Perhaps this is a naive question, but I have been reading about conformal invariance and conformally related metrics and I would like to know if someone can clarify me some concepts on this.

Anti de Sitter space-time (in 3+1 dimensions) is a maximally symmetric Lorentzian manifold with constant negative scalar curvature. The metric is given by $$ds^{2}=-c^{2}(1+k^{2}\,r^2)dt^{2}+\dfrac{1}{(1+k^{2}\,r^2)}dr^{2}+r^{2}d\Omega^{2},$$ where $k^{2}>0$ and $d\Omega$ is the solid angle element in 3d spherical coordinates.

By making the coordinate transformation

$$t=\alpha(\tau), \;\; \rho^{2}=\dfrac{r^{2}}{1+k^{2}r^{2}}$$ so the whole space is now inside a sphere of radius $1/k$, and $\alpha(\tau)$ is a smooth function with no extremal points, one gets the metric

$$ds^{2}=\dfrac{1}{1-k^{2}\rho^{2}}\alpha'^{2}(\tau)\left(-c^{2}d\tau^{2}+\dfrac{1}{\alpha'^{2}(\tau)}\left(\dfrac{1}{1-k^{2}\rho^{2}}d\rho^{2}+\rho^{2}d\Omega^{2}\right)\right)$$ which is conformally equivalent to the FLRW metric with postitive curvature.

What does it mean this result?

  • $\begingroup$ It is not clear to me what are you asking. Is it about what means to be conformal invariant to FLRW or which are te consequences of this? $\endgroup$ May 4, 2017 at 17:35
  • $\begingroup$ both @AlejandroMenaya $\endgroup$ May 4, 2017 at 17:56

1 Answer 1


Two metrics $g,g'$ are conformally related if it exist a function $\omega$ s.t. $g'=e^{\omega}g=\Omega g$. In the case you propose it is clear that $\Omega = \frac{\alpha^\prime(\tau)}{1-k^2\rho^2}$

We can highlight several interesting properties:

  • Conformal transformations are diffeomorphisms. Then the spacetimes are diffeomorphic: they will have same number of holes (in topological sense).Topologically those manifolds are the same. A phisical interpretation: conformal transformations are changes in the reference frame, so the physics in those spacetimes must be equal.
  • Christoffel symbols and curvature tensors between spacetimes are related.
  • Null geodesic in one spacetime maps to null geodesic in the other one (you can see a proof here. Then causal structure is preserved.
  • The Weyl tensor, which roughly counts for gravity outside the sources (interesting for gravitational waves, for example) is conformally invariant.
  • Killing vectors in one spacetime will be conformal killing vectors in the other: Let $\xi$ be a Killing vector for the metric $g$ (i.e. $\mathcal{L}_{\xi}g=0$) then $\mathcal{L}_{\xi}(\Omega g) = \mathcal{L}_{\xi}\Omega g$. Which leads to new currents: Assume $T^{\mu\nu}$ is the energy momentum tensor in a spacetime with a conformal Killing $\xi$. Then $\nabla_{(\mu}\xi{_\nu)} \propto g_{\mu\nu}$. Using equations of motion $\nabla_\mu T^{\mu\nu}=0$ $$ J^\nu = T^{\mu\nu}\xi_\mu\Rightarrow \nabla_\nu J^\nu\propto {T^\nu}_\nu $$ In lots of situations, the field theory we build has conformal symmetric action, which implies a traceless energy-momentum tensor. In this kind of theories, a conformal transformation gives relations between conserved currents in both spacetimes. Furthermore: suppose we define in AdS our theory without an action (which is typical in CFT since we can compute 2 and 3 point correlation functions via conformal symmetry). If the corresponding theory in FLRW does not have a conserved current asociated with the conformal transformation, the theory will present a gravitational anomaly.

Probably there are more interesting facts, but those are the first that popped out of my mind.

  • $\begingroup$ Thanks for the useful information dear @AlejandroMenaya So, what does it mean? That an observer in a AdS space-time, moving through the surface $(\alpha^{-1}(\tau), \rho/\sqrt{1-k^{2}\rho^{2}},\theta,\phi)$ will see space-time as our Universe but shrinked by the conformal factor? $\endgroup$ May 5, 2017 at 13:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.