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$g^{\mu\nu}(x)=\Omega^{2}(x)g'^{\mu\nu}(x)$ 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!

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As for necessary and sufficient conditions for conformal flatness of the given Riemannian metric in dimensions $\ge 4$, the key words are "Weyl curvature tensor". – Misha Mar 18 '14 at 10:38
@Misha: I don't think that this is what the OP is asking. Clearly since both $g$ and $g'$ are flat, their Weyl curvature tensors vanish. What the OP is asking, I think, is what are conditions on $\Omega$ so that both $g$ and $g'$ are flat. – José Figueroa-O'Farrill Mar 18 '14 at 18:33
up vote 4 down vote accepted

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 $\lambda_n$.

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I can only tell something about the positive definite case: In dimension 2, the equation in flat coordinates is the Laplace equation (as in José's answer): if $g$ is flat, than $e^{2f}g$ is flat if and only if $\Delta f=0.$

For manifolds $M$ of higher dimension $n$, you have the powerful Liouville theorem: If $g'$ and $g$ are already flat, you can use flat coordinates $x'\colon U'\subset M\to V'\subset \mathbb R^n$ for $g'$ and $x\colon U\subset M\to V\subset \mathbb R^n$ for $g.$

As $g'$ and $g$ are related by a conformal factor, there is a conformal transformation ($x'\circ x^{-1}\colon V\to V'$ to be explicit) which must be part of a Moebius transformation (i.e. an element of the group generated by translations, homotheties, reflections and inversions on spheres). This group is given by $SO(n,1)$ and one can check by hand which Moebius transformations have the property that the flat (euclidean) metric stays flat. Then, exactly those corresponding conformal factors are possible.

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