2002
DOI: 10.1006/jabr.2001.9084
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On the Cartan–Jacobson Theorem

Abstract: The well-known Cartan-Jacobson theorem claims that the Lie algebra of derivations of a Cayley algebra is central simple if the characteristic is not 2 or 3. In this paper we have studied these two cases, with the following results: if the characteristic is 2, the theorem is also true, but, if the characteristic is 3, the derivation algebra is not simple. We have also proved that in this last case, there is a unique nonzero proper seven-dimensional ideal, which is a central simple Lie algebra of type A 2 , and … Show more

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Cited by 19 publications
(27 citation statements)
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“…The argument above shows then that ker The Z 2 × Z 2 -grading of the Lie algebra g = g(S, S ) constructed above induces a Z 2 gradation given by g0 = g (0,0) ⊕ g (1,0) and g1 = g (0,1) ⊕ g (1,1) . The structure of g0 is given by: Corollary 3.5 Let (S, * , q) and (S , * , q ) be two symmetric composition algebras and let g = g(S, S ) and g0 as above.…”
Section: Remark 32mentioning
confidence: 95%
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“…The argument above shows then that ker The Z 2 × Z 2 -grading of the Lie algebra g = g(S, S ) constructed above induces a Z 2 gradation given by g0 = g (0,0) ⊕ g (1,0) and g1 = g (0,1) ⊕ g (1,1) . The structure of g0 is given by: Corollary 3.5 Let (S, * , q) and (S , * , q ) be two symmetric composition algebras and let g = g(S, S ) and g0 as above.…”
Section: Remark 32mentioning
confidence: 95%
“…The roots in g (0,0) are ±ε i ± ε j , 1 ≤ i < j ≤ 4. Since g (1,0) = S ⊗ S with the natural action of o(4, 4) ∼ = tri(S, * , q) on S; the roots in g (1,0) are ±ε i ± (δ 1 + δ 2 ), 1 ≤ i ≤ 4, and the roots in g (0,1) and in g (1,1) are obtained by applying to the roots in g (1,0) the triality automorphisms on the Dynkin diagram of D 4 that fixes ε 2 − ε 3 and permutes cyclically ε 1 − ε 2 , ε 3 − ε 4 and ε 3 + ε 4 , while substituting δ 1 + δ 2 by δ 1 and δ 2 . As a consequence, the roots in g (0,1) (respectively g (1,1) ) are 1 2 (±ε 1 ± ε 2 ± ε 3 ± ε 4 ) ± δ 1 (resp.…”
Section: Proof From Lemma 33 Tri(s * Q) = T Ss ⊕ Kermentioning
confidence: 99%
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“…is the trivial product by Eqs. (2.27a) and (2.27b), or J = k with n(α) = α 3 for any α ∈ k, because then t (α) = 3α = 0 for any α ∈ k, so that the trace form is trivial (this has to do with the fact that the exceptional Lie algebra of type G 2 is no longer simple in characteristic 3-see, for instance, [3]). Theorem 2.16 and Brown's classification of the Freudenthal triple systems in characteristic 3 [6] yield: ; or (iii) dim T = 2, and hence T is described in item (ii) of Proposition 2.7; or (iv) up to isomorphism, T is the symplectic triple system described in Proposition 2.14.…”
Section: Remark 225mentioning
confidence: 97%
“…Hence there is a bilinear nondegenerate form ( | ) : 1 for any a ∈ W and u ∈W . This bilinear form determines h and allows us to identifyW with the dual W * .…”
Section: Hermitian Typementioning
confidence: 99%