quadratic forms¸ clifford lgebras and spinors

Max-Albert Knus

Quadratic Forms,

Cli�ord Algebras and Spinors

IMECC-UNICAMP

1988

Max-Albert Knus

ETHMathematikR�amistr. 101CH{8092 Z�urichSwitzerland

AMS Subject Classi�cation (1985 Revision): 11E04, 11E16, 11E20, 11E39,11E57, 11E81, 11E88, 15A63, 15A66, 15A90, 20G15, 20G45

Foreword

These notes grew out of a course given at UNICAMP for two months between August and October1987. One aim of the course was to present the basic theory of quadratic forms and Cli�ord alge-bras over a �eld in a characteristic free way. In even dimension we restrict attention to nonsingularforms. In odd dimension nonsingularity implies that the characteristic must be di�erent from 2.An elegant way to avoid this restriction is to work with 1

2{regular forms. Such a form is the or-

thogonal sum of a 1{dimensional form and of a nonsingular form of even dimension. The 1

2{regular

forms behave almost as nicely as nonsingular forms and they are also de�ned in characteristic 2.A typical example in dimension 3 is given by the elements of reduced trace zero in a quaternionalgebra. The notion of 1

2{regularity is due to Kneser (see his G�ottingen lecture notes mentioned

in the bibliography at the end of these notes).Another purpose of the lectures was to give a detailled study of forms of low dimension, low

meaning � 6. The Cli�ord algebra gives much information for these forms. In particular theCli�ord algebra can be used to describe quite explicitly the spin group, the Lie algebra of theorthogonal group and the group of special similitudes of a form. In dimensions 4, 5, and 6 we haveapplied methods developed in joint papers with Parimala and Sridharan. These methods also givea di�erent approach to results of Dieudonn�e (Acta Math. 87, 1952). Moreover they do not dependon the characteristic of the �eld. The main tool is a reduced pfaÆan, introduced independentlyby Fr�ohlich, Jacobson and Tamagawa. Its role is very similar to the role of the reduced norm of aquaternion algebra in the study of forms of dimension 3 and 4. A general theory of the pfaÆan andapplications to quadratic forms of rank 6 over commutative rings can be found in the papers withParimala and Sridharan mentioned above. This theory of forms of rank 6 is related to the theoryof central simple algebras of dimension 16. In particular it yields new proofs of some theorems ofAlbert.

These notes have two appendices. In the �rst one we summarize, in a chart, all the provenresults on the structure of Cli�ord algebras in low dimensions. The second gives some concreteapplications. We introduce spinors, following Chevalley, and compute explicitly the spinor repre-sentations of four Cli�ord algebras occuring in physics: the Pauli algebra, the Minkowski algebra,the Majorana algebra and the Dirac algebra.

Even if this course deals with forms over �elds, it was sometimes convenient (and even nec-essary) to consider forms over commutative rings. Moreover readers, familiar with the techniquesof commutative algebra, will observe that the argument, used many times, to prove a result bypassing to some �eld extension, comes from descent theory.

We make the following conventions. The ground �eld is denoted by K. Vector spaces arealways �nite dimensional. Algebras are associative with 1 (if not explicitly mentioned) and K isidenti�ed with K � 1. Unadorned tensor products are taken over K. The group of units of analgebra is denoted by A�. Maps are written on the left, like functions. We use a unique numberingin each chapter for Examples, Lemmas, Propositions, Theorems and Corollaries. The bibliographyonly contains books and papers mentioned in the notes. We did not try systematically to givecredits for the results or the proofs used.

iv Foreword

I am very grateful to A. Paques, UNICAMP, who made my visit to Campinas possible. Heattended my course with patience, read the notes very carefully and made many suggestions (andcorrections). I also would like to thank P. H�ansli, who helped me a lot in correcting the �nalversion, R. Parimala and R. Sridharan, who read a preliminary version and Elda Mortari for herexcellent script with LATEX. Of course I am still responsible for any mistakes remaining. My visitto UNICAMP was �nanced by grants from FINEP and FAPESP. I thank these institutions andUNICAMP for their support.

Contents

1 Quadratic Forms 1

2 Central Simple Algebras 22

3 Involutions on Central Simple Algebras 35

4 The Cli�ord Algebra 56

5 Invariants of Quadratic Forms 67

6 Special Orthogonal Groups and Spin Groups 79

7 Quadratic Forms of Dimension 2 89



10 The PfaÆan 117



A A Chart of Results 162

B Spinors 166

v

Contents

Foreword

Chapter 1. Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Chapter 2. Central Simple Algebras . . . . . . . . . . . . . . . . . . . . . . . 17

Chapter 3. Involutions on Central Simple Algebras . . . . . . . . 26

Chapter 4. The Cli�ord Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 40

Chapter 5. Invariants of Quadratic Forms . . . . . . . . . . . . . . . . . 48

Chapter 6. Special Orthogonal Groups and Spin Groups . . . 57

Chapter 7. Quadratic Forms of Dimension 2 . . . . . . . . . . . . . . . 65



Chapter 10. The PfaÆan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85



Appendix A. A Chart of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

Appendix B. Spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

Chapter 1

Quadratic Forms

This chapter gives basic facts about quadratic forms which hold over �elds in-

dependently of the characteristic. The proof of Theorem 10 (Witt cancellation) fol-

lows Wagner (see also the notes of Micali{Revoy). Useful references are the books of

Bourbaki, Chevalley, Lam, Scharlau, Baeza, Demazure{Gabriel and Micali{Revoy,

as well as the paper of Bass on Cli�ord algebras.

Let K be a �eld. A quadratic form is a pair (V; q), where V is a �nite dimen-

sional vector space over K and q is a map q : V ! K such that

1) q(�x) = �2q(x); � 2 K; x 2 V ;2) the map bq : V � V ! K de�ned by bq(x; y) = q(x + y) � q(x) � q(y) is

bilinear.

We call bq the polar of q.

Let b : V � V ! K be a symmetric bilinear form on V . The map

h : V ! V � = HomK(V;K); h(x) = b(x; �); x 2 V;

is called the adjoint of b. If we identify V with V �� through the map x 7! x��; x��(f) =

f(x); f 2 V �, we have h�(x) = b(�; x); where h� : V �� ! V � is the transpose of h,

i.e. h�(x��)(y) = x��(h(y)) = h(y)(x). Hence h� = h; since b is symmetric. We say

that b is nonsingular and that the pair (V; b) is a symmetric bilinear space if h is an

isomorphism. A quadratic form q : V ! K is called nonsingular and the pair (V; q)

is called a quadratic space if the polar bq is nonsingular. We denote its adjoint by

hq. If the characteristic of K is not equal to 2, the polar bq determines q, since

q(x) = 12bq(x; x); x 2 V;

2 1. Quadratic Forms

and the theory of quadratic spaces is identical with the theory of symmetric bilinear

spaces. The two theories are quite di�erent in characteristic 2. In these notes we

shall restrict to quadratic spaces.

A morphism of quadratic forms

' : (V; q)! (V 0; q0)

is a K{linear map ' : V ! V 0 such that q0('(x)) = q(x) for all x 2 V . Similarly a

morphism of bilinear forms

' : (V; b)! (V 0; b0)

is a K{linear map ' : V ! V 0 such that b0('(x); '(y)) = b(x; y) for all x; y 2 V .

A morphism ' : (V; b) ! (V 0; b0) (resp. (V; q) ! (V 0; q0)) must be injective if b

(resp. q) is nonsingular. Morphisms, which are isomorphisms, are called isometries.

The group of isometries of (V; b) (resp. (V; q)) is denoted by O(V; b) (resp. O(V; q))

and is called the orthogonal group of (V; b) (resp. (V; q)). An isometry of quadratic

forms induces an isometry of the associated polar forms but not conversely (if char

K = 2), see Example 2.

Let q : V ! K be a quadratic form and K � L a �eld extension. We can

extend the form q to a form qL : L V ! L by putting

qL(Xi

�i vi) =Xi

�2i q(vi) +Xi<j

�i�jbq(vi; vj):

The polar of qLis 1 bq and its adjoint is 1 hq : L V ! L V � = (L V )�. It

is usual to denote qLby 1 q and (LV; 1 q) by L (V; q). Obviously L (V; q)

is nonsingular if, and only if, (V; q) is nonsingular.

Let fe1; : : : eng be a basis of V and let

ai = q(ei); aij = bq(ei; ej); 1 � i; j � n:

The form q is determined by the ai and aij, since

q(x) =Xi

aix2i +

Xi<j

aijxixj for x =Xi

xiei:

1. Quadratic Forms 3

In matrix notation, we have q(x) = �t��, �t denoting the transpose of �, where

� =

[email protected]

1CCA and � =

0BBBB@a1 a12 � � � a1n0 a2 � � � a2n... 0

. . ....

0 � � � 0 an

1CCCCA :

Conversely, for any (n � n)-matrix �, we can de�ne a quadratic form q on V by

putting q(x) = �t�� for x =Pi xiei and � = (x1; : : : xn)

t.The matrix � is not

uniquely determined by q. We have

�t�� = �t�0�; 8� 2 Kn , � = �0 + ; alternating:

We recall that an alternating matrix is a matrix of the form

=

0BBBB@0 c12 � � � c1n

�c12 0 � � � c2n...

.... . .

...�c1n �c2n � � � 0

1CCCCAi.e. is antisymmetric and has zeroes on the diagonal. A matrix is alternating if

and only if = Æ� Æt for some matrix Æ. Thus, a basis fe1; : : : eng of V being �xed,

q is determined by the class [�] of � modulo alternating matrices and each such class

corresponds to a quadratic form. Moreover in each class [�] there is a triangular

matrix. We say that � and �0 are equivalent if [�] = [�0] i.e. if �� 0 is alternating.

If � is as above, the matrix � = (bq(ei; ej)) of the polar is

� =

0BBBB@2a1 a12 � � � a1na12 2a2 � � � a2n...

.... . .

...a1n a2n � � � 2an

1CCCCA = � + �t

and bq(x; y) = �t�� with � = (x1; : : : ; xn)t; � = (y1; : : : ; yn)

t if x =Pxiei and

y =Pyiei. Obviously � depends only on the class [�] of �. Let fe�1; : : : ; e�ng be the

dual basis of V �, i.e. e�i (ej) = Æij. We get

hq(ej) =Xi

aije�i

so that (V; q) is nonsingular if and only if det� 6= 0. We call det� the discriminant of

q with respect to the basis fe1; : : : ; eng and denote it by d(e1; : : : ; en). Let fe01; : : : ; e0ngbe another basis of V and let

ej =Xi

uije0i:


Let � be the matrix (uij) and let � 0 = (bq(e0i; e

0j)). We have

� = �t� 0�:

Let K�2 be the set of squares in K�. The class of d(e1; : : : ; en) modulo K�2 does not

depend on the choice of the basis fe1; : : : ; eng. We denote this class by disc(q) and

call it the discriminant of q. Similarly, if ' : (V; q)! (V 0; q0) is an isometry and

'(ej) =Xi

uije0i

for bases fe1; : : : ; eng of V , resp. fe01; : : : ; e0ng of V 0, we have � = �t� 0�, putting � =

(bq(ei; ej)); � 0 = (bq0(e0i; e

0j)) and � = (uij). The condition � = �t� 0� is necessary

and suÆcient for the existence of an isometry of the bilinear spaces (V; bq); (V0; b0q)

but is not suÆcient for an isometry of the quadratic spaces (V; q) and (V 0; q0). If y =

'(x); x 2 V , let y = Pi yie

0i; x =

Pi xiei; � = (y1; : : : ; yn)

t and � = (x1; : : : ; xn)t,

so that � = ��. Assume that q(x) = �t�� and q0(y) = �t�0� for matrices �, resp. �0.

We have

�t�� = �t�0� = �t�t�0�� for all � 2 Kn:

Therefore � 2 GLn(K) de�nes an isometry if and only if [�] = [�t�0�]. It follows

that isometry classes of quadratic spaces are in 1� 1{correspondence with equiva-

lence classes [�] of matrices � 2Mn(K) such that � + �t is nonsingular.

Remark 1. The interpretation of a quadratic space as the class [�] of a matrix �

modulo alternating matrices was already applied in Klingenberg{Witt. This inter-

pretation has led to important generalizations of the notion of a quadratic space (see

the papers or books by Bak, Bass, Tits, Wall and Scharlau). We shall try to give an

idea of these generalizations by some examples. Let R be a ring with an involution

r 7! r, i.e. an automorphism of the additive group of R such that 1 = 1; rs = s r

and r = r for all r; s 2 R. Further let " = �1 and let � be an additive subgroup of

R such that

(1) fr � "r; r 2 Rg � � � fr 2 R j r = �"rg,(2) for all r 2 R; r�r � �.


Further, let

�"n = f� 2Mn(R) j � =

0BBBB@�1 a12 � � � a1n

�"a12 �2. . .

�"a1n �n

1CCCCA ; aij 2 R; �i 2 �g:

For any matrix � = (aij), we put �� = (aij)

t. It follows from the properties (1) and

(2) of � that

2 �"n ) �� 2 �"n for all � 2Mn(R):

We de�ne now an equivalence relation � � �0 on Mn(R) by

� � �0 , �� 0 2 �"n

and denote the equivalence class by [�]. We say that [�] is an (";�){quadratic

form on Rn and that [�] is nonsingular if � = � + "�� is invertible (observe that �

depends only on the class [�] of �). An isometry [�]! [�0] is by de�nition a matrix

� 2 GLn(R) such that [�] = [��0�]. We remark that an (";�){quadratic form on

Rn is not, as in the classical theory, a quadratic map Rn ! R. But we can associate

to [�] a map

q : Rn ! R=�; q(�) = class of �� mod �:

We now consider di�erent special cases of (";�){quadratic forms. Let �min =

fr � " r; r 2 Rg and �max = fr 2 R j r = �" rg.1) R a �eld with trivial involution, " = 1 and � = �min: quadratic forms,

� = �max: symmetric bilinear forms.

2) R a �eld with trivial involution, " = �1 and � = �min: alternating forms,

� = �max: antisymmetric bilinear forms.

3) R a skew �eld with a nontrivial involution, " = 1 � = �min : even hermitian

forms, � = �max: hermitian forms.

More exotic cases are possible. We give two examples (without claiming that

the corresponding theories have applications!)

4) R a �eld of characteristic 2 with trivial involution, " = 1 (!) and � = (R2;+),

the additive group of squares in R.

5) R a commutative ring with trivial involution, " = �1 and � an ideal of R


such that 2R � �.

We now give an important example of a quadratic space:

Example 2. Let V be a 2{dimensional vector space over K with a basis fe; fg. We

de�ne a quadratic form on V by

q(xe+ yf) = xy or q(�) = �t

0 10 0

!� for � =

xy

!:

The polar bq has matrix (0110), hence (V; q) is nonsingular. We observe that (V; q) '

(V; �q) for any � 2 K�, since 1 00 �

! 0 10 0

! 1 00 �

!=

0 �0 0

!:

The quadratic space (V; q) is called the hyperbolic plane and is denoted by H(K).

We have

O(V; bq) =

8<: x yu v

!2 GL2(K) j

x yu v

!t 0 11 0

! x yu v

!=

0 11 0

!9=;=

( x yu v

!2 GL2(K) j 2xu = 2vy = 0; uy + xv = 1

)so that

O(V; bq) = SL2(K) =

( x yu v

!2 GL2(K) j det

x yu v

!= 1

)

if char K = 2. On the other hand

O(V; q) =

( x yu v

!2 GL2(K) j xu = vy = 0; uy + xv = 1

):

Thus O(V; q) is strictly contained in O(V; bq) in characteristic 2.

Let (V1; q1); (V2; q2) be quadratic forms. We de�ne the orthogonal sum

(V; q) = (V1 ? V2; q1 ? q2)

as V = V1 � V2 and q(x) = q1(x1) + q2(x2) for x = (x1; x2) 2 V; xi 2 Vi: We

have obviously

disc(q1 ? q2) = disc(q1) � disc(q2):


In particular q1 ? q2 is nonsingular if, and only if, q1; q2 are nonsingular. Let (V; q)

be a quadratic form (not necessarily nonsingular) and let U � V be a subspace. We

denote the restriction of q to U by qjU and we set

U? = fx 2 V j bq(x; y) = 0; 8y 2 Ug:

Lemma 3. Let (V; q) be a quadratic form and let U � V be a subspace such that

(U; qjU) is nonsingular. Then

(V; q) = (U; qjU) ? (U?; qjU?):

Proof. It clearly suÆces to show that V = U � U?. Let h : V ! V � be the

adjoint of q and let hjU : U ! U� be its restriction to U . By hypothesis it is an

isomorphism. Let now v 2 V . The restriction of h(v) 2 V � to U� is in the image of

hjU since hjU is an isomorphism. Thus there exists u 2 U such that b(x; u) = b(x; v)

for all x 2 U . This means that v� u is an element of U? or that V = U +U?. The

fact that U \ U? = f0g follows from the fact that qjU is nonsingular.

Important examples of quadratic spaces are hyperbolic spaces. The hyperbolic

planeH(K) was de�ned in Example 2. A quadratic space isometric to the orthogonal

sum of n hyperbolic planes is called a hyperbolic space of dimension 2n and is denoted

by H(Kn). The space H(Kn) is given by the class [(001n0 )], where 1n is the n� n{

identity matrix. There is another, more intrinsic, way to de�ne hyperbolic spaces.

Let V be a vector space of dimension n over K. We de�ne a quadratic form on

H(V ) = V � V �

by

q(x; f) = f(x); x 2 V; f 2 V �:

Choosing a basis fe1; : : : ; eng of V and taking the dual basis in V �, it is easy to see

that H(V ) is isometric to the hyperbolic space H(Kn). Any K{linear isomorphism

' : V ~!V 0 induces an isometry

H(') = ('; '��1) : H(V ) ~!H(V 0):

Further, we have canonical isometries H(V �W ) ' H(V ) ? H(W ).


Let (V; q) be a quadratic space. Then (V;�q) is also a quadratic space. We have

Proposition 4. (V; q) ? (V;�q) ' H(V ).

Proof. If (V; q) is given by the class [�], then q ? �q is given by the class [(�00��)].

We have to �nd an invertible (2n� 2n){matrix � such that��t � 00 ��

!��=�

0 1n0 0

!�:

Such a matrix is given by

� =

1 �1� �t

!�1

(use that ( 0��

�t

0 ) is alternating and that [�] = [�t]).

Corollary 5. Any quadratic space is an orthogonal summand in a hyperbolic space.

Let (V; q) be a quadratic space. An element e 6= 0 2 V is called isotropic if

q(e) = 0 and a subspace U of V is called totally isotropic if qjU is identically zero.

For example the subspace V of the hyperbolic space H(V ) is totally isotropic. Con-

versely, we have

Proposition 6. Let U be a totally isotropic subspace of a quadratic space (V; q).

There exists a morphism ' : H(U) ! (V; q) whose restriction to U is the identity.

In particular (V; q) contains an orthogonal summand isometric to H(U).

Proof. The embedding U ! V induces a surjective map V � ! U�. Composing

with the isomorphism hq : V ~!V � we obtain a surjective map t : V ! U� given by

v 7! bq(v;�)jU . Its kernel is U?, i.e. the sequence

0! U? ! Vt! U� ! 0

is exact. Let W be a direct summand of U? in V such that tjW is an isomorphism

W ~!U�. It follows that the dual map U ! W �, which is given by u 7! bq(u;�)jW is

also an isomorphism. Since U � U?; U \W = f0g and U;W generate a subspace

of V which is a direct sum U �W . The restriction of bq to U �W is given by a bloc


matrix

� =

0 �t1�1 Æ

!

if we choose a basis fe1; : : : ; e2ng of U �W such that fe1; : : : ; eng is a basis of U

and fen+1; : : : ; e2ng is a basis of W . Since � is of the form � + �t; Æ is of the form

� + �t and we can choose for � the matrix

� =

0 0�1 �

!:

If we make a change of basis in U �W given by a bloc matrix � = (10x1), we get

�0 =

0 0�1 �1x + �

!

for the new matrix. This does not change the basis of U . Since �1 is invertible, we

may choose x = ��11 �. Therefore we can assume that � = 0 or � = (0�1

00). We now

replace � by (00�t10 ), which di�ers from � by the alternating matrix ( 0

��1

�t10 ). We

recall that �t1 is the matrix of tjW : W ~!U�. Thus (100

(�t1)�1) de�nes an isomorphism

' : U � U� ~!U �W � V . Since 1 00 ��1

1

! 0 �t10 0

! 1 00 (�t1)

�1

!=

0 10 0

!;

' is a morphism H(U)! V of quadratic spaces and 'jU is the identity.

Corollary 7. Let (V; q) be a quadratic space of dimension 2m. If U � V is a totally

isotropic subspace of dimension m, then (V; q) ' H(U).

Corollary 8. Let (V; q) be a quadratic space and let e 2 V be an isotropic element.

There exists an isotropic element f 2 V such that bq(e; f) = 1. In particular V

contains a hyperbolic plane as an orthogonal summand.

A pair fe; fg of elements of V is called a hyperbolic pair if e; f are isotropic

and bq(e; f) = 1.

Let (V; q) be a quadratic form. We say that x 6= 0 2 V is anisotropic if q(x) 6= 0

and that (V; q) is anisotropic if all x 6= 0 2 V are anisotropic.


Corollary 9. Let (V; q) be a quadratic space. There exists a decomposition

(V; q) = (V0; q0) ? H(Kn)

with (V0; q0) anisotropic.

The decomposition is unique up to isometry, in view of the following Witt can-

cellation theorem and we call n the Witt index of (V; q).

Theorem 10 (Witt cancellation).. Let (V1; q1); (V2; q2) and (V 0; q0) be quadratic

spaces such that (V1; q1) ? (V 0; q0) ' (V2; q2) ? (V 0; q0). Then (V1; q1) ' (V2; q2).

Before proving Theorem 10, we remark that Witt cancellation does not hold

for bilinear spaces in characteristic 2. We have

� =

0B@ 1 0 00 0 10 1 0

1CA =

0B@ 1 1 01 0 11 1 1

1CAt0B@ 1 1 0

1 0 11 1 1

1CAso that 13 and � de�ne isometric bilinear spaces over IF2. But (

10

01) and (01

10) are

not isometric.

Proof of Theorem 10. By Corollary 5 and induction, we may assume that we

have two orthogonal decompositions

(V1; q1) ? H1 = (V; q) = (V2; q2) ? H2

of a quadratic space (V; q) with H1; H2 hyperbolic planes. Let fe1; f1g be a hyper-

bolic pair in H1 and fe2; f2g a hyperbolic pair in H2. In view of the next propo-

sition there is an isometry ' of (V; q) such that '(e1) = e2; '(f1) = f2. Since

(V1; q1) = H?1 and (V2; q2) = H?

2 ; ' is an isometry (V1; q1) ~!(V2; q2).

Proposition 11. Let (V; q) be a quadratic space and fe1; f1g; fe2; f2g be hyper-

bolic pairs in V . There exists an isometry ' of (V; q) such that '(e1) = e2; '(f1) =

f2.

We give some de�nitions before proving Proposition 11. We say that two hy-

perbolic planes H1; H2 in (V; q) are adjacent if there exists a hyperbolic pair fe1; f1g


in H1 and a hyperbolic pair fe2; f2g in H2 with a common element. We say that H1

and H2 are related if there is a �nite chain of adjacent hyperbolic planes connecting

H1 and H2.

Lemma 12. Two hyperbolic planes H1; H2 in a quadratic space (V; q) are always

related.

Proof. Let fe1; f1g; fe2; f2g be hyperbolic pairs in H1; H2. If one of the 4

elements bq(e1; e2); bq(e1; f2); bq(f1; e2); bq(f1; f2) is not zero, say bq(e1; e2), we

replace the pair fe2; f2g of H2 by the pair fe3; f3g; e3 = e2bq(e1; e2)�1; f3 =

f2bq(e1; e2). Then H1; H2 are adjacent to the hyperbolic plane generated by the

hyperbolic pair fe1; e3g. If all of the above 4 elements are equal to zero, then

fe1; f1+e2g; ff2; f1+e2g are hyperbolic pairs generating hyperbolic planes H3 and

H4. The chain H1; H3; H4; H2 shows that H1 and H2 are related.

Proof of Proposition 11. In view of Lemma 12, we may assume that e1 = e2.

Let f = f2 � f1 and let U = Kf . If bq(f; f1) 6= 0, then V = Kf1 � U?. The

map ' : V ! V given by '(�f1 + w) = �(f1 + f2) + w is an isometry such that

'(f1) = f2 and '(e1) = e1. If bq(f; f1) = 0, then q(f) = 0 and f 2 H?1 . Let

v 2 H?1 such that ff; vg is a hyperbolic pair in H?

1 and let H3 be the hyperbolic

plane generated by ff; vg. The linear map u1 : H1 ? H3 ! H1 ? H3 given by

e1 7! e1; f1 7! f1 + f; f 7! f and v 7! v � e1 has the matrix

� =

0BBB@1 0 0 00 1 1 00 0 1 0�1 0 0 1

1CCCAwith respect to the basis fv; f; f1; e1g. Since

�t

0BBB@0 1 0 00 0 0 00 0 0 10 0 0 0

1CCCA � =0BBB@

0 1 1 00 0 0 0�1 0 0 10 0 0 0

1CCCA �0BBB@

0 1 0 00 0 0 00 0 0 10 0 0 0

1CCCA ;we see that u1 is an isometry of H1 ? H3 such that u1(e1) = e1 and u1(f1) = f2.

Since

(V; q) ' H1 ? H3 ? (H1 ? H3)?;


u1 can be extended to an isometry u = u1 ? 1 with the wanted property.

If the characteristic of K is not equal to 2, there is a much simpler proof of

the Witt cancellation theorem. We need the notion of a diagonalizable form. Let

V = Kn with the canonical basis fe1; : : : ; eng and let a1; : : : ; an be elements of K.

We denote by ha1; : : : ; ani the quadratic form q : V ! K given by

q(ei) = ai; i = 1; : : : ; n; bq(ei; ej) = 0; i 6= j:

A quadratic form (V 0; q0) is called diagonalizable if (V 0; q0) ' ha1; : : : ; ani for some

a1; : : : ; an 2 K. Obviously

ha1; : : : ; ani ' ha1i ? � � � ? hani:

Lemma 13. If char K 6= 2, any quadratic form is diagonalizable.

Proof. This is clear if q(v) = 0 for all v 2 V . Assume that q(v1) = a1 6= 0 and

let U = Kv1. Since char K 6= 2; (U; qjU) is nonsingular and the claim follows from

Lemma 3 and induction on the dimension.

Thus to prove Witt cancellation for quadratic spaces over a �eld of characteristic

not equal to 2, it suÆces to show that

(V1; q1) ? hai ' (V2; q2) ? hai ) (V1; q1) ' (V2; q2):

This, in turn, is a consequence of

Proposition 14. Let (V; q) be a quadratic space over a �eld K of characteristic not

equal to 2 and let x; y be elements of V such that q(x) = q(y) 6= 0. There exists an

isometry ' of V such that '(x) = y.

To prove Proposition 14 we need the notion of a re ection. Let u be an

anisotropic element in a quadratic space (V; q) (over any �eld K). The map �u :

V ! V de�ned by

�u(x) = x� bq(x; u)

q(u)u


is called a re ection. Obviously �u(u) = �u and �u = 1 on (Ku)?. We let it as an

exercise to check that �u is an isometry of (V; q) and that � 2u = 1.

Proof of Proposition 14. If q(x) = q(y) we have the equality

bq(x + y; x+ y) + bq(x� y; x� y) = 4bq(x; x):

So either x� y or x + y is anisotropic. If x� y is anisotropic, then �x�y(x) = y. If

x+ y is anisotropic, then �x+y�x(x) = y.

Corollary 15. O(V; q) is generated by re ections if char K 6= 2.

Proof. Since (V; q) is a quadratic space over a �eld K with char K 6= 2, there exists

x 2 V anisotropic. Let ' 2 O(V; q) and let y = '(x). By Proposition 14, there

exists a product of re ections � such that �(y) = x, so (�')(x) = x. Let U = (Kx)?.

We have V = U ? Kx, since x is anisotropic (and char K 6= 2 !) and �'jU is an

isometry of U . By induction on the dimension, �'jU is a product of re ections in

U . These re ections can be extended to re ections on V (by taking the identity on

x). This proves the claim.

Remark 16. Corollary 15 is a weak form of the theorem of Cartan{Dieudonn�e.

The corollary holds also in characteristic 2, except for (V; q) = H(IF2) ? H(IF2),

where the switch of the 2 hyperbolic planes is not a product of re ections. The proof

in the general case is quite complicated (see Dieudonn�e, Sur les groupes classiques).

If the characteristic ofK is not equal to 2, any quadratic space is the orthogonal

sum of 1{dimensional spaces. In general, we have the weaker (and best possible) re-

sult that a quadratic space of even dimension is an orthogonal sum of 2{dimensional

subspaces. For a; b 2 K, let [a; b] denote the quadratic form (K2; q) with

q(e1) = a; q(e2) = b and bq(e1; e2) = 1;

where fe1; e2g is the canonical basis of K2. The form [a; b] is nonsingular if, and

only if, 1� 4ab 6= 0.


Lemma 17. Any quadratic space (V; q) of even dimension 2m is isometric to an

orthogonal sum of m spaces [ai; bi] with 1� 4aibi 6= 0; i = 1; : : : ; m.

Proof. Assume �rst that char K = 2. Since (V; q) is nonsingular, there exist

x; y 2 V such that bq(x; y) = � 6= 0. Replacing y by 1�y, we have bq(x; y) = 1:

Let a1 = q(x); b1 = q(y). Since d(x; y) = 4a1b1 � 1 = 1, we see that: (1) x; y

are linearly independent; (2) the restriction of q to Kx � Ky is nonsingular. The

claim then follows by Lemma 3 and induction on the dimension. If char K 6= 2, it

suÆces to show that a 2{dimensional diagonal space ha; bi; a � b 6= 0, is isometric to

a space [a1; b1] with 1 � 4a1b1 6= 0. Let x1; x2 2 V with q(x1) = a; q(x2) = b and

bq(x1; x2) = 0. Let x3 = x2 +12ax1. We get

bq(x1; x3) = 1; q(x3) = b + 14a

and 1� 4a(b+ 14a) = �4ab 6= 0

so that ha; bi ' [a; b + 14a].

In characteristic 2 a quadratic space cannot have odd dimension, since the

matrix of its polar form is alternating and an (n � n){alternating matrix has zero

determinant if n is odd. To overcome this diÆculty (and since there exist interest-

ing forms in odd dimensions also in characteristic 2 !), we introduce the notion of

12{regular forms (see the notes of Kneser). We �rst de�ne the 1

2{discriminant. For

this we need the following

Lemma 18. Let n be an odd integer and let ai; 1 � i � n; and aij; 1 � i < j � n,

be indeterminates. There exists a polynomial p(ai; aij) 2 ZZ[ai; aij] such that

det

0BBBB@2a1 a12 � � � a1na12 2a2 � � � a2n...

......

a1n a2n � � � 2an

1CCCCA = 2p(ai; aij):

Proof. Since n is odd the determinant on the left hand side is zero modulo 2.

For example, we have

det

0B@ 2a1 a12 a13a12 2a2 a23a13 a23 2a3

1CA = 2(4a1a2a3 + a12a23a13 � a1a223 � a2a

213 � a3a

212):


Let now (V; q) be a quadratic form over K with DimKV = n odd. Let

fe1; : : : ; eng be a basis of V and let ai = q(ei); aij = bq(ei; ej); i < j. We

call the element p(ai; aij), where p is as in Lemma 18, the 12{discriminant of q with

respect to the basis fe1; : : : ; eng and denote it by d0(e1; : : : en). As for the discrimi-

nant, the class of d0(e1; : : : en) modulo K�2 is independent of the choice of the basis

and we denote it by 12disc(q). If (V1; q1) is of odd dimension and (V2; q2) of even

dimension, we have

12disc(q1 ? q1) =

12disc(q1) � disc(q2):

We say that (V; q) is 12{regular if 1

2disc(q) 6= 0. A 1{dimensional form hai is 1

2{

regular if a 6= 0 and q1 ? q2 is12{regular if q1 is

12{regular of odd dimension and q2

is nonsingular of even dimension.

We have the following decomposition theorem for 12{regular forms:

Lemma 19. Let (V; q) be 12{regular of dimension 2m + 1. There exist elements

a0; a1; b1; : : : ; am; bm 2 K with a0 and 1� 4aibi; i = 1; : : : ; m, not zero, such that

(V; q) ' ha0i ? [a1; b1] ? � � � ? [am; bm]:

Proof. If char K 6= 2, the claim follows from Lemma 17 and Lemma 13. Assume

that char K = 2 and that DimKV � 3. Since (V; q) is 12{regular, bq is not identi-

cally zero and there exists x1; y1 2 V with bq(x1; y1) = 1. Then x1; y1 are linearly

independent and q restricted to Kx1 �Ky1 is nonsingular. The claim now follows

by Lemma 3 and induction.

We observe that the notion of 12{regular quadratic forms is also important in

the theory of forms over commutative rings R such that 1262 R but not necessarily

2 = 0 in R. A recent application can be found in the paper of Swan mentioned in

the bibliography.

Remark 20. The decomposition (V; q) = ha0i ? hV 0; q0i with (V 0; q0) nonsingular,

given by Lemma 19, is in fact unique in characteristic 2. The 1{dimensional space

ha0i is the radical of q, i.e.ha0i = fx 2 V j bq(x; y) = 0; 8y 2 V g


and (V 0; q0) is the nonsingular part, i.e. V 0 is generated by all x 2 V with bq(x; z) 6= 0

for some z 2 V .

Chapter 2

Central Simple Algebras

This chapter presents basic results about �nite dimensional algebras. References

are the books of Cohn, Herstein, Jacobson (see the bibliography) or any advanced

algebra textbook. A reference for this chapter as well as for the next two chapters

is the book of Scharlau. Many proofs in this chapter are copied from Scharlau's book.

Let R be a ring. A left or right R{module M is called simple if M 6= 0 and

M has no submodules other than M and 0. The ring R is called simple if it has

no two{sided ideals other than 0 and R. By Schur's lemma the endomorphism ring

A = EndR(M) of a simple module is a skew �eld (i.e. a ring such that any non-zero

element has an inverse). We de�ne the centre Z(R) of a ring as

Z(R) = fx 2 R j ax = xa; 8a 2 Rg:

Obviously R is a Z(R){algebra. For any ring R we denote the ring of all (n � n){

matrices with entries in R by Mn(R). We de�ne the opposite ring Rop of R as

Rop = R as additive group while the product in Rop is de�ned as ao � bo = (ba)o,

where ao is a viewed as an element of Rop.

Lemma 1. Let D be a skew �eld with centre K. The ring Mn(D) is simple. Its

centre is K � 1n, where 1n is the (n� n){unit matrix.

Proof. Let Eij be the matrix with entry 1 in place (i; j) and entries zero otherwise.

Then EijEk` = ÆjkEi`, where Æjk is the symbol of Kronecker. If � is a matrix with a

non-zero entry aij, then a�1ij Eki�Ejk = Ekk. Therefore the two{sided ideal generated

by � contains the unit matrix 1n = E11+� � �+Enn. This proves thatMn(D) is simple.

18 2. Central Simple Algebras

If � = (aij) 2 Z(Mn(D)), the equation Eij� = �Eij implies that aii = ajj for all i

and j and aij = 0 for all i 6= j. Thus � 2 K �1n. This shows that Z(Mn(D)) = K �1n.

We have the following converse:

Lemma 2. Let K be a �eld and A a �nite dimensional simple K{algebra. There

exists a skew �eld D such that A 'Mn(D).

Proof. Let M be a non-zero minimal left ideal of A. Such exist since A is a �nite

dimensional vector space over K. Then M is a simple A{module and EndA(M)

is a skew �eld by Schur's lemma. Let D = EndA(M)op. We view M as a right

D{module putting m � d = d0(m). For any a 2 A, let à : M ! M be given by left

multiplication with a. Since à 2 EndD(M), we have a ring homomorphism

i : A! EndD(M); a 7! à;

which is injective, since A simple. We check that i is surjective. Let x; y 2 M .

The map �y : x 7! xy is an element of D, thus f(xy) = f(x)y for x; y 2 M

and f 2 EndD(M). We claim that i(M) is a left ideal in EndD(M). We have

f � i(x)(y) = f(xy) = f(x)y = i(f(x))(y), hence f � i(x) = i(f(x)) 2 i(M) for

f 2 EndD(M) as claimed. Since A is simple, we have A =MA; so i(A) = i(M)i(A.

Therefore i(A) is a left ideal of EndD(M). Since 1 2 i(A); i is surjective. Finally,

we have EndD(M) 'Mn(D) for some n since M is a �nite dimensional vector space

over D.

Lemma 3. Let D be a skew �eld and A =Mn(D). Then

1) The ideals Li of column matrices:

Li =

0BBBB@0 � � � � 0...

......

0 � � � � 0i� th column

1CCCCA ; i = 1; 2; : : : n

are minimal left ideals of A and Mn(D) ' L1 � � � � � Ln.

2) All simple A{modules, in particular all minimal ideals of A, are isomorphic.

3) Every A{module M 6= 0 which is �nite dimensional as a D{vector space, is

2. Central Simple Algebras 19

a direct sum of simple A{modules.

Proof. 1) Let � =Pj Ejiaj 2 Li and ak 6= 0, then (a�1

k Ekk)� = Eki 2 Li. Hence

the left ideal generated by any non zero element � of Li is Li, so Li is minimal.

2) Obviously all Li are isomorphic as left A{modules. Let N be a simple A{

module. Since A =Pi Li and AN = N , there is i such that LiN 6= 0. The map

Li ! N; � 7! �x, is not the zero map for some x 2 N . Since Li and N are simple

this map is an isomorphism.

3) Let N be a maximal submodule of M such that N � M and N 6= M and

let x 2MnN . Since M=N is simple, M = N +Ax = N +Pi Lix. Here Lix is either

0 or a simple module. There is i such that Lix�

6= N . Since Lix is simple and N

is a maximal submodule of M , M = N � Lix. The claim follows by induction on

DimDM .

Lemma 4. If D; D0 are skew �elds such that Mn(D) 'Mn0(D0), then D ' D0 and

n = n0.

Proof. To show that D ' D0 it suÆces to show that for any simpleMn(D){module

N , there exists an isomorphism Dop ' EndMn(D)(N). Since two simple Mn(D){

modules are isomorphic, we can take for N the column Dn. Since Dn is also a right

D{module, we get a map

' : Dop ! EndMn(D)(Dn); '(d) = rd;

where rd is right multiplication with d. Obviously, ' is a ring homomorphism. Since

D is a skew �eld, ' is injective. We let it as an exercise to check that ' is surjective.

The next result follows from Lemma 2 and Lemma 4.

Theorem 5 (Wedderburn). Let A be a �nite dimensional simple K{algebra.

Then A 'Mn(D), for a suitable skew �eld D. The integer n is uniquely determined

by A and D is determined up to an isomorphism.

Let A be a K{algebra. For any subset B � A, we de�ne the centralizer of B


in A to be

ZA(B) = fa 2 A j ab = ba; 8b 2 Bg:

In particular ZA(A) is the centre of A. Obviously K � ZA(A) = Z(A) and we say

that A is central if K = Z(A). We shall use the abbreviation \c:s: algebra" for a

�nite dimensional central simple K{algebra.

Lemma 6. 1) If A; B are �nite dimensional K{algebras and A0 � A; B0 � B

are subalgebras, then ZAB(A0 B0) = ZA(A

0) ZB(B0). In particular we have

Z(A B) = Z(A) Z(B).

2) If A is a c:s: algebra and B is a simple algebra, then A B is simple.

3) A B is a c:s: algebra if and only if A and B are c:s: algebras.

Proof. 1) The inclusion � is obvious. Let fb1; : : : ; bng be a basis of B and let

x =nXi=1

xi bi 2 ZAB(A0 B0):

We have

(ax1) b1 + � � � = (x1a) b1 + � � � for all a 2 A0

hence xi 2 ZA(A0). Let fc1; : : : ; ckg be a basis of ZA(A0). Then x =Pk

i=1 ci yi; yi 2 B. The same argument shows that yi 2 ZB(B0).

2) Let I 6= 0 be a two{sided ideal of A B. We have to show that I =

A B. Assume �rst that I contains an element a b 6= 0. We have AaA = A,

hence there exist ai; a0i 2 A; 1 � i � t, such that 1 =

Pti=1 aiaa

0i. ThereforePt

i=1(ai 1)(a b)(a0i 1) = 1 b. The same argument applied to b shows that

1 1 2 I. Let now x =Pki=1 ai bi; ai 2 A; bi 2 B be an element of I with k

minimal. Suppose that k > 1. By the above argument we can assume that ak = 1.

By the minimality of k; ak�1 and ak are linearly independent. Since A is central

ak�1 62 Z(A). Let a 2 A such that aak�1 � ak�1a 6= 0 and let y be the element

(a 1)x� x(a 1) = (aa1 � a1a) b1 + � � �+ (aak�1 � ak�1a) bk�1:

The bi are linearly independent by the minimality of k and the last summand is not

zero, hence y 2 I is not zero and has a smaller k. Therefore k = 1 and we are in

the case considered above.


3) One direction is an immediate consequence of 1) and 2) and the other is ob-

vious: if A is not central, AB is not central. If A is not simple, AB is not simple.

An algebra A is c:s: if and only if the opposite algebra Aop is c:s. We have

Lemma 7. Let A be a c:s: algebra. The homomorphism of algebras

' : A Aop ! EndK(A); '(a bo)(x) = axb

is an isomorphism.

Lemma 7 is an immediate consequence of Lemma 6 and of the following (obvi-

ous) result

Lemma 8. Let ' : A ! B be a homomorphism of �nite dimensional K{algebras.

If DimKA = DimKB and A is simple, then ' is an isomorphism.

Let A;B be c:s: algebras over K. We say that A and B are Brauer equivalent

if there exist integers k and ` such that

AMk(K) ' B M`(K):

Let A =Mr(D) and B =Ms(D0). Then A and B are Brauer equivalent if, and only

if, D ' D0. Let [A] be the equivalence class of A. We get a well-de�ned multiplica-

tion by putting [A] � [B] = [AB]. The multiplication is commutative, associative,

has [K] = [Mn(K)] as element 1 and each element has an inverse, since [A][Aop] = 1.

Thus the equivalence classes of c:s: algebras form a group. It is called the Brauer

group of K and is denoted by Br(K).

Example 9. We have Br(K) = f1g for any algebraically closed �eld. In fact, let

D be a central division algebra over K, i.e. a skew �eld D with centre Z(D) = K

and which is �nite{dimensional over K. If K � D and K 6= D, let d 2 DnK. We

get a �eld extension K � K(d) � D which is �nite over K, so d is algebraic over K

and d 2 K. Contradiction.

If K = IR, a classical result of Frobenius says that IR; IC and the quaternions IH


are the only �nite dimensional division algebras over IR. Since IC is commutative, any

c:s: algebra over IR is of the formMn(D) withD = IR or IH. Hence Br(IR) = ZZ=2ZZ.

If A is a K{algebra and a is a unit of A, i.e. an invertible element, then

ia(x) = axa�1; x 2 A;

is an automorphism of A. Such automorphisms are called inner. We now prove a

very useful classical result.

Theorem 10 (Skolem-Noether).. Let A be a c:s: algebra over K and B a �nite

dimensional simple K{algebra. Let �; : B ! A be two algebra homomorphisms.

There exists an inner automorphism ia of A such that = ia Æ �.

Proof. Let �rst A = EndK(V ); V a vector space over K. We can view V as a left

B{module in two di�erent ways, putting b �v = �(b)(v) or b �v = (b)(v). By Lemma

3 these two B{modules are isomorphic so there exists a K{linear isomorphism f :

V ! V such that

f(�(b)(v)) = (b)f(v) for all v 2 V

hence (b) = f�(b)f�1. For the general case, we consider the maps

� 1; 1 : B Aop ! A Aop = EndK(A):

The algebra B Aop is simple by Lemma 6. It then follows from the case A =

EndK(V ) that there exists f 2 A Aop such that

(b) ao = f(�(b) ao)f�1

for all b 2 B; ao 2 Aop. Setting b = 1, we see that f 2 ZAAop(1 Aop) = A 1

(by Lemma 6). Therefore we get f = a 1. The claim �nally is a consequence of

(b) 1 = (a 1)(�(b) 1)(a 1)�1.

Corollary 11. Let A be a c:s: algebra. Every homomorphism A ! A is an inner

automorphism.


Remark 12. Let A be a c:s: algebra over K and let S be a �nite dimensional

commutative K{algebra. We claim that S{automorphisms of B = S A are inner.

This generalization of Corollary 11 will be used later. Let Rad(S) be the Jacobson

radical of S i.e. the set of nilpotent elements of S. The quotient S=Rad(S) is a

product of �eld extensions of K. Let now ' be an automorphism of B and let ' be

the induced automorphism of B = (S=Rad(S)) A. By an obvious generalization

of Corollary 11, ' is inner, ' = iv; v 2 B�. Let

J' = fx 2 B j '(b)x = xb; 8b 2 Bg:

The automorphism ' is inner if, and only if, there exists u 2 J' which is a unit of

B. We get

J' = (S=Rad(S)) J' = J'

= fx 2 B j '(b)x = xb; 8b 2 Bg

and

J' = Sv since ' = iv:

Let v 2 B be such that its image in B is v. By Nakayama's Lemma (see any of

the algebra textbooks in the bibliography or Scharlau's book), v is a unit of B and

J' = Sv. This shows the claim.

Corollary 13. Let A; B; C be c:s: algebras. Then A B ' A C implies that

B ' C.

Proof. Let E = A C and let � be an isomorphism AB ~!A C. We have

C = fx 2 E j x(a 1) = (a 1)x; 8a 2 Ag = ZE(A 1)

�(B) = fy 2 E j y�(a 1) = �(a 1)y; 8a 2 Ag = ZE(�(A 1)):

Let now : A ! E be given by (a) = a 1 and ' : A ! E by '(a) = �(a 1).

By Theorem 10 there exists u 2 E such that = iu Æ '. It follows that iu induces

an isomorphism �(B) ~!C.

Lemma 14. 1) Let K � L be a �eld extension and let A be a K{algebra. Then

A is a c:s: K{algebra if and only if L A is a c:s: L{algebra. In particular K � L


induces a group homomorphism Br(K)! Br(L).

2) Let A be a c:s: K{algebra. There exists a �nite �eld extension K � L such

that L A 'Mn(L). In particular DimKA is always a square.

Proof. 1) follows from Lemma 6 (at least one direction of 1) !).

2) Let K be the algebraic closure of K. By Example 9, we get that K A 'Mn(K). Let � be such an isomorphism. Let fx1; : : : xrg be a basis of A over K,

then

�(1 x`) =Xi;j

�ij`Eij; �ij` 2 K;

where fEij j 1 � i; j � ng, is the canonical basis of Mn(K). Let L be the sub�eld

of K generated by K and the �ij`. Since the �ij` are algebraic over K; L is a �nite

extension of K and � can be viewed as an isomorphism L A ~!Mn(L).

We call a pair (L; �); � : LA ~!Mn(L), a splitting of A. The following result

is a nontrivial sharpening of Lemma 14. We refer to any of the books mentioned at

the beginning of this chapter for a proof (for example Herstein or Scharlau).

Theorem 15. For any c:s: K{algebra A, there exists a splitting (L; �) with K � L

a �nite Galois extension.

We now use the existence of Galois splittings to construct the characteristic

polynomial of a c:s: algebra. Let (L; �) be a Galois splitting of A. If L[X] is the

polynomial ring in one variable X over L, we extend K{automorphisms of L to

K[X]-automorphisms of L[X] by putting g(X) = X for all g 2 G = Gal(L=K). As

for �elds, we have

K[X] = fp(X) 2 L[X] j g(p(X)) = p(X); 8g 2 Gg:

Further, we can extend the isomorphism � : L A ~!Mn(L) to an isomorphism

� : L[X] A ~!Mn(L[X])

by putting �(X 1) = X � 1n. For a 2 A, we de�ne the reduced characteristic

polynomial of a by

�(X; a) = det(�(X 1� 1 a)):


Let �(X; a) = Xn + �n�1Xn�1 + � � � + �0. The coeÆcient (�1)n�0 is called the

reduced norm of a. We denote it by n(a) or nA(a). The coeÆcient { �n�1 is the

reduced trace and it is denoted by tr(a) or trA(a).

Lemma 16. The polynomial �(X; a) is independent of the choice of the Galois

splitting (L; �) and has coeÆcients in K. In particular n(a) 2 K and tr(a) 2 K for

a 2 A.

Proof. Let � : L A ~!Mn(L) and �0 : L0 A ~!Mn(L

0) be two Galois splittings.

Replacing L and L0 by a Galois extension L00 which contains L and L0, we see that

we may assume L = L0. In view of Corollary 11, we have �0 = iu Æ � for some

u 2 GLn(L) so that

det(�0(X 1� 1 a)) = det(u(�(X 1� 1 a))u�1)

= det(�(X 1� 1 a)):

We �nally show that �(X; a) 2 K[X]. For any g 2 Gal(L=K), let g be the K{

automorphism of Mn(L[X]) given by

g(aij) = (g(aij)):

The automorphism �(g 1)��1g�1 of Mn(L[X]) is L{linear, hence there is u 2GLn(L) such that �(g 1)��1g�1 = iu. Then we get

�(1 a) = ug(�(1 a))u�1:

Since det(g(aij)) = g(det(aij)), it follows that �(X; a) = g(�(X; a)) for all g 2 G.

Therefore �(X; a) 2 K[X], as claimed.

The reduced norm and trace satisfy the following formulas

n(ab) = n(a)n(b) tr(a+ b) = tr(a) + tr(b)n(�a) = �nn(a) tr(�a) = �tr(a)

tr(ab) = tr(ba)

for a; b 2 A and � 2 K. Further a 2 A is invertible if and only if n(a) 6= 0. These

properties follow from the corresponding properties of the determinant and the trace

for matrices. Further �(a; a) = 0 for all a 2 A, i.e. the Cayley{Hamilton theorem

holds for the characteristic polynomial.

Chapter 3

Involutions on Central Simple Algebras

Let R be a ring. As already mentioned in Chapter 1, Remark 1, an involution

on R is a map � : x 7! x of R to itself such that

1) a+ b = a+ b 2) ab = ba; 3) a = a:

for all a; b 2 R. A homomorphism ' : (R; �) ! (R0; �0) of rings-with-involution is

a homomorphism ' : R! R0 such that ' Æ � = �0 Æ '.

Let now A be aK{algebra,K a �eld. If � is an involution of A, the restriction of

� to K is an involution �0 of K. If �0 is the identity, we say that � is an involution of

the �rst kind. Hence involutions of the �rst kind are K{linear. If �0 is not the iden-

tity, letK0 be the �xed �eld of �0,K0 = fx 2 K j �0(x) = xg: The extensionK0 � K

is a quadratic Galois extension with Galois group Gal(K=K0) = f1; �0g ' ZZ=2ZZ.

If this case occurs we say that � is an involution of the second kind.

We now study involutions of the �rst kind on c:s: algebras and begin with

the algebra A = Mn(K). For any matrix x, the map x 7! xt is an involution of

Mn(K) of the �rst kind. If � is another such involution, the map x 7! �(xt) is

an automorphism of Mn(K). Therefore, by the Skolem{Noether theorem, there is

u 2 GLn(K) such that

�(xt) = uxu�1 or �(x) = uxtu�1; x 2Mn(K):

The fact that �2 = 1 implies ut = "u, where " = �1. We denote the involution

x 7! uxtu�1 by �u and call " the type of �u. This is well de�ned, since �u = �v

implies u = �v for 0 6= � 2 K. Involutions of type 1 are usually called of orthogonal

type and involutions of type { 1 of symplectic type. If char K = 2 there are no

3. Involutions on Central Simple Algebras 27

di�erences between the two types. In this case it is useful to consider a more special

class of involutions. We say that �u is of even "�type if u is of the fom v + "vt. In

particular �u is of even symplectic type if u is an alternating matrix

u =

0BBBB@0 u12 : : : u1n

�u12 0...

. . .�u1n 0

1CCCCA ;not just a skew{symmetric matrix. We observe that involutions of even symplectic

type can only occur if n is even. This is clear if charK 6= 2. If charK = 2, an

alternating matrix is the matrix of the polar of a quadratic form. Its determinant is

a multiple of 2 if n is odd (Lemma 18 of Chapter 1) hence is zero in characteristic

2. Thus an (n� n){alternating matrix, n odd, is never invertible. Let

Altn(K) = fy 2Mn(K) j y is alternating g = fx� xt j x 2Mn(K)g:

The vector space Altn(K) is of dimension n(n�1)2

over K. Let now � = �u be an

involution of Mn(K) of type ". We call

Alt�n(K) = fx� "�(x) j x 2Mn(K)g:

the space of alternating elements of �. We have for y 2 Alt�n(K),

y = x� "�(x) = x� "uxtu�1 = (xu� "uxt)u�1

= (xu� (xu)t)u�1

and similarly y = u(u�1x� (u�1x)t). Thus

Lemma 1. We have Alt�n(K) = Altn(K) �u�1 = u �Altn(K) if � = �u. In particular

DimKAlt�n(K) = n(n�1)

2:

Let �u; �v be two involutions on Mn(K) and let

' : (Mn(K); �u) ~!(Mn(K); �v)

be an isomorphism of K{algebras{with{involution. By Corollary 11 of Chapter 2,

' = ic; c 2 GLn(K), and c�u(x)c�1 = �v(cxc

�1) for x 2 Mn(K):

A computation shows that v�1cuct is in the centre of Mn(K). Hence there exists

� 2 K� such that

�v = cuct:

28 3. Involutions on Central Simple Algebras

In particular the type of the involution �u is an invariant of the isomorphism class

of (Mn(K); �u). Another consequence is the following

Lemma 2. Let sm be the 2m{alternating matrix with m blocs ( 0�1

10) on the diag-

onal and let v be any invertible alternating matrix. There exists c 2 GL2m(K) such

that v = csmct. In particular (M2m(K); �v) ' (M2m(K); �sm).

Proof. Let V = K2m and let b : V � V ! K be the alternating bilinear form

de�ned by b(�; �) = �tv�; � = (x1; : : : ; x2m)t and � = (y1; : : : ; y2m)

t. Since v is

invertible, there exist �1; �2 2 V such that b(�1; �2) 6= 0 and we can even assume

that b(�1; �2) = 1. The two vectors �1 and �2 are linearly independent. Let now

V1 = K�1 �K�2 and V2 = f� 2 V j b(�i; �) = 0; i = 1; 2g:

We have V = V1 � V2. Let f�3; : : : ; �2mg be a basis of V2 and let d = (dij) be the

matrix of the change of basis, i.e., �j =Pdijei, where fe1; : : : ; e2mg is the canonical

basis of K2m. Putting v0 = (b(�i; �j)), we have

v0 =

s1 00 v00

!and v0 = dtvd:

The matrix v00 is alternating and the claim then follows by induction on the dimen-

sion of V .

We now discuss involutions of the �rst kind on c:s: algebras. Let A be such

an algebra, with involution � : a 7! a, and let (L; �) be a splitting of A. Then

�(1�)��1 is an involution of Mn(L), hence of the form �u; u 2 GLn(L). Let " bethe type of �u. Using that " depends only on the isomorphism class of (Mn(L); �u)

it is easy to check that " is independent of the choice of the splitting (L; �). We call

" the type of the involution �. Again, " depends only on the isomorphism class of

(A; �). In the next Lemma we collect some results whose proofs are straightforward.

Lemma 3. Let A1; A2 be K{algebras with involutions �1( resp. �2).

1) Then �1 �2 is an involution of A1 A2 of the �rst kind if �1; �2 are of the

�rst kind and of second kind if �1; �2 are of the second kind.

2) If �1 has type "1 and �2 type "2, then �1 �2 has type "1"2.


3) If u is a unit of A1 such that �1(u) = �u; � = �1 and if �1 is of type "1,

then iu Æ �1 is an involution of type "1�. Conversely if iu Æ �1 is an involution for

some unit u of A1, then �1(u) = �u.

Let A be a c:s: algebra of dimension n2 with an involution � of type ". We call

the set

Alt�(A) = fx� " �(x) j x 2 Ag

the set of alternating elements of A. Let � : L A ~!Mn(L) be a splitting of A

such that �(1 �)��1 = �u. By Lemma 1, we have �(Alt�(A)) = u � Altn(L), thusAlt�(A) is a vector space of dimension n(n�1)

2over K.

We say that � is of even "{type if there is a splitting (L; �) such that �(1 �)��1 = �u with u = v + "vt 2 GLn(L). This de�nition is in fact independent

of the splitting. If (L0; �0) is another splitting, we may assume that L0 = L and

�0 = ic Æ�, for some c 2 GLn(L). If �0(1�)�0�1 = �u0, with u0 2 GLn(L), we have

ic Æ �u Æ i�1c = �u0 , hence �u

0 = cuct for some unit � of L. We have �u0 = v0 + "v0t,

where v0 = cvct. If " = �1, we say that the involution is of even symplectic type

and if " = 1, we say that the involution is of even orthogonal type. In fact, if the

involution is of even symplectic type, by Lemma 2, we can even choose the splitting

(L; �) such that �(1 �)��1 = �sm with sm = diag(( 0�1

10); : : : ; (

0�1

10)), where

4m2 = DimKA. We further observe that 1 2 Alt�(A) if � is of even symplectic type.

Going over to a splitting of A as above, it suÆces to remark that s�1m �1 2 Alt2m(K).

In the following examples, we now discuss two important classes of K{algebras

with K{linear involutions, the quadratic algebras and the central simple algebras of

dimension 4.

Example 4. Quadratic Algebras. We say that a K{algebra S is quadratic if

DimKS = 2. Let f1; vg be a basis of S. We have

v2 = av + b

for some a; b 2 K. Hence

S ' K[X]=(X2 � aX � b):


It follows, in particular, that a quadratic algebra is always commutative. We de�ne

a K{linear map � : x 7! x of S by

�v + � = �v + � and v = a� v:

It is easy to check that v2 = v2, so � is an involution of S. we have

�v + �+ �v + � = �a+ 2�;

and

(�v + �)(�v + �) = �2vv + ��(v + v) + �2

= �b�2 + a��+ �2:

Therefore x + x 2 K and xx 2 K for all x 2 S. By the following Lemma 6, there

exists exactly one involution � of S with

x + �(x) 2 K and x�(x) 2 K for all x 2 S:

We call it the standard involution of S. We denote x+ x by tr(x) or trS(x) and xx

by n(x) or nS(x). The map n : S ! K is a quadratic form and its polar form

bn(x; y) = n(x + y)� n(x)� n(y)

has the matrix 2 aa �2b

!with respect to the basis f1; vg of S. We say that S is a separable quadratic K{

algebra if n is nonsingular. If K � L is a �eld extension, then S is separable

quadratic over K if and only if L S is separable quadratic over L. Since

�det

2 aa �2b

!= a2 + 4b

is the discriminant of the polynomial X2 � aX � b, S is a separable quadratic K{

algebra if and only if S ' K[X]=p(X) for some separable quadratic polynomial

p(X). Let now S be separable quadratic. Since

n(xy) = xyxy = xyy x = n(x)n(y);

S is a �eld if and only if n is anisotropic. If n is isotropic, it follows from Corollary

8 of Chapter 1 that the quadratic space (S; n) is isometric to H(K). Thus there


exists e; f 2 S such that ee = 0; ff = 0 and ef + fe = 1. It follows from

x+ x = tr(x) 2 K and xx = n(x) 2 K that

x2 � tr(x)x + n(x) = 0; x 2 S:

Let now 1 = �e+�f; �; � 2 K. We have also 1 = �e+�f , hence e = �fe; f = �ef

and ef = 0. It follows that e = �e2. Since e2 � tr(e)e + n(e) = 0, we get e+ e = 1�

and �f = �e. Similarly, we have f = �f 2. Now 1 = ef +fe = �

�f 2+ �

�e2 = 1

�f + 1

�e,

so �� = 1; � = e+ �2f = e+ f = � and �2 = �2 = 1. Finally we get

1 = �2e2 + �2f 2 = e2 + f 2; e2f 2 = 0

and fe2; f 2g is a pair of orthogonal idempotents in S. Therefore

S ' K �K with the involution (�; �) 7�! (�; �):

To summarize, we have shown that a separable quadratic K{algebra S is either a

quadratic �eld extension of K or is isomorphic to K � K. The �rst case occurs if

the norm of S is anisotropic and the second if the norm is isotropic. Finally we

observe that for any separable quadratic K{algebra S; S S ' S � S. An explicit

isomorphism is given by x y 7! (xy; xy); x; y 2 S.

Example 5. Quaternion Algebras. We de�ne a quaternion algebra over a �eld

K to be a c:s: K{algebra A of dimension 4. In view of Theorem 5 of Chapter 2,

A is either a central division algebra over K or A ' M2(K). The reduced norm

is a quadratic form on A, the trace is a linear form on A and the characteristic

polynomial reduces to

�(X; a) = X2 � tr(a)X + n(a):

If A =M2(K), the reduced norm is the determinant. Its polar bdet has the matrix0BBB@0 �1 0 0�1 0 0 00 0 0 10 0 1 0

1CCCAwith respect to the basis e1 = E12; e2 = E21; e3 = E11 and e4 = E22. Therefore det is

a nonsingular quadratic form on M2(K). The subspace U = (KK00) of M2(K) is to-

tally isotropic. Thus we get by Corollary 7 of Chapter 1 (or by a direct computation


!) an isometry of quadratic spaces

(M2(K); det) ' H(U) ' H(K) ? H(K) = H(K2):

For any matrix � = (acbd) 2M2(K), let � = tr(�)� �. We have

� =

d �b�c a

!=

0 1�1 0

! a bc d

!t 0 1�1 0

!�1

:

Thus � 7! � is an involution of even symplectic type of M2(K) such that

� + � = tr(�) 2 K and �� = �� = det(�) 2 K

for all � 2M2(K). By the following Lemma 6, � 7! � is the unique involution � of

M2(K) with the property that � + �(�) 2 K and ��(�) 2 K for all � 2 M2(K).

Let now A be any c:s: K{algebra of dimension 4 and let � : A! A be de�ned by

�(a) = a = trA(a)� a:

Let � : LA ~!M2(L) be a splitting of A. Since trA(a) is the trace of the matrix �(1a), the map �(1 �)��1 is the involution de�ned above for A =M2(K). Therefore

� is an involution of even symplectic type. We call � the standard involution of A.

We have nA(a) = a�(a) = �(a)a since it holds over a splitting and � is the unique

involution of A such that

a+ �(a) 2 K and a�(a) 2 K for all a 2 A:

Let ' : A ~!B be an isomorphism of c:s: algebras of dimension 4 and let �0 = '�A'�1

where �A is the standard involution of A. The map �0 is an involution of B such that

x+ �0(x) 2 K and x�0(x) 2 K for all x 2 B. Hence �0 = �B and ' is automatically

an isomorphism (A; �A) ~!(B; �B) of algebras{with{involutions. It follows that '

induces an isometry (A; nA) ~!(B; nB) of the reduced norms. In particular, a splitting

of A, � : L A ~!M2(L) induces an isometry

L (A; nA) = (L A; nLA) ~!(M2(L); det):

Since (M2(L); det) is nonsingular over L; (A; nA) is nonsingular. We have nA(xy) =

nA(x)nA(y) for x; y 2 A and nA(1) = 1. Thus A is a division algebra if and only if

nA is anisotropic. Since either A is a division algebra or A = M2(K), we see that

nA is isotropic if, and only if, A =M2(K).


Let A be a K{algebra. We say that a K{linear involution � on A is a standard

involution if

a+ �(a) 2 K and a�(a) 2 K for all a 2 A:

(In fact it suÆces to assume that a�(a) 2 K for all a 2 A, since (a + 1)�(a + 1) =

a�(a) + a + �(a) + 1 if A has an element 1).

Lemma 6. A K{algebra A can carry at most one standard involution.

Proof. Let a�(a) = n(a) and a+ �(a) = tr(a). We have

a2 � tr(a)a+ n(a) = 0

in A for any standard involution �. Let now �1; �2 be standard involutions on A

and let tri(a) = a + �i(a); ni(a) = a�i(a). Let fe1; : : : ; eng be a basis of A with

e1 = 1. Since

e2i � tr1(ei)ei + n1(ei)e1 = e2i � tr2(ei)ei + n2(ei)e1

we get tr1(ei) = tr2(ei) for i � 2, hence �1(ei) = �2(ei) for i � 2. On the other

hand, �1(1) = 1 = �2(1), thus, as claimed �1 = �2.

The existence of standard involutions is related with the existence of com-

position algebras. Let A be a �nite dimensional K{vector space with a bilinear

multiplication (a; b) 7! ab which has a neutral element 1, i.e. a � 1 = a = 1 � a. In

the following, we call A an algebra even if the multiplication is not associative. We

say that A is a composition algebra if there exists a nonsingular quadratic form n

on A such that n(ab) = n(a)n(b) for all a; b 2 A. An associative K{algebra with

1, which carries a standard involution �, is a composition algebra for the quadratic

form n(a) = a�(a), since

n(ab) = ab�(ab) = ab�(b)�(a) = a�(a)b�(b) = n(a)n(b):

The following result, which is due to Hurwitz in characteristic not 2, describes all

possible composition algebras. It implies that separable quadratic algebras and c:s:

algebras of dimension 4 are the only examples of associative algebras with a stan-

dard involution such that the corresponding norm is nonsingular.


Theorem 7. Let (A; n) be a composition algebra over K. Then either

1) A = K,

2) A is a separable quadratic K{algebra and n = nA,

3) A is a c:s: K{algebra of dimension 4 and n = nA,

4) A is a separable Cayley algebra. In particular DimKA = 8.

Proof. We give the proof of van der Blij and Springer, which works also in charac-

teristic 2. Let (x; y) be the polar of n, i.e. (x; y) = n(x+ y)� n(x)� n(y). Since byassumption n is nonsingular,

(x; y) = 0; 8y 2 A ) x = 0:

The following formulas are deduced from n(xy) = n(x)n(y) by linearization:

(xy1; xy2) = n(x)(y1; y2)

(x1y; x2y) = n(y)(x1; x2)(3.1)

(x1y2; x2y1) + (x1y1; x2y2) = (x1; x2)(y1; y2):

We must have n(1) = 1. Putting x1 = y2 = x; x2 = z and y1 = 1 in the third

formula of (3.1), we obtain

(x2 � (1; x)x+ n(x) � 1; z) = 0

for all z 2 A. Since n is nonsingular we get

x2 � (1; x)x + n(x) � 1 = 0(3.2)

for all x 2 A. We call (1; x) the trace of x and denote it by tr(x). Let x = tr(x)�x.It is straightforward to check that

xy = y x; x = x and 1 = 1:

Hence x 7! x is an involution on A. Further

xx = xx = n(x); n(x) = n(x); (xy; z) = (y; xz) = (x; zy):(3.3)

We also need the formulas

x(xy) = n(x)y and (yx)x = n(x)y:(3.4)


For the proof of the �rst one, we have

(x(xy); z) = (xy; xz) = (n(x)y; z) for all z 2 A:

The proof of the other one is similar. It follows from (3.4) that x(xy) = (xx)y.

Therefore

x(xy) = x(tr(x)y � xy) = (xtr(x)� xx)y = (xx)y

and similarly (yx)x = y(xx). This shows that A is an alternative algebra. Assume

now that DimK(A) > 1. Since n is nonsingular, it is easy to check that there exists

u 2 A such that the set f1; ug is linearly independent in A and such that the re-

striction of n to B1 = K1�Ku is nonsingular. By (3.2), B1 is a separable quadratic

K{algebra. We have A = B1 ? B?1 . If B

?1 6= 0, the restriction of n to B?

1 is non-

singular and there exists v 2 B?1 such that n(v) 6= 0. We put n(v) = ��; � 2 K.

To conclude we use the following:

Lemma 8. Let B be a proper subalgebra of A such that the restriction of n to B

is nonsingular and let v 2 B? be such that n(v) = �� is not zero. Then B � vB is

a subalgebra of A. We have

n(a+ vb) = n(a)� �n(b) and a + vb = a� vb

for a; b 2 B. Further B is associative and B is commutative if A is associative.

Proof. The fact that B � vB is a subalgebra follows from the formulas

(va)b = v(ba)

a(vb) = v(ab)(3.5)

(va)(vb) = �ba n(v) = �ba

for a; b 2 B. We only check the �rst one. The proof of the others is similar. We

have v = �v, since (v; 1) = 0, and 0 = (v; a) = va + av = �va + av, thus va = av

for a 2 B. Further

((vb)a; z) = (vb; za) = (bv; za) = �(ba; zv):

The last equality follows from the third formula of (3.3), using that (v; a) = 0 for

a 2 B. On the other hand

�(ba; zv) = �((ba)v; z) = (v(ab); z):


This holds for all z 2 A, hence (vb)a = v(ab) as claimed. The formulas for n and

the involution are easy. Let now

n((a+ vb)(c+ vd)) = n(ac + �db+ v(cb+ ad))

= n(ac + �db)� �n(cb+ ad):

On the other hand

n((a+ vb)(c+ vd)) = n(a + vb)n(c+ vd)

= (n(a)� �n(b))(n(c)� �n(d)):

Comparing both expressions and using once more that n is multiplicative, we obtain

(ac; �db) + n(v)(cb; ad) = 0

or

(ac; db) = (ad; cb)

then it follows from (3.3) that

((ac)b; d) = (a(cb); d) for all a; b; c; d 2 B:

Thus (ac)b = a(cb) and B is associative. If A is associative, we have (va)b = v(ab) =

v(ba) and ba = ab. Therefore B is commutative.

We now go back to the proof of Theorem 7. The algebra B2 = B1 � vB1 is an

associative K{algebra of dimension 4. We let it as an exercise to check that B2 is

central simple. If B2 is a proper subalgebra of A, let B3 = B2 + wB2 with w 2 B?2 .

If B3 = A, the formulas (3.5) show (by de�nition!) that A is a Cayley algebra (see

the book of Schafer). If B3 is a proper subalgebra of A, we get B4 = B3 + zB3 with

z 2 B?3 . By Lemma 8, B3 must be associative and hence B2 commutative. This is

not possible.

Let A be a c:s: K{algebra with an involution � of the �rst kind. Since � is

an isomorphism A ~!Aop, the class of A in Br(K) is of order � 2. Conversely any

c:s: K{algebra A such that its class in Br(K) is of order � 2 carries an involution

� of the �rst kind. More precisely:


Theorem 9 (Albert). Let A be a c:s: K{algebra such that 2[A] = 0 in Br(K):

1) There exists an involution � of the �rst kind on A: 2) If the dimension of A is

even, the involution � can be chosen of even orthogonal type or of even symplectic

type.

Proof. We follow the proof given by Saltman. If 2[A] = 0 in Br(K); [A] = [Aop]

so there exists an isomorphism � : A ~!Aop of K{algebras. Let

' : A A ~!EndK(A)

be the isomorphism de�ned by '(a b)(x) = ax�(b). Let now !A be the switch of

A A, i.e. !A(a b) = b a. By the Skolem{Noether theorem, !A = iu for some

u 2 (AA)�. If A = EndK(V ); AA = EndK(V V ) and u : V V ! V V can

be chosen as the switch !V . If � : LA ~!EndL(V ) is a Galois splitting, the element

u = (� �)�1(!V ) 2 L A A is in fact in A A (this can easily be checked by

Galois theory). Thus u 2 (A A)� can be chosen such that (� �)(u) = !V for

any Galois splitting. In particular u2 = 1. Let now : A ! A be given by '(u).

We have

2 = 1 and (ax�(b)) = b (x)�(a) for a; b; x 2 A:

Lemma 10. Let w = (1) 2 A: 1) We have 1 = w�(w) = �(w)w and � 2 = (iw)�1.

2) If there exists a unit a 2 A such that (a) = a, then � = ia Æ � is a K{linear

involution of A.

Proof. 1) We have 1 = (w) = (w � 1) = w�(w). On the other hand 1 =

(1 �w) = ��1(w)w. Thus w is a unit of A with inverse �(w) = ��1(w). Further we

have x = 2(x) = ( (x � 1)) = (w�(x)) = (�(x) � 1) � �(w) =w� 2(x)�(w) = (iw Æ � 2)(x): 2) We have �(xy) = �(y)�(x) since � is an antiau-

tomorphism. We check that �2 = 1. It follows from a = (a) = (1 � a) = �(a)w

that w = a�(a)�1. Thus

�2 = (ia Æ �)2 = ia�(a)�1 Æ � 2 = 1 since � 2 = iw�1:

Proof of Theorem 9. To prove the �rst part, we have, by Lemma 10, to construct

a unit a 2 A such that (a) = a. We may assume that A = Mn(D); D a division

algebra. Since 2[D] = 0 in Br(K) there exists an antiautomorphism � : D ~!D. If


� is an involution, we extend � to A by taking the transpose on Mn(K). If � is

not an involution, we construct as above a map : D ! D such that 2 = 1 and

(ax�(b)) = b (x)�(a) for a; b; x 2 D. Since �2 6= 1; �2 = i�1w for w = (1)

implies that w 62 K. Thus 1 + (1) 6= 0 and a = 1 + (1) is the wanted element a.

If A 'Mn(K); A has an involution given by the transpose.

We now prove the second claim of Theorem 9. By 1) A has an involution �,

say of type ". Let � = �1. If A contains a unit of the form u = z + "��(z), then

�0 = iuÆ� is of type � and in particular, �0 will be of even symplectic type if � = �1.Thus to prove 2) it suÆces to show that for some K{linear involution � of A and

any " = �1, there exists a unit of A of the form u = z + "�(z). Assume that

A = Mn(D); D a division algebra, D 6= K. By 1) D has an involution �1 and it

obviously suÆces to �nd z1 2 D such that z1 + "�1(z1) 6= 0. Since D 6= K; �1 6= 1

so z1 + "�1(z1) = 0 for all z1 2 D is impossible. If A = Mn(K) with n = 2m we

write A =M2(K)Mm(K), take � = �1 �2, where �1 is the standard involution

of M2(K) and �2 is transposition. Then for z = (1000) 1; z + "�(z) = (10

0") 1

is a unit of A.

One can also give conditions for the existence of involutions of the second kind

on c:s: algebras. For this we need the notion of corestriction. Let Z be a separable

quadratic K{algebra with standard involution �0 and let A be a K{algebra. Then

Z A is a Z{algebra and, by Galois theory, A can be identi�ed with the set of

elements y 2 ZA such that (�01)(y) = y. The map ~� = �01 is a �0{semilinear

automorphism of Z A such that ~�2 = 1. Conversely, let B be a Z{algebra with a

�0{semilinear automorphism ~� such that ~�2 = 1. Let

A = fy 2 Bj~�(y) = yg:

Then A is a K{algebra and the multiplication map � : ZA! B; ya 7! ya is an

isomorphism of Z{algebras such that ~� = �(�01)��1. We say that A is descended

from B (and ~�) by Galois descent. Obviously Galois descent also applies to other

structures, like quadratic forms. If B is a Z{algebra, we denote by �0B the algebra

with the Z{action twisted through �0, i.e. y � b = �0(y)b; y 2 Z; b 2 B. The map

~� : B Z �0B ! B Z �0B


given by ~�(b c) = c b is �0{semilinear and the corresponding K{algebra

A = fx 2 B Z �0B j ~�(x) = xg

is called the corestriction of B and is denoted by cor(B). If B has a Z{linear in-

volution �, then cor(B) has an induced K{linear involution cor(�). We can now

formulate a well-known existence criterion for involutions of the second kind:

Theorem 11 (Albert, Riehm).. Let L be a separable quadratic extension of K

and let A be c:s: over L. Then A admits an involution of the second kind if and

only if [cor(A)] = 0 in Br(K).

Theorem 11 will not really be used in the following and we do not prove it. A

proof can be found in the book of Scharlau (p. 309).

Let K � L be a separable quadratic extension and let A be a c:s: L{algebra

with a �xed involution x 7! x� of the second kind. In the next Lemma we describe

all possible other involutions of the second kind on A.

Lemma 12. Let � be an involution of the second kind on A.

1) There exists g 2 A�, with g� = g, such that �(x) = gx�g�1.

2) Let g 2 A� such that g� = g and let �g(x) = gx�g�1. Then �g is an involution

of the second kind and �g = �g0 if and only if g = �g0 for � 2 K�.

Let ' : (A; �g) ~!(A; �g0) be an isomorphism of L{algebras-with-involution.

There exist c 2 A� and � 2 K� such that g0 = �cgc�.

Proof. 1) By Skolem{Noether, there exists g 2 A� such that �(x�) = ig(x). It then

follows from �2 = 1 that g� = �g for some � 2 L such that ��0(�) = 1. By the

following Lemma (which is a special case of Hilbert Theorem 90) there exists � 2 Lsuch that � = ��1�0(�). Replacing g by ��1g, we see that we can assume g� = g.

The claims 2) and 3) are similar to the corresponding claims for involutions of the

�rst kind and we do not check them.

Lemma 13. Let � 2 L be such that ��0(�) = 1. There exists � 2 L such that


� = ��1�0(�).

Proof. If there exists � 2 L such that � = �0(�)+�0(�)� 6= 0, then �0(�)��1 = �. If

not, � 7! ��0(�)� is a K{linear automorphism of L. This can only occur if � = �1(put � = 1!). Let then z be a generator of L such that �0(z) = 1� z. The element

� = 1� 2z is such that �0(�) = �� and � = �1 = ��1�0(�).

Chapter 4

The Cli�ord Algebra

Let (V; q) be a quadratic form over a �eld K. We would like to �nd a K{algebra

C and a K{linear map i : V ! C such that

i(x)2 = q(x) � 1C for all x 2 V:(4.1)

We de�ne the Cli�ord algebra C(V; q) of (V; q) as the K{algebra which is universal

with respect to the property (4.1): A Cli�ord algebra for (V; q) is a K{algebra C to-

gether with a K{linear map i : V ! C verifying (4.1), such that for any K{algebra

A and any K{linear map ' : V ! A with '(x)2 = q(x) � 1A, there exists a unique

K{algebra homomorphism '0 : C ! A such that ' = '0 Æ i.

Lemma 1. For any quadratic form (V; q) over K, there exists up to isomorphism a

unique Cli�ord algebra (C(V; q); i).

Proof. The uniqueness follows from the universal property of the Cli�ord algebra

and we let its proof as an exercise. We prove existence. Let

T (V ) = K � V � V V � � � �

be the tensor algebra of V , let J be the 2{sided ideal of T (V ) generated by all

elements x x� q(x) � 1 for x 2 V and let C = T (V )=J . We claim that C together

with the canonical map

i : V ! TV ! TV=J

is a Cli�ord algebra. Let ' : V ! A be a K{linear map such that '(x)2 = q(x) � 1A.By the universal property of the tensor algebra, ' extends to a unique homomor-

phism ~' : TV ! A of K{algebras such that ~'jV = '. We have ~'(J ) = 0, so ~'

42 4. The Cli�ord Algebra

induces, as claimed, a unique homomorphism '0 : C ! A such that ' = '0 Æ i.

Since a Cli�ord algebra is uniquely determined up to a unique isomorphism we

shall speak in the following of \the" Cli�ord algebra of (V; q) and we shall denote it

by C(V; q) or C(q). From now on we identify K � 1C with K.

We observe that (4.1) implies

i(x)i(y) + i(y)i(x) = bq(x; y); x; y 2 V(4.2)

where bq is the polar of q.

We say that a K{algebra A is ZZ=2ZZ{graded (or simply graded) if A has a direct

sum decomposition A = A0�A1, as K{vector space, such that Ai �Aj � Ai+j; i+ j

mod 2. We have K � 1 � A0. Such a grading on the tensor algebra TV is given by

(TV )0 = K � V V � � � �(TV )1 = V � V V V � � � �

i.e. TV is graded by the degree. The generators x x � q(x) of J have even

degree. Therefore we get a ZZ=2ZZ{grading of C(q), C(q) = C0 � C1; where C0

is generated by the even products of elements of V and C1 by the odd products.

In particular C0 is a subalgebra of C, sometimes called the even Cli�ord algebra,

and i(V ) � C1. By de�nition a graded homomorphism � : A0 � A1 ! A00 � A0

1

of ZZ=2ZZ{graded algebras is an algebra homomorphism such that �(Ai) � A0i. An

immediate consequence of the universal property of the Cli�ord algebra is that any

morphism ' : (V; q)! (V 0; q0) of quadratic forms induces a graded homomorphism

C(') : C(q) ! C(q0). In particular any isometry ' : (V; q) ~!(V 0; q0) induces an

isomorphism C(') : C(q) ~!C(q0). As we shall see later the converse does not hold,

i.e. non isometric forms may have isomorphic Cli�ord algebras.

Lemma 2. Let K � L be a �eld extension. There is a canonical isomorphism

C(L (V; q)) ' L C(V; q) for any quadratic form (V; q).

Proof. The map 1 i : LV ! LC(V; q) induces a homorphism C(L(V; q))!L C(V; q). The homomorphism V ! L V ! C(L (V; q)) yields the inverse

4. The Cli�ord Algebra 43

homomorphism.

We now compute the Cli�ord algebra of an orthogonal sum. For this we need

the graded tensor product A bB of two ZZ=2ZZ{graded K{algebras A = A0�A1 and

B = B0 � B1. As a K{vector space A bB = A B. The product is de�ned for

homogeneous elements a; a0 2 A; b; b0 2 B by

(a b)(a0 b0) = (�1)@(b)@(a0)aa0 bb0;

where @(b) is the degree (0 or 1) of b, and is extended by linearity to arbitrary

elements. The algebra A bB is graded by

(A bB)0 = A0 B0 + A1 B1

(A bB)1 = A1 B0 + A0 B1:

Let A = A0 � A1 be a graded algebra. We call the grading

M2(A)0 =

A0 A1

A1 A0

!and M2(A)1 =

A1 A0

A0 A1

!

of M2(A) the checker-board grading. On M2(K) the checker-board grading is

M2(K)0 =

K 00 K

!and M2(K)1 =

0 KK 0

!:

Let B = B0 � B1 be any ZZ=2ZZ{graded algebra. As an exercise in the theory of

graded algebras (and because it will be used later) we prove the following

Lemma 3. There exists an isomorphism of graded algebras

' :M2(K) bB ~!M2(K)B =M2(B):

Proof. We de�ne '(x 1B) = x 1B for x 2M2(K) and

'(1 (b0 + b1)) =

b0 + b1 00 b0 � b1

!2 M2(B):

Proposition 4. C(q ? q0) ' C(q) bC(q0) as graded algebras.

Proof. Let y = (x; x0) 2 V � V 0. The map y 7! i(x) 1 + 1 i(x0) induces a map

C(q ? q0) ! C(q) bC(q0). On the other hand, the inclusions x 7! (x; 0) 2 V � V 0

and x0 7! (0; x0) 2 V �V 0 extend to algebra homomorphisms C(q)! C(q ? q0) and


C(q0) ! C(q ? q0). These two maps combine to a homomorphism C(q) bC(q0) !C(q ? q0) which is the inverse of the �rst map.

Let fe1; : : : ; eng be a basis of V . In view of the relations (4.1) and (4.2) the 2n

elements 1; i(e1); : : : ; i(en); i(e1)i(e2); : : : ; i(en�1)i(en); i(e1)i(e2)i(e3); : : : ;

i(e1) � � � i(en) form a set of generators of C(V; q) as a vector space over K. A cele-

brated theorem says that they form a basis of C(V; q).

Theorem 5 (Poincar�e{Birkho�{Witt).. If fe1; : : : ; eng is a basis of V , the set

f1; i(ej1) � � � i(ejr); 1 � r � n; 1 � j1 < j2 � � � < jr � ng is a basis of C(V; q).

Proof. If the theorem is true for q1 and q2 it is true for q1 ? q2 by Proposition 4. The

theorem is true if (V; q) = hai because, if v is a generator of V with q(v) = a, then

TV ' K[X]; J ' (X2 � q(v)) and C(q) ' K[X]=(X2 � q(v)) = K � 1�K � i(v):Hence the theorem is true if fe1; : : : ; eng is an orthogonal basis. If the theorem is

true for a given basis, it is clearly true for any basis of V . By Lemma 13 of Chapter

1 the theorem is true for any �eld of characteristic not 2. Let R be a domain of char-

acteristic not 2. By embedding R into its �eld of fractions, we see that the theorem

is true for R. If now K is a �eld of characteristic 2, we can write K as a quotient

R=I with R a domain of characteristic zero (for example a polynomial ring over ZZ).

Let V be the free module with basis a set of symbols fe1; : : : ; eng, so V = V =IV .

We de�ne a quadratic form q on V by choosing elements ai; aij (i 6= j) in R such

that ai + I = q(ei) and aij + I = bq(ei; ej). We have (V ; q)R R=I ' (V; q) and by

Lemma 2 (or better an obvious extension of Lemma 2 for forms over commutative

rings) C(V ; q)R R=I ' C(M; q). The theorem is true for C(V ; q) because R is a

domain of characteristic zero. Therefore it is also true for (V; q).

Remark 6. This proof of the Poincar�e{Birkho�{Witt theorem, which is due to

Kneser, works obviously also for quadratic forms over �nitely generated free R{

modules, R any commutative ring. In fact we have used the notion of quadratic

forms over rings in the proof as well as some basic facts, which are straightforward

generalizations of the corresponding facts for forms over �elds.


Corollary 7. If V has dimension n, then C(V; q) has dimension 2n and C0; C1 have

both dimension 2n�1.

It follows from Theorem 5 that i : V ! C(V; q) is injective. From now on

we shall identify V with its image i(V ) in C(q). We now describe the structure

of the Cli�ord algebras for nonsingular even dimensional forms and 12{regular odd

dimensional forms. We put C = C(V; q); C0 = C0(V; q) and C1 = C1(V; q).

Theorem 8. 1) Let (V; q) be a quadratic space of even dimension 2m over K. Then

C is a c:s: K{algebra, the centre Z(C0) of C0 is a separable quadratic K{algebra.

If Z(C0) is a �eld, C0 is c:s: over Z(C0). If Z(C0) ' K � K, there exists, up to

isomorphism, a unique c:s: K{algebra A such that C0 ' A� A as K �K{algebra.

2) Let (V; q) be 12{regular of odd dimension 2m + 1. Then C0 is c:s: over K,

the centre Z(C) of C is a ZZ=2ZZ{graded quadratic K{algebra Z0 � Z1 such that

Z0 = K and Z1 is generated by an element z with z2 = � � 1; � 2 K�. Further the

multiplication in C induces an isomorphism Z(C) C0 ~!C.

Proof. We �rst prove Theorem 8 in dimension 1 and 2, then by induction for forms

(V; q) = H(K) ? (V 0; q0) and �nally for arbitrary forms.

Case n = 1 : We have (V; q) = hai; a 6= 0. Let v 2 V with q(v) = a. Then

C = K � 1 +Kv; C0 = K � 1; C1 = Kv and the multiplication is given by v2 = a.

The algebra C is commutative, so Z(C) = C.

Case n = 2 : We assume that (V; q) = [a; b]. Let V = Ku � Kv with q(u) =

a; q(v) = b and bq(u; v) = 1. Then C has the basis f1; u; v; uvg with the relations

u2 = a; v2 = b; uv + vu = 1:

To show that C is c:s:, it suÆces to show that L C is c:s: over L for some �eld

extension L of K. Let e = �u + v; � a symbol. We get q(e) = �2a + � + b and

we choose L such that the polynomial X2a + X + b has a root � in L. Then e is

isotropic in L V . By Corollary 8 of Chapter 1, L (V; q) ' [0; 0] and L C is

generated by elements u; v with u2 = 0; v2 = 0 and uv+vu = 1. It is easy to verify


that

u 7! (0010); v 7! (01

00)

induces an isomorphism

L C ~!M2(L):

Hence C is c:s: over K. We have in fact proved that

C(H(K)) ' M2(K);

as graded algebras, ifM2(K) has the checker{board grading. This will be used later.

Let now (V; q) = [a; b] as above. We have C0 = K �1�Kuv and (uv)2 = u(1�uv)v =uv � ab. Thus the element z = uv is a generator of C0 such that

z2 = z � ab:

By Example 4 of Chapter 3, C0 is a quadratic algebra and is separable if and only

if 1 � 4ab 6= 0. Since the discriminant of q with respect to the basis fu; vg is

4ab � 1; Z(C0) = C0 is a separable quadratic K{algebra. In particular Z(C0) is

either a �eld or Z(C0) ' K �K.

This shows that Theorem 8 holds in dimension 1 and 2. We assume now that

(V; q) = H(K) ? (V 0; q0). Let C 0 = C 00 � C 0

1 be the Cli�ord algebra of (V 0; q0). We

get

C 'M2(K) bC 0

by Proposition 4 and the fact that C(H(K)) ' M2(K). Thus by Lemma 3

C 'M2(K) C 0 =M2(C0):

If we use the isomorphism C ~!M2(C0) given by Lemma 3 to identify C withM2(C

0),

then

C0 =

C 00 C 0

1

C 01 C 0

0

!and C1 =

C 01 C 0

0

C 00 C 0

1

!in M2(C

0):

Assume that V has even dimension, so V 0 also has even dimension. By induction,

Theorem 8 holds for (V 0; q0). In particular C is c:s: over K since C 0 is c:s: over K.

We now construct an inner automorphism ofM2(C0) which maps C0 toM2(C

00). Let

x 2 V 0 be anisotropic (such an element exists since q0 is nonsingular). Then x is

invertible in C 0 with inverse x�1 = x�q(x)�1 2 C 01. Left and right multiplication with


x induce isomorphisms C 01 ~!C 0

0 of K{vector spaces. In particular xC 01 = C 0

0; C01x =

C 00 and xC

00x = C 0

0. Thus inner conjugation with the unit (100x) of M2(C

0) maps

C0 =

C 00 C 0

1

C 01 C 0

0

!to

C 00 C 0

1xxC 0

1 xC 00x

!=M2(C

00):

It follows that Z(C0) ' Z(C 00) and that C0 is as described in 1). If (V; q) is

12{regular

of odd dimension, then (V 0; q0) is also 12{regular of odd dimension. We have as above

C 'M2(C0) and C0 'M2(C

00)

so that, by induction on the dimension, C0 is c:s: over K and Z(C) ' Z(C 0) is a

quadratic algebra. Since the isomorphism C 'M2(C0) is an isomorphism of graded

algebras, Z(C) ' Z(C 0) as graded algebras. The last claim Z(C) C0 ~!C is also

clear by induction.

If (V; q) is arbitrary, there exists a �eld extension K � L such that L (V; q) 'H(L) ? (V 0; q0). Therefore Theorem 8 holds for L (V; q). Assume that V has

even dimension. The algebra C is c:s: over K since L C is c:s: over L and Z(C0)

is separable quadratic since Z(L C0) = L Z(C0) is separable quadratic over L.

If Z(C0) is a �eld, it also follows from Theorem 8 that C0 is c:s: over Z(C0). If

Z(C0) = K �K, we only get that there exists an isomorphism of K �K{algebras

C0 ~!A� B; A; B c:s: over K with L A ' L B. To show that A ' B, we use

the following

Lemma 9. Let (V; q) be nonsingular of even dimension. There exists an isomor-

phism

' : Z(C0) C ~!EndC0(C) =M2(C0)

of Z(C0){algebras, where C is viewed as a right C0{module through the multiplica-

tion in C.

Proof. We de�ne '(zc)(x) = cxz for z 2 Z(C0); c; x 2 C. If Z(C0) is a �eld, ' is

an isomorphism since Z(C0)C is a c:s: algebra over Z(C0) and Z(C0)C; M2(C0)

have the same dimension. A similar argument works if Z(C0) = K � K. The fact

that EndC0(C) ' M2(C0) follows from C1 = C0x for x 2 V anisotropic.


We go back to the proof of Theorem 8. If Z(C0) = K � K and C0 = A � B,

we have by the above lemma

M2(A)�M2(B) ' C � C:

Since M2(A); M2(B) and C are simple algebras, it follows that

M2(A) ' C 'M2(B);

hence A ' B by Corollary 13 of Chapter 2.

Finally if V is 12{regular of odd dimension, C0 is c:s: over K since L C0 is

c:s: over L. Further Z(C) is quadratic. Let z = z0 + z1; zi 2 Ci be a generator of

Z(C). The part z0 of degree zero lies in the centre of C0, hence is a scalar, so z1

also generates Z(C) as an algebra. Since z21 6= 0 in L Z(C); z21 6= 0 and Z(C) is

as claimed.

Chapter 5

Invariants of Quadratic Forms

In this chapter and the next the fact that we do not assume char K 6= 2 has a

strong in uence on the presentation of the results. Demazure{Gabriel is a useful ref-

erence for these two chapters. Parts of Proposition 5 are copied from Micali{Revoy.

Let (V; q) be a quadratic form which is either nonsingular of even dimension or

12{regular of odd dimension. Let

Z(V; q) = Z(q) =

(Z(C0(V; q)) if Dim V is evenZ(C(V; q)) if Dim V is odd.

We call Z(q), which is a graded quadratic algebra, the discriminant algebra of (V; q).

The algebra Z(q) is separable if Dim V is even or if Char K 6= 2. Clearly any

isometry (V; q) ~!(V 0; q0) induces a graded isomorphism Z(q) ~!Z(q0). By the proof

of Theorem 8, Chapter 4, we have

Z(H(K) ? q) ' Z(q) and Z(H(K)) ' K �K;

hence Z(H(U)) ' K �K for any hyperbolic space H(U).

Lemma 1. Z(q) = ZC(C0).

Proof. The claim follows from Theorem 8, Chapter 4 and Lemma 6, Chapter 2, if

DimV is odd. Assume that DimV is even. The inclusion � is clear. Let x 2 ZC(C0)

and let x = x0 + x1 be its decomposition in homogeneous parts. Since C0 is ho-

mogeneous, x0 and x1 must lie in ZC(C0). By Proposition 5 of this chapter, we

have a generator z of Z(q) such that x1z + zx1 = x1, so (1 � 2z)x1 = 0. But

(1� 2z)2 = 1+ 4(z2 � z) = 1+ 4r is a unit (Proposition 5), hence x1 = 0. (The use

50 5. Invariants of Quadratic Forms

of Proposition 5 in the proof is not very elegant but is legal since Lemma 1 is not

applied in the proof of Proposition 5 !).

We call the isomorphism class of Z(q) the Arf invariant of (V; q) and denote it

by a(q). We say that the Arf invariant a(q) is trivial if Z(q) ' K �K.

Example 2. Assume that char K 6= 2 and that fe1; : : : ; eng is an orthogonal basis

of (V; q). The element z = e1 � � � en is a generator of Z(q) and

z2 = (�1)n(n�1)2 a1 � � �an = (�1)n(n�1)

2 d(e1; : : : ; en):

Let Z = K � 1 + K � z and Z 0 = K � 1 + K � z0 be two quadratic algebras with

z2 = a and (z0)2 = b. We claim that Z ' Z 0 if and only if a = �2b for some � 2 K�

(assuming char K 6= 2 !). Let ' : Z ~!Z 0 be an isomorphism and let '(z) = �z0 + �.

Since ' is an isomorphism, � 6= 0. We have '(z2) = '(z)2, so a = �2b+2��z0 + �2.

Since 2� 6= 0; � = 0 and a = �2b as claimed. Therefore Z(q) is determined up to

isomorphism by the class of (�1)n(n�1)2 a1 � � �an in K�=K�2 if char K 6= 2. This class

is equal to disc(q) multiplied by the class of 2n(�1)n(n�1)2 in K�=K�2, since disc(q)

is the class of 2na1 � � �an. For any a 2 K�, let [a] denote its class in K�=K�2. The

element

Æ(q) = [(�1)n(n�1)2 ] � disc(q) 2 K�=K�2

is called the signed discriminant of (V; q). It follows from the above discussion that,

for forms of the same dimension, Z(q) ' Z(q0) if and only, if Æ(q) = Æ(q0) (if char

K 6= 2).

The following property of the discriminant algebra follows from Theorem 8 of

Chapter 4 or from Lemma 1 and the fact that C(L (V; q)) ' LC(V; q); C0(L(V; q)) = L C0(V; q):

Lemma 3. For any �eld extension K � L; Z(L (V; q)) ' L Z(V; q).

We now construct a generator z of Z(q), for any characteristic, given orthogonal

decompositions (see Chapter 1)

(V; q) = [a1; b1] ? � � � ? [am; bm] if DimV = 2m;

5. Invariants of Quadratic Forms 51

(V; q) = [a1; b1] ? � � � ? [am; bm] ? ha2m+1i if DimV = 2m+ 1:

We construct z by induction, using the following

Lemma 4. Let (V1; q1) and (V2; q2) be quadratic spaces of even dimensions, and let

Z1 = Z(q1); Z2 = Z(q2). For i = 1; 2, we assume that Zi has a generator zi such

that

z2i = zi + ri; ri 2 K with 1 + 4ri 6= 0

and

zivi + vizi = vi for all vi 2 Vi:

Then Z = Z(q1 ? q2) has a generator z such that

z2 = z + r with r = r1 + r2 + 4r1r2; hence 1 + 4r = (1 + 4r1)(1 + 4r2)

and

zv + vz = v for all v 2 V = V1 � V2:

Proof. Let z = z1 1 + 1 z2 � 2z1 z2 in C(q1 ? q2) = C(q1) bC(q2). We get

z2 = z + r with r = r1 + r2 + 4r1r2. The other claims follow by straightforward

computations.

Let (V; q) = [a; b] and let fx; yg be a basis of V such that q(x) = a; q(y) = b

and bq(x; y) = 1. The discriminant algebra Z(q) = C0 is generated by z = xy

and z2 = z � ab. It is easy to verify that zv + vz = v for all v 2 V . Further

1 � 4ab = �d(x; y) is not zero. Thus we can apply Lemma 4 to quadratic spaces

V1; V2 of type [a; b] or to orthogonal sums of such spaces. Let Vi = [ai; bi] and

(V; q) = (V1; q1) ? : : : ? (Vm; qm):

We choose in each Vi a basis fei; ei+mg such that

q(ei) = ai; q(ei+m) = bi and bq(ei; ei+m) = 1

and put zi = eiei+m; ri = �q(ei)q(ei+m) = �aibi. By Lemma 4 (and induction) the

element

z =mXj=1

(�2)j�1Sj(z1; : : : ; zm);


where Sj is the j � th elementary symmetric function in m letters, is a generator of

Z(q) such that

z2 = z + r; r =mXj=1

4j�1Sj(r1; : : : ; rm)

and zv + vz = v for all v 2 V . Further

1 + 4r =mYi=1

(1 + 4ri) = (�1)md(e1; e1+m; : : : ; em; e2m):

Therefore the class of 1 + 4r in K�=K�2 is the signed discriminant Æ(q) of (V; q).

Let now (V; q) be of odd dimension. We assume that

(V; q) = (V1; q1) ? � � � ? (Vm; qm) ? ha2m+1i = (V 0; q0) ? ha2m+1i

with Vi as above. Let e2m+1 2 V be a basis of ha2m+1i such that q(e2m+1) = a2m+1.

By the even dimensional case, we have a generator z0 of Z(q0) � C0(q0) such that

z02 = z0 + r0 and z0v0 + v0z0 = v0 for all v0 2 V 0. The element

z = e2m+1 (1� 2z0)

of C(ha2m+1i) bC(q0) = C(q) is of degree 1 and commutes with all elements of C0(q),

thus z 2 Z(q). Since

z2 = a2m+1(1 + 4r0) = (�1)md0(e1; : : : ; e2m; e2m+1);

the element z is a homogeneous generator of Z(q). We call the class of

(�1)md0(e1; : : : ; e2m+1) in K�=K�2 the signed 1

2{discriminant of (V; q) and denote it

by 12Æ(q). The element 1

2Æ(q) depends only on the isometry class of (V; q). By the

above discussion, we have for spaces of the same odd dimension

Z(q) ' Z(q0) (as graded algebras), 12Æ(q) = 1

2Æ(q0):

Observe that if (V; q) = [a1; b1] ? � � � ? [am; bm] ? ha2m+1i and char K = 2, then

12Æ(q) = [a2m+1] 2 K�=K�2. Summarizing, we have

Proposition 5. 1) Let (V; q) be a nonsingular quadratic form of even dimension.

There exists an element z 2 C0 such that

i) Z(C0) = K � 1�K � z; z2 = z + r; r 2 K and 1 + 4r 2 K�.

ii) The class of (1 + 4r) in K�=K�2 is the signed discriminant of q.


iii) vz + zv = v for all v 2 V , hence xz + zx = x for all x 2 C1.

2) Let (V; q) be 12{regular of odd dimension. There exists an element z 2 C1 such

that

i) Z(C) = K � 1�K � z; z2 = s; s 2 K�.

ii) The class of s in K�=K�2 is the signed 12{discriminant of q.

The quadratic algebra Z(q) has a unique standard involution �0. We now study

how �0 is related to the structure of the Cli�ord algebra C(q). By the universal prop-

erty of the Cli�ord algebra, there exists a unique K{linear involution � of C(q) such

that �(v) = �v for v 2 V . We call � the standard involution of C(q) (even if � is

not necessarily a standard involution in the sense of Chapter 3). The involution �0

of C which is the identity on V will be called the canonical involution of C. Fur-

ther, let C(�1) be the automorphism of C(q) such that C(�1)(v) = �v for all v 2 V .

Proposition 6. 1) If (V; q) is 12{regular of odd dimension, then �0 is the restriction

of C(�1) to Z(q): 2) Assume that (V; q) is nonsingular of even dimension. Then

i) �0(x)y = yx for all x 2 Z(q) and y 2 C1.

ii) If the dimension of V is congruent to 2 (mod 4), �0 is the restriction of

the standard involution � of C(q).

iii) If the dimension of V is congruent to 0 (mod 4), the standard involution

� of C(q) induces the identity on Z(q).

Proof. Let z 2 Z(q) be as in Proposition 5. The claims follow from Proposition

5, using that �0(z) = �z if the dimension of V is odd and �0(z) = 1 � z if the

dimension of V is even. Observe that we cannot describe �0 if DimV � 0(4).

In characteristic 2, the discriminant does not give any information on q or Z(q),

since Æ(q) = 1 for any nonsingular form. On the other hand the element r 2 K of

Proposition 5 can be used to characterize Z(q). We have

Lemma 7. Let K be a �eld of characteristic 2 and let Z = K � 1 � K � z; Z 0 =

K �1+K �z0 be quadratic K{algebras with z2 = z+r; z02 = z0+r0; r; r0 2 K. Then

Z and Z 0 are isomorphic if and only if there exists � 2 K such that r0� r = �2+�.


Proof. Let ' : Z ! Z 0 be an isomorphism given by '(z) = �z0+�. It follows from

'(z2) = ('(z))2 that �2 = � and �2r0�r = �2+�. Hence � = 1 and r0�r = �2+� as

claimed. Conversely we de�ne an isomorphism ' : Z ! Z 0 by putting '(z) = �z0+�.

The set of elements of K of the form �2 + �; � 2 K is an additive subgroup

of K if char K = 2. We denote it by }(K). Let (V; q) be a quadratic space and let

z 2 Z(q) be a generator as given by Proposition 5, i.e. z2 = z + r; r 2 K. The

class of r in the additive group K=}(K) is the classical Arf invariant of (V; q). We

denote it by �(q). In view of Lemma 7, �(q) is an invariant of the isometry class of

(V; q) and by Lemma 4

�(q1 ? q2) = �(q1) + �(q2):

If (V; q) = [a1; b1] ? � � � ? [am; bm], then

�(q) = a1b1 + � � �+ ambm:

In particular �(q) = 0 for any hyperbolic space (V; q).

The following formulas hold for orthogonal sums:

Lemma 8. 1) If (V2; q2) is nonsingular of even dimension and (V1; q1) is any

quadratic form, then

C(q1 ? q2) ' C((1 + 4r2)q1) C(q2);

where r2 is as in Proposition 5 (for (V2; q2)).

2) If (V2; q2) is12{regular of odd dimension and (V1; q1) is any quadratic form, then

C0(q1 ? q2) ' C(�s2q1) C0(q2);

where s2 is as in Proposition 5 (for (V2; q2)). In particular

C0(h�i ? q) ' C(��q)

for any quadratic form q : V ! K.


Proof. 1) Let z2 and r2 be as in Proposition 5 (for (V2; q2)). The element w2 = 1�2z2satis�es �0(w2) = �w2 and w2

2 = 1 + 4r2, where �0 is the standard involution of

Z(q2). We de�ne a homomorphism

' : C(q1 ? q2) = C(q1) bC(q2)! C((1 + 4r2)q1) C(q2)

by '(v1 1) = v1 w2 and '(1 v2) = 1 v2 for vi 2 Vi (apply Proposition 6,

the universal property of the graded tensor product and the universal property of

the Cli�ord algebra !). Since V1 and V2 are contained in the image of ', ' is an

isomorphism.

2) Let z2 2 Z(q2); s2 2 K be as in Proposition 5 (for q2). We de�ne

: C(�s2q1) C0(q2)! C0(q1 ? q2)

by (1 x) = j2(x), where ji : C(qi) ! C(q1 ? q2) is the canonical embedding,

i = 1; 2; and (v1) = j1(v1)j2(z2) for v1 2 V1. The map is surjective, hence an

isomorphism.

Another important invariant of quadratic forms is the Witt invariant w(q),

which takes values in the Brauer group Br(K) of the �eld K. We de�ne

w(q) =

([C(q)] 2 Br(K) if q is nonsingular of even dimension[C0(q)] 2 Br(K) if q is 1

2-regular of odd dimension.

The Witt invariant obviously depends only on the isometry class of q. Further, by

the proof of Theorem 8, Chapter 4, we have

w(H(K) ? q) = w(q) and w(H(K)) = 1

so w(H(U)) = 1 for any hyperbolic space H(U). We now compute the Witt in-

variant of an orthogonal sum q1 ? q2 with q2 of even dimension. For any class

Æ 2 K�=K�2 and any quadratic form q, the isometry class of Æ � q is well de�ned.

With this in mind, we get from Lemma 8:

Proposition 9. Let q2 be a nonsingular form of even dimension.

1) If q1 is nonsingular of even dimension, then

w(q1 ? q2) = w(Æ(q2)q1)w(q2):


2) If q1 is12{regular of odd dimension, then

w(q1 ? q2) = w(q1)w(�12Æ(q1)q2):

Let (V; q); (V 0; q0) be quadratic forms and let � 2 K�. A K{linear isomorphism

' : V ~!V 0 such that q0('(x)) = �q(x) for all x 2 V is called a similitude with mul-

tiplier �. Two similar forms have isomorphic even Cli�ord algebras:

Lemma 10. Let ' : (V; q) ~!(V 0; q0) be a similitude with multiplier �. There exists

an isomorphism C0(') : C0(q) ~!C0(q0) such that

C0(')(xy) = ��1'(x)'(y) for x; y 2 V:

Proof. Let T 0(V ) be the even part of the tensor algebra of V . The algebra isomor-

phism T 0(') : T 0(V )! T 0(V 0), de�ned by T 0(')(x0� � �x2n) = ��n'(x0) � � �'(x2n),induces the wanted isomorphism.

We describe now another useful isomorphism of Cli�ord algebras. First we need

a de�nition. Let Z be a separable quadratic K{algebra with standard involution

�0 : z 7! z and let � 2 K�. As in Lemma 8 of Chapter 3, we de�ne a K{algebra

A = Z � uZ

of dimension 4 by the multiplication rules

(z1 + uz2)(z3 + uz4) = z1z3 + �z2z4 + u(z2z3 + z1z4):

The algebra A is c:s: of dimension 4, hence a quaternion algebra over K and has

the grading A0 = Z, A1 = uZ. We denote it by (�; Z=K].

Proposition 11. Let (V; q) be nonsingular of even dimension. There exists an

isomorphism of graded algebras

M2(K) C(V; �q) ~!(�; Z(q)=K] C(V; q):

In particular w(�q) = w(q) �w((�; Z(q)=K]).

Proof. Let Z = Z(q) and let '1 : V ! (�; Z=K] C(V; q) be given by '1(v) =

uv; v 2 V: By the universal property of the Cli�ord algebra ' extends to a homo-

morphism ' : C(V; �q)! (�; Z=K]C(V; q) which is injective since C(V; �q) is c:s:


Let A be the commutant of '(C(V; �q)) in (�; Z=K] C(V; q), so A C(V; �q) '(�; Z=K] C(V; q): The algebra A is generated by (Z Z)�0�0 ' K �K and by

u, so A ' (K �K)� u(K�K) ' (�;K�K]. We claim that (�;K�K] 'M2(K):

In fact an isomorphism is induced by K �K ! K 00 K

!and u 7! (01

�0).

Chapter 6

Special Orthogonal Groups and Spin

Groups

Let (V; q) be a quadratic form and let O(q) be its orthogonal group. We denote

by AutgK(C) the group of automorphisms of the Cli�ord algebra C = C0�C1 of (V; q)

as a graded algebra. Any ' 2 O(q) induces an element � = C(') 2 AutgK(C) such

that �V � V . Conversely, let � 2 AutgK(C) be such that �V � V . Then ' = �jVis an automorphism of V such that q('(v)) = '(v)2 = �(v2) = q(v); v 2 V , hence

' 2 O(q). Therefore we can identify O(q) with the subgroup of elements � of

AutgK(C) such that �V � V :

O(q) = f� 2 AutgK(C) j �V � V g:

We assume in the following that (V; q) is either 12{regular of odd dimension or

nonsingular of even dimension. Let Z(q) be the discriminant algebra of (V; q), i.e.

the centre of C if the dimension of V is odd or the centre of C0 if the dimension

of V is even. For any ' 2 O(q) the restriction of ' to Z(q) is an automorphism of

Z(q) as a graded algebra. Thus the restriction de�nes a group homomorphism

� : O(q)! AutgK(Z(q)):

Lemma 1. The group AutgK(Z(q)) is either trivial (if the standard involution �0 of

Z(q) is the identity) or has two elements 1; �0.

Proof. If DimV is even, Z(q) is concentrated in degree zero and has a generator z

such that z2 = z + r for some r 2 K. Let � be a K{automorphism of Z(q) and let

�(z) = �z + �. It follows from �(z2) = �(z)2 that

1 = �+ 2� and �+ r = �2r + �2:

6. Special Orthogonal Groups and Spin Groups 59

Substituting � = 1� 2� in the second equation we get

�2(4r + 1) = �(4r + 1):

By Proposition 5 of Chapter 5, 4r + 1 6= 0, so � = 1 or � = 0 and �(z) = �z + 1 or

�(z) = z. If DimV is odd, Z(q) is generated by an element z of degree 1 such that

z2 = s; s 6= 0 2 K. Thus, if � 2 AutgK(Z(q)), we must have �(z) = �z and �2 = 1.

In view of Lemma 1, we may de�ne a homomorphism O(q) ! ZZ=2ZZ, the

so-called Dickson map, by putting:

Dick(') =

(0 if C(')jZ(q) = 11 if C(')jZ(q) 6= 1:

The kernel of the Dickson homomorphism is called the special orthogonal group and

is denoted by SO(q). This is not the usual de�nition of SO(q) as kernel of the

determinant map, which can only be used if the characteristic is not equal to 2. We

need a di�erent de�nition to include the case of characteristic 2. However the next

lemma shows that the two de�nitions are related. Any ' 2 O(q) is represented by a

matrix (also denoted ') if we �x a basis of V . The determinant of ' is independent

of the choice of the basis. Thus the determinant induces a homomorphism

det : O(q)! K�:

Lemma 2. Assume that (V; q) is either 12{regular of odd dimension or nonsingular

of even dimension. Then

1) det(') = �1 if ' 2 O(q).2) SO(q) � Ker(O(q)

det! f�1g).3) SO(q) = Ker(O(q)

det! f�1g) if (V; q) is 12{regular or char K 6= 2.

Proof. Let fe1; : : : ; eng be a basis of V and let (uij) be the matrix of ' 2 O(q) withrespect to the basis, i.e.

'(ej) =X

uijei:

We have det(uij)2d(e1; : : : ; en) = d(e1; : : : ; en) 6= 0 if q is nonsingular and

det(uij)2d0(e1; : : : ; en) = d0(e1; : : : ; en) 6= 0 if q is 1

2{regular. Thus det(uij)

2 = 1

and the �rst claim is proved. For the second claim, we choose a basis of (V; q) as

in Lemma 17 (resp. 19) of Chapter 1 and choose a generator z of Z(q) as given by

60 6. Special Orthogonal Groups and Spin Groups

Proposition 5 of Chapter 5. Let ' 2 O(q). We claim that

C(')(z) = det(')z + �

for some � 2 K. This is clear if char K = 2, since det(') = 1 by 1) and z 7! z; z 7!z + 1 are the only possible automorphisms of Z(q) by Lemma 1. Assume that char

K 6= 2. We have a unique decomposition

z = �e1 � � � en + terms of lower length

by the P:B:W: theorem. In view of the expression for z computed between Lemma

4 and Proposition 5 of Chapter 5, the coeÆcient � is not zero (since char K 6= 2 !).

We get

C(')(z) = �'(e1) � � �'(en) + terms of lower length

= �det(')e1 � � � en + terms of lower length:

Since, on the other hand, C(')(z) = z + � for some ; � 2 K; 6= 0, we

must have = det(') by the uniqueness of the decomposition z = �e1 � � � en+terms of lower length. The formula C(')(z) = det(')z + � implies 2). We now

check 3) if (V; q) is 12{regular. Since Z(q) is graded and generated by an element

z of degree 1 and since C(') preserves grading, we must have C(')(z) = det(')z.

Thus C(') is the identity on Z(q) if and only if det(') = 1. If char K 6= 2, we

choose an orthogonal basis fe1; : : : ; eng for V . The element w = e1 � � � en is a gen-

erator of Z(q). For any ' 2 O(q), the basis fe0i = '(ei)g is also orthogonal and

C(')(w) = e01 � � � e0n = det(')w. This shows 3) if char K 6= 2.

Example 3. Let (V; q) = H(K). By Example 2 of Chapter 1,

O(q) = f(xu yv) 2M2(K) j xv + uy = 1; xu = 0; yv = 0g

if we identify V with K2 through a basis given by a hyperbolic pair fe; fg. The

element z = ef is a generator of Z(q). We have '(e) = xe + uf; '(f) = ye +

vf if ' = (xuyv): Thus

C(')(z) = (xe+ uf)(ye+ vf) = (xv � uy)z + uy:

If follows that

SO(q) =�

x 00 x�1

!2M2(K); x 2 K�

�' K�:


In characteristic 2, O(q) \ SL2(K) = O(q), so SO(q) is a proper subgroup of

Ker(O(q)det! f�1g).

In the next section we assume that K is a �eld of characteristic not equal to

2. For any quadratic space (V; q) of dimension n, we de�ne the Lie algebra of the

group SO(q) as the set

so(q) = ff 2 EndK(V ) j bq(f(x); y) + bq(x; f(y)) = 0g;

where bq is the polar of q. If q = h1; : : : ; 1i, then so(q) = Altn(K). In particular

DimKso(h1; : : : ; 1i) = n(n�1)2

and the K{vector space so(q) is obviously a Lie algebra

for the Lie product [f; g] = fÆg�gÆf . Since charK 6= 2, any quadratic space (V; q) is

diagonalizable and there exists a �eld extension K � L such that L q ' h1; : : : ; 1i.Thus in general DimKso(q) =

n(n�1)2

and so(q) is a Lie algebra.

Let now [V; V ] be the K{linear subspace of C(q) generated by all [x; y] =

xy � yx; x; y 2 V . It follows from the P.B.W. theorem that DimK[V; V ] =n(n�1)

2.

Further, [V; V ] is a Lie algebra for the product [ ; ] and [[V; V ]; V ] � V . Again this

is easily seen by taking an orthogonal basis of V . Let

: [V; V ]! EndK(V )

be de�ned by (�) = [�; v] = �v � v�; � 2 [V; V ]. We have

bq([�; v]; w) + bq(v; [�; w]) = 0:

This can be checked for � = [x; y] using the de�ning relations of the Cli�ord algebra

and the general case follows by linearity. Thus is a homomorphism of Lie algebras

[V; V ] ! so(q). We claim that is an isomorphism. Since both algebras have the

same dimension, it suÆces to check that is injective. If (�) = 0, then � lies in the

centre of C(q) and in the centre of C0(q) so � 2 K. But K \ [V; V ] = 0. Therefore

the Lie algebra so(q) can be identi�ed with [V; V ] � C0(q). We observe that

[so(q); so(q)] = so(q) if n � 3:

This, again, can be easily checked by taking an orthogonal basis of V and us-

ing the identi�cation so(q) = [V; V ]. It follows that so(q) is always contained in

C0(q)0 = [C0(q); C0(q)]. We shall use this remark later.


We now introduce di�erent subgroups of the group of units of the Cli�ord

algebra C = C0 � C1 of (V; q). These groups will help to describe O(q) and SO(q).

The �eld K can be of arbitrary characteristic. Let u 2 C� be a unit which is

homogeneous , i.e. u 2 C�0 or u 2 C�\C1. We de�ne the graded inner automorphism

igu : C ! C by igu(x) = (�1)@(u)@(x)uxu�1

for x homogeneous, @(u) denoting the degree of u, i.e. @(u) = 0 if u 2 C0 and

@(u) = 1 if u 2 C1. The group

�(q) = fu 2 C�; u homogeneous j igu(V ) � V g

is called the Cli�ord group of (V; q) and the subgroup

S�(q) = fu 2 C�0 j iu(V ) � V g = �(q) \ C0

is the special Cli�ord group.

Let � be the standard involution of C. We de�ne a map

� : C ! C by �(c) = �(c)c; c 2 C:

We have �(v) = �q(v) for v 2 V .

Lemma 4. Assume that (V; q) is nonsingular if the dimension of V is even and

12{regular if the dimension of V is odd. Then �(c) 2 K� for c 2 �(q) and � induces

a group homomorphism

� : �(q)! K�:

Proof. If suÆces to check that �(c) 2 K� for c 2 �(q). The property �(xy) =

�(x)�(y) then is immediate. For any c 2 �(q) and v 2 V , we have igc(v) = ��(igc(v)),since igc(v) 2 V . On the other hand �(igc(v)) = �ig�(c)�1(v), since �(v) = �v. There-fore �(c)cv = v�(c)c for all v 2 V . It follows that �(c)c is in the centre of C0 and in

the centre of C, so �(c)c 2 K�.

Example 5. An element v 2 V belongs to �(q) if and only if q(v) 2 K�, i.e., v is

anisotropic. For if q(v) 2 K�, then

igv(x) = �vxv�1 = �vxvq(v)�1 = (xv � bq(v; x))vq(v)�1

= x� bq(v;x)q(v)

v = �v(x):


Thus igv is the re ection �v.

Since igu(V ) � V for u 2 �(q); igu 2 O(q) and we have a group homomorphism

�(q) ! O(q); u 7! igu. We denote it by �. Since iu is the identity on Z(q) if

u 2 S�(q); � restricts to a homomorphism S� : S�(q)! SO(q).

Proposition 6. The sequences

1! K� ! �(q)�! O(q)! 1

and

1! K� ! S�(q)S�! SO(q)! 1

are exact.

Proof. If igu(v) = v for all v 2 V , then u 2 ZC(C0) = Z(q) by Lemma 1 of Chapter

5. Thus u 2 S�(q) if DimV is even and igu = iu. But then iu(v) = v implies that

u 2 Z(C) = K. If the dimension of V is odd, either u 2 K� or u = �z; z a generator

of Z(q) as given in Proposition 5 of Chapter 5. If u = �z; we get igu(v) = �zvz�1,

thus zv = �vz. But zv + vz = v by Proposition 5 of Chapter 5, thus v = 0 which

is absurd. The surjectivity of � if (V; q) 6' H(IF2) ? H(IF2) follows from Exam-

ple 5 and the fact that O(q) is generated by re ections (Corollary 15 of Chapter

1). The case (V; q) = H(IF2) ? H(IF2) can be checked directly by counting the

order of the groups. We �nally verify that S� is always surjective. Let ' 2 SO(q).We have to show that C(') = iu for some u 2 C�

0 . If DimV is even and Z(q)

is a �eld, then C0 is c:s: over Z(q); C(')jC0 is Z(q){linear so by Skolem{Noether

C(')jC0 = iu; u 2 C�0 . If Z(q) = K �K, we apply Skolem{Noether componentwise

to get C(')jC0 = iu; u 2 C�0 . On the other hand C(') = iv for v 2 C� since C is

c:s:We have ivjC0 = iu, hence u = Æv for Æ 2 Z(q)� and we may assume that v 2 C0.

A similar argument works for the odd dimensional case since Z(q){automorphisms

of C are inner by the generalization of Skolem-Noether mentioned in Remark 12 of

Chapter 2.

Let ' 2 SO(q). By Proposition 6 there exists u 2 S�(q) such that C(') = iu

and by Lemma 4 �(u) 2 K�. We call the class of �(u) in K�=K�2 the spinor norm


of ' and denote it by SN('). The element SN(') does not depend on the choice

of u such that C(') = iu. If iu = iv, then u = �v; � 2 K� and �(u) = �2�(v). Let

now

Spin(q) = fu 2 S�(q) j �(u) = 1g:

Theorem 7. The sequence

1! f�1g ! Spin(q)S�! SO(q)

SN! K�=K�2

is exact.

Proof. Obvious by the de�nition of S� and SN .

Even if its de�nition looks quite complicated, the group Spin(q) is very use-

ful. Moreover, Theorem 7 shows that Spin(q) is a good "approximation" of SO(q).

As we shall see later, for low dimensions, the group Spin(q) is easier to compute

than SO(q). As an algebraic group it is connected and simply connected if q is

of dimension � 3. There are also topological reasons to introduce Spin(q). If

(V; q) = (IRn; h1; : : : ; 1i); n � 3, then Spin(q) is the universal covering of SO(q).

Another classical group associated with a quadratic from (V; q) is its group of

similitudes. We denote it by GO(q). There is an exact sequence

1! O(q)! GO(q)�! K�;

where �('); ' 2 GO(q), is the multiplier of the similitude '. Let fe1; : : : ; eng be abasis of V and let � = (bq(ei; ej)) be the matrix of the polar of q. Further let ' be

the matrix of ' 2 GO(q). We have

't�' = ��

so �n = (det')2. Thus, if n is odd, � is a square, � = �2, and '0 = ��1' is an

isometry. It follows that there exists a direct product decomposition

GO(q) = O(q)�K� if n is odd:

By Lemma 10 of Chapter 5, similitudes of (V; q) induce automorphisms of C0. As-

sume that DimV is even. We say that a similitude ' is special if the induced


automorphism C0(') of C0 is Z(q){linear, i.e. is the identity on the centre of C0.

We denote the group of special similitudes of (V; q) by GO+(q). It contains SO(q)

and K� but is not a direct product of these groups. The computation of GO+(q)

for nonsingular quadratic forms of dimension � 6 over a �eld of characteristic not

equal to 2 was done by Dieudonn�e in a paper of Acta Mathematica 1952 (see the

bibliography). In the next chapters, we shall study forms of dimension � 6 and in

particular reprove Dieudonn�e's results in a characteristic{free way.

Chapter 7

Quadratic Forms of Dimension 2

Let (V; q) be nonsingular of dimension 2 and let C = C0 � C1 be the Cli�ord

algebra of q. We have C1 = V; C0 = Z(q) and C is c:s: of dimension 4. Let � be

the standard involution of C (as a Cli�ord algebra) and �0 the standard involution

of Z(q). By Proposition 6 of Chapter 5, �0 is the restriction of � to C0. Further it is

easy to check, using a basis of C given by the Poincar�e{Birkho�{Witt theorem, that

�(x)x 2 K for all x 2 C. Hence � is the standard involution of C as a quaternion

algebra. Writing an element x 2 C as x = v + y; v 2 C1 = V and y 2 C0, we have

�(x) = �v + �0(y). Thus the reduced norm n of C is given by

n(x) = x�(x) = �q(v) + n0(y);

where n0 is the norm of C0 = Z(q), and we have an orthogonal decomposition

(C; n) ' (V;�q) ? (C0; n0):(7.1)

Proposition 1. Two quadratic spaces of dimension 2 are isometric if and only if

they have the same Witt invariants and the same Arf invariants.

Proof. Let q; q0 be the two forms and let C 0 = C 00 � V 0 be the Cli�ord algebra of

q0. From w(q) = w(q0), we get C ' C 0 and from a(q) = a(q0), we get C0 ' C 00. The

following lemma implies that (C; n) ' (C 0; n) and (C0; n0) ' (C 00; n0), so that the

claim is a consequence of Witt cancellation.

Lemma 2. Let A; B be algebras with standard involutions �A; �B. Any iso-

morphism ' : A ~!B of K{algebras is an isometry of the corresponding norms

nA(x) = x�A(x); nB(y) = y�B(y).

7. Quadratic Forms of Dimension 2 67

Proof. The involution '�A'�1 is a standard involution of B, hence by the unique-

ness of standard involutions '�A'�1 = �B. Thus

nB('(x)) = '(x)�B('(x)) = '(x)'(�A(x)) = '(nA(x)) = nA(x):

Example 3. Let Z be a separable quadratic algebra with standard involution �Z .

We compute the Cli�ord algebra of its norm. By de�nition, we have C(Z; nZ) =

(1; Z=K], where (1; Z=K] is the algebra Z�uZ with the multiplication rules u2 = 1

and xu = u�Z(x). We de�ne a K{linear map

� : Z ! EndK(Z)

by �(x)(y) = x�Z(y). Since �(x)2(y) = nZ(x)y, the map � induces a homomorphism

C(Z; nZ) ! EndK(Z), which is an isomorphism since C(Z; nZ) is c:s: over K. On

the other hand, if z 2 Z is a generator such that z2 = z + r and if we take f1; zg asa basis of Z, it is easy to check that C0(Z; nZ) ' Z. We get

C(Z; nZ) ' EndK(Z) and C0(Z; nZ) ' Z:

Thus, by Lemma 2 above and Lemma 10 of Chapter 5, two quadratic separable

algebras are isomorphic if and only if their norms are similar.

Proposition 4. Let (V; q) be a quadratic space of dimension 2. Then

1) (V; q) is isotropic if and only if Z(q) = C0 ' K �K, i.e. its Arf invariant is

trivial.

2) (V; q) represents 1 if and only if C(q) ' M2(K), i.e. its Witt invariant is

trivial. In particular, C(q) is a division algebra if and only if q does not represent 1.

Proof. 1) Assume that q represents some � 6= 0 2 K and let x 2 V with q(x) = �.

The map C0 ! C1 = V de�ned by c 7! xc is an isometry (C0; �n0) ~!(V; q) since

q(xc) = �n(xc) = �xc�(xc) = �x�(x)n0(c) = q(x)n0(c). Thus the claim follows

from Example 3 and the fact that q is hyperbolic if and only if �q is hyperbolic.

2) Assume that C ' M2(K). Then by Lemma 2,

(C; n) ' (M2(K); det) ' H(K2):

Further, by Proposition 4 of Chapter 1, (V; q) ? (V;�q) ' H(K2), so that (7.1)

and Witt cancellation imply that (V; q) ' (C0; n0). This shows that q represents

68 7. Quadratic Forms of Dimension 2

1. Conversely, if q represents 1, then (V; q) ' (C0; n0) and the claim follows from

Example 3.

Remark 5. The hyperbolic plane H(K) obviously represents any element � 2 K.

Hence if the Arf invariant of a quadratic space of dimension 2 is trivial, its Witt

invariant is trivial. Example 3 shows that the converse is not true.

Proposition 1 can be weakend to the following

Proposition 6. Two quadratic spaces of dimension 2 are similar if and only if they

have isomorphic even Cli�ord algebras, i.e. the same Arf invariants.

Proof. The claim follows from the fact that (V; q) ' (C0; �n0) if q represents the

element �.

We conclude this chapter by computing Spin(q) of a quadratic space (V; q) of

dimension 2. We use freely the notations of Chapter 6. The norm of C as a Cli�ord

algebra is the reduced norm of C, so that �(x) 2 K for all x 2 C. Since V = C1,

we have

S�(q) = C�0

and

Spin(q) = fx 2 C0 j n0(x) = 1g:Lemma 7. Let ' 2 SO(q). Then

1) C(')jC0 = 1C0 .

2) '(x) = ux; x 2 V , for some u 2 C�0 such that n0(u) = 1. Conversely, let

�u : x ! ux; x 2 V and u 2 C0, then �u 2 SO(q) if n0(u) = 1. In particular, the

map u 7! �u is an isomorphism

Spin(q) ~!SO(q):

Proof. 1) Follows from the de�nition of SO(q), since C0 is commutative. We prove

2). Let v 2 V be anisotropic and let q(v) = � 6= 0. We have for x 2 V; C(')(x) =

C(')(��1v2x) = ��1C(')(v)vx since vx 2 C0 and C0 is invariant under C('). The

element u = ��1C(')(v)v lies in C0 and '(x) = C(')(x) = ux. Since the map ' is an


isometry q(x) = q('(x)) = q(ux) = (ux)2 = n0(u)x2 = n0(u)q(x), hence n0(u) = 1.

Conversely let u 2 C�0 with n0(u) = 1, then �u 2 O(q). Since �u(x)�u(y) = u�(u)xy

for x; y 2 V , �u induces the identity on Z(q) and �u 2 SO(q).

The map S� : Spin(q)! SO(q) is given by u 7! iu. We have iu(x) = uxu�1 =

u�0(u)�1x by Proposition 6 of Chapter 5. Since n0(u) = 1; u�0(u

�1)

= u2 and S� can be identi�ed with u 7! �u2 :

Remark 8. By Example 3 any separable quadratic K{algebra is the even Cli�ord

algebra of a quadratic space of dimension 2. We claim that any c:s: K{algebra A

of dimension 4 is the Cli�ord algebra of a quadratic space of dimension 2. As in the

proof of Theorem 7, Chapter 3, we can �nd a separable quadratic K{algebra B � A

and an element u 2 B? � A (? with respect to the reduced norm of A) such that

A = B � uB; u2 = � 2 K� and ub = �B(b)u for b 2 B. By the universal property

of the Cli�ord algebra the map B ! A; b 7! (0; ub); extends to an isomorphism

C(B; �nB) ' A.

Chapter 8


Let (V; q) be a 12{regular form of dimension 3 with Cli�ord algebra C = C0�C1.

By Proposition 5 of Chapter 5, C0 is a c:s: K{algebra of dimension 4, hence a sep-

arable quaternion algebra, the centre Z(q) of C is a quadratic K{algebra generated

by an element z 2 C1 such that z2 = s 2 K and the class of s in K�=K�2 is the

signed 12{discriminant of q.

Example 1. Let A be a separable quaternion algebra (i.e. a c:s: K{algebra of

dimension 4) with standard involution �, trace tr, norm n and let

A0 = fx 2 A j x + �(x) = 0g:We claim that (A0; n), is 1

2{regular of dimension 3. By Remark 8 of Chapter 7, we can

assume that A is the Cli�ord algebra of a space [a; b]. If fx; yg is the correspondingbasis, then A0 has the basis f1 � 2xy; x; yg and (A0; n) ' h4ab � 1i ? [�a;�b] is12{regular, as claimed. The signed 1

2{discriminant of (A0; n) is the class of �1 in

K�=K�2. For A =M2(K), we get

A0 = fx 2M2(K) j tr(x) = 0g ' h�1i ? H(K):

We compute the Cli�ord algebra of (A0;�n). The signed 12{discriminant is trivial.

Let Z be the quadratic K{algebra generated by a symbol ~z such that ~z2 = 1. We

de�ne

' : A0 ! A Z

by '(a) = a ~z for a 2 A0. For any a 2 A, we have a2 � tr(a)a+ n(a) = 0, so that

a2 = �n(a) if a 2 A0 and '(a)2 = �n(a). By the universal property of the Cli�ord

algebra, ' induces a homomorphism (also called ')

' : C(A0;�n)! A Z:


Let fe0; e1; e2g be the basis f1; x; yg (if char K = 2), resp. fxy; x; yg (if char K 6= 2).

The element z = e0(1 � 2e1e2) of Proposition 5, Chapter 5, is mapped to 1 ~z so

that ' maps C0(A0;�n) into A. Since C0(A

0;�n) is c:s: over K; ' restricts to an

isomorphism C0(A0;�n) ' A and is an isomorphism C(A0;�n) ~!A Z. Thus

C(A0;�n) ' A Z; where Z = K[X]=(X2 � 1)

and

C0(A0;�n) ' A:

We now use Example 1 to describe (V; q) in general. Let C(q) = C0 Z(q) and

z 2 Z(q) as above. Since z is of degree 1, the homomorphism � : x 7! zx maps

V into C0. Let � be the standard involution of C as a Cli�ord algebra. Using a

basis of C0 given by the P:B:W: theorem, it is easy to check that x�(x) 2 K for any

x 2 C0. Hence � is the standard involution of C0 as a quaternion algebra. We get

�(zx) = �(x)�(z) = �xz = �zx for x 2 V , so that �(V ) � C 00. Since � is injective

and V; C 00 have the same dimension, � is an isomorphism V ~!C 0

0. Further, since

�n(zx) = (zx)2 = x2z2 = sq(x); � is an isometry

(V; q) ' (C 00;�sn):

We also have V = fx 2 C1 j �(x) + x = 0g since ��1 maps C 00 to the set of elements

x 2 C1 such that �(x) + x = 0. Summarizing we get

Proposition 2. Let (V; q) be a 12{regular form of dimension 3. Then

1) V = fx 2 C1 j x + �(x) = 0g, where � is the standard involution of C.

2) (V; q) ' (C 00;�sn), where C 0

0 = fx 2 C0 j x + �(x) = 0g and s 2 K is such

that [s] = 12disc(q) 2 K�=K�2.

Corollary 3. Two 12{regular forms of dimension 3 are isometric if and only if they

have same Witt invariants and same 12{discriminants. They are similar if and only

if they have same Witt invariants.

Proof. If suÆces to notice that if A; B are two quaternion algebras then, by unique-

ness of standard involutions, any homomorphism of algebras ' : A ! B maps A0

into B0.


Corollary 4. A 12{regular form (V; q) of dimension 3 has trivial Witt invariant if,

and only if, it is isotropic.

Proof. Since q is isotropic if and only if sq is isotropic for any s 2 K�, we can assume

that (V; q) = (A0; n) for some c:s: K{algebra A of dimension 4. Then w(q) = [A] in

Br(K). If [A] = 1, then A = M2(K), and A0 = s`2(K) is isotropic. Conversely, if

(A0; n) is isotropic, (A; n) is isotropic and A =M2(K).

Corollary 5. C0(q) is a division algebra if and only if (V; q) is anisotropic.

We now compute the special Cli�ord group, the spin group and the Lie algebra

of (V; q). Since the norm � of C0 is the reduced norm n of C0; �(x) 2 K for all

x 2 C0. Similarly, we claim that �(x) 2 K for all x 2 C1. Writing x = uz with

u 2 C0, we have �(x) = uz�(uz) = �z2�(u) 2 K. We now claim that iu(V ) � V

for all u 2 C�0 . By Proposition 2, V =

fx 2 C1 j x = ��(x)g: We get �(iu(x)) = �(uxu�1) = ��(u)�1x�(u) = �uxu�1 =

�iu(x) since u�(u) 2 K: This shows that

S�(q) = C�0 and Spin(q) = fu 2 C�

0 j n(u) = 1g:

For any c:s: K{algebra A, we denote the group of elements of Mn(A) of reduced

norm 1 by SLn(A),

SLn(A) = fx 2Mn(A) j nMn(A)(x) = 1g;

so Spin(q) = SL1(C0). By Proposition 2, we have (V; q) ' (C 00;�sn), where [s] =

12Æ(q) 2 K�=K�2, so that

SO(V; q) ' SO(V; sq) ' SO(C 00;�n):

We assume now that (V; q) = (A0;�n) for some quaternion algebra A. It follows

from Example 1 that

C0(V; q) = A and C = A Z

with Z generated by an element z of degree 1 such that z2 = 1. If we identify

V � C1 with zV � C0, then (V; q) is identi�ed with (A0;�n) as a subset of C0 = A.


The map

S� : SL1(A)! SO(A0;�n) = SO(A0; n)

is then given by u 7! iu, where iu is now viewed as an automorphism of C0 (not C).

Similarly

S� : S�(A0; n) = A� ! SO(A0; n)

is identi�ed with u 7! iu; u 2 A�. The map iu is indeed an isometry of A0 since

nA(uxu�1) = nA(x). Proposition 6 of Chapter 6 implies that

SO(q) ' C�0=K

�:

In particular

SO(A0; n) ' A�=K�:

If the 12{regular form (V; q) is anisotropic, C0 is a division quaternion algebra D and

Spin(q) ' SL1(D); S�(q) ' D�:

If (V; q) is isotropic, C0 'M2(K),

Spin(q) ' SL2(K); S�(q) ' GL2(K)

and

SO(q) ' PGL2(K):

Let for example (V; q) = (IR3; h1; 1; 1i) = (IH 0; n). We have S�(q) = IH�. Since

the standard involution of C0 = IH as a Cli�ord algebra is the standard involution of

IH as a quaternion algebra, we get SN(') > 0 for any ' 2 SO(IH 0; n) (by de�nition

SN(') = nIH(u), where C(') = iu). It then follows from Theorem 7 of Chapter 6,

that

S� : Spin(IH 0; n)! SO(IH 0; n)

is surjective. Therefore any ' 2 SO(IH 0; n) can be represented as '(x) = axa�1

with a 2 IH of reduced norm 1.

For the Lie algebra, we get

so(A0; n) = A0 = [A;A];

since so(q) � [C0; C0] and both algebras have the same dimension.

Chapter 9


Let (V; q) be a quadratic space of dimension 4 over K. The Cli�ord algebra

C = C0 � C1 is c:s: of dimension 16, the even Cli�ord algebra C0 is of dimension

8, its centre Z(q) is a separable quadratic K{algebra and C0 is c:s: of dimension 4

over Z(q) if Z(q) is a �eld or C0 ' A� A if Z(q) = K �K, A c:s: of dimension 4

over K.

Example 1. Let A be a quaternion algebra with standard involution a 7! a, norm

n and let � 2 K�. The map

'0 : A!M2(A); '0(a) =

0 a�a 0

!; a 2 A

induces a graded homomorphism ' : C(A; �n)!M2(A), which must be an isomor-

phism by Lemma 8 of Chapter 2. If we identify both algebras through ', we get

C0(A; �n) = A� A; Z(A; �n) = K �K and the standard involution � of C(A; �n)

is given by a bc d

!7!

a ��1c��b d

!:

since its restriction to '0(A) is �1. The canonical involution �0 corresponds to a bc d

!7!

a ��1c�b d

!:

Both involutions � and �0 are of even symplectic type, since they are the tensor

product of the standard involution of A with the involutions x yu v

!7!

x ��1u��y v

!=

1 00 ��

! x yu v

!t 1 00 ��1

!resp.

x yu v

!7!

x ��1u�y v

!;


which are of orthogonal type. Since there is a �eld extension K � L such that

L (V; q) ' H(L2) ' (M2(L); det);

the standard involution and the canonical involution of C(V; q); V a quadratic

space of dimension 4, are of even symplectic type. Since C(A; n) ' M2(A) and

Z(A; n) = K � K, the Witt invariant of (A; n) is the class of A in Br(K) and the

Arf invariant of (A; n) is trivial. Let A; B be quaternion algebras and

: (A; n)! (B; n)

a similitude with multiplier �. By the above computations and Lemma 10 of Chapter

5, induces an isomorphism

0 : A� A = C0(A; n) ~! C0(B; n) = B � B:

Since A and B are c:s: algebras, it follows that A ~! B. This, together with Lemma

2 of Chapter 7, shows that two quaternion algebras are isomorphic if and only if

they have similar norms.

The centre Z(q) of C0 is either a quadratic �eld extension ofK or Z(q) ' K�K.

In the �rst case C0 is a quaternion algebra over Z(q) and in the second C0 ' A�Aas a K�K{algebra (see Theorem 8 of Chapter 4). In both cases C0 has a standard

Z(q){linear involution (the componentwise involution in the second case).

Lemma 2. The standard Z(q){linear involution of C0 as a quaternion algebra is

the restriction to C0 of the standard involution � of C. Further, x�(x) 2 Z(q) for

any x 2 C1.

Proof. There exists a �eld extension K � L such that L (V; q) ' H(L2) '(M2(L); det). Thus by uniqueness of the standard involution, we may assume that

(V; q) = (A; n) for the algebra A = M2(K). The �rst claim then follows from the

formula for the involution given in example 1. For the second claim we may also

assume that (V; q) = (A; n). Let x = (0cb0) be an element of C1 = (0A

A0 ). We have

x�(x) =

��bb 00 ��1cc

!2

K 00 K

!= Z(q);


as claimed.

Let n : C1 ! Z(q) be the quadratic map de�ned by n(x) = x�(x). Even if Z(q)

is not necessarily a �eld, we can view Z(q) (V; q) in an obvious way as a quadratic

space over Z(q).

Lemma 3. The multiplication in C induces an isometry : Z(q) (V; q) '(C1;�n). In particular Z(q) (V; q) is similar to (C0; n).

Proof. Let y 2 Z(q) and v 2 V . Since n(yv) = yv�(yv) = �y2q(v), the map

is a morphism of quadratic spaces. Further is injective because Z(q) (V; q)

is nonsingular. Comparing dimensions shows that is an isomorphism. The last

claim follows by choosing some v 2 V anisotropic and observing that the map

c 7! vc; c 2 C0, is a similitude (C0; n)! (C1; n) with multiplier �q(v) 6= 0.

Assume that the Arf invariant of (V; q) is not trivial, i.e. Z(q) is a quadratic

�eld extension L of K. It follows from Lemma 3 that C0 is a quaternion division

algebra over L if and only if L(V; q) is anisotropic and that C0 ' M2(L) if and only

if L (V; q) is isotropic. On the other hand we claim that L (V; q) is anisotropic if

and only if (V; q) is anisotropic. Assume that L (V; q) is isotropic and that (V; q)

is anisotropic. Let z be a generator of L such that z2 = z + r; r 2 K and let

u+ zv 6= 0 2 L V; u; v 2 V be such that q(u+ zv) = 0. We have

q(u) + rq(v) = 0 and bq(u; v) + q(v) = 0:

Let s = q(v); s 6= 0 since q is anisotropic. We get bq(u; u) bq(u;�v)bq(�v; u) bq(�v;�v)

!=

�2rs ss 2s

!

and det(�2rss

s2s) = �s2(1 + 4r) 6= 0. Therefore u; �v are linearly independent and

qjU ; U = Ku�K(�v), is nonsingular. The form qjU is isometric to (L; sn) where

n is the norm of L, so there exists an orthogonal decomposition

(V; q) ' (L; sn) ? (V 0; q0):

By Example 3 of Chapter 7 and Lemma 10 of Chapter 5, the Arf invariant of (L; sn)

is L. Since L is also the Arf invariant of (V; q); (V 0; q0) must have trivial Arf invari-


ant (consider the cases char K 6= 2 and char K = 2 separately). By Proposition 4

of Chapter 7, (V 0; q0) is isotropic, hence (V; q) is isotropic. This is a contradiction.

Summarizing, we get

Proposition 4. Let (V; q) be a quadratic space of dimension 4 with nontrivial Arf

invariant L. Then

1) C0(q) is a quaternion division algebra over L if and only if (V; q) is anisotropic.

2) C0(q) 'M2(L) if and only if (V; q) is isotropic.

Proposition 5. V = fx 2 C1 j x+ �(x) = 0g.

Proof. Let �0 be the standard involution of Z(q) and let ~� = (�0 1) �1, with

as in Lemma 3. We get

V = fx 2 C1 j ~�(x) = xg

by Lemma 3 (and Galois descent!). But ~�(yv) = �0(y)v = vy = ��(yv) by Propo-

sition 6 of Chapter 5, thus ~� = �� and V = fx 2 C1 j x = ��(x)g as claimed.

We now put Z(q) = Z and identify Z V with C1. Let a; b 2 C0 and x 2 C1.

The map x 7! ax�(b) is a homomorphism

: C0 Z �C0 ! EndZ(Z V ) = Z EndK(V ):

Lemma 6. 1) is an isomorphism and restricts by Galois descent to an isomorphism

: cor(C0) ~!EndK(V ):

2) The involution cor(�) of cor(C0), induced by the standard involution � of

C0, is such that

cor(�) �1(f) = h�1q f �hq; f 2 EndK(V );

where hq : V ~!V � is the adjoint of q and f � is the transpose of f .

Proof. The map is an isomorphism, since it is a map between c:s: algebras of the

same dimension over Z. We prove that cor(C0) ' EndK(V ). Let f = (Pai bi) 2

EndK(V ). It suÆces to check that f = (Pbi ai). Putting x = �(x), we have, by


Proposition 5, f(v) = �f(v) for v 2 V . Since f(v) = Paibi = �P biai = �f(v),

we get f = (Pbi ai) as claimed. To prove 2), we verify that

( (a b))� Æ (1 hq) = (1 hq) Æ (a b)(9.1)

for a; b 2 C0. We have

(1 hq)(x)(y) = bq(x; y) = �(x y + y x)

by Lemma 3, so that (9.1) reduces to

x�(ay b) + ay b �(x) = a xb�(y) + y�(a xb)(9.2)

for a; b 2 C0; x; y 2 C1. The left hand side of (9.2) is equal to tr(xb y a) and the

right hand side to tr(axby), where tr is the reduced trace of C0 as Z{algebra. Both

sides are equal since tr(uv) = tr(vu) for u; v 2 C0.

Theorem 7. Two quadratic spaces of dimension 4 over K are similar if and only if

they have isomorphic even Cli�ord algebras.

Proof. If q; q0 are similar, then C0(q) ' C0(q0) by Lemma 10 of Chapter 5.

Conversely, let ' : C0(q) ~! C0(q0) be an isomorphism of K{algebras and let Z =

Z(q) be the centre of C0(q). Then ' induces an isomorphism Z ~!Z(q0). We view

C 00 = C0(q

0) as a Z{algebra through this isomorphism, so that ' is Z{linear. In view

of the uniqueness of standard involutions, ' is an isomorphism of Z{algebras{with{

involution. By Lemma 6, ' induces an isomorphism : EndK(V )! EndK(V0) such

that Æ = 0 Æ cor('), where : cor(C0) ~!EndK(V ); 0 : cor(C 0

0) ~!EndK(V0)

are as in Lemma 6. Hence is an isomorphism of algebras{with{involution. Fixing

isomorphisms EndK(V ) ' M4(K); EndK(V0) ' M4(K), we can write by Skolem{

Noether, (f) = �f��1 for some K{linear isomorphism � : V ~!V 0. Since is an

isomorphism of algebras{with{involution, we get

�h�1q f �hq�

�1 = h�1q0 (� f �

�1)�hq0

for all f 2 EndK(V ). This implies that � � hq0 = ��hq� for some � 2 K�. Therefore

� is a similitude (V; bq) ~!(V 0; bq0) of the polar forms. To check that � is in fact a

similitude of the quadratic spaces, we observe that


1) the reduced trace tr : C0 ! Z induces an isomorphism

t : HomC0(Z M;C0)! HomZ(Z M;Z);

where we view Z M = C1 as a left C0{module through multiplication in C.

2) t�1 Æ (1 hq)(v)(v) = 1 q(v) for v 2 V .3) The reduced trace commutes with '.

3) follows by uniqueness of the standard involution. 1) and 2) can easily be checked

in the case (V; q) = (A; n) (see Example 1) and the general case follows from the

fact that there exists a �eld extension K � L such that L (V; q) ' H(L2) =

(M2(L); det).

Corollary 8. Two quadratic spaces of dimension 4 are similar if they have the same

Witt and Arf invariants.

Proof. By Lemma 9 of Chapter 4, we have [C0(q)] = [Z(q) C(q)] in Br(Z(q))

(assume that Z(q) is a �eld or do the necessary changes if Z(q) = K �K !). There-

fore C0(q) ' C0(q0) if Z(q) ' Z(q0) and [C(q)] = [C(q0)] in Br(K).

Not any multiplier � can occur in Corollary 8. We have:

Lemma 9. Let (V; q) be a quadratic space of even dimension. Then w(q) = w(�q)

if and only if � = n0(�); � 2 Z(q)� and n0 is the norm of Z(q).

Proof. If � = n0(�), then V ! C(q) given by v 7! �v extends to an isomorphism

C(�q) ~!C(q). If w(q) = w(�q), then w((�; Z(q)=K]) = 1 by Proposition 11 of

Chapter 5. The map

Z(q)! (�; Z(q)=K] = Z(q)� uZ(q)

given by x 7! (0; ux) extends by the universal property of the Cli�ord algebra to an

isomorphism

C(Z(q); �n0) ' (�; Z(q)=K]:

By Proposition 4 of Chapter 7, C(Z(q); �n0) ' M2(K) if and only if �n0 represents


1, i.e. n0 represents �.

Remark 10. Let Z be a quadratic algebra and let B be a quaternion algebra over

Z. By Lemma 6 a necessary condition for B to be isomorphic to the even Cli�ord

algebra of a quadratic space of dimension 4 is that the class of the corestriction of

B in Br(K) is trivial. As shown in Knus{Parimala{Sridharan this condition is also

suÆcient.

Remark 11. As an application of Theorem 7, (and Example 1) we get that a

quadratic space of dimension 4 and trivial Arf invariant is similar to the norm of a

quaternion algebra. This could of course be easily checked directly.

To discuss the structure of C (in particular to give conditions under which C is

a division algebra) we need some results of Chapter 11. We shall see in Proposition

8 of Chapter 11 that C is a division algebra if and only if (Z(q); n0) ? (V;�q) isanisotropic.

We now compute the groups Spin(q), the Lie algebra so(q) and GO+(q) for

quadratic spaces of dimension 4. By de�nition the group Spin(q) is a subgroup of

the group SL1(C0) of units of C0 of Cli�ord norm 1. We claim that in fact

Spin(q) = SL1(C0):

We have to check that for any u 2 SL1(C0); iu(V ) � V . By Proposition 4, V =

fx 2 C1 j x+ �(x) = 0g. We get

�(iu(x)) = �(uxu�1) = ��(u�1)x�(u) = �uxu�1;

since u�(u) = 1. Thus �(iu(x)) = �iu(x) as claimed.

Assume now that (V; q) has trivial Arf invariant. As observed in Remark 11,

(V; q) is similar to the reduced norm of a quaternion algebra A. Therefore, to

compute Spin(q) and SO(q), we may as well assume that (V; q) = (A; n). As usual,

we denote by a 7! �a the standard involution of A. By Example 1,

C0 =

A 00 A

!�M2(A) = C


or C0 = A� A. The involution � maps (a; b) to (a; b). Since

V =�

0 xx 0

!; x 2 A

�;

the condition iu(V ) � V for u 2 C�0 is equivalent to bxa�1 = ax b�1 for all x 2 A.

This, in turn, is equivalent with n(a) = n(b). Thus

S�(q) = f(a; b) 2 A� A j n(a) = n(b)g;

Spin(q) = f(a; b) 2 A� A j n(a) = n(b) = 1g = SL1(A)� SL1(A)

and the homomorphism S� : S�(q) ! SO(q), resp. Spin(q) ! SO(q) is given by

S�(a; b)(x) = bxa�1. If A =M2(K), we get

Spin(H(K2)) = SL2(K)� SL2(K):

If A = IH, the spinor norm of any ' 2 SO(IH; n) is trivial, so that by Theorem

7 of Chapter 6, any ' 2 SO(IH; n) can be represented as '(x) = bxa�1 for a; b

quaternions of norm 1. In general, by Proposition 6 of Chapter 6 and the above

computation of S�(q), any ' 2 SO(A; n) can be written as '(x) = bxa�1 for ele-

ments a; b 2 A of equal norm. We recall that A = M2(K) if (V; q) is isotropic and

A is a division algebra if (V; q) is anisotropic.

If the Arf invariant of (V; q) is not trivial, the centre of C0(q) is a quadratic

�eld extension L of K and C0 is a quaternion algebra over L. Thus C0 is either a

division algebra over L or C0 ' M2(L). By Proposition 4 the �rst case occurs if

(V; q) is anisotropic and the second if (V; q) is isotropic. Therefore

Spin(V; q) ' SL2(L)

if (V; q) is isotropic and

Spin(V; q) ' SL1(B);

with B = C0 a division algebra over L if (V; q) is anisotropic. We claim that

B ' L B0;

B0 a division algebra over K. By Lemma 5, cor(B) is trivial. By the theorem of

Albert{Riehm (Theorem 11 of Chapter 3), B admits an involution � of the second

kind, i.e. such that its restriction to L is the standard involution �0 of L. Now ��


is an involution of B which is L{linear and such that ��(x) � x 2 L for all x 2 B;since ��(x) � �(x) 2 L. In view of the uniqueness of the standard involution �,

we get �� = � or (��)2 = 1. The map ~� = �� : B ! B is �0{semilinear.

By Galois descent, we have B ' L B0 with B0 = fx 2 B j ~�(x) = xg. One

could also construct B0 such that B = L B0 using a basis of V : Assume that

(V; q) = [a; b] ? [c; d] and let fx1; y1; x2; y2g be a corresponding basis of V . Then

z = x1y1 + x2y2 � 2x1y1x2y2

is a generator of L and the elements x1y1; x1x2; y1x2 generate a subalgebra B0 of

dimension 4 over K. It can be checked that L B0 = C0, so B0 is c:s: over K.

The computation of the Lie algebra so(q) is simple. We have so(q) � C 00 =

[C0; C0] in general. Since

C 00 = fx 2 C0 j x+ tr(x) = 0g

is of dimension 3 over Z, we must have

so(q) = C 00 = fx 2 C0 j x + tr(x) = 0g:

We �nally compute the group of special similitudesGO+(q) of a quadratic space

of dimension 4. Let

� : K� � C�0 ! GO(q)

be de�ned by �(�; u)(x) = ��1uxu. Since M = fx 2 C1 j x + �x = 0g (see (4.2.3)),

�(�; a)(x) is a linear automorphism of V for all a 2 C�0 ; � 2 R�. Further it follows

from

q(�(�; a)(x)) = (�(�; a)(x))2

= ��2ax�aax�a= ��2a�0(�(a))q(x)�a= ��2n0(�(a))q(x)

that �(�; a) is a similitude with multiplier ��2n0(�(a)), where �(a) = a�a 2 Z (see

(4.1.1)). We now check that �(�; a) 2 GO+(q). We have for all elements x; y of V

�(�; a)(x)�(�; a)(y) = ��2ax�aay�a= ��2a�0(�(a)xy�a= ��2�(a)�0(�(u)axya

�1

= ��2n0(�(a))axya�1


Since the products xy; x; y 2 V , generate the algebra C0, we see that the extension

C0(�(�; a)) to C0 of the similitude �(�; a) is the inner conjugation ia by a. Thus

C0(�(�; a)) is Z{linear and, by de�nition, �(�; a) is a special similitude. We now

claim that the kernel of � is the subgroup

f(n0(z); z) 2 K� � Z�j z 2 Z�g

of K� � C�0 . It follows from �(�; a)) = 1 that C0(�(�; a)) = ia = 1C0 and a is an

element of the centre Z of C0. We have a = �a for a 2 Z. Thus �(�; a)(x) = x

implies ��1a�0(a)x = ��1n0(a)x = x for all x 2 V . Since V is nonsingular, we get

n0(a) = � as claimed. We �nally claim that � is surjective. Let Z = Z(q). For any

Z{module M , let �0M be the Z{module M with the action of Z twisted through

�0. The subset E of M2(C) consisting of matrices a bc d

!2

�0C0 �0C1

C1 C0

!

is a Z{algebra. The map

Z V ! E; x 7!

0 xx 0

!;

extends by the universal property of the Cli�ord algebra to a homomorphism Z C ! E, which is an isomorphism since both algebras have the same dimension and

C is c:s: over K. We use it to identify Z C with E. The standard involution of

Z C is

� :

a bc d

!7! a cb d

!

since � restricted to Z V = f(0x x0); x 2 Z V g is �1ZV . Let f 2 GO+(q) with

multiplier �. The map 0 �xx 0

!7!

0 ��1f(x)

f(x) 0

!; x 2 V;

extends by the universal property of the Cli�ord algebra to an automorphism ' of

Z C. It is easy to check that the restriction '0 of ' to

Z C0 =

�C0 00 C0

!

is the automorphism C0(1 f). By Skolem{Noether (and Remark 12 of Chap-

ter 2), ' = iu for some u 2 (Z C)�. Since f 2 GO+(q); '0 = C0(1 f) is


Z Z{linear, hence we also have '0 = iv; v 2 (Z C0)�. Since iv = iu on

ZC0 u = rv; r 2 ZZ = (Z00Z) and u 2 (ZC0)

�. Let u = (a00d) 2 (ZC0)

�,

so that f(x) = dxa�1. Since V = fx 2 C1 j x + x = 0g we get dxa�1 = a�1xd for

all x 2 V or cx = xc with c = ad. It follows that cx 2 V for all x 2 V . Thus we get(cx)2 = cxx�c = c�cq(x) 2 K� for all x 2 V , so that c�c 2 K�. Since also �cxc 2 V , wehave c�cxc = �cxc�c, hence xc = �cx for all x 2 V . It follows from xyc = x�cy = cxy that

c 2 Z and, since �c = c for c 2 Z, that xc = cx for all x 2 V . This �nally implies

that c 2 K�, let c = �, so that a = �d�1

and f(x) = ��1dxd for d 2 C0, as claimed.

Summarizing, we get

Theorem 12. Let (V; q) be a quadratic space of dimension 4 with Cli�ord algebra

C = C0�C1. The homomorphismK��C�0 ! GO(q); (�; u) 7! �(�; u)(x) = ��1uxu

induces an isomorphism

K� � C�0=f(n0(�); �); � 2 Z� g ~!GO+(q);

where Z is the centre of C0.

We describe some special cases of Theorem 12. If (V; q) has trivial Arf invariant,

we have GO+(q) ' GO+(A; n) for some quaternion algebra A, since (V; q) is similar

to (A; n). Then

C�0 = A� � Z�; Z� = K� �K� and n0(Z

�) = K�;

so

GO+(q) ' A� � A�=f(�; �); � 2 K�g:

In particular

GO+(q) ' GL2(K)�GL2(K)=f(�; �); � 2 K�g

if further (V; q) is isotropic, i.e. (V; q) = H(K2).

If the Arf invariant of (V; q) is not trivial, Z is a quadratic �eld extension of K and

Z (V; q) is similar to a quaternion algebra (B; n) over Z. Then

GO+(q) = K� � B�=f(n0(z); z); z 2 Z�g;


where B is a division algebra if (V; q) is anisotropic and B = M2(Z) if (V; q) is

isotropic. In this case

GO+(q) = K� �GL2(Z)=f(n0(z); z); z 2 Z�g:

Chapter 10

The PfaÆan

We �rst recall the de�nition and some properties of the pfaÆan of an alternating

matrix and then de�ne a reduced pfaÆan for c:s: algebras. Let xij; 1 � i < j � 2m

be indeterminates and let xji = �xij for i < j; xii = 0; i = 1; : : : ; 2m. The matrix

� = (xij) with entries in ZZ[xij] is called the generic alternating (2m� 2m){matrix.

In view of Lemma 2 of Chapter 3 its determinant is a square in IQ(xij). Since det� 2ZZ[xij] and ZZ[xij] is integrally closed in IQ(xij), det� is the square of a polynomial

pf(�) in ZZ[xij]. The polynomial pf(�) is uniquely determined up to a factor �1.We normalize pf(�) by requiring that pf(sm) = 1 for sm = diag(( 0

�110); : : : ; (

0�1

10)).

In particular, pf(�) is homogeneous of degree m, i.e. pf(��) = �mpf(�) for � 2 K.

For example, we get

pf

0BBB@0 x12 x13 x14

0 x23 x240 x34

0

1CCCA = x12x34 � x13x24 + x14x23:

Let now K be a �eld and ZZ[xij]! K[xij ] be induced by the unique homomorphism

ZZ ! K. Specializing � 7! � for any alternating (2m� 2m){matrix �, we get

det(�) = pf(�)2:

Lemma 1. For any � 2 Alt2m(K) and � 2M2m(K), we have

pf(�t��) = det(�)pf(�):

Proof. It suÆces to verify the claim over ZZ for � and � generic matrices (i.e. with

indeterminates aij for � and uij for � as entries). We have pf(�t��)2 = det(�t��) =

det(�)2pf(�)2 over the polynomial ring ZZ[aij; uij], so that pf(�t��) = �det(�)pf(�)

and the sign � is independent of the choice of � and �. Specializing � to the identity

10. The PfaÆan 87

matrix shows that the sign + must hold.

Let A be a c:s: K{algebra of dimension 4m2 with a K{linear involution � . We

assume that either � is of even symplectic type or that A is isomorphic as an algebra

with involution to the tensor product of two quaternion algebras with the involution

given by the tensor product of the two standard involutions. For example M4(K),

with the involution x 7! xt, is isomorphic to such a tensor product. Let " be the

type of � and let

Alt� (A) = fx� "�(x); x 2 Ag

be the set of alternating elements of A. We claim that there exists a reduced pfaÆan

pfA : Alt� (A)! K such that

pfA(x)2 = nA(x) and pfA(�(a)xa) = nA(a)pfA(x)

for x 2 Alt� (A); a 2 A, nA denoting the reduced norm of A. We �rst consider

the case where � is of even symplectic type. Let � : L A ~!M2m(L) be a Galois

splitting of A. The involution �(1 �)��1 of M2m(L) is of the form �u(x) = uxtu�1

with u 2 GL2m(K) an alternating matrix. In view of Lemma 2 of Chapter 3, we can

modify � by an inner automorphism of M2m(L) in such a way that u is the matrix

sm de�ned above.We get by Lemma 1 of Chapter 3,

�(L Alt� (A)) = Alt�u2m(L)

= sm � Alt2m(L) = Alt2m(L) � s�1m

and we de�ne

pfA(a) = pf(�(1 a)sm) for a 2 Alt� (A):

We �rst check that pfA(a) 2 K of all a 2 Alt� (A). For any g 2 Gal(L=K), let

~g : M2m(L) ! M2m(L) be given by ~g((aij)) = (g(aij)) i.e. ~g acts entrywise. The

map �(g1)��1~g�1 is an L{automorphism ofM2m(L) which respects the involution

�(1 �)��1 of M2m(L). Hence there is c 2 GL2m(L) such that

�(g 1)��1(x) = c~g(x)c�1 for all x 2M2m(L)

and �sm = csmct for some � 2 L�. We have

pfA(a) = pf(�(1 a)sm) = pf(c~g(�(1 a))c�1sm)

88 10. The PfaÆan

since �(g 1)��1(�(1 a)) = �(1 �). On the other hand c�1sm = ��1smct, so

that

pfA(a) = pf(��1c~g(�(1 a))smct)

= ��mdet(c)pf(~g(�(1 a))sm):

We get �m = det(c) by taking the pfaÆan of �sm = csmct and

pf(~g(�(1 a))sm) = pf(~g(�(1 a)sm))

= g(pf(�(1 a)sm)) = g(pfA(a));

since sm is de�ned over K. This shows that pfA(a) 2 K. To verify that pfA is

independent of the choice of the splitting (L; �), it suÆces (as for the characteristic

polynomial) to consider splittings (L; �) and (L; �0) (i.e. the same �eld L). Let

�0 = iy Æ �; y 2 GL2m(L). Since �(1 �)��1 and �0(1 �)�0�1 give the involution

�sm ofM2m(L), we get by the discussion between Lemma 1 and Lemma 2 of Chapter

3, �sm = ysmyt for some � 2 L�. Let now pfA be the pfaÆan de�ned through �

and pf 0A the pfaÆan de�ned through �0. We have

pf 0A(a) = pf(�0(1 a)sm) = pf(y�(1 a)y�1sm)

= pf(��1y�(1 a)smyt) = ��mdet(y)pf(�(1 a)sm)

= pfA(a)

since �sm = ysmyt implies �m = det(y) (take the pfaÆan on both sides). The

formulas

pfA(x)2 = nA(x) and pfA(�(a)xa) = nA(a)pfA(x)

follow from

pfA(x)2 = pf(�(1 x)sm)

2 = det(�(1 x)sm) = det(�(1 x)) = nA(x)

and

pfA(�(a)xa) = pf(sm�(1 a)ts�1m �(1 x)�(1 a)sm)

= pf((s�1m �(1 a)sm)

t�(1 x)sm(s�1m �(1 a)sm))

= det(�(1 a))pf(�(1 x)sm)

= nA(a)pfA(x):

10. The PfaÆan 89

Further we have pfA(�a) = �mpfA(a) for � 2 K and pfA(1) = pf(sm) = 1 (we

observed earlier that 1 2 Alt� (A) if � is of even symplectic type).

We now de�ne the reduced pfaÆan pfA for A = A1 A2; A1; A2 quaternion

algebras and � = �1 �2 the tensor product of the standard involutions �i of Ai.

We observe that � is of orthogonal type since �1; �2 are of symplectic type. Let

�i : L Ai ~!M2(L) be a splitting of Ai; i = 1; 2 (we can choose the same L for

both algebras). Then � = �1 �2 is a splitting of A and �(1 �)��1 is the tensor

product of two copies of the involution

� = (xuyv) 7!

�v�u

�yx

�= s1�

ts�11 :

Thus �(L Alt� (A)) = (s1 s1)Alt4(L) = Alt4(L)(s1 s1)�1, where

s1 s1 =

0BBB@0 0 0 10 0 �1 00 �1 0 01 0 0 0

1CCCAand we de�ne

pfA(a) = pf(�(1 a)(s1 s1)) for a 2 Alt� (A):

As in the case of even symplectic involutions, one checks that pfA(a) 2 K for all

a 2 Alt� (A) and that pfA does not depend on the choice of the splitting �

(� always of the form �1 �2 !) One has to use that det(u1 u2) = (detu1 � detu2)2for ui 2 GL2(L). We leave the details as an exercise.

Let now A be a c:s: algebra of dimension 16 with an involution � which is either

of even symplectic type or A = A1 A2 and � = �1 �2; Ai a quaternion algebra

with standard involution �i; i = 1; 2.

Lemma 2. pfA : Alt� (A)! K is a nonsingular quadratic form on the vector space

Alt� (A) of dimension 6.

Proof. By de�nition of the reduced pfaÆan, it suÆces to consider the case of the

classical pfaÆan on Alt4(K). The formula given at the beginning of this chapter

shows that

(Alt4(K); pf) ' H(K3):

90 10. The PfaÆan

We now construct a \standard involution" on Alt� (A); A as above. We �rst

consider the case A =M4(K). Let � : Alt4(K)! Alt4(K) be de�ned by

�

0BBB@0 x12 x13 x14

�x12 0 x23 x24�x13 �x23 0 x34�x14 �x24 �x34 0

1CCCA =

0BBB@0 �x34 x24 �x23x34 0 �x14 x13�x24 x14 0 �x12x23 �x13 x12 0

1CCCA :We get x�(x) = �(x)x = pf(x) for x 2 Alt4(K). Further � is, in some sense, unique:

Lemma 3. Let : Alt4(K) ! Alt4(K) be a K{linear automorphism such that

(x)x 2 K for all x 2 Alt4(K) (or x (x) 2 K for all x 2 Alt4(K)). There exists

� 2 K� such that = ��.

Proof. Let fEijg; 1 � i; j � 4 be the standard basis of M4(K). Let

fgig; 1 � i � 6 be the basis fhij = Eij �Ejig of Alt4(K) ordered lexicographically.

The condition (g1)g1 2 K implies that (g1) = �g6; � 2 K and (g6)g6 2 K

implies (g6) = �0g1; �0 2 K. The condition (g1 + g6)(g1 + g6) 2 K then implies

� = �0 and the lemma follows by further similar computations.

If A is c:s: of dimension 16 as above, we shall de�ne �A : Alt� (A) ! Alt� (A)

through a splitting � : L A ~!M4(L) as above. We put

�(1 �A(a)) = s2�(�(1 a)s2) a 2 A

in the even symplectic case and

�(1 �A(a)) = (s1 s1)�(�(1 a)(s1 s1))

if A = A1 A2. Let s = s2 (resp. s1 s1). We have

�(1 a)�(1 �A(a)) = �(1 a)s�(�(1 a)s)

= pf(�(1 a)s) = pfA(a);

�(1 �A(a))�(1 a) = s�(�(1 a)s)�(1 a)

= s(�(�(1 a)s)�(1 a)s)s�1

= spf(�(1 a)s)s�1 = pfA(a)

so

�A(a)a = a�A(a) = pfA(a); a 2 Alt� (A):

10. The PfaÆan 91

Further �A is, up to scalars, the unique isomorphism : Alt� (A) ! Alt� (A) such

that (a)a 2 K for all a 2 Alt� (A) (or such that a (a) 2 K for all a 2 Alt� (A)).

The automorphism �A allows us to compute the Cli�ord algebra of (Alt� (A); pfA):

Proposition 4. Let � 2 K�. The map

Alt� (A)!M2(A); x 7!

0 �A(x)�x 0

!induces an isomorphism ' : C(Alt� (A); �pfA) ~!M2(A).

Proof. The existence of ' follows from the universal property of the Cli�ord alge-

bra. It is an isomorphism since C(Alt� (A); pfA) is c:s: and both algebras have the

same dimension.

If we identify C(Alt� (A); pfA) with M2(A), we get C0 = (A00A); C1 = (0A

A0 )

and the standard involution � of C is given by

�

a bc d

!=

d "b"c a

!=

0 "1 0

! d bc a

! 0 "1 0

!where a 7! a = �(a) is the involution of A and " is the type of � . Indeed, the

restriction of � to

V '�

0 �A(x)�x 0

!; x 2 Alt� (A)

�is �1V .

Let A be c:s: over K with an even symplectic involution � and let A[X] =

A K[X] be the polynomial ring in one variable X over A. We extend � to A[X]

by putting �(X) = X. We obviously have

Alt� (A[X]) = Alt� (A)K K[X]

and we get by scalar extension a pfaÆan pfA[X] over A[X] with values inK[X]. Since

1 2 Alt� (A), we can de�ne, for all a 2 Alt� (A), a polynomial ��(X; a) 2 K[X] by

��(X; a) = pfA[X](X � 1� a):

We call ��(X; a) the pfaÆan characteristic polynomial. It is of degree m if A has

dimension 4m2. In particular, if m = 2,

��(X; a) = X2 � pft(a)X + pfA(a)

92 10. The PfaÆan

where pft is a linear form on Alt� (A). We call it the pfaÆan trace of A. We have a

Cayley{Hamilton theorem. i.e. ��(a; a) = 0 for all a 2 Alt� (A).

Example 5. Let B be a quaternion algebra with standard involution b 7! b and let

A =M2(B). The involution

� :

a bc d

!7!

a cb d

!

of A is of even symplectic type since it is the tensor product of the involution x 7! xt

of M2(K) with the standard involution of B. We get

Alt� (A) = f�xb

by

�; x; y 2 K; b 2 Bg:

To compute the pfaÆan, we split B and apply the formula pfA(a) =

pf(�(1 a)sm) (which de�nes pfA(a) !). We get

pfA�xb

by

�= x y � bb:

Let : Alt� (A)! Alt� (A) be de�ned by �xb

by

�=�y�b

�bx

�: Since a (a) = pfA(a),

we must have = �A, so

�A

x bb y

!=

y �b�b x

!:

Further pft(a) = b(a; 1), where b is the polar of pfA, so

pft

x bb y

!= x+ y

(as one would guess!). In particular, we get

a+ �A(a) = pft(a) 2 K for all a 2 Alt� (A):

This strengthens the (heuristic) fact, already mentioned, that �A is a "standard

involution" on Alt� (A).

Chapter 11


At the end of Chapter 10, we have shown that the Arf invariant of a quadratic

space (Alt� (A); �pfA); A a c:s: K{algebra of dimension 16 with an even symplectic

involution and � 2 K�, is trivial. Conversely we claim that

Proposition 1. Let (V; q) be a quadratic space of dimension 6 with trivial Arf

invariant. Assume that (V; q) represents � 2 K�. There exist a c:s: K{algebra A

and an even symplectic involution � on A such that (V; q) ' (Alt� (A); �pfA).

Let C = C0�C1 be the Cli�ord algebra of (V; q) and let e; f be two nontrivial

idempotents generating Z(q) = K �K. Before proving Proposition 1, we describe

V as a subspace of C1. We denote as usual the standard involution of C by �.

Lemma 2. The map V ! V e; x 7! xe, is an isomorphism and

V e = f(x� �(x))e; x 2 C1g = fy � �(y); y 2 C1eg:

Similarly V ' V f and

V f = f(x� �(x))f; x 2 C1g = fy � �(y); y 2 C1fg:

Proof. Since if suÆces to check the claims over an extension K � L, we can assume

that (V; q) = (Alt� (A); pfA) for some c:s: algebra A of dimension 16 with an even

symplectic involution (for example A = M4(K) and �(x) = s2xts�1

2 ). Then by

Proposition 4 of Chapter 10,

C =M2(A); �

a bc d

!=

�(d) ��(b)

��(c) �(a)

!; C0 =

A 00 A

!


and we can choose e = (1000); f = (00

01). The space V is embedded in C as the set

V =�

0 �A(x)x 0

!; x 2 Alt� (A)

�:

All the claims of Lemma 2 can now easily be explicitly checked.

Proof of Proposition 1. The algebra A = C0e (with 1A = e) is c:s: of dimension

16 over K. We shall construct an even symplectic involution � on A such that

(V; q) ' (Alt� (A); �pfA). Replacing q by �q, we may assume that q represents 1.

Let v 2 V be such that q(v) = 1 and let P = vA. The standard involution � of C

maps P to P since

�(vce) = �f�(c)v = ��(c)ve = vc0e; for c0 = �v�(c)v;

by Proposition 5 of Chapter 5. Let � : A ! A be de�ned by �(va) = �v�(a). We

have �(ab) = �(b)�(a) and va = �2(va) = ��(v�(a)) = v� 2(a), so � 2 = 1. Therefore

� is an involution of A and by Lemma 2,

V e = v � Alt� (A):

We claim that the type of � is independent of the choice of v. Let w 2 V with

q(w) = 1 and let w = vu; u 2 A. If � 0 is de�ned by �(wa) = �w� 0(a), we get

�(vua) = �vu� 0(a) = �v�(ua) = �v�(a)�(u):

On the other hand w 2 V so �(vu) = �vu and �(vu) = �v�(u). It follows that

u = �(u) and � 0 = iu Æ� . Thus � 0 is of even symplectic type if and only if � is of even

symplectic type. To compute the type of � we may now assume (as in the proof of

Lemma 2) that (V; q) = (Alt�1(A); pfA), with �1 an even symplectic involution of A,

and we can choose v = 1. For this choice, we get � = �1, thus � is of even symplectic

type. We �nally check that the map � : V ! Alt� (A) de�ned by xe = v�(x) is an

isometry: we have similarly

V f = Alt� (A) � vand we de�ne : Alt� (A)! Alt� (A) by the formula ��1(a)f = (a)v. We get

(a)a = (a)vva = q(��1(a))e for all a 2 Alt� (A);

so by Lemma 3 (and the discussion following Lemma 3) of Chapter 10 , = � ��A for

some � 2 K� and q(x)e = �pfA(�(x)) (recall that e = 1A !). Since �(v) = e; � = 1


and � is an isometry, as claimed.

In view of the next theorem, the algebra A in Proposition 1 is determined up

to isomorphism by the class of similitudes of (V; q).

Theorem 3. Let A and B be c:s: algebras of dimension 16 over K with pfaÆans

pfA and pfB. Then A ' B if and only if the corresponding pfaÆans are similar.

Proof. Let �A be the involution of A and �B the involution of B. Let � : A ~!B

be an isomorphism of algebras and let �0 = ��A��1. Then � is an isomorphism

(A; �A) ~!(B; �0) of algebras with involutions and we get a pfaÆan pf 0B : Alt�0

(B)!K such that � : (Alt�A(A); pfA)! (Alt�

0

(B); pf 0B) is an isometry. Let �0 = iu Æ �B,so

Alt�0

(B) = uAlt�B(B) = Alt�B(B)u�1:

We claim that the map

u�1� : Alt�A(A)! Alt�B(B)

is a similitude of the corresponding pfaÆans. Let

: Alt�B(B)! Alt�B(B)

be de�ned by (y) = �(�A��1(uy))u. We have (y)y = pfA(�

�1(uy)) 2 K for all

y 2 Alt�B(B). Therefore by Lemma 3 of Chapter 10, there exists � 2 K� such that

� = �B and, as claimed, u�1� is a similitude. Conversely, let

' : (Alt�A(A); pfA)! (Alt�B(B); pfB)

be a similitude. By Proposition 4 of Chapter 10, ' induces a graded isomorphism

M2(A) ~!M2(B), where both matrix algebras have the checker{board grading. In

degree zero, we get A� A ' B �B, hence A ' B.

Example 4. Let A = M2(B); B a quaternion algebra with standard involution

b 7! b. The involution

�1 :

a bc d

!7!

d �b�c a

!


of A is the tensor product of the two standard involutions. We get

Alt�1(A) =�

b xy �b

!; b 2 B; x; y 2 K

�and for the corresponding pfaÆan (use a splitting !):

pf1

b xy �b

!= �xy � bb

so that (Alt�1(A); pf1) ' H(K) ? (B;�nB). On the other hand, (see Example 5 of

Chapter 10)

�2 :

a bc d

!7!

a cb d

!is of even symplectic type. We get

Alt�2(A) =�

x bb y

!; b 2 B; x; y 2 K

�and for the corresponding pfaÆan:

pf2

x bb y

!= xy � bb

so that (Alt�2(A); pf2) ' H(K) ? (B;�nB) and both pfaÆans are indeed similar.

Corollary 5. There exists a bijection between similitude classes of quadratic spaces

of dimension 6 with trivial Arf invariant and isomorphism classes of c:s: algebras

of dimension 16, with even symplectic involutions. The correspondence is given by

(V; q) 7! A = C0e, where e is a nontrivial idempotent of the centre of C0. Further

1) A is a division algebra if, and only if, (V; q) is anisotropic.

2) A 'M2(D); D a quaternion division algebra if, and only, if (V; q) has Witt

index 1.

3) A ' M4(K) if, and only if, (V; q) ' H(K3).

The class of A in Br(K) is the Witt invariant of (V; q).

Proof. The �rst claim follows from Lemma 10 of Chapter 5, Proposition 1 and

Theorem 3; the others, then from Example 4.

We now apply Theorem 3 to prove a classical result of Albert. Observe that,

again, we do not assume char K 6= 2.


Theorem 6. Any c:s: K{algebra of dimension 16, whose class in Br(K) is of order

2, is isomorphic to a tensor product of quaternion algebras.

Proof. Let A be c:s: of dimension 16 such that 2[A] = 0 in Br(K). By Theorem 9 of

Chapter 3, there exists an even symplectic involution � onA. Let pfA : Alt�(A)! K

be the corresponding pfaÆan. We have 1 2 Alt�(A) and pfA(1) = 1. We denote by

b : Alt�(A)� Alt�(A)! K the polar of pfA. Since pfA is nonsingular, there exists

z 62 K � 1 such that b(1; z) = 1. The element z is a root of its pfaÆan characteristic

polynomial, so

z2 = z � r; where r = pfA(z);

and z generates a quadratic separable K{algebra Z. With the notations of Chapter

1, we get [1; z] ' (Z; n0) and

(Alt�(A); pfA) ' [1; z] ? [a; b] ? [c; d]

for some quadratic spaces [a; b]; [c; d]. We remark that [a; b] ? [c; d] has Arf invari-

ant Z (compute the discriminant if char K 6= 2 and the classical Arf invariant if char

K = 2). Let now (V1; q1) = [a; b]; (V2; q2) = [c; d], (V; q) = (V1; q1) ? (V2; q2); and

let eC = C(�q) = A1 A2 with A1 = C(�Æ(q2)q1); A2 = C(�q2) (see Lemma 8 of

Chapter 5). The canonical involution �0 of eC is of even symplectic type by Example

1 of Chapter 9 and C0 has centre Z.

Lemma 7. (Alt�0

( eC); pfeC) = (Z; n0) ? (V; q).

Proof. If suÆces to prove Lemma 7 over some �eld extension K � L so that we

can assume that (V;�q) = (A; n), where A is a quaternion algebra with reduced

norm n. By Example 1 of Chapter 9, we get

Alt�0

( eC) = � x bb y

!; x; y 2 K; b 2 A

��M2(A) = C

and pfeC( xb by) = xy � bb as claimed.

By Lemma 7 A and eC = A1 A2 have isometric pfaÆans and by Theorem 3,

A ' A1 A2. This proves Theorem 6. Lemma 7 may also be used to describe the


structure of the Cli�ord algebra of a quadratic space of dimension 4 :

Proposition 8. Let (V; q) be a quadratic space of dimension 4. Then

1) C(q) is a division algebra if, and only, if the space (Z(q); n0) ? (V;�q) isanisotropic.

2) C(q) ' M2(D); D a quaternion division algebra if, and only, if the space

(Z(q); n0) ? (V;�q) has Witt index 1.

3) C(q) 'M4(K) if, and only, if (Z(q); n0) ? (V;�q) ' H(K3):

Proof. By Lemma 7 we have (Alt�0

(C(q)); pfC(q)) ' (Z; n0) ? (V;�q) so that

Proposition 8 follows from Corollary 5.

Example 9. Let A = A1 A2 be a tensor product of quaternion algebras and

let � = �1 �2 be the tensor product of the corresponding standard involutions.

Assume that char K 6= 2 and let A0i = (K � 1)? � Ai for the reduced norms. Then

Alt�(A) = A01 1 + 1 A0

2 ' A01 � A0

2:

We claim that

pfA(x1 + x2) = nA1(x1)� nA2(x2) for xi 2 A0i;

so

(Alt�(A); pfA) ' (A01; nA1) ? (A0

2;�nA2):

It suÆces to check the formula over a �eld extension K � L. Hence we can assume

that A1 =M2(K). By Example 4, we have

Alt�(A) =�

b xy �b

!; x; y 2 K; b 2 A2

�:

Writing b as z + b0; b0 2 A02, we get b = z � b0,

b xy �b

!=

z xy �z

!+

b0 00 b0

!and

pfA

b xy �b

!= �xy � z2 � n(b0) = det

z xy �z

!� n(b0);

as claimed. The quadratic space (A01; nA1) ? (A0

2 � nA2) is called the Albert form

of A1 A2. We denote it by Q(A1; A2). In view of Theorem 3, two tensor prod-

ucts A1 A2 and B1 B2 of quaternion algebras are isomorphic if and only if the


Albert forms Q(A1; A2) and Q(B1; B2) are similar (assuming char K 6= 2). This

result is due to Jacobson. We mention two more results of Albert which are easy

consequences of Theorem 3. We assume char K 6= 2.

1) Let A; B; C be quaternion algebras. If [A][B][C] = 0 in Br(K), then

Q(A;B) is isotropic.

2) Let A; B be quaternion algebras. Then A B ' M2(C) if, and only if,

there exists a quadratic separable algebra Z which is contained in A and B.

1) is clear by Theorem 3. We sketch a proof of 2). If Z � A and Z � B then

(A; nA) ' (Z; n0) ? (V1; q1) and (B; nB) ' (Z; n0) ? (V2; q2) so that (A0; nA)

and (B0; nB) represent a (nonzero) common element. If follows that Q(A;B) is

isotropic. Conversely if AB 'M2(C); Q(A;B) is isotropic and (A0; nA); (B0; nB)

represent a common element � 2 K. If � = 0 A ' M2(K); B ' M2(K) and

the claim is obvious (take Z = K � K). If � 6= 0 and q(v1) = � = q(v2); then

Z = K � 1�K � v1 ' K � 1�K � v2 is the algebra Z.

We now consider quadratic spaces (V; q) of dimension 6 with arbitrary Arf

invariant. Let Z = Z(q) be the centre of the even Cli�ord algebra C0 of (V; q). Let

� be the standard involution of C = C(q) and let

Alt�(C1) = fx� �(x); x 2 C1g:

Lemma 10. The inclusion V ! C1 induces an isomorphism

Z V ~!Alt�(C1):

Proof. Since it suÆces to prove Lemma 10 over a �eld extension K � L, we may

assume that (V; q) = (Alt� (A); pfA) for A a c:s: K{algebra of dimension 16 with an

even symplectic involution � : a 7! a. The claim then follows easily from Example 4.

We now assume that q represents 1, let v 2 V with q(v) = 1, so v2 = 1 in C.

We have �(v) = �v and �1 = iv Æ � is a Z{linear involution of C0 (observe that �

is �0{semilinear, where �0 is the standard involution of Z). We claim that �1 is an

involution of C0 of even symplectic type. The type does not depend on the choice

of v 2 V (such that v2 = 1), so we can take (V; q) = (Alt� (A); pfA) as above and


v = (0110) = 1A. Then

�1

a 00 b

!=

�(a) 00 �(b)

!:

Since � is of even symplectic type, �1 is of even symplectic type. By de�nition of

�1, the maps C1 ! C0 given by �v : x! vx and �v : x 7! xv induce isomorphisms

�v; �v : Alt�(C1) ~!Alt�1(C0):

Let pf1 : Alt�1(C0) ! K be the pfaÆan given by �1 and let �1 : Alt�1(C0) !Alt�1(C0) be such that �1(x)x = x�1(x) = pf1(x) for all x 2 Alt�1(C0). We claim

that �v is an isometry

�v : (Z V; 1 q)! (Alt�1(C0); pf1):

We have (1 q)(x) = �x�(x) for all x 2 Z V = Alt�(C1). Let : Alt�1(C0) !Alt�1(C0) be de�ned by (xv) = �v�(x) for x 2 ZV . We get xv (xv) = �x�(x) =(1q)(x), so = ��1 for some � 2 K�. Putting x = v shows that � = 1; = �1 and

as claimed, �v is an isometry. Further we know that V = fx 2 Z V j �(x) = �xgor, since �(x) = �v�1(xv),

V = fx 2 Z V j v�1(xv) = xg:(11.1)

We are now ready to prove the following

Theorem 11. Let (V; q); (V 0; q0) be quadratic spaces of dimension 6 and let

C0 = C0(q); C00 = C0(q

0) with the standard involutions. Then (V; q) and (V 0; q0) are

similar if and only if C0 and C00 are isomorphic as algebras{with{involution.

Proof. We assume that the centre Z of C0 is a �eld (the other case is similar).

Since ' maps Z to the centre Z 0 of C 00, we view C 0

0 as a Z{algebra through '. Then

' is a Z{linear isomorphism. Let ' : C0 ~!C 00 be an isomorphism as algebras{with{

involution, i.e. such that '� = �'; � the standard involution. We can assume

that V and V 0 represent 1 (since only similitude classes matter), so let v 2 V

with q(v) = 1 and v0 2 V 0 with q0(v0) = 1. Let �1 be the involution iv Æ � of C0

and �01 = iv0 Æ �. Further let pf1 : Alt�1(C0) ! Z, pf 01 : Alt�0

1(C 00) ! Z be the

corresponding pfaÆans and

�1 : Alt�1(C0)! Alt�1(C0); �01 : Alt

�01(C 00)! Alt�

0

1(C 00)


be such that x�1(x) = pf1(x); x0�01(x0) = pf 01(x). The involution �

0 = '�1 '�1 of

C 00 is of even symplectic type and, writing �0 = iu Æ �01; u 2 C�

0 , we get �01(u) = u

or v0�(u)v0 = u. Since '� = � ', it follows that ' iv � '�1 = ' iv '

�1 � = iu iv0 �

so that ' Æ iv = iuv0 Æ ' or '(vxv) = uv0'(x)v0u�1. Replacing x by v x v, we get

'(x) = uv0'(vxv)v0u�1, so there is � 2 Z� such that �uv0 = v0u�1. This, together

with v0�(u)v0 = u, implies that �(u)u 2 Z�, so also u�(u) 2 Z�. The Z{linear

isomorphism C0 ! C 00; x 7! '(x)u maps Alt�1(C0) to Alt�

0

1(C 00) since

'(Alt�1(C0)) = Alt�0

(C 00) = Alt�

0

1(C 00) � u�1

= u � Alt�1(C 00):

Let now : Alt�1(C0)! Alt�1(C0) be de�ned by

(x) = '�1(u�01('(x)u)):

Since

x (x) = '�1('(x)u�01('(x)u)) = pf 01('(x)u) 2 Z;there exists � 2 Z� such that = ��1 and x 7! '(x)u is a similitude

(Alt�1(C0); pf1)! (Alt�0

1(C 00); pf

01)

with multiplier �. We compute �. By the discussion preceeding Theorem 11, the

map x 7! '(xv)uv0 is a similitude Z (V; q)! Z (V 0; q0) with multiplier �. Since

(1 q0)(x0) = �x0�(x0), we get (1 q0)('(xv)uv0) =

('(xv)uv0)�('(xv)uv0) = '(xv)u�(u)�'(xv). Now u�(u) 2 Z and �' = '�, so

(1 q0)('(xv)uv0) = u�(u)(1 q)(x). It follows that � = u�(u) and �(�) = �0(�) =

�, so � 2 K�. We claim that, in fact, x 7! '(xv)uv0 restricts to a similitude

(V; q)! (V 0; q0) with multiplier �. We have by (11.1)

V = fx 2 Z V j v�1(xv) = xg; V 0 = fx0 2 Z V 0 j v0�01(x0v0) = x0g

so we have to check that v0�01('(xv)u) = '(xv)uv0 if x = v�1(xv): By de�nition of

and the fact that � = u�(u), we get

�(u)'�1(xv) = �01('(xv)u) for xv 2 Alt�1(C0):

On the other hand we obtain, using that x = v�1(xv),

�(u)'�1(xv) = �(u)'(vx) = �(u)'(v(xv)v)

= �(u)uv0'(xv)v0u�1 = v0'(xv)u�(u)�(u)�1v0 = v0'(xv)uv0:


Thus

v0'(xv)uv0 = �01('(xv)u) or '(xv)uv0 = v0�01('(xv)u)

as claimed.

Remark 12. The �rst part of Corollary 5 is a consequence of Theorem 11 if

Z = K � K. In fact the algebra C0 is isomorphic to a product A � Aop and the

standard involution � corresponds to the involution �A : (a; b0) 7! (b; a0). Any iso-

morphism of algebras A ! B extends to an isomorphism A � Aop ! B � Bop of

algebras{with{involution. (Observe that the standard involution � of C0 is of the

second kind).

Remark 13. Assume that the Arf invariant of (V; q) is not trivial, i.e. Z = Z(q) is

a �eld. Since

Z (V; q) ' (Alt�1(C0); �pf1)

for some � 2 Z� and some even symplectic involution �1 of C0, it follows from

Corollary 5 that

1) C0 is a division algebra if, and only, if Z (V; q) is anisotropic.

2) C0 ' M2(D); D a quaternion division algebra over L if, and only, if

Z (V; q) has Witt index 1.

3) C0 'M4(Z) if and only if Z (V; q) ' H(Z3).

Remark 14. Let B be a c:s: Z{algebra, Z a separable quadratic �eld extension

of K. The corestriction cor(B) is a c:s: K{algebra and B 7! cor(B) induces a

homomorphism cor: Br(Z) ! Br(K). As shown by Arason for �elds of character-

istic di�erent from 2 or (in a more general context, in particular for �elds of any

characteristic) by Knus{Parimala{Srinivas, the sequence

2Br(K)!2 Br(Z)cor!2 Br(K)

is exact. Here 2Br(K) denotes the subgroup of Br(K) of elements of order 2. Let

now C0 be the even Cli�ord algebra of a quadratic space of dimension 6. Since

the standard involution � is of the second kind, the corestriction of C0 is trivial

(Theorem 11 of Chapter 3) and by the exactness of the above sequence, C0 is of the

form Z B for some c:s: K{algebra B. This is also true for forms of dimension


4 (see Chapter 9) and dimension 2. We believe that the even Cli�ord algebra of a

quadratic space of arbitrary even dimension is of the form C0 = Z(q) B. This is

clear if n � 2 (4) by Proposition 6 of Chapter 5 and if CharK 6= 2 by Lemma 8 of

Chapter 5. Observe that the algebra B is not uniquely determined by the form q

(except if Z(q) = K �K).

Let A be a c:s: Z{algebra of dimension 16 which is isomorphic to the even Cli�ord

algebra C0 of a quadratic space (V; q) of dimension 6 with Arf invariant Z. As we

have seen a necessary condition for such an A is that A has an involution b 7! b�

of the second kind (necessary and suÆcient conditions are given in Knus{Parimala{

Sridharan). Let � be another involution of A of the second kind. By Lemma 12

of Chapter 3, � = �g for g 2 A� such that g� = g (�g denotes the involution

�g(x) = gx�g�1). Further (A; �g) ' (A; �g0) , g0 = �cgc� for � 2 K� and c 2 A�.

In this case we call g and g0 similar. By Theorem 10, the similitude classe of (V; q)

corresponds to the similitude class of g.

Remark 15. Let (V; q) and (V 0; q0) be quadratic spaces of dimension 6 with the

same Arf invariant. We choose as a representative of the Arf invariant the centre

Z of C0(q) and view C0(q0) as a Z-algebra through the choice of an isomorphism

Z ~!Z(C0(q0)). We claim thatC0(q) and C0(q

0) are isomorphicK{algebras if and only

if the quadratic spaces (V; q)Z and (V 0; q0)Z are similar. Let ' : C0(q) ~!C0(q0)

be an isomorphism of K-algebras, so that ' 1 is an isomorphism of the even

Cli�ord algebras of the spaces (V; q) Z and (V 0; q0) Z. Since Z Z ~!Z � Z

( an isomorphism being � : x y 7! (xy; x�0(y)), these spaces have trivial Arf

invariant. Hence, by Corollary 5, they are similar. Conversely, if the quadratic

spaces (V; q) Z and (V 0; q0) Z are similar, there exists an isomorphism of Z-

algebras C0(q)Z ~!C0(q0)Z (see Lemma 10 of Chapter 5). The isomorphism � :

ZZ ~!Z�Z composed with the projection Z�Z ! Z onto the �rst factor induces

a homomorphism of K-algebras C0(q0) Z ! C0(q

0). Let ' be the composition

C0(q)! C0(q) Z ~!C0(q0) Z ! C0(q

0):

The map ' is K-linear and maps the centre Z of C0(q) into the centre Z0 of C0(q

0).

By the uniqueness of the standard involution of quadratic algebras, the restriction

'Z of ' to Z is a morphism (Z; nZ) ! (Z 0; nZ0) of quadratic spaces of the same


dimension. Since 'Z is not the zero map, 'Z is an isomorphism. If we view C0(q0)

as a Z-algebra through 'Z, ' is a Z-homomorphism between c:s: Z-algebras of the

same dimension. Thus ' must be an isomorphism.

We now compute the special Cli�ord group, the spin group and the Lie algebra

of a quadratic space (V; q) of dimension 6. We may assume that q represents 1, so let

v with q(v) = 1. Let u 2 C�0 . With the same notation as in the discussion preceding

Theorem 10, the condition iu(V ) � V implies uxu�1v 2 Alt�1(C0) for all x 2 V , in

particular �1(uxu�1v) = uxu�1v; so uxu�1v = vv�(u�1)vxv�(u)v and �(u)u must

be in the centre of C0. Since �0(�(u)u) = �(�(u)u) = �(u)u, we get �(u)u 2 K.

To simplify notations we put �(u)u = u�(u) = �(u). Further since by (11.1)

V = fx 2 ZV j v�1(xv) = xg; iu(V ) � V is equivalent with v�1(uxu�1v) = uxu�1

for all x 2 V . We have, using that �1(ay�1(a)) = n(a)�1(a)�1�(y)a�1; n the

reduced norm,

v�1(uxu�1v) = v�1(uxv�(u)

�1�1(u))

= v�(u)�1�1(u)�1�(xv)u1n(u)

= �(u)�1�(u)�1v�(xv)u�1n(u)

= �(u)�2n(u)uxu�1:

Thus iu(V ) � V is equivalent with n(u) = �(u)2 and

S�(q) = fu 2 C�0 j n(u) = �(u)2g

Spin(q) = fu 2 S�(q) j �(u) = 1g:

Let Z be a quadratic �eld extension of K. For any c:s: Z{algebra A with an

involution of the second kind �, we de�ne an involution � 7! �� of Mn(A) by

�� = (�(aij))t if � = (aij) 2Mn(A);

i.e. � 7! �� is the tensor product of the involution � of A with the involution

x 7! xt of Mn(Z). We call a matrix g 2Mn(A) hermitian if g� = g and we say that

two hermitian matrices g; g0 are congruent if there exists x 2 GLn(A) such that

g0 = xgx�.


Let g 2Mn(A) be an hermitian matrix. We call the group

GUn(A; g) = fx 2 GLn(A) j xgx� = �g; � 2 Kg

the group of unitary similitudes of g, the group

Un(A; g) = fx 2 GLn(A) j xgx� = gg

the unitary group of g and

SUn(A; g) = fx 2 Un(A; g) j nMn(A)(x) = 1g

the special unitary group of g. With these notations we get

Spin(V; q) = SU1(C0; 1):(11.2)

We consider some special cases. First let

(V; q) ' (Z; nZ) ? (B; �nB);(11.3)

where B is a quaternion algebra over K and � 6= 0 2 K. We have

C(q) ' C(B; �nB) bC(Z; nZ):By Example 1 of Chapter 9, C(B; �nB) is isomorphic to M2(B) with the checker-

board gradation. The standard involution is given by a bc d

!7!

�a ��1�c��b �d

!; a; b; c; d 2 B;

where a 7! �a is the standard involution of B. Further by Example 3 of Chapter 7

C(Z; nZ) = (1; Z=K] ' Z � uZ;

with the multiplication rules xu = u�0(x); u2 = 1 (�0 is the involution of Z) and

the standard involution of C(Z; nZ) corresponds to x+uy 7! �0(x)u�uy; x; y 2 Z.We de�ne a graded isomorphism

: M2(B) b(Z � uZ) ~!M2(B (Z � uZ))

by a bc d

! (x + uy) 7!

a (x + uy) b (x+ uy)c (x� uy) d (x� uy)

!:


We use to identify C(q) with M2(B (Z � uZ)). In particular we get

C0 =

B Z B uZB uZ B Z

!

and it is easy to check that the map a x1 b ux2c ux3 d x4

!7! a x1 b �0(x2)c x3 d �0(x4)

!

is an isomorphism ' : C0(q) ~!M2(B Z). Let � be the standard involution of

C0(q). We get

�

a x1 b ux2c ux3 d x4

!=

�a �0(x1) �c ��1ux3��b ux2 �d �0(x4);

!so

'�'�1

a x1 b x2c x3 d x4

!=

�a �0(x1) ��1�c �0(x3)��b �0(x2) �d �0(x4)

!Thus the transport of � to M2(B Z) is the involution

� � Æ

!7! � � Æ

!�=

1 00 �

! ~� ~�

~ ~Æ

!t 1 00 �

!�1

;

where e� = �a �0(x) for � = a x 2 B Z. In view of (11.2) we get

Spin((B; �nB) ? (Z; nZ)) ' SU2(B Z; diag(1; �));(11.4)

If (V; q) is a quadratic space with Arf invariant [Z] such that

(M; q) Z ' ((B; nB) ? (Z; nZ)) Z

then, by Remark 15, C0(q) ' M2(B Z). The standard involution of C0(q) is of

the form �(x) = gx�g�1 for some hermitian matrix g 2M2(B Z) (i.e. g = g�). In

this case

Spin(V; q) ' fx 2 SL2(B Z) j xgx� = gg ' SU2(B; g):

We observe that g is determined up to similarity by (V; q) (see Remark 15). If , in

(11.4), B =M2(K) (and � = 1), we get

Spin(M; q) ' SU4(Z; s2)

since the standard involution ofM2(K) is � 7! s1�ts�1

1 for � 2M2(K). If i 6= 0 2 Zis an element of trace zero, the matrix is1 2 M2(Z) is hermitian and is congruent

to (0110). In fact we have (01

i0)(

01

10)(

0�{

10) = is1. Thus

Spin(H(K2) ? (Z; nZ)) ' SU4(Z; diag(1; 1; �1; �1)):


Finally we assume that there exists an isomorphism B Z ~!M2(Z) of algebras{

with{involution where the involution on BZ is the tensor product of the standard

involutions and the involution on M2(Z) is the involution

� =

a bc d

!7! �� =

�0(a) �0(b)�0(c) �0(d)

!t:

Then

Spin((Z; nZ) ? (B; �nB)) ' SU4(Z; diag(1; 1; �; �)):

This is the case for B = IH and Z = IC. In fact the isomorphism IH = IC � j IC 'M2( IC) given by

(x� jy) z 7! x �y�y �x

!z

has the wanted property.

We now assume that (V; q) has trivial Arf invariant and that (V; q) represents 1. By

Proposition 1, we can choose (V; q) = (Alt� (A); pfA) for some c:s: algebra A with an

involution � of even symplectic type. Then C0 ' A � A and �(a; b) = (�(b); �(a)).

It follows that

Spin(V; q) ' f(a; �(a)�1); a 2 SL1(A)g ' SL1(A)

and the map S� : Spin(V; q)! SO(V; q) is given by

a 2 SL1(A) 7! S�(a)(x) = �(a)�1xa�1:

If (V; q) is anisotropic, A is a division algebra. If (V; q) has Witt index 1, (V; q) is

similar toH(K) ? (B; nB), B a quaternion division algebra with standard involution

b 7! b (by Example 4). Then A =M2(B) with the involution

�

a bc d

!=

a cb d

!:

So

Spin(V; q) ' SL2(B):

If (V; q) ' H(K3), we have B =M2(K) in the above formula. Therefore A =M4(K)

with the involution �(x) = s2xts�1

2 and

Spin(V; q) ' SL4(K):


By de�nition of pfA, the map � : x 7! s�12 x is an isometry

� : (V; q) = (Alt� (M4(K)); pfM4(K)) ~!(Alt4(K); pf) = H(K3):

Let u 2 Spin(Alt� (M4(K)); pfM4(K)). Since

�S�(u)��1(x) = s�12 (s2u

ts�12 )�1s2xu

�1 = (ut)�1xu�1;

the map S� corresponds through � to the map

SL4(K)! SO(Alt4(K); pf) = SO(H(K3))

de�ned by u 7! (x 7! (ut)�1xu�1).

We now compute the Lie algebra so(q). Let

S(C0; �) = fx 2 C0 j x+ �(x) = 0g:

We obviously have so(q) = [V; V ] � S(C0; �), thus

so(q) = [so(q); so(q)] � S(C0; �)0 = [S(C0; �);S(C0; �)]:

We claim that

so(q) = S(C0; �)0:

It suÆces to verify that DimKS(C0; �)0 = 15. We can assume that (V; q) = (A; pfA)

so

S(C0; �) = f(a;�a); a 2 Ag ' A:

Taking further A = M4(K), we get DimK[A;A] = 15 as claimed. We leave it as an

exercise to compute so(q) for the di�erent cases considered above.

We �nally compute the group GO+(q) of special similitudes of a quadratic space

of dimension 6. Let Z = Z(q) be the centre of C0. Let n0(x) = x�0(x) be the norm

of Z; n the reduced norm of C0 (as Z{algebra), � the standard involution of C and

�(u) = u�(u). Let GU1(C0) = fu 2 C0 j u sigma(u) 2 Z�.

Theorem 16. LetG be the subgroup of Z��GU1(C0) of pairs (z; u) with z�0(z)�1 =

n(u)�(u)�2. There exists an isomorphism

G=f(z2; z�1); z 2 Z�g ~!GO+(q):


Proof. Let C = C0 � C1 be the Cli�ord algebra of (V; q). By Lemma 10, we may

identify Z V with Alt�(C1). For (z; u) 2 Z� � GU1(C0), let �(z; u) : Alt�(C1)!

Alt�(C1) be the map �(z; u)(x) = zux�(u). Since (1 q)(x) = �x�(x); �(z; u) is asimilitude of ZV with multiplier n0(z)�(u)

2 2 K�. We claim that �(z; u) restricts

to a similitude of V if and only if z�0(z)�1 = n(u)�(u)�2. By (11.1) and with

the notations introduced after Lemma 10, we have to check that v�1(zux�(u)v) =

zux�(u) for x 2 V: We get

v�1(zux�(u)v) = �0(z)�1(uxv�1(u)):

It now follows from pf1(uc�1(u)) = n(u)pf1(c) for c 2 C0 that

�1(uc�1(u)) = n(u)�1(u)�1�1(c)u

�1;

hence

v�1(zux�(u)v) = �0(z)n(u)�1(u)�1�1(xv)u

�1

= �0(z)n(u)�(u)�2ux�(u);

using that v�1(xv) = x; �1(u) = vuv and �(u) = u�(u). Thus, as claimed,

zux�(u) 2 V if and only if �0(z)n(u)�(u)�2 = z. The multiplier of �(z; u) is

n0(z)�(u)2. The extension of �(z; u) to C0 is given on elements xy; x; y 2 V , by

C0(�(z; u))(xy) = n0(z)�1�(u)�2zux�(u)xuy�(u) = uxyu�1

(see Lemma 10 of Chapter 5), thus �(z; u) 2 GO+(q). If �(z; u) = 1, then, by the

above formula, u 2 Z and zu2 = 1. Therefore � induces an injective map

�0 : G=f(z2; z�1); z 2 Z�g ! GO+(q):

We check that �0 is surjective. Let

E =

�C0 �C1

C1 C0

!� M2(C):

The map Z V = Alt�(C1)! E; x 7! (0xx0) extends to an isomorphism

Z C ~!E of Z{algebras and we identify Z C with E through this isomorphism.

Let f : V ! V be a special similitude with multiplier �. The map 0 xx 0

!2 Z V 7!

0 ��1f(x)

f(x) 0

!2 E = Z C


extends to an automorphism ' of ZC. As in the proof of Theorem 12 of Chapter

9, one checks that ' = iu with u 2 (Z C0)�. As in Lemma 4, Chapter 6, we

have �(u)u 2 Z�. The standard involution of Z C viewed as an involution of E

corresponds to a bc d

!7�!

�(d) �(b)�(c) �(a)

!:

Since

Z C0 =

�C0 00 C0

!;

the condition �(u)u 2 Z� implies

u =

��1�(d)�1 0

0 d

!

for some d 2 C�0 and � 2 Z�. If follows that (1 f)(x) = �dx�(d). Since if 1 f

descends to f : C ! V , we must have ��0(�)�1 = n(d)�1�(u)2 and f = �(�; d).

As for Spin(V; q), the computation of GO+(q) given in Theorem 16 can be

applied to a lot of cases (see the Appendix A), depending on the type of (V; q). We

only consider the cases of trivial Arf invariant. We may assume that (V; q) represents

1, so (V; q) = (Alt� (A); pfA) for some c:s: K{algebra of dimension 16 with an even

symplectic involution � . We have C0 ' A� A; �(a; b) = (�(b); �(a)),

GU1(C0) = f(��(d)�1; d); d 2 A�; � 2 K�g

so �(u) = � for u = (��(d)�1; d) and

n(u)�(u)�2 = (�2�0(nA(d))�1; nA(d)�

�2) = z�0(z)�1

for z = (1; nA(d)��2), where nA is the reduced norm of A. Projecting Z� �C�

0 onto

the second factor K� � A�, we get

G ' K� � A�

and

GO+(q) ' K� � A�=f(�2; ��1); � 2 K�g:

Chapter 12


The theory of 12{regular quadratic forms of dimension 5 is a by{product of the

theory of quadratic spaces of dimension 6. We begin with an example:

Example 1. Let A be a c:s: K{algebra of dimension 16 with an even symplectic

involution �A and let

Alt�A(A) = fx+ �A(x); x 2 Ag

be the corresponding set of alternating elements. Let pfA : Alt�A(A) ! K be the

reduced pfaÆan. For any a 2 Alt�A(A). we have a pfaÆan characteristic polynomial

��(X; a) = X2 � pft(a)X + pfA(a) (see Chapter 10) and ��(a; a) = 0. To simplify

notations, we put

Alt�A(A) = A+; pft(a) = t+(a); pfA(a) = n+(a)

and

A0+ = fa 2 A+ j t+(a) = 0g:

We check that (A0+; n+) is

12{regular of dimension 5 and compute its Cli�ord algebra.

Let Z be the graded quadratic K{algebra with a generator z of degree 1 such that

z2 = 1. We grade A Z by assuming that the elements of A are of degree 0.

Proposition 2. 1) (A0+;�n+) is 1

2{regular of dimension 5. 2) There exists a

graded isomorphism C(A0+;�n+) ~!A Z. In particular C0(A

+;�n+) ~!A and the

12{discriminant of (A0

+;�n+) is trivial. 3) The standard involution of C(A+;�n+)corresponds to the involution �A �0, where �0 is the standard involution of Z, i.e.

�0(z) = �z.


Proof. If suÆces to prove 1) over an extension K � L so that we can assume that

A = M2(B); B a quaternion algebra with standard involution b 7! b and �A is

given by (see Example 5 of Chapter 10)

�A

a bc d

!=

a cb d

!:

Then

A+ =�

x bb y

!x; y 2 K; b 2 B

�;

n+

x bb y

!= x y � b b; t+

x bb y

!= x + y;

A0+ =

� x bb �x

!; x 2 K; b 2 B

�;

and (A0+;�n+) ' h1i ? (B; nB) is

12{regular. We have a2 = �n+(a) since a satis�es

its pfaÆan characteristic polynomial and t+(a) = 0. Thus the map

' : A0+ ! A Z; '1(a) = a z;

extends, by the universal property of the Cli�ord algebra, to a homomorphism of

graded K{algebras

' : C(A0+;�n+)! A Z:

Let Z(C) be the centre of C(A0+;�n+). We claim that the restriction of ' to Z(C)

is an isomorphism Z(C) ~!Z. By passing to an extension K � L, we may assume

that (A+;�n+) = h1i ? (M2(K); det). Then Z(C) is generated by

z1 = e0(1� 2e1f1)(1� 2e2f2);

where e0 is the generator of h1i and the elements

e1 =

0 01 0

!; f1 =

0 10 0

!; e2 =

1 00 0

!; f2 =

0 00 1

!

are hyperbolic pairs in M2(K) (see Chapter 5). A computation (!) shows that

'(z1) = z, thus as claimed ' : Z(C) ~!Z. It then follows that ' maps C0(A0+;�n+)

to A and is an isomorphism. To prove 3), we observe that the restriction of �A �0to the image of A0

+ is �1.

Let (V; q) be a 12{regular quadratic form of dimension 5. Since there exists

an extension K � L such that L (V; q) ' h1i ? H(K2) ' h1i ? (M2(K); det),


it follows from Proposition 2 that the standard involution of C0(V; q) is of even

symplectic type. Let C0+ be the corresponding set of alternating elements and

n+ : C0+ ! K the reduced pfaÆan. Let z 2 Z(q) be a generator of degree 1

with z2 = s and [s] = 12Æ(q) in K�=K�2 (see Proposition 5 of Chapter 5), and let

� : V ! C0 be the map induced by v 7! zv. By going over to an extension K � L

and applying Proposition 2, we get that � is an isometry

(V; q) ~!(C 00+;�sn+):(12.1)

Theorem 3. 1) Let (V; q) be a 12{regular quadratic form of dimension 5. There ex-

ists � 2 K� and A c:s: of dimension 16 with an even symplectic involution �A such

that (V; q) ' (A0+; �n+). The class of � in K�=K�2 is equal to 1

2Æ(q) and (A; �A) is

unique up to isomorphism of algebras{with{involution. 2) Two 12{regular quadratic

forms (V; q); (V 0; q0) are isometric if and only if 12Æ(q) = 1

2Æ(q0) and C0(q) ' C0(q

0)

as algebras{with{(standard){involution. 3) (V; q); (V 0; q0) are similar if and only if

C0(q) ' C0(q0) as algebras{with{involution.

Proof. 1) We take � = �s and A = C0 with the standard involution.

2) It suÆces to check that an isomorphism ' : (A; �A)! (B; �B) of algebras{with{

involution induces an isometry (A0+; n+) ~!(B0; n+). Clearly ' maps A+ to B+. Let

�A : A+ ! A+ be such that x�A(x) = �A(x)x = pfA(x) and �B : B+ ! B+ such

that x�B(x) = �B(x)x = pfB(x). Let = '�A'�1 : B+ ! B+. Since is a

K{linear isomorphism such that (x)x 2 K for all x 2 B+; = ��B for � 2 K�.

It follows from (1) = 1 that � = 1 and '�A = �B'. Thus ' is an isometry

(A+; n+) ~!(B+; n+). The fact that ' restricts to an isometry (A0+; n+) ~!(B0

+; n+)

follows from the formula x+ �A(x) = t+(x) (see Example 5 of Chapter 10). Finally

3) follows from (12.1).

Remark 4. In view of Proposition 6 of Chapter 7, Corollary 3 of Chapter 8, Theo-

rem 7 of Chapter 9, Theorem 10 of Chapter 11 and Theorem 3 above, two quadratic

forms (V; q); (V 0; q0) which are of dimension � 6 and are nonsingular in even dimen-

sion, resp. 12{regular in odd dimension, are similar if and only if the even Cli�ord

algebras C0(q) and C0(q0) are isomorphic as algebras{with{involution, where the

involutions are the standard involutions.


The next result is due to Tamagawa in the case of (signed) trivial discriminant.

We do not know if it holds for 12{regular forms of dimension 5 in characteristic 2.

Proposition 5. Let K be a �eld of characteristic di�erent from 2 and let (V; q)

be a quadratic space of dimension 5 and signed discriminant [d] 2 K�=K�2. The

following conditions are equivalent:

1) (V; q) represents d.

2) (V; q) ' hdi ? (B; �nB) for some quaternion algebra B and � 2 K�.

3) C0(V; q) 'M2(B) for some quaternion algebra B.

Thus C0 is a division algebra if and only if (V; q) does not represent d.

Proof. 1) implies 2) by Remark 11 of Chapter 9, since hdi? is a quadratic space of

dimension 4 with trivial discriminant (and hence trivial Arf invariant). If 2) holds,

then C0(V; q) ' C(B;�d�nB) by Lemma 8 of Chapter 5 and C(B;�d�nB) 'M2(B)

by Example 1 of Chapter 9. Let now C0(V; q) ' M2(B). By Lemma 8 of Chapter

5, we have

C0(h�di ? q) ' C(dq) ' C0(dq) Z(dq) ' C0(q)� C0(q)

(since Z(dq) ' K � K). On the other hand h�di ? q is of dimension 6 and has

trivial Arf invariant. By Corollary 5 of Chapter 11 (applied to the form h�di ? q

and A = C0(dq)), we get that C0(q) ' C0(dq) 'M2(B) if and only if h�di ? q has

Witt index � 1. Thus C0(q) 'M2(B) implies that

h�di ? q ' h�1i ? h1i ? q0 ' h�di ? hdi ? q0

and

q ' hdi ? q0

by Witt cancellation. As claimed, q represents d.

We now compute the spinor group of a 12{regular form of dimension 5. By

Theorem 3 we may assume that (V; q) = (A0+;�n+) for a c:s: K{algebra A of

dimension 16 with an involution � of even symplectic type. In view of Proposition

2 C0(V; q) ' A and the standard involution on C0 is the involution � of A. For any


u 2 A, let �(u) = u�(u) and n(u) = nA(u). We have Z(q) = K � 1 � K � z with z

of degree 1 such that z2 = 1. Through the map V ! C0; v 7! zv, we can identify

V ' A0+ � C1 with A

0+ as a subspace of A = C0. Any isometry ' of V extends to

an automorphism of A = C0 as an algebra{with{involution, hence to an isometry

of (A+; n+) (or (A+;�n+)), which restricts to ' on A0+. Thus for any u 2 A�; iu

is an isometry of A0+ (i.e. iu(A

0+) � A0

+) if and only if iu(A+) � A+. By Lemma

4 of Chapter 6, �(u) 2 K� if u 2 S�(A0+;�n+). On the other hand, obviously

iu(A+) � A+ if u�1 = ��(u); � 2 K�. Thus

S�(A0+;�n+) = fu 2 A� j u�(u) 2 Kg:

We recall that a matrix g 2Mn(A) is hermitian (with respect to �) if g = g�, where

g� = (�(aij))t if g = (aij):

The group

GUn(A; g) = fx 2 GLn(A) j xgx� = �g; � 2 K�g

is the group of (unitary) similitudes of g (see Chapter 11). With this de�nition,

S�(A0+;�n+) = GU1(A; 1)

and

SO+(A0+;�n+) ' GU1(A; 1)=K

�:

We claim that

Spin(A0+;�n+) = fx 2 A� j x�(x) = 1g:

We have to check that nA(u) = 1 if u�(u) = 1. By passing to an extension K � L,

we may assume that A = M4(K) and �(a) = s2ats�1

2 . It follows from u�(u) = 1

that s2 = us2ut. Taking pfaÆans shows that det(u) = 1 as claimed. The map

Spin(A0+;�n+)! SO(A0

+;�n+)

is given by u 7! (x 7! uxu�1 = ux�(u)). In fact x 7! ux�(u) is an isometry since

n+(ux�(u)) = nA(u)n+(x) and nA(u) = 1.

We consider now some special cases. Let

(V; q) = h1i ? (B; �nB)


for a quaternion algebra B with standard involution b 7! b and � 2 K�. By Lemma

8 of Chapter 5, we have C0(V; q) ' C(B;��nB) and by Example 1 of Chapter 9,

C(B;��nB) 'M2(B) with the standard involution given by

� :

a bc d

!7�!

a ��1c�b d

!=

1 00 �

! a bc d

!t 1 00 �

!�1

:

Therefore

SO+(V; q) ' GU2(B; diag(1; �))=K�

and

Spin(V; q) ' SU2(B; diag(1; �)):

If B =M2(K) and � = 1, then b = s1bts�1

1 , so

S�(h1i ? H(K2)) = fx 2 GL4(K) j xs2xts�12 2 K�g

= GSp4(K)

and

Spin(h1i ? H(K2)) = fx 2 GL4(K) j xs2xts�12 = 1g

= Sp4(K);

if we de�ne the group of symplectic similitudes GSp2m(K) by

GSp2m(K) = fx 2 GL2m(K) j xsmxt = �sm; � 2 K�g:

and the symplectic group Sp2m(K) by

Sp2m(K) = fx 2 GL2m(K) j xsmxt = smg

We �nally show that h1i ? H(K2) can be identi�ed in a natural way with a subspace

of Alt4(K), the set of alternating (4� 4){matrices.

We have

h1i ? H(K2) = (M4(K)0+;�n+)for the involution �(x) = snx

ts�12 , and the map � : x! s�1

2 x is an isometry

� : (M4(K)+;�n+) ~!(Alt4(K);�pf)

by de�nition of n+. Since

M4(K)0+ =�0BBB@

x 0 d �b0 x �c aa b �x 0c d 0 �x

1CCCA 2 M4(K)�;


M4(K)0+ has image

�(M4(K)0+) =�0BBB@

0 �x c �ax 0 d �b�c �d 0 xa b �x 0

1CCCA 2 M4(K)�:

We call matrices in �(M4(K)0+) alternating matrices with pfaÆan trace zero and

denote �(M4(K)0+) by Alt04(K). Let u 2 Spin(M4(K)0+;�n+), since � iu ��1(x) =

s�12 us2xu

�1 = ut�1xu�1, the map

s� : Spin(M4(K)0+;�n+)! SO(M4(K)0+;�n+)

corresponds through � to the map

Sp4(K)! SO(Alt04(K);�pf)

de�ned by u 7! (x 7! ut�1xu�1).

We �nally compute the Lie algebra of (V; q) = (A0+;�n+). Let

S(A) = fx 2 A j x+ �(x) = 0g:

We have so(q) � S(A): Since DimKS(A) = 10, we get

so(A0+;�n+) = S(A) = fx 2 A j x+ �(x) = 0g:

Appendix A

A Chart of Results

We summarize in the following chart all the information we have obtained in

these notes on the structure of Cli�ord algebras of forms of dimension � 6. We

restrict to nonsingular forms, so we assume that char K 6= 2 in odd dimensions. We

use the following notations:

` : the dimension of V� : the Witt index of (V; q)L : a separable quadratic �eld extension of KDm : a central division algebra over K of dimension m2

LDm : a central division algebra over L of dimension m2 induced byscalar extension from a central division algebra over K

Mn(R) : the ring of n� n{matrices over R.

` (V; q) Z(q) C0(q) C(q)2 � = 0; 1 2 q(V ) L L M2(K)

� = 0; 1 62 q(V ) L L D2

� = 1 K �K K �K M2(K)

3 � = 0 K �K D2 D2 �D2

� = 0; �(L q) = 0 L D2 LD2

� = 0; �(L q) = 1 L D2 M2(L)� = 1 K �K M2(K) M2(K)�M2(K)� = 1 L M2(K) M2(L)

A. A Chart of Results 119

` (V; q) Z(q) C0(q) C(q)4 � = 0 K �K D2 �D2 M2(D2)

�((V;�q) ? (L; n)) = 0 L LD2 D4

� = 0; �((V;�q) ? (L; n)) = 1 L LD2 M2(D2)� = 1; 1 2 q(H(K)?) L M2(L) M4(K)� = 1; 1 62 q(H(K)?) L M2(L) M2(D2)� = 2 K �K M2(K)�M2(K) M4(K)

` (V; q) Z(q) C0(q) C(q)

5 Let L = K(pd); d 62 K�2 :

� = 0; 1 62 q(V ) K �K D4 D4 �D4

� = 0; 1 2 q(V ) K �K M2(D2) M2(D2)�M2(D2)� = 0; d 62 q(V ); 1 62 q(L V ) L D4 LD4

� = 0; d 62 q(V ); 1 2 q(L V );�(h1i?) = 0 in L V L D4 M2(LD2)� = 0; d 62 q(V ); 1 2 q(L V );�(h1i?) = 2 in L V L D4 M4(L)� = 0; d 2 q(V );�(L hdi?) = 0 L M2(D2) M2(LD2)� = 0; d 2 q(V );�(L hdi?) = 2 L M2(D2) M4(L)� = 1 K �K M2(D2) M2(D2)�M2(D2)� = 1; �(L q) = 1 L M2(D2) M2(LD2)� = 1; �(L q) = 2 L M2(D2) M4(L)� = 2 K �K M4(K) M4(K)�M4(K)� = 2 L M4(K) M4(L)

` (V; q) Z(q) C0(q) C(q)6 � = 0 K �K D4 �D4 M2(D4)

� = 0; �(L q) = 0 L LD4 ?� = 0; �(L q) = 1 L M2(LD2) ?� = 0; �(L q) = 3 L M4(L) ?� = 1 K �K M2(D2)�M2(D2) M4(D2)� = 1; �((V;�q) ? (L; n)) = 1 L M2(LD2) M2(D4)� = 1; �((V;�q) ? (L; n)) = 2 L M2(LD2) M4(D2)� = 2; 1 2 q(H(K2)?) L M4(L) M8(K)� = 2; 1 62 q(H(K2)?) L M4(L) M4(D2)� = 3 K �K M4(K)�M4(K) M8(K)

It would be interesting to know the structure of the Cli�ord algebra C(q) for ` =

6; � = 0 and Z(q) ' L. In particular one would like to know under which conditions

(on q) C(q) is a division algebra. We observe that the case � = 0; �(L q) = 0

cannot occur over IQ : Let d be the signed discriminant of q, so that L = IQ(pd).

By Meyer's theorem (see Serre) any quadratic space of dimension � 5 over IQ, which

is inde�nite over IR, is isotropic. So d must be negative. Let now x 6= 0 2 V .

The quadratic form qjx? ? dqjIQx is isotropic, again by Meyer's theorem, since it is

120 A. A Chart of Results

inde�nite over IR. Thus there exists y ? x such that q(y) = �dq(x) and L q is

isotropic. The case � = 0; �(L q) = 1 cannot occur over IR but can occur over

IQ. An example is given by the form h1; 1; 1; 1; 1; 7i. Let L = IQ(p�7). If L q is

hyperbolic, then Lh1; 1; 1; 1i is hyperbolic, since Lh1; 7i is hyperbolic. Thus thequaternion algebra (�1; �1

L) is a matrix algebra. This is the case if h1; 1i represents

�1 over L. It would follow that �1 is a square modulo 7.

Appendix B

Spinors

We de�ne spinors as in the book of Chevalley and compute some examples from

physics. We shall only consider nonsingular forms and assume that the character-

istic is not equal to 2. More on applications of Cli�ord algebras to physics can be

read for example in the book of Hermann.

I. Spaces of even dimension.

In view of Theorem 8 of Chapter 4, the Cli�ord algebra C = C(q) of a quadratic

space (V; q) of even dimension is simple. Thus there exists up to isomorphism only

one type of simple C(q){modules or, in the language of representation theory, all

irreducible representations of C(q) are equivalent. If we select such a representation,

� : C(q)! EndK(W ), we call W the space of spinors of (V; q), � the spin represen-

tation of C(q) and the restriction �0 of � to C0 = C0(q) the spin representation of C0

. Thus spinors are elements of a �xed simple C(q){module. The representation �

induces representations of �(q); S�(q) and Spin(q). The algebra C0 is either simple

(if its center is a �eld) or the sum of two simple algebras (if Z(q) ' K � K), see

Theorem 8 of Chapter 4.

Lemma 1. Let � : C ! EndK(W ) be an irreducible representation of C. The

restriction �0 of � to C0 is irreducible if C0 is simple and is the sum of two nonequiv-

alent irreducible representations �+0 : C0 ! EndK(W+), ��0 : C0 ! EndK(W

�) if

C0 is not simple.

122 B. Spinors

Proof. Since all irreducible representations of C are equivalent, we may assume

that W is a minimal left ideal of C and that � is the regular representation, i.e.

�(u)y = uy. Let W 0 6= f0g be a subspace of W of minimal dimension which is

invariant under all operations �0(x); x 2 C0, so that the induced representation

C0 ! EndK(W0) is irreducible. Let x 2 V be anisotropic and let W 00 = �(x)W 0.

Since C1 = xC0 = C0x and C0 = xC1 = C1x, we get �0(C0)W00 = W 00 and

�(C)(W 0 +W 00) = W 0 +W 00. Since � is irreducible, we must have W = W 0 +W 00.

If W 0 \W 00 6= f0g, the minimality of W 0 implies that W 0 \W 00 =W 0. Since W 0 and

W 00 have the same dimension, it follows that W = W 0 = W 00. If W 0 \W 00 = f0g,then W = W 0 �W 00. Thus the representation �0 is either irreducible or the sum of

two irreducible representations. Assume now that C0 is not simple. Then C0 has

two types of nonequivalent irreducible representations (see Theorem 8 of Chapter

4). The representation �0 being faithful, these two types occur in �0. Therefore, as

claimed, �0 is the sum of two nonequivalent irreducible representations W+ = W 0

and W� =W 00. This case occurs if K is algebraically closed. Thus we see, by going

over to the algebraic closure, that, if �0 is the sum of two irreducible representations,

these representations cannot be equivalent. It follows that �0 must be irreducible if

C0 is simple.

The elements of W+; W� are called 12{spinors (if W 0 is not equivalent to W 00)

and the induced representations of C0 are the12{spin representations.

Lemma 2. Let (V; q) be a nonsingular quadratic space (of arbitrary �nite di-

mension) non isometric to H(IF3). Let v 2 V be anisotropic and let M = fx 2V j q(x) = q(v)g. 1) The set M generates as a linear space V . 2) C0(q) is generated

as a K{algebra by all products xy; x; y 2 M . 3) S�(q) is a set of generators of

C0(q). 4) Spin(q) is a set of generators of C0(q).

Proof. 1) Let U be the linear hull of M . Let w 2 V be an anisotropic element and

let �w be the re ection at w. Since �w(v) = v � b(v;w)q(w)

w, we get that either w 2 U

or w 2 U?. Since V has an orthogonal basis we must have V = U ? U?. Assume

now that U? 6= 0. Let v 2 M and w 6= 0 2 M?. For any � 6= 0 2 K, the element

�v + w is isotropic, thus �2q(v) + q(w) = q(v) + q(w) = 0 and �2 = 1. There exists

B. Spinors 123

only one �eld K of characteristic not 2 such that �2 = 1 for all � 6= 0 2 K, namely

IF3. Since V = U ? U?, U and U? are nonsingular. It is then easy to see that

DimKU = DimKU? = 1 and V = H(IF3). 2) and 3) are immediate consequences of

1). We check 4). Since C0(q) is generated by all products of two elements of M , it

is also generated by the products a�1xy with a = q(x) = q(y). But a�1xy belongs

to Spin(q).

Proposition 3. Assume that (V; q) is nonsingular of even dimension and (V; q) 6'H(IF3). The spin representations of S�(q) and Spin(q) are either irreducible or the

sum of two irreducible nonequivalent representations.

Proof. Proposition 3 is an immediate consequence of Lemma 1 and Lemma 2.

II. Spaces of odd dimension.

Since we restrict to nonsingular forms, we must have charK 6= 2. The algebra

C0(q) is c:s:, hence its irreducible representations are all equivalent. We �x one,

which we call the spin representation:

�0 : C0(q)! EndK(W ):

The elements of W are called spinors of (V; q).

Proposition 4. The induced representations �0 : S�(q)! EndK(W );

�0 : Spin(q)! EndK(W ) are irreducible.

Proof. As Proposition 3.

If C(q) is simple, then C(q) = L C0(q); K � L a quadratic �eld extension

and �0 can be extended in an unique way to an irreducible representation of C(q):

1 � : L C0(q) = C(q)! EndL(LW ):

We call this representation the spin representation of C(q).

124 B. Spinors

Proposition 5. If C(q) is not simple, then �0 can be extended to exactly 2 non

equivalent irreducible representations of C.

Proof. If C(q) is not simple, we have Z(q) ' K[x]=(x2 � 1). Let z be a generator

of degree 1 of Z(q) such that z2 = 1. Any u 2 C(q) has a unique representation

u = u1 + u2z; ui 2 C0(q). Since z is in the centre of C(q) the maps

'0 : u 7! u1 + u2 and '00 : u1 � u2

are homomorphisms C ! C0 of K{algebras. The representations �0 = �0 Æ '0 and�00 = �0 Æ '00 extend �0 to C. Conversely let � : C ! EndK(W ) be an irreducible

representation which extends �0. Let = �(z) 2 EndK(W ). We have 2 = 1 and

commutes with all elements �0(C0). Therefore W =W1 �W2, where

W1 = fx 2 W j (x) = xg and W2 = fx 2 W j (x) = �xg:

These spaces are invariant under �0(C0). Since � is irreducible, one is zero and the

other W . Thus = �1 and � is one of the representations �0 or �00.

If C(q) is not simple, the two non equivalent representations of C(q) given by

Proposition 5 are called the spin representations of C(q).

In chapter 6 we identi�ed the Lie algebra so(q) with a subalgebra of C0(q). Thus

a spin representation of C0(q) induces, by restriction, a representation of so(q). We

call this representation the spin representation of so(q).

The chart of Appendix A gives an (abstract) description of spin representations

in low dimensions. To have a concrete description, we need explicit representations

of the Cli�ord algebras, i.e. explicitly given simple C(q){modules. We now construct

such representations for three cases occuring in physics. In some of the examples,

we get representations over the real quaternions IH. Writing any quaternion z as

x + jy; x; y 2 IC, we de�ne an injective map � : IH ! M2(IC) through the regular

representation `a(z) = az. In the following we shall identify IH with its image

�(IH) =�

x �yy x

!; x; y 2 IC

��M2(IC):

B. Spinors 125

In particular we get for the pure quaternions

i =

i 00 �i

!j =

0 �11 0

!and k = ij =

0 �i�i 0

!:

Example 6. Let (V; q) = h1; 1; 1i over IR. The algebra C(q) is the Pauli algebra.

We have (see Appendix A)

C(q) 'M2(IC) and C0(q) ' IH:

Explicit isomorphisms can be constructed using the Pauli matrices:

�1 =

0 11 0

!�2 =

0 �ii 0

!�3 =

1 00 �1

!in M2(IC):

Let fe1; e2; e3g be an orthonormal basis of V . We de�ne

'0 : V !M2(IC)

by ei 7! �i; i = 1; 2; 3. By the universal property of the Cli�ord algebra, '0 extends

to a homomorphism ' : C(q) ! M2(IC). It is easy to check that '0 is a IC{linear

isomorphism if we identify IC with Z(q) through i 7! e1e2e3. The subalgebra C0(q)

is mapped by ' onto the IR{subalgebra of M2(IC) with basis 1 00 1

!; �1�2 =

i 00 �i

!; �1�3 =

0 �11 0

!; �2�3 =

0 ii 0

!:

Thus ' restricts to an isomorphism C0 ~!IH. The isomorphism ' : C ~!M2(IC) is a

spin representation. We construct an explicit representation: let e = 12(1 + e3) 2

C(q). We have '(e) = (1000), so '(C(q)e) = (ICIC

00) and we can choose W = (ICIC

00).

Spinors are pairs (xy) 2 IC2. Since Spin(V; q) = fa 2 C0 j nC0(a) = 1g, ' induces an

isomorphism

Spin(V; q) ~! fa 2 IH j nIH(a) = 1g= SU2(IC) =

� x �yy x

!; x; y 2 IC j det

x �yy x

!= 1

�:

The Lie algebra so(q) = [V; V ] is generated by �1�2; �2�3 and �1�3. Let

su2(IC) =�

x �yy x

!; x; y 2 IC j tr

x �yy x

!= 0

�;

we get '(so(q)) = su2(IC).

126 B. Spinors

Example 7. Let (V; q) = h1;�1;�1;�1i over IR. The algebraC(q) is theMinkowski

algebra. By Appendix A, we have

C(q) 'M2(IH) and C0(q) 'M2(IC):

We construct explicit isomorphisms. Let fe0; : : : ; e3g be an orthogonal basis of V

such that q(e0) = 1; q(ei) = �1 i = 1; 2; 3. Let q1 = h1;�1i and q2 = h�1;�1i, so

C(q1) ~!M2(IR) with e0 7!

0 11 0

!and e1 7!

0 1

�1 0

!;

C(q2) ~!IH with e2 7! i; e3 7! j

(where i; j; k = ij are the pure quaternions in IH) and

: C(q) ~!M2(IR) bIH ~!M2(IR) IH =M2(IH)

(the second isomorphism is as in Lemma 3 of Chapter 4). Explicitely

(e0) =

0 11 0

!; (e1) =

0 1�1 0

!;

(e2) =

i 00 �i

!and (e3) =

j 00 �j

!:

It is not obvious that restricts to an isomorphism C0(q) ~!M2(IC). Using a trick

as in the proof of Theorem 8 of Chapter 4, we replace by ' = iu Æ , whereu = (10

0i ) 2M2(IC). We get

'(e0) =

0 �ii 0

!; '(e1) =

0 �i�i 0

!;

'(e2) =

i 00 �i

!and '(e3) =

j 00 j

!:

Then

'(e0e1) =

�1 00 1

!; '(e0e2) =

0 �1

�1 0

!; '(e0e3) =

0 �kk 0

!

'(e1e2) =

0 �11 0

!; '(e1e3) =

0 kk 0

!; '(e2e3) =

k 00 �k

!and

'(e0e1e2e3) =

�k 00 �k

!;

so ' maps C0(q) toM2(IC), with IC = IR(k). The element e = 1+e1e02

is an idempotent

of C(q) such that '(e) = (1000). So C(q)e is a space of spinors. We have

'(C(q)e) =W =

IH 0IH 0

!:

B. Spinors 127

By the general theory of Chapter 9 ' induces an isomorphism Spin(q) ' SL2(IC).

Further the image of so(q) is

sl2(IC) = fx 2M2(IC) j tr(x) = 0g:

Example 8. Let (V; q) = h1; 1; 1;�1i over IR. The algebra C(q) is called the

Majorana algebra. By our chart

C(q) 'M4(IR) and C0 'M2(IC):

We construct explicit isomorphisms. We decompose q as

q = q1 ? q2 with q1 = h�1; 1i and q2 = h1; 1i

and take corresponding orthogonal bases fe0; e1g; fe2; e3g. We have

C(q1) ~!M2(IR); e0 7!

0 1�1 0

!; e1 7!

0 11 0

!

and

C(q2) ~!M2(IR); e2 7!

0 11 0

!; e3 7!

1 00 �1

!:

These two isomorphisms induce

: C(q) ~! C(q1) bC(q2) 'M2(IR) bM2(IR)

~! M2(IR)M2(IR) =M4(IR);

applying Lemma 3 of Chapter 4. Here again it is not obvious that maps C0(q) to

M2(IC) and we �rst modify to get a better representation (as in Example 7). To

simplify notation, we use the symbols

" =

0 11 0

!; � =

1 00 �1

!; � =

0 1

�1 0

!:

Let now ' = iu Æ , where u = (100") 2 M4(IR) = M2(M2(IR)). We have by

de�nition of

(e0) =

0 1

�1 0

!; (e1) =

0 11 0

!;

(e2) =

" 00 �"

!; (e3) =

� 00 ��

!;

where 1 stands for the (2� 2)-unit matrix. So

'(e0) =

0 "

�" 0

!; '(e1) =

0 "" 0

!;

128 B. Spinors

'(e2) =

" 00 �"

!; '(e3) =

� 00 �

!:

If we identify IC with the subalgebra IR("�) of M2(IR) by putting i = "�, we get

'(e0e1) =

1 00 �1

!; '(e0e2) =

0 �1

�1 0

!; '(e0e3) =

0 i�i 0

!;

'(e1e2) =

0 �11 0

!; '(e1e3) =

0 ii 0

!; '(e2e3) =

i 00 �i

!

and

'(e0e1e2e3) =

i 00 i

!

so ' : C0(q) 'M2(IC) as claimed. The element ~e = 1+e0e12

is an idempotent of C0(q)

such that '(~e) = (1000). Thus

e = ~ee3 + 1

2=

1 + e3 + e0e1 + e0e1e34

is an idempotent of C(q) such that

'(e) =

0BBB@1 0 0 00 0 0 00 0 0 00 0 0 0

1CCCA and '(C(q)e) =

0BBB@IR 0 0 0IR 0 0 0IR 0 0 0IR 0 0 0

1CCCA :Therefore C(q)e is an explicitly given space of spinors for q. As in Example 8, '

induces isomorphisms Spin(q) ~!SL2(IC) and so(q) ~!sl2(IC).

Example 9. Let (V; q) = h�1; 1; 1; 1; 1i over IR. The algebra C(q) is called the

Dirac algebra. By Appendix A,

C(q) 'M4(IC) and C0(q) 'M2(IH):

Let fe0; : : : ; e4g be an orthogonal basis such that q(e0) = �1 and q(ei) = 1; i =

1; : : : ; 4. We decompose q as q1 ? q2 with q1 = h�1; 1i and q2 = h1; 1; 1i, soC(q) ' C(q1) bC(q2). We have, as before, an isomorphism

C(q1) ~!M2(IR); e0 7!

0 1�1 0

!; e1 7!

0 11 1

!

and by Example 6

C(q2) ~!M2(IC); e2 7! �1; e3 7! �2; e4 7! �3;

B. Spinors 129

�i the Pauli matrices. Using Lemma 3 of Chapter 4, we get

: C(q) ~!M4(IC) =M2(M2(IC))

such that

(e0) =

0 1

�1 0

!; (e1) =

0 11 0

!; (e2) =

�1 00 ��1

!;

(e3) =

�2 00 ��2

!and (e4) =

�3 00 ��3

!:

Let a 7! a be the standard involution of M2(IC). The involution

�� :

a bc d

!7! d bc a

!=

0 11 0

! a cb d

! 0 11 0

!

of M2(M2(IC)) is the identity on the images (ei) of ei; i = 0; : : : ; 3. Thus �� is

equal to �0 �1, where �0 is the canonical involution of C(q) (usually we prefer to

work with the standard involution � of the Cli�ord algebra, i.e. the involution which

is �1 on V . But we could not guess how � looks like !).

We now twist by an inner automorphism iu in such a way that the diagram

C(q)iuÆ �! M4(IC) =M2(M2(IC))S S

C0(q)iuÆ �! M2(IH)

is commutative. Let

u =

1 00 �1

!2 GL4(IC);

where �1 = (0110) is the �rst Pauli matrix and let 0 = iu Æ . We get

0(e0) =

0 �1

��1 0

!; 0(e1) =

0 �1�1 0

!; 0(e2) =

�1 00 ��1

!

0(e3) =

�2 00 �2

!and 0(e4) =

�3 00 �3

!:

It follows from Example 6 that 0 maps C0(q) to M2(IH), as claimed. Further the

involution b� = 0�0 0�1 = iu��i�1u is given by b�(x) = u��(u)��(x)(u��(u))�1 or

b� a bc d

!=

0 ��1�1 0

! a cb d

! 0 ��1�1 0

!�1

=

�1d�1 ��1b�1��1c�1 �1a�1

!:

We have �1a�1 = kak�1 for all a 2M2(IC). Thus

b� a bc d

!=

0 �kk 0

! a cb d

! 0 �kk 0

!�1

:

130 B. Spinors

Observe that b� is of symplectic type, as it should be. Since the standard and the

canonical involutions coincide on C0(q), the computation of Spin in Chapter 12

shows that

0(Spin(q)) =�x 2M2(IH)

�� xb�(x) = 1�

=�

a bc d

!2M2(IH)

�� a bc d

! 0 �kk 0

! a cb d

!=

0 �kk 0

!�

so 0 is an isomorphism

Spin(q) ~!SU2(IH; (0k

�k0)):

By formula (12.2) of Chapter 12, we have

Spin(q) ' Spin(�q) ~!SU2(IH; diag(1; �1):

In fact the hermitian matrices (0k�k0) and (�1

001) are congruent since

0 �kk 0

!=

1 1k2�k

2

! 1 00 �1

! 1 �k

2

1 k2

!:

We �nally compute an idempotent of C(q) which generates a simple C(q){module.

We have 0(1+e0e12

) = (1000) 2M2(IC), thus

e = (1 + e0e1

2)(1 + e42

) =1 + e4 + e0e1 + e0e1e4

4

is such that

0(e) =

0BBB@1 0 0 00 0 0 00 0 0 00 0 0 0

1CCCA and '(C(q)e) =

0BBB@IC 0 0 0IC 0 0 0IC 0 0 0IC 0 0 0

1CCCA :

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quadratic forms¸ clifford lgebras and spinors

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