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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=

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

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