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The two-dimensional Lie algebra of infinitesimal dilations and translations of a line has a basis $h,e$ with $[h,e]=e$. Its derived algebra is the abelian line spanned by $e$, but every nontrivial term of its Lower central series of a Lie algebra is that same line. It is solvable and not nilpotent.

For $x$ in the derived algebra of a complex matrix Lie algebra $L$, define $y$ by multiplication by $\overline{\lambda }$ on each generalized eigenspace $V_{\lambda }$ of $x$. Polynomial interpolation and adjoint compatibility of additive Jordan decomposition make $\mathrm{ad}y$ a polynomial in $\mathrm{ad}x$ with zero constant term, so $[y,L]\subseteq [L,L]$. Trace orthogonality of $[L,L]$ and $L$ forces the displayed sum to vanish. Thus $x$ is a nilpotent endomorphism, and the Engel theorem implies solvability. The auxiliary $y$ and the Jordan components are not required to lie in $L$.

A finite-dimensional complex Lie algebra is a Solvable Lie algebra exactly when its derived algebra is a Nilpotent Lie algebra. The forward direction follows from the Lie theorem in the Adjoint representation and lifting nilpotence through its central kernel. The reverse direction follows because the derived series of a Lie algebra after its first term is the derived series of a Lie algebra of the derived algebra.

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.

Over the complex numbers, every finite-dimensional representation of a solvable Lie algebra has a basis in which every representing matrix is upper triangular. The Lie theorem is often stated first as the existence of a common eigenvector in every nonzero finite-dimensional Lie algebra representation of a Solvable Lie algebra. Applying that assertion successively to quotient representations gives an invariant complete flag, and hence the upper triangular form. The same proof works over any algebraically closed field of characteristic zero.

We prove the common eigenvector assertion by induction on $\dim \mathfrak{g}$, writing the action as $av$. The zero Lie algebra is immediate. If $\mathfrak{g}\ne 0$ is solvable, its derived series of a Lie algebra shows that $[\mathfrak{g},\mathfrak{g}]\ne \mathfrak{g}$. Choose a codimension-one ideal of a Lie algebra $\mathfrak{a}$ containing $[\mathfrak{g},\mathfrak{g}]$, and choose $x\notin \mathfrak{a}$. By induction there are $v\ne 0$ and a linear functional $\lambda :\mathfrak{a}\to \mathbb{C}$ such that $av=\lambda (a)v$ for all $a\in \mathfrak{a}$.

Let $W$ be the span of $v,xv,x^{2}v,\dots$. If $m=\dim W$, the first $m$ of these vectors form a basis, and $W$ is $x$-invariant. We claim that for each $a\in \mathfrak{a}$,For $j=0$ this is the definition of $v$. For the induction step, use $ax=xa+[a,x]$ and $[a,x]\in \mathfrak{a}$. Applying the induction hypothesis to both $a$ and $[a,x]$ proves the claim. Consequently $W$ is $\mathfrak{a}$-invariant, and every $a\in \mathfrak{a}$ acts on $W$ by an upper triangular matrix with all diagonal entries $\lambda (a)$.

Since both $x$ and $a$ preserve $W$, the matrix trace of their commutator on $W$ is zero. The claim applied to $[x,a]\in \mathfrak{a}$ givesHere characteristic zero is essential: $m\ne 0$ in the field, so $\lambda ([x,a])=0$. Now the common weight spaceis nonzero and $x$-invariant. Indeed, for $w\in E_{\lambda }$,An endomorphism of a nonzero finite-dimensional complex vector space has an eigenvector, so choose an eigenvector of $x$ in $E_{\lambda }$. It is a common eigenvector for $\mathfrak{g}=\mathfrak{a}+\mathbb{C}x$. This proves the Lie theorem.

For the printed matrices in characteristic $p$, $[x,y]=x$, and there is no common eigenvector. Index the standard basis by $e_0,\dots ,e_{p-1}$. The cyclic entry in the PDF givesThe $p$ diagonal eigenvalues of $y$ are distinct in $k$. Thus every eigenvector of $y$ is a scalar multiple of a single $e_j$. Since $p\ge 2$, $x{e}_j$ is never a scalar multiple of $e_j$, proving the assertion even when $k$ is not an algebraically closed field.

For $1\le j\le p-1$,At the cyclic boundary,Hence $[x,y]=x$. The two-dimensional Lie subalgebra $kx+ky$ has derived algebra $kx$, whose own derived algebra is zero, so it is solvable. It nevertheless has no common eigenvector, including after extending $k$ to its algebraic closure. This is a failure of Lie theorem in positive characteristic. In the proof above, the obstruction is precisely that $\dim W=p$ can vanish as a scalar in $k$.

The derived algebra of a complex solvable Lie algebra is nilpotent. First suppose $\mathfrak{g}\subseteq \mathfrak{gl}(V)$. By the Lie theorem, put every element of $\mathfrak{g}$ in upper triangular form. The diagonal of a commutator of upper triangular matrices is zero, so $\mathfrak{d}=[\mathfrak{g},\mathfrak{g}]$ consists of strictly upper triangular matrices. The Lie algebra of all such matrices is a Nilpotent Lie algebra: if $F_r$ consists of matrices whose entries vanish whenever $j-i<r$, thenTherefore the Lower central series of a Lie algebra of $\mathfrak{d}$ reaches zero. Alternatively, every element of $\mathfrak{d}$ is a nilpotent linear map, and the Engel theorem states that a finite-dimensional Lie subalgebra consisting of nilpotent linear maps is a Nilpotent Lie algebra.

For an abstract complex Solvable Lie algebra $\mathfrak{g}$, apply the preceding result to its Adjoint representation. The Lie algebra $\mathrm{ad}\mathfrak{d}=[\mathrm{ad}\mathfrak{g},\mathrm{ad}\mathfrak{g}]$ is nilpotent. Since $\ker (\mathrm{ad}{|}_{\mathfrak{d}})=\mathfrak{d}\cap Z(\mathfrak{g})$ is central in $\mathfrak{d}$, some term of the Lower central series of a Lie algebra of $\mathfrak{d}$ lies in that central ideal; the next term is zero. Thus $\mathfrak{d}$ itself is nilpotent.

Conversely, every Nilpotent Lie algebra is solvable, since its derived series of a Lie algebra is contained term by term in its Lower central series of a Lie algebra. If $\mathfrak{d}$ is nilpotent, it is therefore solvable, and $\mathfrak{g}/\mathfrak{d}$ is Abelian. More directly, the derived series of a Lie algebra of $\mathfrak{g}$, after its first term, is the derived series of a Lie algebra of $\mathfrak{d}$. Consequently the derived algebra nilpotence criterion is

The form attached to $V$ is $B_V(x,y)={\mathrm{tr}}_V(xy)$. More generally, for a Lie algebra representation $\rho$, the Trace form of a Lie algebra representation isThe unqualified Killing form is the special case of the Adjoint representation,The distinction matters: a Trace form of a Lie algebra representation can be degenerate even when $\mathfrak{g}$ is semisimple, for example on the trivial Lie algebra representation.

The Trace form of a Lie algebra representation is bilinear and symmetric, because $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. It is an invariant bilinear form on a Lie algebra:This follows by expanding both commutators and cyclically permuting factors under the matrix trace. Equivalently,Its radical of a bilinear form is an ideal of a Lie algebra, since if $B_V(y,-)=0$, then $B_V([x,y],z)=-B_V(y,[x,z])=0$. The Killing form is also preserved by every automorphism of a Lie algebra, because the corresponding adjoint operators are conjugate. On a complex finite-dimensional Lie algebra, the Cartan criterion for semisimplicity says that the Killing form is nondegenerate exactly when the Lie algebra is semisimple. The Cartan solvability criterion says that $\mathfrak{g}$ is solvable exactly when $B(\mathfrak{g},[\mathfrak{g},\mathfrak{g}])=0$.

We next construct the sl2 subalgebra associated with a root. Use the root-space decompositionFor $u\in {\mathfrak{g}}_{\beta }$, $v\in {\mathfrak{g}}_{\gamma }$, invariance of the Killing form givesThus $B({\mathfrak{g}}_{\beta },{\mathfrak{g}}_{\gamma })=0$ unless $\beta +\gamma =0$, and $B({\mathfrak{g}}_{\beta },\mathfrak{h})=0$ for nonzero $\beta$. Nondegeneracy of $B$ on $\mathfrak{g}$ now implies that $-\alpha$ is a root and that $B$ pairs ${\mathfrak{g}}_{\alpha }$ and ${\mathfrak{g}}_{-\alpha }$ nondegenerately.

Nondegeneracy of $B{|}_{\mathfrak{h}}$ defines a unique $t_{\alpha }\in \mathfrak{h}$ byChoose $x\in {\mathfrak{g}}_{\alpha }$ and $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)=1$. Their Lie bracket lies in the zero root space, namely $\mathfrak{h}$, andTherefore $[x,y]=t_{\alpha }$.

The essential nonisotropic root lemma is that $\alpha (t_{\alpha })\ne 0$. Suppose instead that it vanished. Then $[t_{\alpha },x]=[t_{\alpha },y]=0$, so $⟨x,y,t_{\alpha }⟩$ would be a Solvable Lie algebra with derived algebra $\mathbb{C}{t}_{\alpha }$. Apply the Lie theorem to its action on $\mathfrak{g}$ by the Adjoint representation. The commutator $\mathrm{ad}{t}_{\alpha }=[\mathrm{ad}x,\mathrm{ad}y]$ is strictly upper triangular in a suitable basis, hence nilpotent. But $t_{\alpha }\in \mathfrak{h}$, so the root-space decomposition makes $\mathrm{ad}{t}_{\alpha }$ diagonalizable. A diagonalizable nilpotent linear map is zero. Thus $t_{\alpha }$ is central in $\mathfrak{g}$. The center of a Lie algebra of a semisimple Lie algebra is zero; equivalently a central element lies in the radical of the Killing form. This forces $t_{\alpha }=0$, contradicting $\alpha \ne 0$.

Writing $c=\alpha (t_{\alpha })$, defineThe root-space decomposition and $[x,y]=t_{\alpha }$ giveThe three vectors are linearly independent because they lie in the distinct summands ${\mathfrak{g}}_{\alpha }$, $\mathfrak{h}$, and ${\mathfrak{g}}_{-\alpha }$. Their span is therefore a copy of the sl2 Lie algebra.

The weight lattice consists of the functionals integral on all coroots. With $H_{\alpha }$ the coroot above, the weight lattice iswhere the fundamental weights satisfy ${\omega }_i(H_{{\alpha }_j})={\delta }_{ij}$ for the simple roots ${\alpha }_j$. Here $P$ lies in the real span of the roots, viewed inside ${\mathfrak{h}}^{\ast }$.

The classification of finite-dimensional sl2 representations says that every finite-dimensional complex sl2 Lie algebra representation is a direct sum of irreducibles $L_m$, $m\in {\mathbb{Z}}_{\ge 0}$, on which the standard $H$ has eigenvalues $m,m-2,\dots ,-m$. Restrict any finite-dimensional Lie algebra representation of $\mathfrak{g}$ to each sl2 subalgebra associated with a root. If $v$ has weight $\lambda$, then $H_{\alpha }v=\lambda (H_{\alpha })v$, so $\lambda (H_{\alpha })$ is an integer. Thus every weight lies in $P$. The same restrictions show that the commuting simple coroots act diagonalizably, justifying the simultaneous weight-space decomposition.

For $\mathfrak{sp}(4)$, the roots are $\pm 2{\epsilon }_1$, $\pm 2{\epsilon }_2$, and $\pm {\epsilon }_1\pm {\epsilon }_2$. Work on ${\mathbb{C}}^4$ with the alternating bilinear form having matrixThe symplectic Lie algebra isUsing the matrix units $E_{ij}$, take the Cartan subalgebraDefine ${\epsilon }_i(a_1{h}_1+a_2{h}_2)=a_i$. A regular diagonal element of $\mathfrak{h}$ has centralizer precisely $\mathfrak{h}$, and every element of $\mathfrak{h}$ acts diagonalizably. Thus it is a Cartan subalgebra. The requested Cartan decomposition is the root-space decompositionChoose positive roots ${\epsilon }_1-{\epsilon }_2$, $2{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$, $2{\epsilon }_1$. The symplectic root sl2 triple are given explicitly byFor the negative root spaces, use the corresponding $Y_{\alpha }$. These eight root vectors, together with $h_1,h_2$, form a basis: the block description above has dimension $4+3+3=10$, and the ten listed vectors are independent. Finally, the matrix unit identityverifies $[X_{\alpha },Y_{\alpha }]=H_{\alpha }$ for every row. The diagonal differences verify $[H_{\alpha },X_{\alpha }]=2X_{\alpha }$ and $[H_{\alpha },Y_{\alpha }]=-2Y_{\alpha }$. Thus each row supplies a basis of the required sl2 subalgebra associated with a root.