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isometries (ψi : Vi ’ Vi ) such that β —¦ (ψ0 , ψ1 ) = ψ2 —¦ β. The triple (π, π, 1C ) is

an isometry

(35.17) Vi , β(x, y) = x y ’ Vi , β(x, y) = x y .


A similitude (ψi : Vi ’ Vi ), with multiplier (»0 , »1 , »2 ) is de¬ned in a similar way.

Observe that the equation »2 = »0 »1 holds for a similitude. The main steps in the
proof of Proposition (??) were to associate to a quadratic space (V, q) of rank 8,
with trivial discriminant and trivial Cli¬ord invariant, a composition of quadratic
spaces V — U1 ’ U2 similar to a composition of type C — C ’ C.
(35.18) Proposition. (1) Compositions of quadratic spaces of dimension 8 are
classi¬ed by H 1 (F, Spin8 ).
(2) Let β012 : V0 —V1 ’ V2 be a ¬xed composition of quadratic spaces of dimension 8.
The action of S3 on the set H 1 (F, Spin8 ) is given by βijk ’ βs(i)s(j)s(j) where the
βijk are de¬ned by
bq0 v0 , β120 (v1 , v2 ) = bq2 β012 (v0 , v1 ), v2 = bq1 β201 (v2 , v0 ), v2
and
β102 (v1 , v0 ) = β012 (v0 , v1 ), β021 (v0 , v2 ) = β201 (v2 , v0 ).
Proof : By the proof of Proposition (??), any composition of 8-dimensional quad-
ratic spaces is isometric over a separable closure of F to the composition
Cs , Cs , Cs , β(x, y) = x y
where Cs is the split Cayley algebra. By Proposition (??), Spin8 is the group of
automorphisms of the composition
Cs , Cs , Cs , β(x, y) = x y Cs , Cs , Cs , β(x, y) = x y .
Consider the representation
ρ : G = GL(Cs ) — GL(Cs ) — GL(Cs ) ’ GL(W )
where
W = S 2 (C— ) • S 2 (C— ) • S 2 (C— ) • HomF (Cs —F Cs , Cs ),
s s s

given by the formula
ρ(±0 , ±1 , ±2 )(f, g, h, φ) = f —¦ ±’1 , g —¦ ±’1 , h —¦ ±’1 , ±2 —¦ φ —¦ (±’1 — ±’1 ) .
0 1 2 0 1
Let w = (n, n, n, m) where m : Cs — Cs ’ Cs is the product m(x, y) = x y in
the para-Cayley algebra Cs . The group scheme AutG (w) coincides with the group
scheme Spin(S, n) in view of Proposition (??). (Observe that by (??) elements t,
t+ , t’ ∈ O(Cs , ns ) such that t(x y) = t’ (x) t+ (y) are already in O+ (Cs , ns ).)
Thus (??) follows from (??). Claim (??) follows from Proposition (??).
488 VIII. COMPOSITION AND TRIALITY


§36. Twisted Compositions
Let L be a cubic ´tale F -algebra with norm NL , trace TL , and let # : L ’ L be
e
the quadratic map such that # = NL ( ) for ∈ L. Let (V, Q) be a nonsingular
quadratic space over L, let bQ (x, y) = Q(x + y) ’ Q(x) ’ Q(y), and let β : V ’ V
be a quadratic map such that
#
(1) β( v) = β(v)
Q β(v) = Q(v)#
(2)
for all v ∈ V , ∈ L. We de¬ne N (v) = bQ v, β(v) and further require that
(3) N (v) ∈ F
for all v ∈ V . We call the datum “ = (V, L, Q, β) a twisted composition over
L with quadratic norm Q, quadratic map β, and cubic norm N . A morphism
∼ ∼
φ : (V, L, Q, β) ’ (V , L , Q , β ) is a pair (t, φ), t : V ’ V , φ : L ’ L , with φ an
’ ’
F -algebra isomorphism, t φ-semilinear, and such that tβ = β t and φQ = Q t.
(36.1) Lemma. Let “ = (V, L, Q, β) be a twisted composition. Then for any » ∈
L— ,
“» = (V, L, »# Q, »β)
is again a twisted composition and, conversely, if (V, L, µQ, »β) is a twisted com-
position, then µ = »# .
Proof : The ¬rst claim is straightforward. If (V, L, µQ, »β) is a twisted composition,
the equality »2 µ = µ# follows from (??), (??). Since (µ# )# = µ# µ2 , it follows
that (»# )2 = µ2 , thus µ = ±»# . Since »2 µ = µ# , multiplying both sides by µ and
using that (»# )2 = µ2 shows that NL/F (»# ) = NL/F (µ). Since L is cubic over F ,
µ = »# .

Observe that the map v ’ »v is an isomorphism “» ’ “ if » ∈ F — . A

morphism

• = (t, φ) : (V, L, »# Q, »β) ’ (V , L , Q , β )

is a similitude with multiplier »# of (V, L, Q, β) with (V , L , Q , β ). For any ¬eld
extension K of F and any twisted composition “ over F , we have a canonical
twisted composition “ — K over K.
Examples of twisted compositions arise from symmetric composition algebras:
(36.2) Examples. (1) Let (S, , n) be a symmetric composition algebra. Let L =
F —F —F , let V = S —L = S —S —S and let Q = (n, n, n). We have (x0 , x1 , x2 )# =
(x1 x2 , x0 x2 , x0 x1 ) and putting
βS (v0 , v1 , v2 ) = (v1 v2 , v2 v0 , v0 v1 )
gives a twisted composition S = (S—L, L, Q, βS ). Condition (??) of the de¬nition of
a twisted composition is equivalent with the associativity of the norm. In particular
βC (v0 , v1 , v2 ) = (v 1 v 2 , v 2 v 0 , v 0 v 1 )
de¬nes a twisted composition associated with the Cayley algebra (C, ). We call C
a twisted Cayley composition. A twisted Cayley composition with L and C split is
a split twisted composition of rank 8.
§36. TWISTED COMPOSITIONS 489


(2) Let (V0 , V1 , V2 ), β012 be a composition of quadratic spaces as in Proposi-
tion (??). If we view V0 • V1 • V2 as a module V over L = F — F — F and
group the βijk to a quadratic map β : V ’ V we obtain a twisted composition over
F — F — F ; any twisted composition over F — F — F is of this form:
(36.3) Proposition. A twisted composition over F — F — F is of the form V0 •
V1 • V2 as in (??.??) and is similar to a twisted composition C for some Hurwitz
algebra (C, n).
Proof : Let V0 = (1, 0, 0)V , V1 = (0, 1, 0)V , V2 = (0, 0, 1)V , so that V = V0 •V1 •V2 .
We ¬rst construct a multiplication on V0 . We use the notations x = (x0 , x1 , x2 ) for
x ∈ L and v = (v0 , v1 , v2 ) for v ∈ V . We have x# = (x1 x2 , x2 x0 , x0 x1 ). Let
β(v) = β0 (v), β1 (v), β2 (v) .
It follows from
β (0, 1, 1)(v0 , v1 , v2 ) = (1, 0, 0)β(v0 , v1 , v2 )
that
β0 (v0 , v1 , v2 ) = β(0, v1 , v2 )
β1 (v0 , v1 , v2 ) = β(v0 , 0, v2 )
β2 (v0 , v1 , v2 ) = β(v0 , v1 , 0).
Furthermore, the F -bilinearity of
β(x, y) = β(x + y) ’ β(x) ’ β(y)
implies that β(0, v1 , v2 ) = β (0, v1 , 0), (0, 0, v2 ) is F -bilinear in the variables v1
and v2 . Thus there are three F -linear maps βi : Vi+1 —Vi+2 ’ Vi where i, i+1, i+2
are taken mod 3, such that
β0 (v1 — v2 ) = β(0, v1 , v2 )
β1 (v2 — v0 ) = β(v0 , 0, v2 )
β2 (v0 — v1 ) = β(v0 , v1 , 0)
and such that
β(v) = β0 (v1 — v2 ), β1 (v2 — v0 ), β2 (v0 — v1 ) .
Since Q is F — F — F -linear, we may write Q(v) = q0 (v0 ), q1 (v1 ), q2 (v2 ) . Condi-
tion (??) of the de¬nition of a twisted composition reads
q0 β0 (v1 — v2 ) = q1 (v1 )q2 (v2 ), q1 β1 (v2 — v0 ) = q2 (v2 )q0 (v0 )
and q2 β2 (v0 — v1 ) = q0 (v0 )q1 (v1 ). This is the ¬rst claim. Linearizing gives
bq0 β0 (v1 — v2 ), β0 (w1 — v2 ) = q2 (v2 )bq1 (v1 , w1 )
bq1 β1 (v2 — v0 ), β1 (w2 — v0 ) = q0 (v0 )bq2 (v2 , w2 )
bq2 β2 (v0 — v1 ), β2 (w0 — v1 ) = q1 (v1 )bq0 (v0 , w0 ).
Similarly, Condition (??) reduces to
bq0 v0 , β0 (v1 — v2 ) = bq1 v1 , β1 (v2 — v0 ) = bq2 v2 , β2 (v0 — v1 ) .
If ν1 , ν2 ∈ F — and e1 ∈ V1 , e2 ∈ V2 are such that ν1 q1 (e1 ) = 1 and ν2 q2 (e2 ) = 1,
then ν1 ν2 q0 (e0 ) = 1 for e0 = β0 (e1 — e2 ). Replacing β by (1, ν2 , ν1 )β and Q by
490 VIII. COMPOSITION AND TRIALITY


(ν1 ν2 , ν1 , ν2 )Q, we may assume that there exists an element e = (e0 , e1 , e2 ) such
that Q(e) = 1. We claim that
β1 (e2 — e0 ) = e1 and β2 (e0 — e1 ) = e2 .
We have for all (v0 , v1 , v2 ) ∈ V
bq1 v1 , β1 (e2 — e0 ) = bq0 e0 , β0 (v1 — e2 )
= bq0 β0 (e1 — e2 ), β0 (v1 — e2 )
= bq1 (v1 , e1 )q2 (e2 ) = bq1 (v1 , e1 )
Since bq1 (x, y) is nonsingular, we must have β1 (e2 —e0 ) = e1 . A similar computation

shows that β2 (e0 — e1 ) = e2 . We de¬ne isometries ±1 : (V0 , q0 ) ’ (V1 , q1 ) and


±2 : (V0 , q0 ) ’ (V2 , q2 ) by ±1 (v0 ) = β1 (e2 — v0 ) and ±2 (v0 ) = β2 (v0 — e1 ) and de¬ne

a multiplication on V0 by
x y = β0 ±1 (x) — ±2 (y) .
We have q0 (x y) = q1 ±1 (x) q2 ±2 (y) = q0 (x)q0 (y) and, for all y ∈ V0 ,
bq0 (y, x e0 ) = bq0 y, β0 β1 (e2 — x) — e2
= bq1 β1 (e2 — x), β1 (e2 — y)
= q2 (e2 )bq0 (x, y) = bq0 (x, y)
Thus x e0 = x, e0 is an identity for and V0 is, by Theorem (??), a Hurwitz
algebra with multiplication , identity e0 and norm q0 . We call it (C, n). Let
C = C — C — C be the associated twisted composition over F — F — F . We check
¬nally that C V via the map (IC , ±1 π, ±2 π) where π is the conjugation in C. It
su¬ces to verify that
β0 (±1 πx — ±2 πy) = x y
β1 (±2 πx — y) = ±1 π(x y)
β2 (x — ±1 πy) = ±2 π(x y).
The ¬rst formula follows from the de¬nition of . For the second we have
bq1 β1 (±2 πx — y), ±1 πz = bq0 y, β0 (±1 πz — ±2 πx)
= bq0 (y, z x)
= bq0 (y x, z)
= bq1 ±1 π(x y), ±1 πz
hence the claim. The proof of the third one is similar.
(36.4) Corollary. For any twisted composition (V, L, Q, β) we have dim L V = 1,
2, 4 or 8.
We observe that the construction of the multiplication in the proof of Propo-
sition (??) is similar to the construction in Proposition (??) or to the construction
given by Chevalley [?, Chap. IV] for forms of dimension 8 of maximal index.
(36.5) Proposition. Let C be a Cayley algebra over F with norm n. The group
scheme Aut(C) of F -automorphisms of the twisted composition C is isomorphic to
the semidirect product Spin(C, n) S3 . In particular the group scheme of automor-
phisms of a split twisted composition of rank 8 is isomorphic to Spin8 S3 .
§36. TWISTED COMPOSITIONS 491


Proof : Let R be an F -algebra such that L — R = R — R — R (the other cases are
let as exercises). Let

p : Aut(C)(R) ’ Autalg (L)(R)
be the restriction map. Since Autalg (L)(R) = S3 , we have to check that p has a
section and that ker p = Spin(n). The permutations ρ : (x0 , x1 , x2 ) ’ (x1 , x2 , x0 )
and : (x0 , x1 , x2 ) ’ (x0 , x2 , x1 ) generate S3 . A section is given by ρ ’ ρ,
ρ(v0 , v1 , v2 ) = (v1 , v2 , v0 ), and ’ , (v0 , v1 , v2 ) = (v 2 , v 1 , v 0 ). Now let t ∈
Aut(C)(R) be such that p(t) = 1, i.e., t is L-linear. Putting t = (t0 , t1 , t2 ), the ti
are isometries of n and the condition tβC = βC t is equivalent to
ˆ ˆ
t0 (v1 v2 ) = t1 (v1 )t2 (v2 )
ˆ ˆ
t1 (v2 v0 ) = t2 (v2 )t0 (v0 )
ˆ ˆ
t2 (v0 v1 ) = t0 (v0 )t1 (v1 ).
In fact any of these three conditions is equivalent to the two others (see Proposi-
tion (??)). By (??) the ti are proper, so that ti ∈ O+ (n)(R) and by Proposition (??)
we have t = (t0 , t1 , t2 ) ∈ Spin(n)(R).

The split exact sequence
p
(36.6) 1 ’ Spin8 ’ Spin8 S3 ’ S3 ’ 1

induces a sequence in cohomology
p1
’ H 1 (F, Spin8 ) ’ H 1 (F, Spin8 S3 ) ’ H 1 (F, S3 ).

(36.7) Proposition. (1) Twisted compositions of dimension 8 over F are classi-
¬ed by the pointed set H 1 (F, Spin8 S3 ).
(2) The map H 1 (F, Spin8 ) ’ H 1 (F, Spin8 S3 ) is induced by β012 ’ (V, β) with
V = V0 • V1 • V2 and β = (β120 , β201 , β012 ) (with the notations of (??)).
(3) For any class γ = [L] ∈ H 1 (F, S3 ) the ¬ber (p1 )’1 ([L]) is in bijection with the
Gal(Fsep /F )
in H 1 F, (Spin8 )γ .
orbits of (S3 )γ
Proof : In view of Proposition (??) any twisted composition over Fsep is a twisted
composition Cs for the split Cayley algebra Cs . Thus (??) will follow from Propo-
sition (??) and Proposition (??) if we may identify Aut(C) with AutG (w) for
some tensor w and some representation ρ : G ’ GL(W ) where H 1 (F, G) = 0, Let
Cs = (V, L, Q, β), let W be the F -space
W = SL (V — ) • SL (V — ) — V • HomF (L — L, L)
2 2


and let G = RL/F GLL (V ) — GL(L) . Then G acts on W as

ρ(±, φ)(f, g, h) = φ —¦ f —¦ ±’1 , ± —¦ g —¦ ±’1 , φ —¦ h —¦ (±’1 — ±’1 ).
We have H 1 F, RL/F GLL (V ) = H 1 L, GLL (V ) by (??), hence H 1 (F, G) = 0
by Hilbert 90. We now choose w = (Q, β, m), where m is the multiplication of L,
getting Aut(C) = AutG (w) and (??) is proved.
(??) follows from the description of H 1 (F, Spin8 ) given in Proposition (??)
and (??) is an example of twisting in cohomology (see Proposition (??)).
492 VIII. COMPOSITION AND TRIALITY


36.A. Multipliers of similitudes of twisted compositions. We assume
in this section that char F = 2, 3. Let “ = (V, L, Q, β) be a twisted composition
over F and let
G = Aut(V, L, Q, β)0 = AutL (V, L, Q, β)
be the connected component of Aut(V, L, Q, β). If L is split and V is of split
Cayley type, the proof of Proposition (??) shows that G = Spin8 , hence G is
always a twisted form of Spin8 and we write G = Spin(V, L, Q, β). If L = F — K
for K quadratic ´tale, and accordingly (V, Q) = (V1 , Q1 ) • (V2 , Q2 ) for (V1 , Q1 )
e
a quadratic space over F and (V2 , Q2 ) a quadratic space over K, the projection

(V, Q) ’ (V1 , Q1 ) induces an isomorphism Spin(V, L, Q, β) ’ Spin(V1 , Q1 ) by

Propositions (??) and (??).
The pointed set H 1 (F, G) classi¬es twisted compositions (V , L, Q , β ) with
¬xed L. Let γ : Gal(Fsep /F ) ’ S3 be a cocycle de¬ning L. By twisting the exact
sequence (??) we see that G = (Spin8 )γ . Let C be the center of G; the center C
of Spin8 ¬ts into the exact sequence
m
1 ’ C ’ µ2 — µ2 — µ2 ’ µ2 ’ 1

(see Lemma (??)) hence, twisting with γ gives an exact sequence
NL/F
1 ’ C ’ RL/F (µ2,L ) ’’’’ µ2,F ’ 1.
Thus
NL/F
H 1 (F, C) = ker L— /L—2 ’’’’ F — /F —2
and the exact sequence (??)
NL/F
#
1 ’ F — /F —2 ’ L— /L—2 ’ L— /L—2 ’’’’ F — /F —2 ’ 1

gives the identi¬cation
H 1 (F, C) = L— /F — · L—2 .
The group H 1 (F, C) acts on H 1 (F, G) through the rule
» · (V, L, Q, β) = (V, L, »# Q, »β), » ∈ L— ,
(36.8)
hence by Proposition (??), (V, L, »# Q, »β) (V, L, Q, β) if and only if the class
— —2 1
» · F · L ∈ H (F, C) belongs to the image of the connecting homomorphism
δF = δ : G(F ) ’ H 1 (F, C)
associated to the exact sequence (recall that char F = 2)
1 ’ C ’ G ’ G ’ 1.
We would like to compute the image of δ. For this we ¬rst consider the case L =
F — K; then H 1 (F, C) = K — /K —2 and, since G = Spin(V1 , Q1 ), G = PGO+ (Q1 )
and δ is the map
S : PGO+ (Q1 )(F ) ’ K — /K —2
de¬ned in (??). The composition of S with the norm map NK/F : K — /K —2 ’
F — /F —2 gives the multiplier map (Proposition (??)). If L is a ¬eld, let ∆ be the
§36. TWISTED COMPOSITIONS 493


discriminant of L, so that L — L = L — L — ∆ (see Corollary (??)). We would like
to compare the images of δF and δL . Since [L : F ] = 3, we have
im δF = NL/F —¦ resF/L (im δF ) ‚ NL/F (im δL ).
By Gille™s norm principle (see the following Remark) we have
NL/F (im δL ) ‚ im δF ,
hence
im δF = NL/F (im δL ).
The group scheme GL is isomorphic to PGO+ (Q), so that
im δL = im S ‚ (L — ∆)— /(L — ∆)—2 = H 1 (L, C)
(since L — L = L — L — ∆). One can check that the diagram
1
NL/F
1
H 1 (F, C)
H (L, C) ’’’
’’


(NL/F )—
(L — ∆)— /(L — ∆)—2 ’ ’ ’ L— /F — · L—2
’’’
commutes. It follows that
im δF = NL/F (im δL ) = NL—K/L (im S)
is the image of the group of multipliers G(Q) in L— /F — · L—2 . We have proved:
(36.9) Theorem. There exists an L-isomorphism (V, L, »# Q, »β) (V, L, Q, β)
if and only if » ∈ F — · G(Q).
(36.10) Remark. The condition » ∈ F — · G(Q) does not depend on β. One can
show that the condition is also equivalent to the fact that the quadratic forms
»# · Q and Q are isomorphic over L. The theorem says that this isomorphism can
be chosen in such a way that it takes » · β to β. This is the hardest part of the
theorem, where we use Gille™s result [?]: it says that the norm NL/F takes R-trivial
elements in im δL ‚ H 1 (L, C) to R-trivial elements of im δF ‚ H 1 (F, C). So it
su¬ces to prove that the image of S in (L — ∆)— /(L — ∆)—2 consists of R-trivial
elements. In facts the following weaker statement is enough: for any element x in
the image of S there is another element y in this image such that y is R-trivial and
the norms of x and y in L— /F — · L—2 are equal. For the proof of this statement
note ¬rst that Q is “almost” a P¬ster form, i.e., there exists a 3-fold P¬ster form q
over L and a ∈ L— such that Q = q + a, ’ad ∈ W (L) where d ∈ F — is such that

∆ = F ( d). By the theorem (??) of Dieudonn´, e
G(Q) = G(q) © NL—∆/L (L — ∆)— = G( a, ’ad ),
hence G(Q) = G(q) © NL—∆/L (L — ∆)— . An element in the intersection is a norm
from (L — ∆)— , i.e., a norm from a quadratic extension of L which splits q, hence
is (up to a square) a norm of a biquadratic composite. Thus
G(Q) = G(q) © NL—∆/L (L — ∆)— = NL—∆/L G(q∆ ) .
We start with x ∈ im S ‚ (L — ∆)— /(L — ∆)—2 . Write z = NL—∆/L (x) ∈
L— /F — · L—2 . We would like to ¬nd y as described above. Since z ∈ G(Q) =
NL—∆/L G(q∆ ) and G(q∆ ) = D G(q∆ ) , we can ¬nd a rational function z(t) ∈
494 VIII. COMPOSITION AND TRIALITY


QL(t) such that z(0) = 1 and z(1) = z (this means that the elements of G(Q) are
R-trivial). Now take any y(t) in the image of
— —2
S : PGO+ QL(t) ’ L — ∆(t) / L — ∆(t)

such that NL—∆(t)/L(t) y(t) = z(t). The element y = y(1)y(0)’1 is by de¬nition
an R-trivial element, is in the image of S and NL—∆/∆(y) = z.
36.B. Cyclic compositions. Twisted compositions of dimension 8 over cyclic
cubic extensions were introduced in Springer [?]. We ¬rst recall Springer™s de¬ni-
tion. Let (L/F, ρ) be a cyclic F -algebra of degree 3 with ρ a generator of the
group Gal(L/F ) = A3 . A cyclic composition is a nonsingular quadratic space
(V, Q) over L, together with an F -bilinear multiplication (x, y) ’ x — y, which is
ρ-semilinear in x and ρ2 -semilinear in y, and such that
Q(x — y) = ρ Q(x) · ρ2 Q(y) ,
(1)
bQ (x — y, z) = ρ bQ (y — z, x) = ρ2 bQ (z — x, y)
(2)
where bQ (x, y) = Q(x+y)’Q(x)’Q(y). Observe that the choice of a generator ρ of
the group Gal(L/F ) is part of the datum de¬ning a cyclic composition. Morphisms
of cyclic compositions are de¬ned accordingly.
Linearizing (??) gives
bQ (x — z, x — y) = ρ Q(x) ρ2 bQ (z, y) ,
(3)
bQ (x — z, y — z) = ρ bQ (x, y) ρ2 Q(x) .
(4)
It then follows that
bQ (x — y) — x, z = ρ bQ (x — z, x — y)
= ρ2 Q(x) bQ (y, z)
= bQ ρ2 Q(x) y, z
so that
(x — y) — x = ρ2 Q(x) y
(5)
and similarly
(6) x — (y — x) = ρ Q(x) y.
Linearizing conditions (??) and (??) gives
(x — y) — z + (z — y) — x = ρ2 bQ (x, z) y,
(7)
(8) x — (y — z) + z — (y — x) = ρ bQ (x, z) y.
By (??) we have bQ (x — x, x) ∈ F , thus (V, L, β, Q) with β(x) = x — x is a twisted
composition. Conversely, we shall see in Proposition (??) that any twisted compo-
sition over a cyclic cubic extension comes from a cyclic composition.
(36.11) Example. Let (S, , n) be a symmetric composition algebra over F and
let L be a cyclic cubic algebra. Let ρ be a generator of Gal(L/F ). It is easy to
check that V = SL = S — L, with the product
x — y = (1 — ρ)(x) (1 — ρ2 )(y)
and the norm Q(x) = (n — 1)(x), is cyclic. Thus, by putting β(x) = x x, we
may associate to any symmetric composition algebra (S, , n) and any cubic cyclic
§36. TWISTED COMPOSITIONS 495


extension L a twisted composition “(S, , L). If, furthermore, L = F — F — F and
ρ1 (ξ0 , ξ1 , ξ2 ) = (ξ1 , ξ2 , ξ0 ), we have a product
β1 (x, y) = x —1 y = (x1 y2 , x2 y0 , x0 y1 )
for x = (x0 , x1 , x2 ) and y = (y0 , y1 , y2 ) in SF —F —F = S — S — S such that
β1 (x, x) = β(x0 , x1 , x2 ) = (x1 x2 , x2 x0 , x0 x1 )
and we obtain a twisted composition as de¬ned in Example (??). By taking ρ2 = ρ2
1
as generator of Gal(F — F — F/F ), we have another product
β2 (x, y) = x —2 y = (y1 x2 , y2 x0 , y0 x1 )
such that β2 (x, x) = β(x). Observe that β1 (x, y) = β2 (y, x).
(36.12) Proposition. Let L/F be a cyclic cubic algebra, let ρ1 , ρ2 be di¬erent
generators of the group Gal(L/F ) = A3 and let (V, L, Q, β) be a twisted composition
over L. There is a unique pair of cyclic compositions βi (x, y) = x —i y over V ,
i = 1, 2, with βi ρi -semilinear with respect to the ¬rst variable, such that βi (x, x) =
β(x). Furthermore we have β1 (x, y) = β2 (y, x).
Proof : There exists at most one F -bilinear map β1 : V — V ’ V which is ρ1 -

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