`Fontaine's rings and p-adic L-functionsPierre 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 p-adic Lfunctions using the theory of (, )-modules of Fontaine. This construction is an adaptation of an idea of Perrin-Riou. 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 p-adic Hodge theory which are used freely in these notes, which means that they are not entirely self-contained. 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ContentsI Classical p-adic L-functions: zeta functions and modular forms 11 The 1.1 1.2 1.3 p-adic zeta function of Kubota-Leopoldt The Riemann zeta function at negative integers . . . . . . . p-adic 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 p-adic 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 p-adic 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 30 311.41.51.61.71.81.9ivCONTENTS 1.9.1 Analytic functions inside C r functions . . . . . . . . . . 31 1.9.2 Distributions of order r . . . . . . . . . . . . . . . . . . 33 1.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 Modular forms 2.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 The upper half-plane . . . . . . . . . . . . . . 2.1.2 Definition of modular forms . . . . . . . . . . 2.1.3 q-expansion 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 p-adic L-functions of modular forms 3.1 L-functions 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 L-functions . . . 3.1.4 Euler products . . . . . . . . . . . . 3.2 Higher level modular forms . . . . . . . . . . 3.2.1 Summary of the results . . . . . . . . 3.2.2 Taniyama-Weil Conjecture . . . . . . 3.3 Algebraicity of special values of L-functions 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 3.4vp-adic L-functions of modular forms . . . . . . . . . . . . . . . 77IIFontaine's rings and Iwasawa theory. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8385 85 85 87 89 91 91 92 95 95 98 100 103 104 1054 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 Euler-Poincar´ 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 Euler-Poincar´ 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 Kubota-Leopoldt 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 7.5.5CONTENTS Tate's normalized trace maps . . . . . . . . . . . . . . 127 Applications to Galois cohomology . . . . . . . . . . . 130 No 2i in Cp ! . . . . . . . . . . . . . . . . . . . . . . . 1318 (, )-modules and p-adic L-functions 133 8.1 Tate-Sen'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 p-adic representations . . . . . . . . . . 140 8.3 Overconvergent (, )-modules . . . . . . . . . . . . . . . . . . 141 8.3.1 Overconvergent elements . . . . . . . . . . . . . . . . . 141 8.3.2 Overconvergent representations . . . . . . . . . . . . . 145 8.3.3 p-adic Hodge theory and (, )-modules . . . . . . . . 147 8.3.4 A map of the land of the rings . . . . . . . . . . . . . . 148 8.4 Explicit reciprocity laws and p-adic L-functions . . . . . . . . 149 8.4.1 Galois cohomology of BdR . . . . . . . . . . . . . . . . 149 8.4.2 Bloch-Kato's dual exponential maps . . . . . . . . . . . 150 8.4.3 The explicit reciprocity law . . . . . . . . . . . . . . . 152 8.4.4 Cyclotomic elements and Coates-Wiles morphisms. . . 154 8.4.5 Kato's elements and p-adic L-functions of modular forms.155Part I Classical p-adic L-functions: zeta functions and modular forms1Chapter 1 The p-adic zeta function of Kubota-Leopoldt1.1 The Riemann zeta function at negative integersWe first recall the definitions of Riemann zeta function and the classical Gamma function:+(s) = (s) =0n-s =n=1 + p(1 - p-s )-1 , if Re (s) &gt; 1. dt , if Re (s) &gt; 0. te-t tsThe -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 KUBOTA-LEOPOLDT Then we have the following formulas: n-s = 1 (s)+e-nt ts0 + + 0 n=1dt , t 1 t = t (s)+ 01 (s) = (s)e-nt s dt1 s dt t . et - 1 tLemma 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)ts0dt , Re(s) &gt; 0 thas 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) &gt; 0, L(f1 , s) = 1 ts 1 [f1 (t) ]|+ - 0 (s) s s(s)+f1 (t)ts+10 (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+10f1(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 -1Note thatf (t) =0Bntn , n!1.2. P -ADIC BANACH SPACES where Bn  Q is the n-th 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)+5f (t)ts-10dt 1 = L(f, s - 1), t s-1so 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+1Theorem 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)pk-1 , then (1 - a1+n1 )(-n1 )  (1 - a1+n2 )(-n2 ) mod pk .1.2p-adic Banach spacesDefinition 1.2.1. A p-adic Banach space B is a Qp -vector space with a lattice B 0 (Zp -module) separated and complete for the p-adic topology, i.e., B0nNlim B 0 /pn B 0 .  -6CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT 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 = p-vB (x) defines a norm on B, such that B is complete for and B 0 is the unit ball.BExample 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))].xZDefinition 1.2.3. A Banach basis of a p-adic 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  0iIwhen i  ; equivalently for any C, the set {i | vp (xi )  C} is a finite set. (ii) vB (x) = inf vp (xi ).iITheorem 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 p-adic Banach spaces with Banach basis (ei )iI and (fj )jJ respectively, then B B is a p-adic Banach space with Banach basis (ei  fj )(i,j)I×J . Thus for all x  B B , x =i,jxi,j ei  fj yj  fjj(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 ).71.31.3.1Continuous functions on ZpMahler's coefficientsWe have the binomial function:  1, x = x(x - 1) · · · (x - n + 1)  n , n! Lemma 1.3.1. vC 0 (x nif n = 0, if n  1.) = 0.x nx 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 [k-1] (x) = f [k] (x + 1) - f [k] (x)and write the Mahler's coefficient an (f ) = f [n] (0). Hence:nf (x) =i=0 n[n](-1)i (-1)ii=0n f (x + n - i), i n f (n - i). ian (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=0an (f )x n,(iii) vC 0 (f ) = inf vp (an (f )).8CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT 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=0anx 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=0an+kx . nHence 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 havepk -19f[pk ](x) = f (x + p ) - f (x) +i=1kk(-1)ipk k f (x + pk - i) + (1 + (-1)p )f (x). ikNow 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.1.3.2Locally constant functions.Choose a z  Cp , such that vp (z - 1) &gt; 0. Then+fz (x) =n=0x (z - 1)n  C 0 (Zp , Cp ). nNote 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=01+n1 2(z - 1)n . z =16 ,z 9-1 =7 , 9the seriesconverges 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) =n1 &gt; 0. (p - 1)pn-1Note 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 =11 pnz -i z xz p =1npn 0if x  pn Zp ; if not.10CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT 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 ), letpk -1fk =i=0f (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 ).1.41.4.1Measures on ZpThe Amice transformDefinition 1.4.1. A measure µ on Zp with values in a p-adic 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 L-vector space, then µ extends by continuity and L-linearity 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) &gt; 0, Aµ (z - 1) =Zpz x µ(x).1.4. MEASURES ON ZP Proof. Since z x = can exchange+ n=011x n(z - 1)nwith normal convergence in C 0 (Zp , Qp ), oneand .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=0bn T n , bn bounded, and bn  Qp } with the valuation v(++bn T n ) =n=0inf nN vp (bn ). Proof. On one hand, for all µ  D0 (Zp , Qp ), write Aµ (T ) = bn (µ) =Zp x nbn (µ)T n , thenµ. Since vC 0 (x nn=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=0bn an (f )(by Mahler'stheorem, an (f )  0) gives a measure µb whose Amice transform is+Aµb (T ) =n=0Tn Zpx µb = n x )= i++T (n=0 i=0nx b i ai ( )) = n+bn T nn=0since an ( Hence+1 if i = n, 0 otherwise.vp (n=0bn 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 KUBOTA-LEOPOLDT 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 functions0 pn -1 i=0fn =f (i)1i+pn Zp  f in C 0 .ZpNow if µ  D0 (Zp , Qp ), set µ(i + pn Zp ) = given by the following &quot;Riemann sums&quot;pn -11i+pn Zp µ. ThenZpf µ isf µ = limZpnf (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 .1.4.2examples 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 .Zpf (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 Zpf (x)g(x)µfor 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 n13the Amice transform is+Axµ =n=0 +TnZpx xµ n x µ+n n+1 x µ n=n=0T n (n + 1)ZpZpd Aµ . dT (ii) Let f (x) = z x , vp (z - 1) &gt; 0. For any y, vp (y - 1) &gt; 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) = p-nz p =1nhence AResi+pn Zp µ (T ) = p-nz pn =1z -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 Zpf (px)µ.Hence+A(µ) (T ) =n=0TnZppx µ = Aµ ((1 + T )p - 1) = (Aµ (T )) nwhere  : T  (1 + T )p - 1 (compare this formula with (, )-modules). We define the action of  by f (x)(µ) =Zpx f ( )µ. p Zp14CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT Then A(µ) = (Aµ ) where (F )((1 + T )p - 1) = 1 F ((1 + T )z - 1). p zp =1The 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(Zpf (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) &gt; 0, then zx  µ =Zp Zpz x µ(x)Zpz y (y),thus Aµ = A Aµ .1.5. THE P -ADIC ZETA FUNCTION151.51.5.1The p-adic zeta functionKummer'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 - 1Proof. This follows from Theorem 1.4.5 and the fact a = (1 + T )a - 1 since a-1a na n=1 a nTn=1 · T 1+ n=21 a-1a nT n-11 + Zp [[T ]] T Zp . Moreover, we have vD0 (a ) = 0.ZpProposition 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).et1 a - at , -1 e -1then fa is in C  on R+ and rapidly decreasing. Hence+ 1 dt L(fa , s) = fa (t)ts = (1 - a1-s )(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 px n a = (Zpd n ) ( dtetx a )|t=0 = (Zpd n (n) ) Aa (et - 1)|t=0 = fa (0). dtCorollary 1.5.3. For a  Z , k  1 (k  2 if p = 2), n1 , n2  k, n1  p n2 mod (p - 1)pk-1 , then vp ((1 - a1+n1 )(-n1 ) - (1 - a1+n2 )(-n2 ))  k.16CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT 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 ).ZpFrom 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)pk-1 and p n1 - n2 is divisible by (p - 1)pk-1 . Remark. The statement is not clean because of x  pZp .1.5.2Restriction to Z . p1 1 Lemma 1.5.4. ( T ) = T . 1 Proof. Let F (T ) = ( T ), thenF ((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 - 1Proposition 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 ( ) a-1 T (1 + T ) T Twhere 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).p17Remark. 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 p-adic continuity, one has to modify  by some &quot;Euler factor at p&quot;. (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.1.5.3Leopoldt's -transform.Lemma 1.5.8. (i) Every x  Z can be written uniquely as x = (x) x , p with (x)  µ(Qp ) = {±1} µp-1 , 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 = [¯]. xnRemark. (i)  is the so-called 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 p-adically 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 Zpusy  (y) = A(i) (us - 1).(i)18CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT Proof. We have (x)i x s (x)Z p=µ(Qp )()i ()i+2pZpx s (x)=µ(Qp )1+2pZpx 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+2pZpf ((x))-1 .Now xs= exp(s log x) = exp(s log uy) = usy and hence ()i x s (x) =1+2pZp µ(Qp )()iZpusy  (-1 · ),µ(Qp )we just set  =µ(Qp )(i)()i  (-1 · ).Definition 1.5.10. p,i (s) = -1 1 - (a)1-i a1-s(x)-i xZ p-sa (x).Proof of Theorem 1.5.7. If n  -i mod p - 1, then p,i (-n) = 1 1 - (a)1-i a 1+n 1 = 1 - (a)1+n a 1+n = (1 - p-n )(-n). The function p,i is analytic if (a)1-i = 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 FUNCTIONS19Remark. (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 &gt; 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) = 0.1.61.6.1C k functionsDefinition.Let f : Zp  Qp be a given function. We define f {0} (x) = f (x) {i} f (x, h1 , · · · , hi ) 1 = (f {i-1} (x + hi , h1 , · · · , hi-1 ) - f {i-1} (x, h1 , · · · , hi-1 )) hi 1 = hj )) ( (-1)i-|I| f (x + h1 · · · hi jII{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]if (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:ivp (f{i}(x, h1 , · · · , hi ))  vC 0 (f ) -j=1vp (hj ).20CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT 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=0p2n an (x), then vp (f (x) - f (y)) = 2vp (x - y). Hencef (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 ) = mini 0ik (x,h1 ,··· ,hi )Zi+1 pinfvp (f {i} (x, h1 , · · · , hi )). nj = n, nj  1}Let L(n, k) = max{j=1vp (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=0ann+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=0an (f )x nand 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 FUNCTIONS211.6.2Mahler'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 Nak0 ,··· ,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)i-|I| gT (x +jI ihj ))I{1,··· ,i}= (1 + T )xx n 1 x x n 1 x-1 n n-1(1 + T )hj - 1 hj j=1{i}  n=0Let 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 11 x0 n · · · ni n0 +···+n =n, 1h1 - 1 hi - 1 ··· . n1 - 1 ni - 122CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT Let{i} Qn,i (x0 , · · · , xi ) = Pn (x0 , x1 + 1, · · · , xi + 1) 1 x0 = n · · · ni n0 n +n +···+n =n, 10n1 ,··· ,ni 1 +1ix1 xi ··· . n1 - 1 ni - 1For all f  C 0 (Zp , Qp ), we have f (x) =+an (f )n=0x n. We denotegi (x0 , · · · , xi ) =n=0an (f )Qn,i (x0 , · · · , xi )if xj + 1 = 0, j  1. We have:+an0 ,n1 -1,··· ,ni -1 (gi ) =n=0an (f )an0 ,n1 -1,··· ,ni -1 (Qn,i )where an0 ,n1 -1,··· ,ni -1 (Qn,i ) =  0    1   n1 ···niiif n =j=0 inj , nj .j=0if 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 =ni+n=0 n0 +n1x1 xi ··· n1 - 1 ni - 1defines 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 FUNCTIONS231.71.7.1locally analytic functionsAnalytic 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=0an T n . Then:(i) If x0  OCp , then f (k) (x0 ) converges for all k andnlim vp (f (k) (x0 )) = . k!(ii) If x0 , x1 are in OCp , then+f (x1 ) =n=0f (n) (x0 ) (x1 - x0 )n n!andf (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 nwhere (i.e.,outside a finite number of xi + mCp ). Proof. (i)f (k) k! +=n=0an+kn+k kT n . Let T = x0 ; since vp (n+k k)  0, vp (xn )  00, 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 = 1n=0 + + n=0an (k=0n (x1 - x0 )k xn-k ) 0 k+(k=0 n=0ann n-k x0 )(x1 - x0 )k = kn=0f (n) (x0 ) (x1 - x0 )n . n!So we can exchange the the roles of 0 and x0 to getnNinf vp (f (n) (x0 ) ) = inf vp (an ). nN n!24CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT (iii) That inf vp (an )  inf vp (f (x)) is clear. As vp (an ) goes to +,nN xOCpvp (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 nNFp [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=0an T , g =n=0nbn T , then f g =n=0ncn T n ,where cn =i=0ai bn-i . Suppose that vp (an ) and vp (bn ) go to infinity when nn n ngoes 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) (xn!) + 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 FUNCTIONS251.7.2Locally 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 SS 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 )m-1 , 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 basisx nProof. The idea is to try to relate the gn = [ pn ]! h (i) First step: For 1  j  ph , we denoten-1 hto the en .n 1 gn,j (x) = gn (-j + p x) = [ h ]! (-j - k + ph x). p n! k=0 If vp (j + k) &lt; 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=0n +n-1#{k : vp (k)  i, 1  k  n}.i=1Since vp (n!) =k=1vp (k) =i=1 hn [ pi ], we have nn vp (n!) - vp ([ h ]!) = p Thus,#{k : vp (k)  i, 1  k  n} =i=1 k=1inf(vp (k), h).nv0 (gn,j ) =k=1 h{0}[inf(vp (j + k - 1), h) - inf(vp (k), h)] ([l=1=j-1 n n+j-1 ] - [ l ] - [ l ]). l p p p26CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT 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 &gt; i, (ii) deg gn,j = m - 1, if j = i, (iii) deg gn,j  m - 1 if j &lt; 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 &gt; i, then j - 1  i. Since [{0} {0}j-1 n n+j-1 ] - [ h ] - [ h ] = m - (m - 1) = 1, h p p pwe have v0 (gn,j )  1, then gn,j = 0. (ii) and (iii):If j  i, writengn,j (x) =k=0ak xk , ak  Zp .The zeros of gn,j are thej+k ,0 ph k  n - 1 and j-1 n+j-1 ] - [ 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 , · · · , m-1 in Zp and m , · · · , n not in Zp . Thenm-1 ngn,j = cl=1(x - l )-1 (1 - l x), (c is a constant ). l=m {0}-1 Since vp (l ) &gt; 0 when l  m, then vp (am-1 ) = vp (c) = v0 (gn,j ). It implies c  Zp . Hence m-1gn,j = cl=1(x - l ).1.8. DISTRIBUTIONS ON ZP It remains to prove v0 (gn,i ) = 0. Sinceh {0} v0 (gn,i ) {0}27=l=1([mph - 1 i-1 mph - i ]-[ l ]+[ ]) pl p pland -[ -i ] = [ i-1 ] + 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 ]!)  (p-1)ph , we have the following theorem: h x is in LA if and only if there n n=0 exists r &gt; 0, such that vp (an ) - rn  + when n  +. Theorem 1.7.10. The function f = an+1.81.8.1Distributions on ZpThe 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) µ|LAh is continuous for all h  N. Set vLAh (µ) = inf (vB (f LAh Zpf 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 KUBOTA-LEOPOLDT Let R+ be the ring of analytic functions defined on D(0, 0+ ) = {x  Cp , vp (x) &gt; 0}. A function f  R+ can be written as f =+ n=0an T n , an  Qpfor all n  N. 1 Let vh = (p-1)ph = vp ( - 1), where  is a primitive ph+1 root of 1.+If F (T ) =n=0bn T n  R+ , we define v (h) (F ) to be v (h) (F ) = v0{vh }(F ) = inf vp (bn ) + nvh .nNThen, 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=0TnZpx µ= n(1 + T )x µ.ZpNote that the last identity in the above definition is only a formal identity here. However, we have Lemma 1.8.3. If vp (z) &gt; 0, thenZp(1 + z)x µ = Aµ (z)Proof. Choose h such that vh &lt; vp (z). Then vp (+zn )  +, [ pn ]! hthereforen=0znx nconverges 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+29x nProof. Let Aµ (T ) =n=0bn T n . Since bn =Zpµ and vp (n!) n , p-1then wehave: 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)phHence Aµ  R+ and v (h) (Aµ )  vLAh (µ). Conversely, for F  R+ , F = vp ([+ + n=0bn T n , then for all h,n n ]!bn ) = vp (bn ) +  . h p (p - 1)phSo f n=0bn an (f ) is a continuous map on LAh . Denote the left hand sidebyZpf 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)p1.8.2Examples of distributions.(i) Measures are distributions and D0  D. (ii) One can multiply a distribution µ  D by g  LA, and one getsd · Axµ = Aµ ,  = (1 + T ) dT ;· Azx µ (T ) = Aµ ((1 + T )z - 1); · AResa+pn Zp µ (T ) = p-nzz -a Aµ ((1 + T )z - 1)pn=130CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT (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=0pnh f (n) (x + y0 ) y - y0 n ( h )  LAh (x) n! pLAh (y),and vLAh ( pnh f (n) (x+y0)n!) goes to +, when n  +. Hence ( f (x + y)µ(x))(y) =Zp ZpfµZpis 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 Ax-1 µ of x-1 µ is a primitive(or called antiderivative) of (1 + T )-1 Aµ , so Ax-1 µ is defined up to 0 ,   Qp (we have x0 = 0).1.8.3Residue at s = 1 of the p-adic zeta function.log(1+T ) . TThe Kubota-Leopoldt distribution µKL given by AµKL (T ) = x µKL =Zp nThend dt d dtn(t=0 n Zpe µKL ) =txd dtnAµKL (et - 1)t=0= Since(t=0ett ) = (-1)n n(1 - n), for all n  N. -11 1 ( ) = and (log(1 + T )) = p log(1 + T ), T T 1 we get (µKL ) = p µKL and xn µKL = (1 - pn-1 )Z p Zpxn µKL = (-1)n n(1 - pn-1 )(1 - n); (x)1-i xZ p 1-sp,i (s) =(-1)i-1 s-1µKL .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 p311 µKL = 1 - p , ( compare withLemma 1.8.6. a+pn Zp µKL = p-n , for all n, for all a  Zp (almost a Haar measure but µ  a = µ). Proof. µKL = p-na+pn Zp z pn =1nz -a AµKL (z - 1) = p-n (1 +z pn =1,z=1log z ), z-1andlog z z-1= 0, if z p = 1, z = 1.1.91.9.1Tempered distributionsAnalytic functions inside C r functionsTheorem 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 hnvC r (f ) = inf (vp (an (f ))-rnlog(1 + n) n log(1 + n) )  vLAh (f )+inf (vp ([ h ]!)-r ). n log p p log pWe have a formula for every a: a a a log(1 + a) vp (a!) = [ ] + · · · + [ h ] + · · ·  - . p p p-1 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 h0bp -1a log(a + 1) - (r + 1) - rh. p-1 log p32CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDTa The function - p-1 + (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) &gt; 0, p let+log x =n=1(-1)n+1 (x - 1)n ; nin 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 so-called 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)n-1 (i) logL x = (x - 1)n , here vp (x - 1) &gt; 0, n n=1 (ii) logL xy = logL x + logL y, (iii) logL = L. Proposition 1.9.3. If r  0, j &gt; r, then xj logL x  C r . Proof. We have+ p-1x logL x =n=0 a=1j1pn 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 DISTRIBUTIONS331.9.2Distributions of order rDefinition 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 Zpf µ) - vC r (f )).Remark. (i) Under the above valuation, Dr (Zp , B) is a p-adic 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 ( hf LAh Zpf 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 (x-a)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 ZpaZp jN,nN(x - a)j µ) - h(j - r))34CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDT Lemma 1.9.6. Suppose F  R+ , F =+ n=0bn 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 nNn ) + rh), (p - 1)phC1 = inf (vp (bn ) + rnNlog(1 + n) ). log pLet h = [ log(1+n) ], then log p vp (bn )  C0 - rh - n log(1 + n)  C0 - r - 2, (p - 1)ph log pn (p-1)phwhich implies C1  C0 - 2. Now, if h is fixed, then C1 - r log(1+n) + log p h (p - 1)p r. Hence, C1 - ris minimal for (1 + n) =n log(p - 1)r log(1 + n) +  C1 - rh - . h log p (p - 1)p log pThus, C0  C1 - r log(p-1)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 &gt; 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)nfor 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,jN35infvp (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 &gt; r - 1, putpn -1 Nfn =i=01i+pn Zp (k=0f (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 ( phk f (h) (i)0ip -1 kN +1infninf vp (pnkf (k) (i) ). k!h!)  vLAh (f ). HencevLAh (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 &gt; 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 KUBOTA-LEOPOLDT SetZpf µ = limn+ vp (ZpZpfn µ, then fn µ), vp ( inf (fn-1 - fn )µ))Zpf µ)  inf(vp (Zpnh vLAh (f ) - rh + (vDr,N (µ) - r). This implies that µ  Dr .1.10SummaryTo 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) =Zpx 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 ) - rnlog(1 + n) ). log p· f  LA if only if there exists r &gt; 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 kN37pkh 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)) &gt; -.n· µ  Dr if and only if vDr (µ) = inf vp (bn (µ)) + r log(1+n) &gt; -. log pn· µ  D if and only if for all r &gt; 0, inf vp (bn (µ)) + rn &gt; -.n +(iv) f =n=0an (f )x n+andfµ = Zpan (f )bn (µ).n=038CHAPTER 1. THE P -ADIC ZETA FUNCTION OF KUBOTA-LEOPOLDTChapter 2 Modular forms2.12.1.1GeneralitiesThe upper half-planeBy 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 + d|2We denote H = {z, Imz &gt; 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 y2is 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40CHAPTER 2. MODULAR FORMS2.1.2Definition of modular formsDefinition 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 dLet Mk (, ) be the set of modular forms of weight k, character  for . Then Mk (, ) is a C-vector 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 ).2.1.3q-expansion 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 &lt;| qM |&lt; 1}.2iz2.1. GENERALITIES41Corollary 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 withM nZ2ny Man = e1 · MM 2f (x + iy)e-M 2-2inx Mdxfor all y. If n &lt; 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 Nan (f )q n , where q(z) = e2iz ,which is called the q expansion of modular forms.2.1.4Cusp 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 ) &gt; 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 e-v (f |k )2y . Theorem 2.1.11. Sk (, ) and Mk (, ) are finite dimensional C-vector 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 ), ).42CHAPTER 2. MODULAR FORMS2.22.2.1The case  = SL2(Z)The generators S and T of SL2 (Z).0 -1 , 1 0 1 n 0 1 1 1 . 0 1Let 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 || &lt; |b|, which implies |µ|  |a|. Then we have µ a  SL2 (Z) and |µ| + || &lt; |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 andCorollary 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| &lt; 1.2.2.2Eisenstein series1 (k) 2 (-2i)k 1  Mk (1), (mz + n)kProposition 2.2.3. If k  3, then Gk  Mk (1), where Gk (z) =m,nand 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)-km,n1 (m az+b cz+d + n)k=m,n1 , ((am + cn)z + (bm + dn))kand (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) =d|n, d1(k) (k) + k-1 (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) =nZCHAPTER 2. MODULAR FORMS1 = (z + n)k^ (l)q llZfor the last identity given by the Poisson summation formula of Fourier transforms, and (by residue computation)+^ (l) =-e-2ilx dx = (x + iy)k0,(-2i)k k-1 l , (k-1)!if l  0, if l  0.It follows that (k) Gk (z) = (k) + (-2i)k m=1+ +ll=1k-1 lmq(k) = (k) + k-1 (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=11 (n)q n is not a modular form, but1 1 (2) 1 ys = lim , 8y 2 (-2i)2 s0 m,n (mz + n)2 |mz + n|2sG is not holomorphic, but G |2  = G , for any   SL2 (Z). 2 2 2 (ii) Let Ek = a0Gk k ) , so that a0 (Ek ) = 1. (G2.2.3The 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)45Figure 2.1: The Fundamental Domain.Proof. Let z0  H, = a b c d  SL2 (Z)0 Since Im (z0 ) = |cz0z+d|2 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 - &lt; 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 |2which 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 + d|2  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 - &lt; 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,46CHAPTER 2. MODULAR FORMSFigure 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.2.2.4Thek 12formula.k 12The following proposition is usually called &quot;the Proposition 2.2.6. Let f  Mk -{0}, then 1 1 v (f ) + vi (f ) + v (f ) + 2 3formula&quot;.vz (f ) =zD-{i,}k . 12Proof. 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 2id log|z|=e-2Man (f )z n = -v (f ),1 0 2i lim1 d log f = - vi (f ), 2 C(i,)2.2. THE CASE  = SL2 (Z)470lim1 2ii1 1 1 d log f = - v (f ) = - v2 (f ) = lim 0 2i 6 6 C(,)d log fC(2 ,)1 ( 2id log f +2 id log f ) =1 2i 1 2i k 2i1 (d log f - d log f (- )) z 2ii=- =-(d log f - d log z k f (z))2 i 2k k dz =- (log i - log 2 ) = . z 2i 12Putting 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 ) - 2a-1 (G6 ) 0does not vanish on D (v () = 1).+Remark. One can prove  = qn=1(1 - q n )24 .2.2.5Dimension 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 &lt; 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48CHAPTER 2. MODULAR FORMSk [ 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) =nZqn2 2=nZein z ,2 = {  SL2 (Z),   I or   S mod 2},  :   {±1}.  () = 1 if   I -1 if   SOne 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}| = 8d|n,4 dd,from which we can deduce that any positive integer can be written as a sum of 4 squares.2.2.6Rationality 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)49in (), compare the coefficients of q and q 2 , we have the following equations: 2 = 2 (1 + 18) = 12921 1 The solution is:  = 240 ,  = 480 . In particular, ,   Q, which implies G4 and G8 have rational q-expansions, 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) =kMk (, A) is an A-algebra.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 dk kis 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(n-1)Mk (SL2 (Z), C)., · · · , n is a basis of Mk (1); if k = 12n+2, Now if k = 12n, G3n , G4 4 3(n-1)+2 3(n-2)+2 G4 G 6 , G4 G6 , · · · , G2 G6 n-1 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 useCHAPTER 2. MODULAR FORMSz 4G (2z) - G ( )  M2 ( , Q) 2 2 2 to deduce(2) 2 Q.Remark. (i) The zeta function  is a special case of L-functions, and (k) are special values of L-functions (i.e. values of L-functions at integers). Siegel used the above method to prove rationality of special values of L-functions for totally real fields. (ii) With a lot of extra work, we can prove integrality results. As (k) k-1 (n)q n , (k) + Gk (z) = (-2i)k n=1 and k-1 (n) =Zp +xk-1 (d|nd ), we have all an (Gk ) are given by measures onZp , therefore a0 (Gk ) is also given by measures. From which we can deduce other constructions of Kubota-Leopoldt zeta functions (the work of Serre, Deligne, Ribet).2.3The 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):]&lt;+Mk (, A) =Let M(A) = Mk (A), then it is an A-algebra. Let Mcong (A) = congruence subgroupM(, 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 G51[G:]&lt;  normallim (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 /11 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, &quot;esquisse d'un programme&quot;). (iii) There are p-adic 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 = Symk-2 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 1an 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  52CHAPTER 2. MODULAR FORMS\ SL2 (Z). Therefore knowing N(, A) is equivalent to knowing M(, A). So it suffices to prove if f=nn0an q n  N (C) =N(, C)and   Aut C, then f   N (C). G3 Let j = a0 (G4 )  = q -1 + · · ·  N(SL2 (Z), Q). 34Proposition 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 33(a-1)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=0Denote Pf (X) =l=0gl X l , where gl  N(SL2 (Z), C), andgl  N(SL2 (Z), C) thanks to the Corollary 2.2.11. We give the proof in two steps.2.4. HECKE OPERATORS53Step 1: Prove that f  is holomorphic on H, by the Proposition 2.3.2. We have n n Pf (X) =l=0Pl (j)X l ,Pf  (X) =l=0Pl (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=01k(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=0k(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 nPf  (f  ) =l=0gl (f  ) =l=0lgl  (f   )l = Pf  (f  )   = 0So f    belongs to the finite set of roots of Pf  , which leads to the required conclusion.2.42.4.1Hecke operatorsPreliminary.Let   G be groups (for example,  = SL2 (Z), G = GL2 (Q)+ ), let x  G, x = {x :   }, x = {x :   }.54CHAPTER 2. MODULAR FORMSLet 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 .iIRemark. (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/]iIimplies i 1xi  (x) =iI iIi 1xi (x -1 ) = (x -1 ) = (x) =iIi 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  =iIi 1xi ,  =jJµj 1yj  A[\G/], then =(i,j)I×Ji µj 1xi yj  A[\G/],and it does not depend on the choices. (ii) (A[\G/], +, ) is an associative A-algebra with 1 as a unit. (iii) If M is a right G-module with G action m  m  g, and  = i 1xi  A[\G/], then for any m  M  , m   = i m  xi does notiIdepend 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.2.4.2Definition 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 OPERATORS55Proof. 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 &lt; 0. Then 0 g = a d , 0 where a &gt; 0. Thus completes the proof of existence. If 1 , 2   satisfies 1 g = then (1 g)(2 g)-1a1 b 1 0 d1a1 a22 g =a2 b 2 0 d2=0a2 b1 -a1 b2 a2 d2 d1 d2 SL2 (Z)This implies a1 = a2 , d1 = d2 , b1 - b2 divisible by d1 . Lemma-definition 2.4.3. For any n  1, Rn = 1( n 0 )  Z[\G/],0 nTn = 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 dRemark. 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 Tpr-1 . (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 d1 aa0ab +bd dd.56CHAPTER 2. MODULAR FORMSAs (n, l) = 1, (a, a ) = 1, (a, d ) = 1. This implies {aa : a|n, 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 haverTpr =i=0 b mod pi1 pr-i b  , Tp = 1( p 0 ) +0 pi 0 11( 1 c )0 pc mod pThenr rTpr Tp =i=0 b mod pi1  pr+1-i0b pi+i=0 b mod pi c mod p1 pr-i pb+pr-i c 0 pi+1r-1=Tpr+1+ Rp (i=0 b mod pi c mod p1 pr-1-i b+pr-1-i  ) = Tpr+1 + pRp Tpr-1 .0 pi(iv) It follows from (i),(ii),(iii).2.4.3Action 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 [ :  ] &lt; +, then  contains some  which is normal in , and [ :  ] &lt; +. (ii) If [ : 1 ] &lt; +, [ : 2 ] &lt; +, then [ : 1  2 ] &lt; +. (iii) If H  H  G, [H : H ] &lt; +, then [H   : H  ] &lt; +. 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) : ] &lt; +,   GL2 (Q)+ , then [SL2 (Z) : SL2 (Z)  -1 ] &lt; +.2.4. HECKE OPERATORS Proposition 2.4.7. Mk (C) =[SL2 (Z):]&lt;+57Mk (, C),Sk (C) =[SL2 (Z):]&lt;+Sk (, C)are stable under GL2 (Q)+ . Proof. For any   , f|k  = f . For   GL2 (Q)+ , we have (f|k )|k (-1 ) = f|k , so f|k  is invariant for the group -1   SL2 (Z). To verify that f|k  is slowly increasing at , write  =    SL2 (Z), then (f|k )(z) = (ad)k-1 d-k (f|k ) then we get the result. Let  = SL2 (Z), G = GL2 (Q)+ ,  =iI a b 0 dfor someaz + b d,i 1i  Z[\G/], we definef|k  =iIi f|k i ,for f  Mk (1) = Mk (C) .The definition is independent of the choice of i . From the general theory, we have (f|k )|k  (z) = f|k (   )(z). If f  Mk (1) (resp. Sk (1)), then f|k   Mk (1) (resp. Sk (1)). Facts: f|k Rn = nk-2 f , and f|k Tn = nk-1 d-k f ( az+b ). dad=n,a1 b mod dProposition 2.4.8. If f =m=0am (f )q m , then am (f|k Tn ) =a1, a|(m,n)ak-1 a mn (f ). a2Proof. For fixed d|n, d  1,b mod dd-k f ( az+b ) = d-k d = d-k = d b mod d m=0 am (f )e2imaz+b dam (f )e2inaz/db mod de2imb/dm=0  1-k m=0 d|m  l=0am (f )e2imaz/d adl (f )q al .= d1-k58 So f|k Tn = nk-1CHAPTER 2. MODULAR FORMSdad=n,a11-k l=0adl (f )q al ,summing the coefficients of q , this gives: am (f|k Tn ) = nk-1 =a1 a|(m,n) k-1m(n/a)1-k a mn (f ) a2aa1 a|(m,n)a mn (f ). a2Corollary 2.4.9. (i) Mk (, Z) and Mk (, Q) are stable under Tn and Rn . (ii) a0 (f|k Tn ) = ak-1 a0 (f ) = k-1 (n)a0 (f ).a|n(iii) a1 (f|k Tn ) = an (f ), therefore f is determined by T - a1 (f|k T ).2.5Petersson scalar product.dxdy = y21 2Lemma 2.5.1.+  1-x2SL2 (Z)\H-1 2dxdy  = &lt; . y2 3Corollary 2.5.2. (i) If [SL2 (Z) : ] &lt; +, 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.59Let 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\Hdxdy y2converges 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 + d|2 i D with |I| = C(). SoiIso 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\Hdxdy 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 yBecause f|k i and g|k 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) = andm,n1 (k) 2 (-2i)km,n1  Mk (1), (mz + n)k 1 (amz + an)k 1 , (cz + d)k1 = (mz + n)k =a=1 (m,n)=1(k) (k) (2i)k \ SL2 (Z)60CHAPTER 2. MODULAR FORMSwhere  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)\H1 k (cz+d)  \ SL2 (Z)  \ SL2 (Z)f (z)y k dxdy y2f (z) Im (z)k dxdy y2f (z)y k dxdy y2  \H  1 f (x + iy)y k-2 dxdy = 0, 0 01 2inx e dx 0where the last equality is because a0 (f ) = 0 and 1. Lemma 2.5.5. (i) For   GL2 (Q)+ , we have= 0 for n f|k , g|k  = (det )k-2 f, g . (ii) Let  = (det )-1 , then f|k , g = f, g|k  . Proof. (i) Choose  such that f, g  Sk () and -1   SL2 (Z), then C(-1 ) f|k , g|k  = (det )2(k-1)-1 \Hf (z)g(z) f (z)g(z)y k dxdy y2yk dxdy 2k y 2 |cz + d|= (det )k-2\H= (det )k-2 C() f, g . (ii) Replace g by g|k -1 , then we get f|k , g = (det )k-2 f, g|k -1 1 = (det )k-2 f, g|k det   = f, g|k  .2.6Primitive formsTheorem 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 FORMS61Proof. (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 f|k , g = = = =0 f|k (1 p 1 2 ), g 0 f|k p 0 , g|k 2 0 1 f|k p 0 , g 0 1 f, g|k p 0 , 0 1thus f|k Tp , g = (p + 1) f|k p 0 , g = f, g|k 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=0an (f )q n  Mk (1) - {0}. If for all n, f|k 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 ) + pk-1 apr-1 (f ) for p prime and r  1. Proof. (i) Since an (f ) = a1 (f|k 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 &quot;Multiplicity 1 theorem&quot;). (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.62CHAPTER 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 = k-1 (n)Gk , we get a basis of Mk (1) of eigenforms. Example 2.6.5. Write =qn=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  (pr-1 ), 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)p-s + p11-2s = 0)as a consequence of the proof of Riemann Hypothesis (Weil Conjecture) for zeta functions of varieties over finite fields.Chapter 3 p-adic L-functions of modular forms3.13.1.1L-functions of modular forms.Estimates for the fourier coefficientsProposition 3.1.1. Let   SL2 (Z) be a subgroup of finite index, let f = an (f )q n  Mk (, C). Then1 n M N(i)  O(nk-1 ),  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 2k kif k  3; if k = 2; if k = 1.1 MM 0y 2 f (x + iy)e-2inx dx,k y.sup\ SL2 (Z)|f|k (z)|.It is finite since [SL2 (Z) : ] &lt; +, and (z) = (z) for   SL2 (Z). 6364CHAPTER 3. P -ADIC L-FUNCTIONS OF MODULAR FORMSLet D be the fundamental domain of SL2 (Z). For any   \ SL2 (Z), there exists C such that, for all z  D, |f|k (z) - a0 (f|k )|  C e- M . Let C = sup C , (z) =2ysup(c,d)=(0,0)ky , |cz+d|2then (z)  C(z)k/2 + B for someB.1 an (f )  e2ny y - 2 M k 1  e2ny y - 2 MM (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 .k1  |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=1Meas({x  [0, M ] :[- log4 y]1 4j y (x + iy) 1 1 })( 4j-1 y )k/2 4j-1 y+4k/2 Meas({x  [0, M ] : (x + iy)  4})  M 4k/2 +j=1 [- log4 y] j=1 [- log4 y] 1 4j 2yM ( 4j-1 y )k/2= M 4k/2 1 + 2y 1-k/2 4j(1-k/2) . 4j(1-k/2) converges, we get an (f ) =[- log4 y] j=1When k  3, let y = 1/M n. Asj=1O(nk-1 ). When k = 2, it is obvious. For k = 1, 2 - y 1/2 &lt; 2, then we get the result.y 1-k/2 4j(1-k/2) &lt;3.1. L-FUNCTIONS OF MODULAR FORMS. Remark. (i) L(f, s) =n=065 0.an (f )n-s converges for Re (s)(ii) If  is a congruence subgroup, f  Sk (), Deligne showed that an (f ) = O(n(k-1)/2+ ), &gt;0in the same theorem mentioned above. Question: What about the noncongruence subgroups?3.1.2Dirichlet series and Mellin transform an . nsDefinition 3.1.2. Let {an }n1 be a sequence in C, the Dirichlet series of (an ) is D(s) =n=1Lemma 3.1.3. If D(s0 ) converges, then D(s) converges uniformly on compact subsets of Re (s) &gt; 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=0an z n , then the maximal open disc of convergenceof 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)n-1 , then D(s) = (1 - 21-s )(s), which converges for Re (s) &gt; 0, absolutely converges Re (s) &gt; 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 &gt; B satisfying, for 0  i  n, (i) (t) = O(tA-i ) O(tB-i ) near 0 near .66 LetCHAPTER 3. P -ADIC L-FUNCTIONS OF MODULAR FORMSMel(, s) :=0(t)tsdt . tThen it is holomorphic on -A &lt; Re (s) &lt; -B, and O(|s|-r ) on -A &lt; a  Re (s)  b &lt; -B. C+i 1 (ii) If r  2, (x) = 2i C-i Mel(, s)x-s ds, for any C with -a &lt; C &lt; -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.3.1.3For f =Modular forms and L-functions n=0 an (f )q n  M2k (1), define an (f ) , ns (s) L(f, s). (2)sL(f, s) =n=1(f, s) =Example 3.1.6. Take f = G2k , we get L(G2k , s) = =a=1  2k-1 (n) = ns n=1 n=1  d2k-1 (ad)-sad=na-sdd=12k-1-s= (s)(s - 2k + 1).Theorem 3.1.7.(i) L(f, s) absolutely converges for Re (s) &gt; 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. L-FUNCTIONS OF MODULAR FORMS.67Proof. (i) The result follows from an (f ) = O(n2k-1 ). (ii) Let (t) = f (it) - a0 (f ), then  is C  on R , and (t) = O(e-2t ) + at . f  M2k (1) implies (t-1 ) = (-1)k t2k (t) + (-1)k a0 (f )t2k - a0 (f ). For Re (s) &gt; 0, we have + -2nt s dt e t t 0 (s) . (2n)s=Then for Re (s) &gt; k,(f, s) = = = =n=1 + 0 + 1 + 1(s) an (f ) (2n)s(t)ts dt t + (t)ts dt + 1 (t-1 )t-s dt t t (t)(ts + (-1)k t2k-s ) dt - a0 (f ) t(-1)k 2k-s+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 inC such that L(s) =n=1cn nsconverges for Re (s) &gt; A, and (s) =(s) L(s) (2)ssatisfy (ii)(a) - (d) of previous theorem, then f (z) :=n=0cn q n  M2k (1).Proof. Since f (z) converges if |q| &lt; 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 &gt; 0. Leton H.(t) = f (it) - c0 =n=1cn e-2nt ,one can check that (s) = Mel(, s). Take c &gt; A, then (t ) - (-1) (t-1 ) t2k c+i c+i 1 = 2i c-i (s)t-s ds - (-1)k c-i (s)ts-2k ds c+i c+i 1 = 2i c-i (s)t-s ds - c-i (2k - s)ts-2k ds c+i 2k-c+i 1 = 2i c-i (s)t-s ds - 2k-c-i (s)t-s ds .k68CHAPTER 3. P -ADIC L-FUNCTIONS OF MODULAR FORMS +  R 2k - c? r rR c 602k-R  - Consider the integral of the function (s)t-s around the closed path . Since (s)  0 on vertical strips, by Cauchy formula,R+ Rlim(s)t-s ds =c+i c-i(s)t-s ds -2k-c+i 2k-c-i(s)t-s ds= 2i ress=0 ((s)t-s ) + ress=2k ((s)t-s ) = 2i(-c0 + (-1)k c0 t-2k ). So (t) - (-1)k (t-1 ) - (-c0 + (-1)k c0 t-2k ) = 0, 2k t(-1)k (-g(it)), t2kby an easy computation, the left hand is just g(it) = 0, which completes the proof.then we get3.1.4Euler productsTheorem 3.1.9. If f =n=0an (f )q n  M2k (1) is primitive, then 1pL(f, s) =1 - ap(f )p-s+ p2k-1-2s.Proof. By Theorem 2.6.2, anm (f ) = an (f )am (f ) whenever (n, m) = 1, soL(f, s) =p r=0apr (f )p-rs .Since, apr+1 - ap apr + p2k-1 apr-1 = 0,3.2. HIGHER LEVEL MODULAR FORMS multiplying by p-(r+1)s , and summing over r from 1 to +, we get  69apr pr=2-rs- ap p-s r=1apr p-rs+p2k-1-2s r=0apr p-rs = 0.Using the fact that a1 = 1, the result follows.3.23.2.1Higher level modular formsSummary of the resultsa b c dFor 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 &quot;old&quot;}. k On Sk (N ), we have the Hecke operators Tn , (n, N ) = 1, f|k Tn = nk-1ad=n,a&gt;1 b mod dd-k faz + b d,and for p | n, the operator 1 f|k Up = pp-1fi=0z+i p.We also have a involution wN given by f|k wN = N - 2 z -k fk-1 Nz.Definition 3.2.2. f  Sk (N ) is called primitive if f  Snew (N ), a1 (f ) = 1 k and f|k Tn = an (f )f , whenever (n, N ) = 1.70CHAPTER 3. P -ADIC L-FUNCTIONS OF MODULAR FORMSTheorem 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 ) + pk-1 apr-1 (f ) = 0. (c) f|k Up = ap (f )f , and this implies apr (f ) = (ap (f ))r for p|N ; (d) There exists f = ±1, such that f|k wN = f f .Theorem 3.2.4. Suppose f =n=1 an q n  Sk (N ) is primitive. Define  N 2sL(f, s) =n=1an , ns(f, s) = (s)L(f, s).Then (i) L(f, s) =p|N1 1-ap p-spN1 ; 1-ap p-s +pk-1-2s(ii) (s) has an analytic continuation to C. And (f, s) = i-k 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 and1 1 ; 1-(p)ap p-s 1-(p)ap p-s +2 (p)pk-1-2s pN p|N  s = (s) D2N L(f  , s) has a analytic continuation(-N )(f  , s) (f  -1 , s) = i-k f G(x) G(-1 )2ix Dwhere 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, thenm=1am q m  Sk (N ) and is primitive.3.3. ALGEBRAICITY OF SPECIAL VALUES OF L-FUNCTIONS713.2.2Taniyama-Weil ConjectureLet  be a finitely generated Z-algebra. Define its Hasse-Weil zeta function  (s) by 1 .  (s) = (1 - |/|-s )  prime in  Conjecture 3.2.6 (Hasse-Weil).  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, Breuil-Conrad-Diamond-Taylor). 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 Hasse-Weil conjecture in this case thanks to theorem 3.2.4. Theorem 3.2.8 (Mordell-Weil). E(Q)  {} Zr(E) finite group.Conjecture 3.2.9 (Birch,Swinnerton-Dyer). ords=1 L(fE , s) = r(E).3.33.3.1Algebraicity of special values of L-functionsModular symbols.Let N  1, f  Sk (N ), P  A[x](k-2) (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, defineirj (f ) =0f (z)z j dz =(j + 1) L(f, j + 1). (-2i)j+1Let Lf be the Z-module generated by rj (f|k ), 1  j  k - 2 and   0 (N )\ SL2 (Z). Then Lf is finitely generated.72CHAPTER 3. P -ADIC L-FUNCTIONS OF MODULAR FORMSi rTheorem 3.3.1. If P  A[x](k-2) , 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 0f (z)P (z)d(z) f|k (z)P|2-k (z)dz,where P|2-k (z) = (cz + d)k-2 P ( az+b )  A[x](k-2) . Take r = a/b, (a, b) = cz+d al-1 al 1, then there exists l =  SL2 (Z) satisfying (a0 , b0 ) = (1, 0), bl-1 bl (an , bn ) = (a, b).i r nf (z)P (z)dz =l=1 nal-1 bl-1 al blf (z)P (z)dz f (z)P (z)dz=l=1 nl (i) l (0) i 0=l=1f|k l (z)P |2-k l (z)dz  A · Lf .Exercise. For N = 1, let L+ (resp. L- ) be the Z-module generated by rj (f ) f f for all odd (resp. even) j. For P  A[X](k-2) , r  Q,  = ±, theni if (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 nsCorollary 3.3.2. (i) Suppose f n=1 M Z for some M . Then L(f, , s) =n=1to C and (f, , j) =(j) ¯ L(f, , j)  Q · Lf , (-2i)jif 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 L-FUNCTIONSnu73Proof. 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 )j-1 dz = u M  Q · Lf . +For (ii), we may assume (n) = e2i M + (-1)j e-2i M , and similarly, (f, , j) = =iu M u Mnunuf (z)(z - f (z)(z -iu j-1 ) dz M u j-1 ) dz M+ (-1)j -i u -Mi u -Mf (z)(z +f (z)(-z -u j-1 ) dz M u j-1 ) dz, Mthen one uses the exercise.3.3.2The resultsTheorem 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 ¯ rk-2 (f )rl (f )  Q f, f , for l odd. This implies +  f f, f f, f   k-2 rk-2 (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.74CHAPTER 3. P -ADIC L-FUNCTIONS OF MODULAR FORMS3.3.3Rankin's methodAssume k = l + j for k, l, j  N. Suppose 1 , 2 : (Z/N Z)×  C× are multiplicative characters. Let+ +f=n=1an q n  Sk (N, -1 ), 1g=n=0bn 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)=11 2 (n)y s+1-k (mz + n)j | mz + n |2(s+1-k) 1 2 (d) · Im (z)s+1-k . (cz + d)j(j) L(1 2 , j + 2(s + 1 - k)) · (-2i)j= a b  \0 (N ) c dD(f, g, s) = L(1 2 , j + 2(s + 1 - k))n=1 s jan b n ¯ ns=(4) (-2i) · f, gGj , 1 2 , s · [SL2 (Z) : 0 (N )]. (s) (j)+ 0 0 1Proof. Using the Fourier expansion, then+n=1an b n ¯ (s) = ns (4)s (s) (4)s (s) = (4)s = = (s) (4)sf (z)g(z) dx · y s f (z)g(z)y s+1 \Hdy ydxdy y2 dxdy y2(f (z)g(z) Im (z)s+1 )0 (N )\H  \ (N )  0f (z) g(z)0 (N )\H  \0 (N )dxdy 1 2 (d) Im (z)s+1-k y k , j (cz + d) y2this implies the Proposition.3.3. ALGEBRAICITY OF SPECIAL VALUES OF L-FUNCTIONS75Theorem 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+k-1 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 ,k-1 = 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 ,k-1  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 = ns1 nZ[ N ]×an ¯ ns bn ns1pN(1 - pp-s )(1 1- p p-s ), p p = 1 (p)pk-1 ,n=11 nZ[ N ]×pN(1 - pp-s )(1- p p-s ), p p = 2 (p)pl-1 ,then D(f, g, s) = an b n ¯ ns 1pNnZ[1 × N](1 - p pp-s )(1- p pp-s )(1- p p p-s )(1 - p p p-s ).76CHAPTER 3. P -ADIC L-FUNCTIONS OF MODULAR FORMSProof. Exercice, noting that apr = ¯r+1 r+1 p - p , p - pbpr =r+1 r+1 p - p . p - pWe give one application here: Corollary 3.3.7. The claim (3.1) holds, i.e. ¯ rk-2 (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=1bn = nsp1 , (1 - p-s )(1 - pl-s-1 )hence D(f, Gl , s) = L(f, s)L(f, s - l + 1). Therefore ¯ L(f, k - 1)L(f, k - l)  Q ·  j+k-1 f, f which implies ¯ rk-2 (f )rk-l-1 (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 p-s )(1 - p,2 p-s ) pN then L(f1  · · ·  fn , s) = pN j1 ,j2 ,··· ,jn {1,2}1 (1 - p,j1 · · · p,jn p-s )(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 L-FUNCTIONS OF MODULAR FORMS773.4p-adic L-functions of modular formsIn 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 - p-s )(1 - p-s ) 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 = pp-1 = pk-1 .fi=0z+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 asabove lemma implies that bnp = bn forbn = -r bpr n , r0.78CHAPTER 3. P -ADIC L-FUNCTIONS 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 p-r 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) = p-my mod pm(y)e-2ixy ,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 p-m 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,k-2] (Zp , Qp )  Qp , ¯ such that for all   LC(Zp , Q), ^ ~ (x)xj-1 µf, = (f , , j), 1  j  k - 1.Zp 1 Moreover, (µf, ) =  µf, , or equivalentlypZpx pµf, =1 µf, .Zp(ii) if vp () &lt; k - 1, then µf, extends uniquely as an element of Dvp () .3.4. P -ADIC L-FUNCTIONS OF MODULAR FORMS79¯ ¯ Proof. (i) The existence of µf, : LP [0,k-2] (Zp , Qp )  Qp is just the linearity ^ of   . The uniqueness is trivial. The second claim follows from pZpx px pj-1µf, =p1 ~ ^ (f , p-1 (px), j) j-11 1~ ^ = (f , , j) =  (x)xj-1 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) = Thenjp-n e-2iax , 0,if x  p-n Zp , = p-n a (pn x) if not.axe2i pn 0,x  Zp , otherwise.(x - a)j µf, =a+pn Zp l=0(-a)lj -n l=0j -n ~ p (f , a (pn x), l + 1) l j nl ~ p (f , a , l + 1). l= Sincei(-1)l~ pnl (f , a , l + 1) = pnl0f (z -a l )z dz = pnif (z)(pn z + a)l dz,- pa nwe getj(-1)ll=0j li(x - a)j µf, = -n pnja+pn Zp - pa nf (z)z j dz  -n pnj Lf .We just pick C = min(vp (~j (f |k ))). r80CHAPTER 3. P -ADIC L-FUNCTIONS OF MODULAR FORMSRemark. (i) If p | N , then  = 0, and  = 0 implies vp() = k-2 &lt; 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 px-1 (x)µf, .In particular, take (x) = x 2 xks- k 2where x t = exp(t log x). Set x 2 -1 xZ pkLp, (f, s) =s- k 2µf, .Proposition 3.4.6. For 1  j  k - 1, Lp, (f  j ) = (1 - Proof. Follows from (i) 1Z = 1Zp - p-1 1p-1 Zp , p ~ (ii) (f , 1Zp , j) = (1 -1 (iii) (µf, ) =  µf, .  ~ (f, j)), pjpj-1  ~ )(1 - j )(f, j).  pRemark. (i) As (f, s) = (f, k - s) and  = pk-1 if p N , then (1 - pj-1  ) = 1 - k-j .  pNote Ep (f, s) = (1 - p-s )(1 - p-s ). Then the Euler factor of the p-adic 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. k-2 (ii) If p | N ,  = 0, the vp () = k-2 . It can happen that  = p 2 , which 2 means Lp, (f, k ) = 0. In this case 23.4. P -ADIC L-FUNCTIONS OF MODULAR FORMS Conjecture 3.4.7 (Mazur-Tate-Teitelbaum Conjecture). k k ~ Lp, (f, ) = LF ont. (f )(f, ). 2 281Here the p-adic L-function is related to 2-dimensional (, 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 Kato-Kurihara-Tsuji, Perrin-Riou, and Stevens, Orton, Emerton with other definitions of the L-invariant. (iii) Mazur, Tate and Teitelbaum have also formulated a p-adic analog of the BSD conjecture. For E/Q an elliptic curve, by Taniyama-Weil, 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 (p-adic 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 Kato-Kurihara-Tsuji, we need another construction of p-adic L-functions via Iwasawa theory and (, )-modules; this construction is the subject of the next part of the course and is based on ideas of Perrin-Riou.82CHAPTER 3. P -ADIC L-FUNCTIONS OF MODULAR FORMSPart II Fontaine's rings and Iwasawa theory83Chapter 4 Preliminaries4.1 Some of Fontaine's ringsThis section is a review of notations and results from Fontaine's course. For details, see Fontaine's notes.4.1.1Rings of characteristic p1 }. 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 xkis 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 valuationCHAPTER 4. PRELIMINARIESvE (x) = vp (x(0) ) and maximal ideal ~ mE + = {x  E + , vE (x) &gt; 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 (¯ ) = p-1 &gt; 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 Ax-Sen-Tate's Theorem), + ~ EK := {x = (xn )  E + , xn  OKn /a, n  n(K)}. ¯ HK = Ker  = Gal(Qp /K ),K =nNKn .K = GK /HK = Gal(K /K)  Z . p4.1. SOME OF FONTAINE'S RINGS Then+ ~+ ~   EK  EK  E + , ¯87K.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 ]&lt;+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. p-n p-n ~K ), i.e., ~ E E (resp. EK ). In particular, E is algebraically closed.nN nN4.1.2(6) SetRings of characteristic 0~ ~ A+ := W (E + ) = W (R), ~ ~ A := W (E) = W (Fr R).~ Every element x  A can be written as+x=k=0pk [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 .88CHAPTER 4. PRELIMINARIES~ ~ There are two topologies in A+ and A: (i) Strong topology or p-adic 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=0p [xk ]) =k=0kp [g(xk )],k(k=0p [xk ]) =k=0kpk [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 p-adic 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 . pak  -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 ] &lt; +, 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 ]&lt;+is the maximal unramified extension of BQp = E. Set B = E ur be the closure of[K:Qp ]&lt;+~ ~ BK in B for the strong topology. Then A = B  Ais the ring of integers OE ur and the residue field of B is A/pA = E. By d Ax-Sen-Tate, 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 semi-linear continuous (for the weak topology) and commuting actions of  and . A (, )-module over BK is a finite dimensional BK -vector space with semi-linear 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 ).90CHAPTER 4. PRELIMINARIESThe 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 pro-cyclic, 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 p-adic L-functions. - DdR (V ), Dcris (V ), Dst (V ).Chapter 5 (, )-modules and Galois cohomology5.1 Galois CohomologyLet M be a topological Zp -module (e.g. a finite module with discrete topology or a finitely generated Zp -module with p-adic topology, or a Fontaine's ring + BdR · · · ), with a continuous action of GK . Let H i (GK , M ) be the i-th 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 1-cocycle 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, gx-x+cg ) = 0, that is, cg = (1-g)x is a coboundary. 1 1- 1 9192CHAPTER 5. (, )-MODULES AND GALOIS COHOMOLOGYTheorem 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 2-i (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 p-adic 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 Mittag-Leffler 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)) .5.2The complex C, (K, V )Assume that K is pro-cyclic (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) pr0  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.93Proof. 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 letg-1 y -1 -1g-1 y - (g - 1)b. -1i+is: as (g) = lim ()ni , y is fixed by HK , weg-1 y = lim (1 +  + · · · +  ni -1 )y. i+ -1y 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) = g-1 (g - 1)z for all g  GK , that is, -1 y = (g - 1)(b - z) for all g. Now b-z  D(V ), because it is fixed by g  HK . Then we have: y = ( -1)(b-z) 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,94CHAPTER 5. (, )-MODULES AND GALOIS COHOMOLOGYhere 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)~, ethey are both in D(V ) and satisfy ( - 1)x = ( - 1)y. Let b = e - e  ~ g-1 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 Hochschild-Serre 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 -1we get D(V ) D(V ) D(V ) ( - 1) = . -1 -1 ( - 1,  - 1)H 2 (GK , V )Remark. (1) The inflation-restriction exact sequence becomes the commutative diagram 0/ H 1 ( , V HK ) K / H 1 (G , V ) K,/ H 1 (H , V K ) K/00/D(V )=1 -1/ H 1 (C, (K, V ))/ ( D(V ) )K-1/0´ 5.3. TATE'S EULER-POINCARE FORMULA.95where 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  -1D(V )Id/ D(V ) Id/0C, : 0/ D(V )/ D(V ) D(V )/ D(V )/0It induces a commutative diagram· H 1 (C, ) ,NNN NNN NN lK (), NNN&amp;H 1 (GK , V )/ H 1 (C · ) , ppp pp ppp wppp lK ( ),where lK () = log () for log (K ) pr(K) on the choice of &quot;.pr(K) Zp . So lK ()c, &quot;does not depend5.35.3.1Tate's Euler-Poincar´ formula. eThe operator .Lemma 5.3.1. (i) {1, , · · · , p-1 } is a basis of EQp over (EQp ); (ii) {1, , · · · , p-1 } is a basis of EK over (EK ), for all [K : Qp ] &lt; +; (iii) {1, , · · · , p-1 } is a basis of E over (E); (iv) {1, [], · · · , []p-1 } 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 )96CHAPTER 5. (, )-MODULES AND GALOIS COHOMOLOGY (iv) To show thatp-1(x0 , x1 , · · · , xp-1 )  A -i=0p[]i (xi )  Ais a bijection, it suffices to check it mod p and use (iii). Definition 5.3.2. The operator  : A  A is defined byp-1(i=0[]i (xi )) = x0 .Proposition 5.3.3. (i)  = Id; (ii)  commutes with GQp . Proof. (i) The first statement is obvious. (ii) Note thatp-1 p-1g(i=0[] (xi )) =i=0i[]i(g) (g(xi )).If for 1  i  p - 1, write i(g) = ig + pjg with 1  ig  p - 1, thenp-1 p-1(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)xif 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 -1x=i=0[]i n (xi )where xi =  n ([]-i x).´ 5.3. TATE'S EULER-POINCARE FORMULA.97Proof. (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 Qp-1F () =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 =1p p-1(z(1 + ))i Fi ((z(1 + ))p - 1)i=0=1 F (z(1 + ) - 1). p zp =1Zp+ Recall D0 (Zp , Qp ) BQp = Qp Zp A+p by µ  Aµ () = Q that (µ) is defined by[]x µ. Recall(x)(µ) =Zpx ( )µ. p pZpFrom 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  {}.98CHAPTER 5. (, )-MODULES AND GALOIS COHOMOLOGYSince 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 )eiNow x  (x) is continuous in AQp , hence  is continuous in D.5.3.2D=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)) &lt; +. 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 haved(x) =i(xi )ei and ei =i=1ai,j (ej ).Let c = inf vE (ai,j ), then we havei,jvE ((x))  c + inf vE ((xi )).i + + From (EQp )  EQp and (¯ p x) =  k (x), we get vE ((x))   ¯k(5.1)vE (x) p. SovE ((x))  c + inf [ivE (xi ) vE (x) ]c+[ ]. p pIf vE (x) &lt; p(c-1) , then vE ((x)) &gt; vE (x). Now D=1 is closed since  is p-1 continuous, and is a subset of the compact set p(c - 1) M := {x : vE (x)  } p-1d k Fp [[¯ ]] · (ei ). ¯ i=1´ 5.3. TATE'S EULER-POINCARE FORMULA.99Hence 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=1bi,j ej . Let c0 = inf vE (bi,j ),i,j d dthend d dx=i=1 dxi (ei ) =i=1 dxij=1bi,j ej =(j=1 i=1xi bi,j )ej .Let yj =i=1xi bi,j , then x =j=1yj ej , and vE (yj )  c0 + vE (x).dFrom (x) =j=1(yj )(ej ), we getvE ((x)) = inf vE ((yj )) = p inf vE (yj )  pvE (x) + pc0 . So, if vE (x) -pc0 p-1+ 1, then vE (n (x))  pn . It implies y =+ ++ i=1i (x)converges in D. Now ( - 1)y =i=0i (x) -i=1i (x) = x-pc0 p-1implies that ( - 1)D contains {x|vE (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100CHAPTER 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.5.3.3The -module D=0 .Z . Let n  0 and n 1 + pn Zp if n  1. p = µp-1 , and n = lim n /n+m . We define  -mIf p = 2, we let 0 = Qp Then 0 = × 1 whereZp [[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. ZZp [[T ]] / A+ Qp /TAQp (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 EULER-POINCARE FORMULA.101Lemma 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 pnnu- 1)In general, for x =k=k0ak  k , then vE (x) = k0 vE (¯ ). Now ¯ +n (x) - x n (¯ )k -  k  ¯ = ak , n (¯ ) -  k=k  ¯ n (¯ ) -   ¯0and 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 (p-1 j (D)) = p-1 i+p j n+1 (D), j=0 j=0p and for n &gt; 1, n+1 = n , so 1 n -1n=1 (1 n+1 -1p-1 + n + · · · + n ), andp-1 Fp {{n }} = Fp {{n+1 }} + · · · + n Fp {{n+1 }}.102CHAPTER 5. (, )-MODULES AND GALOIS COHOMOLOGYSo we can assume n big enough. Recall vE (x) = inf vE (xi ) if x =i ixi ei . We can, in particular, assumevE (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=0x 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 havenvE (f k (x) - k x)  vE (x) + (k + 1)vE (¯ ). + + Let M = EQp e1  · · ·  EQp ed , then f k inducesM/¯ 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 p-1 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 EULER-POINCARE FORMULA. Proof. BK follows from AK by Qp Zp ; and we can reduce AK to AQp .p-1 p-1103Since D=0  (D/p)=0 is surjective, (i=1i (xi ) can be lifted toi=1[]i (^i )), xso we have the following exact sequence: 0 - (pD)=0 - D=0 - (D/p)=0 - 0. Everything is complete for the p-adic topology, so we just have to verify the result mod p, which is in corollary 5.3.12.5.3.4Computation of Galois chomology groups(-1, -1) (-1) pr -(-1) prProposition 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 ) -/0C, : 0/ D(V )/ D(V ) D(V )/ D(V )/0which 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104CHAPTER 5. (, )-MODULES AND GALOIS COHOMOLOGYLet C(V ) = ( - 1)D-1  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 -1since 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 k0. Let gi =n=0n ((¯ k fi )). Then we have: · (gi ) = gi , because ((¯ k fi )) = 0;  · ( - 1)gi = -(¯ k fi )  C.  We can take ei =  k fi . ¯5.3.5The Euler-Poincar´ formula. eTheorem 5.3.17. If V is a finite Zp -representation of GK , then2(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 )GQ105H i (GQp , IndGKp V ).GQSince | 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 - -1M implies that |M =1 | = | -1 |. Now C(V ) -1is a (Fp [[0 ]]/( - 1)) = Fp -moduleof rank dimEQp D(V ) = dimFp V . Hence | C(V ) | = |V |. -15.4Tate's duality and residuesH i (GK , M ) × H 2-i (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 =kZak  k  BQp , define res(xd) = a-1 .The residue of x, denoted by Res(x) is defined as Res(x) = res(x d ). 1+106CHAPTER 5. (, )-MODULES AND GALOIS COHOMOLOGYThe 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 ) od1/ D(V )  D(V )d2/ D(V )d2D (1)  D (1) od1D (1) : C, -1 (Qp , V  (1))5.4. TATE'S DUALITY AND RESIDUES107are 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?).108CHAPTER 5. (, )-MODULES AND GALOIS COHOMOLOGYChapter 6 (, )-modules and Iwasawa theory6.16.1.1i Iwasawa modules HIw (K, V )Projective limits of cohomology groupsIn 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 pn-1 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110CHAPTER 6. (, )-MODULES AND IWASAWA THEORYThus 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 , definei 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 ).6.1.2Reinterpretation in terms of measuresProposition 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 Mittag-Leffler conditions of H i-1 , which are automatic, because of the Finiteness Theorem: /(pn , n - 1)  V is a finite module, so H i-1 (GK , /(pn , n - 1)  V ) is a finite group.I 6.1. IWASAWA MODULES HIW (K, V )111Remark. (i) Recall that D0 (K , V ) is the set of p-adic 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). gHence, 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 ) K1Kn ·µ(1 , ..., i )  VnN.(ii) Let g  GK , , µ  Zp [[K ]], x  V , then g(µ  v) = g µ  gv = ¯µ  gv = g(µ  µ). ¯ gi So  and g commutes, it implies that HIw (K, V ) are Zp [[K ]]-modules.6.1.3Twist by a character (` la Soul´) a eLet  : 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.112CHAPTER 6. (, )-MODULES AND IWASAWA THEORYBy the above exercise, we have a commutative diagram:i HIw (K, V ) i/ H i (K, V ()) IwH i (GK , D0 (K , V ))/ H i (G , D ( , V ())) K 0 KSo i is an isomorphism of cohomology groups. It can be written in a concrete way i : (1 , ..., i )  µ(1 , ..., i ) - (1 , ..., i ) K1Kn ·µ(1 , ..., i )enN.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.6.2i 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  - 0Lemma 6.2.1. Let n = diagram C,n (Kn , V ) : 0n -1 n-1 -1p-1 = 1 + n-1 +, ..., +n-1  Zp [[K ]], the/ D(V )n/ D(V ) nD(V )Id/ D(V ) Id/0C,n-1 (Kn-1 , V ) :0/ D(V )/ D(V ) D(V )/ D(V )/0is 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 /Kn-1 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 )113Remark. (i) The isomorphism1 Exp : HIw (K, V )  D(V )=1is the map that will produce p-adic L-functions. 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 ) = -1D(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).  -nProof. 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 finiteiIset. Let n = max nxi . It implies that ( - 1)M  V , if   n , theniI(n - 1)M = 0, this shows the injectivity.nNSurjectivity: Let (xn )nN  limn (M/n -1). From the proof of injectivity,  - we know that xn is a Cauchy-sequence. 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.114CHAPTER 6. (, )-MODULES AND IWASAWA THEORYi 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 .  - TrV 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 p-divisible 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 ) Kncorp2/ ( D(V ) )n =1 -1 n/00/D(V )=1 n-1 -1p1/ H 1 (GK n-1 , V )/ ( D(V ) )n-1 =1-1/0where 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 )1156.31 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 sub-Zp [[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)/00 / (D(V )/p)=1-1/0Using the exact sequence 0  pV  V  V /p  0116CHAPTER 6. (, )-MODULES AND IWASAWA THEORYand 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 p-torsion ofC(V /p) C(V )/pC(V )is a finite dimensional Fp -vector space, thusis 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 inflation-restriction 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 just1 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 -1It 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 Kubota-Leopoldt zeta function7.1 The module D(Zp(1))=1The 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 map1 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-10 - Zp - (A+p )=1 -  (A+p )=0 - 0. - Q Q Remark. Under the map µ  with support in Z ( = 0) and pZp[]x µ, (A+p )=0 is the image of measures Q µ=0 Zp(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 KUBOTA-LEOPOLDT 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=0n (x) converges, and one check that (y) = y, ( - 1)y = -x.This implies the surjectivity of  - 1.7.2Kummer theory+ ~  = (1, (1) , (2) , ..., (n) , ...)  EQp  E + = R, (1) = 1.Recall thatLet 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 , theng(x) = c(g) , xc(g)  Zpgives a 1-cocycle 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 SERIES119 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 corn+1 n Fn - - -  H 1 (GFn , Zp (1))is commutative, we have a map: 1  : lim Fn  HIw (Qp , Zp (1))  -and1  HIw (Qp , Zp (1)) = Zp · (n )  (lim OFn ).  -7.3Coleman's power seriesTheorem 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. Writep-1() = 1 + pn u for u  Zp . Let x =i=0 p-1xi (1 + n+1 )i , where xi  OFn . Then(x) - x =i=0xi (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 =1f ((1 + T )z - 1).120CHAPTER 7. ZP (1) AND KUBOTA-LEOPOLDT 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 ]], thenk+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) Thenk N(g)((1 + T )p - 1) = 1 + 1 z p =1 k+1 h((1 + T )z - 1) mod 1 ,andz p =1h((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   µp-1 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 havek wn = NFn+k /Fn (wn+k )  1 + 1 OFn for all k,121then wn = 1. This completes the proof. Corollary 7.3.7. N(fu ) = fu ,d where  = (1 + T ) dT .(fu fu )= fu fuProof. 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 =1f ((1 + T )z - 1)(f )((1 + T )p - 1) =1 f ((1 + T )z - 1) p zp =1Then 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 ))) = (logz p =1f ((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 KUBOTA-LEOPOLDT ZETA FUNCTION7.4An 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))=1Theorem 7.4.1. The diagramis 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 = e-a pn - 1 e- pn - 12i 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 TSo fu is nothing but the Amice transform of µa that was used to construct fu p-adic zeta function. S o Exp produces Kubota-Leopoldt zeta function from the system of cyclotomic units.7.5. PROOF OF THE EXPLICIT RECIPROCITY LAW123(iv) The example in (iii) is part of a big conjectural picture. For V a fixed representation of GQ , then conjecturally1 {compatible system of global elements of V } - HIw (Q, V ) 1 - HIw (Qp , V ) - D(V )=1 - - -  p-adic L-functions. --- transf orm Exp AmiceAt 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.7.57.5.1Proof of the explicit reciprocity lawStrategy of proof of Theorem 7.4.1Let u  lim OFn , and g  Cn (g) be the cocycle on GFn by Kummer theory,  - i.e the image of u under the composition of1 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 ) pnfu (). fu (Exp )-1(g) - 1 y - ((g)g - 1)bn , (n ) - 1where 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 KUBOTA-LEOPOLDT 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 - 1in 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 p-k TrFn+k /Fn x  pn-c 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 ) = p-1 TrFn+1 /Fn (h(n+1 )). By hypothesis, (y) = y, we get p-k 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 g-1 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 p2g-1 x n -1log (g) fu y(n ) + (g - 1)bn pn fug-1 x + (g - 1)b  n, g  GFn n - 1Let=log (g) x. log (n )Usinglog((n )) fu y(n ) - (n )  pn-c OFn , n p fu7.5. PROOF OF THE EXPLICIT RECIPROCITY LAW then for every n, y(n ) - fu (n )  pn-c-2 OFn . fu125Let h = y - fu , then (h) = h and h(n )  pn-c-2 OFn . Using the fact fu p-k TrFn+k /Fn OFn+k  OFn and the formula (*), then for every i  N, n  i, h(i ) = p-(n-i) TrFn /Fi (h(n ))  pn-c-2 OFi , thus h(i ) = 0 for every i  N, hence h = 0.7.5.2Explicit formulas for cocylesThis 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 ), pkAs vE (-k (zl )) = we can find k such that vE (-(n+k) (zl ))  -1, ~ Let p = (p, ...)  E + , then for every l ~ [~]Cn (g) = pfor all ln - 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 ) g-1 p· y(n+k ) + (g - 1)bn n p n - 1where bn = ([~]-(n+k) (bn )). p126CHAPTER 7. ZP (1) AND KUBOTA-LEOPOLDT ZETA FUNCTIONk  (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 x-1 x-1 + H : 1 + tBdR  Cp , x  ( ) = ( ),  t recall that t = log(1 + ). We haveH((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) + · · · , thusCn (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 identity1 O p1 CpH(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 have1/pg(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 fu127Using 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 fuWe 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.7.5.3Tate's normalized trace mapsFn .nLet n = (n) - 1, Fn = Qp (n ), F =Lemma 7.5.3. If n  1, x  F , then p-k 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 /Fi-1 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+i-1 (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 KUBOTA-LEOPOLDT ZETA FUNCTION Proof. (i) is obvious.   (ii) is also obvious, since Rn+i-1 (Rn+i (x)) = 0  Rn+i (x)  Yn+i . (iii)  is obvious. For , let x  OFn+k , thenpk -1x=j=0aj (1 + n+k )j ,aj  O Fn .Write j = pk-i j with (j , p) = 1, then Rn (x) = a0 , since p-1 TrFn+i /Fn+i-1 (1 + n+i )j = Thus vp (x) 0  vp (Rn (x)) 0 and vp (Rn+i (x))  Rn+i (x) = (j ,p)=1apn-i 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:   p-1 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) &gt; 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 i-1 = ji-j-1, theni-j-1(uj - 1)-1 = (upi-1 - 1)-1 (1 + uj + · · · + (uj )pi-j-1 -1),so it is enough to treat the case j = i - 1. Let x  OFi  Yi , writep-1x=a=1xa (1 + i )a ,xa  OFi-1 ,7.5. PROOF OF THE EXPLICIT RECIPROCITY LAW write (i-1 ) = 1 + pi-1 v with v  Z , then pp-1129(ui-1 - 1)x =a=1xa (1 + i )a (u(1 + 1 )av - 1).We can check directlyp-1(ui-1 - 1) x =a=1-1xa (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) &gt; 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)) &gt; vp (x) - vp (n ), vp (y) &gt; 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 KUBOTA-LEOPOLDT ZETA FUNCTION7.5.4Applications to Galois cohomologyx  Fn - (  x log ())  H 1 (Fn , Fn )Proposition 7.5.7. (i) The mapinduces 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) &gt; vp (1 ). Using the above proposition (let u = (n )), we get ^ H 1 (Fn , F ()) = ^ F Fn Xn Fn = = . (un - 1) (un - 1) un - 1If 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 n0, we immediately get the result.Recall that the following result is the main step in Ax's proof of the Ax-Sen-Tate 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). g-1 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.7.5.5No 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 Ax-Sen-Tate, 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)Swhere 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 KUBOTA-LEOPOLDT ZETA FUNCTION Then we have g(y) = = 1 [K : Qp ] 1 [K : Qp ] g(x) =S1 [K : Qp ] h xS (x + log (h ))S1 = [K : Qp ](x) +S1 [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 , -kthen we haveChapter 8 (, )-modules and p-adic L-functions8.18.1.1Tate-Sen's conditionsThe 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) &gt; 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 Tate-Sen's conditions for the quadruple (G0 , , , v) are the following three conditions TS1­TS3. (TS1). For all C1 &gt; 0, for all H1  H2  H0 open subgroups, there exists an ~   H1 with v() &gt; -C1 and  () = 1. H2 /H1133134CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONS~ ~ (In Faltings' terminology, /H0 is called almost ´tale.) e (TS2). Tate's normalized trace maps: there exists C2 &gt; 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 ) = gHg-1 ,n g  RH,n = RgHg-1 ,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) - C38.1.2Example : 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 ] &lt; +, 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. TATE-SEN'S CONDITIONS on Ln if n 0 and TrLn /Kn (OLn )  dLn /Kn OKn ,135thus 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 /HLThen for all C1 &gt; 0, we can find x  OL such that vp (TrL /K (x)) is small enough, thus vp () &gt; -C1 . ^ (TS2) and (TS3): By Ax-Sen-Tate, 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 d-1 /Fn and vp (e ) i Kn written asdx=i=1TrK /K (xei )e , ithen inf vp (TrK /F (xei ))  vp (x)  inf vp (TrK /F (xei )) - vp (dKn /Fn ),i iand RHK ,n (x) =dRn (TrK /F (xei ))e , ii=1nm.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.136CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONS8.2Sen'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) &gt; 2C2 + 2C3 for any   G. Set ~ H = G  H0 , then there exists M  GLd () with v(M - 1) &gt; 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 1-cocycle on GK with values in GLd (Cp ), there exists [L : K] &lt; , 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.8.2.1Almost ´tale descent e~ Lemma 8.2.3. If  satisfies (TS1), a &gt; 0, and   U is a 1-cocycle 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 , weSLet 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=0a/2 and moreover(1 - MS )n ,8.2. SEN'S METHOD137-1 ~ 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 getMS - M S =S()(U - U ) =S()U (1 - (U )),thus v(MS - MS ) For any   H, U  (MS ) =Sa + 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 semi-linear action of H, an open i ~ subgroup of H0 . Then H (H, M ) is killed by any x  H with v(x) &gt; 0. Hint: Adapt the proof that if L/K is finite Galois and M is a L-module with semi-linear 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 n-cocycle, let c (g1 , · · · , gn-1 ) =hGal(L/K)g1 · · · gn-1 h()c(g1 , · · · , gn-1 , h),then dc = c.138CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONSTheorem 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 pro-cyclic.inf res8.2.2DecompletionNow 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  &gt; 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 METHOD139Proof. 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) &gt; 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) &gt; v(C) + C3 . By (TS3), this implies v(C) = +, i.e. C = 0. Proof of Proposition 8.2.1. Let   U be a continuous 1-cocycle on G0 with values in GLd (). Choose an open normal subgroup G of G0 such thatGv(V )  v(U2 ) - C2 - C3 .v(V2 - 1) &gt; C3 ,(B) = V1 BV2 ,inf v(U - 1) &gt; 2(C2 + C3 ).By Lemma 8.2.3, there exists M1  GLd (), v(M1 - 1) &gt; 2(C2 + C3 ) such -1 that   U = M1 U (M1 ) is trivial in H = G  H0 (In particular, it has values in GLd (H )).140CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONSNow 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 ) &gt; 2(C2 + C3 ), and by Corollary 8.2.7, there exists M2  GLd (H ) satisfying-1 v(M2 - 1) &gt; 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) &gt; 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 ) = V-1 V  (V ). Apply Lemma 8.2.8 with V1 = V-1 , V2 =  (V ), then we obtain what we want.8.2.3Applications to p-adic representationsProposition 8.2.9. Let T be a free Zp -representation of G0 , k  N, v(pk ) &gt; 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) &gt; 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 8.2.8.141Remark. 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,nH0 ,nDH0 ,n (T ) - DH,n (T ).Example 8.2.10. For  = Cp , let V be a Qp -representation of GK for [K : Qp ] &lt; +, 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 Hodge-Tate if and only if Sen is semi-simple, and the eigenvalues lie in Z. These eigenvalues are the Hodge-Tate weights of V . Remark. For general V , the eigenvalues of Sen are the generalized HodgeTate weights of V . log  , where   Kn , log () = 0. log ()8.38.3.1Overconvergent (, )-modulesOverconvergent elements+Definition 8.3.1. (i) For x =i=0 ikpi [xi ]  A, xi  E = Fr R, k  N, definewk (x) := inf vE (xi ) (One checks easily that wk (x)  vE (),   E, if and only if []x  A+ + pk+1 A).142CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONS(ii) For a real number r &gt; 0, define v (0, r] (x) := inf wk (x) +kNk k = inf vE (xk ) +  R  {±}. r kN rk 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=0pk [xk ]  A, the following conditions are equiv-alent: (i) (ii)++ pk [xk ] converges in BdR ;k=0 + k=0pk 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 = pk-ak yk , then yk  E + . We have p [xk ] =kp pk [p]ak [yk ] = pak (1 + )ak -k [yk ]. p8.3. OVERCONVERGENT (, )-MODULES143 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  p-n .-n ]- A(0, 1]Define A :=r&gt;0 + + A(0, r] = {x  A : -n (x) converges in BdR for n0}.Lemma 8.3.4. x k=0pk [xk ] is a unit in A(0,r] if and only if x0 = 0 andvE ( xk ) &gt; - k for all k  1. x0 r Proof. Exercise. Just adapt the proof of Gauss Lemma. Set 1 p-n A(0,r] , B (0, r] = A(0,r] [ ] = p nN B =r&gt;0endowed 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.144CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONS(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 &lt; r &lt; 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= {nNn 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 &lt; vp (T )  rvE (¯K ) to the ring BK .  Proof. The technical but not difficult proof is again left as an exercise. See Cherbonnier-Colmez 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 &lt; 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]]  BdRn(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.nn8.3. OVERCONVERGENT (, )-MODULES145For 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 . WritedPK (x) =i=0ai ()xi , ai ()  OF [[]].dDefine PK,n (x) = ai (n )xi , i=0 1 p-nthen PK,n (K,n ) = ((PK (K ))) = 0. Then we have vp (PK,n (n ))  vp (PK,n (n )) =and1 1 1 v (PK (¯K )) = n vE (dEK /EQp ) &lt;  if pn rK &gt; 1. n E p p 2pThen 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 &lt; r &lt; rK and pn r  1, -n (AK )  Kn [[t]]  BdR . (0, r]8.3.2Overconvergent representationsD(0, r] := (A(0, r] Zp V )HK  D(V ).Suppose V is a free Zp representation of rank d of GK . LetThis 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 &gt; 0, rV  rK such that  (0, rV ] AK (V ) - D(V ). (0, rV ] DAKBy definition, it is easy to see for all 0 &lt; r &lt; rV , D(0, r] (V ) = AK(0, r] AK(0, rV ]D(0, rV ] (V ).146CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONSProposition 8.3.13. If V is overconvergent, then there exists a CV such that if   K , n() = vp (log(())) and r &lt; inf{p-1 rV , p-n() }, then  - 1 is invertible in D(0, r] (V )=0 and v (0, r] (( - 1)-1 x)  v (0, r] (x) - CV - pn() vE (¯ ). p-1Proof. Write x =i=1[]i (xi ) and adapt the proof of the same statement asn -1in the characteristic p case. One has to use the fact that []ip (0, r] AK if r &lt; p-n and i  Z . p Remark. This applies to (AK )=0 .(0, r]is a unit inTheorem 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 , andn+ (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] KO ? / A(0,1] K,n x; xx x xx xx n n+k n+k(0,1] AK,n+k8.3. OVERCONVERGENT (, )-MODULES147One 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 p-1 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]8.3.3p-adic Hodge theory and (, )-modulesSuppose we are given a representation V , 0 &lt; r &lt; rV and n such that pn r &gt; 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  VK(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148CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONSLet K/Qp be a finite extension, and define BK = {F (K ) : F is a bounded analytic function on 0 &lt; vp (t)  r(F ), r(F ) &gt; 0},  Brig, K = {F (K ) : F is an analytic function on 0 &lt; vp (t)  r(F ), r(F ) &gt; 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 pKN =- Theorem 8.3.16 (Berger). Ford 1 · . vE (¯K ) d log K D (V ) = (B   V )HK =r&gt;0D(0, r] (V ),if V is semi-stable, then Blog, K1 1  K0 Dst (V ) = Blog, K B  D (V ) K t tis an isomorphism of (, N, K )-modules. This implies that Dst (V ) is the invariant under K .8.3.4A map of the land of the ringsThe 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 L-FUNCTIONS149p-adic Hodge o Theory ~ Blog oO_ _ _ -n(, )modules O1 p~ Brig oO _ _  / ~ o B  O   B O A ~ BO  o ~ / A o    ~ / E o  _ _~ A o ~ E omod pE 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 p-adic Hodge theory are in the wrong direction, but overconvergence and Berger's theorem allow us to go backwards.8.48.4.1Explicit reciprocity laws and p-adic L-functionsGalois cohomology of BdRSuppose 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 1-dimensional K-vector 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 )).150CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONSRemark. 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 Hochschild-Serre spectral sequence to conclude. Proposition 8.4.2. Suppose i &lt; j  Z  {±}, then if i  1 or j  0,+ + H 1 (GK , ti BdR /tj BdR ) = 0;if i  0 and j &gt; 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.8.4.2Bloch-Kato's dual exponential mapsLet 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 L-FUNCTIONS151 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 mmH 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 non-triviality 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)) = t-k L K DdR (V ) by an easy computation. Thus, Fil0 (DdR, L (V (k))) = t-k 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)) - t-k L K DdR (V ) is identically 0 for k 0.152CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONS8.4.3The explicit reciprocity lawRecall that1 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 )) -Knµ  H 1 (GKn , V  ).where we write V  , not as V () to distinguish from V (k) = V  k . Then exp (Knk µ)  t-k 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 p-n -n (Exp (µ)) =kZexp (Knk 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) =kZexp (Knk µ).8.4. EXPLICIT RECIPROCITY LAWS AND P -ADIC L-FUNCTIONS153 Proof. (ii) follows from (i) and from the commutative diagram: H 1 (GL2 , V ) - - L2 K DdR (V ) -    Tr K Id corL2 /L1expH 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:nexpg-1 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, p1-n -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 = p-n -n (y) =ii0yi ti ,yi  Kn K DdR (V ),then cg = log (g)y-k t-k(g)i+k - 1 · yi ti + i+k - 1 (n ) i=-k yi ti . ((n )i+k - 1) i=-k= log (g)y-k t-k + (g - 1) So we get expKnk µ = y-k t-k .154CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONS8.4.4Cyclotomic elements and Coates-Wiles 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 thatdfu 1 () = . dT  1 , (1 + 1 )et/p - 1-1 ()-1 = ([1/p ] - 1)-1 = then Exp Qp ((u)) = = =n=11 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 - p-n )(1 - n) (-t)n-1 . (n - 1)! if k  0; if k  -1.+So expQpk (µ) =0, -k-1 , (1 - pk )(1 + k) (-t) (-k-1)!Remark. (i) The map1 lim OFn - {0} - HIw (Qp , Qp (1)) - Qp ,  -u  tk+1 expQpk (u)is the Coates-Wiles 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 non-trivial and are even not de Rham. (iii) dimQp H 1 (GQp , Qp (k)) = 1 if k = 0, 1. Corollary 8.4.6. Non-trivial 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 ] &lt; , prove the same statement.8.4. EXPLICIT RECIPROCITY LAWS AND P -ADIC L-FUNCTIONS1558.4.5Kato's elements and p-adic L-functions of modular forms.Now we come to see the relations with modular forms. Suppose f=n=1an q n  Sk (N ), k  2is 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 ), non-ramified outside N p, such that if N p, for  the 2i 2i arithmetic Frobenius at ( (e pn ) = e pn ), then det(1 - X-1 ) = 1 - a X +k-1X 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 (Faltings-Tsuji-Saito). (i) Vf is a de Rham representation of GQp with Hodge-Tate weights 0 and 1 - k, the 2-dimensional Qp (f )-vector space DdR (Vf ) contains naturally f , and0 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 + pk-1 . If p| N but ap = 0, then Vf is semi-stable 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)),Knthen 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156CHAPTER 8. (, )-MODULES AND P -ADIC L-FUNCTIONS1 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 =expZp(x)xk-1-j · zKato =1 f (f, , j + 1) · k-1-j j! twhere (f, , j + 1)  Q(f, µpn ), f tk-1-j  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 + pk-1 with vp () &lt; k - 1; if p N , let  = ap = 0 (in this case p2 = pk-1 ). In both cases, take  = pk-1 -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, K1  K0 Dst (Vf ) - Blog, K B  D (Vf ). K tTheorem 8.4.10. (i)  (f ) = 0; (ii)  Exp (zKato ) =Zp[]x µf,  (f ) . tk-1Remark. µf,  exists up to now only in the semi-stable case, but zKato exists all the time. So a big question is: How to use it for p-adic L-function?`

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