Say I have a Lie algebra, $\mathfrak{g}$, and some element of the dual $m \in \mathfrak{g}^*$ with the property that $m([u,v])=0$ for all $u,v \in \mathfrak{g}$ ($m$ non-zero). What can be said about the Lie algebra $\mathfrak{g}$? For one it cannot be semi-simple. It is true when $\mathfrak{g}$ is the semi direct product of a one dimensional Lie algebra and $m$ in the corressponding dual vector space (I hope my terminology makes sense). Are there other cases or is a Lie algebra fulfilling these conditions necessarily a semi direct product of a one dimensional Lie algebra with something else?

Unfortunately I don't know enough about Lie groups to know if this is more appropriate for overflow or stack exchange, if I should delete and repost on stack exchange please let me know.


migrated from mathoverflow.net Nov 1 '17 at 11:35

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  • 2
    $\begingroup$ The existence of $m$ is equivalent to $g\neq [g,g]$. Every such algebra is trivially a semidirect product: just take $Ker(m)$ to be the kernel, and and vector $v$ wuth $m(v)\neq 0$ to span the complement. The fact that $Ker(m)$ contains $[g,g]$ makes everything work here. $\endgroup$ – Ehud Meir Oct 18 '17 at 10:30
  • $\begingroup$ You're assuming that $\mathfrak{g}$ is non-perfect. This indeed implies that $m$ is a homomorphism, and $\mathfrak{g}$ is semidirect product of the kernel of $m$ with any complementing line. $\endgroup$ – YCor Oct 18 '17 at 10:31
  • $\begingroup$ PS the question can easily be moved to MathSE, it's better than erasing and repost $\endgroup$ – YCor Oct 18 '17 at 10:32
  • $\begingroup$ @YCor Thanks for your help, I think I don't have enough reputation to move my own question, unfortunately $\endgroup$ – R Mary Oct 18 '17 at 13:43
  • $\begingroup$ No: if the complement vector $v$ is not central then it will not be a direct product. By "trivially" I meant to say that it is easy to describe it as the semidirect product, this semi direct product is not necessarily a direct product. $\endgroup$ – Ehud Meir Oct 18 '17 at 15:30

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