Bracket connection on a Lie group (${\nabla }_{X}Y=\lambda [X,Y]\text{for left-invariant }X,Y$)

Specify the displayed rule on left-invariant vector fields and extend by the Leibniz rule to an affine connection on the Lie group. In a left-invariant vector field frame $L_i$, the extension is ${\nabla }_{L_i}(z^j{L}_j)=L_i(z^j)L_j+\lambda z^j[L_i,L_j]$. Applying the bare bracket formula to arbitrary vector fields instead would violate the Leibniz rule unless $\lambda =1$. With $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity givesfor left-invariant arguments, where $B$ is the Killing form and $\mathrm{Ric}(X,Y)=\mathrm{tr}(Z\mapsto R(Z,X)Y)$. These identities determine the tensor fields everywhere. The endpoints $\lambda =0,1$ have zero curvature, with respectively left- and right-invariant parallel frames, and opposite torsion tensors. The midpoint $\lambda =1/2$ is torsion-free but need not have zero curvature.