Flat translation connections on a Lie group (${\nabla }_X^{-}Y=0,{\nabla }_X^{+}Y=[X,Y]$)

Declare a global left-invariant, respectively right-invariant, frame parallel to obtain two flat affine connections on a Lie group. For left-invariant vector fields, their derivatives are as displayed and their torsion tensors are $-[X,Y]$ and $+[X,Y]$. The tensors vanish on Abelian groups; two genuinely torsionful connections are not asserted there.

Declare a global frame of left-invariant vector fields parallel on a semisimple Lie group. This affine connection is flat and preserves a left-invariant metric tensor because its frame components are constant. If $[e_a,e_c]=C_a{}^b{}_c{e}_b$, its torsion tensor is $T_a{}^b{}_c=-C_a{}^b{}_c$. For fixed $c$, the matrix with entries $T_a{}^b{}_c$ is $\mathrm{ad}{e}_c$. Its trace vanishes because a semisimple Lie algebra is perfect and adjoints of brackets are commutators. The trace of the product for $c,d$ is the Killing form $K_{cd}$, proving both contractions. This flat metric connection differs from the curved torsion-free Levi-Civita connection of the Killing metric.

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 ${\nabla }^0$.