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

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?

• As Jerry pointed out in his answer, $R_{\mu \nu}$ being equal to zero does not imply that flat spacetime is its unique solution. For flat spacetime, you need all the components of the Riemann tensor $R^{\mu}_{\ \nu \lambda \sigma}$ to be zero. – Avantgarde Jun 6 '17 at 23:54
• Yes, I was mistaken about that. But I think it is in some sense not all that core of my question. I have edited the question to circumvent this issue. Kindly have a look. – Dvij D.C. Jun 7 '17 at 5:25
• This still has the problem of it not being clear how any of your variables go to zero, which is going to affect the problem. I don't think there's a way around calculating $g_{ab}$ for arbitrary $D$, and then taking the $D\rightarrow\infty$ limit. – Jerry Schirmer Jun 7 '17 at 14:10
• In particular, setting $\Lambda = 0$ feels deeply wrong, because it is a free parameter of your theory. If anything, this feels more like a proof that anti-de Sitter spaces are necessarily finite dimensional – Jerry Schirmer Jun 7 '17 at 14:23
• @JerrySchirmer It seems to me that I am not setting $\Lambda=0$ by hand. Rather, the large $D$ limit forces it upon us that $\Lambda$ must go to zero. Regarding your suggestion that rather than saying that (anti) de Sitter spaces tend to Minkowskian in the large $D$ limit, we should say that (anti) de Sitter spaces must be finite dimensional seems quite appropriate - but still, I am a bit confused. – Dvij D.C. Jun 8 '17 at 4:57

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 $\Lambda$.

$R_{\mu\nu} = 0$ 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.

• Aren't all spaces with Ricci tensor zero asymptotically Minkowskian? – Dvij D.C. Jun 6 '17 at 20:24
• @Dvij: no. The obvious counterexample is a non-asymptotically flat universe containing only a plane gravitational wave. – Jerry Schirmer Jun 6 '17 at 20:39
• What about non-compact Calabi-Yaus? – Kosm Jun 7 '17 at 2:56
• @JerrySchirmer Yes, I realize I was mistaken about that. But I think it is in some sense not all that core of my question. I have edited the question to circumvent this issue. Kindly have a look. – Dvij D.C. Jun 7 '17 at 5:26

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.

• Thank you for your answer! I have edited my question to circumvent the issues of in what case $R_{\mu\nu}=0$ suggests asymptotically Minkowskian spacetime. Kindly have a look. – Dvij D.C. Jun 7 '17 at 5:28