Monotonicity Formulas for Static Metrics with Non-zero Cosmological Constant

Abstract

In this paper we adopt the approach presented in Agostiniani and Mazzieri (J Math Pures Appl 104:561–586, 2015; Commun Math Phys 355:261–301, 2017) to study non-singular vacuum static space-times with non-zero cosmological constant. We introduce new integral quantities, and under suitable assumptions we prove their monotonicity along the level set flow of the static potential. We then show how to use these properties to derive a number of sharp geometric and analytic inequalities, whose equality case can be used to characterize the rotational symmetry of the underlying static solutions. As a consequence, we are able to prove some new uniqueness statements for the de Sitter and the anti-de Sitter metrics. In particular, we show that the de Sitter solution has the least possible surface gravity among three-dimensional static metrics with connected boundary and positive cosmological constant.

Appendix A: Technical Results

This appendix will be dedicated to the proof of the technical results that we have used in our work. Specifically, we will give a complete proof of Theorems 1.2, 1.5 (for the ease of reference, we have restated them here as Theorem A.1 and Theorem A.7), we will prove an estimate on the static solution near the conformal boundary in the case Λ < 0, and we will state the version of the divergence theorem that we have used in the proofs of Propositions 5.1, 5.2.

Theorem A.1

Let (M, g ₀, u) be a solution of (1.4) satisfying Assumption 1 . The set MAX(u) is discrete (and finite) and

$$\displaystyle \begin{aligned} \liminf_{t\to 1^-} \,U_p(t)\,\,\geq\,\, |{\mathrm{MAX}}(u)|\,|\mathbb{S}^{n-1}|\,,\end{aligned} $$

(A.1)

for every 0 ≤ p ≤ n − 1.

In the proof of this theorem, we will need the following result, that will be proven later.

Proposition A.2

Let (M, g ₀, u) be a solution of (1.4) and let y ₀ ∈MAX(u). Then for every d > 0 it holds

$$\displaystyle \begin{aligned} \liminf_{t\to 1^-} \bigg(\frac{1}{1-t^2}\bigg)^{n-1}\int_{\{u=t\}\cap B_{d}(y_0)} |{\mathrm{D}} u|{}^{n-1} \,{\mathrm{d}}\sigma \,\,\geq\,\, |\mathbb{S}^{n-1}|\end{aligned} $$

(A.2)

We first show how to use this result to prove Theorem A.1.

Proof of Theorem A.1

First we notice that the functions U _p(t) can be written as follows

$$\displaystyle \begin{aligned}U_p(t) \,=\, \bigg(\frac{1}{1-t^2}\bigg)^{\frac{n-1}{2}}\int_{\{u=t\}} \bigg[\frac{|{\mathrm{D}} u|{}^2}{1-u^2}\bigg]^{\frac{p}{2}} \,{\mathrm{d}}\sigma\,.\end{aligned} $$

From formula (1.10) in Theorem 1.1, we have that the term in square bracket is less or equal to 1. Thus, for every p ≤ n − 1, we have

$$\displaystyle \begin{aligned}\bigg[\frac{|{\mathrm{D}} u|{}^2}{1-u^2}\bigg]^{\frac{p}{2}} \,\geq\, \bigg[\frac{|{\mathrm{D}} u|{}^2}{1-u^2}\bigg]^{\frac{n-1}{2}}\,, \end{aligned}$$

hence U _p(t) ≥ U _n−1(t) and, in particular

$$\displaystyle \begin{aligned} \liminf_{t\to 1^-}\, U_{p}(t) \,\geq\, \liminf_{t\to 1^-}\, U_{n-1}(t)\,, \end{aligned} $$

(A.3)

so it is enough to prove the inequality (A.1) for p = n − 1.

Now we pass to analyze the set MAX(u). Suppose that it contains an infinite number of points. Then for each $k\in \mathbb {N}$ we can consider k points in MAX(u). Let 2d be the minimum of the distances between our points. Applying Proposition A.2 in a neighborhood of radius d of each of these points, we obtain

$$\displaystyle \begin{aligned}\liminf_{t\to 1^-} \, U_{n-1}(t) \,\geq\, k\,|\mathbb{S}^{n-1}|\,. \end{aligned}$$

Since this is true for every $k\in \mathbb {N}$, we conclude that $\lim _{t\to 1^-} U_{n-1}(t)=+\infty$ and, using (A.3), we find that $\lim _{t\to 1^-} U_1(t)=+\infty$. But this is impossible, since from the monotonicity of U ₁(t) (stated in Theorem 1.1-(ii)) we know that

$$\displaystyle \begin{aligned}\lim_{t\to 1^-} U_1(t)\,\leq\, U_1(0)\,=\, |\partial M|\,. \end{aligned}$$

Therefore MAX(u) contains only a finite number of points. Repeating the argument above with k = |MAX(u)| we obtain the inequality in the thesis. □

Now we turn to the proof of Proposition A.2, that will be done in various steps. Our strategy consists in choosing a suitable neighborhood of the point y ₀ where we are able to control the quantities in our limit, and then proceed to estimate them.

Notation 1

Here and throughout the paper, we agree that for $f \in {\mathcal C}^{\infty }(M)$ , $\tau \in \mathbb {R}$ and $k \in \mathbb {N}$ it holds

where the J’s are multi-indexes.

Consider a normal set of coordinates (x ¹, …, x ⁿ) in B _d(y ₀), that diagonalize the hessian in y ₀. Note that, since y ₀ is a maximum of u, the derivatives $\partial ^2_{\alpha }u_{|{ }_{y_0}}$ are non positive numbers for all α = 1, …, n, hence it makes sense to introduce the quantities $\lambda _\alpha ^2=-\partial ^2_\alpha u_{|{ }_{y_0}}$ for α = 1, …, n. Since Δu = −nu, we have $\sum _{\alpha =1}^n \lambda _\alpha ^2=n$. In particular, at least one of the λ _α’s is different from zero. We have the following Taylor expansion of u in a neighborhood of y ₀

(A.4)

From (A.4) we easily compute

(A.5)

Now we consider polar coordinates (r, θ ¹, …, θ ⁿ⁻¹), where $\theta =(\theta ^1,\dots ,\theta ^{n-1})\in \mathbb {R}^{n-1}$ are stereographic coordinates on $\mathbb {S}^{n-1}\setminus \{north\ pole\}$.

Lemma A.3

With respect to the coordinates (r, θ) = (r, θ ¹, …, θ ⁿ⁻¹), the metric g ₀ writes as

$$\displaystyle \begin{aligned} g_0\,=\,dr\otimes dr+r^2 g_{\mathbb{S}^{n-1}}+\sigma\, dr\otimes dr+\sigma_i\left(dr \otimes d\theta^i+d\theta^i \otimes dr\right)+\sigma_{ij}\, d\theta^i\otimes d\theta^j\end{aligned} $$

(A.6)

where σ = o ₂(r), σ _i = o ₂(r ²), σ _ij = o ₂(r ³), as r → 0⁺.

Proof

To ease the notation, in this proof we use the Einstein summation convention. It is known that, with respect to the normal coordinates (x ¹, …, x ⁿ) the metric g ₀ writes as

$$\displaystyle \begin{aligned}g_0=(\delta_{\alpha\beta}+\eta_{\alpha\beta})d x^\alpha\otimes d x^\beta\end{aligned} $$

where δ _αβ is the Kronecker delta and η _αβ = o ₂(r) (actually, the term η _αβ can be estimated better, but this is enough for our purposes).

Moreover, it is easy to check that the quantities ϕ ^α  =  x ^α∕r are smooth functions of the coordinates (θ ¹, …, θ ⁿ⁻¹) only, and that

$$\displaystyle \begin{aligned}\delta_{\alpha\beta}\,\frac{\partial\phi^{\alpha}}{\partial\theta^i}\, \frac{\partial\phi^{\beta}}{\partial\theta^j} \, d\theta^i \otimes d\theta^j=g_{\mathbb{S}^{n-1}} \, .\end{aligned} $$

From r ² = δ _αβ x ^α x ^β one also finds the equality δ _αβ ϕ ^α ϕ ^β = 1. Deriving it with respect to θ ⁱ we get

$$\displaystyle \begin{aligned}\delta_{\alpha\beta}\,\frac{\partial\phi^{\alpha}}{\partial\theta^i}\,\phi^{\beta}\,=\,0\, ,\qquad \mbox{for all } i=1,\dots,n-1\end{aligned} $$

We are now ready to compute

$$\displaystyle \begin{aligned} g_0 &=(\delta_{\alpha\beta}+\eta_{\alpha\beta})\, dx^{\alpha}\otimes dx^{\beta}\\ &=(1+\eta_{\alpha\beta}\phi^{\alpha}\phi^{\beta})dr\otimes dr+ (\delta_{\alpha\beta}+\eta_{\alpha\beta})r^2\,\frac{\partial\phi^{\alpha}}{\partial\theta^i}\, \frac{\partial\phi^{\beta}}{\partial\theta^j} \, d\theta^i \otimes d\theta^j\\ &\quad +\eta_{\alpha\beta}\phi^{\alpha}\frac{\partial\phi^{\beta}}{\partial\theta^i}\,r \left(dr \otimes d\theta^i+d\theta^i \otimes dr\right) \\ &=dr\otimes dr+r^2 g_{\mathbb{S}^{n-1}}+\sigma\, dr\otimes dr+\sigma_i\left(dr \otimes d\theta^i+d\theta^i \otimes dr\right)+\sigma_{ij}\, d\theta^i\otimes d\theta^j \, ,\end{aligned} $$

where σ, σ _i, σ _ij are infinitesimals of the wished order. □

We can rewrite formulæ (A.4), (A.5) in terms of (r, θ) as

$$\displaystyle \begin{aligned} u(r,\theta)\,&=\,1\,-\,\frac{r^2}{2}\,\sum_{\alpha=1}^n \big[\,\lambda_\alpha^2\, |\phi^\alpha|{}^2(\theta)\,\big] \,+\,w(r,\theta)\,, \end{aligned} $$

(A.7)

$$\displaystyle \begin{aligned} |{\mathrm{D}} u|{}^2(r,\theta)\,&=\,r^2\sum_{\alpha=1}^n \big[\,\lambda_\alpha^4\, |\phi^\alpha|{}^2(\theta)\,\big]+h(r,\theta)\,. \end{aligned} $$

(A.8)

where , . Moreover, since we know from (A.4) that ∂w∕∂x ^α = o(r) for any α, we have the following estimates on the order of the derivatives of w with respect to (r, θ)

$$\displaystyle \begin{aligned} \frac{\partial w}{\partial r}(r,\theta)\,&=\, \sum_{\alpha=1}^n\Big[\frac{\partial w}{\partial x^\alpha}(r,\theta)\,\frac{\partial x^\alpha}{\partial r}(r,\theta)\Big]\\ &= \sum_{\alpha=1}^n\Big[\frac{\partial w}{\partial x^\alpha}(r,\theta)\,\phi^\alpha(\theta)\Big]\,=\,o(r)\,, \qquad \mbox{as }r\to 0^+{} \end{aligned} $$

(A.9)

$$\displaystyle \begin{aligned} \frac{\partial w}{\partial \theta^j}(r,\theta)\,&=\, \sum_{\alpha=1}^n\Big[\frac{\partial w}{\partial x^\alpha}(r,\theta)\,\frac{\partial x^\alpha}{\partial\theta^j}(r,\theta)\Big]\\ &= \sum_{\alpha=1}^n\Big[\frac{\partial w}{\partial x^\alpha}(r,\theta)\,r\,\frac{\partial\phi^\alpha}{\partial\theta^j}(\theta)\Big]\,=\,o(r^2)\,, \quad \mbox{as }r\to 0^+{} \end{aligned} $$

(A.10)

To estimate the limit in (A.2), we need to rewrite the set {u = t}∩ B _d(y ₀) and the density

$$\displaystyle \begin{aligned}\sqrt{\det({g_0}_{|{}_{\{u=t\}\cap B_d(y_0)}})} \end{aligned}$$

as functions of our coordinates. In order to do so, it will prove useful to restrict our neighborhood B _d(y ₀) to a smaller domain where we have a better characterization of the level set {u = t}. In this regard, it is convenient, for any ε > 0, to define the set

$$\displaystyle \begin{aligned}C_\varepsilon\,=\,\Big\{\theta=(\theta^1,\dots,\theta^{n-1})\in\mathbb{R}^{n-1}\,:\,\sum_{\alpha=1}^n\big[\lambda_\alpha^2\, |\phi^\alpha|{}^2(\theta)\big]>\varepsilon\Big\}\,. \end{aligned}$$

The following result shows that, for t small enough, the level set {u = t}, is a graph over C _ε.

Lemma A.4

For any 0 < ε < 1, there exists η = η(ε) > 0 such that

• (i) the estimates $|w|(r,\theta )<\frac {\varepsilon ^2}{4}\,r^2$ , $|\partial w/\partial r|(r,\theta )<\frac {\varepsilon }{2}\,r$ , |h|(r, θ) < ε ² r ² holds on the whole B _η(y ₀).

• (ii) it holds $\frac {\partial u}{\partial r}(r,\theta )<0$ in (0, η) × C _ε.

• (iii) for every 0 < δ < η, there exists τ = τ(δ, ε) such that for any τ < t < 1, there exists a smooth function r _t : C _ε → (0, δ) such that

  $$\displaystyle \begin{aligned}\{u=t\}\cap B_\delta(y_0)\cap C_\varepsilon\,=\,\{(r_t(\theta),\theta)\,:\,\theta\in C_\varepsilon\}\,. \end{aligned}$$

Proof

Since the functions w, h in (A.7), (A.8) are o(r ²), while ∂w∕∂r is o(r) thanks to (A.9), it is clear that statement (i) is true for some η small enough. Moreover, from expansion (A.7) we compute

$$\displaystyle \begin{aligned}\frac{\partial u}{\partial r}(r,\theta)\,=\,-\,r\,\sum_{\alpha=1}^n \big[\,\lambda_\alpha^2\, |\phi^\alpha|{}^2(\theta)\,\big] \,+\,\frac{\partial w}{\partial r}(r,\theta)\,<\,-\varepsilon\,r\,+\,\frac{\varepsilon}{2}\,r\,=\,-\frac{\varepsilon}{2}\,r\,. \end{aligned}$$

This proves point (ii). To prove (iii), fix t ∈ (0, 1) and consider the function u(r, θ) − t. Since u(r, θ) → 1⁻ as r → 0⁺, we have u(r, θ) − t > 0 for small values of r.

On the other hand, from expansion (A.7) we find

$$\displaystyle \begin{aligned}u(\delta,\theta)-t&=(1-t)-\frac{\delta^2}{2}\,\sum_{\alpha=1}^n \big[\,\lambda_\alpha^2\, |\phi^\alpha|{}^2(\theta)\,\big] \,+\,w(\delta,\theta) \,<\,(1-t)\,-\,\frac{\varepsilon}{2}\,\delta^2\\ &\quad + w(\delta,\theta)\,<\,(1-t)-\frac{\varepsilon}{4}\,\delta^2\,, \end{aligned} $$

and the quantity on the right is negative for any $t>\tau =1-\frac {\varepsilon }{4}\delta ^2$.

Therefore, fixed a θ ∈ C _ε the function r↦u(r, θ) − t is positive for small values of r and negative for r = δ. Moreover from point (ii) we have that $\frac {\partial u}{\partial r}(r,\theta )<0$ for any (r, θ) ∈ (0, δ) × C _ε, hence for any θ ∈ C _ε, there exists one and only one value 0 < r _t(θ) < δ such that (r _t(θ), θ) ∈{u = t}. The smoothness of the function r _t(θ) is a consequence of the Implicit Function Theorem applied to the function u(r, θ). □

As anticipated, Lemma A.4 will now be used to estimate the density of the restriction of the metric g ₀ on {u = t}∩ ((0, δ) × C _ε).

Lemma A.5

There exists 0 < δ < η(ε) such that it holds

$$\displaystyle \begin{aligned}\sqrt{\det({g_0}_{|{}_{\{u=t\}}})}\,(r_t(\theta),\theta) \,\,>\,\, (1-\varepsilon)\,r_t^{n-1}(\theta)\,\sqrt{\det(g_{\mathbb{S}^{n-1}})}\,, \end{aligned}$$

for every θ ∈ C _ε , τ(δ, ε) < t < 1.

Proof

Let r _t be the function introduced in Lemma A.4. Taking the total derivative of u(r _t(θ), θ) = t, we find, for any θ ∈ C _ε

$$\displaystyle \begin{aligned}dr_t\,=\,-\Big[\frac{\partial u}{\partial r}(r_t(\theta),\theta)\Big]^{-1}\,\sum_{j=1}^{n-1}\frac{\partial u}{\partial \theta^j}(r_t(\theta),\theta)\, d\theta^j \,=\,-\,r_t(\theta)\sum_{j=1}^{n-1}\xi_j(\theta)d\theta^j\,, \end{aligned}$$

where

$$\displaystyle \begin{aligned}\xi_j(r_t(\theta),\theta)\,=\,\frac{1}{r_t(\theta)}\,\frac{\frac{\partial u}{\partial \theta^j}}{\frac{\partial u}{\partial r}}(r_t(\theta),\theta)\,. \end{aligned}$$

To ease the notation, in the rest of the proof we avoid to explicitate the dependence of the functions on the variables r _t(θ), θ. In order to compute the restriction of the metric on {u = t}∩ ((0, δ) × C _ε), we substitute the term dr in formula (A.6) with the formula for dr _t computed above. We obtain

$$\displaystyle \begin{aligned}{g_0}_{|{}_{\{u=t\}}}\,=\,r_t^2\, \Big[\xi_i\,\xi_j(1+\sigma)+g_{ij}^{\mathbb{S}^{n-1}}-\frac{\sigma_i}{r_t}\xi_j-\frac{\sigma_j}{r_t}\xi_i+\sigma_{ij}\Big]\,d\theta^i\otimes d\theta^j\,. \end{aligned}$$

Set $\xi =\sum _{j=1}^{n-1}\xi _j d\theta ^j$. We have the following

$$\displaystyle \begin{aligned} \notag \sqrt{\det\Big[\Big(\xi_i\,\xi_j+g_{ij}^{\mathbb{S}^{n-1}}\Big)\,d\theta^i\otimes d\theta^j\Big]}\,&=\,\sqrt{\det\Big(\xi\,\otimes\,\xi\,+\,g_{\mathbb{S}^{n-1}}\Big)} \\ \notag &=\,\sqrt{(1+|\xi|{}_{g_{\mathbb{S}^{n-1}}}^2)\,\det (g_{\mathbb{S}^{n-1}})} \\ {} &\geq\,\sqrt{\det (g_{\mathbb{S}^{n-1}})}\,, \end{aligned} $$

(A.11)

where in the second equality we have used the Matrix Determinant Lemma.

On the other hand, since σ _i = o(r ²) and σ _ij = o(r ³), we deduce that

$$\displaystyle \begin{aligned}\sigma\,\xi_i\,\xi_j-\frac{\sigma_i}{r_t}\xi_j-\frac{\sigma_j}{r_t}\xi_i+\sigma_{ij}=o(r_t)\,, \end{aligned}$$

hence

$$\displaystyle \begin{aligned} \sqrt{\det({g_0}_{|{}_{\{u=t\}}})}\,=\,(1+\omega)\,r_t^{n-1}\,\sqrt{\det\Big[\Big(\xi_i\,\xi_j+g_{ij}^{\mathbb{S}^{n-1}}\Big)\,d\theta^i\otimes d\theta^j\Big]}\,, \end{aligned} $$

(A.12)

with ω = o(1) as r → 0⁺. In particular, we can choose δ small enough so that |ω| < ε on (0, δ) × C _ε. Combining Eqs. (A.11) and (A.12) we have the thesis. □

We also need an estimate of the integrand in (A.2), which is provided by the following lemma.

Lemma A.6

We can choose 0 < δ < η(ε) such that

$$\displaystyle \begin{aligned}\frac{|{\mathrm{D}} u|{}^2}{1-u^2}(r,\theta) \,>\, (1-\varepsilon)\,\sum_{\alpha=1}^n \big[\,\lambda_\alpha^2\, |\phi^\alpha|{}^2(\theta)\,\big] \end{aligned}$$

for every (r, θ) ∈ (0, δ) × C _ε.

Proof

To ease the notation, in this proof we avoid to explicitate the dependence of the functions on the coordinates r, θ. From expansions (A.7) and (A.8) we deduce

Using the Cauchy-Schwarz Inequality we have

$$\displaystyle \begin{aligned} \sum_{\alpha=1}^n \big(\lambda_\alpha^2\,|\phi^\alpha|{}^2\,\big) \,\leq\, \bigg[\sum_{\alpha=1}^n \big(\lambda_\alpha^4\,|\phi^\alpha|{}^2\,\big)\bigg]^{\frac{1}{2}}\,\cdot\,\bigg[\sum_{\alpha=1}^n |\phi^\alpha|{}^2\bigg]^{\frac{1}{2}} \,=\, \bigg[\sum_{\alpha=1}^n \big(\lambda_\alpha^4\,|\phi^\alpha|{}^2\,\big)\bigg]^{\frac{1}{2}}\,, \end{aligned} $$

(A.13)

hence, recalling that $\sum _{\alpha =1}^n (\lambda _\alpha ^2 |\phi ^\alpha |{ }^2)>\varepsilon$ on C _ε, we have also $\sum _{\alpha =1}^n (\lambda _\alpha ^4 |\phi ^\alpha |{ }^2)>\varepsilon ^2$. Therefore, from Lemma A.4-(i) we easily compute

$$\displaystyle \begin{aligned}\bigg|\left(1+\frac{h}{\sum_{\alpha=1}^n (\lambda_\alpha^4\, |\phi^\alpha|{}^2)}\right)\left(1-\frac{2\,w}{\sum_{\alpha=1}^n (\lambda_\alpha^2\, |\phi^\alpha|{}^2)}\right)^{-1}\bigg|\,>\,\frac{1-\delta^2}{1+\frac{\varepsilon\delta^2}{2}}\,. \end{aligned}$$

In particular, we can choose δ small enough so that the right hand side of the inequality above is greater than 1 − ε. Hence, we get

$$\displaystyle \begin{aligned} \frac{|{\mathrm{D}} u|{}^2}{1-u^2} \,\geq\, \frac{|{\mathrm{D}} u|{}^2}{2(1-u)} \,\geq\,(1-\varepsilon)\,\frac{\sum_{\alpha=1}^n \big(\lambda_\alpha^4\, |\phi^\alpha|{}^2\,\big)}{\sum_{\alpha=1}^n \big(\lambda_\alpha^2\, |\phi^\alpha|{}^2\,\big)} \,\geq\, (1-\varepsilon)\,\sum_{\alpha=1}^n \big[\,\lambda_\alpha^2\, |\phi^\alpha|{}^2\,\big] \,, \end{aligned}$$

where in the first inequality we have used that u ≤ 1 on M and in the latter inequality we have used (A.13). □

Now we are finally able to prove our proposition.

Proof of Proposition A.2

For every ε > 0 and 0 < δ ≤ d, we have the following estimate of the left hand side of condition (A.2)

(A.14)

Since u ≤ 1, we have 2∕(1 + u) ≥ 1. Moreover, from (A.7) and Lemma A.4-(i), we obtain

$$\displaystyle \begin{aligned}{}[2(1-u)](r,\theta)&=r^2\,\sum_{\alpha=1}^n\big[\lambda_\alpha^2|\phi^\alpha|{}^2(\theta)\big]\,\Big(1-\frac{w(r,\theta)}{r^2\sum_{\alpha=1}^n\lambda_\alpha|\phi^\alpha|{}^2(\theta)}\Big)\\ &< (1+\varepsilon)\,r^2\,\sum_{\alpha=1}^n\big[\lambda_\alpha^2|\phi^\alpha|{}^2(\theta)\big]\,. \end{aligned} $$

Now fix a δ small enough so that Lemmas A.5, A.6 are in charge. Taking the limit of integrand (A.14) as t → 1⁻ we obtain the estimate

(A.15)

It is clear that the functions converge to as ε → 0⁺, where

$$\displaystyle \begin{aligned}C_0\,=\,\Big\{\theta=(\theta^1,\dots,\theta^{n-1})\in\mathbb{R}^{n-1}\,:\,\sum_{\alpha=1}^n\big[\lambda_\alpha^2\, |\phi^\alpha|{}^2(\theta)\big]\neq 0\Big\}\,\subseteq \,\mathbb{S}^{n-1}\,. \end{aligned}$$

Therefore, taking the limit of (A.15) as ε → 0⁺ and using the Monotone Convergence Theorem, we find

To end the proof, it is enough to show that the set $\mathbb {S}^{n-1}\setminus C_0$ is negligible. But this is clear. In fact, since $\sum _{\alpha =1}^n\lambda _\alpha ^2=n$, there exists at least one integer β such that λ _β ≠ 0. Thus $\mathbb {S}^{n-1}\setminus C_0$ is contained in the hypersurface {ϕ ^β = 0}, hence its n-measure is zero. This proves inequality (A.2) and the thesis. □

This concludes the proof for the de Sitter case. In the anti-de Sitter case we can prove the following analogue of Theorem A.1.

Theorem A.7

Let (M, g ₀, u) be a conformally compact static solution of problem 1.17 satisfying Assumption 2 . Then the set MIN(u) is discrete (and finite) and

$$\displaystyle \begin{aligned}\liminf_{t\to 1^+} \,U_p(t)\,\,\geq\,\, |{\mathrm{MIN}}(u)|\,|\mathbb{S}^{n-1}|\,, \end{aligned}$$

for every p ≤ n − 1.

The proof follows the exact same scheme as the de Sitter case, the only small modifications being in the proof of Lemma A.5 and in the computation (A.15), where we have used the fact that u ≤ 1. This is not true anymore, however, since we are working around a minimum point, we can suppose u < 1 + κ, where κ is an infinitesimal quantity that can be chosen to be as small as necessary. Aside from this little expedient, the proof is virtually the same as the de Sitter case, thus we omit it.

We pass now to the proof of some other results that we have used in our work. The next lemma is useful in order to study the behavior of the static solutions of problem (1.17) near the conformal boundary.

Lemma A.8

Let (M, g ₀, u) be a conformally compact static solution to problem (1.17). Suppose that $1/\sqrt {u^2-1}$ is a defining function, so that the metric g = g ₀∕(u ² − 1) extends to the conformal boundary ∂M. Then

• (i) $\lim _{x\to \bar {x}}(u^2-1-|{\mathrm {D}} u|{ }^2)$ is well-definite and finite for every $\bar x\in \partial M$ ,

• (ii) ∂M is a totally geodesic hypersurface in $(\overline {M},g)$.

Proof

For the proof of this result, it is convenient to use the notations introduced in Sect. 3. Let φ be the function defined by (3.18). By hypothesis, M is the interior of a compact manifold $\overline {M}$ and the metric g is well defined on the whole $\overline {M}$. In particular, the scalar curvature R_g is a smooth finite function at ∂M. Therefore, from Eq. (3.24) we easily deduce that $\lim _{x\to \bar x}u^2(1-|\nabla \varphi |{ }_g^2)$ is well-definite and finite for every $\bar x\in \partial M$. Since

$$\displaystyle \begin{aligned}|\nabla\varphi|{}_g^2=\frac{|{\mathrm{D}} u|{}^2}{u^2-1}\,, \end{aligned}$$

this proves point (i).

To prove statement (ii), we first observe that, since |∇φ|_g = 1 at ∂M (as it follows immediately from point (i)), there exists δ > 0 such that |∇φ|_g ≠ 0 on the whole collar $\mathcal {U}_\delta =\{\varphi <\delta \}$. Therefore, proceeding as in Sect. 3.4, we find a set of coordinates {φ, 𝜗 ¹, …, 𝜗 ⁿ⁻¹} on $\mathcal {U}_{\delta }$, such that the metric g writes as

$$\displaystyle \begin{aligned}g=\frac{d\varphi\otimes d\varphi}{|\nabla\varphi|{}_g^2}+g_{ij}(\varphi,\theta^i,\dots,\theta^{n-1})d\theta^i\otimes d\theta^j\,. \end{aligned}$$

With respect to these coordinates, the second fundamental form of the boundary ∂M = {φ = 0} is

$$\displaystyle \begin{aligned}{\mathrm{h}}^{(g)}_{ij}\,=\,\frac{\nabla^2_{ij}\varphi}{|\nabla\varphi|{}_g}=\nabla^2_{ij}\varphi\,,\qquad \mbox{for } i,j=1,\dots,n-1\,. \end{aligned}$$

On the other hand, from the first equation of problem (3.23), we easily deduce that ∇² φ = 0 on ∂M. This concludes the proof of point (ii). □

Finally, in order to prove the integral identities in Sect. 5, we need an extension of the classical Divergence Theorem to the case of open domains whose boundary has a (not too big) nonsmooth portion. Note that [4, Theorem A.1] is not enough for our purposes, because hypothesis (ii) is not necessarily fulfilled. To avoid problems, we state the following generalization, due to De Giorgi and Federer.

Theorem A.9 ([16, 17, 19, 20])

Let (M, g) be a n-dimensional Riemannian manifold, with n ≥ 2, let E ⊂ M be a bounded open subset of M with compact boundary ∂E of finite (n − 1)-dimensional Hausdorff measure, and suppose that ∂E =  Γ ⊔ Σ, where the subsets Γ and Σ have the following properties:

• (i) For every x ∈ Γ, there exists an open neighborhood U _x of x in M such that Γ ∩ U _x is a smooth regular hypersurface.

• (ii) The subset Σ is compact and $\mathcal {H}^{n-1}(\Sigma ) = 0$.

If X is a Lipschitz vector field defined in a neighborhood of $\overline {E}$ then the following identity holds true

$$\displaystyle \begin{aligned} \int\limits_E {\mathrm{div}} X \, {\mathrm{d}} \mu \, = \, \int\limits_{\Gamma} \langle X \,|\, {\mathrm{n}} \rangle \, {\mathrm{d}} \sigma , \end{aligned} $$

(A.16)

where n denotes the exterior unit normal vector field.

Appendix B: Boucher-Gibbons-Horowitz Method

In this section we discuss an alternative approach to the study of the rigidity of the de Sitter and anti-de Sitter spacetime, which does not require the machinery of Sect. 3. Without the need of any assumptions, this method will allow to derive results that are comparable to Theorems 2.2, 2.6 (case Λ > 0) and Theorems 2.14, 2.18 (case Λ < 0). In the case Λ > 0, the computations that we are going to show are quite classical (see [12, 14]). However, to the author’s knowledge, the analogous calculations in the case Λ < 0 are new.

As usual, we start with the case Λ > 0. Recalling the Bochner formula and the equations in (1.4) we compute

$$\displaystyle \begin{aligned} \notag \Delta|{\mathrm{D}} u|{}^2\,&=\,2\,|{\mathrm{D}}^2 u|{}^2\,+\,2\,{\mathrm{Ric}}({\mathrm{D}} u,{\mathrm{D}} u)\,+\,2\langle{\mathrm{D}}\Delta u\,|\,{\mathrm{D}} u\rangle \\ \notag &=\,2\,|{\mathrm{D}}^2 u|{}^2\,+\,2\,\Big[\,\frac{1}{u}\,{\mathrm{D}}^2 u({\mathrm{D}} u,{\mathrm{D}} u)\,+\,n\,|{\mathrm{D}} u|{}^2\,\Big]\,-\,2n\,|{\mathrm{D}} u|{}^2 \\ {} &=\,2\,|{\mathrm{D}}^2 u|{}^2\,+\,\frac{1}{u}\,\langle{\mathrm{D}}|{\mathrm{D}} u|{}^2\,|\,{\mathrm{D}} u\rangle\,. \end{aligned} $$

(B.1)

Now, if we consider the field

$$\displaystyle \begin{aligned}Y\,=\,{\mathrm{D}}|{\mathrm{D}} u|{}^2\,-\,\frac{2}{n}\,\Delta u\,{\mathrm{D}} u \end{aligned}$$

we can compute its divergence using (B.1).

$$\displaystyle \begin{aligned} {\mathrm{div}}(Y)\,&=\, \Delta|{\mathrm{D}} u|{}^2\,-\,\frac{2}{n}\langle{\mathrm{D}}\Delta u\,|\,{\mathrm{D}} u\rangle\,-\,\frac{2}{n}(\Delta u)^2 \\ &=\,2\,\Big[\,|{\mathrm{D}}^2 u|{}^2-\frac{(\Delta u)^2}{n}\,\Big]\,+\,\frac{1}{u}\,\langle{\mathrm{D}}|{\mathrm{D}} u|{}^2\,|\,{\mathrm{D}} u\rangle\,+\,2\,|{\mathrm{D}} u|{}^2\,. \end{aligned} $$

More generally, for every nonzero $\mathcal {C}^1$ function α = α(u):

$$\displaystyle \begin{aligned} \frac{{\mathrm{div}} (\alpha\, Y)}{\alpha}\,&=\,{\mathrm{div}}(Y)\,+\,\frac{\dot\alpha}{\alpha}\,\langle Y\,|\,{\mathrm{D}} u\rangle \\ &=\,2\,\Big[\,|{\mathrm{D}}^2 u|{}^2-\frac{(\Delta u)^2}{n}\,\Big]\,+\,\Big(\frac{\dot\alpha}{\alpha}+\frac{1}{u}\Big)\Big(\langle{\mathrm{D}}|{\mathrm{D}} u|{}^2\,|\,{\mathrm{D}} u\rangle\,+\,2\,u\,|{\mathrm{D}} u|{}^2\Big)\,. \end{aligned} $$

where $\dot \alpha$ is the derivative of α with respect to u. The computation above suggests us to choose

$$\displaystyle \begin{aligned}\alpha(u)\,=\,\frac{1}{u}\,. \end{aligned}$$

With this choice of α, we have

$$\displaystyle \begin{aligned} {\mathrm{div}}\Big(\frac{1}{u}\,Y\,\Big)\,=\,\frac{2}{u}\,\Big[\,|{\mathrm{D}}^2 u|{}^2-\frac{(\Delta u)^2}{n}\,\Big]\,. \end{aligned} $$

(B.2)

Proposition B.1

Let (M, g ₀, u) be a static solution to problem (1.4). Then, for every t ∈ [0, 1) it holds

(B.3)

Moreover, if there exists t ₀ ∈ (0, 1) such that

(B.4)

then the triple (M, g ₀, u) is isometric to the de Sitter solution.

Remark B.1

Recalling Remark 1, it is easy to realize that the integral on the left hand side of (B.3) is well defined also when t is a singular value of u.

Remark B.2

Note that the right hand side of inequality (B.3) is always nonnegative, as opposed to formula (5.8), where we needed to suppose Assumption 3 to achieve the same result. This is one of the reasons why this approach works without the need to suppose any assumption.

Proof of Proposition B.1

Suppose for the moment that {u = t} is a regular level set. Integrating by parts identity (B.2), we obtain

(B.5)

where n = −Du∕|Du| is the outer g-unit normal of the set {u ≥ t} at its boundary. On the other hand, from the first formula in (3.27) it is easy to deduce that

$$\displaystyle \begin{aligned}\langle Y\,|\,{\mathrm{D}} u\rangle \, = \, 2\left( {\mathrm{D}}^2 u({\mathrm{D}} u,{\mathrm{D}} u)\,-\,\frac{|{\mathrm{D}} u|{}^2\Delta u}{n} \right)\, = \, - 2\left(|{\mathrm{D}} u|{}^3 \, {\mathrm{H}} -\, \frac{n-1}{n}\,|{\mathrm{D}} u|{}^2\,\Delta u\right) . \end{aligned} $$

Substituting in (B.5) proves formula (B.3) in the case where {u = t} is a regular level set.

In the case where t > 0 is a singular value of u, we need to apply a slightly refined version of the Divergence Theorem, namely Theorem A.9 in the Appendix A, in order to perform the integration by parts which leads to identity (B.5). The rest of the proof is identical to what we have done for the regular case. We set

$$\displaystyle \begin{aligned} \begin{array}{rcl} X = \frac{1}{u}\, Y \qquad \mbox{and} \qquad E= \{ u>t \}\, . \end{array} \end{aligned} $$

so that ∂E = {u = t}.

As usual, we denote by Crit(u) = {x ∈ M | Du(x) = 0} the set of the critical points of u, From [30] (see also [27, Theorem 6.3.3]), we know that there exists an open (n − 1)-dimensional submanifold N ⊆Crit(u) such that $\mathcal {H}^{n-1}({\mathrm {Crit}}(u)\setminus N)=0$. Set Σ = ∂E ∩ (Crit(u) ∖ N) and Γ = ∂E ∖ Σ, so that ∂E can be written as the disjoint union of Σ and Γ. We have $\mathcal {H}^{n-1}(\Sigma )=0$ by definition, while Γ is the union of the regular part of ∂E and of N, which are open (n − 1)-submanifolds. Therefore the hypotheses of Theorem A.9 are met, and we can apply it to conclude that Eq. (B.5) holds true also when t is a singular value of u.

To prove the second part, we observe that from (B.3) and (B.4) one immediately gets D² u = ( Δu∕n) g ₀ in {u ≥ t ₀}. Since u is analytic, the same equality holds on the whole manifold M. Now we can use the results in [28] to conclude that (M, g ₀, u) is the de Sitter solution. □

The proposition above is particularly interesting when applied at the boundary ∂M = {u = 0}.

Corollary B.2

Let (M, g ₀, u) be a static solution to problem (1.4). Then it holds

(B.6)

Moreover, if the equality holds then the triple (M, g ₀, u) is isometric to the de Sitter solution.

Proof

First we compute from the equations in (1.4) and formula (3.27), that

$$\displaystyle \begin{aligned}\frac{{\mathrm{H}}\,|{\mathrm{D}} u|}{u}\,=\,-{\mathrm{Ric}}(\nu,\nu)\,, \end{aligned}$$

where ν = Du∕|Du| as usual. In particular, we have H = 0 on ∂M. Hence we can use the Gauss–Codazzi identity to find

$$\displaystyle \begin{aligned}\frac{{\mathrm{H}}\,|{\mathrm{D}} u|}{u}\,=\,\frac{{\mathrm R}^{\partial M}-{\mathrm R}}{2}\,=\,\frac{1}{2}\,\big[\,{\mathrm R}^{\partial M}-\,n(n-1)\,\big]\,. \end{aligned}$$

Substituting t = 0 in formula (B.3) and applying Proposition B.1, we have the thesis. □

Now we turn our attention to the case Λ < 0. Mimicking the computations done in the case Λ > 0, but using the equations in (1.17) instead of the ones in (1.4) we obtain

$$\displaystyle \begin{aligned} {\mathrm{div}}\Big(\frac{1}{u}\,Y\,\Big)\,=\,\frac{2}{u}\,\Big[\,|{\mathrm{D}}^2 u|{}^2-\frac{(\Delta u)^2}{n}\,\Big]\,. \end{aligned} $$

(B.7)

Incidentally, we notice that this equation coincides with the analogous formula (B.2) in the case Λ > 0. We are now ready to state the analogous of Proposition B.1.

Proposition B.3

Let (M, g ₀, u) be a static solution to problem (1.17). Then, for every t ∈ (1, +∞) it holds

(B.8)

Moreover, if there exists t ₀ ∈ (1, +∞) such that

(B.9)

then the triple (M, g ₀, u) is isometric to the anti-de Sitter solution.

Remark B.3

Recalling Remark 1, it is easy to realize that the integral on the left hand side of (B.8) is well defined also when t is a singular value of u.

Proof of Proposition B.3

The proof is almost identical to the proof of Proposition B.3. The only change is that, when we apply the divergence theorem, we need the outer g-unit normal of the set {u ≤ t}, that is n = Du∕|Du| instead of −Du∕|Du|. This is the reason of the different signs in formulæ (B.3), (B.8). □

Now suppose that the manifold M is conformally compact. We would like to use Proposition B.3 to study the behavior of a static solution at the conformal boundary ∂M. In order to simplify the computations and to emphasize the analogy with the case Λ > 0, it will prove useful to suppose that Assumption 2-bis holds. Therefore, from now on we suppose that $1/\sqrt {u^2-1}$ is a defining function, and that $\lim _{x\to \bar x}(u^2-1-|{\mathrm {D}} u|{ }^2)=0$ for every $\bar x\in \partial M$. We are now ready to prove the analogous of Corollary B.2 in the case of a negative cosmological constant.

Corollary B.4

Let (M, g ₀, u) be a conformally compact static solution to problem (1.17) satisfying assumption 2-bis , and let g = g ₀∕(u ² − 1). Then it holds

(B.10)

Moreover, if

(B.11)

where ν _g = Du∕|Du|_g , then the triple (M, g ₀, u) is isometric to the anti-de Sitter solution.

Proof

First we compute from the equations in (1.17) and formula (3.27), that

$$\displaystyle \begin{aligned}\frac{{\mathrm{H}}\,|{\mathrm{D}} u|}{u}\,=\,-{\mathrm{Ric}}(\nu,\nu)\,,\end{aligned} $$

where ν = Du∕|Du| as usual. Therefore, we can rewrite formula (B.8) as

(B.12)

Now we use Eq. (3.16) in order to rewrite the term in the square brackets in the following way

$$\displaystyle \begin{aligned}\big[\,(n-1)u^2-1\,\big]\,\big[\,{\mathrm{Ric}}(\nu,\nu)\,+\,(n-1)\,\big] \,=\, {\mathrm{Ric}}_g(\nu_g,\nu_g)-\bigg(\frac{(n-1)u^2+1}{u^2-1}\bigg)\left(u^2-1-|{\mathrm{D}} u|{}^2\right)\,.\end{aligned} $$

Now it is easy to obtain from inequality (B.12) the following formula

Since $\lim _{x\to \bar x} (u^2-1-|{\mathrm {D}} u|{ }^2)=0$, in particular $|{\mathrm {D}} u|/\sqrt {u^2-1}$ goes to zero as t → +∞. Therefore, taking the limit as t → +∞ of the formula above, we obtain

(B.13)

where ν _g = Du∕|Du|_g. Since ∂M is a totally geodesic hypersurface by Lemma A.8-(ii), from the Gauss–Codazzi equation and formula (3.24) we obtain

$$\displaystyle \begin{aligned}2\,{\mathrm{Ric}}_g(\nu_g,\nu_g)\,=\,{\mathrm{R}}_g-{\mathrm{R}}_g^{\partial M}=\,(n-1)(n-2)-{\mathrm{R}}_g^{\partial M}\,. \end{aligned}$$

Substituting in Eq. (B.13) we obtain formula (B.10).

To prove the rigidity statement, we observe that we can rewrite formula (B.11) as

Now we recall that u [Ric(ν, ν) + (n − 1)] = −H|Du| − (n − 1) Δu∕n and we conclude using Proposition B.3. □