1989
DOI: 10.1007/bf01061314
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Lie algebras, decomposable into a sum of an Abelian and a nilpotent subalgebra

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Cited by 22 publications
(20 citation statements)
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“…In view of the above result it is natural to get results relating the degree of nilpotence of g * with the degree of solvability of g. In this direction we mention that it was proved recently by Andrada-Barberis-Dotti-Ovando [1] (see also [12]) that g is 2-step solvable if it is sum of two abelian subalgebras.…”
Section: G Is Solvable If G * Is Nilpotentmentioning
confidence: 93%
“…In view of the above result it is natural to get results relating the degree of nilpotence of g * with the degree of solvability of g. In this direction we mention that it was proved recently by Andrada-Barberis-Dotti-Ovando [1] (see also [12]) that g is 2-step solvable if it is sum of two abelian subalgebras.…”
Section: G Is Solvable If G * Is Nilpotentmentioning
confidence: 93%
“…In [1], [3], [4], [5], a similar statement is proved under additional restrictions on the nilpotency index of one summand (with fewer restrictions on the characteristic of the ground field). Note that the theorem is no longer true when p = 2 (an appropriate counter example has been constructed in [4]).…”
mentioning
confidence: 83%
“…Note that the theorem is no longer true when p = 2 (an appropriate counter example has been constructed in [4]). We make an essential use of Weisfeiler's results [6] on Lie algebras with a solvable maximal subalgebra which dictates a restriction upon the characteristic of the ground field.…”
mentioning
confidence: 99%
“…(O1) Let g be a real Lie algebra admitting an abelian complex structure. Then g is 2-step solvable [17].…”
Section: Abelian Complex Structures On Lie Algebrasmentioning
confidence: 99%
“…If G is a connected Lie group with Lie algebra g, by left translating J one obtains a complex manifold (G, J) such that left multiplication is holomorphic and, in the bi-invariant case, also right multiplication is holomorphic, which implies that (G, J) is a complex Lie group.Our concern here will be the case when J is abelian. In this case the Lie algebra has abelian commutator, thus it is 2-step solvable (see [17]). However, its nilradical need not be abelian (see Remark 8).…”
mentioning
confidence: 99%