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\begin{document}


\title{O(\textit{N}) Phantom, A Way to Implement $w<-1$ }% Force line breaks with \\

\author{Xin-zhou Li}\email{kychz@shtu.edu.cn}
\author{Jian-gang Hao}
\affiliation{SUCA, Shanghai United Center for Astrophysics,
Shanghai
Normal University, Shanghai 200234, China}%


\date{\today}% It is always \today, today, but you may specify any date with \date.

\begin{abstract}
ABSTRACT: In this paper, we study the cosmological evolution of
imaginary configuration for scalar fields with an O(\textit{N})
symmetry and discuss that it is possible to realize the
equation-of-state parameter $w<-1$. We show that the introduction
of negative kinetic energy of scalar fields could be equivalently
realized by a vector anti-unitary representation of the
O(\textit{N}) group. When the fields are constrained with an
O(\textit{N}) symmetry, they are more stable than that without
symmetry. In the exponential potential, the O(\textit{N}) phantom
field imposes a lower bound for the equation of state $w>-2$. We
also derive the relations between the potential of the field and
the red-shift, known as reconstruction equations.
\end{abstract}

\pacs{04.40.-b, 98.80.Cq, 98.80.Es, 98.80.Hw} \maketitle

\vspace{0.4cm} \textbf{1. Introduction} \vspace{0.4cm}

Recent observation shows that our Universe is made up of two third
of dark energy whose effective equation-of-state parameter may
smaller than $-1$\cite{newobservation,hannestad,melchiorri}. While
the equation of state of conventional quintessence
models\cite{steinhardt}that based on a scalar field and positive
kinetic energy can not evolve to the the regime of $w<-1$.  Some
authors\cite{caldwell1,sahni,parker,chiba,boisseau,schulz,faraoni,maor,onemli,
torres,carroll,frampton,hao,caldwell2,gibbons} investigated a
phantom field model which has negative kinetic energy and can
realize the $w<-1$ in its evolution. Although the introduction of
a phantom field causes many theoretical problems such as the
violation of some widely accepted energy condition and the rapid
vacuum decay\cite{carroll}, it is still very interesting in term
of fitting current observations. Comparing with the other
approaches to realizing the $w<-1$, such as the modification of
Friedman equation, it seems economical.

In our pervious work, we investigated the scalar fields with
O(\textit{N}) symmetry as quintessence\cite{li}, which could be
viewed as the unitary representation of the O(\textit{N}) group
and has a positive kinetic energy. However, if we abandon the
prohibition of negative kinetic energy, as did in the phantom
models, we essentially work in the anti-unitary representation of
the O(\textit{N}) group, in which the corresponding configuration
of fields will be imaginary and thus naturally leads to a negative
kinetic energy term in the lagrangian of the system. we show that
the existence of "angular" motion of the field makes the phantom
fields more stable against the perturbation ( tachyon
instability). Especially, via a specific model in which the
potential is exponentially dependent on the field, we show that
the attractor property of the dynamical system and the
introduction of O(\textit{N}) symmetry leads to a lower bound on
the equation of state $w>-2$. Also, we derive the relation between
the red-shift and the scalar fields and the potential. It turns
out that the potential relates to the red-shift the same way as
that in ordinary scalar fields theory while the variation of the
fields do in a quite different manner.




\vspace{0.4cm} \textbf{2. O(N) Phantom}
 \vspace{0.4cm}



We start from the flat Robertson-Walker metric

\begin{equation}\label{metric}
ds^{2}=dt^{2}-a^{2}(t)(dr^{2}+r^{2}d\alpha^{2}+r^{2}sin^{2}\alpha
d\beta^{2})
\end{equation}

\noindent The Lagrangian density for the Phantom with
 O(\textit{N}) symmetry is

\begin{equation}
L_{\Phi}=\frac{1}{2}g^{\mu\nu}(\partial_{\mu}\Phi^{a})(\partial_{\nu}\Phi^{a})-V(|\Phi^{a}|)
\end{equation}

\noindent where $\Phi^{a}$ is the component of the scalar field,
$a=1,2,\cdots,N$. To make it possess a O(\textit{N}) symmetry, we
write it in the following form

\begin{eqnarray}\label{imag}
\Phi^{1}=iR(t)\cos\varphi_{1}(t)\hspace{4.2cm}\nonumber\\
\Phi^{2}=iR(t)\sin\varphi_{1}(t)\cos\varphi_{2}(t)\hspace{2.85cm}\nonumber\\
\Phi^{3}=iR(t)\sin\varphi_{1}(t)\sin\varphi_{2}(t)\cos\varphi_{3}(t)\hspace{1.5cm}\\
\cdots\cdots\hspace{4cm}\nonumber\\
\Phi^{N-1}=iR(t)\sin\varphi_{1}(t)\cdots\sin\varphi_{N-2}(t)\cos\varphi_{N-1}(t)\hspace{-0.35cm}\nonumber\\
\Phi^{N}=iR(t)\sin\varphi_{1}(t)\cdots\sin\varphi_{N-2}(t)\sin\varphi_{N-1}(t)\nonumber
\end{eqnarray}

\noindent Therefore, we have $|\Phi^{a}|=R$ and assume that the
potential of the O(\textit{N}) phantom depends only on \textit{R}.
The Lagrangian density then take the following form:

\begin{eqnarray}
L_{\Phi}&=-&\frac{1}{2}(\dot{R}^{2}+R^{2}\dot{\varphi}_{1}^{2}+R^{2}\sin^{2}\varphi_{1}\dot{\varphi}_{2}^{2}
+\cdots\hspace{-4cm}\nonumber\\
&+&R^{2}\sin^{2}\varphi_{1}\sin^{2}\varphi_{2}\cdots\sin^{2}\varphi_{N-2}
\dot{\varphi}_{N-1}^{2})-V(R)
\end{eqnarray}

\noindent where the dot denotes the derivative with respect to
$t$. The Eq.(\ref{imag}) is a basis of vector anti-unitary
representation of O(\textit{N}) group, which corresponds to the
generators
\begin{equation}\label{generator}
s_{ab}=e_{ab}-e_{ba}   \hspace{2cm}(a>b)
\end{equation}

\noindent Here, $e_{ab}$ is a matrix with a 1 on the intersection
of the $a$-th row and the $b$-th column and zeros elsewhere. The
multiplication rule for the $e_{ab}$ is $e_{ac}e_{cb}=e_{ab}$,
$e_{ac}e_{db}=0$ $(c\neq d)$. It suffices to know the commutators
\begin{equation}\label{comutator}
[s_{ab},s_{cd}]=\delta_{bc}s_{ad}+\delta_{ad}s_{bc}-\delta_{bd}s_{ac}-\delta_{ac}s_{bd}
\end{equation}
\noindent as one obtain using the multiplication rule for the
$e_{ab}$.

The action for the universe is :

\begin{equation}\label{action}
S=\int d^{4}x\sqrt{-g}(-\frac{1}{16\pi
G}R_{s}-\rho_{\gamma}+L_{\Phi})
\end{equation}

\noindent where $g$ is the determinant of the metric tensor $
g_{\mu\nu}$ , $G$ is the Newton's constant, $R_{s}$ is the Ricci
scalar, and $\rho_{\gamma}$ is the non-relativistic matter density
whose equation of state is $p_{\gamma}=\gamma\rho_{\gamma}$. Using
the metric tensor(\ref{metric}) and action (\ref{action}), we can
obtain the equation of motion for $\varphi_{1}$ , $\cdots $,
$\varphi_{N-1}$ as follows:

\begin{eqnarray}
\ddot{\varphi}_{1}+(2\frac{\dot{R}}{R}+3H)\dot{\varphi}_{1}-\cos\varphi_{1}\sin\varphi_{1}
(\dot{\varphi}^{2}_{2}+\sin^{2}\varphi_{2}\dot{\varphi}^{2}_{3}\hspace{1cm}\nonumber\\+\cdots+\sin^{2}\varphi_{2}
\sin^{2}\varphi_{3}\cdots\sin^{2}\varphi_{N-2}\dot{\varphi}^2_{N-1})=0\hspace{1cm}\\
\ddot{\varphi}_{2}+(2\frac{\dot{R}}{R}+3H+2\cot\varphi_{1}\dot{\varphi}_{1})\dot{\varphi}_{2}
-\cos\varphi_{2}\sin\varphi_{2}(\dot{\varphi}^{2}_{3}+\hspace{0.8cm}\nonumber\\\sin^{2}\varphi_{3}\dot{\varphi}^{2}_{4}
+\cdots+\sin^{2}\varphi_{3}\sin^{2}\varphi_{4}
\cdots\sin^{2}\varphi_{N-2}\dot{\varphi}^{2}_{N-1})=0\hspace{0.5cm}\\
\cdots\cdots\hspace{4cm}\nonumber\\
\ddot{\varphi}_{N-2}+(2\frac{\dot{R}}{R}+3H+2\cot\varphi_{1}\dot{\varphi}_{1}+\cdots+\hspace{2cm}\nonumber\\
2\cot\varphi_{N-3}\dot{\varphi}_{N-3})\dot{\varphi}_{N-2}
-\cos\varphi_{N-2}\sin\varphi_{N-2}\dot{\varphi}^{2}_{N-1}=0\hspace{0.5cm}\\
\ddot{\varphi}_{N-1}+(2\frac{\dot{R}}{R}+3H+2\cot\varphi_{1}\dot{\varphi}_{1}+\cdots+\hspace{2cm}\nonumber\\
2\cot\varphi_{N-2}\dot{\varphi}_{N-2})\dot{\varphi}_{N-1}=0\hspace{1.5cm}
\end{eqnarray}



we have $N-1$ independent first integral of the system of
equations

\begin{eqnarray}
\dot{\varphi}_{1}&=&(\Omega^{2}-\frac{\Omega^{2}_{1}}{\sin^{2}\varphi_{1}})^{\frac{1}{2}}
(a^{3}R^{2})^{-1}\\
\dot{\varphi}_{2}&=&(\Omega^{2}_{1}-\frac{\Omega^{2}_{2}}{\sin^{2}\varphi_{2}})^{\frac{1}{2}}
(a^{3}R^{2}\sin^{2}\varphi_{1})^{-1}
\end{eqnarray}

\begin{eqnarray}
&&\cdots\cdots\nonumber\\
\dot{\varphi}_{N-2}&=&(\Omega^{2}_{N-3}-\frac{\Omega^{2}_{N-2}}{\sin^{2}\varphi_{N-2}})^{\frac{1}{2}}
(a^{3}R^{2}\sin^{2}\varphi_{1}\cdots\nonumber\\\sin^{2}\varphi_{N-3})^{-1}\hspace{-6.3cm}\\
\dot{\varphi}_{N-1}&=&\Omega_{N-2}(a^{3}R^{2}\sin^{2}\varphi_{1}\cdots\sin^{2}\varphi_{N-2})^{-1}
\end{eqnarray}


\noindent where $\Omega$ , $\Omega_{1}$ , $\cdots$ ,
$\Omega_{N-2}$ are $N-1$ independent constants determined by the
initial condition of $\dot{\varphi}_{i}$ , $i=1,2,\cdots,N-1$. The
Einstein equations and the radial equation of scalar fields can be
written as

\begin{equation}
H^{2}=(\frac{\dot{a}}{a})^{2}=\frac{8\pi
G}{3}[\rho_{M}+\rho_{\Phi}]
\end{equation}

\begin{equation}
(\frac{\ddot{a}}{a})=-\frac{8\pi G}{3}[\frac{1}{2}\rho_{M}+2
p_{\Phi}+V(R)]
\end{equation}

\begin{equation}
\ddot{R}+3H\dot{R}-\frac{\Omega^{2}}{a^{6}R^{3}}-\frac{\partial
V(R)}{\partial R}=0
\end{equation}


\noindent where

\begin{equation}\label{rho11}
\rho_{\Phi}=-\frac{1}{2}(\dot{R}^{2}+\frac{\Omega^{2}}{a^{6}R^{2}})+V(R)
\end{equation}

\begin{equation}\label{p11}
p_{\Phi}=-\frac{1}{2}(\dot{R}^{2}+\frac{\Omega^{2}}{a^{6}R^{2}})-V(R)
\end{equation}

\noindent are the energy density and pressure of the $\Phi$ field
respectively, and $H$ is Hubble parameter. The equation-of-state
parameter for the O(\textit{N}) phantom is

\begin{equation}
w =\frac{p_\Phi}{\rho_\Phi}=
\frac{\dot{R}^{2}+\frac{\Omega^{2}}{a^{6}R^{2}}+2V(R)}{\dot{R}^{2}+
\frac{\Omega^{2}}{a^{6}R^{2}}-2V(R)}
\end{equation}

It is clear that the O(\textit{N}) phantom could realize the
equation-of-state parameter $w<-1$ which is equivalent to

\begin{equation}
0<\dot{R}^{2}+\frac{\Omega^{2}}{a^{6}R^{2}}<2V(R)
\end{equation}

\noindent where the term$\frac{\Omega^{2}}{a^{6}R^{2}}$ comes form
the "total angular motion". The most prominent feature of
O(\textit{N}) phantom is that it will not reduce to a cosmological
constant even when the $R$ is spatially uniform and
time-independent.


\vspace{0.4cm} \textbf{3.  Attractor Property and Stability of
O(\textit{N}) Phantom } \vspace{0.4cm}

In this section, we firstly investigate the attractor property of
the O(\textit{N}) phantom field in an exponential potential. To do
so, we rewrite the equations of motion as

\begin{equation}\label{sys1}
\dot{H}=-\frac{\kappa^2}{2}(\rho_\gamma+p_\gamma+\dot{R}^2 +
\frac{\Omega^2}{a^6R^2})
\end{equation}

\begin{equation}\label{sys2}
\dot{\rho_\gamma}=-3H(\rho_\gamma+p_\gamma)
\end{equation}


\begin{equation}\label{sys3}
\ddot{R}+3H\dot{R}-\frac{\Omega^2}{a^6R^3}-\lambda\kappa V(R)=0
\end{equation}

\begin{equation}\label{sys4}
H^2=\frac{\kappa^2}{3}[\rho_\gamma+\frac{1}{2}(\dot{R}^2+\frac{\Omega^2}{a^6R^2})+V(R)]
\end{equation}

Note that the equations of motion for the "angular" parts of the
scalar fields have been integrated by the first integrals and
their contribution is reduced to the terms containing $\Omega^2$
in Eqs.(\ref{sys1}),(\ref{sys3}) and (\ref{sys4}), see
Ref.\cite{li} for details. The potential $V(R)$ is exponentially
dependent on $R$ as $V(R)=V_0\exp(-\lambda\kappa R)$.

Now, we will introduce the following variables to obtain the
autonomous system for the above dynamical system. The variables
could be defined as :

\begin{eqnarray}\label{vari}
x=&&\frac{\kappa}{\sqrt{6}H}\dot{R}\\\nonumber
y=&&\frac{\kappa\sqrt{V(R)}}{\sqrt{3}H}\\\nonumber
z=&&\frac{\kappa}{\sqrt{6}H}\frac{\Omega}{a^3R}\\\nonumber
\xi=&&\frac{1}{\kappa R}\\\nonumber N=&&\log a
\end{eqnarray}



\noindent the equation system(\ref{sys1})-(\ref{sys4}) become the
following autonomous system:

\begin{equation}\label{auto1}
\frac{dx}{dN}=\frac{3}{2}x[\gamma(1+x^2-y^2+z^2)-2(x^2+z^2)]-(3x-\sqrt{6}z^2\xi+\sqrt{\frac{3}{2}}\lambda
y^2)
\end{equation}

\begin{equation}\label{auto2}
\frac{dy}{dN}=\frac{3}{2}y[\gamma(1+x^2-y^2+z^2)-2(x^2+z^2)]-\sqrt{\frac{3}{2}}\lambda
xy
\end{equation}

\begin{equation}\label{auto3}
\frac{dz}{dN}=-3z-\frac{3}{2}z[\gamma(1+x^2-y^2+z^2)-2(x^2+z^2)]-\sqrt{6}xz\xi
\end{equation}

\begin{equation}\label{auto4}
\frac{d\xi}{dN}=-\sqrt{6}\xi^2x
\end{equation}

Also, we have a constraint equation

\begin{equation}\label{constraint}
\Omega_R+\frac{\kappa^2\rho^2_\gamma}{3H^2}=1
\end{equation}

\noindent where
\begin{equation}\label{omegar}
\Omega_R=\frac{\kappa^2\rho^2_R}{3H^2}=y^2-x^2-z^2
\end{equation}

The equation of state for the O(\textit{N}) phantom could be
expressed in term of the new variables as

\begin{equation}\label{equaofstate}
 w_{\Phi}=\frac{p_{\Phi}}{\rho_{\Phi}}=\frac{x^2+y^2+z^2}{x^2-y^2+z^2}
\end{equation}

It is not difficult to prove that the point
$(x,y,z,\xi)=(-\frac{\lambda}{\sqrt{6}},
\sqrt{1+\frac{\lambda^2}{6}}, 0, 0)$ is a stable node of the
autonomous system when $\lambda^2<6$. This corresponds to a late
time attractor solution which is a phantom dominated epoch
$\Omega_R=1$ and an equation of state $w=-\frac{\lambda^2}{3}$.
Unlike the singlet phantom field in exponential\cite{hao}, the
introduction of internal symmetry imposes a lower bound to the
parameter $\lambda^2$ for attractor solution, which equivalent to
the equation of state $w>-2$. This is a very interesting
characteristic of the O(\textit{N}) phantom field.

Next, we study the stability of the O(\textit{N}) Phantom fields
against small perturbation. We work in synchronous gauge and write
the metric perturbation as
\begin{equation}
ds^2=dt^2-a^2(t)(\delta_{ij}-h_{ij})dx^{i}dx^{j}
\end{equation}

\noindent Fourier model of the radial part of the phantom field is
\begin{equation}\label{fourier}
R(t,\textbf{k})=\frac{1}{\sqrt{2\pi}}\int
R(t,\textbf{x})e^{-i\textbf{k}\cdot\textbf{x}}d^3x
\end{equation}

\noindent satisfies the equation of motion

\begin{equation}\label{perturbationeq}
\delta \ddot{R_k}+3H\delta
\dot{R_k}+(k^2+\frac{\Omega^2}{a^6R^4}-V''(R))=-\frac{1}{2}\dot{h}\dot{R}
\end{equation}

\noindent where the $h$ is the trace of $h_{ij}$ and the term
$\frac{\Omega^2}{a^6R^4}$ is resulted from the O(\textit{N})
symmetry. It is necessary to point out that the O(\textit{N})
symmetry is an internal symmetry so that the perturbation of
spacetime does not destroy the symmetry\cite{li}. The effective
mass for the perturbation is
$(k^2+\frac{\Omega^2}{a^6R^4}-V''(R))^{1/2}$, which will become
imaginary for long wave perturbations and thus make the system
unstable. One may avoid this instability by choosing the potential
so as $V''<0$, as suggested in Ref.\cite{carroll}. However, even
if the potential has positive curvature, contrasting to the single
scalar phantom model, the term $\frac{\Omega^2}{a^6R^4}$ in
Eq.(\ref{perturbationeq})make the model more stable than that of
real phantom model. When the potential is chosen as the
exponential potential $V(\phi)=V_0\exp(-\lambda\kappa \phi)$ as we
analyzed in the former part of this section, we will have
$[k^2+\frac{\Omega^2}{a^6R^4}-\lambda^2\kappa^2V_0\exp(-\lambda\kappa
\phi)]^{1/2}$. When the field evolves to its stable attractor
solution, the critical wave number of the perturbation should be
\begin{equation}\label{conditionus}
k_{critical}=\lambda H\sqrt{3+\frac{\lambda^2}{2}}
\end{equation}
\noindent When the wave number is greater than $k_{critcial}$, or
equivalently the wave length is less than the critical wave
length, the instability will not appear. Suppose that current
universe is at the attractor solution epoch, then we may estimate
the critical wave length of the perturbation should be about the
order of $10^{28}$ cm, which is even greater than the radius of
our observable universe. Note that the critical wave number for
the O(\textit{N}) phantom field in exponential potential is the
same as that of phantom field without O(\textit{N})
symmetry\cite{hao}. This is because that the angular contribution
$\frac{\Omega^2}{a^6R^4}$ tends to zero in the attractor regime.
But before the field evolves to the attractor solution, the
O(\textit{N}) phantom model is more stable than the phantom model
without O(\textit{N}) symmetry.

\vspace{0.4cm} \textbf{4. Reconstruction} \vspace{0.4cm}

Now, we will correlate the potential with the observable red-shift
of SNe Ia. To do so, following the earlier study in this
field\cite{turner2}, we introduce the quantity
\begin{equation}
  r(z)=\int^{t_0}_{t(z)}\frac{du}{a(u)}=\int^z_0\frac{dx}{H(x)}
\end{equation}

\noindent which is the Robertson-Walker coordinate distance to an
object at red-shift $z$. Also, we denote
\begin{equation}\label{rhom}
\rho_M=\Omega_M \rho_{crit}=\frac{3\Omega_M H_0^2(1+z)^3}{8\pi G}
\end{equation}

\noindent where $H_0$ is the present Hubble constant, $\Omega_M$
is the fraction of non-relativistic matter to the critical density
and $\rho_{crit}$ is the critical energy density of the universe.

We then readily have
\begin{equation}
  (\frac{\dot{a}}{a})^2=H(z)^2=\frac{1}{(dr/dz)^2}
\end{equation}

\begin{equation}
 \frac{\ddot{a}}{a}=\frac{1}{(dr/dz)^2}+(1+z)\frac{d^2r/dz^2}{(dr/dz)^3}
\end{equation}

\begin{equation}\label{relation}
  \frac{dz}{dt}=-(1+z)H(z)=-(1+z) \frac{dr}{dz}
\end{equation}

\noindent together with Eqs.(\ref{rho11}) and (\ref{p11}), one can
obtain the reconstruction equations as
\begin{equation}\label{vrecons}
V[R(z)]=\frac{1}{8\pi
G}\Big[\frac{3}{(dr/dz)^2}+(1+z)\frac{d^2r/dz^2}{(dr/dz)^3}
\Big]-\frac{3\Omega_M H_0^2(1+z)^3}{16\pi G}
\end{equation}

\begin{equation}\label{Rrecons}
\Big(\frac{dR}{dz}\Big)^2+\frac{\Omega^2}{R^2}(1+z)^4\Big(\frac{dr}{dz}
\Big)^2=\frac{(dr/dz)^2}{4\pi
G(1+z)^2}\Big[\frac{(1+z)(d^2/dz^2)}{(dr/dz)^3}+\frac{3}{2}(1+z)^3
\Big]
\end{equation}
\noindent Eq.(\ref{vrecons}) is the same as those ordinary
quintessence while the Eq.(\ref{Rrecons}) is different in that
there is a sign difference. The right hand side of the
Eq.(\ref{Rrecons}) is positive while it is negative in
conventional O(\textit{N}) quintessence model.


\vspace{0.4cm} \textbf{5. Conclusion and discussion}
\vspace{0.4cm}

Up to now, the observation data do not tell us what should be the
nature of dark energy. But the future observation will be helpful
to determine whether the dark energy is phantom, quintessence, or
cosmological constant. If the equation-of-state parameter $w<-1$
is completely confirmed by observations, then its implications to
fundamental physics would be astounding, since it cannot be
achieved with substance with canonical lagrangian. Phantom field
could be a good candidate for such substance because of its
negative kinetic energy and simplicity. In this paper, we study
 a class of new phantom models, in which the scalar field possesses a O(\textit{N}) internal
 symmetry. The negative kinetic energy can be naturally incorporated by considering the anti-unitary
representation of the O(\textit{N}) group. In a specific model, in
which the potential of the phantom fields is an exponential
potential, we show that there exists an attractor solution when
$\lambda^2<6$ and the attractor solution corresponds to a phantom
energy dominant phase and an equation-of-state parameter
$w=-\frac{\lambda^2}{3}$. Most strikingly, the introduction of
O(\textit{N}) symmetry imposes a restriction to the existence of
the attractor solution $\lambda^2<6$, which accordingly puts a
lower bound to the equation-of-state parameter $w>-2$.



\vspace{0.8cm}

This work was partially supported by National Nature Science
Foundation of China under Grant No. 19875016, and Foundation of
Shanghai Development for Science and Technology under Grant
No.01JC14035.


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