Lecture Notes in Mathematics
Edited by A. Dold and B. Eckmann
914
Max L. Warshauer
The Witt Group of Degree k Maps and Asymmetric Inner Product Spaces
Springer-Verlag Berlin Heidelberg New York 1982 Author
Max L.Warshauer Department of Mathematics, Southwest Texas State University San Marcos, TX ?8666, USA
AMS Subject Classifications (1980): 10 C 05
ISBN 3-540-11201-4 Springer-Verlag Berlin Heidelberg NewYork ISBN 0-387-11201-4 Springer-Verlag New York Heidelberg Berlin
This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under w 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to "Verwertungsgesellschaft Wort", Munich. by Springer-Verlag Berlin Heidelberg 1982 Printed in Germany Printing and binding: Beltz Offsetdruck, Hemsbach/Bergstr. 2141/3140-543210 TABLE OF CONTENTS
INTRODUCTION ...... 1
CONVENTIONS ...... ~ ...... II
Chapter I THE WITT RING ...... 12
i. Setting and notation ...... 12
2. Inner products ...... 14
3. Constructing new inner products out of old ...... 23
4. The symmetry operator ...... 26
5. The Witt equivalence relation ...... 33
6. Anisotropic representatives ...... 41
Chapter II WITT INVARIANTS ...... 47
i. Prime ideals ...... 48
2. Hilbert symbols ...... 51
3. Rank ...... 54
4. Diagonalization and the discriminant ...... 55
5. Signatures ...... 66
Chapter III POLYNOMIALS ...... 70
Chapter IV WITT GROUP OF A FIELD ...... 79
i. Decomposition by characteristic polynomial ...... 79
2. The trace lemma ...... 90
3. Computing Witt groups ...... 100
4. Torsion in W(-k,F) ...... 105
Chapter V THE SQUARING MAP ...... 108
I. Scharlau~s transfer ...... !ii
2. The exact octagon over a field ...... 123 IV
Chapter VI THE BOUNDARY ~..~ ...... , ...... 133
i. The boundary homomorphism ...... 133
2. Reducing to the maximal order ...... 142
3. Computing the local boundary ~(D,P) for the
maximal order ...... 152
4. Computing the cokernel of ~(D} ...... 167
Chapter VII NON MAXIMAL ORDERS ...... 184
i. Traces and canonical localizers ...... 184
2. Normal extensions ...... 192
3. Computing trace for finite fields ...... 198
4. The conductor and T(M~ ...... 203
Chapter VIII THE GLOBAL BOUNDARY ...... 207
i. The coupling invariants ...... ~. 207
2. The boundary is onto ...... 210
Chapter IX A DETAILED ANALYSIS OF THE OCTAGON ...... 226
i. The involutions ...... 226
2. The map I : We(k2,F) § We(-k,F) ...... 229 e
3. The map d s : We(-k,F) § A(F) ...... 238
4. The map Se: We(k,F) § We(k2,F) ...... 241
5. The map me: A(F) + We(k,F) ...... 242
Chapter X THE OCTAGON OVER Z ...... 249
NOTATION ...... 259
REFERENCES ...... 264
INDEX ...... 267 INTRODUCTION
In these notes our goal has been to develop the algebraic machinery for the study of the Witt group W(k,I) of degree k mapping structures and the Witt group of asymmetric inner product spaces. We are particularly interested in their relationship which arises in an exact octagon which is studied both for a field F and for the integers Z We show that this octagon is the appropriate generalization of the Scharlau transfer sequence [!~n 201].
We have tried to develop the properties of these Witt groups in a self-contained and complete manner in order to make this accessible to a larger audience, When references are given they generally are specific including page numbers. However, we should mention two general references. First, The Algebraic Theory of
Quadratic Forms by T.Y. Lam [Lm] develops the Witt group over a field of characteristic not equal to 2, Second, Note____~so_~n th___eeW itt
Classification of Hermitian Innerproduct Spaces over a Ring of Algebraic
Integers by P.E. Conner [C] discusses Hermitian forms and the Witt group over an algebraic number field and ring of integers therein.
Together these should provide any background material the reader might need.
Although the viewpoint we take is entirely algebraic, we would be remiss if we did not mention its topological motivation. Much of this work originated in our efforts to explain and exposit the important work of N.W. Stoltzfus, Unravellin~ the Integral
Knot Concordance Group [Sf-l]. We shall describe this topological connection momentarily. First however we need to describe the algebraic objects at hand.
Our object is to define and study a Witt group W(k,I) of
degree k maps, where k is an integer, D is the underlying ring,
and I is a D-module. This group consists of Witt equivalence
classes of triples (M,B,/) satisfying:
(i) B: M x M + I is an I-valued inner product defined on
the D-module M.
(2) /:M § M is a D-module homomorphism of M satisfying
B(/x,/y) = kB(x,y). We refer to 1 as a map of degree k.
A triple (M,B,/) is metabolic (Witt equivalent to zero) if there
is an /-invariant submodule NC M with N = N ~. Here N ~ is the
orthogonal complement. This enables us to define the Witt
equivalence relation ~ by:
(M,B,/) ~ (MI,BI,/I) iff (M ~ MI, B ~9 -B I, Z ~) /I )
is metabolic. The witt equivalence class of (M,B,/) is denoted
[M,B,/]. The operation of direct sum ~ makes this collection of
equivalence classes into a group W(k,I). The identity is the Witt
equivalence class of metabolic triples.
We also develop the basic properties of asymmetric inner
product spaces (M,B) where no symmetry requirement is placed on the
inner product B. The key to understanding these asymmetric inner products is a "symmetry operator" s satisfying B(x,y) = B(y,sx) for all x,y e M. As above we define the notion of metabolic by
(M,B) is metabolic if there is an s-invariant subspace N C M with N = N ~. This leads us to define Witt equivalence as before, and there results the asymmetric Witt group.
There is a very interesting relationship between this asymmetric
Witt group and the Witt group of degree k maps. This comes from the squaring map S: W(k,D) + W(k2,D) given by [M,B,/] ~ [M,B,/2].
For D = F a field an eight term exact sequence is developed from
S. This octagon involves the Witt group of asymmetric inner products
A(F) just described.
As a special case of the octagon we obtain the transfer sequence of Scharlau, Elman and Lain ~m 20~ and ~,L 23-2~. This appears as an exact octagon in which several terms vanish. We are able to reinter- pret the kernel and cokernel of this transfer sequence. Thus, these
Witt group W(k,F) and A(F) are an appropriate generalization of the classical Witt group W(F), at least in so far as relating to and explai- ning the Scharlau transfer sequence. We should like to prove exactness in the octagon over Z. in order to continue we have a boundary sequence which relates
W(k,Z) to W(k,Q)
O ~ W(k,Z) + W(k,Q) ~ W(k,Q/Z)
When k = • this sequence is short exact. Using this we are able to prove exactness in the octagon over Z when k = • by comparing the octagon to the octagon over Q. AS we have said, in these notes we are developing the algebraic machinery to study the Witt group of degree k maps and the asymmetric Witt group. We should discuss the topological motivation for this work. When k = +i the Witt group W(+I,Z) is the crucial element determining the bordism groups A n of orientation preserving diffeomorphisms of n-dimensional closed oriented smooth manifolds.
Medrano introduced the appropriate Witt invariant as follows.
Let f: M 2n + M 2n be an orientation preserving diffeomorphism of a
closed, oriented, smooth 2n-dimensional manifold M 2n. We consider
then the pair (M2n,f) in A2n. The degree +i mapping triple
associated to this pair is
(Hn (M2n;z)/torsion , B, f*)
where B(x,y) = e. ((x ~ y) ~ [M2n]) , E, is augmentation, [M 2n]
is the fundamental class, ~ is cup product, ~ is cap product,
f* is the induced map on cohomology. If (M2n,f) bounds, this
triple is metabolic. Thus there is an induced homomorphism
I: A2n + W(+I,Z).
The task of computing the bordism groups A2n. was completed
by Kreck [K] who showed that this Witt invariant was essentially
the only invariant for bordism of diffeomorphisms.
More generally, given a closed oriented 2n-dimensional manifold
together with a map 1 of degree k, the corresponding Witt triple
(M,B,I) as above satisfies B(s = kB(x,y) We are thus led to examine the Witt group W(k,Z). The Witt group of asymmetric inner product spaces arises in Quinn's work [Q] on open book decomposition.
The relation of this to the Witt group arising in Kreck's ~ork above
is discussed by Stoltzfus in The Algebraic Relationship Between
Quinn's Invariant of Open Book Decomposition and the Isometric
Structure of the Monodromy [Sf-21. in this he gives a geometric
application of the exact octagon obtained in Chapter X.
The exact octagon (renamed "the eight fold way") has also been
extended to the setting of the surgery obstruction groups by A.Ranicki,
L. Taylor, and B. Williams. For a further discussion of the use
of quadratic forms and the Witt group in topology the reader is
referred to Alexander, Conner, Hamrick, Odd Order Group Actions and
Witt Classification of Inner Products [ACH], and Stoltzfus
Unravelling the Integral Knot Concordance Group [Sf-l~.
We now describe briefly the organization of these notes. Chapter I
lays out the inner product spaces and Witt groups we will be studying.
We continue our study of the Wilt group in Chapter II by describing Witt
invariants which will be used to compute the Witt group in many cases.
These invariants include rank mod 2, discriminant, and signatures.
In Chapter III we study the characteristic and minimal polynomial
of the degree k map l in a degree k mapping structure (M,B,~).
This study is used in Chapter IV where we compute the Witt group
W(k,F) for F a field. This is done by decomposing w(k,F)
according to the characteristic polynomial of s and A(F)
according to the characteristic polynomial of s.
In Chapter V we develop an 8 term exact octagon which relates
W(k,F) to A(F). W+I (k,F) S-~ W+I (k2,F) § W+ 1 (-k,F)
A(F) A(F)
~" w-l(-k , F) + W_I(k2,F ) S§ W_I( k ,F)
S is the squaring map [M,B,/] § [M,B,/2]. We prove again the
Scharlau, Elman, Lam transfer sequence, and see that the exact octagon we develop is its appropriate generalization. In order to study this octagon over Z, we relate W(k,Z) to W(k,Q) by a boundary sequence in Chapter VI.
Now we place an additional requirement on the degree k mapping structure [M,B,/] , namely:
(*) 1 satisfies the monic integral irreducible polynomial f(t] .
The resulting Witt group of triples satisfying the additional requirement (*) is denoted W(k,Z;f) .
The action of l induces a Z[t]/(f(t)) - module structure on
M. To simplify the notation let S = Z[t]/(f(t)). Note that S is
only an order in the Dedekind ring of integers O(E) of the algebraic
,lumber field E = Q[t]/(f(t)) . This order S may not be the maximal
order O(E).
The first step to understanding W(k,Z;f) or W(k,Z;S) (the
same thing) is to study the group w(k,Z;D) for D = O(E) the
maximal order. This group consists of Witt equivalence classes of
inner product spaces (M,B) in which M is a finitely generated
torsion free D-module. This is in contrast to W(k,Z;S) in which
we only insist that the module structure of M lifts to the order
S = Z [t]/(f(t)) .
In Chapter VI we are interested only in W(k,Z;D) and the
resultant boundary sequence for the maximal order. We read this boundary sequence on the Hermitian level, where the - involution on E is given by t + kt -I and t -I + k-lt. One uses the following commutative diagram.
(O) O + H(A-I(D/Z)) § H(E) § H(E/A-I(D/Z))
Ct +t +t
0 + w(k,Z;D) ~ w(k,Q;D) ~(D) w(k,Q/Z;D)
~-l(D/Z) denotes the inverse different of D over Z. The vertical
isomorphisms denoted by t are induced by the trace of E over Q.
Thus the method employed for computing W(k,Z;D) is to study
the corresponding boundary sequence in the isomorphic Hermitian
case. The image of ~(D) is the group H(E/~-I(D/Z)) which is
computed as follows:
t g tr H(E/A-I(D/Z)) § W(k,Q/Z;D) ~ (~ w(k,Fp;D/P) § ~ H(D/P)
Here we sum over all - invariant maximal ideals P in D. The
isomorphism g is induced by selecting a generator I/p for the
p-torsion in Q/Z. The trace map on finite fields from D/P to
F induces the isomorphism tr with the Hermitian groups P H (D/~).
We use the letter M to denote - invariant maximal ideals in
S. In order to study W(k,Z;S) one must use the following
commutative diagram: 3 (D) 0 -~ H(A-I(D/Z)) + H(E) -~ H(E/A-I(D/Z)) % t r t ~(D) ~ t g~ tr 0 + W(k,Z;D) § W(k,Q;D) * W(k,Q/Z;D) + E) w(k,Fp;D/P)§ ~) H(D/P) fl + f2 ~(S) + f3 g tr % tr 0 + W(k,Z;S) § W(k,Q;S) + W(k,Q/Z;S) + E) W(k,Fp;S/M)§ ~) H(s/M) First one computes W(k,Z;D) for the maximal order by going to the Hermitian level and reading ~ (D) in the group ~) H(D/P).
Then one forgets the D-module structure and remembers only the
S-module structure via the maps f. to gain a computation for 1 W(k,Z;S).
Thus in Chapter VII we study non-maximal orders. Let us be explicit here in describing the key problems involved.
At every prime P in D there exists a canonically defined element p(F) in E/A-I(D/Z) with the following properties.
(i) The map of O(E) + E/A-I(D/Z) given by
+ I p(P) induces an embedding of the residue field.
(2) We also have the map Z ~ Q/Z given by
n § n/p which induces an embedding of
F = Z/pZ + Q/Z. P
The element p(P) is canonical in the sense that it makes the following diagram commute.
0 (EyP § E/A -I (D/Z)
%tr %t
F § Q/Z P
The horizontal maps were just described. tr again denotes the map induced by the finite field trace. t denotes the map induced by the number field trace. Thus we see that it is precisely these elements p(P) which determine the isomorphism tr -I o g o t identifying H(E/~-I(D/Z)) with (~ H(D/P). If we wish to use the commutative diagram just P = discussed to compute W(k,Z;S) , we must therefore study those elements p(P). For it is in terms of these elements that one reads the local boundary
~(D,P) : H(E) + H(E/~-I (D/Z)) + ~ H (D/P) + H (D/P) . P =
in such a way as to make our diagram commute. The last map is projection to the pth coordinate.
The first two Sections of Chapter VII are due to Conner. In these we present his theorems which develop the fundamental properties of these canonical localizers p(P). We complete our study of
W(k,Z;S) for non-maximal orders by discussing the finite field trace
tr, and the maps f~. l In Chapter 8 we finish our discussion of the boundary homomor- phism. This includes the notion of coupling from Stoltzfus [Sf-l]
between various ~(D), and a proof that the boundary
2: W(k,Q) § W(k,Q/Z) is onto when k = •
In Chapter IX the terms and maps in the octagon are studied
in detail. This, together with the information about the boundary
map enables us to prove exactness in the octagon over Z in
Chapter X. 10
The idea to study this problem and the possibilities inherent in the program we have undertaken comes from Professor P.E. Conner.
It is thus a pleasure to thank him for his help in this project with- out which the notes would ner have been written. The author feels fortunate to have had the opportunity to study under Professor Conner, whom we thank not only for his invaluable ideas, but also for his patience and understanding.
Further, for many of the ideas herein, he should receive credit.
The author is also grateful to Professor Stoltzfus for numerous conversations throughout this project. We also thank Professor
Dan Shapiro at Ohio State University for his help and suggestions, not only on this paper but also when the author was just beginning to study mathematics. We also express our appreciation to Professors
Cordes and Butts at Louisiana State University; to Professor
A. Liulevicius at the University of Chicago; and to Professor
A. Ross at Ohio State University for their interest in the author at various stages of his career.
Finally, the author wishes to express his gratitude to his wife, Hiroko, and parents Dr. and Mrs. Albert Warshauer who have given their constant support and encouragement. Chapter I THE WITT RING
In this chapter we define the algebraic structures which are to be studied. Section 1 begins by describing the setting for our inner product spaces, and includes a brief discussion of prime ideals and valuations. In Section 2 we develop the elementary properties of these inner product spaces. Section 3 shows how to construct new inner product spaces out of old. In particular the operations of direct sum and tensor product are discussed. These operations later become addition and multiplication in the Witt ring.
Since our inner products need not be symmetric we are led to examine a symmetry operator in Section 4. An inner product space comprises part of the data of a degree k mapping structure which is defined in Section 5. A Witt equivalence relation is then defined on these structures.
Section 6 is concerned with selecting from each Witt equivalence class a certain "anisotropic" representative. We show that in certain cases this representative is unique.
i. Setting and notation
We are interested in developing a Witt classification for triples
(M,B,Z) where B: M x M + Z is a Z - valued inner product and s M § M is a map of degree k. This means B(Ix,ly) = kB(x,y) for all x, y E M. In order to accomplish this one must investigate algebraic number fields and the Dedekind ring of integers in these number fields. Further, we must carefully study the role of non- maximal orders in the Dedekind ring of integers. We begin by des- cribing the setting. 13
Let D be a Dedekind domain [O'M 52] together with an involution
- The quotient field of D is E. The involution - extends to a
Galois automorphism of E and we denote the fixed field of - by F.
It may happen that F = E; in fact this is precisely the case when - is the trivial involution.
We shall also use the symbol O(E) to denote the Dedekind ring of integers D in E. Of course O(F) = O(E) ~ F. Let S be an order in D. This means S is a subring of D, containing i, with the same rank as D [B 88]. We shall be particularly interested in principal orders S of the form S = Z[t,t-l]/(f(t)) = Z[@], where f(t) is a monic, integral irreducible polynomial. Thus S is an order in O(E), where E is the algebraic number field Q[t]/(f(t)). Although our pri- mary interest is in these principal orders in algebraic number fields, the theory we shall develop applies to arbitrary orders in a Dedekind domain.
Let p be a prime ideal in O(E) = D. Then p/~ O(F) = P will denote the corresponding prime ideal in O(F) and p~ S = M will be the corresponding prime ideal in S. The image of p under the
involution - is denoted by p . p is also prime, hence maximal since
D is a Dedekind domain [O'M 49]. Let I be a - invariant frac-
tional ideal in D. Since I factors uniquely into a product of n. prime ideals, I = ~p i I , n i e Z, it follows that I C I if and only 1 if ~ = I.
Associated to a prime P in O(F) or p in O(E) is a discrete non-Archimidean valuation IIp' respectively IIp [O'M i]. F
lies over P, meaning p/~ O(F) = P if and only if I Ip extends I Ip.
Now 1 Ip is a homomorphism from the units in E, E*, onto a
cyclic, multiplicative subgroup of R +. We may thus form OF(P) and
OE(P), the local rings of integers associated to the primes P and 14
P. we have
OE(P% = {w s E: lWIp _< i} .
We also have O(E) = /p%OECP~.
In the local ring of inteaers OE(P) ~ associated to a prime P is a maximal ideal
mCP) = { w s E: lWIp < i} .
In fact m(P) is a principal ideal generated by some element ~ s OE(P).
Any such z generating m(P) is called a local uniformizer. Two such clearly have as quotient a local unit.
It is also useful for what follows to think of the exponential version of I ! p Following [B,S 23] we denote this by Vp .
Vp : E* + Z is given by:
Vp(X) = n means Ixl p = 171 n P
We now have m(~ = {w s E: ~ (w) > 0} If y ~ F*, ~ (y) is the exponent to which the prime ideal P is raised in the factorization of yO(E).
We shall review prime ideals and algebraic number theory results in Chapter II. Our object here has only been to establish notation for the inner product spaces which we are now ready to describe.
2. Inner products
Again D = O(E) is a Dedekin~ domain. We consider pairs M,D which 15 satisfy either:
(a) M is a finitely generated torsion free D-module (and hence projective since D is a Dedekind domain [R-I 85]) with K = I a - invariant fractional ideal in D or
(b) M is a finitely generated torsion D-module with K = E/I where E is the quotient field of D and I is a - invariant fractional ideal in D.
We are interested in studying the D-module HomD(M,K), where
M,K satisfy either (a) or (b) above. The D-module structure of
HomD(M,K) is now given by defining:
df(x) = f(dx) for d s D, x s M, f s HOmD(M,K) .
Definition 2.1 A K-valued inner product space (M,B) over D is a finitely generated D-module M, together with a non-singular bi- additive mapping
B: M• M+ K
satisfying B(dx,y) = B(x,dy) = dB(x,y) for all x,y s M, d ~ D.
B is linear in the first variable, conjugate linear in the second variable. Again, M and K are assumed to satisfy one of the standard assumptions (a) or (b) above.
It is still necessary to say what it means for B to be non-sin- gular.
Definition 2.2 The map B: M • M + K is non-singular 16 provided the adjoint map AdRB: M + HOmD(M,K) is a D-module iso- morphism. By AdRB w_ee are denoting the right adjoint map, namely
AdRB(X) = B(-,x).
The left adjoint map, AdLB is similarly defined by AdLB(X) =
B(x,-). We must conjugate in order to have AdLB(X) e HomD(M,K) , ie. to make AdLB(X) D-linear.
We have left out any symmetry requirements on B. This is taken care of by:
Definition 2.3 An inner product space (M,B) i__ss u Hermitian provided B satisfies B(x,y) = uB(y,x) for all x,y e M, u fixed u e D.
Since B(x,y) = uuB(x,y), it follows that uu = 1 and u is a unit in D of norm 1 [S 60]. We see that 1 Hermitian is the usual notion of Hermitian, while -i Hermitian corresponds to skew-Hermitian.
When the involution - is trivial, meaning the identity, 1
Hermitian is symmetric since B(x,y) = B(y,x). Similarly, we define skew-symmetric, and u symmetric in the case that - is the identity.
Let us now return to study the module M in case (a), namely M is a finitely generated projective D-module.
We form the vector space M ~D E = V over E.
Definition 2.4 The rank of a finitely generated torsion free
D-module M i_~s the dimension of the vector space M | E over E, E being the quotient field of D. 17
Thus viewed M is a D-lattice in V [O'M 209]. Hence, M splits n as a direct sum M = ~9 A. , n = rank M, where each A. is a fractional i= 1 i I ideal in D. In fact, [O'M 212], there is the splitting:
M = AlZ 1 ~ Dz 2 ~ ... ~ Dz n
where A 1 = fractional ideal in D;
A 2 = A 3 = ... = A n = D.
{z i} is a basis for V.
Since M splits as a sum of fractional ideals, and Hom D is additive over direct sums, we are reduced to studying HOmD(A,I), where A is a fractional ideal in D.
Lemma 2.5 Let A, I be fractional ideals in D, with I a
- invariant fractional ideal. Then the map
T : A-II + HomD(A,I )
given by
x + y (x) with T (x)(c) = cx
is a D-module isomorphism. Here the D-module structure on HomD(A,I) is as previously defined.
Proof: First observe that T(x) e HomD(A,I). To see this, note 18 that T(x) is clearly D-linear, and T is a D-module homomorphism since:
T(dx) (C) = c(d--x) = (cd)(~) = T(X) (dc) = (dT(x)) (C).
We must show that T is an isomorphism.
(a) T is i-I: Suppose T(x) = T(y) Then cx = cy, for all c 8 A. Let c ~ 0, and cancel to obtain x = y, hence x = y.
(b) T is onto: Let f ~ HOmD(A,I). We must show f ~ image T .
Tensoring with E, we extend f to HomD(E,E) , where E is the quotient field of D. Since the Lemma is clearly true for E, it follows that f(c) = x0 c, for x0 e E. But x0 c ~ I for all c s A. Hence 2 0 s A-II, and x 0 ~ ~-i~ = ~-iI. Therefore, f(c) = T(x0), and f ~ image T .
Theorem 2.6 Let M be a finitely generated torsion free D-module, and I a - invariant maximal ideal in D. Then there is a canonical
D-module isomorphism
r M ~ HomD(HOmD(M,I)I))
given by
(x) (f) = f-U~Y
Proof: Recall again the module structure on Y E HOmD(HOmD(M,I),I)) is given by: (dy) (f) = y(df) , where (df) (x) = f(dx). By the remarks immediately preceding Lemma 2.5, it suffices to prove the theorem for
M = A a fractional ideal. 19
We apply Lemma 2.5 twice to obtain an isomorphism ~ . 1 T r A = (A-II)I-- + HomD(A-II,I) + HomD(HOmD(A,I),I)). ~ is given by the composition. We have then:
%(x) (f) = 7(x) (m) where T(m) = f, so f(x) = x m
= x m
= f (x)
as claimed. We again observe that r is a D-module isomorphism since:
r (f) = f(dx) = (df) (x) = r (df) = ((de) (x)) (f) []
We now wish to establish this result in case (b), namely when M is a finitely generated torsion D-module, with K = E/I, and I = ~ as usual. In order to do this we simplify matters by recalling the decomposition of finitely generated torsion modules over a Dedekind domain. This is done as follows. Let P be a prime ideal in D, and let M(P) denote the localization of M at P , so
M(P) is a torsion module over the principal ideal domain (p.i.d.)
D(~ .
M(P) = M localized at P = {x E M: P nx = 0 for some n}
We first decompose M as M = ~ M(P). Then using the structure theorem
for modules over a p.i,d., M is isomorphic to a direct sum of cyclic modules with each cyclic module of the form D/~ some P , i s Z.
Since Hom D is additive over direct sums, without loss of generality 2O we may assume M is a cyclic module, M = D/P i.
In order to study HomD(M,E/I) , we begin by simplifying E/I.
Suppose the fractional ideal I factors as
I = Pli... Pk k , where the Pi are maximal ideals in D.
Let ~i be a uniformizer for Pi" Then in the notation above,
I(P) = I localized at p = IS -I where S = D -P . Since I = P11 ... Pk k , it is clear then that
i. I(Pj) = z 3D(p).
We can now simplify K = E/I.
Lemma 2.7 E/I = E/I(P). P
Proof: Define f: E/I § E/I(P) by e + I + ~ (e + I(P)). P P f is clearly well-defined, and a homomorphism.
(a) f is i-i: Suppose f(e + I) = 0. Then e e I(P) for all
P. Thus e E I, by [O'M 46]. Hence, f is i-i.
(b) f is onto: Consider ~ (a i + I(Pi)) s ~ E/I(P). We now i P apply the Strong Approximation Theorem, [O'M 42]. Letting I Ii denote the Pi-adic valuation we can find x e E with Ix - ail i
I I(P i) I i' at the finite set of i when (a) a i 4 I(P i) or (b)
I(P i) @ D(Pi) ; with Ixli ~ 1 otherwise. It follows that f(x + I) = e(ai + I(Pi)),and f is onto. Q l
Notice that the only summand of E/I with P-torsion is E/I(P). 21
Thus, when M = D/P i , we may apply Lemma 2.7 to obtain an isomorphism:
HOmD(M,E/I ) = HOmD(M,E/I(P)).
We further identify E/I(P) ~ E/zJD(P). Here I(P) = ~JD(P), where z is a uniformizer for P as before. Multiplication by z -j then
gives the isomorphism E/zJD(P) = E/D(P)
The module structure of M lifts to D(P) , and we claim there is a D(P)-module isomorphism T: M(P) § HOmD(p) (M(P) ,E/D(P)). In order to define T, it suffices to consider the case M(P) = D/~ i, since
M(F) is a direct sum of cyclic modules. Let x s M be a generator
for D/P i. Then define T(x) by:
T(x) (cx) = c/z I where c E D(P), x e M(P).
As in Lemma 2 .5, T(dx) (cx) = d c/~ l = (dT)(x) (cx) defines T over M.
is clearly an isomorphism. Combining these isomorphisms over
all cyclic module summands of M, we obtain:
Lemma 2.8 Let M be a finitely generated torsion D-module. Then
there is a canonical isomorphism M = HomD(M,E/I).
Exactly as Theorem 2.6 was proved, we now have:
Theorem 2.9 Let M be a finitely generated torsion D-module,
I a - invariant fractional ideal. Then there is a canonical D- module isomorphism: 22
~: M + HomD(HOmD(M,E/I),E/I)
given by
r = f(x) []
For the purposes of the next section, we also need the following
propositions which describe the relationship of | to Hom .
Proposition 2.10 There is a canonical isomorphism
4: HomD(A,HOmD(B,C) + HomD(A~ B, C), where A, B, C are D-modules.
Proof: [R-I 25] Define 9 by (~f)(a~ b) = f(a)(b). The inverse of ~ is given by (~-lg) (a) (b) = g(a~ b) F~
Preposition 2.11 Let X I, X 2, YI" Y2 be finitely generated
projective D-modules. Then
HomD(XI,Y I) Q HomD(X2,Y 2) ~ HomD(Xl| X 2 , Y1 | Y2 )-
The isomorphism is given by
feD g ~ f g , where (f g) (Xl~ x 2) = f(x I) ~ g(x 2) ,
and f g is extended to Xl@ X 2 bilinearly.
Proof: For X i, Yj free, the assertion is clear using bases. 23
Now, if Xi, Yj are projective, then each is a direct summand of a free. Hom and ~ are additive over finite direct sums. Hence the isomorphism for free splits into isomorphisms for the summands.
3. Constructing new inner products out of old
3.1 Direct Sums
Let (M,B) and (MI,B I) be two K-valued inner product spaces. The easiest way to construct a new inner product space is to form the sum.
(M,B) ~ (MI,B I) + (M ~ M I, B ~ B I)
Here (B ~ BI) ((x,y), (z,w)) = B(x,z) + Bl[Y,W). It is clear that
B ~ B 1 is an inner product since the adjoint map
AdR(B B I) : M ~ M 1 + HomD(M ~ MI,K)
splits as AdRB @ AdRB I.
3.2 Tensor Products
The next operation on an inner product space is ~ . Let (M,B) and (MI,B I) be two type (a) inner product spaces. In other words,
M and M 1 are both finitely generated projective D-modules. Assume that:
B: M x M + Y1 and BI: M1 • M1 § Y2 "
We have the adjoint isomorphisms: 24
AdRB: M ~ HomD(M,Y l) and AdRBI: M 1 + HomD(MI,Y2).
Taking the tensor product of these, we obtain by Proposition l.ll
AdR(B O B I) =
AdRB Q AdRBI: M ~ M 1 HomD(M,Y l) ~ Hom D(MI,Y 2)
= H~ | MI' Y1 ~ Y2 )"
This shows that the adjoint of B O B 1 is an isomorphism, and hence
(M ~ MI, B | BI) is a Y1 ~ Y2 - valued inner product space. We can identify Y1 Q Y2 with the product of ideals, YIY2 .
3.3 Scaling an inner product
There is the operation of scaling an inner product.
Let (M,B) be a K-valued inner product space with d e E*. E* denotes the units in E, ie. E* = E - {0}, since E is a field.
Clearly, (M,dB) is a dK-valued inner product space, where
(dB) (x,y) = d B(x,y). We may view this as a special case of tensor product, namely
(M,B) • (D,B d) = (M,dB), where Bd(X,y) = dxy.
3.4 Tensoring with the quotient field of D
Given an inner product space (M,B), with M of type (a), we can form (M,B) ~D E = (M | E, B | i) , where we now denote the extension of B to the quotient field by B O i. When there is no confusion, we will write B | 1 = B. 25
3.5 The discriminant inner product space
Let (M,B) be a D-valued inner product space, and suppose
M = A ~ D~... ~ D with n factors, where n is the rank of M. We form th the n-- exterior power, AnM = A, with inner product
AnB: AnM x AnM + D defined by
AnB(XlA...AXn , y! A ...Ay n) = determinant (aij),
where the matrix (aij) is given by aij = B(xi,xj).
In order to verify that this is an inner product, one again
needs to check that the adjoint AdR(AnB) is an isomorphism, [B 30].
We then call (AnM, AnB) the discriminant inner product space.
Comment 3.6 The adjoint map of B, AdRB or simply Ad B,
Ad B: M + HomD(M,I) is an isomorphism by hypothesis. Taking the
n--th exterior power, An(Ad B) : AnM + A n (Hom D (M,I)) is an isomorphism.
n-i However, HomD(M,I) -~ ~) HomD(D,I) ~) HomD(A,I)= ~) I A-II. i=l i=l
Thus An(HomD(M,I)) ~ I ~ ... I ~ A-II. However, I | ... | ~-i I
is not in general isomorphic to HomD(AnM , I) so that AnB is not in
general non-singular.
However, for I = D, the Dedekind ring of integers,
Ad(AnB) = An(Ad B) : A nM ~ HomD(AnM, D)
will be an isomorphism by the above, and (AnM,AnB) is indeed an inner product space.
We note that M is free as a D-module if and only if the ideal is principal. Thus the discriminant inner product space yields infor- mation about the structure of the D-module M. 26
We may apply the operation of 3.4, tensoring with the quotient field, to the discriminant inner product space. AnB ~ 1 is then multiplication by a fixed x 0 s E. Thus associated with an inner product space (M,B) is a pair (x0,A) , where A = AnM, and x 0 is as described.
This x 0 specifies the adjoint isomorphism,
AdRAnB: A + HomD(A,D) = A-ID, by AnB(al,a2 ) = x0ala 2 ,
with x 0 unique in F*/NE*, where F* is the fixed field F of -, and NE* denotes the multiplicative group of norms of elements in E*.
Hence, x0A = A-ID~ ie. x0AA = D.
4. The symmetry operator
In Definitions 2.1 and 2.2 of an inner product space, and non- singular mapping, we made use of the right adjoint operator Ad R. In this section, we relate the two adjoint operators Ad R and Ad L-
Using Theorems 2.6 and 2.9, this is done as follows.
Theorem 4.1 Let (M,B) be an inner product space of either type
(a) or (b). Let B: M • M + K satisfy B(dx,y) = B(x,dy) = dB(x,y).
Then the right adjoint map AdRB is an isomorphism if and only if the left adjoint map is.
Proof: Let ~ denote the canonical isomorphism
: M +Hom ~Hom ~M,K) ,K) 27 of Theorems 2.6 and 2.9, given by ~(x)(f) = f(x).
Assume AdRB is an isomorphism. We can thus identify
M = HOmD(M,K) via AdRB. Here y e M is identified with B(-,y)
HomD(M,K).
The isomorphism r is now given by:
AdRB r M + HomD(HOmD(M,K),K) § HomD(M,K)
x § Cx where Cx(y) = B(x,y) ,
in other words r is Ad B. The converse follows similarly. D L
Corollary 4.2 Let (M,B) be an inner product space. Then we
can define a unique D-linear isomorphism s: M + M by the equation
B(x,y) = B(y,sx).
Proof: AdLB(X) = ~ s HomD(M,K). Since B is an inner product, we define s(x) by :
AdR(SX) = B(x,-) = B(-,sx).
s is an isomorphism by Theorem 4.1. []
Notation 4.3 We shall reserve the letter s for this map which
is related to the symmetry of B as described above. It is precisely
this map s which enables us to work with non-symmetric inner product
spaces.
Let N be a subspace of M. We say that N is s invariant provided 28
s (N) C N.
Proposition 4.4 N i__{s s invariant if and only if sN = N.
Proof: Sufficiency is clear.
In order to prove necessity, suppose s(N) C N. Then we can form an ascending chain of submodules of M, T. = {m s M: si(m) s N}, 1 Ti+ 1 ~ T i. Since D is Noetherian, and M is finitely generated, M is Noetherian, IS 47]. Hence, this chain terminates. Suppose
Ti= T N for i ! N.
Claim: T 0 = T 1 = ... = T N, and hence sN = N.
It clearly suffices to show that T i ~ Ti+ 1 implies Ti+ 1 ~ Ti+2.
i+l Suppose then that T i ~ Ti+ I. Let m 0 e Ti+ 1 - T i. Then s (m 0)
N and sl(m 0) # N. s is an isomorphism, so there exists mle M with sm I = m 0. Hence, mle Ti+ 2. If mle Ti+l, then m0C T i which is a contradiction. Thus, mls Ti+ 2 - Ti+ I.
We observe that Proposition 4.4 can be proved for any ring R
in place of D, with R not necessarily Noetherian. Namely, let s
be an isomorphism s: M § M of a finitely generated R-modules, with
R not necessarily Noetherian.
Claim: If NCM is a submodule with sN C N, then sN = N.
We sketch the proof using the following form of Nakayama's
Lemma: Assume M is finitely generated. Then M = IM implies
I + Ann(M)= R where Ann(M)denotes the annihilator of M. 29
Consider the diagram :
S 0 + M + M + 0 + + ^
S M/N ~ M/N + 0
It suffices to show ~ is I-i. View M/N as an R[x]-module, where x acts as { Suppose x m 0 = 0. We want to show m 0 = 0. Let I = (x).
Since ~ is onto, M/N = I(M/N). Hence, by Nakayama, above, I + Ann (M/N)
= R[x]. Write 1 = rx + t, where r ~ R[x], t s Ann (M/N). Then
1. m 0 = rx- m 0 + t- m 0 = 0, so m 0 = 0. Thus ~ is i-i. Q
For a subspace N of M, we define
N R = {v 6 M: B(n,v) = 0 for all n E N} .
Thus N R is the kernel of AdRB restricted to N, AdRB: M + HOmD(N,K) .
We call N R the right orthogonal complement of N. The underlying inner product space is understood. Similarly,
N L = {v E M: B(v,n) = 0 for all n s N} .
Now, let N be s invariant. Since sN = N, it follows that
N L = N R. We denote this common orthogonal complement N ~ .
Note 4.5 N ~ is only defined when N i__ss s invariant. 30
Proposition 4.6 If (M,B) is an inner product space of type (a) , meaning M is torsion free, then N L and N R are direct summands of M.
Proof: Consider the exact sequence:
0 § N L § M -~ M/N L -~ 0
It suffices to show that M/N L is torsion free, hence projective over the Dedekind domain D. Then the sequence splits and N L is a summand.
Suppose to the contrary M/N L has torsion. Then there exists x ~ M, d ~ 0, d s D, with x ~ N L and dx e N L. So B(dx,y) = dB(x,y) =
0 for all y ~ N. Thus B(x,y) = 0 for all y e N since D is a domain.
Hence x ~ N L , contradiction. Thus M/N L is torsion free. []
We have already remarked that if N is s invariant, then N L = N R-
The above shows that N L and N R are always summands. These two con- ditions turn out to give the converse.
Proposition 4.7 Let (M,B) be an inner product space with M torsion free. Let N be a summand of M. Then N is s invariant if and only if N L = N R.
Proof: As observed before Note 4.5, necessity is clear. In order to prove sufficiency, consider the two exact sequences:
AdLB 0 § N L + M § HomD(N,K) ~ 0
AdRB 0 § (NL) R ~ M + HOmD(NL,K) + 0 . 31
AdLB and AdRB are onto because N is a summand. Clearly we have
rank N L = rank (HOmD(NL,K)). Thus, by the two sequences above,
rank (NL) R = rank (HOmD(N,K)) = rank N. However, B(ns = 0
for all ns ~ N L, n E N. Thus N C (N L) R" Since N is a summand,
ranks equal, it follows that N = (NL) R. Similarly, N = (NR) L.
We now wish to show that sN C N = (NL) R. So we compute
B(nl,sn) = B(n,ns = 0 for all ns E N L, since N L = N R by hypothesis.
Thus sN C (N L) R = N as desired. [~
Proposition 4.8 N = N L if and only if N = N R.
Proof: Suppose N = N L. Then B(m,n) = 0 for all m, n e N.
Thus, n ~ N R and N C N R. By Proposition 4.6, N = N L implies N is
a summand. Clearly, as in Proposition 4.7, rank N = rank N R, so that
N = N R-
The converse is similar, r3
Theorem 4.1 stated that for an inner product space (M,B),
both AdRB and AdLB are isomorphisms. This enabled us to define the
symmetry operator s, with B-~-x,y) = B(y,sx) . In a like manner, one
can see:
Proposition 4.9 Fixin~ an inner product space (M,B), let
s M + M, be a D-linear operator. Then there is a unique
s M + M, D-linear, with B(x,s = B(s
Notation: Z* is called the adjoint operator of Z, not to be
confused with the adjoint maps AdRB, AdLB previously defined. 32
Proof: For fixed x, we have the map B(x,s s HOmD(M,K).
Since B is non-singular, Ad L is an isomorphism and we can find a unique w s M such that:
B(x,s = B(w,-)
Define s = w. Then B(x,Zy) = B(s ~* is clearly a well- defined, D-linear map with the desired properties. Ad L being an isomorphism shows ~* is unique. C]
One can similarly define a left adjoint operator of s by insisting B(ix,y) = B(x,*s . It then follows that if s s HomD(M,M)
B( (*i) *x,y) = B(x,*iy) = B(s
Since B is non-singular, (~s =s Similarly, *(~*) =~, so that these two * operations are inverses of one another.
Further, we note that
B(s = B(x,s
= B(i*y,sx)
= B(y,s
= B(s-llsx,y)
Thus s = s-lis. Also we have 33
B(x,y) = B(y,sx)
= B (sx, sy)
= B (s*sx,y)
-i -i so that s* = s Similarly *s = s We summarize these remarks in
Theorem 4.10 The correspondences s + Z* and s + *s of the al~ebra of linear operators on M are inverses. They satisfy
~** = s-l~s r s* = *s = s -i
Thus s = s for all s 8 HomD(M,M) if and only if s is central.
An easy calculation also shows (iT)* = T*s so that when s is central the correspondence s + s gives an anti-involution of the algebra of linear operators on M.
5. The Witt equivalence relation
Definition 5.1 Let k b_~e ~iven, k s D. A degree k mapping
structure over D is a triple (M,B,s satisfying:
(a) (M,B) is an inner product space over D .
(b) ~: M + M is a D-linear map satisfying
B(s163 = kB(x,y) for all x,y e M .
is called a map o_ff de~ree k. For all future considerations, we
shall assume henceforth that k e Z.
In the case that M is torsion free, and k ~ 0 , it follows 34 that ~ is non-singular. To see this suppose i(x) = 0. Then
B(Zx, Zy) = kB(x,y) = 0. Since B has values in K = I a
- invariant fractional ideal, we cancel k and conclude B(x,y) = 0 for all y s M. However, B is non-singular, so that x = 0, and
is i-I.
The Witt equivalence relation for degree k mapping structures comes from:
Definition 5.2 A degree k mapping structure (M,B,s is metabolic if there is a D-submodule N C M satisfying:
(a) N is s invariant
(b) N is s invariant
(c) N = N ~
When (M,B,s is metabolic, an N satisfying (a) , (b) and (c) above will be called a metabolizer for M. We shall also refer to the triple (M,B,s as M, when B and s are understood, and speak of M as being metabolic.
The operation of direct sum on inner product spaces extends to degree k mapping structures. The notation:
(V,B,Z) e (W,B', s = (V W,B B',s ~ s
It is clear that ~ ~ s is of degree k with respect to B ~ B'
At this point we can introduce a relation -- on degree k mapping structures by:
(V,B,s ~ (W,B',s when (V ~ W,B ~ -B',s s 35 is metabolic. In what follows, we will show - is an equivalence relation, called the Witt equivalence relation. This agrees with the usual notion of Witt equivalence [M-H] when no s is present.
Notation: W+I(k,K) denotes degree k mapping structures
(M,B,s modulo ~ with values in K; together with the additional requirement that the symmetry operator s is the identity map, so B is symmetric. The underlying ring D is understood.
Emphasis is given to the range of the inner product, namely K.
Similarly, W-I(k,K) is Witt equivalence classes of triples
(M,B,i) having B skew-symmetric.
When there is no s , so that we are taking inner product spaces modulo ~ , without condition (a) of 5.2, we denote the symmetric equivalence classes W+I(K), the skew-symmetric W-I(K). Finally, when no i is present, with no symmetry requirement at all placed on B , the resulting Witt group is denoted A(K).
We have also defined the notion of B being u Hermitian. We write Hu(K) to denote those Witt equivalence classes [M,B] for which B(x,y) = uB(y,x).
In summary, our notation is:
W +I : B symmetric -i W : B skew-symmetric
A : no symmetry requirements on B (B asymmetric)
H u : B u Hermitian
If we write w+l(K), we are thinking of pairs (M,B); if we write
W+I(k,K), we are thinking of triples (M,B,s with B(s163 = kB(x,y). 36
The K means that B: M • M ~ K.
We write A(k,K) to denote the group which consists of triples
(M,B,Z), with no symmetry requirements on B.
If we let k range over Z, we can form a graded ring, with multiplication defined by O ,
A(k,K) • A(k',K') ~ A(kk',KK')
(M,B,I) x (M',B',Z') + (M •D M',B | B',s Q i')
This follows from 3.2.
However, for this paper, we shall only be concerned with the
Abelian group structure arising from direct sum.
Our objective now is to show that is an equivalence relation.
It is easy to see that is reflexive, since (V ~ V,B -B,~ ~ s ) has metabolizer N = {(x,x) : x e V}. Clearly ~ is symmetric.
We must show ~ is transitive.
Again when (M,B,i) is metabolic, we will say M is metabolic, the B, ~ being understood, and write M ~ 0. The following pro- position is clear.
Proposition 5.3 ~ is transitive if and only if H ~ 0 and
M ~ H ~ 0 implies M ~ 0.
We call M stably metabolic if there exists H ~ 0 with
M ~ H - 0. We may then restate Proposition 5.3 as saying ~ is transitive if and only if stably metabolic implies metabolic.
Comment: Once we have show that ~ is transitive, it is also
clear that the following relation, ~ , would have yielded the same 37 relation as Define M 0~ M 1 if and only if there exists
H 0 ~ 0, H 1 ~ 0 with M 0 ~ H 0 isomorphic to M 1 ~ H 1 .
Lemma 5.4 Suppose M is a finitely generated torsion free
D-module. Then (M,B,~) ~ 0 over D if and only i_~f (M,B,I) ~ E ~ 0 over E .
Here E is the quotient field of D (see 3.4).
Proof: Necessity is clear, for if N is a metabolizer for
(M,B,I), then N O D E is a metabolizer for (M,B,s O D E.
To show sufficiency, consider the exact sequence
0 § D + E ~ E/D + 0 .
Tensoring with M, we note that M is embedded into M ~ E as
M | i. Suppose M ~D E has metabolizer N = N ~. Let
N 1 = N /~ (M | i) C M • 1 .
Claim: N 1 = N1J- in M | 1 = M, so that M ~ 0.
To begin with, N 1 is s,s invariant since N is. It also is clear that NIC N1"L. Conversely, if x ~ 1 e N1J', B(x ~ l,y | i) = 0 for all y | 1 E N I. However, if y | r c N, then y ~ 1 e N I, so
B(x ~) l,y ~) r) = rB(x | l,y ~ i) = 0. Hence x | 1 ~ N J- = N.
Thus x O 1 E N I, and NI~C N I. Thus N 1 = N~and M ~ 1 = M ~ 0 o T'I 38
Using Lemma 5.4, we see that to prove ~ is transitive in the case that M is torsion free, we can assume that D = E a field, and K = E .
We thus assume for the rest of the proof of transitivity that
K = E = D a field when M is torsion free, and K = E/I for
M torsion as usual.
In either case K is an injective D-module and we have:
Theorem 5.5 Let (M,B) be an inner product space, with Values in K = E or K = E/I. If N is Z , s invariant, then
N = (N~ ~
Remark 5.6 This is not true for K = I an arbitrary fractional ideal or even K = D
Proof: We have the exact sequence
AdRB 0 § N ~ ~ M § HomD(N,K) 0
AdRB is onto since K is an injective D-module.
Applying the Hom functor, we obtain
0 § Horn D (HornD (N, K) , K) + HornD (M, K) § HornD (N ~,K) §
Again, Ext(HomD(N,K),K) = 0 , since K is injective and the last map is onto. 3g
We can identify HOmD(HOmD(N,K),K) = N by Theorem 2.6. This clearly yields the commutative diagram:
0 § N ~ HomD (M,K) § HOmD (Ni, K) + 0 + + + Ad B 0 § (NI)~ + M +R HomD (N J-, K) § 0
The inner product provides an isomorphism of the middle terms, so by diagram chase [M 50] the inclusion N C (N~) ~ is an isomorphism.
Lemma 5.7 For any two ~ , s invariant submodules R and S of
M, where (M,B) is an inner product space as above, we have:
Cl~ (R+ s)~ = ~ s ~ and
(2) R ~ + S ~ = (R{N S) ~
Proof: (i) follows from the definition of ~ . TO show (2), observe that
(R ~+ S~) ~ = (R~)~fA (SX) ~ = R{NS
Thus, taking , R ~ + S ~ = (R ~ S) ~ O
Lemma 5.8 Let (M,B) be an inne___~r produc t space as above.
Suppose that M ~ 0 with metabolizer N. Let L C M satisfy
L C L ~. Then L + N ~L ~ = (L + N N L~) ~.
Remark 5.9 This Lemma shows how to go from a metabolizer N, and 40 a subspace L C LJ- to another metabolizer, namely L + N ~ L J', which contains the self annihilating subspace L.
Proof: The assumption that L is Z, s invariant is under- stood, in order that L ~ make sense.
We compute using Lemma 5.7:
(L + (N ~ L ~))~= LJ-{~ (N~ LJ-)~ = LJ'{~ (N J- + (L/')~)
= L/'{'~ (N + L) since N ~= N and (LJ')/" = L
= (L~'{'% N) + (LJ'/~ L)
= L + (N ~ LJ-) since LCL/" .
Thus L + (N/~L ~) is also a metabolizer. E3
Theorem 5.10 (Transitivity of ~ ) Let H ~ 0. Then
M ~ H ~ 0 if and only if M ~ 0 .
Proof: Sufficiency is clear. To .show necessity, let N be a metabolizer for M ~ H, and H 0 a metabolizer for H. We embed
H 0 and H into M ~ H as 0 ~ H 0, 0 ~ H respectively. Notice that 0 ~ H 0 C (0 ~ H0 )'L , so that by Lemma 5.8, we may rechoose N
such that 0 ~ H 0 C N.
We review our notation. (M,B), (H,B') and (M ~ H,B ~ B') are the inner product spaces. We will write elements in M ~ H as pairs (x,y) with x e M, y e H.
Let N O = projection of N onto M = {a s M: (a,h) e N for some h} 4~
Claim: N O is a metabolizer for M.
N O is clearly s s invariant since N is, and projection commutes with Z, s on M ~ H.
We first show that N O @ H 0 = N. If! (a,h) s N, we claim h e H 0.
Let (0,h I) s 0 e H 0 N = N ~. Then
(B ~ B')((0,h), (0,hl)) = B(0,0) + B'(h,h I)
= B(a,0) + B' (h,h I)
= (B B')((a,h), (0,hl))
= 0
since N = N ~ . Hence B'(h,h l) = 0 for all h I E H0, so that h ~ H~ = H 0 as claimed. Thus N O ~ H 0 = N.
Clearly, N0 C N~. Conversely, let b s N~. Then by computing as above (b,0) e N = N, so that b s N O . Hence N 0 = N 0 is a metabolizer for M. [-]
Thus ~ is an equivalence relation, and we may form the Witt group consisting of equivalence classes of triples (M,B,s modulo ~.
(M,B,s ~ (MI,BI,s I) provided (M ~ M,B ~ -BI,s @ Z I) ~ 0 .
Notation: [M,B,s will denote the Witt equivalence class of
(M,B, s .
6. Anisotropic representatives
Our final goal of this chapter is to find a representative of
each equivalence class. As long as K = E a field, or E/I in the 42 torsion case, this representative is unique.
We begin by describing the representative we will obtain.
Definition 6.1 A degree k mapping structure (M,B,s i__ss anisotropic if for any s,s invariant D-submodule N of M,
N~N ~ = 0.
Theorem 6.2 Every Witt equivalence class [M,B,Z] has an anisotropic representative.
We prove this theorem by way of a sequence of Lemmas which are of interest in their own right.
Lemma 6.3 Let T be an s,s invariant D-submodule of M, with
T C T~ Then T~/T inherits a quotient degree k mapping structure
(T~/T,B, ~) .
Proof: Let it] denote an element in T~/T . Define
B( [tl], [t2]) = B(tl,t 2)
where tl, t 2 are representatives of [tl], It2] respectively.
is clearly well-defined since T is self-annihilating, ie. T C T .
It is likewise clear that ~, the induced map on T~/T is of degree k with respect to B, and well-defined.
We must show that B is an inner product, ie. that
AdRB: T~/T § HOmD(T~T,K) is an isomorphism. 43
Applying the functor HomD(-,K) to the exact sequence:
0 § T ~ T ~ + T T ~ 0,
we obtain the embedding;
0 + HomD(T~/T,K) + HomD(T~,K).
1_ L Suppose g s HOmD(TTT,K) § ~ ~ HomD(T--,K). We can lift to g: M § K since T is a summand by Proposition 4.6 in the torsion free case and since K is injective in the torsion case.
AdRB: M + HomD(M,K) is an isomorphism. Hence g = B(-,x). g restricted to T equals O, so x c T ~ Thus (x + t) gives the same g for all t ~ T. So we may read x s T2"/T.
This procedure defines a map:
HOmD (T/-/T, K) + T J-/T, namely g § ~ + [x].
The inverse of this map is simply
§ ~ (-, [x] ) .
Hence AdRB is an isomorphism and B is an inner product. D
Lemma 6.4 With the same hypotheses as in Lemma 6.3,
M -T~/T is metabolic.
Proof: In the torsion free case, by Lemma 5.4, we may assume 44
K = E a field. Thus, in any case, there is no loss of generality in assuming the hypotheses of Theorem 5.5, namely that K = E or
K = E/I. Consequently, for N an s s invariant subspace of M, we have N = (N~) m.
We wish to show M ~ -T T ~ 0. So consider N = {(x,x + T) :x E T}.
N is an i, s invariant subspace, and clearly N C N ~. Let
(a,b + T) e N ~, with b e T J-. We compute
(B -B)((a,b + T),(x,x + T)) =B(a,x) - B(b,x) = 0
for all (x,x+T) E N. Thus B(a - b,x) = 0 for all x s T .
Hence, (a - b) c (T~) ~ = T, since by assumption Theorem 5.5 applies. So [b] = [a] - [(a - b)] = [a] in T~/T , and
(a,b + T) = (a,a + T) c N. Therefore, N~C N, and N is a metabolizer for M ~ -T /T.
Lemma 6.4 shows then that M ~ T--/Tt whenever TO T ~. In the torsion free case, we can use that rank M < ~ to conclude, after successive applications of Lemmas 6.3 and 6.4, that M ~ M 0 where M 0 has no Z,s invariant subspace T with TO T ~.
In other words, M 0 is anisotropic.
For the torsion module case, we repeatedly apply Lemmas 6.3 and 6.4 to obtain sequences:
MD T~T~ ...... D T 2 D T I~ T
Since M is Noetherian, the ascending chain condition implies that the sequence {T i} terminates. Hence, it follows the sequence {T i} 45 will also terminate.
Since both chains terminate, we have M ~ Tr/-/Tr ; with
T /T r having no s s invariant submodule Tr+ 1 with Tr+IC Tr+ 1 . r Thus, [M,B,s has an anisotropic representative, TrJ-/Tr , as claimed. This completes the proof of Theorem 6.2.
This anisotropic representative need not be unique for torsion free D-modules [M-H 28]. However, for K = E a field, or K = E/I, we shall see that it is unique.
Theorem 6.5 As long as [M,B,s e w(k,K) satisfies Theorem 5.5,
ie. for K = E or E/I, every Witt equivalence class [M,B,s has a unique anisotropic representative up to isomorphism.
Proof: Suppose (M,B,s ~ (M',B',s with M and M' both anisotropic. Let N C M ~ M' be a metabolizer, with respect
to B G -B'. We will show that N is the graph of an isomorphism
f: M § M' which satisfies B'(f(k),f(y)) = B(x,y), s o f = f o s
and s' o f = f o s. Thus f is an isomorphism between (M,B,s
and (M',B',s
Let A = {a ~ M: there exists a I s M' with (a,a I) e N} .
Claim: Given a e A, then there exists a unique a I e M'
with (a,a I) E N.
For suppose (a,a I) and (a,a 2) e N. Then (0,a I - a 2) e N.
Consider the s s' invariant subspace, M 1 of M' generated by
a I - a 2 . Since N is s ~ s' and s ~ ~' invariant, this subspace 46
M 1 will have (0,M I) C N. Hence MIC M~ since N = N ~ . This is a contradiction to M' being anisotropic unless a I - a 2 = 0 so that a I = a 2 .
Similarly, let B = {a I s M': there exists a e M with
(a,a I) s N} . As above, each a I s B has a unique a 6 M with
(a,a I) ~ N. It follows that N is the graph of a i-i function f .
We claim that A = M and B = M'. To see this we show that
A i = 0 in M, hence (A~)~ = A = M . So let a ~ A , and consider
(a,0) E M ~ M'. Take any (x,y) s N. Then:
(B ~) -B') ((a,0), (x,y)) = B(a,x) = 0 since
a e A J', x e A.
Thus, (a,0) e ~= N. By the first claim, this implies a = 0.
A similar argument shows B = M'. It follows that f: M + M' is an isomorphism.
Let (a,f(a)) e N. Then (la,Z'f(a))s N since N is
~' invariant. Thus, by definition, (f o s (a) = (s o f) (a).
Similarly, (f o s) (a) = (s' o f) (a).
Finally, consider (x,f(x)) and (y,f(y)) ~ N.
(B ~) -B') ((x,f(x)) , (y,f(y)) = 0 so
B(x,y) - B'(f(x),f(y)) = 0 and B(x,y) = B'(f(x),f(y))
as desired. Chapter II WITT INVARIANTS
Having defined the Witt ring we are led to examine invariants which will enable us in many cases to compute at least the group
structure of W(k,F).
Section 1 begins with a preliminary discussion of prime ideals and some results from algebraic number theory. In Section 2 we state the basic properties of Hilbert symbols. The reader should also see O'Meara [O'M] Introduction to Quadratic Forms for a complete exposition.
Following this introduction, we continue by considering
(M,B) a u Hermitian inner product space over a field E with
involution - and fixed field F. B: M • M ~ E satisfies:
B(x,dy) = uB(dy,x) = dB(x,y) for d c E, x,y ~ M .
In Section 3, we discuss the rank mod 2 of M as a Witt group
invariant. Next, in Section 4, we introduce the discriminant in-
variant, the Witt analog of the determinanat for matrices. Thus, we review the matrix representation of B and diagonalization in
order to define this invariant. Section 5 is concerned with the
signature invariants, which arise from the real infinite ramified
primes.
These invariants completely determine the Hermitian group
H+I(E) for E an algebraic number field by Landherr's Theorem.
Notation: F 2 = {0,i} = additive group of Z modulo 2.
(field with two elements)
Z* = ~i,-i} = multiplicative group of units in Z. 48 i. Prime ideals
The setting is as in Chapter 1. Again E is an algebraic number field together with an involution - The fixed field of
- is F . The Dedekind rings of integers in E and F are denoted by O(E) and O(F) respectively. O(F) = O(E)/~ F.
If P is a prime ideal in O(E), then P = p/~ O(F) will
denote the corresponding prime ideal in O(F).
Conversely, if P is a prime ideal in O(F), by going up,
[A,Mc 63] there exists a prime ideal P in O(E) with p /~O(E) = P. In fact, there may be several such prime ideals in
O(E) lying over P. The answer is given by considering PO(E), g e. IS 71]. We factor PO(E) = E pi I The Pi satisfy i=l g pi/~ O(E) = P. Since the extension [E:F] is of degree 2, Z e.f. i= 1 i l = 2, where fi = [O(E)/p i : O(F)/P] is the residue field degree.
We thus obtain the following cases:
i.i Split e = i, f = i, g = 2. In this case PO(E) = PP where p is a prime ideal in O(E) with P ~ P. We say that P
splits in this case.
We may examine the split case in terms of the local completion
of F at P, which we denote F(P).
Write E = F(/o), and suppose /~ satisfies the irreducible
polynomial p/o/F(X) = p(x). Then factor p(x) in F(P) Ix]. The
split case corresponds to p(x) = fl(x) f2(x). The prime spots
Pi dividing P are determined b}, F - monomorphisms 7: E ~ L,
where L is an algebraic closure of F(P). The - involution
interchanges the prime spots, hence, PO(E) = FP. [E(P) :F(P)] = 1
is the local degree. 49
1.2 inert e = i, f = 2. In this case PO(E) = P a prime in
O(E). P = ~ , and we say P remains prime, or is inert.
1.3 Ramified e = 2, f = i. PO(E) = p2, P = P in this case also. We say P ramifies.
In both the inert and ramified cases, p(x) = p/~/F(X) is irreducible in F(P) [x], and the local degree [E(P) :F(P)] equals 2.
This describes the situation for finite primes.
We next consider all embeddings T: F § C, where C is the complex numbers. If y: F + R we call T a real infinite prime.
Otherwise T is called a complex infinite prime. We denote infinite primes by P Our only concern will be with real infinite primes.
Again, since [E:F] = 2, and the characteristic of these fields is 0 (not 2), we may write E = F(/~), for ~ s F* unique up to multiplication by a square in F*. For an infinite prime P , there are two cases:
1.4 Split If P is complex infinite, P is split. If
P is real infinite, and ~ > 0 with respect to the ordering induced by P , we again say P is split. In the case of a real split prime, P , the ordering of F can be extended to E in two distinct ways.
1.5 Ramified If P is a real infinite prime, and a < 0 with respect to the P induced ordering we say P is ramified. In this case, the P ordering of F can be uniquely extended to an embedding of E into C in such a way that the imaginary part of
/~ is positive. 50
Let T denote the extension of P to E. Then T is
equivariant with respect to complex conjugation - This means
there is a commutative diagram:
T E § C
E § C
The map -: E ~ E is the involution, -: C § C is complex
conjugation. There should be no confusion.
Associated to a finite prime P in O(F), or P in O(E),
is a discrete, non-Archimedian valuation I Ip, respectively I Ip"
P lies over P if and only if l Ip extends IIp, as discussed
in Chapter i.
We next describe prime ideals in terms of local uniformizers.
If PO(E) = PP is split, then a local uniformizer ~p ~ OF(P) is also a local uniformizer for both OE(P) and OE(~). Careful, this does not mean P and P induce the same valuation. On the con- trary, OE(P) ~ OE(P). It only says (~) in OE(P) is the unique maximal ideal.
If PO(E) = P is inert, then a local uniformizer Zp for
OF(P) is a local uniformizer for OE(P) also.
If PO(E) = p2 is ramified, then any local uniformizer z of
OE(P) will have norm z~, and ~ is a local uniformizer for
OF(P). This follows since P = P, hence z and ~ are both local uniformizers for OE(P). So vp(z~) = 2 = Vp(~p) Thus
z~ is a local uniformizer for OF(P), when P ramifies. 51
For y ~ F* , we summarize:
(i) If P splits, PO(E) = p~. Vp(y) = ~Tp(y)
= v~(y)
(2) If P is inert, PO(E) = P Vp(y) = vp(y)
(3) If P ramifies, PO(E) = p2 2Vp(y) = vp(y)
This is not true for y e E*.
Associated to a prime p , finite or infinite, lying over a prime P, is the extension of localized completions. The degree
[E(p) :F(p)] is denoted n F . n F = 1 if p is split and 2 otherwise.
2. Hilbert symbols
we begin by recalling the theorem of Hasse.
Theorem 2.1 Let Y c F*. Then y is a norm from E* if and only if Y e F(P)* is a norm from E(P)*, for all p , finite and infinite in E [O'M 186].
This condition is trivial over P split, for then F(P) = E(P).
We should like to rephrase this in terms of Hilbert symbols
[O'M 169]. We now state briefly the salient properties of these symbols.
If a,o s F*, a symbol (a,o)p is defined by: (a,G)p = +I
if and only if a is a norm from F(P) (/o) if and only if there exists x, y e F(P) satisfying ax 2 + ~y2 = +i. 52
In terms of the prime ideals, we summarize.
2.2 If P splits: (a,a)p = +l
2.3 If P is inert: The local degree np = [E(P):F(P)] = 2.
By [O'M 169], evert local unit is a local norm, and the local uniformizer ~ s F(P) is not a local norm.
In terms of Hilbert symbols, for a e F*, we have the following. n Let a = ~ v, for v a local uniformizer and v a local unit.
n (a,u) p = (z,q)p(V,q)
n = (~,~) p n = (-i) Vp (a) = (-1)
2.4 If P is ramified: Again the local degree is 2. As we have seen, in this case we may pick a local uniformizer ~p of
P to be a local norm, namely, ~P = ~P ~P ' where ~p is a uniformizer for E(P). We thus study s local units.
The residue field, OF(P)/m(P) = O(F)/P is isomorphic to the completion OF(P)/m(P). If u is a local unit, then for the following we denote by u I the image of u in the residue field, OF(P)/m(P).
Claim: For P ramified, a local unit u is a local norm, ie. (u,O)p = +i if and only if u I is a square in the residue field. (characteristic # 2) 53
Proof: If u I is a square in the residue field, we may factor the polynomial t 2 - u I = f(t) in the residue field as
(t + /Ul) (t - /Ul) .
We are assumlng the characteristic of the residue field is not
2, so these two factors are relatively prime. Hence, we may apply
Hensel's Lemma, and conclude that t 2 - u factors in the completion
F(P). Thus u has a square root in F(P), and (u,O)p = +i.
Conversely, if (u,O)p = +I, then u is a norm from E(P), say xx = u. We write x as x = wz r, for w a local unit, z a local uniformizer. Then xx = w~( r~ r) = u. Since u is a local unit, r = 0. Thus u is a square in OE(P)/m(F), since the induced involution is trivial there. However, OE(P) /m(P) = OE(P)/m(P), so that u I is a square in the residue field.
2.5 If P is infinite ramified: (a.a~ = -i if and only
if a < 0 and o < 0. This is clear as the completion with respect to P is R.
We restate the Theorem of Hasse in terms of symbols.
Theorem 2.6 y r F* is a norm from E* if and only if
(y,O)p = +i at all primes P , finite and infinite.
We also list the important properties of the Hilbert symbols, in
addition to the discussion above.
2.7 (a,~)p = +i for almost all P since at almost all P ,
a and a are both units [O'M 166]. Almost all means all but
finitely many in this case. 54
2.8 Hilbert Reciprocity H(a,o) = +i p P
2.9 Realization: If s(P) ~ Z* is a function defined for all P satisfying
(i) e(P) = +i if P splits
(2) e(P) = +i at all but finitely many primes
(3) H s(P) = +i P then there is an f s F* with (f,~)p = s(P).
We again refer to O'Meara [O'M 203].
Note: At non'split primes, np = 2, and ~ is not a square in
F(P).
3. Rank
Let [M,B] s H u (E) . We define the rank rood 2 of [M,B] , denoted rk[M,B], by
rk[M,B] = 0 if [M:E] is even.
= 1 if [M:E] is odd.
Here [M:E] is the rank of the vector space M over E. 55
Theorem 3.1 rk: Hu(E) + F 2 is a well-defined group
homomorphism.
Proof: The only problem is to show that rk is well-defined.
So, let [M,B] c H u (E) have [M,B] = 0. Then there is a metabolizer N C M with N = N ~. This yields the exact sequence:
0 + N ~ AdRB § M § HOmE(N,E) + 0
Hence rank M = rank N + rank(HOmE(N,E))
= rank N ~ + rank N
= rank N + rank N
= 2rank N.
Thus [M,B] = 0 implies rk[M,B] = 0. It follows that rk is well-defined.
Clearly rk is additive, so rk defines a group homomorphism. we also note that rk is in fact a ring homomorphism. P1
Corollary 3.2 rk: Hu(I) + F 2 defined as above is a well- defined group homomorphism. Here Hu(I) denotes I-valued u Hermitian inner products on torsion free D-modules.
Proof: Apply I 5.4 .
4. Diagonalization and the discriminant
In order to discuss the discriminant, we must establish some notation. We first pick a fixed basis, {e I .... ,en} for M . 56
n Thus, if x E M, we write x = (al,...,a n) to mean x = Z a e. , a s E. i=l i i i Associated to the inner product B: M • M + E, there is the matrix B' = (bij), where bij = B(ei,ej). If x = (a I ..... an), and y = (bl,...,bn), then in terms of B' we have
= B (x,y) (a I ..... an) B'Ibl 1
k~n/ which we also write as B(x,y) = xB'y t This follows since B is linear in the first variable and conjugate linear in the second.
Now [M,B] e Hu(E). Thus B(ei,e j) = uB(ej,e i) so that b.. = ub It follows that B' satisfies B' = uB 't. 13 31 We now let {e#}x ni=l denote the dual basis to {e i} . e~1 : M + E is defined on a basis of M by e #i (ej) = ~ij' the
Kronecker ~ , and extended linearly to M.
We consider the adjoint map of B, AdRB: M + HomE(M,E) .
AdRB: e i + B(-,ei). We express (AdRB) (e i) as a linear combination of the {e j#} This yields
n (AdRB)e i = Z B(ej e# j=l 'ei) 3
We thus see that the matrix of the adjoint transformation, in terms of the bases {ei}, {e~} is none other than (bij) = B' = (B(ei,ej)).
We can thus state:
Proposition 4.1 Given a bilinear map B: M x M + E, the adjoint AdRB: M + HomE(M,E) is an isomorphism if and only if 57
(B(ei,ej)) is an invertible matrix. []
Next, we wish to relate u Hermitian to 1 Hermitian.
Proposition 4.2 Hu(E) = HI(E)
Proof: Since u satisfies uu = l, by Hilbert's Theorem 90, we can find x I e E with XlXll = u. x I then clearly yields an isomorphism: Hu(E) = HI(E), merely by scaling the inner product -i with x I . In other words, [M,B] e Hu(E) + [M,B I] e HI(E) where
Bl(X,y) = (i/Xl)B(x,y). We must check B 1 is 1 Hermitian.
Bl(X,y) = (i/Xl)B(x,y) = (U/Xl)B(y,x) = (i/Xl)B-q-y,x) = Bl(Y,X).
Conversely, if [M,B I] c HI(E), we must check that B is u Hermitian, where B(x,y) = XlBl(X,y). We compute: B(x,y) =
XlBl(X,y ) = UXlBl(X,y ) = UXlBl(Y,X ) = U(XlBl(Y,X) ) = uB(x,y) []
Remark: We could choose x I = 1 + u.
When the characteristic of E is not 2, we shall see that it
is possible to choose a basis for M so that the matrix B' of B is diagonalized. We prove this first for [M,B] e HI(E).
For [M,B] e Hu(E), we apply Proposition 4.2 and the above, observing that the isomorphism given in 4.2 preserves diagonalization.
Now let [M,B] e HI(E). By the trace lemma, B 1 defined on
M by Bl(X,y) = trE/F ~ B(x,y) is a non-singular symmetric bilinear form on M. Here trE/f denotes the trace map. Since
Bl(X,y) = (I/2) (Bl(X + y,x + y) - Bl(X,X) - BI(y,y) ), and B 1 ~ 0,
it follows that there exists v e M with Bl(V,V) ~ 0. It follows
that B(V,V) ~ 0 also. Extend v to a basis of M, {v,v 2 ..... v n} . 58
Notice that {v,v 2 - (B(v2,v)/B(v,v))v ..... v n - (B(Vn,V)/B(v,v))v }
= {w i} is also a basis for M. The computations:
B(v i - (B(vi,v)/B(v,v))v,v) = B(vi,v} - B(vi,v) = 0
= B(v,v i - (B(vi,v)/B(v,v))v)
show that with respect to {w i} , the matrix of B looks like:
B (v ,v) 0 IZ ~I
Continuing, consider Bl(X,y), for y in the span of {w 2 ..... w n} , x s M. Again, since B 1 is non-singular, we can find x with
0 ~ Bl(X,y) = (i/2) (Bl(X + y,x + y) - Bl(X,X) - Bl(y,y) ). Write n x = Z a.w.. Then it is clear that: i=l i 1
n n Bl(X,y ) = BI( ~ aiwi,Y] + Bl(alwl,Y) = BI( Z aiwi,Y). i=2 i=2
Thus, we can find v e
Proposition 4.3 Given [V,B]e HI(E), there is a basis for V which makes the matrix of B diagonal. (characteristic of E ~ 2) [] 59
Remark 4.4 This also holds for H (E) by applying 4.2. u
Remark 4.5 As long as the involution - , on E is non-trivial, we may prove 4.3 directly even if the characteristic of E is 2.
In order to see this, we must show how to produce a vector v with
B(v,v) ~ 0. Suppose to the contrary that B(v,v) = 0 for all v c M.
Assuming that B is non-singular, so not identically 0, we can find v, w e M with B(v,w) ~ 0. However B(v+w,v+w) = 0 = B(v,w) + B(w,v).
Thus B(v,w) = -B(w,v). Hence, for any a e E,
aB(v,w) = B(av,w) = -B(w,av) = -aB(w,v)
Since B(v,w) ~ 0, a = a for all ac E, and the involution on E is trivial. Contradiction.
Once we have such a vector v, we proceed as in 4.3 to produce an orthogonal basis.
Remark 4.6 Thus, we see that 4.3 holds for Hu(E) , provided we are not in the situation of a trivial involution or a field of characterisitc 2. For [M,B] E Hu(E), where the characteristic of
E is 2, we may write B as a direct sum of 1-dimensional forms and metabolic forms, C0 :)
see [K-I 22].
Diagonalizing an inner product space (M,B) means choosing a 60 basis of M with respect to which the matrix of B is diagonal.
In other words, (M,B) = ~ (Mi,Bi), where M i is a 1-dimensional vector space. It is natural then to compare the matrices of B given by different choices of bases for M.
Suppose that {e I , .... e n} and {fl ..... f~ are two bases of
M. We write E = matrix of B with respect to {e.} , and 1 F = matrix of B with respect to {f.} ] n We may express {e i} in terms of {fj} . Suppose e i = Z c..f. , j=l 13 ] and let C = (cij), C t = transpose of C .
Proposition 4.7 E = CFC t
.th l place Proof: In terms of {e. } , write e. = (0 ..... 1 ..... 0) , 1 l e i. = (0, ..... 1 ..... 0) and e.Ee~13 = eij4 = B(ei,es).~ The ij component of CFC t is likewise given by:
(0 ..... 1 ..... 0) CF~ t /
i th place I .th 3 place
= (Cil ..... Cin)F Cjl = B(Cilf I + + Cinfn , (-/ cjlf I + + Cjnf n) Cjn
= B(ei,e j )
= e. as above. [] 13
We would now like a Witt group invariant corresponding to the determinant of a matrix. This invariant should be independent of the choice of basis, as well as the Witt representative of the given 81
Witt equivalence class.
Let [M,B] s Hl(E). Let B 1 and B 2 denote two different matrices of B. By 4.7, we can write B 1 = CB2Ct, for a non-singular matrix C. Let det B 1 denote the determinant of B I. Then
det B 1 = det C det B 2 det
= det C det B 2 (det C)
Thus, we can read the determinant of B in F*/NE*, since when B 1 is diagonalized, the diagonal elements must be in F* as B is
Hermitian. Unfortunately, this is not a Witt invariant. For example, det B need not be in NE* even when IV,B] = 0 as we see below.
B l>o detB ii
which may not be a norm.
We are thus led to define a corresponding notion:
Definition 4.8 Let B 1 be a matrix with coefficients i_nn E, corresponding to a Hermitian form B . Then dis B = (-l)n(n-l)/2det B !, where n is the dimension of M, is called the discriminant of the inner product space (M,B).
Lemma 4.9 If [M,B] = 0 then dis B s NE*.
Proof: Let N be a metabolizer for M. Let {nl,...,n t} 62 be a basis for N. Extend this to a basis for M, say
{n I, .... nt,nt+ I, .... n2t} With respect to this basis, B has matrix
Interchanging the first t colums with the last t columns, we obtain a matrix
C 0
x
This requires interchanging t 2 columns. Hence B has
t 2 det B = (-i) det C det C = (-l)tdet C ~ det C.
Thus,
dis B = (-l)2t(2t-l)/2(-l)tdetC det
= (-l)t+t(2t-l)det C det
= (-l)t+tdet C det
= det C det C e NE* as claimed. []
It follows that dis is exactly the kind of invariant we seek.
There is still a problem, namely dis is not additive. Hence, we do not obtain a group homomorphism:
HI(E) + F*/NE*. 63
To remedy this problem we invent the group
Q(E) = F*/NE* • F 2 F 2 = {0,i}
[Lm 38]. The binary operation in Q(E) is given by:
ele 2 (dl,e I) (d2,e 2) = ((-i) dld2,e I + e 2)
This is an associative, commutative operation with (1,0) the additive
identity. The additive inverse of (d,e) is ((-l)ed,e).
In fact, Q(E) becomes a ring when one defines multiplication by:
e 2 e 1 (dl,e l) Q (d2,e 2) = (d I d 2 ,ele 2)
The multiplicative identity is (i,i).
Proposition 4.10 The map ~: HI(E) + Q(E) defined by
[M,B] 4 (dis B,rk M) is a group homomorphism.
Proof: Consider [M,B] and [W,B I] in Hl(E). Suppose that rank M = n and rank W = m. We have then:
~([M,B]) = ((-l)n(n-l)/2det B,n)
([W,B I]) = ((-l)m(m-l)/2det Bl,m ) ~([M,B]) ~([W,BI]) = ((-l)nm(-l)n(n-l)/2(-l)m(m-l)/2 det B det Bl,n + m) ~([M,B] + [W,BI]) = ((-i) (n+m) (n+m-l)/2 det B det B I, n + m) 84
But (_l)nm+n(n-l),2/ + m(m-l)/2 = (-i) (n2-n+m2-m+2nm)/2
((n+m) 2-(n+m))/2 = (-i) (n+m) (n+m-l)/2 = (-i)
Thus # indeed gives a well-defined group homomorphism. []
As an exercise, the reader should verify that r is actually a ring homomorphism.
We next consider the kernel of the rank homomorphism rk, which we shall call J. Thus J is the subgroup of HI(E) generated by the even dimensional forms.
Proposition 4.11 HI(E)/J 2 i_~s isomorphic t_~o Q(E).
Proof: J is additively generated by 2-dimensional forms,
Applying ~ to a generator, we obtain:
~
Thus, # induces a map ~ : HI(E)/j2 + Q(E). We now construct an inverse of ~ ,
Define y : Q(E) § HI(E)/j2 by
(a,0) + < l,-a > modulo j2
(a,l) ~ modulo j2 65
It is easy to check that y is a homomorphism, and y o ~ = id
o y = id, where id is the appropriate identity map. []
Proposition 4.11 implies that $ is i-i. Hence,
Corollary 4.12 j2 consists of even dimensional forms [M,B], with dis B = 1 e F*/NE*, ie. det B = (-i) n(n-l)/2 where n = rank m. ~]
Corollary 4.13 Restricting $ t__oo j/j2, we have j/j2 = F*/NE* . ~
In fact, we may think of this as follows: F*/NE* is embedded into Q(E) by: d + (d,0). This is a subgroup of index 2. We may 2 represent the non-identity coset by (I,i). (i,i) = (-1,0). Thus, we have the exact sequence:
q2 1 ~ F*/NE* ~ Q(E) + F 2 § 0
q2 is projection onto the second factor. By the above remarks.
Corollary 4.14 This sequence splits if and only if (i,0) = (-i,0) i__nn Q(E) if and only if -i is a norm in F*/NE*.
Remark 4.15 This defines the discriminant for HI(E). In order to define discriminant for H u (E), we fix an isomorphism:
fxl: Hu(E) + HI(E), by Proposition 4.2
B + B 1 Bl(X, Y) = (i/Xl)B(x,y). 66
Then define dis B = dis fxl B = dis B 1. We must note that this depends on the isomorphism chosen ie. this depends on Xl, where ----1 XlX 1 = U.
Remark 4.16 The discriminant inner product space, Chapter I 3.5, yields the information crucial to the discriminant invariant above, namely the determinant of B. Its advantage is that it generalizes the notion to H(D), for D the Dedekind ring of integers.
5. Signatures
The real infinite primes, P , give rise to the signature invariant which we now discuss. Suppose then that E = F(/o), and o < 0 with respect to P Thus P is an infinite ramified prime.
Lemma 5.1 If x e E, then NE/F(X) > 0 with respect t__oo P
Proof: N=N denotes the norm. Write x = a + b/o . Then E/F N(x) = a 2 - b2o > 0 since o < 0. [~
Let [M,B] e HI(E). By 4.3, we can find a basis {e i} of
M in which B is diagonalized. We can thus write M = X + ~) X- , where B(ei,e i) > 0 for e i e X +, B(ei,e i) < 0 for e i ~ X-.
Now, let v e X +, so v = Za.e.. We compute i I I
B(v,v) = B(Zaiei,Zaiei) = Z B(aiei,aie i)
= ZaiaiB(ei,e i) = ~N(ai)B(ei,ei) > 0
by Lemma 5.1. 67
Similarly, for all v s X-, B(v,v) < 0. We now define:
sgn[M,B] = dim X + - dim X- .
sgn[M,B] is called the signature of [M,B]. In order to show sgn is well-defined, we first need:
Lemma 5 .2 sgn[M,B] is independent of the basis chosen for M.
Proof: Suppose M has two bases, {ei} , {fi } which make B diagonal. Let B I, B 2 be the matrices of B with respect to
{ei}, {fi } .
Consider [M,B I] - [M,B 2] which is Witt equivalent to 0
With respect to the basis {ei,f i} of M ~ M, B 1 ~ -B 2 has matrix
It follows that sgn[M ~ M, B 1 ~ -B 2] = sgn B 1 - sgn B 2 . Thus in order to show sgn B 1 = sgn B2, it clearly is sufficient to show:
Any metabolic space [V,h] has sgn[V,h] = 0 with respect to an arbitrary basis.
So suppose V = X + ~ X- . Let N be a metabolizer for V,
N = N~ We note that n e N implies h(n,n) = 0. Thus, by the remarks preceding this theorem, X+/~ N = 0 = X-/~ N. However,
X+/~ N = 0 implies dim N < dim V - dim X + = dim X- . Similarly, + dim N < dim X + . Now, 2dim N = dim V, so that dim X- and dim X are both > (i/2)dim V. However, dim X + + dim X- = dim V. 68
Thus, dim X + = dim X- = (1/2)dim V, and sgn[V,h] = 0 with
respect to any basis.
In the process of this proof, we have shown:
Corollary 5.3 if [V,h] is metabolic, sgn[V,h] = 0. []
It is thus clear that sgn gives a well-defined Witt-invariant, which is a group homomorphism:
sgn: HI(E) + Z.
It is clear that if [M,B] has finite order, sgn[M,B] = 0,
since every element in Z has infinite order. Thus sgn is non- trivial only on the non-torsion elements in HI(E) .
We finally recall Landherr's Theorem which explicitly computes
HI(E), for E an algebraic number field [Lh]
Landherr's Theorem 5.4 There is an exact sequence:
r 0 + (4Z) + H(E) + Q(E) + 0
[M,B] = (dis B, rk V) .
The kernel of ~, ker~ , is determined by the real infinite ramified primes and the corresponding signatures, each of which is divisible by 4. Here r is the number of real infinite primes.
When r = 0, ~ is an isomorphism. 69
This theorem is important in the boundary computation in
Chapter VI.
By Proposition 4.10, H(E)/J 2 = Q(Z), so we can state:
Corollary 5.5 j2 = 4(Z ~ Z .... Z) [] r times
Remark 5.6 As with the discriminant, we can define a signature
invariant for H u (E). This is done by picking an isomorphism fxl: Hu(E) § HI(E) as in 4.15. We then define
sgn[M,B] = sgn (fxl [M,B]) Chapter III POLYNOMIALS
Given a Witt equivalence class, [M,B,I] in w(k,F) , we
shall decompose it as [M,B,Z] = ~[Mi,Bi,ii], according to the
irreducible factors of the characteristic polynomial of i.
This is the object in Chapter IV.
In this chapter, we lay the groundwork for the above decom- position. This involves a careful study of the characteristic and minimal polynomials of i. These polynomials belong to
K(F) = {p(t) : p(t) is a monic polynomial with non-zero constant term, coefficients in F a field} . We assume throughout this
section that we are working over a field F.
On K(F) we define an involution Tk: K(F) + K(F) . The characteristic and minimal polynomials are shown to be T k fixed.
When p(t) is irreducible and T k fixed, we consider the
field F[t,t-l]/(p(t)) = F(~). It is shown that there is an -i induced involution of F(@) given by @ = k0
This discussion provides the key ingredients for the computations to be made later.
Let (M,B) be an inner product space, and let i: M + M be F-linear. Recall the adjoint, ~* of ~ is defined by the
equation B(v,s = B(s [I 4.9].
Lemma i.i s and s have the same characteristic polynomial
and the same minimal polynomials.
Proof: For any polynomial p(t) , B(p(s = B(v,p(i)w) . 71
Thus, p(s = 0 if and only if p(s = 0, since B is non- singular. The assertion about minimal polynomials follows.
Working over a field, we may view M as the space of n • 1 column matrices and B as an n x n matrix B', [II 4].
B(v,w) = vtB'w, where v t denotes the transpose of v, w denotes the conjugate of w,
W = , W =
n n
is multiplication by an n • n matrix L. To simplify our notation, we identify B with its matrix and write B' = B.
We compute, [(B-iLtB)v]tBw = vtBLB-IBw = vtB(Lw). It follows that L* = B-ILtB = matrix of Z*. Hence, letting det denote the determinant, and I the n x n identity matrix,
det(tI - B-ILtB) = (det B) (det (tl - B-ILtB)) (det B -1)
= det (BtB -I - BB-ILtBB -I)
= det (tI - L t)
= det (tI - L).
The assertion for characteristic polynomials follows. []
Let s be a map of degree k. Then ~ is non-singular, and s is related to s by:
Lemma 1.2 If s has matrix L, and s has matrix L*, then
L* = kL -I . 72
Proof: B(s = B(v,s = B(s163163 = kB(Z-Iv,w) = B(ks
Again, since B is non-singular, it follows that s = k~ -I.
Proposition 1.3 Let s be a map o_ff de~ree k. Then both the minimal and characteristic polynomials o_ff Z satisfy:
tdeg ree P(t)p(t-lk) = a0P(t) ,
where a 0 = constant term of p(t)
Proof: Let x(t) = characteristic polynomial of ~ . Since s is non-singular the constant term of x(t) is non-zero. (Of course the dimension of the vector space M is n, the degree of x(t)). -i -i ~* = k ~ by Lemma 1.2. Thus, by Lemma i.i, s and ks have the same characteristic polynomial. The identity
(-t-iL) (tI-kL -I) = (kt-iI - L)
yields det (-t-iL)x(t) = x(kt -I)
However,
det (-t-lL) = (-l)nt-ndet L
= (-l)nt-n(-l)na0
= t-na0 , a 0 = constant term of x(t).
Thus, t-na0x(t) = x(kt-l), so that tnx(kt -I) = a0x(t) as desired.
In order to check the result for the minimal polynomial p(t) of ~ we again use s = ks -I. Let degree p(t) = m. By Lemma i.i 73 p(k1-1) = 0. Thus, q(t) = a0-1tmp(kt-1 ) is a monic polynomial of degree m = degree p(t) with q(1) = 0. Hence p(t) = a0-1tmp(kt -I) as claimed. []
We continue the study of these polynomials by letting
K(F) = {p(t) : p(t) is a monic polynomial, with constant term a 0 ~ 0} .
n Here, p(t) = Z a t i This is a cancellation semigroup with i=0 i " respect to multiplication of polynomials. Further, any polynomial in K(F) can be uniquely factored into a product of powers of irreducible polynomials in K(F) .
Fer k ~ 0, k e F*, we are led by Proposition 1.3 to introduce an automorphism of period 2 on K(F) by:
Tk: p(t) + tdegp(t)a0-1p(kt-l) = (TkP)(t)
Proposition 1.3 then says that for a degree k mapping structure
(M,B,s both the characteristic and minimal polynomial of s
are T k fixed.
Lemma 1.4 A polynomial p(t) is fixed under T k if and only
if its coefficients satisfy a.k j 0 < j < n = degree p(t). ] = a0an-j ' -- --
Proof: Clear by definition of T k. ~I
We note that a02 = k n.
To summarize, if p(t) s K(F) is T k fixed, exactly one of the
following three cases applies. 74
Type I: deg p(t) = 2n and a 0 = k n.
Thus
= kn-Ja2n - 0 < j < n . a 3 j - _
Type 2: deg p(t) = 2n and a 0 = -k n . Assume char F ~ 2.
Thus
a = -kn-Ja2n - j 0 _< j _< n ,
so that
a = -a = 0 . n n
Note: There is no loss of generality in assuming characteristic F ~ 2 in this case.
Type 3: deg p(t) = 2d + 1.
Thus
k 2d+l = a 0 2 , and k = (a0/ k d )2
Lemma 1.5 If p(t) E K(F) i_{s T k fixed, of degree 2d + i, then -a0k-d is a root of p(t) ~
Proof: Consider p(-a0k-d). The 2j term is a2j(-a0k-d) 2j =
k j However, this 2j term cancels with the 2(d - j) + 1 a2j term since
a2(d_j)+l(-a0k-d)2(d-j)+l = a2(d_j)+l(-a0k-d)(-a0k-d)2(d-J) 75
However,
a2(d_j)+la 0 = a2jk2J, and (-a0k-d) 2 = k.
So the above equals:
= _a2jk2Jk-dkd-J = -a2jkJ
Lemma 1.6 If p(t) ~ K(F) is of type 2, then (t 2 - k) divides p(t). (Characteristic F # 2)
.th Proof: For 0 < j < n, p(/k) will have 3 term a.(/k) j, 3 and (2n - j)th term a2n_j(/k) 2n-j Further,
aj (Jk) j = -kn-Ja2n_j (/k) j = -(/k) 2(n-j)+J a2n-j = -a2n_j (/k) 2n-j
and these terms cancel.
Since char F r 2, a = 0 and /k is a root of p(t). Hence n we can write (t 2 - k)q(t) = p(t) over F(/k). It is clear that q(t) ~ F[t], since t 2 - k and p(t) are. r']
Hence, irreducible polynomials in K(F) which are T k fixed fall into the three following types.
Type i: deg p(t) = 2n and a 0 = k n 78
Type 2: k ~ F** and t 2 - k = p(t), when char F 9 2
Type 3: k e F** and p(t) = t • /k
On F[t,t -I] we introduce the involution: t + kt -I , t -I § k-lt. Denote this by: 7 + ~. See [VI 2].
t_Ii n Let ~" ~ F[t, , say 7 = ~A.t j. Then y = ~ if and only -m 3 if n = m and we have A . = A.k j , 0 < j < n . -3 3 -- -- Suppose, y = ~ and A = I. Then tn7 = p(t) belongs to n K(F) and is a T k fixed polynomial of type i. Conversely, any
T k fixed polynomial of type 1 can be written as tn~ = p(t) , for a unique y = ~ , where 2n = degree p(t).
Continuing, suppose p(t) is a T k fixed polynomial of type 2.
Then by Lemma 1.6, p(t) = (t 2 - k)q(t) . However, T k is multiplicative, and t 2 - k is T k fixed. It follows that q(t) is also a T k fixed polynomial, q(t) has degree 2(n - i) and constant term k n-l.
Hence p(t) = (t 2 - k) q(t) where q(t) is a T k fixed polynomial of type i, or q(t) = i. So we can write p(t) =
(t 2 - k)tn-i 7 = tn(t - kt-l)~ where y = ~ and degree p(t) = 2n.
Finally for type 3, let p(t) have constant term a 0. By
Lemma 1.5, p(t) = (t + a0k-d)q(t). As above we show q(t) is a type 1 T k fixed polynomial of degree 2d or q(t) = i.
Lemma 1.7 If p(t) e K(F) i_ss T k fixed, then the principal
ideal (p(t)) C F[t,t -I] is - invariant.
Proof: We take first the case when p(t) is of type i, say 77 p(t) = tn7 . It follows that (p(t)) = (y) since t is a unit in F[t,t-l]. But T = ~, so that (p(t)) is - invariant.
Next, let p(t) be of type 2. Then by the discussion before the Lemma, p(t) factors as p(t) = tn(t - kt-l)T , where T = T and (t - kt -I) = -~t - kt-l). Clearly then, (p(t)) = ((t - kt-l)y ) is - invariant.
For p(t) of type 3, p(t) = td( t + a0k-d) T, with 7 = ~ .
Now, we compute (t + a0k-d) = kt -I + a0k-d = (t + a0k-d) (a0k-dt-l) since (a0/kd)2 = k. However, a0k-dt-i is a unit in Fit,t-l], which again yields that (p(t) is - invariant. []
We summarize this discussion. Let p(t) be a T k fixed
irreducible polynomial in K(F). Then there are three cases to consider.
Type I: F[t,t-l]/(p(t)) = F(8) is a simple algebraic extension of F together with a non-trivial involution ~ = ~-i Here @
is identified with t. If ~ = t, then kt -I = t so that t 2 = k, t 2 - k = 0 and we are in type 2.
Type 2: F[t,t-l]/(p(t)) = F[t,t-l]/(t 2 - k) = F(/k) for the
case that k ~ F**. The induced involution is /k + k(/k) -I = /k, which is trivial. Note that this is not the involution /k ~ -/k.
Type 3: In this case, k e F**, say f2 = k. The irreducible
polynomial is p(t) = t • f. The field F[t,t-l]/(t • f) = F, by
identifying t with • f. The involution:
+_ f+ k/• f = ~f)2/(+f) = _+f 78 is trivial, and so is the extension.
Finally, in the type 1 situation when the involution - is non- trivial, we wish to describe the fixed field.
Lemma 1.8 In the type 1 situation, the fixed field of
(F(8),-) is F(8 + ks-l).
Proof: There is the embedding F[x] + F[t,t -I] given by x + t + kt -I. We claim that the image of F[x] is the subring of
- fixed elements. n Let 7 e Fit,t-l], y = Z A.t j with A = A.k j, be a typical -n 3 -3 3
- fixed element. Consider %, - An(t + kt -I n. This is still - fixed,
n-i and can be written as ~ B.tJ Continulng inductively, -n+l 3
n = i=0Z a.l (t + kt-l) i = q(t + kt -I) as claimed.
Suppose A n = 1 and y is the image of a monic polynomial,
q(x) , in F[x] .
Claim: If tny = p(t) is irreducible, then so is q(x). For
if q(x) factors as q(x) = ql(x)q2(x), with r = degree ql(x), w = degree q2(x), then
p(t) = tny = [trql(t + kt-l)] [tWq2(t + kt-l)],
SO that p(t) also factors.
We may thus write F[x]/(q(x)) as the - fixed elements in
(f(%),-). Clearly, the minimal polynomial of 8 over F(8 + k8 -I) is x 2 - (6 + ks-l)x + k. Chapter IV WITT GROUP OF A FIELD
We wish to compute the Witt group W(k,F) for F a field.
This is done by decomposing W(k,F) as a direct sum of groups w(k,F;f) according to the irreducible factors f(t) of the charac- teristic polynomial of ~ . We identify each group,
W(k,F;f) = W(k,F;F[t]/(f(t))
by taking anisotropic representatives.
On F[t]/(f(t)) there is an induced involution by Chapter III.
We prove a trace lemma which then enables us to compute these groups w(k,F;F[t]/(f(t)). In this manner then we will have computed W(k,F).
The trace lemma is then used in several cases to compute Witt groups. This computation is valuable for the ensuing chapters.
Finally, we prove a result showing the relation between torsion in W(-k,F) and the number of squares necessary to express k as a sum of squares.
1. Decomposition by characteristic polynomial
Given a degree k mapping structure (M,B, ~, we may view M as a D[t]-module by defining the action of the indeterminate t
to be the same as s By III .1.3, the characteristic polynomial of ~ , p(t), is T k fixed.
Proposition i.i If (M,B,Z) is metabolic, and ~ has charac- teristic polynomial p(t), then p(t) factors as p(t) = f(t)- Tkf(t) 80 for some monic polynomial f(t).
Proof: Let f(t) be the characteristic polynomial of s restricted to N, where N is a metabolizer for M.
We now make HomD(N,K) into a D[t]-module. This is done by defining the action of t by:
Let h e HomD(N,K) Then (t-h) (n) = h(s , where s is the adjoint of i. Viewed thus, AdRB: M + HomD(N,K) is a D[t]-module homomorphism since:
AdRB(t m) = AdRB(Zm) = B(-,Zm) = B(s
= t-B(-,m) = t'AdRB(m) ,
We thus obtain an exact sequence of D[t]-modules:
AdRB 0 + N + M + HomD(N,K) + 0
By definition of the action of t on HomD(N,K), its characteristic polynomial is simply that of Z*IN . We can see this by identifying
N with its dual space, HOmD(N,K) . The action of t on N induced from the corresponding action of t on HomD(N,K) above is then s
Note: In this section we are working over F a field, so that
D = K = F. We have used the notation D, K to follow our previous conventions.
The question arises; what is the characteristic polynomial of
s ? To begin with, by Lemma IIL1.2, ~* = ks -i on M, hence all
the more so on N. We write L 1 as the matrix of ~ restricted to N.
Then the matrix of ~* is kLl -I Now f(t) = characteristic polynomial 81 of s = det (tI - LI). We compute
det(tI - kLl-l) = det (-tL 1 + kI) det (-LI-I)
= det (tI) det (kt-II - L I) det (-LI -I) = t n det (kt-II - L I) det (-LI-I)
= t n f(kt -I) det (-LI -I)
where n = degree f(t) = dimension N. Here det(-Ll-l) is a constant; from which it follows that det (-LI -I) = a0 -I where a 0 is the constant term of f(t), and that Tkf(t) = det (tI - kLl -I) = characteristic polynomial of s .
The exact sequence given, together with the computation given above then yields, by [L-I 402] that p(t) = f(t)-Tkf(t) as claimed.
We continue by forming GK(F), the Grothendieck group associated to K(F). This is the free abelian group generated multiplicatively by the irreducible polynomials in K(F) . T k induces an automorphism of period 2 on GK(F), so we can form H2(C2;GK(F)), denoted simply H2(k;K(F)) [M 122]. This is identified as {f s K(F) :Tkf = f} modulo {g ~ K(F) : g = h-Tkh} This in turn is an F2-vector space with a basis element for each T k fixed irreducible polynomial. We denote this basis by B
Lemma 1.2 The map X : W(k,F) + H2(k,K(F)) given by:
[M,B,s + characteristic polynomial of ~ is an epimorphism.
Proof: X is well-defined by !.i. To see that X is onto it suffices to show that every p(t) s B is in the image of X 82
Given p(t), consider V = F[t,t-l]/(p(t)) = F(9). Let B be given by: B(x,y) = traceF(6)/FX ~ , where - denotes the involution induced on F(@) by e § k@ -I
B is symmetric since trace xy = trace xy. B is clearly non- singular; one may apply the trace lemma 2.1 to be proved or prove it directly.
Define g: V + V by x + @x. We compute:
B(gx,gy) = B(@x,Sy) = trace(@0xy) = trace(kyx) = kB(x,y).
Hence, [V,B,g] e W(k,F). Since 6 satisfies p(t), the minimal polynomial of g is p(t). However, p(t) divides the characteristic polynomial of g , and degree p(t) = degree of characteristic polynomial.
Hence p(t) = characteristic polynomial of g . Q
Given a Witt equivalence class [M,B,s M is a D[t]-module by identifying t with g . We now wish to decompose [M,B,s according to the characteristic polynomial of g. We begin with:
Lemma 1.3 Let p(t) b_~e T k fixed. Then we can factor
r 1 r w s I _s I s k _s k p(t) = Pl "'" Pw ql ql .... qk qk
into irreducible factors, where qi denotes Tkq i. In this decomposition, each Pi(t) i__{s - invariant, ie. T k fixed, and the qi are not - invariant, rl 83
Lemma 1.4 Suppose (M,B,s has characteristic polynomial p(t) = Pl(t) P2(t), with Pl(t), P2(t) relatively prime polynomials, which are both - invariant. Then (M,B,i) = (MI,BI,s I) G (M2,B2,s 2) where Z i has characteristic polynomial Pi(t).
Proof: Let M i = {v e M: pi(s = 0} , B i = BIM" , s = s l l Since (pl,P2) = i, we can find polynomials r(t), s(t) with
Pl(t)r(t) + P2(t)s(t) = i.
Remark 1.5: This statement is false over Z, and is the reason
the decomposition fails over Z.
Hence, if v s M, then pl(s163 + p2(s163 = v. However,
pl(1)r(~)v ~ M 2 since pl(s = 0. It follows that M = M 1 + M 2.
If v s MI/~ M 2 , by the above it is clear that v = 0. It
follows that M is a direct sum of MI, M2, ie. M = M 1 ~ M 2.
We next show that B = B 1 ~ B 2, so that B 1 and B 2 are inner
products.
To begin with, consider p2(s MI+ M 1 . p2(s is clearly l-l,
and hence an isomorphism as we are working over a field. Thus, if
v ~ M I, we may write v = p2(s I. Let w ~ M 2.
B(v,w) = B(P2(s
= B(Vl,P2(s ) = i/k n B(s163163
= i/k n B(invl,a0P2(s )
= 0
where a 0 = constant term of P2(t), since P2(t) is - invariant. 84
Thus, B = B 1 G B 2 as claimed.
Finally, we must show li: M i M l . Note that pi(1) (s =
s (v)) = ~(0) = 0. Hence, s = Z!M. maps M i ~ M i This 1 shows
(M,B,s = (MI,BI,s I) G (M2,B2,s 2)
as claimed. D
Lemma 1.6 Suppose (M,B,Z) has characteristic polynomial r I _r 1 p(t) = characteristic polynomial of Z and P = ql ql ' where ql
is irreducible. Then (M,B,s ~ 0.
Proof: We are assuming ql ~ ql" Since (ql,ql)= i, M will
split as M = M 1 M 2, as in 1.4. We must now examine B.
Let v e M I. As in 1.4, we may write v = q~(s (Vl). Let w e M 1 .
B(V,W) = B(ql(Z)vl,ql(s I) r . r = B(Vl,ql(s ql(s I)
= 0 as e~ us asmarx I0 Since B is non-
singular, dim M 1 = dim M 2 = 1/2 dim M. As in 1.4, M 1 is s invariant.
It follows that M 1 is a metabolizer for (M,B,s 85
r We are almost ready to state the Decomposition Theorem.
First, some notation.
Definition 1.7 Let T be a multiplicative subset of D[t].
Then W(k,K;T), respectively A(K,K;T), denotes Witt equivalence classes in W(k,K) , A(k,K), which are annihilated by T. I_nn particular, when T consists of non-negative powers of a T k fixed irreducible polynomial f, we shall use the notation W(k,K;f).
Theorem 1.8 (The Decomposition Theorem) For F a field
W(k,F) = ~ W(k,F;F) where B denotes the basis of H2(k;K(F)) f E
consisting of T k fixed irreducible polynomials.
Proof: Let [M,B,s e W(k,F). Let p(t) be the characteristic polynomial of s By 1.3, we can factor p as
r I r w s I _s I s k _s k p(t) = Pl "'" Pw (ql ql ) "'" (qk qk )
By induction and 1.4, [M,B,Z] = ~ [Mi,Bi,s i] , where each s has 1 r. S. S, ( I - l characteristic polynomial pi I or qi qi )" By 1.6, those
S. S, [Mi,Bi,s i] with characteristic polynomial (qi I -qi l ) are
Witt ~ 0. This defines a homomorphism L: w(k,F) § ~ W(k,F;f). fEB We must show L is well-defined. So suppose [M,B,I] is
metabolic. Then (e [Mi,Bi,s = 0 where Z i has characteristic
polynomial pi ri We need to show that if [MI,BI,s I] [M2,B2,s 2]
= 0, where Pi(t) is the characteristic polynomial of s and
Pl(t) is relatively prime to P2(t), then M 1 ~ 0 and M 2 ~ 0.
We identify M 1 with M 1 ~ 0 C M 1 M 2. Let H be a metabolizer 86 for M 1 G M 2. Then H is s = s ~ 12 invariant. Further, since p2(s = P2s ~ P2s , p2(Z)HC M I. In fact, since Pl(t) and
P2(t) are relatively prime, it follows that p2(s is a i-i mapp- ing: H {~M 1 + H {~M 1 . Since we are working over a field, p2(s (H ~ M I) = H {~M I. We claim (H ~ M I) = (H /~MI )A" in
M 1 so that M 1 ~ 0 and L is well-defined.
Clearly H ~ M 1 is i I invariant, and H f% MIC (H /~MI )j" .
Let x e (H{~ M1 )A" . We must show x e H f~M I.
To begin with, note that if (hl,h 2) e H, then so is (p2(il)hl,0) since H is Z1 ~ s = ~ invariant. Further, since p2(il) is an isomorphism on H /~ M I, it follows that (hl,0) E H.
Now (x,0) = x ~ (H /~MI )A" . If h = (hl,h 2) e H, it follows that (hl,0) e H~ M I. Thus
(B1 ~ B 2) ((x,0),(hl,h2)) = B((x,0),(hl,0)) = 0
Hence, x s H ~= H, and (H ~ MI)A'C H ~ M I.
L is clearly onto by 1.2.
L is i-i, since if ~[Mi,Bi,i i] has each M i ~ 0 then so too is e[M i] = 0 .
Let us give another interpretation of this isomorphism L.
Let f(t) be a T k fixed irreducible polynomial, so f e B .
Let S = D[t] - (f(t)), and [M,B,s e W(k,K). Then localizing with respect to S, we obtain, (M(S),Bs,s Note that the adjoint map,
AdRBs: M(S) + (HOmD(M,K)) (S) = HomD(S) (M(S),K(S)) 87 is an isomorphism. AdRB S is an isomorphism since localization is an exact functor, [A,Mc 39]. The second isomorphism follows from
[B-2 II 2.7].
M is a torsion D[t]-module. Thus M(S) is annihilated by fi(t), some i. Hence (M(S),Bs,s s) ~ W(k,K;f).
We combine over all f e B , to obtain exactly the L given in
Theorem 1.8. Since L can be viewed as arising from localizing, we shall call L the localization homomorphism.
In fact, as long as we localize at all prime ideals in D, or
D[t], where M is a finitely generated torsion D, or D[t]- module, we obtain such an L.
Theorem 1.9 Let K = F/D. By localizing at all prime ideals p
in D, we obtain an isomorphism:
L: W(k,K) + ~ W(k,K(P);D(P)) P prime in D
Here
K(P) = (F/D) (P) = F/D(P).
Proof: Exactly as in 1.8. []
We should like to describe these pieces W(k,F;f). In order to do this, we need some further notation.
Definition i.i0 Let S be a D-algebra, finitely @enerated as a D-module. Then W(k,K;S) denotes Witt equivalence classes,
[M,B,s in W(k,K) with a compatible S-module structure, meaning 88 there exists r e S with rm = s for all m e M.
We shall be specifically interested in the case S = F[t,t-l]/(f(t)) where f(t) ~ B . For this S, observe that there is an inclusion
j: W(k,F;S) ~ W(k,F;f).
Structures on the left are annihilated by f(t), those on the right are annihilated by some power of f.
Proposition i.ii j is an isomorphism.
Proof: j is clearly 1-1. Let (M,B,Z) be an anisotropic representative of a Witt equivalence class in W(k,F;f). Thus if
N C M, N ~ 0 is s invariant, then N ~ N ~= 0. It follows that
S = N N ~ , and that (M,B,i)=(N,B I ,s (N~,BI,s , where
BI, Z I denote the restrictions of B,s to N,N ~. This is standard linear algebra, see [H 157]. Continuing we can write
M = N 1 ~) N 2 ~) ... ~) N r
as a direct sum of inner product spaces (Ni,Bi,s , where each N i has no non-trivial Z invariant submodules. Such N. are called 1 irreducible.
Let T. = annihilator of N. in F[t]. We want to show T. 1 l l is a maximal ideal in F[t]. Suppose not. Then TiC SiC F[t],
for some ideal S.. l 89
Claim: S.N..~ N.. For if S.N = N., we recall [K-2 50] 1 1 1 1 1 1 Theorem 76: Let R be a ring, I an ideal in R, A a finitely generated R-module satisfying IA = A. Then (l+y)A = 0 for some y s I.
It follows that (i + y)N i = 0 for some Y e S i- Hence,
(i + y) s TiC S i, so I + y s S i. Hence 1 s S i. This contradicts
S i ~ F[t]. Thus SiN i ~ N i.
S.N. @ 0, since S. ~ T.. S.N. is t invariant, ie. Z invariant 1 1 1 1 1 1 as we identify the action of t with ~ , because S. is an ideal. 1 However, we have thus constructed a non-trivial s invariant submodule of N i. This contradicts N i being irreducible. It follows that T is indeed a maximal ideal in Fit]. Thus 1 T. = (f(t)), and j is onto. [] l
Remark 1.12 Proposition 1.11 has shown the two notations given in 1.7 and i.i0 to be redundant. Nonetheless we shall use both. The notation W(k,F;f) is used when we wish to stress the polynomial aspect of the mapping structure, w(k,F;F[t]/(f(t)) i__ss used when we wish to stress the module structure.
Proposition 1.13 Let K = F/D. Then the inclusion
J W(k,K;D/P) + W(k,K(P) ;D(P))
is an isomorphism, where P is a prime ideal in D.
Proof: Same as i.ii. ~1 90
Here W(k,K;D/P) denotes equivalence classes [M,B,s in which M
has a D/p module structure. W(k,K(P) ;D(P)) denotes equivalence
classes in which M has a D(P) module structure.
For F = Q, D = Z, D/p = F n a finite field. Since M is a P vector space over Fpn, B must take its values in the cyclic sub-
group of Q/Z annihilated by p. By the natural choice of generator
for this subgroup, namely l/p, we have W(k,K;D/D = W(k,Fp) .
Proposition 1.14 A(F) decomposes as
A(F; f) = A(F;F[t]/(f(t)) f sB feB
Proof: The proof is exactly like 1.11, where now t acts as
s , the symmetry operator. []
2. The trace lemma
Given f(t) E B , meaning f is irreducible, T k fixed, we form the field F[t,t-l]/(f(t)) = F(8). By III.l.7, (f(t)) is - in- variant, so there is an induced involution on F(@ ) . This involution
is non-trivial in the type 1 situation only. We aim now to identify explicitly the group W(k,F;F[t]/(f(t))). We begin with:
Lemma 2.1 (The trace lemma). Let R be a ~ ring with unit, A an R-al~ebra, E an A-module, and F an R-~. Than there is the followin~ correspondence:
Let <,> : M x M + E be a non-sinqulax bilinear form nver A.
Let t: E + F be an R-linear map, which induces an isomorphism 91
^ t: E § HomR(A,F) , b__yy e + t(- e) = t(e).
Then the map t o <,> : M • M § F is non-singular.
Conversely, if M is an A-module with non-singular form
(,) : M x M + F, ,) R-linear, then there is non-singular form
<,> : M • M + E with t~ = (,). <,> is A-linear.
Further, this correspondence preserves annihilators of sub- modules and the metabolic property provided the R-module structure of
M lifts compatibly to A.
Proof: Part l: Given <,> : M • M § E, and t: E + F, we wish to show ,) = to<,> : M x M + F is non-singular.
Let AdR: M + HomR(M,F ) denote the adjoint of (,). We want to show Ad R is an isomorphism.
Ad R is i-i: Let m ~ 0 be in M. We want to show (-,m) ~ 0.
Since <,> is non-singular, we can find n e M with
Now
<,> is bilinear over A, so
a
Hence,
t(a
Thus, (-,m) ~ 0 as claimed, and Ad R is i-i.
Ad~ is onto: Let f ~ HomR(M,F). For each m e M, define an
A R-linear map A § F by a + f(am). Since t is an isomorphism, this map equals t(-f0(m)) for some f0(m) e E. Now f0 defines 92 an A-linear map f0: M § E. By non-singularity of <,> it follows that f0(m) =
f(m) = t(fo(m)) = t(
so that Ad R is onto as claimed.
Part 2: Let M be an A-module, together with a non-singular
R-linear form (,) : M • M + F. We need to define <,> with to<,> = (,).
Let (-,n O ) e HomR(M,F). As before, for each m e M we can define an R-linear map A + F by a + (am,n0). Again E ~ HOmR(A,F) implies (am,n 0 ) = t(af0(m)) for some unique f0(m) s E. Now define
Let Ad R denote the adjoint of <,> , AdR: M + HomA(M,E)-
A_~dR is i-i: Let m ~ 0 be in M. We want to show <-,m> ~ 0.
By non-singularity of (,), we can find n s M with (n,m) ~ 0.
Hence,
Ad R is onto: Let f 8 HomA(M,E). Then (t o f) s HomR(M,F). By the non-singularity of (,) there exists n 0 E M such that
(t o f) (m) = (m,n0), for all m ~ M. Hence
(ta f) (am) = t(f(am)) = t(af(m)) = (am,n 0 ) ,
so that by definition of <,> we have
The last statement of the theorem follows from the defnitions.
We may extend Lemma 2.1 in a special case.
Lemma 2.2 Suppose that A,E ~iven in 2.1 have a compatible involution -, meaning (a-e)= ae . Suppose also that t: E + F satisfies t(e) = t(e) for all e e E. Then the correspondence of
Lemma 2.1 extends to a correspondence between Hermitian forms
(M,<,>) with values in E, and symmetric forms (M, (,)) with values in F which have a compatible A-module structure, meaning
(ax,y) = (x,ay).
Proof: If <,> is Hermitian,
a
(x,y) = t(
so that (,) is symmetric. Also
(ax,y) = t(
Conversely, let (,) be symmetric. Then
t(a
= t(
However, E = HomR(A,F) via ;, so
Hermitian. [] 94
We recall the identification made at the beginning of Section 2,
F[t]/(f(t)) = F(8). We are now ready to compute W(k,F;F(@)), where
f(t) is T k fixed and irreducible.
Theorem 2.3 W(k,F;F(@)) = H(F(8))
Proof: We apply 2.1 and 2.2, with R = F, and A = E = F(8),
t: E + F is the trace homomorphism.
We must check the non-singularity condition on t, namely that
t: F(e) + HomF(F(e),F) induces an isomorphism.
We must assume F(e) is a finite, separable extension of F.
Thus t(- x) ~ 0 for x ~ 0 [L-I 211]. It follows that t is i-i.
However, F(0) and HomF(F(e),F) are vector spaces over F of the ^ same dimension. Hence t is an isomorphism, so that we may apply
2.1 and 2.2. []
Comment: Clearly, t(e) = t(e), so that 2.2 applies. In our
identification, [V,<,>] e H(F(0)) corresponds to [V,(,),Z]
W(k,F;F(8)). Here to<,> = (,). The map s is recovered from
Hermitian as multiplication by @, Zv = 8v.
We shall reserve the term Hermitian for the case that the
involution is non-trivial. Thus 2.3 is for type 1 polynomials.
For type 2 irreducible, f(t) = t 2 - k. We then read Theorem 2.3
as:
W(k,F;F[t]/((t 2 - k)))= W(F(/k)).
We may thus restate the Decomposition Theorem 1.8, as 95
Theorem 121.4 If k ~ F**, then
W(k,F) = W(F (/k)) H(F[t]/(f (t))) feb f of type 1
If k e F**
W(k,F) = W(F) ~ W(F) ~ H(F[t]/(f(t))) feb f of type 1
Remarks:
(!). The Hermitian terms runs over all irreducible T k fixed polynomials in K(F) of type i. The same field F[t]/(f(t)) = F(@)
= F(o) = F[t]/(g(t)) may be repeated.
(2). If k e F**, the two Witt terms correspond to the two irreducible polynomials of type 3, t + /k, t - /k.
(3). If the characteristic of F is 2, then k e F**, so
W(k,F) = W(F) H(F(@)) feB since
t + /k = t - /k
in this case. Again, the field F(@) may be repeated.
(4). This theorem equally applies to the skew case; simply write
We(k,F) =We(F(/k)) He(F(@)), e = -+ i. feb
Our final goal of this section is to relate this discussion to the asymmetric case. Let [M,B] e A(F). Recall the symmetry operator s: M + M satisfying B(x,y) = B(y,sx). Consequently B(x,y) = B(sx,sy), 96 so that s yields a map of degree 1 on IM,B].
Again, there is the involution on F[t,t -I] induced by T k, with k = I. This extends to F(8) = F[t,t-l]/(f(t)), where f is an irreducible T k fixed polynomial. The induced involution on F(8) is e ~ ~ = e -I
Let S = F[t]/(f(t)) , and consider A(F;S). This denotes structures
[M,B], in which t acts as s the symmetry operator. Since
A(F) = ~) A(F;S) by 1.14, we wish now to compute A(F;S). We do this in two ways.
Theorem 2.5 Let f(t) be an irreducible T k fixed polynomial of type 1 (k = i) . Then
A(F;f) -~ H(F(e)) -~ HeF(e) .
Proof: The idea in this computation is to apply the trace lemma.
We may do this either using trace: F(6) + F, or a scaled trace:
F(6) § F.
Part(a) : Using trace = t: F(e) + F.
Let [M, <,>] ~ HeF(8) and [M,(,)] r A(F;f).
We need to show that <,> is e Hermitian with values in F(8) if and only if (,) is asymmetric with values in F, satisfying
(x,y) = (y,sx) with s being identified with @. We are of course applying the trace lemma with the map t = trace F(@)/F.
Let (,) satisfy (x,y) = (y,ex). Then
t(a
= t
Hence, since t is an isomorphism
is 8 Hermitian.
Conversely, suppose <,> is @ Hermitian. Then
(x,y) = t
= t
as desired.
Thus A(F;f) = Hs(F(8)) , where F[t]/(f(t)) = F(8).
Part (b) : Using the scaled trace. Since ee = I, by Hilbert 90,
there exist u ~ F(8) with uu -I = 8. Let tl: F(8) + F be given by x + trace(x~-l). It is clear that tl is an isomorphism.
Again t denotes trace F(8)/F.
Now suppose [M,<,>] e H(F(8)).
Then
(x,y) = tl
= t
= t
= t
= (y,ex) .
Conversely, suppose (x,y) = (y,Sx).
Then
t(a
= tl
= (uy,a@x) = (u~,ax)
= (uy,ax) = tlua
= t(a
= t(a
Again, t is non-singular, so
Note that we can choose u = 6/(1 + 8) so u = i/(i + e), ----1 and u = 1 + @ . We give both identifications in this theorem since on certain occasions it is more convenient to think of Hermitian forms as giving A(F). The disadvantage is that we must use a scaled trace to make this identification.
We should also give the third identification. Namely, it follows from this theorem that
H(F(6)) = He(F<@)) via: h: <'> + <'>i with h defined by
= u-ly> = (a) If <'>l is e Hermitian, = <~uy,x> 1 = = = and < , > is Hermitian. 99 (b) If <,> is Hermitian, = = [ = @ so that <'>i is 8 Hermitian. Corollary 2.6 There is a commutative triangle of h H(F(@)) + HeF(8) t I ~ J A(F;F[t]/(f(t)) --I isomorphisms, t is trace F(8)/F. t I is trace scaled by u , where u/u = @ h: <'> § <'>l' is defined by Thus, the decomposition theorem reads, Theorem 2.7 A(F) = ~) H (F(O)) -- ~) H(F(e)) f~ B 8 f~ B For F = Q the rationals, or F = F a finite field, Hermitian P of F(@) is well known [Lh] and [M,H] , see Chapter II 5.4 . 100 3. Computing Witt groups We are interested in the group W(k,Z). Let [M,B,s e W(k,Z). Then we may view M as a Z[t,t-l]/(f(t)) module, where f(t) is the characteristic polynomial of s , and t acts as s As has been pointed out, for S = Z[t,t-l]/(f(t)) , the decomposition theorem fails; W(k,Z) ~ ~ W(k,Z;S) Later, we shall measure this failure, [VIII i]. Our next task is to describe these pieces W(k,Z;S), for S = Z[@] above. Thus, let f be a monic,integral T k fixed irreducible polynomial, S = Z(@) = Z[t,t-l]/(f(t)). We begin by describing the maximal ideals in S. Proposition 3.1 The maximal ideals of S are of the form M = (P,g(0)), where g is a monic integral polynomial whose mod p reduction Y is irreducible and y divides the mod p reduction of f, denoted Proof: S/M is clearly a finite field, indeed it embeds into D/P where D is the maximal order, and P/~ S = M Suppose S~{ lies over the prime field Fp. It follows that p ~ M Further, S/M is generated by @i' the image of G in residue field. Let y(t) be the monic irreducible polynomial over F P of 8 1 Let g(t) be a monic integral polynomial whose mod p reduction is y(t). Then clearly g(@) e M and (p,g(0)) = M. Since g is irreducible mod p, g is irreducible. Further, f(@) = 0, so ~(el ) = 0, and y divides ~ as claimed, r-1 101 Remark: M is invariant under the involution - induced by = k@ -I if and only if S/M has a well-defined involution induced by -. Further, S/M has involution - if and only if 7(t), the irreducible polynomial of @i is T k fixed. For if 7 is T k fixed, we have already seen there is an involution induced on Fp(@l) = S/M Conversely, when there is the - involution on Fp(01), 7(01 ) = 7(0 I) = 0. Equating coefficients it follows easily that 7 is T k fixed. We wish to apply the trace lemma to compute W(k,Z;S). Thus we consider the inverse different of S, &-1(S/Z) = I = {x e E: traceE/Q(XS) C Z} Here E is the quotient field of S. Again, there is the - involution on E, and S is a - invariant order. It follows that the inverse different I is a - invariant fractional ideal over S. We may describe I by Euler's theorem [A 92], namely I = Z(0)/(f'(@)),where f' is the derivative of f. We wish to apply the trace lemma with: A = S, E = ~-I(s/z), F= Z, t: E + F, t = trace E/F, R = Z. In order to do this, we must verify that t: &-1(S/Z) § HOmz(S,Z) is an isomorphism. To begin with, the map x + t(-x) is i-i since it is i-i on the quotient fields. Continuing, let h s HOmz(S,Z). Then h = t(-x0), for x 0 e E, since t is an isomorphism on the 102 field level. However his is Z-valued, so that trace(x0S) C Z, and x 0 ~ I. Hence t is onto. Thus, there is an isomorphism between I-valued Hermitian forms, and Z-valued symmetric forms with a compatible S-module structure, meaning (rx,y) = (x,~y) for all r e S. We state this as: Theorem 3.2 The trace lemma yields a__nn isomorphism t,: H( ~-i (S/Z)) + W(k,Z;S) [] The same result naturally holds for asymmetric, and we have: Theorem 3.3 The trace lemma yields an isomorphism HS(ZI(s/z)) = A(Z;S). [] Caution: Using a scaled trace may be impossible since if --i uu = 0 , u e E, and u may not even be in S. We next apply the trace lemma to the trace map t,: E/A-I(s/z) + Q/Z, in other words t, is induced from trace E/Q. There is the induced isomorphism: ;,: E/A-I(s/z) + HOmz(E/A-I(s/Z),Q/Z) . Again, the trace lemma yields, (see Remark 3.9): Theorem 3.4 H(E/A-I(s/z)) = W(k,Q/Z;S) H0(E/A-I(s/z)) = A(Q/Z;S). 103 While we are discussing W(k,Q/Z;S), we continue with the analog of the Decomposition Theorem 1.9. Theorem 3.5 W(k,Q/Z;S) = O W(k,Q/Z;S/M) = _W(k'Fp;S/M) M= M ~ H (S/M) where runs over all - invariant maximal ideals in S. Proof: The proof is exactly as before. Let [M,B,t] s W(k,Q/Z;S). We write this as [Mi,Bi,ti], of irreducible modules. Thus A I = S - annihilator of MI = {r ~ S: rm = 0 for all m e M i} , is a maximal ideal in S. Now we check that A. is - invariant. Observe 0 = Bi(ax,y) = Bi(x,ay) for all a ~ A , x,y e M i- Hence a e A i, for if a ~ A; ay ~ 0 for some ye M i, and 0 = B(x,ay) for all x contradicts the non-singularity of B i. The rest is as before to give the first isomorphism. Now S/M is a finite field, with induced involution since M = M. Of course, this may be the trivial involution. Any finitely generated S/M - module is a finite dimensional vector space whose underlying abelian group is p torsion, where p = char S/M . The second isomorphism then follows by selecting a generator, say I/p for the p torsion in Q/Z. The last isomorphism follows by 2.1. Remark 3.6 A similar theorem holds for A(k,Q/Z;S). Remark 3.7 If the involution on S/M is trivial, the last 104 term is actually W(S/M), Witt of a finite field, which is determined by the cardinality of S/M Let q = cardinality of S/A4 . (a) If q - 1 (mod 4) W(S/M) = Z/2Z (9 Z/2Z (b) If q -= 0 (mod 2) W(S/M) = Z/2Z (c) If q - 3 (mod 4) W(S/M) -- Z/4Z If the involution on S/M is non-trivial, we have Hermitian of a finite field. Here rank is the only invariant [M-H 117]. Remark 3.8 We have thus shown that H(E/I) -~ (9 _H(S/M). M = M In fact, this holds directly, when I = ~ is a - invariant fractional ideal with S= D the underlying ring of integers in E. For if [M,B] e H(E/I), with M a finitely generated torsion D-module, we take anisotropic representatives, decompose into irr- educibles, etc. It follows that [M,B] = (9 [Mi,Bi], where the annihilator of M i, say P , is a - invariant maximal ideal in D. Thus B i takes values in E/I(P). We may identify this with a D/P -valued form [VI 3]. For I = A-I(D/Z), this was done in 3.5. In general, we embed D/P into E/ I(P) by r + P + pr + I(P), where p has valuation vp(I) - i, [VI 3], and obtain an isomorphism between Hermitian forms [M,B] with values in D/P , and Hermitian forms [M,B], where M is D/P -module, with values in E/I (P) . Remark 3.9 The Hermitian groups H(E/~-I(s/z)) are defined because E/~-I(s/z) is an injective S-module, for S an order. One verifies this using the trace induced map from E/~ -I to Q/Z. 105 4. Torsion in W(-k,F) While we are computing Witt groups, it seems natural to mention the torsion in W(-k,F) . Rather surprisingly this is related to the number of squares needed to express k as a sum of squares. Theorem 4.1 If k i_{s a square i_nn F, then W(-k,F) is all 2-torsion. 2 Proof: Suppose k = r , and let [M,B,s e W(-k,F). Then clearly N = {(rx,s : x s M} is a metabolizer for [M,B,s e [M,S,i] = 2[M,B,s [] Theorem 4.2 If k is a sum of two s~uares i__nn F, then W(-k,F) is all 4-torsion. Proof: Suppose k = r 2 + s 2, and let [M,B,z] e W(-k,F). We want to show 4[M,B,s = 0. So consider N C M M ~ M e M defined by: N = subspace generated by {(rx,sx,s (-sy,ry,0,Zy) : x,y 6: M}. It is easy to see that N is a metabolizer, so 4[M,B,s = 0. [-] Theorem 4.3 If k is a sum of four squares i__nn F, then W(-k,F) is all 8-torsion. 2 Proof: Suppose k = r 2 + s 2 + t 2 + u , and let [M,B,s W(-ktF) . As above we produce a metabolizer for 8[M,B,i]. 106 Let N = subspace generated by {(rx,sx,tx,ux,ix,0,0,0), (-sy,ry,-uy,ty,0,s (-tz,uz,rz,-sz,0,0,s (-uw,-tw, sw,rw,0,0,0,s : x,y,z,w s M}. Cearly N = N A is a metabolizer for 8[M,B,s O As long as k is an integer this completes the discussion since every positive integer is a sum of at most 4 squares. A few questions remain however: (I) If k is a sum of 4-squares for example, could W(-k,F) in fact be all 2-torsion? How would one recognize this? (2) In general we have only demanded k E F. Thus there remains open the question of whether one can relate the torsion in W(-k,F) to the minimum number of squares needed to express k as a sum, for arbitrary k. The technique used to prove Theorems 4.1-4.3 is basically to construct an orthogonal design of type (i,i,i .... ) on the independent variables r,s,t .... see [G-S], and use this to construct a metabolizer N. We can thus extend Theorem 4.3 to the case of k being a sum of 8 squares by constructing an orthogonal design using the Cayley numbers. However, by a Theorem of Radon this is absolutely as far as this method will go. The reader is referred to [G-S] for a discussion 107 of orthogonal designs. However, the fact that our method of constructing metabolizers does not generalize still yields no information as to whether Theorems 4.1-4.3 generalize or not. Chapter V THE SQUARING MAP We wish now to study the squaring map S : Ws § WS(k~K), s where K is a field, or Dedekind domain. S is defined by: s [M,B,Z] § [M,B,I 2] . Here s = +i if B is symmetric; E = -1 if B is skew-symmetric. To begin with we study S for K = F a field. We shall relate this to the case of K = Z the integers in the ensuing chapters. We shall derive an exact sequence involving the groups A(F) , and W~(-k,F). The octagon we obtain is: S I I I wl (k,F) + wl (k2,F) ~ wl (-k,F) ml ~ dl A(F) A(F) d I S -i ~ -i -i m_ 1 W-I (-k,F) + W-1 (k2 ,F) § W-I (k,F) me: A(F) + WS(k,F) is defined by: [M,B] + [M ~ M,Bs,%s ] where B s ((x,y), (z,w)) = B(X,W) + sB(z,y) and %s(x,y) = (sks-ly,x) 109 I : We(k2,F) s § WS(-k,F) is defined by: [M,B, 1] [M ~ M, B ~ -kB,s where ~(x,y) = (Zy,x). d : Ws A(F) is defined by: [M,B,I] § [M,B] where B(x,y) = k-iB(x,/y). This octagon is only defined for k ~ 0 in F. When k = 0, the maps do not all make sense. Remark 0.i If k = 0, W(k,F) = W(F). To see this, let [M,B,/] s W(k,F) and suppose (M,B,/) is anisotropic. We claim 1 = 0. Consider the 1 invariant subspace of M generated by: {/M,/2M, .... ~tM}, where t = degree of char polynomial of Z. This subspace is self annihilating since B(/x,/y) = kB(x,y) = 0. It follows that it must be 0 since (M,B,/) is anisotropic. Hence 1M = 0 as claimed. Remark 0.2 For k = • the maps given are well-defined over Z. The key observation is that the image spaces actually are inner products. For the map m , let [W,B] ~ A(Z), with symmetry operator s. Using Theorem I 4.10 we see that s* = s -I, so that s -I exists and s is non-singular. Thus m makes sense over Z, and E is well-defined. S s and I E are clearly well-defined. Let [M,B,Z] ~ We(-k,Z). As above, 1 is non-singular and B(x,y) = k-IB(x,/y); so B is non-singular when k = • 110 In Section 2, we shall prove exactness of this octagon over F a field. The proof given does not work over Z. We shall develop the machinery to study this problem in the following chapters. In Section 1 we motivate the methods used in Section 2 by deriving the transfer sequence of Scharlau, Elman and Lam. In fact, the exact octagon we obtain is a generalization of these maps. The Scharlau, Elman, Lam transfer sequence is an exact octagon with several terms vanishing. We shall prove the sequence below is exact: S I W +I (F (/a) ) + w+l (F) ~ w+l (-a,F;f) m , W +I (F) d ~ W-I (-a,F; f) § 0 § 0 where f(t) = t 2 - a. The term WE(-a,F;f) with f(t) = t 2 - a denotes witt equivalence classes of triples (M,B,Z) with: (a) B e symmetric B: M x M + F (b) B(Ix,Zy) = -aB(x,y) (c) 2. satisfies s = a. It follows that s = -s This is consistent with our previous notation. By the trace lemma, W ~(-a,F;f) = H E(F(/a)). The maps in this sequence are given by: d: W-I (-a,F; f) + W(F) [M,B,Z] + [M,B] where B(x,y) = B(x,z-ly) m: W(F) + W(F (/a)) [M,B] -~ [M,B] ~F F(/a) 111 S: W(F(/a)) ~ W(F) is Scharlau's transfer [M,B] + [M,B] where B(x,y) = tB(x,y) t: F(/a) ~ F is the scaled trace /a [Lm 201] t(x) = trace (~--~ x) I: W(F) + w+l (-a,F; f) [M,B] ~ [M M,~,~] B((x,y) , (u,v)) = B(x,u) aB(y,v) = B ~ -aB((x,y) ,(u,v)) s = (ay,x) We shall begin with the case above, and discuss the Scharlau, Elman, Lam transfer sequence. i. Scharlau's transfer Let F be a field, a ~ F**. Then we can form F(/a). Consider W(F). This is Witt equivalence classes of pairs [M,B], where B: M • M + F is a symmetric inner product. There is themap m: W(F) + W(F/a) given by [M,B] ~ [M,B] ~F F(/a)- Likewise, there is a map S: W(F(/a) ~ W(F) given by [M,B] + [M,B] where B(x,y) = t o B(x,y) with t defined by t(1) = 0, t(/a) = i. This map S is called Scharlau's transfer. It arises from a scaled trace, namely t(x) = traceF(/a)/F(~-/a ~ " x) In [Lm ]01] an exact sequence involving m and S is discussed. m W(F) ~ W(F(/a) ~ W(F). We examine this sequence in a setting which will generalize to our situation. In so doing, we compute the cokernel of S. To begin with, let us identify M ~ FF(/a) with M ~ M as a vector space over F(/a). Thus, if {v i} is a basis for M over 112 F, {(vi,0) } will be a basis for M G M over F(/a) . We must be careful about scalar multiplication. We are viewing ~a as (0,i), so that (i,0) (vi,0) = (vi,0) (0,i) (vi,0) = (0,v i) (i,0) (0,v)z = (0'vi) (0,i) (0,v i) = (avi,0) In other words, scalars from F(/a) are ordered pairs (c,d), c,d s F to be identified with c + d/a. They operate on M ~ M as described above. Under this identification, we can view the map m as defined by: m: [M,B] + [M M,B'] where B' ((xl,Y I) , (x2,Y 2)) = B(Xl,X 2) + aB(Yl,y 2) + [B(Xl,Y 2) + B(x2,Yl)]/a. With these preliminaries, we proceed to define the maps involved in the transfer exact sequence. Let f(t) = t 2 - a, and form w-l(-a,F;f). The maps are defined by: d: w-i (-a,F; f) ~ W(F) [M,B,Z] + [M,B] B(x,y) = B(x,s m: W(F) + W(F/a)) [M,B] + [M,B] ~FF (/a) S: W(F(/a)) + W(F) [M,B] + [M,B] B(x,y) = t o B(x,y) t = scaled trace 113 I: W(F) ~ w+l(-a,F;f) [M,B] ~ [M + M,B,~] = B -aB ~(x,y) = (ay,x) We begin by studying the map d. d: [M,B, s + [M,B] B(x,y) = B (x, s is symmetric since B(x,y) = B(x,s = -B (x,-s = B (-s since B is skew-symmetric = B(y, s since s = -s = ~ (y,x) Lemma I.i ker d = 0. Define I: W-I(F) + w-l(-a,F;f) by [M,B] + [M G M,B, ~ where B((x,y), (u,v)) = B(x,u) - aB(y,v) s = (ay,x). I is clearly a well-defined group homomorphism. We shall show ker d ~ im I. However, W-I(F) = 0, so this will show that kernel d = 0. Let (M,B,s be an anisotropic representative of a Witt class in w-l(-a,F;f), with d([M,B,s = 0. Let N be a metabolizer for [M,B]. Consider [N,BI], where B 1 = B~, the restriction of B to N. B 1 is non-singular since (M,B,s is anisotropic over a field, and N is a metabolizer for B. B 1 is a skew-symmetric inner product on N. Applying I we obtain IN ~ N,BI,s where Bl(X,y) = B 1 ~ -aB I. 114 Define 7: N E) N + M bv (nl,n 2) + n I + 9~n 2. We shall show 7 is an equivariant isomorphism, hence [M,B,s s im I, which completes the proof. Since dim(N ~) N) = dim M, in order to show that ~ is an isomorphism, it suffices to show 7 is 1 - i, since these are vector spaces. So suppose ~(nl,n 2) = n I + In 2 = 0. Then form W = We compute: B(n2,n 2) = B(n2,-s = B(n2,n I) = 0 since N = N ~ with respect to B(n2,s 2) = B(-s = -B(nl,~-in I) = 0 Similarly, B(s = 0, and we see that W is an s invariant sub- space of M with W C W~ This contradicts (M,B,s being anisotropic unless W = 0. Thus n 2 = 0 and since n I + s 2 = 0, n I = 0. Hence 7 is an isomorphism. The following computations show that ~ is equivariant: Bl((nl,n2), (n~,n~)) = Bl(nl,n~) - aBl(n2,n ~) = B(nl,n ~) + B(/n2,/n ~) + B(nl,/n ~) + B(/n2,n ~) (Since N = N~with respect to B) ! = B(n I + /n2,n I + /n~) ! = B(7(nl,n 2) ,~(n~,n2)) o l(nl,n2) = 7(an2,nl) = an 2 + Zn 1 1 o 7(nl,n2) = /(n I + Zn2) = ~n I + Zg"n2 = /n I + an 2. 115 Thus y yields an equivariant isomorphism (N ~ N, BI' ~) ~ (M,B,s It follows that [M,B,I] s im I as was shown. The natural second step is to continue computing kernels. We could do this formally although it would amount to computing the kernel of the 0 mapping: I: W -I(F) + W -l(-a,F;f) where W-I(F) = 0. We shall return to I later, as this "same map" occurs at the end of our octagon. Thus, we next study the cokernel of d, and the map m: W(F) ~ W(F(/a)). As observed previously we may view m as being defined by [M,B] + [M ~ M,B'] where B'((xl,Yl), (x2,Y2)) = B(Xl,X 2) + aB(Yl,y 2) + [B(Xl,Y 2) + B(x2,Yl)]/a. Lemma 1.2 ker m =im d. d m Step i" im d C_ ker m. Suppose [M,B,I] [M,B] + [M 8 M,~']. Consider N = { (s : v ~ M} C M ~) M 116 B'((s (s = B(Zv,Zw) + aB(v,w) + [~(~v,w) + ~(s = B(Zv,w) + aB(v,s + {B(Zv,s + B(s163 aB (v, s + aB (v, ~-lw) + [B (v,-w) + B (-w,v) ] /a = 0 Thus NCN~ rank N = rank M, and rank N + rank N ~= rank M ~ M. Hence rank ~= rank M = rank N, and N = N~ so [M M,B'] = 0. Step 2. ker m ~ im d. Let (M,B) be anisotropic. Suppose m[M,B] = 0. Let N be a metabolizer for [M ~ M,B']. Consider K = {x ~ M: (x,0) ~ N}. B'((x,0), (y,0)) = B(x,y) = 0 for all x,y e N. Thus K C K~ so that K = 0 since (M,B) is anisotropic. Similarly, {x s M: (0,x) s N} = 0, and we can conclude that N is the graph of a 1 - 1 function s M + M, i.e. N = { (s x ~ M}. B' ((ix,x), (iy,y)) = B(ix,ly) + aB(x,y) + [B(s + B(s = 0. Hence B(Zx,s = -aB(x,y). B(Ix,y) = -B(Zy,x) = B(x,-s Thus is of degree -a, and s = -s Further, -aB(x,y) = B(Zx,s = -B(x,s so that s = a. 117 Now we form the space (M,BI,s , where B 1 is defined by Bl(x,y) = B(x,/y). In order to show B 1 is non-singular, we claim: Bl(-,x) = 0 implies x = 0. Proof: Bl(-,x) = 0 = B(-,/x), hence Ix = 0 since B is non-singular. Thus x = 0 since 1 is 1 - I. Since we are over a field, AdRB 1 1 - 1 implies AdRB 1 is an isomorphism, and B 1 is non-singular. B 1 is skew-symmetric since: Bl(X,y) = B(x,/y) = -B(/x,y) = -B(y,/x) = -Bl(Y,X). l is of degree -a with respect to B 1 since: Bl(/x,/y) = B(/x,/2y) = -aB(x,/y) = -aBl(X,y)- Thus [M,BI,/] s w-l(-a,F;f), where f(t) = t2-a. Applying d we obtain [M,BI]. Bl(X,y) = Bl(X,l-ly) = B(x,y). Hence [M,B] s im d as required. Lemma 1.3 ker S =im m. Step I: im m C ker S. Let m([M,B]) = [M ~ M, B']. Applying S we obtain [M ~ M, t o B']. Now consider M ~ 0 C M ~ M. By definition of t o B', we have 1 M ~ 0 C (M ~ 0). However, dim (M ~ 0) = ~ dim (M ~ M) , so that M ~ 0 is a metabolizer, and [M ~ M, t o B'] = 0. Step 2: ker S ~_ im m. Let (M,B) be anisotropic. Suppose S[M,B] = [M,t o B] = 0. Recall t is the scaled trace /a of F(/a) over F, trace (~ -). Let N be a metabolizer for [M,t o B]. 118 If c + d/a a F(/a), where c,d ~ F, we shall call c the F-part, d the /a-part of c + d/a. t: F(/a) + F is given by c + d/a ~ d, projection to the ~%-part. Consider [N,B~] s W(F). Applying m, we obtain IN ~ N,B']. Define y : N ~ N § M by (nl,n 2) ~ n I + /an 2. We shall show that 7 is an isomorphism of (N ~ N,B') with (M,B), and hence ker S C im m. Comment: In order that [N,B~] e W(F), we should again check that B~ is non-singular. This follows as in i.i since (M,B) is anisotropic. An alternate proof can be given since y is an equivariant isomorphism. Since B is non-singular, so is B', and hence so is In order to show y is an equivariant isomorphism, we first show that y is 1 - i, and hence an isomorphism as we are working over a field. Suppose Y(nl,n 2) = n I + /an 2 = 0. Consider B(nl,n I) has /a-part 0 since N = N ~ with respect to B = t o B. However, B(nl,nl) = B(nl,-/an 2) = -/aB(nl,n2). Observe that the /a-part of B(nl,n 2) is 0 also, since nl,n 2 s N = ~. So B(nl,n I) = -/aB(nl,n2) , implying that the F-part of B(nl,n I) is 0 also. Hence B(nl,n I) = 0. This contradicts (M,B) being anisotropic, unless n I = n 2 = 0. Thus y is 1 - i, and hence an isomorphism. We must check that y is equivariant. 119 B' ((nl,n2), (n~,n~,))= B(nl,n {) + aB(n2,n ~) + [B(nl,n ~) + B(n{,n2)] /a t = B(n I + /an2,n I + /an~) = B(y(nl,n2), y(n~,n It follows that (M,B) - (N ~)N,B') as desired. [] Lemma 1.4 ker I = im S. Step 1. im S C ker I. Let S[M,B] = [M,t o B]. Appling I we obtain [M ~) M, (t~ B) ,~]. Let N = {(/av,v) :v ~ M}. N is l invariant since Z(/av,v) = (av,/av). Further it is self-annihilating since (t o B) ((/av,v), (~aw,w)) = (t o B) ((~av,/aw)) - a(t ~ B) (v,w) = a(t o B) (v,w) - a(t o B) (v,w) = 0 Since rank N = { rank (M ~ M), N = N~ and M ~ M ~ 0. Step 2. ker I C im S. Let (M,B) be anisotropic. Suppose I[M,B] = [M G M,B,I] = 0. Let N be a metabolizer for M ~ M. Let K = {x ~ M: (x,0) E N). If x,y s K, B((x,0),(y,0)) = B(x,y) = 0. Since (M,B) is anisotropic, K = {0}. similarly, {x e M: (0,x) s N} = 0. Thus N is the graph of a 1 - 1 function l: M + M. (We need F a field to conclude that Z is an isomorphism of M onto M.) 120 We may write N = {(/v,v),l: M § M}. B((/v,v), (/w,w)) = B(/v,/w) - aB(v~w) = 0 Thus 1 is of degree a. N is 1 invariant, so that (Zv,v) s N implies (av,lv) ~ N. Hence /(/v) = av, and /2 = a. Also B((av,/v), (~w,w)) = B(av,/w) - aB(lv,w) = 0. Thus B(v,/w) = B(/v,w). Hence l* = 1. N is already an F - vector space, with d(/v,v) = (d/(v),dv) for d ~ F. We make N into an F(/a) - vector space by defining /a(/v,v) = (av,lv) = l(/v,v). Now define B 1 on N by Bl((/v,v) , (lw,w)) = B(/v,w) + B(v,w)/a B 1 is symmetric since B(v,/w) = B(lv,w). B 1 is clearly F(/a) - bilinear with the F(/a) - vector space structure defined above. B 1 is an inner product since B is. Now we apply S to [N,BI] , SIN,B1] = [N, t o BI]. Then define y: N ~ M by (/v,v) + v. B(7(/v,v) , 7(/w,w)) = B(v,w) = (t o B I) ((/v,v), (/w,w)). Hence [N,t o B1 ] = [M,B], and [M,B] ~ im S. O 121 Lemma 1.5 I is onto. We define: d: w+l(-a,F;f) + W-I(F) by [M,B,/] + [M,B] where B(x,y) = B(x,/-ly) . B(y,x) = B(y,Z-ix) = B(-/-iy,x) = -B(x,/-ly) = -B(x,y). Hence B is skew-symmetric. (This of course is the same d we have already defined. However, B symmetric yields B skew-symmetric.) Note that W-I(F) = 0, so that kernel d = w+l(-a,F;f). ) Let [M,B,/] ~ w+l(-a,F;f), so [M,B,/] s kernel d. Suppose (M,B,/) is anisotropic. Since [M,B] = 0, we let N be a metabolizer for (M,B). Then N ~ 1N = {0} since (M,B,s is anisotropic. Consider [N,B I] e W(F) , where B 1 = B~. Applying I we obtain [N ~ N,BI,1 ]. We shall show this is isomorphic to (M,B,/). From this it follows that I is onto. Define ~:N ~ N + M by (nl,n 2) § n I + /n 2. Then B(y(nl,n2),~(n~,n~)) = B(n I + /n2,n i + Zn~) = B(nl,n ~) + B(/n2,1n ~) + B(nl,/n ~) + B(/n2,n ~) ( B(nl,/n h) = B(/n2,n {) = 0 since B = 0 on N. = B(nl,n {) - aB(n2,n~)) = Bl((nl,n2) , (n{,n~)). y/(nl,n2) = 7(an2,nl) = an 2 + /n 1 /7(nl,n2) = /(n I + /n 2) = /n I + /2n 2 = /n I + an 2. Thus 7 is equivariant. Again, 7 is an isomorphism since N J% 1N = 0. Hence (M,B,s -~ (N (9 N,BI,~) as claimed. P'] 122 We have thus shown: Theorem 1.5 There is an exact octagon: S I W+I (F (/a)) § W+I (F) + W+I (-a,F; f) m I" w +I (F) d ~, W-I (-a,F; f) + 0 § 0 The map S is Scharlau's transfer, with the other maps as previously described. ['] The proof of Theorem 1.5 just given involved analyzing what actually is an 8 term exact sequence as we shall describe in Theorem 2.1. As one continues with Theorem 2.1, one should note the similarities in its proof to Theorem 1.5 In fact we may restrict the maps in Theorem 2.1 to obtain: S I W+I (a,F; f) + w+l (a2,F;g) + w+l(-a,F;f) m A(F;h) -I W (-a,F;f) + 0 § 0 Here f is the polynomial f(t) = t 2 -- a. h is the polynomial h(t) = t - 1 g is the polynomial g(t) = t - a 123 We then identify the groups: A(F;h) = W+I(F ) since s = 1 the identity map w+l(a,F;f) = W(F(/a)) by the trace lemma, since I = ~, and the induced - involution is trivial. w+l(a2,F;g) = w+l(F) since I = a. If one examines the induced maps under these "trace induced" identifications, then Theorem 2.1 will restrict to the Scharlau transfer as described. 2. The exact octagon over a field. The Scharlau, Elman, Lam transfer sequence has been derived by computing successive kernels. For degree k mapping structures the maps of the Scharlau, Elman, Lam transfer sequence become: d : We (-k,F) § A(F) e [M,B,/] + [M,B] B(x,y) = k-IB(x,/y) b rn A(F) ~ We (k,F) s [M,B] + [M M,Be,s e] where B ((x,y), (z,w)) = B(x,w) +eB(z,y) e 1 (x,y) = (eks-ly,x) s is the symmetry operator for B e S : W E (k,F) ~ W E (k 2 ,F) e [M,B,/] + [M,B,/2] I : We (k2,F) ~ We (-k,F) e [M,B,/] ~ [M ~) M, B -kB,~] s = (s 124 We shall prove: Theorem 2.1 The above maps combine to yield an exact octagon: WI(k,F) §S 1 W1 (k2,F) §I WI(_k,F) m. dl A(F) A(F) d_~1 i_ 1 S_ 1 ~w" m_l W-I (-k,F) + W-I (k2,F) § W-I (k,F) We begin with the map m s . First, we observe that s is of degree k with respect to B E since: BE(Ze(x,Y), s163 = Be((eks-ly,x), (eks-lw,z)) = B (sks-ly,z) + ~/ B(~ks-lw,x) = kB(s-lw,x) + EkB(s-ly,x) = kB(x,w) + EkB(z,y) = k[B(x,w) + eB(z,y)] = kBe((x,y), (z,w)) . m E is well-defined, for if (M,B) ~ 0 has metabolizer N then m e(M,B) has metabolizer N N. Lemlna 2.2 ker Ss = im m s . Proof: Suppose me[M,B] = [M M, Be, le ]. Then M ~ 0 is an 2 le invariant subspace, equal to its own annihilater by the way B e is defined. Thus im m C ker S . E -- E Conversely, suppose [M,B,I] E ker S E. Let N be an 12 invariant subspace of M with N = N J'. N is a metabolizer for 125 Sc[M,B,/] = [M,B, s We assume as usual (M,B,/) is anisotropic. Note: N /~ ZN = 0. This is seen as follows. Let n s NP~ IN, and form N 1 = NIC N~. Thus, since we took (M,B,/) anisotropic, N 1 = 0 and n = 0. For vector spaces then it follows that M = N ~ 1N. Define B 1 on N by Bl(nl,n2) = B(nl,ln2). We must verify that B 1 is an inner product. So consider the adjoint of B I, AdRBl: N ~ HomF(N,F). Since we are working over a field, it suffices to show AdRB 1 is 1 - i. We need k ~ 0, so that 1 is non-singular. Suppose Bl(-,n 2) = 0 = B(-,/n2). Then /n 2 s N = N. However, by the note, N /'% I N = 0. Thus /n 2 = 0. Since 1 is non-singular n 2 = 0. Hence B 1 is non-singular. We may thus form [N,B I] e A(F). Applying m e we obtain [N N, B 1 , ~]. Here s Bi(n!,n 2) = B(nl,~n2) = s I) = ekB(n2,/-inl) = B(n2,s163 )) = Bl(n2,ckl-2nl) Thus s = s -2 for B I, or 12 = eks -I. It follows that ls = (eks-ly,x) = (12y,x) Now define y : N ~ N + M by (nl,n2) § n I + /n 2. Since N~/N = 0, 7 is an isomorphism. We now show that 7 is equivariant, ie. that (N ~ N,Ble,/e) = (M,B,/). It follows that ker S e _C_ im m e as desired. We compute: Ble((x,y), (z,w)) = BI(X,W) + eBl(Z,y) = B(x,/w) + eB(z,/y) + B(x,z) + B(/y,/w) (since B(x,z) = B(/y,/w) = 0 as N = N J') = B(x,lw + z) + B(/y,lw + z) = B(x +/y, z + lw) = B(y(x,y), y (z,w)). 126 Also, 7(ZE(nl,n2)) = ,,,(~ks-ln2,nl) = y(/2n2,n I) = 12n2 + /n I = Z(/n 2 + n I) = Z(Y(nl,n2)). [] Lemma 2.3 ker I s =im S E. Proof: Let [M,B,/2] ~ im S . We first show I ([M,B,Z2]) = 0. E E I [M,B,/] = [M e M, B -kB,/2],- where (~2)- (x,y) = (/2y,x). E Let N = {(x,f-lx): x E M}. N is (2 2 ) invariant. Further, (B e -kB) ((x,s (y,s = B(x,y) - kB(s163 = B(x,y) - B{x,y) = 0. Thus NC N~ rank N = (1/2) rank M, so that N = N ~ and M M ~ 0. Conversely, let [M,B,/] E ker I . So I [M,B,/] = s E [M @ M, B ~ -kB,l] = 0. Let N be a metabolizer for M G M. N is ~ l invariant. Now assume (M,B,/) is anisotropic. We claim that N is the graph, N = {(x,tx) : x E M}, of a 1 - 1 function t: M + M. Consider K = {x e M: (x,0) s N}. If (x,0) s N, then (0,x) e N since N is l invariant. Thus (/x,0) e N, and lxe K. Hence K is 1 invariant. However, if x,y s K, B(x,y) = (B ~ -kB) ((x,0) , (y,0)) = 0, since N = N~ Thus K C K A'. This contradicts (M,B,/) anisotropic unless K = {0}. Similarly, {x e M: (0,x) e N} = 0. Thus N is the graph of a 1 - 1 function t: M + M and we can write N = {(x,tx): x c M}. t maps M onto M since dim N = dim M. On N we define BI,/I by: Bl((x,tx), (y,ty)) = B(x,y). /l(x,tx) = (f(tx),x) /i: N + N since N is l invariant. 127 We compute: BiI/l(x,tx),/l(Y,ty)) = Bl((/(tx),x) , (/(ty),y)) = B(s ,/(ty)) -- k2B(tx,ty) . However, (B ~ -kB) ((x,tx),(y,ty)) = B(x,y) - kB(tx,ty) = 0 since N = N/-. Thus, the above equals kB(x,y) = kBl((x,tx),(y,ty))- In order to have [N,BI,/i] E wl(k,F) , we still must show B 1 is an inner product. Again, it suffices to show that the adjoint is 1 - i. AdRBI : N + HomF (N,F) (x,tx) + BI(-, (x,ty)) Suppose BI(-, (x,tx)) = 0 on N. Then B(-,x) = 0 for all y, when (y,ty) s N. Since we are over a field, B(-,x) = 0 on M. This contradicts (M,B,/) anisotropic unless x = 0. Thus AdRB 1 is 1 - !, and hence an isomorphism. We consider [N,BI,/I]. Define y: (N,BI,I ~) § (M,B,/) by y: (x,tx) + x. ~ is clearly an isomorphism. We claim y is equivariant, so that Se[N,BI,11] = [M,B,/] as desired. Bl((x,tx), (y,ty)) = B(x,y) = B(7(x,tx) ,~(y,ty)) . 7 ol I2(x,tx) = y o /l(/(tx,x) = y o (/x,/(tx)) = lx = 1 o ~(x,tx). [] 128 Lemma 2.4 ker d = im I s s Let Is = [M ~ M, B ~ -kB,l]. Applying d we obtain c [M G M, B ~ -kB] where (B ~ -kB) ((x,y), (u,v)) = (B ~ -kB) ((x,y),l(u,v)) k = I(B ~-kB) ((x,y), (/v,u)). In general, if [M,B,/] ~ W (-k,F) , applying d we obtain s 6 [M,B] where B(x,y) = ~1 B(x,/y) = s B(/y,x) = -s B(y,/-Ix) = 1 S (y,/(/-2(-skx)) = B(y /-2(-s k ' " So s = -s is the symmetry operator for B. Now consider N = M ~ 0 C M ~ M. N is s = -s -2 invariant 1 rank N = ~ rank (M ~ M) and 1 (B e-kB) ((v,0), (w,0)) = ~(B G -kB) ((v,0),(0,w)) = 0. Thus (M e M, B ~ -kB) 0, with metabolizer N. Conversely, suppose [M,B,Z] E ker d s . Let N be an s = -ek/-2 invariant subspace of M with N = N ~ with respect to B, where 1 B(x,y) = ~B(x,/y). Define ~: N ~ N + M by (nl,n 2) § n I + /n 2. We assume (M,B,/) is anisotropic. 129 Claim: y is 1 - Suppose n I + In 2 = 0. Then form K = B(nl,n I) = B(nl,-2~n 2) = (-k) (iB(nl,ln 2)) = -kB(nl,n 2) = 0 J_ as N = N with respect to B. By similar computations, it follows & that K C K . K is l invariant. This contradicts (M,B,/) being anisotropic, unless K = {0}. Thus 7 is 1 - I, and hence an isomorphism. Consider [N,B,I 2] s WS(k2,F). We must show B is an inner product Again, it suffices to show AdRB is 1 - i. So suppose B(-,n) = 0 on N. Note that B(/N,n) = 0 since N = N ~. By the above N ~ f~N = M, so B(-,n) = 0 on M. Since B is an inner product, n = 0 and AdRB is an isomorphism on N. We now apply I s to IN,B,/2]. This yields IN ~) N, B ~) -kB,12]. provides an equivariant isomorphism y: I [N,B,/2] § IN,B,/]. Claim: y E To see this, we compute: (B -kB) ((x,y) , (u,v)) = B(x,u) - kB(y,v) = B(x,u) + B(/y,/v) + B(x,/v) + B(/y,u) = B(x + /y, u + /v) = B(y(x,y),y(u,v)) y o 12(x,y) = y(/2y,x) = /2y + Ix = l(lY + x) = 1 o y(x,y). [] Lemma 12.5 ker m =im d E -E 130 Suppose d_E[M,B,I] = [M,B]. The associated symmetry operator for B is s = -(-s -2. Note: B is -s symmetric. Applying m s to [M,B] , we obtain [M ~ M,BE,/e] where /s(w!,w 2) = (s = (12w2,wl). Consider N C M ~ M defined by N = {(x,/-ix) : x 6 M}. N is l invariant. Further [s (y,/-ly)) = ~(x,/-ly) + 6~(y,/-ix) = ~B(x,y) + s (y,x) = ~B(x,y) + #{-#)~B(~,y) = o. Thus [M ~ M,Bs = 0 and im d_s ker ms Conversely, suppose me[M,B] = [M e M,Bs = 0. Let N be a metabolizer for M M above. Assume (M,B) is anisotropic. Consider K = {x e M: (x,0) e N}. Since N is s invariant (x,0) e N implies (0,x) E N and (sks-lx,0) e N. Also, N is le -I invariant since Be((u,v),/~l(x,y)) = ~Bc(/c(u,v),(x,y)) = 0 for all (u,v) , (x,y) e N. Thus (x,0) s N implies (e/k(sx) ,0) e N. N is a subspace, so (sx,0) e N whenever (x,0) e N. Thus K is s invariant. Further B(x,y) = Be((x,0), (0,y)) = 0 for all x,y s K. This follows since (y,0) e N implies s s (y,0) = (0,y) e N also, and N = N J" with respect to B . Thus K is an s invariant, self s annihilating subspace of M. This contradicts M being anisotropic unless K = 0. Similarly {x e M: (0,x) c N} = {0}. It follows that 131 N is the graph of a 1 - 1 function t: M § M, and we can write N = {(w,tw) : w S M }. -i We now study this map t. First, since N is I s and I s invariant, if (x,y) s N then so is s (x,y) = (~ks-ly,x) and s z~l(x,y) = (y,s/k sx). s Thus, if (y,ty) s N, so is (ty,E/k sy) , so that t(ty) = ~sy. More simply, t 2 = ~s, or Ekt 2 = s. Moreover, B ((x,tx) , (y,ty) = 0 s = B(x,ty) + s B(y,tx) = B(ty,sx) + e B(y,tx) = B(y,t*sx) + s B(y,tx) where t* = adjoint of t = B(y, (t*s + st)x) = 0 Since B is non-slngular, t*s + s t = 0. Thus t* = -sts -I = (-st) (s/k t -2) = -t-I k On M, define an inner product B 1 by Bl(X,y) = kB(x,ty). B 1 is non-singular since B and t are, as usual. B 1 is (-s) symmetric since: Bl(X,y) = kB(x,ty) = kB(ty,sx) = kB(y,t*sx) = kB(y,-stx) = -skB(y,tx) = -SBl(Y,X). Now consider (M,BI,t-I). Bl(t-lx,t-ly ) = kB(t-lx,y) = kB(x,t*-ly) = kB(x,-kty) = (-k) (k)B(x,ty) = (-k) B1 (x,y) 132 Thus [M,BI,t-I] s W-~ (-k,F). Applying d_e we obtain [M,BI]. Bl(x,y) = k-iBl(X,t-ly) : B(x,y) as desired. [] This completes the proof of the exact octagon over a field F, when k ~ 0. The failure of this proof for a Dedekind domain, eg. Z the ring of integers, is the verification that the bilinear maps we are constructing are actually Z-inner products. In order to overcome this difficulty, we use a different approach. Namely, we study a boundary sequence relating W(k,Z) to W(k,Q). This boundary sequence then enables us to study the octagon over Z. Chapter VI THE BOUNDARY Our ultimate goal is to study the octagon over Z, when k = • In order to do this. we relate W(k,Z) to W(k,Q) by means of an exact sequence. In Section i, the boundary homomorphism, 3 , is defined. This enables us to establish an exact sequence: 0 + W(k,Z) § W (k,Q) § W (k,Q/Z) . Next let S = Z[t,t-!]/(f(t)), where f(t) is a T k fixed monic, integral, irreducible polynomial. We have the decomposition W(k,Q) = ~ W(k,Q;f) = ~ W(k,Q;S) . It is thus natural to consider the restriction of 3 to W(k,Q;S). We denote this restriction 3(S). We wish to compute ~(S). This will eventually allow us to analyze ~. The first step is the reduction of the study of ~(S) to the study of $(D): W(k,Q;D) ~ W(k,Q/Z;D) where D is the Dedekind ring of integers in E = Q[t,t-l]/(f(t)). This is done in Section 2. In Section 3, we begin the computation of $ by studying the local case 3(D,P). Here 3(D,P) is the localization of the 3(D) sequence at a - invariant maximal ideal P in D. We use this in Section 4 to compute the cokernel of ~(D). i. The boundary homomorphism We shall construct an exact sequence 0 § W(k,z) + W(k,Q) + W(k,Q/Z). 134 The script W indicates that we have placed a restriction on the minimal polynomial of ~ in the degree k mapping structure [M,B,s This restriction is that the minimal polynomial of Z be a monic integral polynomial. We also should observe that this construction works equally well for Z replaced by D a Dedekind Domain, Q replaced by E the quotient field of D, and Q/Z replaced by E/D. The sequence then reads: i 0 ~ W(k,D) + W (k,E) ~ W (k,E/D) . Note that this construction also applies to the asymmetric case. We similarly will obtain an exact sequence: 0 + A(Z) + A(Q) + A (Q/Z) . If [M,B] ~ A(Q), with symmetry operator s satisfying B(x,y) = B(y,sx), we then require that the minimal polynomial of s be a monic integral polynomial in order for [M,B] to be in script A(Q). The restriction that we have placed on the minimal polynomial of l (respectively s) is dictated by: Proposition i.i A degree k mapping structure (M,B,s over Q contains an s invariant integral lattice A if and only if the minimal polynomial of s is a monic, integral polynomial. 135 Comments: (1) Z is replaced by the symmetry operator s for the asymmetric case, A(Q). (2) All lattices are assumed to be full. [B,S 99] This means that A is a finitely generated Z-submodule of M with rank A = rank M. (3) An integral lattice A is one on which the inner product B is integrally valued. Proof: Necessity. We assume M contains an s invariant integral lattice A. Since BIA is integral the characteristic polynomial of ~ is integral. Since the characteristic polynomial of s is integral, so is the minimal polynomial [L - 1 402]. Sufficiency. Conversely, suppose that the minimal polynomial, f(x}, of ~ is integral. Write m-i m f(x) = a^u + alx + ... + am_~Xx + x . By Lemma II. 1.1, f(s = f(s = 0. Thus, we obtain the identities m-i m-i m =_ g a..~i, (.~,)m=_ E ai(.~*)i i=0 i i=0 We now construct an s s invariant integral lattice. Let {e I ..... en} be a basis for M over Q. Since B(x,y) e Q for x,y ~ M, we can find integers rijk, sij k so that rijkB(s ~ Z and SijkB((~*)kei,e j) e Z for i,j ~ n and k < m. Let d = HrijkSijk . Then clearly {de I ..... den} generates a free Z-module on which B is integrally valued. 136 Let f. = de.. We then define A to be the Z-lattice spanned by l l {fl ..... fn'Zfi'bf2 ..... Zfn's ' "'" ' s .... ~m ifl ' .... s ' s fl .... ~ fn" "'" z*m-lfl' "'" s rn ]" Here m = degree f (x), the minimal polynomial of Z and s The identities for s and (Z)* m , together with ~ * = k show that A is s and Z* invariant. Since B is integral on {fl' .... fn }" it follows that B is integral on A because s has degree k. Thus A is an s ~ invariant lattice as desired. ~] Definitioh 1.2 W(k,Q) (respectivel~ A(Q)) denotes Witt eguivalence classes [M,B,Z] in W(k,Q) for which the minimal polynomial of s (respectively s) is integral. We now define the maps in the boundary sequence: i 0 + W(k,Z) ~ W(k,Q) ~ W(k,Q/Z). If [M,B,s E W(k,Z), i[M,B,s - [M,B,s • Z Q. The boundary homomorphism ~ is more involved. Let [M,B,s e W(k,Q) o By Proposition i.i, we can find an ~ invariant integral lattice L C M. We define the dual lattice of L to be L # = {x [ M: B(x,L) C Z}. We observe that L # ~ L, with rank L # = rank L. Thus L#/L is a finitely generated torsion Z-module. 137 If x,y ~ L~/L, we let x and y be preimageS of these equivalence classes in L #. Let q: Q ~ Q/Z be the quotient map. We define B': L~/L x L#/L + Q/Z by: B' (x,y) = q o B(x,y). We define i': L#/L ~ L#/L by: s = ~(x) We now need to show that [L#/L,B',s '] is an inner product space in W(k,Q/Z). Once this is done, we define ~ by: ~: [M,B,s + [L#/L,B',s To begin with, since L is s invariant so is L #. Thus s is well-defined. The fact that B is integral on L implies that B' is well-defined on L#/L, with values in Q/Z. We must show that B' is non-singular. In order to do this we define h: L % ~ Horn Z [L#/L,Q/Z) by x + B' (x,-) We need to show that h is epic, with kernel L. We shall need: 138 Lemn, a 1.3 For a lattice L, (L%) # = L. Proof: If {v I .... ,Vn} is a basis for L, then a basis for L # is given by {v I # ,...,v n# }, where B(vi,vj# ) = 6.1. It follows ] that a basis for (L#) # is { v I ..... Vn}. We show h is onto. Let g: L#/L + Q/Z be given. Let {Wl, .... w n} be a basis for L #. Then define g:L # + Q as follows. g(w i) is chosen so that: q o 9(w i) = g(wi ) w. E L#/L 1 A The1% extend g linearly to L #. In fact, tensoring with Q, we may assume g e HOmZ(M,Q). Thus since B is an inner product we may write B (Vl,-) = g(-) . However B(Vl,-) is integer valued on L since g(-) = B(Vl,-) and g is integral on L. Thus v I s L #. It is now clear that g(-) = q o 9(-) = q o B(Vl,-) = B' (Vl,-) and h is onto. In order to show that the kernel of h is L, suppose h(w) = 0. So B' (w,-) = 0. Thus for all v E L #, we have B(w,v) e Z. Hence, w ~ (L#) # = L by Lemma 1.3, and the kernel of h is L. 139 Thus, given [M,B,~] s w(k,Q) , we have a method for obtaining an element in W(k,Q/Z). We define: ~ [M,B,s = [L#/L,B',s '] Lemma 1.4 ~ is well-defined. In order to show that ~ is well-defined, we must show that this construction is independent of the lattice chosen, and that is trivial on metabolic forms. If we do this, then ~ preserves Witt-equivalence and gives a homomorphism of Witt groups. (a) Independence of lattice. Let L o be another choice of Z invariant integral lattice. Without loss of generality, Lo C L, for otherwise, we can show that both L o and L give the same result as the ~ invariant integral lattice Lo ~ L. We then have Lo C L C L # ~ Lo # Consider L/L o C Lo#/Lo The annihilator of L/Lo in Lo#/Lo is (L/Lo) -L = L#/Lo . Thus, by 1.6.4, [Lo#/Lo, B~, Z ~] is Witt -equivalent to [(L/Lo)~/(L/Lo), Bo'', s where B~', Z ~' indicate the appropriate induced forms. However, (L/Lo)~-/(L/Lo) = (L#/Lo) /(L/L o) = L#/L. Hence [Lo#/Lo, B~, s = [L#/L,B ', s as was to be shown. (b) We now show ~ (metabolic) ~ 0. So let (M,B,~) ~ 0 with metabolizer N. Let L be an s invariant lattice. Define N = N~ L # . 1 140 Clearly N 1 is [ invariant. We also have the exact sequence 0 ~ N 1 L # ~ L#/N 1 ~ 0. Claim: This sequence splits so that N 1 is a summand of L #. To show this, it clearly suffices to show L#/N 1 is torsion free and hence projective. Suppose to the contrary that there is x E L # , x ~ N I , d ~ 0 with dx E N I . Now dx 6 N 1 ~ N, so B (dx,y) = dB (x,y) = 0 for all Y s N 1 Tensoring with Q, (NI/~ L#) ~ Z Q ~ N. Therefore dB(x,y) = 0 for all y ~ N, and B(x,y) = 0 for y E N. Hence x e N /- = N, x s L # so that x E N f~L # = N I. Contradiction. It follows that L#/N 1 is torsion free, and N 1 is a summand of L # . Now let H = (L#r~ N)/L C L#/L. Clearly H C H/'. Conversely, suppose k s H ~. This means B'(k,h) ~ Z for all h ~ H. Let k be a lift of k to L #. Consider the diagram 141 B(k,-) : L# /~ N + Z L # f B(k,-) extends to L # since L # f~ N = N 1 is a summand. Call this extension f. Then f(-) = B(w,-) since B is non-singular. B(w,x) s Z for x E L #. Thus w E (L#) # = L. Now consider k - w. B(k-w,x) = 0 for x s L# f~ N = N I. Thus k - w e (L#('% N)/', Since L # is a lattice , k - w e N ~ = N. Clearly k - w s L # . Thus k - w s L # /~ N = N I. Now (~--C--Q) s H. However, w 6 L, so w = 0 s H. Thus s H as desired. So H "~ = H and ~(metabolic) ~ 0. Since we have now shown that ~ is well-defined and independent of the choices made, we are ready to prove: Theorem 1.5 The sequence i a 0 + W(k,Z) + W (k,Q) + W(k,Q/Z) is exact. Proof: i is 1 - 1 by Lemma I 5.4. ira i C ker ~. Let [M,B,Z] ~ [M,B,i] ~)Z Q. Choose the lattice M ~ 1 = L in M | Z Q. For this choice of L, we have L = HOmz(L,Z) since B is non-singular on M. Thus L # = L and ~ o i = 0. ker a C im i. Suppose ~([M,B, ~]) = 0. Let H ~ L#/L be a metabolizer for (L#/L,B ', s Let Lo = inverse image of H in L# C M under 142 the projection L # ~ L#/L. Then BIL ~ has values in Z since H is a self-annihilating subspace of L#/L, meaning that B' = 0 e Q/Z on H. L O# = {x ~ M: B(x Lo) C Z}. If x E L~ t projecting to L#/L r ~ H ~ = H Thus x ~ L o Obviously LoC ~, so L o= ~= Ho~,z(Lo,Z), and the adjoint is an isomorphism on L o- L o is Z invariant since H is s invariant. Thus consider [Lo, BILo,s e W(k,Z). Applying i, we obtain [M,B,s since L is a full lattice. Corollary 1.6 The sequence 0 § A(Z) + A(Q) ~ A (Q/Z) is exact. [] 2. Reducin 9 to the maximal order We continue our study of the boundary homomorphism by recalling the computation of W(k,Q) and A!Q). This was done as follows. Let S = Z(@) = Z[t,t-l]/(f (t)) , where f(t) is a monic, integral, irreducible T k fixed polynomial. Let E = Q[t,t-l]/(f(t)). Then, as we have seen, the field E has an involution - induced from T k. The fixed field of - is denoted by F. S is an order in O(E), the ring of integers in E. To simplify our notation, we write O(E) = D. 143 As in Theorem IV 1.8, we have the following computation: ~)(k,Q) -- (~ W(k,Q;f) -- (9 W(k,Q;S) . f s ~ Note: Here we are using the symbol B to denote the collection of T k fixed, monic, integral , irreducible polynomials. This should not be confused with IV 1.8 where the integral requirement is omitted, for W(k,Q). We denote the restriction of Z to W(k,Q;S) by $(S). Lem~a 2.1 There are exact sequences: i ~(S) 0 ~ W(k,Z;S) ~ W(k,Q;S) + W(k,Q/Z;S) i ~(S) 0 ~ A(Z;S) + A(Q;S) + A (Q/Z;S) . These follow from Theorem 1.5, since the S-module structure is preserved by i and ~(S). ~] In order to study and compute the map ~(S) we begin by comparing this with the exact sequence for ~(D) where D is the maximal order. We have the following commutative diagram of forgetful maps. Lemma 2.2 Let fl,f2,f3 be the ma~s which forget the D-module structure and remember only the S-module structure. Then the dia@ram below commutes : 144 i I ~ (D) O ~ W(k,Z;D) ~ w(k,Q;D) W(k,Q/Z;D) +fl +f2 4.f3 i 2 8(S) O + W(k,Z;S) ~ W(k,Q;S) W(k,Q/Z;S) Proof: We begin by remarking that the notation W(k,Q;D) or W(k,Q;S) is somewhat redundant. Indeed W(k,Q;D) = W(k,Q;S) = H(E). in order to verify commutativity recall that ~(D) is defined by choosing an integral D-lattice, L, forming L#/L, etc. However, "by forgetting," the D-lattice L is also an S-lattice, L # is the same, and so consequently is L#/L. PI Thus, we see that we can reduce the study of ~(S) to studying Z(D) where D is the maximal order. However, this reduction is more complicated than might first appear since we still must compute both Z(D) and the maps f.. 1 To see precisely the problems involved in this reduction we continue with the study of ~(D) for D the maximal order. We find it profitable to give the computation for ~(D) on the "Hermitian" level. Thus we recall the following identifications. W(k,Z;D) = H(A-I(D/Z)) where A-I(D/Z) is the inverse different of D over Z W(k,Q;D) = H(E) W(k,Q/Z;D) = H(E/A-I(D/Z)) = ~ H(D/P) 145 In the asymmetric case we likewise had: A(Z;D) = H@ (~-i (D/Z)) A (Q;D) -~ H@ (E) A(Q/Z;D) = H@ (E/~-I (D/Z)) These isomorphisms together with Lemma 2.1 lead us to the exact sequence: i ~(D) O + H (I) ~ H (E) + H (E/I) U U U where I = A-l(D/Z). We should discuss this new boundary map which we continue to call ~(D) since there should be no confusion. Let [M,B] e Hu(E). We first must construct a D-lattice L in M so that BIL takes values in I. In order to do this we follow Proposition i.i. Let {el, .... e n} be a basis for M. Then we can find non-zero integers rijk,Sijk so that rijkB(@kei,ej) e I and SijkB(eei,e j) e I for i,j ~ n and k < m, where m is the degree of the minimal polynomial of 0. Let d = HrijkSij k. Then the lattice L generated by {del,...,den, .... 0mdel,...0mden } is a D-lattice on which B is I-valued as in i.i. Let L # = {v c M: B(v,L) C I}. L # is also a D-lattice. We form L#/L with induced E/I-valued inner product B' as before. Then define ~(D): [M,B] + [L#/L,B']. With ~(D) so defined, the following is clear. Proposition 2.3 ~(D) is well-defined for Hermitian and the followin~ diagram commutes 146 i .~(D) O + H (~-i (D/Z)) ~ H(E) + H (E/~-I (D/Z)) + = t + = t $ = t i ~(D) O + W(k,Z;D) ~ W(k,Q;D) ~ W(k,Q/Z;D) The vertical isomorphisms are induced by trace, t, of E over Q. SO the method we shall employ for computing ~(D) is to study the corresponding boundary for Hermitian. The image of boundary is the group H(E/~-I(D/Z)) = W(k,Q/Z;D) . We recall the computation of this group given in Chapter IV 3.5, 3.8. We had: g tr W(k,Q/Z;D) § ~ W (k,Fp; D/P) § ~) H (D/P) . p=T p=~ Here we sum over all - invariant maximal ideals in D. The 1 isomorphism g is induced by selecting a generator, say ~, for the p-torsion in Q/Z. tr denotes the isomorphism induced by trace on finite fields, D/P over F . P Similar computations apply to W(k,Q/Z;S). We use the letter M to denote a - invariant maximal ideal in S. Proposition 2.4 The diagram below commutes. g tr W(k,Q/Z;D) ~ ~ W(k,Fp;D/p) § ~) H (D/p) p=~ P=P f3 ~ ~tr g tr W(k,Q/Z;S) ~ W (k,Fp; S/M) § ~ H (S/M) M= M M=~ 147 The map tr: H(D/p) ~ H(S/M) where ~ = P ~ S is given by trace on finite fields. Proof: Let [M,B] e H(D/p). Then apply tr to obtain an -i element in W(k,Fp;D/P). Apply the isomorphism g and "forget" to obtain an element in W(k,Q/Z;S). However, tr: D/P ~ F is the same as the composition P tr: D/p § S/P~S ~ Fp. Thus f3 o g-i o tr = g-i o tr o ~ tr, ie. the above diagram commutes. In order to appreciate Proposition 2.4, we collect those - invariant maximal ideals in D which lie over a given - invariant maximal ideal in S. Let T(M) = {P: P is a - invariant maximal ideal in D with P /~ S = M} , where M is a - invariant maximal ideal in S. T(M) may or may not be empty. We shall discuss T(M) further in Chapter VII 4. We now define the local boundary by: ~(D,p) : w(k,Q;D) + H(D/P) by ~(D,P) = q(P) o a(D) where q(P) is the composition of (tr -1 o g) with projection to the P th coordinate in ~ H(D/P). The local boundary 8(S,M) is similarly defined. Theorem 2.5 8(S,M) = ~ tr o a(D,P) P e T(M) Proof: By Lemma 2.2, f3 o a(D) = a(S). Thus, using Proposition 2.4 we need only determine which - invariant maximal ideals P in D 148 lie over ~ in S. These are given by T(M) by definition. Thus, the comm~utative diagram in 2.4 yields: ~(S,M) = ~) tr o q(P) o Z(D) P ~ T(M) = ~) tr o ~(D,P) P e T(M) In order to tabulate the progress we have made and to indicate the problems involved in the computation of ~(S) we record the following diagram. i ~(D) 0 + H(s § H(E) ~ H(E/A-I(D/Z)) # t + t r t i ~ (D) g 0 + W(k,Z;D) + W(k,Q;D) + W(k,Q/Z;D) + ~) W(k,Fp;D/P) -- ~)H(D/P) P=Z + fl + f2 + f3 + tr i ~(S) g 0 + W(k,Z;S) § W(k,Q;S) ~ W(k,Q/Z;S) + 0_ W(k,Fp;S/A~) = (~ H(s/M) M=M We wish to compute the group W(k,Z;S). The idea is to use the above diagram. We compute the group W(k,Z;D) for the maximal order D by using the exact sequence with ~(D). Then we compute the map fl" We first discuss fl" 149 Lemma 2.6 fl i_ss 1 - i. Proof: Clear by diagram chase since i is 1 - 1 and f2 is an isomorphism. F'I In order to compute the cokernel of fl we recall some homological algebra. [M 50] im 8(D) /~ ker f3 = coker fl" Thus we need to compute $(D) and f3" By Proposition 2.4, f3 is determined by two things: (i) The map tr: H(D/P) + H(S/p~S) induced by trace on finite fields. (2) T(A4) = {maximal ideals P in D: P /~ S = M, P = ~} We shall discuss both of these in Chapter VII. In analyzing the group w(k,Z;D) ~ H(A-I(D/Z)) and ~(D) we use the isomorphic Hermitian sequence. (D) O + H (A-I (D/Z)) + H(E) + H (E/A-1 (D/Z)) = @) H (D/P) Again we call ~(D,P) = q(P) o ~(D), where q(P) denotes projection to the P th component H(D/P). We may identify ~(D) with ~ ~(D,P), P so that in essence our task is to compute ~(D,P). It is at this point that we must be very careful however. The isomorphism H(E/A-I(D/Z)) = ~ H(D/P) can be given in any P=~ fashion and will enable us to compute H(A-I(D/Z)) = W(k,Z;D). By this 150 we mean we are free to choose localizers embedding the residue field D/P into E/A-I(D/Z) which specify the isomorphism. However, in order that the isomorphisms determined by our particular choice of localizers be applicable to computing ~(S), the particular choice of iocalizers given must make our previous diagram commute, ie. H(E/&-I(D/Z) ) + @) H (D/P) + t + tr g W(k,Q/Z;D) ~ G) W(k,Fp;D/P) P=~ There are many ways to give the isomorphism H(E/~-I(D/Z)) = ~ H(D/P). As we see from the preceding, there is one canonical isomorphism, namely tr -I o g o t. We shall see in the next section that there is also a convenient way to give this isomorphism in order to read ~(D,P). Thus, we have the following plan. (i) First we study the case of a maximal order. We shall use a convenient choice of localizers to compute ~(D,P). This enables us to compute ~(D,P) and the cokernel of ~(D). (2) In Chapter 7 we discuss non-maximal orders. We must discuss three key parts in this regard. (a) We discuss the canonical loealizers which make the finite field trace and number field trace induced isomorphisms agree. (b) We compute trace induced map tr: H(D/P) ~ H(S/P~ S) for finite fields. (c) We describe the set T(M). 151 We shall next be computing the group W(k,Z;D) = H(A-I(D/Z)) by using the exact sequence from Proposition 2.3. (n) 0 -~ H (A-I (D/Z)) + H(E) + H (E/A-1 (D/Z)) We recall that H(E) for E an algebraic number field, and H(E/A -I) = ~ H(D/P) are known. (Here we abbreviate A-I(D/Z) by A-l). In analyzing the boundary map ~(D), we reduce to the local case by using: Proposition 2.7 Localization at a - invariant, maximal ideal P in D induces the commutative diagram: (D) 0 + H(A-I(D/Z))-- + H(E) + H(E/A -I) = G) H(D/P) P=; q(P) (D, P) 0 + H(A-I(P)) § H(E) § H(E/A-I(P)) -- H(D/P) This diagram also commutes for an order S. Proof: $(D,P) = q(P) o ~(D) by definition. What we show here is that these maps are induced by localization. As long as S is an order in D, maximal or not, S is a finitely generated Z-algebra, and hence Noetherian by [A,M 81]. Thus by [B-2 20], every finitely generated S-module M is finitely presented. Hence [B-2 76] the adjoint isomorphism M + Hom(M,E/A -I) localizes to M(M) + HOms(M) (M(M), (E/A -1 ) (M)), where M is a maximal ideal in S. Thus localization preserves the non-singularity of the forms, and indeed defines a map. 152 In order to identify (E/A -I ) (P) with E/A-I(P), use the exact sequence 0 + A -I ~ E + E/A -I 0, and [A,M 39]. ~] Comment: This re-emphasizes the remarks before Theorem IV 1.9. Corollary 2.8 H(A-I(D/Z)) = /~ H(A-I(D/Z) (P)) P = Proof: H(A -I) = ker D (D) = /~ ker q(P) o 3(D) P= F = ~ ker 3 (D,P) P=~ = /'~ H(A-I (p)) [~ The computation of ~(D,P) will be made in Section 3. 3. Computing the local boundary ~(D,P) for the maximal order. We consider the general case ~(D): H u (E) + H u (E/I) whore I = is a - invariant fractional ideal. Of course, we have in mind I = A-I(D/Z). Following Proposition 2.7, we wish to compute the localization, 3(D,P) of ~(D) at a - invariant maximal ideal P = P in D. Since D will be fixed throughout this section, we simplify our notation of ~(D,P) to ~(P). 153 From 2.7, (E/I) (P) = E/I(P). We now consider E/I(P). (See IV 3.8). We embed the residue field D/P into E/I(P) as follows. Let p ~ E satisfy Vp(0) = Vp(I) i. Then define f: D/P § E/I(P) by r + P + pr + I(P). f is 1 - 1 since pr ~ I(P) implies vp(0r) >_ Vp(I(P)) = vp(I) SO Vp(I) - 1 + Vp(r) _> Vp(I) vp(r) > 1 Hence r g p. For the case when S is an order in E, with I = s we claim that there is a commutative diagram w 0 ~ S/M + E/I(M) +tr +t W 0 ~ F + Q/Z (p) . P Here tr denotes trace on the finite field level, t is induced by trace of E/Q. w is given by the canonical choice of uniformizer in Q/Z(p) annihilated by p, namely w: 1 + (~). In order to see that w exists we proceed as follows. Let A be a finitely generated S-module. Then S is the image of a free 154 f S-module, F l, and we have the exact sequence 0 + kerf § F 1 ~ A+ 0. Of course Fl/ker f= A. This leads to the diagram below, with F 2 = kerf. 0 0 + h 2 + F 2 Z + h + § F 1 Q + h + A Q/Z Given h s HOmz(A,Q/Z), h lifts to hl: F 1 § Q since F 1 is projective. By commutativity, hllF2 = h 2 s HOmz(F2,Z). For finitely generated projective S-modules A, we assert that there is a trace E/Q = t induced isomorphism: HOms(A,E) + HOmz(A,Q) . t(g) = t o g is clearly onto since A is S-projective. In order to see t is i - i, suppose g ~ HOms(A,E). Let a ~ A satisfy g(a) @ 0. Then clearly there exists e e E with t(eg(a)) ~ 0. However S is an order in E so we can write m - e ~ S for some m c Z. Thus, since t is Z-linear, m t(eg (a)) = t(me g(a)) = t (g (mea)) ~ 0. 155 It follows that t o g { 0, and t is an isomorphism. Hence h I s HOmz(FI,Q) may be written uniquely as t o k I where k I s HOmS(FI,E). Further, since t o kll = h 2, we observe that F 2 kll = k 2 s HOms(F2,A-I(s/z)). F 2 Thus k I induces an S-module homomorphism k s HOms(FI/F2,E/~-I(s/z)). Clearly t o k = h. We claim that this k is unique. For suppose t o j = t o k = h. Then t o (j - k) = 0 in Q/Z. Now consider the diagram ^ (j - k) F 1 + E r r (j - k) FI/F 2 E/A -I ^ (j - k) exists since F 1 is S-projective. However, we also have the commutative diagram: t E + Q + r E/A-1 t Q/Z ^ Thus t o (j - k) C Z, from which it follows that we have ^ im(j - k) C A-I (S/Z). Hence j - k - 0 as maps in HOms(FI,E/A-I(s/z)) and j = k is unique. 156 We apply this to the finitely generated S-module S/M, where h: S/M + Q/Z(p) is the Z(p)-module homomorphism h = w o tr. By the above, there exists a unique k = w with t o w = w o tr as claimed. w is evidently determined by where 1 s S/M is taken. Hence is in fact determined by PM for a suitable choice of localizer PM" However, in our computation of D(P) , we shall find it convenient to specify the localizers pp in a different manner. The manner in which we pick these is dictated by our desire to have the boundary computation read by Hilbert symbols. Theorem 3.1 Let [M,B] s Hu(E). We diagonalize B as ----l B = I. If P is over inert, so that ~(P) : Hu(E) § Hu(E/I(P)) + H(D/P) -- {0,i} we have the following formulas. (a) If B = (a,~) = (-l)~(P) then: (d,a)p = (-i) ~(P) (B) 157 (c) If there are no ramified primes, and x I has odd valuation v p(X I) at an even number of primes formulas (a) and (b) above hold. (d) If there are no ramified primes, and x I has odd valuation at an odd number of primes, formulas (a) and (b) are valid at all inert primes except one specified prime P1 over PI" At PI' v p (ax I) = Vp(a) + 1 and we then have 1 ! (a) (a,o) = (-i) ~(P) (b) For B of even rank, formula (b) still holds. II. Ramified primes are divided into two classes: (a) cl (P) = 0 if vF(I) E v p(x I) (mod 2) (b) cl (P) = 1 if Vp(I) E vp(x I) + 1 (mod 2) ~(P) ~ 0 at ramified primes of class 0. ~(P) preserves rank at ramified primes of class i. Note: This determines ~(p) at dyadic ramified primes. Further, under the choice of localizers pp made, Hu(E/I(p)) = W(D/p) reads ~(p) as follows: ~(p) ( Proof: We begin by considering u, with u~ = i. By Hilbert's Theorem 90, there exists x e E* with xx -I = u. Our first task as described in the theorem is to rechoose x appropriately. 158 Thus, we consider vp(x), for P over inert. If at least one prime ramifies, finite or infinite, we can find y E F* with (y,0)p = (-i) vP(x) for all P = P /~ F which are inert, by realization of Hilbert symbols. If there are no ramified primes, finite or infinite, there are two possibilities. (i) If vp(x) is odd at an even number of inert primes, by Realization it is still possible to choose y s F* with (y,0)p = (-i) vP(x) = (-i) vP(y) at all inerts. (2) If vp(x) is odd at an odd number of inert primes, we may find y g F* with (y,o)p = (-i) vP(x) at all inert primes except one specific inert prime, say PI' at which (x) + 1 (y,O)pl (-i) VPl P1 /~ F = PI" ----1 We now rechoose x I = xy. We still have XlX = u since y g F* has yg-i = yy-i = i. Note, however, that x I now has even valuation at all inert primes, with at most one exception as described above. We next describe how to choose the localizers pp. First, at P inert, we choose pp = XlW , where w g F* satisfies vp(w) = vp(I) - vp(x) - i, so vp(pp) = vp(I) - i. This is possible at inert P, since a local uniformizer for P{~ F is also a local uniformizer for P. In order to describe the ramified primes P we begin as follows. We note that the - involution makes the local units in OE(P), denoted OE(P)* into a C2-module. 159 Lemma 3.2 If P = ~ is over inert, then HI(c2;OE(P)*) = i. If P = ~ is over ramified, HI(c2;OE(P)*) = C 2. Proof: Recall that HI(c2;OE(P)*) = {x e OE(P)*:xx = !} modulo {v/J: v [ OE(P)*}. If x is a local unit of norm i, then by Hilbert 90, there exists z c E* with zz -I = x. Write z =~P(Z)v, where ~ is a local uniformizer for P, and v s OE(F)*. If P is over inert, we may choose ~ g F*, so that = ~. Thus zz -I = v~ -I = x, and H 1 is trivial in this case. If P is over ramified, (~-I) is a local unit, and zz--i = (~-i) vp(z)vv--I = x. Thus H 1 is generated by the class of ~-i cl(~-l) Of course (~-i)2 is trivial in H 1 since (z~-l)2 = (~$-i) (~ -i)-I, a quotient of local units. Thus, to complete the proof we need only show that cl(~ -I) is non-trivial in H I. Suppose to the contrary that there is a local unit v with --I -I. -i ---i -i vv = z~ Then ~v = zv , so that ~v is a local uniformizer of OE(F) which lies in F. This is impossible as we are in the ramified case. Hence, cl(~ -I) is non-trivial as claimed. We now consider cl(u) e HI(c2;oE(P)*) where P = ~ is over ramified. By Lemma 3.2, HI(c2;OE(P)*) ~ 0 and we may write cl(u) = cl (~-I) vP(I) - ~ where g = 0 or I. Definition 3.3 If P = ~ is over ramified, P is of class 0 if ~ = O. P is of class 1 if g = i. We observe the following: 160 Lemma 3.4 We may rephrase class as follows: (a) cl(P) = 0 if and only if vp(I) - Vp(X I) (mod 2) (b) cl(P) = 1 if and only if Vp(I) _-- Vp(X I) + 1 (mod 2) Proof: Here Xlx~l = u. Write x I = ~iw, w g OE(P)*. Then cl(u) = cl(XlXi I) = cl(z{-l)i However, cl(~[ -I) generates HI(c2;OE(P)*) # 0, from which the result follows. When P is tamely ramified, we choose z = -[, a skew - uniformizer. When the ramification is wild, any uniformizer will do. Note that z~ is a uniformizer for P /~ F = P, since P is over ramified. Also, vp(z~) = 2. We now choose pp at ramified primes as: Pp = Xl(~)t z with t suitably chosen so that vp(pp) = Vp(I) - 1 if cl(P) = O. pp = Xl(Z~)t with t suitably chosen so that vp(pp) = Vp(I) - i if cl(P) = i. With these choices of localizers made, we now identify the image groups of ~(P), Hu(E/I(P)). Let [V,B] ~ Hu(E/I(P)). By assuming that (V,B) is anisotropic, it follows that the annihilator of the finitely generated OE(P ) - module V(P) is the maximal ideal m(P) 161 in OE(P). Thus V is an OE(P)/m(P) - module. This is equivalently phrased by saying V is a vector space over the residue field 0 E P)/m(P) = D/P. Let x,y E V. Suppose B(x,y) = [a] s E/I(P). Letting z be a uniformizer for P as above, nx = 0 since V is an OE(P)/m(P) - module. Thus ~[a] s I(P). Let a I be a lift of a to E. Since za I E I(P) , it follows that vp(a I) ~ Vp(I) - i. Also B is u Hermitian, so that [a] = u[a] in E/I(P). We may thus write a I - ua I g I(P). We now consider the D/P - valued form on V given by B1 = p~l . B, where the choice of pp has been previously specified. (i) P inert. With a I - ua I = i s I(P), we show B 1 is +! Hermitian. Here pp = XlW where w s F*. -- --i a I - i _-i -i alp P = (~)x I = (~)(~i )w-I -i alw .w -I - -i -i - l = alP F - ipp x I x I = al ppl in D/P. This last follows because Vp(i pp-l) = Vp(i) - Vp(I) + 1 _> i, so that ip'pl s p. This shows there is an isomorphism between 162 Hu(E/I (P)) and H+I (OE (P)/m (P)) given by scaling with pp. Since P is inert, the involution induced on OE(P)/m(P) is non-trivial, and we have true +i Hermitian. (2) The tamely ramified case. As before, we have the form B 1 = p~IB. (a) cl(P) = 0, so pp = Xl(~)t~ where ~ = -~ We now compute as before: ----i a I - i alPp = (____~) (Xl~ (z~)-t ([)-i = (____V_)a I - i (~_) (~)-t (_~)-i x I = (a I - i) (_ppl) = _alppl in OE(P)/m(P). Since P is ramified, we obtain this time an isomorphism between Hu(E/I(P)) = W-I(D/P) = 0. (b) cl(P) = +i, so pp = Xl(Z~)t. The same computation shows al ~i = alppl in D/P, and we have Hu(E/I(P)) = W+I(D/P). (3) The case for wild ramification follows as above. In either case, H (E/I(P) = W(D/P). U With these preliminaries, we are ready to compute ~(P). To begin with, consider a 1-dimensional form in Hu(E). By II 4.15 we may write this as uniquely determined in F/NE/FE. 163 I. We first compute ~(p) for p over inert. We begin by considering ~(P) ( for z a uniformizer for P. It follows that without loss of generality, we may assume either: (a) Vp(aX I) = Vp(I) -2 or (b) Vp(aX I) = Vp(I) -i. This is done by rechoosing a as a(z~) t. Here, recall we may choose s P/~ F = P since this is the inert case. In any case, Vp(Z~) = 2 and z~ is not a uniformizer for P. Thus case (a) or (b) only depends on Vp(a) compared to Vp(I), since by choice Vp(Xl) E 0(mod2) with at most one exceptional prime PI" Now consider the lattice L = P. Since Vp(aX I) = Vp(I) - 1 or Vp(I) - 2, L # = {x E E: B(x,L) ~ I} = {x E E: xaxl~ C I} = {x e E: x e I~ -I (axl)~I} = ~P in case (a) OE(P) in case (b) Thus viewed, in case (a), we clearly get ~(P) In case (b), ~(P) E/I(P) by Again, P is inert so that [OE(P)/m(P) : OF(P)/m(P)] = 2 and the induced involution on the finite field OE(P)/m(P) is non-trivial. Hermitian of a finite field is determined by rank modulo 2. Thus, we have completed our computation of the local boundary on a 1-dimensional form when P is inert. Identifying Hu(E/I(P)) = H+I(D/P) = F 2 = {0,i}, we may summarize this as: 3(P) = (a,a)p. Continuing in the inert case, let B be a form of even rank. Since E is a field, we may diagonalize B as before, B = n(n-l) As in II 4.15, we define the discriminant of B to be (-i) 2 Ha i. Note that this depends on the fixed choice of x I. Adding a hyperbolic form (d,O)p = (a I ... an,O) P = ~ (ai,~) P 1 = H (_i) ~ (P) = H (_I) ~ (P) This completes the inert case. II. The ramified case. As in the inert case, we begin by considering a 1-dimensional form from E, we may assume vp(a) = 0. Note: This does not affect (a,O)p, nor ~(P) Vp(aXl (~) t ) = Vp(Xl (n~) t) = I~ p (I) - 2 p(I) - 1 depending on cl(P) . As in the inert case, we let L = P, and compute L # = P if cl(P) = 0 = O E (P) if cl (P) = 1 Thus, if cl(P) = 0, ~(P) = 0. If cl(P) = i, we obtain the E/I(P) - valued form identify this with the OE(P)/m(P) - valued form on OE(P)/m(P) given -i by viewing a in OE(P)/m(P) : a Witt inner product since P ramifies, so that [OE (P)/re(P) : OF (P)/m(P) ] = l, and the induced involution on the finite field OE(P)/m(P) is trivial. (i) If the characteristic of OE(P)/m(P) = 2, rank is the only invariant, and we are done. (2) If the characteristic of OE(P)/m(P) ~ 2, we must determine if a is a square in the residue field. By II 2.4, a is a square in OE(P)/m(P) if and only if (a,O)p = +i. We continue by letting B be a form of even rank. As before, we diagonalize B, B = 4. Computing the cokernel of ~(D). In this section, we use the computation of the local boundary, ~(P), to compute 2: Hu(E) + Hu(E/I) We also show how to compute H (I), where I = A-I(D/Z). Of course, U H(A-~D/Z) = W(k,Z;D) , and Hu(A-I(D/Z)) = A(Z;D) . Further, the computation of the boundary on the Hermitian level ~ will subsequently be used in the computation of the global boundary 2: W(k,Q) + W(k,Q/Z) in Chapter VIII. In order to describe the boundary homomorphism, the complicated case is described in the next Lemma. Lemma 4.1 Suppose E has involution -, fixed field F as usual. Suppose E/F has no signatures, n_oo dyadic ramified primes, and all ramified primes are of class i. Then we may write the Gollection o__~f ! ramified primes as P1 ~ "'" 'P2t'PI ' .... Pr where the Pi' i = 1 ... 2t have residue fields OE(Pi)/m(pi) = Fq with q H 3 (4) and the ! Pi' i = 1 .... r have residue fields Fq __with q ~ 1 (4) (assuming 2t + r ~ 0). 168 Proof: We wish to show that the number of ramified primes whose residue fields F have q s 3 (4) is even. Let P. be such a q l prime. Then -i is not a square in each OE(Pi)/m(Pi) = Fq [Lm 43]. However at ramified primes the square class in the residue field determines the Hilbert symbol. Thus (-l,o)p. = -i at each such 1 Pi = P'~l F over ramified. Notice that at the other ramified primes P!'l whose residue fields are OE(PI)/m(P i) = Fq with q E 1 (mod 4) , -i is a square in the residue field, so that (-i,o) = +I. Also, P! 1 -i is a local unit at inerts, so that vp(-1) (-l,o)p = (-I) = +i at all inerts. We now apply Hilbert reciprocity, H(-l,q) p = +i. This shows i i that the number of primes with residue fields F with q ~ 3 (4) q elements must be even, since (-l,O)p. = -I only at those primes. l Next we form the group G given by G = Z* ~(F2x ... • x (F2 x ... xF2)r , where 2 t = number of ramified primes at which (-l,O) p = -i, r = number of ramified primes at which (-l,O)p = +i, as given by Lemma 4.1. We write Z* = {i,-i}, F 2 = {0,i}. Z* is designed to keep track of the discriminant and reciprocity, F 2 will take care of ranks. On G we define a multiplication as follows: 169 t I ! (c,a I .... a2t, b I .... b r) ~ (c', a~ .... a2t, b I .... b r) f ((-l) alal "'" + a2ta2tcc ' ! = , a 1 + a i .... a2t + a2t, b 1 + b 1 .... I ,b r + b r ). Note that (I,0, ..., 0) is the identity in G, and that the order of every element divides 4. The purpose of this group G is to describe the cokernel of ~: Hu(E) ~ Hu(E/I) in the special case that there are no signatures, no dyadic ramified primes, and there are ramified primes, all of which are of class 1 in the extension E/F. We assume now that we are in this case. To begin with, recall that Hu(E/I) = ~ H(D/P) ~ W(D/P). P = ~ inert P = ~ ramified Thus any element in H (E/I) can be expressed as a direct sum of U elements ~ [MiBi], where [MiB i] is either in H(D/P) or W(D/P). We define a map h: Hu(E/I) § G by: Let [Mi,B i] s H(D/P). This Hermitian element depends only on the rank modulo 2 of M. over D/P. We define: l h([Mi,Bi]) = ((_i) rank M.l, 0, 0 .... 0) ~ G for these Hermitian summands. Continuing, suppose ~ [Mi,B i] s ~ W(D/P) P P ramified 170 Let d = discriminant of B. 1 1 c = Z (di,O)p P. ramified l l a. = local rank of M. over D/P i i = i, ... 2t 1 1 b i = local rank of M i over D/P i' i = I, ... r Define h: [Mi,B i] + (c, a I .... a2t, b I .... br). Clearly, by P additivity, this defines h as a map h: H u (E/I) + G. Lemma 4.2 h is a homomorphism. Proof: This follows using the product formula for discriminants, II 4.10. The local discriminants satisfy dis ([MI,B I] ~ [M2,B2]) = (_l)rank M 1 rank M 2 dis B 1 dis B 2 We now use the following abbreviations: dis B 1 = d I dis B 2 = d i rank M 1 = a I rank M 2 = a i. At each of the first 2t ramified primes, a a' ! = (_i) alal I ((-I) 1 ldldl ,a)p (dld I , a) p ala ~ = (-i) (d I, a) p (d i, a) p since (-i) is not a square in D/F. At each of the next r ramified primes, 171 ((-i) bibidi d i,O)p : (dldl,o) P = (di,o)p(di,o) P since (-I) is a square in D/P. From these formulas, it clearly follows that h is a homomorphism. FI Lemma 4.3 The image of ~ in H (E/I) is mapped under h to i U the subgroup W 2 of G whose elements are (i,0 .... 0) and (E,I,I .... i) where c is given by ~ = E (-l)vp (I) (same P inert hypotheses as 4.1) Proof: Let ~ [Mi,Bi] G [Mi,B i] be in the image P inert P ramified of 2, say ~[M,B] equals this element in H (E/I). In other words ~(Pi) [M,B] = [Mi,Bi] , where [Mi,B i] E H(D/P i) or W(D/P i) depending on whether P is over inert or ramified. l Case (i) M has even rank, and discriminant d. By Theorem 3.1, ~(P) is read by the Hilbert symbol, (d,0)p = (-i) ~(P) [M'B] Rank is preserved at class 1 ramified primes. Hence by Hilbert reciprocity, h o ~ [M,B] = (i, 0 .... 0) E G. Case (2) M has odd rank, and discriminant d. At inert primes: (d,o) = (_I)~(P) [M,B] + Vp(I) P 172 At ramified primes, the square class of the discriminant of ~(P) [M,B] is determined by the Hilbert symbol (d,o)p. Rank mod 2 is preserved, since by hypothosis all ramified primes are of class I. Hence, (P) [M,B] h o ~ [M,B] = ( H (-i) (d,O)p, i, 1 .... i] V inert P ramified However, by Hilbert reciprocity, (d,O)p = +i. Thus, all P vp(I) H (_i) ~ (P) [M,B] H (d,O)p = ~ (-i) P inert P ramified P inert as claimed. [] Let h: Hu(E/I) + G/W 2 by the composition of h: Hu(E/I) + G with projection G + G/W 2. Corollary 4.4 Again with the hypotheses of Lemma 5.1, there is an exact sequence h 0 + image ~ § Hu(E/I) ~ G/W 2 ~ 0. Proof: By Lemma 4.3, image ~ is contained in the kernel of h. It follows by realization of Hilbert symbols and Theorem 3.1 that image ~ = kernel h. h and consequently h maps onto G by realization of Hilbert symbols. 173 Corollary 4.4 in fact is the proof of part (2) of the following theorem. Theorem 4.5 The cokernel of 2: Hu(E) + Hu(E/I) is given as follows. (i) If there are no ramified primes, finite or infinite, the cokernel of ~ depends on the number of primes at which vp(x I) + vp(I) is odd. If this number is even, the cokernel of ~ is C 2. If this number is odd, ~ is onto. (2) If there are no signatures, dyadic ramified primes, and there are ramified primes, all ramified primes being class I, then there is an isomorphism induced by h: H(E(I)/im~ = cokernel ~ = G/W 2 Thus the cokernel has order 2 2t + r There is an element of order 4 in cokernel ~ if and only if t ~ 0. (3) If there are signatures, dyadic ramified primes, or ramified primes of class 0 the cokernel of ~ is a product of C2's. Its order is determined as follows. Let a be the number of ramified primes of class i. Let b be the number of dyadic ramified primes of class 0. Then the cokernel has order 2 max(a - 1,0) + b Proof: (i) If there are no ramified primes, we must consider two cases. (a) Vp(X I ) is odd at an even number of primes. (b) Vp(X I) is odd at an odd number of primes. 174 In case (a) , we can read 2: Hu(E) ~ Hu(E/I) = ~ H(D/P) by P Theorem 3.1. The formulas for ~(P) are: 8(P) (d,O)p = (-i) ~(P) (B) d = discriminant of B, on even rank forms. Suppose Vp(I) is odd at an even number of primes. Then clearly Vp(X I) + Vp(1) is odd at an even number of primes, since we are in case (a). By realization of Hilbert symbols we may pick a ~ F with (a,O)p arbitrarily specified subject only to reciprocity. Since Vp(I) is odd at an even number of primes, the formulas given determine that ~(P) is non-trivial at an even number of primes. Thus the cokernel has order 2 by realization; and ~ is subject only to the stated restriction. Continuing, if vp(I) is odd at an odd number of primes, Vp(X I) + Vp(I) is odd at an odd number of primes also. Thus by the ~(P) (b) vp(x I) is odd at an odd number of primes. In this case, by Theorem 3.1, we have the previously stated formulas, with one exception P1 at which the Hi!bert symbol is read "backwards." Again, suppose vp(I) is odd at an odd number of primes. This means Vp(X I) + Vp(I) is odd at an even number of primes. However, reciprocity now reads !75 ~(P) Comment: In this instance we are penalized for our choice of x I. Had we chosen x I with Vp(X!) ~ vp(I) (mod 2) almost everywhere, (meaning with at most one exception) the local boundary ~(P) would ~(P) Part (i) of this theorem would have required no special analysis. Indeed, since the cokernel of ~ does not depend on Xl, this provides an alternate proof. However, we find it more natural to give x I with even valuation at inerts. Thus, forms in H+I(E ) do not require a special "Xl" in their diagonalization. (2) As remarked earlier, Part (2) follows from Corollary 4.4. (3) If there are signatures, dyadic ramified primes or tamely ramified primes of class 0, Hilbert symbols determining boundary may be arbitrarily specified, with corrections made at the infinite ramified, dyadic or class 0 primes. Thus, by realization there is a form in H(E) with non-trivial discriminant at prescribed inert and ramified primes of class i. Applying 2, we obtain a form in H(E/I) with prescribed rank at inert primes, and prescribed discriminant at class 1 ramified primes. However, as required by Theorem 3.1, rank is 176 preserved at class 1 ramified primes, with ~(P) = 0 at dyadic ramified primes of class 0. We may thus write H(E/I)/im ~ -- (F 2 • ... • F2)a• 2 ... • F2)b/w via y with y given by: y: ~ [Mi,Bi] § (r I .... ra, s I .... Sb) where r i = rank [Mi,B i] at class 1 ramified primes, dyadic or not. s i = rank [Mi,Bi] at class 0 dyadic ramified primes. W = C 2 subgroup of the product a + b F 2 generated by (1 .... i, 0, 0, ... 0) i = 1 (0 .... 0, 0, 0) This completes the proof. F] Comment: ~(P) = 0 at tamely ramified class 0 primes and also Hu(D/P) = 0. Thus, there is no contribution to the cokernel. By applying Theorem 3.1 we are also able to calculate H u (I). We let J C H(E) be the subgroup of even rank forms. On J /~ H u (I) we may define a local discriminant homomorphism at each dyadic ramified prime: 177 d. : J H (I) + H(E) (dio) Pi Z* 1 U where d is the discriminant of the form in H(E), and (d,O)p. is 1 the Hilbert symbol at P . Since we have restricted the domain of d. l 1 to the even rank forms, the discriminant is multiplicative, as is the Hilbert symbol, so that d. is indeed a homomorphism. 1 Let t be the number of dyadic ramified primes. We then define the total discriminant homomorphism d to be the product of the d i. = H (d i) : J /~ Hu(I) + (Z*) t There are also the infinite ramified primes to take care of. By Chapter II 5, we may define a signature sgni: Hu(E) ~ Z at each infinite ramified prime P~. Combining, we obtain a total signature homomorphism sgn: J ~ H(I) ~ (2Z) r, where r is the number of infinite ramified primes. Recall how the Hilbert symbol was read at an infinite ramified prime, Chapter II 2.5. Let d be the discriminant, P~ an infinite ramified prime. Then (d,~ = • depending on the sign of d in P~. d = (-i) w(w-l)/2 det where w = rank B, det = determinant B. Hence d has signature 2 w(w - i) w- sgnB w - 2w - s~nB (-i) 2 (-1) 2 = (-i) 2 For [M,B] in J /~ Hu(I), w -- 0 mod 2, so that (d,~)~ = (-i) -sgnB/2 = (-i) sgnB/2 178 We thus may state: Theorem 4.6 Hu(I) is determined as follows. For the even rank forms in Hu(I), we have an exact sequence: sgn G d H 0 + J ~Hu(I) § (2Z) r ~ (Z*) t Z* + 0 r sgni/2 t where H = H (-i) H d. is the Hilbert reciprocity map. i = 1 i = 1 i ---- In order for a rank 1 form to exist in H (I) there must be no U class 1 ramified primes. If there are any signatures or class 0 ramified primes, a rank 1 form exists. If there are no ramified primes, a rank 1 form exists if and only if Vp(I) + Vp(X I) E 1 (2) at an even number of primes. Further, if a rank 1 form exists, there is an exact sequence: rank 0 ~ J ~ Hu(I) ~ H(I) + F 2 + 0 This sequence s~ts if and only if -i is a nomn. Proof: J ~ H(I) embeds into J ~ H(E), which by Landherr's Theorem II 5.4 is determined by the discriminant and multisignatures. However, H (I) = /~ ker ~(F), by Corollary 2.8. Now ~ = ~ ~(P) u p = is read by the discriminant at all inert primes of even rank, and at all class 1 tamely ramified primes. Hence (d,O)p must be trivial at all these primes in order for a form to be in the kernel of = ~ ~(P). 179 Thus, by Landherr we obtain sgn d H r t 0 ~ J/~Hu(I). (2Z) ~ (Z*) ~ Z* + 0 is exact. To complete the discussion, we must examine whether a rank 1 form can exist in Hu(I). ~(P) Hilbert symbols. Thus, in this case, there exists a s F/NE/F with 8(P) 1 form. We apply Theorem 3.1 if there are no ramified primes, finite or infinite. In order to have 8(P) ~(P) when Vp(X I) is odd at an even number of primes. This is possible if and only if Vp(I) is odd at an even number of primes, so that Vp(I) + vp(x I) is odd at an even number of primes. Similarly, if Vp(X I) is odd at an odd number of primes, 8(P) = (_l)vp(I) + 1 (a,O)pl in order to have ~ Vp(I) is odd at an even number of primes, so that Vp(I) + Vp(X I) ~ 1 (mod 2) at an even number of primes. Finally, we must discuss the extension rk 0 ~ J ~ Hu(I) ~ H(I) + F 2 + 0 when a rank 1 form exists. By Landherr, and Corollary II 4.14, this sequence splits if and only if -i is a norm in F/NE. [] In this Theorem we have computed H (I). We should note that when there are no rank 1 forms it is still possible to describe exactly what an additive set of generators for Hu(I) looks like, see [W]. Corollary 4.7 If there are no ramified primes, the boundary s__eequence depends on the number of primes at which Vp(X I) + Vp(I) i__ss odd. (a) If this number is even, we have: 0 + Hu(I) + Hu(E) ~ Hu (E/I) + F 2 + 0 (b) If this number is odd, Hu(I) = 0, and 0 + Hu(E) + Hu (E/I) § 0. [] Thus we have shown how to compute Hu(I) , and the boundary homomorphism 2: Hu(E) ~ Hu(E/I). To apply this to I = A-I(D/Z) we 181 clearly only need to know vp(A-I(D/Z)). We can read this for dyadic ramified primes in terms of type as we shall see in 4.12. Let P be a - invariant prime ideal in O(E). Then the - involution makes the local units in OE(P), denoted OE(P)*, into a C2-module. We examine the resulting cohomology group HI(c2;OE(P)*). Lemma 4.8 If P = P i__ss over inert, then HI(c2;OE(P) *) = 1 . If P = ~ is over ramified, HI(c2;OE(P)*) = C 2 Proof: Recall that HI(c2;OE(P) *) = {x e OE(P)*: xx = i} modulo {v/~: v g OE(P)*} . If x is a local unit of norm i, then by Hilbert Theorem 90, there exists z e E with zz = x . We now write z as z = ~VP(Z)v where ~ is a local uniformizer for P and v e OE(P)* If P is over inert, we may choose ~ 6 F* , so that ~ = Thus z~ -I = v~ -I = x and H 1 is trivial in this case. If P is over ramified, (~-i) is a local unit, and zz -I = (~-l)Vp(z) vv--I = x . Thus H 1 is generated by the class of z~--i which we denote cl(z~ -I) Of course (~-i)2 is trivial in H 1 since (~-i)2 = (z~-l) (~ -i)-i a quotient of local units. Thus, to complete the proof we need only show that cl(z~ I) is non-trivial in H 1 Suppose to the contrary that there is a local unit v with vv--i = ~ --i Then ~v -i = zv---i so that zv -i is a local uniformizer of OE(P) which lies in F . This is impossible as we are in the ramified case. Hence, cl(~ -I) is non-trivial as claimed. Definition 4.9 P is of type i provided cl(-l) ~ I e HI(c2;OE(P)*)- P is of type 2 provided cl(-l) = 1 e HI(c2;OE(P)*). We can thus classify type by: 182 Proposition 4.10 P is of type 1 if and only if there exists a local uniformizer ~ c OE(P) with ~ = -z. P is of type 2 if and only if there is a skew unit, u = -u s OE(P)* Proof: cl(-l) ~ 1 if and only if for any local uniformizer z, cl(-l) = cl(z~-l). Hence, for some v e OE(P)*, -v~ -I = z~ -I Replacing n by z~, we obtain a skew uniformizer, cl(-l) = 1 if and only if there is a local unit v with v~ -I = -I. Fl Proposition 4.11 If P is non-dyadic ramified, meaning the characteristic of the residue field O(F)/P ~ 2, then P is of type 1. Proof: Since P is ramified, e = 2 and f = i. Thus O(E)/P = O(F)/P, and the induced involution on O(E)/P is trivial. Suppose P is of type 2. Then by4.10, there is a skew unit v, with v = -v. However, viewing v in the residue field, v = Thus 1 = -i. This is a contradiction unless the characteristic is 2. [] Finally, we wish to examine type in terms of the local different. Let P ramify, say PO(E) = p2. By [A 83], we can find ~ s OE(P) where OE(P) denotes the localized completion of O(E) at P, such that i, ~ forms a basis of OE(P), as an OF(P ) - module. satisfies t 2 - (a + K) t + aK = 0. By [A 92], 2~ - (a + ~) = a - generates the different of the extension. dp Factor ~ - ~ = ~ v, where dp is called the local differential expotent, is a local uniformizer for P, v E OE(P)* We have _dp_ dp (~ - ~) = -(~ - ~) = ~ v = -7 v. 183 Thus,-vv -I = (~z-l) dP and cl(-l) = cl(z~-l) dP . Thus, by definition, Proposition 4.12 Let P be dyadic ramified. Then P is of type 1 if and only if the local differential exPonent dp is odd. Chapter VII NON-MAXIMAL ORDERS In this chapter we discuss the ingredients necessary to compute W(k,Z;S) for S a non-maximal order. First there is the problem of the map w: O(E)/P + E/~-I(D/Z) from Chapter VI 3. w is determined by a localizer p satisfying Vp(0) = Vp(~-l(E/Q)) ' i. However, we must choose a canonical localizer 0(P) in order that the diagram given after Theorem VI 2.5 will commute, and be applicable to our computation. A series of theorems of Conner develop the fundamental properties of these canonical localizers. In Section 2, normal extensions are discussed. The basic question is for a non-dyadic ramified prime P whether P(P) is a local norm. The best result known is the result of Conner, Theorem 2.9, which shows that in some cases it is equivalent to asking whether i/p is a local norm. In Section 3 we compute the trace map for finite fields. In Section 4 we discuss the conductor and the set T(M) = {maximal ideals P in D: P /~ S = M, P = ~}. As previously noted, these are the three crucial steps to under- standing non-maximal orders. i. Traces and canonical localizers In the previous chapter we have studied the boundary map (D) q(p) ~(D,P) : Hu(E) ~ Hu(E/A-I(D/Z)) Hu (D/P) 185 The manner in which this map is read depends on the choice of localizer 0(P), with Vp(0) = Vp(s -I) - i, which embeds D/P into E/s Our object is to study non-maximal orders. To do this requires choosing the isomorphism Hu(E/A-I(D/Z)) = ~ Hu(D/p) in such a way p = that the following diagram commutes: Hu(E/A-I(D/Z)) .... H u (D/P) +t +tr g W(k,Q/Z;D) W(k,Fp; D/P) P--T Thus we are faced with the problem of relating the number field trace t to the residue field trace tr. Again we recall the setting in which we work. We let E be an algebraic number field, with F C E a subfield. O(E) C E and O(F) ~ F are the respective Dedekind rings of algebraic integers. If P is a prime ideal in O(E), then P /~ O(F) = P is the corresponding prime ideal, in O(F) over which P lies. Consider the quotient homomorphism yp: O(E) + O(E)/P onto the finite residue field. Restricting yp to O(F) induces yp: O(F) + O(F)/P + O(E)/P. Our object is to compare the two trace homomorphisms: tE/F: E + F and trEp/F P : O(E)/P ~ O(F)/p where Ep is an abbreviation for O(E)/P the residue field. 186 We consider the diagram: O(E) -~ O(F) +Yp +Tp tr O(E)/P ~ O(F)/P The composition tr o yp is an O(F) - module epimorphism, as is yp. Since O(E) is projective as an O(F) - module, there is an O(F) - module homomorphism T : O(E) § O(F) which makes the diagram above commute. Now consider the relative inverse different ~I(E/F) = HomO(F) (O(E) , O(F)) [A 86]. Since T ~ HOmo(F) (O(E) , O(F)) , using this isomorphism we see that there is an element ~E/F (P) s 4 -1 (E/F) such that tE/F (~E/F(P) I) = ~ (~) . Thus, (*) Yp(tE/F(~E/F (P) l)) = tr(yp(1)) for all I s O(E). 1 2 Now if ~E/FCP) and ~E/F(F) are two elements satisfying (*), then clearly tE/F((~ 1 - u 2) I ) e P for all I e O(E). Thus, using the isomorphism P~-I(E/F) = HOmo(F) (O(E),P) it follows that 1 - 2 c PA -I (E/F) . 187 So although ~E/F(P) is not unique in A-I(E/F), it is unique in s (E/F)/PA -I (E/F) . The following Lemmas of Conner develop the basic properties of these ~E/F(P) . Lemma i.I For any ~E/F(P), ~E/F(P)PP-I C A-I(E/F). Proof: Let e E P and B E p-l. We abbreviate DE/F(P) by ~. Then for any X ~ O(E) tE/F(X ~ ~ $) = B tE/F (I ~ ~) since ~ s p-Ic F. But tE/F(X a ~) s P since yp(l ~) = 0 and yp (tE/F(~ ~ ~)) = tr(yp(~ ~)). Hence ~ tE/F(I ~ ~i) = tE/F(~ ~ ~ B) g O(F) for all X e O(E). Thus by definition of the inverse different, ]J ~ B s A-I(E/F). [] Lemma 1.2 Let (~E/F(P)) be the principal O(E) - ideal generated by u. Then ~p-i is not contained in A-I(E/F). Proof: Suppose the ~P-I C &-1(E/F). Then for all B s p-i and e O(E) tE/F(I ~ B) = B tE/F(X ~) s O(F) SO tE/F(X ~) e (p-l)-i = p. However, we also have 188 7p tE/F(I D) = tr Ep/Fp (%'p(1)) , so trEp/% : Ep + Fp is trivial. Contradiction. [] When F = Q we may use these p as follows. For each prime ideal P C O(E) we obtain a canonically defined element p(p) p-1 = DE/Q (P) E E/A -1(E/Q) where p E Z is the prime over which P lies. Now define a homomorphism O(E) + E/A-1 (E/Q) by l + (lp(P)) . By Lemma i.i this map induces a map w of the residue fields w: O(E)/P + E/A-I(E/Q) since ~E/Q(P) P p-Ic A-I(E/Q). By Lemma 1.2, the induced map is an embedding since ~E/Q(P)p -I ~ A-I(E/Q) . We also have the map Z § Q/Z defined by n + (np-l). This also induces an embedding w: Fp = Z/pZ + Q/Z. Combining w and w we have Theorem 1.3 At every prime P ~ O(E) there is a commutative diagram Q O + Ep = O(E)/P + E/A-I(E/Q) +tr ~t w Q/Z P 189 and w have been described. tr = finite field trace t = map induced from number field trace tE~Q./ Proof: This follows immediately by the definition of ~E/Q(P), since w(1) = Ip(P) = Ip-I~E/Q (p) , so 1 t(Q(1)) - P t (I~E,Q(P))/ = ! tr(1) = w(tr(1)) . I I P Thus we are able to explicitly describe the map w given in Chapter VI 3. As we saw there, the canonical localizer p(P) which defines is the correct choice to make if one wishes to compute W(k,Z;S), where S is a non-maximal order. The reader is referred to the discussion preceding Proposition VI 2.7. The choice we make for localizer p(P) will determine the isomorphism H(E/A-I(E/Q) = (~ H (E/A-1 (E/Q) (P)) p=~ = ~) H(Ep) ~=~ since the embedding Ep + E/A-I(E/Q) (P) defines the isomorphism H(Ep) + H(E/A-I(E/Q) (P)). 190 There are two cases. (i) If P is inert, the isomorphism is independent of localizer, and rank mod 2 is the only invariant. (2) If F is ramified, W(Ep) = H(Ep) = H(E/g-I(E/Q) (P)) since the induced involution is trivial. For P non-dyadic, the identification does depend on the choice of localizer. The localizer we choose therefore will determine the way we read the boundary homomorphism ~{P) : H(E) § H(Ep). We wish to compare the canonical localizer with the localizer given in Theorem VI 3.1. Now different localizers P'QI determine different embeddings Ep + E/&-I(E/Q) (P). The relation between different embeddings is that one is the composition of the other with the map of Ep to itself given by multiplication by p@l -I Since we are only -i interested in this map on H(Ep) , only the norm class of PPl matters. For the localizer in Theorem VI 3.1, this norm class is specified. Thus, in order to understand the canonical localizer above, the question we need to answer is whether @(P) is a local norm. Once we have this information, we can read the boundary homomorphism p(F) with Hilbert symbols, where the identifications come from our canonical localizers. -i Since p(P) = p UE/Q(P), we must study those elements ~E/Q" The following theorems of Conner are fundamental. Theorem 1.4 For any ~E/F(p) we have: (i) vF(~E/F(P)) = vF(P) - 1 + vp(~-l(E/F)) (2) If p ~ P then v~(~E/F(F)) _> v~ P + v~(~-l(E/F)).p 191 Proof: By i.i, ~pp-i ~ ~-I(E/F) . Thus -i v~(~)p + v~(P) - v~(P)p -> v~~p (E/F) . If p ~ p then v~(p) = 0, and (2) follows. -i Again using i.i, we have Vp(~) + 1 - Vp(P) ! Vp A (E/F). However, if vp(~) -Vp(P) Z Vp ~-I(E/F), then ~P-IC ~-I(E/F) which contradicts 1.2. This shows equality and proves (I). We next recall some facts about the different A(E/F), and its inverse A-I(E/F). Let e > 0 be defined by v,(P) = e. P Definition 1.5 If v. (P) = e > 0 is divisibly by the p -r- characteristic of the residue field ch(Fp) then we say P is wildly ramified over P. We recall that if p is wildly ramified v~(~(E/F)) > e. p [S 96]. Definition 11.6 If-- v_(P)p = e > 0 is not divisible by ch(Fp) then we say P is tamely ramified over P. In this case v~(A(E/F)) = e - i. Thus v~(E/F) = e - 1 if and only if P is tamely ramified. Definition 1.7 A prime P C O(F) is tame in E/F if and only if no prime P C O(E) which divides PO(E) i{s wildly ramified. Thus P is tame if and only if for every prime P with v~P = e > 0, e is relatively prime to ch Fp. 192 Theorem 1.8 Let UE/F(P) satisfy YptE/F(~E/F(P) ~) = tr (7 p (~)) Ep/Fp (i) If vp(P) = e > 0 i_{s relatively prime to ch(Fp), then ~E/F(P) s OE(P)* i_ss a loca_____!lunit. (2) If vp(P) = e > 0 is divisible by ch(Fp), then ~E/F(P) is not a local integer a_~t P. (3) If ~ ~ P but v~(P) = e > 0 is relatively prime to ch(Fp) then ~E/F(P) belonqs to the maximal ideal m(~) ~_ OE(P) - (4) I_~f ~ is prime to both PO(E) and A(E/F) then UE/F(P) is a local inteqer i__nn OE(P). Proof: We use the previous remarks about the inverse different together with Theorem 1.4. (i) By 1.4, vp(~) = (e - i) + -(e - i) = 0, so ~ is a local unit. (2) In this case vp(~-l(E/F)) ~ -e. Thus by 1.4, vp(~) ! (e - i) - e = -i, so ~ is not a local integer. (3) By 1.4, v~(~) ~ e - (e - I) = i, so ~ e m(~). (4) By 1.4, v~(~) > 0 so ~ E O (~) is a local integer. [] -- E 2. Normal extensions We continue by letting E/F be a normal extension, and we let G be the Galois group. Again, this section is due to Conner. 193 Theorem 12.1 If g s G and PCO(E) is a Frime ideal then g~E/F(P) = ~E/F(gP). Proof: Let g, be the induced Fp - linear isomorphism induced by g making the diagram below commute. g O(E) + O(E) +7 F %Ygp g* Ep = O(E)/P + O(E)/gp = E gP For I s O(E), tE/F(Ig(u)) = tE/F(g-l(1)~) . Recall the defining equation for ~E/F (gp) trEgp/Fp (Ygp(1)) = 7p tE/F(I~E/F(gP)). We also need the fact that trEp/F P (~) = trEp/Fp(r for all ~ in the Galois group Ep/Fp. We now compute: 7ptE/F (I g(~E/F(P))) = 7ptE/F(g-I(1)~E/F(P)) = trEp/Fp(yp(g-l(1)) ) = trEp/Fp(g,ypg-l(1)) = trEp/F P (yg pC1) ) Thus g(~E/F(P)) = ~E/F(gp) . [] 194 Now let Gp denote the decomposition subgroup at P, ie. all g s G with gP = P. Theorem 2.2 If P C O(E) lies over P, and P is tame in E/F, then for g ~ Gp, gUE/F(P) e m(P). Proof: Since g / Gp, gP ~ P. Thus by 2.1, g~E/F(P) = ~E/F(gP) . We now apply 1.8 to conclude that gu e m(P). [] Theorem 2.3 If PC O(E) lies over P, and P is tame in E/F, then ~E/F(P) E OE(P)* and 7p(~E/F(P)) e Fp. Proof: By 1.8 (I), ~ s OE(P)* If g e G?, then by 2.1, ~E/F(P) - g~E/F(P) e P~-I(E/F). However, P is tame so Vp(pA-I(E/F)) = i, since vp(P) = e and Vp(A-I(E/F)) = 1 - e. Hence 7p(~E/F (P)) = 7p(g~E/F(P) ) = g,yp(~E/F(P)) for all g c Gp. Since Gp maps onto the Galois group of Ep/F P by g + g, it follows that 7p(UE/F(P)) is fixed by all g, in the Galois group of Ep/Fp. Thus 7p(~E/F(P)) e Fp. [] Theorem 12.4 If P lies over a prime P O(F) which is tame i__nn E/F then 7p(~E/F(P))e = 1 where e = Vp(P). Proof: By definition of trace, 195 tE/F(l ~) = Z g(l U), where ~ e O(E). g ~ G Now g(l'~) = g(1) g(~) s OE(P) for all g e G by 2.1 and 1.8. However, for all g { Gp, g(l)g(u) E re(P) by 2.2. Thus Yp tE/F (~) = Z yp(g(l) g(~)) g eGp = Z g, ( Yp(P) ) g, ( Yp(~) ) g eGp But by 2.3, Yp(~) e Fp. Thus, the above equals = 7p (P) Z g,(yp(1)). g e Gp Now if we consider G F + Galois (Ep/Fp) + 1 this has kernel the inertial subgroup whose order is e > 0. Thus by the above YptE/F (X~) = eYp(P) tr E p/F p(Yp(~) ) However by definition u = trEp/F~yp(X)). Thus eyp(~) = 1 as was to be shown. [] Corollary 2.5 Suppose P CO(E) lies over a prime tame in E/F and suppose Gp = G. Let d satisfy de 5 1 (mod ch(Fp)). Then we may choose ~E/F(P) = d. 196 Proof: We compute 7p(tE/F(Id)) = dyp (tE/F (l)) = dyp( ~ g(1) ) g s G = d( Z g.yp(1) ) g s G = de trEp/Fp (yp (i)) = trEp/F p (yp(1)) [] We need one more theorem, which shows that the p behave multiplicatively in towers. This makes no assumption of normality. Theorem 2.6 (Tower Theorem) Let N ~ E -)F, and C o (N) prime ideal with P~O(E) = P, P/~O(F) = P. Then ~N/F(P) = ~N/E(P) ~E/F(P). Proof: We simplify our notation by letting ~i = ~N/E (~) and ~2 = ~E/F (P)" For I e O(N) we now compute: YptN/F(Y~I~2 ) = 7ptE/F(~2 tN/E(I~I)) = trEp/F p (YptN/E (i~ I) ) = trEp/Fp (trN~/Ep (Y~ (I)) = trN~/Fp (y~ (I)) 197 Thus, ~i~2 = ~N/F(P) as desired. [] We recall now that our object has been to determine whether @(P) = p-l~E/Q is a local norm. In particular, this is important precisely when P = ~ lies over a non-dyadic ramified prime. Here we are letting E have involution -, and the fixed field of - is F. Corollary 2.7 I_~f P = ~ lies over a non-dyadic ramified prime p then ~E/Q (F) = d~ F/Q (p) where O < d < ch(Fp) = p and 2d - 1 (mod p). Proof: By the Tower Theorem 2.6, ~E/Q(P) = ~E/F(P)~F/Q(P). By Corollary 2.5, ~E/F(P) = d where 2d ~ 1 (mod p). [] Theorem 2.8 If the non-dyadic ramified prime P CO(F) lies over a rational prime p which is tame in F/Q then d~F/Q(P) i__{s a local norm, ie. the Hilbert symbol (dDF/Q(P),o) P = 1 if and only i_~f 2 Vp(p) is a square in the residue field Fp. Proof: By 2.4 applied to F/Q, yp(~F/Q(P)) e = 1 where e = Vp(p). Also by 2.7, 2d ~ 1 (mod p). Thus 2Vp(p) is a square in the residue field if and only if its reciprocal dyp(~F/Q(P)) is a square. However, by Chapter II 2.4, a local unit is a norm if and only if it is a square in the residue field. FI 198 Theorem 2.9 If the non-dyadic ramified prime PC O(F) lies over a rational prime p which is tame in F/Q then (P(P),O)p = (i/p,O)p provided deg Fp/Fp is even. Proof: (p(P),O)p = (i/p,O)p(d~F/Q(P),o) P. By Theorem 2.8, (dPF/Q(P),o) P = 1 if and only if 2e is a square in the residue field Fp, where e = Vpp. Now consider 2e g Fp. If this is already a square then (d~F/Q(P),o) P = 1 and we are done. If 2e is not a square, then because deg Fp/Fp is even, 2e will be a square in Fp, and again (d~F/Q(P),o) p = i. Admittedly one still needs criZeria to determine whether i/p is a local norm. However, this result of Conner represents a great step foward. What we have attempted to do here is outline the general theory concerning canonical localizers, which are crucial to our under- standing of non-maximal orders. 3. Computing trace for finite fields We next compute the trace induced map tr,: H(E) ~ H(K) where E and K are finite fields. As we have observed, this computation is important in computing W(k,Z;S) for non-maximal orders, if we wish to use the commutative diagram after Theorem VI 2.5. Theorem 3.1 Suppose [E : K] < ~ is an extension of finite fields. E has an involution - which is possibly trivial. Then the homomorphism tr, is 9iven as follows: 199 (1) If - is non-trivial on both E and K, tr,: H(E) ~ H(K) is an isomorphism. (2) If - is trivial on both E and K, tr,: W(E) + W(K) is an isomorphism if [E : K] is odd. If [E : K] is even, tr, is an isomorphism o__nn the fundamental ideal, and has kernel C 2. (3) If - is non-trivial on E, trivial on K, then tr,: H(E) + W(K) i_ss ~i - 1 if p ~ 2 [ 0 if p = 2. where p is the characteristic of the finite field E. Proof: We should observe that since E is finite, K is automatically - invariant, and thus has an involution induced on it. Thus the statements make sense. (1) If [E : K] is even, then K is contained in the fixed field of - by Galois theory, since the fields are finite. Hence in case (I) we may assume [E : K] is odd. We recall that the Hermitian group of a finite field is determined by rank mod 2 [M,H 117]. So let [M,B] e H(E). Then tr,[M,B] has rank equal to [rank M] [E : K]. Hence rank modulo 2 is preserved when [E : K] is odd, so that tr, is an isomorphism in this case as claimed. (2) Let [E : K] be odd, with - trivial on E. By [Lm 193], the composition ~E tr, W(K) + W(E) + W(K) 200 is multiplication by tr, W(K). Thus the composition is an isomorphism. Since all groups have order 4, tr, must be an isomorphism too. Now consider the case [E : K] is even, and - is trivial on E. By Galois theory, E/K factors into a maximal odd order extension and successive extensions of degree 2. [KI: K] = odd [Ki,Ki_ I] = 2 for i > 1 tr,: W(E) ~ W(K) is the composition of the trace maps: tr, tr. W(E) W(K i) ~ W(Ki_ I) ~ ... + W(K 2) ~ W(K I) ~ W(K) To finish the proof of (2) , we need to examine separately each tr,: W(K i) ~ W(Ki_ I) and show that each tr. is an isomorphism on the fundamental ideal, with kernel C 2 when [Ki: Ki_ I] = 2. tr,: W(K I) + W(K) is an isomorphism by the first part of the theorem for [KI: K] = odd. Note: We assume now that the characteristic of E is not 2, for in that case, rank mod 2 determines everything, and (2) is clearly true as stated. We thus consider tr,: W(F(/w)) + W(F). Since the characteristic of E is not 2, any quadratic extension of F is given by adjoining /w, where w K F**. In order to describe a basis for GF(/w ) s F(/w)*/F(/w)** as an F 2 - vector space, we recall the general theory from [G,F]. Let G F = F*/F** have a basis w,b I ..... b n as an F 2 - vector space. 201 Then GF(/w ) has a basis given by {b I ..... b n} together with {xi + Yi/W} as xi,Y i run through distinct square classes 2 2 represented by x i yi w in F*/F** Thus for the case of GF(/w ) where F is a finite field we have two cases: (i) (-i) is a square in F. Then, letting x i = 0, Yi = 1 2 2 x i - yi w = -w = w in F*/F**, so that we may choose /w = g the non-square class in F(~/w). (2) (-i) is not a square in F. In order to find g the non-square class in F(/w) = F(/-I), we must solve the equatlon 2 2 x + y = -I. Then g = x + y/-i is the new square class. We could as well choose g = 1 + x/-l, since 12 - x2(-l) = 1 + x 2 = _y2 = non-square class in F*. Now the Witt group of any finite field is generated by the 1 - dimensional forms , (i) (-i) is a square in F. g = /w. W(F(/w)) is generated by the 1 - dimensional forms and . We compute. 1 /w Here the matrix of tr, 1 /w In this case the discriminant 4w 2 is trivial. I By additivity, tr, ~ ~ 0, and case (2) follows. (2) (-I) is not a square, g = 1 + x/-l. 1 /-i The discriminant 4 is trivial. tr, = 1 ( 2 0> /-i 0 -2 1 /-i The discriminant = non-square /-i -2x - = non-trivial. Again, by additivity tr, (3) Finally, let - be non-trivial on E, and trivial on K. Let F be the fixed field of -. Then E ~ F ~ K. tr,: H(E) + W(K) is the composition tr, tr, H(E) ~ W(F) + W(K) By Jacobson's Theorem [M,H 115], we have the exact sequence 0 + H(E) t~, W(F), and tr, is injective into the fundamental ideal in W(F). By part (2) of this theorem, W(F) t~, W(K), tr, is an 203 isomorphism on the fundamental ideal in W(F). Thus, tr, is 1 - i, tr,: H(E) + W(K). For p = 2, the fundamental ideal is trivial, so that tr, = 0 as claimed. ~] 4. The conductor and T(M) In order to finish describing f3' the last task is to describe which - invariant maximal ideals M in S have - invariant maximal ideals in D lying over them. This is related to the conductor. Definition 4.1 The conductor of D over S is the largest set C which is an ideal in both D and S. We shall need the following theorem. n Theorem 4.2 If A factors as A = Pi where A is prime i = 1 to the conductor C, then n A /~ S = .~ (pi/~ S)- i = 1 Proof: See [G 38]. We shall also need a few results from ideal theory. Lemma 4.3 Let D be a domain, A,B,C ideals in D. Let C be generated by k elements, C = AkC B, and similarly BkcA. 204 Proof: It clearly suffices to show an arbitrary product (a I ... a k) of k elements in A is contained in B. A k is generated by finite sums of such products. Since AC = BC we may write alc i = JZ bjlC j ..... akc k = JE bjk c~3 This system of equations can then be written as 0 = -aic i + bliC 1 + b2ic2 + ... + bkiC k, i = i, ... , k. Solving for c i, using Cramer's rule we obtain ~c i = 0 where A = determinant of coefficient matrix, c. ~ 0 yields ~ = 0. 1 However, the determinant A can be written as • I ... a k) + b where b E B. Thus (a I ... a k) E B as desired. [] Lemma 4.4 If a prime ideal P in D factors as p = A B, then A = D or B = D (D a Dedekind Domain) (A Dedekind Domain has unique factorization). Proof: P = AB clearly implies P ~A or P ~ B. Say P ~A. Then we may write A = PW, W an ideal in D. Hence P = P(WB) = PQ. So P . D = P Q. Since D is Dedekind, P is generated by 2 elements, [O'M], and we may apply Lemma 4.3. D 2 = D C Q. Hence Q = D = WB, so that W = B = D, and p = A. [] Note: This Lemma is also clearly true for an order S. We also recall the following: 205 Theorem 4.5 Let S C_ D be rings, with D integral over S. Let M be a prime ideal in S. Then there exists a prime ideal P i__nn D with P~S = M. Proof: See [A,M 62]. [] With these preliminaries, we are in a position to give a sufficient condition for a - invariant maximal ideal in S to have a unique - invariant maximal ideal in D lying over it. Theorem 4.6 Let M be a - invariant maximal in S. If M is prime to the conductor C, then D has a unique - invariant maximal ideal P with P /~ S = M. Proof: Let M be maximal in S. By Theorem 4.6, we can find P maximal in D with P/~ S = M. We claim that P is unique, and hence - invariant, for clearly ~S = M = M. So suppose P. ~ P 1 with Pi/~S = M. Pi may be ~, or some other maximal ideal in D. w Each such Pi will clearly appear in the factorization of DM = 17 P,. i = 1 z We claim that each P. ~ S = M. Clearly, PiA S ~ M, since l P ~ DM. However, Pi/~ S ~ M implies that P.~ S = S, so that 1 -- 1 1 s P., contradiction. Thus P.~ S = M. l i NOW note that DM ~ S = M. This follows since DM ~ S ~ M; DM C_ Pi and Pi/~ S = M so DM~ S C~ Pi ~ S = M. We now apply Theorem 4.2. Since DM is prime to C, W W DM ~ S = M = H (Pi /~ S) = ~ M. i = 1 i = I However, for M a prime ideal, by Lemma 4.4, w = i, and consequently P is unique. [] 206 For an order S C D, the conductor C ~ 0. For let m # 0 E Z satisfy mD C s. Then (m) generates an ideal in S which is also an ideal in D. Thus, we see that the cardinality of the set T(M) = {P: P ~ S = M, P = ~} is one except possibly for those ideals M which are not prime to C. Since C ~ 0, there are only a finite number of maximal ideals M in S which are not prime to C. Theorems 3.1 and 4.6 then enable us to compute the map f3" The set T(M) must be computed explicitly only at the finite set of ideals not prime to C. Given P in D, we should like to relate D/P to $/P S. Definition 4.7 Let P be a maximal ideal in D, so that P S = M is a maximal ideal in S. [A,Mc 61] We will say S is integrally closed at M if S(M) , the localization of S at M, is integrally closed. D is integrally closed, hence so also is D(P). [A,Mc 62] Thus, S is integrally closed at M if and only if D(P) = S(M). Proposition 4.8 If D(P) = S(M), then D/P = S/M. Proof: D/P = D(P)/m(P) = S(M)/m(M) = S/M. [] Note that D and S have the same quotient field, E, and D(P) = S(M) except at finitely many primes. It follows by 4.8 then that D/P = S/M with only finitely many exceptions also. Chapter VIII THE GLOBAL BOUNDARY Section 1 describes the coupling effect due to Stoltzfus [Sf] between various ~(D). In Section 2 we pnove 2: W(k,Q) + W(k,Q/Z) is onto when k = • This is crucial to studying the octagon over Z. i. The coupling invariants We recall our notation: W(k,K;f) denotes Witt equivalence classes of triples [M,B,s where the characteristic polynomial of s is a power of the T k fixed irreducible polynomial f(x). By taking anisotropic representatives, Proposition IV i.ii identified this as: W(k,K;F) = w(k,K;S) where S = Z[t]/(f(t)). We used the notation ~(S) to denote the restriction of the boundary map to W(k,K;S). In this section we wish to emphasize the polynomial f rather than the module structure. Thus we use the notation of ~(f) to denote the restriction of the boundary map to w(k,K;f). Of course ~(f) is really ~(S). We have the commutative diagram i qf W(k,Z) + W (k,Q) § @) W(k,Q; f) W(k,Q;f) + ~ (f) s(f) W(k,Q/Z;f) We label the composition ~ (f) o qf o i --- s 208 It follows that there is the commutative diagram: ~) e(f) W(k,z) ~ (~ W(k,Q/Z; f) f +i +~i W(k,Q) § W(k,Q/Z) The map ~i just adds up all the terms in W(k,Q/Z). We wish to measure how the various groups W(k,Q/Z;f) couple together. Theorem 1.1 There is an exact sequence i ~ s(f) 0 + W(k,Z; f) + W(k,Z) § ~) W(k,Q/Z ; f) f s feb W(k,Q/Z) ~9 coker ~ (f) . feb Comment: ~ is onto provided 2: W(k,Q) + W(k,Q/Z) is onto. This is the topic of Section 7. Proof: The map ~ is el on the first factor, and the appropriate projection into the cokernel on the other factor. i just adds up terms in W(k,Z), as does ~i" We begin by considering the commutative diagram: 209 0 0 i W(k,Z;f) ~ W(k,Z) f + Jl + i 1 0 + ~ W(k,Q;f) ~ W(k,Q) f i is then clearly 1 - 1 since all the other maps are. To check exactness at W(k,Z), suppose i< ~ Lf,Bf,tf> = Conversely, suppose (~ e (f)) E(f) (M(f)) . Exactly as in the proof of exactness for ~, this yields a self-dual lattice and hence an element in W(k,Z;f) ; which under i is mapped to M(f) . Thus, ker (~ s ~_ im i. Next we show ~ o(~ s = 0. For the first factor, this follows by the commutative diagram el o (~ e(f)) = ~ o i = 0. The other components are also 0 since E(f) = ~(f) o qf o i so that elements in the image of (~ e (f)) are in the image of ~(f), thus 0 in coker ~(f). Conversely, suppose we are given a collection ~[M(f),B(f),t(f)] f of torsion forms in ~W(k,Q/Z;f), which are in the kernel of ~. f Write these as ~[Mf,B,t]. Then each [Mf,B,t] is the trivial element in cokernel of ~(f), hence [Mf,B,t] is in the image of ~(f). So let ~(f) [Mf',B',t'] = [Mf,B,t]. Then ~)[Mf,B,t] = ~9 ~(f) [Mf',B',t'] . f f 210 Applying ~i' ie. adding up in W(k,Q/Z), we obtain: al( ~f [Mf,B,t]) = ~(~) [Mf',B',t']) f However, we are assuming ~i ( ~ [Mf,B,t]) = 0. Thus ~( ~[Mf f',B',t']) = 0. By the exactness of the boundary sequence, Theorem VI 1.5, this implies there exists [M,B,t] s W(k,Z) with i[M,B,t] = (~[Mf ! ,B',t ! ]. f Hence ~ (f) qfi ( [M,B, t] ) = G) ~(f) [M f',B',t'], = •[Mf,B,t] f = tt~ a (f) [M,B,t] f SO that ker~ ~_ im(~ c(f)) r'1 f 2. The boundary is onto In this Section, we derive the results needed to study the octagon over Z. In particular, we will show that 2: W(k,Q) + W(k,Q/Z) is onto when k = • or k = positive prime, k E 2,3(4), or k { 1(8). We also show that ~:A(Q) + A(Q/Z) is onto, and compute the cokernel of ~: W-I(k,Q) + W-I(k,Q/Z) when k = • To begin with, we recall our computations: /- W(k,Q) = ~ ~f s 8 H(Q(0)) ~) W(Q(/k)) k ~ 1 ~ of type 1 ~) H(Q(@)) ~) W(Q) + W(Q) k = 1 211 Also, we had: W(k,Q/Z) -- ~9 W(k,Fp) (~ W(Fp) p~k p {k -- [ ~9 ~9 [H(Fp(O)) ~9 W(F (/k)] p~ f e ~ k ~ F** P P of type 1 W(F ) ~9 W(Fp) ] ~9 W(F 2) ~) W(Fp) . keF ** P Pl k P pM2 p ~ 2 In this decomposition, we sum over all T k fixed irreducible polynomials f(x) of type i, where f has coefficients in F . We P observe that any such f can be lifted (not uniquely) to a T k fixed integral polynomial. To see this write 2n f(x) = Z a.x i where a. s F . l l p i=0 Lift the first n coefficients a i to a i s Z with ai ~ ai (mod p). By Lemma III 1.4, a polynomial g(x) is T k fixed if and only if its coefficients satisfy a.kl i = a0a2n_i, We define g(x) by 2n i g(x) = Z c.x where c. = a. i = 0 ..... n l l l i = 0 c i = a0a2n i k-1 i = n + l, ... , 2n By Lemma III 1.4, g(x) is T k fixed, and clearly the mod p reduction of g(x) is f(x) , since f(x) is T k fixed also. 212 In fact, we should observe that if Q(@) has a non-trivial involution ~ = k0 -I, then the irreducible polynomial of @ over Q is T k fixed. For if 2n i p(x) = Z a x is the monic irreducible polynomial of @, i=O I 2n then p(@) = p(~) = 0. Thus p(k@ -I) = ai(k| = 0. i = 0 2n Multiplying by 0 2n, Z a. (k) i @ 2n-i = 0. Thus i = 0 l 2n a ki@ 2n-i @2n = _ Z l i = 1 a 0 2n-i However, @2n = _ E a.@ i since p(@) = 0. i = 0 z Using the fact that i,@ .... 0 2n-I is a basis for Q(G) over Q since p(x) is irreducible, we may equate coefficients of the two sums for @2n, and obtain k i ai = a2n - i a 0 so that p(x) is T k fixed by Lemma III 1.4. This remark is important when we show the Hermitian elements in W(k,Q/Z) are in the image of boundary. Given H(Fp(@I)), we construct an extension Q(0) of Q, with a non-trivial involution ~ = k0 -I, with the property that the mod p reduction of O is 0 I. By the above it follows that the irreducible polynomial of 0 is T k fixed. Hence H(Q(0)) occurs in the decomposition of W(k,Q). Applying to this H(Q(@)) we show that H(Fp(GI)) is the image of ~, possibly 213 together with some Witt contribution. However, all Witt elements are first shown to be in the image of 2, so that ~ is onto. We begin our study of 3: W(k,Q) ~ W(k,Q/Z) by studying ~IW(Q(/k)). Letting S = Z[t]/(t 2 - k), we previously used the notation ~(S) for ~ IW(Q(/k) ) . Let D denote the ring of integers in Q(/k). By [S 35] D = S for k = • 1 or p, where p is a prime p ~ 2 or p ~ 3 (mod 4). 1 For p H 1 (mod 4). D = {all elements of the form ~ (u + v/k), where u,v s Z with u ~ v (mod 2)}. We now recall the computation for 3(D) given in [M,H 94]. There is an exact sequence: W(Q(/k)) ~ ~ W(D/P) ~ C/C 2 0. P max. in D where C = ideal class group, and r is defined on each generator of W(D/P) by: <~> + ideal class of P modulo C 2. We must be careful. This boundary sequence is for I = D. We are interested in the case I = ~-I(D/Z) of the inverse different. Fortunately, in our case, 4 -1 and D are principal orders, and we may write 4 -1 = D/e, for some ~ E D. We are of course in the special case of a quadratic extension of Q. -I We denote ~' (D) = boundary for I = A ; ~(D) = boundary for I = D in the next Lemma. Lemma 2.1 Scaling by e induces a commutative diagram. 214 (D) W(Q (/k)) ~ W (E/D) 1 1 _-+ _ -_+ -- d ~'(D) W(Q(/k)) + W(E/I) Proof: eA -I = D, so that commutativity follows by definition of ~.D Thus, once we have computed ~(D), we will also have a computation for ~'(D). To begin with we will show C/C 2 is trivial in the stated case. This requires some number theory; we refer to [B,S]. We shall show that for k = • p ; p prime, p ~ 3 (4), ~'(D) = ~(D) = ~(S) is onto. We also show $(S) is onto when k H 1 (8), and compute the cokernel, a C 2 arising from W(F2) , when k E 5 (8). Caution: This Witt piece, W(F 2) , arising in W(k,Q/Z) , k E 5 (8), is thus not in the image of the Witt piece W(Q(/k)) in W(k,Q) . However, we have not shown that this W(F 2) is not the image of a Hermitian piece in W(k,Q). This question is still open. We now aim to prove: Theorem 2.2 C/C 2 is trivial for C the ideal class 9roup i__nn Q(/k), provided k = • or k a positive prime. To set our notation, there are three classes: (I) k K F** (p) remains prime in D. e = 1 f = 2 P D/P = Fp(/k) where P (~ Z = (p) . 215 (2) k s F** (p) splits in D. e = I f = I P D/P i = D/P 2 where pD = PIF2 . (3) p divides k, written plk (p) ramifies e = 2 f = 1 D/P = F where p ~ z = (p). P We follow Borevich-Shafarevich [B,S] in defining: Definition 2.3 Two ideals A and B of D are strictly equivalent if there exists ~ ~ 0 i__nn Q(/k) satisfying NQ(~/k)/Q(e) > 0 and A = B(e). For k < 0, NQ(/k)/Q(a) > 0 always, so that this is the usual definition of equivalence in the ideal class group C. However, if k < 0 and NQ(/k)/Q(E) = +i for all units e, then each ideal class in C will split into two classes equivalent in the strict sense. [B,S 239]. [B,S] calls A,B divisors. For the case of the maximal order D, [B,S 215], divisors correspond in a 1 - 1 fashion with ideals in D. Notation: If P is an ideal, let [F] denote its equivalence class, denotes its strict equivalence class. When all units s have positive norm and k > 0 we can write [P] = ~.~ . Lemma 2.4 [P] is a square in C if and only if there exists an ideal Q ~ [P] with Proof: Sufficiency is clear, since if 2 Conversely, let [P] e C . Then there exists an ideal Q with p = ~ Q2. If NQ(/k)/Q(e) > 0, ~ Thus <~p> ~ Thus to check if [P] is a square, we need only check if either of its strict equivalence classes is a square. By [B,S 246], Theorem 7, ~ Note: We shall use our usual notation in this section for the Hilbert symbol rather than following [B,S]. N'(A) is a positive integer, see [B,S 124]. Remark: (N'(A),D) = +i automotically for p ~ D, and p = ~, P [B,S 242]. We are now ready to examine C/C 2 for k = • k a positive prime. Case I. For k = • 2, D is a principal ideal domain, so that C = 0 = C/C 2. Case II. Let p > 0 be a prime p E 1 (mod 4). Claim: For Q(/p), C/C 2 is trivial. Proof: In this case, equivalence coincides with strict equivalence. If P is a prime ideal, we shall show ~ Let P /~ Z = (q). 217 Case I. (q) is inert. Then N'(p) = q2. (q2,p) = +i. P Case 2. (q) splits, q odd, so N'(P) = q and (~) = +i. Note: (~) is the Legendre symbol. (N' (P),p)p = (q,p)p = (q) (so by Quadratic Reciprocity) = (~) (-i) (p - 1)/2 (q - 1)/2 = + 1 Case 3. (q) ramifies. Again N' p) = q = p. (p,p)p = (p,-p)p(p,-l)p = (~) = (-i) (p - 1)/2 = +i Case 4. (q) = (2) (a) p E 5 (mod 8). (2) is inert. N'(P) = 2 2, and we are done as in Case I. (b) p ~ 1 (mod 8). (2) splits. (2,p)p = (~) = (-i) (p2 - 1)/8 = +I Thus, by Theorem 7 from [B,S], all prime ideals P in C are squares and C = C 2. Case III. Let~ p > 0 be prime, p ~ 3 (mod 4). Claim: C/C 2 is trivial for Q(/p). In this case, each ideal class [P] in C splits into two strict equivalence classes. We may represent these as and , since N(/p) = - p < 0. 218 Let P be a prime ideal in Q(/p). P~ Z = (q). In this case case the discriminant D = 4p. Again, we have 4 cases. Case i. (q) is inert. N' (P) = q2, and (q2,4p)p = +i as before. Case 2. q is odd. (q) splits. So (~) = +i. N(p) = q. We compute (N'(P),4p)pi for Pi = 2 or p. (q,4p)p = (q) = (qP--)(-i) (p - 1)/2 (q - 1)/2 : (-i) (q - 1)/2 (q'4P) 2 = (q'P) 2 = (-i) (p - 1)/2 - (q - 1)/2 = (-i) (q - 1)/2 (see [O'M 206]. If (-i) (q - 1)/2 = -i, P is a strict square. If (-i) (q - 1)/2 = -i, consider /pP ~ [P]. In this case, namely (-i) (q - 1)/2 = -i, we will show is a strict square. Hence, IF] is a square in C. To begin with, N' (/pP) = p q. We compute, (N' (/pP) , 4p) = (pq,p) = (p,p) (q,p) P P P P = (p,-p)p(p,-l)p(-l) (q - 1)/2 = (-I) (p - i)/2(_i) (q - i)/2 = (-i) (-i) = +i (N (/pP), 4p) 2 = (Pq'P) 2 = (P'P) 2 (q'P) 2 = (-I) (p - 1)/2 (p - 1)/2(_i)(q - 1)/2 (p - 1)/2 = (-i) (-i) = +i Thus, again by Theorem 7, we conclude is a strict square. 219 Case 3. P~ Z = (q) = (p), so that (q) ramifies. [p] contains /pP. N'(/pP) = pp = p2. Hence [P] contains a strict square class, namely . Case 4. P~ Z = (2). N' (P) = 2. If p ~ 3 (mod 8), consider /pP ~ [P]. N'(/pP) = 2p. (2p, 4p)p = (2,p) p(p,p)p = (-i) (p2 - i)/8 ~/-i 1 = (-i) (-i) = +i (2p, 4p) = (2p,p) 2 = (2,P) 2(p,p)2 = (-i) (p2 _ i)/8(_i) (p - i)/2 (p - i)/2 = (-I) (-i) = +l Thus [P] s C 2 . If p H 7 (mod 8), N' P) = 2. (2,4p) p = (2,p)p = (2) = (-i) (p2 _ 1)/8 = +i 2 (2,4p) 2 = (2,p) 2 = (_1) p - 1/8 = +1 Again [P] E C 2. We have thus completed the proof of Theorem 2.2 ~'~ Remark: For k a negative prime congruent to 1 modulo 4, Q(/k) also has C/C 2 trivial. The argument is just like the above. 220 It is also possible for one to anaylize Q(/k), for k = p ~ 3 (4) , in which case the above argument fails. Corollary 2.5 For k = +I, p with p --- 2,3 (4), 8'(D) = ~(D) = ~(S) is onto. Proof: Immediate from the boundary sequence and Theorem 2.1, since S = D in this case. Recall S = Z[t]/(t 2 - k). [] For p ~ 1 (4), we apply Proposition VII 4.8. It follows that D/P = S/P{'~ S for P~ Z @ (2). At (2) however, when P/~Z = (2), D/P has 4 elements when p ~ 5 (8), for then (2) is inert and f = 2. For p H 1 (8), (2) splits as PIP2 , f = i, and D/P 1 = D/P 2 = F 2. Thus, ~(S) = tr, o $(P) cannot possibly be onto W(F 2) when p H 5 (8), and is onto when p H 1 (8). We summarize, Corollary 2.6 ~(S) is onto when p ~ 1 (8), and has cokernel C 2 = W(F 2) when p ~ 5 (8). . ~ Thus, in order to show 2: W(k,Q) + W(k,Q/Z) is onto all of the Witt pieces in W(k,Q/Z) , it remains to hit this one last Witt piece, W(F2), when k = p H 5 (8). We thus need to show how to find a Hermitian element in W(k,Q) which under ~ hits W(F 2) e W(k,Q/Z) whenever k = p ~ 5 (8) is a positive prime. This question remains open. 221 Corollary 2.7 For k = • (4) o__rr k E 1 (8), k a positive prime, a: W(k,Q) ~ W(k,Q/Z) is onto all Witt pieces in the decomposi- tion of W(k,Q/Z). Proof: We observe that all Witt pieces in W(k,Q/Z) occur in W(D/P) = W(E/D) E = Q(/k) By Corollaries 2 5, and 2.6 P max in D ' " " ~(S) is onto these Witt pieces. Hence, so is ~ all the more so. [] Corollary 2.8 a: A(Q) § A(Q/Z) is onto all Witt pieces in A(Q/Z). Proof: Same as above, since W(Q) occurs in A(Q). [] Thus ~ restricted to W(Q(/k)) ; ~(D) = ~(S), for S = Z[t]/(t 2- k) is onto when k is a positive prime, k E 1,2,3 (4) or k ~ 1 (8), or k = • By onto, we mean all Witt pieces in W(k,Q/Z) will be in the image of a(S), and hence in the image of ~. For these k, it remains to show ~ is onto. To do this, we must show that all Hermitian pieces in W(k,Q/Z) are in the image of ~. In this inert case, we show ~ is onto by hitting each H(Fp(@I)) separately by a, where 01 satisfies a T k fixed polynomial of type I. First, assume p ~ 2; we will do the case p = 2 last. Let q be a prime, with (q,p) = 1 = (2,p). Suppose | satisfies 2 x - alx + k = 0 over the fixed field of the involution - : 01 + ~i = k01 -I" Here a I e Fp(@ 1 + k@l-l), the fixed field of We write a I = 2b I, which is possible since 2 and p are relatively prime. 2n Let Fp(0 I) = F 2n = finite field with p elements. The fixed P field of - is F n. We shall now construct an extension Q(0) of Q, P 222 together with an involution - : @ ~ k@ -I which is non-trivial. Further, we shall arrange that the monic irreducible polynomial of @ when read mod p is the monic irreducible polynomial of @i" Let F denote the fixed field of - on Q(@). We shall arrange for at least one prime ideal, P in O(F), with P ~ Z = (q), to ramify in Q(@) over F. We then consider the boundary map 3 restricted to H(Q(@)), with H(Q(@)) a direct summand of W(k,Q) . Since there are ramified primes, the cokernel of 3 will be in terms of these. Hence, 31H(Q(@)) will be onto the Hermitian piece H(Fp(@l)), modulo the Witt pieces in W(k,Q/Z). But by 2.7, these Witt pieces have already been shown to be in the image of 3. Thus, H(Fp(@I)) is in the image of 3: W(k,Q) + W(k,Q/Z) as desired. Again, @i satisfies x 2 - 2blX + k = 0. We begin our construction by defining b 2 by the equation: bl 2 = q2b 2 + q + k. Suppose now that b 2 satisfies x m + Cm_l xm-I + ... + c 0 = f2(x) over Fp. We have the following field extensions: Fp C Fp(b2) C Fp(bl) _~ Fp(el). We now choose g2(x) = x TM + dm_l xm-I + ... + d 0, a monic integral irreducible polynomial, with mod p reduction f2(x). We also arrange for the mod q reduction of g2(x) to be irreducible. This is possible by the Chinese Remainder Theorem. Thus, both ideals (p) and (q) remain prime in Q(62), where B2 is a root of g2(x). Next consider the extension of Q(82) given by adjoining a root of x 2 - (q262 + q + k) = 0. Call this extension Q(61). The extension Q(62 ) C Q(8 I) may or may not be proper. In any case, 223 Lemma 2.9 (q) does not ramify i_nn Q(S2 ) ~ Q(81 ) Proof: The different ~ of this extension is the greatest common divisor of the element differents, (f'(~)), where ~ generates Q(SI ) l over Q(82 ) . Hence D divides 2~q2~ 2 + q + k. If (q) ramifies, q divides 4(q262 + q + k), so that q divides k. This is impossible since (q,k) = I. [] 2 Finally, we let E = Q(0), where @ satisfies x - 2~ix + k = 0 over Q(SI). Notice that the mod p reduction of @ is 01. Thus we have an extension of degree 2n, [Fp(@l) : Fp] . There is the fixed field of the involution 0 ~ k| -I given by Q(~I ) . We are adjoining 8 - 4k = 2 to Q(~I ) . By 2 construction, ~i - k = q28 + g = q(q 8 2 + i). This has q-adic 2 valuation I, ie. (q) ramifies in Q(@) over Q. However, also by construction, (q) does not ramify in Q(61). Hence, some prime lying over (q) must ramify in Q(@) over Q(~I ) . Now consider ~(D) : H(Q(0)) + H(E/I). Since there are ramified primes in Q(@) over the fixed field, the cokernel of ~(D) is given in terms of ramified primes. In other words, the term H(Fp(01)) in H(E/I) is in the image of ~(D), modulo ramified primes. Since all the ramified primes have already been shown to be in the image of 2, so also is H(Fp(01)) in the image of ~ as desired. The final construction is to show that H(F2(@I)) is in the image of boundary. Suppose F2(@ I) = F22n, with fixed field F2n. Suppose 2 @i satisfies x - alx + k = 0 over F2n. Suppose a I satisfies fl(x) over F 2. Lift each of these polynomials to Q to obtain Q ~ Q(a) _~ Q(0). 224 Now consider ~(D) : H(Q(0)) + H(E/I) . This time the cokernel will be in terms of ramified primes, or possibly C 2 if there is no ramification. This does not matter, since all cokernel elements are already in the image of 3 by previous work. So modulo these pieces, 3(D) hits H(F2(@I)) as desired. We have thus shown how to hit with 3 a typical Hermitian term H(Fp(@I)) in W(k,Q/Z). This of course works equally well for 3: A(Q) + A(Q(Z)) . We summarize. Theorem 2.10 3: WI(k,Q) § WI(k,Q/Z) is onto when k = • o_~r k a positive prime k E 2,3 (4) o_~r k E 1 (8). Theorem 2.11 3: A(k,Q) § A(k,Q/Z) is onto. in the skew case W-I(k,Q/Z), we need a slight modification. The above argument does show that ~ is onto all inert primes modulo ramified primes. There is in fact only one Witt piece in W-I(k,Q/Z), namely W(F2). Lemma 2.12 W(F 2) _~ W(k,Q/Z) is not in the image of 3. Proof: Consider the commutative diagram of forgetful maps. 3 W-I(k,Q) + W-I(k,Q/Z) +fl +f2 W-I(Q) ~ W-I(Q/Z) = W(F2). 225 fl and f2 are the maps which forget the degree k map in the data of a degree k mapping structure, f2 is the identity: W(F 2) ~ W(F2) . However, W-I(Q) = 0, so that W(F 2) is not in the image of 2. As a consequence, Theorem 2.13 3: s + w-l(k,Q/Z), for k = • or k = prime has cokernel C 2 given by the Witt element W(F 2) i_nn W-I(k,Q/Z) which is not in the image of 3. In order to understand the octagon, and apply the boundary sequences above, it is first necessary to analyze the individual maps in the octagon. We do this next, in terms of the T k fixed polynomials determining the Hermitian pieces. Chapter IX A DETAILED ANALYSIS OF THE OCTAGON We have an exact octagon over a field with typical term We(k,F). We also have analyzed each term WE(k,F) ~ ~ H~(F(0)) ~ We(F(/k)) k ~ F** = ~ HS(F(0)) ~ WS(F) ~ WS(F) k ~ F** In this chapter, we analyze the maps in the octagon using this direct sum decomposition. Each of these Hermitian and Witt summands is determined by a T k fixed irreducible polynomial. In Section i, we classify these polynomials, and discuss several cases that may arise. In Sections 2,3,4,5 we examine the various maps in the octagon. This analysis involves determining the effect of the homomorphisms on rank mod 2, signature, and discriminant. For F an algebraic number field, these invariants determine H (F) by Landherr s Theorem. i. The involutions Recall K(F) = {monic polynomials, non-zero constant term, coefficients in F}. On K(F) , we have,several automorphisms defined. t n (i) TkP(t) = -- p(kt -I) Denote p(t) = Tk(P(t) ) a N t n (2) T-kP(t) = a0 P(-kt-l) p*(t) = T_k(P(t)) (3) ToP(t) = (-l)np(-t) p~ ) = ToP(t) These arise from the corresponding involutions on F[t,t -I] given by, respectively: (i) t + kt -I [ = kt -i 227 (2) t + -kt -I t* = -kt -I (3) t + -t t ~ = -t These involutions, together with the identity, thus determine an action of the Klein 4 group, Z/2Z ~ Z/2Z on Fit,t-l] . We should recall the origin of these involutions. Let [M,B, ~] e W(k,F), with (M,B,Z) anisotropic. Let p(t) be the minimal polynomial of s As we have seen in III 1.3, we may assume p(t) is a T k fixed irreducible polynomial. By III 1.7, the ideal (p(t)) in F[t,t -I] is then - invariant. We then consider F[t,t-l]/(p(t)) = F(Q) . F(@) has an involution - induced by ~ = k0 -I, since (p(t)) was - invariant. Under this involution, [M,B,s is identified with the Hermitian inner product space [M,B'] over F(@) , with t, o B' = B, where t, = trace F(@)/F. We will describe the effect of the maps in the octagon on these Hermitian inner product spaces, and the associated polynomials. To begin with, we need a criteria to determine whether F (@) equals F(02), when F(@) has an involution induced by T k for some k. The answer is given by considering the following cases. We will assume that the characteristic of F is not 2, since F(G) = F(@ 2) when F has characteristic 2. Notation: For the remainder of this chapter, we let p(t) = irreducible polynomial of @ over F 2 q(t) = irreducible polynomial of @ over F. Case i: F(0) ~ F(Q 2) when ~(t) = p(t) and p,(t) : p(t) . 228 Proof: In this case, the ideal (p(t)) is both - and * invariant. Hence F[t,t-l]/(p(t)) = F(@) has the induced involutions: = k@-l, @, = _k@-l, @o = (~), = _@. These are Galois automorphisms of F(@). The fixed field of o is F(@2). As long as o is non-trivial, which happens provided the characteristic of F is not 2, F(@ 2) ~ F(@) by Galois theory. ~'~ Remark: Since F(@ 2) ~ F(@) in this case, we have degree p(t) = 2 degree q(t), and consequently p(t) = q(t2). Case 2: F(0) = F(@ 2) when p(t) = p(t) and p*(t) ~ p(t) . In this case, the o involution is not present. Recall our notation; q(t) is the minimal polynomial satisfied by @2 over F. q(t 2) has @ and -@ as roots. Note that q(t 2) has only even degree terms. The hypothesis p(t) = p(t) and p*(t) 9 p(t) implies p(t) has odd degree terms. Thus, q(t 2) is not irreducible. Hence, we may write (a) q(t 2) = p(t)p(-t)w(t) . This follows since p(t) ~ p(-t), else p*(t) = (p(-t)) = p(t) = p(t), contradiction. Further, degree q(t 2) = 2 degree q(t) < 2 degree p(t). However, by (a), degree q(t 2) > 2 degree p(t). Thus, degree q(t 2) = 2 degree p(t), and w(t) = i, so q(t 2) = p(t) p(-t). Hence, degree q(t) = degree p(t), and F(@) = F(02). O Case 3: F(G) = F(02) when p*(t) = p(t) and p(t) ~ p(t) . 229 Proof: Exactly as above. 2. The map Is: WS(k2,F) + WS(-k,F) We begin with the induction map I s defined by Is: [M,B,Z] + [M ~) M, B ~ -kB,Z] where Z(x,y) = (Zy,x). Consider [M,B'] = [M, [,]] s Hs , where HS(F(o)) embeds into We(k2,F) via t,. To keep our notation consistent with Section i, we write o = @2, and HS(F(o)) = HE(F(@2)) . Here we have F[t,t-l]/(q(t)) = F(@2). Case i: F(@) ~ F(@ 2) F(@ 2) has involution given from T 2' @2 § k2$-2. This extends _i k _ * in two ways to F(@), namely @ § k@ = @, and @ + -k@ -I = $ In fact, this is the way this case is recognized, namely both the - and * involutions are present on F(@) . We identify [M,B'] with [M,B,Z] e W s (k2,F) . B(x,y) = t, o B' (x,y) and s is multiplication by @2. Here B' = [,], so that B = t, o [,]. Mapping over with I s we obtain [M M,B ~ -kB,~]. We wish now to identify the (Hermitian) form we obtain in W~(-k,F) . Recall that ~2(x,y) = (~x,[y) . Hence, ~ acts as a square root of i. Thus, we wish to identify [M M, B ~ -kB,s with an Hermitian form over F(/o) = F(@) . This is a question of how to define an F(C) - vector space structure on M ~ M compatible with the F(3 2) - vector 230 space structure. The following can best be understood by considering M = F(@2), the one dimensional case. The identifications work equally well for M arbitrary. We now write F(@) = F(@ 2) $ F(@2). We are naturally thinking of @ = /~ as the ordered pair (0,i). Addition of ordered pairs is componentwise. Multiplication of pairs is given by (a,b) (c,d) = (ac + bd@ 2, ad + hc). If the dimension [M: F(@2)] = n, we now view M ~ M as a vector space over F(8), with dimension, [M ~ M, F(@)] = n also. A basis for (M ~ M)/F(@) is given by {(vi,0)} where {v i} is a basis of M/F(@2). Scalar multiplication by F(@) = F(@ 2) ~ F(@ 2) is given by (a,b) (vi,0) = (avi,bv i) on basis elements by following the above. These operations extend linearly to make M M into a vector space over F(@) . We obtain the form <,> in HS(F(@)) given by: <(x,y), (z,w)> = i/2([x,z] + -k[y,w] - k@-l[x,w] + @[y,z]) where [,] = B': M • M + F(@2) . One easily checks that this respects the vector space operations given by the identification (x,y) = x + y@, with involution on F(@) given by 0" = -k# -I. We compute tr, o <,> by using trF(O)/F = trF(@2)/F o trF(0)/F(@ 2) Here tr, is the map induced by the appropriate trace, denoted tr. Note, that trF(@)/F(@2 ) (r) = 2r for r K F(@ 2) and 231 -1 trF (@)/F (02 ) (0) = tr 0 = 0. It follows that tr, o <,> = B ~ -kB as desired. We see in this case, that I is identified with a map s I : H~(F(02)) § H~(F(0)). E " (1) I preserves rank. This is clear since [M: F(@2)] = [M M, F(@)], ie. dimF(@2 ) (M) = dimF(0) (M ~ M). (2) Signatures. If k < 0, W(k,F) = ker I is all torsion. Hence, in order that [M, [,]] s ker I there must be no signatures in HE(F(02)). E If k > O, I [M,[,]] ~ H~(F(@)) is in W(-k,F); again this group is all torsion. So there is no signature in the image in this case. (3) Discriminant. Here we must be careful because the discriminant of [M, [,]] is read in F(02 + k20-2)/NF(o2)/F(@2 + k2@ -2) where NF(@2)/F(02 + k20-2) denotes elements in F(@ 2 + k2@ -2) which are the norms of elements from F(G2) , whereas the discriminant of the image, [M ~ M,<,>] is read F(@ - k@-I)/NF(@)/F(G _ k@-l). These may be different groups as the example which follows will show. 232 To summarize, when F(@) is an algebraic number field, Theorem 2.1 Let [M, [,]] e He(F(@2)), and assume F(0 2) @ F(@). Then [M, [,]] e ker I if and only if E (a) M has even rank. (b) M has signature 0 if k < 0. (c) The discriminant of M when read in F(0 - k@-I)/NF(e)/F(@ _ k@-l) must be trivial. Proof: (a) and (b) have already been discussed. To verify (c), we need to calculate the discriminant of <,>. If dim M = n, this is exactly given by (i/2)ndis([,]), by the formula for <,> applied to a 1 - dimensional form, and induction. Hence, (a), (b) , (c) follow by Landherr's Theorem. F'1 An Example. Let o = 0 2 = /-i = i 0 = /i We now have the extensions: Q(i) § Q (~/i) + Q § Q(/i + i//i) This example is for k = -i. 233 The involution on Q(i) is i § k2/i = i/i = -i The involution on Q(/i) is /i + i//i = -i/i By elementary number theory, 3 is not a norm in Q(i)/Q, since 3 is not the sum of 2 squares. However, consider (i - I) + /i in Q(/i). we compute its norm in Q(/i)/Q(/i + I//i), ((i - i) + /i) ((-i - l) + I//i) = 3. Thus 3 becomes a norm. This leads us to consider the following example. Let k = -i, and let M be a 2 - dimensional vector space over Q(i) , with basis el,e 2. With respect to this basis, consider the Hermitian form over Q(i) given by -3~ 1 This 2 - dimensional form has signature 0, and discriminant +3, which is not a norm. ~ , ~ ~ --% If ~l,e2 is a basis for M over Q(i) , e I , le I, e2, ie 2 is a basis for M over Q. We'thus identify the Hermitian form [M, [,]] with the Witt class in W(+I,Q) given by [M,t, o [,],i] where ix = i~, so that with respect to the basis given for M/Q, Z has matrix _% .-% 41 ie~l e 2 ie 2 e I 0 -i 0 0 ...% le 1 1 0 0 0 ~2 0 0 0 -i ...% ie 2 0 0 1 0 We write B = t, o [,]. 234 Next, apply I to [M,B,s to obtain [M ~ M, B G -kB,~]. s This in turn is identified with an Hermitian form over Q(/i) . With respect to the basis (el,0), (e2,0) for M ~ M = V over Q(/i), we obtain the form (<2 0] = <,> -3/2 Since this form has discriminant 3/4, which is a norm, it follows that [V,<,>] = 0 in H(Q(/i)) . In this manner, we see how the norm groups Q(@2 + k2@-2)/NQ(o2)/Q(@2 + k2@2 ) and Q(@ - k@-I)/NQ(@)/Q(@ _ kO -I) give rise to kernel elements for I s . Case II: F(@) = F(@ 2) To begin with, F(@ 2) has the involution given from T k,2 namely 0 2 + k2@ -2. Call this involution ~ , so 9 2 = k2@ -2. Hence ~2@2 = k 2, and (@~)2 = k 2. It follows that G0 = • This gives us two subcases. (a) G = + k@ -I (b) @ = - k@ -I In either case, we begin with [M, [,]] s H(F(@2)) . We embed this into W(k2,F) via t, to obtain [M,B,%] where B = t, o [,] and 235 Z(x) = @2x. Applying I s we obtain [M ~ M, B ~ -kB,i] where ~(x,y) = (@2y,x). (a) In case (a), consider N C M ~ M given by N = {(@x,x): x ~ M N is clearly Z invariant, with rank N = 1/2 rank(M ~ M) . Further, (B (D -kB) ((@x,x) , (@y,y)) = B(Gx,@y) - kB(x,y) = @~B(x,y) - kB(x,y) = kB (x.~y) kB (x,y) = 0 Thus N C N , from which it follows that I [M,B,Z] = 0. This S completes case (a). (b) In case (b), we consider the rank 1 case, M = F(@2) ; the general case follows by diagonalizing. Let ~, = (i,0) and ~ = (0,i) be a basis for M ~ M over F(@ 2) = o. % + e% ~( ~2 ) = @ 2-~e I + 0 -~e 2 Thus viewed, the matrix of s relative to e I , is -.% ._% e I e 2 e% <01 9 2 ) -.% e 2 0 The characteristic polynomial of ~ is det = x 2 - 92 = [x + q) (x - @). 236 The eigenvalues are thus @, -@. We next compute the corresponding eigenvectors. 0 @ x I = @x 1 <1 0 21(Ix 2 \ 0x 2 @2x 2 - @x I = 0 x I - @x 2 = 0 Solving the eigenspace corresponding to eigenvalue @ is generated by fl = (@,i) namely {t~l: t ~ F(@ 2) }. Similarly, the eigenspace corresponding to -@ is {t~ 2 = t(-@,l) : t ~ F(@2) }. Eigenvalue Eigenvector s dimension @ t(@,l) t ~ 0 1 -@ t(-@,l) t ~ 0 1 Expressing each of ~i in terms of the ~i' we have fl = @el + e2 f2 = -@ + e 2. This leads to the change of basis matrix (: :I = L, so L-l~orig L = ~new" where ~new is diagonalized with respect to the basis fl,f2. ~orig denotes ~ with respect to the basis el,e 2 o = [ and ~ = I we obtain With ~new 0 - \l 0 / the diagram: 237 L M(~M -~ M(~M l new L M(~M + M(~M L: (x,y) + (@(x - y) ,x + y) . The question arises: what inner product on M ~ M makes L into an isometry, L: (M ~ M, b, Znew ) § (M G M, B ~ -kB, [orig). We compute: b((x,y) , (z,w)) = (B ~9 -kB) (L(x,y) ,L(z,w)) = (B E) -kB) ((@(x - y) ,x + y) ,(@(z - w) ,z + w) = B(0(x - y),@(z- w)) + -kB(x + y,z + w) = @GB(x - y,z - w) - kB(x + y,z + w) = -kB(x - y,z - w) - kB(x + y,z + w) = -k[B(x,z) + B(y,w) - B(y,z) - B(x,w) + B(x,z) + B(y,w) + B(y,z) + B(x,w)] = -2k[B(x,z) + B(y,w) ] . Thus under the isometry L, we may view (M (~ M, B (~ -kB,Z) -- (M ~9 M, b, kne w ) , where b((x,y),(z,w)) = -2k(B(x,z) + B(y,w)) s = $x, -{iy) Thus, the image of I is in two distinct pieces, H(F(@)) ~ H(F(-O)) in this case. I is given by: c 238 I [M, [,]] § [M,bl] ~) [M,b2] where bl(X,y) = b2(x,y) = -2k[x,y]. It follows that Is[M, [,]] is non-zero if [M, [,]] @ 0. We summarize with: Theorem 2.2 Let [M,[,]] e HeF(G2), and assume F(~ 2) = F(@) . There is an induced involution ~ on F(@ 2) given by ~2 = k~-2. This gives rise to two cases. (a) 0 = k@ -I (b) e = -k@ -z In case (a), [M,[,]] s kernel I In case (b), Is HS(F(e2)) + HE(F(@)) e HI(F(-0))- is 1 - I. [] Finally, we should observe that this analysis works equally well for F a finite field. The only difference is that He(F($)) is determined by rank mod 2 only. In particular, when F has characteristic 2, and k and 2 are relatively prime, WS(k,F 2) = ~ H~(k,F2(@)) ~ W~(F2 ) . We examine the map Is For all Hermitian summands H6(k,F2(0)) , we are in case (a) of Theorem 2.2. Thus I s (Hs ~ 0. We must also check what happens to We(F2). So consider [M,B,~] , where Zx = x. Applying I s , we obtain [M ~ M, B ~ -kB,Z]. A metabolizer is given by N = {(x,x) : x s M}. Thus, when F has characteristic 2, I is identically 0. 3. The map_ d 6 : W s + A(•) . 239 Recall that d is defined by: [M,B,z] § [M,B] where g B(x,y) = k-IB(x,Zy). From V 2.4, the symmetry operator for B, satisfying B(x,y) = B(y,sx) is s = -gkz -2. We consider [M,B'] = [M, [,]] e HE(F(0)), where F(@) = F[t,t-l]/(p(t)) has induced involution 0" = -k@ -I. This embeds into WE(-k,F) via t, and we identify [M,B'] with [M,B,%] ~ WS(-k,F) where B = t, o B', s = @x. Apply d and obtain [M,B]. We must now identify the Hermitian form we obtain. For A(F) this is done by using a scaled trace t I IV 2.5. t I depends on the scaling factor u chosen where uu--i = s = -ek@ -2. tl: F(0 2) + F. Recall that u may be chosen as u = s/(l + s) = -ek0-2/(l + -Ek@ -2) = -Ek/(@ 2 - Ek) and tl(x) = t(xu -I) where t is the usual trace. Since s = -sks -2 = -ek@ -2, we obtain a Hermitian form with values in He(F(@2)). Case i. F(@) ~ F(@ 2) In this case, [M,B] may be identified with the Hermitian form in HE(F(-Ek@-2)) = HS(F(@2)) given by: We then have t I = k-lB(x,@y) = B(x,y) as desired. (h is defined in IV 2.6) 240 We now examine the Witt invariants. We obtain a form in HS(F(@2)). In this case, F(@) ~ F(@2) , so that the rank of M as a vector space over F(@ 2) is twice the rank of M over F(@). We describe the method for obtaining the other invariants by examining the one dimensional case. So assume M = F(8), and [i,i] = d s F(@ - k@-l). Then ~i = i, ~2 = @ is a basis for M over F(@2) . With respect to %,% we examine the matrix of the Hermitian form <,>. We assume u = -s 2 - Ek) . u is given by the involution @* = -k@-l; the , involution extends the - involution. Thus the matrix of <,> is: 1 @ 1 (tr(dk i.u@ ) tr(dk-lu@ .2) -i- *2 @ tr(dk-lu@@ *) tr(dk u@@ )] tr denotes trace F(@)/F(e 2) Using this, one may determine the signature and discriminant of <,>. Case 2. F(@) = F(@ 2) In this case, [M,B'] = [M, [,]] has the same rank as [M,<,>] . Again we examine the 1 - dimensional case, M = F(C), where [I,I] = d. Then <,> is given by: -i-- * _ -1 = d[+sk@/(k 2 - sk@2)] . 241 Thus, the value of the discriminant depends on this factor ekO/(k 2 - ek@2). When k > 0, WS(-k,F) is all torsion, and there are no signatures. When k < 0, we must check that the resulting form [M,<,>] has 0 signature in order that [M, [,]] be in the kernel of d e- Again, we examine We(-k,F2) . We must be in case (2). Rank is the only invariant, so that d e is 1 - i. It is also clear that the Hermitian forms, He(F2(@)) , in We(-k,F 2) are mapped under d e to Hermitian forms. The form [M,B,s with M = F 2, B = , = identity, corresponding to WE(F2), likewise maps under d e to W(F2). These remarks are needed for the computation of the exact octagon over Z to be made later. 4. The map S : W E (k,F) + W e (k 2,F) Recall that S is defined by [M,B,Z] + [M,B,s Let s [M,B'] = [M, [,]] s HE(F(@)), where F(@) = F[t,t-l]/(p(t)) has induced involution ~ = k@ -I. Embed [M,B'] into Ws via t, and identify [M,B'] with [M,B,Z], where B = t, o B', Zx = @x. We apply S and obtain [M,B,Z2]. We wish to identify the Hermitian form we obtain. Case i. F(@) ~ F(@ 2) We clearly obtain the Hermitian form in HE(F(@2)) given by [M,BI], where B 1 = trF(@)/F(@2 ) o B' . The :rank of M over F(@ 2) is twice the rank of M over F($) . 242 In order to examine the other invariants, consider the 1 - dimensional case, M = F(@). A basis for M over F(@ 2) is _~e I = i, ~e 2 = @. Suppose B' = [,] has [i,i] = d e F(@ + k@-l) Then with respect to the basis i, 8 B 1 has matrix: 1 @ = k@ -I @ tr (@d) tr (kd) This is with tr denoting traceF(@)/F(02 ) . Again, this matrix enables one to compute the signature and discriminant invariants. Case 2. F(@) -- F(@ 2) In this case F(@) has involution @ + ~ = k0 -I so that 02 + ~2 = k2@-2 It follows that SE[M,B,s may be identified with the Hermitian form in Hs (@2)) given by [M,B']. In this case S is then E clearly 1 - i. We remark that when the characteristic of F is 2, S s is 1 - 1 by case (2). In particular, S : W(F 2) § W(F2) ; where the non-trivial form in W(F 2) is identified with the form [M,B,Z] in w(k,F 2) given by: M = F 2, B = , i = identity. 5. The map ms: A(F) ~ WS(k,F) 243 m is defined by: [M,B] + [M (B M,B , Z ] , where E ~ E B ((x,y),(z,w)) = B(x,w) + s s (x,y) = (~ks-ly,x). Let [M,B'] = [M,[,]] s Hs where F(@) = F[t,t-l]/(p(t)) has the involution - induced by @ § ~ = @-i We identify [M,B'] with [M,B] e A(F) using a scaled trace, t I. Here the symmetry operator s acts as @, and tl(x) = t(xu-l) , where uu -I = @. As observed before IV 2.6, we may choose u = @/(i + @), so that ----1 u =i + @. Then B = t I o B' The analysis of m e is then similar to I E. The image of [M, [,] ] under Ie can be viewed in He(F(~)) . This is because i 2 = s -I. s There are two cases. Case i. F(@) # FW/Ek@ -[) Case 2. F(@) = F(~k@ -i) Note: Let e = s -I. The involution on F(a) is then ~ k~ -I = ~. So (ek@ -I) ~ k2(ck@-l) -I t and ek$ -I + s @ ~ @-i. In other words, the involution on F(@) extends the - involution on F(@) . Case i. F(@) # F(~) In this case, the dimension M G M over F(Vs ~) is the same as the dimension of M over F(@) . As observed previously, we can 244 make M (9 M into a vector space over F( s~kG -I) = F(e). The idea again is to vfew F(e) = F(@) ~ F(@). We view e as the ordered pair (0,1). Multiplication of ordered pairs is given by identifying (a,b) a + b~ (c ,d) + c + d~ (a,b) (c,d) ac + bd(~ 2) + (ad + bc) ac + bd(sk@ -I) + (ad + bc) (ac + bd(s -I) ,ad + bc). Thus, me: H(F(@)) + H(F(~)) in this case. If M has basis { (v i) } over F(@), M (B M has basis {(vi,0) } over F(~). Consider the form <,>: M ~9 M § F(0,) given by <(x,y) , (Z,W) > = (i/2k) u-l~([x,z] + (k/~) Ix,w] + e[y,z] + k[y,w] 2 Here ~ = 6k@ -I and [,] is the Hermitian form we began with. We must check that <,> is ~ Hermitian. There are the identities: --i u = 1 + @ u--i c~ = (l +@) --i u-l~ = ((i + @)/@)k~/(~k@ -I) = s U A . Hence: <(z,w),(x,y)> = (i/2k)u-l~([z,x] + k[-l[z,y] + ~[w,x] + k[w,y]) 245 = (e/2k) u-le([x,z]- + (k/a) [x,w] + ~[y,z] + k[y,w]) = ~<(x,y) , (z,w) >. - denotes the involution. [a,b] = [b,a] since [,] is Hermitian. Next we compute tr, o <,>, where tr, is traceF(~)/F. trF (a)/F = trF(@)/F o trF(e)/F(@). trF (~)/F (@) <(x,y), (z,w)> : (i/k) u-lk[x,w] + (i/k)u-l(sk@-l) [y,z]. Hence, trF (e)/F <(x,y), (z,w)> = trF(@)/F(u-l[x,w] + u-ls@-l[y,z]) = tl([X,W] + s = B(x,w) + EB(@-ly,z) = B(x,w) + sB(z,y) . Hence, tr, o <,> = B as desired. s We have thus identified the Hermitian form we obtain in the image of m s in this case. Rank mod 2 is clearly preserved, and we read the discriminant and signatures from the extension F(~)/F(e + ke -I) in order to determine if [M, [,]] is in the kernel of m s . Case 2. F(@) = F(/ek@-l) . F(@) has the involution @ ~ ~ = @-i Under this involution, (~2) = ~k@. Also ~ 2 = s -I. Thus, (,~2) (~2) = (a~)2 = (sk) 2 = k 2. 246 Hence, ~ = • This gives two cases: (a) = -k jl (b) - = k j 1 i Case (a) ~ = -k -I Let N = {(ex,x) : x e M}. N is clearly s invariant, since E ~e(x,y) = (a2y,x). Further, N is self-annihilating since: Be((~x,x),(~Y,y)) = B(ex,x) + s = B(ex,y) + B(gx,~Gy) = B(ex,y) + B(se @x,y) . However, 2 = ek@-i , so e ---- (ek@-l) (-i) o ee @ = e(-ke -I) (G -1) = -e. Thus the above equals 0 and N is a metabolizer for [M ~ M,Bs,s163 Thus [M, [,]] is in the kernel of m in this case. E Case (b) --~ = k -i As with IE, we consider the one dimensional case, M = F(@). Let ~i = (I,0) , e 2 = (0,i) be a basis for M M over F(8) = F(~.) . ZE( ~I ) = 0 ~i + 1 - % and Ze( ~2 ) = ~ 2-*el + -Ae 2, so that 9~e has matrix (: :) with respect to e I , e 2 We now diagonalize this matrix, and obtain the diagram: 247 L M~)M + M~)M (: :) (o :) L M~M + M~)M where I, -- /i -~)l is the change of basis matrix. We compute: b((x,y), (z,w)) = BE(L(x,y) ,L(z,w)) = B ((~(x - y) ,x + y) , (e(z - w) ,z + w)) = B(e(x - y),(z + w)) + eB(~(z - w),(x + y)) = B(~x - ~y,z + w) + eB(x + y,0(~) (z - w)) = B(ex,z) + B(ex,w) - B(ay,z) - B(~y,w) + eB(x,e@z) - eB(x,e@w) + sB(y,az@) - eB(y,~w@). However, e = (ek0-1)~ -I = e~ e in this case. Thus, the above becomes = 2e[B(x,z) - B(y,w) ] . We may thus view the image of m as in H e(F(~)) ~ H s in this case, namely m [M,[,]] ~ [M,bI,Z I] @) [M,b2,Z2], where bl(X,y) = 2B(ex,y) 21(x) = ~x b2(x,y) =-2B(~x,y) "%2(x) = -~x. 248 B is an F-valued asymmetric form. We must apply the trace lemma to identify bl,b 2 with <'>i' <'>2 where <'>i and <'>2 correspond with bl,b 2 above, by tr, o <'>i = bl' tr, ~ <'>2 = b2" Then mg: [M, [,]] § [M,<,> I] ~ [M,<,>2] . <'>i: M x M + F(~) is defined by Similarly, = 2u-l~[y,x] = 2eu-le[x,y] = 2e[-l[~x,y] = E trF(~)/F = tl(2[ax,y]) = 2B(ax,y) = bl(X,y) Hence, [M,<,> I] is merely [M, [,]] with the scaling factor 2~ -I , so that m is 1 - 1 in this case. Chapter X THE OCTAGON OVER Z We recall the decompositions We(k,Q/Z) = ~ W(k,Fp) W(k,Fp) A(Q/Z) ~ ~ A(Fp) pXk plk P For plk, the maps in the octagon for W(k,F ) do not make sense, P as these terms W(k,Fp) have k = 0. Therefore, in this section, we assume k = • Hence, by the results for a field, we restate: Lemma i.i There is an exact octagon where k = • S 1 I 1 W(k,Fp) + W(k 2 ,Fp) W(-k,Fp) m I 7 a 1 A(Fp) A(F ) P Z m_1 I_ 1 S_ 1 -1 W-1 (-k,Fp) § W-I (k2,Fp) + W (k,Fp) Proof: "This is the octagon over the field F . Taking the direct P sum over all p, we obtain Theorem 1.2 For k = • there is an exact octagon: 250 S 1 I 1 wl (k,Q/Z) + W1 (k2,Q/Z) W 1 (-k,Q/Z) m 1 dl A(Q/Z) A (Q/Z) m_ 1 d_l~ I_ 1 S_ 1 W-1 (-k, Q/Z) § w-l(k2 ,Q/Z) W -I (k,Q/Z) Although we have yet to prove exactness of the octagon over Z, the homomorphisms nonetheless are defined over Z. It is easy then to check that we have the commutative diagram which follows, i denotes the map • Q, and ~ denotes the appropriate boundary homomorphism. "d_ 1 "d_ 1 "d 1 1 0 § A(Z) A(Q) ~ A(Q/Z) § 0 +m +ml 3 +ml 1 l 0 + WI(k,Z) W(k,Q) ~ W1(k,Q/Z) +S 1 +S 1 ~S 1 0 ~ WI(k2,Z) WI(k2,Q) + W1 (k2Q/Z) +I 1 *Ii ~ii 0 § WI(-k,Z) W1 (-k,Q) + W1 (-k,Q/Z) +d 1 ~dl 3 +d 1 0 + A(Z) A(Q) + A(Q/Z) ~ 0 +m +m -i ~m-i 3 -i 0 § w-l(k,Z) W-I(k,Q) ~ W-I (k,Q/Z) +S -i ~S_I +S_I 0 § w-l(k 2 i Z) w-l(k2,Q) + W-I (k2 ,Q/Z) +I_ 1 +I_ 1 #I_ 1 l 0 + W-I (-k,Z) § W-I (-k,Q) ~ W-I (-k,Q/Z) +d I Cd_l Sd_ 1 251 The last two columns are exact, as are all rows. The problem is that the last three rows are not short exact, as there is the term W(F2) in w-l(k,Q/Z), not in the image of 2. We recall now the diagram chase that would prove exactness of the first column if all the rows were short exact. For simplicity, we label Witt equivalence classes now with symbols x, y, z, u, v, w. To prove: ker S 1 = im m I Let x g WI(k,Z) have x 6 im m I. So we can find y e A(Z) with ml(Y) = x. Now Sl o ml o i(y) = 0 by exactness of the 2 nd column. Hence i o Sl o ml(Y ) = 0 by commutativity. So i(Sl o ml(Y)) = 0. But i is 1 - I, so S 1 o m(y) = Sl(X) = 0. Thus im m I _ ker S I. Pictorially: i y ~ i(y) +m ! +m 1 x ml(i(y) ) ~S 1 ~S 1 i 0 + Sl(X) § 0 Conversely, let x s wl(k,Z) have x a ker S I. The picture below will facilitate reading the proof. 252 ? Z -~ + d_l y § 3y + m 1 m 1 X i(x) -~ 0 4- S 1 0 0 i o SI(X ) = 0, hence S 1 o i(x) = 0 by commutativity. The middle column is exact, so we can find y s A(Q) with ml(Y) = i(x) . Now ml(~y) = (9 o ml) (y) = (3 o i) (x) = 0. Thus, by exactness of the last column, we can find z s W-I(-k,Q/Z) with d_l(Z) = ~y. This is the point that we need 3: w-l(-k,Q) + W-I(-k,Q/Z) is onto. If 9 is onto z, we can find ws W-I(-k,Q) with ~w = z. Then consider (y - d_lW) . 3(y - d_lW) = 3y - 3y = 0. Thus, by exact- ness of the row, we can find v ~ A(Z) with i(v) = y - d_lW. However, (m I o i) (v) : ml(Y) - (m I o d_l) (w) = mlY = ix. Hence, (i o ml) (v) = i(x) . Since i is 1 - i, mlv : x. [] We see that the problem arises from z ~ im 3 going under d_l to Sy, which is in the image of 9. However, we have explicitly calculated what such z must be. Namely, z must arise from W-I(F2) . We recall the computations given in the last chapter. S_ 1 W-I (F2) + W-I(F2) W-I(F2 ) I+-i 0 d W-i (F2) +-i W-i (F2) 253 We may thus conclude that the octagon is exact over Z with one possible exception, the term A(z) d-1 w+l(k,z) ~1 w§ z). We must carefully analyze exactness at w+l(k,Z). To begin with, consider W(F 2) e A(Q/Z) . This is the source of the problem. Consider [V,B] e A(Q), where V = <~i >, and B = [,], with [~l,~l] = 2. We apply ~ to [V,B]. So let L = <~i > be a Z - lattice. L # = <(i/2)~i> and <(I/2)~I,(i/2)~i > = 1/2. It follows that ~[V,B] = W(F2) (meaning 8 ([V,B]) ~ 0 in W(F 2) ). Next, apply m I to [V,B]. Since s = identity is the symmetry operator, we obtain: IV ~ V, Be,[ ]. V ~ V has basis (1,0) = (~i,0) = and (0,i) = (0,~ I) = f2" With respect of ' f2 B s has matrix iI~ ~2 2~ 021 since B ((x,y) , (z,w)) = B(x,w) + eB(z,y) . Z has matrix E 6 since Ze(x,y) = (sks-ly,x). Of course. ~ = +I in this case. 254 Now ~[V (9 V,B ,Z a] = ~ o mI[V,B] = m I o ~[V,B] = ml o d_I(W-I(F2) ) = 0. In fact, we may apply ~ by: Let L be the Z - lattice <~i, ~2 >. Then L # = <(i/2)fl,-~ (i/2) f-~o> A metabolizer for L#/L is N = <(i/2)~ 1 + (I/2)~2>. There is the projection L # q L#/L. Then q-l(N) has basis {(i/2)~ 1 + (I/2)~ 2, (i/2)~ 1 - (I/2)~2}, which we may write as {gl,~2}. This enables us to construct an element in W+I(k,Z), which when tensored with Q yields [V ~9 V,Bs Zs]. The element is W = <~i,~2 > as a Z - module, with inner product l(l 01 ~2 0 - 1 with respect to ~i,~2, and degree k map (k + 1)/2 (-k + 1)/2 1 I(-i + k)/2 (-k - i)/2 ._% -~ with respect to ~i,~2, where ~i = (i/2)fl + (i/2) f2' and ..% ~2 = (1/2) fl - (1/2) f2" This follows since ~ is 0 ..% -~ with respect to fl,f2. We label this element [W,bl,t I] = x where W = <~1,~2 >, 255 bl :/: and, % (-k + i)/2 \ t 1 = ) (-k - l) 12 / We observe that [W,bl,t l] = x has order 2. When k = -1, x has order two since every element in W(-l,Z) has order two. (IV 4) For if [W,B,I] e W(-I,Z), {(x, Zx)} will be an Z ~ Z invariant self-annihilat- ing subspace of [W ~ W, B ~ B, ~ i], hence W ~ W ~ 0. When k = +i, we consider [V ~ V,B,s = y = ix. Since i is one-to-one, 2x = 0 if and only if 2y = 0. With the matrices given, /0 B E = s (i 210 ii Clearly {(r,s,r,-s) : r,s s V} is a metabolizer for (V O V ~) V (~ V, B (B B, s ~ s = 2y, so that 2y = 0 = 2x. Lemma 11.3 x above is not in the ~ of m I, but x is in the kernel of S 1 256 Proof: By construction, x is in the kernel of S 1. This follows since i o Sl(X ) = S 1 o i(x) = S 1 o ml[V,B ] = 0. i is 1 - i, so Sl(X) = 0. The picture below explains the proof that x is not in the image of m I . § ~v +d_ 1 z ~ y + ~y ~ 0 in W(F 2) J I m I + m I i x + i(x) S 1 0 Suppose ml(z) = x. Then m I o i(z) = i o ml(z ) = i(x) . However, ml(Y) = ix also. Thus, ml(Y - i(z)) = 0. By exactness of the middle column, there exists v with d_l(V) = y - i(z) . Now consider ~v. d 1 o ~(v) = ~ o d iv = ~(y - i(z)) = ~y. However, by construction, ~y M 0 in W(F2). The question then is: Can 3v have d_l(~V) = u ~ 0 in W-I(F2)? Clearly ~(v) ~ u as u is not in the image of ~. However, Hermitian summands are mapped under d_l to Hermitian summands by the results of the last chapter. Thus, no such v can exist, and hence x is not in the image of m I. Lemma 1.4 If x I s ker S, then either x 6 im m I or x I - x ~ im m I, where x = [W,bl,t I] a_s described before Lemma 1.3. 257 Proof: The picture below may be useful. w ....~ z 1 d_l Yl § ~ Yl +m 1 + m I i x 1 ix I § 0 + S 1 ~S 1 0 0 Since SlX 1 = 0, S 1 o ix I = 0, and there exists Yl with mlY 1 = ix I by exactness of the middle column. Now consider ~YI" By commutativity, m I o ~Yl = 0, so that by exactness of the last column we can find z I with d_l(Z I) = ~YI" The question is: Is z I in the image of 2? If the answer is yes, we proceed as in the general case and conclude x I is in the image of m I. If the answer is no, consider z I + u. This clearly must be in the image of ~ say ~w = z I + u where u ~ 0 in W -I (F2). Again we proceed as before and conclude x I - x is in the image of m I. We may now state the theorem we have been aiming for: Theorem 1.5 The following octagon is exact. w+l(k,z) Sl w+i(k2 Z) I.i W +I (-k, Z ) m I ~ i/~ ~,d 1 / A(Z) (~) C 2 A(Z) I k2 S 1 W-I(_k,Z) ~-l w-l( ,Z) ~- W-I(k,Z) 258 Proof: Here C 2 denotes the element [W,Bl,tl] = x constructed prior to Lemma 1.3. As we have seen the only question is exactness at W+I (k,Z) . Let x I s im (m I ~ i), where i is the identity on x; so x I = mlY or x I = mlY + x. Applying S I, we obtain Sl(mlY) +SlX = SlmlY by i. 3 = 0 Conversely, let x I e ker S. By 1.4, either x I ~ im m I or x I - x s im m I. In either case, x I is in the image of m I ~ i as desired. Finally, we should remark that adding the term x to A(Z) does not create new kernel elements for m I. This is because x s W+I(k,Z) is not in the image of m I by 1.3. - Remark: The reason no problem occurred with S_l: W-I (F2) + W-I(F2 ) is that neither term is in the image of I. NOTATION This is a list of commonly used symbols and abbreviations. A complete definition and description of each symbol is generally given in the text. This list is intended as an index of symbols. Symbol Description Z The ring of integers Q The rational numbers D A Dedekind domain E The quotient field of D An involution on E F The fixed field of E Units in E E Squares in E NE Norms from E F /NE Group of - fixed elements modulo norms O(E) Dedekind ring of integers in E namely D O(F) Dedekind ring of integers in F O(E) Units in O(E) S An order in D P A prime ideal in O(E) M A prime ideal in S P A prime ideal in O(F) O E (P) Local ring of integers at P ~E(P) Completion of O(E) at P 260 Symbol Description I Fractional ideal in O(E) I(P) I localized at P M(P) M localized at P re(P) The localization of P in OE(P) D/P The residue field O(E)/P, also isomorphic to OE (P)/m (P) ~p or Uniformizer for P Zp Uniformizer for P l lp P-adic valuation on E lip P-adic valuation on F Vp Additive version of Ip Vp Additive version of [p (M,B) Inner product space [M,B] Witt equivalence class of (M,B) (M,B,/) Degree k mapping structure 1 Degree k map [M,B,I] Witt equivalence class of (M,B,I) s Right adjoint operator of l *s Left adjoint operator of l AdRB Right adjoint map of B AdLB Left adjoint map of B N R Right orthogonal complement of N N L Left orthogonal complement of N N ~ Orthogonal complement of N W +I (K) Witt equivalence classes of inner product spaces, (M,B) with B symmetric 261 Symbol Description W+I(k,K) Witt equivalence classes of degree k mapping structures with B symmetric A (K) Witt equivalence classes of inner product spaces, (M,B) with no symmetry requirements s The symmetry operator W(k,K) Degree k mapping structures (M,B,/) under Witt equivalence, with the characteristic polynomial of 1 integral A(K) The characteristic polynomial of s is integral Ann M Annihilator of M Ext Cokernel of Hom functor K(F) Monic polynomials, coefficients in F, nonzero constant term GK (F) Grothendieck group associated to K(F) H 2 (k;K (F)) Cohomology group H2(C2;GK(F)) B Basis for H2(k;K(F)) as an F2-vector space T k Involution on K(F) D(E/F) = ~(E/F) Different of E over F A -I (E/F) Inverse different of E over F NE/F The norm map of E over F C Ideal class group char Characteristic deg Degree dim Dimension det Determinant Dis Discriminant 262 Symbol Description rk Rank ker Kernel im Image e~ Ramification index l f. Residue field degree 1 J Fundamental ideal of even rank forms Witt equivalence relation D End of proof Plk p divides k (ipl,P2) = 1 Pl and P2 are relatively prime t, tr, t, Various trace maps t 1 Scaled trace a Boundary homomorphism L # Dual lattice T(M) {P : P~S = M} sgn Signature F q Finite field with q elements D[t-] Polynomials over D D[t,t -1 ] Finite Laurant series over D C Cyclic group with p elements P (g,~lp Hilbert symbol Legendre symbol Q/Z Quotient as a Z-module W(k,F;f) Witt equivalence classes [M,B,s with f(1) = 0 263 S~ol Description W(k,F;D) Witt equivalence classes with a compatible D-module structure (D) restricted to W(k,F;D) ~ (D,P) = ~(P) The localization of ~(D) at P G F F /F N' CA) Norm of ideal [B,S 124] REFERENCES [A,C,H] Alexander, J.P., P.E. Conner and G.C. Hamrick, Odd Order Group Actions and Witt Classification of Inner Products, Lecture Notes in Math. 625, Springer-Verlag, Heidelberg, Germany (1977). 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[Sf-2] Stoltzfus, N.W., The Al~ebraic Relationship Between Quinn's Invariant of Open Book Decomposition and the Isometric Structure of the Monodromy, (to appear). [v] Vick, J., Homology Theory, Academic Press, N.Y. (1973). [w] Warshauer, M.L., Diagonalization up to witt, Pacific Journal of Mathematics, (to appear). [z,s-1] Zariski, O. and P. Samuel, Commutative Algebra, vol. l, Springer-Verlag, Heidelberg, Germany (1958). [Z,S-2] Ibid. Vol. 2, (1960). INDEX adjoint map 16 adjoint operator 31 anisotropic representatives 42 annihilator N L N R 29 boundary 137 cokernel 173 exact sequence 141 local 147, 156 Chinese remainder theorem 222 class of ramified primes 159 cohomology groups 81, 181 completion 48 conductor 203 decomposition theorem 85, 99 Dedekind domain 13 degree k mapping structure 33 anisotropic 42 map 33 metabolic 34 quotient 42 determinant 61 diagonalization 58 different 183 discriminant inner product space 25 map 61 fundamental ideal J 64 F-part 118 going up 48 Grothendieck 81 Hasse 51 Hensel's lemma 53 Hermitian 16 Hilbert reciprocity 54 symbols 51 Theorem 90 57 HomD(M,K) 15 ideals equivalence class [P] 215 inert 49 infinite 49 maximal of S 100 268 ideals norm of 216 prime 48 ramified 49 split 48 strict equivalence class 215 inner product space 15 discriminant 25 skew Hermitian 16 symmetric 16 u Hermitian 16 integrally closed 206 inverse different i01 involution - 13, 226 T k 73, 226 * 227 irreducible inner product space 88 Isotropic (not anisotropic) J fundamental ideal 64 Jacobson's theorem 202 Landherr's theorem 68 lattice 17 dual 136 full 135 integral 135 local degree 52 local differential exponent 183 local ring of integers 13 local uniformizer 14 localization homomorphism 87 localizer 156,158,188 map of degree k 33 mapping structure 33 metabolic 34 metabolizer 34 module structure of HomD(M,K) 15 Nakayama's lemma 28 non-singular map 15 norm 61 orthogonal complement 29 polynomials characteristic 70 minimal 70 type 74 prime ideals (.see ideals) quotient mapping structure 42 269 ramification index e. 48 ramified 49 l rank 54 realization of Hilbert symbols 54 residue field degree f. 48 Scharlau transfer sequence 112 signature 67 stably metabolic 36 Strong approximation theorem 20 symmetry operator s 27 s invariant 27 tensor product 23 trace 198 lemma 90 scaled 97 transfer, Scharlau 112 type of polynomial 74 of ramified prime 181 uniformizer 14,50 valuation 13,50 Witt equivalence relation 34 Vol. 759: R. 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is strictly equivalent to