Lectures on Some Aspects of P-Adic Analysis

Home , Algebraic number field, Complex number

Lectures on Some Aspects of p-Adic Analysis

By F. Bruhat

Notes by Sunder Lal

No part of this book may be reproduced in any form by print, microﬁlm or any other means without written permission from the Tata Institute of Fundamental Research, Colaba, Bombay-5

Tata Institute of Fundamental Research Bombay 1963

Introduction

These lectures are divided in three parts, almost independent part I is devoted to the more less classical theory of valuated fields (Hensel’s lemma, extension of a valuation, locally compact fields, etc.,) In the second part, we give some recent results about representations of classical groups over a locally compact valuated field. We first recall some facts about induced representations of locally compact groups and representations of semi-simple real Lie groups (in connexion with the theory of “spherical functions”). Afterwards, we construct a class of maximal compact subgroups K for any type of classical group G over a p-acid field and the study of the left coset and double coset modulo K decomposition of G allows us to prove the first results about spherical functions on G. Some open problems are indicated. Part III is devoted to Dwork’s proof of the rationality of the zeta function of an algebraic variety over a finite field. We first need some results (well known, but nowhere published) about analytic and meromorphic functions on an algebraically closed complete valuated field. Then we settle the elementary facts about the zeta function of a scheme (in the sense of Grothendieck) of finite type over Z and we give, following Dwork, the proof of the rationality of these zeta functions

iii

Contents

Introduction iii

I Classical Theory of Valuated Fields 1

1 Theory of Valuations-I 3 1 ...... 3 2 Valuation Rings and Places ...... 7 3 Topology Associated with a Valuation ...... 9 4 Approximation Theorem ...... 10 5 Completion of a ﬁeld with a valuation ...... 12 6 Inﬁnite Series in a Complete Field ...... 14 7 Locally Compact Fields ...... 15 8 Convergent Power Series ...... 16

2 Theory of Valuations -II 19 1 Hensel’sLemma ...... 19 2 Extension of Valuations - Transcendental case ...... 27 3 Residual Degree and Ramiﬁcation Index ...... 29 4 Locally compact Fields ...... 32 5 Extension of a Valuation to an Algebraic Extension.. . . 34 6 GeneralCase ...... 36 7 Complete Algebraic Closure of a p-adic Field ...... 38 8 Valuations of Non-Commutative Rings ...... 39

v vi Contents

II Representations of classical groups over p-adic Fields 43

3 Representations of Locally Compact and Semi-Simple... 45 1 Representations of Locally Compact Groups ...... 45 2 Irreducible Representations ...... 52 3 Measures on Homogeneous spaces ...... 56 4 Induced Representations ...... 57 5 SemiSimpleLieGroups ...... 59

4 Classical Linear Groups over p-adic Fields 63 1 General Deﬁnitions ...... 63 2 Study of GLn(P)...... 65 3 Study of O (ϕ, P)...... 74 n e 4 Representations of p-adic Groups ...... 83 5 SomeProblems ...... 88

III Zeta-Functions 93

5 Analytic Functions over p-adic Fields 95 1 Newton Polygon of a Power-Series ...... 95 2 Zeroes of a power series ...... 96 3 Criterion for the Rationality of power-series ...... 100 4 Elementary Functions ...... 103 5 An Auxiliary Function ...... 106 6 Factorisation of additive characters of a Finite Fields . . 108

6 Zeta-functions 111 1 ...... 111 2 Fields of ﬁnite type over Z ...... 112 3 Convergence of the product ...... 114 4 Zeta Function of a Prescheme ...... 118 5 Zeta Function of a Prescheme over FP ...... 120 6 Zeta Function of a Prescheme over Fq ...... 121 7 Reduction to a Hyper-Surface ...... 125 8 Computation of Nr ...... 126 Contents vii

9 Trace and Determinant of certain Inﬁnite Matrices . . . . 128 10 Meromorphic character of ξV (t) in Ω ...... 131

Part I

Classical Theory of Valuated Fields

Chapter 1 Theory of Valuations-I

In this and the next chapter we give a short account of the classical 1 theory of valuated ﬁelds. Unless otherwise stated by a ring we mean a commutative ring with the unit element 1 and without zero divisors.

Deﬁnition. Let A be a ring and Γ a totally ordered comutative group [1]. A valuation v of the ring A is a mapping from A∗ (the set of non-zero elements of A) into Γ such that

(I) v(xy) = v(x) + v(y) for every x, y in A∗.

(II) v(x + y) ≥ inf(v(x), v(y)) for every x, y in A∗.

We extend v to A by setting v(0) = ∞; where ∞ is an abstract element added to the group Γ satisfying the equation

∞ + ∞ = α + ∞ = ∞ + α = ∞ for α in Γ.

We assume that α < ∞ for every α in Γ. The valuation v is said to be improper if v(x) = 0 for all x in A∗, otherwise v is said to be proper.

The following are immediate consequences of our deﬁnition.

(a) v(1) = 0. For, v(x.1) = v(x) = v(x) + v(1), therefore v(1) = 0

3 4 1. Theory of Valuations-I

(b) If for x in A, x−1 is also in A, we have v(x−1) = −v(x), because 2 v(1) = v(xx−1) = v(x) + v(x−1) = 0 (c) If x is a root of unity, then v(x) = 0. In particular v(−1) = 0, which implies that v(−x) = v(x) (d) Por n in Z (the ring of integers)

v(n) = v(1 + −−− + 1) ≥ inf(v(1)) = 0.

(e) If for x, y in A v(x) , v(y), then v(x + y) = inf(v(x), v(y)). Let us assume that v(x) > v(y) and v(x + y) > v(y). Then v(y) = v(x + y − x) ≥≥ inf(v(x + y), v(−x)) > v(y), which is impossible.

If xi belongs to A for i = n, 1, 2,..., n, then one can prove by in- n duction on n that v( xi) ≥ inf (v(xi)) and that the equality holds if i=1 1≤i≤n there exists only onePj such that v(x j) = inf (v(xi)). In particular if 1≤i≤n n xi = 0 (n ≥ 2) then v(xi) = v(x j) = inf (v(xk)) for at least one pair i=1 1≤k≤n ofP unequal indices i and j. For, let xi be such that v(xi) ≤ v(xl) for i , l. Then v(xi) ≥ inf (v(xk)) = v(x j), which proves that v(xi) = v(x j). 1≤k≤n k,i Obviously we have Proposition 1. Let A be a ring with a valuation v. Then there exists one and only one valuation w of the quotient field K of A which extends v. x It is seen immediately that w = v(x) − v(y) for x, yin A. y ! 3 So without loss of generality we can confine ourselves to a field. The image of K∗ (the set of non-zero elements of field K) by v is a subgroup of Γ which we shall denote by Γv Proposition 2. Let K be a field with a valuation v. Then (a) The set O = {x|x ∈ K, v(x) ≥ 0} is a subring of K, which we shall call the ring of integers of K with respect to the valuation v. (b) The set Y = {x|x ∈ K, v(x) > 0} is an ideal in O called the ideal of valuation v. 1. 5

(c) O∗ = O − Y = {x|x ∈ K, v(x) = 0} is the set of inversible elements of O (d) O is a local ring (not necessarily Noetherian) and Y is the unique maximal ideal of O. We omit the proof of this simple proposition. The ﬁeld k = O/Y is called the residual ﬁeld of the valuation v. It is obvious form the proposition 2 that the valuation v of K which is a homomorphism form K∗ to Γ can be split up as follows

∗ v1 ∗ ∗ v2 v3 K −→ K /O −→ Γv −→ Γ. where v1 is the canonical homomorphism, v2 the map carrying an ele- ∗ ment xO to v(x) and (v3) the inclusion map of Γv into Γ. Deﬁnition. Two valuations v and v′ of a ﬁeld K are said to be equivalent ′ if there exists an order preserving isomorphism σ of Γv onto ΓV such that v′ = σ ◦ v.

From the splitting of the homomorphism v it is obvious that a valua- 4 tion of a field K is completely characterised upto an equivalence by any one of O, or Y . A valuation of a field K is said to be real if Γv is contained in R (the field fo real numbers). Since any subgroup of R is either discrete i.e., isomorphic to a subgroup of integers or dense in R, either Γv is contained in Z or Γv is dense in R. In the former case we say that v is a discrete valuation and in the latter non-discrete. Moreover v is completely determined upto a real constant factor, because if v and v′ are two ′ non-discrete equivalent valuations of K, the isomorphism of Γv onto Γv can be extended to R by continuity, which is nothing but multiplication by a element of R. If v and v′ are discrete and equivalent, the assertion is trivial. If Γv = z we call v a normed discrete valuation. Definition. Let K be a field with a normed discrete valuation v. In K we can find an element π with v(π) = 1. The element π is called a uniformising parameter for the valuation v. 6 1. Theory of Valuations-I

Let K be a ﬁeld with a normed discrete valuation v and O , (0) an ideal in O. Let α = inf (v(x)). Such an α exists because v(x) > 0 x∈O for every x in O. Moreover there exists an element x0 in O such that α 5 v(x◦) = α, because the valuation is discrete. Then O = O x = Oπ For, x 0 x belongs to O ⇐⇒ v(x) ≥ v(x ) ⇐⇒ v( ) ≥ 0 ⇐⇒ x/x belongs 0 x 0 x 0 to O ⇐⇒ x belongs to O x . Since v( o ) = v(x ) − αv(π) = 0, we get 0 πα 0 α α that x0 is in Oπ , conversely π belongs to O x0 is obvious. Therefore O = Oπα. In particular Y = Oπ. In general let v be any valuation of a ﬁeld K. Let O be any ideal of O and HO = {α|α ∈ Γϑ, such that there exists x in O with v(x) = α}. Then the map O → HO is a 1 − 1 correspondence between the set of ideals O in O and the subsets HO of Γv having the property that if α belongs to HO and β belonging to Γv is such that β ≥ α, then β belongs to HO . In particular if Γv is contained in R, then the ideals of O are of one of the two kinds ′ O (i) Iα = {x|x ∈ , v(x) ≥ α}

(ii) Iα = {x|x ∈ O, v(x) >α}

for any α> 0.

Examples. (1) Let Q be the field of rational numbers. For any m in α1 αr Q we have m = ±p1 ··· pr uniquely, where α1,...,αr are in Z and p1,..., pr are distinct primes. If v is any valuation of Q, we have v(m) = r α jv(p j). Therefore it is sufficient to define a valuation for primes in j=1 ZP. We note that for a valuation v there exists atmost one p for which v(p) > 0. If possible let us suppose that there exist two primes p1 and p2 such that v(pi) > 0 for i = 1, 2. 6 Since (p1, p2) = 1, there exist two integers a and b such that ap1 + bp2 = 1. This implies that 0 = v(1) ≥ inf(v(ap1)), v(bp2)) > 0, which is impossible. Thus our assertion is proved, If there does not exist any prime p for which v(p) > 0, then v is improper. For a prime p we define vp(p) = 1 and vp(m) = α, where α is the highest power of p dividing m. It is easy to verify that this is a valuation of Q and any valuation of Q for which v(p) > 0 is equivalent to this 2. Valuation Rings and Places 7 valuation. It is a discrete normed valuation of Q. One can take p as a uniformising parameter and prove that the residual field is isomorphic to Z/(p) (2) Let K be any field, K((x)) the field of formal power series over ∞ r K. For any element f (x) = ar x of K((x)) we define v( f (x)) = t, r=m if at is the first non-zero coeffiPcient in f (x). One an easily verify that v is a normed discrete valuation of K((x)). The ring of integers of the valuation is the ring of formal power series with non-negative exponents and the ideal is the set of those elements in the ring of integers for which the constant term is zero. One can take x as a uniformising parameter.

2 Valuation Rings and Places

This section is added for the sake of completeness. The results men- tioned here will not be used in the sequel.

Remark. Let K be a field with a valuation v and ring of integers O. 7 Then for any x in K, either x belongs to O or x−1 belongs to O. Motivated by this we define A subring A of a field K is called a valuation ring of K if for any x in K either x belongs to A or x−1 belongs to A. In general a ring A is said to be a valuation ring if it is a valuation ring for its quotient field. Proposition 3. A ring A is a valuation ring if and only if the set of principle of A is totally ordered by inclusion. Proof. Let A be a valuation ring. Let Ax and Ay be two proper principle x ideals of A. Consider z = belonging to K the quotient field of A. y Since A is a valuation ring, either z or z−1 belongs A. But this implies that either Ax ⊂ Ay or Ay ⊃ Ax. Therefore the set of principal ideals y is totally ordered conversely let x = , where y and z belong to A and z x , 0, be an element of K which is not in A. x < A implies that y does not belong to Az. But the set of principle ideals of A is totally ordered, therefore we get Az ⊂ Ay implying z = ay for some a in A. But a = x−1, therefore A is a valuation ring. 8 1. Theory of Valuations-I

Corollary. A valuation ring is a local ring. If possible let M1 , M2 be two maximal ideals in a valuation ring A. M1 , M2 implies that there exists x1 ∈ M1, x1 < M2 and x2 ∈ M2, x2 < M1. 8 x1 < M2 =⇒ Ax1 is not contained in M2 which implies that Ax1 is not contained in Ax1. Similarly x2 not belonging to M1 implies that Ax1. But this is impossible, therefore M1 = M2. Proposition 4. A ring A is a valuation ring if and only if A is the ring of a valuation of its quotient ﬁeld K determined upto an equivalence.

Proof. Let M be the unique maximal ideal of the valuation ring A and A∗ = A/M. For x, y in K∗ we deﬁne x ≥ y if and only if x belongs to Ay. It is easy to verify that this relation among the elements of K∗ induces a total order in the group K∗/A∗ and the canonical homomorphism K∗ onto K∗/A∗ is a valuation of K for which the ring of integers is A. The ring of integers of a valuation is a valuation ring has already been proved.

Let k be a ﬁelds. By k ∪∞ we mean the set of elements of k together with an element ∞. We extend the laws of k to (not everywhere deﬁned) laws in k ∪∞ in this way

(i) ∞ + a = a + ∞ = ∞ for a in k∗

(ii) ∞× a = a ×∞ = ∞×∞ = ∞, for a in k∗ 0 ×∞ and ∞ + ∞ are not deﬁned.

Let K be a field with a valuation v and let k = O/Y be the residual fields of v. Then the canonical homomorphism ρ of O onto k extended to K by setting ρ(x) = ∞ for x not in O gives rise to a map of K onto k ∪∞ called a place of K. 9 In general, we define A place of a field K is a mapping ρ form K to k ∪∞ such that

(i) ρ(a + b) = ρ(a) + ρ(b)

(ii) ρ(ab) = ρ(a)ρ(b) 3. Topology Associated with a Valuation 9 for a, b in K and whenever the right hand side is meaningful. It is easy to prove that O = ρ−1(K) is a valuation ring with the maximal ideal Y = ρ−1(0). Thus there exists a 1−1 correspondence between the set of valuation rings and the set of inequivalent places of a ﬁeld (Two places ρ1 and ρ2 of a ﬁeld K carrying K into k ∪∞ and k′ ∪∞) respectively are said to be equivalent if there exists an isomorphism σ of k onto k′ such that ρ2 = σ ◦ ρ1, with σ(∞) = ∞.

3 Topology Associated with a Valuation

Let K be a ﬁeld with a valuation v. For any α ≥ 0 in Γv consider the ideal Iα = {x|x ∈ K, v(x) >α} Then there exists one and only topology on K for which

(1) Iα for diﬀerent α in Γv form a fundamental system fo neighbourhoods of 0.

(2) K is a topological group for addition.

We see immediately that the operation of multiplication in K is continuous in topology. Iα for any α ≥ 0 in Γv is an open subgroup and hence a closed subgroup of K. Thus the residual field k is discrete for 10 the quotient topology. The topology of K is discrete if and only if the valuation v is improper ( if Γv = {o}). In particular K with a discrete and proper valuation is not discrete as a topological space. The topology of K is always Hausdorff, because if x , 0, then x does note belong to Iα with α = v(x), therefore Iαα > 0 = (0) which proves our assertion. α∈Γ Sv ′ Remark 1. If v is not improper, then the ideals Iα for α ≥ o in Γv also constitute a fundamental system of neighbourhoods of 0 for the topology ′ ′ of K. For, Iα and for α> o Iα contains I2α. Remark 2. Let A be a ring a with a decreasing filtration by ideals i.e. there exists a sequence (An)n≥0 of ideals such that An ⊃ An+1 and 10 1. Theory of Valuations-I

AnAm ⊂ Am+n. Then there exists one and only one topology for which A is an additive topological group and (An)n≥o constitute a fundamental system of neighbourhoods of 0. A is a topological ring this topology.

Let M be any ideals of a ring A. Then A can be made into a topo- n logical ring by taking An = M . We call the topology deﬁned by M on A the M− adic topology. In particular the ring of integers of a ﬁeld K which a real valuation v has the M− adic topology for every M = {x/v(x) ≥ α > 0} We shall speak of this topology of K as the M−adic topology. 11 If the valuation v is discrete and normed. We can take α = 1 and M = Y .

Remark 3. If K is a ﬁeld with a real valuation v, then the Y −adic topology completely characterises the valuation upto a constant factor, because x belongs to Y if and only if xn tends to zero as n tends to inﬁnity.

4 Approximation Theorem

For the sake of simplicity we confine ourselves in this section to real valuations though analogous results could be prove for any valuation. In this section we deal with the question whether there exists any connection between various inequivalent valuations of a field. We first prove:-

Lemma 1. Let K be a ﬁeld with two valuations v1 and v2. Then v1 and v2 are inequivalent if an only if O1, the ring of integers of v1, is not contained in O2, the ring of integers of v2.

Proof. If O1 ⊂ O2, then K − O1 contains K − O2 implying Y2 ⊂ Y1 ⊂ O1 ⊂ O2. Therefore Y2 is a prime ideal in O1. Assume Y2 , Y1, then there exists x in Y1 which does not belong to Y2. Since Y2 is an ideal in O Y 1, there exists α> 0 in Γv1 such that 2 contains Iα. Let v1(x) = β. Then for large enough q we have

q v1(x ) = qv1(x) = qβ>α, 4. Approximation Theorem 11

q which means that x belongs to Y2, but Y2is a prime ideal, therefore x belongs to Y2. Hence our assumption is wrong. Therefore Y2 = Y1 and v1 is equivalent to v2. The converse is obvi- 12 ous.

Lemma 2. Let K be a ﬁeld with v1,..., vn(n ≥ 2) proper valuations such that vi is inequivalent to v j for i , j. Then there exists an element z in K such that v1(z) > 0, v2(z) < 0 and vi(z) , 0 for i = 1, 2,..., n.

Proof. We shall prove the results by induction on n. When n = 2, v1 inequivalent to v2implies that O1 is not contained in O2 (lemma 1). There- fore there exists x in O1 and not in O2. Moreover O2 not contained in O1 implies that Y1 is not contained in Y2.

Therefore there exists y in Y1 and not in Y2. Then z = xy is the required element. When n > 2. By induction there exists an element x in K such that v1(x) > 0, v2(x) < 0 and vi(x) , 0 for i = 1, 2,..., n − 1. If vn(x) , 0, we have nothing to prove. If vn(x) = 0, we take an element y with vn(y) , 0. Let z = yxs, s a positive integer. Then for suﬃciently large s, z fulﬁlls the requirements of the lemma.

Theorem 1. Let K be a field with v1,..., vr proper valuations such that vi is inequivalent to v j for i , j. Let Ki be the field K with the topology r defined by vi and ρ the canonical map from K → Ki = P i.e. ρ(a) = i=1 (a, a,..., a). Then ρ(K) = D is dense in P. Q

Equivalently stated if a1,..., ar are any r elements in K, then for every α1,...,αr in R there exists an element x in K such that 13

v(x − ai) >αi for i = 1, 2,..., r.

Proof. The theorem is trivial for r = 1. Let us assume that it is true in case the number of valuations is less then r.

By lemma 2 there exists an elements x in K such that v1(x) > xn 0, v (x) < 0 and v (x) , 0 for 1 ≤ i ≤ r, then y = tends to r i n 1 + xn 12 1. Theory of Valuations-I

0 in K1, to 1 in Kr and to 0 or 1 in others as n tends to inﬁnity. Let the notation be so chosen that ρ(yn) → (0, 0,..., 0, 1,..., 1) as n tends to inﬁnity, 0 occurring in s places where 1 ≤ s ≤ r − 1. Now D is a subspace of P over K, therefore

lt xρ(yn) = lt ρ(xyn) = (0,... 0, x,..., x) n→∞ n→∞ r and (0, 0,..., 0, x, x,..., x) is in D¯ . Consider the product Ki, by i=s+1 r Q induction assumption the diagonal of Ki which is imbedded in D¯ is i=s+1 dense in the product which implies thatQ (0,..., 0, as+1,..., ar) belongs to D¯ for ai in K, s + 1 ≤ i ≤ r. Similarly (a1,a2,..., as, 0,..., 0) belongs to D¯ . But D¯ is a vector space over K, therefore (a1, a2,..., ar) is in D¯ . r Hence Ki = D¯ . i=1 Q Corollary. Under the assumptions of the theorem for α j ∈ Γv j ( j = 1, 2,..., r) there exists x in K such that v j(x) = α j.

14 For α j in Γv j , there exists a j ∈ K such that v(a j) = α j. By approximation theorem there exists an element x in K such that v(x − a j) >α j. By deﬁnition we have v(x) = v(x − a j + a j) = inf v((x − a j), v(a j)) = v(a j) = α j.

5 Completion of a ﬁeld with a valuation

Let K be a field with a valuation v. Since K is a commutative topological group for the topology defined by v, it is a uniform space. Let Kˆ denote the completion K. The composition laws of addition and multiplication can be extended by continuity to Kˆ , for which Kˆ is a topological ring. In fact Kˆ is a topological field, because if Φ is a Cauchy filter on K converging to a , 0, then Φ−1(the image of Φ by the map x → x−1 in K) is a Cauchy filter. For Φ not converging to 0 implies that there exists α ≥ 0 in Γv and a set A in Φ such that v(x) <α for every x in A. Since Φ is a Cauchy filter, for every β in Γv, there exists a set B in Φ contained in A such that v(x − y) > 2α + β for x, y in B. 5. Completion of a field with a valuation 13

Then v(x−1 −y−1) = v(x−1y−1(y− x)) = −v(x)−v(y)+v(y− x) > −α−α+2α+β which implies that Φ−1 is a Cauchy ﬁlter converging to a−1 in Kˆ . The valuation v can also be extended to be valuationv ˆ of Kˆ , in fact it is a ∗ continuous representation of K onto Γv considered as a discrete topological group, so v can be extended as a continuous representationv ˆ of Kˆ∗ in Γ and we getv ˆ(x + y) ≥ inf(ˆv(x), vˆ(y)) by continuity. Moreover 15 O ˆ Oˆ O¯ O ˆ Kˆ (the ring of integers of K) = K = K, since Kˆ is open in K and K ˆ O O O O O¯ is hence in K, Kˆ ∩ K = K is dense in Kˆ , this implies that Kˆ ⊃ K. O¯ O But K ⊃ Kˆ , therefore our result is proved. More generally

Iˆα = x|vˆ(x) > α, x ∈ Kˆ = I¯α = x|v(x) > α, x ∈ K Y Y¯ Y O Y In particular Kˆ = K. We have K = K ∩ Kˆ , so we may identify O /Y with a subset of O /Y , and O /Y is dense in O /Y . But K K Kˆ Kˆ K K Kˆ Kˆ O /Y is discrete, therefore O /Y = O /Y . Kˆ Kˆ Kˆ Kˆ K K Remark. Let K be a field with a real valuation v, with v we associate a map from K to R. We defined for any x in K the absolute value |x| = a−v(x), where a is a real number > 1. The map || satisfies the following properties

(1) |x| = 0 if and only if x = 0

(2) |xy| = |x| |y|

(3) |x + y|≤ sup (|x|, |y|) ≤ |x| + |y|.

The absolute value of elements of K, which deﬁnes the same topology on K as the valuation v.

By Qp we shall always denote the completions of the ﬁeld Q for p-adic valuation and by Zp the ring of integers in Qp. For the absolute 16 value associated to the p-adic valuation. We take a = p so that |x|p = p−v p(x) 14 1. Theory of Valuations-I

6 Inﬁnite Series in a Complete Field

Let K be a complete field for a real valuation v. Since every Cauchy sequence in K has a limit in K, the definition of convergence of infinite series and Cauchy criterium can be given in the same way as in the case of real numbers. However in this case we have the following.

Theorem 2. A family (ui)i∈I of an infinite number of elements of K is summable if and only if ui tends to 0 following the filter of the comple- ments of finite subsets of I.

Proof. The condition is clearly necessary. Conversely for any α in Γv we can ﬁnd a ﬁnite subset J of I such that for i not in J, v(ui) > α, r

then for i1,..., ir not in J we have v ∪i j > α which is nothing but j=1 ! Cauchy Criterium. Hence the family isP summable.

∞ Corollary. Let un be inﬁnite series of elements of K Then the follow- n=0 ing conditions areP equivalent.

∞ (a) un is convergent. n=0 P ∞ (b) un is commutatively convergent. n=0 P (c) un tends to 0 as n tends to inﬁnity.

17 Application. Let K be a complete field for a normed discrete real valuation v,π a uniformising parameter for K, R a fixed system of representatives in O for the elements of the residual field K. Then the series ∞ q rqπ , where rq belongs to R is convergent to an element x in K and q=m converselyP every x in K can be represented in this form in one and only q one way. The series is convergent because v(rqπ ) ≥ q for q , 0 and therefore tends to infinity as q tends to infinity. Conversely by multiplying with a suitable power of π we can take x in O, then there exists a unique r0 ∈ R such that x ≡ r0 (mod Y ). 7. Locally Compact Fields 15

−1 This implies that (x − r0)π is in O. Therefore there exists unique r1 in R such that

−1 (x − x0)π ≡ r1 (mod Y ). 2 or x ≡ r0 + r1π (mod Y ).

Proceeding in this way we prove by induction that

m m+1 x ≡ ro + r1π ··· + rmπ (mod Y ) ∞ m Now it is obvious that the series rmπ , is convergent and that x = r=0 ∞ P q rqπ . The uniqueness of the series is obvious from the construction. q=0 P In particular if, K = QP then any x in Qp can be represented in the ∞ q form rq p , where rq ∈{0, 1, 2 ..., p − 1}. q=m P 7 Locally Compact Fields

In this section we give certain equivalent conditions for valuated fields 18 to be locally compact. Later on we shall completely characterise the locally compact valuated fields. Theorem 3. Let K be a field with a proper valuation v. Then the following conditions are equivalent. (a) K is locally compact.

(b) O is compact.

′ Proof. (a) =⇒ (b). Since (Iα)α∈Γv form a fundamental system of closed ′ neighbourhoods for 0, there exists an α such that Iα is compact. But m ′ = O = O = −1 Iα xo , if v(x0) α, therefore x0 Iα is compact. (b) =⇒ (a) is trivial, as O is a compact neighbourhood of 0. (a) =⇒ (c)K is complete because it is a locally compact commutative group. For any α> 0 in Γv O/Iα is compact because O is compact. 16 1. Theory of Valuations-I

But O/Iα is a discrete space, therefore it contains only a finite number of elements. In particular k = O/Y is finite field. For any β in Γv, 0 <β<α, we have Iα ⊂ Iβ ⊂ O, therefore Iβ/Iα is a nontrivial ideal of O/Iα and distinct elements give rise to distinct ideals. But O/Iα is a 19 finite set, therefore there exist only a finite number of β with 0 <β<α, so we get that

(i) Γv has a smallest positive element

(ii) Γv is Archimedian.

Thus Γv is isomorphic to Z and the valuation v is discrete. (c) =⇒ (b). We shall prove that discreteness of the valuation v and finiteness of k implies that O is precompact, which together with the fact that K is complete implies that O is compact. Let V be any neighbourhood of 0. Since v is discrete, for some n > 0 V contains Y n. We shall show by induction on n that O/Y n is finite for n > 0. The result is true for n = 1; let us assume it to be true for all r < n. We have O/Y n−1 ≃ O/Y n/Y n−1/Y n But O/Y n−1 is finite by induction hypothesis and Y n−1/Y n is finite because it is isomorphic to O/Y , therefore O/Y n is finite. Hence there exist a finite number of elements x1 −−− xr in O r r n such that O ⊂ (xi + Y ) ⊂ (xi + V) and since this is true for every i=1 i=1 neighbourhoodS of 0, O is precompact.S

8 Convergent Power Series

Let K be complete ﬁeld with a real valuation v. Then the power series ∞ n f (x) = an x with coeﬃcients from K is said to be convergent at a n=0 P ∞ n point x of K if the series an x is convergent. It has already been n=0 ∞ P n 20 proved that the series an x converges if and only if n=0 P n v(an x ) = v(an) + nv(x) →∞ as n →∞ (1) 8. Convergent Power Series 17

1 From (1) it is obvious that if take t = lim inf (v(an)), then the series n n f converges for all x which v(x) > −t and does not converge for those x for which v(x) < −t and for those x for which v(x) = −t either the series converges for all x or does not converge at all. The number −t is called the order of convergence of the power series f and the set {x|v(x) > −t} or {x|v(x) ≥ −t, if the series converges at a point x with v(x) = −t} is called the disc of convergence, which we shall denote by D f . If we consider the absolute value associated to v then the radius of convergence is

−1 −t 1/n ρ = a = lim sup(|a|n) n→∞ and D f = {x| |x| < ρ} or {x| |x|≤ ρ}

The mapping x → f (x) from D f to K is continuous because it is a uni- ∞ n form limit of polynomials namely the partial sums of the series an x n=0 in the disc {x|v(x) ≥ −t1, for all t1 > t} or in the disc {x|v(x)P≥ −t} if the series converges on the disc. The classical results about addition and multiplication, ... of power series can be carried over to the power series with coeﬃcient in a complete valuated ﬁeld. For instance ∞ ∞ n n if f (x) = an x and g(x) = bn x are two power series with D f and n=0 n=0 Dg as theirP discs of convergenceP respectively; then if for one x in D f , 21 i ai x belongs to Dg for every i, f (x) also belongs to Dg and we have

∞ r g( f (x)) = cr x , where Xr=0 ∞

cr = bq ai1 ai2 ... aiq , Xq=0 i1+i2X+···+iq=r all the series being convergent.

Remark 1. If k = O/Y is an inﬁnite ﬁeld, then

i inf(v(ai x )) = inf (v( f (y))). i v(y)=v(x) 18 1. Theory of Valuations-I

i For, v( f (x)) ≥ infi(v(ai x )). We get equality, if there does not exist any two terms of the same valuation. In the exceptional case as the series ∞ n as the series any is convergent, we have n=0 j◦P r f (y) = ary + terms of higher valuation, where i◦ ≤ r ≤ j◦ < ∞. r=i◦ and withoutP loss of generality we can assume that v(x) = 0 and j◦ i r infi v(ai x ) = 0. Now v( f (y)) > 0 if and only if ary belongs to r=i◦ j◦ P r Y i.e., if and only if the polynomial a¯ry¯ (the image in k) = 0. But r=i◦ k has inﬁnite number of elements andP the above polynomial not being identically zero has only a ﬁnite number of zeros, therefore there exists 22 atleast one y for which v( f (y)) = 0 and v(x) = v(y). Thus in this case whenever x is in D f and f (y) belongs to Dg for all those y for which v(x) = v(y), we have

i inf v(ai x ) = inf v( f (y)). i v(y)=v(x)

∞ r Then f (g(x)) = cr x with r=0 P ∞

cr = bq av1 ··· avq . Xr=0 v1+−−X+vq=r

Remark 2. Let A be a ring with a topology defined by a decreasing filtration (An)n≥0 of ideals for which A is Hausdorff and complete space. ∞ n Then the formal power series an x converges at x in A if and only n=0 n if an x → 0 as n tends to infinityP and obviously the series converges everywhere in A if and only if an tends to 0 as n tends to infinity. Chapter 2 Theory of Valuations -II

1 Hensel’s Lemma

In this section we give a proof of Hensel’s lemma and deduce certain 23 corollaries which will be used quite often in the following. In this section by a ring we mean a commutative ring with unity (It may have zero divisors). Deﬁnition. Let A be a ring. Two elements x and y in A are said to be strongly relatively prime if and only if Ax + Ay = A i.e. if and only if there exist two elements u and v in A such that ux + vy = 1. In particular if k[x] is the ring of polynomials over a ﬁled k then any two elements in k[x] are strongly relatively prime if and only if they are coprime in the ordinary sense. It is obvious that if xand y are two strongly relatively prime elements in a ring A, then for any z in A x divides y z implies that x divides z.

Lemma 1. Let P and P′ be two polynomials with coeﬃcients in a ring A such that P is monic and P and P′ are strongly relatively prime. Let us assume that degree P = d(P) = s and d(P′) = s′. Then for every polynomial Q in A[x] there exists one and only one pair of polynomials U and V such that Q = UP + VP′ with d(V) < s and for every t > s′, d(Q) < t + s if and only if d(U) < t. 24

19 20 2. Theory of Valuations -II

Proof. The existence of one pair U and Vsuch that Q = UP + VP′ is trivial. If d(V) > s, we write V = AP + B where d(B) < s, which is possible because P is a monic polynomial, so we get

q = (U + A).P + BP′withd(B) < s. 2 Thus we can assume in the beginning itself that d(V) < s. If possible let there exists another pair U′ and V′ such that

Q = U′P + V′P′, d(V′) < s.

Then U′P + V′P′ = UP + VP′ implies that (U − U′)P = (V′ − V)P′. But P and P′ are strongly relatively prime, therefore P divides V′ − V. Since d(V′ − V) < s, V′ − V = 0. This implies that P(U − U′) = 0. As P is monic we must have U = U′. Let d(Q) < t + s. Then d(UP) = d(Q − VP′). But d(V) < s and d(P′) = s′ < t, therefore d(UP) < t + s, which implies that d(U) < t because P is a monic polynomial of degree s. It is obvious that d(V) < t (t > s′) implies that d(Q) < t + s.

Deﬁnition. Let A be a ring, the intersection of all the maximal ideals is called the radical of A and shall be denoted by r(A).

It can be easily proved that any element x of A belongs to r(A) if and only if 1-xy is invertible for all y ∈ A.

25 Lemma 2. Let A be a ring O an ideal in A contained in r(A). Then two polynomials P and P′ in A[x] ane of which (say P) is minic are strongly relatively prime if and only if P¯ and P¯ ′ (the images of P and P′ in A/O[x]) are strongly relatively prime.

Proof. P and P′ are strongly relatively prime implies P¯ and P¯ ′ are strongly relatively prime is obvious.

Suppose that d(P′) = s′ and d(P) = s. Then d(P¯) = d(P) = s, because P is monic. Let E = { f | f ∈ A[x], d( f ) < s + t, for some t > s′}. Then E is a module of ﬁnite type over A. Let E¯ = E/OE, since P¯ and P¯ ′ are strongly relatively prime in A/O[x], E¯′ is generated by the 1. Hensel’s Lemma 21 polynomials XuP¯ and XvP¯ ′ for 0 ≤ u ≤ s. For, by Lemma 1 for every polynomials Q¯ in E¯ there exists one only one pair of polynomials U¯ and V¯ in A/O[X] such that

Q¯ = U¯ P¯ + V¯ ′P¯ ′ d(V¯ ) < t + s.

But d(Q¯) < t + s, therefore d(U¯ ) < t. Thus

u λ U¯ = a¯λX , 0 ≤ u ≤ t Xλ=0 v µ V¯ = b¯µX , 0 ≤ v ≤ s Xµ=0

u v λ µ ′ and Q¯ = a¯λ(X P¯) + b¯µ(X P¯ ). λ=0 µ=0 By a simpleP corollaryP of Nakayama’s lemma (For proof see Algebre by N. Bourbaki chapter 8 section 6) which states that if E is a module 26 of ﬁnite type over a ring A and q and ideal in r(A) then if (a1,..., an) generate E module qE, they generate E also, we get XuP and XvP for 0 ≤ u ≤ t and 0 ≤ v ≤ s constitute a set of generators for E. Therefore

u v = r + k k ′ 1  arX  P  b X  P Xr=0  Xk=0          because 1 belongs to E. Hence P and P′ are strongly relatively prime in A[X]. Let A be a ring with a decreasing filtration of ideals (On)n>0, defin- ing a topology on A for which A is a complete Hausdorff space. If ∞ n f (X) = anX is a power series over A converging everywhere in A n=0 then λn( fP) = sup (i) ai

Hensel’s Lemma. Let A be a ring with a decreasing ﬁltration of ideals (On)n>0. Let A for this topology be a complete Hausdorﬀ space. If ∞ n f (X) = anX is an everywhere convergent power series over A and if n=0 there existP two polynomial ϕ and ψ an A/O1[X] such that

(1) ϕ is monic of degree s

(2) ϕ and ψ are strongly relatively prime

(3) f¯ = ϕψ

27 then there exists one and only pair (g, h) such that

(a) g is a monic polynomial of degree s in A[X] andg ¯ = ϕ.

(b) h is every where convergent power series over A and h¯ = ψ.

Moreover λn(h) = λn( f ) − s. If f is a polynomial then h is a polynomial and g and h are strongly relatively prime.

Proof. Existence We construct two sequences of polynomials (gn) and (hn) an A[X] by induction on n such that

(α) gn is monic of degree s, g¯n = ϕ and

gn + 1 ≡ gn (mod On+1) for n ≥ 0

(β) h¯n = ψ, hn+1 ≡ hn (mod On+1) and

d(hn) = λn+1( f ) − s

(γ) f ≡ gnhn (mod On+1), n ≥ 0

s−1 r s For n = 0, we take go = arX + X if r=0 P s−1 t r s u ϕ = a¯rX + X and ho = buX if Xr=0 Xu=0 1. Hensel’s Lemma 23

t u ψ = b¯uX , with t = d(ψ) = d( f¯) − s = λ1( f ) − s. Xu=0

Let us assume that we have constructed the polynomials g1, g2 ··· gn−1 and h1,..., hn−1 satisfying the conditions (α), (β), and (γ). By lemma (2) gn−1 and hn−1 are strongly relatively prime modulo Oq for every integer q ≥ 1, because gn−1 and hn−1 are strongly relatively prime 28 in A/O1[X] = A/O [X] and O1/Oq is contained in r(A/Oq), every q O1/Og element of O1/Og being nil potent. Therefore by lemma (1) there exist polynomials Xn and Yn in A(X) such that

f − gn−1hn−1 ≡ Yngn−1 + Xnhn−1 (mod On+1) and d(Xn) < s.

But by induction assumption f − gn−1hn−1 ≡ 0 (mod On) therefore 0 ≡ Yngn−1 + Xnhn−1 (mod On). Thus from the uniqueness part of lemma (1) we get Xn ≡ 0 (mod On) and Yn ≡ 0(On). We take gn = gn−1 +Xn and hn = hn−1 +Yn obviously the polynomials gn and hn satisfy the conditions (α), (β) and (γ). Hence we get two sequences of polynomials (gn) and (hn). The respective coefficients of (gn) and (hn) converge as n tends to infinity because of the condition gn+1 ≡ gn (mod On+1) and hn+1 ≡ hn (mod On+1). Therefore lim gn = g is a monic polynomial n→∞ of degree s and lim hn = b is power series over A which converges ev- n→∞ erywhere in A, because h ≡ hn (mod On+1). We see immediately that f = gh h¯ = ψ andg ¯ = ϕ. Moreover λn(h) = d(hn) = λn( f ) − s, because h ≡ hn (mod On+1) =⇒ λn+1(h) = λn+1(hn) = d(hn) ≤ λn+1( f ) − s but f = gh implies that λn+1( f ) ≤ s + λn+1(h), therefore we get our result. If 29 f is a polynomial then λn( f ) is constant for n sufficiently large implying λn(h) is constant for n large, therefore h is a polynomial. Since gn and hn are strongly relatively prime modulo On+1, there exist by lemma (1) polynomials an and bn such that

1 ≡ angn + bnhn (mod On+1), where d(bn) < s and d(an) < d(hn) = λn+1( f ) − s. 24 2. Theory of Valuations -II

Similarly we have polynomials an+1 and bn+1 such that

1 ≡ an+1gn+1 + bn+1hn+1 (mod On+2)

where d(bn+1) < s and d(an+1) < d(hn+1) = λn+2( f ) − s. Combining these two we get

(an+1 − an)gn + (bn+1 − bn)hn ≡ 0 (mod On+1)

Hence by uniqueness if lemma (1) we get an+1 ≡ an (mod On+1) and bn+1 ≡ bn (mod On+1). Since d(bn) < s for every n, we get that lim bn = b is a polynomial, moreover lim an = a is everywhere con- n→∞ n→∞ vergent power series in A; a is a polynomial if f is a polynomial. Hence we get 1 ≡ ag + bh (mod On+1) for every n ≥ 1, which implies that g and h are strongly relatively prime in A [[X]]. Uniqueness. If possible let us suppose that there exists another pair (g′, h′) satisfying the requirements of the lemma. Let V = h − h′ and U = g′ − g. Sinceg ¯ = g¯′ = ϕ and h¯ = h¯′ = ψ, U is in O[X] and V is in O1[[X]]. 30 Let us assume that U belongs to On[X] and V belongs to On[[X]] for all n < m, m > 1. We have

f = gh = g′h′ = (U + g)(V + h) = UV + Uh + gV + gh

which implies that −UV = Uh + gV. But UV is in O2n−2[[X]] (2n − 2 > n, as n > 1), therefore

Uh + gV ≡ 0 (mod On)

Let ρn be the canonical homomorphism from OnA[[X]] onto A/On [[X]]. Obviously we have

ρn(U)ρn(h) + ρn(g)ρn(V) = 0, d(U) < s (1)

But ρn(h) and ρn(g) are strongly relatively prime in AOn[[X]], because they are so in A/O1[[X]] and O1/On is contained in r(A/On), therefore by uniqueness part of lemma (1) we get from (1)

ρn(U) = 0 and ρn(V) = 0 1. Hensel’s Lemma 25

′ ′ This means that V = h − h ≡ 0 (mod On) and U = g − g ≡ 0 (mod On) for every n. But ∩On = 0, because A is a Hausdorff space, therefore U = 0 and V = 0. Hence the uniqueness of g and h is established. Corollary 1. Let K be a complete filed for a real valuation v. Let f (X) = ∞ n anX be an every-where convergent powerseries with coefficient from n=0 OP. Let ϕ and ψ be two polynomials in O/Y [X] = k[X] such that

(1) ϕ is monic of degree s. 31 (2) ϕ and ψ are strongly relatively prime in k[X] (3) f¯ (image of f in k[X])= ϕψ Then there exists one and only one pair g and h such that (1) g ∈ O[X], g is monic of degree s and g¯ = ϕ (2) h ∈ O[X], h converges everywhere in O and h¯ = ψ (3) f = gh. and the radius of convergence of h is the same as that of f . If f is a polynomial, then h is a polynomial. Moreover g and h are strongly relatively prime.

s−1 t r s u Proof. Suppose that ϕ = a¯rX + X and ψ = b¯uX . r=o u=0 P P s−1 t r s u Let ϕ◦ = arX + X , ψ◦u = a¯uX . r=0 u=0 P P Obviously ϕ◦ is monic of degree s and f¯ = ϕ◦ψ◦, which implies that n f − ϕ◦ψ◦ belongs to Y [[X]] i.e., if f − ϕ◦ψ◦ = bnX , then v(bn) > 0 for every n. Let α = inf v(bn),α is obviouslyP strictly positive. Let O n O O O On 1 = {x /x ∈ , v(x) ≥ α}, then ( n)n>0, n = 1 deﬁnes a decreasing ﬁltration on O. Obviouslyϕ ˜ ◦ and ψ˜ ◦ (images of ϕ and ψ in O/O[X]) are strongly relatively prime modulo O1 andϕ ˜ ◦ and ψ˜ ◦ satisfy all the requirements of Hensel’s lemma, therefore there exists one and only one pair (g, h) such that 26 2. Theory of Valuations -II

(i) g is monic polynomial of degree s andg ˜ = ϕ˜ ◦ 32

(ii) h is an every where convergent powerseries in O, h˜ = ψ◦ and λn( f ) − s. (iii) f = gh

Form the choice of ϕ◦ and ψ◦ it is obvious that this pair (g, h) satisfies the conditions (1), (2) and (3) of the corollary. If possible let there exist another pair (g′, h′) satisfying the conditions stated in the corollary. Let g′′ = g−g′, h′′ = h−h′ Sinceg ¯′′ and h¯′′ Y ′ ′′ ′′ O′ are in [x], there exists α > 0 such that g and h are in 1[[x]] where O′ O ′ O 1 = {x|x ∈ , v(x) ≥ α }. Let us take in Hensel’s lemma instead of O′ O O′ On the ideal 1 and the filtration defined by ( n) where n = 1 . But then have two pairs (g, h) and (g′, h′) satisfying the conditions (a), (b), (c) of the lemma, which is not possible, therefore g = g′ and h = h′. If f is a polynomial, the result about radius of convergence is obvious. Let us assume that f is not a polynomial, then λn( f ) tends to infinity v(a ) as n tends to infinity. It has already been proved that t = lim inf i . f i i Since v is a real valuation, for any i we can find an integer k such v(ai) (k − 1)α that(k − 1)α ≤ v(ai) ≤ kα, =⇒ λk( f ). Therefore ≥ i λk( f ) and v(ai) kα lim inf ≥ lim inf i→∞ i i→∞ λk( f )

33 moreover for i = λk( f ) we have kα v(a ) ≥ i . λk( f ) i

Let k →∞, which implies that i = λk( f ) →∞. Then we get kα v(a ) lim inf ≥ i i→∞ λk( f ) i v(ai) nα nα Thus lim inf = lim = lim i→∞ i n→∞ λn( f ) n→∞ λn(h) This proves that t f = tn. 2. Extension of Valuations - Transcendental case 27

Corollary 2. Let K be a complete valuated field with a real valuation v and f a polynomial in O[X]. Then if α in k is a simple root of f¯ , there exists one and only one element a in O such that a is a simple root of f and a¯ = α. Proof. Since α is a simple root of f¯, we have f¯(X) = (X−α) ψ(X) where ψ(α) = 0. Moreover (X − α) and ψ(X) are strongly relatively prime in k[X], (X − α) being a prime element. Therefore form corollary 1, we have in one and only one way f (X) = (X − a)h(X), where h¯ = ψ and a¯ = α. Moreover a is a simple root of f because h(¯a) = h¯(α) = ψ(α) = 0. In particular if K is a locally compact field such that characteristic k , 2, then we shall show that K∗/(K∗)2 is a group of order 4. K locally compact implies that v is discrete and k is a finite. Let π be 34 a uniformising parameter and let C ∈ O∗ = O − Y be an element such that C¯ is not a square in k, such an element exists because k∗/(k∗)2 is of order 2. Then it can be seen that 1, C,π, Cπ represent the distinct cosets in K∗/(K∗)2 and any element in K∗ is congruent to one of them modulo (K∗)2.

2 Extension of Valuations - Transcendental case

In order to prove that a valuation of a field can be extended to an extension field it is sufficient to consider the following two cases: (i) When the extension field is an algebraic extension. (ii) When the extension field is a purely transcendental extension. Proposition 1. Let L = K(X) be a purely transcendental extension of a field K with a valuation v, let Γ′ be any totally ordered group containing ′ Γv. Then for ξ in Γ there exists one and only one valuation ωξn of L extending v such that

n j wξ a j x = inf v(a j) + jξ   0≤ j≤n Xj=0      28 2. Theory of Valuations -II

Proof. It is sufficient to verify that wξ satisfies the axioms of a valuation for K[X]. The axioms aξ(P) = ∞ ⇐⇒ P = 0 and wξ(P + Q) ≥ inf(wξ(P), wξ(Q)) can be easily verified. To prove wξ(PQ) = wξ(P) + n j wξ(Q), where P = a jX , j=0 m P i Q = biX and PQ , 0, we write P = P1 + P2, Q = Q1 + Q2, P1 i=0 P j 35 being the sum of all terms a jX of P such that wξ(P) = v(a j)+ jξ and Q1 i being the sum of those terms biX of Q for which wξ(Q) = v(bi)+iξ. Let r jo and ko be the degree of P1 and Q1 respectively. If P1Q1 = CrX , then we have P

wξ(P1Q1) = v(C jo+ko ) + ξ( jo + ko)

= v(a jo ) + ξ jo + v(bko ) + ξko = wξ(P1) + wξ(Q1). Now

wξ(PQ) = wξ(P1Q1 + P1Q2 + Q1P2 + P2Q2) = wξ(P1Q1),

because the valuation of the other terms is greater than wξ(P1Q1). This implies that

wξ(PQ) = wξ(P1Q1) = wξ(P1) + wξ(Q1) = wξ(P) + wξ(Q).

Corollary. There exists one and only one valuation w of K(X) such that (i) w extends v.

(ii) w(X) = 0.

(iii) The class X¯ of X in kw is transcendental over kv.

The valuation w is the valuation wξ for ξ = 0 and kw is a purely transcendental extension of degree 1 over kv.

It is obvious that wo(i.e. wξ for ξ = 0) satisﬁes (1) and (2) and that ¯ ¯ kwo = kv(Xn).If X were algebraic over kv, then there exists a polynomial n P¯(Y) = a¯ jY j such that at least onea ¯ j , 0 and P¯(X¯) = 0, which means j=0 P 3. Residual Degree and Ramiﬁcation Index 29

n j 36 that P(X) = a jX is in Yw, where at least one a j is not in Yv and all j=0 P a j are in Ov. But this is impossible because w(P(X)) = inf v(a j) = 0. j Conversely let w be a valuation of K(X) satisfying 1), 2), 3). Let P = m i aiX be a polynomial over K. We have to prove that w(P) = inf v(ai). i=0 P m i Let P = aiX be a polynomial over K. We can assume without loss i=0 of generalityP that ai are in Yv and at least one of them is not in Yy, then inf v(ai) = 0. If w(P) > 0, then P¯ = 0 in kw, which implies, that X¯ is i algebraic over kv, which is a contradiction. But we know that

i w aiX ≥ inf{v(ai) + iw(X)} = 0   i Xi     therefore w(P) = inf v(ai). i

3 Residual Degree and Ramiﬁcation Index

Let L be a field and K a subfield of L. Let w be a valuation of L and v the restriction of w on K. Since Yw ∩ K = Yv, the filed kv can be imbedded in the field kw. We shall say the dimension of kw over kv the residual degree of w with respect to v or of L with respect to K. We shall denote it by f (w, v). The index of the group Γv in Γw is called the ramification index of w with respect to v or of L with respect to K and is denoted by e(w, v). If no confusion is possible, we shall denote f (w, y) by f (L, K) and 37 e(w, v) by e(L, K). If e(w, v) = 1, then L is said to be an unramified extension of K. If f (w, v) = 1, L is said to be totally ramified extension of K. Since the group of values and residual field of Kˆ are the same as that of K we have

e(Lˆ, Kˆ ) = e(L, K) and f (Lˆ, Kˆ ) = f (L, K) 30 2. Theory of Valuations -II

Proposition 2. Let L be a filed with a valuation w and let K be a field contained in L and v the restriction of w on K. Then e(L, K) f (L, K) ≤ (L : K) = n, where (L : K) is dimension of L over K. Proof. If n is infinite, the result is trivial. Let us assume that n is a finite number. Let r ≤ e and s ≤ f be two positive integers, then we can find ∗ r elements X1,..., Xr in L such that w(Xi) . w(X j) (mod Γv) for i , j and s elements Y¯1,... Y¯s in kw such that they are linearly independent over kv. Let Y1,... Ys be a system of representatives for Y1,... Ys in O∗ w. Then the elements (XiY j, i = 1,..., r; j = 1, 2,... s) are linearly independent over K. If they are not linearly independent, then there exists elements ai j in K not all 0 such that

ai jXiY j = 0 Xi, j

38 Let α = inf w(ai jXiY j), obviously α is ﬁnite and belongs to Γw. i, j Therefore w(aklXkYl) = α for some k and l. We have

w(ai jXiY j) = w(ai j) + w(Xi) + w(Y j)

= w(akl) + w(Xk) + w(Yl)

if w(ai jXiY j) = α for some i and j .

But w(Y j) = w(Yl) = 0, therefore we get w(Xi) ≡ w(Xk) (mod Γv), which is possible only if i = k. Thus we get

′ aklXkYl + ak jXkY j ≡ 0 (mod O ) (1) Xj,l where O′ = {X/X ∈ L, w(X) >α} −1 −1 Multiplying the congruence (1) with akl Xk we get −1 Y Yl + akl ak jY j ≡ 0 (mod w). Xj,l Therefore

¯ −1 ¯ −1 Yl + (akl ak j)Y j = 0, where akl ak j are in kv. Xj,l 3. Residual Degree and Ramiﬁcation Index 31

But this is impossible, because Y¯1 −−−Ys¯ are linearly independent over kv, therefore (XiY j) are linearly independent over K. Since (L : K) = n, the number of linearly independent vectors in L over K cannot be greater than n. Hence e f ≤ n.

Corollary 3. If L is algebraic over K, then kw is algebraic over kv and Γl/Γk is a torsion group of order ≤ (L : K). The assertion is trivial when (L : K) is finite. When (L : K) is infinite we can write L = Li and kL = ∪kLi, where Li is a finite i∈I algebraic extension of K. S

Then ΓL/ΓK is the union of the quotient groups ΓLi /ΓK for i in I and 39 therefore it is a torsion group. Corollary 4. Suppose that L is algebraic over K, then w is improper if and only if v is improper.

v improper implies that Γv = {0}. Therefore by corollary (3) Γw is a torsion group. But Γw is a totally ordered and abelian group, therefore it consists of identity only. Corollary 5. Let (L : K) be ﬁnite. Then w is discrete if and only v is discrete.

v discrete implies that Γv is isomorphic to Z and (L : k) finite implies Γw/Γv is of finite order. Moreover Γw is Archimedian, because if α and β are in Γw, then nα and nβ where n = order Γw/Γv, are in Γv; therefore there exists an integer q such that qnα > nβ, which shows that qα > β. There exists a smallest positive element in Γw. For, each coset of Γw/Γv has a smallest positive element, the smallest among them is the smallest positive element fo Γw. Hence Γw is isomorphic to z. Corollary 6. If the valuation v on K is discrete, K is complete and (L : K) is finite, then e f = (L : K).

Proof. Let π be a uniformising parameter in L. Let Y¯1,..., Y¯ f be a basis O∗ R of kw over kv and Y1,..., Y f their representatives in w. Let denote a O∗ system of representatives of kv in v . 32 2. Theory of Valuations -II

f 40 Then any element X in Ow can be written in the form αiYi modulo i=1 ′ Yw with αi ∈ R in one and only one way. Let L beP vector space j over K generated by (Yiπ ) for i = 1, 2,..., f and j = 0, 1, 2,..., e − 1. Since L′ is a finite dimensional vectorspace over a complete field K, L′ is complete (for proof see Espaces Vectoriels Topologiques by N. Bourbaki, chapter I section 2) and therefore closed in L. But L′ is dense in L, because for every element X in L and an integer n there exists an ′ element Xn is L such that v(X − Xn) ≥ n e. For sufficiently small n the result is obviously true. Let us assume that it is true for all integers r ≤ n. Since n e is in Γv, there exists an element U in K such that −1 w(U) = v(U) = n e. Therefore U (X − Xn) belongs to Ow and we have −1 U (X − Xn) ≡ αioYi (mod Yw = Owπ) Xi −1 −1 This means that π [(U (X − Xn) − i αioYi] belongs to Ow, therefore P π−1 (U−1(X − X ) − α Y ) ≡ α Y (mod Y )  n io i  i1 i w  Xi  Xi   Proceeding in this way we obtain  −1 e−1 Y e U (X − Xn) ≡ αioYi + ··· + αie−1Yiπ (mod w ) Xi Xi e−1 j Y (n+1)e or (X − Xn) ≡ U  αi jYiπ  (mod w ) Xj=0 Xi     e−1  j 41 Let us take Xn+1 = Xn + U αi jYiπ . " j=0 i # P P ′ Then w(X − Xn+1) ≥ (n + 1)e. Thus L is dense in L and therefore L′ = L. So n = (L : K) ≤ e f . But we know that e f ≤ n, therefore n = e f .

4 Locally compact Fields

Proposition 3. If K is a locally compact filed of characteristic o with a discrete valuation v, then K is a finite extension of Qp where p is 4. Locally compact Fields 33 characteristic of the residual field k.

Proof. Since characteristic K = 0, K contains Q the field of retinal numbers. We see immediately that v is proper, because if v is improper then Q is contained in k which is a finite field by theorem in §7.1 and this is impossible. The restriction of v to Q is vp for some p because p adic valuations are the only proper valuations on Q and this p is the characteristic of k. We have already proved in §7.1 that K is complete, therefore K contains QP. The valuation v on K is discrete, therefore Γv is isomorphic to Z, but Γvp is also isomorphic to Z and is contained in

Γv, therefore = (Γv : Γvp ) is finite. Moreover f = (kv : kvp ) is also finite, because kv is a finite filed. Hence (K : Qp) = e f (by corollary 4 of 42 §3.2) is finite.

Proposition 4. Let K be complete ﬁled for a real proper valuation v such that

(1) characteristic K= characteristic k.

(2) k and all its sub ﬁelds are perfect.

Then there exists a subﬁeld F ⊂ O which is a system of representatives of k in O. Moreover if v is discrete then K is isomorphic to k((x)).

Proof. Let Φ be the family of subfields S of O such that the restriction of ϕ, the canonical homomorphism from O onto k to S is an isomorphism onto a subfield of k. The family Φ , φ, because the prime fields contained in O and k are the same. Obviously Φ is inductively ordered by inclusion, therefore by Zorn’s lemma it has a maximal element F. We shall prove that k = ϕ(F). The field k is algebraic over ϕ(F). If possible let there exist an elementx ¯ in k transcendental over ϕ(F). Let ϕ(x) = x¯, where x is in O, then x is transcendental over F. It is obvious that F(x) is isomorphic to ϕ(F)(x ¯), which contradicts the maximality of F, therefore k is algebraic over ϕ(F). Suppose that ϕ(F), then there exists one elementx ¯ in k and not in ϕ(F). Since ϕ(F) is a perfect field,x ¯ is a simple root of an irreducible monic polynomial P¯ over ϕ(F). Let s s−1 P¯ = X +a¯ s−1X +···+a¯o = (X−x¯)Q¯, where Q¯ is some polynomial 43 over ϕ(F) and Q¯(x ¯) , 0. 34 2. Theory of Valuations -II

By Corollary (4) fo Hensel’s lemma we obtain that the polynomial S s−1 P = X + as−1X + ··· + a0 has a simple root x in O such that ϕ(x) = x¯ and Q is an irreducible polynomial. Therefore F(x) is isomorphic to F[X]/(P). But ϕ(F)(x ¯) is isomorphic to ϕ(F)[X]/(P¯) therefore we see that ϕ is still an isomorphism from F(x) onto ϕ(F)(x ¯). But this is impossible, because F is a maximal element of Φ. Thus our theorem is proved.

When v is discrete, we have seen that every element x in K is of ∞ i the form riπ with ri in F and conversely. Therefore the mapping i=m ∞ P∞ i i riπ → ϕ(ri)X is from K onto k((x)). It is trivial to see that it is i=m i=m anP isomorphism.P Corollary. A non-discrete locally compact valuated field of characteristic p > o is isomorphic to a field of formal power series over a finite field.

Since we have already proved in §7.1 that a locally compact valuated ﬁeld K is complete, its valuation is discrete and k is ﬁnite, our corollary follows from the theorem.

5 Extension of a Valuation to an Algebraic Exten- sion (Case of a Complete Field)

44 Theorem 1. If L is an algebraic extension of a complete ﬁeld K with a real valuation v, then there exists one and only valuation w on L extending v.

Proof. If v is improper w is necessarily improper. So we assume that v is a proper valuation. Suppose that L is a ﬁnite extension of K. If there exists a valuation w on L extending v, then w is unique, because on L any valuation deﬁnes the same topology as that of K(L:K) and the topology on L determines the valuation upto a constant factor and in this case the constant factor is also determined because the restriction of the valuation to K is v. 5. Extension of a Valuation to an Algebraic Extension.. 35

Let L be a Galois extension of K. Then if w is a valuation on L extending v, w o σ for any σ in G(L/K) (the Galois group of L over K) is also a valuation extending v. Therefore by uniqueness of the extension w(x) = Woσ(x) for every x in L. This shows that

v (N(x)) = w( (x)) = w o σ(x) = now(x) L/K Yσ Xσ where (L : K) = n. Thus 1 w(x) = v (N(x)). (1) n L/K

Now suppose that L is any finite extension of K of degree n. We 45 define a mapping w on L by (1) and prove that it satisfies the axioms for a valuation. It is well known that (Bourbaki, algebra chapter V, §8) that if L is the separable closure of K is L, and if p is the characteristic exponent of K (i.e., p = 1 if characteristic K = 0 and p= characteristic K , 0), then n = (L : K) = qpe with q = (L′ : K) and pe = (L : L′). Moreover L is a purely inseparable ′ ′ extension of L , and for each K-isomorphism σi (1 ≤ i ≤ q) of L , in an algebraic closure Ω of K there exists one and only one K-isomorphism of L which extends σi. This extended isomorphism will also be denoted by σi. Then q pe NL/K(x) = σi(x) Yi=1 It is easy to prove that w(x) = ∞ if and only if x = 0 and w(xy) = w(x) + w(y) for x, y in L. To prove that w(x + y) ≥ inf(w(x), w(y)), it is sufficient to prove that w(α) ≥ 0 implies that w(1 + α) ≥ 0 for any r r−1 α in L. We know that if P(X) = X + ar−1X + ··· + ao is the monic n irreducible polynomial of α over K, then N α = (ao) r and P(1 − X) is L/K the irreducible polynomial of 1 + α. Thus

r r n N (1 + α) = (−1) (1 + ar−1 + ··· + (−1) ao) r = bo L/K 36 2. Theory of Valuations -II

So to prove our result we have to show that bo is in O when ao is 46 v(a )n/r in O, because w(α) = o . This will follow from the following n theorem, which completely proves our theorem. Theorem 2. Let K be complete ﬁeld with a real valuation v and x any r r−1 element of an algebraic extension of K. If f (X) = X + ar−1X ··· + ao is the minimum polynomial of x over K, then ao belonging to O implies that all the coeﬃcients of f (X) are in O.

Proof. If possible suppose that all ai are not in O, then v(a j) < 0 for some j, 0 < j < r. Let −α = inf v(a j),α > 0 and j the smallest index such that v(a j) = −α. We have o < j < r. Since α belongs to Γv, there exists an element C in K such v(C) = α. Consider the polynomial k j g = C f (X) = CX + ··· + Ca j X + ··· + C◦a. Because of the choice of j j j, g¯ = ··· + r¯ jX , wherer ¯ j = ca¯ j , 0. Thereforeg ¯ has X as a factor which is a monic polynomial and ifg ¯ = X jψ, then X j and ψ satisfy the requirements of Corollary (4) of Hensel’s lemma, which gives that g is reducible, which is a contradiction. Hence all a j are in O. When L is infinite algebraic extension of K, we can express L = Li where each Li is a finite algebraic extension of K and the family i∈I 47 {SLi}i∈I is a directed set by the relation of inclusion. We define the valuation w for any x in L as w(x) = wi(x) if x is in Li and wi is the extension of v on Li. It is obvious that w is the unique valuation on L extending v.

6 General Case

We shall study now how a valuation of an incomplete field can be extended to its algebraic extension. Let K be field with a valuation v and L an algebraic extension of K. If w is a valuation of L extending v, we can look at the completion Lˆ of L. Lˆ contains L and Kˆ , so it contains a well defined composite extension Mw of L and Kˆ . Then there exist one and only one maximal ideal mw in L ⊗ Kˆ such that the canonical mapping from L ⊗ Kˆ → Mw gives K K an isomorphism from L ⊗ Kˆ /mw onto Mw. So we get a map ϕ from K 6. General Case 37 the set of the valuation w extending v to the set of the maximal ideals of L ⊗ Kˆ . Conversely if we start from a maximal ideal M in L ⊗ Kˆ , then K K the corresponding composite extension M = L ⊗ Kˆ /M is an algebraic extension of K and there one and only one valuation wM of M which extends v and the restriction of wM to L gives a valuation of L extending v. So we get a map ψ from the set of the maximal ideals of L ⊗ Kˆ (or K of the classes of complete extensions) to the set of the valuations of L extending v. Moreover the completion Lˆ of L with respect to wM is exactly Mˆ 48 and the composite extension of L and Kˆ contained in Lˆ is M.So we have ϕ ◦ ψ = I (identity map) Now we have also ψ ◦ ϕ = I, for if w is any valuation of Lthen the valuation wMw is necessarily the same as w by the uniqueness of the extension to M of the valuation v of Kˆ . Hence there exists a 1-1 correspondence between the set of valuations on L extending v and the set of inequivalent composite extensions of L and Kˆ . In particular if (L : K) = n < ∞, then any composite extension of L and Kˆ is complete which means that Lˆ = L ⊗ K/M, where M is some K maximal ideal of L ⊗ Kˆ . K Suppose L is an algebraic extension of an incomplete valuated field K with a valuation v. Let (wi)i∈I be the set of valuations on L extending v. We shall denote by Li the field L with the valuation wi, by ei the ramification index e(Li : K) = e(Lˆ i : Kˆ ) by fi the residual degree f (Lˆ i : Kˆ ) = f (Li : K) and by ni the dimension of Lˆ i over Kˆ . The sequence

0 → r(L ⊗ Kˆ ) → (L ⊗ Kˆ ) → Lˆ i K K Yi∈I is exact, because the radical of L⊗Kˆ is the intersection of all the maximal K ideals of L ⊗ Kˆ , that is of all the kernels of the map L ⊗ Kˆ → Li. If the 49 K K dimension of L over K is ﬁnite we have the following result. 38 2. Theory of Valuations -II

Theorem 3. If L is a finite extension of degree n of a field K with a real valuation v, then there exist, only finitely many different valuations (wi) on L extending v. Moreover we have ni ≤ n and the sequence P 0 → r(L ⊗ Kˆ ) → L ⊗ K → Lˆ i → 0 K K Y is exact.

Proof. We observe that wi is not equivalent to w j for i , j, because wi equivalent to w jmeans that they differ by a constant factor and since their restriction to K is same, we have wi = w j The number of different valuations (wi) on L extending v is finite because the number of inequivalent composite extensions of L and Kˆ is finite. To prove that the sequence is exact, we have to show that the mapping ρ : L ⊗ Kˆ → Lˆ i is surjective.By the approximation K Q theorem of valuations ρ(L) is dense in Lˆ i, therefore ρ(L ⊗ Kˆ ) is dense K Q in Lˆ i, where ρ is the canonical map from L → Lˆ i. But ρ(L ⊗ Kˆ ) is K a finiteQ dimensional vector space over Kˆ thereforeQ it is complete. Hence ρ(L ⊗ Kˆ ) = Lˆ i i.e., ρ is onto. Obviously dim Lˆ i ≤ dim L ⊗ Kˆ over K K Q Q Kˆ ,which means that ni ≤ n. P 50 Corollary. If Kˆ or L is separable over K, then we have ni = n. Kˆ or L separable over K implies that r(Kˆ ⊗ L) = 0P(for proof see K Algebre by N. Bourbaki chapter 8 section 7), therefore ρ is an isomorphism.

7 Complete Algebraic Closure of a p-adic Field

Proposition 5. Let K be a complete ﬁeld with a real valuation v and Ω the algebraic closure of K.Then Ωˆ the completion of Ω by the valuation extending v is algebraically closed. We shall denote the extended valuation also by v. Proof. To prove that Ωˆ is algebraically closed we have to show that any irreducible polynomial f (X) in Ωˆ [X] has a root in Ωˆ . Without loss of 8. Valuations of Non-Commutative Rings 39

O generality we a can assume that f (X) is in Ωˆ [X] and leading coeﬃcient r r−1 of f (X) is 1. Suppose that f (X) = X + ar−1X + ···+ a0 then for every r (m) r−1 (m) integer m there exists a polynomial ϕm(X) = X + b X + ··· + b◦ in r−1 OΩ[X] such that for every x in O , v( f (x) − ϕ (x) > rm Let ϕ (X) = Ωˆ m m r (X − α jm), α jm are in OΩ as ϕm(X) is in OΩ[X]. Then j=1 Q

ϕm+1(α jm) = ϕm+1(α jm) − f (α jm) + f (α jm) − ϕm(α jm) implies that v(ϕm+1(α jm)) > rm r or v(α jm − αtm+1) > rm. Xt=1 2

Therefore there exists a root αtm+1 of ϕm+1(X) such that

v(α jm − αtm+1 ) > m.

So we get a sequence ϕm(X) of polynomials converging to f and 51 a sequence of elements βm converging to β in Ωˆ and each βm is a root of ϕm(X). Since polynomials are continuous functions, we have lim f (βm) = f (β) m→∞ But lim f (βm) = 0, because given integer N > 0, for m > N we m→∞ have v( f (βm)) = v( f (βm) − ϕm(βm)) > rm > N. Hence β is a root of f (X). One can easily prove that the residual field of Ωˆ is the algebraic closure of the residual field of K. In particular if K = Qp, then the ˆ residual field of Ω i.e., kΩˆ is algebraic closure of Z/(P).Thus kΩˆ = ∪Fi, where each Fi is a finite extension of Z/(P)

8 Valuations of Non-Commutative Rings

We deﬁne a valuation of a non-commutative ring A without zero divisors containing the unit element in the same way as of a commutative ring. Almost all the results proved so far about valuated can be carried over to 40 2. Theory of Valuations -II

division rings with valuations with obvious modiﬁcations. WE mention the following facts far illustration. Let L be a division ring with a valuation v.Then

(1) The set OL = x/x ∈ L, v(x) ≥ 0 is a non-commutative ring,which we call the valuation ring of L with respect to the valuation v.

52 (2) YL = x/x ∈ L, v(x) > 0 is the unique two sided maximal ideal of OL.

−1 (3) Any ideal in OL is a two sided ideal. For, v(x yx) = −v(x) + v(y) + −1 v(x) ≥ 0 for x in L and y in OL which means that x yx belongs to OL, therefore yx = xz for some z in OL. Hence OL x = xOL.

(4) The ideals of OL are any one of the two kinds

Iα = x|v(x) >α ≥ 0 ′ Iα = x|v(x) ≥ α> 0

(5) The division ring L is locally compact non-discrete division ring for the valuation v if and only if v is a discrete valuation, L is complete and O/YL is ﬁnite

Regarding the extension of valuations to an extension division ring we prove the following.

Theorem 4. Let P be a division algebra of finite rank over a complete valuated field P with a valuation v such that P is contained in the centre e of P. Then there exists one and only one valuation w of P which extends v. e e Proof. Existence We define N (x) = determinant of the endomorphism P/P e ρxy → xy of P˜, for any x in P˜. 8. Valuations of Non-Commutative Rings 41

1 We shall prove that w(x) = v( N (x) is a valuation of P if r is the r P|P 53 rank of P over P. The axioms we(x) = ∞ if and only ifex = 0 and w(xy) = w(x) + w(y) are obviously true. e To prove w(x + y) ≥ inf(w(x), w(y)) it is sufficient to prove that w(x) ≥ 0 implies w(1 + y) ≥ 0. Let F = P(x) F is clearly a field containing P and P is a vector space over F by left multiplication. The mapping ρ is an F endomorphism. We know that if U is any F-endomorphism x e and Up the P-endomorphism defined by U, then det Up = N (det U) F/P (P˜:F) and we have det ρx = (x) if ρx is considered as an F-endomorphism. Therefore we have 1 w(x) = (P : F) = v( N (x)). r F/P e

Now w(x) ≥ 0 ⇐⇒ vv (N(x)) ≥ 0 =⇒ v(N(1 + x)) ≥ 0, because we F/P F/P have proved this for commutative case.Hence w is a valuation on P˜. Uniqueness. Since P is complete and P is of ﬁnite rank r over P, any valuation deﬁnes the same topology on P˜ as that of Pr. But the topology determines the valuation upto a constant factor, If w1 and w2 are two valuations of P extending v then w1 = Cw2 for some C in P.But restriction of w to w to P is v, therefore C = 1 and w = w . 1 e 2 1 2

Part II

Representations of classical groups over p-adic Fields

Chapter 3 Representations of Locally Compact and Semi-Simple Lie Groups

1 Representations of Locally Compact Groups

In this section we give a short account of some deﬁnitions and results 54 about the representations of locally compact groups.We assume the fundamental theorem on the existence and uniqueness (upto a constant factor) of right invariant Haar measure on a locally compact groups.For simplicity we assume that the locally compact groups in our discussion are unimodular i.e., the Haar measure is both right and left invariant.By L(G) we shall denote the space of continuous complex valued functions with compact support and by L(G, K), where K is some compact set of G,the set of elements of L(G) whose support is contained in K.Obviously we have L(G) = ∪ L(G, K) and L(G, K) is a Banach K⊂G space under the norm f = sup| f (x)|. x∈K The space L(G) can be provided with a topology by taking the direct limit of the topologies of L(G, K). This topology makes L(G) a locally convex topological vector space. Let G be a locally compact group and H a Banach space

45 46 3. Representations of Locally Compact and Semi-Simple...

Deﬁnition 1. A continuous representation U of G in H is a map x → Ux ∈ Hom(H, H) such that

(i) Uxy = Ux ◦ Uy for x; y in G.

(ii) The map H × G → H deﬁned by (a, x) → Ux a is continuous.

55 Deﬁnition. Let H be a Hilbert space.The representation U is said to be Unitary if Ux is a unitary operator on H for every x in H.

Let M(G) be the space of measures on G with compact support.The space M(G) is an algebra for the convolution product deﬁned by

µ ∗ ν( f ) = f (xy) dµ (x) dν (y) Z Z

The space L(G) can be imbedded into M(G) by the map f → µ f = f (x)dx. It is infact a subalgebra of M(G) because µ f ∗ µg = µ f ∗g where

f ∗ g(x) = f (xy−1)g(y)dy. Z

Moreover if v is any element of M(G), then µ f ∗ ν belongs to L(G), because for any g ∈ L(G) we have

(µ f ∗ ν)(g) = x g(xy) f (x)dxdν(y) = dν(y) g(x) f (xy−1)dx. Z Z −1 = µh(g) where h(x) = f (xy )dν(y) Z Thus we deﬁne the convolution of a measure µ and function f ∈ L(G) by setting

(µ ∗ f )(y) = f (x−1y)dµ(x) Z ( f ∗ µ)(y) = f (yx−1)dµ(x) Z 1. Representations of Locally Compact Groups 47

Let U be a representation of G in H. Then U can be extended to M(G) by setting

Uµ(a) = Uxadµ(x) (for a ∈ H,µ ∈ M(G) ZG

Now let H be a Hibert space and U a Unitary representation. 56 Then if µ and ν are any two elements in M(G), we have

hUνUµa, bi = hUxUµa, bidν(x) Z

= hUµa, U −1 bidν(x) Z x

= dν(x) hUyaU −1 bidµ(y) Z Z x

= hUxUya, bidν(x)dµ(y) Z

This means that Uµ∗ν = Uµ ◦ Uν i.e., U is a representation of the algebra M(G).It can be easily veriﬁed that map µ → Uµ is a continuous representation of the algebra M(G). Moreover

∗ hUµa, bi = hUµb, ai

−1 = hUxa, bidµ(x ) Z

∗ −1 This shows that Uµ = Uµ˜ , where dµ˜(x) = dµ(x ). Thus the operator Uµ∗µ˜ on H is Hermitian. In particular we get a representation of L(G) in H given by f →

Uµ f = U f , where

U f (a) = Uxa f (x)dx. ZG We can also get a representation of M(G) by considering regular representations of G i.e., representations G by right or left translations in G in any function space connected with G with some convenient topology, 48 3. Representations of Locally Compact and Semi-Simple...

for instance the space L(G) or L2(G) (the space of square integrable functions). We shall denote by σx the left regular representations and by τx the 57 right regular representations of G i.e., for any function f on G we have

−1 σx( f )(y) = f (x y), τx( f )(y) = f (yx) we have for any µ in M(G)

σµ( f ) = µ ∗ f v v −1 τµ( f ) = f ∗ µ where dµ(x) = dµ(x ) Let K be a compact group, M an equivalence class of (unitary) irreducible representations of K. For any x belonging to G, let Mx = M (Ci j (x)) be the matrix of Mx with respect to some basis of the repre- rm M sentation space. Let rM be the dimension of M and χM = Ci i the i=1 character of M. For any two irreducible unitary representationsP of K we have the following orthogonality relation, M M′ , ′ (1) Ci j ∗ Ckl if M M

M M l M (2) Ci j ∗ Ckl = δ jkCil rM where the value of the convolution product at the unit element e of G is M M M ji(y) M given by Ci j ∗ Ckl (e) = C CKl(y)dx. When we write (1) andR (2) in terms of characters we get ′ (1) χM ∗ χM′ = 0 if M , M l (2) χM ∗ χM = χM rM 58 Obviously we have

M M M (rMχM) ∗ Ci j = Ci j ∗ χMrM = Ci j

M¯ Let LM(K) be the vector space generated by the coeﬃcients Ci j , where M¯ is the complex conjugate representation of M. If f is in L2(G), 1. Representations of Locally Compact Groups 49

N then by Peter-Weyl’s theorem, f = λi jNCi j .Further if rmχM¯ ∗ f = f , i, j,N M¯ P then we have f = λi jMCi j , which means that f belong to LM(K). i, j P M¯ Conversely if f belongs to LM(K), then f = λi jM¯ Ci j . Therefore i, j 2 P rMχM¯ ∗ f = f .Hence f ∈ L (G) is in LM(K) if and only if rMχM¯ ∗ f = f . In this paragraph we give another interpretation of the space LM(K). Deﬁnition. Let M be an irreducible unitary representation of K and U any representation of K in a Banach space H.

We say that an element a ∈ H is transformed by U following M, if a is contained in a ﬁnite dimensional invariant subspace F of H such that the restriction of U to F is direct sum representations of the equivalence class of M. U Let HM = HM = {a |∈ H, a transformed by U following M}. It is 59 easy to verify that HM is a vector space.

2 Proposition 1. LM(K) is exactly the subspace of L (K) formed by the elements which are transformed following M (respectively following M) by the left (respectively right) regular representation of K. = Proposition 2. If U is a representation of K in H, then EM UrM χM is a continuous projection from H → HM. In order to prove the proposition 1 and 2 prove the following results.

M (1) Suppose that ϕ belongs to LM(K), then ϕ = λi jCi, j. For i, j P σ M M −1 M −1 M x ∈ K, we have xCi j (y) = Ci j (x y) = Ci j (x )Ci j (y) Xk M M = (Cki (x))Ck j (y). Xk

So the space E j generated by Ci j, ··· , Cr j(rM = r) is invariant by σ and the restriction of σ to E j is of class M. Therefore LM(K), which σ 2 is direct sum of the E j, is contained in (L (G) ) . M 50 3. Representations of Locally Compact and Semi-Simple...

(2) If ϕ belongs to LM(K) and a belongs to H, then we show that Uϕa belong to HM. We have

UxUϕa = U∈ ∗ϕap = Uσ◦ϕa x x

where εx is the Dirac measure at the point x, and

Uεx b = Uyb dεx(y) = Uxb. ZK

60 This shows that ϕ ∈ LM(K) → Uϕ a ∈ H is a morphism of representation σ and U. Hence Uϕa is transformed by U following M.

(3) If a belongs to HM, then EMa = a. Since a belongs to some ﬁnite dimensional invariant subspace F of H and the restriction of U to F is the direct sum of representation of class M, we can ﬁnd a basis M (e jk) of F such that Uxe jk = Ci j (x)eik K P Let a = λi jei j. Then i, j P

M EM(a) = rM λ jkC jk (x)eikχM(x)dx Z Xi, j,k

M = rm λ jk Ci j (x)χM(x)dx eik Xi,k Xj Z

= λikeik = a. Xi,k

Moreover M −1 M rMCi j (x)χM(x )dx = rMχM ∗ Ci j (e) = δi j R (4) In particular if ϕ belongs to L2(G) it is transformed by σ following M, then

σrM χM ϕ = rM χM ∗ ϕ = ϕ 1. Representations of Locally Compact Groups 51

Therefore ϕ belongs to LM(K). Clearly the results (1)and (4) imply proposition 1.

˙ Since EM EM = U r χM U r χM U r 2 χM ∗ χM M M M

= UrM χM = EM, the proposition (2) is proved by result (3) ′ Similarly we prove that EM · EM′ = 0 for M , M . Thus we get a 61 family of projections EM with EM(H) = HM. The sum HM is direct ′ and is dense in H. It is suﬃcient to prove that if a is a continuousP linear ′ form on H, which is zero on every HM, then ha, a i = o for every a ∈ H. ′ Let us put ϕ(x) = hUxa, a i. Then

M M ′ hϕ, C i = C (y)hUya, a idy. i j Z i j ′ M = hUga, a i with g = Ci j

′ But Uga belongs to HM¯ , therefore we get that ϕ is orthogonal to all M the coeﬃcients Ci j for any M, so ϕ = 0. In particular if U is unitary (for instance the regular representation in 2 L (K)), then the EM are orthogonal projections and H is exactly Hilber- tian sum of the closed subspaces HM. Let G be a locally compact group, K a compact subgroup of G. Sup- pose that U is a continuous representation of G in H and M an equiv- U alence class of unitary representation of K. By HM = HM we shall mean the vector subspace of H consisting of elements which are transformed by the restriction of U to K following M. As in the above case = EM UrMχM is a projection of H to HM. Let

LM(G) = f | f ∈ L(G), f ∗ rMχM = rMχM ∗ f = f

It is easy to prove that LM(G) is a subalgebra of L(G) and the mapping f → rMχM ∗ ∗˙rHχM is a projection from L(G) to LM(G). If f belongs to LM(G) and a belongs to H, then is in HM. If b 62 52 3. Representations of Locally Compact and Semi-Simple...

′ = = belongs to HM, then U f (a) UrM (a)χM ∗ f EM U f a ⇒ U f a is in HM. If b belongs to HM′ , then

U f b = U EM′ b = U f EM EM′ b = 0 f ∗ rM χM

This shows that U is a representation of LM(G) in HM and U f = EMU f EM. Moreover for f ∈ LM(G)

−1 f (y) = rM f (k y)χ (k)dk. Z M K

In particular if M is the identity representation, then χM is constant and f is in LM(G) if and only if

f (y) = rM f (ky)dk = rM f (yk)dk Z Z K K ⇐⇒ f (hyk) = f ( f yk) = f (y).

Such functions are called spherical function on G with respect to K. They can be considered as functions on G/K which are left invariant, provided we write G/K = {K, aK, −−−}

2 Irreducible Representations

In this section we study how we can get some information about the representation of a group G by studying the representation of the algebra LM(G).

Deﬁnition 1. A representation U of a group G in a vector space V is said to be algebraically irreducible if there exists no proper invariant subspace of V.

63 Deﬁnition 2. A representation U of a topological group G in a locally convex space E said to be topologically irreducible if there exists no proper closed invariant subspaces of E. 2. Irreducible Representations 53

Deﬁnition 3. A representation U of a topological group G in a Ba- nach space H is said to be completely irreducible if U(L(G)) is dense in Hom(H, H) in the topological of simple convergence i.e., given an operator T on H and element a1, a2, ··· , ap in H, there exists for every ∈> 0 an element f in L(G) such that

k (U f − T)ai k<∈ for i = 1, 2, ··· , p. It is obvious that (1) ⇒ (2). To prove that (3) ⇒ (2), suppose that F is a proper closed invariant subspace of H. Let a , 0 be any element of F, then for every b in H there exists a T ∈ Hom(H, H) such that T(a) = b. But by definition for every ε > 0 there exists an element f in L(G) such that k U f a − T(a) k< ε. This means that F is dense in H which is a contradiction because F was assumed to be a closed proper subspace of H. The definitions (2) and (3) are equivalent for unitary representation by Von Neumann and all the three representation are equivalent for finite dimension representations. The proof can be found in [9]. The definition (1) implies (3) (for proof see annals of Mathematics, 1954 Godement).

Lemma 1. If U is a completely irreducible representation of G in a Ba- M nach space H, then the representation U of LM(G) is H(M) is also completely irreducible.

Proof. Suppose that T belongs to Hom(H(M), H(M)). Extend T to H ˜ −1 by setting T = T on H(M) and O on EM (0). Obviously T is continuous on H.

Since U is completely irreducible, T˜ can be approximated by U f for 64 ˜ f in L(G) i.e., T = lim U fi . Therefore

˜ EMTEM = lim EMU fi EM = lim UrM χM ∗ fi∗rM χM

= Hence in HM, T lim UrM χM ∗ fi∗rM χM M where rMχM ∗ fi ∗ rMχM is in LM(G). Thus U is completely irreducible. 54 3. Representations of Locally Compact and Semi-Simple...

Let U be a unitary irreducible representation of G in a Hilbert space H. By coefficient of U we means positive definite function hUxa, ai on G. We state without proof the following theorem about the coefficients of unitary representations. Theorem 1. If two irreducible unitary representations have same non- zero coefficient associated to them, they are equivalent. We have seen that the representation U can be extended to the space M(G) and the operator Uµ∗µ˜ for any µ in M(G) is Hermitian. In particular if we take µ = rMχ¯ Mdk, we have µ = µ˜. There fore Uµ∗µ = EM is Hermitian.

Moreover for any f in L(G) and a in HM

hU f a, ai = hU f EMa, EMai = hEMU f EMa, ai

= hU f0 a, ai

65 where f0 = rM χM ∗ f ∗ rMχM belongs to LM(G). Thus if we know nonzero coefficient associated to U M, we know coefficient associated to U as a representation of L(G), which determines coefficient of U as a representation of G. Thus a unitary irreducible representation of G is M M completely characterised by its restriction U to LG(G) if U is not zero. Definition. A set Ω of representations of an algebra A in a vector space is said to be complete if for every nonzero f in A there exists U ∈ Ω such that U f , O. Proposition 3. If there exists a complete set Ω of representations of an algebra A which are of dimension ≤ K (K a fixed integer), then every completely irreducible representation of A in a Banach space is of dimension ≤ k. We first prove a lemma due to Kaplansky. Let A be any algebra. For

x1, ··· , xp in A we deﬁne [x1, ··· , xp] = εσ xσ1 ... xσp where S p is σ∈S P p the set of all permutations σ on 1, 2, ··· , p and εσ is the signature of σ. Obviously if dim A < p, then [x1, ··· , xp] = o for all x1, x2, ··· , xp in A. 2. Irreducible Representations 55

In particular we take A = Mn(C), algebra of n × n matrix with coefficient from C, the field of complex numbers, We define

r(n) = inf(p)such that

[X1, ··· , Xp] = 0, Xi ∈ Mn(C).

Clearly r(n) ≤ n2 + 1. We shall prove that r(n + 1) ≥ r(n) + 2. We have r(n) − 1 elements X1, X2, ··· Xr−1 in Mn(C)(r = r(n)) such 66 that [X1, ··· , Xr−1] , 0, Let Ekh be the canonical basis of Mn(C). Then

n [X1, ··· , Xr−1] = λkhEkh. kX,h=1 , , Since [X1, ··· , Xr−1] 0, there exists k0 and h0 such that λk0h0 0. Let X˜i be the matrix obtained by adding a row and a column of zeros to Xi. Then ˜ ˜ ˜ ˜ X1, ··· , Xr−1Eh0,n+1En+1,n+1 = X1, ··· , Xr−1 Eh0,n+1En+1n+1 h i h ˜ i = λkhEkhEh0 n + 1 Xh,k , = λkh0 Ek,n+1 0. X Thus r(n + 1) ≥ r(n) + 2. Now we prove the proposition. Suppose that r(k) = r and U is a complete irreducible representation of dim > K in a Banach space H. Let F be a subspace of H of dim k + 1. Since r(k + 1) > r(k), there exist operators [A1,..., Ar] in Hom(F, F) such that [A1,..., Ar] , 0. We extend each Ai to the whole space H by deﬁning Ai to be zero on F′, where F′ is any closed subspace such that H is the topological direct sum of F and F′. Suppose that A = lim U , where f ∈ A. We have 1 fi1 i1

, ˜ ˜ ˜ ˜ ˜ ˜ 0 [A1, ··· , Ar] = Aσ1 ,... Aσr = lim[U fi A2, ··· , Ar]. σX∈Sr

˜ ˜ , Therefore there exists f1 in A such that [U f1 A2, ··· , Ar] 0. Repeating 67 this process we obtain that there exist elements f1, ··· , fr in A such that 56 3. Representations of Locally Compact and Semi-Simple...

, U[ f1,··· , fr] = [U f1 , ··· , U fr ] 0. But this is a contradiction because [ f1, ··· , fr] = 0 if [ f1, ··· , fr] , 0, then there exists a V in Ω such that , V[ f1,··· , fr](a) 0 ⇒ , [V f1 , ··· , V fr ](a) 0. But r ≥ rk and dimV ≤ k, therefore [V f1 , ··· ,

V fr ] = 0. Hence dim U < k. Corollary. Let G be a locally compact group, K a compact subgroup, M a class of irreducible unitary representations of K in a Banach space H. If there exists a system Ω of representations of G in a Banach space M such that (i) for every U in Ω, the representation U of LM(G) is of dim ≤ p dim M. Equivalently M occurs atmost p times in each U. (ii) The representations U M for U in Ω form a complete system of representation of algebra LM(G). Then M occurs atmost p times in any completely irreducible representation of G.

3 Measures on Homogeneous spaces

Let G be a locally compact group, dx the right invariant Haar measure and ∆(x) the modular function on G i.e., d(yx) = ∆(y)dx. Let Γ be a closed subgroup of G. We shall denote by ξ,η... the elements of Γ by dξ and δ the Haar measure and the modular function on Γ. It is well 68 known that there exists a right invariant Haar measure on G/Γ if and only if ∆(ξ) = δ(ξ). In general it is possible to ﬁnd a quasi-invariant measure on G/Γ. In order to show the existence, one shows that there exists a δ(ξ) strictly positive continuous function ρ on G such that ρ(ξx) = ρ(x) ∆(ξ) for every x in G and ξ in Γ. Then the measure ρ(x)dx gives rise to a measure dµ(x) on G/Γ such that for any f in L(G) we have

f (x)ρ(x)dx = dµ(x) f (ξx)dξ (1) Z Z Z G G/Γ Γ It is obvious from (1)that ρ(xy) dµ((xy)) = dµ(x) ρ(x) 4. Induced Representations 57

ρ(xy) where depends only on the cosets of x modulo Γ. Thus µ(x) is a ρ(x) quasi-invariant measure on G/Γ. The details could be found in [9].

4 Induced Representations

Let L be a representation of Γ in Hilbert space H. We shall deﬁne two types of induced representation on G given by L.

(1) Assume that L is unitary. Let HL be the spaces of functions f on G such that

(1) f is measurable with values in H. 1/2 (2) f (ξx) = [p(ξ)] Lξ f (x), for ξ ∈ Γ. (3) (ρ(x))−1 k f (x) k2 dx < ∞. GR/Γ Since the function (ρ(x))−1 k f (x) k2 is invariant on the left by Γ, it can be considered as a function on G/Γ. Thus we deﬁne

k f k2= (ρ(x))−1 k f (x) k2 dµ(x) Z G/Γ

It can be proved that HL is a Hilbert space with the scalar product 69

h f, g >, = (ρ(x))−1h f (x), g(x)idµ(x). Z G/Γ

Let UL be the map from G to HL such that

L Ux f (y) = f (xy)

Obviously UL is continuous. Since we have

k UL f k2 = (ρ(x))−1 k f (xy) k2 dµ(x) y Z G/Γ 58 3. Representations of Locally Compact and Semi-Simple...

ρ(xy−1) = (ρ(xy−1))−1 k f (x) k2 dµ(x) Z ρ(x) G/Γ =k f k2

It follows that UL is unitary. We say that UL is the unitary representation induced by L. (ii) Let L be any representation of Γ. Let us suppose that there exists a compact subgroup K of G such that G = ΓK. Let CL be the space of functions f such that (1) f is continuous with values in H. 1

(2) f (ξx) = (ρ(ξ))2 Lξ( f (x)) for ξ ∈ Γ. We define k f k= sup k f (x) k. Clearly CL with this norm is a x∈K Banach space. Again right translation by elements of G give rise to a representation of G in GL. We denote this also by UL. L Let f → restriction of f to K = f0 be the map from the C to C(K) (the set of continuous functions on K with values in H ). The image of L 70 C by this map is the set of elements f0 ∈ C(K) which satisfy condition (2) above for all ξ in Γ ∩ K and x is K. But ρ(ξ) = 1, because ρ is a positive real character of K ∩ Γ, therefore f0(ξx) = f0(x). Through the space CL is identified with a subspace of C(K) yet the representation UL cannot be defined on this subspace. However the restriction of UL to K and the representation induced by the restriction of L to Γ ∩ K are identical. L If L is unitary then f belongs to H if and only if f0 belongs to L2(K). We can choose ρ in such a way that ρ(xk) = ρ(x) for k ∈ K. Since the group K/K ∩Γ is compact homogeneous space, there exits one and only one invariant Haar measure on it. But K/K ∩ Γ is isomorphic to G/Γ therefore with the above choice of ρ, the quasiinvariant measure on G/Γ gives rise to the invariant measure on K/K ∩ Γ. We have

(ρ(x))−1 k f (x) k2 dµ(x) = k f (k) k2 dµ(k) Z Z G/Γ K/K∩Γ 5. Semi Simple Lie Groups 59

= k f (k) k2 dk (A) Z K Our result is obvious from (A).

5 Semi Simple Lie Groups

Let G be a semi simple Lie group worth a faithful representation. We state here two theorems the proof of which could be found in [19]. Theorem 2. The group G has a maximal compact subgroup and all the maximal compact subgroup are conjugates. Theorem 3. Suppose that K is maximal compact subgroup of G, then there exists a connected solvable T of G such that G = TK.

We shall prove the following theorem about completely irreducible 71 representation of G. Theorem 4. Every irreducible representation M of K is contained atmost dim (M) times in every completely irreducible representation of G. Proof. (1) The finite dimensional irreducible representations of G is a vector H is a complete system of representations of L(G). Let x → ρx be a representation of G in a vector space H. We call the function ′ ′ ∗ θ(x) = hρxa, a i where a belongs to H and a belongs to H (the conjugate space of H), a coefficient of the representation. Let V denote the vector space generated by all coefficients of all finite dimensional irreducible representations of G. Since every finite dimensional representation of G is completely reducible, V contains all the coefficients of all finite dimensional representations of G. Let ρ1 and ρ2 be two finite dimensional irreducible representations of G. Then we have

1 ′ 2 ′ 1 2 ′ ′ hρxa1, a1ihρxa2, a2i = hρx ⊗ ρx a1 ⊗ a2, a1 ⊗ a2i showing that V is an algebra. Moreover V is a self adjoint algebra, be- ′ ′ cause if θ(x) = hρa, a i is in V, then θ¯(x) = hρ¯xa¯, a¯ i is also in V. Since G 60 3. Representations of Locally Compact and Semi-Simple...

has a ﬁnite dimensional faithful representation, V separates points i.e., if θ(x) = θ(x′) for every θ in V, then x = x′. Thus Stone- Weierstrass’ approximation theorem every continuous function on G can be approximated uniformly on every compact subset by elements of V. Hence if f is a non-zero elements of L(G), then f (x)g(x)dx = 0 for every element 72 g of C(G) (the set of all continuous functionsR on G), because

′ ρ f = ρx f (x)dx and < ρxa, a > f (x)dx = 0 Z Z

for every a in H and a′ in H∗ and ρ. Therefore f must be =0 (2) The representations of G induced by all characters of T form a complete system for L(G) Let ρ be a ﬁnite dimensional irreducible representation of G and let v t ρ = (ρ)−1) be the representation contragradient to ρ. By Lie’s theorem v [19],the restriction of ρ to T has an invariant subspace of dimension 1, which implies that there exists a vector b′ , 0 in E∗(the conjugate space v of the representation space E of, ρ) such that ρ(t) = b′ = χ(t)b′ for every χ−1 ′ t ∈ T. Consider the mapping a ∈ E → aεc , where a(x) =< ρxa, b >. Since e e ′ r−1 ′ −1 ′ −1 a(tx) =< ρtxa, b >=< ρxa, ρt b >= χ (t) < ρxa, b >= χ (t)a(x),

ae(x) is covariant by left translation. Obviously the map a → a is contin-e uous. Let Uχ−1 be the representation of G induced by χ−1. The mapping ea → a is a morphism of representations ρ and Uχ−1, becausee

e ′ χ−1 ρy a(x) = hρx ρya, b i = a(xy) = Uy (a).

The mapping ae→ a˜ is not zero. If ae, 0, then (ρexa) generates the whole space E because ρ is irreducible, therefore for atleast on x in ′ Ghρxa, b i , 0 ⇒ a , 0. Let f be a non-zero element of L(G). If χ−1 U f = 0. For every χ then ρ f = 0 for every ρ which means the f = 0 by (1). This is a contradiction,e hence our result is proved. 73 (3) We shall show that if χ is a character of T, then M occurs atmost dim (M) times in Uχ. Clearly Uχ/K(restriction of UχtoK) = 5. Semi Simple Lie Groups 61

Uχ/K∩T .But the space of this representation is the space of continuous functions f on K such that

f (tk) = χ(t) f (k) for t ∈ K ∩ T.

Therefore Uχ/K∩T is a subrepresentation of the right regular representa- χ 2 tion of K. Hence (C )M ⊂ LM(K) which is a space of (dimM) . Thus M occurs at most dim(M) times in U. Our theorem follows from (2), (3) and proposition 1.3.

Chapter 4 Classical Linear Groups over p-adic Fields

1 General Deﬁnitions

We shall study the following types of classical linear groups over ﬁeld 74 P or over a division algebra.

(I) (a) GLn(P)- The group of all non-singular n x n matrices with coeﬃcients from P is called the general linear group

(b) PrGLn(P) Let CLn(P) be the centre of the group GLN(P). The group pr GLn(P) = GLn(P)/CLn(P) is called the projective linear group.

(c) SLn(P)-The subgroup of GLn(P) consisting of all the matrices of determinant 1 is called the special linear group or the unimodular group. It can be proved that PrSLn(P) = SLn(P)/C(SLn(P)) is a simple group

(II) -Let E = Pn and ϕ a non-degenerate bilinear form over E

(a) S pn(P)-If ϕ is an alternating form,then the the set of all matrices in GLn(P) which leave this bilinear form invariant is a group called the linear symplectic group. We shall denote the by S pn(P). This group is independent of the choice of the

63 64 4. Classical Linear Groups over p-adic Fields

alternating bilinear form because any two such bilinear forms are equivalent. (b) If ϕ is a symmetric non-degenerate bilinear form, then the set of elements in GLn(P) leaving ϕ invariant is group called the linear orthogonal group. (III) Let P be a separable quadratic extension of P. Let ξ → ξ¯ be the unique nontrivial automorphism of P. If ϕ is a non-degenerate e 75 Hermitian bilinear form over E i.e., ϕ(y, x) = ϕ(x, y), then the e set Un(ϕ, P) of elements of GLn(P) leaving ϕ invariant is a group called the unitary group. (IV) Let P be a division algebra of finite rank over P, such that P is the centre of P. We define GL (P) as in I (a). The group SL (P) can e n n be defined as the kernel of the map σ(determinant of Dieudonne) e ∗ ∗ from GLn(P) to P /C where C is the commutator subgroup of P . (V) Let P be the algebrae of quaternions over P. In this case there exists an involution in P i.e., an anti automorphism of P of order e 2. So we can define as in (3) the group U (ϕ, P) which leaves e n e invariant the bilinear form ϕ over Pn. As in (1) one can define S 0 (ϕ, P) and SV (ϕ, P) and prove that their projective groups n n e are in general simple. Suppose that P is a locally compact p-adic field. All the groups of type (1), (2) and (3) are locally compact, because on Mn(P) (the set of all nχn matrices with coefficients from P) we have the topology of Pn2 and GLn(P) is an open subset of Mn(P) and the groups SLn(P) etc.are closed subgroups of GLn(P). Let us assume that the rank of P over P in (4) is r. Then Mn(P) may be imbedded in M (P), as Pn can be considered as a space of nr e e dimension nr over P, since a matrix is inversible in M (P) if and only if e n it is invertible in M (P), we have nr e GLn(P) = GLnr(P) ∩ Mn(P)

But GLnr(P) is an opene subset of Mnr(P),e therefore GLn(P) is an 76 open subset of M (P). Since M (P) is locally compact, because it has n n e e e 2. Study of GLn(P) 65 e n2 the same topology as the P , GLn(P) is locally compact. Un(ϕ, P) is locally compact, because it is a closed subgroup of GL (P). e e n e e 2 Study of GLn(P) By P we shall mean ae division algebra of ﬁnite rank over P, which is a locally compact valuated ﬁeld, contained in the centre of P. Let O˜ e denote the ring of integers of P e As we have already seen that O˜ is a compact subset of P, there- e fore M (O˜) which is homeomorphic to On2 is compact in M (P˜ ). Let n ne GL (O˜) be the set of elements M (O˜) which are invertible in M (O˜). n n e n ObviouslyGLn(P) contains GLn(O˜). Therefore −1 GLn(Oe˜) = GLn(P) ∩ Mn(O˜) ∩ [GLn(P) ∩ Mn(O˜)]

Since O˜ is open in P, Men(O˜) is open in MNe(P˜). Therefore GLn(O˜) is open in M (O˜). Similarly GL (O˜) is open in GL (P). Moreover n e n n GL (O˜) is closed in M (O˜). For, let (X ) be a sequence of elements in n n p e GLn(O˜) such that Xp tends to X ∈ Mn(O˜) as p tends to inﬁnity. Because O˜ −1 O˜ Mn( ) is compact, we can assume that Xp has a limit Z in Mn( ). But then ZX = XZ = I, therefore X belongs to GLn(O˜). Hence GLn(O˜) is compact. We deﬁne in the following some subgroups of GLn(P), which will be of use later on. e (i) a1 ∗ . Γ= γ =  ..  |γ ∈ GLn(P)           0 an e      where(*) indicates that there may be some non-zero entries. (ii) πα ∗ . T = t =  ..  |t ∈ GLn(P),αi ∈ Z  e     α     0 π      e     ˜  π being a uniformising parametere in P  77 e 66 4. Classical Linear Groups over p-adic Fields

(iii) 1 ∗ . N = n =  ..      1 1        (iv)    a11 0 . D = d =  ..  ai j ∈ P, ai j , 0       0 a    1n  e     (v)     πα1 0 . ∆=  ..  αi ∈ Z e    α    0 π n         We see immediately that T = ∆Neand Γ = DN. Moreover T is a solvable group, Γ is solvable if P is commutative and T (respectively Γ) is a semi direct product of ∆ and N(respectively D and N). e Proposition 1. GLn(P˜) = G = TK, where K = GLn(O˜). Proof. When n = 1, the proposition is trivially true. Suppose that it is true for all GLs(P) for s ≤ n − 1. We shall prove it for GLn(P). Let g = (g ) be an element of G. We can ﬁnd integers (k ) ≤ ≤ such that i j e j1 1 j n e n gi jk j1 = 0 for 2 ≤ i ≤ n Xj=1

= a11 , 0 for i = 1.

78 By multiplying on the right with a suitable element of P we can take atleast one of k to be 1. Let k = (γ ) be a matrix, where γ = j1 i j e i1 ki1 for i = 1, 2,..., n with k ji = 1, γ jr = 0 for r = 2,..., n and the other γi j are so determined that k belongs to K. So we get

a11 ∗ π 0 1 ∗ y 0 gk = = 0 ∗! 0 1! 0 g′! 0 1! e ′ α ∗ where g is n − 1 × n − 1 matrix and a11 = π˜ y, y ∈ O . 2. Study of GLn(P) 67 e But by induction hypothesis g′ = t′k′ where t′ belongs to T ′ and ′ ′ ′ ′ k ∈ K the subgroups T and K deﬁned in GLn−1(P) in the same way as T and K in G Thus we get e 1 ∗ 1 ∗k′−1 1 0 = 0 g′! 0 t′ ! 0 k′!

This implies that

πα 0 1 ∗k′−1 1 o y 0 gk = 0 1! 0 t′ ! 0 k′! 0 1! e = t1k1, t1 ∈ T and k1 ∈ K.

Hence our result follows: We shall now prove an analogue of Elementary divisors theorem. Let A be a ring with unity (but without any other condition). Let us consider the following assertions(where module signiﬁes left module): (I a) any ﬁnitely generated module is isomorphic to a direct sum i=r , ⊕ A/ai, where ai are left ideals with A a1 ⊃···⊃ ar 79 i=1 (I b) Such a decomposition, if it exists, is unique.

(II a) if M is a free module of ﬁnite type and N a ﬁnitely generated submodule of M, there exists a basis e1,..., er and r elements α1,...,αr of A such that αi+1 ⊂ Aαi and such that N is the direct sum of submodules Aαiei.

(II b) if such elements ei and αi exist, the ideal Aαi are independent of the choice of the ei and αi satisfying (II a).

(III a) if g is a m × n matrix with coeﬃcients in A, there exists two m × m and mxn invertible matrices p and q such that d = pgq is a m × n “diagonal” matrix(i.e., di j = for i , j) and αi = dii ∈ Aαi+1.

(III b) if such matrices p and q exist, the ideals Aαi are independent of the choice of p and q (satisfying(III b)). 68 4. Classical Linear Groups over p-adic Fields

It is obvious that (III a) implies (IIa): consider a basis x1,..., xn of M and a system of generators y1,..., ym of N and define the matrix g by −1 y j = gi j xi. Then ei = (q )ik xk is basis of M and the αiei = pikyk generateP N. if a is left NoetherianP then (IIa) implies (Ia), for anyP finitely generated module is a quotient M/N, with M free of finite type and N finitely generated. Conversely,it is obvious that (I a) implies (II b) and (II b) implies (III b). It is well known that all these six assertions are true if A is a commutative principal ideal ring (without zero divisors )(see for instance Bourbaki, Alg., ch VII, §4). We shall now prove the following extension:

80 Theorem. Let A be a ring with unity (but A may be non-commutative and may have zero divisors),which satisﬁes the following conditions:

1) any left or tight ideal is two sided (equivalently, Ax =xA for any x ∈ A

2) the set of the principal ideals is totally ordered by inclusion (hence any ﬁnitely generated ideal is principal).

Then, the assertions (IIIa) and (Ia) are true (hence also (IIa), and (IIIb). If moreover A dis Noetherian (that is if any ideal is principal),then (Ia) is also true Proof of (III a): the result is obviously true form n = m = 1. Assume it is proved for (m − 1) × (n − 1) matrices. Let us consider the ideals A gi j : by(2) they are all contained in one of them, and we can assume without loss of generality, that g ji ∈ Ag11 for any indices i,j. Let gi1 = cig11 for 2 ≤ i ≤ m. By multiplying g on the left by a m×m matrix k where i 0 ··· 0 −c 1 0 ··· 0 k =  2   ···    −cm o ··· 1      we get a matrix kg with (kg)11 = g11 and (kg)i1 = 0 for i ≥ 2. Moreover, the matrix k is invertible. Similarly, using the fact that gi j ∈ g11A. We 2. Study of GLn(P) 69 e ﬁnd a n × n inversible matrix h such that

g11 0 ··· 0 0 kgh =    g′     0      Now, we have just to apply the induction hypothesis to g′ (remember 81 ′ that all the coeﬃcients of g, hence of g belong to Ag11).

Proof. (Ib): more generally, we shall prove that assumption 1) alone implies (Ib).

n m Let M = A/ai = A/bi, with a1 ⊃ a2 ⊃···⊃ an and b1 ⊃ b2 ⊃ i=1 j=1 P P ··· ⊃ bm, ai , A and bi , A for any i. Then m = n and ai = bi for i = 1, 2,..., n.

′ ′ Proof. Let xi (respectively y j) be the canonical generator of A/ai (re- ′ spectively A/b j) and xi (respectively y j) the canonical image of xi (re- n ′ spectively y j) in M. Then y j = ai j xi, where ai j ∈ A and is determined i=1 P m completely modulo ai and therefore modulo ai. Similarly xi = bkiyk, k=1 where bki ∈ A and is completely determined modulo b1. LetP m be a maximal left ideal containing b1. We see immediately that m is a two n n sided ideal and A/m is a division algebra. Since y j = ai j = bkiyk, i=1 i=1 we have P P n ai j ≡ δk j (mod m) Xi=1 But this is possible only when n ≥ m, because if Vm and Vn are two vector spaces over a division ring of dimension m and n respectively such that ϕ and ψ are two linear transformations from Vm to Vn and Vn to Vm respectively. then ϕψ = I implies that ψ is an isomorphism of Vm onto a subspace of Vn. In the same way we get that m ≥ n. Hence m = n. 70 4. Classical Linear Groups over p-adic Fields

If possible let us suppose that ai , bi for some i. Let us suppose that 82 there exists an element a in ai which does not belong to bi. Consider the set aM, it is a submodule of M. Therefore

n n aM = aA/aA ∩ ai = aA/a ∩ ai Xi=1 Xi=1 because every left principal ideal in A is a right principal ideal in A. Let x ∈ A → xa ∈ Aa be a map from A to Aa/aA ∩ ai, its kernel is the set {x|xa ∈ ai} = B. Therefore we get that Aa/aA ∩ ai is isomorphic to A/B. Moreover A/B = (0) if and only if a belongs to ai. Now rank of aM=number of ai such that a does not belong to ai. Since a belongs to ai, a belongs to a j for j ≤ i, therefore rank of a M ≤ n − i. On the other hand rank of aM = number of b j, such that a does not belong to b j. Since a does not belong to bi, rank aM > n − i. Hence we arrive at a contradiction. Thus ai = bi and our result is proved.

Remark. It can be shown that the six assertions (I a ) to (III b) are true if the ring A satisﬁes the conditions 1) and:

1) any ideal is principal;

2) A has no zero divisors.

The proof works exactly as in the commutative case (see Bourbaki, loc. cit.)

Obviously, the ring O˜ of the integers of any valuated non - commu- βi tative field satisfies 1)and 2).Moreover we have in this case dii = (π˜) ∗ with yi ∈ O˜ and 1 ≤ i ≤ r and dii = 0 for i > r. The diagonal n × n matrix y defined by yii = yi for 1 ≤ i ≤ r and yi = 1 for i > r is invert- 83 ible and multiplying d on the right by y−1 and q on the left by y, we get a decomposition g = p′d′q′ where p′ and q′ are invertible and d′ is a diagonal matrix whose diagonal coefficientsπ ˜ βi are positive powers of the uniformising parameter π with β1 ≤ ··· ≤ βr, and the βi are completely determined by these conditions (we used the fact that ideal in O˜ is generated by one and onlye one power of π).

e 2. Study of GLn(P) 71 e Now, let us return to the group G. For any n-tuple of rational integers, α = (α1,...,αn), let dα be the diagonal n × n matrix with diagonal coeﬃcientsπ ˜ αi and let ∆+ be the subset of the subgroup ∆ consisting of the matrices dα with α1 ≤···≤ αn.

Proposition 2. In each double coset KgK modulo K, there exists one and only one element of ∆+.

Proof. Let g = (gi j) be any element of G. Multiply g by a diagonal k matrix (aii), where aii = a , a ∈ P, v(a) > 0 and k is a suﬃciently large ′ integer so chosen that the matrix g = g(aii)belongs to K. Then by the above theorem there exist matrices p′ and q′ in K such that

′ ′ ′ + g(aii) = g = p dβq with dβ ∈ ∆

Let us take αi = βi − kv(a). Then we have g = pdαq with q, p in K and + ′ dα in ∆ . Conversely if g belongs to KdαK. then g belongs to KdβK. But dβ is unique, therefore dα is unique.

Corollary 1. K is a maximal compact subgroup of G.

If possible let H ⊃ K be a compact subgroup of G. Obviously there exists α , 0 such that dα belongs to H. Then

πα1r 0 r . (dα) =  ..  e   0 παnr     rα   If αi , 0, then v(π i → ±∞ as r → ±∞e, which is a contradiction as 84 v is a continuous function form P to R. Hence H = K. Let E be a vector space over P. Let I be a lattice in E i.e., a ﬁnitely e generated O˜ module such that its basis generate E. Since I has no tor- e sion, basis of I is a basis of E. In particular if we take E = Pn and I = On and if we identify G with the group of endomorphisms of E, e then g ∈ K and only if g(I)=I. Moreover if we take any lattice L, then e the subgroup of G which leaves L invariant is a conjugate subgroup of K. 72 4. Classical Linear Groups over p-adic Fields

Let H be a compact subgroup of G. Let e1,..., en be a basis of E. Let J be an O˜-module generated by the elements h(e j), 1 ≤ j ≤ n and h ∈ H. Evidently we have

(1) J is invariant by H

(2) J ⊃ I

(3) The map h → h(e j) is a continuous map from H to E.

But H is compact, therefore the image of H in E by the map defined in (3) is compact and hence bounded. Therefore there exists an integer k such that J ⊂ π−kI, which shows that J is finitely generated, but J ⊃ I, therefore J is generated by a finite set of element which generate E. Hence J is a lattice.e Thus H is contained in a conjugate subgroup of K namely the subgroup of G which leaves J invariant. Hence we have proved the the following.

Corollary 2. Any two maximal compact subgroups of G are conjugates and any compact subgroup of G is contained in a maximal compact subgroup of G.

85 Remark 3. Any double coset KxK, x ∈ G, is a ﬁnite union of left cosets modulo K, because K is open and compact, therefore every double coset and left coset modulo K is open and compact.

We introduce a total ordering in Zn by the lexicographic order i.e., if n α = (α1, ··· ,αn) and β = (β1, ··· , βn) are two elements of Z , then we say that β>α if βi >αi, for the least index i for which βi , αi.

n + Proposition 3. If NdβK ∩ KdαK , φ, where α · β are in Z and dα ∈ ∆ then β ≥ α and NdαK ∩ KdαK = dαK.

Proof. Let ndβ belongs to N dβ K∩ KdαK, where

1 ∗ π˜ β1 0 . . n =  ..  , dβ =  ..      0 1 0π ˜ βn          2. Study of GLn(P) 73 e Then ndβ belongs to KdαK. But ndβ belongs to KdαK if and only α α 1 n α if the invariant factors of ndβ are π,˜ ··· π˜ . Therefore we get thatπ ˜ 1 β dividesπ ˜ i for i = 1, 2, ··· , n. If α1 < β1, our assertion is proved. If α1 = β1, then we multiply the matrix ndβ on the right by a matrix δ, where α α 1 −π˜ 1 a12 ··· −π˜ 1 a1n δ = 0 1 ··· 0  0 0 ··· 1      if   α π˜ 1 a12 ··· a1n 0π ˜ β2 ···∗ nd =   β  ..   .   0 0π ˜ βn      So we get 86 π˜ α1 0 ··· 0 β α  0π ˜ 2 ∗  π˜ 1 0 ndβδ = =  ..  0 g′!  .   0 0π ˜ βn      It is obvious that δ belongs to K. Therefore ndβδ is in KdαK, which means that its invariant factors areπ ˜ α1 , ··· , π˜ αn . Thusπ ˜ α2 , ··· π˜ αn are the ′ ′ invariant factors for g , which implies that g belongs to Kn−1dα − Kn−1 with obvious notations. Our assertion is trivially true forn = 1. If we assume that it is true for all groups GLr(P˜ ) for r ≤ n − 1, we get α ≤ β. But α = β, therefore α ≤ β. We prove the second assertion also by induction on n. For n = 1, it is trivially true. Let us assume that the results is true for all groups GLr(P) for r ≤ n − 1. We have to show that −1 dα ndα belongs to K if ndα belongs to KdαK Let us suppose that

1 a12 ··· a1n 0 1 ∗ n =    ..   .  0 0 0      α Since ndα belongs to KdαKπ˜ 1 divides a1i for i = 2, ··· , n. Obvi- 74 4. Classical Linear Groups over p-adic Fields

ously 1 x12 ··· x1n −1 1 X dα ndα = 0 1 y  = ′   0 g ! 0 0 1     −α1  ′ 87 where X consists of integers xi j = π˜ ai j, and g is a (n − 1 × n − 1) −1 ′ − ′ mat rix of the form dα− n dα and the invariant factors of n dα− are α α ′ π˜ 2 , ··· , π˜ n . Therefore by induction hypothesis g belongs to Kn−1. −1 This shows that dα ndα belongs to K.

3 Study of On(ϕ, P)

In this section we shall prove some of the results of §2 for the group G = On(ϕ, P).The same results can be proved for other such groups of GLn(P) namely SLn(P) etc. with obvious modifications. Throughout our discussion P will denote a locally compact p-adic field such that K = OP|YP has characteristic different from 2.

Deﬁnition 1. Let E be a vector space of dimension n over P. A subspace F ⊂ E is called isotropic with respect to ϕ (a bilinear form as E) if there exists an element x in F such that ϕ(x, y) = 0 for every y in F, in other words the bilinear form when restricted to F is degenerate.

Deﬁnition 2. A subspace F ⊂ E is called totally isotropic with respect to ϕ if the restriction of ϕ to F is zero i.e., ϕ(x, y) = 0 for every x, y in F.

It is obvious from the deﬁnition that the set of totally isotropic subspaces of E is inductively ordered. Therefore there exist maximal totally 88 isotropic subspaces of E. They are of the same dimension, which we call the index of ϕ. If index of ϕ = 0,ϕ is called a non-isotropic form. Witt’s decomposition. Let E1, E2 and E3 be three subspaces of E such that

(1) E = E1 ⊕ E2 ⊕ E3

(2) E1 and E3 are totally isotropic. 3. Study of On(ϕ, P) 75

(3) E1 + E3 is not isotropic.

(4) E2 is orthogonal to E1 + E3 i.e., for x in E2,ϕ(x, y) = o for every y ∈ E1 + E3. It can be proved that for the vector space E = Pn, there exists a Witt decomposition and we can ﬁnd a basis e1, e2, ··· , er of E1, er+1, ··· , er+q of E2 and er+q+1, ··· , en of E3, where 2r + q = n, in such a way that ϕ(ei, e j) = δi,n+1− j for 1 ≤ i ≤ r and r + q < j ≤ n.(I) and that rr+1, ··· , er+q is an orthogonal basis for E2. Clearly the matrix of the bilinear form ϕ with respect to this basis of E is

O O S 0 0 1 Φ= OAO where S = 0 1 0 SOO 1 0 0         and A is a q × q matrix, which is the matrix of ϕ restricted to E2. We shall now completely determine the restriction of ϕ to the non - isotropic part. For simplicity we assume that r = 0 and q = n. Let e1, ··· , eq be an orthogonal basis of E. If x = (x1, ··· , xq) is a point q 2 of E with respect to these basis. Then ϕ(x, x) = ai xi with ai ∈ P. i=1 − P a j ∗ 2 If for i , j is in (P ) , then the vector (o, ··· , a j, ··· , ai, ··· , o) 89 ai is an isotropic vector of ϕ, which is not possible. Therefore ai . a j (mod P∗2), which implies that q ≤ 4. We shall say that two bilinear forms ϕ and ϕ′ are equivalent if there exists a linear isomorphism of the space of ϕ onto the space of ϕ′ and a constant c , 0, such that ϕ′ ◦ λ = ◦ϕ. Then it can be proved that every non-isotropic bilinear form over E is equivalent to one and only one of the following type:

(1) q=4

2 2 2 2 (a) x1 − Cx2 − πx3 + Cπx4

(2) q=3

2 2 2 (a) x1 − Cx2 − πx3 76 4. Classical Linear Groups over p-adic Fields

2 2 2 (b) x1 − Cx2 − Cπx3 (3) q=2

2 2 (a) x1 − Cx2 2 2 (b) x1 − πx2 2 2 (c) x1 − Cπx2 (4) q=1

2 (a) x1 (5) q=o

90 (a) The O-form as where (1, C,π, C π) is a set of representatives of P∗ modulo (P∗)2 as obtained in Corollary 2 of Hensel’s Lemma.

We shall say that a basis ei,..., en is a Witt basis for ϕ if the relations in (I) are satisﬁed and if the restriction of ϕ to E2 has one of the above forms. It is obvious that for ϕ or for a constant multiple fo ϕ, we can always ﬁnd a Witt besides and the matrix of ϕ with respect to a Witt basis is independent of the choice of the Witt basis.

′ Proposition 4. If M = Mq(P) is a matrix such that M AM belongs to ′ Mq(O) (M denotes the transpose of the matrix M and A denotes the matrix of the restriction of ϕ to E2), then M belongs to Mq(O).

Proof. We prove ﬁrst that if for x ∈ E, ϕ(x, x) is in O, then the coordinates of x are in O. Let us assume for instance that q = 4. If possible let v(x1) < 0 and v(x1) ≤ min(v(x2), v(x3), v(x4)). Suppose that 2 2 2 2 2 2 v(x1) = α. Since v(x1 − cx2 − πx3 − cπx4) ≥ 0 we have x1 − Cx2 ≡ o 2r+1 −α 2 −α 2 (mod Y ), where r = max(0,α). Therefore (π x1) − s(π x2) ≡ 0 (mod Y ). But this is impossible, because C is not a square in k. Thus our result is established. The other cases can be similarly dealt with.

′ Let M = (mi j), then M AM = (γi j) where γi j = ϕ(m1i, ··· , mqimq j), ′ If M AM belongs to Mq(O) then γii belongs to O, which implies that 3. Study of On(ϕ, P) 77 mri belongs to O for i, r = 1, 2, ··· , q. It is obvious that it is suﬃcient to assume that only the diagonal elements of M′AM are in O. In the following we shall be dealing with a ﬁxed Witt basis of the 91 space E. We shall adhere to the following notations throughout our discussion.

Ko = G ∩ K, T o = G ∩ T, No = G ∩ N, ∆o = G ∩ ∆+ and π−α1 .  ..     π−αr 0    ◦  .  dα =  1 ..   1   παr     .   ..     πα1      where α = (α1, ··· ,αr) Proposition 5. G = T oKo

Proof. We have already proved that GLn(P) = TK. Therefore g ∈ G implies that g = tk where t and K belong to T and K respectively. We know that det (g) = ±1 and det (k) belongs to O∗. So det (t) belongs to O∗. But det (t) is a power of π, therefore det (t) = 1. Now g belongs ′ ′ to G if and only if g′Φg = Φ i.e., t′Φt = k−1 ΦK−1. Since k−1 Φk−1 ′ belongs to Mn(O), t Φt belongs to Mn(O). Let us suppose that

a1 XZ t = O a2 Y    O O a2     ′ o O a1S a3 ′ ′ ′ ′ then t Φt =  o a2Aa2 X S a3 + a2AY  a′ S a Y′Aa + a SXZ′S a + Y′AY + a′ SZ.  3 1 2 3 3 3     ′ ′ O  This shows that a1S a3 and a3 and a2Aa2 belong to Mn( ). More- 92 78 4. Classical Linear Groups over p-adic Fields

over, we have 1 = det t = (det a1)(det a2)(det a3) and (det a2) and O ′ O (det a1). (det a3) belong to (for, a1S a3 belongs to Mn ( )). So det a2 ∗ belongs to O implying a2 belongs to K. By above proposition we get that the matrix a2 has coefficients from O. We shall find a matrix δ in T ∩ K such that tδ belongs to G. Then g = tK = tδδ−1K implies that δ−1K belongs to Ko and our result will be proved. Multiply the matrix t by the matrices h and h′ on the right. where b 0 0 1 ξ ζ −1 ′ h = 0 a2 0 , h = 0 1 0 0 0 1 0 0 1       −1   a1b a1b + a2 X a1bζ + z ′ we get t h h =  0 1 Y   0 0 a   3    We shall determine the matrices b,ξ and ζ in such a way that t h h′ belongs to G. Now t h h′belongs to G if and only if (t h h′)′Φ(t h h′) = Φ i.e., if and only if the following conditions are satisfied

′ ′ b a1S a3 = S (1) ′ −1 ′ ′ ′ AY + X a2 S a3 + ξ b a1S a3 = 0 (2) ′ ′ ′ ′ ′ ′ ′ a3S a1bζ + a3SZ + y AY + ζ b a1S a3 + Z sa3 = 0 (3) ′ ′ −1 Let us take b = S (a1S a3) . Then h belongs to K ∩ T and the 93 conditions (2) and (3) reduce to

′ −1 ′ ′ AY + X (a2 ) S a3 + ξ S = 0 ′ ′ ′ ′ S ζ + a3SZ + Y AY + ζ S + Z S a3 = 0

′ ′ −1 1 So if we take S ξ = −AY − X a2 S a3 and sζ = − V where V = ′ ′ ′ ′ 2 a3SZ + Y AY + Z S a3, we see that the matrix thh belongs to G. It is obvious that the matrix hh′ belongs to T ∩ K.Hence we get g = ′ ′ −1 thh .(hh ) k = t◦k◦, which proves our result completely. Deﬁnition. Let I be a lattice in E. The O module N(I) generated by the set of elements ϕ(x, y) for x, y in I is called the norm of the lattice I. 3. Study of On(ϕ, P) 79

A lattice I is called a maximal lattice if it is maximal among the lattices of norm N(I). It is easy to see that any lattice of a given norm is contained in a maximal lattice of the same norm. The lattice Io generated by the Witt basis (e1, ··· , en) of E is a maximal lattice of norm n On. Let I be a lattice of norm O containing Io. Let x = xiei be any i=1 element in I. Then ϕ(x, ei) = ±xn+1−i for 1 ≤ i ≤ r and rP+ q < i ≤ n. r+q let y= xiei, since ϕ(y, e j) is an integer for r + 1 ≤ j ≤ r + q, x j is i=r+1 an integerP for r + 1 ≤ q + r. Hence x belongs to Io. Therefore Io is a maximal lattice.

Theorem 2. Let I1 and I2 be two maximal lattices of norm O, then there exists a Witt basis ( f1, f2, ··· , fn) of E and r integers αi ≥···≥ αr ≥ 0, such that (r =index ϕ)

(1) I1 is generated by ( f1, f2,..., fn) 94

(2) I2 is generated by

−α1 −αr αr α1 π f1,..., π fr, fr+1,..., fr+q, π fr+q+1,..., π fn . Proof. We shall prove the theorem by induction on r. When r = 0, ϕ is non-isotropic and there exists only one maximal lattice of norm O which is generated by any witt basis of E. Let us assume that the theorem is true for all bilinear forms of index < r. We ﬁrst prove the following result.

If I is a maximal lattice of norm O and X is an isotropic vector in I such that π−1X does not belong to I, then there exists an isotropic vector X′ ∈ I such that ϕ(X, X′) = 1. If possible let us suppose that the result is not true. Let us assume that ϕ(X, Y) belongs to Y for every Y in I. Then ϕ(π−1X, Y) belongs to O. Consider I′ = I + Oπ−1X. It is a lattice because I′ is ﬁnitely generated O module containing I. Moreover ϕ(Y + απ−1X, Z + βπ−1X) = ϕ(Y, Z) + αϕ(π−1X, Z) + βϕ(π−1X, Y) is an integer for every α, β in O. Therefore norm of I′ is O. But this is a 80 4. Classical Linear Groups over p-adic Fields

contradiction because I is a maximal lattice of norm O. Therefore there exists a vector Y in I such that ϕ(X, Y) belongs to O∗. By multiplying Y by some inversible element of O, we get a vector Y′ in I such that ϕ(X, Y′) = 1. Let us take 1 X′ = Y′ − ϕ(Y′, Y′)X. Obviously ϕ(X, X′) = 1 and ϕ(X′, X′) = O. 2 Now we shall prove the theorem. For every isotropic vector X ∈ I1 95 (respectively I2) let t(X) (respectively u(X)) denote the smallest integer t(x) u(X) such that π X (respectively π X) belongs to I2(respectivelyI1 ). Such an integer exists. because I1 is an O-module of ﬁnite type and I2 gen- t erates E, therefore there exists an integer t such that π I1 ⊂ I2. Thus −1 t(X) ≤ t always. Let X be an isotropic vector in I1 such that π X t(X) −1 does not belong to I1. Then Y = π X belongs to I2 and π Y does −1 not belong to I2. Since π X does not belong to I1, it is obvious that ′ u(Y) = −t(X). By the above result there exists a vector X in I1 such that ϕ(X, X′) = 1 and ϕ(X′, X′) = 0. This shows that π−1X′ does not belong ′ to I1. By the deﬁnition of t(X) and t(X) we get that

′ ′ ϕ πt(X)X,πt(X )X′ = πt(X)+t(X ) ′ Since ϕ(πt(X)X,πt(X)X,πt(X )X) belongs to , we get that 0 t(X) + t(X′) ≥ 0. (1)

′ Similarly there exists an isotropic vector Y in I2 such that ϕ(Y, Y′) = 1and u(Y) + u(Y′) ≥ 0.

′ Let Z = πu(Y )Y′, then t(Z) = −u(Y′) Therefore we get t(X) + t(Z) ≤ 0 (2) 1 obviously Z is isot ropic and π Z does not belong to I1. Therefore there ′ ′ ′ ′ exists a vector Z in I1 such that ϕ(Z, Z ) = 1 and ϕ(Z , Z ) = 0 and t(Z) + t(Z′) ≥ 0 (3)

Let us suppose that the vector X is so chosen that t(X) is of maximum value, which exists because t(X) ≤ t for every X for some integer t. 3. Study of On(ϕ, P) 81

96 Therefore in particular we get t(Z′) ≤ t(X). From (2) and (3) it follows that

t(X) + t(Z) = 0 t(X) + t(Z′) = 0

α Thus we have found two vectors X and Z in I1 such that π 1 X and −α π 1 Z where α1 = t(X) belong to I2 and

ϕ(Z, X) = ϕ(π−t(Z)Y′,πt(X)Y) = 1.

Let F denote the subspace of E orthogonal to the subspace of E generated by the vectors X and Z.Obviously ϕ restricted to F is non - degenerate and its index is r − 1. Moreover I1 = OX ⊕ OZ ⊕ F ∩ I1, because for any a in I1 we have a = λX + µ Z + b, where λ and O belong to ρ and b belongs to F. But ϕ(a, X) = µ, therefore it is an integer, similarly λ is an integer. Thus b belongs to I1 and the assertion is proved. Similarly we have α −α I2 = Oπ 1 X ⊕ Oπ 1 Z ⊕ I2 ∩ F. It can be easily sen that I j ∩ F( j = 1, 2) is a maximal lattice of norm O. Hence by induction hypothesis there exists a Witt basis f2, f3, ··· , fn−1 of F and there exist r-1 integers α2 ≥ −− αr ≥ o such that

(1) f1, f2, ··· , fn−1 generate I1 ∩ F.

−α2 −αr αr α2 (2) π f2,..., π fr, fr+1,..., fr+q, π fr+q+1′ π fn−1 generate I2 ∩ F.

If we take f1 = Z, fn = X and α1 = t(X) we get a Witt basis ( f1, ··· , fn) 97 of E and r integers α1, ··· ,αr satisfying the requirements of the theorem because α2 = t( fn−1) ≤ α1. Corollary 3. The group G acts transitively on the set of lattices of norm O. The mapping g deﬁned by

γ g( fi) = π fi, where

γ = αi for 1 ≤ i ≤ r 82 4. Classical Linear Groups over p-adic Fields

= O for r + 1 ≤ i ≤ r + q = 2r + q − i + 1 for r + q + 1 ≤ i ≤ 2r + q.

leaves Φ invariant. Therefore g belongs to G. Proposition 6. In each double coset of G modulo Ko there exists one o and only one element dα of ∆+. Proof. Let g be any element of G. We shall denote by g itself the automorphism of E with respect to the initial Witt basis(e1, ··· , en). The lattice g(Io) is obviously a maximal lattice of norm O. Therefore by the above theorem we get a Witt basis ( f1, ··· , fn)of E such that

(1) Io is generated by f1, ··· , fn, γ (2) g(Io) is generated by g1, ··· , gn where gi = π fi with

γ as deﬁned in the corollary of above theorem. Let k1 (respectively k2) be the matrix with respect to the basis e1,..., en) (respectively g!, g2, ... gn) of the automorphism k1 (respectively k2) deﬁned by k1(ei) = fi (respectively k2(gi) = g(ei)) for i = 1, 2, ··· , n. We see immediately o that the matrix K1 and K2 are in K . Moreover the matrix of the o automorphism fi → gi with respect to the basis fi is dα where α = (α1,α2, ··· αr).

98 It is obvious that

g(e ) = (k ) g i 2 ji j Xj = (k ) (doα) f 2 ji k j k Xj,k = (k ) (doα) (k ) e 2 ji k j 1 lk l Xj,k,l

0 0 0 Thus we get g = k2 dα k1, which means dα belongs to K gK .The o uniqueness part of the propositional follows from the uniqueness of dα in KxK for x in GLn(P). We introduce a total ordering in Zn which is inverse of the lexicographic ordering. 4. Representations of p-adic Groups 83

r 0 0 Proposition 7. Let α and β be two elements in Z such that dα ∈△+. If 0 0 0 0 0 0 , 0 0 0 0 0 0 N dβ K ∩K dα K φ then β ≥ α. Moreover N dα K ∩ K dα K = 0 0 dα K .

0 0 0 0 0 0 ′ ′ Proof. Since N dβ K and K dα K are contained in N dβK and K dα K respectively with

′ α = (−α1, −α2,..., −αr, 0 ··· 0,αr,αr−1,...,α1) ′ β = (−β1, −β2,..., −βr, 0 ··· 0, βr, βr−1,...,β1)

′ ′ we have N dβ, K ∩ K dα, K , φ. Therefore β ≥ α for the lexicographic ordering introduced in Zn before proposition 3 in this chapter. It is obvious that β ≥ α for the new ordering of Zr. The other assertion follows trivially from the fact that

0 0 0 dαK ∩ G = dαK . 2 4 Representations of p-adic Groups

We prove here an analogue of the theorem about the representations 99 of semisimple Lie Groups in chapter I of this part. We shall give the proof of the theorem for the general linear group GLn(P) = G, though the same theorem could be proved for other classical linear groups with obvious modifications. We shall adhere to the notations adopted in the earlier chapter. Let λ denote a character of T which is trivial on N. Since △ is isomorphic to T/N, λ can be considered as a character of △. Let us λ ∗ assume that U f = 0 for every λ in △ (the group of characters of △) and f ∈ L(G) such that f , 0. We first try to find the condition under which our assumptions are valid. Let ϕ be an element of Cλ (the space of the 1 induced representation of λ). Then ϕ(tx) = (ρ(t)) 2 λ(t)ϕ(x) for x ∈ G and t ∈ T. Moreover

λ λ U f ϕ(e) = ϕ(y) f (y)dy = 0, because U f = 0 (I) ZG 84 4. Classical Linear Groups over p-adic Fields

Since the support of f is a compact set, it intersect only a ﬁnite number ofP double cosets modulo K. Let

S = S ( f ) = α|dα ∈△+, ∩K dα K , φ h X i α = α( f ) = min {β ∈ S ( f )} . β

The set S is a ﬁnite non-empty set because f , 0. Therefore α exists. For any dα in △+ the coset K dα K is a ﬁnite union of left cosets modulo K, the representatives for which could be found in T,because G = TK. Let Iα be the set of left cosets C modulo K such that K dα K = 100 C, where C = t(C)K, t(C) ∈ T. But we know that T = N△, therefore C∈I Sα t(C) = n(C) dγ(C) where n(C) and dγ(C) belong to N and △ respectively. Since n(C) dγ(C) belongs to K dα K proposition 3 implies that γ(C) ≥ α, Thus we get that K dα K = m n(C) K, γ(C) ≥ α and if γ(c) = α, C∈ I S α then C = dα K and we can take t(c) = dα. Let us assume that the right invariant Haar measure on G is such that its restriction to K is = = normalised i.e., k dk 1. Then for any left coset C t(C)K, we have R

f (g)dg = △(t(C)) f (t(C)k) dk ZG ZK and the equation (I) gives

0 = ϕ(y) f (y) dy = △(t(C)) ϕ(t(C)k) f (t(C)k)dk ZG ZK Xβ∈S CX∈Iβ

= σ(t(C)) ϕ0(k) f (t(C)k) dk Xβ XC ZK

1 with σ(t) = [δ(t)△(t)]2 and where ϕ0 denotes the restriction of ϕ to K. We have shown earlier that ϕ0(tx) = λ(t) ϕ0(x) for t ∈ T ∩K = N∩K, but λ(N) = 1, therefore the space Cλ is independent of λ. Moreover there is only one term corresponding to β = α in the summation, since λ for others γ(c) ≥ α. Separating the term for β = α we get U f ϕ(e) = 4. Representations of p-adic Groups 85

1 o σ(dα)2 λ(dα) ϕ (k) f (dα k)dk + Qγ( f,ϕ)λ(dγ) with K γ≥α R P 1 0 Qγ( f,ϕ) = σ(t(C)) 2 ϕ (k) f (t(C)k)dk (II) ZK C∈IβXγ(C)=α n It is obvious that Qγ( f,ϕ) is independent of λ. For every γ ∈ Z , the 101 ∗∗ mapping dγ ∈ △ −→ χγ ∈△ given by χγ(λ) = λ(dγ)is an isomorphism ∗ of the groups △ and △∗ . But the characters of an abelian group are linearly independent, therefore (II) gives us Qγ( f,ϕ) = 0 for every γ and in particular Qα( f,ϕ) = 0. Thus we obtain

ϕ(k) f (dαk)dk = 0, for every ϕ with ϕ(nk) = ϕ(k) for n ∈ N ∩ K. ZK (III) The equation (III) is true for left and right translations of f by elements of K because Uλ = Uλ = UλUλ = 0 and σx f εxn∗ f x f Uλ = Uλ = Uλ Uλ = 0. τx f f ∗εx f x −1 λ So if g(x) = f (k x) for k in K, we have Ug = 0. Obviously S ( f ) = ′ −1 −1 S (g) and α( f ) = α(g). Let Kα = K ∩ dαK dα and Kα = K ∩ dα K dα be two subgroups of K. Now

−1 −1 ϕ(k) f (dα(dα hdα k))dk = ϕ(dα hdα k) f (dαk)dk = 0 ZK ZK

Thus the function k → f (dα k) is orthogonal to all the functions ϕ λ in C = C and their left translates by the elements of Kα, where ϕ is invariant on the left by the elements of N ∩ K. n Lemma. For every α ∈ Z such that dα ∈△+, the subgroup Kα contains N′ ∩ K where N′ is the group consisting of the transpose of elements of N. Proof. By deﬁnition πα1 0 . dα =  ..  with α1 ≤ α2 ≤,..., ≤ αn.    0 παn      86 4. Classical Linear Groups over p-adic Fields

102 −1 αi−α j Let h = (hi j) be an element of K. Then (dα h dα )i j = π hi j which shows that the groups Kα consist of matrix h in K such that α −α ′ π i j hi j is integral. If we take h ∈ N ∩ K, obviously h belongs to ′ Kα. Thus Kα contains N ∩ K. This lemma shows that the groups Kα ′ and Kα are sufficiently big. In addition to the above assumption about f , let us further assume that f belongs to LM(G) where M is some irreducible representation of K. Clearly M is a subrepresentation of left regular representation of K in L2(K). Let E ⊂ L2(K) be an invariant subspace of the left regular representation σ of K such that σ when restricted to E is of class M. Therefore E ⊂ LM(K). Define F(k) = f (dα k). We can assume that F , 0. Since F is transformed following M¯ by the right regular representation of K, F belongs to LM(K). But F is orthogonal to all the functions ϕ in C invariant on the left by the elements of N ∩ K, the left translates of ϕ by the elements of K and the right translates of ϕ by the elements of K. Hence if M satisfies the condition (S ) i.e. The smallest subspace of E invariant by N′ and which contains elements invariant on the left by the elements of N ∩ K is E. Then F is orthogonal to LM(K), 103 because LM(K) is generated by the right translates of E. But this is a contradiction, because F ∈ LM(K). Thus we get the following Theorem 3. The representations Uλ for λ ∈△∗ form a complete system of representations of the algebra LM(G) if the irreducible representation M satisfies the condition (S ). Corollary 1. If M satisfies (S ) then M occurs atmost (dim M) times in any completely irreducible representation of G. Since Uλ for any λ in △∗ when restricted to K is a subrepresentation of the left regular representation of K, C ⊂ LM(K) which is a subspace of dimension (dim M)2, thus M is contained at most (dim M) times in Uλ. Our result follows from proposition 1.3. Corollary 2. The identity representation of K occurs at most once in any completely irreducible representation of G. This follows from Corollary 1 as the identity representation satisfies the condition (S ). 4. Representations of p-adic Groups 87

Corollary 3. If M is the identity representation of K, then the algebra LM(G) is commutative.

The algebra LM(G) has complete system of representations of dim 1. λ Therefore if x and y are any two elements of LM(G), then U (x y) = Uλ(y x) for every λ ∈ △∗, because Uλ is of dimension 1. Therefore Uλ(xy−yx) = 0 for every λ in △∗. But this is possible only if xy−yx = 0 i.e., the algebra LM(G) is commutative. Finally we try to ﬁnd out what are the various representations of K which satisfy the condition (S ). It is obvious that a representation which satisﬁes the condition (S ) when restricted to N ∩ K contains the identity 104 representation of N ∩ K. It is not known whether there exist or not representations of K which when restricted to N ∩ K contain the identity representation but which do not satisfy the condition (S ). However in this connection we have the following result.

Theorem 4. Every irreducible representation M of K which comes from a representation of GLn(O/Y ) and the restriction of which to N ∩ K contains the identity representation of N ∩ K satisﬁes the condition (S ).

It can be easily proved that GLn(O/Y ) is isomorphic to K/H, where H is a normal subgroup K consisting of the matrices (δi j + ai j) where ai j belongs to Y . Therefore a representation of GLn(O/Y ) gives rise to a representation of K.

Remark. We have proved that in the case of real or complex general linear group the representations induced by the unitary characters of T form a complete system of representations of algebra L(G). But in the case of general linear groups over p-adic ﬁelds the representations induced by the characters of △ do not form a complete system. In fact the algebra L(K) is a sub-algebra of L(G), because K is open and compact in G. Therefore if the representations Uλ form a complete system for L(G), their restrictions to K will form a complete system of representations of L(K). But the restriction of Uλ to K is a representation of K induced by the unit character of N ∩ K, therefore by Frobenius reciprocity theorem the irreducible representations of K which occur in 88 4. Classical Linear Groups over p-adic Fields

Uλ are precisely those which when restricted to N ∩ K contain the identity representation. But there exist representations of K for which this property is not satisﬁed.

5 Some Problems

105 For any classical group, we have found a maximal compact subgroup K. If G is the general linear group, it is easy to see that: (i) any maximal compact subgroup is conjugate to K by an inner automorphism; (ii) any compact subgroup is contained in a maximal compact subgroup. (For, let H be a compact subgroup of GL(n, P˜) let e1,..., en be n the canonical basis of P˜ . Let I0 be the O˜-module generated by the ei and let I be the O˜-module generated by the h ei for h ∈ H: because H is compact, the coordinates of the h.ei are bounded and there is an integer −n n ≥ 0 such that I ⊂ π˜ I0. Hence I is a lattice and H is contained in the maximal compact subgroup K1 formed by the g ∈ G such that g.I = I. −1 Moreover, if g ∈ G is such that g.Io = I, then K1 = gKg .) But for the other types of classical groups, it is not known if the results (i) and (ii) are true or not. Actually, one cannot hope that (i) is true: already in SL(n, P), we have only: (i bis) any maximal compact subgroup is conjugate to K by an (not necessarily inner) automorphism. It seems possible that there exist several but a finite number of classes of maximal compact subgroups: for instance, it seems unlikely that the maximal compact subgroup K′ of the orthogonal group 0(n,P) which leaves invariant a maximal lattice of norm P is conjugate to K. But perhaps, any maximal compact subgroup of 0(n, P) is conjugate to K or to K′. It may be noted that (i) and (ii) are not both true in the projective 106 group G = PGL(2, P): a maximal compact subgroup K is the canonical image of GL(2, 0) in G; the determinant defines a map d from G to the quotient group P∗/(P∗)n and the image of any conjugate of K is 5. Some Problems 89 contained in the image D of 0∗ in P∗/(P∗)2. Now, let u be the image 0 π of in G: we have u2 = 1 and d(u) < D. Hence, u generates a 1 0! compact subgroup which is not contained in any conjugate of K. II. It seems very likely that our results about classical groups are valid for any semi-simple algebraic linear group over P(at least if char P = 0). The general meaning of the subgroups N, D, T Γ is clear: N is a maximal unipotent, D is a maximal decomposed torus (a decomposed torus is an algebraic group isomorphic to (P∗)r), which normalised N. Then D can be written as D = △.U, where △ ≈ Zr and U ≈ (0∗)r and we have T = △.N. The subgroup Γ is the normaliser of N. It can be proved (A.Borel,unpublished) that D and N exist in any such G (at least if the base field P is perfect) and are unique, upto an inner automorphism. Now the problems are: (i) define a maximal compact subgroup K;

(ii) prove that G = T.K;

(iii) prove that G = K.△.K and deﬁne △+ (which is certainly related with the Weyl group and the Weyl chambers);

(iv) prove the key Lemma about the intersection Ndα K ∩ KdβK. For (i), the simplest idea is to take a lattice I in the vector space in which G acts, and to put K = {g|g ∈ G, g.I = I}. Then we get a compact subgroup. But it is obvious that K will be maximal and satisfy (ii) and (iii) only if I is conveniently chosen. 107 Assume that char P = 0: then we may consider the Lie algebra G of G and the adjoint representation. Then we can choose a lattice I in G such that [I, I] ⊂ I (in other words, I is a Lie algebra over 0);such a lattice always exists: take a basis G and multiply it by a suitable power of π in such a way that the constants of structure become integral. Now there exist such lattices which are maximal, because [I, I] ⊂ I implies that I is a lattice of norm ⊂ 0 for the Killing form of G . As this form is non-degenerate, it is impossible to get an indeﬁnitely growing sequence 90 4. Classical Linear Groups over p-adic Fields

of such lattices. Hence we can choose such a maximal lattice I and put K = {g|g ∈ G, g.I = I}. But let us look at the compact case: it can be shown that G is compact if any only if the Lie algebra G has no nilpotent elements. In this case, we should have K = G′. So we are led to the following conjectures:

Conjecture 1. there is a unique lattice in G which is a maximal Lie subalgebra over 0;

Conjecture 2. the set I if the X ∈ G such that the characteristic polynomial of the operator ad X has its coeﬃcients in 0, is a Lie subalgebra over 0;

Conjecture 3. (A.Weil): any algebraic simple compact group over a locally compact P-adic ﬁeld of characteristic zero is (up to ﬁnite groups) the quotient of the multiplicative group of a division algebra Q over P by its center.

108 It is easy to prove that (3) implies (2): the Lie algebra G is the quotient of the Lie algebra Q by its center and the X ∈ I are exactly the images of the integers of Q. It is obvious that (2) implies (1), because any Lie subalgebra over 0 is contained in I. Moreover, (3) is true for the classical groups: we have only for compact groups the groups ∗ PGL1(P˜) ≈ P˜ / center and the orthogonal and unitary groups for an anisotropic form; but 01 and 02 are abelian 03 gives the quaternion field, 04 is not simple, etc. But one does not know a general proof of (3). On the other hand, we can look at the “anticompact” case, that is the case of the groups defined by Chevalley in (12). Then the results (i) to (iv) can be proved (for the definition of K and proof of (ii), see Bruhat (10); for (iii) and (iv), my results are not yet published). Then if one can prove one of the above conjectures, one can hope to generalize these results to any semi-simple group by an argument by induction on the dimension of a maximal nilpotent subalgebra of G . III. Extension to the representations of K which do not satisfy the condition (S ). 5. Some Problems 91

This problem is related with the construction of other representations of G: we have seen that the representation Uλ do not form a complete system. Hence, by the Gelfand-Raikov theorem, there certainly exist other irreducible unitary representations of G. We have two indications: first the case of a real semi-simple Lie group G. It seems very likely that to any class of Cartan subgroups H of G, corresponds a series of representations of G, indexed by the characters of H. This has been verified in some particular cases (of.Harish- Chandra and Gelfand-Graev). In particular, assume that there exists a 109 compact Cartan subgroup H: then in many cases (more precisely in the cases where G/K is a bounded homogeneous domain in the sense of E. Cartan (K is a maximal compact subgroup)), we can get irreducible unitary representations of G in the following way: take a character λ of H. take the unitary induced representations Uλ in the space H λ ; this representation is not irreducible. But we have a complex-analytic structure on G/H and we can look at the subspace of H λ formed by the functions which correspond to holomorphic functions on G/H. Then we get an irreducible representation (of (22) or (21). This is in particular true for compact semi-simple Lie groups (after Borel-Well,of (32)). On the other hand, in the case of classical linear groups over a finite field, for instance for the special linear group G with 2, 3 or 4 variables, one knows all the irreducible representations of G and one sees that to each class of Cartan subgroup H, corresponds a series of representations indexed by the characters of H (of Steinberg (33)). But one does not know how exactly this correspondence works. It seems likely that the representation U(λ) associated with character λ of H is a subrepresentation of the induced representation Uλ, and it would be extremely interesting to get a “geometric” definition of U(λ). If one could get such a definition, it would perhaps be possible to generalize it to the algebraic simple linear groups (or at least to the classical groups) over a p-adic field. IV. Study of the algebra of spherical functions Let M be the unity representation of K and Let A be the algebra 110 LM(G): by our results,this is a commutative algebra. It seems possible to determine completely the structure of A. The representations Uλ likely 92 4. Classical Linear Groups over p-adic Fields give all the characters λˆ of A. The λ describe a space isomorphic to a space Cr and the map a → (λˆ(a)) is probably an isomorphism of A onto the algebra of polynomials on Cr which are invariant by the Weyl group of G. (It seems that a recent work by Satake (unpublished) gives a positive answer). V. Computation of the “characters” of the Uλ. The representations Uλ are “in general” irreducible (of (10)). More- over, if f is a continuous function on G, with carrier contained in K, and λ if f belongs to some LM(K), then it is trivial to show that the operator U f if of finite rank, and hence has a trace. The same is obviously true if f is a finite linear combination of translates of such functions. But the space of those f is exactly what I called the space of “regular” functions of G λ (space D(G)) and the map f → Tr U f is a “distribution” on G(o f (10)). A problem is to compute more or less explicitly this distribution (which is the “character” of Uλ. It seems likely that, at least on the open subset of the “regular” elements g of G it is a simple function of the proper values of g (by analogy with the case of complex or real semi-simple Lie groups, of works of Harsih-Chandra and Gelfand-Naimark). Part III

Zeta-Functions

Chapter 5 Analytic Functions over p-adic Fields

Unless otherwise stated K will denote a completed valuated ﬁeld with 111 a real valuation v. We shall adhere to the notations adopted in part I throughout our discussion.

1 Newton Polygon of a Power-Series

∞ i Deﬁnition. Let f (x) = ai x be a power-series over K. Let S be the i=0 set of points Ai = (i, v(aiP)) in the Cartesian plane. The convex hull of S together with the point y = ∞ on the ordinate axis is called the Newton Polygon of the power series f .

It is obvious that the point Ai = (i, v(ai)) lies on the line Y + v(x)X = i i v(ai x ), where v(ai x ) is the intercept cut off by the line on the Y − axis. If the series is convergent at the point x = t then intercepts cut off on the axis of Y by the lines through the points Ai with the slope -v(t) tend to infinity as i tends to infinity. Moreover it can be easily proved that if (mi) is the sequence of slopes of the sides of Newton Polygon of f , v(a ) then (m ) is monotonic increasing and − limm =− lim inf n = ρ( f ) (the i i→∞ i i→∞ n order of convergence of f ).

95 96 5. Analytic Functions over p-adic Fields

2 Zeroes of a power series

∞ i − lim inf v(ai) Let f = ai x be a power series over K. Let ρ( f ) =i→∞ . We i=0 i 112 have alreadyP proved that f is convergent for all points x in K for which v(x) > ρ( f ). Let r be a real number greater than ρ( f ). We shall try to ﬁnd the zeroes of f on the circle v(x) = r. Let us assume that a◦ , 0. (i) If there exists no side of the Newton Polygon of f with slope-r, then there exists there exists one and only term of minimum valuation i i j inf k in ai x . For, if v(x) = r and a = v(ai x ) = v(a j x ) =k v(ak x ), then allP the points Ak are above the line Y + rX = a and AiA j is a side of the Newton Polygon of slope-r. This is contrary to the hypothesis. Thus i v( f (x)) = v(ai x ) for some i and for v(x) = r, which implies that there is no zero of f on the circle v(x) = r. (ii) If there exists a side Ap Aq of slope-r, then there exist at least two terms of minimum valuation. Therefore there may to be a zero of f on the circle v(x) = r. Assume that p < q. Let v(x◦) = r for some x◦ in p q K and c = v(aq x◦ ) = v(aq x◦). Consider the power series

−1 −q i f1(y) = aq x◦ f (x◦y) = biy X Obviously v(bp) = v(bq) = 0, v(bi) ≥ 0 for i , p, q and v(y) = 0 whenever v(x) = r. Hence without loss of generality we can take r = 0, v(ap) = v(aq) = 0, v(ai) > 0 for i < o and i < p and i > q and aq = 1. Therefore

q p p q−p f (x) = x + ··· + ap x = x (x + ··· + ap) where ap , 0

p q−p The polynomials x and (x + −−− + ap) satisfy the requirements of Hensel’s lemma, therefore there exists a monic polynomial g of degree q − p and a power series h, both with coeﬃcients in O, such that

q−p p g = x + ··· + ap, h = x , f = gh

113 and the radius of convergence of h is equal to the radius of convergence q−p of f . Let us assume that g = x + ··· + g◦. Then g0 = ap , 0. Let us further assume that K is an algebraically closed ﬁeld. Then g has q − p 2. Zeroes of a power series 97 zeroes in K which belong obviously to O∗. Moreover h has no zeroes on the circle v(x) = 0. Thus f has exactly q − p zeroes on the circle v(x) = 0 where q − p is the length o the projection of the side of the Newton Polygon of f with slope 0. If λ1, λ2,...,λq−p are the zeroes of q−p f , on v(x) = 0 then f = h · (x − λi). We have also proved that if f is a i=1 Q f (x) power series and λ is its zero on a circle v(x) = r > ρ( f ), then is x − λ also a power series with the same radius of convergence. Regarding the zeroes of f inside the circle v(x) ≥ r we prove the following.

Proposition 1. The power series f has a ﬁnite number of zeroes λ1,..., λk in the disc v(x) ≥ r > ρ( f ) and there exists a power series h such that

k f (x) =i (x − λi) · h(x) with ρ( f ) = ρ(h). Y1=1

Proof. We have proved that f (x) has zeroes on the circle v(x) = ri > ρ( f ) if and only if there exists a side of the Newton Polygon of f of slope −ri. But we know that if (mi) is the sequence of slopes of sides of the lim Newton Polygon of f , then i→∞mi = −ρ( f ). Therefore there exist only a ﬁnite number of sides of the Newton Polygon of slope −r1 < −r < −ρ( f ) i.e., there exists only a ﬁnite number of r1 such that r1 > r > ρ( f ) for which there are zeroes of f (x) on v(x) = r1. Hence the theorem follows.

∞ i If f (x) = ai x is convergent in a disc v(x) > r, then we shall say 114 i=0 that f(x) is analyticP v(x) > r.

Proposition 2. If f (x) has no zeroes in the disc v(x) ≥ r > ρ( f ) in 1 particular f (0) , 0, then the power series is analytic for v(x) > r. f (x) Proof. Let us assume that f (0) = 1. Since f has no zeroes in v(x) ≥ r,there exists no side of the Newton Polygon of f of slope ≤ −r. This v(a ) implies that i ≥ −r for every i. Considering f as a formal power i 98 5. Analytic Functions over p-adic Fields

series over K we get

∞ k ∞ 1 1 = = (−1)k a xi = b x j f 1 + a xi  i  j i Xk=0 Xi=0  Xj=0 i>0   P   k   where b j = (−1) ai1 ··· aik Xk i1+i2X+···+ik= j

k k Therefore v(b j) ≥ inf v(ai1) > − ril = −r j k   i,+−−+ik= j Xl=1  Xl=1   v(b j)   ⇒ > −r. j

1 Hence ρ f ≥ r. Proposition 3. If f is an entire function(i.e., ρ( f ) = −∞) and has no zeroes, then f is a constant. ∞ i Let f (x) = i=0 ai x . As in the proof of the preceding proposition, we see that: P v(a j) ≥−r j for any r.

115 Hence, we have a j = 0 for j ≥ 1. From these propositions, we can deduce the complete structure of entire functions: Weierstrass' Theorem. Let K be an algebraically closed complete field with a real valuation v. Let f be an everywhere convergent power series over K. Then the zeroes of f different from zero form a sequence (λ1, λ2,...,λn,..., ) such that v(λn) is a decreasing sequence which tends to −∞ if the sequence (λn) is infinite and we have

∞ k x f (x) = a◦ x 1 − (1) λ ! Yi=1 i the infinite product being uniformly convergent in each bounded subset of K. Conversely for any sequence (λn) such that v(λn) is a decreasing 2. Zeroes of a power series 99 sequence tending to −∞ as n tends to infinity, the infinite product (1) is uniformly convergent in every bounded subset of K and defines an entire having zeros at the prescribed points λn. Proof. We shall prove the latter part first. Consider

i=1 N x k ϕN(x) = 1 − = akN x , λ ! YN i Xk=0 k 1 where akN = (−1) λi1 λi2 ...λik 1≤i1

∞ ∞ x k ϕ(x) = 1 − = 1 + ak x , where λ ! Yi=1 i Xk=1 1 ak = λi1 ··· λik + lt akn 1≤i1

v(ak) ρk (obviously the series giving a is convergent and ≥ ). There- 116 k k k fore the series ϕ(x)represents an entire function. We have to show that the polynomials ϕN converge to ϕ uniformly on every bounded subset of K. Given two real numbers M and A there exists an integer q such that v(a xk) ≥ M for k ≥ q, for all x with v(x) ≥ Aand for all N,because ρ kN k →∞ as k →∞. This implies that for any N k q v ϕ (x) − a xk ≥ M for v(x) ≥ A. (2)  N kN   Xk=0    Similarly we get  

q v ϕ(x) − a xk ≥ M for v(x) ≥ A. (3)  kN   Xk=0      100 5. Analytic Functions over p-adic Fields

Since akN → aK as N tends to inﬁnity, combining (2) and (3) we get v(ϕ(x) − ϕN (k)) ≥ M for N suﬃciently large. It can be easily proved that the λi are the only zeroes of the function ϕ(x).

Let us denote by f1 the product given by (1). Take a disc v(x) ≥ r. In this disc f (x) has only a finite number of zeroes. Let the zeroes of f in v(x) ≥ r be 0(k times) and λ1, λ2,...λp. Then p x f (x) = xk 1 − g(x) λ ! Yi=1 1 1 where g(x) has no zeroes in the disc v(x) ≥ r. Therefore is analytic in g f1 g1 ∞ x 1 the disc v(x) > r. Consider = = 1 − , where g1 is f g i=p+1 λi g Q f analytic and has no zeroes in the disc v(x) > r. Therefore 1 is analytic f f in the disc v(x) > r and has no zeroes in it. Since it is true for every r, 1 f is a constant function. Hence our theorem is proved. Form the proposition 2, we can derive some properties of the meromorphic functions: ∞ i 117 Definition. A power series ϕ = ai x over a field K is said to be a i=−m meromorphic function in a disc v(Px) ≥ r if and only if there exist two f functions f and g analytic in the same disc such that ϕ = . g In any disc v(x) ≥ r′ > r, g has a finite number of zeroes, therefore g = Pg′ where P is a polynomial and g′ has no zeroes v(x) ≥ r′ which 1 ′ means that g′ is analytic in v(x) > r . Therefore we can can write ϕ = f ′ 1 , where f ′ = f is a convergent power series in v(x) > r′. P g′

3 Criterion for the Rationality of power-series

∞ k Let F be any ﬁeld and f = k=0 ak x an element in F [x] . It can be easily proved that f is a rationalP function if and only if there exists a 3. Criterion for the Rationality of power-series 101

ﬁnite sequences (qi)◦≤i≤h of elements of F at least one of which is non- zero and an integer k such that

anqh + an+1qh−1 + ··· + an+hqo = 0 h+1 for all integers n such that n + h > k. Let us denote by An the determinant of the matrix (an+i+ j)0≤i, j≤h.

Lemma 1. The power series f is a rational function if and only if there h+1 exists integer h and n◦ such that An = 0 for all v > n◦. Proof. It is obvious that the condition is necessary. We shall prove that the condition is sufficient by induction on h. When h = 0, we have an = 0 for n sufficiently large. Therefore f is actually a polynomial. h+1 Let us assume that An = 0 for n > n◦. Moreover we may assume h h that An , 0 for infinitely many n, because if An = 0 for n large then by h+1 induction hypothesis we get that f is a rational function. Since An = 0 118 2 h h h h , for n > n◦, An An+2 = An+1 . So it follows that An 0 for n > n◦. Consider the following system of linear equations

Er = an◦+r x1 + an◦+1+r x2 + ··· + an◦+h+r xh+1 = 0 for r = 0, 1, 2,...

For any q ≥ n◦ the system qof the h if h equations Eq, Eq+1,... Eq+h−1 h is of rank h (because Aq ,P0). So has a unique solution upto a constant ′ factor. But the system q of the h + 1 equations Eq,..., Eq+h is also of h , h+1 ′ rank h (because Aq+1 P0 and Aq = 0) and therefore q and q+1 on ′ the hand and qand q + 1 on the other hand have theP same solution.P

Thus any solutionP of Pq is a solution of q+1 and any solution of no is a solution of Eq for qP≥ n◦. Thus we haveP found a ﬁnite sequenceP (xi) such that an◦+r x1 + ··· + an◦+h+r xh+1 = 0 for r ≥ 0. Hence f is a rational function.

i Theorem 1. Let f (x) = ai x be a formal power series with coeﬃcients in Z. Let R and r be twoP real numbers such that (1) Rr > 1

(2) f considered as a power series over the ﬁeld of complex numbers is 119 holomorphic in the disc |x| < R. 102 5. Analytic Functions over p-adic Fields

(3) f considered as a power series over Ωp(the complete algebraic clo- ′ ′ sure of Qp) is meromorphic in the disc |x| ≤ r with r > r. (where ||p) is the absolute value associated to vp). Then f is a rational function.

Proof. We can assume that R ≤ 1, because R > 1 implies that f is a polynomial and we have nothing to prove. Moreover r > 1, because ′ Rr > 1. Since f is meromorphic in |x|p ≤ r , there exist two functions g g and h analytic in |x|p ≤ r such that f = h . If necessary by multiplying f by a suitable power of x we can assume that f has no pole at x = 0 and hence that h is polynomial with h(0) = 1. Let

∞ k i i g gi x and h = hi x Xi=0 Xi=0

gn+k = anhk + an+1hk−1 + ··· an+k−1h1 + an+k (1) By Cauchy’s inequality we obtain the following

−s (1) |as|≤ MR

−s (2) |gs|≤ Nr

−s By taking R and r smaller if necessary we assume that |as| ≤ R and −s |gs|p ≤ r for s > s◦. Let

an an+1 ··· an+k an+m a a a a Am+1 = n+1 n+2 n+k+1 n+m+1 n ··· ··· ··· ···

a a a ··· a n+m n+m+1 n+m+k n+2m

where m > k. 120 The equation (1) gives

an an+1 ··· an+k−1 gn+k gn+m a a a a ··· Am+1 = n+1 n+2 n+k n+k+1 n ··· ··· ··· ··· ···

an+m an+m+1 an+m+k−2 gn+m−k r2+2m

4. Elementary Functions 103

Obviously for n > so we have m+1 −(n+2m) m+1 |An |≤ (m + 1)!(R ) m+1 −n m−k+1 and |An |p ≤ (r ) m+1 because |an|p ≤ 1 for every n. If An , 0, then m+1 m+1 −2m(m+1) kn −n(m+1) m+1 −k −n 1 ≤ |An ||An |≤ (m + 1)!R r [Rr] = k1[(R r) r ] m+1 −k Let m be so chosen that (Rr) r > 1. Then there exists an integer n◦ such that for n > n◦ m+1 m+1 |An ||An | < 1. m+1 This is a contradiction. Therefore A = 0 for n > n◦. Hence f is a rational function.

Corollary. If f is a power series over Z such that f has a non-zero radius of convergence considered as series over the complex number ﬁeld is meromorphic in Ωp, then f is a rational function.

4 Elementary Functions

We consider the convergence of the exponential logarithmic and binom- 121 inal series in this section. We assume that the ﬁeld K is of characteristic 0 and the real valuation v on Q induces a p-adic valuation. ∞ xn The exponential series e(x) = . Converges in the disc v(x) > n=0 n! 1 P and in the domain of convergence v(e(x) − 1) = v(x). Let n = p − 1 r r r+1 a◦ + a1 p + ··· + ar p where p ≤ n ≤ p and 0 ≤ ai ≤ p − 1. One can easily prove that n n n − S v(n!) = + −−− + = n " p# " pr # p − 1 r where S n = ai Therefore i=0 P v( 1 ) −1 S ∴ n! = + n n p − 1 n(p − 1) 104 5. Analytic Functions over p-adic Fields

1 S log n v n! −1 But n ≤ + 1 . Therefore lim = . Hence the p − 1 log p ! n→∞ n p − 1 1 v(xn) series e(x) converges for v(x) > . If v(x) = 1 , then = p − 1 p−1 n! 1 < ∞ whenever n is a power of p. Thus the series does not con- p − 1 1 verge for v(x) = . The latter part of the assertion is trivial. We p − 1 see immediately that e(x + y) = e(x).e(y) and e(x) has no zeroes in the domain of convergence. ∞ yk We deﬁne log (1 + y) = (−1)k−1 as a formal power series over k=1 k 122 K. We shall show that the seriesP log(1 + y) converges for v(y) > 0 and 1 v(log(1 + y)) = v(y) for v(y) > we have p − 1

(−1)nyn v = nv(y) − v(n) n ! log n (−1)nyn But v(n) ≤ therefore v tends to inﬁnity as n →∞ when- log p n ever v(y) > 0. On the other hand v(n) = 0 if (n, p) = 1, therefore the 1 series is not convergent for v(y) ≤ 0. For n > 1 and v(y) > , it can p − 1 (−1)n−1yn easily proved that v > v(y), which proves our last assertion. n 1 Moreover for v(x) > we have the equalities p − 1

e(log(1 + x)) = 1 + x (1) log(e(x)) = x (2)

Let 1 G = x|x ∈ K, v(x) > ( p − 1) 1 G = x + 1|x ∈ K, v(x) > ( p − 1) 4. Elementary Functions 105

∗ be subgroups of K+ (the additive group of K) and K respectively. The mapping x → e(x) is an isomorphism of G onto G′, the inverse of which is the mapping 1 + x → log(1 + x). In fact the mapping 1 + y → log(1+7) is a homomorphism of the group 1+YΩ(Ω begin the complete algebraic closure of K) into the subgroup of Ω+, where v(y) > 0.It is not 1 an isomorphism because it ζ is a p−th root of unity, then v(ζ−1) = p − 1 and log ζ = 0. ∞ We deﬁne (1 + Y)Z = h(m, Z)Ym = e(Z log(1 + Y)) where m=0 Z(Z − 1) ··· (Z − m P+ 1) h(m, Z) = as a formal power series in the vari- m! ables Y and Z over K. Since h(m,Z) is a polynomial in Z, we can substitute for Z any element of K to get a power series in the one variable Y.

Proposition 4. For any element t in K the power function (1+x)t deﬁned 123 1 1 above is analytic for v(x) > (respectively for v(x) > −v(t)+ ) p − 1 p − 1 t if v(t) ≥ 0 (respectively if v(t) < 0) Moreover if t belongs Zp, then (1+ x) is analytic for v(x) > 0. Proof. When v(t) < 0 m − 1 v(h(m, t)) = m(v(t)) − v(m!) ≥ mv(t) − p − 1 v(h(m, t)) 1 Therefore lim inf = v(t) − Hence (1 + x)t is analytic m→∞ m p − 1 1 in v(x) > − v(t). Similarly one can prove the convergence when p − 1 v(t) ≥ 0.

Let t be in Zp. Then h(m, t) is a p-adic integer. Suppose that v(m!) + 1 = α, then there exists an element km in Z such that k t ≡ km (mod p ) Therefore

t(t − 1) ... (t − m + 1) ≡ km(km − 1) 106 5. Analytic Functions over p-adic Fields

or h(m, t) ≡ h(km, m) (mod p).

But h(km, m) is a rational integer, therefore v(h(m, t)) ≥ 0. From this our assertion follows easily.

5 An Auxiliary Function

Throughout our discussion Fq shall denote a finite field consisting of q elements. Let us consider the infinite product

m m−1 T P − T T p − T p m m F(Y, T) = (1 + Y)T (1 + Y P) P (1 + Y P ) p (1)

The product is well deﬁned as formal power series in two variables Y and T over Q. Clearly (1) is convergent in Q Y, T . Expressing F(Y, T) as a power series over Q T and Q Y we obtain ∞ m F(Y, T) = Bm(T)Y , d(Bm(T)) ≤ m mX=0 ∞ m = αm(Y)T , mX=0

124 where αm(Y) is a power series, the terms being of degree ≥ m.

Lemma 2. The coeﬃcients of F(T, Y) are p-adic integers.

Lemma 3. If F is an element of Q [Y, Z] such that F(0, 0) = 1, then F (F(Y, Z))p ﬃ belongs to Zp [Y, Z] if only if the coe cients of p p are in pZp F(Y , Z ) excepts for the ﬁrst.

i j Proof of Lemma 3. Let us suppose that F(Y, Z) = 1 − ai jY Z , then i+ j>0 P (F(Y, Z))p G = = F xF where F(Y p, Z p) 1 2 5. An Auxiliary Function 107

r i j p r i j F1 = 1 − p ai jy Z + ··· + (r )(−1)  ai jY Z  i+Xj>0 i+Xj>0     p  p i j + ··· + (−1)  ai jY Z  . i+Xj>0    ∞ k  pi p j F2 = 1 +  ai jY Z  Xk=1 i+Xj>0      i j If G = 1 + bi jY Z , then i+ j>0 P bi j = −pai j+ (terms of the form pX polynomials in a with rational integers coeﬃcients with

∞

r + s < i + j) + ai1 j1 ... aik jk Xk=1 X ′ i1 + ··· + ik = i ′ j1 + ··· + jk = j

ir + jr > 0 ∞ +(−1)p ap j a j ··· a j i + ··· + i = i′ i1 1 i2 2 ik k 1 k Xk=1 ′ j1 + ··· + jk = j

ir + jr > 0 where the last two sums appear only if i and j are divisible by p and in 125 this case pi′ = i, p j′ = j.

Assume that bi j belongs to pZp for i + j > 0. We shall prove that ai j are in Zp by induction. Obviously a00 is in Zp. Assume that ars ∈ Zp for r + s < i + j; then in the formula giving bi j all the terms except p perhaps −pai j. But a − a belongs to pZp if a belongs to Zp, therefore pai j belongs to pZp and ai j belongs Zp. The other part of the assertion is trivial 108 5. Analytic Functions over p-adic Fields

Proof of Lemma 2.

− T pm − T pm 1 ∞ pT pm pm−1 p (1 + Y) (1 + Y ) (F(Y, T)) m=1 = Q − F(Y p, T p) T pm − T pm 1 ∞ m−1 (1 + Y p)T p (1 + Y pm ) p m=2 Q (1 + Y)p T = "(1 + Y p)# ∞ T = + k a p bkY   Xk=1      where bk are p-adic integers. Moreover

∞ T ∞ ∞ m + k = m k 1 p bkY  h(m, T)p  bkY   Xk+1  mX=0 Xk=1          v(pm) m − 1 F(Y, T) p But ≥ m − > 0, therefore − 1 has its m! p − 1 F(Y p, T p) coefficients in pZp. Thus by lemma (3) the coefficients of F(Y, T) are p-adic integers. 126 One deduces from lemma (2) that F(y, t) is analytic for v(t) ≥ 0 and v(y) > 0, because if v(t) ≥ 0, then v(Bm(t)) ≥ 0 because Bm(t) is a ∞ m polynomial with coefficients from Zp. Therefore the series Bm(t)y m=0 converges for v(y) > 0. P

6 Factorisation of additive characters of a Finite Fields

Ps Let Rs = x | x ∈ Ωp = Ω, x = x . We have the canonical map from R2 to Fps namely the restriction on the canonical homomorphism of OΩ 6. Factorisation of additive characters of a Finite Fields 109 onto kΩ. In order t prove that this map is bijective, it is sufficient to prove s that is surjective ; because both Rs and Fps have p elements. Ifx ¯ , 0 is ps−1 ps−1 in Fps , thenx ¯ −1 = 0 andx ¯ is a simple root of the polynomial X − 1. Therefore by Hensel’s lemma there exists an element α belonging to ps−1 Ω such thatα ¯ = x¯ and α − 1 = 0, which proves that α is in R2 and the mapping is onto. Infact the canonical homomorphism of OΩ onto ∞ kΩ when restricted to R = R2 is an isomorphism onto kΩ. Finally s=1 Hensel’s lemma shows that RS1 is contained in Qp. Let Us = Qp(Rs). Clearly Us is a Galois extension of Qp and the Galois group is cyclic generated by the automorphism σ : ρ → ρp, s where ρ is a primitive p − 1 th root of unity. Moreover Us is an unramified extension of Qp, because [Us; Qp] = [Fps ; Fp]. If we take ∞ U = Us, then the completion of U is the maximum unramified ex- s=1 tensionS of Qp in Ω and σ is called the Frobenius automorphism of U. If ′ t is an elements is R2, then

′ ′ s−1 Tr t′ = t′ + t p + ··· + t p Us / Qp

Tr t′ belongs to Zp. Thus the function (1 + Y) is analytic for v(y) > 0. 127 ′ Let t be the representative of t ∈ Fp2 in R2. If y belongs y belongs Tr t′ to YΩ then (1 + y) belongs to Ω. We shall choose y in such a way Tr t′ that mapping t → (1 + y) is a character of the additive group of Fps . Obviously for any u and v in Fps we have

′ ′ ′ (u + v) ≡ u + v (mod YΩ) ′ ′ ′ ′ Tr(u + v ) ≡ Tr u + tr v (mod YΩ) ′ ′ ≡ Tr u + Tr v (mod pZp) because Tr u′ is a p-adic integer. Therefore

′ ′ ′ (1 + y)tr(u+v) = (1 + y)Tr u (1 + y)Tr v (1 + y)a,

p where a belongs to pZp. Let us take 1 + y = ζ where ζ = 1 and ζ , 1. ′ It follows that (1 + y)a = 1. Thus the mapping u → ζTr u is a character a of FPs . We shall show that it is a non -trivial character. Firstly, ζ = 1 if 110 5. Analytic Functions over p-adic Fields

and only if a belongs to pZp proved that a already belongs to Zp. For by 1 choice of y we have v(y) = > 0 and p − 1

ζa = (1 + y)a = 1 + ay + ··· + h(m, a)ym + ···

Since a is p-adic integer, v(h(m, a) ≥ 0 and hence v(h(m, a)ym ≥ 2 2 2 for m ≥ 2, (a+y)a , 1 if v(ay) < . Therefore v(ay) ≥ , p − 1 p − 1 p − 1 1 which implies that v(a) ≥ < 0, thus a belongs to pZ . But the p − 1 p ′ canonical image of Tr u in Fp is the trace of u as an element of Fps ′ ′ over Fp, therefore there exists as least one u such that Tr u is not in Tr u′ pZp. Hence the mapping u → ζ is a non-trivial character of Fps . By deﬁnition of the product F(Y, T) we have

− u′pm − u′pm 1 ′ m m F(y, u′) = (1 + y)u ··· (1 + yp ) p + − ′ m u′pm 1 −u p ′ m m F(y, u′p) = (1 + y)u p ··· (1 + yp ) −p + − + − u′pm s 1 − u′pm s 2 s−1 ′pg−1 m m F(y, u′p ) = (1 + y)u ··· (1 + yp ) p

s 128 Since u′p = u′, by multiplying these identities we get

s−1 r ′ F(y, u′p ) = (1 + y)Tr u Yr=0

′ s−1 Thus ζTr u = ϕ(u′pk ) where ϕ(T) = F(ζ − 1, T), is the splitting of k=0 additive charactersQ of Fps which we shall require later. Chapter 6 Zeta-functions

It is well known that the Riemann zeta function ζ(s) = (1 − p−s)−1, 129 p where p runs over all prime numbers, is absolutely convergentQ for Re s >. We can generalise this definition for any commutative ring with unit element . In the case of ring of integers p is nothing but the gener- ating element of the maximal ideal (p) and it is also equal to the number of elements in the field Z/(p). Motivated by this we define for any commutative ring A with identity −s −1 ζA(s) = (1 − N(M) ) (I) YM where M runs over the set of all maximal ideals of A and N(M) is the number of elements in the field A/M. But in general N(M) is not finite and even if N(M) is finite the produce (I) is not convergent, therefore we have to put some more restrictions on the ring. In the following we shall prove that if A is finitely generated over Z i.e., if there exist a finite number of elements x1,..., xk in A such that the homomorphism from

Z X1,..., Xk to A which sends Xi to xi is surjective, then N(M) is finite and the infinite product (I) is absolutely convergent fot Re s > dim A, where the dimension of A is defined as follows. Definition. If A is an integral domain, the dimension of A is the transcendence degree (respectively transcendence degree +1) of the quotient

111 112 6. Zeta-functions

ﬁeld of A over Z/(p) (respectively Q) if characteristic of A is p (respec- 130 tively 0). In the general case dim A is the supremum of the dimension of the rings A/Y where Y is any minimal prime ideal.

It can be proved that dimension of A is equal to the supremum of the lengths of strict maximal chains of prime ideals. Before proving the convergence of the zeta function we give some examples of ﬁnitely generated rings of over Z.

1. The ring Z is ﬁnitely generated over itself.

2. Any ﬁnite ﬁeld Fq.

3. The ring of polynomials in a ﬁnite number of variables over Fq i,e., the ring Fq[X1,..., Xk]

4. The ring Fq[X1,..., Xr]/U , where U is any prime ideal of Fq[X1, ..., Xr]. This is the set of regular functions defined over Fq on the variety V defined by the ideal U affine space.

5. Let K be any algebraic number ﬁeld. The ring of integers A in K is ﬁnitely generated over Z.

6. Let V be an affine variety defined over the algebraic number field K and let O ⊂ K[X1,..., Xr] be the ideal of V. Then the ring of regular functions on V i.e., K[X1,..., Xr]/O is not finitely generated over Z. But the ideal O is generated by the ideal O0 = O ∩ A[X1,..., Xr] of the ring A[X1,..., Xr] and we can associate to V the quotient ring A[X1,..., Xr]/O which is obviously finitely generated over Z. It is to be noted that this ring is not intrinsic and depends on the choice of the coordinates in Kr

2 Fields of ﬁnite type over Z

We shall require the following lemma in the course of our discussion. 131 Normalisation lemma of Noether. Let K be a field. Let R and S be subrings of K containing a unit elements such that S is finitely generated 2. Fields of finite type over Z 113 over R. Then there exists an elements a , 0 in R and a finite number pf element X1,..., Xr in S such that

1. X1,..., Xr are algebraically independent over the quotient ﬁelds of R.

−1 2. Any elements of S is integer over R[a , X1,..., Xr].

Proposition 1. Let K be a field. Let R be a subring of K and L the quotient field of R. If K as a ring is finitely generated over R, then (K : L) is finite and there exists an element a in R such that L = R[a−1].

We first prove the following: If a field K is integral over a subring R then R is a field. Let x be any element of R, then x−1 belongs to K and therefore satisfies an equation

n n−1 X + a1X + ··· + an = 0, ai ∈ R

This implies that x−1 is a polynomial in x over R. But R[x] = R, therefore x−1 belongs to R. Hence R. Hence R is a field Proof of proposition 1. Since K is finitely generated over R, by the normalisation lemma, there exists an element a , 0 in R and a finite family (x1,..., xr) in K algebraically independent over L such that K −1 is integral over R[a , x1,..., xr]. By the remark above it follows that −1 R[a , x1,..., xr] is a field. But x1,..., xr are algebraically independent over L, therefore r = 0 and L = R[a−1]. Since K is finitely generated and integral over L, (K : L) is finite.

Proposition 2. If a commutative ring A is ﬁnitely generated over Z, then WN(M) is ﬁnite for any maximal ideal M of A.

Proof. Since A is finitely generated over Z, the field K = A/M is finitely 132 generated over Z. If characteristic of K is zero then K contains Z. There- fore by proposition (1) Q = Z(a−1) for some a , 0 and a in Z, which is impossible. Thus characteristic of K is p and by proposition (1) K is a finite extension of Fp, hence K is a finite field. 114 6. Zeta-functions

3 Convergence of the product

Proposition 3. The infinite product ζA(s) is a absolutely convergent for Re s > dim A and uniformly convergent for Re s > dim A + ε for every ε> 0. Proof. We shall prove the result by induction on r = dim A. If r = 0 then A is a finite field. Let us assume that A = Fq. Then 1 ζ (s) = A 1 − q−s is a meromorphic function in the plane with a simple pole at s = 0. Let us assume that the result is true for all those rings which are finitely generated over Z and dimension of which are less than r. Before proving the result for rings of dimension r we prove the following result.

Let A be a ﬁnitely generated ring over Z and B = A[X], the ring of polynomials in one variable over A, then ζB(s) = ζA(s − 1) in a suitable domain of convergence. In fact if ζA(s) is convergent for Re s > x, then ζB(s) is convergent for Re s > x + 1. If dim A = 0, then A = Fq for some q and B = Fq[X]. Since he maximal ideals in B are generated by irreducible polynomials, which can be assumed to be monic, we get

sd(p) −1 ζB(s) = (1 − q ) YP

133 where P runs over the set of monic irreducible polynomials over A. In order to prove the absolute convergence of ζB(s), it is suﬃcient to prove the convergence of the inﬁnite series

σ S = q−d(P) where s = σ + it XP

Since the number of monic polynomials of degree r is qr, we have

∞ S = |q−d(P)|σ ≤ qr|q−r|σ XP Xr=1 3. Convergence of the product 115

∞ = q(1 − σ)r Xr=1 Obviously the series S is convergent if 1 − σ< 0 i.e., σ> 1. More- over in this domain 1 ζ (s) = (Q a monic polynomial inB) B qsd(Q) XQ ∞ ∞ qk 1 1 = = = qsk qk(s−1) 1 − q1−s Xk=0 Xk=0 Hence ζB(s) = ζA(s − 1). Now let the dimension of A be arbitrary and B = A[X]. We shall denote by Spm(B) the set of maximal ideals of B. For any M in Spm(B), M ∩ A is in Spm(A), because A/M ∩ A, being a subring of the finite field B/M, is a field. Let π denote the mapping M ∈ Spm(B) −→ M ∩ A ∈ Spm(A). It can be easily proved that the set π−1N and Spm(A/N[X]) are isomorphic, where N is any maximal ideal of A. Therefore

−s −1 ζB(s) = [1 − (N(M)) ] M∈YSpm(B) = (1 − N(M)−s)−1 N∈Spm(Y A) M∈Yπ−1(N) (s) = ζA/M[X] N∈Spm(Y A) 134 N But A/ is a ﬁnite ﬁeld, therefore ζA/N[X](s) = ζA/N(s − 1) So we get

ζB(s) = (ζA/N(s − 1)) N∈Spm(Y A) = (1 − N(N)1−s)−1 N∈Spm(Y A) 116 6. Zeta-functions

= ζA(s − 1). 1 It follows that ζ (s) ,..., = and ζ ,..., = ζ (k − s) Fq [X1 Xk] 1 − qk−s Z[X1 XK] Z where ζZ is nothing but the Riemann zeta function. Now we shall prove our main proposition. Assume that A is an integral domain. Let K be the quotient field and R the prime ring of A. Since A is finitely generated over R, by the normalisation lemma we have the following: (1) If characteristic A = p , 0, then there exist r elements x1, x2,..., 135 xr in A such that A is integral over R[x1,..., xr], where x1,..., xr are algebraically independent over R = Fp. (ii)- If characteristic A = 0, then there exits an element a in R = Z and r − 1 elements x1,..., xr−1 −1 in A such that every element of A is integral over Z[a , x1,... xr−1] and the elements x1,..., xr−1 are algebraically independent over Q. We get r elements in the first case and r − 1 elements in the second case because r is the dimension of A which is equal to the transcendence degree of K over Fp or transcendence degree of K over Q + 1 according as the characteristic of A is non-zero or not. It can be proved that A (re- ′ −1 spectively A = A(a )) is a finite module over B = Fp[x1,..., xr] ( re- ′ −1 spectively B = Z[a , x1,..., xr−1]) and the mapping π from Spm(A) → Spm(B) (respectively from Spm(A′) → Spm(B′)) is onto. Let A (respectively A′) be generated by k elements as a B (respectively B′) module. We shall prove that π−1(N) for any N in Spm(B) ( respectively in Spm(B′)) has at most k elements. Let C = A/AN. It is an algebra of rank t ≤ k over B/N. Since π−1(M) is isomorphic to Spm(A/AN) it is sufficient to prove that C has at most k maximal ideals. This will follow from the following.

Lemma II. Let A be any commutative ring with identity and (Ui1≤i≤m a ﬁnite set of prime ideals in A such that

A = Ui + U j for i , j

m Then the mapping θ : A → P =i AUi is surjective i=1 Q 3. Convergence of the product 117

m Proof. It is suﬃcient to prove that 1 = ai where ai belongs to U j for 136 i=1 j , i because if (t1,..., tm) is any elementP of P, then m ′ ′ θ ti ai = (t1,..., tm), where ti is a representative of ti in A. i=1 ! P If m = 2, the result is obvious i.e., 1 = a1 + a2 where a1 is in O2 and a2 is in O1. Let us assume that it is true for less than m ideals. m−1 Then 1 = vi where vi ∈ O j for 1 ≤ j ≤ m − 1 and j , i. Since i=1 A = Oi + Om, weP have 1 = xi + yi for 1 ≤ i ≤ m − 1 with xi ∈ Om and m−1 m−1 Yi ∈ Oi. Clearly xivi + viyi = 1. i=1 i=1 P P m−1 Let us take ui = vi xi for i ≤ i ≤ m − 1 and um = yivi, then i 1 m P ui = 1 and ui ∈ O j for j , i. i=1 P

Let M1, M2,..., Mt be any ﬁnite set of distinct maximal ideals of C. t t Then by lemma (1) C/i Mi is isomorphic to ⊕ C/Mi(⊕ indicates the i=1 i=1 direct sum). Thus t ≤ k.T Assume that the characteristic of A is 0. Let M be any maximal ideals of A. If a does not belong to M, then MA[a−1] is a maximal ideal in A[a−1], because A[a−1]/MA[a−1] is isomorphic to A/M. If a belongs to M, then M contains one and only one prime Pi occurring in the unique factorisation of a and the set of maximal ideals which contains pi. is α1 αt isomorphic to Spm(A/piA). Therefore if a = p1 ,..., pt , then

t ζA(s) = ζA[a−1](s) ζ (s) Yi=1 A/piA

But dim A/piA < dim A, therefore inorder to prove the convergence 137 of ζA(s) it is suﬃcient to consider ζA[a−1](s). We have

M −s −1 ζA[a−1](s) = (1 − (N ) ) N Spm(YB′) N∈Y−1π(N) 118 6. Zeta-functions

Since N(M) ≥ N(M), we get

−σ −σ |NM| ≤ k |NN| ≤ kζz(r − σ − 1) N∈Spm(X B) N∈πX−1(M) N∈Spm(X B′)

Therefore ζA[a−1](s) is convergent for ℜs > dim A. If characteristic A = p, then we get −σ N −σ |NM| ≤ k |N | ≤ kζFp (r − s) N∈Spm(X B) M∈Xπ−1(N) N∈Spm(X B′) which gives the same result as above, Now we have to prove our theorem in the general case(A is not an integral domain). But we shall prove in the next § a more general result.

4 Zeta Function of a Prescheme

Let A be a commutative ring with unity. We shall denote by Sp (A) the set of all prime ideals of A. On Sp(A) we deﬁne a topology by classifying the sets F(O) as closed sets, where

F(O) = {Y |Y ⊃ O, Y ∈ Sp (A)}.

and O is any ideal in A. This topology is referred to as the Jacobson Zariski topology. It is obvious that in this topology a point is closed if and only if it is a maximal ideal of A. We associate with every point Y of Sp(A) a local ring A namely the ring of quotient of A with respect to 138 the multiplicatively closed set A − Y . On O the sum of all these local rings we deﬁne a sheaf structure by giving“suﬃciently many” sections. For any a, b, ∈, A we consider the open subset

V(b) = {Y |Y ∈ Sp(A), Y = b}. a a For any Y ∈ v(b), the, fraction , is an element of AY . Then b Y b a the mapping Y → gives a section S (a, b) of O. The pair (X, O) bY together with the sheaf of local rings O is called an aﬃne scheme, where X = Sp(A). 4. Zeta Function of a Prescheme 119

Definition. Let (X, O) be a ringed space. We say that X is a prescheme if every point has an open neighbourhood which is isomorphic as a ringed space to Sp(A) for some ring A. Such a neighbourhood is called an affine neighbourhood. We shall assume that the pre-scheme X satisfies the ascending chain condition for open sets, then X is quasi-compact and it can be written as the union of a finite number of affine open sets Xi. We shall denote by Ai the ring such that Xi is isomorphic to Sp(Ai). Then the ring Ai is Noetherian and has a finite number of minimal prime ideals Yi j. Each prime ideal of Ai contains a Yi j and Xi = Sp(Ai) is the union of the si j = Sp(Ai j) (with Ai j) = Ai/Yi j) , each S i j being a closed subset of Xi and the Ai j being integral domains. Moreover the residue field of the local ring associated to a point x ∈ S i j is the same for the sheaf of the scheme X and for the sheaf of the scheme Sp(Ai j) We define the dimension of X as the maximum of the dimensions of the rings Ai(or of the rings Ai j). It can be proved that if X is irreducible (i.e. if X cannot be represented as union of two proper closed subsets). then Ai = dim A j for i , j. A prescheme S is a finite type over Z if there exists a decomposition of S into a union of a finite number of open affine sets Xi such that each 139 Ai, the ring associated to Xi, is finitely generated over Z. It can be proved that the same is true for any decomposition into a finite number of affine open sets. In particular, a ring A is finitely generated over Z if and only if the scheme Sp(A) is of finite tyte over Z and an open prescheme of S is also of finite type over Z. Let S be a prescheme of finite type over Z. A point x ∈ S is closed if and only if the residue field of the local ring of x is finite (we shall denote by N(x)) the number of elements of this field). In particular, if ′ S = UXi , then a point x ∈ Xi is closed in S if and only if it is closed in Xi Now we define the ζ-function of S by: −s −1 ζS (s) = (1 − (N(x)) ) Y where x runs over the set of closed points of S . It is clear that if S = S P(A), then ζS = ζA. As above, we can write S as a union of a finite number of subsets S i, each S i being affine open subset, with 120 6. Zeta-functions

S i = Sp(Ai), where Ai is an integral domain ﬁnitely generated over Z. Then it is obvious that:

ζS i ζsi∩S j∩S k ··· ) π ! i< j

ζS i∩S j ··· i< j ! Q Now we shall prove the following generalisation of the Theorem 1 bis:- The ζ function of a prescheme S of ﬁnite type over Z is convergent for Re s > dim S . Of course, theorem 1 bis implies theorem 1. Assume we have proved the theorem 1 bis for prescheme of dimension < dim S . Then we get as in the preceding § the convergence of ζA for any integral domain A ﬁnitely generated over Z of dimension ≤ dim S , and in particular the

140 convergence of the ZS i . After (I), we have just to prove this: if U (resp. F) is an open (resp. closed) subset of X = Sp(A) (with dim A ≤ dim S ), then ζU∩F is convergent for Re(s) > dim S . But let G = X − U; we have: ζ∪∩F = ζF/ζF∩G and F ∩ G is closed in X. Hence we have just to prove the convergence of ζF. But F is deﬁned by an ideal of A and F = Sp(A/O) and ζF = ζA/O . If O = {0}, we have ζF = ζA and if O , {0} then the minimal prime ideals of A/O give non trivial prime ideals of A and we have dim A/O < dim S : the induction hypothesis ensures the convergence of ζF. Hence we have completely proved the theorems 1 and 1 bis

5 Zeta Function of a Prescheme over FP

Let S be a prescheme over Z of ﬁnite type. We have a canonical map from a prescheme S to Sp(Z) given by π(x) = characteristics of the residue ﬁeld of local ring of x for any x in S . Suppose that π(x) = p for every x in S . In this case each Ai is of characteristic p and the canonical map from Z into Ai can be factored through FP. In this case we say that the prescheme S is over FP. 6. Zeta Function of a Prescheme over Fq 121

Let S be a prescheme of ﬁnite type over FP. Then the residue ﬁeld k(x) of the local ring associated to a closed point x is of characteristic P for every x in S . Therefore k(x) = FPd(x) where d(x) is a strictly positive integer. thus −sd(x)−1 ζS (s) = 1 − p xY=x∈S Let us take t = p−s. Then

d(x)−1 ζS (s) = 1 − t = ζS (t). xY=x∈S

The function ζ˜s(t) is also called a zeta function on S . It is absolutely 141 convergent in the disc |t| < P− dim(s). we have

∞ ∞ kd(x) h ζ˜s(t) = t = aht xY=x∈S Xk=0 Xh=0 with a0 = 1 and an ∈ Z. The end of these lectures will be devoted to the proof of the following theorem (Dwork’s theorem):

Theorem 1. The function ζ˜S (t) of a prescheme S of ﬁnite type over FP is a rational function of t.

6 Zeta Function of a Prescheme over Fq

In order to prove Dwork’s theorem it is sufficient to prove it for an affine scheme and open sets of an affine scheme because of the equation (1). Then we have to look at thezeta function of a ring A finitely generated over FP. Such a ring can be considered as the quotient of FP[X1,..., Xk] by some ideal O and we can associate to A the variety V defined by O in k K where K is the algebraic closure of FP. It may be noted that V is not necessary irreducible. We shall call ζA the zeta function of the variety V. f More generally we consider a variety V over Fq, where q = p . The variety V is completely determined by the ring 122 6. Zeta-functions

A = Fq[X1,... Xn]/O ∩ Fq[X1,..., Xn] where O is an ideal in K[X1, ..., Xn] generated by O0 = Fq[X1,..., Xn] ∩ O K being the algebraic closure of Fq. We deﬁne

ζv = ζA and ζ˜v = ζ˜A.

142 For every maximal ideal M of Fq[X1,... Xn] there exists a maxi- ′ mal ideal M in K[X1,..., Xn] such that Fq[X1,..., Xn] ∩ M = M. But n Spm(K[X1,....Xn]) is isomorphic to K , therefore a maximal ideal M of n Fq[X1,..., Xn] is determined by one point x of K . Moreover this point x belongs to V if and only if M ⊃ O However this correspondence be- n tween the maximal ideals of Fq[X] and the points of K is not one-one. So we want to ﬁnd the condition when two points x and y of Kn correspond to the same maximal ideal of Fq[X1,..., Xn] = Fq[X]. Let Mx and My be the maximal ideals of K[X] corresponding to x = (x1,..., xn) and y = (y1,..., yn) respectively such that Mx ∩ Fq[x] = My ∩ Fq[x]. It is obvious that Fq[X]/Mx ∩ Fq[x] = F[x]/My ∩ Fq[x] is isomorphic to Fq[x1,..., xn] = Fq f for some f > 0. We shall show that the necessary and suﬃcient condition that Mx ∩ Fq[X] = My ∩ Fq[X] is that there exists an element σ in G(Fq f /Fq) such that σ(x) = y. For n = 1 the existence of σ is trivial. Let us assume that there exists a σ in G(Fq f /Fq) such that σ(xi) = yi for i = 1, 2,..., r − 1 for ≤ n. Let σ(x j) = z j for j ≥ r. Let P(x) be the polynomial of zr over Fq(y1,..., yr−1). Then P(y1,..., yr−1zr) = 0, which gives on applying σ the equation P(x1,..., xr−1, yr) = 0. There- fore P is in My ∩ Fq[X] i.e., P(y1,... yr−1, yr) = 0. Thus yr and zr are conjugate over Fq(y1,..., yr−1). Let τ be the automorphism of K over Fq(y1,..., yr−1) such that τ(ar) = yr. Then τoσ is an element of G(Fq f /Fq) such that τoσ(xi) = yi for i = 1, 2,..., r. Our result fol- 143 lows by induction. The converse is trivial. Hence we see that if M is a f maximal ideal of Fq[X] containing O with N(M) = q , then there exist n exactly f points conjugate over Fq, in K ∩ V and f = (Fq(x) : Fq) if n and only if f is the smallest integer such that x belongs to (Fq f ) . Let

n N f = number of points in V ∩ (Fq f ) n n J f = number of points in V ∩ (Fq f f ) − U (V ∩ (Fq f ) ) q f ′< f 6. Zeta Function of a Prescheme over Fq 123

f I f = number of maximal ideals of A of norm q .

We have proved that J f = f I f . By deﬁnition of the ζ- function of V we have

−s −1 ζV (s) = ζA(s) = (1 − (nM) ) M∈YSpm(A) = (1 − q−s f (M))−1 M∈YSpm(A) where f (M) is deﬁned by the equation N(M) = q f (M) So we see that we can substitute t = q−s in the zeta function (and not only t = p−s as in the general case) and get a new zeta function.

∞ f (M) −1 f −I f ζV (s) = (1 − t ) = (1 − t ) = ζ˜v,q(t) M∈YSpm(A) Yf =1 Therefore ∞ f Log ζ˜v,q(t) = −I f log(1 − t ) Xf =1 ∞ ∞ tk f = I f k Xf =q Xk=1 J tk f = f f k Xf Xk ∞ tn = J  f  n Xn=1 Xf /n    ∞    tn  = N n n Xn=1 144 ∞ tn Thus ζ˜v,q(t) exp Nn , where Nn is the number of points of V n=1 n ! n P in Fq. We have already seen that this is a power series with integral coeﬃcients. 124 6. Zeta-functions

′ Theorem 1 . ζ˜V,q(t) is a rational function of t.

We shall show that in order to prove the rationality of ζÃ(t) where −s −s t = q , it is sufficient to prove the rationality of ζÃ(t) where t = p . Since ζ˜v,q(t) and ζ˜v(t) are both convergent in a neighbourhood of the origin, we have f f ζ˜v,q(t ) = ζ˜v(t) with q = p . Let µ be any f -th root of unity. Then

f f f ζ˜v(µt) = ζ˜v,q(µ t ) = ζ˜v,q(t ) = ζ˜v(t)

If we have n k bkt ζ˜ (t) = k=0 v n k Pk=0 Ckt then also P

n k k µ( bkµ t ζ˜ (t) = k=0 v n k k Pµ Pk=0 Ckµ t n k k P Pbk( µ )t = k=0 µ n k k Pk=0 Ck(Pµ µ )t k f P ≤ ≤ Pbk f t = 0 k [n/ f ] k f P0≤k≤[n/ f ] Ck f t P 145 because µk = 0 if k . 0 (mod f ) µ ThusP we get

k f bk f t 0≤k≤[n/ f ] ζ˜ (t) = P = ζ˜ (t f ) v k f v,q Ck f t 0≤k≤[n/ f ] P k bk f t 0≤k≤[n/ f ] i.e., ζ˜ (t) = P V,q k Ck f t 0≤k≤[n/ f ] P

Hence ζV,q˜(t) is a rational function of t. 7. Reduction to a Hyper-Surface 125

7 Reduction to a Hyper-Surface

We shall show that to prove our theorem it is sufficient to consider the zeta function of a hypersurface V defined by a polynomial P(X1,..., Xn) r in FP[X1,..., Xn]. We know that we can write V = Vi where each Vi i=1 T i is a hyper surface. Let E be any subset of {1, 2...., r} and VE = Vi. iǫE T Let NV (respectively NVE ) be the number of points of V(respectively VE) in any field FPn . We now prove that

1 NV = (−1) + n(E)NVE (I) XE where n(E) is the number of elements in E. Let any point x in V belong to k hypersurface Vi where 1 ≤ k ≤ r. Then x appears l times in the right hand side of equation (I), where

r−k k r−k k r−k s+1 I = C0 C1 − C0 C2 + C1 + ··· + (−1) k r−k k k r−k Cs + C1 Cs−1 + ··· + Ch Cs−h + ··· + ··· ∞ ∞ = r − k (−1)h+t−1Ck Ct  h Xt=0 Xh=1  ∞   t−1 r−k  = (−1) Ct Xt=0 146 Thus I = 0 or 1 according as r < k or r = k. Hence the equality (I) is established. This proves that

˜ ˜ (−1)1+n(E) ζV (t) = [ζVE (t)] (2) YE This proves that it is enough to prove theorem 1 for a hypersurface. Let V be a hypersurface deﬁned by the polynomial P(X1, X2,... Xn) in FP[X1,... Xn]. Let B be any subset of {1, 2,..., n}. Let

WB = {x|x ∈ V, xi = 0 for i not in B} 126 6. Zeta-functions

= | ∈ = UB x x WB, xi 0  Yi∈B    It is obvious that V is union of disjoint subsets WB − UB where B runs over all the subsets of {1, 2,..., n}. Hence the zeta function of V is the product of the zeta functions of the varieties (WB − UB) and the theorem 1 will be a consequences of the following lemma.

Lemma 2. Let P be a polynomial in FP[X1,..., Xn]. then the zeta func- n tion of the open subset deﬁned by xi = 0 in the hyper surface W i=1 deﬁned by P is a rational function. Q

8 Computation of Nr

147 We shall adhere to the following notation throughout our discussion.

x = (x1...., xn+1), xi ∈ FPr .

α = (α1,...,αn+1),αi ∈ Z. α α1 αn+1 x = x1 ··· xn+1 |α| = α1 + α2 + ··· + αn+1.

Let X be any additive character of Fpr . Then we have

X (UP(X1,..., Xn)) = 0 if P(x1,..., xn) , 0

UX∈FPr r = p if P(x1,..., xn) = 0

Therefore

r X (UP(x1,..., Xn)) = p Nr ∈ ∗ U∈F r x1XFpr Xp r r n where p Nr = (p − 1) + n + 1X (xn+1P(x1,..., xn)) ∈ ∗ x X(FPr ) 8. Computation of Nr 127

Let α xn+1P(x1,..., xn) = aα x Xα where only a ﬁnite number of aα are nonzero. Then

α X (xn+1P) = X (aα x ). Yα r r n α Therefore p Nr = (p − 1) + X (aα,..., x ) ∈ X∗ n+1 Yα x (Fpr )

r−1 k We take the character X deﬁned by X (t) = ϕ(t′p ). where 148 k=0 ′ ′ Q t ∈ Rr such that t = t and ϕ(y) = F(ζ − 1, y), ζ being a primitive p-th root of unity. Thus from equation (1) we get

r−1 r r n α pk p Nr = (p − 1) + ϕ(bαξ ) ∈ X∗ n+1 Yα Yk=0 x (FPr ) R∗ R where ξi = xi,ξi ∈ r , bα = aα and bα belongs to 1. Let

r−1 α pk G(§) = ϕ(bα§ ) and Gr(§) = G(ξ ). Y kY0 Then r r n p Nr = (p − 1) + Gr(ξ) ∗ n+1 ξ∈(XRr ) We have already proved that G(ξ) is analytic for ξ integral. There- fore α Gr(ξ) = grαξ α∈XZn+1 Then

r r n p Nr = (p − 1) + G(ξ) ∗ n+1 ξ∈(XRr ) r n α = (p − 1) + grα ξ n+1 ∗ n+1 α∈XZ ξ∈(XRr ) 128 6. Zeta-functions

n+1 = (pr − 1)n + g ξαi rα  i  Xn+1 Yi=1 Xi α∈Z     αi . r But ξi = 0 if αi 0 (mod p − 1) i P r r = p − 1 if αi ≡ 0 (mod p − 1)

Therefore

r r n r n+1 p Nr = (p − 1) + grα (p − 1) α=X(pr−1) r n r n+1 = (p − 1) + gpr α−α(p − 1) (I) Xα 149

9 Trace and Determinant of certain Inﬁnite Matri- ces

Let K be any ﬁeld and A = K [X1,..., Xn+1] be the ring of formal α power series in n + 1 variables over K. Let H = hαX by any element α of A. We deﬁne an operator TH on A as follows P

′ ′ ′ TH(H ) = HH for every H inA.

For any integer r we deﬁne an operator λr Such that

λ a Xα = a Xα. r  α  rα Xα Xα     It can be easily proved that these two operators are continuous for the topology given by the valuation on A deﬁned earlier. Let us set ΓH,r = λr ◦ TH. It is obvious that the monomials constitute a topological basis of A and the operator ΓH,r has a matrix (γαβ) with respect to this ′ ′ basis, where γαβ = hrα−β. It is trivial to see that THH = TH ◦ TH for any 9. Trace and Determinant of certain Inﬁnite Matrices 129

′ ′ ′ two elements H and H of A and λrr′ = λr ◦ λr for any two integers r and r′. Moreover we have

s Γ = S ◦ S −1 H,r λr TH·H(r)····Hr where H(r)(X) = H(Xr). In order to prove the above identity it is suﬃcient to prove that the action of the two sides is the same on the monomials. We have β TH0λr(X ) = 0 if β is not a multiple of r

β β TH ◦ λr(X ) = TH(X r ) if β is a multiple of r

α+ β rα + β = hαX r = hαX X X r rα + β with the convention that coeﬃcient of X r = 0 if r does not divide 150 β. Therefore

T ◦ λ (Xβ) = λ h Xrα Xβ) H r r  α  Xα    (r)  = λr ◦ TH Thus

2 ΓH,r = λr ◦ TH ◦ λr ◦ TH

= λr ◦ λr ◦ TH(r) ◦ TH

= λr2 ◦ TH,H(r)

Let us assume that we have proved that

s Γ = S ◦ s−1 H,r λr TH◦H(r)◦...◦H(r ) Then

s+1 s rS −1 ◦ ΓH,r =ΓH,r ◦ ΓH,r = λrS TH◦H(2) ···◦ H ◦ λr TH

= λrS TH◦TH(2) TH(rS −1) ◦ λr ◦ TH 130 6. Zeta-functions

= λrS +1 TH ◦ TH(r) ◦···◦ TH(rs)

s We see immediately that ΓH,r is an operator of the same type as ΓH,r s s ′ s ′ (r) (rS −1) namely ΓH,r =ΓH′ ◦ r′ where r = r and H = HH ... H .

151 Lemma 3. Let us assume that K =Ω the complete algebraic closure of f QP and r = p . Let us further assume that the coeﬃcients hα tend to 0 s s as |α| tends to inﬁnity. Then the series Tr(ΓH,r) = (ΓH,r)αα giving the α s α P trace of ΓH,r with respect to the basis (X ) is convergent and we have

1 S −1 Tr(Γs ) = = H(ξ) ...... H ξr H,r (rs − 1)n+1 RX∗ n+1 ξ∈( f s)

β Proof. For any monomial X in K [X1,..., Xn+1] β α+β ΓH,r(X ) = λr ◦ hαX Xα α+β = hαr X Xα

Therefore the matrix of the operator ΓH,r with respect to the basis β (X ) is (γαβ) with γαβ = hrα−β and Tr(ΓH,r) = hrα−α. But hα tends to α 0 as | α | tends to inﬁnity, therefore the series P hrα−α is convergent in α K. We have already proved that P

n+1 H(ρ) = (r − 1) hrα−α Xr−1 Xα≥0 ρ=1

Therefore 1 T (Γ ) = H(ρ) r H,r (r − 1)n+1 ρrX−1=1

s Hence our lemma is proved for s = 1 for s > 1, ΓH,r is of the same type as ΓH,r.Thus our lemma is completely established. 10. Meromorphic character of ξV (t) in Ω 131

s s n s n+1 s 152 Corollary. p Ns = (p − 1) + (p − 1) Tr Γ where Γ=ΓG,p we have already proved that

s−1 s s n pk p Ns = (p − 1) + G(ξ ) ∗ n+1 ξ∈(XRs) Yk=0 Therefore the corollary follows from the lemma.

10 Meromorphic character of ξV(t) in Ω

We have seen that

∞ ts ζ˜ (t) = exp N , where V  s s  Xs=1    n  n+1  n n−i s(i−1) n + 1 n+1−i s(i−1) s Ns = (−1) p + (−1) p Tr Γ i! i ! Xi=0 Xi=0 Therefore ∞ ∞ N ts n n (pi−1t)s s = (−1)n−i s i! s Xs=1 Xs=1 Xi=0 ∞ + n 1 n + 1 (pi−1t)s = (−1)n+1−i Tr Γs i ! s Xs=1 Xi=0 ∞ ∞ N ts n (pi−1t)s n + 1 exp s = exp (−1)n+1−i  s   " s # i ! Xs=1  Yi=0 Xs=1        n+1 ∞  (−1)n+1−i    (pi−1t)s  n + 1 = exp Tr Γs " s # i ! Yi=0 Xs=1 n n+1 n−i n n−i n + 1 = (1 − pi−1t)(−1) ) ∆(pi−1t)(−1) ) i! i ! Yi=0 Yi=0 ∞ ts where ∆(t) = exp − Tr Γs s=1 s ! P 132 6. Zeta-functions

So in order to prove that ζ˜v(t) is meromorphic in Ω, it is sufficient to 153 prove that ∆(t) is every where convergent in Ω. If Γ were a finite matrix, then its trace is well defined. If the order of the matrix is N, then

N s s Tr Γ = λi are the eigen values of Γ. Xi=1 Moreover

∞ N ts N ∆(t) = exp − λs = (1 − tλ )  s i  i  Xi=1 Xs=1  Yi=1   = det(I − tΓ) 

∞ m If Γ is an inﬁnite matrix, we deﬁne det(I − tΓ) = dmt , where m=0 P d = (−1)m ε γ γ ...γ i m σ i1 iσ(1) im σ(m) 1≤Xi1<

εσ being the signature of any permutation σ in sm. Then for Γ=ΓG,p we get

m dm = (−1) , εσγα1ασ(1) ...γαmασ(m)αi being distinct. Xαi 1X≤i≤m σX∈sm

Let us assume that there exists a constant M such that v(gα) ≥ M | α |. Then

v(γαβ) = v(gpα−β) ≥ M | pα − β | ≥ M(p | α | − | β |)

We consider one term of the series giving dm

m m = v  γα j γσ( j) v(γα j ασ( j)) Yj=1  Xj=1     10. Meromorphic character of ξV (t) in Ω 133

≥ M p | α j | − | ασ( j) | Xj

≥ M(p − 1) α j Xj 154 Now there exist only a ﬁnite number of indices αi such that their length | α | is less than some constant, therefore the series dm converges. m Moreover we get v(dm) ≥ M(p − 1) inf α j where inﬁmum is taken j=1 P m over all the sequence α1,...,αm. Let ρm = inf | α j |. Now let us j=1 n+1 P order the sequence of indices α ∈ Z+ in such a way that | αi |<| αi+1 |, m then we have ρm = | αi | and we see immediately that i=1 P m ρm lim = αi = ∞ m→∞ m Xi=1 m

v(d ) Therefore m tends to inﬁnity as m tends to inﬁnity. Hence we get m the following lemma.

α Lemma 4. If an element G = gαX satisﬁes the condition n+1 α∈PZ+

then the series det(I − tΓ) with Γ=ΓG is well deﬁned as an element of Ω [t.] and is an every where convergent power series in Ω. It is evident from the above discussion that if we prove that

α (i) The function G deﬁned by = πϕ(aαξ ) satisﬁes the condition (C) 155

∞ ts Tr Γs (ii) The formal power series exp − and det(I − tΓ) are s=1 s ! identical. P 134 6. Zeta-functions

Then ∆(t) is every where convergent in Ω which implies that ζ˜v(t) is meromorphic in Ω. We have already proved the result (ii) when Γ is a ﬁnite matrix. Let Γh denote the matrix of ﬁrst h rows and columns of Γ. Then ∞ tsTrΓs det(I − tΓ ) = exp − h h s Xs=1 ∞ h m = dmt mX=0

where dh = (−1)m γ ...γ being an element of S . m i1iσ(1) imiσ(m) m ≤i1≤i2<...

h m dm − dm = (−1) γα1 ασ(1) ...γαmασ(m) α1X,...,αm σX∈S m m − (−1) γα1 ασ(1) ...γαmασ(∗) αXi≤h σX∈S m

h 156 Obviously v(dm −dm) tends to inﬁnity as h tends to inﬁnity. Similarly one can prove that

v Tr Γs − Tr Γs = v g ... g h  pα1−α2 pαs−α1  αX1...αs      − g ... g  pα1−α2 pαs−α1  α ...αX≤h   1 s    tends to infinity as h tends to infinity. In order to prove that the α function G satisfies (1) it is sufficient to prove that each term ϕ(aαξ ) of 10. Meromorphic character of ξV (t) in Ω 135 the product satisfies (1). We have

ϕ(t) = F(ζ − 1, t)

∞ m m But F(Y, t) = Am(Y)t with Am(Y) = Y Bm(Y) and Bm(Y) belongs m=o to O [Y] . ThereforeP

α α m α m ϕ(aαξ ) = (ζ − 1) Bm(ζ − 1)(aαξ ) mX=0 α β = hβξ . Xβ=0

Thus hβ = 0 if β , αm

β β α α = (ζ − 1) Bβ/α (ζ − 1)aα . which shows that |β| 1 1 1 v(hβ) ≥ = |β| |α| p − 1 p − 1 |α|!

β/α because Bβ/α (ζ − 1)aα is of positive valuation. Hence G satisﬁes (I). We have proved that ζ˜v(t) is convergent in a disc |t| <δ< 1 as 157 a series of complex numbers and is meromorphic in the whole of Ω, therefore by the Criterion of rationality proved earlier we obtain that ∼ ζv(t) is a rational function of t. Hence the lemma 2 of §7 is completely proved and also the theorem 1.

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