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Write $D=\mathrm{ad}y$. The generalized eigenspace decomposition of the linear map $D$ isfor any sufficiently large $N$. Because $D$ is a derivation, the generalized-eigenspace bracket lemma givesConsequently $L_{0,y}$ is a Lie subalgebra.

The set $I(K)=\{x\in L:[x,K]\subseteq K\}$ is the normalizer of a Lie subalgebra $K$. Certainly $L_{0,y}\subseteq I(L_{0,y})$. Conversely, if $x\in I(L_{0,y})$, then $y\in L_{0,y}$ givesOn the direct sum of the nonzero generalized eigenspaces, $D$ is invertible. Hence the nonzero-eigenvalue component of $x$ vanishes, and

Now let $K$ be a Lie subalgebra containing $L_{0,y}$. Since $y\in K$, the subspace $K$ is $D$-invariant. The generalized zero eigenspace of the induced map on $L/K$ is the image of $L_{0,y}$, hence is zero. If $x\in I(K)$, then $Dx=[y,x]\in K$, so $x+K$ lies in that zero eigenspace. Thus $x\in K$ and

A nilpotent Lie algebra is one whose lower central serieseventually reaches zero. Suppose $L$ is nilpotent and $K⊊L$. Choose the least $r\ge 2$ for which ${\gamma }_r(L)\subseteq K$. Then ${\gamma }_{r-1}(L)⊈K$, and anysatisfies $[x,K]\subseteq [L,{\gamma }_{r-1}(L)]={\gamma }_r(L)\subseteq K$. Therefore $x\in I(K)\setminus K$, proving the normalizer condition for a nilpotent Lie algebra

It remains to prove the converse needed here. The Engel lemma states that if a finite-dimensional Lie algebra of linear maps consists of nilpotent maps, then the maps have a common nonzero vector in their kernels. To prove it, induct on the dimension of the algebra. For a maximal proper subalgebra $H$, induction applied to the action of $H$ on $L/H$ produces $x\notin H$ with $[H,x]\subseteq H$. Thus $H$ is an ideal of codimension one. Induction also gives a nonzero common kernelThe ideal property makes $W$ invariant under $L$; a nilpotent representative of a basis of $L/H$ has a nonzero kernel on $W$, yielding a vector killed by all of $L$.

Apply the lemma to the Adjoint representation. It produces a nonzero element of the center of a Lie algebra. Induction on $\dim L$, followed by passage to the quotient by this center, proves Engel theorem: if every $\mathrm{ad}y$ is nilpotent, then $L$ is nilpotent. The hypothesis $L_{0,y}=L$ says exactly that every $\mathrm{ad}y$ is nilpotent, so

For this Heisenberg Lie algebra,because $c$ belongs to the center. Thus its lower central series of a Lie algebra is $\mathfrak{g},\mathbb{C}c,0$, so $\mathfrak{g}$ is a two-step nilpotent Lie algebra.

The radical of a bilinear form of the Killing form is the ideal of a Lie algebra $\{x:B(x,y)=0\text{ for all }y\}$. It is zero for a complex semisimple Lie algebra. Every element of the center of a Lie algebra lies in this radical.

For a bi-invariant Riemannian metric on a Lie group, the induced inner product on its Lie algebra makes every adjoint operator skew-adjoint. The Levi-Civita connection of a bi-invariant metric gives $R(X,Y)Z=-\frac{1}{4}[[X,Y],Z]$. Thus $K(X,Y)=\frac{1}{4}\Vert [X,Y]{\Vert }^2$ for orthonormal $X,Y$, andThe nullspace of this quadratic form is exactly the center of a Lie algebra. A zero center therefore gives positive Ricci curvature, uniformly bounded below by a positive constant times the metric through left invariance and compactness of the unit sphere in the Lie algebra.

Each term is the preimage of the center of a Lie algebra after quotienting by the preceding term. Equivalently, $Z_{i+1}=\{x:[\mathfrak{g},x]\subseteq Z_i\}$. A Lie algebra is nilpotent exactly when this series reaches the whole algebra after finitely many steps. For the Heisenberg Lie algebra, the series is $0\subset \mathbb{C}c\subset \mathfrak{g}$.