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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.

Work over $\mathbb{C}$. Decompose $V$ as the direct sum of the generalized eigenspaces $V_{\lambda }$ of $X$. Define $X_s$ to be $\lambda I$ on $V_{\lambda }$ and set $X_n=X-X_s$. Then $X_s$ is a diagonalisable endomorphism, $X_n$ is a nilpotent endomorphism, and they commute. For uniqueness, any commuting decomposition $X=S+N$ has $S$ and $N$ commuting with $X$, hence preserving each $V_{\lambda }$. Decompose $V_{\lambda }$ further into eigenspaces of $S$. On a nonzero such space with eigenvalue $\sigma$, the operator $X=\sigma I+N$ has only the eigenvalue $\sigma$, so $\sigma =\lambda$. Since $S$ is diagonalizable, $S=\lambda I$ throughout $V_{\lambda }$. Thus $S=X_s$ and $N=X_n$. This is the Additive Jordan decomposition.

The Chinese remainder theorem for the pairwise coprime polynomials $(t-\lambda {)}^{r_{\lambda }}$ also gives a polynomial $p$ with $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^{r_{\lambda }})$, where $r_{\lambda }$ is the largest Jordan block size. Consequently $X_s=p(X)$ and $X_n=X-p(X)$. This polynomial description shows that both parts preserve every $X$-invariant subspace.

On $\mathrm{Hom}(V_{\lambda },V_{\mu })$ the operator $\mathrm{ad}{X}_s$ is the scalar $\mu -\lambda$, so it is diagonalizable. The nilpotence of commutation by a nilpotent endomorphism makes $\mathrm{ad}{X}_n$ nilpotent. They commute, since $[\mathrm{ad}{X}_s,\mathrm{ad}{X}_n]=\mathrm{ad}[X_s,X_n]=0$. Uniqueness of the Additive Jordan decomposition therefore gives the adjoint compatibility of additive Jordan decomposition:Now let $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ be a complex semisimple Lie algebra and $X\in \mathfrak{g}$. Since the semisimple part of $\mathrm{ad}X$ is a polynomial in $\mathrm{ad}X$, it preserves $\mathfrak{g}$. Thus $[X_s,\mathfrak{g}]\subseteq \mathfrak{g}$. By the Weyl complete reducibility theorem, the Adjoint representation of $\mathfrak{g}$ on $\mathrm{End}(V)$ has a decomposition $\mathrm{End}(V)=\mathfrak{g}\oplus M$ into invariant subspaces. Write $X_s=y+z$ with $y\in \mathfrak{g}$ and $z\in M$. For $a\in \mathfrak{g}$, the vector $[z,a]=[X_s,a]-[y,a]$ lies in $\mathfrak{g}$, and invariance of $M$ puts it in $M$ as well. Hence $[z,a]=0$.

Decompose $V={⨁}_j{U}_j$ into Irreducible Lie algebra representations. Each $U_j$ is preserved by $X_s=p(X)$ and by $y$, hence by $z$. The Schur lemma makes $z{|}_{U_j}=c_{j}I$. A semisimple Lie algebra is a perfect Lie algebra, so every representing element of $\mathfrak{g}$ has zero trace on every $U_j$. Also $\mathrm{tr}(X_s{|}_{U_j})=\mathrm{tr}(X{|}_{U_j})$, because $X_n$ is nilpotent there. It follows that $c_j\dim U_j=\mathrm{tr}(z{|}_{U_j})=0$. In characteristic zero, $c_j=0$, so $z=0$. Therefore $X_s\in \mathfrak{g}$ and $X_n=X-X_s\in \mathfrak{g}$. This proves that semisimple matrix Lie algebras are closed under additive Jordan decomposition, including representations with repeated isomorphic irreducible summands.

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.

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.

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.

Let ${\alpha }_i^{\vee }=2{\alpha }_i/({\alpha }_i,{\alpha }_i)$ be the coroots. With a chosen set of simple roots, the root lattice and weight lattice areThe fundamental weights ${\omega }_j$ are defined by $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$, and form an integral basis of $P$. We use the Cartan matrix convention $A_{ij}=⟨{\alpha }_j,{\alpha }_i^{\vee }⟩$.

For the A2 root system, inversion of its Cartan matrix yields the A2 fundamental weights and weight lattice:Equivalently ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$. The root lattice has index three in the weight lattice, since the change-of-basis matrix has determinant three. For a planar realization takeThe weight lattice is triangular. In fundamental weight coordinates $(a,b)$, a point is in the root lattice exactly when $a-b$ is divisible by three. The following sketch marks both bases and the sublattice:

The integers called Dynkin indices here are the Dynkin labels of the highest weight:For finite-dimensional Irreducible Lie algebra representations, they are nonnegative integers. All weight coordinates below are these Dynkin label coordinates, not simple-root coordinates.

A weight-string enumeration algorithm gives the set of weights without their weight multiplicities. Begin with $\Lambda$ and process known weights by increasing height below $\Lambda$. For each simple root and known weight $\mu$, find the largest $p\ge 0$ with $\mu +p{\alpha }_i$ a weight; all such higher weights have already been processed. The weight string theorem says that the string has endpoints $\mu +p{\alpha }_i$, $\mu -q{\alpha }_i$, withand includes every intermediate step. Append $\mu -{\alpha }_i,\dots ,\mu -q{\alpha }_i$ and repeat until no new weights appear. This terminates in finite dimension and supplies all weights; every nonhighest weight can be reached by simple-root lowering. A string can contain contributions from several sl2 summands, so the resulting set does not by itself determine weight multiplicities.

For the explicit calculation we use the following general facts: finite-dimensional representations of a complex semisimple Lie algebra are completely reducible by the Weyl complete reducibility theorem; the symmetric powers of the defining sln representation are irreducible of highest weight $m{\omega }_1$; and weights in a tensor product of Lie algebra representations add with multiplicities multiplied. We also use the highest-weight representation classification: each finite-dimensional irreducible has a unique dominant integral highest weight, and a nonzero vector killed by all simple-root raising operators supplies an irreducible summand with that highest weight in a completely reducible module. For $A_2$, the Weyl dimension formula specializes toFor $A_2\cong {\mathfrak{sl}}_3(\mathbb{C})$, the defining fundamental representation has weights $(1,0),(-1,1),(0,-1)$. Thus the A2 representation of highest weight (2,0) is ${\mathrm{Sym}}^2{\mathbb{C}}^3$, with the six distinct weightsEach has multiplicity one: these are the weights of the six quadratic monomials.

For the tensor square of the A2 representation of highest weight (2,0), add every ordered pair of elements of $W_6$. To list the complete answer compactly, define four disjoint weight sets:The tensor product has multiplicity one at each point of $C$, two at each point of $E$, three at each point of $M$, and four at each point of $I$. These fifteen distinct weights account for $3+12+9+12=36$ states.

We next identify the irreducible summands rather than just their dimensions. Split the tensor square of the six-dimensional space into its symmetric power and exterior power, of dimensions $21$ and $15$. The square of a highest-weight vector in the symmetric part has highest weight $(4,0)$, giving $R(4,0)$ of dimension $15$.

Let $v_a,v_b$ be vectors of weights $a=(2,0)$ and $b=(0,1)=a-{\alpha }_1$. In the exterior part $v_a\wedge v_b$ has weight $(2,1)$ and is killed by both simple-root raising operators: raising $v_b$ along ${\alpha }_1$ gives a multiple of $v_a$, whose wedge with itself is zero, and the other raising actions vanish. Thus it is a highest-weight vector. The Weyl dimension formula gives $\dim R(2,1)=15$, so the entire exterior part is this irreducible summand.

The remaining symmetric part has dimension six. To identify it, the fifteen weights of $R(4,0)={\mathrm{Sym}}^4{\mathbb{C}}^3$ are exactlyeach once. Subtract them from the unordered-pair weights of ${\mathrm{Sym}}^{2}R(2,0)$. The residual weights are $M\cup I$, each once, with highest weight $(0,2)$. They are the negatives of $W_6$, so this six-dimensional summand is $R(0,2)$. ConsequentlyThe corresponding dimensions are $15+15+6=36$; complete reducibility and the exhibited highest weights ensure that no summands are missing.

The full weight multiplicities in the A2 tensor square of highest weight (2,0) are summarized below. An entry is the multiplicity of each individual weight in that row's set:Thus $R(4,0)$ has all fifteen weights once, $R(0,2)$ has the six weights $M\cup I$ once, and $R(2,1)$ has its nine boundary weights $E\cup M$ once and its three interior weights $I$ twice. Only the three interior weights of $R(2,1)$ are degenerate among the irreducible components.

Choose a positive root system $R^{+}$ for a complex semisimple Lie algebra, with Weyl group $W$ and Weyl vector $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$. For a dominant integral weight $\lambda$, let $L(\lambda )$ denote its finite-dimensional irreducible representation, and use formal symbols with $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. Its formal character of a weight module is $\mathrm{ch}L(\lambda )={\sum }_{\mu }\dim L(\lambda {)}_{\mu }{e}^{\mu }$. The Weyl character formula saysHere $\det (w)=(-1{)}^{\ell (w)}$, the sign of the product of simple root reflections. The Weyl denominator formula isThe quotient is initially interpreted in the fraction field of the group algebra of the weight lattice; the theorem makes it a finite character.

Use the real Lie-algebra convention of the question, with a gauge covariant derivative $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The gauge field strength is its curvature:Write $a_{\mu }={\delta }_X{A}_{\mu }/ϵ=-{\partial }_{\mu }X+[X,A_{\mu }]$. To first order in $ϵ$,Substitute $a_{\mu }$. The mixed second derivatives of $X$ cancel. The terms containing first derivatives of $X$ cancel in pairs. The remaining terms areThe Jacobi identity combines the last two into $[X,[A_{\mu },A_{\nu }]]$. ConsequentlyThe transformation is homogeneous even though the connection transformation contains an inhomogeneous derivative term.

For a finite-dimensional Lie algebra, define its Adjoint representation by ${\mathrm{ad}}_X(Y)=[X,Y]$ and its Killing form byIt is a symmetric bilinear form by cyclicity of the trace. The Jacobi identity gives ${\mathrm{ad}}_{[Z,X]}=[{\mathrm{ad}}_Z,{\mathrm{ad}}_X]$. Set $A={\mathrm{ad}}_Z$, $B={\mathrm{ad}}_X$, $C={\mathrm{ad}}_Y$. ThenThis proves the invariant bilinear form on a Lie algebra property, without assuming simplicity or nondegeneracy.

For definiteness use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and take a real compact semisimple Lie algebra as the gauge algebra. Its positive internal metric is $B=-\kappa$. A Killing-form Yang-Mills Lagrangian with the coupling absorbed into the connection isThe spacetime metric is unchanged by internal gauge transformations. HenceThus gauge invariance follows directly from invariance of the Killing form. Other overall conventions are possible, but the energy sign must be checked rather than inferred from a prefactor in isolation.

For physical kinetic terms the internal form must be real, nondegenerate and positive definite after choosing the overall sign. If ${\mathcal{E}}_i=F_{0i}$ and ${\mathcal{B}}_i=\frac{1}{2}{ϵ}_{ijk}{F}_{jk}$, the above convention has Lagrangian $[B({\mathcal{E}}_i,{\mathcal{E}}_i)-B({\mathcal{B}}_i,{\mathcal{B}}_i)]/(2g_{\mathrm{YM}}^2)$ and physical energy densityThe canonical Hamiltonian density also contains the nondynamical multiplier and a spatial divergence. With ${\Pi }_i={\mathcal{E}}_i/g_{\mathrm{YM}}^2$ and $D_i{\Pi }_i={\partial }_i{\Pi }_i+[A_i,{\Pi }_i]$, integration by parts using the invariant bilinear form on a Lie algebra givesThe Gauss law constraint in gauge theory sets $D_i{\Pi }_i=0$. If the boundary flux vanishes, or the appropriate boundary contribution is included, the integrated physical Hamiltonian is the positive energy displayed above.

An indefinite internal form would give gauge-field polarizations with opposite kinetic signs. A degenerate form would fail to supply a kinetic term for some directions. The compactness criterion from the Killing form says that negative-definiteness of the Killing form of a real finite-dimensional algebra is equivalent to compact semisimplicity; thus the pure Killing-form construction selects compact semisimple real forms. One cannot use a complex-bilinear Killing form on arbitrary complex field components as if it were a positive Hermitian metric.

This does not prohibit Abelian gauge theories. A compact Abelian factor has zero Killing form, so it needs a separately chosen positive invariant bilinear form on a Lie algebra, rather than the Killing form. More generally an algebra with a positive invariant metric on a Lie algebra is compact reductive, namely a direct sum of a compact semisimple algebra and an Abelian center. A noncompact group can also share the same compact Lie algebra through global covering choices in an Abelian factor; positivity is a statement about the algebra and internal metric, not by itself a classification of global gauge-group topology. Quantum matter anomalies and global restrictions would require additional input; no matter content is specified here.

For the finite matrix transformation, let $g(x)\in SU(N)$ and regard $g$ as a multiplication operator. The identity ${\partial }_{\mu }{g}^{-1}=-g^{-1}({\partial }_{\mu }g)g^{-1}$ givesTaking commutators of these differential operators cancels the adjacent multiplication operators $g^{-1}g$, yieldingThis argument keeps the derivatives acting on test fields and avoids treating $D_{\mu }$ as just a matrix.

There is a normalization issue in the printed last paragraph. The standard Killing form defined through the adjoint trace on $\mathfrak{su}(N)$ isnot simply $\mathrm{tr}(XY)$. For example, with $N=2$ and $X=Y=-i{\sigma }_3/2$, the defining trace is $-1/2$, whereas the adjoint trace is $-2$. The printed trace formula can be used as a rescaled invariant form, with the constant absorbed into the gauge coupling; it has exactly the invariance needed here. For either normalization,by cyclicity. This proves finite Yang-Mills gauge transformation invariance, with the healthy sign chosen for anti-Hermitian gauge fields. The normalization discrepancy is not a failure of gauge invariance.