Abdenacer Makhlouf - Algebra and Applications 1

For any and, as and , we conclude that x ∙ e 1= e 2∙ x , and in the same way, x ∙ e 2= e 1∙ x , for any .

But x = 1 ∙ x = e 1∙ x + e 2∙ x , and e 2∙ ( e 1∙ x )= (1 − e 1) ∙ ( e 1∙ x ) = 0 = e 1∙ ( e 2∙ x ). It follows that splits as with

For any , n( u ) = n( e 1∙ u ) = n( e 1)n( u ) = 0, so , and too, are totally isotropic subspaces of paired by the norm. In particular, , and this common value is either 0, 1 or 3, depending on being 2, 4 or 8. The case of = 0 is trivial, and the case of = 1 is quite easy (and subsumed in the arguments below). Hence, let us assume that is a Cayley algebra (dimension 8), so .

For any and , using [2.5]we get

Hence, is orthogonal to both and , so it must be contained in . Also .

Besides,

so that , and also . But for any and , we have

and the analogues for v ∙ u . We conclude that

Now, for linearly independent elements , let with n( u 1, v ) ≠ 0 = n( u 2, v ). Then the alternative law gives ( u 1∙ u 2)∙ v = −( u 1∙ v )∙ u 2 +u 1∙ ( u 2∙ v + v ∙ u 2) = −n( u 1, v ) u 2≠ 0, so that u 1∙ u 2≠ 0. In particular, , and the same happens with .

Consider the trilinear map:

This is alternating because x ∙2= 0 for any by the Cayley–Hamilton equation, and n( x ∙ y , y ) = −n( x , y ∙ y ) = 0. It is also non-zero, because , so that .