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.


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

  • $\begingroup$ Indeed that would be a way to explain it. However when working in 2d in string theory, we often say that a conformal transformation is equivalent to a making a difféomorphism following the same transformation, and the rescaling the metric back. So at least in 2d, it would seem that all conformal transformations are diffeos. Could you maybe provide a counter-example in 4d ? $\endgroup$ – Frotaur Jun 27 at 18:51
  • $\begingroup$ @Frotaur: You refer to counterexamples in your own question. $\endgroup$ – Ben Crowell Jun 27 at 18:52
  • $\begingroup$ 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 ? $\endgroup$ – Frotaur Jun 28 at 0:34

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