Arithmetic of Quadratic Forms (Cambridge Tracts in Mathematics)

CAMBRIDGE TRACTS IN MATHEMATICS General Editors B. BOLLOBAS, P. SARNAK, C. T. C. WALL 106 Arithmetic of quadratic for...

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CAMBRIDGE TRACTS IN MATHEMATICS

General Editors B. BOLLOBAS, P. SARNAK, C. T. C. WALL

106

Arithmetic of quadratic forms

YOSHIYUKI KITAOKA Department of Mathematics, Nagoya University

Arithmetic of quadratic forms

AMBRIDGE

UNIVERSITY PRESS

PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE

The Pitt Building, Trumpington Street, Cambridge CB2 1RP, United Kingdom CAMBRIDGE UNIVERSITY PRESS

The Edinburgh Building, Cambridge CB2 2RU, UK http://www.cup.cam.ac.uk 40 West 20th Street, New York, NY 10011-4211, USA http://www.cup.org 10 Stamford Road, Oakleigh, Melbourne 3166, Australia © Cambridge University Press 1993

This book is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1993 First paperback edition (with corrections) 1999

Typeset in Computer Modern 10/12pt, in AMS-T

[EPCJ

A catalogue record for this book is available from the British Library ISBN 0 521 40475 4 hardback

ISBN 0 521 64996 X paperback Transferred to digital printing 2003

Contents

Preface

vii

Notation

ix

Chapter 1 General theory of quadratic forms 1.1 Symmetric bilinear forms 1.2 Quadratic forms 1.3 Quadratic forms over finite fields 1.4 The Clifford algebra 1.5 Quaternion algebras 1.6 The spinor norm 1.7 Scalar extensions

1 1

3 12

20 24 29 32

Chapter 2 Positive definite quadratic forms over R 2.1 Reduction theory 2.2 An estimate of Hermite's constant

33 33 42

Chapter 3 Quadratic forms over local fields 3.1 p-adic numbers 3.2 The quadratic residue symbol 3.3 The Hilbert symbol 3.4 The Hasse invariant 3.5 Classification of quadratic spaces over p-adic number fields

47 47 54 56 60

Chapter 4 Quadratic forms over Q 4.1 Quadratic forms over Q

64 64

52

Contents

vi

Chapter 5 Quadratic forms over the p-adic integer ring 5.1 Dual lattices 5.2 Maximal and modular lattices 5.3 Jordan decompositions 5.4 Extension theorems 5.5 The spinor norm 5.6 Local densities Chapter 6 Quadratic forms over Z 6.1 Fundamentals 6.2 Approximation theorems 6.3 Genus, spinor genus and class 6.4 Representation of codimension 1 6.5 Representation of codimension 2 6.6 Representation of codimension > 3 6.7 Orthogonal decomposition 6.8

The Minkowski-Siegel formula

70 70 71 79 86 92 94 129 129 134 147 151 157 164 169 173

Chapter 7 Some functorial properties of positive definite quadratic forms 189 7.1 Positive lattices of E-type 190 7.2 A fundamental lemma 199 7.3 Weighted graphs 217 7.4 The tensor product of positive lattices 222 7.5 Scalar extension of positive lattices 239 Notes References

250

Index

269

263

Preface

The purpose of this book is to introduce the reader to the arithmetic of quadratic forms. Quadratic forms in this book are mainly considered over the rational number field or the ring of rational integers and their completions. It is of course possible to discuss quadratic forms over more general number theoretic fields and rings. But it is not hard to understand the general theory once readers have learned the simpler theory in this book. This book is self-contained except for the following: The theory of quadratic fields is used to prove the first approximation theorem in Chapter 6. In Chapter 7, the relative theory is treated and so the use of algebraic number fields is inevitable. Dirichlet's theorem on prime numbers in arithmetic progressions is also used without proof. Several exercises are given. The answers are known anyway. Some of them are easy and some of them are not. Some problems are also given in the Notes. Problems mean questions for which I do not know the answer and want to know it. They are very subjective and arbitrary. I do not say that they have citizenship in the world of quadratic forms. Inheriting pioneering works of Eisenstein, Smith, Minkowski, Hardy and others, Hasse established the so-called Minkowski-Hasse principle on quadratic forms and Siegel gave the so-called Siegel formula and other great works. Some of them were formulated from the view point of algebraic groups, and stimulated and developed them. One of the achievements of Eichler was to introduce the important concept of spinor genus. Kneser showed that group-theoretic consideration was fruitful. Their method gave a unified perspective to sporadic results. Now there seem to be no famous

viii

Preface

outstanding problems. But it does not mean that there remains nothing to search out. The important view point is again to look at quadratic forms as diophantine equations. Although it may be hard, it must be interesting and deep.

It must be emphasized that this book does not cover even all of the arithmetic part of quadratic forms, of course, and I did not touch Witt rings, hermitian forms, analytic theory, etc. At first I intended to give a brief introduction to analytic theory, in particular the theory of modular forms, but I gave up as the book was already too long. I say only that analytic theory is a fertile part of quadratic forms. I wish to thank Mr. D. Tranah, who gave me the chance to write this book, and Professor M. Kneser who gave useful suggestions and finally I am cordially grateful to Professor J. S. Hsia who examined in full detail the first version of the manuscript mathematically and linguistically.

Notation

Z and Q denote the ring of integers and the field of rational numbers, respectively, Zp, Qp are their p-adic completion, and 1R is the field of real numbers. A:= B means that A is defined by B and sometimes B by A.

For any two sets X, Y, X C Y denotes the inclusion, so X = Y is possible.

X\Y denotes Ix EX For a set Z, #Z denotes the cardinality of Z. Almost all elements of a set X means all but a finite number of elements of X. For a commutative ring R, M,n,n(R) is the set of m x n matrices with entries in R, and we put Mm(R) := Mm,m(R). lm denotes the unit matrix of degree m. GLm(R) is the group of all invertible elements of Mm(R). For X E Mm,n, tX means the transposed matrix of X, and for matrices X, Y, we put Y[X] := tYXY if it is defined. An) means For square matrices Al, , An, diag(Al, )

If the product x o y is defined for x E X, y E Y, then S o T denotes

{sotI sES,teT} for SCX,TCY.

x

Notation

Let F be a field and R a ring such that R generates F, and V be a vector space over F and M be a finitely generated submodule in V over R. A submodule N over R of M is called primitive if N = FN fl M. A matrix X E Mm,,,,n(R) is said to be primitive if X can be extended to a matrix in GL,,,,(R) or GLn(R), that is every elementary divisor of X is in R" if R is a principal ideal domain.

1

General theory of quadratic forms

Throughout this book, rings R are commutative and contain the unity 1, and modules M over R are finitely generated with 1m = m for m E M. In this chapter, we give some basics about quadratic forms which include the so-called Witt theorem, Clifford algebra and quaternion algebra. Besides them, the theory of quadratic forms over finite fields is outlined. It is useful for readers to get used to how to deal with quadratic forms and also to their applications.

1.1 Symmetric bilinear forms Let M be a module over a ring R and b a mapping from M x M to R satisfying the conditions b(x, y) = b(y, x) for x, y E M b(rx + sy, z) = rb(x, z) + sb(y, z) for r, s E R, x, y, z E M. We call b a symmetric bilinear form and the pair (M, b) or simply M a symmetric bilinear module over R. When R is a field, we often use "space" instead of "module". (1) (2)

If M is free and {v}1 is a basis of M, then we write

M=(A) for A = (b(v2,vj)). For another basis {uz} 1, there is a matrix T = (t2j) such that (u1, , u,,,) = (v1, , vn)T, t2j E R, det T E R", and

2


1

(b(ui, uj)) = (b(Ek tkivk, Eh thjvh)) _ (>k,h tkib(vk, vh)thj) = tTAT = A[T] holds and so we have M = (A) = (A[T]). Thus det A(R" )2 is independent of the choice of a basis and is uniquely determined by M, and we denote it by

dM and call it the discriminant. (It is defined only for free modules.)

For a non-zero symmetric bilinear space U over a field F, we call U regular if d U 54 0, and it is easy to see that

U is regular if b(x, U) = 0, then x = 0 HomF(U, F) = {y H b(x, y) I x E U} e=> for a basis {ui} of U, there is a subset {vi} of U such that b(ui, vj) _ Si,j (Kronecker's delta).

(We use "regular" for spaces only!) For a subset S of a symmetric bilinear module M over a ring R, we put

S'={xEMjb(x, s)=0 for sES}. For symmetric bilinear modules M, M1,

,

M,,,, over a ring R such that

M = Ml ® ®M., b(Mi, Mj) = 0 if i $ j, we call M the orthogonal sum of M1,

,and , Mn

write

M=MI When M has a basis {vi} such that M = Rv1 L

1 Rvn, {vi} is

called an orthogonal basis of M. By the above notation, M =Li (ai) with ai = b(vi, vi). For a symmetric bilinear space U = U1 L L Urn over a field F, d U = d U1 ... d U,,,, is clear and so U is regular if and only if every Ui is regular.

Proposition 1.1.1. Let U be a symmetric bilinear space over a field F and V a subspace of U. If V is regular, then U = V L V1 holds. Proof. Since V is regular, we have V fl V1 = {x E V I b(x, V) = 0} = {0} and every linear mapping f E HomF (V, F) is given by x --> b(x, y) for y E V. For u E U, x --> b(x, u) is a linear mapping from V to F. Therefore,

there is an element y E V such that b(x, u) = b(x, y) for x E V, which implies u - y E This means U = V + V1 and then U = V ® V1 = V1.

VLV1.

1.2

Quadratic forms

3

Proposition 1.1.2. Let U be a symmetric bilinear space over a field F and V a subspace of U. Then V fl U' _ {O} if and only if HomF(V, F) _ {x H b(x, y) y E U}, and then dim V1 = dim U - dim V. Proof.

Suppose v fl Ul = {0}. Take a subspace W(D V) such that

U = W ® Ul. If w E W satisfies b(w, W) = {0}, then b(w, U) = {0} and hence w E U1 fl w = {0}. Thus W is regular, and for f E HomF(V, F) we extend it to HomF(W, F) by f (WO) = 0 with W = Wo ® V. Because of regularity of W, there is an element y in W so that f (x) = b(x, y) for x in W and especially in V. If, conversely v fl Ul {0}, then we take a basis {v2} of V such that v1 E V fl U1. A linear mapping f defined by f (v1) = 1, f (vi) = 0 (i > 2) is not of form f (x) = b(x, y), since b(v1, U) = 0. Thus the former part has been proved. Suppose V fl U1 = {0} and for a basis {vi}m IT, of V we can take a subset {u2}m 1 of U such that b(vi, uj) = Sid for 1 < i, j < m . Then {ui}m 1 is linearly independent. We define a linear mapping

f:U->Uo:=®m1Fu2 by f (u) = >i b(u, vi)ui. Then f is surjective by virtue of f (u2) = u2 and ker f = V j- is clear. Thus we have dim U - dim V -L = dim Uo = dim V. The proof shows

Corollary 1.1.1. In Proposition 1.1.2, v fl U1 = {0} if and only if V is contained in some regular subspace of U.

1.2 Quadratic forms Let M be a module over a ring R and q a mapping from M to R which satisfies the conditions (i) q(ax) = a2q(x) for a E R, x E M, (ii) b(x, y) := q(x + y) - q(x) - q(y) is a symmetric bilinear form.

We call the pair (M, q) or simply M a quadratic module over R, q a quadratic form and b the associated symmetric bilinear form. When R is a field, we often use "space" instead of "module". Putting x = y, we have b(x, x) = 2q(x)

for x E M.

If 2 is in R', then we can associate a symmetric bilinear form 1 b(x, y) _ B(x, y). Then B(x, x) = q(x) and conversely for a given symmetric bilinear form B(x, y), q(x) := B(x, x) is a quadratic form and q(x+y) -q(x) -q(y) =

4

1


2B(x, y). Thus "quadratic module" and "symmetric bilinear module" are equivalent if 2 E R'. In this section and the next we will associate b(x, y) with M, but after 1.4, the fields are of characteristic 0, and we will prefer the bilinear form B instead of b. The difference between them is minor and the choice depends on the individual's taste. Considering a quadratic module M as a symmetric bilinear module with b(x, y) (B(x, y) after section 1.4), the notations and terminology 1, M (A), d M and regular remain meaningful.

If 2 is a unit, then 1 and "regular" are independent of the choice of B(x, y) or b(x, y). But d M differs by 2' with n = rank M, since M = (A) or (2-1A) for A = (b(x, y)), according to b(x, y) or B(x, y), respectively. For a quadratic module M, we put

RadM:=Ix EM1I q(x)=0}. This is a submodule of M and if 2 E R', then Rad M = M-L.

Theorem 1.2.1. Let U be a quadratic space over a field F. If ch F

2,

then we have

where the U,'s are regular and 1-dimensional. If ch F = 2, then we have

1RadU, where the V is are regular and 2-dimensional, the Wz 's are 1-dimensional

and 0 < t < [F' : (F" )2] and

1RadU. Proof. Suppose ch F # 2. If there is an element ul E U such that q(ul) # 0, then b(ul, ul) = 2q(ul) 0 and so U1 = Ful is regular. Proposition 1.1.1 implies U = U1 1 Ui . Repeating this, we have U = U1 1 1 Ur -L Ur+1 where U1, , Ur are regular and 1-dimensional and q(x) = 0 for all x E Ur+1, which implies b(Ur+l, Ur+1) = 0 and hence Ur+1 C U. Decomposing x E Ul along the above orthogonal decomposition of U, U1 C Ur+1 is easy to see.

Next, suppose ch F = 2. If x, y E U satisfies b(x, y) # 0, then V1 = F[x, y] is regular with d V1 = -b(x, y)2, noting b(x, x) = b(y, y) = 0. Hence we have U = V1 1 V L. Repeating this, we have U = V1 1 1 V 1 Uo

1.2

Quadratic forms

5

, V3 are regular and 2-dimensional and b(U1, UO) = 0. As above, Uo = U1 holds. We take a direct sum decomposition where V 1 ,

U-L =W1ED ®Wt ®RadU where the Wi's are 1-dimensional. Since b(U1, U1) = 0, U1 = W1 1

1

Wt I Rad U holds. Put Wi = Fwi. Since wi c U1 but wi Rad U, q(wi) # 0 holds. If q(wi)(F")2 = q(wj)(F")2 for i j, then q(wi) = a2q(wj) for a E F" holds and this means q(wi - awl) = 0. Hence we have a contradiction wi - awl E Rad U. Thus the q(wi)'s give distinct representatives of F" /(FX )2. Let M, N be quadratic modules over a ring R. If a linear mapping a from

M to N satisfies that a is injective and q(a(x)) = q(x) for x E M,

we call a an isometry from M to N and say that M is represented by N and write

a:M'-->N. When a(M) = N, we write

a:M=N, and say that M and N are isometric. The group of all isometries from M to itself is denoted by O(M).

Suppose that R is a subring of a field F and generates F. For an Rsubmodule M of a quadratic space V over F satisfying FM = V, we denote by

O+(M):_{aEO(M)I deter=1 } where det a is defined by det T for a matrix T with

(a(vl), ... , a(vn)) _ (vi, ... v.)T for a basis {vi} of V.

For an isometry a from M to N, it is easy to see

b(a(x),a(y)) = b(x,y) for X, y E M.

Conversely, an injective linear mapping a from M to N which satisfies b(a(x), a(y)) = b(x, y) is an isometry if 2 E R", since q(x) = 2-'b(x, x). For quadratic modules M, N and a linear mapping a from M to N with

q(a(x)) = q(x) for x E M, a(x) = 0 yields x E M' since b(x, M) _ b(a(x), a(M)) = {0}. If, moreover M is a regular quadratic space, then or is an isometry. Now we give an important example of isometry.

6

1


Proposition 1.2.1. Let M be a quadratic module over a ring R. For an element x in M with q(x) E R", we put Tx(y):= y - b(x,y)q(x)-lx for y E M. Then Tx is an isometry from M to M and satisfies

7-X (X) = -x, Tx (y) = y for y E x1, and Tr = id. Proof. Tx(x) = -x and Tx(y) = y for y E x1 are easy. 7-x2(y)

= 7-x(y)

- b(x,y)q(x)-1Tx(x) = y

implies Tr = id and the bijectivity of Tx. Finally

q(Tx(y)) = q(y) + b(x,y)2q(x)-2q(x) - b(x,y)q(x) lb(x,y) = q(y) implies that Tx is an isometry. Tx is called a symmetry. If R is a field, then the determinant of a transformation Tx is -1. The following theorem of Witt type is due to Kneser.

Theorem 1.2.2. Let R D R be rings and P a proper ideal of R satisfying R = R" UP and R" n P = 0. Let L, M, N, H be R-submodules of a quadratic module U over R such that they are finitely generated over R, M, N are free R-modules and L D M, N, H. Suppose that (1)

q(H) C R, b(L, H) C R,

(2a)

HomR(M, R) _ {x F-, b(x, y) I y E H}, HomR(N, R) _ {x'--* b(x, y) I y E H},

(2b)

and o : M = N is an isometry such that (3)

or (x) = x mod H

for x E M. Then a can be extended to an element of O(L) which satisfies (4)

v=id on H1

and (3) for every x in L. Proof. For z in H with q(z) E R", the symmetry

Tz(x) = x - b(x,z)q(z)-'z satisfies three properties: rz(L) = L because of the property (1), secondly

(3) for x E L and (4) by definition of T. For a submodule J of H, we denote by S(J) a subgroup of O(L) generated by Tx (x E J, q(x) E R'), and then every element in S(J) satisfies the condition (3) for x E L and the condition (4). Note that the quotient ring R := R/P is a field. First, we will prove a preparatory assertion:

1.2

Quadratic forms

7

Assertion 1. We assume, moreover (5)

#R>2=q(x)ER" forsomexEH,

(6)

#R = 2

q(x) E R" and b(x, H) C P for some x E H.

Then a is a restriction of an element of S(H). We prove this by induction of the rank of M. Suppose rank M = 1 and M = Rm, N = Rn, n = a(m), and put, by (3)

h=n - mEH.

(7)

Then q(h) = 2q(n) - b(n, m) = 2q(m) - b(n, m) and (7), (1) imply

q(h) = b(n, h) = -b(m, h) E R.

(8)

If q(h) E R", then Th(m) = m - b(h, m)q(h)-1h = n by (8), (7), and we complete the proof in the case of rank M = 1. Suppose

q(h) E P.

(9)

First, we show that if there is an element f in H satisfying q(f ), b(f, m), b(f, n) E R',

(10)

then the proof in the case of rank M = 1 is completed. Supposing the existence of such f, we put g := n -Tf(m); then (7), (10) imply

g=h+b(f,m)q(f)-lf E H,

(11) and

q(g) = q(h) + b(f, m)2q(f)-2q(f) + b(f, m)q(f)-lb(h,f)

(12)

= q(h) + b(f, m)b(f, n)q(f)-1 E R"

by (9), (10). Moreover, we have T9(n) = n - b(n,g)q(g)-1g

= n - (b(n, h) + b(f, m)q(f)-1b(n, f))q(g)-1g

= n - (b(n,h) + q(g) - q(h))q(g)-1g

=n-g by (8) = Tf (m)

by definition of g.

by (12)

by (11)

8

1


Thus o is a restriction of rg-f e S(H). It remains to show the existence of f. We denote a vector space H/PH over R by H and an element of H, R represented by x E H, y E R by x, y, respectively. By virtue of (1), H becomes a quadratic space over R by q(x) := q(x) for x E H. This is well defined by (1). Put ml = {x E H I b(x, in) = 0}, n1 = {x E H I b(x, n) = 0}.

The condition (10) is equivalent to q(H \ (m1 U n')) 54 0. From (8), (9) follows m,1 fl n1 D h, and by virtue of (2) there exist h,,,, hn E H such that b(m, hm) = 1 and b(n, hn) = 1. Thus dim m1 = dim n1 = dim ft - 1. Suppose (13)

For (14)

q(H \ (1n1 U n1)) = 0.

IF-1- fl F)1- and yEft \(m1Un1),we have, by (13) q(ax + y) = a2q(x) + ab(x,9) + q(y) = 0 for a E R.

If OR > 2, then (14) implies (15)

q(x) = b(x ) =

0.

Note that h E in-J- n n1, and the vectors which are not in the union of two hyperplanes rn1, n1- span the whole space if the number of elements of the coefficient field is greater than 2. Putting x := h in (15), we have b(h, H) = 0 and then m1 = n1 by (7). From this with (15) it follows that q(n1) = q(H \ m1) = 0 and so q(H) = 0. This contradicts the assumption (5).

Suppose Of? = 2 and put

H1 := m1 fl n1 fl H1 and H2 :_ (H \ (yn1 U n1)) n fl-L;

then (14) implies q(Hi) = q(H2) = 0, noting b(H1iH2) = 0. For x E H, we have

x E m1 n H1

b(x, 7n) = 0, b(-;t, H) = 0

b(x, n) = 0, b(x, H) = 0 by (7)

,# 2En'nH',

1.2

Quadratic forms

9

and hence H1 = m1 n H1 =,h1 n H1, and then

H2=(H\m1)n(H\n1)nH1=H1\Hi. This means q(H1) = q(H1) U q(H2) = 0 which contradicts (6). Thus we have completed the proof in the case of rank M = 1. Suppose r = rank M > 1. Let {m1, , mr} be a basis of M and we take hi E H such that b(mi, hj) = Sid by virtue of the assumption (2). It is easy to see (16)

H=®iRhi®K,K:=M--nH.

From the assumption (5), (6) there is an element h E H or H1- such that q(h) # 0 according to OR > 2 or Of? = 2. Changing a basis if necessary, we may suppose

h E Rhr + K.

(17)

(,)i=1 r- Rmi, there is r E S(H) Applying the inductive assumption to Mo such that r = Q on Mo. Taking r-1(N), r-1v instead of N, v, they satisfy conditions (2), (3), because of T(H) = H and a remark at the beginning of the proof. Hence we may suppose

(18)

o,(mi)=mi for1 2, and wl := y - q(y)vl satisfies q(wi) = 0 and b(vl, wl) = 1. This means that Wl := F[vl, wl] is a hyperbolic plane. Thus, in particular the case of dim V = n = 1 has been completed. Suppose n > 1. Since V is totally isotropic, b(V, V) = 0 and {V2, , vn} C Wi holds. Since

((W,'))' := {x E Wi I B(x, Wi) = 0} C U1 because of U = Wl I Wi , we have

F[v2... , vn] n ((Wi ))1 C V fl U1 = 0, and hence by the assumption of the induction, there is a totally isotropic subspace W2 in Wi with dim W2 = n - 1 such that F[v2i , vn] ® W2 is a hyperbolic space. So W := Fwl ® w2 is what we want. From Corollary 1.2.1 it follows that all maximal hyperbolic subspaces in a quadratic space U have the same dimension. A half of the dimension of a maximal hyperbolic subspace is called the Witt index of U which we write ind U. By virtue of the above proposition, ind U is equal to the dimension of a maximal totally isotropic subspace if U is regular.

Corollary 1.2.4. Let U be a regular quadratic space over a field F and V = Fv a 1-dimensional quadratic space with q(v) = -a E F". Then a is represented by U, i.e. a E q(U) if and only if U 1 V is isotropic.

Proof. Suppose q(x) = a for x E U; then x + v E U I V is isotropic. Conversely, suppose that U 1 V is isotropic and u + by is an isotropic vector for u E U, b E F. If b # 0, then q(b-lu) = -q(v) = a. If b = 0, then u is isotropic. Hence the regular space U contains a hyperbolic plane and then q(U) = F D a.

12

1


Proposition 1.2.3. Let U be a 2-dimensional regular quadratic space over a field F with ch F 2. Then assertions (i), (ii), (iii) are equivalent; (i) (ii)

U is isotropic; U is a hyperbolic plane;

(iii)

d U = -1 (in F" /(F" )2).

(iii) is Proof. (i) (ii) follows from the previous proposition. (ii) clear. Suppose that (iii) holds. Since ch F # 2, there is an orthogonal

basis {u1iu2} of U and q(ul)q(u2) = -a2 for some a E F". Hence ul + q(ul)a-1u2 is an isotropic vector of U. Exercise 1. Prove Sylvester's law of inertia, that is for a quadratic space U over the field R of real numbers, there is a basis {v2} satisfying q(E aivi) _ E?" 1 a, 1 a2n+i, where rn and n are dependent only on U.

Exercise 2. Let V be a quadratic space over a field F. If V = V1 1 Rad V = V2 1 Rad V, then V1 is isometric to V2. If ch F = 2, then this does not hold for V' instead of Rad V. Namely, suppose that ch F = 2 and that H is a hyperbolic plane, V is a regular 2-dimensional quadratic space; then V I Fu -- H I Fu if q(u) E q(V). (Hint: Note that for a basis {v1, v2} of V with q(vi) = q(u), F[u + v1, v2] is regular and isotropic and hence hyperbolic.)

Exercise 3. Let W, H be quadratic modules over a ring R and suppose that H is a hyperbolic plane, that is H has a basis {hl, h2} over R satisfying

q(aih1+a2h2)=a1a2.ForwEWandvEV:=WJ H, we put E,,,(v) := v + b(v, hl)w - b(v, w)h1 - q(w)b(v, hl)h1.

E,,, is an isometry of V such that E,,,(h1) = h1 and = E,,,,E,,,2. If an isometry a of V satisfies a(hl) = hl, then v-1Ew = id on H for some w E W. This is called a Siegel-Eichler transformation.

1.3 Quadratic forms over finite fields Throughout this section, let F be a finite field Fq with q elements.

Let F = F(9) be a unique quadratic extension of F; then the norm N and the trace tr from F' to F are surjective mappings. For x, y E F', tr(xy) = N(x + y) - N(x) - N(y) is clear, where y denotes the conjugate of y over F. If N(x) = 0 for x E F', then x = 0 holds, and tr(xF') = 0, for x E F, implies x = 0. Thus (F', N) is an anisotropic regular quadratic space over F.

1.3

Quadratic forms over finite fields

13

Proposition 1.3.1. Let U be a 2-dimensional anisotropic quadratic space over a finite field F. Then U = (F', N) holds, and hence U is regular and q(U) = F. Proof. For a basis Jul, u2} of U and x E F, we have

q(xul + u2) = q(ui)x2+b(ui,u2)x + q(u2) # 0 since U is anisotropic. Hence there exists a E F such that

q(x1u1 + x2u2) = q(ul)(x1 - ax2)(x1 - ax2) = q(u1) N(xi - ax2) for xl, x2 E F. Taking /3 E F' with N(/3) = q(ul), a mapping x1u1+x1u2 (3(x1 - ax2) gives an isometry from U onto (F', N).

Corollary 1.3.1. A regular 2-dimensional quadratic space over F is isometric to a hyperbolic plane or an anisotropic space (F', N). Proof. Unless a quadratic space is anisotropic, then it is isotropic. Hence Propositions 1.2.2 and 1.3.1 yield immediately the corollary. Proposition 1.3.2. A quadratic space U over a finite field F with dim U > 2 is isotropic.

Proof. Suppose that U is anisotropic. Take a 2-dimensional subspace V;

then V is regular and q(V) = F by virtue of the previous proposition. For a non-zero vector x E V1(¢ {0}), there is a vector v E V = (F',N) such that q(v) = -q(x) ( 0). Thus v + x is isotropic. This contradicts our assumption on U.

Corollary 1.3.2. An orthogonal sum V of two isometric 2-dimensional regular quadratic spaces over F is a hyperbolic space. Proof. Suppose that an orthogonal sum V of two 2-dimensional aniso-

tropic quadratic spaces is not a hyperbolic space. Since dim V = 4 > 2, V is isotropic by Proposition 1.3.2. Then it is an orthogonal sum of a hyperbolic plane and an anisotropic 2-dimensional quadratic space (F', N). From Corollary 1.2.3 it follows that a hyperbolic plane is isometric to (F', N), which is a contradiction. Exercise 1. Suppose ch F # 2. Since a regular 2-dimensional quadratic space represents 1, every regular quadratic space V is isometric to (1) 1

1 (1) 1 (d V),

and hence it is classified by dim V and d V (E F" /(F" )2).

14

1


Suppose ch F = 2, and let V =1i 1 F[ei, fi] be a regular quadratic space over F with b(ei, fi) = 1 for all i. Then A(V)

i-i

q(ei)q(fi) mod P

depends only on V, where P := {x2 + x I x E F}. We call A(V) the Arf invariant of V. (It is not necessary to confine ourselves to finite fields.) (Hint: First, verify the case of n = 1. Suppose V L--L

1 H 1 (12 n,+i (F', N))

where H is a hyperbolic plane; then A(V) = (n - m)A((F', N)) holds and

hence, noting A((F', N)) # 0, we have A(V) = 0 if and only if V is a hyperbolic space.)

Theorem 1.3.1. A regular quadratic space over a finite field F which is not isometric to a hyperbolic space is isometric to an orthogonal sum of a hyperbolic space and an anisotropic p2-dimensional space (= (F', N)), or an orthogonal sum of a hyperbolic space and a 1-dimensional regular quadratic space. Here the latter case can occur only when ch F # 2.

Proof. If a regular quadratic space U over F is of dim U > 2, then Proposition 1.3.2 and Proposition 1.2.2 imply that U contains a hyperbolic plane

H. Applying a similar argument to H-L we have, finally, U = V 1 W for a hyperbolic space V and W with dim W < 2. By Proposition 1.2.2 and Proposition 1.3.1, W is a hyperbolic plane or isometric to (F', N) if dim W = 2. If ch F = 2, then dim U is even by Theorem 1.2.1 and hence dim W = 1 does not happen. We put, for U with dim U even, X(U) = 1 if U is a hyperbolic space, and

X(U) = -1, otherwise. For an odd-dimensional regular quadratic space

U, we put X(U) = 0. Then it is easy to see that X(U 1 H) = X(U) if H is a hyperbolic space, and X(U 1 V) = -X(U) if V is an anisotropic 2-dimensional space and U is regular, since an orthogonal sum of two 2dimensional anisotropic spaces is a hyperbolic space. Let us evaluate the number r(V, W) of isometries from V to W. For a quadratic space U over F and t E F, we put a(t, U) := O{x E U I q(x) = t}.

If Fu is a quadratic space with q(u) = t # 0, then r(Fu, U) = a(t, U) is obvious.

1.3


15

Lemma 1.3.1. For a regular quadratic space U over F with dim U = u we have a(0, U) =

qu-1

-

x(U)q'/2-1 + X(U)qu/2.

Proof. We use induction on u. If u = 1, then a(0, U) = 1 is easy, since the regularity of U with u = 1 implies ch F # 2, and q(U) # {0}. Suppose u = 2. If U is anisotropic, then a(0, U) = 1 and X(U) = -1 imply the assertion. If U is a hyperbolic plane, then a(0, U) = #{x1, x2 E F I x1x2 = 0} = 2q - 1

which is the assertion. Suppose dim U > 3 and decompose U as U = H 1 Uo, where H is a hyperbolic plane, and we take a hyperbolic basis {h1, h2}. Writing u = a1h1 + a2h2 + UO for u E U, uo E Uo, we have

a(0, U) = #{al, a2 E F, uo E UO a1a2 + q(uo) = 0}

= (2q - 1)a(0, Uo) +

(q - 1)a(t, Uo) tEFx

= (2q - 1)a(0, Uo) + (q -

_ (q -

1)qu-2 +

1)(qu-2 - a(0,

Uo))

qa(0, Uo).

By the induction hypothesis, we complete the proof. Lemma 1.3.2. Let U be a regular quadratic space over F with even dimen-

sion. For t E F", there is an automorphism f of U such that q(f(u)) = tq(u) for u E U, and a(t, U) = a(1, U) = q1-1(1 - X(U)q-u/2) holds with u:= dim U.

Proof. Let H be a hyperbolic plane with hyperbolic basis {hl, h2}. For t E F", f (h1) = h1, f (h2) = th2 is an automorphism of H with q(f (h)) = tq(h) for h E H. If K is an anisotropic 2-dimensional quadratic space, then identifying K with F' at the beginning and taking an element k E F' such that q(k) = t, f (x) = kx is an automorphism of K with q(f (x)) = tq(x) for x E K. By virtue of Theorem 1.3.1 we obtain the existence of an automorphism f of U. Then

a(t, U) = a(1, U) = (q - 1)-1 > a(t, U) tEFx

= (q - 1)-1(qu - a(0, U)) =

qu-1

-

X(U)q'/2-1

from the previous lemma. For a regular quadratic space (U, q), we denote a new quadratic space (U, -q) by U(-1). Let U be a regular quadratic space such that dim U = 1

16

1


if chF 0 2, dim U = 2 if chF = 2; then U L

U(-1)

is hyperbolic, by Proposition 1.2.3 if chF # 2, Corollary 1.3.1 and 1.3.2 if chF = 2. Thus, for a regular quadratic space U, U L U(-1) is hyperbolic and hence X(U L

U(-1))

= 1.

Theorem 1.3.2. Let V, W be regular quadratic spaces over a finite field F(= Fq) and put v := dim V, w := dim W and suppose v < w. Then the number r(V, W) of isometries from V to W is equal to qvw-v(v+l)/2(1- X(W)q-'/2)(I + X(V L W(-1))q(v-w)V2) X

fi

(1 - q-e).

w-v+1<e<w-1 e:even

Proof. Let r'(V,W) be the number given by the above formula; we show

r(V, W) = r'(V, W) by induction on v = dim V. We note first that for regular quadratic spaces U, U1, U2, we have

r(U L U1i U L U2) = E r(Ul,

r(U, U L U2)r(Ul, U2)

o: U- +U±U2

since o(U)' = U2 follows from Corollary 1.2.3.

It is easy to verify r'(U L U1, U L U2) = r'(U, U L U2)r'(Ul, U2) for dim U = 1 if ch F # 2, a hyperbolic plane U if ch F = 2. To show r (V, W) = r' (V, W), we have only to prove r (V, W) = r' (V, W) for 12, and for 2-dimensional spaces V if dimensional spaces V if ch F ch F = 2, noting that a regular quadratic space is an orthogonal sum of 1-dimensional subspaces if ch F # 2, or an orthogonal sum of hyperbolic planes and at most one anisotropic 2-dimensional subspace if ch F = 2.

Claim. If ch F

2, then r(V, W) = r'(V, W) in the case of v = dim V = 1.

Proof. We have only to prove a(t, W) = r'(V, W) if V = Fv, q(v) = t(O 0). If w = dim W is even, then a(t, W) = q"i-1(1-X(W)q-"'/2) follows from Lemma 1.3.2 and it is equal to r'(V, W). Suppose w = dim W = 1; then we know X(V L W(-1)) = 1 #: d V = V = W, and r (V, W) = 2 or 0 depending on whether V = W or dW V W respectively. Hence r (V, W) = r' (V, W) follows. Suppose that w is odd and put W = Fw1 L WO. Then from a(t, W) = U{x E F, wo E Wo I x2q(w1) + q(wo) = t}

it follows that a(t, W) = ga(1,W0) if x2q(wi)

t for all x E F, and

a(t, W) = 2a(0, Wo) + (q - 2)a(1, Wo)

1.3


17

if x2q(wi) = t for some x E F. Here we note that t # 0. Suppose x2q(wi) t for all x E F; then Fwl V, and hence V 1 (Fwi)(-1) is anisotropic. Thus

x(V L and

a(t,W) =

qw-1(1

W(-1)) = -x(Wo-1))

+x(V L

_ -x(Wo) W(-1))q(1-w)/2) = r'(V,W).

Suppose x2q(wi) = t for some x E F; then Fwl = V and V L is hyperbolic. Thus x(V L W(-1)) = x(Wo) follows. Finally -)C(Wo)q(w-1)/2-1 + x(Wo)q(w-i)/2)

a(t, W) = 2(qir-2

+ (q -

(Fwi)(-1)

2)qw-2(1

-

X(l'Vo)q-(w-1)/2)

= qw-1(1 + x(Wo)q(1-w)l2) = r'(V, W).

Thus we have completed the proof in the case of ch F

2.

Claim. If ch F = 2, then r(V, W) = r' (V, W) in the case of v = dim V = 2.

Proof. Let U be a hyperbolic plane and {u1, u2} a hyperbolic basis. By virtue of the fact that {isotropic vectors in U} = {tui, tut I t E F" }, an

isometry a of U is a(ui) = tui,a(u2) = t-1u2 or a(ui) = tu2,a(u2) _ t-1u1. Hence we have r(U, U) = 2(q - 1) = r'(U, U). Let K be an anisotropic 2-dimensional quadratic space identified with (F', N) as explained at the beginning of this section. If a is an isometry

of K with a(1) = 1, then a is a linear isomorphism of F' over F. Now q(a(x)) = x and b(1, a(x)) = b(1, x) imply N(a(x)) = x and tra(x) = trx, and hence the generator 9 with F' = F(9) is transformed by a to a conjugate. Thus a is either the identity or the conjugate mapping. Hence

an isometry a of K is a(x) = ax or ax for some a E K with N(a) = 1, and r(K, K) = 2a(1, K) = 2q(1 + q-1) = r'(K, K). If v = w = 2 and V W, then V 1 W(-1) is not a hyperbolic space and hence x(V L W(-1)) = -1 and r(V, W) = 0 = r'(V, W) follows. Thus we have completed the proof of w = 2. Suppose w > 4 and put

W=HLWo where H is a hyperbolic plane with hyperbolic basis {h1i h2}. If V is a hyperbolic plane, then Corollary 1.2.1 shows that every isotropic vector in W is transformed to h1 by an isometry of W. We have

r(V, W) = (a(0, W) - 1) {w E W I b(h1, w) = 1, q(w) = 0}.

18

1


By Lemma 1.3.1, we know

a(O, W) - 1 =

qw-1(1

+X(W)gl-w/2)

- X(W)q-w/2)(1

On the other hand, we have O{w E W I b(hi, w) = 1, q(w) = 0} = O{w = x1h1 + x2h2 + wo E W I x1, x2 E F, wo E Wo, b(hi, w) = 1, q(w) = 0} = O{x1 E F, wo E Wo I x1 + q(wo) = 0} = qw-2 and hence +X(W)q'-w/2) = r'(V,W).

r(V, W) = q2w-3(1 - X(W)q-"2)(1

Suppose that V is an anisotropic 2-dimensional quadratic space with q(xiv1 + x2v2) = ax2 + bx1x2 + cx2 for X1, x2 E F; then r(V, W) equals

a(a, W)#{w E W I b(h1 + ah2, w) = b, q(w) = c} = a(1,W)#{w = a1 h1 + a2h2 +wo I a1, a2 E F, wo E Wo, a2 + aal = b, ala2 + q(wo) = c} = a(1, W)#{ al E F, wo E Wo I q(wo) = aai + bal + c} = a(1, W)a(1, Wo)q, since aai + bal + c = q(alvl + v2)

0 for al E F.

Thus we have

r(V, W) =

q2w-3(1

- x(W)q'2)(1 - X(Wo)q(2-w)l2) = r'(V, W),

noting that X(Wo) = X(W(-1)) = -X(V 1 W(-1)), which completes the proof in the case of ch F = 2 and so the proof of Theorem 1.3.2.

0

Proposition 1.3.3. Let V, W be quadratic spaces over F and suppose that V is regular and represented by W. Put W = Wo I W1 and v = dim V, w = dim Wo, r = dim Rad W. Then we have

r(VIW)=

r(V, Wo)gv"' vdimW-v(v+1)/2 g

w-v even e:even

(1 - q-e)

if W -L = Rad W, 1 W. if W j-

Proof. Let {vi} be a basis of V. Every isometry o, from V to W is given by o, (vi) = q(vi) + wi for an isometry rl from V to W1 and wi E Rad W where

1.3


19

Wl is defined by W = Wl I Rad W. Hence we have r(V, W) = r(V, Wi)q"' and the first equation is clear. Note that W' = Rad W if ch F 2. Next we suppose ch F = 2, W -L Rad W. From (F" ) 2 = F" follows W' = Fwo + Rad W, q(wo) = 1 by virtue of Theorem 1.2.1. We have only to show

r(V,WI) = q(w+l)v-v(v+l)/2

fJ w-v+2<e<w

(1 - q-e)

e:even

We may suppose Wl = Wo 1 Fwo. We use induction on v. Since V is regular, v is even. Suppose v = 2 and q(vi) = ai, b(vl,V2) = 1. r(V, Wl) = O{(wl, W2) E Wl x Wl I q(wi) = ai, b(wl,w2) = 1}

is clear. If q(wl) = q(wi) = al and b(wl,Wi) # 0, b(wi,Wi) 0 for w1, wi E W1, then they are transformed by an isometry of Wl by virtue of Corollary 1.2.1. Hence

r(V, Wl) = #{x E Wl I q(x) = al, b(x, Wl) 54 0} x #{x E Wl I q(x) = a2, b(x, wl) = 1}

where wl E Wl is any vector such that q(wl) = al, b(wl,Wi) # 0. Let us evaluate J{x E Wl q(x) = al, b(x,WI) # 0}. Since q(Fwo) = F, there is, for an element y E Wo, a unique f E F such that q(y + f wo) = q(y) + f2 = al; moreover b(y + fwo, Wl) = b(y, Wo) 0 is equivalent to y 0 0. Hence, writing x = y + f wo (y E WO), it follows easily that I

#{x E Wi I q(x) = al, b(x, WI) Next, let us evaluate

0} = qw - 1.

O{x E Wl I q(x) = a2, b(x, wl) =1}

and put wl := yo+awo, yo E Wo, a E F. Suppose q(wl) = al, b(wl, Wl) j4 0. For x = y + cwo (y E WO, c E F), we have b(x, wl) = b(yo, y), and q(x) = q(y) + c2 = a2 uniquely determines c by y. Thus we have O{x E Wl I q(x) = a2, b(x, wl) = 1} _ {y E Wo I b(y, yo) = 1} =

qw-l.

Hence r(V,W1) = (q"' - 1)q'-l is what we desired for v = 2. If v > 2, then we decompose V as V = Vo I Vl with dim Vo = 2 and use r(V,WI) = r(Vo,Wl)r(Vl,VL in Wl), (regarding Vo C WI) and the induction hypothesis while noting that dim(V1 in Wl)1 = 1.

20

1


1.4 The Clifford algebra Henceforth, unless mentioned otherwise, the characteristic of fields is different from 2 and a ring is embedded in a field. We replace a quadratic form q by Q, the associated bilinear form b by B := 1b. Thus we have, for a quadratic form Q and the associated bilinear form B,

Q(x + y) - Q(x) - Q(y) = 2B(x, y), Q(x) = B(x, x). A quadratic module (M, Q) is considered as a symmetric bilinear module with respect to B. So, M = (A) means (B(vi, vj)) = A for a basis {v2} of M.

For a quadratic space V over a field F, we put

T(V) :_

®m=° ®m V, ®°V := F, ®1V := V,

where ® denotes the tensor product over F. For u := ul ® ® um E ®"'V, v := v1 ® ®vn E ®nV (ui, vj E V), we define a product of u and v by

u®V:= ul ®...®um ®vi ®...®vn E

®m+nV

This is well defined and T(V) is called the tensor algebra of V. Denoting by I (V) the two-sided ideal of T (V) generated by v ® v - Q(v) (v E V), we put

C(V) := T(V)/I(V), which is called the Clifford algebra. For simplicity, an element of C(V) which is represented by v1 0 . . . 0 vm (v2 E V) is denoted by v1 v,,,,. By definition, we have, in C(V) (1)

J vEV

rv2=Q(v)

U, v E V = uv, + vu = 2B(u, v), uv = -vu if B(u, v) = 0.

Proposition 1.4.1. Let V be a quadratic space over F and A an algebra over F. If f is a linear mapping from V to A with f (v)2 = Q(v) for v E V, then there is a homomorphism g from C(V) to A such that f (v) = g(v) for v E V.

Proof. We define an algebra homomorphism g' from T(V) to A by 9 '(vi

®... ®vm) = .f (vi) ... f(vm) for v2 E V.

Then kerg' is a two-sided ideal of T(V) containing v 0 v - Q(v) for v E V and hence kerg' contains I(V). Thus g' induces a homomorphism g from C(V) = T(V)/I(V) to A satisfying g(v) = g(v) = f (v) for v E V.

1.4

The Clifford algebra

21

Corollary 1.4.1. For isometric quadratic spaces V, W over F, Clifford algebras C(V), C(W) are isomorphic.

Proof. Let o be an isometry from V on W; then o induces a linear mapping from V to C(W) by x '--> u (x) E C(W). Hence there exists a homomorphism f from C(V) to C(W) satisfying f (x) = cr(x) for x E V. Similarly, there exists a homomorphism g from C(W) to C(V) satisfying g(w) = a-1(w) for w E W. Thus g f (resp. f g) is the identity mapping on elements in C(V) (resp. C(W)) represented by elements in V (resp. W) and hence on C(V) (resp. C(W)) respectively.

Theorem 1.4.1. Let V be a quadratic space over F with an orthogonal. basis {vi} 1. Then {vi' vn^ ei = 0 or 1} is a basis of the Clifford algebra C(V). In particular dimC(V) = 2n. I

Proof. By virtue of (1), the Clifford algebra is spanned by vl' vn, (el = 0 or 1) as a vector space and hence dimC(V) < 2n. We must show the

converse inequality. Let F2 be a prime field with 2 elements and K = F2 an n-dimensional vector space over F2, and moreover let A be a 2n-dimensional vector space whose basis are elements of K. When we regard an element f in K as a basis vector of A, we will write [f]. So, note [fl + f2] [fi] + [f2] for fl, f2 E K. We introduce a product in A by [fl] [f2] =

where f, = (al, .

xEF, aEF2

(-1)ajb,

fJ 1 O(V) with image = 0+(V) (gvg-1)2 = gv2g-1 = Q(v) is or O(V). (Hint: Q(cr(v)) = (u(v))2 = obvious. If v = 1, then g E Co (V) is in the centre of C(V). This implies g E F. If v =,T1 ...Tv_ E 0+(V), then u(x) _ (v-,,,, ... v1)-1x(v,,,.... v1) _ v1)-1.) gxg-1 with g = Proposition 1.6.3. Let V be a regular isotropic quadratic space over a field F. Then 0'(V) is generated by TTTy for x, y E V satisfying Q(x) _ Q(y) 0 0. Proof. Let SI be the subgroup of O(V) which is generated by TrTy for (v,,,....

x, y E V satisfying Q(x) = Q(y) # 0. Noting QTXU-1 = Toixi for a E O(V), Sl is a normal subgroup of 0'(V). Let V = H 1 W where H is a hyperbolic

plane with a basis {e1, e2} such that Q(e1) = Q(e2) = 0, B(e1, e2) = 1 . Let u = TT1 TT,n E 0'(V). We take yi E H such that Q(xi) = Q(yi) for i = 1, , n. Then or- Ty1 ...Tyn E Q (1)1

1.6 The spinor norm

31

follows from TyjTvj E SZ and the normality of Il in 0'(V).

Put 77

Tyn; then 1J W = id since y2 E H, and 77' 77TH E O'(H) holds by 0(77) = 0(77Iw)0(77IH) = 0(70 from Proposition 1.6.2. Put 77'(el) = ale, + a2e2, 77'(e2) = b1e, + b2e2; then a1a2 = b1b2 = 0 and r7' E O+(H) imply 77'(el) = ale,, 17'(e2) = a-11 e2 and hence 77' = Tel-e2Te,-ale2 is easily seen. Therefore 17 = Te, _e2Te, _aje2 on V since they are equal on H and W, and 0(77') = a, and 0(77') E (F")2 implies a, = a2 for some a E F' x + y and xy are continuous from F x F to F and x '--> x-1 is also continuous from F" to F" . For a vector space V over F with dim V = n, we introduce a topology on V, identifying V = Fn. The topology is independent of the identification.

Proposition 1.6.5. Let F be a topological field such that (F" )2 is open in F". For a regular quadratic space V over F, the spinor norm 0 : O(V) F" /(F" )2 is continuous. Proof. We have only to show that if an isometry or is sufficiently close to

the identity mapping, then 0(a) = 1 in F"/(F")2. We use induction on dim V. If dim V = 1, then a = ±1 and so or = 1 if a is close to 1. Hence 0 is continuous. Suppose that dim V > 2 and {vi} is an orthogonal basis of V. Let or be an isometry of V close to 1, and put a(vi) = vi + >j aijvj; then the aij are close to 0, and hence

Q(a(vl) + vi) = (2 + al1)2Q(v1) + EaijQ(vj)

j#i is close to 4Q(v1)(54 0), since Q(v1) # 0 and 2+all is close to 2. Hence we

have Q(a(vl) + v1) = 4Q(vl)a2 for a E F", since (F' )2 is open. Putting T := TQ(v,)+v we have

T(v1) = v1 - 2B(vi, or(vi) + vl)Q(a(vl) + 111)-l(a(vl) + vi)

_ -a(vl), 0(T) = Q(a(vl) + v1) = Q(vl) in F"/(F" )2,

32

1


and

T (vi) = vi - 2B (vi, o (vi) + vl)Q(o,(vl) + vl)-1(o,(vl) + vi) is close to vi for i > 2 since u(vi) + v1 is close to 2v1. Hence 7-or(vi) and vi are close for i > 2 and so To(vi) = -v1 yields that is also close to

1 on vi . From the induction hypothesis, it follows that °(T°-I -) = 1 and then we have 9(0.)

= 0(T2o') = B(T)B(Ta) =Q(vl)e(TQ) =Q(vi)O(-1 on Fvl)O(Tojvi ) = Q(vl)2 = 1 in F"/(F')'

0

1.7 Scalar extensions Let R' D R be commutative rings and (M, B) be a symmetric bilinear module over R. Denote R' ®R M by R'M; then we can define a symmetric bilinear form B' by B'(rx, sy) = rsB(x, y) for r, s E R' and x, y E M. The new symmetric bilinear module (R'M, B') is called the scalar extension of (M, B) and usually we use the same letter B instead of B. For a quadratic module (M, Q), we associate the symmetric bilinear module (M, B) with B(x, x) = Q(x) and take the scalar extension (R'M, B) to define the scalar extension of (M, Q) as (R'M, Q) which is associated with (R'M, B). Hence for ai E R', xi E M, we have

Q ( aixi) = B (E aixi, E aixi) = E aiaj B(xi, xj ). i,j

If M has a basis over R, then M is identified with Rn for some n by using a basis of M over R, and R'M is identified with R'n through the identification of M and Rn; thus the scalar extension is simply extending the domain R of variables to R', keeping the corresponding matrix.

2 Positive definite quadratic forms over IR

The aim of this chapter is to give a reduction theory of quadratic forms over R. Let P be the set of all positive definite matrices of degree m and denote GL,,,,(Z) by r. Then I' acts discontinuously on P by p -- p[g] (p E P, g E I'). An important problem is to determine the fundamental domain P/r, that is to give a "canonical" matrix in the orbit p[g] (g E t). This was done by Minkowski for general m and the set of canonical matrices is called Minkowski's domain. Unless there is a need for an exact fundamental

domain, Siegel's domain is more useful, especially for analysis on P. So we introduce Siegel's domain instead of Minkowski's. As an application, we prove the finiteness of the class number of (not necessarily positive definite) quadratic forms. In Section 2.2, we give a better estimate due to Blichfeldt of Hermite's constant introduced in Section 2.1. The proof will be an introduction to the theory of geometry of numbers. We need it in the last chapter.

2.1 Reduction theory In this section, we describe a reduction theory of positive definite symmetric

matrices. Let m be an integer larger than 1 and put G = G(m) :=GL,,,.(IR),

A = A(m) :={diag(al,

,

a,,,,)

a2 > O for i = 1,

,

m},

34

2

Positive definite quadratic forms over R

N = N(m) :=the subgroup of all upper triangular matrices with all diagonals being 1 in G, K = K(m) := the subgroup of all orthogonal matrices in G For positive numbers t, u, we define subsets At, Nu as follows:

At:={diag(al, ,a.) EAlai/ai+l ai(1

- hi2 t)

2

> (3/4)ai

follows.

Lemma 2.1.3. Let g E G; then there exists an element 'yo E IF such that O(g'yo) = min.yEr 0(g'Y) and g'yo E S2/v,1/2.

Proof. We use induction on m. We can choose 'y' E IF such that O(gy') _ min.yEr 0(gry), and write

g'y' = kan

where k E K, a = diag(al, a2i

) E A, n E N, and moreover we may

assume n E N112 by virtue of Lemma 2.1.1 and (1). First suppose m = 2; by the previous lemma, we have al /a2 < 2/v/3-. Putting 'Yo = 'y', we have gryo = kan E S2/,/-3- 1/21 and O(gyo) = O(gy') is the minimum. This completes the case of m = 2. Suppose m > 3. k-lgy' = an is of the form

b E G(m - 1). b

Applying the assumption of the induction, we write

bx = k'a'n'

2 Positive definite quadratic forms over IR

36

where x E I'(m -1), k' E K(m -1), a' = diag(a'1, .. , a',,_1) E A2/f(m 1), n' E N112 (M - 1). Then

/ g

:=k-lgry C 1

\

l x)\ =

an x) =an('

*...*

al 0

bx

/)

0 1

0

0

0

0

0

0

Noting 0(g) = 0(gry') by (1) and the minimality of it, Lemma 2.1.2 implies al/a1 < which yields diag(al, a') E A2/,/-3- with a' E A2/V'3-(m - 1). Now applying Lemma 2.1.1 to the final matrix in (2) we can take n E NZ such that gn E S2/,V'5 1/2. Thus, putting 'Yo = 7' (1

0

/

n we have g-yo =

kgn E S2/ 1/2, and 'yo is a required matrix in F, since/ O(g-yo) = 0(g) =

0 0(g'y') by (1). Now we apply the reduction theory of GL, ,,(R) to the reduction theory of quadratic forms. Denote by P = P(m) the set of all positive definite symmetric matrices of degree m and put

St,u=St,u(m):={a[n]IaEAt,nENu}, which is called Siegel's domain.

Theorem 2.1.1. P = U,yEr S4/3,1/2['Y] and /µm, := sup

p) m pEP (d mmp

< (4/3)im-1)/2,

where min p denotes the minimum of p[x] (0 # x E Zm).

Proof. For p E P, write p = tgg (g E G), and making use of the previous lemma, we can write g'yo = kan (yo E F, k E K, a E A2/ f, n E N112). Moreover we may assume O(g-yo) G 0(gry) for all ry E F. Then p[yo] = a2[n] E 54/3,1/2 is clear. This implies the first assertion. To prove the inequality, we show first, minxEZm\{o} IgxI 5

(4/3)(--1)/4Idetgi l/m.

Reduction theory

2.1

37

Since the left-hand side is attained by the primitive vector x, it is equal to min.yEr Igyel = min,yEr 0(9'Y) _ 4(g'Yo) = a1,

where we put a = diag(ai,

,

am). Now

al/ai = (al/a2)...(ai-1/ai) < (2/v'3-)'-'

implies m f1(2/v3)'-1 = (2/v)m(m-1)/2

a1 / det a < i=2

This is what we want, noting det g = f det a. Now we can complete the proof of the theorem as follows: minp =minxEZ-\{o} t(gx)gx = minxEZm\{o} Igxl2

i

i =(I

aixil +:I nij I aixj 1)2. I

j>i

,

xm)

2 Positive definite quadratic forms over llt

38

Noting I aixil
i c2,i as c2. From the formula of the inverse matrix it follows that n-1 E Nu, for all n E Nu for an appropriate u' > 0. Hence a[n-1] E St,,,, and applying the above for St,u, instead of St,u, there is a constant cl such that a[n-1][x] < ci la[x] for all x E IIt'. Replacing x by nx, we have a[x] < ci 1p[x], which yields the first inequality. F o r f = t (0, , 0,1, 0, , 0), we have p[f] = pii, a[f] = ai and hence the latter inequality in the assertion.

Theorem 2.1.3. Put S = St,u for t, u > 0 and let d be a positive number. Then there is only a finite number of m x m integral matrices x such that 0 < I det xI < d and S f1 S[x] # 0. Proof. We use induction on m. Suppose that for x E Mm(Z) with 0 < det x < d there is a matrix p such that p = a[n], p[x] = a'[n'] (a, a' E At, n, n' E Nu). (i) We assume first that we can write

x=

(X1

I (x1 E Mmi (Z), X2 E Mm2 (Z), X3 E Mmi,m2 (Z)); 2 X2

then putting

a = diag(al, a2), (al E At (mi), a2 E At (M2)),

n= I 0j

n2 I (n1 E Nu(ml),n2 E Nu(m2),n3 E Mml,M2(R)),

Reduction theory

2.1

39

we have

p = a[n] = ( (3)

_ (a10

1]

0

a2)

a2[n2]/

[(ni

/ (0

0

n11n3

O L

p[x] - ((al[n'ffxl] Oa2[n0][x2]

0

n11n3

[(1

(nixi)-1(nix3 +nsx2) /J

0

We write similarly

a' = diag(ai, a'2), n' =

n', 0

n3 n2

where a?, nZ have the same degree as ai, ni, and a'[n'] has the similar decomposition to (3). Noting the matrix identity

1 01

(0 c) [ (0 1) ]

(t(bf) 1

b[f]11+c)

,

we have from p[x] = a'[n'] that (4)

(ai[ni])[xi] = ai[ni], a2[n2][x2] = a2[ni] 1 n3 . (nix,) 1(nix3 + n3x2) = n'1

The last equality gives (5)

x3 = x1n1

n3 - n1 n3x2

Ifm=2, then we have ml=m2=1,n1=n' =n2=n2=1,In3l,In3l ah(i)

by (7)

jai

by h(i) > i.

by xh(i),k(i) 54 0

Thus we have

(9)

ai r ai.

For d1 := det x, and p' = p[x] implies p'[d1x-1] = dip, but p', dip are also in St,u, and it is easy to see that d1x-1 does not have a decomposition as in (i). By a similar argument, we have for i < m - 1, (10)

ai >- ai.

det p' and det p' >- det p, det p = fj ai and det p' = fl a'i Since det p imply the validity of (9), (10) for i = m. Then, for i < m (8), (10) and (6) with h(i) > i imply (11)

ai > ah(i) >- ah(i) >

ai+1.

2.1

Reduction theory

(6), (9), (10) and (11) show that al,

, a,,,,,

ai,

41 .

. ,

a;,, have the same

magnitude up to constants dependent only on m, t, u, d and then (7) implies that I xji l - 1, which shows the finite possibility of x. Let q be a not necessarily positive definite symmetric regular matrix of degree m over R. If a positive definite symmetric matrix p satisfies q-1 [p] = q,

then p is called a majorant of q. The existence of a majorant is shown as follows: write q = d[g] for d = diag(l,,,,, -1n), g E GL,, (R) and put p = tgg; then q-11p] = (g-1d-1tg-1)[tgg] = d-1[g] = q. It is easy to see that if p is a majorant of q, then p[u] is a majorant of q[u] for every u E GLm(]R).

Corollary 2.1.1 (finiteness of class number). Let d be a non-zero integer and put Qd = {q E M,,, (7d) I tq = q, det q = d}. We define an equivalence relation qi - q2 by q1 = q2[u] for u E GLm(Z). Then the number of equivalence classes in Qd is finite.

Proof. Put J = (xij) where xi.1 = 1 if i + j = m + 1 and xi9 = 0 if i+j

m + 1. J2 is the identity matrix. We show St u C JSt,u, J

Let a = diag(a1,

,

for some u'.

a,,,,) E At, n E Nu; then

J(a[n])-'J = t(JtnJ)-1(Ja-1J)(JtnJ)-1, and

Ja-1 J = diag(aM-1, ... , ai 1) E At,

and JtnJ E Nu implies E Nun for some u'. Thus Ja[n]-1J E St,u, follows, which is what we want. Let q E Qd and p be a majorant of q. Choose g E GL,,,,(Z) such that p[g] E S413,112 by Theorem 2.1.1, and a positive number u > 1/2 such that S4/s,112 C JS413,,J. Then q-1[p] = q (JtnJ)-1

implies p[g] = q[g](p[g])-1q[g] E 54/3,1/2 n q[g]S4/3 1/2q[g] C S4/3,u n (q[g]J)S4/3,u(Jq[g])

By the previous theorem, the number of possible integral matrices x with I det xI = d such that S4/3,u f1 S4/3,u[x] 0 is finite. It follows that Jq[g], and hence also q[g], is in a finite set.

42

2

Positive definite quadratic forms over lR

There is an easier method to get the corollary. The following is enough. For given natural numbers m and d, there is a constant c satisfying that for any integral symmetric regular m x m matrix A with I det Al < d, there exists a non-zero vector x E Zm such that I A[x] I < c. Indeed, we know the following:

Exercise (Hermite). min{IA[x]I I x E Zm, x

0}
0, e,'q E Zp ). Multiplying a/3 = 0 by have pm+n = 0 in Z p. Since 7L is injectively embedded in Zp, this is a contradiction. By the corollary, we can define the quotient field Qp of 7Lp which is called the field of p-adic numbers. Thus a non-zero p-adic number x can be written uniquely as pte (m E Z, e E 7LP) and we put

ordpx:=m

and

lxlp:=p_m.

We set loIP := 0

by definition. Then it is easy to see that Ix + y1p < max(jx1p, jylp) with equality holding if x I p# I y I p and

xEZ, x E p7Lp

I xEZP

if and only if if and only if if and only if

Jx1p 0 there exists a number c(e) such that d(xn, xm) < c for n, m > c(e). If xn = 0 for infinitely many n, then it is easy to see that xn converges to 0. Hence we may suppose that xn = 0 for finitely many n.

50

3

Quadratic forms over local fields

Removing those n with Xn = 0, we may assume xn 0 0 for n > 1. Put xn = pa" En (an E Z, En E Z). If there is a subsequence an- such that an, > ... > anm > ... - -oo, then putting anm = bm, Enm = ym, d(xnm, xnm+k) = Ipbm /m -

pbm+k

11m+k I =

p-bm.+k

which contradicts the property of the Cauchy sequence. Thus there exists an integer a such that an > a and we have only to prove that xnp a(E z P) converges. Hence we may suppose Xn E Zp. We write

xn = {xn o, ... , EXn,jP, ... } i=0

with 0 < xn,i 0, we take a number c(p-m) such that d(xk, Xh) < p-' if k, h > c(p-""). Then d(Xk, Xh) < p-m means Xk - Xh - 0 mod pm and hence Xk,i = Xh,i for i < m. So Xk,i depends only on i if k is sufficiently large, k > c(p-m) suffices. We denote it by bi and put m

y = {bo,...

,Ebip',... } E S.

i=o Since

m

m

Ym - Xk,m = E bip2 - E xk,iPi = 0 i=0

i=0

for a sufficiently large k, that is y - xk E p zp, {xn} converges to y. Next we show that Zp is compact. Let m

yn = {yn,0, ... , E yn,ip", ... } E Zp (0 < Yn,i < p) i=0

There is an infinite subsequence of yn such that yn,0 = ao for some ao (0 < ao < p), and there is an infinite subsequence in this subsequence such that yn,i = a1 for some a1. Repeating this, there exists a sequence {ai}°O0 (0 < ai < p) such that there exist infinitely many Yn satisfying yn,i = ai for 0 < i < a n y given integer. Hence {ao, , E'0 aipi, } is an accumulation point in Zp. Since I E2 1 xil p < max, 2.

Suppose p = 2; we have only to show the existence of X, Y E N with X, Y, XY V (F')2 to verify [N : (F')2] > 4. Noting x2 -cy2, we can take

X=2(=12-c.12),Y=5(=12-c 22)

if c = -1;

ifc=-3; X=5(-02-c 12mod8),Y=-2(=12- C. 12)

ifc=3;

X=-2(=02-c.12),Y=-1(=12-c. 12) X=2(=02-c.12),Y=3(=12-C.12)

if c = 2; 12)

X=3.2(=02-c.12),Y=7(=12-c.12)

ifc=-2; ifc=3.2; ifc=-3.2,

which exhaust all possibilities of c. Thus (3) has been proved. Let us prove (6). Let a, b E ZP; then ax 2 + by2 = z2 has a non-trivial solution in 7L/pZ, regarding a, b E Z/pZ by reduction modulo p. If z 0, then we regard x, y E Z C Z. and applying Theorem 3.1.4, ax 2 + by2 = zo

for some zo E ZP which means (a, b)F = 1. If z = 0, then -ab-1 is a

3 Quadratic forms over local fields

56

square in (7L/p7L)2 and applying the same theorem, -ab-1 = w2 holds for some w E 7Lp and then a 12 +b. w2 = 0 holds, which implies (a, b)F = 1. If, for a E 7Lp (a, p) F = 1, that is ax 2 +py2 = z2 has a non-trivial solution, then multiplying by a power of p, we may assume x, y, z E Z, and one of them is in 7LP . Then z E 7LP follows necessarily. Hence (a, p)F = 1 for

a E Z implies that a is a square in Z/pZ by reduction modulo p, and so (P) = 1. Conversely (P) = 1 implies a =_ z2 mod p and then by Theorem 3.1.4, a E F 2 and so (a,p)F=1. Let us prove (7). Let a E Z2. Suppose (a, 2)F = 1; then axe +2 y2 = z2 has a non-trivial solution in Q2 and we may assume x, y, z E Z2 and one of them is in 7L2 . Then x, z E Z2 holds, and a + 2y2 . 1 mod 8 since x2 z2 - 1 mod 8. Hence a = 1 mod 8 if y E 2Z2, and=- -1 mod 8 if y E 7L2 . Thus (2, a)F = 1 implies a ±1 mod 8. Conversely if a - ±1 mod 8, then ±a - 1 mod 8 and so ±a = x2 for x E Z2 . Hence a 12 + 2 0 = x2 if

a = x2, and a 12 + 2 x2 = x2 if a = -x2 which mean (a, 2) F = 1. Thus (2, a)F = 1 if and only if a = ±l mod 8. Next, for a, b E Z', we show (a, b) F = 1 if and only if a or b 1 mod 4. Noting a unit in Z2 (- 1 mod 8) is a square in Z2, we may assume a, b = ±1, ±3. First we show (a, b) F = 1 if a or b - 1 mod 4. We may assume

a - 1 mod 4, i.e. a = 1 or -3. (1, b) = 1 is clear and so we may assume

a = -3. Noting -7, -15 E (Z2 I)2 from -7 =_ -15 =_ 1 mod 8, we have 2 , -3 12 + 3 12 = 0, -3 12 - 3.22 = -152, which -3 12 -2 2 = yield (-3, -1)F = (-3,3)F = (-3,-3)F = 1. Hence a or b =- l mod 4 implies (a, b)F = 1.

Conversely suppose (a, b)F = 1 for a, b E Z'. Then axe + by2 = z2 has a non-trivial solution, where x, y, z E Z2 and one of them is in 7L2. If z E 2Z2, then axe+by2 - 0 mod 4 implies x, y E Z', and so a+b - 0 mod 4.

Thus a - 1 orb - 1 mod 4. If z E Z', then axe + bye - 1 mod 8, which implies that only one of x, y is in 7L2'2 . If, for example x E Z', then y E 2Z2

and we have a = 1 mod 4. Thus we have proved that (a, b) F = 1 yields a or b = 1 mod 4. (8) easily follows from (5), (6), (7) using (3).

To prove (9), we may assume, by using (3) and the fact that (a, a) _ (a, -1), that either a, b are different primes, or else a may be a prime or -1 and b = -1. Now (9) follows from (6), (7) and (2), (3), (4) in Theorem 3.2.1

3.4 The Hasse invariant Let the field F be a local field JR or @p as in the previous section. We define an important invariant S(V) for a regular quadratic space V over

3.4

Hasse invariant

57

F, which is called the Hasse invariant.

Theorem 3.4.1. Let V - (al) 1

I (an) be a regular quadratic space

over F. Then n

S(V) := J (ai,aj)F = fl(ai, II aj)F 1 3. Suppose m = 2; } is an orthogonal basis then {alul + a2u2i a2Q(u2)ul - a1Q(ul)u2i u3, whose value of v is equal to 1, and hence it is linked to {ui}. Thus we have completed the proof of the lemma. It should be remarked that the definition of Hasse's invariant is sometimes defined by replacing < by 5, we have only to show that a quinary quadratic space is isotropic. Suppose dim U = 5 and U is anisotropic. For a E d U, (a) ® U is also anisotropic and d((a) (& U) = a6 = 1 in F"/(F")2. Hence we may assume d U = 1. The discriminant of every quaternary subspace of U is equal to 1 in F" /(F" )2 by the case of m = 4. Hence we have U =15 (1). If p # 2, then S(13 (1)) = 1 = (-1, -1) and so 13 (1) is isotropic. This is a contradiction. If p = 2, then -7 E (Q2 )2 and 12+12+12 +22 + y / 2 = o show that 15 (1) is isotropic, which is again a contradiction. Hence, U is isotropic.

The above proof for m < 4 shows that the theorem is also valid for R.

3 Quadratic forms over local fields

62

Corollary 3.5.1. For a regular quaternary quadratic space U over Qp,

Q(U) = Q, holds.

Proof. For a E QP, U 1 (-a) is a regular quinary quadratic space and hence it is isotropic, which yields Q(U) E) a.

Corollary 3.5.2. If p # 2 and ai E Z (i = 1, 2, 3), then the quadratic space (al) 1 (a2) 1 (a3) over Q is isotropic.

Proof. The Hasse invariant is equal to 1 = (-1, -1) by (6) in Theorem 3.3.1.

Theorem 3.5.2. Let U, V be regular quadratic spaces over Q p. Then U V if and only if dim U = dim V, d U = d V and S(U) = S(V).

Proof. The " only if " part is clear. We use induction on m dim U to prove the " if " part. If m = 1, then U = V follows from U = (d U) (d V) = V.

Suppose m = 2 and put W = U 1 ((-1) ® V); then d W = 1 and S(W) = S(U)S((-1) ® V)(d U, d V)

= S(U)S(V)(-1,dV)(-1,-1)(dU,dV) = (-1,-1), using S(U) = S(V), d U = dV. Hence W is isotropic by the previous theorem, and there exist u E U, v E V (u # 0, v # 0) such that Q(u) _ Q(v) 34 0. Thus

U = (Q(u)) 1 (Q(u) d U) = (Q(v)) 1 (Q(v) d V) = V holds.

Suppose m > 3 and put W = U 1 ((-1) ® V); then dim W > 5 implies that W is isotropic and so we can take non-zero vectors u E U, v E V such that Q(u) = Q(v) 0. Write U = (Q(u)) 1 Uo, V = (Q(v)) 1 Vo. Then dim Uo = dim Vo and d Uo = d Vo are obvious, and S(U) = (Q(u), Q(u))S(Uo) (Q(u), d Uo)

and a similar formula for V implies S(Uo) = S(Vo). By the induction hypothesis, we have Uo = Vo and then U = V.

3.5

Classification of quadratic spaces over p-adic number fields

63

Corollary 3.5.3. Let U be a regular quaternary anisotropic quadratic space over QZ,. Then U = (1) 1 (-S) 1 (p) 1 (-Sp) where S E Z such that (p, S) = -1, (6, Z ) = 1. In particular, d U = 1, S(U) = -(-1, -1). Proof. By Theorem 3.5.1, U is anisotropic if and only if d U = 1 and S(U) = -(-1, -1), which are satisfied for the given form in the assertion. And from the theorem it follows that they are unique up to isometry if they exist. Hence we have only to prove the existence of S in the assertion.

If p = 2, then S = 5 satisfies (2, S) = -1, (S, Z) = 1 by (7) of Theorem 3.3.1. If p > 2, then we have only to take an integer S such that (P) = -1 by (6) of the same theorem. Thus the existence of S has been proved.

Corollary 3.5.4. Let U, V be regular quadratic spaces over Q2v with dim U < dim V. Then U is represented by V if and only if dim V - dim U >

3 or V=U

V = U I (d U d V)

if dim V = dim U, if dim V =dim U + 1,

V=UIH

if dimV=dimU+2 and dV=-dU,

where H is the hyperbolic plane.

Proof. If dim V - dim U = 0 or 1, then the assertion is clear. Suppose dim V - dim U = 2. If d V d U = -1, then the assertion is clear by Proposition 1.2.3. Suppose d V d U -1; then we take a, b E Qp such

that ab d U = d V and put W = U 1 (a) 1 (b). Clearly U is represented by W. Moreover dim V = dim W and d V = d W. Therefore S(W) = S(U) (d U, ab) (a, ab) (b, b)

= S(U)(dU,dUdV)(a,dUdV)(adUdV,-1) = S(U) (d U, d U d V) (d U d V, -1)(a, -dUdV).

Since - d U d V # 1 in QP /(Q2, )2, we can choose a such that S(W) _ S(V). Then W = V and so U is represented by V. Suppose dim V - dim U > 3. We have only to prove the case of dim V dim U= 3. Take a E Qp such that a d U d V 0 -1; then

dim V - dim(U 1 (a)) = 2 and dV d(U 1 (a)) # -1 are obvious and so by the above argument, U 1 (a) (and hence U) is represented by V.

4 Quadratic forms over Q

4.1 Quadratic forms over Q In this chapter, we prove the so-called Minkowski-Hasse theorem and the existence of a quadratic space over Q with prescribed local behaviours. Both are generalized over algebraic number fields. In number theory, we often say that Hasse's principle holds when some assertion over Q is true if and only if it is true over all local fields. It comes from the Minkowski-Hasse theorem. We denote by Sl the set consisting of all prime numbers and the formal symbol oo. We call an element of Sl a place (of Q); a prime number is called a finite place and oo is called the infinite place. For a quadratic space U over Q and place v E S2, we write U, for the scalar extension Q2vU of the quadratic space U, where we put Q2... = R. We do not use the letters p or q for the infinite place.

Theorem 4.1.1. Let U be a regular quadratic space over Q. Then U is isotropic if and only if Uv is isotropic for every place v E Q.

Proof. The " only if "-part is clear. We prove the converse by induction on dim U. If m := dim U = 1, then there is nothing to do, since a regular quadratic space with dim = 1 is anisotropic. Suppose m = 2 and put U = (a) 1 (b) (a, b E (Q'). Since Uv is isotropic, -ab E (QZx )2 for every place v E SZ by Proposition 1.2.3. Hence we have

4.1

Quadratic forms over Q

65

-ab > 0, taking v = oo and ordp(-ab) is even for v a prime p. Thus -ab is a square in Q', that is U is isotropic. Suppose m = 3; then U is isotropic if and only if (c) ® U is isotropic for c E 0'. Hence we may suppose

U = (1) 1 (-a) 1 (-b) for a, b E Q'. Moreover multiplying a, b by a square, we may assume that a and b are integers. We prove the case of m = 3 using induction on IaI + Ibl.

We note that (1) 1 (-a) 1 (-b) is isotropic if and only if x2 - aye = b has a solution. If Ial + Ibi = 2, then a, b = ±1, and since U,,,) is isotropic, a or b must be positive. Hence a or b = 1 and then U is isotropic.

Suppose Ial + IbI > 2 and Ial _< Ibi; then we have Ibi > 2. Moreover, dividing by a square if necessary, we may assume a and b are square-free integers. If a = 1, then U is obviously isotropic. Hence we assume a # 1. First we show that c2 = a mod b is soluble with c E Z. Since b is a product of different primes up to sign, we have only to show c2 - a modp is soluble for a prime divisor p of b. Since U, is isotropic, x2 -ay2 -bz2 = 0 has a nontrivial solution over Q2p and we may assume x, y, z E 7Lp and one of them is in 7Lp . Now b 0- 0 mod p2 implies y E 7Lp and then a - (xy-1)2 mod p follows, where xy-1 E 7Lp. Thus w2 - a mod p is soluble and so is w2 = a mod b. Hence there exists an integer c such that c2 = a + bd for some d E Z and we may assume Icl < b/2. Then IdI = Ice - aI/IbI 4 and put

U= V I W, V= (a,) 1 (a2), W= (a3) I ... 1 (am),

66

4


where we may assume all ai are integers, and put d = 2a1 am. Since Uv is isotropic for every place v, there exists b E QX such that by E Q(Vv) n -Q(Wv)

(1)

for every place v. Considering U(-1) := (-1) ® U if necessary, we may assume b,,, > 0. Write by = cppe(p)

(cp E 7LP', e(p) E 7L, e(p) > 0) and put

c = [J pe(p). pid

Now we need Dirichlet's prime number theorem:

Theorem. Every arithmetic progression a + kb, k = 1, 2, with a and b relatively prime contains infinitely many primes. Proof. For the proof see [BS]. Since cp(cp e(p))-1 is in Z for pid, there is a sufficiently large prime q such that q = cp(cp-e(p))-1 modp3 for every prime divisor p of d by using the above theorem. Put b = qc.

Then b/bp = qcp-1cp e(p) - 1 modp3 for pfd implies b/bp E (Z )2 for pfd, and hence (1) yields b E Q(Vv) n -Q(Wv)

(2)

for vId, and b,,. > 0 yields that (2) is valid for v = oo. If a prime p does not divide qd, then all ai and b are in 7Lp and then V 1 (-b) and W 1 (b) are isotropic over Qp by virtue of Corollary 3.5.2 and hence (2) holds for such a prime p. Thus (2) holds for all places v # q, which means that quadratic

spaces V' := V 1 (-b), W' := W 1 (b) over Q are isotropic over Qv for v # q. Let us show that Vq and Wq are also isotropic. V is a ternary isotropic quadratic space if v # q. Using (9) in Theorem 3.3.1 and 3.5.1, we have

H(-1, -1)Q = (-1, -1)@9

S(Vv) =

S(Vq) _ v#q

v,{q

which implies that Vq is isotropic. If m > 5, then dim W > 3 and ai E Z imply that Wq and hence Wq is isotropic by Corollary 3.5.2. Suppose that

m = 4; then dim W' = 3 and Wy is isotropic for the same reason that Vq is. Thus we have proved that V; and W; are isotropic for all places. Since dim V', dim W' < dim U, we can apply the induction hypothesis to conclude that Wand W' are isotropic which means b E Q(V), -b E Q(W). This means that U = V 1 W is isotropic.


4.1

67

Corollary 4.1.1. Let U be a regular quadratic space over Q and a E Qx. Then a E Q(U) if and only if a E for all places v. Proof. We have only to apply the theorem to U 1 (-a). Corollary 4.1.2. Let U, V be regular quadratic spaces over Q. Then U is represented by V if and only if Uv is represented by V, for all places v. Proof. We only have to show the " if " part. We use induction on dim U. If dim U is 1, then it is just the previous corollary. Write U = (a) 1 U' for C Q(V,) holds for all places v and then a E Q(V) a E @"; then a E

holds. Hence we can write V = (a) 1 V. By virtue of Corollary 1.2.3, U y VT, implies U;

* V; for all places v. The induction hypothesis tells

usU' -+ V'andsoU --+ V. Corollary 4.1.3. For regular quadratic spaces U and V over Q, U = V if and only if U, = V, for all places. Proof. This is a trivial corollary of the previous one.

Corollary 4.1.4. Let U and V be regular quadratic spaces over Q with

dimV-dimU>3. ThenU -* V if U.+VV. Proof. From Corollary 3.5.4 it follows that Up - Vp for every prime p, and then Corollary 4.4.2 yields the result.

Theorem 4.1.2. Let n be a natural number and U(v) be a regular quadratic space over Q with dim U(v) = n for all places. Then there exists a regular quadratic space U over Q such that U = U(v) for all places v if and only if the following three conditions hold: (i) There exists a non-zero rational number a such that

dU(v) = a in QX /(Q )2 (ii)

for all places v. S(U(v)) = 1 for all but finitely many places.

(iii) fv S(U(v)) = 1. Proof. If there exists a regular quadratic space U over Q such that U,, U(v) for all places, then (i) is clear and (ii), (iii) follow from (6), (9) in Theorem 3.3.1. Now suppose (i), (ii) and (iii). If n = 1, then we have only to put

U = (a). Suppose n > 2. Define a subset S of Il by

S = {2, oo} U {p I a Z } U {p I S(U(p)) = -1}.

68


4

S is a finite set. For v E S, we take azJ E Q(U(v))

(a # 0) and write ap = bppe(p) (bp E Z P', e(p) E 7L) for a prime p E S. Put c = sgn(aoo) fi pe(p) PE S

where sgn(a... ) = a,,,/Ia,,,,I and p runs over finite places in S; then is in 7LP' for p E S and using Dirichlet's theorem as in the proof of Theorem 4.1.1, we can take a sufficiently large prime number q such that q - bp(cp e(p))-1 modp3 for a prime p E S, bp(cp-e(P))-1

which implies qc/ap - bppe(p)/ap - 1modp3 since c/ap E Z'. Thus we have qc/ap E (Q )2 for p E S, p oo, and qc/a. > 0 is clear. So we have shown

qc/a E (Q,1 )2

(iv)

for v E S.

Suppose n = 2 and put U = (qc) 1 (qca); then d U = a = d U(v) in

1 U(v) holds by (iv). v /((Qv )2 by (1), and for v E S, U, = If p S and p # q, then qc, qca E Z implies S(Up) = 1 = S(U(p)) and hence Up = U(p). So,

S(Uq) = 11 S(U,,) = 11 S(U(v)) = S(U(q)) v56q

v#q

implies Uq = U(q) which completes the case of n = 2. Suppose n > 3. We claim first qc E Q(U(v))

that is, U(v) I (-qc) is isotropic over Qv for all places v. If n > 4, then it follows from Theorem 3.5.1 for finite places and from (iv) for the infinite place.

Suppose n = 3. It is clear for v E S by (iv). Suppose p

d(U(p) 1 (-qc)) = -aqc # 1 in @P /(@p )2,

S. If

4.1

Quadratic forms over (Q

69

then U(p) 1 (-qc) is isotropic by Theorem 3.5.1. Next suppose d(U(p) 1 (-qc)) = -aqc = 1 in Q /((QX )2; then

S(U(p) 1 (-qc)) = S(U(p)) (d U(p), -qc)(-qc, -qc)

= (a, -qc)(-qc, -qc) _ (-aqc, -qc) = 1 = (-1, -1) by p 2. Hence U(p) 1 (-qc) is isotropic by Theorem 3.5.1 Thus we have verified that U(v) 1 (-qc) is isotropic for all places v and the claim is proved. Therefore, qc E Q(U(v)) and we can write U(v) = (qc) 1 W(v) for every place v. Then d W (v) = aqc is clear. From S(U(v)) = (qc,qc)(qc,aqc)S(W(v))

it follows that the conditions (ii), (iii) hold for W(v) in place of U(v). Induction then gives the existence of a quadratic space W over Q such that W = W(v) for all places v. We have only to put U = (qc) 1 W.

5 Quadratic forms over the p-adic integer ring

Throughout this chapter, p is a prime number, and R and F denote the p-adic integer ring Z. and the p-adic number field Q., respectively. We denote by (a) the principal ideal aR for a E F. An important property of R is that R is a principal ideal domain, that is R is a commutative ring without zero-divisors and every ideal of R is principal. Then it is known that every (finitely generated) module M over R is isomorphic to R/Il ®. . ® R/I,,, where the Ik are ideals of R satisfying C I,, which are uniquely determined by M. Therefore if M is Il C torsion-free, i.e. ax = 0 (a E R, x E M) implies a or x = 0, then M has a basis over R. Let V be a vector space over F. A submodule L over R in V is called a lattice on V if L is finitely generated over R and contains a basis of V over

F. Then there exists a basis {vi} of V over F such that L = ®iRvi. We can define the discriminant dL of L by det(B(vi,vj)) x (R")2. This does not conflict with the definition in Chapter 1. By a regular quadratic lattice over R, we mean a lattice on some regular quadratic space over F. If K is a torsion-free (finitely generated) module over R, then K is a lattice on FK. In this sense, if there is no confusion, we often refer to a submodule M of possibly smaller rank as a regular (sub)lattice when FM is regular.

5.1 Dual lattices Let L be a regular quadratic lattice over R, i.e. a lattice on a regular

5.2

Maximal and modular lattices

71

quadratic space V over F. We define the dual lattice LO of L by

LO:={xEVI B(x,L)CR}. Let {vi} 1 (n := dim V) be a basis of L over R and {ui} be the dual basis

of {vi}, i.e. (B(vi,uj)) = ln. Then it is easy to see La = Rut ®... ® Run and hence

(Ll)a = L.

Put (v1, ... , vn) = (u1, ... , un)T

T E GLn(F). Then 1n = (B(ui, vj)) = (B(ui, Ek tkjuk)) = (B(ui, uj))T implies (1)

for

(2) T = (B(ui, uj))-1 = tT and then (B(vi, vj)) = tT(B(ui, uj))T = (B(ui, uj))-1 = T by (1) and (2).

Thus we have

detT=det(B(vi,vj)) =dL, dLO = (dL)-1 and Lp - ((B(vi,vj))-1). If B(L,L) C R, then LO

L and [LO

:

L] = pordp det T = pordp d L If

L = J 1 K, then La = JO 1 Kb trivially, where Ja and Ka are dual lattices of J and K on FJ and FK respectively. For lattices L, M on V, it is easy to see LDM c Ma, (L + M)a = La n Ma, Ltt + Ma = (L n M)l. For a submodule M of V, Ma is defined in FM as above if FM is regular.

5.2 Maximal and modular lattices Let L be a regular quadratic lattice over R and V = FL. We define the scale s(L) and the norm n(L) by s(L) = B(L, L), n(L) = { > aiQ(xi) I ai E R, xi E L}. Noting B(x, x) = Q(x), 2B(x, y) = Q(x + y) - Q(x) - Q(y), we have 2 s(L) C n(L) C s(L). Hence n(L) = s(L) if p 0 2, and n(L) equals either s(L) or 2 s(L) if p = 2. Let {vi} be a basis of L over R; then let us recall that we write L = (A) for A = (B(vi,vj)). If a -'A E GLn(R) (n = rankL) for some a E F", then

L is said to be an aR(or (a))-modular lattice. If a E R", then it is called unimodular. For an (a)-modular lattice L, s(L) = (a) (= aR) is clear and n(L) equals either (a) or (2a), and n(L) = (2a) if and only if all diagonals of A are in (2a).

For L = M 1 N, L is (a)-modular if and only if M and N are (a)modular.

72

5

Quadratic forms over the p-adic integer ring

Proposition 5.2.1. Let L be a quadratic lattice over R. Then L is (a)modular if and only if LO = a-1L. In particular, L is unimodular if and only if La = L.

Let {vi} be a basis of L, and {ui} be the dual basis,

Proof.

i.e.

(B(vi,uj)) = In. Then by the previous section, {ui} is a basis of LO and (vi,...

vn) = (ul ... ,un)(B(vi,vj)) and (B(ui,uJ)) = (B(vi,vj))-'.

If L is (a)-modular, then (B (vi, vj)) = aA for A E GLn(R) and hence (ul

. . .

, un) (vii =. .

.

,4J-n)(B(vi) vj))-1 = (v1, .

.

.

,

vn)a-1A-1

which means La = a-1L. If LO = a-1L, then

v,)a-1C for some C E GLn(R), which implies

In = (B(ui, vj)) = a-1(B(j: ckivk, vi)) = a-1tC(B(vi, vi)) k

for C = (cij) and so (B(vi, vj)) = atC-1. Thus L is (a)-modular. Proposition 5.2.2. Let L be a regular quadratic lattice over R and M be a submodule of L and suppose that M is (a) -modular. Then L = M 1 N holds for a submodule N if and only if B(M, L) C s(M). In particular, s(L) = s(M) implies L = M 1 M1. Proof. The "only if" part is clear. Let us prove the "if" part. For X E L, the mapping m H a-1B(m, x) is a homomorphism from M to R by virtue of B(M, L) C s(M) = (a). Hence there is an element y E Ma = a-1M such that a-1B(m, x) = B(m, y) for m E M, which implies x - ay E M1 fl L

because ay E M C L. Thus x E M 1 (M1 fl L) and hence L C M 1 (M1 fl L) C L holds and we have only to put N = M1 fl L. For a regular quadratic lattice L over R, L is called (a)-maximal if and only if n(L) C (a) and for any lattice M containing L properly, n(M) (a) holds.

Suppose that L is (a)-maximal and n(L) C (a) but n(L) # (a); if n(L) C (p2a), then p-1L D L and n(p 1L) C (a), which contradict the (a)-maximality of L. Hence for an (a)-maximal quadratic lattice L, we have n(L) = (a) or (pa). The following two theorems are most fundamental.

5.2


73

Theorem 5.2.1. Let L be a lattice on a regular anisotropic quadratic space V. L is (a)-maximal if and only if L = {x E V I Q(x) E (a)}.

Proof. First we claim that M := {x E V I Q(x) E (a)} is a module over R. Let x, y E V satisfy Q(x), Q(y) E (a), x 0, y # 0. Suppose 2B(x, y) (a); Then we define an integer n > 1 by (2B(x, y)pf) = (a). It is easy to see Q(x)Q(y)B(x, Y)-2 E (a)2(4p2na-2) _ (4p2n).

Hence by virtue of Theorem 3.1.4, 1 - Q(x)Q(y)B(x, y)-2 is a square of some element c E R" and then Q(x)Q(y) - B(x, y)2 = -(cB(x, y))2 0. It means that a binary subspace F[x, y] (C V) is a hyperbolic plane and hence it is isotropic, which contradicts the assumption. Thus we have 2B(x, y) E (a) and then Q(x + y) E (a), meaning x + y E M. Thus M is a module.

Next we show that M is finitely generated. Let {vi} be a basis of V over F such that vi E M for all i. Put K = ®Rvi and let N be a finitely generated submodule such that K C N C M. Then

s(N) C 2 n(N) C (2) implies (d N) C (2) m (m := dim V). Since d K = [N : K] 2 d N implies

R D ([N: K]2) = (dK (dN)-') D

(dK)(a)-_,

[N : K] is bounded and hence M is finitely generated over R and (a)maximal.

Let L be an (a)-maximal lattice; then L C M is clear, and then the maximality of L and n(M) C (a) imply L D M and hence L = M. Theorem 5.2.2. Let L be a lattice on a regular quadratic space V over F. If L is (a)-maximal, then there exist hyperbolic planes Hi and an anisotropic subspace VO such that

V=LiHi LVo, L=1i(LnHi)L(LnV0), LnHi=(2(01

0)) and L n V0 is (a)-maximal. In particular, (a)-maximal lattices on V are mutually isometric.

To prove this, we prepare with three lemmas.

74

5


Lemma 5.2.1. Let L be a regular quadratic lattice over R with rank L = m. If n(L) C (a) and ((2a-1)' d L) = R or pR, then L is (a)-maximal.

Proof. Let M be a lattice on FL such that M D L and n(M) c (a). Then we have d L = [M: L]2 d M and (d M) C s(M)m c (2-1 n(M))'m' C (a/2)m. Thus ((2a-1)m'dL)

= ([M :

L]2(2a-1)m'dM)

C [M : L]2R

and so by the assumption we have [M : L] = 1, i.e. M = L. Therefore L is (a)-maximal.

Corollary 5.2.1. If L is unimodular with n(L) = 2R, then L is (2)maximal.

Proof. We have only to put a = 2 in the lemma.

Lemma 5.2.2. For c E F" and a lattice L on the hyperbolic plane H, the following three conditions are equivalent: (i)

L- ((0 c));

(ii) L is (c) -modular and n(L) C (2c); (iii) L is (2c)-maximal.

Proof. (i) = (ii) is clear. (ii) (iii) follows from the previous lemma, putting a = 2c. Let us prove (iii) (i). Let {e1, e2} be a basis of L such that Q(e1) = 0, B(e1, e2) = a, Q(e2)=b for a, b E F.

Note that a E s(L) C 2-'n(L) C (c) and b E n(L) C (2c). We show (a) = (c). If (a)

(c), then a E (cp) holds, and then for x1, x2 E R we have

Q(xip-1e1 + x2e2) = 2x1x2p 1a + x2 2b E (2c).

It means L C R[p'e1i e2], L 0 R[p-1e1i e2] and n(R[p-1e1i e2]) C (2c), which contradicts the maximality of L. Thus we have shown (a) = (c) and

can put c = au (u E R'). Then b(2a)-1 E R and so L = R[uel, e2 b(2a)-1e1] = (0 c)) which completes the proof. Note that c in (i)

clan

be replaced by cu for u E R'.

5.2


75

Lemma 5.2.3. Let L be an (a)-maximal lattice on an isotropic regular quadratic space V over F. For an isotropic primitive element x in L, there exists an isotropic element y E L such that L = R[x, y] I R[x, y]1, B(x, y) = a/2 and R[x, y] is (a)-maximal.

Proof. We show B(x, L) = (a/2). From the definition follows B(x, L) C s(L) C 2-1 n(L) C (a/2). So, we assume B(x, L) C (ap/2); then Q(w + p-1x) = Q(w) + 2p-1B(w, x) E (a)

for w E L. Hence L C L+p 1Rx, L# L+p-1Rx and n(L+p-1Rx) c (a), which contradicts the maximality of L. Thus we have shown B(x, L) =

(2-1a), and we take z E L such that B(x, z) = 2-1a and put y = z a-1Q(z)x E L. Then

M:=R[x,y](2-via 20-'a )) and so M is (2-1a)-modular and hence B(M, L) c s(L) c (2-1a) = s(M). By Proposition 5.2.2, we have L = M 1 M-'-. By the previous lemma, M is (a)-maximal. Using Lemma 5.2.3 repeatedly, we have the orthogonal decomposition of L in Theorem 5.2.2. Since L is (a)-maximal, L n Vo is also (a)-maximal. The number of hyperbolic planes is ind V and then by Corollary 1.2.3 the isometry class of an anisotropic subspace Vo is uniquely determined by V. Then Theorem 5.2.1 yields the class of (a)-maximal lattices which are uniquely determined up to isometry. Thus we have completed the proof of Theorem 5.2.2. We know the following stronger assertion.

Exercise 1. Let V be an isotropic regular quadratic space over F. For maximal lattices L, N whose norms are not necessarily equal, show there exists a hyperbolic

plane H C V such that

L = (LnH) 1 (LnH1), N = (NnH) 1 (NnH-i). Exercise 2. For an R-maximal lattice M, ((102

1//2)) 1 M is also R-maximal.

But in general, the orthogonal sum of R-maximal lattices is not R-maximal.

5

76


Exercise 3. Let V be the 4-dimensional anisotropic quadratic space over F and L be the R-maximal lattice on it. Prove L

(1)1(-S)1(p)L(-6p)

ifp

0112

ifp=2,

112))1((2 2))

2,

where 6 E R" such that the quadratic residue symbol (P) = -1.

Lemma 5.2.4. Let V be a regular quadratic space over F. Then for a submodule L over R with n(L) C (a), there exists an (a)-maximal lattice on V which contains L. Proof. Let {vi}m 1 be a basis of V over F such that a subset {vi} 1 is a basis of L over R (n < m). We put M = ®i 1Rva ®%' -n,+1 Rptvi

for an integer t. If t is sufficiently large, then n(M) C (a). If a lattice N on V satisfies N D M and n(N) C (a), then d M = [N : M]2 d N and dN E s(N)m C (2-1a)-. Hence [N: M]2R = d M(d N)-1R E) d M (2-1a)-and so [N : M] is bounded. Therefore there exists a maximal lattice K with respect to inclusion such that K D M and n(K) C (a). Such a lattice K is an (a)-maximal lattice. Theorem 5.2.3. Let L and M be regular quadratic lattices over R and suppose that rank L + 3 < rank M, M is (a) -maximal and n(L) C (a). Then L is represented by M.

Proof. Put V := FL, W := FM; then by virtue of Corollary 3.5.4, V is represented by W and hence there exists a module K (c W) isometric to L. By the previous lemma, there exists an (a)-maximal lattice N on W which contains K. From Theorem 5.2.2, it follows that M and N are isometric, and hence L is represented by M. Now we give a few facts about unimodular lattices. Theorem 5.2.4. Suppose p 2. For a unimodular lattice L, we have

Since n(L) = s(L) = R, there is an element v1 E L such that Q(v1) E R', and then Proposition 5.2.2 implies L = Rvi 1 Rvi, where vi is unimodular. Repeating this, L has an orthogonal basis, that is L Proof.

(e1) 1 ...

I

(E,,), ei E R". Put M = (1) 1

1 (1) 1 (d L) with

rank M = rank L. Hasse invariants of both L and M are 1 by (6) in Theorem

3.3.1, and then by Theorem 3.5.2 we have FL = FM, and L and M are R-maximal by Lemma 5.2.1. Theorem 5.2.2 yields L = M.

5.2


77

Corollary 5.2.2. Suppose p : 2 and let K be a unimodular lattice of rank K > 3. Then Q(K) = R holds. Proof. Put

L=(I

0 I)1(1)1

1(1)1(-dK)

with rank L = rank K. By the theorem, K = L and hence we have

R=Q((0

0)))cQ(K)cR.

Lemma 5.2.5. Let p = 2 and L be a unimodular lattice over R. Then L is an orthogonal sum of unimodular lattices of rank 1 and binary unimodular lattices with norm (2). Proof. Since s(L) = R, we have n(L) = R or (2).

If n(L) = R, then

there exists an element v E L such that Q(v) E R" and Proposition 5.2.2 yields L = Rv L v1. Suppose n(L) = (2); since there exist v1, v2 E L such that B(vl, v2) = 1 by virtue of s(L) = R, M := R[vl, v21 is unimodular by Q(vi) E (2) (i = 1, 2). The same proposition implies L = M L M1. Repeating this, we complete the proof. Lemma 5.2.6. Let p = 2 and L be a binary unimodular lattice with n(L) _ (2).

If L is isotropic, then L = ((1 0) ), d L = -1, S(FL) = -1 and Q(L) = (2).

If L is anisotropic, then L =

((1 2) ), d L = 3, S(FL) = 1 and

Q(L) = {0} U {x E R I ord2 x- 1 mod 2}.

Proof. L is (2)-maximal by virtue of Corollary 5.2.1. If L is isotropic, then the assertion follows from Lemma 5.2.2. Suppose that L is anisotropic;

then L = ((21 a

1)) for some a, b E R. If ab =- 0 mod 2, then

dL = -(1 - 4ab)(Rx)2 = _(Rx)2 and so FL is isotropic, which is a contradiction. Thus ab - 1 mod 2 and so d L = 3. Put r := 1 - a - b E R" ; then are + r + b - 1 mod 8 is easy. Since Q(L) E) 2(ar2 + r + b), we have 2 E Q(L). Hence

FL =(2)1(2.3)=((1

2))

78


5

holds over F, and since L and ((2 implies L =

2

I

2

)) are (2)-maximal, Theorem 5.2.2

I ). It remains to show

1

Q(L)={xERIx=0or ord2 x is odd}. For x E F", x E Q(FL) e-* FL I (-x) is isotropic S(FL 1 (-x)) = -1 by Theorem 3.5.1, and S(FL 1 (-x)) = (-3, x)Q2 = (-1)n for n = ord2 x. Thus 0 # x E Q(FL) # ord2 x is odd. Thus the proof is complete. Theorem 5.2.5. Let p/= 2 and L be a unimodular lattice. If n(L) = 2R,

then L = H or H 1 (I copies of (I 0

2 I) where H denotes an orthogonal sum of

2

1 )). If n(L) = R, then L has an orthogonal basis.

Proof. Suppose n(L) = 2R; then L is (2)-maximal by Corollary 5.2.1. By Lemma 5.2.5 L is an orthogonal sum of binary unimodular lattices with norm=(2), which are isometric to

(0 0)) or ((2 2)) by Lemma 5.2.6.

To complete the proof in the case of n(L) = (2), we have only to show 1

((1

2))_L((2 1

2)) _

((O1

0))

((1 1

0)

By Corollary 5.2.1 both are (2)-maximal, and hence we have only to prove that they are isometric over F by virtue of Theorem 5.2.2. It is easily done by comparing discriminants and Hasse invariants.

Next suppose n(L) = R; then by Lemma 5.2.5 L has an orthogonal component of rank 1. Hence we have only to show that M := R[vi, v2] L Rv3 with Q(vi) = Q(v2) = 2a (= 0 or 2), B(vl, v2) = 1, Q(v3) E R" has an orthogonal basis. It is easy to see M = R[vi + v3] L R[vi - 2av2, v2 - `v%(v3)-lv3],

where both orthogonal components are unimodular and the norm of the second one is R. Hence by the proof of Lemma 5.2.5, the second lattice has an orthogonal basis and so M has an orthogonal basis.

Proposition 5.2.3. Let p = 2 and ei E R" (i = 1, 2, 3). Then (El) L (E2) L (E3) = ((

2a 1

1

2a)) L

(E)

5.3 Jordan decompositions

79

fore E R" and a = 0 or 1. If a = 1, then it is anisotropic. Proof. Let L = Rv1 J Rv2 1 Rv3 and Q(v2) = ei; then M := R[vl + v2i v2 + v3]

e1 + e2 (

E2

e2

1

e3)(2a

E2 +

2a)

1

for a = 0 or 1 by Lemma 5.2.6. By Proposition 5.2.2, we have L = M 1 M1. When a = 1, L is anisotropic by Lemma 5.2.6. In the above proof, if El + E2 0 mod 4, then M is isotropic by discriminant consideration, and then a = 0 holds by Lemma 5.2.6.

Proposition 5.2.4. If p = 2 and L is unimodular with rank L > 5, then L=

((0

\

1

I) 1 (some lattice).

Proof. If n(L) = 2R, then Theorem 5.2.5 implies the assertion. Otherwise,

L has an orthogonal basis {vi} (Q(vi) = ci). If ei + Ej = Omod4 for some i, j, then M := R[vi + vj, vj + vk] =

((0 1)) for k 0 i, j by the

note just before the proposition, and M splits L. If ei + ej - 2 mod 4, i.e. Ei - ej mod 4 for all i, j, then

M:=R[vl+v2+v3+v4i v2+v3+v4+v5]^-( (0 1

1) 0

)

and L is split by M.

5.3 Jordan decompositions Let L be a regular quadratic lattice over R. If s(L) = n(L), then there exists an element u E L such that (Q(u)) = s(L) and by Proposition 5.2.2 we have

L = Ru 1 u1. If s(L) # n(L), i.e. n(L) = 2s(L) and p = 2, then there exist v, w E L such that Q(v), Q(w) E 2 s(L) and (B(v, w)) = s(L), and then M := R[v, w] is an s(L)-modular lattice and by the same proposition,

L = M 1 M1 holds. Thus repeating this argument, L is an orthogonal sum of modular lattices and we can write

where Li is s(Li)-modular and s(Li) D D s(Lt), where every inclusion is proper, i.e. s(Li) s(L2+1) This decomposition is called a Jordan decomposition, and each Li is called a Jordan component. In general, a Jordan decomposition is not unique. The following is fundamental.

5

80


Theorem 5.3.1. Let L be a regular quadratic lattice over R, and L = Ll 1 1 Lt = K1 I 1 Ku be Jordan decompositions. Then we have (i) t = u, (ii) s(Li) = s(Ki), and rankLi = rankKi for 1 < i < t (iii) n(Li) = n(Ki) for 1 < i < t, (iv) n(Li) = n(L) and s(Li) = s(L). Proof. For a regular quadratic lattice M over R, we put

M(a) :_ {x E M I B(x, M) C (a)} = aMa fl M C M. Then s(M(a)) C (a)

(1)

is clear. Suppose that M is (s)-modular; then M(a) = as-1M fl m by Proposition 5.2.1. Hence when (a) D (s), M(a) = M is (s)-modular, and when (a) C (s), M(a) = as-1M is (a2s-1)-modular. Thus, for an (s)modular lattice M,

M(a) is (a)-modular if and only if (s) = (a), and (s) (a) implies s(M(a)) C (pa). I Lt(a), from (1) and (*) it follows that Since L(a) = Li(a) I s(L(a)) = (a) if and only if (a) = s(Li) for some i. Thus the set of s(Li) (1 2. Thus we have n(Li) = n(L) = n(Ki), which is the assertion (iii). L/pL becomes a symmetric bilinear space over Z/pZ by B(x, y) mod pR and L1 induces a regular subspace and Ij>2 Lj induces (L/pL)' and rank L1 = rank L - dim(L/pL) and the similar assertion about the decomposition L =1 Ki. Thus we have rank L1 = rank K1. s(Li) = s(L) is clear. Finally n(Li+1) C s(Li+1) C ps(Li) C n(Li)

implies n(L) = n(Li).


81

Theorem 5.3.2. Suppose p 2. For a regular quadratic lattice L over R, let L = L1 1 1 Lt = K1 1 1 Kt be Jordan decompositions. Then Li = Ki holds for every i.

Put s(Li) = s(Ki) _ (a); then as in the proof of the previous theorem L(a) = Li 1 = Ki 1 become Jordan decompositions. Hence we may assume i = 1 and s(Li) = s(Ki) = R by scaling. Then Proof.

L/pL becomes a quadratic space over Fp =: Z/pZ. While L1 and K1 induce regular subspaces, both 1j>2 Lj and 1j>2 Kj become Rad(L/pL).

Thus Li/pLi - Ki/pKi over Fp and implies dL1 = dK1 in F 1. Let L = L1 1 be a Jordan decomposition; we may suppose s(Li) = R by scaling and then s(L) = R and L1 is unimodular. Let o, E O(L), and in Theorem

1.2.2putR=R,P=(p)andU=H=L, M=L1i N=a(Li),and

q(x) = Q(x), b(x, y) = 2B(x, y). Then conditions (1), (2), (3) and (5) are satisfied, and therefore there is a T E O(L) such that T is a product of

82

5


symmetries in O(L) and r = o on L1. Thus r-1o = id on L1 and hence T-1vILi E O(Li) is a product of symmetries by the induction hypothesis. A symmetry in O(L-) induces canonically a symmetry in O(L) which is the identity on L1. Thus r-1o is a product of symmetries in O(L), so or is a product of symmetries in O(L). Henceforth, we study the case of p = 2, but we do not necessarily assume

p=2.

Theorem 5.3.4. Let L be a regular quadratic lattice over R and M, N be submodules of L. For a E F', we put La

La

{x E L I B(x,L) C (a)}, {x E La I Q(x) C (2a)}.

Suppose that M and N satisfy (3)

HomR(M, R) _ {x H a-1B(x, y) I y E La}, { HomR(N, R) _ {x a-'B (x, y) I y E La}.

If or : M = N satisfies

a(x) - x mod La for x E M,

(4)

then or can be extended to an isometry of L which satisfies

o(x) - x mod La

for x E L.

Proof. It is easy to see that La and La are submodules of L. Putting

h:=o(x)-xELa for xE M, Q(o(x)) = Q(h) + 2B(h, x) + Q(x) implies Q(h) = -2B(h, x) E (2a) and hence h E La. Thus we may replace La in the condition (4) by La. Putting H := La and taking scaling by (2a)-1, we introduce a new quadratic form

q(x) = (2a)-1Q(x), and then b(x, y) := q(x + y) - q(x) - q(y) = a 1B(x, y)

and Q(H) C (2a) yield

b(H, L) = a-1B(H, L) C R and q(H) C R. Applying Theorem 1.2.2 to

R=R,P=(p),U=L, we complete the proof.


83

Lemma 5.3.1. Let L be a regular quadratic lattice over R and suppose L = M I M1 = N J N1 where M and N are unimodular with n(M) _ n(N) = (2). Then M1 = Ni holds if M and N are isometric. Proof. We apply the previous theorem for a = 1. To do it, we must verify

the conditions (3), (4). Since La D M, N and s(La) C s(L') C (a) = R, Proposition 5.2.2 implies La = M I * = N I * , where * is an appropriate lattice, and then the condition (3) in Theorem 5.3.4 is satisfied. For an isometry v : M - N, let us verify the condition (4). For x E M and y E L, we have

B(v(x) - x, y) = B(v(x), y) - B(x, Y) E B(N, L) - B(M, L) = R

and hence o(x) - x E La for x E M, which is nothing but the condition (4). Thus the theorem implies that o can be extended to an isometry of L, and we have proved the lemma.

Theorem 5.3.5. Let L = M 1 M1 = N I N1 be a regular quadratic lattice over R and suppose that M and N are isometric and for every Jordan component K of M, n(K) = 2 s(K). Then M1 = Ni. Proof. By scaling, Lemma 5.3.1 holds if M and N are isometric, modular and n(M) = 2 s(M). Applying it repeatedly to each Jordan component of M and N in the theorem, we complete the proof.

Theorem 5.3.6. Suppose p = 2. Let L = M I M1 = N I N1 be a regular quadratic lattice over R. Suppose that M = N - (E) for E E R", s(M1) =

s(Ni), n(Mi) = n(Ni) and either s(Mi) C (2) or s(Mi) = n(Mi) = R. Then M1 - N1.

Proof. Put M := Rv. First suppose s(Mi) C (2). We apply Theorem 5.3.4 for a = 2; then La = 2M I M1 = 2N I N1 and La D 2M, 2N. Hence (3) is obviously satisfied. Putting v(v) = by + w (b E R, w E M1) for an isometry o : M = N, we have Q(o (v)) = b2Q(v) + Q(w) and (b2 - 1)Q(v) = -Q(w) E s(Mi) C (2). Since Q(v) E R", we have b E R". Thus, for y E L

B(v(v) - v, y) = B((b - 1)v + w, y) = (b - 1)B(v, y) + B(w, Y) E (b - 1)B(M, L) + B(M1i L) C (2)

because s(Mi) C (2). This means v(v) - v E La, which is the condition (4). By the theorem, or is extended to an isometry of L and then M1 - N1.

Next suppose s(Mi) = n(Mi) = R; then s(L) = R. We apply the same theorem for a = 1; then clearly La = L and hence the condition (4) is also clear for v : M - N. We have only to verify the condition (3). Now

84

5


n(Mi) = R implies that there exists x c- M1 such that Q(x) E R". Now v + x E L1 follows from Q(v + x) = Q(v) + Q(x) - 0 mod 2. By

B(v, v + x) = Q(v) E R", the condition (3) is verified for M and similarly for N.

Corollary 5.3.2. Let L be a regular quadratic lattice over R and L = M1 1 M2 1 = N1 1 N2 1 ... be Jordan decompositions. If M1 = N1, then Mi ^_' Ni holds.

Proof. By scaling, we may assume that M1 (= N1) is unimodular. If n(Mi) = n(Ni) = (2), then we have only to apply Lemma 5.3.1. If n(Mi) = n(Ni) = R, then by Theorem 5.2.5 M1 and N1 have an orthogonal basis.

Noting rankM1 = rankN1i n(MI) = n(Ni) and s(Mi) = s(Ni) C (2) by Theorem 5.3.1, Theorem 5.3.6 yields the corollary inductively on rank M1.

Theorem 5.3.7. Let V be a regular quadratic space over F. Suppose that L and M are unimodular lattices on V with n(L) = n(M). Then L = M holds.

Proof. First, we note that d L = d M since they are units and equal to d V in F/(F" )2. If p # 2, then Theorem 5.2.4 completes the proof. Assume p = 2. Now we claim, first that if K is a unimodular lattice on a hyperbolic space, then

(e)IK--(e)I(I(I0 ' )))forEERx. Let us prove it by induction on rank K. Suppose rank K = 2. Using either Proposition 5.2.3 if n(K) = R, or Lemma 5.2.6 if n(K) = 2R, we have

(e)IK(1

2))1(-3e)or0

1))1(e).

Now

1)))={xERIx=0or ord2x is odd}

2

in Lemma 5.2.6 yields that

((1

2)) 1 (-3e) is anisotropic. Since K

is isotropic, it is a contradiction and so (e) 1 K = (I

1

2 )) 1 (e). If

rank K > 4, then by Proposition 5.2.4, (e) 1 K = (1 ((0

2

I )) 1 H,


85

where rank H = 1 or 3 and H is unimodular. If rank H = 1, then we have H = (e), comparing discriminants. If rank H = 3, then H is unimodular and isotropic, comparing Witt indices. Moreover H has an orthogonal rthogonal basis by Theorem 5.2.5 and then Proposition 5.2.3 implies H = (I Thus we have proved the claim.

I) 1 (e).

\\\

If n(L) = n(M) = (2), then Theorem 5.2.5 completes the proof. Hence

we may assume n(L) = n(M) = R; then L and M have an orthogonal basis. Applying the above to K = M 1 L(-1) and L 1 V-1), we have

L1(M1L(-1))=L1(lrankL ((0 0))),

M1 (L 1L(-1))MI(lrankL ((0 0))) Since the left-hand sides are isometric, we have L = M by Lemma 5.3.1.

Proposition 5.3.2. Let N be a maximal lattice isometric to the orthogonal sum of n copies of (

(1

10) ) and M be a regular quadratic lattice over R

with rank M = n and s(M) C 2R. Then there is a primitive submodule MO

in N such that MO = M and Mo = M(-1). Proof. By using a Jordan splitting of M, we may assume that either n = 1

and M = (a) or n = 2 and M = (2b (l2 c 1) ), where the latter occurs only when p = 2 and c = 0 or 1. Suppose M = (a) and N = R[e, f] with Q(e) = Q(f) = 0, B(e, f) = 1. Put v = e + a/2 f; then Q(v) = a and

v1 = R[e - a/2 f] = (-a).

/ 2c

Next, suppose p = 2 and M = R[zl, z2] = (2b 1 R[ul,

,

2c)) and N =

u4] such that Q(E xjui) = 2(xlx2 + x3x4). Putting V2 := ul +2 b U2 - 2bu3 + U4, wl := ul + 2bu2, w2 := 2bu2 + u3 + 2bu4, v1 := U1,

then

M1 := R[vl, v21 - (2b (1 M2

R[wl, w2]

(2

b

01) )>

2

1

1

2

)

86

5


are clear. M1, M2 are primitive and

Mi = R[ul - 2bu3, 26u3 + u41 = (1 M2 = R[ul -

26u2

Ob

- 2bu4, u3 - 26u4] ti

_26 2b+1

) = Mi

_26+1

-2b

-2b -2b+1

)-

M2-1)

Now, we have only to take MO = Ml and M2 for c = 0 and 1 respectively.

0 Proposition 5.3.3. Let N be a regular quadratic lattice over R with s(N) C R and let L be a regular submodule of N. Then d Ll divides d N d L, and the number of submodules of N which are isometric to L and are not mutually transformed by isometries of N is finite.

Proof. For x E N, z H B(x, z) is a homomorphism from L to R and hence

there exists y E La such that B(x,z) = B(y,z) for z E L. Define the mapping 0 from N to La by O(x) = y. Since ker 0 = L-1- and O(x) = x for

x E L, ¢-1(L) = L 1 L1 and implies [N: L 1 L1] = [¢(N) : L], which divides [La : L]. Then

ordp(dLdL1) = ordl,([N : L 1 L1]2 dN) < ord,([LO : L]2 dN) implies 0 < ordp d L1 < ordp(d N d L) from which it follows, upon considering the various Jordan decompositions of L1 with s(L1) C R, that there are only a finite number of isometry classes of L1. For a given K which is one of the possibilities for L-1-, there are only finitely many lattices

isometric to N which contain L 1 K. This completes the proof.

5.4 Extension theorems In this section, we give some results of Witt type theorems. The following is essentially due to Eichler.

Theorem 5.4.1. Let L be a regular quadratic lattice over R, and M and N be isometric submodules of L. We put M := FM n L and N := FN n L and suppose

M/M-N/N=R/(pal)®...®R/(p°"`) for 0 < al
2 or p = 2 and rank L > 3 holds.

Proof. By scaling, we may suppose that L is unimodular. First suppose p 0 2 and rank L > 2. Let a E RI; then by Theorem 5.2.4, L

(a)J (adL)J

If v E L gives a basis of (a), then Tv E O(L) and 9(Tv) = a. Thus we have 0(0+ (L)) D R".

Suppose p = 2. Let M = (I

is

2c I) for c = 0 or 1; then Lemma 5.2.6

2R". For a E R", we can take v E M such that Q(v) _ 2a, and then Tv E O(M) is easy to see. 9(T,) = 2a yields 9(0+(M)) D R" . L is split by M for c = 0 or 1 either in the case of (i) or (ii) with n(L) = 2s(L) by Theorem 5.2.5, and in the case of (ii) with n(L) = s(L) by Proposition 5.2.3. Thus we have 9(0+(L)) D 9(0+(M)) D R". implies Q(M)

Proposition 5.5.5. If L is unimodular with rank L -> 2 and p 0 2, then 0(0+ (L)) = R" (F" )2. Proof. By Theorem 5.3.3, O(L) is generated by symmetries. Let Tv E O(L)

and we may assume that v is a primitive element in L since Tav = Tv

for a E F". Now r, (x) = x - 2B(x,v)Q(v)-lv E L for x E L implies B(L, v)Q(v)-' E R, and then Proposition 5.2.2 implies L = Rv 1 v1. Thus Q(v) E R" and hence 9(0+(L)) C R" (F" )2. The converse inclusion follows from the previous proposition.

94

5


5.6 Local densities In this section, we introduce the notion of local density, which is an important invariant of representations of quadratic forms over R. By Siegel's theorem, which will be proved in the next chapter, it turns out to be important for representations of quadratic forms over Z. First, we show the following fundamental lemma.

Lemma 5.6.1. Let S and T be regular symmetric matrices over R of degree s and t respectively. Put

Et(R) := {B = tB E Mt(R) I all diagonals of B E (2)},

and take an integer h such that phT-1 E Et(R). For G E Ms,t(R) and integers r, e > 0, we put Apr (T, S; G, pe) :_

{X E M8,t(R)modp' I S[X] -Tmodp"Et(R),X - Gmodpe}. If r > h + max(e,1), then (pr+1)t(t+1)/2-st#Ap,r+, (T, S; G, pe) = (pr)t(t+1)/2-st#Apr (T, S; G, pe) holds.

Proof. Since

S[x + pry] = S[x] + pr(txSy + tySx) + p2rS[y] for x, y E Ms,t(R), the set Apr (T, S; G, pe) is well-defined if r > max(e, l). Since T is integral, the integer h is non-negative and so the sets Apr, Apr+1

are well-defined for r > h + max(e, 1). Let {Ti}a 1 be a complete set of representatives of symmetric matrices T' mod pr+1 Et (R) such that

T' -TmodprEt(R). The number a of representations is equal to [Et(R) : pEt(R)] = pt(t+1)/z We define the mapping 0 from the disjoint union of Apr+1 (Ti, S; G, p') to Apr (T, S; G, pe) by

O(X) = X mod pr. This is well-defined. For X E Apr (T, S; G, pe) and Y E Ms,t (R), we put

X' = X + prY; then S[XI] = S[X] + pr(tX SY + tySX) + p2'S[Y] - S[X] - T mod prEt (R)

5.6 Local densities

95

and hence S[X'] __ Tj modp"'+lEt(R) for some j because of the definition of the set {Ti}. Thus 0 is surjective and the number of inverse images is pSt and so pst#Apr (T, S; G, pe) = 1: OApr+1(Ti, S; G, pe).

Now let us show OApr+1 (Ti, S; G, pe) = OApr+i (T, S; G, pe).

There exists

Uj

E

GLt (R) such that

= T [Ui] and Ui

Ti

1 mod p' T-1Mt(R), applying Corollary 5.4.4 to A T, B := Ti and h r - 1. Since r > h + e, p'rT-1 - 0 mod pe holds and hence Ui - 1 mod pe. Thus we have Apr+1(Ti, S; G, pe) = OApr+1(T, S; G, pe). Hence we have pst #Apr (T, S; G, pe) _

OApr+1(Ti , S; G, pe) = pt

(t+l) /2 jApr+1(T S; G, pe) , ,

which implies the assertion.

Lemma 5.6.2. Let S, T, G and Apr (T, S; G, pe) be those in Lemma 5.6.1. For a non-negative integer a, we have #FApr+a

(paT,pa S+. G, pe) = past#Apr (T,

S; G,pe) for r > h + max(e, 1).

Proof. It is easy to see #Apr+a (paT, pa S; G, pe )

=#{X E M3,t(R) mod p'+aM8,t(R) I pas[X]

paT mod pr+aEt(R),

=

X = Gmodpe} =O{X E M8,t(R) modpr+aM8 t(R) I S[X] __ T mod p'Et(R), X - G mod pe } =past IApr

(T, S; G, pe )

0 Lemma 5.6.3. Let S and T be regular symmetric matrices over F of degree s and t respectively, and G E Ms,t(R). Put Bpr (T, S; G, pe )

{X E Ms,t(R) mod pr'S-1Ms,t(R) I S[X] - T mod pr'Et(R),

X - Gmodpe}.

96


5

Let h' be an integer such that ph' S-1 E ES (R). Then Bpr (T, S; G, pe) is well-defined for r > h' + e. For an integer a and a sufficiently large r, we have OBpr+° (paT, pa S; G, pe) = OBpr (T, S; G, pe)

Moreover if S and T are integral and r > h' + max(e, 1), then we have #Apr (T, S; G, pe) = pt ordp (det S) OBpr (T, S; G, pe).

Proof. For X, Y E M,,t(R), it is easy to see

S[X +

prS-1Y]

= S[X] + pr(tXY + tYX) + p2rS-1 [Y] S[X] modprEt(R) for r > h',

and if r > h' + e, then prS-1Ms,t(R) C pe(ph'S-1)M,,t(R) C peM,,t(R)

Hence for r > h' + e, Bpr (T, S; G, pe) is well-defined. For an integer a and a sufficiently large integer r, we have #Bpr+" (paT, pa S; G, pe)

=#{X E M8,t(R) mod prS-1Ms,t(R) I paS[X] - paT mod pr+aEt(R), X - G mod pe } =IiBpr (T, S; G, pe)

Suppose that S and T are integral and r > h' + max(e, 1); then OApr(T,S;G,pe) = #Bpr (T, S;

G,pe)[S-1M8,t(R)

: Ms,t(R)]

is easy to derive. It gives the required equation.

Lemma 5.6.4. Let S and T be regular symmetric matrices over F of degree s and t respectively, and G E M8,t(R). Then for a sufficiently large

r, (pr)t(t+1)/2-sty}Bpr an (T, S; G, pe) is independent of r. Proof. Let a be

integer such that paT and paS are integral; then for a

sufficiently large r, the previous lemma yields #Bpr (T, S; G, pe) = NBp+o. (paT, pa S; G, pe) ordp(det(p"S))OApr+a

= p-t

(paT, pa S; G, pe)

Lemma 5.6.1 implies the assertion. Before the definition of our local density, we give one more lemma, which connects ours with Siegel's original.

5.6 Local densities

97

Lemma 5.6.5. Let S and T be regular symmetric matrices over R of degree s and t respectively and G E M8,t(R). Put Ap,. (T, S; G, pe) :_ {X E Ms,t (R) mod prM8,t (R) I S[X] - T mod pr,

X-Gmodpe}. Then we have

OA',r (T, S; G, pe) = 2t52 NApr (T, S; G, pe)

for a sufficiently large r where 6 is Kronecker's delta function. Proof. When p 2, Et(R) is simply {A = to E Mt(R)}. Hence A'pr = Apr holds. So we assume p = 2. Let {Ti} be a complete set of representatives of integral symmetric matrices T' mod 2r+1 such that T' - T mod 2rEt(R). The number of them is

[2-Et(R) : {x E 2r+1Mt(R) I x = tx}] =

2t(t-1)/2

Define a mapping y

: LJjA2r+1(Ti, S; G, 2e) - A2r (T, S; G, 2e)

by O(X) - X mod 2r. For X E A2r (T, S; G, 2e) and Y E Ms,t (R), we have

S[X + 2rY] = S[X] + 2r(tXSY + tYSX) + 22rS[Y] = S[X] mod 2r Et (R) =Tmod2rEt(R).

Thus X + 2rY is in A2r}1(Ti, S; G, 2e) for some i and so 0 is surjective and the number of inverse images is 2'. Using Corollary 5.4.4 as in the proof of Lemma 5.6.1, we have

OX, (Ti, S; G, 2e) _

OA2r+1 (T, S; G, 2e)

for a sufficiently large integer r. Thus 2t(t-1)l2#A2r+1 (T, S; G, 2e)

=

OA2r+1 (Ti, S; G, 2e)

= 2stOA2r (T, S; G, 2e)

=

28t+t(t+l)/2-st#A2r+1 (T, S; G, 2e)

holds by Lemma 5.6.1. This completes the proof.

98

5


Let us define the local density. Let M and N be regular quadratic lattices over R with rank M = m and rank N = n and {mi }, {n2 } be bases of M, N respectively. Let g be a homomorphism from M to N; then and (1) nn)G for someG E Mn,m(R) (g(m1),... ,g(mm)) = (nl,..

Putting

T = (B(mi, mj)) and S = (B(ni, nj)), the dual basis of N is given by (ni, , nn)S-1. Then the set (2)

BPy(M,N;g,pe) {u : M - N/prNO I Q(v(x)) = Q(x) mod 2p', v(x) g(x) mod peN for x E M} is canonically identified with Bpr (T, S; G, pe) through matrix representation. Hence by Lemma 5.6.4, the following definition of the local density is well-defined.

ip(M, N; g,pe)

:=pm,ordp dr N lim(pr)m(m+1)/2-mnOBP. oo

(M, N; g,pe)

=pmord, d N lim (pr)m(m+1)/2-mny}Bpr (T, S; G, pe)

r-oo

lim(pr)m(m+1)/2-mn#Apr (T, S; G, pe),

oo

where in the last row S and T are supposed to be integral and we used Lemma 5.6.3. When e = 0, the additional condition v(x) - g(x) mod peN is trivially satisfied and we put Apr (T, S) := Apr (T, S; G, p°), A'pr (T, S)

A'pr (T, S; G, p(),

/3 (M, N)

/P(M,N;g,p°)

Through matrix representation,

Apr (M, N) := {v : M - N/prN I Q(v(x)) - Q(x) mod 2pr for x E M} is identified with Apr (T, S) if S(M), S(N) C R. Siegel's original definition of local density is ap(S T) := 2-6m," lim

r-* oo

(pr)m(m+1)/2-mnA,r(T S) H

2m62,P-5m,,,/3P(M, N)

:= ap(M, N). Note that the positions of S and T are reversed. When we adopt the matrix notation S[X] = T, it is natural that S is on the left, but when we use the lattice notation M -> N, it is natural that S is on the right.

5.6 Local densities

99

Proposition 5.6.1. Let M, N, g, T, S, G be those as above. Then (i)

/3 (M,N;g,pe) = (pr)m(m+l)/2-mnOAp, (T, S; G, pe)

= moHd

v(x) - g(x)

M - N/prN I Q(v(x)) - Q(x) mod 2pr, peN for x E M}

if S and T are integral and r > h + max(e, 1) for an integer h satisfying phQ(MO) C (2), (ii)

/3 (M, N; g,

pe) = 2-m62,p (pr)mm(m+1)2-mnOAPr(T,

S; G, pe)

if S and T are integral and r is sufficiently large, (jp(M(p°),N(pa);g,pe) = pam(m+1)/2/p(M,N;g,pe)

(iii)

if S and T are integral and a > 0. where M(pa) = (paT) and N(pa) = (paS) are scalings of M and N by pa. Proof. The assertions (i), (ii) and (iii) follows from Lemma 5.6.1, 5.6.5 and 5.6.2 respectively.

Exercise. Show 3p (M, N; g, pe)

to N such that v(x)

0 if and only if there exists an isometry v from M g(x) modpeN for x E M. In particular, rank M >

rank N implies /3p (M, N; g, pe) = 0.

It is not easy to evaluate local densities. But there are several cases in which we know the explicit formula. For example, if M is unimodular with

n(M) C (2), and N is a regular quadratic lattice with n(N) C (2), then Proposition 5.6.1 implies /3p(M, N) =

pm(m+1)/2-mn#Ap(T, S)

= pm(m+1)/2-mng{ isometries from M to N

where M and N are regular quadratic spaces over R/(p) defined by (MI pM, Q) and (NIpN, Q) respectively. If, moreover N is unimodu2 number of isometries is given in Theorem a lar with n(N) C (2), then the

1.3.2.

For regular quadratic lattices M, N over R and a homomorphism g from

M to N, we still keep (1) and (2). We define an integer h' such that ph S-1 E En(R), i.e. ph Q(N1) C (2),

100

5


which implies phi Na C N as in the proof of Corollary 5.4.2. Applying Theorem 5.4.2 to L := M, G := NO, k > h' + max(e, 1), (p-k)mn-m(m+l)/2O{Q : M -> N/pkNO I * (k)}

is independent of k, where the condition *(k) is:

*(k) : Q(a(x)) - Q(x) mod 2pk, a(x) = g(x) mod peN for x E M and v induces an injective mapping from M/pM to N/pN. Hence we can define the primitive local density dp(M, N; g, pe) by

li(pk)m(m+1)/2-mn#{a : M --> N/pkN I

* (k)}

_ (pro)m(m+1)/2-mnpmordPdN{a : M -p N/prONO I * (r0)}

for rO := h' + max(e, 1). When e = 0, we denote dp(M, N; g, pe) by dp(M, N). HH

If N is unimodular with n(N) C (2), then dp(M, N) is the product of pm(m+1)/2-mn with the number of isometries from M/pM to N/pN, where quadratic forms are defined by 1Q. If s(M), s(N) C R, then we have, for integers e > 0, r > h' + max(e, 1), #{X E Apr (T, S; G, pe) I X : primitive }

*(r) = [N# : NJ

}

{o, : M -* N/prNO I * (r)}.

If the three sets are well-defined, then the equalities are clear. The first is well-defined if r > max(e, 1) as at the beginning of the proof of Lemma 5.6.1. The second is the translation of the first to lattices. The third one is well-defined if p'NO C pN, prNO C peN and r > h', which, in order, correspond to the injectivity, a(x) - g(x) modpeN and the dependence of Q(a(x)) only on a mod p"NO, as at the beginning of the proof of Lemma 5.6.3. Then the three sets are well-defined and the equalities hold if r > h' + max(e, 1). Thus we have dp(M,N;g,pe) = li m(pr)m(m+1)/2-mn#{X E Apr (T, S; G, pe) I X : primitive}.

We note that rO in the above is independent of M and hence there is a positive number cN dependent only on N such that dp(M, N; g, pe) > CN if it is not zero, and that dp(M, N; g, pe) = 0 unless there exists an isometry

a : M -* N such that a(M) is primitive in N and a(x) - g(x) mod peN for x E M. Now we give two reduction formulas. The following is due to Siegel.

5.6 Local densities

101

Theorem 5.6.1. Let M and N be regular quadratic lattices over R, then we have

/32(M, N) = E [M' : M]m+1-ndP(M,, N), FMDM'DM

where m = rank M < n = rank N. The summation on M' is finite. Proof. We still keep (1) and (2). The property similar to (iii) in Proposition 5.6.1 holds for primitive local density since I{Q :

MV) , N(P°)/prN(P°) I * (r)}

=I{a:M->N/pr'N I *(r-a)} = pamn,}{Q : M --> N/pr-a'N I * (r - a)} and hence we may suppose that S and T are integral by scaling. Suppose r > ordp det T

and

S[X] - T mod prEn.(R)

for

X E Mn,m(R).

Then we will show that rank X = m and writing X = X1 K by virtue of elementary divisor theory, where X1 E Mn,m(R) is primitive and K E Mm(R), GLm(R)K is uniquely determined by X modpr and 2 ordp det K < ord, det T < r.

(3)

Now S[X] - T mod prE,,,(R) implies det S[X] - det T mod pr, which implies det S[X] # 0 mod pr by r > ordp det T, and hence rank X = m. Then

det S[X] = det S[XiI(det K)2 - det Tmod pr implies (3) and

prK-1 - Omodp. Moreover (3) and S[X1] - T[K-1] modprEn,(R)[K-1] imply that T[K-1] is integral. Now suppose

X'-Xmodpr

and X' = X'K' is a similar decomposition. Then

X'K'K-1 = X'K-1 = XK-1 + (X' - X)K-1 =X1+(X'-X)K-1 -X1modp

implies that X'K'K-1 is integral and hence Xi is primitive. Similarly = K(K')-1

is integral since is integral and thus

K'K-1 (K'K-1)-1

5

102


E GL.,,,.(R) follows. Thus GL,n(R)K has been uniquely determined by X mod pr. Next, let {Ji}h 1 be a complete set of representatives of K'K-1

{prJ[K-1] I J E E.m(R)}modprE,n(R)

for a regular integral matrix K satisfying (3). Then Ji E En(R) is clear. Put M(T, S; K)

{X E Mn,,n(R) mod prMn,,n,(R)K I S[X] Tmodp'E,n(R), XK-1 : integral and primitive},

and

Mi(T, S; K)

{X E Mn,,n(R) mod p'Mn,m(R) I S[X] T[K-1] + Ji modp'E,n(R), X : integral and primitive},

and define a mapping 0 from M(T, S; K) to uiMi(T, S; K) by O(X) = XK-1. We will show that 0 is bijective. Both sets are clearly well-defined and 0 is well-defined. If X E Mi (T, S; K), then

S[XK] - T + Ji[K] modp'E,n(R) - T modprE,n(R). Thus XK E M(T, S; K) and q(XK) = X, which yields that 0 is surjective. The injectivity is obvious. Thus we have

#M(T,S;K) _

#MM(T,S;K)

Next, we show that tMM(T, S; K) is independent of i. Put

f = ordp det K.

Then writing Ji = prJ[K-1] for J E E,,,,(R),

Ji = pr-2f J[f K-1] - 0 mod pr-2f E., (R). Applying Corollary 5.4.4 to A := T[K-1] and B := T[K-1] +Ji for a large integer r there is a matrix U E GL,n(R) such that T [K-1] + Ji = T [K-1] [U].

5.6 Local densities

103

Hence we have

#Mi(T, S; K) = #{X E Apr(T[K-1], S) I X : primitive }. Since

#M(T, S; K) = pnf #{X E Apr (T, S) I XK-1 integral and primitive}, we have

#{X E Apr (T, S) I XK-1 integral and primitive} = p-'f #M(T, S; K)

= p n f E #Mi (T, S; K) = p-nf h#{X E Apr (T [K-1], S) IX primitive}. Let us evaluate the number h. Put K = UK1V (U, V E GLm(R)) and Kl = diag(pf', ... , pfm ); then h = #({prJ[K-1] I J E E,, (R)) mod Pr E,,, (R)

_ [Em(R)[Ki

since f =

1]

: Em(R)] = p(m+l)f

fi. Thus we have obtained

#{X E Apr(T,S) I XK-1 integral and primitive} =p(m-n+1)ordpdetK#{X E Apr (T[K- 1], S) I X primitive}. Therefore we have, for a sufficiently large integer r #Apr(T,S)

= E p(m-n+1) ordp det K#{X E Apr (T [K-1], S) I X primitive}, K

where K ranges over regular matrices in GLm(R)\Mm(R) such that T[K-1] is integral. Considering the correspondence between K and the lattice

M' defined by (ml, , mm)K-1, we have [M' : M] = pordp det x Thus we have obtained the equation in the assertion. If dp(M', N) 0, then there is an isometry o, : M' y N. Hence dM' C (s(N))m holds and d M = [M' : M] 2 d M' implies

0 < ordp[M' : M]2 = ordp(dM/dM') < ordp(dM(s(N))-m'). Thus the summation on M' is finite. Now let us give another reduction formula.

5

104


Let M, N be regular quadratic lattices over R with rank M = m and rank N = n respectively and S(M), S(M) C R; let K be a submodule of N isometric to M. For a basis {mi} of M, we put Cp,(M,N;K) := {a : M --* N/p'N I Q(o(x)) = Q(x) mod 2pr and where the condition (**) is (**): there exists an isometry ri E O(N) such that 77(K) = R[Q'(ml), where o''(mi) is an element of N which represents o,(mi) mod p'N. Let us verify that the condition (**) is independent of the choice of o'(mi) for a sufficiently large r, which means that Cpr (M, N; K) is well-defined for a sufficiently large integer r. To do it, we have only to show the following: then for any given xi E N, there is an isometry

77/ :K=K":=R[vl+prxl,... Ivm+prxm] with r7' E O(N) when r is sufficiently large. This is simply Corollary 5.4.5. Thus Cpr (M, N; K) is well-defined.

Let us give Cpr (M, N; K) in terms of matrices. Let {ki} and {ni} be bases of K and N respectively and put

(ki, ... km) _ (nl, ... nn)Y (Y E Mn,m.(R) ), (Q (?nl), ... D'(mm)) = (nl, ... nn) X (X E Mn m(R) ), ,rl(km)) _ (o'(ml) ... (nl, , nn)XG for GE GLm(R) (ri(n1), ... , i7(nn)) _ (nl, ... nn)A (A E GLn(R) ) and finally

S :_ (B(ni, nj)) and T :_ (B(mi, mj)). Then Q(o,(x)) - Q(x) mod 2p' is equivalent to S[X] - T mod prEm(R), 77 E O(N) means S[A] = S, and lastly r7(K) = R[Q'(ml), (i7(kl),...

,

o'(mm)], i.e.

,i7(km))=(o'(m1),... ,Q (?nm))`-T

implies

(ni, ... , nn)AY = (i7(ni), ... , i1(nn))y (71(k1),... ,rl(km)) _ (nl,... nn)XG

5.6 Local densities

105

and hence AY = XG. Thus Cpr (M, N; K) is, in terms of matrices, Cpr(T, S; Y) = {X E Mn,,m,,,(R) mod prMn,m (R) I S[X] = T mod prEm(R), S[A] = S, AY = XG for some A E GL,, (R) and G E GL,,,,(R)}.

Note that this set is well-defined when r is sufficiently large by the above lattice-theoretic consideration, that is the condition on X is independent of the choice of X mod prMn,m,(R).

Lemma 5.6.6. Keeping the above, ,3p (M, N; K) :_

(pr)m(m+l)/2-mn0Cpr (M, N; K)

is independent of r if r is sufficiently large.

Proof. As above, let {Ti} be a complete set of representatives of symmetric matrices T' mod p' +1 E,,,, (R) such that T' - T mod p''Em (R). The number of representatives is pm(m+l)/2 We define a mapping 0 from LJZCpr+l (Ti, S; Y) to Cpr (T, S; Y) by O(X) = X mod p'. Obviously 0 is well-defined and surjective. Let X E Cpr (T, S; Y) and Y E Mn,,m (R); regarding X E Mn,m(R), X + p'Y gives the same element of Cpr (T, S; Y) and X + pry is in Cpr+l (Ti, S; Y) for some i. Thus the number of the inverse images is pmn. By virtue of Corollary 5.4.4, there exists a unimodular matrix Ui E GL ,,,(R) such that Ti = T[Ui]. Hence OCpr+l (Ti, S; Y) = #Cpr+i (T, S; Y),

and so

OCpr+1 (T, S; Y) = pnm-m(m+l)/2y}Cpr (T, S; Y).

0

Lemma 5.6.7. Suppose M = Ml I M2, mi := rank Mi > 0. Then we have

JApr (M, N) _ OCpr (Ml, N; Ki) i=1

x 0{0'2 E Apr(M2iN) I B(Ki,v2(M2)) -0modpr} for a sufficiently large r, where the set {Ki} is a complete set of submodules of N isometric to Ml which are not transformed into each other by

106

5


isometries of N. If M2 = 0, then the number of 0'2 in the above equation should be 1.

Proof. First, we define a mapping 0 from Ap, (M, N) to 0 (N) as follows: For al E APT (Ml, N) we consider that the image of al is in N, taking any fixed representatives. Then by virtue of Corollary 5.4.2, M1 is isometric to al(Mi) when r is sufficiently large. Hence there exists a E O(N) such

that a(a1(Mi )) = Ki for some i. We fix any such a for al and put O(al) = a ((=- O(N)). Let a E Ap, (M, N) and put

al = 01M1 E Apr(MiiN)

and

0`2 = cb(al)(a1M2) E Ap, (M2,N)

Now B(M1 i M2) = 0 implies B(a(Mi), a(M2)) - O modpr and hence B(O(al)a(Ml),O(oi)a(M2)) = Omodpr. Noting 0(aIMja(M1) = Ki for some i, we have B(Ki, 0'2 (M2)) - 0modp"". Now we define a mapping

rl: Apr(M,N) -+u2=1 Cp-(Mi,N;Ki) x {0-2 E Apr(M2iN) IB(Ki,a2(M2)) __ 0modpr}

by 77(o) = (al, 0'2) as above. The injectivity of rl is obvious. For a given (al, 0`2), we define or E Ap, (M, N) by aI M1 = al and or I M2 = O(al)-1a2. Then 7(a) = (al, a2) is clear.

Lemma 5.6.8. 010`2 E Apr(M2iN) I B(Ki,a2(M2)) __ 0modp'"}

=([K":Ki]/[N:Ki1Ki where r is sufficiently large.

Proof. First we show

{x EN I B(Ki,x)-Omod pr}=prKp IK2 K. The left-hand side obviously contains the right-hand side. Let x E N satisfy B(K2, x) - 0 mod pr. Write x = x1 + x2 (x1 E FKi and X2 E FK2 ). Then B(K2, xi) = B(Ki, x) __ 0 mod p' implies x1 E p'KO C Ki C N and hence x2 = x - x1 E N n FKZ = KZ . Thus the above equality has been proved and we have

0{a2 E Apr(M2iN) I B(Ki,a2(M2)) - 0modpr} _ #{a2 E Ap- (M2, N) 1 a2 (M2) C prKi 1 KZ }

_ 01a2 : M2 - prKo I Ki modprN I Q(92(x)) = Q(x) mod 2pr

forxEM2}.

5.6 Local densities

107

Here we fix any positive integer a such that paK# C Ki and then the above is equal to [N: paKP 1 Ki ]-m2{U2 : M2 -> prKq I Ki modpr(paKd 1 K--L) Q(v2(x)) - Q(x) mod 2pr for x E M2}. Also, we write a2(x)

_ 1(x)+72(x),

'yl (x) E prKl mod pr+aKa,

Ki mod prKK . Since Q(prKf) = p2rQ(K$), we have for a sufficiently large r, 'y2 (x) E

Q('y1(x)) E Q(PrKP) C (2pr) and hence the condition Q(a2(x)) - Q(x) mod 2pr is equivalent to Q('y2(x)) - Q(x) mod 2pr. Thus the number required is equal to [N : PaKa _- Ki ]-m2

[Pf Kq :

pr+aKb]m2

x 01-y2: M2 - Ki mod prKi I Q(ry2(x)) - Q(x) mod 2pr for -t E M2} = [N : paKa 1 Ki ]-m2Pam1m2NApr(M2,Ki ), since rank Ki = rank M1 = ml. On the other hand, we have paml [N : paK# 1 Ki ]-1

=Paml[N Ki 1 Ki ]-1[Ki 1 Ki : paK# 1 Ki ]-1 =Pam'[N Ki 1 Ki ]-1[Ki : paKd]-1 =[Ka pa Kt] [N : Ki 1 Ki ]-1[Ki : paKp]-1

=[Ka:Ki][N:Ki1Ki ]-1, which completes the proof. Summarizing, we have the second reduction

Theorem 5.6.2. Let M = M1 1 M2 and N be regular quadratic lattices over R and mi = rank Mi > 0. Let {K2} be a complete set of submodules of N isometric to M1 which are not transformed into each other by isdmetries

of N. Moreover we assume S(M), S(N) C R. Then we have ap(M, N) _

([M1 : M1]/[N : Ki I Ki ])m2/3p(Mi, N, Ki)Qp(M2, Ki )

If m2 = 0, then we put Op (M2, KZ) = 1.

Proof. Since Ki = M1i we have [Ka : Ki] = [Mi : M1]. For

m=rankM=m1+m2i m(m+ 1)/2 - mn = (mi(ml + 1)/2 - min) + (m2(m2 + 1)/2 - m2(n - ml)) holds and then the above lemmas complete the proof.

5

108


Corollary 5.6.1. In Theorem 5.6.2, assume that all submodules of N isometric to M1 are transformed into each other by O(N). Then we have ,3p(M, N) = ([Mi : M1]/[N : K 1 K1])m2,3p(M1, N)lip(M2, K1), where K is a submodule of N isometric to M1.

Proof. We note that Corollary 5.4.2 implies M = R[o,'(ml), v'(m,,,)] in the definition of Cp, (M, N; K) when r is sufficiently large, and so if the submodules of N isometric to M are mutually transformed by O(N), then ,3p (M, N; K) =,3p (M, N). The corollary is now obvious. 0 Using this corollary, we can evaluate the local density j3p(N, N) (see [Ko], [W3]).

Theorem 5.6.3. Let N =1jEZ Mj be a regular quadratic lattice over R, where Mj is either (pj)-modular or {0}, and Mj # {0} occurs only for finitely many integers j, and write Mj = Mp3 for the scaling of a unimodular lattice Nj by pi. Put

w :_ E jnj{T nk + (nj + 1)/2}, j k>j

nj := rankNj = rank Mj, m

P(m) :_ fl(1 - p-2x) (= 1 if M=O). =1

With, for a regular quadratic space U or U = {0} over 7L/p7L, 0

x(U)

1

-1

if dim U is odd, if U is a hyperbolic space, otherwise,

and for a unimodular lattice M over R with n(M) = 2 s(M), we define a regular quadratic space M (M/pM, over7L/p7L, and put 2Q)

X(M)

X(M)

(I) The case of p # 2. Op (N, N) = 23p'PE, where

8 : =0 0 1 Mj 0 {0}},

P:= [JP([nj/2]),

j

E:= fl (1 + x(Nj)p-n'l2)-1, j:MjO0

5.6 Local densities

109

and [x] denotes the largest integer which does not exceed x. (II) The case of p = 2.

(32(N, N) = 2--qPE. We define q, P, E as follows: For a unimodular lattice M over 7L2i we say that M is even if n(M) = (2), odd otherwise; for the sake of convenience, {0} is also called even. With

10 qj:=

nj nj + 1

if Nj is even, if Nj is odd and Nj+1 is even, if Nj and Nj+1 are odd,

we put

We write for a unimodular lattice M, M = M(e) I M(o), where M(e) is even and M(o) is either odd or {0} with rank M(o) < 2. Then we put

P := fl P(2 rank Nj(e)). 7

Defining Ej by

a (1 + x(Nj (e))2-ank Nj(e)/2)

if both Nj_1 and Nj+1 are even and unless Nj(o) - (el) I (e2) with El - e2 mod 4, otherwise,

1

2

we put

E:=fEj-.1 Before giving the proof, we note that qj depends on Nj and N,+1, and that qj = 0 if Nj = {0}. Moreover Ej depends on Nj_1, Nj and Nj+1 and

Ej =1

if

Nj_1=Nj =Nj+1={0}

110

5


but

Ej

I 1

2

if

either

(Nj_1 = Nj = 0 and Nj+1 is even) or (1V1_1 is even and Nj =Nj+1={0}), (Nj_1 = Nj = {0} and Nj+1 is odd) or (Nj_1 is odd and Nj = Nj+1 = {0}).

Let us give a sketch of the proof. We use induction on rank N. First we note that our formula satisfies the condition (iii) in Proposition 5.6.1. The values except w are invariant by

scaling. Thus we may assume No # {0} and Nj = 0 if j < 0 by scaling. Let us denote the formula for the local density given in the Theorem by ,(3'; we will verify /3 = 0'.

Suppose p # 2; when rank N = 1, the Theorem is easy and since every submodule isometric to No is transformed to No by an isometry of N, by virtue of Proposition 5.2.2 and Theorem 5.3.2, applying Corollary 5.6.1 to M0 I Ms-, we have /3 (N, N) =,3p (Mo, N),3p(Mo , MO L).

Proposition 5.6.1 implies

/p(Mo, N) = pno(no+1)/2-nonr(Mo, N)

where n = rank N, and M0,N are quadratic spaces (Mo/pMo, 2Q),

(N/pN, 2Q) over Z/pZ respectively and the function r is the number of all isometries from Mo to N. Using Proposition 1.3.3, we have ap(Mo, N) =

pno(1-no)/2r(Mo,

M0).

The formula for r(Mo, Mo) in Theorem 1.3.2 shows

/p(Mo, N) = /3 (N, N)/1p(Md , MO-L) = 2(1- x(Mo)p no/2)P([(no -1)/2}) and hence Op (N, N)10p(Mo , Mo) =)3p(Mo, N) = ap(N, N)l0p(Mo , MO L)

Hence the induction hypothesis yields /3 (N, N) = /'' (N, N).

5.6 Local densities

111

Suppose p = 2. It is easy to verify ,3 = 0' if N is even unimodular or N = (c) for e E Z', using Proposition 5.6.1. The following is the fundamental reduction formula needed to prove the theorem: (3)

/32 (N, N)

32(Mo(e), N)/32(Mo(e)', Mo(e)1)

if Mo(e) 54 {0},

,32((el),N)/2((E2) L MO L, (e2) 1 MO-L)

if MO = (e1)'L (E2),

,32((e), N),32(Mo-, Mo)

if Mo = (E),

using Theorem 5.3.5 for the first and Theorem 5.3.6 for the others to apply Corollary 5.6.1. We will verify the following three formulas: ,32(Mo(e), N) _ /32(N, N)/02(Mo(e)1, Mo(e)1) if Mo(e) (4)

,32((f1), N)

= /3 (N, N) /,32' ((e2) I. Mo , (E2) L Mo )

if 32((E), N)

0,

MO = (El) L (E2),

_ 32(N, N)/,6z(Mo , Mo) if Mo = (e),

Once they have been proved, our theorem is proved by induction on rank N as follows:

If Mo(e) 0 0, then use the first equations of (3) and (4). So suppose Mo(e) = 0. If no = 1, then use the third equations of (3) and (4). If no = 2, then use the second equations of (3) and (4). (i) The case of Mo(e) # {0}. We must verify /32(Mo(e), N) = /3 (N, N)/,32(Mo(e)1, Mo(e)1)

= 2qo-goP(rankMo(e)/2)E' 1EoEi(E-IEoE1)-',

(5)

where qo, Ez mean the invariants qo, Ei for Mo(e)1. Putting rank Mo(e) = k and rank N = n, we have by Proposition 5.6.1 ,32(Mo(e), N)

= 2k(k+l)/2-kn {o.:Mo(e)-+N/2NI Q(a(x))Q(x) mod 4 for xEMo(e)} Put

M:={xENIQ(x)-0mod 2} N Mo(e) 1 2Mo(o) L Mo

if Mo is even, if rank Mo(o) = 1,

Mo(e) L (7L2(v1 + v2] + 2Mo(o)) 1 Mo

if MO (0) = Z2 [V1,

with B(vl, v2) = 0,


5

112

Since x E Mo(e) implies Q(x) - 0 mod 2, we have

02(Mo(e),N) = 2k(k+1)/2-kny}{a:Mo(e)+M/2NjQ(o.(x))

Q(x)mod4 for x E

Mo(e)},

and

M/2N N/2N if MO is even, (Mo(e)/2Mo(e))1(Mo /2Mo) if rank Mo(o) = 1, (Mo(e)/2Mo(e)) -L (El + E2) I (Mo /2Mo) i Mo(o) (El) -L (E2) Thus 02(Mo(e), N) if rank Mo(o) 1, =2k(k+l)/2-kn r(Mo(e),Mo(e)1Mo) I r(Mo(e),Mo(e)1(2(E1+E2))1MO) if rank Mo(o) = 2,

where for a quadratic lattice K over 712 with n(K) C 2712, K denotes the quadratic space (K/2K, 1Q) over 71/271 and r(U, V) is the number of isometries from U to V.

Put

1

U := Mo(e) 1 Mo and V := U 1 ( 2 (El + E2)); then Ul = Me , V1 = Mo 1 (2(el + E2)) follow at once from 1

2

(Q(x + y) - Q(x) - Q(y)) = B(x, y),

and Rad U = Ul if and only if N1 is even. Lastly Rad V = V1 if and only if E1 0 E2 mod 4 and N1 is even. Now we can verify (5) using Theorem 1.3.2 and Proposition 1.3.3. (ii) The case of Mo = (E) (E E 712 ). In this case we must verify 02(Mo, N) _ 02' (N, N)/,a2(Mo , MO J-)

=

(6)

2qo-qOEl

1EoEi(E-1EoE1)

where the dashed invariants are those for Mo . Putting m := rank MOL n - 1, we have by Proposition 5.6.1 02(Mo, N)

= (22)1-(1+m.)#{v : Mo -p N/4N I Q(r(x)) - Q(x) mod 8 for x E Mo} = 2-2m,4{a mod 4, x E Mo /4Mo a2E + Q(x) - E mod 8}

5.6 Local densities

113

(here we note Q(x) - 0 mod 2 implies a - 1 mod 2 and so a2 - 1 mod 8)

= 21-2m {x E Mp /4MO I Q(x) - Omod8} = 21-m#{x E Mo /2Mo I Q(x) = 0 mod 8} (by s(MM) C 2Z2) = 21-n'1-n2#{x E M1/2M1 I M2/2M2 I Q(x) - Omod8} = 21-n1-n2#{x E N1/2N1 1 N22)/2N22) I Q(x) - Omod4}.

We note from the above that /32(Mo, N) = 2 if M1 = M2 = 0, and so (6) is true in that case.

(ii.a) Suppose that N2 and N1 are odd. Putting N1 =1 (ui), N2 =1 (vi), we have 02(Mo, N) n2

n1

= 21-n1-n2 O{ai, bj mod 2

uia2 + 2 > vib2 = 0 mod 4} i=1

=21-n1-n2

i=1

vib2 - 9 E uia2 mod 2}

{bj mod 2 vlb1 a.i mod 2

i>2

=2 1-ni-n2 2 n2 -'Ofaimod2l Euia2 - 0mod2} =2 -n' 2 ni-1 = 2-1 which yields (6).

(ii.b) Suppose that N2 is odd but N1 is even; then we have, putting N2 = 1 (vi), 032(Mo, N)

= 21-n1-n2µ{x E Ni /2N1, bi mod 2 Q(x) + 2 =

21-n1-n2µ{x

E Nl/2N1, bi mod 2 I vlb2

vib2 = 0 mod 4}

-2-1Q(x) + E vib2 mod 2} i>2

which yields (6).

Suppose that N2 is even hereafter in the case of Mo = (e); then we have

/32(Mo,N) = 21-'#{x E N1/2N1 I Q(x) - 0mod4}.


5

114

(ii.c) If N1 is even, then 32 (Mo, N)

= 21-n1#{x E N1/2N1 1 1Q(x) - Omod2}

= 21-n1(2n1-1 - x(Nl)2n1/2-1 + X(Nl)2n1/2) X(N1)2-ni/2 =1+

by Lemma 1.3.1

which yields (6). (ii.d) Suppose N1 is odd and N1 (o) = (u), then we have i32(Mo, N) = 21-n1 #{x E Nl (e)/2N1(e), a mod 2 I Q(x) + a2u - O mod 4}

= 21-n1 #{x E Nl (e)/2N1(e) I Q(x) - O mod 4} X(Ni(e))2(nl-1)/2-1 +X(N1(e))2(n,-1)/2) = 21-n1(2n1-2 by Lemma 1.3.1 = 2-1(1 +

X(N1(e))2-(nl-1)/2),

which yields (6).

(ii.e) If N1(o) = (u1) 1 (u2), then we have 32(Mo, N)

=21-n1#{al,a2mod2,xENl(e)/2N1(e) I ula2+u2a2+Q(x) -0mod4} =21-n1#{xENi(e)/2N1(e) 1 1Q(x) _- Omod2}

(when a1 - a2 - 0 mod 2 ) 2 (ul + u2) mod 2}

+ 21-nl a{x E N1 (e)/2N1(e) 1 1 Q(x)

(when al - a2 - 1 mod 2 ) 2(2 ni-3 = 21-n1 2n1-2

( 2-1(1 Sl 2-1

-X(Nl(e))2(nl-2)/2-1 +x(N1(e))

if u1

u2 mod 4,

if u1

u2 mod 4,

+X(Ni(e))2-(nl-2)/2)

if ul

u2 mod 4,

if u1

u2 mod 4,

which yields (6) and completes the case of rank MO = 1. (iii) The case of MO = (e1) 1 (e2) (ei E Z2')-

2(nl-2)/2)

5.6 Local densities

115

As noted, we have only to verify the second equality in (4). Putting m := rank MM = n - 2, Proposition 5.6.1 yields 02 ((El), N) =(22)1-(m+2)O{Q : (El) -> N/4N I Q(u(x)) = Q(x) mod 81

=2-2-2m0{a,bmod4,x E MOL /4Mo I a2El +b2E2+Q(x) - El mod8} =2-1-2my}{b mod 4,x E Mo /4Mo I b2E2 + Q(x) = 0 mod 8} a = 1 mod 2) (when

+2- 1-2, Oja mod 4,x E Mo /4Mo I a2E1 + Q(x) = El - E2 mod 8} (when a = 0 mod 2) =2-1-2m(0{x E MOL /4Mp I Q(x) = Omod4} + #{x E Mo /4Mo I Q(x) - El - E2 mod 4}) =2-1-'rn($${x E Mp /2Mp 1 1 Q(x) = 0 mod 2}

+ b{x E Mo /2Mo I 2Q(x) = 2 (El - E2) mod 2}) 2-1

if

{ 2-'p{x E Mo /2Mo I ZQ(x) = 0 mod 2} if 2-1 if E1 0-E2mod4, 1 if El = E2 mod 4 and Nl is even, 2-1 if El = E2 mod 4 and N, is odd,

El

E2 mod 4,

El = E2 mod 4

which yields the second equality of (4) and completes the proof of the theorem. Let us study quantitative properties of local densities. They may suggest something global and indeed play a substantial role in global representation problems.

Theorem 5.6.4. Let M, M', N and N' be regular quadratic lattices over R and m = rank M = rank M' < n = rank N = rank N'. Then we have: (a) /P(prM,prN) = pr(m+1)mfjp(M N). (b) If N C N', then )3p(M (c)

N) < [N: N]m/p(M, N').

If N C N', prN' C N, then Op(M N'):5 p rm(m+l) [N : prN']m Qp(p"M, N).


116

5

(d)

If M C M', then Op (M, N) :5 [MI: M]n---',3p(M, N).

(e)

If N' C N and ??(M) C N' for every isometry rl : M -+ N, then Op (M, N) = [N : N']-m/3p(M, N').

In particular, if N is maximal and anisotropic, and M '-+ N, then /3p(pM, N) =

pm(m+l-n)/3p(M N).

Proof. (a) follows from (iii) in Proposition 5.6.1. By virtue of (a), we may assume that the scales of M, M', N, N' are in R in the rest of the proof. (b): Let i : N/ptN -* N'/pt N' be the canonical inclusion mapping. For

0 E Apt (M, N) := {v : M - N/ptN I Q(o (x)) - Q(x) mod 2pt }, i o 0 is clearly contained in Apt (M, N'). #Apt (M, N')

#{i o 0 1 -0 E Apt (M, N)}

is also clear. Noting that a linear mapping 0 : M -> NlptN satisfies i o = 0 if and only if 0(M) C p'N' fl N, we have

{0 : M - N/ptN I i o0 = 0} = O{V : M -; (ptN' f1 N)/ptN} _ [ptN' f1 N : ptN]m < [N' : N]m. Grouping elements 0 of Apt (M, N) by the inverse image of have OApt (M, N) < [N: N]m#Apt (M, N'). (c) is proved by using (a) and (b) as follows:

-p i o 0, we

/3p(M, N') = p-rm(m+1)/33p(prM,prN') < p-rm(m+l) [N : prN']m/3p(prM, N).

(d): By Theorem 5.6.1, we have /3p(M,N)

_

E [Mo

:

M]m+1-ndp(Mo, N)

FMDMOJM

E ([M0 : FMDMoDM' =[M' :

M'][M'

M]m+l-n/3p(M', N).

: M])m+1-ndp(Mo,N)

5.6 Local densities

117

(e): Regarding o E Apt (M, N) as a linear mapping from M to N and then applying Corollary 5.4.2, there exists an isometry i : M y N such that 77(M) = o(M) if t is sufficiently large. By the assumption, a(M) = 77(M) C N' holds and then we have

At (M, N) = #{o : M - N'/ptN I Q(o(x)) - Q(x) mod2pt} = [N : N']'Of a : M -p N'/ptN' I Q(o(x)) - Q(x) mod 2pt} = [N : N']-m#Apt (M, N'). Suppose that M y N and N is maximal anisotropic; then n(M) C n(N). For an isometry Ti : pM -* N, n(77(pM)) = n(pM) C n(pN) holds and then Theorem 5.2.1 yields r7(pM) C pN. Applying the above, we have l-3P(pM, N) = [N : pN]-m,3P(pM,pN) = p-mn+m(m+1) p(M, N).

0 Theorem 5.6.5. Let M and N be regular quadratic lattices over R and m := rank M < n := rank N. Suppose M is represented by N, then the following are true: (a) If a submodule Mo of N is isometric to M, then ,3P(M, N) > c[FMo fl N : (b)

Mo]m+l-n

where c is a positive number dependent only on N. If ind FN > m and /3P (M, N) # 0, then (3P(M, N) > cl,

(c)

where cl is a positive number dependent only on N. If n > 2m + 1, then /3P(M, N) < c2 minM0 [Mo :

M]2m-n

where c2 is a positive number dependent only on N, and MO ranges over lattices on FM such that Mo D M and dd(Mo, N) 0.

Proof. (a): Put M' := FMo fl N and applying (d) in the previous theorem, we have ,13P(M, N) _ /3P(Mo, N) > [MI:

> [M' :

Mo]m+1-n/3P(M',

Mo]m+l-ndd(M', N).

N)

118

5


Since M' is primitive in N, dp(M', N) 0 and hence by the remark before Theorem 5.6.1, there is a positive number c dependent only on N such that dp(M', N) > c. (b): Since Op (M, N) # 0, we may assume M C N, and we use induction on rank N. We fix any maximal lattice No in N. Suppose n(M) C n(No); then by virtue of Theorem 5.2.2 and Proposition 5.3.2 there is a submodule MO isometric to M which is primitive in No. Then

[FMonN : Mo] = [FMonN : FMo n No] [FMo n No : Mo]

= [FMonN: FMonNo] < [N: No] and (a) implies /3p(M, N) > c[N :

No]-+1-'a

since indFN > m implies n > 2m > m + 1, i.e. m + 1 - n < 0.

Suppose n(M) D n(No); then let M = M1 1 splitting with s(M1) D

1 Mt be a Jordan

D s(Mt), where all s(Mi) are different. We note

that M1 is modular and Ml C M C N implies n(N) D n(Mi) = n(M) D n(No). Since the number of the isometry classes of modular lattices with n(N) D norm D n(No), rank < m is finite, by Proposition 5.3.3 a modular submodule M' of N with n(M') D n(No) is transformed by an isometry of N to one of a finite set S of submodules of N. S depends only on N. By Theorem 5.6.2 we have /3p(M, N)

=E([Mi:M1]/[N:K2IKi ])rankM-rank

Ka ),

where Ki are a complete set of submodules of N isometric to Mt which are not transformed into each other by isometries of N and hence we may assume Ki E S. Thus we have Op (M, N) > c Ei /3p(Mi , KZ) for a positive number c dependent only on N. We show ind K2 > rank MjL. Put FN =

U, FM=V, FM1 = W, V =WI X and U= W I Y; then we want to see ind Y > dim X. We decompose W, Y as W = Wt I W2 I Hk, Y = Yt 1 Y2 I Hh, where k = indW, h = indY, W1 = Y(-1), Hi is a hyperbolic space of dim = 2i and W2 I Y2 is anisotropic. Then the assumption ind U - dim V > 0 implies

0< k + h + dim Wt - dim W - dim X < h - dim X, i.e. ind Y > dim X. Then the induction assumption implies /3p(Mi , K2) > c(K2) for some positive number c(KZ) dependent only on Ki , which completes the proof.

5.6 Local densities

(c): We may suppose Op (M, N)

that M' D M and dp(M', N)

119

0. Then there is a lattice M' such 0 by Theorem 5.6.1. We can put r =

min ordp [M' : M] where M' ranges over lattices on FM such that M' D M and dp(M', N) # 0. By Theorem 5.6.1, we have [MI: M]m+l-ndp(M', Op (M, N) _ N) FM3M'3M

r

where A(m, s) is the number of modules with rank m over R which contains a given module with index ps. Let us evaluate the number A(m, s). Put L = R[ul... , um] C L= R[vl, ... vm] and [L' : L] = ps. By (vl, ... , vm) =

(ul, , un)a (a E GL,,,.(F)), L' corresponds to aGL,(R) bijectively and [L' : L] = ps means a-1 E Mm(R) and det a-l E p8R". Thus we have A(m, s) = #(GLm(R)\{b E Mm(R) I det b E p8R" }) and taking upper triangular matrices as representatives of the left cosets, rj ps;(i-1) Then s;>o we can get A(m, s) _ el+...+sm -s

m

A(m

s)p-(m-1)8

< E ripsi(i-m) si>Oi=1

< E fl p-si = (1 _ p 1)m-1 ei_o i<m i<m

is easy. Thus we have ,dp(M, N)
y and a is an isometry}. Take a vector z in V such that b(y, z) = 1; then Fq[y, z] = Fq[y, z - q(z)y]

is a hyperbolic plane. Let us evaluate the number of isometries

a: U1--->y1

5.6 Local densities

121

such that for a E Fq, w E U1, the mapping Q : ax + w - ay + o, (w) is an isometry from Fx .1 U1 to V. Write V = Fq[y,z] 1 Vo;

then y1- = Fqy 1 Vo is clear. For an isometry o from U1 to y1, we write o ,(w) = f (w)y + 77(w), f (w) E Fq, r7(w) E Vo. Clearly q(w) = q(o,(w)) = q(77(w)). Suppose that & is an isometry and r7(w) = 0. Then &(-f (w)x +

w) = 0 implies w = 0 and i is injective and so an isometry from Ul to Vo. Conversely for any f E Hom(Ul, Fq) and 77 : Ul '- Vo, the mapping w --> f (w)y+t7(w) is an isometry from Ul to yl such that & is an isometry

from Fx 1 Ul to V. Hence we have r(U, V) = qu-lr(Fgx, V)r(Ul, Vo). Applying the induction hypothesis, we complete the proof.

Lemma 5.6.10. Let p be an odd prime, and U, V be quadratic spaces over a finite field Fp with p elements. Suppose that U -+ V and V is regular. Write U = Uo 1 U1 and put t = dim Uo, u = dim U, v = dim V. Suppose u < v and v > 2. Then we have a

pu(u+l)/2-uvr(U, V) =11(1 ± p-r' ) i=1

1+X(W)p-1

ifv-2u+t=0, ifv-2u+t=2,

1

otherwise,

12 x

where 2 < ri < b and a, b are smaller than a number dependent only on v, W is defined by V = Uo 1 W, and X is as in the previous lemma. Proof. If u = t, then U is regular and Theorem 1.3.2 yields our assertion. Hence we assume

u>t

(7)

Since V is regular, by Corollary 1.2.3 we have

r(U, V) = r(Uo, V)r(U1, W) and then by Theorem 1.3.2 and the previous lemma, we have

r(U,V) =ptv-t(t+1)/2(1

-

x

(1 - p-') x p(u-t)(v-t)-(u-t)(u-t+1)/2

JJ

X(V)p-v/2)(1

+ X(Uo _- V

(-1))p(t-v)/2)

v-t+l<e dim U', yielding

v -t=dimW > 2dimU1 =2(u-t).

(9)

Denoting by P the product of factors 1 ± p-a for r(U,V) such that a is greater than 1, we have (10)

pu(u+1)/2-uvr(U, V)

_ p(1 + X(Uo 1 V(-1))p(t-v)/2) 11 - p (v-t-2(u-t)+1) (v-t-2(u-t)+2)/2)X X (1

if vtmod2,

- X(W)P-

(1 + X

(1+

X(W)p-(v-t-2(u-t))/2) X(W)p-(v-t-2(u-t)+2)/2

1

if v - t mod 2,

if v - t = 2(u - t) >4, otherwise,

Sl 1

where the last factor comes from the case of i = 1 and v - t mod 2, which

implies u - t > 1 + i = 2. Since Uo 1 0-1) = Uo 1 U0(-1) 1 W(-1) and Uo 1 U0(-1) is a hyperbolic space,

X(Uo 1 V(-1)) = X(W(-1)) = X(W).

Suppose v - t = 2(u - t) i.e. v - 2u + t = 0; then (8) and (9) imply that W is hyperbolic and then X(Uo -L

0-1)) = X(W) = 1.

By the assumption v - t = 2(u - t) implies (t - v)/2 = u - v < 0. If u - v = -1, then we have t - v = -2, u - t = 1 and then (10) implies pu(u+1)/2-uvr(U, V)

= P(1 +p-1)(1 -p-1) 2 = 2P(1 -p 2).

Ifu-v_ -1, then u > v - 2 = 2u - t which contradicts (7). Thus u + 1 - v < -2 and then we complete the proof

of the case v-2u+t= 2. Suppose v - t - 2(u - t) > 3, i.e. v - 2u + t > 3; then u > t implies (t -v)/2 < (t - v + 2u- 2t)/2 < -1.5, and other powers of p in (10) are less than -1. Thus we have completed the proof of this final case and hence also the proof of the lemma. 0 Let U, V be regular quadratic spaces over Q and M, N be lattices on U and V respectively, that is M, N are finitely generated modules over Z and they contain a basis of U, V respectively. We denote quadratic lattices

ZpM, ZpN over Zp by Mp, Np. We study the infinite product of local densities. It is known that the behaviour of such products over all primes suggests something global in some cases through the Siegel formula (see the next chapter), although it is not fully satisfactory. Let M, N be as above and rank M = m < rank N = n, and define two infinite sets of primes

P(N) :={p p P(M, N) :={p p

2, NP is unimodular}, 2, Np and Mp are unimodular},

For a prime p E P(M, N), we have by Proposition 5.6.1 Op(Mp,Np)

=pm(m+1)/2-mn#{v : MP -> Np/pNN I Q(v(x)) - Q(x) mod 2p}.

Denoting a quadratic space (Mp/pMp, 2Q) over Z/pZ by Mp and defining similarly Np, we have /3p(Mp, Np) =

pm(m+1)/2-mn#{a

: RP - Np},

124

5


and then Theorem 1.3.2 yields /P(MP, NP)

_ (1-XP(NP)p-n/2)(1+XP(MP 1 Np-1))p(m-n)/2)

II

(1 -p-e),

n-m+l<em+2, n=m+1>2 orifn=m#2 with Mp=Np for every

to infinity.

p E P(M, N), then there exist positive numbers c1i C2 dependent only on it such that

cl < a'(M, N) < c2. (iii) If n = m + 2 and - d M d N is not a square in Q, then there exist positive numbers c3i c4 dependent only on n such that

c3fl(1-gyp(-dMdN)p 1)-1 1, we have

[J(1 - p-a) < fl (1 - p-a) < 1, p

1< pES

pES

f[(1_p-2a),

[1(1+p-a)=

pES

fJ(1-p-a)-1 1. Thus fpes(1 +p-a) and Hpcs(1 - p-a) are bounded from above and below by positive numbers independent of S but dependent on a. Let q be a natural number greater than 1 and X be a homomorphism from (7L/q7L)'< to C', and define L(s, X) := fl(1- X(p)p')-'P

where we put X(p) = 0 if p divides q. If the real part of s is greater than 1, then L(SIX) is absolutely convergent and L(s, X) has a meromorphic continuation on C. If X 1, then the infinite product f p(1 is convergent and it is equal to L(1, X) (see §109 in [Lan]). Thus to prove the proposition we can neglect factors 1 ± p-a (a > 1) in the formula of X(p)p-1)-1

,3p (Mp, Np) and we have only to take care over the factor

Ep := E(1 - Xp(N)p n /2)(1 + Xp(M 1 N( 1))P(m n)l2) where E = 1 if n > m, and = 2 if n = m. If moreover n/2 or (n - m)/2 > 1, then the corresponding factor in Ep can be replaced by 1. We note that Xp(N) = 0 if rank N is odd, and is equal to gyp((-1)n/2 dN) otherwise. The case in which we can neglect the factor 1 - Xp(N)p-n/2 is that either n is N(-1))p(--n)/2) odd or n > 2. The case in which the factor E(1+Xp(M 1 can be neglected is that m + n is odd, n - m > 2 or m = n with Mp = Np. Thus the case (ii) is over. Let us consider the case (iii). Then

(-1)(-+n)/2dMdN(-1) = -dMdN and

EP = (1 - Xp(N)p-n/2)(1

(1 - Xp(N)p-" /2)(1

- p2)(1 - gyp(- d M d N)p-1)-1

implies this case. In the case of (iv), Ep = 1 - gyp(- d N)p-1 and this case is over.

In the case of (i) Ep = (1 - Xp(N)p-n/2)(1 + p-1) and this completes the proof of the case (i) and so the proof of the proposition. In (iv), - d N E (Qx )2 if and only if the infinite product diverges to 0.

5

126


Corollary 5.6.2. Let M, N be lattices on regular quadratic spaces over Q

and n = rank N, m = rank M. If n > 2m + 3 and MP y NP for every prime p, then there exist positive numbers c1i c2 dependent only on N such that cl < 11 ap(MP, NP) < c2. P

Proof. Using (ii) in the last proposition, we have only to consider the products of aP(MP, NP) for p V P(M, N). Since n _> 2m + 3 implies

ind QPN > m, (b) in Theorem 5.6.5 implies OP(MP, NP) > cl

for some positive number cl dependent only on N. Suppose p E P(N) but p P(M, N); then p 2 and NP is unimodular. By Lemma 5.2.1, NP is Zr-maximal and hence there is a primitive submodule M' isometric to MP in NP because of Theorem 5.2.2, ind NP > m and Proposition 5.3.2. Theorem 5.6.1 implies OP(MP, NP) > dd(M', NP) = pm(m+l)/2-mnr(MP/pMP, NP/pNP),

where r is the number of isometries from (MP/pMP, to (NP/pNP, 1Q). Then Lemma 5.6.10 together with the consideration 2Q) of the infinite product 1±p-a (a > 1) given at the beginning of the proof of the last proposition of would yield 3P(MP, NP) > c2 pEP(N)

pVP(M,N)

for some positive constant c2 dependent only on n. Thus the estimate from below has been proved.

Let us estimate from above. Using the estimate of the proof of (c) in Theorem 5.6.5, we have j3 (MP, NP)
0

max

@pMDM'DMp

>o

dP (M', NP) EPs(m+l-n) A(m, s), s>0

fl psi(i-1). Hence m

ps(m+l-n)A(m, s) = 11

E

i=1si>O

m

psi(m-n+i) = fl(1 _ pm-n+i)-1 i=1

5.6 Local densities

127

and m - n + i < m - (2m + 3) + m < -3. Hence the product of on all p is convergent. If p E P(N), then Np is unimodular and dp(M', Np) is equal to pm(m+1)/2-mnr(U, V), putting U = M'/pM', V = Np/pNp. Using Lemma 5.6.10, the product of /3 (Mp, Np) for p E P(N) is convergent. For p V P(N), (c) of Theorem 5.6.5 implies Es>ops(m+1-n)A(m, s)

the existence of a positive number cp dependent only on Np such that i3p(Mp, Np) < cp, which completes the proof.

The estimate from below in this corollary induces Theorem 6.6.2 via an analytic approach explained in the notes of the next chapter, although only the case of m = 1, 2 is proved. The estimate from below in the corollary does not hold in general when n = 2m + 2, and the theorem is false when n = 4, m = 1. What is the global implication of the behaviour of fp ap(Mp, Np)?

Exercise 1. Let n = 2m+2, let U, V be regular quadratic spaces over Q with dim U =

m, dim V = n and let M, N be lattices on U, V, i.e. finitely generated modules containing bases of U, V respectively. Suppose that there is a primitive submodule in Np isometric to 7M7, for every prime p. Show

fli3 (Mp,Np)>c 11 (1-p-1), P

pA n(M)

where c is a positive number dependent only on N. (Hint: By (a) in Theorem 5.6.5 and Proposition 5.6.2, fp,Qp(Mp, Np) > cl (> 0) where p ranges over {p I Np is either not unimodular or both Np and Mp are unimodular}. If Np is unimodular but Mp is not so, then use 3p(Mp, Np) > dp(Mp, Np) and Lemma 5.6.10.)

Exercise 2. Let n = 2m + 1, m > 1, and let U, V be regular quadratic spaces over Q with dim U = in, dim V = n and M, N be lattices on U, V respectively. Suppose that there is a primitive submodule in Np isometric to Mp for every prime p. Show

f /p(Mp, Np) > crj(1 + cpp-1), pEP(N)

p

where c is a positive number dependent only on N, ep = 0, ±1 and p in the right-hand side ranges over {p I Np is unimodular but Mp is not so}. (Hint: Use Lemma 5.6.10 and Proposition 5.6.2.)

5

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Exercise 3. Let n = 2m + 2 or 2m +1 and let N be a regular quadratic lattice over R with ind FN = m -1. For a regular submodule M of N with rank M = m, show /3p(ptM, N) --+ 0 as t -> oo. (Hint: Use (c) in Theorem 5.6.5.)

Exercise 4. Let Fq be a finite field with q elements. Let U = Uo I Rad U be a quadratic space over Fq and Hk be the hyperbolic space of dim = 2k over Fq. Suppose that U is represented by Hk. Then show q u(u+l)/2-2kur(U,

Hk)

=(1 - q-k)(1 +

(1 - q2i-2k)

X(Uo)qu-d/2-k)

1 0 for every non-zero x E V, then V is called positive definite and a lattice on it is called a positive (definite quadratic) lattice. If V,, is negative definite, i.e. Q(x) < 0 for every nonzero x E V.., then V is called negative definite and a lattice on it is called a negative (definite quadratic) lattice.

If V... is isotropic, then V is called indefinite and a lattice on it is called an indefinite (quadratic) lattice . We note that if V is indefinite and dim V > 5, then V is isotropic by Theorem 4.4.1 and 3.5.1. However the

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Quadratic forms over 7L

indefiniteness of V in the case of dim V < 4 does not always imply that V is isotropic. For a vector space V over Q, we denote by Vp the scalar extension Qp®QV

of V to the vector space over @p as in Chapter 4 and for a lattice L on V, we denote 7Lp ®z L by L. Through a basis {v2} of L over 7L, we can identify L = 7Ln, V = (Qn, Lp = Z , Vp = (P canonically. Thus Lp is the closure of L in Vp, and we have L = flp(V fl Lp), where p runs over all primes, since 7L = flp(Q l7Lp).

Let us recall that a place is a prime number or the formal symbol oo as defined in Chapter 4.

Lemma 6.1.1. Let S be a finite set of primes and v a place not in S. For a given ap E ip for p E S, there is a rational number a E Q such that a and ap are sufficiently close in Qp for p E S, and for every prime p V SU {v} a is contained in 7Lp. Moreover if v = oo, then a can be positive and sufficiently large, and if v is a prime, then a is sufficiently close to ate, a real number given in advance.

Proof. First we assume v = oo. Write ap = bppe(p) (bp E Zp , e(p) E Z) for p E S. Put c = fpES peip> and take a sufficiently large integer t. We can choose a sufficiently large integer d such that d - bp(cp-e(P))-1 modpt7Lp

for p E S. Then dc - bpp6(P) mod cptZp for p E S. Putting a := dc, we have a - ap mod cpt7Lp for p E S. This a has the required properties. Next suppose that v = q is a prime and a..) E 1R is given. Let h be a natural number so that qh - 1 mod (11pES p) t. If a , = 0, then the rational number aq-'h is the required number for the integer as constructed above and a sufficiently large integer m. Suppose a,, # 0; for sgn(a,,.)ap instead of ap we construct c, d as above, letting t satisfy c(flpES)t > la,,. I. We put

a:= sgn(a,,)c(d + xn(T7 p)t)q-hmn pEs

for x

qhm

- 1 and integers m, n. We can take an integer n so that

0 < log(a/a,,) = log(([J p)t + dx-n)I aoo l

-1- n log(ghm/x)

PES

< log((ghm/x),

which implies 1 < a/a00 < ghm/x. By virtue of 41pES p)t > ja... I and qhm > x, we have n > 0 and then a is the desired number. The following basic correspondence between global and local lattices is due to Eichler.

6.1

Fundamentals

131

Theorem 6.1.1. Let V be a finite-dimensional vector space over Q, and L be a lattice on it. For each prime p, let M(p) be a lattice on Vp. Then there is a lattice M on V such that Mp = M(p) for every prime p if and only if Lp = M(p) for almost all primes. When this is the case, there is only one such lattice M, which is given by M := flp(V fl M(p)).

Proof. Let {vi}? 1 be a basis of L. Suppose that there is a lattice M on V such that Mp = M(p) for all primes. Let {ui}? 1 be a basis of M and (u,, , un.) = (vi, , vn)A for A E GLn(Q). Then for almost all primes p, A, A-1 E GLn(Zp) holds and for such a prime p, {ui} and {vi} span the same lattice over Zp, i.e. Lp = Mp = M(p). Conversely suppose that LP = M(p) for almost all primes p. Let S be a finite set of primes p such that M(p) # Lp and put q = ripES p. We take

a large natural number t such that qtLp C M(p) for p E S. Fix a set of the representatives {wp,j,} of M(p)lgtLp for p E S and for each collection J = {jp}PES we can take wj E V such that

wj - wp,1p E qtLp for every prime p E S, and wj E LP for p 0 S by using the last lemma, considering the coordinate representation by a basis of L. Denote by M the lattice spanned by qtL and wt's. Then by

the method of the construction, we have Lp = Mp = M(p) for p 0 S, and moreover M C M(p) in Vp and hence Mp C M(p). Let us show Mp = M(p) for p E S. For x E M(p) (p E S), we can take a prescribed representative wp,jp such that x E wp,jp + qtLp. Take any representative

jp' for p' E S with p' # p; then for a collection J of jp (p' E S) we have x E wp,jp + qtLp = wi + qtLp C Mp, since M contains wj and qtL. Thus M(p) C Mp and hence Mp = M(p) holds for a prime p E S. Thus we have completed the proof of Mp = M(p) for all primes. The uniqueness of M follows from m = flp(V fl Mp) = flp(V n M(p)). Let us introduce some important notions about quadratic lattices. Let V be a quadratic space over Q and L, M be lattices on V. We say that L and M are in the same class if L = ar(M)

for some

o E O(V).

The set of lattices on V in the same class as L is denoted by cls(L).

In the above, when we replace O(V) by O+(V), we write the proper class and cls+(L) instead of the class and cls(L).


6

132

We define the genus gen(L) of L to be the set of all lattices M on V such that for every prime p Mp = o-p(Lp)

for some

ap E O(Vp).

By Proposition 5.3.1, we can replace 0(Vp) by O+(Vp) in the definition.

The spinor genus spn(L) of L is defined by the set of lattices M such that for 71 E O(V) and vp E O'(Vp)

,l(M)p = a (Lp)

for every prime p.

Here 0'(Vp) is the kernel of the spinor norm from O+(Vp) to @P I/( )2 When we replace the condition q E 0(V) by ri E 0+ (V), we say the proper spinor genus and use spn+(L). Let us introduce some groups. We denote the set of all bijective linear mappings of a vector space V over Q into itself by GL(V). Taking a basis of V, GL(V) is isomorphic to GLn(Q) for n := dim V. The adele group X)2.

GL(V)A is the subset of g = in the product of fl,, GL(,) where v runs through all places such that gpLp = Lp

for almost all primes p

for any fixed lattice L on V. This definition is independent of the choice of a lattice L since for another lattice M, Lp = Mp holds for almost all primes (g = g). When p. GL(V) is canonically embedded in GL(V)A by g -+ dim V = n, V is isomorphic to Qn and GL(V)A is canonically identified with g = such that g,, is in GLn(Qi,) for every place v and gp is in GLn(7Lp) for almost all primes p. When n = 1, GL(V)A is {

I gv E Qv and gp E 7LP for almost all primes p}

which is called the idele group of Q and denoted by Q'. As above, Q' is canonically embedded in @A by x H where x = x for all places. For a quadratic space V over Q, the adelization groups OA(V) ), OA(V ), OA(V )

are defined by the set of g =

in GL(V)A such that g, E

and 0'(V,) for all places v respectively. Since V is a topological space through V, = @v , GL (V) = GL, ,(Q,) is also a topological space. Then we take as bases of open sets of GL(V)A all the subsets which are of form 11 U, x II GLn(7Lp) vES

pOS

6.1

Fundamentals

133

for which S is a finite set of places containing oo and U is an open set in GLn(Q1) for v E S. This topology is called the restricted direct product topology. Now the topology on groups OA (V), OA (V), O'y (V) is induced by restriction. Let us consider the action of GL(V)A on lattices. For g = E GL(V)A and a lattice L on V, we put gL := np(V n gpLp).

By definition of GL(V)A, gpLp = Lp for almost all primes p, and by Theorem 6.1.1, np(V n gpLp) becomes a lattice M on V with Mp = gpLp for all p. We put O+9(L) = {g E 0,+4(V) I gL = L}.

Using the adele groups, we can write

cls+(L) = O+(V)L, cls(L) = O(V)L, gen(L) = OA(V)L = OA+(V)L, spn(L) = O(V)O'4(V)L, spn+(L) = O+(V)O'q(V)L Then it is easy to see

cls+(L) C cls(L) C spn+(L) C spn(L) C gen(L). Two symmetric regular matrices A and B E Mn(Q) are said to be in the same genus if there exists gp E GLn(Zp) such that A = B[gp] for every prime p and A = B[g,,] for g,,,, E GLn(IR). This is compatible with our definition. For quadratic spaces U = (A), V = (B), the above assumptions mean UT, = VT, for all places and then Corollary 4.1.3 yields U = V.

The fundamental result on the number of the classes in gen(L) is the following.

Theorem 6.1.2. Let V be a regular quadratic space over Q, and L be a lattice on V. Then the number of the different proper classes in gen(L) is finite.

Proof. Take an integer a such that s(L(')) C Z. Since the number of the different proper classes in gen(L) is the same as in gen(L(')), we may assume s(L) C Z. If M E gen(L), then s(Mp) = s(Lp) for every prime p

implies s(M) = s(L) c Z. If L = (A), M = (B) for symmetric matrices A, B, then A, B are integral matrices and M E cls+(L) if and only if A = B [T] for some T E SLn (Z). Noting [GLn (Z) : SLn (Z)] = 2, Corollary 2.1.1 asserts the finiteness of the number of equivalence classes in question.

0

6

134


Theorem 6.1.3. Let U, V be quadratic spaces over Q and L, M be lattices on U, V respectively. If there is an isometry or, : L * M for every place v, then L y N holds for some N E gen(M). Here Lv, MT, mean U,,, V for V = 00.

Proof. The existence of Qv induces Uv y V, for every place v and so U V. Hence we may assume U C V and av E O(V,), and so ap(LP) C Mp. We define a lattice N on V by Np.

Mp

if LP C MP,

vp 1(MP)

otherwise.

This is well-defined because LP C Mp holds for almost all primes p. Then LP C Np is clear for all primes p and hence L C N E gen(M) holds.

6.2 Approximation theorems In this section, we give two approximation theorems which are needed later.

Theorem 6.2.1. Let V be a regular quadratic space over Q with m dim V > 2, and suppose that V is not a hyperbolic plane and that V,,, (L,. (1)) 1 (13 (-1)) (r + s = m). Suppose that the following data are given: (a) (b)

a lattice M on V, a finite set S of primes p such that S 5) 2 and Mp is unimodular if

(c) (d)

-11,p, ... xn P E Mp for p E S.

integers r', s' with 0 0, then we

6.2 Approximation theorems

135

put r" = r', s" = s'. Otherwise, put r" = s' and s" = r'. Applying the theorem to V(a), there is a vector x E M' for which (i)1 x and x1,p are sufficiently close in Vp for p E S(a), (ii)1 for p S(a), aQ(x) E 7Ly with precisely one exception p# q, where aQ(x) E qZ4 , and

(iii)1 aQ(x) > 0 (resp. < 0) if r" = 1, s" = 0 (resp. r" = 0, s" = 1). The conditions (i)1, (iii)1 imply immediately (i), (iii) for V. For p E S(a)\S, Q(x1) and Q(xl,p) are sufficiently close by (i)1. Since Q(xl,p) E Z P', Q(x1) is also in Z P'. For p = q (0 S(a)), aQ(x) E qZ is nothing but Q(x) E g7Lq . Thus the conditions (i), (ii), (iii) have been verified for V. Let {vl, v2} be an orthogonal basis of V; then

Q(alvl + a2v2) = a2Q(vl) + a2Q(v2) = Q(vl)(a2 + Q(vl)-1Q(v2)a2). Since V is not a hyperbolic plane, -Q(vl)Q(v2) V Q2 holds and hence

F :_ Q( -Q(vl)Q(v2)) is a quadratic extension of Q and V is isometric to the scaling of F by Q(v1), where F is considered as a quadratic space over Q whose quadratic form is given by the norm N from F to Q. By the above reduction, we may assume V = (F, N). We denote by RF the maximal order of F. Put

S':=SU{pIMp

(RF)p}.

For p E S'\S, p 54 2 and Mp is unimodular and hence we can choose xl,p E MP such that Q(x1,p) E Z P'. The assertions (i), (ii), (iii) for S follow obviously from that for S'. Thus we may assume that Mp = (RF)p for p S. Now, applying Lemma 6.1.1 to coordinates of x1,, which are expressed in terms of an orthogonal basis of F, we take y E F sufficiently close to x1,p for p E S, y E Mp for p S, and we can choose y so that

N(y) = Q(y) > 0 (resp. < 0) if r' = 1, s' = 0 (resp. r' = 0, s' = 1) since for an element u of an orthogonal basis {v1} such that Q(u) > 0 (resp. Q(u) < 0) for r' = 1 (resp. r' = 0) we can make a coefficient of y at u sufficiently large. Decompose the principal ideal (y) = yRF as

for ideals m, n of F such that the prime divisor p` appears in ?a if and only if p lies above a prime number in S. Since y E Mp = (RF)p for p S, n is an integral ideal. Put P := rjpES p and take a sufficiently large integer

t. Let us consider the ray class group mod Pt in F. It is known that the

6

136

Quadratic forms over Z

ray class group represented by n contains infinitely many prime ideals 4 of degree 1 (hence N(4) is a prime S). Hence there exists z E F such that

4 = nz, z - 1 mod Pt(RF)p for p E S and N(z) > 0.

Put x := yz. Then the conditions (i), (iii) are obviously satisfied. In particular x E MP holds for p E S, and x = yz E m4 implies x E (RF)P = MP for p S. Thus x is in M. The second condition follows from Q(x) = N(x) = N(yz) = ± N(m4) = ± N(m) N(4), since N(4) is a prime 0 S and the primes in N(m) are in S. Thus the case of n = 1 and m = 2 has been completed.

Suppose n = 1 and m = dim V > 3. Take any prime q S. Then q 54 2 and Mq is unimodular. By Theorem 5.2.4, there is a basis {v2} of Mq such

with B(vl, v2) = 1, Q(vl) = Q(v2) = 0

that M. = 7Lq[V1i V21 L Zgv3 1 .

and Q(v3) E 7Lq . We choose xl E M for which xl and x1,, are sufficiently close in Vp for p E S, and xl and vl +qv2 are sufficiently close in Vq. Then Q(xi) and Q(vi + qv2) = 2q are close in Qq and hence Q(xi) E g7Lq . Put

T:= {p: prime IpVSand Q(xl)V7LP}. Q(x1) E qZ implies Q(xi) # 0 and then T is a finite set of primes and q E T. If p E T, then p is odd and MP is unimodular and so Q(MP) D Z .

We take x2 E V by Lemma 6.1.1 such that Q(x2) > 0 (resp. < 0) if

r' = 1, s' = 0 (resp. r' = 0, s' = 1), x2 E MP and Q(x2) E Z

for

p E T and lastly x2 and v3 are sufficiently close in Vq. Taking a natural number a such that axe E M and p does not divide a for p E T, we put M' := 71[x1, axe] C M. Then Q(xl)Q(ax2) -B(xl, axe) is sufficiently close in Qq to 2ga2Q(v3) - B(vi + qv2i av3) = 2ga2Q(v3) E g7Lq and hence M' is regular and d Mq E g7Lq . Thus QM' is binary and not hyperbolic. Put

dM'VZp} and then S' := S U T U U is a finite set. Put

xlP

(x1 x2

if pESUU, if p E T.

Since MP' C MP, d M' E Z and MP is unimodular for p S', MP' is unimodular for p V S'. Applying the above case of n = 1, m = 2 to xi,P and MP, there is an element x E M' C M such that (i)2

x and x'1 P are sufficiently close in VP for p E S',

(ii)2 Q(x) E ZP for p Q(x) E h7Lh ,

S' with precisely one exception p = h, where

6.2

Approximation theorems

137

(iii)2 Q(x) > 0 (resp. < 0) if r' = 1, s' = 0 (resp. r' = 0, s' = 1). Let us show that x is what we want. If p E S, then x and x'1,p = xl, and xl and x1,p are sufficiently close in Vp respectively and hence so are x and x1,p. If p E T, then (i)2 yields that x and x2 are sufficiently close in Vp, and Q(x2) E 7LP . Hence Q(x) E 7LP for p E T. If p E U, then x and xl are sufficiently close in Vp, and p T and the definition of T imply Q(xl) E Z', hence Q(x) E 7Lp . Thus we have proved that if p E S'\S, Q(x) E 7LP then (ii)2 implies the condition (ii) and we have completed the proof of the case n = 1 and m > 3. Now suppose 1 < n < m. We use the induction hypothesis twice. First, we apply it to xl,p, , xn-l,p, S and M. Then there exist (1)

xl ... , xn-1 E M

such that (i)3 xi and xi,p are sufficiently close in Vp for 1 < i < n - 1, p E S, (11)3 det(B(xi, xj))1 0. Let us show the claim (q). For p V S U {h}, Mp and Np are unimodular

by (b) and (ii)3, and hence Mp = Np 1 N1 and then NP C Wp yields MP = Np . Thus MP is unimodular for p V S U {h}. Let us recall that (3) is valid even for p = h, and then the given condition Xn,p E Mp implies zn,p E M, for p E S U {h}. (iii)3 implies that RU 1 (6) y RV and hence lRW = (6) 1 .

Now we can apply the induction hypothesis to M', S U {h}, zn,p unless W is a hyperbolic plane. Vh D Mh and Uh D Nh imply d Vh E Z (@h )2 and d Uh E h7Lh (@h )2 and then dWh = d Vh d Uh E h7Lh (@h) 2

which means that W is not a hyperbolic plane. Thus we get the claim (#) by the induction hypothesis. Putting xn

yn+zn,

. . . , xn} is what we want. To verify the condition (i), we have only to check that xn and Xn,p are sufficiently close for p E S by (i)3. By the choice of yn, yn and Yn,p are sufficiently close for p E S U {h} and then (3) and (i)4 yield that Xn and xn,p are sufficiently close for p E SU{h}, which implies the condition (i).

we will show { x 1 ,

6.2


139

Let us verify xi E M. By (1), we have only to consider xn. xn,p E Mr holds for p E S by the original assumption and for p = h by choice. Since xn and xn,p are sufficiently close for p E S U {h} as above, xn E Mp holds for p E S U {h}. Suppose p 0 S U {h}. Then we get xn = yn + zn E Np 1 MP = Mp as above and then xn E Mp for every prime p, i.e. xn E M. Let us verify the condition (ii). Since N = Z[xi, , xn_111 xn and Xn,h are close for h, and Nh + Zhxn,h = Lh is unimodular,

det(B(xi,xj)) EZh.

det(B(xi,x;))(Z )2=d(Np+Zp[xn]) = d(Np + Z [yn + zn]) = d(Np + Zp [zn] )

since yn E Np. Moreover zn E W and QN = U = W1 imply

det(B(xi,xj)) = dNp Q(zn) for p S U {h} and then (ii)3i(ii)4 imply the condition (ii). (iii) follows from IR[xl, , xn] = RN + 1Rxn = RU 11Rzn, (iii)3 and (iii)4. Thus we have completed the proof of the theorem. The next theorem is a strong approximation theorem for O'(V). To state the theorem, we recall the topology of End(V) for an n-dimensional vector

space V over a local field F. Fix any basis {vi} of V, and then we can identify End(V) with Mn(F), and the topology of Mn(F) is transferred to End(V). It is independent of the choice of a basis once it has been fixed, because f E End(V) is close to the zero mapping if and only if , vn)(fi,,j) are the coordinates fi,j defined by (f (vi), , f (Vn)) = (v1, close to zero, and the base change induces (fi,j) -> g(fi,j)g-1 for some g E GLn(F). The following strong approximation theorem is due to Eichler. Theorem 6.2.2. Let V be a regular quadratic space over Q with dim V > 3

and suppose that V, is isotropic for some place v; (v may be finite or infinite). Let L be a lattice on V and S a finite set of primes with v S. Given op E O'(Vp) for p E S, there is an isometry a E O'(V) such that

v(Lp) = Lp for every prime p 0 S U {v} and or and op are sufficiently close in End(Vp) for p E S. To prove the theorem, we need several preparatory lemmas.

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140


Lemma 6.2.1. Let V be a regular quadratic space over Q and S a finite set of places including the infinite place oo. Given isometries av E O+(VV)

for v E S, there are vectors xi,

, x2n E V such that av and r,1

Tx2n

are sufficiently close for v E S, where Tx denotes the symmetry with respect to X.

Proof. Put av = Tx1 (v) Tx2n ivi for xi (v) E V. Since the order of a symmetry is 2, we may suppose that n is independent of v E S. By Lemma 6.1.1 we can choose xi E V such that x2 and xi(v) are sufficiently close in Vv for v E S. Then Txi and Tx,(v) are sufficiently close, and hence av and Tx2n(v) are, for v E S. Tx1(v)

Lemma 6.2.2. Let W be a regular quadratic space over Q with dim W > 3, S a finite set of places and v a place not in S. For a lattice K on W, there is an integer r dependent on K such that

(i) r E Z for every prime p E S and (ii) if a rational number c is represented by W and c E rzp fl Q(Kp) for every prime p # v, then there is an element w E W for which Q(w) = c and w E Kp for every prime p # v.

Proof. Enlarging S, we may assume that a prime p different from v is in

S if p = 2 or Kp is not unimodular. Let {Ki}h 1 be a complete set of representatives of the classes in gen(K). We show first that Ki can be chosen so that (Ki)p = Kp for every prime p E S. By the definition of the genus and Proposition 5.3.1 there is an isometry ai,p E O+(Wp) such that

ai,p((Ki)p) = Kp. By the last lemma, there is ai E O+(W) such that ai and ai,p are sufficiently close for p E S. As representatives we have only to take ai(K2) again. Thus we may assume (Ki)p = Kp for 1 < i < h, p E S. Now we take an integer s such that

sKi C K

sEZp

for every i and

for pES,

and put r := s2. The condition (i) is satisfied. Suppose that c is a rational number as in the condition (ii). If p E S, then p 54 v and hence c/r = c/s2 E Q(Kp/). If p S and p # v, then p 2 and Kp is unimodular, and

then Kp = (

0

1 I) 1 (some lattice). Since c/r E Zr,, c/r is represented

by (I ' 1 )) and hence by K. Thus c/r E Q(Kp) if a prime p is not the place v. If v is a finite prime q, then c/r q2t E Q(Kq) for a large integer t since c is represented by W. Thus c/r or c/r q2t is represented by Kp for every prime p if v = oo or q respectively. By Theorem 6.1.3, there exists

6.2 Approximation theorems

141

an element x in Ki for some i such that Q(x) = c/r or c/r q2t for v = 00 or q, respectively. Put w = sx or sq-tx for v = oo or q, respectively; then Q(w) = c holds. If v = oc, then w = sx E sKi C K. If v = q, then w = sq-tx E sq-tKi c q-'K. Thus w is what we want. Lemma 6.2.3. Let V be a regular quadratic space over Q with dim V > 4 for which V is isotropic for some place v, L a lattice on V and T a finite set of prime numbers with v T. Suppose that a non-zero rational number a r= Q(V) and zp E Vp (p E T) satisfy the conditions (a) a = Q(zz) for every prime p E T, (b) a E Q(Lp) for every prime p T. Then there is an element z E V satisfying the conditions (i) z and zp are sufficiently close in Vp for every prime p E T, (ii) z E Lp for every prime p T U {v}, (iii) Q(z) = a.

Proof. Scaling by a-1, we may suppose a = 1 without loss of generality. v, and T1 U {v} p implies p# 2 and that LP is unimodular. For a prime p E T1 \ T, we take zp E Lp for which Q(zp) = 1 by (b). Hence the conditions (a), (b) are satisfied for T1. If the conclusions are true for T1 instead of T, then they imply the assertion for T. So we may assume (c) if p V T U {v}, then p 0 2 and Lp is unimodular. If V(,, is isotropic and v is a prime number q, then we put T' := T U {q}; by the condition (b) for T, there exists an element zQ E L. with Q(zq) = 1 and hence (a) holds for T' instead of T. If the lemma has been verified for T' and v = oo, then the conclusion for T' implies it for T. Thus we may

We take a finite set T1 of primes such that T1 D T, T1

assume (d) if V,, is isotropic, then v = oo. Moreover, taking (rIpETp)-1L for a large integer t instead of L, we may

assume (e)

zpELpfor pET.

Fix any element x E V with Q(x) = 1, and take an isometry op E O+(Vp)

such that oa(x) = zp for p E T. By Lemma 6.2.1, there is an isometry a E O+(V) for which a and ap are sufficiently close for p E T. Put y = o(x); then Q(y) = 1, and y (= o(x)) and zp (= op(x)) are sufficiently close for p c- T. Then we have by (e) (f)

yELpforpET.

We choose an integer b by Lemma 6.1.1 such that (g)

b and 1 are sufficiently close in Q, for p E T and u := by E Lp for

pVTU{v}. Then we have, with (f) and (g) (h) u E Lp for p 0 v and Q(u) = V.

6

142

Put W


u-1- and determine a lattice K on W by fKp (LnW)p LpnWp if pET, Kp C ptLp for a sufficiently large t if p E T.

Obviously K C L holds. Put

Tb:={pIb0ZP andp

v}.

Then Tb is a finite set and T n Tb = 0 since b E Z for p E T by (g). If p 0 Tb U {v}, then (h) implies Q(u) = b2 E Zp , and if moreover p 0 T, then Kp = LP n (u1)p is unimodular since u E Lp by (h) and Lp is unimodular by (c). Therefore Corollary 5.2.2 yields (k) Q(Kp) = Zp for p 0 T U Tb U {v}.

Applying the last lemma to W, S := T U Tb, v, there is an integer r such

that

r E Z for p E T U Tb, and for a rational number c E Q(W) with

(1)

c E rzpnQ(Kp) for every p# v, there is an element w E W for which Q(w)=c and w E Kp for every p v.

Put

TT:={pI r0Zp andp#v}.

Then the first part of (1) implies (m) T, Tb, Tr and {v} are disjoint. We claim (n)

there is a rational number d such that d E Zp if p# v, and d and 1 are sufficiently close in Qp for p E T, and moreover 1 - (bd)2 E Q(W) and

1-(bd)2 ErZpnQ(Kp)

if p#v.

We will come back to the proof of this later; but first we complete the proof of the lemma with its help. Applying (1) to c := 1 - (bd)2, there is an element w E W such that Q(w) = 1 - (bd)2 and w E Kp for p # v. We show that

z:=du+w

is what we want. For P E T, z - zp = (dby - zp) + w is sufficiently close to 0, since b, d and 1 are sufficiently close, by (g) and (n), y and zp are sufficiently close and w E Kp c ptLp, by (j). Thus condition (i) holds. If p TU{v}, then (h) and K C L imply z = du+w E dLp+Kp C Lp, which is condition (ii). Lastly, Q(z) = d2Q(u) + Q(w) = (bd)2 + (1 - (bd)2) = 1 is condition (iii).

It remains to verify the claim (n). First we construct dp E Zp for p E T U Tb U Tr which satisfies the local version of (n):

6.2


143

(n') 0 74 1 - (bdp)2 E r7Lp fl Q(Kp) and dp E Zp hold for p e T U Tb U Tr, and dp and 1 are sufficiently close for p E T. Let p E T; taking a non-zero number ep E Q(Kp) that is sufficiently close to 0, we put dp = b-1(1- ep/2). By (g), dp is sufficiently close to 1 in @p. Clearly 1 - (bdp)2 = 1 - (1 - ep/2)2 = ep(1 - ep/4) E ep(Z )2.

It implies (bdp)2 # 1 and 1 - (bd1,)2 E Q(Kp). By (1) we have r E Z and hence 1 - (bdp)2 E r7Lp. Thus (n') has been proved in this case. Let p E Tr; then p Tb and so b E Zp . Choose dp E 7Lp such that dp ( b-1) and b-1 are sufficiently close in Q,. Then 1-(bdp)2 is sufficiently close to 0 and hence it is in r7Lp. By (k) and (m), Q(Kp) = 7Lp holds and then the assertion (n') is true in this case. Let p E Tb; we put dp

0. Since (1) implies r E 7Lp

1 = 1 - (bdp)2 E r7Lp.

It remains to show 1 E Q(Kp). Suppose that v1 = fy for f E 42p' is primitive in L. By (c), Lp is unimodular and hence Q(vi) = f2 E Zr,. Suppose f E Z P'. Since v1 = f b-1u and (j) imply that Kp = vi in Lp, Kp is unimodular and then Q(Kp) E) 1, since rank Kp = dim V -1 > 3 and p 2. If f E p7Lp, then Q(v1) E p7Lp holds. Since Lp is unimodular, B(vi, Lp) = 7Lp and hence there is an element v2 E LP such that B(v1, v2) = 1. Then 7Lp [v1 v2] is unimodular and so is 7Lp [v1 V2]1 in Lp which is contained in Kp = (u1)p in Lp. Thus Q(Kp) D Q(7Lp[U1iv2]1) E) 1 holds since rank Zp[v1iv2]1 > 2 and p 54 2, and we have verified the assertion (n'). Let us verify assertion (n). First, suppose v = oo; then by Lemma 6.1.1, we can choose a large number d E Q such that 1 - (bd)2 is negative, and d i

i

and dp are sufficiently close in @p for p E T U Tb UTr and d E 7Lp otherwise. Since dp E 7Lp for p E T U Tb U Tr by (n'), d E 7Lp there; but d E 7Lp for

p 0 T UTb UTr implies that d is an integer. By (n'), d and 1 are sufficiently close for p E T. For p E T U Tb U Tr, 1 - (bdp)2 E (r7Lp fl Q(Kp))\{0}

which is open, and d and dp are sufficiently close. Therefore 1- (bd)2 E r7Lp fl Q(Kp) holds for p E T U Tb U Tr. If p T U Tb U Tr, then b, d E 7Lp and so by (m) and (k) 1 - (bd)2 E 7Lp = r7Lp fl Q(Kp).


6

144

It remains to show 1 - (bd)2 E Q(Woo). By the way of the construction, V,,, = (1) 1 W,, and V,,, is isotropic since v = oo. Thus Q(W... ) contains a negative number and hence 1 - (bd)2 E Next suppose v := q 34 oo; for a sufficiently large integer t, we put dq := q-t and let dp be the integer in 7Lp in (n'). We choose a rational number d' such that d' and dp are sufficiently close for p E T U Tb UTr U {q}

and d E 7Lp otherwise. Take a sufficiently large integer m for which qm and 1 are sufficiently close in Qp for p E T U Tb U Tr and 1 - (bd)2 > 0 for d := d'q-m. Then d and dp are sufficiently close for p E T U Tb U Tr. (n') implies that d' E 7Lp if p E T U Tb U Tr, and hence d E 7Lp if p q. Thus d E 7Lp if p q, which is the first condition in (n). Since dp and 1 are sufficiently close for p E T by (n'), d and 1 are too, which is the second part of (n). For p E T U Tb U Tr 1 - (bd)2 E (r7Lp fl Q(Kp)) \ {0}

follows from (n'). If p

T U Tb U Tr U {q}, then

r7Lp fl Q(Kp) = 7Lp E) 1 - (bd)2

follows from (k), b E 7Lp and d E 7Lp. Thus 1 - (bd)2 E r7Lp fl Q(Kp) if

p # v = q which is the last condition in (n). Since V. = (1) 1 W. is isotropic, Q(Wq) E) -1 holds and then 1 - (bd)2 = -(bd)2(1 - (bd')-2g2m) E Q(Wq).

By (d), V,, = (1) 1 W,, is anisotropic and so Q(W,,) = {x > 0} E) 1-(bd)2. Thus 1-(bd)2 is represented locally by W and so 1-(bd)2 E Q(W) which proves (n) and so the lemma.

Proof of Theorem 6.2.2 with dim V > 4. We keep the notions in the theorem. We divide the proof into three steps. Suppose up = TxPTy,,, Q(xp) = Q(yp) (xp, yp E Vp) for every p E S. We take any vector x such that x and xp are sufficiently close for p E S, and x E LP otherwise, using Lemma 6.1.1, and take rip E O(Vp) such that Since x and xp are sufficiently close for p E S, Q(x) 0 follows yp = from Q(xp) 0. Choose a finite set S' of primes such that S' fl (SU{v}) = 0, and that p S' U S U {v} implies that p 2, Q(x) E Zp, Tx Lp = LP and LP is unimodular. Since Q(x) and Q(yp) ( 0) are sufficiently close for p E S, we have Q(x) = ccQ(yp) for some cp E 7LP . Put (i)

gp(xp).

(5)

zp

.

I cpyp x

for p E S, for p E S ' .

6.2


145

S' U S U {v}, then p 2 and Lp is unimodular and hence Q(x) E 7Lp C Q(Lp). Applying the last lemma to zp,T = S' U S and 0 a Q(x) E Q(V), there is a vector z E V with Q(z) = Q(x) such that If p

z and zp are sufficiently close for p E S' U S,

(6)

zELP for

p E S, then o = Txrz and Qp = rx,Typ are sufficiently close since x and xp, z and zp = cpyp are sufficiently close. If p E S', then or = Tx rz and Tx = id are sufficiently close since by (5), (6) x and z are sufficiently close, and hence a(Lp) = Lp. If p V S' U S U {v}, then TxLp = Lp follows from

the definition of S'. On the other hand, z E LP and Q(z) = Q(x) E Z imply r,, E O(Lp) since Lp is unimodular. Thus a(Lp) = Lp holds for p V S' U S U {v}, and a is what we want. (ii)

Suppose that ap = TX1 PTY1

P

Tx,, PTy, P, Q(xi p) = Q(yi p) for every

p E S.

In this case, we may assume that r is independent of p E S, since the order of symmetries is 2. Applying (1) to Tx PTy P, we complete this case. (iii) General case. Since Vp is isotropic for almost all primes, we may enlarge S by adding

anisotropic primes where we make vp the identity mapping. So we may assume that Vp is isotropic if p S. Put 2r

vp := 7-x1

P

... Tx2,

P

With

II Q(xi,p) = 1, i=1

and assume that r is independent of p E S. Take xi,

, x2r_1 E V

satisfying (a) xi and xi,p are sufficiently close for p E S and 1 < i < 2r - 1, and so are f?r i 1 Q(xi) and 11 r i 1 Q(xi,p) ( 0) for p E S. Hence there is

a unit e'' such that 2r-1

2r-1

Q(x2r,p)-i = 11 Q(xi,p) = ep fi Q(xi), i=1

i=1

and eP is sufficiently near 1. Therefore there is a unit ep such that eP = and ep is sufficiently close to 1 for p E S.

146

6


We claim the existence of a vector x2r E V which satisfies 2r-1

Q(x2r)

_

Q(xi)-1

i=1

and (b)

x2r, X2r,p are sufficiently close for p E S.

Put 2r-1

(7)

a:= rl

Q(xi)-1

i=1

First let us verify that a is locally represented by V. For p E S, a = Q(Epx2r,p) E Q(Vp) is clear. Since Vp is isotropic for p 0 S, a is represented by Vp for p V S. If V,,, is isotropic, then a is also represented by V. If V"

is anisotropic, then aQ(x2r_1) = 11zr12Q(xi)-1 > 0 implies a E Q(V,,,,). Thus a is locally represented by V and hence a is represented by V, that is there is a vector w E V with (8)

Q(w) = a.

Take rip E O+(Vp) such that 77p (w) = Epx2r,p for p E S and by Lemma 6.2.1

there is an isometry rt E O+(V) such that r) and qp are sufficiently close for p E S. Put X2r := r/(w); then Q(x2r) = fl?T i 1 Q(xi)-1 by (7) and (8), and x2r = 7](W), Epx2r,p = r/p(w) are sufficiently close for p E S. Since ep and 1 are sufficiently close, x2r and x2r,p are so. Thus we have verified the claim.

Put Since Vp is isotropic for p E S', Tx, Tx2, is a product of TxTy (where x, y E Vp, Q(x) = Q(y) 54 0) by Proposition 1.6.3. Hence by virtue of (ii) there exists an isometry al E O'(V) such that of and the identity are sufficiently close for p E S, and a1 and Tx,

-r X2, are sufficiently close for

p E S' and al (Lp) = Lp for p 0 S U S' U {v}. Now a := ai 1Tx1 Tx2r is what we want, since xi and xi,p are sufficiently close for p E S and 1 4. Suppose now dim V = 3. By scaling, we may assume d V = 1, and then V is identified with the set of all pure elements of some quaternion algebra C as in 1.5. Then

C=@®V and forx=a+wEC(aEQ,wEV)weputx:=a-w

6.3 Genus, spinor genus and class

147

and N(x) = xJr-. N is a quadratic form on C and the restriction on V is the original one on V. Put L' = 7L 1 L and take a finite set T ( v) of primes such that T D S and if p T U {v}, then Lp' = L'' is unimodular and a subring of C. By Proposition 1.6.4 there exists zP E CP such that ap(x) = zpxzp 1

and

9(ap) = N(zp)

for the given up for p E S. By N(zz) = 9(a,) E (@P )2, we put N(zp) _ P, ap E Q Q. Taking aP 1zp instead of zp, we may assume ap(x) = zpxzz 1,

N(zp) = 1.

For p E T \ S, we put zp = 1. Applying Lemma 6.2.3 to a lattice L' on C with a = 1, there is a vector z E C such that N(z) = 1, z and z, are sufficiently close for p E T and z E L' for p T U {v}. We define an isometry a E O(C) by a(y) = zyz-1 By Proposition 1.5.4, a E 0+ (C) holds and a(1) = 1 implies aIv E 0+ (V), which then by Proposition 1.6.4 yields a(x) = vxv-1 for x E V, v E C and

9(al v) = N(v). Then a(x) = vxv-1 = zxz-1 for x E C. Thus z-1v is in the centre of C, that is v = az for some a E Q. Thus 9(al v) = N(v) = N(az) = a2 and so a1 v E 0'(V). Now we regard ap as an isometry of C. If p E S, then z and zp are sufficiently close. Hence or and ap are sufficiently close, as are ajvp and aaIvv for p e S, using a(1) = aa(1) = 1. If p E T\S, then or and id are sufficiently close since z and zp = 1 are so and hence or I v, is sufficiently close to id, which yields aI vp (Lp) = Lp. Suppose p 0 T U {v}; then by the assumption on T, LIP = L'P is unimodular and a ring. Then z E L, and N(z) = 1 imply L'' = Zpz 1 (some lattice); hence r-. (L') = LP'

-zxz

and z-1 = z E L'p = L, and so z-1 E L,. Thus Tz(x) yields P

P

P

P

P

P

P

Therefore alvv(Lp) C L, holds since Lp is the orthogonal complement of 1, and the theorem has been proved.

6.3 Genus, spinor genus and class In this section, we generalize the concepts of genus, spinor genus and class given in 6.1. Let V be a regular quadratic space over Q and S a finite set

6

148


of places with oo E S. Put Zs := {a E Q I a E Z for p L on V, we can associate an S-lattice LS by

S}. For a lattice

n (VnLp) PVS

which is equal to Ut>o(fPES p)-tL and is a finitely generated module over ZS. In other words, an S-lattice L is a submodule of V such that LP = VP if p E S, and LP is a lattice on Vp if p S. If S = {oo}, then an S-lattice is nothing but a lattice defined as before. We define genus, spinor genus and class for S-lattices in the same way as for the standard lattice:

gens(L) := OA(V)L = OA(V)L, spns(L) := O(V)O'y(V)L, spns (L) := O+(V)O'A(V)L, clss(L) := O(V)L, clss (L) := O+(V)L, where g = (gv) E OA(V) acts on the set of S-lattices L by

gL:= n (V n gpLp)

ps

We define the stabilizer group of L by {g E O+4(V) I g(L) = L}

OA+(L)

={g EO,+4(V) Igp(Lp)=LpforpVS} If S = loo}, then the above definitions are the same as those given in 6.1. The following is due to Kneser.

Theorem 6.3.1. Let L be an S-lattice on a regular quadratic space V over Q; then the number gs (L) of the different proper spinor genera spns in gens(L) is given by

[QA : fl Q

B(O+(V))8(O,+q (L))]

if dim V > 3.

VES

Proof. First we note that OA (V) contains all commutators in O,+4(V); hence OA (V) is a normal subgroup of O+4 (V) and O+9 (V) /OA (V) is commutative.

Suppose spns (M) = spns (N) for M, N E gens (L). Write M = gL, N = hL for g, h E O+9 (V). By definition, we have

O+(V)O'A(V)gL = O+(V)OA(V)hL.

6.3

Genus, spinor genus and class

149

Thus g-1h E 0+(V)OA' (V)O+A(L) and gs (L) = [0A +(V) : 0+(V)0A(V )Oa (L)1. Now we consider the spinor norm 0: 0,+g (V) -+ QA, where we identify the image with its inverse image by @A --+ @a/((QA)2. This norm induces (1)

0: 0a(V)/0+(V)O+(L)OA(V)

-' QA/(rj w

-

9(0+(V))O(OA(L)))

VES

Note that the v-components of the above factor groups are trivial for v E S.

Let us show the surjectivity. Let x = (xv) E QA with x = 1 for v E S. Since dim V _> 3, Proposition 5.5.1 guarantees the existence of gp E O+(Vp) such that 9(gp) E) xp and moreover Proposition 5.5.4 implies the existence

of gp E 0+(Lp) such that 9(gp) 9 xp if xp E Zp and Lp is unimodular. Hence g := (gv) with g, = id for v E S is in GL(V)A, i.e. g E 0A+(V). Thus the surjectivity has been shown. Next suppose 9(g) E 11

Q . 9(0+(V))9(OA(L))

VES

Taking h E 0+(V), k E 0,+4 (L), we may suppose

9(ghk) C fj @v x II (Q )2. vES pS Moreover we may assume (ghk)v = 1 if v E S, since the v-component of 0,+4(L) is 0+(Vv) for v E S. Then 9(ghk) C (@A)2, i.e. ghk E 0'A (V). Thus the injectivity has been proved.

Corollary 6.3.1. gs (L) is a power of 2. Proof. First, we note that the mapping 9 at (1) in the proof of the theorem is injective even for dim 2. Since 9(0,+q(L)) D (@A)2, the injectivity of the mapping 0 at (1) in the proof of the theorem yields gs (L) is a power of 2 if it is finite. If S = {oo}, then g+ (L) is finite by Theorem 6.1.2. Suppose

S # {oo}. Take any lattice Lp on Vp for every prime p E S, and put L':= npo-(V n Lp). For a lattice M on V, we associate an S-lattice MS by MS :=

n Mp). Then

cls(M) -> clss(MS), spn+(M) - spns (Ms), gen(M) -' gens(Ms) are well-defined and we have a surjection from cls in gen(L') to clss in gens(L), since for N E gens(L), N' = npES(V n Lp) npgs (V n Np) E gen(L') corresponds to N. Hence the finiteness of the number of cls in gen(L') implies that the number gs of spns in gens(L) is finite. Let us determine 9(0+(V)).

150


6

Proposition 6.3.1. Let V be a regular quadratic space over Q with dim V > 3. Then we have

9(O+(V)) -

J {a E

a > 0} if V,,,, is anisotropic,

l (QX

otherwise.

Proof. Let a E Qx and suppose a > 0 if V... is anisotropic. Put 6

j -1 l

1

if Voo is positive definite,

otherwise.

If Vp is anisotropic for a prime p, then we take by E Qp such that 6abp d V

(Q )2 and 6bp d V ¢ (Qpx )2. Take 0 < b E (Q such that b and by are sufficiently close in Qp for a prime p for which Vp is anisotropic. Note that the number of primes for which Vp is anisotropic is finite by Theorem 3.5.1.

If there is no anisotropic prime, then we put b = 1. Then V 1 (6b) and V 1 (bab) are isotropic at every place by the same theorem. Thus they are isotropic and hence -6b, -6ab E Q(V). Taking x, y E V such that Q(x) = -6b, Q(y) = -bob, 9(rxTy) = a(@x)2. Thus 9(0+(V))

I {a Etnx J a > 0} if V... is anisotropic, ll Q otherwise.

The converse inclusion is clear.

Corollary 6.3.2. Let L be an S-lattice on a regular quadratic space V over Q with dim > 3. If

0(0+(Lp))

Zp

for every prime p V S, then gens(L) = spns (L).

Putting Q+ _ Proof. By the assumption 9(OA(L)) contains {a E Q I a > 0}, we know Qa =1R> x @+ x jlZp and then Qx C fi Q x @+ x 9(OA(L)) C 11 `Qv x 9(O+(V))9(O+(L)) C QxA* A vES

vES

Thus Theorem 6.3.1 implies the corollary.

Remark. Let LP = Ll 1 1 Lt be a Jordan splitting. If either rank Li > 2 (resp. 3) for p # 2 (resp. p = 2) for some i, or Lp is maximal and rank Lp > 3, then 9(Lp) D Z by Proposition 5.5.4 and 5.5.3. Thus if s(L) C 7L and d L is odd and square-free for a ({oo}-)lattice L, then gen(L) = spn+(L) if rank L > 3. Next let us give a sufficient condition for spn+ = cls+, spn = cls.

6.4

Representation of codimension 1

151

Theorem 6.3.2. Let V be a regular quadratic space over Q with dim V > 3, and S be a finite set of places with S E) {oo}. Suppose that there is a place v in S such that V is isotropic. Then for every S-lattice L, we have spns(L) = clss(L) and spns(L) = cls,+s(L).

Proof. Suppose K E spns(L); then there exist isometries p E 0(V), ap E 0'(V,) such that p(Kp) = ap(Lp)

for p

S.

Put T := {p V S I u(Kp) Lp}, which is a finite set. Then by Theorem 6.2.2, there is an isometry or E 0'(V) such that a(Lp) = Lp for p 0 S U T, and or and up are sufficiently close for p E T. Therefore a(Lp) =

LP = u(Kp) for p 0 S U T and a(Lp) = ap(Lp) = p(Kp) for p E T. Thus we have a(Lp) = p(Kp) for p 0 S and so K E clss(L). Therefore spns(L) = clss(L) holds. If we take p E 0+(V) in the above, then we obtain spns (L) = clss (L).

Remark. The most important case is S = {oo}. For example, if L is a lattice on an indefinite regular quadratic space V over Q with dim V > 3 and s(L) C Z,

and d L is odd and square-free, then we have gen(L) = cls+(L) by the above two theorems.

Exercise 1. Let V be a regular quadratic space over Q of dim > 3 and L a lattice on V and suppose that V9 is isotropic for a prime q and 9(0+(Lp)) D 7LP for p # q. Prove for every K E gen(L), there is an isometry p E 0+(V) such that p(Kp) = Lp for p 54 q. (Hint: Put S = {oo, q} and apply Corollary 6.3.2 and Theorem 6.3.2.)

Exercise 2. Let L be a unimodular lattice on an indefinite quadratic space V over i.e. s(L) C 7L and d L = ±1. Prove: If s(L) = 7L, then L is isometric to (1 (1)) 1 (1 (-1)). (i) (ii) If s(L) = 2Z, then L is isometric to an orthogonal sum of E8, E(-1) and (1 0 1 I ), where E8 is the unique lattice which is positive def\\\

inite, with rank 8 and norm 2Z. (The explicit form of E8 is given in 7.1 of the next chapter.) (Hint: Use the above remark if rank L > 3.)

6.4 Representation of codimension 1 In this section we consider the representation problem of codimension 1.

6

152

Let L = Li 1


1 Lk be a Jordan splitting of a regular quadratic

lattice L over Zp, that is Li is modular and ordp s(Li) < Then we put tp(L)

=

(a1,...,a,,a2,...

a2 ...

ak,...

< ordp s(Lk). ak),

where ai is defined by pai7Lp = s(Li) hence ai < < ak and the number of ai in the above sequence is rank Li. Note tp(L) is independent of the particular Jordan decomposition. If L is unimodular, then tp(L) = (0, , 0). For two ordered sets a = (a1, , an), b = (bi, , bn), we define the or-

dering a < b by either ai = bi for i < k and ak < bk for some k < n or ai = bi for all i. Lemma 6.4.1. Let U be a regular quadratic space over @p and L a lattice on it. Then L contains a submodule M such that (i)

rank M = rank L - 1, d M # 0 and M is a direct summand of L as

(ii)

a module, and any lattice L' on U containing M with d L' = d L and tp(L') > tp(L) is equal to L.

Proof. We use induction on dim U. First, we assume that L is unimodular and has an orthogonal basis {vi} 1 (n := dim U); then put

M = Zp[vl, ... vn_1]. The condition (i) is clearly satisfied. Let L' be the lattice in (ii). Since L is unimodular, tp(L) = (0, , 0), and tp(L') > tp(L) implies that every coordinate of tp(L') is non-negative and hence s(L') C Zp. Since M is unimodular and L' D M, L' = M 17Lpavn for some a E Qp by Proposition 5.2.2. Then d L = d L' = Q(avn) d M implies a2Q(vn) E 7 L p and so a E Z .

Thus L' = L. Next, we assume that L is unimodular but has no orthogonal basis; then we have p = 2 and L =1i tp(Lp) for every prime p. If there is an isometry u from M to L', then U' = U and extending u to an isometry

154

6


from U to U', we have L' = u(L) or uTr (L) where x is an element such that U = QM 1 Qx, and u(L) # urx(L). Proof. Let L' be the given lattice in the theorem. Since M y L',

U'=QL'-QM1(dMdL')=QM1 (dMdL)-U. Hence we may suppose that U' = U and L' D M, taking u-1(L) instead of L'. We will show L' = L or TE(L). Take an orthogonal basis {w,} i of Mq such that Q(wi) E Z for i < n - 1 and Q(wn_1) E qZQ, by virtue of (iii). Since L. is unimodular, N := Zq [wl, , wn_21 1 is also unimodular and wn_1 is primitive in N. We take an element wn E N such that {wn_11 wn} is a basis of N. Let us see that QqN is hyperbolic. Since N is unimodular and Q(wn-1) E qZq, we have B(wn_1i wn) E 7Lq , then d N = Q(wn-1)Q(wn)

- B(wn-1, wn)2

_ -B(wn-1, wn)2(1 - Q(wn-1)Q(wn)B(wn-1, wn)-2) and

Q(wn-1)Q(wn)B(wn-1, wn)-2 E qZq

imply that dN E -(Q )2 and QqN is hyperbolic. By Lemma 5.2.2, there is a basis {el, e2} of N such that Q(el) = Q(e2) = 0, B(el, e2) = 1. Put wn_1 = a1e1 + a2e2 (ai E Zq); then Q(wn-1) = 2a1a2 E qZq . Multiplying

ei by a unit and renumbering, we may suppose wn_1 = el + ege2 for some c E Z'. Now tq(L'q) _> tq(Lq) implies s(LQ) C Zq and then dL' _ d L implies that L'q is also unimodular. Since Lq contains Mq, there is a unimodular submodule Kq such that L''q =1i t2(L2), dL'2 = dL2 and 2q E Q(L2), then t2(L2) = t2(L2), s(L2) = s(L2) and L2 is unimodular. If L2 has no orthogonal basis, then n(L'2) = 2 s(L2), dL2 = dL2 yield L'2 = L2. Suppose L2 has an orthogonal basis. Then L'2 = (e1) 1 (e2) f o r ei E Z . Now dL2 = d L'2 implies El E2 =_ 2qc - b2 mod 8 and so e1 e2 =_ -1 mod 4. On

the other hand, 2q E Q(L2) implies that elx1 + e2x2 = 2q has a solution xi in Z2. This is impossible because eie2 -1 mod 4. Thus the second condition has been verified. Clearly Mp ?' (Q(v1)) = (2q) satisfies the third condition and we have completed the case that U is a hyperbolic plane.

Suppose that U is not a hyperbolic plane. Let MP be a characteristic submodule of Lp for p E S, and {x1,p, , x,,-,,p} be a basis of M. Applying Theorem 6.2.1, there are vectors x1i , xn_1 E L such that (i) xi and xi,p are sufficiently close for 1 < i < n - 1, p E S and (ii) det(B(xi, xj)) E Z for p S with precisely one exception p = q, where det(B(xi,xj)) E qZ . Then put M := Z[xl, , xn_1]. By Corollary 5.4.5, there is an isometry

o'p E O(Lp) such that Mp = vp(Mp') for p E S, and hence Mp is also characteristic of Lp. The other conditions are clear.

Ci

Remark. In general, the codimension two analogue of a global characteristic sublattice can not exist. The intuitive reason is this: the orthogonal complement of a sublattice of codimension two is two-dimensional, and the class number of a binary quadratic lattice is in general greater than 1. Corollary 6.4.1. Let {Li}m 1 be a set of lattices on quadratic spaces Vi

such that rank Li = n, d Li = d (1 < i < m) and they are not mutually isometric. Then there is a submodule M of Lj for some j such that rank M = n - 1 and M is not represented by Li for every i $ j. P r o o f . Let S be a finite set of primes such that 2 E S and (Li)p is unimodular f o r every i and p V S. Put S = {p1 i , pr} and define a subset Aj of {Li}Z"1 inductively as follows:

Al {Li I tpl (Li) < tpl (Lk) for k = 1, 2, , m}, A,+1 :_ {Li E Aa I tpj+l (Li) < tp3+1(Lk) for every Lk E Aj}.

6.5 Representation of codimension 2

Then Al D

157

D Ar is obvious. Take a lattice Lj in Ar, and let M be

a submodule satisfying the conditions in Theorem 6.4.1 as constructed by Proposition 6.4.1. Assume that M is represented by Li. Then V = QLi = QM 1 M) = V. Since Lj E Ar C Al, tP, (Lj) < tp, (Li). Furthermore

since Mp, is characteristic to (Lj)P (Lj)p, = (Li)P1. Thus t,,(Li) = tp, (Lj) and Li is in Al, and Lj E Ar C A2 implies tp2 (Lj) < tp2 (Li) since Li E Al. Thus we have tP2 (Li) = tP2 (Li) similarly to the above. Repeating this argument, we have Li E Ar which means tp(Li) = tp(Lj) for p E S. Thus Theorem 6.4.1 yields Li = Lj.

6.5 Representation of codimension 2 In this section, we show the following fact due to Kneser, Hsia and Schulze-

Pillot whose proof is group-theoretic. As an abbreviation, we say that K is represented by gen(L) (resp. spn+ (L)) if K is represented by a lattice in gen(L) (resp. spn+(L)).

Theorem 6.5.1. Let V = W 1 U be a regular quadratic space over Q and suppose dim U = 2. Let L, K be lattices on V, W respectively and suppose L D K. Then K is represented by either every spn+ or exactly a half of the spn+ in gen(L). The latter case occurs if and only if the following three conditions hold: (i) (ii)

- d U (QX )2; 9(O,+g (L)) C N :_ fv N n Q 1, where

N,,:={xEQZv I (x,-dU)@ =1}, (iii) putting 9(Lp; Kp) := {9(cp) C Q2p I W E O+(Vp) with Kp C cp(Lp)},

9(Lp; Kp) C Np

for every prime p. We need several lemmas. We take 6 E QX such that 6 E d U.

Lemma 6.5.1. Suppose -6

(Q')'; then we have [QA : Q2" N] = 2.

Proof. Putting

f

p 7Lp

, we have Qx = Q2" 7L 1R , where

R':={aEIR" ja>O} is identified with elements in Q2' with all finite components being 1. We define the mapping cp : QA -* {±1} by V(x) = rjv(x, -6)@ for x =

158

6


E Q'. It is well-defined since for almost all odd primes p, xp, -S E ZP and hence (xv, -S)@, = 1. Since -6 0 (Qx)2, there is a place v such that

-b

(@v )2. By (8) in Theorem 3.3.1, co is surjective. From (9) in the same theorem ker cp D QX `Let. Hence ker cp D QX N. Suppose x E ker cp. We must verify x E QX N. , v2,,, be the places such that (x,,, -S)@ = -1. vi,

2, vi for 1 < i < 2n and 6, x7, E 7LP hold. Put ep = ordp xp and h = Ilpes pe'. We take a

Let S be a finite set of primes such that for p 0 S, p

prime q such that q - 17h-1xP modpt for S, where t is a sufficiently large

integer and q = 1 if all places vi are finite, -1 otherwise. Let us show qr]h-1x E N, i.e. (g17h-lx,,, -S)(Q = 1. If a prime p is not in S and p # q, then p # 2 and g17h-lxp and b are in Z , and so (g17h-lxp, -S)Qp = 1. If p E S, then g17h-lxp is a square in @PX and so (g17h-1xp, -S)Qp = 1. If all places vi are finite, then (g17h-1x,,., -5)R = (x,,,,, -S)R = 1. If some place

vi is infinite, then q = -1 and (x,,,,, -S)R = -1 and so x,,, < 0. Thus (gi7h-lxD, -S)R = 1. Finally we have (q?7h-1xq, -6)(Q9

= fJ(g17h-lxv, -S)(Qv = 1. vOq

Therefore we have shown gr7h-lx E N and completed the proof.

Lemma 6.5.2. [Q : N9(O+(V))] = 2[@X : 0(0+(V))] if -6 V (Q')'. Proof. The left-hand side is equal to [QA : NQX][N@X : N9(O+(V))]

= 2[NQX : 99(O+(V))]

= 2[@X : QX n 99(O+(V))]

If V,, is isotropic, then 0(O+(V)) = QX by Proposition 6.3.1 and then QX n N9(O+(V)) = QX. So in this case the assertion has been verified. Suppose that V,, is anisotropic; we have only to show (QX n N9(O+(V)) = 9(O+(V)) (= {a E QX 1 a > 0}).

The left-hand side obviously contains the right-hand side. Suppose b E QX n 90(0+ (V)) and write b = cx, c E QX,c > 0,x E N. We have only `too prove b E 9(O+(V)). Since V,,, is anisotropic and dim U = 2, we have d U > 0 and so 6 > 0 follows; then

x(, EN.={yERX I (y,-S)R=1}=IR

.

Thus x. > 0 and then b > 0, which means b E 0(O+(V)) by Proposition 6.3.1.


6.5

159

Lemma 6.5.3. Suppose -6 V (@X )2. Then 0(0+q(L))1RX C NB(O+(V)) t* o(O,+q(L))1RX c N.

If we remove R'< from both sides, it is still valid. Proof. is clear. Let us prove =. Suppose that the right-hand side is false. Then there is a place v such that the v-component of 9(O,+9(L))1R" is

not in N. So we choose a v-component x of 9(O,+4 (L))1RXwith x V N. We define x = (xv,) E QA by x,, := 1 if u v. Since x E 9(O++(L))1RX, we

write x = ay, a E 9(O+(V)), y E N. For a place u # v, xv = 1 implies 1 = ayu and so a = y,-,' E N,,,, i.e. (a, -6)@u = 1 for u # v. On the other hand, f 1v (a, -S)@ = 1 yields (a,

fl (a,

u#v

Thus a E Nv and so xv = ayv E Nv. This is a contradiction. The case without 1RX is quite similar.

Lemma 6.5.4. Suppose that -6 0 (QX)2 and V,,. is anisotropic. Then [N6(O+(V))O(O+(L))RX : NB(O+(V))l = 2

if and only if [NO(O+(V))O(Oa(L))1RX : 1V9(O+(V))9(Oa(L))] = 2 and

N9(O+(V))6(O,+9(L)) = NB(O+(V)).

Proof. The "if" part is clear. Let us verify the converse. If the conclusion is false, then we have 1V6(O+(V))6(Oa(L))1RX =1V0(O+(V))9(Oa(L)) and hence

]RX

c N O(O+(V))O(O+(V,,.)) = N 9(O+(V,,,,)).

Since V,,. is anisotropic, 9(O+(V,,,)) = ]R+ and 6 > 0, and so

N,,,, _{aE1RX I (a,-6)R=1}=R+. This is a contradiction.

0

160

6


Lemma 6.5.5. Suppose -6

Then

[Qi : 1V9(O+(V))9(OA(L))1R"] < 2, and the equality holds if and only if

9(OA(L)) C N.

Proof. Put i:= [Q i : N9(O+(V))9(Oa(L))1R"], j:= [N9(O+(V))9(Oa(L))1R" : NB(O+(V))]. Then by Lemma 6.5.2, we have 2[Qx : 9(O+(V))] = ij. If V,, is isotropic, then by Proposition 6.3.1 we have Qx = 9(O+(V))

and so ij = 2. In particular we have i < 2. i = 2 if and only if j = 1, which is equivalent to 9(OA(L))1Rx C N9(O+(V)). Then Lemma 6.5.3 9(OA(L))1RX C N. Noting 0(0+(V,,)) = R", we get implies i = 2 the assertion. Suppose that V,,, is anisotropic; then Proposition 6.3.1 implies ij = 4. By Lemma 6.5.4 i = 2, i.e. j = 2, if and only if

9e(O+(V))9(O+(L)) = 99(O+(V)) and

k:= [N9(O+(V))O(Oa(L))1R" : N9(O+(V))9(O++(L))] = 2. k is clearly equal to [1R" :1R" n 1V9(O+(V))9(O,+q (L))]

where 1R" is identified with elements in Q with the component being 1 for every prime. Since V,,,, is anisotropic, 6 > 0 hence N,,, = R +x, 9(O+(V)) = {a E Qx I a > 0} by Proposition 6.3.1 and the oo-part of 9(OA(L)) is equal to

Thus

1R" n N9(O+(V))B(Oa(L))=1R+

and hence k = 2 is automatically satisfied if V... is anisotropic. Thus i = 2 is equivalent to N9(O+(V))9(O (L)) = N9(O+(V)), which means 9(O++(L)) C 99(O+(V)). By Lemma 6.5.3, this is equivalent to 0(O,+y (L)) C N. Lastly we must verify

i < 2. Otherwise i = 4 and so j = 1, which implies 1R'< C N,,9(O+(V)). Since V,, is anisotropic, we get 6 > 0, N,,,, =1R+ and 9(O+(V)) = {a E Qx I a > 0}. Thus the above inclusion does not hold.

6.5


161

Lemma 6.5.6. Put HL

OA(U)O+(V)O''A(V)OA(L),

where OA(U) is canonically embedded in OA(V) by assuming that O+(U,) acts trivially on W,,. If -6 then OA(V) = HL. If -6 (Qx)z then (1)

OA(V)IHL = QZA/NO(O+(V))0(OA(L))]RX,

where the isomorphism is induced by 0.

Proof. First, we suppose -6 E (Q2" )2; then U is a hyperbolic plane since 6 E d U. Take g E OA(V). Since 0(Ov (UU)) = Qv and for a lattice L' on U 0(O+(L'P)) D 7GP holds for almost all p by Proposition 5.5.5, we have 0(OA(U)) = @A and hence there is an isometry h E OA(U) such that 0(gh) = (@A)2, that is gh E 0'4(V). Noting O'9(V) is a normal subgroup of 0+(V), this case has been shown.

Next suppose -6 (Q > )2. Let us show first 0(OA(U)) = N. Write U = (b) 1 (b6) (b E Q"). Then 0(O+(UU)) = 0(O+((1) 1 (6))) holds. If Uv is isotropic, then -6 E (QZ,x )2 hence 0(O+(UV)) = QZv , and

N = {a E Qv I (a, -6)@v = 1} = Q2v

.

If U is anisotropic, then 0(O+(UU)) = {a2 + b26 E Q2v I a, b E Qv} since

(a2 + b26)(A2 + B26) = (aA - bB6)2 + (aB + Ab)26.

On the other hand, x E N if and only if (x, -6)Q = 1, which means a2x - b26 = c2 for a, b, c E Qv which are not 0 at the same time. It is equivalent to x = e2 + d26 for e, d E Q2v. Thus we have 0(O+(Uv)) = NV,

and then 0(OA(U)) = N follows from Proposition 5.5.5 as above. So 0 implies a homomorphism of the left-hand side and the right-hand side of (1). Since dim V > 3, Proposition 5.5.1 and 5.5.5 yield the surjectivity. Let us show the injectivity. Suppose that

0(f) c N0(O+(V))0(OA(L))IR"

for f E OA(V). Multiplying by an element of OA(U)O+(V)OA(L), we may assume 0(f) C 1Rx (Q2A)2. Put 0(f) = r(QA)2 for r E 1RX. If V,,,, is anisotropic, then 0(O+(V,)) = 1R and so r > 0. Thus 0(f) = (Q )2 holds and hence f E O'4 (V). If V... is isotropic, then 0(0+(V,,,,)) = 1R'< holds, and therefore OA(L) contains x = (xv) such that xv = 1 if v # 00

and 0(xv)={aEJRIa0,then 0(f)=(Q2A)2. If r < 0, then 0(x f) = -r(QA)2 =

Thus 0 is injective.

162

6


Lemma 6.5.7. Put i :_ [OA(V) : HL].

Then i = 1 or 2, and i = 2 if and only if -S V (Q')' and B(OA(L)) C N,

Proof. By the last lemma, -S E (QX )2 implies i = 1. So we may assume -S (QX)2. Then, (1) in Lemma 6.5.6 and Lemma 6.5.5 imply the assertion.

Proof of Theorem 6.5.1. First let us see that 0A(V)/O+(V)OA(V)OA(L) acts transitively and faithfully on spn+ in gen(L). Take L1, L2 E gen(L) and g E 0+(V). If L1 E spn+(L2), then L1 = qj(L2) where 17 E 0+(V)

and j E 0' (V). Now g(L1) = i7(?7-1gqg-1)(gjg-1)g(L2) ri-1ggg-1, gjg-1

means g(L1) E spn+g(L2) since acts on spn+ in gen(L).

E 0' (V). Thus 0,+4(V)

Next, g(L1) E spn+(L1) means g(L1) = r7j(Ll) for 77 E 0+(V) and j E 0' (V). Writing L1 = k(L) for k E 0A(V), this means (ijjk)-lgk E OA(L). Thus gE

?7jkOA(L)k-1

C 0+(V)0'A(V)OA(L)

since 0A (V) contains the commutator subgroup of 0A +(V), ), i.e. the sub-

group which fixes spn+(Li) is 0+(V)O'A(V)OA(L). The transitivity follows from the fact that 0+(V) acts transitively on gen(L). Thus the claim is verified.

For j E 0A(V), we have 0A(jL) = jOA(L)j-1 C 0' (V)OA(L) and hence HiL C HL. Similarly we have HL = Hj-ljL C HiL and so HL = HjL. Suppose that the lattice K on W in the theorem is represented by jL, that is there is an isometry v : K -+ jL. Extending it, we may assume Q E O(V). We will show that K is also represented by a lattice in

spn+(j'jL) for j' E HL. Since HL = HJL = 0'-10A(U)0'0+(V)OA(V)OA(jL)

we can write j' = jlj2j3j4 where

ji E

QOA(U)v-1

j2 E 0+(V), j3 E 0' (V), j4 E 0A(jL)

Then

j' = j2(j2 1j1j2j1 1)(jlj3j1 1)jlj4 E O+(V)O'A(V)jlj4

6.5


163

and so spn+(j'j(L)) E) jlj4j(L) = jlj(L). Here j(L) D o-(K) implies j(L) D v(K) I N for some binary lattice. Then

jij(L) D ji(o(K) I N) = jlo,(K 1o,-1(N)) = a(Q-1j1Q)(K 1 or-'(N)) D a(K), since a-1 j1o is a trivial isometry on W. Thus K is represented by jl j(L) E spn+(j'j(L)) if j' E HL and K is represented by jL. By Lemma 6.5.7, we have [O,+4(V) : HLI G 2. Thus K is represented by either every spn+ or exactly a half of the spn+ in gen(L). The latter occurs if and only if [O,+4 (V) : HL] = 2 and there exists some L' in gen(L) such that K is not represented by spn+(L'). [O,+4(V) : HL] = 2 is equivalent to (i) and (ii) by Lemma 6.5.7. So we have only to verify that the condition (iii) does not hold if and only if every spn+ in gen(L) represents K, assuming [O,+q (V) : HLI = 2. Before the proof of this, we note that

Np = 0(0+ (Up)) C O(Lp; Kp) since an isometry cp E 0+ (Up) is extended by VjwP = id and then cp E O+(Vp) and cp(Kp) = Kp C Lp. Therefore the condition (iii) is equivalent to O(Lp; Kp) = Np for every prime p. From now on we assume [O,+4(V) : HLI = 2. First we suppose that K is represented by every spn+ in gen(L). We must show O(Lp; Kp) Np for some prime p. By Lemma 6.5.6, there exists cp E O,+9 (V) such that (2)

e((p) ¢

98(0+(V))e(O+(L))Rx.

By the assumption, there is a lattice L' E spn+(cpL) such that K y L. Multiplying cp by an element of O+ (V )OA (V ), we may assume K C cpL without loss of generality. By the definition of O(Lp; Kp), we have, for every prime p (3)

8(cp) C O(Lp; Kp),

where cpp is a p-component of V. On the other hand, from (2) it follows that there is a prime q such that O(cpq) ¢ Ng8(O+(V))8(O+(Lq)), which yields N. 54 O(Lq; Kq) by (3).

Conversely we assume 8(Lq; Kq) # Nq for some prime q. Since Nq C 8(Lq; Kq) as above, 8(Lq; Kq) 0 Nq and hence there is cpq E O+(Vq) such that cpq(Lq) D Kq and 8(cpq) 0 Nq. We extend cpq to cp E O,+4(V) by

putting cp = id on V, for every place v 0 q. Then ca(L) D K is clear and 8(cp) ¢ N. Let us see (4)

8(cP) 0 98(O+(V))

164

6


If 9(p) c 99(O+(V)), then there is a rational number b such that bO(co) C 9. For v q, b E N is clear by the definition of cpT,. Thus (b, -6)Q = 1 holds for v q, and then (b, -6)Qq = 1 holds, which means b E Nq. Thus b E N and O(W) C N, which contradicts O(cpq) Nq. Next we show (5)

O(W)

90(0+(V))e(O+(L))1R".

If (5) is false, then 6(p) C N9(O+(V))O(O+(L))1Rx, which implies a9(V) C NB(OA(L))1R < for some a E B(O+(V)). Hence a6(pp) C NpO(O+(Lp)) for every prime p. Since cps = id, a9(V0) C e(O+(V))(1R")2 C

Hence we have C N9(0,+9(L)) and O(V) C N6(O+(V))O(O,+g(L)). Then Lemma 6.5.7 implies 9(p) C NB(O+(V)) which contradicts (4). Thus (5) is true and we have cp HL by (1) in Lemma 6.5.6. Thus we have proved that W(L) in spn+((p(L)), which is not in the orbit of spn+(L) by HL, represents K. Hence every spn+ in gen(L) represents K.

Are there any diophantine-theoretic (not group-theoretic) results like [BoSP] in the codimension two case?

6.6 Representation of codimension >3 In the indefinite case, we have a satisfactory result. Theorem 6.6.1. Let V, W be regular quadratic spaces over Q with dim V+ 3 < dim W and L, M lattices on V, W respectively. Suppose that LP is represented by Mp for every prime p, V,,, is represented by W,,. and W is indefinite. Then L is represented by M.

Proof. By Corollary 4.4.2 V is represented by W. So we may assume V C W. By the assumption, there is an isometry ap from Lp to Mp, where we may assume ap E O(Wp). Multiplying by some isometry in O(Vp1) from

the right if necessary, we may assume ap E 0'(Wp) by Proposition 5.5.1. Applying Theorem 6.2.2 to W, M, there is an isometry or E 0'(V) such that

fa(Mp) = Mp if Lp C Mp a is sufficiently close to ap if Lp 0 Mp. Hence we have a(Lp) C Mp for every prime p. Thus a(L) C M. In the definite case, this Hasse principle does not hold in general. At present we can only show the following result.

6.6 Representation of codimension > 3

165

Theorem 6.6.2. Let W be a positive definite quadratic space over Q and N a lattice on it. Then there is a positive constant c(N) such that a lattice M on a positive definite quadratic space V over Q is represented by N provided that

dim W > 2 dim V + 3, min(M) := min{Q(x) I x E M, x 0 0} > c(N) and Mp is represented by Np for every prime p. We need several preparatory lemmas.

Lemma 6.6.1. Let IN be the set of non-negative integers; we introduce a , yk) if xi < yi for all partial order in Nk defined by (x1, , xk) < (yl, i. Then every subset S of Nc contains only finitely many minimal elements.

Proof. We use induction on k. The assertion is trivial for k = 1. Write x = (p(x), xk) E N k

with p(x) E Nk-1, and put SS,, = {x E Nk-1 I (x, n) E S}.

Let Mn, M be the sets of the minimal vectors of Sn,, U°n°-_OSn respectively.

By the induction assumption, Mn and M are finite sets. For y' E M we fix any element y E S satisfying y = (y', yk) and denote by M' the finite set of such y. Put m := max{yk y E M'}. Suppose that x E S is minimal. Then from the definition p(x) E SXk there exists y E M' such that p(y) < p(x). If xk > Yk, then x > y and so the minimality of x implies x = y E M'. Suppose xk < yk (< m). Since x is minimal, p(x) is so in I

Sxk, i.e. p(x) E Mxk. Thus X E M' U (Un 0(Mn, n))

0 Lemma 6.6.2. Let Np be a regular quadratic lattice over 7Lp of rank Np =

n > m. Then there are only finitely many regular submodules Mp(j) of rank M(j) = m such that every regular submodule Mp of rankMp = m of Np is represented by some Mp(j).

Proof. By scaling we may assume s(Np) C 7Lp. Let Mp be a regular quadratic lattice over 7Lp of rank M = m and Mp =1i_1 Li be a Jordan splitting. Since Li is modular, there is an integer bi E N such that p-biLi

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166


is unimodular or pZ modular. Since there are only finitely many nonisometric lattices over Z, of unimodular or pZ modular lattices of fixed rank, there are only finitely many possibilities for the collection t, rank Li , t). Fix one of them and and the isometry class K i of p b`Li (i = 1, , bt). By the last lemma, there are only consider the corresponding (b1, finitely many minimal ones. It is clear that if Mp =J Li, MP =12=1 L2, %

and bi
3 and U. is isotropic for a prime q. Let L be a lattice on U such that LP is maximal for every prime p. Then there is a natural number s such that L represents every positive definite quadratic lattice H for which Hp is represented by g8Lp for every prime p.

Proof. Let {Li} be a complete set of representatives of classes in gen(L). We put S = {q, oo} and denote by the S-lattice LZ

np,49(U n (L2)p)

Then Theorem 6.3.2 and Corollary 6.3.2 imply cls+(L') = spn+(Li) = gens(Li) = gens(L'). Thus there is an isometry Qi such that (o (Li))p = LP for every prime p # q. Hence as a set of representatives of classes in gen(L), we may assume (Li)p = Lp for p # q. Then for a sufficiently large integer s, we have g8Li c L. For a lattice H in the theorem, H is represented by gsLi for some i by Theorem 6.1.3, noting that {gsLi} is a complete set of representatives of the classes in gen(g8L). Thus H is represented by L.

Lemma 6.6.4. Let L, q, s be those in the last lemma, m an integer satisfying rank L > m + 3 and K a lattice on a positive definite quadratic space over Q. Then there is a positive constant c such that K 1 L represents a lattice M = Z[vl, , v,,,,] if Mp is represented by (K 1 g8L)p for every prime p and (B(vi,vj)) > cl,n. Proof. Let S be a finite set of primes such that 2, q E S and for p V S, Kp and LP are unimodular, and fix a natural number r such that (1)

pr s(Kp) C n(gsLp)

if

p E S.

6.6

Representation of codimension > 3

167

Choose vectors vh E K (i = 1, , t) such that for any given h = 1, h < t) for which xl,p, , Xm,p E Kp, there is some h (1 (2)

vh - xi p mod prKp

for i= l

m and for all p c S.

We choose a positive number c such that (3)

clm > (B(vh, v ))

f o r h = 1, . . , t.

Let M be a given lattice in the lemma. By the first condition on it, there exist xi,p E Kp, yi,p E gsLp such that (4)

B(vi, vj) = B(xi,p, xj,p) + B(yi,p, yj,p)

for all primes p.

For some h satisfying (2) for these xi,p, we put (5)

A := (B(vi, vj))

- (B(vh, vj ))

All entries of A are rational and A = (B(vi, vj)) - Clm + Cl,.,.,, - (B(vh, vj ))

implies that A is positive definite by virtue of the second condition on M and (3). By the choice of h, we can write xi,p = vh + p7'zi,p for some zi,p E Kp for i = 1, , m, p E S. Then (4) implies A =(B(xi,p, xj,p)) + (B(yi,p, yj,p))

- (B(vh, vv ))

=(B(yi,p, yj,p)) + (prB(vi , zj,p) + prB(zi,p, vj) + p2rB(zip zj,p))

By (1) and vh, zi,p E Kp, the (i, j) entry of A is congruent to B(yi,p, yj,p) modn(gsLp) for p E S. Then denoting by

H:=Z[ul,...,u,] a quadratic module defined by (B(ui, uj)) := A, yi,p E g8Lp yields n(Hp) C n(g8Lp) for p E S. Next, for p V S, vi E M, vh E K and n(Mp) C n(Kp I qsLp) = 7Lp imply n(Hp) C n(Mp) + n(Kp) C 7Lp = n(g5Lp)

by (5). Thus we have proved n(Hp) C n(gsLp) for every prime p. Since

dim QL - dim QH = rank L - m > 3


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168

and g8Lp is maximal for every p, Hp is represented by gsLp for every prime p by Theorem 5.2.3. Since A is positive definite, (Q2H),, is positive definite and then by the last lemma, H is represented by L. Thus for some wi E L,

(B(wi,wj)) = A and hence

(B(vi,v.7)) = (B(vh,vj )) + (B(wi,wj)) This means that M is represented by K 1 L. Proof of Theorem 6.6.2. Let N be the lattice on W in the theorem. Let S be a finite set of primes such that 2 E S, Np is unimodular for p S and Nq is unimodular for some prime q ( 2) in S. We will construct a set of submodules K(J), L(J) of N as in Lemma 6.6.4 and show that a given M in the theorem satisfies the condition in Lemma 6.6.4 for some J. Put m = dim V. For each p E S, we choose finitely many submodules Mp(jp) of rank = m in Np according to Lemma 6.6.2. To each collection J = (jp)pES, we take a submodule K(J)

of rank = m in N satisfying the conditions K(J)p = Mp(jp) for p E S and d K(J) E 7LP or p7LP for p V S by Theorem 6.2.1 and Corollary 5.4.5. We construct a submodule L(J) of

rank L(J) = rank N - m (> m + 3) in {x E N I B(x, K(J)) = 0} by local conditions as follows: We put

L(J)p = K(J)p in Np

for p V S.

Since s(K(J)p) C 7Lp and d K(J)p E 7Lp U p7LP , Proposition 5.3.3 implies d L(J)p C 7Lp u p7Lp . Thus L(J)p is 7Lp maximal if p S. For p E S, we

take any maximal submodule of K(J)P as L(J)p. From Proposition 5.5.3 and Corollary 6.3.2, it follows that gen(L) = spn(L). We show that L(J)q is isotropic. If rankL(J)q > 5, then L(J)q is isotropic. Otherwise, we have rank L(J)q = 4, m = 1, rank N. = 5. By the assumption on q ( 2), Nq is unimodular. Hence we have (7)

Nq =

((

0 ))-L((o '))-L (some lattice). 1

If L(J)q is anisotropic, QqN D QqL(J)q contains a 4-dimensional anisotropic subspace and hence indQ2gN = 1, which contradicts (7). Thus L(J)q is isotropic. So, we have constructed submodules K(J), L(J) of N such that rank K(J) = dim V, L(J) is locally maximal, especially 7Lp-maximal if p V S, L(J)q is isotropic, and finally every submodule M(p) in Np with rank M(p) = dim V is represented by K(J)p for p E S for some J.

6.7 Orthogonal decomposition

169

Let M be a lattice of rank m of a positive definite quadratic space over Q such that Mp is represented by Np for every p. Suppose p 0 S; then Np is unimodular. Hence n(Mp) C Z, holds. Since L(J)p is Zp maximal and rank L(J)p > m + 3, M, is represented by L(J)p = (gSL(J))p by Theorem 5.2.3 for every J where s is the maximal integer among those integers s given in Lemma 6.6.3 for L(J). For some J, Mp is represented by K(J)p for all p E S. Thus Mp is represented by (K(J) 1 g8L(J))p for every prime p. By Lemma 6.6.4, there is a constant c(J) such that M is represented by

K(J) 1 L(J) C N if (B(vi,vj)) > c(J)1,,,, for some basis {vi} of M. Put c' := maxi c(J). By the reduction theory in Chapter 2, there is a basis {vi} of M such that (B(vi,vj)) E 54/3,1/2 (the Siegel domain; see Chapter 2), and then (B(vi, v7)) > cdiag(Q(vi), ... , Q(vm)) for some positive constant c by Theorem 2.1.2. If Q(v) is sufficiently large, then we have (B (vi, vj)) > c'l,,,, and hence M is represented by N. This completes the proof.

Exercise. Let K, L be regular quadratic lattices over Z with rank L > rank K + 3. If K is represented by gen(L), show K is represented by spn+(L).

6.7 Orthogonal decomposition If a quadratic module L has a non-trivial orthogonal decomposition L = L1 1 L2 (L1

0, L2

0),

then L is called decomposable. Otherwise it is called indecomposable. The aim of this section is to show the following two facts.

Theorem 6.7.1. Let V be a positive definite quadratic space over Q. If L = L1 L 1 Lt is an orthogonal splitting into indecomposable submodules of a lattice L on V, then the set of {Li} is uniquely determined by L.

Theorem 6.7.2. Let V be an indefinite quadratic space over Q. Then there is no indecomposable lattice on V if dim V > 12.

The first is due to Eichler and the second is due to Watson. Proof of Theorem 6.7.1.

Let L be a lattice on V. By scaling of V if necessary, we may assume

s(L) C Z. An element x in L is called reducible if there are non-zero elements y, z E L such that x = y + z and B(y, z) = 0. Otherwise x is


6

170

called irreducible. Let S be the set of all irreducible elements of L. Since V is positive definite and s(L) C 7L, x = y + z (y # 0, z # 0) and B(y, z) = 0 imply 1 < Q(y), Q(z) < Q(x) for x, y, z E L. Thus every non-zero element

in L is a sum of irreducible elements, and S spans L. We introduce an equivalence relation - in S. For x, y E S, we write x - y if there is a chain of elements zl,

,Zr in S such that z1 = x, Zr = y and B(zi, zi+1)

0 for

, r - 1. Let {Si}i1 be the equivalence classes of this equivalence relation. Note that B(Si,S,) = 0 if i # j. Let Li be the submodule spanned by Si. Then B(Li, Lj) = 0 if i j. Since S spans L, we have i = 1,

L =1i= Li. Thus t is a finite number. Let L =-L 1 Ki be another orthogonal decomposition with Ki indecomposable. Take an element x in Sk. Since x is irreducible, x is in some Kj. For an element y in Sk, there is a chain z1, ,Zr E Sk such that z1 = X, Zr = y, B(zi, zi+l) 0 0 (i = 1, . . , r-1). Hence x = zl, z2i , Zr = y E Kj follows. Thus Sk is contained in some Kj. Therefore Lk C Kj holds. Put Cj = {k I Lk C Kj}; then L =1'_1 (1kEc3 Lk) C1'-1 Kj = L, which yields Kj =1kECj Lk. Since Kj is indecomposable, #Cj = 1 follows for all j. This is the assertion of the theorem. To prove Theorem 6.7.2, we divide the proof into several steps.

Lemma 6.7.1. Let K be a regular quadratic lattice over 7Z with n rank K. Then there is a splitting K = K1 1 K2 such that d K1 E (Q< )2 and moreover rank K1 = 2 if p # 2 and n > 5, rank K1 = 2 or 4 if p = 2 and n > 8.

Proof. First suppose p # 2. Let us recall (Q /(Q )2) = 4 and K has an orthogonal splitting 171 (yi) (,1i E Qp ). For n > 5, there is a pair (i, j) such that rlirlj E (@p )2. We have only to take (,qi) 1 (rid) as K1. Next suppose p = 2 and write

K =J i= 1

(,3,)

1 (_ -2 i=1(2a`

(?

))) 1 (lit 1 (2bi

(11 2

))),

where mi > 0 and rank K = n = m1 + 2(m2 + m3) > 8. For any given five different classes in 422 /(@2 )2, we can choose four classes whose product is in (Q2 X)2 by Corollary 3.1.4. Thus we may assume m1 < 4 and the product /3j 13 is not in (Q2 )2 for any i # j. If m2 > 2, then we can take ((2a1

(0 o))

o))) as K1. Thus we may assume m2 and

1 (2a2 (0

similarly m3 < 1. Now n > 8 implies ml = 4, m2 = m3 = 1. If the product of two elements of Ni (1 < i < 4) is in -(Q2 X)2 or 3(Q22 )2, then

6.7

Orthogonal decomposition

171

taking an orthogonal sum with (2a1 (0 ')) or (2b1 (1 1) ), we get a required submodule K1 with rank K1 = 4. Thus it remains

/to

show the non-existence of the subset S (represented by /3i) in Q 21 /(@2 )2 with S = 4

such that the product of any two distinct elements of S is different from ±1, 3 in Q22 /(Q21 ) 2 and the product of all elements is not a square. Write S = {si}4 1. If ord2 81 - ord2 s2 - ord2 83 mod 2, then ord2 $182 - ord2 s1 53 - 0 mod 2

and 81 S2 and 51S3 are different classes in Q / (Q22) 2. One of them must be equal to ±1 or 3 mod(QQ )2. This is a contradiction. Thus we may assume

ord2 Si - 0 mod 2 (i = 1, 2), ord2 Si - 1 mod 2 (i = 3,4) without loss of generality. Since S1s2 and S3S4 # f1, 3 mod(Q2 )2, {s1, s2} = {1, 5} or {3, 7} and {33i s4} = {2, 2.5} or {2.3, 2 7}. However this yields 4

fl si E (Q2 )2. i=1

This contradicts the assumption on S and then completes the proof of the lemma.

Lemma 6.7.2. Let W be an isotropic regular quadratic space over JR with dim W > 11, and r be 2,4,6 or 8. Suppose

W (1s (1))

-- (-Lt (-1))

with s > t. Then there is a subspace Wo C W such that d Wo = 1, the Hasse invariant of Wo = 1 and dim Wo = r. Proof.

Since s > t ands + t > 11, we haves > 6. If r < 6, then put

Wo =1,. (1). Suppose r = 8. Ifs > 8, then put Wo =1T (1). Ifs = 6 or 7, then t > 4 and so put WO = (14 (1)) 1 (14 (-1)). Lemma 6.7.3. Let L be a lattice on the quadratic space given in Theorem 6.7.2. Then there is a decomposable lattice M E gen(L). Proof. By assumption, dim V > 12 and V... is isotropic. Scaling of V by -1 if necessary, we may assume V,,,, = (13 (1)) 1 (1t (-1)) with s > t.

Suppose that there is an integer r = 2,4,6 or 8 such that there is a splitting for every prime p

(Cl)

Lp = Lp,l 1 Lp,2

with

rank L,1 = r, dLp,1 E (Qp ) 2.

172

6


If LP is unimodular, then so is Lp,1i hence the Hasse invariant SP(QpLp,l) = 1 under the additional assumption p 0 2. Thus Sp(QpLp,1) = 1 for almost all primes, and s,,,, := rjSp(QpLp,l) P

is well-defined. Next, we suppose

s. = 1.

(C2)

First we show that the assumptions (Cl), (C2) give a required lattice M. By the last lemma, there is a subspace Wo of V... such that dim Wo = r, d Wo = 1 and S,, (WO) = 1. Thus by Theorem 4.4.2 there is a quadratic space U over Q such that Up = Q,Lp,1 i U. = Wo. We may assume U C V. Taking any lattice K on U and N on U-'-, we construct a new lattice M by local conditions as follows; we put

MP :=KPINP

if

Kp1Np=LP.

This is true for almost all primes. Suppose Kp 1 Np LP. By the definition of U, there is a lattice KP on Up isometric to L,,,1. Therefore there is an isometry qp E O(Vp) such that ijp (Lp,1) C Up. By using it, we put MP

j,(LP) = j,(LP,1) I j,(LP,2)

if

Kp 1 Np # Lp.

Then Kp, gp(Lp,1) and NP, gp(Lp,2) define the lattices M1, M2 on U, U1 respectively, M = Ml 1 M2 E gen(L) is clear.

It remains to verify the conditions (Cl) and (C2) for r = 2,4,6 or 8. Applying Lemma 6.7.1 to LP, we get a splitting LP = Lp',1 J- LP,2 1 (some lattice) with rank Lp,i = 2 or 4, d L' ,j E (Q )2

and moreover rank Lp,1,rank Lp,2 are independent of p when p runs over all primes. Put si = fp Sp(QpL'P i) (i = 1, 2); then 3152 - fJ SP(QP(LP,1 I Lp 2))

holds. Hence at least one of Si, 82 and 3152 is equal to 1, and so taking L, 1i L, 2 or L'P 1 1 L'P 2 as Lp,1i the conditions (Cl) and (C2) are satisfied.

Proof of Theorem 6.7.2.

6.8


173

Let L be a lattice on V and M = M1 1 M2 a decomposable lattice in gen(L) as given in Lemma 6.7.3. We may assume rank M2 > 3. Let Qp be an isometry from Lp to Mp. Since dim QpM2 > 3, there is an isometry ?p E O(QpM2) such that considering rlp = id on QpM1i gpup E O'(Vp) holds. Putting up = rlpvp, we have up c- O'(Vp) and

up(Lp) = (M1)p 1 (some lattice). We define a lattice N on V by

if Lp = (M1)p 1 (M2)p,

Lp

Np

1up(Lp)

otherwise.

Then N E spn+L and Np = (M1)p I (some lattice) for every prime p. Thus N = M1 1 (some lattice) follows. Since V,,, is isotropic, Theorem 6.3.2 implies spn+ = cls+ and hence N = L.

6.8 The Minkowski-Siegel formula In this section we prove the definite case of the so-called Minkowski-Siegel

formula by Siegel's original proof without its analytic part. But we will refer to some analytic ideas, one of which will convince readers who are familiar with the theory of automorphic forms. For a lattice K on a positive definite quadratic space over Q, we put E(K) := OO(K)

and

w(K) :_ E

E(Ki)-1

where Ki runs over a complete set of representatives of isometry classes in gen(K). w(K) is called the weight of K.

Theorem 6.8.1. Let M, N be lattices on positive definite quadratic spaces U, V and put m := rank M, n := rank N and

( 1/2 Em.n '

S

if either n=m+1 or

n = m > 1,

otherwise.

1

Then we have

w(M)-1 = (i)

em

m-1 m7rm(m+1)/4 TT r((m ii=1o

x 11 ap(Mp, Mp), p

- i)/2))-1(d M)-(,n+l)/2


6

174

(ii)

w(N)-1

r( M, Ni) M-1

= em

nirm(2n-m+1)/4

11 I'((n - 2)/2))-1(d N)-m/2(d

M)(n-m-1)/2

i=0

x fl ap(Mp, Np), P

(iii)

w(N)-1

Ni)

N'Z) M-1

Em

n.7rm(2n-m+1)/4 fl

r((n -

i)/2))-1(dN)-m/2(dM)(n-m-1)/2

i=O

x fJ DP(Mp, Np), P

where Ni runs over a complete set of representatives of the isometry classes

in gen(N), r(M, Ni) is the number of the isometrics from M to Ni, t is the ordinary gamma function, rpr(M, Ni) is the number of isometrics 0' from M to Ni such that v(M) is primitive in Ni and finally Dp(Mp, Np) = 2m62,P-Sm,ndP(MP NP).

We note that (i) is a special case of (ii), by putting N = M. We call the left-hand side of (ii) (resp.(iii)) the mean value m(M, N) (resp. mpr(M, N)) of the (resp. primitive) representation of M by N. We need several lemmas. Consider the following situation M.

(*): Let the ring R and the field F be either Z and Q, or Z, and (p, and let N, M, K be quadratic lattices over R such that FN = FM I FK is regular, M is a primitive submodule of N, and lastly K = proj(N) where proj is the projection mapping from FN to FK along FN = FM ® FK.

We put m:= rank M, n:= rank Nandk:=n-m= rank K. Lemma 6.8.1. Under the condition (*), there is a unique linear mapping cp from K -> FM/M such that N = M + R[x + W(x) I x E K]. Conversely,

let 0 be a linear mapping from K to FM/M. Then M is primitive in No = NN(M, K) := M + R[x + O(x) I x E K] and projNk = K, and No = N if and only if 0 = V.

Proof. Let {vi} be a basis of K. By the assumption K = proj(N), there are elements V (vi) E FM such that (O(vi) + vi E N. That is cp gives a linear

6.8


175

mapping from K to FM. Let x be an element of N, proj(x) is in K and hence proj(x) = E aivi for some ai E R. Then

x - (E aivi + co(J: aivi)) = (x - proj(x)) - cp(j: aivi) E FM fl N = M. Thus

x E M+ R[y + cp(y) I y E K], and so

N C M + R[y + cp(y) IyEK] CN. So the existence of cp has been shown. Suppose that two linear mappings cpi : K -* FM/M satisfy N = M + R[y + cpi(y) I y E K]. Then for x E K,

x + cpi(x) E N (i = 1, 2) implies p1(x) - p2(x) E Nn FM = M. Thus cpl and cp2 determine the same mapping from K to FM/M. For bases {vi}, {ui} of M, K, {vi} U {ui + ¢(ui)} is a basis of N4,. So M is primitive

in No andprojN4,=K. Lemma 6.8.2. Under the condition (*), d N= d M d K, s(K) C s(N)2 s(MO) + s(N) and [KO : K# fl K] is bounded from above by a constant dependent only on M, N. Moreover #cp(K) is bounded by some constant depending only on M, N where cp is the linear mapping in the last lemma. If M and N are

unimodular, then N = M I K. Proof.

We take a basis {vi} 1 and {ui}m1 of K and M such that

(co(ve), , cp(vk)) = (ql lul, q, 0, , 0) for some r and for posi, q,. or a power qi of the prime p according to R = Z or tive integers ql,

Z, respectively. Put (1)

N = R[ql lul, ...

, q,-

lur] + N.

Then N = M + R[y + cp(y) I y E K] implies (2)

N = R[ql lul, ... qr 121r, ur+l, ... um] 1 K.

Hence d M d K = (ql . . . q,.)2 d N. Since M is primitive in N, the basis {ui}T 1 of M can be extended to a basis of N, and hence (1) implies [N : N] = ql . . . q,.. Thus we have d M d K = d N. Next vi + cp(vi) E N for vi E K implies B(M, (p(vi)) = B(M, vi + cp(vi)) C B(M, N) and hence q, 1B(M, ui) = B(M, cp(vi)) C B(M, N) for i < r. Thus (3)

ui E q B(M, N)MO

for

i < r.

176

6


Hence qi is bounded from above and so the boundedness of #V(K) is clear. On the other hand B(vi, vj) + B((p(vi), co(vj)) = B(vi + Co(vi), vj + co(vj)) E s(N)

yields B(vi, vj) E s(N) if i or j > r, otherwise

B(vi, vj) E -B(qi 1ui, q 'uj) + s(N) C B(M, N)2 s(Mo) + s(N) C s(N)2 s(Mo) + s(N).

Thus s(K) C s(N)2 s(MO) + s(N) holds. Let a be a natural number such that a2(s(N)2 s(MO) + s(N)) C R. Then we have s(aK) c R and [KO: KO fl K] _ [a2(aK)0 : a2(aK)0 fl aK] < [a2(aK)O : a2((aK)O fl aK)] = [(aK)O : aK]

_ (d(aK)/d((aK)a))1/2 which is dependent only on M and N. Suppose that M and N are unimodular; then (3) implies ui E qiM for i G r and hence qi = 1. Thus N = N = M 1 K follows from (1) and (2). For a linear mapping cp from K to FM/M, we put N. = NW(M, K)

M+R[x+cp(x)IxEK]. Lemma 6.8.3. We keep the condition (*). Denote by 4 the set of the linear mappings cp from K to FM/M such that there is an isometry or N, = N and by I the set of the isometries or : M y N such that o(M) is primitive in N and projo(M)LN = K, where proj,(M)L denotes the :

projection to Fo(M)-L. Then the mapping (cp, a) H O M from LJWEP{O' : N} to I is surjective, and Z;((cp1, a1)) = ((cp2, a2)) if and only if V1 = cp2a and of = a2(id IM I a) as isometries on FN for a E 0 (K). :

Proof. For cp E -1), let o be an isometry from N. on N. Since M is primitive in N,,, o(M) is so in N. Moreover we have o(K) = o(projM-LNW) = projo(M)-L N.

Thus l; is well-defined. Let us show the surjectivity. Let rl E I and put H = proj,n(M) -L N; then by the definition of I, there is an isometry

a:K -H. Put o := r) 1 a. M = o-1(77(M)) is primitive in o,-'(N), since rl(M) is so in N. Now K = o-' (H) is the projection of o-1 (N) to FM' = FK, since

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177

H is the projection of N to Fq(M)L. The condition (*) is satisfied for M, K, a-1(N). Applying Lemma 6.8.1 to a-1(N) instead of N, we have a-1(N) = N. for some co : K -+ FM/M. Since ((cp, a)) = aI M =,q, the mapping 1; is surjective.

Next suppose az N (i = 1,2) and t;((cpli a1)) _ ((cp2i a2)); then a := a21a1 : NN1 = N,,2 and al m = id. Thus a(ML) = M1 and so a(FK) = FK. Considering the projection of the both sides to FK, we have :

a(K) = K. Put aIK = a; then a E O(K) and a21a1 = a = id IM I a. Thus al = a2(id IM 1 a) and for x E K, a(x + p1(x)) = a(x) +W1(X) = a(x) + cp2(a(x)) + (c' - cp2a)(x) implies

('P1 - cp2a)(x) = a(x + (P1(x)) - (a(x) + P2(a(x))) E NN2 n FM = M. Thus cp1 = cp2a mod M.

Conversely cp1 = cp2a and a1 = a2(id IM 1 a) for a E O(K) imply obviously ((coi, al)) = alI M = 0'21M = ((cc2, a2))

Lemma 6.8.4. Keep the condition (*) with R = Z and assume that QN is positive definite. Then E(N)-1#{a

: M' N I a(M) is primitive in N and proj,(M)-L N = K} = K --> QM/M I N}.

Proof. Using the notation of Lemma 6.8.3, we have

#1= E(K)-1 x 04) x E(N), which is our assertion.

Lemma 6.8.5. Keep the condition (*) with R = 74, and let q be a sufficiently large power of p. Then in Lemma 6.8.3, (4)

e(((pl, al))(x)

mod

O

for x E M

if and only if

al =3a2(idIM Ia) on FN,

901 =cp2a

for a E O(K) and ,(3 E O(N) satisfying /3(x) - x mod qNl for x E N. Proof. Suppose (4) for Vi E 4D and ai :

N. Then

a1(x) - a2 (x) mod qNO

178

6


for x E M and hence all ai (x) =- x mod for x E M. al l al (M) is primitive in N,2 = a2a1(NN1) since M is so in N,1. Then applying Theorem 5.4.1 to a2-1a1 : M = a21a1(M) in NN2, there is an isometry y E 0(N,2) such that 771M = a2 1alI M and (5)

77(x) =- x mod

for

xE

Then,7-1a21a1 : NN1 = N,,2 and y-1a21a1I M = id. Taking the projection

on FK = FM' of N, we have r7-la2'oi(K) = K. Define a E O(K) by a := 17_1x2 10.11 K; then 71-1a21a1 = idm 1 a. Put /3 := a2r7a21 E 0 (N); then a1 = i3a2(id IM 1 a) and (5) implies 3(y) - y mod qN0 for y E N. To complete the proof of the " only if " part, it remains to verify cpl = V2a. For x E K, we have 77-1a21a1(x + p1(x)) = a(x) + cci(x) = a(x) + 02(a(x)) + (cPi - W2a) (x),

77-10121a1(x+'Pi(x)) E rl-10.2 1a1(N,1) =

a(x) + co2(a(x)) E Nw2.

Thus we have (cci - c02a)(x) E N,12 n FM = M, and so coi = cP2a as mappings from K to FM/M. Thus the " only if " part has been proved. The " if " part is clear because of ((coi, 0.i)) = aiI M

Lemma 6.8.6. We keep the notation of Lemma 6.8.3 and assume R = Z,. Then we have, for a sufficiently large power q of p, j,b X O(O(N)/ ^'q) = O(I/ ^'q)#(O(K)l ^'q);

where for a1, a2 E O(H) for H = N or K; a1 -q a2 is defined by al 0.21(x) - x mod qHa

for

x E H;

for

x E M.

for 171, 112 E I, 771 -q 772 is defined by

1h (x) - q2 (x) mod qNa

Proof. Put A := UwE41({a : NP = N}/ -q),

where for ai : Nw = N, a1 -q a2 is defined by

0.10.21

OA = #'D #(O(N)/ ,,,q)

-q id IN. Then

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179

is clear. Consider the mapping ' from A to I/ -q by (cp, a) c-, aI M. This is well-defined. Since ' is surjective by Lemma 6.8.3, we have only to show

that the number of inverse images is #(O(K)/ -q). By the last lemma, '((W1, al)) = t;'((cp2i a2)) if and only if

al = 0a2(id I M 1 a),

cpi = cp2a

for a E O(K), 3 E O(N) with 0 -q id IN. SO '((cpl, ai)) _'((cP2> a2)) means al -q a2(id IM 1 a), cpl = cp2a. On the other hand, a E O(K) acts on A by (V, o,) H (cpa, a(id I M 1 a)), and so putting G := {a E O(K) I cpa = cp, a(id IM I a) -q a}; the number of inverse images is the number of elements in an orbit by O(K) and so equal to [O(K) : G]. Thus we have only to verify G = {a E O(K) I a -q id IK}. First let us show qNN n K = qKt.

(6)

For y E FK, y is in the left-hand side e=> y E K and B(q-ly, NN) = B(q-ly, K) C Z, y E K n qKa = qKa since q is sufficiently large.

Suppose a -q id IK. Let a : N, = N f o r c p E $. For a E O(K), put Ti = a(id IM I a)a-1; then a(id I M 1 a) -q a

'q(x) - x mod qN0 f o r x E N q (id I M 1 a)(y) - y mod qN0 for y E Np

(id IM I a)(y+z+cp(z)) y+ z+ cp(z) mod gNN for y E M, z E K

eza(z)-zmodgN0 for zEK '=>a(z)-zEgKa for zEK by(6)

aa-gid1K To complete the proof, we have only to show that a -q id IK implies cp = cpa. By Lemma 6.8.2 [Kt : K# n K] is bounded by a number dependent

only on M, N. Thus cp(gKO) c q[K# : KO n K]

cp(KO n K) c q[KO : KO n K]

cp(K) C M,

and so a -q id IK implies cp((l - a)(K)) C cp(gKO) C M, i.e. cp = cpa. Conversely suppose a E G; then a(id IM I a) (x + y + cp(y)) - a(x + y + cp(y)) mod qNd

for x E M, y E K, and aa(y) - a(y) mod qNO for y E K, which yields a(y) - y E qa-'(N)# n K = qKO by (6). Thus a -q id IK.

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180


Lemma 6.8.7. Continuing the last lemma, we have #,D _qm(m+1)/2-mnpn ordp d N+(m-n) ordp d K/P (N, N)-l/p(K, K)

x#{v : M -> N/qNt I Q(v(x)) - Q(x) mod 2q, a(M) is primitive in N, and projo(M)LN = K}. Proof. Since for a linear mapping a : N --4 N such that

Q(a(x)) - Q(x) mod 2q,

there exists an isometry a' E 0 (N) such that a'(x)

a(x) mod qNa, by applying Theorem 5.4.2 to L = N, G = Na, k = ordp q, we see the canonical

mapping from O(N)/ -q to {a : N -> N/qNO I Q(a(x)) - Q(x) mod 2q} is bijective. Thus

#(O(N)/ -q) = p-nordydNgn2-n(n+l)/2ap(N,N) Similarly O(O(K)/ -q) q) =

p-(n-m) ordp d Kq(n-m)(n-m-1)/20p(K K).

By virtue of the last lemma, we have only to verify #(I/ ^+q) is the number of linear mappings v : M -* N/qNO such that

Q(a(x)) - Q(x) mod 2q,

a(M) is primitive in N and proj,(M)-LN = K. Here we note that the conditions are independent of taking representatives of the images in N (keeping the linearity of a), because the first and second are clear and the third follows from Corollary 5.4.5. So to complete the proof, we have only to show that for a linear mapping a : M --4 N/qNa such that Q(v(x)) =Q(x) mod 2q, a(M) is primitive and proj,(M)L N = K, there is an element 71 E I such that ii(x) - or(x) mod qNa for x E M. Applying Theorem 5.4.2 to a, G = Na, there is an isometry r) : M -+ N such that rj(x) =or(x) mod qNa for x E M. It is clear that r1(M) is primitive in N since a(M) is so. By Corollary 5.4.5, there is an isometry ry E O(N) such that yq(M) = a (M). So proj,,

and

I.

(M)-LN = proj,y,q(M)-LN = K,

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181

Lemma 6.8.8. Keep the condition (*) with F Q, and let S be a finite set of primes such that if p S, then Mp, Np are unimodular. Then {cp : K -> QM/M I N, E gen(N)} and

lj{WP:Kp-QPMPIMp I NWP(MP,KP)=Np} PES

are bijective.

Proof. First, suppose that cpp is given for p E S. Let {vz}k 1 and be bases of K and M respectively; making use of them we define A(cpp) E Mm,k(Qp) by (oP(v1), ... , cop(vk)) = (u1, ... um)A(cop)

and take a matrix AE Mm,k(Q) such that A - A(cpp) mod Mm,k(Zp)

1 A E Mm,k(Zp)

p E S, otherwise. for

Define the mapping cp by

('P(vl), ... , cO(vk)) :_ (u1, ... um)A.

We show N, E gen(N). For p E S ((co -

cp)(vl),...

(gyp

- OP)(vk))

_ (u1 ... , um)(A - A(Wp)) E (MP, ... , Mp) implies (cp - (pp) (K) C MP and

(NN4)P = MP+Zp[x+y(x) I X E Kp] = MP+Zp[x+cc (x) I x E Kp] = Np. If p S, then MP and Np are unimodular and so Lemma 6.8.2 yields that NP = Mp 1 Kp. By the definition of A, (Nw)p = MP + Z [x + W(x) I x E Kp] = 1llp 1 Kp,

and so we have (N,,)p = Np. Thus we have proved N. E gen(N). Conversely a linear mapping

cp : K -- QM/M such that N,, E gen(N) induces canonically cop : Kp - QpMp/Mp such that NWP (Mp, Kp) = Np. If p S, then N,,p and Np equal Mp 1 Kp by Lemma 6.8.2 and in particular cop(Kp) C Mp. The correspondence co --+ {cop} gives the inverse of the above correspondence. Thus we have proved the lemma.

6

182


Lemma 6.8.9. Under the condition (*) with p # 2 and R = 7Lp, assume that M, N are unimodular. Then we have b4 = 1 and in Lemma 6.8.7 we can remove the condition on a that a(M) is primitive in N and projo(M).L N = K.

Proof. Take cp E fi; then there is an isometry 77 : NN = N. Since M and

N are unimodular, N,, = M 1 M1. For x E K, x + cp(x) E N. implies V(x) E M. It means V(K) C M. Thus (D contains only one linear mapping

from K to FM/M. Moreover cp(K) C M implies N,, = M 1 K. Suppose a : M - N/qNO such that Q(v(x)) - Q(x) mod 2q. We assume that v is a linear mapping to N. Let us show that the other two conditions on a in the assertion of Lemma 6.8.7 are automatically satisfied. By Corollary 5.4.2, there is an isometry /3 : M' N such that a(M) =,3(M). Since M is unimodular, /3(M) is so and hence N = /3(M) 1 *. Thus a(M) = /3(M) is primitive in N. Since p 2, /3(M) is transformed to M by O(N) by Corollary 5.4.1. Thus

projo(M)LN = projmLN = K.

0 Proof of Theorem 6.8.1. We denote by p(M) the quotient of the left-hand side by the right-hand side of (i). First let us show p(M) = 1 if m = rankM = 1. Since both sides of (i) do not change under scaling by Proposition 5.6.1, we may assume

M = (1). Then gen(M) = cls(M) and w(M) = 2-1 are clear. For an odd prime p, by Proposition 5.6.1

ap(M, M) = 2-1/p(M, M) = 2-1pAp(1,1) = 2-10{x mod p I x2 - 1 mod p}

=1. If p = 2, then a2(M, M) = /32(M, M) = tA4(1,1)

=#{xmod4Ix21mod8} =2. Since t(1/2)

we get p(M) = 1. Now we assume that the equation in (i) holds for any m < n and suppose m < n. When Ni runs over a complete set of representatives of isometry


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183

classes in gen(N) as in the theorem, we have rpr(M,NZ) E(NZ)

=>>E(Nj)-1O{v : M'- Ni I o(M) is primitive in Ni and K

i

projQ(M)LNZ = K},

where K runs over all possible isometry classes (that is K is a lattice on

a positive definite quadratic space such that QN = QM I @K) d N = dMdK, s(K) C s(N)2 s(MO)+s(N) and rankK = n-m by Lemma 6.8.2; then by Lemma 6.8.4

(Rl) I=

E(K)-1#{cp : K - QM/M I NN(M, K) - Nil i

K

_ E E(K)-10{cp : K ---> QM/M I NW(M, K) E gen(N)} K

=E

E(K)-1 [I

K

µµ

N{(o

mm

: KP - QPMPI MP I N,pp (MP, KP) - NP},

pES

by Lemma 6.8.8 where S is a finite set of primes such that if p MP, Np are unimodular. Now Lemma 6.8.7 implies

S, then

1 E E(K) -1 fi {qm(m+1)/2-mnpn ordp d N+(m-n) ordp d K K

pES u

qq

X I3P(NP, NP)-1NP(Kp, KP)H{O : MP -a NP/qNP I Q(o(x))

Q(x) mod 2q, o (MP) is primitive in NP, and projo(Mp)LNp - Kp}} where q is a sufficiently large power of p,

_ K

11

{qm(m+1)/2-mnpnordp d N+(m-n) ordp d K

pES

X ,3P(NP, NP)-'OP(KP,

N{U : MP -+ NN/gNp I Q(o (x)) KP)y

Q(x) mod 2q, v(MP) is primitive in Np, and projo(Mp)L NP = Kp}}

where K runs over a complete set of representatives of possible genera. Since

l3P(MP,Np) =

2m62,P+6.,,,,,

ap(MP, NP)>

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184


we have /3p(Np, Np)-1,3p(Kp, Kp) = 2m62,par(Nr, Np)-'ap(Kp, Kp).

By Proposition 5.6.2 and the remark in the proof, the infinite products fp ap(Np, Np), fJp ap(Kp, Kp) converge. Using Lemma 6.8.9, we have

l =1:K w(K) (fl p

x [f

ap(Kp,Kp))(dK)m-n(dN)n2mHap(Np Np)

1

p {qm(m+1)/2-mn#{v

: Mp -> Np/qNP I Q(v(x)) - Q(x) mod 2q,

p

o(Mp) is primitive in Np and projo(MP)±Np = Kp}} Using the induction hypothesis for K, we have K)m-n

w(K) H op (Kp, Kp) (d p

n-m-1

=E

11 h((n - m -

i)/2)(dK)(m+1-n)/2

i=o

where e = 1/2 if n - m > 1, 1 otherwise. Thus 1 is equal to n-m-1 e-17r-(n-m)(n-m+1)/4

TT

I'((n - m - i)/2)2m(d N)n f ap(Np, Np)

ii=o

x

1

p

fjp(ordy dK)(m+1-n)/2gm(m+l)/2-mn

K p

x #{Q : MP -p Np/qNN I Q(o (x)) - Q(x) mod 2q,

v(Mp) is primitive in Np, and projo(Mp)1Np = Kp}.

Since QN = QM I QK, the isometry class of QK is unique. By Theorem 6.1.1, K runs over a complete set of representatives of possible genera if and only if Kp runs over a complete set of representatives of possible isometry classes for p E S, noting d Np = d Mp d Kp and s(Kp) C s(Np)2 s(MM) + s(NN).


6.8

185

Finally we have

n-m-1 l=E-17f-(n-m)(n-m+1)/4 TT r((n-m-i)/2)2m(dN)nJap(NP p

ii=10

x

NP)-1

lip(ordp d N-ordr, d M)(m+1-n)/2gm(m+1)/2-mn P

x Of o, : Mp - NP/qNP I Q(v(x)) - Q(x) mod 2q, v(Mp) is primitive in Np} n-m-1 _ E-1.-(n-m)(n-m+1)/4 r(( n - m - i)/2)2n'(dN)(m+l+n)/2

11

ii=1o

x (d

M)(n-m-1)/2 11

ap(Np, Np)-1 f p-m ordr d Ndp(M, N)

P

P

n-m-1

_ E-17r-(n-r)(n-m+1)/4 ri r((n - in -

i)/2)(dM)(n-m-1)/2

i=O

x (d

N)(n-m+1)/2

fl ap(NN, NP)-1 fl DP(MP, NP). P

P

By the definition of p, we have n-1 p(N)-127r-n(n+1)/4

ap(Np, Np) =

fi r'((n - i)/2)(d

N)(n+1)/2W(N)-i

i=0

P

and hence

m-1 =p(N)2-1E-1,,m(2n-m+l)/4 11

l/w(N)

I'((n -

i)/2)-1

i=0

x (d

M)(n-m-1)/2(d N) -m/2 TT DP(MP, NP) P

Thus we have p(N) = 1 if and only if (iii) is true when m < n. Here we claim that p(N) is also the quotient of the left-hand side of (ii) by the right-hand side of (ii). To see it, we note that r(M, Ni) =

E rpr(M', Ni) QMJM'DM

since, considering v : M y Ni as an isometry from QM to QNi, it induces

v : m':= Qm no,-' (Ni) - Ni

Quadratic forms over 71

6

186

with Q(M') being primitive in Ni, and that locally aP(MP, (Ni)p) = and

(d

[MP : L, Q,,M,DM,,jM,,

MP1m+1-nDP(M,1,

(NN)P)

MP]m+1-n = (d MM)(n-m-1)/2.

MP)(n-m-1)/2[Mp :

Since the correspondence between global and local lattices are given by Theorem 6.1.1, we get the claim. In particular, (ii) follows from (iii). Thus denoting by p(M) the quotient of the left-hand side by the righthand side of (i), we have proved that, if rank K = 1, then p(K) = 1, and if p(K) = 1 with rank K < n, then p(N) is the quotient of the left-hand side by the right-hand side of (ii) and (iii) for m < n; so the ratio is independent of M. It remains to verify p(N) = 1 but we give only the basic ideas of the proof.

By scaling, we may assume s(N) C 2Z. Denoting the left-hand side and right-hand side of (ii) by L(M, N) and R(M, N) respectively, we have shown L(M, N) = p(N)R(M, N). For T E C with imaginary part of r > 0, we put 0(T, N) : =

e(Q(x)T) xEN 00

_

r((k), N)e(kr), k=0

where e(z) denotes e21riz and

e(,r, N) = w(N)-1

E, (Ni)-10(T, Ni)

00

=1+

m((k), N)e(k-r). k=1

On the other hand, we put 00

E(r, N) := 1 + e1,nirn/2I'(n/2)-1(d N) -1/2 E kn/2-1 11 aP((k), NP)e(kr). k=1

P

Then we have proved that O(T, N) - p(N)E(r, N) is a constant number. Siegel showed that for m > 4 (7)

H(N, a/b)(bT - a) -m/2

E(-r, N) = 1 + a,b

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187

where a, b run over relatively prime integers with b > 0 and

H(N a/b)/ :_l(i/b)m/2(dN)-1/2 / /

xEN/bN

e(a

Q(x))

Then O(T, N) and E(T, N) are modular forms (see the definition in the Notes) and if a constant number is a modular form, then it is identically

zero. Thus O(T, N) = p(N)E(T, N) holds. Take limy on both sides with T = iy. Comparing both sides, we have p(N) = 1. For m < 4, (7) is not absolutely convergent. But, when we multiply each term by IbT - aj-s it is absolutely convergent for real part of s sufficiently large, and it has an analytic continuation on the whole s-plane. We can then define E(T, N) to be the value at s = 0. This is an idea of Hecke. We need a modification in the case m = 3. Siegel's original idea is as follows. The Dirichlet series D(s) corresponding to 9(7-, N) is (through the Mellin transform) defined by cc

E m((k), N)k-s; k=1

it is holomorphic for Res > n/2 and has a simple pole at a := n/2 with the residue (dN)-1/2irn/2r(n/2)-1

b :=

By the Tauberian theorem,

lira x-a

X-00

m((k), N) = b. k<x

But, for technical reasons, Siegel considered instead the Dirichlet series 00

m((k), N) ka-1

k=1

Then it has a simple pole at s = 1 and (8)

lim x-1 E m((k),

N)k-(a-1)

= b.

k<x

These follow from classical analytic number theory. Replacing m((k),N) by the right-hand side of (ii), Siegel proved the formula similar to (8) and got p(N) = 1.

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188


Remark. If we accept the existence of a unimodular Haar measure on OA (V ), it

is not hard to see that the partial sum w3(K) = EK E(Ki)-' where Ki runs over a complete set of representatives of isometry classes in spn(K) and is dependent only on gen(K). Note that in the equation (Ri),

#{cp:K-Q2M/MI N,,Espn(N)} depends only on spn(K). To see this, let K' = vrl(K) for v E O(QK), 77 E

0'A (QK). Putting a = r) = id on QM,

NN E spn(N).

Thus the correspondence cp'- cpr/ lv 1 gives the invariance of the number

in question. Then restricting Ni in spn(N) and grouping the isometry classes, the second line of (Rl) becomes

ws(K)

(9) K

#{cp : K' -> Q2M/M I N,(M, K') E spn(N)} K'

where K runs over the different genera and K' runs over the different spinor

genera in gen(K). If either n - m > 3 or n - m = 2 and OA(V) = OA(W)O+(V)O'' (V)O+(N)

where V = U 1 W (cf. Lemma 6.5.7), then the inner sum on K' is dependent only on gen(N), since, for v E OA(V), we can take r) E 0+(Q1K), a E O+(V), ry E 0A' (V), 0 E O+(N) such that o = -yrj(a,3)-1. Then NW,,- (M, rjK') = riN,, (M, K') E spn(riN) = spn(oN).

Therefore (9) depends only on gen(N), considering the correspondence Thus the left-hand sides of (ii), (iii), restricting K', 0 - rqK', Ni E spn(N), are dependent only on gen(N) if n - m > 3 or the above condition is satisfied for when n - m = 2.

7 Some functorial properties of positive definite quadratic forms

In this chapter, we investigate functorial properties of positive definite quadratic forms with respect to the tensor product and scalar extension. Let L, M and N be quadratic modules over 71 and RK the maximal order of an algebraic number field K. Some of the fundamental problems are:

Does L®M=L®Nimply M=N? What can we say about M, N when RKM = RKN, for example, is

M=N?

When we consider these in the category containing indefinite quadratic forms, it is not interesting. For example, let M, N be unimodular positive definite quadratic modules over Z with the same rank and n(M) = n(N) =

271, then L ® M = L ® N holds for L = (1) 1 (-1). So the condition L ® M = L ® N says nothing. If the field K is not totally real, RKM is isotropic at an infinite imaginary place of K and hence the isometry class of RKM is nothing but its spinor genus (by the generalization of Theorem 6.3.2). Therefore RKM = RKN holds. So the above problems are meaningless. However if we confine ourselves to the category of positive definite quad-

ratic forms and totally real algebraic number fields, the answer seems affirmative. At least there is no counter-example so far. Let us give some results in this chapter. In this chapter, by a positive lattice we mean a lattice on a positive definite quadratic space over Q. Hence, if L is a positive lattice, then for a basis {ei} of L over 71, B(ei,ej) E Q and the matrix (B(ei,ej)) is positive definite.

190

Functorial properties of positive definite quadratic forms

7

RK denotes the maximal order of an algebraic number field K. For quadratic spaces U, V over Q, we defined the tensor product U V after Proposition 1.5.4. It is easy to see, taking orthogonal bases of U, V, that if U, V are positive definite quadratic spaces, then U ® V is also

positive definite. We introduce a quadratic form on the tensor product L ® M of positive lattices L, M by its restriction from QL ® QM. Since for bases {ei}, { fi} of L, M, respectively, {ei 0 fj } is a basis of L 0 M, we

see that L ® M is a positive lattice if L, M are positive lattices, and the corresponding matrix is

B((ei 0 fj, ek 0 fh)) = (B(ei, ej)) 0 (B(fi, fj)) (tensor product of matrices). We recall that B(xi ®yi, x2 ®y2) = B(xl, x2)B(yi, y2) for xi E L, yi E M. In 7.1, we introduce the notion of a positive lattice of E-type, which at the present time is the most convenient one for our problem. Sections 7.2 and 7.3 are preparatory. In 7.4 and 7.5, the main results in this chapter are given.

7.1 Positive lattices of E-type For a positive lattice L, we put min(L) = min{Q(x) I x E L, x

0},

M(L) ={x E L I Q(x) = min(L)j.

M(L) is obviously a finite set. An element of M(L) is called a minimal vector of L. For positive lattices L, M, we have Q(x 0 y) = Q(x)Q(y) and so if x E .M (L), y E M (M), then

min(L ® M) < Q(x 0 y) = Q(x)Q(y) = min(L) min(M).

(1)

Definition. A positive lattice L is of E-type if and only if .M(L (&M) C {x ®y jx E L, y E M} for any positive lattice M holds.

Note that elements of L 0 M cannot necessarily be written as x 0 y. Elements x ® y in L 0 M are called split. Positive lattices of E-type play an essential role in this chapter. The first basic result is

7.1

Positive lattices of E-type

191

Lemma 7.1.1. Let L, M be positive lattices. Then we have min(L ® M)

(2)

min(L) min(M),

and then by (2), min(L (D M) = min(L) min(M). Thus X E .M (L), y E .M (M) follow and so

min(L ® M) = min(L) min(M), M(L ®M) = {x ®y I x E .M(L), y E .M(M)}. Thus (3) has been proved.

(i) is obvious. We prove (ii); let N be a positive lattice. Let v = x + y (x E

L 0 N, y E M 0 N) be a minimal vector of (L I M) ®N. Since Q(v) _ Q(x) + Q(y) and Q(v) is minimal, x = 0 or y = 0, i.e. v = y E M ®N or = x E L 0 N. Suppose that v is in L ® N for example; then

min((L I M) ® N) = Q(v) > min(L ®N) > min((L 1 M) ®N), and

min((L I M) (9 N) = Q(v) = min(L 0 N) and

VEM(L®N)=M(L)0M(N) follow. Hence, v is split in L 0 N C (L I M) 0 N. Thus L I M is of E-type.

192

7


Next let u be a minimal vector of L 0 M ® N; then

u=z®w (zEM(L),wEM(M®N)) follows from (3) and w is split since M is of E-type. Hence u is split in (L 0 M) 0 N. This means that L 0 M is of E-type. Let us prove the third assertion. For a positive lattice N, we have min(L) min(N) = min(L (9 N)

< min(M ®N) < min(M) min(N) by (2) = min(L) min(N) by the assumption. Thus we have min(M ® N) = min(L ® N). Therefore a minimal vector v of M ® N is also a minimal one of L 0 N and so it is split in L 0 N. Write v = x®y, x E M(L), y E A4 (N); then y is primitive and so {y} is extended a basis of N and hence x ® y E M & N implies x E M. Thus v is split in

M ON. We note that if L is a positive lattice of E-type, then a scaling of L is also of E-type by (ii).

Lemma 7.1.2. ,ar < Vfr- holds for 2 < r < 43, where µr is the Hermite constant in Chapter 2. Proof. By virtue of Theorem 2.2.1, we have only to show

(µr 2+2 14/3.

7.1


193

Hence drawing the graph of 3z3(x2 f'(x))', it is negative for x > 8 and hence x2 f'(x) is monotonically decreasing there and f'(8) < 0 implies that f'(x) < 0 for x > 8. Then, f (43) > 0 implies f (x) > 0 for 8 < x < 43.

Lemma 7.1.3. Let A, B be positive definite symmetric matrices of degree n; then trAB > n(IAIIBI)1/n Proof. Write B = D[T] for a diagonal matrix D = diag(d1i , dn) and an orthogonal matrix T. Denoting by a1, , an the diagonal entries of TAtT, we have

trAB =trTAtTD = Eaidi > n(fl(aidi))1/n =nI BI1/n(jl ai)1/n > nIBI1/nITAtTI1/n =nIBI1/nJAI1/n,

where we used the fact that the arithmetic mean is greater than or equal to the harmonic mean and that for a positive definite matrix c = (cij), jJ cii > det c.

Theorem 7.1.1. If L is a positive lattice with rank L < 43, then L is of E-type.

Proof. Let M be a positive lattice and v = i=1 xi ® yj be a minimal vector of L ® M and assume that r is minimal among these expressions; then {x1i , xr}, {y1, , Yr} are linearly independent and so L1 := 7L[x1, ... xr], Ml

Z[yl, ... , yr]

are positive lattices of rank = r. Then we have min(L1) min(M1) > min(L) min(M) > min(L ® M) = Q(v)

= B(v, v) = E B(xi, xj)B(yi, yj ) i,j

(4)

= tr(B(xi, xj)) (B(yi, yj)) > r(dL1 dM1)1/r > rµr 2 min(Li) min(M1),

which yields r < p2 and so by Lemma 7.1.2, r = 1, that is, v is split.

Corollary 7.1.1. Put cn := minl cn for some constant c is known. Hence c,,, tends to zero.

Lemma 7.1.4. If n > 40, then t1n < n/6. Proof. As in the proof of Lemma 7.1.2, we have only to show h(x)

log x/6 - log {21r-11'(2 + x/2)21x}

= f (x) - 2-1 log x + log(x/6) > 0 for x > 40, where f is the function defined in the proof of Lemma 7.1.2. It is easy to see that x2h'(x) > 0 for x > 40 and h(40) > 0, which complete the proof.

Theorem 7.1.2. Let L be a positive lattice such that s L C Z and min(L) < 6; then L is of E-type.

Proof. Let M be a positive lattice and v = ? 1 xi ® yi be a minimal vector of L ® M with r being minimal; then as in the proof of Theorem 7.1.1, we have, since B(xi, xj) E Z,

min(L ®M) = Q(v) > r(det(B(xi, xj)) det(B(yi, yj)))1/r > r(det(B(yi, yj)) 1/r On the other hand, min(L (D M) < min(L) min(M) < 6 min(Mi), where we put M 1 := Z[y1i , yr]. Thus we get r(dM1)1/r < 6 min(Mi) and hence r/6 < µr From the previous lemma follows r < 40 and then Theorem 7.1.1 yields that M1 is of E-type and hence v E L 0 Ml is split, since v is a minimal vector of L ® M and hence L ® M1. min(M1)(dMl)-1/r
4) i=1

i=1

be the lattice spanned by ±ei±ej (1 < i # j < n) in R'. Note rank Dn = n and d Dn = 4.

7.1


195

Denote by E8 the lattice spanned by D8 and a E$ 1 ei. Then rank E8 = 8 and d E8 =1. Denote by E7 the orthogonal complement of 2 Ei 1 ei in E8. Then s

8

xi=0}

I

i=1

i=1

is isometric to the orthogonal complement in E8 of any minimal vector of E8. Note rank E7 = 7 and d E7 = 2. Denote by E6 the orthogonal complement of 7L[el +e8i i E8 1 ei] in E8. Then s i=1

is isometric to the orthogonal complement in E8 of any submodule of E8 isometric to A2. We have rank E6 = 6 and d E6 = 3. For these lattices L, it is easy to see that L is spanned by M(L) and is an indecomposable positive lattice with s(L) = 7L, min L = 2. Conversely, such lattices are isometric to one of the above lattices, by using the theory of classification of root systems of Lie algebras. Anyway, by Theorem 7.1.2 they are of E-type. Next we show

Theorem 7.1.3. Let F be a finite abelian extension of Q, and RF the ring of integers in F. We introduce the symmetric bilinear form BF by BF (x, y) := trF/@ xy where y denotes the complex conjugate of y. Then (RF, BF) is a positive lattice of E-type.

To prove it, we need several lemmas. Since F is abelian, Q(x) BF (x, x) = tr IxI2 = E I a(x) I2 where a runs over the Galois group Gal(F/Q); so RF is clearly a positive lattice.

Lemma 7.1.5. minRF = [F : Q] and M(RF) are roots of unity in F. Proof. Put n := [F : Q]; then for 0 # x E RF, we have Q(x) = E la(x)I2 > n(fl Ia(x)I2)1/Th =

n(NF/(Q(x))2/' > n,

where NF/@ is the norm from F to Q. Now Q(1) = n is clear and hence minRF = n. If Q(x) = n for x E RF, then in the above inequality Ia(x)I is independent of a so Ia(x)I = 1 for all a E Gal(F/Q). Thus x is a root of unity since x E RF.


7

196

Lemma 7.1.6. Let p be a prime and L := Z[ul, , up_1] a quadratic lattice defined by B(ui, uj) = -1 if i # j, B(ui, ui) = p - 1 for 1 < i, j < p - 1. Then L is a positive lattice of E-type.

Proof. Let M be a positive lattice, and for x =

ui ® wi E L ® M (ui E

L, wi E M), we have

Q(x) =

B(ui, uj)B(wi, wj ) i,j

=(p - 1)

Q(wi) - E B(wi, wj ) i#j

=(p - 1)

Q(wi) + E Q(wi - wj) - (p - 2)

Q(wi)

i<j

Q(wi)+T, Q(wi-wi)

(5)

i<j

Hence Q(x) _> 0 and moreover Q(x) = 0 yields wi = 0, so x = 0. Thus L ® M is a positive lattice and hence L is also, taking M = (1). It is clear that min(L ® M) < (p -1) min M. To show the converse inequality, we may assume w1 i , Wk 0, wk+1 = = wp_1 = 0 (1 < k (p - 1) min M. Thus we have min(L ®M) = (p - 1) min M. Suppose x E .M (L ® M); then all elements wi, wi - wj except for the above

p - 1 elements should be zero. Thus w1 - wk = 0 and wi - wj = 0 for i = 2 < j, so w1 = wk, w2 = = wp_1. If k = 1, then by definition of k, w2 = = wp_1 = 0 and so x = u1 0 w1 is split. If k > 1, then w1 = wk = w2 and so w1 = = wp_1 follows. This means that x = > ui ®w1 is split. Thus L is of E-type.

Lemma 7.1.7. Let p be a prime and n > 2 and put F = Q(C), where S is a primitive pn-th root of unity. Then RF is a positive lattice of E-type.

Proof. It is well known that RF has a basis vi :_(i-1, i-1,1 < i < pr-1(p-1). Then B(vi, vj) = trF/(Q ('-j and trF/@ rm = S

i mod pn

S-im

- [ (ipm = imod pn- i

pn-1(p - 1) if pnIm, _pn-1 if pn-11m,pn%m , 0

Let L be a positive lattice in the previous lemma. Let

if pn-1 %m.


7.1

197

be a positive lattice defined by (B(wi,wj)) = pn-llpn-,. We define a basis {zi}of L(D M by ub+1®Wa(i=a+bpri-1,1 Q(ui) - E(E laid l)Q(ui) iES

iES iES i#i

> (minM)(s - (s - 1)s/n) > min M).

198

7


Exercise 2. Let L be a positive lattice such that L = (a[n]), where

and n = (nij) is an upper triangular matrix with all diagonals being 1. If ai > a1 for all i, show L is of E-type. (Hint: For a positive lattice M and x = ei 0 ui E M(L (D M), define k = 0. Then show by Uk # 0 and Uk+1 = min L min M > min(L ® M) = Q(x)

= tr(a (B(ui,uj))[tn]) E aibii + akQ(uk)

a, Q(uk) = Q(el ® uk),

i 1. Then (2) implies

Q(vi(D M) =o'(vi®L®...®L(9 M') (3)

=L&...L®vs-i(1)(vi)®L0...0L®N' 01-1(vi (9 N) =Q-'(vi ®L®... ®L(9 N')

(4)

=L

L®ai1(vi)®L0...0L®M'

where Q3-,(1)(vi) (resp. vi 1(vi)) is on the s-1(1)-th (resp. s(1)-th) coordinate. Here, denoting by Ni, Mi positive lattices such that L®Ni, L®MM are the right-hand side modules of (3), (4), respectively, we have (3') (4')

01-1 (L ®Ni) =vi ® M

r(L ®Mi) =vi ®N,

and then rank Ni = rank M/ rank L = rank N/ rank L = rank Mi, and

Mi={mEMI v(L0m)Cvi®N}, Ni={nENI o,-'(L ®n)Cvi®M}.

7.2 A fundamental lemma

201

Thus the condition (B2) is satisfied. From (3'), (4') it follows that

a(vi®Mi) =a(vi®MnL®Mi) C L®Ninvi®N=v2®NN._ and hence (B3). Now let M, N be (not necessarily indecomposable) positive lattices and

suppose a : L ® M = L ® N with the condition (1). Let M

=1

Yj be decompositions into indecomposable lattices. By the condition (Al), L ® Xj, L ® Yj are indecomposable. Hence by virtue of 1

Theorem 6.7.1, we have k = h and may assume a(L 0 Xj) = L 0 Yj, changing suffices if necessary. We can apply the above to Xj, Yj, a. Putting

Xji={xEXj I o,(L(D x)Cvi 0Yj}, Yji ={y E Yj

I

o--' (L (9 y) C vi 0 Xj},

Mi={mEM I a(L(9 m) Cvi®N}, Ni ={n E N I o 1(L 0 n) C vi 0 M}, we show Mi =1j Xji, Ni =1j Y. The left-hand sides obviously contain

the right-hand sides. Suppose m = Exj E Mi (xj E Xj). For z E L,

o,(z®m)Ea(z0xj)Evi ®Nyields v(z0xj)Cvi 0Yj forallzEL.

Hence xj E Xji holds and so Mi C1j X. Similarly Ni C1j Yji, which is what we wanted. Since rank Mi = E j rank Xji = E j rank Yj i = rank Ni and rank Xji = rank X j / rank L, we have rank Mi = rank M/ rank L and similarly rank Ni = rank N/ rank L, which implies the condition (B2). (B3) follows from v(Q(vi ® Xji)) C Q(vi ®Yji) and Mi =1j Xji, Ni =1j Y. Next we show (B) (A). We need several lemmas. In the rest of this section, L denotes an indecomposable positive lattice such that the assertion

(B) is true. If rank L = 1, then we have nothing to do. So we assume rank L > 1.

Lemma 7.2.1. Letting M, N, Mi, Ni, a, vi be those in (B), we have M N, or (L ®Mi) = vi ®N and o, 1(L ®Ni) = vi ®M. In particular, rank M = rank N > rank L. Proof. By virtue of (B2), the primitiveness of vi implies that of Mi, Ni in M, N, respectively, that is we have (5)

L®Mi=@(L®MM)nL®M, L®Ni=QZ(L®Ni)nL®N.

(B2) yields a(L ® Mi) C vi 0 N and rank Mi = rank N/ rank L and hence o,(Q2(L 0 Mi)) = Q(vi ® N). From (5) follows

a(L®Mi) =a(Q(L®MM)) na(L®M) (6)

=Q2(vi®N)nL®N=vi®N,

202

7


and similarly

or-' (L 0 Ni) = vi 0 M.

(6')

Now dim@(vi ® Mi) = dimQ(vi ® Ni) follows from (B2) and then (B3) implies v(Q(vi (9 Mi)) = Q(vi 0 Ni), which, since v, ® Mi, vi ® Ni are primitive in L 0 M, L 0 N, respectively, implies

v(vi®Mi)= Q(Q(vi0Mi)nL®M)=Q(vi®NN)nL®N = vi 0 N.,. Hence we can define an isometry pi : Mi

Ni by

a(vi (9 m) = vi ®µi(m)

(7)

We will construct an isometry p : M = N by making use of pi. We first show

B(a, b) = B(pi(a), pj (b)) for a E Mi, b E Mj .

(8)

Since

B(vi, vj)B(a, b) =B(vi (& a, vj 0 b)

=B(vi 0 p (a), vj0 i (b))

by (7)

=B (vi, vj) B (pi (a), p (b) )

(8) holds if B(vi, vj) 0. Suppose B(vi, vj) = 0; then

B(L, L)B(Mi, Mj) =B(L ®Mi, L ®Mj) = B(vi ®N, vj®N) =B (vi, vj)B(N, N) = 0,

by (6)

and then B(Mi, Mj) = 0 follows. Similarly,

B(L, L)B(pi(Mi), pj (Mj)) = B(L, L)B(Ni, Nj) =B(L ®Ni, L (DNj) = B(vi ®M, vj(&M) by (6') =B(vi, vj)B(M, M) = 0.

Thus we have B(µi(MM), pj(Mj)) = 0 and hence B(µi(Mi), pj(Mj)) _ B(Mi, Mj) = 0 if B(vi, vj) = 0 which completes the proof of (8).


By (Bl), we may assume that a subset {v1,

203 ,

vn} is a basis of QL,

changing suffices if necessary; then by (6), ®i 1Mi is a submodule of M of finite index and n

QM=(@QMZ. i=1

Hence by virtue of (8), we can define an isometry µ : QM = QN by µ(E mi) = E lµi(mi) for mi E QMi. We have only to show µ(M) = N. Let lei} be a basis of L and ei = Eji aijvj, vi = Eh-1 biheh, aij E Q, bih E Z; then

n

E bikakj = bij

(9)

k=1

For m E M, we write m = ' 1 mi, mi E QMi, and a(vj ®mi) vi ® nji, nji E QN by (6); then (7) yields nii = µi(mi). Now we have

a(ek ® m) =a(E akjvj ®E mi) = E akjvi ® nji j

i

i,j

_ 1: eh ®(E akjbihnji) h

i,j

From a(ek (9 m) E L ® N follows Ei j akjbihnji E N. Summing them up with k = h, (9) implies >i nii E N and so p(m) = E µi(mi) = E nii E N. Thus we have shown µ(M) C N. The assumption L ® M = L ® N implies

dM=dNandso p(M)=N.

0

The proof shows the following, putting ui = vi for 1 < i < n.

Corollary 7.2.1. If a subset {uj} 1 of {vi}i=1 is a basis of QL, then a(ui ® m) = ui ®µ(m) for m E Mui for some isometry y : M = N and p(Mui) = Nui Lemma 7.2.2. Let p be a prime number and let A be a matrix in Mn (7L) satisfying A 1n mod p. If Ak = In fork > 1, then A = In or A = , ±l)W-1 for W E GLn(7L). The latter case occurs only W diag(±1,

when p=2. Proof.

Let the order of A be pr'h (h, p) = 1, and put B = AP'; then

Bh=lnand B-1nmodp.

Suppose h # 1, i.e. B 0 In and define an integer m (> 1) by B 1n modpm but B # In modpm+l Since (h, p) = 1,

1n = B = (1n + (B - 1n))h = 1n + h(B - ln) modpm+l

204

7


gives B - 1n modpm+1, contradicting the defining property of m. There-

fore we get B = In andh=1. Suppose p 0 2. Putting C AP'-1, we have C - 1n mod p, Cr = 1, C ln. Writing C= 1n + ptD (t > 1,D E Mn (7L), D # 0 mod p), P-1

1n =CP = 1n+Pt+1D+E (8p)(ptD)s+ptpDp s=2

-1n + pt+1 D mod pt+2 yields D - 0 mod p which is a contradiction. Thus we have completed the case ofp54 2. Suppose p = 2 and r > 2; then putting F := A2' 2, F - In mod 2 and the order of F is 4. Writing F = 1n + 2G, we have F2 1n mod 4. Defining D by the same equation as in the case of p # 2 for C := F2, we have t > 2 and then 1n = C2 = 1n + 2t+1D + 22tD2 - 1n + 2t+1Dmod2t+2 yields

D - 0 mod 2, again a contradiction. Thus we have r < 1, i.e. A2 = In. Put L = 7Ln(column vectors) and let A act on L canonically. Set Lf = {x E L I Ax = ±x}. Then the identity x = (x + Ax)/2 + (x - Ax)/2 gives L = L+ ® L_ since Ax - x mod 2L. Hence we have only to take W as a base change matrix.

Lemma 7.2.3. Let K, M, N be positive lattices and suppose that K is indecomposable and there exist submodules M2, N' in M, N such that

[M M'] 1. We will show that M = Mk 1 (Mk)1 under these assumptions. Once this has been proved, we can then apply the induction assumption to

a:K®(Mk)1=K®(Nk)1, which completes the proof.

Noting that it = ul is primitive, choose a basis {ui} of M such that B(u1 i ui) = 0 for i > 3. If B(ul, u2) = 0, then M = Mi 1 (M,') -L and this

contradicts the assumption [M : M'] > 1. Thus we have B(ul, u2) # 0. Considering a-1, we can choose a similar basis {vi} of N and hence for a basis of {ui} of M and {vi} of N, we have (11)

( B(u1iu2)#0,B(ui,ui)=0for i>3, Sl

B(vi, v2)

-

0, B(vi, vi) = 0 for i > 3.

206

7


Take elements xi, yi E K such that

> B(xi, yi) = 1,

(12)

by virtue of the assumption s K = Z. For m E M, by (12)

B(ul, m) = E B(xi ®u1, yi ®m) E B(K ®u1, K ®M) is clear and hence B(ul, M) C B(K (& u1, K ® M) follows. The converse inclusion is clear, and so B(ul, M) = B(K ® u1, K ® M) and similarly

B(vi, N) = B(K ® vi, K ® N). Hence we have

B(ul, M) = B(K ®ul, K ®M) = B(v(K (9 ul), o(K (9 M))

= B(K®v1,K(9 N) = B(vi,N), which implies

(Q(ui),B(ui,u2)) = (Q(vi),B(vi,v2))

by (11).

Since Q(ul) = Q(vi), the reduced denominator of B(ul,u2)/Q(ul) and B(vl, v2)/Q(vl) are the same; we denote it by r. If r = 1, then B(ul, U2 B(ul, u2)Q(ul)-lul) = 0 and u2 - B(ul, E M implies M = Zu1 1 ui . This contradicts [M : M'] > 1 in (10) and so r > 1 holds. Let u2)Q(ul)_lu1

U1

-Z[x>u3,...],v1

=Z[y,v3i...]

where

x := r(u2 - aul), y := r(v2 - a'vi) for a := B(ul, u2)/Q(ul), a' := B(vl, v2)/Q(vl), and put x=

mi (mi E M'), i>2

y=

ni (ni E Ni). i>2

Fix any prime p dividing r. Then a/a' is a p-adic unit, and for z E K, we have

0`(Z ® u2) = a(z ® (r-lx + aul)) = r-lo,(z ® x) + av(z (9 ul) = r-1 r(z ® x) + az ® v1 = r io,(z (9 x) + az ® aj-1(v2

- r-ly)

= (a/a')z ®v2 + r-1(E o% (z) (D µi (mi) - a/a' > z ® ni) i>2

i>2


207

and then r-1(E ai(z) ® µi(mi) - a/a E z (D ni) = a(z ® u2) - (a/a )z ® v2 i>2

i>2

EZp(K®N)nQ(K®vi) c7p(K®v1 ). Thus we get, taking the k-th component, (13)

ak(z) ® lµk(mk) - (a/a') z ®nk E r7Lp(K (9 Nk) for z E K.

Taking a basis {wi} of Nk such that nk = akwl, /Lk(mk) = bkwl + CkW2

for ak,bk,ck E Z, (13) means (bkok(z) - (al a) akz) 0 W1 + ckak(z) 0 w2 E r7Lp(K ® Nk) for z E K, and hence

bkvk(z)-(a/a')akz modr7LpK, ckak(z)Er7LpK for z E K. We show that ak $ ± id yields ak - 0 mod r7Lp. Suppose ak # 0 mod r7Lp; then it implies ak(z) - cz mod p7LpK for c = (a/a') ak/bk E 7Lp by (14). Making use of (12), 1 = E B(ak (xi), ak (yi)) = c2 mod p and hence c (14)

±1 mod p. Thus we get ak (z) - ±z mod p7LpK where the sign is determined

by c and hence it is independent of each z E K. Since the order of ak is

finite, the previous lemma implies ak = ± id or K = K+ 1 K_ where Kf = {z E K I akz = ±z}. These are contradictions. Thus we have shown ak - 0 mod r7Lp. Then (14) implies bk = ck = 0 mod r7Lp and so nk,pk(mk) E r7LpNk. Since p is any prime number dividing r, we have nk,µk(mk) E rNk, which implies Mk E rMk C rM. Then

u2 - r-1mk = r-1x + aul - r-1mk = r-1 E mi + aul E M n Q(Mk)1 = (Mk)1. i>2 i#k

With it, implies 1 Mk 1 WD 1 E) ul, u2 - r - mk, u3,

and hence Mk I (Mk)' E) ui for all i since r-1mk E Mk. This means M = Mk I (Mk) 1, since {ui} is a basis of M. This completes the proof of the lemma as we noted before.

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Lemma 7.2.4. Let K, X, Y be positive lattices and suppose that K is indecomposable and o, : K ® X = K ® Y. Then there exist submodules Mo, M

of X and No, N of Y such that: (i) Mo, M (resp. No, N) are primitive submodules in X (resp. Y) (unless they are {0}) such that

[X: Mo I M] < oo, [Y: No I N] < oo,

a(K®Mo)=K®No, o,(K®M)=K®N. (ii)

There exist decompositions Mo =1i Mo,i, No =1i No,i such that for every i

o, =ai®/3i : K®Mo,i =K®No,i for ai E O(K) and ,Qi : Mo,i - N0,i. (iii)

In particular, MO - No. a(K ®m) = K ®n for m E M, n E N implies m = 0, n = 0.

Proof. Let ml,

, mr be mutually orthogonal elements of X such that a (K ®mi) = K ®ni for some ni E Y, and we assume that r is the maximal number. Then putting

M := {ml, ... , mr}1, N := In,, ... , nr}1, Mo := M1 and No := N1, (i), (iii) are clear, since

and =K®Q[ni,... ,nr]nK®Y=K®No.

(ii) follows from Lemma 7.2.3.

Lemma 7.2.5. Let M, N be indecomposable lattices and suppose M®N = K1 I K2 (K1 0, K2 0). Then the isometry a of M ® N defined by a = id on K1, _ - id on K2 is not contained in O(M) ® O(N).

Proof. Suppose a = a ®p (a E O(M), p E O(N)). a - id mod 2(M ® N) implies (15)

a(x)®µ(y) -x®ymod2(M®N) for x E M,y E N.


209

We take a primitive vector v1 of N, and if µ(v1) 42v1, then we choose v2 E N such that {v1, v2} is a basis of Q[vl, p(v1)] fl N. Hence there are integers a, b such that p(v1) = av1 + bv2, which is also valid even in the case of p(vi) E @v1. Then (15) implies

a(x) 0 µ(v1) - x ® v1 = (aa(x) - x) 0 v1 + ba(x) 0 v2 E 2(M 0 N). Since {vi, v2} can be extended to a basis of N, it yields aa(x) = x mod 2M for x E M. Thus a = 1 mod 2 and or (x) = x mod 2M. By virtue of Lemma 7.2.2, there is a decomposition M = M1 ® M2 such that

a=id on M1ia=-id on M2. For xi E Mi (i = 1, 2), B(xl, X2) = B(a(xl), a(x2)) = -B(xl, x2) is clear and this yields that M1 and M2 are orthogonal with each other. The indecomposability of M yields M1 or M2 = 0, i.e. a = ± id on M. Similarly

we have µ = ± id on N and then a = f id on M ® N which contradicts the definition of a.

Lemma 7.2.6. L ® L is indecomposable and O(L ® L) is generated by O(L) and the interchanges of L.

Proof. Let a E O(L ®L). In Lemma 7.2.4 we put K = X = Y = L, and let M, Mo, N, No be those there. We show L = Mo or M. Suppose M = 0; then (i) in Lemma 7.2.4 implies L = Mo. Suppose M 0. By (i), we have or : L ® M = L 0 N, and by (iii) we can apply Lemma 7.2.1 to a : L ® M = L ® N, we have rank M > rank L. Since M is a primitive submodule of L, M = N = L holds. Thus we have shown L = Mo or M. If L = Mo, then (ii) yields a = a ®,Q for a,,3 E O(L) since L is indecomposable.

If L = M, then Mo = 0 and applying Lemma 7.2.1 by (iii) there exists a submodule Mi # 0 and vi E L such that a(L (9 Mi) = vi ® L. Comparing ranks, we see that rank Mi = 1. Thus we have shown a(L ®u) = v ®L for some u, v E L, u 0, v # 0. Let p E 0(L 0 L) be the isometry defined by p(x (D y) = y ®x (x, y E L). Then

pa(L(9 u)=L®v follows.

Applying the above to pa : L 0 L = L 0 L, and denoting by

M', Mo', N', No the corresponding submodules in Lemma 7.2.4 for pa, we have Mo $ 0 so Mo = L, and pa E O(L) ®O(L) as above. Thus O(L ®L) is generated by O(L) and the interchange isometry p. Next suppose that L ® L is decomposable, i.e.

L®L=K11K2iK1 00, K2 $0,

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7


and define an isometry a E O(L ®L) by a = id on K1, the previous lemma and the above, we have

id on K2. By

a = µ(f3 ®Q)

(16)

for 0, v E 0(L) and the above interchange isometry µ. a2 = id implies, for

x, y E L, x 0 y = a2(x ® y) = µ(l3 ® a)(o(y) 0l3(x)) = ai3(x) 0,3a(y) Hence x ® L = o-,3(x) ® L follows thus oi3(x) = ±x for x E L and so o-/(x) - x mod 2L for x E L. Since L is indecomposable, taking -a if necessary, we may assume (17)

a/3 = id on L.

Let {vi} be a basis of L and put x = a(vi) 0 vj for i 54 j (we are assuming

rank L > 1); then x = (x + a(x))/2 + (x - a(x))/2, and (x + a(x))/2 E K1i(x - a(x))/2 E K2

follows from a2 = 1 and a - id mod 2(L 0 L) and then (16), (17) imply (o (vi) 0 vj + a(vj) (9 vi) /2 = (x + a(x))/2 E K1. This contradicts that {v(vh) 0 vk I h, k} is a basis of L 0 L.

Lemma 7.2.7. Let m be a natural number. Then if O(®mL) is generated by O(L) and interchanges of L, and if ®m-1L is indecomposable, then OtmL is indecomposable.

Proof. By the previous lemma, we may assume m > 3. We assume ®mL = K1 L K2, Ki 0 0 (i = 1, 2) and define an isometry a E O(®mL) by (18)

a = id on K1i = - id on K2.

From the assumption, we can write a = (®6i)/.µ, where Qi E O(L) and µ(x1 0 . 0 Xm) = xµ(1) ® 0 xµ(m) (xi E L). Here we identified the isometry µ and the permutation of the set {1, , m}. Now a2 = 1 implies x1 ®... ®xm, =a2(x1 ®... (& xm) =a(61(xµ(1)) ®... 0 6m(xµ(m))) =61(°p(1)(xµ2(1))) ® ... 0 6m(6Fi(m)(xFi2(m))),

which implies p2 = id. If µ(1) = 1, then a c O(L) ® O(®m-1L) which contradicts Lemma 7.2.5 by the assumption that ®m-1L is indecomposable.

If µ(1) = 2, then µ(2) = 1 so a E O((D2L) 0 O(Om-2L) follows. It also


211

contradicts Lemma 7.2.5, since ®2L is indecomposable by Lemma 7.2.6. Put j = p(1) > 3 and define an isometry i by 'q(xl®x2®...®xi (9 ...)=x2®xi ®...®xl®..

where all elements except for the first, the second and the j-th coordinates are fixed. Then we have

r1_1arl(xl®...®X-)=17-1a(x2®xj (D ...®xl®...) =77-101(x1) ®... ®Qj (x2) ®...

(aj (x2) is on the j-th coordinate) =a.7 (x2) ®al (x1) ®.. .

and hence q-lcij E O(®2L) ® 0((gm-2L). Since r1-1all = id on r1-1(Kl) and is -id on r1-1(K2) by (18), we have a contradiction to Lemma 7.2.5 as above.

Lemma 7.2.8. Let K, X, Y be positive lattices. Suppose that K is indecomposable with rank K > 1 and a : K ® X = K ® Y, and that there exist submodules Mo, M of X and No, N of Y satisfying the conditions (i), (ii), (iii) below. Then X = Mo L M, Y = No L N and X = Y. (i)

Mo, M (resp. No, N) are primitive submodules of X (resp. Y) (unless they are {0}) such that

[X:Mo1M] 0 for all t E Q, we get (min L) 2 = Q(x)Q(y) > B(x, y)2 and so 0 < [x, y] < 1. If [x, y] = 1,

then B(x, y) = ±Q(x) and so Q(tx + y) = (t ± 1)2 min L, which yields

x = ±y. Thus M(L)/ ± 1 becomes a weighted graph. We call it the weighted graph induced by a positive lattice L. Let L be the submodule of L spanned by the elements of M (L). We show that L is indecomposable if and only if the graph M (L) / ± 1 is connected.

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7


If L = L1 1 L2, L1 # 0, L2 0, then x e M(L) is in L1 or L2. Hence putting Gi {x E M(L) I x E Li}, we find M(L) = G1 U G2. If, for instance G1 = 0, then M(L) = G2 C L2 and then L C L2. This is a contradiction. Thus G1 0 and similarly G2 0. By definition of L1, L2, [G1, G2] = 0 is clear. Thus M(L)/ ± 1 is not connected. Conversely suppose M (L)/ f 1 is not connected; then there exist subsets H1, H2 of .M (L) such that

Hi = -Hi, M(L) = H1 U H2, B(H1,H2) = 0. Let Mi be the submodule of L spanned by the elements of Hi; then M1 1 M2 =L and so L is not indecomposable.

Lemma 7.3.1. Let A = lei, , en}, A', B, C be weighted graphs and suppose a : A x B = A' x C. Take and fix any element b E B. We define fi E A', ci E C, gij E A, bij E B by

o (ei, b) = (fi, ci), o (gij, bij) = (fi, cj) for 1 < i, j < n. Then b # bij yields [ei, ej] = 0.

Proof. Put aij = [ei, ej]; then the property (ii) means aij = aji, and

(1)

aij =[ei, ej] = [(ei, b), (ej, b)] = [(fi, ci), (fj, cj)] =[fi, fj] [ci, cj].

We must show that b # bij implies aij = 0. Choose 1 < u, m < n with b # bun,, such that aum > aij if b 54 bij. We have only to conclude the contradiction, supposing au,,,, # 0. Put gum = eP, bum, = b'; then o, (eP, b') = (fu, cm), b'

(2)

b.

aiP[b, b'] = [ei, eP] [b, b'] _ [(ei, b), (er, b')]

(3) (4) (5)

= [(fi, ci), (fu, cm)] = [fi, ful [ci, cm].

if fi = fu (by (3)) amp[b,b'] = [fm, fu] (putting i = min (3)). aiP[b, b'] = [ci, cm]

Hence we have

aum = [fu, fm] [cu, cm] =aupamp[b,b' ]2

fP0fu

by (1)

by (4), (5)

7.3

Weighted graphs

219

Suppose fu = fp; then (2) implies v(ep, b') _ (fp, c,,,,). If b = bp,,,,,, then it implies o (gp,,,,, b) = a(g3,,,,, bp,..) = (fp, c,,,,) = v(ep, b') and so b = b' which contradicts (2). Thus we have b bp,,,, and then the definition of u, m implies aum, > apm, which yields 0 < aum = aupamp [b, b']2

by (6)

< aupaum [b, b 1]2 < aum.

Hence we get [b, b'] = 1, contradicting (2). Thus we have shown (7). If bup # b, then aup au.n and so 0 < aum = aupamp[b, b']2 < aumamp [b, b/ ]

by (6)

2

aum

which also induces [b, b'] = 1 contradicting (2). Thus we have (8)

bup = b

and

a(gup,b) = (fu,cp)

Now, we define t by gup = et; then a(et, b) = (ft, ct) and (8) yield (9)

fu=ft, cp=ct;

then (4) implies (10)

atp[b, b'] _ [ct, cm].

Substituting i = p to (3), we have (11)

[b,b'] =

[fp,fu][cp,cm.].

Thus we get [cp, cm.] = [ct, cm]

=atp[b,b']

by (9) by (10)

= atp [fp, fu] [cp, cm]

by (11)

and [cp,cm](1 - atp[fp, fu]) = 0. If [cp,cm] 0, then atp[fp, fu] = 1 and hence [fp, fu] = 1, which contradicts (7). If [cp, cm] = 0, then [b, b'] = 0 by (11) and (6) induces aum. = 0. This contradicts the assumption aum, 0, which completes the proof.

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7

Lemma 7.3.2. Let A, A', B and C be weighted graphs and suppose that A = lei, , en} is connected and a : A x B = A' x C. Fix any element b E B. Putting v(ei, b) = (fi, ci), we have A =' {o,(ei, b) 1 < i < n} = {(fi, cj) 1 < i, j < n}. Proof. For 1 < i < n, we put Ci = {ck I fi = fi}. Clearly ci E Ci; denote by C' the connected component of Ci containing ci. Here we note that x E C', y E Ci and [x, y] 0 0 yield y E C. We first show that lei, ej] _ [fi, fj] [ci, cj] 0 induces Ci = Cj. Suppose lei, ej] = [fi, fj][Ci, cj] 0 0. Then by Lemma 7.3.1, we have o, (et, b) = (fi, cj) for some t. Hence a(et, b) _ (12)

(ft, ct) implies cj = ct, fi = ft and so ct E Ci. Now lei, ej] 54 0 implies [ci, cj]

0 and hence C? E) cj since ci E Ci and cj = ct E Ci. Thus we get

(cje)C'nCjI

0.

Next we show that x E C' n c,,, y E Cj' and [x, y] 0 imply y E C. Let x and b be above; then by definition x E Ci and y E Cj imply (13)

x = c.,, fu = fi, y = ck and fk = fj for some u, k.

They give [eu, ek] = [(eu, b), (ek, b)]

_ [(Jf, Cu), (fk, Ck)]

= [fu, fk] [Cu, Ck] _ [fi, J j] [x, y] by (13),

which yields [en, ek] # 0 by virtue of the assumptions lei, ej]

0 and

[x, y] # 0. Applying Lemma 7.3.1 to [en, ek] # 0, we have v(e3, b) = (fu, ck)

for some s. Then v(es, b) _ (fs, cs) implies, by virtue of (13), fs = fu = fi and hence y = ck = cs E Ci.

(14)

Now [y, x] # 0 by the assumption, x E C' and (14) imply y E C', which is what we wanted. As Cj' is connected, for any fixed yl E Cj, there exist y2i , ym E Ca such that [yi, yi+11 0 0

(i = 1, 2, ... , m - 1), ym = cj (E Cj ).

As proved as above, cj E C' n Cj, and applying what we have proved to x = cj, y = ym_1, we have ym_1 E CZ n Cj; repeating this argument we get ym_2, , yi E C' n C,'. Thus C.,' C Ci. Changing i and j, we have the converse inclusion and have proved C' = Cj' if lei, ej] 0 0. Since A is connected, we get Cl' = = Cn. For 1 < s, t < n, we have ct E Ct = C,' C Cs and then by definition of Cs we find ct = ck, fk = fs for some k. Finally, we get (f y, CO = (fk, Ck) = v(ek, b), which completes the proof of the lemma.

7.3

Weighted graphs

221

Lemma 7.3.3. Let A, B and C be connected weighted graphs and suppose

a : A x B = A x C. If there exist bo E B, co E C such that a(x, bo) = (f (x), co) for any x E A, where f is a mapping from A to A, then f is an isometry from A to A and there exists an isometry g : B = C such that a(x, y) = (f (x), g(y)) Proof. Since A is a finite set and or is injective, f is injective and hence bijective. For a, a' E A, [a, a'] = [(a, bo), (a', bo)] = [f (a), f (a')] implies that f is an isometry. Composing f -1 x id with or, we may assume f = id. We consider a subset Bo of B having the property (14)

a(a, b) = (a, g(b)) for a E A, b E Bo,

where g is a mapping from Bo to C. Now Bo = {bo} is such a subset by hypothesis. We let Bo be a maximal subset among all subsets having this property. Once Bo = B is shown, then g is clearly an isometry and the proof is finished. Suppose B0 0 B. We have only to show the existence of b' V B0 such that

a(x, b') = (x, c) for every x E A and some c E C, which would contradict the maximality of B0. We put Co := g(Bo), and m = max [b, b']. bEBO VOBO

Considering a-1 if necessary, we may assume (15)

m> max[c,c']. CECO 'OCO

Since B is connected, m > 0. Fix b E Bo, b' define h(x) E A, c(x) E C for x E A by

Bo such that m = [b, b']. We

a(x, b') = (h(x), c(x)).

Let us show that h(x) = x and c(x) is independent of x E A, which contradicts the maximality of Bo will complete the proof. If c(x) E CO, then c(x) = g(bi) for some b1 E B0, and from a(x, b') = (h(x), g(b1)) = a(h(x), b1)

we get b' = b1i contradicting the property b'

Bo. Thus c(x)

m = [b, b'] = [(x, b), (x, b')] _ [(x, g(b)), (h(x), c(x)] = [x, h(x)] [g(b), c(x)] [x, h(x)lm by g(b) E Co, c(x) Co and (15) < M.

Co, and

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Now m > 0 implies [x, h(x)] = 1 so x = h(x). Thus we have

a(x, b') = (x, c(x)) for x E A. If x, y E A and [x, y] # 0, then [x, y] = [(x, b'), (y, b')] = [(x, c(x)), (y, c(y))] = [x, y] [c(x), c(y)]

yields [c(x), c(y)] = 1, i.e. c(x) = c(y). Since A is connected, it means that c is constant on A. This contradicts the maximality of Bo and completes the proof. We say that a weighted graph A is indecomposable unless A - B x C for weighted graphs B, C with OB > 1, #C > 1.

Theorem 7.3.1. Let Ai (1 1, OBE > 1 for all i, j, and

assume or : fi'_' 1 Ai = f 1 Bj. Then we have m = n with or a product of , m}. ai : Ai = B8(i) for some permutation s of {1, Proof. Changing suffices and taking v-1 if necessary, we may suppose

OA1 > 0& OB3 for all i, j. For each i > 2, take and fix an element ei E Ai. Changing suffices if necessary, we may suppose that the projection of o(Al, e2, , e,,,,) to B1 contains at least two distinct elements by the assumption that OA1 > 1. Applying Lemma 7.3.2 to A = A1, B = fli>2 A,, A'=B1 and C = [Ii>2 B, we have (16)

Al-a(Al,e2i...,em.)=AoxCo (AoCB1,CoC11Bi) i>2

By hypothesis, #Ao > 1 and the assumption that Al is indecomposable implies #Co = 1. Therefore OA1 = #Ao < OB1 yields OA1 = OB1 and hence

AO = B1 since #A1 > #Bj. So (16) induces an isometry from Al to B1. Applying Lemma 7.3.3 to v : Al x B = B1 x C, we have or = f x g for

f:A1-B1i g:f Ai-fJB., i>2

j>2

The theorem is now inductively proved.

7.4 The tensor product of positive lattices For a positive lattice L, L denotes a submodule of L spanned by M(L). In general, rank L < rank L.

7.4

The tensor product of positive lattices

223

Lemma 7.4.1. Let L be a positive lattice of E-type such that [L : L] < oo and L is indecomposable. Then L is indecomposable, and if v : L M = L ® N for positive lattices M, N, then for in E M(M), we have a(L ®m) = F ®G for some primitive submodules F C L, G C N satisfying min F = min L, min G = min N. Proof. Suppose L = L1 1 L2; then any minimal vector of L is in L1 or L2-

Put Mi = fv E M(L) I vELi} for i=1,2;then ,M(L)=M1UM2 and for a submodule Ki of Li spanned by Mi we have L = K1 I K2. Since L is indecomposable, K1 or K2 = 0, which means .M1 or .M2 = 0. Thus M(L) C L2 or L1 and hence L C L2 or L1. Since [L : L] < oo means rank L = rank L, we have L1 or L2 = 0. Therefore L is indecomposable.

Since L is of E-type, o,(M(L) (9 M(M)) = M(L) ® .M (N) and hence v canonically induces an isometry from A x B to A x C as weighted graphs where A = .M(L)/ ± 1, B = .M(M)/ ± 1, C = .M(N)/ ± 1. Since L is indecomposable, A is connected. Applying Lemma 7.3.2, we have a(M(L)®m) = {x®y I x E F1, y E G1} for some subsets F1 C .M(L), G1 C M(N). Denoting by F', G' the submodules spanned by F1, G1, we have v(L®m) = F'®G'. Put F = QF'f1L, G = QG'f1N; then F®G is a primitive submodule of L®N and [F®G : F'®G'] < oo. Also L®m is a primitive submodule of L ® M and [L ® m : L ® m] < oo. Then a(L (9 m) = F' O G' induces a(L ® m) = F ® G. Now F' C F C L and F' contains the subset F1 of ,M (L) and so min L = min F, similarly min G = min N. Corollary 7.4.1. Let L be a positive lattice of E-type such that [L : L] < 00 and L is indecomposable. If L = L1 ® L2 for positive lattices L1, L2, then L1, L2 are of E-type and [Li : Li] < 00 and Li is indecomposable for i = 1, 2. Proof. We define an isometry v E O(L1 ® L2 (& L2) by

a(x(Dy®z) =x®z®y (x E Ll,y,z E L2). For m E M(L2), a(L (D m) = (L1 ® m) ® L2, and applying Lemma 7.4.1,

we have min(Li ® m) = min(L). From (iii) of Lemma 7.1.1 it follows that L1 ® m is of E-type and so L1 is of E-type. Similarly L2 is of E-

type. Hence M(L) = M(L1) ® .M(L2) and then L = Ll ® L2. Now the indecomposability of L and the property [L : L] < oo induce those of L,, L2For a positive lattice L, we say that L is multiplicatively indecomposable if L = K1®K2 for positive lattices K1, K2 implies rank K1 or rank K2 = 1.

Theorem 7.4.1. Let L be a positive lattice of E-type such that [L : L] < oo, L is indecomposable and L is multiplicatively indecomposable. Then for any indecomposable positive lattice X,

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(i) (ii)

L ® X is indecomposable,

(iii)

if X = ®tL 0 X' and X' is not divisible by L, then O(L ® X) is

L ® X = L 0 Y yields X = Y for a positive lattice Y, generated by O(L), O(X') and interchanges of L.

Proof. If rank L = 1 then the assertion is clear. Assume rank L > 1. We have only to prove that L is indecomposable and L satisfies the condition (B) in the Fundamental Lemma in 7.2. Lemma 7.4.1 shows the indecomposability of L. To prove the condition (B), we use induction on rank M for the lattice

M in the condition. Put .M (L) = {±vz}and for positive lattices M, N

assumeo:L®M'=L®N and thatv(L®m)=Lon(mEM,nEN) yields m = 0, n = 0. Now {v2} satisfies the condition (B1). If rank M = 1, then this condition is not satisfied and hence we have rank M > 1. Put

M1 := M1, M2 :_ (M1)1,

N' := N1 N2 :_ (N1)1. Then M', N' (i = 1, 2) are primitive submodules; moreover QM2 = QM, and @N2 = QN are clear. Since L is of E-type, we have Q(L®M) = L®N, which implies (1)

v(L®M')=L®N' (i=1,2),[M:M11M2] rank ®iEsMi >_ (#S) rank M/ rank L

by (5). By the choice of S, OS = rank L is valid and then the inequality in (5) becomes equality.

Similarly we get rank Nj = rank N/ rank L which is nothing but the condition (B2). Moreover (4) implies

QMj = @M3l 1 QM and QNj = QN 1 QN

.

Then (3) implies a(Q(vj 0 Mj)) = Q(vj 0 Nj) and (B3) is proved. Next we assume M1 = 0; then N' = 0 and [M : M] < oo, [N : N] < oo. Note that the situation is similar for a-1 : L 0 N = L 0 M. We begin by showing that for m E M(M) the following holds: (6)

a(L®m)Cw®N forsomewEM(L).

By virtue of Lemma 7.4.1, there exist, for m E M(M), submodules F C L, G C N satisfying a(L®m) = F®G, min F = min L and min G = min N. Since L is multiplicatively indecomposable, either rank F or rank G must be one. If rank G = 1, then putting G = Zg we have a(L®m) = FOG C L®g, and Q(g) = min G = min N = min M = Q(m) implies d(L ®m) = d(L ®g) and so a(L ® m) = L ® g, which contradicts the assumption. Therefore we

get rank F = 1, and putting F = Zw, min L = min F yields w E .M (L) which is simply (6).

For uEM(L)and Mu:=fm EMIa(L(D m)Cu®N}, we show (7)

u®Nco(L®Mu)Cu®N.

For n E M(N), write o,-'(u ®n) = v ®m (v E .M(L),m E M(M)); then by virtue of the above a(L 0 m) C v' 0 N holds for v' E M(L).

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Then a(v 0 m) = u ® n yields v' = ±u and hence m E Mu. Thus we have u ® n = a (v ® m) E v(v (9 Mu) C v(L ® Mu) c u ® N, which immediately yields (7).

The assumption [N : N] < oo then implies rank L rank Mu = rank N. Similarly we have rank L rank Nu = rank M which yields the condition (B2).

In the above, m E Mu implies

min(Mu) = min(M) and M(Mu) C M(M). Similarly, by considering v-1 : L ®N = L ®M, we have .M (Nu) C .M (N), where Nu := In E N I o,-'(L on) C u ®M}. Starting with u E .M (L), n E

.M(Nu) C M(N), we put or-'(u ®n) = v ®m for v E .M(L), m E M(M). Then n E Nu implies o,-'(L ®n) C u ®M and hence a-1(u ®n) = v ® m gives v = ±u and v(L (& m) C u 0 N by (6), i.e. m E Mu. Thus we have or-' (u ®M(Nu)) C u ®M(Mu) and v- 1(u ®Nu) C u ®Mu.

(8)

Since min(L) min(M) = min(u ® N) = min(u ® N), taking submodules spanned by minimal vectors in (7) we have u ® N C or (L ® Mu) C u ®N. Hence a(L (D Mu) = u ®N, and then o (L 0 Mu) C u ® N implies

[N: NJ =[u®N:u®NJ = [u®N: a(L0Mu)][v(L®Mu) : u®N] =[u®N:a(L(9 Mu)][o(L(9 Mu):0(L®Mi)]. Now [N : N] < oo implies that [L ®Mu : L ®Mu] < oo so [Mu : Mu] < oo.

Similarly [Nu : Nu] < oo is valid and then (8) yields Qu-1(u 0 Nu) C @(u 0 Mu). The condition (B2) proved above implies rank Nu = rank Mu and so @o-1(u (9 Nu) = Q(u 0 Mu), which yields the condition (B3).

Example 1. Let L be one of the positive lattices An,, D, E6, E7, E8 in Example 1 in 7.1. Then L satisfies the assumptions in Theorem 7.4.1. Except for L being multiplicatively indecomposable, the other conditions are obvious.

Suppose L = K 0 H for positive lattices K, H. Assuming rank K > 1 and rank H > 1, we will induce a contradiction. Let us assume it. By Corollary 7.4.1, K, H are of E-type and [K : K], [H : H] < oo. Since

K is of E-type, we get L = k (-9H = K 0 H and then L = L yields

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227

K = K, H = H. By scaling, we may assume min K = 1, min H = 2. For e E .M (K), u, v E M(H), we have B(u, v) = Q(e)B(u, v) = B(e ® u, e (D v) E B(L, L) = Z.

If B(u, v) is even for all u, v E ,M (H), then H = H yields s H C 2Z which

implies H = Zu 1 ul for u E M(H). This means that H and hence L = K ® H is decomposable, which is a contradiction. Therefore B(u, v) is odd for some u, v E .M (H). Since B(u, V)2 < Q(u)Q(v) = 4, B(u, v) = ±1 and hence we may assume B(u, v) = 1 for some u, v E M (H). For

e, f E .M (K), we have B(e, f) = B(e 0 u, f 0 v) E B(L, L) = Z. Then K = K yields s K C Z. Since Q(e) = 1, K = 71e el holds and it implies the decomposability of L. Thus we have shown that L is multiplicatively I

indecomposable.

Example 2. Let L be the lattice in Exercise 1 in 7.1 and use the notation there. Now L is of E-type and L = L. If the weighted graph induced by L is connected, then L is indecomposable. If L = K ® H and L is indecomposable, then

Corollary 7.4.1 implies that K, H are of E-type and .M (K) ® M(H) =' .M(L). Hence the indecomposability of the weighted graph induced by L implies that L is multiplicatively indecomposable. Thus L satisfies the assumption of Theorem 7.4.1 if and only if the weighted graph induced by L is connected and indecomposable. Among the conditions on L in Theorem 7.4.1, the multiplicative indecomposability is necessary to the assertion (iii). The author does not know

whether other conditions are really necessary or not. When rank L = 2, the following gives a satisfactory answer.

Theorem 7.4.2. Let E be an indecomposable binary positive lattice. Then the assertion (A) is true for L := E in the Fundamental Lemma.

By the Fundamental Lemma, we have only to verify the assertion (B). We need several lemmas. First note that E is of E-type by Theorem 7.1.1; we may assume min E = 1 without loss of generality by scaling of E by (min E)-1. We fix a basis lei, e2} of E such that

Q(el) = 1, a:= Q(e2) > 1, b := B(el, e2) and 0 < b < 1/2. This is possible since B(el, ±e2 + nel) = ±B(el, e2) + n. If a = 1, then .M (E) = {±el,±e2} and so this case is contained in the previous theorem.

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We may assume hereafter

Q(ei) = 1, a = Q(e2) > 1, 0 a > 1 for m E Z.

Hence if mei + nee E M(E), then n = 0 must be valid and then m = ±1, i.e. M(E) = {±e1}. Lemma 7.4.2. Let M be a positive lattice with min M = 1. Then we have, for f E M (M)

{zEE®MI B(z,ei®f)=b,Q(z) 1, and Q(z) < a yields, by (9), a > Q(z) > b2 + (a - b2)Q(x2) > a, and hence (10)

Q(z) = a, Q(x2) = 1, Q(xi + bx2 - b f) = 0, i.e. x1 + bx2 = b f.

Since M is positive definite, we have B(x2i f)2 _< Q(x2)Q(f) = 1, and then, by (10) (11)

Q(xi) = b2Q(f - x2) = b2(2 - 2B(f, X2)):5 4b2 < 1.

The assumptions minM = 1 and x1 E M yield Q(xi) = 1 or 0 by (11). If

Q(xi) = 0, then x1 = 0 and x2 = f by (10), i.e. z = e2 ® f. This is the second case.

Suppose Q(xi) = 1; then (11) implies b = 1/2 and B(f, x2) _ -1. Hence Q(f +x2) = 0 and x2 = -f are valid. From (10) we have x1 = b(f -x2) = f and so z = (el - e2) 0 f. 0

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229

Lemma 7.4.3. Let M, N be positive lattices with min M = min N = 1. Assume we have an isometry or : E 0 M = E O N such that a(E 0 m) = E O n (m E M, n E N) implies m = 0. Then for a given fl E .M (M) there exist f2 E M, 91, 92 E N such that

a(el O fl) = el O gl, a(e2 O fl) = el 092, a(el O f2) = e2 091Proof. Since E is of E-type, a(.M(E) OM (M)) =.M(E) O.M(N) is valid and hence there exists gl E M(N) such that a(el (9 fl) = el O gi. Then B(a(e2 O fl), el 091) = b, Q(a(e2 0 fl)) = a are clear. From Lemma 7.4.2, we have o,(e2 ®fi) = el ®g2 for 92 E N, e2 091 or (el - e2) ®gl. The latter two yield or (E 0 fl) = E 0 gi contradicting the assumption. Thus we have a(e2Ofl) = elOg2. Applying a similar argument to o- I : EON = EOM, and g1 E .M(N), we have a-1(e2 O 91) = el ®f2 for f2 E M. Lemma 7.4.4. O(E 0 E) is generated by O(E) and the interchange of E. Proof. Let a E O(E(&E). Suppose that there exist x(# 0), y(# 0) E E such that a(EOx) = EOy. Then a(EOx1) = EOy1 and, applying Lemma 7.2.3

to K = M = N = E, we have aEO(E)®O(E).Suppose a(EOx)=EOy yields x = y = 0; applying Lemma 7.4.3 to M = N = E, fl = el, there exist 91,92 E E such that a(el 0 el) = el O gi, a(e2 (D el) = el 0 g2. Hence a(E 0 el) C el 0 E is valid and so a(E 0 el) = el 0 E. Letting µ E O(E 0 E) be the interchange of E we have ua(E 0 el) = E 0 el. By the above argument, we have µa E O(E) 0 O(E).

Lemma 7.4.5. Let M, N be positive lattices isometric to E and suppose that an isometry a : EOM = EON satisfies a(EOm) = EOn only when m = 0, n = 0. Then there exist decompositions M = M1®M2, N = Nl ®N2 such that, for j = 1, 2

a(E®M;)=ej ON,a-'(E(9 NS)=ej OM,a(ej (& M3)=ej ONj. Proof. We may suppose M = N = E. The proof of the previous lemma shows a(xoy) = al (y)®a2(x) (x, y E E) for some isometries al, a2 E O(E). Noting .M (E) = {±el}, clearly a(E (D el) = el 0 E. Thus we have only to put Ml = Ze1i M2 = Zal 1(e2), N1 = Ze1 and N2 = Za2(e2)

Lemma 7.4.6. The assertion (B) for E follows from the following assertion (C): Let M, N be positive lattices and suppose that an isometry or : EOM

E ®N satisfying a(E O m) = E O n (m E M, n E N) implies m = 0. Then there exist submodules Ml, Nl of M, N respectively such that M, = N, = (a scaling of E) and a(E (g Ml) = E O N1. Proof. Let M, N be positive lattices with or : E 0 M = EON and suppose that or (E (D m) = M O n (m E M, n E N) yields m = 0. Applying the

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7


assertion (C) repeatedly, there exist submodules Ml, , Mt, N1, , Nt of M, N respectively such that Mi = Ni = ECM (which denotes the scaling

ofEbyci), a(E®Mi)=E®Ni and (12)

(B1) is satisfied for vi := ei (i = 1, 2). By virtue of Lemma 7.4.5, there exist decompositions, Mi = Mil E Mil, Ni = Nil ®Ni2 such that, for 1 < i < t, j = 1, 2

(13) a(E®Mij) = vj

®Ni,or-1(E®Nij)

= vj ®Mi,a(vj ®Mii) = vj ®Nij.

Putting

M3 ={mEMI a(E(9 m)Cvj ®N},

Nv, ={nENIa-1(E®n)Cvj®M}, Now Mv; D EaMij, Nv, D EiNij follow from (13), and hence rank Mv, > t, rank Nvi > t. On the other hand, Mvl nMv2 = {0}, Nvl nNv2 = {0} yield rank Mvl E M,2 > 2t, rank Nvl E Nv2 > 2t. Since (12) implies rank M = rank N = 2t, we have rank Mv3 = rank Nvj = t for j = 1, 2, which yields (B2) and QMv; = Q(EiMij),QNv3 = @((DiNij). Hence

a(Q(vj ® Mv,)) = a(Q(vj ® ((DiM5))) = Q(vj 0 (EiNNj)) = Q(v7 ®Nv, ).

(by (13))

Thus (B3) is valid.

Lemma 7.4.7. Let K be a positive lattice with min K = 1. Then a submodule H of E ®K isometric to E is either E ®f (f E M (K)) or el ®Ko (Ko C K). Proof. Let {u, z} be a basis of H such that Q(u) = 1, B(u, z) = b, Q(z) = a.

Since E is of E-type, min(E 0 K) = 1 and hence u E M(E (9 K) = .M (E) ® .M (K). Thus we can put u = el ®f for f E .M (K). Then from Lemma 7.4.2 it follows that z = el Ox (x E K, B(f, x) = b, Q(x) < a), e2 ® f or (el - e2) ®f. Thus, H is either el 0 7L[ f, x] or E ® f . Hereafter we make the assumption (*):

for positive lattices M and N, an isometry a : E 0 M = E ® N satisfying a(E 0 m) = E 0 n(m E M, n E N) implies m = 0, and

moreover min M = 1 and hence min N = 1. We have only to show that the assertion (C) in Lemma 7.4.6 holds under (*) which we assume henceforth.

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231

Lemma 7.4.8. For x E M(M),y E M(N), there are submodules M' C M, N' C N such that a(E 0 x) = el 0 N', a(el ® M') = E 0 y. Proof. Since min M = min N = 1 by assumption, Q(x) = Q(y) = 1 implies o, (E ®x) = a-1(E ® y) = E. Hence Lemma 7.4.7 we have the assertion, noting that the assumption that or (E 0 m) = E ® n is valid only for m = 0.

Lemma 7.4.9. Let N' be a submodule of N containing n E A4(N) and being isometric to E. Then either a-1(el ®N') = E ® m (m E M(M)) or = el ® M' (M' C M); the former is possible for at most one N' when n is given.

Proof.

Since E = a-1(el ® N') C E 0 M, from Lemma 7.4.7 we have

a-1(el 0 N') = E 0 m (m E M(M)) or el 0 M' (M' C M). Suppose that there exist submodules Ni (i = 1, 2) of N such that n E Ni = E and a-1(el ® Ni) = E ® mi (mi E M(M)); then E ® ml fl E ® m2 E) a-1(el ® n) implies ml = fm2 since m1, m2 are primitive, and hence we have a-1(el ®N1) = a-1(el ®N2), i.e. N1 = N2.

For ml E M(M), we put a(el 0 ml) = el 0 n1 (ni E M(N)). From Lemma 7.4.7 and the assumption (*) we have a(E®ml) = elONo for some submodule No of N. Define a basis {nl, n2} of No by a(ei ®ml) = el ®ni

(i = 1,2); then Q(nl) = 1, B(nl, n2) = b and Q(n2) = a are obvious. Similarly we define a submodule Mo of M by or' (E ®nl) = el 0 Mo and a basis {m1, m2} of Mo by a-1(ei ®n1) = el ®mi (i = 1, 2); ml is already given. Then the following are clear. (14)

Q(ml) = Q(ni) = 1,B(ml,m2) = B(nl,n2) = b, Q(m2) = Q(n2) = a,

(15)

a(el ®ml) = el ®ni,

(16)

a(e2 ®ml) = el ®n2, a(el ®m2) = e2 ®nl, a(e2 ®m2) =el 0 x1 +e2®x2 for some xiEN (i=1,2).

(17) (18)

Once we have xl = 0, x2 = n2i we obtain a(E®7L[ml, m2]) = E®7L[nl,n2] and 7L[ml, m2] = 7L[nl, n2] = E, which is nothing but the assertion (C) in Lemma 7.4.6 and we complete the proof of Theorem 7.4.2. We shall verify xl = 0, x2 = n2 below.

Lemma 7.4.10. We have B(ni,xi) = 0, B(nl, x2) = b, x2 0, B(n2i xl + bx2) = ab, Q(xl + bx2) + (a - b2)Q(x2) = a2 and Q(x2) < a. Proof. From the inner product of (15) and (18) follows (19)

b2 = B(nl, xl) + bB(nl, x2).

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From the inner product of (17) and (18) we have

ab = bB(nl, xi) + aB(nl, x2).

(20)

Similarly, (16) and (18) imply (21)

ab = B(n2, xl) + bB(n2, x2) = B(n2, xl +bx2).

(18) implies

(22)

a2 = Q(xi) + 2bB(xi, x2) + aQ(x2) = Q(xi + bx2) + (a - b2)Q(x2).

Then (19), (20) yield B(ni, x1) = 0, B(nl, x2) = b which implies x2 0 0 by b 54 0. It remains to show Q(x2) < a. We know the inequality B(n2, xl + bx2)2 < Q(n2)Q(xl + bx2)

and substituting (21), (14), (22) into it, we get a2b2 < a(a2-(a-b2)Q(x2)), which yields immediately Q(x2) < a.

Lemma 7.4.11. cr '(el ® x2)

el ® M.

Proof. Suppose (23)

a-i(ei ®x2) = ei ®y for some y E M.

Then we will show Q(x2) = b2(< 1), which contradicts x2(0 0) E N and min N = 1, and hence we complete the proof. So, let us verify Q(x2) = b2.

Put (24)

or-'(e2 ®x2) = ei ®yi + e2 ®y2 (yZ E M).

The inner product of (15) and (23), and Lemma 7.4.10 imply (25)

B(ml, y) = B(nl, x2) = b,

and (16), (23) and (25) give (26)

B(n2, x2) = bB(ml, y) = V.

From (15), (24) and Lemma 7.4.10 we have (27)

B(ml, yi) + bB(ml, y2) = bB(nl, x2) = b2,

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233

and (16), (24) and (26) give

bB(ml, yi) + aB(ml, y2) = bB(n2, x2) = b3, which implies with (27)

B(mi, yi) = b2, B(mi, y2) = 0.

(28)

From (23) and (24) we find Q(x2) = Q(y), aQ(x2) = Q(yi) + 2bB(yi, Y2)+ aQ(y2) = Q(yi + bye) + (a - b2)Q(y2), bQ(x2) = B(y, yi) + bB(y, y2) = B(y, yi + by2). Those yield b2Q(x2)2 = B(y, yi + by2)2 < Q(y)Q(yi + bye) = Q(x2)(aQ(x2) - (a - b2)Q(y2)),

which immediately gives, since Q(x2) 34 0 Q(y2)

(29)

Q(x2)-

From (18) and (23) we find bB(m2i y) = B(xl, x2) + bQ(x2). Since, on the other hand B(m2, Y) = B(el (9 m2, el ®y) = B(e2 ®nl,el ®x2)

(by (17), (23))

= b2

by Lemma 7.4.10, we have (30)

B(xi, x2) = bB(m2, y) - bQ(x2) = b3 - bQ(x2)-

The equalities

B(yl, m2) + bB(y2, m2) = aB(nl, x2) = ab, bB(yl, m2) + aB(y2, m2) = bB(xl, x2) + aQ(x2) = b4 + (a - b2)Q(x2), follow from (17), (24) and Lemma 7.4.10 and by (18), (24) and (30) respectively. Solving them, we have (31)

B(yi, m2) = b(a + b2) - bQ(x2), B(y2, M2) = Q(x2) - b2.

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If m2 = Y2, then (31) and Lemma 7.4.10 imply Q(m2) = Q(x2)-b2 < a-b2,

which contradicts Q(m2) = a in (14) by b 54 0. So, m2 0 y2, giving Q(y2 - m2) > min M = 1. From (29), (31) and (14) we have 1 < Q(y2 - m2) = Q(y2) - 2B(y2, M2) + Q(m2) < Q(x2) - 2(Q(x2) - b2) +a= a+ 2b2 - Q(X2)-

Thus we get, with (29) Q(y2) < Q(x2) < a + 2b2 - 1.

(32)

Now we put v(el ® Y2) = el ® w1 + e2 ® w2 (wi E N). Then Q(y2) = a - b2 if w2 # 0. Hence w2 # 0 yields Q(wi + bw2) + (a - b2)Q(w2)

a - b2 < Q(y2) < a + 2b2 - 1 by (32); so 3b2 > 1 which contradicts 0 1/2 and so b = 1/2. Then the right-hand side is equal to a in the above inequality and al = ±1, al + a2 = 0, a2 = ::F1. Then we have

n2 = al(nl - x2) and b = B(nl,n2) = a,B(nl,nl - x2) = al(1 - b) by Lemma 7.4.10. Since b = 1/2, we have al = 1 and then n2 = n1 - x2. It follows from (35) that x1 = 2(n2 - x2) = nl/2 - x2. Thus we have ni/2 = xl + x2 E N contradicting the primitiveness of n1 in N. Suppose a2 = 0; then we get n2 = a1n1 + a2x2 = alnl contradicting the linear independence of n1, n2.

Thus we have al = 0 and then n2 = a2x2. Then b = B(nl, n2) = a2B(nl, x2) = alb

(by (14)) (by Lemma 7.4.10)

yields a2 = 1 which is what we wanted. Next we give a result on the uniqueness of the decomposition.

0

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Theorem 7.4.3. Let Li (1 1, [Li

:

Li] < oo, Li is indecomposable and Li is

multiplicatively indecomposable.

Let Mi (1 1 and suppose that ®Z_" L®= ®i 1Mi. Then we have m = n and changing suffices if necessary, or = ®vi, ai : Li = M°`4 (scaling of Mi by ai). Proof. Put L = ®m 1Li; then L is of E-type by Lemma 7.1.1, and L = ®Li. Since the Li's are of E-type, [Li : Li] < oo implies [L : L] < oo. Next we show L is indecomposable. By virtue of (iii) in Lemma 7.1.1, Li is of Etype. If L®= ®Kij for multiplicatively indecomposable positive lattices Kid, then applying Corollary 7.4.1 repeatedly to Li, Ki3 = Kid is of E-type and indecomposable. Taking Kid as L in Theorem 7.4.1, we obtain the indecomposability of L = ®Li = ®ijK13. Now we can apply Corollary 7.4.1 to the above L to show Mi is of E-type, [Mi Mi] < oo and Mi is indecomposable; so Li, Mj satisfy the same properties. Hence, without loss of generality we may assume by scaling that min(Li) = min(Mi) = 1. :

Changing suffices and considering or-' if necessary, we may further assume (36)

(37)

rank L1 > rank Li, rank Mj

(1 2Mj

such that v(ei®x) = ai(e1')®o2(x) for x E ®i>2Li. Suppose B(ei, ei) Let ei E M(Li) (i > 2); then (40) gives Q2(e2(9 ...0 em) =S0'2(e2®...®em) for 6 =f1.

0.

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Then B(ei, ei) = B(el ®e2

®... (& e,,,,, ei ®e2

®... ®e,,,,)

B(ei, ei)SQ(e2 ®... ® e,,,,) = SB(ei, ei)

so S = 1. Thus we have a2 = 0-2 on ®i>2M(Li) and hence on ®i>2Li if

B(ei,ei) # 0. As L1 is indecomposable, for any given e1,ei E M(Ll), there exist z1 = e1, z2, ,Zr = ei(zi E M(Li)) such that B(zi, zi+1) # 0 for i = 1, , r - 1. Hence 0`2 is independent of the choice of el E M(L1). So we have proved a(x 0 y) = a,(x) 0 a2(y) for x E M(Li), y E ®i>2Li and then al : L1 = M1, 0'2 : ®i>2Li ®j>2Mj imply or = o-1 0a2 on L10(®i>2Li). Our theorem is now proved inductively.

0 Corollary 7.4.2. Let Li be those in Theorem 7.4.3 and assume that if Li is isometric to a scaling of Lj, then Li - Lj. Then O((&Li) is generated by O(Li) and interchanges of isometric components. Proof. We have only to put Mi = Li in the theorem.

Remark. Lattices in Example 1 satisfy the assumptions of Theorem 7.4.3.

Exercise 1. Let Li be binary indecomposable lattices with min(Li) = 1; show O((DLi) is generated by O(Li) and interchanges of isometric components. (Hint: Show that the binary submodule in ®Li with minimal discriminant is of the form el 0 . . . 0 ej_1 0 Lj 0 ej+l ®. . . for ei E M(Li). Theorem 7.4.2 then yields that ®Li determines the components Li. Suppose without loss of generality d L1 < d Li for i > 1 and L1 = = Lk 99- Li for i > k.

If ®i>kLi = L1 ® K for some K then min K = 1 so ®i>kLi contains a submodule Li isometric to L1 and hence the minimal discriminant of the binary submodules in ®i>kLi is equal to d Li for some i > k and is greater than or equal to d Li = d L1. By the choice of L1, we have d L1 = d Li, which contradicts the assumption on k. Therefore (&i>kLi is not divisible by L1. Now use the condition (A3).)

Exercise 2. We say a positive lattice L is IR-type if for any positive lattice M, u ® v (u E M(L), v E M(M)) is irreducible, that is u®v = x+y and B(x, y) = 0 (x, y E L 0 M) imply x or y = 0. Show the following.

7.5 Scalar extension of positive lattices (a)

239

If tµt 2 > 1/2 for Hermite's constant utt for t < T, then a positive lattice L with rank L < T is of IR-type. (T > 114 follows from

Theorem 2.2.1.) If L is an indecomposable positive lattice of IR-type and [L : L] < 00, then L OM is indecomposable for any indecomposable positive lattice M. (Hint: Imitate [Ki8].) (b)

7.5 Scalar extension of positive lattices Assuming elementary theory of algebraic number fields we treat the following problem: Let L and M be quadratic modules over Z and F be an algebraic number

field with the maximal order RF. Then does RFL = RFM imply L = M? As we noted, this does not make sense unless L and M are positive lattices and F is a totally real algebraic number field. But under these conditions the problem becomes interesting. In this section we give some partial answers. The maximal order of an algebraic number field F is denoted by RF. Lemma 7.5.1. Let F be a totally real finite Galois extension of Q. Then the following are equivalent: (i) (ii)

For positive lattices L and M, a : RFL = RFM yields or(L) = M. Let G be a finite group in GLn(RF) such that g(A) := (g(aij)) E G, with A (aid) E G, holds for every g E Gal(F/Q). Then G C GLn (7L) .

Proof.

(i) = (ii) Let G be as in (ii) and put P = >AEG tAA. The as-

sumption {g(A) I A E G} = G for every g E Gal(F/Q) implies we have

g(P) = P for every g E Gal(F/QZ). Since tAA is positive definite for a real regular matrix A, P is positive definite with rational entries. Now we can define a positive lattice L = 7L[ul, , un] by (B(ui, uj)) = P. Using tAPA = P for A E G, the mapping a defined by

(a(ul), ... , a(un)) = (ul, ... , un)A becomes an isometry of RFL. Applying (i) to L = M, a, we have or (L) = L,

that is A E GLn(Z). (ii) = (i). Let L, M, or be as in (i). We define an isometry

rl E O(RF(L 1 M)) by 77 = a on RFL and v-1 on RFM. Let {ui}m 1 be a basis of L 1 M over Z, and define a matrix A = A,1 for q E O(RF(L 1 M)) by (1)

(77(ul), ... gl(um)) = (ul, ... , um)A.

240

7


Then the following are clear: A E GL n(RF), (B (u2, uj)) [A] = (B(uti, u;))

(2)

Put G := {A I A satisfies (2)}; since B(ui, uj) E Q, g(A) E G for g E Gal(F/Q), A E G. Moreover (B(ui, uj)) is positive definite and F is totally real. Therefore G is a finite group, and then (ii) implies A E G C GLm(Z). This means g(L I M) = L 1 M, and so a(L) C (L 1 M) f1 RFM, yielding

a(L) C M, and similarly a-1(M) C L. Thus we get a(L) = M. We note that for a linear algebraic group G defined over Q, the points G(RF) of G with coordinates in RF satisfy gG(RF) = G(RF) for g E Gal(F/(Q).

Lemma 7.5.2. Let K/Q be a finite Galois extension, p, q different rational primes, and p, 4 prime ideals of K which lie on p, q respectively. Let G be a finite subgroup of GLn(RK) such that G is fixed by Gal(K/(Q) as a set where a E Gal(K/Q) acts entrywise. If g E G, g - in mod 4, then ag = g for every a in the inertia group of p. Proof. First we claim that if g - In mod 4 for g E G, then the order of g is a power of q. Suppose not. Then considering gq' (- In mod 4)

instead of g, we may assume that the order h of g is greater than 1 and is relatively prime to q. We consider the completion field Kq of K by 4, and denote a prime element of Kq by 7r. Write g = In + 7rrgl where r > 1 by assumption and gl is an integral matrix over Kq such that gl # 0 mod ir. yields Then In = 9h = In + Eh=1 h

h7rrgi = -

h

G2) zrrjg1 = 0 mod 7r 2r

j=2 3

and hence gl - 0 mod 7rr which contradicts the defining property of gl. Thus the claim has been verified. Take an element a of the inertia subgroup

1'o (p) :_ {a E Gal(K/Q) I ax - xmodp for every x E RK}

and g E G with g - In mod

We put go = a(g)g-la(g)-1g E G. Then

go - In mod 4 follows from g In mod 4, and a E Fo(p) implies a(g) g mod p and hence go - In mod p. Applying the above claim to go and p, 4, we see the order of go is a power of both p and q. Since p and q are relatively prime, go ='n, and hence a(g) and g-1 commute. Therefore the order of a(g)g-1 is a power of q since the order of g is so. On the other hand, a(g)g-1 - 1n mod p implies the order of a(g)g-1 is a power of p. Thus a(g)g-1 = In, which is what we want.

7.5 Scalar extension of positive lattices

241

Lemma 7.5.3. Let K/Q be a finite Galois extension and denote by G a finite subgroup of GLn(RK) which is invariant under the action of Gal(K/(Q) as a set. If the following are satisfied, then G is in GLn(Z). (i) GfGLn(RF) C GLn(7L) is valid for every proper subfield F which is a Galois extension of Q. (ii) In K/QQ, at least two different rational primes p and q ramify. Proof. We may suppose q 0 2. Take a prime ideal 4 of K which lies above q and take g E G and a E Fo(4), the inertia subgroup of 4; then v(g)g-1 = In mod 4, and by the previous lemma, is fixed by elements of the inertia groups of prime ideals lying above p. Since such elements generate a normal subgroup H, we have a(g)g-1 E GLn(RF) where F is the subfield of K corresponding to H. Since p ramifies, F K and using the condition (i) we have a(g)g-1 E GLn(7L). Since o-(g)g-1 - In mod 4, we have a(g)g-1 In mod q and then by virtue of Lemma 7.2.2 it follows a(g)g-1 = ln. Thus we have shown that a(g) = g is valid for g E G, and o, E Fo(4) for every prime ideal 4 lying above q. Therefore v(g) = g is valid for g E G and every

element a of the normal subgroup H' generated by inertia subgroups of prime ideals lying above q. Since q ramifies, H' 0 {1} and then using the condition (i), we get g E GLn(ZL).

Lemma 7.5.4. Let K/Q be a totally real finite Galois extension such that only one rational prime ramifies and the Galois group Gal(K/Q) is nilpotent. Then Gal(K/(Q) is cyclic. Proof. We use induction on [K Q. If [K : Q2] = 1, there is nothing to do. Let p be a ramified prime. Put F = Gal(K/Q) and denote the centre of r by Z. Since r is nilpotent, Z {1} and F/Z is nilpotent. By the induction hypothesis, F/Z is cyclic and then F is abelian. Since p is a unique prime which ramifies in the totally real abelian extension K/Q, K is contained in Q(( + (-1) where S is a primitive pn-th root of unity for some n. It is known that Q2(( + (-1) is a cyclic extension of Q, and hence K/QZ is cyclic.

0 Lemma 7.5.5. Let L be a positive lattice and K a totally real algebraic We introduce a symmetric bilinear form BK(x, y) on RK by BK(x, y) = trK/Q(xy). If (RK, BK) or L is of E-type, then z E RKL number field. satisfying

Q(z) = min{Q(y) I y E RKL, 0 0 Q(y) E Q} is already contained in L.

Proof. We show first M((RK, BK)) = {fl}. For a non-zero v E RK, we have

BK(v,v) =trK/Qv2 > [K Q](NK/(Q

v2)1/[K:Q]

> [K Q.

7

242


then all conjugates of v2 are equal hence v2 = 1 and so v = ±1. Since K is totally real, (RK, BK) is clearly a positive lattice. We define a bilinear form t on RKL by B(x, y) = trK/Q B(x, y) for x, y E RKL. Then If BK(V, v) = [K

for a, b E RK, x, y E L, we have

B(ax, by) = trK/Q B(ax, by) = trK/Q(ab) B(x, y) where we use the same letter B for bilinear forms on L and RKL. Thus we get (RKL, B) = (RK, BK) ® L. Since (RK, BK) or L is of E-type,

M((RK, BK) (9 L) = M((RK, BK)) ® M(L) = {±1} ® M(L) and hence M(RKL, B) = .M (L) follows. Let z be as in the lemma. Since Q(z) E Q, we have Q(z) < min L. Then

B(z, z) = trK/@ Q(z) = [K: Q]Q(z) < [K: 42] min L = min((RK, BK) (9 L) = min(RKL, B)

and hence z E .M(RKL, B) = M(L) C L.

Theorem 7.5.1. Let K/Q be a totally real extension. Then for positive lattices L and M, a : RKL '-' RKM yields a(L) = M provided that rank L = rank M < 43 or (RK, BK) defined in the previous lemma is of E-type.

Proof. We note that L is of E-type by Theorem 7.1.1 if rank L < 43. Take xl E RKL such that

Q(xl) = min{Q(y) I y E RKL, 0# Q(y) E Q};

then by the previous lemma, we have x1 E L and similarly a(x1) E M. Hence a induces a : RK(X -) = RK(a(xl)'). Repeating the similar argument, we obtain x1, , x... E L such that L D 7x1 1 1 Zxn, (finite index) and a(xi) E M. Hence we get a(QL) = QM and so

a(L) =a(RKLnQL)=RKMn@M=M. 0


243

Theorem 7.5.2. Let K/Q be a totally real nilpotent Galois extension. Then for positive lattices L, M a : RKL = RKM yields or (L) = M.

Proof. We use induction on [K : Q]. If [K : Q] = 1, then we have nothing to do. By virtue of Lemma 7.5.1, we have only to verify the condition (ii) there. Noting that every subfield of K which is a Galois extension of Q is a nilpotent extension of Q, Lemma 7.5.3 completes the proof of the theorem if at least two different rational primes ramify in K/Q. Hence we may suppose that the only one prime can ramify; then by Lemma 7.5.4, K is a cyclic extension and from Theorem 7.1.3, (RK, BK) is of E-type. Theorem 7.5.1 completes the proof of the theorem.

Exercise 1. Let m be a natural number > 1. Show there exists a finite set Em of algebraic numbers with E,,,, fl Q = 0 depending only on m such that if K is a totally real algebraic number field with [K : (Q] = m and K fl Em,,, = 0, then v : RKL = RKM yields v(L) = M for positive lattices L, M. (Hint. Let x E RKL satisfy Q(x) = min{Q(y) I RKL 3) y # 0, Q(y) E Q}

and write x = E2"_l ciu2(ci E RK, ui E L) where (B(u2, uj)) = a[n] E S413,112 (a Siegel domain) for a diagonal matrix a and an upper triangular

matrix n with diagonals being 1. Then Q(x) = a[nc] < al holds for c = t(cl, , and a = diag(al, , a,,,,). Since Q(x) E Q, the inequality yields ai x (conjugate of i-th entry of nc)2 < al. Hence conjugates of entries

of nc are bounded and similarly for c, and so they are roots of a finite number of polynomials over Z. Denote by E,,,, the set of non-rational roots. Then c is a rational vector, that is x E L; finally use the argument of Theorem 7.5.1.)

Exercise 2. Let E be a totally real Galois extension over Q and suppose that all Sylow subgroups of the Galois group Gal(E/Q) are cyclic. Show O(REL) = O(L) holds for any positive lattice L. (Hint. See [BK].)

Exercise 3. Let K1, K2 be totally real algebraic number fields with their discriminants being relatively prime. Suppose that O(RK;L) = O(L) holds for every positive lattice and i = 1, 2. Show 0(RK1K2L) = O(L) holds for every positive lattice where K1K2 denotes a composite field of K1, K2. (Hint. RK1K2 = RK1 ® RK2 if the discriminants of K1, K2 are relatively prime.)

The next result is concerned with indecomposability under the scalar extension.

244

7


Theorem 7.5.3. Let E/F be a Galois extension of totally real algebraic number fields such that F is the unique unramified (over F) intermediate subfield of E. Let V be a quadratic space over F such that V, = F ® V is positive definite for the completion F of F at every infinite place v of F. Let L be a lattice on V, that is FL = V an L is finitely generated over RF. If L is indecomposable, then REL is also indecomposable.

Proof. Before the proof, we recall three facts:

(F1) L, REL are positive definite for every infinite place, and then the generalization of Theorem 6.7.1 holds, that is if

for indecomposable submodules Li, Mj, then m = n and {Li 1 < i < n} = {Mi 11 < i < n}. It is also true for REL. (F2) Let U be a vector space over an algebraic number field K, and N a finitely generated submodule of U over RK. Then there exist u1,

, U. E U

such that N = RKU1® . ® RKU,._1® Au, where A is an integral ideal of K such that NK/@(A) := [RK : A] is relatively prime to any given natural number. (F3) The integral ideal of F generated by (det where a1, , a, E

RE (n = [E : F]) and xU) means the j-th conjugate of x, is called the relative discriminant d(E/F) of EIF, and ElF is unramified if and only if d(E/F) is RF. Let us begin the proof of the theorem.

Let L be a lattice on V. Suppose that L is indecomposable but REL is decomposable. We have only to prove the existence of the intermediate

field K such that K # F and K is unramified over F, which contradicts the assumption. Write

(m>1)

(3)

where Li's are indecomposable RE-modules for i = 1,

, m. We make

G:= Gal(E/F) act on EV by g(av) = g(a)v

for

a E E, v E V and g E G.

We note that for w E EV, g(w) = w for all g E G yields w E V. For x

aivi, y = E biwi (ai, bi E E, vi, wi E V), we have

g(B(x, y)) =g(J: aibjB(vi, wj)) i, j

_ 1: g(ai)g(bj)B(vi, wj) = B(g(x), g(y)) i,j


245

Hence we have REL = g(REL) = g(L1) I. 1 g(L,,,,) for g E G. Since Li is an RE-module, g(Li) is so for g E G. By (3),

{g(Li) 11< i< m} _ {Li 11< i< m}. We show that G acts transitively on {Li 1 < i < m}. Otherwise, changing suffices if necessary we may suppose that G acts on {Li 1 < I

i 0}, and T runs over all symmetric integral non-negative matrices of size m. Then the theta function f (z) := 0,,,, (z, S) is a modular form of degree m, weight n/2, that is it satisfies the following three properties: (i) f (z) is holomorphic on Hm. (ii) f (z) satisfies (f I

(a

where I

d ))(z) := det(cz + d) -n/2 f ((az + b)(cz + d)-1) = r f (z), ac

d

I

is in

Sp(m,Z) := {M E SL(2m,Z) I tMJM = J}

for J = (?m

), with c =- 0 mod q where q := 41SI.

(a, b, c, d

are m x m matrices and cz + d is always regular in this case.) s; is a complex number with I K I = 1 dependent on a, b, c, d. (iii) For any ME Sp(m,Z), (f IM) (z) has a Fourier expansion

E h(T, M)exp(2iri trTz/q) T

for some complex number h(T, M) where T runs over all m x m nonnegative symmetric matrices such that the scale s((T)) C Z.

Notes

256

A function which satisfies these three conditions is called a modular form of weight n/2, degree m and level q. Then Problem 6 is equivalent to

Problem 6'. When does the theta function 8,,,,(z, S) determine the isometry class of (S) ? Corollary 6.4.1 shows that this is true if det S is fixed and n = m + 1. Is it always true with n = m + 1 ? This is true when n = 2, m = 1. A related problem is

Problem 7. Let Si be integral positive definite matrices of degree n and assume that quadratic lattices (Si) are not mutually isometric. When are 6,,,,(z, Si) linearly independent ?

This is true if n = m + 1 and det Si are independent of i by virtue of Corollary 6.4.1. Schulze-Pillot taught me that this is false when n = m + 2 and (Si) runs over a complete set of representatives of the classes in the

genus of L = (1) 1 (172) 1

1 (172k) (2k = n). It is based on the

Remark in §8 and

"2

0(O+(Lp)) _

(mX

17 )2

ZZP11

for p = 2, for p = 17,

forp#2,17,

Because, these imply that gen(L) contains two different spinor genera, and Siegel's weighted sum of theta functions in a spinor genus is independent of the choice of a spinor genus. How about the case when (Si) runs over a fixed spinor genus ? Incidentally it is a problem to generalize the Siegel formula for spinor genus (c.f. Remark in §8 and [SP21). The number of isometry classes of positive definite integral matrices with det = D, degree = n is bounded by Din-1i/2 from below and by D(n-1)/2+E for any positive number a from above up to a constant, which is a corollary

of Theorem 6.8.1 [Ki2], [Bi]. On the other hand, the main contribution in the dimension formula of the above modular forms is supposed to be

Ia d E Sp(m,Z) c - 0 mod q}. So, the main part (up to a constant) of (this should be [Sp(m,Z) : Po(q)], which is almost equal to If the the volume of the fundamental domain of H,,,, by ro(q)

I

I

q'(-+l)/2.

number of non-isometric quadratic forms is greater than the dimension of the modular forms, then we get a linear relation of theta functions. Hence, if n - 1 > m(m + 1), then we will get ample linear relations among these

theta functions. What happens in the case of n - 1 < m(m + 1) ? For example, as a refinement of Problem 6',

Notes on Chapter 6

257

Problem 8. If n - 1 < m(m + 1), then does the theta function 9,,,(z, S) determine the isometry class of S ? If n - 1 > m(m + 1), then are there non-isometric positive definite quadratic forms which have the same theta function ? When m = 1, the problem is solved affirmatively [Shio], [CS3], [Schi],

[Schi2]. A finite number of examples such that non-isometric positive definite quadratic lattices have the same theta function is known when n = 4, m = 1 ([Schi], [Shio]). Are there infinitely many such examples and if so, what characterizes them ?

Problem 9. Let m be a natural number. Then, what is the smallest number n which satisfies the following? There exists a positive definite quadratic lattice N of rank N = n so that any (possibly with finitely many exceptions) quadratic lattice M of rank M = m which is locally represented by N is represented by N.

Of course, if gen(N) = cls(N), then the answer is affirmative. But it is known that the number of such N is finite, and if n = 3 and m = 1, then there are only a finite number of lattices N which do not have exceptions above. For recent developments, see the article of M-H. Kim in the Proceedings of the International Conference on Integral Quadratic Forms and Lattices in Korea (1998).

Suppose that N is a quinary positive definite quadratic form with Q(N) C 2Z and dN = 20. Then gen(N) contains two classes N1, N2 whose quadratic forms are:

Q1 /(> xivi) =2 (x1 + x2 + x3 + x4 + 2x5 - x1x2 - x2x3 - X3X4), Q2

xi Ui) = 2 (x1 + x2 + x3 + x4 + 3x5 - x1 x2 - x1 x3 - x1 x4

- x1x5 + x2x5 + x3x5).

For these, the exceptional binary quadratic forms seem to be finite. To show the existence of an isometry from M to N, it is important to study the totality of all the numbers r(M, N) of isometries from M to N. Because the generating function 9,,,.(z, S) is a modular form, we can make use of the theory of modular forms. If n > 4m + 4, then we can give the asymptotic formula for r(M, N) and in particular r(M, N) > 0 under the assumption of the local representability and min M being sufficiently large. Lastly, we briefly refer an analytic approach to Theorem 6.6.2. Let S

be a positive definite integral matrix of degree n and {Si} a complete representative of the classes in the genus of S. Put f (z) := 0" (z, S) w(S)-1 Ei E(Si)-16m(z, S). Then f (z) is a modular form for which h(0, M) = 0 for every M E Sp(m, Z) for h in the third condition of the

258

Notes

definition of a modular form. f (z) is a so-called cusp form if m = 1,

but it is not so if m > 1. Put a(T) := w(S)-1 Ei E(Si)-lr(Si,T), and f (z) = > b(T)exp(2iri tr Tz). A basic strategy is to show b(T)/a(T) --> 0, which implies an asymptotic formula for r(S, T) = a(T) (l + b(T)/a(T)). When m = 1, we get another proof of Theorem 6.6.2. This works even for n = 3, 4. The case of n = 4 = 2m + 2 is due to Kloosterman by introducing so-called Kloosterman's sums, and to Eichler by solving Ramanujan's conjecture. The case of n = 3 = 2m + 1 involves Linnik-Selberg's conjecture [DS]. So, Theorem 6.6.2 is still valid with some modifications in the

caseofm=1,n=3,4. The case of m = 2 and n > 2m + 3 = 7 is also successful and we have another proof of Theorem 6.6.2, and a partial result

for n = 6 = 2m + 2. But we do not have a complete solution in this case

nor any results for n = 5 = 2m + 1. In the case of m > 3 we do not have a complete analytic proof of Theorem 6.6.2. This strategy is hard (in particular for the case n < 2m + 2) but very attractive [Ki21]. Here we give a more general problem than Problems 2 and 3. Let N be a positive quadratic lattice with rank N = n. Let m be a natural number. For a positive definite quadratic lattice M with rank M = m, there is a matrix p = a[n] such that M = (p) where p = a[n] E S4/3,12 is a decomposition as in Chapter 2 with a E A4/3 and n E N112. For any bounded domain in 54/3,1/2, there are only finitely many positive definite integral matrices, and for each M it can be checked whether M is represented by N or not, although it may take much time even by using a computer. So, we want to find a good path to the boundary. Here, that a path tends to the boundary means a1 = minM -> oo, al/a2 , or am_1/am -p 0. Note that 0, N112 is a compact set. Let us call a path asymptotically good if and only if we have an asymptotical formula of the representation number r(M, N) when M tends to the boundary along the path, and let us call a path good if and only if we have r(M, N) > 0 when M tends to the boundary along the path. Now we spell out:

Problem 10. What is a(n) (asymptotically) good path for N? From this view point, Theorem 6.6.2 shows that min M -> oo is a good path if n > 2m + 3 and it is an asymptotically good path if n > 7 for m = 2 [Kill]. Probably min M - oo is an asymptotically good path if n > 2m+3.

When n = 4 and m = 1, t -- p oo for M = (t) is an asymptotically good path with the restriction ap(M, N) > , for any fixed positive number k and any prime p f o r which M p is anisotropic. The condition c (M, N) -. 0 gives an example of the invalidness of Theorem 6.6.2. So, the condition

c (M, N) > K(> 0) may mean that M does not tend to the boundary at a finite prime p. When n = 6, m = 2, the path M = (tB) t -> oo with (t, dN) = 1 is an asymptotically good path [Ki17], and a1 = min M -> 00 and a1 /a2 < al 30.2 is a good path. [HJ] gives also a good path as referred

Notes on Chapter 6

259

above. Theorem 6.6.2 gives a wide path, but others are rather narrow. The author does not know any example of a good but not asymptotically good path. We may find a counter-example to Theorem 6.6.2 in the case of n < 2m + 2 in the sequence of M such that c (M, N) -> 0 or the infinite product fly, ap (M, N) -; 0. Siegel's formula for indefinite quadratic forms is valid with some modifications. The orthogonal group of an indefinite quadratic form N is an infinite group unless QN is a hyperbolic plane. So, the left hand sides of the equalities of Theorem 6.8.1 are meaningless. However, it holds under some modifications. This involves some geometry, which we do not pursue. For modern approaches to Siegel's theory, see [We2], [S]. A generalization of Siegel's formula given in [Bi], [Ki15] is false when the ranks are equal. Siegel's formula, Theorem 6.8.1 implies that the number of the different classes in gen(N) is almost the same as 2w(N) when rank N is fixed [Ki2], [Ki3], [Bi]. But we do not know much when rank N varies. For example,

Problem 11. Let Nn be a unimodular positive definite lattice with rank Nn = n. Is the class number of N,,, almost the same as 2w(Nn) ? There are many lattices in gen(N,,) with only trivial isometries if n is large [Ba]. But it is not known how many lattices have only trivial isometries if n = 32 and n(Nn) = 2Z . Finally, let us introduce a problem related to congruent numbers. A positive rational number n is a congruent number if and only if there exist positive rational numbers x, y, z satisfying x2 + y2 = z2 and xy/2 = n. The

problem is: Let n be a square-free and odd (respectively, even) natural number. When does 20{(x, y, z) E Z3In = 2x2+y2+32x2} = #{(x, y, z) E Z In = 2x2+y2+8z2} (respectively, 20{(x,y, z) E Z3In/2 = 4x2 + y2 + 32x2} _

#{(x, y, z) E 7131 n/2 = 4x2 + y2 + 8z2})

hold? ([Kob] ) Other comments:

Theorem 6.3.1 shows how to calculate the number of proper spinor genera in a genus. In fact, there is a natural isomorphism between the quotient groups QA/ MIES Q,1, 9(O+(V))9(OA(L)) and ter group has in its denominator an open subgroup containing the principal ideles. This fact leads to a class field perspective for the arithmetic of spinor genera. For further details see [EH]. QA/QXo(OA(L)). The lat

260

Notes

Theorem 6.5.1 has now been extended to all codimensions (including codimension zero). In particular, the set of spn+ C gen(L) representing K has the algebraic structure of a homogeneous space with respect to a subgroup of the adele group 0,+9(V). See [HSX], [H2].

Chapter 7 The results in §1 show that there exists ample positive lattices of E-type. However, an example which is not of E-type is known at rank = 292 by Steinberg (see p. 47 in [MH].)

Problem 1. Improve the number 43 in Theorem 7.1.1, and construct and study more examples not of E-type. Related to §5,

Problem 2. Let E be a finite Galois extension over Q and suppose that the complex conjugation induces an element of the center of the Galois group Gal(E/Q). Then (RE, BE) (BE(x, y) := trE1Q(xy)) is a positive lattice. Is it of E-type? So far, there is no counter-example.

Let us give several problems related to the tensor product. The basic one is

Problem 3. Is there any positive lattice L which does not satisfy the condition (A) in the Fundamental Lemma? In particular, what about the case of rank L = 3 ?

Problem 4. Is there any example for which the uniqueness of decomposition does not hold? Even for L = M 0 N with rank M = 2, rank N = 3, it is not known whether M and N are determined uniquely by L up to scaling or not.

Problem 5. Characterize lattices which do not satisfy the condition (A), if they exist. As we saw, Lemma 7.5.5 is critical. We propose a natural problem:

Problem 6. Is Lemma 7.5.5 true without the assumption that (RK, BK) or L is of E-type ? Problem 7. If a positive lattice L satisfies the condition (A) in the Fundamental Lemma, does RFL satisfy the similar condition over RF? More generally, does the Fundamental Lemma hold over RF for a totally real algebraic number field F and lattices which are positive definite at every infinite place of F?

Notes on Chapter 7

261

As we will see below, the existence of unramified extensions seems to be crucial.

Problem 8. When a totally real algebraic number field F does not have any totally real unramified extension, is there any example of L which is positive definite at every infinite place such that L does not satisfy the condition (A)? In particular, to what extent can one generalize the results in this chapter to Q(/5)? (It is known that it has no unramified extension.)

Problem 9. F = Q( 10) has a non-trivial totally real unramified extension (the maximal (not necessarily abelian) unramified totally real extension is F(/), which the author learned from Prof. Washington). (As we will see below, some of the results in this chapter are false in general.) For a Galois extension of totally real algebraic number fields E/F, look for any examples independent of the unramified extensions for this special field.

Problem 10. If L is a multiplicatively indecomposable positive lattice, is RFL so for a totally real algebraic number field F? In particular, what about the lattices in Example 1 in §1? Here we considered the scalar extension of positive definite quadratic forms to totally real algebraic number fieds. But it is also possible to do so for positive definite Hermitian forms [Ki10], and the problem is formulated as

Problem 11. Let K be a Galois extension over Q with Galois group r and let RK be the maximal order of K. Let G be a finite r-stable subgroup of GL,, (RK), that is for g = (gij) E G and v E r, we have (a(gij)) E G. G acts on RI canonically. Are there a direct sum decomposition Z = E 1Li such that for g E G there exist roots of unity ei(g) E K (i = 1, , k) and a permutation s(g) of {1, , k} satisfying ei(g)gLi = Ls(g)i? If the answer to this is affirmative, then for positive lattices L, M and a Galois extension K/Q with the complex conjugation being in the centre of Gal(K/@), a : RKL = RKM as Hermitian forms yields L = M, and if K does not contain roots of unity except ±1, we have a(L) = M. So, we can take the superfluous assumption off Theorems 7.5.1 and 7.5.2. The answer to problem 11 is affirmative if Gal(K/Q) is nilpotent [KS], n = 2 [Ki23] or n = 3 [Su]. In this chapter, we considered the case that the base ring is the rational

integer ring Z. If the base ring is not Z, then the answers to problems are negative in general. Let us explain it. Let E/F be an unramified Galois extension of totally real algebraic number fields with Galois group Put V F[G] (group ring) and we define an inner prodG=

uct B(,) by B(g, g') = Sg,g' (Kronecker's delta) for g, g' E G. Then V is positive definite at all infinite places of F. We extend the operation of

262

Notes

G to EV = E[G] by g(>2 aigi) = >2 g(ai)ggi (g E G, ai E E), and put L' :=1gEG REg, L :_ {>29EG g(a)g a E RE} where RE is the maximal order of E. Then for al, , a,, E RE we have >29EG g(an)g E L and t(>gEG g(ai)g, ... , >2gEG g(af.)g) = (gi (ai))t(gi, ... , g,,). Since E/F I

is unramified, det(gi(aj))2 generates RF as an ideal. Hence

L'=REL holds. Let us show that L is indecomposable. Suppose L = Ll 1 L2

0), then L' = REL1 1 REL2 and since REL1 is an orthogonal sum (L1 of REg (g E G) by a generalization of Theorem 6.7.1, GREL1 contains G and so GREL1 = L'. However, G operates trivially on L and so L' _ GREL1 = REL1. This means L2 = 0, that is L is indecomposable. Put M =1gEG RFg; then REM = L' = REL but M L since M is decomposable. Let (RE, BE) be the quadratic lattice over RF with BE(x, y) = trEIF xy; then R'E := (RE, BE) is positive definite at every infinite place and taking trace of REM = REL, we have R'E (&M = R'E ®L. On the other hand, R'E E) a '--> >2gEG g(a)g E L gives an isometry of R'E and L. Thus we have an indecomposable lattice L and a decomposable one

M such that L®L=L®M but LAM. Problem 12. In the above example, we constructed lattices L, M and N (:= L) such that L ®M = L ®N does not imply M = N, making use of the existence of unramified extensions of the base field. Is there any such example independent of the existence of unramified extensions of the base field ?

Let E/F be an unramified Galois extension of totally real algebraic number fields; then as above, we can construct an indecomposable lattice L

over RF such that REL is decomposable and V := FL is a quadratic space over F which is positive definite at every infinite place of F. Write REL = Ll 1 L2 (Li 0 0) and define an isometry or E O(REL) by o = id. on L1, = -id. on L2. Since L is indecomposable, v O(L) holds. Hence O(REL) = O(L) is false. If conversely there is a special isometry or E O(REL) such that Q2 =

1,a(x) - x mod 2REL for X E REL and or 0 ±1, then REL = {x E REL I Qx = x} 1 {x E REL I Qx = -x} is clear and by virtue of Theorem 7.5.3, there exists an intermediate field K(# F) which is unramified over F. So a natural but rather vague problem is Problem 13. What is the role of the existence of an unramified extension of a field F when we consider the problems in this chapter?

References

[AK] G.E.Andrews and Y.Kitaoka, A result on q-series and its application to quadratic

forms, J. of Number Theory 34 (1990), 54-62. [A] T.Asai, The class numbers of positive definite quadratic forms, Japan. J. Math. 3 (1977), 239-296. [Ba] E.Bannai, Positive definite unimodular lattices with trivial automorphism groups,

Memoirs of A.M.S. 85 (1990). [Bar] H.-J.Bartels, Zur Galoiskohomologie definiter arithmetischer Grappen, J. Reine

Angew. Math. 298 (1978), 89-97. [BK] H.-J.Bartels and Y.Kitaoka, Endliche arithmetische Untergruppe der GL",, J. Reine Angew. Math. 313 (1980), 151-156. [Bi] J.Biermann, Gitter mit kleiner Automorphismengruppe in Geschlechtern von Z-Gittern mit positiv-definiter quadratischen Formen, Dissertation Gottingen (1981). [B1] F. van der Blij, On the theory of quadratic forms, Ann. of Math. 50 (1949), 875-883.

[Bor] A.Borel, "Introduction aux groups arithmetiques," Hermann, 1969. [Bou] N.Bourbaki, "Algebra, Ch.9," Hermann, 1959. [BoSa] S.Bocherer and F.Sato, Rationality of certain formal power series related to local densities, Comment. Math. Univ. St. Paul. 36 (1987), 53-86. [BoSP] S.Bocherer and R.Schulze-Pillot, On a theorem of Waldspurger and on Eisenstein series of Klingen type, Math.Ann. 288 (1990), 361-388. [BS] Z.I.Borevich and I.R.Shafarevich, "Number theory," Academic Press, 1966.

[Cl] J.W.S.Cassels, "An introduction to the geometry of numbers," SpringerVerlag, 1959. [C2]

, "Rational quadratic forms," Academic Press, 1978.

[CEJ] W.K.Chan, D.Estes and M.Jochner, Representations of codimensions > 3 by definite quadratic forms, J. of Number Theory 71 (1998), 81-85.

References

264

[CS1] J.H.Conway and N.J.A.Sloane, "Sphere packings, lattices and groups," Springer-Verlag, 1988. [CS2I]

, Low-dimensional lattices. I. Quadratic forms of small deter-

minant, Proc. R. Soc. London A 418 (1988), 17-41. , Low-dimensional lattices. II. Subgroups of GL(n, Z), [CS2II] Proc. R. Soc. London A 419 (1988), 29-68. [CS2111]

,

Low-dimensional lattices. III. Perfect forms, Proc. R.

Soc. London A 418 (1988), 43-80. [CS21V]

,

Low-dimensional lattices. IV. The mass formula, Proc.

R. Soc. London A 419 (1988), 259-286. [CS2V]

, Low-dimensional lattices. V. Integral coordinates for integral

lattices, Proc. R. Soc. London A 426 (1989), 211-232. [CS3]

, Four-dimensional lattices with the same theta series, Inter-

nat. Math. Res. Notices (1992), 93-96. [D] L.E.Dickson, "Studies in the theory of numbers," Chelsea, 1930. [DS] W.Duke and R.Schulze-Pillot, Representation of integers by positive ternary quadratic forms and equidistribution of lattice points on ellipsoids, Invent. Math 99 (1990), 49-57.

[E] M.Eichler, "Quadratische Formen and orthogonal Gruppen," SpringerVerlag, Berlin, 1974. [EH] D.Estes and J.S.Hsia, Spinor genera under field extension.IV Spinor class fields,

Japanese J. Math. 16 (1990), 341-350. [G] L.J.Gerstein, Splitting quadratic forms over integers of global fields, Amer.J.Math.

91 (1969), 106-133. [Hi] Y.Hironaka, On a denominator of Kitaoka's formal power series attached to local

densities, Comment. Math. Univ. St. Paul. 37 (1988), 159-171. [Hi2]

, Spherical functions and local densities on hermitian forms, to

appear in Journal of Math. Soc. Japan 51 (1999). [Hi3]

, Local zeta functions on hermitian forms and its application to

local densities, J. of Number Theory 71 (1998), 40-64. [H1] J.S.Hsia, Representations by sinor genera, Pac. J. Math. 63 (1976), 147-152. [H2] J.S.Hsia, Arithmetic of indefinite quadratic forms, Proceedings of conference on

Integral quadratic forms and lattices in Korea (1998). [HI] J.S.Hsia and M.I.Icaza, Effective version of Tartakowsky's theorem.

[HJ] J.S.Hsia and M.Jochner, Almost strong approximations for definite quadratic spaces, Inventiones Math. 129 (1997), 471-487. [HKK] J.S.Hsia, M.Kneser and Y.Kitaoka, Representations of positive definite quadratic forms, J. Reine and Ang. Math. 301 (1978), 132-141. [HSX] J.S.Hsia, Y.Y.Shao and F.Xu, Representations of indefinite quadratic forms, J. reine angew. Math. 494 (1998), 129-140. [Jl] M.Jochner, Representation of positive definite quadratic forms with congruence and

primitive conditions. II, J. of Number Theory 50 (1995), 145-153. , Representation of positive definite quadratic forms with congruence and primitive conditions. III. [JK] M.Jochner and Y.Kitaoka, Representation of positive definite quadratic forms with [J2]

congruence and primitive conditions, J. of Number Theory 48 (1994), 88-101. [Kal] H.Katsurada, A certain formal power series of several variables attached to local densities of quadratic forms. I, J. of Number Theory 51 (1995), 169-209. [Ka2]

, An explicit formula for Siegel series.

References

265

[Kil] Y.Kitaoka, On the relation between the positive definite quadratic forms with the

same representation numbers, Proc. of Japan Acad. 47 (1971), 439-441. [Ki2]

, Two theorems on the class number of positive definite quadratic forms,

Nagoya Math.J. 51 (1973), 79-89. [Ki3]

, On the asymptotic behaviour of Brauer-Siegel type of class numbers of

positive definite quadratic forms, Proc. of Japan Akad. 49 (1973), 789-790. [Ki4] , Representations of quadratic forms and their application to Selberg's zeta functions, Nagoya Math. J. 63 (1976), 153-162. [Ki5]

,

Class numbers of quadratic forms over real quadratic fields, Nagoya

Math. J. 66 (1977), 89-98. [Ki6]

, Scalar extension of quadratic lattices, Nagoya Math. J. 66 (1977),

139-149. [Ki7]

, Scalar extension of quadratic lattices II, Nagoya Math. J. 67 (1977),

159-164. , Tensor products of positive definite quadratic forms, Gottingen Nachr. (1977), 1-14.

[Ki8] [Ki9]

Modular forms of degree n and representation by quadratic forms,

,

Nagoya Math. J. 74 (1979), 95-122. [Kil0] , Finite arithmetic subgroups of GLn II, Nagoya Math. J. 77 (1980), 137-143. [Kill] , Modular forms of degree n and representation by quadratic forms II, Nagoya Math. J. 87 (1982), 127-146. , A note on local densities of quadratic forms, Nagoya Math. J. 92 (1983), 145-152.

[Ki12] [Ki13]

, Fourier coefficients of Eisenstein series of degree 3, Proc. of Japan

Akad. 60 (1984), 259-261. [Kil4]

, Local densities of quadratic forms and Fourier coefficients of Eisen-

stein series, Nagoya Math. J. 103 (1986), 149-160. [Ki15]

, Modular forms of degree n and representation by quadratic forms IV,

Nagoya Math. J. 107 (1987), 25-47. [Ki16]

, Local densities of quadratic forms, Investigations in Number The-

ory (Advanced Studies in Pure Mathematics) 13 (1988), 433-460. [Ki17]

, Modular forms of degree n and representation by quadratic forms V,

Nagoya Math. J. 111 (1988), 173-179. [Ki18]

, Some remarks on representations of positive definite quadratic forms,

Nagoya Math. J. 115 (1989), 23-41. [Ki19]

, A note on representation of positive definite binary quadratic forms by

positive definite quadratic forms in 6 variables, Acta Arith. LIV (1990), 317-322. [Ki20]

,

"Lectures on Siegel modular forms and representation by

quadratic forms," Springer-Verlag, Heidelberg, 1986. [Ki21]

, Representation of positive definite quadratic forms and analytic num-

ber theory, to appear in English, in "Sugaku," Iwanami-Shoten, Tokyo, 1991, pp. 115-127. [Ki22] , Finite arithmetic subgroups of GLn. III, Proc. Indian Acad. Sci. 104 (1994), 201-206. [Ki23] , Finite arithmetic subgroups of GLn. V, Nagoya Math. J. 146 (1997), 131-148. [Knl] M.Kneser, Darstellungsmasse infefiniter quadratischer Formen, Math.Zeit. 77 (1961), 188-194.

References

266 [Kn2]

,

Witts Satz fur qiiadratische Formen fiber lokalen Ringen, Nachr. die

Akad.der Wiss.Gottingen,Math.-Phys.II 9 (1972), 195-203. [Kn3] , "Quadratische Formen," Vorlesungs-Ausarbeitung, Gottingen, 1973/74. [Kn4]

,

Representations of integral quadratic forms, Canadian Math.Soc.

(1983), 159-172. In "Quadratic and Hermitian forms". [Kob] N.Koblitz, Introduction to elliptic curves and modular forms, Springer-Verlag,

Graduate Texts in Mathematics 97 (1984). [Ko] O.Korner, Die Masse der Geschlechter quadratischer Formen vom Range < 3 in quadratischen Zahlkorpern, Math.Ann. 193 (1971), 279-314.

[KS] Y.Kitaoka and H.Suzuki, Finite arithmetic subgroups of GLn. IV, Nagoya Math. J. 142 (1996), 183-188. [La] T.Y.Lam, "The algebraic theory of quadratic forms," Benjamin, 1973. [Lan] E.Landau, "Primzahlen I," Chelsea, 1953. [Le] C. G.Lekkerkerker, "Geometry of Numbers," North-Holland, AmsterdamLondon, 1969. [M] M.Mazur, Finite arithmetic subgroups of GLn.

[MH] J.Milnor and D.Husemoller, "Symmetric bilinear forms," SpringerVerlag, Berlin-Heidelberg-New York, 1973. [MS] K.A.Morin-Strom, A Witt theorem for non-defective lattices, Can. J. Math 30 (1978), 499-511.

[0] O.T.O'Meara, "Introduction to quadratic forms," Springer-Verlag, Berlin, 1971.

[P] M.Peters, Darstellungen durch definite ternare quadratische Formen, Acta Arith.

XXXIV (1977), 57-80. [S] F.Sato, Siegel's main theorem of homogeneous spaces, Comment. Math. Univ. St.

Paul 41 (1992), 141-167. [Scha] W.Scharlau, "Quadratic and Hermitian forms," Springer-Verlag, Berlin, 1985. [Schil] A.Schiemann, Ein Beispiel positiv definiter quadratischer Formen der Dimension

4 mit gleichen Darstellungszahlen, Arch. Math. 54 (1990), 372-375. , Ternary positive definite quadratic forms are determined by

[Schi2]

their theta series, Math. Ann. 303 (1997), 507-517. [Shim] G.Shimura, Confluent hypergeometric functions on tube domains, Math. Ann

260 (1982), 269-302. [Shio] K.Shiota, On theta series and the splitting of S2(I'o(q)), J. Math. Kyoto Univ.

31 (1991), 909-930. [Si] C.L.Siegel, "Gesammelte Abhandlungen," Springer-Verlag, Berlin, 1966. [SP1] R.Schulze-Pillot, Darstellung durch Spinorgeschlechter terndrer quadratischer Formen, J. of Number Theory 12 (1980), 529-540. [SP2]

, Darstellungsmasse von Spinorgeschlechtern terndrer quadratis-

cher Formen, J. Reine and Ang. Math. 352 (1984), 114-132. [SP3]

, A linear relation between theta series of degree and weight 2, in

"Number theory, Lect. Notes in Math.," Springer-Verlag, Berlin-HeidelbergNew York, 1989, pp. 197-201. [Su] H.Suzuki, Finite arithmetic subgroups of GL3.

[W1] G.L.Watson, "Integral quadratic forms," Cambridge Univ. Press, London, 1960. [W2]

, Quadratic diohantine equations, Philos. Trans. Roy. Soc. London

Ser.A 253 (1960/61), 227-254.

References [W3]

267

, The 2-adic density of a quadratic form, Mathematika 23 (1976),

94-106.

[Wel] A.Weil, "Discontinuous subgroups of classical groups," University of Chicago, 1958. [We2] A.Weil, Sur la theorie des forms quadratiques, in "collected works," SpringerVerlag, Berlin-Heidelberg-New York, 1979, pp. 471-484. [Wi] E.Witt, Theorie der quadratischen Formen in beliebigen Korpern, J. Reine and

Ang. Math. 176 (1937), 31-44. [Y] T.Yamazaki, Jacobi forms and a Maass relation for Eisenstein series, J. Fac. Sci.

Univ. Tokyo 33 (1986), 295-310. [Yan] T.Yang, An explicit formula for local densities of quadratic forms, to appear in

J. of Number Theory.

Index

adele group, 132 anisotropic, 10 Arf invariant, 14 associated symmetric bilinear form, 3 canonical involution, 22 characteristic submodule, 153 class, 131 Clifford algebra, 20 connected, 217 connected component, 217 decomposable, 169 discriminant, 1, 70 divisible, 199 dual lattice, 71

finite place, 64 genus, 132

Hasse invariant, 56 Hermite's constant, 37 Hilbert symbol, 54 hyperbolic basis, 11 hyperbolic plane, 11 hyperbolic space, 11

indecomposable, 169, 222 indefinite, 129

indefinite (quadratic) lattice, 129 infinite place, 64 irreducible, 170 isometric, 5 isometry, 5, 217 isotropic, 10 Iwasawa decomposition, 34 Jordan component, 79 Jordan decompositon, 79 Legendre symbol, 52 lattice, 70, 129 linked, 57 local field, 47 local density, 94

majorant, 41 maximal, 72 minimal vector, 190 modular, 71 multiplicatively indecomposable, 223 negative definite, 129 negative (definite quadratic) lattice, 129

idele group, 132

norm, 25, 71

270

not divided, 199 orthogonal basis, 2 orthogonal sum, 2 p-adic integer, 48 p-adic number, 47 place, 64 positive definite, 129 positive (definite quadratic) lattice, 129

positive lattice, 190 primitive local density, 100 proper class, 131 proper spinor genus, 132 pure, 24

quadratic form, 3 quadratic module, 3 quadratic residue symbol, 52 quaternion algebra, 24 reducible, 169 regular, 2

Index regular quadratic lattice, 70, 129 represented, 5 scalar extension, 32 scale, 71 scaling, 28 second Clifford algebra, 22 Siegel domain, 34 Siegel-Eichler transformation, 12 spinor genus, 132 spinor norm, 29 split, 190 symmetric bilinear form, 1 symmetric bilinear module, 1 symmetry, 6

tensor algebra, 20 tensor product, 27 totally isotropic, 10 trace, 25 weight, 173 weighted graph, 217

Witt index, 11

Arithmetic of Quadratic Forms (Cambridge Tracts in Mathematics)

Arithmetic of quadratic forms

Arithmetic of quadratic forms

Invariants of Quadratic Differential Forms (Cambridge Tracts in Mathematics)

Divisors (Cambridge Tracts in Mathematics)

Convexity (Cambridge Tracts in Mathematics)

Divisors (Cambridge Tracts in Mathematics)

Sets of Multiples (Cambridge Tracts in Mathematics)

Compositions of Quadratic Forms (De Gruyter Expositions in Mathematics)

Invariants of quadratic differential forms

Invariants of Quadratic Differential Forms

Introduction to quadratic forms

The Lebesgue Integral (Cambridge Tracts in Mathematics)

The Lévy Laplacian (Cambridge Tracts in Mathematics)

Matrices and Quadratic Forms

Hopf Algebras (Cambridge Tracts in Mathematics)

Simple Noetherian Rings (Cambridge Tracts in Mathematics)

Quasi-Frobenius Rings (Cambridge Tracts in Mathematics)

The Lebesgue Integral (Cambridge Tracts in Mathematics)

Simple Noetherian Rings (Cambridge Tracts in Mathematics)

Proximity Spaces (Cambridge Tracts in Mathematics 59)

Forcing Idealized (Cambridge Tracts in Mathematics)

Totally Positive Matrices (Cambridge Tracts in Mathematics)

Multiple Forcing (Cambridge Tracts in Mathematics 88)

Totally Positive Matrices (Cambridge Tracts in Mathematics)

Forcing Idealized (Cambridge Tracts in Mathematics)

Asymptotic Expansions (Cambridge Tracts in Mathematics)

Fixed Point Theorems (Cambridge Tracts in Mathematics)

Quadratic Forms Over Semilocal Rings

Quadratic forms in infinite dimensional vector spaces

Synthetic Geometry of Manifolds (Cambridge Tracts in Mathematics 180)

Arithmetic of Quadratic Forms (Cambridge Tracts in Mathematics)