CA Z = B —Z CA B.

The restriction σ |CA Z preserves both factors, hence it decomposes as

σ |C A Z = „ — σ |C A B .

Since the degree of CA B is odd, the involution σ |CA B is orthogonal by (??). There-

fore, it follows from (??) that the involution σ |CA Z has the same type as „ . Propo-

sition (??) shows that σ and σ |CA Z have the same type and completes the proof

in the case where char F = 2.

Suppose next that char F = 2. If „ is unitary, then it induces a nontrivial

automorphism of order 2 on Z, hence there exists z ∈ Z such that z + „ (z) = 1.

For every extension σ of „ to A we have that z + σ (z) = 1, hence 1 ∈ Alt(A, σ )

and σ is symplectic by (??). Similarly, if „ is symplectic then 1 ∈ Alt(B, „ ) hence

also 1 ∈ Alt(A, σ ) for every extension σ of „ . Therefore, every extension of „ is

symplectic.

Suppose ¬nally that „ is orthogonal and that Z/F is separable. As above, we

have CA B = B —Z CA B. If deg CA B is even, then (??) shows that we may ¬nd an

involution θ1 on CA B of orthogonal type and an involution θ2 of symplectic type.

By (??), the involution „ — θ1 (resp. „ — θ2 ) is orthogonal (resp. symplectic). It

follows from (??) that every extension of this involution to A has the same type.

If deg CA B is odd, the same arguments as in the case where char F = 2 show

that every extension of „ is orthogonal.

If the subalgebra B ‚ A is not simple, much less is known on the possibility

of extending an involution from B to A. We have however the following general

result:

(4.17) Proposition. Let A be a central simple algebra with involution of the ¬rst

kind. Every element of A is invariant under some involution.

Proof : Let a ∈ A and let σ be an arbitrary orthogonal involution on A. Consider

the vector space

V = { x ∈ A | σ(x) = x, xσ(a) = ax }.

It su¬ces to show that V contains an invertible element u, for then Int(u) —¦ σ is an

involution on A which leaves a invariant.

Let L be a splitting ¬eld of A. Fix an isomorphism AL Mn (L). The existence

of g ∈ GLn (L) such that g t = g and gat = ag is shown in Kaplansky [?, Theorem 66]

(see also Exercise ??). The involution „ = Int(g) —¦ t leaves a invariant, and it is

orthogonal if char F = 2. By (??), there exists an invertible element u ∈ AL such

§5. QUADRATIC FORMS 53

that „ = Int(u) —¦ σL and σL (u) = u. Since „ leaves a invariant, we have u ∈ V — L.

We have thus shown that V —L contains an element whose reduced norm is nonzero,

hence the Zariski-open subset in V consisting of elements whose reduced norm is

nonzero is nonempty. If F is in¬nite, we may use density of the rational points to

conclude that V contains an invertible element. If F is ¬nite, we may take L = F

in the discussion above.

Note that if char F = 2 the proof yields a more precise result: every element is

invariant under some orthogonal involution. Similar arguments apply to involutions

of the second kind, as the next proposition shows.

(4.18) Proposition. Let (B, „ ) be a central simple F -algebra with involution of

the second kind of degree n and let K be the center of B. For every b ∈ B whose

minimal polynomial over K has degree n and coe¬cients in F , there exists an

involution of the second kind on B which leaves b invariant.

Proof : Consider the F -vector space

W = { x ∈ B | „ (x) = x, x„ (b) = bx }.

As in the proof of (??), it su¬ces to show that W contains an invertible element.

Mn (L) — Mn (L)op .

By (??), we may ¬nd a ¬eld extension L/F such that BL

Fix such an isomorphism. Since the minimal polynomial of b has degree n and

coe¬cients in F , its image in Mn (L) — Mn (L)op has the form (m1 , mop ) where

2

m1 , m2 are matrices which have the same minimal polynomial of degree n. We

may then ¬nd a matrix u ∈ GLn (L) such that um2 u’1 = m1 . If µ is the exchange

involution on Mn (L) — Mn (L)op , the involution Int(u, uop ) —¦ µ leaves (m1 , mop )

2

invariant. This involution has the form Int(v) —¦ „L for some invertible element

v ∈ W — L. If F is in¬nite, we may conclude as in the proof of (??) that W also

contains an invertible element, completing the proof.

If F is ¬nite and K F — F , the arguments above apply with L = F . The

remaining case where F is ¬nite and K is a ¬eld is left to the reader. (See Exer-

cise ??.)

§5. Quadratic Forms

This section introduces the notion of a quadratic pair which is a twisted ana-

logue of quadratic form in the same way that involutions are twisted analogues of

symmetric, skew-symmetric or alternating forms (up to a scalar factor). The full

force of this notion is in characteristic 2, since quadratic forms correspond bijec-

tively to symmetric bilinear forms in characteristic di¬erent from 2. Nevertheless

we place no restrictions on the characteristic of our base ¬eld F .

As a preparation for the proof that quadratic pairs on a split algebra EndF (V )

correspond to quadratic forms on the vector space V (see (??)), we ¬rst show

that every nonsingular bilinear form on V determines a standard identi¬cation

EndF (V ) = V —F V . This identi¬cation is of central importance for the de¬nition

of the Cli¬ord algebra of a quadratic pair in §??.

5.A. Standard identi¬cations. In this subsection, D denotes a central di-

vision algebra over F and θ denotes an involution (of any kind) on D. Let V be

a ¬nite dimensional right vector space over D. We de¬ne9 a left vector space θ V

9 Note that this de¬nition is consistent with those in § ?? and § ??.

54 I. INVOLUTIONS AND HERMITIAN FORMS

over D by

θ

V = { θv | v ∈ V }

with the operations

θ

v + θ w = θ (v + w) and ± · θ v = θ v · θ(±)

for v, w ∈ V and ± ∈ D. We may then consider the tensor product V —D θ V which

is a vector space over F of dimension

dimF V —F V

dimF V —D θ V = = (dimF V )2 dimF D.

dimF D

Now, let h : V — V ’ D be a nonsingular hermitian or skew-hermitian form on V

with respect to θ. There is an F -linear map

•h : V —D θ V ’ EndD (V )

such that

•h (v — θ w)(x) = v · h(w, x) for v, w, x ∈ V .

(5.1) Theorem. The map •h is bijective. Letting σh denote the adjoint involution

on EndD (V ) with respect to h, we have

σh •h (v — θ w) = δ•h (w — θ v) for v, w ∈ V ,

where δ = +1 if h is hermitian and δ = ’1 if h is skew-hermitian. Moreover,

TrdEndD (V ) •h (v — θ w) = TrdD h(w, v) for v, w ∈ V

and, for v1 , v2 , w1 , w2 ∈ V ,

•h (v1 — θ w1 ) —¦ •h (v2 — θ w2 ) = •h v1 h(w1 , v2 ) — θ w2 .

Proof : Let (e1 , . . . , en ) be a basis of V over D. Since h is nonsingular, for i ∈

{1, . . . , n} there exists a unique vector ei ∈ V such that

1 if i = j,

h(ei , ej ) =

0 if i = j,

and (e1 , . . . , en ) is a basis of V over D. Every element x ∈ V —D θ V therefore has

a unique expression of the form

n

ei aij — θ ej

x= for some aij ∈ D.

i,j=1

The map •h takes the element x to the endomorphism of V with the matrix

(aij )1¤i,j¤n (with respect to the basis (e1 , . . . , en )), hence •h is bijective. Moreover,

we have

n n

TrdEndD (V ) •h (x) = TrdD (aii ) = TrdD h(ej , ei aij ) ,

i=1 i,j=1

hence in particular

TrdEndD (V ) •h (v — θ w) = TrdD h(w, v) for v, w ∈ V .

For v, w, x, y ∈ V we have

h x, •h (v — θ w)(y) = h(x, v)h(w, y)

§5. QUADRATIC FORMS 55

and

h •h (w — θ v)(x), y = h wh(v, x), y = θ h(v, x) h(w, y),

hence σh •h (v — θ w) = δ•h (w — θ v).

Finally, for v1 , w1 , v2 , w2 , x ∈ V ,

•h (v1 — θ w1 ) —¦ •h (v2 — θ w2 )(x) = v1 h(w1 , v2 )h(w2 , x)

= •h v1 h(w1 , v2 ) — θ w2 (x),

hence •h (v1 — θ w1 ) —¦ •h (v2 — θ w2 ) = •h v1 h(w1 , v2 ) — θ w2 .

Under •h , the F -algebra with involution EndD (V ), σh is thus identi¬ed with

V —D θ V endowed with the product

(v1 — θ w1 ) —¦ (v2 — θ w2 ) = v1 h(w1 , v2 ) — θ w2 for v1 , v2 , w1 , w2 ∈ V

and the involution σ de¬ned by

σ(v — θ w) = δw — θ v for v, w ∈ V ,

where δ = +1 if h is hermitian and δ = ’1 if h is skew-hermitian. We shall refer

to the map •h in the sequel as the standard identi¬cation of EndD (V ), σh with

(V —D θ V, σ). Note that the map •h depends on the choice of h and not just on the

involution σh . Indeed, for any ± ∈ F — ¬xed by θ we have σ±h = σh but •±h = ±•h .

The standard identi¬cation will be used mostly in the split case where D = F .

If moreover θ = IdF , then θ V = V , hence the standard identi¬cation associated to

a nonsingular symmetric or skew-symmetric bilinear form b on V is

∼

(5.2) •b : (V —F V, σ) ’ EndF (V ), σb ,

’ •b (v — w)(x) = vb(w, x),

where σ(v — w) = w — v if b is symmetric and σ(v — w) = ’w — v if b is skew-

symmetric.

(5.3) Example. As in (??), for all integers r, s let — : Mr,s (D) ’ Ms,r (D) be the

map de¬ned by

t

(aij )— = θ(aij ) .

i,j i,j

Let V = Dr (= Mr,1 (D)) and let h : Dr — Dr ’ D be the hermitian form de¬ned

by

h(x, y) = x— · g · y for x, y ∈ Dr ,

where g ∈ Mr (D) is invertible and satis¬es g — = g.

Identify Mr (D) with EndD (Dr ) by mapping m ∈ Mr (D) to the endomorphism

x ’ m · x. The standard identi¬cation

∼

•h : Dr —D Dr ’ Mr (D)

’

carries v — w ∈ D r —D Dr to the matrix v · w — · g, since for all x ∈ D r

v · w— · g · x = vh(w, x).

56 I. INVOLUTIONS AND HERMITIAN FORMS

5.B. Quadratic pairs. Let A be a central simple algebra of degree n over a

¬eld F of arbitrary characteristic.

(5.4) De¬nition. A quadratic pair on A is a couple (σ, f ), where σ is an involution

of the ¬rst kind on A and f : Sym(A, σ) ’ F is a linear map, subject to the

following conditions:

(1) dimF Sym(A, σ) = n(n + 1)/2 and TrdA Skew(A, σ) = {0}.

(2) f x + σ(x) = TrdA (x) for all x ∈ A.

Note that the equality x + σ(x) = x + σ(x ) holds for x, x ∈ A if and only

if x ’ x ∈ Skew(A, σ). Therefore, condition (??) makes sense only if the reduced

trace of every skew-symmetric element is zero, as required in (??).

If char F = 2, the equality TrdA Skew(A, σ) = {0} holds for every involution

of the ¬rst kind, by (??), hence condition (??) simply means that σ is of orthogonal

type. On the other hand, the map f is uniquely determined by (??) since for

1 1

s ∈ Sym(A, σ) we have s = 2 s + σ(s) , hence f (s) = 2 TrdA (s).

If char F = 2, Proposition (??) shows that condition (??) holds if and only

if σ is symplectic (which implies that n is even). Condition (??) determines the

value of f on the subspace Symd(A, σ) but not on Sym(A, σ). Indeed, in view

of (??), condition (??) simply means that f is an extension of the linear form

Trpσ : Symd(A, σ) ’ F . Therefore, there exist several quadratic pairs with the

same symplectic involution.

(5.5) Example. Let „ be an involution of orthogonal type on A. Every element

a ∈ A such that a+„ (a) is invertible determines a quadratic pair (σa , fa ) as follows:

let g = a + „ (a) ∈ A— and de¬ne

σa = Int(g ’1 ) —¦ „ and fa (s) = TrdA (g ’1 as) for s ∈ Sym(A, σa ).

From (??), it follows that σa is orthogonal if char F = 2 and symplectic if char F =

2. In order to check condition (??) of the de¬nition of a quadratic pair, we compute

fa x + σa (x) = TrdA (g ’1 ax) + TrdA g ’1 ag ’1 „ (x)g for x ∈ A.

Since ag ’1 „ (x) and „ ag ’1 „ (x) = xg ’1 „ (a) have the same trace, the last term on

the right side is also equal to TrdA xg ’1 „ (a) , hence

fa x + σa (x) = TrdA xg ’1 a + „ (a) = TrdA (x).

We will show in (??) that every quadratic pair is of the form (σa , fa ) for some

a ∈ A such that a + „ (a) is invertible.

We start with a couple of general results:

(5.6) Proposition. For every quadratic pair (σ, f ) on A,

1

f (1) = deg A.

2

Proof : If char F = 2, we have f (1) = 1 TrdA (1) = 1

deg A. If char F = 2, the

2 2

proposition follows from (??) since f (1) = Trpσ (1).

(5.7) Proposition. For every quadratic pair (σ, f ) on A, there exists an element

∈ A, uniquely determined up to the addition of an element in Alt(A, σ), such that

f (s) = TrdA ( s) for all s ∈ Sym(A, σ).

satis¬es + σ( ) = 1.

The element

§5. QUADRATIC FORMS 57

Proof : Since the bilinear reduced trace form on A is nonsingular by (??), every

linear form A ’ F is of the form x ’ TrdA (ax) for some a ∈ A. Therefore,

extending f arbitrarily to a linear form on A, we may ¬nd some ∈ A such that

f (s) = TrdA ( s) for all s ∈ Sym(A, σ). If , ∈ A both satisfy this relation, then

TrdA ( ’ )s = 0 for all s ∈ Sym(A, σ), hence (??) shows that ’ ∈ Alt(A, σ).

Condition (??) of the de¬nition of a quadratic pair yields

TrdA x + σ(x) = TrdA (x) for x ∈ A.

Since TrdA σ(x) = TrdA xσ( ) , it follows that

TrdA + σ( ) x = TrdA (x) for x ∈ A,

hence + σ( ) = 1 since the bilinear reduced trace form on A is nonsingular.

(5.8) Proposition. Let „ be an orthogonal involution on A. Every quadratic pair

on A is of the form (σa , fa ) for some a ∈ A such that a + „ (a) ∈ A— . If a, b ∈ A

are such that a + „ (a) ∈ A— and b + „ (b) ∈ A— , then (σa , fa ) = (σb , fb ) if and only

if there exists » ∈ F — and c ∈ Alt(A, „ ) such that a = »b + c.

Proof : Let (σ, f ) be a quadratic pair on A. By (??), there exists an invertible

element g ∈ A— such that σ = Int(g ’1 ) —¦ „ and g ∈ Sym(A, „ ) if char F = 2, and

g ∈ Alt(A, „ ) if char F = 2. Moreover, (??) yields an element ∈ A such that

σ( ) + = 1 and f (s) = TrdA ( s) for all s ∈ Sym(A, σ). Let a = g ∈ A. Since

„ (g) = g and σ( ) + = 1, we have a + „ (a) = g, hence σa = σ. Moreover, for

s ∈ Sym(A, σ),

fa (s) = TrdA (g ’1 as) = TrdA ( s) = f (s),

hence (σ, f ) = (σa , fa ).

Suppose now that a, b ∈ A are such that a + „ (a), b + „ (b) are each invertible

and that (σa , fa ) = (σb , fb ). Writing g = a + „ (a) and h = b + „ (b), we have

σa = Int(g ’1 ) —¦ „ and σb = Int(h’1 ) —¦ „ , hence the equality σa = σb yields g = »h

for some » ∈ F — . On the other hand, since fa = fb we have

TrdA (g ’1 as) = TrdA (h’1 bs) for s ∈ Sym(A, σa ) = Sym(B, σb ).

Since Sym(A, σb ) = Sym(A, „ )h and g = »h, it follows that

»’1 TrdA (ax) = TrdA (bx) for x ∈ Sym(A, „ ),

hence a ’ »b ∈ Alt(A, „ ), by (??).

This proposition shows that quadratic pairs on A are in one-to-one correspon-

dence with equivalence classes of elements a + Alt(A, „ ) ∈ A/ Alt(A, „ ) such that

a + „ (a) is invertible, modulo multiplication by a factor in F — .

In the particular case where A = Mn (F ) and „ = t is the transpose involution,

the elements in A/ Alt(A, „ ) may be regarded as quadratic forms of dimension n,

by identifying a + Alt(A, „ ) with the quadratic form q(X) = X · a · X t where

X = (x1 , . . . , xn ). The matrix a + at is invertible if and only if the corresponding

quadratic form is nonsingular (of even dimension if char F = 2). Therefore, quad-

ratic pairs on Mn (F ) are in one-to-one correspondence with equivalence classes of

nonsingular quadratic forms of dimension n modulo a factor in F — (with n even if

char F = 2).

58 I. INVOLUTIONS AND HERMITIAN FORMS

(5.9) Example. Suppose that n = 2 and char F = 2. Straightforward compu-

tations show that the quadratic pair on M2 (F ) associated to the quadratic form

aX 2 + bXY + cY 2 is (σ, f ) with

a11 a12 a22 a12

σ =

a21 a22 a21 a11

and

a11 a12

= a11 + ab’1 a12 + cb’1 a21 .

f

a21 a11

The observation above explains why quadratic pairs on central simple algebras

may be thought of as twisted forms of nonsingular quadratic forms up to a scalar

factor. We next give another perspective on this result by relating quadratic forms

on a vector space V to quadratic pairs on the endomorphism algebra EndF (V ).

Let V be a vector space of dimension n over F and let q : V ’ F be a quadratic

form on V . We recall that bq is the polar symmetric bilinear form of q,

bq (x, y) = q(x + y) ’ q(x) ’ q(y) for x, y ∈ V ,

which we assume to be nonsingular. This hypothesis implies that n is even if

char F = 2, since the bilinear form bq is alternating in this case. We write simply

σq for the adjoint involution σbq on EndF (V ) and

(5.10) •q : V —F V ’ EndF (V ), •q (v — w)(x) = vbq (w, x)

for the standard identi¬cation •bq of (??). Under this identi¬cation, we have

(V — V, σ) = EndF (V ), σq ,

where σ : V — V ’ V — V is the switch.

(5.11) Proposition. There is a unique linear map fq : Sym EndF (V ), σq ’ F

such that

fq —¦ •q (v — v) = q(v) for x, y ∈ V .

The couple (σq , fq ) is a quadratic pair on EndF (V ). Moreover, assuming that

dimF V is even if char F = 2, every quadratic pair on EndF (V ) is of the form

(σq , fq ) for some nonsingular quadratic form q on V which is uniquely determined

up to a factor in F — .

Proof : Let (e1 , . . . , en ) be a basis of V . The elements •q (ei —ei ) for i = 1, . . . , n and

•q (ei —ej +ej —ei ) for i, j ∈ {1, . . . , n}, j = i, form a basis of Sym EndF (V ), σq =

•q Sym(V — V, σ) . De¬ne

fq •q (ei — ei ) = q(ei ), fq •q (ei — ej + ej — ei ) = bq (ei , ej )

n

and extend by linearity to a map fq : Sym EndF (V ), σq ’ F . For v = e i ±i ∈

i=1

V we have

— e i ±2 +

fq —¦ •q (v — v) = fq —¦ •q 1¤i¤n ei 1¤i<j¤n (ei — ej + ej — ei )±i ±j

i

2

= 1¤i¤n q(ei )±i + 1¤i<j¤n bq (ei , ej )±i ±j = q(v),

hence the map fq thus de¬ned satis¬es the required condition. Uniqueness of fq is

clear, since Sym EndF (V ), σq is spanned by elements of the form •q (v — v).

§5. QUADRATIC FORMS 59

Since the bilinear form q is symmetric, and alternating if char F = 2, the

involution σq is orthogonal if char F = 2 and symplectic if char F = 2. To check

that (σq , fq ) is a quadratic pair, it remains to prove that

fq x + σ(x) = TrdEndF (V ) (x) for x ∈ EndF (V ).

Since both sides are linear, it su¬ces to check this formula for x = •q (v — w) with

v, w ∈ V . The left side is then

fq —¦ •q (v — w + w — v) = fq —¦ •q (v + w) — (v + w) ’ fq —¦ •q (v — v)

’ fq —¦ •q (w — w)

= bq (v, w)

and the claim follows, since (??) shows that bq (v, w) = bq (w, v) = Trd •q (v — w) .

Suppose now that (σ, f ) is an arbitrary quadratic pair on EndF (V ). As shown

in the introduction to this chapter (and in (??)), the involution σ is the adjoint

involution with respect to some nonsingular symmetric bilinear form b : V — V ’ F

which is uniquely determined up to a factor in F — . Use the standard identi¬cation

•b (see (??)) to de¬ne a map q : V ’ F by

q(v) = f —¦ •b (v — v) for v ∈ V .

From the de¬nition, it is clear that q(v±) = q(v)±2 for ± ∈ F . Moreover, for v,

w ∈ V we have

q(v + w) ’ q(v) ’ q(w) = f —¦ •b (v — w) + f —¦ •b (w — v)

= f •b (v — w) + σ •b (v — w) .

Since (σ, f ) is a quadratic pair, the right side is equal to

TrdEndF (V ) •b (v — w) = b(w, v) = b(v, w).

Therefore, q is a quadratic form with associated polar form b, and it is clear that

the corresponding quadratic pair (σq , fq ) is (σ, f ). Since b is uniquely determined

up to a factor in F — , the same property holds for q.

For later use, we give an explicit description of an element ∈ EndF (V ) satis-

fying property (??) for the quadratic pair (σq , fq ). It su¬ces to consider the case

where char F = 2, since otherwise we may take = 1 . 2

(5.12) Proposition. Let (V, q) be a nonsingular quadratic space of even dimension

n = 2m over a ¬eld F of characteristic 2 and let (e1 , . . . , en ) be a symplectic basis

of V for the bilinear form bq , i.e., a basis such that

bq (e2i’1 , e2i ) = 1, bq (e2i , e2i+1 ) = 0 bq (ei , ej ) = 0 if |i ’ j| > 1.

and

Set

m

= •q e2i’1 — e2i’1 q(e2i ) + e2i — e2i q(e2i’1 ) + e2i’1 — e2i ∈ EndF (V ).

i=1

(1) The element satis¬es tr( s) = f (s) for all s ∈ Sym EndF (V ), σq .

(2) The characteristic polynomial of equals

Pc (X) = X 2 + X + q(e1 )q(e2 ) · · · X 2 + X + q(e2m’1 )q(e2m ) ,

hence

m m(m’1)

s2 ( ) = q(e2i’1 )q(e2i ) + .

i=1 2

60 I. INVOLUTIONS AND HERMITIAN FORMS

Proof : It su¬ces to check the equation for s of the form •q (v —v) with v ∈ V , since

n

these elements span Sym EndF (V ), σ . If v = i=1 ei ±i , we have bq (e2i’1 , v) = ±2i

and bq (e2i , v) = ±2i’1 , hence

•q (v — v) =

m

•q i=1 (e2i’1 — v)±2i q(e2i ) + (e2i — v)±2i’1 q(e2i’1 ) + (e2i’1 — v)±2i’1 .

It follows that

m

tr •q (v — v) = bq (v, e2i’1 )±2i q(e2i ) + bq (v, e2i )±2i’1 q(e2i’1 )

i=1

+ bq (v, e2i’1 )±2i’1

n m

±2 q(ei )

= + ±2i’1 ±2i = q(v).