# Björling problem for zero mean curvature surfaces in the three-dimensional
light cone

Joseph Cho  Institute of Discrete Mathematics and Geometry, TU Wien, Wien,
1040, Austria<EMAIL_ADDRESS>,  So Young Kim  Department of
Mathematics, Korea University, Seoul, 02841, Republic of Korea
<EMAIL_ADDRESS>,  Dami Lee  Department of Mathematics, Indiana
University, Bloomington, IN, 47405, USA<EMAIL_ADDRESS>,  Wonjoo Lee
Department of Mathematics, Korea University, Seoul, 02841, Republic of Korea
<EMAIL_ADDRESS>and  Seong-Deog Yang  Department of Mathematics, Korea
University, Seoul, 02841, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

We solve the Björling problem for zero mean curvature surfaces in the three-
dimensional light cone. As an application, we construct and classify all
rotational zero mean curvature surfaces.

###### Key words and phrases:

Zero mean curvature surfaces, Björling representation

###### 2020 Mathematics Subject Classification:

Primary 53A10; Secondary 53B30.

##  1\. Introduction

The _classical Björling problem_ [bjorling_integrationem_1844] poses the
following question: Suppose that a curve
$\gamma\mathrel{\mathop{\ordinarycolon}}I\to\mathbb{E}^{3}$ and a unit vector
field $N$ along $\gamma$ in the Euclidean 3-space $\mathbb{E}^{3}$ are given
so that

$\dot{\gamma}\cdot N=0,$

where $\dot{\phantom{\gamma}}$ denotes the differentiation with respect to
$u\in I$, and $\cdot$ is the standard Euclidean inner product. The goal is to
find a minimal surface $X$ which contains $\gamma$ and whose unit normal along
$\gamma$ is the prescribed $N$. Due to the Weierstrass representation of
minimal surfaces [weierstrass_untersuchungen_1866], this problem can be solved
via analytic extensions of $\gamma$ and $N$ to the complex plane:

$X(z)=\int_{u_{0}}^{z}(\dot{\gamma}(w)-iN(w)\times\dot{\gamma}(w))\operatorname{d\\!}{w}$

where $\times$ denotes the standard Euclidean cross product. The curve and the
prescribed normal together are called the _Björling data_. Such Björling type
problems are well studied across various types of surfaces in various space
forms (see, for example, [alias_bjorling_2003, asperti_bjorling_2006,
brander_bjorling_2010, brander_bjorling_2018, dussan_bjorling_2017,
kim_spacelike_2011, kim_prescribing_2007, yang_bjorling_2017]).

The Björling data can be slightly modified by noting that
$\mathcal{L}\mathrel{\mathop{\ordinarycolon}}=N\times\dot{\gamma}$ is
perpendicular to $\dot{\gamma}$ and of the same length to $\dot{\gamma}$. Thus
if $\mathcal{L}$ is prescribed then one can recover $N$ via
$N=\frac{\dot{\gamma}\times\mathcal{L}}{|\dot{\gamma}\times\mathcal{L}|}$,
allowing us to adopt $\gamma$ and $\mathcal{L}$ instead of $\gamma$ and $N$ as
Björling data. Geometrically, the vector field $\mathcal{L}$ of the Björling
data is equivalent to the prescription of the tangent vector field of the
surface that is perpendicular to $\dot{\gamma}$ along $\gamma$. This change in
viewpoint of the prescribed data for the Björling problem proved to be useful
for obtaining the _(singular) Björling representation_ [kim_prescribing_2007]
for maxfaces [umehara_maximal_2006] and generalized timelike minimal surfaces
[kim_spacelike_2011] in Lorentz $3$-space, and also the Björling
representation for zero mean curvature surfaces in isotropic $3$-space
[seo_zero_2021].

Our goal of the paper is to solve the Björling problem for zero mean curvature
surfaces in the $3$-dimensional light cone $\mathbb{Q}^{3}_{+}$. In the
similar cases of other quadrics of Lorentz $4$-space, namely, the hyperbolic
$3$-space $\mathbb{H}^{3}(-1)$ and de Sitter $3$-space
$\mathbb{S}^{3}_{1}(1)$, the tangent space is isomorphic to either the
Euclidean $3$-space or the Lorentz $3$-space, respectively. Thus one can use
the cross product in the tangent space to obtain Björling representations
[yang_bjorling_2017] in an analogous manner to the Euclidean case. However,
the tangent space to $\mathbb{Q}^{3}_{+}$ is isomorphic to isotropic
$3$-space, so that it does not have a cross product structure. Therefore, for
the Björling problem in $\mathbb{Q}^{3}_{+}$, we use the alternative viewpoint
and assume that the given Björling data consists of a spacelike analytic curve
$\gamma$, together with a spacelike vector field $\mathcal{L}$ along the
curve, and find the zero mean curvature surface which contains $\gamma$ and
has $\mathcal{L}$ as a tangent vector field along $\gamma$.

The paper is structured as follows: After reviewing the basic geometry and
surface theory of $\mathbb{Q}^{3}_{+}$ in Section 2, we describe the process
to solve the Björling problem for a given spacelike analytic curve with
prescribed tangent vector field in Section 3, culminating in the _Björling
representation for zero mean curvature surfaces in $3$-dimensional light cone_
(see Theorem 3.8). As an application, we construct and classify (see Theorem
4.4) rotationally invariant zero mean curvature surfaces in
$\mathbb{Q}^{3}_{+}$ in Section 4. Furthermore, many surfaces admitting
Weierstrass representations in various space forms with indefinite metric can
be extended across lightlike lines (see, for example,
[akamine_reflection_2021, akamine_reflection_2022, akamine_space-like_2019,
fujimori_analytic_2022, fujimori_zero_2015-1, umehara_hypersurfaces_2019]); we
give an example of such surface in $\mathbb{Q}^{3}_{+}$ in Section 4.5.

##  2\. Preliminaries

We first briefly review the geometry of three-dimensional light cone, and the
surface theory within. For a detailed description, see [liu_surfaces_2007,
liu_hypersurfaces_2008, liu_representation_2011].

###  2.1. Hermitian matrix model of three-dimensional light cone

Let $\mathbb{L}^{4}$ denote the Lorentz $4$-space, with inner product

$\langle(t_{1},x_{1},y_{1},z_{1}),(t_{2},x_{2},y_{2},z_{2})\rangle=-t_{1}t_{2}+x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2}.$

The Lorentz $4$-space can be identified with the set of $2\times 2$ Hermitian
matrices $\mathrm{Herm}(2,\mathbb{C})$ via

$(t,x,y,z)\sim\begin{pmatrix}t+z&x+iy\\\ x-iy&t-z\end{pmatrix}$

where we will abuse notation between vectors and matrices. Then for any
$V,W\in\mathrm{Herm}(2,\mathbb{C})\cong\mathbb{L}^{4}$,

$\langle V,W\rangle=-\frac{1}{2}\left(\det{(V+W)}-\det{V}-\det{W}\right),$

so that

$|V|^{2}\mathrel{\mathop{\ordinarycolon}}=\langle V,V\rangle=-\det{V}.$

We note here that the symmetric bilinear form $\langle\cdot,\cdot\rangle$ is
well-defined for all $A\in\mathrm{M}(2,\mathbb{C})$.

Defining

$f_{0}\mathrel{\mathop{\ordinarycolon}}=\begin{pmatrix}0&0\\\
0&-2\end{pmatrix},\quad
f_{1}\mathrel{\mathop{\ordinarycolon}}=\begin{pmatrix}0&1\\\
1&0\end{pmatrix},\quad
f_{2}\mathrel{\mathop{\ordinarycolon}}=\begin{pmatrix}0&i\\\
-i&0\end{pmatrix},\quad
f_{3}\mathrel{\mathop{\ordinarycolon}}=\begin{pmatrix}1&0\\\
0&0\end{pmatrix},$

we see that $\\{f_{0},f_{1},f_{2},f_{3}\\}$ form an asymptotically orthonormal
basis of $\mathbb{L}^{4}\cong\mathrm{Herm}(2,\mathbb{C})$.

In the Hermitian model, hyperbolic $3$-space $\mathbb{H}^{3}(-1)$,
$3$-dimensional light cone $\mathbb{Q}^{3}_{+}$, and de Sitter $3$-space
$\mathbb{S}^{3}_{1}(1)$ can be defined as quadrics in $\mathbb{L}^{4}$ via

$\displaystyle\mathbb{H}^{3}(-1)$
$\displaystyle\mathrel{\mathop{\ordinarycolon}}=\\{X\in\mathrm{Herm}(2,\mathbb{C})\mathrel{\mathop{\ordinarycolon}}\langle
X,X\rangle=-1,\operatorname{tr}{X}>0\\},$ $\displaystyle\mathbb{Q}^{3}_{+}$
$\displaystyle\mathrel{\mathop{\ordinarycolon}}=\\{X\in\mathrm{Herm}(2,\mathbb{C})\mathrel{\mathop{\ordinarycolon}}\langle
X,X\rangle=0,\operatorname{tr}{X}>0\\},$ $\displaystyle\mathbb{S}^{3}_{1}(1)$
$\displaystyle\mathrel{\mathop{\ordinarycolon}}=\\{X\in\mathrm{Herm}(2,\mathbb{C})\mathrel{\mathop{\ordinarycolon}}\langle
X,X\rangle=1\\},$

respectively. Note that for any $X\in\mathbb{Q}^{3}_{+}$, there is some
$F\in\mathrm{SL}(2,\mathbb{C})$ such that

$X=F\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}F^{\star}=Ff_{3}F^{\star}$

where $F^{\star}$ denotes the conjugate transpose of $F$.

When we visualize surfaces in $\mathbb{Q}^{3}_{+}$, we will use the following
stereographic projection

$\mathbb{Q}^{3}_{+}\ni(t,x,y,z)\mapsto\left(\frac{x}{1+t},\frac{y}{1+t},\frac{z}{1+t}\right).$

Then $\mathbb{Q}^{3}_{+}$ is identified with
$\\{(a,b,c)\in\mathbb{R}^{3}\mathrel{\mathop{\ordinarycolon}}0<a^{2}+b^{2}+c^{2}<1\\}$.

###  2.2. Surface theory and Weierstrass-type representation

Let $D$ be a two-dimensional simply-connected domain, and suppose
$X\mathrel{\mathop{\ordinarycolon}}D\to\mathbb{Q}^{3}_{+}$ is a spacelike
immersion, that is, the induced metric on the tangent plane of $X$ at every
point $p\in D$ is Riemannian. Then for conformal coordinates $(u,v)\in D$ with
complex structure given via $z=u+iv$, suppose that the first fundamental form
is given by

$\operatorname{d\\!}{s}^{2}=\phi^{2}(\operatorname{d\\!}{u}^{2}+\operatorname{d\\!}{v}^{2})=\phi^{2}\operatorname{d\\!}{z}\operatorname{d\\!}{\bar{z}}$

for some $\phi\mathrel{\mathop{\ordinarycolon}}D\to\mathbb{R}^{\times}$.
Furthermore, we have $\langle X,X_{u}\rangle=\langle X,X_{v}\rangle=0$ so that
there exists a unique lightlike
$n\mathrel{\mathop{\ordinarycolon}}D\to\mathbb{L}^{4}$ such that

(2.1) $\langle n,n\rangle=\langle n,X_{u}\rangle=\langle
n,X_{v}\rangle=0,\quad\langle n,X\rangle=1.$

Such $n$ is called the _Gauss map_ of $X$, and the coefficients of the second
fundamental form is then computed in terms of $n$ via

$L\mathrel{\mathop{\ordinarycolon}}=-\langle X_{u},n_{u}\rangle,\quad
M\mathrel{\mathop{\ordinarycolon}}=-\langle X_{u},n_{v}\rangle=-\langle
X_{v},n_{u}\rangle,\quad N\mathrel{\mathop{\ordinarycolon}}=-\langle
X_{v},n_{v}\rangle.$

Hence, the shape operator $S$ is

$S=\phi^{-2}\begin{pmatrix}L&M\\\ M&N\end{pmatrix}$

with the (extrinsic) Gaussian curvature $K$ and mean curvature $H$ given by

$K=\det{S},\quad H=\frac{1}{2}\operatorname{tr}S.$

Those surfaces $X$ with $H\equiv 0$ will be referred to as _zero mean
curvature surfaces_.

We recall the Weierstrass-type representation for zero mean curvature surfaces
in the three-dimensional light cone [liu_representation_2011, Theorem 3.2]
(see also [pember_weierstrass-type_2020, § 4.2] and [seo_zero_2021, Theorem
39]):

######  Fact 2.1.

Any zero mean curvature surface
$X\mathrel{\mathop{\ordinarycolon}}D\to\mathbb{Q}^{3}_{+}$ can locally be
represented as

$X=Ff_{3}F^{\star}=F\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}F^{\star}$

where $F\mathrel{\mathop{\ordinarycolon}}D\to\mathrm{SL}(2,\mathbb{C})$
satisfies

$\operatorname{d\\!}{F}F^{-1}=\begin{pmatrix}G&-G^{2}\\\
1&-G\end{pmatrix}\Omega$

for some meromorphic function
$G\mathrel{\mathop{\ordinarycolon}}D\to\mathbb{C}$ and holomorphic $1$-form
$\Omega\mathrel{\mathop{\ordinarycolon}}D\to\mathbb{C}$ such that
$G^{2}\Omega$ is holomorphic. The pair $(G,\Omega)$ is called the _Weierstrass
data_.

##  3\. Solution of the Björling problem

Using the Weierstrass-type representation in Fact 2.1, we will now solve the
Björling problem for zero mean curvature surfaces in $\mathbb{Q}^{3}_{+}$.
Unlike the cases of hyperbolic $3$-space $\mathbb{H}^{3}(-1)$ and de Sitter
$3$-space $\mathbb{S}^{3}_{1}(1)$, the Björling data we consider will be an
analytic curve together with a tangent vector field due to the lack of cross
product structure in the tangent space of $\mathbb{Q}^{3}_{+}$.

###  3.1. The conformality condition for the Björling data

For an interval $I$ with parameter $u\in I$, let
$\gamma\mathrel{\mathop{\ordinarycolon}}I\to\mathbb{Q}^{3}_{+}$ be a spacelike
analytic curve and $\mathcal{L}$ be an analytic spacelike vector field along
$\gamma$ such that

(3.1)
$\langle\dot{\gamma},\dot{\gamma}\rangle=\langle\mathcal{L},\mathcal{L}\rangle,\quad\langle\dot{\gamma},\mathcal{L}\rangle=0,\quad\langle\gamma,\mathcal{L}\rangle=0.$

Geometrically, the conditions (3.1) ensure that $\mathcal{L}$ will be a
tangent vector field; thus, we refer to (3.1) as the _conformality condition_.
Our goal is to find a zero mean curvature surface
$X\mathrel{\mathop{\ordinarycolon}}D\to\mathbb{Q}^{3}_{+}$ which contains
$\gamma$ and which has $\mathcal{L}$ as a tangent vector field along $\gamma$.

Note that if there is a such a zero mean curvature surface $X$, then

(3.2)
$\operatorname{d\\!}{X}=\operatorname{d\\!}{F}f_{3}F^{\star}=\operatorname{d\\!}{F}F^{-1}Ff_{3}F^{\star}=\begin{pmatrix}G&-G^{2}\\\
1&-G\end{pmatrix}\Omega X=\mathrel{\mathop{\ordinarycolon}}\mathbf{\Omega}X.$

Thus, from $\gamma$ and $\mathcal{L}$, we will construct such $1$-form
$\mathbf{\Omega}$.

######  Remark 3.1.

An important distinction from the cases of $\mathbb{H}^{3}(-1)$ and
$\mathbb{S}^{3}_{1}(1)$ is that $X\in\mathbb{Q}^{3}_{+}$ is not invertible
since $\det X=0$. Thus, instead of considering $\operatorname{d\\!}{X}X^{-1}$
as in the cases of $\mathbb{H}^{3}(-1)$ and $\mathbb{S}^{3}_{1}(1)$, we
consider the equation as in (3.2).

For Björling data $\gamma$ and $\mathcal{L}$ satisfying conformality condition
(3.1), let $\gamma(u)=X(u,0)$ and
$\mathcal{L}(u)\mathrel{\mathop{\ordinarycolon}}=X_{v}(u,0)$. Then on one
dimensional domain $I$, (3.2) implies that we must solve for
$\mathbf{\Omega}=\mathrel{\mathop{\ordinarycolon}}\tilde{\Omega}\operatorname{d\\!}{u}$
satisfying

(3.3)
$\Lambda(u)\mathrel{\mathop{\ordinarycolon}}=\frac{1}{2}(\dot{\gamma}(u)-i\mathcal{L}(u))=X_{z}(u,0)=\tilde{\Omega}(u)X(u,0)=\tilde{\Omega}(u)\gamma(u).$

We claim that there is a unique $\tilde{\Omega}$ which solves (3.3).

Note that $\det{\Lambda}=0$ because of (3.1), and that $\det{\gamma}=0$ since
it is a curve in $\mathbb{Q}^{3}_{+}$. By standard theory we know that

$\det{\tilde{\Omega}}=\operatorname{tr}{\tilde{\Omega}}=0.$

Therefore

$\mathbf{\Omega}=\begin{pmatrix}G&-G^{2}\\\ 1&-G\end{pmatrix}\Omega$

for some functions $G$ and $1$-form $\Omega$. We rewrite (3.3) as

(3.4) $\begin{pmatrix}\Lambda_{11}&\Lambda_{12}\\\
\Lambda_{21}&\Lambda_{22}\end{pmatrix}=\begin{pmatrix}G\tilde{\Omega}&-G^{2}\tilde{\Omega}\\\
\tilde{\Omega}&-G\tilde{\Omega}\end{pmatrix}\begin{pmatrix}\gamma_{11}&\gamma_{12}\\\
\gamma_{21}&\gamma_{22}\end{pmatrix}.$

By examining $(1,1)$ and $(2,1)$ components of (3.4), we see that $G=G_{1}$
and $\tilde{\Omega}=\tilde{\Omega}_{1}$ where

(3.5)
$G_{1}=\frac{\Lambda_{11}}{\Lambda_{21}},\quad\tilde{\Omega}_{1}=\frac{\Lambda_{21}^{2}}{\Lambda_{21}\gamma_{11}-\Lambda_{11}\gamma_{21}}.$

On the other hand, by examining $(1,2)$ and $(2,2)$ components of (3.4), we
see that $G=G_{2}$ and $\tilde{\Omega}=\tilde{\Omega}_{2}$ where

(3.6)
$G_{2}=\frac{\Lambda_{12}}{\Lambda_{22}},\quad\tilde{\Omega}_{2}=\frac{\Lambda_{22}^{2}}{\Lambda_{22}\gamma_{12}-\Lambda_{12}\gamma_{22}}.$

It is immediate to see $G_{1}=G_{2}$ from the fact that $\det{\Lambda}=0$.
Then

$\displaystyle\tilde{\Omega}_{1}$
$\displaystyle=\frac{\Lambda_{21}^{2}}{\Lambda_{21}\gamma_{11}-\frac{\Lambda_{12}\Lambda_{21}}{\Lambda_{22}}\gamma_{21}}=\frac{\Lambda_{21}\Lambda_{22}}{\Lambda_{22}\gamma_{11}-\Lambda_{12}\gamma_{21}},$
$\displaystyle\tilde{\Omega}_{2}$
$\displaystyle=\frac{\Lambda_{22}^{2}}{\Lambda_{22}\gamma_{12}-\frac{\Lambda_{11}\Lambda_{22}}{\Lambda_{21}}\gamma_{22}}=\frac{\Lambda_{21}\Lambda_{22}}{\Lambda_{21}\gamma_{12}-\Lambda_{11}\gamma_{22}}.$

Hence, it is enough to show that the denominators are the same.

For this matter, we multiply (3.4) with the adjunct of $\gamma$ from the right
to obtain

$\begin{pmatrix}\Lambda_{11}&\Lambda_{12}\\\
\Lambda_{21}&\Lambda_{22}\end{pmatrix}\begin{pmatrix}\gamma_{22}&-\gamma_{12}\\\
-\gamma_{21}&\gamma_{11}\end{pmatrix}=\begin{pmatrix}G\tilde{\Omega}&-G^{2}\tilde{\Omega}\\\
\tilde{\Omega}&-G\tilde{\Omega}\end{pmatrix}\begin{pmatrix}\det{\gamma}&0\\\
0&\det{\gamma}\end{pmatrix}.$

However, we have that $\det{\gamma}=0$ since it is a curve in
$\mathbb{Q}^{3}_{+}$. Therefore, the left hand side is a zero matrix. In
particular, its trace is 0; hence

$\Lambda_{11}\gamma_{22}-\Lambda_{12}\gamma_{21}-\Lambda_{21}\gamma_{12}+\Lambda_{22}\gamma_{11}=0,$

implying that $\tilde{\Omega}_{1}=\tilde{\Omega}_{2}$.

Now, the expressions for $\tilde{\Omega}_{1}$ and $\tilde{\Omega}_{2}$ in
(3.5) and (3.6) implies that the Björling data $\gamma$ and $\mathcal{L}$
satisfying (3.1) must additionally satisfy

(3.7) $\gamma_{11}\Lambda_{21}-\gamma_{21}\Lambda_{11}\neq 0.$

Unfortunately, the conformality conditions (3.1) alone does not guarantee
this; one can even construct explicit examples where the expression (3.7)
vanishes for some Björling data satisfying the conformality conditions.

###  3.2. The orientability condition for the Björling data

To give a geometric interpretation of the condition (3.7), we define the
following notion.

######  Definition 3.2.

For two linearly independent spacelike vectors $U,V\in\mathbb{L}^{4}$ spanning
a Riemannian subspace $W$ of $\mathbb{L}^{4}$, let $\\{e_{1},e_{2}\\}$ be an
orthonormal basis of $W$ such that there is some
$F\in\mathrm{SL}(2,\mathbb{C})$ with $Ff_{i}F^{\star}=e_{i}$ for $i=1,2$. The
_signed area of $U$ and $V$_, denoted by $\mathrm{SA}(U,V)$, is defined via

$U\wedge V=\mathrel{\mathop{\ordinarycolon}}\mathrm{SA}(U,V)e_{1}\wedge
e_{2}.$

An important feature of the signed area is the fact that it is invariant under
_orientation-preserving isometries_ of $\mathbb{Q}^{3}_{+}$. On the other
hand, the square of the signed area (or the squared area) can be calculated
using the inner product of the ambient Lorentz $4$-space:

######  Lemma 3.3.

For linearly independent spacelike vectors $U,V$ spanning a Riemannian
subspace $W$, we have

$\mathrm{SA}(U,V)^{2}=\langle U,U\rangle\langle V,V\rangle-\langle
U,V\rangle^{2}.$

###### Proof.

Let $\\{e_{1},e_{2}\\}$ be an orthonormal basis $W$ such that
$Ff_{i}F^{\star}=e_{i}$ for $i=1,2$ with $F\in\mathrm{SL}(2,\mathbb{C})$.
Writing

$U=ae_{1}+be_{2},\quad V=ce_{1}+de_{2}$

for some constants $a,b,c,d\in\mathbb{R}$, we calculate that

$\mathrm{SA}(U,V)e_{1}\wedge e_{2}=(ae_{1}+be_{2})\wedge(ce_{1}+de_{2})=(ad-
bc)e_{1}\wedge e_{2}.$

On the other hand, we have

$\langle U,U\rangle\langle V,V\rangle-\langle
U,V\rangle^{2}=(a^{2}+b^{2})(c^{2}+d^{2})-(ac+bd)^{2}=(ad-bc)^{2},$

giving us the desired conclusion. ∎

The next lemmata give additional geometric criteria for the Björling data to
satisfy (3.7) in terms of the signed area.

######  Lemma 3.4.

Given Björling data $\gamma$ and $\mathcal{L}$ satisfying the conformality
condition, the signed area of $\dot{\gamma}$ and $\mathcal{L}$ does not
vanish.

###### Proof.

Suppose for contradiction that $\mathrm{SA}(\dot{\gamma},\mathcal{L})=0$. Then
direct calculations show that
$\mathcal{\mathcal{L}}=\alpha\gamma\pm\dot{\gamma}$ for some $\alpha$. Thus

$\pm\langle\dot{\gamma},\dot{\gamma}\rangle=\langle\dot{\gamma},\alpha\gamma\pm\dot{\gamma}\rangle=\langle\dot{\gamma},\mathcal{L}\rangle=0,$

which is a contradiction since $\dot{\gamma}$ is spacelike. ∎

######  Lemma 3.5.

Given Björling data $\gamma$ and $\mathcal{L}$ satisfying the conformality
condition, $G$ and $\Omega$ are well-defined if and only if the signed area of
$\dot{\gamma}$ and $\mathcal{L}$ is negative.

###### Proof.

Fix any $u_{0}\in I$ throughout the proof. Since signed area is invariant
under orientation-preserving isometries, we may assume without loss of
generality that $\gamma=2f_{3}$, so that any vector in
$T_{\gamma}\mathbb{Q}^{3}_{+}$ takes the form

$\ell\gamma+xf_{1}+yf_{2}$

for some $\ell,x,y\in\mathbb{R}$. By the conformality conditions (3.1) on the
Björling data, we have that $\dot{\gamma},\mathcal{L}\in
T_{\gamma}\mathbb{Q}^{3}_{+}$; therefore, there are some constants
$a,b,c,d,e,f\in\mathbb{R}$ such that

$\dot{\gamma}=a\gamma+bf_{1}+cf_{2},\quad\text{and}\quad\mathcal{L}=d\gamma+ef_{1}+ff_{2}.$

Since
$\Lambda\mathrel{\mathop{\ordinarycolon}}=\frac{1}{2}\left(\dot{\gamma}-i\mathcal{L}\right)$,
we calculate that

$\displaystyle\gamma_{11}\Lambda_{21}-\gamma_{21}\Lambda_{11}=(b-f)-i(c+e),$

so that

(3.8) $\displaystyle|\gamma_{11}\Lambda_{21}-\gamma_{21}\Lambda_{11}|^{2}$
$\displaystyle=b^{2}+c^{2}+e^{2}+f^{2}+2(ce-bf)$
$\displaystyle=|\dot{\gamma}|^{2}+|\mathcal{L}|^{2}+2(ce-bf)$
$\displaystyle=2(|\dot{\gamma}|^{2}-(bf-ce)).$

Now since $\gamma$-component does not affect the signed area, we note that

$\mathrm{SA}(\dot{\gamma},\mathcal{L})=\mathrm{SA}(bf_{1}+cf_{2},ef_{1}+ff_{2}).$

However, we can also calculate that

$\mathrm{SA}(bf_{1}+cf_{2},ef_{1}+ff_{2})f_{1}\wedge
f_{2}=(bf_{1}+cf_{2})\wedge(ef_{1}+ff_{2})=(bf-ce)f_{1}\wedge f_{2},$

allowing us to deduce that

$\mathrm{SA}(\dot{\gamma},\mathcal{L})=bf-ce.$

On the other hand, Lemma 3.3 and the conformality conditions (3.1) imply that

$\mathrm{SA}(\dot{\gamma},\mathcal{L})^{2}=|\dot{\gamma}|^{2}|\mathcal{L}|^{2}-\langle\dot{\gamma},\mathcal{L}\rangle^{2}=|\dot{\gamma}|^{4}$

so that

$bf-
ce=\mathrm{SA}(\dot{\gamma},\mathcal{L})=\begin{cases}|\dot{\gamma}|^{2},&\text{if
}\mathrm{SA}(\dot{\gamma},\mathcal{L})>0,\\\ -|\dot{\gamma}|^{2},&\text{if
}\mathrm{SA}(\dot{\gamma},\mathcal{L})<0.\end{cases}$

Thus, using (3.8) allows us to conclude that

$|\gamma_{11}\Lambda_{21}-\gamma_{21}\Lambda_{11}|^{2}=\begin{cases}0,&\text{if
}\mathrm{SA}(\dot{\gamma},\mathcal{L})>0,\\\ 4|\dot{\gamma}|^{2},&\text{if
}\mathrm{SA}(\dot{\gamma},\mathcal{L})<0.\end{cases}$

Therefore, $G$ and $\Omega$ are well-defined if and only if
$SA(\dot{\gamma},\mathcal{L})<0$. ∎

######  Remark 3.6.

The signed area condition of Lemma 3.5 encodes the fact that the orientation
of $\dot{\gamma}$ and $\mathcal{L}$ must be correct. For this reason, we call
the signed area condition, the _orientability condition_.

######  Remark 3.7.

The orientability condition is unnecessary for the Björling data in other
quadrics such as $\mathbb{H}^{3}(-1)$ and $\mathbb{S}^{3}_{1}(1)$; the crucial
difference arises in the definition of the (lightlike) Gauss map used to
define the second fundamental form. In the case of $\mathbb{H}^{3}(-1)$ and
$\mathbb{S}^{3}_{1}(1)$, the Gauss map $n$ and the position vector $X$ must
satisfy $\langle n,X\rangle=0$; however, in the case $\mathbb{Q}^{3}_{+}$, we
have that $\langle n,X\rangle=1$, so that one cannot switch the signs on the
Gauss map freely.

We summarize our results in the next main theorem of our paper, the _Björling
representation for zero mean curvature surfaces in the $3$-dimensional light
cone_:

######  Theorem 3.8.

Consider $\mathbb{Q}^{3}_{+}$ as a subset of $\mathrm{Herm}(2,\mathbb{C})$
identified with $\mathbb{L}^{4}$. Given a spacelike analytic curve
$\gamma\mathrel{\mathop{\ordinarycolon}}I\to\mathbb{Q}^{3}_{+}$ and an
analytic vector field $\mathcal{L}$ satisfying the conformality condition

$\langle\dot{\gamma},\dot{\gamma}\rangle=\langle\mathcal{L},\mathcal{L}\rangle,\quad\langle\dot{\gamma},\mathcal{L}\rangle=0,\quad\langle\gamma,\mathcal{L}\rangle=0,$

and the orientability condition

$\mathrm{SA}(\dot{\gamma}(u),\mathcal{L}(u))<0\qquad\text{for all }u\in I,$

there exists a unique zero mean curvature surface $X$ such that

$X(u,0)=\gamma(u),\qquad X_{v}(u,0)=\mathcal{L}(u)$

for all $u\in I$. It is given by $X=Ff_{3}F^{\star}$ where $F\in
C^{\omega}(\mathcal{U}\subset\mathbb{C},\mathrm{SL}(2,\mathbb{C}))$ is a
solution to

$\operatorname{d\\!}{F}F^{-1}=\text{ the analytic extension of
}\mathbf{\Omega},\qquad F(u_{0})f_{3}F(u_{0})^{\star}=\gamma(u_{0})$

where $u_{0}$ is an element of the domain of $\gamma$ and
$\mathbf{\Omega}=\mathrel{\mathop{\ordinarycolon}}\tilde{\Omega}\operatorname{d\\!}{u}$
is the solution of

$\Lambda=\tilde{\Omega}\gamma,\qquad\text{where}\quad\Lambda(u)\mathrel{\mathop{\ordinarycolon}}=\frac{1}{2}(\dot{\gamma}(u)-i\mathcal{L}(u))\in
M(2,\mathbb{C}).$

##  4\. Rotational zero mean curvature surfaces in $\mathbb{Q}^{3}_{+}$

In this section we construct and classify rotationally invariant zero mean
curvature surfaces, which we call _catenoids_ of the $3$-dimensional light
cone. The Björling problem is especially apt for the construction of such
surface as the initial curve can be selected as the orbit of a single point
under rotations of $\mathbb{Q}^{3}_{+}$.

First, we define rotational surfaces in $\mathbb{Q}^{3}_{+}$ analogously to
the definition given in hyperbolic spaces [do_carmo_rotation_1983, Definition
2.2] (see also [abe_constant_2018, Definition 3.1]):

######  Definition 4.1.

Let $P^{k}$ denote a $k$-dimensional subspace of $\mathbb{L}^{4}$. Choosing
some $P^{2}$ and $P^{3}\supset P^{2}$ such that
$P^{3}\cap\mathbb{Q}^{3}_{+}\neq\varnothing$, let $\mathrm{O}(P^{2})$ denote
the set of orthogonal transformations that leave $P^{2}$ fixed. For a regular
curve $\gamma$ in $P^{3}\cap\mathbb{Q}^{3}_{+}$ that does not meet $P^{2}$, we
call the orbit of $\gamma$ under the action of $\mathrm{O}(P^{2})$ a
_rotational surface generated by $\gamma$_. Furthermore, we say that the
rotational surface is

  * •

an _elliptic_ rotational surface if the induced metric on $P^{2}$ is
Lorentzian,

  * •

a _parabolic_ rotational surface if the induced metric on $P^{2}$ is
degenerate, and

  * •

a _hyperbolic_ rotational surface if the induced metric on $P^{2}$ is
Riemannian.

The following facts will be useful in recovering the explicit parametrizations
of the catenoids from the rotationally invariant Björling data:

######  Fact 4.2 ([yang_bjorling_2017, Lemma 4.2]).

For $\nu\in\mathbb{C}\setminus\\{0\\}$,

$F_{0}(z)\mathrel{\mathop{\ordinarycolon}}=\frac{1}{2}\begin{pmatrix}z^{1/2}&0\\\
0&z^{-1/2}\end{pmatrix}\begin{pmatrix}\sqrt{\nu}+\sqrt{\nu}^{-1}&\sqrt{\nu}-\sqrt{\nu}^{-1}\\\
\sqrt{\nu}-\sqrt{\nu}^{-1}&\sqrt{\nu}+\sqrt{\nu}^{-1}\end{pmatrix}\begin{pmatrix}z^{-\nu/2}&0\\\
0&z^{\nu/2}\end{pmatrix}$

is a solution to

(4.1) $\operatorname{d\\!}{F}F^{-1}=\begin{pmatrix}G&-G^{2}\\\
1&-G\end{pmatrix}\Omega\quad\text{with}\quad
G\mathrel{\mathop{\ordinarycolon}}=z,\quad\Omega\mathrel{\mathop{\ordinarycolon}}=\lambda\frac{\operatorname{d\\!}{z}}{z^{2}},\quad\lambda\mathrel{\mathop{\ordinarycolon}}=\frac{1-\nu^{2}}{4}.$

where $\lambda\in\mathbb{C}\setminus\\{\frac{1}{4}\\}$. Furthermore,
$\tilde{F}\mathrel{\mathop{\ordinarycolon}}=F_{0}R$ also solves (4.1) where
$R\in\mathrm{SL}(2,\mathbb{C})$ controls the initial condition.

######  Fact 4.3 ([yang_bjorling_2017, Lemma 4.2]).

For $\beta\in\mathbb{C}\setminus\\{0\\}$, let

$F_{1}(z)\mathrel{\mathop{\ordinarycolon}}=\begin{pmatrix}1&z\\\
0&1\end{pmatrix}\begin{pmatrix}\beta^{-1/2}&0\\\
0&\beta^{1/2}\end{pmatrix}\begin{pmatrix}\cos{\beta z}&-\sin{\beta z}\\\
\sin{\beta z}&\cos{\beta z}\end{pmatrix}\begin{pmatrix}\beta^{1/2}&0\\\
0&\beta^{-1/2}\end{pmatrix}$

is a solution to

(4.2) $\operatorname{d\\!}{F}F^{-1}=\begin{pmatrix}G&-G^{2}\\\
1&-G\end{pmatrix}\Omega\quad\text{with}\quad
G\mathrel{\mathop{\ordinarycolon}}=z,\quad\Omega\mathrel{\mathop{\ordinarycolon}}=\beta^{2}\operatorname{d\\!}{z}.$

Furthermore, $\tilde{F}\mathrel{\mathop{\ordinarycolon}}=F_{1}R$ also solves
(4.2) where $R\in\mathrm{SL}(2,\mathbb{C})$ controls the initial condition.

###  4.1. Elliptic catenoids

Consider an elliptic rotation

$R^{\mathrm{E}}(u)=\begin{pmatrix}e^{iu}&0\\\ 0&e^{-iu}\end{pmatrix}$

applied to the point $\begin{pmatrix}1&1\\\
1&1\end{pmatrix}\in\mathbb{Q}^{3}_{+}$ to obtain an elliptic circle

$\gamma(u)=R^{\mathrm{E}}(u)\begin{pmatrix}1&1\\\
1&1\end{pmatrix}R^{\mathrm{E}}(u)^{\star}=\begin{pmatrix}1&e^{2iu}\\\
e^{-2iu}&1\end{pmatrix}.$

Note that $\langle\dot{\gamma},\dot{\gamma}\rangle=4$ and that
$\gamma(0)=\begin{pmatrix}1&0\\\ 1&1\end{pmatrix}f_{3}\begin{pmatrix}1&0\\\
1&1\end{pmatrix}^{\star}$.

To complete the Björling data, we first note that any $\mathcal{L}$ satisfying
the conformality condition must be of the form

$\mathcal{L}_{\pm}(u)=f(u)\gamma(u)\pm 2e_{3}$

for any function $f$. Now, we can check that $\mathcal{L}_{+}=f\gamma+2e_{3}$
fails the orientability condition, that is,

$\Lambda_{21}\gamma_{11}-\Lambda_{11}\gamma_{21}=0,$

while $\mathcal{L}_{-}=f\gamma-2e_{3}$ satisfies the orientability condition.

Thus we take $\mathcal{L}=\mathcal{L}_{-}$, and note that for rotationally
invariant surfaces, we must have that $\mathcal{L}$ is generated by the
elliptic rotation under consideration, i.e.

$\mathcal{L}(u)=R^{\mathrm{E}}(u)\mathcal{L}(0)R^{\mathrm{E}}(u)^{\star}.$

Thus, we find that

$f(u)\gamma(u)-2e_{3}=\mathcal{L}(u)=R^{\mathrm{E}}(u)\mathcal{L}(0)R^{\mathrm{E}}(u)^{\star}=f(0)\gamma(u)-2e_{3},$

and we have $f(u)=f(0)=\mathrel{\mathop{\ordinarycolon}}a$ is a constant
function so that

$\mathcal{L}(u)=a\gamma(u)-2e_{3}.$

Considering $\gamma$ and $\mathcal{L}$ as the Björling data, we use the
Björling representation in Theorem 3.8 to calculate that

$G=\frac{a-2}{a+2}e^{2iu},\quad\text{and}\quad\Omega=\frac{-i(a+2)^{2}}{8}e^{-2iu}\operatorname{d\\!}{u}.$

To obtain explicit parametrizations, we analytically extend $G$ and $\Omega$
and change the variables via

$\frac{a-2}{a+2}e^{2iw}=\mathrel{\mathop{\ordinarycolon}}z.$

Then

$G=z,\quad\Omega=\lambda\frac{\operatorname{d\\!}{z}}{z^{2}},\quad\lambda\mathrel{\mathop{\ordinarycolon}}=\frac{4-a^{2}}{16},$

allowing us to use Fact 4.2 to compute the explicit parametrizations for the
elliptic catenoid:

$X^{\mathrm{E}}=\begin{pmatrix}e^{(a-2)v}&e^{2iu+av}\\\
e^{-2iu+av}&e^{(a+2)v}\end{pmatrix}=R^{\mathrm{E}}(u)e^{av}\begin{pmatrix}e^{-2v}&1\\\
1&e^{2v}\end{pmatrix}R^{\mathrm{E}}(u)^{\star}$

where $w=u+iv$ (see also Figure 1).

Figure 1. Elliptic catenoids in $3$-dimensional light cone with
$a=\frac{3}{2}$ on the left, and $a=4$ on the right, where the initial given
curve is highlighted.

###  4.2. Hyperbolic catenoids

Given a one parameter group of hyperbolic rotations

$R^{\mathrm{H}}(u)=\begin{pmatrix}e^{u}&0\\\ 0&e^{-u}\end{pmatrix},$

the curve

$\gamma(u)\mathrel{\mathop{\ordinarycolon}}=R^{\mathrm{H}}(u)\begin{pmatrix}1&1\\\
1&1\end{pmatrix}R^{\mathrm{H}}(u)^{\star}=\begin{pmatrix}e^{2u}&1\\\
1&e^{-2u}\end{pmatrix}$

is a hyperbolic circle. Note that $\langle\dot{\gamma},\dot{\gamma}\rangle=4$
and that $\gamma(0)=\begin{pmatrix}1&0\\\
1&1\end{pmatrix}f_{3}\begin{pmatrix}1&0\\\ 1&1\end{pmatrix}^{\star}$.

To find the Björling data, we find that any $\mathcal{L}$ satisfying the
conformality condition must take the form

$\mathcal{L}_{\pm}(u)=f(u)\gamma(u)\pm f_{2},$

for any function $f$. Since $\mathcal{L}_{+}$ fails the orientability
condition, we set $\mathcal{L}=\mathcal{L}_{-}$. We also have that
$\mathcal{L}$ is generated by the rotation under consideration; thus,

$\mathcal{L}(u)=b\gamma(u)\pm f_{2},$

for some constant $b$.

Now, we calculate the Björling data as

$G=\frac{b+2i}{b-2i}e^{2u},\quad\text{and}\quad\Omega=\frac{1}{8}(b-2i)^{2}e^{-2u}\operatorname{d\\!}{u}.$

Analytically extending $G$ and $\Omega$ and letting

$\frac{b+2i}{b-2i}e^{2w}=z,$

so that

$G=z,\qquad\Omega=\lambda\frac{\operatorname{d\\!}{z}}{z^{2}},\qquad\lambda=\frac{4+b^{2}}{16},$

we can use Fact 4.2 to compute the explicit parametrizations for hyperbolic
catenoid:

$X^{\mathrm{H}}=\begin{pmatrix}e^{2u+bv}&e^{(b+2i)v}\\\
e^{(b-2i)v}&e^{-2u+bv}\end{pmatrix}=R^{\mathrm{H}}(u)e^{bv}\begin{pmatrix}1&e^{2iv}\\\
e^{-2iv}&1\end{pmatrix}R^{\mathrm{H}}(u)^{\star}$

where $w=u+iv$ (see Figure 2).

Figure 2. Hyperbolic catenoid in the $3$-dimensional light cone with parameter
$b=\frac{3}{2}$, where the initial given curve is highlighted.

###  4.3. Parabolic catenoids

Finally, we consider the parabolic rotation

$R^{\mathrm{P}}(u)=\begin{pmatrix}1&u-1\\\ 0&1\end{pmatrix}$

applied to the point $\begin{pmatrix}1&1\\\ 1&1\end{pmatrix}$ to obtain a
parabolic circle

$\gamma(u)=R^{\mathrm{P}}(u)\begin{pmatrix}1&1\\\
1&1\end{pmatrix}R^{\mathrm{P}}(u)^{\star}=\begin{pmatrix}u^{2}&u\\\
u&1\end{pmatrix}.$

Note that $\langle\dot{\gamma},\dot{\gamma}\rangle=1$ and that
$\gamma(1)=\begin{pmatrix}1&0\\\ 1&1\end{pmatrix}f_{3}\begin{pmatrix}1&0\\\
1&1\end{pmatrix}^{\star}$.

We note that any $\mathcal{L}$ satisfying the conformality condition must
satisfy

$\mathcal{L}_{\pm}(u)\mathrel{\mathop{\ordinarycolon}}=f(u)\gamma(u)\pm f_{2}$

for some function $f$, and direct calculation tells us that $\mathcal{L}_{-}$
fails the orientability condition. Thus, we choose
$\mathcal{L}=\mathcal{L}_{+}$, and the fact that $\mathcal{L}$ must generated
by the rotation under consideration allows us to deduce that

$\mathcal{L}(u)=c\gamma(u)+f_{2}.$

for some constant $c$.

Using $\gamma$ and $\mathcal{L}$ as the Björling data, we calculate that

$G=u+\frac{2i}{c},\quad\text{and}\quad\Omega=\frac{c^{2}}{4}\operatorname{d\\!}{u}.$

Analytically extending $G$ and $\Omega$ and making change of coordinates so
that

$w+\frac{2i}{c}=z,$

we have

$G=z,\qquad\Omega=\frac{c^{2}}{4}\operatorname{d\\!}{z}.$

This allows us to use Fact 4.3 to obtain explicit parametrizations for the
parabolic catenoids:

$X^{\mathrm{P}}=\begin{pmatrix}e^{cv}(u^{2}+v^{2})&e^{cv}(u+iv)\\\
e^{cv}(u-iv)&e^{cv}\end{pmatrix}=R^{\mathrm{P}}(u+1)e^{cv}\begin{pmatrix}v^{2}&iv\\\
-iv&1\end{pmatrix}R^{\mathrm{P}}(u+1)^{\star}$

where $w=u+iv$ (see Figure 3).

Figure 3. Parabolic catenoid in $3$-dimensional light cone with parameter
$b=\frac{1}{2}$, where the initial given curve is highlighted.

###  4.4. Classification of rotationally invariant zero mean curvature
surfaces

Any circle in $\mathbb{Q}^{3}_{+}$ is congruent to one of the circles we
constructed as orbits of points under rotations up to homotheties and
isometries. Thus we conclude:

######  Theorem 4.4.

Any rotationally invariant zero mean curvature surfaces in
$\mathbb{Q}^{3}_{+}$ must be a piece of one of the following surfaces (given
with its respective Weierstrass data):

  * •

elliptic catenoid
$(G=\frac{a-2}{a+2}e^{2iw},\Omega=\frac{-i(a+2)^{2}}{8}e^{-2iw}\operatorname{d\\!}{w})$

  * •

hyperbolic catenoid
$(G=\frac{b+2i}{b-2i}e^{2w},\Omega=\frac{(b-2i)^{2}}{8}e^{-2w}\operatorname{d\\!}{w})$,
or

  * •

parabolic catenoid
$(G=w+\frac{2i}{c},\Omega=\frac{c^{2}}{4}\operatorname{d\\!}{w})$

up to homotheties and isometries of $\mathbb{Q}^{3}_{+}$.

###  4.5. Additional example with analytic extensions

As in the parabolic catenoid case, let us take the parabolic circle as the
initial curve, i.e.

$\gamma(u)=R^{\mathrm{P}}(u)\begin{pmatrix}1&1\\\
1&1\end{pmatrix}R^{\mathrm{P}}(u)^{\star}=\begin{pmatrix}u^{2}&u\\\
u&1\end{pmatrix}.$

We have seen that $\mathcal{L}=f\gamma+f_{2}$ for any function $f$ satisfies
both the conformality condition and the orientability condition.

Now, if we take

$f(u)=\frac{c}{u}$

for $u\in(0,\infty)$, then we have $\mathcal{L}=\frac{c}{u}\gamma+f_{2}$ is
not generated by the rotation under consideration; thus, the resulting surface
will not be rotationally invariant. We can still calculate that

$G=\frac{c+2i}{c}u,\qquad\Omega=\frac{c^{2}}{4u^{2}}\operatorname{d\\!}{u}.$

Analytically extending $G$ and $\Omega$ and letting

$\frac{c+2i}{c}w=\mathrel{\mathop{\ordinarycolon}}z,$

so that

$G=z,\qquad\Omega=\lambda\frac{\operatorname{d\\!}{z}}{z^{2}},\qquad\lambda=\frac{c(c+2i)}{4},$

we can use Fact 4.2 to compute the explicit parametrizations:

$X=e^{cv}\begin{pmatrix}e^{2u}&e^{u+iv}\\\ e^{u-iv}&1\end{pmatrix},$

where $w=e^{u+iv}$ (see Figure 4).

Figure 4. A parabolic catenoid and a non-rotational zero mean curvature
surface sharing the same initial curve, where the initial given curve is
highlighted (on the left); the non-rotational zero mean curvature surface
drawn over bigger domain (on the right).

If we change parameters so that $\tilde{u}=e^{u}$, then we obtain

$\tilde{X}=e^{cv}\begin{pmatrix}\tilde{u}^{2}&e^{iv}\tilde{u}\\\
e^{-iv}\tilde{u}&1\end{pmatrix}$

so that

$\langle\tilde{X}_{v},\tilde{X}_{v}\rangle=e^{2bv}\tilde{u}^{2}.$

This implies $L(v)\mathrel{\mathop{\ordinarycolon}}=\tilde{X}(0,v)$ is a
lightlike curve in $\mathbb{Q}^{3}_{+}$. In fact,

$\displaystyle L(v)$ $\displaystyle=\tilde{X}(0,v)=\begin{pmatrix}0&0\\\
0&e^{cv}\end{pmatrix}$ $\displaystyle=\begin{pmatrix}e^{\frac{-cv}{2}}&0\\\
0&e^{\frac{cv}{2}}\end{pmatrix}\begin{pmatrix}0&0\\\
0&1\end{pmatrix}\begin{pmatrix}e^{\frac{-cv}{2}}&0\\\
0&e^{\frac{cv}{2}}\end{pmatrix}^{\star}$
$\displaystyle=R^{\mathrm{H}}(\tfrac{-cv}{2})\begin{pmatrix}0&0\\\
0&1\end{pmatrix}R^{\mathrm{H}}(\tfrac{-cv}{2})^{\star},$

and thus $L$ is a lightlike circle. Therefore, we have that $\tilde{X}$ is an
analytic extension of $X$ across the lightlike circle $L$ (see Figure 5).

Figure 5. Analytic extension of the non-rotational zero mean curvature surface
across a lightlike circle (on the top); closer look at the analytic extension
near the lightlike circle (on the bottom). On all figures, the lightlike
circle is highlighted.

Acknowledgements. The last author gratefully acknowledges the support from NRF
of Korea (2017R1E1A1A03070929 and NRF 2020R1F1A1A01074585).

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