Large $D$ limit of (Anti) de Sitter Space is Minkowskian Space?

Question

As is well known, the solution of the vacuum Einstein equations with a non-zero cosmological constant, $G_{\mu\nu}+\Lambda g_{\mu\nu}=0$, is an asymptotically (anti) de Sitter space based on the sign of the cosmological constant.

Now, taking the trace of these vacuum Einstein equations, one reads $R - \dfrac{1}{2}RD + \Lambda D = 0$. Or, equivalently, $R=\dfrac{2\Lambda D}{D-2}$. Now, thus, in the large $D$ limit, $R=2\Lambda$. Thus, the Einstein equations read as $R_{\mu \nu} - \dfrac{1}{2}(2\Lambda) g_{\mu\nu} + \Lambda g_{\mu \nu} = 0$ in this large $D$ limit. Or, $R_{\mu \nu}=0$. Thus, $R=0$ and, consequently, $\Lambda=0$. This means that the Schwarzschild (anti)de Sitter metric (a solution of $G_{\mu\nu}+\Lambda g_{\mu\nu}=0$) which is asymptotically (anti) de Sitter approaches a Schwarzschild metric which is asymptotically Minkowskian in the large $D$ limit - because the cosmological constant approaches zero in this limit. In this sense, the large $D$ limit of an (anti) de Sitter space appears to be a Minkowskian space.

Is this result true? If it is then can someone mathematically show (without reference to the physical arguments presented here) how an (anti) de Sitter space can approach a Minkowskian one as the number of spacetime dimensions approaches infinity?

Answer 1

The illusion of a vanishing cosmological constant is created because of the erred approximation procedure. Since $R=\dfrac{2\Lambda D}{D-2}$, what one should (and can at best) claim in the large $D$ limit is that $R= 2\Lambda \bigg( 1+ \mathcal{O} \bigg(\dfrac{1}{D}\bigg)\bigg)$.

Thus, the Einstein equations give $R_{\mu\nu} - \Lambda g_{\mu\nu} \bigg( 1+ \mathcal{O} \bigg(\dfrac{1}{D}\bigg)\bigg)+\Lambda g_{\mu\nu}=0$.

Thus, $R_{\mu\nu}=\Lambda g_{\mu\nu} \bigg(\mathcal{O}\bigg(\dfrac{1}{D}\bigg)\bigg)$.

Now, we don't know how $\Lambda$ scales with $D$ (or even $g_{\mu\nu}$ for that matter) and thus, it is inappropriate to claim that $R_{\mu\nu}$ goes to zero in the large $D$ limit. What one can claim is that $R_{\mu\nu}$ goes as $\dfrac{1}{D}$ times $\Lambda g_{\mu\nu}$ in the large $D$ limit. Without $R_{\mu\nu}$ going to zero, neither $R$ can be claimed to go to zero nor the $\Lambda$.

Answer 2

$R_{\mu\nu} = 0$ does not give the Minkowski spacetime as a unique solution. In particular, it tells you nothing about the value of the Weyl tensor $C_{\mu \nu \alpha \beta}$. Also, more concretely, Schwarzschild spacetimes are Ricci flat (and you can find some parallels between cosmologies and schwarzschild spacetimes)

I haven't particularly studied anti-de Sitter spaces in the case of infinite dimensions, but the above does not prove that they are identically Minkowski.

Answer 3

Asymptotic flatness / Asymptotic Minkowskian is a bit of a tricky issue in terms of how one defines it. The general criteria more/less are:

1. The existence of a global timelike Killing vector field.

Now, consider a spacetime $(\hat{M}, \hat{g}_{ab})$, where $\hat{M}$ is some spacetime manifold. Then, a spacetime is called asymptotically simple if: 2. There exists a function $\omega \geq 0$ such that $g_{ab} = \omega^2 \hat{g}_{ab}$, 3. On the boundary of the spacetime, $\omega = 0$, and $\omega_{,a} \neq 0$, 4. Every null geodesic intersects this boundary at two points.

Now, if we associate the metric $g_{ab}$ with the Einstein field equations, the existence of these conditions implies that the spacetime is asymptotically flat / asymptotically Minkowskian. The Schwarzschild metric for example obeys such properties and is asymptotically flat. Related to your question, I don't think the adS metric even in the asymptotic sense behaves the same way as asymptotically flat spacetimes, because the conformal boundary is quite different than typical asymptotically flat spacetimes.