that since Γ is a congruence subgroup there exists some N ∈ Z such that Γ(N Γ(N )) Γ. −1 −1 − 1 − 1 Then, γ Γ(N Γ(N ))γ γ Γγ , but Γ(N Γ(N )) SL2 (Z) and so γ Γ(N Γ(N ))γ = Γ(N Γ(N ). ). Thus, Thus, Γ(N Γ(N )) γ Γγ and so γ Γγ −1 is a congruence subgroup as desired.
Solution
This exercise proves Proposition 1.2.4. Let Γ be a congruence subgroup of SL2 (Z), thus containing Γ(N Γ(N )) for some N , N , and suppose that the function f : H → C is holomorphic and weight-k weight- k invariant under Γ. Suppose Suppose
Exercise 1.2.6.
∞
also that in the Fourier expansion f (z ) = constants C constants C and r.
n an q N the coefficients for n > 0 satisfy |an |
Cn r for some positive
n=0
i) Show that for any z = x + iy ∈ H , ∞
|f ( f (z )| |a0 | + C
nr e
2πny
−
N
(1)
n=1 2πty
−
Changing n to a nonnegative real variable t, show that the continuous version g (t) = tr e N of the summand rN rN increases monotonically on the interval 0, and then decreases monotonically on , ∞ . Usin Usingg this, this, 2πy 2πy represent all but two terms of the sum in (1) as unit-wide boxes under the graph of g and consider the missing terms individually to establish the estimate (where in this exercise C 0 and C can denote different constants in different places)
∞
|f ( f (z )| C 0 + C
0
C
1 g (t) dt + r y
∞
tr e−t dt. dt. This last integral is a gamma function yr+1 0 integral, integral, to be introduced formally in Section 4.4, but at any rate it converges at both ends and is independent of y . In sum, After a change of variables the integral takes the form
f (z )| C 0 + |f (
C yr
as y → ∞ . ii) For every α ∈ SL2 (Z), the transformed function (f [ f [α]k )(z )(z ) is holomorphic and weight-k weight- k invariant under α−1 Γα and therefore has a Laurent expansion (f [ f [α]k )(z )(z ) =
n an q N
n∈Z
To show that the Laurent series truncates from the left to a power series it suffices to show that lim (f [ f [α]k )(z )(z )q N N = 0
qN →0
If α fixes ∞ then this is immediate from the Fourier series of f itself. itself. Otherwise Otherwise show that the transforme transformed d −k function (f [α]k )(z )(z ) = (cz + d) f ( f (α(z )) satisfies r −k lim |(f [ f [α]k ])(z ])(z )q N |q N N | C lim (y N |)
qN →0
Recalling that q N N = e to complete the proof.
2πi(x+y) N
qN →0
, show that y = −C log( C log(|q N N |), and use the fact that polynomials dominate logarithms
i) The equation ( 1) follows immediately by applying the triangle inequality, the fact that |an | Cn r , and πny that exp exp 2πinz = exp Re 2πinz = exp −2N . N N
Solution
r
2πty
2πty
πyt To establish the claim about g(t) we differentiate to find that g (t) = −2N e N + rtr−1 e N , or g (t) = 2πty πyt Nr t. e N tr−1 −2N + r . Since the multiplier is always positive, this is negative precisely when r 2πyt N or 2πy Nr Nr Thus, we see that g is decreasing precisely when t ∈ , ∞ , and thus g must be increasing on 0, as 2πy 2πy desired. −
∞
∞
g (n) =
n=1
−
Nr we may use Riemann sums to estimate that 2πy
Using the fact that the sign change occurs at ∞
−
g (n)
0
n=0
N r 1 g (t) dt + r (2πye (2πye)) y r
(2)
(the first equality being true because g (0) = 0). In particular, particular, choosing choosing the partition partition {[n − 1, n] : n ∈ N} and using Nr Riemann left hand Riemann sums for intervals lying in 0, 2πy , using right hand Riemann sums on intervals lying
Nr Nr Nr inside , ∞ , and bounding the sum on the partition containing by g 2πy 2πy 2πy the maximum g takes, and the interval is measure 1) we get ( 2). Thus, we see that ∞
g(n) C
∞
0
n=1
1 g (t) + r y
=
N r 1 (since this is r (2πye (2πye)) y r
N r where C = max 1, . Thus, we see that (2πye (2πye))r
∞
f (z )| |a0 | + C |f (
n=1
g (n) |a0 | + CC
∞
0
1 g (t) dt + r y
Thus, taking C 0 = |a0 | and C = CC (where, as was stated, the C means different things on the right and left hand sides) we get our desired inequality.
ii) We have the estimation estimation f [α]k (z )q N j (α, z)| |f [ N | = | j(
−k
f (α(z ))q ))q N j (α, z )| |f ( N | j(
−k
C C 0 + r | j( j (α, z )2r | q N N y
Im(z Im(z ) . No Now, w, lettin lettingg q N N → 0 is j (α, z)|2 | j( equivalent to letting y → ∞ . Now, from this it’s easy to see that | j( j (α, z)| is asymptotically C y for some constant C (really it’s the c entry of α). Thus, we see that where we used the previous part and made use of the fact that Im( α(z )) =
lim |f [ f [α]k (z )q N j (α, z )| N | | j(
qN →0
−k
C C 0 + r | j( j (α, z)|2r q N N y
C | j( j (α, z)|2r−k q N N qN →0 y r = lim Dy r−k q N N = lim
qN →0
where where the first equalit equality y is via the (obvio (obvious) us) fact fact that lim | j( j (α, z)|−k C 0 q N second fact fact follo follows ws N = 0, and the second qN →0
from the asymptotic for | j( j (α, z )| with D = CC . Now, by definition definition we have have that q N N = exp
|q N N | = e
2πiy
−
N
and so |y | = A log |q N N | where A is a constant. Thus, r −k r −k | lim Dy r−k q N |q N |q N N | = lim |y | N | = A lim log |q N N | N | = 0 qN →0
r −k because |q N . N dominates log |q N N |
qN →0
qN →0
2πi (x+iy ) N
and so
