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In view of (??), this shows: σ(g)(w) = c’1 · w · c, so
χ(c) = σ(g)’1 = g.


By extending scalars to a splitting ¬eld of A, it follows from the proposition
above and (??) that χ(c) ∈ O+ (A, σ, f ) for all c ∈ “(A, σ, f ). The commutative
diagram of (??) has an analogue for algebras with quadratic pairs:
(13.15) Proposition. For brevity, set C(A) = C(A, σ, f ). Let Z be the center
of C(A) and let σ denote the canonical involution on C(A). For all c ∈ “(A, σ, f ) we
have σ(c)c ∈ F — , hence “(A, σ, f ) ‚ Sim C(A), σ . The map χ and the restriction
to O+ (A, σ, f ) of the canonical map
C : GO(A, σ, f ) ’ AutF C(A), σ
¬t into a commutative diagram with exact rows:
χ
O+ (A, σ, f )
1 ’ ’ ’ F— ’ ’ ’
’’ ’’ “(A, σ, f ) ’’’
’’ ’’’ 1
’’
¦ ¦ ¦
¦ ¦ ¦
C

Int
1 ’ ’ ’ Z — ’ ’ ’ Sim C(A), σ
’’ ’’ ’ ’ ’ AutZ C(A), σ
’’ ’’’ 1
’’
Proof : All the statements follow by scalar extension to a splitting ¬eld of A and
comparison with (??), except for the surjectivity of χ onto O+ (A, σ, f ).
To prove this last point, recall the isomorphism ν of (??) induced by the C(A)-
bimodule structure on B(A, σ, f ):

ν : ι C(A)op —Z C(A) ’ EndA—Z B(A, σ, f ) .

This isomorphism satis¬es
ι op
xν( c1 —c2 )
= c1 — x · c2 for x ∈ B(A, σ, f ), c1 , c2 ∈ C(A).
For g ∈ O+ (A, σ, f ), it follows from (??) that C(g) is the identity on Z, hence β(g)
is an A — Z-endomorphism of B(A, σ, f ), by (??). Therefore, there exists a unique
element ξ ∈ ι C(A)op —Z C(A) such that ν(ξ) = β(g).
In the split case, (??) shows that ξ = ι (c’1 )op — c, where c ∈ “+ (V, q) is such
that χ(c) = g. Since the minimal number of terms in a decomposition of an element
of a tensor product is invariant under scalar extension, it follows that ξ = ιcop — c2
1
for some c1 , c2 ∈ C(A). Moreover, if s is the switch map on ι C(A)op —Z C(A),
de¬ned by
op
s(ιcop — c ) = ιc — c for c, c ∈ C(A),
then s(ξ)ξ = 1, since ξ = ι (c’1 )op — c over a splitting ¬eld. Therefore, the elements
c1 , c2 ∈ C(A)— satisfy
c1 c2 = » ∈ Z with NZ/F (») = 1.
180 III. SIMILITUDES


By Hilbert™s Theorem 90 (??), there exists »1 ∈ Z such that » = »1 ι(»1 )’1 .
(Explicitly, we may take »1 = z + »ι(z), where z ∈ Z is such that z + »ι(z) is
invertible.) Then ξ = ι (c’1 )op — c3 for c3 = »’1 c2 , hence
3 1

xβ(g) = c’1 — x · c3 for x ∈ B(A, σ, f ).
3
In particular, for x = 1b we obtain by (??):
c’1 — 1b · c3 = g b .
3
This shows that c3 ∈ “(A, σ, f ) and χ(c3 ) = g.
(13.16) Corollary. Suppose deg A ≥ 4; then
“(A, σ, f ) © Z = { z ∈ Z — | z 2 ∈ F — };
thus “(A, σ, f ) © Z = F — if char F = 2, and “(A, σ, f ) © Z = F — ∪ z · F — if
char F = 2 and Z = F [z] with z 2 ∈ F — .
Proof : Since χ is surjective, it maps “(A, σ, f ) © Z to the center of O+ (A, σ, f ). It
follows by scalar extension to a splitting ¬eld of A that the center of O+ (A, σ, f )
is trivial if char F = 2 and is {1, ’1} if char F = 2 (see Dieudonn´ [?, pp. 25, 45]).
e

Therefore, the factor group “(A, σ, f ) © Z /F is trivial if char F = 2 and has two
elements if char F = 2, proving the corollary.
(13.17) Example. Suppose Q is a quaternion F -algebra with canonical involution
γ. If char F = 2, let σ = Int(u) —¦ γ for some invertible pure quaternion u. Let F (u)1
be the group of elements of norm 1 in F (u):
F (u)1 = { z ∈ F (u) | zγ(z) = 1 }.
As observed in (??), we have O+ (Q, σ) = F (u)1 . On the other hand, it follows
from the structure theorem for Cli¬ord algebras (??) that C(Q, σ) is a commutative
algebra isomorphic to F (u). The Cli¬ord group “(Q, σ) is the group of invertible
elements in C(Q, σ). It can be identi¬ed with F (u)— in such a way that the vector
representation χ maps x ∈ F (u)— to xγ(x)’1 ∈ F (u)1 . The upper exact sequence
of the commutative diagram in (??) thus takes the form
1’γ
1 ’ F — ’’ F (u)— ’’ F (u)1 ’ 1.

Similar results hold if char F = 2. Using the same notation as in (??) and (??),
we have O+ (Q, γ, f ) = F ( )1 , “(Q, γ, f ) F ( )— , and the upper exact sequence of
the commutative diagram in (??) becomes
1’γ
1 ’ F — ’’ F ( )— ’’ F ( )1 ’ 1.

The extended Cli¬ord group. In this subsection, we de¬ne an intermediate
group „¦(A, σ, f ) between “(A, σ, f ) and Sim C(A, σ, f ), σ , which covers the group
PGO+ (A, σ, f ) in the same way as “(A, σ, f ) covers O+ (A, σ, f ) by the vector rep-
resentation χ. This construction will enable us to de¬ne an analogue of the spinor
norm for the group PGO+ (A, σ).
The notation is as above: (σ, f ) is a quadratic pair on a central simple algebra A
of even degree over an arbitrary ¬eld F . Let Z be the center of the Cli¬ord algebra
C(A, σ, f ). Since (??) plays a central rˆle almost from the start (we need injectivity
o
of the canonical map C for the de¬nition of χ in (??) below), we exclude the case
of quaternion algebras from our discussion. We thus assume
deg A = n = 2m ≥ 4.
§13. QUADRATIC PAIRS 181


We identify PGO(A, σ, f ) with AutF (A, σ, f ) by mapping g · F — to Int(g), for
g ∈ GO(A, σ, f ). By (??) and (??), the canonical map C induces an injective
homomorphism C : PGO+ (A, σ, f ) ’ AutZ C(A, σ, f ), σ . Consider the following
diagram:
PGO+ (A, σ, f )
¦
¦
C

Int
Sim C(A, σ, f ), σ ’ ’ ’ AutZ C(A, σ, f ), σ .
’’
(13.18) De¬nition. The extended Cli¬ord group of (A, σ, f ) is the inverse image
under Int of the image of the canonical map C:
„¦(A, σ, f ) = { c ∈ Sim C(A, σ, f ), σ | Int(c) ∈ C PGO+ (A, σ, f ) }.

Thus, „¦(A, σ, f ) ‚ Sim C(A, σ, f ), σ , and there is an exact sequence
χ
1 ’ Z — ’ „¦(A, σ, f ) ’ PGO+ (A, σ, f ) ’ 1
(13.19) ’
where the map χ is de¬ned by
if Int(c) = C(g), with g ∈ GO+ (A, σ, f ).
χ (c) = g · F —
If char F = 2, the group „¦(A, σ, f ) = „¦(A, σ) may alternately be de¬ned by
„¦(A, σ) = { c ∈ C(A, σ)— | c · c(A) · c’1 = c(A) },
since the Z-automorphisms of C(A, σ) which preserve c(A) are exactly those which
are of the form C(g) for some g ∈ GO+ (A, σ), by (??). We shall not use this
alternate de¬nition, since we want to keep the characteristic arbitrary.
For c ∈ “(A, σ, f ), we have Int(c) = C(g) for some g ∈ O+ (A, σ, f ), by (??),
hence “(A, σ, f ) ‚ „¦(A, σ, f ). Our ¬rst objective in this subsection is to describe
“(A, σ, f ) as the kernel of a map κ : „¦(A, σ, f ) ’ Z — /F — .
The multiplier map µ : GO(A, σ, f ) ’ F — induces a map
µ : PGO+ (A, σ, f ) ’ F — /F —2 ,
since µ(±) = ±2 for all ± ∈ F — . This map ¬ts into an exact sequence:
µ
π
O+ (A, σ, f ) ’ PGO+ (A, σ, f ) ’ F — /F —2 .
’ ’
(13.20) Lemma. The kernel of the map µ —¦ χ : „¦(A, σ, f ) ’ F — /F —2 is the sub-
group Z — · “(A, σ, f ). In particular, if F is algebraically closed, then, since µ is
trivial, „¦(A, σ, f ) = Z — · “(A, σ, f ).
Proof : For c ∈ “(A, σ, f ), we have χ (c) = g · F — for some g ∈ O+ (A, σ, f ), hence
µ—¦χ (c) = 1. Since ker χ = Z — , the inclusion Z — ·“(A, σ, f ) ‚ ker(µ—¦χ ) follows. In
order to prove the reverse inclusion, pick c ∈ ker(µ—¦χ ); then χ (c) = g·F — for some
g ∈ GO+ (A, σ, f ) such that µ(g) ∈ F —2 . Let µ(g) = ±2 for some ± ∈ F — . Then
±’1 g ∈ O+ (A, σ, f ), so there is an element γ ∈ “(A, σ, f ) such that χ(γ) = ±’1 g.
We then have
Int(γ) = C(±’1 g) = C(g) = Int(c),
hence c = z · γ for some z ∈ Z — .
182 III. SIMILITUDES


We now de¬ne a map
κ : „¦(A, σ, f ) ’ Z — /F —
as follows: for ω ∈ „¦(A, σ, f ), we pick g ∈ GO+ (A, σ, f ) such that χ (ω) = g · F — ;
then µ(g)’1 g 2 ∈ O+ (A, σ, f ), hence there exists γ ∈ “(A, σ, f ) such that χ(γ) =
µ(g)’1 g 2 . By (??) it follows that Int(γ) = C(g 2 ) = Int(ω 2 ), hence
ω2 = z · γ for some z ∈ Z — .
We then set
κ(ω) = z · F — ∈ Z — /F — .
To check that κ is well-de¬ned, suppose g ∈ GO+ (A, σ, f ) also satis¬es χ (ω) =
2
C(g ). We then have g ≡ g mod F — , hence µ(g)’1 g 2 = µ(g )’1 g . On the other
hand, the element γ ∈ “(A, σ, f ) such that χ(γ) = µ(g)’1 g 2 is uniquely determined
up to a factor in F — , by (??), hence the element z ∈ Z — is uniquely determined
modulo F — .
Note that, if char F = 2, the element z is not uniquely determined by the
condition that ω 2 = z · γ for some γ ∈ “(A, σ, f ), since “(A, σ, f ) © Z — = F — (see
(??)).
(13.21) Proposition. The map κ : „¦(A, σ, f ) ’ Z — /F — is a group homomor-
phism and ker κ = “(A, σ, f ).
Proof : It su¬ces to prove the proposition over an algebraic closure. We may thus
assume F is algebraically closed; (??) then shows that
„¦(A, σ, f ) = Z — · “(A, σ, f ).
For z ∈ Z — and γ ∈ “(A, σ, f ) we have κ(z · γ) = z 2 · F — , hence κ is a group
homomorphism. Moreover, ker κ consists of the elements z · γ such that z 2 ∈
F — . In view of (??), this condition implies that z ∈ “(A, σ, f ), hence ker κ =
“(A, σ, f ).
Our next objective is to relate κ(ω) to µ —¦ χ (ω), for ω ∈ „¦(A, σ, f ). We need
the following classical result of Dieudonn´, which will be generalized in the next
e
section:
(13.22) Lemma (Dieudonn´). Let (V, q) be a nonsingular even-dimensional quad-
e
ratic space over an arbitrary ¬eld F . Let Z be the center of the even Cli¬ord al-
gebra C0 (V, q). For every similitude g ∈ GO(V, q), the multiplier µ(g) is a norm
from Z/F .
Proof : The similitude g may be viewed as an isometry µ(g) · q q. Therefore,
the quadratic form q ⊥ ’ µ(g) · q is hyperbolic. The Cli¬ord algebra of any
form 1, ’± · q is Brauer-equivalent to the quaternion algebra21 (Z, ±)F (see for
instance [?, (3.22), p. 47]), hence Z, µ(g) F splits, proving that µ(g) is a norm
from Z/F .
Continuing with the same notation, and assuming dim V ≥ 4, consider ω ∈
„¦ EndF (V ), σq , fq ‚ C0 (V, q) and g ∈ GO+ (V, q) such that χ (ω) = g · F — . The
preceding lemma yields an element z ∈ Z — such that µ(g) = NZ/F (z) = zι(z),
where ι is the nontrivial automorphism of Z/F .
21 We use the same notation as in § ??.
§13. QUADRATIC PAIRS 183


(13.23) Lemma. There exists an element z0 ∈ Z — such that κ(ω) = (zz0 )’1 · F —
2

and, in C(V, q),
’1
ω · v · ω ’1 = ι(z0 )z0 z ’1 g(v) for v ∈ V .
Proof : For all v ∈ V , we have in C(V, q)
2
z ’1 g(v) = z ’1 ι(z)’1 g(v)2 = µ(g)’1 q g(v) = q(v),
hence the map v ’ z ’1 g(v) extends to an automorphism of C(V, q), by the universal
property of Cli¬ord algebras. By the Skolem-Noether theorem, we may represent
this automorphism as Int(c) for some c ∈ C(V, q)— . For v, ∈ V , we then have
Int(c)(v · w) = z ’1 g(v) · z ’1 g(w) = µ(g)’1 g(v) · g(w),
hence (??) shows that the restriction of Int(c) to C0 (V, q) is C(g). Since g is a proper
similitude, it follows from (??) that Int(c) is the identity on Z, hence c ∈ C0 (V, q)— .
Moreover, Int(c)|C0 (V,q) = Int(ω)|C0 (V,q) since χ (ω) = g · F — , hence c = z0 ω for
some z0 ∈ Z — . It follows that for all v ∈ V ,
’1 ’1
ω · v · ω ’1 = z0 c · v · c’1 z0 = ι(z0 )z0 z ’1 g(v).
Observe next that
Int(c2 )(v) = z ’2 g 2 (v) for v ∈ V ,
since c commutes with z. If γ ∈ “+ (V, q) satis¬es χ(γ) = µ(g)’1 g 2 , then
γ · v · γ ’1 = µ(g)’1 g 2 (v) for v ∈ V ,
hence γ ’1 zc2 centralizes V . Since V generates C(V, q), it follows that
γ ≡ zc2 ≡ zz0 ω
2
mod F — .
By de¬nition of κ, these congruences yield κ(ω) = (zz0 )’1 · F — .
2


The main result of this subsection is the following:
(13.24) Proposition. The following diagram is commutative with exact rows and
columns:
χ
O+ (A, σ, f )
1 ’ ’ ’ F — ’ ’ ’ “(A, σ, f ) ’ ’ ’
’’ ’’ ’’ ’’’ 1
’’
¦ ¦ ¦
¦ ¦ ¦π

χ
1 ’ ’ ’ Z — ’ ’ ’ „¦(A, σ, f ) ’ ’ ’ PGO+ (A, σ, f ) ’ ’ ’ 1
’’ ’’ ’’ ’’
¦ ¦
¦κ ¦µ

NZ/F
Z — /F — F — /F —2 .
’’’
’’
Proof : In view of (??) and (??), it su¬ces to prove commutativity of the lower
square. By extending scalars to a splitting ¬eld of A in which F is algebraically
closed, we may assume that A is split. Let (V, q) be a nonsingular quadratic space
such that (A, σ, f ) = EndF (V ), σq , fq .
Fix some ω ∈ „¦(A, σ, f ) and g ∈ GO+ (V, q) such that χ (ω) = g · F — . Let
z ∈ Z — satisfy µ(g) = NZ/F (z). The preceding lemma yields z0 ∈ Z — such that
κ(ω) = (zz0 )’1 · F — . Then
2

NZ/F κ(ω) = NZ/F (zz0 )’1 · F —2 = µ(g) · F —2 .
2
184 III. SIMILITUDES


To conclude our discussion of the extended Cli¬ord group, we examine more
closely the case where deg A is divisible by 4. In this case, the map κ factors through
the multiplier map, and the homomorphism χ factors through a homomorphism
χ0 : „¦(A, σ, f ) ’ GO+ (A, σ, f ).
We denote by µ : „¦(A, σ, f ) ’ Z — the multiplier map, de¬ned by
µ(ω) = σ(ω)ω for ω ∈ „¦(A, σ, f ).

The element µ(ω) thus de¬ned belongs to Z — since „¦(A, σ, f ) ‚ Sim C(A, σ, f ), σ .
(13.25) Proposition. Suppose deg A ≡ 0 mod 4. For all ω ∈ „¦(A, σ, f ),
κ(ω) = µ(ω) · F — .
The Cli¬ord group can be characterized as
“(A, σ, f ) = { ω ∈ „¦(A, σ, f ) | µ(ω) ∈ F — }.
Proof : The second part follows from the ¬rst, since (??) shows that “(A, σ, f ) =
ker κ.
To prove the ¬rst part, we may extend scalars to an algebraic closure. For
ω ∈ „¦(A, σ, f ) we may thus assume, in view of (??), that there exist z ∈ Z — and
γ ∈ “(A, σ, f ) such that ω = z · γ. We then have κ(ω) = z 2 · F — . On the other
hand, since deg A ≡ 0 mod 4 the involution σ is of the ¬rst kind, by (??), hence
µ(ω) = z 2 µ(γ). Now, (??) shows that µ(γ) ∈ F — , hence

µ(ω) · F — = z 2 · F — = κ(ω).



Another interesting feature of the case where deg A ≡ 0 mod 4 is that the
extended Cli¬ord group has an alternate description similar to the de¬nition of the
Cli¬ord group. In the following proposition, we consider the image Ab of A in the
Cli¬ord bimodule B(A, σ, f ) under the canonical map b : A ’ B(A, σ, f ).
(13.26) Proposition. If deg A ≡ 0 mod 4, then
„¦(A, σ, f ) = { ω ∈ Sim C(A, σ, f ), σ | σ(ω) — Ab · ω = Ab }.

Proof : Let ω ∈ Sim C(A, σ, f ), σ . We have to show that Int(ω) = C(g) for some
g ∈ GO+ (A, σ, f ) if and only if σ(ω) — Ab · ω = Ab .
Assume ¬rst that ω ∈ „¦(A, σ, f ), i.e., Int(ω) = C(g) for some g ∈ GO+ (A, σ, f ).
To prove the latter equality, we may reduce by scalar extension to the split case.
Thus, suppose that (V, q) is a nonsingular quadratic space of dimension divisible
by 4 and (A, σ, f ) = EndF (V ), σq , fq . Under the standard identi¬cations asso-
ciated to q we have Ab = V — V ‚ V — C1 (V, q) (see (??)), hence it su¬ces to
show
σ(ω) · v · ω ∈ V for v ∈ V .
Let ι be, as usual, the nontrivial automorphism of the center Z of C(A, σ, f ) =
C0 (V, q) over F , and let z ∈ Z — be such that µ(g) = zι(z). By (??), there is an
element z0 ∈ Z — such that κ(ω) = (zz0 )’1 · F — and
2

’1
ω · v · ω ’1 = ι(z0 )z0 z ’1 g(v) for v ∈ V .
§13. QUADRATIC PAIRS 185


The canonical involution „ of C(V, q) restricts to σ = „0 on C0 (V, q); by (??)
this involution is the identity on Z. Therefore, by applying „ to each side of the
preceding equation, we obtain
’1
σ(ω)’1 · v · σ(ω) = g(v)ι(z0 )z0 z ’1 = z0 ι(z0 )’1 ι(z)’1 g(v) for v ∈ V ,
hence
’1
σ(ω) · v · ω = z0 ι(z0 )ι(z)g ’1 (v)µ(ω) for v ∈ V .
By (??), we have µ(ω) · F — = κ(ω), hence µ(ω) = ±(zz0 )’1 for some ± ∈ F — . By
2

substituting this in the preceding relation, we get for all v ∈ V
’1
σ(ω) · v · ω = ±z0 ι(z0 )ι(z)ι(zz0 )’1 g ’1 (v) = ±NZ/F (z0 )’1 g ’1 (v).
2


Since the right-hand term lies in V , we have thus shown σ(ω) — Ab · ω = Ab .
Suppose conversely that ω ∈ Sim C(A, σ, f ), σ satis¬es σ(ω) — Ab · ω = Ab .
Since b is injective, there is a unique element g ∈ A such that
σ(ω) — 1b · ω = g b .
(13.27)

We claim that g ∈ GO+ (A, σ, f ) and that Int(ω) = C(g). To prove the claim,
we may extend scalars to a splitting ¬eld of A; we may thus assume again that
(A, σ, f ) = EndF (V ), σq , fq for some nonsingular quadratic space (V, q) of dimen-
sion divisible by 4, and use the standard identi¬cations A = V — V , Ab = V — V ‚
B(A, σ, f ) = V — C1 (V, q). Since B(A, σ, f ) is a left A-module, we may multiply
each side of (??) by v — w ∈ V — V = A; we thus obtain
b
σ(ω) — (v — w)b · ω = (v — w) —¦ g for v, w ∈ V .
Since (v — w) —¦ g = v — σ(g)(w), it follows that, in C(V, q),
σ(ω) · w · ω = σ(g)(w) for w ∈ V .
Since ω is a similitude of C(A, σ, f ), by squaring this equation we obtain
NZ/F µ(ω) q(w) = q σ(g)(w) for w ∈ V .
This shows that σ(g) is a similitude, hence g ∈ GO(V, q), and


(13.28) µ σ(g) = µ(g) = NZ/F µ(ω) .
Moreover, for v, ∈ V ,
ω ’1 · (v · w) · ω =µ(ω)’1 σ(ω) · v · ω µ(ω)’1 σ(ω) · w · ω
’1
=NZ/F µ(ω) σ(g)(v) · σ(g)(w).
By (??) it follows that
ω ’1 · (v · w) · ω = µ(g)’1 σ(g)(v) · σ(g)(w) for v, w ∈ V ,
hence, by (??),
ω · (v · w) · ω ’1 = µ(g)’1 g(v) · g(w) = C(g)(v · w) for v, w ∈ V .
This equality shows that Int(ω)|C0 (V,q) = C(g), hence g is proper, by (??), and the
proof is complete.
186 III. SIMILITUDES


(13.29) De¬nition. Suppose deg A ≡ 0 mod 4. By using the description of the
extended Cli¬ord group „¦(A, σ, f ) in the proposition above, we may de¬ne a ho-
momorphism
χ0 : „¦(A, σ, f ) ’ GO+ (A, σ, f )
mapping ω ∈ „¦(A, σ, f ) to the element g ∈ GO+ (A, σ, f ) satisfying (??). Thus, for
ω ∈ „¦(A, σ, f ) the similitude χ0 (ω) is de¬ned by the relation
σ(ω) — 1b · ω = χ0 (ω)b .
The proof above shows that
χ (ω) = χ0 (ω) · F — for ω ∈ „¦(A, σ, f );
moreover, by (??), the following diagram commutes:
χ0
„¦(A, σ, f ) ’ ’ ’ GO+ (A, σ, f )
’’
¦ ¦
¦ ¦µ
µ

NZ/F
Z— F —.
’’’
’’
Spinor norms. Let (σ, f ) be a quadratic pair on a central simple algebra A
of even degree over an arbitrary ¬eld F .
(13.30) De¬nition. In view of (??), we may de¬ne a homomorphism
Sn : O+ (A, σ, f ) ’ F — /F —2
as follows: for g ∈ O+ (A, σ, f ), pick γ ∈ “(A, σ, f ) such that χ(γ) = g and let
Sn(g) = σ(γ)γ · F —2 = µ(γ) · F —2 .
This square class depends only on g, since γ is uniquely determined up to a factor
in F — . In other words, Sn is the map which makes the following diagram commute:
χ

<< . .

. 25
( : 75)



. . >>