= Solution
A <Lie algebra representation> on a <vector space> $V$ is a linear map $R:\mathfrak g\to\operatorname{End}(V)$ preserving the <Lie bracket>:
$$
R([X,Y])=[R(X),R(Y)].
$$
The <Adjoint representation> acts on $V=\mathfrak g$ by $\operatorname{ad}_X(Y)=[X,Y]$. The <Jacobi identity> gives
$$
[\operatorname{ad}_X,\operatorname{ad}_Y]=\operatorname{ad}_{[X,Y]}.
$$
If $\mathfrak g$ is nonabelian, some $[X,Y]$ is nonzero, so this <Adjoint representation> is not the <trivial Lie algebra representation>. \b[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> is
$$
\boxed{\kappa(X,Y)=\operatorname{tr}(\operatorname{ad}_X\operatorname{ad}_Y).}
$$
It is a <symmetric bilinear form> by cyclicity of the <matrix trace>. To prove its invariance, write $A=\operatorname{ad}_X$, $B=\operatorname{ad}_Y$, $C=\operatorname{ad}_Z$. Then
$$
\begin{aligned}
\kappa([Z,X],Y)+\kappa(X,[Z,Y])&=\operatorname{tr}([C,A]B+A[C,B])\\
&=\operatorname{tr}(CAB-ABC)=0.
\end{aligned}
$$
This 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 $\operatorname{ad}_X$ maps $\mathfrak g$ into $I$ and vanishes on $I$. Every $\operatorname{ad}_Y$ preserves $I$. Therefore $\operatorname{ad}_X\operatorname{ad}_Y$ has zero diagonal blocks relative to a basis adapted to $I$, and
$$
\kappa(X,Y)=0\qquad(X\in I,\ Y\in\mathfrak g).
$$
Nondegeneracy 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, proving
$$
\boxed{\kappa\text{ nondegenerate}\quad\Longrightarrow\quad\mathfrak g\text{ semisimple}.}
$$
This 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 form>
$$
K=\{X:\kappa(X,Y)=0\text{ for all }Y\}
$$
is 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. Thus
$$
\boxed{\mathfrak g\text{ simple and }\kappa\not\equiv0\quad\Longrightarrow\quad\kappa\text{ nondegenerate}.}
$$
The computations in the two following parts illustrate both a nondegenerate compact example and a degenerate algebra with an abelian ideal.
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