Read colmez.pdf text version
Fontaine's rings and padic Lfunctions
Pierre Colmez C.N.R.S. Institut de Math´matiques de Jussieu e
2
i These are notes from a course given at Tsinghua University during the fall of 2004. The aim of the course was to explain how to construct padic Lfunctions using the theory of (, )modules of Fontaine. This construction is an adaptation of an idea of PerrinRiou. The content of the course is well reflected in the table of contents which is almost the only thing that I modified from the notes taken and typed by the students Wang Shanwen, Chen Miaofen, Hu Yongquan, Yin Gang, Li Yan and Hu Yong, under the supervision of Ouyang Yi, all of whom I thank heartily. The course ran in parallel to a course given by Fontaine in which the theory of (, )modules was explained as well as some topics from padic Hodge theory which are used freely in these notes, which means that they are not entirely selfcontained. Also, as time ran short at the end, the last chapter is more a survey than a course. For a bibliography and further reading, the reader is referred to my Bourbaki talk of June 2003 published in Ast´risque 294. e
ii
Contents
I Classical padic Lfunctions: zeta functions and modular forms 1
1 The 1.1 1.2 1.3 padic zeta function of KubotaLeopoldt The Riemann zeta function at negative integers . . . . . . . padic Banach spaces . . . . . . . . . . . . . . . . . . . . . . Continuous functions on Zp . . . . . . . . . . . . . . . . . . 1.3.1 Mahler's coefficients . . . . . . . . . . . . . . . . . . 1.3.2 Locally constant functions. . . . . . . . . . . . . . . . Measures on Zp . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 The Amice transform . . . . . . . . . . . . . . . . . . 1.4.2 examples of measures on Zp and of operations on measures. . . . . . . . . . . . . . . . . . . . . . . . . . . . The padic zeta function . . . . . . . . . . . . . . . . . . . . 1.5.1 Kummer's congruences. . . . . . . . . . . . . . . . . . 1.5.2 Restriction to Z . . . . . . . . . . . . . . . . . . . . . p 1.5.3 Leopoldt's transform. . . . . . . . . . . . . . . . . C k functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6.1 Definition. . . . . . . . . . . . . . . . . . . . . . . . . 1.6.2 Mahler's coefficients of C r functions. . . . . . . . . . . locally analytic functions . . . . . . . . . . . . . . . . . . . . 1.7.1 Analytic functions on a closed disk. . . . . . . . . . . 1.7.2 Locally analytic functions on Zp . . . . . . . . . . . . Distributions on Zp . . . . . . . . . . . . . . . . . . . . . . . 1.8.1 The Amice transform of a distribution. . . . . . . . . 1.8.2 Examples of distributions. . . . . . . . . . . . . . . . 1.8.3 Residue at s = 1 of the padic zeta function. . . . . . Tempered distributions . . . . . . . . . . . . . . . . . . . . . iii . . . . . . . . . . . . . . . . . . . . . . . 3 3 5 7 7 9 10 10 12 15 15 16 17 19 19 21 23 23 25 27 27 29 30 31
1.4
1.5
1.6
1.7
1.8
1.9
iv
CONTENTS 1.9.1 Analytic functions inside C r functions . . . . . . . . . . 31 1.9.2 Distributions of order r . . . . . . . . . . . . . . . . . . 33 1.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
2 Modular forms 2.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 The upper halfplane . . . . . . . . . . . . . . 2.1.2 Definition of modular forms . . . . . . . . . . 2.1.3 qexpansion of modular forms. . . . . . . . . . 2.1.4 Cusp forms. . . . . . . . . . . . . . . . . . . . 2.2 The case = SL2 (Z) . . . . . . . . . . . . . . . . . . 2.2.1 The generators S and T of SL2 (Z). . . . . . . 2.2.2 Eisenstein series . . . . . . . . . . . . . . . . . 2.2.3 The fundamental domain for SL2 (Z) . . . . . k 2.2.4 The 12 formula. . . . . . . . . . . . . . . . . . 2.2.5 Dimension of spaces of modular forms. . . . . 2.2.6 Rationality results. . . . . . . . . . . . . . . . 2.3 The algebra of all modular forms. . . . . . . . . . . . 2.4 Hecke operators . . . . . . . . . . . . . . . . . . . . . 2.4.1 Preliminary. . . . . . . . . . . . . . . . . . . . 2.4.2 Definition of Hecke operators: Rn , Tn , n 1. . 2.4.3 Action of Hecke operators on modular forms. . 2.5 Petersson scalar product. . . . . . . . . . . . . . . . . 2.6 Primitive forms . . . . . . . . . . . . . . . . . . . . . 3 padic Lfunctions of modular forms 3.1 Lfunctions of modular forms. . . . . . . . . 3.1.1 Estimates for the fourier coefficients . 3.1.2 Dirichlet series and Mellin transform 3.1.3 Modular forms and Lfunctions . . . 3.1.4 Euler products . . . . . . . . . . . . 3.2 Higher level modular forms . . . . . . . . . . 3.2.1 Summary of the results . . . . . . . . 3.2.2 TaniyamaWeil Conjecture . . . . . . 3.3 Algebraicity of special values of Lfunctions 3.3.1 Modular symbols. . . . . . . . . . . . 3.3.2 The results . . . . . . . . . . . . . . 3.3.3 Rankin's method . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . .
39 39 39 40 40 41 42 42 43 44 46 47 48 50 53 53 54 56 58 60 63 63 63 65 66 68 69 69 71 71 71 73 74
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
CONTENTS 3.4
v
padic Lfunctions of modular forms . . . . . . . . . . . . . . . 77
II
Fontaine's rings and Iwasawa theory
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
85 85 85 87 89 91 91 92 95 95 98 100 103 104 105
4 Preliminaries 4.1 Some of Fontaine's rings . . . . . . . . . . 4.1.1 Rings of characteristic p . . . . . . 4.1.2 Rings of characteristic 0 . . . . . . 4.2 (, )modules and Galois representations. 5 (, )modules and Galois cohomology 5.1 Galois Cohomology . . . . . . . . . . . . 5.2 The complex C, (K, V ) . . . . . . . . . 5.3 Tate's EulerPoincar´ formula. . . . . . . e 5.3.1 The operator . . . . . . . . . . . 5.3.2 D=1 and D/(  1) . . . . . . . 5.3.3 The module D=0 . . . . . . . . 5.3.4 Computation of Galois chomology 5.3.5 The EulerPoincar´ formula. . . . e 5.4 Tate's duality and residues . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . groups . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
6 (, )modules and Iwasawa theory i 6.1 Iwasawa modules HIw (K, V ) . . . . . . . . . . 6.1.1 Projective limits of cohomology groups 6.1.2 Reinterpretation in terms of measures . 6.1.3 Twist by a character (` la Soul´) . . . a e i 6.2 Description of HIw in terms of D(V ) . . . . . 1 6.3 Structure of HIw (K, V ) . . . . . . . . . . . . . 7 Zp (1) and KubotaLeopoldt zeta function 7.1 The module D(Zp (1))=1 . . . . . . . . . . 7.2 Kummer theory . . . . . . . . . . . . . . . 7.3 Coleman's power series . . . . . . . . . . . 7.4 An explicit reciprocity law . . . . . . . . . 7.5 Proof of the explicit reciprocity law . . . . 7.5.1 Strategy of proof of Theorem 7.4.1 7.5.2 Explicit formulas for cocyles . . . . . . . . . . . . . . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
109 . 109 . 109 . 110 . 111 . 112 . 115 117 . 117 . 118 . 119 . 122 . 123 . 123 . 125
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
vi 7.5.3 7.5.4 7.5.5
CONTENTS Tate's normalized trace maps . . . . . . . . . . . . . . 127 Applications to Galois cohomology . . . . . . . . . . . 130 No 2i in Cp ! . . . . . . . . . . . . . . . . . . . . . . . 131
8 (, )modules and padic Lfunctions 133 8.1 TateSen's conditions . . . . . . . . . . . . . . . . . . . . . . . 133 8.1.1 The conditions (TS1), (TS2) and (TS3) . . . . . . . . . 133 8.1.2 Example : the field Cp . . . . . . . . . . . . . . . . . . 134 8.2 Sen's method . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 8.2.1 Almost ´tale descent . . . . . . . . . . . . . . . . . . . 136 e 8.2.2 Decompletion . . . . . . . . . . . . . . . . . . . . . . . 138 8.2.3 Applications to padic representations . . . . . . . . . . 140 8.3 Overconvergent (, )modules . . . . . . . . . . . . . . . . . . 141 8.3.1 Overconvergent elements . . . . . . . . . . . . . . . . . 141 8.3.2 Overconvergent representations . . . . . . . . . . . . . 145 8.3.3 padic Hodge theory and (, )modules . . . . . . . . 147 8.3.4 A map of the land of the rings . . . . . . . . . . . . . . 148 8.4 Explicit reciprocity laws and padic Lfunctions . . . . . . . . 149 8.4.1 Galois cohomology of BdR . . . . . . . . . . . . . . . . 149 8.4.2 BlochKato's dual exponential maps . . . . . . . . . . . 150 8.4.3 The explicit reciprocity law . . . . . . . . . . . . . . . 152 8.4.4 Cyclotomic elements and CoatesWiles morphisms. . . 154 8.4.5 Kato's elements and padic Lfunctions of modular forms.155
Part I Classical padic Lfunctions: zeta functions and modular forms
1
Chapter 1 The padic zeta function of KubotaLeopoldt
1.1 The Riemann zeta function at negative integers
We first recall the definitions of Riemann zeta function and the classical Gamma function:
+
(s) = (s) =
0
ns =
n=1 + p
(1  ps )1 , if Re (s) > 1. dt , if Re (s) > 0. t
et ts
The function has the following properties: (i) (s+1) = s(s), which implies that has a meromorphic continuation to C with simple poles at negative integers and 0. (ii) (n) = (n  1)! if n 1.
1 (iii) (s)(1  s) = sin(s) , which implies that (s) is an entire(or holomorphic) function on C with simple zeros at n for n N. 1 (iv) ( 2 ) = .
3
4CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT Then we have the following formulas: ns = 1 (s)
+
ent ts
0 + + 0 n=1
dt , t 1 t = t (s)
+ 0
1 (s) = (s)
e
nt s dt
1 s dt t . et  1 t
Lemma 1.1.1. If f : R+ C is a C function on R+ , rapidly decreasing (i.e., tn f (t) 0 when t + for all n N), then L(f, s) = 1 (s)
+
f (t)ts
0
dt , Re(s) > 0 t
has an analytic continuation to C, and L(f, n) = (1)n f (n) (0). Proof. Choose a C function on R+ , such that (t) = 1 for t [0, 1] and (t) = 0 for t 2. Let f = f1 + f2 , where f1 = f , f2 = (1  )f . Then 0 f2 (t)ts dt is t holomorphic on C, hence L(f2 , s) is also holomorphic and L(f2 , n) = 0 = (n) f2 (0). Since, for Re (s) > 0, L(f1 , s) = 1 ts 1 [f1 (t) ]+  0 (s) s s(s)
+
f1 (t)ts+1
0 (n)
dt t
= L(f1 (t), s + 1) = (1)n L(f1 , s + n), we get analytic continuation for f1 and hence for f , moreover, L(f, n) = L(f1 , n) = (1)n+1 L(f1
+ (n+1)
, 1)
= (1)n+1
0
f1
(n+1)
(t)dt = (1)n f1 (0) = (1)n f (n) (0).
(n)
We now apply the above lemma to the function f (t) =
t . et 1
Note that
f (t) =
0
Bn
tn , n!
1.2. P ADIC BANACH SPACES where Bn Q is the nth Bernoulli number with value: 1 1 1 B0 = 1, B1 =  , B2 = , B3 = 0, B4 =  , B5 = 0 · · · 2 6 30 Since f (t)  f (t) = t, we have B2k+1 = 0 if k 1. Now : (s) = 1 (s)
+
5
f (t)ts1
0
dt 1 = L(f, s  1), t s1
so we obtain the following result. Theorem 1.1.2. (i) has a meromorphic continuation to C. It is holomorphic except for a simple pole at s = 1 with residue L(f, 0) = B0 = 1. (ii) If n N, then (n) = (1)n (n+1) 1 L(f, n  1) = f (0) n+1 n+1 Bn+1 = (1)n Q n+1 Bn+1 =  if n 2 . n+1
Theorem 1.1.3 (Kummer). If p does not divide the numerators of (3), (5), · · · , (2  p), then the class number of Q(up ) is prime to p. Remark. This theorem and a lot of extra work implies Fermat's Last Theorem for these regular primes. We will not prove it in these notes, but we will focus on the following result, also discovered by Kummer, which plays an important role in the proof. Theorem 1.1.4 (Kummer's congruences). Let a 2 be prime to p. Let k 1. If n1 , n2 k such that n1 n2 mod (p  1)pk1 , then (1  a1+n1 )(n1 ) (1  a1+n2 )(n2 ) mod pk .
1.2
padic Banach spaces
Definition 1.2.1. A padic Banach space B is a Qp vector space with a lattice B 0 (Zp module) separated and complete for the padic topology, i.e., B0
nN
lim B 0 /pn B 0 . 
6CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT For all x B, there exists n Z, such that x pn B 0 . Define vB (x) = sup {n : x pn B 0 }.
nN{+}
It satisfies the following properties: vB (x + y) min(vB (x), vB (y)), vB (x) = vp () + vB (x), if Qp . Then x B = pvB (x) defines a norm on B, such that B is complete for and B 0 is the unit ball.
B
Example 1.2.2. (i) B = Cp = Qp , B 0 = OCp , vB (x) = [vp (x)] Z; (ii) The space B = C 0 (Zp , Qp ) of continuous functions on Zp . B 0 = 0 C (Zp , Zp ) is a lattice, and vB (f ) = inf vp (f (x)) =  because Zp is comxZ pact. (iii) Let B = C 0 (Zp , Cp ), B 0 = C 0 (Zp , OCp ); vB (f ) = inf [vp (f (x))].
xZ
Definition 1.2.3. A Banach basis of a padic Banach space B is a family (ei )iI of elements of B, satisfying the following conditions: (i) For every x B, x = xi ei , xi Qp in a unique way with xi 0
iI
when i ; equivalently for any C, the set {i  vp (xi ) C} is a finite set. (ii) vB (x) = inf vp (xi ).
iI
Theorem 1.2.4. A family (ei )iI of elements of B is a Banach basis if and only if (i) ei B 0 for all i; (ii) the set (ei )iI form a basis of B 0 /pB 0 as a Fp vector space. ¯ Proof. We leave the proof of the theorem as an exercise. Let B and B be two padic Banach spaces with Banach basis (ei )iI and (fj )jJ respectively, then B B is a padic Banach space with Banach basis (ei fj )(i,j)I×J . Thus for all x B B , x =
i,j
xi,j ei fj yj fj
j
(xi,j Qp , xi,j 0 as (i, j) )
= =
i
(yj B, yj 0 as j ) (zi B , zi 0 as i ).
ei zi
1.3. CONTINUOUS FUNCTIONS ON ZP Exercise. C 0 (Zp , Cp ) = Cp C 0 (Zp , Qp ).
7
1.3
1.3.1
Continuous functions on Zp
Mahler's coefficients
We have the binomial function: 1, x = x(x  1) · · · (x  n + 1) n , n! Lemma 1.3.1. vC 0 (
x n
if n = 0, if n 1.
) = 0.
x n
x Proof. Since n = 1, vC 0 ( n ) 0. n x If x N, then n N implies vp ( x vp ( n ) 0 because N is dense in Zp .
) 0. Hence for all x Zp ,
For all f C 0 (Zp , Qp ), we write f [0] = f, f [k1] (x) = f [k] (x + 1)  f [k] (x)
and write the Mahler's coefficient an (f ) = f [n] (0). Hence:
n
f (x) =
i=0 n
[n]
(1)i (1)i
i=0
n f (x + n  i), i n f (n  i). i
an (f ) =
Theorem 1.3.2 (Mahler). If f C 0 (Zp , Qp ), then (i) lim vp (an (f )) = +,
n
(ii) For all x Zp , f (x) =
n=0
an (f )
x n
,
(iii) vC 0 (f ) = inf vp (an (f )).
8CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT Proof. Let Then
= {a = (an )nN : an Qp bounded}, v (a) = inf nN vp (an ).
.
· f a(f ) = (an (f ))nN is a continuous map from C 0 (Zp , Qp ) to and v (a(f )) vC 0 (f ).
· The space 0 = {(an )nN : an 0, as n } is a closed subspace of and B = {f : a(f ) 0 } is a close subspace of C 0 (Zp , Qp ). · For all a
0 , +
fa =
n=0
an
x n
C 0 (Zp , Qp )
because the series converges uniformly. Moreover, vC 0 (fa ) v (a) x+1 x x and as n+1  n+1 = n ,
+ [k] fa
=
n=0
an+k
x . n
Hence we have: ak (f ) = f [k] (0) = ak , which implies a(fa ) = a. · f a(f ) is injective. Since a(f ) = 0 implies f (n) = 0 for all n N. Hence f = 0 by the density of N in Zp . Now for f B, a(f ) 0 implies f  fa(f ) = 0 because a(f  fa(f ) ) = a(f )  a(f ) = 0 and a is injective. So f B implies that f satisfies (ii). Moreover, since v (a(f )) vC 0 (f ) = vC 0 (fa(f ) ) v (a(f )), (iii) is also true. It remains to show that: Claim: B = C 0 (Zp , Qp ). (a) First proof. We have a lemma: Lemma 1.3.3. If f C 0 (Zp , Qp ), then there exists k N such that vC 0 (f [p ] ) vC 0 (f ) + 1.
k
1.3. CONTINUOUS FUNCTIONS ON ZP Proof. We have
pk 1
9
f
[pk ]
(x) = f (x + p )  f (x) +
i=1
k
k
(1)i
pk k f (x + pk  i) + (1 + (1)p )f (x). i
k
Now vp ( pi ) 1, if 1 i pk 1 et vp (1+(1)p ) 1. Since Zp is compact, f is uniformly continuous. For every c, there exists N , when vp (x  y) N , we have vp (f (x)  f (y)) c. It gives the result for k = N . First proof of the Claim. Repeat the lemma: for every c = vC 0 (f ) + k, there exists an N , such that vC 0 (f [N ] ) c. Hence, for all n N , vp (an (f )) c.
1.3.2
Locally constant functions.
Choose a z Cp , such that vp (z  1) > 0. Then
+
fz (x) =
n=0
x (z  1)n C 0 (Zp , Cp ). n
Note k N, fz (k) = z k . So we write, fz (x) = z x and we have z x+y = z x z y . Example 1.3.4. (i) z 2 =
n=0
1
+
n
1 2
(z  1)n . z =
16 ,z 9
1 =
7 , 9
the series
converges in R to 4 , and converges in Q7 to  4 . 3 3 (ii) If z is a primitive pn th root of 1, then vp (z  1) =
n
1 > 0. (p  1)pn1
Note that z x+p = z x for all x, then z x is locally constant( constant mod pn Zp ). The characteristic function of i + pn Zp is given by 1i+pn Zp (x) = since zx =
z pn =1
1 pn
z i z x
z p =1
n
pn 0
if x pn Zp ; if not.
10CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT Lemma 1.3.5. The set of locally constant functions LC(Zp , Qp ) B. Proof. By compactness of Zp , a locally constant function is a linear combination of 1i+pn Zp z x , z µp , thus a linear combination of z x . But an (z x ) = (z  1)n goes to 0, when n goes to , hence z x B. Lemma 1.3.6. LC(Zp , Qp ) is dense in C 0 (Zp , Qp ). Proof. For every f C 0 (Zp , Qp ), let
pk 1
fk =
i=0
f (i)1i+pk Zp .
Then fk f in C 0 because f is uniformly continuous. Second proof of the Claim. By the above two lemmas, LC(Zp , Qp ) B C 0 (Zp , Qp ), B is closed and LC(Zp , Qp ) is dense in C 0 (Zp , Qp ), hence B = C 0 (Zp , Qp ).
1.4
1.4.1
Measures on Zp
The Amice transform
Definition 1.4.1. A measure µ on Zp with values in a padic Banach space B is a continuous linear map f Zp f (x)µ = Zp f (x)µ(x) from C 0 (Zp , Qp ) to B. Remark. (i) If L Cp is a closed subfield and B is an Lvector space, then µ extends by continuity and Llinearity to C 0 (Zp , L) = L C 0 (Zp , Qp ). (ii) We denote D0 (Zp , B) the set of the measure on Zp with values in B, then D0 (Zp , B) = D0 (Zp , Qp ) B. Definition 1.4.2. The Amice transform of a measure µ is defined to be the map: + x x µ Aµ (T ) = (1 + T ) µ(x) = Tn µ. Zp Zp n n=0 Lemma 1.4.3. If vp (z  1) > 0, Aµ (z  1) =
Zp
z x µ(x).
1.4. MEASURES ON ZP Proof. Since z x = can exchange
+ n=0
11
x n
(z  1)n
with normal convergence in C 0 (Zp , Qp ), one
and .
Definition 1.4.4. The valuation on D0 is vD0 (µ) = inf (vp (
f =0 Zp +
f µ)  vC 0 (f )).
Theorem 1.4.5. The map µ Aµ is an isometry from D0 (Zp , Qp ) to the set {
n=0
bn T n , bn bounded, and bn Qp } with the valuation v(
+
+
bn T n ) =
n=0
inf nN vp (bn ). Proof. On one hand, for all µ D0 (Zp , Qp ), write Aµ (T ) = bn (µ) =
Zp x n
bn (µ)T n , then
µ. Since vC 0 (
x n
n=0
) = 0 by Lemma 1.3.1, x ) vD0 (µ) n
+
vp (bn (µ)) vD0 (µ) + vC 0 ( for all n, hence v(Aµ ) vD0 (µ).
On the other hand, if (bn )nN is bounded, f
n=0
bn an (f )(by Mahler's
theorem, an (f ) 0) gives a measure µb whose Amice transform is
+
Aµb (T ) =
n=0
T
n Zp
x µb = n x )= i
+
+
T (
n=0 i=0
n
x b i ai ( )) = n
+
bn T n
n=0
since an ( Hence
+
1 if i = n, 0 otherwise.
vp (
n=0
bn an (f )) min(vp (bn ) + vp (an (f )))
n
min(vp (bn )) + min(an (f ))
n n
= v(
bn T ) + vC 0 (f )
n
= v(Aµ ) + vC 0 (f ). Thus vD0 (µb ) v(Aµ ). Then we have v(Aµ ) = vD0 (µ).
12CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT By Lemma 1.3.6, we know that locally constant functions are dense in C (Zp , Qp ). Explicitly, for all f C 0 (Zp , Qp ), the locally constant functions
0 pn 1 i=0
fn =
f (i)1i+pn Zp f in C 0 .
Zp
Now if µ D0 (Zp , Qp ), set µ(i + pn Zp ) = given by the following "Riemann sums"
pn 1
1i+pn Zp µ. Then
Zp
f µ is
f µ = lim
Zp
n
f (i)µ(i + pn Zp )
i=0
(1.1)
Note that vp (µ(i + pn Zp )) vD0 (µ). Theorem 1.4.6. If µ is an additive bounded function on compact open subsets of Zp (by compactness of Zp is a finite disjoint union of i + pn Zp for some n), then µ extends uniquely as a measure on Zp via (1.1). Proof. Since µ is an additive function on compact open subsets, µ is linear on locally constant functions. And µ is bounded, hence µ is continuous for vC 0 . As the locally constant functions are dense in C 0 (Zp , Qp ), we have µ as a measure on Zp .
1.4.2
examples of measures on Zp and of operations on measures.
Example 1.4.7. Haar measure: µ(Zp ) = 1 and µ is invariant by translation. We must have µ(i + pn Zp ) = p1 which is not bounded. Hence, there exists n no Haar measure on Zp . Example 1.4.8. Dirac measure: For a Zp , we define a by f (a). The Amice transform of a is Aa (T ) = (1 + T )a .
Zp
f (x)a =
Example 1.4.9. Multiplication of a measure by a continuous function. For µ D0 , f C 0 , we define the measure f µ by g · fµ =
Zp Zp
f (x)g(x)µ
for all g C 0 .
1.4. MEASURES ON ZP (i) Let f (x) = x, since x x n = (x  n + n) x n = (n + 1) x x +n , n+1 n
13
the Amice transform is
+
Axµ =
n=0 +
Tn
Zp
x xµ n x µ+n n+1 x µ n
=
n=0
T n (n + 1)
Zp
Zp
d Aµ . dT (ii) Let f (x) = z x , vp (z  1) > 0. For any y, vp (y  1) > 0, then = (1 + T ) y x (z x µ) =
Zp Zp
(yz)x µ = Aµ (yz  1)
which implies that Azx µ (T ) = Aµ ((1 + T )z  1). (iii) The restriction to a compact open set X of Zp : it is nothing but the z i z x , multiplication by 1X . If X = i + pn Zp , then 1i+pn Zp (x) = pn
z p =1
n
hence AResi+pn Zp µ (T ) = pn
z pn =1
z i Aµ ((1 + T )z  1).
Example 1.4.10. Actions of and . For µ D0 , we define the action of on µ by f (x)(µ) =
Zp Zp
f (px)µ.
Hence
+
A(µ) (T ) =
n=0
Tn
Zp
px µ = Aµ ((1 + T )p  1) = (Aµ (T )) n
where : T (1 + T )p  1 (compare this formula with (, )modules). We define the action of by f (x)(µ) =
Zp
x f ( )µ. p Zp
14CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT Then A(µ) = (Aµ ) where (F )((1 + T )p  1) = 1 F ((1 + T )z  1). p zp =1
The actions and satisfy the following properties: (i) = Id; (ii) (µ) = 0 µ has a support in Z ; p (iii) ResZ (µ) = (1  )µ. p The map is very important in the theory of (, )modules. Example 1.4.11. Action of . Let = Gal(Qp (µp )/Qp ). Let : Z p be the cyclotomic character. For and µ D0 , we let µ be given by f (x)µ =
Zp Zp
f (()x)µ.
One can verify that Aµ (T ) = Aµ ((1 + T )()  1) = (Aµ (T )) for (T ) = (1 + T )()  1. (Compare this formula with (, )modules.) For all , commutes with and . Example 1.4.12. Convolution µ. Let , µ be two measures, their convolution µ is defined by f (x) µ =
Zp Zp
(
Zp
f (x + y)µ(x))(y).
Here we have to verify y Zp f (x + y)µ(x) C 0 , which is a direct consequence of the fact f is uniformly continuous. Let f (x) = z x , vp (z  1) > 0, then zx µ =
Zp Zp
z x µ(x)
Zp
z y (y),
thus Aµ = A Aµ .
1.5. THE P ADIC ZETA FUNCTION
15
1.5
1.5.1
The padic zeta function
Kummer's congruences.
Lemma 1.5.1. For a Z , there exists a measure a D0 such that p Aa =
Zp
(1 + T )x a =
1 a  . T (1 + T )a  1
Proof. This follows from Theorem 1.4.5 and the fact a = (1 + T )a  1 since a1
a n
a
n=1 a n
Tn
=
1 · T 1+
n=2
1 a1
a n
T n1
1 + Zp [[T ]] T
Zp . Moreover, we have vD0 (a ) = 0.
Zp
Proposition 1.5.2. For every n N, Proof. For a R , for T = et  1, let + fa (t) = Aa (T ) =
xn a = (1)n (1  a1+n )(n).
et
1 a  at , 1 e 1
then fa is in C on R+ and rapidly decreasing. Hence
+ 1 dt L(fa , s) = fa (t)ts = (1  a1s )(s) (s) 0 t n n fa (0) = (1) L(fa , n) = (1)n (1  a1+n )(n) n The identity fa (0) = (1)n (1  a1+n )(n) is algebric, so is true for all a, hence even on Z . Thus p
x n a = (
Zp
d n ) ( dt
etx a )t=0 = (
Zp
d n (n) ) Aa (et  1)t=0 = fa (0). dt
Corollary 1.5.3. For a Z , k 1 (k 2 if p = 2), n1 , n2 k, n1 p n2 mod (p  1)pk1 , then vp ((1  a1+n1 )(n1 )  (1  a1+n2 )(n2 )) k.
16CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT Proof. The left hand side LHS = vp ((1  a1+n1 )(n1 )  (1  a1+n2 )(n2 )) is vp ( (xn1  xn2 )a ) vD0 (a ) + vC 0 (xn1  xn2 ).
Zp
From the proof of Lemma 1.5.1, vD0 (a ) = 0, thus LHS vC 0 (xn1  xn2 ). It suffices to show vC 0 (xn1  xn2 ) k. There are two cases: If x pZp , then vp (xn1 ) k and vp (xn2 ) k since n1 , n2 k. If x Z , vp (xn1  xn2 ) k because (Z/pk Z) has order (p  1)pk1 and p n1  n2 is divisible by (p  1)pk1 . Remark. The statement is not clean because of x pZp .
1.5.2
Restriction to Z . p
1 1 Lemma 1.5.4. ( T ) = T . 1 Proof. Let F (T ) = ( T ), then
F ((1 + T )p  1) =
1 1 p zp =1 (1 + T )z  1
+
1 = ((1 + T )z)n p zp =1 n=0
+
=
n=0
(1 + T )pn =
1 . (1 + T )p  1
Proposition 1.5.5. (a ) = a . Proof. We only need to show the same thing on the Amice transform, but Aa (T ) = 1 a 1 1  =  a · a ( ) a1 T (1 + T ) T T
where a is the inverse of a by : Z , i.e., (a ) = a. Since and p 1 1 a commutes and ( T ) = T , we have (Aa ) = 1 1  aa ( ) = Aa . T T
1.5. THE P ADIC ZETA FUNCTION Corollary 1.5.6. (i) ResZ (a ) = (1  )a = (1  )a , p (ii) Z xn a = Zp xn (1  )a = (1)n (1  an+1 )(1  pn )(n).
p
17
Remark. The factor (1  pn ) is the Euler factor of the zeta function at p. Theorem 1.5.7. For i Z/(p  1)Z (or i Z/2Z if p = 2), there exists a unique function p,i , analytic on Zp if i = 1, and (s  1)p,1 (s) is analytic on Zp , such that p,i (n) = (1  pn )(n) if n i mod p  1 and n N. Remark. (i) If i 0 mod 2, then p,i = 0 since (n) = 0 for n even and 2; (ii) To get padic continuity, one has to modify by some "Euler factor at p". (iii) Uniqueness is trivial because N is infinite and Zp is compact. (iv) The existence is kind of a miracle. Its proof relies on Leopoldt's transform.
1.5.3
Leopoldt's transform.
Lemma 1.5.8. (i) Every x Z can be written uniquely as x = (x) x , p with (x) µ(Qp ) = {±1} µp1 , if p = 2, if p = 2 and x 1 + 2pZp .
(ii) (xy) = (x)(y), xy = x y . Proof. If p = 2, it is obvious. n If p = 2, (x) = lim xp = [¯]. x
n
Remark. (i) is the socalled Teichm¨ller character ; u (ii) x = exp(log(x)); (iii) xn = (x)n x n , here x n is the restriction to N of x s which is continuous in s, (x)n is periodic of period p  1, which is not padically continuous. Proposition 1.5.9. If is a measure on Z , u = 1 + 2p, then there exists a p (i) measure on Zp (Leopoldt's transform) such that (x)i x s (x) =
Z p Zp
usy (y) = A(i) (us  1).
(i)
18CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT Proof. We have (x)i x s (x)
Z p
=
µ(Qp )
()i ()i
+2pZp
x s (x)
=
µ(Qp )
1+2pZp
x s 1 · (x),
where is such that ( ) = . We have a isomorphism : 1 + 2pZp Zp log(x) . log(u)
x y= Then f (y) (1 ) =
Zp 1+2pZp
f ((x))1 .
Now x
s
= exp(s log x) = exp(s log uy) = usy and hence ()i x s (x) =
1+2pZp µ(Qp )
()i
Zp
usy (1 · ),
µ(Qp )
we just set =
µ(Qp )
(i)
()i (1 · ).
Definition 1.5.10. p,i (s) = 1 1  (a)1i a
1s
(x)i x
Z p
s
a (x).
Proof of Theorem 1.5.7. If n i mod p  1, then p,i (n) = 1 1  (a)1i a 1+n 1 = 1  (a)1+n a 1+n = (1  pn )(n). The function p,i is analytic if (a)1i = 1, which can be achieved if i = 1. If i = 1, there is a pole at s = 1. (x)i x n a (x)
Z p
(x)n x n a (x)
Z p
1.6. C K FUNCTIONS
19
Remark. (i) A theorem of Mazur and Wiles (originally the Main conjecture of Iwasawa theory) describes the zeros of p,i (s) in terms of ideal class groups of Qp (µpn ), n N. (ii) Main open question: For i 1 mod 2, can p,i (k) = 0, if k > 1 and k N? The case k = 1 is known. In this case, p,i (1) is a linear combination ¯ with coefficients in Q× of log of algebraic numbers, hence by transcendental number theory (Baker's theorem), p,i (1) = 0.
1.6
1.6.1
C k functions
Definition.
Let f : Zp Qp be a given function. We define f {0} (x) = f (x) {i} f (x, h1 , · · · , hi ) 1 = (f {i1} (x + hi , h1 , · · · , hi1 )  f {i1} (x, h1 , · · · , hi1 )) hi 1 = hj )) ( (1)iI f (x + h1 · · · hi jI
I{1,··· ,i}
One notes that f {i} is the analogue of the usual derivation in C(R, C). In fact, if f : R C is in C k and i k, define f {i} by the above formula, then f {i} (x, h1 , · · · , hi ) =
[0,1]i
f (i) (x + t1 h1 + · · · + ti hi )dt1 · · · dti ,
hence f {i} is continuous and f {i} (x, 0, · · · , 0) = f (i) (x). Definition 1.6.1. A function f : Zp Qp (or Cp ) is in C k if f {i} can be extended as a continuous function on Zi+1 for all i k. p Remark. If f C 0 and h1 , · · · , hi = 0, then we have:
i
vp (f
{i}
(x, h1 , · · · , hi )) vC 0 (f ) 
j=1
vp (hj ).
20CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT Example 1.6.2. The definition of C k here is different than the usual case.
+
Here is an example. For all x in Zp , x =
n=0 +
pn an (x) with an (x) {0, 1, · · · , p
1}. Let f (x) =
n=0
p2n an (x), then vp (f (x)  f (y)) = 2vp (x  y). Hence
f (x) = 0 for all x Zp , thus f is in C in the usual sense. But f is not C 2 in our case. In fact, let (x, h1 , h2 ) = (0, pn , pn ) and ((p  1)pn , pn , pn ), here p = 2, we have: f {2} (0, pn , pn ) = 0; f {2} ((p  1)pn , pn , pn ) = p  p2 . We define a valuation on C k functions by: vC k (f ) = min
i 0ik (x,h1 ,··· ,hi )Zi+1 p
inf
vp (f {i} (x, h1 , · · · , hi )). nj = n, nj 1}
Let L(n, k) = max{
j=1
vp (nj ), i k,
x is a Banach basis of C k . n Exercise. there exists a Ck , such that for all n 1, log n log n k  Ck L(n, k) k . log p log p Corollary 1.6.4. The following three conditions are equivalent: Theorem 1.6.3 (Barsky). pL(n,k)
+
(i)
n=0
an
n+
x n
Ck,
(ii) lim vp (an )  k log n = +, log p (iii) lim nk an  = 0.
n+
Definition 1.6.5. If r 0, f : Zp Qp is in C r if
+
f=
n=0
an (f )
x n
and nr an (f ) 0 when n +. C r becomes a Banach space with the valuation: log(1 + n) vC r (f ) = inf {vp (an )  r }. nN log p
1.6. C K FUNCTIONS
21
1.6.2
Mahler's coefficients of C r functions.
We need M¨hler's Theorem in several variables to prove Barsky's theorem. a Let g(x0 , x1 , · · · , xi ) be a function defined on Zi+1 . We define the action p [k] j on g by the following formula: j g(x0 , · · · , xi ) = g(x0 , · · · , xj + 1, · · · , xi )  g(x0 , · · · , xi ), j = j j · · · j , k times . We set ak0 ,··· ,ki (g) = 0 0 · · · i i g(0, · · · , 0). Recall that C 0 (Zi+1 , Qp ) = C 0 (Zp , Qp ) p ··· C 0 (Zp , Qp ).
[k ] [k ] [k] [1] [1] [1] [1]
Theorem 1.6.6 (M¨hler). If g is continuous on Zi+1 , then ak0 ,··· ,ki (g) 0 a p when (k0 , · · · , ki ) and we have the following identity: g(x0 , · · · , xi ) =
k0 ,··· ,ki N
ak0 ,··· ,ki (g)
x0 xi ··· k0 ki
(1.2)
Conversely, if ak0 ,··· ,ki 0, then the function g via equation (1.2) is continuous on Zi+1 , ak0 ,··· ,ki (g) = ak0 ,··· ,ki , and p vC 0 (g) = inf vp (ak0 ,··· ,ki ). Proof of Theorem 1.6.3. Let gT (x) = (1 + T )x , then we have: gT (x, h1 , · · · , hi ) =
{i}
1 ( h1 · · · hi
(1)iI gT (x +
jI i
hj ))
I{1,··· ,i}
= (1 + T )x
x n 1 x x n 1 x1 n n1
(1 + T )hj  1 hj j=1
{i} n=0
Let Pn =
. Since
=
and gT (x, h1 , · · · , hi ) =
Pn (x, h1 , · · · , hi )T n ,
{i}
we have the following formulas:
{i} Pn (x0 , h1 , · · · , hi ) =
n0 +n1 i n1 ,··· ,ni 1
1 x0 n · · · ni n0 +···+n =n, 1
h1  1 hi  1 ··· . n1  1 ni  1
22CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT Let
{i} Qn,i (x0 , · · · , xi ) = Pn (x0 , x1 + 1, · · · , xi + 1) 1 x0 = n · · · ni n0 n +n +···+n =n, 1
0
n1 ,··· ,ni 1 +
1
i
x1 xi ··· . n1  1 ni  1
For all f C 0 (Zp , Qp ), we have f (x) =
+
an (f )
n=0
x n
. We denote
gi (x0 , · · · , xi ) =
n=0
an (f )Qn,i (x0 , · · · , xi )
if xj + 1 = 0, j 1. We have:
+
an0 ,n1 1,··· ,ni 1 (gi ) =
n=0
an (f )an0 ,n1 1,··· ,ni 1 (Qn,i )
where an0 ,n1 1,··· ,ni 1 (Qn,i ) = 0 1 n1 ···ni
i
if n =
j=0 i
nj , nj .
j=0
if n =
If f is in C k , i k, then gi is continuous on Zi+1 , thus p an0 +n1 +···+ni (f ) 0. n1 · · · ni Conversely, if
+
an0 +n1 +···+ni (f ) 0, then n1 · · · ni an0 ,n1 ,··· ,ni (f ) x0 n1 · · · ni n0 +···+n =n
i
+
n=0 n0 +n1
x1 xi ··· n1  1 ni  1
defines a continuous functions Gi on Zi+1 . But Gi = gi on Ni+1 , hence p Gi = gi , xj + 1 = 0, for all j 1,hence f is in C k .
1.7. LOCALLY ANALYTIC FUNCTIONS
23
1.7
1.7.1
locally analytic functions
Analytic functions on a closed disk.
+
Lemma 1.7.1. Let (an )nN with an in Cp be a sequence such that vp (an ) when n , let f =
n=0
an T n . Then:
(i) If x0 OCp , then f (k) (x0 ) converges for all k and
n
lim vp (
f (k) (x0 )) = . k!
(ii) If x0 , x1 are in OCp , then
+
f (x1 ) =
n=0
f (n) (x0 ) (x1  x0 )n n!
and
f (n) (x0 ) inf vp ( ) = inf vp (an ); nN nN n! (iii) inf vp (an ) = inf vp (f (x)) and vp (f (x)) = inf vp (an ) almost everynN xOCp n
where (i.e.,outside a finite number of xi + mCp ). Proof. (i)
f (k) k! +
=
n=0
an+k
n+k k
T n . Let T = x0 ; since vp (
n+k k
) 0, vp (xn ) 0
0, we get (1) and also vp ( (ii)
+ + +
f (n) (0) f (k) (x0 ) ) inf vp (an ) = inf vp ( ). nN nN k! n!
f (x1 ) = =
an x n = 1
n=0 + + n=0
an (
k=0
n (x1  x0 )k xnk ) 0 k
+
(
k=0 n=0
an
n nk x0 )(x1  x0 )k = k
n=0
f (n) (x0 ) (x1  x0 )n . n!
So we can exchange the the roles of 0 and x0 to get
nN
inf vp (
f (n) (x0 ) ) = inf vp (an ). nN n!
24CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT (iii) That inf vp (an ) inf vp (f (x)) is clear. As vp (an ) goes to +,
nN xOCp
vp (an ) reaches its infimum at some n0 N. So we can divide everything by an0 and we may assume that inf vp (an ) = 0. Let f (T ) = f (T ) mod mCp
nN
Fp [T ]. If x OCp doesn't reduce mod mCp to a root of f , then f (x) = 0, equivalently, vp (f (x)) = 0.
+ + +
Corollary 1.7.2. Let f =
n n=0
an T , g =
n=0
n
bn T , then f g =
n=0
n
cn T n ,
where cn =
i=0
ai bni . Suppose that vp (an ) and vp (bn ) go to infinity when n
n n n
goes to infinity, then vp (cn ) goes to infinity and inf vp (cn ) = inf (an ) + inf (bn ). Definition 1.7.3. For x0 Cp , r R, we define D(x0 , r) = {x Cp , vp (x  x0 ) r}. Definition 1.7.4. A function f : D(x0 , r) Cp is analytic if it is sum of its Taylor expansion at x0 or equivalently, if lim (vp ( f (n) (x0 ) ) + nr) = +. n!
0)
n+ {r}
We define vx0 (f ) = inf n (vp ( f
(n) (x
n!
) + nr).
Proposition 1.7.5. If the function f : D(x0 , r) Cp is analytic, then (i) For all k N, f (k) is analytic on D(x0 , r),
{r} v x0 (
f (k) (x0 ) {r} ) + kr vx0 (f ) k!
and goes to + if k goes to +. (ii) f is the sum of its Taylor expansion at any x D(x0 , r). {r} (iii) vx0 (f ) = inf vp (f (x)). (iv)
{r} vx0 (f g) xD(x0 ,r) {r} = vx0 (f )
+ vx0 (g).
{r}
Proof. If r Q, one can choose Cp , such that vp () = r. Let F (x) = f (x0 + x), x OCp . Apply the previous lemma, we can get the result. If r Q, choose rn decreasing with the limit r, rn Q. Use D(x0 , r) = / n D(x0 , rn ) and the case r Q, we get the result.
1.7. LOCALLY ANALYTIC FUNCTIONS
25
1.7.2
Locally analytic functions on Zp .
Definition 1.7.6. Let h N be given. The space LAh (Zp , Qp ) is the space of f whose restriction to x0 +ph Zp is the restriction of an analytic function fx0 on {h} D(x0 , h), for all x0 Zp . The valuation of the space is vLAh = inf vx0 (fx0 ),
x0 S
S be any set of representations of Zp /ph Zp . (Use above proposition to prove that this does not depend on S.) Lemma 1.7.7. LAh is a Banach space. Moreover, let x+i en = 1i+ph Zp ( h )m1 , n = mph  i, m 1, 1 i ph , p then en 's are a Banach basis of LAh . Theorem 1.7.8 (Amice). The functions [ pn ]! h of LAh .
x n
, n N are a Banach basis
x n
Proof. The idea is to try to relate the gn = [ pn ]! h (i) First step: For 1 j ph , we denote
n1 h
to the en .
n 1 gn,j (x) = gn (j + p x) = [ h ]! (j  k + ph x). p n! k=0 If vp (j + k) < h, then vp (j  k + ph x) = vp (j + k), for all x in OCp . If / vp (j + k) h, then vp (j  k + ph x) h with equality if x Fp Fp . So, we get
{0} v0 (gn,j )
n = vp ([ h ]!)vp (n!)+ inf(vp (j+k), h) = p k=0
n +
n1
#{k : vp (k) i, 1 k n}.
i=1
Since vp (n!) =
k=1
vp (k) =
i=1 h
n [ pi ], we have n
n vp (n!)  vp ([ h ]!) = p Thus,
#{k : vp (k) i, 1 k n} =
i=1 k=1
inf(vp (k), h).
n
v0 (gn,j ) =
k=1 h
{0}
[inf(vp (j + k  1), h)  inf(vp (k), h)] ([
l=1
=
j1 n n+j1 ]  [ l ]  [ l ]). l p p p
26CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT As [x + y] [x] + [y], we have v0 (gn,j ) 0, for all 1 j ph . So, we have vLAh (gn ) 0. (ii) Second step: we need a lemma Lemma 1.7.9. Let n = mph  i, gn,j Fp [x], then: (i) gn,j = 0, if j > i, (ii) deg gn,j = m  1, if j = i, (iii) deg gn,j m  1 if j < i. The lemma implies the theorem: gn can be written in terms of the en , multiplying by an invertible upper triangular matrix. Now use the fact that xn is a Banach basis if and only if xn is a basis of LA0 /pLA0 over Fp . h h Proof of Lemma 1.7.9. (i) If j > i, then j  1 i. Since [
{0} {0}
j1 n n+j1 ]  [ h ]  [ h ] = m  (m  1) = 1, h p p p
we have v0 (gn,j ) 1, then gn,j = 0. (ii) and (iii):If j i, write
n
gn,j (x) =
k=0
ak xk , ak Zp .
The zeros of gn,j are the
j+k ,0 ph
k n  1 and j1 n+j1 ]  [ h ] = m  1. h p p
#{zeros in Zp } = #{k : vp (j + k) h} = [
Let {i : 1 i m  1} be the set of the roots with 1 , · · · , m1 in Zp and m , · · · , n not in Zp . Then
m1 n
gn,j = c
l=1
(x  l )
1 (1  l x), (c is a constant ). l=m {0}
1 Since vp (l ) > 0 when l m, then vp (am1 ) = vp (c) = v0 (gn,j ). It implies c Zp . Hence m1
gn,j = c
l=1
(x  l ).
1.8. DISTRIBUTIONS ON ZP It remains to prove v0 (gn,i ) = 0. Since
h {0} v0 (gn,i ) {0}
27
=
l=1
([
mph  1 i1 mph  i ][ l ]+[ ]) pl p pl
and [ i ] = [ i1 ] + 1, we get the result. a a Let LA = {locally analytic functions on Zp }. Because Zp is compact, LA = LAh and is an inductive limit of Banach spaces. So (i) A function : LA B is continuous if and only if LAh : LAh B is continuous for all h. (ii) A sequence fn f converges in LA if and only if there exists h, such that for all n, fn LAh and fn f in LAh . 1 1 Since n vp ([ pn ]!) (p1)ph , we have the following theorem: h x is in LA if and only if there n n=0 exists r > 0, such that vp (an )  rn + when n +. Theorem 1.7.10. The function f = an
+
1.8
1.8.1
Distributions on Zp
The Amice transform of a distribution.
Definition 1.8.1. A distribution µ on Zp with values in B is a continuous linear map f Zp f µ from LA to B. We denote the set of distributions from LA to B by D(Zp , B). Remark. (i) µLAh is continuous for all h N. Set vLAh (µ) = inf (vB (
f LAh Zp
f µ)  vLAh (f )).
Then vLAh is a valuation on D(Zp , B) for all h, and D(Zp , B) is complete for the Fr´chet topology defined by vLAh , h N which means that µn goes to µ e if and only if vLAh (µn  µ) + for all h. (ii) D(Zp , B) = D(Zp , Qp ) B. From now on, we will denote D(Zp , Qp ) by D.
28CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT Let R+ be the ring of analytic functions defined on D(0, 0+ ) = {x Cp , vp (x) > 0}. A function f R+ can be written as f =
+ n=0
an T n , an Qp
for all n N. 1 Let vh = (p1)ph = vp (  1), where is a primitive ph+1 root of 1.
+
If F (T ) =
n=0
bn T n R+ , we define v (h) (F ) to be v (h) (F ) = v0
{vh }
(F ) = inf vp (bn ) + nvh .
nN
Then, for F, G R+ , v (h) (F G) = v (h) (F ) + v (h) (G). We put on R+ the Fr´chet topology defined by the v (h) , h N. e Definition 1.8.2. The Amice transform of a distribution µ is the function:
+
Aµ (T ) =
n=0
Tn
Zp
x µ= n
(1 + T )x µ.
Zp
Note that the last identity in the above definition is only a formal identity here. However, we have Lemma 1.8.3. If vp (z) > 0, then
Zp
(1 + z)x µ = Aµ (z)
Proof. Choose h such that vh < vp (z). Then vp (
+
zn ) +, [ pn ]! h
therefore
n=0
zn
x n
converges to (1 + z)x in LAh .
Theorem 1.8.4. The map µ Aµ is an isomorphism of Fr´chet spaces e from D to R+ . moreover, v (h) (Aµ ) vLAh (µ) v (h+1) (Aµ )  1.
1.8. DISTRIBUTIONS ON ZP
+
29
x n
Proof. Let Aµ (T ) =
n=0
bn T n . Since bn =
Zp
µ and vp (n!)
n , p1
then we
have: vp (bn ) = vp (bn )  vLAh ( x x ) + vLAh ( ) n n n x ) = vLAh (µ)  vp ([ h ]!) vLAh (µ) + vLAh ( n p n vLAh (µ)  = vLAh (µ)  nvh . (p  1)ph
Hence Aµ R+ and v (h) (Aµ ) vLAh (µ). Conversely, for F R+ , F = vp ([
+ + n=0
bn T n , then for all h,
n n ]!bn ) = vp (bn ) + . h p (p  1)ph
So f
n=0
bn an (f ) is a continuous map on LAh . Denote the left hand side
by
Zp
f µ, this defines a distribution µ D. Moreover, vLAh (µ) = inf vp ([ n n ]!bn ) inf vp ([ h+1 ]!bn ) h nN nN p p n (h+1) inf (vp (bn ) + )  1 = vLAh (Aµ )  1. h+1 nN (p  1)p
1.8.2
Examples of distributions.
(i) Measures are distributions and D0 D. (ii) One can multiply a distribution µ D by g LA, and one gets
d · Axµ = Aµ , = (1 + T ) dT ;
· Azx µ (T ) = Aµ ((1 + T )z  1); · AResa+pn Zp µ (T ) = pn
z
z a Aµ ((1 + T )z  1)
pn
=1
30CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT (iii) one gets actions , , with the same formulas than on measures. (iv) Convolution of distributions: If f LAh and for all y y0 + ph Zp ,
+
f (x + y) =
n=0
pnh f (n) (x + y0 ) y  y0 n ( h ) LAh (x) n! p
LAh (y),
and vLAh ( p
nh f (n) (x+y
0)
n!
) goes to +, when n +. Hence ( f (x + y)µ(x))(y) =
Zp Zp
fµ
Zp
is well defined, Aµ = A Aµ . (v) The derived distribution: µ dµ given by Zp f dµ = Zp f µ. Easy to check Adµ (T ) = log(1+T )Aµ (T ). µ can't be integrated because log(1+T ) = 0 if T =  1, µp . (vi) Division by x, the Amice transform Ax1 µ of x1 µ is a primitive(or called antiderivative) of (1 + T )1 Aµ , so Ax1 µ is defined up to 0 , Qp (we have x0 = 0).
1.8.3
Residue at s = 1 of the padic zeta function.
log(1+T ) . T
The KubotaLeopoldt distribution µKL given by AµKL (T ) = x µKL =
Zp n
Then
d dt d dt
n
(
t=0 n Zp
e µKL ) =
tx
d dt
n
AµKL (et  1)
t=0
= Since
(
t=0
et
t ) = (1)n n(1  n), for all n N. 1
1 1 ( ) = and (log(1 + T )) = p log(1 + T ), T T 1 we get (µKL ) = p µKL and xn µKL = (1  pn1 )
Z p Zp
xn µKL = (1)n n(1  pn1 )(1  n); (x)1i x
Z p 1s
p,i (s) =
(1)i1 s1
µKL .
The integral is analytic in s by the same argument as for measures.
1.9. TEMPERED DISTRIBUTIONS Proposition 1.8.5. lims1 (s  1)p,1 (s) = lims1 (s  1)(s) = 1). Proof. It follows from the following lemma.
Z p
31
1 µKL = 1  p , ( compare with
Lemma 1.8.6. a+pn Zp µKL = pn , for all n, for all a Zp (almost a Haar measure but µ a = µ). Proof. µKL = pn
a+pn Zp z pn =1
n
z a AµKL (z  1) = pn (1 +
z pn =1,z=1
log z ), z1
and
log z z1
= 0, if z p = 1, z = 1.
1.9
1.9.1
Tempered distributions
Analytic functions inside C r functions
Theorem 1.9.1. For all r 0, LA C r . Moreover there exists a constant C(r) depending on r, such that for all h N and for all f in LAh , vC r (f ) vLAh (f )  rh  C(r). Proof. Since vLAh (f ) = inf (vp (an (f ))  vp ([ pn ]!)), we have h
n
vC r (f ) = inf (vp (an (f ))r
n
log(1 + n) n log(1 + n) ) vLAh (f )+inf (vp ([ h ]!)r ). n log p p log p
We have a formula for every a: a a a log(1 + a) vp (a!) = [ ] + · · · + [ h ] + · · ·  . p p p1 log p Write n = ph a + b, 0 b ph  1, then we have vC r (f )  vLAh (f ) inf (vp ([ = n log(1 + n) ]!)  r ) h n p log p log(aph + b + 1) inf (vp (a!)  r ) aN log p h
0bp 1
a log(a + 1)  (r + 1)  rh. p1 log p
32CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT
a The function  p1 + (r + 1) log(a+1) of a is bounded above, we just let C(r) log p be its maximum.
Observe that the function log is well defined on Z . First if vp (x  1) > 0, p let
+
log x =
n=1
(1)n+1 (x  1)n ; n
in general, if x = (x) x , let log x = log x . If x = p, let log p = 0. By the formula log xy = log x + log y, log is well defined in Qp  {0}. This log is the socalled Iwasawa's log, or log0 . However, we can define the value at p arbitrarily. For L Qp , define logL p = L, then logL x = log0 x + Lvp (x). Theorem 1.9.2. Choose a L in Cp . Then there exists a unique logL : C p Cp satisfying: + (1)n1 (i) logL x = (x  1)n , here vp (x  1) > 0, n n=1 (ii) logL xy = logL x + logL y, (iii) logL = L. Proposition 1.9.3. If r 0, j > r, then xj logL x C r . Proof. We have
+ p1
x logL x =
n=0 a=1
j
1pn a+pn+1 Zp xj logL x.
Let fn,a = 1pn a+pn+1 Zp xj logL x. We have to prove the sum converges in C r . On pn a + pn+1 Zp , we have xj logL x = (x  pn a + pn a)j logL (pn a + (x  pn a)) x  pn a x  pn a = pnj (a + p n+1 )j (logL pn a + log0 (1 + p n+1 )). p p a So fn,a LAn+1 , vLAn+1 (fn,a ) nj. Use the previous theorem, we get vC r (fn,a ) nj  r(n + 1)  C(r) and it goes to +.
1.9. TEMPERED DISTRIBUTIONS
33
1.9.2
Distributions of order r
Definition 1.9.4. Let r 0 and B be a Banach space. A distribution µ D(Zp , B) is a distribution of order r if f Zp f µ is a continuous map from C r (Zp , Qp ) to B. We denote the set of distributions of order r by Dr (Zp , B). We define a valuation on Dr (Zp , B) by vDr (µ) = infr (vp (
f C Zp
f µ)  vC r (f )).
Remark. (i) Under the above valuation, Dr (Zp , B) is a padic Banach space and Dr (Zp , B) = Dr (Zp , Qp ) B. We denote Dr (Zp , Qp ) by Dr . (ii) Dtemp = Dr =set of tempered distributions. (iii) Since LAh C r , and for f LAh , vC r (f ) vLAh (f )  rh  C(r), we get, for µ Dr LA , h vLA (µ) = inf (vp ( h
f LAh Zp
f µ)  vLAh (f )) vDr (µ)  rh  C(r).
Theorem 1.9.5. µ D, the following are equivalent: (i) µ Dr i.e. µ can be extended by continuity to C r . x (ii) There exists a constant C, such that vp ( Zp n µ) C  r log(1+n) , for log p all n. (iii) There exists a constant C, such that vp ( a+ph Zp (xa)j µ) C +h(j  r), for all a Zp , j N, h N. (iv) There exists a constant C, such that vLAh (µ) C rh, for all h N. Remark. It follows that vDr (µ) = is equivalent to vDr . Proof. (i) (ii) is just the definition of vDr . (iii) (iv) is true by the definition of LAh (with some C). Remains to prove (ii) (iv). We have v (h) (Aµ ) vLAh (µ) v (h+1) (Aµ )  1, hence the proof is reduce to the following lemma with F = Aµ . inf (vp (
a+ph Zp
aZp jN,nN
(x  a)j µ)  h(j  r))
34CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT Lemma 1.9.6. Suppose F R+ , F =
+ n=0
bn T n , the following are equivalent:
(i) there exists C, such that v (h) (F ) C  rh, for all h N, (ii) there exists C , such that vp (bn ) C  r log(1+n) for all n. log p Proof. Let C0 = inf (v (h) (F ) + rh) = inf (inf (vp (bn ) +
hN hN nN
n ) + rh), (p  1)ph
C1 = inf (vp (bn ) + r
nN
log(1 + n) ). log p
Let h = [ log(1+n) ], then log p vp (bn ) C0  rh  n log(1 + n) C0  r  2, (p  1)ph log p
n (p1)ph
which implies C1 C0  2. Now, if h is fixed, then C1  r log(1+n) + log p h (p  1)p r. Hence, C1  r
is minimal for (1 + n) =
n log(p  1)r log(1 + n) + C1  rh  . h log p (p  1)p log p
Thus, C0 C1  r log(p1)r . log p For N 0, let LP [0,N ] be the set of the locally polynomial functions of degree no more than N on Zp . Theorem 1.9.7. Suppose r 0, N > r  1. If f Zp f µ is linear function from LP [0,N ] to a Banach space B , such that there exists C, vp (
a+pn Zp
(x  a)j µ) C + (j  r)n
for all a Zp and n, j N, then µ extends uniquely to an element of Dr . Remark. (i) Let r = 0, N = 0, we recover the construction of measures as bounded additive functions on open compact sets.
1.9. TEMPERED DISTRIBUTIONS (ii) We define a new valuation on Dr vDr,N (µ) =
aZp nN,jN
35
inf
vp (
a+pn Zp
(x  a)j µ)  n(j  r),
then vp ( Zp f µ) vLAh (f ) + vDr,N (µ)  rn for all f LP [0,N ] LAh ; (iii) The open mapping theorem in Banach spaces implies that vDr,N is equivalent to vDr . Proposition 1.9.8. If f LA, r 0, N > r  1, put
pn 1 N
fn =
i=0
1i+pn Zp (
k=0
f (k) (i) (x  i)k ) LP [0,N ] , k!
then fn f in C r . Hence LP [0,N ] is dense in C r . Proof. There exists h, such that f LAh . We assume n h, then vLAh (f  fn ) = f LAh implies vp ( p
hk f (h) (i)
0ip 1 kN +1
infn
inf vp (pnk
f (k) (i) ). k!
h!
) vLAh (f ). Hence
vLAh (f  fn ) vLAh (f ) + (N + 1)(n  h). Then vC r (f  fn ) vLAh (f  fn )  rn  C(r) vLAh (f )  C(r)  (N + 1)h + (N + 1  r)n +, because N + 1  r > 0. Proof of Theorem 1.9.7. The proposition implies the uniqueness in the theorem. We only need to prove the existence. We show that if f LAh , then limn Zp fn µ exists: vp (
Zp
(fn+1  fn )µ) vLAn+1 (fn  fn+1 ) + vDr,N (µ)  r(n + 1) inf(vLAn+1 (f  fn ), vLAn+1 (f  fn+1 )) + vDr,N (µ)  r(n + 1) vDr,N (µ) + vLAh (f )  r(h  1) + (n  h)(N + 1  r) +.
36CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT Set
Zp
f µ = limn+ vp (
Zp
Zp
fn µ, then fn µ), vp ( inf (fn1  fn )µ))
Zp
f µ) inf(vp (
Zp
nh
vLAh (f )  rh + (vDr,N (µ)  r). This implies that µ Dr .
1.10
Summary
To summarize what we established: (i) We have the inclusions: C 0 C r LA LAh D0 Dr D LA . h Now, if f is a function on Zp and µ is a linear form on polynomials, then we have: n n f an (f ) = (1)i f (n  i) i i=0 µ bn (µ) =
Zp
x µ n
(ii) For f a function, · f C 0 if only if vp (an (f )) + and vC 0 (f ) = inf vp (f (x)) = inf vp (an (f )).
xZp n
· f C r if only if vp (an (f ))  r log(1+n) + and log p vC r (f ) = inf vp (an (f )  r
n
log(1 + n) ). log p
· f LA if only if there exists r > 0 such that vp (an (f ))  rn +. LA is not a Banach space; it is a compact inductive limit of Banach spaces.
1.10. SUMMARY · f LAh if and only if vp (an (f ))  vp ([ pn ]!) + and h vLAh (f ) = inf inf vp (
xZp kN
37
pkh f (k) (x) n ) = inf (vp (an (f ))  vp ([ h ]!)). n h! p
(iii) For µ a distribution, · µ D0 if and only if vD0 (µ) = inf vp (bn (µ)) > .
n
· µ Dr if and only if vDr (µ) = inf vp (bn (µ)) + r log(1+n) > . log p
n
· µ D if and only if for all r > 0, inf vp (bn (µ)) + rn > .
n +
(iv) f =
n=0
an (f )
x n
+
and
fµ = Zp
an (f )bn (µ).
n=0
38CHAPTER 1. THE P ADIC ZETA FUNCTION OF KUBOTALEOPOLDT
Chapter 2 Modular forms
2.1
2.1.1
Generalities
The upper halfplane
By SL2 we mean the group of 2 × 2 matrices with determinant 1. We write SL2 (A) for those elements of SL2 with entries in a ring A. In practice, the ring A will be Z, Q, R. b Let = a d in SL2 (R), z in C  { d }, let z = az+b , then c c cz+d Im(z) = (ad  bc) Imz Im(z) = . 2 cz + d cz + d2
We denote H = {z, Imz > 0} the upper half plane. It is stable under z z and one can verify (1 2 )z = 1 (2 z). Proposition 2.1.1. The transform action z z defines a group action of SL2 (R) on H. Proposition 2.1.2.
dxdy y2
is invariant under SL2 (R).
i (hint : dx dy = 2 dz dz and z z is holomorphic.)
Definition 2.1.3. Let f : H C be a meromorphic function and = a b c d be in SL2 (R). If k in Z, we define the weight k action of SL2 (R) by (f k )(z) = (cz + d)k f (z). Exercise. (f k 1 )k 2 = f k 1 2 . 39
40
CHAPTER 2. MODULAR FORMS
2.1.2
Definition of modular forms
Definition 2.1.4. Let be a subgroup of SL2 (Z) of finite index, is a finite order character of (i.e. () µN ). f : H C is a modular form of weight k, character for , if: (i) f is holomorphic on H; (ii) f k = ()f , if ; (iii) f is slowly increasing at infinity, i.e. for all \SL2 (Z), there exists C() and r() such that  f k (z)  y r() , if y C(). Definition 2.1.5. is a congruence subgroup if (N ) = Ker (SL2 (Z) SL2 (Z/N Z)) for some N in N. Example 2.1.6. 0 (N ) = { a b c d SL2 (Z) : c 0 mod N } (N ).
Any character : (Z/N Z) C extends to a congruence character : 0 (N ) C ( a b ) (d). c d
Let Mk (, ) be the set of modular forms of weight k, character for . Then Mk (, ) is a Cvector space.
0 Remark. (i) If 1 1 = I and (I) = (1)k , then Mk (, ) = 0; 0 (ii) f Mk (, ), g SL2 (Z), f k g Mk (g 1 g, g ) where g () = (gg 1 ).
2.1.3
qexpansion of modular forms.
Lemma 2.1.7. If is a subgroup of finite index of SL2 (Z) and : C is of finite order, then there exists M in N  {0}, such that 1 M and 0 1 ( 1 M ) = 1. 0 1 Proof. We can replace by Ker and assume = 1. There exists n1 = n2 , 1 such that 1 n1 and 0 n2 have the same image in \SL2 (Z), then M = 0 1 1 n1  n2  satisfy the condition. For M N{0}, let qM (z) = e M . Then z qM (z) gives a holomorphic bijection M Z\H D = {0 < qM < 1}.
2iz
2.1. GENERALITIES
41
Corollary 2.1.8. If f Mk (, ), then there exists M = 0, M N, such ~ that f (z + M ) = f (z). Thus there exists f holomorphic on D , such that ~ f (z) = f (qM ). ~ ~ Now f has a Laurent expansion f (qM ) = an q n with
M nZ
2ny M
an = e
1 · M
M 2
f (x + iy)e
M 2
2inx M
dx
for all y. If n < 0, when y , the right hand side goes to 0, so an = 0. Hence we get the following result. Proposition 2.1.9. If f is in Mk (, ), there exists M N  {0}, and 1 elements an (f ) for each n M N, such that f=
1 n M N
an (f )q n , where q(z) = e2iz ,
which is called the q expansion of modular forms.
2.1.4
Cusp forms.
Definition 2.1.10. (i) v (f ) = inf{n Q, an (f ) = 0} 0 and we say that f has a zero of order v (f ) at . We say that f has a zero at if v (f ) > 0. (ii) A modular form f is a cusp form if f k has a zero at for all in \SL2 (Z). We denote Sk the set of cusp form of weight k. Sk (, ) Mk (, ). Remark. If f is a cusp form, then f is rapidly decreasing at since  (f k )(z) = O ev (f k )2y . Theorem 2.1.11. Sk (, ) and Mk (, ) are finite dimensional Cvector spaces with explicit formulas for the dimensions( if k 2). Remark. k, Mk (, ) = M () is an algebra. The study of Mk (, ) for congruence subgroup and congruence characters (Ker congruence subgroup ) can be reduced to the study of Mk (0 (N ), ) for a simple group theoretic reason. From now on, we write Mk (N, ) = Mk (0 (N ), ), Sk (N, ) = Sk (0 (N ), ).
42
CHAPTER 2. MODULAR FORMS
2.2
2.2.1
The case = SL2(Z)
The generators S and T of SL2 (Z).
0 1 , 1 0 1 n 0 1 1 1 . 0 1
Let Mk (1) = Mk (SL2 (Z), 1), Sk (1) = Sk (SL2 (Z), 1). Let S= It is easy to verify Tn =
1 So Sz =  z , T n z = z + n.
T =
for any n Z.
Proposition 2.2.1. (i) If (a, b) = 1, then there exists n = n(a, b), (a0 , b0 ) = (1, 0), (a1 , b1 ) = (0, 1), · · · (an , bn ) = (a, b), such that al al+1 bl bl+1 (ii) SL2 (Z) = S, T . Proof. (i) We prove it by induction on a + b. If a + b = 1, one can do it by hand: I= 1 0 , S= 0 1 0 1 , S2 = 1 0 1 0 , S3 = 0 1 0 1 . 1 0 SL2 (Z) for any l.
If a + b 2, there exists µ, Z, such that bµ  a = 1, and  < b, which implies µ a. Then we have µ a SL2 (Z) and µ +  < a + b. b Therefore the conclusion is obtained by the inductive assumption. b (ii) Let = a d SL2 (Z), there exists n = n(a, b), (a0 , b0 ) = (1, 0), c (a1 , b1 ) = (0, 1), · · · (an , bn ) = (a, b), such that l = As 1 = I and
1 l+1 l =
al al+1 bl bl+1
SL2 (Z) for any l.
nl 1 1 0
= T nl S 3 ,
then =
1 (l+1 l )1 T, S .
2.2. THE CASE = SL2 (Z)
+
43 an q n , where q = e2iz , then f Mk (1) if and
Corollary 2.2.2. Let f =
n=0 +
only if the following two conditions hold: (i)
n=0 1 (ii) f ( z ) = z k f (z).
an q n converges if q < 1.
2.2.2
Eisenstein series
1 (k) 2 (2i)k 1 Mk (1), (mz + n)k
Proposition 2.2.3. If k 3, then Gk Mk (1), where Gk (z) =
m,n
and means the summation runs over all pairs of integers (m, n) distinct from (0, 0). Proof. As mz + n min(y, y/z) sup(m, n), the series converges uniformly on compact subsets of H and is bounded at . b Let = a d SL2 (Z), since c (cz + d)k
m,n
1 (m az+b cz+d + n)k
=
m,n
1 , ((am + cn)z + (bm + dn))k
and (m, n) (am + cn, bm + dn) is a bijection of Z2  {(0, 0)}, it follows that Gk k = Gk . Proposition 2.2.4. Gk (z) = where s (n) =
dn, d1
(k) (k) + k1 (n)q n , (2i)k n=1
+
ds , and k is even (if k is odd, Mk (1) = 0, since  I
SL2 (Z)). Proof. (k) (k) Gk (z) = (k) + k (2i) (2i)k
+
Ak (mz),
m=1
44 where Ak (z) =
nZ
CHAPTER 2. MODULAR FORMS
1 = (z + n)k
^ (l)q l
lZ
for the last identity given by the Poisson summation formula of Fourier transforms, and (by residue computation)
+
^ (l) =

e2ilx dx = (x + iy)k
0,
(2i)k k1 l , (k1)!
if l 0, if l 0.
It follows that (k) Gk (z) = (k) + (2i)k m=1
+ +
l
l=1
k1 lm
q
(k) = (k) + k1 (n)q n . k (2i) n=1
+
Remark. (i) G2 (z) = it is almost one. Let G (z) = G2 (z) + 2
(2) (2) + (2i)2
+ n=1
1 (n)q n is not a modular form, but
1 1 (2) 1 ys = lim , 8y 2 (2i)2 s0 m,n (mz + n)2 mz + n2s
G is not holomorphic, but G 2 = G , for any SL2 (Z). 2 2 2 (ii) Let Ek = a0Gk k ) , so that a0 (Ek ) = 1. (G
2.2.3
The fundamental domain for SL2 (Z)
Theorem 2.2.5. Let D denotes the shadows in Figure 1.1. Then it is a fundamental domain for PSL2 (Z). Moreover, the stabilizer of z D is  {I} if z = i, ;  {I, S} if z = i;  {I, T S, (T S)2 } if z = .
2.2. THE CASE = SL2 (Z)
45
Figure 2.1: The Fundamental Domain.
Proof. Let z0 H, = a b c d SL2 (Z)
0 Since Im (z0 ) = cz0z+d2 tends to zero, as (c, d) tends to infinity, there exists 0 such that Im (0 z0 ) is maximal. There exists a unique n such that:
1 1  < Re (0 z0 ) + n . 2 2 Let 1 = T n 0 , then Im (1 z0 ) = Im (0 z0 ) Im (S1 z0 ) = Im (1 z0 ) 1 z0 2
which implies 1 z0  1. Therefore D contains a fundamental domain. If z1 , z2 D, and there exists SL2 (Z), such that z1 = z2 , we want to show z1 = z2 . By symmetry, we may assume Im (z2 ) Im (z1 ). If Im (z2 b = a d , Im (z2 ) cz2 +d)2 implies cz2 + d2 1. As Im (z2 ) 23 , we have c c 1, d 1. It remains only finite number of cases to check. If c = 0, then d = ±1, and is the translation by ±b. Since 1 1  < Re (z1 ), Re (z2 ) , 2 2 this implies b = 0, and = ± I. If c = 1, the fact z2 + d 1 implies d = 0 except if z2 = , in which case we can have d = 0, 1. The case d = 0 gives z2  1, hence z2  = 1; on the other hand, SL2 (Z) implies b = 1, hence z1 = z2 = a  1/z2 D,
46
CHAPTER 2. MODULAR FORMS
Figure 2.2: The Route C(M, ) of Integration.
which implies a = 0, and z1 = z2 = i. The case z2 = , and d = 1 gives 1 a + b + 1 = 0 and z1 = z2 = a  1 = a + D, which implies a = 0 and z1 = z2 = . If c = 1, we have similar argument as c = 1. This completes the proof of the Theorem.
2.2.4
The
k 12
formula.
k 12
The following proposition is usually called "the Proposition 2.2.6. Let f Mk {0}, then 1 1 v (f ) + vi (f ) + v (f ) + 2 3
formula".
vz (f ) =
zD{i,}
k . 12
Proof. Apply Cauchy residue formula to d log f over the path showed in Figure 1.2. As M +, and 0, we have: 1 2i 1 M + 2i lim d log f =
C(M,) zD{i,}
vz (f ),
d log f = lim 
C (M ) M +
1 2i
d log
z=e2M
an (f )z n = v (f ),
1 0 2i lim
1 d log f =  vi (f ), 2 C(i,)
2.2. THE CASE = SL2 (Z)
47
0
lim
1 2i
i
1 1 1 d log f =  v (f ) =  v2 (f ) = lim 0 2i 6 6 C(,)
d log f
C(2 ,)
1 ( 2i
d log f +
2 i
d log f ) =
1 2i 1 2i k 2i
1 (d log f  d log f ( )) z 2
i
i
= =
(d log f  d log z k f (z))
2 i 2
k k dz = (log i  log 2 ) = . z 2i 12
Putting all these equations together, we get the required formula. Corollary 2.2.7. G4 has its only zero on D at z = , G6 has its only zero on D at z = i. = (( G4 3 G6 2 1 ) ( ) ) 1 = q + · · · M12 (1) a0 (G4 ) a0 (G6 ) 3a0 (G4 )  2a1 (G6 ) 0
does not vanish on D (v () = 1).
+
Remark. One can prove = q
n=1
(1  q n )24 .
2.2.5
Dimension of spaces of modular forms.
Theorem 2.2.8. (i) Mk (1) = 0, if k is odd or k = 2. (ii) dim Mk (1) = 1, if k = 0 or k is even and 2 < k 10. In this case Mk (1) = C · Gk (We have G0 = 1). (iii) Mk+12 (1) = C · Gk+12 · Mk (1). Proof. If f Mk+12 , then f= a0 (f ) Gk+12 + g, a0 (Gk+12 )
where g Mk (1), because does not vanish on H, v () = 1 and v (f  a0 (f ) G ) 1. a0 (Gk+12 ) k+12
48
CHAPTER 2. MODULAR FORMS
k [ 12 ], k [ 12 ] + 1,
Corollary 2.2.9. If k is even, dimC Mk (1) =
k 2 mod 12, if not.
Remark. Finite dimensionality of spaces of modular forms has many combinatorical applications. For example, let (z) =
nZ
q
n2 2
=
nZ
ein z ,
2
= { SL2 (Z), I or S mod 2}, : {±1}. () = 1 if I 1 if S
One can check that dim M2 ( , ) 1, 4 M2 ( , ), and 4G (2z)  2 G ( z ) M2 ( , ), so we have 2 2 3(2)(2) 4 z , 4G (2z)  G ( ) = 2 2 2 (2i)2 hence {(a, b, c, d) Z4 : a2 + b2 + c2 + d2 = n} = 8
dn,4 d
d,
from which we can deduce that any positive integer can be written as a sum of 4 squares.
2.2.6
Rationality results.
As M8 (1) and M10 (1) are of dimension 1, we have a0 (G8 )G2 = a0 (G4 )2 G8 , a0 (G10 )G4 G6 = a0 (G4 )a0 (G6 )G10 . 4 Let = Substituting G4 = + q + 9q 2 + · · · , G8 = + q + 129q 2 + · · · (4) (8) (4), = (8). 4 (2i) (2i)8 ()
2.2. THE CASE = SL2 (Z)
49
in (), compare the coefficients of q and q 2 , we have the following equations: 2 = 2 (1 + 18) = 1292
1 1 The solution is: = 240 , = 480 . In particular, , Q, which implies G4 and G8 have rational qexpansions, and (4) Q, (8) Q. 4 8 1 Exercise. a0 (G6 ) =  504 , which implies (6) 6
Q.
Let A be a subring of C, let Mk (, A) = {f Mk (), an (f ) A, for all n}, then M(, A) =
k
Mk (, A) is an Aalgebra.
Theorem 2.2.10. (i) M(SL2 (Z), Q)  Q[X, Y ], where X = G4 , Y = G6 . (ii) M(SL2 (Z), C) = C M(SL2 (Z), Q). Proof. If k fk = 0, where fk Mk (SL2 (Z), C), then for any z, for any a b SL (Z), we have (cz + d)k fk (z) = 0. Therefore (Xz + Y )k fk (z) 2 c d
k k
is identically zero because it (as a polynomial in X and Y ) has too many zeros. Hence fk (z) = 0, which implies that M(SL2 (Z), C) =
k 3(n1)
Mk (SL2 (Z), C).
, · · · , n is a basis of Mk (1); if k = 12n+2, Now if k = 12n, G3n , G4 4 3(n1)+2 3(n2)+2 G4 G 6 , G4 G6 , · · · , G2 G6 n1 is a basis of Mk (1), and so on, 4 = aG3 + bG2 , a, b Q. As G4 , G6 M(SL2 (Z), Q), this proves both 6 4 results. Corollary 2.2.11. Let f Mk (1), Aut(C), then f = (k) Mk (1). Moreover, (2i)k Q if k is even and k 4. an (f ) q n
Proof. The first assertion is a direct consequence of Theorem 2.2.10 (ii). For any Aut(C), we have G  Gk = a0 (Gk )  a0 (Gk ) Mk (1). k This implies a0 (Gk ) = a0 (Gk ) for any Aut(C), therefore a0 (Gk ) Q.
50 Remark. When k = 2, we can use
CHAPTER 2. MODULAR FORMS
z 4G (2z)  G ( ) M2 ( , Q) 2 2 2 to deduce
(2) 2
Q.
Remark. (i) The zeta function is a special case of Lfunctions, and (k) are special values of Lfunctions (i.e. values of Lfunctions at integers). Siegel used the above method to prove rationality of special values of Lfunctions for totally real fields. (ii) With a lot of extra work, we can prove integrality results. As (k) k1 (n)q n , (k) + Gk (z) = (2i)k n=1 and k1 (n) =
Zp +
xk1 (
dn
d ), we have all an (Gk ) are given by measures on
Zp , therefore a0 (Gk ) is also given by measures. From which we can deduce other constructions of KubotaLeopoldt zeta functions (the work of Serre, Deligne, Ribet).
2.3
The algebra of all modular forms.
an q n Mk (, C), an A, n N .
Let A be a subring of C, let Mk (A) =
[SL2 (Z):]<+
Mk (, A) =
Let M(A) = Mk (A), then it is an Aalgebra. Let Mcong (A) =
congruence subgroup
M(, A).
Theorem 2.3.1. (i) If f M(C), and Aut(C), then f M(C). (ii) M(C) = C Q M(Q) = C Q M(Q). (iii) Let Q = Aut(M(Q)/ M(SL2 (Z), Q)), GQ = Gal(Q /Q), then we have an exact sequence: 1
/ SL2 (Z) / Q w / GQ /1
2.3. THE ALGEBRA OF ALL MODULAR FORMS. where G
51
[G:]< normal
lim (G/), and GQ Q is induced by the action on Fourier 
coefficients. (iv) Mcong (Qab ) is stable by Q , and
^ Aut(Mcong (Qab )/ M(SL2 (Z), Q))  GL2 (Z).
Moreover, we have the following commutative diagram: 1
/ SL2 (Z) / SL (Z) ^ 2 / Q v / GQ / ^ Z /1
1
r / GL (Z) ^ 2
/1
^ ^ ^ where GQ Z is the cyclotomic character, GL2 (Z) Z is the determinant 0 ^ ^ map, and Z GL2 (Z) maps u to 1 u . 0 ^ Remark. (i) SL2 (Z) is much bigger than SL2 (Z). (ii) We can get an action of GQ on SL2 (Z) by inner conjugation in Q . This is a powerful way to study GQ (Grothendieck, "esquisse d'un programme"). (iii) There are padic representations of GQ attached to modular forms (by Deligne) for congruence subgroups. They come from the actions of GQ on H1 (SL2 (Z) , W ), where W = Symk2 Vp Zp Zp [SL2 (Z)/], Vp is Q2 with p actions of Q through GL2 (Zp ) and are cut out using Hecke operators on these spaces. Proof of Theorem 2.3.1 (i). Let N(, A) denote the set of holomorphic functions f : H C satisfying the following conditions: (a) for any , f (z) = f (z), (b) for any \ SL2 (Z), f =
nn0 (,f ) n 1 Z M 1
an q n , and an A for any n.
As S12 (SL2 (Z), Q) does not vanish on H, 12 S1 (SL2 (Z), , Q), where k : SL2 (Z) µ12 . Let 0 = Ker . If f Mk (, A),  12 f N( 0 , A). If f N(, A), k f M12k (, A), where k + n0 (, f ) 0 for any
52
CHAPTER 2. MODULAR FORMS
\ SL2 (Z). Therefore knowing N(, A) is equivalent to knowing M(, A). So it suffices to prove if f=
nn0
an q n N (C) =
N(, C)
and Aut C, then f N (C). G3 Let j = a0 (G4 ) = q 1 + · · · N(SL2 (Z), Q). 3
4
Proposition 2.3.2. (i) N(SL2 (Z), Q) = Q[j], N (SL2 (Z), C) = C[j]. (ii) j : SL2 (Z)\H C is bijective. 3 if SL2 (Z) (iii) j(z)  j() has a zero at z = of order e() = 2 if SL2 (Z)i, 1 otherwise. (iv) j(i), j() Q. , · · · , a is a basis of M12a (SL2 (Z), Q). Proof. (i) Note that G3a , G4 4 (ii) and (iii): For any C, f = (j  ) · M12 (SL2 (Z), C), with v (f ) = 0. As D = SL2 (Z)\H, and 1 1 z (f ) + i (f ) + (f ) = 1, 2 3
3(a1)
zD{,i}
we can deduce the required results. (iv) G4 () = 0, G6 (i) = 0. Let f N(, C), Pf (X) =
\ SL2 (Z)
(X  f ) N(SL2 (Z), C)[X] C((q))[X]
Pf (X) =
\ SL2 (Z) n
(X  (f ) ) N(SL2 (Z), C)[X] C((q))[X] gl X l , Pf (X) =
n l=0
Denote Pf (X) =
l=0
gl X l , where gl N(SL2 (Z), C), and
gl N(SL2 (Z), C) thanks to the Corollary 2.2.11. We give the proof in two steps.
2.4. HECKE OPERATORS
53
Step 1: Prove that f is holomorphic on H, by the Proposition 2.3.2. We have n n Pf (X) =
l=0
Pl (j)X l ,
Pf (X) =
l=0
Pl (j)X l .
The roots of Pf are the f 's, where \ SL2 (Z). They are holomorphic on H. The roots of Pf are multivalued holomorphic functions on H. In order to prove that are single valued, it suffices to show there is no ramification. Let be an arbitrary element in H. we have, around , n distinct formal solutions
+
al,k ()(j  j()) e()
k=0
1
k
(1 l n)
of Pf (X) = 0 as (j  j()) e() is a local parameter around by Proposition 2.3.2. Let H satisfies j( ) = j() , then we have e( ) = e(). Therefore
+
al,k () (j  j( )) e( ) ,
k=0
k
(1 l n)
are n distinct formal solutions around . It follows that there is no ramification around , for any . Hence the roots of Pf are holomorphic on H. In particular, f is holomorphic on H. Step 2: Prove that there exists SL2 (Z) of finite index, such that f = f for any . For any SL2 (Z),
n n
Pf (f ) =
l=0
gl (f
) =
l=0
l
gl (f )l = Pf (f ) = 0
So f belongs to the finite set of roots of Pf , which leads to the required conclusion.
2.4
2.4.1
Hecke operators
Preliminary.
Let G be groups (for example, = SL2 (Z), G = GL2 (Q)+ ), let x G, x = {x : }, x = {x : }.
54
CHAPTER 2. MODULAR FORMS
Let A be a ring, define A[\G/] to be the set of : G A satisfying the following two conditions: (i) (x) = (x) = (x), for all x G, . (ii) There exists a finite set I such that = i 1xi .
iI
Remark. (i) We impose xi to be distinct in \G, in this situation, the decomposition is unique, i 's are unique. (ii) For any , 1xi (x) = 1xi (x 1 ). So = i 1xi A[\G/]
iI
implies i 1xi (x) =
iI iI
i 1xi (x 1 ) = (x 1 ) = (x) =
iI
i 1xi (x)
Therefore there exists a permutation: : I I, and for any i I, there exists i , such that (i) = i , xi = i x(i) . Proposition 2.4.1. (i) If =
iI
i 1xi , =
jJ
µj 1yj A[\G/], then
=
(i,j)I×J
i µj 1xi yj A[\G/],
and it does not depend on the choices. (ii) (A[\G/], +, ) is an associative Aalgebra with 1 as a unit. (iii) If M is a right Gmodule with G action m m g, and = i 1xi A[\G/], then for any m M , m = i m xi does not
iI
depend on the choices of xi . Moreover, m M , m(1 2 ) = (m1 )2 , m (1 + 2 ) = (m 1 ) + (m 2 ). Proof. Exercise, using the previous remark. Remark. If = 1, then A[\G/] = A[G] is commutative if and only if G is commutative.
2.4.2
Definition of Hecke operators: Rn , Tn , n 1.
Let G = GL2 (Q)+ , = SL2 (Z). Lemma 2.4.2. Let g G M2 (Z), then there exists a unique pair (a, d) b N  {0}, and b Z unique mod dZ, such that g = a d 0
2.4. HECKE OPERATORS
55
Proof. Let g = , there exists µ, Z, such that (µ, ) = 1, and µ + = 0. And there exists x, y Z, such that x  µy = 1, Let x y x y b 0 = µ if x + y 0; 0 =  µ if x + y < 0. Then 0 g = a d , 0 where a > 0. Thus completes the proof of existence. If 1 , 2 satisfies 1 g = then (1 g)(2 g)
1
a1 b 1 0 d1
a1 a2
2 g =
a2 b 2 0 d2
=
0
a2 b1 a1 b2 a2 d2 d1 d2
SL2 (Z)
This implies a1 = a2 , d1 = d2 , b1  b2 divisible by d1 . Lemmadefinition 2.4.3. For any n 1, Rn = 1( n 0 ) Z[\G/],
0 n
Tn = 1{gM2 (Z),det g=n} Z[\G/]. Proof. Left and right invariance come from det gg = det g det g . And Lemma 2.4.2 implies Tn = 1( a b ) , so get the finiteness needed.
ad=n,a1 b mod d 0 d
Remark. If p is prime, Then Tp = 1( p 0 ) by elementary divisors for prin0 1 ciple ideal domains. Theorem 2.4.4. (i) For any n 1 and l 1, Rn Rl = Rnl = Rl Rn , Rn Tl = Tl Rn . (ii) If (l, n) = 1, Tl Tn = Tln = Tn Tl . (iii) If p is prime and r 1, Tpr Tp = Tpr+1 + pRp Tpr1 . (iv) Let TZ be the subalgebra of Z[\G/] generated by Rn and Tn (n 1). It is a commutative algebra. Proof. (i) It is trivial. (ii) We have Tn Tl =
ad=n,a1 a d =n,a 1 b mod d b mod d
1 aa
0
ab +bd dd
.
56
CHAPTER 2. MODULAR FORMS
As (n, l) = 1, (a, a ) = 1, (a, d ) = 1. This implies {aa : an, a l} = {a : a nl}. Therefore in order to show Tn Tl = Tnl , it suffices to verify that {ab + bd } is a set of representatives of Z/(dd )Z, where b is a set of representatives of Z/dZ, b is a set of representatives of Z/d Z. It suffices to show the injectivity under the mod dd Z map. If ab1 + b1 d ab2 + b2 d , then b1 b2 mod d , so b1 = b2 , which leads to the required conclusion. (iii) We have
r
Tpr =
i=0 b mod pi
1 pri b , Tp = 1( p 0 ) +
0 pi 0 1
1( 1 c )
0 p
c mod p
Then
r r
Tpr Tp =
i=0 b mod pi
1 pr+1i
0
b pi
+
i=0 b mod pi c mod p
1 pri pb+pri c
0 pi+1
r1
=T
pr+1
+ Rp (
i=0 b mod pi c mod p
1 pr1i b+pr1i ) = Tpr+1 + pRp Tpr1 .
0 pi
(iv) It follows from (i),(ii),(iii).
2.4.3
Action of Hecke operators on modular forms.
The following two propositions are exercises in group theory. Proposition 2.4.5. Assume G are groups. Then (i) If [ : ] < +, then contains some which is normal in , and [ : ] < +. (ii) If [ : 1 ] < +, [ : 2 ] < +, then [ : 1 2 ] < +. (iii) If H H G, [H : H ] < +, then [H : H ] < +. Proposition 2.4.6. (i) Suppose GL2 (Q)+ , and N N such that N , N 1 M2 (Z), then 1 SL2 (Z) SL2 (Z) (N 2 ) := SL2 (Z) (1 + N 2 M2 (Z)). (ii) If [SL2 (Z) : ] < +, GL2 (Q)+ , then [SL2 (Z) : SL2 (Z) 1 ] < +.
2.4. HECKE OPERATORS Proposition 2.4.7. Mk (C) =
[SL2 (Z):]<+
57
Mk (, C),
Sk (C) =
[SL2 (Z):]<+
Sk (, C)
are stable under GL2 (Q)+ . Proof. For any , fk = f . For GL2 (Q)+ , we have (fk )k (1 ) = fk , so fk is invariant for the group 1 SL2 (Z). To verify that fk is slowly increasing at , write = SL2 (Z), then (fk )(z) = (ad)k1 dk (fk ) then we get the result. Let = SL2 (Z), G = GL2 (Q)+ , =
iI a b 0 d
for some
az + b d
,
i 1i Z[\G/], we define
fk =
iI
i fk i ,
for f Mk (1) = Mk (C) .
The definition is independent of the choice of i . From the general theory, we have (fk )k (z) = fk ( )(z). If f Mk (1) (resp. Sk (1)), then fk Mk (1) (resp. Sk (1)). Facts: fk Rn = nk2 f , and fk Tn = nk1 dk f ( az+b ). d
ad=n,a1 b mod d
Proposition 2.4.8. If f =
m=0
am (f )q m , then am (fk Tn ) =
a1, a(m,n)
ak1 a mn (f ). a2
Proof. For fixed dn, d 1,
b mod d
dk f ( az+b ) = dk d = dk = d
b mod d m=0
am (f )e2im
az+b d
am (f )e2inaz/d
b mod d
e2imb/d
m=0 1k m=0 dm l=0
am (f )e2imaz/d adl (f )q al .
= d1k
58 So fk Tn = n
k1
CHAPTER 2. MODULAR FORMS
d
ad=n,a1
1k l=0
adl (f )q al ,
summing the coefficients of q , this gives: am (fk Tn ) = nk1 =
a1 a(m,n) k1
m
(n/a)1k a mn (f ) a2
a
a1 a(m,n)
a mn (f ). a2
Corollary 2.4.9. (i) Mk (, Z) and Mk (, Q) are stable under Tn and Rn . (ii) a0 (fk Tn ) = ak1 a0 (f ) = k1 (n)a0 (f ).
an
(iii) a1 (fk Tn ) = an (f ), therefore f is determined by T  a1 (fk T ).
2.5
Petersson scalar product.
dxdy = y2
1 2
Lemma 2.5.1.
+ 1x2
SL2 (Z)\H
1 2
dxdy = < . y2 3
Corollary 2.5.2. (i) If [SL2 (Z) : ] < +, then dxdy = C(), 2 y 3
\H
¯ ¯ where C() = [PSL2 (Z) : ], is the image of in PSL2 (Z). + (ii) If GL2 (Q) such that 1 SL2 (Z), then C(1 ) = C(). Proof. (i) Since dxdy is invariant under the action of , the integral is well y2 defined. Put {i } be a family of representatives of \ SL2 (Z), then \H = i (D) up to sets of measure 0 (maybe have overlap in SL2 (Z)i SL2 (Z)). (ii) Since \H = 1 \H , the two integrals are the same by the invariance of dxdy . y2
2.5. PETERSSON SCALAR PRODUCT.
59
Let f, g Sk (C), choose SL2 (Z) of finite index such that f, g Sk (, C). Proposition 2.5.3. f, g := 1 C() f (z)g(z)y k
\H
dxdy y2
converges and is independent of the choice of . Proof. For , we have f (z) = (cz + d)k f (z), Im (z) = g(z) = (cz + d)k g(z), Im z . cz + d2 i D with I = C(). So
iI
so f (z)g(z)y k is invariant under . Now \H =
if also satisfy that f, g Sk ( , C), then f, g Sk ( , C), and 1 C() f (z)g(z)y k
\H
dxdy dxdy 1 = f (z)g(z)y k 2 2 C( ) ( )\H y y dxdy 1 = f (z)g(z)y k 2 . C( ) \H y
Because fk i and gk i are exponentially decreasing as y on D, f, g converges. Remark. In fact, we can choose one modular form and one cusp form, and the integral will still converge. Proposition 2.5.4. For f Sk (1), we have Gk , f = 0. Proof. By definition, Gk (z) = and
m,n
1 (k) 2 (2i)k
m,n
1 Mk (1), (mz + n)k 1 (amz + an)k 1 , (cz + d)k
1 = (mz + n)k =
a=1 (m,n)=1
(k) (k) (2i)k
\ SL2 (Z)
60
CHAPTER 2. MODULAR FORMS
where denotes the subgroup of SL2 (Z) consisting of all upper triangular 1 , f . We have matrices. So we just compute (cz+d)k
\ SL2 (Z) 1 ,f (cz+d)k
= = = =
\ SL2 (Z)
SL2 (Z)\H SL2 (Z)\H
1 k (cz+d) \ SL2 (Z) \ SL2 (Z)
f (z)y k dxdy y2
f (z) Im (z)k dxdy y2
f (z)y k dxdy y2 \H 1 f (x + iy)y k2 dxdy = 0, 0 0
1 2inx e dx 0
where the last equality is because a0 (f ) = 0 and 1. Lemma 2.5.5. (i) For GL2 (Q)+ , we have
= 0 for n
fk , gk = (det )k2 f, g . (ii) Let = (det )1 , then fk , g = f, gk . Proof. (i) Choose such that f, g Sk () and 1 SL2 (Z), then C(1 ) fk , gk = (det )2(k1)
1 \H
f (z)g(z) f (z)g(z)y k dxdy y2
yk dxdy 2k y 2 cz + d
= (det )k2
\H
= (det )k2 C() f, g . (ii) Replace g by gk 1 , then we get fk , g = (det )k2 f, gk 1 1 = (det )k2 f, gk det = f, gk .
2.6
Primitive forms
Theorem 2.6.1. (i) If n 1, then Rn and Tn are hermitian. (ii) The eigenvalues of Tn are integers in a totally real field. (iii) Sk (1) has a basis of common eigenvectors for all Tn , n 1.
2.6. PRIMITIVE FORMS
61
Proof. (i) It is trivial for Rn . Since TZ is generated by Rp and Tp for p prime, it suffices to consider Tp . Let M2 (Z), det = p, then there exist 1 , 2 SL2 (Z) such that = 1 p 0 2 , then 0 1 fk , g = = = =
0 fk (1 p 1 2 ), g 0 fk p 0 , gk 2 0 1 fk p 0 , g 0 1 f, gk p 0 , 0 1
thus fk Tp , g = (p + 1) fk p 0 , g = f, gk Tp . 0 1 (ii) Sk (SL2 (Z), Z) is a lattice in Sk (1) stable under Tn , so det(XI  Tn ) Z[X], so the roots are algebraic integers, and real since Tn is hermitian. (iii) Tn s are hermitian, hence they are semisimple. Since the Tn commute to each other, by linear algebra, there exists a common basis of eigenvectors for all Tn .
+
Theorem 2.6.2. Let f =
n=0
an (f )q n Mk (1)  {0}. If for all n, fk Tn =
n f , then (i) a1 (f ) = 0; (ii) if f is normalized, i.e. a1 (f ) = 1, then an (f ) = n , for all n, and (a) amn (f ) = am (f )an (f ) when (m, n) = 1. (b) ap (f )apr (f ) = apr+1 (f ) + pk1 apr1 (f ) for p prime and r 1. Proof. (i) Since an (f ) = a1 (fk Tn ) = a1 (n f ) = n a1 (f ), if a1 (f ) = 0, then f = 0. (ii) The first assertion is obvious, and the other two follow by the same formulae for the Rp , Tp . Definition 2.6.3. f Sk (1) is called primitive if a1 (f ) = 1 and f is an eigenform for all Hecke operators. Theorem 2.6.4. (i) If f, g are primitive with the same set of eigenvalues, then f = g. (called "Multiplicity 1 theorem"). (ii) The primitive forms are a basis of Sk (1). Proof. (i) Apply (i) of the previous theorem to f  g, since a1 (f  g) = 0, so f = g.
62
CHAPTER 2. MODULAR FORMS
(ii) By (iii) of Theorem 2.6.1, there exists a basis of primitive forms. For any two distinct such forms f and f , then there exist n and = such that f k Tn = f, f k Tn = f,
then f, f = f k Tn , f = f, f k Tn = f, f , so f, f = 0. Therefore one has to take all the primitive forms to get a basis of Sk (1). Remark. Since (Gk )k Tn = k1 (n)Gk , we get a basis of Mk (1) of eigenforms. Example 2.6.5. Write
=q
n=1
(1  q n )24 =
n=1
(n)q n ,
where (n) is Ramanujan's function. Then (mn) = (m) (n), (p) (pr ) = (pr+1 ) + p11 (pr1 ), if (m, n) = 1, if p is a prime, n 1.
Proof. Since S12 (1) = C · , and is stable by the Tn , is an eigenform of Tn with eigenvalue (n). Remark. In 1973, Deligne proved Ramanujan's conjecture that  (p) 2p11/2 ( Re (s) = 11/2, if 1  (p)ps + p112s = 0)
as a consequence of the proof of Riemann Hypothesis (Weil Conjecture) for zeta functions of varieties over finite fields.
Chapter 3 padic Lfunctions of modular forms
3.1
3.1.1
Lfunctions of modular forms.
Estimates for the fourier coefficients
Proposition 3.1.1. Let SL2 (Z) be a subgroup of finite index, let f = an (f )q n Mk (, C). Then
1 n M N
(i) O(nk1 ), an (f ) = O(n log n), O( n), (ii) an (f ) = O(nk/2 ), if f Sk (). Proof. We have that an (f ) = e2ny y  2 Define (z) = y 2
k k
if k 3; if k = 2; if k = 1.
1 M
M 0
y 2 f (x + iy)e2inx dx,
k
y.
sup
\ SL2 (Z)
fk (z).
It is finite since [SL2 (Z) : ] < +, and (z) = (z) for SL2 (Z). 63
64
CHAPTER 3. P ADIC LFUNCTIONS OF MODULAR FORMS
Let D be the fundamental domain of SL2 (Z). For any \ SL2 (Z), there exists C such that, for all z D, fk (z)  a0 (fk ) C e M . Let C = sup C , (z) =
2y
sup
(c,d)=(0,0)
k
y , cz+d2
then (z) C(z)k/2 + B for some
B.
1 an (f ) e2ny y  2 M k 1 e2ny y  2 M
M (x + iy)dx 0 M (C(x + iy)k/2 0
+ B)dx.
1 If C = 0, take y = M n , then we get (ii). We now need to evaluate M 0 1 1 Let y 1 (in application, y = M n ), then (x + iy) y . Let j N. If (x + iy) 41j y, there exists (c, d) such that c2 y 2 + (cx + y)2 4j y 2 , hence there exist c, d Z, such that
(x + iy) 2 .
k
1 c 2j , Now
cx + d 2j y.
Meas({x [0, M ] : d, s.t.cx + d 2j y}) 2j+1 yM, so Meas({x [0, M ] : (x + iy)
M 0 1 }) 4j y
4j 2yM , and
(x + iy)k/2 dx
[ log4 y]
j=1
Meas({x [0, M ] :
[ log4 y]
1 4j y
(x + iy)
1 1 })( 4j1 y )k/2 4j1 y
+4k/2 Meas({x [0, M ] : (x + iy) 4}) M 4k/2 +
j=1 [ log4 y] j=1 [ log4 y] 1 4j 2yM ( 4j1 y )k/2
= M 4k/2 1 + 2
y 1k/2 4j(1k/2) . 4j(1k/2) converges, we get an (f ) =
[ log4 y] j=1
When k 3, let y = 1/M n. As
j=1
O(nk1 ). When k = 2, it is obvious. For k = 1, 2  y 1/2 < 2, then we get the result.
y 1k/2 4j(1k/2) <
3.1. LFUNCTIONS OF MODULAR FORMS. Remark. (i) L(f, s) =
n=0
65 0.
an (f )ns converges for Re (s)
(ii) If is a congruence subgroup, f Sk (), Deligne showed that an (f ) = O(n(k1)/2+ ), >0
in the same theorem mentioned above. Question: What about the noncongruence subgroups?
3.1.2
Dirichlet series and Mellin transform
an . ns
Definition 3.1.2. Let {an }n1 be a sequence in C, the Dirichlet series of (an ) is D(s) =
n=1
Lemma 3.1.3. If D(s0 ) converges, then D(s) converges uniformly on compact subsets of Re (s) > Re (s0 ). Proof. One can assume s0 = 0, then use Abel's summation. Corollary 3.1.4. There exists a maximal half plane of convergence (resp. absolute convergence).
Remark. (i) if f (z) =
n=0
an z n , then the maximal open disc of convergence
of f is the maximal open disc of absolute converge, and also is the maximal open disc of center 0 on which f can be extended analytically. (ii) Let an = (1)n1 , then D(s) = (1  21s )(s), which converges for Re (s) > 0, absolutely converges Re (s) > 1 and can be extended analytically to C. (iii) In general you can't extend D(s) outside its half plane of absolute convergence, but for D(s) coming from number theory, it seems that you can always extend meromorphically to C (Langlands program). We review some basic facts about Mellin transform: Proposition 3.1.5. (i) Let : R C be in C r , and suppose there exist + A > B satisfying, for 0 i n, (i) (t) = O(tAi ) O(tBi ) near 0 near .
66 Let
CHAPTER 3. P ADIC LFUNCTIONS OF MODULAR FORMS
Mel(, s) :=
0
(t)ts
dt . t
Then it is holomorphic on A < Re (s) < B, and O(sr ) on A < a Re (s) b < B. C+i 1 (ii) If r 2, (x) = 2i Ci Mel(, s)xs ds, for any C with a < C < B. Proof. (i) The first assertion is clear. For the second, use Mel(, s) = (1)r 1 Mel((r) , s + r). s(s + 1) · · · (s + r  1)
^ ^ (ii) Mel(, C + it) = C (t), where C (x) = (ex )eCx , and C is the Fourier transform of C . Then use Fourier inversion formula.
3.1.3
For f =
Modular forms and Lfunctions
n=0
an (f )q n M2k (1), define an (f ) , ns (s) L(f, s). (2)s
L(f, s) =
n=1
(f, s) =
Example 3.1.6. Take f = G2k , we get L(G2k , s) = =
a=1 2k1 (n) = ns n=1 n=1
d2k1 (ad)s
ad=n
a
s
d
d=1
2k1s
= (s)(s  2k + 1).
Theorem 3.1.7.
(i) L(f, s) absolutely converges for Re (s) > 2k;
(ii) (a) (f, s) has a meromorphic continuation to C; (b) (f, s) is holomorphic except for simple poles at s = 0 of residue a0 (f ) and 2k of residue (1)k a0 (f ); (c) (f, 2k  s) = (1)k (f, s); (d) (f, s) goes to zero at in each vertical strip.
3.1. LFUNCTIONS OF MODULAR FORMS.
67
Proof. (i) The result follows from an (f ) = O(n2k1 ). (ii) Let (t) = f (it)  a0 (f ), then is C on R , and (t) = O(e2t ) + at . f M2k (1) implies (t1 ) = (1)k t2k (t) + (1)k a0 (f )t2k  a0 (f ). For Re (s) > 0, we have
+ 2nt s dt e t t 0 (s) . (2n)s
=
Then for Re (s) > k,
(f, s) = = = =
n=1 + 0 + 1 + 1
(s) an (f ) (2n)s
(t)ts dt t + (t)ts dt + 1 (t1 )ts dt t t (t)(ts + (1)k t2ks ) dt  a0 (f ) t
(1)k 2ks
+
1 s
,
()
since the first term is holomorphic for all s C, this gives (a) and (b). Replacing s by 2k  s in (), we get (c). (d) follows from integration by part. Theorem 3.1.8 (Hecke's converse theorem). Let (cn )nN be a sequence in
C such that L(s) =
n=1
cn ns
converges for Re (s) > A, and (s) =
(s) L(s) (2)s
satisfy (ii)(a)  (d) of previous theorem, then f (z) :=
n=0
cn q n M2k (1).
Proof. Since f (z) converges if q < 1, it is holomorphic on H. Obviously f (z + 1) = f (z), we just have to verify 1 g(z) = f ( )  z 2k f (z) = 0 z It suffices to prove that g(it) = 0 for t > 0. Let
on H.
(t) = f (it)  c0 =
n=1
cn e2nt ,
one can check that (s) = Mel(, s). Take c > A, then (t )  (1) (t1 ) t2k c+i c+i 1 = 2i ci (s)ts ds  (1)k ci (s)ts2k ds c+i c+i 1 = 2i ci (s)ts ds  ci (2k  s)ts2k ds c+i 2kc+i 1 = 2i ci (s)ts ds  2kci (s)ts ds .
k
68
CHAPTER 3. P ADIC LFUNCTIONS OF MODULAR FORMS + R 2k  c
? r r
R c 6
0
2k
R  Consider the integral of the function (s)ts around the closed path . Since (s) 0 on vertical strips, by Cauchy formula,
R+ R
lim
(s)ts ds =
c+i ci
(s)ts ds 
2kc+i 2kci
(s)ts ds
= 2i ress=0 ((s)ts ) + ress=2k ((s)ts ) = 2i(c0 + (1)k c0 t2k ). So (t)  (1)k (t1 )  (c0 + (1)k c0 t2k ) = 0, 2k t
(1)k (g(it)), t2k
by an easy computation, the left hand is just g(it) = 0, which completes the proof.
then we get
3.1.4
Euler products
Theorem 3.1.9. If f =
n=0
an (f )q n M2k (1) is primitive, then 1
p
L(f, s) =
1  ap
(f )ps
+ p2k12s
.
Proof. By Theorem 2.6.2, anm (f ) = an (f )am (f ) whenever (n, m) = 1, so
L(f, s) =
p r=0
apr (f )prs .
Since, apr+1  ap apr + p2k1 apr1 = 0,
3.2. HIGHER LEVEL MODULAR FORMS multiplying by p(r+1)s , and summing over r from 1 to +, we get
69
apr p
r=2
rs
 ap p
s r=1
apr p
rs
+p
2k12s r=0
apr prs = 0.
Using the fact that a1 = 1, the result follows.
3.2
3.2.1
Higher level modular forms
Summary of the results
a b c d
For N 2, define 0 (N ) = SL2 (Z) : c 0 (mod N ) .
and write Sk (0 (N )) = Sk (N ). Exercise. If DM N , f Sk (M ), let fD (z) = f (Dz), then fD Sk (N ). Such a form is said to be old if M = N . Definition 3.2.1. Snew (N ) = {f Sk (N ) : f, g = 0, g "old"}. k On Sk (N ), we have the Hecke operators Tn , (n, N ) = 1, fk Tn = nk1
ad=n,a>1 b mod d
dk f
az + b d
,
and for p  n, the operator 1 fk Up = p
p1
f
i=0
z+i p
.
We also have a involution wN given by fk wN = N  2 z k f
k

1 Nz
.
Definition 3.2.2. f Sk (N ) is called primitive if f Snew (N ), a1 (f ) = 1 k and fk Tn = an (f )f , whenever (n, N ) = 1.
70
CHAPTER 3. P ADIC LFUNCTIONS OF MODULAR FORMS
Theorem 3.2.3. (i) The primitive forms are a basis of Snew (N ). k (ii) If f is primitive, then Q({an (f )}, n N) is a totally real number field, an (f ) are integers, and f is primitive for all Aut(C). (iii) If f is primitive, then (a) anm (f ) = an (f )am (f ) if (n, m) = 1, (nm, N ) = 1; (b) For p N , apr+1  ap (f )apr (f ) + pk1 apr1 (f ) = 0. (c) fk Up = ap (f )f , and this implies apr (f ) = (ap (f ))r for pN ; (d) There exists f = ±1, such that fk wN = f f .
Theorem 3.2.4. Suppose f =
n=1
an q n Sk (N ) is primitive. Define N 2
s
L(f, s) =
n=1
an , ns
(f, s) = (s)
L(f, s).
Then (i) L(f, s) =
pN
1 1ap ps
pN
1 ; 1ap ps +pk12s
(ii) (s) has an analytic continuation to C. And (f, s) = ik f (f, k  s); (iii) More generally, if (D, N ) = 1, : (Z/DZ) C is a character of conductor D. Then (a) f = an (n)q n Sk (N D2 , 2 );
n=1
(b) L(f , s) = (c) (f , s) to C and
1 1 ; 1(p)ap ps 1(p)ap ps +2 (p)pk12s pN pN s = (s) D2N L(f , s) has a analytic continuation
(N )
(f , s) (f 1 , s) = ik f G(x) G(1 )
2ix D
where G() is the Gauss sum G() =
x(Z/DZ)
(x)e
.
Theorem 3.2.5 (Weil's Converse Theorem). Conversely, if (am )m1 satisfy (b) and (c) of condition (iii) of the above theorem for all of conductor D,
(D, N ) = 1, then
m=1
am q m Sk (N ) and is primitive.
3.3. ALGEBRAICITY OF SPECIAL VALUES OF LFUNCTIONS
71
3.2.2
TaniyamaWeil Conjecture
Let be a finitely generated Zalgebra. Define its HasseWeil zeta function (s) by 1 . (s) = (1  /s ) prime in Conjecture 3.2.6 (HasseWeil). has a meromorphic continuation to C. Let E : y 2 = x3 + ax2 + bx + c, a, b, c Q be an elliptic curve, E = Z[x, y]/(y 2  x3  ax2  bx  c) be its coordinate ring, which is a finitely generated algebra over Z. Theorem 3.2.7 (Wiles, BreuilConradDiamondTaylor). There exists a unique NE and fE S2 (NE ) which is primitive, such that E (s  1) L(fE , s)
while means up to multiplication by a finite numbers of Euler factors. Remark. This proves HasseWeil conjecture in this case thanks to theorem 3.2.4. Theorem 3.2.8 (MordellWeil). E(Q) {} Zr(E) finite group.
Conjecture 3.2.9 (Birch,SwinnertonDyer). ords=1 L(fE , s) = r(E).
3.3
3.3.1
Algebraicity of special values of Lfunctions
Modular symbols.
Let N 1, f Sk (N ), P A[x](k2) (polynomials of degree k  2) with i A C a subring. For r Q, the integral r f (z)P (z)dz converges because f is exponentially small around i and r. These integrals are called modular symbols. For 0 j k  2, define
i
rj (f ) =
0
f (z)z j dz =
(j + 1) L(f, j + 1). (2i)j+1
Let Lf be the Zmodule generated by rj (fk ), 1 j k  2 and 0 (N )\ SL2 (Z). Then Lf is finitely generated.
72
CHAPTER 3. P ADIC LFUNCTIONS OF MODULAR FORMS
i r
Theorem 3.3.1. If P A[x](k2) , r Q, then Proof. For SL2 (Z),
(i) (0)
f (z)P (z)dz A·Lf C.
f (z)P (z)dz = =
i 0 i 0
f (z)P (z)d(z) fk (z)P2k (z)dz,
where P2k (z) = (cz + d)k2 P ( az+b ) A[x](k2) . Take r = a/b, (a, b) = cz+d al1 al 1, then there exists l = SL2 (Z) satisfying (a0 , b0 ) = (1, 0), bl1 bl (an , bn ) = (a, b).
i r n
f (z)P (z)dz =
l=1 n
al1 bl1 al bl
f (z)P (z)dz f (z)P (z)dz
=
l=1 n
l (i) l (0) i 0
=
l=1
fk l (z)P 2k l (z)dz A · Lf .
Exercise. For N = 1, let L+ (resp. L ) be the Zmodule generated by rj (f ) f f for all odd (resp. even) j. For P A[X](k2) , r Q, = ±, then
i i
f (z)P (z)dz 
r r
f (z)P (z)dz A · L . f ¯ an q n , : Z Q is constant mod (n) an has an analytic continuation ns
Corollary 3.3.2. (i) Suppose f
n=1
M Z for some M . Then L(f, , s) =
n=1
to C and (f, , j) =
(j) ¯ L(f, , j) Q · Lf , (2i)j
if 1 j k  1. ¯ (ii) If N = 1 and (x) = (1)j (x), then (f, , j) Q · L , if f 1 j k  1.
3.3. ALGEBRAICITY OF SPECIAL VALUES OF LFUNCTIONS
nu
73
Proof. we may assume (n) = e2i M for some 0 u M  1 because such functions form a basis, then
(s) L(f, , s) (2)s
= =
+ 0 + 0
2i nu 2ny s dy M e y y n=1 an e u s dy f ( M + iy)y y ,
this proves the first assertion of (i) as f is exponentially small around i u and M .
u (f, , j) = 0 f ( M + iy)(iy)j d(iy) iy i u f (z)(z  M )j1 dz = u M Q · Lf . +
For (ii), we may assume (n) = e2i M + (1)j e2i M , and similarly, (f, , j) = =
i
u M u M
nu
nu
f (z)(z  f (z)(z 
i
u j1 ) dz M u j1 ) dz M
+ (1)j 
i u M
i u M
f (z)(z +
f (z)(z 
u j1 ) dz M u j1 ) dz, M
then one uses the exercise.
3.3.2
The results
Theorem 3.3.3. If f is primitive, then there exist + and  C, if f f ¯ : Z Q (mod M Z), 1 j k  1, (x) = (1)j (x), ¯ then (f, , j) Q · . f Proof. We prove the case N = 1, = 1. We shall prove that ¯ rk2 (f )rl (f ) Q f, f , for l odd. This implies + f f, f f, f k2 rk2 (f ) L(f, k  1) (3.1)
where stands for equality up to multiplication by an algebraic number. The method to show (3.1) is the Rankin's method in the following section.
74
CHAPTER 3. P ADIC LFUNCTIONS OF MODULAR FORMS
3.3.3
Rankin's method
Assume k = l + j for k, l, j N. Suppose 1 , 2 : (Z/N Z)× C× are multiplicative characters. Let
+ +
f=
n=1
an q n Sk (N, 1 ), 1
g=
n=0
bn q n Ml (N, 2 ).
So f (z) = 1 (d)(cz + d)k f (z), 1 Let 1 (j) Gj,1 2 ,s (z) = · · 2 (2i)j = We have Proposition 3.3.4.
+
g(z) = 2 (d)(cz + d)l g(z).
N m (N,n)=1
1 2 (n)y s+1k (mz + n)j  mz + n 2(s+1k) 1 2 (d) · Im (z)s+1k . (cz + d)j
(j) L(1 2 , j + 2(s + 1  k)) · (2i)j
= a b \0 (N ) c d
D(f, g, s) = L(1 2 , j + 2(s + 1  k))
n=1 s j
an b n ¯ ns
=
(4) (2i) · f, gGj , 1 2 , s · [SL2 (Z) : 0 (N )]. (s) (j)
+ 0 0 1
Proof. Using the Fourier expansion, then
+
n=1
an b n ¯ (s) = ns (4)s (s) (4)s (s) = (4)s = = (s) (4)s
f (z)g(z) dx · y s f (z)g(z)y s+1
\H
dy y
dxdy y2 dxdy y2
(f (z)g(z) Im (z)s+1 )
0 (N )\H \ (N ) 0
f (z) g(z)
0 (N )\H \0 (N )
dxdy 1 2 (d) Im (z)s+1k y k , j (cz + d) y2
this implies the Proposition.
3.3. ALGEBRAICITY OF SPECIAL VALUES OF LFUNCTIONS
75
Theorem 3.3.5. (i) D(f, g, s) admits a meromorphic continuation to C, which is holomorphic outside a simple pole at s = k if l = k and 1 2 = 1 ¯ (ii) if f is primitive, g Ml (N, 2 , Q), then ¯ D(f, g, k  1) Q · j+k1 f, f . Proof. As D(f, g, s) = f, gGs , (i) we have to prove the same statement for Gs , which can be done by computing its Fourier extension. The pole comes from the constant Fourier coefficients. (ii) For the case N = 1,1 = 2 = 1 and j 3, then Gj,1 2 ,k1 = Gj , we are reduced to prove ¯ f, gGj Q f, f . Let fi , i I be a basis of Sk (1) of primitive forms, with f1 = f . As ¯ ¯ gGj Mk (1, Q), we can write gGj = 0 Gk + i i fi , with i Q. Since Gk , f = 0, Then f, gGj = 1 f, f . Remark. The general case can be treated in the same way, once we prove that ¯ Gj,1 2 ,k1 Mj (N, 1 2 , Q) (if j = 2 or 1 2 = 1). Proposition 3.3.6. If
+
f, fj = 0, if j = 1,
n=1 +
an ¯ = ns bn = ns
1 nZ[ N ]
×
an ¯ ns bn ns
1
pN
(1  p
ps )(1 1
 p ps )
, p p = 1 (p)pk1 ,
n=1
1 nZ[ N ]
×
pN
(1  p
ps )(1
 p ps )
, p p = 2 (p)pl1 ,
then D(f, g, s) = an b n ¯ ns 1
pN
nZ[
1 × N
]
(1  p p
ps )(1
 p p
ps )(1
 p p ps )(1  p p ps )
.
76
CHAPTER 3. P ADIC LFUNCTIONS OF MODULAR FORMS
Proof. Exercice, noting that apr = ¯
r+1 r+1 p  p , p  p
bpr =
r+1 r+1 p  p . p  p
We give one application here: Corollary 3.3.7. The claim (3.1) holds, i.e. ¯ rk2 (f )rl (f ) Q f, f , for l odd. Proof. Let f Sk (1) be primitive, k given. For l even, let g = Gl , then
+
n=1
bn = ns
p
1 , (1  ps )(1  pls1 )
hence D(f, Gl , s) = L(f, s)L(f, s  l + 1). Therefore ¯ L(f, k  1)L(f, k  l) Q · j+k1 f, f which implies ¯ rk2 (f )rkl1 (f ) Q f, f . Remark. In the general case, L(Gj , 1 2 , k  1, s) (s)L(1 2 , s  l + 1). If f1 , f2 , · · · , fn are primitive forms Sk (Ni ) for Ni  N . Write 1 L(fi , s) = , (i) (i) (1  p,1 ps )(1  p,2 ps ) pN then L(f1 · · · fn , s) =
pN j1 ,j2 ,··· ,jn {1,2}
1 (1  p,j1 · · · p,jn ps )
(1) (n)
.
One has the following conjecture: Conjecture 3.3.8 (Part of Langlands Program). L(f1 · · · fn , s) has a ¯ meromorphic continuation to C, and is holomorphic if fi = fj , for all i = j. Remark. Rankin's method implies the above conjecture is OK for n = 2. The case for n = 3 is due to Paul Garrett. The case for n 4 is still open.
3.4. P ADIC LFUNCTIONS OF MODULAR FORMS
77
3.4
padic Lfunctions of modular forms
In the following, we assume f Sk (N ) is primitive. Definition 3.4.1. + (x) = 1 ((x) + (x)),  (x) = 1 ((x)  (x)). 2 2 Then (f, + , j) (f,  , j) ¯ ~ (f, , j) = + Q (1)j (1)j+1 f f ¯ if : Z Q and 1 j k  1. ¯ ¯ Fix an embedding Q Qp . The function L(f, s) has an Euler product L(f, s) = 1 ¯ , E (s) Q[ E (s) prime
s
], deg E (s) 2.
Write Ep (s) = (1  ps )(1  ps ) and assume = 0. Then = 0 if and only if p  N . Set f (z) = f (z)  f (pz). Lemma 3.4.2. f k Up = f in all cases. Proof. It is clear if p  N as in the case = 0. If p N , then + = ap , and f k Tp = ( + )f , thus 1 f k Up  f = p
p1
= pk1 .
f
i=0
z+i p
 f (z + i)  f (z) + f (pz)
=  ( + )f (z) + fk Tp = 0.
If we write f = all n. Define bn for n
+ n n=1 bn q , the 1 Z p as
above lemma implies that bnp = bn for
bn = r bpr n , r
0.
78
CHAPTER 3. P ADIC LFUNCTIONS OF MODULAR FORMS
¯ Take LCc (Qp , Q) a locally constant function with compact support and let an (n) s . L(f, , s) = n 1 nZ[ p ] ¯ If has support in pr Zp , then (x) = 0 (pr x) for 0 : Z Q constant mod m p Z for some m.Then L(f, , s) = r prs L(f, 0 , s) which implies ¯ ¯ ¯ ~ (f, , j) Q Qp , for all LCc (Qp , Q). ¯ Definition 3.4.3. Assume LCc (Qp , Q) and is constant modulo pn Z. The discrete Fourier transform of is ^ (x) = pm
y mod pm
(y)e2ixy ,
for m n  vp (x), where xy Qp Qp /Zp Q/Z. This definition does not depend on the choice of m n  vp (x). ^ Exercise. (i) is constant mod pm Zp if and only if has support in pm Zp . ^ ^ (ii) (x) = (x). ^ (iii) For a Qp , let a (x) = (ax), then a (x) = pvp (a) x .
a
¯ ¯ Theorem 3.4.4. (i) There exists a unique µf, : LP [0,k2] (Zp , Qp ) Qp , ¯ such that for all LC(Zp , Q), ^ ~ (x)xj1 µf, = (f , , j), 1 j k  1.
Zp 1 Moreover, (µf, ) = µf, , or equivalently
pZp
x p
µf, =
1
µf, .
Zp
(ii) if vp () < k  1, then µf, extends uniquely as an element of Dvp () .
3.4. P ADIC LFUNCTIONS OF MODULAR FORMS
79
¯ ¯ Proof. (i) The existence of µf, : LP [0,k2] (Zp , Qp ) Qp is just the linearity ^ of . The uniqueness is trivial. The second claim follows from
pZp
x p
x p
j1
µf, =
p
1 ~ ^ (f , p1 (px), j) j1
1 1~ ^ = (f , , j) =
(x)xj1 µf, .
Zp
(ii) One needs to show there exists a constant C, such that vp (
a+pn Zp
(x  a)j µf, ) C + (j  vp ())n,
for all a Zp , n N, j k  2. Note that 1a+pn Zp (x) = for a (x) = Then
j
pn e2iax , 0,
if x pn Zp , = pn a (pn x) if not.
ax
e2i pn 0,
x Zp , otherwise.
(x  a)j µf, =
a+pn Zp l=0
(a)l
j n l=0
j n ~ p (f , a (pn x), l + 1) l j nl ~ p (f , a , l + 1). l
= Since
i
(1)l
~ pnl (f , a , l + 1) = pnl
0
f (z 
a l )z dz = pn
i
f (z)(pn z + a)l dz,
 pa n
we get
j
(1)l
l=0
j l
i
(x  a)j µf, = n pnj
a+pn Zp  pa n
f (z)z j dz n pnj Lf .
We just pick C = min(vp (~j (f k ))). r
80
CHAPTER 3. P ADIC LFUNCTIONS OF MODULAR FORMS
Remark. (i) If p  N , then = 0, and = 0 implies vp() = k2 < k  1, 2 hence µf, exists by the above Theorem. (ii) If p N , then vp (), vp () 0. Since vp () + vp () = k  1, at least one of µf, or µf, always exists. In the case vp () = k  1, then + = ap (f ) is a unit. This case is called the ordinary case. The conditions are not strong enough for the uniqueness of µf, , as we can add the (k  1)th derivative of any D0 . (iii) In the case = = 0, we do not understand what happens. Definition 3.4.5. Let : Z C be a continuous character. Set p p Lp, (f , s) =
Z p
x1 (x)µf, .
In particular, take (x) = x 2 x
k
s k 2
where x t = exp(t log x). Set x 2 1 x
Z p
k
Lp, (f, s) =
s k 2
µf, .
Proposition 3.4.6. For 1 j k  1, Lp, (f j ) = (1  Proof. Follows from (i) 1Z = 1Zp  p1 1p1 Zp , p ~ (ii) (f , 1Zp , j) = (1 
1 (iii) (µf, ) = µf, . ~ (f, j)), pj
pj1 ~ )(1  j )(f, j). p
Remark. (i) As (f, s) = (f, k  s) and = pk1 if p N , then (1  pj1 ) = 1  kj . p
Note Ep (f, s) = (1  ps )(1  ps ). Then the Euler factor of the padic Lfunction is actually the product of one part of the Euler factor for L(f, s) and one part of the Euler factor for L(f, k  s). This is a general phenomenon. k2 (ii) If p  N , = 0, the vp () = k2 . It can happen that = p 2 , which 2 means Lp, (f, k ) = 0. In this case 2
3.4. P ADIC LFUNCTIONS OF MODULAR FORMS Conjecture 3.4.7 (MazurTateTeitelbaum Conjecture). k k ~ Lp, (f, ) = LF ont. (f )(f, ). 2 2
81
Here the padic Lfunction is related to 2dimensional (, N )filtered modules D with N = 0 and Fil0 D = D, Fil1 D = D. For the pair (, ) as in Fontaine's course, where is the eigenvalue of and is the parameter associated to the filtration, is our and is our LF ont. . The conjecture is proved by KatoKuriharaTsuji, PerrinRiou, and Stevens, Orton, Emerton with other definitions of the Linvariant. (iii) Mazur, Tate and Teitelbaum have also formulated a padic analog of the BSD conjecture. For E/Q an elliptic curve, by TaniyamaWeil, it is associated to a primitive form f S2 (N ). Set Lp, (E, s) = Lp, (f, s) if it exists, which is the case if E has either good reduction (hence p N ) or multiplicative reduction (hence p  N, p2 N ) mod p. Conjecture 3.4.8 (padic BSD Conjecture). ords=1 Lp, (E, s) = Kato showed that ords=1 Lp, (E, s) rank E(Q), rank E(Q) + 1, if p N or = 1; if p  N and = 1. rank E(Q), rank E(Q) + 1, if p N or = 1; if p  N and = 1.
(iv) To prove Kato or KatoKuriharaTsuji, we need another construction of padic Lfunctions via Iwasawa theory and (, )modules; this construction is the subject of the next part of the course and is based on ideas of PerrinRiou.
82
CHAPTER 3. P ADIC LFUNCTIONS OF MODULAR FORMS
Part II Fontaine's rings and Iwasawa theory
83
Chapter 4 Preliminaries
4.1 Some of Fontaine's rings
This section is a review of notations and results from Fontaine's course. For details, see Fontaine's notes.
4.1.1
Rings of characteristic p
1 }. p
(1) Cp is the completion of Qp for the valuation vp with vp (p) = 1. a = {x Cp , vp (x)
~ (2) E + is the ring R in Fontaine's course. By definition ~ E + := {x = (xn )nN  xn Cp /a, xp = xn , n} n+1 ~ is a ring of characteristic p with an action of GQp . For x = (xn ) E + , for every xn , pick a lifting xn OCp , then ^
k+
lim (^n+k )p := x(n) OCp x
k
is a canonical lifting of xn such that ~ E + = {x = (x(n) )nN  x(n) OCp , (x(n+1) )p = x(n) , n} with the addition and multiplication by (x + y)(n) = lim (x(n+k) + y (n+k) )p ,
k+
k
(xy)(n) = x(n) y (n) .
85
86 ~ E + is a valuation ring with valuation
CHAPTER 4. PRELIMINARIES
vE (x) = vp (x(0) ) and maximal ideal ~ mE + = {x E + , vE (x) > 0}. ~ (3) Choose once for all ~ = (1, (1) , · · · ) E + , (1) = 1.
Then (n) is a primitive pn th root of 1 for all n. Set ~ =  1 E +. ¯
p We know that vE (¯ ) = p1 > 0. From now on, : GQp Z will be the cyclotomic character. The action p of GQp on is given by +
g() =
(g)
=
k=0
(g) k . ¯ k
(4) In the following, without further specification, K Qp will be a finite extension of Qp . Denote by k = kK its residue field. Set Kn = K((n) ), Set F K= the maximal unramified extension of Qp inside K, F K = the maximal unramified extension of Qp inside K . Set ¯ GK = Gal(Qp /K), and (5) For every K, let ~ ~+ ~ EK := {x = (xn ) E + , xn OK /a, n} = (E + )HK (by AxSenTate's Theorem), + ~ EK := {x = (xn ) E + , xn OKn /a, n n(K)}. ¯ HK = Ker = Gal(Qp /K ),
K =
nN
Kn .
K = GK /HK = Gal(K /K) Z . p
4.1. SOME OF FONTAINE'S RINGS Then
+ ~+ ~ EK EK E + , ¯
87
K.
We set
+ ~ ~+ ~ ~ EK := EK [¯ 1 ] EK := EK [¯ 1 ] E = E + [¯ 1 ] = Fr R
with valuation vE (¯ k x) = vE (x)  kvE (¯ ). The following Theorem is the topics in the last section of Chapter 2 of Fontaine's Notes. ¯ ~ Theorem 4.1.1. (i) E is a field complete for vE with residue field Fp , ring ~ + and GQp acts continuously with respect to vE . of integers E (ii) EF = kF ((¯ )) if F/Qp is unramified. In general, EK is a totally ramified extension of EF of degree [K : F ], + thus a local field of characteristic p, with ring of integers EK and residue field kF . (iii) E = EK is a separable closure of EQp , is stable under GQp
[K:Qp ]<+
and Gal(E/EK ) = HK . So HQp acts continuously on E for the discrete topology. ~ ~ (iv) E (resp. EK ) is the completion of the radical closure of E (resp. pn pn ~K ), i.e., ~ E E (resp. EK ). In particular, E is algebraically closed.
nN nN
4.1.2
(6) Set
Rings of characteristic 0
~ ~ A+ := W (E + ) = W (R), ~ ~ A := W (E) = W (Fr R).
~ Every element x A can be written as
+
x=
k=0
pk [xk ]
~ while xk E and [xk ] is its Teichm¨ller representative. u As we know from the construction of Witt rings, there are bijections ~ = ~ A+ (E + )N , ~= ~ A (E)N .
88
CHAPTER 4. PRELIMINARIES
~ ~ There are two topologies in A+ and A: (i) Strong topology or padic topology: topology by using the above bijec~ ~ tion and the discrete topology on E + or E. A basis of neighborhoods of 0 k ~ are the p A, k N. (ii) Weak topology: topology defined by vE . A basis of neighborhoods of ~ 0 are the pk A + [¯ n ]A+ , k, n N. ~ The commuting actions of GQp and on A are given by
+ + + +
g(
k=0
p [xk ]) =
k=0
k
p [g(xk )],
k
(
k=0
p [xk ]) =
k=0
k
pk [xp ]. k
~ ~ ~ ~1 (7) B := A[ p ] is the fraction field of A. B is complete for the valuation vp , ~ ~ its ring of integers is A and its residue field is E. For the GQp and actions, ~ A=1 = Zp , ~ ~ ~ AHK = W (EK ) := AK , (8) Set = []  1, t = log[] = log(1 + ). The element [] is the padic analogue of e2i . The GQp  and  actions are given by ( + 1) = ( + 1)p , ( + 1) = ( + 1)(g) . (9) Set ~ A+p := Zp [[]] A+ Q which is stable under and GQp . Set 1 ~ AQp := Zp [[]][ ] A while stands for completion under the strong topology, thus AQp = {
kZ
~ B =1 = Qp , ~ ~ 1 ~ B HK = AK [ ] := BK . p
ak k  ak Zp , lim vp (ak ) = +}.
k
1 Set BQp := AQp [ p ], then BQp is a field complete for the valuation vp , with ring of integers AQp and residue field EQp .
4.2. (, )MODULES AND GALOIS REPRESENTATIONS.
89
~ Moreover, if [K : Qp ] < +, B contains a unique extension BK of BQp ~ whose residue field is EK , and AK = BK A is the ring of integers. By uniqueness, BK is stable under and GK acting through K . The field E ur = BK
[K:Qp ]<+
is the maximal unramified extension of BQp = E. Set B = E ur be the closure of
[K:Qp ]<+
~ ~ BK in B for the strong topology. Then A = B A
is the ring of integers OE ur and the residue field of B is A/pA = E. By d AxSenTate, B HK = BK , AHK = AK . Remark. If K is a uniformising parameter of EK , let K AK be any ¯ lifting. Then AK = {
kZ k ak K ak OF , lim vp (ak ) = +}. k
~ Remark. In the above construction, the correspondence  is obtained by making bijective and then complete, where = (EK , E, AK , A, BK , B).
4.2
(, )modules and Galois representations.
Let K be a fixed finite extension over Qp , let = K . Definition 4.2.1. (i) A (, )module over AK is a finitely generated AK module with semilinear continuous (for the weak topology) and commuting actions of and . A (, )module over BK is a finite dimensional BK vector space with semilinear continuous (for the weak topology) and commuting actions of and . (ii) A (, )module D/AK is ´tale (or of slope 0) if (D) generates D as e an AK module. A (, )module D/BK is ´tale (or of slope 0) if it has an AK lattice e which is ´tale, equivalently, there exists a basis {e1 , · · · , ed } over BK , such e that the matrix of (e1 ), · · · , (ed ) in e1 , · · · , ed is inside GLd (AK ).
90
CHAPTER 4. PRELIMINARIES
The following theorem is similar to Theorem 1.5.9 in §1.5.4 of Fontaine's Notes. Theorem 4.2.2. The correspondence V  D(V ) := (A Zp V )HK is an equivalence of categories from the category of Zp representations (resp. Qp resp) of GK to the category of ´tale (, )modules over AK (resp. e BK ), and the Inverse functor is D  V (D) = (A AK D)=1 . Remark. (i) K is essentially procyclic, so a (, )module is given by two operators and commuting relations between them. For example, if D/AK is free of rank d, let U be the matrix of for = K , let P be the matrix of , then U (P ) = P (U ), U, P GLd (AK ). (ii) We want to recover from D(V ) the known invariants of V :  H i (GK , V ); we shall do so in the coming lectures. We will also recover the Iwasawa modules attached to V and thus give another construction of padic Lfunctions.  DdR (V ), Dcris (V ), Dst (V ).
Chapter 5 (, )modules and Galois cohomology
5.1 Galois Cohomology
Let M be a topological Zp module (e.g. a finite module with discrete topology or a finitely generated Zp module with padic topology, or a Fontaine's ring + BdR · · · ), with a continuous action of GK . Let H i (GK , M ) be the ith cohomology groups of M of continuous cohomology. Then: H 0 (GK , M ) = M GK = {x M : (g  1)x = 0 g GK }; {c : GK M continuous, g1 cg2  cg1 g2 + cg1 = 0, g1 , g2 GK } H 1 (GK , M ) = {c : g (g  1)x, for some x M } To a 1cocycle c, we associate a GK module Ec such that 0 M Ec N 0 where Ec Zp × M as a Zp module and GK acts on Ec by g(a, m) = (a, gm + cg ). One can check easily g1 (g2 (a, m)) = g1 (a, g2 m + cg2 ) = (a, g1 g2 m + g1 cg2 + cg1 ) = g1 g2 (a, m). ^ Ec is trivial if and only if there exists 1 Ec , such that g ^ = ^ for all g, i.e. 1 1 ^ = (1, x), g ^ ^ = (0, gxx+cg ) = 0, that is, cg = (1g)x is a coboundary. 1 1 1 91
92
CHAPTER 5. (, )MODULES AND GALOIS COHOMOLOGY
Theorem 5.1.1 (Tate's Local Duality Theorem). Suppose K is a finite extension of Qp . Let M be a Zp [GK ]module of finite length. Then: (i) H i (GK , M ) = 0 for i 3; H i (GK , M ) is finite if i 2.
2
(ii)
i=0
H i (GK , M )(1) = M [K:Qp ] ; Hom(H i (GK , M ), Qp /Zp ).
i
(iii) H 2i (GK , Hom(M, µp ))
We will give a proof using (, )module (Herr's thesis). Remark. (i) If M is a finitely generated Zp module with padic topology, then M lim M/pn M , and H i (GK , M ) lim H i (GK , M/pn M ).   Z! Not tautological, the proof uses finiteness of (i) to ensure MittagLeffler conditions. (ii) If V is a Qp representation of GK , let T V be a Zp lattice stable by GK . Then H i (GK , V ) Qp H i (GK , T ). Corollary 5.1.2. If V is a Qp representation of GK . Then:
2
(i)
i=0
(1)i dimQp H i (GK , V ) = [K : Qp ] dimQp V ;
(ii) H 2 (GK , V ) = H 0 (GK , V (1)) .
5.2
The complex C, (K, V )
Assume that K is procyclic (Qp Z ), is a topological generator of K . p This assumption is automatic if p 3, or if K Q(µ4 ) when p = 2. Let V be a Zp  or Qp representation of GK . Set D(V ) = (A Zp V )HK .
· Definition 5.2.1. The complex C, (K, V ) = C, (K, V ) is (1, 1) (1) pr (1) pr
0 D(V )    D(V ) D(V )    1   D(V ) 0.       2 It is easy to see C, (K, V ) is really a complex (as , commute to each other). We shall denote the complex by C · (V ) if no confusion is caused. We
5.2. THE COMPLEX C, (K, V ) have H 0 (C · (V )) = {x D(V ), (x) = x, (x) = x}, {(x, y) : (  1)x = (  1)y} H 1 (C · (V )) = , {((  1)z, (  1)z) : z D(V )} D(V ) , H 2 (C · (V )) = (  1,  1) H i (C · (V )) = 0, for i 3. Theorem 5.2.2. H i (C, (K, V )) H i (GK , V ) for all i in N.
93
Proof. We have the following exact sequence (which can be proved by reducing mod p): 0 Zp A  A 0,  here A = OE ur in Fontaine's course. d (1) i = 0: For x D(V )=1 , since D(V ) = (A Zp V )HK , we have D(V )=1 = (A=1 Zp V )HK = V HK , and (V HK )=1 = V GK . (2) i = 1: Let (x, y) satisfy the condition (  1)x = (  1)y. Choose b (A Zp V )HK , (  1)b = x. We define the map: g GK cx,y (g) = while the meaning of let
g1 y 1 1
g1 y  (g  1)b. 1
i+
is: as (g) = lim ()ni , y is fixed by HK , we
g1 y = lim (1 + + · · · + ni 1 )y. i+ 1
y This is a cocycle with values in V , because g (g  1)( 1  b) is a cocycle, and (  1)cxy (g) = (g  1)x  (  1)(g  1)b = 0, which implies that cxy (g) D(V )=1 = V . Injectivity: If cxy = 0 in H 1 (GK , V ), then there exists z V , cxy (g) = g1 (g  1)z for all g GK , that is, 1 y = (g  1)(b  z) for all g. Now bz D(V ), because it is fixed by g HK . Then we have: y = ( 1)(bz) and x = (  1)(b  z), hence (x, y) equal to 0 in H 1 (C · (V )). Surjectivity: If c H 1 (GK , V ), we have:
0 V  Ec  Zp 0,
94
CHAPTER 5. (, )MODULES AND GALOIS COHOMOLOGY
here Ec = Zp × V , e Ec 1 Zp and ge = e + cg for g cg representing c. We have: 0 D(V )  D(Ec )  AK 0, here D(Ec ) A Ec and e D(Ec ) 1 AK . Let ~ x = (  1)~, e y = (  1)~, e
they are both in D(V ) and satisfy (  1)x = (  1)y. Let b = e  e ~ g1 A Zp Ec . Then cx,y (g) = 1 y  (g  1)b = cg and (  1)(b) = x. (3) i general: from the exact sequence: 0 Zp A  A 0,  tensoring with V and taking the cohomology H i (HK , ), we get 0 V HK D(V )  D(V ) H 1 (HK , V ) 0,  because A V (A/pi ) as HK modules and H i (HK , E) = 0, if i 1, so H i (HK , A V ) = 0 for all i 1. Hence H i (HK , V ) = 0 for all i 1. By the HochschildSerre Spectral Sequence for 1 HK GK K 1, we have H i (K , H j (HK , V )) H i+j (GK , V ). When j or i 2, the cohomology vanishes. So we have: H q (GK , V ) = 0, if q 3 H 2 (GK , V ) H 1 (K , H 1 (HK , V )). Since H 1 (HK , V ) =
D(V ) , 1 1 1
we get D(V ) D(V ) D(V ) (  1) = . 1 1 (  1,  1)
H 2 (GK , V )
Remark. (1) The inflationrestriction exact sequence becomes the commutative diagram 0
/ H 1 ( , V HK ) K / H 1 (G , V ) K
,
/ H 1 (H , V K ) K
/0
0
/
D(V )=1 1
/ H 1 (C, (K, V ))
/ ( D(V ) )K
1
/0
´ 5.3. TATE'S EULERPOINCARE FORMULA.
95
where the map H 1 (C, (K, V )) ( D(V ) )K is given by sending (x, y) to the 1 image of x. 1 (2) Let be another generator of K , we have 1 (Zp [[K ]]) and a commutative diagram: C, : 0
/ D(V )
1 1
/ D(V )
1 1
D(V )
Id
/ D(V )
Id
/0
C, : 0
/ D(V )
/ D(V )
D(V )
/ D(V )
/0
It induces a commutative diagram
· H 1 (C, ) ,
NNN NNN NN lK (), NNN&
H 1 (GK , V )
/ H 1 (C · ) , ppp pp ppp wppp lK ( ),
where lK () = log () for log (K ) pr(K) on the choice of ".
pr(K) Zp . So lK ()c, "does not depend
5.3
5.3.1
Tate's EulerPoincar´ formula. e
The operator .
Lemma 5.3.1. (i) {1, , · · · , p1 } is a basis of EQp over (EQp ); (ii) {1, , · · · , p1 } is a basis of EK over (EK ), for all [K : Qp ] < +; (iii) {1, , · · · , p1 } is a basis of E over (E); (iv) {1, [], · · · , []p1 } is a basis of A over (A). Proof. (i) Since EQp = Fp ((¯ )) with =  1, we have (EQp ) = Fp ((¯ p )); ¯ (ii) Use the following diagram of fields, note that EQp /(EQp ) is purely inseparable and (EK )/(EQp ) is separable: EK EQp (iii) Because E = EK . (EK )
(EQp )
96
CHAPTER 5. (, )MODULES AND GALOIS COHOMOLOGY (iv) To show that
p1
(x0 , x1 , · · · , xp1 ) A 
i=0
p
[]i (xi ) A
is a bijection, it suffices to check it mod p and use (iii). Definition 5.3.2. The operator : A A is defined by
p1
(
i=0
[]i (xi )) = x0 .
Proposition 5.3.3. (i) = Id; (ii) commutes with GQp . Proof. (i) The first statement is obvious. (ii) Note that
p1 p1
g(
i=0
[] (xi )) =
i=0
i
[]i(g) (g(xi )).
If for 1 i p  1, write i(g) = ig + pjg with 1 ig p  1, then
p1 p1
(
i=0
[]
i(g)
(g(xi ))) = ((g(x0 )) +
i=1
[]ig ([]jg g(xi ))) = g(x0 ).
Corollary 5.3.4. (i) If V is a Zp representation of GK , there exists a unique operator : D(V ) D(V ) with ((a)x) = a(x), (a(x)) = (a)x
if a AK , x D(V ) and moreover commute with K . (ii) If D is an ´tale (, )module over AK or BK , there exists a unique e operator : D D with as in (i). Moreover, for any x D,
pn 1
x=
i=0
[]i n (xi )
where xi = n ([]i x).
´ 5.3. TATE'S EULERPOINCARE FORMULA.
97
Proof. (i) The uniqueness follows from AK (AK ) (D) = D. For the existence, use on A V D(V ). D(V ) is stable under because commutes with HK , commutes with K since commutes with GK . (ii) D = D(V (D)), thus we have existence and uniqueness of . The rest is by induction on n. Example 5.3.5. Let D = AQp A+p = Zp [[]] be the trivial (, )module, Q here [] = (1 + ). Then for x = F () A+p , (x) = F ((1 + )p  1). Write Q
p1
F () =
i=0
(1 + )i Fi ((1 + )p  1),
then (F ()) = F0 (). It is easy to see if F () belongs to Zp [[]], Fi () + + belongs to Zp [[]] for all i. Then (EQp ) EQp = Fp [[]]. Hence (A+p ) Q A+p . Consequently, is continuous for the weak topology. Q Moreover, we have: 1 ((F )) = F0 ((1 + )  1) = p zp =1
p p1
(z(1 + ))i Fi ((z(1 + ))p  1)
i=0
=
1 F (z(1 + )  1). p zp =1
Zp
+ Recall D0 (Zp , Qp ) BQp = Qp Zp A+p by µ Aµ () = Q that (µ) is defined by
[]x µ. Recall
(x)(µ) =
Zp
x ( )µ. p pZp
From the above formula, we get, using formulas for Amice transforms, A(µ) () = (Aµ )(). Proposition 5.3.6. If D is an ´tale module over AK , then is continuous e for the weak topology. Proof. As AK is a free AQp module of rank [K : Qp (µp )], we can assume K = Qp . Choose e1 , e2 , · · · , ed in D, such that D = (AQp /pni )ei , ni N {}.
98
CHAPTER 5. (, )MODULES AND GALOIS COHOMOLOGY
Since D is ´tale, we have D = (AQp /pni )(ei ). Then we have the following e diagram: D (AQp /pni )(ei ) xi (ei )
/
/D / (AQ /pni )ei p
(xi )ei
Now x (x) is continuous in AQp , hence is continuous in D.
5.3.2
D=1 and D/(  1)
Lemma 5.3.7. If D is an ´tale module over EQp , then: e =1 (i) D is compact; (ii) dimFp (D/(  1)) < +. Proof. (i) choose a basis {e1 , · · · , ed }, then {(e1 ), · · · , (ed )} is still a basis. + Set vE (x) = inf i vE (xi ) if x = i xi (ei ), xi EQp . We have
d
(x) =
i
(xi )ei and ei =
i=1
ai,j (ej ).
Let c = inf vE (ai,j ), then we have
i,j
vE ((x)) c + inf vE ((xi )).
i + + From (EQp ) EQp and (¯ p x) = k (x), we get vE ((x)) ¯
k
(5.1)
vE (x) p
. So
vE ((x)) c + inf [
i
vE (xi ) vE (x) ]c+[ ]. p p
If vE (x) < p(c1) , then vE ((x)) > vE (x). Now D=1 is closed since is p1 continuous, and is a subset of the compact set p(c  1) M := {x : vE (x) } p1
d
k Fp [[¯ ]] · (ei ). ¯
i=1
´ 5.3. TATE'S EULERPOINCARE FORMULA.
99
Hence D=1 is also compact. (ii)  1 is bijective on D/M from the proof of (i). We only need to prove that M/((  1)D M ) is finite, equivalently, that (  1)D contains {x : vE (x) c } for some c .
d
(xi ) can be written uniquely as (xi ) =
j=1
bi,j ej . Let c0 = inf vE (bi,j ),
i,j d d
then
d d d
x=
i=1 d
xi (ei ) =
i=1 d
xi
j=1
bi,j ej =
(
j=1 i=1
xi bi,j )ej .
Let yj =
i=1
xi bi,j , then x =
j=1
yj ej , and vE (yj ) c0 + vE (x).
d
From (x) =
j=1
(yj )(ej ), we get
vE ((x)) = inf vE ((yj )) = p inf vE (yj ) pvE (x) + pc0 . So, if vE (x)
pc0 p1
+ 1, then vE (n (x)) pn . It implies y =
+ +
+ i=1
i (x)
converges in D. Now (  1)y =
i=0
i (x) 
i=1
i (x) = x
pc0 p1
implies that (  1)D contains {xvE (x)
+ 1}.
Proposition 5.3.8. If D is an ´tale module over AK (resp. over BK ), e then: (i) D=1 is compact (resp. locally compact); (ii) D/(  1) is finitely generated over Zp (resp. over Qp ). Proof. We can reduce to K = Qp . BK follows from AK by Qp Zp . So we consider D over AQp . (i) Note that D=1 = lim(D/pn D)=1 . From the previous lemma we have  D/pn D is compact by easy induction on n. So D=1 is compact. (ii) The quotient (D/(  1))/p (D/p)/  1 is finite dimensional over Fp . We have to check that
100
CHAPTER 5. (, )MODULES AND GALOIS COHOMOLOGY if x = (  1)yn + pn Zp for all n, then x (  1)D.
If m n, ym  yn (D/pn )=1 , which is compact, we can extract a sequence converging mod pn . Thus we can diagonally extract a sequence converging mod pn for all n. Then yn converges to y in D and x = (  1)y.
5.3.3
The module D=0 .
Z . Let n 0 and n 1 + pn Zp if n 1. p = µp1 , and n = lim n /n+m . We define 
m
If p = 2, we let 0 = Qp Then 0 = × 1 where
Zp [[n ]] = lim Zp [n /n+m ] = D(n , Zp ).  If n 1, let n be a topological generator of n . So n = n p . The correspondence Zp [[n ]] o n  1 o is just the Amice transform. Then Zp [[0 ]] = Zp [ ] Zp [[1 ]], Zp {{n }} := (Zp [[n ]][(n  1)1 ]) Zp {{0 }} = Zp [ ] Zp {{1 }}. Modulo p, we get Fp {{n }} EQp as a ring.
Z
Zp [[T ]]
/ A+ Qp /
T
AQp (as a ring),
Remark. Zp [[0 ]] D0 (0 , Zp ) D0 (Z , Zp ) (A+p )=0 . So (A+p )=0 is p Q Q a free Zp [[0 ]]module of rank 1. This a special case of a general theorem which will come up later on. Lemma 5.3.9. (i) If M is a topological Zp module (M = lim M/Mi ) with  a continuous action of n (i.e. for all i, there exists k, such that n+k acts trivially on M/Mi ), then Zp [[n ]] acts continuously on M ; (ii) If n  1 has a continuous inverse, then Zp {{n }} also acts continuously on M .
´ 5.3. TATE'S EULERPOINCARE FORMULA.
101
Lemma 5.3.10. (i) If n 1, vE (n (¯ )  ) = pn vE (¯ ); ¯ (ii) For all x in EQp , we have vE (n (x)  x) vE (x) + (pn  1)vE (¯ ). Proof. Since () = 1 + pn u, u Z , we have p n (¯ )  =n (1 + )  (1 + ) = (1 + )((1 + )p ¯ ¯ ¯ ¯ ¯ =(1 + )((1 + )  1) . ¯ ¯ Then we get (i).
+ u pn
nu
 1)
In general, for x =
k=k0
ak k , then vE (x) = k0 vE (¯ ). Now ¯
+
n (x)  x n (¯ )k  k ¯ = ak , n (¯ )  k=k ¯ n (¯ )  ¯
0
and vE (
n (¯ )k  k ¯ ) (k  1)vE (¯ ). n (¯ )  ¯
Proposition 5.3.11. Let D be an ´tale (, )module of dimension d over e EQp . Assume n 1, (i, p) = 1. Then (i) induces i n (D) ()i n (D); (ii) n 1 admits a continuous inverse on i n (D). Moreover if {e1 , · · · , ed } is a basis of D, then: Fp {{n }}d  n (D) (1 , · · · , d )  1 i n (e1 ) + · · · + d i n (ed ) is a topological isomorphism. Proof. (i) is obvious. Now, remark that (ii) is true for n + 1 implies (ii) is true for n, since i n (D) = i n (p1 j (D)) = p1 i+p j n+1 (D), j=0 j=0
p and for n > 1, n+1 = n , so 1 n 1
n
=
1 (1 n+1 1
p1 + n + · · · + n ), and
p1 Fp {{n }} = Fp {{n+1 }} + · · · + n Fp {{n+1 }}.
102
CHAPTER 5. (, )MODULES AND GALOIS COHOMOLOGY
So we can assume n big enough. Recall vE (x) = inf vE (xi ) if x =
i i
xi ei . We can, in particular, assume
vE (n (ei )  ei ) 2vE (¯ ), it implies vE (n (x)  x) vE (x) + 2vE (¯ ) for all n x D (as vE (n (x)  x) vE (x) + (p  1)vE (¯ ) for all x EQp ). Now (n ) = 1 + pn u, u Z , p so n (i n (x))  i n (x) = i (ip u n (n (x))  n (x)) = i n (iu n (x)  x). So we have to prove x f (x) = iu n (x)  x has a continuous inverse on D, and D is a Fp {{f }}module with basis {e1 , · · · , ed }. Let = iu 1; iu Z , p f f so vE () = vE (¯ ). Then vE ( (x)  x) vE (x) + vE (¯ ). It implies has an inverse + f (1  )n and vE (g(x)  x) vE (x) + vE (¯ ). g= n=0
x x So f has an inverse f 1 (x) = g( ) and vE (f 1 (x)  ) vE (x). By induction, for all k in Z, we have
n
vE (f k (x)  k x) vE (x) + (k + 1)vE (¯ ).
+ + Let M = EQp e1 · · · EQp ed , then f k induces
M/¯ M
k M/k+1 M
k M/¯ k+1 M. ¯
So f k Fp [[f ]]e1 · · ·f k Fp [[f ]]ed is dense in k M and is equal by compactness. ¯ Corollary 5.3.12.  1 has a continuous inverse on D=0 , and D=0 is a free Fp {{0 }}module with basis {(e1 ), · · · , (ed )}. Proof. Copy the proof that (ii) for n + 1 implies (ii) for n in the previous p1 proposition, using 1 = 0 . Proposition 5.3.13. If D is an ´tale (, )module over AK or BK , then e  1 has a continuous inverse on D=0 .
´ 5.3. TATE'S EULERPOINCARE FORMULA. Proof. BK follows from AK by Qp Zp ; and we can reduce AK to AQp .
p1 p1
103
Since D=0 (D/p)=0 is surjective, (
i=1
i (xi ) can be lifted to
i=1
[]i (^i )), x
so we have the following exact sequence: 0  (pD)=0  D=0  (D/p)=0  0. Everything is complete for the padic topology, so we just have to verify the result mod p, which is in corollary 5.3.12.
5.3.4
Computation of Galois chomology groups
(1, 1) (1) pr (1) pr
Proposition 5.3.14. Let C, be the complex 0 D(V )     D(V ) D(V )    1   D(V ) 0.         2 Then we have a commutative diagram of complexes C, : 0
/ D(V )
Id
/ D(V )

D(V )
Id
/ D(V )

/0
C, : 0
/ D(V )
/ D(V )
D(V )
/ D(V )
/0
which induces an isomorphism on cohomology. Proof. Since ()(  1) =  1 and commutes with (i.e. = ), the diagram commutes. is surjective, hence the cokernel complex is 0. The kernel is nothing but 0  0  D(V )=0  D(V )=0  0,  it has no cohomology by Proposition 5.3.13. Theorem 5.3.15. If V is a Zp or a Qp representation of GK , then C, (K, V ) computes the Galois cohomology of V : (i) H 0 (GK , V ) = D(V )=1,=1 = D(V )=1,=1 . D(V ) (ii)H 2 (GK , V ) (1,1) . (iii) One has an exact sequence 0  D(V )=1  H 1 (GK , V )  1 (x, y)  D(V ) 1 x
=1 1
 0
104
CHAPTER 5. (, )MODULES AND GALOIS COHOMOLOGY
Let C(V ) = (  1)D1 D=0 , the exact sequence 0  D(V )=1  D(V )=1  C(V )  0 induces an exact sequence 0  D(V )=1 D(V )=1 C(V )    0 1 1 1
since C(V )=1 (D=0 )=1 = 0. Proposition 5.3.16. If D is an ´tale (, )module of dimension d over e =1 is a free Fp [[0 ]]module of rank d. EQp , then C = (  1)D Proof. We know: · C D=0 , it implies C is a Fp [[0 ]]module of rank less than d; · C is compact, because D=1 is compact; · So we just have to prove (see proposition 5.3.11 and corollary 5.3.12) that C contains {(e1 ), · · · , (ed )}, where {e1 , · · · , ed } is any basis of D over EQp . Let {f1 , · · · , fd } be any basis. Then n (¯ k fi ) goes to 0 when n goes to +
+
if k
0. Let gi =
n=0
n ((¯ k fi )). Then we have:
· (gi ) = gi , because ((¯ k fi )) = 0; · (  1)gi = (¯ k fi ) C. We can take ei = k fi . ¯
5.3.5
The EulerPoincar´ formula. e
Theorem 5.3.17. If V is a finite Zp representation of GK , then
2
(V ) =
i=0
H i (GK , V )(1) = V [K:QP ] .
i
5.4. TATE'S DUALITY AND RESIDUES Proof. From Shapiro's lemma, we have H i (GK , V )
GQ
105
H i (GQp , IndGKp V ).
GQ
Since  IndGKp V  = V [K:Qp ] , we can assume K = Qp . Given an exact sequence 0 V1 V V2 0, then (V ) = (V1 )(V2 ) and V  = V1 V2  from the long exact sequence in Galois Cohomology, thus we can reduce to the case that V is a Fp representation of GK . Then we have: H 0  = D(V )=1,=1 ; C(V ) D(V )=1 · · 1 1 D(V ) . H 2  =  (  1,  1) H 1  =  D(V ) 1
=1
;
So H 0 H 2 H 1 1 =  C(V ) 1 , because D(V )=1 and D(V ) are finite groups, 1 1 and for a finite group M , the exact sequence: M 1  0 0  M =1  M  M  1
M implies that M =1  =  1 . Now C(V ) 1
is a (Fp [[0 ]]/(  1)) = Fp module
of rank dimEQp D(V ) = dimFp V . Hence  C(V )  = V . 1
5.4
Tate's duality and residues
H i (GK , M ) × H 2i (GK , M (1))  Qp /Zp .
Let M be a finite Zp module. We want to construct a perfect pairing By using Shapiro's lemma, we may assume K = Qp . Definition 5.4.1. Let x =
kZ
ak k BQp , define res(xd) = a1 .
The residue of x, denoted by Res(x) is defined as Res(x) = res(x d ). 1+
106
CHAPTER 5. (, )MODULES AND GALOIS COHOMOLOGY
The map Res : BQp Qp maps AQp to Zp , thus it induced a natural map BQp /AQp Qp /Zp . Proposition 5.4.2. Res((x)) = Res(x); Res((x)) = ()1 Res(x) Proof. Exercise. Let D be an ´tale (, )module over AQp , denote D = HomAQp (D, BQp /AQp ), e let x D , y D, denote x, y = x(y) BQp /AQp . Then (x), (y) = ( x, y ), (x), (y) = ( x, y ) determines the (, )module structure on D . Set [x, y] := Res( x, y ) Qp /Zp . The main step is following proposition. Proposition 5.4.3. (i) The map x (y [x, y]) gives an isomorphism from D to D (V ) = Homcont (D, Qp /Zp ). (ii) The following formulas hold: [x, (y)] = [(x), y] [(x), y] = ()1 [x, 1 (y)]. Corollary 5.4.4. Let V (1) = HomZp (V, (Qp /Zp )(1)), then D(V (1)) = D (1). Now the two complexes C, (Qp , V ) : D(V ) D (V ) o
d1
/ D(V ) D(V )
d2
/ D(V )
d2
D (1) D (1) o
d
1
D (1) : C, 1 (Qp , V (1))
5.4. TATE'S DUALITY AND RESIDUES
107
are in duality, where d1 z = ((  1)z, (  1)z), d2 (x, y) = (  1)x  (  1)y, d 1 z = ((  1)z , ( 1  1)z ), d2 (x , y ) = ( 1  1)x  (  1)y , and the duality map in the middle given by [(x, y), (x , y )] = [x , x]  [y , y]. One can check that the images are closed. Therefore their cohomology are in duality. For details, see Herr's paper in Math Annalen (2001?).
108
CHAPTER 5. (, )MODULES AND GALOIS COHOMOLOGY
Chapter 6 (, )modules and Iwasawa theory
6.1
6.1.1
i Iwasawa modules HIw (K, V )
Projective limits of cohomology groups
In this chapter we assume that K is a finite extension of Qp and GK is the Z ¯ Galois group of K/K. Then Kn = K(µpn ) and n = Gal(K /Kn ) = n p if n 1 (n 2 if p = 2) where n is a topological generator of n . We pn1 choose n such that n = 1 . The Iwasawa algebra Zp [[K ]] is isomorphic to Zp [[T ]] with the (p, T )adic topology by sending T to  1. We have Zp [[K ]]/(n  1) = Zp [Gal(Kn /K)]. Furthermore Zp [[K ]] is a GK module: let g GK and x Zp [[K ]], then gx = g x, where g is the image of g in K . By the same way, GK acts on ¯ ¯ Zp [Gal(Kn /K)]. Using Shapiro's Lemma, we get, for M a Zp [GK ]module, H i (GKn , M )  H i (GK , Zp [Gal(Kn /K)] M ), with the inverse map given by (1 , ..., i )
gGal(Kn /K)
gCg (1 , ..., i )  (1 , ..., i ) Cid (1 , ..., i ) .
109
110
CHAPTER 6. (, )MODULES AND IWASAWA THEORY
Thus we have a commutative diagram: H i (GKn+1 , M )   H i (GK , Zp [Gal(Kn+1 /K)] M )  cor H i (GKn , M )    H i (GK , Zp [Gal(Kn /K)] M ) One can check that the second vertical arrow is just induced by the natural map Gal(Kn+1 /K) Gal(Kn /K). Definition 6.1.1. (i) If V is a Zp representation of GK , define
i HIw (K, V ) = lim H i (GKn , V )  n
while the transition maps are the corestriction maps. (ii) If V is a Qp representation, choose T a stable Zp lattice in V , then define i i HIw (K, V ) = Qp Zp HIw (K, T ).
6.1.2
Reinterpretation in terms of measures
Proposition 6.1.2. H i (GK , Zp [[K ]] V ) HIw (K, V ). = i Proof. The case of Qp follows from the case of Zp by using Qp Zp . Now assume that V is a Zp representation of GK . By definition, = Zp [[K ]] = lim Zp [[K ]]/(n  1),  it induces the map : H i (GK , V ) W
/ lim H i (GK , /(n  1) V ) = H i (K, V ) Iw  WWWWW WWWWW WWWWW WWWWW +
lim H i (GK , /(pn , n  1) V ) 
The surjectivity is general abstract nonsense. The injectivity of implies the injectivity of ; to prove that of , it is enough to verify the MittagLeffler conditions of H i1 , which are automatic, because of the Finiteness Theorem: /(pn , n  1) V is a finite module, so H i1 (GK , /(pn , n  1) V ) is a finite group.
I 6.1. IWASAWA MODULES HIW (K, V )
111
Remark. (i) Recall that D0 (K , V ) is the set of padic measures from K to V : Zp [[K ]] V D0 (K , V ), v v, = where is the Dirac measure at . Let g GK , µ D0 (K , V ); the action of GK on D0 (K , V ) is as follow: (x)(gµ) = g(
K K
(¯x)µ). g
Hence, for any n N, the map H i (GK , Zp [[K ]] V ) H i (GKn , V ) (translation of Shapiro's lemma) can be written in the following concrete way: (1 , ..., i ) µ(1 , ..., i )  (1 , ..., i )
K
1Kn ·µ(1 , ..., i ) V
nN
.
(ii) Let g GK , , µ Zp [[K ]], x V , then g(µ v) = g µ gv = ¯µ gv = g(µ µ). ¯ g
i So and g commutes, it implies that HIw (K, V ) are Zp [[K ]]modules.
6.1.3
Twist by a character (` la Soul´) a e
Let : K Q be a continuous character. It induces a transform p D0 (K , V ) D0 (K , V ), For Zp [[K ]], we have · (µ) = ( · )( · µ). Indeed, it is enough to check it on Dirac measures. In this case · (1 2 v) = (1 2 )1 2 v = ( · 1 )( · 2 ) v. Recall that Zp () = Zp · e , where, if g GK , then ge = (¯)e . Define g V () = V Zp (). Exercise. The map µ D0 (K , V ) ( · µ) e D0 (K , V ) is an isomorphism of Zp [GK ]modules. µ · µ.
112
CHAPTER 6. (, )MODULES AND IWASAWA THEORY
By the above exercise, we have a commutative diagram:
i HIw (K, V ) i
/ H i (K, V ()) Iw
H i (GK , D0 (K , V ))
/ H i (G , D ( , V ())) K 0 K
So i is an isomorphism of cohomology groups. It can be written in a concrete way i : (1 , ..., i ) µ(1 , ..., i )  (1 , ..., i )
K
1Kn ·µ(1 , ..., i )e
nN
.
It is an isomorphism of Zp modules. Warning: i is not an isomorphism of Zp [[K ]]modules, because i (x) = ( · )i (x): there is a twist.
6.2
i Description of HIw in terms of D(V ) 0.
i Remark. HIw (K, V ) = limnn H i (GKn , V ), so we can always assume n  0
Lemma 6.2.1. Let n = diagram C,n (Kn , V ) : 0
n 1 n1 1
p1 = 1 + n1 +, ..., +n1 Zp [[K ]], the
/ D(V )
n
/ D(V )
n
D(V )
Id
/ D(V )
Id
/0
C,n1 (Kn1 , V ) :
0
/ D(V )
/ D(V )
D(V )
/ D(V )
/0
is commutative and induces corestrictions on cohomology via H i (C,n (Kn , V ))  H i (GKn , V ). Proof. n is a cohomological functor and induces TrKn /Kn1 on H 0 , so it induces corestrictions on H i . Theorem 6.2.2. If V is a Zp or Qp representation of GK , then we have: i (i) HIw (K, V ) = 0, if i = 1, 2. 1 2 (ii) HIw (K, V ) D(V )=1 , HIw (K, V ) D(V ) , and the isomorphisms are = = 1 canonical.
I 6.2. DESCRIPTION OF HIW IN TERMS OF D(V )
113
Remark. (i) The isomorphism
1 Exp : HIw (K, V ) D(V )=1
is the map that will produce padic Lfunctions. Let's describe (Exp )1 . Let y D(V )1 , then (  1)y D(V )=0 . There exists unique xn D(V )=0 satisfying that (n  1)xn = yn , then we can find bn A V such that (  1)bn = xn . Then g log (n ) pn (g  1) y  (g  1)bn (n  1)
( gives a cocycle on GKn with values in V , and log pn n ) does not depend on n. Denote by ,n (y) H 1 (GKn , V ) the image of this cocycle, then 1 (Exp )1 : y  (· · · , ,n (y), · · · )nN HIw (K, V )
doesn't depend on the choice of n . 2 (ii) We see that D(V ) is dual to D(V (1))=1 = V (1)HK , so HIw (K, V ) = 1
D(V ) 1
= (V (1)HK ) .
Before proving the theorem, we introduce a lemma. Lemma 6.2.3. If M is compact with continuous action of K , then M lim(M/n  1). 
n
Proof. We have a natural map from M to limn (M/n  1).  Injectivity: let V be an open neighborhood of 0. For all x M , there exists nx N and Ux x, an open neighborhood of x such that ( 1)x V for Knx and x Ux . By compactness, M = Uxi , where I is a finite
iI
set. Let n = max nxi . It implies that (  1)M V , if n , then
iI
(n  1)M = 0, this shows the injectivity.
nN
Surjectivity: Let (xn )nN limn (M/n 1). From the proof of injectivity,  we know that xn is a Cauchysequence. Because M is compact, there exists x = lim xn . We have xn+k  xn = (n  1)yk for all k 0, as M is compact, there exists a subsequence of yk converging to y, passing to the limit, we get x  xn = (n  1)y. This shows the surjectivity.
114
CHAPTER 6. (, )MODULES AND IWASAWA THEORY
i Proof of Theorem 6.2.2. HIw (K, V ) is trivial if i 3 and the case of Qp follows from Zp by Qp Zp . For i = 0, 0 HIw (K, V ) = lim V GKn .  Tr
V GKn is increasing and V , as V is a finite dimensional Zp module, the sequence is stationary for n n0 . Then TrKn+1 /Kn is just multiplication by p for n n0 , but V does not contain pdivisible elements. This shows that limTr V GKn = 0.  D(V For i = 2: H 2 (GKn , V ) = (1,n)1) . The corestriction map is induced by Id on D(V ), thus D(V ) D(V ) 2 HIw (K, V ) = lim (n  1) = 1  1 by Lemma 6.2.3, as D(V ) is compact (and even finitely generated over Zp ). 1 For i = 1: we have commutative diagrams: 0
/
D(V )=1 n 1
/ H 1 (G , V ) Kn
cor
p2
/ ( D(V ) )n =1 1
n
/0
0
/
D(V )=1 n1 1
p1
/ H 1 (GK n1 , V )
/ ( D(V ) )n1 =1
1
/0
where p1 (¯) = y , p2 ((¯, y )) = x, for any x, y D(V ). Using the functor lim, y ¯ x ¯ ¯  we get: 0
/ lim

D(V )=1 n 1
/ lim H 1 (GK , V ) n

/ lim( D(V ) )n =1 . 1

) Because D(V )=1 is compact, by Lemma 6.2.3 we have D(V )=1 lim D(V 1 . n  1 By definition, HIw (K, V ) = lim H 1 (GKn , V ). The same argument for showing  0 HIw (K, V ) = 0 shows that lim( D(V ) )n =1 = 0. So we get 1  1 D(V )=1  HIw (K, V ).
=1
1 6.3. STRUCTURE OF HIW (K, V )
115
6.3
1 Structure of HIw (K, V )
Recall that we proved that if D is an ´tale (, )module of dim d over EQp , e then C = (  1)D=1 is a free Fp [[Qp ]]module of rank d. The same proof shows that if n 1, i Z , C n (D) is free of rank d over Fp [[n ]]. p Corollary 6.3.1. If D is an ´tale (, )module of dimension d over EK , e then C is a free Fp [[K ]]module of rank d · [K : Qp ]. Proof. Exercise. Hint: D is of dimension d · [HQp : HK ] over EQp and [K : Qp ] = [GQp : GK ] = [Qp : K ][HQp : HK ]. Proposition 6.3.2. If V is a free Zp or Qp representation of rank d of GK , then (i) D(V )=1 is the torsion subZp [[K 1 ]]module of D(V )=1 . (ii) We have exact sequences: 0  D(V )=1  D(V )=1  C(V )  0.  and C(V ) is free of rank d · [K : Qp ] over Zp [[K ]] (or over Qp Zp Zp [[K ]]). Corollary 6.3.3. If V is a free Zp representation of rank d of GK , then 1 the torsion Zp [[K 1 ]]module of HIw (K, V ) is D(V )=1 = V HK , and 1 HIw (K, V )/V HK is free of rank d · [K : Qp ] over Zp [[K ]]. Proof of Proposition 6.3.2. D(V )=1 = V HK is torsion because it is finitely generated over Zp , so (ii) implies (i). To prove (ii), we have to prove C(V )/pC(V ) is free of rank d · [K : Qp ] over Fp [[K ]]. Consider the following commutative diagram with exact rows 0
/ D(V )=1 / (D(V )/p)=1 / D(V )=1
1 1
/ C(V ) / C(V /p)
/0
0
/ (D(V )/p)=11
/0
Using the exact sequence 0 pV V V /p 0
116
CHAPTER 6. (, )MODULES AND IWASAWA THEORY
and apply the snake lemma to the vertical rows of the diagram above, we have the cokernel complex is p  torsion of D(V ) C(V /p) D(V ) p  torsion of 0. (  1) (  1) C(V )/pC(V )
D(V ) (1)
Note that the ptorsion of
C(V /p) C(V )/pC(V )
is a finite dimensional Fp vector space, thus
is also a finite dimensional Fp vector space, hence C(V )/pC(V ) is a Fp [[K ]]lattice of C(V /p), but C(V /p) is a free Fp [[K ]]module of rank d · [K : Qp ] by Corollary 6.3.1. Remark. (i) The sequence 0 D(V )=1 D(V )=1 C(V ) 0 is just the inflationrestriction exact sequence 0 H 1 (K , V HK ) H 1 (GK , V ) H 1 (HK , V )K 0. (ii) Let 0 V1 V V2 0 be an exact sequence, then the exact sequence 0 D(V1 ) D(V ) D(V2 ) 0 and the snake lemma induces 0 D(V1 )=1 D(V )=1 D(V2 )=1 By Theorem 6.2.2, this is just
1 1 1 0 HIw (K, V1 ) HIw (K, V ) HIw (K, V2 ) 2 2 2 HIw (K, V1 ) HIw (K, V ) HIw (K, V2 ) 0.
D(V1 ) D(V ) D(V2 ) 0. 1 1 1
It can also be obtained from the longer exact sequence in continuous cohomology from the exact sequence 0 Zp [[K ]] V1 Zp [[K ]] V Zp [[K ]] V2 0.
Chapter 7 Zp(1) and KubotaLeopoldt zeta function
7.1 The module D(Zp(1))=1
The module Zp (1) is just Zp with the action of GQp by g GQp , x Zp (1), g(x) = (g)x. We shall study the exponential map
1 Exp : HIw (Qp , Zp (1)) D(Zp (1))=1 .
Note that D(Zp (1)) = (A Zp (1))HQp = AQp (1), with usual actions of and , and for , (f ()) = ()f ((1 + )()  1), for all f () AQp (1). Proposition 7.1.1. (i) A=1 = Zp · Qp (ii) We have an exact sequence:
1
(A+p )=1 . Q
1
0  Zp  (A+p )=1  (A+p )=0  0.  Q Q Remark. Under the map µ with support in Z ( = 0) and p
Zp
[]x µ, (A+p )=0 is the image of measures Q µ=0 Z
p
(A+p )=0 = C(Zp ) = (  1)Zp [[Qp ]]. Q Zp [[Qp ]] can be viewed as measures on Qp Z , and µ (  1)Zp [[Qp ]] = p means Zp µ = 0. It implies that C(Zp ) is free of rank 1 over Zp [[Qp ]] which is a special case of what we have proved. 117
118CHAPTER 7. ZP (1) AND KUBOTALEOPOLDT ZETA FUNCTION Proof. (i) We have proved (A+p ) A+p , Q Q vE x vE ((x)) [ ], p 1 1 ( ) = , if x EQp .
+ These facts imply that  1 is bijective on EQp /¯ 1 EQp and hence it is also bijective on AQp / 1 A+p . So Q
(x) = x x 1 A+p . Q (ii) We know that (  1)A+p A+p For x (A+p )=0 , then Q Q Q n (x) n ()A+p 0 if n . Q
+
Hence y =
n=0
n (x) converges, and one check that (y) = y, (  1)y = x.
This implies the surjectivity of  1.
7.2
Kummer theory
+ ~ = (1, (1) , (2) , ..., (n) , ...) EQp E + = R, (1) = 1.
Recall that
Let n = (n)  1, Fn = Qp (n ) for n 1. Then n is a uniforming parameter of Fn , and NFn+1 /Fn (n+1 ) = n , OFn+1 = OFn [n+1 ]/((1 + n+1 )p = 1 + n ).
~ For an element a Fn , choose x = (a, x(1) , ...) E. This x is unique up to u with u Zp . So if g GFn , then
g(x) = c(g) , x
c(g) Zp
gives a 1cocycle c on GFn with values in Zp (1). This defines the Kummer map:
: Fn  H 1 (GFn , Zp (1)) a  (a).
7.3. COLEMAN'S POWER SERIES
119
By Kummer theory, we have H 1 (GFn , Zp (1)) = Zp · (n ) (OFn ). The diagram Fn+1   H 1 (GFn+1 , Zp (1))  NF /F cor
n+1 n
Fn    H 1 (GFn , Zp (1))
is commutative, we have a map:
1 : lim Fn HIw (Qp , Zp (1)) 
and
1 HIw (Qp , Zp (1)) = Zp · (n ) (lim OFn ). 
7.3
Coleman's power series
Theorem 7.3.1 (Coleman's power series). Let u = (un )n1 lim(OFn ){0}  (pour les applications NFn+1 /Fn ), then there exists a unique power series fu Zp [[T ]] such that fu (n ) = un for all n 1. Lemma 7.3.2. (i) If x OFn , Gal(Fn+1 /Fn ), then (x)  x 1 OFn+1 . (ii) NFn+1 /Fn x  xp 1 OFn+1 . Proof. It is easy to see that (i) implies (ii) since [Fn+1 : Fn ] = p. Write
p1
() = 1 + pn u for u Zp . Let x =
i=0 p1
xi (1 + n+1 )i , where xi OFn . Then
(x)  x =
i=0
xi (1 + n+1 )i ((1 + 1 )iu  1) 1 OFn+1 .
+ ¯ ¯ Corollary 7.3.3. u = (¯p , u1 , ..., un , ...) EQp , where un is the image of ¯ u1 ¯ un mod 1 .
Definition 7.3.4. Let N : OF1 [[T ]] OF1 [[T ]] such that N(f )((1 + T )p  1) =
z p =1
f ((1 + T )z  1).
120CHAPTER 7. ZP (1) AND KUBOTALEOPOLDT ZETA FUNCTION Lemma 7.3.5. (i) N(f )(n ) = NFn+1 /Fn (f (n+1 )), (ii) N(Zp [[T ]]) Zp [[T ]], (iii) N(f )  f 1 OF1 [[T ]], k (iv) If f OF1 [[T ]] , k 1, if (f  g) 1 OF1 [[T ]], then
k+1 N(f )  N(g) 1 OF1 [[T ]].
Proof. (i) The conjugates of n+1 under Gal(Fn+1 /Fn ) are those (1 + n )z  1 for z p = 1, this implies (i). (ii) Obvious, is just Galois theory. (iii) Look mod 1 , because z = 1 mod 1 , we have N(f )(T p ) = f (T )p . k (iv) We have N( f ) = N(f ) , so we can reduce to f = 1 and g = 1 + 1 h. g N(g) Then
k N(g)((1 + T )p  1) = 1 + 1 z p =1 k+1 h((1 + T )z  1) mod 1 ,
and
z p =1
h((1 + T )z  1) is divisible by p.
+ Corollary 7.3.6. (i) If u EQp and vE (¯) = 0, then there exists a unique ¯ u gu Zp [[T ]] such that N(gu ) = gu and gu (¯ ) = u. ¯ k+1 k (ii) If x 1 + 1 OFn+1 , then NFn+1 /Fn (x) 1 + 1 OFn .
Proof. (i) Take any g Zp [[T ]] such that g(¯ ) = u, then g Zp [[T ]] , by (iv) ¯ of Lemma 7.3.5, Nk (g) converges in g + 1 Zp [[T ]] and gu is the limit. k (ii) There exists f 1 + 1 OF1 [T ] such that x = f (n+1 ). Then use (i) and (iv) of Lemma 7.3.5. Proof of Theorem 7.3.1. The uniqueness follows from the fact that 0 = f Zp [[T ]] has only many finitely zeros in mCp (Newton polygons). k Existence: let u = (un ), write un = n un , where k Z and µp1 do not depend on n, and un 1 + mFn . Then NFn+1 /Fn un+1 = un . If for all n, fu (n ) = un , let fu = T k fu , then fu (n ) = un . Thus we are reduced to the case that un 1 + mFn for all n. By (i) of Corollary 7.3.6, we can find gu Zp [[T ]] for u. We have to check ¯ that gu (n ) = un for all n = 1. Write vn = gu (n ). Then N(gu ) = gu , by (i) of Lemma 7.3.5, implies that NFn+1 /Fn (vn+1 ) = vn ; and gu (¯ ) = u implies ¯ vn that vn = un mod 1 for all n. Let wn = un , then we have NFn+1 /Fn (wn+1 ) = wn and wn 1 + 1 OFn .
7.3. COLEMAN'S POWER SERIES By (ii) of Corollary 7.3.6, we have
k wn = NFn+k /Fn (wn+k ) 1 + 1 OFn for all k,
121
then wn = 1. This completes the proof. Corollary 7.3.7. N(fu ) = fu ,
d where = (1 + T ) dT .
(
fu fu )= fu fu
Proof. By (i) of Lemma 7.3.5, we have N(fu )(n ) = NFn+1 /Fn (fu (n+1 )) = fu (n ), for all n, thus N (fu ) = fu . Using the formula ( log f ) = (log N(f )), we immediately get the result for . As for the proof of this last formula, we know that (N(f )(T )) = N(f )((1 + T )p  1) =
z p =1
f ((1 + T )z  1)
(f )((1 + T )p  1) =
1 f ((1 + T )z  1) p zp =1
Then we have two ways to write (log (N(f ))) (log (N(f ))) = p ( log N(f ))( = p ) N(f ) N(f ) = p( ) = p( )((1 + T )p  1) N(f ) N(f ) = p( log N(f ))((1 + T )p  1), (log (N(f ))) = (log
z p =1
f ((1 + T )z  1))
= =
(1 + T )zf ((1 + T )z  1) f ((1 + T )z  1) z p =1 f f ((1 + T )z  1) = p( )((1 + T )p  1) f f z p =1
= p(( log f ))((1 + T )p  1), hence the formula.
122CHAPTER 7. ZP (1) AND KUBOTALEOPOLDT ZETA FUNCTION
7.4
An explicit reciprocity law
/ H 1 (Qp , Zp (1)) lim(OFn  {0}) Iw  QQQ mm QQQ mmmm QQQ m QQ mmm u fu () QQ( vmmm Exp fu D(Zp (1))=1
Theorem 7.4.1. The diagram
is commutative. Remark. (i) The proof is typical of invariants defined via Fontaine's rings: easy to define and hard to compute. (ii) For another example, let X/K be a smooth and projective variety, then i i DdR (Het (X × K, Qp )) = HdR (X/K). ´ The proof is very hard and is due to Faltings and Tsuji. (iii) Let a Z such that a = 1, (a, p) = 1. The element un = ea pn  1 e pn  1
2i 2i
Q(µpn )
is a cyclotomic unit in OQ(µpn ) (whose units are called global units). Then un Fn = Qp (µpn ), un = a (n ) , 1 (n )
where b Qp such that (b ) = b. From NFn+1 /Fn (n+1 ) = n , one gets NFn+1 /Fn (un+1 ) = un ( commutes with norm), thus u = (un ) lim OFn .  Obviously the Coleman power series fu = (1 + T )a  1 , (1 + T )1  1 fu a 1 =  . fu (1 + T )a  1 T
So fu is nothing but the Amice transform of µa that was used to construct fu padic zeta function. S o Exp produces KubotaLeopoldt zeta function from the system of cyclotomic units.
7.5. PROOF OF THE EXPLICIT RECIPROCITY LAW
123
(iv) The example in (iii) is part of a big conjectural picture. For V a fixed representation of GQ , then conjecturally
1 {compatible system of global elements of V }  HIw (Q, V ) 1  HIw (Qp , V )  D(V )=1    padic Lfunctions.  transf orm Exp Amice
At present there are very few examples representation of GQ for which this picture is known to work. The Amice transform works well for Zp (1), because 1 improves denominators in , and A=1 A+p can be viewed as measures. Qp Q In general, to use the properties of , we will have to introduce overconvergent (, )modules.
7.5
7.5.1
Proof of the explicit reciprocity law
Strategy of proof of Theorem 7.4.1
Let u lim OFn , and g Cn (g) be the cocycle on GFn by Kummer theory,  i.e the image of u under the composition of
1 lim(OFn  {0})  HIw (Qp , Zp (1))  H 1 (GFn , Zp (1)). 
Let y D(Zp (1))=1 = A=1 (1), let g Cn (g) be the image of y under the Qp composition of   1 D(Zp (1))=1 = A=1 (1)    HIw (Qp , Zp (1))  H 1 (GFn , Zp (1)). Qp We need to prove that Cn = Cn for all n implies y = For Cn , we have Cn (g) = log (n ) pn
fu (). fu (Exp )1
(g)  1 y  ((g)g  1)bn , (n )  1
where bn A is a solution of (  1)bn = ((n )n  1)1 (  1)y, we know that (  1)y A=0 . The exact value of bn is not important. Qp (0) (k) (0) ~ For Cn , choose xn = (xn , ..., xn , ...) E + such that xn = un . Let un = [xn ], then ~ g(~n ) u = []Cn (g) . un ~
124CHAPTER 7. ZP (1) AND KUBOTALEOPOLDT ZETA FUNCTION Proposition 7.5.1. Assume n 1. (i) There exists k Z, bn OCp /pn such that p2 Cn (g) = p2 log (n ) g  1 y(n+k ) + (g  1)bn · pn n  1
in OCp /pn . (ii) There exists k Z, bn OCp /pn such that p2 Cn (g) = p2 in OCp /pn . Proposition 7.5.2. There exists a constant c N, such that for all n and for all k, if x OFn+k , b OCp satisfy vp then pk TrFn+k /Fn x pnc OFn . We shall prove Proposition 7.5.1 in the next no , and Proposition 7.5.2 in the third no . We first explain how the above two propositions imply the theorem: If h() = (h()), then h(n ) = p1 TrFn+1 /Fn (h(n+1 )). By hypothesis, (y) = y, we get pk TrFn+k /Fn (y(n+k )) = y(n ), n, k (). log (n ) fu y(n+k )  (n ) , b = bn  bn . n p fu By Proposition 7.5.1, and the hypothesis Cn (g) = Cn (g), we get x = p2 g1 x + (g  1)b = p2 (Cn (g)  Cn (g)) = 0. n  1 The first equality is because for every x Fn , Proposition 7.5.2, we get p2
g1 x n 1
log (g) fu y(n ) + (g  1)bn pn fu
g1 x + (g  1)b n, g GFn n  1
Let
=
log (g) x. log (n )
Using
log((n )) fu y(n )  (n ) pnc OFn , n p fu
7.5. PROOF OF THE EXPLICIT RECIPROCITY LAW then for every n, y(n )  fu (n ) pnc2 OFn . fu
125
Let h = y  fu , then (h) = h and h(n ) pnc2 OFn . Using the fact fu pk TrFn+k /Fn OFn+k OFn and the formula (*), then for every i N, n i, h(i ) = p(ni) TrFn /Fi (h(n )) pnc2 OFi , thus h(i ) = 0 for every i N, hence h = 0.
7.5.2
Explicit formulas for cocyles
This no is devoted to the proof of Proposition 7.5.1 (0) (i) Recall that = []  1, ( pm [xm ]) = pm xm and () = 1  1 = n ~ 0. Let n = n () A+ , then n = [1/p ]  1, (~n ) = n . Write ~ ~
+
bn =
l=0
~ pl [zl ], where zl E. As Cn (g) Zp , we have (n+k) Cn (g) = Cn (g), for all n and k. 1 vE (zl ), pk
As vE (k (zl )) = we can find k such that vE ((n+k) (zl )) 1, ~ Let p = (p, ...) E + , then for every l ~ [~]Cn (g) = p
for all l
n  1.
~ n  1, p · (n+k) (zl ) E + . We have ~
log (n ) (g)g  1 [~] · p y(~n+k ) + [~]((g)g  1)(n+k) (bn ). p n p (n )n  1
~ ~ ~ Both sides live in A+ + pn A, reduce mod pn and use : A+ /pn OCp /pn , then [~] p and p pCn (g) = p log (n ) g1 p· y(n+k ) + (g  1)bn n p n  1
where bn = ([~](n+k) (bn )). p
126CHAPTER 7. ZP (1) AND KUBOTALEOPOLDT ZETA FUNCTION
k (ii) Write u = (n )(vn ), where vn are units OFn . So we just have to prove the formula for (n ) and (vn ). Thus we can assume vp (un ) 1. Let x1 x1 + H : 1 + tBdR Cp , x ( ) = ( ), t recall that t = log(1 + ). We have
H((1+x )(1+y )) = H(1+(x +y )+ 2 x y ) = (x +y ) = H(1+x )+H(1+y ), thus H(xy) = H(x) + H(y).
1 p Write un = [(un , un , ...)], we have ~
g(~n ) u un ~
= []Cn (g) = 1 + Cn (g) + · · · , thus
Cn (g) = H(
g(~n ) u ). un ~
We know un = fu (n ) and (~n ) = un , then u (fu (~n )) = fu ((~n )) = fu (n ) = un = (~n ). u
(~ So, if we set an = fuun ) , then (an ) = 1. ~n ~ We know that [~]an A+ since vp (un ) 1. Then we get H(an ) p because of the following identity
1 O p1 Cp
H(an ) = and because =
1 ~
[~]an  [~] p p 1 [~]an  [~] p p = · , [~] p [~]~1 p
~ is a generator of Ker in A+ as Ker , and 1 1 , so vE (¯ ) = (1  )vE (  1) = 1. 1 p
= ¯ Then we have
1/p
g(fu (~n )) fu ((1 + n )(g)  1) ~ = fu (~n ) fu (~n ) fu ((1 + n )(1 + ) pn  1) ~ = fu (~n ) fu (g)  1 =1+ (~n ) · + terms of higher degree in , fu pn
(g)1
7.5. PROOF OF THE EXPLICIT RECIPROCITY LAW hence H g(fu (~n )) (g)  1 fu = (n ). · fu (~n ) pn fu
127
Using formula fu (~n ) = un an , we get ~ Cn (g) = H g(~n ) u g(fu (~n )) g(an ) =H H un ~ fu (~n ) an (g)  1 fu (n )  ((g)g  1)H(an ). = · pn fu
We conclude the proof by multiplying p2 , noticing that (g) = 1 mod pn , so (g)  1 exp(log (g))  1 log (g) = = mod pn ; n n n p p p set bn = p2 H(an ), we get the result.
7.5.3
Tate's normalized trace maps
Fn .
n
Let n = (n)  1, Fn = Qp (n ), F =
Lemma 7.5.3. If n 1, x F , then pk TrFn+k /Fn x does not depend on k such that x Fn+k . Proof. Use the transitive properties of the trace map and the fact [Fn+k : Fn ] = pk . Let Rn : F  Fn be the above map. Denote Yi = {x Fi , TrFi /Fi1 x = 0}. Lemma 7.5.4. (i) Rn commutes with Qp , is Fn linear and Rn Rn+k = Rn .
+
(ii) Let x F , then x = Rn (x)+
i=1
Rn+i (x), where Rn+i (x) = Rn+i (x)
Rn+i1 (x) Yn+i and is 0 if i 0. (iii) Let k Z, then vp (x) kvp (n ) if and only if vp (Rn (x)) and vp (Rn+i (x)) kvp (n ) for every i N.
kvp (n )
128CHAPTER 7. ZP (1) AND KUBOTALEOPOLDT ZETA FUNCTION Proof. (i) is obvious. (ii) is also obvious, since Rn+i1 (Rn+i (x)) = 0 Rn+i (x) Yn+i . (iii) is obvious. For , let x OFn+k , then
pk 1
x=
j=0
aj (1 + n+k )j ,
aj O Fn .
Write j = pki j with (j , p) = 1, then Rn (x) = a0 , since p1 TrFn+i /Fn+i1 (1 + n+i )j = Thus vp (x) 0 vp (Rn (x))
0 and vp (Rn+i (x)) Rn+i (x) = (j ,p)=1
apni j (1 + n+i )j
(1 + n+i )j , 0,
if p  j if (j, p) = 1. 0.
By Fn linearity we get the result. Remark. In the whole theory, the following objects play similar roles: p1 TrFn+1 /Fn = 0 Yi . Lemma 7.5.5. Assume that j i  1 and j 2. and assume j is a generator of j . Let u Qp . If vp (u  1) > vp (1 ), then uj  1 is invertible on Yi . Moreover if x Yi , vp (x) kvp (n ), then vp ((uj 1)1 x) kvp (n ) vp (1 ).
p Proof. If i1 = j
ij1
, then
ij1
(uj  1)1 = (up
i1  1)1 (1 + uj + · · · + (uj )p
ij1 1
),
so it is enough to treat the case j = i  1. Let x OFi Yi , write
p1
x=
a=1
xa (1 + i )a ,
xa OFi1 ,
7.5. PROOF OF THE EXPLICIT RECIPROCITY LAW write (i1 ) = 1 + pi1 v with v Z , then p
p1
129
(ui1  1)x =
a=1
xa (1 + i )a (u(1 + 1 )av  1).
We can check directly
p1
(ui1  1) x =
a=1
1
xa (1 + i )a . (u(1 + 1 )av  1) vp (1 ).
Moreover, if vp (x) 0, then vp ((uj  1)1 x)
Proposition 7.5.6. Assume n 1, u Q and vp (u  1) > vp (1 ), then p (i) x F can be written uniquely as x = Rn (x) + (un  1)y with Rn (y) = 0, and we have vp (Rn (x)) > vp (x)  vp (n ), vp (y) > vp (x)  vp (n )  vp (1 ).
^ ^ (ii) Rn extends by continuity to F , and let Xn = {x F , Rn (x) = 0}. ^ Then every x F can be written uniquely as x = Rn (x) + (un  1)y with y Xn and Rn (x) Fn , and with the same inequalities vp (Rn (x)) Proof. (i) As
+
vp (x)  vp (n ),
vp (y)
vp (x)  vp (n )  vp (1 ).
x = Rn (x) +
i=1 +
(un  1)((un  1)1 Rn+i (x)).
we just let y =
i=1
(un  1)1 Rn+i (x).
(ii) By (i), we have vp (Rn (x)) vp (x)  C, so Rn extends by continuity ^ to F ; the rest follows by continuity. ^ Remark. (i) The maps Rn : F  Fn are Tate's normalized trace maps. (ii) they commutes with Qp (or GQp ). ^ (iii) Rn (x) = x if x F and n 0, hence Rn (x) x if x F and n .
130CHAPTER 7. ZP (1) AND KUBOTALEOPOLDT ZETA FUNCTION
7.5.4
Applications to Galois cohomology
x Fn  ( x log ()) H 1 (Fn , Fn )
Proposition 7.5.7. (i) The map
induces isomorphism
^ Fn H 1 (Fn , Fn ) H 1 (Fn , F ).
^ (ii) If : Fn  Q is of infinite order, then H 1 (Fn , F ()) = 0. p Proof. If n 0 so that vp ((n )  1) > vp (1 ). Using the above proposition (let u = (n )), we get ^ H 1 (Fn , F ()) = ^ F Fn Xn Fn = = . (un  1) (un  1) un  1
If u = 1, we get (n  1)Fn = 0. If u = 1, we get Fn /(u  1)Fn = 0. For n small, using inflation and restriction sequence, as Gal(Fn+k /Fn ) is ^ finite, and F () is a Qp vector space, we have ^ H 1 (Gal(Fn+k /Fn ), F ()Fn+k ) = 0, then we get an isomorphism
^ ^ H 1 (Fn , F ())  H 1 (Fn+k , F ())Gal(Fn+k /Fn ) .
H 2 = 0,
From the case of n
0, we immediately get the result.
Recall that the following result is the main step in Ax's proof of the AxSenTate theorem (cf. Fontaine's course). Proposition 7.5.8. There exists a constant C N, such that if x Cp , if H GQp is a closed subgroup, if for all g H, vp ((g  1)x) a for some a, then there exists y CH such that vp (x  y) a  C. p The following corollary is Proposition 7.5.2 in the previous section. Corollary 7.5.9. For x OF , if there exists c OCp such that ^ vp Then we have vp (Rn (x)) n  C  1(or 2). g1 x  (g  1)c n  1 n, for all g GFn .
7.5. PROOF OF THE EXPLICIT RECIPROCITY LAW Proof. By assumption, we get vp ((g  1)c) n, g HQp = Ker ,
131
^ then by Ax, there exists c F such that vp (c  c ) Take g = n , then vp (x  (n  1)c ) n  C. n  C. As Rn n = n Rn = Rn , we get n  C  vp (1 )  vp (n ),
vp (Rn (x)) = vp (Rn (x  (n  1)c )) hence the result.
7.5.5
No 2i in Cp !
Theorem 7.5.10. (i) Cp does not contain log 2i, i.e. there exists no x Cp satisfies that g(x) = x+log (g) for all g GK , where K is a finite extension of Qp . (ii) Cp (k) = 0, if k = 0. Proof. (i) If K = Qp , if there exists such an x, by AxSenTate, we get HQ ^ x F = Cp p . Then we have: Rn (g(x)) = g(Rn (x)) = Rn (x) + log (g). Because Rn (x) Fn , it has only finite number of conjugates but log (g) has infinitely many values, contradiction! Now for K general, we can assume K/Qp is Galois, let y= 1 [K : Qp ] (x)
S
where S are representatives of GQp /GK . For g GQp , we can write g = h for h GK and S. From this we get log (h ) = [K : Qp ] log (g).
S
132CHAPTER 7. ZP (1) AND KUBOTALEOPOLDT ZETA FUNCTION Then we have g(y) = = 1 [K : Qp ] 1 [K : Qp ] g(x) =
S
1 [K : Qp ]
h x
S
(x + log (h ))
S
1 = [K : Qp ]
(x) +
S
1 [K : Qp ]
log (h )
S
= y + log (g). Then by the case K = Qp , we get the result. (ii) If 0 = x Cp (k), then g(x) = (g)k x. Let y = g(y) = y + log (g), which is a contradiction by (i).
log x , k
then we have
Chapter 8 (, )modules and padic Lfunctions
8.1
8.1.1
TateSen's conditions
The conditions (TS1), (TS2) and (TS3)
Let G0 be a profinite group and : G0 Z be a continuous group homop morphism with open image. Set v(g) = vp (log (g)) and H0 = Ker . ~ Suppose is a Zp algebra and ~ v :  R {+} satisfies the following conditions: (i) v(x) = + if and only if x = 0; (ii) v(xy) v(x) + v(y); (iii) v(x + y) inf(v(x), v(y)); (iv) v(p) > 0, v(px) = v(p) + v(x). ~ ~ Assume is complete for v, and G0 acts continuously on such that ~ v(g(x)) = v(x) for all g G0 and x . ~ Definition 8.1.1. The TateSen's conditions for the quadruple (G0 , , , v) are the following three conditions TS1TS3. (TS1). For all C1 > 0, for all H1 H2 H0 open subgroups, there exists an ~ H1 with v() > C1 and () = 1.
H2 /H1
133
134
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
~ ~ (In Faltings' terminology, /H0 is called almost ´tale.) e (TS2). Tate's normalized trace maps: there exists C2 > 0 such that for all open subgroups H H0 , there exist n(H) N and (H,n )nn(H) , an ~ increasing sequence of sub Zp algebras of H and maps ~ RH,n : H  H,n satisfying the following conditions: ~ (a) if H1 H2 , then H2 ,n = (H1 ,n )H2 , and RH1 ,n = RH2 ,n on H2 ; (b) for all g G0 , g(H,n ) = gHg1 ,n g RH,n = RgHg1 ,n g;
(c) RH,n is H,n linear and is equal to Id on H,n ; ~ (d) v(RH,n (x)) v(x)  C2 if n n(H) and x H ; (e) lim RH,n (x) = x.
n+
(TS3). There exists C3 , such that for all open subgroups G G0 , H = G H0 , there exists n(G) n(H) such that if n n(G), G/H and ~ v() = vp (log ()) n, then  1 is invertible on XH,n = (RH,n  1) and v((  1)1 x) for x XH,n . ~ Remark. RH,n RH,n = RH,n , so H = H,n XH,n . v(x)  C3
8.1.2
Example : the field Cp
~ Theorem 8.1.2. The quadruple ( = Cp , v = vp , G0 = GQp and =the cyclotomic character) satisfies (TS1), (TS2), (TS3). Proof. (TS1): In Fontaine's course, we know that for any Qp K L such that [L : Qp ] < +, then vp (dLn /Kn ) 0 as n +. The proof showed that vp ((n )  n ) 0 as n +, where n is a uniformizer of Ln and Gal(Ln /Kn ) = Gal(L /K ) when n 0. We also have TrL /K = TrLn /Kn
8.1. TATESEN'S CONDITIONS on Ln if n 0 and TrLn /Kn (OLn ) dLn /Kn OKn ,
135
thus TrL /K (OL ) contains elements with vp as small as we want. Take x x OL and let = TrL /K (x) , then
() = TrL /K () = 1.
HK /HL
Then for all C1 > 0, we can find x OL such that vp (TrL /K (x)) is small enough, thus vp () > C1 . ^ (TS2) and (TS3): By AxSenTate, CHK = K , let HK ,n = Kn , and p k RHK ,n = p TrKn+k /Kn on Kn+k . If K = Qp , RHK ,n = Rn , that's what we did in last chapter. We are going to use what we know about Rn . For G = GK , then H = HK , choose m big enough such that for any n m, vp (dKn /Fn ) is small and [K : F ] = [Kn : Fn ] = d. Let {e1 , ..., ed } be a basis of OKn over OFn and {e , ..., e } be the dual basis of Kn over Fn 1 d for the trace map (x, y) TrKn /Fn (xy). This implies that {e , ..., e } is a 1 d vp (dKn /Fn ) are small. Any x K can be basis of d1 /Fn and vp (e ) i Kn written as
d
x=
i=1
TrK /K (xei )e , i
then inf vp (TrK /F (xei )) vp (x) inf vp (TrK /F (xei ))  vp (dKn /Fn ),
i i
and RHK ,n (x) =
d
Rn (TrK /F (xei ))e , i
i=1
n
m.
So use what we know about Rn to conclude. Remark. By the same method as Corollary 7.5.7, we get ^ = (i) H 1 (, K ) K, where the isomorphism is given by x K  ( x log ()). ^ (ii) H 1 (, K ()) = 0 if is of infinite order.
136
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
8.2
Sen's method
~ Proposition 8.2.1. Assume verifying (TS1), (TS2) and (TS3). Let ~ U be a continuous cocycle from G0 to GLd (). If G G0 is an open normal subgroup of G0 such that v(U  1) > 2C2 + 2C3 for any G. Set ~ H = G H0 , then there exists M GLd () with v(M  1) > C2 + C3 such that  V = M 1 U (M ) satisfies V GLd (H,n(G) ) and V = 1 if H. ~ Example 8.2.2. Example of Sen: For the case = Cp , for U a 1cocycle on GK with values in GLd (Cp ), there exists [L : K] < , such that U is cohomologous to a cocycle that which is trivial on HL and with values in GLd (Ln ) for some n. The proof of Proposition 8.2.1 needs three Lemmas below. It is technical, but the techniques come over again and again.
8.2.1
Almost ´tale descent e
~ Lemma 8.2.3. If satisfies (TS1), a > 0, and U is a 1cocycle on H open in H0 and v(U  1) a for any H, ~ then there exists M GLd () such that a , v(M 1 U (M )  1) a + 1. 2 Proof. The proof is approximating Hilbert's Theorem 90. Fix H1 H open and normal such that v(U  1) a + 1 + a/2 for ~ satisfies (TS1), we can H1 , which is possible by continuity. Because ~ find H1 such that v(M  1) v() a/2,
H/H1
() = 1. ()U , we
S
Let S H be a set of representatives of H/H1 , denote MS = have MS  1 =
S + 1 MS
()(U  1), this implies v(MS  1) =
n=0
a/2 and moreover
(1  MS )n ,
8.2. SEN'S METHOD
137
1 ~ so we have v(MS ) 0 and MS GLd (). If H1 , then U  U = U ((U )  1). Let S H be another set of representatives of H/H1 , so for any S , there exists H1 and S such that = , so we get
MS  M S =
S
()(U  U ) =
S
()U (1  (U )),
thus v(MS  MS ) For any H, U (MS ) =
S
a + 1 + a/2  a/2 = a + 1.
()U (U ) = M S .
Then
1 1 MS U (MS ) = 1 + MS (M S  MS ), 1 with v(MS (M S  MS )) a + 1. Take M = MS for any S, we get the result.
Corollary 8.2.4. Under the same hypotheses as the above lemma, there ~ exists M GLd () such that v(M  1) a/2, M 1 U (M ) = 1, H.
Proof. Repeat the lemma (a a+1 a+2 · · · ), and take the limits. ~ ~ ~ Exercise. Assume satisfies (TS1), denote by + = {x v(x) 0}. Let + ~ M be a finitely generated module with semilinear action of H, an open i ~ subgroup of H0 . Then H (H, M ) is killed by any x H with v(x) > 0. Hint: Adapt the proof that if L/K is finite Galois and M is a Lmodule with semilinear action of Gal(L/K), then H i (Gal(L/K), L) = 0 for all i 1. Let L such that TrL/K () = 1. For any c(g1 , · · · , gn ) an ncocycle, let c (g1 , · · · , gn1 ) =
hGal(L/K)
g1 · · · gn1 h()c(g1 , · · · , gn1 , h),
then dc = c.
138
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
Theorem 8.2.5. (i) The map x  (g x log (g)) gives an isomorphism K H 1 (GK , Cp ). (ii) If : GK K Q is of infinite order, then H 1 (GK , Cp ()) = 0. p Proof. Using the inflation and restriction exact sequence 0  H 1 (K , Cp ()HK )  H 1 (GK , Cp ())  H 1 (HK , Cp ())K . by the above exercise, H 1 (HK , Cp ())K = 0, then the inflation map is actu^ ally an isomorphism. We have Cp ()HK = K (), and use Corollary 7.5.7. In fact ^ K = H 1 (K , K ) = H 1 (K , K) = Hom(, K) = K · log , the last equality is because K is procyclic.
inf res
8.2.2
Decompletion
Now recall that we have the continuous character: G0  Z , H0 = Ker . p is complete for v, with continuous action of G0 . H is an open subgroup of H0 , and we have the maps:RH,n : H H,n . By (TS2), v(RH,n (x)) v(x)C2 ; and by (TS3), v(( 1)1 x) v(x)C3 , if RH,n (x) = 0 and vp (log ()) n. We can use these properties to reduce to something reasonable. Lemma 8.2.6. Assume given > 0, b 2C2 + 2C3 + , and H H0 is open. Suppose n n(H), G/H with n() n, U = 1 + U1 + U2 with U1 Md (H,n ), v(U1 ) b  C2  C3 U2 Md (H ), v(U2 ) b. Then, there exists M GLd (H ), v(M  1) b  C2  C3 such that M 1 U (M ) = 1 + V1 + V2 , with V1 Md (H,n ), v(V1 ) b  C2  C3 ), V2 Md (H ), v(V2 ) b + .
8.2. SEN'S METHOD
139
Proof. Using (TS2) and (TS3), one gets U2 = RH,n (U2 ) + (1  )V , with v(RH,n (U2 )) v(U2 )  C2 , Thus, (1 + V )1 U (1 + V ) = (1  V + V 2  · · · )(1 + U1 + U2 )(1 + (V )) = 1 + U1 + (  1)V + U2 + (terms of degree 2) Let V1 = U1 + RH,n (U2 ) Md (H,n ) and W be the terms of degree 2. Thus v(W ) 2(b  C2  C3 ) b + . So we can take M = 1 + V, V2 = W . Corollary 8.2.7. Keep the same hypotheses as in Lemma 8.2.6. Then there exists M GLd (H ), v(M  1) b  C2  C3 such that M 1 U (M ) GLd (H,n ). Proof. Repeat the lemma (b b+ b+2 · · · ), and take the limit. Lemma 8.2.8. Suppose H H0 is an open subgroup, i n(H), G/H, n() i and B GLd (H ). If there exist V1 , V2 GLd (H,i ) such that v(V1  1) > C3 , then B GLd (H,i ). Proof. Take C = B  RH,i (B). We have to prove C = 0. Note that C has coefficients in XH,i = (1  RH,i )H , and RH,i is H,i linear and commutes with . Thus, (C)  C = V1 CV2  C = (V1  1)CV2 + V1 C(V2  1)  (V1  1)C(V2  1) Hence, v((C)  C) > v(C) + C3 . By (TS3), this implies v(C) = +, i.e. C = 0. Proof of Proposition 8.2.1. Let U be a continuous 1cocycle on G0 with values in GLd (). Choose an open normal subgroup G of G0 such that
G
v(V ) v(U2 )  C2  C3 .
v(V2  1) > C3 ,
(B) = V1 BV2 ,
inf v(U  1) > 2(C2 + C3 ).
By Lemma 8.2.3, there exists M1 GLd (), v(M1  1) > 2(C2 + C3 ) such 1 that U = M1 U (M1 ) is trivial in H = G H0 (In particular, it has values in GLd (H )).
140
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
Now we pick G/H with n() = n(G). In particular, we want n(G) big enough so that is in the center of G0 /H. Indeed, the center is open, since in the exact sequence: 1 H0 /H G0 /H G/H 1, G/H Zp × (finite), and H0 /H is finite. So we are able to choose such a n(G). Then we have v(U ) > 2(C2 + C3 ), and by Corollary 8.2.7, there exists M2 GLd (H ) satisfying
1 v(M2  1) > C2 + C3 and M2 U (M2 ) GLd (H,n(G) ).
Take M = M1 · M2 , then the cocycle V = M 1 U (M ) a cocycle trivial on H with values in GLd (H ), and we have v(V  1) > C2 + C3 and V GLd (H,n(G) ). This implies V comes by inflation from a cocycle on G0 /H. The last thing we want to prove is V GLd (H,n(G) ) for any G0 /H. Note that = as is in the center, so V (V ) = V = V = V (V ) which implies (V ) = V1 V (V ). Apply Lemma 8.2.8 with V1 = V1 , V2 = (V ), then we obtain what we want.
8.2.3
Applications to padic representations
Proposition 8.2.9. Let T be a free Zp representation of G0 , k N, v(pk ) > 2C2 + 2C3 , and suppose G G0 is an open normal subgroup acting trivially on T /pk T , and H = G H0 . Let n N, n n(G). Then there exists a unique DH,n (T ) T , a free H,n module of rank d, such that: (i) DH,n (T ) is fixed by H, and stable by G; (ii) H,n DH,n (T )  T ; (iii) there exists a basis {e1 , . . . , ed } of DH,n over H,n such that if G/H, then v(V  1) > C3 , V being the matrix of .
8.3. OVERCONVERGENT (, )MODULES Proof. Translation of Proposition 8.2.1, by the correspondence representations of G0 H 1 (G0 , GLd ()). For the uniqueness, one uses Lemma 8.2.8.
141
Remark. H0 acts through H0 /H (which is finite) on DH,n (T ). If H,n is ´tale e (H0 /H) over H0 ,n (the case in applications), and then DH0 ,n (T ) = DH,n (T ) , is locally free over H0 ,n (in most cases it is free), and H,n
H0 ,n
DH0 ,n (T )  DH,n (T ).
Example 8.2.10. For = Cp , let V be a Qp representation of GK for [K : Qp ] < +, T V be a stable lattice. Then DSen,n (V ) := DHK ,n (T ) is a Kn vector space of dimension d = dimQp V with a linear action of Kn . Sen's operator is defined as follows: Sen = It is easy to see: Proposition 8.2.11. V is HodgeTate if and only if Sen is semisimple, and the eigenvalues lie in Z. These eigenvalues are the HodgeTate weights of V . Remark. For general V , the eigenvalues of Sen are the generalized HodgeTate weights of V . log , where Kn , log () = 0. log ()
8.3
8.3.1
Overconvergent (, )modules
Overconvergent elements
+
Definition 8.3.1. (i) For x =
i=0 ik
pi [xi ] A, xi E = Fr R, k N, define
wk (x) := inf vE (xi ) (One checks easily that wk (x) vE (), E, if and only if []x A+ + pk+1 A).
142
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
(ii) For a real number r > 0, define v (0, r] (x) := inf wk (x) +
kN
k k = inf vE (xk ) + R {±}. r kN r
k r
(iii) A(0, r] := {x A : lim vE (xk ) +
k+
= lim wE (xk ) +
k+
k r
= +}.
Proposition 8.3.2. A(0, r] is a ring and v = v (0, r] satisfies the following properties: (i) v(x) = + x = 0; (ii) v(xy) v(x) + v(y); (iii) v(x + y) inf(v(x), v(y)); (iv) v(px) = v(x) + 1 ; r (v) v([]x) = vE () + v(x) if E; (vi) v(g(x)) = v(x) if g GQp ; 1 (vii) v (0, p r] ((x)) = pv (0, r] (x). Proof. Exercise.
+
Lemma 8.3.3. Given x
k=0
pk [xk ] A, the following conditions are equiv
alent: (i) (ii)
+
+ pk [xk ] converges in BdR ;
k=0 + k=0
pk xk converges in Cp ;
(0)
(iii) lim (k + vE (xk )) = +;
k+
(iv) x A(0, 1] . Proof. (iii) (iv) is by definition of A(0, r] . (ii) (iii) is by definition of + vE . (i) (ii) is by the continuity of : BdR Cp . So it remains to show (ii) (i). ~ Write p = (p, p1/p , · · · ) E + , then = [p]p is a generator of Ker A+ . We know ak = k + [vE (xk )] + if k +. Write xk = pkak yk , then yk E + . We have p [xk ] =
k
p p
k
[p]ak [yk ] = pak (1 + )ak k [yk ]. p
8.3. OVERCONVERGENT (, )MODULES
143
Note that pk (1 + p )ak k pak m A+ + (Ker )m+1 for all m. Thus, ak + + implies that pk [xk ] 0 BdR /(Ker )m+1 for every m, and therefore also in + + BdR by the definition of the topology of BdR . + Remark. We just proved A(0,1] := BdR A, and we can use
n : A(0,p
+ to embed A(0,r] in BdR , for r pn .
n ]
 A(0, 1]
Define A :=
r>0 + + A(0, r] = {x A : n (x) converges in BdR for n
0}.
Lemma 8.3.4. x
k=0
pk [xk ] is a unit in A(0,r] if and only if x0 = 0 and
vE ( xk ) >  k for all k 1. x0 r Proof. Exercise. Just adapt the proof of Gauss Lemma. Set 1 pn A(0,r] , B (0, r] = A(0,r] [ ] = p nN B =
r>0
endowed with the topology of inductive limit, and B (0, r] ,
again with the topology of inductive limit. Theorem 8.3.5. B is a subfield of B, stable by and GQp , both acting continuously. B is called the field of overconvergent elements. We are going to prove elements of D(V )=1 are overconvergent. Definition 8.3.6. (i) B = B B, A = A B (so B is a subfield of B stable by and GQp ), A(0, r] = A(0, r] B. (ii) If K/Qp is a finite extension and {A , B , A , B , A(0, r] , B (0, r] }, (0, r] define K = HK . For example AK = A(0, r] AK . (iii) If {A, B, A , B , A(0, r] , B (0, r] }, and n N, define K,n = n (K ) B.
144
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
(0, r]
We now want to make AK more concrete. Let F K be the maximal ¯ unramified extension of Qp , K be a uniformizer of EK = kF ((¯K )), PK ¯ + EF [X] be a minimal polynomial of K . Let PK AF [X] (note that A+ = ¯ F ¯ OF [[]]) be a lifting of PK . By Hensel's lemma, there exists a unique K AK such that PK (K ) = 0 and K = K mod p. If K = F , we take K = . ¯ Lemma 8.3.7. If we define rK = 1, (2vE (dEK /EQp ))1 , if EK /EQp is unramified, otherwise . for all 0 < r < rK .
then K and PK (K ) are units in AK
(0, r]
Proof. The proof is technical but not difficult and is left to the readers. Proposition 8.3.8. (i) AK = {
nN n an K : an OF , lim vp (an ) = +}; n
(ii)
(0, r] AK
= {
nN
n an K
: an OF , lim (vp (an ) + rnvE (¯K )) = +}.
n
So f f (K ) is an isomorphism from bounded analytic functions on the (0, r] annulus 0 < vp (T ) rvE (¯K ) to the ring BK . Proof. The technical but not difficult proof is again left as an exercise. See CherbonnierColmez Invent. Math. 1998. Corollary 8.3.9. (i) AK is a principal ideal domain; (0, r] (ii) If L/K is a finite Galois extension, then AL is an ´tale extension e (0, r] of AK if r < rL , and the Galois group is nothing but HK /HL . Define n = n (), K,n = n (K,n ). Proposition 8.3.10. (i) If pn rK 1, (K,n ) is a uniformizer of Kn ; + (ii) K,n Kn [[t]] BdR . Proof. First by definition
+ n = [1/p ]  1 = (n) et/p  1 Fn [[t]] BdR
n
(0, r]
(for [1/p ] = (n) et/p : the value of both sides is (n) , and the pn th power of both side is [] = et (recall t = log[])). This implies the proposition in the unramified case.
n
n
8.3. OVERCONVERGENT (, )MODULES
145
For the ramified case, we proceed as follows. By the definition of EK , K,n = (K,n ) is a uniformizer of Kn mod a = 1 {x : vp (x) p }. Write n be the image of K,n in Kn mod a. So we just have to prove K,n Kn . Write
d
PK (x) =
i=0
ai ()xi , ai () OF [[]].
d
Define PK,n (x) =
ai (n )xi , i=0 1 p
n
then PK,n (K,n ) = ((PK (K ))) = 0. Then we have vp (PK,n (n )) vp (PK,n (n )) =
and
1 1 1 v (PK (¯K )) = n vE (dEK /EQp ) < if pn rK > 1. n E p p 2p
Then one concludes by Hensel's Lemma that K,n Kn . For (ii) , one uses Hensel's Lemma in Kn [[t]] to conclude K,n Kn [[t]].
+ Corollary 8.3.11. If 0 < r < rK and pn r 1, n (AK ) Kn [[t]] BdR . (0, r]
8.3.2
Overconvergent representations
D(0, r] := (A(0, r] Zp V )HK D(V ).
Suppose V is a free Zp representation of rank d of GK . Let
This is a AK module stable by K . As for , we have : D(0, r] (V )  D(0, p
1 r]
(0, r]
(V ).
Definition 8.3.12. V is overconvergent if there exists an rV > 0, rV rK such that (0, rV ] AK (V )  D(V ). (0, rV ] D
AK
By definition, it is easy to see for all 0 < r < rV , D(0, r] (V ) = AK
(0, r] AK
(0, rV ]
D(0, rV ] (V ).
146
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
Proposition 8.3.13. If V is overconvergent, then there exists a CV such that if K , n() = vp (log(())) and r < inf{p1 rV , pn() }, then  1 is invertible in D(0, r] (V )=0 and v (0, r] ((  1)1 x) v (0, r] (x)  CV  pn() vE (¯ ).
p1
Proof. Write x =
i=1
[]i (xi ) and adapt the proof of the same statement as
n 1
in the characteristic p case. One has to use the fact that []ip (0, r] AK if r < pn and i Z . p Remark. This applies to (AK )=0 .
(0, r]
is a unit in
Theorem 8.3.14 (Main Theorem). (i) All (free Zp or Qp ) representations of GK are overconvergent. (ii) D(V )=1 D(0, rV ] (V ). Sketch of Proof. (ii) is just because improves convergence. (i) follows from Sen's method applied to = A(0, 1] , v = v (0, 1] , G0 = GK , HK,n = n (AK ). Now we show how to check the three conditions. (TS1). Let L K Qp be finite extensions, for = [¯L ]( then for all n, (n ()) = 1,
HK /HL (0, 1]
HK /HL
([¯L ]))1 ,
and
n+ (0,1]
lim v (0, 1] (n ()) = 0.
(TS2). First HK ,n = AK,n . Suppose pn rK 1. We can define RK, n by the following commutative diagram: RK,n : A(0,1] K
O ? / A(0,1] K,n x; xx x xx xx n n+k n+k
(0,1] AK,n+k
8.3. OVERCONVERGENT (, )MODULES
147
One verifies that n n+k n+k does not depend on the choice of k, using the fact = Id. Then the proof is entirely parallel to that for Cp with in the role of p1 TrFn+1 /Fn and n+k in the role of n+k . (TS3). For an element x such that RK,n (x) = 0, write
+
x=
i=0
RK,n (x), where RK,n (x) (n+i+1) ((AK
(0,p(n+i+1) ] =0
)
).
Then just apply Proposition 8.3.13 on (AK )=0 . (0,1] Now Sen's method implies that there exists an n and a AK,n module (0, 1] DK,n A(0, 1] V such that A(0, 1] A(0, 1] DK,n  A(0, 1] V.
K,n
(0,p(n+i+1) ]
(0, 1]
Play with (TS3) just like Lemma 8.2.8, one concludes that DK,n n (D(V )) n (0, 1] and n (DK,n ) D(0, p ] (V ). We can just take rV = n.
(0, 1]
8.3.3
padic Hodge theory and (, )modules
Suppose we are given a representation V , 0 < r < rV and n such that pn r > 1. Then we have
+ n (D(0, r] (V )) BdR V  Cp V
and
(0, r]
n (AK ) Kn [[t]]  Kn . So we get the maps n : Kn A(0, 1] D(0, r] (V )  Cp V
K
(8.1)
and
+ n : ti Kn [[t]] A(0, r] D(0, r] (V )  ti BdR V, i Z.
K
(8.2)
Theorem 8.3.15. There exists an n(V ) N such that if n n(V ), then we have (i) the image of n in (8.1) is exactly DSen, n (V ); (ii) Fili DdR (V ) = (Im n )K in (8.2) for all i; (iii) DdR (V ) = Kn ((t)) A(0, r] D(0, r] (V ) K .
K
148
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
Let K/Qp be a finite extension, and define
BK = {F (K ) : F is a bounded analytic function on 0 < vp (t) r(F ), r(F ) > 0}, Brig, K = {F (K ) : F is an analytic function on 0 < vp (t) r(F ), r(F ) > 0}
(this last ring is the Robba ring in the variable K ), and
Blog, K = Brig, K [log K ]. Extend , K by continuity on Brig, K , and set
(log K ) = p log K + log
(K ) , p K (K ) (log K ) = log K + log K
where log (K ) BK and log (K ) Brig, K . Let K p
K
N = Theorem 8.3.16 (Berger). For
d 1 · . vE (¯K ) d log K
D (V ) = (B V )HK =
r>0
D(0, r] (V ),
if V is semistable, then
Blog, K
1 1 K0 Dst (V ) = Blog, K B D (V ) K t t
is an isomorphism of (, N, K )modules. This implies that Dst (V ) is the invariant under K .
8.3.4
A map of the land of the rings
The following nice picture outlines most of the objects that we have discussed till now and that we shall have to discover more about in the future.
8.4. EXPLICIT RECIPROCITY LAWS AND P ADIC LFUNCTIONS149
padic Hodge o Theory ~ Blog o
O
_ _ _
n
(, )modules O
1 p
~ Brig o
O _ _ / ~ o B O
B O A
~ BO o
~ / A o
~ / E o _ _
~ A o ~ E o
mod p
E where
B+ O 7 8 dR 7 77 7 77 ~ + 777 / B+ 77 Blog O st O 77 77 77 77 / B + B+ ~ 77 cris rig O 77 7 / Cp ~+ BO O / OCp ~ A+ / OC /p ~ E+ p _ _ _
~+ Brig =
n
+ n (Bcris ),
~+ Blog =
n
+ n (Bst ).
Note that most arrows from (, )modules to padic Hodge theory are in the wrong direction, but overconvergence and Berger's theorem allow us to go backwards.
8.4
8.4.1
Explicit reciprocity laws and padic Lfunctions
Galois cohomology of BdR
Suppose K is a finite extension of Qp . Recall that we have the following: Proposition 8.4.1. For k Z, then (i) if k = 0, then H i (GK , Cp (k)) = 0 for all i (ii) if k = 0, then H i (GK , Cp ) = 0 for i 2, H 0 (GK , Cp ) = K, and H 1 (GK , Cp ) is a 1dimensional Kvector space generated by log H 1 (GK , Qp ). (i.e, the cup product x x log gives an isomorphism H 0 (GK , Cp ) H 1 (GK , Cp )).
150
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
Remark. This has been proved for i 1. For i 2, H i (HK , Cp (k)) = 0 by using the same method as for H 1 . Then just use the exact sequence 1  HK  GK  K  1 and HochschildSerre spectral sequence to conclude. Proposition 8.4.2. Suppose i < j Z {±}, then if i 1 or j 0,
+ + H 1 (GK , ti BdR /tj BdR ) = 0;
if i 0 and j > 0, then x x log gives an isomorphism
+ + H 0 (GK , ti BdR /tj BdR )( + + K)  H 1 (GK , ti BdR /tj BdR ).
Proof. Use the long exact sequence in continuous cohomology attached to the exact sequence 0  ti+n Cp (
+ + + + Cp (i + n))  ti BdR /tn+i+1 BdR  ti BdR /ti+n BdR  0,
+ + and use induction on j i (note that in the base step j = i+1, ti BdR /tj BdR = Cp (i)), and Proposition 8.4.1 to do the computation. This concludes for the case where i, j are finite. For the general case, one proves it by taking limits.
8.4.2
BlochKato's dual exponential maps
Let V be a de Rham representation of GK , we have BdR Qp V BdR K DdR (V ) = H 0 (GK , BdR V ) = and H 1 (GK , BdR V ) = H 1 (GK , BdR K DdR (V )) = H 1 (GK , BdR ) K DdR (V ). So we get an isomorphism DdR (V )  H 1 (GK , BdR V );
x x log .
8.4. EXPLICIT RECIPROCITY LAWS AND P ADIC LFUNCTIONS151 Definition 8.4.3. The exponential map exp is defined through the commutative diagram: exp : H 1 (GK , V )
RRR RRR RRR RRR R( / DdR (V ) mmmm6 mm mmm mmm mmmmmmmm vm m mm
H 1 (GK , BdR V )
Proposition 8.4.4. (i) The image of exp lies in Fil0 DdR (V ). (ii) For c H 1 (GK , V ), exp (c) = 0 if and only if the extension Ec 0  V  Ec  Qp  0, is de Rham as a representation of GK . Proof. (ii) is just by the definition of de Rham. For (i), c H 1 (GK , V ) + implies c = 0 H 1 (GK , (BdR /BdR ) V ). But x x log gives an isomorphism
+ + DdR (V )/ Fil0 (DdR (V ))(= H 0 (GK , (BdR /BdR )V ))  H 1 (GK , (BdR /BdR )V )).
So exp (c) = 0 (mod Fil0 ) Remark. exp is a very useful tool to prove the nontriviality of cohomology classes. Now suppose k Z, L is a finite extension of K. Then V (k) is still de Rham as a representation of GL . Define DdR, L (V (k)) := H 0 (GL , BdR V (k)) = tk L K DdR (V ) by an easy computation. Thus, Fil0 (DdR, L (V (k))) = tk K Filk DdR (V ) and this is 0 if k 0. So for every k Z, for L/K finite, exp : H 1 (GL , V (k))  tk L K DdR (V ) is identically 0 for k 0.
152
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
8.4.3
The explicit reciprocity law
Recall that
1 HIw (K, V )  H 1 (GK , Zp [[K ]] V ) = H 1 (GK , D0 (K , V )).
If : K Q is a continuous character, for n N, p µ H 1 (GK , D0 (K , V )) 
Kn
µ H 1 (GKn , V ).
where we write V , not as V () to distinguish from V (k) = V k . Then exp (
Kn
k µ) tk Kn K DdR (V )
and is 0 if k 0. Recall also that we have the isomorphism Exp : H 1 (K, V )  D(V )=1 , that D(V )=1 D(0, rV ] (V ) and that there exists n(V ) such that n (D(0, rV ] (V )) Kn ((t)) K DdR (V ), for all n n(V ). Now denote by TrKn+k /Kn = TrKn+k ((t))/Kn ((t)) Id : Kn+k ((t))DdR (V ) Kn ((t))DdR (V ). Theorem 8.4.5 (Explicit Reciprocity Law). Let V be a de Rham represen1 tation of GK and µ HIw (K, V ). (i) If n n(V ), then pn n (Exp (µ)) =
kZ
exp (
Kn
k µ).
(ii) For n N, n + i n(V ), then Exp Kn (µ) := TrKn+i /Kn p(n+i) (n+i) (Exp (µ)) does not depend on i, and Exp Kn (µ) =
kZ
exp (
Kn
k µ).
8.4. EXPLICIT RECIPROCITY LAWS AND P ADIC LFUNCTIONS153 Proof. (ii) follows from (i) and from the commutative diagram: H 1 (GL2 , V )   L2 K DdR (V )  Tr K Id cor
L2 /L1
exp
H 1 (GL1 , V )   L1 K DdR (V )  where L1 L2 are two finite extensions of K. For (i), suppose y = Exp (µ), x D(V ), and x(k) is the image of x in D(V (k)) = D(V )(k) (Thus, (x(k)) = (x)(k) and (x(k)) = ()k (x)(k)). The integral K k µ is represented by the cocycle:
n
exp
g1 log (n ) · y(k)  (g  1)b n p n  1 where b A V is the solution of g cg = (  1)b = (n  1)1 (  1)(y)(k) . From y D(0, rV ] (V )=1 one gets (  1)y D(0, p and then (n  1)1 (  1)y D(0, p
1n 1 r V
]
(V )=0
n ]
(V )=0 .
Thus b A(0, p ] V . This implies that n (b) and n (y) both converge + in BdR V . Then cg = n (cg ) differs from log (n ) g  1 · · n (y)(k) pn n  1 by the coboundary (g  1)(n (b)). Therefore, they have the same image in + H 1 (GKn BdR V (k)). Write cg = pn n (y) =
ii0
yi ti ,
yi Kn K DdR (V ),
then cg = log (g)yk t
k
(g)i+k  1 · yi ti + i+k  1 (n ) i=k yi ti . ((n )i+k  1) i=k
= log (g)yk tk + (g  1) So we get exp
Kn
k µ = yk tk .
154
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
8.4.4
Cyclotomic elements and CoatesWiles morphisms.
n lim OFn , 1+n n1  T . Then we have 1+T 1 (u) HIw (Qp , Qp (1)),
Let K = Qp , V = Qp (1), u = the Coleman power series fu =
Exp ((u)) = (1 + T ) Note that
dfu 1 () = . dT 1 , (1 + 1 )et/p  1
1 ()1 = ([1/p ]  1)1 = then Exp Qp ((u)) = = =
n=1
1 1 1 TrQp (1 )/Qp 1 ()1 = t/p  1 p p zp =1, z=1 e et 1 1 1 1 t t/p  · t/p = · t  t/p 1 p e 1 t e 1 e 1 (1  pn )(1  n) (t)n1 . (n  1)! if k 0; if k 1.
+
So exp
Qp
k (µ) =
0, k1 , (1  pk )(1 + k) (t) (k1)!
Remark. (i) The map
1 lim OFn  {0}  HIw (Qp , Qp (1))  Qp , 
u tk+1 exp
Qp
k (u)
is the CoatesWiles homomorphism. (ii) Since (1 + k) = 0 if k 1 is even, the above formula implies that the extensions of Qp by Qp (k + 1) constructed via cyclotomic elements are nontrivial and are even not de Rham. (iii) dimQp H 1 (GQp , Qp (k)) = 1 if k = 0, 1. Corollary 8.4.6. Nontrivial extensions of Qp by Qp (k) are not de Rham if k 0 is odd. Exercise. (i) Prove that this is also true for k 1 even by taking a general element of D(Qp (1))=1 . (ii) For [K : Qp ] < , prove the same statement.
8.4. EXPLICIT RECIPROCITY LAWS AND P ADIC LFUNCTIONS155
8.4.5
Kato's elements and padic Lfunctions of modular forms.
Now we come to see the relations with modular forms. Suppose f=
n=1
an q n Sk (N ), k 2
is primitive. So Q(f ) = Q(a1 , · · · , an , · · · ) is a finite extension of Q, and Qp (f ) = Qp (a1 , . . . , an , . . . ) is a finite extension of Qp . Theorem 8.4.7 (Deligne). There exists a representation Vf of GQ of dimension 2 over Qp (f ), nonramified outside N p, such that if N p, for the 2i 2i arithmetic Frobenius at ( (e pn ) = e pn ), then det(1  X1 ) = 1  a X +
k1
X 2.
Remark. A Qp (f )representation V of dimension d is equivalent to a Qp representation of dimension d · [Qp (f ) : Qp ] endowed with a homomorphism Qp (f ) End(V ) commuting with GQ . Therefore, Dcris (V ), Dst (V ), DdR (V ) are all Qp (f )vector spaces. Theorem 8.4.8 (FaltingsTsujiSaito). (i) Vf is a de Rham representation of GQp with HodgeTate weights 0 and 1  k, the 2dimensional Qp (f )vector space DdR (Vf ) contains naturally f , and
0 k i DdR (Vf ) = DdR (Vf ), DdR (Vf ) = 0, DdR (Vf ) = Qp (f )f if 1 i k  1.
(ii) If p N , then Vf is crystalline and det(X  ) = X 2  ap X + pk1 . If p N but ap = 0, then Vf is semistable but not crystalline and ap is the eigenvalue of on Dcris (V ); if ap = 0, then Vf is potentially crystalline.
1 Remark. If V is a representation of GK , µ HIw (K, V ),
k µ H 1 (GKn , V (k)),
Kn
then this is also true for aK k µ for all a K and for n : K Zp being constant modulo Kn .
K
(x)k µ, with
156
CHAPTER 8. (, )MODULES AND P ADIC LFUNCTIONS
1 Theorem 8.4.9 (Kato). There exists a unique element zKato HIw (Qp , Vf ) (obtained by global methods using Siegel units on modular curves), such that if 0 j k  2, is locally constant on Z Qp with values in Q(f ), then p =
exp
Zp
(x)xk1j · zKato =
1 f (f, , j + 1) · k1j j! t
where (f, , j + 1) Q(f, µpn ), f tk1j Fil0 (DdR Vf (k  1  j)) .
Our goal is to recover Lp, (f, s) from zKato (recall Lp, is obtained from µf, Dvp () (Zp ) before). We have Exp (zKato ) D(Vf )=1 , but the question is how to relate this to Dcris (Vf ), Dst (Vf ). If p  N , let be a root of X 2  ap X + pk1 with vp () < k  1; if p N , let = ap = 0 (in this case p2 = pk1 ). In both cases, take = pk1 1 . Thus, , are eigenvalues of on Dst (Vf ). Assume = (which should be the case for modular forms by a conjecture). Define =  to be the projection on the eigenspace in Dst (Vf )  and extend it by Blog, K linearity to
Blog, K
1 K0 Dst (Vf )  Blog, K B D (Vf ). K t
Theorem 8.4.10. (i) (f ) = 0; (ii) Exp (zKato ) =
Zp
[]x µf,
(f ) . tk1
Remark. µf, exists up to now only in the semistable case, but zKato exists all the time. So a big question is: How to use it for padic Lfunction?
Information
164 pages
Report File (DMCA)
Our content is added by our users. We aim to remove reported files within 1 working day. Please use this link to notify us:
Report this file as copyright or inappropriate
737769