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An abelian Lie group has commutative multiplication. Its Lie algebra is an abelian Lie algebra, and its Lie exponential map is a group homomorphism from the additive Lie algebra. Conversely, an additive Lie exponential map forces the identity component of a Lie group to be abelian, but does not force all disconnected components to commute.

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 abelian Lie algebra $kI\subseteq \mathrm{End}(V)$, for $V\ne 0$, is a nilpotent Lie algebra but contains the nonnilpotent identity map. Thus nilpotence of the abstract Lie algebra does not imply that a particular Lie algebra representation consists of nilpotent endomorphisms. Engel theorem requires nilpotence of every represented endomorphism to obtain strict upper triangularity.

The correspondence in part (a) respects Lie brackets. If the Lie algebra is abelian, any two left-invariant vector fields satisfy $[X,Y]=0$ everywhere, not just at the identity. Part (b) therefore gives $d\omega (X,Y)=0$ for every such pair.

At a point $g$, the map $(d{L}_g{)}_e:T_{e}G\to T_{g}G$ is an isomorphism. Thus every tangent vector at $g$ is the value of a left-invariant vector field. The two-form $d\omega$ consequently vanishes on every pair of tangent vectors at every point, provingNo connectedness assumption on the Lie group is needed: the argument uses its left translations and its abelian Lie algebra.

First consider finite-dimensional Lie algebra representations. The operators representing an abelian Lie algebra commute. They have a common eigenvector: take an eigenspace for the first operator of a basis of the algebra, observe that every remaining operator preserves it, and repeat inside this nonzero space. The resulting line is an invariant subspace, so an irreducible representation must be one-dimensional. Its action has the form $xv=\lambda (x)v$ for a linear functional $\lambda$. Conversely, every such functional defines a Lie algebra representation, since both the Lie brackets in the algebra and the commutators of its scalar operators vanish. Distinct functionals give nonisomorphic representations. Thus the irreducibles are precisely the characters $\lambda \in {\mathfrak{g}}^{\ast }$.

Even without a finite-dimensional assumption on the module, this conclusion for a finite-dimensional abelian Lie algebra follows from the Weak Hilbert Nullstellensatz: its universal enveloping algebra is the polynomial algebra $\mathrm{Sym}(\mathfrak{g})$. A simple module over a commutative algebra is its quotient by a maximal ideal, and every maximal ideal of this polynomial algebra over $\mathbb{C}$ has residue field $\mathbb{C}$.

Indecomposable does not imply irreducible. For the one-dimensional abelian Lie algebra $\mathbb{C}x$, let $x$ act on ${\mathbb{C}}^2$ by the Nilpotent Jordan blockThe line spanned by the first coordinate vector is an invariant subspace, so the module is reducible. If it were a direct sum of two nonzero submodules, both would have dimension one; nilpotence would force $N$ to act as zero on both, contrary to $N\ne 0$. Thus it is an indecomposable representation.

For a finite-dimensional module $V$, put $d=\dim V$ and define its generalized weight spaces intrinsically byTo obtain the decomposition, choose a basis $x_1,\dots ,x_r$ of $\mathfrak{g}$, split into generalized eigenspaces of $\rho (x_1)$, and successively split each piece by $\rho (x_2),\dots ,\rho (x_r)$. All the pieces are submodules because the acting operators commute. On one resulting piece, $\rho (x_j)$ has a unique eigenvalue ${\lambda }_j$. The commuting operators can be simultaneously upper triangularized by repeating the common-eigenvector argument on quotients. Hence on this piece every $\rho (x_j)-{\lambda }_{j}I$ is strictly upper triangular. Any linear combination is also strictly upper triangular, and its $d$th power vanishes. Extending ${\lambda }_j$ linearly therefore identifies the piece with $V^{\lambda }$. Different tuples give disjoint pieces; if a component has a different tuple from $\lambda$, at least one $\rho (x_j)-\lambda (x_j)I$ is invertible there. Consequentlywith only finitely many nonzero summands. Since the definition uses every $x$ rather than a chosen basis, this generalized-weight decomposition for a nilpotent Lie algebra is canonical. Its summands need not be irreducible, as the Jordan block example already shows.

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 $\dim V=0$.

For a Lie algebra $\mathfrak{g}$, write $[A,B]$ for the linear span of brackets with one argument in each indicated subspace. The three definitions areThese are respectively an Abelian Lie algebra, a solvable Lie algebra and a nilpotent Lie algebra. The second and third sequences are the derived series of a Lie algebra and the lower central series of a Lie algebra.

An Abelian Lie algebra has ${\gamma }_2=0$, and a nilpotent Lie algebra is a solvable Lie algebra: induction gives ${\mathfrak{g}}^{(j)}\subseteq {\gamma }_{j+1}$. Thus all the implications are generated byNone of the reverse implications holds. The Heisenberg Lie algebra with basis $x,y,z$ and $[x,y]=z$, all other basic brackets zero, is nonabelian but has ${\gamma }_2=\mathbb{C}z$, ${\gamma }_3=0$. The two-dimensional affine Lie algebra of the line with $[h,e]=e$ has ${\mathfrak{g}}^{(1)}=\mathbb{C}e$ and ${\mathfrak{g}}^{(2)}=0$, but ${\gamma }_j=\mathbb{C}e$ for every $j\ge 2$. It is solvable and neither nilpotent nor abelian. These two examples answer all six ordered-pair comparisons.

The Lie theorem says that a finite-dimensional Lie algebra representation of a finite-dimensional solvable Lie algebra over an algebraically closed field of characteristic zero has a common eigenvector whenever its representation space is nonzero. Equivalently it admits an invariant complete flag, or simultaneous upper triangularization. Here the field may be taken to be $\mathbb{C}$; these hypotheses are essential.

To prove the common-eigenvector assertion, induct on $\dim \mathfrak{g}$. The zero algebra is immediate. Since $\mathfrak{g}$ is nonzero and solvable, its derived algebra is proper. Choose a codimension-one Lie algebra ideal $\mathfrak{h}$ containing it, and write $\mathfrak{g}=\mathfrak{h}\oplus \mathbb{C}x$. The Lie algebra $\mathfrak{h}$ is solvable, so induction supplies $v\ne 0$ and a linear functional $\lambda :\mathfrak{h}\to \mathbb{C}$ with $hv=\lambda (h)v$.

Consider the finite-dimensional cyclic subspace $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Until the first linear dependence, these powers form a basis. The identityand $[h,x]\in \mathfrak{h}$ show by induction, simultaneously for all $h\in \mathfrak{h}$, that $W$ is $\mathfrak{h}$-invariant and that $h$ is upper triangular on it with every diagonal entry $\lambda (h)$. It is also $x$-invariant by construction. HenceThe trace of a commutator is zero, and characteristic zero gives $\lambda ([x,h])=0$.

The simultaneous eigenspace $E_{\lambda }=\{w:hw=\lambda (h)w\text{ for every }h\in \mathfrak{h}\}$ is nonzero and $x$-invariant, sinceOver an algebraically closed field, $x{|}_{E_{\lambda }}$ has an eigenvector, which is therefore a common eigenvector for all of $\mathfrak{g}$. This finishes induction. Apply the same assertion to the quotient representation by its invariant line, and then to successive quotients. A basis adapted to the resulting complete flag gives the stated simultaneous triangularization of a Lie algebra representation, completing the proof of the Lie theorem.

A nilpotent Lie algebra is a solvable Lie algebra, so the inclusion $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ satisfies the Lie theorem and is upper triangular in a suitable basis, for finite-dimensional complex $V$.

This is insufficient to prove the Engel theorem. Its matrix version starts with a Lie subalgebra of nilpotent endomorphisms and concludes that they are simultaneously strictly upper triangular; its abstract version concludes nilpotence from nilpotence of every Adjoint representation endomorphism. Merely upper triangular matrices can have nonzero diagonal entries: the one-dimensional algebra $\mathbb{C}{I}_V$ is an Abelian Lie algebra and a nilpotent Lie algebra, but acts by a nonnilpotent identity matrix. This is nilpotent Lie algebras need not act nilpotently. Moreover, in the abstract Engel theorem nilpotence of the algebra is a conclusion, so assuming it first to invoke the Lie theorem would be circular. Abstract nilpotence and nilpotence of each representing matrix are different conditions.

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:

Let $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $\mathfrak{g}=\mathbb{C}(I+N)\subseteq {\mathfrak{gl}}_2(\mathbb{C})$. This one-dimensional abelian Lie algebra is not a semisimple Lie algebra. For $X=I+N$, its Additive Jordan decomposition is $X_s=I$ and $X_n=N$. Neither belongs to $\mathfrak{g}$: comparing diagonal entries and then the upper-right entry rules out both scalar multiples. Therefore nonsemisimple matrix Lie subalgebras need not be closed under Additive Jordan decomposition.

Take the Lie subalgebra $\mathfrak{b}\subseteq {\mathfrak{gl}}_2(\mathbb{C})$ of all upper-triangular matrices. It is a solvable Lie algebra with a nonzero abelian ideal of a Lie algebra, hence is not a semisimple Lie algebra. For every $X\in \mathfrak{b}$, the polynomial description $X_s=p(X)$ in the Additive Jordan decomposition keeps $X_s$ upper triangular; then $X_n=X-X_s$ is upper triangular too. Thus the upper-triangular matrix algebra is closed under Additive Jordan decomposition despite being nonsemisimple. A diagonal matrix algebra gives an even simpler abelian Lie algebra example.

Suppose first that the left-invariant 1-form $\alpha$ is closed. The scalar function $\alpha (l_{\xi })$ is constant for every $\xi$, because both the form and vector field are left-invariant. Cartan's magic formula therefore givesPart b says that $g\mapsto R_g^{\ast }\alpha$ is locally constant. Since $G$ is connected, it is constant, and its value at the identity is $\alpha$. Hence $R_g^{\ast }\alpha =\alpha$ for every $g$, so $\alpha$ is bi-invariant.

Conversely, if $\alpha$ is bi-invariant, then all these Lie derivatives vanish. Cartan's formula and the constancy of $\alpha (l_{\xi })$ give ${\iota }_{l_{\xi }}d\alpha =0$. The left-invariant vector fields span every tangent space, so $d\alpha =0$.

Connectedness cannot be omitted. The orthogonal group $O(2)$ has an abelian Lie algebra, so every left-invariant 1-form is closed by the Maurer-Cartan equation in a Lie-algebra basis. Conjugation by a reflection acts as $-1$ on its one-dimensional Lie algebra, so a nonzero left-invariant 1-form is not right-invariant. This is the standard obstruction recorded by the closed left-invariant 1-form criterion.