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For the Adjoint representation of a Lie algebra, the operator associated with $x\in L$ isThe Jacobi identity gives $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$, so this really is a Lie algebra representation. Define the lower central series of a Lie algebra by $L^1=L$ and $L^{r+1}=[L,L^r]$. A nilpotent Lie algebra is one for which $L^{c+1}=0$ for some integer $c$. If this holds, ${\mathrm{ad}}_x$ carries $L^r$ into $L^{r+1}$, and consequently $({\mathrm{ad}}_x{)}^{c}L=0$. This proves one implication without any condition on the field.

For the converse, we prove the linear form of the Engel theorem: if $A\subseteq \mathrm{End}(V)$ is a finite-dimensional Lie algebra consisting entirely of nilpotent linear maps, with $V\ne 0$ finite-dimensional, then there is a nonzero vector annihilated by all of $A$. This argument works over any field.

Induct on $\dim A$. The assertion is immediate when $A=0$. First observe that if $b^m=0$, then its commutator action on $\mathrm{End}(V)$ is nilpotent, becauseLet $B$ be a proper Lie subalgebra of $A$. The commutator action of $B$ on $A/B$ therefore consists of nilpotent linear maps. Its image has dimension at most $\dim B<\dim A$, so induction supplies a nonzero coset $a+B$ annihilated by $B$. Thus $[B,a]\subseteq B$, and the normalizer $N_A(B)=\{a\in A:[a,B]\subseteq B\}$ strictly contains $B$. This is the Engel normalizer lemma.

Choose $B$ maximal among proper Lie subalgebras. Its normalizer must be $A$, so $B$ is a Lie algebra ideal. Moreover $\dim (A/B)=1$: otherwise the inverse image of a one-dimensional subalgebra of $A/B$ would lie strictly between $B$ and $A$. By induction,Since $B$ is an ideal, $W$ is $A$-invariant: $b(av)=a(bv)+[b,a]v=0$. Take $x\in A\setminus B$. Its restriction to $W$ is nilpotent and hence has a nonzero kernel. A nonzero vector in that kernel is annihilated by $B$ and $x$, and therefore by all of $A$. This completes the induction.

Apply this result successively to $V$, then its quotient by a common annihilated line, and so on. We obtain a complete flag $0=V_0\subset V_1\subset \cdots \subset V_n=V$ with $A{V}_i\subseteq V_{i-1}$. Equivalently every element of $A$ is strictly upper triangular in one common basis; any product of $n$ such operators is zero.

Now take $V=L$ and $A=\mathrm{ad}(L)$. The hypothesis supplies exactly the nilpotence needed above. Every iterated bracket of length $n+1$, where $n=\dim L$, is an application of $n$ adjoint operators and vanishes. These brackets span $L^{n+1}$, giving $L^{n+1}=0$. The zero algebra is already nilpotent. Thus $L$ is nilpotent if and only if every ${\mathrm{ad}}_x$ is nilpotent. The common flag is essential: separate nilpotence of unrelated operators would not imply that their mixed products vanish.

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.

Let $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$ be a finite-dimensional Lie algebra representation. The form denoted $B_V$ isIt is the Trace form of a Lie algebra representation; the Killing form $B$ without a subscript is specifically the case $V=\mathfrak{g}$, $\rho =\mathrm{ad}$. The distinction matters: the form of the trivial representation cannot detect whether the algebra is semisimple.

Linearity of $\rho$ and trace proves bilinearity, and cyclicity of the trace gives symmetry. With $A=\rho (X)$, $D=\rho (Y)$ and $C=\rho (Z)$, the representation identity givesThus it is an invariant bilinear form on a Lie algebra, and in particular the Adjoint representation preserves the Killing form.

The Cartan solvability criterion has two useful formulations. For a finite-dimensional complex Lie algebra $\mathfrak{g}$,Its matrix version says that a Lie subalgebra $L\subseteq \mathfrak{gl}(V)$ is solvable precisely when $\mathrm{tr}(xy)=0$ for all $x\in [L,L]$ and $y\in L$. We prove the matrix version first, with ordinary trace in $V$.

If $L$ is solvable, the Lie theorem makes all its matrices upper triangular. Their commutators are strictly upper triangular, so multiplying such a matrix by an upper triangular one still has zero diagonal and hence zero trace. This proves the easy direction.

Conversely assume the trace-orthogonality condition and fix $x\in [L,L]$. We show that all eigenvalues of $x$ vanish. Use its Additive Jordan decomposition $x=x_s+x_n$ with $[x_s,x_n]=0$. On the generalized eigenspace $V_{\lambda }$ of $x$, define an auxiliary endomorphism $y$ to be $\overline{\lambda }I$. This $y$ need not be in $L$; we only need control of its commutator with $L$.

On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}{x}_s$ acts by $\lambda -\mu$ and $\mathrm{ad}y$ by $\overline{\lambda }-\overline{\mu }$. Choose a polynomial $q$ with $q(0)=0$ and $q(\lambda -\mu )=\overline{\lambda -\mu }$ at the finitely many distinct differences. Polynomial interpolation supplies it because equal differences have equal conjugates. Therefore $\mathrm{ad}y=q(\mathrm{ad}{x}_s)$.

The adjoint compatibility of additive Jordan decomposition identifies $\mathrm{ad}{x}_s$ as the semisimple part of $\mathrm{ad}x$. Elementary Jordan–Chevalley decomposition gives $\mathrm{ad}{x}_s=p(\mathrm{ad}x)$ for a polynomial $p$ with $p(0)=0$. Hence $\mathrm{ad}y$ is a polynomial in $\mathrm{ad}x$ with zero constant term. Since $\mathrm{ad}x$ maps $L$ into $[L,L]$ and preserves that Lie algebra ideal, we obtainNo assumption that $x_s$, $x_n$ or $y$ belongs to $L$ was made.

Write $x={\sum }_i[a_i,b_i]$ with $a_i,b_i\in L$. Cyclicity and the assumed orthogonality now giveOn the other hand the nilpotent parts have zero trace on each generalized eigenspace, soThus all $\lambda$ are zero and $x$ is a nilpotent endomorphism. This is the Conjugate-spectrum proof of Cartan solvability.

The permitted Engel theorem, in the form needed here, states: a finite-dimensional Lie subalgebra of endomorphisms in which every element is nilpotent has a nonzero vector annihilated by all its elements, and iteration on quotients makes every element simultaneously strictly upper triangular. Apply it to $[L,L]$. The strictly upper triangular algebra is nilpotent, so $[L,L]$ is a nilpotent Lie algebra and therefore solvable. Since $L/[L,L]$ is abelian, the derived series of a Lie algebra of $L$ terminates too. This proves the matrix criterion.

Apply it to $L=\mathrm{ad}\mathfrak{g}$. The condition on $B$ is exactly the matrix condition on $L$. Thus $\mathrm{ad}\mathfrak{g}$ is solvable. The kernel of $\mathrm{ad}$ is the center of a Lie algebra, which is abelian, so $\mathfrak{g}$ is solvable as well: once the derived series maps to zero it is central, and its next term vanishes. Conversely a solvable $\mathfrak{g}$ has solvable adjoint image, proving the abstract Cartan solvability criterion in both directions.

A finite-dimensional Lie algebra is a semisimple Lie algebra when it has no nonzero solvable Lie algebra ideal, equivalently its solvable radical is zero. Let $R=\mathrm{rad}B=\{x:B(x,\mathfrak{g})=0\}$. Invariance makes $R$ an ideal. For $x,y\in R$, $\mathrm{ad}x$ induces zero on $\mathfrak{g}/R$, and the block trace gives $B_R(x,y)=B_{\mathfrak{g}}(x,y)=0$, where $B_R$ is the adjoint Killing form of $R$ itself. The Cartan solvability criterion therefore makes $R$ solvable. If $\mathfrak{g}$ is semisimple, $R=0$.

Conversely suppose $B$ is a nondegenerate bilinear form. If a nonzero solvable ideal existed, its last nonzero derived term $A$ would be an abelian ideal of $\mathfrak{g}$. For $a\in A$ and $x\in \mathfrak{g}$, the operator $\mathrm{ad}a\mathrm{ad}x$ has image in $A$ and is zero on $A$, so its square and its trace are zero. Thus $B(A,\mathfrak{g})=0$, contradicting nondegeneracy. This is Abelian ideals lie in the radical of the Killing form. We conclude the Cartan criterion for semisimplicity:

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.

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

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

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

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

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

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

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

For an arbitrary finite-dimensional complex Lie algebra $L$, a Cartan subalgebra is a nilpotent Lie subalgebra $H$ equal to its own normalizer. When $L$ is semisimple, this is equivalently a maximal abelian subalgebra consisting of elements that act semisimply in the Adjoint representation.

Choose such an $H$. Simultaneous diagonalization gives the root-space decompositionThe nonzero weights $\alpha \in H^{\ast }$ are the roots. The restriction of the Killing form $B$ to $H$ is a nondegenerate bilinear form, so each $\alpha$ corresponds to a unique $t_{\alpha }\in H$ with $\alpha (h)=B(t_{\alpha },h)$. On the real span of these $t_{\alpha }$, the restriction of $B$ supplies a positive-definite inner product after choosing the standard real form. The sl2 subalgebra associated with a root givesand shows that these reflections preserve the finite set $\Phi$. Thus the roots form a finite reduced crystallographic root system, whose Weyl group is generated by these reflections.