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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

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