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An alternating trilinear form is a scalar-valued map, linear in three arguments, which vanishes whenever two arguments agree. In characteristic zero, interchanging two arguments negates its value, while cyclic permutation preserves it. For a symmetric bilinear form $B$ on a Lie algebra, the tensor $B([x,y],z)$ is alternating exactly when $B$ is an invariant bilinear form on a Lie algebra.

A real Lie algebra is compact when it is the Lie algebra of some compact Lie group. Equivalently, it has a positive-definite invariant bilinear form on a Lie algebra. It is a direct sum of an abelian center and a compact semisimple Lie algebra. For its semisimple part, the compactness criterion from the Killing form is negative-definiteness of the Killing form. This definition concerns the algebra; a global group with that algebra can still have a noncompact central covering direction.

For Lie-algebra-valued fields and a constant invariant bilinear form on a Lie algebra, the connection terms in the displayed identity cancel. Indeed, invariance gives $\kappa ([A,Y],Z)+\kappa (Y,[A,Z])=0$. Ordinary integration by parts then yields $\int \kappa (D_{\mu }Y,Z)=-\int \kappa (Y,D_{\mu }Z)$ when boundary terms vanish, for example for compactly supported variations. This preserves covariance in variational calculations of Yang-Mills equations.

In $\mathcal{L}=-g_{ab}{F}_{\mu \nu }^a{F}^{b\mu \nu }/4$, only the symmetric part of $g$ contributes. A constant symmetric bilinear form gives local gauge invariance precisely when it is an invariant bilinear form on a Lie algebra, because the gauge field strength transforms in the Adjoint representation. A nondegenerate kinetic term requires a nondegenerate bilinear form; positivity of the Hamiltonian is an additional condition. For a compact simple real Lie algebra, $-\kappa$ is positive definite, and a positive multiple supplies the usual positive-energy convention with Minkowski signature $(+---)$.

For the Killing form or another invariant bilinear form on a Lie algebra,The connection terms cancel by invariance. Thus, for compactly supported variations, a gauge covariant derivative can be moved to the other factor in an integral with a minus sign, just as an ordinary derivative.

// Target: quantum-field-theory.bigb

An adjoint scalar has adjoint covariant derivative $D_{\mu }\varphi ={\partial }_{\mu }\varphi +[A_{\mu },\varphi ]$. A symmetric invariant bilinear form on a Lie algebra contracts the gauge field strength and scalar kinetic terms to give gauge invariance. On a compact real form the positive internal metric is $-\kappa$, because the Killing form itself is negative definite; complex scalar components require the associated Hermitian pairing. The Killing form is degenerate on Abelian factors, where another invariant metric is needed for a nondegenerate kinetic term.

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

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

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

For a finite-dimensional Lie algebra, the Killing form is defined using the Adjoint representation on the algebra itself, rather than the given Lie algebra representation $d$:The Jacobi identity gives ${\mathrm{ad}}_{[X,Y]}=[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]$. Using the cyclic property of the trace,The same trace identity makes the Killing form symmetric. For a basis $(t_a)$, the lowered structure constants of a Lie algebra are $c_{abc}=\kappa ([t_a,t_b],t_c)$. Antisymmetry of the Lie bracket gives $c_{abc}=-c_{bac}$, while invariance gives $c_{abc}=\kappa (t_a,[t_b,t_c])=-c_{acb}$. The two adjacent transpositions generate every permutation, proving the antisymmetry of Killing-lowered structure constants: $c_{abc}$ is totally antisymmetric, even if the Killing form is degenerate.

A semisimple Lie algebra has no nonzero solvable ideal; equivalently, its solvable radical is zero, or, in characteristic zero, its Killing form is nondegenerate. A real compact Lie algebra is the algebra of some compact Lie group, equivalently an algebra admitting a positive-definite invariant bilinear form on a Lie algebra. For a real semisimple Lie algebra the compactness criterion from the Killing form says precisely that $\kappa$ is negative-definite. A compact algebra with a nonzero abelian center has a degenerate Killing form, so the semisimplicity qualification matters.

For a semisimple Lie algebra, let ${\kappa }^{ab}$ denote the inverse matrix of ${\kappa }_{ab}=\kappa (t_a,t_b)$. Its Casimir operator in $d$ isThis contraction is independent of the chosen basis. To prove centrality directly, put $[X,t_b]=A^a{}_b{t}_a$. Invariance gives $A^{\mathsf{T}}\kappa +\kappa A=0$, hence $A{\kappa }^{-1}+{\kappa }^{-1}{A}^{\mathsf{T}}=0$. The Lie algebra representation property and the product rule for a commutator now giveIndeed, the coefficient of $d(t_c)d(t_b)$ in this expression is $(A{\kappa }^{-1}+{\kappa }^{-1}{A}^{\mathsf{T}}{)}^{cb}$. Thus $[d(X),C]=0$ for every $X$. No irreducibility of $d$ is required.

The matrix units obey $E_{ij}{E}_{pq}={\delta }_{jp}{E}_{iq}$. Consequently the general linear Lie algebra $\mathcal{M}={\mathfrak{gl}}_n(\mathbb{R})$ hasFor its Killing form, write $L_X(Z)=XZ$ and $R_X(Z)=ZX$, so that ${\mathrm{ad}}_X=L_X-R_X$. Left multiplication acts identically on each of the $n$ columns, and right multiplication identically on each of the $n$ rows. Thus $\mathrm{Tr}(L_X{L}_Y)=\mathrm{Tr}(R_X{R}_Y)=n\mathrm{tr}(XY)$. The diagonal coefficient of $L_X{R}_Y$ on $E_{ij}$ is $X_{ii}{Y}_{jj}$, giving $\mathrm{Tr}(L_X{R}_Y)=\mathrm{tr}X\mathrm{tr}Y$; the other mixed trace is identical. This proves the Killing form of the general linear Lie algebra:The scalar matrices form a nonzero central abelian ideal, so $\mathcal{M}$ is not semisimple for any $n\ge 1$. More explicitly, the radical of the Killing form is $\mathbb{R}I$, since vanishing against every $Y$ forces $nX-(\mathrm{tr}X)I=0$ by nondegeneracy of the trace pairing. For $n\ge 2$, take $H=E_{11}-E_{22}$. Then $\kappa (H,H)=4n>0$, whereas a positive-definite invariant inner product would make every adjoint map skew and hence give $\kappa (X,X)\le 0$. Therefore $\mathcal{M}$ is not compact when $n\ge 2$. For the allowed exceptional value $n=1$, the abstract abelian algebra $\mathbb{R}$ is a compact Lie algebra, being isomorphic to the algebra of $U(1)$; the particular matrix group $G{L}_1(\mathbb{R})$ is nevertheless noncompact. This distinguishes compactness of an algebra from that of a chosen group realizing it.

For the final algebra, the printed basis specifies strictly upper triangular matrices. If $i<j$ and $p<q$, the first term in $[E_{ij},E_{pq}]$ can be nonzero only when $i<j=p<q$, and then $E_{iq}$ is strictly upper triangular; the second can be nonzero only when $p<q=i<j$, and then $E_{pj}$ is strictly upper triangular. Bilinearity proves closure under the Lie bracket. To compute its own Killing form, grade $E_{ij}$ by the positive integer $j-i$. Bracketing with any basis element increases that degree whenever the result is nonzero. In a basis ordered by degree, all adjoint maps are strictly triangular, as are their products. Every such product has zero trace, so the Killing form of a nilpotent Lie algebra vanishes here. Therefore the subspace denoted by a prime in the PDF isThis is the defined radical of the Killing form, not the derived algebra; for $n=1$ both $\mathcal{L}$ and this radical are zero.

For a finite-dimensional Lie algebra, write $A_X=\mathrm{ad}X$ for its Adjoint representation. The Killing form is the symmetric bilinear formThe Jacobi identity gives $A_{[X,Y]}=[A_X,A_Y]$. Using the cyclic property of the trace,Thus it is an invariant bilinear form on a Lie algebra. If $[T_a,T_b]=f_{ab}{}^c{T}_c$, the adjoint matrix has entries $(A_{T_a}{)}^c{}_b=f_{ab}{}^c$. Its trace product yields the structure constants of a Lie algebra formulaA semisimple Lie algebra has zero solvable radical, equivalently no nonzero solvable ideal. In characteristic zero, the Cartan criterion for semisimplicity characterizes it by nondegeneracy of its Killing form. A real compact Lie algebra is the Lie algebra of a compact Lie group, equivalently one admitting a positive-definite invariant inner product. For a real semisimple algebra, the compactness criterion from the Killing form says compactness is equivalent to negative-definiteness of $\kappa$. An abelian compact-type algebra has zero Killing form, so this last characterization specifically concerns the semisimple case.

Let $R_a=\mathrm{ad}(T_a)$, where $\mathrm{ad}(X)Y=[X,Y]$. In the given basis, the Adjoint representation of a Lie algebra isThe Jacobi identity gives $[\mathrm{ad}X,\mathrm{ad}Y]=\mathrm{ad}[X,Y]$, so $[R_a,R_b]=f^c{}_{ab}{R}_c$, as required for a Lie algebra representation. The Killing form is the symmetric Trace form of a Lie algebra representationCyclicity of the trace proves its invariance directly:In components this is precisely $\boxed{{\kappa }_{dc}{f}^d{}_{ab}+{\kappa }_{bd}{f}^d{}_{ac}=0}$, so the Killing form is an invariant bilinear form on a Lie algebra.

Choose the infinitesimal gauge transformation convention $\delta \varphi =\omega \varphi$, with $\omega (x)={\omega }^a(x)t_a$, and write $A_{\mu }=A_{\mu }^a{t}_a$. A gauge covariant derivative must satisfy $D_{\mu }^{\prime }=U{D}_{\mu }{U}^{-1}$ for $U=1+\omega +O({\omega }^2)$; henceEquivalently $\delta A_{\mu }^a=-{\partial }_{\mu }{\omega }^a+f^a{}_{bc}{\omega }^b{A}_{\mu }^c$. Expanding $\delta (D_{\mu }\varphi )$ makes the cancellation transparent:Thus $\boxed{\delta (D_{\mu }\varphi )={\omega }^a{t}_a{D}_{\mu }\varphi }$: it transforms in the same representation as $\varphi$. In adjoint covariant derivative notation, $\delta A_{\mu }=-D_{\mu }^{\mathrm{ad}}\omega$.

The commutator of two gauge covariant derivatives defines the gauge field strength:ThusConjugating the derivative commutator gives $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$, so the gauge field strength transforms homogeneously in the Adjoint representation:Even if the matrices $t_a$ in a particular matter representation are not faithful, the displayed component formula defines the Lie-algebra-valued curvature independently of that representation.

Only $g_{(ab)}$ contributes to the Yang-Mills theory Lagrangian, since $F_{\mu \nu }^a{F}^{b\mu \nu }$ is symmetric in $a,b$. Thus take $g_{ab}$ to be a constant real symmetric bilinear form. Its infinitesimal variation isArbitrary local field strengths and gauge parameters give the necessary and sufficient invariant gauge kinetic form conditionAn antisymmetric part is unconstrained but contributes nothing. A nondegenerate kinetic term additionally needs a nondegenerate bilinear form; positive energy additionally needs a positive internal metric in the signature $(+---)$ convention. Neither extra condition follows solely from gauge invariance. For a compact simple real Lie algebra, $-\kappa$ is positive definite and is the standard choice up to a positive factor.

For the intended complex-simple setting, the Killing form is nondegenerate. Consequently write any invariant bilinear form on a Lie algebra uniquely as $g(X,Y)=\kappa (SX,Y)$. Invariance of both forms impliesfor every $X,Y,Z$, so $S$ commutes with the Adjoint representation. A subspace invariant under all adjoint maps is an ideal, hence simplicity makes the adjoint representation irreducible. By Schur lemma, $S=cI$, giving the uniqueness of an invariant bilinear form on a simple Lie algebra:The same conclusion holds for compact simple real algebras by complexification, with $c$ real; nondegeneracy excludes $c=0$, and positive energy selects $c<0$ in the stated convention.

There is a genuine real-simple exception to uniqueness of the Killing form if “simple” is read as an arbitrary real simple algebra. Regard ${\mathfrak{sl}}_2(\mathbb{C})$ as real. It is real simple: its complexification consists of two complex simple factors exchanged by conjugation, so a conjugation-stable ideal is zero or the whole algebra. Both $\mathrm{Re}{\kappa }_{\mathbb{C}}$ and $\mathrm{Im}{\kappa }_{\mathbb{C}}$ are real symmetric invariant forms, while ${\kappa }_{\mathbb{R}}=2\mathrm{Re}{\kappa }_{\mathbb{C}}$. With $H=\mathrm{diag}(1,-1)$, the Killing form of the special linear Lie algebra gives ${\kappa }_{\mathbb{C}}(H,H)=8$ and ${\kappa }_{\mathbb{C}}(H,iH)=8i$. Therefore the imaginary part cannot be a real multiple of ${\kappa }_{\mathbb{R}}$. The uniqueness conclusion requires the customary complex-simple or compact-real interpretation.

Let $B$ be the given invariant bilinear form on a Lie algebra. Choose a basis $x_1,\dots ,x_d$ and its $B$-dual basis $x^1,\dots ,x^d$, so $B(x_i,x^j)={\delta }_i^j$. The Casimir element isThe tensor ${\sum }_i{x}_i\otimes x^i$ is the inverse tensor of $B$, so it is independent of the chosen basis. Multiplication in the universal enveloping algebra therefore makes ${\Omega }_B$ independent of that choice as well.

For $z\in \mathfrak{g}$, write $[z,x_i]={\sum }_j{a}_{ji}{x}_j$. Invariance means $B([z,x],y)+B(x,[z,y])=0$, whence $[z,x^i]=-{\sum }_j{a}_{ij}{x}^j$. Thereforeby interchanging $i,j$ in the second sum. Since $\mathfrak{g}$ generates its universal enveloping algebra, ${\Omega }_B$ is central. This argument does not require adding a symmetry assumption to the given bilinear form.

For the sl2 Lie algebra, takeThe cyclic identity for the trace gives $B([Z,X],Y)+B(X,[Z,Y])=0$. The form is nondegenerate, since $B(e,f)=1$, $B(h,h)=2$, and all the other pairings except $B(f,e)$ are zero. The dual of $(e,f,h)$ is $(f,e,h/2)$, soBy the classification of finite-dimensional sl2 representations, an irreducible of dimension $n$ has a highest-weight vector $v$ with $ev=0$ and $hv=mv$, where $m=n-1$. Since $ef=fe+h$, one obtainsCentrality and the Schur lemma then give the answer on the entire representation:This Casimir eigenvalue for sl2 depends on normalization. Choosing the Killing form instead, $K=4B$, scales the dual basis and the Casimir element by $1/4$, giving $(n^2-1)I/8$.

A finite-dimensional complex Lie algebra is a semisimple Lie algebra when its solvable radical is zero, equivalently when it has no nonzero solvable ideals. Its Killing form is the symmetric bilinear formThe cyclic trace identity and ${\mathrm{ad}}_{[z,x]}=[{\mathrm{ad}}_z,{\mathrm{ad}}_x]$ giveThus the Killing form is an invariant bilinear form on a Lie algebra. In particular its radical $R=\{x:B_L(x,L)=0\}$ is an ideal.

We supply the trace argument needed for nondegeneracy rather than assuming the Cartan criterion for semisimplicity. The matrix form of the Cartan solvability criterion says: if $A\subseteq {\mathrm{End}}_{\mathbb{C}}(V)$ and $\mathrm{tr}(uv)=0$ for every $u\in [A,A]$, $v\in A$, then $A$ is solvable. To prove this direction, fix $u\in [A,A]$. Let $V={⨁}_{\lambda }{V}_{\lambda }$ be its generalized eigenspace decomposition, and define $s$ to act on $V_{\lambda }$ by the scalar $\overline{\lambda }$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, the semisimple part of ${\mathrm{ad}}_u$ has eigenvalue $\lambda -\mu$, whereas ${\mathrm{ad}}_s$ acts by $\overline{\lambda }-\overline{\mu }=\overline{\lambda -\mu }$. Polynomial interpolation on the finitely many eigenvalues, with all derivatives through the sizes of the nilpotent blocks set to zero, therefore givesHere the same difference always has the same conjugate, so the interpolation is consistent; the prescribed zero derivatives remove every nilpotent block. Since ${\mathrm{ad}}_u(A)\subseteq [A,A]$, it follows that $[s,A]\subseteq [A,A]$.

Write $u={\sum }_j[a_j,b_j]$. By cyclicity and the trace hypothesis,On the other hand, $\mathrm{tr}(us)={\sum }_{\lambda }\dim (V_{\lambda })|\lambda {|}^2$. Thus all eigenvalues of $u$ vanish, and every element of $[A,A]$ is nilpotent. The Engel theorem makes $[A,A]$ nilpotent as a Lie algebra, hence solvable; $A/[A,A]$ is abelian, so $A$ is solvable. This is the Conjugate-spectrum proof of Cartan solvability.

Apply this to the Killing radical $R$. For $x,y\in R$, the adjoint actions preserve $R$ and act as zero on $L/R$, because $R$ is an ideal. Computing traces in a basis adapted to $R$ givesThe matrix Lie algebra ${\mathrm{ad}}_R(R)$ consequently satisfies the trace criterion and is solvable. Its kernel is $Z(R)$, an abelian ideal; a central extension of a solvable algebra is solvable. Thus $R$ is solvable. This proves the reusable assertion that the Killing radical is a solvable ideal. Semisimplicity forces $R=0$, and hence $B_L$ is nondegenerate.

A Cartan subalgebra is a nilpotent subalgebra $H$ which is self-normalizing:For a complex semisimple algebra this is equivalently a maximal toral subalgebra. The nilpotent, self-normalizing definition permits a proof of the restricted nondegeneracy without first assuming the toral characterization.

Use the generalized-weight decomposition for a nilpotent Lie algebra for the adjoint action of $H$:For completeness, the stability underlying this decomposition follows directly from nilpotence of $H$. For fixed $h$, every ${\mathrm{ad}}_h$ on $H$ is nilpotent. If $T=\rho (h)$ and $S=\rho (y)$ in a finite-dimensional representation, then $({\mathrm{ad}}_T{)}^{r}S=0$ for sufficiently large $r$. The identityshows that $S$ preserves each generalized eigenspace of $T$. Starting with a basis of $H$, refine these primary decompositions successively; all summands remain $H$-invariant. Each resulting summand has only one eigenvalue for each basis element. The Lie theorem triangularizes the action on that summand, so those eigenvalues extend to a single linear character $\lambda$ on all of $H$. This proves the displayed decomposition and nilpotence of all shifted operators there.

Since $H$ is nilpotent, $H\subseteq L^0$. The space $L^0$ is a subalgebra: repeated use of the derivation rule for ${\mathrm{ad}}_h$ shows that the bracket of two generalized zero-eigenvectors is another such vector. If $L^0/H\ne 0$, the adjoint action of $H$ on this quotient consists entirely of nilpotent maps. The Engel theorem supplies a nonzero coset $x+H$ with $[H,x]\subseteq H$, contradicting self-normalization. Therefore the zero generalized weight space of a Cartan subalgebra is exactly $H$.

Finally, $B_L(H,L^{\lambda })=0$ when $\lambda \ne 0$. Choose $h\in H$ with $\lambda (h)\ne 0$ and put $T={\mathrm{ad}}_h$. On $H$, a power $T^m$ vanishes. On $L^{\lambda }$, $T$ is invertible. For $u\in H$ and $v\in L^{\lambda }$, write $v=T^{m}w$; invariance givesIf $u\in H$ is orthogonal to $H$, it is now orthogonal to every summand of $L$, so nondegeneracy of $B_L$ implies $u=0$. The restriction $B_L{|}_{H\times H}$ is nondegenerate. This establishes the nondegeneracy of the Killing form on a Cartan subalgebra for every Cartan subalgebra, without requiring a chosen root basis.

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:

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.

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.