The Case of Equality for the Spacetime Positive Mass Theorem

Abstract

The rigidity of the spacetime positive mass theorem states that an initial data set (M, g, k) satisfying the dominant energy condition with vanishing mass can be isometrically embedded into Minkowski space. This has been established by Beig–Chruściel and Huang–Lee under additional decay assumptions for the energy and momentum densities $\mu$ and J. In this note, we give a new and elementary proof in dimension 3 which removes these additional decay assumptions. Our argument uses spacetime harmonic functions and Liouville’s theorem. We also provide an alternative proof based on the Killing development of (M, g, k).

Appendices

Appendix A: The Fundamental Theorem of Hypersurfaces

Proof of Proposition 3.1

We follow the proof of [18, p. 100]. Let U be a compact subset of M. We construct the metric $\bar{g}=-dt^2+g_t$ on $(-\varepsilon ,\varepsilon )\times U$ by prescribing

$$\begin{aligned} \partial _t g_t(\partial _i,\partial _j)&=2{\bar{\nabla }}^2_{ij} t, \end{aligned}$$

(55)

$$\begin{aligned} {\bar{g}}|_{t=0}&=g, \end{aligned}$$

(56)

$$\begin{aligned} \partial _t ({\bar{\nabla }}^2_{ij}t)-({\bar{\nabla }}^2t)^2_{ij}&= 0, \end{aligned}$$

(57)

$$\begin{aligned} {\bar{\nabla }}^2_{ij} t|_{t=0}&= k_{ij}, \end{aligned}$$

(58)

where $({\bar{\nabla }}^2t)^2_{ij}={{\bar{g}}}^{kl}({\bar{\nabla }}^2_{ik}t)({\bar{\nabla }}^2_{jl}t)$. We will use Roman letters $\{i,j,k,l\}$ to denote indices tangential to M. By standard ODE existence theory there exists a small $\varepsilon >0$ such that we can solve the above equation for $t\in (-\varepsilon ,\varepsilon )$. Next, we take a cover $\{U_i\}$ of M. According to the asymptotics of (M, g, k), there exists a uniform $\varepsilon >0$ for each $U_i$. Therefore, we can patch together the above construction and (M, g) can be embedded in $((-\varepsilon ,\varepsilon )\times M,{\bar{g}})$ with the second fundamental form k.

To verify the flatness of ${{\bar{g}}}$ we proceed exactly as in [18]. It suffices to verify that the curvatures ${\bar{R}}_{tijt}$, ${\bar{R}}_{ijkl}$ and ${\bar{R}}_{tijk}$ are vanishing. Observe that $\langle {\bar{\nabla }}t,{\bar{\nabla }} t\rangle =-1$ implies ${\bar{\nabla }}_i{\bar{\nabla }}_t t =0$. Combining this with Eq. (57) yields

$$\begin{aligned} 0&=\partial _t({\bar{\nabla }}^2_{ij}t)-({\bar{\nabla }}^2 t)^2_{ij} \end{aligned}$$

(59)

$$\begin{aligned}&={\bar{\nabla }}_t{\bar{\nabla }}_i{\bar{\nabla }}_j t+({\bar{\nabla }}^2 t)^2_{ij} \end{aligned}$$

(60)

$$\begin{aligned}&= {\bar{\nabla }}_i{\bar{\nabla }}_t{\bar{\nabla }}_j t-{\bar{R}}_{tijt}+({\bar{\nabla }}^2 t)^2_{ij} \end{aligned}$$

(61)

$$\begin{aligned}&=\partial _i({\bar{\nabla }}_t{\bar{\nabla }}_j t)-{\bar{\nabla }}^2t(\partial _t,{\bar{\nabla }}_i\partial _j)-{\bar{\nabla }}^2t(\partial _j,{\bar{\nabla }} _i\partial _t)-{\bar{R}}_{tijt}+({\bar{\nabla }}^2 t)^2_{ij} \end{aligned}$$

(62)

$$\begin{aligned}&=-{\bar{R}}_{tijt}. \end{aligned}$$

(63)

Since ${\bar{R}}_{tijt}=0$, ${\bar{\nabla }}_t \partial _t=0$ and ${\bar{\Gamma }}_{ti}^t=0$, we obtain

$$\begin{aligned} \partial _t({\bar{R}}_{tijk})&=({\bar{\nabla }}_t{\bar{R}})_{tijk}+{\bar{R}}_{tljk}{\bar{\Gamma }}_{ti}^l +{\bar{R}}_{tilk}{\bar{\Gamma }}_{tj}^l+{\bar{R}}_{tljk}{\bar{\Gamma }}_{tk}^l \end{aligned}$$

(64)

$$\begin{aligned}&=({\bar{\nabla }}_j{\bar{R}})_{titk}+({\bar{\nabla }}_k{\bar{R}})_{tijt} +{\bar{R}}_{tljk}{\bar{\Gamma }}_{ti}^l+{\bar{R}}_{tilk}{\bar{\Gamma }}_{tj}^l+{\bar{R}}_{tljk}{\bar{\Gamma }}_{tk}^l \end{aligned}$$

(65)

$$\begin{aligned}&=\partial _j({\bar{R}}_{titk})-{\bar{R}}_{litk}{\bar{\Gamma }}_{jt}^l-{\bar{R}}_{tilk}{\bar{\Gamma }}_{jt}^l+ \partial _k({\bar{R}}_{tijt})-{\bar{R}}_{lijt}{\bar{\Gamma }}_{kt}^l-{\bar{R}}_{tijl}{\bar{\Gamma }}_{kt}^l \end{aligned}$$

(66)

$$\begin{aligned}&\qquad +{\bar{R}}_{tljk}{\bar{\Gamma }}_{ti}^l+{\bar{R}}_{tilk}{\bar{\Gamma }}_{tj}^l+{\bar{R}}_{tljk}{\bar{\Gamma }}_{tk}^l \end{aligned}$$

(67)

$$\begin{aligned}&=-{\bar{R}}_{litk}{\bar{\Gamma }}_{jt}^l-{\bar{R}}_{tilk}{\bar{\Gamma }}_{jt}^l-{\bar{R}}_{lijt} {\bar{\Gamma }}_{kt}^l-{\bar{R}}_{tijl}{\bar{\Gamma }}_{kt}^l +{\bar{R}}_{tljk}{\bar{\Gamma }}_{ti}^l\nonumber \\&\qquad +{\bar{R}}_{tilk}{\bar{\Gamma }}_{tj}^l+{\bar{R}}_{tljk}{\bar{\Gamma }}_{tk}^l. \end{aligned}$$

(68)

According to the Codazzi equation, ${\bar{R}}_{tijk}|_{t=0}=0$, and thus ${\bar{R}}_{tijk}=0$. Next, we compute

$$\begin{aligned} \partial _t({\bar{R}}_{ijkl})&=({\bar{\nabla }}_t {\bar{R}})_{ijkl}+{\bar{R}}_{sjkl}{\bar{\Gamma }}_{ti}^s +{\bar{R}}_{iskl}{\bar{\Gamma }}_{tj}^s+{\bar{R}}_{ijsl}{\bar{\Gamma }}_{tk}^s+{\bar{R}}_{ijks}{\bar{\Gamma }}_{tl}^s \end{aligned}$$

(69)

$$\begin{aligned}&=(\nabla _k {\bar{R}})_{ijtl}+(\nabla _l {\bar{R}})_{ijkt} +{\bar{R}}_{sjkl}{\bar{\Gamma }}_{ti}^s+{\bar{R}}_{iskl}{\bar{\Gamma }}_{tj}^s +{\bar{R}}_{ijsl}{\bar{\Gamma }}_{tk}^s+{\bar{R}}_{ijks}{\bar{\Gamma }}_{tl}^s \end{aligned}$$

(70)

$$\begin{aligned}&=-{\bar{R}}_{ijsl}{\bar{\Gamma }}_{kt}^s-{\bar{R}}_{ijks}{\bar{\Gamma }}_{lt}^s +{\bar{R}}_{sjkl}{\bar{\Gamma }}_{ti}^s+{\bar{R}}_{iskl}{\bar{\Gamma }}_{tj}^s +{\bar{R}}_{ijsl}{\bar{\Gamma }}_{tk}^s+{\bar{R}}_{ijks}{\bar{\Gamma }}_{tl}^s \end{aligned}$$

(71)

$$\begin{aligned}&={\bar{R}}_{sjkl}{\bar{\Gamma }}_{ti}^s+{\bar{R}}_{iskl}{\bar{\Gamma }}_{tj}^s. \end{aligned}$$

(72)

According to the Gauss equations, ${\bar{R}}_{ijkl}|_{t=0}=0$, and thus ${\bar{R}}_{ijkl}=0$. Therefore, ${{\bar{M}}}$ is flat which implies together with $M\cong {\mathbb {R}}^3$ that ${{\bar{M}}}$ is a subset of Minkowski spacetime. $\square$

Appendix B: Killing Development

Another way to prove rigidity for the spacetime PMT is to construct a spacetime using spacetime harmonic function, and demonstrating that this spacetime is Minkowski space. For this purpose, we define on ${\tilde{M}}^4={\mathbb {R}}\times M^3$ the Lorentzian metric

$$\begin{aligned} \tilde{g}=2\text {d}\tau \text {d}u+g, \end{aligned}$$

(73)

where $\tau$ is the flat coordinate on the ${\mathbb {R}}$-factor. This so-called Killing Development is motivated by [3, 11], though we note that the Killing Development in [3, 11] was obtained from three, rather than a single vector field. Since $M^3\cong {\mathbb {R}}^3$, we have ${\tilde{M}}^4\cong {\mathbb {R}}^4$, and thus it suffices to show that ${\tilde{g}}$ is flat. The flatness of ${\tilde{g}}$ follows essentially from the Gauss and Codazzi equations computed in Sect. 3. We present here another approach which has the advantage that it does not require the additional regularity assumptions $g\in C^3(M^3)$ and $k\in C^2(M^3)$ used in Lemma 3.4, and therefore establishes Theorem 1.2 in full generality.

We first claim that we can write

$$\begin{aligned} g=(|\nabla u|^{-2}+a^2+b^2) \text {d}u^2+2a\text {d}u\text {d}x_1+2b\text {d}u\text {d}x_2+\text {d}x_1^2+\text {d}x_2^2, \end{aligned}$$

(74)

for some functions $a,b\in C^2(M^3)$. This essentially follows from the flatness of the level sets of u, but let us elaborate more on this construction:

To write g in the above form, we need to define globally defined coordinates $x_1,x_2$. To do so, we begin with introducing global polar coordinates. Given some point $p_0\in M^3$, let $\Gamma :(-\infty ,+\infty )\rightarrow M^3$ be the integral curve through $p_0$ with respect to the vector field $\nabla u$. We define the function $\rho (p)=d(p,\Gamma \cap \Sigma _{u(p)})$, where d denotes the distance within the level set $\Sigma _{u(p)}$. Since $u\in C^3(M)$ and $|\nabla u|\ne 0$, the second fundamental form of $\Sigma _{u(p)}$ is $C^1$. On each level set $\Sigma _t$ of u, we can write the metric $g_{\Sigma _t}$ as $d\rho ^2+\rho ^2d\theta ^2$. We would like g to have globally such a form, i.e. we need to define an angle function $\theta (p)\in [0,2\pi )$ for any $p\in M^3\backslash \Gamma$. To uniquely determine $\theta (p)$, we fix another point $p_1\in M^3$ not contained in the image $\text {im}(\Gamma )$. Let $\Gamma _1:(-\infty ,\infty )\rightarrow M^3$ be the integral curve through $p_1$ with respect to the vector field $\nabla u$. Since $|\nabla u|\ne 0$, we have $\text {im}(\Gamma )\cap \text {im}(\Gamma _1)=\varnothing$. We set $\theta (\Gamma _1)=0$. Thus, the Lorentzian metric ${\tilde{g}}$ can be written in the form

$$\begin{aligned} {\tilde{g}}= & {} 2\text {d}\tau \text {d}u+(|\nabla u|^{-2}+a_0^2+\rho ^{-2}b_0^2)\text {d}u^2+2a_0\text {d}u\text {d}\rho +2b_0\text {d}u\text {d}\theta \nonumber \\&+\,\text {d}\rho ^2+\rho ^2\text {d}\theta ^2 \end{aligned}$$

(75)

for some functions $a_0,b_0\in C^2(M^3\backslash \Gamma )$, where the $C^2$ regularity follows from the second fundamental form being $C^1$. Finally, we change coordinates via $x_1=\rho \cos \theta$, $x_2=\rho \sin \theta$ and set

$$\begin{aligned} a=a_0 \cos \theta - b_0\rho ^{-1}\sin \theta ,\quad b=a_0\sin \theta +b_0\rho ^{-1}\cos \theta \end{aligned}$$

(76)

to obtain

$$\begin{aligned} \tilde{g}=2\text {d}\tau \text {d}u+(|\nabla u|^2+a^2+b^2)du^2+2a\text {d}u\text {d}x_1+2b\text {d}u\text {d}x_2+\text {d}x_1^2+\text {d}x_2^2 \end{aligned}$$

(77)

as desired.

In $(\tau ,u,x_1,x_2)$ coordinates, the inverse metric ${\tilde{g}}^{-1}$ is given by

$$\begin{aligned} \tilde{g}^{-1}=\begin{bmatrix} -|\nabla u|^{-2} &{}1 &{} -a &{}-b \\ 1&{}0&{}0&{}0 \\ -a&{}0&{}1&{}0 \\ -b&{}0&{}0&{}1 \end{bmatrix}. \end{aligned}$$

(78)

Therefore, we have

$$\begin{aligned} {\tilde{\nabla }} u=\tilde{g}^{u i}\partial _i=\partial _\tau . \end{aligned}$$

(79)

Moreover, the null vector ${{\tilde{\nabla }}} u=\partial _\tau$ is covariantly constant, i.e. ${\tilde{\nabla }}^2 u=0$. Thus, $({\tilde{M}}^4,{\tilde{g}})$ is a pp-wave. See [4] for a more detailed discussion of such spacetimes. Therefore, we have on $(M^3,g,k)$

$$\begin{aligned} 0={\tilde{\nabla }}^2_{ij}u|_{TM^3}=\nabla ^2_{ij}u+\text {II}_{ij}{\hat{N}}(u)=(\text {II}_{ij}-k_{ij})|\nabla u|, \end{aligned}$$

(80)

where $N=|\nabla u|(-|\nabla u|^{-2}\partial _\tau +\partial _u-a\partial _1-b\partial _2)$ is a time-like unit normal vector. Thus, the second fundamental form $\text {II}$ of $(M^3,g)\subset ({\tilde{M}}^4,{\tilde{g}})$ is given by k.

The vector fields $\{\partial _1,\partial _2,\partial _u,\partial _\tau \}$ form a frame of $T{\tilde{M}}^4$ and $\{\nabla u,\partial _1,\partial _2\}$ form an orthogonal frame of $TM^3$. Using Mathematica, we obtain that the only non-vanishing Ricci curvature terms of ${\tilde{g}}$ are given by

$$\begin{aligned} {\widetilde{{\text {Ric}}}}(\partial _u,\partial _{1})&=\frac{1}{2}(-a_{x_2x_2}+b_{x_1x_2}), \end{aligned}$$

(81)

$$\begin{aligned} {\widetilde{{\text {Ric}}}}(\partial _u,\partial _2)&=\frac{1}{2}(a_{x_1x_2}-b_{x_1x_1}), \end{aligned}$$

(82)

$$\begin{aligned} {\widetilde{{\text {Ric}}}}(\partial _u,\partial _u)&=\frac{1}{2}(a_{x_2}-b_{x_1})^2-\frac{1}{2}\Delta _{{\mathbb {R}}^2}(|\nabla u|^{-2}+a^2+b^2)+a_{ux_1}+b_{ux_2}. \end{aligned}$$

(83)

Taking the trace of ${\widetilde{{\text {Ric}}}}$, we have $\tilde{R}=0$, then $\mu ={\widetilde{{\text {Ric}}}}({N},{N})$ and $J={\widetilde{{\text {Ric}}}}({N},\cdot )$. The identity $\langle J,\partial _1\rangle =\langle J,\partial _2\rangle =0$ yields ${\widetilde{{\text {Ric}}}}({N},\partial _1)={\widetilde{{\text {Ric}}}}({N},\partial _2)=0$. Combining this with $\mu \ge 0$, we obtain ${\widetilde{{\text {Ric}}}}(\partial _u,\partial _u)\ge 0$. The equation ${\widetilde{{\text {Ric}}}}({N},\partial _1)={\widetilde{{\text {Ric}}}}({N},\partial _2)=0$ also implies

$$\begin{aligned} a_{x_2x_2}=b_{x_1x_2} \quad \text {and}\quad a_{x_1x_2}=b_{x_1x_1}. \end{aligned}$$

(84)

Thus, $\psi :=a_{x_2}-b_{x_1}$ only depends on u. Hence, there exists a function l such that $a=x_2\psi (u)+l_{x_1}$ and $b=-x_1\psi (u)+l_{x_2}$. Inserting this into Eq. (83), we obtain

$$\begin{aligned} \Delta _{{\mathbb {R}}^2}\left( \frac{1}{2}|\nabla u|^{-2}+\frac{1}{2}l_{x_1}^2+\frac{1}{2}l_{x_2}^2+l_{x_1} x_2\psi (u)-l_{x_2}x_1\psi (u)-l_u\right) \le 0. \end{aligned}$$

(85)

Next, we define

$$\begin{aligned} F(u,x_1,x_2):=\frac{1}{2}|\nabla u|^{-2}+\frac{1}{2}l_{x_1}^2+\frac{1}{2}l_{x_2}^2+l_{x_1} x_2\psi (u)-l_{x_2}x_1\psi (u)-l_u. \end{aligned}$$

(86)

Another computation and the fact that ${{\tilde{\nabla }}} u$ is covariantly constant, yield that the only non-vanishing Riemann curvature terms of ${\tilde{g}}$ in the frame $\{\partial _1,\partial _2,\partial _\tau ,\nabla u \}$ are given by

$$\begin{aligned} \tilde{R}(\nabla u,\partial _1,\partial _1,\nabla u)&= R(\nabla u,\partial _1,\partial _1,\nabla u) +k(\nabla u,\nabla u)k(\partial _1,\partial _1)-k^2(\nabla u,\partial _1) \end{aligned}$$

(87)

$$\begin{aligned}&=-|\nabla u|^4F_{x_1x_1}, \end{aligned}$$

(88)

$$\begin{aligned} \tilde{R}(\nabla u,\partial _2,\partial _2,\nabla u)=&R(\nabla u,\partial _1,\partial _1,\nabla u) +k(\nabla u,\nabla u)k(\partial _2,\partial _2)-k^2(\nabla u,\partial _2) \end{aligned}$$

(89)

$$\begin{aligned}&=-|\nabla u|^4F_{x_2x_2}, \end{aligned}$$

(90)

$$\begin{aligned} \tilde{R}(\nabla u,\partial _1,\partial _2,\nabla u)&=-|\nabla u|^4F_{x_1x_2}. \end{aligned}$$

(91)

According to Theorem 4.2 in [11], we have $|\nabla u|= 1+O_1(|x|^{-\tau })$. Combining this with the asymptotics for g and k in (1), we obtain $F_{x_ix_j}=O(|x|^{-\tau -2})$, where $i,j=1,2$. Therefore, we can follow the proof of Lemma 3.5 to conclude that F is a linear function with respect to $x_1$, $x_2$. Thus, ${\tilde{g}}$ is flat which finishes the proof.