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If $\mathfrak{g}$ is a solvable Lie algebra with center $Z(\mathfrak{g})$, can we decompose it over its center? Writing $\mathfrak{g}=Z(\mathfrak{g})\oplus\mathfrak{h}$ where $\mathfrak{h}$ is a Lie subalgebra?

Bernard
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Mohammadreza Bidar
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    For $\mathfrak{g}$ arbitrary (solvability plays no role), such a decomposition exists if and only if $Z(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]={0}$. This often fails, including in the solvable case, e.g., in Dietrich's example. Another example: for $\mathfrak{g}$ nilpotent non-abelian this always fails. – YCor Dec 21 '19 at 12:32

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In general a solvable Lie algebra does not decompose as a Lie algebra direct sum with the center. Take the $3$-dimensional Heisenberg Lie algebra $L$ with basis $(x,y,z)$ and $[x,y]=z$, $Z(L)=\langle z\rangle$. If you consider this sum just as direct sum of vector spaces, then we always have $$\mathfrak{g}=Z(\mathfrak{g})\oplus \mathfrak{h}=Z(\mathfrak{g})\oplus\mathfrak{g}/Z(\mathfrak{g}).$$

Dietrich Burde
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