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