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How it can be proved, that every four dimensional nilpotent Lie algebra contains a three dimensional abelian ideal?

Smolin Vlad
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  • To directly prove this, show that a nilpotent derivation of the 3-dimensional Heisenberg Lie algebra has a kernel of dimension $\ge 3$. – YCor Jan 08 '20 at 16:40

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Every $4$-dimensional nilpotent Lie algebra over a field $K$ of characteristic zero is either abelian or isomorphic to the direct sum $\mathfrak{n}_3(K)\oplus K$, where $\mathfrak{n}_3(K)$ denotes the $3$-dimensional Heisenberg Lie algebra, or to the filiform nilpotent Lie algebra $\mathfrak{f}_4(K)$ with adapted basis $(e_1,\ldots e_4)$ and brackets determined by $[e_1,e_i]=e_{i+1}$ for $i=2,3$. This is well known and not difficult to show. Obviously, both algebras contain an abelian ideal of dimension $3$. For example, $\langle e_2,e_3,e_4\rangle$ is an abelian ideal in $\mathfrak{f}_4(K)$.

References on this site:

Four dimensional nilpotent real Lie algebras

Dietrich Burde
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