# Co-positivity of tensors and Stability conditions of CP conserving two-
Higgs-doublet potential111This work was supported by the National Natural
Science Foundation of P.R. China (Grant No.12171064), by The team project of
innovation leading talent in chongqing (No.CQYC20210309536) and by the
Foundation of Chongqing Normal university (20XLB009)

Yisheng Song Corresponding author. School of Mathematical Sciences, Chongqing
Normal University, Chongqing 401331 P.R. China, Email:
<EMAIL_ADDRESS>
###### Abstract

We give how to calculate the vacuum stability conditions of scalar potential
for two Higgs doublet model with explicit CP conservation. Moreover, the
analytical sufficient and necessary conditions are obtained. The argument
methods are first used to study the tensor with two parameters.

Keywords: Co-positivity, Fourth order tensors, Homogeneous polynomial, 2HDM

##  1 Introduction

The stability model of multi-Higgs potential is very noticeable in particle
physics community, and such a model itself was first proposed by Lee [1] for
the two-Higgs-doublet model (for short, 2HDM) in 1973. Weinberg [2] gave a
general model of multi-Higgs potential in 1976. It was studied in hundreds of
papers since then, one of the simplest extensions of the standard model is the
2HDM. There are many studies on the bounded from below (for short, BFB) or the
vacuum stability conditions of the 2HDM potential, including CP conservation
and CP violation. These results are different kinds of analytical conditions
of the vacuum stability of such a potential, for examples, 2HDM with CP
conservation in Refs. [3, 4, 5, 6, 7, 8, 9], the most general 2HDM in Refs.[6,
3], 2HDM with CP conservation and CP violation in Ref. [10, 5, 6, 11], 2HDM
handled numerically [12] and others references that are no cited here. For the
tree-level metastability bounds of the most general 2HDM see Ref.[13].
Recently, H. Bahl et al. [14] presented an analytically sufficient condition
of the vacuum stability of 2HDM with CP violation. However, there are not a
simple analytical expression in the vacuum stability conditions even for 2HDM
until now. In this paper, we will try our best to give a argument technique to
solve this problem, and provide a simple analytical expression (Theorem 1) of
the vacuum stability for 2HDM with CP conservation.

It is well-known that for the 2HDM with explicit CP conservation, all
couplings of the Higgs potential are real [10, 13, 1, 3]. The scalar potential
of such a 2HDM with Higgs doublets $\Phi_{1}$ and $\Phi_{2}$ is

$\displaystyle V_{H}(\Phi_{1},\Phi_{2})=$ $\displaystyle
V_{2}(\Phi_{1},\Phi_{2})+V_{4}(\Phi_{1},\Phi_{2})$ (1) $\displaystyle
V_{2}(\Phi_{1},\Phi_{2})=$ $\displaystyle
m_{11}^{2}|\Phi_{1}|^{2}+m_{22}^{2}|\Phi_{2}|^{2}-m_{12}^{2}(\Phi_{1}^{*}\Phi_{2}+\Phi_{2}^{*}\Phi_{1})$
$\displaystyle V_{4}(\Phi_{1},\Phi_{2})=$
$\displaystyle\Lambda_{1}|\Phi_{1}|^{4}+\Lambda_{2}|\Phi_{2}|^{4}+\Lambda_{3}|\Phi_{1}|^{2}|\Phi_{2}|^{2}$
$\displaystyle+\Lambda_{4}(\Phi_{1}^{*}\Phi_{2})(\Phi_{2}^{*}\Phi_{1})+\frac{\Lambda_{5}}{2}[(\Phi_{1}^{*}\Phi_{2})^{2}+(\Phi_{2}^{*}\Phi_{1})^{2}]$
$\displaystyle+\Lambda_{6}|\Phi_{1}|^{2}(\Phi_{1}^{*}\Phi_{2}+\Phi_{2}^{*}\Phi_{1})$
$\displaystyle+\Lambda_{7}|\Phi_{2}|^{2}(\Phi_{1}^{*}\Phi_{2}+\Phi_{2}^{*}\Phi_{1}),$

where $\Phi^{*}$ is Hermitian conjugate of $\Phi$. The stability (the BFB) of
the scalar potential in the 2HDM is considered only the non-negativity of the
quartic part $V_{4}$ [10], i.e., $V_{4}(\Phi_{1},\Phi_{2})\geq 0.$

In this paper, with the help of the corresponding theory and methods of higher
order tensors, we mainly present the sufficient and necessary conditions of
the vacuum stability for the 2HDM potential with explicit CP conservation.
That is, our main result is the following:

######  Theorem 1.

Let $\Lambda_{1}>0$, $\Lambda_{2}>0$. Then $V_{4}(\Phi_{1},\Phi_{2})\geq 0$ if
and only if

$\displaystyle(1)$
$\displaystyle\Lambda_{6}=\Lambda_{7}=0,\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq
0,\Lambda_{3}+\Lambda_{4}-|\Lambda_{5}|+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0;$
$\displaystyle(2)$ $\displaystyle\Delta\geq
0,\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0,$
$\displaystyle|\Lambda_{6}\sqrt{\Lambda_{2}}-\Lambda_{7}\sqrt{\Lambda_{1}}|\leq
2\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})+2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}},$
$\displaystyle\mbox{(i)
}-2\sqrt{\Lambda_{1}\Lambda_{2}}\leq\Lambda_{3}+\Lambda_{4}+\Lambda_{5}\leq
6\sqrt{\Lambda_{1}\Lambda_{2}},$ $\displaystyle\mbox{(ii)
}\Lambda_{3}+\Lambda_{4}+\Lambda_{5}>6\sqrt{\Lambda_{1}\Lambda_{2}}\mbox{ and
}$
$\displaystyle|\Lambda_{6}\sqrt{\Lambda_{2}}+\Lambda_{7}\sqrt{\Lambda_{1}}|\leq
2\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})-2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}},$

where
$\Delta=4(12\Lambda_{1}\Lambda_{2}-12\Lambda_{6}\Lambda_{7}+(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})^{2})^{3}-(72\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})+36\Lambda_{6}\Lambda_{7}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})-2(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})^{3}-108\Lambda_{1}\Lambda_{7}^{2}-108\Lambda_{6}^{2}\Lambda_{2})^{2}$.

##  2 2HDM potential and Real tensors

In order to show our main result, we need turn the polynomal
$V_{4}(\Phi_{1},\Phi_{2})$ about two complex variable into a 4th order
symmetric real tensor, and then, use some conclusions of tensors to prove our
conclusion.

Let $\phi_{k}$=$|\Phi_{k}|$, the modulus of $\Phi_{k}$ for $k=1,2.$ Then

$\Phi_{1}^{*}\Phi_{2}=\phi_{1}\phi_{2}\rho e^{i\theta}\mbox{ and
}\Phi_{2}^{*}\Phi_{1}=\phi_{1}\phi_{2}\rho e^{-i\theta},$

here $i^{2}=-1$ and $\rho\in[0,1]$ is the orbit space parameter [7, 11]. So,
we have

$\displaystyle V_{4}(\Phi_{1},\Phi_{2})=$
$\displaystyle\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+\Lambda_{3}\phi_{1}^{2}\phi_{2}^{2}+\Lambda_{4}\rho^{2}\phi_{1}^{2}\phi_{2}^{2}$
$\displaystyle+\frac{\Lambda_{5}}{2}(\phi_{1}^{2}\phi_{2}^{2}\rho^{2}e^{2i\theta}+\phi_{1}^{2}\phi_{2}^{2}\rho^{2}e^{-2i\theta})$
$\displaystyle+\Lambda_{6}\phi_{1}^{2}(\phi_{1}\phi_{2}\rho
e^{i\theta}+\phi_{1}\phi_{2}\rho e^{-i\theta})$
$\displaystyle+\Lambda_{7}\phi_{2}^{2}(\phi_{1}\phi_{2}\rho
e^{i\theta}+\phi_{1}\phi_{2}\rho e^{-i\theta})$ $\displaystyle=$
$\displaystyle\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+\Lambda_{3}\phi_{1}^{2}\phi_{2}^{2}+\Lambda_{4}\rho^{2}\phi_{1}^{2}\phi_{2}^{2}$
$\displaystyle+\Lambda_{5}\rho^{2}\phi_{1}^{2}\phi_{2}^{2}\cos
2\theta+2\Lambda_{6}\phi_{1}^{3}\phi_{2}\rho\cos\theta+2\Lambda_{7}\phi_{2}^{3}\phi_{1}\rho\cos\theta$
$\displaystyle=$
$\displaystyle\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+(\Lambda_{3}+\Lambda_{4}\rho^{2}-\Lambda_{5}\rho^{2})\phi_{1}^{2}\phi_{2}^{2}$
$\displaystyle+2\Lambda_{5}\rho^{2}\phi_{1}^{2}\phi_{2}^{2}\cos^{2}\theta+2\Lambda_{6}\rho\phi_{1}^{3}\phi_{2}\cos\theta+2\Lambda_{7}\rho\phi_{2}^{3}\phi_{1}\cos\theta.$

Let $x=\cos\theta$. Then $x\in[-1,1]$ and

$\displaystyle V_{4}(\Phi_{1},\Phi_{2})=$
$\displaystyle\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+(\Lambda_{3}+\Lambda_{4}\rho^{2}-\Lambda_{5}\rho^{2})\phi_{1}^{2}\phi_{2}^{2}$
(2)
$\displaystyle+2\Lambda_{5}\rho^{2}\phi_{1}^{2}\phi_{2}^{2}x^{2}+2\rho(\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2})\phi_{1}\phi_{2}x.$

This defines a 4th-order 2-dimensional symmetric tensor
$\mathcal{A}(\rho,x)=(a_{ijkl})$ with a parameter $\rho$ and $x$:

$\displaystyle a_{1111}=$ $\displaystyle\Lambda_{1},a_{2222}=\Lambda_{2},$ (3)
$\displaystyle a_{1122}=$
$\displaystyle\frac{1}{6}[\Lambda_{3}+\Lambda_{4}\rho^{2}+\Lambda_{5}\rho^{2}(2x^{2}-1)],$
$\displaystyle a_{1112}=$ $\displaystyle\frac{1}{2}\Lambda_{6}\rho x,\
a_{1222}=\frac{1}{2}\Lambda_{7}\rho x.$

So the vacuum stability (or the BFB) of the system $V_{H}(\Phi_{1},\Phi_{2})$
may turn into the co-positivity of a 4th-order tensor $\mathcal{A}(\rho,x)$.

The positive definiteness and the co-positivity of a 4th order symmetric
tensor are applied to deal with the vacuum stability conditions of the
particle physics model in Ref. [7]. Recently, In Refs. [19, 16, 17, 18, 15],
the distinctly sufficient conditions were given for the co-positivity of 4th
order 3-dimensional symmetric tensors, which may be used to the vacuum
stability conditions of scalar potential of the particle physics model.

##  3 Co-positivity of Matrices and Tensors

The co-positivity of a matrix $M=(m_{ij})$ is used to verify the vacuum
stability of the particle physics model in Refs. [7, 8]. A symmetric matrix
$M=(m_{ij})$ is co-positive if the quadratic form $x^{\top}Mx\geq 0$ for all
non-negative vectors $x\in\mathbb{R}^{n}$. The co-positivity of a $2\times 2$
symmetric matrix $M=(m_{ij})$ was showed in Ref. [20, Lemma 2.1], also see
Hadeler [21, Theorem 2] and Nadler [22, Lemma 1] for more details. A $2\times
2$ symmetric matrix $M=(m_{ij})$ is co-positive if and only if

$m_{11}\geq 0,m_{22}\geq 0\mbox{ and }m_{12}+\sqrt{m_{11}m_{22}}\geq 0.$

The co-positivity of a symmetric tensor is tried to test the vacuum stability
of the particle physical model in Ref. [7]. A $m$th-order $n$-dimensional
symmetric tensor $T=(t_{i_{1}\cdots
i_{m}})(i_{j}=1,2,\ldots,n,j=1,2,\ldots,m)$ is co-positive [24, 25, 26, 27,
28] if the quartic form $T{\bf x}^{m}\geq 0$ for all non-negative vectors
${\bf x}\in\mathbb{R}^{n}$, where ${\bf x}=(x_{1},x_{2},\cdots x_{n})^{\top}$
and

$T{\bf x}^{m}={\bf x}^{\top}(T{\bf x}^{m-1})=\sum_{i_{1}\cdots
i_{m}=1}^{n}t_{i_{1}\cdots i_{m}}x_{i_{1}}\cdots x_{i_{m}},$

$T{\bf x}^{m-1}=(y_{1},y_{2},\cdots,y_{n})^{\top}$ is a vector with its
entiries

$y_{k}=(T{\bf x}^{m-1})_{k}=\sum_{i_{2}\cdots i_{m}=1}^{n}t_{ki_{2}\cdots
i_{m}}x_{i_{2}}\cdots x_{i_{m}}.$

Let $f(x,y)$ be a quartic homogeneous real polynomial about two variables
$x,y$,

$f(x,y)=a_{0}x^{4}+a_{1}x^{3}y+a_{2}x^{2}y^{2}+a_{3}xy^{3}+a_{4}y^{4}.$ (4)

Then it gives a 4th-order 2-dimensional symmetric tensor $T=(t_{ijkl})$ with
its entires,

$t_{1111}=a_{0},$ $t_{2222}=a_{4},$ $t_{1122}=\displaystyle\frac{1}{6}a_{2},$
$t_{1112}=\displaystyle\frac{1}{4}a_{1},$
$t_{1222}=\displaystyle\frac{1}{4}a_{3}$,

Assume that $a_{0}>0$ and $a_{4}>0$. In Ref. [16], the co-positivity of the
above tensor $T$ was proved (also see Refs. [30, 29]).

######  Lemma 1.

[16, Lemma 3.1] Let $a_{0}>0$ and $a_{4}>0$. Then $f(x,y)\geq 0$ for all
$x\geq 0,\ y\geq 0$ if and only if

(1) $\Delta\leq 0$ and $a_{1}\sqrt{a_{4}}+a_{3}\sqrt{a_{0}}>0;$ or

(2) $a_{1}\geq 0$, $a_{3}\geq 0$ and $2\sqrt{a_{0}a_{4}}+a_{2}\geq 0;$ or

(3) $\Delta\geq 0$, $|a_{1}\sqrt{a_{4}}-a_{3}\sqrt{a_{0}}|\leq
4\sqrt{a_{0}a_{2}a_{4}+2a_{0}a_{4}\sqrt{a_{0}a_{4}}}$ and

(i) $-2\sqrt{a_{0}a_{4}}\leq a_{2}\leq 6\sqrt{a_{0}a_{4}}$,

(ii) $a_{2}>6\sqrt{a_{0}a_{4}}$ and

$a_{1}\sqrt{a_{4}}+a_{3}\sqrt{a_{0}}\geq-4\sqrt{a_{0}a_{2}a_{4}-2a_{0}a_{4}\sqrt{a_{0}a_{4}}}$,

where
$\Delta=4(12a_{0}a_{4}-3a_{1}a_{3}+a_{2}^{2})^{3}-(72a_{0}a_{2}a_{4}+9a_{1}a_{2}a_{3}-2a_{2}^{3}-27a_{0}a_{3}^{2}-27a_{1}^{2}a_{4})^{2}$.

##  4 Vacuum stability conditions

In this section, we mainly give the vacuum stability conditions of the 2HDM
with explicit CP conservation. That is, how to logically establish our main
result, Theorem 1.

The quartic part (2) of such a CP conserving two-Higgs-doublet potential may
be rewritten as follow

$\displaystyle V_{4}(\Phi_{1},\Phi_{2})=$ $\displaystyle
2\Lambda_{5}\rho^{2}\phi_{1}^{2}\phi_{2}^{2}x^{2}+2\rho(\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2})\phi_{1}\phi_{2}x$
$\displaystyle+\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+(\Lambda_{3}+\Lambda_{4}\rho^{2}-\Lambda_{5}\rho^{2})\phi_{1}^{2}\phi_{2}^{2}.$

If $\phi_{1}>0,\ \phi_{2}=0$ (or $\phi_{1}=0,\ \phi_{2}>0$), then
$V_{H}(\Phi_{1},\Phi_{2})=\Lambda_{1}\phi_{1}^{4}$ (or
$\Lambda_{2}\phi_{2}^{4}).$ Without loss of generality, we may assume
$\Lambda_{1}>0$, $\Lambda_{2}>0$, $\phi_{1}>0,\ \phi_{2}>0$ in the sequel. Let

$A=2\Lambda_{5}\rho^{2}\phi_{1}^{2}\phi_{2}^{2},\
B=2\rho(\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2})\phi_{1}\phi_{2},$ (5)
$C=\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+[\Lambda_{3}+(\Lambda_{4}-\Lambda_{5})\rho^{2}]\phi_{1}^{2}\phi_{2}^{2}.$
(6)

Then, $V_{4}(\Phi_{1},\Phi_{2})$ may be seen as a quadratic function $f(x)$
about a variable $x\in[-1,1]$,

$f(x)=V_{H}(\Phi_{1},\Phi_{2})=Ax^{2}+Bx+C.$ (7)

So, if $A>0$, then as you can see from the image below, the function $f(x)$
attains its minimum $\frac{4AC-B^{2}}{4A}$ at a point
$x=-\frac{B}{2A}\in[-1,1]$ or attains its minimum $A+C\pm B$ at a point $x=\pm
1$ ($-\frac{B}{2A}\notin[-1,1]$). At that time, the graph-like of $f(x)$ is as
shown below:

$-2$$-1$$1$$2$$1$$2$$x$$y$$2A>B>-2A$$-\frac{B}{2A}<-1$$-\frac{B}{2A}>1$

Figure 1: Graph of $f(x)$ ($A>0$)

If $A\leq 0$, then $f(x)$ attains its minimum $A-B+C$ or $A+B+C$ at a point
$x=-1$ ($B\geq 0$) or $x=1$ ($B<0$). At that time, the graph-like of $f(x)$ is
as shown below:

$-2$$-1$$1$$2$$0.5$$1$$1.5$$2$$x$$y$$B<0$$B>0$

Figure 2: Graph of $f(x)$ ($A<0$)

So the following conclusions is easy to be obtained. For detail proof, see
Appendix.

######  Proposition 2.

$V_{H}(\Phi_{1},\Phi_{2})\geq 0$ if and only if

$\begin{cases}4AC-B^{2}\geq 0,-2A\leq B\leq 2A,\\\ A-B+C\geq 0,\\\ A+B+C\geq
0.\end{cases}$

######  Proposition 3.

$4AC-B^{2}\geq 0$ and $-2A\leq B\leq 2A$ for all $\phi_{1}\geq 0,\
\phi_{2}\geq 0$ if and only if $\Lambda_{6}=0,\Lambda_{7}=0,\Lambda_{5}\geq
0,$

$\Lambda_{3}+\Lambda_{4}-\Lambda_{5}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq
0,\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0.$

It follows from the equations (5) and (6) that

$\displaystyle A\pm B+C=$ $\displaystyle
2\Lambda_{5}\rho^{2}\phi_{1}^{2}\phi_{2}^{2}\pm
2\rho(\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2})\phi_{1}\phi_{2}$
$\displaystyle+\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+[\Lambda_{3}+(\Lambda_{4}-\Lambda_{5})\rho^{2}]\phi_{1}^{2}\phi_{2}^{2}$
$\displaystyle=$
$\displaystyle(\Lambda_{4}+\Lambda_{5})\phi_{1}^{2}\phi_{2}^{2}\rho^{2}\pm
2(\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2})\phi_{1}\phi_{2}\rho$
$\displaystyle+\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+\Lambda_{3}\phi_{1}^{2}\phi_{2}^{2}$
$\displaystyle=$ $\displaystyle a\rho^{2}\pm b\rho+c,$

where

$a=(\Lambda_{4}+\Lambda_{5})\phi_{1}^{2}\phi_{2}^{2},\
b=2(\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2})\phi_{1}\phi_{2},$ (8)
$c=\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+\Lambda_{3}\phi_{1}^{2}\phi_{2}^{2}.$
(9)

For $\rho\in[0,1],$ Let

$s(\rho)=a\rho^{2}+b\rho+c\mbox{ and }t(\rho)=a\rho^{2}-b\rho+c.$ (10)

So, if $a>0$, then the function $s(\rho)$ reaches its minimum
$\frac{4ac-b^{2}}{4a}$ at a point $\rho=-\frac{b}{2a}\in[0,1]$, or attains its
minimum $c$ (or $a+c+b$) at a point $\rho=0$ (or $1$)
($-\frac{b}{2a}\notin[0,1]$). At that moment, the graph-like of quadratic
function $s(\rho)$ is illustrated by the following image:

$-2$$-1$$1$$2$$1$$2$$3$$x$$y$$0>b>-2a$$-\frac{b}{2a}<0$$-\frac{b}{2a}>1$

Figure 3: Graph of $s(\rho)$ ($a>0$)

If $a\leq 0$, then $s(\rho)$ attains its minimum $c$ or $a+b+c$ at a point
$\rho=0$ ($-\frac{b}{2a}\geq\frac{1}{2}$) or $\rho=1$
($-\frac{b}{2a}\leq\frac{1}{2}$); if $a=0$, then $s(\rho)$ attains its minimum
$c$ or $b+c$ at a point $\rho=0$ ($b\geq 0$) or $\rho=1$ ($b\leq 0$). The
graph-like of $s(\rho)$ is as shown below:

$-2$$-1$$\frac{1}{2}$$1$$2$$0.5$$1$$1.5$$2$$x$$y$$-\frac{b}{2a}>\frac{1}{2}$$-\frac{b}{2a}<\frac{1}{2}$$x=\frac{1}{2}$

Figure 4: Graph of $s(\rho)$ ($a<0$)

Similarly, if $a>0$, then the function $t(\rho)$ reaches its minimum
$\frac{4ac-b^{2}}{4a}$ at a point $\frac{b}{2a}\in[0,1]$, or attains its
minimum $c$ (or $a+c+b$) at a point $\rho=0$ (or $1$)
($\frac{b}{2a}\notin[0,1]$). The graph-like of $t(\rho)$ is the image below:

$-2$$-1$$1$$2$$1$$2$$3$$x$$y$$0<b<2a$$\frac{b}{2a}<0$$\frac{b}{2a}>1$

Figure 5: Graph of $t(\rho)$ ($a>0$)

if $a<0$, then $t(\rho)$ attains its minimum $c$ or $a-b+c$ at a point
$\rho=0$ ($\frac{b}{2a}\geq\frac{1}{2}$) or $\rho=1$
($\frac{b}{2a}\leq\frac{1}{2}$); if $a=0$, then $t(\rho)$ attains its minimum
$c$ or $-b+c$ at a point $\rho=0$ ($b\leq 0$) or $\rho=1$ ($b\geq 0$). The
graph-like of $t(\rho)$ is the following:

$-1$$\frac{1}{2}$$1$$2$$0.5$$1$$1.5$$2$$x$$y$$\frac{b}{2a}>\frac{1}{2}$$\frac{b}{2a}<\frac{1}{2}$$x=\frac{1}{2}$

Figure 6: Graph of $t(\rho)$ ($a<0$)

So the following conclusions is easy to be obtained. For detail proof, see
Appendix.

######  Proposition 4.

(1) $A+B+C\geq 0$ if and only if

$\begin{cases}\frac{4ac-b^{2}}{4a}\geq 0,-2a\leq b\leq 0;\\\ c\geq 0;\\\
a+b+c\geq 0.\end{cases}$

(2) $A-B+C\geq 0$ if and only if

$\begin{cases}\frac{4ac-b^{2}}{4a}\geq 0,0\leq b\leq 2a;\\\ c\geq 0;\\\
a-b+c\geq 0.\end{cases}$

######  Proposition 5.

(1) $c\geq 0$ if and only if

$\Lambda_{1}\geq 0,\Lambda_{2}\geq
0,\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0.$

(2) $\dfrac{4ac-b^{2}}{4a}\geq 0$ and $-2a\leq b\leq 0$ if and only if
$\Lambda_{4}+\Lambda_{5}\geq 0$,

$\Lambda_{6}=0,\Lambda_{7}=0,\Lambda_{1}\geq 0,\Lambda_{2}\geq
0,\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0.$

(3) $\dfrac{4ac-b^{2}}{4a}\geq 0$ and $2a\geq b\geq 0$ if and only if
$\Lambda_{4}+\Lambda_{5}\geq 0$,

$\Lambda_{6}=0,\Lambda_{7}=0,\Lambda_{1}\geq 0,\Lambda_{2}\geq
0,\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0.$

Next we show $a\pm b+c\geq 0$ for all $\phi_{1}\geq 0,\ \phi_{2}\geq 0$.

######  Proposition 6.

$a-b+c\geq 0$ for all $\phi_{1}\geq 0,\ \phi_{2}\geq 0$ if and only if

(1) $\Delta\leq 0$,
$\Lambda_{6}\sqrt{\Lambda_{2}}+\Lambda_{7}\sqrt{\Lambda_{1}}<0$;

(2) $\Lambda_{6}\leq 0$, $\Lambda_{7}\leq 0$,
$\Lambda_{3}+\Lambda_{4}+\Lambda_{5}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0$;

(3) $\Delta\geq 0$,
$|\Lambda_{6}\sqrt{\Lambda_{2}}-\Lambda_{7}\sqrt{\Lambda_{1}}|\leq
2\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})+2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}}$
and

(i)
$-2\sqrt{\Lambda_{1}\Lambda_{2}}\leq\Lambda_{3}+\Lambda_{4}+\Lambda_{5}\leq
6\sqrt{\Lambda_{1}\Lambda_{2}}$,

(ii) $\Lambda_{3}+\Lambda_{4}+\Lambda_{5}>6\sqrt{\Lambda_{1}\Lambda_{2}}$ and

$\Lambda_{6}\sqrt{\Lambda_{2}}+\Lambda_{7}\sqrt{\Lambda_{1}}\leq
2\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})-2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}}$,

where
$\Delta=4(12\Lambda_{1}\Lambda_{2}-12\Lambda_{6}\Lambda_{7}+(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})^{2})^{3}-(72\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})+36\Lambda_{6}\Lambda_{7}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})-2(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})^{3}-108\Lambda_{1}\Lambda_{7}^{2}-108\Lambda_{6}^{2}\Lambda_{2})^{2}$.

###### Proof.

From the equations (8) and (9), it follows that

$\displaystyle a-b+c=$
$\displaystyle\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})\phi_{1}^{2}\phi_{2}^{2}$
$\displaystyle-2\Lambda_{6}\phi_{1}^{3}\phi_{2}-2\Lambda_{7}\phi_{2}^{3}\phi_{1}$
$\displaystyle=$ $\displaystyle
f(\phi_{1},\phi_{2})=a_{0}\phi_{1}^{4}+a_{1}\phi_{1}^{3}\phi_{2}+a_{2}\phi_{1}^{2}\phi_{2}^{2}+a_{3}\phi_{1}\phi_{2}^{3}+a_{4}\phi_{2}^{4},$

where $a_{0}=\Lambda_{1}$, $a_{1}=-2\Lambda_{6}$,
$a_{2}=\Lambda_{3}+\Lambda_{4}+\Lambda_{5}$, $a_{3}=-2\Lambda_{7}$,
$a_{4}=\Lambda_{2}$. Then an application of the co-positivity of quartic form
(4) (Lemma 1) yields that the sufficient and necessary conditions of the
inequality $a-b+c\geq 0$ is

(1) $\Delta\leq 0$,
$-2\Lambda_{6}\sqrt{\Lambda_{2}}-2\Lambda_{7}\sqrt{\Lambda_{2}}>0$;

(2) $-2\Lambda_{6}\geq 0$, $-2\Lambda_{7}\geq 0$,
$\Lambda_{3}+\Lambda_{4}+\Lambda_{5}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0$;

(3) $\Delta\geq 0$,
$|2\Lambda_{6}\sqrt{\Lambda_{2}}-2\Lambda_{7}\sqrt{\Lambda_{1}}|\leq
4\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})+2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}}$
and

(i)
$-2\sqrt{\Lambda_{1}\Lambda_{2}}\leq\Lambda_{3}+\Lambda_{4}+\Lambda_{5}\leq
6\sqrt{\Lambda_{1}\Lambda_{2}}$,

(ii) $\Lambda_{3}+\Lambda_{4}+\Lambda_{5}>6\sqrt{\Lambda_{1}\Lambda_{2}}$ and
$-2\Lambda_{6}\sqrt{\Lambda_{2}}-2\Lambda_{7}\sqrt{\Lambda_{1}}\geq-4\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})-2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}}$.

The required conclusions follow. ∎

Similarly, the following conclusion is easy to be showed.

######  Proposition 7.

$a+b+c\geq 0$ for all $\phi_{1}\geq 0,\ \phi_{2}\geq 0$ if and only if

(1) $\Delta\leq 0$,
$\Lambda_{6}\sqrt{\Lambda_{2}}+\Lambda_{7}\sqrt{\Lambda_{1}}>0$;

(2) $\Lambda_{6}\geq 0$, $\Lambda_{7}\geq 0$,
$\Lambda_{3}+\Lambda_{4}+\Lambda_{5}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0$;

(3) $\Delta\geq 0$,
$|\Lambda_{6}\sqrt{\Lambda_{2}}-\Lambda_{7}\sqrt{\Lambda_{1}}|\leq
2\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})+2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}}$
and

(i)
$-2\sqrt{\Lambda_{1}\Lambda_{2}}\leq\Lambda_{3}+\Lambda_{4}+\Lambda_{5}\leq
6\sqrt{\Lambda_{1}\Lambda_{2}}$,

(ii) $\Lambda_{3}+\Lambda_{4}+\Lambda_{5}>6\sqrt{\Lambda_{1}\Lambda_{2}}$ and

$\Lambda_{6}\sqrt{\Lambda_{2}}+\Lambda_{7}\sqrt{\Lambda_{1}}\geq-2\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})-2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}}$.

Combining Propositions 4 (1) with Propositions 5 and 7, the following results
are easy to obtain.

######  Proposition 8.

$A+B+C\geq 0$ for all $\phi_{1}\geq 0,\ \phi_{2}\geq 0$ if and only if
$\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0$ and

$\displaystyle(1)$
$\displaystyle\Lambda_{6}=\Lambda_{7}=0,\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq
0,\Lambda_{4}+\Lambda_{5}\geq 0;$ $\displaystyle(2)$ $\displaystyle\Delta\leq
0,\Lambda_{6}\sqrt{\Lambda_{2}}+\Lambda_{7}\sqrt{\Lambda_{1}}>0;$
$\displaystyle(3)$ $\displaystyle\Lambda_{6}\geq 0,\Lambda_{7}\geq
0,\Lambda_{3}+\Lambda_{4}+\Lambda_{5}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0;$
$\displaystyle(4)$ $\displaystyle\Delta\geq 0,$
$\displaystyle|\Lambda_{6}\sqrt{\Lambda_{2}}-\Lambda_{7}\sqrt{\Lambda_{1}}|\leq
2\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})+2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}},$
$\displaystyle\mbox{(i)
}-2\sqrt{\Lambda_{1}\Lambda_{2}}\leq\Lambda_{3}+\Lambda_{4}+\Lambda_{5}\leq
6\sqrt{\Lambda_{1}\Lambda_{2}},$ $\displaystyle\mbox{(ii)
}\Lambda_{3}+\Lambda_{4}+\Lambda_{5}>6\sqrt{\Lambda_{1}\Lambda_{2}}\mbox{ and
}$
$\displaystyle\Lambda_{6}\sqrt{\Lambda_{2}}+\Lambda_{7}\sqrt{\Lambda_{1}}\geq-2\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})-2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}}.$

Simultaneously applying Proposition 5 and Proposition 6 to Propositions 4 (2),
the following results are easy to obtain.

######  Proposition 9.

$A-B+C\geq 0$ for all $\phi_{1}\geq 0,\ \phi_{2}\geq 0$ if and only if
$\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0$ and

$\displaystyle(1)$
$\displaystyle\Lambda_{6}=\Lambda_{7}=0,\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq
0,\Lambda_{4}+\Lambda_{5}\geq 0;$ $\displaystyle(2)$ $\displaystyle\Delta\leq
0,\Lambda_{6}\sqrt{\Lambda_{2}}+\Lambda_{7}\sqrt{\Lambda_{1}}<0;$
$\displaystyle(3)$ $\displaystyle\Lambda_{6}\leq 0,\Lambda_{7}\leq
0,\Lambda_{3}+\Lambda_{4}+\Lambda_{5}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0;$
$\displaystyle(4)$ $\displaystyle\Delta\geq 0,$
$\displaystyle|\Lambda_{6}\sqrt{\Lambda_{2}}-\Lambda_{7}\sqrt{\Lambda_{1}}|\leq
2\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})+2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}},$
$\displaystyle\mbox{(i)
}-2\sqrt{\Lambda_{1}\Lambda_{2}}\leq\Lambda_{3}+\Lambda_{4}+\Lambda_{5}\leq
6\sqrt{\Lambda_{1}\Lambda_{2}},$ $\displaystyle\mbox{(ii)
}\Lambda_{3}+\Lambda_{4}+\Lambda_{5}>6\sqrt{\Lambda_{1}\Lambda_{2}}\mbox{ and
}$
$\displaystyle\Lambda_{6}\sqrt{\Lambda_{2}}+\Lambda_{7}\sqrt{\Lambda_{1}}\leq
2\sqrt{\Lambda_{1}\Lambda_{2}(\Lambda_{3}+\Lambda_{4}+\Lambda_{5})-2\Lambda_{1}\Lambda_{2}\sqrt{\Lambda_{1}\Lambda_{2}}}.$

By combining Propositions 2 with Propositions 3, 8 and 9, we easily showed our
main result, Theorem 1.

###### The proof of Theorem 1.

It follows from Propositions 2 and 3 that $V_{H}(\Phi_{1},\Phi_{2})\geq 0$ if
and only if

$\displaystyle\Lambda_{6}=\Lambda_{7}=0,\Lambda_{5}\geq 0,$ $\displaystyle\
\Lambda_{3}+\Lambda_{4}-\Lambda_{5}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0,$
(11) $\displaystyle\ \Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0;$
$\displaystyle A-B+C\geq 0;$ $\displaystyle A+B+C\geq 0.$

Propositions 8 (3) and 9 (3) mean that

$\Lambda_{6}=\Lambda_{7}=0,\Lambda_{3}+\Lambda_{4}+\Lambda_{5}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq
0;$

Propositions 8 (1) and 9 (1) mean that

$\Lambda_{6}=\Lambda_{7}=0,\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq
0,\Lambda_{3}+\Lambda_{4}+\Lambda_{5}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0,$

The above several inequalities together is equivalent to

$\Lambda_{6}=\Lambda_{7}=0,\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq
0,\Lambda_{3}+\Lambda_{4}-|\Lambda_{5}|+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0.$

This obtain (1) of Theorem 1.

The two inqualities $A-B+C\geq 0$ and $A+B+C\geq 0$ imply that there is
contradiction between Proposition 8 (2) and Proposition 9 (2), and so, it can
not hold.

By Propositions 8 (4) and 9 (4), (2) of Theorem 1 follows easily. ∎

##  5 Conclusions

In summary, we prove the analytical sufficient and necessary conditions of the
vacuum stability conditions of 2HDM potential with explicit CP conservation.
At the same time, for a 4th-order 2-dimensional symmetric tensor
$\mathcal{A}(\rho,x)=(a_{ijkl})$ (3) with two parameters $\rho\in[0,1]$ and
$x\in[-1,1]$, the co-positivity is proved. Which first gives a argument
methods to study the tensor with two parameters.

##  6 Some remarks

When $\lambda_{6}=\lambda_{7}=0$, Theorem 1 coincides with the corresponding
conclusion of Refs. [3, 11, 9] and other references not cited here.

Recently, Bahl et al. [14] gave a stronger sufficient conditions (Eqs. (43)
and (44)) of the vacuum stability of CP conserving 2HDM potential with the
constraint $\Lambda_{6}\sqrt{\Lambda_{2}}=\Lambda_{7}\sqrt{\Lambda_{1}}$ using
Ulrich and Watson’s result (Eq.(30))[30]. In this paper, we use the optimizing
version (Lemma 1) of Ulrich and Watson’s result to yield the analytic
expression of the vacuum stability. For the more details about Lemma 1 see
Refs. [16, 29].

In terms of the eigenvalues of a $4\times 4$ matrix, Ivanov [6] gave the
sufficient and necessary conditions of the vacuum stability of CP conserving
2HDM potential, but not analytic. Our conclusions is analytic in this paper.

Our work also first applies the successive removal parameter method to study
the properties of a symmetric tensor with multi-parameters, which deserves
further study and perfectness.

## Data Availablity Statement

This manuscript has no associated data or the data will not be deposited.
[Authors’ comment: This is a theoretical study and there are no external data
associated with the manuscript.]

## Appendix

###### The proof of Propositions 3.

From the equation (5), it follows that

$\displaystyle
B-2A=2(\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2}-2\Lambda_{5}\rho\phi_{1}\phi_{2})\rho\phi_{1}\phi_{2},$
$\displaystyle
B+2A=2(\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2}+2\Lambda_{5}\rho\phi_{1}\phi_{2})\rho\phi_{1}\phi_{2},$

and hence,

$\displaystyle B-2A\leq
0\Leftrightarrow-\Lambda_{6}\phi_{1}^{2}-\Lambda_{7}\phi_{2}^{2}+2\Lambda_{5}\rho\phi_{1}\phi_{2}\geq
0,$ $\displaystyle B+2A\geq 0\Leftrightarrow\ \
\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2}+2\Lambda_{5}\rho\phi_{1}\phi_{2}\geq
0.$

Thus, $B-2A\leq 0$ is equivalent to the co-positivity of a matrix $M=(m_{ij})$
with its entries
$m_{11}=-\Lambda_{6},m_{22}=-\Lambda_{7},m_{12}=m_{21}=\Lambda_{5}\rho.$ From
the co-positivity of $2\times 2$ matrix, it follows that

$B-2A\leq 0\Leftrightarrow\Lambda_{6}\leq 0,\Lambda_{7}\leq 0\mbox{ and
}\Lambda_{5}\rho+\sqrt{\Lambda_{6}\Lambda_{7}}\geq 0.$

Similarly, we also have

$B+2A\geq 0\Leftrightarrow\Lambda_{6}\geq 0,\Lambda_{7}\geq 0\mbox{ and
}\Lambda_{5}\rho+\sqrt{\Lambda_{6}\Lambda_{7}}\geq 0.$

Thus, the inequalities $-2A\leq B\leq 2A$ imply that
$\Lambda_{6}=0,\Lambda_{7}=0,\Lambda_{5}\geq 0$, i.e., $B=0,A\geq 0$, and so,
the inequality $4AC-B^{2}\geq 0$ means that

$C=\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+[\Lambda_{3}+(\Lambda_{4}-\Lambda_{5})\rho^{2}]\phi_{1}^{2}\phi_{2}^{2}\geq
0.$

which is equivalent to the co-positivity of $2\times 2$ matrix $M=(m_{ij})$
with its entries,

$m_{11}=\Lambda_{1},\ m_{22}=\Lambda_{2},\
m_{12}=m_{21}=\frac{1}{2}[\Lambda_{3}+(\Lambda_{4}-\Lambda_{5})\rho^{2}].$

That is,
$\Lambda_{3}+(\Lambda_{4}-\Lambda_{5})\rho^{2}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq
0.$ Therefore, this is equivalent to

$\displaystyle\Lambda_{3}+\Lambda_{4}-\Lambda_{5}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq
0,$ $\displaystyle\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0.$

The required conclusion follows. ∎

###### The proof of Propositions 5.

(1) It is obvious that

$c=\Lambda_{1}\phi_{1}^{4}+\Lambda_{2}\phi_{2}^{4}+\Lambda_{3}\phi_{1}^{2}\phi_{2}^{2}\geq
0,$

is equivalent to

$\Lambda_{1}\geq 0,\Lambda_{2}\geq
0,\Lambda_{3}+2\sqrt{\Lambda_{1}\Lambda_{2}}\geq 0.$

(2) It follows from the equations (8) that

$b+2a=2(\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2}+(\Lambda_{4}+\Lambda_{5})\phi_{1}\phi_{2})\phi_{1}\phi_{2},$
$2a-b=2(-\Lambda_{6}\phi_{1}^{2}-\Lambda_{7}\phi_{2}^{2}+(\Lambda_{4}+\Lambda_{5})\phi_{1}\phi_{2})\phi_{1}\phi_{2},$

and hence,

$\displaystyle b+2a\geq 0\Leftrightarrow$
$\displaystyle\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2}+(\Lambda_{4}+\Lambda_{5})\phi_{1}\phi_{2}\geq
0,$ $\displaystyle b\leq 0\Leftrightarrow$
$\displaystyle\Lambda_{6}\phi_{1}^{2}+\Lambda_{7}\phi_{2}^{2}\leq 0.$

So, we have

$\displaystyle b+2a\geq 0\Leftrightarrow$ $\displaystyle\Lambda_{6}\geq
0,\Lambda_{7}\geq
0,(\Lambda_{4}+\Lambda_{5})+2\sqrt{\Lambda_{6}\Lambda_{7}}\geq 0,$
$\displaystyle b\leq 0\Leftrightarrow$ $\displaystyle\Lambda_{6}\leq
0,\Lambda_{7}\leq 0,$

and hence, $\Lambda_{4}+\Lambda_{5}\geq 0$, $\Lambda_{6}=0,\Lambda_{7}=0$,
i.e., $b=0$. Clearly, if $a\neq 0$, then

$\frac{4ac-b^{2}}{4a}\geq 0\Leftrightarrow c\geq 0.$

Similarly, (3) is also showed easily. ∎

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MPP-2022-57

Dynamical Cobordism of a Domain Wall
and its Companion Defect 7-brane

Ralph Blumenhagen, Niccolò Cribiori, Christian Kneißl and Andriana Makridou

_Max-Planck-Institut für Physik (Werner-Heisenberg-Institut),
Föhringer Ring 6, 80805 München, Germany _

###### Abstract

Starting from an already known solution in the literature, we study the
dynamical cobordism induced by the backreaction of a non-supersymmetric,
positive tension domain wall in string theory. This could e.g. be a non-BPS
D8-brane of type I or a $\overline{D8}/O8$ stack of a non-supersymmetric type
IIA orientifold. The singularities which typically appear indicate either an
inherent inconsistency of this background or the required presence of a
suitable defect, as predicted by the cobordism conjecture. We provide evidence
that this end-of-the-world $7$-brane is explicitly described by a new kind of
non-isotropic solution of the dilaton-gravity equations of motion.
Intriguingly, on the formal level this solution turns out to be closely
related to the initial solution for the non-supersymmetric domain wall.

##  1 Introduction

One of the most fundamental swampland conjectures [1, 2, 3] is the absence of
global symmetries in quantum gravity [4, 5]. The notion of symmetry can be
extended to include higher-form global symmetries [6], which should also
vanish. Thus, taking topological global charges into account, led to the
recent proposal that the cobordism group in quantum gravity needs to be
trivial [7]. Starting with a non-trivial cobordism group, the induced global
symmetry can either be gauged or broken. Its breaking requires the
introduction of defects, which have to cancel the cobordism charge. We refer
to [8, 9, 10, 11, 12, 13, 14, 15] for more recent developments related to the
cobordism conjecture.

In the series of papers [16, 17, 18], it has been argued and supported by many
examples that, from the effective supergravity point of view, certain non-
trivial cobordism groups are related to dynamical tadpoles or more general
scalar potentials. This is the case in the presence of running solutions of
the equations of motion that feature end-of-the-world (ETW) branes at finite
space-time distance, where in addition some scalar fields go to infinite
distance in field space. This picture was dubbed Local Dynamical Cobordism,
since it is based on the interplay between the dynamics of the scalars and the
breaking of cobordism symmetries, while it can be given a local description in
large classes of models. It is the purpose of the current work to continue
along these ideas and provide one more non-trivial example. We will give
evidence that breaking a certain global cobordism symmetry requires the
existence of a new object in the effective theory, which can be detected by a
novel kind of non-isotropic solution of the dilaton-gravity equations of
motion.

Concretely, we consider the backreaction of a gauge neutral, non-
supersymmetric 9-dimensional domain wall carrying only a positive tension and
coupling to the dilaton like a D-brane, i.e. with the factor $\exp(-\Phi)$ in
its action. Such a domain wall arises e.g. as the non-BPS D8-brane of type I
string theory or as a local R-R tadpole-free $16\times\overline{D8}+O8^{++}$
stack in a certain non-supersymmetric orientifold, which is the T-dual of the
Sugimoto model [19]. In short, we will address such an object as a neutral
domain wall.

Fortunately, this backreaction problem has already been addressed in 2000 by
Anamaria Font and one of us [20]. Similar to the solution for the backreacted
original Sugimoto model [21], it was shown in [20] that the gravity and
dilaton equations of motion of the T-dual setup do not admit a solution with
nine-dimensional Poincaré symmetry (along the longitudinal directions of the
brane), but induce a spontaneous compactification on a transversal circle
times a non-trivial longitudinal direction. In [20], two explicit solutions
were presented, where for the first the longitudinal direction was non-
compact, whereas for the second it was a finite size interval.

In this work, we revisit these solutions in light of the recent developments
around (dynamical) cobordism. It was already realized in [20] that in such
solutions there appear additional singularities, where e.g. the string
coupling diverges. At that time this was a rather puzzling issue, but in view
of the improved understanding of cobordism symmetries in quantum gravity, it
might just be pointing to the required existence of an ETW defect 7-brane,
curing the singularity (and breaking the symmetry) and making the whole
solution consistent. Let us also mention that the possible presence of
8-branes in the original Sugimoto model has been proposed in [22], although
from a different perspective.

In section 2, we first review the two solutions of [20] for the backreaction
of such a neutral domain wall and show that the so called solution II+ fits
especially well into the recent discussion of Local Dynamical Cobordism [16,
17, 18]. Analyzing the behavior close to the singularity, we find that it is
of logarithmic type, something one would indeed expect from a codimension-two
object. This is strong indication for the presence of ETW 7-branes at the
endpoints of the finite size interval.

As a new development, in section 3 we construct the (local) solution of the
dilaton-gravity equations of motion around the potentially present
ETW-7-brane. Since this object will also just carry tension, we cannot expect
to find a familiar solution, preserving both 8D Poincaré and 2D rotational
symmetry. Given that we do not want to break the 8D Poincaré symmetry (and be
probably forced to introduce more defects of lower dimensionality), we make an
ansatz for a solution breaking the 2D rotational symmetry. It turns out that
the resulting equations of motion do admit a solution that has a striking
similarity to the neutral domain wall we started from. The ETW 7-brane turns
out to have positive tension as well and shows the expected logarithmic
singularities. The coupling to the dilaton is $\exp{(-2\Phi)}$ in string
frame, which does not correspond to any known (to us) codimension-two object
in string theory. It would thus be interesting to confirm the presence of this
candidate new object in string theory with an independent analysis.

##  2 Backreacted Domain Wall

We consider a neutral domain wall configuration in ten dimensions carrying
positive tension $T$ and being located at the position $r=0$ in the
transversal directions. Since it carries no other gauge charge, at leading
order its supergravity action assumes the form

$\begin{split}S={\frac{1}{2\kappa_{10}^{2}}}\int
d^{10}x\sqrt{-G}\left(\mathcal{R}-{\frac{1}{2}}(\partial\Phi)^{2}\right)-T\int
d^{10}x\sqrt{-g}\,e^{{\frac{5}{4}}\Phi}\,\delta(r),\end{split}$ (2.1)

where $G_{MN}$ denotes the metric in ten dimensions and $g_{\mu\nu}$ the
metric on the nine-dimensional worldvolume of the brane. Moreover, one has
$\kappa_{10}^{2}=\ell_{s}^{8}/(4\pi)$, with $l_{s}$ the string length. Note
that this is actually the same action as considered in [20]. The only slight
difference is the presence of a single source instead of two. Indeed, the
second source would sit at $r=R$ and it is here taken into account by
extending periodically the solution beyond $r\in[-\frac{R}{2},\frac{R}{2}]$.

The resulting gravity equation of motion reads

$\begin{split}\mathcal{R}_{MN}-{\frac{1}{2}}G_{MN}\mathcal{R}-{\frac{1}{2}}&\left(\partial_{M}\Phi\partial_{N}\Phi-{\frac{1}{2}}G_{MN}(\partial\Phi)^{2}\right)=\\\
&\hskip
113.81102pt-\lambda\,\delta_{M}^{\mu}\delta_{N}^{\nu}\,g_{\mu\nu}\,\sqrt{\frac{g}{G}}\,e^{{\frac{5}{4}}\Phi}\,\delta(r),\end{split}$
(2.2)

with $\lambda=\kappa_{10}^{2}T$. In addition, there is only the dilaton
equation of motion

$\begin{split}\partial_{M}\left(\sqrt{-G}\,G^{MN}\,\partial_{N}\Phi\right)={\frac{5}{2}}\lambda\sqrt{-g}\,e^{{\frac{5}{4}}\Phi}\,\delta(r)\,.\end{split}$
(2.3)

These are non-linear equations that share some features with the backreaction
of BPS-branes but, since they do not involve any $p$-form gauge field, they
are actually of non-supersymmetric type and in general much harder to solve.

###  2.1 Two solutions breaking 9D Poincaré symmetry

These equations do not admit a solution preserving 9D Poincaré invariance.
However, in [20] a solution was found that preserved 8D Poincaré invariance
featuring a single non-trivial longitudinal direction $y$. The general ansatz
for the metric was

$\begin{split}ds^{2}=e^{2{\cal A}(r,y)}ds_{8}^{2}+e^{2{\cal
B}(r,y)}(dr^{2}+dy^{2}),\end{split}$ (2.4)

with a separated dependence of the warp factors $\mathcal{A}$, $\mathcal{B}$
and the dilaton $\Phi$ on the coordinates $r$ and $y$, i.e.

$\begin{split}&{\cal A}(r,y)=A(r)+U(y)\,,\quad{\cal B}(r,y)=B(r)+V(y)\,,\\\
&\Phi(r,y)=\chi(r)+\psi(y)\,.\end{split}$ (2.5)

For such a separation of variables, by redefining $r$ and $y$ the ansatz (2.4)
is actually the most general one.

The equations of motion lead to five a priori independent equations. The one
related to the variation $\delta G^{\mu\nu}$ is

$\begin{split}\Big{(}7A^{\prime\prime}+28(A^{\prime})^{2}+B^{\prime\prime}+{\textstyle\frac{1}{4}}(\chi^{\prime})^{2}\Big{)}+&\Big{(}7\ddot{U}+28(\dot{U})^{2}+\ddot{V}+{\textstyle\frac{1}{4}}(\dot{\psi})^{2}\Big{)}\\\\[2.84544pt]
&\hskip
85.35826pt=-\lambda\,e^{B+V}\,e^{{\frac{5}{4}}\Phi}\,\delta(r)\,.\end{split}$
(2.6)

The prime denotes the derivative with respect to $r$ and the dot the
derivative with respect to $y$. For the two variations $\delta G^{rr}$ and
$\delta G^{yy}$, we obtain

$\begin{split}&\Big{(}28(A^{\prime})^{2}+8A^{\prime}B^{\prime}-{\textstyle{\frac{1}{4}}}(\chi^{\prime})^{2}\Big{)}+\Big{(}8\ddot{U}+36(\dot{U})^{2}-8\dot{U}\dot{V}+{\textstyle{\frac{1}{4}}}(\dot{\psi})^{2}\Big{)}=0,\\\\[5.69046pt]
&\Big{(}8A^{\prime\prime}+36(A^{\prime})^{2}-8A^{\prime}B^{\prime}+{\textstyle\frac{1}{4}}(\chi^{\prime})^{2}\Big{)}+\Big{(}28(\dot{U})^{2}+8\dot{U}\dot{V}-{\textstyle\frac{1}{4}}(\dot{\psi})^{2}\Big{)}\\\\[2.84544pt]
&\hskip 170.71652pt=-\lambda\,e^{B+V}\,e^{{\frac{5}{4}}\Phi}\,\delta(r)\,,\\\
\end{split}$ (2.7)

and for the off-diagonal $\delta G^{ry}$

$\begin{split}&-8A^{\prime}\dot{U}+8B^{\prime}\dot{U}+8A^{\prime}\dot{V}-{\textstyle\frac{1}{2}}\chi^{\prime}\dot{\psi}=0\,.\end{split}$
(2.8)

Finally, the dilaton equation of motion becomes

$\begin{split}\Big{(}\chi^{\prime\prime}+8A^{\prime}\chi^{\prime}\Big{)}+\Big{(}\ddot{\psi}+8\dot{U}\dot{\psi}\Big{)}={\frac{5}{2}}\lambda\,e^{B+V}\,e^{{\frac{5}{4}}\Phi}\,\delta(r)\,.\end{split}$
(2.9)

Note that taking the sum of the two equations in (LABEL:eomyyrr) gives the
simpler equation

$\begin{split}8\Big{(}A^{\prime\prime}+8(A^{\prime})^{2}\Big{)}+8\Big{(}\ddot{U}+8(\dot{U})^{2}\Big{)}=-\lambda\,e^{B+V}\,e^{{\frac{5}{4}}\Phi}\,\delta(r)\,.\end{split}$
(2.10)

One first solves these equations in the bulk and then implements the
$\delta$-source via a jump of the first derivatives
$A^{\prime},B^{\prime},\chi^{\prime}$ at $r=0$. We proceed completely
analogously to [20], so that we can keep the presentation short and
essentially just describe the two solutions found there.111In fact, in this
new approach we did not find any further relevant solution, either. In
appendix A we present a physically less interesting solution we have
identified.

#### Solution I

From (2.10) one learns that both expressions in the brackets need to be
constant (away from the sources). This forces one of the two functions $A(r)$
or $U(y)$ to be of trigonometric type and the other one in general of
hyperbolic type. It turns out that all equations of motion and boundary
conditions are satisfied by the following functions

$\begin{split}A(r)=B(r)&={\frac{1}{8}}\log\Big{|}\sin\left[{\textstyle
8K(|r|-{\textstyle{\frac{R}{2}}})}\right]\Big{|},\\\\[2.84544pt]
\chi(r)&=-{\frac{3}{2}}\log\Big{|}\tan\left[4K(|r|-{\textstyle{\frac{R}{2}}})\right]\Big{|}+\phi_{0},\\\\[2.84544pt]
U(y)&=-K\,y,\,\qquad\psi(y)=V(y)=0\,,\end{split}$ (2.11)

where the integration constant $\phi_{0}$ is related to the string coupling
constant. Moreover, the parameters appearing have to satisfy

$\begin{split}\cos(4KR)={\frac{3}{5}}\,,\qquad
e^{{\frac{5}{4}}\phi_{0}}=3\left(\frac{5}{2}\right)^{\frac{1}{8}}\frac{K}{\lambda}\sim\frac{1}{\lambda
R}\,.\end{split}$ (2.12)

The second relation nicely reflects that the compactness of the $r$ direction
is a consequence of the backreaction of the neutral domain wall. Indeed,
sending the string coupling to zero, the compact space decompactifies.

Hence, the direction transversal to the neutral domain wall is spontaneously
compactified on a topological $S^{1}$ of a size related to $R$.222Another
possibility would be to consider the $r$-direction to be an interval of finite
proper size proportional to $R$. Since there is no distinguished singular
point in the $y$-direction, in this way one would be led to a pair of ETW
8-branes at $r=-R/2$ and $r=R/2$. However, an 8-brane is not consistent with
the singularities (2.13) of the solution at these points and is also not
expected from the cobordism argument (which suggests an ETW 7-brane).
Moreover, since there are really periodic trigonometric functions appearing in
the solution (2.11), we consider this as a clear indication that the
$r$-direction is circular. However, as opposed to the upcoming solution II,
the proper length of the $y$-direction is infinite so that there are no walls
of nothing in this solution (equivalently, walls of nothing are infinitely far
away and thus cannot be captured by the given effective description). In other
words, we necessarily ask for the presence of a coordinate of finite proper
length in order to give an interpretation in terms of local dynamical
cobordism. Moreover, the solution as it stands is inconsistent in the sense
that at $r=\pm R/2$ there appear logarithmic singularities in the warp factors
and the dilaton

$\begin{split}A(r)=B(r)\sim{\frac{1}{8}}\log\rho\,,\qquad\chi(r)\sim-\frac{3}{2}\log\rho,\end{split}$
(2.13)

with $\rho=|r|-R/2$. This leads to curvature singularities, where the string
coupling diverges. The appearance of a singularity at $\rho=0$ was already
observed in [20], but in those early days it could not be clarified what it
was indicating. We will come back to this issue after discussing Solution II.

#### Solution II

Let us now review the more involved Solution II to the equations of motion.
The equations in the bulk still admit three free parameters, $\alpha,K,R$,
which are further restricted by implementing the 8-brane boundary conditions
at $r=0$. Eventually, it was found that the $r$-dependent solutions satisfying
the proper jump conditions at $r=0$ are

$\begin{split}A(r)&={\frac{1}{8}}\log\Big{|}\sin\left[{\textstyle{8K}}(|r|-{\textstyle{\frac{R}{2}}})\right]\Big{|},\\\\[2.84544pt]
\chi(r)&={\frac{\alpha^{\pm}}{8}}\log\Big{|}\sin\left[{\textstyle{8K}}(|r|-{\textstyle{\frac{R}{2}}})\right]\Big{|}\mp
2\log\Big{|}\tan\left[{\textstyle{4K}}(|r|-{\textstyle{\frac{R}{2}}})\right]\Big{|}+\phi_{0},\\\\[2.84544pt]
B(r)&={\frac{\mu}{8}}\log\Big{|}\sin\left[{\textstyle{8K}}(|r|-{\textstyle{\frac{R}{2}}})\right]\Big{|}\mp{\frac{\alpha^{\pm}}{8}}\log\Big{|}\tan\left[{\textstyle{4K}}(|r|-{\textstyle{\frac{R}{2}}})\right]\Big{|},\end{split}$
(2.14)

with $r\in[-{\frac{R}{2}},{\frac{R}{2}}]$. The parameters appearing above are
defined as

$\begin{split}\mu={\frac{\alpha^{2}}{16}}+{\frac{5\alpha}{4}}+1\,,\qquad{\rm
for}\quad\alpha^{\pm}=-4(5\mp 4\sqrt{2})\,,\end{split}$ (2.15)

where the latter are the positive and the negative root of

$\begin{split}\alpha^{2}+40\alpha-112=0\,.\end{split}$ (2.16)

Moreover, consistency of the boundary conditions implies

$\begin{split}\cos(4KR)=\pm{\frac{16}{\alpha^{\pm}+20}}={\frac{1}{\sqrt{2}}},\end{split}$
(2.17)

with the minimal solution $K=\pi/(16R)$. The integration constant $\phi_{0}$
is related to the string coupling constant and has to satisfy

$\begin{split}e^{{\frac{5}{4}}\phi_{0}}={\frac{\pi}{\lambda
R}}e^{-B-{\frac{5}{4}}(\chi-\phi_{0})}\Big{|}_{r=0}={\frac{\pi}{\lambda
R}}\sqrt{2}\left(\sqrt{2}-1\right)^{2\sqrt{2}}\,.\end{split}$ (2.18)

Again, we have a compact direction parametrized by $R$ which becomes non-
compact as the string coupling is sent to zero. The solutions for the
$y$-dependent functions are a bit simpler and read

$\begin{split}U(y)&={\frac{1}{8}}\log\Big{(}\cosh\left[8K\,y\right]\Big{)}\,,\quad\psi(y)=\alpha
U(y)\,,\quad V(y)=-{\frac{5\alpha}{4}}U(y)\,,\end{split}$ (2.19)

i.e. all three functions are proportional. Notice also that we have chosen one
integration constant such that the solution is symmetric around $y=0$. We will
denote the present two solutions as II±, to indicate that they refer to the
roots $\alpha^{\pm}$ respectively. As we will see shortly, only Solution II+
leads to a finite size of the $y$-direction. For this solution, in figure 1 we
display the three functions depending on $r$, showing their characteristic
features.

\begin{picture}(0.0,0.0)\end{picture}

Figure 1: For Solution II+, the three functions $A(r),\chi(r)$ and $B(r)$ are
shown for $R=1$ in a double period $r\in[-R,R]$.

One can see that by construction they all have a kink at the location $r=0$ of
the neutral domain wall, but exhibit a singularity at $r=\pm R/2$. Introducing
the coordinate close to the singularity, $\rho=r-R/2$, and expanding
$\sin[8K(|r|-{\frac{R}{2}})]\simeq{8K}\rho$, the behavior of these three
functions close to $\rho=0$ is

$\begin{split}A(\rho)\simeq{\frac{1}{8}}\log\rho\,,\quad\chi(\rho)\simeq{\frac{1}{8}}(\alpha^{+}-16)\log\rho\,,\quad
B(\rho)\simeq{\frac{1}{8}}(\mu-\alpha^{+})\log\rho\,.\end{split}$ (2.20)

Since $(\alpha^{+}-16)<0$ while $(\mu-\alpha^{+})>0$, for $\rho\to 0$ the warp
factors $A$ and $B$ go to zero while the string coupling $g_{s}=\exp(\Phi)$
goes to infinity, i.e. we are facing a strong coupling singularity.

###  2.2 The geometry of the solution

In this section, we would like to discuss the stringy geometry of Solution II
in more detail. For this reason, we mainly work in the string frame, while we
will mention some results in the Einstein frame as well.333We recall that the
string frame metric is obtained by multiplying the Einstein frame metric (2.4)
with $\exp((\Phi-\phi_{0})/2)$.

We start by studying the geometry along the $y$-direction. In string frame the
proper length (at fixed $r$) is

$\begin{split}L_{y}=\int_{-\infty}^{\infty}dy\,e^{V(y)+{\frac{1}{4}}\psi(y)}=\int_{-\infty}^{\infty}dy\,\Big{(}\cosh\left[{\textstyle{\frac{\pi}{2R}}}y\right]\Big{)}^{-{\frac{\alpha}{8}}}\,.\end{split}$
(2.21)

By inspection of the integral, one can see that this length can only be finite
for positive $\alpha$. In this respect, Solution II- is thus similar to
Solution I: in both cases the length is infinite. Evaluating the integral for
Solution II+, one finds instead

$\begin{split}L_{y}=\frac{2R}{\sqrt{\pi}}\frac{\Gamma\left(\frac{\alpha^{+}}{16}\right)}{\Gamma\left({\frac{\alpha^{+}}{16}}+{\frac{1}{2}}\right)}\approx
4.7\,R\,.\end{split}$ (2.22)

We note that in the Einstein frame the situation is qualitatively similar:
Solutions I and II- exhibit an infinite proper length $L_{y}$, while Solution
II+ a finite one ($\approx 3.9R$).

Since (2.22) is finite, there are two end-of-the-world walls at finite
distance from one another (and thus within the same effective description). In
view of the cobordism conjecture and by taking the logarithmic singularity at
$|r|=R/2$ into account, it is natural to assume that there will be two
corresponding ETW 7-branes located in the $(r,y)$-plane at the two
distinguished positions

$\begin{split}{\rm ETW}_{1}:\ (r,y)=(R/2,-\infty)\,,\qquad{\rm ETW}_{2}:\
(r,y)=(R/2,+\infty)\,.\end{split}$ (2.23)

In the following, we focus on Solution II+. We can compute the string-frame
proper length in the $r$-direction (at fixed $y$). It is given by

$\begin{split}L_{r}=\int_{-R/2}^{R/2}dr\,e^{B(r)+{\frac{1}{4}}\chi(r)}=2\int_{-R/2}^{0}dr\,\frac{\left(\sin\left[{\textstyle\frac{\pi}{2R}}\left(r+\frac{R}{2}\right)\right]\right)^{\frac{\mu}{8}+\frac{\alpha^{+}}{32}}}{\left(\tan\left[{\textstyle\frac{\pi}{4R}}\left(r+\frac{R}{2}\right)\right]\right)^{\frac{\alpha^{+}}{8}+{\frac{1}{2}}}}\approx
2.1\,R\,\end{split}$ (2.24)

and, despite a singularity of the integrand at $r=\pm R/2$, it comes out
finite. In the Einstein frame it is finite as well $(\approx 0.9R)$. The area
of the compact $(r,y)$ space is then simply (in string frame)

$\begin{split}\text{Area}=\int drdy\,\sqrt{G_{rr}G_{yy}}=L_{r}\,L_{y}\approx
9.9\,R^{2}\,\end{split}$ (2.25)

and once again is a finite quantity.

Actually, the length $L_{r}$ also depends on the coordinate $y$ due to the
$y$-dependent part of the warp factor and similarly $L_{y}$ depends on $r$. We
discuss briefly these behaviors below. First, let us consider how the length
of the circle in the $r$-direction gets warped with the $y$-coordinate

$\begin{split}R(y)=e^{V(y)+{\frac{1}{4}}\psi(y)}\,R=e^{-\alpha^{+}U(y)}\,R\underset{y\to\pm\infty}{\longrightarrow}0\,.\end{split}$
(2.26)

The behavior in the Einstein frame is qualitatively similar,
$R(y)=e^{-\frac{5}{4}\alpha^{+}U(y)}R$. As for $L_{y}$, close to the
singularity at $r=R/2$ it is multiplied with the exponential of

$\begin{split}B(r)+{\frac{1}{4}}\chi(r)\simeq{\frac{1}{8}}\left({\frac{(\alpha^{+})^{2}}{16}}+{\frac{\alpha^{+}}{2}}-3\right)\log\rho\approx-0.16\log\rho,\end{split}$
(2.27)

indicating that the proper length of the interval diverges at $\rho\to 0$.
Notice that, on the contrary, in the Einstein frame the warp factor becomes

$\begin{split}B(r)\simeq{\frac{1}{8}}\left({\frac{(\alpha^{+})^{2}}{16}}+{\frac{\alpha^{+}}{4}}+1\right)\log\rho\approx
0.26\log\rho,\end{split}$ (2.28)

and thus the proper length goes to zero as $\rho\to 0$.

Therefore, at the presumed location of the ETW 7-branes, in string frame the
size of the circle in the $r$-direction tends to zero, while the length of the
interval in the $y$-direction goes to infinity, such that the area stays
finite. This means that topologically this space is the unreduced suspension
of the circle, $S(S^{1})=S^{2}$. In figure 2, we provide a schematic
representation of the solution in the string frame. he two gray circles on the
left and on the right hand side of the upper figure actually have zero size
and should be considered as points. This is shown explicitly in the lower
figure.

Figure 2: Schematic view of the string frame geometry.

8-brane$r$$r=0$$r=R/2$$y$ETW 7-braneETW 7-brane

As we have seen, in the Einstein frame the distance of the two ETW 7-branes at
$r=R/2$ goes to zero, which means that the space in the $(r,y)$-plane is
topologically the reduced suspension $\Sigma(S^{1})=S^{2}$.

We would like now to relate our setup and the general picture of Local
Dynamical Cobordism [16, 17, 18]. Recall that for the distance $\Delta$ to the
ETW-defects and their tension $\mathcal{T}$, the following scaling behavior

$\begin{split}\Delta^{-n}\sim{\cal T}\end{split}$ (2.29)

was proposed in [16], with $n$ some order one parameter. For the present
example, one can indeed express $\Delta$ in terms of the tension ${\cal
T}\sim\lambda\exp({\frac{5}{4}}\phi_{0})$ of the neutral domain wall

$\begin{split}\Delta\sim L_{y}\sim{\cal T}^{-1},\end{split}$ (2.30)

where we used (2.18). This means that in our case we have the exponent $n=1$.
It was also proposed [17, 18] that (in units where $\kappa_{10}=1$ and in
Einstein frame) the distance $\Delta$ and the scalar curvature ${\mathcal{R}}$
scale with the distance $D$ in field space as

$\Delta\sim e^{-{\frac{\delta}{2}}D}\,,\qquad|{\mathcal{R}}|\sim e^{\delta D}\
$ (2.31)

for a suitable parameter $\delta$. This is reminiscent of the swampland
distance conjecture [23], as it means that close to the end-of-the-world the
field goes to infinity with a logarithmic behavior.

We will check that a relation of the type (2.31) holds in our setup, both in
string and Einstein frame, for a certain value of $\delta$ different in the
two frames. Consider the ETW 7-brane located at $y=+\infty$ and $|r|=R/2$.
Since the proper length of the circle in the $r$-direction goes to zero at
$y=+\infty$, all trajectories specified by a fixed $r$ and $y\to+\infty$ do
reach the ETW 7-brane location. Note that e.g. trajectories with a fixed
finite value of $y$ and $r\to R/2$ do not reach the ETW 7-brane location.
Therefore, for the string-frame distance close to the location of the ETW
brane, we can approximate

$\begin{split}\Delta\sim\int_{y}^{\infty}dy^{\prime}\,e^{-{\frac{\pi\alpha}{16R}}y^{\prime}}\sim
e^{-{\frac{\pi\alpha}{16R}}y}\,.\end{split}$ (2.32)

Expressing this in terms of the field distance of
$\psi(y)\sim{\frac{\pi\alpha}{16R}}y$ and its canonical normalized field
$D=\psi(y)/\sqrt{2}$, one finds444Actually, we are neglecting the
$r$-dependence in $\Delta$. By including it, we get a power law correction of
the type (in string frame) $\Delta\sim\rho^{(1-2\sqrt{2})}e^{-\sqrt{2}D}$
yielding $\delta=2\sqrt{2}$.

$\begin{split}\Delta\sim e^{-\sqrt{2}D}\,.\end{split}$ (2.33)

A similar calculation in the Einstein frame gives $\Delta\sim
e^{-\frac{5\pi\alpha}{64}y}\sim e^{-\frac{5}{4}\sqrt{2}D}$ and thus
$\delta=\frac{5}{2}\sqrt{2}$. It remains to check whether the second relation
in (2.31) is satisfied for these values of $\delta$. For the general warped
ansatz of the 10D metric (2.4), the Ricci scalar is

$\begin{split}{\mathcal{R}}=-e^{-2{\cal B}}\Big{(}16\,\Box{\cal
A}+72\,(\nabla{\cal A})^{2}+2\,\Box{\cal B}\Big{)}\end{split}$ (2.34)

giving in the Einstein frame $|\mathcal{R}|\sim e^{-2V(y)}\sim
e^{\frac{5}{2}\sqrt{2}D}$, while in the string frame

$\begin{split}|{\mathcal{R}}|\sim
e^{-2\left(V(y)+{\frac{1}{4}}\psi(y)\right)}\sim e^{2\sqrt{2}D}\,.\end{split}$
(2.35)

The relations (2.31) are thus satisfied in our setup.

###  2.3 Cobordism interpretation

Comparing the three solutions, we observe that only Solution II+ features an
interval of finite proper length. As we will discuss in the next section, this
behavior matches precisely with the presence of a 7-brane solution curing the
logarithmic singularity and capping off the $y$-direction. Such an ETW-brane
configuration can be actually predicted by the requirement of a vanishing
cobordism group proposed in [7], as we are going to argue below.

We recall that a trivial cobordism group, $\Omega_{k}=0$, corresponds to the
statement that any compact $k$-dimensional quantum gravity background is the
boundary of a $(k+1)$-dimensional compact manifold. This is equivalent to the
requirement that (certain) higher-form global symmetries should be absent in
the $(d-k)$-dimensional effective theory. Thus, the global symmetry must
either be gauged or broken. In the latter case, one needs to introduce a
$(d-k-1)$-defect, which is a domain wall in the effective theory.

As was pointed out in this context in [14], also non-trivial K-theory classes
$K^{-k}({\rm pt})\neq 0$ correspond to global symmetries, which are deeply
related to cobordism by a generalization of the classic Conner-Floyd theorem
[24, 25]. In fact, K-theory symmetries are always gauged [26] and their
charges appear in the same charge neutrality condition together with the
cobordism invariants, which lie in the image of the Atiyah-Bott-Shapiro
orientation. In many cases, this has been identified with familiar tadpole
cancellation conditions in string theory [14].

In our setup, we always have a spontaneous compactification on a topological
$S^{1}$. A natural cobordism group to consider would then be $\Omega_{1}^{\rm
Spin}=\mathbb{Z}_{2}$. The choice of Spin structure is dictated by the fact
that we want to have fermions in our effective theory. Since the group is not
trivial, we have a global symmetry, which must be either broken or gauged.
However, only breaking the symmetry would predict a 7-brane defect.

As was argued in [14], for one of the two string realizations of the
9-dimensional neutral domain wall that we are aware of, namely the type I non-
BPS D8-brane, the breaking is not expected. In this case, the symmetry has to
be gauged. Indeed, it is known that type I (non-BPS) Dp-branes are classified
by KO-theory [27, 28]. In particular, a non-BPS D8 brane is associated to
KO-1(pt)$\,=\\!\mathbb{Z}_{2}$. Via the Atiyah-Bott-Shapiro orientation, this
group is indeed isomorphic to $\Omega_{1}^{\rm Spin}$. However, as proposed in
[14], in this peculiar case the cobordism invariant contribution to the
$\mathbb{Z}_{2}$-valued tadpole cancellation condition vanishes (mod 2). Thus,
gauging the global $\mathbb{Z}_{2}$ symmetry we started with requires an even
number of non-BPS D8 brane charges to cancel the tadpole on the KO-theory
side.

Therefore, we are led to the proposal that Solution I or II- describes the
backreacted non-BPS D8-brane, classified by KO-1(pt). These solutions do not
have an end-of-the-world brane at finite distance, but still always feature a
log-singularity. We now interpret this as an inconsistency telling us that one
should not consider a single such non-BPS D8-brane on a compact space $S^{1}$,
in agreement with the aforementioned gauging.

For the other string realization of the 9-dimensional neutral domain wall,
namely the $\overline{D8}/O8$ stack of a non-supersymmetric type IIA
orientifold, a natural interpretation is that instead the cobordism symmetry
$\Omega_{1}^{\rm Spin}=\mathbb{Z}_{2}$ is broken. Indeed, since we are not in
type I, we cannot make use of the aforementioned relation between cobordism
and K-theory and there is no obvious motivation why the cobordism charge
should be gauged. On the other hand, breaking such a symmetry predicts the
existence of 7-brane defects which can be nicely identified with the ETW-
branes of Local Dynamical Cobordism [16, 17, 18]. In our interpretation, this
situation would correspond to Solution II+, where in fact the singularities
are at finite distance and spacetime closes off.555In this picture, it is also
tempting to identify the topological $S^{1}$ as the generator of
$\Omega_{1}^{\rm Spin}$. As we have seen, the size of this $S^{1}$ shrinks to
zero at the positions of the two ETW-branes and thus the boundary conditions
for the fermions should be periodic.

##  3 The ETW 7-brane

The cobordism conjecture suggests that the singularity we have seen for
Solution II+ can be cured by introducing an appropriate pair of ETW 7-branes.
What kind of 7-branes could they be? To clarify this point, we move a step
forward and try to construct a new 7-brane solution of the dilaton-gravity
equations of motion. Close to its core, this solution should show the features
that we need to close-off the singularity found for Solution II+.

Thus, we are looking for a 7-brane solution to the equations of motion (2.2)
and (2.3) that preserves 8D Poincaré symmetry, that has $\log\rho$
singularities close to its core ($\rho\to 0$) and, as figure 2 suggests, that
is non-isotropic in the two transversal directions. Indeed, recall that there
is one direction in which the length scale vanishes and another direction in
which it goes to infinity. The tension of the brane and the dependence on the
dilaton are not yet known, but we are going to determine them.

###  3.1 Solution breaking rotational symmetry

Guided by the aforementioned constraints, we make a non-isotropic ansatz for
the Einstein-frame metric

$\begin{split}ds^{2}=e^{2\hat{\cal A}(\rho,\varphi)}ds_{8}^{2}+e^{2\hat{\cal
B}(\rho,\varphi)}(d\rho^{2}+\rho^{2}d\varphi^{2}),\end{split}$ (3.1)

with a separated dependence of the warp factors and the dilaton on the radial
coordinate $\rho$ and the angular coordinate and $\varphi$, i.e.

$\begin{split}&\hat{\cal
A}(\rho,\varphi)=\hat{A}(\rho)+\hat{U}(\varphi)\,,\quad\hat{\cal
B}(\rho,\varphi)=\hat{B}(\rho)+\hat{V}(\varphi)\,,\\\
&\hat{\Phi}(\rho,\varphi)=\hat{\chi}(\rho)+\hat{\psi}(\varphi)\,.\end{split}$
(3.2)

The hat is used here to distinguish the various quantities from the similar
ones employed in the 9-dimensional neutral domain wall solution.

The equation of motion are analogous to the ones we have seen in the previous
section. The one related to the variation $\delta g^{\mu\nu}$ is

$\begin{split}\Big{(}7\hat{A}^{\prime\prime}+7{\frac{\hat{A}^{\prime}}{\rho}}&+28(\hat{A}^{\prime})^{2}+\hat{B}^{\prime\prime}+{\frac{\hat{B}^{\prime}}{\rho}}+{\textstyle{\frac{1}{4}}}(\hat{\chi}^{\prime})^{2}\Big{)}\\\\[2.84544pt]
+&{\frac{1}{\rho^{2}}}\Big{(}7\ddot{\hat{U}}+28(\dot{\hat{U}})^{2}+\ddot{\hat{V}}+{\textstyle{\frac{1}{4}}}(\dot{\hat{\psi}})^{2}\Big{)}=-\hat{\lambda}\,e^{a\hat{\Phi}}\,{\frac{1}{2\pi\rho}}\delta(\rho)\,,\end{split}$
(3.3)

where as mentioned above the tension $\hat{\lambda}$ and the parameter $a$
(giving the 7-brane dependence on the dilaton) are left undetermined for the
moment. The prime denotes the derivative with respect to $\rho$ and the dot
the derivative with respect to $\varphi$. For the two variations $\delta
g^{\rho\rho}$ and $\delta g^{\varphi\varphi}$ we obtain

$\begin{split}&\Big{(}8{\frac{\hat{A}^{\prime}}{\rho}}+28(\hat{A}^{\prime})^{2}+8\hat{A}^{\prime}\hat{B}^{\prime}-{\textstyle{\frac{1}{4}}}(\hat{\chi}^{\prime})^{2}\Big{)}+{\frac{1}{\rho^{2}}}\Big{(}8\ddot{\hat{U}}+36(\dot{\hat{U}})^{2}-8\dot{\hat{U}}\dot{\hat{V}}+{\textstyle{\frac{1}{4}}}(\dot{\hat{\psi}})^{2}\Big{)}=0,\\\\[5.69046pt]
&\Big{(}8\hat{A}^{\prime\prime}+36(\hat{A}^{\prime})^{2}-8\hat{A}^{\prime}\hat{B}^{\prime}+{\frac{1}{4}}(\hat{\chi}^{\prime})^{2}\Big{)}+{\frac{1}{\rho^{2}}}\Big{(}28(\dot{\hat{U}})^{2}+8\dot{\hat{U}}\dot{\hat{V}}-{\frac{1}{4}}(\dot{\hat{\psi}})^{2}\Big{)}=0\,,\\\
\end{split}$ (3.4)

and for the off-diagonal $\delta g^{\rho\varphi}$

$\begin{split}&8{\frac{\dot{\hat{U}}}{\rho}}-8\hat{A}^{\prime}\dot{\hat{U}}+8\hat{B}^{\prime}\dot{\hat{U}}+8\hat{A}^{\prime}\dot{\hat{V}}-{\frac{1}{2}}\hat{\chi}^{\prime}\dot{\hat{\psi}}=0\,.\end{split}$
(3.5)

The dilaton equation of motion becomes

$\begin{split}\Big{(}\hat{\chi}^{\prime\prime}+{\frac{\hat{\chi}^{\prime}}{\rho}}+8\hat{A}^{\prime}\hat{\chi}^{\prime}\Big{)}+{\frac{1}{\rho^{2}}}\Big{(}\ddot{\hat{\psi}}+8\dot{\hat{U}}\dot{\hat{\psi}}\Big{)}={2a}\lambda\,e^{a\hat{\Phi}}\,{\frac{1}{2\pi\rho}}\delta(\rho)\,.\end{split}$
(3.6)

Note that taking the sum of the two equations in (LABEL:eomyyrr2) gives the
simple equation

$\begin{split}8\Big{(}\hat{A}^{\prime\prime}+{\frac{\hat{A}^{\prime}}{\rho}}+8(\hat{A}^{\prime})^{2}\Big{)}+{\frac{8}{\rho^{2}}}\Big{(}\ddot{\hat{U}}+8(\dot{\hat{U}})^{2}\Big{)}=0\,.\end{split}$
(3.7)

All these equations are similar to the ones from the previous section, albeit
featuring some extra $1/\rho$-terms. For instance, they contain the 2D
Laplacian in polar coordinates

$\begin{split}\Box\,{F}(\rho,\varphi)={\frac{1}{\rho}}\partial_{\rho}(\rho\,\partial_{\rho}F)+{\frac{1}{\rho^{2}}}\partial_{\varphi}^{2}F\,.\end{split}$
(3.8)

We find it quite remarkable that these equations admit a three-parameter bulk
solution which is also similar to the one from the previous section. Now it is
the $\rho$-dependent functions that are of hyperbolic type

$\begin{split}\hat{A}(\rho)&={\frac{1}{8}}\log\left(\cosh\Big{[}8\hat{K}\log\Big{(}{\frac{\rho}{\rho_{0}}}\Big{)}\Big{]}\right)\,,\\\\[2.84544pt]
\hat{\chi}(\rho)&=\hat{\alpha}\hat{A}(\rho),\\\\[2.84544pt]
\hat{B}(\rho)&=-\log\Big{(}{\frac{\rho}{\rho_{0}}}\Big{)}+\left({\textstyle{\frac{\hat{\alpha}^{2}}{32}}-{\frac{7}{2}}}\right)\hat{A}(\rho)\,,\end{split}$
(3.9)

where at this stage $\hat{\alpha}$, $\hat{K}$ and the dimensionful parameter
$\rho_{0}$ are still undetermined. Notice the additional
$\hat{f}(\rho)=-\log(\rho/\rho_{0})$ term in the solution for the warp factor
$\hat{B}$. The angle dependent solutions are instead

$\begin{split}\hat{U}(\varphi)&={\frac{1}{8}}\log\big{|}\cos(8\hat{K}\varphi)\,\big{|},\\\\[2.84544pt]
\hat{\psi}(\varphi)&={\frac{\hat{\alpha}}{8}}\log\big{|}\cos(8\hat{K}\varphi)\,\big{|}\pm
2\log\Big{|}\tan\left(4\hat{K}\varphi+{\frac{\pi}{4}}\right)\Big{|},\\\\[2.84544pt]
\hat{V}(\varphi)&=-\left({\textstyle{\frac{\hat{\alpha}^{2}}{32}}-{\frac{9}{2}}}\right)\hat{U}(\varphi)+{\frac{\hat{\alpha}}{16}}\hat{\psi}(\varphi)\,,\end{split}$
(3.10)

where due to the periodicity of the cosine-function we have
$\varphi\in[0,\frac{\pi}{4\hat{K}}]$. The sign in the second line is in fact a
free parameter, so that the solution comes in a pair. Flipping such a sign is
the same as shifting the argument of the $\tan$-function by $-\pi/2$, which is
the same as flipping the sign of $\hat{K}$. We denote this pair as ETW
$7^{\pm}$-branes and from now on we invoke the freedom to choose $\hat{K}>0$.

Finally, each of the quantities in the solutions (3.9) and (3.10) can be
shifted by arbitrary integration constants, which we omitted for convenience.
We will see below that matching with our desired delta source configuration
can be achieved by fixing one of these constants.

###  3.2 Brane sources

Given the bulk solution, we would like now to study the behavior at the
boundary. To this end, we include the $\delta$-function source on the right
hand side of the equations of motion.

First, we notice that $\hat{f}=\log\rho$ is the 2D Green’s function satisfying

$\begin{split}\Box_{\rho}\hat{f}\equiv\hat{f}^{\prime\prime}+{\frac{\hat{f}^{\prime}}{\rho}}={\frac{1}{\rho}}\delta(\rho)=2\pi\delta^{2}(\vec{y})\,.\end{split}$
(3.11)

To understand where potential $\delta$-functions can appear, we do not
initially specify $\hat{A}(\rho)$. Rather, we introduce just the restricted
ansatz

$\begin{split}\hat{\chi}(\rho)&=\hat{\alpha}\hat{A}(\rho)\,,\qquad\hat{B}(\rho)=\underbrace{-\log(\rho/\rho_{0})}_{\tilde{B}(\rho)}+\left({\textstyle{\frac{\hat{\alpha}^{2}}{32}}-{\frac{7}{2}}}\right)\hat{A}(\rho),\end{split}$
(3.12)

together with the solution (3.10) for the angle dependent functions, and we
insert them into the equations of motion to obtain

$\begin{split}\delta
G^{\mu\nu}&:\quad\phantom{ii}\Box_{\rho}\tilde{B}+\left({\textstyle{\frac{\hat{\alpha}^{2}+112}{32}}}\right)\left(\Box_{\rho}\hat{A}+8(\hat{A}^{\prime})^{2}-{\frac{8\hat{K}^{2}}{\rho^{2}}}\right)=\,^{\prime\prime}\delta(\rho)^{\prime\prime}\,,\\\
\delta G^{\varphi\varphi}&:\quad
8\,\left(\Box_{\rho}\hat{A}+8(\hat{A}^{\prime})^{2}-{\frac{8\hat{K}^{2}}{\rho^{2}}}\right)=\,^{\prime\prime}\delta(\rho)^{\prime\prime}\,,\\\
\delta\Phi&:\quad\hat{\alpha}\,\left(\Box_{\rho}\hat{A}+8(\hat{A}^{\prime})^{2}-{\frac{8\hat{K}^{2}}{\rho^{2}}}\right)=\,^{\prime\prime}\delta(\rho)^{\prime\prime}\,,\\\
\end{split}$ (3.13)

with the $\delta G^{\rho\rho}$ and $\delta G^{\rho\varphi}$ gravity equations
of motion leading to no source term. Here, with
${}^{\prime\prime}\delta(\rho)^{\prime\prime}$ we indicate potential source
terms with generic coefficients.

Clearly, the term $\Box_{\rho}\tilde{B}$ in the first equation generates the
source term in (3.3) for $a=0$ and $\hat{\lambda}=2\pi$. The second potential
contribution on the right hand side of (3.13) is related to $\hat{A}$, which
enters always with the specific combination

$\begin{split}\hat{A}^{\prime\prime}+{\frac{\hat{A}^{\prime}}{\rho}}+8(\hat{A}^{\prime})^{2}-{\frac{8\hat{K}^{2}}{\rho^{2}}}=0\,,\qquad{\rm
for}\ \rho\neq 0\,.\end{split}$ (3.14)

Due to $\hat{A}(\rho)\simeq-{\hat{K}}\log\rho/\rho_{0}$ for $\rho\ll\rho_{0}$
this could lead to another two-dimensional $\delta$-source. However, from
(3.13) one would deduce that this source must come from an 8-brane wrapping
the $\varphi$ direction. Another possibility is that this term is zero even at
the core, $\rho=0$, so that the right hand side of the last two equations in
(3.13) is everywhere identically vanishing. This is preferable in our case, as
it would avoid the introduction of yet another 8-brane.

To see that this possibility can indeed occur, notice that the first two terms
in (3.14) recombine into $\Box_{\rho}\hat{A}$ and thus definitely give a
$\delta^{2}$-term so that integrating over a disc of small radius
$\varepsilon_{0}$ yields

$\begin{split}\int_{D_{\varepsilon_{0}}}d\rho d\varphi\,\rho\
\Box_{\rho}\hat{A}=-{2\pi\hat{K}}\,.\end{split}$ (3.15)

Therefore, this needs to be cancelled by the last two terms in (3.14), in
order to avoid the appearance of additional sources. Note that all four terms
scale like $\pm 1/\rho^{2}$ close to the core. To obtain the desired
cancellation, we recall that we still have the freedom to choose arbitrary
integration constants in the ansatz, as mentioned at the end of the last
section. Actually, after a redefinition of the coordinates $x_{\mu}$ and
$\rho$, we are left with one physical integration constant that is usually
identified with $\hat{\phi}_{0}$. Hence, we supplement our ansatz for
$\hat{A}(\rho)$ by

$\begin{split}\hat{A}(\rho)={\frac{1}{8}}\log\left(\cosh\Big{[}8\hat{K}\log\Big{(}{\frac{\rho}{\rho_{0}}}\Big{)}\Big{]}\right)+{\frac{\hat{\phi}_{0}}{\hat{\alpha}}}\,,\end{split}$
(3.16)

where we have added explicitly such an integration constant. Then, we can
write

$\begin{split}8(\hat{A}^{\prime})^{2}-{\frac{8\hat{K}^{2}}{\rho^{2}}}&=8\left(\hat{A}^{\prime}-{\frac{\hat{K}}{\rho}}\right)\left(\hat{A}^{\prime}+{\frac{\hat{K}}{\rho}}\right)\\\
&\simeq-{\frac{16\hat{K}}{\rho}}\,\partial_{\rho}\\!\left(\hat{A}(\rho)+\hat{K}\log\left({\frac{\rho}{\rho_{0}}}\right)\right),\end{split}$
(3.17)

so that integrating over a disc of small radius $\varepsilon_{0}$ and invoking
Stoke’s theorem one gets

$\begin{split}\int_{D_{\varepsilon_{0}}}d\rho
d\varphi\,\rho\,\left(8(\hat{A}^{\prime})^{2}-{\frac{1}{8\rho^{2}}}\right)=-\frac{32\pi\hat{K}}{\hat{\alpha}}\hat{\phi}_{0}\,.\end{split}$
(3.18)

We conclude that the behavior of the peculiar expression (3.14) at the core
depends on the value of an integration constant. Without specifying this
parameter, the expression is ambiguous. For the specific value
$\hat{\phi}_{0}=-\hat{\alpha}/16$, the term (3.18) cancels against (3.15) and
(3.14) vanishes everywhere. In this case, the only physical $\delta$-source
term on the left hand side of (3.13) arises from the contribution $\tilde{B}$,
which can be reproduced by a $7$-brane localized at $\rho=0$ with Einstein-
frame action

$\begin{split}S_{7}=-T_{7}\int
d^{10}x\,\sqrt{-g}\;\frac{\delta(\rho)}{2\pi\rho}\,.\end{split}$ (3.19)

Since there is no contribution in the dilaton equation, we need to set $a=0$
and choose $\hat{\lambda}=\kappa_{10}^{2}T_{7}=2\pi$. Note that for a 7-brane
$\hat{\lambda}$ is dimensionless.

###  3.3 Comparison to Solution II

Let us compare the behavior of the solution close to the core, $\rho=0$, with
the singular geometry (2.20) of Solution II+. The parameter capturing the
local behavior close to the ETW-wall (string frame) is the factor
$\delta=2\sqrt{2}$ in the scaling relation (2.31). We now require that our ETW
7-brane solution scales in the same way close to its core.

The $\rho$-dependent functions (3.9) close to $\rho=0$ read

$\begin{split}\hat{A}(\rho)&\simeq-{\hat{K}}\log\rho\,,\quad\hat{\chi}(\rho)\simeq-{\hat{K}\hat{\alpha}}\log\rho\,,\\\
\hat{B}(\rho)&\simeq-{\hat{K}}\left({\textstyle{\frac{\hat{\alpha}^{2}}{32}}-{\frac{7}{2}}+{\frac{1}{\hat{K}}}}\right)\log\rho\,.\end{split}$
(3.20)

Then, the string-frame distance to the core behaves as

$\begin{split}\Delta\sim\int_{0}^{\rho}d\rho^{\prime}\,e^{\hat{B}(\rho^{\prime})+{\frac{1}{4}}\hat{\chi}(\rho^{\prime})}\sim\rho^{-\hat{K}\left({\frac{\hat{\alpha}^{2}}{32}}+{\frac{\hat{\alpha}}{4}}-{\frac{7}{2}}\right)}\end{split}$
(3.21)

and for $\hat{K}>0$ it comes out finite only if
${\frac{\hat{\alpha}^{2}}{32}}+{\frac{\hat{\alpha}}{4}}-{\frac{7}{2}}<0$. This
distance can be expressed in term of the canonically normalized field
$D=\hat{\chi}/\sqrt{2}$ as

$\begin{split}\Delta\sim\exp\left({\textstyle{\frac{\sqrt{2}}{\hat{\alpha}}}}\left({\textstyle{\frac{\hat{\alpha}^{2}}{32}}+{\frac{\hat{\alpha}}{4}}-{\frac{7}{2}}}\right)D\right),\end{split}$
(3.22)

so that, recalling the scaling relation (2.31), we get $\delta=2\sqrt{2}$
(with $\Delta$ finite) precisely for

$\begin{split}\hat{\alpha}=\alpha^{+}=4(4\sqrt{2}-5)\,.\end{split}$ (3.23)

As for the scalar curvature, we find close to $\rho=0$ and in string frame

$\begin{split}|{\mathcal{R}}|\sim{\frac{1}{\rho^{2}}}e^{-2\left(\hat{B}(\rho)+{\frac{1}{4}}\hat{\chi}(\rho)\right)}\sim\rho^{2\hat{K}\left({\frac{\hat{\alpha}^{2}}{32}}+{\frac{\hat{\alpha}}{4}}-{\frac{7}{2}}\right)}\sim
e^{2\sqrt{2}D},\end{split}$ (3.24)

which is also consistent with the scaling behavior (2.31).666We also checked
that the scaling relations (2.31) are obeyed in Einstein-frame. In this case
we found $\delta=\frac{5}{2}\sqrt{2}$, which consistently leads to
$\hat{\alpha}=\alpha^{+}$, as well. Thus, close to the core, the coordinate
$\rho$ in the 7-brane solution plays the role of the coordinate $y$ (or better
$\Delta$) in the original domain wall solution, for $y\to\pm\infty$.
Similarly, the $\varphi$-coordinate for the ETW 7-brane solution (3.10) is
related to the periodic $r$-coordinate of the domain wall (2.14) via
$8\hat{K}\varphi=8K(|r|-3R/2)={\pi\over 2R}|r|-{3\pi\over 4}$. Therefore,
depending on $\hat{K}$, the former domain wall coordinate $r$ parametrizes a
segment of the $\varphi$ circular coordinate. For convenience we choose the
value $\hat{K}=1/8$ from now on, so that $\varphi$ has periodicity $2\pi$.
Hence, we conclude that close to the core the non-isotropic ETW 7-brane
solution has the right properties to close-off the singularity found in the
neutral domain wall solution of the previous section.

It is instructive to compare the set of parameters in of Solution II+ and of
the ETW 7-brane solution. Initially, for the neutral domain wall the bulk
solution admitted five parameters,

$\begin{split}\alpha,\,K,\,R,\,e^{\phi_{0}},\,\lambda\,.\end{split}$ (3.25)

Implementing the boundary conditions from the domain wall and requiring a
finite size of the spontaneously compactified longitudinal direction fixed
$\alpha=4(4\sqrt{2}-5)$ and led to the two conditions

$\begin{split}K={\frac{\pi}{16R}}\,,\qquad
e^{{\frac{5}{4}}\phi_{0}}\sim{\frac{1}{\lambda R}}\,.\end{split}$ (3.26)

Finally, we were left with two unconstrained parameters like e.g. the radius
$R$ and the overall scale of the string coupling constant
$g_{s}=e^{\phi_{0}}$. In a concrete string theory setting, the tension would
also be fixed, of course. Note that for large radius $R$ the string coupling
becomes small so that one expects to have good control over the employed low
energy effective action, which is just dilaton gravity in this case.

For the 7-brane, we also started with five parameters in the bulk (even if
$\hat{\phi}_{0}$ was introduced explicitly afterwards),

$\begin{split}\hat{\alpha},\,\hat{K},\,\rho_{0},\,e^{\hat{\phi}_{0}},\,\hat{\lambda}\,,\end{split}$
(3.27)

but the boundary condition fixed the tension $\hat{\lambda}=2\pi$ and gave
rise to one relation

$\begin{split}\phi_{0}\sim\hat{\alpha},\end{split}$ (3.28)

so that one is left with three free parameters. Requiring that the 7-brane
closes off the singularities present in the neutral domain wall solution (i.e.
that it is really the ETW-brane) fixed the parameter
$\hat{\alpha}=\alpha^{+}$. Hence, at this stage we are also left with two free
parameters, namely $\hat{K}$ and the radial scale $\rho_{0}$. In contrast to
the domain wall solution II+, we now face a fixed value of the string coupling
constant $g_{s}=\exp(-\hat{\alpha}/16)\approx 0.85$, which is at the boundary
of having control over the utilized low energy effective action. However, the
dilaton-gravity solution per se is valid for arbitrary values of
$\hat{\alpha}$ and consequently $g_{s}$. Therefore, we can retain perturbative
control and are thus confident that the solution captures some physical
features of the ETW 7-brane.

###  3.4 Geometry of the ETW 7-brane

Let us now discuss the geometry around the ETW $7^{\pm}$-branes in more
detail. First, we display the $\rho$-dependent functions (3.12) with (3.16) in
figure 3 below.

\begin{picture}(0.0,0.0)\end{picture}

Figure 3: The three functions $\hat{A}(\rho),\hat{\chi}(\rho)$ and
$\hat{B}(\rho)$. Here we have chosen $\rho_{0}=1$ and
$\phi_{0}=-\hat{\alpha}/16$.

The figure reflects the just discussed $\log\rho$ behavior close to the core,
but except from this no further singularities appear. Let us also determine
the finite proper length of the radial direction in string frame, which is (we
always set $\hat{K}=\textstyle\frac{1}{8}$ below)

$\begin{split}L_{\rho}&=\int_{0}^{\infty}d\rho\,e^{\hat{B}(\rho)+{\frac{1}{4}}\hat{\chi}(\rho)}\\\
&=\int_{0}^{\infty}d\rho\;{\frac{\rho_{0}}{\rho}}\,\Big{(}\cosh\Big{[}\log\Big{(}{\frac{\rho}{\rho_{0}}}\Big{)}\Big{]}\Big{)}^{{\frac{1}{8}}\left({\frac{\hat{\alpha}^{2}}{32}}+{\frac{\hat{\alpha}}{4}}-{\frac{7}{2}}\right)}\approx
7.38\,\rho_{0}\,.\end{split}$ (3.29)

Consider now the more intriguing angular dependence, which we display in
figure 4.

\begin{picture}(0.0,0.0)\end{picture}

Figure 4: The three functions $\hat{U}(\varphi),\hat{\psi}(\varphi)$ and
$\hat{V}(\phi)$ for the ETW $7^{-}$ brane. For the $7^{+}$ brane the plots are
just shifted by $\pi$.

First, we see that the solution is non-isotropic, with a singular behavior at
the two angles $\varphi_{1}=\pi/2$ and $\varphi_{2}=3\pi/2$. At $\varphi_{1}$
the string coupling goes to the zero whereas at $\varphi_{2}$ it diverges.
Next, we compute the proper length of the angular direction in string frame,
which is

$\begin{split}L_{\varphi}&=\int_{0}^{2\pi}d\varphi\,e^{V(\varphi)+{\frac{1}{4}}\psi(\varphi)}\\\
&=\int_{0}^{2\pi}d\varphi\,\Big{|}\cos\varphi\Big{|}^{{\frac{1}{8}}\left({\frac{\hat{\alpha}^{2}}{32}}+{\frac{\hat{\alpha}}{4}}+{\frac{9}{2}}\right)}\Big{|}\tan\left({\frac{\varphi}{2}}+{\frac{\pi}{4}}\right)\Big{|}^{\pm{\frac{1}{8}}(\hat{\alpha}+4)}\approx
1.07\,(2\pi)\,.\end{split}$ (3.30)

Thus, both proper lengths $L_{\rho}$ and $L_{\varphi}$ are finite but the
area, $A=\tilde{L}_{\rho}L_{\varphi}$, is not just the product of the two, as
it involves the divergent integral

$\begin{split}\tilde{L}_{\rho}&=\int_{0}^{\infty}d\rho\,\rho\,e^{\hat{B}(\rho)+{\frac{1}{4}}\hat{\chi}(\rho)}\to\infty\,.\end{split}$
(3.31)

Hence, the ETW 7±-brane solution is non-compact.

In order to better visualize what happens to the geometry, in figure 5 we
provide a contour plot of the warp factor
$\exp(V(\varphi)+{\frac{1}{4}}\psi(\varphi))$.

ETW $7^{-}$ braneETW $7^{+}$ brane

Figure 5: Contour plot of the $\varphi$-dependent warp factor
$\exp(V(\varphi)+{\frac{1}{4}}\psi(\varphi))$ in string frame. For a more
intuitive depiction, we have chosen $\varphi=0$ along the negative vertical
axis.

This picture is very suggestive, as one can literally see the structure of the
ETW-branes. There is one weak coupling direction ($\varphi=\mp\pi/2$ for the
$7^{\pm}$ brane), where the length scale goes to zero and geometry disappears
into nothing. Along the opposite strong coupling direction, the length scale
goes to infinity. This reflects the expected divergent behavior
(2.27).777Doing the same computation in Einstein-frame, as already visible in
the rightmost plot in figure 4, along both opposite directions the radial
length scale goes to zero. This is consistent with the behavior (2.28).  Note
that for each $\varphi\neq\mp\pi/2$ (for the $7^{\pm}$ brane), the string
coupling diverges as one radially approaches the core at $\rho=0$.

To summarise, we believe we found a very promising candidate for an explicit
solution to the leading order string equations of motion describing a defect
$7^{\pm}$-brane that can close-off the singularities from the backreacted non-
supersymmetric domain wall solution II+. As a new characteristic feature,
these ETW-branes have a non-isotropic geometry around them, thus breaking the
rotational symmetry in the transversal directions familiar from supersymmetric
BPS brane solutions. They have positive tension
$\hat{\lambda}=\kappa^{2}_{10}T_{7^{\pm}}=2\pi$ and in string frame are
governed by an action

$\begin{split}S=-T_{7^{\pm}}\int
d^{10}x\sqrt{-g}\,e^{-2\Phi}\,\delta^{2}(\vec{r})\,.\end{split}$ (3.32)

To our knowledge, no 7-brane object of this type has so far appeared in the
string theory literature.

##  4 Conclusions

In this work, we performed an analysis of previously known classical solutions
of dilaton-gravity equations of motion describing neutral domain walls of
positive tension, which are necessarily non-supersymmetric. They could
describe either a non-BPS D8-brane of type I or a $\overline{D8}/O8$ stack of
a non-supersymmetric type IIA orientifold. Among the three solutions, one
featured a spontaneous compactification on a finite size interval with two
end-of-the-world walls. We showed that this Solution II+ fits nicely into the
framework of dynamical cobordism, recently proposed in [16, 17, 18]. As a new
aspect, we explored a new type of solution of dilaton-gravity equations of
motion. Such a solution has the potential to describe the two ETW 7-branes
that, according to the cobordism conjecture, should exist in order to close-
off the inconsistent background containing just the neutral domain wall. A
novel feature is the non-isotropic nature of this 7-brane solution.

As for future directions, we can point out a couple of open questions, even if
some of them might just be artifacts of the effective dilaton-gravity action
we are using. First, what is the role of the other two (also singular)
solutions that do not admit ETW walls? Since they have a non-compact
longitudinal direction and are of topology $S^{1}\times\mathbb{R}$, it is
conceivable that they are of higher “energy” and as such are therefore not
stable solutions. Eventually, they might decay to Solution II+. On the
contrary, as we argued in section 2.3, it could also be that they are on the
same level and thus their logarithmic singularity indicates that the global
symmetry related to the domain wall charge can only be gauged instead of
broken by ETW 7-branes.

Second, we have noticed a compelling mathematical relation between the
spontaneously compactified original bulk solution for the domain wall and the
one-dimension-higher non-isotropic ETW 7-brane solution. In the first case,
the longitudinal Poincaré symmetry was spontaneously broken, while in the
second case it was the transverse rotational symmetry. It would be interesting
to study whether this is just an accident of this specific example or whether
there exists a whole family of such pairs of solutions in e.g. higher
codimensions.

Finally, given that our analysis hints at the existence of a possible new
(non-supersymmetric) 7-brane in string theory, it would be clearly important
to confirm this result with an independent method. Similarly, new string
theory defects were predicted in [7] and they still call for a better
understanding.

##### Acknowledgments:

We would like to thank R. Angius, I. Basile and S. Raucci for discussions. The
work of N.C. is supported by the Alexander-von-Humboldt foundation.

##  Appendix A Additional formal solution

In this appendix, we collect a new, though arguably not physically
interesting, codimension-one solution. The solution in the bulk is similar to
Solution I and given by

$\displaystyle A(r)$
$\displaystyle=\frac{1}{8}\log\Big{|}\sin\left[{\textstyle
8K(|r|-{\textstyle{\frac{R}{2}}})}\right]\Big{|},$ (A.1) $\displaystyle B(r)$
$\displaystyle=-\frac{7}{16}\log\Big{|}\sin\left[{\textstyle
8K(|r|-{\textstyle{\frac{R}{2}}})}\right]\Big{|},$ $\displaystyle\chi(r)$
$\displaystyle=\phi_{0},$ $\displaystyle U(y)$ $\displaystyle=\pm Ky,$
$\displaystyle V(y)$ $\displaystyle=\pm\frac{9}{2}Ky,$ $\displaystyle\psi(y)$
$\displaystyle=0.$

The jump conditions then fix

$\cot{(4KR)}=0,$ (A.2)

which admits the minimal solution $K=\pi/(8R)$. However, this means that the
jumps in $A^{\prime}(r)$ and $B^{\prime}(r)$ at $r=0$ are vanishing separately
so that also the right hand side must be vanishing. This implies $\lambda
e^{\frac{5}{4}\phi_{0}}\sim\cot{(4KR)}=0$ and thus the string coupling
vanishes. Therefore, this solution does not really describe the backreaction
of a positive tension object.

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