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For a finite-dimensional Lie algebra over a field of characteristic zero, Cartan's criterion characterizes a solvable Lie algebra by vanishing of its Killing form on $\mathfrak{g}\times [\mathfrak{g},\mathfrak{g}]$, and a semisimple Lie algebra by nondegeneracy of that form. The Cartan solvability criterion and Cartan criterion for semisimplicity state the two distinct tests.

A complete flag in an $n$-dimensional vector space is a nested sequence of vector subspaces with $\dim V_i=i$. An action preserves such a flag exactly when its matrices are upper triangular in a basis adapted to the flag. The Lie theorem supplies an invariant complete flag for every complex finite-dimensional Lie algebra representation of a solvable Lie algebra.

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.

In a field of characteristic $p$, take a basis $e_0,\dots ,e_{p-1}$, and let $x{e}_j=e_{j-1}$ with indices modulo $p$, while $y{e}_j=j{e}_j$. Then $[x,y]=x$, so $kx+ky$ is a solvable Lie algebra, but the distinct eigenvalues of $y$ force any common eigenvector to be a coordinate vector, and $x$ cyclically permutes those vectors. There is consequently no common eigenvector, even over an algebraically closed field.

The solvable Lie algebra $\mathbb{C}x\oplus \mathbb{C}y$, with $[x,y]=y$, has an infinite-dimensional Irreducible Lie algebra representation on $\mathbb{C}[t]$: $xf(t)=tf(t)$ and $yf(t)=f(t-1)$. Indeed $[x,y]f=yf$. An invariant subspace is an ideal $p(t)\mathbb{C}[t]$ because it is invariant under multiplication by $t$. Shift invariance gives $p(t)\mid p(t-1)$, hence $p(t-1)=p(t)$ and $p$ is constant. This shows why the Lie theorem needs finite-dimensional representations, even when the algebra itself is finite-dimensional.

For a root of a root system of a complex semisimple Lie algebra, define $t_{\alpha }$ by $B(t_{\alpha },h)=\alpha (h)$ using the Killing form on its Cartan subalgebra. Then $\alpha (t_{\alpha })\ne 0$. Otherwise vectors $x\in {\mathfrak{g}}_{\alpha }$, $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)=1$ would generate a solvable Lie algebra with $[x,y]=t_{\alpha }$ central. The Lie theorem would make $\mathrm{ad}{t}_{\alpha }$ nilpotent, while the root-space decomposition makes it diagonalizable, forcing $t_{\alpha }$ into the zero center of a Lie algebra.

Take the semidirect product of a Lie algebra and a modulewhere $k^2$ is the defining representation of the sl2 Lie algebra. To see directly that this is a Lie algebra, realize $(X,u)$ as the block matrix $\left(\begin{array}{cc}X& u\\ 0& 0\end{array}\right)$. These matrices are closed under the operator commutator, whose Jacobi identity holds in every associative algebra.

Using the standard sl2 triple $e,h,f$, its Lie brackets show $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]={\mathfrak{sl}}_2$: indeed $h=[e,f]$, $e=[h,e]/2$, and $f=-[h,f]/2$. Also $h=\mathrm{diag}(1,-1)$ is invertible on $k^2$. For any $u\in k^2$, put $v=h^{-1}u$; thenIt follows that the derived algebra contains both ${\mathfrak{sl}}_2\oplus 0$ and $0\oplus k^2$, so $[L,L]=L$: this is a perfect Lie algebra.

However, $0\oplus k^2$ is a nonzero ideal of a Lie algebra with zero Lie bracket. It is therefore a solvable Lie algebra and is contained in the solvable radical of $L$. A semisimple Lie algebra has zero solvable radical, soThis illustrates the perfect semidirect product criterion: the Lie brackets internal to the first summand generate it, while its action generates the second summand.

For this diamond Lie algebra, every bracket lies in $⟨c,p,q⟩$, and each of these three vectors occurs as a bracket. Thus its derived algebra is exactly ${\mathfrak{g}}^{(1)}=⟨c,p,q⟩$. Inside that subalgebra the only nonzero bracket is a multiple of $[p,q]=c$, so ${\mathfrak{g}}^{(2)}=\mathbb{C}c$. Centrality of $c$ then gives ${\mathfrak{g}}^{(3)}=0$. HenceIn particular it is a solvable Lie algebra, although its lower central series of a Lie algebra does not terminate: $[\mathfrak{g},⟨c,p,q⟩]=⟨c,p,q⟩$. Solvability does not imply nilpotence.

Define a symmetric bilinear form $B$ by pairing $c$ with $d$ and $p$ with $q$, each with value one, and setting all other pairings of basis vectors to zero. Its Gram matrix in the order $(c,d,p,q)$ isThe determinant is one, so this is a nondegenerate bilinear form.

For invariance, put $T(x,y,z)=B([x,y],z)$. The nonzero values on distinct basis triples are generated byand the reversed triples have value minus one. Every value involving $c$, or a repeated basis vector, vanishes. Thus $T$ is an alternating trilinear form, explicitly $d^{\ast }\wedge p^{\ast }\wedge q^{\ast }$. In particular $T(x,y,z)=T(y,z,x)$. By symmetry of $B$, this reads $B([x,y],z)=B(x,[y,z])$, the defining identity for an invariant bilinear form on a Lie algebra.

Suppose a finite-dimensional Lie algebra representation $\rho$ realized this form as $B(x,y)=\mathrm{tr}(\rho (x)\rho (y))$. The Lie theorem says that every finite-dimensional representation of a complex solvable Lie algebra admits a basis in which all representing matrices are upper triangular. A commutator of upper triangular matrices has zero diagonal, hence is strictly upper triangular. Therefore $\rho ([\mathfrak{g},\mathfrak{g}])$ is strictly upper triangular, and its product with any representing matrix still has zero diagonal. It follows thatThis is the general fact that trace forms of solvable Lie algebras annihilate the derived algebra. Here $c\in [\mathfrak{g},\mathfrak{g}]$, so such a trace form would have $B(c,d)=0$, contradicting the constructed value one. Thus the nondegenerate invariant form exists, but no finite-dimensional representation has it as its trace form. Finite dimensionality is implicit in the ordinary representation trace; no trace on arbitrary infinite-dimensional endomorphisms is being assumed.

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 in finite dimension over $\mathbb{C}$; the characteristic-zero real version follows by complexification. The Cartan solvability criterion isIts semisimplicity consequence, also often called Cartan's criterion, is $\mathfrak{g}$ is semisimple if and only if its Killing form is nondegenerate. We prove both forms.

The auxiliary results used are as follows. The Lie theorem simultaneously triangularizes every finite-dimensional representation of a complex solvable Lie algebra. The Engel theorem says that a matrix Lie algebra consisting entirely of nilpotent endomorphisms has a common annihilated nonzero vector and a basis making all its matrices strictly upper triangular; in particular it is a nilpotent Lie algebra. Nilpotent means that the lower central series terminates, which implies that the derived series terminates as well. Finally, the Additive Jordan decomposition writes any complex endomorphism uniquely as $x=x_s+x_n$, where $x_s$ is diagonalizable, $x_n$ is nilpotent and they commute. On a generalized eigenspace $V_{\lambda }$, the two parts are $\lambda I$ and $x-\lambda I$, and both are polynomials in $x$. On endomorphisms, $\mathrm{ad}{x}_s$ is diagonalizable, $\mathrm{ad}{x}_n$ is nilpotent, and these are the Jordan parts of $\mathrm{ad}x$: the first assertion follows by splitting into maps between eigenspaces, and the second by expanding repeated commutators with the nilpotent $x_n$.

First prove the linear trace version: if $L\subseteq \mathfrak{gl}(V)$ andthen $L$ is solvable. Fix $x\in [L,L]$ and let $V=⨁V_{\lambda }$ be its generalized eigenspace decomposition. Define $y$ to act on $V_{\lambda }$ by the scalar $\overline{\lambda }$. We do not assume that $y$, $x_s$ or $x_n$ belongs to $L$.

On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}x$ has the form $(\lambda -\mu )I+N$ with $N$ nilpotent, whereas $\mathrm{ad}y$ is multiplication by $\overline{\lambda -\mu }$. Choose a polynomial $q$ with $q(d)=\bar{d}$ for every occurring eigenvalue difference $d$, and with all its derivatives of positive order up to the relevant nilpotency index zero at $d$. Such a polynomial exists by Hermite interpolation, or the Chinese remainder theorem for the pairwise coprime powers of $(t-d)$. At $d=0$ its value is zero, so $q(0)=0$. Expanding $q(dI+N)$ now givesBecause $x\in L$ and $[L,L]$ is an ideal, every positive power of $\mathrm{ad}x$ sends $L$ into $[L,L]$. The zero constant term therefore gives $[y,L]\subseteq [L,L]$.

Express $x$ as a sum of commutators ${\sum }_j[a_j,b_j]$ with $a_j,b_j\in L$. Cyclicity of trace and the hypothesis implyBut on $V_{\lambda }$, $xy$ has trace $\dim (V_{\lambda })|\lambda {|}^2$, since the nilpotent part has trace zero. Henceforcing every eigenvalue of $x$ to vanish. Thus every member of the derived algebra $[L,L]$ is a nilpotent endomorphism. By the Engel theorem, $[L,L]$ is nilpotent and therefore solvable; its derived series terminates, and adjoining the initial term $L$ shows that $L$ is solvable.

Conversely, if $L$ is solvable, the Lie theorem makes it upper triangular. Commutators of upper-triangular matrices have zero diagonal, and multiplying such a commutator by an upper-triangular matrix still has zero diagonal. Therefore the trace condition holds. Apply this equivalence to $L=\mathrm{ad}\mathfrak{g}$. Its trace pairing is the Killing form, and solvability of its image is equivalent to solvability of $\mathfrak{g}$: the kernel is its abelian centre, so a terminating derived series in the quotient terminates after at most one more step in $\mathfrak{g}$. This proves the solvability criterion.

For the semisimplicity criterion, let $R$ be the radical of the Killing form. Invariance makes $R$ a Lie algebra ideal. For $x,y\in R$, their adjoint actions on $\mathfrak{g}/R$ vanish, so computing trace in a basis adapted to $R$ gives $B_R(x,y)=B_{\mathfrak{g}}(x,y)=0$. The solvability criterion applied to $R$ shows it is a solvable ideal. Thus a semisimple Lie algebra has $R=0$ and a nondegenerate Killing form.

Conversely, any nonzero solvable ideal has a last nonzero term $A$ in its derived series; this is an abelian ideal of $\mathfrak{g}$. For $x\in A$, $y\in \mathfrak{g}$, the map $\mathrm{ad}x\mathrm{ad}y$ sends $\mathfrak{g}$ into $A$ and vanishes on $A$, so has trace zero. Hence $A\subseteq R$. A nondegenerate Killing form therefore rules out nonzero solvable ideals, exactly the definition of semisimplicity.

For classification, nondegeneracy of the Killing form provides the duality and root-space pairings used to extract a reduced crystallographic root system from a semisimple Lie algebra. Classification then reduces to the finite Dynkin diagrams and reconstruction of the Lie algebra from its root data.

The derived series of a Lie algebra isThe algebra is a solvable Lie algebra if ${\mathfrak{b}}^{(m)}=0$ for some $m$. Here a bracket of subspaces means the linear span of their Lie brackets.

The Lie theorem states that a nonzero finite-dimensional representation of a complex solvable Lie algebra has a common eigenvector; equivalently, its operators can be simultaneously upper triangularized. Applied to an Irreducible Lie algebra representation, the common eigenvector gives a nonzero invariant line, which must be the entire module. A scalar action $\rho (x)=\lambda (x)$ respects the Lie bracket exactly when $\lambda ([x,y])=0$. Thus, in the finite-dimensional setting,Every functional displayed gives a one-dimensional irreducible, and two such modules are isomorphic exactly when their functionals agree.

The finite-dimensional qualification is essential here, although the printed wording does not repeat it. For example, let $[x,y]=y$ and act on $\mathbb{C}[t]$ by $xf(t)=tf(t)$ and $yf(t)=f(t-1)$. These operators satisfy $[x,y]f=yf$. A nonzero submodule stable under $x$ is an ideal $p(t)\mathbb{C}[t]$. Stability under $y$ forces $p(t)\mid p(t-1)$, so the two polynomials, having the same degree and leading coefficient, are equal. In characteristic zero a polynomial invariant under a nonzero shift is constant. The module is therefore an infinite-dimensional simple module for the two-dimensional affine Lie algebra. It rules out extending the one-dimensional classification to unrestricted representations.

Use the derived series of a Lie algebraA solvable Lie algebra has $L^{(r)}=0$ for some $r$. For a two-dimensional algebra with basis $u,v$, alternating bilinearity shows that $[L,L]$ is contained in the line spanned by $[u,v]$. A one-dimensional Lie algebra is abelian, so $[[L,L],[L,L]]=0$. Every two-dimensional complex Lie algebra is solvable, with derived length at most two. This dimension argument actually works over any field with the alternating definition of a Lie bracket.

There is a genuine qualification in the last request: the assertion about all Irreducible Lie algebra representations is true for finite-dimensional representations. Without that restriction, the infinite-dimensional simple module for the two-dimensional affine Lie algebra is a counterexample. Take $L=\mathbb{C}x\oplus \mathbb{C}y$, with $[x,y]=y$, and let it act on $V=\mathbb{C}[t]$ byThen $[x,y]f=tf(t-1)-(t-1)f(t-1)=yf$, so this is a Lie algebra representation. Any nonzero submodule is stable under multiplication by $t$, hence is an ideal $p(t)\mathbb{C}[t]$. Stability under $y$ implies $p(t)\mid p(t-1)$. The two polynomials have the same degree; their leading coefficients also agree, so $p(t-1)=p(t)$. In characteristic zero this forces $p$ to be constant. Thus $V$ is infinite-dimensional and irreducible, while $L$ is solvable. The literal unrestricted assertion is false.

Here is a proof of the intended finite-dimensional assertion, including the essential Lie theorem argument. We prove that a nonzero finite-dimensional representation $\rho :L\to \mathrm{End}(V)$ of a complex solvable algebra has a common eigenvector, by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise $[L,L]\ne L$; choose a hyperplane $K$ containing $[L,L]$. It is a solvable ideal, and $L=K\oplus \mathbb{C}x$. Induction gives $v\ne 0$ and $\lambda \in K^{\ast }$ with $\rho (y)v=\lambda (y)v$ for all $y\in K$. Write $X=\rho (x)$ and suppress $\rho$ on elements of $K$.

Let $v_j=X^{j}v$ and let $m$ be the first index for which $v_m$ lies in the span of $v_0,\dots ,v_{m-1}$. The space $W=\mathrm{span}(v_0,\dots ,v_{m-1})$ is $X$-invariant. Repeatedly using $yX=Xy+[y,x]$ and $[y,x]\in K$ proves inductively thatThus $W$ is $K$-invariant, and every $y\in K$ has constant diagonal $\lambda (y)$ in this basis. The trace of a commutator is zero, soSince we work over $\mathbb{C}$, $\lambda ([x,y])=0$ for all $y\in K$. Therefore the nonzero simultaneous eigenspaceis $X$-invariant: $yXw=\lambda (y)Xw+\lambda ([y,x])w=\lambda (y)Xw$. A complex linear operator on a nonzero finite-dimensional space has an eigenvector. An eigenvector of $X{|}_{V_{\lambda }}$ is consequently a common eigenvector for $L$.

Its span is a one-dimensional submodule. If $V$ is irreducible, that span must be all of $V$. Hence every finite-dimensional irreducible representation of a finite-dimensional complex solvable Lie algebra has dimension one. Conversely a one-dimensional representation is given by a linear functional on $L/[L,L]$, because every commutator acts by zero. Passing repeatedly to quotients also gives the simultaneous triangularization of a Lie algebra representation. Neither the eigenvector step nor the trace argument extends to the infinite-dimensional counterexample above.

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.

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.

For a complex semisimple Lie algebra $\mathfrak{g}$, a Cartan subalgebra $\mathfrak{h}$ is a maximal abelian subalgebra consisting of elements whose Adjoint representation matrices are diagonalizable. Equivalently it is a nilpotent self-normalizing Lie subalgebra. The root-space decomposition iswhere the roots are the nonzero weights of that Adjoint representation.

We use these properties of the Killing form $B$: it is nondegenerate on $\mathfrak{g}$ and on $\mathfrak{h}$, it is invariant, distinct root spaces are orthogonal unless their roots sum to zero, and $B$ pairs ${\mathfrak{g}}_{\alpha }$ nondegenerately with ${\mathfrak{g}}_{-\alpha }$. Define $t_{\alpha }$ by $B(t_{\alpha },h)=\alpha (h)$, and choose $e\in {\mathfrak{g}}_{\alpha }$, $f\in {\mathfrak{g}}_{-\alpha }$ with $B(e,f)=1$. Since weights add, $[e,f]\in \mathfrak{h}$, and invariance givesThe nonisotropic root lemma shows $\alpha (t_{\alpha })\ne 0$. Here is its short proof: if this number were zero, the span of $e,f,t_{\alpha }$ would be a solvable Lie algebra with $t_{\alpha }$ central. Apply the Lie theorem to its action on $\mathfrak{g}$. The commutator $\mathrm{ad}{t}_{\alpha }=[\mathrm{ad}e,\mathrm{ad}f]$ would be strictly upper triangular and hence nilpotent. But $t_{\alpha }\in \mathfrak{h}$ makes it diagonalizable. It would therefore be zero, putting $t_{\alpha }$ in the zero center of a Lie algebra of $\mathfrak{g}$, contrary to its definition. SetThenso their span is the sl2 subalgebra associated with a root.

The abstract reduced crystallographic root system axioms are as follows. In a finite-dimensional real inner product space $E$, the set $R$ is finite, consists of nonzero vectors and spans $E$; for every $\alpha \in R$, one has $R\cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$; the root reflection $s_{\alpha }(\beta )=\beta -2(\beta ,\alpha )\alpha /(\alpha ,\alpha )$ preserves $R$; and $2(\beta ,\alpha )/(\alpha ,\alpha )$ is an integer for every pair of roots. We verify the positive-definite real form as well as these axioms, rather than assuming the complex Killing form is already positive.

First $R$ is finite and has no zero element by its definition. The roots span ${\mathfrak{h}}^{\ast }$ over $\mathbb{C}$: any $h\in \mathfrak{h}$ annihilated by all roots commutes with the whole root-space decomposition and is central, hence zero. Therefore the $t_{\alpha }$, and also the $H_{\alpha }$, span $\mathfrak{h}$ over $\mathbb{C}$. For every root $\beta$, the classification of finite-dimensional sl2 representations applied to the Adjoint representation of the root subalgebra gives $\beta (H_{\alpha })\in \mathbb{Z}$. Put ${\mathfrak{h}}_{\mathbb{R}}={\mathrm{span}}_{\mathbb{R}}\{H_{\alpha }:\alpha \in R\}$. Every root takes real values on this space, andThe strict inequality follows because the roots span ${\mathfrak{h}}^{\ast }$. This also shows that the complexification of ${\mathfrak{h}}_{\mathbb{R}}$ injects into $\mathfrak{h}$: an equality $h_1+i{h}_2=0$ with real $h_j$ would contradict positivity of $B(h_1,h_1)$ and $B(h_2,h_2)$. Since its complex span is all of $\mathfrak{h}$, it is a real form. Moreover $B(H_{\alpha },H_{\alpha })=4/\alpha (t_{\alpha })>0$, so $t_{\alpha }$ is a real scalar multiple of $H_{\alpha }$. The dual inner product thus makes $E={\mathrm{span}}_{\mathbb{R}}R$ a Euclidean space, as in the Euclidean subspace of a Cartan subalgebra.

To prove reducedness without assuming it, consider the root-subalgebra modulewith absent root spaces understood to be zero. Its $H_{\alpha }$ weights are even, so every nontrivial Irreducible Lie algebra representation in it has even positive highest weight and one-dimensional weight-zero space. The action of $X_{\alpha }$ on its weight-zero space $\mathfrak{h}$ has image exactly $\mathbb{C}{X}_{\alpha }$, of dimension one. Therefore there is precisely one nontrivial irreducible summand, the already embedded adjoint ${\mathfrak{sl}}_2$ module of highest weight $2$. Thus $\dim {\mathfrak{g}}_{\alpha }=1$ and $2\alpha \notin R$. If $\beta =c\alpha$ is any root on the same real line, the integral numbers $\beta (H_{\alpha })=2c$ and $\alpha (H_{\beta })=2/c$ have product $4$. Hence $c$ is one of $\pm \frac{1}{2},\pm 1,\pm 2$; the half and double cases are excluded by applying the preceding argument to the appropriate root. This proves the root-space reducedness lemma and $R\cap \mathbb{R}\alpha =\{\pm \alpha \}$.

For a root $\beta$ not parallel to $\alpha$, the sum of root spaces ${⨁}_{j\in \mathbb{Z}}{\mathfrak{g}}_{\beta +j\alpha }$ is stable under the root ${\mathfrak{sl}}_2$. Its integer $H_{\alpha }$ weights are symmetric under sign in each irreducible summand. Therefore the weight $-\beta (H_{\alpha })$ also occurs, at the root $\beta -\beta (H_{\alpha })\alpha$. The Killing form normalization givesso this root is exactly $s_{\alpha }(\beta )$. For $\beta =\pm \alpha$ the reflection just swaps the two roots. This proves reflection invariance and the Cartan integer condition. All axioms of the reduced crystallographic root system have now been verified.

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.