# Anti de Sitter and FLRW metrics

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?

• 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? – Alejandro Menaya May 4 '17 at 17:35
• both @AlejandroMenaya – Ernesto Lopez Fune May 4 '17 at 17:56

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.

• 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? – Ernesto Lopez Fune May 5 '17 at 13:45