Lie derivative of a metric with nonmetricity (${\mathcal{L}}_{\xi }{g}_{ab}={\xi }^c{\nabla }_c{g}_{ab}+g_{cb}{\nabla }_a{\xi }^c+g_{ac}{\nabla }_b{\xi }^c$)

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 $g=e^{2x}d{x}^2$, 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.