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The three-dimensional Heisenberg Lie algebra has a basis $a,b,c$ with $[a,b]=c$ and $c$ central. It is a two-step <Nilpotent Lie algebra>.

Heisenberg Lie algebra

A vector subspace $I\subseteq\mathfrak g$ is an <ideal of a Lie algebra> when $[\mathfrak g,I]\subseteq I$. A <Nilpotent Lie algebra> is one whose lower central series
$$
\mathfrak g=\gamma_1(\mathfrak g),\qquad
\gamma_{r+1}(\mathfrak g)=[\mathfrak g,\gamma_r(\mathfrak g)]
$$
eventually vanishes.

By the <Engel theorem>, the operators $\operatorname{ad}_x$ for a nilpotent complex Lie algebra can be represented simultaneously by strictly upper triangular matrices. Their products are strictly upper triangular and have zero trace. Hence
$$
\boxed{\kappa(x,y)=0\quad\text{for every }x,y\in\mathfrak g}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 102 / 1 / b / Solution

For this <Heisenberg Lie algebra>,
$$
[\mathfrak g,\mathfrak g]=\mathbb Cc,
\qquad
[\mathfrak g,\mathbb Cc]=0,
$$
because $c$ belongs to the <Center of a Lie algebra>[center]. Thus its <Lower central series of a Lie algebra> is $\mathfrak g,\mathbb Cc,0$, so $\mathfrak g$ is a two-step <Nilpotent Lie algebra>.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 102 / 1 / i / Solution

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A <Lie algebra> is nilpotent when its <Lower central series of a Lie algebra>[lower central series] eventually becomes zero.

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 Incoming links: Nilpotent Lie algebra