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Nov 12, 2015 at 15:03 history edited Dietrich Burde CC BY-SA 3.0
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Nov 12, 2015 at 9:34 comment added Dietrich Burde Any abelian ideal of $L=\mathfrak{s}\ltimes {\rm rad}(L)$ lies of course in the solvable radical ${\rm rad}(L)$. This may already be helpful. However, also the algorithm to compute the Levi decomposition has to be taken into account.
Nov 11, 2015 at 12:49 comment added user47116 Thank you. It's rather interesting that for solvable Lie algebra the question is easier. How do you think: could Levi decomposition be useful in general case? Since, it decomposes Lie algebra into semi-direct product of radical ideal and semi-simple algebra...
Nov 10, 2015 at 20:15 history edited Dietrich Burde CC BY-SA 3.0
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Nov 10, 2015 at 20:05 history answered Dietrich Burde CC BY-SA 3.0