Proof on metric tensor identity

Question

What is the proof for this proposition: $$\frac{1}{2} g^{mi} \partial_k g_{mi} =\frac{1}{2g} \partial_k g $$

Answer 1

There is a formula for the derivative of an invertible matrix given by $$\partial_k \det(A) = \det(A) \operatorname{Tr}(A^{-1} \partial_k A).$$ I would prefer not to prove this here, so see this post by Terry Tao for a proof of this formula. In our situation, $(g_{ij})$ is an invertible matrix with $(g_{ij})^{-1} = (g^{ij})$ and $\det(g_{ij}) = g$, so the formula translates to \begin{align*} \partial_k g & = g \operatorname{Tr} (g^{im} \partial_k g_{mj}) \\ & = g g^{im} \partial_k g_{mi} \\ & = g g^{mi} \partial_k g_{mi}. \end{align*} Hence $$\frac{1}{g} \partial_k g = g^{mi} \partial_k g_{mi}.$$