Some believe that if $L$ is a nilpotent Leibniz algebra and $N$ is a nilpotent ideal such that $N\subset Z^l(L)$ and $L/N$ is nilpotent then $L$ is nilpotent. In this theorem 3.1 I read a proof of this and I think is not true. I want a counterexample to show that this is not true.
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Why do you not think it is true? Is there some problem with the proof? – Tobias Kildetoft Feb 08 '17 at 11:55
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@TobiasKildetoft:yes there is a problem in proof.we have $L^{m+n}\supseteq [L^m,L^n]$ but in this proof use converse – pink floyd Feb 08 '17 at 12:40
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http://link.springer.com/content/pdf/10.1007/978-94-011-5072-9_1.pdf you can read this – pink floyd Feb 08 '17 at 12:43
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No, I cannot read that as it is behind a paywall and I am not at a university with access right now. Anyway, you should include the specific issues you have with the proof in the question itself. – Tobias Kildetoft Feb 08 '17 at 12:44
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@TobiasKildetoft:llemma:for any $i$,$j\in \mathbb Z [L^i,L^j]\subseteq L^{i+j}$ this article prove this – pink floyd Feb 08 '17 at 12:52
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But having inclusion in one direction does not rule out equality in special cases. – Tobias Kildetoft Feb 08 '17 at 12:57
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Let us continue this discussion in chat. – pink floyd Feb 08 '17 at 13:04
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Theorem 3.1 seems to say that L is nilpotent if and only if L/N^2 is nilpotent. – David Towers Feb 08 '17 at 13:07
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But that isn't what the Theorem is saying. It talks about L factored by the derived subalgebra of N (which is the Frattini subalgebra of N). – David Towers Feb 08 '17 at 13:19
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@DavidTowers:I want to prove theorem ,but I have problem where it use equation $L^{m+n}=[L^m,L^n]$ – pink floyd Feb 08 '17 at 13:27
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@DavidTowers:the proof of this theorem is same to theorem 2 of "Some characterizations of nilpotent lie algebras cHAo".I think the proof of this theorem have problem. – pink floyd Feb 08 '17 at 13:52
1 Answers
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This result follows from Theorem 5.5 of Barnes, D. (2011). Some theorems on Leibniz algebras. Comm. Algebra 39(7):2463–2472.
David Towers
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