# How is the Ricci scalar of a conformally flat metric non-zero?

I am having trouble with a contradiction arising from some computation, and I cannot figure out at which point I make a mistake.

Consider a conformally flat metric $$g_{\mu\nu}=e^{2\phi}\eta_{\mu\nu}$$. Then, the ricci scalar of $$g$$ is not always 0, depending on $$\phi$$, as can be seen for example from the formulas here.

However, let us not consider a "conformal" change of variables $$x'^\rho(x)$$ such that, in the new coordinates, the metric is rescaled as such : $$g'_{\mu\nu} = e^{-2\phi}g_{\mu\nu} = \eta_{\mu\nu}$$. In other words, identify the conformal transformation which rescales the metric by $$e^{-2\phi}$$, and apply it as a change of variables.

However, since scalars remain unchanged under a change of variables (more precisely under a diffeomorphism), we should have that $$R[g] = R[g'] = R[\eta] = 0$$ (where $$R[g]$$ is the ricci scalar of the metric $$g_{\mu\nu}$$.

Now this is a problem since we saw that $$R[g]$$ need not be $$0$$ even if g is conformally flat. Thus the contradiction.

I don't understand at which step in my reasoning I have made a mistake. The only possibility that I see is that there is such conformal transformation that rescales the metric by $$e^{-2\phi}$$, but this seems very restrictive looking again at this.

I am sure that the mistake I made it's much more elemental, but I can't figure it out.

• I see what you mean, but I don't think that I agree. At least for the conformal transformations I know, they are always invertible, as well as $C^\infty$, so they should belong to diffeomorphisms. physics.stackexchange.com/questions/116853/… for example seem to agree. But maybe I am still missing something ? – Frotaur Jun 28 at 0:34