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If $\mathfrak{a}$ is an abelian ideal of a Lie algebra, then ${\mathrm{ad}}_x{\mathrm{ad}}_y$ maps $\mathfrak{g}$ into $\mathfrak{a}$ and vanishes on $\mathfrak{a}$ whenever $x\in \mathfrak{a}$. Its trace is zero, so $\kappa (\mathfrak{a},\mathfrak{g})=0$. Every nonzero solvable ideal has a last nonzero term in its derived series of a Lie algebra, which is such an abelian ideal. Consequently a nondegenerate Killing form forces the solvable radical to vanish.

The derivations of a Lie algebra form a Lie subalgebra of $\mathrm{End}(\mathfrak{g})$ with the commutator bracket. The identity $[D,\mathrm{ad}x]=\mathrm{ad}(Dx)$ makes the inner derivations an ideal of a Lie algebra. For a complex semisimple Lie algebra, the Weyl complete reducibility theorem and the vanishing center of a Lie algebra imply $\mathrm{Der}(\mathfrak{g})=\mathrm{ad}\mathfrak{g}$.

The derived algebra is the linear span of all Lie brackets $[x,y]$ in a Lie algebra. The Jacobi identity makes it an ideal of a Lie algebra. Its quotient is the largest abelian Lie algebra quotient of $\mathfrak{g}$.

The direct sum $\mathfrak{g}\oplus \mathfrak{h}$ has componentwise bracketThe two summands are commuting ideals.

If $L$ is a finite-dimensional Lie algebra of nilpotent endomorphisms, every proper Lie subalgebra $M$ is strictly contained in its normalizer of a Lie subalgebra. The Adjoint representation of $M$ on $L/M$ consists of nilpotent maps by nilpotence of commutation by a nilpotent endomorphism. In the inductive proof of Engel theorem, the lower-dimensional Engel lemma gives a nonzero class $y+M$ with $[M,y]\subseteq M$. A maximal proper $M$ is consequently an ideal of a Lie algebra of codimension one.

The real Lie algebra $\mathfrak{iso}(2)$ is a semidirect product of planar translations with infinitesimal rotations. One orientation convention is $[J,E_1]=-E_2$, $[J,E_2]=E_1$, $[E_1,E_2]=0$. Its translation space is a two-dimensional abelian ideal of a Lie algebra.

An inner derivation of a Lie algebra is ${\mathrm{ad}}_x:y\mapsto [x,y]$. The Jacobi identity makes this a derivation of a Lie algebra. For every derivation of a Lie algebra $D$, $[D,{\mathrm{ad}}_x]={\mathrm{ad}}_{Dx}$, so the inner derivations form an ideal of a Lie algebra in the derivation Lie algebra. The kernel of $x\mapsto {\mathrm{ad}}_x$ is the center of a Lie algebra.

Conjugation by every element of $G$ preserves a normal Lie subgroup $H$. Its differential therefore preserves $\mathfrak{h}=T_{e}H$. Differentiate that invariance along $\exp (tX)$ to obtain $\mathrm{ad}(X)\mathfrak{h}\subseteq \mathfrak{h}$, proving that $\mathfrak{h}$ is an ideal of a Lie algebra. Closed subgroups have a canonical Lie-subgroup structure; an arbitrary abstract subgroup need not.

The normalizer of a Lie subalgebra $H\subseteq L$ isIt is the largest Lie subalgebra of $L$ in which $H$ is an ideal of a Lie algebra.

For an ideal of a Lie algebra $I$ and nondegenerate Killing form, invariance makes $I^{\perp }$ an ideal. An element of $I\cap I^{\perp }$ commutes with $I$, so the intersection is an abelian ideal. Abelian ideals lie in the radical of the Killing form, forcing the intersection to vanish. Thus $\mathfrak{g}=I\oplus I^{\perp }$ as commuting ideals. Iterating minimal nonzero ideals gives the direct-sum formulation of a semisimple Lie algebra.

Let ${\hat{T}}_a={\mathrm{ad}}_{T_a}$ on the Lie algebra itself. The Jacobi identity says, for every $X$,Therefore $[{\hat{T}}_a,{\hat{T}}_b]=c_{ab}^c{\hat{T}}_c$, proving that the displayed structure-constant matrices form the Adjoint representation of a Lie algebra.

The bilinear form ${\kappa }_{ab}=\mathrm{tr}({\hat{T}}_a{\hat{T}}_b)$ is the Killing form. It is symmetric by cyclicity of trace. Its invariance follows by expanding the two commutators:In components this is preciselyEquivalently, the adjoint action preserves the bilinear form, so it is an invariant tensor.

The invariant abelian subalgebra is an abelian ideal of a Lie algebra $\mathfrak{a}$. Assume it is nonzero, as required for the conclusion. For $X\in \mathfrak{a}$ and $Y\in \mathfrak{g}$, the map ${\mathrm{ad}}_X{\mathrm{ad}}_Y$ sends $\mathfrak{g}$ into $\mathfrak{a}$, because it is an ideal. It vanishes on $\mathfrak{a}$, because ${\mathrm{ad}}_Y$ preserves the ideal and ${\mathrm{ad}}_X$ kills it. Relative to $\mathfrak{a}$ and a vector-space complement, this map has zero diagonal blocks, so its trace is zero. ThusThis proves that abelian ideals lie in the radical of the Killing form. Choose a nonzero $X=x^a{T}_a$ in the ideal. Then ${\kappa }_{ba}{x}^a=0$ for every $b$, so the coordinate vector $x$ is a nonzero null eigenvector of the Killing matrix. The nonzero-ideal qualification cannot be dropped: the zero ideal is abelian even in a semisimple algebra with nondegenerate Killing form.

Now suppose $\kappa$ is invertible and let $C_{\kappa }={\kappa }^{bc}{t}_b{t}_c$ in any Lie algebra representation. In matrix notation invariance is $A_a^T\kappa +\kappa A_a=0$, where $(A_a{)}^d{}_b=c_{ab}^d$. Multiplying by the inverse form gives $A_a{\kappa }^{-1}+{\kappa }^{-1}{A}_a^T=0$. HenceNo commutativity of the representation matrices has been assumed. This proves the centrality of the quadratic Casimir operator formed with the inverse Killing form. Schur lemma makes it a scalar on every finite-dimensional complex irreducible representation.

For SU(2), make the normalization explicit. In a Hermitian angular-momentum basis with $[J_a,J_b]=i{ϵ}_{abc}{J}_c$, the structure-constant matrices obeyThus $C_{\kappa }=\frac{1}{2}{\sum }_a{J}_a^2$. In the compact anti-Hermitian basis $T_a=-i{J}_a$, the brackets have real structure constants ${ϵ}_{abc}$ and the Killing form of the SU(2) Lie algebra is instead $-2{\delta }_{ab}$; the inverse contraction gives the same $C_{\kappa }$. This sign change is a basis convention, not a different spectrum.

In the complex highest-weight basis $(H,E^{+},E^{-})$ used earlier, the same Killing form and its inverse areConsequentlyOn a highest vector of weight $m=2j$, the ladder calculation gives $E^{+}{E}^{-}v=mv$ and $E^{-}{E}^{+}v=0$. Its Casimir eigenvalue is thereforeCentrality gives the same value throughout the irreducible multiplet. This is the Killing-normalized SU(2) quadratic Casimir; the familiar angular-momentum operator ${\sum }_a{J}_a^2$ has eigenvalue $j(j+1)$ and is twice this inverse-Killing-form contraction.

The derived series of a Lie algebra is defined recursively by ${\mathfrak{g}}^{(0)}=\mathfrak{g}$ and ${\mathfrak{g}}^{(r+1)}=[{\mathfrak{g}}^{(r)},{\mathfrak{g}}^{(r)}]$, where the bracket denotes the linear span of all brackets of pairs of elements. A solvable Lie algebra is one whose derived series of a Lie algebra terminates at zero:The solvable radical $\mathrm{rad}\mathfrak{g}$ of a finite-dimensional Lie algebra is its largest solvable ideal of a Lie algebra. A semisimple Lie algebra has no nonzero solvable ideals of a Lie algebra, equivalentlyThese are definitions; identifying semisimplicity with a direct sum of simple ideals additionally uses the structure theorem in characteristic zero. The calculations and representation theorems below are over $\mathbb{C}$, as in the complex representation setting of this paper.

Work over $\mathbb{C}$ with finite-dimensional Lie algebras and Lie algebra representation spaces. The adjoint form of Engel theorem isHere a nilpotent Lie algebra has terminating lower central series, ${\gamma }_1\mathfrak{g}=\mathfrak{g}$, ${\gamma }_{j+1}\mathfrak{g}=[\mathfrak{g},{\gamma }_j\mathfrak{g}]$. We prove the stronger linear statement: if $V\ne 0$, $L\subseteq \mathfrak{gl}(V)$ and every member of $L$ is a nilpotent endomorphism, then there is a nonzero $v\in V$ killed by all of $L$, and $L$ can be simultaneously represented by strictly upper triangular matrices.

For the proof of Engel theorem by induction and normalizers, first establish a useful nilpotence fact. If $x^m=0$, left and right multiplication by $x$ on $\mathrm{End}(V)$ commute. Expanding their difference givesFor $N=2m-1$, every summand contains a power of $x$ at least $m$, so $(\mathrm{ad}x{)}^{2m-1}=0$. Restrictions and induced quotient maps remain nilpotent.

Now prove the common-kernel statement, the Engel lemma, by induction on $\dim L$. The zero algebra is immediate. Choose a maximal proper Lie subalgebra $M$. Its action on $L/M$ by commutators consists of nilpotent maps by the preceding fact. Applying the lower-dimensional induction hypothesis to its image gives a nonzero coset $y+M$ with $[M,y]\subseteq M$. HenceThe normalizer of a Lie subalgebra is a Lie subalgebra by the Jacobi identity. Maximality implies $N_L(M)=L$, so $M$ is an ideal of a Lie algebra. Moreover $L/M$ has no proper nonzero Lie subalgebra, and every one-dimensional subspace is a Lie subalgebra. Thus $\dim L/M=1$.

Induction also provides a nonzero common annihilatorIt is stable under $L$: for $m\in M$, $x\in L$, $v\in W$,because $[m,x]\in M$. Choose $x$ representing a basis of $L/M$. The restriction of the nilpotent endomorphism $x$ to the nonzero space $W$ has nonzero kernel. A vector in this kernel is killed by both $M$ and $x$, hence by all of $L$. This finishes the induction. When an action is not faithful, induction is applied to its image, whose dimension is no larger than $\dim M$; no faithfulness assumption is hidden.

Starting with the common-kernel line, apply the same statement to each successive quotient Lie algebra representation space. It constructs a complete flag of invariant subspacesAn adapted basis represents every member of $L$ by a strictly upper triangular matrix. Every product of $d$ such matrices is zero, so every iterated commutator with $d$ entries is zero as well.

Finally, if every $\mathrm{ad}x$ on $\mathfrak{g}$ is nilpotent, apply the linear statement to the Adjoint representation. With $d=\dim \mathfrak{g}$, every product of $d$ adjoint operators vanishes, so ${\gamma }_{d+1}\mathfrak{g}=0$. Conversely, if ${\gamma }_{c+1}\mathfrak{g}=0$, then $(\mathrm{ad}x{)}^{c}y=0$ for all $x,y$, proving the other implication. This proves Engel's theorem and its simultaneous-triangularization form. The hypothesis concerns every element of the algebra, not merely a chosen set of generators.

For $X\in L$, define the Adjoint representation of a Lie algebra by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity gives, for every $Z$,Thus $\mathrm{ad}$ preserves brackets and is a Lie algebra representation. Its Killing form isUsing the representation identity and the cyclic property of the trace,The Killing form is also symmetric by cyclicity.

In a basis $T_a$, write $[T_a,T_b]=c_{ab}{}^d{T}_d$. The lowered structure constants satisfyThe first expression is antisymmetric in $a,b$ and the second in $b,c$. These transpositions generate all permutations, so $c_{abc}$ is totally antisymmetric.

An invariant subspace of $L$ under its Adjoint representation of a Lie algebra is exactly an ideal of a Lie algebra: $[L,I]\subseteq I$. A simple Lie algebra has no nonzero proper ideal, proving the adjoint representation is irreducible.

Put $D_X=d(X)$. The same trace calculation proves invariance of the Trace form of a Lie algebra representation:Because the $D_X$ are anti-Hermitian, $H$ is a real symmetric bilinear form. The Killing form of a simple compact Lie algebra is negative definite. Define a real linear map $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance of both forms implies that $T$ commutes with every ${\mathrm{ad}}_Z$, while symmetry makes $T$ self-adjoint for the positive inner product $-\kappa$. By the spectral theorem, its real eigenspaces are invariant under the Adjoint representation of a Lie algebra. Irreducibility forces a single eigenvalue, so $T=\mu I$ and $H=\mu \kappa$. This also avoids the extra care needed when applying the complex form of the Schur lemma to a real vector space.

The kernel of $d$ is an ideal of a Lie algebra. It cannot be all of $L$: a trivial representation of dimension greater than one is reducible. Simplicity therefore makes $d$ faithful. For every nonzero $X$,Hence the positive trace index for a compact simple Lie algebra givesin the adapted basis.

For the cubic trace tensor of a Lie algebra representation, the identityexpands toSet $Y=T_d$, $X=T_a$, $Z=T_b$, $W=T_c$. Cyclicity of $B$ converts $B_{\ell bc}$ to $B_{bc\ell }$ and $B_{a\ell c}$ to $B_{ca\ell }$. Thus the invariance identity for a cubic trace tensor is

For the final contraction, it is important to keep the specified index-raising convention. In the adapted basis write $C_{ab}{}^c=c_{ab}{}^c$. Then $c_{abc}=-C_{ab}{}^c$, while raising the last two indices with ${\kappa }^{ab}=-{\delta }^{ab}$ gives $c_a{}^{mn}=-C_{am}{}^n$. Total antisymmetry and the definition of the Killing form implyOnly the antisymmetric part of $B$ in $m,n$ contributes, soConsequently the Killing-normalized contraction of a cubic trace tensor isThe negative sign printed in the last requested identity is incompatible with raising indices by the inverse Killing form. The opposite sign is obtained if one instead defines the lowered structure constants using the positive Euclidean metric, a different convention from the one specified.

For a concrete check, take the two-dimensional representation of $\mathfrak{su}(2)$ with $T_a=-i{\sigma }_a/(2\sqrt{2})$, where the ${\sigma }_a$ are Pauli matrices. ThenTheir contraction is $+{\delta }_{ab}/8$, confirming the positive sign with $\mu =1/4$ and providing a counterexample to the printed sign.

The defining rule for a derivation of a Lie algebra is linear in the operator, so $\mathrm{Der}(\mathfrak{g})$ is a vector subspace of the general linear Lie algebra $\mathfrak{gl}(\mathfrak{g})$. Its Lie bracket is the operator commutator. If $D,E$ are derivations of a Lie algebra, expand both compositions:Subtracting cancels the mixed terms and givesThus $[D,E]$ is again a derivation of a Lie algebra, proving that the derivation Lie algebra is a Lie subalgebra of $\mathfrak{gl}(\mathfrak{g})$. This calculation works in every characteristic of a field.

The Jacobi identity states $[x,[y,z]]=[[x,y],z]+[y,[x,z]]$. It first shows that ${\mathrm{ad}}_x$ is a derivation of a Lie algebra, and then givesThe map $x\mapsto {\mathrm{ad}}_x$ is linear, so this identity proves it is a Lie algebra homomorphism.

Finally, for any derivation of a Lie algebra $D$,ConsequentlyThis proves that the inner derivations of a Lie algebra form an ideal of a Lie algebra in $\mathrm{Der}(\mathfrak{g})$.

Use a real Lie algebra convention $[T_a,T_b]=f_{ab}{}^c{T}_c$. Its Adjoint representation of a Lie algebra acts on $\mathfrak{g}$ itself:Linearity is immediate. The Jacobi identity givesso $[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]={\mathrm{ad}}_{[X,Y]}$, the defining Lie algebra representation condition. In the basis $T_b$, its matrix entries areThus the adjoint generators are the structure constants arranged as matrices. The representation has kernel equal to the center of a Lie algebra; it need not be faithful for an arbitrary algebra.

The Adjoint representation of a Lie group is ${\mathrm{Ad}}_g(Y)=gY{g}^{-1}$. It satisfies ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$, preserves the identity, and sends inverses to inverse matrices. Differentiating $e^{tX}Y{e}^{-tX}$ gives $[X,e^{tX}Y{e}^{-tX}]$, soConsequently the positive adjoint exponentials furnish the representation on elements $g=e^X$ and their products, and the intrinsic conjugation action defines it globally.

The inverse-conjugation formula in the PDF needs a minus adjoint exponent with this standard definition. Its first-order term is $Y-[X,Y]$, whereas $e^{+{\mathrm{ad}}_X}Y$ has first-order term $Y+[X,Y]$. For example $X=T_1,Y=T_2$ in a nonabelian algebra with $[T_1,T_2]\ne 0$ already distinguishes the two. The inverse conjugation and adjoint antirepresentations identity explains the group-order issue as well: $F(g)={\mathrm{Ad}}_{g^{-1}}$ satisfies $F(gh)=F(h)F(g)$. Retaining the printed inverse conjugation as a left action without this order reversal would not give an ordinary group representation.

The Killing form is the symmetric bilinear formTo prove degeneracy for a non-semisimple Lie algebra, take its nonzero solvable radical and the last nonzero member $\mathfrak{a}$ of its derived series of a Lie algebra. This is a nonzero abelian ideal of a Lie algebra. For $x\in \mathfrak{a}$, ${\mathrm{ad}}_x$ maps $\mathfrak{g}$ into $\mathfrak{a}$ and kills $\mathfrak{a}$. Every ${\mathrm{ad}}_y$ preserves $\mathfrak{a}$. Hence ${\mathrm{ad}}_x{\mathrm{ad}}_y$ maps the full space into $\mathfrak{a}$ and has zero restriction there, so its trace is zero. Thus $B(x,y)=0$ for every $y$. This is the abelian ideals lie in the radical of the Killing form argument, and establishes a nonzero kernel and vanishing determinant.

For a compact real semisimple Lie algebra, the adjoint action is unitary in an invariant positive inner product. Its infinitesimal generators are skew-Hermitian, givingStrictness follows because the adjoint kernel is the zero center. This explains the compactness criterion from the Killing form: “strictly negative” means negative-definite, not that every entry of its matrix is negative.

In the first three-generator example, take columns to be images of the basis $(X,Y,H)$. Direct use of the brackets givesTaking traces of products yields the Killing form for cyclic three-generator bracketsIt is nondegenerate, hence semisimple by the preceding degeneracy result, but it has mixed signature and is not compact. An explicit realization iswhich span the real traceless two-by-two matrices and satisfy exactly these brackets.

Changing the sign of $[X,H]$ changes the first and third adjoint matrices towhile ${\mathrm{ad}}_Y$ is unchanged. Nowthe negative-definite compact case. The basis $X=-i{\sigma }_1$, $Y=-i{\sigma }_2$, $H=-i{\sigma }_3$ realizes it as the real special unitary Lie algebra $\mathfrak{su}(2)$.

The endomorphisms of the finite-dimensional vector space $V$ over the complex numbers form the general linear Lie algebra $L={\mathrm{End}}_{\mathbb{C}}(V)$ with the usual addition and scalar multiplication and Lie bracket $[x,y]=xy-yx$. This commutator is bilinear and antisymmetric, and expanding the six products verifies the Jacobi identity.

For a Lie subalgebra $\mathfrak{g}=L_1$, being an abelian Lie algebra means $[\mathfrak{g},\mathfrak{g}]=0$. A nilpotent Lie algebra has ${\gamma }_1=\mathfrak{g}$, ${\gamma }_{j+1}=[\mathfrak{g},{\gamma }_j]$ eventually zero; a solvable Lie algebra has ${\mathfrak{g}}^{(0)}=\mathfrak{g}$, ${\mathfrak{g}}^{(j+1)}=[{\mathfrak{g}}^{(j)},{\mathfrak{g}}^{(j)}]$ eventually zero. These are the lower central series of a Lie algebra and derived series of a Lie algebra, respectively. Nilpotence is a condition on the Lie bracket, and does not require every member to be a nilpotent endomorphism: a nonzero scalar multiple of the identity spans an abelian Lie algebra.

A flag of a vector space is an increasing chain of vector subspaces. The flag we construct is a complete flag, $0=V_0\subset V_1\subset \cdots \subset V_d=V$, where $\dim V_i=i$, and each $V_i$ is an invariant subspace for $\mathfrak{g}$. We first prove the common-eigenvector assertion in the Lie theorem, by induction on $\dim \mathfrak{g}$; the zero algebra is immediate. For nonzero solvable $\mathfrak{g}$, its derived algebra is proper, so there is a codimension-one ideal of a Lie algebra $\mathfrak{k}$ containing $[\mathfrak{g},\mathfrak{g}]$. Write $\mathfrak{g}=\mathfrak{k}\oplus \mathbb{C}x$. By induction there are $v\ne 0$ and a linear functional $\lambda$ on $\mathfrak{k}$ such that $kv=\lambda (k)v$ for every $k\in \mathfrak{k}$.

Let $W$ be the cyclic subspace spanned by $v,xv,x^{2}v,\dots$. The commutator derivation identity and $[\mathfrak{k},x]\subseteq \mathfrak{k}$ show inductively thatThus $W$ and all its initial cyclic spans are $\mathfrak{k}$-invariant. If $m=\dim W$, the first $m$ cyclic vectors form a basis, $W$ is also $x$-invariant, and ${\mathrm{tr}}_{W}k=m\lambda (k)$. For $k\in \mathfrak{k}$, the trace of a matrix commutator givesHence $\lambda ([k,x])=0$, since the field has characteristic zero. The nonzero common weight spaceis $x$-invariant: $k(xw)=x(kw)+[k,x]w=\lambda (k)xw$. The restriction of $x$ to $V_{\lambda }$ has an eigenvector, because $\mathbb{C}$ is an algebraically closed field. This is a common eigenvector for $\mathfrak{g}$. Its line is invariant, and repeating the argument on the quotient vector space gives the complete invariant flag. Equivalently, this proves simultaneous triangularization of a Lie algebra representation.

In a basis adapted to this complete flag, every member of $\mathfrak{g}$ is upper triangular, so every member of its derived algebra is strictly upper triangular. Products of $d$ strictly upper triangular matrices vanish, and each iterated Lie bracket of $d$ such matrices is a sum of these products. Consequently the derived algebra is a nilpotent Lie algebra. We may therefore take $L_2=[L_1,L_1]$: it is an ideal, and the quotient Lie algebra $L_1/L_2$ is abelian. This also covers $\dim V=0$.

A finite-dimensional Lie algebra over the complex numbers $L$ is a semisimple Lie algebra when its solvable radical is zero, equivalently when it has no nonzero solvable ideals. Its Killing form isThe cyclic property of the trace makes this bilinear form symmetric and gives its invariance of a bilinear form on a Lie algebra:It follows that $R=\{x:B_L(x,L)=0\}$ is an ideal of a Lie algebra. For $x\in R$, $\mathrm{ad}x$ induces the zero map on $L/R$. Therefore, for $x,y\in R$, the matrix trace splits over the invariant subspace $R$ and the quotient to give $B_R(x,y)=B_L(x,y)=0$. In particular $B_R([R,R],R)=0$. The Cartan solvability criterion implies that $R$ is solvable. Since $L$ is semisimple, $R=0$: the Killing form is nondegenerate.

For completeness, the trace step in the Cartan solvability criterion is precisely the mechanism of the previous solution. For a complex matrix Lie algebra $\mathfrak{a}$ with $\mathrm{tr}(xy)=0$ for $x\in [\mathfrak{a},\mathfrak{a}]$, $y\in \mathfrak{a}$, set $U=[\mathfrak{a},\mathfrak{a}]$ and $W=\mathfrak{a}$. If $\beta \in M$ and $x=[a,b]$, then $\mathrm{tr}(x\beta )=\mathrm{tr}(a[b,\beta ])=0$, since $[b,\beta ]\in U$. Linearity and the trace orthogonality nilpotence lemma show that every member of $U$ is nilpotent. The Engel theorem makes $U$ nilpotent and hence $\mathfrak{a}$ solvable. Apply this to $\mathfrak{a}=\mathrm{ad}R$; the kernel of this Adjoint representation of a Lie algebra is the abelian center of $R$, so $R$ is solvable as claimed.

For an arbitrary complex Lie algebra, a Cartan subalgebra $H$ means a nilpotent Lie algebra that is self-normalizing: $N_L(H)=\{x:[x,H]\subseteq H\}=H$. This definition does not assume that $H$ is abelian. We prove that it is abelian when $L$ is semisimple.

Use the generalized-weight decomposition for a nilpotent Lie algebra for the action of $H$ on $L$. Its zero generalized weight space isWe have $H\subseteq L^0$, since $H$ is nilpotent. If $L^0/H\ne 0$, the Engel theorem gives a nonzero coset annihilated by every $h\in H$. Its representative $x$ satisfies $[H,x]\subseteq H$, contradicting $N_L(H)=H$. Thus $L^0=H$.

For a nonzero generalized weight $\lambda$, choose $h_0\in H$ with $\lambda (h_0)\ne 0$. The operator $\mathrm{ad}{h}_0$ is invertible on $L^{\lambda }$ and nilpotent on $H$. For $h\in H$, $z\in L^{\lambda }$, write $z=(\mathrm{ad}{h}_0{)}^{m}w$ with $m$ large enough that $(\mathrm{ad}{h}_0{)}^{m}h=0$. Invariance of the Killing form givesOn the other hand, $H$ is solvable, so the Lie theorem triangularizes its action on $L$. For $a,b,h\in H$, the matrix $\mathrm{ad}[a,b]$ is strictly upper triangular, while $\mathrm{ad}h$ is upper triangular. Thus $B_L([H,H],H)=0$. Together with $L=H\oplus {⨁}_{\lambda \ne 0}{L}^{\lambda }$, this yields $B_L([H,H],L)=0$. Nondegeneracy gives $[H,H]=0$.

Finally, if $x$ commutes with $H$, it normalizes $H$, hence lies in $H$. Any abelian subalgebra containing $H$ consists of such elements. Thus $H$ is a maximal abelian subalgebra, indeed $C_L(H)=H$.

A nilpotent endomorphism $x$ satisfies $x^N=0$ for some positive integer $N$. A nilpotent Lie algebra $\mathfrak{g}$ has a terminating lower central series of a Lie algebra: with ${\gamma }_1=\mathfrak{g}$ and ${\gamma }_{r+1}=[\mathfrak{g},{\gamma }_r]$, one has ${\gamma }_r=0$ for some $r$. These are different conditions: the first concerns a particular linear map, whereas the second concerns repeated Lie brackets.

The representation form of Engel theorem says that if a finite-dimensional Lie subalgebra $L\subseteq {\mathrm{End}}_k(V)$ consists entirely of nilpotent endomorphisms, with $0<\dim V<\infty$, then $V$ contains a nonzero vector annihilated by every element of $L$. Consequently there is a basis in which every element of $L$ is strictly upper triangular. No assumption on the characteristic of a field or algebraic closure is required.

We prove the common-vector assertion by induction on $\dim L$, simultaneously for every nonzero finite-dimensional representation space. The zero algebra is immediate. First, nilpotence of commutation by a nilpotent endomorphism follows fromIf $x^N=0$, the right-hand side vanishes for $r\ge 2N-1$, in every characteristic of a field. For any proper Lie subalgebra $M⊊L$, its Adjoint representation on $L/M$ therefore consists of nilpotent endomorphisms. Its image has dimension at most $\dim M<\dim L$, so the inductive assertion gives a nonzero class $y+M$ annihilated by $M$. Equivalently $y\notin M$ and $[M,y]\subseteq M$. Thus the Engel normalizer lemma gives $M⊊N_L(M)$.

Choose a maximal proper Lie subalgebra $M$. Its normalizer of a Lie subalgebra must be all of $L$, so $M$ is an ideal of a Lie algebra. Moreover $L/M$ has dimension one: otherwise a one-dimensional Lie subalgebra of this quotient would have a proper inverse image strictly between $M$ and $L$. By induction the space $W=\{v\in V:Mv=0\}$ is nonzero. It is invariant under $L$, since $m(xw)=x(mw)+[m,x]w=0$. Write $L=M+kx$. The restriction of the nilpotent endomorphism $x$ to $W$ has a nonzero kernel, and any vector in that kernel is annihilated by all of $L$. This completes the induction. Apply the assertion repeatedly to $V/kv$ to obtain a complete invariant flag $0=V_0\subset V_1\subset \cdots \subset V_d=V$ with $L{V}_j\subseteq V_{j-1}$. A basis adapted to that flag makes every matrix strictly upper triangular.

A nilpotent abstract Lie algebra need not act by nilpotent matrices. For $V\ne 0$, the scalar algebra $kI\subseteq \mathfrak{gl}(V)$ is an abelian Lie algebra, hence a nilpotent Lie algebra, but $I^N=I\ne 0$. No change of basis makes $I$ strictly upper triangular. The failed implication is precisely the distinction described in nilpotent Lie algebras need not act nilpotently.

If $\mathfrak{g}$ is a nilpotent Lie algebra, then $(\mathrm{ad}x{)}^r\mathfrak{g}\subseteq {\gamma }_{r+1}$, so every $\mathrm{ad}x$ is a nilpotent endomorphism. Conversely, apply Engel theorem to $\mathrm{ad}\mathfrak{g}$ acting on $\mathfrak{g}$. Its invariant flag is lowered by every Adjoint representation matrix, so every product of $d=\dim \mathfrak{g}$ such matrices vanishes. Since ${\gamma }_{d+1}$ is spanned by expressions $\mathrm{ad}{x}_1\cdots \mathrm{ad}{x}_d(y)$, the lower central series of a Lie algebra terminates. ThereforeFinally suppose $k$ is an algebraically closed field. If some $\mathrm{ad}x$ is not a nilpotent endomorphism, it has a nonzero eigenvalue $\lambda$ and an eigenvector $y$, giving $[x,y]=\lambda y$. The vectors $x,y$ are linearly independent and span a nonabelian two-dimensional Lie subalgebra. Conversely, a Lie subalgebra of a nilpotent Lie algebra is nilpotent because its lower central series of a Lie algebra lies termwise in the ambient series. A nonabelian two-dimensional Lie algebra has a basis $u,v$ with $[u,v]=v$: choose $v$ spanning the nonzero derived algebra and rescale a complementary vector. Its Adjoint representation has the nonzero eigenvalue $1$, so it is not nilpotent. This proves the two-dimensional subalgebra criterion for Lie algebra nilpotence: