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Dear forum,

I would like to ask if $H(m)$ is the Heisenberg Lie algebra of dimension $2m+1$ and $M$ is an ideal of $H(m)$. Can we say that $M$ has a complement in $H(m)$?

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  • $\begingroup$ Can you clarify the meaning of "complement" here? What happens when you take $M$ to be the 1-dimensional center? $\endgroup$
    – Jim Humphreys
    Commented May 20, 2013 at 12:26
  • $\begingroup$ Hint: what happens if $M$ is the centre of $H(m)$? $\endgroup$
    – user91132
    Commented May 20, 2013 at 12:28
  • $\begingroup$ If $M\neq 0$, then I guess $H(m)/M$ is a free Lie algebra and so $0\rightarrow M\rightarrow H(m)\rightarrow H(m)/M\rightarrow 0$ is split so the result follows. $\endgroup$
    – Francesco
    Commented May 20, 2013 at 12:47
  • $\begingroup$ In fact, I think $H(m)/M$ is abelian Lie algebra. $\endgroup$
    – Francesco
    Commented May 20, 2013 at 12:49
  • $\begingroup$ But Im not sure in this way!!! $\endgroup$
    – Francesco
    Commented May 20, 2013 at 13:04

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As pointed out in the comments, if $M$ would have a complement, then the exact sequence $$ 0 \rightarrow M \rightarrow H(m)\rightarrow H(m)/M \rightarrow 0 $$ would be split. But this is not true in general. If it were true for $M=Z(H(m))$, then $H(m)$ were abelian ($M$ and $H(m)/M$ abelian), a contradiction.

For an answer to a similar question, see Lie algebras and complements.

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