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let $A$ be a leibniz algebra over an algebrically closed field $K$ of characteristic $0$. I think if $A$ has basis $e_i$ ,$f_j$ , for $i=1,2,...,m$ and $j=1,2,...,n$, with $e_if_j=\lambda_{ij}f_j$, either $e_if_j=-f_je_i$, or $f_je_i=0$, and all other products between basis elements equal to $0$. is solvable leibniz algebra.

is it true? I want to this statement to levi theorem

thanks for help

pink floyd
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  • Could you write a meaningful sentence? Is the question whether every Leibniz algebra possessing a basis with the prescribed property is solvable? – YCor Nov 28 '16 at 08:57
  • If all other products are zero then the square of every element is zero, which implies anti-commutativity in characteristic zero, so the second possibility doesn't occur unless the lambda's are zero. So A is a solvable Lie algebra. – David Towers Nov 29 '16 at 10:30
  • @DavidTowers :thanks – pink floyd Nov 29 '16 at 15:36

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