= Lie derivative of a metric with nonmetricity
{c}
{title2=$\mathcal L_\xi g_{ab}=\xi^c\nabla_cg_{ab}+g_{cb}\nabla_a\xi^c+g_{ac}\nabla_b\xi^c$}
For any <torsion-free connection>, expand the coordinate <Lie derivative of a tensor field> and cancel the symmetric connection terms to obtain the displayed identity. If the connection is <metric-compatible>, lowering the vector index commutes with the <covariant derivative> and it becomes $\mathcal L_\xi g_{ab}=2\nabla_{(a}\xi_{b)}$. Symmetry of the connection alone does not suffice: with $g=e^{2x}dx^2$, zero connection coefficients and $\xi=\partial_x$, the two expressions $\mathcal L_\xi g$ and $2\nabla_x\xi_x$ are $2e^{2x}$ and $4e^{2x}$, respectively.
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