I'm working on an exercise from Geometric Relativity by Dan A. Lee, but things didn't go well:
Following Lee's hint, I was trying to use Exercise 1.12 and view $\color{red}{g}$ as the background metric ($\bar g$ in Exercise 1.12). In Exercise 1.12, I've shown that the Ricci curvature of a Riemannian metric $g$ and the Ricci curvature of another Riemannian metric $\bar g$ are related by $$R_{ij}=\bar{R}_{ij}+(\overline{\nabla}_k W_{ij}^k-\overline{\nabla}_j W_{ki}^k)+(W_{k\ell}^k W_{ij}^\ell-W_{j\ell}^k W_{ik}^\ell).\tag{1}$$ Please see Relating the Ricci curvatures of two Riemannian metrics to know about (1).
Now I have $$R_{g_t}=g_t^{ij}\bar{R}_{ij}+g_t^{ij}(\overline{\nabla}_k W_{ij}^k-\overline{\nabla}_j W_{ki}^k)+g_t^{ij}(W_{k\ell}^k W_{ij}^\ell-W_{j\ell}^k W_{ik}^\ell)\tag{2}$$ and $$\left.\frac{d}{dt}\right|_{t=0}R_{g_t}=\lim_{t\to 0}\frac{R_{g_t}-R_g}{t}=\lim_{t\to 0}\frac{R_{g_t}-g^{ij}\bar{R}_{ij}}{t}.\tag{3}$$ To evaluate the difference quotient in (3), I write $$R_{g_t}\simeq (g^{ij}+t\dot{g}^{ij})\bar{R}_{ij}+(g^{ij}+t\dot{g}^{ij})(\overline{\nabla}_k W_{ij}^k-\overline{\nabla}_j W_{ki}^k)+(g^{ij}+t\dot{g}^{ij})(W_{k\ell}^k W_{ij}^\ell-W_{j\ell}^k W_{ik}^\ell)\tag{4}$$ for small $t$. I'm not sure if the linear approximation of $g_t$ at $t=0$ is legitimate, but I can't imagine what it would be like if there were any other ways. Now I have $$\frac{R_{g_t}-g^{ij}\bar{R}_{ij}}{t}\simeq \dot{g}^{ij}\bar{R}_{ij}+(\frac{g^{ij}}{t}+\dot{g}^{ij})(\overline{\nabla}_k W_{ij}^k-\overline{\nabla}_j W_{ki}^k)+(\frac{g^{ij}}{t}+\dot{g}^{ij})(W_{k\ell}^k W_{ij}^\ell-W_{j\ell}^k W_{ik}^\ell)\tag{5}$$ for small $t\neq 0$. If the above is true, all we have to do now is take its limit at $t=0$. But after seeing that $$\lim_{t\to 0}\dot{g}^{ij}\bar{R}_{ij}=\dot{g}^{ij}\bar{R}_{ij}=\langle\mathrm{Ric}_g,\dot{g}\rangle_g,\tag{6}$$ I began to hesitate about the computation because I didn't have a minus sign to compensate (6) for the formula to be derived in Exercise 1.18. Why's that? What should I do next? Thank you.
Remark 1. I'm not worried about $\dot{g}^{ij}(\overline{\nabla}_k W_{ij}^k-\overline{\nabla}_j W_{ki}^k)$ and $\dot{g}^{ij}(W_{k\ell}^k W_{ij}^\ell-W_{j\ell}^k W_{ik}^\ell)$ in (5) because these terms contribute nothing as indicated by the hint.
Remark 2. The hint says that in any orthonormal frame $\{v_i\}_i$, the double divergence of $\dot{g}$ is given by $$\mathrm{div}_g\mathrm{div}_g\dot{g}=\sum_{i,j}\dot{g}_{ij;ij},$$ which I have just checked. And in the same frame, I also checked that $$\begin{align} \Delta_g\mathrm{tr}_g\dot{g}&=\mathrm{div}_g\mathrm{grad}(g^{ij}\dot{g}_{ij})\\ &=\sum_k\left(v_k(v_k(\sum_i \dot{g}_{ii}))+\sum_j v_j(\sum_i\dot{g}_{ii})\overline{\Gamma}_{kj}^k\right) \end{align}$$ and that $$\langle\mathrm{Ric}_g,\dot{g}\rangle_g=\sum_{i,j}\bar{R}_{ij}\dot{g}_{ij}.$$ These local expressions follow from a straightforward computation. The crux of my attempt really lies in the expansion of those $W$ terms. Originally, I planned on taking it by employing $$W_{ij}^k=\frac{1}{2}g_t^{k\ell}\left(\overline{\nabla}_i(g_t)_{\ell j}+\overline{\nabla}_j(g_t)_{i\ell}-\overline{\nabla}_\ell(g_t)_{ij}\right),$$ another result obtained previously (see The components $W_{ij}^k$ of the difference $W=\nabla-\overline{\nabla}$ between two Levi-Civita connections), but the algebra really hurts.
