= Solution
For $X\in L$, define the <Adjoint representation of a Lie algebra> by $\operatorname{ad}_X(Y)=[X,Y]$. The <Jacobi identity> gives, for every $Z$,
$$
[\operatorname{ad}_X,\operatorname{ad}_Y]Z
=[X,[Y,Z]]-[Y,[X,Z]]
=[[X,Y],Z]=\operatorname{ad}_{[X,Y]}Z.
$$
Thus $\operatorname{ad}$ preserves brackets and is a <Lie algebra representation>. Its <Killing form> is
$$
\kappa(X,Y)=\operatorname{Tr}(\operatorname{ad}_X\operatorname{ad}_Y).
$$
Using the representation identity and the <cyclic property of the trace>,
$$
\begin{aligned}
\kappa([X,Y],Z)
&=\operatorname{Tr}([\operatorname{ad}_X,\operatorname{ad}_Y]\operatorname{ad}_Z)\\
&=\operatorname{Tr}(\operatorname{ad}_X[\operatorname{ad}_Y,\operatorname{ad}_Z])
=\kappa(X,[Y,Z]).
\end{aligned}
$$
The <Killing form> is also symmetric by cyclicity.
In a basis $T_a$, write $[T_a,T_b]=c_{ab}{}^dT_d$. The lowered <structure constants> satisfy
$$
c_{abc}=\kappa([T_a,T_b],T_c)=\kappa(T_a,[T_b,T_c]).
$$
The first expression is antisymmetric in $a,b$ and the second in $b,c$. These transpositions generate all permutations, so \b[$c_{abc}$ 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 \b[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>:
$$
H([X,Y],Z)=\operatorname{Tr}([D_X,D_Y]D_Z)
=\operatorname{Tr}(D_X[D_Y,D_Z])=H(X,[Y,Z]).
$$
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 $\operatorname{ad}_Z$, 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 $X$,
$$
H(X,X)=\operatorname{Tr}(D_X^2)
=-\operatorname{Tr}(D_X^\dagger D_X)<0.
$$
Hence the <positive trace index for a compact simple Lie algebra> gives
$$
\boxed{H_{ab}=-\mu\delta_{ab},\qquad\mu>0}
$$
in the adapted basis.
For the <cubic trace tensor of a Lie algebra representation>, the identity
$$
\operatorname{Tr}([D_Y,D_XD_ZD_W])=0
$$
expands to
$$
B([Y,X],Z,W)+B(X,[Y,Z],W)+B(X,Z,[Y,W])=0.
$$
Set $Y=T_d$, $X=T_a$, $Z=T_b$, $W=T_c$. 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
$$
\boxed{c_{da}{}^\ell B_{bc\ell}
+c_{db}{}^\ell B_{ca\ell}
+c_{dc}{}^\ell B_{ab\ell}=0.}
$$
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> imply
$$
\sum_{m,n}C_{am}{}^nC_{rm}{}^n=-\kappa_{ar}=\delta_{ar},
\qquad
c_a{}^{mn}c_{mn}{}^r=-\delta_a{}^r.
$$
Only the antisymmetric part of $B$ in $m,n$ contributes, so
$$
\begin{aligned}
c_a{}^{mn}B_{mnb}
&=\frac12c_a{}^{mn}\operatorname{Tr}([d(T_m),d(T_n)]d(T_b))\\
&=\frac12c_a{}^{mn}c_{mn}{}^rH_{rb}
=-\frac12H_{ab}.
\end{aligned}
$$
Consequently the <Killing-normalized contraction of a cubic trace tensor> is
$$
\boxed{c_a{}^{mn}B_{mnb}=+\frac{\mu}{2}\delta_{ab}.}
$$
The 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 $T_a=-i\sigma_a/(2\sqrt2)$, where the $\sigma_a$ are <Pauli matrices>. Then
$$
\kappa_{ab}=-\delta_{ab},\quad
H_{ab}=-\frac14\delta_{ab},\quad
c_a{}^{mn}=-\frac1{\sqrt2}\epsilon_{amn},\quad
B_{mnb}=-\frac1{8\sqrt2}\epsilon_{mnb}.
$$
Their contraction is $+\delta_{ab}/8$, confirming the positive sign with $\mu=1/4$ and providing a counterexample to the printed sign.
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