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3 votes
1 answer
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Example of a supersolvable Lie group/algebra whose nilradical does not have a complement

What is an example of a real solvable simply-connected Lie group $G$ whose nilradical does not have a complement (that is, $G$ is not a semidirect product of the nilradical and another subgroup)? Is ...
2 votes
0 answers
60 views

Is there always a purely real representative for a metrized solvable Lie group?

Alekseevski proves for Heintze groups (a special class of solvable Lie groups) that any such group admits a (left-invariant) metric which is isometric to a purely real Heintze group (again equipped ...