Principal Component Analysis (PCA) Quadratic Forms Multivariate
QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC 2 … · respectively binary quadratic and ternary...
Transcript of QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC 2 … · respectively binary quadratic and ternary...
QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC 2
by
Lawrence Ervin Gosnell
Thesis submitted to the Graduate Faculty of the
Virginia Polytechnic Institute and State University
in partial fulfillment of the requirements for the degree of
APPROVED:
C. W. Patty
C. J. fa~ry
MASTER OF SCIENCE
in
Mathematics
E. A. Brown, Chairman
G. W. /rofts
R. J. tS'aigle
June, 1973
Blacksburg, Virginia
ACKNOWLEDGEMENTS
I would like to thank
under whose direction this thesis was written, for his
advice and encouragement.
ii
TABLE OF CONTENTS
INTRODUCTION . . . . . . . . . . . . . . . . . . . 1
I. QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC :; 2 • • • • • • • • • • • • • • • • • • • • • 4
II. QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC 2 • • • • • • • • • • • • • • • • 8
III. THE SYMPLECTIC GROUP . . . . . . . . . . . . IV. EQUIVALENCE OF FORMS . . . . . . . . . . . .
BIBLIOGRAPHY . . . . . . . APPENDIX
VITA
ABSTRACT
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24 26
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INTRODUCTION
By a form we shall mean a homogeneous polynomial
with coefficients in a given field. A form is linear if,
for each term of the polynomial, the sum of the powers
of the indeterminants is one. A form is quadratic if
the sum is two. In an n-ic form, the sum of the powers
is n. A form is unary if it involves only one
indeterminant, binary if it involves two indeterminants,
and n-ary if it involves n indeterminants. Thus 3x + y
- 4z is a ternary linear form over the integers.
Similarly, 4x2 + xy - 2y2 and 2xn + x2yn-2 + yn + zn are
respectively binary quadratic and ternary n-ic forms.
Quadratic forms (in fact forms in general) have
been extremely important in the study of number theory.
The two-square, three-square, and four-square problems
are merely questions of the representations of primes 2 2 2 2 2 2 2 2 2 by the forms x + y , x + y + z , and x + y + z + w
respectively. In addition, the properties of binary
quadratic forms over the reals play a large part in the
solutions of these problems. Fermat's Last Theorem
can be stated in terms of forms by saying that the
ternary n-ic form xn + yn ± zn never represents zero
with x, y, z ~ O, n > 2. For more on arithmetic use of
forms see Jones [4].
1
2
Quadratic forms over the integers have been
studied since the time of Gauss. However, as the
twentieth century began the idea of generalization· of many
mathematical concepts became popular. Among the concepts
generalized to arbitrary settings was that of quadratic
forms over the integers. But in generalization problems
arose. With standard definitions, forms over fields of
characteristic 2 had to be ruled out. An authoritative
and often cited source for quadratic forms over fields of
characteristic ~ 2 is the 1937 paper of Ernst Witt [7].
The characteristic 2 case, after some slight adjustments,
turned out to be non-trivial and some new concepts were
developed. One of the most important of these is the
Arf Invariant, first introduced in 1940 by Cahit Arf [l].
In this paper we shall investigate quadratic forms
over fields of characteristic 2. Some sources for back-
ground material on these forms are Artin [2], Dieudonne
[3], and Kaplansky [5]. In Section I we will outline the
generalization of quadratic forms to fields of
characteristic ~ 2. In Section II we complete the necessary
adjustments to make the characteristic 2 case non-trivial.
In passing, we shall mention in Section III some results
on the symplectic group. Finally, in Section IV we shall
investigate the properties of quadratic forms over fields of
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characteristic 2; in particular, we study conditions for
equivalence of such forms.
I. QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC ~ 2
The extension of quadratic forms over the real
numbers to quadratic forms over arbitrary fields of
characteristic ~ 2 is standardly done in the following
manner. Let V be an n-dimensional vector space over a
field F of characteristic ~ 2. In our discussions we shall
limit ourselves to finite dimensional spaces. We then
define a map ( , ): VxV+F such that for x, y, z £ V and
a, S £ F:
1) (x,y) = (y,x)
2) (ax+ Sy,z) = a(x,z) + S(y,z).
This map is called the symmetric bilinear form on V (or
sometimes just the bilinear form on V). This map is very
much like an inner product on V and is called such by
Artin [2] and Kaplansky [5]. We shall follow their example
in referring to the map ( , ). Based upon this map we
define a second map, the quadratic form, Q:V+F by
Q(x) = (x,x).
It is immediate that for x, y, xi £ V and a, S, a1 £ F,
Q has the following properties:
1) Q(ax) = a2Q(x)
2) Q(ax + Sy) = a2Q(x) + s2Q(y) + 2aS(x,y)
4
5
3)
Rewriting 2), we find
(x,y) = l/2(Q(x + y) - Q(x) - Q(y))
which gives a clear indication of some of the difficulties
to arise in quadratic forms over fields of characteristic
2.
We associate a given quadratic form Q for an n-
dimensional space V with a matrix M = (aij) by choosing a
base for V, say {x1 , ... , xn}' and letting (xi, xj) = aij.
Since ( , ) is symmetric, M will also be symmetric and the
matrix derived in the same way by choosing a different base,
say {y1 , ... , yn} will be equivalent to Min the usual
way. So their determinants will be equal up to square
factors. To bypass this problem, we define the determinant
of V (sometimes called the determinant of Q) to be the
canonical image of the determinant of the matrices defined
above in {O} U (F/F2 ) where F is the multiplicative group
of non-zero elements of F. The determinant of V, written
det V or det Q, is then well-defined. It is also possible
to go in the other direction. If V is an n-dimensional
vector space over a field F of characteristic .J 2 and
M = (aij) any nxn symmetric matrix with entries in F, then
we can define a symmetric bilinear form ( , ) on v by
defining (xi, xj) = aij where {xl, . . . , xn} is a base for
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V over F. In either case we say V has the matrix M in the
base {x1 , . , xn}. A vector space V with such a map
Q and its associated map ( , ) is called a quadrafic space.
If det V = 0 we say V is singular. Otherwise V is non-
singular.
A linear map from one quadratic space V into another
quadratic space W which preserves the quadratic form is
called a representation. In other words, a linear map
a:V+W is a representation if for all x E V, QV(x) = QW(crx)
where QV is the quadratic form on V and QW the quadratic
form on W. Whenever there is no confusion as to which
quadratic space and quadratic form Q we are considering,
we shall suppress all subscripts. If a : V+W is a one-to-
one representation into (respectively onto), we call a
an isometry into (respectively onto). If there is an
isometry a from V onto W we say V and W are isometric and
write v~w. If V=W and Qv=Qw, the set of isometries of V
onto V form a group under composition of functions, which
we call the orthogonal group of V for a particular form Q.
The structure of this group is of some interest and has
been investigated in detail [6]. A primary use for
isometries is in discussing the equivalence of forms over
a quadratic space V. If Q1 and Q2 are two different forms
over two quadratic spaces V and W respectively, then Q1 is equivalent to Q2 , written Q1~Q2 , if there is an isometry
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of V onto W. If V=W, this is the same as requiring the
existence of a change of base transformation which
preserves the quadratic form.
In searching for equivalent quadratic forms many
invariants have been found. Among these are:
1) the dimension of the space
2) the range of the form
3) the determinant of the space
4) the spinor norm
5) the Hasse Algebra.
We shall be interested in seeing which of these
invariants carry over to the case where F has characteristic
2. For a more thorough treatment of quadratic forms over
a field of characteristic ~ 2, see chapters 4 and 5 of
O ' Meara [ 6 ] •
II. QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC 2
Hereafter, any field F will have characteristic 2
unless otherwise specified. Let V be an n-dimensional
vector space over F. Using the standard derivations of
( , ) and Q, the fact that F has characteristic 2 causes
difficulties and ambiguities almost from the beginning.
A quadratic space V is called non-trivial if there are at
least two elements x,y in V such that (x,y) f O. As a
first step to proving the existence of an orthogonal base,
a base {x1 , ... , xn} such that (xi, xj) = 0 for if j,
it is usually necessary to prove the existence of a vector
x in V such that Q(x) f 0. Once this is established,
orthogonalization easily follows. The usual proof goes
as follows:
Theorem: A non-trivial quadratic space V contains
a vector x such that Q(x) f O.
Proof: Suppose Q(x) = O for every x E V. Then
Q(x+y) = O for all x, y E V. So 2(x,y) = Q(x+y) - Q(x)
- Q(y) = o.
Now if V is over a field of characteristic f 2 we have the
implication (x,y) = 0 for all x, y E V and so the form on
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Vis trivial, contrary to hypothesis. But if V is over a
field F of characteristic 2 we can draw no such conclusion,
for, given any o. e: F, 2o. = O. So since (x,y) is in F,
2(x,y) = 0 provides absolutely no information about (x,y)
and we cannot reach the contradiction necessary to prove
the theorem.
It therefore becomes apparent that if we hope to
have a non-ambiguous entity to work with we shall have to
make some adjustments in our definitions of ( , ) and Q.
However, we still wish to retain our relationships with
square symmetric matrices. We therefore pick a base
{x1 , .. . , xn} for V, a symmetric nxn matrix M = (aij)
and make the following definition. For x = o.1x1 + ...
+ o.nxn with o.ie:F, define Q:V+F by
Q( x) = Ei~j o.i o.j aij.
Then for o. e: F and x, y, yi e: V, these properties follow:
1) Q(o.x) = o.2Q(x)
2) Q(x+y) = Q(x) + Q(y) + (x,y) defines a symmetric
bilinear form on V.
Induction yields: n n
Q( 1: yi) = 1: Q(y.) + Ei<J.(yi,yJ.). i=l i=l 1
We also retain for our chosen base {x1 , ... , xn}, the
relationships (xi,xj) = aij for i ~ j and Q(xi) = aii"
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Now for each x E V,
which implies
Q(x+x) = 4Q(x) = 0
0 = Q(x+x)
= Q(x) + Q(x) + (x,x)
= (x,x).
So (x,x) = 0 for each x E V. This makes the form we have
defined on V what is called an alternate form. Over a
field of characteristic ~ 2 this form would also be skew-
symmetric, i.e. (x,y) = -(y,x). But in a field of
characteristic 2, -1 = 1 and skew-symmetry degenerates
into symmetry.
If (x,x) = 0 for every x E V, it does not necessarily
follow that Q(x) = 0 for every x E V. Let V be a 2-dimen-
sional space with base {x1 ,x2 } and let
M = Then defining things as above (x1 ,x2 ) = Q(x1 ) = Q(x2 ) = 1.
But we have seen that (x1 ,x1 ) must be 0 which shows that
Q(x) and (x,x) are not always the same, as they were in the
case of fields of characteristic ~ 2. Consider the matrix
N of Vin the base {x1 ,x2}.
N = ~ ) . The two matrices we have associated with V are not only
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different, but also their determinants differ modulo
squares. So in order to avoid confusion, if we speak of a
matrix associated with a form we will mean the matrix of
inner products. Instead of defining the quadratic form
from a matrix we will define the form from a homogeneous
quadratic polynomial. For example, we will consider the
form discussed above as defined by the polynomial x2 + xy
+ y2 rather than the matrix
0 i) We choose the polynomial x2 + xy + y2 since every element
z of V can be written in the form ax1 + Sx2 with a, S £ F
and so 2 2 . Q(z) = a Q(x1 ) + S Q(x2 ) + aS(x1 ,x2 )
= a 2 + aS + s2 .
Substituting x and y for a and S respectively to indicate
that a and S can take on any values in F, we arrive at the 2 2 polynomial x + xy + y . In this way we continue to have
the determinant of the space defined as the determinant of
the matrix of inner products modulo squares, and it is still
an invariant.
We now return to the question of orthogonalization.
If a non-trivial quadratic space with the alternate form
defined above has an orthogonal base, then the matrix of
inner products consists entirely of zeros which implies
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the form is trivial, which is a contradiction. It is possible
to define non-alternate forms over fields of characteristic
2 so that orthogonalization is still possible. But these
forms closely parallel the characteristic ¥ 2 case and are
of little interest here. Thus, whenever we speak of a
quadratic space over a field of characteristic 2 we shall
mean a space with the alternate form defined above.
As before, a representation is a linear function
with certain preservative properties. There is, however,
some ambiguity in the literature concerning the exact
definition of a representation in the characteristic 2
case. A representation can be defined by:
1) Q(x) = Q(crx) for all x E V
2) (x,y) = (crx,cry) for all x, y E V.
As seen in O'Meara [6], these two definitions are equivalent
over fields of characteristic ¥ 2. But in the case of
characteristic 2, this is not necessarily the case. Consider
a 2-dimensional vector space V over F with a quadratic form
derived from the polynomial xy. That is, V has a base
{x1 ,x2} such that Q(x1 ) = Q(x2 ) = 0 and (x1 ,x2 ) = 1. Then
define a : v~v by
cr(x1 ) = x1 + x2 cr(x2 ) = x2 .
Extending a linearly we get a linear map from V to V. Let
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axl + ax2 and yxl + ox2 be arbitrary members of v where
a, a, y, o £ F. Then (ax1 + ax2 , yx1 + ox2 ) = ao + Sy.
Also
(cr(axl + ax2),cr(yxl + ox2)) = (a(xl + x2) + ax2,
y(x1 + x2 ) + ox2 )
= ao(xl + x2' x2)
+ Sy(x2 , x1 + x2 )
= ao + Sy.
This implies that (x,y) = (crx,cry) for all x, y £ V. In
contrast, Q(x1 ) = 0 and
Q(crx1 ) = Q(x1 + x2 )
= Q(xl) + Q(x2) + (xl,x2)
= 1.
So the two definitions are not equivalent here.
From the general definition of Q,
Q(x + y) = Q(x) + Q(y) + (x,y)
and
Q(cr(x + y)) = Q(crx) + Q(cry) + (crx,cry).
If Q(z) = Q(crz) for all z £ V, then
Q(x + y) = Q(crx +cry).
Hence (x,y) = (crx,cry) for all x, y £ V. So preservation of
Q implies the preservation of ( , ) while the converse does
not always hold. We shall define a representation as a
linear map between quadratic spaces which preserves the
bilinear form. Properties which hold under this definition
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will hold under the other definition. In addition, our
definition has the advantage of agreeing (at least in
spirit) with the ideas of representation and isometry in
analysis and topology in which ( , ) is considered as an
inner product.
Now we determine conditions which are necessary and
sufficient, in addition to the preservation of ( , ), for
the preservation of Q. Let V be an n-dimensional quadratic
space with base {x1 , ... , xn}. Then for each x £ V, n
x = E aixi with the ai £ F. Then Q(x) = Q(crx) for all i=l
x £ V if and only if n
Q( E aix.) i=l l
if and only if
n = Q( E ai crxi)
i=l
But a is a representation, so
Ei<jaiaj(xi,xj) = Ei<jaiaj(crxi,crxj).
This implies Q(x) = Q(crx) for all x £ V if and only if for
all ai £ F
n 2 E ai Q(x.) =
i=l l
if and only if for each xi in the base for V
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Q(xi) = Q(crxi).
And so we have the following theorem:
Theorem 2.1: Let V be an n-dimensional quadratic
space over F with base {x1 , ... , xn}' Wan m-dimensional
quadratic space over F, and cr V+W a representation. Then
Q(x) = Q(crx) for each x E V if and only if Q(xi) = Q(crxi)
for each xi in the base for V.
Since a representation cr is a linear transformation,
if cr maps V to V, then cr can be represented as a transforma-
tion matrix between two matrices of V. We shall call the
determinant of the matrix associated with cr the determinant,
det cr, of cr.
A quadratic space V is said to have an orthogonal
splitting into subspaces vl' . , V m if V = V 1 EB . . .
EB Vm and (Vi,V.) = 0 for i "F j. m J
We write v = v1 .L. . . .l.
Vm or V = J_ v. and define ..L <P vi = {O}. A subspace u i=l J.
of V s:elits V if there is a subspace W of V such that
V = U ~ W. Let v1 , ... , Vm be quadratic spaces with
associated bilinear forms which we represent by ( , )i.
Then V = V1 (;9 •
by de fining ( ,
. )
• (:9 v m VxV+F
m ( E xi, i=l
can be made into a quadratic space
such that for xi' Yi E Vi
m m E yi) = E (xi,yi)i.
i=l i=l
16
Since each element of V can be written uniquely as a sum
of elements of the Vi, this map is well-defined. Also with
the bilinear form V = V1 .J. ••• ..l Vm since (Vi,vj·) = 0
for i F j and the bilinear form is unique. In addition,
if Vi has the matrix Mi, then V has the matrix
M =
M1 • • • 0 • M
2
0 • • • M
..L W , and m ai : Vi+Wi is a representation for each i=l, ... , m,
then a:V+W defined by
a(x) m
= a( E x.) = i=l l
m r a.x.
l l i=l
is a representation. We write a = a 1 .l. ... ..1.. am. If
Let U be a subspace of V. Then we define the orthogonal
complement U* of U to be {x E V:(x,U) = O}. We call V* the
radical of V and write V* = rad V. For a subspace U of V,
rad U = Un U*. Vis said to be regular if rad V = {O}.
For the following let V = v1 + . . . + Vm with
(Vi,Vj) = O for i ~ j.
Lemma 2.2: rad V =rad V1 + ... +rad Vm.
17
Proof: Let x be an element of rad V. Then m
x = I: xi with xi E Vi. For i = 1, ... , m: i=l
since (xj,Vi) = 0 for i ~ j. So
(x1 ,Vi) = (x,V1 ) = 0
since (x, V) = o. Thus each xi E rad vi and so x E rad v1
+ . . . + rad v m . Then
m Now let x = I: xi with each xi
i=l
m (x, V) = ( I: xii/)
i=l m
= ( . E xi, V l + . . . i=l
+ v ) m
E rad vi.
+(x,V). m m
All other terms drop out since (xi,Vj) = O for i ~ j.
But xi E rad Vi so (xi,Vi) = O and thus (x,V) = o. So
x E rad V. Therefore rad V = rad v1 + ... + rad Vm.
Corollary 2.2.1: V is regular if and only if all
the Vi are regular.
Proof: {O} = rad V = rad v1 + ... + rad Vm
implies rad Vi = {O} for each i.
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Rad Vi = {O} implies
rad V = rad v1 + .•. + rad Vm
= {0} + ..• + {O}
= {0}.
So V is regular if and only if all the Vi are regular.
Lemma 2. 3: If V is regular then V = V1 ..1.. ••• ..L Vm.
Proof: V = V1 + ... + Vm and (Vi,Vj) = 0 for i ~ j,
so it suffices to show that V = V1 $ ..• W Vm. In other
words, we need only show that if
0 = x = x1 + • + x m with xi E Vi, then xi= 0 for each i = 1, .. , m. So
let 0 = x1 + •.. + xm. Then for each i,
0 = (O,Vi)
= (xl + . . + x m' Vi)
= (xi, Vi)
since (xj,Vi) = 0 for i ~ j. Thus xi E rad vi for each
But since Vis regular, Corollary 2.2.1 says xi = 0 for
each i.
i.
We now wish to relate the regularity of a quadratic
space to its matrix of inner products. A matrix Mis non-
singular if and only if the determinant of M is different
from O.
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Theorem 2.4: Let V be an n-dimensional quadratic
space with base {x1 , ... , xn}. Let M be the matrix of
inner products in this base. Then V is regular if and only
if M is non-singular.
Proof: We shall prove the contrapositive, that V
is not regular if and only if M is singular. Suppose V
is not regular. Then there is an x E V, x ~ O, such that
(x,V) = O. Then
x = cx1x1 +
with cxi E F and not all cxi = 0.
O = (x,x1 )
. + ex x n n
Then for each i,
= cxl(xl,xi) + ... + cxn(xn,xi)
which implies that the rows of Mare linearly dependent.
So M is singular. The converse is proved by reversing the
steps above.
This theorem makes it reasonable to refer to a regular
quadratic space as non-singular and to a non-regular
quadratic space as singular.
Theorem 2.5: If U is a regular subspace of the
quadratic space V, then V = U .J.. U*. If V = U ..L W is
another splitting, then W = U*.
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Proof: (U,U*) = 0 since (x,U) = 0 for each x E U*.
Un U* =. {O} since rad U = UnU* and rad U =. {O}. It remains
only to show that V = U + U*. Let {u1 , ... , un}.be a
base for U. Then we must find a1 E F and u* E U* such
that
x = a1u1 + ... + anun + u*.
Let M = (aij) be the matrix of inner products for U. Then
calculating (x,ui) for each i, we get the following system
of equations:
al(ul,ul) + . . . + an(un,ul) + (u*,u1 ) = (x~u1 ) . . al(ul,un) + . . . + an(un,un) + (u*,un) = ( X, Un).
But (ui,u*) = 0 for each i, so we reduce to the following
system of n equations inn unknowns a1 , ... , an:
alall + · · · + analn = (ul~x) .
where aij = (u1 ,uj). But U is regular so M = (a1j) is non-
singular and the system has a unique solution. Now if
V = U -1.. W, then (U,W) = 0. So W<U*. But V=U J.. U* implies
the dimension of U* equals the dimension of W. So W = U*.
In passing we also note that if U is a regular subspace of
V, then U** = U.
Suppose there is a subspace U of V such that
V = U@ rad V. Then V = U ~rad V since (U,rad V) = O. If
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such a splitting exists, we call it a radical splitting.
Unless rad V = V or rad V =·{a}, it is not necessarily
true that U is unique. Also V = U .J.. rad V implies
rad V = rad U ..J.. rad (rad V).
But
rad (rad V) = rad V
since
(rad V, rad V) = (rad V, V) = O.
So
rad V = rad U .J... rad V
and thus rad U = {0}. Therefore U is regular. We now
proceed to use radical splittings to show we can limit our
examination of quadratic spaces to regular spaces.
Lemma 2.6: If V and W are quadratic spaces with V
regular and cr:v~w a representation, then cr is an isometry.
Proof: Let x E ker cr. Then
(x, V) = (crx,crV)
= ( 0 ,crV)
= o. So x E rad V. But V is regular so x = 0 and cr is one-to-
one.
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Theorem 2.7 (Witt): Let the quadratic spaces V and
v1 have radical splittings U J_ rad V and u1 -L. rad v1 respectively. Then
1) There is a representation a:V+V1 if and only
if there is a representation y:U+U1 .
2) V ~ v1 if and only if U ~ u1 with rad V ~ rad v1 .
Proof: 1) Let y:U+U1 be a representation. Define
T:rad V+rad vl by
T(X) = 0
for all x £ rad V. Then T is a representation and cr ..L T:V+V1 is also a representation.
Let a:V+V1 be a representation. Define y:U+U1 by
cr(x) = y(x) + z
for x £ U, y(x) £ u1 , and z £rad V1 . Then for x, y £ U
(x,y) = (ax,cry)
= (yx + z1 ,yy + z2 )
= (yx,yy) + (yx, z2 ) + (z1 ,yy)
+ (zl,z2)
= (yx,yy).
So y is a representation.
2) Let y:U+U1 and T:rad+V rad v1 be isornetries onto.
Then y ..L T:V+V1 is also an isornetry onto.
23 .
Let cr:V+V1 be an isometry onto. Of necessity, cr carries
rad v onto rad v1 . So rad V ~ rad v1. Now, from part 1)
there is a representation y:U+U1 . But u is regular, so by
Lemma 2.6 y is an isometry. Since rad v ~ rad v1 , y must
be onto and so u ~ u1 .
This theorem does much for us. First, by letting
V = v1 we can see that although the "U" in a radical
splitting may not be unique, it is unique up to isometry.
Second, it allows us to limit our discussions to regular
spaces, which we shall do from this point on. Note that
this does not rule out the possibility of singular sub-
spaces, for a regular space can easily have singular
subspaces.
III. THE SYMPLECTIC GROUP
The set of all isometries of an n-dimensional
quadratic space V onto itself forms a group under
composition of functions. In the case of characteristic
t- 2, the structure on the quadratic space is an
orthogonal geometry and so we call this group the
orthogonal group, On(V), of V. In the characteristic 2
case our quadratic space has a symplectic geometry
structure and hence we call the group of isometries of
V onto itself the symplectic group, Spn(V), of y. As
Artin [2] does a thorough development of the symplectic
group, we shall state the major results here without
proof and then pass on.
Let V be a regular quadratic space. For each
y E v, define cry:v~v by
cry(x) = x + c(x,y)y
where c is a fixed element of F. Then cry is an isometry
onto. If c = o, cr = 1 y v for each y E V. If c t- o, then
cry leaves x fixed if and only if (x,y) = 0. So cry fixes
the elements of (Fy)*. An isometry of the form cry for
some y E V is called a symplectic transvection in the
direction of y.
We now state the following theorems, which
characterize Spn(V).
24
25
Theorem 3.1: Every element of Spn(V) has
determinant 1. The group Spn(V) is generated by the
symplectic transvections.
Theorem 3.2: The center of Spn(V) consists of lv·
These theorems characterize Spn(V) and show that
every isometry from an n-dimensional quadratic space V
onto itself is the composition of isometries of the form
cry(x) = x + c(x,y)y
for some y £ V. Also, if Spn(V) has more than one
element, then Spn(V) is non-commutative.
IV. EQUIVALENCE OF QUADRATIC FORMS
Prior to discussing equivalence of forms, w~ must
develop a little more information about the structure of
quadratic spaces over fields of characteristic 2. Let
V be an n-dimensional quadratic space. We define a
vector x E V to be isotropic if (x,x) = 0 and anisotropic
if (x,x) ~ 0. A space is totally isotropic if (x,y) = 0
for every x, y E V. At times, in the characteristic ~ 2
case, a vector x E V is called isotropic if Q(x) = O.
In this case, since Q(x) = (x,x) the two definitions are
equivalent. In characteristic 2, this is not the case.
Suppose V is a quadratic space in which every vector is
isotropic. If x being isotropic means Q(x) = 0, then for
any x, y E V,
0 = Q(x + y)
= Q(x) + Q(y) + (x,y)
= (x,y).
So V is totally isotropic. But if being isotropic means
(x,x) = O, then every vector in the 2-dimensional quadratic
space with the mat.rix
M = (~ 6) is isotropic. To see this, let {x1 ,x2} be the base for V
in which it has the matrix M; then for ax1 + ax2 an
26
27
arbitrary element of V;
= 0
and (x1 ,x2 ) = 1. So Vis not totally isotropic. So in
the characteristic 2 case, the two possible definitions
are not equivalent. We therefore follow our precedent
and adopt the weaker definition. That is, x E V is
isotropic if (x,x) = 0. Since our form is alternate
(x,x) = 0 for every x E V, and so every vector in a
quadratic space over a field of characteristic 2 is
isotropic.
A space having the matrix
M = is clearly isotropic. A quadratic space which has the
matrix M in some base is called a hyperbolic plane. Thus
all hyperbolic planes are regular, isotropic, and have
determinant 1. Since all hyperbolic planes have the
same matrix, they are all isometric.
Theorem 4.1: Every regular 2-dimensional quadratic
space V is a hyperbolic plane.
Proof: Vis regular so Vis not totally isotropic.
28
and (x1 ,x2 ) = a F O. Otherwise Vis totally isotropic.
Then {x1 ,x2/a} is also a base for V and
= l/a(a)
= 1.
So V has the matrix
M = G 6) in the base {x1 ,x2/a}. So Vis a hyperbolic plane.
A quadratic space V is universal if Q(V) = F. If
H is a hyperbolic plane over a field of characteristic ~
2, then H is universal. We exhibit an example of a non-
universal hyperbolic plane in the following lemma. We
shall make future use of this example.
Lemma 4.2: Let V be a 2-dimensional quadratic space
over z 2 (a) where z2 = {O,l} and a is transcendental over
Z2· Suppose the form Q on V is defined by the polynomial x2 + xy + y2 Then a is not an element of Q(V).
Proof: Suppose there is a z E V so that Q(z) = a.
This means there are x, y E z 2 (a) such that x2 + xy + y2
= a. Then
29
2 since if x = O, then y = a and la e z2 (a) which is impossible.
Similarly y ~ O. Also x I y since 2 x = (y + a)/(x + y).
Now since x, ye z2 (a), x = f (a) grar
h(a) y = JTciT
where f(a), g(a), h(a), j(a) E z2[a] and g(a) ~ 0 ~ j(a).
Then
(r(a)\ 2 + f(a)h(a) + (h(a)) 2 = lit'CiT) g(a)j(a) JTCiT a
Finding a common denominator and letting
f(a) = f(a)j(a)
h(o:) = h(a)g(a)
g(a) = g(a)j(a),
we reduce to the following equation:
f(a) 2 + f(a)h(a) + h(a) 2 = ag(a) 2 .
Also f(a) I h(a) so f(a) 2 ~ h(a) 2 and g(a) 2 ~ 0. So let
P(x) f(x) 2 + f(x)h(x) + h(x) 2 + - 2 = xg( x) .
Then P(x) is a polynomial in z2 [x] with a as a root. We
need only show that P(x) is not the zero polynomial to
arrive at the desired contradiction. We have
30
f(x) n + + f o = x . . . g(x) m + + = x . . . go h(x) k + + ho = x . . .
with f 0, go, ho E z2 . Since these polynomials are over a
field of characteristic 2, squaring them just squares
each term. So f(x) 2 , g(x) 2 , and h(x) 2 have first terms 2n 2m 2k x , x , and x respectively and all powers of x in each
polynomial are even. Then xg(x) 2 has only odd powers of
x, and so f(x)h(x) must cancel all the even powers of
f(x) 2 + h(x) 2 ~ 0. The first term of f(x)h(x) is xn+k.
Suppose n<k. Then
2n<n+k<2k.
So nothing cancels the term x2k and P(x) ~ O. A similar
argument holds if k<n.
This leaves only the case n = k. Then
2n = n+k = 2k
and so the highest even power of P(x) is x2n + xn+k + x2k = x2k
Again P(x) ~ 0. But if P(x) ~ O, then a is algebraic over
z2 , which is a contradiction. Therefore a is not an
element of Q(V).
We have seen that not all hyperbolic planes are
universal. However any forms equivalent to the zero form
and any forms over a perfect field are universal.
31
We now proceed to obtain the closest possible analogue
to an orthogonal base. To do this we first need a lemma.
Lemma 4.3: Every regular quadratic space Vis
split by a hyperbolic plane.
Proof: Let x be any vector of V, x ~ O. Then x is
isotropic. Since Vis regular, there is a y E V such that
(x,y) = a ~ 0. Then H = Fx + Fy
is a regular isotropic 2-dimensional subspace of V. So
H is a hyperbolic plane, and since H is regular, H splits
V by Theorem 2.5.
Theorem 4.4: Every regular n-dimensional quadratic
space V has a splitting V = H1 .1... ••• ...L.Hr where n = 2r and
each Hi is a hyperbolic plane. Furthermore, this splitting
is unique up to isometry.
Proof: Since V is regular, by Lemma 4.3, V is split
by a hyperbolic plane, say H1 . So
V = H1 J... H1 *. Then since Vis regular, Corollary 2.2.1 implies that H1* is regular and of dimension n-2. We then proceed by
induction to derive
V = H1 ..l.. ••• ..l.Hr.
2r = n since at each step, H1* is regular and so split by
a hyperbolic plane or H1* = {O}. Since all hyperbolic
32
planes are isometric, this splitting is unique up to
isometry.
Thus if V is a regular n-dimensional quadratic space,
then V has a base {x1 ,y1 , ... , xr"~} where 2r = n. Then
V has the nxn matrix
M =
0 1 • • • 1 0
0 1 1 0
0
·o 1 0 • . • 1 0
in this base. We call such a base a symplectic base. This
result can easily be extended to regular quadratic spaces
with alternate forms over fields of characteristic ~ 2
(see Kaplansky [5]).
M =
The matrix then assumes the form:
0 1 • • • 0 -1 0
·o 1 0 ... -1 0 .
Note that Theorem 4.4 implies that all regular quadratic
spaces with alternate forms over any field are even-
dimensional. Also every regular quadratic space over a
field of characteristic 2 has determinant 1.
We have now reached the point where we can begin to
consider the equivalence of quadratic forms. Let V and v1 be regular quadratic spaces with forms Q and Q1 respectively.
33
We shall say that Q is equivalent to Q1 , or what will mean
the same thing, V is equivalent to v1 , if there is an isometry
a from V onto v1 such that Q(x) = Q(crx) for each r E v. We
notice that the statement that v is equivalent to vl is the
same thing as the statement that V is isometric to v1 under the stronger definition of isometry discussed in
Section II. If Q is equivalent to Q1 , we write Q~Q1 or
v~v1 . If Q and Q1 are different forms on the same
quadratic space V, then the statement that Q~Q1 is the
same as the statement that there is a change of base for V
so that Q agrees with Q1 with respect to the new base .
That is, let {x 1, . , xn} be a base for V; then
and
If Q~Q1 , we must be able to find a new base {y1 , ... , yn}
for V such that
and if ( , ) and ( , )1 are the bilinear forms associated
with Q and Q1 respectively, then
(yi,yj) = (xi,xj)l.
Let V be a 2-dimensional quadratic space. Then Q on
V is defined by some binary quadratic polynomial, say
34
ax2 + bxy + cy2 with a, b, c E F, b ~ O. This means, if
{xl,x2} is a base for v and axl + ax2 an arbitrary element
of V, then 2 2 Q(axl + ax2) = aa + baa + ca .
Now (x1 ,x2 ) = b and {x1 ,x2/b} is also a base for V. In
fact, since
= l/b(b)
= 1,
{x~2/b} is a symplectic base for V and
Q(x2/b) = (l/b) 2Q(x2 )
= c/b2
= d.
Then if yx1 + ox2/b is an arbitrary element of V1 then
Therefore any binary quadratic form is equivalent to a form
defined by a polynomial with 1 as the coefficient of the
xy term. We shall therefore consider only forms generated 2 2 by polynomials of the form ax + xy + by . We shall
abbreviate this form by [a,b]. The form generated by xy,
we symbolize by [O,O] and call the zero form.
Now suppose V is regular and n-dimensional. Then
V is an orthogonal sum of hyperbolic planes and, since the
35
hyperbolic planes are orthogonal, the form on V is the
orthogonal sum of the forms on the hyperbolic planes.
That is,
where [ai,bi] is the form on Hi and V = H1 _L. . • ..L Hr'
2r = n. Then Q is generated by the polynomial 2 + + 2 + 2 + x2y2 + b2y2
2 + alxl xlyl blyl a2x2 . .
All other terms disappear due to orthogonality.
.
In discussing equivalence of forms, we wish to find
a system of invariants. Since we can have only a
symplectic base for a quadratic space V, the spinor norm
and Hasse Algebra are no longer invariants, as we lack the
orthogonal base prerequisite to the construction of the
necessary Clifford Algebras and Hasse Algebras [6]. From
our definition of equivalence, the dimension of the quadratic
space and the range of the quadratic form are useful
invariants. The determinant of V is also an invariant but
since every regular quadratic space has determinant 1, it
is of no use to us here. So we study an analogue to the
determinant, developed by Cahit Arf [l], called the Arf
Invariant. In doing this study, we follow the development
of Kaplansky [5]. Throughout this development, all forms
36
considered will be regular forms on quadratic spaces over
fields of characteristic 2.
Lemma 4.5: [a,b] J_ [c,d] ~ [c,d] ...L [a,b].
Proof: We have a symplectic base {x1 ,y1 ,x2 ,y2}
for V such that,
and
and
Q(x1 ) = a
Q(yl) = b
Q(x2 ) = c
Q(y2) = d
(xl,yl) = (x2,y2) = 1
Q(u1 ) = c
Q(v1 ) = d
Q(u2 ) = a
Q(v2 ) = b
(u1 ,v1 ) = (u2 ,v2 ) = 1
37
Let
ul = x2
vl = Y2
u2 = xl
v2 = Y1·
Then [a,b] 1... [c,d] "' [c,d] J_[a,b].
From the symmetry of the form, it is clear that
[a,b] "' [b,a]. If necessary, this fact can be proven in
the style of Lemma 4.5. In fact, proof of equivalence
using the definition will follow this method, and hence
will be completed in slightly less detail. Also, whenever
we prove two forms equivalent, we shall implicitly assume
they are forms on the same quadratic space.
Lemma 4.6: [a,b] ..L [c,d]"' [a+ c,d] J_ [a,b + d].
with
Q(ul) = a
Q(vl) = b
Q(u2) = c
Q(v2) = d.
38
Then {u1 + u2 ,v2 ,u1 ,v1 + v2} is also a symplectic base
for V and
= a + c + O
= a + c.
Similarly
Q(v1 + v2) = b + d.
So [a,b] _L [c,d] "' [a + c,d] J_ [a,b + d].
Lemma 4.7: If Q is a regular binary quadratic form
which represents zero non-trivially, then Q"' [O,O].
Proof: Since Q represents zero non-trivially, there
is a non-zero vector u E V such that Q(u) = O. Complete
u to a symplectic base {u,v}. Then (u,v) = 1 and Q(v) = a.
Then {u,w} where w = v + au is also a symplectic base
since
Also
(u,w) = (u,v + au)
= (u,v) + a(u,u)
= 1 + 0
= 1.
Q(w) = Q(v + au)
= Q(v) + a2Q(u) + a(v,u)
39
= o. So Q rv [O,O].
Corollary 4.7.1: [a,O] rv [O,O] rv [O,S] for all
a, a E F.
As a further example of this technique of proving
equivalence, we offer the following lemma:
Lemma 4.8: 2 [a,a] rv [a ,l] for any a E F.
Proof: {u,v} is a symplectic base for V such that
(u,v) = 1 and Q(u) = Q(v) = a. Then {s,t} where
s = au + (a + l)v
and
t = u + v
is also a base for v. (s,t) = (au + (a + l)v, u + v)
= a(u,u) + a(u,v) + (a +
+ (a + l)(v,v)
= 0 + a + (a + 1) + 0
= 1.
So {s,t} is a symplectic base for V.
Q(s) = Q(au + (a + l)v)
l)(v,u)
40
= a2Q(u) + (a + 1) 2Q(v) +
a(a + l)(u,v)
= a2 (a) + (a + 1) 2 (a) + a2 + .a
= a3 + a3 + a + a 2 + a
= 2 a .
Q(t) = Q(u + v)
= Q(u) + Q(v) + (u,v)
=a+a+l
= 1. 2 So [a,a] ~ [a ,l].
group.
Since F is a field, it is also an additive abelian
Let P = {x2 + x : x E F}. Then since F has
characteristic 2,
which implies P is a subgroup of F. What we shall be
interested in is the factor group
F0 = F/P.
Lemma 4.9: If [a,b] ~ [c,d] then ab+ P = cd + P.
Proof: V is a quadratic space with symplectic base
{u,v} such that Q(u) = a, Q(v) = b, and V has a second
symplectic base {s,t} such that Q(s) = c, Q(t) = d.
Suppose
Then the matrix
41
s = au + Sv
t = yu + ov.
represents the transformation from· {u,v} to.{s,t}. Now
1 = (s,t)
= (au + Sv,yu + ov)
= ay(u,u) + ao(u,v) + Sy(v,u) + So(v,v)
= ao + Sy
= det A.
So we know that A can be written as a product of elementary
matrices. So there are three cases we need consider:
Case 1. s = u, t = av. Then
1 = (s,t)
= (u,av)
= a(u,v)
= Cl. •
So t = v.
Case 2. s = v, t = u. Then Q(s)
surely ab = ba.
Case 3. s = u + av, t = v. Then
d = Q(t) = b
and
c = Q(s)
= b, Q(t) = a, and
Then
42
= Q(u + av)
= Q(u) + a2Q(v) + a(u,v) 2 = a + a b + a.
ab + cd = ab + (a + a2b + a)b
= ab + ab + Cl2b2 + ab
= (ab) 2 + ab e: p.
So ab + P = cd + P.
In the binary case, we will call the canonical
image of ab in Fa the Arf Invariant of the form [a,b].
The preceding lemma makes it reasonable to call this an
invariant. We would now like to extend the Arf Invariant
to an arbitrary regular n-dimensional quadratic space in
the following manner. If V is a 2r-dimensional quadratic
space, then a regular quadratic form on V has the form
[a1 , b1 ] J_ • • • .....L [ar, br]. We define the Arf Invariant
of this form to be the canonical image in Fa of a1b1 +
... + arbr. Before proving this to be an invariant, we
reduce our calculations to the 4-dimensional case.
Lemma 4.la: Let V be a regular n-dimensional
quadratic space with two symplectic bases {u1 ,v1 , ... ,
ur,vr} and {s1 ,t1 , ... , sr' tr}. Then it is possible
43
to pass from the first base to the second base by a
sequence of changes each affecting only two pairs of
ui, vi. Such changes we shall call dyadic changes·.
Proof: We shall prove the lemma by induction on r
where 2r = n is the dimension of V.
The case r = 2 is trivial.
Assume the lemma is true for r = k and let V have dimension
2k + 2. Then V = H1 ..L .
base {ui,vi}. Then
with ai,bi E Hi.
Case 1. Suppose for some i,
ai = bi = o.
-1.Hk+l where each Hi has a
Then without loss of generality,
Let W = H1 ...L ••• ..l.Hk. Then s 1 , t 1 E W and we can
complete s 1 , t 1 to a base for W and by induction move that
base to the base for W, {s 1 ,t1 ,u2 ,v2 , ... , uk,vk} by a
series of dyadic changes. Thus we can move from
{ul,vl, ... , uk+l'vk+l} to {sl,tl,u2,v2, .,uk+l'vk+l}
by a series of dyadic changes. Let Z = H2 ...L • • • _L Hk+l.
44
Then Z has bases {u2 ,v2 , ... , uk+l'vk+l} and.{s 2 ,t2 ,
... , sk+l'tk+l} and by induction we can move from one
to the other by a series of dyadic changes. And ·so
the lemma follows.
Case 2. Using the orthogonality of the Hi, we note that,
1 = (s1 ,t1 )
= (al,bl) + · · · + (ak+l'bk+l)
implies that at least one of the (ai,bi) ~ O. We may
assume (a1 ,b1 ) = a ~ O. Write D = H1 i. H2 and define
q = bl + b2.
Suppose (p,q) = o ~ o. Then (p,q/o) = 1 and.{p,q/o}
form a symplectic base for a hyperbolic plane which can
be completed to a symplectic base for D. A dyadic change
is possible from the base {u1 ,v1 ,u2 ,v2} to this new
base.
Then
sl = P + ~3 + · · · + ak+l
tl = q + b3 + . . . + bk+l
and we reduce to Case 1.
Suppose (p,q) = O. Then (a2 ,b2 ) = a and there is a dyadic
change from the base {u1 ,v1 ,u2 ,v2} to the symplectic base
{a1 + a 2 , b1/a, a2/a, b1 + b2 }. There is a third
symplectic base in which a 1 + a 2 and b1 + b2 are elements
45
of the same hyperbolic plane of D. After a dyadic change
to this base, we again reduce to Case 1. And so the lemma
is proved.
Theorem 4.11: The Arf Invariant is an invariant.
Proof: By Lemma 4.10, we need only consider the case
where Vis a 4-dimensional regular quadratic space. Let
V = H1 J_ H2 have a symplectic base· {u1 ,v1 ,u2 ,v2} and let
[a,b] ..L [c,d] be the form on V in this base. So the Arf
Invariant of this form is ab + ed. Let s 1 , t 1 start a
second symplectic base and, as in the preceding lemma,
write
We note that from Lemma 4.9, once we have a symplectic
base for a hyperbolic plane, we may replace it with any
other symplectic base for that hyperbolic plane without
affecting the Arf Invariant. Also if H is a 2-dimensional
regular quadratic space and x, y E H such that (x,y) = 1,
then {x,y} is a symplectic base for H.
Case 1. (a1 ,S1 ) = 1. Then {a1 ,s1} is a symplectic base
for Hi and so we can replace it with {u1 ,v1 }. Suppose
a2 = 0. Then if s2 = O,
sl = ul
tl = vl.
46
We then complete s 1 , t 1 to a base for V by choosing
s2 = u2
t2 = v2.
Then the Arf Invariant is ab + ed. If $2 ~ O, then we
can complete s2 to a base for H2 and then replace s2 by u2 . Then
sl = ul
tl = vl + u2
and we complete sl' tl to a symplectic base for V by
choosing
s2 = u2
t2 = ul + v2.
Then
Q(sl) = a
Q(tl) = b + c
Q(s2) = c
Q(t2) = a + d.
So the Arf Invariant is
a(b + c) + c( a + d) = ab + ac + ca + cd
= ab + ed.
47
and since cx2 I- 0 we can complete it to a symplectic base
for H2 and then replace cx 2 by u2 . Then
t 1 = v1 + yu 2 .
We then complete s 1 , t 1 to a symplectic base for V by
choosing
Then
e = Q(sl) = a + c
f = Q(tl) = b + y2c
g = Q(s2) = c
h Q(t2) 2 + y + b + d = = Y a
and ef + gh is our representative for the Arf Invariant
and
(ab + cd) + (ef + gh) = ab + cd + (a+ c)(b + y2c)
+ c(y2a + y + b + d)
= ab + cd + ab + ay 2c + cb
+ y2c2 + cy2a + cy + cb + cd
= (yc) 2 + ye E P.
48
So (ab + cd) + P = (ef + gh) + P. A symmetric argument
proves the case (a2 ,S 2) = 1.
Case 2. (a1 ,e1 ) = y ;i O, 1. Then (a2 ,e2 ) = 1 + y· ~ O, 1.
So {a1 ,e1/y,a2 ,e2/(l +y)} is a symplectic base for V.
Changing bases as before, we get the symplectic base
{s1 ,t1 ,s2 ,t2 } such that
Then
t 1 = yv1 + (1 + y)v2
s 2 = (1 + y)u1 + yu2
Q(s 1 ) = a + c
Q(tl) = y 2b + (1 + y) 2d
Q(s 2 ) = (1 + y) 2a + y 2c
Q(t2) = b + d.
Here our representative of the Arf Invariant is
(a+ c)(y2b + (1 + y) 2d) + ((1 + y)2a + y2c) (b
= ay2b + a(l + y) 2d + cy2b + c(l + y) 2d + (1 +
ab + (1 + 2 y) ad + y2cb + y2cd 2 + (1 + y) 2cd + (1 + y) 2ab + y2cd = y ab
= ab(y2 + (1 + y)2) + cd(y2 + (1 + y)2)
= (ab + cd)(y2 + (1 + y)2)
+ d) y)2
49
= (ab+ cd)(y2 + 1 + y2 )
= ab + ed.
So the Arf Invariant is invariant.
We now know that, if Vis any n-dimensional regular
quadratic space over a field of characteristic 2 with two
quadratic forms Q and Q1 , then Q ~ Q1 implies Q and Q1
have both the same determinant and the same Arf Invariant.
Over certain fields of characteristic 2, we get a stronger
result. A given form on a quadratic space V is equivalent
to itself no matter what base is chosen for V. So if V
has the form [a,b] in the symplectic base {u,v} and a is
a square in V, then by replacing {u,v} with {u/la,la v}
we get an equivalent .form [l,c] where c = ab. This result
leads to the following lemma and theorem and also gives
us a hint as to the location of non-equivalent forms with
the same Arf Invariant.
Lemma 4.12: If a+ b E P, then [l,a] ~ [l,b].
Proof: a + b E P implies 2 b = a + z + z
for some z E F. Let {u,v} be the symplectic base in which
the space has the form [l,a]. Then'{s,t} is also a base
where
50
s = u
t = zu + v.
Then
(s,t) = (u,zu + v)
= z(u,u) + (u,v)
= 1
and {s,t} is a symplectic base. Hence
Q(s) = 1
Q(t) = z 2Q(u) + Q(v) + z(u,v) 2 + a + z = z
= b.
So [l,a] ~ [l,b].
Theorem 4.13: .If V is a regular n-dimensional
quadratic space with two forms Q = [a1 ,b1 ] _L ••• _L
[ar,br] and Q1 = [c 1 ,d1 J J_ .•• J_ [cr,dr], 2r = n, and
.each ai and ci is a square in F, then Q~Q1 if and only if
their Arf Invariants are equal.
Proof: The necessity is a direct result of Theorem
4.11. For the sufficiency, we rely on the discussion
preceding Lemma 4.12 and repeated use of Lemma 4.6. Then r
[a1 ,b1 J J_ .•. J_[ar,br]~[l, I: a.bi] J_ [O,O] ..l.. i=l J.
•.. J_ [O,O]
and
Since
51
r [c1 ,d1 J J_ .•• J_ [cr,dr],\,[1, E cidi] _L[O,O] J_
i=l ••• _L [O,O].
r E aibi +
i=l
r .E cidi i=l
e: p' r r
then by Lemma 4.12, [l, E aib.J~[l, E c.di]. i=l l. i=l l.
Corollary 4.13.1: Two forms on a regular quadratic
space over a perfect field of characteristic 2 are
equivalent if and only if their Arf Invariants are equal.
Proof: This follows from the theorem since in a
perfect field of characteristic 2, every element is a square.
Every finite field and every algebraically closed
field is perfect, so we now have a set of necessary and
sufficient conditions for equivalence over every finite
or algebraically closed field of characteristic 2. Consider
the perfect field z2 = {O,l}. Then the four possible
regular binary quadratic forms on a 2-dimensional quadratic
space V over z2 are [O,O], [O,l], [l,O], and [1,1]. From
Corollary 4.13.1,
since each has Arf Invariant O. Also
52
[l,l]~[O,O]
since 1 ~ O. Therefore the only regular binary quadratic
forms over z2 up to equivalence are [1,1] and [O,OJ.
Expanding this we see that any regular form on a quadratic
space V over z2 is equivalent to an orthogonal sum of
copies of [1,1] and [O,O]. By Lemma 4.6,
[1,1] ~ [l,l]'V[O,l] _L [l,O]
'V[O,O] _J_ [O,O].
So a regular form over z2 is an orthogonal sum of copies
of [O,O] plus (possibly) one copy of [1,1]. For a
listing of all binary quadratic forms over certain other
finite fields of characteristic 2, see the Appendix.
We now show that the Arf Invariant is not always
sufficient for equivalence of forms. Consider the field
z2 (a) where a transcendental over z2 . We showed in
Lemma 4.2 that the form [1,1] over z2 (a) does not represent
a. That is, there is no vector u in V, a regular binary
quadratic space with base {s,t} over z2 (a), such that
Q(u) = a. The form [1,1] has Arf Invariant 1. The form
[a,l/a] also has Arf Invariant 1. If {u,v} is the
symplectic base for V in which V has the form [a,l/a],
then Q(u) = a. So [l,l]~[a,l/a] over z2(a) since there is
a transformation of {s,t} into another base· {w,z} such
that Q(w) = a. So the Arf Invariant is not always
53
sufficient. Notice that z2 (a) is not perfect since la is not an element of z2 (a).
Returning to the binary case, we notice that over
z2 no form [a,$] with a~ 0 ~ a is equivalent to [O,O].
Over z2 , the only form of this type is [1,1]. It is only
natural to ask when it is possible to extend this result
to other fields of characteristic 2. However, the
answer to our question is "never," as the following
lemmas show.
Lemma 4.14: Let V be a regular quadratic space over
a field F of characteristic 2; suppose there is an a E F
such that a ~ O, 1. Then [a, a + l]~[O,O].
So
Proof: Let {u,v} be a symplectic base for V. Then
Q(u) = a
Q(v) = a + 1
(u,v) = 1.
Q(u + v) = Q(u) + Q(v) + (u,v)
= a + (a + 1) + 1
= o. So by Lemma 4.7, [a,a + 1]~[0,0].
54
We note also that any forms [a,(z2 + z)/a] or
[Sz,(z + l)/aJ are equivalent to [O,O] where a, a E F,
a ~ 0 ~ a, and z E F. We observe that a field F of
characteristic 2 contains an a ~ O, 1 if and only if F
Corollary 4.14.1: If F is a field of characteristic
2, and a, a E F such that a ~ 0 ~ a, then [a,aJ~[O,O] for
all such a, a E F if and only if F = z2 .
Restating the corollary, we get the following.
Corollary 4.14.2: For any field F of characteristic
2 such that F ~ z2 , there are a, a E F, a ~ 0 ~ a such that
[a,aJ~[o,oJ.
We would now like to develop a set of necessary and
sufficient conditions for two quadratic forms over any
field of characteristic 2 to be equivalent. We will also
develop a technique to determine the equivalence of binary
quadratic forms.
Theorem 4.15 (Witt's Cancellation Theorem): If
V = U ..LU* and v2 = W J.. W* and U ~ W, then U* ~ W* 1 .
55
provided U and W are regular and v1 = v2 .
Proof: It is sufficient to prove the theorem in the
case that U and W are hyperbolic planes, since each regular
space is a sum of hyperbolic planes and all hyperbolic
planes are isometric. So let U < V1 and W < V2 be hyperbolic
planes with V1 = V2 . Then
V l = U .l H1 _L • • • .J_ Hn J_ I
and
V 2 = W .l. K1 .l.. • • • J_ Km .J.. J
where I and J are totally isotropic and Hi = Ki for each
i = 1, ... , min (m,n). We claim m = n and I= J. Now
and
V 2 = ( W ...L K1 _L • • • _J_ Km) _J_ J
are radical splittings of v1 and v2 respectively. Then
since V1 = V2 , Theorem 2.7 implies
and
I = J.
Hence 2n + 2 = 2m + 2 and m = n. We have now proved our
claim. But
56
and
with I= J and Hi= Ki for each i = 1, ... , n. So
U* = W*.
Theorem 4.16: Let q_, u2 , w1 , w2 be regular
quadratic spaces such that ul "' wl and for vl = ul _l_ u2
and V2 = W1 _L W2 , V1 = V2 . Let a 1 :u1~w1 be the isometry
which makes u1 "'w1 and let a2 :u2~w2 be the isometry proven
to exist in Theorem 4.15, and suppose v1 "' v2 by a = a 1 J_
a2 • Then u2 "' W2 by cr 2 .
Proof: Since u2 "' W2 it is sufficient to prove
that for all u2 £ u2
Q(u2) = Q(a2u2).
For each x £ V, x can be represented uniquely by
x = u1 + u2
where u1 £ lJ_, and u2 £ u2 . Notice that by picking
appropriate x £ V, we can involve every u2 £ u2 in an
expression of this type. By the definition of cr,
cru1 = cr1u1
Then
57
Q(x) = Q(ul + u2)
= Q(ul) + Q(u2) + (ul,u2)
and
Q(crx) = Q(cru1 + cru2 )
= Q(crlul + cr2u2)
= Q(crlul) + Q(cr2u2) + (crlul,cr2u2) ·
But since V1 'V V2 by cr,
Q(x) = Q(crx)
and since u1 'V w1
Q(ul) = Q(crlul).
So
Q(u2) + (ul,u2) = Q(cr2u2) + (crlul,cr2u2)
= Q(cr2u2 ) + (cru1 ,cru2 ).
But since a is an isometry
(ul,u2) = (cru1 ,cru2 )
and so
Q(u2) = Q(cr2u2)
Hence u2 'V w2.
Corollary 4.16.1: [a,b]'V[c,d] if and only if
[a,b] J.. [c,d]rv[O,O] J_ [O,O].
58
Proof: [a,b]~[c,d] if and only if
[a,b] l. [c,d]~[a,b] J_ [a,b]
~[a + a,b] J_ [a,b + b]
~[O,b] J_ [a,O]
~co,oJ J. co,oJ.
Corollary 4.16.2: Let Q1 = [a1 ,,b1 J _l ••• _L [ar,br]
and Q2 = [c1 ,d1 ] ..L .•• _L [cr,dr]. Then Q1 ~ Q2 if and
only if Q1 ..L Q2 is equivalent to the orthogonal sum of 2r
copies of [O,O].
Proof: This follows by repeated use of Lemma 4.6 and
Corollary 4.16.1.
Theorem 4.17: Let Q1 = [a,b], Q2 = [c,d], and
Q = Q1 ..LQ2 . Then Q is a regular form on a 4-dimensional
quadratic space V. If there are u1 , w1 E V,, u1 F 0 F w1 ,
u1 ~ w1 , such that
Q(u1 ) = Q(w1 ) = Q(u1 + w1 ) = O,
then Q ~ [ 0, 0] J_ [ 0, 0].
Proof: (u1 ,w1 ) = Q(u1 + w1 ) + Q(u1 ) + Q(w1 ) = O·
Now there is a v if (Fu1 + Fw1 ) such that (u1 , v) = a F 0.
Otherwise V is totally isotropic which is impossible. If
59
(v,w1 ) = O, then {u1 ,v1 } where v1 = v/a is a symplectic
base for a hyperbolic plane H1 < V such that (w1 ,H1 ) = 0.
If (v,w1 ) = S ~ O, then.{u1 ,v1 } is still a symplectic base
for H1 , but we must replace w1 with
w * 1 = w1 + Su1/a.
Then
Q(w1*) = Q(wl) + (S/a) 2Q(u1 ) + (S/a)(w1 ,u1 )
= 0
and
Q(w1 * + u1 ) = Q(w1 + (1 + (S/a))u1 )
= Q(w1 ) + (1 + (S/a)) 2Q(u1 )
+ (1 + (S/a))(u1 ,w1 )
= a. So w1* satisfies the hypothesis in conjunction with u1 .
We have shown (u1 ,w1 *) = 0 so
(v1,w1*) = (v/a,wl + Sul/a)
2 = l/a(v,w1 ) + Sia (v,u1 )
= l/a(S) + S/a2 (a)
= o. From this point on, we identify w1* with w1 . * So w1 e: H1 which is also a hyperbolic plane and we complete w1 to a
symplectic base.{w1 ,z} for H1* such that Q(z) = 0 using
60
Lemma 4.7. Similarly we get a base.{u1 ,v*} for H1 such
that Q(v*) = O. Then.{u1 ,v*,w1 ,z} is a symplectic base
for V such that
So [a,b] J_ [c,d]'V[O,O] J_ [O,O].
Corollary 4.17.1: Let Q = [a,b] _l_ [c,d] be a form
on a quadratic space V. Then [a,b]'V[c,d] if there are
u, w E V, u ~ w, u ~ 0 ~ w, such that
Q(u) = Q(w) = Q(u + w) = o.
We will now illustrate the use of Corollary 4.17.1
by verifying the results of Lemma 4.8. Let V be a 2 4-dimensional quadratic space with the form [a,a] J_ [a ,l].
Then V has a base.{u1 ,vpu2 ,v2} such that for a, B, y, o E F
Q(au1 + Sv1 + yu2 + ov2 ) = aa2 + aB + aB 2 + a 2y 2
+ yo + o2
If a = O, then
Q(au1 + Sv1 + yu2 + ov2 ) = aB + yo
and
If a ~ O, let
Then
and
Also
and
Hence
61
Q(x) = a(l) 2 + (1)(0) + a(0) 2 + a2 (1) + ·(l)(a) + a2
= a + a2 + a + a2
= 0
Q(y) = a(a2 ) + (a)(a) + a(a2 ) + a2 (0) + (O)(a) + a2
= a3 + a2 + a3 + a2
= 0.
Q(x + y) = a(a + 1) 2 + (a + l)(a) + a(a) 2
+ a2(1) + (1)(0) + 02
a3 + a + a2 + a + a3 + 2 = a
= o. 2 [a,a]"'[a ,l].
We would now like to characterize the non-equivalent
forms over certain fields F of characteristic 2. If F is
finite, then cr:F~P defined by 2 cr(x) = x + x
is a homomorphism of F onto P with kernel {O,l}. Hence
[F:P] = 2 by La Grange's Theorem and F0 has exactly 2
elements, and so over a finite field there are exactly
2 non-equivalent forms.
62
Lemma 4.18: If Fis a finite field, 1 E P if and k only if F has 2 elements where k is even.
Proof: Every finite field F of characteristic 2
has 2k elements. 1 E P if and only if for some x E F
x2 + x = 1
if and only if
x2 + x + 1 = 0 2 has a solution. x + x + 1 = 0 has a solution if and only
if x3 -1 = 0 has a solution other than 1 if and only if
F contains a primitive cube root of unity if and only if
F has order 2k where k is even.
Corollary 4.18.1: If V is a binary quadratic space k over a finite field F of order 2 where k is odd, then
[O,O] and [1,1] are the nonequivalent forms one V.
When k is even, we have shown that 1 E P and we
cannot get such a characterization. Attempts at further
characterizations in the cases where k is even or F is an
infinite field are recommended. We close with one final
set of necessary and sufficient conditions for
equivalence of binary quadratic forms.
63
Lemma 4.19: Let V be a regular binary quadratic space
over a finite field F with binary quadratic forms Q and Q1 .
Then Q ~ Q1 if and only if either both Q and Q1 represent
zero non-trivially or neither Q nor Q1 represent zero
non-trivially.
Proof: Clearly if a form is equivalent to [O,O],
it represents zero non-trivially. Suppose Q ~ Q1 and Q
represents zero non-trivially. Then by Lemma 4.7 Q ~ [O,O] which implies Q1 ~ [O,O] and Q1 represents zero
non-trivially. If Q does not represent zero non-trivially
and Q1 does, then Q1 ~ [O,O] which implies Q ~ [O,O]
which is a contradiction. Conversely if Q and Q1 both
represent zero non-trivially, then they are both
equivalent to [O,O], and hence equivalent. If neither
represents zero non-trivially, then neither is equivalent
to [O,O]. But V has only 2 non-equivalent forms, so
Q ~ Ql.
BIBLIOGRAPHY
1. Arf, Cahi t, "Unterschungen i.iber quadratische Formen in Korpern der Charakteristik 2 (Teil I)," J. filr reine und angew. Math., Vol. 183 (1940), pp.-148-167 .- --
2. Artin, Emil, Geometric Algebra, Interscience Publishers, Inc., New York, 1957.
3. Dieudonne, J., La geometrie des groupes classiques, Ergebnisse der Mathematik und ihrer Grenzgebiete, Neue Falge, Heft 5, Springer, Berlin, 1955.
4. Jones, Burton, The Arithmetic Theory of Quadratic Forms, Carus Monograph no. 10, The Mathematical Association of America, New York, 1950.
5. Kaplansky, Irving, Linear Algebra and Geometry; ~ Second Course, Allyn and Bacon, Inc., Boston, 1969.
6. O'Meara, O. T., Introduction to Quadratic Forms, Springer, Berlin, 1963.~
7. Witt, Ernst, "Theorie der quadratischen Formen in beliebigen Korpern," J. fur reine und angew. Math., Vol. 176(1937)~ pp:-31-44.
64
APPENDIX
65
Table I
~l
0 0 1
1 1 0
p = {0}
F - F 0 -
Possible Forms (up to symmetry)
[O,O]
[O,l]
[1,1]
66
f:-:1
0 0 0
1 0 0
ae Arf Invariant
0 0
0 0
1 1
67
Table II
F = {O,l,a,b}
+ 0 1 a b 0 1 a b
0 0 1 a b 0 0 0 0 0
1 1 0 b a 1 0 1 a b
a a b 0 1 a 0 a b 1
b b a 1 0 b 0 b 1 a
p = {O,l}
Fa = {P, a + P}
Possible Arf Invariants - O, a
Possible Forms al3 Arf Invariant (uE to s;ymmetr;y)
[O,O] 0 0
[0,1] 0 0
[O,a] 0 0
[O,b] 0 0
[1,1] 1 0
[l,a] a a
[l,b] b a
[a,a] b a
[a,b] 1 0
[b ,b] a a
68
Table III .. ,
F = {O,l,a,b,c,d,e,f}
+ 0 1 a b c d e f 0 1 a b c d e f
0 0 1 a b c d e f 0 0 0 0 0 0 0 0 0
1 1 0 c f a e d b 1 0 1 a b c d e f
a a c 0 d 1 b f e a 0 a b c d e f 1
b b f d 0 e a c 1 b 0 b c d e f 1 a
c c a 1 e 0 f b d c 0 c d e f 1 a b
d d e b a f 0 1 c d 0 d e f 1 a b c
e e d f c b 1 0 a e 0 e f 1 a b c d
f f b e 1 d c a 0 f 0 f 1 a b c d e
P =·{o,a,b,d}
F0 = {P, 1 + P}
Possible Arf Invariants - O, 1
Possible Forms (uE to s;:tmmetr;:t) a.a Arf Invariant
[O,O] 0 0
[O,l] 0 0
[O,a] 0 0
[O,b] 0 0
[O,c] 0 0
[O ,d] 0 0
[O,e] 0 0
69
Table III (cont)
Possible Forms (uE to s~mmetr~) as Arf Inva:riant
[O,f] 0 0
[1,1] 1 1
[l,a] a 0
[l,b] b 0
[l,c] c 1
[l,d] d 0
[l,e] e 1
[l,f] f 1
[a,a] b 0
[a,b] c 1
[a,c] d 0
[a,d] e 1
[a;e] f 1
[a,f] 1 1
[b,b] d 0
[b' c] e 1
[b ,d] f 1
[b ,e] 1 1
[b, f] a 0
[c,c] f 1
[c,d] 1 1
[c ,e] a 0
70
Table III 9cont)
Possible Forms (u;e to s~mmetr~::) a(3 Arf Invariant
[c,f] b 0
[d,d] a 0
[d,e] b 0
[d,f] c 1
[e ,e] c 1
[e, f] d 0
[f ,f] e 1
The vita has been removed from the scanned document
QUADRATIC FORMS OVER FIELDS OF CHARACTERISTIC 2
by
Lawrence Ervin Gosnell
(ABSTRACT)
This thesis is concerned with the study of
quadratic forms over fields of characteristic 2. First,
we consider the extension of quadratic forms to fields
of characteristic I 2. Then we make the adjustments
necessary to make the characteristic 2 case non-trivial,
and investigate the structure of quadratic spaces.
We define equivalence of quadratic forms and
investigate a set of invariants. We develop a set of
necessary and sufficient conditions for equivalence
and give a characterization of non-equivalent forms
over certain finite fields.