That is, if a manifold $(M, \hat{g})$ is Einstein and $\hat{g}_{ij} = \cfrac{\delta_{ij}}{F^2}$ (where $F: M \to \mathbb{R}$ is positive and differentiable), then $M$ has constant sectional curvature. Some computations give:
$$
\begin{aligned} K_{i j} &=\frac{R_{i j i j}}{g_{i i} g_{j j}}=R_{i j i j} F^{4} \\ &=\left(-\sum_{\ell} f_{\ell}^{2}+f_{i}^{2}+f_{j}^{2}+f_{i i}+f_{j j}\right) F^{2} \end{aligned}
$$
where $
\cfrac{\partial}{\partial x_{i}}, \cfrac{\partial}{\partial x_{j}}
$ are orthogonal. But how can I relate this to the condition that $\text{Ric}(X, Y) = \lambda g(X, Y)$ (i.e, $M$ is einstein)?
I'm aware we could replace "$\hat{g}_{ij} = \cfrac{\delta_{ij}}{F^2}$" to $\hat{g}_{ij} = \cfrac{g_{ij}}{F^2}$ where $g$ is the metric of some other manifold with constant sectional curvature, but I'm trying to prove this weaker result first.
Best Answer
First, recall the standard decomposition of curvature $$ R=\frac{\text{scal}}{2n(n-1)} (g\odot g) + (\text{Ric}_0\odot g)+W $$ where $R$ is the Riemann curvature tensor, $\text{scal}$ is the scalar curvature, $\text{Ric}_0:=\text{Ric}-\frac{\text{scal}}{n} g$ is the traceless-Ricci, $W$ is the Weyl curvature tensor, and $\odot$ is the Kulkarni-Nomizu product of two (0,2)-tensor as a (0,4)-tensor.
The Einstein condition $\text{Ric}=\lambda g$ gives the traceless Ricci vanishes (and constant scalar curvature), and $M$ being conformally flat means $W=0$. So $R$ is a constant multiple of $g\odot g$, i.e. constant sectional curvature.