Algebra and Applications 1. Abdenacer Makhlouf

      Then (, ∙, n) is a composition algebra.

      In Elduque and Pérez (1997), one may find examples of infinite-dimensional composition algebras of arbitrary infinite dimension, which are even left unital.

      2.3.2. Isotropic Hurwitz algebras

      Assume now that the norm of a Hurwitz algebra (, ∙, n) represents 0. That is, there is a non-zero element such that n(a) = 0. This is always the case if and is algebraically closed.

      With a as above, take such that , so that n(a ∙ b, 1) = 1. Also n(a ∙ b) = n(a)n(b) = 0. By the Cayley–Hamilton equation, the non-zero element e₁ := a ∙ b satisfies , that is, e₁ is an idempotent. Consider too the idempotent , and the subalgebra generated by e₁. (1= e₁ + e₂).

      For any and, as and , we conclude that x ∙ e₁ = e₂ ∙ x, and in the same way, x ∙ e₂ = e₁ ∙ x, for any .

      But x = 1 ∙ x = e₁ ∙ x + e₂ ∙ x, and e₂ ∙ (e₁ ∙ x)= (1 − e₁) ∙ (e₁ ∙ x) = 0 = e₁ ∙ (e₂ ∙ x). It follows that splits as with

      For any , n(u) = n(e₁ ∙ u) = n(e₁₎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(u₁, v) ≠ 0 = n(u₂, v). Then the alternative law gives (u₁ ∙ u₂)∙ v = −(u₁ ∙ v)∙ u₂ +u₁ ∙ (u₂ ∙ v + v ∙ u₂) = −n(u₁, v)u₂ ≠ 0, so that u₁ ∙ u₂ ≠ 0. In particular, , and the same happens with .

      Consider the trilinear map:

      This

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