= Solution
For a <Lie group> $G$, <Left translation on a Lie group> and <right translation on a Lie group> are the <diffeomorphisms> $L_g(h)=gh$ and $R_g(h)=hg$. Their differentials identify the tangent space at the identity with every tangent space. Translating a basis $e_a$ of the <Lie algebra> gives global <left-invariant vector fields> $E_a(g)=(dL_g)_e e_a$ and <right-invariant vector fields> $F_a(g)=(dR_g)_e e_a$. Left and right translations commute. If $[e_a,e_b]=c^c{}_{ab}e_c$, then
$$
[E_a,E_b]=c^c{}_{ab}E_c,\qquad [F_a,F_b]=-c^c{}_{ab}F_c.
$$
The opposite sign for the right-invariant frame is essential.
Define an <affine connection> $\nabla^-$ by declaring the $E_a$ parallel. Explicitly, $\nabla^-_X(f^aE_a)=X(f^a)E_a$. Its <curvature form of a connection> vanishes, since successive derivatives of the component functions satisfy the vector-field commutator identity. Its <torsion tensor> is
$$
T^-(E_a,E_b)=\nabla^-_{E_a}E_b-\nabla^-_{E_b}E_a-[E_a,E_b]=-[E_a,E_b].
$$
Similarly, declaring the $F_a$ parallel gives another flat <affine connection> $\nabla^+$, with $T^+(F_a,F_b)=-[F_a,F_b]$. In a left-invariant frame the latter connection obeys
$$
\nabla^+_XY=[X,Y].
$$
To see this, express $E_a=F_{\operatorname{Ad}_g e_a}$. Along $g\exp(t e_b)$, differentiation of $\operatorname{Ad}_{g\exp(t e_b)}e_a$ gives $\operatorname{Ad}_g[e_b,e_a]$, producing the displayed bracket. Consequently $T^+(X,Y)=[X,Y]$ for left-invariant $X,Y$. These are the <flat translation connections on a Lie group>. They have opposite torsion; if $G$ is Abelian, both torsion tensors vanish and the two connections coincide. The phrase “with torsion” does not mean nonzero torsion for Abelian groups.
The average $\nabla^0=(\nabla^-+\nabla^+)/2$ is again an <affine connection> and satisfies $\nabla^0_XY=\frac12[X,Y]$ for left-invariant fields. Its <torsion tensor> is zero. With the curvature convention $R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$, the <Jacobi identity> gives
$$
\begin{aligned}
R^0(X,Y)Z
&=\frac14\bigl([X,[Y,Z]]-[Y,[X,Z]]\bigr)-\frac12[[X,Y],Z]\\
&=-\frac14[[X,Y],Z].
\end{aligned}
$$
Define the <Ricci tensor> by $\operatorname{Ric}(X,Y)=\operatorname{Tr}(Z\mapsto R(Z,X)Y)$. Since $[[Z,X],Y]=\operatorname{ad}_Y\operatorname{ad}_X Z$, this gives
$$
\boxed{R^0(X,Y)Z=-\frac14[[X,Y],Z],\qquad
\operatorname{Ric}^0(X,Y)=-\frac14K(X,Y),}
$$
where $K(X,Y)=\operatorname{Tr}(\operatorname{ad}_X\operatorname{ad}_Y)$ is the <Killing form>. Thus the <canonical torsion-free connection on a Lie group> is generally curved. Its exact flatness criterion is $[[\mathfrak g,\mathfrak g],\mathfrak g]=0$: Abelian and two-step nilpotent groups are exceptions. A vanishing <Ricci tensor> alone does not prove flatness.
For a <semisimple Lie algebra>, the <Killing form> is nondegenerate. Its adjoint invariance gives $K([X,Y],Z)+K(Y,[X,Z])=0$, so translating it defines a <bi-invariant pseudo-Riemannian metric> $g_K$. For left-invariant fields,
$$
(\nabla^0_Xg_K)(Y,Z)=-\frac12\bigl(K([X,Y],Z)+K(Y,[X,Z])\bigr)=0.
$$
A metric-compatible <torsion-free connection> is uniquely the <Levi-Civita connection>; therefore $\nabla^0$ is its Levi-Civita connection. Moreover,
$$
\boxed{\operatorname{Ric}(g_K)=-\frac14g_K.}
$$
The special feature is a canonical nondegenerate metric and an <Einstein manifold> structure. The metric need not be positive definite: noncompact semisimple groups generally require indefinite signature. For compact semisimple groups, using $-K$ gives a <bi-invariant Riemannian metric> with $\operatorname{Ric}=\frac14g_{-K}$ instead.
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