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The average of the two flat translation connections on a Lie group is a torsion-free connection. For left-invariant vector fields, the Jacobi identity gives $R(X,Y)Z=-\frac{1}{4}[[X,Y],Z]$ with $R(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$. Contracting gives $\mathrm{Ric}(X,Y)=-\frac{1}{4}K(X,Y)$ for the Killing form $K$. It is flat exactly when all double Lie brackets vanish, including the case of a two-step nilpotent group. For a semisimple Lie algebra, $K$ is nondegenerate and supplies a bi-invariant indefinite metric whose Levi-Civita connection is ${\nabla }^0$.

Define the raw Killing form byFor 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 $[e_i,e_j]={\epsilon }_{ijk}{e}_k$, $K_{ij}=-2{\delta }_{ij}$ 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 givewhose 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 toTake $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. The Jacobi identity givesTaking the trace of $Z\mapsto R(Z,X)Y=-\mathrm{ad}Y\mathrm{ad}X(Z)/4$ yieldsThus 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: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 isFor each fixed $c$, antisymmetry gives $T_a{}^b{}_c=C_c{}^b{}_a$, so this is the matrix of $\mathrm{ad}{e}_c$. The torsion contractions of a left-parallel Lie-group connection therefore areThe 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.

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 $E_a(g)=(d{L}_g{)}_e{e}_a$ and right-invariant vector fields $F_a(g)=(d{R}_g{)}_e{e}_a$. Left and right translations commute. If $[e_a,e_b]=c^c{}_{ab}{e}_c$, thenThe opposite sign for the right-invariant frame is essential.

Define an affine connection ${\nabla }^{-}$ by declaring the $E_a$ parallel. Explicitly, ${\nabla }_X^{-}(f^a{E}_a)=X(f^a)E_a$. Its curvature form of a connection vanishes, since successive derivatives of the component functions satisfy the vector-field commutator identity. Its torsion tensor isSimilarly, 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 obeysTo see this, express $E_a=F_{{\mathrm{Ad}}_g{e}_a}$. Along $g\exp (t{e}_b)$, differentiation of ${\mathrm{Ad}}_{g\exp (t{e}_b)}{e}_a$ gives ${\mathrm{Ad}}_g[e_b,e_a]$, 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 ${\nabla }_X^{0}Y=\frac{1}{2}[X,Y]$ for left-invariant fields. Its torsion tensor is zero. With the curvature convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity givesDefine the Ricci tensor by $\mathrm{Ric}(X,Y)=\mathrm{Tr}(Z\mapsto R(Z,X)Y)$. Since $[[Z,X],Y]={\mathrm{ad}}_Y{\mathrm{ad}}_{X}Z$, this giveswhere $K(X,Y)=\mathrm{Tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$ 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,A metric-compatible torsion-free connection is uniquely the Levi-Civita connection; therefore ${\nabla }^0$ is its Levi-Civita connection. Moreover,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 $\mathrm{Ric}=\frac{1}{4}{g}_{-K}$ instead.