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4 the right side of (5.17) is equal to _ 44/7)1 - 2sn s-2
1- ( 2
— s)(1 + Or' — 2 (2 + 20nL(E„, 2 — s) n (5.20)
-___1(2 ( -172) ns -21— s)(80 1-si,(E„, 2 — s). On the other hand, if we bring the term (1/2)s over to the right in the functional equation (5.11)-(5.12) in the theorem, we find that what we want
s 87
§5. The Hasse—Weil L-function and its functional equation
to prove is: n - sr(s)L(E„, s) = (
—2 ) (N/2)s(27) 5-2 ( \/N) 2-sr(2 n
(--n2)
(N/4 )' - sns -2 1-(2
s)L(En , 2 — s)
s)L(En , 2 — s).
And this is precisely (5.20). Thus, to finish the proof of the theorem for odd n, it remains to prove the lemma. PROOF OF LEMMA. First suppose that in + m2 i is not divisible by I + i. This is equivalent to saying that m i and m 2 have opposite parity, i.e., their sum is odd. Now as a, b range from 0 to 4n, the Gaussian integer a + bi runs through each residue class modulo (2 + 2i)n exactly twice. Each time gives the same value of x,i'(a + bi), since Li' (a + bi) depends only on what a + bi is modulo n' = (2 + 2i)n. But meanwhile, the exponential terms in the two summands have opposite sign, causing the two summands to cancel. To see this, we observe that if a l + b 1 i and a2 + b2 i are the two Gaussian integers in different residue classes modulo 4n but the same residue class modulo (2 + 2i)n, thei, a l + b 1 (a2 + b2 (2 + 2i)n (mod 4n), and so (2nil4n)m.ya1,b1)-(a2,b2)) = e(27cil4rOm.(2n,2n) = e rri(nt 1 +m 2 ) = _1 . e This proves the first part of the lemma. Now suppose that m 1 + m 2 i = (1 + i)x. Note that in (a, b) = m l a + m2 b = Re((m, tn 2 i)(a + bi)) = Re((1 i).5(a + hi)). Hence, the exponential term in the summand in Sm is tli(ic(a + bi)), where •
i/f(x) def --=-- e211'" (with n' = (2 + 2i)n). Since x„' is a primitive character modulo (2 + 2i)n (see Proposition 3), we can apply Problem 9(a)—(b) of §11.2. Since the summation in (5.19) goes through each residue class modulo (2 + 2i)n twice, we have
S„, 2
E
x (a + bi)111(.t(a + hi))
a+biEgil/( 2 +2 0,8
2 k/r1(:)g(Z;1) = 2 Xax)9(4). This proves the lemma, and hence the theorem (except for some slight modifications in the case of even n, which will be left as an exercise).
In the problems we shall outline a proof of the analogous theorem in the case of an elliptic curve, namely y2 = x 3 + 16, which has complex -multiplication by another quadratic imaginary integer ring, namely Z[w], where co + -PO. There is one additional feature which is needed because we end up summing not over [i], which can, be thought of as 1 2 , but rather over a lattice which is the image of 1 2 under a certain 2 x 2-matrix.
II. The Hasse-Weil L-Function of an Elliptic Curve
88
So we have to apply Poisson summation to a function much like the function in this section, but involving this matrix. We conclude this section by mentioning two references for a more general treatment of the theory of which we have only treated a few special cases. First, in C. L. Siegel's Tata notes [Siegel 1961] (see especially pp. 60-72) one finds L-functions whose summand has the form 1 e
27r im- u
P(m + y) (Q[ni + Y]) s+g12 '
where nier, u, ye Er, Q is the matrix of a positive definite quadratic form, and P is a "spherical polynomial with respect to Q of degree g". The case we needed for L(Eii, s) was: n = 2, Q = ( ?), P(x 1 , x 2) = x l + ix 2 , g = 1. In Problem 8 below we have the case Q = ( 1,2 if), P(x 11 x2) =4 (a) + 1 /2)x1 ± (w12 + 1) x2 (where w = —1/2 ± i012), g = 1. In [Lang 1970, Chapters XIII and XIV], two approaches are given to this topic. In Ch. XIII, the approach we have used (originally due to Hecke) is applied to obtain the functional equation for the Dedekind zeta-function of an arbitrary number field. This is a generalization. of Problems 2 and 6 below. However, the case of more general Hecke L-series is not included in that chapter. A quite different approach due to J. Tate—using Fourier analysis on p-adic fields—is given in Ch. XIV of Lang's book. PROBLEMS 1. Finish the proof of the theorem for n even. 2. (a) Find a functional equation for 0(1) = E„„ z2e - orn 12, t > O. (b) The Dedekind zeta-function of a nucigiber field K is defined as follows: CK(S) SZ E(Ni)_s,
where the sum is over all nonzero ideals I of the ring of integers of K. This sum converges for Re s> I (see [Borevich and Shafarevich 1966, Ch. 5, §1]). Let K = 04 Prove that CI< (s) is an entire function except for a simple pole at s = 1 with residue n/4, and find a functional equation relating CK (s) to CK (1 — s). 3. For u and y in R 2, let
0:(t) . y def
e2itim-ve-ntim+u12
I-a
1
t > 0.
mEZ 2
Find a functional equation relating 0:(t) to 0!..,(f). 4. (a) In the situation of Proposition 11, express the Fourier transform of (w in terms off(y) for any nonnegative integer k. (b) Suppose that k is a nonnegative integer, ue R 2 is fixed with u it V,, t > 0 is fixed, and w = (1, 0 E C 2. What is the Fourier transform of g(x) = ((x + u) • oke -ntix+u1 2 ? (c) With k, u, t, w as in part (b), define:
-
89
§5. The Hasse-Weil L-function-and its functional equation
(On
Ou, k(t) =
+
. ok e -ntIm+ui
2 5
m e 22
o u,k (o=
E (m . oke2nim•ue -ntimi2
•
m € 22
r Find a functional equation relating 0 k (t) to 0"(+). (d) Suppose that / is a fixed ideal of Z[i], and x: (Z[i]//)*
C* is a nontrivial
character. Outline (without computing the details) how one would prove that the function Exe zro xkx(x)(Nx)_s extends to an entire function of s (if x were trivial, there would be a simple pole at s = 1), and satisfies a functional equation relating its value at s to its value at k + 1 s. (e) 'Explain why any Hecke L-series for Z[i] is essentially of the form in part (d).
5. Let f
C, fe
(a) For MeGL.(R), let Mt denote the transpose matrix, and let M* = (M - V. Let g(x) = f(Mx). Express 4(y) in terms of f(y). (b) Let L be a lattice in fi??; equivalently, L is of the form L = MZ", where M E GL(), i.e., L is obtained by applying some matrix M to the elements in the standard basis lattice Z. Let L' be the "dual lattice" defined by: L' = {ye Rnix • ye Z for all xe L}. First show that L' is a lattice. Then prove a functional equation relating ExeL f(x) to EyE d(y).
f.
6. Let w= - 1+ (a) Let M ((., 3 2 ). Show that the lattice L = Zkoj C = R2 is L = MZ 2 . What are,M* and L'? Show that L', considered as a lattice in C, is twice the "different" of Z[co], which is defined as {xee(co)ITr(xy) E Z for ally E Z [C0] 1. Here Tr x =
/x+.t- =2Rex. (1)) Prove that E xelfroj
e'"2
t- e -(40t)ixt2 2r_ 2s IN/ 3 xEzz(0,1
(c) Let K = Q(co). Prove that CK(s) extends meromorphically onto the complex sPlane with its only pole a simple pole at s = 1. Find the residue at the pole, and derive a functional equation for C K(S). (d) With K as in part (c), prove the identity: CK(s) = C(s)L(, s), where x is the nontrivial character of (Z0Z)*. (e) Use part (d), the theorem in §11.4, and Problem 5(e) of §11.4 to give another proof of the functional equation in part (c).
7. Let E be the elliptic curve y2 = x3 + 16. Let w = 4- 10. (a) Show that the reduction mod p of y2 = x3 + 16 is an elliptic curve over Fp if and only if p # 2, 3. For such p, recall that the Euler fador at p is (1 — 2a p p + p 2s)— where (1 — 2ap T + pr) is the numerator of Z(EIEF;; T). Show that for p # 2, 3 this factor is cxdpeg P N p- -1 )
where the product -is over all (one or two) prime ideals P of Z[co] dividing (p) and ot`iteg P is the unique generator of P such that acileg P I (mod 3). (I)) When p = 3, define the Euler factor at 3 to be simply 1, as we did for En : y2 = x3 — 112 .x when pl2n. When p = 2, at first glance it looks like we again have
90
IL The Hasse—Weil L-Function of an Elliptic Curve
y2 =--- x3 , and so we might be tempted to take 1 for the Euler factor at 2 as well. However, this is wrong. When one defines L(E, s), if there exists a 0-linear change of variables that takes our equation E: y2 = x3 + 16 in FD4 to a curve C whose reduction mod p is smooth, then we are obliged to say that E has good reduction at p and to form the corresponding Euler factor from the zeta-function of C over Fir In Problem 22 of §II.2, we saw that y2 = x3 + 16 is equivalent over Q to y 2 + y = x 3, a smooth curve over F2 whose zeta-function we computed. Show that the Euler factor at p = 2 is given by the same formula as in part (a). (c) For x E 7L[(.1)] prime to 3, let x(x) = — ce)i be the unique sixth root of I such that x(x) 1 (mod 3). Let x(x) = 0 for x in the ideal (,./ — 3). Show that L(Es)_ !
xEli
i (Nx)
(d) Let 11/(X) =
e(27c43)Tr(x/W3)
def
where Tr x = x + ï = 2 Re x. Verify that tif(x) is an additive character of Z[w]13 satisfying the condition in Problem 9 of §II.2 (i.e., that it is nontrivial on = IxEztco)/3 X(X)11/(X), where any larger ideal than (3)). Find the value of g(x, x is as in part (b).
8. (a) Let w = (1, co), u e — Z 2 , and t > 0 be fixed. What is the Fourier transform of g(x) = (x + u) • w e -ntl(x+to.wI 2 ? (b) Let 6,i(t) = E2 On), where g(x) is as in part (a). Find the Mellin transform 4)(s) of E x(a + bco)0.(i). tr-. a13,b/3) 0 0}. It is important to note that any g e SL2 (ER) preserves H, i.e., Im z > 0 implies lm gz> O. This is because
Im gz = Im
(az + 0(c-if + d) = az + b icz + df -2 Im(adz + bcZ). = Im Icz + dI2 . cz + d
But Im(adz + bd) = (ad — bc)Im z = Im z, since det g = 1. Thus,
lm gz = icz + d1 -2 Im z
a b for g= c d e SL2(R).
/( 1.2)
Thus, the group SL2 (R) acts on the set H by the transformations (1.1). The subgroup of SL2 (I1) consisting of matrices with integer entries is, by definition, SL2 (Z). It is called the "full modular group", and is sometimes denoted F. We shall denote f d=ef r/+./.. (Whenever we have a subgroup G . denote G1+1 if G contains — ./; if —I0G, we set of SL 2 (R), we shall let 6 6= G.) This group f = s12 (7L)/-Fi acts faithfully on H. It is one of the basic groups arising in number theory and other branches of mathematics. Besides F = SL 2 (7L), certain of its subgroups have special significance. Let N be a positive integer. We define
f(N)7 {(ac b) d ESL 2 (Z)la -=. d -74. 1 (mod N), b :÷-_. c E-2 0 (mod MI . (
1.3
)
This is a subgroup of r = SL 2 (Z), actually a normal subgroup, because it is the kernel of the group homomorphism from F to SL 2 (7L/NZ) obtained by reducing entries modulo N. In other words, r(N) consists of 2 x 2 integer matrices of determinant 1 which are congruent modulo N to the identity matrix. Note that F(1) = F. We shall later see that F(N) is analogous to the subsemigroup 1 + NZ c: Z consisting of integers congruent to 1 modulo N. 1"(N) is called the "principal congruence subgroup of level N". - Notice that F (2) = F(2)/ + I, whereas for N > '2 we have f(N) = because —1 # 1 (mod N), and so — I is not in r(N). A subgroup of F (or of r) is called a "congruence subgroup of level N" if it -contains F(N) (or F(N), if we are considering matrices modulo + I). Notice that a congruence subgroup of level N also has level N' for any multiple N' of N. This is.because F(N) = r(N/). We say that a subgroup of r (or of f) is a "congruence subgroup" if there exists N such that it is a congruence subgroup of level N. Not all subgroups of F are congruence
III. Modular Forms
100
subgroups, but we shall never have occasion to deal with non-congruence subgroups. For our purposes the most important congruence subgroups of F are:
)er ic 0 (mod N)};
170 (N) Fi (N)
c d
a {(c ci)e Fo (N)ia —= 1 (mod MI.
(1.4) (1.5)
bd )e It is easy to check that these are, in fact, subgroups. Note that if F1. (N), then since ad — be = 1 and Nic, it follows that ad 1 (mod N), and hence d 1 (mod N). We sometimes abbreviate the definitions (1.3)—(1.5) as follows:
T(N) =()) mod NI; 01
ro(N) =
0*
1* ) mod N}; I', (N) = {( 01
) mod N},
where * indicates the absence of any congruence condition modulo N. Whenever a group acts on a set, it divides the set into equivalence classes, where two points are said to be in the same equivalence class if there is an element of the group which takes one to the other. In particular, if G is a subgroup of F, we say that two points z 1 , z2 e H are "G-equivalent" if there exists g e G such that z2 = gz 1 . Let F be a closed region in H. (Usually, F will also be simply connected.) We say that F is a "fundamental domain" for the subgroup G of F if every z eH is G-equivalent to a point in F, but no two distinct points z 1 , z2 in the interior of F are G-equivalent (two boundary points are permitted to be
G-equivalent). The most famous example of a fundamental domain is shown in Fig.
:
F def = {z e —1 :5_ Re z
and Izi 1 } .
(1.6)
Proposition 1. The region F defined in (1.6) is a fundamental domain for F.
PROOF. The group F = SL2 (1) contains two fractional linear transformations which act as building blocks for the entire group:
T
1 1) : G1 0 —1
S
i 0
kf(
z + 1; (1.7)
: z —11z.
§1. SL 2 (Z) and its congruence subgroups
101
1 2
Figure III.!
To prove that every zeH is F-eqiiivalent to a point in F, the idea is to lf it use translations Ti to move a point z inside the strip —1 < Re z lands outside the unit circle, it is in F. Otherwise use S to throw the point to bring it inside the strip, outside the unit circle, then use a translation and continue in this way until you get a point inside the strip and outside the unit circle. We now give a precise proof. Let zeH be fixed. Let r be the subgroup of F generated by S and T (we shall soon see that actually F = + F'). If y = ( db ) e F', then Im yz = Im ziicz + dI 2 by (1.2). Since c and d are integers, the numbers Ia . + di are bounded away from zero. (Geometrically, as c and d vary through all integers, the complex numbers cz + d run through the lattice generated by 1 and z; and there is a disc around 0 which contains no nonzero lattice point.) Thus, there is some y = (c' bd) E r such that Im yz is maximal. Replacing y by Ty for some suitable j, without loss of generality we may suppose that yz is in the strip —I< Re yz < But then, if yz were not in F, i.e., if we had lyzi Im yz, which contradicts our choice of y e F' so that Im yz is maximal. Thus, there exists y e such that yz E F. We now prove that no two points in the interior of F are F-equivalent. We shall actually prove a more precise result. Suppose that z 1 , z 2 e F are F-equivalent. Here we are not supposing that z 1 and z2 are necessarily distinct or that they are necessarily in the interior of F. Without loss of generality, suppose that Im z2 Im z1 . Let y = er, e r be sticil that z2 = 1. Since z 1 is in yz i . Since Im z2 Im z 1 , by (1.2) we must have icz i + F and d is real (in fact, an integer), it is easy to see fromzFig. III.1 that this 2. This leaves the cases :i) c 7-7-- 0, d = +1; inequality cannot hold if (ii) c = +1, d = 0, and z 1 is on the unit circle; (iii) c d = .+1 and z 1 = + In casel(i), either y or —y ; (iV)C = —d= +1 and z, +
102
in. Modular Forms
is a translation V; but such a y can take a point in F to another point in F only if it is the identity or if j = + I and the points are on the two vertical boundary lines Re z = +1. In case (ii), it is easy to see that y = + TS with a = 0 and 2 1 and z2 on the unit circle (and symmetrically located with respect . In case to the imaginary axis) or with a = ± 1 and 2 1 = z 2 = ± + 1), and if such a map takes z 1 €F to 22 eF (Hi), y can be written as + we must have a = 0 and z2 = 2 1 = — + `ffs or else a = 1 and z2 = 2 1 + 1 = + Ç3. Case (iv) is handled in the same way as case (Hi). We conclude that in no case can 2 1 or z2 belong to the interior of F, unless +y is the identity and 22 = 2 1 . This proves the proposition. In the course of the proof of Proposition 1, we have established two other facts. Proposition 2. Two distinct points 2 1 , z 2 on the boundary of Fare F-equivalent only if Re 2 1 = ±4 and 22 = 21 + 1, or if 2 1 is on the unit circle and 22 =
In the next proposition, we use the notation Gz. for the "isotropy subgroup" of an element z in a set on which the group G acts: by definition, Gz =
{g eGigz = z}. Proposition 3. If z e F, then
rz +Iexcept
in the following three cases .•
(i) r= ± {/, S } if z = ; (ii) F = ± {I, ST, (ST) 2 } if z = co = + (Hi) F= {I, TS, (TS) 2 } if z = —63 = + Both Propositions 2 and 3 follow from the second part of the proof of Proposition 1. (ST) 3 = —I, and (TV = — I. Thus, in F = Notice that S 2 = SL 2 (7L)/+ I the elements S, ST, TS generate cyclic subgroups of order 2, 3, 3, respectively; and these subgroups are the isotropy subgroups of the elements i, (1), — el-3, respectively. These points in H with nontrivial isotropy subgroups are called "elliptic points". Another by-product of the proof of Proposition 1 is the following useful fact. Proposition 4. The group = SL,(Z)/+I is generated by the two elements S and T (see (1.7)). In other words, any fractional linear transformation can be written as a "word" in S (the negative-reciprocal map) and T (translation by 1) and their inverses. PROOF. As in the proof of Proposition 1, let F' denote the subgroup of f generated by S and T. Let z be any point in the interior of F (e.g. z = 21). Let g be an element of F. Consider the point gz EH. In the first part of thé
103
§1. SL 2 (Z) and its congruence subgroups
proof of Proposition 1, we showed that there exists y e I"' such that y(gz) e F. But since z is in the interior of F, it follows by Propositions 1 and 3 that yg = + I, i.e., g = e r. This shows that any g e r is actually (up to a sign) in r. The proposition is proved. Thus, any element in Î can be written in the form Sal T b, s a, T b2 Sairi, where all of the a bi are nonzero integers, except that we allow a l and/or b1 to be zero; and, since S 2 = —I, we may suppose that all of the ai equal 1, except that a l = 0 or 1. We may also use the identity (ST) 3 —1 to achieve a further simplification in some cases. We now return to the fundamental domain F for F = SL 2 (Z). Recall that in Chapter I §4 we also had a fundamental domain, in that case a parallelogram H c c for the lattice L. In that case the group was L, the action of g EL on a point z E C was simply g(z) = g + z. Every z e C is L-equivalent to a point in H, and no two points in the interior of fl are L-equivalent. In that situation we found it useful to glue together the boundary of fl by identifying L-equivalent points. We obtained a torus, and then we found that the map zi--+ (0(4 p'(z)) gives an analytic isomorphism from the torus CIL to the elliptic curve y2 = 4.r 3 g2 (L)x g 3 (L) (see §1.6). In our present situation, with the group F acting on the set H with fundamental domain F, it is also useful to identify r-equivalent points on the boundary of F. Visually, we fold F around the imaginary axis, gluing the 23 and + iy and the point eti° to e 2 i(112-0) for y > ..rpoint + iy to < O < I-. The resulting set F with its sides glued is in one-to-one correspondence with the set of f-equivalence classes in H, which we denote FVf. (The notation F\l/ rather than H/F is customarily used because the group f acts on the set H on the left.) In Chapter I we saw how useful it is to "complete" the picture by including a point or points at infinity. The same is true when we work with rvi. Let ii denote the set H {co} u G. That is, we add to H a point at infinity (which should be visualized far up the positive imaginary axis ; for this reason we sometimes denote it ico) and also all of the rational numbers on the real axis. These points {co} u Q are called "cusps". It is easy to see that r permutes the cusps transitively. Namely, any fraction a/c in lowest terms can be completed to a matrix (ca e r by solving ad — bc = 1 for d and b; 'this matrix takes co to a/c. Hence all rational numbers are in the same F-equivalence class as co. If r/ is a subgroup of r, then F' permutes the cusps, but in general not transitively. That is, there is usually more than one r' equivalence class among the cusps {co} u O. We shall see examples later. By a "cusp of F'" we mean a F'-equivalence class of cusps. We may choose any convenient representative of the equivalence class to denote the cusp. Thus, we say that "f has a single cusp at oo", where oo can be replaced by any rational number ,) a/c in this statement. We extend the usual topology on H to the set 11 as follows. First, a funda,
-
-
III. Modular Forms
104 00
iC
a Nc
0 a lc
Figure 111.2
mental system of open neighborhoods of co is Nc = {zEHIIM Z >C} L.) {CO} for any C > 0. Note that if we map H to the punctured open unit disc by sending z ,-.4 q d= ef e27tiz
,
(1.8)
and if we agree to take the point co E,171 to the origin under this map, then Nc is the inverse image of the open disc of radius e'Rc centered afthe origin, and we have defined our topology on I/ v {co } so as to Make this map (1.8) continuous. The change of variables (1.8) from z to q plays a basic role in the theory of modular functions. We use (1.8) to define an analytic structure on Hu {po } . In other words, given a function on H of period 1, we say that it is meromorphic at oo if it can be expressed as a power series in the variable q having at most finitely many negative terms, i.e. it has a Fourier expânsion of the form
f(z) =
E an e 2n1nz = E anqn, neZ
(1.9)
neZ
in which an = 0 for n 14 sk(r) =
118
III. Modular Forms
AMk-12(r
) (i.e., the cusp forms of weight k are obtained by multiplying
modular forms of weight k 12`by the function A(z)). (e) Mk (f) = Sk (F) CEk for k > 2. PROOF. Note that for a modular form all terms on the left in (2.21) are nonnegative.
(a) Letfe Mo (F), and let c be any value taken by f(z). Then f(z) c eMo (f) has a zero, i.e., one of the terms on the left in (2.21) is strictly positive. Since the right side is 0, this can only happen if f(z) — c is the zero function. (b) If k < 0 or k 2, there is no way that the sum of nonnegative terms on the left in (2.21) could equal k/12. (c) When k = 4, 6, 8, 10, or 14 we note that there is only one possible way of choosing the v(f) so that (2.21) holds: for k = 4, we must have v,,,(f ) = 1, all other vp(f)= 0; for k = 6, we must have vi(f)= I, all other vp(f)= 0; for k = 8, we must have v.(f) = 2, all other vp(f)= 0; for k = 10, we must have v0(f ) = vi(f)= 1, all other vp(f)= 0; for k = 14, we must have v »(f ) = 2, vi (f ) = 1, all other v(f) = O. Let f1 (z), f2 (z) be nonzero elements of Mk (z). Sincefi (z) and f2 (z) have the same zeros, the weight zero modular function f1 (z)/f2 (z) is actually a modular form. By part (a), f1 and f2 are proportional. Choosing f2 (z) E k (z) gives part (c). (d) For fe Si (r) we have v,„(f ) > 0, and all other terms on the left in (2.21) are nonnegative. Notice that when k 12 and f = A, (2.21) implies that the only zero of A(z) is at infinity. Hence, for any k and any Je Sk (F), the modular functionfl A is actually a modular form, i.e.,f/A E Mk _ 2(F). This gives us all of the assertions in part (d). (e) Since Ek does not vanish at infinity, given fe Mk (f) we can always subtract a suitable multiple of Ek so that the resulting f cEk eMk (f) vanishes at infinity, i.e., f cEk e sk (r). We now prove that any modular form for I" is a polynomial in E4, E6 (see Problem 4 of §I.6 for a different proof of this fact). Proposition 10. Any feAl k (r) can be written in the form
f(z) =
ci,jE4 (2) iE6' (Z) i .
(2.27)
4i+ 6j=k
PROOF. We use induction on k. For k = 4, 6, 8, 10, 14 we note that E4, E6, E4 E6 , E4 E6, respectively, is an element of Mk (r), and so, by Proposition 9(c), must span Mk (f). Now suppose that k = 12 or k> 14. It is clearly possible to find i and j such that 41 4- 6/ = k, in which case E‘it Ei Emk(r). Given fe Mk(r), by the same argument as in the proof of Proposition 9(e)
119
§2. Modular forms. for SL 2 (1)
we can find ceC such that f we can write f in the form
—
cE14ie Sk (F). By part (d) of Proposition 9, -
. f= cElEi + AA . = cELE-4 +
(270 1 2 1728
(E43 — E 62 )fl , s
where fi e Mk-12( 1") , Applying (2.27) to fi by the induction assumption a (with k replaced byjc — 12), we obtain the desired polynomial forf.
The kinvariant.We now define a very important modular function of weight zero:
E4 (Z) 3 2 1728 1728g2 (z) 3 Jkz litf A (z) — E4(z) 3 E6 (Z) •,
(by (2.14)-(2.15))
(2.28)
-
Proposition 11. The function j gives a bijection from FV-1- (the fundamental
dom qin with r-eguivakni sides identified and the point at infinity included) and the Riemann sphere N = C u lool. In the proof of Proposition 9(d) we saw that A(z) has a simple zero at infinity and no other zero. Since g 2 does not vanish at infinity, this means that j(z) has a simple pole at infinity and is holomorphic on H. For any C E C the modular form 1728g3 — cA eM 12 (F) must vanish at exactly one point P El"\H, because when k = 12 exactly one of the terms on the left in (2.21) is strictly positive. Dividing by A, we see that this means that j(z) — c ---z-- 0 for exactly one value of ze F\l/. Thus, j takes co to oo and on o F\li is a bijection with C. PROOF.
Proposition 12. The modular functions of weight zero for F are precisely the
rational functions off. PROOF. A rational function of j(z) is a modular function of weight zero (see Remark 3 at the beginning of this section). Conversely, suppose that f(z) is a modular function of weight zero for F. If zi are the poles of f(z) in rvy, counted with multiplicity, then f(z)•11 i (j(z) j(zi)) is a modular function of weight 0 with no poles in H, and it suffices to show that such a function is a rational function of f. So, without loss of generality we may assume thatf(z) has no poles in H. We can next multiply by a suitable power of A to cancel the pole off(z) at oo. Thus, for some k we will have A(z)Y(z) e M/2k (F). By Proposition 10, we can write f(z) as a linear combination of modular functions of the form EE6ilY (where 4i + 6j = 12k), so it suffices to show that such a modular function is a rational expression in j. Since 4i + 6j is divisible by 12, we must have i = 3i0 divisible by 3 and j :----- 2j0 divisible by 2. But it is easy to check that EPA and EPA are each of the form aj + b, by (2.28); and Ero.Eljo/A k is a product of such factors. This o proves the proposition. —
120
III. Modular Forms
The definition (2.28) is not the only way one could have defined the invariant. It is not hard to see that any ratio of two non-proportional modular forms of weight 12 would have satisfied. Propositions 11 and 12. But j(z) has the additional convenient properties that: its pole is at infinity, i.e., it is holomorphic on H; and its residue at the pole is 1, as we easily compute from (2.28) that the q-expansion off starts out !-1 + • • . Recall that in Chapter I we used the Weierstrass p-function and its derivative to identify the analytic manifold C/L with an elliptic curve in Pt. Similarly, in our present context we have used the j-invariant to identify 1"\ii as an analytic manifold with the Riemann sphere PL Proposition 12 then amounts to saying that the only meromorphic functions on the Riemann sphere are the rational functions. Thus, Proposition 12 is analogous to Proposition 8 in §1.5, which characterized the field of elliptic functions as the rational functions of x = p(z), y =
A loose end from Chapter I. In Chapter I we defined an elliptic curve over C to be a curve given by an equation of the form y2 = f(x), where f(x) is a cubic with distinct roots. We then worked with curves whose equations were written in the form
= 4x 3 — g2 (L)x — g 3 (L)
(2.29)
for some lattice L. It is not hard to see that a linear change of variables will bring an equation y2 = f(x), where f(x) has distinct roots, into the form
y 2 = 4x3 Ax B
with
A 3 — 27B2 0 0.
(2.30)
But in order to write such a curve in the form (2.29) we need to show that a lattice L can always be found such that g2 (L)= A, g 3 (L) z--- B. This was not proved in Chapter I, but it is easy to prove now using modular forms. In what follows we shall write a lattice L= {mo t + nco2 }, where z = c1) 1 1 o2 e H, as follows: L = 2Lz = {nviz n2 }. Here 2 = (0 2 . Thus, any lattice L is a complex multiple of (i.e., a rotation plus expansion of) a lattice of the form L. Proposition 13. For any A, BE C such that A 3 0 27B 2 there exists L =AL z
such that
g 2 (L)
A;
(2.31) (2.32)
PROOF. It follows immediately from the definition 1- 4 g 2 (Lz), and similarly g 3 (2L) = 1 6 g 3 (L3. -
of g 2 that g2 (2Lz)=
By (2.14), we can restate the proposition: there exist 2 and z such that E4 (z) = 24(3/4n4)A and E6 (z) = 26(27/80)B. Let a = 3A14E4,b = 27B/8e. Then the condition A 3 0 27B2 becomes a3 b2 . Suppose we find a value of z such that E6 (z) 2/E4(z) 3 = b 2/a3 . Then choose 2 so that E4 (z) = 24a,
121
§2. Modular forms for SL 2 (7Z)
in which case we must have
E6(Z) 2 = b2E4 (z) 31a3 = 212 b 2,
i.e.,
E6 (z) = +16b.
If we have + in the last equation, then our values of z and 2 have the required properties (2.31)42.34 If we have —, then we need only replace 1 by So it remains to find a value of z which gives the right ratio E6 (z) 2/E4(z) 3 . Now EVE43. = I — 1728/i, by (2.28). Since j(z) takes all finite values on H, and b2/a3 0 I, we can find a value of z such that b 2 la' = 1 — 17281j(z) = E6 (z) 21E4 (z) 3 . This completes the proof. Thus, there was no loss of generality in Chapter I in taking our elliptic curves to be in the form (2.29) for somelattice L.
The Dedekind eta function and the product formula for A(z). We conclude this section by proving the functional equation for the Dedekind a-function, from which Jacobi's product formula for A(z) will follow. The function q(z), z e H, is defined by the product -
ri(z) = e 2 ' 24
(1 11.- kJ. n=1
e2ninz).
(2.33)
(In Problem 7 of §II.4 we gave another definition, and another proof of the functional equation. In the problems below we shall see the equivalence of the two definitions.) Proposition 14. Let .1- denote the branch of the square root having nonnegative
realpart. Then
n(— 1/z) = fzTig(z).
(2.34)
PROOF. The product (2.33) clearly converges to a nonzero value for any ze H, and defines a holomorphic function on H. Suppose we show that the logarithmic derivatives of the left and right sides of (2.34) are equal. Then (2.34) must hold up to a multiplicative constant; but substituting z --= i shows that the constant must be 1. Now the logarithmic derivative of (2.33) is
ty(z) 2ni t n(z) — 24
co
24
E
ne 2. ninz
n= 1 1 — e 2irin )
If we expand each term in the sum as a geometric series in qn (0? = e27'), and then collect terms with a given power of q, we find that til(z)
_27ri
?Az) — 24
— 24
o-1 (n)qn) =
(2.35)
Meanwhile, the logarithmrc derivative of the 'relation (2.34) that we want is:
122
III. Modular Forms --,
rii (— 1 I z) z _2 ti(— 11z)
(2.36)
2z
Using (2.35), we reduce (2.36) to showing that E2 ( — liz)z -2 =
12 + E2(Z), 2niz 0
and this is precisely Proposition 7. Proposition 15.
(2/0 -12 A(z) -z--- q fl (1 — q") 2 4 ,
q = elniz
(2.37)
.
n=1
PROOF. Let f(z) be the product on the right of (2.37). The function f is holomorphic on H, periodic of period 1, and vanishes at infinity. Moreover, f(z) is the 24-th power of q(z), by definition; and, raising both sides of (2.34) to the 24-th power, we find that f(-11z) = z i2f(z). Thus, f(z) is a cusp form of weight 12 for r . By Proposition 9(d), f(z) must be a constant multiple of d(z). Comparing the coefficient of q in their q-expansions, we conclude that o f(z) equals (270' 2 A(z). Notice the role of the Eisenstein series E2 (z) in proving the functional equation for ri(z), and then (2.37). The 17-function turns up quite often in the study of modular forms. Some useful examples of modular forms, especially for congruence subgroups, can be built up from ri(z) and functions of the form ri(Mz). The q-expansion in (2.37) is one of the famous series in number theory. Its coefficients are denoted T (n ) and called the Ramanujan function of n, since it was Ramanujan who proved or conjectured many of their properties: Go co
E
efq t(n) q n d=
n=1
fi (1 — qn)24•
n=1
Among the properties which will be shown later: (1) t(n) is multiplicative (t(nm) = t(n)t(m) if n and m are relatively prime); (2) t(n)-.=- 0-11 (n) (mod 691) (see Problem 4 below); (3) t(n)/n 6 is bounded. Ramanujan conjectured a stronger bound than (3), namely: F rom < n i 1/20_o (n) (where o (n) is the number of divisors of n). The Ramanujan conjecture was finally proved ten years ago by Deligne as a consequence of his proof of the Weil conjectures. For more discussion of t(n) and references for the proofs, see, for example, [Serre 1977] and [Katz 1976a]. -0
PROBLEMS
1. Prove that for k > 4: Ek (z) = 4
E rn,nER (rn,n)=1
On: + nr k .
123
§2. Modular forms for SL 2 (7L) 2. What is E2 (0?
3. (a) Show that El = E8 E4E6 = E10, and E6E8 = E14 (I)) Derive relations expressing a 7 in terms of a3 ; a9 in terms of a 3 and as ; and a 13 in terms of as and u7 . 4. (a) Show that E12 — El = di, with c = (270- 12 26 35 7 2/691. (b) Derive an expression for t(n) in terms of a„ and 475 . (c) Prove that t(n) a 1 1 (n) (mod 691). 5. Prove that E4 and E6 are algebraically independent; in particular, the polynomial in Proposition 10 is unique. 6. (a) Prove the identity
oo
n aX n
cic
E
JO X n
n1 n= 1 (b) Let a 1 (mod 4) be an integer greater than L Show that Ea 4. 1 (i) = 0, and prove the following sequence of summation formulas: co
E1 e
n=
/1/504, no
B +1 = 1/264,
t, —
2(a + 1) a
1/24,
u = 5;
= 9; a = 13, etc.
a
7. Lef(z) be a modular form of weight k for F. Let
1 .f (z) —k E2 (z) f(z). g (z) =12 2n Prove that q(z) is a modular form of weight k if and only iff(z) is a cusp form.
2 for F, and that it is a cusp form
. 8. (a) Prove that E6 = E4 E2 and E8 = E6E2 — (b) Derive relations expressing a5 in terms of al and a3 , and o-7 in terms of a l and C5 .
9. Recall the following definitions and relations from Problems 4 and 7 in §11.4. Let x be a nontrivial even primitive Dirichlet character mod N. Define GO
0 (X, 1 ) =
E An)
2
Re 1>0.
n= 1
Then OCX, = (N 2 0 -1129(X) 0 ( 0-5 1/N 2 t). Now let x be the character mod 12 such that x(+ 1) = 1, x( + 5) = —1, and define (z) =-- 0(x, — I 12) for z H. Then Pi( — 1/z) = Vz/i (a) - Prove that Ti(z + 1) = e 224 - ) and that ri 24 e Si 2 (F). (b) Prove that i)(z) = (c) Write the equality in part (b) as an identity between formal power series in q . (Note. This identity is essentially Euler's "pentagonal number theorem." For a discussion of its combinatoric meaning and two more proofs of the identity, see [Andrews 1976, Corollaries 1.7 and 2.9].) 10. Find the j-invariant of the elliptic curve in Problem 3(b) of §1.2, which comes from the generalized congruent number problem. What is j in the classical case = 1?
124
In. Modular Forms Prove that there is no other value of 2 e15 for which the corresponding j is an integer. It is known that an elliptic curve with nonintegralf cannot have complex multiplication.
§3. Modular forms for congruence subgroups Let y = ( ) e f = SL 2 (1), let f(z) be a function on ii =1/uQu {co} with values in C u {co } , and let k e Z . We introduce the notation . f J[y]k to denote the function whose value at z is (cz_+ d) -1R(az + b)I(cz + d)). We denote the value off f [y] k at z by f(z)l[y] k :
f(Z)I Mk d:7--:f (CZ
+
dr kf(YZ) for
y
= 7 a b) ) e r. c dj
(3.1)
More generally, let GL(Q) denote the subgroup of GL 2 (Q) consisting of matrices with positive determinant. Then we define
f(z)l[y] k d=ef (det y)142 (cz ± d) -if(yz)
for
y = (ac bd e G LI" (0). )
(3.2) For example, if y = (f; °a ) is a scalar matrix, we have f i[y] k .--.--- f unless a < 0 and k is odd, in which case f j[yi k = f. Care should be taken with this notation. For example, by definition f(2z)i[y] k means f l[y] k evaluated at 2z, i.e., (2cz + d)Y(Paz + NI (2cz + d)). This is not the same as g(z)I[y] k for g the function defined by g (z) = f(2z); namely, g (z)l[y] k = (cz + d)- k g (yz) = (cz + d)_"f(2(az + b)I (cz + d )) . With this notation, any modular function of weight k for f satisfies fl[y]k =-- f for all y e I". Some functions are invariant under DI for other y e GL (Q). For example, recall the theta-function defined in §4 of Chapter II: 0(t) = EnE2 e'n2 for Re t > O. We saw that it satisfies the functional equation 0(1) = r 112 0(11t) (where \r- is the branch such that .-If = 1 ) . We define a function of z e H, also called the theta-function, by setting 0(z) = 0(— 2iz), i.e., —
0(z) = E
e2
_E
ne2
Substituting
—
qn2 for
ZEI-19
q = e2niz.
(3.3)
neZ
2iz for t in the functional equation fox : 0, we have 0(z) = (2z/0 -112 0( — 1/4z).
(3.4)
Squaring both sides and using the notation (3.2), we can write
° III= _ ie 2
for
_. = I E GLI(Q). Y (— 4 0 )
(3.5)
125
§3. Modular forms for congruence subgroups
_Notice that
fiEv1v21,=(fiEv1l,)1[Y2]k
for y 19 y2 e GLI(G).
(3.6)
To see this, write (det y) k12 (cz + d ) k in the form (dyzIdz) k12 , and use the chain rule. We are now ready to define modular functions, modular forms, and cuspforms for a congruence subgroup r c F. Let qN denote e 2z1z .
Definition. Let f(z) be a meromorphic function on H, and let I' r be a congruence subgroup of level N, i.e., r' r(N). Let k EL. We call f(z) a modular function of weight k for r if
fi[y] h =1 for all
y e r',
(3.7)
and if, for any y0 eJT = SL2 (1), f(z)l[yo]k has the form Ea,A
with
an = 0 for n a ne n= no
= (a/d)'42
E a e 27ribnAIN nan Nd n = no
n
briggrd, = nE = ano
where
10
bn
if ci,11 n;
(ald)ka e 2 ninbladN ania
if a
in.
This proves the lemma.
Li
We now turn to the proof of the proposition. The first assertion in part (c) follows from Lemma 1. Now suppose that fe Mk (r). Then for a y'a e r" (where y' e I') we have (f Mk)! [a' y' = (fi [yl k)i = f [a]k ; • in addition, for yo e F we have (f [Œ]k)1 [YO] k = fi [ayo ] k , and the condition (3.8) holds for f [a] k , by Lemma 2. Thus,f1 e Mk (F"). Iff vanishqs at all of the cusps, then so doesfl[a] k , by Lemma 2. To obtain the last assertion in Proposition 17(a), we write a = ?), g = N - k2fl[a]k , and note that a-l Fa n F = Fo (N). The values at the two cusps co and 0 come from the above formula for bn with n = 0 and a replaced by ( ?) at the cusp co and by (i:4,1 Sa Iv) ) at the cusp O. This completes the proof of Proposition 17(a). (1)) Let = e2 , and let g =EN x i (j)tj be the Gauss sum. Then co (1 Nil
N —1
f 1 (z) =
E
n=0
1=0
N v=0 co
N —1 = Ar IT
anqn
(v) X1 (1V)
1,v= 0
iv
E ane2 irtn(z—vIN) n= 0
N —1 = n
N
E0
(v)fiz — vIN).
Now let y = bd ) e F0 (MN 2). We want to examine fxa (yz). 'Let yv denote k0 1 . Then for each y and y', 0 < y y' < N, we compute ,
yory;; 1 =
— cvIN b + (v'a vd)IN cvv'IN 2) d + cv'IN
•
129
§3. Modular forms for congruence subgroups r
If y' is chosen for each y so that v'a -=-- vd mod IV (such a choice is unique, because a and d are prime to N), then we have yvyy,7, 1 E ro (M and so ),
fy
4 1 (yz)=--A ,.
N-1
E -ii(v)ftyvvy
1 1, z)
n N-1
=
N v=o
y i (v)x(d)(cy,,z + d + cv'IN) kAy,,z) a
=
N-1
X(d)(ez + d) k --- E -ii(of(z N v. 0
- v' IN).
But )71 (v) = -ii (036. (a)h-Ci (d) = X1(c1) 2 i1(V). Thus, a
xi fti (yz) = xxi(d)(cz + d) k ----'IV
N-1
E i-c-- ,(0ftz - v' IN) = xxi(d)(cz + d)kfx, (z),
V=0
and so 4, i has the right transformation formula to be in Mk (MN 2 , XXI)The cusp conditions are verified by the same method as in the proof of part (a) of the proposition. Namely, for all y E F,fx ,(z)1 [y] k is a linear combination a off(z)l [m ]k, and so the cusp conditions follow from Lemma 2. The next proposition generalizes Proposition 9(a) in the last section. Like Proposition 9(a), it is useful in proving equality of two modular forms from information about their zeros. Proposition 18. Mo (r) = C for any congruence subgroup
r c F. That is,
there are no non-constant modular forms of weight zero. Letfe Mo (r), and let a = f(z 0) for some fixed zo e H. Let r = U air be a disjoint union of cosets, and consider g d=ef ri (f 1[000 — a), i.e., PROOF.
g(z) =
ri (f(Œ
-1 z)
— a).
(3. 1 1)
Then g(z) is holomorphic on H, and it satisfies (3.8), because f does. M oreover; given y f - we have gl [y-1 ] 0 = II (f i [(yai) 1 0 — a). But since f 1[(4-10 does not change if a is replaced by another element ay' E ar, and since left multiplication by y permutes the cosets ai r, it follows that { f I[(y air1 ] 0 } is merely a rearrangement of { f i[oci 9 0 } . Thus, g l[y- 1 0 = g, and we conclude that g e Mo (T). By Proposition 9(a), g is a constant. Since the term in (3.11) corresponding to the coset ir isf(z) — a, it follows that for z = z0 the product (3.11) includes a zero factor. Thus, g = O. Then one of th`e factors in (3.11) must be the zero function (since the meromorphic functions on H form a field). That is, f(cci l z) — a = 0 for some j and for all z e IL Replacing z by aj z, we have: f(z) = a for all ze H, as claimed. o As an example of the applications of Proposition 18, we show that for any positive integers N and k (with k even) such that k(N + 1) =--- 24, a
-
130
HI. Modular Forms
cusp-form of weight k for ro(N) must be a multiple of (n(z)ri(Nz)) k . Here n(z) = e 2riz/24 nn (1 _ qn) , q = e2, as in §2.
Proposition 19. Let f(z) be a nonzero element of sk (ro(N)), where N = 2, 3, 5, or 11 and k = 8, 6, 4, or 2, respectively, so that k(N + 1) = 24. Then f(z) is a constant multiple of g(z) aTf (ti(z)ri(Nz)) k . By Proposition 17(a), g(z) N+1 ---= A(z)A(Nz) is an element of S24( r.0 (N)). In addition, g(z) N+1 is nonzero on H, since A(z) 0 0 on H. At infinity, g(z)' = q N+1 n (1 _ q n)24(1 _ en\) 24 has a zero of order N + 1 in its q-expansion. At the cusp zero, we write -24z12A (z)(zIN) 12A(zIN) oz , , N+1 I[S]24 = z -246,(-1/z),A(—Nlz) = z ) = N -12 q r1 ii 0 _ 4)240 _ q 11 1/1)2 4 , PROOF.
which has a zero of order N + 1 in its qN-expansion. On the other hand, since fesk (ro (N)), its q-expansion at oo is divisible by q, and the 'isexpansion of fl[S] k is divisible by qAi. Now (PO', as a ratio of two elements of S24 (f0 (N)), is a modular function of weight zero for Fo (N). Since g(z) 0 0 on H, this ratio is holomorphic on H. Moreover, the qexpansion off ' is divisible at least by q N+1 , and at zero the qN-expansion in g' and the 4 +1 in is divisible at least by q"- . Hence, the g N+1 1[S] 24 are 'canceled, and the ratio is holomorphic at the cusps, i.e., ( .fig ) v +1 e Mo (fo (N)). By Proposition 18, ( flg) N+1 is a constant, and hence fig is also a constant. This completes the proof. 0
ei +1
Notice that we did not actually prove that g(z) ---= (ri(z)ri(Nz)) k is in sk (ro(N)), unless we can be assured that there exists a nonzero element feS k (ro (N)). The same proof, for example, would tell us that any nonzero Je S3 (r0(7)) must be a constant multiple of (n(z)v(7z)) 3 ; but s3(r0(7)) = 0, since 3 is odd and — /e I-10 (7). However, for the values of N and k in Proposition 19 it can be shown that dim sk(ro(N)) = 1 (see Theorem 2.24 and Proposition 1.43 in Chapter 2 of [Shimura 1971]). Thus, Proposition 19 is not vacuous. For N = 2, k = 8 we can see this directly, since we know that T and ST 2S ------ ( 1 _?) generate r0 (2) (see Problem 13(b) of §III.1).
Proposition
20. (q(z)ri(2z)) 8 E S8(1-0(2)).
Clearly, g(z) = (ri(z)q(2z)) 8 is holomorphic on H. Its q- expansion at oo is: ( e 2iriz/24±2ni2424)8 n 0 ...... qn)8 (1 _ q2.)8 = q ri 0 _ qn)8 (1 _ q 2n) s . Using the relation ?A-1/z) =--- V zlig(z), we easily see that g(z) vanishes at the cusp 0 as well. It remains to show invariance under [(1 _?)] 8 . Set a = (2 1). Then (1 _?) = lane, and PROOF.
g (z)l[a] 8 = 24 (20 -8 0/( — 1/2z)q( — 1/z)) 8 = (2z2) -4(V2z/i n (2z) / ti(z)) 8 = (r/(z)t/(2z)) 8 =-- g(z).
131
§3. Modular forms for congruence subgroups
Since g[T]8 = g, and Ha is always trivial, it follows that g is invariant under la Ta, as desired.
Eisenstein series. Let N be a positive integer. Let a = (al , a2) be a pair of integers modulo N. We shall either consider the ai as elements of ZINZ or as 'integers in the range 0 < ai 3 to be either odd or even. Proposition 21. Gkcj-mod
N
mk(r(N .
)) and Gr a 2 ) mod N e mk (r i (N) ).
PROOF. First, the series (3.12) is absolutely and uniformly convergent for z in any compact subset of H, because k > 3 (see, for example, Problem 3 in §1.5). Hence G(z) is holomorphic in H. Now let y = (ca bd) e F. Then G(z)I [y] k =
E
(cz + dyk
mr, a mocIN
E
k ( az + b mi cz-+ d + m2)
t(p"
mz-Eamod N
W" 1"" + m2 c)z + ()nib + rn2c1))k
Let m' = (m i a + m2 c, mb + m 2 d) (m i m2 )( ca bd) =m1). Note that modulo N we have m' ay. Let a' = ay (reduced modulo N). Then the maps mi--m/ = my, and m' = m'y' give a one-to-one correspondence between pairs m e V with m a mod N and pairs m' e Z 2 with m' a' mod N. This means that the last sum above is equal to Em' Ea' mod NOnfiz + (z). Thus, = Gkay mod N
Gg-m°dN
for y e re
(3.13)
If 'y e F(N), then by definition y I mod N, and so ay a- a mod N. Thus, (3.13) shows that G: is invariant under [y] k for iy e F(N). Similarly, if y (N) and a l = 0, we have (O a2)y (0, a2) mod N, and so Gra2) m0d N is ,
132
HI. Modular Forms
invariant under [y]k for y e f 1 (N). It remains to check the holomorphicity condition at the cusps. But [yoik permutes the G" dN for yo e F, by (3.13). Hence it suffices to show that each G: is finite at infinity. But
E
lim GAz) =
z-4.iao
if a l 0 0;
0
E
mEarnedN,m1=0
n"
if a 1 = O.
n-rta 2 mod N
The sum / n" over n L--s a2 mod N converges because k >_ 3. It is essentially a "partial" zeta-function. More precisely,
± (_ l) CCk.` - a2( ) 9 where Ca(k)i-Q. c Gr,a 2)(co ) _ a,(k) ts1 n.s.---amodN
This completes the proof of the proposition. As a special case of (3.13), if we set y = —! we have
GI-(2 = (— Div. This can also be seen directly from the definition (3.12). Thus, for example, G: = 0 if k is odd and 2a--= 0 mod N. It is now not hard to construct modular forms for any congruence subgroup F', f = I"' D f(N), out of the Eisenstein series G:d rno N . For fixed a, the elements y E F' permute the Eisenstein series G:, where a' ranges over the orbit of a under the action of r, i.e.,
g' Gel ra foiverl. Let r = #(41') be the number of elements in the orbit. If F(Xi , .. . , X,.) is any homogeneous symmetric polynomial in r variables with total degree d, and we set Xi = Gri (where ayi rubs through the orbit ar'), then F(Gr', . . . , Ge ') is easily seen to be a modular form of weight kd for F'. For example, taking F to be X1 + - • - + Xr or X1 . . . Xr , we have E Ge
,„2 , mod
N(z )
€ mk (ro(N)) ;
a2 e (7L /NZ)* Gi(c 0 , a2) mod AT( 2.) e mo(N),,
n
(3.16) ( Fo(N)).
a2 e(ZINZ)*
We now compute the qN-expansion for G:. We shall be especially interested in the cases when a l = 0 or a2 =---- O. Recall the formula used in deriving the q-expansion for Gk in the last section (see the proof of Proposition 6):
1 k = (..._ ok--12c(k).___ k 'E i• e2nijz nE ez (z + n) Bk j = 1
for
k > 2,
z e H.
(3.17) Let
133
§3. Modular forms for congruence subgroups
c
cf
a
=
NkBk
{0 k d7f Ca2(k) (— 1) k r i2 (k)
if a / 0 0; if a / = O.
(3.18) Then
G:(z) =
14,
E + ni 1 :4a 1 mod E N,m i *0 m2 Ea 2 mod N
(
E
1)"
-k
N + n)
m i Ea, mod N,m i >0 neZ
+ N -k
in2) –k
m iz + a2
E
=b+N
iz +
E
(m i z - a2
m 1 E–a 1 modN,m i >0 nEZ
=
00
E
ck(
ik_ie2Rifi(rniz+a2)/N)
m i Ea ' mod N j=1 m i >0 03
E
(—ok
m1=--- – al mod m >0
e2iriAmiz–a2)1N)
E N j=1
In these computations we had to split up the sum into two parts, with m i replaced by m 1 for m i negative, because in applying (3.17) with z replaced by (m / z ± a2)I N we need (rn i z + a 2)IN E II, i.e., m 1 > O. Now let -
qN = e2nizIN
crf e2mmi
(3.19)
Then G(z) =
E
b + ck
00 E i k„ va2q.km 1
m 1 E-a l mod N j=1
(3.20)
( _ ok
co
E
E
m i s--. –a i mod N j=1 m >0
To find the coefficient 91; of q?k, it remains to gather together terms with = n. We shall only do this in the cases when a l = 0 and a2 = O. As a result, we have the following proposition. Proposition 22. Le t ck ,bt k ,,q N be as in (3.18) (3.19). For k > 3 let Gklm°dN (z) -
a mod be the Eisenstein series (3.12). Then the qN- expansion of G N
G:(z) = bt k + E
(3.21)
tfirT07;
n=1
Can be computed from (3.20). If a = (a 1 , 0), then for n > 1 = Ck(
E iln n/jE-..-`. a l mod N
j k, +( _ ok
E jjn
nijE –a l mod N
(3.22)
134
HI. Modular Forms
If a = (0, a2), then for n __ I
bNa n,k = ck E ik—i (Va 2 ± ( ..._1)k—i(2 2) . (3.23)
er, k = 0 if N )(n;
jln
Thus, for a = (0, a 2) co
q = e2niz .
Ge ,a 2)(z) = b(00:ka2 ) ± ck 1 E i k, (Va 2 ± (____ 01,: ja 2) q n 1 n=1i
(3.24)
Proposition 23. If 2a a- (0, 0) mod N and k is odd, then G laT md N = O. Other-
wise, Ge'a2) is nonzero at co, and Ge" 9) has a zero of order min(a i , N — a 1 ), where we are taking al in the range 0 N/2. n=c, ( a2 + nN) k (N — a 2 + nN)k)t< 0 I
Finally, we look for the first possible value of n in (3.22) for which either sum in (3.22) is nonzero. That value is n = min(a i , N — a l ), where we have the possible value/ = 1 in one of the two sums. Thus, b' °) = +ck for n = 0 min(a / , N — a 1 ). This completes the proof.
- "dN can be As an application, we show that a certain product of G31" expressed in terms of the n-function. We shall use this result in the next section, where we give Hecke's proof of the transformation formula for 0(z). Recall the Weierstrass p-function and its derivatives from Chapter I: So(z; w 1
,
(0
1 1 1 • 2) = —2z + „iI ,„ ez l (z + nu°, + nw 2) 2 (n2C01 + no) 2 ) 2 '
sg (z co 1 , co 2) = —2 E '
(z
;
+ m, u) + nco2)- 3
;
m,neR
(0 1 9
(0 2) =
E
(— l )"(k — 1)!
(z
+ imo, + nw 2) -k ,
k > 3.
m,neZ
If a 0 (0, 0), we can express Gk'llmd N (Z) in terms of gp (k -2) as follows: G: mod N (z)
—k a i z + a., .`+ Ira + n) N m,nEZ
= N _k z
( -- Ok = Nk (k — l)!
(k_ 2 )
al z + a2 ;
(3.25)
Z,
1) . This is the value of p (k-2) for the lattice L, = linz + n1 at a point of order
135
§3. Modular forms for congruence subgroups
N modulo the lattice, namely (a1 z + a2)/N. Thus, the Eisenstein series for r(N) are related to the values of the derivatives of p at division points. Now suppose that k = 3. By (3.25), G(z) can vanish only if 0 1 ((a1 z + a2)IN ; z, 1) = O. But pi vanishes only at half-lattice points (see the end of §1.4). If 2a ET: (0, 0) mod N, i.e., if 2(a 1 z + a2)IN is a point in L z , then G39- is the zero function, by Proposition 23. Otherwise, oVa l z + a2)IN ; z, 1) is nonzero. We have proved Proposition 24. If 2a # 0 mod N, then Grd N (z) V: lr ) for z-e H.
Let p be an odd prime. We now define 12- 1
h(z) =-- n Gl o,a2 ,modp(z).
(3.26)
a 2 =1
Proposition 25. h(z)e M3(p ... 1)(1- 0( p)), its only zero is a (p 2 — 1)/4 —fold zero at 0, and it is a constant multiple of (TIP(z)lri(pz)) 6 . The first part is the special case of (3.16) when N = p, k = 3. h(z) is nonzero on H by Proposition 24, and at infinity by Proposition 23. To find its order of zero at 0, we examine the qp-expansion of PROOF.
hl[S]3 (p _i ) =
P— 1 P-1 [I Gr 2 ,0)m04 p n Gl o 2,modp i[s h= ,a
a 2 =1
a2=1.
by (3.13). According to Proposition 23, the first power of qp which appears in Gr2 ' 0)rmdP is qpinin("2 4'- '2 ). Thus, the order of zero of h(z) at 0 is (1) - 1)12
12 - 1
E min(a 2 , p --: a2) = 2 a 2 =1
E
a2 = (p2 — 1)/4.
a2 =1
Now set 172(z) == (re(z)li(pa. Then î(z) 4= A(z)P I A(pz) is holomorphic and nonzero on H, since A(z) is holomorphic and nonzero on H. Because A(z) e S12 (r) c S12 (1-0 (p)) and A(pz) e Si2 (ro(p)) by Proposition 17(a), it follows that h(z) 4 is a modular function of weight 12(p — 1) for Fo (p). Its q-expansion at infinity is q p no ...... q n) /q p no _ e p) 24 _ n „, _ q n) p,o 24" hence 1i(z) 4 is holomorphic and nonzero at co. At the cusp 0 we have I 2(p-1) • f — , at 11z)P-IA(—plz) kz)4 1[S] 1 2(p-1) = z— = z-12(p-1)Z12PA(z)P1((zIp)12A(zip)) = p'2 A(z)P I A(z Ip) = p 12 q p no .....
q .) 24p/qpn(1
qnp)24 1
which has leading term p 1242-1 . Thus, both h4 and ii 4 are elements of M12(p_ 0(1"0 (p)) with no zero except for a (p2 — 1)-ordér zero at O. Hence their ratio is a constant by Proposition 18. But (///h )4 = const implies that hiii = const. This concludes the proof of Proposition 25. 0
136
III. Modular Forms ,
Notice that, by (3.15), we have some duplication in the definition (3.26) of 12( 2). That is, if we define ( p— 1)12 f(z) = fi , Gr,a2) mod p(z ) ,
(3.27)
a2 =1
we find, by (3.15), that
h(z) = (
0
( p-1 )/2f(z)2.
'
—
(3.28)
Proposition 25 then tells us that the square of the ratio off(z) to 07 P(z)11(pz)) 3 is a constant. Hence, the ratio of those two functions must itself be a constant, and we have proved Proposition 26. The function f(z) defined in (3.27) is a constant multiple of
(r1P(z)M(pz)) 3 . Because each of the GV'a2) in (3.27) is in M3 (1", (p)) by Proposition 21, it follows that fe M3(p _ 0/2 (1"1 (p)). However, unlike 144 f(z) is not, strictly speaking, a modular form for the larger group Fo (p). Proposition 27. Let f(z) be defined by (3.27), and let y = Ca D e Fo (p). Then
f l[y] 3( P — 1)/2 = (1,)f, where (i) is the Legendre symbol (which is +1, depending on whether or not d is a square modulo p). PROOF.
Since (0, a 2 ) (° )2L---_ (0, da2) mod p, it follows by (3.13) that
f I [y] 3,p - 1 „ 2 =
(p-1.)/2
fi Gr ,da 2
)
mod p .
a 2 =1
But by (3.15), the terms in this product are a rearrangement of (3.27), except that a minus sign is introduced every time the least positive residue of da2 modulo p falls in the range (p + 1)/2, (p + 3)/2, . . . , p 1. Let nil be the number of times this occurs. Thus, fl[Y]3(p012 = ( 1)ndf. According to Gauss's lemma, which is an easily proved fact from elementary number theory (see, for example, p. 74 of [Hardy and Wright 1960 ]) , we have (— 1)d = (1). This proves the proposition. 0 —
—
The transformation formula in Proposition 27 is an example of a general relationship between modular forms for r1 (N) and "twisted-modular" forms for Fo (N), called "modular forms with character". We now discuss this relationship. We start with some very general observations. Suppose, that F" is a subgroup of F', and f(z) is a modular form of weight k for the smaller group F" but not necessarily for the bigger one. Then for y e I' we can at least say thatf i[y - l k only depends on the coset of y modulo F". That is, if y" e r", then fl [(yy") - 1 = fi[y- lk . Now suppose that the subgroup F" is normal in F'. To every coset yF" associate the linear map fF--> f l[y- lk which takes an element in Mk(F") to Mk (F") by Proposition 17 (with F" and y -1 in place of F' and a; note that yF"y -1 n F = r", since y e F' and F" is normal in F'). This gives us a group
137
,
§3. Modular forms for congruence subgroups
-
1
homomorphism p from F'/F" to the linear automorphisms of the vector space Mk(F"), because for yi , y2 e F' , P(Yi Y2): fE.-fl{(yiy2) - 9k = (fl[Ynk)i[Yilk = POO (p(y2)f).
In other words, we have what is called a "representation" of the group
r'/r" in the vector space mk(r"). If : F'/F" -- C* is a character of the quotient group, then define Mk(F', x) to be the subspace of Mk(F") consisting of modular forms on which the representation p acts by scalar multiplication by x, i.e.,
mk(r, x) a tfe Afk(r")1p(y)f= lLe ---,
x(y)f for all ye
r}.
If F'/]T" happens to be abelian, then according to a basic fact about representations of finite abelian groups, Mk(F") decomposes into a direct sum of Mk(F', x) over the various characters x of FYI'. We shall soon recall the simple proof of this fact in the special case that will interest us. We apply these observations to the case F" = ri (N), F' = Fo (N). Since ri (N) is the kernel of the surjective homomorphism from FO (N ) to (11 N 1)* that takes (ca db to d, it follows that Fi (N) is a normal subgroup of Fo (N) with abelian quotient group isomorphic to (ZINZ)* (see Problems 1-2 of §III.1). Let x be any Dirichlet character modulo N, i.e., any character of (Z/NZ)*. In this context the subspace Mk(ro(N), x) Mk(ri (N)) is usually abbreviated Mk(N, x). That is, )
Mk (N, x) ,1---3 ffe Mk (Fi (N))I f l[y] k = x(d)f for y = (
a 11 ) e ro c_ d
(N4 (3.29)
In particular, if x = 1 is the trivial character, then Mk (N, 1) =
mk(ro(N)).
Proposition 28. Mk (f i (N)) = 10Mk (N, x), where the sum is over all Dirichlet /characters modulo N. As mentioned before, this proposition is actually a special case of the basic fact from representation theory that any representation of a finite abelian group decomposes into a direct sum of characters. However, we shall give an explicit proof anyway. First, any function f that satisfies f 1[y] k = x(d)f for two distinct characters x must clearly be zero; hence, it suffices to show that anyfe Afk (Fi (N)) can be written as a sum of functions fx E x). Let , PROOF.
Mk(N,
1,
fx = 0(:AT)
d
E
k(d)f I[Ydlo
where yd is any element of Fo (N) with lower-right entry congruent to d mod N. We check that for y = (` S)e fo (N)
fx I Hit =--- 4)(1N) d Zz)* TCfrOf f Fy 1 which, if we replace dd' by d as the variable of summation, is easily seen
138
HI. Modular Forms
to equal x(d)f, i.e.,ft e Mk (N, x). Finally, we sum theft over all characters x modulo N, reverse the order of summation over d and x, and obtain:
Efx =
1
_
Ex(d) E N) x d. ( zINE) * o(
fiETL,
which is equal to f, because the inner sum is 1 if d = 1 and 0 otherwise. o Thus, f can be written as a sum of functions in Mk (N, x), as claimed.
Notice that Mk (N, x) = 0 if x has a different parity from k, i.e., if (— 1) 0 (-1)k . This follows by taking y = —I in the definition (3.29) and recalling that f i[ nk = ( 1)k. f For example, as an immediate corollary of Proposition 28 and the pre7 ceding remark, we have —
—
Proposition 29.
AO , 1), k even; 4 AM r 1( )) = I ./IA(4 , x), k odd, where 1 denotes the trivial character and x the unique nontrivial character , modulo 4. Notice that the relationship in (3.29) is multiplicative in y; that is, if it holds for y l and y2 , then it holds for their product. Thus, as in the case of modular forms without character, to show that f(z) is in Mk (N, x) it suffices to check the transformation rule on a set of elements that generate F o (N). As another example, we look at 0 2 (z) = (Inez qn2)2 , whose n-th qexpansion coefficient is the number of ways n can be written as a sum of two squares. Proposition 30. 0 2 e M1 (F 1 (4)) ---.--- Aft (4,
x), where x(d) = (
-
1)("w 2 .
the transformation rule for —I, T, and ST 4S = (-1 _7), which generate To (4) (see Problem 13 of §1117.1). This is immediate for T, since 02 has period F. Next, the relation fl[ n i = f= x( 1)f holds for anyf, by definition. So it remains to treat the case ST4S. Let PROOF. It suffices to verify
—
—
—
(0 —1 CtIV d=ef V
and
aN
aN-1 = —laN , N
so that
o) '
(a b) -1 c d ccN
(
=
—c/N\ — Nb a ). d
(3.30)
We write ST4S = —a4 Tcc4-1 = ia4 Tec4 , and use the relationship 0 2 1 [cc4j i _j® 2 (see (3.5)) to obtain 021[574511 z..= 021[Œ4T,,4]1 =_._ — 021[Ta4]1 = —i®21 [°44]1 [Œ4]i = (Recall that the scalar matrix il acts trivially on all functions, i.e., [1/4], identity.)
6111•••18 ■
-
139
§3. Modular forms for congruence subgroups
To finish the proof of the proposition, we must show the cusp condition, i.e., that 0 2 1[701 1 is finite at infinity for all yo e T. But the square of 0 2 1[y0] 1 is 04 I [To] 2 , and it will be shown in Problem 11 below that 04"e M2 (ro (4)) ; in particular, this means that CO'f[yo ] 2 , and hence also Col [yo]i , is finite at o infinity. This completes the proof. The spaces Mk (N, x) include many of the most important examples of modular forms, and will be our basic object of study in several of the sections the follow. We also introduce the notation Sk (N, x) to denote the subspace of cusp forms: Sk (N, x) j=f2Mk (N, x) n S k (r i (N)).
The Mel/in transform of a modular form. Suppose that f(z) = E an 4 (where q Iv ......_ e 2niz1N ) is a modular form of weight k for a congruence subgroup F' of level N. Further suppose that Ia n ' = 0(n') for some constant c ER, i.e., that an/ne is bounded as n --> co. It is not hard to see that the qN-expansion coefficients for the Eisenstein series G:m" N have this property with c = k — 1 + e for any r, > O. For example, in the case F' = F, the coefficients are a constant multiple of o-k _ i (n), and it is not hard to show that ak _ 1 (n)I rl k—i+e 0 as n ---÷ cc . We shall later show that, if f is a cusp form, we can take c = lc/2 + E. It has been shown (as a consequence of Deligne's proof of the Weil conjectures) that one can actually do better, and take c = (k — 1)/2 + E. In Chapter II we saw that the Mellin transform of OW = E e -'2 and certain generalizations are useful in investigating some important Dirichlet series, such as the Riemann zeta-function, Dirichlet L-functions, and the Hasse-Weil L-function of the elliptic curves En : y2 = x 3 — n2x. We now look at the Mellin transform for modular forms. Because we use a variable z in the upper half-plane rather than t (e.g., t = — 2iz), we define the Mellin transform by integrating along the positive imaginary axis rather than the positive real axis. The most important case is F' = F1 (N). For now we shall also assume that f(co) z.-- O. Thus, let f(z) = E,7=1 an qn e Mk (Fi (N)). (Recall that since Te F1 (N), we have an expansion in powers of q = e27riz rather than qN .) We set ioo f(z)zs' dz. (3.31) def —0
We now show that if f(z) = E,T, i anq" with lan i = 0(d), then the integral g(s) defined in (3.31) converges for Re s> c 4- 1: \ ico
.
JO
oo
it°
f(z)zs -i dz = E an n =1
J0
2 dz zse ninz ______ z
— E0° an (— 27r1 1n y) n=1
foe
dt tse-r 7
(where t = —27rinz)
0
00
(-270 -sr(s) E an n -s
(see (4.6) of Ch. II)
n=1
(3.32)
140
111. Modular Forms
(where the use of I" in the gamma-function F(s) has no relation to its use in the notation for congruence subgroups; but in practice the use of the same letter F should not cause any confusion). Since iann-si = o(nc-Res\), this last sum is absolutely convergent (and the interchanging of the order of integration and summation was justified). - Inœ=0 42Ane mk (r,(N)) has la c) 0 0, we replace f(z) by f(z) — ac, If f(z) =----in (3.31). In either case, we then obtain g(s) = (-2ni)_sr(s)LAs), where co co ;le Lf (s) d=ef and for Re s > c + 1, if f(z) =
E
E
n=1
n=0
(3.33)
with Ian ' = 0(d).
In addition to their invariance under [y]k for y el",(N), many modular forms also transform nicely under [acia, where ŒN = ( ) as in (3.30). It will be shown in the exercises that, for example, any function in Mk(N, x) for z a real character (i.e., its values are +1) can be written as a sum of two functions satisfying f1[aN]k= Ci-kf
C 7--- 1 or —1,
(3.34)
where one of the functions satisfies (3.34) with C = 1 and the other with C = —1. An example we already know of a function satisfying (3.34) is 02 : the relation (3.5) is a special case of (3.34) with k = 1, C = 1, N = 4. We now show that if (3.34) holds, then we have a functional equation for the corresponding Mellin transform g (s) which relates g (s) to g (lc — s). For simplicity, we shall again suppose that f(z)= I an sq" with ao = 0. We can write (3.34) explicitly as follows, by the definition of [aid', : f(-1INz) = CN -42 (—iNz) kf(z).
(3.35)
In (3.31), we break up the integral into the part from 0 to il\FN and the part from i//i to ix. We choose ifIKT because it is the fixed point in H of ŒN : zF.--+ —1INz. We have
ico g(s) =
fib.rsT
= •••■•••■■•
••••••■••
dz
.fiz)z s--z-
—f
ioo
111Vz)
f(— 1 INz)(— 11Nz) 3d( — 1INz iw-st
ioo (f(z)zs ± f(-1INz)(-11Nz)s)—dz z fiwN ico ) —dz (f(z)zs ± i k CN - wf(z) ( — 1 /Nz)s—k
. f ibliq
Z
,
because of (3.35). In the first place, this integral converges to an entire function of s, because f(z) decreases exponentially as z -4 ioo. That is, because the lower limit of integration has been moved away from zero, we no longer have to worry about the behavior of the integrand near O. (Compare with the proof of Proposition 13 in Chapter II, where we used a similar technique to find a
141
§3. Modular forms for congruence subgroups
rapidly convergent series for the critical value of the Hasse-Weil L-function ; see-the remark following equation (6.7) in §II.6.) Moreover, if we replace s by k s in the last integral and factor out ik CN -42 (- NY , we obtain:
(i'C'N k/2 (- N)sf(z)z's
g(k s) = i k CN -k2 (-
= eCN -k12 (-
f(z)zs)
dz
ioo I (Pc CAT -kl2f(z)(- 11Nz)s -k +f(z )e)
= i k CA T -142 (- N)sg(s), because the last integral is the same as our earlier integral for g(s). This equality can be written in the form
(- i\ N) 5g(s) = C(- iffV)- k- sg(k s). Thus, by (3.32)-(3.33), if we define A(s) for Re s> c + 1 by
(3.36)
A(s) = (- i Kr)g(s) = ( N 12n)sr(s)L f (s),
we have shown that A(s) extends to an entire function of s, and satisfies the functional equation
(3.37)
A(s) = CA(k s).
An,
As an example of this result, we can take f(z) = A(z)e S 1 which satisfies (3.35) with N = 1, k = 12, C = 1. Then A(z) = Z t(n)q", 6.(s) = z,f_ -c(n)n -- , and A(s) = (2n) -sr(s)LA(s) satisfies the relation : A(s) =
A(12 The derivation of (3.37) from (3.34) indicates a close connection between Dirichlet series with a functional equation and modular forms. We came across Dirichlet series with a functional equation in a very different context in Chapter II. Namely, the Hasse-Weil L-function of the elliptic curve : y2 = .x3 - n 2 x satisfies (3.36)-(3.37) with k = 2, N = 32n' for n odd and 16n2 for 'n even, C = for n odd and ( ) for n even (see (5.10)(5.12) in Ch. II). We also saw that the Hasse-Weil L-function of the elliptic curve y2 = x3 + 16 satisfies (3.36)-(3.37) with k = 2, N = 27, C = I (see Problem 8(d) of 11.5). So the question naturally arises: Can one go the other way? Does every Dirichlet series with the right type of functional equation come from some modular form, i.e., is it of the form Lf(s) for some modular form f? In particular, can the Hasse-Weil L-functions we studied in Chapter II be ,obtained by taking the Mellin transform of a suitable modular form of weight 2? That is, if we write L(E„, s) in the form E:=1 bm ars (see (5.3) in Ch. II), is E b niqm the q-expansion of a weight two modular form? Hecke [1936] and Weil [1967] showed that the answer to these questions is basically yes, but with some qualifications. We shall not give the details,
142
W. Modular Forms ,
which are available in [Ogg 1969 ], but shall only outline the situation and state Weil's fundamental theorem on the subject. Suppose that L(s)= I anti' satisfies (3.36)7(3.37) (and a suitable hypothesis about convergence). Using the "inverse Mellin transform", one can reverse the steps that led to (3.36)-(3.37), and find that f(z)= E anqn satisfies (3.34). For now, let us suppose that N = A2 is a perfect square, and that C = ik (k even). Then, iff(z) satisfies (3.35), it follows that ft (z) cre--f f(z I A) = 1 anqr,1 satisfies: f1 (— 11 z) = z kfi (z). Thus, fi is invariant under [S] k and [T] k , and hence is invariant under the group generated by S and V. Hecke denoted that group 0(2). We have encountered the group 0(2) before. In this way one can show, for example, that L(Eino , s) corresponds to a modular form (actually, a cusp form) of weight 2 for 0(8n 0). Unfortunately, however, Hecke's groups 0(A) turn out not to be large enough to work with satisfactorily. In general, they are not congruence r(2) is an exception.) subgroups. (6(2) But one can do much better. Weil showed, roughly speaking, that if one has functional equations analogous to (3.36)-(3.37) for enough "twists" E x(n)ann -s of the Dirichlet series Z anti', then the corresponding qexpansion is in mk(ro(N)). we now give a more precise statement of Weil's theorem. Let xo be-a fixed Dirichlet character modulo N (x o is allowed to be the trivial character). Let x be a variable Dirichlet character of conductor m, where m is either an odd prime not dividing N, or else 4 (we allow m = 4 only if N is odd). By a "large" set of values of m we shall mean that the set ' _iv} jez , contains at least one m in any given arithmetic progression {u+ where u and y are relatively prime. According to Dirichlet's theorem, any such arithmetic progression contains a prime; thus, a "large" set of primes is one which satisfies (this weak form of) Dirichlet's theorem. By a "large" set of characters x we shall mean the set of all nontrivial x modulo m for a "large" set of m. Let C = ±1, and for any x of conductor m set
C, --z--• CX0(m)( — N)g(X)igW,
(3.38)
where g(x) = E.7,_, x(j)e 27'iiiN is the Gauss sum. Given a q-expansion f(z)= Ef=o anqn, q = e'iz, for which lan i = 0(d), we define Lf (s) by (3.33) and A(s) by (3.36), and we further define CO
Lf (x, s) =
E x(n)anirs ;
A(y, s) = (m \ I-A 1 72n)r (s)Lf (x, s).
(3.39)
n=1
Weil's Theorem. Suppose that f(z)=E„œ_ o an qn, q = e2', has the property that Ian ' =.-- 0(nc), ce R. Suppose that for C =1 or —1 the function A(s) defined by (3.36) has the property that A(s)-1- a o (lls --I 1 (k — 4) extends
143
§3 Modular forms for congruence subgroups
to an entire function which is bounded in any vertical strip of the complex plane, and satisfies the functional equation A(s) CA(k s). Further suppose that for a "large" set of characters x of conductor m (in the sense explained above), the function A(z, s) defined by (3.39) extends to an entire function which is bounded in any vertical strip, and satisfies the functional equation A(z, s) = Cx A(, k s), with C, defined in (3.38). Then feM k (N, x 0), and f satisfies (3.34). lf, in addition, L1 (s) converges absolutely for Re s> k s for some e> 0, then fis a cusp form. -
One can show that the Hasse—Weil L-functions of the elliptic curves in Chapter II satisfy the hypotheses of Weil's theorem (with zo = 1). The same techniques as in the proof OC the theorem in §I1.5 can be used io show this. However, one must consider the Hecke L-series obtained in (5.6) of Ch. II by replacing 2,7(I) by the character jc",(.0z(Ni1) with z any Dirichlet character modulo ni as in Weil's theKm. For example, if we do this for L(E, , s), where E1 is the elliptic curve, 1=- x 3 x, we can conclude by Weirs theorem that
fE i (z) = q
2q5 3 (i) 6q' 3 + 2q' 7
E
bni e
(3.40)
rn>25
(see (5.4) of Cu. 10 is a cusp form of . weight two for 1'0 (32). If we form the q-expansion corresponding to the L-series of E : y 2 = n 2x, namely, fEn (z) = z n(m)k nqin, it turns out that fEn e M2 (F0(32n 2)) for n odd and fEn eM2 (F0 (16n 2)) for n even. Note that when n 1 mod 4, so that zn is a character of conductor n, this is an immediate consequence of the fact that fE , M2 (f0 (32)), by Proposition 17(b). More generally, it can be shown that the Hasse—Weil L- function for any elliptic curve with complex rrfultiplication satisfies the hypotheses of Weirs theorem with k = 2, and so corresponds to a weight two modular form (actually, a cusp form) tor FAN). (N is the so-called "conductor" of the elliptic curve.) -Many elliptic curves without complex multiplication are also known to have this property. In fact, it idtS conjectured (by Taniyama and Weil) that every elliptic curve defined over Vivo rational numbers has L-function which satisfies Weil's theorem for some N. Geometrically, the cusp forms of weight two can be regarded as holornorphic differential forms on the Riemann surface F0(N)V-1 (i.e., the fundamental domain with FAN} equivalent boundary sides identified aï: the cusps included). The TaniyamaWeil conjecture then can be shown to take the form: every elliptic curve over G can be obtained as a quotient of the Jacobian of some such Riemann surface. For more information about the correspoK dence between modular forms and Dirichlet series, see [Hecke 1981] ,teil 1967], [Ogg 1969 ], and
[Shimura 197f].
144
III. Modular Forms
PROBLEMS
L Let a e GL1. (0), and let g(z) ---= f(az). Let y = (cz bd )e F. Notice ihat f(az)l[y] h was defined to be (caz + d) kf(yaz), which is not the same as g(z)l[y]k = (cz ± d)' f(ayz). Show that if a = (g ?), i.e., if az =-- nz, then g(z)1[T]k -----.ficaaccrx - lk = f(nz)IEGin "Mk. be a congruence subgroup of r of level N, and denote r; = {y E FITS = . 5 } for sEQ v {co}. Let s= Œ 1 CC, 0: E F. (a) Prove that ars'OE -1 z-_- (Œria -1 )co . (b) Show that there exists a unique positive integer h (called the "ramification index" of r at s) such that
2. Let
r'
(1) in the case —.ter
• r; = + aoe t
trinEz a;
(ii) in the case —1.0r either r; = 0,-1 {T hrilf ne za; Or
r; =
nela•
Show that h is a divisor of N. (c) Show that the integer h and the type (I, Ha, or III)) of s does not depend on the choice of a E r with s = a' co ; and they only depend on the F'-equivalence class of s. (d) Show that if a -1 oc is of type I or Ha andfe /cr.), thenfl[a -l ] k has a Fourier expansion in powers of qh . A cusp of r is called "regular" if it is of type I or Ha; it is called "irregular" if it is of type lib. (e) Show that if a-1 co is an irregular cusp, and femk (r), then fl[a- i]k has a Fourier expansion in powers of q2,, in which only odd powers appear if k is odd and only even powers appear if k is even. If k is odd, note that this means that to show that fE Mk (r) is a cusp form one need only check the q-expansions at . the regular cusps.
3. Let h be any positive integer, and suppose 2hIN, N 4. Let F' be the following level N congruence subgroup: F' = f(a, ,1!1 ) --a-. (1 -4)i mod N for some j}. Show that co is a cusp of type Jib. 4. (a) Show that F1 (N) has the same cusps as ro(N) for N = 3, 4. (b) Note that —.10 ri (N) for N> 2. Which of the cusps of r'1 (3) and 1T1 (4), if any, are irregular?
5. Find the ramification indices of F' at all of its cusps when: (a) F' = Fo(p) (p a prime); (b) F' = Fo(p2); (c) F' = F(2). 6. Prove that if r a F is a normal subgroup, then all cusps have the same ramification index, namely
[Toe : ±r].
7. (a) Show that any weight zero modular function for F' c r satisfies a polynomial of degree [F: r ] over the field C( j) of weight zero modular functions for T.
145
§3. Modular forms for congruence subgroups
(1)) Show that, if r is a normal subgroup, and if f(z) is F'-invariant, then so is - f(Œz) for any °ter. Then show that the field of weight zero modular functions for r' is a galois field extension of C( j) whose galois group is a quotient of In practice (e.g., if r is a congruence subgroup), it can be shown that the galois group is equal to f/F'.
8. Prove the following identities, which will be useful in the problems that follow and in the next section, by manipulation of power series and products in q = e2 1ni :
(a) 0(z) + e(z + = 20(4z); E2 (z) = 48Z odd n> 0 61 (n)q n ; (b) E2 (z + 2k p4 n crk-i(n)q n (c) Ek(Z) — ( 1 + Pk-1 )Ek(PZ) ± Pk-1 Ek(P 2 Z) = Bk (k 2, p prime); (d) E2 (z) 3E2(2z) + 2E2 (4z) =-1-(E2 (z) E2(z +1)); (e) q(z + -21) = e21ti/4 8 7/ 3(2z)/ri (z)ri(4z). 9. Prove that if k is even and f(z) has period one and satisfies f(— 114z) = (— 4z 2)42 then fl[y]k = f for all ye To (4). f(z),
10. (a) Prove that 18 (4z)/I 4 (2z)e M2 (FM)), and find its value at each cusp. 67 + 1). (13) For a E Z prove that E2(ST'Sz) = (az + 1)2 E2 (z) — 4(az -214-(E2 (z) — 3E2 (2z) + 2E2(4z)) = Zoddn> 0 ai (n)qn by Problem (c) Let F(z) 8(c). Prove that F(z)E m2(1T0(4)), and find its value at each cusp. (d) Prove that F(z) = ri8 (4z)M 4(2z). Then derive the identify q
(1 n;=.- 1
q4n)4 (
1+
q 2 r74.
E odd
0.
1
(n) q .
n> 0
(e) Give a different proof that —24F(z) = RE2 (z) — E2(z + 1/2)) is in M2 (1-0(4)) by proving that, more generally, E2 (z) — 141,7E7:01 E2 (z + j/N) is in M2 (r0 (N 2)).
11. (a) Prove that 0(z)4 em2 (r0 (4)), and find its value at each cusp. (13) Show that 0(z)4 and F(z) (see preceding problem) are linearly independent. (c) Prove that te 0 (2z)/q 8 (z)r/8 (4z)em 2 (r0 on, and find its value at each cusp. (d) Prove that 0(z) = (e) Prove that 40(z) = 2n1124172 (z +. D/r/(2z). 12. Let N = 7 or 23, and let k = 24/(N + 1). Let x be the Legendre symbol x(n) Prove that any nonzero element of Sk(N, x) must be a constant multiple of (ri(z)ri(Nz)) k 13. Using Propositions 25-27, prove that (r/(z)r/(3z)) 6 eS6 (fo (3)) and (r/(z)n(7z)) 3 E S3 (7, x) where x(n) = 14. Let 4 (z) = En , z e izn2 = 0(z/2). Let x be the unique nontrivial character of 0(2)/ F(2) (which has 2 elements). Show that 04 e M2 (6(2), x). 15. Let fe Mk (N, x), and set ŒN =( 1). (a) Prove that f i[ŒN]k E Mk(N1 3(), and that the mapfl-4.f I kNI is an isomorphism of vector spaces from Mk (N, x) to Aik (N, 7). Prove that the square of this map (i.e., where one uses it to go from M/Or, x) to Mk(N, je) and then again to go , from ./4k (N, k) back to Mk(N7 X)) is the map (-1)k on Mk(Ar7 x)(b) 1f x = X, i.e., if x takes only the values +1, then prove that Mk(N, = Mk+ (N, X) 0 MC (N, X), where Mk ± (N, X) crf ffe,Mk(N, X)Ifi[ŒN]k = In other words, any modular form in Mk(Ne, x) can be written as a sum of one
146
-
Ill. Modular
FOrms
which is fixed under ik [aN]k and one which is taken to its negative by i rL.aN_,k• (c) Let N = 4. In Problem 17(d) below, we shall see that 04 and Fspan M2 (r0 (4)) = M2(4, 1) (where F = En odd ci(n)q n as in Problem 10, and 1 denotes the trivial character in M2 (4, 1)). Assuming this, find the matrix of [a4 ] 2 in the basis 04, F; show that M2+ (4, 1) and M2- (4, 1) are each one-dimensional; and find a basis for M2 (r0 (4)) consisting of eigenforms for r04.1 2 If you normalize these eigenvectors by requiring the coefficient of q in their q-expansions to be 1, then they are uniquely determined. 8 2 : + Enœ, i o-k . . i (n)qn. Express L1(s) (see = — 16. (a) For k 2 even, let f(z) the definition in (3.33)) in terms of the Riemann zeta-function. (1)) Write an Euler product for Lf(s). (c) Let fx(z)= E;f)= 1 (4_ 1 (n)x(n)q" for a Dirichlet character x. Write an Euler product for Li(s).
17. Let F(2) be the fundamental domain for F(2) constructed in §1 (see Fig. 111.3). Then
F' = ŒF(2), where a = (2) 1), is a fundamental domain for F0 (4) = al- (2)a-1 (see Problem 10 in §III.1). The boundary of F' consists of: two vertical lines extending from ( 3 + i0)/4 and from (1 + LA/4 to infinity ; two arcs of circles of radius 1, one centered at and one centered at 0; the arc of the circle of radius -4- and center -4- which extends from 0 to (9 + i.A/28; and the arc of the circle of radius and center A which extends from (9 + LA/28 to 1. Consider F0 (4)-equivalent points on the boundary of F' to be identified. (a) Find all elliptic points in F' (i.e., points which are F-equivalent to i or co = (-1 + 0)14 Which are on the boundary and which are in the interior of F' ? (b) Let f(z) be a nonzero modular function of weight k (keZ even) for 1-0 (4). Let v(f) denote the order of zero or pole of f(z) at the point P. At a cusp P = Œ 1 co, we define v(f) to be the first power of qh with nonzero coefficient in the Fourier expansion of fl[a - lk (where h is the ramification index; see' , Problem 2 above). Prove that: Ep€F, v(f) = k/2, where the summation is over all points in the fundamental domain F', including the three cusps, but taking only one point in a set of F0 (4)-equivalent boundary points (e.g., { + iy, 14- iy} or { + + + (c) Describe the zeros of 0(z)4 and F(z) (see Problems 10-11 above). (d) Prove that 04 and F span Af2 (r0 (4)). (e) Prove that Mk(F0 (4)) = 0 if k < 0, and it contains only the constants if k = 0. (f) Prove that for k = 2,k o a nonnegative even integer, any fEMk (F0 (4)) can be written as a homogeneous polynomial of degree k0 in F and O. (g) Prove that S6 (F0 (4)) is one-dimensional and is spanned by 0 8 F 1604F 2 . (h) Prove that ,j 12(2z) S6 (F0(2)), but that 2 (2z) E S6 (F0(4)). Then conclude that q 12 (22.) = 08 F— 1604F 2 . (i) Prove that for k = 2k0 6, any fe Sk (F0(4)) can be written as a homogeneous polynomial of degree k 0 in F and 04 that is divisible by 04F(04 — 16F). 18. (a) IffE Mk , (N, x i ) and g E Mk2 (N, X2), show that fg e Mk, ÷k2 (N, Xi X2)• (b) Let x be the unique nontrivial character of (/4)*. Show that any element of M1 (4, x) is a constant multiple of 0 2 . (c) With x as in part (b), find a formula for dim Sk (FI (4)) = dim Sk(4, (d) Let f(z) = (ii(z)q(2z)) 8 , and let g(z) = f(2z). Show that S8 (1"1 (4)) is spanned by f and g.
§4. Transformation formula for the theta-function
,
j 47
19. Let I' c r be a congruence subgroup, with r _-= Uai r, so that F' = Ucc.T' F is a fundamental domain for rt. Let j; ---- f I [ocr ' ],, forfe Mk (r). Suppose thatfe Sk (r), the ramification index of r' at the cusp so that fi (z) = Z__, anA, where h si = aj i oo. In particular,f(z) = I nc°,_ i anqii: at the cusp co. (a) Show that there exists a constant C independent off and x such that Ifi(x +
(b) (c) (d) (e) (f)
i y)I Lç CC-2 "mi for y> e.
Let g(z) = (Im z) k12 1f(z)f. Show that g(yz) = Om z)"12 1f(z)l[y] k i for y E F. Show that g(z) ref Om zr 14(2)1 is bounded on F. Show that g(z) is bounded on F'. Show that g(z) is bounded on H. Show that for any fixed y: 1fh im-27rirox+iyvh dx. an = - fi x ±
h o
(g) Show that there exists a constant C1 such that for all y:
Ian ! < ciy-ki2 e21oty/h . (h) Choosing y = 1/n in part (g), show that for C2 = C1 e2irm : lan i < C2 n k12 . (i) Show that la1n -42 is similarly bounded for each j.
§4. Transformation formula for the theta-function We first define some notation. Let d be an odd integer, and let c be any integer. The quadratic residue symbol () is defined in the usual way when d is a (positive) prime number, i.e., it equals 0 if dic, 1 if c is a nonzero quadratic residue modulo d, and — 1 otherwise. We extend this definition to arbitrary odd d as follows. First, if g.c.d.(c, d) > 1, then always (i) = O. Next, if d is positive, we write d as a product of primes d=rlipi (not necessarily distinct), and define (-) = II i (j-i). If d = 4 1 and c = 0, we adopt the i convention that (K) = 1. Finally, if d is negative, then we define (f-i ) = ( t) if c > 0 and () = —(1`6) if c < O. It is easy to check that this quadratic residue symbol is bimultiplicative, i.e., it is multiplicative in c if d is held fixed and multiplicative in d if c is held fixed. It is also periodic with period d.when d is positive : (cP) = (-) if d> O. However, one\ must be careful, because periodicity fails when d is negative and c + d and c have different signs: (q4-) = —() if e> O> c + d. This is because of our convention that ( -s) = — (t'il) when both c and d are negative. On the other hand, this convention ensures that the usual formula (2-) = (— 1) (" )/2 holds whether d is positive or negative. Next, we adopt the convention that „ii: for zE C always denotes the branch whose argument is in the interval ( — n/2, n/2]. We next define Ed for d odd by: -
ser = -Ni(l), i.e.,
148
HI. Modular Forms
11
if d d
1 mod 4; 3 mod 4. •
(4.1)
With these definitions we finally define the "automorphy factor" j(y, z), which depends on y = db) e F0 (4) and z e H:
AY,
crlf
c; 1 NiCZ + d
for
y
(
a b c d)er°(4),
z e H.
(4.2)
Inez q n2 = Inez e Irtizn 2 . Recall our definition of the theta-function 0(z) = The purpose of this section is to prove the following theorem, following Hecke [1944 ].
Theorem. For y eF0(4) and z e H 0 (yz) =j(y, z)0 (z),
(4.3)
where Ay, z) is defined by (4.2). Notice that the square of Ay, z) is ( 1-)(cz d), and so the square of the equality is precisely what we proved in Proposition 30. Thus, for fixed y e F0 (4) the ratio of the two sides of (4.3) is a holomorphic function of z e H whose square is identically 1. Thus, the ratio itself is + 1. The content 1, i.e., that j(y, z) has the right sign. of the theorem is that this ratio is Simple as that sounds, the theorem is by no means trivial to prove. At first, it might seem sensible to proceed as in the proof of Proposition 30, proving that (4.3) holds for generators of F0 (4). However, then we would have to show that the expression j(y, z) in (4.2) has a certain multiplicative property which ensures that, if (4.3) holds for y1 and y2 , then it must hold for Vi y2 . But that is a mess to try to show directly. We shall, in fact, conclude such a property for j(y, z) as a consequence of the theorem (see Problem 3 at the end of this section). In proving the theorem, it turns out to be easier to work with the function 4 (z) def = 0(42) = Ine z ein2z , which we encountered in Problem 14 of §111.3. This function satisfies : 4(T 2z) = 4)(z) (obvious from the definition) and izO(z) (immediate from (3.4)). Hence, 4)(yz) has a transforma0(Sz) = tion rule for any y in the group 0(2) generated by + T 2 , S. It is because 0(2) is such a large group—having only index 3 in F—that it is sometimes easier to work with 4). The corresponding group under which 0(z) = a= ?), has a transformation rule is a- '0(2)a (see Problem 1 of §III.3). But cc -1 4:5(2)crt is not contained in I" = SL2 (7L); its intersection with f is the subgroup F0 (4) of index six in F. Thus, we can work with a "larger" subgroup of F (i.e., its index is smaller) if we work with 4)(z) rather than 0(z). The next lemma gives an equivalent form of the theorem in terms of 4)(z).
Lemma 1. The theorem follows if we prove the following transformation formula for 4)(z):
149
§4. Transformation formula for the theta-function
ck(yz) = i" -c)12- 6 \I — i(cz for y r such that y =(
)
+ d)(1)(z)
(4.4)
(mod 2) with d 0 O.
Suppose that we havé the transformation rule for O(z). We must show that 0 satisfies (4.3) for all y = (ac ' ) e F0 (4). We first note that if c = 0, then (4.3) holds trivially, since in that case 0(yz) = 0(z), while = 1 (since d = +1). So in what follows we suppose that j(y, z) = c O. )E F0 (4) with c Owe write For any y = PROOF.
,
0 (7z) =
n az
a) = (y`( — 1 /2z)), 1/2z) — c/2
+.1 _ /4, (2b(— 112z) —
where y' = (2,7 .42) and y' m- (_? (13) (mod 2), because 41c. We apply (4.4) = (1)(), with y' in place of y and — 1/2z in place of z. Using the fact that we obtain
en
0
(yz) = i( -d)/2 d2) (
— i(d( —
112z)
c12)4)( - 112z).
2i0(z) by (3.4). The product Next, we have 4)( — 1/2z) = 0( — 1/4z) = of the two square root terms is + Vcz + d (note that, because of our convention on the branch of the square root, we have .fx,./j; =± xy; for example, — 1 = — Nib. But since the three functions — i(d( — 112z) — c12), — 2iz, and Vcz + d are all holomorphic on H, the + must be the same for all z; so it suffices to check for any one value of z, say z = i. But in that case „I— 2iz = „/2, and .\/.7-c.f; = +\41)cy always holds when y is positive real. Thus, the product of the two square root terms is Vcz + d, and we have C) (yz) = i( 1 -d)/2
2
_c d d
/c.z +
N
d0(z).
1 --d)/2(-- 2) -= To complete the proof of Lemma 1, it remains to check that i( k d which we easily do by considering the cases d 1, 3, 5, 7 (mod 8). —
t'd
The remainder of this section is devoted to proving (4.4). For a fixed odd prime p, let us denote
= n '(z)/(pz),
(4.5)
where n (z) is the Dedekind eta-function, as in §2 and §3. According to i.e., tfr3 is a modular Propositions 26-27, we have 1/.3 Em3,p_ iofp, form for ro (p) with character x(d) = (9,-). This is the basic tool which will be used to prove (4.4). But it will take several lemmas to relate tir 3 and 0.
Lemma 2. (P(pz)/(z) = 131/(z)/0 2 (z + 2 'I) •
(4.6)
150
III. Modular Forms
By Problem 11(e) in the preceding section, with z replaced by we find that the left side of (4.6) is equal to
PROOF.
by
e -2/ri/24 1,1 2 (Pz2+1 )/n ,
nz
,hfz)
,
V)IrKz)y = e(CP -
(e - 27ri/24 1,1 2(z -
11/24)2irt
-42
and
(p.z2f-1) .:_
02(z12-1) n 2 09 .12-1 ) -
But since q(z + 1) = e224q(z), and p((z + 1)/2) = (pz + 1)/2 + 91, it follows that the last term on the right is e-( P -1)2ni124. This proves (4.6).
Lama 3. Le t - p be an odd prirne, let y = 0 3(2 1 1). Then fl[y] 3(p - 1)12 =
cl D E 0(2) n fo(p), and let f(z)=
().!
Let a =( ), so that Iii((z -I- 1)/2) = tfr(az). Then 0((yz + 1)/2) = kayz) = ig(aya-i )ocz). Now for y as in the lemma, we have PROOF.
ŒyŒ 1 =
+ c + d — a — 012) 2c d c
(Note that b+d a c is divisible by 2 because y e 6(2)) Hence, by Propositions 26-27, we have —
—
yz + 1) = (d_ c)
3 (
2
(2caz + d
—
c) 3( P-1)J2tp 3(otz) z+1 2 )'
.( 4)(cz + d)3(P-1)1203(
because d
—
c
d (mod p ) . This is the relation asserted in the lemma.
Lemma 4. Let p be an odd prime, let y = 4(pz)I0P(z). Then gl[y](l _p)/2
bd)
e 0(2) nTo(P), and let g(z)=
=
We first claim that g8 transforms trivially under y. Let a = (g ?), and let y' = yo -1 . Then both y and y' are in 6(2), and we can use Problem 14 of the preceding section to compute PROOF.
g8(z)i [y1441 p) =
(CZ
± d)4 P-1 )0 8(y'ocz)14) 8P(yz)
_ (Œz ± d)-40 8('Y'az) (cz + d)-41'08 P(yz) _ 0 8(Œz) —
8
8P (Z) = g (z),
as claimed. Meanwhile, the ninth power of g transforms under y by (11,), as we see by raising both sides of (4.6) to the 9th power and using Propositions 26-27 and Lemma 3 (here f(z) is as in Lemma 3):
99 1 [7] 9(1 -p)/2 = (0 91f6 )1[7]9(1-p)/2 = (IA 31 [11(13-1)/2)3 — 4-415)1fr 3)3 — (d)a9 • i) ( V) 6 ( J. I [7] 3( p-1)/2) 6
151
§4 . \Transformation formula for the theta-function
faking the quotient of these two relationships gives
gibli-p)/2- := g9 ibil 9( 1 - p)/2/g 8 I [T] 4( 1 -p)
=
0
p)g .
(
r
The next lemma generalizes Lemma 4 by replacing the prime p by an arbitrary positive odd number n. Lemma 5. Let n be
a positive odd integer, let y ,:--- (a, bd)e 0(2) n ro (n), and let
g(z) = 4)(nz)10n(z). Then gibli-no= ()g PROOF. We write n =pi - • . p, as a product of primes (not necessarily distinct), and we use induction on the number r of prime factors. Lemma 4 is the case r = 1. Now suppose we know Lemma 5 for n; we shall prove the corresponding equality for a product n'rz---- np of r + I primes. We write
0(n'z) _ cknaz) N(pz)Y çbn'(z) — O n(OLZ)0 P(Z) ) 1
(4.7)
where a = (6 ?). For y = (a, be') e Fo (n) n 0(2) we have y' = aya-1 = fan) n 0(2), and so, by the induction assumption,
4)(nocyz)10"(Œyz) = 4)(ny'az)10"(y'az) = (1:7-1 ) (az + d)
(
c 7p bdi E
0(nocz)b4?(ocz)
= (-nd)(cz + d)" -n)I2 0 onezve(ŒZ). In addition, by Lemma 4, we have
d 4)(pyz)10P(yz) =(- )(cz -I- d)(1-P)12 4)(pz)10P(z). P Combining these two relations, we see that replacing z by yz in (4.7) has the effect of multiplying by
. ,Gd)
(CZ
+ d)" -n)12 (()(CZ + d) (1-P)12) n = (1)( 4)(CZ ± n p P
d , (n ')
d) (1- np)I2
(cz + d)"-n1)12.
This completes the induction step, and the proof of the lemma.
0
We are now ready to prove (4.4). We first note that.both sides of (4.4) remain unchanged if y is replaced by —y (see Problem 2 below). Hence, without loss of generality we may suppose that y = (:. bd ) F... (_? D mod 2 with c > O. We now apply Lemma 5 with n replaced by the positive odd integer c,
obtainir
152
III. Modular Forms
(1)0(cC(7YzZ) =
()(CZ
± d ) 1 -c)/2 O(CZ) CV(Z)
(4.8) •
The ratio that ultimately interests us is 4)(yz)10(z). Solving for this in (4.8) gives
(0
(cczz)) . Yzz) )c = ( d c ) ( c z 4_ d) c- 1 0 04)(Y
: (
(4.9)
)
On the other hand, we have (using: ad
1 = bc)
—
cyz . c az + b ,... a cz+d
I cz±d i
Since a is even and 4) has period 2, this means that
Ccyz) = 0 ( cz + 1 d) = V —i(cz + d)(k(cz + d) (4.10) = \I —i(cz + d)0(cz). Combining (4.9) and (4.10) gives 4)(Yz)
0(z)
)c = (--1)(cz + d) — 1 ) /2.\/ — i(cz + d). c
(4.11)
(
Meanwhile, we saw in Problem 14 of the preceding section that 0 8 is invariant under Ed4 for y E (5(2), i.e.,
8k = 0)(v/
(CZ
+
VI'
for any k e Z.
, (4.12)
We now raise both sides of (4.11) to the c-th power and divide by (4.12), where k is chosen so that c2 = 8k + 1. (Since c is odd, of course c2 :1-=._ 1 mod 8.) The result is: (KYz) = (-1-)(cz + d)cc -112-41c(— i(cz + d))(` -l v2V — i(cz + d).
0(z)
c
But c(c — 1)/2 — 4k + (c
—
1)/2 = (c2
—
1)/2 — 4k = O. Hence,
0(yz) = (y_),_ i.,,c_ i ,i2 , . y It v — i(cz c‘ 0(z)
+ d),
which is the transformation formula (4.4) that we wanted to prove. This concludes the proof of the main theorem as well. 0
The transformation formula for the theta-function is similar to the transformation formula for a modular form of weight k if we take k =1, i.e., except for a power of i the "automorphy factor" is (cz + d) 1/2 . In the next chapter we shall see that there is a general theory of modular forms whose weight is a half-integer, and the transformation formula for 0(z) plays a fundamental role in describing such functions.
153 -
§5., The modular interpretation, and Hecke operators
PROBLEMS
,
1. Prove that the generalized quadratic residue symbol (i) as defined in this section satisfies the following form of quadratic reciprocity: if c and d are both odd integers, then
( _ occ-tud-io()
if c or d is positive; (f) = _(...... o(c-i)(d- 1W4() if c and d are negative.
2. Show directly that both sides of (4.4) remain unchanged if y is replaced by — y. 3. (a) Show directly (using (3.4)) that the theorem (4.3) holds for the generators T, and ST'S of 1-0 (4). (b) Show that the theorem would follow from part (a) if one could show that
Ace13, z) = Act, &M g, z) for all a, /3e r0(4).
—
I,
(4.13)
(c) Conversely, show that the theorem proved in this section implies the relation
(4.13).
§5. The modular interpretation, and Hecke operators A basic feature of modular forms is their interpretation as functions on lattices. More precisely, we consider the most important cases of a congruence subgroup F': F' = F, F(N), F 0 (N) or F(N). (Of course, F = fl (1) = F0 (1) = F(1), so everything we say about the cases F 1 (N), F0(N) or F(N) will apply to F if we set N = 1.) By a "modular point" for F' we mean : (j ) for F' = f: a lattice L c C; (ii) for F' = F1 (N): a pair (L, t), where L is a lattice in C, and t e CIL, is a point of exact order N; (iii) for F' = F0(N): a pair (L,- S), where L is a lattice in C, and S c C1L is a cyclic subgroup of order N, i.e., S = Zt for some point te CIL, of exact order N. _ (iv) for f' = F(N): a pair (L, { t 1 , t2 }), where 1 1 , 1 2 e CIL have the property that every tekLIL is of the form t = mi l ± nt 2 , i.e., t i , t 2 form a basis for the Points of order N (in particular, t i and t must each have exact, , order N). Given a lattice L, in general there will be several modular poir.. s of the form (L, t), (L, S), or (L, {l i , 1 2 )). However, when N = 1, there is only one modular point corresponding to each L, and we identify it with the modular point L for F. Let k e Z. In each case (i)-(iv), we consider complex-valued functions F on the set of modular points which are of "weight k" in the following sense. If we scale a modular point by a nonzero complex number A, then the value of F changes by a factor of A - k . That is, for 2 e C* we consider AL =
154
III.
P.111 e LI, /it e C , S = Pik ES} c weight k if for all E C *
Modular Forms
AL. Then F is defined to be of
case (i) F(2L) = A'F(L) for all modular points L; = ATY(L, t) for all modular points (L, 1); case (ii) case (hi) F(2L, /IS) = /1-1`F(L, S) for all modular points (L, S); ) t2 }) = .1-kF(L, t 1 , t2 }) for all modular points case (iv) F(AL, (L, { t 1 , t2}). An example of such a function of weight k is Gk (L) =
E
(k > 2 even).
(5.1)
0*IeL
Notice that any function F in case (0, such as Gk, automatically gives a function for the other groups; for example, by setting F(L, t) = F(L). Given a function F of weight k, we define two corresponding functions P and f as follows. fl(co) is a complex-valued function on column vectors = ( T) such that co 1 ko 2 E H; f(z) is a function on the upper, half-plane H. Let 2L. be the lattice spanned by co l and co2 , and let L, be the lattice spanned by -z and 1. Given F as above, we define case (i) P(c )) = F(L); case (ii) P(w) = F(L., (o 2IN); case (iii) P(co) = F(L., Zco 21 N); case (iv) P(co) = F(L., {w 1 /N, w 2/N}). In all cases we define f(z) = P(f). Thus, for example, the function f(z) that corresponds to Gk (L) (see (5.1)) is the Eisenstein series we denoted Gk (z) in §2 (see (2.5)). For y = (ca e F = SL 2 (Z), we define the action of y on functions of co Rya)), where yoi is the usual multiplication of a column by the rule 'AO vector by a matrix.
r, r,
(N), FAN) or f(N). The above Proposition 31. Let k E Z, and let r = association of F with P and f gives a one-to-one correspondence between the following sets of complex-valued functions: (1) F on modular points which
have weight k; (2) P on column vectors co which are invariant under y for y E r and satisfy 1-7(.1co)= .1.-k .P(w); (3) f on H which are invariant under [Y]k for y e
PROOF. We shall treat case (ii), and leave the other cases as exercises. Suppose y E 1"1 (N) and F is a weight k function of modular points (L, We first compute: P4- (yco) = F (1,
cco ± cico 2 -
because Law1+bc02.cw1+do)2 = L ) mod L (since y E F1 (N)). We also have
= F(L., w2/N) Aco), y e F) and (co), + dco 2)/.N
co2/N
§5._ The modular interpretation, and Hecke operators
155
-
-
\
P(co) =. F(LAw , Aco 2 IN) = /1 -4 F(L., co 2IN) = Next, we have ^
f(yz)= F (L (az+b)1(cz+d) , ,
I
Tv
)
= (ez ±
clY c F (Laz+b,cz+d,
ez + d N
)1
because F has weight k. But the lattice spanned by az + b and cz + d is Lz ; and (cz + d)I N.=..--- k mod Lz ; hence
f(yz) =--- (cz + d)k F (L z ,
NI) = (cz ± d)kf(z).
-
Thus, the P and feorresponding to F have the properties claimed. To show the correspondence in the other direction, given f(z) we define P(co) to be coi kfico //co 2); and given P we define F(L, t) to be Aco), where co is chosen to be any basis of L such that co2/N a t mod L. One must first check that the definition of F makes sense (i.e., that such a basis co exists), and that the definition of F is independent of the choice of such a basis co. The first point is routine, using the fact that t has exact order N in C/L, and the second point follows immediately because any other such basis must be of the form yco with y e F1 (N). It is also easy to check that, oncefis invariant under [y]k for y e F1 (N), it follows that F is invariant under y and has weight k; and that, ifFhas weight k, then so does the corresponding F. The construction going from F to P to f and the construction going from f to P to F are clearly inverse to one another. This concludes the proof. o We say that F is a modular function/ modular form/ cusp form if the corresponding f is a modular function/ modular form/ cusp form as defined in §3. We now discuss the Hecke operators acting on modular forms of weight k for r1 (N). We could define them directly on f(z) E Mk (1"1 (N)). However, the definition appears more natural when given in terms of the corresponding functions F on modular points. Let ..r denote the 0-vector space of formal finite linear combinations of modular points, i.e., 2' = OCIeL is the direct sum of infinitely many one-dimensional spaces, one for each pair (L, t), where L is any lattice in C and t E C1L is any point of exact order N. A linear map T: 2' --0 2' can be given by describing the image TeL,, = E. an ep.n of each basis element; here {P.} are a finite set of modular points. For each positive integer n we define a linear map Tn : 2) --* ..r by the following formula giving the image of the basis vector eL , t : 1 Tri (eL, r) = — 4'5' ev,t, n
(5.2)
where the summation is over all lattices L' containing L with index n such that (L', t) is a modular point. (Here for t E Ca, we still use the letter t to /
156
Ill. Modular Forms
denote the image of t modulo the larger lattice L'.) In other words, L' IL C CIL is a subgroup of order n, and t must have exact order N modulo the larger lattice L' as well as modulo L. The latter condition means that the only multiples Zt which are in L' IL are the multiples INt which are in L. = I", this condition disappears, and we sum In the case N = 1, i.e., = n. The condition on t is also empty if over all lattices L' with [L' : n and N have no common factor. To see this, suppose that g.c.d.(n, N)= 1, and suppose that N't c L'. Then the order of N't in L'IL divides N (because N'Nt E L) and divides n (because # L'IL=n), and so divides g.c.d.(N,n)=1. Thus N't E L. Notice that the sum in (5.2) is finite, since any lattice L' in the sum must because each element of L' IL has order be contained in L = dividing n = # L' IL. Thus, each L' in the sum corresponds to a subgroup of order n in iLIL ,-z-!,(11nZ) 2 . Note that T1 = 1 = the identity map. Next, for any positive integer n prime to N we define another linear map T7, „ :2—*Y by
1 Tn,n (eL,t ) = --ye( 1/n)L,t•
(5.3)
Note that t has exact order N modulo L because g.c.d.(N, n) = 1. Again we are using the same letter t to denote an element in C/L, and the corresponding element in CAL. It is easy to check the commutativity of the operators ,
Tni , ni Tn2,n2
Tnin2,nin2 = T,g2,„2 Tni ,ni
Tn,n Tm = Tni Tn, n .
(5.4)
It is also true, but not quite so trivial to prove, that the T.'s commute with one another for different m's. This will follow from the next proposition. Proposition 32. (a) If g.c.d.(m, n) = 1, then T„,„= T.T, 7 ; in particular, 2",,, and Ti, commute. (to) Ifp is a prime dividing N, then Tpt = 7. (c) If p is a prime not dividing N, then for 1> 2
Ti = Tpi--1Tp — pTp1- 2l
,
.
(5.5)
PROOF. (a) In the sum (5.2) for Tmn , the L' correspond to certain subgroups S' of order mn in namely, those which have trivial intersection . with the subgroup It CIL. Since g.c.d.(m, n)= 1, it follows that any such S' has a unique subgroup S" of order n; if L" L is the lattice corresponding to S", then S'/S" gives a subgroup of order m in ,-1,-,- L"/L". Both S" and S'/S" have nontrivial intersection with Zt. Conversely, given S"=L"ILc -1-iLIL of order n and a subgroup S' =L'IL",c kL"/L" of order m, where both subgroups have trivial intersection with It, we have a unique subgroup L'IL c ALIL of order mn with nontrivial intersection with Zt. This shows
157
§5. The modular interpretation, and Hecke operators
thatthe modular points that occur in Tm7,(eL ,,)= ï E eL ,,, and in Tm (Tn(e L. )) =-1-ITm (er ,,,) are the same. (b) By induction, it suffices to show that Tpi-i Tp = Tpt for I> 2. Let t' =It. Then Tp /(eL , t) = p -' lev ,„ where the summation is over all L' L LIL has order p i and does not contain t'. Notice that such that L' IL c L' IL must be cyclic, since otherwise it would contain a (p, p)-subgroup of pT iLIL. There is only one such (p, p)-subgroup, namely f,L1L, and t' E since pt' = Ni e L. Once we know that L' IL must be cyclic, we can use the same argument as in part (a). Namely, for each L' that occurs in the sum for Tp r(e L ,,) there is a unique cyclic subgroup of order pin L' IL; the corresponding lattice L" occurs in the sum for Tp (e"), and L' is one of the lattices that occur in 7;i -i(eL „,,). This shows the equality in part (b). (c) Since p,t N, the condition about the order of t in C/L' is always fulfilled. We have 1,i -iTp (eL,,)= p -1 IL ,, EL, e L,.„ where the first summation is over all lattices LI' such that S" = L" IL has order p, and the second summation is over all L' such that S' = L' IL" has orderp'. On the other hand, Tpt (eL. ,) = poe 1 EL, eL .,,, where the summation is over all L' such that L' IL has order IA Clearly, every L' in the inner sum for 1p r--1 Tp is an L' of the form in the sum for Tpi, and every L' in the latter sum is an L' of the form in the former sum. But we must count how many different pairs L", L' in the double sum lead to the same L'. First, if L' IL is cyclic, then there is only one possible L". But if L'1L is not cyclic, i.e., if L'IL D g1L, then L" can be an arbitrary lattice such that L" IL has order p. Since there are p 1 such lattices (for example, they are in one-to-one correspondence with the points on the projective line over the field of p elements), it follows that there are p extra times that eL ,,, occurs in the double sum for Tpt-iTp . Thus,
Tpl(eL , t) = 7p l-i 7p (e 1 , 1)
p
,t• L'(1/p)L
[L':( lip)Li=pI- 2
But
1 T 1-2T (e )= — 2 TP I-2e (1/p)L, t =p 14,t P, P
[L':( 1/P)Li
e
t
This concludes the proof of part (c). If n = prŒr is the prime factorization of the positive integer n, then Proposition 32(a) says that 71 = Tpf, Tpft . Then parts (b)-(c) show that each Tpcp is a polynomial in Tpi and Tpi , pi . It is easy to see from this and (5.4) that all of the Yin's commute with each other. Thus, the operators T,,, (n a positive integer prime to N) and Tm (m Any positive integer) generate a commutative algebra AP of linear maps from 2 to 2; actually, eit9 is generated by the Tp , p (p,I1 N a prime) and the Tp (p any prime). There is an elegant way to summarize the relations in Proposition 32 as formal power series identities, where the coefficients of the power series
III. Modular Forms
158 J
are elements in follows:
Ye.
First, for p
v.:el
IN,
Tp 1X 1 =--
we can restate Proposition 32(b) as
1 I — TX
PIN'
(5.6)
i.e., (I TpIX 1)(1 — Tp X) = 1 in le[[X]]. This follows from Proposition 32(b) because, equating coefficients, we see that the coefficient of X' is — Tri-iTp . Similarly, for p,I 'N, part (c) of Proposition 32 is equivalent to the identity 1 pN, (5.7) i TplX 1 = Tp X + pTp, pil: ' 1 i.--, 0
7;,X + pTp, p X 2 and equate i.e., if we multiply both sides of (5.7) by 1 coefficients of powers of X, we see that (5.7) is equivalent to the equalities —
Tp — Tp = 0,
Ti = 1,
Tpt — Tpr-iTp + pTpi-2 T„,p for / > 2.
To incorporate part (a) of Proposition 32, we introduce a new variable s by putting X = p - s for each p in (5.6) or (5.7). We then take the product of (5.6) over p with pIN and (5.7) over all p with p,11 N: co I 7 1-2s I -H fI E p1P -Is = fl piN 1 — TpP s p-f1■1 1 — TpP -s + Tp,pP all p 1 ,-- 0 But, by part (a) of the proposition, when we multiply together the sums on the left in this equality, we obtain / To', where the sum is over all positive integers n. The proof is exactly like the proof of the Euler product for the Riemann zeta-function. We use the factorization n = p71 • • • A.'', and the « p r-ars). We hence conclude that relation: T„n -3 = (Tpli p -l Œts) • - • (T Pr' oo
To' = n=1
ri piN 1 —
1 11 I E 12s• - • T ,PP TpP s pp' 1 — TpP - +P
(5.8)
For d an integer prime to N, let [d]: Y -÷ 2 be the linear map defined on basis elements by Me ', , =-- eL di • Note that di has exact order N in C/L, because g.c.d.(d, N) =--- 1. A lso nc;te that [d] depends only on d modulo N, i.e., we have an action of the group (Z/ NZ)* We now consider function's F on modular points and the corresponding functions f(z) on H. Again we suppose that we are in the case I' = 1"1 (N). If T: 21 ---* 21 is a linear map given on basis elements by equations of the form T(e La ) = 1 an e pa , then we have a corresponding linear map (which we also denote T) on the vector space of complex-valued functions on modular points: TAL, t) =EaF(P). For example,
[d]F(L, t) ------ F(L, dt) T„,„F(L, t) = n -20, t) n T„F(L, t) = 1 E F(L1 , t), n L,
(here g.c.d.(d, N) = 1);
(g.c.d.(n, N) = 1);
(5.9)
159
§5. The modular interpretation, and Hecke operators
where the last summation is over all modular points (L', t) such that [L' = n, as in (5.2). Proposition 33. Suppose that F(L, t) corresponds to a function f(z) on H which is in Mk (Ti (N)). Then [d]F, TF, and TF also correspond to functions (denoted [d]f, Tf, and Tnf) in Mk (ri (N)). If f is a cusp form, then so are [d]f, Tn.nf, and Tnf. Thus, [d], 7, and Tn may be regarded as linear maps on mk(ri (N)) or on sk(ri (N)). In this situation, let x be a Dirichlet character modulo N. Then Je MAN, x) if and only if [d]F = x(d)F, i.e., if and only if
F(L, dt) = x(d)F(L, t) for de (ZI NZ)* .
(5.10)
PROOF. To show that [d]f, T„, „And Tjare invariant under [y] k for y e F1 (N), by Proposition 31 it suffices to show that [d]F(L, t), TF(L, t), and T„F(L, t) have weight k, i.e., [d]F(AL, At) =--- A -k [d]F(L, 1), TF(AL, At) =--A -k T„,nF(L, t), and TnF(AL, At) = A-k TnF(L, t); but this all follows trivially from the definitions. We next check the condition at the cusps. Note that if a = e GL (Q) and iff(z) corresponds to F(L, t), then f(z)l[a] k = (det Œ) 2 (CZ d) -k F (L„z , -110
= (det 00'12 F (1-,..z+b,cz+d, cz In particular, if ŒE F, then this equals F(Lz , (cz d)/N). Next, for each de(ZINZ)*, choose a fixed ad e T such that ad (cr) )mod N. (This is possible, because g.c.d.(d, N) = 1, and the map F SL 2 (7L/N7L) is surjective, by Problem 2 of §1II.1.) We then have .f(z) I [act]k
F (45 / c-v )= [d] F
(L
Z
[d]f(z),
i.e., [d]f = fif ad_ik• 1 Thus, for 70 E F we check (3.8) for [d]f as follows: [d],f i[Ydk = f I [6dYdk , which has a q-expansion of the required type, i.e., [d]fsatisfies (3.8) iffdoes. Similarly, we find that 7,J(z) = n-2 F(L,L,, h) = nk -2 F(Lz , ;v2-) = n k- 2 [I ] f(z) , so this case has already been covered. We next consider the cusp condition for T„f(z), which is a sum of functions of the form 1,7), where L' is a lattice containing L 2 with index n. We take such an L' and let (co l c02) be a basis for L'. Since Lz L' with index n, there is a matrix t with integral entries and determinant n such that (7) = TO) (co denotes the column vector with entries co l and c02). We can choose a set T of Such t (independently of z) such that
Tf(z) =
E F(L., ( z ) n t€ , 9 N
We now consider each F(L., h), where (I) = t(: 21). We find a yer such that yr 71 ,1,(g ',7,) with zero lower-left entry and a, b, d integers with ad = n
III. Modular Forms
160 -
(it is an easy exercise to see that this can be done). The lattice spanned by CO = T -1 (7) is the same as that spanned by y() = -,vg ",)(7). Thus, a
F(Lto I ) = F (I,( az+ bjln,d In , -9 N = ci c F (L(az + b )1d , 1 , -N'— N = ak [a] f((az + 1))1d). But if f(z) satisfies (3.8), then so does f((az + 12)1d), and [a] also preserves (3.8), as shown above. This proves the cusp condition for Tnf. Finally, let y d E To (N) be any element with lower-right entry d. As shown above, f(z)l[yj k = F(Lz , kJ) = [d]F(Lz , Ntf). Thus, f INk corresponds to [diF, and so ,f,ifL.,vd_iI k T his completes 0 the proof of the proposition. We saw before (Proposition 28) that a function Je Mk(Ti (N)) can be written as a sum of functions in Mk (N, x) for different Dirichlet characters x. Thus, using the one-to-one correspondence in Proposition 31, we can write a modular form F(L, t) as a direct sum of F's which satisfy (5.10) for various x. Proposition 34. The operators 7,, and 7 commute with [d], and preserve the space of F(L, t) of weight k which satisfy (5.10). If F(L, t) has weight k and
satisfies-(5.10), then T„F = nk-2 x(n)F.
PROOF. That the operators commute follows directly from the definitions.
Next, if [d]F = x(d)F, it follows that [d] TF = T[d] F = x(d)TF and [d] TF = T[d]F = x(d)7,F. This is just the simple fact from linear algebra that the eigenspace for an operator [d] with a given eigenvalue is preserved under any operator which commutes with [d]. Finally, if F(L, t) satisfies (5.10), then 1;,,nF(L, t) = n-2 F(L, t) = nk-2 F(L, nt) =
0
nk-2 [n]F(L, 0 = nk-2 X(n)F(1-1 1).
If we translate the action of Tn , T. „ and [d] from functions F(L, t) to functions f(z) on H, then Proposition 34 becomes Proposition 35. Tn and T„„ preserve Alk(N, X), and also Sk (N, x). For 4 f€ Mk (N, X) the action of T„,:, is given by T„,„f = nk-2 x(n)fs Proposition 36. The operators 7
Mk (N, x) satisfy the formal power series
identity a,.
E nr,-- 1
Tun - s = jj 0 - Tpp -s + x(p)p k-i — 2s‘)—1 .
(5.11)
all p
PROOF. We simply use (5.8) and observe that for
p)( N we have Tp, pf = p k-2 x(p)f, while for .1)1 N the term on the right in (5.11) becomes (1 — Tpp -s)l because x(p) = 0. 0
161
§5. The modular interpretation, and Hecke operators
As a special case of these propositions, suppose we take x to be the trivial character of (ZIN1)*. Then Mk (N, -x) = Mk(Fo (N)), and the modular forms fe Mk(N, x) correspond to F(L, 1) on which [d] acts trivially, i.e., F(L, = F(L, dt) for all de (ZINZ)*. Such F(L, 1) are in one-to-one correspondence with functions F(L, S), where S is a cyclic subgroup of order N in C/L. Nainely, choose t to be any generator of S, and set F(L, S) = F(L, t). The fact that F(L, t) = F(L, dt) means that it makes no difference which generator of S is chosen. Conversely, given F(L, S), define F(L, t) = F(L, S1), where = Zt is the subgroup of C/L generated by t. So we have just verified that functions of modular points in the sense of case (iii) at the beginning of this section correspond to modular forms for ro (N). We now examine the effect of the Hecke operator Tm on the q-expansion at oo of a modular form f(z) E Mk(N, x). That is, if we write f(z) = E an q" and Tmf(z) = Zbq, q = e'r iz, we want to express b„ in terms of the an . We first introduce some notation. IffE C [Pin f Za n e, we define ;
ben
;
Um f=
(5.12)
an gnim,
where the latter summation is only over n divisible by m. Note that U1 = identity, and Um o J' the identity, while V„,o Un, is the map on power series which deletes all terms with n not divisible by ni. Suppose that f(z) = E anqn, q = e2niz, converges for ze H. Then we clearly have:
Vmf(z) = f(mz);
Umf(z) =
Proposition 37. Let f(z) = E,T=0 anqn , q = e 2 E,T_.. 0 kg". Then
bn= apn
X(P)P
m -1
m i=0
f
+j m
E).
(5.13)
, fe Mk (N, x), and let 7,f(z)
k-1. a
nip9
(5.14)
where we take x(p) = 0 i piN and we take anip = O fn is not divisible by p. In other words,
Tp =
x(p)p k-1 Vp on
Mk (N, x).
(5.15)
We have Tpf(z) = F(L' , k), where F is the function on modular points which corresponds to f and the sum is over all lattices L' containing Lz with index p such that ki has order N modulo L'. Such L' are contained in the lattice -,15Lz generated by 1, and +„ and the lattices of index p are in one-toone correspondence with the projective line Np over the field of p elements Fp = ZIpl. Namely, the point in 1131-p with homogeneous coordinates (a 1 , a2 ) corresponds to the lattice generated by Lz and (a l z a 2)/p. Thus, there are p + 1 possible L' corresponding to (1, j) for j = 0, 1, . . . , p — 1 and (0, 1). If ON, then all p ± 1 of these lattices L' are included; if piN, then the last lattice (generated by Lz and must be omitted, since k has order 1, in that case. Note that the lattice generated by Lz and (z Alp is L(z+i)/p . Thus, if pi N we have 7,f(z) = t, h) = Upf(z). If PROOF.
162
III. Modular Forms
ON, then we have the same sum plus one additional term corresponding
to the lattice generated by
s pz . Thus, in that case 4 and 1,-; this lattice is f,L
p:N) 1) = Upf(z) + plc' F (Lpz, N
Tpf(z) = Upf(z) + ;I'01 , pz,
= Upf(z) + p k-1 x(p)F (L pz , Tv1 -) = Upf(z) + pk-1 x(p)f(pz),
which is (5.15). The expression (5.14) for the bn follows directly from (5.15) 0 if we use the expressions (5.12) for the operators Vp and U. As a consequence of Proposition 37 we have the factorization
1 — 7X+ x(p)p k-1 X 2 = (1 — UpX)(1 — x(p)p k-1 VpX),
(5.16)
where both sides are regarded as polynomials in the variable X whose coefficients are in the algebra of operators on the subspace of C [[q]] formed by the q-expansions of elements f(z) E MOT, X). To see (5.16), note that the equality of coefficients of X is precisely (5.15), while the coefficients of X' agree because U po Vp = 1. Proposition
38.
We have the following formal identity : co
co
E
ru n — s
n=1
= E x (n)nk —i
i Un).
n=1
or, equivalently,
Tn =
(5.17)
n=1
E x(d)d' Vd
(5.18)
0 Unid.
din
PROOF. By (5.11) and (5.16) we find that the left side of (5.17) is equal to
Fl ((1 — upp — s)(1 — x(p)pk -1 VpP —s
)) -1 •
P
Since the Up and Vp do not commute, we must be careful about the order of the factors. Moving the inverse operation inside the outer parentheses, we reverse the order, obtaining
ri (0 -
x(p)pk - i vpp - syi (1
- Up').
P
Note that Up , and Vp2 do commute for p 1 0 p2 , as follows immediately from (5.12). This enables us to move all of the (1 — U p - s)_l terms to the right past any (1 — X(P2)Pr i Vp,P2-s) -1 term for /32 0 p. This gives us separate products with the Vp's and with the Up's :
1
1
Il l - x(p)pk -1 Vpp - s n 1 - Upp - s: p
p
We now expand each term in a geometric series and use the fact that V. = Vn, o Vn and U. = Una o U. The result is (5.17). 0
163
§5. the modular interpretation, and Hecke operators
Proposition 39. Under the conditions of Proposition 37, if Tmf(z) = E(:,°_, 0 b,,qn,
then
x(d)dk-i
Z
bn =
a mnid 2
(5.19)
'
dig-c.c1-(m,n)
PROOF. According to
(5.18), we have
Tm Z (lair = n= 0
,
E x(d)d k-1 Vd o Umid E an q n n= 0
dim
= E x(d)d" 1 E dim
mldfn
a qd2 n
ron .
If we set r = d2 n1m, the inner sum becomes I arm1d2gr with the sum taken over all r divisible by d. Replacing r by n and gathering together coefficients o of q' , we obtain the expression (5.19) for the n-th coefficient. Most of the most important examples of modular forms turn out to be eigenvectors ("eigenforms") for the action of all of the Tm on the given space of modular forms. Iff E Mk(N, z) is such an eigenform, then we can conclude a lot of information about its q-expansion coefficients.
is an eigenform for all of the operators Tm with eigenvalues Am , m = 1, 2, . . . : Tmf = Anil'. Let am be the q-expansion coefficients: f(z) =Em",, o am qm. Then am = Amai for m = 1, 2, . ... In addition, al 0 0 unless k = 0 and f is a constant function. Finally, if ao 0 0, then A m is given by the formula Proposition 40. Suppose that f(z) E Alk (N,
A m ------
)()
E x(d)d'.
(5.20)
dim
Using (5.19) with n = 1, we find that the coefficient of the fi rst power of q in Tmf is am . If Tmf = Am!, then this coefficient is also equal to A m a 1 . This proves the first assertion. If we had a l = 0, then it would follow that all am = 0, and f would be a constant. Finally, suppose that a() 0 O. If we compare the constant terms in Tmf = 2.f and use (5.19) with n = 0, we obtain: A m ao = 1)0 = Eon x(d)d' i ao . Dividing by a() gives the formula for PROOF.
0
Am .
Iff is an eigenform as in Proposition 40 (with k 0), then we can multiply it by a suitable constant to get the coefficient of g‘ equal to 1: a l = 1. Such an eigenform is called "normalized". In that casé, Proposition 40 tells us that am = Am is simply the eigenvalue of Tm . If we then apply the operator identity (5.11) to the eigenform f, we obtain identities for the g-expansion coefficients of f. Namely, applying both sides of (5.11) to a normalized eigenform fE Mk(N, x), we have: CO
E an n=1
= fio _ app_s +x(p)?,,,sy i
•
III. Modular -Forms
164
1. Let N = 1, and let k > 4 be an even integer. Suppose that is an eigenform for all of the T., and suppose that a o 0 0 f= auir G Mk and f is normalized (i.e., a l = 1). Then Proposition 40 says that f= (10 + Eo-k _ i (n)qn. But recall the Eisenstein series Ek = 1 — Lik-1(1)qn (see (2.11)). Thus, f and --IkEk are two elements of Mk (r) which differ by a constant. Since there are no nonzero constants in Mk (F), they must be equal, i.e., a o =— Bk/2k. Therefore, up to a constant factor, any weight k eigenform for r which is not a cusp form (i.e., a o 0 0) must be Ek. Conversely, it is not hard to show that Ek is actually an eigenform for all of the Tm . This can be done using the q-expansion and (5.14) (it suffices to show that it is an eigenform for the Tp , since any Tm is a polynomial in the operators Tp for p prime). Another method is to use the original definition of Tm on modular points, applied to Gk (L) = - o*rel.r k . 2. Let N = 1, k = 12. Since S12 (1") is one-dimensional, spanned by (2n) -12 A(z) = E;;°=1 t(n)q" (see Propositions 9(d) and 15), and since the L preserve Sk (F), it follows that f= Et(n)q" is an eigenform for L. It is normalized, since t(1) = 1. Then Proposition 40 says that Tmf = t(m)f for all in. Thus, the relation (5.11) applied to f gives co 1 (5.21) -c(n)n - s = sp11—2s• EXAMPLES.
(r)
E
allp 1 — t(P)P—
n=1
This Euler product for the Mellin transform off = E t(n)q' is equivalent to the sequence of identities (see Proposition 32):
t(mn) = t(m)t(n)
for in, n relatively prime;
T(P i) = t(p 1 )c(p) P 1 'T(P 1-2 )Ramanujan conjectured that in the denominator of (5.21) the quadratic 1 — t(p)X p' X 2 (where X = p -s) has complex conjugate reciprocal roots ap and a-14p ; equivalently, 1 — t(p)X pll X 2 = (1 — ap X)(1 — FipX), i.e., t(p) = ap + p with lap ! = p l 112 From this it is easy to conclude that IT(p)i 2/3' 1/2 , and, more generally, IT(n)! < 0- 0 (n)n 1112 (see the proof of Proposition 13 in §II.6). The Ramanujan conjecture and its generalization, the Ramanujan-Petersson conjecture for cusp forms which are eigenforms for the Hecke operators, were proved by Deligne as a consequence of the Weil conjectures (see [Katz 1976a]). The Euler product (5.21) is reminiscent of the Hasse-Weil L-series for elliptic curves. For more information on such connections, see [S4imura 1971]. 3. Let N = 4, z(n) = (4) = (— 1)(11-1)12 for n odd. We saw that Mi (N, x) is one-dimensional and is spanned by .0 2 . If we apply Proposition 40 to e2 = + q + • - • + 21,qm + • • • 9 we find that Am = (-L -;1 ), where the sum is over odd d dividing m. For example,
Ap 1
,—1
-r
—
p
)
10 ifp 3 (mod 4); 2 if pE:-. 1 (mod 4).
(5.22)
165
§5. The modular interpretation, and Hecke operators
Since 2„, is one-fourth the number of ways m can be written as a sum of two squares, we have recovered the well-known fact that p can be so expressed in eight ways if p 1 (mod 4) and in no way if p E-: 3 (mod 4). That fact is usually proved using factorization in the Gaussian integers Z[i]. The Gaussian field 0(i) has discriminant 4, and corresponds to the character x of Z having conductor 4: x (n) = 6-,1). Let 1 denote the trivial character of (Z/41)*, and let p denote the two-dimensional representation ( ) of (Z/4Z)*. Then we can write (5.22) in the form A p = Tr p(p). This turns out to be a very special case of a general fact: a normalized weight-one eigenform in M l (N, x) has its q-expansion coefficients determined by the trace of a certain two-dimensional representation of a certain galois group. For more information about this, see [Deligne and Serre 1974].
Another approach to Hecke operators. Let F1 and some group G.
F2
be two subgroups of
Definition. F1 and F2 are said to be "commensurable" if their intersection has finite index in each group: [F1 F/ n F2] < 00 , [F2 : F1 n F2] < 00 . Let r be a congruence subgroup of F SL 2 (1), and let a e G = G.L4 - (0). Then r and a' F'a are commensurable. To see this, suppose r F (N ). By Lemma 1 in the proof of Proposition 17, we have r(ND) for some D. Let F" = a-l r'a F(ND) and F' n ara F' n cc Pa. Then F" r(ND), otr"oc— i r (ND). Thus, [r F"] [F: ['WD)] < cc, and also [a' F'a : F"] = [F' : aF"a -1 ] [F: F(ND)]
BASIC EXAMPLE.
, S +) • A"(N, V , S +)\c A"(N, V , S +),
(5.24)
and that A l (N, S' , S +) is a group. A l (N, 5' , S +) is clearly a congruence subgroup of I", since it contains r(N/), where N' is the least common multiple of M and N (recall S+ = MZ). Here are our familiar examples:
ro (N ) = A l (N, (7L1 NZ)* , Z);
ri (N) = A l (N , {1 , Z); }
I" (N) = A 1 (N, {1} , NZ). Definition. Let r be a congruence subgroup of r, and let a e GL I (0). Let 1"" = r n a -1 1-"a, and let d z---.- [f': r"], I"' = U1=1 1""yj. Let f(z) be a function on H which is invariant under [y]k for y e r. Then , f(Z)
Proposition
d
I Crlarik ,-;-,. E f(z)f[aynk.
(5.25)
i---1
42. f(z)l[rall k does not change if a is replaced by any other
representative a' of the same double coset : Fair = Far. Nor does it depend on the choice of representatives y.; of r/ modulo r". If Je mk (r), then fi[railk e
mk(n.
PROOF. We first prove the second assertion, that is, that
(5.25) is unchanged
if y.; is replaced by ylyi, where 7; e r". Since r" a-l ra, we have 7; = a-l .pia for some -,i e r. Then f l[ay; y:]k = flP jayAk =f[ŒY5]k, because f IDA, =f. Next,. we observe that, by Proposition 41, the sum on the right in (5.25) can be written Ef(z)I[ai] k , where- the ai are any elements such that rar = U rai . It is then immediate that the definition depends only on the double coset Far and not on the choice of representative a. Finally, suppose that fe Mk (r). If y e r', then (f f[rocrlk)([y]k = Ef lEotyAk = f [Par ], since right multiplication by y just rearranges the right cosets ray]. If f
I
167
§5. The modular interpretation, and Hecke operators
satisfies the cusp conditions, then so does fl[ayA k for each j, by Lemma 2 in the proof of Proposition 17. This completes the proof of Proposition 42. CI
Definition. Let r = A 1 (N, S", S+ ), and let n be a positive integer. Let fe Mk (r). Then
7f
(k12)-1
Ef I [rank,
(5.26)
where the sum is over all double cosets of F' in An(N, Sx , S t ). By Proposition 42, we have Tnfe mk (r). _ Equivalently, we can define Tn f
= n(k/2)-1 E
.,...k, fi[al where Pai runs through the right cosets of ri in An(N, S x , S). Proposition 43. In the case
r = ri ( N), the definition
(5.27)
(5.26) agrees with our
earlier definition of the Hecke operators T n . PROOF. Let A" = An (N , {1}, z). For each a e (Z1 NZ)* we fix a
GF
such that
Lemma. (5.28)
A" = U fi (N)øa (" disjoint
0 dt
where the disjoint union is taken over all positive a dividing n and prime to N, and for each such a we set d = nla and take b = 0, 1, .. . , d — 1. The terms on the right in (5.28) are clearly contained in A. Suppose the union were not disjoint. Then for some a', b', d' = nid we would have Yi aa (g :11) = y2 o-„,q; ct;:), and so F would contain the matrix (g byl; ':,:y1= (a/81' 4,); then a' = a, d' = d, and so (g ,'0,) = ab('; db :) for some j, i.e., b = b' + jd. If 0 < b, b' < d, this means that b = b'. To prove the lemma, it remains to show that any a = (ae: n e A" is in one of the terms on the right in (5.28). Choose g, h relatively prime so that go' + hc' =- 0, and complete the row g, h to a matrix y = ge il; ) e F. Then ya has determinant n and is of the form (8' :). Replacing y by +(13,. by if necessary, without loss of generality we may suppose that ya = (g ) with a> 0, ad = n, 0 b < d. Then, considered modulo N, the equality a = y -1 (g bd ) gives (1) ,7*) r_.=_- ( 1,!; :)(g 1,;), i.e., y-1 is of the form Yi aa for some yi e 1"1 (M. Thus, a = y1 a,,(0` bd) e r-i1 (N)cra(8 !,), as desired. This completes the proof of the lemma. o PROOF OF LEMMA.
(
,We now prove the proposition. Let Tr be the linear map, on Mk (Fi (N)) defined by (5.26) (or (5.27)), and let Tir d be the earlier definition. Since Aik(Fi (N)) = C i x Mit(N , x), it suffices to show that T:ew = T:k1 on Mk (N, z).
168
Iii.
Modular Forms
Let Je Mk (N, x). By the lemma, we have T: ewf = w k/2)-1 Ef i pia (g bA k. Since f Mk (N, x) we have f I [aa ],, = x(a)f. Also, for each a> 0 and d ad = n we have d-1
E f(z)IN
b=0
bdnk
with
= 11 nki2 cl-kf(az d± 1 b=0 = n ki2 ai-k Va 0 dUdf(z),
by (5.13). Thus, Tnnewf = (k/2)- 1
Ex(a)n ka d i
—k va 0 udf.
an
(The sum is over a > 0 dividing n and prime to N; but the term x(a) ensures . that the terms with g.c.d.(a, N) > 1 will drop out.) That is,
T:ew z---
E x (a)ak - i va o u„,a =
Told ,
all;
by (5.18). This proves Proposition 43.
to
In many situations, the definition of Tn in terms of double cosets is more convenient than the definition in terms of modular points. For example, we shall use this definition below to show that Tu is a Hermitian operator with respect to the Petersson scalar product. Moreover, the double coset approach can be generalized to situations where no good interpretation in terms of modular points is known. We shall see an example of this in the next chapter, where we shall define Hecke operators on forms of half-integral weight. For a more detailed treatment of the double coset approach in a more general context, see [Shimura 1971].
The Petersson scalar product. Suppose we have an integral over some region in the upper half-plane and make the change of variable z F.+ az = where a = (ca 1:1) E GL I (0). If we have a term dxdyly 2 in the integrand, this does not change under such a change of variable. To see this, we compute:
, daz ....._ = det a dz (cz + d) 2 '
az
Im _ det a Im z — Icz + dr .
(5.29)
In general, if we make a differentiable change of complex variable z 1 = u(z), then the area element dxdy near z is multiplied by lut(z)1 2 (see Fig. 111.5). Thus, interpreting the first equality in (5.29) in terms of the real variables x = Re z and y = Im z, we see that an element of area near z is expanded by a factor idazIdzi 2 = (det a) 2/Icz + dr. Thus, dxdyly 2 is invariant under the change of variables, by (5.29).
Proposition 44. Let F' c F be a congruence Subgroup, let F' c H be any fundamentardomain for F', and define
169
§5._ The modular interpretation, and Hecke operators
Figure 111.5
I .7v
dxdy
11(r1)
cfff
F"
(5.30)
2 '
Then 0 (a) The integral (5.30) converges and is independent of the choice of Fi . (b) [t.- : i-4 1 = griht(r). (e) If a e GLI (0) and cc-1 ra c F, then [r: rf] = [f: a-1 f' al PROOF. First let [r: 11 = d, let f = U1=1 (04'9 and take F' = U ai l FNotice that the integral for p(F) converges, because
f dxdy f 1/2 JF
< I Y2 .1
y'ydx d :=
2 Vi
—1/2J srS/2
If for each j we make the change of variables zi--+ aj z, we find that ja7 1 F dXdY1Y2 = IF dxdyIy 2 ; hence, for our choice of F' we find that the 1 integral in (5.30) converges to dp(F). Suppose we chose a different fundamental domain F1 for F'. Then we divide F1 into regions R for which there exists a e F' with aR c F', and again we use the invariance of dxdyly 2 under zl---+az to show that the integral over R c F1 and the integral over the corresponding region aR in F' are equal. Finally, to show part (c), we note that a' F' is a fundamental domain for a-i ra. Since
f
dxdy _ f dxdy
Gt_ir y2
r y2
,
it follows that p(a - ir a) = p(r), and so part (c) follows from part (b).
a
mk(r).
Now suppose that f(z) and g(z) are two functions in We consider the function f(z)g(z)y k , where the bar denotes Complex conjugation and y = Im z. If we replace the variable z by az for a e GL I (0), we obtain: f(az)g(az)y k (det allez + di 2 )k, by (5.29). But this is just (f(z)l[a] k) (g(z)i[a] k)y k . Thus, the effect of the change of variable is to replace f by f[Œ]k and g by g I [a]k .
rf = r
Definition. Let be a congruence subgroup, let F' be a fundamental domain for F', and let f, g e Mk (F'), with at least one of the two functions f,
170
in Modular Forms
g a cusp form. Then we define
f
o dyxd2y . (5.31) i;g> (if [f : fl < p i" It is immediate from this definition that is linear in fand antilinear in g (i.e., < f, cg> = 5 - (f, g>), it is antisymmetric (i.e., = ), and also < f, f> > 0 for f 0 O. These are the properties one needs in order to have a hermitian scalar product. We shall return to this later. 1
...z .
75 g yk
Proposition 45. The integral in (531) is absolutely convergent, and does not depend on the choice of F'. If F" is another congruence subgroup such that f, g emk (r"), then the definition of is independent of whether f, g are considered in Mk (F) or in mk (r").
n
Remark. The term lir : in (5.31) is needed in order to have the second assertion in the proposition. The requirement that f or g be a cusp form is needed to get convergence of the integral, as we shall see. First take F' = H ail F, where f = U Œif-/- In each region cT i F make the change of variables which replaces z by cc1 1 z, thereby transforming the integral into (z)i[ail]kg(zacc-i]. , dxdy
PROOF.
Efj f)
)
F
K
•T
y
2 •
Sincef, g G mk (r), we can writef l[ct,7 ' ],, = Es13.--.0 and, 91[01.r 1 ]k = Ec:= 0 bnd, with either an = 0 or bn = 0, since f or g is a cusp form. Because IqN I = le2 niziNi= e -21rY1N and y > 1 in F, and because an and b„ have only polyno. mial growth (see Problem 19 in §3), it is not hard to see that the integral is absolutely convergent. Next, if F1 is another fundamental domain, we proceed as in the proof of Proposition 44, comparing SR f(z)g(z)y'dxdy with SŒR f(z)g(z)? -2 dxdy, where R is a subregion of F1 and al?, with a e F', is the corresponding subregion of F'. Making the change of variable z 1- az and using the fact that f I NI, = f, giNk = g, we find that the two integrals are equal. Finally, suppose f, ge Amr"). First suppose that F" c F'. Writing f' = f ", F" =U67'F', we have
U1&
1
0-4 : f 1 fr,
— k dxdy f(z)g(z)y 2 Y
.
=
1
-
1
[I':
E ri"] s
l f(z)1[6i-l ]kg(z)1[6i-l ]Yk k dxdy y F'
[f : ft] But all of the summands on the right are equal, sincefi[bi] k = 1 gi[61- 1k = g; and there are [r': i---"] values of!. We thus obtain the right side of (5.31). If F" s* rt, we simply set r"f = F" r' F', and show that (5.31) and (5.31) with F".in place of F' are each equal to (5.31) ü7' - "I in place of F'. This completes the proof of Proposition 45. 0
171
§5. The modular interpretation, and Hecke operators
Proposition 46. Let f, g e Mk (r) with f or g a cusp form. Let a e GL (Q). Then
(5.32)
<j" g> = .
PROOF. Let F" = F' n ar'a'. Then f, g e Mk(f"), and fi[a]k , gl[a]k E Mk(GC 1 F"a) by Proposition 17(a). Here a -1 1""a = a' Fla n F'. Let F" be a fundamental domain for r", and take o( 1 F" as a fundamental domain for a -l F"a. Then the right side of (5.32) is
1
,c_ iF„ [Ï: .4 : a - r"a] f
kdxdy 1 f(z)1[ 0 ]9(z) I [Œ]k.Y
r-
I
-
fr,f(Z)g (z)
„kdxdy . y2
-,/
--Since [r: a-1 1-- "a] = [r: t- "] by Proposition 44(c), we obtain the left side 0 of (5.32). Now for a E GL I (0) we note that a can be multiplied by a positive scalar without affecting [a]k. So without loss of generality we shall assume in what follows that a = ( 'c'i ) has integer entries. Let D = ad — bc = det a, and set a' = Da- .1 _tic -ab‘) Then [aœl ]k = [alk . (
.
Proposition 47. With f, g, a as in Proposition 46,
= .
(5.33)
In addition, depends only on the double coset of a modulo F'. PROOF. If in (5.32) we replace g by gi{a -l j k and replace F' by r n then Proposition 46 gives (5.33). Now suppose we replace a by )7, ay2 in < f I [a], , g>. We obtain = = , because f is invariant under [yi jk and g is invariant under [Y 1 ]k' since
o
yi , 7;' e r'.
Proposition 48. Let r = IVN), A" = dn(N, {1}, Z) (see (5.23)). Let f, g e Mk (r) with for g a cusp form. For each de (ZINZ)* , fix some ord er such that ad ---- ( '' (2, ) mod N. Let n be a positive integer prime to N. Then
= . In particular, iffe M k (N, x) ., then = An) for n prime to N. PROOF.
If a = C JD e A" then a .7-_-- (1) bn) mod N, CL' (!)- -bnin Mod N. Thus, o-n a' e An. By definition,
(g ---i b) mod N, and
)
,,f= 77 k/2)-1 Efl[ r, r /], 7 (
where the sum is over the distinct double cosets of F' = A' in A". Let dOE denote the number of right cosets in F'ar. Then dOE = [r: r n a -1 ra] by Proposition 41. By ( ',), (5.25) and the second assertion in Proposition 47,
we have
-
172
111. Modular Forms
g>, =n(14/2)-1 E cice as a scalar product on mk (r,(N)), because makes sense only if f or g is a cusp form. Instead, one can use the explicit construction of Eisenstein series in §III.3 above, directly constructing eigenforms. Then Mk(I7i (N)) can be written as the orthogonal direct sum of Sk(fii (N)) and a space spanned by Eisenstein series which are eigenforms. (In fact, an intrinsic definition of an Eisenstein series, which generalizes to many other situations, is: a modular form which is orthogonal to all cusp forms.) The dimension of the space of Eisenstein series turns out to be the number r of regular cusps of 1"1 (N). Roughly speaking, in order for a modular form to be a cusp form it must satisfy r conditions (vanishing of the constant term in the qN-expansion), one at each regular cusp; and so Sk(I-1 (N)) has codimension r in Mk (fl (N)). For more information on Eisenstein series, see, for example, [Gunning 1962 ]. PROBLEMS
1. Prove Proposition 31 in cases (iii)—(iv),
I"' = Fo (N), F(N).
2. (a) Let ŒN = 1). We saw in Problem 15 of §III.3 that [ŒN]k preserves Mk (Fo(N)). Prove that the Hecke operator T. commutes with ŒN for n prime to N. (b) Let F =Enodd cr (n)q", 04 = En anqn M2 (F0(4)) (where a the number of ways n can be written as a sum of four squares). Write the matrix of T2 in the basis 04, F, and find a basis of normalized eigenforms for T2. (c) Show that 1 2 does not commute with a4 (see Problem 15(c) in §III.3). (d) Suppose n is odd. Since T. commutes with cx4 and T2, it preserves each eigenspace in part (b) and each eigenspace in Problem 15(c) of §III.3. Show that the operator T. on the two-dimensional space M2 (F0(4)) is simply multiplication by ai (n). (e) Derive the following famous formula (see, for example, Chapter 20 of [Hardy and Wright 1960 ]) for the number of ways n can be written as a sum of four squares:
an
=
8a, (n) 12401 (n0)
for n odd; for n= 2rn0 even, 2,1'n0 .
3. If JE Mk (N, x), show that f(0o) = 0 implies 7„f(00) = 0, but that f(s) = necessarily imply 7"„f(s) = 0 for other cusps s (give an example).
0 does not
4. (a) Show that if feMk (N, x) and g(z) = f(Mz), then gE Mk (MN, x), where 2( is defined by xi(n) = x(n mod N) for ne(ZIMNZ)*. Let T. be the Hecke operator
§5. The modular interpretation, and Hecke operators
175
considered on Mk(N, x), and let Ts' be the Hecke operator considered on Mk (MN, x'). Suppose n is prime tor M. Show that forfeMk (N, x) c: Mk (MN, x') and g(z) = f(Mz)E Mk (MN, X') we have: 7 f= 7;,f ; '
g(z) = (Tf)(Mz).
(b) Let f(z) = (q(z)q(2z)) 8 , g(z) = f(2z) E S8 (1-0 (4)) (see Problem 18(d) of §111.3). Show that for odd n, T„ acts on the two-dimensional space S8 (F0(4)) by multiplication by a scalar. (c) Find the matrix of T2 acting on s8 (r0(4)) in the basis f, g ; then find the basis of normalized eigenforms for all of the Tn .
5. Let f, g E Mk (N, i) with f or g a cusp form. Let AN. Note that Tp = Up . Show that <J, Vp lj> 6. Let fi = (270- 2 4 A2 5 = (a) 2A "6 9 where (20-12 d q11„(1 qn) 24 and E6 =1 - 504 05 (n)q"• We know that S24 (1") is two-dimensional and is spanned by fi and f2 (see §111.2). (a) Give a simple reason why fi is not an eigenform for the Hecke operators. (b) With the help of a calculator, find the matrix of the Hecke operator T2 in the basis fi , f2 (c) Express in terms off, andf2 the basis for s24(r) consisting of normalized eigenforms for all the 7. (d) Express in terms off, and f2 the cusp form whose n-th q-expansion coefficient is the trace of 7,, on S24(1"). (e) Find Tr T3 from part (d), and also directly by computing matrix entries.
In this problem one encounters fairly large numbers; for example, the coefficients in part (c) are in Q(V144169). While Proposition 51 guarantees the existence of a basis of normalized eigenforms, it does not guarantee that it will always be easy to find it computationally.
7. Let (L, t) be a modular point for Fi (N). Let a e W(N, {1}, z), and let y e (N). For each a and y, consider the lattice L' = !-ay.L. Show that L' is a lattice which contains L with index n, that L' depends only on the right coset of F'1 (N) in An(N, {1 } , Z) which contains ay, and that t has order N in C/L'. Show that the L' in (5.2) are in one-to-one correspondence with the right cosets of Fi on in An(N, { 1 } , Z). Use this to give another proof of Proposition 43 by comparing (5.9) and (5.27). 8. Let fe Mk (Fo(p)). (a) Show that yo = 1 and yi = (1). -1.;), 0 <j < p, are right coset representatives for F modulo Fo (p). (b) Let Tr( f) be defined by: Tr(f) = liP, 0fl[y j k . Show that Tr(f)eMk(F). (c) If f happens to be in man, then show that Tr(f) = (p + 1)f and Trtff [(p) l )] k) = ki2 7;f
CHAPTER IV
Modular Forms of Half Integer Weight
Let k be a positive odd integer, and let A = (k 1)/2. In this chapter we shall look at modular forms of weight k/2 = A + 1/2, which is not an integer but rather half way between two integers. Roughly speaking, such a modular form f should satisfy Maz + 13)1(cz + d)) = (cz + d)+112f(z) for (ca ,i) in f = SL2 (7L) or some congruence subgroup I' c T. However, such a simple—
minded functional equation leads to inconsistencies (see below), basically because of the possible choice of two branches for the square root. A subtler definition is needed in order to handle the square root properly. One must introduce a quadratic character, corresponding to some quadratic extension of Q. Roughly speaking, because of this required "twist" by a quadratic character, the resulting forms turn out to have interesting relationships to the arithmetic of quadratic fi elds (such as L-series and class numbers). Moreover, recall that the Hasse—Weil L-series for our family of elliptic curves E„ : y2 = x 3 — tex in the congruent number problem involved "twists" by quadratic characters as n varies (see Chapter II). It turns out that the critical values L(E„, 1) for this family of L-series are closely related to certain modular forms of half-integral weight. One classical reason why modular forms were studied is their use in investigating the number of ways of representing an integer by a quadratic form: m =41,_. 1 A fi nin, = [nir A[n], where A = [Ail] is a giiren symmetric matrix, [n] is a column vector and [n]' the corresponding row vector. For example, the number of ways m can be represented as a sum of k squares is equal to the coefficient am in the q-expansion of O": k
q EnJ =7-. Eam es i ek = j=11 i gni2 = 1 n i= n 1 , - - • , nk= — —
co
co
In Chapter III we saw that for k even Ok e Mk/2 (4,Xic—Ii), where x_.. / is the
177
§1. Definitions and examples , /
character modulo 4 defined by x_ 1 (n) = (-741) = (-1)(n -1)12 . More generally, for k even one can use In q tn"Ini to construct modular forms of weight k/2. The properties of modular forms are then useful in studying the number of representations m = [n]fil[n]. For example, in Problem 2 of §III.5 we used the action of the Hecke operators on M2 (1"0 (4)) to derive a simple formula for the_number of ways an integer can be written as a sum of four squares. For more information about the connection between quadratic forms in k variables and modular forms of weight k/2, see, for example, [Gunning 1962] and [Ogg 1969]. It is natural to ask whether a similar theory exists for quadratic forms in an odd number of variables. This would be a theory of modular forms of weight k/2 where k is an odd integer. One would want ek for k odd to be an example of such a form. Early investigators of representability of an integer as a sum of an odd number of squares—Eisenstein, and later G. H. Hardy understood the desirability of such a theory of modular forms of halfintegral weight. But it was only recently—starting with Shimura's 1973 Annals paper—that major advances have been made toward a systematic theory of such forms which rivals in elegance and beauty the much older theory of forms of integral weight. In this chapter we shall first present the basic definitions and elementary properties of forms of half-integral weight, largely following [Shimura 1973a, 1973b], but with slightly different notation. We shall discuss examples in some detail. But when we come to the fundamental theorems of Shimura and Waldspurger, we shall not give the proofs, which go beyond the scope of an introductory text. Rather, we shall motivate those theorems using the examples. Finally, we shall conclude by returning to the congruent number problem and Tunnell's theorem, which we used in Chapter I to motivate our study of elliptic curves.
§1. Definitions and examples We shall always take the branch of the square root having argument in (-7r/2, ir/2]. Thus, .sii. is a holomorphic function on the complex plane with the negative real axis ( — co, 0] removed. It takes positive reals to positive reals, complex numbers in the upper half-plane to the first quadrant, and complex numbers in the lower half-plane to the fourth quadrant. For any integer k, we define zk12 to mean Whenever we have a transformation rule such as f(yz) = (cz + d) kf(z) (where y = (ca jt ), yz = (az + b)I(cz + d)), we call the term (cz + d)k_ the "automorphy factor". It depends on y and on z. That is, an automorphy factor J(y, z) for a nonzero- functionf has the property thatf(yz) = Ay z)f(z) 4 for-z e //And y in some matrix gr-oup. Because ,
178
iv. Modular Forms of Half Integer Weight
f(aliz) _ f(aliz) . Agz) .f(z) f(13z) f(z) it follows that any automorphy factor Ay, z) must satisfy
J(oefl, z) = f(Œ, /3z) J(fl, z).
For example, Ay, z) = cz + d for y = ( ,171)e F = SL 2 (7L) satisfies (1.1). If — 1 = G1 Pi ) is in our matrix group, then J(y, z) must also satisfx: J(— 1, z) = 1. Another example of an automorphy factor is Ay, = j(y, z) = (ff)ed-1 \ cz + d, y = bd )e F0 (4), which is the automorphy factor for 0(z) = qn2 , q = e2iriz (see §1II.4). Suppose that, in defining modular forms of weight k/2 for k an odd integer, we try the most obvious thing: we look for functions f satisfying f(yz) = (cz + d) kl2f(z). However, Ay, = (cz + d ) ci2 cannot be an automorphy factor of a nonzero function for any congruence subgroup. F' c F and any odd integer k. To see this, suppose r f(N), N> 2. Let a = iN_N), ?). Then a, fl E F', and (1.1) would require the k-th power of the following equality of holomorphic functions on H:
AN 2 2N)z +1 — N = \41--Nzl(— Nz + 1) + 1 — N • \I —Nz + I.
N
(1.2)
Clearly, the square of (1.2) holds; thus (1.2) itself holds up to a sign. Since both expressions in the radicals on the right are in the lower half-plane, the right side is the product of two complex numbers in the fourth quadrant; Ne H. but the left side is in the first quadrant, since (N 2 — 2N)z + 1 Hence, (1.2) is off by a factor of — 1, and (1.1), which is the k-th power of (1.2), is also off by —1 if k is odd. Thus, we cannot have Ay, z) = (cz + 0 42 . The natural way out of this difficulty is to force (1.1) to hold by simply defining the automorphy factor J(y, z) to be the k-th power of
j(y, , z) cr:f 0 (y z) I 0 (z) = (-c-
d
1 1 \I cz- + d
for
r
y = (a hi\ c d)
E
o
(4). (1.3)
That is, for a congruence subgroup F' c F0 (4), one defines a modular form of weight k/2 to be a holomorphic function on H which transforms like the k-th power of 0(z) under any fractional linear transformation in I"' (and which is holomorphic at the cusps, in a sense to be explained below). Recall that (fi) is the Legendre symbol, defined for a positive odd prime d as the usual quadratic residue symbol, then for all positive odd d by multiplicativity, and finally for negative odd d as (-) if c> 0 and — (-4-) if c < O. (Also, (4---) = 1.) Further recall that we define ed = 1 if d -a- 1.(mod 4) and Ed = i if dE--_-: — 1 (mod 4). Thus, e j . X - i (d) = (— 1)(d-1)12 (note that this holds for negative as well as positive d). Thus, if we define f(z)i[y] k/2 = j(y, z) kflyz) for y e F0 (4) and k odd, we shall require a modular form of weight k/2 for F' to be fixed by [y] k/2 for y e f'. As in the case of integer weight, we shall want to define [Œ]k/2 for arbitrary matrices a e GL I (0) with rational entries and positive determinant.
179
§1. Definitions and examples \
But since the automorphy factor j(y, z) is defined only for ye r0 (4), we have no preferred branch of the square root for arbitrary a e GLI(0). This circumstance requires us to work with a bigger group than GOO), a group G that contains two "copies" of each a EGLI (G)., one corresponding to each branch of the square root of cz + d. The example of the autornorphy factor Ay, z), which is a square root of (-7i)(cz + d), shows that we also should allow square roots of —(cz + d). Thus, we shall actually define G to be a four-sheeted covering of GOO). We now give the precise definition of the group G. Let Tc C denote the group of fourth roots of unity: T :.--- {,± 1, + i} . We now define G to be the set of all ordered pairs (a, 0(z)), where a = (a, 1:) e GL(Q) and 4)(z) is a holomorphic function on H such that
2 qt)(z) =
i
cz + d
Met a
\
for some t e T2 = { +1 } . Tliat is, for each a e GL() and for each fixe'd t = +1, there are two possible elements (a, +0(z))EG. We define the product of two elements of G as follows:
(1.4)
(al 0(z))(fil P(z)) = (Œ)6, 0(13z)0(z))-
Remark. We could define G in the same way but with T defined to be the group of all roots of unity or even (as in [Shimura 19734) the group of all complex numbers of absolute value 1. However, for our purposes we only need fourth roots of unity. Proposition 1. G
is a group under the operation (1.4).
PROOF. We first check closure, i.e., that the right side of (1.4) belongs to G. If a = (a, :11), fi = (eg f,), we have
(0(f3z)tP(z)) 2 = t i (ctiz + d)t 2 (gz + h)t\i'det a det /3
-=.-- t 1 t2 (c(ez +f) + d(gz + h))/Vdet afl = 1 1 12 ((ce +- dg)z + cf + dh)IVdet a/3,
■
which shows that (4, (P(Az)/i(z))e G. Associativity is immediate. It is also obvious that ( ( ), 1)EG serves as the identity element. Finally, to find an inverse of (a, 4)(z)), where a = (, bd ), we write D. = det a, a' = Da - I = ( cr, -ab), and look for (a( 1 , lit(z)) such that: OW 1 z)0(z) = ,1. That is, we set 0(z) = 110(or'z), and then must check that (a', 1//(z)) e G. But
tit(z) 2 = 1 1 ( ( dzb + d)/) = —1 f:6(—cz + a)/(ad — bc) a t —
.
1 t
4.
a )/„Met a -1
D D
which is of the reqUirVI
,
form. This completes the proof.
o
180
Iv. Modular Forms of Half Integer Weight
We have a homomorphism
P: G which simply projects onto the first part of the pair: (a, c/)(z)))--> a. The kernel of P is the set of all (1, 4)(z)) E G. Since 0(z) 2 = t for (1, 0(2))€G, it follows that 0(z) must be a constant function equal to a fourth root of unity. Thus, if we map T to G by I (1, t),_ we have an exact sequence 1
in other words, the kernal of P is the image of T under t (1, t). Similarly, if a = aa for a E 0, the inverse image of a under P is the set of all pairs (a, t) for le T, since in that case we again find that 4)(z) 2 must» be +1. We let G' = P --- 1(r) be the set of all pairs (a, (z)) e G such that Œ € r = SL 2 (Z). G' is clearly a subgroup. If = (a, 4)(z)) e G, we shall sometimes use the notation z to mean the same as az for z e H. For = (a, 0(z)) e G and any integer k, we define an operator []k/2 on functions J on the upper half-plane H by the rule f(z)
k12 f(ŒZ) 0(Z)
k•
(1.5)
This gives an action of the group G on the space of such functions, i.e., we have (f(z)1F1_Ik12., 6 L.- , 2_11 10 = f(Z) I [1 2]k/2 because of (1.4). L.,, A Now let ry be a subgroup of F0(4). Then j(y, , z) is defined for elements y e (see (1.3)). We define d--ff
r'
(Y, AY, Z ))1Y Er/.
(1.6)
Clearly is a subgroup of G (because of (4.13) in §III.4) which is isomorphic to rt under the projection P. Let L denote the map from T0 (4) to G which takes y to 5./ cfff (71./(79 Z)) e G.
Then P and L are mutually inverse isomorphisms from r0 (4) to r0 (4). We call L a "lifting" or "section" of the projection P. In our notation we are using tildes to denote this lifting from 1T0 (4) to G:
L:
(Y1 .1(Y1 z));
P: 6' .1-(Y z ) i---° Y. Suppose that f(z) is a meromorphic function on H which is invariant under where I"' is a subgroup of finite index in F0 (4). We now []ki2 for Y e describe what it means for f(z) to be meromorphic, holomorphic, or vanish at a cusp seQu tool. We first treat the cusp co. Since r has finite index in F0 (4) and hence in f, its intersection with
± fi
j)1 \O 1/jjz
181
§1. Definitions and examples
'Mil must, be of the form f+ a 'Di} if — 1 e I"' and either {a ini} or 1( — z) if — I rf. (We always take h > O.) Since [ -2-'1 ] 42 is the identity, and j((1) = 1, in all cases f is invariant under h, and so has an expansion in powers of gh e2h. As in the case of integer weight, we say that f is meromorphic at oo if only finitely many negative powers of qh occur, we say thatfis holomorphic at oo if no negative power of qh occurs, and we define f(co) to be the constant term when f is holomorphic at co. , Now suppose that se Q L.) fool. Let s = a co, Œ E F, and let =--- (a, 0(z)) be any element of G 1 which projects to a: P( ) = a. Define rsi = e I ys = sl, and let Gal = It/ e G cx) = co} . We have
i1 j)' {(± ( 0 because Foe ={ +(,1) 1)}, and the 4)(z) for ( ) must be a contant function I, as noted above. Now ir is contained in G' and fixes co, i.e., G. Moreover, P gives an isomorphism from Ç-1-rs' to a' F;a Ç. Since F' has finite index in F, it follows that a-i Fs'a is of one of the forms { + nj}, {(t) 1)-11, or ((— Mil. Thus, for some t E T we have Gsp
1 , = {(-±(i h)
iE z
tY} •
h> 0 is determined by requiring a n to be a generator of Given s ana a-1 1a modulo + 1; then t is determined, since P is one-to-one on e I. i.e., we can find t by applying the lifting map L to + c Proposition 2. The element ( ( ';), t) e Gcol depends only on the r-equivalence class of the cusp s. First suppose that c is replaced by another element (a l 4 (z)) e G 1 such that s = oo. Then e G' fixes co, and so it is of the form t i ). Then (±(,!,
PROOF.
Ofgi (M) -1 (± -1 rg)(M) ) t
(
(-±-(j)
hiM i RG
—1 1)
But (() f), t) commutes with ((4 t), so conjugating by affect(( 1), t). Now suppose that s l = ys = ('2c)oo, where y e F', = (y, that r;1 = yr;y — ', and so f;1 = ,pr;53-1 . Thus,
± GO -
=± and weagain obtain the same (0
)g
does not
z)) E f". Note
±
t). This completes the proof.
182
iv.
Modular Forms of Half Integer Weight
We are now ready to define meromorphicity/holomorphicity/vanishing at Let s = the cusp s of r". Suppose f is invariant under DI/2 for .2. E we have e and set g = fiN142 . Then for any element +
fiEgL i2= f1R1q 2= g..
g [
t)] k/2 :
That is, g is invariant under [(a) I
g(z) = g(z)
h
01
t-kg(z + h). t
Icf2
where r = 0, 14 , 1, or Then e -2nirzlh g(z) is invariant under z 1 * z + h, and so we obtain a Fourier series expansion e'i"filg(z) = Eanq'hi, Write i k = —
nign-1-79/11. g(z) = Eane2
We say that f is meromorphic at s if an = 0 for all but finitely many n < 0, and that f is holomorphic at s if an = 0 for all n < O. Iff is holomorphic at s, g(z). Automatically f(s) = 0 if r 0, since in that we define f(s) case the first term is ao e'z'fit. If r = 0, then f(s) = ao . It is easy to see that these definitions depend only on the F'-equivalence class of s, i.e., g = kI2, where y E f' and c oo = s (see flMic /2 may be replaced by g = f the proof of Proposition 2, and also the proof of Proposition 16 in §III.3). It should be noted, however, that there is some indeterminacy in the value f(s) of a modular form at a cusp. Namely, if we replace e G l by ((4 7), t ig, with t t e T, then the effect is to multiply g = fiM k/2 by CI k . If k/2 is a halfinteger, this may alter the value by a power of i; if kl2 is an odd integer, then f(s) may be defined only up to + I (compare with the proof of Proposition 16); and if k/2 is an even integer, then f(s) is well defined in all cases. Given a cusp s of I"' and an integer k, we say that s is k irregular if r 0, and we say that s is k regular if r = 0, i.e., if t k = I. Thus, iff is holomorphic at the cusps, it automatically vanishes at all k-irregular cusps. Note that whether a given cusp s of F' is k-regular or k-irregular depends only on k modulo 4. That is, if t = ± i, then the cusp is k-irregular unless k/2 is an even integer; if t = 1, then it is k-irregular unless k/2 is an integer; and if t = I, then it is always k-regular. In the case when /02 is an odd integer, the terminology agrees with the definition of regular and irregular cusps in Problem 2 of §1.11.3. -
-
—
Definitions. Let k be any integer, and let F' c F0(4) be a subgroup of finite index. Let f(z) be a meromorphic function on the upper half-plane H which is invariant under [nk/2 for all j) E ff. We say that f(z) is a /nodular function of weight 1(12 for r/ if f is meromorphic at every cusp of I'. We say that such an f(z) is a modular form and write Je Mk/2 (r/), if it is holomorphie on H and at every cusp. We say that a modular form f is a cusp form and write fe Slog"), if it vanishes at every cusp.
183
§1. Definitions and examples
rouso c To (4).
Let x be x) denote the subspace of
Now let N be a positive multiple of 4, so that
a character of (ZPVZ)*. We let Mk/2 (ro (N), -44/2(r (N)) consisting off such that for y = (c" bd) e FAN)
(1.7) fiEflka = x(cOf. We define Ski2 (ro (N), x) = Sql2 (f i (N)) n 402 (f" 0 (N), x). The exact same argument as in the integer weight case shows that
Mk/2 (f'l (N)) = Ox Mk/g0 (MI x). Furthermore, it follows immediately from the definitions that if x i and X2 are Dirichlet characters modulo N, then E Mki/2 ( Ir0 (N),
xi)
(i = 1
5
2)
implies
f1 f2
J J
-
(IC
j+k2)12(f 0( N ) , Z1Z2) .
(1.8)
Notice that for any k e Z, we have Mk/2(fro (N), x) = 0 if x is an odd character, as we see by substituting — 1 for y in (1.7): f[((1 Wko f = x( Of For example, since the trivial character x= 1 is the only even character of (Z/41)*, this means that
Mk/2 (fAl (4)) = Mk/2 (to (4), 1) = mk,2(t0(4))If 4IN, recall that x_i denotes the character modulo N defined by x_ i (n) = ( 1) n-0/2 for n (ZI NZ)* . Notice that in our definitions k is any integer, not necessarily odd. For k even, let us compare the above definitions of Mk/2 (0, Sic/2(t '), Mk/2 (f-0 (N) X), Ski2(to(N), x) with the definitions of Mk/2(F /), S12 (r'), Mk/2 (N, )01 Sk12 (N, x) in Chapter III. First, for any = (a, (z)) e G, cc = (a, ',;), 4i(z) 2 = t(CZ j SO d)/.,./det cc, note that f(z)IM k/2 = 4)(Z)- kfoŒ) = rk/2f(z)i [Œ1k12 that Mk/2 for kl2 e Z differs only by a root of unity from the operator [a]k/2 defined in the last chapter. It immediately follows that for kl2e Z the cusp —
condition defined in the last chapter is equivalent to the cusp condition defined above. There is a slight difference, however, between the condition that f be invariant under [y]k/2 for y = eic r" and the condition that f be invariant under []k/2. Namely, firin-111/2 =i ( y, z) - kf(yz) = (fri)(cz dn -ki2f0z)) (7-61 1 )1q2f [Y]k/2 . Thus, if 102 is odd, then M k12 differs from [y],02 by x_, (d). From this discussion we conclude Proposition 3. Let 41N, kl2e Z. Then Mk 1 2 (f‘O(N) ,
= mk12(1V, eix),
sk1 2 r-0(N), X) = Ski2(N (
It is now not hard to describe the structure of Mki2 (ro(4)). If we have a polynomial P(X1 , , X„,)e C[X, , . . . , X.] and assign "weights" wi to Xi, then we say that a monomial ri Xi has weight w = Z ni wi , and we say that a polynomial P has pure weight w if every monomial which occurs in P with nonzero coefficient has weight w.
184
iv. Modular Forms of Half Integer Weight
Proposition 4. Let 00(z) = , qn2 , F(z) = Z n>0 odd at OW q = e2ni2 Assign weight 1/2 to O and weight 2 to F. Then Mk/2(f0(4)) is the space of all polynomials in C[0, F] having pure weight kl2. PROOF. In Chapter III we found that for k/2 e Z, the space Mko (N, consists of polynomials in 0 and F having pure weight k/2 (see Problems 17-18 of §III.3). This gives Proposition 4 when k/2 E Z. Next, by the definition of [13] k/2 , we have OID1 1/2 = Co for y e f0 (4). 0 is holomorphic at the cusps, because we checked holomorphicity of 0 2 at the cusps, and if O had a singularity at s, so would 0 2 . Thus, Cs € M112 (f0 (4)). It now follows by (1.8) that any polynomial in O and F of pure weight kl2 is in Mk/2 (f0(4)). Conversely, suppose that fe Mk/2 (Î o (4)), where k is odd. Then f 0 is in aPk+1)/4 e2P(0 , F) , M(k + 1)/2 ( r0(4)), and so can be written in the formf where a E C is a constant which equals O if 4)(1 k + 1 and P(0, F) has pure weight (k — 1)/2. Then f OP(0, F) = aF(')1410 e Mk/2 (t0 (4)). But because 0 vanishes at the cusp —1, while F does not, this form satisfies the holomorphicity condition at --12- only if a = O. Thus, f= OP(0, F), and the proof is complete. Corollary. Dim Mk12(f210( 4 )) = 1 + [k/4]. PROBLEMS
1. Let y = '1)e ro (4), and let p = ( ( ), m -14). Check that p e G. (a) Compute pp'. (la) If ye f0 (4m), then y/ = 7)y ?) -1 is in r0(4). Compare piip -1 with .ç)i . Show that they are always the same if m is a perfect square, but that otherwise they differ by a sign for certain y e F0(4m). (
2. Let y =
'De r0(4), and let p = N '14,1i). Check that p E G. (a) Compute (b) Suppose that 41N and y e fo(N). Then yl = (g, 1)y( ,( ,, 1) -1 E ro (N). Compare pijp - i with 55 f .
3. Find h and t for each cusp of r0(4). 4. Let r c f0 (4) be a subgroup of finite index. Let /02 e Z. Show that if c I-, (4), then Mk/2 (f") = Mk/2 (r). Otherwise, let x be the unique nontrivial character of n r 1 (4); show that m„12(r) Mk/2(r', e2 ) in the notation of 011.3 (see the discussion following Proposition 27). 5. (a) Describe Ski2 (r0 (4)), and find its dimension. (b) Show that for k 5, the codimension of Ski2 (t0 (4)) in Mk/2 (r0(4)) is equal to the number of k-regular cusps. (c) Find an element E anqn in S1 312 (r0 (4)) such that a l = I and a2 = 0 (there is only one).
6. Let 4IN, and define xiv by kv (d) = (1) if g.c.d.(N, d) = 1, xN(d) = 0 if g.c.d.(N, d) > 1. (a) Show that xN is a Dirichlet character modulo N.
. §2. Eisenstein series of half integer weight for r0(4)
185
(b) Let p be as in Problem 2 above, let x be an arbitrary Dirichlet character modulo N, and suppose that fe Mkigo(N), z). Show that fj[P]ko e Mk12(f-0 (N), i.x ki ) •
ro (4 ).
7. Find the value of 0 G Milgo (4)) at the three cusps of
§2.
Eisenstein series of half integer weight for
r0(4)
Recall the functions Gk (z) and Ek (z) in Mk (F) that we defined in §I11.2 for k > 4 an even integer. We set Gk (z) = E(niz + n) -k , where the sum is over m, n e Z not both zero; and then Ek (z) was obtained by dividing by Gk (ioo) = Mk). Alternately, we can obtain Ek (z) by the same type of summation Z (mz + n) -k if we sum only over pairs m, n which have no common factor (see Problem 1 of 011.2): -
Ek (z) -_-_-_ E
i
(2.1)
Om + n)k'
where the sum is over m, n e Z with g.c.d.(m, n) = 1 and only one pair taken from (m, n), ( in, n). In §111.2 we obtained the g-expansion —
—
Ek (z) = I —
2k '
1 =....... e 2itiZ
4
k n=1
We now want to give an analogous construction, with k replaced by a half-integer k/2 and F replaced by F0(4). To do this, we give a more intrinsic description of the sum (2.1). Notice that if we complete the row (m, n) to form a matrix y = (4,1 bn) e F, which can be done because g.c.d.(m, n) = 1, then the summand in (2.1) is the reciprocal of the automorphy factor Jk (y,9 z) for functions in Afk (F). That is, such functions f satisfy : f(yz) = .1k (y, z)f(z) for .4(y, z) = (mz + n)". In (2.1) we are summing litik(y, z) over all equivalence classes of y e F, where Yi ?--, y2 if y1 and y2 have the same bottom row up to a sign, i.e., if y2 = + a. Dy1 for some +G; D E For, . In other words, we may regard ik (y, z) as a function on the set Fcc\F of right cosets, and then rewrite (2.1) as follows: E k(Z) =
E Jk (y, zy l 9 y E rcAr
k > 4 even.
(2.2)
It is now possible to give a proof of the invariance of Ek (z) under [yjk for Yi e F which easily generalizes to other such sums of automorphy factors. Namely, note that, by the definition of [Y1]k, we have
Ek(z)I [Yi lc ------- .4 0'19 zY
= Jkoi,
l Ek(vi z)
zri
E Jk(y,
y e roAr
yizr
= E Jktyvi, zr, ye
foAr
because of the general relation (1.1) satisfied by any automorphy factor.
iv. Modular Forms of Half Integer Weight
186
But right multiplication by yl merely rearranges the right cosets F.y. Thus, the last sum is just a rearrangement of the sum in (2.2) for Ek. Since the sum is absolutely convergent for k > 2, we conclude that Ek I [yj k = Ek. We now mimic this procedure for half integral weight k/2 and the group f0 (4). In order to have absolute convergence, we must assume k/2 > 2, i.e.,
k > 5. Note that we obviously have F0 (4). = F. = { + ( - 1)}. Definition. Let
k
be an odd integer,
k > 5.
E
Eko (Z) =
Ay,
z)_ ".
(2.3)
7 e roD\ro( 4)
) e F0 (4) for each As representatives y of F.VE0 (4) we take one matrix (in, n) with 41m, g.c.d. (m , n) = 1 (in particular, n is odd), and we may specify n> 0 so that we get only one of the pairs + (m, n). Thus (a m
b n
a b Ek/2 (Z) = , .i E m,n e Z, 4Im,n> 0, g.c.d.(m,n)= 1 (( m n) zY k Substituting the definition of j(y, z) and noting that (-7:) -k = (v), since k is odd, we obtain Elo (z)
=
E
(—m)Einc(inz
41m, n> 0 g.c.d.(m,n)=1
n
+ n)
2,
k > 5 odd.
(2.4)
mk/2 (r0(4)) now follows by the exact same argument The fact that Ek/2 - — . as in the case of the Eisenstein series of integral weight for 1" which we considered above. Namely, Ek/2(z)i [MI0 = j(yi , z)_ kEk12 (y i z), which just gives a rearrangement of the sum (2.3). Finally, the cusp condition is routine to check, so we shall leave that as an exercise (see below). Unlike the case of 1", where there is only one Eisenstein series of each weight, Ek/2 (z) has a companion Eisenstein series Fki2 (z), defined by Fk/2 = Ek/2
That there are two basic Eisenstein series for each half-integer k/2 is related to the fact that F0 (4) has two regular cusps (see Problem 3 of §1 and the discussion at the very end of §III.5). To see that Fkl2 e Mk/2 (f0(4)), we apply Problem 6 of §1 in the case N = 4, x = 1. To write Fk12 more explicitly, we compute
Fk/2 (z)= (2z) -4/2 Ek/2( — 1/4z ) =
242
(4z)kI2
m)r_k
4Im1 . n> 0 g .c.d .(nz, n)= 1
I
n - n (— ml4z + n) k/2
1 = 2k/2 "S" (M) ek 4-4 n " (— m + 4nz)k12'
187
§2. Eisenstein series of half integer weight for fo (4)
where we have used the fact that n > 0 to write ,i4z \i—m/4z + n = —m + 4nz. We now replace m by —4m, so that the sum becomes a summation over all me/ with g.c.d.(4,n, n) = 1, i.e., n odd and g.c.d.(m, n) = 1. We obtain
1 n J F(nz + m)kI2 • IC k
E m,ne 7
Fic/2(Z) =
n>Oodd
(2.5)
g.c.d.(m,n)=1
We next compute the g-expansions of Ei02 and Fki2 . We proceed much as in the computation of the q-expansion of Eit in §111.2, except that there are added complications. The derivation of the g-expansion for Fie/2 (z) turns out to be slightly less tedious than for Ek12 (z). Hence, we shall give that derivation, and omit the analogous proof of the g-expansion formula for Ek/2 (z). To compute the g-expansion of Fie/2 (z), we start with the following relation, which was proved in Problem 8(c) of §11.4 for any a> 1, Z E H (where we have replaced a, s by z, a here): co 00 6,2nit2. E e—nial2r(a)-1 (2.6) E (z + h) ° (Note: for z e H we define z" = ea log z, where log z denotes the branch having imaginary part in (0, n) for z e H.) Since the sum (2.5) for Fk12 (z) is absolutely convergent, we may order the n I and write m = —j + nh, terms as we please. If we let j =-- 0, 1, . h eZ, we obtain Fk I2(z)
n>0 odd
= 2 —k/2
(nz
E
2-142
0 j 0 odd -0 <j 0 odd we let 26.(f) denote (7). Thus, if j is a prime, it is L(J) which tells us whether j splits, remains prime, orj.amifies in CP(jn) (i.e., this depends on whether xpi(j) equals 1, — 1, or 0, respectively). There is a unique character, also denoted x„„ which has 1 mod 4 and conductor 14ml if m 2 or 3 mod 4 and conductor iml if m which agrees with zni (j) = C9 for j > 0 odd. We can express x„, in terms of the usual Legendre symbol as follows:
188
IV. Modular Forms of Half Integer Weight
if m
E---
1 mod 4;
if m -a-- —1 mod 4; if m = 2m0 za-- 2 mod 4, where we define (51--) = ( —1)( i-1)12 for j odd and (il) = 0 for j even, and (4) = (— 0°2-1)/8 for ] odd and () = 0 for j even. i Note that Z.( — 1) = 1
if m > 0,
and ;Cm( — 1) = —1
if m < 0. (2.8)
Recall that we de fi ne A = (k — 1)/2.
has q-expansion coefficient bl which for I squarefree is equal to L(4_ 1)).1 , I — A) times a factor that depends only on A and on 1 mod 8, but not on I itself. This factor is given in (2.16) below. Proposition 5. The Eisenstein series
Fic/2
We first prove a lemma.
Lemma. Let n = no n?, where no is squarefree. Then for I squarefree, the inner sum in (2.7) vanishes unless n 1 11, and f n i ll it equals
1) ---an(no ,Ino p(ndn i ,
(2.9)
wherits to 0 if n i is not squarefree and equal to Al5bius function (equal (-1)" if n 1 is the product of r distinct primes). First, if n 1 = 1, then (Tia) has conductor n = no . If / has a common factor d with n, then, combining terms which differ by multiples of nld, we see that the inner sum in (2.7) is zero. On the other hand, if lealnlr, then making the change of variables f = —/j leads to (-7,-,E)E (De'afin . This Gauss sum is well known to equal snjti (see, e.g., §5.2 and §5.4 of [Borevich and Shafarevich 1966]). This proves the lemma when n 1 = 1. We next show that the inner sum is zero if g.c.d.(no , n 1 ) = d> 1. Note that PROOF OF LEMMA.
1 (4)
if g.c.d.( j, n) = I,
[0
otherwise;
and it follows that (1,) --z-- (--/--d. 2 ) • Combining terms with j which differ by multiples of nId2 and usingnthe fact that c/2 4'/ (since 1 is squarefree), so that If , e -27ti1( j+j'n1d2 )/n = 0, we find that the inner sum is zero in this case.
189
to(4)
§2. Eisenstein series of half integer weight for
Now suppose that n 1 > 1 is prime to n o . Let p, run through all primes dividing n,: n, = 11/4v. Set qv = pv; and let bv(j) = 0 if pd./ and (5,,( j) = I if pv )fr f. Then (-,4) = (54) II,, (VD. Any JE ZinZ can be written uniquely in the form
j =ion?. + >in/q,
0 -...ily 0, is the discriminant of a quadratic field and h) is the corresponding character, then the ID1-th g-expansion coefficient of 1412 is equal to L(x D , 1 — 2). IAl2 ra C( 1 — 22)(Eko
This proposition is due to H. Cohen [1975 ]. (His notation is slightly different from ours.) It can be viewed as a prototype for the theorem of Waldspurger-Tunnell which we shall discuss later. The WaldspurgerTunnel! theorem exhibits a modular form of weight 3- (think of A as being equal to 1 in Proposition 6) for F0(128) such that the square of its n-th g-expansion coefficient is a certain nonzero factor times L(E„, 1), where L(E, s) is the Hasse-Weil L-function of an elliptic curve E (see §I1.5) and E. is the elliptic curve y2 = x3 — n 2 x. Thus, the g-expansion coefficients of this particular form of half integral weight are closely related to all of the critical values 4E., 1) which we encountered in Chapter IL If we were allowed to take 2 = 1 in Proposition 6, then the ID1-th qexpansion coefficient would be L(XD, 0), which differs by a simple nonzero
194
IV. Modular Forms of Half Integer Weight
factor from L(x D , 1) (because of the functional equation), and is essentially equal to the class number of the quadratic imaginary field CP(NA (here = 1 or — 40. We will say more about the case A = 1 below. For another proof of Proposition 6, using so-called "Jacobi forms," see [Eichler and Zagier 1984]. We now discuss some interesting consequences of Proposition 6. Since = any element 1-4/2 e -M - /c/2(T0(4)) can be expressed as a polynomial in - 00 qn2 and F = En> 0 odd al OW (see Proposition 4), it follows that there are formulas expressing L(x D , 1 — .1.) in terms of i (n) and the number of ways n can be expressed as a sum of r squares. We illustrate with the simple case k = 5, /1 = 2. in this case M512 (t0 (4)) is spanned by CO and OF, and a comparison of the constant coefficients and the coefficients of the first power of q shows that H5/2 — 1 100 5 — *OF (see the exercises below). This yields the following proposition. —
Proposition 7. Let sr(n) denote the number of ways n can be written as a sum of r squares thus 0 5 =--- E s5 (n)q". Let D be the discriminant of a real quadratic field, and let XD be the corresponding character. Then 1
L(xD, 20s5(D) 1 = 17l)
D- j2 odd
Many other relations of this sort can be derived, in much the same way as we used the structure of Mk (F) in the exercises in §111.2 to derive identities between various ar (n) . For details, see [Cohen 1975 ]. These relations can be viewed as a generalization to A > 1 of the so-called "class number relations" of Kronecker and Hurwitz. (For a statement and proof of the class number relations, see, for example, Zagier's appendix to [Lang 1976] and his correction following the article [Zagier 1977 ]. ) The class number relations correspond to the case À. = 1, i.e., L(x D , 1 — A) = L(XD, 0). But A = 1 falls outside the range of applicability of Proposition b. To get at the class number relations from the point of view of forms of half integral weight, one must study a more complicated function H3/2 which transforms under F0(4) like a form of weight but which is not analytic. The IDI-th coefficient of H3/2 is essentially the class number of the imaginary quadratic field O(Ji75). The study of 113/2 is due to Zagier [19754 In some sense, the situation with 1/312 is analogous to the situation with E2 in §111.2. In both cases, when we lower the weight just below the minimum weight necessary for absolute convergence, problems arise and we no longer have a true modular form. The study of E2 and H312 is much subtler than that of Ek and 14/2 for k> 2 and k/2 > 2, respectively. But in both cases the Eisenstein series, though they fail to be modular forms, have a special importance: we saw that the functional equation for E2 led to the functional equation for the discriminant form A(z) e S1 2(n ; and H312 has as its coefficients the class numbers of all imaginary quadratic fields.
§2. tisenstein series of half integer weight for
r0 (4)
195
We next examine the quite different question of how the nonsquarefree q=expansion coefficients of Ek12, Flo, and 14/2 can be determined. Again we shall give the details only for Fk12 and shall omit the similar computations
for- Ek12 .We proceed one prime at a time. That is, for each prime p we express bi in terms of bio , where 1 = el° and p2 4". Finally, combining the different primes p, we shall be able to express 1)1 in terms of bio when / = 44 and 10 is squarefree. This information can be conveniently summarized in the form of an Euler product for Er, bioliti- s, as we shall see later. The easiest case is p = 2. Suppose that / = 4v 1o , 44'10 . We return to our earlier computation of 1)1 in (2.7). The general formula (2.7) is valid for all /, not necessarily squarefree. If we replace 1 by 4/ in (2.7), the double sum does not change, because replacing I by 4/ is equivalent to replacing j in the inner sum by 4f, which runs through as j does, since n is odd. Thus, 410-1 9 and so for 1 = 4v/0 we have ba, = = 2(k--2)v bio. (2.21) This case p = 2 can be summarized as follows: for any 10 , co
E
v=o
1
4v2-sv = /71 0
(2.22)
0 1 — 2k -2- s
Comparing coefficients of 2-" on both sides of (2.22) shows that (2.22) is equivaleni to (2.21). (Note that we do not actually need to assume that 44(49 .) Now suppose that p is odd. Let 1 p=2v/o, where p2 V ° . We again use (2.7) and divide into two cases. Case (i ) pi 10 . Let 10 =pio • We write the double sum in (2.7) in the form n> Ø odd Sri, where Sn is the term in (2.7) corresponding to n. Let n = nop', where p2 41 no . We fix n o and look at the sum of the Sn over all n = n0p 21 , h = ct, u 1, 2, . . . . Note that k/2 = no-kap -hk . we 6, = 8,. and nnow evaluate the inner sum overje ilni. First suppose that Ono . As representatives j of i mod ni we choose iop 2h ji no as j0 ranges from 0 to no — 1 and j1 ranges from 0 top 21 — 1. With p fixed we introduce the 'notation
=
if pij;
(2.23)
I if p Then for h > I we have p 2h
no (11) =
5(j) (it
)
;;
6(/1) (k) •
Also, e -21rign =
p I + 2(v-h)
Thus, the term Sn in (2.7) corresponding to n =-- nop
is the product of
Iv.
196
E
k – 142 e no no
(-/0 nn-
—
0 3 be an odd integer, A = (k — 1)/2, 4IN, x be a Dirichlet character modulo N. Let f(z) = 4°_, a ne' e Sic12( f0(N), x) be an eigenform for 7p 2 for all primes p with corresponding eigenvalue A p : 1,2f = Apf. Define a function g(z) = 4'1_ 1 be 2 " by the formal identity co 1 E b a n -s = 11 (4.1) + x(p)2pk-2— 2s • n=1
alip 1 — A pp —s
Then g e Mk-i(N I , x2) for some integer N' which is divisible by the conductor of x2 . If k > 5, then g is a cusp form.
Notice that the definition (4.1) of the b„ is equivalent to the following: (1) b 1 = 1; (2) bp = 2, for all primes p; (3) bpv = 2bp v-i — x(p) 2pk-2 bp v-2 for y >. 2; (4) b,,,,, = bn,b, if m and n are relatively prime. In Shimura's original theorem, the determination of the level N' of g was a little complicated. However, it has since been shown ([Niwa 1975]) that
-
§4. The theorems of Shimura, Waldspurger, Tunnel!, and the congruent number problem 213 one can always take N' ----z NI2. It should also be noted that Shimura actually - proved a somewhat more general theorem, applying to f which are not necessarily eigenforms for all of the Tp2. As a simple numerical ex-ample, suppose N =--- 4, x is trivial. The first nonzero cusp form of half integer weight occurs when k = 9, 11 = 4 (see Problem 5 of §IV.1). Up to a constant multiple, it is f = ORO' — 16F) =--E awl", where we have chosen f so that a l ---7. 1. (By Problem 17(h) of §I11.3 this f is also equal to q 12 (2z)/0 3 (z).) Clearly, f is an eigenform for all 7p 2, because S912 (r0 (4)) is one-dimensional. Then Shimura's theorem holds with N' = 2. Now Ss (f0 (2)) is one-dimensional and spanned by the normalized form g(z) = (n(z),1(2z)) 8 = ql-In"=1 (1 — q") 8 (1 — err ), 8 = E bnqn (see Propositions 19 and 20 in §III.3) ; hence this g(z) must be the g(z) in the theorem. It is now easy to relate the coefficients bin of g to the coefficients an off: Namely, using (3.7) with /0 = / 1 =-- 1, = 4, and noting that a1 = 1, x' = x (this is the trivial character mod 2, which equals 1 on odd numbers and 0 on even numbers), we obtain: bA =
=..- a P 2 + p3
if
p > 2;
b2 = a,.
(4.2)
While (4.2) follows immediately from Shimura's theorem, it is nevertheless quite a remarkable numerical identity : the p-th coefficient in q n (1 — q")8 (1 — q 2") 8 is equal to p 3 plus the p 2 -th coefficient in q 11 (1 — 47 2n) 1 2/(E q)3! Like many numerical relations that follow from the theory of modular forms, this fact looks rather outlandish when stated in this elementary form without the theoretical context. If we have a fixed set of linearly independent forms f in Sko (r0 (N), x) which satisfy the hypotheses of Shimura's theorem, then we can extend the Shimura map by linearity to the subspace of Skj2 (r0 (N), x) spanned by them. Note that the image gi off is always a normalized eigenform in Sk_ 1 (N/2, x 2). If we take another set {f'} of forms which satisfy Shimura's theorem and are also a basis for the same space as thef (for example, if we multiply each f by a scalar), the Shimura map is clearly affected. When we refer to the image of a single eigenform under the Shimura map, we shall always mean the normalized eigenform g in Shimura's theorem. But if we have a space of modular forms with fixed basis f of eigenforms with Shimura(f) = gi , then we define Shimura(/ aifi) = I aigi , which is not necessarily a normalized eigenform. So our meaning of "image" or "preimage" under the Shimura map depends upon the context. In general, it is possible for several different fe Ski2 (r0 (N), x) to go to the same g e Mk_ , (ro (N), x 2). For instance, there might be anf' e Sk12(fO(N) , x') which corresponds to g, where x' is a character different from x which has the same square: x 2 = x'2 . However, when N = 4, Kohnen [1980] described the situation more precisely. We now describe his main result. Let Mk+12 (r0 (4)) denote the subspace of Mk/2 (r0 (4)) consisting of f(z) = Eane for which an = 0 whenever (— 1)'n - -. 2 or 3 modulo 4. It is certainly a priori possible that there are no nonzero forms f with this property, which
214
IV. Modular Forms of Half Integer Weight i
requires that "half" its coefficients vanish. However, we know from Proposition 9 that 14/2 is such a form. Also, it is easy to check (using Proposition 17(a) of §I11.3 and (1,.8) of §IV.1) that 0(z) f(4z) e /1/42 (r0 (4)) for any fE M(k-1)12(SLi(Z)). It turns out that Aik2 (ro (4)) is the direct sum of the one-dimensional space spanned by the Eisenstein series 14/2 and the subspace Sk+12 (4)) of cusp forms:
(ro
,
/
sk412(r0(4)) = {f=
E anqn E Skigo( 4))1an = 0 if (-1)'n -m 2 or 3 mod 41. (4.3)
It is not hard to show that Mk+/2 (f0(4)) is preserved by all of the Hecke operators Tp2 except for T4. According to Shimura's definition of T4, one has T4 I ant? = E a4„q". If we look back at §IV.2, we see that Flo is an eigenform for T4 (with eigenvalue 2 2 , see (2.21)), but Ilk/2 is not. In fact, it is easy to see that T4Hk/20 Alk+i 2 (ro (4)). For this reason, Kohnen modifies' T4, and defines a slightly different operator TI so that the m-th coefficient of T: 1 anqn for (-1)2m -a 2 or 3 mod 4 is zero and for (— l )m a 0 or 1 mod 4 is equal to
a4„, + x(_..1)Am(2)2'ia,, + 2k-211"m14• With Shimura's definition, since we have x(p) = 0 when pIN even if x is the trivial character on (ZINZ)*, the second and third terms vanish in
a4m + XX(-1)Ani( 2) 22-1 ain + X(4) 2k -2am/4. Thus, Kohnen's modification is to replace the trivial character x by the map which takes the value 1 (and never 0) on all numbers including those not prime to N. It is T-4,- rather than T4 which preserves M 2 (F0(4)) and Sk4i2 (f0 (4)). Note that Hk/2 is an eigenform for T: with eigenvalue 2 2 . Kohnen further shows that M4-2 (r0 (4)) has a basis of eigenforms for all of the Tp2 (p 0 2) and for n which is unique up to permutation of the elements and scalar multiplication. There is no obvious way to normalize an eigenform f= E anqn ; for example, we cannot necessarily multiply by a scalar to get a l = 1, since a l = 0 for all fe 47/2 (f0 (4)) if .3. is odd. But one can require that the coefficients all lie in as small a field extension of 0 as possible. It turns out that the images g of these eigen-basis forms Je Sk4i2(r0(4)) under the correspondence in Shimura's theorem are all contained in Sk_i(F), r = 51,2 (Z) (the results of Shimura-Niwa only guarantee that they are in Sk-1( r0(2))) ; they are all distinct; and they form a basis for Sk_i(r) consisting of normalized eigenforms for all of the Hecke operators T„ acting on Sk _ 1 (F). (In particular, dim Sk41-2(r0(4)) = dim Sk_i (F).) Thus, if we take each of our basis elements for S4-2(r0(4)) to its image under the Shimura map—and take - p I/42 to the normalized Eisenstein-series -i---1-1 —and then extend by linearity to all of Afk4i2 (r0 (4)), we obtain an isomorphism from M4-2 (r0 (4)) to Mk_i (F) (and from Sk,2 (r0 (4)) to Sk _i(fl). This isomorphism commutes
§4. The theorems of Shimura, Waidspurger, Tunnell, and the congruent number problem
215
with the Hecke operators, in the sense that:
Mk4i2(r0 (4)) TP2
f
g E Mk -1(r)
I I
TP
and
nI
g
I
T2
T44-fi--+ T2 g
Ti,g
The basic method of Shimura's proof of his theorem was to use Weil's theorem which we discussed briefly at the end of §III.3. Weil's theorem says that if E bnn and its "twists" E bn O(n)n' for certain Dirichlet characters each satisfy the right type of functional equation relating the value at s I s, then g =. E bn iqn e Mic _i(Fo(N /), x 2). But the proof to the value at k that all of these functional equations are satisfied is not easy ; about twenty pages of [Shimura 1973a] are devoted to an investigation of delicate analytic properties of the Dirichlet series corresponding to g and its twists. Shimura's correspondence seems rather roundabout. It says : take the q-expansion of a suitable Je Sk/2 (r0 (N), x); look at the q-expansion coefficients a„ as n varies over integers with fixed squarefree part /0 , and form a Dirichlet series from them which turns out to have an Euler product ; then take the part of this Euler product which is independent of /0 , and expand it and finally, go from this new Dirichlet into a new Dirichlet series Eb n series to the q-expansion E bn qn, which will be your modular form of integral weight. After Shimura's paper appeared, people started looking for a more conceptual, less roundabout construction of the Shimura correspondence. Certain more direct, analytic constructions were given by Shintani [1975] and Niwa [1975 ]. In addition, group representation theory was found to provide a conceptual explanation of this correspondence (see [Gelbart 1976], [Flicker 1980]). Moreover, the use of representation theory has led to striking new results about forms of half integer weight, especially in the work of J.-L. Waldspurger. Using representation theory, Waldspurger [1980, 1981] proved a remarkable theorem establishing a close connection between critical values of 1 e 21 and the coefficients in 4-series for a modular form g of weight k the q-expansion of a form f of half integer weight 10 which corresponds to g under the Shimura map. Roughly speaking, the theorem' says that the critical value is equal to the square of a corresponding q-expansion coefficient times a nonzero factor which can be explicitly described. Waldspurger's general result is complicated to state, so we shall \ only describe what it says in two particular situations. As mentioned before, Kohnen [1980] showed that the Shimura map gives an isomorphism —
—
—
ra s2 (f(4)) Shimu --° Sk-i (n.
(4.4)
Here S 2 (r0 (4)) is defined in (4.3). Let g(z)=-- E bonqn e Sk _ i (f) be a normalized eigenform for all of the Hecke operators, and let xi, be the character corresponding to the quadratic field of discriminant D. Suppose that
216
IV. Modular Forms of Half Integer Weight
(— 1)'D > 0, i.e., the quadratic field is real if A = (k — 1)/2 is even and it is imaginary if 2 is odd. Recall that Lg (ID, s) denotes the analytic continuation of Enap - i XD(n)bnn -8 (which can be shown to converge absolutely if Re s> k12). Let f(z) = Zan g?' e S4-2 (r0 (4)) be the unique preimage of g under the Shimura map (4.4), i.e., g - 7---- Shimura(f ). Let )122Di.
Lig (XD, /1.) = (11)1) (k — 1)! k2 , then the mapping is easily seen to be a character of (Z[i]/D*; if k 2 > k 1 , then def X(X) =f i(OXk (NX)k l is a character of (Z[i]/j)*. In, for example, the case 1( 1 > k 2 , the Irecke L-series is then .1 "E 0 x e ZVI 2(X)(N1 xrs E xk x(x)(NV2 - s. 5. (a) Make the change of variables x' = Mx to obtain 4(y) = Sure-216M 2 Y f(e) idde :`,;41 — -frfiFifif(M *y), since Ats i x' • y = x' • M*y. (b) L' M*Z" ; let g(x) =f(Mx). Then rr g(m) = I mEz nj(rt) ide!mi EyL,./(y) by part (a). Exez.f(x) = 6. (a) M* = -A-(1 ?); considered in C, L' = Note that Tr xy = 2x - y, where on the left x and y are considered as elements of C, and on the right as elements of R2 . (b) In Problem 5(b), let f(x) = e - '11'12 , g(x) = f(Mx) with M as in part (a); then apply Poisson summation. (c) Let 0(t) be the sum on the left in part (b), and let 4)(s) = 1`2,,rst 5(0(t) 1) 1L + 0(0(0 —')1'-. For Re s > I show th a t
Answers, Hints, and References for Selected Exercise
230
4)(s) = (20)sa + 1 10 + n -s1"(s)6CK(s). The residue at s == 1 is n/3f."3. Finally, replacing t by 4/3t in the integrals for ck(s) leads to the relation: 0(s) = (2/j) 21 4(1 s), and this leads to A(s) = A(1 s) for A(s)rf (07270s1"(s)4(s). (d) Use the Euler product form K(S) fl p(1 (INP) -s) -1 . For ea prime ideal P of norm p 1 (mod 3), the contribution from P and P is (1 - sr 2 ; for P = (p), where p 2 (mod 3), the contribution is (1 - p -2T' = (1 (l - X(P)PT"' ; and for P (NI -3) the contribution is (1 - X(3) 3-T"' (since x(3) =-- 0). So the Euler product is the Euler product for “s) times the Euler product for L(x, s). (e) Multiplying the functional equations for C(s) and for L ( c, s) gives A(s) = A(1 s) for A(s) = n.-42 r()C(s)(3/7 ) 42 r((s + 1)/2)L(x, = ("jinn K(s)r OW as + 1)/2) = const -0/27r)sr(s)C K(s) by (4.4). 7. See Problems 20-22 in §II.2; (d) g(x, tfr) = 3. 8. (a) By part (1) of Proposition 9, j(y) equals e'Y times the Fourier transform of (x w)e - nox'12 ; note that x w = Mx • (1, i) with M as in Problem 6(a); then proceed De zniu- y e--(47rooly•(-1„), (I)) Use as in Problem 6(b) to obtain 4" (y) = 34t y . n-sr (s)6L(E, s). (d) Replacing t by IF in the Problem 7(c) to obtain 0(s) = eh where the integrals in 4 (s), one obtains 4)(s) = (-1)s -1 Ex(a + ba))14T 2t summation is over 0 a, b < 3 ; u = (a13, b13); and ) e2g1"•rne-4tInl'( '' 1) I 2 . Then for Re (2 s) > 3/2 use (4.6), OU(t) = E„, z2 tn • ( Probrm 2 of §11.2 (note that u • in = iTr((a + bco)(- &m 1 m 2)1i0)), and the evaluation of the Gauss sum in Problem 7(d). The result is: 0(s) = (4/3)5-1 370 -2 1-(2 s)6L(E, 2 s). Equating this with the expression in part (b) and t 12
1/
collecting terms gives the desired result.
01.6 2. The function f(s) = A(1 + s) is even in the first case (so that its Taylor expansion at s 0 has only even powers of s) and odd in the second case. 3. (c) Use: •(3-1, and 2ap = ap 'ercp .
5-6. n
first few nonzero kn,„ 2 b 1 = 1, b 5 = 2, b9 = -3 3 b 1 = 1, b5 = 2, b 13 = 6, b17 = -2 10 = 1, b9 = -3, b 13 = 6, , l7 = -2
1,(E„, 1)
0.92707 1.5138 1.65
7. You want 14 4. 1 1 to be less than c/2, i.e., c/2 > 4(1---
remainder estimate
0.00027 1R25 1 0.00123 1R29 1.ç.. 0.289 e m+iy,fir ; for 1R131 ...ç.
N' e - gml`51- , so choose M> N ' 1og(8,1Thre) -71-NTAT; log n, i.e. (2n N/2/7r)log n for n odd, (2n/n)log n for n even. 8. (a) Use Problem 3 to find the b„,,,,. (b) See Problem 7 of §I.1. large n the right side is asympotic to
1. Te 1-1 (N) c ro (N ), but STS -1 F0 (N); hence, neither I-, (N) nor Fo (N) is normal in r. 2. (a) Given Ac SL 2(ZI1VZ), let A be any matrix with A = A mod N. Find B, CESL 2 (Z) such that BAC is diagonal: BAC = (g ?,), where ad =' det A 1 (mod N). It suffices to find A= re 'SI) with determinant 1, since then SL 2 (Z), and B -1 B -1 (.4 1)0C -1 = A 71(mod N). To find A so that 1 = det 1 = 1 + ((ad - 1)/N + xd)N - zN 2 , first find x so that xd -(ad
1)/N (mod N), and then choose z so as to make det /i= 1.
231
_Answers, Hints, and References for Selected Exercises
3. (a) (q 2 — 1)(q 2 — q); (b) q(q 2 — 1). 4. (b) Since the kernel in part (a) has p4(`'' ) elements and #GL2 (7L/pZ) = (p 2 — 1)(p2 — p), it follows that #GL 2 (11peZ)= p4e-3(p2 1)(p — 1). (c) Divide the answer in (1)) by (Ape) to get p 3e-2( p2 _ 1) - 5. Use the Chinese remainder theorem. 6. N 3 rIppia — P -2 ). 7. N 3 flpiN (1 — p -2); N; N11,0,(1 — p -1 ); N 2 1-1piN (1 — 1) -1 ); Nri piN (1 + if') 8. The image of F(N) under conjugation by (/?; 1) is the subgroup of F0 (N 2) consisting of matrices whose upper-left entry is r--- 1 (mod N). 10. and 14(b). See Figures A.1—A.2. 12. Besides F and F(2), there are four, namely, the preimages of the three subgroups of order 2 in S3 and the one subgroup of order 3 under the map 1" —0 SL2 (Z/2Z)•:-., S3. (1) r0(2) = {(t, :) mod 2), F0(2) = the right half of F(2); (ii) r(2) = {(: :) mod 2 } , F°(2) = Fu T -1 Fu SF; (iii) 0(2)-.----. {(ccf bd) E..- I or S mod 2 } , fundamental domain = Fu T' Fu T' SF; (iv) {C. bdr-1- I, ST or (ST) 2 mod 21, fundamental domain = Fu T' F. 13. (a) This can be proved using the fundamental domain, as in the proof of Proposition 4 in the text. Here is another method. Let G denote the subgroup of 0(2) generated by +S, T2 . Clearly G C 0(2). Conversely, write g E (g1(2) as a word of the form +Sal T 1'IST b2 • • • STN, where al = 0 or 1 and bi # 0, j = 1, . . . , / — 1. Use induction on / to show that g E G. We work mod +1, so that S 2 = (ST) 3 = 1. Without loss of generality we may suppose a l = 0, bi # 0, since we can always multiply g on the left or right by S without affecting whether g e G. For the same reason we may suppose that b 1 ------- 4 = 1, since T 2 e G. Note that I= I or 2 is impossible, since T,TSTO 0(2). If / > 2, write g = TSP2 - - • . Since (STS)(TST) = 1, we have TST = (STS) - ' = ST' S, and so T 2 Sg = TSTb2 -1 S • • • , which is just like g but with b2 replaced by b2 — 1. If b2 > 0, use induction on b2 to finish the proof. If b2 < O, write ST'g = ST -1 ST b2 - • • = TSr2 +1 - - - , and again use induction on 'bd. (b) Use: r°(2) = F0(2) = STO(2)(ST) -1 . (c) Let G be the subgroup of generated by T2 and ST'S. Since G c= r(2), it suffices to show that r: G]= 6, e.g., that any "word" in S and T can be multiplied on the left by elements of G to obtain one of the elements cx; used in the text for coset representatives. Use S 2 = (ST) 3 = 1 and induction as in part (a). (d) Use part (c) and the isomorphism in Problem 8. 14. (e) Fo( p) is bounded by the vertical lines above l and —1; arcs of circles with diameter [0, 2] and [— 2/(2p — 1), 0]; and arcs of circles with diameter [ —1/(k — 2), —1/4
r
[
(The centers of the boundary arcs are at
—1 /2, 0, 1/6, 4/10.) %
1/2 Figure A.1
232
Answers, Hints, and References for Selected Exercises
I
1 ....Pw
...
■ ......
.......
4....
•as, .....
1
0
1/2
Figure A.2. Two possible fundamental domains for F0 (4):
I. Let F(2) be the fundamental domain for F(2) in the text. Then ŒF(2), where a = (3 1), is a fundamental domain for F0(4) = Œr(2)a -1 (see Fig. A-1). IL U__ / ocT'F is a fundamental domain for F0(4), where F is the fundamental domain for r = SL 2 (Z) given in the text, and ai are coset representatives for r modulo F0 (4). Here we have taken: a l = 1, a2 = S, a 3 = T'S, a4 = 7-2 S, a5 = 7-3 S, a 6 = ST'S. (In Fig. A.2, the number j labels Œ' F; the solid boundary arcs are centered at 1, —4, —i; the dotted arcs are centered at 0, —1, —4, --i, —4; the point P is (-7 + i0)/26.)
k = 3, . . . ,p. For example, F0 (3) is the union of the regions marked 1, 2, 3 and 4 in Figure A.2. 15. See Problems 12 and 14(c) for counterexamples. 16. (b) (i) 2; (ii) 2; (iii) 2; (iv) 1.
§111.2 1. In the sum for Gk, group together indices m, n with given g.c.d. 2. Substitute z = i in Proposition 7 to obtain: E2 (i) = 3/7r. 3. (a) Since both sides of - each equality are in a one-dimensional ilik (F), by Proposition 9(c), it suffices to check equality of constant terms. (b) Equate coefficients of q" on both sides of the ' equalities in part (a) to obtain: o-7 (n) = o.3 (n) + 120 ET.1 o-3 (j)o-3 (n — j); 110-9 (n) = —100.3 (n) + 21a5 (n) + 5040 ri',::,1 o-3 (j)o-5 (n — j); 613 (n) = 21a5 (n) — 20a7 (n) + 10080E7:4 a.5 ( j)a7 (n — j). 4 • (a) Since the left side is in S1 2 (f), by Proposition 9(d) it suffices to check equality of coefficients of q. (b)t(n) = 76556 o-1 / (n) + -a5 (n) — 0.5 ( j)o.5 (n —j). (c) Consider part (b) modulo 691. 5. Let F(X, Y) be an irreducible polynomial satisfied by X ----- E4 , Y = E6. Substituting z = i in E4 (z), E6 (z) leads to a contradiction, since E6 (i) = 0, E4 (i) * 0. 6. (b) Use part (a) with x = e- 2tt = e2ni(i), along with the relation 0 = 2(2, /)/3„, / E„.,. / (i) — 2(a1+ i)Ba+ 1 — Efz/ sria (n)q n with q = e2gi(i). '7. Use Proposition 7 and the derivative of the identity f(— 1 fz) = z kf(z). 8. (a) By Problem 7, the right sides are in m6(r) and m8(r), respectively. Now proceed as in Problem 3(a). (b) 210-5 (n) = I0(3n — 1)(73 (n) + ai (n) + 240E7:1 al (j)a3 (n — j); 20a7 (n) = (42n — 21)a 5 (n) — ai (n) + 504 E.7:1 ai (j)a5 (n — h. 9. (a) Use the fact
233
Answers, Hints, and References for Selected Exercises
that n2 1 (mod 24) if n is prime to 12. (b) Use Proposition 9(d) to show that their 24 ±1 mod 12 q — q n) = 24-th powers are equal. (c) Ens 15 mod 12 q (n2 -1)/24 .
10. j= 256(.1 -2 + I + 12 ) 3/(.1. + .1 - 1) 2 ;j= 1728 if A = I; if A = alb in lowest terms (with a and b positive) and jeZ, that means that a4 b4(a2 + 132) 2 divides 256(a4 a2 b2 + b 4) 3, but since a, b and a2 + 112 each have no common factor with a4 + a2b2 + b 4, that means that a2 b2 (a2 + b2) divides 16, and now it's easy to eliminate all possibilities except a = b = 1. 1
§111.3 4. (a) Since F0 (N) = +F/ (N) in those cases, there is no difference between F0 (N)-equivalence and fl (N)-equivalence. (b) —1/2 is an irregular cusp for F0 (4).
5. Group
ro(p)
Cusp Index
o
co
f(2)
ro(P2)
—1Ikp, k =1, 1
p
,p 1
0
oo 2
2
—
1 2
7. (a) See the proof of Proposition 18; replace a by X in (3.11). 8. (a) Replacing z by z + 1/2 in Z e21riz"2 gives Z(-1)"e'" 2 ; meanwhile, the right-hand side is 2 E e2R1z(2n)2. (b) works the same way. (c) On the left side, the constant term is clearly zero; the coefficient of qn for p ,t n is (-2/c/Bk)ak_ 1 (n); if pin but p21e n, then the coefficient is —2k/B k times 0-k _ 1 (n)— (1 + p k-i )o- (n1p) = 0; if n = pmn o with and Ono , then the coefficient of q" is —2k/B k times (pmn o) —(1 + pk-i )o-k _ i (pm -1 n o) p ic-1 0.k_, ( pm— 2no)
>1
( 1 +Pk-i )crk i(Pm-1 ) +Pk-i crk 1(P m-2)) = O. (d) Use parts (b) and (c) with p = k = 2. (e) Rearrange the infinite product of (1 _ e2icicz -1/2)n) = 1 Ong') by writing 1 + qn = (1 — qn), getting
ak---1010)(0i 1(Pm1) -
—
-
-
(where 11 denotes 11(1 — qn)). But this equals nn2even even Mall n twice an even 11 4in • 9. See the proof of Proposition 30. 10. (a) Since n8 (z)ri8 (2z)ES8(F0 (2)) by Proposition 20, to show invariance of ri 8 (4z)/n 4 (2z) under [y] 2 for y E F0(4) it suffices to show invariance of (4z)e(z)71 4(2z) under [y] 10 for y e r0(4) ; now use Problem 9 to show this. At the cusp 00 5 we see that r/8 (4z)/q4(2z) = q11(1 q4") 8/(1 q2")4 has a first order zero; to lin even • rintwice an oddinnodd
find q(0) we apply [5 ]2 : Z -2 q 8 ( - 41z)17/ 4(— 21z) = z -2 (z/4) 4 n8 (z 14)1( —(z/2) 2 114(42)) = —411(1 — q) 8/( 1 q , which approaches --1-4 as z cc. To find the value at the cusp —1, which is equivalent to the cusp = T(— 1/2), we can apply [(I ?)] 2 , as follows: —
/78 (2z + 1)-2
(
4z
(2z + 1) 4
2z + 1) = (2z + 1)' ( 2z ) 114 2z + 1
4z (2z +
64z 2
( 1 4z
1\ 2,14
2z ) ti
=
8
8
1 77 2° (2z) 16 ri 8 (z)1 8 (4z)'
1)
1 __
2z
)
24
2Z1 ) (-
e —26
)4
—1\ 4z )1 8
1
by Problem
z1) e-2ni/6 4 (
I 2z
8(e)
234
Answers, Hints, and References for Selected Exercises
which approaches as z oo
(b) E2 (ST -aSz) = E2 (S(- a -))= (a + E2( - a - 12") + Ir(- a - +), but E2 (--a - = E2 (-1), so this equals (a + 71) 2 (z 2E2(z) + + TN- a - = (az + 1) 2122 (z) - 6-7- (az + 1). (c) Obviously F1[7] 2 = F. Using Problems 1 and 10(b), we have -24F(z)1[ST -4 S] 2 = E2 (z)1[ST -4 S] 2 3E2(2z)l[5T -2 S] 2 + (E2 (2z) + (42+10 + 2(E2 (4z) 2E2(4z)I[ST-1 S12 = E2 (Z) + (42+1) (42+1)) -24F(z). We now look at the cusps. First, we clearly have F(co) ----= O. In evaluating FI[S] 2 , ignore terms which approach zero as z co ; then, using Proposition 7, obtain F(z)1[S] 2 "rd — 24E2(Z) — 3 (1E2 (Z/ ) 2(- E2(44)) -) = + we similarly ignore terms which approach zero: This is F(0). To find F( + 2(2z ± 1 ) -2 E2(1÷– 24F(z)i[ST -2 S] 2 = E2(z)I[ST -2 Sh 7 3E2 (2z)I[ST f 2) by Problem 1 above, and, by Proposition 7 and part (b), neglecting terms which = D2 E2(-ii approach zero, this becomes E2 (z) - 3E2(2z) + 2(2z + 1) -2 ( E2 (Z) — 3E2(2z) + z 2 E2 — ± 6E2 — "1i) --- 4E2( I)) by Problem 8(d); again applying Proposition 7 and neglecting small terms, we obtain E2(z) - 3E2 (2z) + -13- ( - 16E2 (4z) + 24E2 (2z) - 4E2 (z)) - t - 3 + -I ( -16 + 24 - 4) = (d) Since the only zero of ri 8 (4z)f 4(2z) is a -4, so that F( 1) = M 1) = simple zero at co, it follows by Proposition 18 that F(z)101 8 (4z)M 4(2z))€1110(F0 (4)) is en) = 1 + t )•(e)- • First show a constant. To get the identity, write (1 - 17 4n)/( that A(z + ilN)eS12m(ro(N 2 )), using the same type of argument as in the proof of Proposition 17(b). Then set f(z) = 6.(z) N 111A(z + jIN), and take the logarithmic derivative of the equality f(yz) = f(z) for y eFo (N2). 11. (a) Prove invariance under [A 2 for y E r0(4) as in Proposition 30. At the cusps, clearly 04(co) = 1; at 0 we have 04 (z)1[5 ] 2 = 2 2 04 ( - 1/(4z/4)) = z -2 (- z 2 /4)04(z/4), which approaches -I as z oo ; at .1- we have 04 (z)1[ST -2 S] 2 = (2z + 1) 2 04(z/(2z + 1)) = (2z + 1)2®4(_ 1/(4( - 1/4z - 1/2))) = (2z) 2 ® 4(- 1/4z - 1/2) by (3.4), but by Problem 8(a) this equals (2z) -2 (20( 1/z) - E(- 1/4z))4 = -1(2(20 -112 0(z/4) 0(z))4 = (0(z/4) - 0(z)) 4, which approaches zero as z co. (b) Use: 04(oo) 0 = F(co). (c) One can proceed directly, as in Problem 10(a). Alternately, write it as re (2z) and use Propositions 17 and 20 and Problem 10(a). An even 6.(2z) —
-
(
-
-
-
-
en\
.-
(n(z)g(2z))8778(4z ) ,
easier method is as follows. Letfi (z) 714(2z)M 8 (4z), and let f2(z) re ° (2z)117 8 (z)e (4z). In the solution to Problem 10(a), we saw that for cc = ST -2 S = ?) we havef2 = 16f1 1[4 2 . Sincefi is invariant under [y] 2 for y e F0 (4), it follows thatf2 is invariant under a -1 1-0 (4)a, which is easily checked to be equal to F0 (4); finally, since a keeps 0 fixed and interchanges the'equivalence classes of cusps oo and -1, we see thatf2(0) = 1 6f1(0) = - 1,f2( 16f, ( co) = o, f2(œ) = 145fi(—) = 1. (d) Same procedure as 10(d). (e) Use Problem 8(e). 12. Follow the proof of Proposition 19. 13. Write (q(z)11(3z)) 6 = . (z)(n(3z)17/ 3 (z)) 6 , (1(z)1(7z)) 3 = (z)(q(7 z)1,7 7 (z)) 3 . 14. Check the generators T2 and S, using (3.4) to get 0(Sz) = firk/Az); to check the cusp -1 = (TST)' cc, write 04(z)I[TST] 2 = 04(z/2)1[(1 ?)] 2 (see Problem 1). 15. (a) For y = e 170 (N) note that ocNyaV = (- `kb -`,,1N)e Fo (N), and so (f I [aN]01[Y]k = (f 1 [OENYŒV ]k)I1ŒNik = x(a)f [ŒN]k. But X(a) = TO). (b) For fe Mk (N, x) write f f+ ' + f , where f = ± efl[aN]k). (c) By 10(d), +.1q + - = + F; matrix is F(z)I[Œ4]2 = -- *4 8 (z)M 4 (2z) = (1 116); M; -(4, 1)' = C-10 4, M;(4, 1) = -12-404): 16. (a) E ak_ i (n)n -s = Zfrii _l_„1 -1 (fin)7 = Z Ent 5 = C(S 7 k)(s).
235
Answers, Hints, and References for Selected Exercises
I: 1/2
0 Figure A.3
Sc
SB
Figure A.4
s 4. pk-1-2s)l . (b) L1(s) = n[(1 — p)(1 — Pk-1-5)r =14( 1 — ak-i(P)P(c) L1 x (s) = 11 19[0 — X( P)P - s)( 1 —
= np( 1 — X(P)ak-i(P)P -s + X(P) 21)4-1-25) -1 . 17. (a) Any point r'-equivalent to i must be F0 (4)-equivalent to one of the points aj-1 i, where 1" = Uï... 1 airo (4). In this way, find that i, (2 + i)15 e Int F' are F-equivalent to i; co and (5 + i.11)/14E Int F' are F-equivalent to co ;. the two F0 (4)-equivalent boundary points (— I + 012 and (3 + 0/10 are F-equivalent to i; and no boundary points are F-equivalent to co. (b) Follow the proof of Proposition 8, but with slightly more involved computations. Note that the three "corners" (-3 + i. 1 ) /4, (1 + i0)/4, (9 + i0)/28 are all F0 (4)-equivalent, and the sum of the angles at the three corners is 360 0 . To illustrate the elements in the integration around the cusps and along the circular arcs, let us compute (27ri)-1 f f ' (z)dz If(z) over the contour ABCDEF pictured in Fig. A-3, where we suppose that f(z) has no zeros or poles on the contour (but may have a zero or pole at the cusp 0), and we ultimately want the limit as E --+ O. Here BC is a circular arc of radius e centered at O. The element a l = (1 ?) E F0 (4) takes the arc from 'A to O to the arc from D to O. The image of B is very close to C, and so we can use the integral from D to a l B to approximate the integral from D to C. First, we use S to take the arc BC to a horizontal line between Re z = —3 and Re z = 1. (see Fig. A.4). By (2.24), we have
f : fj
.
z z) )
dz =--- is: f l i dz — k f c dz . B
Z
But the latter integral approaches 0 as E -4 0 (since the angle of the are BC approaches zero), while the first integral on the right'approaches —v o(f ) (as in the proof of Proposition 8 , but we use the map 44 = e21644 into the unit disc). Next, by
236
Answers, Hints, and References for Selected Exercises \
(2.24), we have
= _ k f B dz j A z+ 1/4
fBf/(z)dz +fDr(z)dz,, l'Brkl(tz fœlitLikIdZ
J 4 f(z)
c f(z)
JA f(z)
--, _k J /
0
j Go f(z)
dz z+ 1/4
= -k
I,
1/4
dz
.
— 2 -1-isrS)/4 Z
Similarly, using a2 = a 1) e r0(4) to take the arc DE to the arc FE, we find that the sum of the integrals over those two arcs is equal to - k times the integral of l• from (2 + i0)/28 to i. These two integrals are evaluated by taking in z between the limits of integration, where the branch is determined by following the contour. As a result, we find that the sum of the two integrals is - k times the logarithm of
1/4
1/4 --1 0)/4 (2 + i.0)/28 (-2+
•
where, if we keep track of the contour, we see that we must take ln( - 1) = - ni. Thus, (20 -1 times the sum of these two integrals is equal to k/2. For a systematic treatment of formulas for the sum of orders of zero of a modular form for a congruence subgroup, see [Shimura 1971, Chapter 2]. (c) The only zero of 04 is a simple zero at the cusp -1; the only zero of F is a simple zero at co. (d) If fE M2 (ro(4)), apply part (b) to the element f - f(cc)0 4 - 16f( - - -1) Fe M2 (ro (4)), which is zero at -1 and co, and hence is the zero function. (e)-(f) See the proof of Proposition 9. (g) Look at the value at each cusp of a012 + b08 F + c04 F2 + dF3 : f(cc) = a, f(-1) = dI163 , so a = d= 0; then f= 1(0) = (--4 - ) (-4)( -I), so c ------ -16b. (h) Let f(z) = ql 2 (2z). Since f(z) 2 = A(2z)e M12( 1-0( 2 )), we immediately have vanishing at the cusps. Let Œ = (_.1 ?). By writing 2az = -1/(1 - 1/2z), show that f i[a]6 = -f, and so fit M6 (r'0(2)). However, r0(4) is generated by T and a2. Next, since s6(r0 (4)) =C(08 F- 1604F2), to check the last equality in part (h), it suffices to check equality of the coefficient of q. 18. (b) Iffe M 1 (4, X), by subtracting off a multiple of ®2 we may suppose that f(co) = 0; then M2 (4, 1)f 2 1.--- a?q2 + - • - . But a multiple zero at co Would contradict Problem 17(b), unless f is the zero function. (c) First show that, for k _._. 5 odd, Sk (4, x) = 0 -2 Sk+I (4, 1); then, by Problem 17( 1 ), dim Sk (4, X k) --= [(k - 3)12] (where [ ] is the greatest integer function). (d) Use Proposition 20, Problem 17 (with F' = F0 (2), a = CI 7)), and part (c).
§111.4 2. Show that V- i(cz + d)/V - 4- cz - d) = rsg" (it suffices to check for z = 0), and then divide into four cases, depending on the sign of c and d. 3. (b)-(c) Use the (. 0 a/3z) 0(43z) 0(i3z) relation 0(z) - 0(/3z) 0(z) .
§111.5 2. (a) By (5.27) and (5.28), Tn(f i[ŒN]k) = d k/2" Ea,bf 1[ŒNcra(c; ba) ]k . Let , Œa ,b = anaN ajg riDoc-IV ENV, tl}, Z), and show that the Œ,b are in distinct cosets of r;(N) in An(N, {1}, Z); and hence form a set of representatives; so, by (5.27),
Answers, Hints, and References for Selected Exercises
237
(Tnf)i[aidic = n(42 " Ea,bfl[cca,baN]k = n (W2)-1 Ef i[cr naN0-.(8 !)I beCaUSe fl[Crac = f. (b) TF = 0, T204 = 04 + 16F, so {F, F + A.Ø 4} is a normalized eigenbasis. (c) (7'2 F)i[Œ4] 2 = 0 # T2(FI[Œ4]2)- (d) Any eigenvector of T2 or [M k must be an eigenvector of T„ ; this includes F,iF +2-40 4 , C04, and so all of M2(r0( 4 )); by Proposition 40 applied to F, we have An = al (n) (n odd). (e) Let 04 (z) = Eanqn. For n odd, a, = (n)a i = 8u1 (n) by Proposition 40. For n = 2no twice an odd number, compare coefficients of qn0 in T2 04 = 04 + 16F to get an = a 16a1 (n0) = 24d1 (n0). For n = 2n 1 divisible by 4, compare coefficients of qi_'1 in T2 04 = 04 + 16F to get a„ = an, +13 = 24o-1 (n0) by induction. 3. The first part follows from (5.19) with n = 0; a counterexample for other cusps is in Problem 2(b), since Co4 + 16F does not vanish at s = —1/2. 4. (a) Both 1,, f and Tnf are equal to n(142 ". Eflk(401 M k over the same a, b, d, because g.c.d. (n, M) = 1. Note that g = M -42 f 1[01 M k . Thus, AP/2 T g = n(42)-1 E f 1[( ?)o-jg 'la, where ?,) mod MN. Set aa in the sum for T,, f equal to (411 ?)aj ig -?)'. Then = ((k/2I Ef f[aa(1 ?)(sSI !)O' ?)--I]ol oof ?)],= ( 7nf)1[0o1 ?ilk, because in (f; bia4) = (11 ?)(iO4 ?)--1 the entry bM runs through a complete set of residues mod d. (b) Since s8(r0 (2)) = Cf, f is an eigenform for Ta ;.then by part (a), g is an eigenform with the same eigenvalue. (c) T2 9 = U2 V2 f = f; T2 f = —8f by a comparison of coefficients. Hence {f, 8g +f} is the normalized eigenbasis. 5. = < p0/ 2)-1f1[( )] kl g> = Ef: , by (5 . 33) . ii)] Since gl[(t; li)]k = g(pz j) = e2 g(pz), we have = p(f, Pk-1 g(pz)> = p k . 6. (a) It has a l = O. (b)f, =q2 48q3 + 8 . 27 . 5q4 — 64 .5 . 47q5 + 4 . 9 - 5 .17 - 47q 6 + ••-;
(11■1
[
12 = q
24. 43q 2 + 4.9.49.139e + 64.171337e + • ; T2f1 = q + 1080q 2 + • • • =f2 + 2112f; T2 f2 = —24 . 43q +(64 .171337 + 2 23)q2 + - - = —1032f2 + 29367211(0 Tr T2 = 1080, Det T2 = — 2 ". 3 2 • 2221; eigenvalues = 540 + 12V144169; normalized eigenforms aref2 + (131 ± J144169) 12f1. (d) 2f2 + 3144fi . (e) From part (d): 2(4 . 9 -49 139) — 48 . 3144 = 89 5-23 41. Alternately, compute T3f1 = —48f2 + 36 - 2619f1 5 T3 f2 = 36 . 681112 *f (we don't care about *), so Tr 7'3 = 36(2619 + 6811) = 8 .9 .5 . 23 - 41. 8. In case of difficulty, see §3.1(c) of [Serre 1973 ].
NV.
= «73 ?), n2-114) ((: ()cil " d) ((141n) m 114) = ((;„, ()V czlm + d) by (1.4). (b) yi = (J„, mdb), so "j), = 1), ()E;i 1 Niczim - + d)= ?), (5)). Thus, they are equal if m is a square mod d. For m not a perfect square, to find y for which = 135" p -1 (1, —1), it suffices to find d prime to 4m such that (7) = —1, since one can then find y of the form Lam 'De ro (4m). To do this, first let m = 24mm 0i, where s = 0 or 1 and m o is squ-arefree. Choose d Ez- 1 mod 8 and d do mod mo , where do is chosen so that
1. (a)
(t) = —1. Then one easily checks that (7) = (ia) = (4) = — 1, as desired. 2. Replacing y by —y, if necessary, without loss of generality we may suppose that b> 0 or else b == 0 and a> 0. (a) = (ev N 114Vinf 1:1), (DE» cz + d)-
((-1 1r), —iNi/4,/-z-)=((z ;-,),N14-,iect:::-.:(5)evN cz + d)((_? le) Note that \I(az + b)I(cz + d) - cz + d = q i Vaz + b with 1 1 = —1 if a .< 0' and c 0, th = 1 otherwise. Thus,
238
.
Answers, Hints, and References for Selected Exercises
pi'P-1 z--- (( 2Nb - r), ti 1N '14( E V 1/ a ( — 11Nz)+ b(—iN 114\li))= , 0 4vb 'IN), q i (D8 d-- ' — Nbz + a). (b) We have .•j3 1 = ((-daNb - `,;IN ), (411)--)e c-1- 1 \I — Nbz + a). Since ad — bc = 1 and 41c, we have a 2_ ---: d mod 4, and so ea = ed. Thus, it remains to compare ( 1%16 ) with Ili a We suppose c > 0 (an analogous argument gives the same result if c < 0). Then
r). Now
- fivb ) = ( 04)(7) = (--&) — 1, and (4-) = (÷), so that (-=-(N (N fr (1b )( ) = )( 1 71 — ) sgn a =--- (Dth. Thus, yi = pj3p -1 ( ( ?), (1)). Again ,-, and p,- p -1 are equal if N is a perfect square; otherwise there exist y for which the two differ by (1, — 1). 3. At oo, p = 1, h = 1, t = I. At 0, take p = ((? I), N[Z-), so p" = ((_? 16), —i N/i), and we need f 0 (4) p(( in, Op = 7), -iiv-tiz+ h . VZ) = ((rh1 ), 1 ,I z + h) (( 4 L), -il)= ( (( .7), —it,lhz — I). This is in fo(4) if /i = 4, 1= 1. Finally, at the cusp —I take a = (_1 ?), p = ((A ?), V —2z + 1), so p" = (0, V2z + 1). Since aa Da' er0 (4), we can take h = 1. To find t, compute p(( 1), 0 p' — ((_,1 4), :J-4z — 1), which is j((_ _1), z) provided that ! = i. S. (a) Ski2(f.0(4)) = 0 if k < 9• For k ,.._ 9, it consists of elements of the form OF(I6F — 04)P(0, F), where P is a polynomial of pure weight (k — 9)/2. Thus, for k .... 5, dim Sk2 (r0 (4)) = [(k — 5)14]. (b) Since ! = 1 at the cusps co and 0, there are always those two regular cusps. Since 1 = i at the cusp —4, that cusp is k-regular if and only if ik = 1, i.e., 41k. But I + [k/4] — [(k — 5) 14] = 2 if k ,... 5, 4,1'k and = 3 if k ._._ 5, 41k, as call be verified by checking for k = 5, 6, 7, 8 and then using induction to go from k to k + 4. (c) ®F(® 4 — 16F)(0 4 — 2F). 6. (a) If d'E_-- d mod N and Nj4 is odd, then (1) = (11..) = ( 1 r//4 - 1)(d-I W2( Ndi4 ) = N If N/4 is even, then d' -2E d mod 8, and the proof works (_._ 1)(NI4-1)(d' -1)12(4) = .ur) the same way, with the additional observation that (i) = (-,÷). (b) The proof that the h
(
.. .
cusp condition holds for f 1[4,12 is just like the analogous part of the proof of Proposition 17 in §I11.3. Now let y = C. 'Del-004 and let yl = ( -dbN -eir). By 4. Problem 2 above, we have p53 p-1 = (1, xpl(d))i, 1 . Thus, (f i [P]42) i Vilk/2 "4"--(f 1 [ei'P -1 ]ki2)1[P]k/2 = A(d)(fi Di ik/2)1[P]ko z--- X ki (d);((a)f 1[P]ka. But since ad ÷ .- --_-. 1 mod N, we have x(a) = .k(d). Thus, (f i [P]k2)1[A42 = )---X il v(d)f I [P],42, as desired. 7. 0(co)- = 1, ®(— 1/2) = 0, c(0) = (1 - 012. §IV.2 1. E, 2 (co) = 1, Eki2( 0) = Ekt2 (— 1/2) = 0; Fic/2(0) = (if)_k ; 12(00) = F142(— 1/2) = O. 5. If one uses (2.16) and (2.19) rather than (2.17) and (2.20), then the solutions a and , fl to the resulting 2 x 2 equations involve x, which depends on 1. 6. By Proposition 8, it suffices to show this for squarefree. In that case use (2.16) and . (2.19) to evaluate the /-th coefficient of Ekg + (1 ± ik)2 -k12 F142. 7. lika = ((I — 22) + ( I — 2)q + - . . ; C(1 — A) = 0 if and only if 2 .._ 3 is odd. 9. Use Problem 1 above and Problem 7 of §1V.1 to find a and b so that Co" — aEk12 — bFkg vanishes at the cusps. a = 1, b = (1 + 0'2-k12. e5 = £512 _ (1 + i)/N/2 F512, 0 7 = E712 + (1 — i)/V2 F7/2 .
t
§IV.3 2. The computation is almost identical to that in Problem I in §IV.1. 3. (c) By the lemma in Proposition 43 in §111.5, right coset representatives for ri (N) modulo fl (N) n cC i r'1 (N)Œ, where a = (01 p, are ah = (01 pb,), 0 -._ b < pv, By Problems 1(b)
239 -,
Answers, Hints, and References for Selected Exercises
and 3(b), we use the ab to get representatives for r,(N) modulo ri (N) n r-v i ri (N)ÇV. Since ab = (01 :Oa ), we have Tp v f(z) = p“k14- ' ) • Ebfl[(( (i) :r), Pvil • ( ( i), 1 )L2 z ---- p f( z,-?) ------ E pin ane2xinziPv. The same argument works for Ti v, since the corresponding t is trivial even when 8,f'N. 4. (3.5) gives bn for half integer weight 102; in the case of integer weight k, the formula (5.19) of
§111.5 with m .„.. p 2 gives b„ ...._ ap,n+ x(p)pk-i an+ x ( p 2)p 2k- 2 a nfp 2 . If k is
formally replaced by 1c/2 here, the middle term on the right differs by )( NI ).1„(p)j) from the middle term on the right in (3.5). 5. Let y = (ca 1:1)e ro (N) be such that p2 11). Then for yi e 1", (N) we have .1g2:iiij A'i.. = 1:p2 'g p 21 42 :1) I 53.i : By Problem 2, we have i,2).)„.-2 1 = with Vi = ( p4j2c bliiii 2) e ro(N). Let t y' yiy E1"1 (N). Then D
(fi [f1'1 (M4411001/2)1[342 = Efl[42'Plii)/ki2 i
= EA Dl
pl2;(jk/2
-----
X(d)
Zil Rp2 Mk/2 •
But one checks that the correspondence ri (N)421`)i i-- f1 (N) p2ti permutes the right cosets in r, 0\042 f-, (N), and so this equals x(d)fl[f-i (N),2 fi (N)L2 . (Verify that if ti and Tx are in the same right coset of 1-1 (N) n a - -1 r 1 (N)a in T1 (N), where a =. (01 :2), then so are y; = rciy and yi, = yt.r y'.) Thus, Tp 2 .T1[342 = X(d)72f it remains to note that Fo (N) is generated by 1"1 (N) and y = (ea e fo (N) for which p2 1b. Hence 1p2 fl[fl ki2 = x(d)Tp2 f for all y = (ca dI' ) e ro(N).
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Index
Addition law, 7, 29-35 Affine variety, 52 coordinate ring of, 55 Algebraic geometry, 25-26
Algorithm for congruent number problem, 4, 5, 222 semi-algorithm, 5 Automorphy factor, 148, 152, 153, 177-178
Bernoulli
numbers, 110 polynomials, 54 Bezout's theorem, 32 Birch-Swinnerton-Dyer conjecture, 3, 46, 9093, 218, 221, 222 Branch points, 21, 25
Character additive, 57, 62 conductor, 67 Dirichlet, 62, 75 of group, 61 multiplicative, 57, 62 primitive, 62 quadratic, 82, 176, 187-188, 191-192
/
trivial, 57 Class number, 176, 194, 218 relations, 194 Coates, J., 92 -Wiles theorem, 92, 96, 221 Commensurable subgroups, 165 Complex multiplication, 42, 50, 92, 124, 143, 222 Conductor of character, 67 of elliptic curve, 143 Congruence
subgroup, 99-100 principal, 99 zeta-function, 51, 52 Of elliptic curve, 53, 59 Congruent
number, 1, 3, 5, 46, 70, 92, 221-222 number problem, 1, 2, 4, 221-222 generalized, 8, 123-124, 223-224 Coordinate ri ng, 55 Critical value, 90, 95, 193, 215, 216, 217 Cusp, 103, 106, 108, 126 condition, 125-126, 180-182 form, 108, 117-118, 125, 127, 155, 182 irregular, 144, 182 regular, 144, 1-74, 182
246
Cyclotomic fields, 37
Dedekind eta-function, 78, 121, 122 zeta-function, 56, 88, 89 Deligne, P., 53, 122, 164 111-112, 122, 164 Diagonal hypersurface, 56 Different. 89 Dilogarithm, 76, 78 Diophantus, 1 Dirichlet L-series, 75, 188, 190, 193 series, 80, 141 and modular forms, 140-143 theorem (primes in arithmetic progression), 45, 142 Discriminant modular form, 111-112, 12"1 , 164 of polynomial, 26 Double coset, 165, 204 Doubly periodic, 14
Eigenforms for Hecke operators, 163, 173-174, 201 Euler product, 163 half integer weight, 210-211, 214 normalized, 163 for involution of M2(r0(4)), 146 Eisenstein, F., 177 Eisenstein series, 109-110, 123, 154, 164, 174, 185 of half integer weight, 186-188, 193 Euler product, 199-201 of level N, 131-134 L-function of, 146 normalized, 111, 122 Elementary divisor theorem, 202 Elliptic curve, 9, Il addition law, 7, 29-35 additive degeneracy, 36 complex multiplication, 42, 50, 92, 124, 143, 222 over finite fields, 40-41, 43 inflection points, 13, 35, 41 Legendre form, 224 multiplicative degeneracy, 36 points of order N, 21, 36, 38-40 rank, 44, 46, 51, 91 torsion subgroup, 36, 43.-44, 49-50 Weierstrass form, 24, 26, 33, 120
Index
functions, 14-16, 18, 25 integrals, 27-29, 217 point, 102, 146 -ri(z), 78, 121, 122 Euclid, I
Fermat, 2, 96 curve, 56 last theorem, 2, 5 Fields cyclotomic, 37 of division points, 37 finite, 40 Fourier transform, 71, 83 for finite group, 76 Fractional linear transformation, 98, 102 Functional equation Dedekind eta-function, 78, 121 Dedekind zeta-function, 88, 89 Dirichlet L-series, 77-78 Hasse-Weil L-series, 81, 84, 90, 91 L-series of modular forms, 140-143, 216 Riemann zeta-function, 73-74 theta-functions, 73, 76-78, 85, 88-89, 124 Fundamental domain, 100, 103, 105-107, 146, 231-232 parallelogram, 14
0(X), 107, 142, 148 Galois action on division points, 37-38, 42, 50 Gamma-function, 70-71 Gauss lemma (on quadratic residues), 136 sums, 56, 62, 67-68, 188 Gaussian integers, 14, 41, 42, 65, 165 General linear group, 38, 98 Genus, 53.-54 Good reduction, 43, 90 Grassmannian, 55 Greenberg, R., 222 Gross, B. H., 93, 222
Hardy, G. H., 177 Hasse-Davenport relation, 60, 62-63, 70 Hase-Weil L-function, 3, 61, 64, 75, 79, 81, 84, 90, 141 and modular forms, 143 Hecke, E., 88, 141, 142 character, 81 L-series, 81
247
index
Hecke, E. (cont.) operators, 155, 156, 158, 167, 202 algebra of, 157, 210 Euler product, 158, 160 in half integer weight, 168, 201, 206207, 210 Hermitian, 168, 172 on q-expansion, 161, 163, 207 trace, 175 via double cosets, 167, 168, 202 Heegner, K., 92-93 Homogeneous polynomial, 10, 12, 52 Hurwitz, A., 194 Hypergeometric series, 29 ,-
Irregular cusp, 144, 182 Isotropy subgroup, 102
Jacobi. K., 112 forms, 194 sums, 56, 57, 61 triple product, 219 j-invariant, 105, 119-120, 123-124
Kohnen, W., 214 plus-subspace, 213-214 -Shimura isomorphism, 201, 213-216 -Zagier theorem, 216 Kronecker, L., 194
Lattice, 14, 21, 89, 153 dual, 89 Legendre form of elliptic curve, 224 symbol, 60, 65, 82, 136, 147, 178, 187-188 Level, 99 L-function Dirichlet, 75, 77-78, 188, 190, 193 Hasse-Weil, 3, 61, 64, 75, 79, 81, 84, 90, • 141 of modular form, 140-143, 216 Linear group general, 38, 98 special, 98 Line at infinity, 10-11 Liouville's theorem, 15
Me llin transform, 71, 84, 85, 139
inverse, 142 Möbius function, 188 Modular curve, 91 form, 96-97, 108-109, 117-118, 125, 127, 155 with character, 127, 136-139, 183 and Dirichlet series, 140-143 Euler product, 163 of half integer weight, 3, 176, 178, 182 weight one, 165 function, 108, 119, 125, 127, 155, 182 group SL 2(Z), 99 point, 153 Mordell theorem, 43-44 -Weil theorem, 44 Multiplicity one, 173-174, 214
New form, 174 Niwa, S., 212, 214, 215
Pentagonal number theorem, 123 Petersson scalar product, 168, 169-170, 216 Points at infinity, 10, 11, 13, 103 on elliptic curve, 11, 12 Poisson summation, 72, 83 Projective completion of curve, 11 line, 11, 12, 98, 105 plane, 10 dual plane, 12 space, 11 variety, 52 Pythagoras, 1 Pythagorean triple, 1-2, 3, 5, 8 primitive, 5, 6-7
_ q-expansion, 104, 109, 125, 126 Quadratic character, 82, 176, 187-188, 191-192 form, 176-177, 221 reciprocity-, 66, 82, 153 residue symbol, 60, 65, 82, 136, 147, 178, 187-188
Ramanujan, S., 122 conjecture, 122, 164 -Petersson conjecture, 164 T(n), 122. f23, 164 RamiPe • - • •icx, 144
248
Rank of elliptic curve, 44, 46, 51, 91 Reduction mod p, 43 Regular cusp, 144, 174, 182 Representation, 137, 165 theory, 215 Residue field, 43, 55-56 theorem, 15, 30, 116 Riemann sphere, 10, 21, 98, 105, 119, 120 surface, 53, 54, 143 zeta-function, 27, 51, 64, 70, 73 Root. number, 70, 84, 92, 97
Shimura, G., 177, 215 map, 200-201, 213-215 theorem, 212-213, 215' Shintani, T., 215 Smooth curve, 9, 13, 52-53 Special linear group, 98
Tangency, order of, 13 Taniyama, Y., 91, 143 -Weil conjecture, 91, 143 Tate, J., 88 -Shafarevich group, 218, 222 T(n), 122, 123, 164 Theta-functions, 72, 76-78, 84, 88-89, 96-97, 124, 176-177 Hecke transformation formula, 148 Torsion subgroup, 36-37, 43 ill Torus,1452 Tunnel!, J., 1, 200, 217, 219 theorem, 1, 3, 4, 6, 193, 217, 219-222 Twisting, 81, 127, 142, 176
IndeJ
Type of finite abelian group, 40
Upper half-plane, 99
Variety affine, 52 projective, 52
Waldspurger, J.-L., 215 theorem, 193, 200, 215, 220 Weierstrass p-function, 16-18, 21, 134 derivatives of, 17, 18, 20, 21, 22 differential equation of, 21, 22, 24 form of elliptic curve, 24, 26, 33, 120 Weight, 108, 153-154 half integer, 176 one, 165 of polynomials, 183 Weil, A.,, 44, 91, 141, 142, 143 conjectures, 53-54, 91, 122, 139, 164 parametrization, 91 -Taniyama conjecture, 91 theorem, 142-143, 215 Wiles, A., 92
Zagier, D., 93, 194, 222 Zeta-function congruence, 51, 52 Dedekind, 56, 88, 89 partial, 132 Riemann, 27, 51, 64, 70, 73
