qσ (x + ±y) = qσ (x) + ±qσ (x, y) + ±2 qσ (y) = 0

has at most two solutions and F has more than three elements. For this choice

of ±, we have

qσ (x + ±y)

pσ (x + ±y) = pσ (x + ±y).

qσ (x + ±y)

By linearity of pσ and pσ , it follows that

qσ (x + ±y) q (x + ±y)

pσ (x) + ± σ

pσ (x) + ±pσ (y) = pσ (y).

qσ (x + ±y) qσ (x + ±y)

§16. BIQUATERNION ALGEBRAS 247

On the other hand, we also have

qσ (x) q (y)

pσ (x) + ± σ

pσ (x) + ±pσ (y) = pσ (y),

qσ (x) qσ (y)

hence

qσ (x) q (x + ±y) q (y)

=σ =σ

qσ (x) qσ (x + ±y) qσ (y)

since pσ (x) and pσ (y) are linearly independent.

To conclude, observe that Skew(A, σ) has a basis (ei )1¤i¤6 consisting of in-

vertible elements: this is clear if σ is the involution θ de¬ned above, and it follows

for arbitrary σ since Skew(A, σ) = v · Skew(A, θ) if σ = Int(v) —¦ θ. Denoting

» = qσ (e1 )qσ (e1 )’1 , the argument above shows that

qσ (ei )

»= for i = 1, . . . , 6,

qσ (ei )

hence pσ (ei ) = »pσ (ei ) for i = 1, . . . , 6. By linearity of pσ and pσ , it follows that

pσ (x) = »pσ (x) for all x ∈ Skew(A, σ).

(16.23) Proposition. Let pσ be a non-zero linear endomorphism of Skew(A, σ)

such that xpσ (x) ∈ F for all x ∈ Skew(A, σ), and let

qσ (x) = xpσ (x) ∈ F for x ∈ Skew(A, σ).

The quadratic form qσ is nonsingular; for x ∈ Skew(A, σ) we have qσ (x) = pσ (x)x,

and qσ (x) = 0 if and only if x ∈ A— . Moreover, Skew(A, σ), qσ is an Albert

quadratic space of A.

Proof : Proposition (??) shows that pσ is a scalar multiple of an endomorphism

satisfying (??.??) and (??.??); therefore, pσ also satis¬es these conditions. It fol-

lows that qσ (x) = pσ (x)x for all x ∈ Skew(A, σ) and qσ (x) = 0 if and only if x is

invertible.

In order to show that qσ is nonsingular, we again consider a decomposition of A

into a tensor product of two quaternion subalgebras, so

A = Q 1 —F Q2

and set θ = γ1 — γ2 , which is the tensor product of the canonical involutions on Q1

and Q2 . As observed in the proof of (??), we have

Skew(A, σ) = (Q0 — 1) • (1 — Q0 ),

1 2

and we may consider the endomorphism pθ of Skew(A, σ) de¬ned by

pθ (x1 — 1 + 1 — x2 ) = x1 — 1 ’ 1 — x2

for x1 ∈ Q0 and x2 ∈ Q0 . Denoting qθ = xpθ (x) for x ∈ Skew(A, σ), we then have

1 2

qθ (x1 — 1 + 1 — x2 ) = x2 ’ x2 = ’ NrdQ1 (x1 ) + NrdQ2 (x2 ),

1 2

hence qθ is a nonsingular quadratic form.

If v ∈ Sym(A, θ) © A— is such that σ = Int(v) —¦ θ, (??) shows that there exists

» ∈ F — such that pσ (x) = »vpθ (xv) for all x ∈ Skew(A, σ). Then qσ (x) = »qθ (xv),

hence multiplication on the right by v de¬nes a similitude

∼

Skew(A, σ), qσ ’ Skew(A, θ), qθ

’

with multiplier ». Since qθ is nonsingular, it follows that qσ is also nonsingular.

248 IV. ALGEBRAS OF DEGREE FOUR

To complete the proof, consider the map

i : Skew(A, σ) ’ M2 (A)

de¬ned by

0 pσ (x)

i(x) = for x ∈ Skew(A, σ).

x 0

We have i(x)2 = qσ (x) for all x ∈ Skew(A, σ), hence the universal property of

Cli¬ord algebras shows that i induces an F -algebra homomorphism

i— : C Skew(A, σ), qσ ’ M2 (A).

(Compare with (??).) Since qσ is nonsingular, it follows that the Cli¬ord algebra

C Skew(A, σ), qσ is simple, hence i— is injective. It is also surjective by dimension

count, hence (??) shows that Skew(A, σ), qσ is an Albert quadratic space of A.

(16.24) Example. Let A = (a1 , b1 )F — (a2 , b2 )F . If θ = γ1 — γ2 is the tensor prod-

uct of the canonical involutions on the quaternion algebras (a1 , b1 )F and (a2 , b2 )F ,

the computations in the proof of (??) show that one can take

and qθ (x1 — 1 + 1 — x2 ) = x2 ’ x2

pθ (x1 — 1 + 1 — x2 ) = x1 — 1 ’ 1 — x2 1 2

for xi ∈ (ai , bi )0 , i = 1, 2. Therefore, qθ has the following diagonalization:

F

qθ = a1 , b1 , ’a1 b1 , ’a2 , ’b2 , a2 b2 .

(Compare with (??).)

We now list a few properties of the endomorphism pσ de¬ned in (??).

(16.25) Proposition. Let pσ be a non-zero linear endomorphism of Skew(A, σ)

such that xpσ (x) ∈ F for all x ∈ Skew(A, σ), and let qσ : Skew(A, σ) ’ F be the

(Albert) quadratic map de¬ned by

qσ (x) = xpσ (x) for x ∈ Skew(A, σ).

(1) For all a ∈ A— and x ∈ Skew(A, σ), we have qσ axσ(a) = NrdA (a)qσ (x) and

pσ axσ(a) = NrdA (a)σ(a)’1 pσ (x)a’1 .

(2) There exists some dσ ∈ F — such that

(a) qσ (x)2 = dσ NrdA (x) for all x ∈ Skew(A, σ);

(b) p2 = dσ · IdSkew(A,σ) ;

σ

(c) dσ · F —2 = disc σ;

(d) pσ (x), pσ (y) = dσ [x, y] for all x, y ∈ Skew(A, σ) where [ , ] are the Lie

brackets (i.e., [x, y] = xy ’ yx).

Proof : Consider θ, pθ , qθ as in (??). The relation

qθ (x)2 = NrdA (x) for x ∈ Skew(A, σ)

is easily proved: extending scalars to an algebraic closure, it su¬ces to show that

for any 2 — 2 matrices m1 , m2 of trace zero,

2

det(m1 — 1 + 1 — m2 ) = det(m1 ) ’ det(m2 ) .

This follows by a computation which is left to the reader. It is clear that p2 = θ

IdSkew(A,θ) , and (??) shows that disc θ = 1. For x1 , y1 ∈ (a1 , b1 )0 and x2 , y2 ∈

F

§16. BIQUATERNION ALGEBRAS 249

(a2 , b2 )0 , we have

F

[x1 — 1 + 1 — x2 , y1 — 1 + 1 — y2 ] = [x1 — 1, y1 — 1] + [1 — x2 , 1 — y2 ]

= [x1 — 1 ’ 1 — x2 , y1 — 1 ’ 1 — y2 ],

hence pθ (x), pθ (y) = [x, y] for x, y ∈ Skew(A, σ). Therefore, the properties in (??)

hold with dθ = 1. The relations

and pθ axθ(a) = NrdA (a)θ(a)’1 pθ (x)a’1

qθ axθ(a) = NrdA (a)qθ (x)

for all a ∈ A— and x ∈ Skew(A, σ) follow by the same arguments as in (??) and

(??). The proposition is thus proved for σ = θ.

For an arbitrary orthogonal involution σ, there exists v ∈ A— such that σ =

Int(v) —¦ θ. The proof of (??) then yields » ∈ F — such that

pσ (x) = »vpθ (xv) and qσ (x) = »qθ (xv)

for all x ∈ Skew(A, σ). For a ∈ A— and x ∈ Skew(A, σ), we then have

qσ axσ(a) = »qθ axvθ(a) and pσ axσ(a) = »vpθ axvθ(a) .

Since property (??) is already proved for θ, it follows that

qσ axσ(a) = » NrdA (a)qθ (xv) = NrdA (a)qσ (x)

and

pσ axσ(a) = » NrdA (a)vθ(a)’1 pθ (xv)a’1

= » NrdA (a)σ(a)’1 vpθ (xv)a’1

= NrdA (a)σ(a)’1 pσ (x)a’1 ,

proving(??).

To complete the proof, we show that the properties in (??) hold with dσ =

2

» NrdA (v).

First, we have for x ∈ Skew(A, σ) that

qσ (x)2 = »2 qθ (xv)2 = »2 NrdA (v) NrdA (x)

and

p2 (x) = »vpθ »vpθ (xv)v .

σ

Since θ(v) = v, we may use property(??) for θ to rewrite the right side as

»2 v NrdA (v)v ’1 p2 (xv)v ’1 = »2 NrdA (v)x.

θ

Therefore, (??) and (??) hold. Since

disc σ = NrdA (v) · disc θ = NrdA (v) · F —2 ,

by (??), we also have (??). Finally, to establish (??), observe that by linearizing

the relations

qσ (x) = xpσ (x) = pσ (x)x

we get

bqσ (x, y) = xpσ (y) + ypσ (x) = pσ (x)y + pσ (y)x for x, y ∈ Skew(A, σ).

In particular,

bqσ pσ (x), y = pσ (x)pσ (y) + yp2 (x) = p2 (x)y + pσ (y)pσ (x)

σ σ

250 IV. ALGEBRAS OF DEGREE FOUR

for x, y ∈ Skew(A, σ). In view of (??), it follows that

pσ (x)pσ (y) + dσ yx = dσ xy + pσ (y)pσ (x),

hence

pσ (x)pσ (y) ’ pσ (y)pσ (x) = dσ (xy ’ yx)

for x, y ∈ Skew(A, σ), proving (??).

Alternately, properties (??) and (??) can be established by comparing σ with

a symplectic involution instead of θ (see the proof of (??)), and using (??). Details

are left to the reader.

With the notation of the preceding proposition, we have for all x ∈ Skew(A, σ)

qσ pσ (x) = p2 (x)pσ (x) = dσ xpσ (x) = dσ qσ (x),

σ

hence pσ is a similitude of Skew(A, σ), qσ with multiplier dσ . The group of simil-

itudes of Skew(A, σ), qσ can be described by mimicking (??).

(16.26) Proposition. The proper similitudes of Skew(A, σ), qσ are of the form

x ’ »’1 axσ(a)

where » ∈ F — and a ∈ A— .

The improper similitudes of Skew(A, σ), qσ are of the form

x ’ »’1 apσ (x)σ(a)

where » ∈ F — and a ∈ A— .

The proof is left to the reader.

We now give another point of view on the linear endomorphism pσ and the

Albert form qσ by relating them to the Cli¬ord algebra C(A, σ).

√

Let K = F ( disc σ) and let ι be the nontrivial automorphism of K/F . Recall

from §?? that we may identify

(A, σ) = NK/F (Q, γ)

for some quaternion K-algebra Q with canonical involution γ. The quaternion

algebra Q is canonically isomorphic as an F -algebra to the Cli¬ord algebra C(A, σ).

Recall also that there is a Lie algebra homomorphism n : L(Q) ’ L(A) de¬ned by

™

n(x) = ι x — 1 + ι 1 — x for x ∈ Q.

™

This homomorphism restricts to a Lie algebra isomorphism

∼

n : Q0 ’ Skew(A, σ).

™ ’

(16.27) Proposition. Let ± ∈ K — be such that ι(±) = ’±. The linear endomor-

phism pσ which makes the following diagram commutative:

n

™

Q0 ’ ’ ’ Skew(A, σ)

’’

¦ ¦

¦ ¦ pσ

±·

n

™

Q0 ’ ’ ’ Skew(A, σ)

’’

(where ±· is multiplication by ±) is such that xpσ (x) ∈ F for all x ∈ Skew(A, σ).

The corresponding Albert form qσ satis¬es:

qσ n(x) = trK/F (±x2 ) for x ∈ Q0 .

™

§16. BIQUATERNION ALGEBRAS 251

Let s : Q0 ’ K be the squaring map de¬ned by s(x) = x2 . Then qσ is the Scharlau

transfer of the form ± · s with respect to the linear form trK/F : K ’ F :

qσ = (trK/F )— ± · s .

Proof : It su¬ces to prove that

n(x)n(±x) = trK/F (±x2 ) for x ∈ Q0 .

™ ™

This follows from a straightforward computation:

n(x)n(±x) = (ι x — 1 + 1 — x) ι (±x) — 1 + 1 — ±x

™ ™

= ι(±x2 ) + (±x2 ) + ι (±x) — x + ι x — ±x.

Since ι(±) = ’±, the last two terms in the last expression cancel.

Continuing with the notation of the proposition above and letting dσ = ±2 ∈

F — , we obviously have p2 = dσ IdSkew(A,σ) and dσ · F —2 = disc σ, and also

σ

pσ (x), pσ (y) = dσ [x, y] for x, y ∈ Skew(A, σ)

since n is an isomorphism of Lie algebras. We may thus recover the properties

™

in (??.??).

Conversely, Proposition (??) shows that the linear endomorphism pσ can be

used to endow Skew(A, σ) with a structure of K-module, hence to give an explicit

description of the Cli¬ord algebra C(A, σ).

We may also derive some information on quaternion algebras over quadratic

extensions:

(16.28) Corollary. For a quaternion algebra Q over an ´tale quadratic exten-

e

sion K/F , the following conditions are equivalent:

(1) Q is split by some quadratic extension of F ;

(2) Q (a, b)K for some a ∈ F — and some b ∈ K — ;

(3) ind NK/F (Q) = 1 or 2.

Proof : It su¬ces to prove the equivalence of (??) and (??) for non-split quaternion

algebras Q, since (??) and (??) are clearly equivalent and the three conditions

trivially hold if Q is split. We may thus assume that the squaring map s : Q0 ’ K

is anisotropic. Condition (??) then holds if and only if the transfer (tr K/F )— ± ·s

is isotropic where ± ∈ K is an arbitrary nonzero element of trace zero. By (??)

and (??), (trK/F )— ± · s is an Albert form of NK/F (Q); therefore, by Albert™s

Theorem (??), this form is isotropic if and only if condition (??) holds.

In the special case where K = F — F , the corollary takes the following form:

(16.29) Corollary (Albert [?]). For quaternion algebras Q1 , Q2 over F , the fol-

lowing conditions are equivalent:

(1) there is a quadratic extension of F which splits both Q1 and Q2 ;

(2) Q1 (a, b1 )F and Q2 (a, b2 )F for some a, b1 , b2 ∈ F — ;

(3) ind(Q1 —F Q2 ) = 1 or 2.

The implication (??) ’ (??) may be reformulated as follows:

(16.30) Corollary. Let Q1 , Q2 , Q3 be quaternion algebras over F . If Q1 —Q2 —Q3

is split, then there is a quadratic extension of F which splits Q1 , Q2 , and Q3 .

252 IV. ALGEBRAS OF DEGREE FOUR

Proof : The hypothesis means that Q1 — Q2 is Brauer-equivalent to Q3 , hence its

index is 1 or 2. The preceding corollary yields a quadratic extension of F which

splits Q1 and Q2 , hence also Q3 .

Finally, we relate the preceding constructions to the generalized pfa¬an de¬ned

in §??. Let Z(A, σ) be the center of the Cli¬ord algebra C(A, σ). Let ι be the

nontrivial automorphism of Z(A, σ)/F and let Z(A, σ)0 be the space of elements

of trace zero, so

Z(a, σ)0 = { z ∈ Z(A, σ) | ι(z) = ’z }.

By (??), the generalized pfa¬an is a quadratic map

π : Skew(A, σ) ’ Z(A, σ)0 .

Our goal is to show that this map can be regarded as a canonical Albert form

associated to σ.

Let C(A, σ)0 be the space of elements of reduced trace zero in C(A, σ). Since

C(A, σ) is a quaternion algebra over Z(A, σ), we have x2 ∈ Z(A, σ) for x ∈

C(A, σ)0 . We may then de¬ne a map • : C(A, σ)0 ’ Z(A, σ)0 by

•(x) = x2 ’ ι(x2 ) for x ∈ C(A, σ)0 .

(16.31) Lemma. Let c : A ’ C(A, σ) be the canonical map. For x ∈ Skew(A, σ),

1

π(x) = • 2 c(x) .

Proof : We may extend scalars to an algebraic closure, and assume that

(A, σ) = M4 (F ), t = EndF (F 4 ), σq

where q(x1 , x2 , x3 , x4 ) = x2 +x2 +x2 +x2 . We use q to identify EndF (F 4 ) = F 4 —F 4

1 2 3 4

as in (??). Let (ei )1¤i¤4 be the standard basis of F 4 . A basis of Skew(A, σ) is given

by

hij = 1 (ei — ej ’ ej — ei ) for 1 ¤ i < j ¤ 4,

2

and we have, by (??),

π( xij hij ) = (x12 x34 ’ x13 x24 + x14 x23 )e1 · e2 · e3 · e4

1¤i<j¤4

for xij ∈ F . On the other hand,

1 1

2 c( xij hij ) = xij ei · ej ,

1¤i<j¤4 1¤i<j¤4

2

and a computation shows that

•( xij hij ) = (x12 x34 ’ x13 x24 + x14 x23 )e1 · e2 · e3 · e4 .

1¤i<j¤4

Theorem (??) yields a canonical isomorphism

(A, σ) = NZ(A,σ)/F C(A, σ), σ ;

moreover, using the canonical isomorphism as an identi¬cation, the map

n : C(A, σ)0 ’ Skew NZ(A,σ)/F C(A, σ), σ

™

1

Skew(A, σ) ’ C(A, σ)0 (see (??)). Therefore, the lemma

is the inverse of 2c:

yields:

π n(x) = •(x) = x2 ’ ι(x2 ) for x ∈ C(A, σ)0 .

™

§17. WHITEHEAD GROUPS 253

Let ± ∈ Z(A, σ)0 , ± = 0. According to (??), the map qσ : Skew(A, σ) ’ F de¬ned

by

for x ∈ C(A, σ)0

qσ n(x) = n(x)n(±x)

™ ™ ™

is an Albert form of A.

(16.32) Proposition. For all a ∈ Skew(A, σ),

qσ (a) = ±π(a).

Proof : It su¬ces to prove that

n(x)n(±x) = ± x2 ’ ι(x2 ) for x ∈ C(A, σ)0 .

™ ™

The left side has been computed in the proof of (??):

n(x)n(±x) = ι(±x2 ) + ±x2 .

™ ™

§17. Whitehead Groups

For an arbitrary central simple algebra A, we set

SL1 (A) = { a ∈ A— | NrdA (a) = 1 }.

This group contains the normal subgroup [A— , A— ] generated by commutators

aba’1 b’1 . The factor group is denoted

SK1 (A) = SL1 (A)/[A— , A— ].

This group is known in algebraic K-theory as the reduced Whitehead group of A.

It is known that SK1 (A) = 0 if A is split (and A = M2 (F2 )) or if the index

of A is square-free (a result due to Wang [?], see for example Pierce [?, 16.6] or

the lecture notes [?]). In the ¬rst subsection, we consider the next interesting case

where A is a biquaternion algebra. Let F be the center of A, which may be of

arbitrary characteristic. Denote by I k F the k-th power of the fundamental ideal

IF of the Witt ring W F of nonsingular bilinear spaces over F , and let Wq F denote

the Witt group of nonsingular even-dimensional quadratic spaces over F , which

is a module over W F . We write I k Wq F for I k F · Wq F , so I k Wq F = I k+1 F if

char F = 2. Our objective is to de¬ne a canonical injective homomorphism

± : SK1 (A) ’ I 3 Wq F/I 4 Wq F,

from which examples where SK1 (A) = 0 are easily derived.

In the second subsection, we brie¬‚y discuss analogues of the reduced Whitehead

group for algebras with involution in characteristic di¬erent from 2.

17.A. SK1 of biquaternion algebras. Although the map ± that we will

de¬ne is canonical, it is induced by a map ±σ whose de¬nition depends on the choice

of a symplectic involution. Therefore, we start with some general observations on

symplectic involutions on biquaternion algebras.

Let A be a biquaternion algebra over a ¬eld F of arbitrary characteristic, and

let σ be a symplectic involution on A. Recall from §?? the linear endomorphism

of Symd(A, σ) de¬ned by x = Trpσ (x) ’ x, and the quadratic form

Nrpσ (x) = xx = xx for x ∈ Symd(A, σ).

254 IV. ALGEBRAS OF DEGREE FOUR

As in §??, we let

Symd(A, σ)0 = { x ∈ Symd(A, σ) | Trpσ (x) = 0 },

and we write Symd(A, σ)— = Symd(A, σ) © A— for simplicity. For v ∈ Symd(A, σ)—

and x ∈ A, we have σ(x)vx ∈ Symd(A, σ), because if v = w + σ(w), then σ(x)vx =

σ(x)wx + σ σ(x)wx . We may therefore consider the quadratic form ¦v : A ’ F

de¬ned by

¦v (x) = Trpσ σ(x)vx for x ∈ A.

(17.1) Proposition. For each v ∈ Symd(A, σ)— , the quadratic form ¦v is a

scalar multiple of a 4-fold P¬ster form. This form is hyperbolic if Trpσ (v) = 0.

Moreover, if σ is a hyperbolic symplectic involution, then ¦v is hyperbolic for all

v ∈ Symd(A, σ)— .

Proof : Suppose ¬rst that σ is hyperbolic. The algebra A then contains an isotropic

right ideal I of reduced dimension 2, i.e., dimF I = 8. For x ∈ I we have σ(x)x = 0,

hence for all v = w + σ(w) ∈ Symd(A, σ)— ,

¦v σ(x) = Trpσ xvσ(x) = TrdA xwσ(x) = TrdA wσ(x)x = 0.

Therefore, σ(I) is a totally isotropic subspace of A for the form ¦v , hence this form

is hyperbolic.

For the rest of the proof, assume σ is not hyperbolic, and let V ‚ Symd(A, σ) be

a 3-dimensional subspace containing 1 and v, and not contained in Symd(A, σ)0 . By

(??), there is a decomposition of A into a tensor product of quaternion subalgebras,

so that

(A, σ) = (Q1 , σ1 ) —F (Q2 , γ2 )

where σ1 is an orthogonal involution, γ2 is the canonical involution, and v ∈

Sym(Q1 , σ1 ). In view of the computation of the bilinear form T(Q1 ,σ1 ,v) in (??), the

following lemma completes the proof:

(17.2) Lemma. Suppose (A, σ) = (Q1 , σ1 ) —F (Q2 , γ2 ), where σ1 is an orthogonal

involution and γ2 is the canonical (symplectic) involution. We have Sym(Q1 , σ1 ) ‚

Symd(A, σ) and, for all v ∈ Sym(Q1 , σ1 )— ,

¦v = T(Q1 ,σ1 ,v) · NrdQ2

where NrdQ2 is the reduced norm quadratic form on Q2 .

Proof : Let ∈ Q2 be such that + γ2 ( 2 ) = 1. For all s ∈ Sym(Q1 , σ1 ), we have

2 2

s—1=s— + σ(s — 2 ),

2

hence s — 1 ∈ Symd(A, σ) and

Trpσ (s — 1) = TrdA (s — 2) = TrdQ1 (s) TrdQ2 ( 2 ) = TrdQ1 (s).

Let v ∈ Sym(Q1 , σ1 )— . For x1 ∈ Q1 and x2 ∈ Q2 , we have

σ(x1 — x2 )v(x1 — x2 ) = σ1 (x1 )vx1 — γ2 (x2 )x2 = σ1 (x1 )vx1 — 1 NrdQ2 (x2 ).

Therefore,

¦v (x1 — x2 ) = Trpσ σ1 (x1 )vx1 — 1 NrdQ2 (x2 ) = TrdQ1 σ1 (x1 )vx1 NrdQ2 (x2 ),

hence

¦v (x1 — x2 ) = T(Q1 ,σ1 ,v) (x1 , x1 ) NrdQ2 (x2 ).

§17. WHITEHEAD GROUPS 255

To complete the proof, it remains only to show that the polar form b¦v of ¦v is the

tensor product T(Q1 ,σ1 ,v) — bNrdQ2 .

For x, y ∈ A, we have

b¦v (x, y) = Trpσ σ(x)vy + σ σ(x)vy = TrdA σ(x)vy ,

hence, for x = x1 — x2 and y = y1 — y2 ,

b¦v (x, y) = TrdQ1 σ1 (x1 )vy1 TrdQ2 γ2 (x2 )y2 .

Since TrdQ2 γ2 (x2 )y2 = bNrdQ2 (x2 , y2 ), the proof is complete.

The de¬nition of ±σ uses the following result, which is reminiscent of Hilbert™s

theorem 90:

(17.3) Lemma. Suppose σ is not hyperbolic. For every u ∈ Symd(A, σ) such that

Nrpσ (u) = 1, there exists v ∈ Symd(A, σ)— such that u = vv ’1 . If u = ’1, the

element v is uniquely determined up to multiplication by a factor in F — .

Proof : We ¬rst prove the existence of v. If u = ’1, we may take for v any unit in

Symd(A, σ)0 . If u = ’1, let v = 1 + u. We have

vu = (1 + u)u = u + Nrpσ (u) = v,

hence v satis¬es the required conditions if it is invertible. If v is not invertible, then

vv = 0, since vv = Nrpσ (v) ∈ F . In that case, we also have vvu = 0, hence v 2 = 0

since vu = v. It follows that Trpσ (v) = 0, and by (??) we derive a contradiction

with the hypothesis that σ is not hyperbolic. The existence of v is thus proved.

Suppose next that u = ’1 and v1 , v2 ∈ Symd(A, σ)— are such that

u = v1 v1 ’1 = v2 v2 ’1 .

Then

u + 1 = (v1 + v1 )v1 ’1 = (v2 + v2 )v2 ’1 .

Since vi + vi = Trpσ (vi ) ∈ F for i = 1, 2, these equations together with the

hypothesis that u = ’1 show that Trpσ (v1 ) = 0 and Trpσ (v2 ) = 0. They also yield

Trpσ (v1 )v2 = Trpσ (v2 )v1 ,

hence v1 and v2 di¬er by a nonzero factor in F .

Now, consider the following subgroup of F — — A— :

“ = { (», a) ∈ F — — A— | »2 = NrdA (a) }.

For (», a) ∈ “, we have ’»’1 σ(a)a ∈ Symd(A, σ), since 1 ∈ Symd(A, σ), and

Proposition (??) shows that Nrpσ ’»’1 σ(a)a = 1. Therefore, if σ is not hyper-

bolic, the preceding lemma yields v ∈ Symd(A, σ)— such that

vv ’1 = ’»’1 σ(a)a.

(17.4)

If »’1 σ(a)a = 1 (i.e., a ∈ GSp(A, σ) and » = µ(a)), we have v = ’v, hence

Trpσ (v) = 0. Proposition (??) then shows that ¦v is hyperbolic. If »’1 σ(a)a = 1,

the element v is uniquely determined up to a factor in F — . Therefore, the quadratic

form ¦v ∈ I 3 Wq F is also uniquely determined up to a factor in F — , and its class

in I 3 Wq F/I 4 Wq F is uniquely determined.

We may therefore set the following de¬nition:

(17.5) De¬nition. Let ±σ : “ ’ I 3 Wq F/I 4 Wq F be de¬ned as follows:

256 IV. ALGEBRAS OF DEGREE FOUR

(1) If σ is hyperbolic, let ±σ = 0.

(2) If σ is not hyperbolic, let

±σ (», a) = ¦v + I 4 Wq F

where v ∈ Symd(A, σ)— satis¬es (??).

In particular, the observations above show that ±σ µ(g), g = 0 for g ∈ GSp(A, σ).

If L is an extension ¬eld of F over which σ is hyperbolic, every quadratic form

¦v for v ∈ Symd(A, σ)— becomes hyperbolic over L by (??), hence the de¬nition

above is compatible with scalar extension.

Observe that the group SL1 (A) embeds in “ by mapping a ∈ SL1 (A) to (1, a) ∈

“. Our goal is to prove the following theorem:

(17.6) Theorem. The map ±σ de¬ned above is a homomorphism. Its restriction

to SL1 (A) does not depend on the choice of the symplectic involution σ; letting

± : SL1 (A) ’ I 3 Wq F/I 4 Wq F denote this restriction, we have ker ± = [A— , A— ],

hence ± induces an injective homomorphism

± : SK1 (A) ’ I 3 Wq F/I 4 Wq F.

The rest of this subsection consists of the proof, which we break into three

parts: we ¬rst show that ±σ is a homomorphism, then we investigate the e¬ect of

a change of involution, and ¬nally we determine the kernel of ±σ .

±σ is a homomorphism. If σ is hyperbolic, then ±σ is clearly a homomor-

phism since we set ±σ = 0. Throughout this part of the proof, we may thus assume

that σ is not hyperbolic. Let (», a), (» , a ) ∈ “. In order to show that

±σ (», a) + ±σ (» , a ) = ±σ (»» , aa ),

we ¬rst reduce to the case where a, a ∈ Symd(A, σ) and » = Nrpσ (a), » =

Nrpσ (a ).

(17.7) Lemma. For (», a) ∈ “ and g ∈ GSp(A, σ),

±σ (», a) = ±σ µ(g)», ga = ±σ »µ(g), ag .

Proof : Let v ∈ Symd(A, σ)— be subject to (??); since

»’1 µ(g)’1 σ(ga)ga = »’1 σ(a)a,

the quadratic form ¦v represents ±σ (», a) as well as ±σ µ(g)», ga , hence the ¬rst

equation is clear. To prove the second equation, we calculate

’µ(g)’1 »’1 σ(ag)ag = g ’1 vg · g ’1 v ’1 g.

’1

By (??), the last factor on the right side is equal to g ’1 vg , hence

±σ »µ(g), ag = ¦g’1 vg + I 4 Wq F ∈ I 3 Wq F/I 4 Wq F.

For x ∈ A, we have σ(x)g ’1 vgx = µ(g)’1 σ(gx)vgx, hence

¦g’1 vg (x) = µ(g)’1 ¦v (gx).

This equation shows that the quadratic forms ¦g’1 vg and ¦v are similar, hence

±σ »µ(g), ag = ±σ (», a).

(17.8) Lemma. For all (», a) ∈ “, there exist a similitude g ∈ GSp(A, σ) and

units u, v ∈ Symd(A, σ)— such that

(», a) = µ(g), g · Nrpσ (u), u = Nrpσ (v), v · µ(g), g .

§17. WHITEHEAD GROUPS 257

Proof : By (??), there exists some u ∈ Symd(A, σ)— such that uu ’1 = »’1 σ(a)a.

For g = au’1 we have

σ(g)g = u’1 σ(a)au’1 = » Nrpσ (u)’1 ∈ F — ,

hence g ∈ GSp(A, σ) and µ(g) Nrpσ (u) = ». We thus get the ¬rst decomposition.

In order to get the second, it su¬ces to choose v = gug ’1.

For (», a), (» , a ) ∈ “, we may thus ¬nd g, g ∈ GSp(A, σ) and u, v ∈

Symd(A, σ)— such that

(», a) = µ(g), g · Nrpσ (u), u ,

(» , a ) = Nrpσ (v), v · µ(g ), g ,

hence also

(»» , aa ) = µ(g), g · Nrpσ (u), u · Nrpσ (v), v · µ(g ), g .

From (??), it follows that

±σ (», a) = ±σ Nrpσ (u), u , ±σ (» , a ) = ±σ Nrpσ (v), v

and

±σ (»» , aa ) = ±σ Nrpσ (u) Nrpσ (v), uv .

Therefore, in order to show that

±σ (»» , aa ) = ±σ (», a) + ±σ (» , a ),

it su¬ces to show that

(17.9) ±σ Nrpσ (u) Nrpσ (v), uv = ±σ Nrpσ (u), u + ±σ Nrpσ (v), v .

We have thus achieved the desired reduction.

If u ∈ Symd(A, σ)0 , then ’ Nrpσ (u)’1 σ(u)u = 1, hence

±σ Nrpσ (u), u = ¦1 + I 4 Wq F.

Therefore, if u and v both lie in Symd(A, σ)0 , the right side of (??) vanishes. The

left side also vanishes, since uv ∈ GSp(A, σ) and µ(uv) = Nrpσ (u) Nrpσ (v). For

the rest of the proof of (??), we may thus assume that u and v are not both in

Symd(A, σ)0 .

Consider a 3-dimensional subspace V ‚ Symd(A, σ) which contains 1, u and v,

and is therefore not contained in Symd(A, σ)0 . By (??), there is a decomposition

of A into a tensor product of quaternion subalgebras, so

(A, σ) = (Q1 , σ1 ) —F (Q2 , γ2 )

where σ1 is an orthogonal involution, γ2 is the canonical involution, and u, v ∈

Sym(Q1 , σ1 ), hence also uv ∈ Q1 . For x ∈ Q1 , we have Prd2 1 ,x = PrdA,x , hence

Q

Prpσ,x = PrdQ1 ,x and therefore Nrpσ (x) = NrdQ1 (x). To prove (??), it now su¬ces

to prove the following lemma:

(17.10) Lemma. Suppose A decomposes into a tensor product of quaternion alge-

bras stable under the symplectic involution σ, which may be hyperbolic:

(A, σ) = (Q1 , σ1 ) — (Q2 , γ2 ),

where σ1 is an orthogonal involution and γ2 is the canonical (symplectic) involution.

For all x ∈ Q— ,

1

±σ NrdQ1 (x), x = NrdQ1 (x), disc σ1 · NrdQ2 .

258 IV. ALGEBRAS OF DEGREE FOUR

Indeed, assuming the lemma and letting θ = disc σ1 · NrdQ2 , the left-hand

side of (??) is then NrdQ1 (uv) · θ, while the right-hand side is NrdQ1 (u) · θ +

NrdQ1 (v) · θ. The equality follows from the congruence

mod I 2 F.

NrdQ1 (u) + NrdQ1 (v) ≡ NrdQ1 (u) NrdQ1 (v)

Proof of (??): Suppose ¬rst that σ is hyperbolic. We then have to show that

the quadratic form NrdQ1 (x), disc σ1 · NrdQ2 is hyperbolic for all x ∈ Q1 . For

v ∈ Sym(Q1 , σ1 )— , Proposition (??) and Lemma (??) show that the quadratic form

T(Q1 ,σ1 ,v) · NrdQ2 is hyperbolic. In view of (??), this means that

NrdQ1 (vs), disc σ1 · NrdQ2 = 0 in Wq F

for all v ∈ Sym(Q1 , σ1 )— and all s ∈ Q— such that σ1 (s) = s = ’γ1 (s), where γ1

1

is the canonical involution on Q1 . In particular (for v = 1), the quadratic form

NrdQ1 (s), disc σ1 · NrdQ2 is hyperbolic. Adding it to both sides of the equality

above, we get

for all v ∈ Sym(Q1 , σ1 )— .

NrdQ1 (v), disc σ1 · NrdQ2 = 0

To complete the proof in the case where σ is hyperbolic, it now su¬ces to show

that Sym(Q1 , σ1 )— generates Q— . For all x ∈ Q— , the intersection

1 1

Sym(Q1 , σ1 ) © x Sym(Q1 , σ1 )

has dimension at least 2. Since the restriction of NrdQ1 to Sym(Q1 , σ1 ) is a non-

singular quadratic form, this intersection is not totally isotropic for NrdQ1 , hence

it contains an invertible element s1 ∈ Sym(Q1 , σ1 )— . We have s1 = xs2 for some

s2 ∈ Sym(Q1 , σ1 )— , hence x = s1 s’1 is in the group generated by Sym(Q1 , σ1 )— .

2

For the rest of the proof, assume that σ is not hyperbolic. Let σ1 = Int(r) —¦ γ1

for some r ∈ Skew(Q1 , γ1 ) F ; thus, r2 ∈ F — and disc σ1 = r2 · F —2 .

If σ(x)x = NrdQ1 (x), then x ∈ GSp(A, σ) and µ(x) = NrdQ1 (x), hence

±σ NrdQ1 (x), x = 0. On the other hand, the condition also implies σ1 (x) = γ1 (x),

hence x commutes with r. Therefore, x ∈ F [r], and NrdQ1 (x), disc σ1 is meta-

bolic. The lemma thus holds in this case.

Assume ¬nally that σ(x)x = NrdQ1 (x), and let

w = 1 ’ NrdQ1 (x)’1 σ(x)x = 1 ’ σ(x)γ1 (x)’1 ∈ Q1 ,

so that w ’ NrdQ1 (x)’1 σ(x)x = w. Since σ is not hyperbolic, the same arguments

as in the proof of (??) show that w is invertible, hence

±σ NrdQ1 (x), x = ¦w + I 4 Wq F.

By (??), we have ¦w ≡ T(Q1 ,σ1 ,w) ·NrdQ2 mod I 4 Wq F . Moreover, we may compute

T(Q1 ,σ1 ,w) by (??): since wγ1 (x) = γ1 (x) ’ σ1 (x) ∈ Q— satis¬es

1

σ1 wγ1 (x) = wγ1 (x) = ’γ1 wγ1 (x) ,

we may substitute wγ1 (x) for s in (??) and get

T(Q1 ,σ1 ,w) ≡ NrdQ1 w2 γ1 (x) , disc σ1 ≡ NrdQ1 (x), disc σ1 mod I 3 F.

§17. WHITEHEAD GROUPS 259

Change of involution. We keep the same notation as above, and allow

the symplectic involution σ to be hyperbolic. For x ∈ Symd(A, σ)0 , we have

x2 = ’ Nrpσ (x) ∈ F . We endow Symd(A, σ)0 with the restriction of Nrpσ or,

equivalently for the next result, with the squaring quadratic form sσ (x) = x2 (see

§??).

(17.11) Proposition. Every proper isometry of Symd(A, σ)0 has the form x ’

gxg ’1 for some g ∈ GSp(A, σ). For every g ∈ GSp(A, σ), one can ¬nd two or four

anisotropic vectors v1 , . . . , vr ∈ Symd(A, σ)0 such that

µ(g), g = Nrpσ (v1 ), v1 · · · Nrpσ (vr ), vr in “.

Proof : The proposition readily follows from (??). We may however give a short

direct argument: for all v ∈ Symd(A, σ)0 anisotropic, computation shows that the

hyperplane re¬‚ection ρv maps x ∈ Symd(A, σ)0 to ’vxv ’1 . The Cartan-Dieudonn´ e

theorem shows that every proper isometry is a product of an even number of hy-

perplane re¬‚ections, and is therefore of the form

’1

’1

x ’ (v1 · · · vr )x(vr · · · v1 )

for some anisotropic v1 , . . . , vr ∈ Symd(A, σ)0 . Since

σ(v1 · · · vr ) · v1 · · · vr = v1 · · · vr ∈ F — ,

2 2

the element v1 · · · vr is in GSp(A, σ), and the ¬rst part is proved.

To prove the second part, observe that for g ∈ GSp(A, σ) and x ∈ Symd(A, σ)0

we have gxg ’1 = µ(g)’1 gxσ(g), hence by (??) and (??),

Nrpσ (gxg ’1 ) = µ(g)’2 NrdA (g) Nrpσ (x) = Nrpσ (x).

Therefore, the map x ’ gxg ’1 is an isometry of Symd(A, σ)0 . If this isometry

is improper, then char F = 2 and x ’ ’gxg ’1 is proper, hence of the form x ’

’1

for some g ∈ GSp(A, σ). In that case g ’1 g anticommutes with every

g xg

element in Symd(A, σ)0 . However, using a decomposition of (A, σ) as in (??), it

is easily seen that no nonzero element of A anticommutes with Symd(A, σ)0 . This

contradiction shows that the isometry x ’ gxg ’1 is proper in all cases. By the

Cartan-Dieudonn´ theorem, it is a product of an even number r of hyperplane

e

re¬‚ections with r ¤ 5, hence we may ¬nd anisotropic v1 , . . . , vr ∈ Symd(A, σ)0

such that

’1

gxg ’1 = (v1 · · · vr )x(vr · · · v1 ) for x ∈ Symd(A, σ)0 .

’1

Since Symd(A, σ)0 generates A, it follows that g ’1 (v1 · · · vr ) ∈ F — . Multiplying v1

by a suitable factor in F — , we get g = v1 · · · vr ; then

2 2

µ(g) = v1 · · · vr = Nrpσ (v1 ) · · · Nrpσ (vr ).

Now, let „ be another symplectic involution on A. Recall from (??) the 3-fold

P¬ster form jσ („ ) uniquely determined by the condition

mod I 3 Wq F.

jσ („ ) ≡ Nrpσ ’ Nrp„

Since jσ („ ) ≡ ’jσ („ ) ≡ Nrp„ ’ Nrpσ mod I 3 Wq F , we have jσ („ ) j„ (σ).

(17.12) Proposition. For all (», a) ∈ “,

±σ (», a) + ±„ (», a) = » · jσ („ ) + I 4 Wq F = » · j„ (σ) + I 4 Wq F.

260 IV. ALGEBRAS OF DEGREE FOUR

Proof : If σ and „ are hyperbolic, the proposition is clear since ±σ = ±„ = 0

and jσ („ ) jσ (σ) = 0 in Wq F by (??), since all the hyperbolic involutions are

conjugate. We may thus assume that at least one of σ, „ is not hyperbolic. Let

us assume for instance that σ is not hyperbolic, and let „ = Int(u) —¦ σ for some

u ∈ Symd(A, σ)— .

We consider two cases: suppose ¬rst that Trpσ (u) = 0. Lemma (??) and

Proposition (??) show that “ is generated by elements of the form Nrpσ (v), v ,

with v ∈ Sym(A, σ)— . Therefore, it su¬ces to prove

±σ Nrpσ (v), v + ±„ Nrpσ (v), v = Nrpσ (v) · jσ („ ) + I 4 Wq F

for all v ∈ Symd(A, σ)— . Let V ‚ Symd(A, σ) be a 3-dimensional subspace con-

taining 1, u and v. Since Trpσ (u) = 0, we have27 V ‚ Symd(A, σ)0 . By (??), there

is a decomposition

(A, σ) = (Q1 , σ1 ) —F (Q2 , γ2 )

where Q1 is the quaternion subalgebra generated by V and σ1 = σ|Q1 is an orthog-

onal involution. By (??), we have

±σ Nrpσ (v), v = Nrpσ (v), disc σ1 · NrdQ2 +I 4 Wq F.

Since „ = Int(u) —¦ σ and u ∈ Q1 , the algebra Q1 is also stable under „ , hence

(A, „ ) = (Q1 , „1 ) —F (Q2 , γ2 )

where „1 = Int(u) —¦ σ1 . The involution „1 is orthogonal, since u ∈ Sym(Q1 , σ1 ) and

TrdQ1 (u) = Trpσ (u) = 0. Therefore, by (??) again,

±„ Nrpσ (v), v = Nrpσ (v), disc „1 · NrdQ2 +I 4 Wq F.

On the other hand, we have disc „1 = NrdQ1 (u) disc σ1 by (??), hence

±σ Nrpσ (v), v + ±„ Nrpσ (v), v = Nrpσ (v), NrdQ1 (u) · NrdQ2 +I 4 Wq F.

This completes the proof in the case where Trpσ (u) = 0, since the proof of (??)

shows that

NrdQ1 (u) · NrdQ2 = Nrpσ (u) · NrdQ2 = jσ („ ).

Consider next the case where Trpσ (u) = 0. We then compare ±σ and ±„ via a

third involution ρ, chosen in such a way that the ¬rst case applies to σ and ρ on

one hand, and to ρ and „ on the other hand. Speci¬cally, let t ∈ Symd(A, σ)

Symd(A, σ)0 be an invertible element which is not orthogonal to u for the polar

form bNrpσ . Let

ξ = bNrpσ (u, t) = ut + tu = 0.

Since u = ’u, this relation yields tutu’1 = Nrpσ (t) ’ ξtu’1 . Letting s =

ut’1 , we have s ∈ F since Trpσ (t) = 0 while Trpσ (u) = 0, and s’2 = ξs’1 ’

/

Nrpσ (t) Nrpσ (u)’1 ∈ F . Let ρ = Int(t) —¦ σ, hence „ = Int(s) —¦ ρ since u = st. We

/

have s ∈ Symd(A, σ)t’1 = Symd(A, ρ), and Trpρ (s) = 0 since s2 ∈ F . Therefore,

/

27 If

char F = 2, then V ‚ Symd(A, σ)0 even if Trpσ (u) = 0, since Trpσ (1) = 2 = 0. The

arguments in the ¬rst case thus yield a complete proof if char F = 2.

§17. WHITEHEAD GROUPS 261

we may apply the ¬rst case to compare ±„ and ±ρ , and also to compare ±ρ and ±σ ,

since t ∈ Symd(A, σ)0 . We thus get

/

±σ (», a) + ±ρ (», a) = » · jσ (ρ) + I 4 Wq F,

±ρ (», a) + ±„ (», a) = » · jρ („ ) + I 4 Wq F

for all (», a) ∈ “. The result follows by adding these relations, since jσ (ρ) + jρ („ ) ≡

jσ („ ) mod I 3 Wq F .

(17.13) Corollary. For all a ∈ SL1 (A),

±σ (1, a) = ±„ (1, a).

Proof : This readily follows from the proposition, since 1 = 0 in Wq F .

In view of this corollary, we may de¬ne a map ± : SL1 (A) ’ I 3 Wq F/I 4 Wq F

by

±(a) = ±σ (1, a) for a ∈ SL1 (A),

where σ is an arbitrary symplectic involution on A.

(17.14) Example. Let Q be a quaternion F -algebra with canonical involution γ

and let A = M2 (Q) with the involution θ de¬ned by

q11 q12 γ(q11 ) ’γ(q21 )

θ = .

q21 q22 ’γ(q12 ) γ(q22 )

This involution is symplectic (see (??)), and it is hyperbolic since

q11 q12

I= q11 , q12 ∈ Q

q11 q12

is a right ideal of reduced dimension 2 such that θ(I) · I = {0}. Therefore, ±θ = 0

and ± = 0.

If „ is another symplectic involution on A, Proposition (??) yields

±„ (», a) = » · jθ („ )

for all (», a) ∈ “. More explicitly, if „ = Int(u) —¦ θ with u ∈ Symd(A, θ)— , it follows

from (??) and (??) that jθ („ ) = Nrpθ (u) · NrdQ , hence

±„ (», a) = », Nrpθ (u) · NrdQ .

Kernel of ±. We continue with the notation above. Our objective is to deter-

mine the kernel of the homomorphism ±σ : “ ’ I 3 Wq F/I 4 Wq F ; the kernel of the

induced map ± : SL1 (A) ’ I 3 Wq F/I 4 Wq F is then easily identi¬ed with [A— , A— ].

(17.15) Lemma. Suppose the symplectic involution σ is hyperbolic and let U be an

arbitrary 2-dimensional subspace in Symd(A, σ). For every u ∈ Symd(A, σ), there

exists an invertible element x ∈ A— such that TrpA σ(x)ux = 0 and xσ(x) ∈ U .

/

Note that we do not assume that u is invertible, so the quadratic form x ’

TrpA σ(x)ux may be singular.

Proof : Since σ is hyperbolic, the index of A is 1 or 2, hence A M2 (Q) for

some quaternion F -algebra Q. Consider the involution θ on M2 (Q) de¬ned as in

(??). Since all the hyperbolic involutions are conjugate by (??), we may identify

(A, σ) = M2 (Q), θ . We have Nrpθ Nrp0 (u) 0 = Nrpθ (u), hence Witt™s theorem

θ

1

on the extension of isometries (see Scharlau [?, Theorem 7.9.1]) shows that there

262 IV. ALGEBRAS OF DEGREE FOUR

is an isometry of Sym M2 (Q), θ , Nrpθ which maps u to Nrp0 (u) 0 . Composing

θ

1

with a suitable hyperplane re¬‚ection, we may assume that this isometry is proper.

By (??), it follows that there exist a ∈ A— and » ∈ F — such that

Nrpθ (u) 0

θ(a)ua = » .

0 1

Let b, c ∈ Q— be such that NrdQ (c) = 1 and c ’ 1 ∈ Q— , and let x = a 11 ∈ A.

b bc

Then

1 ’γ(b) Nrpθ (u) 0 11

θ(x)ux = »

’1 γ(bc) 0 1 b bc

Nrpθ (u) ’ NrdQ (b) Nrpθ (u) ’ NrdQ (b)c

=»

’ Nrpθ (u) + NrdQ (b)γ(c) ’ Nrpθ (u) + NrdQ (b)

hence Trpσ θ(x)ux = 0. On the other hand, x is invertible since it is a product of

invertible matrices:

10 10 10 11

x=a .

’1

0b 11 0c 01

0 γ(c’1)γ(b)

Finally, we have xθ(x) = a θ(a), hence xθ(x) ∈ U if and only if

b(1’c) 0

0 γ(c ’ 1)γ(b)

∈ a’1 U θ(a)’1 .

b(1 ’ c) 0

Since b is arbitrary in Q— , it is clear that we can choose b such that this relation

does not hold.

The following result holds for an arbitrary symplectic involution σ:

(17.16) Lemma. If x1 , x2 ∈ Symd(A, σ) F satisfy Trpσ (x1 ) = Trpσ (x2 ) and

Nrpσ (x1 ) = Nrpσ (x2 ), then there exists some g ∈ GSp(A, σ) such that gx1 g ’1 = x2 .

Proof : The hypothesis yields

Nrpσ (ξ + ·x1 ) = Nrpσ (ξ + ·x2 ) for ξ, · ∈ F,

hence the 2-dimensional subspace of Symd(A, σ) spanned by 1, x1 is isometric to

the subspace spanned by 1, x2 . By Witt™s theorem, the isometry which maps 1 to

1 and x1 to x2 extends to an isometry f of Symd(A, σ), Nrpσ , and this isometry

may be assumed to be proper. By (??), there exist » ∈ F — and g ∈ A— such that

f (x) = »’1 gxσ(g) for all x ∈ Symd(A, σ). Since f (1) = 1, we have g ∈ GSp(A, σ)

and » = µ(g), hence

x2 = f (x1 ) = gx1 g ’1 .

(17.17) Proposition. For all (», a) ∈ “ such that ±σ (», a) = 0, there exist g ∈

GSp(A, σ) and x, y ∈ A— such that

(», a) = µ(g), g · (1, xyx’1 y ’1 ).

Proof : Let v = 1 ’ »’1 σ(a)a ∈ Symd(A, σ). If v = 0, then a ∈ GSp(A, σ) and

» = µ(a), so we may take g = a and x = y = 1. For the rest of the proof, we may

thus assume that v = 0.

Claim. There exists some y ∈ A— such that Trpσ σ(y)vy = 0. Moreover, if σ is

hyperbolic, we may assume that y ∈ GSp(A, σ) and ay ∈ GSp(A, σ).

/ /

§17. WHITEHEAD GROUPS 263

This readily follows from (??) if σ is hyperbolic, since we may ¬nd y ∈ A— such

that Trpσ σ(y)vy = 0 and assume moreover that yσ(y) does not lie in the subspace

spanned by 1 and a’1 σ(a)’1 , hence y, ay ∈ GSp(A, σ). If σ is not hyperbolic, then

/

the proof of (??) shows that v is invertible and satis¬es vv ’1 = ’»’1 σ(a)a, hence

also ±σ (», a) = ¦v + I 4 Wq F . By hypothesis, ±σ (», a) = 0, hence there exists some

y0 ∈ A such that ¦v (y0 ) = 0. If y0 is invertible, the claim is proved with y = y0 .

If y0 is not invertible, then we have Nrpσ σ(y0 )vy0 = NrdA (y0 ) Nrpσ (v) = 0

2

and Trpσ σ(y0 )vy0 = 0, hence σ(y0 )vy0 = ’ Nrpσ σ(y0 )vy0 = 0. Since σ is

anisotropic, this relation implies σ(y0 )vy0 = 0 by (??), hence also v ’1 σ(y0 )vy0 =

0, showing that the involution Int(v ’1 ) —¦ σ is isotropic. Since this involution is

symplectic, it is then hyperbolic by (??). Therefore, by (??) we may ¬nd y1 ∈ A—

such that Trpσ Int(v ’1 ) —¦ σ(y1 )v ’1 y1 = 0, i.e.,

Trpσ v ’1 σ(y1 )y1 = 0.

For y = v ’1 σ(y1 ) ∈ A— , we then have

Trpσ σ(y)vy = Trpσ y1 v ’1 σ(y1 ) = 0

and the claim is proved.

In view of the de¬nition of v, we derive from Trpσ σ(y)uy = 0 that

(17.18) » Trpσ σ(y)y = Trpσ σ(y)σ(a)ay .

On the other hand, we also have

Nrpσ »σ(y)y = »2 NrdA (y)

and

Nrpσ σ(y)σ(a)ay = NrdA (a) NrdA (y)

by (??), hence

(17.19) Nrpσ »σ(y)y = Nrpσ σ(y)σ(a)ay .

If σ(y)y ∈ F and σ(ay)ay ∈ F (which may be assumed if σ is hyperbolic), we

/ /

may then apply (??) to get a similitude g0 ∈ GSp(A, σ) such that

’1

σ(ay)ay = »g0 σ(y)yg0 .

Multiplying on the left by σ(y)’1 σ(g0 ) and on the right by g0 y ’1 , we derive from

the preceding equation

σ(ayg0 y ’1 )ayg0 y ’1 = »µ(g0 ).

Therefore, the element g1 = ayg0 y ’1 is in GSp(A, σ), and µ(g1 ) = »µ(g0 ). We then

have

’1 ’1

a = (g1 g0 )(g0 yg0 y ’1 ),

’1

which yields the required decomposition with g = g1 g0 and x = g0 .

To complete the proof, we examine the cases where σ is not hyperbolic and one

of the inclusions σ(y)y ∈ F or σ(ay)ay ∈ F holds.

If σ(y)y = µ ∈ F — , we derive from (??) and (??):

Nrpσ σ(ay)ay = »2 µ2 .

Trpσ σ(ay)ay = 2»µ and

264 IV. ALGEBRAS OF DEGREE FOUR

Therefore, the element b = σ(ay)ay ’ »µ satis¬es Trpσ (b) = Nrpσ (b) = 0, hence

b = 0 since σ is not hyperbolic. We thus have σ(ay)ay = »µ and σ(y)y = µ, hence

a ∈ GSp(A, σ) and µ(a) = ». We may take g = a and x = y = 1 in this case.

Similarly, if σ(ay)ay = ν ∈ F — , then (??) and (??) yield

Trpσ σ(y)y = 2»’1 ν Nrpσ σ(y)y = »’2 ν 2 ,

and

and the same argument as above shows that σ(y)y = »’1 ν. Again, we get that

a ∈ GSp(A, σ) and µ(a) = », hence we may choose g = a and x = y = 1.

Our next goal is to show that all the elements of the form µ(g), g (1, xyx’1 y ’1 )

with g ∈ GSp(A, σ) and x, y ∈ A— are in the kernel of ±σ .

(17.20) Lemma. ±σ NrdA (x), x2 = 0 for all x ∈ A— .

Proof : If σ is hyperbolic, the result is clear since ±σ = 0. For the rest of the proof,

we may thus assume that σ is not hyperbolic. Let v ∈ Symd(A, σ)— be such that

vv ’1 = ’ NrdA (x)’1 σ(x)2 x2 , so that ±σ NrdA (x), x2 = ¦v + I 4 Wq F . We thus

have to show that ¦v is hyperbolic.

If NrdA (x)’1 σ(x)2 x2 = 1, then Trpσ (v) = 0 and the result follows from (??).

If NrdA (x)’1 σ(x)2 x2 = 1, the proof of (??) shows that we may assume v = 1 ’

NrdA (x)’1 σ(x)2 x2 . We then have

¦v (x’1 ) = Trpσ σ(x)’1 x’1 ’ NrdA (x)’1 Trpσ σ(x)x .

Since Nrpσ xσ(x) = xσ(x)xσ(x), we get by (??) that

σ(x)’1 x’1 = NrdA (x)’1 xσ(x),

hence ¦v (x’1 ) = 0. Since every isotropic P¬ster form is hyperbolic, it follows that

¦v + I 4 Wq F = 0.

The main result in this part of the proof of Theorem (??) is the following:

(17.21) Proposition. Embedding GSp(A, σ) and SL1 (A) in “ by mapping g ∈

GSp(A, σ) to µ(g), g and a ∈ SL1 (A) to (1, a), we have

ker ±σ = GSp(A, σ) · [A— , A— ].

Proof : In view of (??), it su¬ces to show that ±σ µ(g), g = 0 for all g ∈ GSp(A, σ)

and ±σ (1, xyx’1 y ’1 ) = 0 for all x, y ∈ A— .

The ¬rst relation is clear, either from the de¬nition of ±σ or from (??), since

±σ is a homomorphism. The second relation follows from the preceding lemma and

the equality

(1, xyx’1 y ’1 ) = NrdA (xy), (xy)2 · NrdA (x)’1 , (y ’1 x’1 y)2 · NrdA (y)’1 , y ’2 .

We proceed to determine the kernel of the induced map

± : SL1 (A) ’ I 3 Wq F/I 4 Wq F.

(17.22) Corollary. ker ± = [A— , A— ]. More precisely, every element in ker ± is a

product of two commutators.

§17. WHITEHEAD GROUPS 265

Proof : The preceding proposition shows that [A— , A— ] ‚ ker ±. To derive the

converse inclusion from (??), it su¬ces to show that every element g ∈ Sp(A, σ)

is a commutator. By (??), there is an involution which leaves g invariant, hence

there exists some x ∈ A— such that xσ(g)x’1 = g. Since σ(g) = g ’1 , it follows

that x(1 + g)x’1 g = 1 + g. Therefore, if 1 + g is invertible, g is a commutator:

g = x(1 + g)’1 x’1 (1 + g).

If g = ’1 and char F = 2, we have g = iji’1 j ’1 , where (1, i, j, k) is a quaternion

basis of any quaternion subalgebra of A. The proof is thus complete if A is a division

algebra. If ind A = 2, we still have to consider the case where 1 + g generates a

right ideal of reduced dimension 2. Denoting by Q a quaternion division algebra

which is Brauer-equivalent to A, we can ¬nd a representation A = M2 (Q) such that

either

’1 0

g=

0±

for some ± ∈ Q such that NrdQ (±) = 1 or

’1 β

g=

0 ’1

for some pure quaternion β = 0: see Exercise ??. In the ¬rst case, we have

± = ±1 ±2 ±’1 ±’1 for some ±1 , ±2 ∈ Q— since every quaternion of reduced norm 1

1 2

is a commutator (see Exercise ??); hence g = xyx’1 y ’1 where

i0 j0

x= ,y = if char F = 2,

0 ±1 0 ±2

and

10 10

x= ,y = if char F = 2.

0 ±1 0 ±2

In the second case, if char F = 2, pick a quaternion γ = 0 which anticommutes

with β. Then g has the following expression as a commutator:

’β 2 /2 β ’1 γ ’1

β γ 0 1/2 0

g= .

β ’1 γ ’1

0 β 0 γ 0 0

If char F = 2, pick γ ∈ Q— such that 1 + γ ’1 β = 0. Then

1 + γ ’1 β (1 + γ ’1 β)’1

1γ 0 1 ’γ 0

g= .

01 0 1 01 0 1

We leave the case where A is split as an exercise (see Exercise ??).

The proof of Theorem (??) is now complete.

(17.23) Example. Let A = (a, b)F — (c, d)F be an arbitrary biquaternion algebra

over a ¬eld F of characteristic di¬erent from 2. Suppose F contains a primitive

fourth root of unity ζ. We then have ζ ∈ SL1 (A), and we may compute ±(ζ) as

follows.

Let (1, i, j, k) be the quaternion basis of (a, b)F , viewed as a subalgebra of

A, and let σ be the symplectic involution on A which restricts to the canonical

involution on (c, d)F and such that σ(i) = i, σ(j) = j and σ(k) = ’k. By de¬nition,

266 IV. ALGEBRAS OF DEGREE FOUR

’1

±(ζ) = ±σ (1, ζ) = ¦1 + I 5 F , since 11 = ’σ(ζ)ζ. A diagonalization of the form

¦1 can be derived from (??):

¦1 ’a, ’b · c, d .

Since ζ is a square root of ’1 in F , we have ’a ≡ a mod F —2 and ’b ≡ b mod F —2 ,

hence

±(ζ) = a, b, c, d + I 5 F.

It is then easy to give an example where ±(ζ) = 0 (hence SK1 (A) = 0): we may

start with any ¬eld F0 of characteristic di¬erent from 2 containing a primitive

fourth root of unity and take for F the ¬eld of rational fractions in independent

indeterminates a, b, c, d over F0 .

17.B. Algebras with involution. The group SK1 discussed in the ¬rst sub-

section is based on the linear group SL1 . Analogues of the reduced Whitehead

group are de¬ned for other simple algebraic groups (see for instance Platonov-

Yanchevski˜ [?, p. 223]). We give here some basic results for algebras with involution

±

and refer to [?] and to Yanchevski˜ [?] for further results and detailed references.

±

We assume throughout that the characteristic of the base ¬eld F is di¬erent

from 2. For any central simple F -algebra A with involution σ we set Sym(A, σ)— for

the set of symmetric units. The central notion for the de¬nition of analogues of the

reduced Whitehead group is the subgroup Σσ (A) of A— generated by Sym(A, σ)— .

It turns out that this subgroup is normal and depends only on the type of σ, as the

following proposition shows.

(17.24) Proposition. (1) The subgroup Σσ (A) is normal in A— .

(2) If σ, „ are involutions of the same type (orthogonal, symplectic or unitary) on

A, then Σσ (A) = Σ„ (A). More precisely, any σ-symmetric unit can be written as

the product of two „ -symmetric units and conversely.

(3) If σ is orthogonal, then Σσ (A) = A— . More precisely, every unit in A can be

written as the product of two σ-symmetric units.

Proof : (??) This readily follows from the equation

asa’1 = asσ(a) σ(a)’1 a’1

for a ∈ A— and s ∈ Sym(A, σ).

(??) Since σ and „ have the same type, (??) or (??) shows that „ = Int(u) —¦ σ

and Sym(A, „ ) = u · Sym(A, σ) for some u ∈ Sym(A, σ)— . The last equation

shows that every element in Sym(A, „ ) is a product of two σ-symmetric elements.

Interchanging the rˆles of σ and „ shows that every element in Sym(A, σ) is a

o

product of two „ -symmetric elements.

(??) By (??), every unit x ∈ A— is invariant under some orthogonal involu-

tion „ . Let „ = Int(u) —¦ σ for some u ∈ Sym(A, σ)— ; the equation „ (x) = x yields

σ(xu) = xu, hence x = (xu)u’1 is a decomposition of x into a product of two

symmetric units.

If A is a division algebra, we may also prove (??) by the following dimension

count argument due to Dieudonn´ [?, Theorem 3]: since dim F Sym(A, σ) > 1 deg A,

e 2

we have Sym(A, σ) © x Sym(A, σ) = {0} for all x ∈ A . If s1 , s2 ∈ Sym(A, σ)—

—

are such that s1 = xs2 , then x = s1 s’1 , a product of two symmetric units.

2

The same kind of argument yields the following result for arbitrary involutions:

§17. WHITEHEAD GROUPS 267

(17.25) Proposition (Yanchevski˜ Let (A, σ) be a central division algebra with

±).

involution over F . Every nonzero element x ∈ A decomposes as x = zs with

s ∈ Sym(A, σ)— and z ∈ A— such that zσ(z) = σ(z)z.

Proof : We have A = Sym(A, σ) • Skew(A, σ) and 1 ∈ Skew(A, σ), hence dimension

count shows that x Sym(A, σ) © F ·1•Skew(A, σ) = {0}. Therefore, one can ¬nd

s0 ∈ Sym(A, σ)— , » ∈ F and z0 ∈ Skew(A, σ) such that xs0 = » + z0 . The element

z = » + z0 commutes with σ(z) = » ’ z0 , and satis¬es x = zs for s = s’1 .

0

To investigate further the group Σσ (A), we now consider separately the cases

where the involution is unitary or symplectic.

Unitary involutions. Let (B, „ ) be a central simple algebra with unitary

involution over a ¬eld F , and let K be the center of B.

(17.26) Proposition (Platonov-Yanchevski˜ Suppose B is a division algebra.

±).

’1 ’1 — —

Every commutator xyx y ∈ [B , B ] is a product of ¬ve symmetric elements.

In particular, [B — , B — ] ‚ Σ„ (B).

Proof : If x ∈ Sym(B, „ )— , the formula

xyx’1 y ’1 = x yx’1 „ (y) „ (y)’1 y ’1

shows that xyx’1 y ’1 is a product of three symmetric elements. For the rest of the

proof, we may thus assume that x ∈ Sym(B, „ ), hence

1

dimF Sym(B, „ ) + F · x = 1 + dimF B.

2

Therefore, Sym(B, „ ) + F · x © Sym(B, „ )y = {0}, and we may ¬nd s1 ∈

Sym(B, „ ), » ∈ F and s2 ∈ Sym(B, „ )— such that

s1 + »x = s2 y.

If s1 = 0, then xyx’1 y ’1 = s2 ys’1 y ’1 , and we are reduced to the case where

2

x ∈ Sym(B, „ )— . We may thus assume that s1 ∈ B — . A direct computation shows

that xyx’1 y ’1 is a product of ¬ve symmetric elements:

xyx’1 y ’1 = f1 f2 f3 f4 s2

where

f1 = xs’1 „ (x), f3 = (1 + »xs’1 )s’1 „ (1 + »xs’1 )

f2 = „ (x)’1 s1 x’1 ,

2 1 1 1

and f4 = „ (1 + »xs’1 )’1 (1 + »xs’1 )’1 .

1 1

The group

UK1 (B) = B — /Σ„ (B)

is the unitary Whitehead group of B. The preceding proposition shows that this

group is a quotient of K1 (B) = B — /[B — , B — ]. We may also consider the group

Σ„ (B) = { x ∈ B — | NrdB (x) ∈ F — },

which obviously contains Σ„ (B). The factor group

USK1 (B) = Σ„ (B)/Σ„ (B)

is the reduced unitary Whitehead group of B. The following proposition is an

analogue of a theorem of Wang:

268 IV. ALGEBRAS OF DEGREE FOUR

(17.27) Proposition (Yanchevski˜ If B is a division algebra of prime degree,

±).

USK1 (B) = 0.

Proof : We ¬rst consider the case where deg B is an odd prime p. Let x ∈ Σ„ (B)

and NrdB (x) = » ∈ F — . Then NrdB (»’1 xp ) = 1, hence »’1 xp ∈ [B — , B — ] since

SK1 (B) = 0, by a theorem of Wang [?] (see for example Pierce [?, 16.6]). It then

follows from (??) that xp ∈ Σ„ (B). On the other hand, we have NrdB „ (x)’1 x =

1, hence, by the same theorem of Wang, „ (x)’1 x ∈ [B — , B — ]. Therefore, x2 =

x„ (x) „ (x)’1 x ∈ Σ„ (B). Since p is odd we may ¬nd u, v ∈ Z such that 2u+pv = 1;

then

x = (x2 )u (xp )v ∈ Σ„ (B)

and the proposition is proved in the case where deg B is odd.

If B is a quaternion algebra, Proposition (??) shows that Σ„ (B) is the Cli¬ord

group of NK/F (B, γ), where γ is the canonical involution on B. On the other hand,

by (??) there is a canonical isomorphism NK/F (B) = EndF Sym(B, „ ) , hence

Σ„ (B) is generated by Sym(B, „ ). To make this argument more explicit, consider

the map i : Sym(B, „ ) ’ M2 (B) de¬ned by

0 γ(x)

i(x) = for x ∈ Sym(B, „ ).

x 0

Since i(x)2 = NrdB (x), this map induces an F -algebra homomorphism

i— : C Sym(B, „ ), NrdB ’ M2 (B)

which is injective since Cli¬ord algebras of even-dimensional nonsingular quadratic

spaces are simple. The image of i— is the F -subalgebra of invariant elements under

the automorphism ± de¬ned by

a11 a12 γ —¦ „ (a22 ) γ —¦ „ (a21 )

± = for aij ∈ B,

a21 a22 γ —¦ „ (a12 ) γ —¦ „ (a11 )

since ± —¦ i(x) = i(x) for all x ∈ Sym(B, „ ). Under i— , the canonical gradation of

C Sym(B, „ ), NrdB corresponds to the checker-board grading, hence i— restricts

to an isomorphism

γ —¦ „ (b) 0

∼

i— : C0 Sym(B, „ ), NrdB ’

’ b∈B B.

0 b

Under this isomorphism, the special Cli¬ord group is mapped to Σ„ (B). From the

Cartan-Dieudonn´ theorem, it follows that every element in “+ Sym(B, „ ), NrdB

e

is a product of two or four anisotropic vectors, hence for every b ∈ Σ„ (B) there

exist x1 , . . . , xr ∈ Sym(B, „ ) (with r = 2 or 4) such that

γ —¦ „ (b) 0 0 γ(x1 ) 0 γ(xr )

= ... ,

0 b x1 0 xr 0

hence

b = x1 γ(x2 ) or b = x1 γ(x2 )x3 γ(x4 ).

This shows that Σ„ (B) = Σ„ (B), since Sym(B, „ ) is stable under γ.

§17. WHITEHEAD GROUPS 269

Symplectic involutions. Let (A, σ) be central simple algebra with symplectic

involution over F . In view of (??), the reduced norm of every element in Sym(A, σ)

is a square. Let

R(A) = { a ∈ A— | NrdA (a) ∈ F —2 };

we thus have Σσ ‚ R(A) and we de¬ne, after Yanchevski˜ [?, p. 437],

±

K1 Spin(A) = R(A)/Σσ (A)[A— , A— ].

Note that R(A) is in general a proper subgroup of A— : this is clear if A is a

quaternion algebra; examples of degree 4 can be obtained as norms of quaternion

algebras by (??), since the equality R(A) = A— implies that the discriminant of

every orthogonal involution on A is trivial.

For every a ∈ A— , Proposition (??) shows that there is an involution Int(g) —¦ σ

which leaves a invariant. We then have σ(a) = g ’1 ag, hence

a2 = aσ(a) (g ’1 a’1 ga) ∈ Σσ (A)[A— , A— ].

This shows that K1 Spin(A) is a 2-torsion abelian group.

(17.28) Proposition (Yanchevski˜ K1 Spin(A) = 0 if deg A ¤ 4.

±).

Proof : Suppose ¬rst that A is a quaternion algebra. Let a ∈ R(A) and NrdA (a) =

±2 with ± ∈ F — ; then NrdA (±’1 a) = 1. Since SK1 (A) = 0 (see Exercise ??), we

have ±’1 a ∈ [A— , A— ], hence

a = ±(±’1 a) ∈ Σσ (A)[A— , A— ].

Suppose next that deg A = 4. Recall from (??) the F -algebra isomorphism

∼

i— : C Sym(A, σ), Nrpσ ’ M2 (A).

’

which maps x ∈ Sym(A, σ) to x x ∈ M2 (A). Under this isomorphism, the gra-

0

0

dation of the Cli¬ord algebra corresponds to the checker-board grading of M 2 (A),

and the canonical involution which is the identity on Sym(A, σ) corresponds to the

involution θ on M2 (A) de¬ned by

a11 a12 σ(a22 ) ’σ(a12 )

θ = .

a21 a22 ’σ(a21 ) σ(a11 )

Moreover, the special Cli¬ord group “+ Sym(A, σ), Nrpσ is mapped to the group

»σ(a)’1 0

» ∈ F — , a ∈ A— and NrdA (a) = »2

“= ‚ GL2 (A).

0 a

The map “ ’ A— which carries

»σ(a)’1 0

∈“

0 a

to a ∈ A— maps “ onto R(A). From the Cartan-Dieudonn´ theorem, it follows that

e

+

every element in “ Sym(A, σ), Nrpσ is a product of two, four, or six anisotropic

vectors in Sym(A, σ), hence for every

»σ(a)’1 0

∈“

0 a

one can ¬nd x1 , . . . , xr ∈ Sym(A, σ)— , with r = 2, 4, or 6, such that

»σ(a)’1 0 0 x1 0 xr

= ··· .

0 a x1 0 xr 0

270 IV. ALGEBRAS OF DEGREE FOUR

Therefore, every a ∈ R(A) can be written as a = x1 x2 . . . xr’1 xr for some x1 , . . . ,

xr ∈ Sym(A, σ)— .

(17.29) Corollary. Let (D, σ) be a central division F -algebra with involution of

degree 4. Any element of the commutator subgroup [D — , D— ] is a product of at

most

(1) two symmetric elements if σ is of orthogonal type,

(2) six symmetric elements if σ is of symplectic type,

(3) four symmetric elements if σ is of unitary type.

Proof : The claim follows from (??) if σ is orthogonal, from (??) if σ is symplectic

and from (??) if σ is unitary.

Exercises

1. Let Q be a quaternion algebra over an ´tale quadratic extension K of a ¬eld F of

e

arbitrary characteristic. Show that the inverse of the Lie algebra isomorphism

∼

n— : c N(Q) ’ Q of (??) maps q ∈ Q to c ι q —q0 + ι q0 —q ’TrdQ (q)ι q0 —q0 ,

™ ’

where q0 ∈ Q is an arbitrary quaternion such that TrdQ (q0 ) = 1.

2. Let Q1 , Q2 be quaternion algebras over a ¬eld F of characteristic 2, with

canonical involutions γ1 , γ2 , and let (A, σ, f ) = (Q1 —Q2 , γ1 —γ2 , f— ). Consider

Symd(A, σ)0 = { x ∈ Symd(A, σ) | Trpσ (x) = 0 }. Suppose V1 , V2 are 3-

dimensional subspaces such that Symd(A, σ)0 = V1 + V2 and that the products

v1 v2 with v1 ∈ V1 and v2 ∈ V2 span the kernel of f . Show that V1 and

V2 are the spaces Q0 and Q0 of pure quaternions in Q1 and Q2 . Conclude

1 2

that Q1 and Q2 are uniquely determined as subalgebras of A by the condition

(A, σ, f ) = (Q1 — Q2 , γ1 — γ2 , f— ).

Hint: Show that if v1 = q11 + q21 ∈ V1 and v2 = q12 + q22 ∈ V2 with

q11 , q12 ∈ Q0 and q21 , q22 ∈ Q0 , then v1 v2 = v2 v1 and f (v1 v2 ) = [q11 , q12 ] =

1 2

[q21 , q22 ].

3. (Karpenko-Qu´guiner [?]) Let (B, „ ) be a central simple algebra with unitary

e

involution of degree 4. Let K be the center of B. Show that

(B, „ ) = (Q1 , „1 ) —K (Q2 , „2 )

for some quaternion subalgebras Q1 , Q2 if and only if the discriminant algebra

D(B, „ ) is split.

Hint: If D(B, „ ) is split, use Theorem (??) to represent (B, „ ) as the

even Cli¬ord algebra of some quadratic space. For the “only if” part, use

Propositions (??) and (??).

4. Suppose char F = 2. Extensions of the form F [X, Y ]/(X 2 ’ a, Y 2 ’ b) with a,

b ∈ F — are called biquadratic. Show that for every central simple F -algebra

of degree 6 with orthogonal involution σ of trivial discriminant, there exists an

´tale biquadratic extension of F over which σ becomes hyperbolic. Deduce that

e

every central simple F -algebra of degree 4 is split by some ´tale biquadratic

e

extension of F . (This result is due to Albert [?, Theorem 11.9].)

5. Let A be a biquaternion F -algebra. Suppose A is split by an ´tale extension

e

of the form K1 — K2 , where K1 , K2 are ´tale quadratic F -algebras. Show that

e

EXERCISES 271

there exist a1 , a2 ∈ F — such that

A (K1 , a1 )F — (K2 , a2 )F .

Hint: Re¬ne the argument used in the proof of Proposition (??).

6. Let σ, „ be distinct symplectic involutions on a biquaternion F -algebra A. Show

that dimF Symd(A, σ)©Symd(A, „ ) = 2. Show also that there is a quaternion

algebra B ‚ A over some quadratic extension of F such that σ|B = „ |B is the

conjugation involution, and that the algebra B is uniquely determined by this

condition.

7. Let σ, „ , θ be symplectic involutions on a biquaternion F -algebra A. Show that

the invariants of these involutions are related by jσ („ ) = j„ (σ) and jσ („ ) +

j„ (θ) + jθ (σ) ∈ I 3 Wq F . Use this result to show that if σ and „ are conjugate,

then jσ (θ) = j„ (θ).

8. Let σ be a symplectic involution on a biquaternion F -algebra A. Let

Symd(A, σ)0 = { x ∈ Symd(A, σ) | Trpσ (x) = 0 }

and let sσ : Symd(A, σ)0 ’ F be the squaring map. Show that ind(A) ¤ 2 if

and only if sσ is a subform of some (uniquely determined) 3-fold P¬ster form

πσ . Suppose these conditions hold; then

(a) show that (A, σ) has a decomposition

(A, σ) = M2 (F ), σ1 —F (Q, γ)

for some quaternion algebra Q with canonical involution γ and some or-

thogonal involution σ1 on M2 (F ), and that πσ = disc σ1 · NrdQ ;

(b) for θ a hyperbolic involution on A, show that πσ = jθ (σ);

(c) show that G(A, σ) = G(πσ ) = Sn(sσ ).

Hint: The equality G(A, σ) = Sn(sσ ) follows from (??) and G(A, σ) =

G(πσ ) follows from (??) and (??).

9. Suppose char F = 2. Let K/F be an ´tale quadratic extension with non-

e

—

trivial automorphism ι and let δ ∈ K be such that ι(δ) = ’δ. Let (V, q)

be an odd-dimensional quadratic space over K with trivial discriminant and

let ζ ∈ C(V, q) be an orientation of (V, q). De¬ne an F -linear map i : V ’

M2 NK/F C0 (V, q) by

’δ ι (x · ζ) — 1 ’ 1 — (x · ζ)

0

i(x) = ι

(x · ζ) — 1 + 1 — (x · ζ) 0

for x ∈ V . Show that the map i induces an F -algebra isomorphism:

∼

i— : C V, (trK/F )— ( δ · q) ’ M2 NK/F C0 (V, q) .

’

Use this result to give a direct proof of the fact that if Q is a quaternion K-

algebra and s : Q0 ’ K is the squaring map on the space of pure quaternions,

then Q0 , (trK/F )— ( δ · s) is an Albert quadratic space of NK/F (Q).

10. Suppose char F = 2. Let Q1 , Q2 be quaternion F -algebras with canonical in-

volutions γ1 , γ2 and let (A, θ) = (Q1 , γ1 ) —F (Q2 , γ2 ). De¬ne a linear endomor-

phism p on Skew(A, θ) = (Q0 —1)•(1—Q0) by p(x1 —1+1—x2) = x1 —1’1—x2

1 2

and a quadratic form q : Skew(A, θ) ’ F by q(x) = xp(x). Consider another

pair of quaternion F -algebras Q1 , Q2 and (A , θ ) = (Q1 , γ1 ) —F (Q2 , γ2 ), and

de¬ne p , q on Skew(A , θ ) as p, q were de¬ned on Skew(A, θ). Show that

∼

for every isomorphism f : (A, θ) ’ (A , θ ) there exists some » ∈ F — such

’

272 IV. ALGEBRAS OF DEGREE FOUR

∼

that f ’1 —¦ p —¦ f = »p, hence f restricts to a similitude (Skew(A, θ), q) ’ ’

Skew(A , θ ), q with multiplier ».

Hint: Use (??).

This exercise is inspired by Knus-Parimala-Sridharan [?, Theorem 3.4] and

Wadsworth [?]. It shows that the forms q and q are similar without using the

fact that they are Albert forms of A and A .

11. Let σ be a symplectic involution on a biquaternion algebra A. Show that

the invariant jσ of symplectic involutions and the homomorphism ±σ : “ ’

I 3 Wq F/I 4 Wq F are related by

±σ Nrpσ (v), v = jσ Int(v) —¦ σ ’ Nrpσ (v) · Nrpσ +I 4 Wq F

for all v ∈ Symd(A, σ)— .

12. Suppose char F = 2. Let A be a biquaternion division algebra and let x ∈

SL1 (A). √

(a) Let σ be an arbitrary symplectic involution and let L = F ( a) be a

quadratic extension of F in Sym(A, σ) which contains σ(x)x. Recall from

(??) that one can ¬nd b, c, d ∈ F — such that A (a, b)F — (c, d)F . Show

that L contains an element y such that σ(x)x = yy ’1 where is the

nontrivial automorphism of L/F , and that

±(x) = NL/F (y), b, c, d + I 5 F.

(b) Suppose x is contained in a maximal sub¬eld E ‚ A which contains an

√

intermediate quadratic extension L = F ( a). Recall from (??) that A

(a, b)F — (c, d)F for some b, c, d ∈ F — . Show that L contains an element y

such that NE/L (x) = yy ’1 where is the nontrivial automorphism of L/F ,

and that

±(x) = NL/F (y), b, c, d + I 5 F.

13. Let Q be a quaternion F -algebra. Show that every element in SL1 (Q) is a

commutator, except if F = F2 (the ¬eld with two elements).

Hint: Argue as in (??). If q ∈ SL1 (Q) and 1 + q is not invertible and

∼

nonzero, then show that there is an isomorphism Q ’ M2 (F ) which identi¬es

’

q with ’1 ’1 . Then q = xyx’1 y ’1 with x = 1 ’1 and y = ’1/2 1 if

1 0

0

0 ’1 0

»+1 0 1 » if char F = 2 and » = 0, 1.

char F = 2, with x = 0 » and y = 0 1

14. Let Q be a quaternion division F -algebra and let (V, h) be a nonsingular her-

mitian space of dimension 2 over Q with respect to the canonical involution,

so that EndQ (V ), σh is a central simple algebra of degree 4 with symplectic

involution. Let g ∈ Sp(V, h) be such that IdV + g is not invertible, and let

v1 ∈ V be a nonzero vector such that g(v1 ) = ’v1 .

(a) If v1 is anisotropic, show that for each vector v2 ∈ V which is orthogonal

to v1 , there is some ± ∈ Q such that g(v2 ) = v2 ± and NrdQ (±) = 1.

(b) If v1 is isotropic, show that for each isotropic vector v2 ∈ V such that

h(v1 , v2 ) = 1, there is some β ∈ Q such that g(v2 ) = v1 β ’ v2 and

TrdQ (β) = 0.

15. Let b be a nonsingular alternating bilinear form on a 4-dimensional F -vector

space V and let g ∈ Sp(V, b) be such that IdV + g is not invertible.

(a) Show that ker(IdV + g) = im(IdV + g)⊥ .

(b) If the rank of IdV + g is 1 or 2, show that V = V1 • V2 for some 2-

dimensional subspaces V1 , V2 which are preserved under g, hence g can be

EXERCISES 273

represented by the matrix

« «

’1 1 0 0 ’1 1 0 0

¬ 0 ’1 0 0· ¬ 0 ’1 0 0·

g1 = ¬ · or g2 = ¬ ·.

0 0 ’1 0 0 0 ’1 1

0 0 0 ’1 0 0 0 ’1

Use Exercise ?? to conclude that g is a commutator if F = F2 . If char F =

2, show that g1 = a1 b1 a’1 b’1 and g2 = a2 b2 a’1 b’1 , where

1 1 2 2

« «

1010 1000

¬0 1 0 0 · ¬ ·

· , b1 = ¬0 1 0 0· ,

a1 = ¬0 0 1 0 0 1 1 0

0001 0001

« «

0010 1000

¬0 0 0 1· ¬ ·

· , b2 = ¬0 1 0 0· .

a2 = ¬

1 0 0 0 0 0 1 1

0100 0001

(c) If IdV + g has rank 3, show that g can be represented by the matrix

«

’1 1 0 0

¬ 0 ’1 1 0·

¬ ·

0 1

0 ’1

0 0 0 ’1

(with respect to a suitable basis).

If char F = 2, let

«

’3 ’8 22 ’12

¬’3 ’9 23 ’12·

y=¬ ·.

’3 ’9 22 ’11

’4 ’5 16 ’8

Show that (x’1)3 (x+1) is the minimum and the characteristic polynomial

of both matrices gy and y. Conclude that there is an invertible matrix x

such that gy = xyx’1 , hence g is a commutator. If char F = 2, let

«

0100

¬1 0 0 1·

z=¬ ·

1 0 1 1 .

1110

Show that x4 + x3 + x2 + x + 1 is the minimum and the characteristic

polynomial of both matrices gz and z. As in the preceding case, conclude

that g is a commutator.

16. Let A be a biquaternion F -algebra with symplectic involution σ. Write simply

V = Symd(A, σ), V 0 = Symd(A, σ)0 and q = Nrpσ , and let q 0 be the restriction

of q to V 0 . Using (??), show that there is a canonical isomorphism SL1 (A)

Spin(V, q) which maps the subgroup [A— , A— ] to the subgroup

Spin (V, q) = Spin(V 0 , q 0 ) · [„¦, “+ (V 0 , q 0 )]

where „¦ = „¦ EndF (V ), σq , fq is the extended Cli¬ord group. Deduce that

the subgroup Spin (V, q) ‚ Spin(V, q) is generated by the subgroups Spin(U )

for all the proper nonsingular subspaces U ‚ V .

274 IV. ALGEBRAS OF DEGREE FOUR

Hint: Use the proof of (??).

17. Suppose char F = 2 and F contains a primitive 4th root of unity ζ. Let A =

(a, b)F — (c, d)F be a biquaternion division F -algebra.

(a) Show that if A is cyclic (i.e., if A contains a maximal sub¬eld which is

cyclic over F ), then a, b, c, d is hyperbolic.

(b) (Morandi-Sethuraman [?, Proposition 7.3]) Suppose d is an indeterminate

over some sub¬eld F0 containing a, b, c, and F = F0 (d). Show that A is

cyclic if and only if a, b, c is hyperbolic.

Hint: If a, b, c is hyperbolic, the following equation has a nontrivial

solution in F0 :

a(x2 + cx2 ) + b(y1 + cy2 ) ’ ab(z1 + cz2 ) = 0.

2 2 2 2

1 2

√ √ √√

Let e = 2 c(ax1 x2 + by1 y2 ’ abz1 z2 ) ∈ F0 ( c). Show that F0 ( c)( e) is

cyclic over F0 and splits (a, b)F (√c) , hence also A.

Notes

§??. If char F = 2, an alternative way to de¬ne the canonical isomorphism

N —¦ C(A, σ) (A, σ) for (A, σ) ∈ D2 (in (??)) is to re¬ne the fundamental re-

lation (??) by taking the involutions into account. As pointed out in the notes

for Chapter ??, one can de¬ne a nonsingular hermitian form H on the left A-

submodule B (A, σ) ‚ B(A, σ) of invariant elements under the canonical involution

ω and show that the canonical isomorphism ν of (??) restricts to an isomorphism

of algebras with involution:

∼

NZ/F C(A, σ), σ ’ EndA B (A, σ), σH

’

where Z is the center of the Cli¬ord algebra C(A, σ). Since deg A = 4, dimension

count shows that the canonical map b : A ’ B(A, σ) induces an isomorphism of A-

∼

modules A ’ B (A, σ). Moreover, under this isomorphism, H (a1 , a2 ) = 2a1 σ(a2 )

’

for a1 , a2 ∈ A. Therefore, b induces a canonical isomorphism

∼

(A, σ) ’ EndA B (A, σ), σH .

’

Similarly, in Theorem (??) the canonical isomorphism D —¦ C(A, σ) (A, σ) for

(A, σ) ∈ D3 can be derived from properties of the bimodule B(A, σ). De¬ne a left

C(A, σ) —Z C(A, σ)-module structure on B(A, σ) by

(c1 — c2 ) u = c1 — u · σ(c2 ) for c1 , c2 ∈ C(A, σ), u ∈ B(A, σ).

If g ∈ C(A, σ) —Z C(A, σ) is the Goldman element of the Cli¬ord algebra, the map

C(A, σ) —Z C(A, σ) ’ B(A, σ) which carries ξ to ξ 1b induces an isomorphism of

left C(A, σ) —Z C(A, σ)-modules

∼

[C(A, σ) —Z C(A, σ)](1 ’ g) ’ B(A, σ).

’

We thus get a canonical isomorphism

»2 C(A, σ) EndC(A,σ)—Z C(A,σ) B(A, σ).

Using (??), we may identify the right-hand side with A —F Z and use this identi¬-

cation to get a canonical isomorphism D C(A, σ), σ A.

In the proofs of Theorems (??) and (??), it is not really necessary to consider

the split cases separately if char F = 0; one can instead use results on the extension

NOTES 275

of Lie algebra isomorphisms in Jacobson [?, Ch. 10, §4] to see directly that the

™1 1 ™™ 1 ™

1

canonical Lie algebra isomorphisms n—¦ 2 c, 2 c—¦n, »2 —¦ 2 c and 2 c—¦»2 extend (uniquely)

to isomorphisms of the corresponding algebras with involution (see Remarks (??)

and (??)).

The fact that a central simple algebra of degree 4 with orthogonal involution

decomposes into a tensor product of stable quaternion subalgebras if and only if

the discriminant of the involution is trivial (Corollary (??) and Proposition (??))