††institutetext: 1School of Physics, Sun Yat-Sen University, Guangzhou 510275,
P. R. China

# Vacuum stability of the type II seesaw leptogenesis from inflation

Chengcheng Han1<EMAIL_ADDRESS>Sihui Huang1
<EMAIL_ADDRESS>Zhanhong Lei1<EMAIL_ADDRESS>
(2020.06.15)

###### Abstract

Recently it has been found that introducing a triplet Higgs to the standard
model could provide a feasible leptogenesis to generate the baryon asymmetry
of our universe, providing that the inflation is driven by the mixing state of
the triplet Higgs and SM Higgs. In this work, we survey the viable parameter
space satisfying the vacuum stability and perturbativity in this model. We
find that the introduction of the triplet Higgs would also ameliorate the
problem of the Higgs vacuum instability. We present two representative
parameter regions where the origin of neutrino masses, baryon asymmetry of the
universe as well as inflation can be explained while keeping consistent with
the condition of vacuum stability and perturbativity.

††preprint:

##  1 Introduction

The Standard Model(SM), a theory for strong and electroweak interactions, is
regarded as a landmark of modern particle physics. However, there remain
several unsolved problems in particle physics and cosmology, such as the
origin of the cosmic inflation Brout:1977ix ; Sato:1980yn ; Guth:1980zm ;
Linde:1981mu ; Albrecht:1982wi , neutrino masses, and baryon asymmetry of our
universe. Numerous models have been promoted to solve one or two of the above
problems but rarely solve all of them at once. Interestingly, recently the
authors found that the type II seesaw model Magg:1980ut ; Cheng:1980qt ;
Lazarides:1980nt ; Mohapatra:1980yp  could provide a simple framework to
address all three problems simultaneously Barrie:2021mwi ; Barrie:2022cub .
The idea is that the neutrino masses are generated by the traditional type II
seesaw mechanism where the neutrino masses are provided by the vacuum value of
the neutral part of the triplet Higgs, and the inflation is driven by the
mixing state of the triplet Higgs and SM Higgs doublet in the early universes,
and the baryon asymmetry is generated by the Affleck-Dine mechanism
Affleck:1984fy ; Dine:1995kz  during the inflation.

However, since the model contains many additional scalars, it is important to
examine whether such a model satisfies various theoretical constraints,
particularly, the stability of the vacuum and the perturbativity of the
parameters. Actually, the standard model Higgs itself already suffers from the
instability problem due to the second minimum of the potential at the high
energy scale Buttazzo:2013uya ; Han:2018yrk . It would be interesting to study
whether this problem can be addressed by the introduction of the triplet
Higgs. Although there is already a lot of work concerning the vacuum stability
and perturbativity in the framework of type II seesaw Arhrib:2011uy ;
Chun:2012jw ; BhupalDev:2013xol ; Bonilla:2015eha ; Haba:2016zbu ;
Moultaka:2020dmb , many of them focus on these problems around the electroweak
scale. However, in the model of type II seesaw triplet leptogenesis
Barrie:2021mwi ; Barrie:2022cub , the triplet Higgs also plays a role of
inflaton. If a second vacuum develops at a higher energy scale, we could
possibly either worry about the transition of our vacuum into the second
vacuum or that we keep staying in the second vacuum from the beginning. The
reason for the latter statement is that after the inflation, there exists an
oscillation stage of the triplet Higgs and SM Higgs doublet Barrie:2021mwi ;
Barrie:2022cub , and then there is a high chance that the oscillation would
end at the second vacuum. One may wonder whether the thermal effect during the
radiation stage following reheating could save this situation. However, it is
believed that the reheating temperature could not be much higher than $\sim
10^{14}$ GeV Sfakianakis:2018lzf ; Barrie:2021mwi . If the second minimum
develops at a higher scale, we would still keep staying in the second vacuum.
In this paper, to simplify our analysis we require the electroweak vacuum to
be absolutely stable and no second vacuum develops at any intermediate scale
between the electroweak scale and Planck scale.

In addition, to generate the baryon asymmetry, the inflation should be driven
by a mixing state of the triplet Higgs and the SM Higgs doublet, providing
additional conditions on the viable parameter space. Therefore it would be
intriguing to give a full analysis of the parameter space satisfying the
vacuum stability and perturbativity, keeping the above three problems being
resolved.

The paper is organized as follows. In Sec. 2 we overview the idea of the
leptogenesis from Higgs inflation. In Sec. 3 we summarize the theoretical
requirement of vacuum stability as well as the necessary condition for Higgs
inflation with successful leptogenesis. We show our numerical result in Sec. 4
and we draw the conclusion in Sec. 6.

##  2 Leptogenesis from Higgs inflation

In this section, we briefly overview how leptogenesis could occur if the
inflation is provided by a mixture of triplet Higgs and SM Higgs doublet. We
will first show the condition of such mixture to drive the inflation and then
present the Affleck-Dine mechanism to generate the baryon asymmetry during the
inflation stage.

###  2.1 Inflation from Higgs

The Type-II seesaw mechanism introduces a triplet scalar $\Delta$ to SM, which
transforms as $(1,3,2)$ under the $SU(3)_{c}\times SU(2)_{L}\times U(1)_{Y}$
gauge group and carries a lepton number $l_{\Delta}=-2$. Under tensor
representation of $SU(2)_{L}$, the triplet $\Delta$ and SM Higgs field can be
written as:

$\displaystyle\Delta=\left(\begin{matrix}\Delta^{+}/\sqrt{2}&\Delta^{++}\\\
\Delta^{0}&-\Delta^{+}/\sqrt{2}\end{matrix}\right)~{}~{}~{}H=\left(\begin{matrix}h^{+}\\\
h\end{matrix}\right)~{}.$ (1)

The potential of the Higgs fields is given by:

$\displaystyle V(H,\Delta)$ $\displaystyle=$
$\displaystyle-m^{2}_{H}H^{\dagger}H+m_{\Delta}^{2}\text{Tr}(\Delta^{\dagger}\Delta)+\lambda_{H}(H^{\dagger}H)^{2}+\lambda_{1}(H^{\dagger}H)\text{Tr}(\Delta^{\dagger}\Delta)$
(2)
$\displaystyle+\lambda_{2}(\text{Tr}(\Delta^{\dagger}\Delta))^{2}+\lambda_{3}\text{Tr}(\Delta^{\dagger}\Delta)^{2}+\lambda_{4}H^{\dagger}\Delta\Delta^{\dagger}H+[\mu(H^{T}i\sigma^{2}\Delta^{\dagger}H)$
$\displaystyle+\frac{\lambda_{5}}{M_{p}}(H^{T}i\sigma^{2}\Delta^{\dagger}H)(H^{\dagger}H)+\frac{\lambda_{5}^{\prime}}{M_{p}}(H^{T}i\sigma^{2}\Delta^{\dagger}H)(\Delta^{\dagger}\Delta)+h.c.]~{}.$

where the terms in the bracket break the $U(1)_{L}$ symmetry and we also
include the Planck suppressed operators. Even if the high dimension operators
have no effect on the low energy physics, they could dominate the $U(1)_{L}$
breaking terms during inflation. The Yukawa term becomes

$\displaystyle\mathcal{L}_{Yukawa}=\mathcal{L}^{SM}_{Yukawa}-\frac{1}{2}Y_{\nu
jk}\bar{L}^{c}_{j}i\sigma^{2}\Delta L_{k}+h.c.~{}.$ (3)

However, the recent CMB observation already exclude the possibility of the
potential being a simple polynomial function. The reason is that the limit of
tensor to scalar ratio around 0.056 indicates that the potential of the
inflaton should be flat enough at which inflation happens. Fortunately, it is
known that in Higgs inflation a non-minimal coupling of Higgs field to gravity
can flatten the potential and result in a Starobinsky type model. Therefore we
extend the model with both H and $\Delta$ taking the non-minimal coupling to
gravity, and then the Lagrangian is as follows:

$\displaystyle\begin{aligned}
\frac{\mathcal{L}}{\sqrt{-g}}=&-\frac{1}{2}M^{2}_{P}R-\xi_{H}H^{\dagger}HR-\xi_{\Delta}\text{Tr}(\Delta^{\dagger}\Delta)R\\\
&-g^{\mu\nu}(D_{\mu}H)^{\dagger}(D_{\nu}H)-g^{\mu\nu}\text{Tr}(D_{\mu}\Delta)^{\dagger}(D_{\nu}\Delta)-V(H,\Delta)+\mathcal{L}_{Yukawa}~{}.\end{aligned}$
(4)

To simplify our analysis, we only focus on the neutral components of $\Delta$
and $H$. The relevant lagrangian becomes:

$\displaystyle\begin{aligned}
\frac{\mathcal{L}}{\sqrt{-g}}=&-\frac{1}{2}M^{2}_{P}R-\xi_{H}|h|^{2}R-\xi_{\Delta}|\Delta^{0}|^{2}R-(\partial_{\mu}h)^{2}-(\partial_{\mu}\Delta^{0})^{2}\\\
&-V(h,\Delta^{0})-[\frac{1}{2}Y_{\nu
ij}\bar{\nu}^{c}_{i}\nu_{j}\Delta^{0}+h.c.]+\cdots~{},\end{aligned}$ (5)

where $V(h,\Delta^{0})$ is defined as

$\displaystyle V(h,\Delta^{0})$ $\displaystyle=$ $\displaystyle-
m_{H}^{2}|h|^{2}+m_{\Delta}^{2}|\Delta^{0}|^{2}+\lambda_{H}|h|^{4}+\lambda_{\Delta}|\Delta^{0}|^{4}+\lambda_{H\Delta}|h|^{2}|\Delta^{0}|^{2}$
(6) $\displaystyle-\left(\mu
h^{2}{\Delta^{0}}^{*}+\frac{\lambda_{5}}{M_{p}}|h|^{2}h^{2}{\Delta^{0}}^{*}+\frac{\lambda^{\prime}_{5}}{M_{p}}|\Delta^{0}|^{2}h^{2}{\Delta^{0}}^{*}+h.c.\right)+...~{},$

where
$\lambda_{\Delta}=\lambda_{2}+\lambda_{3},~{}\lambda_{H\Delta}=\lambda_{1}+\lambda_{4}$.
The Yukawa term of lepton and $\Delta^{0}$ provides a Majorana mass term for
neutrino once $\Delta$ obtains a non-vanishing vacuum expectation value.

We parameterize the $h$ and $\Delta^{0}$ by the polar coordinates:
$h\equiv\frac{1}{\sqrt{2}}\rho_{H}e^{i\eta}$ and
$\Delta^{0}\equiv\frac{1}{\sqrt{2}}\rho_{\Delta}e^{i\theta}$. Then the
Lagrangian can be written as:

$\displaystyle\frac{\mathcal{L}}{\sqrt{-g}}=-\frac{1}{2}(M_{P}^{2}+\xi_{H}\rho_{H}^{2}+\xi_{\Delta}\rho_{\Delta}^{2})R-\frac{1}{2}(\partial_{\mu}\rho_{H})^{2}-\frac{1}{2}(\partial_{\mu}\rho_{\Delta})^{2}-V(h,\Delta^{0})+\cdots~{}.$
(7)

Considering $\xi_{H}\rho_{H}^{2}+\xi_{\Delta}\rho_{\Delta}^{2}\gg 1$ during
inflation, we can perform a weyl transformation:

$\displaystyle
g_{E\mu\nu}=\Omega^{2}g_{\mu\nu},~{}~{}\Omega^{2}=1+\frac{\xi_{H}\rho_{H}^{2}}{M_{P}^{2}}+\frac{\xi_{\Delta}\rho_{\Delta}^{2}}{M_{P}^{2}}~{}.$
(8)

Then Lagrangian is translated from Jordan frame into Einstein frame:

$\displaystyle\frac{\mathcal{L}}{\sqrt{-g_{E}}}=-\frac{1}{2}M_{P}^{2}R_{E}-\frac{1}{2\Omega^{2}}\left((\partial_{\mu}\rho_{H})^{2}+(\partial_{\mu}\rho_{\Delta})^{2}\right)-3M_{P}^{2}(\partial_{\mu}\text{log}\Omega)^{2}-\frac{V(h,\Delta^{0})}{\Omega^{4}}+\cdots~{}.$
(9)

Subsequent derivations in this section mainly follows Lebedev:2011aq . Here we
use the natural unit $M_{p}=1$. Then from Eq. 8 and Eq. 9 we get:

$\displaystyle\mathcal{L}_{kin}=-\frac{3}{4}\left(\partial_{\mu}\text{log}(\xi_{H}\rho_{H}^{2}+\xi_{\Delta}\rho_{\Delta}^{2})\right)^{2}-\frac{1}{2(\xi_{H}\rho_{H}^{2}+\xi_{\Delta}\rho_{\Delta}^{2})}\left((\partial_{\mu}\rho_{H})^{2}+(\partial_{\mu}\rho_{\Delta})^{2}\right)~{}.$
(10)

We can redefine the fields:

$\displaystyle\chi=\sqrt{\frac{3}{2}}\text{log}(\xi_{H}\rho_{H}^{2}+\xi_{\Delta}\rho_{\Delta}^{2}),~{}~{}\kappa=\frac{\rho_{H}}{\rho_{\Delta}}~{}.$
(11)

Plugging Eq. 11 into Eq. 10 and requiring a large non-minimal coupling
$\xi_{H},\xi_{\Delta}\gg 1$, we get:

$\displaystyle\mathcal{L}_{kin}=-\frac{1}{2}(\partial_{\mu}\chi)^{2}-\frac{1}{2}\frac{\xi_{H}^{2}\kappa^{2}+\xi_{\Delta}^{2}}{(\xi_{H}\kappa^{2}+\xi_{\Delta})^{3}}(\partial_{\mu}\kappa)^{2}~{}.$
(12)

For large field value, the quartic terms dominate the potential, thus the
$\mu$ term in $V(h,\Delta^{0})$ can be ignored. Considering a large $\chi$,
the potential of the scalar fields can be written as

$\displaystyle
U(\rho_{H},\rho_{\Delta})=\frac{V}{\Omega^{4}}=\frac{\lambda_{H}\kappa^{4}+\lambda_{H\Delta}\kappa^{2}+\lambda_{\Delta}}{4(\xi_{H}\kappa^{2}+\xi_{\Delta})^{2}}~{},$
(13)

with the following minima:

$\displaystyle(1)~{}2\lambda_{H}\xi_{\Delta}-\lambda_{H\Delta}\xi_{H}>0,~{}2\lambda_{\Delta}\xi_{H}-\lambda_{H\Delta}\xi_{\Delta}>0,~{}\kappa=\sqrt{\frac{2\lambda_{\Delta}\xi_{H}-\lambda_{H\Delta}\xi_{\Delta}}{2\lambda_{H}\xi_{\Delta}-\lambda_{H\Delta}\xi_{H}}}~{},$
(14)
$\displaystyle(2)~{}2\lambda_{H}\xi_{\Delta}-\lambda_{H\Delta}\xi_{H}>0,~{}2\lambda_{\Delta}\xi_{H}-\lambda_{H\Delta}\xi_{\Delta}<0,~{}\kappa=0~{},$
(15)
$\displaystyle(3)~{}2\lambda_{H}\xi_{\Delta}-\lambda_{H\Delta}\xi_{H}<0,~{}2\lambda_{\Delta}\xi_{H}-\lambda_{H\Delta}\xi_{\Delta}>0,~{}\kappa=\infty~{},$
(16)
$\displaystyle(4)~{}2\lambda_{H}\xi_{\Delta}-\lambda_{H\Delta}\xi_{H}<0,~{}2\lambda_{\Delta}\xi_{H}-\lambda_{H\Delta}\xi_{\Delta}<0,~{}\kappa=0,\infty~{}.$
(17)

In scenario 1, the inflation is dominated by $\Delta^{0}$ and $H$, which is
just the case we are interested in. Thus we get our first constraint on the
parameters to ensure the inflation is driven by a mixing state of the triplet
Higgs and SM doublet Higgs, i.e,

$\displaystyle
x=\frac{\xi_{H}}{\xi_{\Delta}},~{}2\lambda_{H}-x\lambda_{H\Delta}>0,~{}2x\lambda_{\Delta}-\lambda_{H\Delta}>0~{}.$
(18)

Note that a necessary condition for vacuum stability is that
$\lambda_{\Delta}>0,\lambda_{H}>0$. However, $\lambda_{H\Delta}$ could be
either positive or negative. Obviously, if $\lambda_{H\Delta}<0$, the above
condition can be automatically satisfied, while for the case of
$\lambda_{H\Delta}>0$, a dedicated study should be preformed and we leave the
details of discussion in Sec. 4.

###  2.2 Leptogensis from Higgs inflation

we can simplify our analysis by defining the inflaton as $\varphi$, through
the following relations to the polar coordinate fields,

$\displaystyle\rho_{H}=\varphi\sin\alpha,~{}\rho_{\Delta}=\varphi\cos\alpha~{},$
$\displaystyle\xi\equiv\xi_{H}\sin^{2}\alpha+\xi_{\Delta}\cos^{2}\alpha~{}.$
(19)

The Lagrangian in Eq. (7) is now given by,

$\displaystyle\frac{\mathcal{L}}{\sqrt{-g}}=-\frac{M_{p}^{2}}{2}R-\frac{\xi}{2}\varphi^{2}R-\frac{1}{2}g^{\mu\nu}\partial_{\mu}\varphi\partial_{\nu}\varphi-\frac{1}{2}\varphi^{2}\cos^{2}\alpha~{}g^{\mu\nu}\partial_{\mu}\theta\partial_{\nu}\theta-V(\varphi,\theta)~{},$
(20)

where

$\displaystyle
V(\varphi,\theta)=\frac{1}{2}m^{2}\varphi^{2}+\frac{\lambda}{4}\varphi^{4}+2\varphi^{3}\left(\tilde{\mu}+\frac{\tilde{\lambda}_{5}}{M_{p}}\varphi^{2}\right)\cos\theta~{},$
(21)

and

$\displaystyle m^{2}=m^{2}_{\Delta}\cos^{2}\alpha-m_{H}^{2}\sin^{2}\alpha~{},$
$\displaystyle\lambda=\lambda_{H}\sin^{4}\alpha+\lambda_{H\Delta}\sin^{2}\alpha\cos^{2}\alpha+\lambda_{\Delta}\cos^{4}\alpha~{},$
$\displaystyle\tilde{\mu}=-\frac{1}{2\sqrt{2}}\mu\sin^{2}\alpha\cos\alpha~{},$
$\displaystyle\tilde{\lambda}_{5}=-\frac{1}{4\sqrt{2}}(\lambda_{5}\sin^{4}\alpha\cos\alpha+\lambda^{\prime}_{5}\sin^{2}\alpha\cos^{3}\alpha)~{}.$
(22)

To fit the current CMB observation, we need roughly
$\frac{\xi}{\sqrt{\lambda}}\approx 5\times 10^{4}$. The Affleck-Dine mechanism
will be realised in our scenario through the motion of the dynamical field
$\theta$. The size of the generated lepton asymmetry will be determined by the
size of the non-trivial motion induced in $\theta$ sourced by inflation.
During inflation, $m\ll\varphi$, which means that the quartic term in the
Jordan frame potential dominates the inflationary dynamics. The lepton number
asymmetry generated by the motion of the triplet Higgs phase $\theta$ at the
end of inflation,

$\displaystyle{n_{L}}_{\textrm{end}}$ $\displaystyle=$ $\displaystyle
Q_{L}\varphi^{2}_{\textrm{end}}\dot{\theta}_{\textrm{end}}\cos^{2}\alpha$ (23)
$\displaystyle\simeq$ $\displaystyle-
Q_{L}\tilde{\lambda}_{5}\varphi^{3}_{\textrm{end}}\sin\theta_{\rm
end}/\sqrt{3\lambda}~{}.$

After inflation, the lepton number density is roughly red-shifted by $a^{3}$.
To generate the correct baryon asymmetry of our univsere, we need
${n_{L}}_{\textrm{end}}\sim 10^{-16}$. For $\lambda\sim\mathcal{O}(0.1)$,
$\varphi_{\textrm{end}}\sim 1$ in Planck unit, we just need
$\tilde{\lambda}_{5}\sim 10^{-15}$.

##  3 Vacuum stability from type II seesaw

In spite of the intriguing motivation, this mechanism suffers from constraints
of theoretical consistency including vacuum stability, perturbativity, and the
condition of inflation. The vacuum stability condition requires that the
potential of Eq. 2 does not develop a second vacuum between TeV scale and the
Planck scale, otherwise our electroweak vacuum would be meta-stable and may
transit to another vacuum. In the large field limit, we only need focus on the
quartic term of the potential. These conditions can be summarized as Eq. 24,
with detailed derivation in appendix A.

$\displaystyle\begin{aligned}
C_{1},C_{2},C_{3},C_{4},C_{5}>0~{}\text{and}~{}[C_{6}>0~{}\text{or}~{}C^{\prime}_{6}>0]~{}.\end{aligned}$
(24)

where

$\displaystyle C_{1}$ $\displaystyle=\lambda_{H}~{},$ (25) $\displaystyle
C_{2}$ $\displaystyle=\lambda_{2}+\lambda_{3}~{},$ (26) $\displaystyle C_{3}$
$\displaystyle=\lambda_{2}+\frac{1}{2}\lambda_{3}~{},$ (27) $\displaystyle
C_{4}$
$\displaystyle=\lambda_{1}+2\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3})}~{},$
(28) $\displaystyle C_{5}$
$\displaystyle=\lambda_{1}+\lambda_{4}+2\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3})}~{},$
(29) $\displaystyle C_{6}$
$\displaystyle=|\lambda_{4}|\sqrt{\lambda_{2}+\lambda_{3}}-2\lambda_{3}\sqrt{\lambda_{H}}~{},$
(30) $\displaystyle C_{6}^{\prime}$
$\displaystyle=2\lambda_{1}+\lambda_{4}+\sqrt{(8\lambda_{H}\lambda_{3}-\lambda_{4}^{2})(2{\lambda_{2}}/{\lambda_{3}}+1)}~{}.$
(31)

Note that the above conditions are should be satisfied at any energy scale
between TeV and $M_{p}$. The perturbativity condition requires that all
couplings should be not much beyond $O(1)$. Here we require all the couplings
smaller than $\sqrt{4\pi}$.

Since inflation in our context requires a flat direction along the mixing of
$h$ and $\Delta^{0}$, the condition in scenario 1 of Eq. 14 should also be
satisfied, which can be rewritten as:

$\displaystyle C_{7}>0,~{}C_{8}>0~{},$ (32)

where

$\displaystyle~{}C_{7}$ $\displaystyle=$ $\displaystyle
2\lambda_{H}-x\lambda_{H\Delta}~{},$ (33) $\displaystyle~{}C_{8}$
$\displaystyle=$ $\displaystyle 2x\lambda_{\Delta}-\lambda_{H\Delta}~{},$ (34)
$\displaystyle x$ $\displaystyle=$
$\displaystyle\frac{\xi_{H}}{\xi_{\Delta}}~{}.$ (35)

We stress that the conditions of $C_{7}>0$ and $C_{8}>0$ should be satisfied
only at around Planck scale.

In our calculation we utilize SARAH Staub:2013tta  to derive the 2-loop
renormalization group equations for the relevant couplings. The 1-loop result
is extracted in appendix C and we find it consistent with the result from
Bonilla:2015eha .

##  4 Numerical results

In our calculation, the free parameters include three gauge coupling
parameters $g_{1},~{}g_{2},~{}g_{3}$, five quartic coupling parameters
$\lambda_{H},~{}\lambda_{1},~{}\lambda_{2},~{}\lambda_{3},~{}\lambda_{4}$ and
yukawa coupling of top quark and leptons with the triplet Higgs. The threshold
effect of $\mu$ parameter can be ignored because it should be much smaller
than GeV to avoid the wash out of the lepton asymmetry Barrie:2021mwi ;
Barrie:2022cub . Here we set the triplet Higgs mass to be 1 TeV, then all the
SM parameters firstly have to run from the top mass scale to TeV scale, then
all the couplings run into Planck scale. During the intermediate scale, the
vacuum stability condition has been checked and the inflation conditions
$C_{7}$ and $C_{8}$ are only imposed at Planck scale. Since all the couplings,
particularly the Higgs self-coupling is sensitive to the top mass, here we set
the top mass to be 172 GeV and Higgs mass to be 125 GeV. All the SM model
couplings at the top mass scale are derived following the formulas from
Buttazzo:2013uya .

Before going to the numerical results, we give some analysis on the preference
of parameter space. First of all, to realize the mixing of the triplet Higgs
and SM Higgs doublet, we have the following relation,

$x=\frac{\xi_{H}}{\xi_{\Delta}},~{}2\lambda_{H}-x\lambda_{H\Delta}>0,~{}2x\lambda_{\Delta}-\lambda_{H\Delta}>0~{}.$
(36)

Since the vacuum stability requires $\lambda_{H}>0,\lambda_{\Delta}>0$, if
$\lambda_{H\Delta}<0$ the above relation can be satisfied automatically. If
$\lambda_{H\Delta}>0$, we need both $2\lambda_{H}-x\lambda_{H\Delta}>0$ and
$~{}2x\lambda_{\Delta}-\lambda_{H\Delta}>0$, imposing additional requirement
for the parameter space.

In addition, it is known that the SM Higgs develops a second vacuum at the
scale around $10^{10-11}$ GeV. However, due to the introduction of the new
scalars, it is possible to remove the second vacuum by changing RGE running of
SM Higgs self-coupling. From the RGE of $\lambda_{H}$

$\displaystyle(4\pi)^{2}\frac{d\lambda_{H}}{d\log\mu}$ $\displaystyle=$
$\displaystyle\frac{27}{200}g_{1}^{4}+\frac{9}{20}g_{1}^{2}g_{2}^{2}+\frac{9}{8}g_{2}^{4}+3\lambda_{1}^{2}+3\lambda_{1}\lambda_{4}+\frac{5}{4}\lambda_{4}^{2}-\frac{9}{5}g_{1}^{2}\lambda_{H}-9g_{2}^{2}\lambda_{H}$
(37) $\displaystyle+24\lambda_{H}^{2}+12\lambda_{H}y_{t}^{2}-6y_{t}^{4},$

we find that there are additional contributions from the term of
$3\lambda_{1}^{2}+3\lambda_{1}\lambda_{4}+\frac{5}{4}\lambda_{4}^{2}$. Note
that this term is always positive and it is possible to lift the $\lambda_{H}$
away from negative at a high energy scale if $\lambda_{1},\lambda_{4}$ are
large enough. We find that there are two typical scenarios, one is that the
couplings of $\lambda_{1}$, $\lambda_{4}$ are small and $\lambda_{H}$ can be
lift to be positive but keeps small at Planck scale($\mathcal{O}(0.01)$), and
the other case is that $\lambda_{1}$, $\lambda_{4}$ are relatively large and
final $\lambda_{H}$ is significantly lifted and becomes rather large at Planck
scale($\mathcal{O}(1)$). In the following we will present the parameter space
for these two typical scenarios.

(a) $x=1$ (b) $x=1/3$

Figure 1: Regions surviving the conditions in the parameter space for
$\lambda_{2}=0.15,\lambda_{3}=0.1,y_{\nu}=0$ at TeV scale.

In Fig. 1 we show the parameter region of a small $\lambda_{H}$ predicted at
$M_{p}$ scale. Here we fixed $\lambda_{2}=0.15,\lambda_{3}=0.1$ at 1 TeV
scale. For the yukawa coupling from the triplet Higgs and leptons, we assume
that it is small enough and has no effect on the RGE running of other
parameters and we will discuss its effect later. From left to right, $x$ is
fixed to be $1,1/3$ respectively. Note that the the condition $C_{2}>0$ and
$C_{3}>0$ can be easily satisfied by set up. The constraints from other
conditions are also shown in combination. Each of the combined conditions
divide the parameter region into two parts and only one of them is allowed, as
indicated by the allowed region(red). Different choices of $x$ has small
effect and it affects slightly condition $C_{7}>0$. The reason is that
$\lambda_{H}$ is small at high energy scale and at the boundary of the
$C_{7}>0$ the $2\lambda_{H}-x\lambda_{H\Delta}$ approaches 0, therefore
varying the $x$ value will slightly change the boundary due to the smallness
of $\lambda_{H}$. We note here that if the initial value of $\lambda_{2}$
becomes smaller, the parameter space would fast vanish. The reason that is
that if we change $\lambda_{2}$ into a smaller value, both $\lambda_{1}$ and
$\lambda_{4}$ would becomes larger and the condition $C_{7}>0$ would be easily
violated.

Figure 2:  Running of couplings from TeV scale to $M_{p}$, with
$\lambda_{1}=-0.2,~{}\lambda_{2}=0.15,~{}\lambda_{3}=0.1,~{}\lambda_{4}=-0.05$
at TeV scale.

Here we give one benchmark in this parameter space. At TeV scale, the input
parameters are fixed as follows:
$\lambda_{1}=-0.2,~{}\lambda_{2}=0.15,~{}\lambda_{3}=0.1,~{}\lambda_{4}=-0.05$.
At $M_{p}$ scale the parameters become
$\lambda_{1}=-0.14,~{}\lambda_{2}=0.97,~{}\lambda_{3}=0.33,~{}\lambda_{4}=0.093$.
Now we find
$\lambda_{H}=0.011,~{}\lambda_{\Delta}=1.3,~{}\lambda_{H\Delta}=-0.046$. For
the case $x=1$, the mixing angle of the Higgs and triplet Higgs during
inflation and the effective $\lambda$ can be found to be

$\displaystyle\tan=\alpha=6.27,~{}~{}\lambda=9.95\times 10^{-3}~{}.$ (38)

To check the consistence of the vacuum stability, in Fig. 2 we show the RGE
running of the vacuum stability conditions for this benchmark point. It is
easy to find all the conditions are satisfied. Even if $C_{6}^{\prime}$ is
negative at certain scale, $C_{6}$ is always keeping positive.

In Fig. 3 we show the parameter region with $\lambda_{H}$ is $\mathcal{O}(1)$
at $M_{p}$ scale. Now we fixed $\lambda_{2}=0,\lambda_{3}=0.1$ at TeV scale.
Again here we assume that the yukawa coupling is small enough and $x$ is fixed
to be $1,1/3$ respectively. It shows that the survived region is mainly
determined by conditions $C_{5}>0$, $C_{7}>0$ and $C_{8}>0$. We also impose a
perturbativity condition that all the parameters should be smaller than
$\sqrt{4\pi}$. Given the relatively large initial values of
$\lambda_{1}~{}\text{and}~{}|\lambda_{4}|$ in Fig. 3 (a), the perturbativity
condition also matters. For a smaller $x$ as in Fig. 3 (b), condition
$C_{7}>0$ is relaxed but condition $C_{8}>0$ is strengthened, but the boundary
of $C_{5}>0$ remains unchanged while the boundaries of $C_{7}>0$ and $C_{8}>0$
shift to left. For the case of $x=3$, we did not identify any parameter region
since the condition $C_{7}>0$ becomes too strong.

(a) $y_{\nu}=0,x=1$ (b) $y_{\nu}=0,x=1/3$

Figure 3:  Regions surviving the conditions in the parameter space for
$\lambda_{2}=0,\lambda_{3}=0.1,y_{\nu}=0,x=1,~{}\frac{1}{3}$ at TeV scale.

Again in Fig. 4 we show one benchmark point in the region of Fig. 3 for the
RGE running of the vacuum stability conditions. The related parameters are
fixed at
$\lambda_{1}=0.7,~{}\lambda_{2}=0,~{}\lambda_{3}=0.1,~{}\lambda_{4}=-0.8$ at
TeV scale. We find that at $M_{p}$ scale
$\lambda_{1}=2.70,~{}\lambda_{2}=0.64,~{}\lambda_{3}=0.35,~{}\lambda_{4}=-1.32$
and $\lambda_{H}=0.8,~{}\lambda_{\Delta}=0.99,~{}\lambda_{H\Delta}=1.38$. For
$x=1$, we have
$2\lambda_{H}-\lambda_{H\Delta}=0.22,~{}2\lambda_{\Delta}-\lambda_{H\Delta}=0.60$.
Then we can calculate the mixing angle and effective quartic coupling during
inflation is

$\text{tan}~{}\alpha=1.65,~{}~{}\lambda=0.77~{}.$

Figure 4:  Running of couplings from TeV scale to $M_{pl}$ scale, with
$\lambda_{1}=0.7,~{}\lambda_{2}=0,~{}\lambda_{3}=0.1,~{}\lambda_{4}=-0.8$.

We stress the importance of imposing the vacuum stability conditions between
low scale and $M_{p}$ since it is possible that the vacuum instability happens
at an intermediate scale. To illustrate this point, in Fig. 5 we show a
particular case where the vacuum stability conditions are satisfied at low
scale and at Planck scale, but a second minimum develops at the scale around
$10^{13}$ GeV. The couplings are fixed to be
$\lambda_{1}=-0.2,~{}\lambda_{2}=0.1,~{}\lambda_{3}=0.1,~{}\lambda_{4}=-0.06$
at TeV scale, and it clearly shows that vacuum stability conditions are
violated between the scale of $10^{4}-10^{13}$ GeV.

Figure 5:  Running of couplings from Electroweak scale to $M_{pl}$ scale, with
$\lambda_{1}=-0.3,~{}\lambda_{2}=0.15,~{}\lambda_{3}=0.1,~{}\lambda_{4}=0.1$.

###  4.1 Effect from yukawa couplings of triplet Higgs with leptons

In this subsection we discuss the effect of yukawa couplings from triplet
Higgs with leptons. Since we have 6 independent yukawa couplings, to simplifiy
our analysis, however, we only require one of them could be sizable and all
the others are too small to have significant effect. In Fig. 6 we show the
effect of yukawa couplings.

As shown in Appendix. C, the yukawa coupling $y_{\nu}$ contributes to the RGE
of $\lambda_{H\Delta}$ from the term $\lambda_{H\Delta}y_{\nu}^{2}$. In the
case of $\lambda_{2}=0.1,\lambda_{3}=0.1$, since $\lambda_{H\Delta}<0$ at
almost all scales below $M_{pl}$, a non-vanishing yukawa coupling reduces the
value of $\lambda_{H\Delta}$ at $M_{p}$, and then condition 7 can be relaxed
slightly as shown in the left panel of Fig. 6. In the case of
$\lambda_{2}=0.1,\lambda_{3}=0.1$, the $\lambda_{1}$ evolves with a positive
and relatively large value. With the $\lambda_{1}y_{\nu}^{2}$ term in the RGE
of $\lambda_{1}$, the value of $\lambda_{1}$ becomes larger at higher energy
scale and it is easily beyond the perturbativity at $M_{p}$. Thus the
perturbativity condition is strengthened with non-zero $y_{\nu}$. We can see
that the survived region shrinks in as shown in the left panel of Fig. 6
comparing with the plot without the contribution of yukawa coupling.

Notice that when $y_{\nu}\gtrsim 0.7$, it will hit the Landau pole before
$M_{p}$ and such case should be avoided.

(a) $\lambda_{2}=0.1,\lambda_{3}=0.1$ (b) $\lambda_{2}=0,\lambda_{3}=0.1$

Figure 6:  Regions surviving the conditions in the parameter space for
$y_{\nu}=~{}0.6,x=1$.

##  5 Phenomenological implications

In the previous section we identify the parameter regions consistent with
theoretical conditions. In this section we will show the low energy
implications of the parameter space. After electroweak symmetry spontaneous
breaking, the triplet and SM Higgs field both obtain a vacuum expectation
value,

$\Delta=\left(\begin{matrix}0&0\\\
v_{t}/\sqrt{2}&0\end{matrix}\right)~{},~{}~{}H=\left(\begin{matrix}0\\\
v_{d}/\sqrt{2}\end{matrix}\right)~{}.$ (39)

Then the type II seesaw model predicts 7 massive particles:
$H^{\pm\pm},~{}H^{\pm},~{}H^{0},~{}A^{0},~{}\text{and}~{}h$.

As in Ref. Ashanujjaman:2021txz , for $v_{t}\ll v_{d}$, which is the region we
are interested in, the mass squared difference of these scalar particles can
be deduced,

$m^{2}_{H^{\pm\pm}}-m^{2}_{H^{\pm}}\simeq
m^{2}_{H^{\pm\pm}}-m^{2}_{H^{0}/A^{0}}\simeq\frac{\lambda_{4}}{4}v_{d}^{2}~{}.$
(40)

Thus, $\lambda_{4}$ can be used to constrain the mass splitting of
$H^{\pm\pm}$ and $H^{\pm}$, as shown in Fig. 7 (a) and Fig. 7 (b), which are
related to the parameter regions in Fig. 1 and Fig. 3, respectively.

We see that for a mass of doubly-charged Higgs around TeV, the mass splitting
of doubly-charged Higgs and charged Higgs is generally below 10 GeV, thus the
cascade decay of the doubly charged Higgs is highly suppressed. In addition,
the vacuum value of the triplet Higgs is less than 10 keV, thus most of the
doubly charged Higgs would decay into dileptons, providing a unique signature
to test the model.

(a) $\lambda_{2}=0.15,\lambda_{3}=0.1,y_{\nu}=0$ (b)
$\lambda_{2}=0,\lambda_{3}=0.1,y_{\nu}=0$

Figure 7:  The mass splitting of $H^{\pm\pm}$ and $H^{\pm}$ predicted in the
case of Fig. 1 and Fig. 3 for $x=1$.

##  6 Conclusions

In this work we investigate the parameter space of type II seesaw model
consistent with the vacuum stability and perturbativity, while solving the
problems of neutrino masses, baryon asymmetry of the universe and inflation. A
successful leptogenesis requires that the inflation is driven by the mixing
state of the SM Higgs doublet and the triplet Higgs, providing additional
constraints on the parameter space. After combining these conditions, we give
two typical viable parameter regions where the SM Higgs self-coupling could be
large $\mathcal{O}(1)$ or small $\mathcal{O}(0.01)$.

##  7 Acknowledgements

We acknowledge Binglong Zhang for his helpful discussions. C. H. is supported
by the Guangzhou Basic and Applied Basic Research Foundation under Grant No.
202102020885, and the Sun Yat-Sen University Science Foundation.

## References

  * (1) R. Brout, F. Englert and E. Gunzig, _The Creation of the Universe as a Quantum Phenomenon_ , _Annals Phys._ 115 (1978) 78.
  * (2) K. Sato, _First Order Phase Transition of a Vacuum and Expansion of the Universe_ , _Mon. Not. Roy. Astron. Soc._ 195 (1981) 467.
  * (3) A. H. Guth, _The Inflationary Universe: A Possible Solution to the Horizon and Flatness Problems_ , _Phys. Rev. D_ 23 (1981) 347.
  * (4) A. D. Linde, _A New Inflationary Universe Scenario: A Possible Solution of the Horizon, Flatness, Homogeneity, Isotropy and Primordial Monopole Problems_ , _Phys. Lett. B_ 108 (1982) 389.
  * (5) A. Albrecht and P. J. Steinhardt, _Cosmology for Grand Unified Theories with Radiatively Induced Symmetry Breaking_ , _Phys. Rev. Lett._ 48 (1982) 1220.
  * (6) M. Magg and C. Wetterich, _Neutrino Mass Problem and Gauge Hierarchy_ , _Phys. Lett. B_ 94 (1980) 61.
  * (7) T. P. Cheng and L.-F. Li, _Neutrino Masses, Mixings and Oscillations in SU(2) x U(1) Models of Electroweak Interactions_ , _Phys. Rev. D_ 22 (1980) 2860.
  * (8) G. Lazarides, Q. Shafi and C. Wetterich, _Proton Lifetime and Fermion Masses in an SO(10) Model_ , _Nucl. Phys. B_ 181 (1981) 287.
  * (9) R. N. Mohapatra and G. Senjanovic, _Neutrino Masses and Mixings in Gauge Models with Spontaneous Parity Violation_ , _Phys. Rev. D_ 23 (1981) 165.
  * (10) N. D. Barrie, C. Han and H. Murayama, _Affleck-Dine Leptogenesis from Higgs Inflation_ , _Phys. Rev. Lett._ 128 (2022) 141801 [2106.03381].
  * (11) N. D. Barrie, C. Han and H. Murayama, _Type II Seesaw leptogenesis_ , _JHEP_ 05 (2022) 160 [2204.08202].
  * (12) I. Affleck and M. Dine, _A New Mechanism for Baryogenesis_ , _Nucl. Phys. B_ 249 (1985) 361.
  * (13) M. Dine, L. Randall and S. D. Thomas, _Baryogenesis from flat directions of the supersymmetric standard model_ , _Nucl. Phys. B_ 458 (1996) 291 [hep-ph/9507453].
  * (14) D. Buttazzo, G. Degrassi, P. P. Giardino, G. F. Giudice, F. Sala, A. Salvio et al., _Investigating the near-criticality of the Higgs boson_ , _JHEP_ 12 (2013) 089 [1307.3536].
  * (15) C. Han, S. Pi and M. Sasaki, _Quintessence Saves Higgs Instability_ , _Phys. Lett. B_ 791 (2019) 314 [1809.05507].
  * (16) A. Arhrib, R. Benbrik, M. Chabab, G. Moultaka, M. C. Peyranere, L. Rahili et al., _The Higgs Potential in the Type II Seesaw Model_ , _Phys. Rev. D_ 84 (2011) 095005 [1105.1925].
  * (17) E. J. Chun, H. M. Lee and P. Sharma, _Vacuum Stability, Perturbativity, EWPD and Higgs-to-diphoton rate in Type II Seesaw Models_ , _JHEP_ 11 (2012) 106 [1209.1303].
  * (18) P. S. Bhupal Dev, D. K. Ghosh, N. Okada and I. Saha, _125 GeV Higgs Boson and the Type-II Seesaw Model_ , _JHEP_ 03 (2013) 150 [1301.3453].
  * (19) C. Bonilla, R. M. Fonseca and J. W. F. Valle, _Consistency of the triplet seesaw model revisited_ , _Phys. Rev. D_ 92 (2015) 075028 [1508.02323].
  * (20) N. Haba, H. Ishida, N. Okada and Y. Yamaguchi, _Vacuum stability and naturalness in type-II seesaw_ , _Eur. Phys. J. C_ 76 (2016) 333 [1601.05217].
  * (21) G. Moultaka and M. C. Peyranère, _Vacuum stability conditions for Higgs potentials with $SU(2)_{L}$ triplets_, _Phys. Rev. D_ 103 (2021) 115006 [2012.13947].
  * (22) E. I. Sfakianakis and J. van de Vis, _Preheating after Higgs Inflation: Self-Resonance and Gauge boson production_ , _Phys. Rev. D_ 99 (2019) 083519 [1810.01304].
  * (23) O. Lebedev and H. M. Lee, _Higgs Portal Inflation_ , _Eur. Phys. J. C_ 71 (2011) 1821 [1105.2284].
  * (24) F. Staub, _SARAH 4 : A tool for (not only SUSY) model builders_ , _Comput. Phys. Commun._ 185 (2014) 1773 [1309.7223].
  * (25) S. Ashanujjaman and K. Ghosh, _Revisiting type-II see-saw: present limits and future prospects at LHC_ , _JHEP_ 03 (2022) 195 [2108.10952].

##  Appendix A A Derivation of the Vacuum Stability Conditions

In this section, we calculate the vacuum stability conditions for a potential
including the SM Higgs doublet and triplet Higgs. All the results can be found
in Bonilla:2015eha ; Moultaka:2020dmb  and here we just add some details. Here
we consider the potential from Eq. 2 where high dimension operators are
neglected,

$\displaystyle V(H,\Delta)=$
$\displaystyle-m^{2}_{H}H^{\dagger}H+m_{\Delta}^{2}\text{Tr}(\Delta^{\dagger}\Delta)+[\mu(H^{T}i\sigma^{2}\Delta^{\dagger}H)+h.c.]$
(41)
$\displaystyle+\lambda_{H}(H^{\dagger}H)^{2}+\lambda_{1}(H^{\dagger}H)\text{Tr}(\Delta^{\dagger}\Delta)+\lambda_{2}(\text{Tr}(\Delta^{\dagger}\Delta))^{2}$
$\displaystyle+\lambda_{3}\text{Tr}(\Delta^{\dagger}\Delta)^{2}+\lambda_{4}H^{\dagger}\Delta\Delta^{\dagger}H+\cdots~{}.$

It can be seen that in the large field limit the potential $V$ mainly depends
on the following four items:$H^{\dagger}H$,
$(\text{Tr}(\Delta^{\dagger}\Delta))^{2}$, $\text{Tr}(\Delta^{\dagger}\Delta)$
and $H^{\dagger}\Delta\Delta^{\dagger}H$. Take
$\text{Tr}(\Delta^{\dagger}\Delta)\neq 0$ and define $r$, $\zeta$ and $\xi$ as
the following non-negative dimensionless quantities,

$H^{\dagger}H\equiv r\text{Tr}(\Delta^{\dagger}\Delta)~{},$ (42)
$\text{Tr}(\Delta^{\dagger}\Delta)^{2}\equiv\zeta[\text{Tr}(\Delta^{\dagger}\Delta)]^{2}~{},$
(43)
$H^{\dagger}\Delta\Delta^{\dagger}H\equiv\xi\text{Tr}(\Delta^{\dagger}\Delta)(H^{\dagger}H)~{}.$
(44)

For all scalar masses are positive, the potential does not classically drop to
another minimum, but this still leaves the possibility of tunneling to a
deeper minimum. In order for this not to happen, we must ensure that $V$ is
bounded, which is equivalent to requiring that the quartic part of the
potential in Eq. 41, $V^{(4)}$, must be positive for all non-zero field
values.

$\frac{V^{(4)}}{[\text{Tr}(\Delta^{\dagger}\Delta)]^{2}}=\lambda_{H}r^{2}+\lambda_{1}r+\lambda_{2}+\lambda_{3}\zeta+\lambda_{4}\xi
r~{}.$ (45)

For $V^{(4)}>0$ and $[\text{Tr}(\Delta^{\dagger}\Delta)]^{2}>0$, we have

$\frac{V^{(4)}}{[\text{Tr}(\Delta^{\dagger}\Delta)]^{2}}>0~{},$ (46)

so we can rewrite Eq. 45 as follow,

$f(r)\equiv ar^{2}+br+c>0~{},$ (47)

where
$a=\lambda_{H},b=\lambda_{1}+\lambda_{4}\xi,c=\lambda_{2}+\lambda_{3}\zeta$.
Obviously, Eq. 47 is regarded as a one-dimensional quadratic inequality.

(I) $b>0$

For $f(r)>0$ and $r\in(0,\infty)$, the parabola opens up, $a>0$, which is

$\lambda_{H}>0~{}.$ (48)

When $r\rightarrow 0$, we have $c>0$ and

$\lambda_{2}+\lambda_{3}\zeta>0~{}.$ (49)

In this case, Eq. 46 is apparently satisfied.

(II)$b<0$

Likewise, $a>0$ and $c>0$ still hold. At this time, due to $f(r)>0$, the
minimum must also satisfy this condition. For $f^{\prime}(r)=0$, it can be
obtained,

$2\lambda_{H}r+\lambda_{1}+\lambda_{4}\xi=0~{},~{}r=-\frac{\lambda_{1}+\lambda_{4}\xi}{2\lambda_{H}}~{}.$
(50)

Substitute the minimum point back to Eq.45,

$-(\lambda_{1}+\lambda_{4}\xi)^{2}+4\lambda_{H}(\lambda_{2}+\lambda_{3}\zeta)>0~{}.$
(51)

Simplified available

$-2\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3}\zeta)}<\lambda_{1}+\lambda_{4}\xi<2\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3}\zeta)}~{}.$
(52)

Since the case where $b$ is less than zero is considered here, it is only
necessary to consider that

$-2\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3}\zeta)}<\lambda_{1}+\lambda_{4}\xi~{},$
(53)

which is

$\lambda_{1}+\lambda_{4}\xi+2\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3}\zeta)}>0~{}.$
(54)

To sum up, we can make a summary and redefine Eq. 49 and Eq. 54,

$\displaystyle\lambda_{H}>0~{},$ (55)
$\displaystyle\lambda_{2}+\lambda_{3}\zeta\equiv F_{1}(\zeta)>0~{},$ (56)
$\displaystyle\lambda_{1}+\lambda_{4}\xi+2\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3}\zeta)}\equiv
F_{2}(\xi,\zeta)>0~{}.$ (57)

Since the value range of $\zeta$ is $2\xi^{2}-2\xi+1\leqslant\zeta\leqslant
1$, which is proofed in Appendix. B, and $F_{2}(\xi,\zeta)$ is monotonic about
$\zeta$ and $\xi$, we can take $\zeta=2\xi^{2}-2\xi+1$ and rewrite Eq. 55 as
the function $F(\xi)$

$F(\xi)=\lambda_{1}+\xi\lambda_{4}+2\sqrt{\lambda_{H}[\lambda_{2}+\lambda_{3}(2\xi^{2}-2\xi+1)]}~{}.$
(58)

Since $\xi\in[0,1]$ and $F(\xi)>0$, there are $F(0)>0$ and $F(1)>0$, i.e.

$F(0)=\lambda_{1}+2\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3})}>0~{},$ (59)
$F(1)=\lambda_{1}+\lambda_{4}+2\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3})}>0~{}.$
(60)

If $F(\xi)$ is a monotonic function, the above two conditions are sufficient.
On the contrary, it is necessary to find its minimum point. So we need to take
the first derivative of $F(\xi)$,

$F^{\prime}(\xi)=\lambda_{4}+\frac{2\lambda_{H}\lambda_{3}(2\xi-1)}{\sqrt{\lambda_{H}[\lambda_{2}+\lambda_{3}(2\xi^{2}-2\xi+1)]}}~{},$
(61)

whose second derivative is

$F^{\prime\prime}(\xi)=\frac{2\lambda_{H}^{2}\lambda_{3}(2\lambda_{2}+\lambda_{3})}{{\lambda_{H}[\lambda_{2}+\lambda_{3}(2\xi^{2}-2\xi+1)]}^{3/2}}~{}.$
(62)

It can be seen from the above formula that the sign of $F^{\prime\prime}(\xi)$
is the same as that of $\lambda_{3}$. If $\lambda_{3}>0$ and
$F^{\prime}(0)<0,~{}F^{\prime}(1)>0$, the minimum falls in $(0,1)$. Otherwise,
the minimum value is $F(0)$ or $F(1)$, and Eq. 59 with Eq. 60 is sufficient to
ensure $F(\xi)>0$. For $F^{\prime}(0)<0$ and $F^{\prime}(1)>0$, we have

$\displaystyle
F^{\prime}(0)=\lambda_{4}-\frac{2\lambda_{H}\lambda_{3}}{\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3})}}<0~{},$
(63) $\displaystyle
F^{\prime}(1)=\lambda_{4}+\frac{2\lambda_{H}\lambda_{3}}{\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3})}}>0~{}.$
(64)

Combining Eq. 63 and Eq. 64 gives

$2\lambda_{3}\sqrt{\lambda_{H}}>|\lambda_{4}|\sqrt{\lambda_{2}+\lambda_{3}}~{}.$
(65)

That is to say, if and only if
$2\lambda_{3}\sqrt{\lambda_{H}}>|\lambda_{4}|\sqrt{\lambda_{2}+\lambda_{3}}$,
we should calculate the minimum of $F(\xi)$ in $(0,1)$. Taking zero for Eq. 61
and solving the equation, the extreme point can be obtained as

$\xi_{0}=\frac{1}{2}-\frac{\lambda_{4}}{2\lambda_{3}}\sqrt{\frac{\lambda_{3}(2\lambda_{2}+\lambda_{3})}{8\lambda_{H}\lambda_{3}-\lambda_{4}^{2}}}~{}.$
(66)

Substituting $\xi_{0}$ back into Eq. 58, the $F(\xi)>0$ condition can be
obtained

$\displaystyle F(\xi_{0})$
$\displaystyle=\lambda_{1}+\frac{1}{2}\lambda_{4}+4\lambda_{H}\sqrt{\frac{\lambda_{3}(2\lambda_{2}+\lambda_{3})}{8\lambda_{H}\lambda_{3}-\lambda_{4}^{2}}}-\frac{\lambda_{4}^{2}}{2\lambda_{3}}\sqrt{\frac{\lambda_{3}(2\lambda_{2}+\lambda_{3})}{8\lambda_{H}\lambda_{3}-\lambda_{4}^{2}}}$
(67)
$\displaystyle=\lambda_{1}+\frac{1}{2}\lambda_{4}+\frac{1}{2}(8\lambda_{H}\lambda_{3}-\lambda_{4}^{2})\sqrt{\frac{2\frac{\lambda_{2}}{\lambda_{3}}+1}{8\lambda_{H}\lambda_{3}-\lambda_{4}^{2}}}$
$\displaystyle=\lambda_{1}+\frac{1}{2}\lambda_{4}+\frac{1}{2}\sqrt{(8\lambda_{H}\lambda_{3}-\lambda_{4}^{2})(2\frac{\lambda_{2}}{\lambda_{3}}+1)}>0~{}.$

In summary, the vacuum stability conditions are obtained as follow

$\lambda_{H},\lambda_{2}+\lambda_{3},2\lambda_{2}+\lambda_{3},\lambda_{1}+2\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3})},\lambda_{1}+\lambda_{4}+2\sqrt{\lambda_{H}(\lambda_{2}+\lambda_{3})}>0~{},$
(68)

and

$2\lambda_{3}\sqrt{\lambda_{H}}\leqslant\lvert\lambda_{4}\rvert\sqrt{\lambda_{2}+\lambda_{3}}~{}\text{or}~{}\lambda_{1}+\frac{1}{2}\lambda_{4}+\frac{1}{2}\sqrt{(8\lambda_{H}\lambda_{3}-\lambda_{4}^{2})(2\frac{\lambda_{2}}{\lambda_{3}}+1)}>0~{}.$
(69)

##  Appendix B B Proof of the properties of $\xi$ and $\zeta$

###  B.1 $0\leq\xi\leq 1$

We already know

$\Delta=\left(\begin{matrix}\Delta^{+}/\sqrt{2}&\Delta^{++}\\\
\Delta^{0}&-\Delta^{+}/\sqrt{2}\end{matrix}\right),~{}\Delta^{\dagger}=\left(\begin{matrix}\Delta^{-}/\sqrt{2}&\Delta^{0}\\\
\Delta^{--}&-\Delta^{-}/\sqrt{2}\end{matrix}\right)~{}.$ (70)

From the above formula, we can easily find

$\text{Tr}\Delta=0,~{}\text{Tr}\Delta^{\dagger}=0~{}.$ (71)

For any $2\times 2$ matrices $M$ and $N$, we have

$MN+NM=\bm{1}(\text{Tr}MN-\text{Tr}M\text{Tr}N)+M\text{Tr}N+N\text{Tr}M~{},$
(72)

where $\bm{1}$ is identity matrix, from which follows immediately

$\Delta\Delta^{\dagger}+\Delta^{\dagger}\Delta=\bm{1}\times\text{Tr}\Delta\Delta^{\dagger}~{}.$
(73)

Then multiply this by $H^{\dagger}$ on the left and right by $H$ to get

$H^{\dagger}\Delta\Delta^{\dagger}H+H^{\dagger}\Delta^{\dagger}\Delta
H=H^{\dagger}H\text{Tr}\Delta\Delta^{\dagger}~{}.$ (74)

Due to $H^{\dagger}\Delta^{\dagger}\Delta H\geq 0$, we can obtain

$H^{\dagger}H\text{Tr}\Delta\Delta^{\dagger}\geq
H^{\dagger}\Delta\Delta^{\dagger}H~{}.$ (75)

According to the definition of $\xi$, it is easy to know

$\xi\equiv\frac{H^{\dagger}\Delta\Delta^{\dagger}H}{H^{\dagger}H\text{Tr}\Delta\Delta^{\dagger}}\leq
1~{}.$ (76)

Since $H^{\dagger}H\text{Tr}\Delta\Delta^{\dagger}$ and
$H^{\dagger}\Delta\Delta^{\dagger}H$ are positive, $\xi$ must be trivially
greater than zero. Finally the two boundary values 0 and 1 are effectively
reached respectively when $H^{\dagger}\Delta=0$ and $\Delta H=0$. Thus

$0\leq\xi\leq 1~{}.$ (77)

###  B.2 $\frac{1}{2}\leq\zeta\leq 1$

Using Eq. 70 and considering their product

$\Delta\Delta^{\dagger}=\left(\begin{matrix}{\lvert\frac{\Delta^{+}}{\sqrt{2}}\rvert}^{2}+{\lvert\Delta^{++}\rvert}^{2}&\frac{\Delta^{+}\Delta^{0}}{\sqrt{2}}+\frac{\Delta^{++}\Delta^{-}}{\sqrt{2}}\\\
\frac{\Delta^{-}\Delta^{0}}{\sqrt{2}}-\frac{\Delta^{--}\Delta^{+}}{\sqrt{2}}&{\lvert\frac{\Delta^{+}}{\sqrt{2}}\rvert}^{2}+{\lvert\Delta^{0}\rvert}^{2}\end{matrix}\right)~{}.$
(78)

First, trace Eq. 78 and then square it, we have

$(\text{Tr}\Delta\Delta^{\dagger})^{2}=({\lvert\frac{\Delta^{+}}{\sqrt{2}}\rvert}^{2}+{\lvert\Delta^{++}\rvert}^{2})^{2}+({\lvert\frac{\Delta^{+}}{\sqrt{2}}\rvert}^{2}+{\lvert\Delta^{++}\rvert}^{2})^{2}+2({\lvert\frac{\Delta^{+}}{\sqrt{2}}\rvert}^{2}+{\lvert\Delta^{++}\rvert}^{2})({\lvert\frac{\Delta^{+}}{\sqrt{2}}\rvert}^{2}+{\lvert\Delta^{0}\rvert}^{2})~{}.$
(79)

Furthermore, square Eq. 78 and then find the trace to get

$\text{Tr}(\Delta\Delta^{\dagger})^{2}=({\lvert\frac{\Delta^{+}}{\sqrt{2}}\rvert}^{2}+{\lvert\Delta^{++}\rvert}^{2})^{2}+({\lvert\frac{\Delta^{+}}{\sqrt{2}}\rvert}^{2}+{\lvert\Delta^{++}\rvert}^{2})^{2}+2[\frac{\Delta^{+}\Delta^{0}}{\sqrt{2}}+\frac{\Delta^{++}\Delta^{-}}{\sqrt{2}}][\frac{\Delta^{-}\Delta^{0}}{\sqrt{2}}-\frac{\Delta^{--}\Delta^{+}}{\sqrt{2}}]~{}.$
(80)

Finally, subtract Eq. 79 and Eq. 80,

$\begin{aligned}
(\text{Tr}\Delta\Delta^{\dagger})^{2}-\text{Tr}(\Delta\Delta^{\dagger})^{2}&=2({\lvert\frac{\Delta^{+}}{\sqrt{2}}\rvert}^{2}+{\lvert\Delta^{++}\rvert}^{2})({\lvert\frac{\Delta^{+}}{\sqrt{2}}\rvert}^{2}+{\lvert\Delta^{0}\rvert}^{2})\\\
&-2[\frac{\Delta^{+}\Delta^{0}}{\sqrt{2}}+\frac{\Delta^{++}\Delta^{-}}{\sqrt{2}}][\frac{\Delta^{-}\Delta^{0}}{\sqrt{2}}-\frac{\Delta^{--}\Delta^{+}}{\sqrt{2}}]\\\
&=2\text{Det}\Delta\Delta^{\dagger}\end{aligned}~{}.$ (81)

Using $\text{Det}\Delta\Delta^{\dagger}\equiv|\text{Det}\Delta|^{2}\geq 0$ and
Eq. 81, it is easily to note that
$(\text{Tr}\Delta\Delta^{\dagger})^{2}\geq\text{Tr}(\Delta\Delta^{\dagger})^{2}$.
According to the definition of $\zeta$, we have

$\zeta\equiv\frac{\text{Tr}(\Delta\Delta^{\dagger})^{2}}{(\text{Tr}\Delta\Delta^{\dagger})^{2}}\leq
1~{},$ (82)

and $\zeta$ must be positive because $(\text{Tr}\Delta\Delta^{\dagger})^{2}$
and $\text{Tr}(\Delta\Delta^{\dagger})^{2}$ are both greater than zero. But in
fact, $\zeta$ cannot go below 1/2. To see this, we will rewrite $\zeta$ by
$M_{1}^{2}$ and $M_{2}^{2}$, the two (real and positive) eigenvalues of
$\Delta\Delta^{\dagger}$,

$\zeta=\frac{M_{1}^{4}+M_{2}^{4}}{(M_{1}^{2}+M_{2}^{2})^{2}}~{},$ (83)

and then divide it up and down by $M_{2}^{4}$, we can easily read the function
$\zeta(x)=(1+x^{2})/(1+x)^{2}$ where $x\equiv M_{1}^{2}/M_{2}^{2}\geq 0$ to
show that it has a minimum of $\zeta=1/2$ when $x=1$. Similarly, we calculated
that $\zeta\leq 1$ and reaches 1 at $x\rightarrow 0$ or $x\rightarrow\infty$.
Therefore,

$\frac{1}{2}\leq\zeta\leq 1~{}.$ (84)

###  B.3 Correlation between $\xi$ and $\zeta$

We first perform a general gauge transformation $H\rightarrow\mathcal{U}H$,
$\Delta\rightarrow\mathcal{U}\Delta\mathcal{U}^{\dagger}$, where $\mathcal{U}$
is any element of $SU(2)_{L}\times U(1)_{Y}$. Since $\mathcal{U}$ is unitary
and $\Delta\Delta^{\dagger}$ is hermitian, we can always find a gauge
transformation that diagonalizes $\Delta\Delta^{\dagger}$ for any given field
$\Delta$. According to the definition of $\xi$ and a series of calculation,
$\xi$ reads

$\xi=\frac{M_{2}^{2}|\tilde{\phi^{0}}|^{2}+M_{1}^{2}|\tilde{\phi^{+}}|^{2}}{(M_{1}^{2}+M_{2}^{2})(|\tilde{\phi^{0}}|^{2}+|\tilde{\phi^{+}}|^{2})}~{},$
(85)

where ${\rm diag}\\{M_{1},M_{2}\\}$ is the diagonalized matrix of $\Delta$ and
the tilde denotes the components of the transformed doublet
$\tilde{H}=\mathcal{U}H$. Furthermore, we define

$\displaystyle
c_{\Delta}^{2}\equiv\frac{M_{1}^{2}}{M_{1}^{2}+M_{2}^{2}},~{}s_{\Delta}^{2}\equiv\frac{M_{2}^{2}}{M_{1}^{2}+M_{2}^{2}}~{},$
(86) $\displaystyle
c_{H}^{2}\equiv\frac{|\tilde{\phi^{+}}|^{2}}{|\tilde{\phi^{0}}|^{2}+|\tilde{\phi^{+}}|^{2}},~{}s_{H}^{2}\equiv\frac{|\tilde{\phi^{0}}|^{2}}{|\tilde{\phi^{0}}|^{2}+|\tilde{\phi^{+}}|^{2}}~{},$
(87)

they scan all their allowed domains
$c_{\Delta}^{2}(\text{or}s_{\Delta}^{2}),c_{H}^{2}(\text{or}s_{H}^{2})\in[0,1]$.
Using these definitions, we can rewrite $\xi$ and $\zeta$ as

$\displaystyle\xi=$
$\displaystyle\frac{M_{1}^{2}\lvert\tilde{\phi^{+}}\rvert^{2}+M_{2}^{2}\lvert\tilde{\phi^{0}}\rvert^{2}}{(M_{1}^{2}+M_{2}^{2})(\lvert\tilde{\phi^{+}}\rvert^{2}+\lvert\tilde{\phi^{0}}\rvert^{2})}$
(88) $\displaystyle=$ $\displaystyle
c_{\Delta}^{2}c_{H}^{2}+s_{\Delta}^{2}s_{H}^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{2}[(c_{\Delta}^{2}+s_{\Delta}^{2})(c_{H}^{2}+s_{H}^{2})+(c_{\Delta}^{2}-s_{\Delta}^{2})(c_{H}^{2}-s_{H}^{2})]$
$\displaystyle=$ $\displaystyle\frac{1}{2}(1+c_{2\Delta}c_{2H})~{},$
$\displaystyle\zeta=$
$\displaystyle\frac{M_{1}^{4}+M_{2}^{4}}{(M_{1}^{2}+M_{2}^{2})^{2}}=(\frac{M_{1}^{2}}{M_{1}^{2}+M_{2}^{2}})^{2}+(\frac{M_{2}^{2}}{M_{1}^{2}+M_{2}^{2}})^{2}$
(89) $\displaystyle=$ $\displaystyle
c_{\Delta}^{4}+s_{\Delta}^{4}=\frac{1}{2}[(c_{\Delta}^{2}+s_{\Delta}^{2})^{2}+(c_{\Delta}^{2}-s_{\Delta}^{2})^{2}]$
$\displaystyle=$ $\displaystyle\frac{1}{2}(1+c_{2\Delta}^{2})~{},$

where we have defined $c_{2H}=c_{H}^{2}-s_{H}^{2}$,
$c_{2\Delta}=c_{\Delta}^{2}-s_{\Delta}^{2}$ with their range of variation
$c_{2H}\in[-1,1]$, $c_{2\Delta}\in[-1,1]$. Combining Eq. 88 and Eq. 89, and
then eliminating $c_{2\Delta}^{2}$, we can obtain

$2\xi^{2}-2\xi+\frac{c_{2H}^{2}+1}{2}=c_{2H}^{2}\zeta~{}.$ (90)

In other words, we have

$2\xi^{2}-2\xi+\frac{1}{2}=(\zeta-\frac{1}{2})c_{2H}^{2}~{}.$ (91)

Since the value of $c_{2H}^{2}\in[0,1]$, it is easy to get

$\zeta\geq 2\xi^{2}-2\xi+1~{}.$ (92)

It is obvious to find that $2\xi^{2}-2\xi+1\geq\frac{1}{2}$ with
$\xi\in[0,1]$, and as mentioned earlier, $\zeta\in[\frac{1}{2},1]$, so we can
obtain

$2\xi^{2}-2\xi+1\leqslant\zeta\leqslant 1~{}.$ (93)

##  Appendix C C 1-loop renormalization group equations

The 1-loop renormalization group equations are extracted below. For
simplicity, among the SM Yukawa couplings, we only retain the top coupling.
Similarly, we take the normal ordering assumption of neutrino mass spectrum
and neglect the contribution of the first 2 generation neutrino Yukawa
coupling. Here $t=\log\mu$.

$\displaystyle(4\pi)^{2}\frac{dg_{1}}{dt}=$
$\displaystyle\frac{47}{10}g_{1}^{3}~{},$
$\displaystyle(4\pi)^{2}\frac{dg_{2}}{dt}=$
$\displaystyle-\frac{5}{2}g_{2}^{3}~{},$
$\displaystyle(4\pi)^{2}\frac{dg_{3}}{dt}=$ $\displaystyle-7g_{3}^{3}~{},$
$\displaystyle(4\pi)^{2}\frac{dy_{t}}{dt}=$
$\displaystyle-\frac{17}{20}g_{1}^{2}y_{t}-\frac{9}{4}g_{2}^{2}y_{t}-8g_{3}^{2}y_{t}+\frac{9}{2}y_{t}^{3}~{},$
$\displaystyle(4\pi)^{2}\frac{dy_{\nu}}{dt}=$
$\displaystyle-\frac{9}{10}g_{1}^{2}y_{\nu}-\frac{9}{2}g_{2}^{2}y_{\nu}+2y_{\nu}^{3}~{},$
$\displaystyle(4\pi)^{2}\frac{d\lambda_{H}}{dt}=$
$\displaystyle\frac{27}{200}g_{1}^{4}+\frac{9}{20}g_{1}^{2}g_{2}^{2}+\frac{9}{8}g_{2}^{4}+3\lambda_{1}^{2}+3\lambda_{1}\lambda_{4}+\frac{5}{4}\lambda_{4}^{2}-\frac{9}{5}g_{1}^{2}\lambda_{H}-9g_{2}^{2}\lambda_{H}$
$\displaystyle+24\lambda_{H}^{2}+12\lambda_{H}y_{t}^{2}-6y_{t}^{4}~{},$
$\displaystyle(4\pi)^{2}\frac{d\lambda_{1}}{dt}=$
$\displaystyle\frac{27}{25}g_{1}^{4}-\frac{18}{5}g_{1}^{2}g_{2}^{2}+6g_{2}^{4}-\frac{9}{2}g_{1}^{2}\lambda_{1}-\frac{33}{2}g_{2}^{2}\lambda_{1}+4\lambda_{1}^{2}+16\lambda_{1}\lambda_{2}+12\lambda_{1}\lambda_{3}$
$\displaystyle+6\lambda_{2}\lambda_{4}+2\lambda_{3}\lambda_{4}+\lambda_{4}^{2}+12\lambda_{1}\lambda_{H}+4\lambda_{H}\lambda_{4}+6\lambda_{1}y_{t}^{2}+\lambda_{1}y_{\nu}^{2}~{},$
$\displaystyle(4\pi)^{2}\frac{d\lambda_{2}}{dt}=$
$\displaystyle\frac{54}{25}g_{1}^{4}+15g_{2}^{4}+2\lambda_{1}^{2}-24g_{2}^{2}\lambda_{2}+28\lambda_{2}^{2}-\frac{36}{5}g_{1}^{2}(g_{2}^{2}+\lambda_{2})+24\lambda_{2}\lambda_{3}$
$\displaystyle+6\lambda_{3}^{2}+2\lambda_{1}\lambda_{4}+2\lambda_{2}y_{\nu}^{2}~{},$
$\displaystyle(4\pi)^{2}\frac{d\lambda_{3}}{dt}=$
$\displaystyle\frac{72}{5}g_{1}^{2}g_{2}^{2}-6g_{2}^{4}-\frac{36}{5}g_{1}^{2}\lambda_{3}-24g_{2}^{2}\lambda_{3}+24\lambda_{2}\lambda_{3}+18\lambda_{3}^{2}+\lambda_{4}^{2}+2\lambda_{3}y_{v}^{2}-y_{\nu}^{4}~{},$
$\displaystyle(4\pi)^{2}\frac{d\lambda_{4}}{dt}=$
$\displaystyle\frac{36}{5}g_{1}^{2}g_{2}^{2}-\frac{9}{2}g_{1}^{2}\lambda_{4}-\frac{33}{2}g_{2}^{2}\lambda_{4}+8\lambda_{1}\lambda_{4}+4\lambda_{2}\lambda_{4}+8\lambda_{3}\lambda_{4}+4\lambda_{4}^{2}+4\lambda_{H}\lambda_{4}$
$\displaystyle+6\lambda_{4}y_{t}^{2}+\lambda_{4}y_{\nu}^{2}~{}.$