ON CONGRUENCES BETWEEN MODULAR FORMS

RICHARD LAWRENCE TAYLOR

A DISSERTATIONPRESENTED TO THE FACULTYOF PRINCETON UNIVERSITYIN CANDIDACY FOR THE DEGREEOF DOCTOR OF PHILOSOPHY.RECOMMENDED FOR ACCEPTANCEBY THE DEPARTMENT OFMATHEMATICS

JUNE 1988

Copyright by Ri hard Lawren e Taylor 1988All Rights Reserved

To John and Mary

Abstra tAn elementary lemma on group ohomology is proved. Applied to ertain arithmeti subgroups of real lie groups this leads to a general method of establishing ongruen esbetween modular forms of di�erent weights. We apply this to establish the existen e of ertain p-adi families (Hida families) of Siegel modular forms and of modular forms forGL2 over an imaginary quadrati �eld. We also use these methods to show that the exis-ten e of ertain Galois representations one expe ts to be atta hed to Siegel modular forms orresponding to holomorphi dis rete series would imply their existen e for Siegel modularforms orresponding to limit of holomorphi dis rete series.

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ContentsAbstra t 1Introdu tion 4

1 A Group Cohomologi al Lemma 71.1 The Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Some Appli ations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Some Congruen es between Siegel Modular Forms 192.1 Introdu tion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Review of Siegel Modular Forms . . . . . . . . . . . . . . . . . . . . . . . . 212.3 Relation to Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.4 Some Lemmas on He ke Operators . . . . . . . . . . . . . . . . . . . . . . . 352.5 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443 p-adi Families of Siegel Modular Forms 493.1 Introdu tion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.2 Review of Siegel Modular Forms II . . . . . . . . . . . . . . . . . . . . . . . 513.3 Some Lemmas on Eisenstein and Theta Series . . . . . . . . . . . . . . . . . 643.4 The General Strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 753.5 Conje tural Appli ations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

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4 Modular Forms over an Imaginary Quadrati Field 964.1 Introdu tion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 964.2 Review of Cohomology Groups and Automorphi Forms . . . . . . . . . . . 994.3 Cohomology of the Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . 1094.4 Change of Level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1154.5 Change of Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1194.6 Cy lotomi Hida Families . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1224.7 Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

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Introdu tionThis work is on erned with ertain methods of Hida for produ ing ongruen es betweenmodular forms. We generalise some of these methods and apply them to the problem of onstru ting Galois representations atta hed to modular forms.Hida in a series of papers (see [Hi1℄ and the referen es ited therein) has onstru ted ertain p-adi families of ellipti modular forms. To des ribe these results �x an odd primep, an integer N and a Diri hlet hara ter � de�ned modulo Np. Fix Q a � Q a p andQ a � C . Let O be the integers of a �nite extension of Q p in whi h � is valued and let� = O[[T ℄℄. By a �-adi form we mean a formal power series P1n=0 anqn with an 2 � su hthat for all pairs (k; �) of an integer k � 2 and a hara ter � : (1 + pZ)=(1 + prZ) ! Q a �,P1n=0 an(�(1 + p)(1 + p)k � 1)qn is the Fourier expansion of an ellipti modular form ofweight k, level Npr and hara ter �!�k�, where ! denotes the Tei hmuller hara ter. If�0 is a �nite extension of � we de�ne the spa e of �0-adi forms to be the spa e of �-adi forms tensored with �0. One an de�ne an a tion of the He ke operators on �-adi forms ompatible with spe ialisation. One an also de�ne a notion of ordinary modularforms, both at the �nite and at the �-adi level. Roughly speaking the spa e of ordinaryforms over a p-adi ring of integers, or over �, is the largest spa e on whi h the a tion ofUp = [�0(Np?)0� 1 00 p

1A�0(Np?)℄ is invertible, and its omplement is that on whi h Upis p-adi ally nilpotent. Hida's main results are that the spa e of �-adi ordinary forms ofgiven level N and hara ter � is a �nite torsion free �-module, and moreover any ordinaryeigenform of the He ke algebra of weight k, level Npr and hara ter �!�k� (with � as4

above) an be lifted to an ordinary �0-adi eigenform of the He ke algebra (for some �nite�0=�). Thus there are very systemati families of ongruen es between ordinary eigenforms.This work has been generalised by Hida ([Hi2℄) and Wiles ([Wi℄) to Hilbert modularforms. In this thesis we shall give generalisations to Siegel modular forms (in hapter 3)and to modular forms over imaginary quadrati �elds (in hapter 4). Our entral te hniqueis a method of omparing modular forms of di�erent weights via group ohomology. Thisseems to be a very general method and we give an exposition of it in hapter 1. It is ageneralisation of te hniques of Shimura ([Sh1℄) and Hida ([Hi1℄). It is based on the fa tthat if G is an algebrai group, P a paraboli subgroup, M1 and M2 G(Z)-modules and �a ongruen e subgroup of G(Z) su h that � mod pr is ontained in P (Z=prZ) then theremay be non-trivial �-morphisms between M1 (Z=prZ) and M2 (Z=prZ) whi h allow usto ompare the �- ohomology of the modules M1 and M2.The easiest type of dedu tion to draw from this method is that the dimension of the spa eof ordinary modular forms for a given group � is bounded independently of the weight as theweight varies over some in�nite set. This together with a suitable modular form ongruentto one modulo p is enough to produ e a lot of ongruen es between ordinary eigenforms ofdi�erent weights. This is arried out in hapter 2 for Siegel modular forms. In the ase ofGSp4 it is applied to the problem of asso iating Galois representations to modular forms.Spe i� ally to eigenforms of \weight (n1; n2)" with 2 � n1 � n2 one expe ts to be able toasso iate ertain four dimensional Galois representations. If 3 � n1 then one hopes to beable to �nd these representations in the ohomology of ertain Shimura varieties. Howeverfor the ase n1 = 2 no su h method is expe ted to exist. We show how the result forordinary eigenforms of weight (2; n2) (n2 � 2) would follow from the result for forms ofweight 3 � n1 � n2.In hapter 3 we organise these ongruen es into \Hida families". However we restri t tothe ase of parallel weight (k; :::; k) and even genus. We follow the method of Wiles ([Wi℄)based on �nding enough �-adi forms by writing down �-adi Eisenstein series, multiplyingthese by suitable modular forms and spe trally de omposing the result. This requires thatone already has suitable bounds on the dimension of various spa es of ordinary modular5

forms involved. This again follows from our results in hapter 1. In the ase of Siegelmodular forms the al ulations required for this method be ome very messy.Finally in hapter 4 we onsider the ase of imaginary quadrati �elds. Here we are notable to multiply together modular forms, so we work ex lusively with the orresponding ohomology groups. Our results are not as sharp as we would like due to torsion in thehomology groups. As a byprodu t of our method we an in fa t exhibit torsion in thehomology of ertain sheaves (of \non-parallel weight", so that the torsion free part of the uspidal part of the �rst homology vanishes) on quotients of hyperboli 3-spa e by ertaindis rete groups.It is a pleasure to a knowledge the in uen e of the work of Hida [Hi1℄ and that of Wiles[Wi℄ on this thesis. I have also enjoyed and bene�ted from many dis ussions with FredDiamond and Mi hael Larsen. Finally I would like to express my great gratitude to myadvisor Andrew Wiles for his onstant help and en ouragement.This work was partially supported by a Sloane Foundation Do toral Dissertation Fel-lowship.

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Chapter 1A Group Cohomologi al Lemma1.1 The LemmaWe shall make repeated use of a ertain argument to ompare the group ohomology ofdi�erent modules, and hen e ertain spa es of modular forms of di�erent \weight". Weshall des ribe this method abstra tly as it seems to be appli able rather generally. It is anextension of some ideas of Shimura ([Sh2℄) and of Hida (see [Hi1℄).If � is a semi-group, � a subgroup and M a �-module then the ohomology groupsH�(�;M) may be onsidered as the image of M under the right derived fun tors of the�xed point fun tor N 7! N� from �-modules to abelian groups. If �1 and �2 are twosubgroups of � and if g 2 � is su h that [�2 : �2 \ g�1g�1℄ is �nite then there is a naturaltransformation: [�2g�1℄ : H�(�1; ) �! H�(�2; )determined as being the unique su h transformation ompatible with the boundary homo-morphisms and whi h oin ide in degree zero with:M�1 �! M�2m 7�! P igmwhere: �2 =a i(�2 \ g�1g�1)

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or equivalently: �2g�1 =a ig�1These are usually alled He ke operators, and the spe ial ase �1 � �2 and g = 1 is alled orestri tion and denoted or.If �1, �2 are two groups and M1, M2 are modules over �1 and �2 respe tively and, ifmoreover, � : �2 ! �1 and � :M1 !M2 satisfy:�((� )m) = (�m)for all 2 �2 and m 2M1; then there is an indu ed map:

(��; ��) : H�(�1;M1) �! H�(�2;M2)It may be de�ned as the unique map fun torially extending:� :M�11 �!M�22Returning to the situation in the last paragraph we may fa tor [�1g�2℄ as:

H�(�1;M) ( �g�1 ;g�)�! H�(�2 \ g�1g�1;M) or�! H�(�2;M)where g�1( ) = g�1 g.We shall now introdu e a slight generalisation of He ke operators. Let �1, �2 again besubgroups of � and g 2 � be su h that [�2 : �2 \ g�1g�1℄ < 1. Let M1, M2 be modulesfor hg;�1i and �2 respe tively. Let � : gM1 !M2 be a map of �2 \ g�1g�1-modules. Thenwe an de�ne a map: [�2g�1℄� : H�(�1;M1) �! H�(�2;M2)to be the omposite:

H�(�1;M1) ( �g�1 ;(�Æg)�)�! H�(�2;\g�1g�1;M2) or�! H�(�2;M2)If we set M1 =M2 and � = Id then we re over the normal He ke operators.8

Lemma 1.1 Let � be a semi-group; �1 � �2 subgroups of �; g 2 � with [�1 : �1 \g�2g�1℄ <1;M1 (resp. M2) a module for h�1; gi (resp. h�2; gi); and j :M1 !M2 a h�2; gimorphism su h that j : gM1 �! gM2. Then there exists I : H�(�2;M2)! H�(�1;M1) su hthat:1. If there exist elements i 2 �1 su h that �1g�1 =` ig�1 and �1g�2 =` ig�2 thenI Æ j� = [�1g�1℄.2. If there exist elements Æi 2 �1 su h that �1g�2 = (�2g�2)q(`�2Æig�2) and jÆigM1 =0 then j� Æ I = [�2g�2℄.The onditions in 1) and 2) are automati ally satis�ed if �1 = �2.Proof:Set I = [�1g�2℄jj�1gM1 . The �rst part is easy, it follows from the ommutativity ofthe following diagram:H�(�1;M1) ( �g�1 ;g�)�! H�(�1 \ g�1g�1; gM1) �! H�(�1 \ g�1g�1;M1) or�! H�(�1;M1)j� # j� # & res & res " orH�(�2;M2) ( �g�1 ;g�)�! H�(�1 \ g�2g�1; gM2) j��= H�(�1 \ g�2g�1; gM1) �! H�(�1 \ g�1g�1;M1)(The only slight problem is the right hand triangle, but working in the ategory of h�1; gi-modules we need only he k this in degree zero.)For the se ond part we must he k that the following diagram is ommutative:H�(�1;M1) j��= H�(�1 \ g�2g�1; gM1) �! H�(�1 \ g�2g�1;M1) or�! H�(�1;M1)( �g�1 ; g�) " j� & # j� # j�H�(�2;M2) ( �g�1 ;g�)�! H�(�2 \ g�2g�1;M2) or�! H�(�2;M2)The left hand side is easy, the only problem is to he k that the two omposite mapsH�(�3; N)! H�(�2;M1) in:H�(�3; N) �! H�(�3;M1) or�! H�(�1;M1)# j� # j�H�(�3 \ �2;M2) or�! H�(�2;M2)are equal, where �3 = �1 \ g�2g�1 and N = gM1. (If �1 = �2 this also is easy.) In fa t weshall prove this under the following assumptions, whi h are learly valid in our ase:9

�3 � �1 and �2 � �1 are subgroups of �, M1 is a �1 module, N is a �3-submoduleof M1, M2 is a �2-module and j : M1 ! M2 is a morphism over �2 su h that �1 =(�2�3)q (`�2Æi�3) with jÆiN = 0.To prove this let � 2 Zn(�3; N) and let 1; :::; n 2 �2. Let �2�3 = `hk;0�3 and�2Æi�3 =`hi;k;0�3 with ea h hk;0 and hi;k;0 in �2, so that jhi;k;0N = 0 and �2 =`hk;0(�2\�3). Then the image of � in Hn(�2;M1) by the lower route is represented by:( 1; :::; n) 7�!Xhk;0j Æ �(h�1k;0 1hk;1; :::; h�1k;n�1 nhk;n)where hk;l is de�ned by h�1k;l�1 lhk;l 2 �3 and hk;l = hk0;0 for some k0. Moreover if we de�nehi;k;l in the similarly, then the image of � by the upper route is represented by:( 1; :::; n) 7�! jPhk;0�(h�1k;0 1hk;1; :::; h�1k;n�1 nhk;n)+jPhi;k;0�(h�1i;k;0 1hi;k;1; :::)=Phk;0j Æ �(h�1k;0 1hk;1; :::; h�1k;n�1 nhk;n)1.2 Some Appli ations

The basi idea in the appli ation of this lemma is that if G is a redu tive group, P a paraboli subgroup, L a Levi omponent of P , A the split omponent of its entre, �P (N) the inverseimage of P (Z=NZ) under G(Z) ! G(Z=NZ), M a G module with weights � � X�(A),�0 2 � a lowest weight with respe t to the partial order orresponding to P , and � 2 X�(A)is su h that �:� � 0 for all � 2 � with equality if and only if � = �0; then one an de�nea map j : M(Z=NZ) ! M�0(Z=NZ) of �P (N)-modules su h that j : �(N)M(Z=NZ) ��!�(N)M�0(Z=NZ) (here M�0 denotes the �0 eigenmodule). By our lemma we then see thatH�(�P (N);M(Z=NZ)) depends up to the a tion of [�P (N)�(N)�P (N)℄ only on the a tionof L onM�0(Z=NZ). No doubt this an be formalised in this generality, but we shall simplytreat several examples.Example 1.1

We shall prove:10

Theorem 1.1 Fix a prime p and an extension of the p-adi valuation on Q to Q a (i.e.Q a � Q a p ) and an integer N . Fix also a onstant C. Then the sum of the dimensions of theeigenspa es in Sk(�1(N)) for the He ke operator Tp = [�1(N)0� 1 00 p1A�1(N)℄ for whi hthe orresponding eigenvalue has p-adi valuation less than C is bounded independently ofk. Note that we shall here employ the standard notation for ellipti modular forms. Weshall also use standard fa ts about them without omment. See for example [Sh2℄.Proof:We �rst redu e to the ase pjN . So suppose p6 jN and without loss of general-ity k > C2 + 1. If f 2 Sk(�0(N); �) is an eigenvalue for Tp with eigenvalue ap wherevalp (ap) < C then the equation X2�apX+�(p)pk�1 has a root � with valp (�) = valp (ap)and a root � with valp (�) > C, and f(z) � �f(pz) 2 Sk(�1(Np)) is an eigenve tor for[�1(Np)0� 1 00 p

1A�1(Np)℄ with eigenvalue �. Moreover if f1; :::; fr 2 Sk(�1(N)) are lin-early independent and if �1; :::; �r 2 C then the fun tions fi(z) � �ifi(pz) are linearlyindependent in Sk(�1(Np)). The desired redu tion now follows at on e.Thus assume pjN . By a theorem of Ei hler and Shimura it will do to establish the theo-rem withH1(�1(N); Sk�2((Q a p )2)) in pla e of Sk(�1(N)), and Tp = [�1(N)0� p 00 11A�1(N)℄.(Here Sn denotes the nth symmetri power.) In fa t it will do to onsiderH1(�1(Npr); Sk�2((Q a p )2)) for any r � 0 (be ause pjN implies that Tp ommutes withrestri tion from �1(N) to �1(Npr)). Let Bk(Q a p ) denote the sum of the eigenspa es ofTp in this ohomology group, whi h have p-adi valuation less than C. Also let Bk =Bk(Q a p )\H1(�1(Npr); Sk�2(Z2p))TF (where TF indi ates the torsion free quotient). ThenBk(Q a p ) = Bk Zp Q a p . Take M = H1(�1(Npr); Sn((Z=prZ)2)) for some �xed hoi e ofn > C and for r = n(n+ 1). Then for k � n+ 2 we have a natural proje tion map:

j : Sk�2((Z=prZ)2) �! Sn((Z=prZ)2)11

as �1(Npr)-modules. If moreover g = 0� pn 00 11A we have:

j : gSk�2((Z=prZ)2) ��! gSn((Z=prZ)2)and so by our lemma 1.1 we know that the kernel of:Bk=prBk ,! H1(�1(Npr); Sk�2((Z=prZ)2)) j��!Mis killed by Tnp . Thus r(rkBk)� valp (detTnp ) � valp (#M) so that r(rkBk) � valp (#M) +nC(rkBk) and rkBk � valp (#M) as desired.Before giving further examples we re all the notion of \Hida idempotent". If M is a Zpmodule with EndZp(M) a �nite Zp-module (for example ifM or Hom(M; Q p=Zp) is a �niteZp-module) and if U :M !M is an endomorphism then there exists a unique idempotenteU 2 Zp [U ℄ � EndZp(M) su h that:1. U is invertible on eUM2. U is topologi ally nilpotent on (1� eU )M3. eU = limr!1 U r!4. if U ommutes with another operator T so does eUIfM 0 is a se ond su h module with an operator U 0 and T :M !M 0 is su h that TU = U 0Tthen eU 0T = TeU . These results are all easy onsequen es of the dis ussion in [MW2℄(se tion 4). If A is a Zp-algebra we an think of eU 2 EndA(M A). Moreover if M and Uare de�ned over Z, say M =M0 Z Zp then eU 2 EndR(M0 Z R) where R = Zp \ Q a . Inparti ular if we �x Q a � C and Q a � Q a p we may think of eU 2 EndC (M0 C ). Exa tlysimilar results hold with Zp repla ed by the ompletion of the integers of any number �eldat any �nite prime.Note that we may dedu e from the above example the following result of Hida:Corollary 1.1 Fix a prime p, an integer N and embeddings Q a � C and Q a � Q a p . Lete be the idempotent asso iated to the He ke operator Tp (de�ned as above) on Sk(�1(N)).Then dim eSk(�1(N)) is bounded independently of k.12

We now onsider our se ond:Example 1.2We re all some fa ts about Sp2g. Fix a maximal torus T in Sp2g(C ) onsisting ofdiagonal matri es in the standard representation into GL2g. Fix X�(T ) �= Zg by setting~n = (n1; :::; ng) : (diag(�1; ::; �g ; ��11 ; ::; ��1g )) = �n11 :::�ngg . Also �x a Borel B onsisting ofmatri es of the form: 0BBBBBBBBBBBBBBBBBBB�

� 0 : : : 0 � : : �� � 0 : :... . . . ... : :� � : : : � � : : �0 : : 0 � : : : � �: : ... . . . ...: : 0 � �0 : : 0 0 : : : 0 �

1CCCCCCCCCCCCCCCCCCCAThen the roots � of T on sp2g, the Lie algebra of Sp2g, are:� �ij the ve tor onsisting of zeroes ex ept for 1 in the ith pla e and �1 in the jth pla e(i 6= j)� �ij the ve tor onsisting of zeroes ex ept for 1 in the ith and jth pla es (i > j)� �ii the ve tor onsisting of zeroes ex ept for 2 in the ith pla e� ij = ��ij (i � j)With respe t to B the positive roots �+ are the �ij with i > j and the �ij . Note that:

��ij = 0� �ii � �jj 00 �jj � �ii1A

��ij = 0� �ii + �jj 00 ��ii � �jj1A

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��ii = 0� �ii 00 ��ii1A

� ij = ���ijwhere �ij is the g � g matrix with one in the ith row and jth olumn and zeroes elsewhere.Also sp2g(C ) has a Chevalley basis onsisting of the following elements:Xij = 0� �ij 00 ��ij

1A (i 6= j)Yij = 0� 0 �ij + �ji0 0

1A (i > j)Yii = 0� 0 �ii0 0

1AZij =t Yij (i � j)together with ��(i+1)i for i = 1; :::; g � 1 and ��11.We shall let UZ denote the Z-subalgebra of the universal enveloping algebra of sp2g(C )generated by elements of the form Xnijn! , Y nijn! and Znijn! . If V is an sp2g(C ) module then byan admissible latti e L � V we shall mean a Z latti e preserved by UZ. Then it is knownthat any �nite dimensional irredu ible sp2g(C ) module ontains an admissible latti e, forexample UZv for any lowest weight ve tor v, and moreover that any admissible latti e isequal to the sum of its interse tions with the weight spa es of T in V .Let R denote Z or Z=NZ . Let L be an admissible latti e in VL where �L : Sp2g(C ) !GLVL . We an de�ne GL(R) to be the subgroup of GLLR generated by elements ofthe form expXij , expYij and expZij . We shall let PL(R) denote the subgroup of GL(R)generated by the expXij and the expYij , and SL(R) the one generated by the expXij . Itis known that if L1 and L2 are as above with ker�L1 � ker�L2 then there is a unique mapGL1(R) ! GL2(R) taking expW 2 GL1(R) to expW 2 GL2(R) for W equal to any Xij,Yij or Zij . (See [Sg℄.) We see that this map takes PL1(R) to PL2(R) and SL1(R) to SL2(R).In parti ular we see that GL(R), PL(R) and SL(R) depend only on ker�L up to anoni alisomorphism. 14

If �L is the standard (faithful) 2g dimensional representation of Sp2g(C ) then GL(R) =Sp2g(R), PL(R) = P (R) the subset of matri es of the form 0� A B0 tA�11A with A 2SL2g(R) and SL(R) = SLg(R) the subset of these matri es with B = 0. We expe tthat this is well known, but know no referen es. Brie y one shows by performing row and olumn operations that Sp2g(R) is generated by matri es of the following forms:

1. 0� 1� �ij 00 1 + �ij1A = exp(�Xij) for i 6= j

2. 0� 1g �ij + �ji0 1g1A = exp(Yij) for i > j

3. 0� 1g �ii0 1g1A = exp(Yii)

4. 0� 1g 0�ij + �ji 1g1A = exp(Zij) for i > j

5. 0� 1g � �ii � �jj � �ji + �ij 00 1g � �ii � �jj � �ji + �ij1A= exp(Xij) exp(�Xji) exp(Xij) for i 6= j

6. 0� 1g � �ii �ii��ii 1g � �ii1A = exp(Yii) exp(�Zii) exp(Yii)

that P (R) is generated by matri es of types 1), 2), 3), and 5); and that SLg(R) is generatedby those of type 1) and 3).Let:�1(N) = 8<:0� A BC D

1A 2 Sp2g(R) j C � 0 mod N detA � 1 mod N9=;

15

Then if L is any admissible latti e we have maps:Sp2g(Z) �! GL(Z=NZ)S S�1(N) �! PL(Z=NZ) �! SL(Z=NZ) ompatible with the map L ! L Z=NZ . In parti ular �1(N) !! SLg(Z=NZ) whereexpXij 7! expXij and expYij 7! 0.Re all that the irredu ible representations of Sp2g(C ) are parametrised by ve tors ~n 2X�(T ) with 0 � n1::: � ng. We shall denote the set of su h ve tors X�(T )+. Let V~n denotethe Sp2g(C ) module parametrised by ~n, and give it a GSp2g(C ) a tion by letting �12g a tby �j~nj, where j~nj =Pni. Note that if ~a 2 X�(T ) is a weight of T on V~n then 0� �1g 00 1g1A

a ts on the orresponding weight spa e V ~a~n as � 12P(ni+ai) and that 12P(ni + ai) 2 Z�0 . Inparti ular if � 2 Z then 0� �1g 00 1g1A preserves any admissible latti e.Now hoose v~n 2 V~n a lowest weight ve tor. Set L~n = UZv~n, an admissible latti e inV~n. Let V 0~n =LV ~a~n , where the sum is taken over weights ~a with P(ni + ai) = 0. Also letU 0Z be the subalgebra of UZ generated by the elements Xnijn! and L0~n = U 0Zv~n. Then L0~n � V 0~n.In fa t it is known that L~n is spanned over Z by ve tors of the form Qi>j Xaijijaij ! Qi�j Y bijijbij ! v~nand so we see that L0~n = L~n \ V 0~n and this is a dire t summand of L~n (for an element of theabove form lies in V 0~n if and only if bij = 0 for all i; j).Fix a positive integer N . L~n (Z=NZ) is a �1(N)-module and this a tion fa torsthrough PL~n . We an also make L0~n (Z=NZ) a �1(N) module through the map �1(N)!SL~n(Z=NZ). I laim that with these a tions the proje tion map:

j : L~n (Z=NZ) �! L0~n (Z=NZ)is a map of �1(N)-modules. But it will do to show that:� j(Xijv) = Xijj(v)� j(Yijv) = 0 16

and these are both lear. Moreover if g = 0� N1g 00 1g1A then:

j : g(L~n (Z=NZ)) ��! g(L0~n (Z=NZ))Thus if � � �1(N) is of �nite index, and if U denotes the He ke operator [�g�℄ we see thatour proposition implies that there exists:

I : H�(�; L0~n (Z=NZ)) �! H�(�; L~n (Z=NZ))su h that I Æ j� = U and j� Æ I = U . It is well known that these ohomology groupsare �nitely generated abelian groups and so if N = p a prime we an asso iate a Hidaidempotent e to U . Then we have that:

j� : eH�(�; L~n (Z=NZ)) ��! eH�(�; L0~n (Z=NZ))Now onsider SLg � GSp2g by A 7! 0� A 00 tA�1

1A. Then we see from the fa t thatV 0~n = L0~n C that V 0~n is an irredu ible slg(C )-module of heighest weight depending onlyon the (g � 1)-tuple (n1 � n2; :::; n1 � ng). Thus if ~m 2 Zg with 0 � m1 � ::: � mg andmi�m1 = ni�n1 for i = 2; :::; g then we have an isomorphism of slg(C )-modules V 0~n �! V 0~msu h that v~n 7! v~m, and so L0~n �! L0~m preserving the a tion of the Xij . Thus eH�(�; L~nFp)depends up to anoni al isomorphism only on the (g � 1)-tuple (n2 � n1; :::; ng � n1). Wededu e:Theorem 1.2 Let p be a prime, � � �1(p) a subgroup of �nite index. Fix Q a � Q a p andQ a � C . Then we an asso iate a Hida idempotent e to the a tion of [�0� p1g 00 1g

1A�℄on H�(�; V~n) for ~n 2 X�(T )+, and dim eH�(�; V~n+m~t) is bounded independently of m � 0,where ~t = (1; :::; 1) 2 Zg .Proof:Set ~m = ~n+m~t. Then it will do to show that dim eH�(�; L~mQ p) is so bounded.But we have seen that dim eH�(�; L~m Fp) is so bounded and we have that:

17

� eH�(�; L~m Zp) Fp ,! eH�(�; L~m Fp)� dim eH�(�; L~m Zp) Q p = dim eH�(�; L~m Q p)so the result follows. (The �rst embedding omes from the long exa t sequen e orrespond-ing to 0! Zp p! Zp ! Fp ! 0.)

18

Chapter 2Some Congruen es between SiegelModular Forms2.1 Introdu tionIn this hapter we are on erned with showing how, starting with a Siegel modular form f oflow weight whi h is an eigenform of the He ke operators on a ertain ongruen e subgroupof �0(p) (those matri es in Sp2g(Z) ongruent to 0� � �0 �

1A mod p) whi h is ordinary in thesense that it is an eigenvalue of a ertain He ke operator Up (see se tion 2.2) with eigenvaluea p-adi unit; we an �nd a series of eigenforms of heigher weight whose eigenvalues underthe He ke operators tend to those of the �rst form p-adi ally. We apply this to show howstandard onje tures about the existen e of p-adi Galois representations orresponding tosu h forms of genus two, if true for high weight (where one hopes to �nd the representationsin ertain p-adi ohomology groups) would also be true for ordinary forms of low weight.To explain our results more pre isely re all that given g integers 0 � n1 � ::: � ngwe may onsider Siegel modular forms of genus g and of weight ~n = (n1; :::; ng). Theseare holomorphi fun tions on Siegel modular spa e valued in the irredu ible representationof GLg(C ) with heighest weight ~n and whi h have ertain transformation properties. If12g(g + 1) � n1 then su h forms orrespond to automorphi representations of GSp2g(A )19

whi h are holomorphi dis rete series at in�nity. In this ase one expe ts ( onje turally) tobe able to asso iate to ertain su h modular forms (those whi h are eigenforms of a He kealgebra) a system of 2g dimensional l-adi Galois representations, and to be able to �ndthese representations in the ohomology of ertain sheaves on anoni al models of ertainquotients of Siegel modular spa e. In the ase n1 = 12g(g+1)� 1 su h forms orrespond toautomorphi representations of GSp2g(A ) whi h are limit of holomorphi dis rete series atin�nity. One still expe ts to be able to asso iate Galois representations to su h forms, butone an no longer expe ts to be able to �nd them geometri ally.We shall show how to redu e the se ond ase to the �rst in the ase of \ordinary"forms of genus two. The restri tion on the genus is probably not important but it simpli�essome of the arguments and genus two is the ase of most interest for us. An eigenform on a ongruen e subgroup � of �0(p) is alled ordinary at p if its eigenvalue for the He ke operatorUp = [�0� 1g 00 p1g1A�℄ is a p-adi unit. An eigenform of genus two for a ongruen esubgroup � of Sp4(Z) whi h is dense in Sp4(Zp) is alled ordinary at p if a ertain quarti polynomial Qp(X) (see 2.4) asso iated to f has distin t roots the ratio of no two of whi hequals p and one of whi h is a p-adi unit (in the ase n1 � 2 this is ertainly true if Qp(X)has two distin t roots whi h are p-adi units).Our method is as follows. One an write down a theta series � of weight (p�1; :::; p�1)whi h is ongruent to one modulo p. Multiplying a form f by high p powers of � produ esvery ongruent forms of heigher weight. This along with some ba kground on Siegel modularforms is dis ussed in se tion 2.2. Unfortunately this does not seem to be enough for theappli ations to Galois representations, the problem being that if one starts with an eigenformof the He ke operators one does not obtain an eigenform highly ongruent to it. To over omethis problem in the ase of a ongruen e subgroup ontained in �0(p) one uses the fa t thatthe number of ordinary eigenforms on � of level ~n+m(1; :::; 1) is bounded independently ofm. This is proved by embedding the usp forms in a ertain ohomology group and usingthe results of the �rst hapter to relate these as m varies. From this it is not diÆ ult tosee (by an argument of Wiles using Fitting ideals) that we an lift f�pm to an eigenform

20

of weight ~n+ pm(p� 1)(1; :::; 1) with eigenvalues ongruent to those of f modulo pm+1C forsome C independent of m, as we wanted. This argument is dis ussed in se tion 2.3. Theappli ation to Galois representations is dis ussed in se tion 2:5. In se tion 2.4 we showhow to onstru t from an eigenform � on a general ongruen e subgroup whi h is densein Sp4(Zp) an eigenform on � \ �0(p) whi h is also an eigenform for Up. We use this togeneralise the results about Galois representations to ongruen e subgroups not ontainedin �0(p).2.2 Review of Siegel Modular FormsFix an integer g � 1. Let GSp2g(R) denote the set of � 2 GL2g(R) su h that:

�0� 0 1g�1g 01A t� = �(�)0� 0 1g�1g 0

1Afor some � : GSp2g(R)! R�. Let Sp2g(R) be those elements of GSp2g(R) in the kernel of�. Let G(R) = GSp2g(R), G1 = G(R ), G+1 = ��1R�>0 � G1, G(Q )+ = G(Q ) \ G+1, andA (resp. A f ) denote the adeles (resp. �nite adeles) of Q . Let U1 denote the group of:0� A B�B A

1A 2 GL2g(R )su h that AtB is symmetri and AtA+BtB is a nonzero s alar, or:U1 = �(�; �) 2 GLg(C ) � C � j (�; �)� = (�; �)where (�; �)� = (� (t� )�1; � ) ( denoting omplex onjugation), and the orresponden eis given by: � = 0� A B�B A

1A 7�! (A� iB; �(�))Thus U1 is a real form of GLg(C ) � C � . It is in fa t the g � g unitary similitudes.Let Z = Zg denote the set of symmetri omplex g�g matri es x+p�1y with y positivede�nite. Then G+1 a ts on Z by 0� A BC D

1A : z ! (Az +B)(Cz +D)�1. If z0 = (p�1)1g21

then � 7! �z0 gives a bije tion G+1=U1 �! Z. We de�ne a map J : G+1�Z ! GLg(C )�C �by: 0�� = 0� A BC D1A ; z1A �! (Cz +D; �(�))

so that: J(��; z) = J(�; �z)J(�; z)If � is a �nite dimensional representation of GLg(C ) � C � on a omplex ve tor spa e Vthen we set J� = � Æ J : G+1 � Z ! Aut(V ).If U � G(A f ) is an open ompa t subgroup we let S�(U) denote the spa e of fun tions� : G(Q )nG(A ) ! V su h that:� �(guu1) = �(u1)�1�(g) for all g 2 G(A ), u 2 U and u1 2 U1� if h 2 G(A ) then the fun tion:fh : Z �! V�z0 7�! J�(�; z0)�(h�)where � 2 G+1, is holomorphi . (It is easily he ked that this fun tion is well de�ned.)� RN(Q )nN(A ) �(nh)dn = 0 where N is the unipotent radi al of any proper paraboli subgroup and where dn is any invariant measure on N(Q )nN(A )We set S� = SS�(U) as U ranges over open ompa t subgroups. Then G(A f ) a ts on S� onthe right by (�jg)(h) = �(hg�1), and S�(U) = SU� . We de�ne similarlyM�(U) by omittingthe last ondition (assuming g > 1, whi h is the only ase we shall be on erned with asthe ase g = 1 is well known).Similarly if � � G+1 is a dis rete subgroup we set S�(�) to be the set of holomorphi fun tions f : Z ! V su h that:� f j = f for all 2 �� lim�!+1(f j )0� z 00 i�

1A = 0 for all 2 G(Q )+ and all z 2 Zg�122

where for 2 G+1 we de�ne:(f j )(z) = J�( ; z)�1f( z)Similarly we may de�ne M�(�) by dropping the last ondition (if g > 1).Assume U � G(A f ) is an open ompa t subgroup su h that (�U)Q �R�>0 = A � . Thenif we set �U = U \G(Q )+ we have an isomorphism:

S�(U) �= S�(�U )given by: � 7�! (f� : �z0 7! J�(�; z0)�(�))and inversely by: f 7�! (�f : u� 7! J�(�; z0)�1f(�z0))where � 2 G+1, u 2 U and 2 G(Q ). The se ond map is well de�ned as, by the strongapproximation theorem and our assumption on U , we have that G(A ) = G(Q )UG+1 . Ifh 2 G(A f ) and h = u with u 2 U and 2 G(Q )+ then:�f jh = �hj and f�jh = f�j Now let U and U 0 be open ompa t subgroups of G(A f ) and let g 2 G(A f ), then wede�ne a He ke operator: [UgU 0℄ : S�(U) �! S�(U 0)� 7�! P�jgiwhere UgU 0 = `Ugi. If U and U 0 also satisfy the ondition of the last paragraph we anthink of [UgU 0℄ : S�(�U ) ! S�(�U 0). It is given by f 7! P f j i where gi = ui i with i 2 G(Q )+ and ui 2 U . Equivalently we may write g = u with 2 G(Q )+ and u 2 U ,and then the i's may be de�ned by �U �U 0 =`�U i.It is well known that if V is irredu ible there is an inner produ t on V , say h ; i, su hthat: h�(�)v1; �(�)v2i = �(�)�hv1; v2i

23

for all � 2 U1 and v1; v2 2 V , and where � depends only on � (in fa t �(x1g; x2) = x�).Now de�ne an inner produ t on S�(U) by:h�1; �2i = ZG(Q)nG(A )=UU1 h�1(g); �2(g)ijj�(g)jj��dgwhere dg is an invariant measure on G(Q )nG(A )=UU1 and jj:jj : Q �nA � ! R�>0 byx 7! Q jxvjv. This is easily he ked to be well de�ned. Moreover the adjoint of g isjj�(g)jj��g�1 (if the two measures are normalised orre tly) and the adjoint of [UgU 0℄ isjj�(G)jj��[U 0g�1U ℄.We now introdu e some spe i� representations �. We re all �rst some multilinearalgebra. Let 0 � n1 � ::: � ng be integers, and set ~n = (n1; :::; ng) and j~nj = n1 + ::: + ng.Also let ~t = (1; :::; 1). Sj~nj, the symmetri group on j~nj letters a ts on Mj~nj. There is anelement in Z[Sj~nj℄ (unique up to �1) satisfying:� 2 = � for some s alar �� =P�2Sj~nj Æ�� where Æ� = 0, 1 or �1� � = if � preserves the sets f1; :::; n1g, fn1 + 1; :::; n1 + n2g, et .� � = (�)� (where (�)� denotes the sign of sigma) if � preserves sets of the formfnij�1 + j; nij�1+1 + j; :::; j~nj � ng + jg where ij is the least index su h that nij � j(where n0 = 0).We an think of 2 End(Mj~nj). Let N~nM denote Mj~nj. Then N~n ommutes withlo alisation and if � : M ! N is linear we get a linear map ~n(�) : N~n(M) ! N~n(N).If R is any ring this gives a natural a tion of GLg(R) on N~n(Rg), and if R is a �eld of hara teristi zero this representation is known to be irredu ible (see [We℄).Let W~n denote the GLg(R)�R� moduleN~n(Rg) with the above a tion of GLg(R) andon whi h R� a ts via � 7! � 12g(g+1)�j~nj. Let �~n denote the orresponding representation,let W~n denote W~n(C ) and drop the � when �~n is used as a sub- or super-s ript.If R is a ring denote the ring of formal power series:Xh2symm�g(Z)�0 ahqh24

in an indeterminate q by R[[q℄℄g. Here symm�g(Z) denotes the semigroup of g�g symmetri integral matri es with even diagonal entries, and a supers ript � 0 (resp. > 0) indi atesthose whi h are positive semi-de�nite (resp. positive de�nite). If M is an R-module wede�ne M [[q℄℄g = R[[q℄℄g M .If f 2M�(�) there is an integer N depending only on � su h that f(z+Nh) = f(z) forh 2 symmg(Z). Thus we have a Fourier expansion:f(z) = Xh2symm�g(Z)�0 ah(f) exp(�p�1N�1tr(hz))and so we get an embedding M�(�) ,! W~n[[q1=N ℄℄g. f 2 S�(�) if and only if ah(f j ) = 0for all 2 G(Q )+ and for all deth = 0. If R � C we de�ne M~n(�; R) = M�~n(�) \W~n(R)[[q1=N ℄℄g. We de�ne S~n(�; R) similarly, and for U an open ompa t subgroup ofG(A f )with (�U)Q �R�>0 = A � we de�ne S~n(U;R) to be S~n(�U ; R), and similarly forM~n(U;R).Lemma 2.1 Let � � G(Q )+ be a dis rete ongruen e subgroup, then there is a �nite abelianextension K=Q su h that S~n(�) = S~n(�;OK) C , and similarly M~n(�) =M~n(�;OK) C .Proof:It will learly do to show the following:1. M~n(�) is �nite dimensional2. M~n(�) =M~n(�; Q ab) C3. S~n(�) = S~n(�; Q ab) C4. if f 2M~n(�; Q ab) then there exists 0 6= C 2 Q ab with Cf 2M~n(�;OQab )1) is well known. If M~n denotes the union over all ongruen e subgroups � of M~n(�) thenShimura has proved that M~n =M~n(Q ) C and that G(Q )+ preserves M~n(Q ab) (see [Sh3℄).2) and 3) follow from this. Finally it will do to establish 4) in the spe ial ase when �is equal to the set of matri es in Sp2g(Z) whi h are ongruent to 12g modulo N for someN > 3.It is known (see for example [Fa2℄) that there is a separated s hemeM and a prin ipallypolarised abelian s heme A=M of relative dimension g together with an isomorphism � :25

(�gN � (Z=NZ)g ) �ZM �! A[N ℄ taking the standard pairing (�gN � (Z=NZ)g )2 ! �N tothe Weil pairing A[N ℄2 ! �N (i.e. a level N stru ture), with the property that if A=S isa prin ipally polarised abelian s heme with level N stru ture, then there is a unique mapS !M su h that A �= A�M S and the polarisation and level N stru ture on A ome bypulling ba k that on A. If A=S is an abelian s heme, let !A=S denote the dire t image ofthe sheaf of relative di�erentials A=S. Then !A=S is quasi oherent on S, and if T ! Sthen !A�T=T is the inverse image of !A=S. (These are easy from the de�nition of and thefa t that A ! S is quasi ompa t.) In parti ular !A=S is the inverse image of !A=M underthe anoni al map S !M, and we get a map !A=M(M)! !A=S(S).Now onsider Z[[q1=N ℄℄. There is an h 2 symm�g(Z) su h that if f = qh and R =Z[[q1=N ℄℄[f�1℄, then it is known (the Mumford onstru tion, but see [Fa2℄) that there is anabelian s heme A=spe R with a anoni al isomorphism !A=R �= Rg. Thus we obtain a map!A=M(M)! Rg. Moreover if ~n is as above we get a map (N~n !A=M)(M)!N~n(Rg). Ifwe tensor over Z with C we get a ommutative diagram:(N~n !A=M)(M) �! N~n(Rg)# #(N~n !AC =MC )(MC ) �! N~n(R C )gIt is further known that (N~n !AC =MC )(MC ) =M~n(�) and that the map (N~n !AC =MC )(MC )! N~n(C [[q1=N ℄℄g[f�1℄g) is just the normal q-expansion. Finally as !A=M is quasi oher-ent and C is at over Z, N~n(!AC =MC (MC )) = N~n(!A=M(M)) C . Thus the image ofN~n(!A=M(M)) in M~n(�) spans M~n(�) and ea h element has a Fourier expansion with oeÆ ients in Z as desired.Corollary 2.1 With the notation as in the lemma, if � = �U for U � QGSp2g(Zl) anopen ompa t subgroup normalised by iQZ�l where i : G m ! GSp2g by t 7! 0� 1g 00 t1g1A,then we may take K = Q .Proof:It will do to show that S~n(�; Q ab) is stable under the a tion of Gal(Q ab=Q ), andsimilarly for M~n(�; Q ab). We shall only treat the �rst ase, whi h is marginally harder.26

Shimura (see [Sh3℄) de�nes an a tion of G(A f ) on M~n(Q ab) su h that in parti ular everyfun tion is stabilised by an open subgroup, G(Q )+ has its normal a tion, and if t 2 QZ�lthen f it = f (t�1;Qab=Q) , where ( ; Q ab=Q ) is the Artin symbol. Let f 2 S~n(�; Q ab) andt 2QZ�l , we must show that f it 2 S~n(�; Q ab).Firstly let � 2 G(Q )+ and h 2 symm�g(Z)�0 with deth = 0. Then we an �nd u 2stabG(A f )(f), � 2 G(Q )+ and s 2QZ�l su h that i(t)� = u�i(s), so we see that:ah(f i(t)j�) = ah(fu�i(s)) = ah(f j�)(s�1;Qab=Q) = 0Thus f i(t) 2 S~n(Q ab). Se ondly let � 2 �, then we an �nd � 2 G(Q )+ , s 2 QZ�l andu 2 W = fx 2 QG(Zl)jx � 12g mod Ng � stabG(A f )(f) \ U for some N , su h thati(t)� = u�i(s). Then we see that � 2 Sp2g(Z):f1; i(�1)g and �(u)s = �t. Without loss ofgenerality we may assume s = t, �(u) = 1 and � 2 Sp2g(Z). In fa t in this ase:� 2 Sp2g(Z) \W:i(t)Ui(t)�1� Sp2g(Z) \ U = �and so: f i(t)j� = fu�i(t) = (f j�)i(t) = f i(t)and we are done.We shall now introdu e some parti ular He ke operators of spe ial importan e for us. Fixa rational prime p and onsider an open ompa t subgroup U = U1�U2 �Ql 6=pGSp2g(Zl)�GSp2g(Zp) satisfying:� U � 0� 1g 00 (QZ�l )1g

1A� there is an integer r = r(U) � 1 su h that U2 is the set of matri es 0� A BC D

1A 2GSp2g(Z) with C � 0 mod pr and (A mod pr) and (D mod pr) lying in some �nite setof possibilities.Let N 0 = N 0(U) be the smallest positive integer su h that U ontains all elements ofQGSp2g(Zl) whi h are ongruent to one modulo N 0, and write N 0(U) = N(U)pr (or if no27

onfusion an arrise N 0 = Npr. Then we de�ne:Up = [U 0� 1g 00 p1g

1AU ℄ = [U 0� 1g 00 �p1g1AU ℄

where �p denotes a uniformizer in Zp . Then we have:Lemma 2.2 Let U be as des ribed above. Then S~n(U) = S~n(U;Z) C and the He keoperator Up preserves S~n(U;Z). In fa t (P ahqh=N )jUp = P aphqh=N . Thus we an de�nea Hida idempotent e on S~n(U) as in se tion 1.2.Proof:Let X be a set of representatives for symmg(Z) modulo p, su h that ea h X 2 Xis ongruent to zero modulo N . Then:U 0� 1g 00 p1g

1AU = aX2X U0� 1g X0 p1g

1Aas follows easily from the fa t that if 0� A BC D

1A 2 U there is an X 2 X with B �AX mod p (A is invertible modulo p) and from the equality:0� 1g 00 p1g1A0� A BC D

1A = 0� A p�1(B � AX)pC D � CX1A0� 1g X0 p1g

1ANow note that for h 2 symm�g(Z):XX2X exp(�p�1N�1p�1tr(hX)) = 8<: p 12g(g+1) if h 2 p symm�g(Z)0 otherwise(If h 2 p symm�g(Z) this is lear. If not pi k Y 2 X with 2Np6 jtr(hY ), and then:PX2X exp(�p�1N�1p�1tr(hX)) = (PX2X exp(�p�1N�1p�1tr(hX)))exp(�p�1N�1p�1tr(hY ))and the result follows.) Now:(P ah exp(�p�1N�1tr(hz)))jUp= �~n(p1g; p)�1P ah exp(�p�1N�1p�1tr(hz))PX exp(�p�1N�1p�1tr(hX))= p� 12g(g+1)Ppjh ahp 12g(g+1) exp(�p�1N�1p�1tr(hz))= P aph exp(�p�1N�1tr(hz))

28

Keep the notation of the lemma. Then Z = Z(U) = ((Z=N 0Z)�)g a ts on S~n(U) by(a1; :::; ag) 7! �a = diag( ~a1; :::; ~ag ; ~a1�1; :::; ~ag�1) where ~ai 2 QZ�l with ( ~ai)l equal to 1 ifl 6 jN 0 and ai if ljN 0. We an de ompose S~n(U) =L�2 �Z S~n(U)�. Let R = Q ab \ OQa p , andS~n(U;�;R) = S~n(U)� \ S~n(U;R). It follows from the results of Shimura dis ussed abovethat S~n(U)� = S~n(U;�; U) R C . Let T = T(U) be the abstra t double oset algebra overZ generated by the operators [UxU ℄ where x 2 M2g(Zl) \ GSp2g(Q l) for all primes l 6 jN 0.It is known that T is ommutative (see for example [A2℄). Moreover T a ts on S~n(U), andea h element T 2 T a ts as a normal operator (i.e. it ommutes with its adjoint). Thusthe elements of T an be simultaneously diagonalised. The a tion of T ommutes with thatof Z, and if r(U) � 1 these both ommute with the a tion of Up. I laim that T preservesS~n(U;�;R).To see this let x 2 M2g(Zl) \ GSp2g(Q ) then we an write UxU = `Uxi wherexi 2M2g(Zl) \GSp2g(Q l) is of the form:0BBBBBBBBBBBBBBBBBBB�

a1 0 : : : 0 � � : : : �� a2 0 � � �... . . . ... ... . . . ...� � : : : ag � � : : : �0 0 : : : 0 ba�11 � : : : �0 0 : : : 0 0 ba�12 : : : �... . . . ... ... . . . ...0 0 : : : 0 0 0 : : : ba�1g

1CCCCCCCCCCCCCCCCCCCA(see for instan e [A2℄). We may further suppose that b and ea h ai lies in lZ � Zl . Then onsider an element x0i 2M2g(Z) \GSp2g(Q ) de�ned by:� x0i lies in the same Borel as xi was required to lie in� x0i � diag(a1; :::; ag ; ba�11 ; :::; ba�1g ) mod N 0� x0i � xi mod l?

29

Then for ? large enough Uxi = U�a�1x0i so that f jxi = �(a�1)f jx0i and x0i learly preservesR[[q1=N ℄℄g.Finally we introdu e a parti ular modular form whi h we shall need later.Lemma 2.3 Fix a rational prime p, and set:�0(p) = 8<:0� A BC D

1A 2 Sp2g(Z) ������ C � 0 mod p9=;Then there is an element � 2M(p�1)~t(�0(p);Z) with a0(�) = 1 and pjah(�) for all h 6= 0.Proof:This argument is due to Hida in the ase g = 1.Let Qn denote the (n� 1)� (n� 1) matrix:0BBBBBBBBBBBB�

2 �1 0 : : 0�1 2 �1 00 �1 2 0: : :: : :0 0 0 : : 2

1CCCCCCCCCCCCAthen Qn 2 symm�n�1(Z) and an easy indu tion on n shows that detQn = n and hen e inparti ular Qn is positive de�nite. Let:�1 = XX2M(p�1);g(Z) exp(�p�1tr(tXQpXz))Then theorems 2 and 3 of [AM℄ imply that � = �21 2 M(p�1)~t(�0(p);Z). It is lear thata0(�) = a0(�1)2 = 1. Now let � be the (p� 1)� (p� 1) matrix:0BBBBBBBBBBBBBBB�

0 0 0 : : 0 �11 0 0 0 �10 1 0 0 �1: : :: : :0 0 0 0 �10 0 0 : : 1 �1

1CCCCCCCCCCCCCCCA30

It has hara teristi polynomial Xp�1 + ::: + 1 and its eigenvalues are the nontrivial pthroots of one. Thus Z=pZ a ts on M(p�1);g(Z) by m : X 7! �mX and the orbits are eitherf0g or have ardinality p. It is easily he ked that t�Qp� = Qp, so that tXQpX is onstanton su h orbits. Then pjah(�1) for h 6= 0, from whi h the result follows.Corollary 2.2 Let U be an open ompa t subgroup of G(A f ) satisfying the onditions de-s ribed before lemma 2.2. Let R = Q ab \ OQa p and � be a hara ter on Z(U). Then thereis a map: im : eS~n(U;�;R) ,! eS~n+(p�1)pm�1~t(U;�;R)su h that for all h 2 symm�g(Z) ah(imf) � ah(f) mod pm and infa t for all T 2 T(U) wehave ah((imf)jT ) � ah(f jT ) mod pm. (Re all that ~t = (1; :::; 1).)

Proof:Set im(f) = e(�pmf). Then �pmf � f mod pm, so U r!p (�pmf) � U r!p f mod pm andhen e e(�pmf) � ef = f mod pm.Also if T 2 T re all that f jT =P �if jx0i where �i is a root of unity and x0i 2 M2g(Z) \GSp2g(Q ) is as des ribed in the dis ussion after lemma 2.2. But if j~n denotes the a tion asfor modular forms of weight ~n, then:(X ahqh=N )j~n+a~tx0i = �ai (X ahqh=N )j~nx0ifor some integer �i depending only on x0i. Thus:im(f)j~n+(p�1)pm�1~txi � im(f j~nxi) mod pmand so im(f)jT � im(f jT ) mod pm as desired.

2.3 Relation to CohomologyOur aim in this se tion is to relate our spa es of automorphi forms to ertain ohomol-ogy groups and use this to prove an analogue of theorem 1.2 for automorphi forms. Wethen apply this to �nd ongruen es between eigenforms of the He ke operators in di�erentweights. 31

We �x some notation:sp2g = 8<:0� A BC �tA

1A 2M2g������ tB = B; tC = C9=;k = 8<:0� A B�B A

1A 2M2g������ tB = B; tA = �A9=;p = 8<:0� A BB �A

1A 2M2g������ tB = B; tA = A9=;a = 8<:0� 0 ��� 0

1A 2M2g������ � is diagonal9=;Then sp2g = k � p, k(R ) is the Lie algebra of K1 = U1 \ Sp2g(R ), and a � k � sp2g is aCartan subalgebra. We �x a(C )0 �= C g by ~x : diag(�1; :::; �g) 7! p�1Pxi�i. Then if Tis the maximal torus of Sp2g(C ) onsidered in example 1.2, and t(C ) is its Lie algebra, wemay onjugate t(C ) to a(C ) su h that the map:Zg �= X�(T ) � t(C )0 ��! a(C )0 �= C gis the anoni al in lusion. Thus the roots of a on sp2g are just the ve tors �ij (i 6= j), �ij(i � j) and ij (i � j) as in example 1.2. The roots � of a on k are the �ij , and those(�n) on p are the �ij and the ij . Choose the same order we hose in example 1.2. LetW denote the Weyl group of sp2g and Wn the subset of elements w su h that �+ � w�+where �+ = � \ �+. Let U denote the universal enveloping algebra of sp2g(C ), and Z(U)its entre. Then the homomorphisms Z(U)! C are parametrised by a(C )0=W (the Harish-Chandra parametrisation). We shall denote this orresponden e � $ ��. We now restatea spe ial ase of theorem 10 of [Fa1℄ in a slightly di�erent notation:Let:� � � G+1 be a torsion free dis rete subgroup� w 2Wn of length l(w)� � be the basis orresponding to �+, and ÆG = (1; :::; g) half the sum of thepositive roots 32

� � 2 a(C )0 be a dominant (integral) weight with respe t to �� W (�) denote the irredu ible K1 module with highest weight �� C� be the spa e of uspidal C1 fun tions f : �nSp2g(R ) !W (�) su h thatf(hk) = k�1f(h) for all h 2 Sp2g(R ) and k 2 K1� C��� be the subspa e of C� that transform by �� under the a tion of Z(U)� V (�) be the irredu ible Sp2g(C ) module of heighest weight �� V(�) the sheaf on �nZ de�ned by setting V(�)(U) to be the set of C1fun tions f : ~U ! V (�) su h that f( z) = f(z) for all z 2 Z and 2 �,and where ~U is the pre-image of U under the map Z ! �nZ� H�P denote the image of the ohomology of ompa t support in the oho-mology, or equivalently the kernel of the map from the ohomology to the ohomology of the boundary of the Borel-Serre ompa ti� ationthen: C��+ÆGw(�+ÆG�w�1ÆG) ,! H l(w)P (�nZ;V(�)) ,! H l(w)(�; V (�))Moreover the He ke operator [�g�℄ orresponds to the He k operator [�g�1�℄,for g 2 Sp2g(R ) for whi h these operators make sense.Now take w : (x1; :::; xn) 7! (�xn; :::;�x1). Note that ÆG � w�1ÆG = (g + 1)~t. LetÆK = 12(1�g; 3�g; :::; g�1) be half the sum of the elements of �+ . Then the representation�~n of K1 � U1 de�ned in the last se tion has heighest weight w~n, whi h is dominantwith respe t to w�+. Thus Z(U) a ts on S~n(�) via �w~n+2wÆK�wÆG = �w~n+ÆG so that for� torsion free and for ~n with g + 1 � n1 � ::: � ng:S~n(�) ,! C�~n+w�1ÆGw~n ,! H 12g(g+1)P (�nZ;V(~n� (g + 1)~t)),! H 12g(g+1)(�; V (~n� (g + 1)~t))and so for any dis rete �:

S~n(�) ,! H 12g(g+1)(�; V (~n� (g + 1)~t))33

(We may hoose a normal subgroup �0 of �nite index whi h is torsion free and then take �=�0invariants, using the in ation restri tion sequen e on the right.) Moreover if g 2 G(Q )+ thenthe a tion of [�g�℄ on S~n(�) orresponds to that of [�{g�℄ on H 12g(g+1)(�; V (~n� (g+1)~t)),where {g = 0� 0 �1g1g 01A tg0� 0 �1g1g 0

1A�1. To see this we need only he k that for� 2 R�>0 [��12g�℄ a ts on both by �j~nj�g(g+1), whi h is easy. Now we have:Proposition 2.1 Let U � G(A f ) be an open ompa t subgroup satisfying the onditionsstated before lemma 2.2 with r(U) � 1, let ~n be su h that 0 � n1::: � ng and let e denotethe Hida idempotent. Then there is a onstant C su h that:

dim eS~n+m~t(U) < Cfor all m � 0, where ~t = (1; :::; 1).

Proof:Without loss of generality we may restri t to m � g + 1. Then by theorem 1.2we an hoose C su h that dim eH 12g(g+1)(�U ; V (~n+m0~t)) < C for all m0 � 0. Now �x m.Also hoose a �nite Z module M � C su h that:S~n+m~t(�U ;Z) ,! H 12g(g+1)(�U ; L~n+(m�g�1)~t)TF MThen: eS~n+m~t(�;Z) ,! e(H 12g(g+1)(�U ; L~n+(m�g�1)~t)TF M)= (eH 12g(g+1)(�U ; L~n+(m�g�1)~t)TF )M� eH 12g(g+1)(�U ; V~n+(m�g�1)~t)and so dimS~n+m~t(�U ;Z) < C and we are done.We an now dedu e our �rst main result:

Theorem 2.1 Let U � G(A f ) be an open ompa t subgroup satisfying the onditions statedbefore lemma 2.2 with r(U) � 1, let ~n be su h that 0 � n1 � ::: � ng, let R = Q ab\OQa , andlet e denote the Hida idempotent. Let f 2 eS~n(U)� be an eigenform for T with eigenvaluesgiven by � : T ! R. Then we an �nd fm 2 eS~n+am~t(U)� su h that:34

� fm is an eigenvalue for T with eigenvalues �m : T ! R� am !1 as m!1� supT j�(T )� �m(T )jp ! 0 as m!1Proof:We may assume f 2 eS~n(U;�;R) and that if � 2 Q ab and �f 2 S~n(U;�;R) then� 2 R. Then by orollary 2.2 we an �nd f 0m 2 eS~n+am~t(U;�;R) with fmjT � f jT mod pr0mfor all T 2 T, where am and r0m tend to in�nity with m. Let Tm denote the image of T inEnd(eS~n+am~t(U;�;R)). Then we get �0m : Tm !! R=pr0mR with �0m � � mod prm . Call itskernel Im. Let C be the bound from the last proposition (proposition 2.1). Then we an �ndless than C fun tions hm;i 2 eS~n+am~t(U;�;R) whi h span eS~n+am~t(U;�;R) and su h thatea h hm;i is an eigenform for Tm with eigenvalues given by �m;i say. Then Vm = �Rhm;i is afaithful Tm module. Thus, if Fitt denotes the Fitting ideal (see, for example, the appendixof [MW1℄), FittTm (Vm) = 0 and so:0 = FittTm=Im(Vm=ImVm)= Qi FittR=pr0mR(R=�m;i(Im))= Qi �mi(Im) � R=pr0mRThus for some i, valp(�m;iIm) � r0m=C. Let fm = hm;i and we are done.

2.4 Some Lemmas on He ke OperatorsThe dis ussion in the other se tions is prin ipally on erned with \ordinary" forms, i.e.modular forms in the image of the Hida idempotent e a ting on a spa e of modular forms foran open ompa t subgroup ontained in U0(p), the set of matri es 0� A BC D

1A in QG(Zl)with C � 0 mod p. It is often more interesting to onsider a modular form f on an open ompa t subgroup ontaining G(Zp). In this se tion we give a riterion for ef 2 S(U\U0(p))not to vanish. In fa t we shall only treat the ase of genus two. The al ulations are alreadyvery messy, and this is the ase of prin ipal interest for us.35

In the ase of genus one the answer to this question is very easy. If f 2 Sk(�1(N)),with N prime to p, is an eigenform for Tp and Sp, say with eigenvalues ap and dp, thenef 2 Sk(�1(Np)) is non-zero if and only if one root of the equation Q(X) = X2�apX+pdpis a p-adi unit. To prove this (at least in the ase that this polynomial has distin t roots)one writes down two forms f1 and f2 in Sk(�1(Np)) whi h are both eigenforms for Up witheigenvalues the roots of Q(X), and su h that f is a linear ombination of the two (and nota multiple of either one separately).For the rest of this hapter we assume that g = 2. Also let N be an integer and U theopen ompa t subgroup of QGSp4(Zl) onsisting of matri es ongruent to 0� 12 00 �121Amodulo N . Let p be a prime not dividing N . Then we de�ne He ke operators:

� T (p) = Tp =26666664U0BBBBBB�

1 0 0 00 1 0 00 0 �p 00 0 0 �p1CCCCCCAU

37777775� Tp2 =

26666664U0BBBBBB�

1 0 0 00 1 0 00 0 �2p 00 0 0 �2p

1CCCCCCAU37777775

� Rp =26666664U0BBBBBB�

1 0 0 00 �p 0 00 0 �2p 00 0 0 �p1CCCCCCAU

37777775� Sp =

26666664U0BBBBBB�

�p 0 0 00 �p 0 00 0 �p 00 0 0 �p

1CCCCCCAU37777775

� T (p2) = Tp2 +Rp + Sp 36

where �p denotes a uniformiser in Zp . Also let Qp(X) be the formal polynomial whose oeÆ ients are He ke operators given by:X4 � TpX3 + (T 2p � T (p2)� p2Sp)X2 � p3TpSpX + p6S2pRe all the following formulae (see [A2℄):� T 2p � T (p2)� p2Sp = pRp + p(p2 + 1)Sp = (p+ 1)�1(pT 2p � pTp2 � (p4 � 1)Sp)� pRp = T 2p � T (p2)� p(p2 + p+ 1)Sp� pTp2 = (p+ 1)T (p2) + p2(p+ 1)Sp � T 2pAlso re all that S~n is a dire t sum, L�, of irredu ible admissible representations � =N�l of GSp4(A f ). If p6 jN and if �U 6= (0) then �p is spheri al and so is the unique spheri alirredu ible subquotient of some unrami�ed prin ipal series representation. (See [C℄ for thisand the fa ts quoted below about su h representations.)We �rst dis uss the a tion of He ke operators on unrami�ed prin ipal series representa-tions, and then we apply these results to spa es of usp forms. Fix the Borel:

B =0BBBBBB�� 0 � �� � � �0 0 � �0 0 0 �

1CCCCCCAand the maximal torus T onsisting of diagonal matri es. Des ribe unrami�ed hara terson T (Q p) by triples (�1; �2; ) of unrami�ed hara ters on Q �p , where:(�1; �2; ) : diag(�; �; ���1; ���1) 7�! �1(�)�2(�) (�)In parti ular let Æ denote the hara ter taking this matrix to j�2�4��3jp. By the unrami�edprin ipal series orresponding to (�1; �2; ) we mean the representation on the spa e oflo ally onstant fun tions � : GSp4(Q p) ! C satisfying �(bh) = ((�1; �2; )Æ 12 )(b)�(h) forall b 2 B(Q p) and h 2 GSp4(Q p); where the a tion is given by (�g)(h) = �(hg�1). Weshall denote this representation by �(�1; �2; ) and its irredu ible spheri al subquotient37

by �(�1; �2; ). Note this a tion is twisted from that in [C℄. Then it is easy to see that�(�1; �2; )GSp4(Zp) = C��1 ;�2; where:�(bk) = ��1;�2; (bk) = ((�1; �2; )Æ 12 )(b)

for b 2 B(Q p) and k 2 GSp4(Zp). That this is a good de�nition follows from the Iwasawade omposition. Then we an ompute that:� �jTp = p 32 (p�1)(1 + �1(p�1) + �2(p�1) + �1�2(p�1))�� �jSp = (p�2)�1�2(p�1)�� �jRp = p2 (p�2)(�1(p�1) + �2(p�1) + �1�2(p�1) + �21�2(p�1) + �1�22(p�1))�� (p�2)�1�2(p�1)�These follow from the oset de ompositions (see [A2℄):� GSp4(Zp)diag(1; 1; p; p)GSp4(Zp) =`GSp4(Zp)� as � runs over the matri es:{0BBBBBB�

1 0 x y0 1 y z0 0 p 00 0 0 p1CCCCCCA for x; y; z = 0; :::; p � 1

{0BBBBBB�

p 0 0 0�i 1 0 z0 0 1 i0 0 0 p1CCCCCCA for i; z = 0; :::; p � 1

{0BBBBBB�

1 0 x 00 p 0 00 0 p 00 0 0 11CCCCCCA for x = 0; :::; p � 1

38

{0BBBBBB�

p 0 0 00 p 0 00 0 1 00 0 0 11CCCCCCA

� GSp4(Zp)diag(p; p; p; p)GSp4(Zp) = GSp4(Zp)diag(p; p; p; p)� GSp4(Zp)diag(1; p; p2; p)GSp4(Zp) =`GSp4(Zp)� as � runs over the matri es:{0BBBBBB�

1 0 x y0 p py 00 0 p2 00 0 0 p1CCCCCCA for x = 0; :::; p2 � 1 and y = 0; :::; p � 1

{0BBBBBB�

p 0 0 pyi 1 y z0 0 p pi0 0 0 p21CCCCCCA for i; y = 0; :::; p � 1 and z = 0; :::; p2 � 1

{0BBBBBB�

p 0 0 00 p2 0 00 0 p 00 0 0 11CCCCCCA

{0BBBBBB�

p2 0 0 0�pi p 0 00 0 1 i0 0 0 p1CCCCCCA for i = 0; :::; p � 1

{ 0� p12 B0 p121A where B runs over non-zero symmetri 2 � 2 integral matri esmodulo p whi h satisfy detB � 0 mod pThen we easily on lude that:

39

� �j(pRp + p(p2 + 1)Sp) = p3 (p�2)(�1(p�1) + �2(p�1) + 2�1�2(p�1) + �21�2(p�1) +�1�22(p�1))�� �jQp(X) = (X�p 32 (p�1))(X�p 32 �1(p�1))(X�p 32 �2(p�1))(X�p 32 �1�2(p�1))�Now let � denote the subgroup of elements of GSp4(Zp) whi h are ongruent to a matrixof the form 0� � �0 �1A modulo p. Then GSp4(Q p) =`41B(Q p)wi�, where:

w1 = 14 w3 =0BBBBBB�

1 0 0 00 0 0 10 0 1 00 �1 0 01CCCCCCA

w2 =0BBBBBB�

0 0 1 00 1 0 0�1 0 0 00 0 0 11CCCCCCA w4 =

0BBBBBB�0 0 1 00 0 0 1�1 0 0 00 �1 0 0

1CCCCCCAThus �(�1; �2; )� has a basis onsisting of fun tions f1; f2; f3; f4 where fi is supported onB(Q p)wi� and where fi(wi) = 1. We shall represent the fun tion P�ifi by the row ve tor(�1; :::; �4), so in parti ular � is represented by (1; 1; 1; 1). We shall al ulate the matrixrepresenting the a tion of the He ke operator Up = [�diag(1; 1; p; p)�℄ with respe t to thisbasis. It is easy to see that it is represented by (fj jUp(wi)), and we laim that this is:

p 12 (p�1)0BBBBBB�

p p� 1 p� 1 p� 10 p�1(p�1) (p� 1)�1(p�1) (p� 1)�1(p�1)0 0 p�2(p�1) (p� 1)�2(p�1)0 0 0 p�1�2(p�1)1CCCCCCA

To see this �rst note that:(fj jUp)(wi) =XX �1�2 2(p�1)fj0�wi0� p12 X0 12

1A1A40

where X runs over 2� 2 integral matri es modulo p. To al ulate these values we write:wi0� p12 X0 12

1A = b(i;X)wk(i;X) (i;X)where b 2 B(Q p), 2 � and k = 1; 2; 3 or 4. Then:

fj0�wi0� p12 X0 121A1A = 8<: ((�1; �2; )Æ 12 )(b(i;X)) if j = k(i;X)0 otherwiseThe b's, 's and k's are given by the following formulae:

� w10� p12 X0 121A = 0� p12 X0 12

1Aw1� If x 6� 0 mod p, say ax � 1 mod p then:

w20BBBBBB�

p 0 x y0 p y z0 0 1 00 0 0 11CCCCCCA =

0BBBBBB�p 0 �a �ay0 p �ay z � ay20 0 1 00 0 0 1

1CCCCCCAw10BBBBBB��a 0 1p(1� ax) 0ay 1 yp (1� ax) 0�p 0 �x y0 0 0 1

1CCCCCCA

� w20BBBBBB�

p 0 0 y0 p y z0 0 1 00 0 0 11CCCCCCA =

0BBBBBB�1 0 0 0y p 0 z0 0 p �y0 0 0 1

1CCCCCCA� If z 6� 0 mod p, say az � 1 mod p, then:

w30BBBBBB�

p 0 x y0 p y z0 0 1 00 0 0 11CCCCCCA =

0BBBBBB�p 0 x� ay2 �ay0 p �ay �a0 0 1 00 0 0 1

1CCCCCCAw10BBBBBB�

1 �ay 0 yp(1� az)0 �a 0 1p(1� az)0 0 1 00 �p �y �z1CCCCCCA

41

� If y 6� 0 mod p, say ay � 1 mod p, then:w30BBBBBB�

p 0 x y0 p y 00 0 1 00 0 0 11CCCCCCA =

0BBBBBB�1 0 0 0a p 0 a2x0 0 p �a0 0 0 1

1CCCCCCAw20BBBBBB�

0 a 1p(ay � 1) 0�a a2x axp (ay � 1) 1p(1� ay)p 0 x y0 �p �y 01CCCCCCA

� w30BBBBBB�

p 0 x 00 p 0 00 0 1 00 0 0 11CCCCCCA =

0BBBBBB�p 0 x 00 1 0 00 0 1 00 0 0 p

1CCCCCCAw3� If X is invertible modulo p, say Y X � 12 mod p then:

w40� p12 X0 121A = 0� p12 �Y0 12

1Aw10� �Y 1p(12 � Y X)�p �X1A

� If xz = y2 and z 6� 0 mod p, say az � 1 mod p, then:w40BBBBBB�

p 0 x y0 p y z0 0 1 00 0 0 11CCCCCCA =

0BBBBBB�1 0 0 0�az p 0 a0 0 p az0 0 0 1

1CCCCCCAw20BBBBBB�

1 ya 1p(ay2 � x) yp (1� az)0 a 0 1p(1� az)0 0 1 00 �p y z1CCCCCCA

� If x 6� 0 mod p, say ax � 1 mod p, then:w40BBBBBB�

p 0 x 00 p 0 00 0 1 00 0 0 11CCCCCCA =

0BBBBBB�p 0 a 00 1 0 00 0 1 00 0 0 1

1CCCCCCAw30BBBBBB�

a 0 1p(1� ax) 00 1 0 0�p 0 x 00 0 0 11CCCCCCA

� w40� p12 00 121A = 0� 12 00 p12

1Aw442

From these results an easy al ulation gives the matrix for the a tion of Up. Note that inparti ular the eigenvalues of Up on �(�1; �2; )� are p 32 (p�1), p 32�1 (p�1), p 32�2 (p�1)and p 32�1�2 (p�1).Finally we are in a position to dedu e the main result of this se tion:Lemma 2.4 Let f 2 S~n(U), where U is an open ompa t subgroup of QGSp4(Zl) onsist-ing of all matri es ongruent to 0� 12 00 �121A modulo some integer N ; be an eigenve torof the He ke operators Sl, Tl and T (l2) for all primes l 6 jN , say f jT = �(T )f . Fix a primep6 jN . Assume that �(Qp(X)) has distin t roots the ratio of no two of whi h is p. Then if �is a root of �(Qp(X)) we an �nd a non-zero form f 0 2 S~n(U \U0(p)) whi h is an eigenformof the He ke operators Sl, Tl and T (l2) for l 6 jNp with eigenvalues given by �, and su h thatf 0jUp = �f 0. In parti ular if � is a p-adi unit then ef 0 = f 0.Proof:We may write S~n =L� with � =N�l being irredu ible admissible representa-tions of G(A f ). We may assume without loss of generality that f 2 �U for some �. Then�p is the spheri al subquotient of an unrami�ed prin iple series representation �(�1; �2; ),where fp 32 (p�1); p 32 �1(p�1); p 32 �2(p�1); p 32 �1�2(p�1); g is the set of roots of �(Qp(X)).By our assumption that �(Qp(X)) has distin t roots, (�1; �2; ) is regular (i.e. its onju-gates under the Weyl group are distin t, i.e. �1 6= �2; �1; �2; �1�2 6= 1) and satis�es the ondition of theorem 3.10 of [C℄, namely:

(�1; �2; )(��11 ; ��12 ; �1�2)9=;8>>>>>><>>>>>>:

diag(p�1; p; p; p�1)diag(p; 1; p�1; 1)diag(1; p; 1; p�1)diag(p; p; p�1; p�1)9>>>>>>=>>>>>>; 6= p

i.e. �2��11 (p) 6= p�1, �1(p) 6= p�1, �2(p) 6= p�1 and �2�1(p) 6= p�1. Thus �(�1; �2; ) isirredu ible and so the result now follows from the above dis ussion.Remark: If �(Qp(X)) has two distin t roots whi h are p-adi units and if 2 � n1 � n2then it ertainly satis�es the ondition of the lemma.43

Before �nishing this se tion we give one further omputation. If H is the He ke algebraof double osets in GSp4(Zp)nGSp4(Q p)=GSp4(Zp) and if H0 is the He ke algebra of double osets in GL2(Zp)nGL2(Q p)=GL2(Zp) then there is an inje tion: H ,! H0[x�1℄ (see forexample [Fa2℄). An easy al ulation shows that this map is given expli itly by:[GSp4(Zp)gGSp4(Zp)℄ 7�! xval� (g)XGL2(Zp)ai

where GSp4(Zp)gGSp4(Zp) = `GSp4(Zp)0� ai bi0 �(g)ta�1i1A. Thus if we let tp denote

[GL2(Zp)0� 1 00 p1AGL2(Zp)℄ and sp denote [GL2(Zp)p12GL2(Zp)℄ we have that:

� Tp 7�! (p3 + ptp + sp)x� Sp 7�! spx2� Rp 7�! (p3tp + sptp + (p2 � 1)sp)x2� Qp(X) 7�! (X � spx)(X � p3x)(X2 � ptpxX + p3spx2)These follow from the de ompositions given above. In parti ular we have that Qp(spx) = 0.2.5 Main ResultsWe ontinue to assume that g = 2. We �x an integer N and we �x U to be the subgroupof QGSp4(Zl) onsisting of matri es ongruent to 0� 12 00 12

1A modulo N . We �x also aprime p not dividing N . We shall let U 0 denote any subgroup of the form:8<:x 2 U ������x �0� �i �0 �t��1i

1A mod pr for some i9=;where r � 1 and f�ig � GL2(Z=prZ) is some set. Also let TM denote the double osetalgebra over Z generated by the He ke operators Tl, Sl and T (l2) for all primes l 6 jM . Thenwe have:

44

Proposition 2.2 Let U 0 be as above and 3 � n1 � n2. Then there is a onstant C su hthat if f 2 eS~n+m~t(U 0) (re all that ~t = (1; 1)) is an eigenform of the ring of He ke operatorsTNp , say f jT = �(T )f ; then there is an integer a � 1, a �nite extension E=Q p su h thata[E : Q p ℄ < C, and a ontinuous representation:� : Gal(Q a =Q ) �! GLa(OE)su h that � is unrami�ed outside Np and su h that if l 6 jNp is a prime then �(Ql)(Frobl) = 0.Proof:We need only onstru t a representation valued in GLa(E), be ause as � is on-tinuous and Gal(Q a =Q ) is ompa t � will stabilise some ompa t OE module spanning Ea,whi h then must be free.In se tion 2.3 we de�ne a representation V (~n + (m � 3)~t) of GSp4=Z. This gives us alo ally onstant etale p-adi sheafVp(~n+(m�3)~t) on a ertain smooth modelMM=Z[ 1M ℄ ofZ=�M , whereM = Npr+1 and �M denotes the set of matri es in Sp4(Z) whi h are ongruentto one modulo M . (See [Fa2℄.) Let W denote H3et(MM � spe Q ;Vp(~n+ (m� 3)~t)). ThenTNp , Up and Gal(Q a =Q ) all a t on W and these a tions ommute with one another. Thea tion of Gal(Q a =Q ) is unrami�ed outside Np and if l 6 jNp then Ql(Frobl) = 0. For theseassertions we refer the reader to [Fa2℄. The operator Up is not treated there, but an betreated in an exa tly analogous manner. Note also that theorem one of the se tion \He keOperators and Frobenius" in [Fa2℄ implies that Ql(Frobl) = 0 be ause we have seen at theend of se tion 2.4 that Ql(xsl) = 0.Now onsider (eW )TNpE, where E is the �nite extension of Q p generated by the imageof � and has a TNp a tion via �. We must show that 1 � dimQp (eW )TNpE � C, for C some onstant independent ofm and f . However (eW )TNpE �= eH3(�M ; V (~n+(m�3)~t))TNpEand so dimQp (eW ) TNp E � dimQp eW � C for some onstant C as in theorem 1.2.Moreover if we �x E ,! C ompatible with the p-adi valuation on Q a � C and if we letC (�) denote the one dimensional omplex TNp module with the a tion via �, then we seethat: ((eW )TNp E) C �= eH3(�M ; V (~n+ (m� 3)~t))TNp C (�)� eS~n(U 0)TNp C (�) 6= (0)45

and we are done.The following onje tural re�nements of this proposition are well known:Conje ture 2.1 Let U 0 be as above and 3 � n1 � n2. Then there is a onstant C su hthat if f 2 eS~n+m~t(U 0) is an eigenform of the ring of He ke operators TNp , say f jT =�(T )f ; then there is a �nite extension E=Q p su h that [E : Q p ℄ < C, and a ontinuousrepresentation: � : Gal(Q a =Q ) �! GSp4(OE)su h that � is unrami�ed outside Np and su h that if l 6 jNp is a prime then �(Ql)(Frobl) = 0.Conje ture 2.2 Let U 0 be as above and 3 � n1 � n2. Then there is a onstant C su hthat if f 2 eS~n+m~t(U 0) is an eigenform of the ring of He ke operators TNp , say f jT =�(T )f ; then there is a �nite extension E=Q p su h that [E : Q p ℄ < C, and a ontinuousrepresentation: � : Gal(Q a =Q ) �! GSp4(OE)su h that � is unrami�ed outside Np and su h that if l 6 jNp is a prime then Frobl has hara teristi polynomial �(Ql).Now we an state the main result of this hapter:Theorem 2.2 Let U and U 0 be as above and let 2 � n1 � n2. Let � : ((Z=Npr)�)2 ! C �be a hara ter. Assume one of the following is true:1. f 2 eS~n(U 0)� is an eigenform of TNp , say f jT = �(T )f2. f 2 S~n(U)� is an eigenform of TN , say f jT = �(T )f , and �(Qp(X)) has distin troots no quotient of two of whi h equals p and one of whi h is a p-adi unit.3. f 2 S~n(U)� is an eigenform of TN , say f jT = �(T )f , and �(Qp(X)) has two distin troots whi h are p-adi unitsThen there is a �nite extension E=Q p , an integer a � 1 and a ontinuous representation:� : Gal(Q a =Q ) �! GLa(OE)46

whi h is unrami�ed outside Np and su h that if l 6 jNp is a prime then �(Ql)(Frobl) = 0.Proof:By lemma 2.4 of se tion 2.4 the se ond two ases redu e at on e to the �rst ase.So onsider the �rst ase. We may �nd forms fm 2 eS~n+bm~t(U 0)� whi h are eigenvalues ofTNp , say fmjT = �m(T )f , and su h that as m goes to in�nity bm !1 and supTNp j�m(T )��(T )jp ! 0. (Lemma 2.1 of se tion 2.3.) Then for any positive integer s we may �nd apositive integer as, a �nite extension Es=Q p with as[Es : Q p ℄ < C (where C is as in the lastproposition, but for (n1 + 1; n2 + 1)) and a ontinuous representation �s : Gal(Q a =Q ) !GLas(OEs=ps) whi h is unrami�ed outside Np and su h that �(Ql)(Frobl) = 0. Howeverthere are only �nitely many extensions E : Q p of degree less than C, so we may assumethat Es = E and as = a are independent of s.We shall re ursively de�ne in�nite subsets It � It�1 � T su h that if s1; s2 2 It then�s1 � �s2 mod pt. This is possible as (�s mod pt) fa tors through Gal(K=Q ) for someGalois extension K=Q unrami�ed outside Np and of degree bounded independently of s. Itis known that there are only �nitely many su h extensions and so there is a �nite Galoisextension L=Q through whi h all the (�s mod pt) fa tor. Now there are only �nitely manymaps Gal(L=Q ) ! GLa(OE=pt) so for in�nitely many s 2 It�1 the maps (�s mod pt) mustbe equal as desired. Now we de�ne � : Gal(Q a =Q ) ! GLa(OE) by the requirement that�(�) � �s(�) mod pt for all � 2 Gal(Q a =Q ) and all t and all s 2 It. This is easily he kedto be a good de�nition of a morphism with the desired properties.If we assume the onje tures mentioned above the same method gives the following onje tural strengthening of this theorem:Theorem 2.3 Assume onje ture 2.1 (resp. 2.2). Let U and U 0 be as above and let 2 �n1 � n2. Let � : ((Z=Npr)�)2 ! C � be a hara ter. Assume one of the following is true:1. f 2 eS~n(U 0)� is an eigenform of TNp , say f jT = �(T )f2. f 2 S~n(U)� is an eigenform of TN , say f jT = �(T )f , and �(Qp(X)) has distin troots no quotient of two of whi h equals p and one of whi h is a p-adi unit.47

3. f 2 S~n(U)� is an eigenform of TN , say f jT = �(T )f , and �(Qp(X)) has two distin troots whi h are p-adi unitsThen there is a �nite extension E=Q p and a ontinuous representation:� : Gal(Q a =Q ) �! GSp4(OE)whi h is unrami�ed outside Np and su h that if l 6 jNp is a prime then �(Ql)(Frobl) = 0(resp. Frobl has hara teristi polynomial �(Ql(X))).

48

Chapter 3p-adi Families of Siegel ModularForms3.1 Introdu tionIn this hapter we develop a theory of Hida families for Siegel modular forms of even genus.To explain our results �x an odd prime p, an even positive integer g and embeddingsQ a � C , Q a � Q a p . The assumption that g be even is probably not essential. It is madebe ause our omputations with Eisenstein series and Rankin's method are dependent on theparity of g, and be ause g = 2 is the ase of greatest interest for us. Let Mk(N;�) denotethe spa e of Siegel modular forms of genus g, weight k, level N and hara ter �. Thisspa e has a Z[�℄ stru ture for some root of unity � and this allows us to de�ne Mk(N;�;A)for any algebra A ontaining enough roots of unity so that we an onsider � as valued inA. Any element of Mk(N;�;A) has a formal Fourier expansion P ah exp(�i tr(hz)) withah 2 A and where h runs over the set S of integral, positive semi-de�nite g � g symmetri matri es with even diagonal entries. If pjN we shall let Up denote the He ke operator[�0(N)0� 1g 00 p1g

1A�0(N)℄ where �0(N) is de�ned in se tion 3.2. If O denotes the integersof a suitably large extension of Q p we de�ne MÆk (N;�;O) to be the largest submodule ofMk(N;�;O) on whi h Up is an automorphism. It is in fa t a dire t summand, whi h we49

shall all the ordinary part. We an arry this notion over to other settings, for examplewe an de�ne MÆk (N;�; C ) using the embeddings Q a � C and Q a � Q a p . Note that iff 2 Mk(N;�) is an eigenform of Up, say f jUp = apf , then f 2 MÆk (N;�) if and only if apis a p-adi unit. We shall also de�ne in se tion 3.2 a ertain ommutative ring of He keoperators TN whi h a ts semi-simply on Mk(N;�) and ommutes with the a tion of Up.Now let N be prime to p. Let � be the power series ring O[[T ℄℄ with O as above. Bya �-adi form of level N and hara ter � (de�ned modulo Np), we shall mean a formalexpansion: Xh2S ahqhwith oeÆ ients in � and su h that for all but �nitely many pairs (k; �) with k � g+1 and:� : (Z=Np?Z)� ! (1 + pZ)=(1 + p?Z) ! Q a �

we have that: Xh2S ah(�(1 + p)(1 + p)k � 1) exp(�i tr(hz))is the Fourier expansion of an element of MÆk (Np?; �!�k�; Q a p ). We shall denote the spa eof su h forms MÆ(N;�). We an de�ne an a tion of TNp on MÆ(N;�) ompatible withspe ialisation.Our �rst main theorem states thatMÆ(N;�) is a �nite free � module. The main point ofthe proof is that dimMÆk (Np; �) is bounded independently of k. The se ond main theoremstates that if f 2 MÆk (Np?; �!�k�) is an eigenve tor of TNp , say f jT (n) = �(n)f , then we an �nd an integerM (divisible byN) and a form F 2 MÆ(M;�)R (R the integers of someextension of the �eld of fra tions of �) whi h is an eigenve tor for TMp , say F jT (n) = �(n)F ,su h that �(n) � �(n) modulo some prime of R lying above (1 + T � �(1 + p)(1 + p)k.To prove this we follow a method of Wiles (see [Wi℄). One �rst writes down some �-adi Eisenstein series, one then multiplies them by a ertain form of low weight (we use a thetaseries) and uses Rankin's method to show that if f is of high enough weight it will o urin the spe tral de omposition of this produ t. For this we must use our �rst theorem. To

50

extend the result to all weights k one shows that if f is an ordinary eigenform of weight k we an �nd ordinary eigenforms arbitrarily ongruent to f with weights of the form k+(p�1)p?.In the last se tion we show how to use this theory to rederive some results of the last hapter about Galois representations. We also show how the standard onje tures on theexisten e of Galois representations one expe ts to asso iate to Siegel modular forms wouldimply similar results about �-adi representations.3.2 Review of Siegel Modular Forms IIWe set up a theory of Siegel modular forms in a more lassi al setting than the last hapter.This is better suited to the purposes of this hapter.We shall �x throughout an odd prime p and embeddings Q a ,! C and Q a ,! Q a p . Weshall also �x a positive integer g. For our main results we must assume that g is even, andwe shall in fa t always assume this ex ept in se tions one and four where we shall need to onsider all g to make a ertain indu tion argument work.We shall let Zg denote the Siegel spa e of genus g, that is:Zg = fx+ iy jx; y 2 symmg(R ); y > 0gHere we write a > 0 if a is a positive de�nite symmetri real matrix. Also if A is a ringsymmg(A) denotes the module of g � g symmetri matri es over A. Moreover symm�g(A)will denote the sub-module whose diagonal entries are in 2A and we shall use the supers ript� 0 to denote the sub-semigroup of positive semi-de�nite elements when this makes sense.A ouple more notes on notation. We shall use �ij to denote the g � g matrix that hasone at the interse tion of the ith row and jth olumn and zeroes elsewhere, and 1g to denotethe g � g identity matrix. We shall let � : GSp2g ! G m denote the hara ter su h that0��0� A BC D

1A1A 1g = AtD �BtC. Also if R is an integral domain FR will denote its �eldof fra tions.51

The group GSp+2g(R ) a ts on Zg by:0� A BC D1A : z 7�! (Az +B)(Cz +D)�1

If f : Zg ! C is a fun tion, k is a non-negative integer and 2 GSp+2g(R ) we de�ne atransform of f by by the formulae:� f jk (z) = �( ) gk2 j( ; z)�kf( z)� j0�0� A BC D

1A ; z1A = det(Cz +D)If � is a subgroup of Sp2g(Z) of �nite index we de�ne a spa e of modular forms, denotedMk(�), to be the set of holomorphi fun tions f : Zg ! C su h that f jk = f for all 2 �.If g = 1 we must supplement these onditions with a growth ondition, but this is wellknown. If we wish to indi ate the genus we shall write M (g)k (�). In this work we shall be on erned with two spe ial ongruen e subgroups of Sp2g(Z) whi h we shall denote:� �0(N) = f0� A BC D

1A 2 Sp2g(Z) jC � 0 mod Ng� �1(N) = f0� A BC D

1A 2 Sp2g(Z) jC � 0 mod N; detD � 1 mod NgWe shall de ompose: Mk(�1(N) =M� Mk(N;�)as � runs over hara ters � : (Z=NZ)� ! Q a �, and where Mk(N;�) denotes the spa eof holomorphi fun tions f : Zg ! C su h that f jk = �( )f for all 2 �0(N). Here� : �0(N)! Q a � by 0� A BC D

1A 7! �(detD). Again we need a growth ondition if g = 1.Any element of Mk(�1(N)) has a Fourier expansion:X ah(f) exp(�i tr(hz))52

as h runs over symm�g(Z)�0 . If R is a ring we shall denote by R[[q℄℄g the ring of formalpower series P ahqh as h runs over the semigroup symm�g(Z)�0 , and where ah 2 R. Withthis notation we see that: Mk(�1(N)) ,! C [[q℄℄gf 7�! P ah(f)qhIf R � C is a sub-Z-module we de�ne Mk(�1(N); R) to be those elements of Mk(�1(N))whose Fourier expansion lies in R[[q℄℄g � C [[q℄℄g . It is a result of Shimura thatMk(�1(N); C )=Mk(�1(N);Z)ZC and so for any Z-module R we may onsistently de�neMk(�1(N); R) =Mk(�1(N);Z) Z R. Similarly if O� denotes the extension of Z generated by the im-age of � then Mk(N;�;O�) O� C = Mk(N;�) so if R is any O� module we an de�neMk(N;�;R) = Mk(N;�;O�) O� R. Note also that this implies that Aut(C =Q ) a ts onMk(�1(N)) by: f�(z) =X ah(f)� exp(�i tr(hz))where � 2 Aut(C =Q ). In fa t Mk(N;�)� =Mk(N;� Æ �).If f is a modular form of genus g for some ongruen e subgroup � of Sp2g(Z) we de�nea modular form �(f) of genus g � 1 by:�(f)(z) = lim�!+1 f 0� z 00 i�

1A� is alled the Siegel operator. Its a tion on Fourier expansions is given by:

�(f) =X ah0(f) exp(�i tr(hz))where h0 = 0� h 00 1

1A. We an embed Sp2(g�1) into Sp2g by:0� A BC D

1A 7�!0BBBBBB�

A B1 0C D0 11CCCCCCA

53

If � is a ongruen e subgroup of Sp2g set �(�) = � \ Sp2(g�1). Then:� :M (g)k (�) �!M (g�1)k (�(�))and: � :M (g)k (N;�) �!M (g�1)k (N;�)We all a modular form of weight k uspidal if for all 2 Sp2g(Z) we have that �(f jk ) =0. We denote the subspa e of Mk(�) onsisting of usp forms by Sk(�). Similarly we usethe notation Sk(N;�). The rationality results des ribed above remain true for usp forms.If Sp2g(Z) = qI�Æi then we have a left exa t sequen e:0! S(g)k (�)!M (g)k (�)!MI M (g�1)k (�(Æ�1i �Æi))and similarly for Mk(N;�).If f 2 Sk(�) and g 2Mk(�) we an de�ne an inner produ t hf; gi� by:Z�nZg f(z)g(z)(det y)k dzwhere z = x + iy and dz = (det y)�g�1Q1�����g dx�� dy��. Then if 2 GSp+2g(R ) wehave that hf jk ; gjk i �1� = hf; gi�. Similarly if f 2 Sk(N;�) and g 2Mk(N;�) we set:hf; giN = Z�0(N)nZg f(z)g(z)(det y)k dzIf U denotes an operator on Sk(N;�) we shall let U� denote its adjoint with respe t tothis inner produ t. It will be onvenient to introdu e a slight variant variant of this innerprodu t. Set: WN = 0� 0 �1gN1g 01A

then for f 2 Sk(N;�) and g 2Mk(N;�) we de�ne:(f; g)N = hf jWN ; g iNwhere denotes omplex onjugation. This is a C bilinear form whi h restri ted to Sk(N;�)2is non-degenerate. If U is an operator on Sk(N;�) we shall let U y denote its transpose withrespe t to this pairing. 54

We shall now re all some fa ts about the theory of He ke operators (see [A2℄ for details).If g 2 GSp+2g(Q ) and g � 0� � �0 �1A mod N we de�ne the He ke operator:

[�0(N)g�0(N)℄ :Mk(N;�) �!Mk(N;�)by: f 7�! �(g) gk2 � g(g+1)2 XI �0(gi)f jkgiwhere:� �0(N)g�0(N) = q�0(N)gi� �00� A BNC D

1A = �(detA) this being given the value 0 if (detA;N) 6= 1.Lemma 3.1 If f1 2 Sk(N;�), f2 2Mk(N;�), g = 0� A BNC D

1A 2 GSp+2g(Q ) andA;B;C;D 2Mg�g(Z) then:h(f1j[�0(N)g�0(N)℄)jWN ; f2iN = hf1jWN ; f2j[�0(N)g��0(N)℄iwhere: g� = 0� N�11g 00 1g

1A tg0� N1g 00 1g1A = 0� tA tCN tB tD

1AIn parti ular if g is as above and if further:

f j[�0(N)g�0(N)℄ = (f j[�0(N)g�0(N)℄) then [�0(N)g�0(N)℄y = [�0(N)g�0(N)℄.

55

Proof:Let �0(N)g�0(N) = q�0(N)g i with i = 0� Ai BiCi Di1A 2 �0(N). Then:

h(f1j[�0(N)g�0(N)℄)jWN ; f2iN= Pi �(g i)?�(AAi)[�0(N) : �0(N) \ �1i g�1�0(N)g i℄�1hf1jWN (W�1N g iWN ); f2i�0(N)\ �1i g�1�0(N)g i= �(g)?�(A)[�0(N):�0(N)\g�1�0(N)g℄Pihf1jWN (W�1N (g�)�1; �(Di)f2j �i i�0(N)\g�1�0(N)g= �(g)?�(A)hf1jWN (g�)�1; f2i�0(N)\g�1�0(N)gwhere ? = gk2 � g(g+1)2 . Similarly:hf1jWN ; f2j[�0(N)g��0(N)℄iN = �(g�)?�(A)hf1jWN ; f2jg�ig��1�0(N)g�\�0(N)and so the lemma follows.Let �0(N) be the semigroup of elements in GSp+2g(Q ) \M2g(Z) with (N;det ) = 1and � 0� � �0 �1A mod N . Let T�N denote the He ke algebra whi h is spanned over Zby the double �0(N) osets ontained in �0(N). Then T�N a ts on Mk(N;�) and thisa tion preserves Sk(N;�). Moreover if N jM then T�M ,! T�N and if N and M have thesame prime fa tors this is an isomorphism. This map is also ompatible with the in lusionMk(N;�) ,! Mk(M;�). Any element T of T�N an be written as Pni�0(N)gi with gi =0� Ai Bi0 Di

1A and then:ah(f jT ) =Xni�(gi)� g(g+1)2 �(detAi)(detAi)k exp(�i tr(hA�1i Bi))aDihA�1i (f)where the sum is taken over those i su h that DihA�1i 2 symm�g(Z). In parti ular T�Npreserves Mk(N;�; Q a ). It is also known that Sk(N;�) has a basis of eigenforms for T�N ,whi h are orthogonal with respe t to the Petersson inner produ t. We see that:� If M �Mk(N;�) is preserved by T�N so is M? � Sk(N;�)� If U is an operator on Sk(N;�) whi h ommutes with the a tion of T�N then U� also ommutes with T�N . 56

For n a positive integer prime to N let T (n) 2 T�N denote the sum of all �0(N) double osets in �0(N) on whi h � takes the value n. We let TN denote the subring of T�N generatedby these operators. We have that:ah(f jT (n)) = n� g(g+1)2 �(ng)ngkPd1jd2j:::dg ja dg1dg�12 :::dgPD �(detD)�1(detD)�kan�1DhtDwhere the se ond sum is taken over a set of representatives for:GLg(Z)nGLg (Z)

0BBB� d1 . . . dg1CCCAGLg(Z)

In parti ular TN preservesMk(N;�;O�;}) for } any prime ofO� dividing N . Also if T 2 TNthen T y = T . This follows from lemma 3.1 and the fa t that �0(N)� = �0(N).For all these results we refer the reader to [A2℄. We also note that any eigenform of T�Nis also an eigenform of ea h algebra Lg0;l(N) (l 6 jN) onsidered in [A1℄ (this follows from [A2℄theorem 4.1.8).If pjN we shall also onsider the operators Upr = [�0(N)0� 1g 00 pr1g1A�0(N)℄. We listsome properties of Upr :Lemma 3.2 Let pjN and let � be de�ned modulo N . Then Up 2 End(Mk(N;�)) satis�es:1. if N jM then the a tion of Up is ompatible with Mk(N;�) ,!Mk(M;�)2. ah(f jUpr) = aprh(f)3. Upr = U rp4. UpMk(Np; �) �Mk(N;�)5. Up ommutes with the a tion of TN6. U yp = Up7. if T 2 T�N then some power of Up ommutes with T .57

Proof:1. and 2. follow from the fa t that:�0(N)0� 1g 00 pr1g

1A�0(N) = q�0(N)0� 1g B0 pr1g1A

where B runs over any set of representatives for the mod pr ongruen e lasses of symmg(Z).3) follows from 2). 4) follows from the fa t that:�0(Np)0� 1g 00 p1g

1A�0(N) = q�0(Np)0� 1g B0 p1g1A

where B is as above. 5) follows from 2) and the formula for ah(f jT (n)), and 6) followsfrom 2) and lemma 3.1. We do not prove these oset de ompositions here. The proof isessentially the same as that given in lemma 3.4 below.7) follows from these de ompositions and the fa ts that if 0� A B0 D1A 2 �0(N), � =

�(0� A B0 D1A), X � 0 mod � is a symmetri integral matrix, and pr � 1 mod � then:

0� 1g X0 pr1g1A0� A B0 D

1A = 0� A B0 D1A0� 1g ��1tD((1� pr)C +XD)0 pr1g

1Aand that X 7! ��1tD((1� pr)B +XD) is a permutation of the modpr ongruen e lassesof symmetri integral matri es.Note that in parti ular Up preserves Mk(N;�;O�).We now want to de�ne the Hida idempotent asso iated to Up whi h will be essential inwhat follows. First re all the following lemma:Lemma 3.3 Let O denote the integers of a �nite extension of Q p andM a �nite O-module.Let U be an operator on M , then there exists a unique idempotent eU in EndO(M) whi h ommutes with U and su h that U is an automorphism on eM and is topologi ally nilpotenton (1�e)M . Moreover e = limr!1 U r!. IfM 0 and U 0 satisfy orresponding onditions and if� :M !M 0 is su h that �U = U 0� then �e = e�.

58

If O0 is a number �eld, M0 a �nite O0-module, } a prime of O0 above p and U 2EndO0(M0) then there is a ring R ontained in a number �eld with O0 � R � O0;} and anidempotent eU 2 EndR(M0 R) su h that eU onsidered as an element of EndO0;}(M0 O0;}) oin ides with the idempotent asso iated to U above.In parti ular in these ir umstan es we an think of e 2 EndQa (M0 Q a ), as we have�xed Q a � Q a p .We shall be interested in the ase M0 =Mk(N;�;O�) with pjN and U = Up. We shalldenote by e the orresponding idempotent, whi h we shall all the Hida idempotent. We anthink of e a ting on Mk(N;�;O�;}) or on Mk(N;�). In either ase we have the followingproperties:� If N jM the a tion of e is ompatible with Mk(N;�) ,!Mk(M;�)� eMk(Npr; �) = eMk(N;�)� e ommutes with the a tion of T�N� ey = eFinally we must study the operator Up and the Hida idempotent e in a slightly di�erentsetting. Let N now be an integer prime to p.Now let � denote one of the following:� �0 = 8<:0� A BC D

1A 2 Sp2g(Z) jB � 0 mod N; C � 0 mod Np;A � D � 1 mod N9=;

� �1 = 8<:0� A BC D1A 2 �0 j detD � 1 mod p9=;� �1(Npr)Any element of Mk(�) has a Fourier expansion P ah(f)qh where h runs over elements ofN�1symm�g(Z). The theory of rationality arries over exa tly to this situation. We de�ne

59

operators Upr to be the He ke operators asso iated to the double osets �0� 1g 00 p1g1A�.Then we have:Lemma 3.4 With the notation as above:1. Upr = U rp2. ah(f jUp) = aph(f)3. Up preserves Mk(�;Z)4. the a tion of Up is ompatible with the in lusions:Mk(�1) � Mk(Np; �)[Mk(�0) � Mk(Np; 1)and: Mk(�1(Npr)) �Mk(Npr; �)Proof:These all follow from the fa ts that:

�0� 1g 00 p1g1A� = q�0� 1g B0 p1g

1Awhere B runs over a set of representatives for the mod pr ongruen e lasses of symmg(Z)ea h hosen � 0 mod N . This de omposition follows from the following equations:� 0� 1g B0 pr1g

1A = 0� 1g 00 pr1g1A0� 1g B0 1g

1A� 0� 1g X0 pr1g

1A0� 1g Y0 pr1g1A�1 = 0� 1g p�r(X � Y )0 1g

1A

60

� If 0� A BpC D1A 2 Sp2g(Z) then tDB 2 symmg(Z) and if further X 2 symmg(Z) withX �t DB mod pr and X � 0 mod N then:0� 1g 00 pr1g

1A0� A BpC D1A = 0� A p�r(B � AX)pr+1C D � pCX

1A0� 1g X0 pr1A

where B � AX � A(tDB �X) � 0 mod pr.In parti ular we an introdu e the Hida idempotent in this ontext, and it will be om-patible with the other ontexts onsidered earlier. We �nish this se tion with a simpli� ationof the ondition that an ordinary modular form is uspidal:Lemma 3.5 Let N be prime to p. Let � (or �(g)) denote either �0 or �1 as de�ned above.Let �(g�1) denote the orresponding group in genus g � 1. Then there is an integer m su hthat for all k � g(g+1)2 we have a left exa t sequen e:0! eSk(�(g))! eMk(�(g))! eMk(�(g�1))m

Proof:Let �nSp2g(Z) = qI�Æi, so that we have an exa t sequen e:0! Sk(�)!Mk(�) �!MI Mk(�(Æ�1i �Æi))

Let J be the subset of I onsisting of those i for whi h �Æi � �0(p). Let � Q a be the�nite Z[�p ℄-module generated by the oeÆ ients f jÆi where f 2 Mk(�;Z) and i 2 I. Thenwe make the following laims:1. f 2Mk(�;Z) and �(f) 2LInJMk(�(Æ�1i �Æi); pr) implies that �(f jUp) 2LIMk(�(Æ�1i �Æi); pr+1)2. f 2Mk(�) and � 2 �0(p) implies that ��1�� = � and that f j�jUp = f jUpj�3. f 2Mk(�;Z) and �(f) 2LInJMk(�(Æ�1i �Æi); pr) implies that �(f jUp) 2LInJMk(�(Æ�1i �Æi); p(r + 1))61

4. The following sequen e is left exa t:0! eSk(�)! eMk(�)!MJ Mk(�(g�1))5. � :Mk(�(g))!Mk(�(g�1)) is ompatible with the a tion of Up.Note that from 2), 4) and 5) the lemma follows at on e. Also note that 5) follows easilyfrom the des ription of Up in terms of its a tion on Fourier expansions; and that 3) and 4)follow easily from 1) and 2).First we prove 1). Note that for � = 0� A BC D

1A 2 Sp2g(Z):f jUpj� = p gk2 � g(g+1)2 XX f j0� A+XC B +XDpC pD

1Aand that: 0� A+XC B +XDpC pD

1A = 0� (A+XC)E�1 B0pCE�1 D01A0� E Y0 ptE�1

1Awith:� 0� (A+XC)E�1 B0pCE�1 D0

1A 2 Sp2g(Z)� 0� E Y0 ptE�1

1A 2M2g�2g(Z)� E upper triangular(That we an �nd su h a de omposition an be easily dedu ed from proposition 1.3.7 of[A2℄.) We now onsider two ases:1. p6 jdetE, i.e. E 2 GLg(Z)In this ase:0� A+XC B +XDpC pD

1A = 0� A+XC B0tE�1pC D0tE�11A0� 1g E�1Y0 p1g

1A62

with 0� A+XC B0tE�1pC D0tE�11A 2 �0(p). Thus:

�0�f j0� A+XC B +XDpC pD1A1A = 0

2. pjdetEIn this ase:p gk2 � g(g+1)2 f j0� A+XC B +XDpC pD

1A =pk� g(g+1)2 (p�1 detE)kP ah exp(�i tr(p�1hY tE)) exp(�i tr(p�1 tEhEz))where: f j0� (A+XC)E�1 B0pCE�1 D0

1A =X ah exp(�i tr(hz))In either ase �(f jUp) 2 LIMk(�; pr+1) as E upper triangular implies thattE�10� h 0

1AE�1 = 0� h0 01A.

Finally for laim 2), let X � 0 mod N and 0� A BpC D1A 2 �0(p). Then:0� A BpC D

1A0� 1g X0 p1g1A0� A BpC D

1A�10� 1g AXtA� AtB0 p1g1A�1

= 0� AtD �BtCp2 � pAXtC �AtB +BtAp+AXtAp(CtD �DtCp� CXtCp) p(�CtB +DtA+ CXtA)1A0� 1g p�1(AtB � AXtA)0 p�11g

1A� 2 Sp2g(Z) as AtD(AtB � AXtA)� AtB +AXtA � 0 mod p� � 0� A0 �0 D0

1A mod p and if detA � detD � 1 mod p then detA0 � detD0 � 1 modp. 63

� � 0� 1g 00 1g1A mod N

The result now follows as X 7! AXtA� AtB is a permutation of symmg(Z=pZ).3.3 Some Lemmas on Eisenstein and Theta SeriesIn this se tion we olle t some results that we need on erning Eisenstein series, theta seriesand Rankin's method. These results are based on work of Andrianov and of Shimura. Firstwe introdu e some Eisenstein series:� E(z; s; k; �; b) = (det 2y)sP�(detD) det(Cz+D)�kjdet(Cz+D)j�2s where z = x+iy,and where the sum runs over 0� A BC D

1A representing the osets �0n�0(b).� E�(z; s; k; �; b) = E(�1z ; s; k; �; b)z�k� eEk;�;b(z) =Ph 2 symm�g(Z)h > 0 ah exp(�i tr(hz)) with:

ah = L(1 + g2 � k; �̂�1�h)0B�Ql 2 P (h)l 6 jb Ml;h(�(l)lk )1CA (deth)k� g+12 f g2�kg��h�(g��h)Ql 6 jf��hl j b (1�g��h(l)l g2�k)where{ e is the primitive hara ter orresponding to { if is a primitive hara ter f denotes its ondu tor and�( ) = Xx2(Z=f Z)� (x) exp(2�ixf )

denotes the orresponding Gauss sum{ �h = �h� g2 where �h is the hara ter orresponding to Q (pdeth)=Q and � is the hara ter orresponding to Q (p�1)=Q64

{ P (h) is a �nite set of primes depending only on Q �h, as de�ned in x10 of [Fe℄.{ for l 2 P (h) Ml;h(X) 2 Z[X℄ is a polynomial depending only on l and Q �h asde�ned in x10 of [Fe℄Lemma 3.6 Assume k � g2 + 1 and that � is a hara ter modulo b, with �(�1) = (�1)k.Assume also that either k > g2 + 2 or �2 6= 1. Then:

C ~Ek;�;b(z) = E�(z; 0; k; �; b) j [b℄ 2Mk(�0(b); �)with C 6= 0 a onstant.Proof:E�(x+ iy; s; k; �; b) =Pbh2symm�g(Z) ah(y; s) exp(�i tr(hx)) whereah(y; s) has been al ulated by Shimura and Feit. In parti ular ah(y; s) is �nite (resp. zero)at s = 0 if Ah(s) is �nite (resp. zero) at s = 0, where, if h has r+ positive eigenvalues, r�negative eigenvalues and rank r = r+ + r�:Ah(s) = �g�r(k� g+12 +2s)�g�r�(k+s)�g�r+(s) �L(k + 2s; �)Q g2i=1 L(2k � 2i+ 4s; �2)��1

: 8<: L(k � g + r2 + 2s; ��h)Q g�r2i=1 L(2k � 2g + r � 1 + 2i+ 4s; �2)Q g�r�12i=0 L(2k � 2g + r + 2i+ 4s; �2)a ording as r is even or odd. Here �t(s) = Qt�1i=1 �(s � i2). (See x10 of [Fe℄.) The onlyterms that ontribute a zero or pole at s = 0 are:� �g�r+(s)�1 whi h ontributes a zero of order the integer part of g�r++12� the L-series in the numerator whi h an ontribute at most a simple pole.Thus Ah(s) has a pole at s = 0 only if g = r+ and the L-series in the numerator have apole, i.e. k = 1+ g2 , ��h = 1, whi h is a ase we have ex luded. Moreover Ah(0) = 0 unlessg = r+ or one of the following hold:� g = r+ + 1, k = g2 + 1, �2 = 1� g = r+ + 2, k = g2 + 2, ��h = 165

� g = r+ + 2, k = g2 + 1, �2 = 1However our assumptions ex lude these last three possibilities. Thus we have:E�(x+ iy; 0; k; �; b) = Xh 2 symm�g(Z)h > 0 ab�1h(y; 0) exp(�i tr(hx))

where using [Fe℄ x10 and the fun tional equation for L-series asso iated to Diri hlet har-a ters: ah(y; o) = 2?�?i?(det b)? L�1ahe�� tr(hy)with ea h ? a rational number independent of h and with:� = �g(k)�1��k+�� g22 ��1 ��1�k+�+ g22 �� L = L(k; �)Q g2i=1 L(2k � 2i; �2)� � = 0 or 1 and � � k � g2 mod 2, so that ��h(�1) = (�1)�.From this the desired equality follows at on e. Moreover we see that E�(z; 0; k; �; b) j [b℄is holomorphi . Finally it lies in Mk(�0(b); �) be ause it is equal to E(z; 0; k; �; b) jWb andE(z; s; k; �; b) transforms under �0(b) by ��1.We now introdu e some theta series. Let be a hara ter modulo r and let Q 2symm�g(Z), Q > 0. We shall let s(Q), the step of Q, denote the smallest positive integersu h that s(Q)Q�1 2 symm�g(Z). Now set:

�Q; (z) = XN2Mg�g(Z) (detN) exp(�i tr(tNQNz))We �rst re ord some transformation properties of �Q; . The proofs are based on work ofAndrianov. In fa t part one is due to him in the ase primitive, but unfortunately wedon't think this method goes over exa tly to the ase of imprimitive.Lemma 3.7 1. �Q; (z) 2 M g2 (�0(r2s(Q)); �Q) where �Q is a hara ter of order twomodulo s(Q) with �Q(�1) = (�1) g2 .

66

2. If is primitive modulo r then �Q; jWr2s(Q) = C�s(Q)Q�1; �1 for some onstant C.Proof:Following Andrianov we introdu e for T 2 Mg�g(Z) with QT � 0 mod s(Q) atheta series:

�(z;Q : T ) = XN2Mg�g(Z) exp(�i tr(t(N + s(Q)�1T )Q(N + s(Q)�1T )z))Then: �Q; (z) = XM2Mg�g(Z)modr (detM)�(z; r2Q : rs(Q)M)Now from x1.3.3 of [A2℄ we know that:

1. If = 0� A BC D1A 2 �0(s(Q)) then:

�(z;Q : T ) j g2 = �Q(detD) exp(�i tr(s(Q)�2AtBtTQT ))�(z;Q : TA)2. �(z;Q : T ) j Ws(Q) = C1PQU � 0 mod s(Q)U mod s(Q) exp(2�i tr(s(Q)�2 tTQU))�(z; s(Q)Q : U)with C1 a non-zero onstant independent of T .Thus if = 0� A BC D

1A 2 �0(r2s(Q)) then:�Q; j g2 (z) = P (detM)�Q(detM) exp(�i tr(AtBtMQM))�(z; r2Q : rs(Q)MA)= �Q (detD)P (det(MA))�(z; r2Q : rs(Q)MA)= �Q (detD)�Q; (z)(Here we have used the fa t that (AtB)(tMQM) has tra e an even integer.)Now assume is primitive. For A 2Mg�g(Z) set:�A( ) = XM2Mg�g(Z=rZ) (detM) exp(2�ir tr(MA))

We shall assume for the moment the following two fa ts:67

� A 2 GLg(Z=rZ) implies that �A( ) = (detA)�1�1( )� A 62 GLg(Z=rZ) implies that �A( ) = 0.Then: �Q; jWr2s(Q) = C1P (detM)PQT � 0 mod s(Q)T mod r2s(Q) exp(2�ir tr(tM QTs(Q)))�(z; r4s(Q)Q : T )= C1PT � QTs(Q) ( )�(z; r4s(Q)Q : T )= C1�1( )PT �1(det QTs(Q))�(z; r4s(Q)Q : T )but: �(z; r2Q : T ) =XN exp(�i tr(zt(r2QN + QTs(Q))s(Q)Q�1(r2QN + QTs(Q))))so that:�Q; jWr2s(Q)(z) = C1�1( )P �1(detX)PN exp(�i tr(zt(r2QN +X)s(Q)Q�1(r2QN +X)))where X = X(T ) = QTs(Q) and where T runs over elements of Mg�g(Z) modulo r2s(Q) withQT � 0 mod s(Q).Now let I be the right ideal in Mg�g(Z) generated by r2Q. Then the sets fT modr2s(Q) j QT � 0 mod s(Q)g and fX mod Ig are in bije tion by the maps:T 7�! s(Q)�1QTs(Q)Q�1X � XThus: �Q; jWr2s(Q)(z) = C1�1( )PXmodI �1(detX)PN�XmodI exp(�i tr(ztNs(Q)Q�1N))= C1�1( )�s(Q)Q�1; �1Now we must return to the two laims about the sums �A( ). The �rst is easy on makinga hange of variable. For the se ond we shall show the existen e of B 2 GLg(Z=rZ) su h thatBA = A and (detB) 6= 1. From this one sees easily that �A( ) = (detB)�A( ) and so

68

�A( ) = 0. To �nd su h a B let M be the submodule of (Z=rZ)g generated by the olumnsof A. What we require is � 2 Aut((Z=rZ)g ) with � jM= Id and with (det�) 6= 1. Howeverwe an pi k a basis e1; :::; eg of (Z=rZ)g su h that M = (�1e1; :::; �geg) with (�1; r) 6= 1.Then let � be represented by : 0BBBBBBBBB�� 0 : : 00 1 0: : :: : :0 0 : : 1

1CCCCCCCCCAwith � � 1 mod r(�1;r) and (�) 6= 1. This is the desired � and ompletes the proof of thelemma.We now introdu e a formal q-expansion whi h looks like an Eisenstein series:

E�;b;s(k; �) = X(deth; p1) = psh 2 symm�g(Z)h > 0ahqh

where:� b is prime to p� � : (Z=ptZ)� hi�! (1 + pZ)=(1 + ptZ) �! Q a �� �(Z=bpZ)� �! Q a � and � = �(p)�(p) orresponding to (Z=bpZ)� �= (Z=bZ)� �(Z=pZ)�and where:ah = (deth)� g+12 f g2̂�(p)�(p)h �(p)�(f ^�(p)�(p)h )�(�̂(p)�(p)h )L(p)(1 + g2 � k; ^��1�!�k�h)��1(f ^�(p)�(p)h )hf ^�(p)�(p)h i�k�(p�s deth)hp�s dethikQl 2 P (h)l 6 jbp Ml;h(�(l)�(l�1)hli�k)Ql 6 jf ^�(p)�(p)hl j b (1� l g2g��h(l)�(l�1)hli�k)69

Here we have used the notation �h = ��ph where �ph is de�ned modulo a number prime to pand where � = !s p�12 . Also L(p) denotes the L-series with the Euler fa tor at p removed.Now we prove that these are not too far o� from being modular forms.Lemma 3.8 Let b be prime to p, � : (Z=bpZ)� ! Q a �, � = �(p)�(p) as above with�(p) 2 6= 1, �(�1) = 1. Let � : (Z=pt1Z)� hi! (1 + pZ)=(1 + pt1Z) ! Q a � also be asabove. Let Q 2 symm�g(Z) with (detQ; p1) = pt2 . Set t = max(t1; t2 + 1). Also let : (Z=rZ)� ! Q a � be su h that s(Q)r2 j bpt and p j r. Finally assume that k � g2 + 1.Then: CU tp(�Q; E�;b;t2(k; �)) = U tp(�Q; �2!�2k ~Ek;���1!k;bp)2 Mk+ g2 (�0(bpt); ��Q�!�k)with C a non-zero onstant.We �rst separate out a part of the proof:Lemma 3.9 Let � be a hara ter de�ned mod pt1 and let Q 2 symm�g(Z) with Q > 0 and(detQ; p1) = pt2 . Then if t = max(t1; t2 + 1) and if A =PAhqh is a formal q-expansion:�(p�t2 det(�Q))U tp(�Q; �2A) = U tp(�Q; A�)where A� =P(deth;p1)=pt2 (p�t2 deth)Ahqh.Proof:We look at the oeÆ ient of qh on both sides. On the left hand side we get:Xg+tnQn=pth�2 (detn)Ag�(p�t2 det(�Q))while on the right hand side we get:Xg+tnQn=pth (detn)Ag�(p�t2 det g)This proves the lemma.Proof of lemma three:First note that: �Q; �2!�2k 2Mk(�0(bpt); �Q�2!�2k)

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and that as �(p) 2 6= 1 and k � g2 + 1 we have:~Ek;���1!k 2Mk(�0(bpt); ���1!k)Thus we need only show that the laimed identity holds (formally). However:�!�k(detQ)U tp(�Q; �2!�2k ~Ek;���1!k) = U tp(�Q; E1)= �(p�t2 detQ)U tp(�Q; E2)where:� E1 =P(deth; p1) = pt2h 2 symm�g(Z)h > 0bhqh

� E2 =P(deth; p1) = pt2h 2 symm�g(Z)h > 0�(p�t2 deth)bhqh

� bh = f g2�k̂�(p)���1!k�(p)(f ^�(p)���1!k)�( ^��1(p)��!�k)ah:8>>>><>>>>:

1 f ^�(p)���1!k = p2m�(p)h (p) f ^�(p)���1!k = p2m�1A�(p)h (p) ��(p)!k��1 = 1with m a positive integer� A� = 1��(p)(p)p g2�k1��(p)(p)�1pk� g2�1� � = ! p�12Here we have used the fa t that if �1, �2 are primitive hara ters with oprime ondu torsthen �(�1�2) = �(�1)�(�2)�1(f�2)�2(f�1). Note that A� 6= 1 or 0 as �(p) 2 6= 1 and alsothat: �(p)h (p) = Æ:�(p�t2 deth)with Æ = �(�1) g2 if p � 1 mod 4 and Æ = �(�1) g2+t2 if p � 3 mod 4.Thus we see that: 71

� If f ^�(p)���1!k is an even power of p and is 6= 1 then:E1 = f g2�k̂�(p)���1!k�(p)(f ^�(p)���1!k)�( ^�(p)���1!k)E�;b;t2(k; �)

� If f ^�(p)���1!k is an odd power of p then:E2 = Æf g2�k̂�(p)���1!k�(p)(f ^�(p)���1!k)�( ^�(p)���1!k)E�;b;t2(k; �)

� If ^�(p)���1!k = 1 then:E1(A�1Æ +A�1�Æ) +E2(A�1Æ �A�1�Æ) =2f g2�k̂�(p)���1!k�(p)(f ^�(p)���1!k)�( ^�(p)���1!k)E�;b;t2(k; �)

In the �rst two ases we on lude at on e the desired equality. In the last ase we on ludethat: A�1�(p�t2 detQ)Æ�!�k(p�t2 detQ)U tp(�Q; �2!�2k ~Ek;���1!k;bp) =f g2�k̂�(p)���1!k�(p)(f ^�(p)���1!k)�( ^�(p)���1!k)U tp(�nQ; E�;b;t2(k; �))and again we are done.We shall now �x a usp form f in Sk(�0(q); �). Assume that f has Fourier expansionPh2symm�g(Z)h>0 ah(f)qh. Also if Q 2 symm�g(Z), Q > 0 and : (Z=rZ)� ! Q a � weintrodu e a Diri hlet series:DQ; (s) = L(s+ g2 ; ��Q)Q g2�10 L(2s+ 2i; 2�2)PM2SLg(Z)nM+g�g(Z) (detM)aMQtM (f)(detM)1�k�sLemma 3.10 DQ; (s) = P1n=1 dnns with jdnj � Cng+ 12 log n for some onstant C. In par-ti ular it is onvergent for Re s > g + 32 .Proof:We shall establish the following laims:1. jah(f)j � C1(deth) k22. #f[M ℄ 2 SLg(Z)nM+g�g (Z)j detM = ng � C2ng� 12 log n

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From these two it at on e follows that:XM2SLg(Z)nM+g�g(Z) aMQtM (f)(detM)1�k�s =X annswith janj � C1C2(detQ) k2ng+ 12 log n, and hen e that:jdnj � Pn=n1n2 jan1 jPn2=n3m21:::m2g2 1� C1C2(detQ) k2ng+ 12 (log n)Pn=n1n2 n�g2 Pn2=n3m21:::m2g2 1� C1C2(detQ) k2ng+ 12 (log n)�Pnn3=1 n�g3 � �Pn1 m�2g� g2� Cng+ 12 log nWe now prove the two laims that we made. For the �rst re all that:

ah(f) = ZX2symmg(R)=symmg (Z) f(X + iY ) exp(��i tr(zh))where z = X + iY and Y is onstant. Thus from the bound given on page 335 of [Sm℄ wesee that: jah(f)j � C3(detY )� k2 exp(� tr(Y h))for any Y 2 symmg(R ) with Y > 0. In parti ular if h 12 denotes the positive de�nite squareroot of h, putting U = h 12Y h 12 we see that:

jah(f)j � C3(deth) k2 (detU)� k2 exp(� trU)for any U > 0, as required.For the se ond laim, we see that:#f[M ℄ 2 SLg(Z)nM+g�g (Z)jdetM = ng = Pn=n1:::ng 1n2n23:::ng�1g� ng�1Pn=mng �Qg�1i=1 Pmj=1 j�i�� Cng�1(log n)#fm jmjn; m > 0g� 2Cng�1(log n)n 12Lemma 3.11 Let f , Q, be as above with:� is primitive modulo r 73

� qjs(Q)r2� all prime divisors of qs(Q) divide r� �(�1) = (�1)k� detQ 6= 0� either k � g + 1 and ( �)2 6= 1, or k � g + 3then: DQ; (k � g) = C(f; �s(Q)Q�1; �1 ~Ek� g2 ;� �Q;b)bwhere C 6= 0 and b = s(Q)r2.Proof:A ording to proposition 2.3 of [AK℄:

DQ; (s) = 1 s2 (s)�1L(s)hf; �Q; � E(z; �s2 + g2 � k2 ; k � g2 ; � �Q; b)ibwith 1 6= 0 6= 2, (s) = Qg1 �( s+k�i2 ), L(s) = L(s + g2 ; �Q�)Q g2�10 L(2s + 2i; �2 2) andRe s suÆ iently large. By proposition 2.4 of [AK℄ the equation remains valid whenever allthe terms remain de�ned. The result now follows from lemmas one and two and the fa tsthat:� hA;Bi = hAjWb; BjWbi� � �Q(�1) = (�1)k+ g2� ~E k;�;b = (�1) g2�(�1) ~Ek;��1;b (as �(�) = �(�1)�(��1) ).

Lemma 3.12 Let f be a usp form in Sk(�0(q); �) with k � 2g+2, whi h is an eigenvalueof the He ke operators and su h that ef = f . Let Q 2 symm�g(Z), Q > 0 be su h thataQ(f) 6= 0. Let pt2 = (p1; s(Q)g detQ�1). Choose r su h that it is divisible by p2 and byall prime divisors of s(Q) and su h that qjr2s(Q). Write s(Q)r2 = bpt1 with b prime to p.Let T 2 Tr with Tf 6= 0.74

Then we an hoose a primitive hara ter modulo r su h that (�1) = (�1)k�(�1)and (� )2 6= 1. Write ��Q = �!k��1 with � a hara ter modulo bp, ! the Tei hmuller hara ter and � : (Z=pt1Z)� hi! (1 + pZ)=(1 + pt1Z) ! Q a �. Then:(f; Te(�s(Q)Q�1; �1!2k��2E�;b;t2(k � g2 ; �)))bpt1 6= 0

where e(�E) is de�ned as U�tp eU tp(�E) for t � max(t1; t2 + 1).Proof:We know from Lemma 3 that the left hand side is a non-zero multiple of:

(f; Te(�s(Q)Q�1; �1 ~Ek� g2 ;� �Q;bp))bpt1(Tef; �s(Q)Q�1; �1 ~Ek� g2 ;� �Q;bp)bpt1whi h is itself a non-zero multiple of DQ; (k� g). However we know from [A1℄ that in this ase DQ; (s) has an Euler expansion of degree 2g + 1. The result thus follows from thefollowing fa t:Let D(s) = P dnns be a Diri hlet series with jdnj � Cna. Assume that for Re ssuÆ iently largeD(s) =QpQp(p�s)�1 with Qp a polynomial of bounded degree.Then D(s) is non-zero and onvergent for Re s > a+ 1.3.4 The General StrategyThis method of onstru ting Hida families is due to Wiles (see [Wi℄).Fix a positive integer N and a hara ter � : (Z=NpZ)� ! Q a �. Let O denote theintegers in a �nite extension of Q p ontaining all the '(N) roots of unity (here ' is Euler'sphi-fun tion), and set � = O[[T ℄℄. Also let � denote the set of pairs (k; �) where k is aninteger � g + 1 and � : (1 + pZp) ! Q a � is a hara ter of �nite order, so that we maythink of � as a hara ter:

(Z=prZ)� hi�! (1 + pZ)=(1 + prZ) �! Q a �We shall denote the smallest possible hoi e of r by r(�).

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Now for I an in�nite subset of � set:MI(N;�;�) =fF 2 �[[q℄℄g j F jT=(�(1+p)(1+p)k�1) 2Mk(�0(Npr(�)); �!�k�;O�) for (k; �) 2 Igwhere O� denotes the integers of the �eld FO(�(1 + p)). Also set:M(N;�;�) =fF 2 �[[q℄℄g j F jT=(�(1+p)(1+p)k�1) 2Mk(�0(Npr(�)); �!�k�;O�)for all but �nitely many (k; �) 2 �gi.e. M(N;�) = [I of �nite omplement in�MI(N;�)Note that:� � 2 �, F 2 �[[q℄℄g and �F 2M implies that F 2 M� if F 2 M then there is a � 2 � su h that �F 2 M�.� MI=(1+T ��(1+p)(1+p)k)M ,!Mk(�0(Npr(�)); �!�k�;O�) if �� I is �nite and(k; �) 2 I.Before giving examples of elements of M(N;�), or as we shall write �-adi forms, wemust introdu e Hida's idempotent in this ontext. First we de�ne an a tion of Up on M ompatible with spe ialisation by setting:Up(Xh Ahqh) =Xh Aphqh

Then we have:Lemma 3.13 Let I � � be in�nite. Then1. there is a unique operator e onMI ompatible with Hida's idempotent under spe ial-isation2. eF = limr!1 U r!p F 76

3. Up is invertible on eMI .Proof:We have the spe ialisation map:� :MI ,! M(k;�)2IMk(�0(Npr(�)); �!�k�;O�)

The image is ertainly preserved by Up. First we shall show that the image is also losed.For this let Fj 2 MI , �(Fj)!QI gi as j !1. Then if � : �!QI O� is the spe ialisationmap we have that �(ah(Fj)) ! Q ah(gi) as j ! 1. However � is ompa t and so �� is losed in QI O�. Thus we an �nd bh 2 � su h that �bh =Q ah(gi). Then it is easily seenthat P bhqh 2MI and that �(P bhqh) =QI gi.Thus the operator e = limr!1 U r!p preserves �MI and this proves the �rst two parts. Itis easily seen that Up is inje tive on eMI . Finally if F 2 eMI then forG = limr!1 U r!�1p F 2MI , whi h exists as limr!1 U r!�1p �F exists, UpG = eF = F . This proves the last part also.Corollary 3.1 The Hida operator de�ned in the lemma satis�es the following properties:1. e2 = e2. if t ommutes with Up then it also ommutes with e3. if I � J the a tion of e is ompatible withMJ �MI4. e extends to an operator onM with the same properties.

These are all easy.We are now in a position to onstru t some examples of �-adi forms:Example 3.1 Let� b be prime to p� � : (Z=bpZ)� ! Q a � be a hara ter with:{ �(�1) = 1{ � = �(p)�(p) its de omposition into \at p" and \ away from p" parts77

{ �(p) 2 6= 1� Q 2 symm�g(Z), Q > 0� (detQ; p1) = pt2� N and r be su h that pjr, s(Q)r2jN and N = bpt3Then there is a �-adi form GN;Q; ;� 2 M�(b; ��Q) su h that:GN;Q; ;�jT=(�(1+p)(1+p)k�1) = e(�Q; E�;b;t2(k � g2 ; �))in the notation of se tion two.Proof:For t � max(t3; t2 + 1) set It = f(k; �) 2 � j r(�) � tg. Then we an de�neG(t) 2MIt to be U tp(�Q; P(deth; p1) = pt2h 2 symm�g(Z)h > 0

Bh(T )qh) where:Bh(T ) = (deth)�(g+ 12 )fg^�(p)�(p)h �(p)�(f ^�(p)�(p)h )�(�̂(p)�(p)h )

(1 + T )logp(f�1̂�(p)�(p)h p�t2 deth)G��1�h!� g2 ( (1+T )(1+p)g � 1)Ql 2 P (h)l 6 jbp Ml;h(�(l)hli g2 (1 + T )� logp l)Ql 6 jf ^�(p)�(p)hljb (1� l g2 hli g2g��h(l)(1 + T )� logp l)and where:� x = !(x)(1 + p)logp x for x 2 Z�p� G�(�(1 + p)(1 + p)n � 1) = L(p)(1� n; ��!�n) for � as above, n > 0 an integer and�(�1) = 1.That Gt is in fa t inMIt follows from lemma 3.8 of se tion 3.3. Then U�tp eG(t) 2MItand U�tp eG(t)jT=(�(1+p)(1+p)k�1) = e(�Q; E�;b;t2(k � g2 ; �)) again by lemma 3.8 of se tion

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3.3. In parti ular if t0 > t then: U�tp eG(t) 2 MItk [U�t0p eG(t0) 2 MIt0and so G = U�tp eG(t) 2 Tt0�tMIt0 =M� and has the desired spe ialisations.Next we re ord two te hni al results about �-adi forms:Lemma 3.14 1. eM(Np; �) = eM(N;�)2. If O0 denotes the integers of a �nite extension of O and �0 denotes O0[[T ℄℄ then:M(N;�;�)� �0 =M(N;�;�0)

These are both straightforward. We an use part two to de�neM(N;�;R) for any �-algebraR to beM(N;�;�)� R.We now prove one of the main theorems on �-adi forms:Theorem 3.1 MÆ(N;�) is a �nite free �-module.Proof:We may and shall assume that N is prime to p (by the last lemma). Note thatMÆ(N;�) is ertainly torsion free over �.Step 1 There is a onstant C su h that dim eMk(�1(Np)) � C for all k.Let: �(g) = f0� A BC D1A 2 Sp2g(Z) jC � 0 mod Np; B � 0 mod N;

A � D � modN; detA � 1 mod pgThen it will do to show that dim eMk(�(g)) is bounded independently of k. However we knowfrom proposition 2.1 that for ea h g0 there is a onstant Cg0 su h that dim eSk(�(g0)) < Cg0 .The result now follows by indu tion using lemma 3.5 of se tion 3.2.Step 2 If C is as in step 1 then MÆ(N;�) does not ontain more than C �-linearlyindependent elements.79

Assume that F0; :::; FC were linearly independent elements ofMÆ(N;�). Then we an�nd h0; :::; hC in symm�g(Z) su h that the matrix (ahi(Fj)) is non-singular. Then for ksuÆ iently large:� Fj jT=((1+p)k�1) 2 eMk(�1(Np);O)� det(ahi(Fj jT=((1+p)k�1))) 6= 0For su h k the Fj jT=((1+p)k�1) are (C+1) linearly independent elements of eMk(�1(Np);O), whi h is a ontradi tion.Step 3MÆ(N;�) is a ompa t �nitely generated �-module.We an hoose F1; :::; Fr a maximal set of linearly independent elements of MÆ(N;�),and we an hoose h1; :::; hr in symm�g(Z) with � = det(ahi(Fj)) 6= 0. Then we laim that�MÆ(N;�) � hF1; :::; Fri�. For if F =P bjFj with bj 2 F� then we have the non-singularset of equations: ahi(F ) =X bjahi(Fj)for \unknowns" bj . As the ahi(F ) and the ahi(Fj) are in �, bj 2 ��1�.Step 4 In parti ular MÆ(N;�) =MÆI(N;�) for some subset I of � with �nite omple-ment. Thus for almost all pairs (k; �) 2 � we have that:MÆ(N;�)=(T � �(1 + p)(1 + p)k)MÆ(N;�) ,!Mk(�1(Npr(�));O�)Thus the theorem follows from the following lemma:

Lemma 3.15 Let M be a ompa t �-module and }i an in�nite olle tion of height oneprimes su h that M=}iM is a �nite torsion free Zp-module thenM is a �nite free �-module.Proof:Let Oi = �=}i. Then M=}iM is a �nite free Oi-module. Let M=}iM �= Orii . Letr = ri0 = min ri. Then by Nakayama's Lemma �r �!! M , and Ori !! (M=}iM) for all i.Thus r = ri for all i and Ori �! (M=}iM). Thus if ~� 2 �r is su h that �(~�) = 0 then ~� 2 }rifor all i and so ~� = ~0. That is � is an isomorphism.

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Corollary 3.2 There exists � 2 � su h that:MÆ(N;�) �MÆ�(N;�) � �MÆ(N;�)Proof:If F 2 M(N;�) then we know that we an �nd � 2 � so that �F 2 M�(N;�), thisis enough asMÆ(N;�) is �nitely generated over �.We shall now de�ne an a tion of the He ke ring TNp on M(n; �) (or MI(N;�)) bysetting: F jT (n) = n� g(g+1)2 �(ng)(1 + T )g logp nPd1jd2j:::jdg jn dg1:::dgPD �(detD)�1(1 + T )� logp detDan�1DhtDwhere the se ond sum is taken over a set of representatives for:GLg(Z)nGLg (Z)

0BBB� d1 . . . dg1CCCAGLg(Z)

This is ompatible with spe ialisation. We an similarly de�ne an a tion of T�Np onM(N;�) O Q a p whi h is ompatible with spe ialisation, by using the formula in se tion3.2.Lemma 3.16 TNp a ts semi-simply onM(N;�; F�).Proof:It will do to show that ea h T (n) 2 TNp a ts semi-simply, be ause TNp is ommuta-tive. Let T (n) have hara teristi polynomial P (X), and let Q(X) 2 �[X℄ be the produ tof the distin t irredu ible fa tors P (X). It will do to show that Q(T (n)) a ts as 0 on Mor equivalently on M=(1 + T � �(1 + p)(1 + p)k)M for in�nitely many (k; �). Howeverfor almost all (k; �) 2 �, M=(1 + T � �(1 + p)(1 + p)k)M ,! M. For su h a pair (k; �)let �P (X) and �Q(X) 2 O�[X℄ denote the redu tion of P and Q. Let q be de�ned from �Pthe same way Q was de�ned from P . Then q(X)j �Q(X). Moreover q(T (n)) = 0 as �P (X) isthe hara teristi polynomial of T (n) on a subspa e of Mk(Np?; ��!�k) on whi h T (n) isknown to a t semi-simply. Thus �Q(T (n)) = 0 and we are done.Before proving our se ond main theorem about lifting eigenforms we state and prove analgebrai lemma whi h we shall require: 81

Lemma 3.17 Let R denote the integers of a �nite extension of Q p . LetM be a �nite torsionfree R module. Let T be a ommutative ring of operators on M . Let MFR =LI Vi whereT a ts on Vi by a hara ter �i : T ! R. Let a : M ! R be a linear form, x 2 M and� : T ! R be su h that:� (�(t)� t)x 2 pAM� valp a(x) � B� rkM � CThen for some i:� valp (�(t)� �i(t)) � A�BC for all t 2 T� a 6� 0 on ViProof:Let J = fi 2 I j ajVi 6� 0g. Let � denote the proje tion of M toLJ Vi. Then� (�(t)� t)�(x) 2 pA�M� valp a(�(x)) � B� rk�M � CThus we may assume that a 6� 0 on all Vi. Let M 0 =LM \ Vi. Let I � T and I 0 � R bethe annihilators of x in M=pAM . Then T=I �= R=I 0 via �. Then we have that:FittT (M 0) = 0and so FittT=I(M 0=IM 0) = 0and so FittR=I0(L(M \ Vi)=�i(I)(M \ Vi)) = 0and so Q�i(I)dimVi � I 0 � pA�Band so for some i, valp �i(I) � A�BCas desired. (Here Fitt denotes the Fitting ideal. For some of its properties see (for example)the appendix of [MW1℄.)82

Theorem 3.2 Let f 2 eSk(Nps; ��!�k;O) be an eigenform of the He ke algebra TNp , sayf jT (n) = �(n)f . Let Q 2 symm�g(Z), Q > 0 be hosen su h that aQ(f) 6= 0. Let M be hosen prime to p su h that N jM and su h that if l (6= p) is any prime dividing s(Q), sayl� jjs(Q), then l�+2ijjM with i some positive integer. Then there exists a �nite extensionof F�, with integers R say, and a prime } of R above (1 + T � �(1 + p)(1 + p)k) andF 2M(M;�;R) su h that:� F jT (n) = �(n)F with �(n) 2 R� �(n) � �(n) mod }for all n su h that T (n) 2 TMp .Proof:We shall �rst show the result for suÆ iently large k and then dedu e it for all k.Choose k0 su h that:� k0 � 2g + 2� M =MI for some I ontaining (k; �) for all k � k0 and for all �.Step 1 If k � k0 then we an �nd a non-zero f 0 = MÆ(M;�)=(1 + T � �(1 + p)(1 +p)k)MÆ(M;�) with f 0jT (n) = �(n)f 0 for all T (n) 2 TMp .We may assume f is an eigenve tor for T�Mp . Then we an �nd T 2 TMp with f jT =�f 6= 0 and su h that TMk(Mpt; ��!�k) is an eigenspa e for TMp . Then hoose r and tsu h that (Nps)js(Q)r2 = Mpt1 and p2jr and hoose a primitive hara ter modr with (�1) = �(�1) and (�(p) (p))2 6= 1. Set � = �2!�2k�Q. Then we see from example 3.1and lemma 3.12 and se tion 3.3 that:

0 6= (f; TeGMpt1 ; s(Q)Q�1; �1!2k��2; �jT=(�(1+p)(1+p)k�1))Mpt1Thus f 0 = TeGMpt1 ; s(Q)Q�1; �1!2k��2; �jT=(�(1+p)(1+p)k�1) is non-zero and will do.Step 2 If k � k0 then the theorem is true.Choose R the integers in a �nite extension of F� su h that TMp is diagonalisable onMÆ(M;�; FR). Let e1; :::; er 2MÆ(M;�;R) be eigenve tors of TMp whi h span83

MÆ(M;�; FR). Let �i : TMp ! R be the eigenvalue orresponding to ei. Let U =LReiand let T � EndR(M(M;�;R)) be the R-subalgebra generated by TMp . Then U is afaithful T module. Let I � T be the annihilator of f 0. Then arguing as in the proofof lemma 3.17 we see that there is a prime } above (1 + T � �(1 + p)(1 + p)k) su h thatFittR=}(U=IU) = 0. ThenQ�i(I) � } and so for some i �i(I) � }, i.e. for all T (n) 2 TMp�i(T (n)) � �(T (n)) mod }.Step 3 The theorem is true for all k.We have seen (proposition 2.1) that there is a onstant C su h thatdimMÆl (Npt; ��!�k; FO) < C for all l. Let valp (aQ(f)) = B. Let � be the theta se-ries de�ned in lemma 2.3. Then �prf 2 Mk+(p�1)pr(Npt; ��!�k;O). Moreover �prf �f mod pr+1, and so e(�prf) � f mod pr+1 and:(�prf)jk+(p�1)prT (n) � f jkT (n)� �(n)f� �(n)(�prf) mod pr+1as ak+(p�1)pr � ak mod pr+1 for all a. Thus also ((�prf)je)jT (n) � �(n)((�prf)je) mod pr+1.Thus by the lemma proved just before this theorem we see that for r > B we an �ndfr 2Mk+(p�1)pr(Npt; ��!�k;O) su h that:� frjT (n) = �r(n)fr� �r(n) � �(n) mod p[ r+1�BC ℄� aQ(fr) 6= 0Now for r suÆ iently large this implies that we an �nd Fr 2 MÆ(M;�;R) (whereR denotesthe integers in an extension of � su h that TMp is diagonalisable on MÆ(M;�; FR)) su hthat:� FrjT (n) � �r(n)Fr� �r(n) � �(n) mod (}r; p[ r+1�BC ℄)� }r is a prime of R above (1 + T � �(1 + p)(1 + p)k+(p�1)pr).84

Now there are only �nitely many hoi es for �r and so there is an in�nite set S ofpositive integers su h that for r 2 S �r = �. Then:�(n) � �(n) mod \S (}r; p[ r+1�BC ℄)We laim that: \S (}r; p[ r+1�BC ℄) \ � � (1 + T � �(1 + p)(1 + p)k)from whi h the theorem would follow. However this in lusion follows from the two fa ts:� (}r; p[ r+1�BC ℄) \ � = (p[ r+1�BC ℄; (1 + T � �(1 + p)(1 + p)k+(p�1)pr))� (p[ r+1�BC ℄; (1+ T ��(1+ p)(1+ p)k+(p�1)pr)) � (p[ r+1�BC ℄; (1+ T ��(1+ p)(1+ p)k))The �rst of these is easy and the se ond not mu h harder. (In general if R1 � R2 are tworings with prime ideals }1 and }2 where }1 = } 2, and if a 2 R1 then (a; }2) = (a; }1). Toprove this one redu es at on e to the ase R1, R2 integral domains and }1 = }2 = 0, inwhi h ase it is obvious.)3.5 Conje tural Appli ationsThroughout the rest of this paper we shall restri t to the ase g = 2, i.e. toGSp4. This is justfor de�niteness and be ause g = 2 was the ase of interest for us. For other even g exa tlysimilar results should hold with � : Gal(Q a =Q ) ! GSp4 repla ed with � : Gal(Q a =Q ) !GLN for suitable N , and with a suitable hara teristi polynomial. We shall also make freeuse of the following onje ture:Conje ture 3.1 Let f 2Mk(N;�) be an eigenform of the He ke algebra TN , say f jT (n) =�(n)f . Assume k � 3. Then there is a ontinuous semi-simple representation:� : Gal(Q a =Q ) �! GSp4(Q a p )unrami�ed outside pN and su h that the hara teristi polynomial of Frobl for l 6 jNp isgiven by:X4 � �(l)X3 + (�(l)2 � �(l2)� l2(k�2)�(l2))X2 � l2k�3�(l2)�(l)X + l4k�6�(l4)

85

Our aim in this se tion is to show that these onje tures would imply similar results forordinary �-adi forms and for ordinary eigenforms of weight two. We �rst prove a generalresult on lifting group representations:Proposition 3.1 Let R be an integral domain, f}igi2I be an in�nite set of prime idealssu h that the interse tion of any in�nite subset is zero. Let Oi denote a �nite extension ofR=}i ontaining its integral losure, and assume that FOi is of hara teristi zero. Let � bea group and �i : �! GLN (Oi) a semi-simple representation. For ea h x 2 � suppose thereexists Tx 2 R with Tx � tr(�ix) mod }i.Then there is an in�nite subset J � I, S an integral domain �nite over R[f�1℄ for somef 2 R, a semi-simple representation � : � ! GLN (S) and for ea h i 2 J a prime Pi over}i, a �eld Li � FOi, a map S=Pi ,! Li su h that:S=Pi ,! LiS SR=}i � Oi ommutes and su h that � mod Pi is onjugate to �i as a representation into GLN (Li).We break up the proof into a series of lemmas.Lemma 3.18 Let K be a �eld of hara teristi zero, let A � MN (K) be a split semisim-ple sub-algebra and let H � GLN (K) be its normaliser. Let fa1; :::; arg and fb1; :::; brgbe subsets of A ea h of whi h spans A and assume that for s = 1; :::; 4 we have thattr(Qsj=1 aij ) = tr(Qsj=1 bij ) for all s-tuples (i1; :::; is) 2 f1; :::; rgs. Then there is an el-ement n 2 H with nain�1 = bi for all i.Proof:A is of the formL�M �(K)a� and without loss of generality we may assume(m��)�;�=1;:::;a� embeds asLm�b��� , whereP� a�b� � = N . Also assume that the notationis su h that for � 6= �0 (b�; �) 6= (b�0 ; �0).Let e�� denote the idempotents orresponding toour de omposition of A into simple algebras. Let e�� = P���iai and set f�� = P���ibiThen we see that:86

� e2�� = e�� ) tr(e2�� � e��)aj = 0 8j) tr(f2�� � f��)bj = 0 8j) f2�� = f��� e�� entral ) tr(e��aiaj) = tr(aie��aj) 8i; j) tr(f��bibj) = tr(bif��bj) 8i; j) f�� entral� � = rk (tr(e��aiaj))i;j ) �rk (tr(f��bibj))i;j) � = dim f��A� tre�� = �b� ) trf�� = �b�From this we see that for ea h � ff��g is a permutation of fe��g. We an onjugatethe bi's by an element of H su h that for ea h �; � f�� = e��. Thus we may withoutloss of generality assume that this equation holds. Now �x �; � and set a0i = e��ai andb0i = e��bi. Then we an onsider a0i and b0i 2 M �(K) su h that for s = 1, 2 or 3 and(i1; :::; is) 2 f1; :::; rgs tr(Qsj=1 a0ij ) = tr(Qsj=1 b0ij ). Moreover we need only show that we an �nd n0 2 GL �(K) with n0a0in0�1 = b0i for all i.Let �kl =P�klia0i and set Ækl =P�klib0i. Then as above we an show that Ækk form a setof ommuting idempotents ea h of tra e one and su h thatP Ækk = 1. Thus by onjugationby some element of GL � we may assume that without loss of generality �kk = Ækk. Thenwe also see that �kkÆkl = Ækl and Ækl�ll = Ækl so that Ækl = �kl�kl for some �kl 2 K. Then�kl�lm = �km and �kk = 1 so there exist �k 2 K� su h that �kl = �k��1l . Thus we an �nda diagonal matrix n0 in GL � su h that after onjugation by n0 we may assume �kl = Ækl forall k and l. Then it is easily seen that a0i = b0i for all i.Lemma 3.19 Let the assumptions be as in the theorem, but assume further that � is gen-erated by x1; :::; xr and that for ea h i if Ai is the subalgebra of GLN (Ki) generated by Im�ithen Ai is spanned by f�i(xj)g. Then the on lusions of the proposition hold, ex ept thatwe do not yet laim that � is semi-simple.

87

Proof:Without loss of generality we may assume that � is the free group on x1; :::; xr.Ai is semi-simple as it has a faithful semi-simple module. We may assume that Ai is splitover FOi . Then after onjugation and dis arding some i we may assume that for all iAi = A FOi where A �=LM �(Z)a� and a ,!MN (Z) by:(x��) 7�! �xb���where the pairs (b�; �) are distin t for distin t � and where P a�b� � = N .Let = f(ai) 2 Ar j the ai spanAg, so that is a subvariety of rP a� 2� dimensionalaÆne spa e de�ned over Q . Then by the last lemma we have a map: �! A Mgiven by (ai) 7! (trQsj=1 aij ) taken over s = 1; :::; 4 and all (i1; :::; is) 2 f1; :::; rgs. Considerthe point T 2 A M (R) de�ned by: T = (TQ xij )Then T mod }i is in the image of (FOi) for all i. Thus T is in the image of (F a R ),be ause the image of is de�ned by some polynomial equalities and inequalities. Thus we an �nd X1; :::;Xr in AF a R whi h span AF a R and whi h satisfy trQs1Xij = TQ xij forall (i1; :::; is) as above. In fa t all the Xi lie in some S=R as des ribed in the proposition.Dis ard the �nite number of }i whi h ontain f and assume (as we may ) that FR=FS isGalois. Choose Pi over }i and hoose Li su h that S=Pi andKi both embed in Li over R=}i.Then (Xj mod Pi) 2 (S=Pi) � (Li) and �(Xj mod Pi) = (TQ xij mod }i). Thus by thelast lemma there is n in the normaliser of A in GLN (Li) with n(Xj mod Pi)n�1 = �i(xj)for all j and so we are done.Proof of proposition 3.1Let A be de�ned as in the proof of the last lemma. Pi k a distinguished index, say 0.Let x1; :::; xr 2 � be su h that f�0(xj)g span AFO0 . Then rk (Txjxk) � dimA and so witha �nite number of ex eptions whi h we may dis ard rk tr�i(xj)�i(xj) � dimA. However iffajg is a �nite subset of A F then rk tr(aiaj) � dimA with equality if and only if theyspan A. Thus we see that we may assume that the �i(xj) span A FOi for all i 2 I.88

Now let � be the subgroup of � generated by the xi. Then we by the last lemma thatwe have a map � : � ! (A S)� � GLN (S) with the notation as in the proposition and�ij� � � mod Pi. Now de�ne: A(S0) ��! S0rfor S0 any S algebra by: a 7�! (tr a�(xi))This is a linear map and the image 0(S0) is thus de�ned by the vanishing of ertain linearforms with oeÆ ients in FS . By inverting some element of S we may assume that � isinje tive and that 0 is de�ned over S. Then for 2 � let t = (T xi). By redu tion mod }ifor in�nitely i we see that t 2 0(S) and hen e there exists a unique �( ) with ��( ) = t .Then �( ) � �i( ) mod Pi for all i 2 I. Thus we have � : �! A S, � � �i mod Pi for alli 2 I. It follows at on e that � is a representation.Finally we repla e � by �ss, then at almost all }i:(�ss mod }i) = (� mod }i)ss = �iCorollary 3.3 Let � be as in se tion 3.4. Let }i be an in�nite set of distin t height oneprimes. For ea h i let: �i : Gal(Q a =Q ) �! GLN (Q a p )be ontinuous representations, unrami�ed outside some integer N . For l 6 jN let l(X) 2 �[X℄be moni of degree N and su h that:

l(X) � har�i(Frobl)(X) mod }iwhere hara(X) denotes the hara teristi polynomial of a. Then there exists R the integersof a �nite extension of F� and:� : Gal(Q a =Q ) �! GLN (FR)su h that (X) = harFrobl(X) for all l 6 jN .

89

Proof:We �rst show that the partial map from Gal(L=Q ) where L denotes the maximalextension of Q unrami�ed outside N :Frobl 7�! l(X)

is ontinuous. Pi k any positive integer n and pi k some l 6 jN . Then we would like toshow that for l0 with Frobl0 suÆ iently lose to Frobl we have that l � l0 mod mn wherem is the maximal ideal of �. But TI;Z�0(}i; ps) = 0 and so by ompa tness there exist}1; :::; }r and s1; :::; sr su h that Tr1(}i; ps) � mn. Then hoose Ui open neighbourhoodsof Frobl su h that for l0 with Frobl0 2 Ui �i(Frobl) � �i(Frobl0) mod psi . Then for l0 withFrobl0 2 U = TUi l � l0 mod mn as desired.Thus we an extend uniquely to a ontinuous map:Gal(Q a =Q ) �! Gal(L=Q ) �! �[X℄

Then � � har�i(�) mod }i for all � 2 Gal(Q a =Q ) by ontinuity. Now apply the proposi-tion and we �nd a representation � : Gal(Q a =Q ) ! GLN (FR) for some appropriate R with� mod Pi onjugate to �i for in�nitely many i. Thus har�(Frobl) � har�i(Frobl) � l(X) forin�nitely many height one primes, so � is the desired representation.Before proving our last two main theorems we prove one further lemma.Lemma 3.20 Let F be a �eld, � a group, � : � ! GSp4(F ) � GL4(F ). Then �ss :� ! GL4(F ) preserves a non-degenerate symple ti form, so we may onsider �ss : � !GSp4(F ).Proof:Let 0 6= V � F 4 be a simple � submodule. Let hi denote the symple ti form.Then we are in one of the following ases:� V = F 4 and �ss = �� dimV = 2, F 4 = V � V ?. In this ase we an hoose a basis of F 4 with respe t to

90

whi h hi is represented by 0� 0 12�12 01A and � is either of the form:

� 7�!0BBBBBB�

a� 0 b� 00 a0� 0 b0� � 0 d�0 0� 0 d0�

1CCCCCCAin whi h ase �ss = � or it is of the form:

� 7�!0BBBBBB�

a� 0 b� �� e� � � � 0 d� �0 0 0 f�1CCCCCCA

when �ss is easily seen to preserve hi.� dimV = 2, V ? = V . In this ase we an hoose a basis of F 4 su h that hi isgiven by 0� 0 12�12 01A and for all � �(�) = 0� A� B�0 D�

1A. Then �ss is given by� 7! 0� A� 00 D�

1A and also preserves hi.� dimV = 1, V ? � V , dimV ? = 3. In this ase we an hoose a basis of F 4 withrespe t to whi h hi is represented by 0� 0 12�12 0

1A and � has the form:� 7�!

0BBBBBB�e� � � �0 a� � b�0 0 f� 00 � � d�

1CCCCCCAThen �ss also preserves hi.

91

Theorem 3.3 Assume the onje ture 3.1. Let F 2 MÆ(N;�;R), with N prime to p andR the integers of some �nite extension of F�, be an eigenform of the He ke algebra TN , sayF jT (n) = �(n)F . Then there is a �nite extension L of FR and a ontinuous representation:� : Gal(Q a =Q ) �! GSp4(L)whi h is unrami�ed outside Np and su h that for l 6 jNp, a prime, Frobl has hara teristi polynomial:

X4 � �X3 + (�(l)2 � �(l2)� l�1�(l))X2 � �(l)�(l)X + �(l)2where �(l) = l�3�(l2)(1 + T )2 logp l.

Proof:Combining onje ture 3.1 and the orollary to the last proposition we at on ededu e the existen e of su h a representation into GL4(L). We need only show that itpreserves a non-degenerate symple ti form. Let G denote the Zariski losure of Im�, andGÆ the onne ted omponent of the identity. Let G =` �( i)GÆ and let � = f 2 � j �( ) 2GÆ and tr�( ) 6= 0g. Then GÆ is the Zariski losure of �. Now what we require is a matrixA 2M4(L) and � : f1; :::; rg ! f�1g su h that:1. (tr�( ))�( )At�( ) = (tr�( �1)A for all in �2. �( i)At�( i) = �ipdet �( i)A3. tA = �A4. detA 6= 0where pdet �( i) is some �xed square root of det �( i) whi h we may assume lies in L. Itis easy to he k that a solution to these equations in S=Pi for in�nitely many height oneprimes Pi of some �nitely generated extension of the integers of L ontained in L impliesthe existen e of su h a solution in S.( More pre isely let r� be the dimension of the spa e of matri es A satisfying on-ditions 1), 2) and 3). Let e(�)1; :::; e(�)r� be a basis of the spa e of su h matri es. That92

det(P e(�)i�i) 6� 0 may be expressed as Fr�(e(�)i)(X) 6= 0 for some polynomial Fr�(e(�)i(X).Thus we have a solution if and only if r� > 0 and Fr�(e(�)i)(X) 6= 0. Then for some � thereexists a solution of 1)-4) for in�nitely many height one primes Pi. This implies r� > 0 andFr�(e(�))(X) 6= 0.)Theorem 3.4 Assume onje ture 3.1. Let f 2MÆ2 (Npr; �;O) be an eigenform of the He kealgebra TNp , say f jT (n) = �(n)f . (O the integers in a �nite extension of Q p .) Then thereis a multiple M of N and a ontinuous representation:

� : Gal(Q a =Q ) �! GSp4(Q a p )whi h is unrami�ed outside Mp and su h that if l 6 jMp then Frobl has hara teristi poly-nomial: X4 � �(l)X3 + (�(l)2 � �(l2)� �(l2))X2 � l�(l2)�(l) + l2�(l4)Proof:By theorem 3.2 we an �nd an M as above; hara ters and � su h that is de�ned modulo Mp, � is of p power order and is de�ned modulo a power of p and� = !�2�; R the integers of a �nite extension of F�; and F 2MÆ(M; ;R) an eigenformfor the ring of He ke operators TMp , say F jT (n) = �(n)F , su h that � � �(n) mod P withP a prime of R above (1 + T � �(1 + p)(1 + p)2)). Then by the last theorem we an �nd a ontinuous representation: � : Gal(Q a =Q ) �! GSp4(L)with L a �nite extension of F�. Moreover it is unrami�ed outside Mp and for l 6 jMp the hara teristi polynomial of Frobl is ongruent modulo a prime (P 0 say) above (1+T��(1+p)(1 + p)2) to the polynomial des ribed in the statement of the theorem. As Gal(Q a =Q )is ompa t we an �nd a �nite R0 (the integers of L) L � L4 su h that L R0 L = L4 andwhi h is preserved by the Galois a tion. Then LP 0 is free over R0P 0 and we an hoose a

93

basis with respe t to whi h the symple ti form is given by:0BBBBBB�0 0 �1 00 0 0 �2��1 0 0 00 ��2 0 0

1CCCCCCAwhere �1j�2. We may assume that p�1 and p�2 lie in R0, and that �1 = 1. Let e1; e2; e01; e02be the orresponding basis. We laim that L0 = he1; �� 122 e2; e01; �� 122 e02i is also preserved byGal(Q a =Q ).Let (aij) denote an element of Im � with respe t to the basis e1; e2; e01; e02. Then:

(aij)0BBBBBB�

0 0 1 00 0 0 �21 0 0 00 �2 0 01CCCCCCA (aji) = �

0BBBBBB�0 0 1 00 0 0 �21 0 0 00 �2 0 0

1CCCCCCAwith � a unit in R0P 0 . From this we see that:� a11a33 � a13a31 = �� a11a23 � a13a21 � 0 mod �2� a11a43 � a13a41 � 0 mod �2� a21a33 � a23a31 � 0 mod �2� a21a43 � a23a41 � 0 mod �2� a31a43 � a23a41 � 0 mod �2and so: a23� = a33(a11a23 � a13a21) + a13(a33a21 � a23a31) � 0 mod �2so that a23 � 0 mod �2. Similarly a21 � a43 � a41 � 0 mod �2. Thus (aij) also preservesthe latti e L0 as we wanted to show.

94

Now redu tion gives us:� : Gal(Q a =Q ) �! Aut(L0 R0P 0=P 0; hi) � GSp4(Q a p )a ontinuous representation unrami�ed outside Mp with the hara teristi polynomials ofthe desired forms.

95

Chapter 4Modular Forms over an ImaginaryQuadrati Field4.1 Introdu tionIn this hapter we onsider Hida families for GL2 over an imaginary quadrati �eld. As abyprodu t we are led to a method to exhibit torsion in the �rst homology groups of ertainsheaves on the 3-manifolds asso iated to su h forms.To explain our results let K be an imaginary quadrati �eld whi h we shall assume has lass number one (though this is almost ertainly unne essary) and let O denote its ringof integers. We �x Ka � C and Ka � C p . Let p be an odd rational prime whi h splitsin K, and � the prime of K above p orresponding to Ka � C p . By an ordinary uspidaleigenform of \weight" n, levelM , and hara ter � : (O=MO)� ! (Ka )�, we shall mean aneigenvalue of the He ke operators Tn a ting on the orresponding spa e of usp forms (seese tion 4.2, but note that \weight n" is the \weight" orresponding to a sheaf of dimension(n + 1)2, and di�ers by a shift of two from the normal terminology in the ase of ellipti modular forms.) with the eigenvalue of Tp prime to �. We shall write M = Npr, with N

96

prime to p and de ompose � = �a � � y orresponding to (O=MO)� = A�B where:A = f� 2 ((1 +NpO)=(1 +NprO)) j ��� = 1g � (O=NpO)�B = f� 2 ((1 +NpO)=(1 +NprO)) j � = ��g(the anti- y lotomi and y lotomi parts respe tively). We shall onsider �-adi eigenformswhi h interpolate ordinary uspidal eigenforms. In the simplest ase an �-adi eigenform oflevel N (prime to p) and hara ter � (an anti- y lotomi hara ter of de�ned modulo Nprfor some r) is a olle tion am(T ) 2 O�[[T ℄℄ for ea h m 2 O su h that for ea h y lotomi hara ter and non-negative rational integer n:fam((1 + T )� (1 + p)2n (1 + p))gare the eigenvalues of the He ke operators Tm a ting on an ordinary uspidal eigenform ofweight n, level Nps (some s) and hara ter � !�n where ! is the Tei hmuller lifting of thenorm map (O=pO)� ! (Z=pZ)� to a map (O=pO)� ! Z� . In general we must allow theai to lie in a �nite extension of O�[[T ℄℄.Our main theorem (theorem 4.1) states that if we �x n then there are only �nitely many�-adi eigenforms of this level. If moreover we �x an anti- y lotomi hara ter � then allbut �nitely many ordinary uspidal eigenforms of level Nps (any s), weight n (any n) and hara ter �!n (any y lotomi hara ter ) lift to an �-adi eigenform. It would be ni eto strengthen this result to say that any ordinary uspidal eigenform lifted to a unique�-adi eigenform. The problem here is torsion in the orresponding homology groups (seethe omments following theorem 4.1).Su h results have been proved by Hida (and Wiles) for modular forms over totally real�elds, however as our modular forms are not analyti we an not multiply them togetherto produ e new ones with good ongruen e properties, so we have to rely ompletely onthe ohomology. Our argument falls into two parts. In se tion 4.4 we use the in ationrestri tion long exa t sequen e to \ hange level". In se tion 4.5 we relate di�erent weights.Both are a hieved by developing ideas of Hida ([Hi1℄).For example in se tion 4.7 we prove that if p is as above and p6 j�(p) where � is Ramanu-jam's fun tion ( i.e. �(z) =P �(n)e2n�iz is the uspidal ellipti modular fun tion of weight97

12 for SL2(Z)), and if n1 6= n2; n1; n2 > 10; n1+n2 > 20+r; and n1 � n2 � 10 mod pr(p�1)then H1(SL2(O); Sn1;n2) and H1 usp(SL2(O); Sn1;n2)have torsion of exponent divisible by pr. (Here Sn1;n2 is the SL2(O)-module whi h is thethe tensor produ t of the nth1 symmetri power of O2 with the natural SL2(O) a tion andthe nth2 symmetri power of O2 with SL2(O) a tion twisted by omplex onjugation. Alsosee se tion 4.2 for the meaning of \ usp" in this ontext.)The rest of the paper is organised as follows. Se tion 4.2 ontains some analyti resultswe need. Se tion 4.3 is somewhat te hni al and is needed to show that our use of thein ation restri tion sequen e respe ts the uspidal ohomology. (Torsion prevents us usingthe analyti theory of Eisenstein series whi h Hida used in [Hi1℄.) In se tion 4.6 we ompletethe proof of our main theorem and show how to onstru t some examples.We should mention that while all the theory goes through in the ase of a prime inertin K, we la k any examples in that ase, so that theory may be va uous.NotationMost of the notation used is either standard or explained in the text. If F is a �eld F a will denote its algebrai losure. C p will denote the ompletion of Q a p and O p the ring ofelements of non-negative valuation in C p . If � is a prime in a number �eld we shall use C �and O � for the orresponding notions. If A is a ring with ideal I then we shall set:� �(I) = f� 2 SL2(A) j � � 0� 1 00 1

1A mod Ig� �1(I) = f� 2 SL2(A) j � � 0� 1 �0 1

1A mod Ig� �0(I) = f� 2 SL2(A) j � � 0� � �0 �

1A mod Ig98

We hope the ontext will always make it lear whi h ring we are talking about.By a o�nite Zp module we shall mean the Pontriagin dual of a �nite Zp module.4.2 Review of Cohomology Groups and Automorphi FormsThroughout let K be an imaginary quadrati �eld of lass number one and let O denote itsring of integers. The assumption that the lass number is one is almost ertainly unne essarybut it simpli�es the notation. Fix also an odd rational prime p, whi h is unrami�ed in Kand a prime � of O lying above p. Note that these onditions imply that the only unit ofO ongruent to 1 modulo � is 1 itself. Let�denote omplex onjugation.For any pair of non-negative integers n1; n2 we have a free(n1 + 1)(n2 + 1) -dimensional O module with an a tion of GL2(O) (or in fa t of M2(O)).Itmay be expli itly des ribed as Sn1(O2)�Sn2(O2) where Sn denotes the n-th symmetri power (i.e. the maximal symmetri quotient of the n-th tensor power) and where 2GL2(O)a ts on the �rst O2 in the natural fashion and on the se ond via � . We will denote this mod-ule Sn1;n2 . If A is an O module Sn1;n2(A) will denote Sn1;n2OA. In parti ular Sn1;n2(C ) an be thought of as the irredu ible �nite dimensional representations of Lie group SL2(C ).When we need to take an O-basis we shall always take the natural basis with respe t towhi h = 0� a b d

1A 2M2(O) a ts as:0BBBBBBBBBBBB�

an1M n1an1�1bM : : : bn1Man1�1 M (an1�1d+ (n1 � 1)an1�2 )M : : : bn1�1dM: : : :: : : :: : : : n1M n1 n1�1dM : : : dn1M

1CCCCCCCCCCCCA

99

where M denotes the blo k:0BBBBBBBBBBBB�

�an2 n2�an2�1�b : : : �bn2�an2�1� (�an2�1 �d+ (n2 � 1)�an2�2� ) : : : �bn2�1 �d: : : :: : : :: : : :� n2 n2� n2�1 �d : : : �dn2

1CCCCCCCCCCCCAWe shall be interested in the ohomology of ongruen e subgroups� < SL2(O) with oeÆ ients in Sn1;n2(A).When A = C these groups an be studied an-alyti ally. More pre isely let Z denote \the quaternion upper half plane" or \hyperboli 3-spa e", that is to say:fquaternions z = x+ ykjx 2 C and y 2 R>0gThen SL2(C ) a ts on Z and in fa t on Z � Sn1;n2(A) by:

= 0� a b d1A : (z; v) 7! ((az + b)( z + d)�1; v)

If � is torsion free, �nZ is a smooth manifold with a sheaf ~Sn1;n2(A) onsisting of �-invariant se tions of Z � Sn1;n2(A). Then it is known that H�(�; Sn1;n2(A)) =H�(�nZ; ~Sn1;n2(A)). In the ase A = C this group is well studied. See for example Harder[Ha1℄,[Ha2℄ for the following results.There is a ompa t manifold with boundary �nZ and an embedding �nZ ,! �nZ whi his a homotopy equivalen e (the Borel-Serre ompa ti� ation). The sheaves ~Sn1;n2(A) extendto �nZ in su h a way that H�(�nZ; ~Sn1;n2(A)) �= H�(�nZ; ~Sn1;n2(A)). We have a naturalmap: H�(�nZ; ~Sn1;n2(A)) �! H�(�(�nZ); ~Sn1;n2(A))We shall denote the kernel of this map by H� usp(�nZ; ~Sn1;n2(A)) and the image byH�Eis(�nZ; ~Sn1;n2(A)).We have that: 100

� H i usp(�nZ; ~Sn1;n2(C )) = 0 unless n1 = n2 and either i = 1 or 2� H1 usp(�nZ; ~Sn;n(C )) �= H2 usp(�nZ; ~Sn;n(C )) �= Sn(�; C )� H0Eis(�nZ; ~Sn1;n2(C )) = 0 unless n1 = n2 = 0 when it is equal to C� dim H1Eis(�nZ; ~Sn1;n2(C )) = 12 dim H1(�(�nZ); ~Sn1;n2(C ))� H2Eis(�nZ; ~Sn1;n2(C )) = H2(�(�nZ); ~Sn1;n2(C )) unless n1 = n2 = 0 in whi h ase it isof odimension oneHere Sn(�; C ) = ��U(�)f where U(�) is the losure of � in SL2 of the �nite adeles of Kand where the sum is taken over all uspidal automorphi representations � = �f �1 ofSL2 over K with �1 the prin ipal series representation of SL2(C ) orresponding to the hara ter 0� a �0 a�11A 7! � ajaj�2(n+1).This set up may be des ribed in terms of group ohomology as follows. De�ne a (�-) usp to be a �- onjuga y lass of Borel subgroups of SL2(K). For B su h a Borel we willdenote by [B℄ (or if ne essary [B℄�) the orresponding usp. Set �B = � \B. The onne ted omponents of the Borel-Serre ompa ti� ation are in one-to-one orresponden e with the usps. Set: H��(�; Sn1;n2(A)) =M[B℄ H�(�B ; Sn1;n2(A))This appears to depend on the hoi e of Borel whi h represents ea h usp, however if 2 �then � B �1 = �B �1 and we get a anoni al isomorphism � : H�(�B; Sn1;n2(A)) ��! H�(� B �1 ; Sn1;n2(A))Using the fa ts that �B is its own normaliser in � and that if M is aG-module and g 2 G then the map g� : H�(G;M)! H�(G;M) indu ed by onjugationby g on G and by translation by g on M is the identity; we see further that:� � : H�(�B; Sn1;n2(A)) ��! H�(� B �1 ; Sn1;n2(A)) is independent of the hoi e of 2 � onjugating B to B �1

101

� The diagram: H�(�B ; Sn1;n2(A))res%H�(�; Sn1;n2(A)) # �res& H�(� B �1 ; Sn1;n2(A)) ommutes.We will identify the groupsH�(�B; Sn1;n2(A)) for B representing a usp [B℄ and write simplyH�(�[B℄; Sn1;n2(A)). Restri tion gives a well de�ned mapH�(�; Sn1;n2(A))! H��(�; Sn1;n2(A))and we have a ommutative diagram:H�(�; Sn1;n2(A)) res! H��(�; Sn1;n2(A))ko k oH�(�nZ; ~Sn1;n2(A)) ! H�(�(�nZ); ~Sn1;n2(A))We shall use H� usp(�;M)andH�Eis(�;M) in the obvious way.If we use the group ohomology the analyti des ription ofH� usp(�; Sn1;n2(C )) given above remains true for � with torsion. To see this hoose �� �of �nite index and without torsion. The results for � follow from those for � be ause, aswe are working over a �eld of hara teristi 0, the Serre-Hos hild spe tral sequen e impliesthat the � ohomology is just the �n� invariant part of the � ohomology.We now want to des ribe the a tion of the He ke operators on these various spa es.For this we will work in the ategory of (M2(O) \GL2(K))-modules. It is easily seen byabstra t nonsense that for � a group ontained in the semi-group (M2(O) \GL2(K)) thatthe ohomology fun tors H�(�; ) de�ned on (M2(O) \GL2(K))-modules an be thought ofas the right derived fun tors of the �xed point fun torM 7!M� and as su h are a universalÆ-fun tors. This will be very helpful in he king that diagrams ommutes. When we an onsider the maps as spe ial instan es of natural transformations between su h universalÆ-fun tors, it will do to he k the ommutativity only in degree zero.102

To des ribe the He ke operators let g2M2(O), det g 6= 0 and let �1;�2 be ongru-en e subgroups of SL2(O). Then we have [�2 : �2 \ g�1g�1℄ <1 and so we an de�ne amap [�2g�1℄ : H�(�1;M)! H�(�2;M) ,or more pre isely a natural transformation betweenÆ-fun tors [�2g�1℄ : H�(�1; )! H�(�2; ) as follows:H�(�1;M) g��! H�(�2 \ g�1g�1;M) or�! H�(�2;M)where the �rst map is indu ed by the ompatible maps :M g�! M�1 onjugation � �2 \ g�1g�1and the se ond map is orestri tion. One an he k straightforwardly that [�2g�1℄ onlydepends on the double oset �2g�1 and not on the parti ular hoi e of g. To des ribeexpli itly the a tion of [�2g�1℄ in degree zero and one, assume that �2 = q i(�2 \ g�1g�1)(whi h is easily seen to be equivalent to �2g�1 = q ig�1). Then� [�2g�1℄ :M�1 !M�2 by m 7!P( ig)m� [�2g�1℄ : H1(�1;M)! H1(�2;M) is indu ed by sending a �1- o y le � to the�2- o y le Æ 7!P( ig)�(( ig)�1Æ( jig)) where ji is the unique index su h that �1i Æ ji 2 g�1g�1.We an des ribe the He ke operators on a topologi al level by onsidering the diagram:�2 \ g�1g�1nZ. &�1nZ �2nZ indu ed by Zg�1 . & IdZ ZThis gives rise to: H�(�2 \ g�1g�1nZ; ~M)% & transferH�(�1nZ; ~M) H�(�2nZ; ~M )whi h is exa tly the He ke operator [�2g�1℄. (Here ~M is the sheaf onstru ted from Mexa tly as ~Sn1;n2 was from Sn1;n2 .) 103

We also want to de�ne their a tion on the ohomology of the boundary. The topologi alpi ture shows us how to do this. For ea h usp [C℄ of �2 \ g�1g�1 we have a map:H�(�1 [g�1Cg℄;M) g��! H�((�2 \ g�1g�1)[C℄;M) or�! H�(�2 [C℄;M)(the orestri tion map is de�ned be ause [�2 [C℄ : (�2 \ g�1g�1)[C℄℄ is less than[�2 : �2 \ g�1g�1℄). Summing these maps over [C℄ we get:[�2g�1℄ : H��(�1;M) �! H��(�2;M)A ertain amount of are is needed to keep tra k of the identi� ation we are making ofH�(�B;M) for di�erent Borels B representing the same usp. One also has to he k thatit is well de�ned up to these identi� ations. Moreover one an easily he k that it dependsonly on the double oset �1g�2 and that it is ompatible with the restri tions maps fromH�(�i;M) and the previous de�nition at this level. From this we see that He ke operatorspreserve the uspidal and Eisenstein omohology. For example to he k the ompatibilitywith our previous de�nition we must he k thatL[B℄H�(�1 [A℄;M) ! L[C℄H�(�1 [g�1Cg℄;M) g�! L[C℄H�((�2 \ g�1g�1)[C℄;M)- " "H�(�1;M) g�! H�((�2 \ g�1g�1);M)and L[C℄H�((�2 \ g�1g�1)[C℄;M) or! L[C℄H�(�2 [C℄;M) ! L[B℄H�(� [B℄;M)" %H�((�2 \ g�1g�1);M) or! H�(�2;M) ommute. The �rst diagram is easy. That the pentagon ommutes follows from thefollowing fa t,whi h it suÆ es to he k in degree zero:Assume � � � with �nite index and � � C andM is a �-module. Let C1; :::; Csbe representatives of the �- onjuga y lasses of �- onjugates of C. ThenLH�(� \ Ci;M) or!LH�(Ci;M) onjugation! H�(C;M)" res " resH�(�;M) or�! H�(�;M) ommutes.

104

To onsider the a tion of the He ke operators analyti ally we must modify our analyti des ription slightly. Let U be an open ompa t subgroup of GL2(Af ) (here Af denotes the�nite adeles of K) su h that � = U \GL2(K) � SL2(O). Then we have:� Sn(�; C ) = ��Uf where the sum is taken over all uspidal automorphi representa-tions � = �f �1 of GL2(K) with �1 the prin ipal series representation of GL2(C ) orresponding to the hara ter 0� a �0 b1A 7! � ajaj�n+1 � jbjb �n+1 jabj�n.

� H1Eis(�nZ; ~Sn1;n2(C )) = Gn1;n2(�; C ) =L�IndGL2(Af )B0(Af ) f�U where the sum is takenover all He ke hara ters: = f 1 : 0� A� �0 A�

1A! C �for whi h 1 : 0� a �0 b

1A 7! a�a�n2b�(n1+1). (The indu tion here is not the usualunitary indu tion but that in Harder [Ha2℄, whi h explains the slight dis repan yfrom the uspidal ase above.)Now assume U1; U2 are as above and that g 2 GL2(K), then we get a map:[U1g�1U2℄ : Sn(�1; C ) = ��U1f �! ��U2f = Sn(�2; C )by v 7! R(U1g�1U2) v �f (x) dx where the Haar measure dx is normalised so that RU1 dx = 1. If�2g�1 = q ug�1 then U1g�1U2 � qU1g�1 �1u . If this is in fa t an equality then [U1g�1U2℄and [�2g�1℄ : Sn(�1; C ) ! Sn(�2; C ) oin ide. This follows for example from results inHarder [Ha2℄. Similar results hold for the Eisenstein ohomology.Finally in this se tion we onsider some spe ial He ke operators. Let T be the abstra t ommutative ring over Z generated by the symbols Tn for n 2 O n f0g. For the rest of thisse tion we onsider only ongruent subgroups of the form:

f0� a b d1A 2 SL2(O) j � 0 modN ; a � d � 1 modM g

105

whereM j N . For su h a group �, Tn 7! [�0� n 00 11A�℄ gives an a tion of T on H�(�;M) .(To see this is a good de�nition one must he k that these operators ommute. By abstra tnonsense we an he k this in degree zero where the problem redu es to showing that if�0� n 00 1

1A� = q�� and �0� m 00 11A� = q�� then the osets ��� and ��� oin ide.For this we refer the reader to Shimura [Sh2℄.) Note that we an apply the remarks of thelast paragraph to des ribe these He ke operators analyti ally, taking:

U = f0� a b d1A 2Yv GL2(Ov) j � 0 mod N ; d � 1 modMg

(Atleast if 1 is the only unit of O ongruent to 1 mod M .) Moreover we an fa torise[U10� n�1 00 11AU2℄ into a produ t of lo al operators, whi h are the identity for primes notdividing n.If �1 > �2 are both of the above form de�ned by N1;M1;N2;M2 and if n is not divisibleby primes dividing N1 but not N2, then we have a ommutative diagram:H�(�1;M) Tn�! H�(�1;M)res # # resH�(�2;M) Tn�! H�(�2;M)and so if N1 and N2 have the same prime fa tors we see that the restri tion map is T -equivariant (or as we shall write \He ke equivariant"). (This need only be he ked indegree zero where it follows be ause we an �nd i su h that �1 = q i(�1 \ g�1g�1) and�2 = q i(�2 \ g�2g�1) - see Shimura [Sh2℄.)We have an a tion of (O=NO)� �= �0(N)=�1(N) on H�(�1(N);M) by onjugation. Weshall �x �0(N)=�1(N) �! (O=NO)� by 0� a b d

1A 7! d.Then it is again easily he ked thatthis a tion ommutes with that of T , and that if N j N 0 then (O=NO)� !! (O=N 0O)� is ompatible with res : H�(�1(N);M)! H�(�1(N 0);M).We shall be parti ularly interested in the He ke operators T� for � dividing some power ofp. Thus we re all the de omposition for � now of the form106

f0� a b d1A 2 SL2(O) j � 0 modNpr ; a � d � 1 modNpsg with r � 1 and r � s and for� as above: �0� � 00 1

1A� = q0� � u0 11A�

as u runs over any set of representatives for ongruent lasses of O mod �. If � also dividessome power of p we have also:[�0� � 00 1

1A�℄[�0� � 00 11A�℄ = [�0� �� 00 1

1A�℄Most essentially we have Hida's idempotent. In general to an operator T on a �nite (or o�nite) Zp-module H we an de�ne an idempotent eT in EndZp(H) ommutes with T su hthat T is an automorphism of eTH and is topologi ally nilpotent on (1� eT )H (see Mazurand Wiles [MW2℄). In fa t eT = limr!1T r!. If S : H ! H 0 is a morphism between �niteZp-modules and T; T 0 are operators on H and H 0 respe tively su h that ST = T 0S then Srestri ts to a map S : eTH ! eT 0H 0. In parti ular for the He ke operator Tp a ting on a�nite Zp-module H we will denote the orresponding idempotent simply e, all it Hida'sidempotent and write HÆ for eH, whi h we will all the ordinary part of H. If p = ��� in Othen we an de�ne e� and e�� similarly orresponding to T� and T�� and we have e�e�� = e.In most of the following we will restri t to the ordinary parts of modules, and this will beessential for our arguments. Any v 2 H whi h is an eigenve tor of Tp with eigenvalue ap-adi unit will be preserved by e. We see that almost all of the dis ussion of this se tiongoes over to ordinary parts in the obvious fashion, but one should be aware that restri tiondoes not usually map (for example) H�(�1(N);M)Æ to H�(�1(Np);M)Æ when p6 jN . (Thisis \be ause we are going from no p in the level to p in the level".)Although Hida's idempotent is initially de�ned only in the p-adi setting we an some-times onsider it in more general situations. Fix embeddings Ka � C � and Ka � C . Inparti ular we have �xed an extension of the �-adi valuation to Ka . Now if H is a �nite

107

torsion free O-module with an a tion of Tp then we have:e 2 O�[Tp℄ � EndO�(H�)\ \K�[Tp℄ � EndK�(H K�)[ [K[Tp℄ � EndK(H K)and there is a �nite extension L=K ontained in K� su h that e 2 L[Tp℄ and hen e 2EndC (H C ).Then we have that:� eSn(�1(Npr); C ) =L(e�Upp ) (�pf )Up� eGn1;n2(�1(Npr); C ) =L�e IndGL2(Kp)B0(Kp) p�Up V pfwhere: Up = f0� a b d1A 2 GL2(Op) j � 0 mod Npr; d � 1 mod Nprg

Up = f0� a b d1A 2Yv 6 jpGL2(Ov) j � 0 mod N; d � 1 mod Ng

here � = �p �pf �1 runs over uspidal automorphi representations as before, where = p� pf � 1 is also as des ribed above, and where V pf does not matter very mu h. Ifp = ��� in O then we have moreover, e�Upp = e��U�� e���U���� . It is known that for v a primeabove p with �Uvv 6= 0 and for r > 0:� �v super uspidal implies that ev�Uvv = 0� �v = �( ; j : j) implies that ev is unde�ned if i is unrami�ed and (Nv) 12 (v) haspositive �-adi valuation. Otherwise the dimension of ev�Uvv is 1 or 0 a ording towhether is unrami�ed and (Nv) 12 (v) is a �-adi unit, or not.� �v = �( 1; 2) prin ipal series implies that ev is unde�ned if i is unrami�ed and(Nv) 12 i(v) has positive �-adi valuation for some i. Otherwise ev�Uvv has dimension108

0, 1 or 2 orresponding to the number of i (i = 1; 2) whi h are unrami�ed and have(Nv) 12 i(v) a �-adi unit.From this one an show that if p is inert in K then eGn1;n2(�1(Npr); C ) = 0 and thatif p = ��� splits in K then:eGn1;n2(�1(Npr); C ) =MV pfwhere the sum is taken over hara ters = p� pf� 1 as des ribed above with the added ondition that if p : 0� a �0 b1A 7! 1(a) 2(b) then 1 is unrami�ed at � and 2 at ��. Inparti ular for r � 1, dim eGn1;n2(�1(Npr); C ) is independent of n1 and n2 (as 1 is the onlyunit ofO ongruent to 1 mod (Npr)), and dim eGn1;n2(�1(Npr); C ) � dimGn1;n2(�1(N); C ).This is be ause the p part of the representation orresponding to is the prin ipalseries representation oming from �1 and �2 where in the inert ase, up to roots of unity,p�1(p) = pn2+1 and p�2(p) = pn1+1. While in the split ase:� p 12�1(�) = ��n2+1

� p 12�2(�) = �n1+1� p 12�1(��) = �n2+1� p 12�2(��) = ��n1+1

4.3 Cohomology of the BoundaryIn this se tion we want to des ribe the ordinary part of the ohomology of the boundary.We shall let M denote a �nite or o�nite Zp-module with a ontinuous a tion of the mul-tipli ative semi-group M2(O)\GL2(K) in the p-adi topology. � will denote a ongruen esubgroup of SL2(O) of the form:

f0� a b d1A 2 SL2(O) j � 0 mod Npr; d � 0 mod Np?g

109

For B a Borel we shall let ~�B denote the unipotent radi al of �B. Then ~�B � �B andWB = �B=~�B embeds into the roots of unity in K and so [�B : ~�B℄ is prime to p. Thus wehave: res : H�(�B ;M) ��! H�(~�B;M)WB(from the Hos hild-Serre spe tral sequen e).It is known that there is a bije tion:�nSL2(O)=B0 ! f�� uspsg 7�! B0 �1where B0 = 8<:0� � �0 �

1A9=;. Then �1~� B0 �1 = �1� \0� 1 O0 11A and we an think of~�B as an ideal in O. As B is its own normaliser we an easily dedu e that this ideal dependsonly on the usp [B℄ and not on its representation B0 �1.We shall write ~�[B℄ for this ideal.Similarly we an legitimately write W[B℄ and H�(~�[B℄;M). Then we have anoni ally:

H��(�;M) �=M[B℄ H�(~�[B℄;M)W[B℄ �=M[B℄ H� t(~�[B℄; p;M)W[B℄Here the \ t " indi ates that we are using ontinuous ohomology, ~�[B℄; p denotes the losureof ~�[B℄ in Op and the latter isomorphism follows be ause there is a bije tion between ~�[B℄ o y les (or oboundaries) and ontinuous ~�[B℄; p o y les ( oboundaries). We shall dropthe \ t " from our notation in future. In the ase p = ��� is split in O we see that:

H�(~�[B℄;M) = H�(~�[B℄; �;M)�H�(~�[B℄; ��;M)and that the W[B℄ a tion preserves this de omposition. Thus we an write:

H��(�;M) = H��(�;M)� �H��(�;M)��We shall introdu e the following de�nitions for a usp [B℄, and forv a prime above p:� [B℄ is v-unrami�ed if ~�[B℄; v = Ov.110

� [B℄ is v-�rst lass if [B℄ has a representative B0 �1 with 2 SL2(O) and �0� � �0 �1A mod vs for all s ( or equivalently for s = r)(1).

� [B℄ is v- third lass if [B℄ has a representative B0 �1 with = 0� a � �1A where( ; v) = 1. In this ase any representative has this form(2) and we an hose a repre-sentative with a � 0 mod vs for any s(1).� [B℄ is v-se ond lass if it is neither v-�rst nor v-se ond lass.Here the results marked (1) follow as � � �1(Npr) and those marked (2) as � � �0(p). Any�rst lass usp is unrami�ed, as follows easily from the formula:0� a b d

1A0� 1 �0 11A0� a b d

1A�1 = 0� 1� a � a2�� 2� 1 + a �1A

and the fa t (1).We shall prove:Proposition 4.1 If v is any prime above p and � and M are as above then we have aninje tion: evH��(�;M)v ,!M[B℄ H�(~�[B℄; v;M)where the sum is over v-�rst lass usps.Before proving this we shall draw the orollary that will be of use to us later:Corollary 4.1 If �1 � �2 are as � above and M is as above then the restri tion map:eH��(�1;M) res,! H��(�2;M)is inje tive (on the ordinary part).Proof: It will do to show that for ea h prime v above pevH��(�1;M)v res,! H��(�2;M)v111

Then onsidering the ommutative diagram:evH��(�1;M)v res�! H��(�2;M)res # # resL[B℄ v� lass 1H�(~�1[B℄; v;M) res! L[C℄H�(~�2[C℄;M)the proposition tells us that the left hand verti al arrow is inje tive and so we need onlyshow that the lower horizontal arrow is inje tive. But above ea h v-�rst lass usp [B℄ of�1 there lies a v-�rst lass usp [C℄ of �2, as we see at on e from the de�nition, andH�(~�1[B℄; v;M) ��! H�(~�2[C℄; v;M)

as ~�1[B℄; v = Ov = ~�2[B℄; v.We now turn to the proof of the proposition. We shall assume p = ��� (whi h is the aseof real interest and slightly harder than the inert ase), and that v = � (v = �� is exa tlythe same).Fix n su h that n > r and �n � 1 mod N ��r and set g = 0� �n 00 11A. We shall showthat:� If x 2 ker(H��(~�;M)� ! L[B℄ lass 1H�(~�[B℄; �;M)) then[�g�℄x 2L[B℄ lass 3H�(~�[B℄; �;M)W[B℄ and this latter spa e is preserved by [�g�℄.� [�g�℄ :L[B℄ lass 3H�(~�[B℄; �;M)W[B℄ !L[B℄ lass 3H�(~�[B℄; �;M)W[B℄ is topologi allynilpotent.from whi h the proposition follows easily. To prove the se ond assertion it will do to takeM of �nite ardinality. Pi torially this all amounts to [�g�℄ a ting as follows:

112

uspsthird lass uspsse ond lass usps�rst lass???6 �

topologi allynilpotentLet [B℄ be a �- usp with B = ÆB0Æ�1, Æ = 0� a � �

1A 2 SL2(O), we want to examine the[B℄- omponent of [�g�℄. Firstly the � \ g�g�1 usps above [B℄ are exa tly [ �1i B i℄ where� =`�B i(�\ g�g�1) or (as we see after a small al ulation) �g� =`�B ig�. Thus the usps [C℄ su h that [�g�℄ gives a non-zero map from H�(�[C℄;M) to H�(�[B℄;M) are repre-sented by [g�1 �1i B ig℄, or equivalently by [g�1u Bgu℄ where gu = 0� �n u0 11A as u varies over ongruen e lasses mod �n and without loss of generality u � 0 mod N ��r. Expli itly [C℄is represented by "B0"�1 with " = 0� a�u �m ��n�m �

1A 2 SL2(O) for somem. Thus if [B℄ is notthird lass( i.e. if �j ) then we see that � 6 j(a � u ) so that m = 0 and [C℄ must be lass one,whi h is our �rst laim.For the se ond assertion onsider [B℄ a third lass usp, whi h we an write [ÆB0Æ�1℄with Æ = 0� �2na b d1A where � 6 jd. Then if [C℄ is a third lass usp giving rise to [B℄ wemust have u � 0 mod �n ( u as above) and so we may take u = 0 and C = "B0"�1 with

113

" = 0� �na b �nd1A. Then � � 0 mod N ��r and � � d mod �r implies that:

"0� 1 �0 11A 2 �1(Npr)

Thus [B℄ = [C℄ and the map:[�g�℄ : M[B℄ lass 3H�(~�[B℄; �;M) �! M[B℄ lass 3H�(~�[B℄; �;M)

splits up as a dire t sum of maps:H�(�1�;M) g��! H�(�3�;M) or�! H�(�2�;M)

where:� �1 = ~�"B0"�1 = 8<:0� 1� �na � �2na2�� 2� 1 + �na �

1A 2 � j � 2 O9=;� �2 = ~�ÆB0Æ�1 = 8<:0� 1� �2na � �4na2�� 2� 1 + �2na �

1A 2 � j � 2 O9=;� �3 = �2 as �2 � g�g�1The map g� is indu ed by the ompatible maps M !M , m 7! gm and �3 ,! �1 by:0� 1� �2na � �4na2�� 2� 1 + �2na �1A 7�! 0� 1� �na (�n�) �2na2(�n�)� 2(�n�) 1 + �na (�n�)

1Aand or redu es to the identity.Thus we may des ribe [�g�℄ : H�(~�[B℄; �;M) ! H�(~�[B℄; �;M) as the map indu ed by ompatible maps M !M by m 7! m and ~�[B℄; � ! ~�[B℄; � by � 7! �n�. Then if M is �nite(as a set), for some a, res : H�(~�[B℄; �;M)! H�(�an~�[B℄; �;M) is zero, and so by our abovedes ription T a�n is also zero, whi h is what we wanted to show.

114

4.4 Change of LevelThe entral result of this se tion is:Proposition 4.2 LetN 2 O be prime to p, let r � s � 1 and let M be a Zp-module. SetGr;s = (1 +NpsZ)=(1 +NprZ). Theni) res : eH�(�1(Nps);M) ��! eH�(�1(Nps) \ �0(Npr);M)If further eM�1(Npr) = 0 thenii) res : eH1(�1(Nps) \ �0(Npr);M) ��! eH1(�1(Npr);M)Gr;sand hen eiii) eH1(�1(Nps);M) ��! eH1(�1(Npr);M)Gr;sThe modules M = Sn1;n2(A) satisfy eM�1(Npr) = 0 for A any O�-module.Proof: Let us establish the notation:

�s = �1(Nps)S� = �1(Nps) \ �0(Npr)5�r = �1(Npr)Then �=�r �= Gr;s.i) I laim the following diagram ommutes:H�(�s;M) res�! H�(�;M)T r�sp # . # T r�spH�(�s;M) res�! H�(�;M)where the diagonal arrow is given by the He ke operator [�s0� pr�s 00 1

1A�℄. From this theresult would follow at on e as Tp is invertible on the ordinary part of any module. Furtherit suÆ es to he k the ommutativity in degree 0 by abstra t nonsense, and here it followsfrom the existen e of u su h that:115

� �s0� pr�s 00 11A�s = q u�s

� �s0� pr�s 00 11A� = q u�

� �0� pr�s 00 11A� = q u�

(see Shimura [Sh2℄).ii) We look at the in ation-restri tion sequen e:0! H1(Gr;s;M�r) inf! H1(�;M) res! H1(�r;M)Gr;s t! H2(Gr;s;M�r)Let Tp and hen e e a t on these groups by giving them their normal a tion on the middleterms and letting them a t on the outer terms through their a tion on M�r = H0(�r;M).Assuming for a minute that these a tions are ompatible we see that there is an exa tsequen e: 0 �! eH1(�;M) res�! eH1(�r;M)Gr;s �! 0as desired.To he k the ompatibility assertion let�denote the map:�=�r ��! Gr;s0� a b d

1A 7�! dLet also gu be su h that:

�r0� p 00 11A�r = qgu�r and �0� p 00 1

1A� = qgu�For ; Æ 2 � let v = v(u) and w = w(u) be the unique indi es su h that g�1u gv andg�1v Ægw 2 �. It is easily he ked that g�1u gv = (and that g�1v Ægw = Æ). Then:a) Let � 2 Z1(Gr;s;M�r). Then for 2 �r:116

� inf(Tp�)( ) = (P gu�) � (Tp inf �)( ) =P gu�(g�1u gv)and so inf(Tp�) = Tp(inf �).b)Let x 2 H1(�r;M)Gr;s . Then there exists � 2 C1(�;M) su h that:� � j�r2 Z1(�r;M) and represents x� d� 2 Z2(Gr;s;M�r) � C2(�;M) and represents t(x)(see Hos hild-Serre [HS℄).Tpt(x) is represented by ( ; Æ) 7! P gu(�( Æ) � �(Æ) � �( )). Moreover onsider 2C1(�;M) de�ned by ( ) =P gu�(g�1u gv). Then j�r2 Z1(�r;M) and represents Tpx.Moreover(d )( ; Æ) = Pu gu�(g�1u Ægw)� Pu gv�(g�1v Ægw)�Pu gu�(g�1u gv)= Pu gu ��(g�1u gvg�1v Ægw)� (g�1u gv)�(g�1v Ægw)� �(g�1u gv)�= Pu gu(d�)(g�1u gv; g�1v Ægw)= Pu gu(d�)( ; Æ) as g�1u gv = and g�1v Ægw = Æ= Pu gu(�( Æ)� �(Æ)� �( ))Thus (d ) 2 Z2(Gr;s;M�r) and represents Tpt(x), i.e. t(Tpx) = Tpt(x).For the �nal assertion that eSn1;n2(A)�r = 0 we shall make an arbitrary extension of Tpto all of Sn1;n2(A) and show that:

TpSn1;n2(A) � pSn1;n2(A)from whi h it follows that eSn1;n2(A) and hen e eSn1;n2(A)�r vanish. Choose a set ofrepresentatives fug for ongruen e lasses of O mod p, and set:Tpm =Xu

0� p u0 11A :m

117

Then:

Tpm � Pu

0BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB�

0 : : : : : : : 0 un1 �un20 : : : : : : : 0 un1 �un2�1: : :: : :0 : : : : : : : 0 un10 : : : : : : : 0 un1�1�un2: : :: : :0 : : : : : : : 0 un1�1: : :: : :: : :0 : : : : : : : 0 1

1CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCA

m mod �

� 0m mod �as for any hara ter � : (O=�O)� ! (O=�O)� one has:Xu2(O=�O)� �(u) = 0. Q.E.D.Now de�ne HÆn1;n2 = lim! reH1(�1(Npr); Sn1;n2(K�=O�)). Then HÆn1;n2 is an O�[[G℄℄-module where G = lim Gr with Gr = (O=NprO)�. Let Hr = ker(G!! Gr) then what wehave just shown amounts to:(HÆn1;n2)Hr = eH1(�1(Npr); Sn1;n2(K�=O�)) (r � 1)If�denotes the Pontriagin dual this is the same as:( �HÆn1;n2)Hr = eH1(�1(Npr); Sn1;n2(O�))We have the following de ompositions for G:G = Gtor �H1 = (O=NO)� �H0 = (O=NO)� � (O=pO)� �H1Moreover omplex onjugation a ts on H1 and we have H1 = H+1 �H�1 where � refer to the orresponding eigenspa es for omplex onjugation. We an hoose isomorphisms Zp ! H�1118

by � 7! u��. Then O�[[G℄℄ �= O�[[H1℄℄[Gtor℄ �= O�[[T+; T�℄℄[Gtor℄the latter map being given by u� $ (T�+1). O�[[H1℄℄ is a omplete lo al noetherian ring,its maximal ideal orresponds to (�; T+; T�)�O�[[T+; T�℄℄.Corollary 4.2 �HÆn1;n2 is a �nitely generated O�[[H1℄℄-module.Proof: By Nakayama's lemma it will do to show that �HÆn1;n2 is ompa t and that ( �HÆn1;n2)H1 =( �HÆn1;n2=(T+; T�) �HÆn1;n2) is �nitely generated over O�. The proposition and the fa t that�HÆn1;n2 = lim eH1(�1(Npr); Sn1;n2(O�)) redu es this to the well known fa t thatH1(�1(Npr); Sn1;n2(O�)) is a �nitely generated O�-module.From the orollary of se tion 4.3 we have for r � s � 1 a ommutative diagram withexa t rows:0 ! eH1 usp(�1(Nps);M) ! eH1(�1(Nps);M) ! eH1Eis(�1(Nps);M) ! 0# #o #0 ! eH1 usp(�1(Npr);M)Gr;s ! eH1(�1(Npr);M)Gr;s ! eH1Eis(�1(Npr);M)Gr;swhere the right hand verti al arrow is an inje tion and where M = Sn1;n2(K�=O�). Thusthe left hand verti al arrow is an isomorphism, and if we set HÆn1;n2 usp to belim! eH1 usp(�1(Npr); Sn1;n2(K�=O�)) then we see that:� HÆn1;n2 usp ,! HÆn1;n2� HÆn1;n2 usp is a �nitely generated O�[[H1℄℄-module.� (HÆn1;n2 usp)Hr = eH1 usp(�1(Npr); Sn1;n2(K�=O�)) for r � 1.4.5 Change of WeightFor n1; n2 integers set:�n1;n2 : G = (O=NO)� �H0 ! H0 = O�p ! O�p ! O��� 7! �n1 ��n2

119

This extends to a homomorphism �n1;n2 : O�[[G℄℄ ! O� and to a homomorphism ~�n1;n2 :O�[[G℄℄! O�[[G℄℄ de�ned by sending h 2 G to �n1;n2(h)h. ThenO�[[G℄℄ ~�n1;n2�! O�[[G℄℄�n1;n2 & . �0;0O� ommutes. ForM an O�[[G℄℄-module we de�ne a twistM(�n1;n2) to be the same underlyingtopologi al abelian group but with a new O�[[H0℄℄ a tion de�ned by:h:m = ~�n1;n2(h)mOur aim is to prove:Proposition 4.3 There is a anoni al He ke equivariant isomorphism:

HÆn1;n2 ��! HÆ0;0(�n1;n2)It also restri ts to an isomorphism on the uspidal parts.Before proving this we note a ouple of orollaries. From now on we will use simply HÆto denote HÆ0;0.Corollary 4.3 Let I be the losed ideal of O�[[H1℄℄ generated byfh��n1;n2(h) j h 2 Hrg then �HÆ=I �HÆ �= eH1(�1(npr); Sn1;n2(O�)), and a similar statementholds for the uspidal part.This is lear.Corollary 4.4 �HÆ usp is a torsion O�[[H1℄℄-module.Proof: Let Ir denote the losed ideal generated by fh� �0;1(h) j h 2 Hrg and let R denoteO�[[H1℄℄ and M denote �HÆ. Then we have that M=IrM is p-torsion and that R=Ir is freeof p-torsion. IfM were not a torsion R-module we would have an inje tion R �,!M . Let Ndenote the submodule fn 2M j 9r 2 Rwith rn 2 Im�g. Then we an de�ne � : N ! FR

120

su h that � Æ � = IdR and there is then a non-zero onstant � 2 R su h that �Im � � R.Then we have for ea h r:M=IrM - N=IrN !! � Im�=Ir� Im�!! (� Im� + Ir)=Ir - ((�) + Ir)=Ir ,! R=IrNow M=IrM is p-torsion and thus ((�) + Ir)=Ir is p-torsion and so in fa t zero. Thus� 2 Ir 8r whi h implies � = 0 a ontradi tion.We now turn to the proof of the proposition. We have that:HÆn1;n2 = lim! eH1(�1(Npr); Sn1;n2(O=�rO))where the ith term on the right hand side is an O=�rO[(O=Npr)�℄-module. Call this ring�r, then O�[[G℄℄ = lim �r and this is ompatible with the above dire t limit. �n1;n2 redu esto a map �r ! O=�rO.Let j : Sn1;n2(O=�rO) �! O=�rO0BBBBBB�x0::xn1n2

1CCCCCCA 7�! xn1n2where we hoose a basis of Sn1;n2(O=�rO) as des ribed at the start of se tion 4.2. It is amap of modules over the semi-group:

� = f0� a b d1A 2M2(O) j ad� b 6= 0; � 0 mod pr; d � 1 mod prg

It thus indu es a He ke equivariant map:j� : H�(�1(Npr); Sn1;n2(O=�rO)) �! H�(�1(Npr);O=�rO)and for r � s � 1:H�(�1(Nps); Sn1;n2(O=�rO)) j��! H�(�1(Nps);O=�rO)res # # resH�(�1(Npr); Sn1;n2(O=�rO)) j��! H�(�1(Npr);O=�rO)121

ommutes. It is easily he ked that j� gives a map of �r-modules:H�(�1(Npr); Sn1;n2(O=�rO)) �! H�(�1(Npr);O=�rO)(�n1;n2)(In fa t it suÆ es to he k that for � 2 (O=prO)� we havej� Æ � = �n1��n2(� Æ j�) and su h an equality need only be he ked in degree zero where itreally is easy.)Finally we know from lemma 1.1 that j� is an isomorphism on ordinary parts as j :gSn1;n2(O=�rO) �! g(O=�rO) for g = 0� pr 00 11A.

4.6 Cy lotomi Hida FamiliesFirst let us introdu e a de�nition of an eigenform ( of the He ke operators having weightn, level Npr, and hara ter � : (O=NprO)� ! Ka �) suited to our purposes. First notethat it is easy to see that there are natural bije tions between the homomorphisms of ea hof the following forms:� � : T(H1(�1(Npr); Sn;n(K�=O�))�)! O �� � : T(H1(�1(Npr); Sn;n(O�))�)! O �� � : T(H1(�1(Npr); Sn;n(K�))�)! C �� � : T(H1(�1(Npr); Sn;n(K�))�)! C �� � : T(H1(�1(Npr); Sn;n(O�))�)! O �� � : T(H1(�1(Npr); Sn;n(K�=O�))�)! O �� � : T(H1(�1(Npr); Sn;n(C ))�)! C ( using our embedding of Ka into C � and C andthe fa t that the eigenvalues of the He ke operators are algebrai )We will all su h a homomorphism an eigenform of weight n, level Npr , and hara ter�. We will all it ordinary (respe tively uspidal) if it fa tors through the He ke algebra

122

a ting on the ordinary (respe tively uspidal) part of the ohomology. Thus the uspidaleigenforms orrespond to homomorphisms:� � : T(Sn(�1(Npr); C )�)! CLet G;H1; et . be as in the last se tion. Let � = O�[[H+1 ℄℄ and let R denote the integersin the algebrai losure of F�. If � : Gtor �H� ! O �� is a �nite hara ter we shall meanby an (ordinary) �-adi eigenform of level N and hara ter � a homomorphism:� : T(HÆ �) �! R

or equivalently: � : T( �HÆ �) �! RWe shall all it uspidal if it fa tors through HÆ� usp.For : H1 ! O �� a �nite hara ter and n a positive integer set: n : H1 �! O ��h 7�! h2n (h)and denote also its extension to a map �! O � by n. Let } ;n be the kernel of n and Qa prime of R above } ;n. Also let ! : (O=pO)� ! �(O�) denote the unique lifting of thenorm map (O=pO)� ! (Z=pZ)� ( i.e. the Tei hmuller hara ter). Then we have:T( �HÆ�) �! R# #T(eH1(�1(Npr); Sn;n(O�))� !�n) = T( �HÆ�=} ;n �HÆ�) � ! R=Q � O �where the verti al arrows are surje tive and the map a ross the bottom making the diagram ommute exists and is unique. (It exists be ause if I is the kernel of the verti al map onthe left then I � p} ;n (as �HÆ� is �nitely generated) and p} ;n � ker(T( �HÆ�) ! R=Q)be ause } ;n is ontained in this kernel.) This just says that for ea h ; n;Q as above �redu es modulo Q to a unique ordinary eigenform of weight n, level Npr for suÆ ientlylarge r, and hara ter � !�n. A similar statement is true for uspidal �-adi eigenforms.We will prove a partial onverse to this.123

Theorem 4.1 Fix N 2 O. Then :1. There are only �nitely many uspidal �-adi eigenforms of level N .2. If we also �x a �nite hara ter � : Gtor�H� ! O �� then for almost all ordinary eigen-forms � of level Npr, weight n and hara ter � for some r; n and a y lotomi hara ter( i.e. : (1 + pO=1 + prO) ! O �� and (�) = (��)) there is a �-adi eigenform �of level N , a hara ter �!n and a prime Q of R above } ;n su h that � redu es to �modulo Q.3. For all but �nitely many �nite hara ters � : Gtor �H� ! O �� there are only �nitelymany uspidal eigenforms of weight n, level Nps, and hara ter � as s; n and vary, with y lotomi .Remark It would be very ni e to be able to strengthen part 2 to assert that we ouldlift any ordinary uspidal eigenform to a unique �-adi one. This is false if one doesnot restri t to the uspidal part as the ordinary Eisenstein series ome in two variable p-adi families. However the examples of uspidal forms oming either by base hange fromGL2=Q or by theta series from grossen hara ters on a quadrati extension of K �t ni elyinto y lotomi families. One ould prove this strengthening if one knew that �HÆ usp wasa free �-module. This latter statement is equivalent to eH1 usp(�1(Np); Sn;n(O�)) beingtorsion free for in�nitely many n. This is a onsequen e of the following result, whi h inturn follows easily from Nakayama's lemma and unique fa torisation:If �i 2 � are in�nitely many distin t primes (all di�erent from p) and M is a ompa t �-module then the following are equivalent:� M is a free �-module� for all i M=�iM is p-torsion free.Proof: Note that part 3 follows from 1 and 2.1) First note that �HÆ usp � is a �nite torsion �-module for all but �nitely many � and that124

there are no �-adi eigenforms for su h �. (Infa t O�[[G℄℄ has only �nitely many heightone primes P su h that �HÆ uspP 6= 0 (su h a prime is a minimal element of the support of�HÆ usp) and so only �nitely many su h that O�[[G℄℄=P ,! �HÆ usp=P �HÆ usp.) For the �nite setof � for whi h �HÆ usp � is not a torsion �-module it is �nite and hen e T( �HÆ usp �) is a �nite�-algebra so there are only �nitely many �-adi eigenforms of hara ter �.2) Let T = T( �HÆ�!m ) for some m = 0; : : : p � 2. Then for n � m mod (p � 1)T !! T(H1(�1(Npr); Sn;n(O�))� ), and an ordinary eigenform � of weight n hara ter� an be thought of as a map T �! O � . If ker � is not a minimal prime ideal then thereis a prime P su h that ker � stri tly ontains P and su h that T =P ,! R and so � an belifted. But T has only �nitely many minimal prime ideals (as it is �nite over an integraldomain) and so there are only �nitely many maps T �! O � whose kernel is a minimal primeof depth 2. This implies that there are only �nitely many eigenforms � as in the theoremof weight n � m mod (p� 1) whi h an not be lifted to a �-adi eigenform, whi h at on eimplies the result.Finally in this se tion we would like to give some examples to show that our theory isnot va uous. Although one ould des ribe these examples very pre isely we shall ontentourselves with an existen e theorem.Proposition 4.4 Assume that p = ��� is split in K then:1. Let N 2 Z be prime to p and su h that there is an ordinary uspidal eigenform ofweight k � 2 and level Npr for some r for GL2=Q . Then there is a uspidal �-adi eigenform over K of level N .2. If � : (O=prO)� ! C �� is an anti y lotomi hara ter(i.e. �(��) = �(�)�1 for � 2 (1 + pO) ) then there is an integer N prime to pand a hara ter 0� : (O=NO)� ! C �� and a �-adi eigenform over K of level N and hara ter �0�.Proof: 1) This omes from base hange.More pre isely if the assumptions of the proposition hold then we know from the workof Hida (see for example [Hi1℄) that as r varies there are in�nitely many ordinary uspidal125

eigenforms of level Npr and weight k for GL2=Q . Hen e there are in�nitely many uspidalautomorphi representations � = �p �pf �1 of GL2=Q su h that:� �1 is the dis rete series representation whose in�nitessimal hara ter has Harish-Chandra parameter (12 ; 32 � k) 2 C 2 = X�(T )C where we identify X�(T ) with Z2 by(n;m) : 0� a 00 b1A 7! anbm.

� �p is either a prin ipal series or spe ial representation �( 1; 2) with (say) 1 unram-i�ed and 1(p) a p-adi unit.Let ~� = ~�p ~�pf ~�1 denote the base hange of � to GL2=K. Then ~�1 is the irredu ibleprin ipal series orresponding to the hara ter:0� a �0 b1A 7�! � ajaj�n+1 jaj�n� �bjbj�n+1 jbj�n

(To he k this one swit hes to the orresponding representations of the Weil groups WRand WC . �1 orresponds to:WR = C �of1; g �! GL2(C )by 7�! 0� 0 �11 01A

C � 3 z 7�! 0� z 12 �z 32�k 00 z 32�k�z 121A

and so ~�1 is the representation orresponding to :WC = C � �! GL2(C )z 7�! 0B� � zjzj�k�1 jzj2�k 00 � �zjzj�k�1 jzj2�k

1CAwhi h orresponds to the Langlands' quotient of the prin ipal series des ribed above, whi his in fa t irredu ible.)

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Thus (e~�Upp )� (~�pf )Up ,! eH1; usp(�1(Npr); Sk�2;k�2(C )) where:Up = f0� a b d

1A 2 GL2(Op) j � 0 mod Npr ; d � 1 mod NprgUp = f0� a b d

1A 2Yv 6 jpGL2(Ov) j � 0 mod N ; d � 1 mod NgMoreover �Upp 6= 0 (resp. (�pf )Up 6= 0) implies that ~�Upp 6= 0 (resp. (~�pf )Up 6= 0). As p is splitfor v j p ~�v = �p and ev ~�v = e�p 6= 0.In summary we have for ea h k � 2 a uspidal automorphi representation ~� = ~�p ~�pf ~�1 su h that:

0 6= (e~�Upp )� (~�pf )Up ,! eH1; usp(�1(Npr); Sk�2;k�2(C ))Thus by theorem 4.1 we are done.2) This omes from using the Weil lifting.More pre isely let L=K be a quadrati extension whi h is Galois over Q and in whi h� and �� split. Let �1; �2; ��1; ��2 be the embeddings L ,! Ka , with the �rst two extendingthe identity on K and with ��i(�) = �i(�). Let11; 12 be the in�nite pla es orrespondingto �1 and �2; and let v1; v2; v1; v2 be the primes above p orresponding to �1; �2; ��1; ��2.We an �nd an integerM prime to p, all whose prime divisors split from K to L, and whi his su h that, for ea h n � 0, there are grossen hara ters �1; �2 of l su h that:� �11 : C � � C � �! C �(a; b) 7�! � ajaj�n+1 jaj�n � �bjbj�n+1 jbj�n� ondu tor(�1) divides M� �2 is a �nite hara ter� ondu tor(�2) divides M(v2)1v11� �2 v2 � �2 v1 : (Z�p )2 ! C � equals �� � �� : (Z�p )2 ! C �

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(This is possible be ause for � in a subgroup of �nite index in the units of L, �i� 2 R>0 fori = 1; 2 and so �11(�) = 1.) In this ase p 12�1 v1(v1) and p 12�1 v2(v2) are v1-adi units.Then let # denote the Weil lifting of �1�2 to an automorphi representation of GL2=K.Then:� # is uspidal as �111 6= �112� #1 is the prin ipal series representation orresponding to:0� a �0 b1A 7�! � ajaj�n+1 jaj�n� �bjbj�n+1 jbj�n

� #� is the prin ipal series representation �(�1; �2) with �1 unrami�ed, p 12�1(�) a v1-adi unit and �2jO�� = ��jO��� #�� is the prin ipal series representation �(�01; �02) with �01 unrami�ed, p 12�01(��) a v1-adi unit and �02jO��� = ���jO���� (#pf )Up 6= 0 where Up is as above with N =M2Thus: 0 6= (e#Upp ) (#pf )Up ,! eH1(�1(Npr); Sn;n(C ))with Up also as above, for some r. As this is true for a single value of N and for all n � 0we are done by theorem 4.1.4.7 TorsionLastly we shall use our results to exhibit torsion in the �rst homology group of ertainsheaves. First we shall re all brie y what is obvious. There are four things we might lookat whi h fall by Pontriagin duality into two pairs:1. torsion in H1 with oeÆ ients in Sn1;n2(O) ! la k of p-divisibility in H1 with oeÆ ients in Sn1;n2(K=O).

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2. torsion in H1 with oeÆ ients in Sn1;n2(O) ! la k of p-divisibility in H1 with oeÆ ients in Sn1;n2(K=O)The obvious way to look for torsion is to onsider the long exa t sequen e orrespondingto: 0 �! Sn1;n2(O) ��! Sn1;n2(O) �! Sn1;n2(O=�O) �! 0In ohomology this gives for � a ongruen e subgroup:0 �! Sn1;n2(O=�O)� �! H1(�; Sn1;n2(O))� �! 0

whi h in some sense des ribes expli itly the torsion in this ase. However in homology weget: H2(�; Sn1;n2(O)) �! H2(�; Sn1;n2(O=�O)) �! H1(�; Sn1;n2(O))� �! 0whi h seems to be very little help as H2 is as mysterious as H1. (Dually if we look at0! O=�O ! K=O �! K=O ! 0 we again see that we get an answer for ii) but not for i).)Our methods allow us to say something about ase i) in the ase n1 6= n2.Theorem 4.2 Fix n1; n2 with n1 6= n2, and supposedim eH1; usp(�1(Npr);O�) > 0, whi h is ertainly the ase if there is a �-adi eigenformof level N . Then for t � s � r with n1 � n2 � 0 mod pt�s, H1(�1(Nps); Sn1;n2(O�)) andH1; usp(�1(Nps); Sn1;n2(O�)) have torsion of exponent at least pt.Notes: 1) Examples of values of N and p for whi h we an apply the theorem areprovided easily by proposition 4.4 of the last se tion.2) If H1; usp(�1(Nps); Sn1;n2(O�)) has torsion of exponent of order pt then the same istrue for the relative homology H1(�1(Nps)nZ; �(�1(Nps)nZ); ~Sn1;n2(O�)) as we see fromthe exa t sequen e:: : : �! H1(�(�nZ); ~M) �! H1(�nZ; ~M) �! H1(�nZ; �(�nZ); ~M) �! : : :

3) These are not the most pre ise results that an be proved by these methods but givea good indi ation of the type of question that an be treated.129

4) We ould dedu e a less pre ise theorem from our general theory using ommutativealgebra, but we shall go ba k to the method of proof as it yields more pre ise informationmore easily.Proof: Without loss of generality r = s � 1.We have from the proof of theorem 4.1 an isomorphism:j� : eH1(�1(Npt); Sn1;n2(O=�tO)) ��! eH1(�1(Npt);O=�tO)As n1 � n2 � 0 mod pt�r we have that xn1�xn2 � 1 mod �t8x 2 Gt;r = (1 + �rO)=(1 + �tO) then this map is Gt;requivariant. In this ase propo-sition 4.2 tells us that:eH1(�1(Npr); Sn1;n2(O=�tO)) ��! eH1(�1(Npr);O=�tO)The long exa t sequen e orresponding to:

0! Sn1;n2(O=�tO)! Sn1;n2(K�=O�) �t! Sn1;n2(K�=O�)! 0and the fa t that eSn1;n2(K�=O�)�1(Npr) = 0 (see the last part of proposition 4.2) implythat: eH1(�1(Npr); Sn1;n2(O=�tO)) �= eH1(�1(Npr); Sn1;n2(K�=O�))�tand hen e that:

eH1(�1(Npr); Sn1;n2(K�=O�))�t �= eH1(�1(Npr);K�=O�)�tor dualising:eH1(�1(Npr); Sn1;n2(O�))O=�tO �= eH1(�1(Npr);O�)O=�tOThe theorem now follows at on e, re alling thatdim eH1Eis(�1(Npr); Sn1;n2(K�)) = dim eH1Eis(�1(Npr);K�). The same arguments ap-ply to the uspidal parts.We an somewhat extend these results on torsion to the ase in whi h p does not dividethe level. The ru ial result will be: 130

Proposition 4.5 Let N be prime to p; n1, n2, n positive rational integers with n1; n2 > n;n1 + n2 � 2n+ r; n1 � n2 � n mod pr. Then we have a map of �1(Npr)-modules:j : Sn1;n2(O=�rO) �!�! Sn;n(O=�rO)su h that if I = [�1(N)0� pr 00 1

1A�1(Npr)℄j�1� then j� Æ I = Tpr and so:I : eH�(�1(Npr); Sn;n(O=�rO)) ,! H�(�1(N); Sn1;n2(O=�rO))This map preserves the uspidal parts.

Proof:This follows from lemma 1.1 of se tion 1.1 be ause if g = 0� pr 00 11A then�1(N)g�1(Npr) = �1(Npr)g�1(Npr) q`�1(Npr)gi�1(Npr) where ea h gi is of the form0� � �0 p�

1A (see Shimura [Sh2℄). The fa t that this preserves the uspidal part an easillybe he ked as in se tion 4.2. From this we dedu e:Corollary 4.5 Under the same assumptions as the proposition we have:eH�(�1(Npr); Sn;n(O=�rO)) ,! H�(�1(N); Sn1;n2(K�=O�))This map preserves uspidal parts.Proof:This follows from the long exa t sequen e orresponding to:0! Sn1;n2(O=�rO)! Sn1;n2(K�=O�) �r! Sn1;n2(K�=O�)! 0and the fa t that: Sn1;n2(K�=O�)�1(N) = Sn1;n2(K�=O�)SL2(O) = 0Corollary 4.6 Let N be prime to p and n1 6= n2; n positive rational integers with n1; n2 >n; n1 + n2 � 2n+ r; n1 � n2 � n mod pr. If dim eH1 usp(�1(Npr); Sn;n(C )) > 0 thenH1; usp(�1(N); Sn1;n2(O�)) and H1(�1(N); Sn1;n2(O�))have torsion of exponent divisible by �r. 131

Proof:This follows as in the proof of theorem 4.2 using the fa t thatdim eGn;n(�1(Npr); C ) � dimGn1;n2(�1(N); C ) whi h was noted at the end of se tion 4.2.For example taking n = 10 we �nd:Example 4.1 Assume that p splits in K and p6 j�(p) where � is Ramanujam's fun tion ( i.e.p is ordinary for �(z) = P �(n)e2n�iz, the uspidal ellipti modular fun tion of weight 12for SL2(Z)), and if n1 6= n2; n1; n2 > 10; n1+n2 > 20+r; and n1 � n2 � 10 mod pr(p�1)then H1(SL2(O); Sn1;n2) and H1 usp(SL2(O); Sn1;n2)have torsion of exponent divisible by pr.

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