Algebra and Applications 1. Abdenacer Makhlouf

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      and this has cyclic symmetry:

image

      THEOREM 2.7.– Let (image, ∗, n) be an eight-dimensional symmetric composition algebra. Then:

      Spin image.

      Moreover, the set of related triples (the set on the right hand side) has cyclic symmetry.

      The cyclic symmetry on the right-hand side induces an outer automorphism of order 3 (trialitarian automorphism) of Spin(image, n). Its fixed subgroup is the group of automorphisms of the symmetric composition algebra (image, ∗, n), which is a simple algebraic group of type G2 in the para-Hurwitz case, and of type A2 in the Okubo case if char image.

      The group(-scheme) of automorphisms of an Okubo algebra over a field of characteristic 3 is not smooth (Chernousov et al. 2013).

      At the Lie algebra level, assume image, and consider the associated orthogonal Lie algebra

image

      The triality Lie algebra of (image, ∗, n) is defined as the following Lie subalgebra of image (with componentwise bracket):

image

      Note that the condition d0(xy) = d1(x) ∗ y + xd2(y) for any x, image is equivalent to the condition

image

      for any x, y, image. But n(xy, z) = n(yz, x) = n(zx, y). Therefore, the linear map:

image

      is an automorphism of the Lie algebra image.

      THEOREM 2.8.– Let (image, ∗, n) be an eight-dimensional symmetric composition algebra over a field of characteristic ≠ 2. Then:

       – Principle of local triality: the projection map:

image

      is an isomorphism of Lie algebras.

       – For any x, , the triple

image

      belongs to image, and image is spanned by these elements. Moreover, for any a, b, x, image:

image

      PROOF.– Let us first check that tx, yimage:

image

      and hence

image

      Also image and image (adjoint relative to the norm n), but RxLx = n(x)id, so RxLy + RyLx = n(x, y)id and hence image, so that image, and image too. Therefore, image.

      Since the Lie algebra image is spanned by the σx, y ’s, it is clear that the projection π0 is surjective (and hence so are π1 and π2). It is not difficult to check that ker π0 = 0, and therefore, π0 is an isomorphism.

      Finally, the formula [ta, b, tx, y] = tσa, b(x), y + tx, σa, b(y) follows from the “same” formula for the σ’s and the fact that π0 is an isomorphism. □

      Given two symmetric composition algebras (image, ∗, n) and image, consider the vector space:

image

      where image is just a copy of image (i = 0, 1, 2) and we write image, image instead of image and image for short. Define now an anticommutative bracket on image by means of:

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