Octonions, Jordan algebras, and exceptional groups

= x + (
Consequently, G D is 3-dimensional and connected.

Proof. Use equation (2.1) with u

= id and the associativity of D.

0

G is contained in the stabilizer of e in the orthogonal group O(N), which also leaves the orthogonal complement e.l in CK invariant. We will need that restriction of the stabilizer of e in SO(N) to e.l yields an isomorphism with the rotation group SO(Nl ), where Nl is the restriction of N to e.l. In characteristic =I 2 this is an almost trivial matter, but in characteristic 2 there are complications due to the fact that then e E e.l. For the purposes of the present chapter we could do with less, but in the next chapter we will use the full isomorphism result. Let F be the stabilizer of e in O(N), and SF = F n SO(N). These are closed subgroups of O(N) and SO(N), respectively. We denote by F and SF the corresponding groups over k. If t E F, it leaves e.l invariant and its restriction to e.l is an element of the orthogonal group O(Nt}. If char(k) = 2 the quadratic form Nl is degenerate. Then O(Nt} and O(N1 ) are defined to be groups of linear isometries of e.l, as in §1.1. In this case we put SO(Nt} = O(Nl), SO(Nt} = O(Nt}. Proposition 2.2.2 The restriction homomorphism

is an isomorphism of algebraic groups which induces an isomorphism SF SO(Nt}.

Proof. It is evident that e is a homomorphism of algebraic groups. If char(k) =I 2, define for u E SO(Nl ) its extension e(U) to C by e(U)(e) = e. Clearly, e: SO(Nt} - SF is a homomorphism of algebraic groups and e = e- l , so e is an isomorphism of algebraic groups. The proposition follows. In characteristic 2 the extension of u E O(Nt} to an element of SF cannot be defined in this way: then e E e.l, whereas we have to define the image of a vector outside e.l. We will follow a different approach in this case. So let char( k) = 2. Note that in this case the restriction Nl is a quadratic form having defect 1, i.e. dim V.l = 1 (cf. [Che 54, § 1.2] or [Dieu, Ch. I, § 16]). Every element of O(Nt} is a product of orthogonal transvections or reflections Sa, N(a) =I 0 of the form (1.1) (see [Dieu, Ch. II, § 11]). The algebraic group O(Nl ) is connected; this can be seen in the following manner. The restriction of the polynomial function N - 1 to e.l is irreducible (over the algebraic closure K), since otherwise Nl would be a square of a linear polynomial, which would vanish on a 6-dimensional subspace of e.l. It would

28

2. The Automorphism Group of an Octonion Algebra

follow that there were 6-dimensional totally isotropic subspaces with respect to the nondegenerate quadratic form N in 0, which is impossible. Hence S = {x E el.I N(x) = I} (over K) is an irreducible variety. The morphism

s -+ O(Nl ), a

H

Sa,

maps S onto an irreducible set of generators of O(Nl ) containing the identity, so O(Nl ) is a connected algebraic group (see [Hu, § 7.5J or [Sp 81, Prop. 2.2.6]). The quadratic form N on 0 is nondegenerate, so here we have a rotation subgroup SO(N). Rotations are now orthogonal transformations with Dickson invariant 0; every rotation is a product of an even number of reflections (see [Dieu, Ch. II, § 10]). The restriction homomorphism

is surjective, since every orthogonal transformation of el. leaves e invariant and can be written as a product of reflections Sa with a E el.. Its kernel has order 2, it is the subgroup generated by the reflection Se. This can be seen by using a symplectic basis el = e, ... , es of 0 as in (1.30); the ei with i # 5 form a basis of el.. Since Se is not a rotation, the restriction homomorphism SF -+ O(Nl ) is an isomorphism of abstract groups. Clearly, e is a homomorphism of algebraic groups. To show that it is an isomorphism of algebraic groups, it suffices by [Sp 81, Cor. 5.3.3J, to prove that the Lie algebra homomorphism de : L(SF) -+ L(O(Nl)) is an isomorphism. The Lie algebra L(SF) is contained in the space S of the linear transformations t : OK -+ OK satisfying

{x,tx)=O (xEOK)andte=O. To see this we work over the ring of dual numbers K[eJ, e2 = 0 (see [Bor, Ch. I, 3.20]). The elements of L(SF) are linear maps t of OK with te = 0 such that id + et, acting on x + ey E K[eJ ® 0, satisfies N«id + d)(x + ey)) = N(x + ey). It follows that (x, tx) = O. Writing this out for the matrix (O!ij ) of t with respect to the above symplectic basis, we find 43 independent equations: O!i,j+4

+ O!j,iH = 0

=0 O!l,; = 0

O!i,H4

(i # j, indices mod 8), (i # 1), (allj).

So S has dimension 21. As L(SO(Nl» also has dimension 21, we have S = L(SO( N l The Lie algebra homomorphism de acts on the matrices of L(SF)

».

2.2 Connectedness and Dimension of the Automorphism Group

29

by deleting the fifth row and the fifth column. Since aj,5 = a1,j+4 for j '" 1, = aj,l for j '" 1 and a1,1 = as,s = 0, de is injective and hence is an 0 isomorphism. The last point is clear.

a5,j+4

Proposition 2.2.3 The automorphism group G of CK is a connected algebmic group of dimension 14.

Proof. Consider the set of special pairs in CK (see Def. 1.7.4). Since K is algebraically closed, special pairs certainly exist. For any two special pairs there exists a linear isometry t of C K fixing e and mapping one pair to- the other, by Witt's Theorem; clearly, we may assume that t is a rotation. So the group SF operates transitively on the set of special pairs. Fix one special pair a, b with a, bE DK, a quaternion subalgebra of CK. The stabilizer H of a, b in SF is isomorphic to the special orthogonal group SO(N2 ) of a five-dimensional quadratic form N 2 , under the isomorphism e of SF with SO(N1)' If char(k) '" 2, this isomorphism is restriction to (Ke61 Ka $ Kb).L; that this is an isomorphism of algebraic groups is seen as in the first paragraph of the proof of Prop. 2.2.2. If char(k) = 2, we restrict in two steps: first to (Ke61Ka).L (same argument as in the characteristic '" 2 case), from there to (Ke $ Ka).L n b.L = (Ke $ Ka $ Kb).L (argument as in the characteristic 2 case in the proof of Prop. 2.2.2). The homogeneous space X = SF /H can be identified as a set with the set of all special pairs. Counting dimensions we find dim X

= dimSF/H = dimSO(N1) -

dimSO(N2 )

= 21-10 = 11.

From Cor. 1.7.5 we infer that G operates transitively on the set of special pairs, i.e., on X. Since e, a, band ab form a basis of DK, the stabilizer in G of the special pair a, b is G D. This implies that X ~ G/GD' So another dimension count yields dimG

= dim X + dimG D = 11 + 3 = 14.

As a homogeneous space of a rotation group, X is an irreducible algebraic variety. Since GD is connected, it follows that G is connected (see [Sp 81, Ex. 5.5.9 (1)]). 0 The transitivity of G on special pairs implies transitivity on the set of quaternion subalgebras, so Corollary 2.2.4 The automorphism group G of CK acts tmnsitively on the set of quaternion subalgebras.

(This also follows from Cor. 1.7.3, since over an algebraically closed field K all quaternion algebras are isomorphic to M 2 (K).) We remarked at the beginning of this chapter that G is contained in O(N). The connectedness of G implies that it must be contained in the

30

2. The Automorphism Group of an Octonion Algebra

connected component of the identity in O(N), which is SO(N). This implies the following corollary. Corollary 2.2.5 The automorphism group G is a subgroup of the rotation

group SO(N).

2.3 The Automorphism Group is of Type G 2 In this section we are going to further determine the structure of the algebraic group G. We view the split octonion algebra CK as an algebra of pairs (x, y) with x, y E DK, the algebra of 2 x 2 matrices over K, and with N«x, y» = det(x) - det(y)j see § 1.8. We begin by describing a maximal torus in G. To this end we consider (2.2) for the particular case where c and p are diagonal matrices. We denote the 2 x 2 diagonal matrix diag(lt, 1t- 1 ) with It E K* by c/(,.

Lemma 2.3.1 The group T consisting of the automorphisms

t.x,,,, : x + ya f-+ c.xxC~1

+ (c",yc~1)a (x, y E DK)'

(2.3)

with A, J,L E K*, is a 2-dimensional torus in G. Every element of G that commutes with all elements of T lies in T, so T is a maximal torus. Hence G has rank 2. Proof. For a E DJ. we take (0, e) with e the 2 x 2 identity matrix. Notice that (x,O)a = (0, x). As a basis of CK we take (ell'O), (e12'0), (e21'0), (e22'0), (0, ell), (O,e12), (0,e2t), (O,e22), where eij denotes the 2 x 2 matrix with (i, j)-entry equal to 1 and the other entries O. The matrix of t.x,,,, with respect to this basis is

Observe that L.x,_", = t.x,,,,, so the obvious parametrization by A and J,L is not one-to-one. We can reparametrize via = AJ,L and 11 = AJ,L-l; this gives an isomorphism of algebraic groups from (K*)2 onto T, viz.,

e

This shows that T is a 2-dimensional torus. We now prove that if t E G commutes with all t.x,,,,, then it lies in T. Any such t leaves the eigenspaces of every t.x,,,, invariant. From (2.4) we see that these eigenspaces are

2.3 The Automorphism Group is of Type G2

31

if we choose A, J.I. E K* in such a way that t).,~ has seven distinct eigenvalues. This implies that t must leave DK invariant, so it has the form given by (2.2):

with c,p E DK, det(c) = det(p) = 1. Since t has to leave the eigenspaces of every t).,~ invariant, its restriction to DK , i.e., the mapping

has to leave K e12 and K e21 invariant. This implies that ce12 E K e12C and ce21 E K e21 c, so c = c, for some ~ E K*. In a similar way, the invariance of all K eij under the mapping

implies that p is diagonal. Thus we have proved that t K*.

= t,,~ for some ~,K. E 0

It is easy now to determine the center of G.

Lemma 2.3.2 The center of G consists of the identity only. Proof. Since a central element commutes with all elements of the maximal torus T, it must lie in T, so it has the form

using the notation of the previous lemma. It must commute with all automorphisms t that leave DK invariant, which by (2.2) have the form

t : x + ya

t--+

cxc- 1 + {pcyc- 1)a

(x, y E D K )

with C,p E DK, N(p) = 1. This implies that t).,~IDK has to commute with all inner automorphisms of DK, the full 2 x 2 matrix algebra over K. This can only happen if >.2 = 1, so t).,~IDK = id. Since this holds for every quaternion subalgebra of CK and since every element of CK is contained in a quaternion 0 subalgebra by Prop. 1.6.4, we conclude that t).,~ = id. In the next theorem, C may be an arbitrary composition algebra. In fact, the proof for octonion algebras uses the case of quaternion algebras, which we therefore have to settle first.

Theorem 2.3.3 Let C be any composition algebra over k. The only nontrivial invariant subspaces of CK under the action of Aut{CK) are Ke and el..

32

2. The Automorphism Group of an Octonion Algebra

Proof. If dim( CK) = 2, the result is trivial if K has characteristic ::j: 2. If char(K) = 2, we have Ke = e.L. Let a ¢ Ke with (e,a) = 1; set N(a) = o. The element e+a satisfies the same conditions, so by Prop. 1.2.3 it satisfies the same minimum equation as a. It follows that there exists an automorphism which carries a to e + a, so a cannot span another invariant subspace. Next we deal with the case that CK is of dimension 4, so C K = M(2, K). Every automorphism of this algebra is inner by the Skolem-Noether Theorem (see § 2.1). Let V be an invariant subspace of CK and let t E M(2, K) be a diagonal matrix with distinct nonzero eigenvalues. Then the eigenspaces of the linear map x f-+ txt- l of M(2,K) are Ken + Ke22, Ke12, Ke21 (where the eij are as in the proof of Lemma 2.3.1). So V is spanned by vectors of the form oen + f3e22 and mUltiples of el2 and e21. If el2 E V then for all E K

>.

10) 1 el? (10) ->. 1 ( >.

= (->. _>.2 >.1)

lies in V, whence e21 E V, en - e22 E V. It follows that then V :::> e.L. If oen + f3e22 E V and 0 ::j: f3 then similar arguments give that e12, e21 E V. So if el2 and e21 do not lie in V, we must have V = K e. This proves the Theorem for quaternion algebras. Finally, let CK be an octonion algebra over K. Consider any subspace V of CK that is invariant under Aut(CK), and any quaternion subalgebra D 1 • Let VI = V n DI. Every automorphism of DI can be extended to an automorphism of CK by Cor. 1. 7.3. So VI is invariant under Aut(Dt}, whence VI = 0, K e, e.L n DI or D1 . Let D2 be another quaternion suba.lgebra of CK; call V n D2 = V2 • By Cor. 2.2.4 there exists an automorphism cp of CK carrying DI to D 2. We must have cp(V1) = V2. Consequently, V2 = 0, Ke, e.L n D2 or D2 according to whether VI = 0, Ke, e.L n DI or D 1. Since by Prop. 1.6.4 every element of CK is contained in a quaternion subalgebra, it 0 follows that V = 0, Ke, e.L or CK. Focussing attention on octonion algebras again, we derive the following important corollary to the above theorem. Corollary 2.3.4 The automorphism group G of the octonion algebra CK over K has a faithful irreducible representation. Proof. If char(K) ::j: 2, the 7-dimensional representation of Gin e.L is faithful and irreducible. If char(K) = 2, this is not true anymore, since in that case K e is an invariant subspace contained in e.L. But then the representation of G in the 6-dimensional quotient space e.L / K e is irreducible, and it is not hard to verify that it is faithful. 0 We can now identify G. Theorem 2.3.5 The algebraic group G defined by an octonion algebra C is a connected, simple algebraic group of type G2 •

2.4 Derivations and the Lie Algebra of the Automorphism Group

33

Proof. We continue to work in the algebra CK over the algebraically closed field K. From Prop. 2.2.3 and Lemma 2.3.1 we know that G is a connected linear algebraic group of dimension 14 and rank 2. By the previous Corollary G has a faithful irreducible representation. This implies that G is reductive (see [Sp 81, Ex. 2.4.15]). As the center of G is trivial by Lemma 2.3.2, G is semisimple. From Prop. 2.2.3, Lemma 2.3.1 and [Sp 81, 8.1.3J we deduce that the two-dimensional root system of G has 12 elements. It then must be irreducible (the only reducible two-dimensional root system has 4 elements), and must be of type G 2 (see [Sp 81, 9.1J or [Bour, p.276]). 0 In Prop. 2.4.6 we will see that k is a field of definition of the algebraic group G = Aut(C) if C is an octonion algebra over k.

2.4 Derivations and the Lie Algebra of the Automorphism Group Let C be an algebra over k with identity element e. A derivation of C is a linear map of C such that

d(xy) = x.d(y)

+ d(x).y (x, y E C).

Taking x = y = e it follows that d( e) check shows that their commutator

[d, d'J

= O. If d, d'

= dod' -

are derivations an easy

d' 0 d

is also one. It follows that the derivations of C form a Lie algebra Der(C) or Derk(C), Our aim is to prove that if C is an octonion algebra, DerK{C) is the Lie algebra L{G), where G is as in the previous section. (From this it will follow that G is defined over k, see Prop. 2.4.6). For more on derivations of octonion algebras see [Ja 71, §2J. Let C be an arbitrary composition algebra. To deal with Der{ C) we use the doubling process of § 1.5. Let D and a be as in [loc.cit.]. If d E Der{C) there are linear maps do and d 1 of D such that for xED

d(x)

= do(x) + d1{x)a.

Lemma 2.4.1 (i) do E Der(D). (ii) For X,y E D we have d1{xy)

= d1{x)y + d1{y)x.

Proof. This follows by a straightforward computation, using the multiplica,tion rules (1.22) and the definition of a derivation. 0 If D is a two- or four-dimensional composition algebra denote by S the space of linear maps d 1 of D with the property of part (ii) of the lemma.

34

2. The Automorphism Group of an Octonion Algebra

Lemma 2.4.2 dimS

~

2

if dimD = 2

and dimS ~ 8

if dimD = 4.

Proof. As in the case of derivations, we have d 1 (e) = 0 for d 1 E S. If dim D = 2 this implies the stated inequality. If dim D = 4 then D is generated by two elements a,b (see Cor. 1.6.3), and d 1 E S is completely determined by d1(a) and d1(b). The inequality follows. 0

C and D being as before, denote by Dero(C) the space of derivations whose restriction to D is zero.

Lemma 2.4.3 dimDero(C) :5 1 (respectively, :5 3) if dimC = 4 (respectively, 8). Proof. Let d E Dero (C). Then d is determined by da. From the relations = A, ax = xa (x E D) we deduce that

a2

(da)a + a(da)

= 0,

(da)x - x(da)

= O.

(2.5)

Write da = b + ca. From the second relation we deduce that bx + xb = 0 for all xED. If dim D = 2 then D is commutative, and (x + x)b = (x, e)b = 0 for all xED, whence b = O. If dimD = 4 then

b(xy)

= -(yx)b = -y(xb) = y(bx) = -(by)x = -b(yx).

So b annihilates all elements of the form xy + yx. We can conclude that b = 0 by producing an invertible element of this form, and it suffices to do this over the algebraic closure K. But then D is isomorphic to the algebra of 2 x 2-matrices over K, and we may take x = e12, Y = e21. We have now shown that da = ca and from the first relation (2.5) we find that c + C = O. The lemma follows. 0

Lemma 2.4.4 Der(C) = 0 if C is two-dimensional and dimDer(O) :5 3 (respectively, :5 14) if C has dimension 4 (respectively, 8). Proof. If dim C

= 2, then C = k[aJ. Using (1. 7) we find that if d is a derivation 2a(da) - (a, e)da

= O.

If char k ~ 2 we may assume that (a, e) = 0 and otherwise that (a, e) = 1. In both cases da = 0 and d = O. Now let dim C > 2. We use the doubling process. D being as before, it follows from Lemma 2.4.1 that we have an exact sequence 0-. Dero(C) -. Der(O) -. Der(D) $ S. Hence dim Der(C) ~ dim Dero(C)

+ dim Der(D) + dim S.

2.5 Historical Notes

35

If dimC = 4 using Lemmas 2.4.2 and 2.4.3 we find that dim Der(C) :5 3. (In fact we have equality since, as is easily seen, the derivation algebra contains the three-dimensional space of inner derivations x ....... ax - xa. But we are mainly interested in the octonion case.) If C is an octonion algebra the same 0 argument gives the desired inequality.

We can now prove the results announced in the beginning.

Proposition 2.4.5 Let C be an octonion algebra. Then dim Derk(C) and DerK(C) is the Lie algebra olG.

= 14

Proof. Since DerK(C) = K ®k Derk(C) and dimL(G) = dimG = 14 (see Th. 2.3.5) it suffices to prove the second point. Let V = End(CK®KCK, CK) be the space of K-linear maps ofCK®KCK to CK. Denote by /I. the multiplication map x®y ....... xy. The algebraic group of automorphisms G = Aut(CK) is the stabilizer of /I. in the algebraic group GL(CK), under the obvious action of GL(CK) on V, and the Lie algebra L(G) consists of derivations of CK (see [Hu, p.77Jj one can also see this by working with the ring of dual numbers K[f], as before in the proof of Prop. 2.2.2). Since dimL(G) = dimG = 14 (by Prop. 2.2.3), it follows from Lemma 2.4.4 that we have the desired equality. 0 As a consequence of the proposition we have:

Proposition 2.4.6 Let C be an octonion algebra over k. Then the automorphism group G is defined over k. Proof. We use the notations of the previous proof. V(k) = Endk(C ®k C) is a k-structure on V in the sense of [Sp 81, §l1J, /I. E V(k) and the algebraic group GL( CK) is defined over k (by [loc.cit., 2.1.5]). Let cp : G -+ V be the morphism 9 ....... 9./1.. It is defined over k. To prove that the stabilizer G is defined over k it suffices by [loc.cit, 12.1.2J to prove that the kernel of the differential dcpe has dimension 14. But this kernel is precisely Der( C K ), as follows from [Hu, p. 77J. An application of the previous Proposition proves 0 the assertion.

2.5 Historical Notes

E. Cartan remarked in rCa 14J that the compact real exceptional simple Lie group of type G2 can be realized as the automorphism group of the real octonionsj see [Fr 51J for an explicit proof. A similar result for Lie algebras of type G 2 over arbitrary fields of characteristic zero was proved independently by N. Jacobson [Ja 39J and E. Bannow (Mrs E. Witt) [BaJ. Generalization to groups over arbitrary base fields was a fairly obvious idea. L.E. Dickson defined analogs of the Lie group G 2 over arbitrary fields without referring to octonion algebras - and proved simplicity over finite fields

36

2. The Automorphism Group of an Octonion Algebra

as early as 1901, after he had dealt with the classical groups; see [Di 01a] and [Di 05]. C. Chevalley refers to this work by Dickson in his Tohoku paper [Che 55]. In [Ja 58], Jacobson made an extensive study of the groups of automorphisms of octonion algebras over fields of characteristic not two, and proved they are simple if the algebra is split, so in particular over algebraically closed fields and over finite fields.

3. Triality

In this chapter we deal with algebraic triality in the group of similarities and in the orthogonal group O(N) of the norm N of an octonion algebra C, and with the related triality in the Lie algebras of these groups, usually called local triality. Geometric triality on the quadric N(x) = 0 in case N is isotropic will be left aside; the reader interested in the subject may consult [BlSp 60] and [Che 54, Ch. IV]. The results are proved here in all characteristics. In the existing literature, local triality has only been dealt with in characteristic not two. It turns out that for local triality in characteristic two one cannot use the Lie algebra L(SO( N)) of the orthogonal group, but has to pass to the Lie algebra of the group of similarities, which is one dimension higher, or to a subalgebra of L(SO(N)) of co dimension one. The proof we give for the characteristic two case is different from that in the other cases. Algebraic triality defines two outer automorphisms of the projective similarity group of N, which generate a group of outer automorphisms isomorphic to the symmetric group S3. Further, we derive a characterization of the automorphism group of C within the group of rotations that leave the identity element of C invariant (cf. Cor. 2.2.5). We use triality to give an explicit description of the spin group of the norm of C, and we describe the outer automorphisms induced by triality in this spin group. Finally, we prove that the corresponding algebraic group is defined over the base field of C. Bear in mind that C will always be an octonion algebra in this chapter. In the first section we present material related to quadratic forms that we need in the remainder of the chapter.

3.1 Similarities. Clifford Algebras, Spin Groups and Spinor Norms Let N be a nondegenerate quadratic form on a vector space V (of finite dimension) over a field k. We recall from § 1.1 that a similarity of V with respect to N is a linear transformation t : V - V such that

N(t(x))

= n(t)N(x)

(x E v)

38

3. Triality

for some n(t) E k*, called the multiplier of t. Then (t(x), t(y») = n(t){ x, y) (x, y E V), so t is bijective. The similarities form a group GO(N) (or GO(N, V) if there is danger of confusion), called the similarity group of N, and n is a homomorphism GO( N) - k*. The kernel of n is the orthogonal group O(N). If t is a similarity, so is At for A E k*, and n(At) = A2 n(t). Since we will often have to deal with similarities up to a nonzero scalar multiple, it is useful to introduce the homomorphism

where k*2 denotes the subgroup of squares in k*. We call v(t} the square class of the multiplier of t. Now consider the case that N is the norm of an octonion algebra C. The left multiplication la by an element a with N(a} :F 0 is a similarity with n(la} = N(a), and so is the right multiplication Ta. Any t E GO(N} can be written as t = lat' with a = t(e) and t' E O(N). It follows that every similarity is a product of a left multiplication and a number of reflections. Here we recall the convention of § 1.1 that for char(k) = 2 we understand by a reflection an orthogonal tmnsvection. The remainder of this section is devoted to a digression on Clifford algebras, spin groups and the spinor norm. We will, to a large extent, only give an exposition of the definitions and results we need, referring to the literature for proofs: [Ar, Ch. V, § 4 and § 5J, [Che 54, Ch. IIJ, [Dieu, Ch. II, § 7 and § 10J, [Ja 80, § 4.8J, [KMRT, § 8J. Consider again an arbitrary vector space V of finite dimension n over a field k with a nondegenerate quadratic form N. In the tensor algebra T(V) = k ffi V ffi (V ®k V) ffi ...

consider the two-sided ideal IN generated by the elements x ® x - N(x} (x E V). The Clifford algebra of N is the algebra CI(N} = T(V}/IN. (In the literature Clifford algebras are commonly denoted by a C, followed or not by something within parentheses. We use the notation CI to avoid confusion with the C denoting a composition algebra.) The canonical map of V, considered as a subspace of T(V), into CI(N) is injective; one identifies V with its image in CI(N}. For x, y E CI(N} we denote their product in CI(N} by x 0 Y, to avoid confusion with the product in C if CI(N) is the Clifford algebra of the norm N on an octonion algebra C. As an algebra, CI(N} is dearly generated by V, and one has the relations

x o2 = N(x} and x 0 y + Y 0 x

= (x, y)

(X,YEV),

using the notation x o2 = x 0 x. It follows that every x E V with N(x) :F 0 has an inverse in CI(N}, viz. x o - 1 = N(X)-lX. If el , e2, ... , en is a basis of V, a basis of CI( N) is formed by the elements

3.1 Similarities. Clifford Algebras, Spin Groups and Spinor Norms

39

so dimCI(N) = 2n. The elements x 0 y with x,y E V generate a subalgebra CI+(N) of CI(N), which has dimension 2n - 1 and which is called the even Clifford algebra of N. This even Clifford algebra can also be described as follows. Take the even tensor algebra

r;

= IN n T+(V) is generated by the elements x ® x - N(x) and The ideal u ® x ® x ® v - N(x)(u ® v) (u, v,X E V). Now CI+(N) = T+(V)jlt. The Clifford group of N is the group r(N) of invertible u E CI(N) such that u 0 V 0 u o - l = V. The even Clifford group is r+(N) = r(N) n CI+(N). For u E r(N), we define tu as the restriction to V of conjugation by u: tu : V ..... V,

u o- l .

X t-+ U 0 X 0

This is an orthogonal transformation of V, since

N(uoxouO-l) =uoxouo-louoxouo- l =N(x). Clearly, tuov = tutv for for x E V,

U,V

E

r(N). If u E V with N(u)

i: 0,

then we have

x 0 u o - l = N(U)-lU 0 X 0 U = N(u)-l( -x 0 u o2 + (x, u )u) = -x + N(u)-l( x, U )u, u

0

so u E r(N) and tu = -suo Every rotation is an even product of reflections sal Sa2 ... Sa2r' SO it is of the form tu with u = al 0 a2 0"·0 a2r E r+(N), all ai E V, N(ai) i: O. All elements of r+(N) are ofthis form, and u E r+(N) is determined up to a factor in k" by t u , since the intersection of the center of CI(N) with r+(N) is k". In other words, there is an exact sequence

1 ..... k" ..... r+(N) ~ SO(N) ..... 1,

(3.1)

where X denotes the homomorphism u t-+ tu. The main involution L of CI(N) is the anti-automorphism of order 2 defined by

For u =

al 0··· 0

i: 0),

a2r E r+(N), (ai E V, N(ai)

N(u) = u 0 L(U) = N(al)'" N(a2r) It is easily verified that N(uou')

= N(u)N(u')

N: r+(N) ..... k", u

t-+

define E

k".

and N(>.u)

N(u),

= >.2N(u). So

40

3. Triality

is a homomorphism. Its kernel is called the spin group Spin(N). Notice that the spin group is contained in the even Clifford algebra Cl+(N). Since u E r+(N) is determined by tu up to a nonzero scalar factor, it makes sense to define the homomorphism q :

SO(N)

-+

k* /k*2, tu 1-+ N(u)k*2.

One calls q(t) the spinor nonn of t E SO(N). If t = sal'" Sa2r with all ai E V, N(ai) 1= 0, then q(t) = N(al)'" N(a2r)k*2. The kernel of q is the reduced orthogonal group O'(N). The homomorphism X of (3.1) maps Spin(N) onto O'(N). Let us consider all these objects over the algebraic closure K of k. The vector space V with the quadratic form N is replaced by VK = K ®k V with the extension of N to VK, which we sometimes denote by NK. Clearly, Cl(NK) = K ®k Cl(N). The invertible elements of Cl(NK) form an algebraic group, of which r(NK) is a closed subgroup. (From the fact that r(NK) is defined over k - though we will not prove that here - it follows that r(N) is the group of k-rational points of r(NK).) Similarly, we have the algebraic group r+ (NK), a closed subgroup of r( N K), and a closed subgroup Spin( N) of r+(NK)' The exact sequence (3.1) (with k replaced by K) is an exact sequence of algebraic groups, and N: r+(NK) -+ K* is a homomorphism of algebraic groups. Every rotation has spinor norm lover K since K* = K*2, so O'(N) coincides with SO(N) over an algebraically closed field. (Notice that, in general, O'(N) is not the group of rational points of an algebraic group.) The algebraic group Spin(N) is connected provided dim V ~ 2; we show this in a similar way as in the proof of Prop. 2.2.2 for SO(N). The polynomial function N - 1 on VK is irreducible, so S = {x E V I N(x) = I} is an irreducible algebraic variety in V. The morphism S xS

-+

Spin(N), (x, y)

1-+

X 0

y,

maps the irreducible variety S x S onto an irreducible set of generators of Spin(N) containing the identity, so this must indeed be a connected algebraic group (cf. [Hu, § 7.5] or [Sp 81, Prop. 2.2.6]). (If dim V = 1, then Spin(N) =

{±1 }.) We have a homomorphism of algebraic groups 11" : Spin(N) -+ SO(N), where 11" is the restriction of X to Spin(N) (so 1I"(al 0 a2 0 '" 0 a2r) = Sal Sa2 ... sa2r)' where Spin(N) is the simply connected covering group of SO(N) (cf. [Sp 81, 10.1.4]). The homomorphism 11" is a separable isogeny with kernel of order 2 if char(K) 1= 2; if char(K) = 2, 11" is an inseparable isogeny with kernel { 1 }. We will discuss this in detail for the case that N is the norm on an octonion algebra C, though in fact the argument will work for the general case. We assume that k = K is algebraically closed. Then N is the unique (up to equivalence) form of maximal Witt index in dimension 8. We denote the corresponding spin group by Spin(8), etc.

3.1 Similarities. Clifford Algebras, Spin Groups and Spinor Norms

41

We have to determine explicitly the Lie algebra L(8pin(8)) of 8pin(8). To this end we introduce again the ring of dual numbers K[c] (as in the proof of Prop. 2.2.2), extend N to a quadratic form on G[c] = K[c] ®K G and consider the corresponding Clifford algebra K[c] ®K CI(N). The Lie algebra of r+ (NK) consists of elements U E CI+ (NK) such that (1

+ cu) 0 G[c] 0 (1 + eut- l

~ G[c],

that is, (1 Since (1

+ eu) 0 (x + cy) 0 (1 + eu)o-l

+ eu)O-l = 1 -

cU, this leads to

x+cy+c(uoX-XOU)EG[c] Since x

(x, y E G).

E G[c]

(X,YEG).

+ cy E G[c], we get the condition (x E G).

uox-xoUEG

(3.2)

Now first let char(K) "# 2. Choose an orthonormal basis el, ... , eg of G, write as a linear combination of these and U = I:'Lo I: fril ... i2Ieil 0·' ·oei21' where the second sum is taken over all sequences ib ... ,i 21 with 0::; i l < ... < i 21 ::; 8. One easily sees that (3.2) holds if and only if U = fro + Ei<j frijei oej. Since Spin(8) consists of the z E r+(NK) with ZOt(z) = 1, we need for L(Spin(8)) the extra condition (1 + eu) 0 t(1 + eu) = 1, which yields U + t(u) = O. Since u + t(u) = 2fro, we get fro = 0, so L(8pin(8)) is contained in the space S of the elements u = I:i<j frijei 0 ej. Since dimS = 28 = dimL(Spin(8)), we have L(Spin(8) = S. The Lie action of u on G, x t-+ U 0 X - X 0 u, has matrix ( 'Yij )l
4

(3.3) Lfri,iH = O. i=l The Lie action of U = fro + I:i<j frijei 0 ej satisfying (3.3) on GK, x t-+ U 0 X - X 0 u, has matrix ( 'Yij )l
(1 ::; i, j ::;

= 'YjH,iH = fri,jH

'Yi,iH = 'YH4,i = 0 =

= 'YH4,i = frH4,jH

frij

4),

(1::; i < j::; 4),

'Yi,jH 'YH4,j

'Yj,iH

=

(1 ::; i ::;

4),

(1 ::; i < j :5

4).

42

3. Triality

It follows that the kernel of d1r is K (considered as a subspace of CI(NK». Its image is a 27-dimensional subalgebra of the 28-dimensional Lie algebra L(80(8», because of condition (3.3). Thus we see that 1r is an inseparable isogeny if char(K) = 2 (cf. [Sp 81, § 9.6]).

3.2 The Principle of Triality The following theorem is the central result of this chapter.

Theorem 3.2.1 (Principle of Triality) Let C be an octonion algebra over a field k, with norm N. (i) The elements t1 E GO(N) such that there exist t2, t3 E GO(N) with (x,y E

CJ

(3.4)

form a normal subgroup SGO(N) of index 2, called the special similarity group. If (ft, t2, t3) and (t~, t~, t~) satisfy (3.4), then so do (fttJ., t2t~, t3t~) and (tll,t2"1,ta 1). (ii) Ifft E GO(N), then there exist t2, t3 E GO(N) such that (x,y E

CJ

(3.5)

if and only ift1 ¢ SGO{N). (iii) The elements t2 and t3 in (3.4) and in (3.5) are uniquely determined by t1 up to scalar factors ,\ and ,\-1 in k*, respectively. For tl = la, the similarities t2 = lara and t3 = la-1 = N(a)-1lo. satisfy (3.4), and similarly with t1 = r a, t2 = r a-1 = N(a)-lrr;., t3 = lara; for tt = Sa, (3.5) is satisfied by t2 : y 1--+ -N(a)-1afj, and t3 : x 1--+ xa; for tt = SaSb we can take t2 = N(a)-llal" and t3 = N(b)-lrar" to satisfy (3.4). Iftl E SGO(N), then so are t2and t3. (iv) Iftl. t2, t3 are as in (3.4) or (3.5), then n(tl) = n(h)n(t3). (v) Let t E GO(N); it is a product of a left multiplication by an invertible element of C and an orthogonal transformation t'. Then t E SGO(N) if and only if t' is a rotation. (vi) If tt, t2, t3 are bijective linear transformations of C that satisfy (3.4) or (3.5), then they are necessarily similarities with respect to N.

Proof. Call t1 E GO(N) even or odd if there exist t2, t3 E GO(N) such that (3.4) or (3.5) holds, respectively. (a) If two triples (tb t2, t3) and (t~, t~, t~) of invertible linear transformations satisfy (3.4), then so does their product (t1ti, t2t~, t3t~), since

If the first triple satisfies (3.4) and the second one (3.5), then their product satisfies (3.5). If the first triple satisfies (3.5) and the second one (3.4), then

3.2 The Principle of Triality

43

(tlt~, t2t~, t3t2) satisfies (3.5), and this last triple satisfies (3.4) if both initial triples satisfy {3.5}. Finally, if (t17 t2, t3) satisfies (3.4), then the same is true for {tIl, t2"l, tal}. For, substituting in (3.4) t2"I{X) for x and t 1{y) for y yields

a

from which we infer

(x,y E c). So the product of two even or two odd similarities is even and that of an even and an odd similarity is odd, while the inverse of an even similarity is even. (b) We show now that a similarity cannot be both even and odd. For assume that {t17 t2, t3} satisfies (3.4) and at the same time (tt, t 2, t~) satisfies (3.5). Then, by the results of (a), (id, U2, U3) with U2 = t2tal and U3 = t~t2"l would satisfy {3.5}. Thus,

(x,y E C). Taking y = e we get x = u2{e)u3{x), so U3 = la with a x = e we get U2 = Tb with b = u3{e)-I. Thus,

xy For x

= (yb){ax)

= u2{e)-lj

taking

(x,y E C).

= y = e this yields e = ba, so b = a-I. If we take y = a, we get (x E C).

xa =ax

This means that a belongs to the center of C, so a = Ae for some A E k by Prop. 1.9.1, and b = A-Ie. Thus we arrive at the relation

xy = yx

(x,y E C).

Since C is not commutative, we have a contradiction. (c) Every similarity is a product of a left multiplication by an invertible element and a number of reflections, so to show that we have either (3.4) or (3.5), it suffices by (a) to consider the cases tl = la and tl = Sa for a E C with N(a) ::f O. For tl = la we deduce from the second Moufang identity (1.14) that

a{xy) = (axa)(a-Iy)

(x,y E C).

Hence if we take t2 = laTa and t3 = la-I, then (t17t2,t3) satisfies (3.4). If we conjugate the above relation and then replace a by a, fi by x and x by y, we get {xy)a = (xa-l){aya) (x,y E C),

44

3. Triality

Next consider a reflection

xa + ax, Sa(x)

=X-

Sa :

X 1-+ X - N(a)-l(x,a}a. Since (x,a}e

=

X - N(a)-laxa = -N(a)-laxa.

Using the first Moufang identity (1.13) we get:

sa(xy) = -N(a)-la(jjx)a = -N(a)-l(ajj)(xa)

(x,y E C),

so if we take tl = Sa, t2 : y 1-+ -N(a)-lajj, and t3 : X 1-+ xa, we have a triple of similarities which satisfies (3.5). Combining this with the results of (a) and (b), we find the statements (i), (ii) and (v) of the Theorem and the statements in the second and third sentence of (iii). The normality of SGO(N) in GO(N) follows from the fact that it has index 2. The relation in (iv) is easily proved by taking norms on both sides of (3.4) and (3.5). (d) We now prove the uniqueness of t2, t3 up to a scalar factor. Assume (tl. t2, t3) and (h, t~, t~) both satisfy (3.4), then with Ui = tilt~ (i = 1,2) we have by (a),

(x,y E C). Taking y = e we find X = U2(x)u3(e), so U2 = similarly U3 = lb with b = u2(e)-1. Hence

xy = (xa)(by)

Ta

with a = u3(e)-1, and

(x,y E C).

Substitution of X = Y = e yields ab = e, so b = a-l. Replacing y by ay we find

x(ay)

= (xa)y

(x,y E C).

By Prop. 1.9.2, a = >.e for some>. E k, so b = >.-le. This implies that t~ = >'t2 and t~ = >.-lt3. Now let (tl, t2, t3) satisfy (3.5). Define similarities Ui by Ui(X) = ti(x). Then (Ul. U2, U3) satisfies (3.4). The uniqueness of U2 and U3 up to factors >. and >.-1, respectively, implies the same for t2 and t3. (e) To prove (vi), finally, consider arbitrary bijective linear transformations tl, t2, t3 of C that satisfy (3.4). Taking y = e in (3.4), we get

(X E C). As tl is bijective, there exists X E C such that N(tl(X)) i= OJ since N(tt(x)) = N(t2(X))N(t3(e)), we see that N(t3(e)) i= 0, so t3(e) is invertible. Hence t2 = Tatl with a = t3(e)-1. By substituting x = e in (3.4), we find in a similar way that t3 = lbtl with b = t2(e)-l. We can now rewrite (3.4) in the following form:

(x,y E C).

(3.6)

3.3 Outer Automorphisms Defined by Triality

45

From tl(e) = t2(e)t3(e) it follows that tl(e) is invertible. Set c = tl(e)-l. The similarity ti = le can be completed to a triple of similarities (ti, t 2,t satisfying (3.4) as we saw in (c) above. Then the product triple (ti tb t 2t2, tat3) also satisfies (3.4) as we saw in (a). Now titl(e) = lctl(e) = e, so for the proof of (vi) we may as well assume that tl(e) = e. Taking x = y = e in (3.6), we find ab = e. Taking norms of both sides of (3.6), we get, since N(a)N(b) = N(e) = 1,

a)

N(tl(XY))

= N(tl(X))N{tl{Y))'

This means that the nondegenerate quadratic form N on C defined by N(x) = N(tl(X)) (x E C) permits composition. This implies by Cor. 1.2.4 that N = N, so ft is orthogonal. Then t2 and t3 must be similarities, since each of them is a product of tl and a similarity. A similar proof shows that bijective linear transformations tb t2, t3 that satisfy (3.5) must be similarities. 0 The elements of SGO(N) are called proper similarities, the other similarities improper. For tl E GO(N) we fix the notation t2 and t3 by the rule that the triple (tb t2, t3) satisfies either (3.4) or {3.5)j such a triple (tb t2, t3) is said to be related. Bear in mind that t2 and t3 are determined by tl up to factors oX and oX -t, respectively. Equation (3.4) is called the first form of triality and (3.5) the second form of triality. Consider a related triple (tb t2, t3)j if tl E SO(N), it satisfies the first form of triality, and hence t2, t3 E SGO(N). With the aid of the spinor norm we can determine when we can take t2 and t3 to be orthogonal. Proposition 3.2.2 For tl E SO(N), the similarities t2 and t3 such that (tl' t2, t3) satisfies (3.4) have square class of multiplier V(ti) = O'(tt}. Hence t2 and t3 can be taken to be orthogonal if and only if O'(tl) = 1; in that case they are both rotations. Proof. The equality V(ti) = O'(tt} follows from the case tl = SaSb in (iii) of Th. 3.2.1. We have also seen there that t2, t3 E SGO(N), so if we take them to be orthogonal, they are rotations by (v) of the same theorem. 0 Remark 3.2.3 If k* = k*2, which is for instance the case if k is algebraically closed, every rotation has spinor norm 1. Then for every tl E SO(N) one can find t2, t3 E SO(N), unique up to a common factor ±1, such that (tb t2, t3) satisfies the first form of triality (3.4), so the Principle of Triality holds in this form for SO(N) over algebraically closed fields, more generally over fields in which every element is a square.

3.3 Outer Automorphisms Defined by Triality For t E GO(N) we define

i E GO(N)

by

£(x) = n(t)-lt(x)

(x E C).

46

3. Triality

Lemma 3.3.1 For t E GO(N) we have n(f)

= n(t)-l

(x E C, N(x)

and

-10).

Further,

tu=£fJ. so t

1-+

£= t,

and

£ is an involutory automorphism of GO(N). o

Proof. Easy verifications.

Lemma 3.3.2 Let tb t2, ta E SGO(N). If (h, t2, ta) satisfies the first form of triality, then so do (t2, tl, £a), (ta, £2, h), (£1, £a, £2), (£2, ta, it) and (ia, £1, t2). Proof. From we infer for N(y)

-10,

N(y)tl(X) = t2(xy)N(y)ta(y-l)

= N(y)t2(xy)(£a(y))-I.

Hence

(x,y

E

C, N(y)

-10).

Working over K, we obtain a polynomial equality which is valid on a Zariski open subset of CK x CK. It must hold for all x, y E CK and in particular for x, y E C. So (t 2 , t l , £a) is a related triple. In a similar way one proves that (ta, £2, tl) is related. Thus, if we interchange in a related triple satisfying the first form of triality the first and the second (or the third) component and replace the remaining component t by i, we get another related triple satisfying the first form of triality. Applying this operation a few times in a suitable way, we get the other three related triples. 0

Corollary 3.3.3 1ft is a rotation, then u(f) = u(t). Iftl is a rotation with U(tl) = 1 and (tl, t2, ta) is a related triple of rotations, then U(t2) = U(t3) = 1.

Proof. If t is a rotation, then £(x) = t(x)j since conjugation is the negative of the reflection sel £ is a rotation with the same spinor norm as t. The second 0 statement follows by combining the above lemma with Prop. 3.2.2. In dealing with similarities of the octonion algebra C up to a scalar factor, it is natural to interpret them as transformations of the set lID(C)( k) of krational points of the seven-dimensional projective space lID (C) = (C\ { 0 }) / K* . These transformations form the projective similarity group PGO(N) =

3.3 Outer Automorphisms Defined by Triality

47

GO(N)jk*. The image of SGO(N) in PGO(N) under the natural projection is the projective special similarity group PSGO(N) = SGO(N)jk*. For t E GO(N), we identify its image tk* in PGO(N) with the projective transformation it induces in IP(C), and we denote both by ttl. Every ttl] E PSGO(N) uniquely determines [t2], [t3] E PSGO(N) such that (tl> t2, t3) is a related triple. Proposition 3.3.4 The mappings a,{3,e: PSGO(N)

--+

PSGO(N),

a : [ttl t-+ [t2], {3 : ttl] t-+ [i3] , e : ttl t-+ til, are automorphisms of PSGO(N). The automorphisms a and {3 generate a group S of outer automorphisms of PSGO(N), which contains e, and which is isomorphic to the symmetric group 83 • Proof. That a and {3 are homomorphisms follows from part (i) of Th. 3.2.1 and Lemma 3.3.1, and their bijectivity follows from Lemma 3.3.2. From Lemma 3.3.2 it also follows that

and {3 :

ttl]

t-+

[i3]

t-+

[i2] t-+ [tl],

[t2]

t-+

[t3]

t-+

[itl t-+ [t2]'

This implies e = a{3. Further one derives the relations

denoting by 1 the identity automorphism. These are the defining relations for the symmetric group 8 3 , if one takes as generators (12) and (123). So there exists a homomorphism r: 8 3 --+ S, (12)

t-+

a and (123)

t-+

{3.

The only nontrivial normal subgroup of 8 3 is the alternating group A 3 . Since (123) E A3 is not mapped onto 1 by r, we see that r must be an isomorphism. There remains the proof that all elements i- 1 of S are outer automorphisms. Let A be the quotient of the group of automorphisms of PSGO(N) by the group of inner automorphisms. We have to show that the homomorphism 8 3 --+ A induced by r is injective. If not, its kernel would contain the alternating group A3 , and {3 would be an inner automorphism. If this were the case, there would exist u E SGO(N) such that

(tl E SGO(N)). If tl = la, then {3([tl]) = [i3] with t3 = N(a)-lla by Th. 3.2.1 (iii). Writing this out we see that for each a E C with N(a) i- 0 there is Aa E k such that

48

3. Triality

(x E C).

N(a)u(ax) = Aau(x)a Taking x

(3.7)

= u- 1 (a- 1 ) we find (a

E

C, N(a) f:. 0).

(3.8)

Taking norms of both sides of (3.7) we get

N(a)2n(u)N(a)

= A~n(u)N(a),

from which we infer A~ = N(a)2, so Aa = ±N(a) fora E C,N(a) f:. O. We may assume that k is algebraically closed. Since a 1--+ Aa is a polynomial function on C by (3.8) and since Ae = 1, we must have Aa = N(a) if N(a) f:. O. By Zariski continuity this must hold for all a E C. Using Prop. 1.9.2 we conclude that e E k* e. Consequently, f3 = Inn( u) = id, which is a contradiction. 0

Remark 3.3.5 Since c E S, it is an outer automorphism of PSGO(N). One can extend c to an automorphism ttl 1--+ til of PGO(N), which is an inner automorphism since in GO(N) we have

i

= n(t)-lcte- 1 ,

where e is the similarity x 1--+ x. Notice that e satisfies (3.5), the second form of triality, so e
3.4 Automorphism Group and Rotation Group of an Octonion Algebra Let C be an octonion algebra over k. We saw in § 2.2 that the algebraic group G is a closed subgroup of the stabilizer SF of e in SO(N), which also leaves el. invariant, and that restriction of elements of SF to el. defines an isomorphism (} of SF onto SO(Nl); see Prop. 2.2.2. We will henceforth identify the elements of SF with their (}-images in SO(Nl)' and conversely identify every t E SO(Nt} with its extension (}-I(t) to a rotation fixing e. Thus, we have identified G with a closed subgroup of SO(Nt}. Similarly, we identify G = Aut(C) with a subgroup of SO(Nl)' For x,y E C, both f:. 0, we denote k*x E IP(C)(k) by [x], and sometimes write [x].[y] for [xy]. For tl E SO(N1 ) there exist similarities t2, t3, unique up to scalar factors A and A-I, respectively, such that (tlJ t2, t3) satisfies the first form of triality (3.4). We define a mapping Ll from SO(N1 ) to IP(C)(k) by Ll: SO(Nt} -+ IP(C)(k), tl 1--+ [t2(e)]. By Lemma 3.3.2, (t2, tlJ i3) also satisfies the first form oftriality. This implies i3(e) = t2(e) since tl(e) = e. Using all this, we find for the map Ll:

3.4 Automorphism Group and Rotation Group of an Octonion Algebra

49

Thus,

= [t](Ll(u».Ll(t) (t,u E SO(Nl». E Aut(C), then tl = t2 = t3, so Ll(tt} = [e]. This necessary condition is Ll(tu)

If tl also sufficient as we see in the following proposition.

Proposition 3.4.1 For t E SO(N1 ) we have t E Aut(C) if and only if .1(t) = tel. Proof. It remains only to prove the "if" part. If tl E SO(Nl) and Ll(tl) = [e], we may assume t2(e) = e. From (3.4) with x = e it follows that tl(Y) = t3(Y) for all y. Then t3(e) = tl(e) = e, hence taking y = e in (3.4) yields that tl(X) = t2(X) for all x. Thus we have found that tl = t2 = t3, which means that tl E Aut(C). 0

If tt E SO(Nd, t2(e) has nonzero norm. We claim that every element of C with nonzero norm can be obtained in this way.

Lemma 3.4.2 For every e E C with N(e) =F 0 there exists tl that t2(e) = e.

E SO(N1 )

such

Proof. We try tl = SaSb with a, bE e.L. Then tl(e) = e, so tl E SO(Nl)' Pick any a E C with (a,e) = (a,e) = 0 and N(a) =F O. Take b = ).aewith)' E k* still to be chosen. Then N(b) =F 0 and

(b,e)

= ).(ae,e) = ).(e,a) = -).(e,a) = o.

By Th. 3.2.1 (iii), we can take t2 = lalr,. Then

t2(e)

= ab = -ab = -).aac =

).aac = ).N(a)e.

So if we take). = N(a)-l, we have t2(e) = e as desired.

o

We define an action T of SO(N1) in P(C)(k) by T(t) = a([t]) with a as in Prop. 3.3.4, so T(tt} = [t2]' If T(t) = [id] , then ttl = lid] since a is an automorphism; this implies that t = id since t E SO(Nl)' Thus, T is faithful. For k = K we have obtained an action of SO(Nd on P(C). Consider in P(C) the Zariski open subset 0= {[e] I N(e) =F O}.

It is the complement of a projective quadric which is defined over k. Lemma 3.4.2 shows that T(SO(Nt}) operates transitively on the set O(k) of k-rational points of O. According to Prop. 3.4.1, t E SO(Nt} is an automorphism of C if and only if T(t) fixes tel, so G = Aut(C) is the stabilizer of tel. Working over K we see that T induces a bijective morphism T of algebraic varieties from the homogeneous space SO(N1)/G onto O. Counting dimensions yields

50

3. Triality

dimSO(N1)/G

= 21- dimG =

dim 0 = dimP(C) = 7

(use [Sp 81, Th. 5.1.6]), from which we derive again that dimG = 14. By Prop. 2.4.6 we know that G is defined over k. The group SO(N1) is the rotation group of a 7-dimensional quadratic form, it is a quasisimple algebraic group of type Ba (see [Sp 81, 17.2.1 and 17.2.2]) and G is a closed algebraic subgroup which is an exceptional simple group of type G2 • We have found a bijective morphism r of SO(N1)/G onto a Zariski open subset 0 of the 7-dimensional projective space P(C), which induces a bijection between SO(Nl)/G(k) and O(k) ~ P(C)(k). If the norm N is anisotropic, i.e., if C is an octonion division algebra, then O(k) = P(C)(k). The bijective morphism r : SO(N1)/G -+ 0 is even an isomorphism of algebraic varieties. According to [Sp 81, 5.3.2] it suffices to show that the tangent map (dr)e is surjective. To prove this, we consider the map

.1 : SO(N1) -+ 0, t

t-+

T(t)[e].

The surjectivity of (dr)e will follow if we can show that (d.1)e is surjective. Pick a E e.L, N(a) = 1. The map «e.L\{ O})/ K*)

nO -+ «ae.L\{ 0 })/K*) n 0, [x]

t-+

[ax]

= [ax],

is an isomorphism of algebraic varieties which factors through .1, for

The image of (d.1)e must therefore contain ae.L, which we can identify with the tangent space to 0 at [a]. This shows that dre is surjective. 0

3.5 Local Triality If k = K is an algebraically closed field, the Principle of Triality holds already for the algebraic group SO(N), as we noticed in Rem. 3.2.3. It has an analog in the Lie algebra L(SO(N)) of this algebraic group, the Principle of Local Triality, at least in characteristic not two. In characteristic two the situation is more complicated. In the first instance we have to work in the Lie algebra L(SGO(N)) of the special similarity group, or in the quotient by its one-dimensional center. For local triality in the Lie algebra L(SO(N)) in characteristic two we have to restrict to a subalgebra M thereof which has co dimension 1j in the next section we will see that M is the commutator subalgebra of L(SO(N)). Before we formulate and prove local triality, we have to discuss the Lie algebras involved here and, in particular, to find suitable generators for them.

3.5 Local Triality

51

The similarity group GO(N) is the algebraic group consisting of the invertible linear transformations t which satisfy the equations

N(t(x))

= N(t(e))N(x)

Notice that for a similarity t with respect to the norm N on C the multiplier

n(t) equals N(t(e)) (cf. § 3.1). ,The special similarity group SGO(N) is the identity component of GO(N). We denote its Lie algebra by Lo. To find it one can use the ring of dual numbers K[cJ. The method of the proof of Prop. 2.2.2 gives that Lo consists of the linear maps t of CK satisfying the condition

(x,t(x)) = (,e,t(e))N(x) We denote by Lo the Lie algebra of linear maps t of C satisfying the same condition. Such a t is called a local similarity, and the factor (e, t(e)) its local multiplier. Examples of local similarities are the left and right multiplications le and re for C E C, because by (1.3)

(x,ex) = (e,c)N(x), and similarly for the right multiplications. The subalgebra Ll = L(SO(N)) is given by the equations

(x,t(x))

=0

The similarly defined subalgebra of Lo is L 1 • It consists of the alternating linear transformations. These are also skew symmetric:

(t(x),y) + (x,t(y)) = 0

(x,y

E

C).

Examples of alternating transformations are

ta,b: C

-+

C,

X

1-+

(x,a)b - (x,b)a,

with a, b E Cj further, the la with (a, e) = 0, the ra with (a, e) = 0, the me = le - re for all c E C, and finally the lalb - lbla and rarb - rbra for all a, bE C, as one easily verifies.

Lemma 3.5.1 If tl and t2 are local similarities of C, then their Lie commutator [tlo t2J = tlt2 - t2tl is alternating. Proof. Let tl and t2 have local multipliers Al and A2, respectively. The relation

(x E C) leads to

52

3. Triality

(x E C). After cancelling terms on both sides of the equality sign we find

(x E c). Now write down this relation with t1 and t2 interchanged, and subtract that from the above relation. Then we find (x, [t1' t2JX) = 0 (x E C). 0

Lemma 3.5.2 (i) L1 is genemted as a vector space by the ta,b with a, bE C. (ii) If char(k) = 2, then L1 is genemted as a Lie algebm by {la, rb, me Ia, bE e.L, C E C}, and Lo is genemted as a Lie algebm by {la, rb Ia, bE C}. Proof. For t E Lit define Nt = {x E C It(x) = o}. For a = t(e) we have (e,a) = o. Pick bE C with (e,b) = -1. Then ta,b{e) = a = tee), so t' = t - ta,b maps e to 0, i.e., e E Nt'. Next assume we have tELl with e E Nt #- C. Pick u E C, u ¢ Nt. For a = t(u) we have (u,a) = 0, and for x E Nt,

(x,a)

= (x,t(u») + (t(x),u) = o.

Choose b E Nl with (u, b) = -1. Then

ta,b{X) = (x,a)b- (x,b)a ta,b{U) = (u,a)b- (u,b)a

=0 = t{u).

(x E Nt),

Hence for t' = t - ta,b we have Nt' ;2 Nt and u E Nt', so Nt' #- Nt. By induction we arrive at statement (i). Notice that after the first step we have only used transformations ta,b with a, b E e.L. For the proof of (ii), consider any tELl. Take a = tee), then (a,e) = o. For t' = t - la we have t'(e) = 0, so e E Nt'. Continuing as in the previous paragraph we find that L1 is generated as a vector space by the transformations la and ta,b with a, b E e.L. In order to prove (ii) it therefore suffices to express these ta,b in left and right multiplications by elements of e.L and transformations me with c E C. Let a, b E e.L. From a = -a we infer that -ax + xa = (x, a )e, so ax = xa - (x, a )e, and similarly xb = bx - (x, b}e. Using this, we find:

[la, rb](x)

= a(xb) -

(ax)b = a{bx) - {xa)b + (x, a)b - (x, b)a = a(bx) - (xa)b + ta,b(X) (x E C),

so

(x E C). Since the associator in C is alternating,

(3.9)

3.5 Local Triality

(xa)b - x(ab)

= (ab)x -

a(bx)

53

(x E C),

so

a(bx)

+ (xa)b =

(ab)x + x(ab)

(x

E

C).

Now assuming that char(k) = 2, we substitute this into (3.9), thus getting

ta,b(X)

= [la, rb](x) + (ab)x + x(ab) = [la, rb](x) + (ab)x - (ab)( x, e) + x(ab) = [la, rb](x) + (ab)x + x(ab).

(x, e ) (ab)

Hence we find which shows that L1 is generated as a Lie algebra by the la, rb and stated in (ii). For Lo, finally, fix any dEC with (e, d) = 1; then by (1.4),

(x,dx)

= (e,d)N(x) = N(x)

me

as

(x E C).

If t E Lo has local multiplier A, then t - Ald E L 1. Since that Lo is generated by aUla and rb.

me

= le -

r e , we see 0

As in the last paragraph of the above proof we see that in any characteristic Lo is spanned by L1 and ld for some d rt el.. If char(k) :/;2, we can take d = !e, but in characteristic 2 we have le E L 1 . Thus, Lemma 3.5.3 (i) dim Lo = dim L1 + 1. (ii) If char(k) :I 2, then Lo = L1 $ kid, a direct sum of Lie algebras. If char(k) = 2, then kid eLI. For later use, we formulate another consequence of Lemma 3.5.2, which is of a technical nature and which easily follows from the proof of part (ii) of that lemma. Lemma 3.5.4 Assume char(k) = 2. Let M be the subalgebra of Ll generated as a Lie algebra by the la and rb with a, b E el., and let d be some element of C with (d,e) = 1. Then L1 = M + kmd. We will see later that md rt M. We are now ready to formulate and prove the result on local triality. Theorem 3.5.5 (Principle of Local Triality) Let C be an octonion algebra with norm N over a field k. (i) If char(k) :I 2, then for every tl E L1 there exist unique t2, t3 E L1 such that (3.10) (x,y E Cj.

The mappings iJ 2 : h ...... t2, and iJ3 : tl ...... t3, respectively, with tit t2, t3 satisfying (3.10), are Lie algebra automorphisms of L 1 .

54

3. Triality

For tl

= ta,b,

(3.10) is satisfied by t2

= l(lbla -lal,;)

and t3 = l(rbrii -

rari».

(ii) If char(k) = 2, then for every tl E Lo there exist t2, t3 E Lo such that (3.10) holds. These t2, t3 are unique up to a change h 1-+ t2 + Aid, t3 1-+ t3 + Aid, for some A E k. The mappings {)2 : tl

+ k id

1-+

t2

+ kid,

and {)3: tl

+ k id

1-+

t3

+ kid,

with tl. t2, t3 satisfying (3.10), are Lie algebm automorphisms of Lo/ kid. (3.10) is satisfied by the triples tl = la, t2 = la + r a, t3 = la and tl = rb, t2 = rb, t3 = lb + rb· Proof. Let N be the set of tl E Ll (or E Lo in characteristic 2) for which there exist t2, t3 E Ll (E Lo, respectively) such that (3.10) is satisfied. If (tl ,t~, t~) also satifies (3.10), then clearly so does (Atl + J,Lti, At2 + J,Lt~, At3 + J,Lt~) for A, J,L E k. FUrther,

ht! (xy) = tl(t~(X)Y + xt~(y»

= tl(t~(X)Y) + tl(Xt~(y» = t2t~(X)Y + t~(X)t3(Y) + t2(X)t;(y) + xt3t~(y) Interchanging the ti and the yields

t~

(x,y E C).

and then subtracting from the above relation

(x,y

E

C).

This shows that N is a Lie sub algebra of L1 (or Lo in characteristic 2). Hence, to prove the Theorem it suffices to prove that N contains a set of generators of L1 or Lo, respectively. First, we consider the char(k) '" 2 case for L 1. We use the ta,b as generators.

ta,b(XY)

= (xy, a}b - (xy, b}a = (y,xa}b- (y,xb}a = (by)(xa) + (b(ax»y - (ay)(xb) - (a(bx»y = (b(ax) - a(bx»y + (by)(xa) - (ay)(xb).

(3.11)

From the definition of ta,b it is immediate that ta,b(Z) = ta,i)(.z), so using the above expression we get

ta,b(XY) = tii,i)(Yx) = x((ya)b - (yb)a)

+ (ay)(xb) -

(by)(xa).

Adding (3.11) and (3.12) and dividing by 2 yields

ta,b(XY)

= 21 (b(ax) -

1 a(bx»y + 2x((ya)b - (yb)a).

(3.12)

3.5 Local Triality

55

Just as we noticed before Lemma 3.5.1, !(lbla -lalii) and !(TbTa - TaTii) are elements of L l . This proves existence in case (i). For existence in case (ii), we take as generators of Lo the La and Tb. Since the associator is symmetric in characteristic 2,

a(xy)

+ (ax)y = x(ay) + (xa)y,

so

= (ax + xa)y + x(ay). Hence (3.10) is satisfied by tl = la, t2 = la + Ta and t3 = La. tl = Tb, t2 = Tb and t3 = lb + Tb· a(xy)

Similarly with

Now about uniqueness. Assume that the triples (tll t2, t3) and (tll t~, t~) both satisfy (3.10). Then U2 = t2 - t2 and U3 = t3 - t~ satisfy

(x,y E C). Taking y = e we find U2(X) = xa with a = -u3(e), while with x = e we find U3(y) = -u2(e)y = -ay. Then

(xa)y - x(ay) = 0

(X,yEC).

By Prop. 1.9.2 this implies that a = >.e for some>. E k, so U2 = -U3 = >. id. If char(k) =1= 2, this transformation does not belong to Ll if>. =1= 0, so in that case t2 and t3 are uniquely determined by tl' The statements about the automorphisms t'J2 and t'J 3 are consequences of the first paragraph of this proof. 0 A triple (tl' t2, t3) of local similarities which satisfies (3.10) is said to be related. The following result is immediate from the first paragraph of the proof of the Theorem combined with Lemma 3.5.1. Corollary 3.5.6 If (tll t2, t3) and (ti, t 2,t a) are related triples of local similarities, then the triple of alternating transformations ([tl' til, [t2' t2l, [t3, t3J) is related. Remark 3.5.7 In characteristic i= 2, the Principle of Local Triality holds in Lo as well. Then tl determines t2 and t3 up to adding opposite mUltiples of the identity. Taking tl = id as an extra generator besides the ta,b, the triple tl = id, t2 = id, t3 = 0 satisfies (3.10). Assume char(k) = 2. Then the principle of local triality does not hold in L l . For consider tl = md = ld + Td E Ll for some dEC with (d, e) = 1. Since the associator is invariant under cyclic permutation, we have

d(xy) so

+ (dx)y =

x(yd) + (xy)d,

56

3. Triality

d(xy)

+ (xy)d =

(dx)y + x(yd).

The Principle of Local Triality in Lo says that t2 = ld + Aid and ta = r d + Aid are the only elements in Lo such that the triple (t1' t2, ta) satisfies (3.10). But none of these t2 and ta are in L1 if d ¢ e.l. One way out is to formulate a Principle of Local Triality for L1 in combination with two other subalgebras L2 and La of Lo which we define as follows: Li is the subalgebra containing k id such that Li/k id = f}i(LI/k id) for i = 2,3. For t1 E L1 one can then find t2 E L2 and ta E La such that (3.10) holds, and these are unique up to adding to both of them one and the same multiple of the identity. Another approach is to restrict to a subalgebra of L 1 • Let M be the Lie sub algebra of L1 generated by all la and rb with a, b E e.l. We saw in Lemma 3.5.4 that L1 = M +kmd, so L2 = M +kld and La = M +krd. Since ld and r d have local multiplier (e, d) = 1, they do not belong to L1' On the other hand, if t1 EM, then t2, ta E M by Th. 3.5.5. Hence md ¢ M, and Lb L2 and La are three distinct Lie algebras which have M in common, while the sum of any two of them is Lo. Recall that dimL l = 28, hence dimM = 27. Thus we have proved the following theorem.

= 2. Let M be the Lie subalgebra of Ll generated as a Lie algebra by all La and r a with a E e.l, and let d be a fixed element of C with d ¢ e.l. (i) dimM = 27, L1 = M E9 k(ld + rd), L2 = M E9ld and La = M E9 rd. Further, Li n Lj = M and Li + Lj = Lo for 1 $ i f; j $ 3. (ii) For tt E Ll there exist t2 E L2 and ta E L3 such that (3.10) holds, and these are unique up to adding a common multiple of the identity to them. (iii) For tl E M there exist t2, ta E M, unique up to adding a common multiple of the identity, such that (3.10) holds. For t1 E L1I T1 ¢ M such t2 and ta in L1 do not exist.

Theorem 3.5.8 Assume char(k)

Lemma 3.3.2 about permutations of related triples of proper similarities has a local analog. First some notation. For a local similarity t, the local similarity £ is defined by

lex) = t(x)

(x E C).

£ and t

have the same local multiplier. The mapping t t-+ £ is an involutory automorphism of the Lie algebra L o, and it maps Ll onto itself.

Lemma 3.5.9 Assume tb t 2, ta E Ll if char(k) f; 2, and tb t2, ta E M if char(k) = 2. If the triple (tt, t2, ta) is related, then so are the triples (t2' tb £a), (ta,t2' tl), (£11 £a, £2), (£2, ta, £1) and (£a, £1, t2)' In the characteristic 2 case, t E M implies

£ EM.

Proof. Substitution in (3.10) of xy for x and of jj for y yields (x,y E C),

3.5 Local Triality

(x,y

E

57

C).

Multiplying all the terms in the latter equation on the right by y, we get

(x,y

E

C).

Now

«xy)ta(y))y = «xy)ia(y»y = -«xy)y)ia(y) + (y, ia(y) )xy (by Lemma 1.3'.3 (iv)) = -N(y)xia(Y) (since ia ELl). Thus we get

(x,y E C, N(y)

1= 0),

and by Zariski continuity this holds for all x, y E C, so (t2' t1, ia) is related. Further,

i1 (xy)

= t 1 (yx) ~-:---,-~

= t2(Y)X + yta(x) =

from which it follows that

Xi2(y) + i3(X)Y

(x,y

E

C),

(ilo ia, i2 ) is related. In particular, we see that

i1 E M if t1 EM. The other three cases follow by applying the above results.

0

In § 3.1 we considered the projection 11 : Spin(N) ---. SO(N), which sends a2 0 ... 0 a2r to Sal sal' .. Sal r ' We saw that in the characteristic 2 case d1f maps L(Spin(N)) onto a 27-dimensional sub algebra of L1' We will see in Cor. 3.6.7 that this image is M (defined as in Th. 3.5.8). Using this result we will then derive that M is the commutator subalgebra [L1' L 1], which provides a characterization of M that does not depend on a special set of generators. The fact that, in characteristic 2, M is the commutator subalgebra of L1 could also be derived in a different way. From Cor. 3.5.6 and Th. 3.5.8 one infers that [Lo, Lo] ~ M, so M is a 27-dimensional ideal in L 1. The latter is a Lie algebra of type D4 of intermediate type, i.e., belonging to the orthogonal group and not to the simply connected nor to the adjoint group. From general results about Lie algebras of classical type in nonzero characteristic it follows that M must be the commutator subalgebra [Llo L1] (see [Ho 78, Th. (8.20)] or [Ho 82, Th. (2.1)]). These general results also tell us that L1 has a 1dimensional center in this case; in the following lemma we prove this directly, and also that L1 has trivial center in the other characteristics.

a1

0

58

3. Triality

Lemma 3.5.10 (i) If char(k) '" 2, the center of Ll is trivial, and the center of Lo as well as the centralizer of Ll in Lo is kid. (ii) If char(k) = 2, then L o, Ll and M have center kid, and this is also the centralizer in Lo of each of these Lie algebras. Proof. Let t E Lo commute with aliia and Ta for a E e.l. From t1a follows that

= lat it

t(ax) = at(x) Taking x = e we get that t(a) = au for a E e.l, where u by r a we find that t( a) = ua for a E e.l. Thus,

t(x)

= t(e). Replacing la

= xu = ux

(3.13)

If char(k) '" 2, we can drop the restriction x E e.l in (3.13), since this equation also holds for x = e. This implies that u is central in C, so by Prop. 1.9.1 u = Ae for some A E k. Hence t = Aid; this is an element of L o, but not of L 1· Now assume char(k) = 2. Then (3.13) and Prop. 1.2.3 together imply

(x, u)e + (u, e)x

=0

Since x need not be a multiple of e, we see that (u, e) = 0, and hence (x, u) = 0 for all x E e.l. This proves that u = Ae for some A E k. Finally, consider x ¢ e.l. Choose a E e.l n x.l with N(a) '" O. Then ax E e.l, since (ax,e) = (x,a) = (x,a) =0. Hence

t(x)

= t(a- 1 (ax))

= N(a)-lt(a(ax))

= N(a)-lat(ax) = N(a)-la(Aax) = AX,

so t = Aid. This is contained in M if char(k) = 2 by Th. 3.5.8 (iii).

0

3.6 The Spin Group of an Octonion Algebra We consider again the rotation group SO(N) of the norm N of an octonion algebra Cover k. The phenomenon of triality makes it possible to give another description of the simply connected covering of SO(N). In fact, we will construct an algebraic group RT( C) which we show to be isomorphic to the spin group Spin(N) of § 3.1. We let the group SO(N)3 = SO(N) x SO(N) x SO(N) act componentwise on the vector space

v = C 3 = {(XbX2,X3) IXi

E C, i

= 1,2,3}.

In SO(N)3 we consider the subgroup RT(C) of related triples, i.e., RT(C) consists of the triples that satisfy the first form of triality:

3.6 The Spin Group of an Octonion Algebra

59

Recall that for related triples (tl. t2, t3) of rotations we must have all ti E O'(N), the reduced orthogonal group. By (i) of Th. 3.2.1, RT(C) is closed under componentwise multiplication and taking inverses. A rotation tl with CT(tl) = 1 can be written as

with ai, bi E C, Ili N(ai)N(b i ) = 1. According to Th. 3.2.1 (iii) the corresponding rotations t2, t3 such that (tl, t2, t3) is related are given by (t2,t3) = ±(lallbl ···la)br,ra1rbl ···rarrbJ·

We can get the plus sign here by replacing al by -al, if necessary. Hence, RT( C) consists of the elements

where ai, bi E C, Ili N(ai)N(b i ) = 1. We denote RT(CK ) by RT(C). Over the algebraic closure K of k the reduced orthogonal group O'(N) coincides with SO(N), so RT(C) is a closed subgroup of the algebraic group SO(N)3. We will see in Prop. 3.7.1 below that the algebraic group RT(C) is defined over k. It then will follow that RT(C) is the group of rational points RT(C)(k). Proposition 3.6.1 RT(C) is a connected algebraic group, and

is a surjective homomorphism of algebraic groups which has a kernel of order 2 if char(k) =I 2 and which is bijective if char(k) = 2. The group RT(C) is mapped onto SO(N) by f!l.

Proof. It is obvious that f!l is a surjective homomorphism of algebraic groups. Since t2 and t3 are determined by tl up to a common factor ±1, ker(l?l) = {1, 1- }, with 1 and 1- denoting (1,1,1) and (1, -1, -1), respectively. Note that 1- = 1 if char(k) = 2. To prove the connectedness of RT(C) , consider in CK the set S = {x E CK I N(x) = I}; this is an irreducible variety as we remarked already in the proof of the connectedness of Spin(N) in § 3.1, hence so is S x S. The elements (SaSb' lalb' rarb) with N(a) = N(b) = 1 generate RT(C). There is a morphism of algebraic varieties

60

3. Triality

Its image under 'Y is an irreducible subvariety of RT( C) which contains the identity and generates RT(C), hence RT(C) is connected (see [Hu, § 7.5] or [Sp 81, Prop. 2.2.6]). The last point is clear. 0 To find the Lie algebra L(RT(C», we work again over the ring K[e] of dual numbers. We must have the triples (tb t2, t3) such that (1+et1, 1+et2, 1+et3) is a related triple of rotations in CK[e]. This leads to triples of ti E L1 which satisfy t1(XY)

= t2(X)Y + xt3(Y)

Thus we find in characteristic

(x,y E C).

1= 2:

Recall that, in this case, t1 E L1 uniquely determines t2 and t3. So the righthand side has dimension dimL 1 = 28 = dimRT(C), and it follows that the inclusion must be an equality. If char(k) = 2, the components ti of a related triple of alternating transformations must lie in M. We obtain that

Now t1 determines the couple (t2, ta) modulo k(l, 1). Again, the inclusion is an equality. Corollary 3.6.2 l>1 is a separable isogeny if char(k) isogeny with kerdl>l of dimension 1 if char(k) = 2. Proof. If char(k)

1= 2,

and an inseparable

1= 2, then dl>l : L(RT(C»

-t

L1, (tb t2, t3)

1-+

t1,

is bijective, so l>1 is a separable isogeny (see [Sp 81, Th. 4.3.7]). If char(k) = 2, then dl>l has a I-dimensional kernel, viz. k(O, 1, 1), so then l>1 is an inseparable 0 isogeny (see [Sp 81, § 9.6]). We see that the situation of RT(C) with its projection l>1 onto SO(N) is the same as that of Spin(N) and its projection 'If' onto SO(N) (treated at the end of § 3.1). Hence RT(C) must be the simply connected covering of SO(N) and is therefore isomorphic to Spin(N), at least in characteristic 1= 2. Rather than exactly figuring out the situation in characteristic 2 (with root systems and all that), we prefer to exhibit directly an isomorphism between RT( C) and Spin(8) in all characteristics. Proposition 3.6.3 There is an isomorphism of algebraic groups cp: Spin(N)

defined by

-+

RT(C)

3.6 The Spin Group of an Octonion Algebra

61

. where ai, bi E CK, I1i N(ai)N(bi ) = 1. It commutes with the projections of Spin(N) and RT(C) on SO(N) and induces an isomorphism of Spin(N) onto RT(C). Proof. We follow a detour via the even tensor algebra of C,

and the algebra End( C) of k-linear transformations of C into itself. .The bilinear transformation

CxC

End(C), (a, b)

-+

t-+

lal;;,

defines a linear transformation

C ®k C

-+

End(C), a ® b t-+ lal;;.

This can be extended to an algebra homomorphism

CPT : T+(C)

-+

End(C)

with CPT(a ® b) = lali)

for

a, bE C.

rJ.

CPT maps the ideal in T+ (C) generated by the elements x ® x - N (x) and u®x®x®v - N(x)(u®v) (u, v,X E C) onto 0, so it can be factored through the even Clifford algebra CI+(N) = T+(C)/TJ. of C, that is to say, we find an algebra homomorphism

cpl:CI+(N)-+End(C)

with cpl(aob)=lali)

for

a,bEC.

for

a, bE C.

In a similar way we define an algebra homomorphism

CP2 : Cl+(N)

-+

End(C)

with CP2(a 0 b) = raTi)

We determine the kernels of CPl and CP2. First assume char(k) i= 2. The center of CI+ (N) has dimension 2 and is spanned by 1 and z = el 0 e2 0 .•• 0 es where el = e, e2, . .. ,es is a standard orthogonal basis of C (see (1.29». Since zo2 = 1, the elements Ul = (1 + z) and U2 = (1 - z) are orthogonal central idempotents. The algebra CI+(N) is the direct sum of two simple two-sided ideals, viz., h = Ul 0 CI+(N) and 12 = U2 0 CI+(N), both of dimension 64 (see [Che 54, Th. 11.2.3 and Th. II.2.4]). Since dimCI+(N) = 128 and dim End( C) = 64, the kernel of CPl must have dimension at least 64. Since CPl i= 0, ker(cpl) is either h or h It follows that CPl(Ul) = or CPl(U2) = 0, so CPl(Z) = 1 or -1. In the former case we replace es by -es, so we may assume CPl(Z) = -1. This implies ker(CPl) = h. We have

4

4

°

62

3. Triality

By conjugating we obtain for i > 1, this implies

r~l T~2 ••• ri!7 Ti!s

= -1. Since el = el

and fi

= -ei

Therefore, ker(
(
E CI+{N)),

is injective and hence an isomorphism since both algebras have dimension 128. Working over K we obtain a homomorphism of algebraic groups 1jJ : RT(C) ..... SO(N) x SO(N), (tl. t2, ta)

1-+

(t2' ta).

RT' (C) = 1jJ(RT(C)) is a closed subgroup of SO(N) x SO(N). For (t2' ta) E RT'(C), the element t1 E SO(N) such that (tl. t2, ta) satisfies the first form of triality is unique, since (3.4) implies

Hence 1jJ : RT( C) ..... RT' (C), has an inverse 'IjJ-1 , which is a morphism. Thus, 'IjJ induces an isomorphism of algebraic groups between RT(C) and RT'(C). From Th. 3.2.1 (iii) it follows that RT'(C) consists of the elements of the form

Since Spin(N) consists of the elements

the algebra isomorphism (
7'1 : (tl' t2, ta) ..... (iI, i a, i2), 7'2 : (tb t2, ta) ..... (ta, t2, tt), 7'a : (tl' t2, ta) ..... (t2' tl, ia). By Th. 3.2.1 (i) and Lemma 3.3.2 these are automorphisms of RT(C). We have similar automorphisms of RT(C).

.s

Proposition 3.6.4 7'2 and 7'a generate a group of automorphisms of RT( C) which is isomorphic to the symmetric group Sa. The nontrivial elements of.s are outer automorphisms. Proof. We have 7'? = 7'~ = id and 7'27'a7'2 = 7'a7'27'a = 7'1. It follows that S consists of id, 7'1. 7'2, 7'a, 7'27'a, 7'a7'2, so has order 6. Assume char(k) # 2. The central elements (1, -1, -I), (-1,1, -1) and (-1, -1, 1) of RT(C) are permuted by the elements of S. This defines an isomorphism o(S onto Sa. An inner automorphism induces the trivial permutation of the center, so the nontrivial elements of S must be outer automorphisms. The above argument breaks down in characteristic 2, since then the above central elements coincide. In that case we work over K, and use the center Z of L(RT( C))). This center consists of the related triples (tb t2, ta) with all ti belonging to the center of M, so by Lemma 3.5.10,

Z = { (A, JL, A+ JL) IA, JL

E

K}.

The above automorphisms 7'i induce in L(RT( C)) automorphisms d7'; which are described by the same formulas as the 7';, but with tl, t2 and ta denoting this time elements of LI (see also Lemma 3.5.9). So they induce permutations of the elements (1,1,0), (1,0,1) and (0,1,1) of Z. Thus, we get an isomorphism of onto Sa again. Since the inner automorphisms of RT( C) act trivially on Z, the nontrivial elements of S are outer automorphisms. 0

.s

Remark 3.6.5 RT(C) ~ Spin(8) is an algebraic group of type 0 4 , The quotient of Aut(RT(C)) by the group of inner automorphisms Inn(RT(C)) is isomorphic to the automorphism group of the Oynkin diagram of 04, which is Sa (see [Hu, § 27.41 or [Stei, § 10]). The above proposition gives an explicit splitting Aut(RT(C)) = S.Inn(RT(C)), with S ~ Sa.

64

3. Triality

We have three representations

(!i

of RT( 0) in OK, given by

Proposition 3.6.6 The representations equivalent.

{}i

are irreducible and pairwise in-

Proof. The image of (}i is SO(N), which acts irreducibly in OK. If char(k) =F 2, then the kernels of {}l. {}2 and {}a are < (1, -1, -1) >, < (-1,1,-1) > and < (-1,-1,1) >, respectively. Since these kernels are distinct, the representations can not be equivalent. In the case that char(k) = 2, we seek refuge in the Lie algebra L(RT(O» again. There, d{}l. d{}2 and d{}a have kernels k(O, 1, 1), k(l, 0,1) and k(l, 1, 0), respectively. These being distinct, the representations are not equivalent. 0 Composing the (}i with the isomorphism cP : Spin(S) = Spin(N) --t RT( 0) of Prop. 3.6.3, we get three surjective homomorphisms 1I"i of Spin(S) onto SO(S), which yield inequivalent representations in O. Prop. 3.6.3 enables us to write these out explicitly. We will see that 11"1 is the same as the homomorphism 11" we considered at the end of § 3.1. For these explicit computations we need to fix a nontrivial idempotent z in the center of CI+(N, K). If char(k) =F 2, we choose a standard orthogonal basis el, e2, ... ,es in OK (see (1.29», and take z = el ° e2 ° ... ° es. A straightforward computation shows that CPl(Z) = -1 and CP2(Z) = 1. (Hint: in the proof of Prop. 3.6.3 it was shown that CPl (z) = ±1 and that CP2{Z) = -CPl(Z), so it suffices, e.g., to check that CP2{z)(e) = e, which is easy using (1.S), Moufang identities and Lemma 1.3.3.) The image of L{Spin{S» under d1l"i is Ll = L{SO(N» in characteristic =F 2, as follows from Th. 3.5.5. If char(k) = 2, we use a standard symplectic basis el = e, e2, . .. , es (as in (1.30» with all N(ei) = 1) and take z = 2::=1 ei 0 ei+4' Then CPl(Z) = 1 and CP2(Z) = O. (Hint: from the proof of Prop. 3.6.3 we know already that CPl (z) = 0 or 1, and that CP2{Z) = CPl (z) + 1, and one easily checks that CPl{z)(e) = e in this case.) As we saw at the end of § 3.1, the Lie algebra L(Spin(S» consists of the elements u = O!o+ 2:i<j O!ijeioej with O!O,O!ij E K,

2::=1 O!i,i+4 = O. Its image under dcp is L(RT(O» = {(tl. t2, ta) E Mal (tl. t2, ta) related}. From Th. 3.5.S we infer that d{}i{L{RT{ 0))) find:

=M

for i

= 1,2,3. We thus

Corollary 3.6.7 The homomorphisms of algebraic groups SO(S) which are defined by

1I"i :

Spin(S)

--t

3.7 Fields of Definition

1r2(al 0 b1 0 · · · 0 ar 0 br } = lalh'l .. ·la)iir , 1r3(al 0 b1 0

... 0

ar 0 br }

= r al rii l ... rarriir (ai, bi E C,

65

IT N(ai)N(bi } = 1), i

are surjective. If char(k) i= 2, they are separable isogenies with kernels of orner 2, viz., < -1 >, < -z > and < z >, respectively. If char(k) = 2, then the 1ri are bijective, inseparable isogeniesi the kernels of d1rl, d1r2 and d1ra are the subspaces k I, k(1 + z} and kz of L(Spin(8», respectively, and the d1ri all have image M. We can now prove the result announced after Lemma 3.5.9.

Lemma 3.6.8 (i) If char(k) = 2, then M = [M, M] (ii) [L(Spin(8», L(Spin(8»] = L(Spin(8».

= [L1I L1] = [Lo, Lo].

Proof. We know already that [Lo, Lo] ~ M (see Cor. 3.5.6 and Th. 3.5.8), hence [M,M] ~ [Ll,L 1 ] ~ [Lo,Lo] ~ M. Therefore, to get (i) it suffices to prove that [M, M] = M. This will follow from [M,MI = M. Since M = d1rl(L(Spin(8))), it suffices to prove (ii). We do this by picking generators of L(Spin(8» and expressing these as sums of commutators. As generators we take 1, the ei 0 ej with 1 ~ i < j ~ 8, j i= i + 4, and the elements ei 0 ei+4 + ej 0 ejH with 1 ~ i < j ~ 4, using the notations introduced just before the above corollary. In the following commutator relations, which are easily verified, the indices i, j, k are distinct and 1 ~ i,j,k ~ 4.

hence

This also enables us to write 1 as a sum of commutators. Further,

lei 0 ek,ek+4 0 ej] = ei 0 ej, lei 0 ek, ek+4 0 ej+4] = ei 0 ej+4, [ei+4 0 ek, ek+4 0 ej+4] = ei+4 0 ej+4. Thus we see that all generators of L(Spin(8» are in the commutator subal0 gebra.

3.7 Fields of Definition The results of the foregoing sections enable us to prove that the algebraic group RT( C) is defined over k.

66

3. Triality

Proposition 3.7.1 Let C be an octonion algebra O1Jer k. The algebraic group RT(C) of related triples of rotations is defined O1Jer k. Proof. We proceed as in the proof of Th. 2.4.6. Let A = G L( CK )3 , the product of three copies of GL( CK). This is an algebraic group which is defined over k. Further, V = EndK(CK®KCK,CK) is a vector space over K with k-structure V(k) = End(C ®k C, C). In a similar way we have the vector space Q(C)(K) over K of quadratic forms on CK, which has Q(C)(k) as a k-structure. The multiplication map J.1. : x ® y 1-+ xy lies in V(k) and in Q(C)(k) we have the norm N of C. On the vector space

x=

V

X

Q(C)(K)3

we have an action of A defined by

(tllt2,t3): X

-+

X, (v, qll Q2,q3) 1-+ (til OVO(t2®t3),QI ott,Q2ot2,Q30t3)'

Now notice that til 0 J.1. 0 (t2 ® t3) = J.1. if and only if the triple (tb t2, t3) satisfies the first form of triality (3.4). Therefore, the stabilizer G~ of x = (J.1., N, N, N) E X consists of the triples (tt, t2, t3) of orthogonal transformations that satisfy (3.4), i.e., of the related triples of rotations. Hence, this stabilizer is just the group RT(C). By [Sp 81, 12.1.2] the stabilizer will be defined if the kernel of the differential dcpe has dimension 28 = dimRT(C), where cp(a) = a.x (a E A). Arguments like those used after the proof 3.6.1 show that this is indeed the 0 case. By Prop. 3.6.3 we identify RT(C) with Spin(N). The previous Proposition then gives a structure of algebraic group over k on Spin(N).

3.8 Historical Notes The study of triality originated with the treatment of geometric triality between points, spaces of one kind and spaces of the other kind on a quadric in complex seven-dimensional projective space, by E. Study [Stu] and E. Cartan rCa 25]. Octonions appeared in this connection in papers by F. Vaney [Va] and E.A. Weiss [Weiss]. Local triality in Lie algebras of type D4 was studied by H. Freudenthal [Fr 51] for the real and complex case. F. van der Blij and T.A. Springer [BlSp 60] treated algebraic triality over arbitrary fields and local triality over fields of characteristic not two. N. Jacobson [Ja 64b], [Ja 71] dealt with local triality in characteristic not two, and with algebraic triality in [Ja 60]. The results on local triality in characteristic two presented here are new, to our knowledge. In the literature one often finds the Principle of Triality formulated in a different way: for any similarity t, there exist similarities tl and t2 such

3.8 Historical Notes

67

that either tl{XY) = t{X)t2{Y) for all X,Y E Cor tl{XY) = t{y)t2{X) for all x, Y E C, and similarly with local trialityj this convention goes back to Cartan rCa 25J. The formulations used here (and also in [Ja 64bJ and [Ja 71]) are more convenient in applications and make some proofs simpler. Lemmas 3.3.2 and 3.5.9 ensure that the different formulations are in fact equivalent.

4. Twisted Composition Algebras

Twisted composition algebras are somewhat similar to composition algebras, and the two notions are strongly interrelated. On the other hand, twisted composition algebras have close connections with Jordan algebras, which we will deal with in the next chapter. In fact, the motivation to set up this theory comes from exceptional Jordan algebras and twisted algebraic groups of type D4. In particular, the notion of reduced twisted composition algebra comes from reduced Jordan algebras. We refer to Ch. 6 for the details. In this chapter, k will denote a field and I a separable cubic field extension of k. The normal closure of lover k is I' = l(d), where d satisfies a separable quadratic equation over k (see, e.g., [Ja 74, Th. 4.13 and ex. 3 in § 4.8]). If char(k) :f; 2, we can take d = ..fIj, i.e., either of the square roots of the discriminant D of lover k . We set k' = k( d). So either I is a Galois extension of k with cyclic Galois group of order three (a cubic cyclic extension), and then I' = I and k' = k, or I' and k' are quadratic extensions of I and k, respectively, and I' is a cubic cyclic extension of k'. We identify [' with k'®k1. Fix a generator (f of Gal(l' jk'). We view it also as an element of Gal(l' jk) and as a k-isomorphism of I into I'. We denote by T the element of Gal(l' jk) whose fixed field is I if I' :f; l, and T = id if l' = I. We denote by Nl/k and Trl/k norm and trace function of the extension I of k. We first introduce twisted composition algebras for the case that the separable cubic extension I of k is normal, so a cubic cyclic extension; we call algebras of this type "normal twisted composition algebras" . The basic theory of these algebras will be developed in § 4.1. In the next section we generalize this to the case that the separable cubic extension I of k is not normal; we then get the general twisted composition algebras. In both cases we have a vector space with an additional operation; this is binary in the normal case (a product), and unary in the nonnormal case (a squaring operation). There will be close relations between the two types. In the theory of normal twisted composition algebras the characteristic of the field k can be arbitrary, but when dealing with the nonnormal twisted composition algebras, more precisely with their relations to the normal algebras, we have to exclude characteristic two and three. For the applications to Jordan algebras in Ch. 6 this is not a problem, since there a restriction to characteristic not two or three will be in force anyway.

70

4. Twisted Composition Algebras

To study the automorphism group of a twisted composition algebra in

§ 4.4, we also have to consider twisted composition algebras over a split cubic extension of a field K, i.e., over K EB K EB K; we develop that theory in § 4.3. But in all other sections we work over a separable cubic field extension 1 of

k. In the last sections we discuss explicit constructions of twisted composition algebras, involving connections with cubic central simple algebras.

4.1 Normal Twisted Composition Algebras We first deal with the case of a cubic cyclic field extension 1 of k, so l' = 1 and k' = k. Definition 4.1.1 A vector space F over 1 provided with a nondegenerate quadratic form N, and associated bilinear form ( , ), is said to be a normal twisted composition algebra over land q if there is a k-bilinear product * on F which satisfies the following three conditions. (i) The product x * y is q-linear in x and q2-linear in y, that is

(Ax) * y x * (AY)

= q{A)(x * y), = q2{A)(x * y)

(x, y E F,

A E l);

(ii) N{x * y) = q(N(X))q2(N(y)) (x, Y E F); (iii) (x*y,z) =q{(y*z,x))=q2{(z*x,y)) (X,y,zE F). One calls N the norm of the algebra. The normal twisted composition algebra is also denoted by (F, *, N), or simply by F. A bijective l-linear mapping t : Fl ~ F2 between normal twisted composition algebras Fl and F2 over 1 satisfying t{x * y) = t(x) * t(y) is called an isomorphism. We will see in Cor. 4.1.5 that the norm of a normal twisted composition algebra is already determined by the linear structure and the product *, and that an isomorphism preserves the norm. If A E k* we define for x,y E F

It is immediate from the definitions that (F, *>., N>.) is a normal twisted composition algebra, denoted by F>.. We say that F>. is an isotope of F, and that F and F>. are isotopic. . It is easy to see that F>.p. = (F>.)p. (A, J1. E l*). From condition (iii) it follows that

4.1 Normal Twisted Composition Algebras

T(x)

= (x*x,x)

71

(x E F)

is invariant under a, so T(x) E k for all x E F. Thus, T is a cubic form over k in 3 diml F variables. In the following two lemmas we give a number of useful identities.

Lemma 4.1.2 In any normal twisted composition algebm F we have for

X,y,Z,w E F, (x*z,y*z) =a«(x,y))a 2 (N(z)), (x*z,x*w) =a(N(x))a 2 «(z,w)), (x * z, y * w) + (x * w,y * z) = a«(x, y))a 2 «( Z,W )).

(4.1) (4.2) (4.3)

Proof. From (ii) in the definition we infer

N«x + y) * (z

+ w)) = a(N(x + y))a2 (N(z + w)).

Write this out and, using (ii), cancel all terms which contain only two of the variables; this yields a relation we call (*). Taking w = 0 in (*), we get (4.1) and taking y = 0 we get (4.2). Now cancel all terms in (*) containing only three variables by using (4.1) and (4.2), then what remains is precisely 0 equation (4.3). Lemma 4.1.3 In any normal twisted composition algebm F we have for

x,y,z

E

F, x * (y * x) = a(N(x))y, x * (y * z) + z * (y * x) = a( (x, z) )y, (x * y) * x = a 2 (N(x))y, (x*y)*z+(z*y)*x=a 2 «(x,z))y, (x * x) * (x * x) = T(x)x - N(x)(x * x).

(4.4) (4.5) (4.6) (4.7) (4.8)

Proof. Since (by (iii) ofDef. 4.1.1) (x*(y*x),u) =a 2 «(u*x,y*x)) = a(N(x))(u,y) (by (4.1)) = a(N(x))(y,u) for all u E F, we get (4.4). Linearizing this equation, we find (4.5). The following two equations are proved in the same way. Finally, to prove (4.8) take y = x and z = x * x in (4.5):

x * (x * (x * x)) + (x * x) * (x * x) Since x * (x * x) = a(N(x))x by (4.4), we get

= T(x)x.

72

4. Twisted Composition Algebras

(x * x) * (x * x)

= T(x)x - x * (O'(N(x»x) = T(x)x -

N(x)(x * x).

o Consider for a E F the left multiplication l: : F - F, x is 0'2-linear, and the right multiplication F - F, x 1-+ O'-linear. From (4.4) and (4.6) we infer:

r: :

1-+

X

a * x, which

* a,

which is

Lemma 4.1.4 If a E F has N(a) '" 0, then l: and r: are invertible, with

inverses r N (a)-l a and IN(a)-l a ' respectively. The following result is also immediate from (4.4).

Lemma 4.1.5 The norm N on a normal twisted composition algebra F over land 0' is determined by the linear structure over l and the product * on F. Isomorphisms of normal twisted composition algebras are norm preserving. There exists a close relationship between normal twisted composition algebras and ordinary composition algebras; in fact, either kind can be related to the other one. First, consider a normal twisted composition algebra F over land 0'. Pick a, bE F with N(a)N(b) '" O. Define

xy = (a * x) * (y * b). This product is I-bilinear, and the norm N satisfies:

N(xy)

= N«a * x) * (y * b)) = O'(N(a * x)0'2(N(y * b» = 0'2(N(a»N(x)N(y)0'(N(b))

= >.N(x)N(y) with>' = 0'2(N(a»0'(N(b» '" O. Take as new norm Iv = >'N. This obviously permits composition (see Def. 1.2.1). An identity element is

as a straightforward computation using (4.4) and (4.6) shows. We have thus obtained a structure of composition algebra on F. As a consequence one finds using Th. 1.6.2: Proposition 4.1.6 A normal twisted composition algebra over l can only

have dimension 1, 2, 4 or 8 over 1. From the considerations below it will follow that in each of the dimensions 1, 2, 4 and 8 there do exist normal twisted composition algebras. We will mainly be interested in such algebras of dimension 8; we also call these nor-

mal twisted octonion algebras.

4.1 Normal Twisted Composition Algebras

73

We further exploit the connection between ordinary composition algebras and normal twisted composition algebras. Let F be as before. Denote by N(F)* the set of nonzero values of the norm N and by M(N) the group of multipliers of similarities of N (see 1.1). Proposition 4.1.7 M(N) is a a-stable subgroup of l* and N(F)* is a coset )'M(N) , with NI/k(.\) E M(N). Proof. Let N be as before. Clearly, M(N) coincides with the similar group M(N), which is the set of nonzero values N(x) (see the beginning of the proof of Theorem 1.7.1). It follows from what we established above that if a, bE F, N(a)N(b) i- 0 we have

M(N)

= 0'2(N(a))0'(N(b))N{F)*,

from which we conclude (using that M(N) contains (l*)2) that N(F)* is a coset of M (N) and also that M (N) is the set of nonzero elements of the form 0'2{N{a))0'{N(b))N(c) (a, b, c E F). This implies that M{N) is a-stable. Taking a = b = c we see that NI/k(.\) E M{N) for all .\ E N(F)*. 0 We now give a construction of a normal twisted composition algebra from a composition algebra Cover k. As before, 1 is cubic cyclic field extension of k and 0' a generator of the Galois group. Let Nc be the norm of C. Extend the base field to l: F = l®kC, extend Nc to a quadratic form over 1 on F, denoted by N, and similarly for conjugation and the product. Define a a-automorphism cp of F by

cp(e ® x) = O'{e) ® x.

= O'(N(z)) for z E F. Define (4.9) x * y = cp(x)cp2(y) (x, Y E F). A straightforward computation shows that (F, *, N) is a normal twisted comNotice that N{cp(z))

position algebra. For the verification of point (iii) in Def. 4.1.1, use (1.IO). In this verification one sees that it is necessary to take the conjugates of x and y in definition (4.9) of the product * ; without conjugation, things go wrong. We denote this twisted composition algebra by F{C). Definition 4.1.8 A normal twisted composition algebra F over land 0' is said to be reduced if there is a composition algebra Cover k and .\ E l* such that F is isomorphic to the isotope F(Ch,. Proposition 4.1.9 (i) If the reduced normal twisted composition algebras F = F( C», and F' = F( C'h,.' over land 0' are isomorphic, then the composition algebras C and C' are isomorphic. (ii) The normal twisted composition algebras F{C)A and F{C)N over land 0' have the same automorphism group.

74

4. Twisted Composition Algebras

Proof. (i) An isomorphism from F onto F' preserves the norm by Lemma 4.1.5, hence the norms Ne of C and Ne' of C' are similar over l. It follows (see the first paragraph of the proof of Th. 1.7.1) that Ne and Ne' are equivalent over l. By a result of Springer (see [Sp 52, p. 1519,b] or [Lam, p. 198]), they must be similar over k, so by Th. 1.7.1 C and C' are isomorphic. (ii) The condition that a linear bijection is an isomorphism is invariant under 0 isotopy. . We will develop several criteria for a normal twisted composition algebra to be reduced. The following theorem is the first result in this line.

Theorem 4.1.10 Let F be a normal twisted composition algebm over ,. The

following conditions are equivalent. (i) F is reduced. (ii) T represents zero nontrivially, i.e., there exists x i= 0 in F such that T{x) = (x * x, x) = O. (iii) There exists x i= 0 in F such that x * x = ).X for some). E l. Proof. (i) => (ii). Assume F is reduced. We may assume that F = F{C». as before. We write ( , ) and ( , )e for the bilinear forms associated with N and Ne, respectively. For x, y E C we have x * y = ).xy. Pick x E C, x i= 0, (x,e)e = 0, then x = -x and x 2 = -Ne(x)e by (1.7). Using (1.10) we find

(x * x,x)

= ).(x2 ,x)e = -).Nc(x)(e,x)e = 0,

which proves (ii). (ii) => (iii). If T represents 0 nontrivially, we pick y E F, y f; 0, such that T(y) = O. By (4.8),

(y

* y) * (y * y) =

-N(y)y * y.

We take x = y if Y * Y = 0, and x = y * y if y * y f; OJ in either case, x f; 0 and x * x = ).x for some). E l. So (iii) holds. (iii) => (i). Under the assumption of (iii), we have to construct a composition algebra Cover k such that F ~ F{C),\ for some). E l*. We divide the proof into a number of steps. Pick x E F, x i= 0, such that x * x = Ax with), E l. We first introduce e E F which is going to play the role of identity element in C. (a) If N(x) = J.I. f; 0, we put e = x. Using (4.6) we see that

a 2 (J.I.)x It follows that a 2 (J.I.) properties

= a 2 (N(x»x =

(x * x) * x = ).a().)x.

= ).a().), so J.I. = a().)a 2 ().) e * e =).e

N(e)

= J.I. =

and), f; O. Hence e has the

with), E " ). f; 0,

a().)a

2

().)

i=

O.

(4.10)

(4.11)

4.1 Normal Twisted Composition Algebras

75

(b) Assume now N(x) = O. By (4.8),

(x *x) * (x *x) Now either x

= T(x)x = (x*x,x)x = >'(x,x)x = O.

* x = 0, or y = x * x =I 0 satisfies y * y = 0 and N(y)

= N(x * x) = a(N(x))a 2 (N(x)) = O.

So we may as well assume that we have x =I 0, N(x) = 0 and x * x = O. Since x is contained in a hyperbolic plane (see § 1.1), there exists y E F with N(y) = 0 and (x,y) = -1. Using (4.5) we find:

(z E F),

(4.12)

since (x, y) = -1. Hence

F=x*F+y*F. The sets x * F and y * Fare subspaces of the vector space F. They are totally isotropic for N, since

N(x * u) = a(N(x))a 2 (N(u)) = 0

(UEF),

and similarly for y * F. It follows that both subspaces have dimension ~ ~ dim F (see § 1.1 again). Together they span F, so they must both have dimension equal ~ dim F and we have a direct sum decomposition

Consider the right multiplication r; in F. If z*x = 0, then z = -x* (Z*y) by (4.12), so kerr; ~ x* F. By (4.6), on the other hand, r;(xu) = (xu) *x = a 2 (N(x))z = O. Hence kerr; = x * F. In the same way we derive from (4.7) the following analog of (4.12): (x

* z) * y + (y * z) * x = -z

(z E F).

(4.13)

With the same arguments as above one derives:

For the left multiplication

l;

one derives using (4.13) and (4.4),

We define a = x * y and b = y * x. Using the identities of Lemma 4.1.3 one verifies the following relations, where a = (y, x * y) = (y, a ).

76

4. Twisted Composition Algebras

x * a = -x, a * x = 0, x * b = 0, b * x = -x, a*a=b+(1(a)x, b* b = a + ax, a * b = 0, b * a = (12(a)x.

(4.14) (4.15) (4.16) (4.17) (4.18) (4.19)

°

From (4.14) and (4.15) we infer that a f. and b f. 0, respectively. Using (4.3) and the relations in Def. 4.1.1 one further sees

(4.20) (4.21) (4.22)

(a,b)=I, N(a) = N(b) = 0, ( a, x) = (b, x) = o. In this case we put e

= (12(a)x + a + b. We see that

* = «(12(a)x + a + b) * «(12(a)x + a + b)

e e

= -ax + b + (1(a)x - (1(a)x + (12(a)x + a + ax = (12(a)x + a + b = e and

N(e) = N«(12(a)x + a + b) = (12(a)2 N(x) + N(a) = (a, b) = 1.

+ N(b) + (12 (a) ( x, a) + (12(a)( x, b) + (a, b)

Again, e satisfies equations (4.10) and (4.11), this time with ,x = tJ. = 1. (c) The next step towards the construction of a composition algebra C is the definition of conjugation in F and of a (1-linear mapping

x = -x + (x, e )N(e)-le Then - is a linear map, e = e, N(x) Further, we have for x E F,

(x E F).

= N(x), so (x,y) = (x,y),

x * e = -x * e + (x * e,e)N(e)-le = -x*e+(1«(e*e,x))N(e)-le = -x * e + (1(,x)(1( (e, x) )N(e)-le = -x*e+(1«(e,x))(,x(12(,x))-1(e*e) = (-x+ (x,e)N(e)-le) *e = x*e,

(4.23) and

x = x.

i

r 4.1 Normal Twisted Composition Algebras

77

and similarly e * x = e * x.

Notice that if F = F(C)., then x define the a-linear mapping


* e = A
A- 1 (X * e).

(4.24)

We show that

* e) 1 = A- a(A)-1( -(e*e)*x+a 2((e,x})e) = A- 1 ( -e*x+A- 1a(A)-1a2((e,x})e*e) =A- 1(e*x).
* e) 1 = A- a(A)-1((e*x) *e) = A- 1a(A)-1a 2(N(e»x = x. = A-1 (-A--:1'-"'(e-*-x""")

Further,

(x,y E F).

(4.25)

We show that this product defines a structure of composition algebra over l on the vector space Fj see Def. 1.2.1. First, for x E F,

and in a similar way we find that ex = x. To prove that the norm N permits composition up to a scalar factor we first show that

N(
N(xy) = N(A-1(
78

4. Twisted Composition Algebras

Hence, if we define

(x E F),

IV(x) = 1-'-1 N(x)

(4.27)

IV permits composition. Thus we have proved that the vector space F with the new, bilinear, product and the norm IV is a composition algebra over l; we denote this by 6. The conjugation - we defined in step (c) is the normal conjugation in this composition algebra 6 with respect to the bilinear form of its norm N. SO xy = yx, xx = N(x), etc. (e) In this step we prove that cp is a l1-automorphism of the composition algebra 6.

cp(xy)

= oX- 1(xy * e) =

oX- 1(yx * e)

= oX- 111(oX)-1((cp2(y) * cp(x)) = oX -ll1(oX)-l ( - (e * cp(x))

* e)

* cp2(y) + u 2((cp2(y), e) )cp(x))

* * ( - cp2 (y) + (cp2 (y), e )N (e) -1 e) ) = oX -lu(oX)-l (( e * cp(x)) * cp2(y)) = oX -lU(oX) -1 ( (oXcp2( cp(x))) * cp2(y)) = oX-I (cp2(cp(x)) * cp2(y)) = oX-I u( oX) -1 ( ( e cp( x))

(x, y E 6).

= cp(x)cp(y) From (4.26) it follows that

(x E 6).

(4.28)

= {x E 61 cp(x) = x}. is a vector space over k such that 6 ~ l ®k C

(4.29)

N(cp(x)) = u(N(x)) (f) Define

C

Then C (see [Sp 81, 11.1.6]). From the multiplicativity of cp it follows that C is closed under multiplication, and (4.28) implies that N(x) E k for x E C. Since C contains the identity element e, it is a composition algebra over k. To verify condition (4.9), we compute

oX(cp(X)cp2(Y)) = cp2(cp(x)) * cp(cp2(y)) =x*y (X,yEF). It follows that F £:! F( Ck

o

The above theorem enables us to prove, for some special fields k, that every normal twisted octonion algebra over a cubic cyclic extension field l of k is reduced, namely, if the cubic form T in 24 variables over k represents 0

4.2 Nonnormal Twisted Composition Algebras

79

nontrivially. This is so in the following two cases. (i) k a finite field.

A theorem of Chevalley [Che 35, p. 75] (also in [Gre, Th. (2.3)], [Lang, third ed., 1993, p. 214, ex. 7]), [Se 70, § 2.2, Th. 3] or [Se 73, p. 5]) implies that every cubic form in more than three variables over a finite field represents 0 nontrivially. (ii) k a complete, discretely valuated field with finite residue class field. Every cubic form in more than nine variables over such a field represents 0 nontriviallyj see [Sp 55, remark after Prop. 2], or also [De] or [Le]. If k is an algebraic number field, then also every normal twisted octonion algebra over a cubic cyclic extension 1 of k is reducedj see the end of 4.8. This will follow from a further study: in §§ 4.5 and 4.6 we will thoroughly explore the structure of normal twisted octonion algebras, especially of those whose norm N is isotropic, leading to another criterion (Th. 4.8.1) for normal twisted octonion algebras to be reduced. But first we introduce general twisted composition algebras in the following section, and in the next section we will identify the automorphism groups of reduced, normal or nonnormal, twisted composition algebras as twisted forms of an algebraic group of type D4·

4.2 Nonnormal Twisted Composition Algebras In the present section the separable cubic extension field 1 of k is not necessarily normal. Notations are as fixed in the introduction to this chapter. Suppose l' f l. Then O'{A) ¢. 1 if A E l, A ¢. k, so we cannot carryover the definition of a normal twisted composition algebra in Def. 4.1.1 to a vector space F over l. However, 0'{A)0'2(A) E l. If F' = A' ®l F is a normal twisted composition algebra over l' and if x*x happens to lie in F for X E F, then also (AX) * (AX) E F. This suggests that instead of the product * we must consider the square x*2 = X * x. Thus we are led to the the following definition. For A E 1 we write A*2 = 0'(A)0'2(A).

Definition 4.2.1 Let 1be a separable cubic extension of k and 0' a nontrivial k-isomorphism of 1into its normal closure l'. By a twisted composition algebra over land 0' we understand a vector space F over 1 provided with a unary operation *2 : F - t F, called the squaring operation, and a nondegenerate quadratic form N, called the norm, with associated bilinear form ( , ), such that the following four axioms are satisfied: (i) (AX) *2 = A*2 X *2 (A E l, X E F)j (ii) f: F x F - t F defined by

80

4. Twisted Composition Algebras

(x,y E F)

is k-bilinearj (iii) N(x *2) = N(x) *2 (x E F)j (iv) T(x) = (X*2,X) E k (x E F). We use the notation (F, *2, N), or simply F, for such a twisted composition algebra. An isomorphism of twisted composition algebras Fl and F2 as above is a bijective l-linear map t : Fl -+ F2 such that t(x*2) = t(x) *2 for x E Fl. Notice that in this definition it is allowed that the cubic extension llk be Galois, so l' = I and k' = k. If F is a twisted composition algebra as before, and >. E 1*, we define the isotope F>. to be (F, *'2, N>.), where x*'2 = >.(x *2) and N>.(x) = >. *2 N(x). It is immediate that FA verifies the axioms of a twisted composition algebra.

The relations between normal and nonnormal twisted composition algebras are given in the following two propositions. Proposition 4.2.2 (i) If l is cubic cyclic over k and (F, *, N) is a normal twisted composition algebm over I, then F with the same norm N and squaring opemtion defined by x *2 = X *X (x E F) is a twisted composition algebm over land 0". (ii) Let char{k) f:; 2,3. If I is cubic over k, but not necessarily Galois, and (F, *2, N) is a twisted composition algebm over I and (7, then F' = I' ®l F with the extension of N to a quadmtic form over l' on F' carries a unique structure of normal twisted composition algebm over I' and (7 such that x *2 = x*x for x E F. (For uniqueness it is necessary that the extension of the isomorphism (7 of l into l' to an automorphism of I' is given.) (iii) If Fl and F2 are twisted composition algebms over land (7 with char(l) ::/: 2,3 and F{ and F2, respectively, are their extensions to normal twisted composition algebms over I' as in (ii), then an l-linear bijection t : Fl -+ F2 is an isomorphism of twisted composition algebms if and only if its I' -linear extension is an isomorphism of normal twisted composition algebms between F{ and F~.

Proof. (i) being obvious, we tackle (ii). Extend the k-bilinear mapping f : F x F -+ F as in (ii) of Def. 4.2.1 to a k'-bilinear mapping F' x F' --+ F'. Pick a basis ell ... ,en of F over lj this is also a basis of F' over l'. Write n

f(x, y) =

L
(x,y E F')

i=l

with symmetric k'-bilinear
j

.,

4.2 Nonnormal Twisted Composition Algebras

81

2

'X,y)

=

(>. E l', X,y

LO'j(>')?p"j(X,Y)

E

F')

I

j=O

with unique k'-bilinear mappings ?p"j : F' x F'

-+

I!

l'. Thus,

2

f(>'x, y) =

I: O'j (>.)fJ(x, y)

"

(>. E l', X,y

E

F')

j=O

with unique k'-bilinear fJ : F' x F' -+ F'. Repeating this argument, we find unique k'-bilinear mappings 9"j : F' x F' -+ F' such that 2

f(>'x, JJY) = L

(>',JJ E l', x,y E F').

O"(>.)O'j (JJ)9"j(X, y)

(4.30)

',j=O

From the symmetry of f we infer, using Dedekind's Theorem again, that

9i,i(X, y)

= 9i,i(Y, x)

(X,y E

F')

(4.31)

>.

and JJ, we

(a, f3 E l', x, Y E F')

(4.32)

for 1 ::::; i, j ::::; n. By considering f (>.ax, JJf3y) as a function of find with Dedekind from (4.30) that

for 1 :$ i,j ::::; n. Using this relation and the fact that f(x,x) = 2X*2, we rewrite condition (i) of Def. 4.2.1 in the form

0'(>.)0'2(>.)f(x, x) = f(>'x, >.x) 2

= L O"(>.)O'j (>')9i,j(X, x)

(>. E l', x E F').

i,j=O

Linearizing in >., we get for >., JJ

E

l', x

E

F'

2

(0'(>.)0'2(JJ)

+ 0'2(>')0'(JJ))f(x,x) =

L (O'i(>.)O'j(JJ)

+ O'j (>.)O'i(JJ))9i,j (x, x)

',j=O 2

=L

O'i (>.)O'i (JJ)(9i,j (x, x)

+ 9j,,(X,x)).

i,i=O By Dedekind this implies

9i,i(x,x)

+ 9i,i(X,X) = 0

(x E F', (i,j) 1= (1,2),(2,1)).

If char(k) 1= 2, it follows by (4.31) that 9i,i(x, x) = 0, so 9i,i is antisymmetric for (i,j) 1= (1,2), (2, 1). From this we derive, using the symmetry of f, (4.30) and (4.31),

82

4. Twisted Composition Algebras

f(x,y)

1 = 2(f(x,y) + f(y,x)) =

1 2(91,2(X,y) + 92,l(X,y) + 91,2(y,X) + 92,l(y,X))

= 91,2(X, y)

+ 91,2(Y, x)

(x, y E F').

Hence if we define

x * y = 9l,2(X, y)

(x,y

E

F'),

we have a k'-bilinear product on F' which by (4.32) satisfies condition (i) of Def. 4.1.1 and such that x *2 = X*x for x E F. The uniqueness of the product * is obvious from the proof. Extend the norm N on F to a quadratic form over l' on F'. In condition (iii) of Def. 4.2.1 we replace x by AX + p,y + vz + (!w, with x, y, Z, w E F and A, p" v, e E k. Writing this as a polynomial in A, p" v and e and equating the terms with Ap,ve, we find

(f(x, y), f(z, w)} + (f(x, z), f(y, w)} + (f(x, w), f(y, z)} = 0'( (x, y) )0'2( (z, w}) + 0'( (z, w) )0'2( (x, y}) + 0'«X,z})0'2«y,W}) + 0'«y,w})0'2«x,z}) + 0'( (x, w) )0'2( (y, z}) + 0'( (y, z) )0'2( (x, w}). Here we use that k has more than four elements. The above relation is fourlinear over k, so it remains valid if we extend k to k'. Hence it also holds for x,y,z,w E F'. Replace x,y,z,w in the above relation by AX,p,y,vz,(!W, respectively, with x, y, z, w E F' and A, p" v, eEL'. In the relation we thus obtain, the terms with 0'(A)0'2(p,)0'(v)0'2(e) on either side must be equal by Dedekind, so

(x * y, z * w) + (x * w, z * y) = 0'( (x, z) )0'2( (y, w}). Replacing z by x and w by y yields the validity of condition (ii) of Def. 4.1.1 for F'. Applying a similar argument to (iv) of Def. 4.2.1, viz., trilinearization over k and then extension of this field to k', yields that T(x) E k' for all x E F'j here we use char(k) i 3. Replacing x by AX + p,y + vz and using Dedekind then proves that condition (iii) of Def. 4.1.1 holds for F'. This completes the proof of part (ii) of the Proposition. As to (iii), let t : Fl -+ F2 be an isomorphism of twisted composition algebras. Denote its l'-linear extension also by t. On F{, t-l(t(x) * t(y)) is a product for a normal twisted composition algebra which extends the squaring operation x I-t x *2 on Fl. since t- 1(t( x) *2) = x .. 2 for x E Fl. By uniqueness in (ii), t-1(t(x) * t(y)) = x * y for x, y E F{, that is, the extension t is an isomorphism of normal twisted composition algebras. The converse is obvious. 0

4.2 Nonnormal Twisted Composition Algebras

83

If P is a twisted composition algebra over 1 with char(l) 1= 2,3, then the normal twisted composition algebra F' over l' determined by F as in part (ii) of the above proposition will be called the normal extension of F. If 1 is a cubic cyclic extension of k of characteristic 1= 2, 3, a twisted composition algebra over l may be identified with the normal twisted composition algebra it determines. The restriction to fields of characteristic 1= 2, 3 is not too much of a nuisance; the theory of twisted composition algebras is set up in view of applications to Jordan algebras (see Ch. 5 and 6), and there we need the same restriction on the characteristic. It is clear that if two twisted compositions algebras are isotopic, the same holds for their normal extensions. Corollary 4.2.3 Let char(l) 1= 2,3. The norm N of a twisted composition algebra F over 1 is uniquely determined by the linear structure and the squaring operation *2. Isomorphisms of twisted composition algebras preserve the norm. A twisted composition algebra can only have dimension 1, 2, 4 or 8 over l. Proof. Let F' be the normal extension of F. In the proof of part (ii) of the above proposition, the product * on F' is determined by the squaring operation *2 on F and the linear structure; the norm plays no role there. By Lemma 4.1.5, the norm on F' is determined by the product and the linear structure. This proves the first statement. The second one is proved in a similar way. The last statement follows from Prop. 4.1.6 on the dimensions of twisted composition algebras. 0 As in the normal case, we speak in the case of dimension 8 about twisted octonion algebras. If F' is a normal twisted composition algebra over l' and l' 1= l, how do we find twisted composition algebras F over l such that F' is the normal extension of F ? The following proposition gives an answer to this question. Recall that T is the generator of Gal(k' /k). Proposition 4.2.4 Assume char(k) ::f. 2,3 and l' ::f. l, so [l' : k] = 6. Let F' be a normal twisted composition algebra over l' . (i) If F' is the normal extension of a twisted composition algebra over land 0', then there exists a unique bijective T-linear endomorphism '1.1. of F' satisfying 2 '1.1. = id and (x,y E F') (4.33) u(x * y) = u(y) * u(x)

such that F = Inv(u) = {x E F'I u(x) = x}. (ii) Conversely, for any '1.1. as in (i), Inv(u) is a twisted composition algebra over l which has F' as its normal extension. (iii) Every '1.1. as in (i) satisfies N(u(x))

= r(N(x))

(xEF').

84

4. Twisted Composition Algebras

(iv) If F = Inv(u) and Fl = Inv(ul) are twisted composition algebras over 1 and 0' which both have F' as their normal extension, then F ~ Fl if and only if there exists an automorphism t of F' such that Ul = tut- 1 , and every isomorphism: F ~ Fl extends to such an automorphism. In particular, Aut(F)

= {t!F !t E Aut(F'), tu = ut }.

Proof. (i) Identify F' with l' ®l F. The transformation u = r ® id is bijective r-linear with u 2 = id and F = Inv(u). To prove (4.33), consider for >',J.I. E 1 and x,y E F, z = (>.x) * (J.l.y) + (J.l.y) * (>.x) E F, so z is invariant under u. By the r-linearity of u,

Since

TO'

= 0'2r and since>. and J.I. are r-invariant, u(z) = z implies 0'2(>.)0'(J.I.)U(X * y)

+ 0'(>.)0'2(J.I.)u(y * x) =

* (J.l.y) + (J.l.y) * (>.x) = 0'(>.)0'2(J.I.)(u(X) * u(y» + 0'2(>.)0'(J.I.)(u(y) * u(x). (>.x)

By Dedekind's Theorem, u(x * y)

= u(y) * u(x) (x,y E F). Since l' = k' ®k 1 we have F' = I' ®l F = k' ®k F. Using the r-linearity of u we see that 4.33 holds. (iii) Apply u to both sides of equation (4.4) and use (4.6). (ii) F = Inv(u) is a vector space over Inv(r) = 1of the same dimension as the dimension of F' over l' (see, e.g., [Sp 81, 11.1.6]). If x E F, then u(x * x) = u(x) * u(x) = x * x, so x * x E F, and further N(x) = N(u(x) = r(N(x», so N(x) E 1. It is straightforward now that F with x· 2 = X * x and the restriction of N as norm is a twisted composition algebra which has F' as its normal extension. (iv) Let s : Fl -+ F2 be an isomorphism of twisted composition algebras over 1 and 0' which both have F' as their normal extension. Let Fi = Inv(ui) (i = 1,2). Define t : F' -+ F' as the I'-linear extension of s. From

(x E F) we derive as in the proof of part (i) above that t(x

* y) =

t(x)

* t(y)

(x,y E

F').

From t(Fl) = F2 and it follows U2 = tUlrl. This implies the "only if" part 0 of (iv). The "if" part is immediate.

4.2 Nonnormal Twisted Composition Algebras

85

A r-linear mapping u as in the above proposition is called an involution of F'. If F = Inv( u), then u is said to be the involution associated with F. Let F be any twisted composition algebra over l, with norm N. We have a (partial) analogue of Prop. 4.1.7. Let, as before, N(F)* to be the set of nonzero values of Non F, and M(N) the multiplier group of N. Proposition 4.2.5 N(F)* is

~

coset >..M(N) with Nl / k (>..)

E

M(N).

Proof. We represent F as in the preceding Proposition, via F' and u. We proceed as in the proof of Prop. 4.1.7, with a E F and b = a. We obtain a structure of composition algebra C'on F ' , with norm N = N(a) *2 N, and identity element e = (N(a) *2)-1 a *2 E F. Moreover, we have u(xy) = u(y)u(x) for x, y E C. Let v = Se 0 u. Then v is a r-linear automorphism of C' with fixed point set F. Now C' induces on F a structure of composition algebra C, with norm Nip. It follows that N(F)* = N(a) *2 M(N), and NI/k(N(a)) = N(a) *2 N(a) E (N(a) *2)2M(N) = M(N). 0 Corollary 4.2.6 (i) If>.. is as in the proposition, then N(F>.)* = M(N). (ii) If FIJ ~ F, then J1, E k* M(N).

Proof. The first point follows from the last equality of the proof. If FIJ. ~ F, then N(FjI.)* = N(F)*. Since the multiplier groups of Nand NjI. are the same, it follows from the proposition that J1, *2 E M(N). But then J1, = NI / k (J1,)(J1, *2)-1 E k* M(N), proving the second point. 0 It can be shown that if F is a normal twisted octonion algebra over land a (and charl :/: 2,3) the converse of (ii) is also true, see [KMRT, Th. (36.9)J. The proof is rather delicate. In view of the close connection between normal and nonnormal twisted composition algebras, one can expect properties of the former to be inherited by the latter. We give one identity, to be used later in Lemma 4.1.3.

Lemma 4.2.7 In a twisted composition algebra F over a field of characteristic :/: 2,3, the following identity holds for x E F, a E l: (ax+x *2) *2 = (T(x) -aN(x) +TrI/k(aN(x)))x+ (a *2 -N(x))x*2. (4.34) Proof. In the normal case this follows from the formulas of Lemma 4.1.3. If F is nonnormal, work in the normal extension. 0 It would be rather natural to call a nonnormal twisted composition algebra

F reduced if its normal extension F' is so. We prefer an apparently stronger definition; we will see in Th. 4.2.10 that these two definitions are in fact equivalent. If C is a composition algebra over k, we have over l' the normal twisted composition algebra F(k' ®k C). Its underlying vector spaces is

86

4. Twisted Composition Algebras

F'

= l' ®k' (k' ®k C) = l' ®k C.

On F' we have the T-linear automorphism u with u(e ®k x) = T(e) ® x (e E l',x E C). Let F = Inv(u). It is straightforward to check that F' and u are as in Prop. 4.2.4. By that Proposition we obtain a structure of twisted composition algebra on F. We denote this twisted composition algebra by F(C). Definition 4.2.8 A twisted composition algebra F over a field l of characteristic # 2, 3 is said to be reduced if there exist a composition algebra Cover k and A E l* such that F is isomorphic to the isotope F(C)>.. If l is cubic cyclic over k, then this boils down to the definition of 4.1.8. If

F(Ch is a nonnormal reduced twisted composition algebra over l, then its normal extension is F(C')>. with C' = k' ®k C. If F = F(C)>. is a reduced nonnormal composition algebra for land u, then F = (l ®k e) EB (lv'D ®k Co),

where D is the discriminant of lover k (so T( /15) = -/15) and Co = e.L in C. We have the following formulas for the squaring operation and the norm E l, x E Co) : in F (where

e,,,,

(e ® e + ",v'D ® x) *2 = A(U(e)q2(e) - Dq(",)q2(",)Nc(x)) ® e -AvD(u(e)q2(",) + u2(e)u(",)) ® x, (4.35) N(e ® e + ",v'D ® x) = U(A)q2(A)(e

+ ",2 DNc(x)).

(4.36)

In Prop. 4.1.9 (ii) we saw that the automorphism group of a reduced normal twisted composition algebra F(C)>. is independent of A. The same holds for reduced nonnormal twisted composition algebras. Proposition 4.2.9 The automorphism group of the twisted composition algebra F = F(Ch does not depend on A. Proof. By Prop. 4.2.4, Aut(F) = {tiF It E Aut(F') , tu = ut}, where F' is the normal twisted composition algebra F(C'h with C' = k' ®k C. Now Aut(F') is independent of A, and the same holds for action on it of the involution u. 0 Th. 4.1.10 on the characterization ofreduced normal twisted composition algebras carries over to the following result for the general case. Theorem 4.2.10 Let F be a twisted composition algebra over l, char(l) # 2, 3, and let F' be its normal extension. The following conditions are equivalent. (i) F is reduced. (ii) F' is reduced. (iii) T represents 0 nontrivially on F, i.e., there exists x # 0 in F such that T(x) = (X*2,X) = O. (iv) There exists x # 0 in F such that x *2 = AX for some A E l.

4.2 Nonnormal Twisted Composition Algebras

87

Proof. We may assume that F is nonnormal. If F is reduced, then F' = F(G'», with G' = k' ®k G and G as in Def. 4.2.8, so (i) implies (ii). If F' is reduced, T represents zero nontrivially on F' by Th. 4.1.10, hence so it does on F by the lemma below. Thus, (ii) implies (iii). If (iii) holds, then (iv) follows by the same argument as in the proof of Th. 4.1.10. Finally, assume (iv) holds. We follow the lines of the proof of the implication (iii) :::::} (i) in Th. 4.1.10, with some adaptationsj we refer to that proof as to ''the old proof". Choose x E F such that x *2 = AX with A E 1. If N(x) :I 0, we take e = x as in part (a) of the old proof. If N(x) = 0, we follow part (b) of that proof. We may assume that x :I 0, N(x) = 0 and x * x = O. Pick y E F with N (y) = 0 and (x, y) = -1. Then F' = x * F' + Y * F'. We take again e = u 2(a)x + x * y + y * x with a = (y, x * y). Notice that

and that x * y + y * x = (x + y) *2 _X*2 _y*2 E F, so e E F. In either case, whether N(x) equals zero or not, we have found e E F satisfying equations (4.10) and (4.11). Now proceeding as in step (c) and (d) of the old proof, one arrives at an l'-bilinear product and a new norm N which define on F' a structure of composition algebra over l' with identity element ej we call this 6. As in step (e) of the old proof, we show that cp (as in the old proof) is a u-automorphism of 6 of order 3 which commutes with conjugation. Let u be the involution of F' associated with F. Straightforward computations show that u commutes with conjugation and that ucp = cp 2u. From this it follows that u is a r-linear anti-automorphism of 6. We define by u". = cp and

(x E F')

an isomorphism f.! f-4 u" of Gal(l' /k) onto a subgroup of the group of semilinear automorphisms of 6 such that each u" is a l>-automorphism. As in step (f) of the old proof one proves that

G = Inv( u q If.! E Gal(l' /k) ) with the restriction of N as norm is a composition algebra over k which has 0 the properties required by Def. 4.2.8. Thus, F is reduced. We still owe the reader the lemma we used in the beginning of the above proof. Lemma 4.2.11 Let U be any vector space over a field k, T a cubic form on U, and k' a quadratic extension of k. 1fT represents zero nontrivially on U' = k' ®k U, then so it does on U itself

Proof. Pick a basis 1, e of k' over k and write the elements of U' as x + ey with x, y E U. Let T(x + ey) = 0 for some x + ey :I o. If y = 0 or if y :I 0 and T(y) = 0, we are done. So let T(y) :I o. Consider the cubic polynomial

88

4. Twisted Composition Algebras

T(x + Xy)

E k[X]. This has a root e in the quadratic extension k' of k, so it must have a root a in k itself. Hence T(x + ay) = O. If x + ay f. 0, we are done. If x + ay = 0, we find

T(x + ey)

= T( -ay + ey) = (e -

a)3T(y) f. 0,

o

a contradiction.

The following lemma will be used later. Let F be an arbitrary twisted composition algebra. The notations are as in Def. 4.1.1. Lemma 4.2.12 There exists a E F such that the following conditions are

satisfied. (i) T(a) = (a· 2,a) f.0; (ii) a and a· 2 are linearly independent over l; (iii) the restriction of ( , ) to the two-dimensional subspace La + La· 2 is nondegenerate, or equivalently (provided (ii) holds), T(a)2 - 4 NI/k(N(a)) f.

O. If N is isotropic, there exists isotropic a with T{a) f. O. Such an a also satisfies the" conditions (ii) and (iii); moreover, a· 2 is isotropic and satifies (i), (ii) and (iii), and we have (a .2).2 = T{a)a and T(a .2) = T{a)2. Proof. If a and a .2 are linearly independent, ( , ) is degenerate on La $la .2 if and only if Le., if

T(a)2 - 4N I/ k(N(a)) = 0, so indeed the two conditions in (iii) are equivalent, provided (ii) holds. If F is not reduced, every nonzero a E F satifies (i) and (ii) by Th. 4.2.10. If N is isotropic, we choose a f. 0 with N(a) = 0, then (iii) also holds. For anisotropic N we argue as follows: if char(k) f. 2, then no vector can be orthogonal to itself, so the bilinear form ( , ) is nondegenerate on any subspace; if char(k) = 2 and a f. 0, then

T(a)2 - 4NI/k(N(a))

= T(a)2 f. O.

Now assume F reduced: F = F(Ch., for an octonion algebra Cover k and some ,X E I·. Let D be a two-dimensional composition subalgebra of C. If we have a E D satisfying (i) and (ii) then (iii) must hold, too. For a E C we have

a*2

and

= 'xCi2 =

'x(-a+ (a,e)e)2 2 = 'x{a - 2( a, e)a + (a, e )2e) =,x( - (a,e)a+ ((a,e)2 - N(a))e),

i

4.3 Twisted Composition Algebras over Split Cubic Extensions

T(a)

89

= (a*2,a) = N,/ k (A)(a,e)((a,e)2 -3N(a)).

The conditions (i) and (ii) together are equivalent to the following four conditions:

atj.ke, (a,e);lO, (a,e)2;lN(a), (a,e)2;l3N(a). If the restriction of N to D is isotropic, we pick a E D with (a, e) = 1 and N(a) = OJ this satisfies our conditions. If k is finite we may assume this to be the case. So we can now assume that k is infinite and N is anisotropic on D. The four conditions require a E D to lie outside a finite number of lines in D. Since k is infinite, such a exist. This proves the first part of the Lemma. Finally, let a with N(a) = 0 satisfy T(a) ;l O. Since (a, a *2) = T(a) ;l 0, (ii) must hold. Further, (iii) holds. From (4.34) we infer that (a *2) *2 = T(a)a, so T(a *2) = T(a)2 ;l OJ further, N(a *2) = o. Hence a *2 is isotropic and satisfies (i), so also (ii) and (iii). 0

At the end of § 4.1 we gave some examples of fields over which every normal twisted composition algebra is reduced. These fields have the same property for arbitrary twisted composition algebras. This is clear for the case of finite fields: since every finite extension of a finite field is Galois, every twisted composition algebra over a finite field is normal. Over a complete, discretely valuated field with finite residue class field every twisted composition algebra is reducedj in the nonnormal case this follows from the above theorem by the same argument as used at the end of § 4.1 for the normal case.

4.3 Twisted Composition Algebras over Split Cubic Extensions In this section we generalize the notion of twisted composition algebra to the situation where the cubic field extension Ilk is replaced by a direct sum of three copies of a field. This generalization will be used in the next section to determine the automorphism groups of eight-dimensional twisted composition algebras. These will turn out to be twisted forms of groups of type D4 · The motivation for the generalization lies in the following situation. Consider as in § 4.1 a normal twisted composition algebra F = (F, *, N) over a cubic cyclic extension field I of k. Let K be an extension field of lj this is a splitting field of lover k, i.e., an extension of k such that L = K®kl ~ K$K$K. See Ch. I, § 16 (in particular ex. 2) in [Ja 64aJ. Denote the three primitive idempotents in L by elo e2 and e3, so el = (1,0,0), etc. The action of the Galois automorphism u of lover k is extended K-linearly to L, i.e., as id ®u; it then induces a cyclic permutation of the primitive idempotents, say,

90

4. Twisted Composition Algebras

O"(ei) = ei-l (indices to be taken mod 3) (cf. the proof of Th. 8.9 in [Ja 80]). The group < 0" > plays the role of "Galois group" of Lover K. Extend the vector space F over 1to the free module FK = K ®k F over L, and the norm N, which is a quadratic form over l, to a quadratic form over L on FK, also denoted by N. Finally, extend the k-bilinear product * on F to a K-bilinear product * on FK. The conditions (i), (ii) and (iii) of Def. 4.1.1 remain valid on FK. (Since (ii) involves polynomials of degree 4 over k, one has difficulties if k does not have at least five elements; these difficulties are avoided if one views N as a polynomial function on FK which is defined over k.) We thus arrive at a notion of twisted composition algebra over the split cubic extension L of K; we formalize this in the following definition.

Definition 4.3.1 Let K be any field. By the split cubic extension of K we understand the K -algebra L = K €a K €a K. Call its primitive idempotents

el = (I, 0, 0), e2 = (O, 1,0) and e3 = (O, 0,1). Fix the K-automorphism 0" of L by O"{ei) = ei-l (i = 1,2,3 mod 3). A twisted composition algebra over L and 0" is a free L-module F provided with a K-bilinear product * and a non-degenerate quadratic form N over L (this notion being defined in the obvious manner) such that the conditions (i), (ii) and (iii) of Def. 4.1.1 hold.

If 1 with char{l) i 2,3 is a cubic field extension of k which is not normal and F is a twisted composition algebra over l, we have a normal extension F' = l' ®IF, which we identify with k'®kF. If now K is any field extension of l', we have again K®kl = K®k,l' = K€aK€aK (with obvious identifications of tensor products), and FK = K ®k F = K ®k' F' with the twisted composition algebra structure induced by that on F' is again a twisted composition algebra over the split extension L = K €a K €a K of K. The 3-cyclic group generated by 0" plays the role of "Galois group" of L over K. The formulas of Lemmas 4.1.2 and 4.1.3 remain valid in the situation of Def. 4.3.1; we will, in fact, only need (4.4) and (4.6). We now turn to an explicit determination of the structure of a twisted composition algebra F over L = K €a K €a K. It will turn out that F is the direct sum of three copies of a composition algebra Cover K, with the product * in F determined in a specific way by the product in C. Put Fi = eiF for i = 1,2,3; these are vector spaces over K, and we have a direct sum decomposition

of vector spaces over K. The formulas in (i) of Def. 4.1.1 with oX = ei show that (i=1,2,3). Fi * F ~ Fi+2 and F * Fi ~ Fi+l It follows that for 1 $ i, j $ 3,

Fi * Fj = 0

(j i i + 1)

and Fi * Fi+l ~ Fi+2.

r 4.3 Twisted Composition Algebras over Split Cubic Extensions

91

For x E Fi we have

So we can define Ni : Fi

-+

K by

It is readily verified that Ni is a nondegenerate quadratic form on the vector space Fi over K for i = 1,2,3. We denote the associated K-bilinear form by Ni (, ). Formulas (ii) and (iii) of Def. 4.1.1 yield

Ni+2(X * y) = Ni(x)Ni+l(Y) Ni(y * z, x) = Ni+l (z * x, y)

(x E Fi , Y E Fi+d, (4.37) (x E F" y E FHl , Z E FH 2) (4.38)

for i = 1,2,3. Put C = Fl' Take a E F3 with N 3(a) ::I 0 and bE F2 with N2(b) ::I O. In the same way as in Lemma 4.1.4 we derive that the K-linear map /2 : C -+ F2, x ...... a * x, is bijective, and similarly for /3 : C -+ F3, x ...... X * b. We define a K-bilinear product on C by

xy = (a * x) * (y * b) = /2(x)

* /3(Y)

(x,y E C).

(4.39)

Using (4.4) and (4.6) one sees that e = N3(a)-l N2(b)-l(b * a) is an identity element for this multiplication. Putting No(x) = N3(a)N2(b)Nl(X) (x E C), we conclude from (4.37) that

No(xy) = No{x)No{y)

(x,y E C).

Thus we have obtained on C a structure of composition algebra over K; we denote this by Ca,b(F). Set It = id : C -+ Fl' We have constructed a K-linear bijection

f

= (h,/2,/3): C$C$C

-+

F, (Xl,X2,X3) ...... (h(Xl),/2(X2),/3(X3»

(4.40) with !i(Xi) E Fi . Notice that F determines the norm of C up to a K*multiple, hence it determines C up to isomorphism by Th. 1.7.1. We encountered a similar situation in the proof of Prop. 4.1.6 (which we gave before stating the proposition itself). In the present case, too, F can have dimension 1, 2, 4 or 8 over L. Notice that the composition algebra structure on C determines the quadratic form N and the product * on F, provided the K-linear bijections /2 : C -+ F2 and /3 : C -+ F3 as well as N 3{a) and N2(b) are given; to prove this, use (4.37), (4.38), (4.4) and (4.6). From any composition algebra Cover K we can construct a twisted composition algebra Fs{C) over L = K $ K $ K. As an L-module we take Fs(C) = C $ C $ C, the product and the norm are defined by

92

4. Twisted Composition Algebras

(x,y,z) * (U,v,w) = (yw,zii.,xy), N«x, y, z» = (No(x), No(Y), No(z», with No denoting the norm of C. One easily verifies the conditions of Def. 4.1.1, keeping in mind that the "Galois automorphism" q acts on L by 0'«0:, fj, 'Y» = (fj, 'Y, 0:). Taking a = (0,0,1) and b = (0,1,0) we reconstruct C from F.(C) as described above: C = Ca,b(F.(C». If we start from an arbitrary twisted composition algebra F over the split cubic extension L of K, and construct the composition algebra C = Ca,b(F) with the aid of a E F3 and bE F2 with N 3(a) = N2(b) = 1 (provided these exist), then F ~ F.(C).

4.4 Automorphism Groups of Twisted Octonion Algebras Let F be a twisted octonion algebra, either normal over the cubic cyclic extension field l of k, or nonnormal over the separable but not Galois cubic extension field l of k; in the latter case we assume char k '" 2,3. Denote the automorphism group of F by Aut(F). An automorphism of a twisted composition algebra FK over a split cubic extension L of K is, of course, an L-linear bijection that preserves *; as in Lemma 4.1.5 one sees that it also leaves N invariant. The group of these automorphisms is denoted by Aut(FK)' If we take for K an algebraic closure of l', the automorphisms of FK = K ®k F form a algebraic group G. Now let K again be any extension field of l'. Let u be an automorphism of FK ; the L-linearity of u implies that it stabilizes every F,. Let Ui be the restriction of u to Fi; it is a K-linear bijection. As in the preceding section, let C be the octonion algebra Ca,b(F) over K defined by a E F3 and bE F2 with N(a)N(b) i=- 0. We have the linear bijection of (4.39)

From (4.39) we infer that

(x,y E C).

(4.41)

Define t = (tl' t2, t3) E GL(C)3 by t = I-Iou 0 I, i.e., ti = li- 1 0 Ui 0 IiSince u preserves N, all ti lie in O(No). According to (4.41) and (4.39), they must satisfy the condition

(x,y

E

C)

(remember that 11 = id). This means that (tt, t2, t3) is a related triple of rotations of C (cf. the Principle of Triality, Th. 3.2.1), necessarily of spinor norm 1 (cf. § 3.2, in particular Prop. 3.2.2 and Cor. 3.3.3). Conversely, any

4.4 Automorphism Groups of Twisted Octonion Algebras

93

related triple of rotations (h, t2, t3) (necessarily of spinor norm 1) of 0 defines an automorphism u of FK as above. Thus we get an isomorphism between Aut(FK) and the group RT(O) of related triples of rotations of OK (see § 3.6):

ep : G

-+

RT(O), u

f-+

,-1 0 U 0

f.

(4.42)

Taking first K = l', we obtain a composition algebra 0 over l'. Then taking for K the algebraic closure of l', we see that the algebraic groups G and RT(OK) are isomorphic. Now by Prop. 3.7.1 the group RT(OK) is defined over l'. Also, the isomorphism I of 4.40 is defined over l'. We thus obtain on G a structure of algebraic group over l'. By Prop. 3.6.3, RT(OK) is isomorphic to the spin group Spin(No). Thus we have shown the following result. Proposition 4.4.1 G is an algebraic group over l' which is isomorphic to Spin(8). We will see in Th. 4.4.3 that G is defined over k. In the rest of this section we take for K a separable closure ks of k containing l'. Then O/(No) = SO(No)· For the case char(k) =I 2 this follows from the fact that ks contains all square roots of its elements, so all spinor norms are 1. In all characteristics, one can use a Galois cohomology argument, see [Sp 81, §12.3J. The Galois group Gal(ks/k) (which we understand to be the topological Galois group if ks has infinite degree over k) acts on ks®kl and on Fk, = ks ®kF by acting on the first factor. It permutes the idempotents ei and the components Fi; thus we have a homomorphism a : Gal(ks/k) -+ S3 such that 1'(ei) = ea(-y)(i) Gal(ks/k»).

hE

Lemma 4.4.2 a(Gal(ks/k» has order 3 ill is Galois overk, and order 6 il 1 is not Galois over k, so it equals the degree 01 l' over k. Proof. The invariants of Gal(ks/k) in kS®kl form the field k®kl = I, so every idempotent ei is displaced. This implies that a(Gal(ks/k» has order at least 3. The idempotents exist already in l'®kl, which is invariant under Gal(ks/k), so their permutation is in fact accomplished by the action of Gal(l'/k), that is, a can be factored through a homomorphism a ' : Gal{l'/k) -+ S3. If l' = I, then Gal(l'/k) has order 3, so then la(Gal(ks/k»1 = 3. If l' =I I, then Gal(l' /k) 9:! S3 and the kernel of a' in Gal(l'/k) is a normal subgroup of order at most 2, so it consists of the identity only, whence la(Gal(ks/k»1 = 6. 0 Over ks the octonion algebra Ok, = ks ®k 0 is split, so we may replace C = Ca,b(F) by the split octonion algebra. Thus we get from (4.42) an isomorphism, which we also call ep, of G(ks ) onto RT(O)(ks), where 0 is the split octonion algebra over I. For l' E Gal(ks/k) we have the conjugate isomorphism 'Yep = l' 0 ep 0 1'-1. Then z(1') = 'Yep 0 ep-1 is an automorphism of RT(O)(ks). This defines a nonabelian 1-cocycle of Gal(ks/k) with values in RT(C)(ks) (see [Sp 81, § 12.3]).

94

4. Twisted Composition Algebras

The automorphism z(-y) acts on RT(C)(ka) as

z('Y) : (tb t2, ta)

1-+

(t~(-y)(l)' t~("')(2)' t~(-y)(a»,

where a is the homomorphism of Gal(ka/k) into Sa as in the above lemma, and tj is the image of tj under some inner automorphism (depending on j) of SO(No). In case char(k) '" 2, tj = tj if tj = ±1, so then z(-y) permutes the central elements (1, -1, -1) etc. of RT(C)(k.) in the same way as a(-y-l) does; if char(k) = 2, a similar argument with central elements of the Lie algebra of RT works (see Prop. 3.6.4 and its proof). It follows that the image of z(-y) in Aut(RT(C»/Inn(RT(C» ~ Sa is a(-y-l). G is a ks-form of RT(C) which can be obtained by twisting RT(C) by the co cycle z (see [Sp 81, § 12.3.7]); it is defined over k. From Lemma 4.4.2 we infer that it is of type a0 4 or 604 (see [Ti]) according to whether the cubic extension 1/ k is Galois or not. Thus we have proved the following theorem.

Theorem 4.4.3 G is defined over k. If 1 is a cubic cyclic extension of k and F a normal twisted octonion algebra over l, then G is a twisted k-form of the algebraic group Spin(8) of type a0 4 • If 1 is cubic but not Galois over k with char(k) '" 2 or 3, and F is a nonnormal twisted octonion algebra over l, then G is a twisted k-form of Spin(8) of type 60 4 •

4.5 Normal Twisted Octonion Algebras with Isotropic Norm In this section, F will be a normal twisted octonion algebra over a cubic cyclic field extension 1 of k (and we will omit the adjective "normal" when speaking about twisted octonion algebras). From now on we fix a E F that satisfies the three conditions of Lemma 4.2.12; if N is isotropic, we moreover assume a isotropic. We take E to be the orthogonal complement of la $la * a, E

= {x E CI (x,a) = (x,a*a) = O}.

Lemma 4.5.1 E * a ~ E and a * E

~

(4.43)

E.

Proof. If x E E, then x * a E E, for

( x * a, a)

= 0"( ( a * a, x }) = 0,

and by (4.1), Similarly for a * x.

o

This allows us to define the O"-linear transformation t in E by t : E - E, x

1-+

X

* a.

(4.44)

r

4.5 Normal Twisted Octonion Algebras with Isotropic Norm

95

Lemma 4.5.2 The transformation t : E - E satisfies

t 2(x)=-(a*a)*x (XEE), 3 t (x) = -T(a)x - a * (a * (a * x))

(x E E),

(4.45) (4.46)

and t

6

+ T(a)t 3 + NI/k(N(a)) = O.

Proof. We compute, using Lemma 4.1.3,

2

t (x)

= (x*a)*a =

-(a*a)*x, t (x) = -«a*a)*x)*a = (a * x) * (a * a) - 0'2 ( ( a * a, a ) )x = -T(a)x - a * (a * (a * x)) (since (a * x,a) 3

= 0'2«(a * a,x}) = 0),

so

t 3 (x) Apply

t3

+ T(a)x + a * (a * (a * x)) =

O.

to both sides of this equation: 6

t (x)

+ T(a)t 3 (x) + [{(a * (a * (a * x))) * a} * a] * a = O.

Since (a * y) * a = 0'2(N(a))y, the last term on the left hand side equals NI/k(N(a))x, sO we get the formula. 0 We call a twisted composition algebra isotropic if its norm is isotropic. From now on we assume that F is an isotropic twisted octonion algebra (normal, as is the convention now). We take a as in Lemma 4.2.12 with N(a) = O. We define and

(4.47)

Using Lemma 4.1.3 one easily verifies e1 * e1 = A1 e2, e2 * e2 = A2e1, e1 * e2 = e2 * e1 = 0,

with Al

= T(a)

= All. Further, = N(e2) = 0 and

(4.48) (4.49) (4.50)

E k* and A2

N(eI)

(el' e2)

= 1.

(4.51)

A straightforward computation now yields T(~lel

+ ~2e2) =

Al NI/k(6)

+ A2 NI/k(6).

(4.52)

Notice that replacing a by T(a)-la * a, which also satifies the conditions in Lemma 4.2.12 and is isotropic, amounts to interchanging el and e2 and also Al and A2.

96

4. Twisted Composition Algebras

Define D = lei EB le2 and E = Dl.. The restriction of ( , ) to D and the restriction to E are both nondegenerate. Define as in (4.44) the u-linear transformations

(4.53) From (4.45) we infer that

t~(x) = -AieHl

*x

(x E E, indices mod 2).

Take Ei = ti(E). Trivially, ti(Ei ) ~ E i . Since N{ei) isotropic, hence they have dimension ~ 3. By (4.7),

(4.54)

= 0, both Ei are totally

* x) * el + (el * x) * e2 = u2{( el, e2 ))x = x. t~+l (E) ~ E, it follows that E = El + E 2. Since dim Ei

(e2

Since ei * E = we must have a direct sum decomposition:

~ 3,

with both Ei having dimension 3. Using (4.7) again, we see for x E E,

(x * el) Since e2

* e2 + (e2 * el) * x =

2

u {(x, e2) )el = O.

* el = 0 by (4.50), we find that t2tl = O. Similarly, tlt2 = O. Hence

and therefore

ti{Ei ) It follows that Ei

= Ei

(i=fti),

(4.55)

(i = 1,2).

(4.56)

= t~{E), so by (4.54) (4.57)

From (4.46) we know that

t~{x) = -AiX - ei By (4.4), ei

* Ei =

ei

* (E * ei) =

* (ei * (ei * x)).

0, so (4.58)

It follows that 1I"i = -AHlt~ is the projection of E on E i • Since El and E2 are totally isotropic and ( , ) is nondegenerate on E, the Ei are in duality by the isomorphism

where

it: E2

-+

l, x

1-+

(u,x),

4.5 Normal Twisted Octonion Algebras with Isotropic Norm and similarly E2 --

Ei. For Xi

97

E Ei we have

(4.59) for

(tl(XI), t2(X2))

= (Xl * el,X2 * e2) = O"«(el * (X2 * e2),xt}) = 0"«(XI,X2 )),

since by (4.54) el * (X2 * e2) = ->'lt~(t2(X2)) = ->'lt~(X2) = X2. This means that t2 = (ti)-l and vice versa. We now compute Xl * X2 and X2 * Xl for Xi E E i . We write Xi as Xi = ti(Zi) = Zi * ei with Zi E Ei and find

Xl * X2 = (Zl * el) * (Z2 * e2) = -«Z2 * e2) * ed * Zl

Xl * X2

+ 0"2 ( (Zl' Z2 * e2) )el

= 0"2( (tIl (xd, X2 ))el = 0"«( Xl, t2(X2) ))el

Similarly for X2

(by (4.7)).

(by (4.59)).

* Xl. Thus, (4.60)

Next consider

since el

X

* Y for x, y EEl. We have

* El = -Alt~(Ed = o.

Further,

(X * y,e2) since EI * e2

= O"«(y * e2,x)) = 0,

= t2(Ed = o. So X * Y E E.

Using (4.5) we find

tl(X) * tl(Y) = (x * el) * (y * el) = -el * (y * (x * el)), since (x

* el, el) = o. Also, y * (x * el) = -el

* (x * y).

Now using (4.54) and (4.58), we find

tl(X) * tl(Y)

= -Alt~( -Alt~(X * y)) = -Alt2(X * y),

and similarly with t2 and x, y E E 2 . Thus,

(4.61)

98

4. Twisted Composition Algebras

An immediate conclusion is that (4.62)

To describe the multiplication of elements of Ei it is convenient to introduce a bilinear product

(i=I,2) by defining

Xi /\ Yi

= t;l(Xi) * ti(Yi)

(4.63)

From (4.61) it is immediate that (4.64)

ti(Xi) /\ ti(Yi) = -Aiti+l(Xi /\ Yi) This wedge product is alternating. For if X E El, then

X /\ x

= (e2 * x) * (x * el) (by (4.54) and (4.58)) = -«x * et} * x) * e2 (by (4.7), since (e2,x * eI) = 0) = N(x)(el * e2) (by (4.6)) = 0 (since N(x) = 0, or el * e2 = 0),

and similarly for x E E2' By linearizing one finds (4.65)

x /\y = -Y /\x

For this wedge product we can prove the formulas that are well known for the vector product (cross product) in three-space.

Lemma 4.5.3 For Xi, Yi E E i ,

(i =

1,2), one has

(Xl /\ Yl) /\ X2 = (Xl, X2 )Yl - (Yl, X2 )Xl, (X2 /\ Y2) /\ Xl = (X2, Xl )Y2 - (Y2, Xl }X2, (Xl/\Yl,X2/\Y2) = (Xl,X2)(Yl,Y2) - (Xl,Y2)(X2,yI).

Further, Xi /\ Yi = 0 for all Yi (or for all Xi) implies Xi tively), (i = 1,2).

=0

(Yi

= 0,

Proof. From (4.63) we get

(Xl/\ Yl) /\ X2

= t2"l(t}l(Xl) * tl(Yl)) * t2(X2) = -Al(t}2(Xl) * yt} * t2(X2) (by (4.61)) = (tl (xt) * Yl) * t2(X2) (by (4.58).)

Now by (4.7) and (4.59),

(4.66) (4.67)

(4.68) respec-

4.6 A Construction of Isotropic Normal Twisted Octonion Algebras

(tl (Xl) * Yl) * t2(X2)

99

+ (t2(X2) * yt} * tl (xt) = u 2((tl (Xl), t2(X2) ))Yl = (Xb X2 )Yb

and by (4.59) and (4.60),

t2(X2) * Yl = U«(tl(Yl),t2(X2) ))e2 = U2«(YbX2 ))e2. Using these relations, we find

(tl(Xl) * yt} * t2(X2) = -(t2(X2) * Yl) * tl(Xl) + (Xb X2 )Yl = -(Yb X2 )(e2 * tl(Xl)) + (Xb X2 )Yl = -( YbX2 )Xl + (XbX2 }Yl (by (4.54) and (4.58)). This proves the first formula, and the same argument leads to the second one. For the third formula we proceed as follows:

(Xl'\ Yb X2 t\ Y2) = (tl 1(Xl) * tl(Yl), t2"1(X2) * t2(Y2)) 2 = (-A2)( -Al)( t2(t 1 (xl) * Yl), tl(t2"2(X2) * Y2)) = U«(tl(Xl) * Yl,t2(X2) * Y2)) = (Y2, (tl (Xl) * Yl) * t2{X2) ). Now (tl (Xl) * Yl) * t2(X2) was computed above; substituting that expression we find formula (4.68). The last statement of the Lemma is easily derived from (4.66) and ~~.

0

Define an alternating trilinear function ( , , ) on Ei by

(X,y,z) = (x,yt\z). From 4.59 and 4.64 we obtain (4.69)

4.6 A Construction of Isotropic Normal Twisted Octonion Algebras We maintain the convention that all twisted composition algebras are normal. The analysis made in the previous section leads to a construction of isotropic twisted octonion algebras, to be described in the present section. This construction will yield all such algebras. We first discuss some generalities. Let V be a three-dimensional vector space over the field l and V' its dual space. The bilinear pairing between V

100

4. Twisted Composition Algebras

and V' is denoted by ( , ). There is a vector product 1\ on V with values in V', and one on V' with values in V with the following three properties (where x, y E V, x', y' E V') :

(x 1\ y) 1\ x' (x' I\y') I\x (x' I\y',xl\y)

= (x,x')y - (y,x' }x, = (x,x')y' - (x,y')x', = (x,x')(y,y') - (x,y'}(y,x').

(4.70) (4.71) (4.72)

Using these properties, one easily verifies that x 1\ y is alternating bilinear on V, nonzero if x and yare linearly independent, and that (x, y, z) = (x, yl\z) is an alternating trilinear function on V (Le., invariant under even permutations of the variables and changing sign under odd permutations); similarly on V'. On a three-dimensional space, an alternating trilinear function is unique up to a nonzero factor, viz., it is a multiple of the determinant whose columns are the coordinate vectors of the three variables with respect to some fixed basis. It follows that the vector products on V and on V' are unique up to multiplication by some a E l* and a- 1 , respectively. If t : V -+ V is a a-linear transformation, where a is an automorphism of l, then another alternating trilinear function of x, y, Z is a- 1{(t{x), t{y), t{z»)). Hence it is a multiple of (x, y, z). We define det{t) E l by

(t{x), t{y), t{z») = det{t)a{ (x, y, z)

(X,y,zEV).

If we replace the wedge product x 1\ yon V by ax 1\ y, so ( , , ) bya( , , ), then det(t) changes to aa(a)-1 det(t). We call det{t) the determinant of t with respect to the given choice of the wedge product (or the choice of the alternating trilinear form). Let t' be the inverse adjoint transformation of t in V', Le.,

(t{x),t'{x'») =a{(x,x')

(x E V, x' E V').

t' is also a-linear. Lemma 4.6.1 For x, y E V we have

t{x) 1\ t{y)

= det{t)t'{x 1\ y).

(4.73)

Moreover, det{t') = det{t) -1. Proof. The first formula follows from the equations

( z, (t') -1 (t{x)

1\ t{y»

) = a 2 { ( t{z), t{x)

1\ t{y)

)

= a2 «t{z), t{x), t{y) ) =

a 2 {det{t» ( z, x 1\ y). Similarly, we have for x', y' E V'

t'{x')

1\ t'{y')

= det(t')t(x' 1\ y').

4.6 A Construction of Isotropic Normal Twisted Octonion Algebras

101

Using (4.70) and the definition of det (t) we see that

(t(x)

/I. t(y)) /I.

(t(Xl)

/I. t(yt)) =

det(t)t«x /I. y) /I. (Xl

/I.

yt)),

which by what we already proved equals

It follows that

t'(x')

/I.

t'(y')

= det(t)-lt(x' /I. y'),

where x' = X /I. y, y' = Xl /I. Yl. Since any element of V' is a wedge product of elements of V, the last formula holds for arbitrary x', y' E V'. The second 0 assertion of the lemma follows. Now assume that l is, as before, a cubic cyclic extension of the field k, and that u is a generator of the Galois group. Assume that t is a ulinear transformation of V such that t 3 = - A with A E k*. We also assume that the vector product on V is such that det(t) = -A. This can always be arranged. For if not, then Nl/k( -A -1 det(t)) = 1, as one sees by computing (t 3(x), t 3(y), t 3(z)} in two different ways. By Hilbert's Theorem 90 there exists a E l such that det(t) = -Aa-lu(a). Replacing the vector product x /I. y on V by ax /I. y changes the determinant of t to det(t) = -A. Our assumption determines the vector product on V up to a multiplicative factor J.L E k* and the vector product on V' up to J.L- l . For the inverse adjoint transformation t' we have t,3 = _A- l and det(t') = _A-I. V and t are the ingredients of the construction of a (normal) twisted composition algebra F(V, t). We are guided by the results of the preceding section. Taking (with the notations of 4.5) V = El, V' = E 2 , t = tt the definition of F(V, t) is explained by the formulas of that section. We define F = F(V, t) = l EEll EEl V EEl V' and put el = (1,0,0,0), e2 = (0,1,0,0). We define a product * in F as follows:

+ 6e2 + x + x') * (77lel + 772e2 + y + y') = + u( (x, t'(y')} nel + {Au(6)u 2(77l) + u( (t(y), x'} ne2 + u(6)t- l (y) + u 2(77l)t(X) + t'(x') /I. (t')-l(y') +

(6el

P- l u(6)U2(772)

u(6)(t')-1(y')

+ u 2(772)t'(X') + t(x) /I. Cl(y).

(4.74)

for ei,77i E l, x,y E V, x',y' E V'. We further define the norm N on F by

(4.75) for

ei E l, x E V and x' E V'.

Theorem 4.6.2 With this product and norm F(V, t) is an isotropic twisted octonion algebra, and all such algebras are of this form. For z = 6el +6e2 + x + x' the cubic form T(z) = (z * z, z) is given by

102

4. Twisted Composition Algebras

T(z) = ANI/k(~l) + A-1 NI/k(~2) + TrI/k(~la( (t(X), X'})) TrI/k(~2a( (X, t ' (X') }))

+

+ (X, t(X), rl(X) } + (X', t ' (X'), t l - l (X' ) }.

(4.76)

Proof. It is clear that the product * is a-linear in the first variable and a 2 _ linear in the second variable. The verification of the other two requirements for a twisted composition algebra (cf. Def. 4.1.1), viz.,

N(x * y)

= a(N(x))a2(N(y))

(x,y E

F)

and

(x,y, Z E F), and the computation of T are straightforward, so we omit them. That we obtain all twisted composition algebras with isotropic norm in this way, follows from the analysis in the preceding section. 0

4.7 A Related Central Simple Associative Algebra We continue to consider an isotropic twisted octonion algebra F = F(V, t). Notations and conventions are as in the preceding section. We introduce the associative algebra Dover k consisting of all transformations of V of the form

(We write 1 for the identity here.) We show that this is a cyclic crossed product. (For crossed products, see [AI 61, Ch. V), [ArNT, Ch. VIII, §§ 4 and 5} or [Ja 80, §§ 8.4 and 8.5}.) The notation D is used since in the most important case for us it is a division algebra. Lemma 4.7.1 The elements 1, t and t 2 form a basis of Dover 1, and D is isomorphic to the cyclic crossed product (1, a, -A), so it is a central simple algebra of degree 3 over k. If A E Nl/k(l*), this crossed product is isomorphic to the algebra M3(k) of 3 x 3 matrices over k, and if A ¢ Nl/k(l*), it is a division algebra.

Proof. If ~O+6t+~2t2 =0

with

~i E

1, then for all

X E

V and ." E 1 we have

"'~f)X + a("')~lt(X)

+ a2("')~2t2(x)

=

o.

Since the automorphisms 1, a, a 2 are linearly independent over l, it follows that all ~i are zero. So 1, t, t 2 is a basis of Dover 1. The cyclic crossed product (l, a, -A) is the algebra generated by 1 and an element u such that u 3 = -A and u~ = a(~)u (~ E l). It has dimension 3

4.7 A Related Central Simple Associative Algebra

103

over 1. Clearly, there is a homomorphism (1,0', -.\) - D sending u to t and extending the identity map of l. Since D has dimension 3 the homomorphism is bijective, hence is an isomorphism. The crossed product is known to be isomorphic to M3(k) if -.\ E N,/kW) and to be a division algebra if -.\ ¢ NZ/k(l*). We may omit the minus sign, since NZ/k(-l) = -1. 0

Lemma 4.7.2 There exists Vo E V such that V

= D.vo.

Proof. If D is a division algebra, we can pick any nonzero Vo E V. For then D.vo is a nine-dimensional subspace of V over k, which must coincide with

V. If D is not a division algebra we have D 9'! M3(k). As a D-module, V is isomorphic to the direct sum of three copies of the simple module k 3 of M3(k). But then the D-module V is isomorphic to D, viewed as a left module over itself. We can then take for Vo the image in V of any invertible element of D. 0 On the central simple algebra Dover k, one has the reduced norm, see [AI 61, Ch. VIII, § 11], [Schar, Ch. 8, § 5] or [Weil, Ch. IX, § 2]. It is the unique polynomial function on D which upon extension of k to a splitting field of D becomes the determinant. It is multiplicative, Le., ND(UV) = ND(U)ND(V) for U,V E D, and 11. is invertible if and only if ND(U) =1= O. The following lemma gives a simple characterization of the reduced norm.

Lemma 4.7.3 If A is a central simple algebra of degree n over k, then the reduced norm NA is the unique homogeneous polynomial function of degree n on the vector space A over k with NA(l) = 1 which satisfies the conditions: x E A is invertible if and only if NA(X) =1= 0 and there exists a homogeneous polynomial map P : A - A of degree n - 1 such that

Proof. For A = Mn(k) it is known that x E A is invertible if and only det(x) -# 0, and then X-I = det(x)-I adj(x), where adj(x) is the adjoint matrix of x, i.e., the matrix whose entries are the cofactors of x (see, e.g., [Ja 74, § 2.3]). As to the uniqueness of NA, let N~ with N~(I) = 1, in combination with pI also satisfy the conditions. Then N~ (x) -# 0 if and only if det(x) -# 0, and

det(x)pl(x) =

adj(x)N~(x)

(x E A,

det(x)

-# 0).

Since det is an irreducible polynomial in the entries of the matrix x (see, e.g., [Ja 74, Th. 7.2]), either det divides all entries of adj (viewed as a matrix with polynomial entries), or det divides N'.4. The first case being absurd, we must have N~ = det.

--~~--~-------------------,

104

4. Twisted Composition Algebras

In the case of an arbitrary central simple algebra A, work with a Galois splitting field m of A, i.e., with a Galois extension m of k such that m®kA ~ Mn(k). We have a reduced norm NA over m. For any u E Gal(m/k), let U(NA) denote the polynomial obtained by the action of u on the coefficients of NA. Then U(NA) also satisfies the conditions for a reduced norm on m®kA, so by the uniqueness of this, U(NA) = NA. This means that NA has its coefficients in k, and hence its restriction to A is a reduced norm on A. The uniqueness of the reduced norm on A is immediate from the fact that its extension to m ®k A is a reduced norm on the latter algebra. 0 With the aid of the above lemma it is not hard to compute the reduced norm on D explicitly. (The proof can, in fact, be adapted to any crossed product.)

Lemma 4.7.4 The reduced nonn ofu ND(U)

= Nl/k(~O) -

= ~o +~lt +~2t2

in Dis

,X Nl/k(~l) + ,X2 Nl/k(~2) +,X Trl/k(~ou(6)u2(~2»'

This can also be written as ND(U) Au

=

= det(Au),

where Au is the matrix 2 eo -'xU(e2) -'xu (ed) 6 u(eo) -'xU2(e2) . ( 6 u(6) u2(eo}

Proof. An element '1.1. = ~o + 6t + e2t2 has an inverse if and only if the right multiplication by '1.1., i.e., v 1-+ VU, is bijective. This right multiplication is a linear transformation over l which has matrix Au as above. Hence '1.1. is invertible if and only if det(Au) # 0, and then '1.1.- 1 is the solution v of Auv = 1, so of the form det(Au)-l P(u). A straightforward computation yields that det(Au} = Nl/k(eO) - ,X Nl/k(~l) +,X2 NI/k(6) +,X TrI/k(~Ou(edu2(e2»' So det(Au) E k and it is a cubic polynomial in coordinates over k. P is a 0 map D -. D that is quadratic in coordinates over k. In exactly the same way as we did with D we introduce the associative algebra D' of l-linear combinations of 1, t' and t,2, acting on V'. This has the same properties as D except that t'3 = -,X -1, so in the formulas for the reduced norm one must replace ,X by ,X -1. We call D' the opposite algebra of D, a name which is justified by the fact that D and D' are anti-isomorphic as we will see in the following Lemma.

Lemma 4.7.5 The mapping D -. D',

'1.1.

= eo + elt + e2t2

1-+

'1.1.'

= eo -

'xU(~2)t' - 'xU2(~1)t/2

is an anti-isomorphism of D onto D'. It preserves the reduced nonn: ('1.1.

E

D).

4.8 A Criterion for Reduced Twisted Octonion Algebras. Applications

105

Proof. Both statements are verified by straightforward explicit computation in the coordinates ~o, 6, 6. Notice that we can also write u' = ~o + (t,)-16 +

(t,)-26.

0

We now return to the twisted composition algebra F(V, t) of §4.6.

Lemma 4.7.6 (i) The nonzero values of the reduced norm ND on D form a subgroup N(D)* of k*. We have N(D')* = N(D)*. (ii) For v E V we have T(v) :/: 0 if and only if V = D.v. Then T(u.v) = ND(U)T(v) for u E D, so the nonzero values of T on V form a coset of N(D)* in k*. (iii) Similarly, the nonzero values of T on V' form a coset of N (D)* in k* . Proof. ND is multiplicative, and ND(U) :/: 0 if and only U is invertible, so the nonzero values of ND form a subgroup of k*. The second point of (i) follows from the preceding lemma. According to Th. 4.6.2, the value of T( v) for v E V is T(v) = _A- 1 ( v, t(v), t 2(v)). Pick Vo E V such that V

= D.vo

(cf. Lemma 4.7.2). Write

Then

T(v) = det(X)T(vo), where X is the matrix which expresses v, t(v), t 2(v) in vo, t(vo), t 2(vo}: X =

(

~o -AO'(6) -AO'2(~1}) O'(~o) -AO' 2(6) .

6 6

0'(6)

a 2 (eo)

From Lemma 4.7.4 we infer that det(X) = ND(U) with U = ~o + ~lt + ~2t2. Thus we find that T(v) = T(vo)ND(U). We see that T(v) :/: 0 if and only if u is invertible, which implies the first assertion of (ii). This proves (ii). (iii) is proved in the same way, also using (i). 0

4.8 A Criterion for Reduced Twisted Octonion Algebras. Applications In Th. 4.1.10 we gave criteria for a normal twisted composition algebra to be reduced. The following theorem can be viewed as a sharpening of part (ii) of Th. 4.1.10. Let F = F(V, t) be as in Th. 4.6.2. Notations and conventions are as before.

106

4. Twisted Composition Algebras

Theorem 4.8.1 The isotropic twisted octonion algebra F is reduced if and only if there exists x E V, x # 0, and U ED such that T(x) = ND(U). Proof. First assume there exists x E V, x # 0, and U ED such that T(x) = ND(U), If T(x) = 0, then F is reduced by Th. 4.1.10. If T(x) = ND(U) # 0, then by Lemma 4.7.6 ND(U) E T(a)N(D)* for some a E V, so T(a) E N(D)*. It follows that there exists y E V with T(y) = 1. Now (y * y, y) = T(y) = 1, whereas (y, y) = 0 (since N is identically zero on V by (4.75», so y and y*y are linearly independent. Hence z = y + Y * y # 0 and

z *z

= (y + y * y) * (y + Y * y) = y * y + Y * (y * y) + (y * y) * y + (y * y) * (y * y) = y * y + q(N(y»y + q2(N(y»y + T(y)y - N(y)y * y = y* y+y = z.

Again we conclude by Th. 4.1.10 that F is reduced. Conversely, assume F reduced. We are going to show the existence of x and U with the required properties. Since F is reduced there exists by Th. 4.1.10 a nonzero z = 6el + 6e2 + x + x' E F with ~i E l, x E V and x' E V' such that z * z = az for some a E l. By Th. 4.6.2 this amounts to saying that the following system of equations has a nontrivial solution

(6, {2,X, x', a):

+ q«x,t'(x'») = 06

(4.77)

= 06 q(6)rl(x) + q2(el)t(X) + t'(x') /I. (t')-l(X') = ax q(6)(t')-l(x') + q2(6)t'(x') + t(x) /I. t-l(x) = ax'.

(4.78)

A-lq(~2)q2({2)

Aq(6)q2(el) +q«t(x),x'»

(a) We first consider the case that there is a solution with x

T(x)

(4.79) (4.80)

# O. We compute

= (x, t(x) /I. t-l(x».

Using (4.80) we get

T(x)

= a{ x, x') -

q(6)( x, (t')-l(x'» - q2(~2)( x, t'(x'».

Applying q2 to both sides of (4.77) and then multiplying the result by q2(~2) yields Further,

q(6)( x, (t')-l(x'» = q(6)q2( (t(x), x'» = q(a66) - A Nl/k(6)

With these formulas we find

(by (4.78»).

4.8 A Criterion for Reduced Twisted Octonion Algebras. Applications

107

T(x) = ANl/k(el) + A-I Nl/k(6) - 0-(066) - 0-2(OeI6) + o( x, x'}. (4.81) If 0 = 0, then T(x) = ND(U) with U = -6t + A- 16t 2 by Lemma 4.7.4. For the rest of case (a) assume that 0 =f 0. By (4.11) we have N(z) 0-(0)0-2 (0). Since N(z) = ele2 + (x, x') we find that

=

(x, x') = 0-(a)0-2(0) - ele2. Inserting this into in (4.81) we find

T(x) = Nl/k(O)

+ ANl/k(6) + A-I Nl/k(6) -

Tr 1/ k(066) = ND(U),

where U = 0 - 0-2(el)t + A- 10-(6)t 2. (b) Now assume we have a solution (6, e2, 0, x', 0) of the system of equations (4.77)-(4.80) with 6 =f 0. From (4.77) we infer that also =f and =f 0. Multiplying the opposite sides of (4.77) and (4.78), we get

0

°

el

OA- 1 Nl/k(e2) = oAN1/k(6),

1

A- 2 = Nl/k(elei ), A = Nl/k(Aelei

1

).

This implies that the cyclic crossed product D is isomorphic to the matrix algebra M3(k). Hence ND(u) runs over all elements of kif U runs over D, so for any nonzero x E V there is U ED with T(x) = ND(U). (c) Finally, let there be a nonzero solution of the form (el,O,O,X',o). By (4.78) we have = 0. By (4.79), t'(x') 1\ e-1(x') = 0. This is only possible if t'(x') = ex' for some eEL. Since t' is o--linear and t'3 = -A-I, we find by computing t,3(x') that A-I = Nl/k( -e). So again A E Nl/k(l*) and we can complete the proof as in case (b). 0

el

We mention, in particular, the following consequence, which we will use in Ch. 8. Corollary 4.8.2 If F is not reduced, then D is a division algebm.

Proof. If D is not a division algebra, then D ~ M3(k). In that case the reduced norm ND takes all values in k, so for any nonzero x E V there exists u ED such that ND(U) = T(x). Hence F is reduced by Th. 4.8.1. 0 As an application of the above theorem we can give another class of special fields k over which all twisted octonion algebras are reduced. We assume char(k) 1= 2,3. Theorem 4.8.3 Assume that k has the following property: If A is a ninedimensional centml simple algebm A whose center k' is either k or a quadmtic extension of k, then N A (A) = k'. Let Jurtherl be a cubic extension of k. Then every twisted octonion algebm over land 0- (as in Def. 4.2.1) is reduced.

108

4. Twisted Composition Algebras

Proof. Let F be a twisted octonion algebra over I and 0'. If the norm N of F is isotropic, we take k' = k, and if N is anisotropic, we choose a quadratic extension k' of k which makes N isotropic. Extend I to a cubic cyclic extension l' of k', with 0' (extended to a k'-automorphism of I') as generator of Gal(l' jk'). In either case, F' = k' ®k F is a twisted octonion algebra over I' of the form described in Th. 4.6.2. The reduced norm of l'[t] takes all values in k'; if l'lt] is a division algebra over k', this follows from the assumption we made about the reduced norm on nine-dimensional division algebras over k', and if l'[t] ~ M3(k'), this is always the case. By the previous theorem, F' is reduced. By Th. 4.1.10 this is equivalent to the fact that the cubic form T represents zero nontrivially on F'. By Lemma 4.2.11, T already represents 0 zero nontrivially on F. That implies that F is reduced. Here are some examples of fields with the property of the theorem. (i) k an algebraic number field If D is a central simple algebra over a field k, then the nonzero reduced norms of elements of D form a subgroup N(D)* of k*. Let SLD be the norm one group of D, i.e. the group of elements of DK with reduced norm 1. This is a an algebraic group over k. Then the quotient group k* jN(D)* can be identified with the Galois cohomology set Hl(k, SLD) (see [Se 64, Ch. III, § 3.2]). The reduced norm map will be surjective if and only if the Galois cohomology set is trivial. That this is indeed the case if k is an algebraic number field and D is nine-dimensional follows from the Hasse principle (see [loc.cit., § 4.7, Remarque 1]). This shows that a number field k has the property ofthe previous theorem. Hence over such a field any twisted composition algebra is reduced. (ii) As in case (vi) of § 1.10 perfect fields with cohomological dimension :::; 2 also have the required property, as follows from [loc.cit., Ch. III, § 3.2J. Examples are finite fields and p-adic fields.

4.9 More on Isotropic Normal Twisted Oct onion Algebras This section gives some complements to the material of § 4.5 and § 4.6. We prove two somewhat technical lemmas that will be used in Ch. 8. Let F be an isotropic normal twisted octonion algebra and consider an isotropic element a E F with T(a) f OJ then a and a * a are linearly independent and the restriction of ( , ) to La EB la * a is nondegenerate (see Lemma 4.2.12). The corresponding subspaces E = (la $la * a).L as in (4.43) and Ei = ti(E) (i = 1,2) with ti as in (4.53) will now be denoted by E(a) and Ei(a}, since we are going to vary a:

4.9 More on Isotropic Normal Twisted Octonion Algebras

109

E(a) = (la EB la * a)l., E1(a) = {x*a!(x,a) = (x,a*a) =O}, E2(a) = {x * (a * a) ! (x, a) = (x, a * a) = O}. By (4.57), we also have

E1(a) = {(a*a)*x!(x,a) = (x,a*a) =O}, E2(a) = {a * x! (x, a) = (x, a * a) = O}. Since (a * a) * (a * a) = T(a)a by (4.8), Ei(a * a) = Ei+1(a). By (4.4) and (4.6), a * (a * a) = (a * a) * a = 0, hence we can also write

E1 (a) = {x * a ! (x, a * a) = 0 }, E2(a) = {a*x! (x,a*a) = O}. Recall that E1 (a) and E2(a) are totally isotropic subspaces which are in duality with respect to ( , ). Lemma 4.9.1 Let a,b E F be isotropic with T(a)T(b)

f. O. Then b E E1(a)

if and only if a E E2(b). Proof. Let b E E1(a). By (4.56), we may assume that b = x*a with x E E1(a). Pick z E E2(a) with (x, z) = 1. According to (4.62), b * b E E2(a), so (z, b * b) = O. From (4.7) it follows:

b * z = (x * a) * z = -(z * a) * x + u 2( (x, z) )a. By (4.55), z * a = O. Hence a = b * z E E2(b). The proof of the converse implication is similar.

o

Lemma 4.9.2 Assume again a, bE F to be isotropic with T(a)T(b) f. O. If a * b = 0, then E2(a) n E 1(b) f. O. Proof. We may assume that F = F(V, t) is as in Th. 4.6.2 with e1 = a, e2 = T(a)-1a * a, V = E1 (a) and V' = E2(a) (see the part of § 4.5 beginning with equation (4.47)). Using the multiplication rule (4.74) it is straightforward to see that a * b = 0 implies that b = o:e2 + v, with 0: E k, v E V. If v = 0, then b is a nonzero multiple of e2 and E1(b) = E2(a), proving the lemma in that case. So we may assume that v f. O. Then

b * b = >. -1u(0:)u 2(0:)e1 Take

Z

E

V, with (z, t(v)

+ u(0:)C 1(v) + t(v) 1\ C 1(v).

1\ r1(v)) =

0, z ¢ kt(v), then

(z,b* b) = (z,t(v) and z * b is a nonzero multiple of t(z)

E2(a) n E1(b).

1\

t- 1(v)) = 0,

1\ t- 1 (v)

which lies in V'

n E 1 (b)

= 0

110

4. Twisted Composition Algebras

4.10 Nonnormal Twisted Octonion Algebras with Isotropic Norm In this section we briefly discuss analogues of the results of § 4.5 and § 4.6 in the case of a nonnormal twisted octonion algebra. We use the notations of Def. 4.2.1. So l is a non-cyclic cubic extension of k. Assume that char(k) ::I 2,3. Let F be a twisted composition algebra over I and (7 and assume that the norm N is isotropic. The normal extension F' = l' ®l F of F, introduced in Prop. 4.2.2, is an isotropic normal twisted composition algebra over l', and U = T ® id defines a T-linear anti-automorphism of F', by Prop. 4.2.4. Take a E F with the properties of Lemma 4.2.12. Then a and a *2 are fixed by u. We carry out the analysis of § 4.5 for a, with F' instead of F. Notations being as in that section, we have for x E F'

u(x * a)

= a * u(x),

u(x * a *2)

= a *2 *u(x).

It follows that for x E Ei

u(x * ei)

= ei * x.

Using (4.54) we see that u induces a T-linear bijection Ei -... Ei+l' From (4.58) we find that for x E Ei (4.82) and from (4.63) it then follows that

U(Xi

1\ Yi) =

U(Yi)

1\ U(Xi)

(Xi, Yi

E

Ei).

(4.83)

Next, (4.60) implies that

(Xl, t2(X2) ) = (7T( (U(X2), (t2 0 U)(Xl) )), and using (4.82)

(Xl, X2 ) = (7T( ( (u 0 t2"l }(X2), (t2 0 U)(Xl)) ) = (7T( (tl (U(X2)), t2( U(Xl)) )), whence

(U(X2), U(Xl))

= T( (Xl, X2))'

(4.84)

From (4.83) and (4.84) we deduce that for X, y, z E Ei

(U(X),U(y),U(z))

= -T((X,y,z)),

where the alternating trilinear form ( , , ) is as at the end of § 4.5. The properties of U which we just established indicate how to modify the construction of § 4.6 in order to deal with nonnormal twisted composition algebras. The notations are as in the beginning of the chapter. Assume we are given a normal twisted composition algebra F' = .1'(V, t) over l' and (7.

4.10 Nonnormal Twisted Octonion Algebras with Isotropic Norm

111

Definition 4.10.1 A hermitian involution of (V, t) is a r-linear bijection t : V -+ V' with the following properties: (i) (x,t{y»)=r«(y,t(x»)) (X,yEV). (ii) tot = (t')-l 0 t. (iii) t-1(x 1\ y) = t(y) 1\ t(x). Using (4.72) one finds by the same kind of argument as used in the proof of Lemma 4.6.1 that we also have (iii)' t{Xll\y') = t-1(yl) I\t-l{X') (X/,y' E V'). The next theorem is the analogue of 4.6.2 for nonnormal twisted composition algebras.

Theorem 4.10.2 Let t be a hermitian involution of (V, t). Then U{elel + 6e2 + x + x') = r(6)el + r{e2)e2 + t-1(x' ) + t{x) defines an involution of F'. The fixed point set Inv( u) is an isotropic nonnormal twisted composition algebra over land (J. All such algebras are of this form. Proof. The proof that u is an involution (as defined after the proof of Prop. 4.2.4) is straightforward. The second point then follows from Prop. 4.2.4. The last point is a consequence of what was established in the beginning of this section. 0 The properties of t of 4.10.1 can be reformulated. Putting

h(x, y) = (x, t{y») (x, y E V),

(4.85)

it follows from property (i) that h is a nondegenerate hermitian form on the l'-vector space V, relative to r. Then, using (ii),

h(t(x), y) = (t(x), t,(y) ) = a( (x, (t/)-l(t(y))))

=

(J(

(x, t(t(y) )

=

a(h(x, t(y))) , which explains the adjective "hermitian" in Def. 4.10.1. Using the three properties we also find

(t(x),t(y),t,(z») = (t-1(yl\x),t,{z») = r«(z,yl\x) = -r«(x,y,z), from which we see that det(t) = -1, the determinant being defined as in § 4.6. Conversely, given a nondegenerate hermitian form on V, there is a unique r-linear map t such that (4.85) holds. The requirements that t be hermitian relative to h and that det( t) = -1 give conditions equivalent to those of Def. 4.10.1. Let D be the cyclic crossed product {l', (J, -).), see § 4.7. Its center is k'. It is immediate that

112

4. Twisted Composition Algebras

defines an involution of the second kind of D, i.e. an anti-automorphism of D which induces a nontrivial automorphism on the center of D. (In the present case this is T.) Finally, we notice that by property (ii) we have t 3 = ,,-1 0 (t,)-3 0 ". Since 3 t and (t,)-3 both are scalar multiplication by ->., we conclude that now T(>') = >.. We will say that in this case the involution of the second kind of D is hermitian.

4.11 Twisted Composition Algebras with Anisotropic Norm In this section we will review analogues Th. 4.6.2 for the case of twisted composition algebras with anisotropic norm. We need a complement to Lemma 4.2.12, which we first establish. For the moment, F is an arbitrary twisted composition algebra, as in Def. 4.2.1. The restriction char(k) i 2,3 remains in force .. Assume that b E F has the properties (i), (ii), (iii) of Lemma 4.2.12 and that N(b) i o. In particular D(b) = T(b)2_4NI/k(N(b» i o. The polynomial with coefficients in k (4.86) has two distinct roots and put

eand .,."

which are nonzero. Assume that they lie in k

a = (e-.,.,)-1{N{b)-1eb+b*2), a'

= N(b)-1b-a = (.,.,-e)-I(N(b)-I.,.,b+b*2).

Lemma 4.11.1 a is isotropic and (a, a') = N(b)-1. We have a *2 = -N{b).,.,-1 a" T(a) = Moreover, a has the properties of Lemma 4.2.12.

_.,.,-1.

Similar results hold for a'.

e

Proof. That a is isotropic follows by a direct computation, using that is a root of (4.86). We have (a, b) = (e _.,.,)-1 (2e+T(b» = 1, since T(b) = -e-T/. Hence, a being isotropic, (a, a') = (a, N(b)-1b - a) = N(b)-1. The formula for a *2 follows from (4.34), using that NI/k(N(b» = N(b)N(b) *2 = e.,.,. The 0 remaining assertions are easy. Now assume that the norm N is anisotropic on F. Then k is infinite. We also assume that (with the notations of Prop. 4.2.5) N(F)* = M(N). By Cor. 4.2.6 this can be achieved by replacing F by an isotope. Lemma 4.11.2 There exists bE F with N{b) = 1 having the three properties of Lemma 4.2.12.

Proof. As N is anisotropic, k and l are infinite. View F as a vector space over k. By Lemma 4.2.12 there exists c E F with T(c) i 0, U(c) = T(C)2 4NI/k(N(c» i 0 and N(c) i o. View T and U as homogeneous polynomial

4.11 Twisted Composition Algebras with Anisotropic Norm

113

functions of respective degree 3 and 6, and N as a homogeneous quadratic mapping of F to 1 = k3 . Let K be an algebraic closure of k. By homogeneity, there is dE K ®k F with T(d) f= 0, U(d) f= 0, N(d) = 1. Let Q be the variety in K ®k F defined by the equation N(x) = 1. The set 0 of x E Q with T(x) f= 0, U(x) f= 0 is an open subset of Q which is nonempty, by what we just saw. By our assumptions there exists bo E F with N(b o) = 1. If x E F, N(x) f= 0, then x = bo - (N(X))-l( bo, x}x E Q, as a straightforward calculation shows. Since k is infinite, we can choose x E F such that N(x) f= 0 and x E O. Then b = x has properties (ii) and (iii) of Lemma 4.2.12, and N(b) = 1. Also, if we had b*2 = ~b, then since N(b) = 1 we had = 1 and T(b) = (~b, b) = 2~, whence the contradiction U(b) = o. Hence property (i) also holds. 0

e

Choose b as in the preceding lemma. Since N is anisotropic the polynomial (4.86) has no roots in k. We have to distinguish several cases, which we briefly discuss. Case (A). F is a normal twisted composition algebra. Let k1 be the quadratic extension of k generated by the roots ~ and .,., of (4.86). Notice that now ~.,., = 1. Denote by T1 the nontrivial automorphism of kdk. Then it = k1 ®k 1 is a Galois extension k which is cyclic of degree 6. Viewing (T and T1 as elements of its Galois group, (TTl is a generator of that group. Now F1 = II ®l F is a normal twisted composition algebra over hand (T, with an isotropic norm. We perform the analysis of § 4.5 for F 1 , with a as in Lemma 4.11.1. Then e1 = a, and by the lemma e2 = a'. Denote by v the T1-linear map T1 ® id of F1 = II ® F. Then v is an automorphism of F1 and F is the space of invariants Inv(v). Moreover v(x * ei) = v(x) * ei+1 (x E FI,i = 1,2). With the notations of 4.5, v induces a T1-linear bijection Ei --+ Ei+I. As in § 4.10 we find

V(Xi

1\ Yi)

= V(Xi) 1\ V(Yi),

(V(X2),V(XI)) =T1«X},X2)). These formulas indicate how to modify the construction of 4.6 in the present case. Assume given a twisted composition algebra FI = F(VI , t) over l1 and (T. Definition 4.11.3 A unitary involution of (V}, t) is a T1-linear bijection L1 : V1 --+ V{ with the following properties:

(i) (X,L1(Y)}=T«y,L1(X))) (X,yEV1). (ii) L1 ot = t' 0 L1. (iii) L11(x 1\ y) = L1(Y) 1\ L1(X).

114

4. Twisted Composition Algebras

We also have (iii)' "l(X' /\ y'}

= "ll(y'} /\ "ll(x'}.

Theorem 4.11.4 Let "I be a unitary involution of (VI! t). Then V(6el + 6e2 +x + x'} = Tl(~2}el + Tl(~1}e2 - £ll(x'} - £l(X) defines a Tl-linear automorphism of Fl. The fixed point set Inv(v) is a normal twisted composition algebra over I and 17, with N(F)* = M(N). All anisotropic normal twisted composition algebras with the last property are of this form. Proof. The proofs of the assertions about v and F are straightforward. The last point follows from what was established in the beginning of this section. 0 Again, there is a reformulation of the properties of Def. 4.11.3. Define

Then h is a nondegenerate hermitian form on VI, relative to Tl. We now have

h(t(x), t(y))

= 17(h(x, y)),

which explains the adjective ''unitary''. Also, det( "1) = -1. The cyclic crossed product occurring in the present case is D = (It, 17, e), where ~ is as before. Notice that T/ = Tl(e} = ~-1. The central simple algebra Dover kl has the involution of the second kind

We now call the involution unitary. Case (B). F is a nonnormal twisted composition algebra and the normal composition algebra F' is anisotropic. In this case (4.86) has no root in k'. Let kl and Tl be as in Case (A). Now li = kl ®k l' is a Galois extension of k whose group is 8 2 X 8 3 , We view 17, T and Tl as automorphisms of Ii. Put F{ = kl ®k F'. By 4.6.2 we may assume that is of the form F{ = .r(V1 , t}, where VI is a vector space over Ii and t is 17-semilinear. By 4.10.2 we have a T-linear hermitian involution t on (VI, t) and an involution on Vb whereas by case (A) we have a Tl-linear unitary involution on (VI, t) and an automorphism v of F{. Then F

= Inv(u, v} = Inv(u} n Inv(v}.

In the present case the cyclic crossed product D is a central simple algebra over the field ki = kl ®k k'. It has two commuting involutions of the second kind: a T-linear one which is hermitian, and a Tl-linear one which is unitary. Case (C). F is a nonnormal twisted composition algebra and (4.86) has roots in k'.

r

i

4.12 Historical Notes

115

We can now take ~ and TJ in k', hence a E F'. Proceeding as in 4.10 we have u(x * ei) = eHl * x (x E E i ),

u 0 ti = til

0

u.

It follows that u defines 7'-linear bijections t and t' of V and V', respectively. Moreover, (4.83) holds, and we have the counterpart of (4.84)

Assume again that F' = F(V, t), with V a vector space over l'. Then we have a 7'-linear automorphism t of order 2 such that

Moreover, t' being as in the last equation,

t'{x 1\ y) = t(x) 1\ t(y) (x, y E V). With these notations,

defines a 7'-linear involution of F', such that F is the fixed point set Inv(u). Any twisted composition algebra of Case (C) can be obtained in this way. The cyclic crossed product D = (l', 0', -A) has the unitary involution of the second kind

4.12 Historical Notes Twisted composition algebras were introduced by T.A. Springer in the normal case, with a view to a good description of nonreduced Albert algebras (see Ch. 6). The theory was first exposed in a course at the University of Gottingen in the summer of 1963 (see [Sp 63]). The generalization to the nonnormal case is due to F.D. Veldkamp (in an unpublished manuscript). Independently, it was also given in [KMRT, §36].

5. J-algebras and Albert Algebras

In this chapter we discuss a class of Jordan algebras which includes those that are usually named exceptional central simple Jordan algebras or Albert algebras. Our interest in Albert algebras is motivated by their connections with exceptional simple algebraic groups of type E6 and F 4, a topic we will deal with in Ch. 7. They also playa role in a description of algebraic groups of type E7 and Eg , but we leave that aspect aside. We will not enter into the general theory of Jordan algebras, but use an ad hoc characterization of the algebras under consideration by simple axioms, which are somewhat reminiscent of those for composition algebras. We call these algebras J-algebras, since they are in fact a limited class of Jordan algebrasj see Remark 5.1.7. In this and the following chapters, fields will always be assumed to have characteristic =F 2,3. The assumption characteristic =F 3 is for technical reasons and could possibly be removed. However, characteristic =F 2 is essential for our approach to Jordan algebras as (nonassociative) algebras with a binary product. If one wants to include all characteristics, it is necessary to use quadratic Jordan algebras as introduced by K. McCrimmon [MCClj see also [Ja 69], [Ja 81] or [Sp 73].

5.1 J-algebras. Definition and Basic Properties Let k be a field with char(k) =F 2,3 and let C be a composition algebra over k. For fixed 'Yi E k*, let A = H(C;'Yb'Y2,'Y3) be the set of ('Yb'Y2,'Y3)-hermitian 3 x 3 matrices x

= h(6,6,6jcl,C2,C3) = ('Y2J~lC3 C2

-f: _ 6

3C2

'Yl:7

'Y3

)

(5.1)

'Y2 Cl

with ~i E k and Ci E C (i = 1,2,3) j here - denotes conjugation as in § 1.3. We define a product in A which is different from the standard matrix product: (5.2) where the dot indicates the standard matrix product and the square is the usual one with respect to the standard product (which coincides with the

118

5. J-algebras and Albert Algebras

square with respect to the newly defined product). This multiplication is not associative. Together with the usual addition of matrices and multiplication by elements of k, it makes A into a commutative, nonassociative k-algebra with the 3 x 3 identity matrix e as identity element. We introduce a quadratic norm Q on A by

Q(x) =

i tr(x

2

)

= 2(~? + ~~ + ~~) + 'Yil'Y2N(Cl) + 'Yl1'Y3N(c2) + 'Y2"1'YIN(C3)

for x

(5.3)

= h(6,6,6;c1,C2,C3) E A, and the associated bilinear form (x, y) = Q(x + y) - Q(x) - Q(y) = tr(xy) (x,y E A).

This bilinear form is nondegenerate. Of special interest is the case that C is an octonion algebra; we then call A = H(C; I'll 1'2, 1'3) an Albert algebra. More generally, A is called an Albert algebra if k' ®k A is isomorphic to such a matrix algebra H(C';'Y1I'Y2,'Y3) for some field extension k' of k and some octonion algebra C' over k'. In Prop. 5.1.6 we will prove the relation x 2(xy) = x(x 2y), which is typical for Jordan algebras. The reader will have no difficulty in verifying the following three rules: Q(x 2) = Q(X)2 if (x, e) = 0, (5.4)

(xy,z) = (x,yz), 3 Q(e) = 2'

(5.5) (5.6)

We will, conversely, consider a class of algebras with a quadratic norm Q that satisfies (5.4), (5.5) and (5.6). This class will turn out to contain, besides the algebras related to the algebras A = H(C; 1'1, 1'2, 1'3) introduced above, one other type of algebras; see Prop. 5.3.5, the remark that follows it, and the classification in Th. 5.4.5. Definition 5.1.1 Let k be a field of characteristic :F 2,3. A J-algebra over k is a finite-dimensional commutative, not necessarily associative, k-algebra A with identity element e together with a nondegenerate quadratic form Q on A such that the conditions (5.4), (5.5) and (5.6) are satisfied. Q is called the norm of A, and the associated bilinear form ( , ) will often be called the inner product. A J-subalgebra is a nonsingular (with respect to Q) linear subspace which contains e and is closed under multiplication. An isomorphism t : A - A' of J-algebras over k is a bijective linear transformation which preserves multiplication: t(xy) = t(x)t(y) (x, YEA).

Remark 5.1.2 It will be shown in Prop. 5.3.10 that in a J-algebra of dimension > 2 the norm Q is already determined by the linear structure and the product, and the same holds for the cubic form det that will be introduced in Prop. 5.1.5. As a consequence, an isomorphism necessarily leaves Q and det invariant in dimension > 2.

r

5.1 J-algebras. Definition and Basic Properties

119

We begin the study of J-algebras with a lemma that gives a linearized version of (5.4). Lemma 5.1.3 If (x, e)

= (y, e) = (z, e) = (u, e) = 0,

then

2(xy, zu )+2( XZ, yu )+2(xu, yz) = (x, y)( Z, u )+( x, z)( y, u )+( x, u)( y, z). Proof. By substituting AX + J1.y + /JZ + (!U for x in (5.4), writing both sides out as polynomials in A,J1.,/J and (!, and equating the coefficients of AJ1./J{! on either side, we immediately get the formula. Here we use that the degree of the polynomials is 4 and that Ikl > 4. 0 Proposition 5.1.4 If A is a J-algebm over a field k and 1 is any extension field of k, then 1®k A, with the extension of the product and the quadmtic

form, is a J-algebm over l. Proof. The linearized version of (5.4) that we proved in the Lemma is in fact equivalent to (5.4) itself. This multilinear version clearly also holds in 1®k A. Similarly for (5.5). 0 By the Hamilton-Cayley Theorem, every element in a 3 x 3 matrix algebra over a field satisfies a cubic equation. It is conceivable that a similar result holds in a "matrix algebra" H(C;'Yl''f2,'Y3). In fact, it does in all J-algebras. Proposition 5.1.5 Every element x in a J-algebm A satisfies a cubic equa-

tion x3

1

(x, e )x2 - (Q(x) - 2(x, e )2)x - det(x)e

-

= 0,

(5.7)

called its Hamilton-Cayley equation. Here det is a cubic form on A. Proof. With the aid of equation (5.5) we derive from Lemma 5.1.3 (with Z = u = x)

(x 3

-

Q(x)x,y)

=0

(x,y

E

A, (x,e)

= (y,e) = 0).

Since ( , ) is nondegenerate, this implies that

x3

-

Q(x)x

= lI:(x)e

(x

E A,

(x,e) =

0),

(5.8)

where II: is a cubic form with values in k. We can write any x E A as x = x' + x, e)e with (x', e) = O. Substitution of x x, e)e in equation (5.8) yields after some computation

l(

x3

-l(

-

1

(x,e)x2 - (Q(x) - 2(x,e)2)x - det(x)e = 0

where det is a cubic form on A with values in k. If det(x)

i= 0, then

(x

E

A), o

5. J-algebras and Albert Algebras

120

satisfies xx- 1 = ej we will come back to this in Lemma 5.2.3. The polynomial

Xx(T)

= T3 -

1

(x, e )T2 - (Q(x) - 2(x, e )2)T - det(x)

(5.9)

is called the characteristic polynomial of x, and det(x) the determinant of Xj the cubic form det is called the determinant function on A, or just the determinant of A. By taking the inner product of the left hand side of (5.7) with e, one finds 1

0= (x3,e) - (x,e){x 2,e) - (Q(x) - 2{x,e)2)(x,e) - det(x)(e,e)

= (x 2,x) -

1

3Q(x)(x,e) + 2(x,e)3 - 3det(x).

Hence

(5.10) Notice that det( e) = l. With the aid of (5.10) one easily computes det(x) for an element x h(6,6,6jCl,C2,C3) ofH(Cj'Yl,1'2,1'3) as in (5.1): det(x)

=

= ~1~26 -1'311'2~lN(cd-1'111'3~2N(C2)-1'211'1~3N(C3)+N(CIC2' C3)' (5.11)

Here N ( , ) denotes the bilinear form associated with the norm N on C. The cubic form det uniquely determines a symmetric trilinear form ( , , ) with (x,x,x) = det(x). We have

6{ x, y, z) = det(x + y + z) - det(x + y) - det(y + z) - det(x + z)+ det(x)

+ det(y) + det(z).

We derive some consequences of the Hamilton-Cayley equation. Replacing x by x + y + z in (5.7) yields 1

(X+y+z)3_(X+y+z, e )(x+y+z)2_(Q(x+y+z)-2(x+y+z,e )2)(X+Y+z) = det(x + y + z)e. Collecting in both sides terms which are linear in each of the variables x, y and z, we obtain the following formula.

x(yz) + y(xz) + z(xy) = {x,e)yz + (y,e)xz + (z,e)xy +

5.1 J-algebras. Definition and Basic Properties

1

121

1

2((y,z} - (y,e}(z,e})x + 2((x,z) - (x,e}(z,e})y + 1

2((x,y) - (x,e}(y,e})z+3(x,y,z}e.

(5.12)

Replacing z by x in this equation we find

2x(xy) + x 2y = 2( x, e }xy + (y, e }x2+ 1

((x,y) - (x,e}(y,e})x + (Q(x) - 2(x,e}2)y + 3(x,x,y}e. (5.13) From equation (5.12) one easily derives a formula that expresses the symmetric trilinear form associated with det in the inner product and the product. Namely, take the inner product of either side of (5.12) with e, apply condition (5.5) several times and use (e, e) = 3. After rearrangement and dividing by 3 one finds:

3(x,y,z}

= (xy,z) -

1

1

1

2(x,e}(y,z} - 2(y,e}(x,z} - 2(z,e}(x,y}+ 1

2(x,e}(y,e}(z,e}.

(5.14)

We can now prove the Jordan identity. Proposition 5.1.6 In any J-algebm A the Jordan identity holds:

x 2(xy) = x(x 2y). Proof. It suffices to prove that x 2(xy) and x(x 2y) have equal inner products with any z E A. In view of (5.5) this amounts to showing that (xy,x 2z) = (xZ,x2y) (x,y,z E A). (5.15) This relation is immediate from (5.5) if y = e or z = e, so it suffices to prove it for the case (y, e) = (z, e) = O. Under these assumptions, we take the inner product of either side of equation (5.13) with xz and, using (5.5), find the relation

2(x{xy),xz} + (x2y,xz) = 2(x,e}(xy,xz} + (x,y}(x,xz)+ 1

Q{x)(y,xz} - 2(x,e}2(y,xz} +3(x,x,y}(x,z}. Replacing 3( x, x, y) in the right hand side by the expression that follows from equation (5.14), we arrive at the formula

(x2y,xz) = -2(x(xy),xz) +2(x,e}(xy,xz} + (x,y}(x,xz)+ 1

Q{x)(y,xz} - 2(x,e}2(y,xz} + (x,xy}(x,z) - (x,e}(x,y}(x,z) for (y, e) = (z, e) = O. It is straightforward to verify that the right hand side of this equation is symmetric in y and z. So the left hand side is symmetric in y and z, too, which just amounts to (5.15). 0

122

5. J-algebras and Albert Algebras

Remark 5.1.7 A commutative algebra over a field k of characteristic =F 2 in which the Jordan identity holds is called a (commutative) Jordan algebm. So the above proposition says that every J-algebra is a Jordan algebra. A consequence of the Jordan identity is power associativity: xmxn = xm+n (m,n ~ 1); see, e.g., [Ja 68, Ch. I, Th. 8J or [Schat, Ch. IV, §1, p. 92J. For Jalgebras, power associativity follows more easily, as we show in the following corollary. Corollary 5.1.8 For any x in a J-algebra A, the subalgebm k[xJ genemted by x is a homomorphic image of k[TJ/Xz(T), where XZ is the chamcteristic polynomial of x (see equation (5.9)). Consequently, A is power associative. Proof. By substituting x for y in the Jordan identity we find that X 2 X 2 = x4. By the Hamilton-Cayley equation (5.7), every element of k[xJ can be written in the form ~oe + 6x + 6x2 (but notice that e, x and x 2 need not be linearly independent). The product of two such elements is associative since (x'xm)xn = xl+m+n for l,m,n ~ 2, as follows from the Jordan identity. In other words, k[xJ is the homomorphic image of the associative algebra k[TI/Xx(T). 0 Remark 5.1.9 The word "sub algebra" above is meant in the sense of the theory of nonassociative algebras, so as a linear subspace containing e and closed under multiplication. For a J-subalgebra we also required in Def. 5.1.1 that the restriction of the norm Q to it is nondegenerate; we will see in Prop. 5.3.8, that this need not be the case with k[x}.

5.2 Cross Product. Idempotents With the aid of the symmetric trilinear form { , , } associated with det and the bilinear inner product we introduce a cross product x that will be used frequently in future computations: in a J-algebra A, we define x x y (x, YEA) to be the element such that

(x x y,z) = 3(x,y,z)

(z E A).

{5.16}

The cross product is evidently symmetric. In the following lemma we express it in terms of the ordinary product, and collect some formulas that will be useful in later computations. Lemma 5.2.1 The following formulas hold for the cross product. (i) x x Y = xy - ~{x,e}y - ~(y,e}x - ~(x,y}e + ~(x,e}(y,e}ej (ii) x(x x x) = det(x)e; (iii) (Xl XX2) xy = !(XIX2}y-!XI (X2Y)-!X2(XIy)+i{ Xl, Y }x2+i{ X2, Y}XI; (iv) (x x x) x (x x x) = det(x}x;

r

t

5.2 Cross Product. Idempotents

+ 4(x X z) X (y X u) + 4(x X u) X (y X z) = 3( x, y, z)u + 3( x, y, u)z + 3( x, z, u)y + 3( y, z, u )x; 4x X (y X (x X x)) = (x,y)x X x + det(x)y; det(x X x) = det(x)2.

(v) 4(x (vi) (vii)

123

X

y)

X

(z

X

u)

Proof. (i) By equation (5.14), (x X y, z) equals the inner product with z of the right hand side of the formula in (i), for all z. Hence (i) holds. (ii) Using the above formula to compute x X x, one finds

x(x

X

x) = x 3

1 (x,e)x2 - (Q(x) - "2(x,e)2)x

-

(by the Hamilton-Cayley equation (5.7)).

= det(x)e

(iii) A straightforward computation using (i) and equation (5.14) yields (x

X

x)

X

y

= x2y -

1 "2(y,e)x 2

-

(x,e)xy +

1 1 1 2 3 "2(x,e)(y,e)x - "2(Q(x) - "2(x,e) )y - "2(x,x,y)e. Using equation (5.13), one reduces this to

(x

1

X

x) x y = "2x2y - x(xy)

1

+ "2(x,y)x.

Linearizing this one obtains the formula in (iii). (iv) Replace Xl and X2 by x, and y by x X x in the formula of (iii). Then using (ii), the Hamilton-Cayley equation and power associativity, one easily gets the result. (v) This follows by linearizing the previous formula. (vi) and (vii) In (v), replace x, z and u by x X x; this yields

4[(xxx) x (x xx)] x [yx (x xx)]

= 3(xx x, xxx, y )xxx+det(xxx)y.

(5.17)

The left hand side equals 4det(x)x x (y x (x x x)) by (iv). Further,

3(x x x,x x x,y) = (x x x) x (x x x),y) = det(x)(x,y), by (5.16) and (iv). Replacing y by x x x in this formula we get

det(x x x) = det(x)2, i.e., (vii) holds. Using these last two equations to replace terms in the right hand side of (5.17), we find

= det(x)( x, y)x x x + det(x)2y. This yields formula (vi) for det(x) i= 0; by continuity for the Zariski topology 4 det(x)x x (y x (x x x))

over an algebraic closure of k it holds everywhere (cf. the end of the proof of Prop. 3.3.4). 0 An important role in developing the theory of J-algebras will be played by idempotent elements, i.e., elements u such that u 2 = U.

-

124

-------

-

-~~~~~~~-

5. J-algebras and Albert Algebras

Lemma 5.2.2 Ifu 1= 0, e is an idempotent in A, then det(u) = 0 and Q(u) = ! or Q(u) = 1; in addition, {u, e} = 2Q(u). F'urther, e - '1.£ is an idempotent withu(e-u) =0, {u,e-u} = 0 andQ(e-u) = ~ -Q(u). So if A contains an idempotent 1= O,e, then it contains an idempotent '1.£ with Q(u) =!. Proof. The Hamilton-Cayley equation reads for an idempotent (1 - {u, e} - Q(u) Since an idempotent and

1= 0, e can

'1.£:

1

+ 2{ '1.£, e }2)u = det(u)e.

not be a multiple of e, we find det(u) = 0

1 - ('1.£, e) - Q( '1.£)

1

+ 2('1.£, e ) 2 = o.

(5.18)

Using (5.5) we see that 1 1 2 1 Q(u) = 2(u,u} = 2('1.£ ,e} = 2(u,e}, so ('1.£, e)

= 2Q('1.£). Substituting this in equation (5.18), we get 3

Q(u)2 - 2 Q (u)

This yields Q(u)

1

+ 2 = O.

= ! or Q(u) = 1. The rest of the proof is straightforward. 0

!

An idempotent '1.£ with Q(u) = is called a primitive idempotent. This name is in agreement with the usual terminology. For let '1.£ be an idempotent with Q(u) = and suppose we could decompose

4,

Then we would have

and (Ui' e) = 2Q(Ui) = 1 or 2, which leads to a contradiction. On the other hand we will see in Prop. 5.3.7 that in most cases e - u is a sum of two orthogonal primitive idempotents. We next prove an addition to Prop. 5.1.5.

Lemma 5.2.3 Let A be a l-algebra. For x E A, there exists x-I E A having the properties xx- l = e and x(x-1y) = x-l(xy) (y E A) if and only det(x) 1=

o.

Such an element x-I is unique, viz.,

5.3 Reduced J-algebras and Their Decomposition

125

Proof. If det(x) '" 0, then

satisfies xx- 1 = e by the Hamilton-Cayley equation (5.7). From the Jordan identity it follows that x(x-1y) = x-1(xy) (y E A). If z satisfies xz = e and x(zy) = z(xy) (y E A), then x 2z = x and x 3 z = x 2. Hence by the Hamilton-Cayley equation,

1 det(x)z = x 2 - (x,e)x - (Q(x) - 2(x,e)2e, which shows uniqueness. Now suppose X-I would exist for x '" 0 with det(x) X

1 2 "7' (x,e)x - (Q(x) - 2(x,e)2)e =

= O. Then

o.

Taking the inner product with e we find that Q(x) = ~(x,e)2. Hence x 2 = (x,e)x. Then X = X ( X -1) X = X -1 X 2 = ( x, e ) e. It follows that x is a nonzero multiple of e. Since det( e) = 1 we obtain a contradiction. 0 The element X-I as in the Lemma is called the }-inverse or just inverse of x. It is not an inverse in the general sense of nonassociative algebras as we defined at the end of § 1.3, since x- 1(xy) = y need not hold for y ¢ k[x].

5.3 Reduced J-algebras and Their Decomposition A J-algebra is said to be reduced provided it contains an idempotent", 0, e. By Lemma 5.2.2 it also contains a primitive idempotent. We consider a reduced J-algebra A over k and fix a primitive idempotent u in A. Define

E

= (ke EB ku).L = {x E A I (x, e) = (x, u) = 0 }.

The restriction of Q to keEBku is nondegenerate, hence the same holds for E. From x E E we infer that ux E E, for (ux,e) = (x,u) = 0 and (ux,u) = (x, u) = o. So we can define the linear transformation t : E --+ E,

x t--+ UX.

(5.19)

126

5. J-algebras and Albert Algebras

Lemma 5.3.1 t is symmetric with respect to { , }, and t 2

E=Eo$EI where Ei

= !t.

We have

with Eo 1. Ell

= {x E Elt(x) = ~iX}.

Proof. Let x, y E E. The symmetry of t follows from property (5.5):

{t(x),y}

= {ux,y} = {x,uy} = (x,t(y)}.

Using Lemma 5.2.2, we get from equations (5.13) and (5.14)

2u(ux) + ux

= 2ux,

from which t 2 = !t follows. This implies that the possible eigenvalues of t are o and and the symmetry of t implies that the eigenspaces are orthogonal 0 and span E.

!,

Eo and EI are called, respectively, the zero space and half space of u. The restrictions of Q to Eo and to EI are nondegenerate. Lemma 5.3.2 The following rules hold for the product in E. (i) For x, y E Eo, 1 xy = '2{x,y}(e - u).

(ii) For x, y EEl, xy

1 = 4{x,y}(e+u) +xoy

withxoy E Eo.

(iii) If x E Eo and y EEl, then xy EEl. Proof. First we derive a formula for arbitrary x, y E E. Replacing z by u in equation (5.12) we get

u(xy) + x(uy) + y(ux)

1 = xy + '2( x, y}u + 3( u,x, y}e

(x,YEE).

Now 3{ u, x, y) = (u, x x y) and in the latter expression we replace x x y by the expression given in Lemma 5.2.1. This leads to the equation

u(xy) +x(uy)+y(ux)

1 = xy+ 2{ x,y )u+({ ux, y) -

1

2{x, y»e

(x,y

E

E).

(5.20) For x, y E Eo this equation reduces to 1

1

u(xy - 2{x,y)e) = xy - 2{x,y)e.

5.3 Reduced J-algebras and Their Decomposition The only elements z of A satisfying the relation uz u, so

127

= z are the multiples of

1 XY - 2"{x,y}e = IW

for some

K,

E

k. Taking inner products with u, we get 1

(xy,u) - 2"(x,y}(e,u} Since (xy,u) = (x,uy) = 0, we find

xy

K,

= K,(u,u}.

= -!(x,y). Hence

1

= 2"(x,y}(e -

u),

which proves (i). Next, we consider x, y EEl. We define x

1 x 0 y = xy - 4( x, y }(e + u)

(x,y

E

y by

0

Ed.

(5.21)

Using that (xoy,u) = !(x,y) we find that (xoy,u) = O. Likewise, (x 0 y, e) = O. Hence x 0 Y E E. From equation (5.20) we infer that u(xy) = !(x,y)u, so

u(x 0 y)

= u(xy -

1

= u(xy) -

4(x,y}(e + u))

1

2"(x,y)u = o.

Hence x 0 y E Eo. This proves (ii). Finally, let x E Eo and y EEl· Then

(xy,u)

= (y,ux)

= 0 and

so xy E E. Equation (5.20) yields u(~y) proves that xy EEl'

(xy,e) = (x,y) = 0,

+ !xy = xy, so u(xy)

= !xy. This

0

In the following lemma we collect a number of formulas involving products between elements of Eo and E 1 .

Lemma 5.3.3 The following formulas hold for x, xl. X2 E Eo and y EEl. (i) x(xy) = ~Q(X)Yi

Xl(X2Y) + X2(X1Y) = ~(Xl.X2 )Yi yo xy = ~Q(Y)Xi (y 0 y)y = ~Q(Y)Yi (Yl 0 Y2)Y3 + (Y2 0 Y3)Yl + (Y3 0 Yl)Y2 ~(Y3'Yl )Y2i (vi) Q(xy) = ~Q(X)Q(Y)i (vii) Q(y 0 y) = ~Q(y)2. (ii) (iii) (iv) (v)

=

~ (Yl. Y2 )Y3

+ ~ (Y2, Y3 )Yl +

128

5. J-algebras and Albert Algebras

Proof. (i) For x E Eo and y E El we have by (5.13),

2x(xy) + x 2y = Q(x)y + 3( x, x, y }e. According to equation (5.16) and Lemma 5.2.1,

3(x,x,y}

= (x,x x y) = (x,xy) = 0,

since x E Eo and xy EEl. By Lemma 5.3.2 (i), x 2 = Q(x)(e - u). So we get 1

2x(xy) + Q(x)y - 2 Q(x)y = Q(x)y, from which (i) follows. (ii) follows from (i) by linearizing. (iii) Interchanging x and y in formula (5.13) and using (5.16) and Lemma 5.2.1 again, we find

xy2 + 2y(xy) = Q(y)x + (x, y2 }e. Using y2

= yo y + !Q(y)(e + u), we get 1

x(y 0 Y + 2Q(y)(e + u)) + 2y(xy)

= Q(y)x + (x, y2 }e.

By Lemma 5.3.2 (i),

x(yoy)

1 1 1 1 Q(y)(e+u)}(e-u) = 2(x,y2}(e-u). = 2(x,yoy}(e-u) = 2(x,y2_ 2

Substituting this into the above formula and rearranging terms, we get 111

2y(xy) - 2{x,y2}u

= 2 Q(y)x + 2(X,y2 }e.

By (5.5) this yields 1

y(xy) - 4(y,x y )(e+u)

= 41Q(y)x,

from which (iii) follows. (iv) The Hamilton-Cayley equation (see Prop. 5.1.5) for y E E reads

y3 _ Q(y)y _ det(y)e = o. Now for y E Eb 3det(y)

= (y,y x y) = (y,y2 _ Q(y)e) = (y,y2) = 1

(y, yo y - 2Q(y)(e + u)) = 0, so the Hamilton-Cayley equation for y E El becomes

5.3 Reduced J-algebras and Their Decomposition

129

y3 = Q(y)y. From this we derive

(y 0 y)y

= (y2 -

1 2Q(y)(e + u))y

= Q(y)y -

1 1 2Q(Y)Y - 4Q(Y)Y

1

= 4 Q (y)y,

thus obtaining (iv). (v) is obtained by linearizing. (vi) From Lemma 5.1.3 we derive

4Q(xy) Computing x 2 and

y2

4Q(xy)

+ (X 2,y2) = 2Q(x)Q(y).

with the aid of Lemma 5.3.2, we get 1

+ 2Q(x)Q(y)( e - u, e + u} = 2Q(x)Q(y).

Since (e - u, e + u) = 2, we arrive at the formula of (vi). (vii) By (vi) and (iv),

1 1 1 4 Q(y 0 y)Q(y) = Q«y 0 y)y) = Q(4 Q (y)y) = 16 Q (y)3, so Q(y 0 y) = lQ(y)2 follows for Q(y) '# O. By Zariski continuity (over an algebraic closure of k), this holds for all y EEl. 0 Statement (i) in the above lemma can be interpreted in terms of Clifford algebras (cf. § 3.1). This will be used later on. Corollary 5.3.4 Let CI(Qj Eo) be the Clifford algebra of the restriction of Q to Eo. The map t.p : Eo - End(E1 ) defined by

t.p(x)(y)

= 2xy

can be extended to a representation ofCI(Qj Eo) in Eb i.e., a homomorphism of Cl(Qj Eo) into End(El). Proof. By (i) in the Lemma,

so the extension of t.p to a homomorphism of the tensor algebra T(Eo) into End(Ed respects the defining relations for CI(Qj Eo). 0 Proposition 5.3.5 Let A be a reduced J-algebra. Consider a primitive idempotent u in A, and let E, Eo and El be as before, with respect to u. (i) Eo = 0 if and only if A is 2-dimensionalj then A = ku ED k(e - u), an

orthogonal direct sum. (ii) If El = 0, then

130

5. J-algebras and Albert Algebras

A

For A, A', J.L, J.L'

E

k, x, x'

E

= kU$k(e-u) $

Eo.

Eo product and nonn given by

(AU + J.L(e - u) + x)(A'u + J.L'(e - u) + x') = 1

= AA'U + (J.LJ.L' + 2q(x,x'»)(e and

u) + J.LX' + p.'x,

1 Q(A + J.L(e - u) + x) = "2A2 + J.L2 + q(x) ,

where q is a nondegenerate quadratic fonn on Eo with associated bilinear fonn q( , ). Conversely, for any vector space Eo (possibly 0) with a nondegenerate quadratic fonn q, the above fonnulas define a J-algebra A. Proof. If Eo = 0, then yo y = 0 for y E Ell so Q(y) = 0 by Lemma 5.3.3 (iv). Since the restriction of Q to EI is nondegenerate, this implies EI = O. Hence A is the 2-dimensional algebra ku $ k(e - u). If EI = 0, then A is an orthogonal direct sum of vector spaces:

A

= ku $

k(e - u) $ Eo.

The product is determined by Lemma 5.3.2 (i), if we take for q the restriction of Q to Eo. Conversely, any vector space Eo with a nondegenerate quadratic form q yields a J-algebra A of dimension equal to dim Eo + 2 as above; it is straightforward to verify the axioms (5.4)-(5.6). 0 We call a J-algebra as in (li) of the above proposition a J-algebra of quadratic type. Such a J-algebra A is closely related to the Jordan algebra of the quadratic form q as in [Ja 68, p. 14J; in fact, A is the algebra direct sum of a one-dimensional algebra ku and the subalgebra k(e - u) $ Eo, the latter being the Jordan algebra of q with e - u as identity element (and also a J-algebra if we multiply Q by ~).

Lemma 5.3.6 If EI :f= 0, hence also Eo :f= 0, then there exists Xl E Eo with Q(XI) = In fact, one can take Xl = Q(y)-ly 0 Y for any y E EI with Q(y) :f= 0; then XIY =

i.

iy·

Proof. Since the restriction of Q to EI is nondegenerate, there exists y E EI such that Q(y) :f= O. Then Xl = Q(y)-ly 0 Y E Eo. By Lemma 5.3.3 (vii), Q(XI) = From (iv) of that same lemma we infer XIY = 0

1.

h.

In the discussion after Lemma 5.2.2 we claimed that in most cases e - u is a sum of two orthogonal primitive idempotents if u is a primitive idempotent. We can now prove the precise result.

5.3 Reduced J-algebras and Their Decomposition

131

Proposition 5.3.7 If A is a reduced J-algebra and u is a primitive idempotent in A, then e - u is a sum of two orthogonal primitive idempotents unless A = kuEDk(e - u) ED Eo and Q does not represent 1 on Eo. The latter condition is independent of the choice of u such that the corresponding half space EI is zero. Proof. If EI # 0, then there exists Xl E Eo with Q(XI) = ~ by the above lemma. Then !(e - u) + Xl and !(e - u) - Xl are primitive idempotents with sum e - u. If EI = 0, then

A = ku ED k(e - u) ED Eo.

+ J.L(e - u) + al with al E Eo, and b = e - u - a, so b = ->.u + (1 - J.L)(e - u) - al and a + b = e - u. One easily verifies that a and b are both idempotent if and only if>. = 0, J.L = ! and Q(al) = ~. Then indeed Q(a) = Q(b) = ! and ab = 0. The restriction of Q to ku ED k(e - u) is independent of u, so by Witt's Consider a = >.u

Theorem the same is true for the restriction to the orthogonal complement &. 0 We saw in Prop. 5.1.5 that every element satisfies a cubic equation, the Hamilton-Cayley equation. We compare this with the minimum equation. Let a E A and denote by rna its minimum polynomial. So k[a) ~ k[T)frna(T) and rna divides Xa'

Proposition 5.3.8 The polynomials rna and Xa have the same roots in a common splitting field. Hence rna = Xa if Xa has three distinct roots in a. For a f/ ke, the restriction of the norm Q of A to k[a) is non degenerate if and only if not all roots of Xa are equal. If Xa has a root in k, then k[a) contains a primitive idempotent if and only if not all roots of Xa are equal. Proof. Upon replacing k by a splitting field of Xa, we may assume that Xa splits in k. If dim k[a) = 1, then a = >.e for some>. E k. Then rna(T) = T - >., and one easily computes that Xa(T) = (T - >.)3. Next assume dim k[a) = 2. If rna(T) = (T - >.)2, consider X = a - >.e. This satisfies x2 and X # 0, so rnx(T) T2. It follows that Xx(T) = T3 - (x,e}T2, so Q(x) = !(x,e}2. But Q(x) = !(x,x) = !(x 2,e) = 0, hence also (x, e) = 0. This implies that (a, e) = 3>. and Q(a) = ~>.2. Since det(a) = >.3, we find that Xa(T) = (T - >.)3. It is easily verified that in k[a] = k[x] there is no idempotent # e, and that the restriction of Q to this subspace is degenerate. If rna(T) = (T - >')(T - J.L) with>' # J.L, then

=

°

k[a]

=

~

k[T)f(T - >.)(T - J.L)

~ kED

k,

so k[a] contains orthogonal idempotents u and e - u; we may assume u to be primitive. Q is nondegenerate on k[a] (see Lemma 5.2.2). If a = au+,8(e-u),

---

132

--------------

5. J-algebras and Albert Algebras

then ma(T) = (T-a)(T-{3), so a = A and {3 = f.t, or a = f.t and (3 = A. One easily computes that Xa(T) = (T - a)(T - (3)2, which is (T - A)(T - f.t)2 or (T - A)2(T - f.t). Now let dim k[a] = 3. Then ma = Xa. If Xa has three distinct roots, then k[a] ~ k Ee k Ee k, i.e., k[a] is spanned by three idempotents Ul, U2, U3 with UiUj = 0 for i :f:. j. These must be orthogonal with respect to Q since (Ui' Uj ) = (e, UiUj ) = 0 if i :f:. j. Hence the restriction of Q to k[a] is nondegenerate. If Xa(T) = (T - A)(T - f.t)2 with A :f:. f.t, then k[a] ~ k $ k[x] for some x :f:. 0 with x 2 = o. So k[a] contains. an idempotent. The restriction of Q to k[x] is degenerate, and since the two components k and k[x] in the direct sum decomposition k[aJ = kEek[x] are ideals generated by orthogonal idempotents, the restriction of Q to k[a] is degenerate. If Xa(T) = (T - A)3, then k[a] = k[x] for some x with x2 :f:. 0 and x 3 = o. With arguments as in the case ma(T) = (T - A)2 treated above, one sees that the restriction of Q to the subspace kx $ kx 2 is identically zero and that this subspace is orthogonal to ke. Thus we find that the restriction of Q to k[a] is degenerate. It is also straightforward to verify that k[x] contains no idempotents :f:. e. Finally, suppose no longer that k is necessarily a splitting field of Xa, but that Xa has a root in k. Then either all three roots lie in k, and then the statement about the existence of a primitive idempotent in k[a] follows from the above analysis. Or there are two distinct roots in a quadratic extension I of k which are not in k itself. Then k[a] ~ kEel, which contains an idempotent. 0 Corollary 5.3.9 If k is algebraically closed and dimk A> 2, then mx = Xx for x in a nonempty Zariski open subset of A.

Proof. We first construct a primitive idempotent Ul EA. To this end, we pick an element x with (x, e) = 0 and Q(x) :f:. O. Then

Xx(T) = T3 - Q(x)T - det(x)

with Q(x):f:. 0,

which does not have three equal roots. By Prop. 5.3.8, k[x] contains a primitive idempotent Ul. Prop. 5.3.7 implies that e - Ul is the sum of two orthogonal primitive idempotents U2 and U3, since Q represents all values on the subspace (kUl $ k(e - ut})J. if k is algebraically closed. An element y = 17lUl + 172 U 2 + 173 U 3 with three distinct 17i has characteristic polynomial Xy with three distinct roots, viz., the 17i' so then Xy = my. The x E A such that Xx has three distinct roots are characterized by the fact that the discriminant 0 of Xx is not zero, so these form a Zariski open set. In Rem. 5.1.2 we indicated that the norm Q and the cubic form det on a J-algebra are determined by the algebra structure, provided it is of dimension > 2. We will now prove this.

,r

5.4 Classification of Reduced J-algebras

133

Proposition 5.3.10 On a l-algebra A over k with dimk(A) > 2, the quadratic form Q satisfying conditions (5.4)-(5.6) and the cubic form det are determined by the algebra structure of A, i.e. the structure of vector space over k and the product. Hence for an isomorphism t : A --+ A' of l-algebras over k of dimension> 2 we have: Q'(t(x» = Q(x) and det'(t(x» = det(x)

(x E A). Proof. It suffices to prove this for algebraically closed k. There is a nonempty Zariski open subset S of A such that Xx is the minimum polynomial of x for XES. The coefficients of Xx determine det(x) and Q(x). So the polynomials det and Q are determined on S and therefore on all of A. 0 Remark 5.3.11 If dimk(A) = 2 and A is reduced, then there are two orthogonal idempotents '1.1.1 and '1.1.2 such that A = kUl EB kU2 (cf. Prop. 5.3.5). One of these idempotents is primitive and the other is not: Q(u) = and Q(e-u) = 1 for '1.1. = '1.1.1 or '1.1.2. For x = ~u+17(e-u) we have Q(x) = ~~ +17 2 ; with the Hamilton-Cayley equation one finds det(x) = ~172. Hence in this case there are two possibilities for Q and det, i.e., these are not determined by the algebra structure. Corollary 5.3.12 Ifdet(x) -10 then x has a l-inverse x- l and det(x- l ) =

1

det(x)-l. Proof. For the first point see Lemma 5.2.3. It suffices to prove the equality for algebraically closed k. We first assume that dimk(A) > 2. Let V be the Zariski open set {x E A I det(x) -I O}, and W the Zariski open set on which mx = Xx (see Cor. 5.3.9). On V n W, Xx(T) = T3 ... - det(x) is the unique cubic polynomial which has x as a root, and similarly for Xx-1 (T) = T3 ... - det(x- l ). But k[x] is associative, so x- l is also a root of - det(x)-lT 3Xx(T- l ) = T3 ... - det(x)-l, so det(x- l ) = det(x)-l for x E V n W. By Zariski continuity, the relation holds on all of V. Now let dimk(A) = 2. For x = ~u + 17(e - '1.1.) as in the above remark, det(x) = ~172. If ~17 -I 0, then det(x- l ) = ~-117-2 = det(x)-l. 0

5.4 Classification of Reduced J-algebras We continue with the determination of the structure of reduced J-algebras. This will lead to the result that besides the J-algebras of quadratic type, which we found in Prop. 5.3.5, there is only one other type of reduced Jalgebra, viz., the matrix algebras H( C; I'll 1'2, 1'3) we introduced at the beginning of § 5.1; see Th. 5.4.5. We again fix a primitive idempotent '1.1. and assume that El -10. Further we fix Xl E Eo with Q(Xl) = ~.

134

5. J-algebras and Albert Algebras

Lemma 5.4.1 Consider the linear mapping

(i) s is symmetric with respect to { , } and S2 = l~' (ii) EI = E+ $ E_ with E+ .1 E_, where E+ and E_ are the eigenspaces of s for the eigenvalues and respectively. (iii) If dim Eo > 1, then both E+ :/= 0 and E_ :/= O. (iv) If dim Eo = 1, then EI = E+ or EI = E_.

1

1,

Proof. (i) The symmetry of s follows.from (5.5). The second statement of (i) is a consequence of Lemma 5.3.3 (i). (ii) From (i) it follows that s has eigenvalues and and that EI is the orthogonal direct sum of the corresponding eigenspaces E+ and E_. (iii) Pick x E Eo with {x, Xl} = 0 and Q(x) :/= O. If E+ :/= 0, pick y E E+ with Q(y) :/= O. By Lemma 5.3.3 (ii),

1

XI(XY)

-1,

1

= -X(XIY) = -4 xy ,

so xY E E_. Further, Q(xy) :/= 0 by Lemma 5.3.3 (vi), so xy :/= O. Hence E+ :/= 0 implies E_ :/= O. Similarly, E_ :/= 0 implies E+ :/= O. (iv) Suppose E+ :/= 0 and E_ :/= O. Pick y E E+ with Q(y) :/= 0 and Z E E_ with Q(z) :/= O. By Lemma 5.3.3 (iii),

yo y Hence (y

0

= 4y 0 XlY = Q(Y)Xl'

y)z = -1Q(y)z. By Lemma 5.3.3 (v), 1 = 41 Q (y)z + 4( y, z )y.

2(y 0 z)y + (y 0 y)z Since (y, z) = 0, we find

2(y 0 z)y which shows that yo z

= 21 Q (y)z,

:/= O. On the other hand,

{XbYoZ}

= (XI,YZ)

1

= {XIY,Z} = 4{y,z} =0.

This implies that dim Eo > 1. Now (iv) follows.

o

In case (iv) of the above lemma we may assume that EI = E+; otherwise, we take -Xl instead of Xl. We can now replace u by another primitive idempotent, viz., u' = !(e - u) - Xl: one easily verifies that indeed u,2 = u' and Q(u') = Further, it is straightforward that A is an orthogonal direct sum

!.

A = ku' $ k( e - u') $ E~,

5.4 Classification of Reduced J-algebras

135

where Eo = k( ~u' + xd $ E+ satisfies u' Eo = o. This shows that A is a J-algebra of quadratic type if dim Eo = 1j see Prop. 5.3.5. This being dealt with, we sharpen our assumptions: A is a reduced Jalgebra with a primitive idempotent u such that dim Eo > 1 and El :f= o. Then, by (iii) of the above lemma, El = E+ E9 E_ with both E+ :f= 0 and E_ :f= o. We will see that under these assumptions A is isomorphic to a Jalgebra H(Cj 1'1, 1'2, 1'3) of Hermitian 3x3 matrices over a composition algebra C, as introduced in the beginning of § 5.1. To start with, we define the vector space

C

= xt n Eo = {x E Eo I (x, xd = O}.

In Prop. 5.4.4, C will be given the structure of a composition algebraj before that, we prove two technical lemmas. Lemma 5.4.2 For y+, Z+ E E+ and y_, z_ E E_ we have:

(i) (ii) (iii) (iv) (v) (vi) (vii)

y+ 0 Z+ = ~(y+, z+ )Xlj y_ 0 z- = y_, z_ )Xlj y+ 0 y_ E Cj (y+ oy_)y+ = iQ(y+)y-j (y+ oy_)y_ = iQ(y-)y+j Q(y+) = Q(y_) = 0 ify+ 0 yQ(y+ 0 y_) = ~Q(y+)Q(y_).

-!(

= 0 and y+ :f= O:f= y_j

Proof. (i) Using Lemma 5.3.3 (iii), we find y+ oy+ = 4y+ OX1Y+ = Q(Y+)Xl. By linearizing we find the result for y+ 0 Z+. Similarly for (ii). (iii) (Xl.Y+ 0 y-) = (Xl,Y+Y-) = (X1Y+,y-) = i(y+,y-) = O. (iv) By Lemma 5.3.3 (v), 2(y+

0

y-)y+

+ (y+ 0 y+)y- =

1 4 Q (y+)y-

1

+ 4(Y+'Y- )y+.

Using (i) and the fact that E+ 1- E_, we easily get the result. The proof of (v) is similar. (vi) This is immediate from (iv) and (v). (vii) By Lemma 5.3.3 (vi),

Using (iv) we find

so Q(y+

1

0

y-) = 4 Q(y+ )Q(y_),

if Q(y+) :f= o. By continuity for the Zariski topology (over an algebraic closure of k), the relation holds everywhere. It is not hard, by the way, to prove (vii)

136

5. J-algebras and Albert Algebras

for Q(y+) lemma:

= 0 directly with the aid of Lemma 5.3.3

(i), and (iv) of the present

o Now we fix a+ E E+ and a_ E E_ with Q(a+)Q(a_)

i= O.

Lemma 5.4.3 The maps e+ : C

-+

E+, x

1-+

4Q(a_)-lxa_,

e_ : C

-+

E_, x

1-+

4Q(a+)-lxa+,

are linear isomorphisms. Their inverses are, respectively, E+ E_

-+ -+

C, y+ C, y_

a_ 1-+ a+ 1-+

0 0

y+ y_

= a_y+, = a+y_.

Proof. From (v) of the above lemma one sees that

is injective, and from Lemma 5.3.3 (iii) that it is surjective. That e+ is its inverse, follows from either of these. Similarly for e_ with (iv) of the above 0 lemma instead of (v). We now make C into a composition algebra. Proposition 5.4.4 The vector space C with the product <> defined by x<>x' = e+(x)e_(x') = 16Q(a+)-lQ(a_)-1(xa_)(x'a+)

{x,x' E CJ

and with the norm N defined by

(x E CJ is a composition algebra.

Proof. We have to verify the conditions of Def. 1.2.1. First, e identity element for the multiplication:

= a+a_

is an

by Lemma 5.4.2 (iv), hence

= e+(x)e_(e) = 4Q(a_)-1(xa_)a_ = x (x E C) by Lemma 5.3.3 (iii), and similarly e <> x = x. Using Lemma 5.4.2 (vii) and x <> e

Lemma 5.3.3 (vi), we derive

5.4 Classification of Reduced J-aigebras

Q(xox')

137

= Q(e+(x)e_(x')) 1

=

"4 Q(e+(x))Q(e_(x'))

=

"41