$g^{\mu\nu}(x)=\Omega^{2}(x)g'^{\mu\nu}(x)$ is a conformal transformation.
If $g'^{\mu\nu}$ is flat, what kind of $\Omega(x)$ is choosed can make $g^{\mu\nu}$ flat.
We can think about any dimension $n$ and any signature.
In special case, $g'^{\mu\nu}=diag(-1,1,1,1)$.
Thanks!



In what condition is a conformal flat manifold flat?
By looking at how the Riemann curvature tensor changes under a conformal transformation (see, Besse *Einstein manifolds*, 1.159 (b)), and writing $\Omega = e^{-f}$, you will find that $g$ is flat if and only if
$$
\nabla df - df \circ df + \tfrac12 |df|^2 g’ = 0~,
$$
where $df\circ df$ is the symmetric product of $df$ with itself, $g’$ is the metric and not the inverse metric and $\nabla$ is the Levi-Civita connection of the reference flat metric $g’$. If you prefer, in flat coordinates, the equation is
$$
\partial\_a \partial\_b f - \partial\_a f \partial\_b f + \frac12 \partial\_c f \partial^c f g’\_{ab} = 0~.
$$
You can take traces now and depending on the dimension ($n$) you find an equation of the type
$$
\Delta f = \lambda\_n \partial\_a f \partial^a f~,
$$
for some constant $\lambda\_n$.


