and the proof is reduced to checking the assertion for the products v v, v x, x v

and x x, where it is easy.

For symmetric compositions of dimension 2, we do not have to assume in Propo-

sition (??) that F contains a contains a primitive cube root of unity since a sepa-

rable alternative algebra A of degree 3 and dimension 3 is ´tale commutative and

e

1

the multiplication reduces to a b = ab ’ 3 TA (ab). Thus (??) and (??) imply:

(34.28) Proposition. Let F be a ¬eld of characteristic di¬erent from 3 and let

(S, , n) be a symmetric composition algebra over F of dimension 2. Then S

(L0 , ) for a unique cubic ´tale F -algebra L. The algebra S is para-quadratic if and

e

only if L is not a ¬eld.

To obtain a complete description of symmetric compositions in dimension 4

and 8 we may take the Structure Theorem (??) for cubic alternative algebras into

account:

(34.29) Theorem (Elduque-Myung). Let F be ¬eld with char = 3 which contains

a primitive cube root of unity. There exist equivalences of groupoids

Scomp4 ≡ Hurw 4 ≡ Hurw 4 ≡ A1

and

1 1

Scomp8 ≡ Hurw 8 Oku ≡ G2 A2 .

However we did not prove (??) and we shall give an alternate proof of these

equivalences. Observe that this will yield, in turn, a proof of Theorem (??)!

We postpone the proof of (??) and begin with an example:

(34.30) Example. Let V be a 2-dimensional vector space over F . We view ele-

ments of EndF (V • F ) of trace zero as block matrices

φ v End(V ) V

∈

V—

f ’ tr(φ) F

The product of such two blocks is given by

φ v φ v φ—¦φ +v—¦f φ(v ) ’ tr(φ )v

· =

f ’ tr(φ) f ’ tr(φ ) f —¦ φ ’ tr(φ)f f (v ) + tr(φ) tr(φ )

where (v —¦ f )(x) = vf (x). With the multiplication (x, y) ’ x y de¬ned in (??)

1

and the quadratic form n = ’ 3 SEndF (V •F ) ,

SV = 1 C EndF (V • F ) = EndF (V • F )0 ,

is a symmetric composition algebra and

’1V 0 End(V ) V

e= ∈

V—

0 2 F

is a nontrivial idempotent. By Proposition (??), (??), the map

•(x) = e (e x) = bn (e, x)e ’ x e

is an automorphism of SV , of order 3, such that SV , reduces to the Petersson

algebra SV • . The corresponding Z/3Z-grading is

S2 = V — .

S0 = EndF (V ), S1 = V,

In particular we have dimF S0 = 4. The converse also holds (Elduque-P´rez [?,

e

Theorem 3.5]) in view of the following:

§34. SYMMETRIC COMPOSITIONS 475

(34.31) Proposition. Let F be a ¬eld of characteristic not 3 which contains a

primitive cube root of unity ω. Let (S, , n) be a symmetric composition algebra

of dimension 8 with a nontrivial idempotent e. Let •(x) = e (e x) and let

S = S0 • S1 • S2 be the Z/3Z-grading of (S, , n) de¬ned by • (see (??)). If

dimF S0 = 4, there exists a 2-dimensional vector space V such that (S, ) S V • =

1

C End(V • F ) .

Proof : Since dimF S0 = 4 we must have dimF S1 = dimF S2 = 2, and S1 , S2 are

maximal isotropic direct summands of S by Lemma (??). Let (x1 , x2 ) be a basis

of S1 and let (f1 , f2 ) be a basis of S2 such that bn (xi , fj ) = δij . Since

(xi + fi ) u (xi + fi ) = (xi + fi ) u (xi + fi ) = u

holds for all u we have

(1) xi (u xi ) = 0 = (xi u) xi

(2) fj (u fj ) = 0 = (fj u) fj

and

(3) xi (u fi ) + fi (u xi ) = u

for all u ∈ S0 . Thus, by choosing u = e and using that

x = ’ωe x = ’ω 2 x e, e = ’ω 2 e f

f = ’ωf

for x ∈ S1 , resp. f ∈ S2 , we see that xi xi = 0 = fj fj . Moreover x1 x2 =

0 = x2 x1 and f1 f2 = 0 = f2 f1 , so S1 S1 = 0 = S2 S2 . Since u xi ∈ Si ,

u fj ∈ S2 for u ∈ S0 , (??) implies that

(f1 x1 , f1 x2 , x1 f1 , x1 f2 )

generates S0 , hence is a basis of S0 and S1 S2 = S0 = S2 S1 . We now de¬ne an

F -linear map ψ : S ’ M3 (F ) on basis elements by

« « « «

001 000 0 00 000

0 0 0, x2 ’ 0 0 1, f1 ’ 0 0 0, f2 ’ 0 0 0

x1 ’

000 000 3 00 030

and

1

xi fj ’ µxi fj + (1 ’ µ)fj xi ’ tr(xi fj ),

3

1

fj xi ’ µfj xi + (1 ’ µ)xi fj ’ tr(xi fj )

3

where multiplication on the right is in M3 (F ) and µ = 1’ω . We leave it as a

3

∼

(lengthy) exercise to check that ψ is an isomorphism of composition algebras S ’

’

C M3 (F ) = SV with V = F 2 .

1

Proof of Theorem (??): Let (S, , n) be a symmetric composition algebra. Since

twisted forms of Okubo algebras are Okubo and twisted forms of para-Hurwitz

algebras of dimension ≥ 4 are para-Hurwitz, we may assume by Lemma (??) that

S contains a nontrivial idempotent. Let S = S0 • S1 • S2 be the grading given

by Lemma (??). Assume ¬rst that dim S = 4. If dim S0 = 4, then S = S0

is para-Hurwitz; and if dim S0 = 2, S is para-Hurwitz by Proposition (??). If

dim S = 8 and dim S0 = 2, S is para-Hurwitz by Proposition (??); if dim S = 8

and dim S0 = 4, S is Okubo by Proposition (??).

476 VIII. COMPOSITION AND TRIALITY

34.D. Alternative algebras with unitary involutions. To overcome the

condition on the existence of a primitive cube root of unity in F , one considers

separable cubic alternative algebras B over the quadratic extension K = F (ω) =

F [X]/(X 2 + X + 1) which admit a unitary involution „ .

(34.32) Proposition. Let K be a quadratic separable ¬eld extension of F with

conjugation ι. Cubic separable unital alternative K-algebras with a unitary involu-

tion „ are of the following types:

(1) L = K — L0 for L0 cubic ´tale over F and „ = ι — 1.

e

(2) Central simple associative algebras of degree 3 over K with a unitary involution.

(3) Products K — (K — C) where C is a Hurwitz algebra of dimension 4 or 8 over F

and „ = (ι, ι — π) where π is the conjugation of the Hurwitz algebra C.

Proof : A cubic separable unital K-algebra is of the types described in Theorem

(??). (??) then follows by Galois descent and (??) follows from Proposition (??)

(the case of a Cayley algebra being proved as the quaternion case).

Assume that F has characteristic di¬erent from 3 and that K = F (ω) is a

¬eld. Let 2Sepalt m (3) be the groupoid of alternative F -algebras (B, „ ) which are

separable cubic of dimension m over K and have unitary involutions. The generic

polynomial of degree 3 on B with coe¬cients in K restricts to a polynomial function

on Sym(B, „ ) with coe¬cients in F . Let

Sym(B, „ )0 = { x ∈ Sym(B, „ ) | TB (x) = 0 }.

on Sym(B, „ )0 as in (??):

We de¬ne a multiplication

x y = µxy + (1 ’ µ)yx ’ 1 TB (yx)1.

(34.33) 3

1

The element x y lies in Sym(B, „ )0 since ι(µ) = 1 ’ µ. Let n(x) = ’ 3 TB (x2 ).

A description of Sym(B, „ ) (and of ) in cases (??) and (??) is obvious, however

less obvious in case (??).

(34.34) Lemma. Let C be a Hurwitz algebra over F , let K be a quadratic ´tale

e

F -algebra, and let B be the alternative K-algebra B = K — (K — C) with unitary

involution „ = (ι, ι — π). Then:

(1)

Sym(B, „ )0 = TC (z), ξ — z + ι(ξ) — π(z) z∈C

where ξ is a (¬xed ) generator of K such that TK (ξ) = ’1.

(2) If K = F (ω) where ω is a primitive cube root of unity, the map z ’ ω — z +

∼

ι(ω) — π(z) is an isomorphism (C, ) ’ Sym(B, „ )0 , .

’

Proof : (??) We obviously have

Sym(B, „ )0 ⊃ TC (z), ξ — z + ι(ξ) — π(z) z∈C ,

hence the claim by dimension count, since the map

Sym(B, „ )0 ’ TC (z), ξ — z + ι(ξ) — π(z) z∈C

given by z ’ ξ — z + ι(ξ) — π(z) is an isomorphism of vector spaces.

(??) follows from (??) and Proposition (??).

§34. SYMMETRIC COMPOSITIONS 477

(34.35) Theorem. Let F be a ¬eld of characteristic not 3 which does not contain

a primitive cube root of unity ω and let K = F (ω). For any cubic separable alterna-

tive K-algebra with a unitary involution „ , the F -vector space Sym(B, „ )0 , is a

symmetric composition algebra. Conversely, for any symmetric composition algebra

(S, ), the unital alternative K-algebra B = K · 1 • (K — S) with the multiplication

xy = (1 + ω)x y ’ ωy x + bn (x, y) · 1, x·1=1·x =x

for x, y ∈ B 0 , admits the unitary involution (ι, ι — 1S ) and the functors

C : (B, „ ) ’ Sym(B, „ )0 ,

2

,

2

A : (S, ) ’ B = K · 1 • (K — S), (ι, ι — 1)

de¬ne an equivalence of groupoids

2

Sepalt m+1 (3) ≡ Scomp m

for m = 2, 4 and 8.

Proof : To check that Sym(B, „ )0 , is a symmetric composition algebra over F ,

we may assume that K = F — F . Then B is of the form (A, Aop ) and „ (a, bop ) =

(b, aop ) for A separable alternative of degree 3 as in the associative case (see (??)).

By projecting on the ¬rst factor, we obtain an isomorphism

Sym(B, „ )0 A0 = { x ∈ A | TA (x) = 0 },

and the product on A0 is as in (??). Thus the composition Sym(B, „ )0 , is

0

isomorphic to (A , ) and hence is a symmetric composition.

Conversely, if (S, ) is a symmetric composition algebra over F , then (S, ) — K

is a symmetric composition algebra over K and (K · 1 • K — S, ·) is a K-alternative

algebra by Theorem (??). The fact that (ι, ι — 1) is a unitary involution on B

follows from the de¬nition of the multiplication of B. That 2 C and 2A de¬ne an

equivalence of groupoids then follows as in (??).

(34.36) Corollary. Let (S, ) be a symmetric composition algebra with norm n.

The following conditions are equivalent:

(1) S contains a nontrivial idempotent.

(2) the cubic form N (x) = bn (x x, x) is isotropic.

(3) the alternative algebra 1A(S) (if F contains a primitive cube root of unity,

otherwise 2A(S)) has zero divisors.

Proof : (??) ’ (??) By Proposition (??) the map •(x) = e (e x) is an auto-

morphism of (S, ) of order ¤ 3 and the corresponding subalgebra S0 (see Lemma

(??)) has dimension at least 2. For every nonzero x ∈ S0 with bn (x, e) = 0 we have

x e = e x = ’x, so that bn (x x, x) = ’bn (x x, e x) = n(x)bn (x, e) = 0.

The implication (??) ’ (??) is Lemma (??).

We check that (??) ” (??): If S is Okubo, then 2A(S) is central simple with

zero divisors and NrdB (x) = bn (x x, x) for x ∈ Sym(B, „ )0 , hence the claim in

this case. Since K — K — S always has zero divisors we are left with showing that

N is always isotropic on a para-Hurwitz algebra. Take x = 0 with bn (x, 1) = 0;

then bn (x x, x) = bn ’n(x)1, x = 0.

In the following classi¬cation of Elduque-Myung [?, p. 2487] “unique” always

means up to isomorphism:

478 VIII. COMPOSITION AND TRIALITY

(34.37) Theorem (Classi¬cation of symmetric compositions). Let F be a ¬eld of

characteristic = 3 and let (S, ) be a symmetric composition algebra over F .

(L0 , ) for a unique cubic ´tale F -algebra L. The

(1) If dimF S = 2, then S e

algebra S is para-quadratic if and only if L is not a ¬eld.

(2) If dimF S = 4, then S is isomorphic to a para-quaternion algebra (Q, ) for a

unique quaternion algebra Q.

(3) If dimF S = 8, then S is either isomorphic to

(a) a para-Cayley algebra (C, ) for a unique Cayley algebra C,

(b) an algebra of the form (A0 , ) for a unique central simple F -algebra A of

degree 3 if F contains a primitive cube root of unity, or

(c) S is of the form Sym(B, „ )0 , for a unique central simple F (ω)-algebra

B of degree 3 with an involution of the second kind if F does not contain a

primitive cube root of unity ω.

Proof : (??) was already proved in Proposition (??). Let K = F (ω). By Theo-

Sym(B, „ )0 , for some alternative K-algebra B with

rem (??) we have (S, )

a unitary involution „ . By Proposition (??) cases (??.??) and (??.??) occur if B

is central simple over K, and cases (??) and (??.??) occur when B K • K — C

for some Hurwitz algebra C of dimension 4 or 8. In view of Lemma (??), we must

have in the two last cases (S, ) (C , ) so that (S, ) is a para-Hurwitz algebra,

as asserted.

We call symmetric composition algebras as in (??.??) or (??.??) Okubo algebras

(the case where ω lies in F is not new!), we denote the corresponding groupoids

by 1Oku (when K = F — F ), 2Oku (when ω ∈ F ) and we set Oku = 1Oku

2

Oku. Assume that F does not contain a primitive cube root of unity ω and

let 2A2 be the full subgroupoid of A2 whose objects are algebras of degree 3 over

F (ω) with involution of the second kind. Since we have equivalences of groupoids

1

Oku ≡ 1A2 and 2Oku ≡ 2A2 , we also call symmetric composition in iOku symmetric

compositions of type iA2 .

(34.38) Corollary. Let F be ¬eld of characteristic di¬erent from 3. There exist

equivalences of groupoids

Scomp4 ≡ Hurw 4 ≡ Hurw 4 ≡ A1

and

1 2

Scomp 8 ≡ Hurw 8 Oku ≡ G2 A2 .

A2

34.E. Cohomological invariants of symmetric compositions. Three co-

homological invariants classify central simple algebras (B, „ ) of degree 3 with uni-

tary involutions „ (see Theorem (??)): f1 ∈ H 1 (F, µ2 ) (which determines the cen-

ter K), g2 ∈ H 2 F, (µ3[K] (which determines the K-algebra B) and f3 ∈ H 3 (F, µ2 )

(which determines the involution). We have a corresponding classi¬cation for sym-

metric compositions:

(34.39) Proposition. Let F be a ¬eld of characteristic not 2, 3. Dimension 8

symmetric compositions of

(1) type G2 are classi¬ed by one cohomological invariant f3 ∈ H 3 (F, µ2 ).

(2) type 1A2 (if ω ∈ F ) by one invariant g2 ∈ H 2 (F, µ3 ).

§35. CLIFFORD ALGEBRAS AND TRIALITY 479

(3) type 2A2 by two cohomological invariants g2 ∈ H 2 (F, Z/3Z) and f3 ∈ H 3 (F, µ2 ).

Proof : The claims follow from Theorem (??) and the corresponding classi¬cations

of Cayley algebras, resp. central simple algebras. For (??), one also has to observe

that if K = F (ω), then the action on µ3 twisted by a cocycle γ de¬ning K is the

usual action of the Galois group. Thus (µ3 )γ = Z/3Z.

Observe that a symmetric composition algebra S of type 1A2 with g2 = 0

comes from a division algebra A over F . However, its norm is always hyperbolic

(see Example (??)), hence f3 (S) = 0. On the other hand, in view of the following

example, there exist composition algebras of type 2A2 with invariants g2 (S) = 0

and f3 (S) = 0.

(34.40) Example. Over R there are no compositions of type 1A2 but there exist

compositions of type 2A2 with K = C and B = M3 (C). There are two classes of

involutions of the second type on M3 (C), the standard involution x ’ „ (x) where

„ (x) is the hermitian conjugate, and the involution Int(d) —¦ „ : x ’ d„ (x)d’1 ,

with d = diag(’1, 1, 1), which is distinguished. Since tr(x2 ) > 0 for any nonzero

hermitian 3 — 3 matrix x, the norm on Sym(B, „ )0 is anisotropic. The restriction

of the norm to Sym B, Int(d) —¦ „ 0 is hyperbolic. Observe that f3 Sym(B, „ )0 = 0

and f3 Sym B, Int(d) —¦ „ 0 = 0.

§35. Cli¬ord Algebras and Triality

35.A. The Cli¬ord algebra. Let (S, ) be a symmetric composition algebra

of dimension 8 over F with norm n. Let C(n) be the Cli¬ord algebra and C0 (n)

the even Cli¬ord algebra of (S, n). Let „ be the involution of C(n) which is the

identity on V . Let rx (y) = y x and x (y) = x y.

(35.1) Proposition. For any » ∈ F — , the map S ’ EndF (S • S) given by

0 »x

x’

rx 0

induces isomorphisms

∼

±S : C(»n), „ ’ EndF (S • S), σn⊥n

’

and

∼

±S : C0 (»n), „ ’ EndF (S), σn — EndF (S), σn ,

’

of algebras with involution.

Proof : We have rx —¦ x (y) = x —¦ rx (y) = »n(x) · y by Lemma (??). Thus the

existence of the map ±S follows from the universal property of the Cli¬ord algebra.

The fact that ±S is compatible with involutions is equivalent to

bn x (z y), u = bn z, y (u x)

for all x, y, z, u in S. This formula follows from the associativity of n, since

bn x (z y), u = bn (u x, z y) = bn z, y (u x) .

The map ±S is an isomorphism by dimension count, since C(n) is central simple.

480 VIII. COMPOSITION AND TRIALITY

Let (V, q) be a quadratic space of even dimension. We call the class of the

Cli¬ord algebra [C(V, q)] ∈ Br(F ) the Cli¬ord invariant of (V, q). It follows from

Proposition (??) that, for any symmetric composition (S, , n) of dimension 8, the

discriminant and the Cli¬ord invariant of (S, »n) are trivial. Conversely, a quadratic

form of dimension 8 with trivial discriminant and trivial Cli¬ord invariant is, by the

following Proposition, similar to the norm n of a Cayley (or para-Cayley) algebra C:

(35.2) Proposition. Let (V, q) be a quadratic space of dimension 8. The following

condition are equivalent:

(1) (V, q) has trivial discriminant and trivial Cli¬ord invariant,

(2) (V, q) is similar to the norm n of a Cayley algebra C.

Proof : This is a classical result of the theory of quadratic forms, due to A. P¬ster

(see for example Scharlau [?, p. 90]). As we already pointed out, the implication

(??) ’ (??) follows from Proposition (??). We include a proof of the converse

which is much in the spirit of this chapter, however we assume that char F = 2.

The idea is to construct a Cayley algebra structure on V such that the corresponding

norm is a multiple of q. This construction is similar to the construction given in

Chevalley [?, Chap. IV] for forms of maximal index. Let

∼

± : C(q), „ ’ EndF (U ), σk

’

be an isomorphism of algebras with involution where „ is the involution of C(V, q)

which is the identity on V . Let be a nontrivial idempotent generating the center

of C0 (q). By putting U1 = ±( )U and U2 = ±(1 ’ )U , we obtain a decomposition

(U, k) = (U1 , q1 ) ⊥ (U2 , q2 )

such that ± is an isomorphism of graded algebras where EndF (U1 • U2 ) is “checker-

board” graded. For any x ∈ V , let

0 ρx

±(x) = ∈ EndF (U1 • U2 )

»x 0

so that »x ∈ HomF (U2 , U1 ) and ρx ∈ HomF (U1 , U2 ) are such that »x —¦ρx = q(x)·1U1

and ρx —¦»x = q(x)·1U2 . Let ˆi : Ui ’ Ui— be the adjoints of qi , i.e., the isomorphisms

∼

b ’

induced by bi = bqi . We have

’1 t

ˆ1 ˆ1

b 0 0 ρx b 0 0 ρx

=

ˆ2 ˆ2

»x 0 »x 0

0 b 0 b

hence

ˆ1 —¦ ρx = »t —¦ ˆ2 and ˆ2 —¦ »x = ρt —¦ ˆ1

b xb b xb

or, putting ρx (u2 ) = ρ(x, u2 ) and »x (u1 ) = »(x, u1 ), we obtain maps

» : V — U 1 ’ U2 and ρ : V — U2 ’ U1

such that

b1 ρ(x, u2 ), u1 = b2 u2 , »(x, u1 ) .

If we set u2 = »(x, u1 ) we then have

b2 »(x, u1 ), »(x, u1 ) = b1 bq (x)u1 , u1

so that, since we are assuming that char F = 2,

q2 »(x, u1 ) = q(x)q1 (u1 )

§35. CLIFFORD ALGEBRAS AND TRIALITY 481

for x ∈ V and u1 ∈ U1 . Similarly the equation q1 ρ(x, u2 ) = q(x)q2 (u2 ) holds for

x ∈ V and u2 ∈ U2 . By linearizing the ¬rst formula, we obtain

b2 »(x, u1 ), »(x, v1 ) = q(x)b1 (u1 , v1 )

and

b2 »(x, u), »(y, u) = bq (x, y)q1 (u)

for x, y ∈ V and u1 , v1 ∈ U1 . By replacing q by a multiple, we may assume that

q represents 1, say q(e) = 1. We may do the same for q1 , say q1 (e1 ) = 1. We then

have q2 (e2 ) = 1 for e2 = »(e, e1 ). We claim that ρ(e, e2 ) = e1 . For any u1 ∈ U1 , we

have

b1 ρ(e, e2 ), u1 = b2 e2 , µ(e, u1 )

= b2 »(e, e1 ), »(e, u1 )

= b1 (e1 , u1 )q(e) = b1 (e1 , u1 ).

Since q1 is nonsingular, ρ(e, e2 ) equals e1 as claimed. The maps s1 : V ’ U1 and

s2 : V ’ U2 given by s1 (x) = ρ(x, e2 ) and s2 (y) = »(y, e1 ) are clearly isometries.

Let

x y = s’1 » x, s1 (y) for x, y ∈ V .

2

We have

q(x y) = q2 » x, s1 (y) = q(x)q1 s1 (y) = q(x)q(y),

x e = s’1 » x, s1 (e) = s’1 »(x, e1 ) = x

2 2

and

bq (v, e y) = b2 »(v, e1 ), µ(x, e1 ) = bq (v, e)

for all v ∈ V , so that e y = y. Thus V is a composition algebra with identity

element e. By Theorem (??) (V, ) is a Cayley algebra.

As an application we give another proof of a classical result of the theory of

quadratic forms (see for example Scharlau [?, p. 89]).

(35.3) Corollary. Let (V, q) be an 8-dimensional quadratic space with trivial dis-

criminant and trivial Cli¬ord invariant. Then (V, q) is hyperbolic if and only if it

is isotropic.

Proof : By Proposition (??), q is similar to the norm of a Cayley algebra, so that

the claim follows from Proposition (??).

35.B. Similitudes and triality. Let (S, , n) be a symmetric composition

algebra of dimension 8 over F . In view of Proposition (??) any similitude t of n

induces an automorphism C0 (t) of C0 (n) and t is proper, resp. improper if C0 (t) re-

stricts to the identity of the center Z of C0 (n), resp. the nontrivial F -automorphism

ι of Z. Let GO+ (n) be the group scheme of proper similitudes of n and GO’ (n)

the set of improper similitudes. The “triality principle” for similitudes of (S, n) is

the following result:

482 VIII. COMPOSITION AND TRIALITY

(35.4) Proposition. Let t be a proper similitude of (S, n) with multiplier µ(t).

There exist proper similitudes t+ , t’ of (S, n) such that

µ(t+ )’1 t+ (x y) = t(x) t’ (y),

(1)

µ(t)’1 t(x y) = t’ (x) t+ (y)

(2)

and

µ(t’ )’1 t’ (x y) = t+ (x) t(y).

(3)

Let t be an improper similitude with multiplier µ(t). There exist improper simili-

tudes t+ , t’ such that

µ(t+ )’1 t+ (x y) = t(y) t’ (x),

(4)

µ(t)’1 t(x y) = t’ (y) t+ (x)

(5)

and

µ(t’ )’1 t’ (x y) = t+ (y) t(x).

(6)

The pair (t+ , t’ ) is determined by t up to a factor (µ, µ’1 ), µ ∈ F — , and we have

µ(t+ )µ(t)µ(t’ ) = 1.

Furthermore, any of the formulas (??) to (??) (resp. (??) to (??)) implies the

others. If t is in O+ (n), the spinor norm Sn(t) of t is the class in F — /F —2 of the

multiplier of t+ (or t’ ).

Proof : Let t be a proper similitude with multiplier µ(t). The map S ’ EndF (S•S)

given by

0 1 0

t(x)

•(t) : x ’ = ±S t(x)

’1

0 µ(t)’1

µ(t) rt(x) 0

is such that (•(t)(x))2 = µ(t)’1 n t(x) = n(x), and so it induces a homomorphism

∼

•(t) : C(n) ’ EndF (S • S).

’

By dimension count •(t) is an isomorphism. By the Skolem-Noether Theorem, the

s0 s1

automorphism •(t) —¦ ±’1 of EndF (S • S) is inner. Let •(t) —¦ ±’1 = Int s3 s2 .

S S

Computing ±’1 —¦ •(t) on a product xy for x, y ∈ S shows that ±’1 —¦ •(t)|C0 =

S S

C0 (t). Since t is proper, C0 (t) is Z-linear. Again by Skolem-Noether we may write

s0 0

±S —¦ C0 (t) —¦ ±’1 = Int 0 s . This implies s1 = s3 = 0 and we may choose s0 = s0 ,

S 2

s0 0

s2 = s2 . We deduce from •(t)(x) = Int —¦ ±S (x) that

0 s2

= s0 x s’1 and µ(t)’1 rt(x) = s2 rx s’1

t(x) 2 0

or

s0 (x y) = t(x) s2 (y) and s2 (y x) = µ(t)’1 s0 (y) t(x), x, y ∈ S.

The fact that C0 (t) commutes with the involution „ of C0 (n) implies that s0 , s2 are

similitudes and we have µ(s0 ) = µ(t)µ(s2 ). Putting t+ = µ(s0 )’1 s0 and t’ = s2

gives (??) and (??).

To obtain (??), we replace x by y x in (??). We have

µ(t+ )’1 n(y)t+ (x) = t(y x) t’ (y).

Multiplying with t’ (y) on the left gives

µ(t+ )’1 n(y)t’ (y) t+ (x) = t(y x)µ(t’ )n(y).

§35. CLIFFORD ALGEBRAS AND TRIALITY 483

By viewing y as “generic”(apply (??)), we may divide both sides by n(y). This

gives (??).

If t is improper, then C0 (t) switches the two factors of Z = F — F and, given t,

we get t+ , t’ such that µ(t+ )t+ (x y) = t(y) t’ (x).

Formulas (??) and (??) follow similarly.

Conversely, if t satis¬es (??), then C0 (t) switches the two factors of Z = F — F ,

hence is not proper. This remark and the above formulas then show that t+ , t’ are

proper if t is proper. To show uniqueness of t+ , t’ up to a unit, we ¬rst observe

that t+ , t’ are unique up to a pair (r + , r’ ) of scalars, since

t+ 0

±S C0 (t)±’1 = Int .

S 0 t’

Replacing (t+ , t’ ) by (r+ t+ , r’ t’ ) gives

µ(t+ )(r+ )’1 t+ (x y) = r’ t’ (x) t(y) = µ(t+ )’1 r’ t+ (x y).

’1

This implies r+ = r’ . We ¬nally check that Sn(t) is the multiplier of t+ (or t’ )

for t ∈ O+ (n). The transpose of a linear map t is denoted by t— . Putting ±S (c) =

(t+ , t’ ) and writing ˆ : S ’ S — for the isomorphism induced by bn , we have

b

— — — —

±S c„ (c) = (t+ˆ’1 t+ ˆ t’ˆ’1 t’ ˆ = (ˆ’1 t+ ˆ + , ˆ’1 t’ ˆ ’ )

b b, b b) b bt b bt

— —

(since c„ (c) ∈ F ). Then the claim follows from t+ ˆ + = µ(t+ )ˆ and t’ ˆ ’ =

bt b bt

’ˆ ’1 —2 — —2

µ(t )b, since Sn(t) = Sn(t ) = c„ (c)F ∈ F /F . The other claims can be

checked by similar computations.

(35.5) Corollary. For any pair », »+ ∈ D(n), the set of values represented by n,

there exists a triple of similitudes t, t+ , t’ such that Proposition (??) holds and

such that », »+ are the multipliers of t, resp. t+ .

Proof : Given » ∈ D(n), let t be a similitude with multiplier », for example t(x) =

u x with n(u) = », and let t+ , t’ be given by triality. If t is replaced by ts

with s ∈ O+ (n), the multiplier of t will not be changed and the multiplier of t+

will be multiplied by the multiplier µ(s+ ) of s+ . By Proposition (??) we have

µ(s+ )F —2 = Sn(s). Since n is multiplicative, Sn O+ (n) ≡ D(n) mod F —2 and

we can choose s (as a product of re¬‚ections) such that Sn(s) = »+ µ(t+ )’1 F —2 ,

hence the claim.

Using triality we de¬ne an action of A3 on PGO+ (n)(F ) = PGO+ (n): Let

[t] be the class of t ∈ GO+ (n)(F ) modulo the center. We put θ + ([t]) = [t+ ] and

θ’ ([t]) = [t’ ] where t+ , t’ are as in Proposition (??).

(35.6) Proposition. The maps θ + and θ’ are outer automorphisms of the group

PGO+ (n). They satisfy (θ + )3 = 1 and θ+ θ’ = 1, and they generate a subgroup of

Aut PGO+ (n) isomorphic to A3 .

Proof : It follows from the multiplicativity of the formulas in Proposition (??) that

the maps θ+ and θ’ are group homomorphisms. The relations among them also

follow from (??). Hence they are automorphisms and generate a homomorphic

image of A3 . The fact that θ + is not inner follows from Proposition (??).

We shall see that the action of A3 is, in fact, de¬ned on the group scheme

PGO+ (n) = GO+ (n)/ Gm . For this we need triality for Spin(n).

484 VIII. COMPOSITION AND TRIALITY

35.C. The group Spin and triality. The group scheme Spin(S, n) is de-

¬ned as

Spin(S, n)(R) = { c ∈ C0 (n)— | cSR c’1 ‚ SR and c„ (c) = 1 }

R

for all R ∈ Alg F . The isomorphism ±S of (??) can be used to give a nice description

of Spin(S, n) = Spin(S, n)(F ).

(35.7) Proposition. Assume that char F = 2. There is an isomorphism

{ (t, t+ , t’ ) | t, t+ , t’ ∈ O+ (S, n), t(x y) = t’ (x) t+ (y) }

Spin(S, n)

such that the vector representation χ : Spin(S, n) ’ O+ (S, n) corresponds to the

map (t, t+ , t’ ) ’ t. The other projections (t, t+ , t’ ) ’ t+ and (t, t+ , t’ ) ’ t’

correspond to the half-spin representations χpm of Spin(S, n).

+

Proof : Let c ∈ Spin(n) and let ±S (c) = t0 t0 . The condition cxc’1 = χc (x) ∈ S

’

for all x ∈ S is equivalent to the condition t (x y) = χc (x) t’ (y) or, by Propo-

+

sition (??), to χc (x y) = t’ (x) t+ (y) for all x, y ∈ S. Since by Proposition (??)

we have

ˆ’1 t+ —ˆ

b b 0

±S „ (c) = ˆ’1 t’ —ˆ

0 b b

where ˆ : S ’ S — is the adjoint of bn , the condition c„ (c) = 1 is equivalent to

∼

b ’

— —

t+ ˆ + = ˆ and t’ ˆ ’ = ˆ i.e., the t± are isometries of b = bn , hence of n since

bt b bt b,

char F = 2. They are proper by Proposition (??). Thus, putting

T (S, n) = { (t, t+ , t’ ) | t+ , t, t’ ∈ O+ (S, n), t(x y) = t’ (x) t+ (y) },

c ’ (χc , t+ , t’ ) de¬nes an injective group homomorphism φ : Spin(S, n) ’ T (S, n).

It is also surjective, since, given (t, t+ , t’ ) ∈ T (S, n), we have (t, t+ , t’ ) = φ(c) for

+

±S (c) = t0 t0 .’

From now on we assume that char F = 2. The isomorphism (??) can be de¬ned

on the level of group schemes: let G be the group scheme

t, t+ , t’ ∈ O+ (S, n)(R), t(x y) = t’ (x) t+ (y) .

(t, t+ , t’ )

G(R) =

∼

(35.8) Proposition. There exists an isomorphism β : G ’ Spin(S, n) of group

’

schemes.

Proof : By de¬nition we have

t+ 0

(±S — 1R )’1 = c ∈ Spin(S, n)(R),

0 t’

so that ±S induces a morphism

β : G ’ Spin(S, n).

Proposition (??) implies that β(Falg ) is an isomorphism. Thus, in view of Propo-

sition (??), it su¬ces to check that dβ is injective. It is easy to check that

Lie(G) = { (», »+ , »’ ) ∈ o(n) — o(n) — o(n) | »(x y) = »’ (x) y + x »+ (y) }.

On the other hand we have (see §??)

Lie Spin(S, n) = { x ∈ S · S ‚ C0 (S, n) | x + σ(x) = 0 } = [S, S]

where multiplication is in C0 (S, n) (recall that we are assuming that char F = 2

here) and the proof of Proposition (??) shows that that dβ is an isomorphism.

§35. CLIFFORD ALGEBRAS AND TRIALITY 485

Identifying Spin(S, n) with G through β we may de¬ne an action of A3 on

Spin(n):

(35.9) Proposition. The transformations θ + , resp. θ’ induced by

(t, t+ , t’ ) ’ (t+ , t’ , t), (t, t+ , t’ ) ’ (t’ , t, t+ )

resp.

3

are outer automorphisms of Spin(S, n) and satisfy the relations θ + = 1 and

’1

θ+ = θ’ . They generate a subgroup of Aut Spin(S, n) isomorphic to A3 . Fur-

thermore Spin(S, n)A3 is isomorphic to Autalg (C), if S is a para-Cayley algebra C,

and isomorphic to Autalg (A), resp. to Autalg (B, „ ), for a central simple algebra A

of degree 3, resp. a central simple algebra (B, „ ) of degree 3 with an involution of

second kind over K = F [X]/(X 2 + X + 1), if (S, ) is of type 1A2 , resp. of type 2A2 .

Proof : Let R be an F -algebra. It follows from the multiplicativity of the formulas

+ ’

of Proposition (??) that the maps θR and θR are automorphisms of Spin(S, n)(R).

The relations among them also follow from (??). They are outer automorphisms

since they permute the vector and the two half-spin representations of the group

Spin(S, n)(R) (since char F = 2, this also follows from the fact that they act non-

trivially on the center C, see Lemma (??)). Now let (t, t+ , t’ ) ∈ Spin(S, n)A3 (R).

We have t = t+ = t’ and t is an automorphism of SR . Conversely, any auto-

morphism of (S, n)R is an isometry and, since ±(x y) = ±(x) ±(y), ± is proper

by (??).

Let Spin8 = Spin(V, q) where V is 8-dimensional and q is hyperbolic.

(35.10) Corollary. (1) There exists an action of A3 on Spin8 such that SpinA3 8

is split of type G2 .

(2) There exists an action of A3 on Spin8 such that SpinA3 = PGU3 (K) where

8

2

K = F [X]/(X +X +1). In particular, if F contains a primitive cube root of unity,

there exists an action of A3 on Spin8 such that SpinA3 = PGL3 .

8

Proof : Take A = F — Cs , resp. A = M3 (K) in Proposition (??).

As we shall see in Proposition (??), the actions described in (??) and (??) of

Corollary (??) are (up to isomorphism) the only possible ones over Fsep .

Let again (S, n) be a symmetric composition of dimension 8 and norm n. In

view of (??) (and (??)) the group scheme Spin(S, n) ¬ts into the exact sequence

χ

1 ’ µ2 ’ Spin(S, n) ’ O+ (n) ’ 1

’

where χ is the vector representation, i.e.,

(χc )R (x) = cxc’1 for x ∈ SR , c ∈ Spin(S, n)(R).

Let χ : Spin(S, n) ’ PGO+ (n) be the composition of the vector representation χ

with the canonical map O+ (n) ’ PGO+ (n). The kernel C of χ is the center

of Spin(n) and is isomorphic to µ2 — µ2 . The action of A3 on Spin(S, n) restricts

to an action of A3 on µ2 — µ2 . We recall that we are still assuming char F = 2.

(35.11) Lemma. The action of A3 on C µ2 — µ2 is described by the exact

sequence

1 ’ C ’ µ2 — µ2 — µ2 ’ µ2 ’ 1

where A3 acts on µ2 —µ2 —µ2 through permutations and the map µ2 —µ2 —µ2 ’ µ2

is the multiplication map.

486 VIII. COMPOSITION AND TRIALITY

Proof : In fact we have

(35.12)

CR = {(1, 1, 1), = (1, ’1, ’1), = (’1, 1, ’1), = = (’1, ’1, 1)},

0 1 2 01

hence the description of C through the exact sequence. For the claim on the action

of A3 , note that θ+ = (θ’ )’1 maps i to i+1 with subscripts taken modulo 3.

Observe that the full group S3 acts on C and that Aut(C) = S3 .

In view of (??) (and (??)) we have an exact sequence

χ

1 ’ C ’ Spin(S, n) ’ PGO+ (n) ’ 1.

(35.13) ’

(35.14) Proposition. There is an outer action of A3 on PGO+ (n) such that the

maps in the exact sequence (??) above are A3 -equivariant.

Proof : The existence of the A3 action follows from Proposition (??), Lemma (??)

and the universal property (??) of factor group schemes. The action is outer, since

it is outer on Spin(S, n). Observe that for F -valued points the action is the one

de¬ned in Proposition (??).

Let (C, , n) be a para-Cayley algebra over F . The conjugation map π : x ’

Z/2Z

x can be used to extend the action of A3 to an action of S3 = A3

∼

+

on Spin(n) and PGO (n): Let ±C : C(C, n) ’ EndF (C • C), σn ⊥ n be the iso-

’

morphism of Proposition (??). The conjugation map x ’ x is an isometry and

since ±C C(π)±’1 = Int π π , π is improper. For any similitude t with multiplier

0

0

C

µ(t), t = µ(t) πtπ is a similitude with multiplier µ(t)’1 and is proper if and only

’1

if t is proper. Proposition (??) implies that

µ(t)’1 t(x y) = t+ (x) t’ (y)

(1)

µ(t’ )’1 t’ (x y) = t(x) t+ (y)

(2)

µ(t+ )’1 t+ (x y) = t’ (x) t(y)

(3)

hold in (C, ) if t is proper. Let θ + and θ’ be as de¬ned in (??) and (??). We

further de¬ne for R an F -algebra, θ([t]) = [t] for [t] ∈ PGO+ (n)(R), θ(t, t+ , t’ ) =

(t, t’ , t+ ) for (t, t+ , t’ ) ∈ G(R) Spin(n)(R) and θ( 0 ) = 0 , θ( 1 ) = 2 , θ( 2 ) = 1

for i as in (??).

(35.15) Proposition. (1) Let G be Spin(n) or PGO+ (n). The maps θ, θ +

and θ’ are outer automorphisms of G. They satisfy the relations

3 ’1

θ+ = 1, θ+ = θ’ θ+ θ = θθ’ ,

and

and they generate a subgroup of Aut(G) isomorphic to S3 . In particular Out(G)

S3 and Aut(G) PGO+ (n) S3 .

(2) The exact sequence

χ

1 ’ C ’ Spin(C, n) ’ PGO+ (C, n) ’ 1

’

is S3 -equivariant.

Proof : The proof is similar to the proof of Proposition (??) using the above formu-

las (??) to (??). In (??) the action of S3 on C is as de¬ned in (??). The fact that

S3 is the full group Out(G) follows from the fact that the group of automorphisms

of the Dynkin diagram of Spin(C, n), which is of type D4 , is S3 (see §??).

§35. CLIFFORD ALGEBRAS AND TRIALITY 487

Observe that for the given action of S3 on Spin(C, n) we have

Spin(C, n)S3 = Spin(C, n)A3 = Autalg (C).

(35.16)

The action of S3 on Spin8 induces an action on H 1 (F, Spin8 ). We now describe the

objects classi¬ed by H 1 (F, Spin8 ) and the action of S3 on H 1 (F, Spin8 ). A triple of

quadratic spaces (Vi , qi ), i = 1, 2, 3, together with a bilinear map β : V0 — V1 ’ V2

such that q2 β(v0 , v1 ) = q0 (v0 )q1 (v1 ) for vi ∈ Vi is a composition of quadratic

spaces. Examples are given by Vi = C, C a Cayley algebra, β(x, y) = x y, and by

∼

Vi = C, β(x, y) = x y. An isometry ψ : (V0 , V1 , V2 ) ’ (V0 , V1 , V2 ) is a triple of

’

∼

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 -

semilinear with respect to the ¬rst variable and ρ2 -semilinear with respect to the

second variable and such that β1 (x, x) = β(x) for all x ∈ V : the di¬erence β of

two such maps would be such that β(x, x) = 0, hence β(x, y) = ’β(y, x). This

is incompatible with the semilinearity properties of β. Over a separable closure

Fsep of F such a map exists, since L — Fsep is split and, by Proposition (??),

VFsep is of Cayley type. Thus, by descent, such a β1 resp. β2 exists. Furthermore

x —i y = βi (x — y) satis¬es the identities of a cyclic composition, since it does

over Fsep . The last claim follows from Example (??).

It follows from Proposition (??) that the twisted composition (S — L, L, Q, β)

is independent of the choice of a generator of Gal(L/F ).

(36.13) Proposition. Cyclic compositions over F are classi¬ed by the pointed set

H 1 (F, Spin8 Z/3Z).

Proof : In view of Propositions (??) and (??) any cyclic composition over a sepa-

rable closure Fsep of F is isomorphic to a Cayley composition with multiplication —

as described in Example (??) and ρ either given by (x0 , x1 , x2 ) ’ (x1 , x2 , x0 ) or

(x0 , x1 , x2 ) ’ (x2 , x0 , x1 ). It then follows as in the proof of Proposition (??) that

the group of automorphisms of the cyclic composition

C — (F — F — F ), n ⊥ n ⊥ n, F — F — F, ρ, — R

is isomorphic to Spin(ns )(R) Z/3Z. The claim then follows by constructing a

representation G ’ GL(W ) such that Spin(ns ) Z/3Z = AutG (w) as in the proof

of (??), (??). We let this construction as an exercise.

Let (V, L, Q, ρ, —) be a cyclic composition. We have a homomorphism

p : Aut(V, L, ρ, —) ’ Aut(L) = A3

induced by restriction. Assume that p is surjective and has a section s. Then s

de¬nes an action of A3 on (V, L, Q, ρ, —).

496 VIII. COMPOSITION AND TRIALITY

(36.14) Proposition. Let F be a ¬eld of characteristic not equal to 3. For any

faithful ρ-semilinear action of A3 on (V, L, Q, ρ, —), V0 = V A3 carries the structure

of a symmetric composition algebra and V = V0 — L. In particular, if dimL V = 8,

then p has only two possible sections (up to isomorphism) over F sep .

Proof : The ¬rst claim follows by Galois descent and the second from the fact

that over Fsep we have only two types of symmetric composition algebras (The-

orem (??)).

(36.15) Corollary. Datas (V, L, Q, ρ, —), p, s where (V, L, Q, ρ, —) is a cyclic com-

position, are classi¬ed either by H 1 (F, G — A3 ) where G is of type G2 , if the sec-

tion s de¬nes a para-Hurwitz composition, or by H 1 F, PGU3 (K) — A3 where

K = F [X]/(X 2 + X + 1).

Proof : This corresponds to the two possible structures of symmetric composition

algebras over Fsep .

Let (V, L, ρ, —) be a cyclic composition of dimension 8. We write ρ V for the

L-space V with the action of L twisted through ρ and put ρ Q(x) = ρ Q(x) . Let

x (y) = x — y and rx (y) = y — x.

(36.16) Proposition. The map

0 2

x

∈ EndL (ρ V • ρ V ),

x’ x∈V

rx 0

extends to an isomorphism of algebras with involution

2

∼

±V : C(V, Q), „ ’ EndL (ρ V • ρ V ), σQ

’

2

where σQ is the involution associated with the quadratic form Q = ρ Q ⊥ ρ Q. In

particular ±V restricts to an isomorphism

2

∼

C0 (V, Q), „ ’ EndL (ρ V ), σρ Q — EndL (ρ V ), σρ2Q .

’

Proof : The existence of ±V follows from the universal property of the Cli¬ord

2

algebra, taking the identities (x — y) — x = ρ Q(x)y and x — (y — x) = ρ Q(x)y into

account. It is an isomorphism because C(V, Q) is central simple over F . The

fact that ± is compatible with involutions is a consequence of the formula (??)

above.

Proposition (??) is a twisted version of Proposition (??) and can be used to

deduce analogues of Propositions (??) and (??). The proofs of the following two

results will only be sketched.

(36.17) Proposition. Let t be a proper similitude of (V, Q), with multiplier µ(t).

There exist proper similitudes u, v of (V, Q) such that

µ(v)’1 v(x — y) = u(x) — t(y),

µ(u)’1 u(x — y) = t(x) — v(y),

µ(t)’1 t(x — y) = v(x) — u(y)

for all x, y ∈ V . If t ∈ RL/F O+ (V, Q) (F ) is such that Sn(t) = 1, then u, v can

be chosen in RL/F (O+ (V, Q))(F ).

§36. TWISTED COMPOSITIONS 497

2

Proof : We identify C(V, Q) with EndL (ρ V • ρ V ), σQ through ±V . The map

0 2

tx

∈ EndL (ρ V • ρ V )

x’ ’1

µ(t) rtx 0

extends to an automorphism t of C(V, Q) whose restriction to C0 (V, Q) is C0 (t).

Thus we may write t = Int u v where u , v are similitudes of Q and

0

0

’1

u 0 0 u 0 0

x tx

= ’1

0 v rx 0 0 v µ(t) rtx 0

or

and v (y — x) = u (y) — t(x)µ(t)’1 .

u (x — y) = t(x) — v (y)

Putting u = u µ(u )’1 and v = v and using that

2

µ(u)’1 = µ(u ) = ρµ(t) ρ µ(v)

gives the ¬rst two formulas. The third formula follows by replacing x by y — x in

the ¬rst one (compare with the proof of Proposition (??)).

(36.18) Proposition. Assume that char F = 2. For R ∈ Alg F we have

RL/F Spin(V, Q) (R)

3

{ (v, u, t) ∈ RL/F O+ (V ) (R) | v(x — y) = u(x) — t(y) },

the group A3 operates on Spin(V, Q), and Spin(V, Q)A3 is a group scheme over F

such that RL/F [Spin(V, Q)A3 ]L = Spin(V, Q).

Proof : Proposition (??) follows from Proposition (??) as Proposition (??) follows

from Proposition (??).

The computation given above of the Cli¬ord algebra of a cyclic composition can

be used to compute the even Cli¬ord algebra of an arbitrary twisted composition:

(36.19) Proposition. Let “ = (V, L, Q, β) be a twisted composition. There exists

an isomorphism

±V : C0 (V, Q) = C EndL (V ), σQ ’ ρ EndL (V ) — ∆

where ∆(L) is the discriminant algebra of L and ρ is a generator of the group

Gal L — ∆(L)/L A3 .

Proof : By Proposition (??) there exists exactly one cyclic composition (with re-

spect to ρ) on (V, L, Q, β)—∆ and by Proposition (??) we then have an isomorphism

ρ2

∼

±V —∆ : C0 (V, Q) — ∆ ’ EndL—∆ ρ (V — ∆) — EndL—∆

’ (V — ∆)

of L — ∆-algebras with involution. By composing with the canonical map

C0 (V, Q) ’ C0 (V, Q) — ∆

2

and the projection ρ EndL (V ) — ∆ — ρ EndL (V ) — ∆ ’ ρ

EndL (V ) — ∆ we

obtain a homomorphism of central simple algebras

±V : C0 (V, Q) = C EndL (V ), σQ ’ ρ EndL (V ) — ∆

which is an isomorphism by dimension count.

498 VIII. COMPOSITION AND TRIALITY

36.C. Twisted Hurwitz compositions. In this section we ¬rst extend the

construction of a twisted composition C — L given in Example (??) for L cyclic to

an arbitrary cubic ´tale algebra L and a para-Hurwitz algebra (C, , n). We refer

e

to ?? for details on cubic ´tale algebras.

e

Let ∆ be the discriminant (as a quadratic ´tale algebra) of L, let ι be the

e

generator of Gal(∆/F ) and let C be a Hurwitz algebra over F . Since L — ∆ is

cyclic over ∆, by Proposition (??) there exists a cyclic composition (x, y) ’ x — y

on C — L — ∆ for each choice of a generator ρ of Gal(L — ∆/∆). The automorphism

˜ = π — 1 — ι of C — L — ∆ is ι-semilinear and satis¬es ˜2 = 1 and ˜(x — y) = ˜(y) —˜(x)

ι ι ι ι ι

for x, y ∈ C — L — ∆. Let

V = { x ∈ C — L — ∆ | ˜(x) = x }

ι

be the corresponding descent (from L—∆ to L). Since ˜(x—y) = ˜(y)—˜(x), the map

ι ι ι

β(x) = x—x restricts to a quadratic map β on V . The restriction Q of n—1—1 to V

takes values in L and is nonsingular. Thus (V, L, Q, β) is a twisted composition.

We write it “(C, L). A twisted composition similar to a composition “(C, L) is

called a twisted Hurwitz composition over L. If C is a Cayley algebra we also say

that “(C, L) is a twisted composition of type G2 . The underlying quadratic space

(V, Q) of “(C, L) is extended from the quadratic space (V0 , Q0 ) over F with

V0 = { x ∈ C — ∆ | π — ι(x) = x }

and Q0 is the restriction of Q to V0 . The space (V0 , Q0 ) is called the ∆-associate

of (C, n) by Petersson-Racine [?] (see also Loos [?]) and is denoted by (C, N )∆ . In

[?] a K-associate (U, q)K is attached to any pointed quadratic space (U, q, e) and

any ´tale quadratic algebra K:

e

(U, q)K = { x ∈ U — K | π — ιK (x) = x }

where π is the re¬‚ection with respect to the point e. For any pair of quadratic ´tale

e

algebras K1 , K2 with norms n1 , n2 , (K1 , n1 )K2 = (K2 , n2 )K1 is the ´tale quadratic

e

algebra

K1 — K2 = { x ∈ K1 — K2 | (ι1 — ι2 )(x) = x }.

Recall that K — K F — F.

(36.20) Lemma. Let (U, p) be a pointed quadratic space. There are canonical

isomorphisms

(U, p)K1 (U, p)K1 —K2 (U, p)F —F (U, p).

and

K2

Reference: See Loos [?].

In what follows we use the notation [a] for the 1-dimensional regular quadratic

form q(x) = ax2 (and, as usual, a for the bilinear form b(x, y) = axy, a ∈ F — ).

(36.21) Lemma. Let (C, n) be a Hurwitz algebra (with unit 1) and let (C 0 , n0 ) =

1⊥ be the subspace of (C, n) of elements of trace zero. We have, with the above

notations, (V0 , Q0 ) = (C, n)∆ and :

(1) If char F = 2, V0 = F · 1 • C 0 , Q0 = [1] ⊥ δ — (C 0 , n0 ).

(2) If char F = 2, let u ∈ C, ξ ∈ ∆ with TC (u) = 1 = T∆ (ξ) and let w =

u — 1 + 1 — ξ. Then V0 = F · w • C 0 , Q0 (w) = n(u) + ξ 2 + ξ, bQ0 (w, x) = bn (x, u)

and Q0 (x) = n(x) for x ∈ C 0 .

§36. TWISTED COMPOSITIONS 499

Proof : Claims (??) and (??) can be checked directly. The ¬rst claim then follows

easily.

(36.22) Proposition. Two twisted Hurwitz compositions “(C1 , L1 ) and “(C2 , L2 )

are isomorphic if and only if L1 L2 and C1 C2 .

Proof : The if direction is clear. We show the converse. If the compositions

“(C1 , L1 ) and “(C2 , L2 ) are isomorphic, we have by de¬nition an isomorphism

∼

L1 ’ L2 and replacing the L2 -action by the L1 -action through this isomorphism,

’

we may assume that L1 = L2 = L. In particular the quadratic forms Q1 and Q2

are then isometric as quadratic spaces over L. By Springer™s theorem there exists

an isometry (C1 , n1 )∆ (C2 , n2 )∆ . Since ∆ — ∆ F — F , (C1 , n1 ) (C2 , n2 )

follows from Lemma (??), hence C1 C2 by Theorem (??).

Let G be the automorphism group of the split Cayley algebra. Since G =

SpinS3 (see ??), we have a homomorphism

8

G — S3 ’ Spin8 S3 .

(36.23) Proposition. Twisted compositions of type G2 are classi¬ed by the image

of H 1 (F, G — S3 ) in H 1 (F, Spin8 S3 ).

Proof : A pair (φ, ψ) where φ is an automorphism of C and ψ is an automorphism

of L de¬nes an automorphism “(φ, ψ) of “(C, L). The map

H 1 (F, G — S3 ) = H 1 (F, G) — H 1 (F, S3 ) ’ H 1 (F, Spin8 S3 )

corresponds to [C] — [L] ’ [“(C, L)].

The following result is due to Springer [?] for L cyclic and V of dimension 8.

(36.24) Theorem. Let (V, L, Q, β) be a twisted composition and let N be the cubic

form N (x) = Q x, β(x) , x ∈ V . The following conditions are equivalent:

(1) The twisted composition (V, L, Q, β) is similar to a twisted Hurwitz composition.

(2) The cubic form N (x) is isotropic.

(3) There exists e ∈ V with β(e) = »e, » = 0, and Q(e) = »# .

Proof of (??) ’ (??): We may assume that (V, L, Q, β) is a Hurwitz composition.

The element x = c — 1 — ξ ∈ C — L — ∆ lies in V if tC (c) = 0 = t∆ (ξ). For such

an element we have β(x) = c2 — 1 — ξ 2 ∈ L. Since tC (c) = 0, c2 = nC (c) and so

N (x) = 0 if char F = 2. If char F = 2, x = 1 — 1 — 1 lies in V , β(x) = x, and

N (x) = bQ (x, x) = 0.

For the proof of (??) ’ (??) we need the following lemma:

(36.25) Lemma. Any element x of a twisted composition (V, L, Q, β) satis¬es the

identity

β 2 (x) + Q(x)β(x) = N (x)x.

Proof : Since over the separable closure Fsep any composition is cyclic, it su¬ces to

prove the formula for a cyclic composition and for β(x) = x—x. In view of Formulas

(??) and (??) (p. ??), we have

(x — x) — (x — x) + (x — x) — x — x = ρ bQ (x — x, x) x

hence the assertion, since (x—x)—x = ρ2 Q(x) x (and the product — is ρ-semilinear

with respect to the ¬rst variable).

500 VIII. COMPOSITION AND TRIALITY

Proof of (??) ’ (??): First let x = 0 be such that β(x) = 0 and N (x) = 0. If

Q(x) = 0 we take e = β(x) and apply Lemma (??). If Q(x) = 0, we replace x by

β(x) and apply (??) to see that our new x satis¬es Q(x) = 0 and β(x) = 0. Thus

we may assume that Q(x) = 0 and β(x) = 0. Let y ∈ V be such that Q(y) = 0 and

bQ (x, y) = ’1. Let

β(u, v) = β(u + v) ’ β(u) ’ β(v).

We claim that

(36.26) β β(x, y) = β(x, y) + f · x

for some f ∈ F . We extend scalars to L — ∆ and so assume that β(x) = x — x is

cyclic. It su¬ces to check that f ∈ ∆ © L. Since β(x, y) = x — y + y — x, we have

β β(x, y) = (x — y + y — x) — (x — y + y — x).

Applying formulas (??) and (??) (p. ??), we ¬rst obtain

(y — x) — x + (x — x) — y = (y — x) — x = ’x

x — (x — y) + y — (x — x) = x — (x — y) = ’x

since β(x) = x — x = 0 and bQ (x, y) = ’1. Applying again formulas (??) and (??)

then give

(x — y) — (x — y) = ρ bQ (x — y, y) x + y — x

(x — y) — (y — x) = 0

(y — x) — (x — y) = ρ bQ (y — x, y) x

(y — x) — (y — x) = ρ2 bQ (y, y — x) x + x — y

and

β β(x, y) = β(x, y) + f · x

with

f = ρ bQ (x — y, y) + ρ bQ (y — x, y) + ρ2 bQ (y, y — x)

= ρ2 bQ β(y), x + bQ β(y), x + ρ bQ β(y), x = TL—∆/∆ bQ β(y), x

hence f ∈ ∆ © L. Let µ ∈ L be such that TL (µ) = f ; we claim that e = µx + β(x, y)

satis¬es β(e) = e:

β(e) = β(µx) + β β(x, y) + β µx, β(x, y)

= β(x, y) + f · x + β µx, β(x, y) .

To compute β µx, β(x, y) we assume that β(x) = x — x is cyclic. We have:

β µx, β(x, y) = (µx) — (x — y + y — x) + (x — y + y — x) — (µx)

= ρ(µ)[x — (x — y + y — x)] + ρ2 (µ)[(x — y + y — x) — x]

= ’ρ(µ)x ’ ρ2 (µ)x = (µ ’ f )x.

This implies that β(e) = β(x, y) + µx = e. The relation »# = Q(e) follows from

β(e) = »e and Lemma (??), since replacing β(e) by »e we see that »# »e+Q(e)»e =

2Q(e)»e.

§36. TWISTED COMPOSITIONS 501

Proof of (??) ’ (??): We ¬rst construct the Hurwitz algebra C. Replacing β by

the similar composition »’1 β, we may assume that β(e) = e. This implies that

Q(e) = 1. Let ∆ be the discriminant algebra of L. We also call β the extension

β — 1 of β to L — ∆. Let ρ1 , ρ2 be the generators of Gal(L — ∆/∆) and let ι be

the generator of Gal(∆/F ), so that (putting ι = 1 — ι on L — ∆ and V — ∆), we

have ρ2 = ρi+1 and ιρi = ρi+1 ι. Let βi (x, y) = x —i y be the two unique extensions

i

of β to L — ∆ as cyclic compositions with respect to ρi . By uniqueness we have

(see Examples (??) and (??))

β1 (x, y) = β2 (y, x) and ιβ1 (x, y) = β2 (ιx, ιy).

Let π : V ’ V be the hyperplane re¬‚ection

π(x) = x = ’x + bQ (e, x)e

∼

as well as its extension to V — ∆ and let •i : V — ∆ ’ V — ∆ be the ρi -semilinear

’

map given by •i (x) = βi (x, e).

(36.27) Lemma. The following identities hold for the maps •i :

(1) •i π = π•i

•2 = •i+1

(2) i

(3) ι•i ι = •i+1

Proof : (??) We have β1 (x, e) = ’β1 (x, e) + ρ1 bQ (e, x) e so that

β1 (x, e) = β1 (x, e) ’ bQ e, β1 (x, e) e

’ ρ1 bQ (e, x) e + bQ e, ρ1 bQ (e, x) e e = β1 (x, e)

(??) By (??)

β1 β1 (x, e), e = β1 β1 (x, e), e = ρ2 bQ (e, x) e ’ β1 (e, x) = β1 (e, x)

using that β1 β1 (x, y), z + β1 β1 (z, y), x = ρ2 bQ (x, z) y, a formula which follows

from (??) (p. ??).

(??) follows from ιβ1 (x, y) = β2 (ιx, ιy).

We next de¬ne a multiplication γ1 on V — ∆ by

γ1 (x, y) = β1 •2 (x), •1 (y) .

(36.28) Lemma. The multiplication γ1 satis¬es the following properties:

(1) Q γ1 (x, y) = Q(x)Q(y)

(2) γ1 (x, e) = γ1 (e, x) = x

(3) γ1 (x, y) = γ1 (y, x)

(4) ιγ1 (x, y) = γ1 (ιy, ιx)

(5) γ1 •1 (x), •1 (y) = •1 γ1 (x, y)

Proof : (??) We have

Q γ1 (x, y) = Q β1 •2 (x), •1 (y) = ρ1 Q •2 (x) ρ2 Q •1 (y)

= ρ1 ρ2 Q(x) ρ2 ρ1 Q(y) = Q(x)Q(y)

(??) We have γ1 (x, e) = β1 β2 (x, e), e = x and similarly γ1 (e, x) = x.

502 VIII. COMPOSITION AND TRIALITY

(??) In view of (??) and (??) γ1 is a Hurwitz multiplication with conjugation

x ’ x and (??) holds for any multiplication of a Hurwitz algebra, see (??).

(??) We have

ιγ1 (x, y) = ιβ1 •2 (x), •1 (y) = β2 ι•2 (x), ι•1 (y)

= β2 •1 ι(x), •2 ι(y) = β1 •2 ι(y), •1 ι(x) = γ1 (ιy, ιx)

(??) Using (??) we have

•1 γ1 (x, y) = β1 γ1 (y, x), e = β1 β1 •2 (y), •1 (x) , e

= ρ2 bQ •2 (y), e •1 (x) ’ β1 β1 e, •1 (x) , •2 (y)

= β1 β1 e, •1 (x) , bQ •2 (y), e e ’ β1 β1 e, •1 (x) , •2 (y)

= β1 β1 e, •1 (x) , •2 (y) = β1 •2 •1 (x), •2 (y)

= γ1 •1 (x), •1 (y)

We go back to the proof of Theorem (??). It follows from Lemma (??) that

γ1 ιπ(x), ιπ(y) = ιπ γ1 (x, y)

Since ιπ•1 = •2 ιπ, the automorphisms {•1 , ιπ} of V — ∆ de¬ne a Galois action

1

of the group Gal(L — ∆/F ) = S3 on the Hurwitz algebra V — ∆. Let C be the