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

Question

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 now 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 isn't such conformal transformation that rescales the metric by $e^{-2\phi}$, except when the $\phi$ in question does not affect the Ricci scalar, but this seems very restrictive looking again at this.

Answer 1

What you've proved is that not every conformal transformation can be represented as a diffeomorphism.

Answer 2

$\newcommand{\co}{\nabla} \newcommand{\lam}{\lambda}$ Ok, so after considering the case of flat space, it appears that my hunch was actually correct, although I am still looking for a general proof.

Let me elaborate : consider the metric $g_{\mu\nu}=e^{2\phi}\eta_{\mu\nu}$ as in the question. If we compute the Ricci scalar, we obtain the following : $$ e^{2\phi} R_g = R_\eta - 2(n-1)(\co^\mu\co_\mu\phi +1/2(n-1)\co_\mu\phi \co^\mu\phi) $$

$n$ here is the dimensionality of the manifold we consider. The covariant derivatives on the right are associated to the initial metric, namely $\eta_{\mu\nu}$ and so are simply partial derivatives. Now, as argued in the question, a conformal transformation can in particular be seen as a diffeomorphism. This means, in particular, that it must be true that $R_g = R_\eta = 0$. In other words, this is a constraint that $\phi$ must satisfy if there exists a conformal transformation with this scale factor. This property should in principle be provable, assuming $\phi$ indeed comes from a conformal coordinate change, but I haven't been able to find a general proof.

However, in our case at hand of flat space it is easy to prove. The conformal group for flat space (here for n>2) is well known. It contains the Poincaré group which are simply isometries, and trivially satisfy $R_g= R_\eta=0$. We also have the dilations, $x' = \lam x$, but for these we have $e^{2\phi} = \lam^2$, so $\phi$ is constant and we can see from the Ricci equation that we also have $R_g=0$.

The only non-trivial case to check is the special conformation transformations, namely $x' = {x'}^{\mu} = \frac{x^{\mu}-b^{\mu}x^2}{1-2(b.x)+b^2x^2}$. We can check that for this change of coordinate we have $e^{-\phi} = \Lambda(x) = (1-2b.x + b^2x^2) \Leftrightarrow \phi = -\ln(\Lambda(x))$. We can then check explicitly that we indeed have $R_g = R_\eta =0$, by computing using the Ricci transformation formula. I don't do it here because it is straightforward, although tedious.

I will keep looking to see if I find a way to show that, given a conformal transformation of scale factor $e^{2\phi}$, then

$$ (e^{2\phi}-1)R = -2(n-1)(\co^\mu \co_\mu \phi+1/2(n-1)\co_\mu\phi\co^\mu\phi)$$

Where $R$ and $\co_\mu$ are associated to the starting metric.