Emergence of Apparent Horizon in Gravitational Collapse

Abstract

We solve Einstein vacuum equations in a spacetime region up to the “center” of gravitational collapse. Within this region, we construct a sequence of marginally outer trapped surfaces (MOTS) with areas going to zero. These MOTS form a marginally outer trapped tube (apparent horizon). It emerges from a point and is smooth (except at that point). In the proof we employ a scale critical trapped surface formation criterion established by An and Luk and a new type of quasilinear elliptic equation is studied. One of the main conclusions in this paper proves a conjecture of Ashtekar on black hole thermodynamics. And the spacetimes constructed here could also be viewed as (non-spherically symmetric) generalizations of the well-known Vaidya spacetime.

Appendices

Appendix A. Ricci Curvature of M

On each MOTS: $(1-R(\underline{u},\omega ), \underline{u}, \omega )$, here we derive a lower bound for ${\text{ Ric}_{M}}(\nabla R, \nabla R)$.

For the 2-dimensional manifold M, we have

$$\begin{aligned} R_{ij}=K'g_{ij}, \end{aligned}$$

where $K'$ is the Gaussian curvature of M respect to frames $e_a', e_b', e_3', e_4'$. Gauss equation gives

$$\begin{aligned} K'=-\rho '+\frac{1}{2} \hat{\chi }'\cdot \hat{\underline{\chi }}'-\frac{1}{4} \text{ tr }\chi ' \text{ tr }\underline{\chi }'. \end{aligned}$$

Let $F_a:=\Omega (\nabla _a R)$ and $F=F^c e_c$. Recall that

$$\begin{aligned} e'_b=e_b-F_b e_3, \quad e'_3=e_3, \quad e'_4=e_4-2F+|F|^2 e_3. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \underline{\chi }'(e_a', e_b')= & {} g(D_{e_a'}e_3, e_b')=g(D_{e_a-F_a e_3}e_3, e'_b)\\= & {} g(D_{e_a}e_3, e'_b)-F_a g(D_{e_3}e_3, e'_b)\\= & {} g(D_{e_a}e_3, e_b-F_b e_3)=g(D_{e_a}e_3, e_b)-F_b g(D_{e_a}e_3, e_3)\\= & {} \underline{\chi }(e_a, e_b). \end{aligned}$$

It follows

$$\begin{aligned} \text{ tr }\underline{\chi }'=\text{ tr }\underline{\chi }, \quad \quad \hat{\underline{\chi }}'=\hat{\underline{\chi }}. \end{aligned}$$

In the same fashion, we have

$$\begin{aligned} \chi '= & {} \chi _{ab}-(\nabla _a F_b+\nabla _b F_a)+\nabla _3(F_a F_b)-(\zeta _b+\eta _b)F_a-(\zeta _a+\eta _a)F_b\\&+|F|^2\underline{\chi }_{ab}-F_b F^c\underline{\chi }_{ac}-F_a F^c \underline{\chi }_{bc}-4\underline{\omega }F_a F_b. \end{aligned}$$

Since

$$\begin{aligned} \nabla _3 F_a=\nabla _3 (\Omega \nabla R)=-\Omega \underline{\chi }\cdot \nabla R-2\underline{\omega }\Omega \nabla R, \end{aligned}$$

contracting with metric, we then arrive at

$$\begin{aligned} \text{ tr }\chi '=\text{ tr }\chi -2\Omega \Delta R-4\Omega \eta \cdot \nabla R-4\Omega ^2\hat{\underline{\chi }}_{bc}\nabla ^b R\nabla ^c R-\Omega ^2\text{ tr }\underline{\chi }|\nabla R|^2-8\Omega ^2\underline{\omega }|\nabla R|^2. \end{aligned}$$

Furthermore notice that on M, we have $\text{ tr }\chi '=0$. Hence,

$$\begin{aligned} \hat{\chi }'_{ab}= & {} \chi '_{ab}-\frac{1}{2}\text{ tr }\chi ' g_{ab}=\chi '\\= & {} \chi _{ab}-(\nabla _a F_b+\nabla _b F_a)+\nabla _3(F_a F_b)-(\zeta _b+\eta _b)F_a-(\zeta _a+\eta _a)F_b\\&+|F|^2\underline{\chi }_{ab}-F_b F^c\underline{\chi }_{ac}-F_a F^c \underline{\chi }_{bc}-4\underline{\omega }F_a F_b\\= & {} \chi _{ab}-\nabla _a \Omega \nabla _b R-\Omega \nabla _a \nabla _b R-\nabla _b \Omega \nabla _a R-\Omega \nabla _b\nabla _a R\\&-\Omega ^2 \underline{\chi }_{ac}\nabla ^c R \nabla _b R-2\underline{\omega }\Omega ^2 \nabla _a R \nabla _b R-\Omega ^2 \underline{\chi }_{bc} \nabla ^c R \nabla _a R-2\underline{\omega }\Omega ^2 \nabla _b R \nabla _a R\\&-\Omega (\zeta _b+\eta _b)\nabla _a R-\Omega (\zeta _a+\eta _a)\nabla _b R+\Omega ^2\underline{\chi }_{ab}|\nabla R|^2\\&-\Omega ^2\underline{\chi }_{ac}\nabla _b R \nabla ^c R-\Omega ^2\underline{\chi }_{bc} \nabla _a R \nabla ^c R-4\Omega ^2\underline{\omega }\nabla _a R \nabla _b R. \end{aligned}$$

For $\rho '$, we have

$$\begin{aligned} \rho '= & {} {}\frac{1}{4} R(e'_4, e'_3, e'_4, e'_3)\\= & {} {}\frac{1}{4} R(e_4-2F^c e_c+|F|^2e_3, e_3, e_4-2F^b e_b+|F|^2e_3, e_3)\\= & {} \rho +2\Omega \underline{\beta }_b \nabla ^b R+\Omega ^2 \underline{\alpha }_{bc}\nabla ^c R \nabla ^b R. \end{aligned}$$

Therefore, on M we conclude

$$\begin{aligned} K'=&-\rho -2\Omega \underline{\beta }_b \nabla ^b R-\Omega ^2 \underline{\alpha }_{bc}\nabla ^c R \nabla ^b R\\&+\frac{1}{2}\hat{\underline{\chi }}^{ab}\chi _{ab}-\hat{\underline{\chi }}^{ab}\nabla _a \Omega \nabla _b R-\Omega \hat{\underline{\chi }}^{ab}\nabla _a\nabla _b R-\frac{1}{2} \Omega ^2 \hat{\underline{\chi }}^{ab}\underline{\chi }_{ac}\nabla ^c R \nabla _b R\\&-\Omega ^2\underline{\omega }\underline{\chi }^{ab}\nabla _a R \nabla _b R-\frac{1}{2} \Omega ^2\hat{\underline{\chi }}^{ab}\underline{\chi }_{bc}\nabla ^c R\nabla _a R-\Omega ^2\underline{\omega }\hat{\chi }^{ab}\nabla _a R \nabla _b R\\&-\frac{1}{2}\Omega \hat{\underline{\chi }}^{ab} (\zeta _b+\eta _b)\nabla _a R-\frac{1}{2}\Omega \hat{\underline{\chi }}^{ab} (\zeta _a+\eta _a)\nabla _b R+\frac{1}{2}\Omega ^2|\nabla R|^2 \hat{\underline{\chi }}^{ab}\underline{\chi }_{ab}\\&-\frac{1}{2}\Omega ^2\hat{\underline{\chi }}^{ab}\underline{\chi }_{ac}\nabla _b R \nabla ^c R-\frac{1}{2}\Omega ^2 \hat{\underline{\chi }}^{ab}\underline{\chi }_{bc}\nabla _a R\nabla ^c R-2\Omega ^2\underline{\omega }\hat{\underline{\chi }}^{ab}\nabla _a R \nabla _b R. \end{aligned}$$

Here

$$\begin{aligned} -\rho +\frac{1}{2}\hat{\underline{\chi }}^{ab}\chi _{ab}=-\rho +\frac{1}{2}\hat{\underline{\chi }}^{ab}\hat{\chi }_{ab}=-\check{\rho }=\frac{\underline{u}a}{2R^3}f(\omega )\cdot [1+o(1)]>0, \end{aligned}$$

and

$$\begin{aligned} 2| K'+\rho -\frac{1}{2}\hat{\underline{\chi }}^{ab}\chi _{ab} |\le \frac{\underline{u}a^{\frac{1}{2}}}{R}\cdot \frac{|\nabla R|^2}{R^2}+\frac{\underline{u}a^{\frac{1}{2}}}{R^2} \cdot |\nabla ^2 R|+\frac{\underline{u}a}{R^3}\cdot \frac{\underline{u}a^{\frac{1}{2}}}{R}\cdot |\nabla R|. \end{aligned}$$

Hence

$$\begin{aligned} 2K'= & {} 2\bigg ( K'+\rho -\frac{1}{2}\hat{\underline{\chi }}^{ab}\chi _{ab}\bigg )+2\bigg ( -\rho +\frac{1}{2}\hat{\underline{\chi }}^{ab}\chi _{ab}\bigg )\\\ge & {} -2|K'+\rho -\frac{1}{2}\hat{\underline{\chi }}^{ab}\chi _{ab}|+2\bigg ( -\rho +\frac{1}{2}\hat{\underline{\chi }}^{ab}\chi _{ab}\bigg )\\\ge & {} -\frac{\underline{u}a^{\frac{1}{2}}}{R}\cdot \frac{|\nabla R|^2}{R^2}-\frac{\underline{u}a^{\frac{1}{2}}}{R^2} \cdot |\nabla ^2 R|-\frac{\underline{u}a}{R^3}\cdot \frac{\underline{u}a^{\frac{1}{2}}}{R}\cdot |\nabla R|. \end{aligned}$$

Therefore, we conclude

$$\begin{aligned} 2{\text{ Ric}_{M}}(\nabla R, \nabla R)= & {} 2K'|\nabla R|^2\nonumber \\\ge & {} -\frac{\underline{u}a^{\frac{1}{2}}}{R}\cdot \frac{|\nabla R|^4}{R^2}-\frac{\underline{u}a^{\frac{1}{2}}}{R^2} \cdot |\nabla ^2 R|\cdot |\nabla R|^2 \nonumber \\&-\frac{\underline{u}a}{R^3}\cdot \frac{\underline{u}a^{\frac{1}{2}}}{R}\cdot |\nabla R|^3. \end{aligned}$$

(A.1)

Appendix B. Construction of Initial Data Along $H_0^{[0,\delta ]}$

Goal of This Section. By four steps, we will construct initial data along $H_0^{[0,\delta ]}$ such that for any $\underline{u}\in (0,\delta ]$ we have

$$\begin{aligned} \int _0^{\underline{u}}|\hat{\chi }|^2(0,\underline{u}', \omega )d\underline{u}'=f(\underline{u}, \omega )\underline{u}a, \quad \quad \text{ with } \quad \quad \frac{20}{21}\le f(\underline{u}, \omega )\le \frac{22}{21}. \end{aligned}$$

(B.1)

Remark 13

In the below, we will focus on achieving (B.1). But with a similar argument, for any large positive constant c, we could also find initial data such that

$$\begin{aligned} \int _0^{\underline{u}}|\hat{\chi }|^2(0,\underline{u}', \omega )d\underline{u}'=f(\underline{u}, \omega )\underline{u}a, \quad \quad \text{ with } \quad \quad 1-\frac{1}{c}\le f(\underline{u}, \omega )\le 1+\frac{1}{c}\qquad \end{aligned}$$

(B.2)

by rescaling initial data satisfying

$$\begin{aligned} \int _0^{\delta }|\hat{\chi }|^2(0,\underline{u}', \omega )d\underline{u}'=f(\delta ,\omega ) \delta a,\quad \quad \text{ with } \quad \quad 1-\frac{1}{c}\le f(\delta , \omega )\le 1+\frac{1}{c}. \end{aligned}$$

We use (B.2) in Section 10.1.

Remark 14

A first trying to achieve (B.1) is to require

$$\begin{aligned} \frac{20}{21}\cdot a \le |\hat{\chi }|^2(0, \underline{u}', \omega )\le \frac{22}{21}\cdot a \end{aligned}$$

(B.3)

for any $\omega \in {\mathbb {S}}^2$, where $0\le \underline{u}' \le \delta$. If we have (B.3), then (B.1) follows. However, by a topological argument, on any fixed $S_{u,\underline{u}}$, traceless two tensor $\hat{\chi }_{ab}$ must vanish on at least one point. This first trying fails. In the below, we will give a more sophisticated approach to achieve (B.1).

Step One We choose a smooth function $\hat{\chi }_0(\underline{u}, \omega )$. And we require that for fixed $\omega$, $\hat{\chi }_0(\underline{u}, \omega )\in C_{c}^{\infty }([0,\delta ])$ in $\underline{u}$ variable and for all $\omega \in {\mathbb {S}}^2$,

$$\begin{aligned} \int _0^{\delta }|\hat{\chi }_0|^2(\underline{u}', \omega )d\underline{u}'=\delta a. \end{aligned}$$

Step Two We choose a number $\frac{99}{100}$ with property

$$\begin{aligned} \frac{20}{21}\le \frac{99}{100} \le 1 \le \frac{100}{99} \le \frac{22}{21}. \end{aligned}$$

With this number, we decompose $(0,\delta ]$ into dyadic pieces

$$\begin{aligned} (0,\delta ]=\bigcup _{k=0}^{+\infty }\left[ \left( \frac{99}{100}\right) ^{k+1}\delta , \left( \frac{99}{100}\right) ^k \delta \right] . \end{aligned}$$

In $[(\frac{99}{100})^{k+1}\delta , (\frac{99}{100})^k \delta ]$, we let

$$\begin{aligned} |\hat{\chi }|^2(0,\underline{u}, \omega ):=|\hat{\chi }_0|^2\left( \left[ \left( \frac{100}{99}\right) ^{k+1}\underline{u}-\delta \right] \cdot 99, \omega \right) . \end{aligned}$$

(B.4)

Thus,

$$\begin{aligned}&\int _{\left( \frac{99}{100}\right) ^{k+1}\delta }^{\left( \frac{99}{100}\right) ^{k}\delta }|\hat{\chi }|^2(0,\underline{u}', \omega )d\underline{u}'\\&\quad =\int _{\left( \frac{99}{100}\right) ^{k+1}\delta }^{\left( \frac{99}{100}\right) ^{k}\delta }|\hat{\chi }_0|^2\left( \left[ \left( \frac{100}{99}\right) ^{k+1}\underline{u}'-\delta \right] \cdot 99, \omega \right) d\underline{u}'\\&\quad =\left( \frac{99}{100}\right) ^{k+1}\cdot \frac{1}{99}\cdot \int _{\left( \frac{99}{100}\right) ^{k+1}\delta }^{\left( \frac{99}{100}\right) ^{k}\delta }|\hat{\chi }_0|^2\left( \left[ \left( \frac{100}{99}\right) ^{k+1}\underline{u}'-\delta \right] \cdot 99, \omega \right) d \left( \left[ \left( \frac{100}{99}\right) ^{k+1}\underline{u}'-\delta \right] \cdot 99 \right) \\&\quad =\left( \frac{99}{100}\right) ^{k+1}\cdot \frac{1}{99}\cdot \int _0^{\delta }|\hat{\chi }_0|^2(\underline{u}', \omega )d\underline{u}'\\&\quad =\left( \frac{99}{100}\right) ^{k+1}\cdot \frac{1}{99}\cdot \delta a\\&\quad =\left( \frac{99}{100}\right) ^k\cdot \frac{1}{100}\cdot \delta a. \end{aligned}$$

And for fixed $\omega$, $|\hat{\chi }|^2(0,\underline{u}, \omega )\in C_c([(\frac{99}{100})^{k+1}\delta , (\frac{99}{100})^{k}\delta ])$ in $\underline{u}$ variable. Furthermore, we have

$$\begin{aligned}&\sum _{k=0}^{+\infty }\int _{\left( \frac{99}{100}\right) ^{k+1}\delta }^{\left( \frac{99}{100}\right) ^{k}\delta }|\hat{\chi }|^2(0,\underline{u}', \omega )d\underline{u}'\\&\quad =\sum _{k=0}^{+\infty } \left( \frac{99}{100}\right) ^k\cdot \frac{1}{100}\cdot \delta a\\&\quad =\frac{\frac{1}{100}\cdot \delta a}{1-\frac{99}{100}}=\delta a. \end{aligned}$$

Step Three For any $\underline{u}\in (0,\delta ]$, there exists $N_0\in {\mathbb {N}}$ such that

$$\begin{aligned} \left( \frac{99}{100}\right) ^{N_0+1}\delta \le \underline{u}\le \left( \frac{99}{100}\right) ^{N_0}\delta . \end{aligned}$$

We thus have

$$\begin{aligned} \int _0^{\left( \frac{99}{100}\right) ^{N_0+1}\delta }|\hat{\chi }|^2(0, \underline{u}', \omega )d\underline{u}' \le \int _0^{\underline{u}}|\hat{\chi }|^2(0, \underline{u}', \omega )d\underline{u}' \le \int _0^{\left( \frac{99}{100}\right) ^{N_0}\delta }|\hat{\chi }|^2(0, \underline{u}', \omega )d\underline{u}'. \end{aligned}$$

On one side

$$\begin{aligned}&\int _0^{\underline{u}}|\hat{\chi }|^2(0, \underline{u}', \omega )d\underline{u}' \\&\quad \le \int _0^{\left( \frac{99}{100}\right) ^{N_0}\delta }|\hat{\chi }|^2(0, \underline{u}', \omega )d\underline{u}'\\&\quad =\frac{\left( \frac{99}{100}\right) ^{N_0}\cdot \frac{1}{100}\cdot \delta a}{1-\frac{99}{100}} =\left( \frac{99}{100}\right) ^{N_0}\cdot \delta a\\&\quad = \frac{100}{99}\cdot \left( \frac{99}{100}\right) ^{N_0+1}\cdot \delta a\le \frac{100}{99}\underline{u}a \le \frac{22}{21}\underline{u}a. \end{aligned}$$

On the other side

$$\begin{aligned}&\int _0^{\underline{u}}|\hat{\chi }|^2(0, \underline{u}', \omega )d\underline{u}' \\&\quad \ge \int _0^{\left( \frac{99}{100}\right) ^{N_0+1}\delta }|\hat{\chi }|^2(0, \underline{u}', \omega )d\underline{u}'\\&\quad =\frac{\left( \frac{99}{100}\right) ^{N_0+1}\cdot \frac{1}{100}\cdot \delta a}{1-\frac{99}{100}} =\left( \frac{99}{100}\right) ^{N_0+1}\cdot \delta a\\&\quad =\frac{99}{100}\cdot \left( \frac{99}{100}\right) ^{N_0}\cdot \delta a \ge \frac{99}{100}\underline{u}a\ge \frac{20}{21}\underline{u}a. \end{aligned}$$

Putting the above inequalities together, for any $\underline{u}\in (0, \delta ]$ we have

$$\begin{aligned} \int _0^{\underline{u}}|\hat{\chi }|^2(0,\underline{u}', \omega )d\underline{u}'=f(\underline{u}, \omega )\underline{u}a, \quad \quad \text{ with } \quad \quad \frac{20}{21}\le f(\underline{u}, \omega )\le \frac{22}{21}. \end{aligned}$$

Step FourWith initial data $|\hat{\chi }|^2(0, \underline{u}, \omega )$ prescribed along $H_0^{(0,\delta ]}$, we need to further check the hyperbolic part. Here we need to note that $\partial _{\underline{u}} \hat{\chi }$ and $\alpha$ are very large and tend to $+\infty$ as $\underline{u}\rightarrow 0$. However, in [3] by a method of renormalization we avoid using $\alpha$. Replace $\delta$ by $\underline{u}$ in [3], for any $\underline{u}\in (0, \delta ]$, we consider the region $(u,\underline{u}')\in [b\underline{u}a^{\frac{1}{2}}, 1]\times [0,\underline{u}]$. All the arguments in [3] hold for this region. Then let $\underline{u}\rightarrow \delta$. With initial data $|\hat{\chi }|^2(0, \underline{u}, \omega )$ prescribed along $u=0$ and Minkowski data prescribed along $\underline{u}=0$, we then have the existence of Einstein vacuum equation in the whole colored region above. And we have a sequence of $M_{\underline{u}}$, of which the radius $\rightarrow 0$ as $\underline{u}\rightarrow 0$.

Remark 15

Another way to construct arbitrary small MOTS is using the following initial data along $u=0$: fix any large natural number $N_1$, we then pick up another natural number $N_2$ such that $1\ll N_1 \ll N_2$ . For $\underline{u}\le (\frac{99}{100})^{N_2+1}$, Minkowski data are prescribed along $u=0$; for $\underline{u}\ge (\frac{99}{100})^{N_2+1}$ we prescribe $|\hat{\chi }|^2(0, \underline{u}, \omega )$ as in Step One to Step Three above. Since $N_1\ll N_2$, for $\underline{u}\ge (\frac{99}{100})^{N_1+1}$ it still holds

$$\begin{aligned} \int _0^{\underline{u}}|\hat{\chi }|^2(0,\underline{u}', \omega )d\underline{u}'=f(\underline{u}, \omega )\underline{u}a, \quad \quad \text{ with } \quad \quad \frac{20}{21}\le f(\underline{u}, \omega )\le \frac{22}{21}. \end{aligned}$$

Then the smallest MOTS is of radius $(\frac{99}{100})^{N_1+1}\cdot a$. Since $N_1$ could be any large positive integer, the radius of the smallest MOTS could be arbitrary small.

Appendix C. Construction of Initial Data Along $H_0^{[0,\delta ]}$: Part II

In this part, we provide examples of initial data satisfying the following two requirements at the same time:

• 1. The initial data along $H_0^{[0,\delta ]}$ satisfy

  $$\begin{aligned}&\int _0^{\delta }|\hat{\chi }_0|^2(\underline{u},\omega )d\underline{u}=f(\delta ,\omega )\delta a, \,\, \\&\quad \text{ with } \,\, 1- \frac{1}{c}\le f(\delta , \omega )\le 1+\frac{1}{c}, \,\, \text{ where } \,\, 1\ll c \ll b \le a^{\frac{1}{2}}. \end{aligned}$$

• 2. Moreover, we require that $|\hat{\chi }_0|^2(\underline{u}, \omega )$ is almost a constant out of two small discs $\{D^1_{\underline{u}}\subset {\mathbb {S}}^2\}$ and $\{D^2_{\underline{u}}\subset {\mathbb {S}}^2\}$ :¹⁵

  $$\begin{aligned}&0\le |\hat{\chi }_0|^2(\underline{u},\omega )\lesssim a, \quad \text{ for } \quad \omega \in D^1_{\underline{u}}\cup D^2_{\underline{u}},\\&|\hat{\chi }_0|^2(\underline{u}, \omega )=[1+o(1)]\cdot a, \quad \text{ for } \quad \omega \in {\mathbb {S}}^2\setminus \big (D^1_{\underline{u}}\cup D^2_{\underline{u}}\big ) \quad \text{ with } \quad |o(1)|\le 1/c,\\&\iint _{D^1_{\underline{u}}\cup D^2_{\underline{u}}}1\cdot d\omega \lesssim \frac{1}{c^2}. \end{aligned}$$

We now prescribe characteristic initial data. We follow some basic calculations in Chapter 2 of [15]. For $\underline{u}\le 0$, we prescribe Minkowskian initial data. For $0 \le R \le 1$, the surface $S_{1-R, 0}$ on the boundary $\underline{H}_0$ of Minkowskian region, is the sphere of radius R in the hyperplane $t=-R$:

$$\begin{aligned} |x|=R, \quad \quad |x|=\sqrt{x_1^2+x_2^2+x_3^3}. \end{aligned}$$

We use coordinates $(x_1, x_2, x_3)$ to define stereographic coordinates $(\theta _1, \theta _2)$ on $S_{0,0}$. We thus have two stereographic charts, the north polar chart and the south polar chart.

Denote $q_2=(0,0,-1)$ to be the south pole. The domain of the north polar chart is then $U_1=S_{0,0} \backslash q_2$. And the chart is the mapping of $U_1$ onto ${\mathbb {R}}^2$ by $(x_1, x_2, x_3)\in U_1 \rightarrow (\theta _1, \theta _2)\in {\mathbb {R}}^2$:

$$\begin{aligned} \theta _1=\frac{2x_1}{1+x_3}, \quad \quad \theta _2=\frac{2x_2}{1+x_3}. \end{aligned}$$

Similarly, for the south polar chart:

$$\begin{aligned} \theta _1=\frac{2x_1}{1-x_3}, \quad \quad \theta _2=\frac{2x_2}{1-x_3}. \end{aligned}$$

In both charts, the standard metric on $S_{0,0}$ is

$$\begin{aligned} \mathring{g}_{AB}(\theta _1, \theta _2)=\frac{\delta _{AB}}{(1+\frac{1}{4}(\theta _1^2+\theta _2^2))^2}. \end{aligned}$$

The transformation from north polar coordinates to south polar coordinates is

$$\begin{aligned} (\theta _1, \theta _2)\rightarrow f(\theta _1, \theta _2)=\frac{(4\theta _1, 4\theta _2)}{\theta _1^2+\theta _2^2}. \end{aligned}$$

Note that $f=f^{-1}$. We define $(\theta _1', \theta _2')=f(\theta _1, \theta _2)$ and hence $(\theta _1, \theta _2)=f(\theta _1', \theta _2').$

We then prescribe initial data along $H_0$. The induced metric $g|_{S_{0,\underline{u}}}$ can be expressed in the form

$$\begin{aligned} g|_{S_{0,\underline{u}}}=(\phi |_{S_{0,\underline{u}}})^2 \hat{g}|_{S_{0,\underline{u}}}, \end{aligned}$$

where $\hat{g}|_{S_{0,\underline{u}}}$ satisfies that $\Phi ^*_{\underline{u}}\hat{g}|_{S_{0,\underline{u}}}$, a metric on $S_{0,0}$, has the same area form as $\mathring{g}|_{S_{0,0}}$:

$$\begin{aligned} d\mu _{\Phi ^*_{\underline{u}}\hat{g}|_{S_{0,\underline{u}}}}=d\mu _{\mathring{g}|_{S_{0,0}}}. \end{aligned}$$

In another word,

$$\begin{aligned} \sqrt{\det \hat{g}(0,\underline{u},\theta _1, \theta _2)}=W^2(\theta _1, \theta _2), \quad \text{ where } \quad W(\theta _1, \theta _2)=\frac{1}{1+\frac{1}{4}(\theta _1^2+\theta _2^2)}. \end{aligned}$$

Thus, in both stereographic charts $\hat{g}$ is given by

$$\begin{aligned} \hat{g}_{AB}(0,\underline{u},\theta _1, \theta _2)=W^2(\theta _1, \theta _2)\,m_{AB}(0,\underline{u}, \theta _1, \theta _2), \end{aligned}$$

where m satisfies $\det m=1$. Set

$$\begin{aligned} O_{CA}(\theta _1, \theta _2)=\delta _{CA}-\frac{2\theta _C \theta _A}{\theta _1^2+\theta _2^2} \quad \text{ and } \quad \tilde{O} \quad \text{ its } \text{ transpose }. \end{aligned}$$

It is straight forward to check

$$\begin{aligned} \tilde{O}O=I, \quad \quad \tilde{O}=O, \quad \quad \det {O}=-1. \end{aligned}$$

From Chapter 2 in [15], we also have

$$\begin{aligned} m(0, \underline{u}, \theta _1, \theta _2)=\tilde{O}(\theta _1, \theta _2)\,m'(0,\underline{u},\theta _1', \theta _2')\,O(\theta _1, \theta _2), \end{aligned}$$

which we write as

$$\begin{aligned} m=\tilde{O}m' O. \end{aligned}$$

(C.1)

Now the matrix m at a given point on $H_0$ is a 2-dimensional positive definite symmetric unimodular matrix. Such a matrix has the form $m= \begin{bmatrix} Z+X &{} Y \\ Y &{} Z-X \end{bmatrix},$ where $Z^2-X^2-Y^2=1$ and $Z\ge 0$. Use exponential map, in north polar chart we can express:

$$\begin{aligned} m(0, \underline{u}, \theta _1, \theta _2)=\exp \Psi (\underline{u}, \theta _1, \theta _2), \end{aligned}$$

where $\Psi$ is a symmetric trace-free 2-dimensional matrix. And $\Psi (\underline{u}, \theta _1, \theta _2)$ is the free data we can prescribe along $H_0$. In the below, we will prescribe $\Psi (\underline{u}, \theta _1, \theta _2)$ in the north polar chart and $\Psi '(\underline{u}, \theta _1', \theta _2')$ in the south polar chart such that (C.1) is satisfied.

Consider the standard spherical coordinates $\{\theta , \phi \}$ on ${\mathbb {S}}^2$, where $0\le \theta \le \pi$ and $0\le \phi <2\pi$. Assume $\theta =0$ and $\theta =\pi$ are corresponding to the north pole and south pole, respectively. For constant c, we assume $1\ll c \ll b \le a^{\frac{1}{2}}$.

In the following, we will prescribe $\Psi (\underline{u}, \theta _1, \theta _2)$ and $\Psi '(\underline{u}, \theta _1', \theta _2')$ such that for fixed $\underline{u}$, in $\{\theta , \phi \}$ coordinates, we have

• $|\hat{\chi }|^2_0(\underline{u}, \theta , \phi )=0$ for $|\phi -{2\pi \underline{u}}/{\delta }|\le \pi /2c$ and $|\theta -\pi /2|\le \pi /2c$,

• $|\hat{\chi }|^2_0(\underline{u}, \theta , \phi )\le a$ for $|\phi -{2\pi \underline{u}}/{\delta }|\le \pi /c$ and $|\theta -\pi /2|\le \pi /c$,

• $|\hat{\chi }|^2_0(\underline{u}, \theta , \phi )=0$ for $|{\pi }-\phi +{2\pi \underline{u}}/{\delta }|\le \pi /2c$ and $|\phi |\le \pi /2c$,

• $|\hat{\chi }|^2_0(\underline{u}, \theta , \phi )\le a$ for $|{\pi }-\phi +{2\pi \underline{u}}/{\delta }|\le \pi /c$ and $|\phi |\le \pi /c$,

• $|\hat{\chi }|^2_0(\underline{u}, \theta , \phi )=a$ otherwise.

In another word we hope, for fixed $\underline{u}$, it holds $|\hat{\chi }|^2_0(\underline{u}, \theta , \phi )=a$ for most points on ${\mathbb {S}}^2$, except for $(\theta , \phi )$ lying in two small discs (with radius $\pi /c$) centered at P and Q. Here P, Q have coordinates $\theta =\pi /2, \phi =2\pi \underline{u}/\delta$ and $\theta =\pi /2, \phi ={\pi }-2\pi \underline{u}/\delta$, respectively. Within these two small discs, it holds $|\hat{\chi }|^2_0(\underline{u}, \theta , \phi )\le a$.

Fix the north pole. For any point $P\in {\mathbb {S}}^2$ there is a $1-1$ correspondence between its stereographic coordinates $(\theta _1(P), \theta _2(P))$ and its spherical coordinates $(\theta (P), \phi (P))$. In particular, we could write

$$\begin{aligned} (\theta , \phi )=(\theta (\theta _1, \theta _2), \phi (\theta _1, \theta _2)). \end{aligned}$$

For simplicity, in the below we omit the explicit forms of $(\theta (\theta _1, \theta _2), \phi (\theta _1, \theta _2))$.

To prescribe $\Psi (\underline{u},\theta _1, \theta _2)$ in the north polar chart, for $\sqrt{\theta ^2_1+\theta ^2_2}<20$ we set

$$\begin{aligned} \Psi (\underline{u},\theta _1, \theta _2)=\sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}\cdot \Psi _0(\underline{u}, \theta _1, \theta _2). \end{aligned}$$

For fixed $\theta$, we require $x(\theta , \cdot )$ to be a periodic function with period $2\pi$ and it will be constructed later. Let

$$\begin{aligned} \Psi _0(\underline{u}, \theta _1, \theta _2)= \begin{bmatrix} 1&{}0\\ 0&{}-1 \end{bmatrix}, \quad \text{ for } \quad \underline{u}>0. \end{aligned}$$

We also define

$$\begin{aligned} m(\underline{u},\theta _1, \theta _2)=\exp (\Psi (\underline{u},\theta _1, \theta _2))=\exp \big (\sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}\cdot \Psi _0(\underline{u}, \theta _1, \theta _2)\big ). \end{aligned}$$

And for $\underline{u}>0$, we have

$$\begin{aligned} m(\underline{u},\theta _1, \theta _2)= \begin{bmatrix} \exp \big (\sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}\big )&{}0\\ 0&{}\exp \big (-\sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}\big ) \end{bmatrix}, \\ m^{-1}(\underline{u},\theta _1, \theta _2)= \begin{bmatrix} \exp \big (-\sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}\big )&{}0\\ 0&{}\exp \big (\sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}\big ) \end{bmatrix}. \end{aligned}$$

Applying

$$\begin{aligned} \frac{d}{d\underline{u}}e^{X(\underline{u})}=\int _0^1 e^{\tau X(\underline{u})}\frac{dX(\underline{u})}{d\underline{u}}e^{(1-\tau )X(\underline{u})}d\tau , \end{aligned}$$

we get

$$\begin{aligned}&{}\frac{\partial }{\partial \underline{u}}m(\underline{u},\theta _1,\theta _2)=\int _0^1 \exp \begin{bmatrix} \sqrt{2} \tau x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}&{}0\\ 0&{}-\sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}\end{bmatrix}\\&\qquad \times \bigg (\sqrt{2}\cdot \frac{\partial x}{\partial \phi }(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot \frac{-2\pi }{\delta }\cdot a^{\frac{1}{2}}\bigg ) \begin{bmatrix} 1&{}0\\ 0&{}-1 \end{bmatrix}\\&\qquad \times \exp \begin{bmatrix} \sqrt{2} (1-\tau )\cdot x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}&{}0\\ 0&{}-\sqrt{2} (1-\tau )\cdot x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}\end{bmatrix}d\tau \\&\quad =\int _0^1 \exp \begin{bmatrix} \sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}&{}0\\ 0&{}-\sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}\end{bmatrix}\\&\qquad \times \bigg (\sqrt{2}\cdot \frac{\partial x}{\partial \phi }(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot \frac{-2\pi }{\delta }\cdot a^{\frac{1}{2}}\bigg )\begin{bmatrix} 1&{}0\\ 0&{}-1 \end{bmatrix} d\tau \\&\quad =\sqrt{2}a^{\frac{1}{2}}\begin{bmatrix} \exp \bigg (\sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}\bigg )&{}0\\ 0&{}-\exp \bigg (-\sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}\bigg ) \end{bmatrix}\\&\qquad \times \bigg (\frac{\partial x}{\partial \phi }(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot \frac{-2\pi }{\delta }\bigg ). \end{aligned}$$

From Chapter 2 in [15], it holds

$$\begin{aligned} \hat{\chi }_{AB}=\frac{1}{2}\phi ^2\frac{\partial \hat{g}}{\partial \underline{u}}=\frac{1}{2}\phi ^2 W^2(\theta _1, \theta _2)\frac{\partial }{\partial \underline{u}}m_{AB}(\underline{u},\theta _1, \theta _2), \end{aligned}$$

and

$$\begin{aligned} e:=\frac{1}{2} |\hat{\chi }|^2_{g}=&{}\frac{1}{2} (g^{-1})^{CA}\hat{\chi }_{AB}(g^{-1})^{BD}\hat{\chi }_{DC}\\ =&{}\frac{1}{8} (\hat{g}^{-1})^{CA}\frac{\partial \hat{g}_{AB}}{\partial \underline{u}}(\hat{g}^{-1})^{BD}\frac{\partial \hat{g}_{DC}}{\partial \underline{u}}\\ =&{}\frac{1}{8} (m^{-1})^{CA}\left( \frac{\partial m}{\partial \underline{u}}\right) _{AB} (m^{-1})^{BD} \left( \frac{\partial m}{\partial \underline{u}}\right) _{DC}. \end{aligned}$$

For our concrete example, we have

$$\begin{aligned} e= & {} {}\frac{1}{8} (m^{-1})^{CA}\left( \frac{\partial m}{\partial \underline{u}}\right) _{AB} (m^{-1})^{BD} \left( \frac{\partial m}{\partial \underline{u}}x\right) _{DC}\nonumber \\= & {} {}\frac{1}{8}\cdot \bigg (\sqrt{2}\cdot \frac{\partial x}{\partial \phi }(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot \frac{-2\pi }{\delta }\cdot a^{\frac{1}{2}}\bigg )^2 \cdot \text{ tr }\begin{bmatrix} 1&{}0\\ 0&{}1 \end{bmatrix}\nonumber \\= & {} {}\frac{1}{2} a\cdot \bigg (\frac{\partial x}{\partial \phi }(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot \frac{-2\pi }{\delta }\bigg )^2. \end{aligned}$$

(C.2)

We now prescribe the function $\partial x/\partial \phi (\theta , \phi )$. Let $\partial x/\partial \phi (\theta , \phi )$ to be $2\pi$-periodic with respect to $\phi$ and require

$$\begin{aligned} |\frac{\partial x}{\partial \phi }|(\theta ,\phi )\left\{ \begin{array}{ll} =0 &{} \text{ if } -\frac{\pi }{2c}\le \phi \le \frac{\pi }{2c} \text{ and } -\frac{\pi }{2c}\le \theta -\frac{\pi }{2}\le \frac{\pi }{2c};\\ =0 &{} \text{ if } {\pi }-\frac{\pi }{2c}\le \phi \le {\pi }+\frac{\pi }{2c} \text{ and } -\frac{\pi }{2c}\le \theta -\frac{\pi }{2}\le \frac{\pi }{2c};\\ \le {\delta }/{2\pi } &{} \text{ if } -\frac{\pi }{c}\le \phi \le \frac{\pi }{c} \text{ and } -\frac{\pi }{2c}\le \theta -\frac{\pi }{2}\le \frac{\pi }{2c};\\ \le {\delta }/{2\pi } &{} \text{ if } {\pi }-\frac{\pi }{c}\le \phi \le {\pi }+\frac{\pi }{c} \text{ and } -\frac{\pi }{2c}\le \theta -\frac{\pi }{2}\le \frac{\pi }{2c};\\ ={\delta }/{2\pi } &{} \text{ otherwise }. \end{array} \right. \end{aligned}$$

Then in the north polar chart $\sqrt{\theta _1^2+\theta ^2_2}<20$ we have

$$\begin{aligned}&\frac{1}{2}a\cdot \left( \frac{\partial x}{\partial \phi }\left( \theta , \phi -\frac{2\pi \underline{u}}{\delta }\right) \cdot \frac{-2\pi }{\delta }\right) ^2 \nonumber \\&\quad \left\{ \begin{array}{ll} =0 &{} \text{ if } -\frac{\pi }{2c}\le \phi -\frac{2\pi \underline{u}}{\delta } \le \frac{\pi }{2c} \text{ and } -\frac{\pi }{2c}\le \theta -\frac{\pi }{2}\le \frac{\pi }{2c};\\ =0 &{} \text{ if } {\pi }-\frac{\pi }{2c}\le \phi -\frac{2\pi \underline{u}}{\delta } \le {\pi }+\frac{\pi }{2c} \text{ and } -\frac{\pi }{2c}\le \theta -\frac{\pi }{2}\le \frac{\pi }{2c};\\ \le {a}/{2} &{} \text{ if } -\frac{\pi }{c}\le \phi -\frac{2\pi \underline{u}}{\delta } \le \frac{\pi }{c} \text{ and } -\frac{\pi }{2c}\le \theta -\frac{\pi }{2}\le \frac{\pi }{2c};\\ \le {a}/{2} &{} \text{ if } {\pi }-\frac{\pi }{c}\le \phi -\frac{2\pi \underline{u}}{\delta } \le {\pi }+\frac{\pi }{c} \text{ and } -\frac{\pi }{2c}\le \theta -\frac{\pi }{2}\le \frac{\pi }{2c};\\ ={a}/{2} &{} \text{ otherwise }. \end{array} \right. \end{aligned}$$

(C.3)

Next we extend $\Psi (\underline{u}, \theta _1, \theta _2)$ to the south polar chart. In $\frac{1}{5}<\sqrt{\theta _1^2+\theta _2^2}<20$,¹⁶ we require $\Psi '(\underline{u}, \theta _1', \theta _2'):=O(\theta _1', \theta _2')\Psi (\underline{u}, \theta _1, \theta _2)O(\theta _1, \theta _2)$. Since $m(\underline{u}, \theta _1, \theta _2)=\exp \big (\Psi (\underline{u}, \theta _1, \theta _2)\big )$ and $m'(\underline{u}, \theta _1', \theta _2')=\exp \big (\Psi '(\underline{u}, \theta _1', \theta _2')\big )$, (C.1) follows. Denote $\tilde{\lambda }:=\sqrt{2} x(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot a^{\frac{1}{2}}.$ We calculate

$$\begin{aligned} e'=&{}\frac{1}{8} (m'^{-1})^{CA}\left( \frac{\partial m'}{\partial \underline{u}}\right) _{AB} (m'^{-1})^{BD}\left( \frac{\partial m'}{\partial \underline{u}}\right) _{DC}\\ =&{}\frac{1}{8} {O^{C}}_{C'}(m^{-1})^{C'A'}{O_{A'}}^{A}{O_{A}}^{A''}\left( \frac{\partial m}{\partial \underline{u}}\right) _{A''B'}\cdot \\&{O^{B'}}_{B}{O^{B}}_{B''}(m^{-1})^{B''D'}{O_{D'}}^{D}{O_{D}}^{D''}\left( \frac{\partial m}{\partial \underline{u}}\right) _{D''C''}{O^{C''}}_{C}\\ =&{}\frac{1}{8} {O^{C}}_{C'}(m^{-1})^{C'A''}\left( \frac{\partial m}{\partial \underline{u}}\right) _{A''B''}(m^{-1})^{B''D''} \left( \frac{\partial m}{\partial \underline{u}}\right) _{D''C''}{O^{C''}}_{C}\\ =&{}\frac{1}{8}\cdot \bigg (\sqrt{2}\cdot \frac{\partial x}{\partial \phi }(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot \frac{-2\pi }{\delta }\cdot a^{\frac{1}{2}}\bigg )^2\\&\times {O^{C}}_{C'}\begin{bmatrix} e^{-\tilde{\lambda }}&{}0\\ 0&{}e^{\tilde{\lambda }}\end{bmatrix}^{C'A''}\begin{bmatrix} e^{\tilde{\lambda }}&{}0\\ 0&{}-e^{-\tilde{\lambda }}\end{bmatrix}_{A''B''}\begin{bmatrix} e^{-\tilde{\lambda }}&{}0\\ 0&{}e^{\tilde{\lambda }}\end{bmatrix}^{B''D''}\begin{bmatrix} e^{\tilde{\lambda }}&{}0\\ 0&{}-e^{-\tilde{\lambda }}\end{bmatrix}_{D''C''} {O^{C''}}_{C}\\ =&{}\frac{1}{8}\cdot \bigg (\sqrt{2}\cdot \frac{\partial x}{\partial \phi }(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot \frac{-2\pi }{\delta }\cdot a^{\frac{1}{2}}\bigg )^2 \cdot \text{ tr }\begin{bmatrix} 1&{}0\\ 0&{}1 \end{bmatrix}\\ =&{}\frac{1}{2} a\cdot \bigg (\frac{\partial x}{\partial \phi }(\theta , \phi -\frac{2\pi \underline{u}}{\delta })\cdot \frac{-2\pi }{\delta }\bigg )^2. \end{aligned}$$

Note that by (C.2) e and $e'$ take the same value at $(\theta _1, \theta _2)$ with $\frac{1}{5}<\sqrt{\theta _1^2+\theta _2^2}<20$. And in view of (C.3), when the points are close to $\sqrt{\theta _1^2+\theta _2^2}=\frac{1}{5}$ and $\sqrt{\theta _1^2+\theta _2^2}=20$, e and $e'$ are a/2. We then extend $m'(\underline{u}, \theta _1', \theta _2')$ as a smooth 2-covariant S tensorfield to the south pole chart $\sqrt{\theta _1'^2+\theta _2'^2}<20$ and require

$$\begin{aligned} e'=\frac{1}{8} (m'^{-1})^{CA}\left( \frac{\partial m'}{\partial \underline{u}}\right) _{AB} (m'^{-1})^{BD} \left( \frac{\partial m'}{\partial \underline{u}}\right) _{DC}=\frac{a}{2} \quad \text{ for } \quad \sqrt{\theta _1'^2+\theta _2'^2}\le \frac{1}{5}. \end{aligned}$$

With the above constructions we find $|\hat{\chi }_0|^2(\underline{u}, \omega )$: it is a constant out of two small discs $\{D^1_{\underline{u}}\subset {\mathbb {S}}^2\}$ centered at P and $\{D^2_{\underline{u}}\subset {\mathbb {S}}^2\}$ centered at Q:

$$\begin{aligned}&0\le |\hat{\chi }_0|^2(\underline{u},\omega )\lesssim a, \quad \text{ for } \quad \omega \in D^1_{\underline{u}}\cup D^2_{\underline{u}}, \\&|\hat{\chi }_0|^2(\underline{u}, \omega )=a, \quad \text{ for } \quad \omega \in {\mathbb {S}}^2\setminus \big (D^1_{\underline{u}}\cup D^2_{\underline{u}}\big ), \\&\iint _{D^1_{\underline{u}}\cup D^2_{\underline{u}}}1\cdot d\omega \lesssim \frac{1}{c^2}. \end{aligned}$$

At the same time, for fixed $\theta$ and fixed $\phi$ we let $\underline{u}$ vary. Then these two discs $D^1_{\underline{u}}$ and $D^2_{\underline{u}}$ rotate (at a speed $2\pi /\delta$) along $\theta =\pi /2$. For $0<\underline{u}\le \delta$, there are at most two intervals of length $\delta /c$ such that $0\le |\hat{\chi }_0|^2\le a$, for all the rest $\underline{u}\in (0,\delta ]$, we have $|\hat{\chi }_0|^2=a$. This implies that along $H_0^{[0,\delta ]}$, $(\hat{\chi }_0)_{ab}(\underline{u},\omega )$ also satisfies

$$\begin{aligned} \int _0^{\delta }|\hat{\chi }_0|^2(\underline{u},\omega )d\underline{u}=f(\delta ,\omega )\delta a, \,\, \text{ with } \,\, 1-\frac{1}{c}\lesssim f(\delta , \omega )\lesssim 1+\frac{1}{c}, \end{aligned}$$

where $1\ll c \ll b \le a^{\frac{1}{2}}$.