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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.

A real Lie group is a finite-dimensional real smooth manifold with a group structure whose multiplication and inversion are smooth. Its tangent space at the identity element $e$ is the vector space $\mathfrak{g}=T_{e}G$ of velocities of smooth curves through $e$. For the general linear group ${\mathrm{GL}}_n(\mathbb{R})$ it identifies with $M_n(\mathbb{R})$. The matrix exponential and local matrix logarithm areThe exponential is globally defined, and each matrix $\exp X$ is invertible with inverse $\exp (-X)$. The logarithm here is a local map near $I$; it is not a globally defined inverse on every real invertible matrix. The differential of $\exp$ at zero is the identity, and the two power series are inverse locally.

For a general $G$, the assumed Exponential map of a Lie group with $d{\exp }_0=I$ has a smooth local inverse by the inverse function theorem, giving a local exponential chart. Define the Lie bracket from local group commutators byFor small $s,t$ the group commutator lies in that chart. More explicitly, the map $F(u,v)=\log (\exp u\exp v\exp (-u)\exp (-v))$ is smooth and has $F(u,0)=F(0,v)=0$. Its mixed second differential at $(0,0)$ is a bilinear map of $X,Y$, proving bilinearity. Swapping the two group elements inverts the commutator, and $\log (g^{-1})=-\log g$ near $e$. Thus $F(v,u)=-F(u,v)$ and $[Y,X]=-[X,Y]$. In the matrix group, expansion to the mixed term $st$ gives the familiar commutator $XY-YX$.

For $g\in G$, let $c_g(h)=gh{g}^{-1}$ and define the Adjoint representation of a Lie group by ${\mathrm{Ad}}_g=d(c_g{)}_e\in \mathrm{GL}(\mathfrak{g})$. It is smooth and satisfies ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$. Its derived representation is $\mathrm{ad}=d(\mathrm{Ad}{)}_e:\mathfrak{g}\to \mathrm{End}(\mathfrak{g})$. Naturality of the Exponential map of a Lie group under conjugation givesAt $t=0$, the derivative of the logarithm of the commutator is ${\mathrm{Ad}}_{\exp (sX)}Y-Y$: the derivative of multiplication at $(e,e)$ adds tangent vectors, and $d{\log }_e=I$. Differentiate in $s$ to concludeTo prove the Jacobi identity without assuming it in the bracket construction, first establish that the differential of a Lie group homomorphism preserves Lie brackets. For a homomorphism $f:G\to H$, the allowed exponential identity gives ${\log }_H(f({\exp }_{G}u))=d{f}_e(u)$ locally. Apply this identity to the group commutator and take the mixed derivative to obtain $d{f}_e([X,Y])=[d{f}_{e}X,d{f}_{e}Y]$. In particular $f=\mathrm{Ad}:G\to \mathrm{GL}(\mathfrak{g})$ givesApplying both sides to $Z$ yields $[[X,Y],Z]=[X,[Y,Z]]-[Y,[X,Z]]$. Rearranging with antisymmetry provesNeither injectivity of $\mathrm{ad}$ nor the existence of a global logarithm was used.

Finally let $H$ be a normal Lie subgroup of $G$, with its corresponding Lie algebra $\mathfrak{h}\subseteq \mathfrak{g}$. For each $g\in G$, conjugation restricts to $H$, so its differential preserves $\mathfrak{h}$: ${\mathrm{Ad}}_g\mathfrak{h}=\mathfrak{h}$. Differentiate along $g=\exp (tX)$ to get $[X,Y]=\mathrm{ad}(X)Y\in \mathfrak{h}$ for every $X\in \mathfrak{g}$ and $Y\in \mathfrak{h}$. Thus the Lie algebra of a normal Lie subgroup is an ideal of a Lie algebra. The assertion concerns a subgroup carrying the corresponding Lie-subgroup structure, for example any closed subgroup; no arbitrary abstract subgroup is being assigned a tangent space.

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.

Schur's lemma says that a nonzero intertwiner between irreducible representations is an isomorphism; over $\mathbb{C}$, every endomorphism of a finite-dimensional irreducible representation is scalar. To prove the Schur lemma, let $T:V\to W$ be a intertwining operator between irreducible representations. Its kernel and image of a linear map are invariant. If $T\ne 0$, irreducibility forces $\ker T=0$ and $\mathrm{im}T=W$. This proves the first assertion over any field. In particular the endomorphisms of an irreducible representation form a division ring.

When $V$ is finite-dimensional over an algebraically closed field, an endomorphism $T$ has an eigenvalue $\lambda$. The endomorphism $T-\lambda I$ has nonzero kernel. By the first assertion it must be zero, so $T=\lambda I$. Consequently, for complex finite-dimensional irreducible representations, the space of intertwining operators has dimension zero for nonisomorphic representations and dimension one for isomorphic representations.

Every finite-dimensional representation of a complex semisimple Lie algebra is completely reducible. We prove the Weyl complete reducibility theorem using the allowed Casimir operator facts, without assuming a splitting in advance. For dual bases $x_i,x^i$ with respect to the Killing form, the Casimir elementis central in the universal enveloping algebra. Hence its Casimir operator $C_M={\sum }_i{\rho }_M(x_i){\rho }_M(x^i)$ commutes with the action on every Lie algebra representation $M$ and is compatible with subrepresentations, quotient representations, and intertwining operators. On the trivial Lie algebra representation it acts by zero. On every nontrivial finite-dimensional Irreducible Lie algebra representation it acts by a nonzero scalar.

For clarity, the last fact can be expressed by the Casimir eigenvalue formula: on an irreducible with dominant integral weight $\lambda$, the scalar is $(\lambda ,\lambda +2\rho )$, with the inner product induced by the Killing form and $\rho$ the half-sum of positive roots. On the real span of the weights this inner product is positive definite, and the scalar is positive for $\lambda \ne 0$. For a semisimple Lie algebra with several simple factors the scalars add, so a nontrivial representation still gives a nonzero scalar. These are properties of the Casimir operator being used here.

First establish Casimir splitting of a trivial quotient. Supposeis a short exact sequence of finite-dimensional Lie algebra representations, with trivial quotient. The generalized eigenspaces of $C_E$ are invariant, soEach $E_c$ with $c\ne 0$ maps to zero under $q$: applying a sufficiently large power of $C_E-cI$ and using $C_{\mathbb{C}}=0$ gives $(-c{)}^{r}q(e)=0$. Thus $q(E_0)=\mathbb{C}$.

Take a composition series of a module of $E_0$. The Casimir operator is nilpotent on $E_0$, so its scalar on every irreducible composition factor is zero. The stated Casimir operator property makes every such factor trivial. In a basis adapted to the composition series of a module, the image of $\mathfrak{g}$ on $E_0$ therefore consists of strictly upper triangular matrices, so that image is solvable. The allowed fact that a semisimple Lie algebra acts trivially on every one-dimensional representation implies that $\mathfrak{g}$ is a perfect Lie algebra: otherwise a nonzero linear functional on $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ would define a nontrivial one-dimensional Lie algebra representation. Thus $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, and its image is consequently a perfect Lie algebra too. A perfect Lie algebra that is also a solvable Lie algebra is zero, since its derived series of a Lie algebra is constant until it vanishes. Hence $\mathfrak{g}$ acts trivially on $E_0$. Choose $e\in E_0$ with $q(e)=1$. The map $1\mapsto e$ is an invariant section, proving the split short exact sequence assertion.

Now let $W\subseteq V$ be any nonzero subrepresentation. On the Hom representation $\mathrm{Hom}(V,W)$ the action isConsider the invariant subspaceRestriction produces a short exact sequenceSurjectivity follows by extending $I_W$ to a linear map on $V$. The quotient is trivial, because $I_W$ commutes with the action on $W$. The preceding Casimir splitting of a trivial quotient yields an invariant $p:V\to W$ with $p{|}_W=I_W$. Thusand $\ker p$ is invariant. The zero subrepresentation also has a complement. Repeatedly splitting off an irreducible subrepresentation now gives a direct sum of irreducibles, proving the Weyl complete reducibility theorem.

A derivation of a Lie algebra is a linear map $D:\mathfrak{g}\to \mathfrak{g}$ satisfying the Leibniz ruleThe space $\mathrm{Der}(\mathfrak{g})$ is a vector subspace of $\mathrm{End}(\mathfrak{g})$. Equip it with the commutator $[D,E]=DE-ED$. Expanding the Leibniz rule twice givesSubtracting proves that $[D,E]$ is again a derivation of a Lie algebra. Antisymmetry and the Jacobi identity hold for the commutator in every associative endomorphism algebra, so this defines the derivation Lie algebra.

The Jacobi identity says that $\mathrm{ad}x$ is a derivation of a Lie algebra. Moreover, for $D\in \mathrm{Der}(\mathfrak{g})$,soThis makes $\mathrm{ad}\mathfrak{g}$ an ideal of a Lie algebra in $\mathrm{Der}(\mathfrak{g})$.

Finally suppose $\mathfrak{g}$ is semisimple. Let $\mathfrak{g}$ act on $\mathrm{Der}(\mathfrak{g})$ by $x\cdot D=[\mathrm{ad}x,D]$. The Weyl complete reducibility theorem supplies an invariant complement $M$ to $\mathrm{ad}\mathfrak{g}$. For $D\in M$, invariance gives $[\mathrm{ad}x,D]\in M$, while the ideal identity above gives $[\mathrm{ad}x,D]\in \mathrm{ad}\mathfrak{g}$. Their intersection is zero, so $\mathrm{ad}(Dx)=0$ for every $x$. The center of a Lie algebra of a semisimple Lie algebra is zero, hence $Dx=0$ for every $x$, and $M=0$. ThereforeEvery derivation of a Lie algebra is inner, and the element giving it is unique because the center of a Lie algebra vanishes.

For the Euclidean motion Lie algebra in two dimensions, use the ordered basis $(J,E_1,E_2)$ and keep the printed rotation signs. The Adjoint representation matrices areOnly the square of ${\mathrm{ad}}_J$ has a nonzero trace. Hence the Killing form of the planar Euclidean motion Lie algebra has matrixIts radical of the Killing form is $\mathrm{span}\{E_1,E_2\}$, precisely the translation subspace. This is a nonzero abelian ideal of a Lie algebra, so the Lie algebra is neither simple nor semisimple. In fact its first derived series of a Lie algebra term is this translation ideal and its next is zero. Thus the degeneracy and the failure of semisimplicity agree with the structural proofs above; the adjoint action nevertheless remains nontrivial.

The Pauli matrix commutator identity givesfor the anti-Hermitian basis of the SU(2) Lie algebra. In the ordered basis $(T^1,T^2,T^3)$ the Adjoint representation matrices, with columns recording the images of basis elements, areTheir squared matrix traces are $-2$, and the mixed products have zero trace. Therefore the Killing form of the SU(2) Lie algebra isIt is negative definite and nondegenerate, so this real Lie algebra is semisimple. It is also simple: the displayed bracket is the three-dimensional cross product; if an ideal of a Lie algebra contains nonzero $x$, the vectors $[y,x]$ span the plane perpendicular to $x$, and together with $x$ span the whole algebra. This confirms that the simplicity implication applies with a nonzero Killing form.

A Lie algebra representation on a vector space $V$ is a linear map $R:\mathfrak{g}\to \mathrm{End}(V)$ preserving the Lie bracket:The Adjoint representation acts on $V=\mathfrak{g}$ by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity givesIf $\mathfrak{g}$ is nonabelian, some $[X,Y]$ is nonzero, so this Adjoint representation is not the trivial Lie algebra representation. It is the required nontrivial representation of dimension $\dim \mathfrak{g}$. Nontriviality does not require its homomorphism to be injective.

For the finite-dimensional algebras here, with normalization one, the Killing form isIt is a symmetric bilinear form by cyclicity of the matrix trace. To prove its invariance, write $A={\mathrm{ad}}_X$, $B={\mathrm{ad}}_Y$, $C={\mathrm{ad}}_Z$. ThenThis also gives $\kappa ([X,Y],Z)=\kappa (X,[Y,Z])$, the equivalent invariant bilinear form on a Lie algebra identity.

For the unheaded structural requests, a simple Lie algebra is nonabelian and has no ideals of a Lie algebra other than zero and itself. A semisimple Lie algebra has no nonzero solvable ideal of a Lie algebra; in finite dimension over $\mathbb{R}$ or $\mathbb{C}$ this is equivalently a direct sum of simple Lie algebras.

Suppose first that the Killing form is nondegenerate. If $I$ is an abelian ideal of a Lie algebra and $X\in I$, then ${\mathrm{ad}}_X$ maps $\mathfrak{g}$ into $I$ and vanishes on $I$. Every ${\mathrm{ad}}_Y$ preserves $I$. Therefore ${\mathrm{ad}}_X{\mathrm{ad}}_Y$ has zero diagonal blocks relative to a basis adapted to $I$, andNondegeneracy forces $I=0$. If a nonzero solvable ideal of a Lie algebra existed, its last nonzero derived series of a Lie algebra term would be a nonzero abelian ideal of a Lie algebra, again impossible. Hence the solvable radical is zero, provingThis argument proves the needed implication rather than assuming the Cartan criterion for semisimplicity.

One can also see the direct-sum formulation explicitly through orthogonal ideal splitting for a nondegenerate Killing form. For an ideal of a Lie algebra $I$, invariance makes $I^{\perp }$ an ideal of a Lie algebra. If $X\in I\cap I^{\perp }$, then $\kappa ([X,Y],Z)=\kappa (X,[Y,Z])=0$ for $Y\in I$, $Z\in \mathfrak{g}$, so $[X,I]=0$. Thus $I\cap I^{\perp }$ is an abelian ideal of a Lie algebra, and must vanish. Consequently $\mathfrak{g}=I\oplus I^{\perp }$ and the two summands commute. Select a minimal nonzero ideal; it is nonabelian, and any ideal inside it is an ideal of $\mathfrak{g}$ because the complementary summand commutes with it. It is therefore simple. The restricted form is the remaining summand’s own Killing form, because the two ideals commute. Repeating the splitting there terminates in a direct sum of simple ideals.

Conversely the radical of the Killing formis an ideal of a Lie algebra by the invariance just proved. For a simple Lie algebra it is either zero or the whole algebra. The permitted hypothesis that $\kappa$ is not identically zero excludes the second alternative. ThusThe computations in the two following parts illustrate both a nondegenerate compact example and a degenerate algebra with an abelian ideal.

For a finite-dimensional complex Lie algebra $\mathfrak{g}$, the following three conditions are equivalent descriptions of a semisimple Lie algebra.

First, its solvable radical is zero. An ideal of a Lie algebra is a linear subspace $I$ satisfying $[\mathfrak{g},I]\subseteq I$. A solvable Lie algebra is one whose derived series, defined by $I^{(0)}=I$ and $I^{(r+1)}=[I^{(r)},I^{(r)}]$, eventually becomes zero. The solvable radical is the largest solvable ideal of a Lie algebra; its vanishing is equivalent to the absence of nonzero solvable ideals.

Second, the Killing formis a nondegenerate bilinear form: if $\kappa (u,v)=0$ for every $v$, then $u=0$. Here $\mathrm{ad}u(v)=[u,v]$ is the Adjoint representation, and $\mathrm{tr}$ is the trace. This equivalence is the Cartan criterion for semisimplicity.

Third, $\mathfrak{g}$ is a direct sum of Lie algebras that are nonabelian simple Lie algebras. A simple Lie algebra is nonabelian and has no ideals of a Lie algebra other than zero and itself. Thus the three equivalent criteria are

The derived series gives a direct test, with no structural theorem needed. By bilinearity and antisymmetry of the Lie bracket, every bracket in $\mathfrak{h}$ is a multiple of $z$, and $z=[x,y]$ actually occurs. ThereforeConsequently this Heisenberg Lie algebra is a nonzero solvable Lie algebra. It is an ideal of a Lie algebra in itself, since $[\mathfrak{h},\mathfrak{h}]\subseteq \mathfrak{h}$. Its solvable radical is therefore all of $\mathfrak{h}$, contradicting the first criterion in part (a). Hence

The Dynkin diagram in part (b) is a three-node chain, hence $D_3=A_3$. By the classification of complex semisimple Lie algebras by their root systems, this gives the Isomorphism between so6 and sl4:There is also a concrete realization. On $U={\mathbb{C}}^4$, choose a nonzero alternating four-form $\mathrm{vol}$. The six-dimensional exterior square ${\Lambda }^{2}U$ carries the nondegenerate symmetric bilinear form $B$ specified byIt is symmetric because exchanging two pairs entails four transpositions, and it is nondegenerate because each basis bivector pairs nontrivially with its complementary pair. The special linear Lie algebra acts by $A(u\wedge v)=Au\wedge v+u\wedge Av$. Its action on ${\Lambda }^{4}U$ is multiplication by $\mathrm{tr}A=0$, so it preserves $B$. We obtain a nonzero Lie algebra homomorphism ${\mathfrak{sl}}_4\to \mathfrak{so}({\Lambda }^{2}U,B)$. Its kernel is an ideal of a Lie algebra, hence zero because ${\mathfrak{sl}}_4$ is a simple Lie algebra. Both sides have dimension $15$, so this homomorphism is a Lie algebra isomorphism.

Work over $\mathbb{R}$ or $\mathbb{C}$, in finite dimension and characteristic zero. The Killing form of a Lie algebra is the trace form of its Adjoint representation:It is bilinear and symmetric by the cyclic property of the trace. It is an invariant bilinear form on a Lie algebra, since the Jacobi identity gives ${\mathrm{ad}}_{[x,y]}=[{\mathrm{ad}}_x,{\mathrm{ad}}_y]$ and henceEquivalently, $\kappa ([z,x],y)+\kappa (x,[z,y])=0$. A Lie algebra automorphism preserves the form, because it conjugates the adjoint matrices. Its radical of a bilinear form ${\mathfrak{r}}_{\kappa }=\{x:\kappa (x,\mathfrak{g})=0\}$ is an ideal of a Lie algebra by invariance. The center of a Lie algebra lies in this radical, so the form vanishes for an abelian algebra. On a direct sum of ideals the summands are orthogonal and the form restricts to their own Killing forms.

A short argument proves the requested implication without assuming the Cartan criterion for semisimplicity. Let $\mathfrak{a}$ be an abelian ideal of a Lie algebra, $x\in \mathfrak{a}$, and $y\in \mathfrak{g}$. The map ${\mathrm{ad}}_x$ takes $\mathfrak{g}$ into $\mathfrak{a}$ and vanishes on $\mathfrak{a}$, while ${\mathrm{ad}}_y$ preserves $\mathfrak{a}$. Therefore ${\mathrm{ad}}_x{\mathrm{ad}}_y$ has image in $\mathfrak{a}$ and is zero on $\mathfrak{a}$. In a basis extending a basis of $\mathfrak{a}$, both diagonal blocks are zero, so its trace vanishes. This proves that abelian ideals lie in the radical of the Killing form.

If the solvable radical $\mathfrak{r}$ were nonzero, its derived series of a Lie algebra would have a last nonzero term $\mathfrak{a}$. The Jacobi identity makes every derived term an ideal of $\mathfrak{g}$, and the last one is abelian. Thus $0\ne \mathfrak{a}\subseteq {\mathfrak{r}}_{\kappa }$, contradicting nondegeneracy. We obtainThe converse holds as well in characteristic zero; together these implications are the Cartan criterion for semisimplicity.

For a complex simple Lie algebra, that criterion gives nondegeneracy on $\mathfrak{g}$. Let $\mathfrak{h}$ be a Cartan subalgebra. Use the standard root-space decompositionFor $h,h^{\prime }\in \mathfrak{h}$ and a root vector $e_{\alpha }$, choose $t\in \mathfrak{h}$ with $\alpha (t)\ne 0$. Invariance givesThus $\mathfrak{h}$ is orthogonal to every nonzero root space. If $h\in \mathfrak{h}$ is also orthogonal to $\mathfrak{h}$, it is orthogonal to all of $\mathfrak{g}$ and hence is zero. This establishes nondegeneracy of the Killing form on a Cartan subalgebra:The same invariance calculation shows that $\kappa ({\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta })=0$ unless $\alpha +\beta =0$. Opposite root spaces are therefore paired nondegenerately. Tracing the adjoint action on the root-space decomposition yieldssince the root spaces of a complex semisimple algebra are one-dimensional and its adjoint action on $\mathfrak{h}$ is zero.

The relevant Euclidean subspace of a Cartan subalgebra iswhere the $h_{{\alpha }_i}$ are the simple coroots, normalized by ${\alpha }_i(h_{{\alpha }_i})=2$. The standard Euclidean property is that $\kappa$ is real and positive definite on this space. The displayed trace formula explains it: $\kappa (h,h)$ is a sum of real squares, and the roots span ${\mathfrak{h}}^{\ast }$, so all squares vanish only for $h=0$. Via the nondegenerate form, the roots can consequently be regarded as vectors in a real Euclidean normed vector space. The induced dual inner product makes root reflections orthogonal and allows root lengths and angles to be encoded by Cartan integers and the Dynkin diagram.

This positivity is on a specified real subspace; the complex Killing form is a bilinear form, not a positive Hermitian inner product. On a compact real form the Killing form is instead negative definite. For example the Killing form of the special linear Lie algebra gives $\kappa (X,Y)=4\mathrm{tr}(XY)$ for ${\mathfrak{sl}}_2$, so $\kappa (L_0,L_0)=8$ but $\kappa (i{L}_0,i{L}_0)=-8$.