I am having issues with the Lie derivative of the metric tensor with respect to a basis vector. I know we can prove that the Lie derivative of the metric tensor with respect to the vector $$\boldsymbol{\xi}$$ satisfies \begin{align} \mathcal{L}_{\boldsymbol{\xi}} g_{\alpha\beta} &= \xi^\mu \partial_\mu g_{\alpha\beta} + g_{\mu\beta} \partial_\alpha \xi^\mu + g_{\alpha\mu} \partial_\beta \xi^\mu \\ &= \xi^\mu \nabla_\mu g_{\alpha\beta} + g_{\mu\beta} \nabla_\alpha \xi^\mu + g_{\alpha\mu} \nabla_\beta \xi^\mu \\ &= g_{\mu\beta} \nabla_\alpha \xi^\mu + g_{\alpha\mu} \nabla_\beta \xi^\mu \end{align} which can be verified simply by subtracting and using the condition of torsion free metric and metric compatibility.
The problem comes when I substitute the vector $$\boldsymbol{\xi}$$ with the basis vector $$\boldsymbol{e_\gamma}$$. Clearly, $$\boldsymbol{e_\gamma} = \delta^\lambda_\gamma \boldsymbol{e_\lambda}$$, so $$(\boldsymbol{e_\gamma})^\mu = \delta^\mu_\gamma$$. Substituting this back in the equations of the Lie derivative, I get from the first line $$\begin{equation} \mathcal{L}_{\boldsymbol{e_\gamma}} g_{\alpha\beta} = \partial_\gamma g_{\alpha\beta} \end{equation}$$ and from the last line, however, $$\begin{equation} \mathcal{L}_{\boldsymbol{e_\gamma}} g_{\alpha\beta} = 0 \end{equation}$$ since the (covariant) derivatives of $$\delta^\mu_\gamma$$ are zero. But $$\partial_\gamma g_{\alpha\beta}=0$$ cannot be true in general. Where did my argument go wrong?
The covariant derivative of $$\delta_{\gamma}^{\mu}$$ is not zero in general. For example, in the $$(R,\theta)$$ plane polar coordinates, the covariant derivative of $$1e_r + 0e_{\theta}$$ in the direction of $$e_{\theta}$$ is surely not zero. In this case: $$\Gamma^{\theta}_{r\theta} \neq 0$$