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For $X\in L$, define the Adjoint representation of a Lie algebra by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity gives, for every $Z$,Thus $\mathrm{ad}$ preserves brackets and is a Lie algebra representation. Its Killing form isUsing the representation identity and the cyclic property of the trace,The Killing form is also symmetric by cyclicity.

In a basis $T_a$, write $[T_a,T_b]=c_{ab}{}^d{T}_d$. The lowered structure constants satisfyThe first expression is antisymmetric in $a,b$ and the second in $b,c$. These transpositions generate all permutations, so $c_{abc}$ is totally antisymmetric.

An invariant subspace of $L$ under its Adjoint representation of a Lie algebra is exactly an ideal of a Lie algebra: $[L,I]\subseteq I$. A simple Lie algebra has no nonzero proper ideal, proving the adjoint representation is irreducible.

Put $D_X=d(X)$. The same trace calculation proves invariance of the Trace form of a Lie algebra representation:Because the $D_X$ are anti-Hermitian, $H$ is a real symmetric bilinear form. The Killing form of a simple compact Lie algebra is negative definite. Define a real linear map $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance of both forms implies that $T$ commutes with every ${\mathrm{ad}}_Z$, while symmetry makes $T$ self-adjoint for the positive inner product $-\kappa$. By the spectral theorem, its real eigenspaces are invariant under the Adjoint representation of a Lie algebra. Irreducibility forces a single eigenvalue, so $T=\mu I$ and $H=\mu \kappa$. This also avoids the extra care needed when applying the complex form of the Schur lemma to a real vector space.

The kernel of $d$ is an ideal of a Lie algebra. It cannot be all of $L$: a trivial representation of dimension greater than one is reducible. Simplicity therefore makes $d$ faithful. For every nonzero $X$,Hence the positive trace index for a compact simple Lie algebra givesin the adapted basis.

For the cubic trace tensor of a Lie algebra representation, the identityexpands toSet $Y=T_d$, $X=T_a$, $Z=T_b$, $W=T_c$. Cyclicity of $B$ converts $B_{\ell bc}$ to $B_{bc\ell }$ and $B_{a\ell c}$ to $B_{ca\ell }$. Thus the invariance identity for a cubic trace tensor is

For the final contraction, it is important to keep the specified index-raising convention. In the adapted basis write $C_{ab}{}^c=c_{ab}{}^c$. Then $c_{abc}=-C_{ab}{}^c$, while raising the last two indices with ${\kappa }^{ab}=-{\delta }^{ab}$ gives $c_a{}^{mn}=-C_{am}{}^n$. Total antisymmetry and the definition of the Killing form implyOnly the antisymmetric part of $B$ in $m,n$ contributes, soConsequently the Killing-normalized contraction of a cubic trace tensor isThe negative sign printed in the last requested identity is incompatible with raising indices by the inverse Killing form. The opposite sign is obtained if one instead defines the lowered structure constants using the positive Euclidean metric, a different convention from the one specified.

For a concrete check, take the two-dimensional representation of $\mathfrak{su}(2)$ with $T_a=-i{\sigma }_a/(2\sqrt{2})$, where the ${\sigma }_a$ are Pauli matrices. ThenTheir contraction is $+{\delta }_{ab}/8$, confirming the positive sign with $\mu =1/4$ and providing a counterexample to the printed sign.

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.

In the ordered basis $(K_1,K_2,J)$, the three Adjoint representation matrices areTheir cross-product traces vanish, while their squared traces are $-2\lambda$, $-2\lambda$ and $-2{\lambda }^2$. ConsequentlyThis is the Killing form for cyclic three-generator brackets with constants $(1,\lambda ,\lambda )$.

For $\lambda >0$, the form is nondegenerate and negative-definite. The algebra is semisimple and compact, isomorphic to the SU(2) Lie algebra. For $\lambda <0$, it is nondegenerate with signature $(2,1)$, so the algebra is semisimple and noncompact, isomorphic to ${\mathfrak{sl}}_2(\mathbb{R})$.

For $\lambda =0$, the only nonzero bracket is $[K_1,K_2]=J$, with $J$ central. This is the Heisenberg Lie algebra, which is nilpotent and neither semisimple nor compact. Compactness is ruled out even though the Killing form is merely degenerate: $A_{K_1}$ is nonzero and nilpotent, whereas in a compact Lie algebra every adjoint map is skew-adjoint for a positive inner product. Skew-adjoint nilpotent operators vanish.

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.

For a nontrivial finite-dimensional anti-Hermitian Lie algebra representation of a compact Lie algebra that is simple, its Trace form of a Lie algebra representation is a positive multiple of the Killing form. Define $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance makes $T$ commute with the Adjoint representation of a Lie algebra. Symmetry makes $T$ self-adjoint for $-\kappa$, so its eigenspaces are ideals. Simplicity implies $T=\mu I$. The kernel of the representation is an ideal and hence vanishes; therefore $H(X,X)=-\mathrm{Tr}(d(X{)}^{†}d(X))<0$ for nonzero $X$, proving $\mu >0$.

// Target: lie-theory.bigb

On a finite-dimensional positive inner-product space, a skew-adjoint nilpotent operator is zero. In an orthonormal basis, $\Vert A{\Vert }_F^2=\mathrm{tr}(A^{\ast }A)=-\mathrm{tr}(A^2)=0$, since a nilpotent matrix has zero trace and so does its square. The Frobenius norm therefore vanishes. In particular, a nonzero nilpotent adjoint map rules out a compact Lie algebra.