= Bracket connection on a Lie group
{title2=$\nabla_XY=\lambda[X,Y]\quad\text{for left-invariant }X,Y$}
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 $\nabla_{L_i}(z^jL_j)=L_i(z^j)L_j+\lambda z^j[L_i,L_j]$. Applying the bare bracket formula to arbitrary vector fields instead would violate the <Leibniz rule> unless $\lambda=1$. With $R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$, the <Jacobi identity> gives
$$
T(X,Y)=(2\lambda-1)[X,Y],\qquad R(X,Y)Z=\lambda(\lambda-1)[[X,Y],Z],\qquad \operatorname{Ric}(X,Y)=\lambda(\lambda-1)B(X,Y),
$$
for left-invariant arguments, where $B$ is the <Killing form> and $\operatorname{Ric}(X,Y)=\operatorname{tr}(Z\mapsto R(Z,X)Y)$. 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.
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