= Solution
Define the raw <Killing form> by
$$
\boxed{K(X,Y)=\operatorname{Tr}(\operatorname{ad}X\operatorname{ad}Y).}
$$
For a finite-dimensional real or complex <Lie algebra> in characteristic zero, it is nondegenerate exactly when the algebra is a <semisimple Lie algebra>. For a real compact semisimple algebra it is negative definite. Consequently the positive compact Killing metric is $g=-K$: for example on $\mathfrak{su}(2)$ with $[e_i,e_j]=\varepsilon_{ijk}e_k$, $K_{ij}=-2\delta_{ij}$ and $-K$ gives a <Riemannian metric> on $SU(2)$. An overall minus sign in the term “Killing metric” is common here; raw trace $K$ itself is negative definite in this example.
For a <Lorentzian metric> use $\mathfrak{sl}(2,\mathbb R)$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. Direct traces give
$$
K_{(h,e,f)}=\begin{pmatrix}8&0&0\\0&0&4\\0&4&0\end{pmatrix},
$$
whose <metric signature> is $(2,1)$. For a non-Abelian degenerate example, the three-dimensional <Heisenberg Lie algebra> has $[x,y]=z$ central. Every product of two adjoint maps is zero, so $K=0$.
For any adjoint-invariant nondegenerate metric on a <Lie group>, the <Koszul formula> on left-invariant fields simplifies to
$$
2g(\nabla_XY,Z)=g([X,Y],Z),\qquad
\boxed{\nabla_XY=\frac12[X,Y].}
$$
Take $R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$. The <Jacobi identity> gives
$$
R(X,Y)Z=\frac14\bigl([X,[Y,Z]]-[Y,[X,Z]]\bigr)-\frac12[[X,Y],Z]
=-\frac14[[X,Y],Z].
$$
Taking the trace of $Z\mapsto R(Z,X)Y=-\operatorname{ad}Y\operatorname{ad}X(Z)/4$ yields
$$
\boxed{\operatorname{Ric}(X,Y)=-\frac14K(X,Y).}
$$
Thus the metric $g=K$ is a <Killing-form Einstein metric> with Einstein constant $-1/4$. With $g=-K$ the same connection has constant $+1/4$. This proves the <Einstein manifold> assertion for every semisimple group, including products when the unscaled Killing form is used on all factors.
There is also a different, flat metric-preserving connection. Declare the global left-invariant frame parallel:
$$
D_X(Y^ae_a)=X(Y^a)e_a.
$$
Its curvature is zero because applying two derivatives to coefficient functions and subtracting their commutator gives zero. Its metric derivative vanishes since $g_{ab}$ is constant in this frame. Its <torsion tensor> on invariant fields is
$$
\boxed{T(e_a,e_c)=-[e_a,e_c],\qquad T_a{}^b{}_c=-C_a{}^b{}_c.}
$$
For each fixed $c$, antisymmetry gives $T_a{}^b{}_c=C_c{}^b{}_a$, so this is the matrix of $\operatorname{ad}e_c$. The <torsion contractions of a left-parallel Lie-group connection> therefore are
$$
\boxed{T_a{}^a{}_c=\operatorname{Tr}(\operatorname{ad}e_c)=0,\qquad
T_a{}^b{}_cT_b{}^a{}_d=K_{cd}.}
$$
The zero trace follows because a semisimple algebra is perfect: every element is a sum of brackets, and the adjoint of a bracket is a matrix commutator with zero trace. If the metric convention is $g=-K$, the second contraction is $-g_{cd}$. These <flat translation connections on a Lie group> are not the torsion-free <Levi-Civita connection>; nonzero torsion permits flatness alongside the curved Killing metric.
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