= Solution
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> $d$:
$$
\kappa(X,Y)=\operatorname{Tr}_{\mathcal L}(\operatorname{ad}_X\operatorname{ad}_Y),\qquad \operatorname{ad}_X(Z)=[X,Z].
$$
The <Jacobi identity> gives $\operatorname{ad}_{[X,Y]}=[\operatorname{ad}_X,\operatorname{ad}_Y]$. Using the <cyclic property of the trace>,
$$
\begin{aligned}
\kappa([X,Y],Z)&=\operatorname{Tr}(\operatorname{ad}_X\operatorname{ad}_Y\operatorname{ad}_Z-\operatorname{ad}_Y\operatorname{ad}_X\operatorname{ad}_Z)\\
&=\operatorname{Tr}(\operatorname{ad}_X[\operatorname{ad}_Y,\operatorname{ad}_Z])=\kappa(X,[Y,Z]).
\end{aligned}
$$
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>: \b[$c_{abc}$ 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 $d$ is
$$
\boxed{C=\kappa^{ab}d(t_a)d(t_b).}
$$
This 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 give
$$
[d(X),C]=\kappa^{ab}\bigl(A^c{}_a d(t_c)d(t_b)+A^c{}_b d(t_a)d(t_c)\bigr)=0.
$$
Indeed, the coefficient of $d(t_c)d(t_b)$ in this expression is $(A\kappa^{-1}+\kappa^{-1}A^{\mathsf T})^{cb}$. Thus \b[$[d(X),C]=0$ 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> $\mathcal M=\mathfrak{gl}_n(\mathbb R)$ has
$$
\boxed{[E_{ij},E_{pq}]=\delta_{jp}E_{iq}-\delta_{qi}E_{pj},\qquad c_{ij,pq}{}^{mn}=\delta_{jp}\delta_i^m\delta_q^n-\delta_{qi}\delta_p^m\delta_j^n.}
$$
For its <Killing form>, write $L_X(Z)=XZ$ and $R_X(Z)=ZX$, so that $\operatorname{ad}_X=L_X-R_X$. Left multiplication acts identically on each of the $n$ columns, and right multiplication identically on each of the $n$ rows. Thus $\operatorname{Tr}(L_XL_Y)=\operatorname{Tr}(R_XR_Y)=n\operatorname{tr}(XY)$. The diagonal coefficient of $L_XR_Y$ on $E_{ij}$ is $X_{ii}Y_{jj}$, giving $\operatorname{Tr}(L_XR_Y)=\operatorname{tr}X\operatorname{tr}Y$; the other mixed trace is identical. This proves the <Killing form of the general linear Lie algebra>:
$$
\kappa(X,Y)=2n\operatorname{tr}(XY)-2\operatorname{tr}X\operatorname{tr}Y,\qquad \boxed{\kappa_{ij,pq}=2n\delta_{jp}\delta_{iq}-2\delta_{ij}\delta_{pq}.}
$$
The scalar matrices form a nonzero central abelian ideal, so \b[$\mathcal M$ is not semisimple for any $n\ge1$]. More explicitly, the <radical of the Killing form> is $\mathbb RI$, since vanishing against every $Y$ forces $nX-(\operatorname{tr}X)I=0$ by nondegeneracy of the trace pairing. For $n\ge2$, 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 $\kappa(X,X)\le0$. Therefore \b[$\mathcal M$ is not compact when $n\ge2$]. 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 $GL_1(\mathbb R)$ 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 is
$$
\boxed{\mathcal L'=\mathcal L,\qquad \dim\mathcal L'=\frac{n(n-1)}2.}
$$
This is the defined <radical of the Killing form>, not the derived algebra; for $n=1$ both $\mathcal L$ and this radical are zero.
Back to article page
