Inverse of a metric under variation

Question

Given a fixed metric $g_{\mu\nu}$, its variation by a small amount could be written as: $$g_{\mu\nu}+h_{\mu\nu}$$ or equivalently as: $$g_{\mu\nu}+\delta(g_{\mu\nu}).$$ The given metric has the property that: $$g_{\mu\alpha}g^{\alpha\nu}=\delta^\nu_\mu,$$ and that $g_{\mu\nu}=g_{\nu\mu}$, which applies also to the metric after variation $g+h$.

Question: What is the relationship between $\delta g^{\mu\nu}$ and $h^{\mu\nu}$? This is actually proving that for arbitrary tensors: $$A_{\mu\nu}B^{\mu\nu}=A^{\mu\nu}B_{\mu\nu}.$$ How to prove this?

Answer 1

Consider the disturbed metric $\widetilde{g}_{\mu\nu}=g_{\mu\nu}+h_{\mu\nu}$ with $|h_{\mu\nu}|<1$, meaning the pertubation is small and we can use $g$ to raise and lower indices, then $\widetilde{g}^{\mu\nu}=g^{\mu\nu}-h^{\mu\nu}$ with $|h^{\mu\nu}|<1$. The sign change arises since we want to have $\widetilde{g}_{\lambda\mu}\widetilde{g}^{\mu\nu}=\delta_\lambda^\nu$ as well as $g_{\lambda\mu}g^{\mu\nu}=\delta_\lambda^\nu$. Neglecting the pertubation in second order, we have: \begin{equation} \widetilde{g}_{\lambda\mu}\widetilde{g}^{\mu\nu} =(g_{\lambda\mu}+h_{\lambda\mu})(g^{\mu\nu}-h^{\mu\nu}) =\delta_\lambda^\nu -\underbrace{g_{\lambda\mu}h^{\mu\nu}}_{=-h_\lambda^\nu} +\underbrace{h_{\lambda\mu}g^{\mu\nu}}_{=h_\lambda^\nu} +\mathcal{O}(h^2). \end{equation} Therefore we need to put $\delta g^{\mu\nu}=-h^{\mu\nu}$. Furthermore, we have: \begin{equation} A_{\mu\nu}B^{\mu\nu} =A^{\kappa\lambda}g_{\kappa\mu}g_{\lambda\nu}B^{\mu\nu} =A^{\kappa\lambda}B_{\kappa\lambda}. \end{equation}