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\begin{document}
\begin{titlepage}

\begin{flushright}
\\
KUNS-1754\\
\today
\end{flushright}

\vspace{4ex}

\begin{center}
{\large \bf
$E_6$ Unification, Doublet-Triplet Splitting \\
and  Anomalous $U(1)_A$ Symmetry
}

\vspace{6ex}

\renewcommand{\thefootnote}{\alph{footnote}}
Nobuhiro Maekawa\footnote
{e-mail: maekawa@gauge.scphys.kyoto-u.ac.jp
}
and 
Toshifumi Yamashita\footnote{
email: yamasita@gauge.scphys.kyoto-u.ac.jp
}

\vspace{4ex}
{\it Department of Physics, Kyoto University, Kyoto 606-8502, Japan}\\
\end{center}

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\vspace{6ex}

%--------------------<<   abstract   >>--------------------
\begin{abstract}
We propose a natural Higgs sector in $E_6$ grand unified theory
(GUT) with anomalous $U(1)_A$ gauge symmetry. In the scenario, 
the doublet-triplet splitting can be realized
while the proton decay via dimension 5 operator is suppressed. 
The gauge coupling unification is also realized without finetuning.
The GUT scale is generally lower than the usual GUT scale 
$2\times 10^{16}$ GeV, the proton decay via dimension 6 operator
can be seen in near future experiment.
The lifetime of nucleon in the model is roughly estimated as 
$\tau_p(p\rightarrow e^+\pi^0)\sim 3\times 10^{33}$ years. 
It is shown that the Higgs sector is compatible with the matter sector
proposed by one of the authors, 
which reproduces the realistic quark and lepton mass matrices including
bi-large neutrino mixing angle. Combining the Higgs sector
and the matter sector, we can obtain the totally consistent $E_6$ GUT.
The input parameters are only 8 integer anomalous $U(1)_A$ charges  
(+3 for singlet Higgs) for 
the Higgs sector and 3 (half) integer charges for the matter sector. 



\end{abstract}

\end{titlepage}

%--------------------<<   section    >>--------------------
\section{Introduction}
Recently in a series of papers
\cite{maekawa,maekawa2,maekawa3} a scenario to construct a realistic 
 GUT  has been 
proposed within $SO(10)$ group. In the scenario, anomalous $U(1)_A$
gauge symmetry,\cite{U(1)} whose anomaly is cancelled by Green-Schwarz
mechanism,\cite{GS} plays a critical role and it has many interesting
features:
1) the interaction is generic, namely, all the interactions, which are
allowed by the symmetry, are introduced. Therefore, once we fix the field
contents with their quantum numbers(integer), all the interactions are
determined except the coefficients of order one;
2) it naturally solves the so-called 
doublet-triplet (DT) splitting problem, \cite{DTsplitting} 
using Dimopoulos-Wilczek(DW) mechanism
\cite{DW,BarrRaby,Chako,complicate};  
3) it reproduces the  realistic structure of quark and lepton 
mass matrices including neutrinos bi-large mixing,\cite{SK}
 using
Froggatt-Nielsen (FN) mechanism\cite{FN};  
4) the  anomalous $U(1)_A$ explains the hierarchical structure of 
the symmetry breaking scales  and the masses of 
heavy particles; 5) all the fields except those of the minimal 
SUSY standard model (MSSM) can become heavy;
6) the gauge couplings are unified just below the usual GUT scale 
$\Lambda_G\sim 2\times 10^{16}$ GeV;
7) in spite of the lower GUT scale, the proton decay via dimension six 
operator $p\rightarrow e^+\pi^0$ is still within experimental limit and 
we expect to observe proton decay  in near future;
8) the cutoff scale is lower than the Planck scale;
9) the $\mu$ problem is also solved.

An extension of the above $SO(10)$ model 
to $E_6$ gauge group has been done
 in the analysis of fermion masses\cite{BM}
and it has been found that $E_6$ is more economical in the sense that 
we have only to introduce  minimal matter fields 
three ${\bf 27}$s  for 3-family fermions instead of the matter contents 
of three ${\bf 16}$ plus one ${\bf 10}$ in $SO(10)$ case. 
Moreover, the charge assignment for realizing bi-large
neutrino mixing automatically satisfies the condition for weakening 
the flavor changing neutral
current (FCNC). Namely, the right-handed down quark and the left-handed
lepton of the first and second generations belong to a single multiplet
${\bf 27}$ thanks to the twisting family structure.\cite{Bando}

In the Higgs sector of $E_6$ unification, however, 
the situation is not so simple. The Higgs fields 
$\bar \Phi({\bf \overline{27}})$ and $\Phi({\bf 27})$ are needed to
break $E_6$ into $SO(10)$, in addition to the Higgs fields 
$A({\bf 78})$, whose VEV breaks $SO(10)$ into 
$SU(3)_C\times SU(2)_L\times SU(2)_R\times U(1)_{B-L}$, 
$\bar C({\bf \overline{27}})$ and  $C({\bf 27})$, whose VEVs break
$SU(2)_R\times U(1)_{B-L}$ into $U(1)_Y$.
The other VEVs 
$\VEV{\bar C}$, $\VEV{C}$, $\VEV{\bar\Phi}$ and $\VEV{\Phi}$ generally
destabilize the DW type VEV 
$\VEV{{\bf 1}_A}=\VEV{{\bf 16}_A}=\VEV{{\bf \overline{16}}_A}=0$, 
$\VEV{{\bf 45}_A}=\tau_2\times {\rm diag}
(v,v,v,0,0)$, which is required to realize
the DT splitting. Therefore we may have to kill 
the interaction between $A$ and $\bar\Phi\Phi$ as well as between
$A$ and $\bar CC$. However, if we forbid
the interactions between these Higgs fields in the superpotential,
then pseudo Nambu Goldstone (PNG) fields appear. 
Since the VEVs $|\VEV{\Phi({\bf 1,1})}|=|\VEV{\bar \Phi({\bf 1,1})}|$ break
$E_6$ into $SO(10)$, the Nambu-Goldstone (NG) modes 
${\bf 16}$, ${\bf \overline{16}}$ and ${\bf 1}$ of $SO(10)$ appear.
On the other hand, the VEVs 
$|\VEV{C({\bf 16,1})}|=|\VEV{\bar C({\bf \overline{16},1})}|$ also break 
$E_6$ into other $SO(10)'$, so 
${\bf 16}$, ${\bf \overline{16}}$ and ${\bf 1}$ of $SO(10)'$ again appear
as NG modes. In the language of usual $SU(5)$,
these NG modes represent as $({\bf 10}+{\bf \bar 5}+{\bf 1})$, 
$({\bf \overline{10}}+{\bf 5}+{\bf 1})$ and ${\bf 1}$.
The VEV of the adjoint field 
$\VEV{{\bf 45}_A}=\tau_2\times {\rm diag}(v,v,v,0,0)$
breaks $E_6$ into $SU(3)_C\times SU(2)_L\times SU(2)_R\times U(1)_{B-L}
\times U(1)_{V'}$,
the NG modes by this breaking are ${\bf 16}$ and ${\bf \overline{16}}$
 of $SO(10)$ and $({\bf 3,2})_{\frac{1}{6}}+
({\bf \bar 3,1})_{-\frac{2}{3}}+({\bf 3,2})_{-\frac{5}{6}}+h.c.$
of the standard gauge group 
$SU(3)_C\times SU(2)_L\times U(1)_Y$. 
Since the part of these modes are absorbed by Higgs mechanism,
the remaining pseudo Nambu-Goldstone modes become
${\bf 10+\overline{10}}+2\times {\bf 5}+2\times{\bf \bar 5}+4\times {\bf 1}$ 
of $SU(5)$
and $({\bf 3,2})_{\frac{1}{6}}+({\bf \bar 3,1})_{-\frac{2}{3}}+h.c.$
of the standard gauge group. If all these PNG modes have only tiny masses
around the weak scale, then not only coupling unification is destroyed but
also the gauge couplings are diverging below the GUT scale.
Therefore, we have to give these PNG fields superheavy masses.
However, in order to give them superheavy masses, we have to introduce
some interaction between Higgs fields, and this requirement is opposite
one to stabilize the DW type of VEV. Such a conflict is similar to that
in $SO(10)$ case. 

This paper aims a unified description of Higgs sector in $E_6$ model, where
the above problem and the DT splitting problem are solved.
It may seem that $SO(10)$ unified models are already appealing.
However,  if we further proceed to $E_6$ we have more  advantages
in addition to the natural FCNC suppression as follows:
\begin{enumerate}
\item  The FN field 
is naturally provided  as an composite operator $\VEV{\bar \Phi\Phi}$,
where $\Phi$ and $\bar \Phi$  are  needed to break 
$E_6$ to $SO(10)$;
\item The usual doublet Higgs field $H$ is already included 
in the field 
$\Phi$.
\item In the Higgs sector the condition 
for the unification of gauge 
coupling constants automatically provides the  ``R parity" 
in terms of anomalous $U(1)_A$ naturally and so we do not have to 
introduce additional R parity.
\end{enumerate}
Moreover, we can construct a totally consistent and realistic
$E_6$ GUT scenario by combining this Higgs sector and 
the matter sector.


After explaining how the vacuum in  Higgs sector is determined
by anomalous $U(1)_A$ charges (section 2) and 
making a quick review of $SO(10)$ 
model(section 3), we shall explain how the above good features in Higgs 
sector are naturally obtained in the $E_6$ unification (section 4) and
a totally consistent $E_6$ GUT scenario(section 5).  


\section{Vacuum Determination}
Here we explore some general structures of VEVs which are 
determined from the superpotential of Higgs sector. 
The Higgs sector is the most unknown part and usually the VEVs of 
Higgs fields are introduced as input parameters, since 
general forms of potential  has  been usually too arbitrary. 
However anomalous $U(1)_A$ provides us with very strong constraints 
to the superpotential  $W$  and thus dictates the scales of the system 
in a quite definite way.   
In our analysis, the supersymmetry (SUSY) is essential because the 
observation of the following vacuum structure comes   
from the analytic property of SUSY theory. 

Let us study the simplest case 
where  all the fields
are gauge singlet fields 
$Z_i^\pm$ ($i=1,2,\cdots n_\pm$)
 with charges $z_i^\pm$ ($z_i^+>0$ and $z_i^-<0$).
Through this paper  we use a unit in which the cutoff $\Lambda=1$ and denote 
all the
superfields with uppercase letters and their anomalous $U(1)_A$ charges
with the corresponding lowercase letters.

We show that none of the fields with positive anomalous
$U(1)_A$ charge acquires non-zero VEV if the FN
mechanism
\cite{FN}
acts effectively in the vacuum.
From the $F$-flatness conditions of the superpotential,
we get $n=n_++n_-$ equations  plus one $D$-flatness condition,
\begin{equation}
 F_{Z_i}\equiv\frac{\partial  W}{\partial Z_i}=0, \qquad D_A=g_A
      \left(\sum_i z_i |Z_i|^2 +\xi^2 \right)=0,
\label{eq:fflat}
\end{equation}
where 
$\xi^2$
is the coefficient of Fayet-Iliopoulos (FI) $D$-term.
\footnote{
In weakly coupled Heterotic string theory, this FI $D$-term can be
induced by stringy loop correction;
$\xi^2=\frac{g_s^2\tr Q_A}{192\pi^2}$.}
The above equations may seem to be over determined. However,
 the $F$-flatness
conditions are not independent because of the gauge invariance:
\begin{equation}
\frac{\partial  W}{\partial Z_i}z_iZ_i=0.
\label{constraint}
\end{equation}
Therefore, generically a SUSY vacuum with $\VEV{Z_i}\sim \Lambda$ 
exists
(Vacuum A),
because the  coefficients of the above  are generically
of order 1. 
However, if $n_+\leq n_-$, we can choose another vacuum (Vacuum B)
with $\VEV{Z_i^+}=0$, which automatically satisfy the $F$-flatness
conditions $F_{Z_i^-}=0$ since they contain
at least one field with positive charge. Then the $\VEV{Z_i^-}$
are determined by the $F$-flatness conditions
$F_{Z_i^+}=0$ with the constraint
(\ref{constraint}) and the $D$-flatness condition $D_A=0$.
Note that if $\xi<1$,
the VEVs of $Z_i^-$ are less than the
cutoff scale. 
This can lead to the FN mechanism.
At this stage, among fields $Z_i^-$, we define the FN field 
$\Theta$ as the field 
whose VEV mainly compensates the FI parameter $\xi$.
If we fix the normalization of $U(1)_A$ charge  so that
$\theta=-1$,
then from $D_A=0$ the VEV of the FN field $\Theta$ is determined  as 
\begin{equation}
\VEV{\Theta}\equiv\lambda\sim \xi,
\label{lambda}
\end{equation}
which breaks $U(1)_A$ gauge symmetry. 
On the other hand,
other VEVs are determined by the $F$-flatness conditions with 
respect to  $Z_i^+$ as
$\VEV{Z_i^-}\sim \lambda^{-z_i^-}$ which is shown below.
Since $\VEV{Z_i^+}=0$, it is sufficient to examine the terms linear
in $Z_i^+$ in the superpotential in order to determine 
$\VEV{Z_i^-}$. Therefore, in general
the superpotential to determine the VEVs can be written
\begin{eqnarray}
W&=&\sum_i^{n_+}W_{Z_i^+}, \label{W}\\
W_{Z_i^+}&=& \lambda^{z_i^+}Z_i^+\left(1+\sum_j^{n_-}\lambda^{z_j^-}Z_j^-
+\sum_{j,k}^{n_-}\lambda^{z_j^-+z_k^-}Z_j^-Z_k^-+\cdots \right) \nn \\
&=&\tilde Z_i^+\left(1+\sum_j^{n_-}\tilde Z_j^-
+\sum_{j,k}^{n_-}\tilde Z_j^-\tilde Z_k^-+\cdots \right),
\end{eqnarray}
where $\tilde Z_i\equiv \lambda^{z_i}Z_i$.
The $F$-flatness conditions of the $Z_i^+$ fields require
\begin{equation}
\lambda^{z_i^+}\left(1+\sum_j\tilde Z_j^-+\cdots\right)=0,
\end{equation}
which generally lead to solutions $\tilde Z_j^-\sim O(1)$
if these $F$-flatness conditions determine the VEVs.
Thus the F-flatness condition requires
\begin{equation}
   \VEV{ Z_j^-} \sim O(\lambda ^{-z_j^-}).
\label{VEV}
\end{equation}
Note that if there is another field $Z_i^-$ which has larger charge
than the FN field $\Theta$, then the VEV of $Z_i^-$ becomes
larger than $\xi$, that is inconsistent with $D_A=0$.
Therefore it is natural that the field with the 
largest negative
charge becomes the FN field.
Note that if $n_+=n_-$, generically all the VEVs of $Z_i^-$ are
fixed, therefore there appears no flat direction in the potential.
So in this case there is no massless field.
On the other hand, if $n_+<n_-$, generally the $n_+$ equations of
$F$-flatness and $D$-flatness conditions do not determine all the
VEVs of $n_-$ fields $Z_i^-$. Therefore, there are flat directions
in the potential, producing some massless fields.
Thus, if we want to realize the case with no massless mode
in the Higgs sector, $n_+=n_-$ must be imposed in Higgs
sector.\footnote{
This rough argument of number counting is based on an assumption 
that the Higgs sector has no other structure by which the freedom 
of $F$-flatness conditions is reduced. Such a situation occurs easily 
 by imposing some symmetry, for example, $Z_2$ parity or R parity.
}

If the Vacuum A is selected, the anomalous $U(1)_A$ gauge symmetry
is broken at the cutoff scale, and the FN mechanism does not act.
Therefore, we cannot know the existence of the $U(1)_A$ gauge symmetry
from the low energy physics. On the other hand, if the Vacuum B is
selected, the FN mechanism acts effectively and we can recognize
the signature of the $U(1)_A$ gauge symmetry from the low energy
physics. Therefore, it is reasonable to assume that Vacuum B is
selected in our scenario, in which the $U(1)_A$ gauge symmetry
plays an important role in the FN mechanism. The VEVs of
the fields $Z_i^+$ vanish, which guarantees that the SUSY zero 
mechanism\footnote{
Note that if total charge
of an operator is negative, the $U(1)_A$ invariance and analytic property 
of the superpotential forbids 
the operator in the superpotential since the field $\Theta$ with negative 
charge cannot compensate for the negative total charge of the operator 
(SUSY zero mechanism). } acts effectively.



Up to now we have examined the VEVs of only singlet fields.
Here we consider the case where there are non-trivial representation 
of the gauge group. 
The same discussion can be applied if we take a set of the independent gauge 
invariant operators 
instead of the gauge singlet fields $Z_i$. The gauge invariant
operator $O$ with negative charge $o$ has
non-vanishing VEV $\VEV{O}\sim \lambda^{-o}$ if the $F$-flatness
conditions determine the VEV. For example, let us introduce fundamental
representation $C({\bf 27})$ and $\bar C({\bf \overline{27}})$ of $E_6$. 
The VEV of the gauge singlet operator $\bar CC$ is estimated as
$\VEV{\bar CC}\sim \lambda^{-(c+\bar c)}$. The essential difference
appears in the $D$-flatness condition
of $E_6$ gauge theory, which requires 
\begin{equation} 
\left|\VEV{C}\right|=\left|\VEV{\bar C}\right|\sim \lambda^{-(c+\bar c)/2}.
\end{equation}
Note that these VEVs are also determined by the anomalous $U(1)_A$
charges but they are different from the naive expectation
$\VEV{C}\sim\lambda^{-c}$.
This is because the $D$-flatness condition strongly constraints 
the VEVs of non-singlet fields.
One more important $D$-flatness condition is
\begin{equation}
D_A=g_A\left(\xi^2+\Sigma_i\phi_i|\Phi_i|^2\right)=0.
\end{equation}
The argument of the singlet fields cannot apply directly to the case of the
non-trivial representation fields $\Phi_i$.
For example, if we adopt the fields $\Phi({\bf 27})$ and 
$\bar \Phi({\bf \overline{27}})$ of $E_6$ with the charges
 $\phi+\bar \phi=-1$, then the above $D$-flatness condition of
 the anomalous $U(1)_A$ gauge symmetry requires
$\xi^2+\phi|\Phi|^2+\bar \phi |\bar \Phi|^2=0$. The $D$-flatness 
condition of $E_6$ gauge group leads to $|\Phi|=|\bar \Phi|$, therefore
we obtain $\xi^2+(\phi+\bar \phi)|\Phi|^2=0$, namely, 
$|\VEV{\Phi}|=\left|\VEV{\bar \Phi}\right|=\xi$. 
In this case, since $\bar \Phi\Phi$ plays the same role as $\Theta$,
the unit of hierarchy becomes $\VEV{\bar \Phi\Phi}=\lambda\sim \xi^2$, which
is different from the previous relation (\ref{lambda}).
It means that even if $\xi$ has
milder hierarchy, the unit of hierarchy becomes stronger.

Of course we can determine the VEVs of the singlet operators
from the same superpotential as (\ref{W}),
replacing the singlet fields $Z_i$ by a set of independent gauge singlet 
operators. This is actually  not easy to be found.
However, the situation can be simplified, if all the fields 
$\Phi_i^+$(including non singlet) 
with positive charges have
vanishing VEVs.
We can obtain the superpotential to determine the VEVs as
\begin{equation}
W=\sum_i^{n_+}W_{\Phi_i^+},
\end{equation}
where $W_X$ denotes the terms linear in the $X$ field.
Note that generally fields with positive charges can have non-vanishing
VEVs, even if all the gauge singlet operators with positive charges have
vanishing VEVs. For example, if we take $\phi=-3$ and $\bar \phi=2$,
then the gauge singlet operator $\bar \Phi\Phi$ can have non-vanishing VEV,
that means that $\bar \Phi$ with positive charge $\bar \phi=2$ has 
non-vanishing VEV. In such cases, it is not guaranteed that 
the $F$-flatness conditions of fields with negative charges are
automatically satisfied. Therefore we have to take account of the 
superpotential $W(\bar \Phi)$ which includes the positive charged fields 
$\bar \Phi$ with non-vanishing VEV.
We shall see such examples later.

In summary, 
\begin{enumerate}
\item
Gauge singlet operators with positive total charge have vanishing VEVs,
in order that the FN mechanism acts effectively. This guarantees that 
the SUSY zero mechanism works well.
\item
The singlet operator $\Theta$ with the largest negative charge becomes the 
FN field. When the singlet operator is just a singlet field, 
the VEV $\VEV{\Theta}\sim \xi$, which is determined from
$D_A=0$. When the singlet operator is a composite operator 
$\Theta\sim \bar \Phi\Phi$, the VEV 
$\VEV{\Theta}=\xi^2$.
\item
The $F$-flatness conditions of singlet operators with positive 
charges determine the 
VEVs of singlet operators $O$ with negative charges $o$ as
$\VEV{O}\sim \lambda^{-o}$, while the $F$-flatness conditions of the singlet
operators with negative charges are automatically satisfied.
When the operator is a composite operator $O\sim \bar CC$, the
$D$-flatness condition requires 
$\left|\VEV{C}\right|=\left|\VEV{\bar C}\right|\sim 
\lambda^{-(c+\bar c)/2}$.
\item
If the number of the independent singlet operators(moduli) with 
positive charges equals to that of the fields with negative charges, 
generically no massless field appears.
\item
General superpotential to determine the VEVs is expressed 
as $W=\sum_i W_{O_i^+}$, where $W_{O_i^+}$ which is linear in the independent
singlet operator $O_i^+$ with positive charges.
When all the fields $\Phi^+$ (including non singlet) with positive charges
have vanishing VEVs, the superpotential can be written
$W=\sum_i^{n_+}W_{\Phi_i^+}$. If some of the positive charged fields 
$\bar \Phi$ have
non-vanishing VEVs, the superpotential $W_{NV}$ must be added, which includes
only the fields with non-vanishing VEVs.
\end{enumerate}

\section{Doublet-triplet splitting in $SO(10)$ GUT}
Here we make a quick review of $SO(10)$  unified scenario 
proposed by one of the authors
\cite{maekawa,maekawa3}.


\subsection{Alignment and DT splitting}
The contents of the Higgs sector in $SO(10)\times U(1)_A$ are 
listed in Table I. 
\vspace{3mm}
\begin{center}
Table I. The typical values of anomalous $U(1)_A$ charges are listed.

\begin{tabular}{|c|c|c|} 
\hline
                  &   non-vanishing VEV  & vanishing VEV \\
\hline 
{\bf 45}          &   $A(a=-1,-)$        & $A'(a'=3,-)$      \\
{\bf 16}          &   $C(c=-3,+)$        
                  & $C'(c'=2,-)$      \\
${\bf \overline{16}}$&$\bar C(\bar c=0,+)$ 
                  & $\bar C'(\bar c'=5,-)$ \\
{\bf 10}          &   $H(h=-3,+)$        & $H'(h'=4,-)$      \\
{\bf 1}           &$\Theta(\theta=-1,+)$,$Z(z=-2,-)$,
                  $\bar Z(\bar z=-2,-)$& $S(s=3,+)$ \\
\hline
\end{tabular}

\vspace{5mm}
\end{center}
Here the symbols $\pm$ denote $Z_2$ parity.
The adjoint Higgs
field $A$, whose VEV 
$\VEV{A({\bf 45})}_{B-L}=\tau_2\times {\rm diag}
(v,v,v,0,0)$ breaks $SO(10)$ into
$SU(3)_C\times SU(2)_L\times SU(2)_R\times U(1)_{B-L}$.
 This Dimopoulos-Wilczek form of the VEV plays an
important role in solving the DT splitting problem.
The spinor Higgs fields 
$C$ and $\bar C$ break $SU(2)_R\times U(1)_{B-L}$ into $U(1)_Y$ 
by developing $\VEV{C}(=\VEV{\bar C}=\lambda^{-(c+\bar c)/2}$).
The  Higgs field $H$ contains usual $SU(2)_L$ doublet.
The gauge singlet operators $A^2$, $\bar CC$ and $H^2$ must
have negative total anomalous $U(1)_A$ charges
to obtain non-vanishing VEVs
as discussed 
in the previous section.
On the other hand, in order to give masses to all the Higgs fields,
we have introduced the same number of fields with positive charges\footnote{
Strictly speaking, since some of the Higgs fields are eaten by 
Higgs mechanism, in principle, less number of positive fields can give 
superheavy masses
to all the Higgs fields. Here we do not examine the possibilities.},
$A^\prime$, $C^\prime$, $\bar C^\prime$ and
$H^\prime$.
This is, in a sense, a minimal set of Higgs contents.

Following the general argument in the previous section,
the superpotential required by determination of the VEVs can be written
\begin{equation}
W=W_{H^\prime}+ W_{A^\prime} + W_S + W_{C^\prime}+W_{\bar C^\prime}
+W_{NV}.
\end{equation}
Here $W_X$ denotes the terms linear in the positive charged field
$X$, which has vanishing VEV.  Note, however, that terms 
including two
fields with vanishing VEVs like 
$\lambda^{2h^\prime}H^\prime H^\prime$
give contributions to the mass terms but not to the VEVs.
All the terms in $W_{NV}$ contains only the fields with non-vanishing 
VEVs. In the typical charge assignment, 
it is easily checked that they do not play a significant role in our 
argument since they
do not include the products of only the neutral components under the
standard gauge group. In the following argument, for simplicity, we
neglect the terms which do not include the products of only
neutral components under the standard gauge group like
${\bf 16}^4$, ${\bf \overline{16}}^4$, ${\bf 10\cdot 16}^2$,
 ${\bf 10 \cdot \overline{16}}^2$ and ${\bf 1\cdot 10}^2$, 
 even if these terms
are allowed by the symmetry.\footnote{ 
 It is easy to include these terms in
our analysis. Then these terms can give some constraints 
to the other vacua than the standard vacuum, but not to the standard vacuum.
}

We now discuss the determination of the VEVs.
If $-3a\leq a^\prime < -5a$,
the superpotential $W_{A^\prime}$ is in general
written as
\begin{equation}
W_{A^\prime}=\lambda^{a^\prime+a}\alpha A^\prime A+\lambda^{a^\prime+3a}(
\beta(A^\prime A)_{\bf 1}(A^2)_{\bf 1}
+\gamma(A^\prime A)_{\bf 54}(A^2)_{\bf 54}),
\end{equation}
where the suffixes {\bf 1} and {\bf 54} indicate the representation 
of the composite
operators under the $SO(10)$ gauge symmetry, and $\alpha$, $\beta$ and 
$\gamma$ are parameters of order 1. Here we assume 
$a+a^\prime+c+\bar c<0$
to forbid the term $\bar C A^\prime A C$, which destabilizes the 
DW form of the VEV $\VEV{A}$. 
The $D$-flatness requires the VEV
$\VEV{A}=\tau_2\times {\rm diag}(x_1,x_2,x_3,x_4,x_5)$, and the $F$-flatness
of the $A^\prime$ field requires
$x_i(\alpha\lambda^{-2a}
+(2\beta-\frac{\gamma}{5})(\sum_j x_j^2)+\gamma x_i^2)=0$, 
which gives only two solutions $x_i^2=0$, 
$-\frac{\alpha}{(1-\frac{N}{5})\gamma+2N\beta}\lambda^{-2a}$. 
Here $N=0$ - 5 is the number of $x_i \neq 0$ solutions.
The DW form is obtained when $N=3$.
Note that the higher terms $A^\prime A^{2L+1}$ $(L>1)$ are forbidden by
the SUSY zero mechanism. If they are allowed, 
the number of possible VEVs other than the DW form
becomes larger, and thus it becomes less natural to obtain the DW form. 
This is a
critical point of this mechanism, and the anomalous $U(1)_A$ gauge 
symmetry
plays an essential role to forbid the undesired terms.
In this way, the scale of the VEV is automatically
determined by the anomalous $U(1)_A$ charge of $A$, as noted in the 
previous section.


Next we discuss the $F$-flatness condition of $S$, which determines
the scale of the VEV $\VEV{\bar C C}$. 
$W_S$ is given by
\begin{equation}
W_S=\lambda^{s+c+\bar c}S\left((\bar CC)+\lambda^{-(c+\bar c)}
+\sum_k\lambda^{-(c+\bar c)+2ka}A^{2k}\right)
\end{equation}
if 
$s\geq -(c+\bar c)$.
Then the $F$-flatness condition of $S$ implies $\VEV{\bar CC}\sim 
\lambda^{-(c+\bar c)}$, and the $D$-flatness condition requires 
$|\VEV{C}|=|\VEV{\bar C}|\sim \lambda^{-(c+\bar c)/2}$.
The scale of the VEV is determined only by the charges of 
$C$ and $\bar C$ again.
If we take $c+\bar c=-3$, then we obtain the VEVs of the fields 
$C$ and $\bar C$
as 
$\lambda^{3/2}$, which differ from the expected values $\lambda^{-c}$ 
and
$\lambda^{-\bar c}$ if $c\neq \bar c$.

Next, we discuss the $F$-flatness of $C^\prime$ and 
$\bar C^\prime$,
which realizes the alignment of the VEVs $\VEV{C}$ and $\VEV{\bar C}$
and imparts masses on the PNG fields. 
This simple mechanism
was proposed by Barr and Raby.
\cite{BarrRaby}
We can easily assign anomalous $U(1)_A$ charges which allow the 
following superpotential:
\begin{eqnarray}
W_{C^\prime}&=&
       \bar C(\lambda^{\bar c^\prime +c+a}A
       +\lambda^{\bar c^\prime +c+\bar z}\bar Z)C^\prime, \\
W_{\bar C^\prime}&=&
       \bar C^\prime(\lambda^{\bar c^\prime +c+a} A
       +\lambda^{\bar c^\prime +c+z}Z)C.
\end{eqnarray}
The $F$-flatness conditions $F_{C^\prime}=F_{\bar C^\prime}=0$ give
$(\lambda^{a-z} A+Z)C=\bar C(\lambda^{a-\bar z} A+\bar Z)=0$. 
Recall that the VEV of $A$ is 
proportional to the $B-L$ generator $Q_{B-L}$ as 
$\VEV{A}=\frac{3}{2}vQ_{B-L}$, and that
 $C$, ${\bf 16}$, is decomposed into 
$({\bf 3},{\bf 2},{\bf 1})_{1/3}$, 
$({\bf \bar 3},{\bf 1},{\bf 2})_{-1/3}$, 
$({\bf 1},{\bf 2},{\bf 1})_{-1}$ and $({\bf 1},{\bf 1},{\bf 2})_{1}$ 
under
$SU(3)_C\times SU(2)_L\times SU(2)_R\times U(1)_{B-L}$.
Since $\VEV{\bar CC}\neq 0$, 
 $Z$ is fixed to be $Z\sim -\frac{3}{2}\lambda v Q_{B-L}^0$, 
where $Q_{B-L}^0$ is
the $B-L$ charge of the component of $C$ which has non-vanishing VEV. 
Once the VEV of $Z$ is determined, 
no other component fields can have non-vanishing VEVs 
because they have different charges $Q_{B-L}$. 
If  the $({\bf 1},{\bf 1},{\bf 2})_1$ field obtains
a non-zero VEV (therefore, $\VEV{Z}\sim -\frac{3}{2}\lambda v$), then the 
gauge group 
$SU(3)_C\times SU(2)_L\times SU(2)_R\times U(1)_{B-L}$ is broken to the 
standard gauge group. Once the direction of the VEV $\VEV{C}$ is 
determined, the VEV $\VEV{\bar C}$ must have the same direction 
because of the $D$-flatness
condition. Therefore, $\VEV{\bar Z}\sim -\frac{3}{2}\lambda v$.

Finally the $F$-flatness condition of $H^\prime$ is examined. 
$W_{H^\prime}$ is
written
\begin{equation}
W_{H^\prime}=\lambda^{h+a+h^\prime}H^\prime AH. 
\end{equation}
The $F_{H^\prime}$ leads to the vanishing VEV of the triplet
Higgs $\VEV{H_T}=0$. 
All VEVs have now been fixed.

There are several terms which must be forbidden for the stability of the 
DW mechanism. For example, $H^2$, $HZH^\prime$ and $H\bar Z H^\prime$
induce a large mass of the doublet Higgs, 
and the term $\bar CA^\prime A C$ would destabilize the DW form of 
$\VEV{A}$.
We can easily forbid these terms using the SUSY zero mechanism.
For example, if we choose
$h<0$, then $H^2$ is forbidden, and if we choose $\bar c+c+a+a^\prime<0$, 
then
$\bar CA^\prime A C$ is forbidden. 
Once these dangerous terms are forbidden
by the SUSY zero mechanism, higher-dimensional terms which also become
dangerous (
for example, 
$\bar CA^\prime A^3 C$ and $\bar CA^\prime C\bar CA C$) are automatically
forbidden.
This is also an attractive point of our scenario. 

In the end of this subsection, we would like to explain how to determine
the symmetry and the quantum numbers in the Higgs sector to realize 
DT splitting.
It is essential that the dangerous terms are forbidden by SUSY zero mechanism
and the necessary terms must be allowed by the symmetry. The dangerous terms
are
\begin{equation}
H^2, HH',HZH', \bar CA'C, \bar CA'AC, \bar CA'ZC, A'A^4, A'A^5.
\end{equation}
On the other hand, the terms required to realize DT splitting are
\begin{equation}
A'A, A'A^3, HAH',\bar C'(A+Z)C, \bar C(A+Z)C', S\bar CC.
\end{equation}
Here we denote both $Z$ and $\bar Z$ as $Z$.
In order to forbid $HH'$ while $HAH'$ is allowed, we introduce $Z_2$ parity.

Of course, the above conditions are necessary but not sufficient. 
As in the next subsection, we have to write down the mass matrices of 
Higgs sector to know whether an assignment truly works well or not. 


\subsection{Mass spectrum of Higgs sector}
Under $SO(10)\supset SU(5) \supset SU(3)_C\times SU(2)_L\times U(1)_Y$,
the spinor ${\bf 16}$, vector ${\bf 10}$ and the adjoint ${\bf 45}$
are classified in terms of the fields  $Q({\bf 3,2})_{\frac{1}{6}}$,
$U^c({\bf \bar 3,1})_{-\frac{2}{3}}$, $D^c({\bf \bar 3,1})_{\frac{1}{3}}$,
$L({\bf 1,2})_{-\frac{1}{2}}$, $E^c({\bf 1,1})_1,N^c({\bf 1,1})_0$,
$X({\bf 3,2})_{-\frac{5}{6}}$ and their conjugate fields, and
$G({\bf 8,1})_0$ and $W({\bf 1,3})_0$;
\begin{eqnarray}
{\bf 16}&= &
\underbrace{[Q+U^c+E^c]}_{\bf 10}+\underbrace{[D^c+L]}_{\bf \bar 5}
+\underbrace{N^c}_{\bf 1}, \nn \\
{\bf 10}&= &
\underbrace{[D^c+L]}_{\bf \bar 5}+\underbrace{[\bar D^c+\bar L]}_{\bf 5},
\label{class}\\
{\bf 45}&= &
\underbrace{[G+W+X+\bar X+N^c]}_{\bf 24}
+\underbrace{[Q+U^c+E^c]}_{\bf 10}
+\underbrace{[\bar Q+\bar U^c+\bar E^c]}_{\bf \overline{10}}
+\underbrace{N^c}_{\bf 1}. \nn
\end{eqnarray}

In the followings, we study how mass matrices of the above fields
are determined, by taking an example of typical charge assignment
shown in Table I.
For the mass terms, we must
take account of not only the terms in the previous section but also
the terms that contain two fields with vanishing
VEVs. 

First we examine the mass spectrum of ${\bf 5}$ and ${\bf \bar 5}$ of
$SU(5)$. 
Considering the additional terms
$\lambda^{2h^\prime} H^\prime H^\prime$,
$\lambda^{c^\prime+\bar c^\prime}\bar C^\prime C^\prime$, 
$\lambda^{c'+c+h'}C'CH'$, 
$\lambda^{\bar c'+z+\bar c+h}Z\bar C'\bar CH$,
$\lambda^{\bar c'+\bar c+h'}\bar C'\bar CH'$ and
$\lambda^{2\bar c+h'}\bar C^2H'$,
we write the mass matrices $M_I$ ($I=D^c(H_T),L(H_D)$),
whose elements should read the mass of $I$ component of
${\bf \bar 5(10)}$ or ${\bf \bar 5(16)}$ and
$\bar I$ component of ${\bf 5(10)}$ or ${\bf 5(\overline{16})}$:
\begin{equation}
M_I=\bordermatrix{
\bar I\backslash I&10_H(-3) & 16_{C}(-3) & 
                  10_{H^\prime}(4)&16_{C^\prime}(2) \cr
10_H(-3) & 0 & 0 & \lambda^{h+h^\prime +a}\VEV{A}  & 0 \cr
\overline{16}_{\bar C}(0)& 0 & 0 & \lambda^{h'+2\bar c}\VEV{\bar C} 
& \lambda^{\bar c+c^\prime} \cr
10_{H^\prime} (4)& \lambda^{h+h^\prime +a}\VEV{A} & 0& \lambda^{2h^\prime} 
 & \lambda^{h'+c'+c}\VEV{C} \cr
\overline{16}_{\bar C^\prime}(5) &  
\lambda^{h+\bar c'+\bar c}\VEV{\bar C} 
 & \lambda^{c+\bar c^\prime} 
 & \lambda^{h'+\bar c'+\bar c}\VEV{\bar C}
  & \lambda^{c^\prime+\bar c^\prime} \cr},
\end{equation}
where the vanishing components are due to SUSY zero mechanism and
we denote the typical charges in parentheses.
It is worth while examining the general structure of the mass matrices.
The first two columns and rows are for the fields with non-vanishing
VEVs which have smaller charges,  and the last two columns and rows are 
for the fields with vanishing VEVs which have larger charges.
Therefore, it is useful to divide the matrices into four $2\times 2$ 
matrices as
\begin{equation}
M_I=\left(\matrix{ 0 & A_I \cr B_I & C_I }\right).
\end{equation}
It is easily seen 
that the ranks of $A_L$ and $B_L$ are reduced to one
when the VEV $\VEV{A}$ vanishes.
It means that
the rank of $M_L$ is reduced, and actually it becomes three.
On the other hand, the ranks of $A_{D^c}$ and $B_{D^c}$ remain two
because the field $A$ becomes non-zero value on $D^c$.
Therefore DT splitting is realized.
The mass spectrum of $L$ is easily obtained as
$(0,\lambda^{2h'},\lambda^{\bar c+c'},\lambda^{\bar c'+c})$.
The massless modes of doublet Higgs are estimated as
\begin{equation}
5_H, \bar 5_H+\lambda^{h-c+\frac{1}{2}(\bar c-c)}\bar 5_C.
\label{mixing}
\end{equation}
The elements of
matrices $A_I$ and $B_I$  become generally larger than the elements of 
the matrices $C_I$ because the total anomalous $U(1)_A$ charge of 
the corresponding pair of fields in $A_I$ and $B_I$ becomes smaller 
than that in $C_I$.
Therefore, the mass spectrum of $D^c$ is essentially estimated by the matrices
$A_I$ and $B_I$ as 
$(\lambda^{h+h'},\lambda^{h+h'},\lambda^{\bar c+c'},\lambda^{\bar c'+c})$.
It is obvious that to develop the proton decay, we have to pick up the element
of $C_I$. Since the element is generally smaller than the mass scale of $D^c$,
the proton decay can be suppressed.
The effective colored Higgs mass is estimated as
$(\lambda^{h+h^\prime})^2/\lambda^{2h^\prime}=\lambda^{2h}$, 
which is larger than the cutoff scale, because
$h<0$. 



Next we examine the mass matrices for the representations 
$I=Q,U^c$ and $E^c$,
which are contained in the {\bf 10} of $SU(5)$,
%Like the superpotential previously discussed, 
where the additional terms
$\lambda^{2a^\prime}A^\prime A^\prime$, 
$\lambda^{c^\prime+\bar c^\prime}\bar C^\prime C^\prime$,
$\lambda^{c^\prime+a^\prime+\bar c} \bar CA^\prime C^\prime$ and
$\lambda^{\bar c^\prime+a^\prime+c} \bar C^\prime A^\prime C$
must be taken into account.
The mass matrices are written
\begin{equation}
M_I=\bordermatrix{
\bar I\backslash I&45_A(-1) &16_{ C}(-3)& 45_{A^\prime}(3) &
16_{C^\prime}(2) \cr
45_A(-1) &0& 0 & \lambda^{a^\prime+a} \alpha_I  & 
\lambda^{\bar c+c^\prime+a}\VEV{\bar C} \cr
\overline{16}_{\bar C}(0)&0 & 0 & 0 & \lambda^{\bar c+c^\prime}\beta_I \cr
45_{A^\prime}(3) &\lambda^{a+a^\prime} \alpha_I & 0& \lambda^{2a^\prime}  & 
\lambda^{\bar c+c^\prime+a^\prime}\VEV{\bar C} \cr
\overline{16}_{\bar C^\prime}(5) &\lambda^{c+\bar c^\prime+a}
\VEV{C} &\lambda^{c+\bar c^\prime}\beta_I &
\lambda^{c+\bar c^\prime+a^\prime}\VEV{C} &
 \lambda^{c^\prime+\bar c^\prime}\cr},
\label{mass10}
\end{equation}
where $\alpha_Q=\alpha_{U^c}=0$ and $\beta_{E^c}=0$ because
there are Nambu-Goldstone modes in breaking 
$SO(10)\rightarrow SU(3)_C\times SU(2)_L\times SU(2)_R\times U(1)_{B-L}$
and $SU(2)_R\times U(1)_{B-L}$, respectively.
Defining $2\times 2$ matrices as in $I=L,D^c$ case,
it is obvious that the ranks of $A_I$ and $B_I$ reduce.
Thus for each $I$, the $4\times 4$ matrices $M_I$ have one 
vanishing eigenvalue, which corresponds to the Nambu-Goldstone mode eaten
by the Higgs mechanism. The mass spectrum of the remaining three
modes is ($\lambda^{c+\bar c^\prime}$, $\lambda^{c^\prime+\bar c}$,
$\lambda^{2a^\prime}$) for the color-triplet modes $Q$ and $U^c$, and
($\lambda^{a+a^\prime}$, 
$\lambda^{a+a^\prime}$,
$\lambda^{c^\prime+\bar c^\prime}$) or 
($\lambda^{c+\bar c^\prime+a}\VEV{C}$, 
$\lambda^{c^\prime+\bar c+a}\VEV{\bar C}$,
$\lambda^{2a^\prime}$) for the color-singlet modes $E^c$. 


The adjoint fields $A$ and $A^\prime$ contain
two $G$, two $W$ and two pairs of $X$ and $\bar X$, 
whose mass matrices $M_I(I=G,W,X)$ is given by
\begin{equation}
M_I=\bordermatrix{
\bar I\backslash I &    45_A(-1)       &        45_{A'}(3)           \cr
45_A(-1)    &     0        & \alpha_I\lambda^{a+a'}  \cr
45_{A'}(3) & \alpha_I\lambda^{a+a'} & \lambda^{2a'}  \cr}.
\end{equation}
Two $G$ and two $W$ acquire masses $\lambda^{a^\prime+a}$.
Since $\alpha_X=0$, one pair of $X$ is massless, which is eaten
by Higgs mechanism. However, the other pair has a rather light mass of
$\lambda^{2a^\prime}$.


\subsection{Gauge unification and proton decay}
In the minimal SUSY $SU(5)$ GUT, the proton stability is not compatible
with the success of the gauge coupling unification.
\cite{Goto}
The proton stability requires that the colored Higgs mass is
larger than $10^{18}$ GeV, which destroys the coupling unification
because it has no other tuning parameter. Of course, if we introduce
other superheavy particles with smaller masses than the GUT scale,
we may recover the gauge coupling unification by tuning their masses. 
However, unless we
have some mechanism which controls the scale of the masses, generically
some finetuning is required. 
In the $SO(10)$ GUT with DW mechanism, the proton stability can be
realized in the mass structure
\begin{equation}
M_I=\left(\matrix{0 & \VEV{A} \cr \VEV{A} & m \cr}\right) (I=L, D_c),
\end{equation}
if $\VEV{A}^2/m_{D^c}>10^{18}$ GeV. However, in order to realize the 
condition $\VEV{A}^2/m_{D^c}>10^{18}$ GeV, the mass scale of 
the additional doublet Higgs $m$ becomes smaller than the GUT scale
$\VEV{A}$, which generally destroys the success of the gauge coupling
unification. Of course, it may be possible to realize
the gauge coupling unification by tuning the other scale, $\VEV{C}$,
or the mass scales of superheavy particles.
Unless we have no mechanism which controls these scales, however, 
such situation cannot explain why the gauge couplings meet
at a scale $\Lambda_G\sim 2\times 10^{16}$ GeV in the minimal SUSY
standard model (MSSM).

In our scenario, 
once we determine the anomalous $U(1)_A$ charges, the mass spectrum
of all superheavy particles and other symmetry breaking scales are
 determined, and hence we can examine whether 
the running couplings from the low energy scale meet at the unification
scale or not. When the initial values of the gauge coupling constants
are replaced by the usual GUT scale $\Lambda_G\sim 2\times 10^{16}$ GeV and
the unified gauge coupling, the condition for the gauge coupling 
unification can be converted into the relation between the charges and
the ratio of the cutoff scale $\Lambda$ 
to the usual GUT scale $\Lambda_G\sim 2\times 10^{16}$ GeV:
\begin{equation}
\frac{\Lambda_G}{\Lambda}\sim \lambda^{-\frac{h}{7}}
                         \sim \lambda^{\frac{h}{8}},
\end{equation}
that leads to 
\begin{eqnarray}
\Lambda&\sim& \Lambda_G, \\
h&\sim &0.
\end{eqnarray}
Here we have used the renormalization group up to one loop approximation.
It is non-trivial that
in the relation, all the charges except that of the Higgs doublet are
cancelled out. 
If we simply have taken $\Lambda=\Lambda_G$ and $h=0$, the model would be
suffering from the proton decay via dimension 5 operator because
the effective colored Higgs mass becomes 
$\lambda^{2h}\Lambda=\Lambda_G<<10^{18}$ GeV. 
However, we have to take negative $h$ to forbid the Higgs mass term
$H^2$. Therefore we would like to know how large negative charge
$h$ can be adopted in our scenario.
To obtain realistic quark and lepton mass matrices
including bi-large neutrino mixing, the maximal value of $h$ is $-3$.
In that case, the effective colored Higgs mass becomes
$\lambda^{2h}\Lambda=\lambda^{-6}\Lambda_G\sim 10^{22}$ GeV, which
is much larger than the experimental limit. 
Note that even for such a small value of $h$, 
the coupling unification can be realized,
using the ambiguities of order one coefficients. 
Since the unified scale becomes $\lambda^{-a}\Lambda$ is
just below the scale $\Lambda_G$,  the proton decay via dimension 6 
operator can be seen in near future.
We will return to this point in the next section.

Once we fixed the anomalous $U(1)_A$ charges, the fact that the gauge 
couplings meet at the usual GUT scale in the MSSM is non-trivially related 
with the 
result that the gauge couplings of the GUT with anomalous $U(1)_A$ gauge
symmetry  almost meet at the GUT
scale $\lambda^{-a}\Lambda$ in our scenario. Therefore, this GUT scenario can
explain why the gauge couplings meet at a scale in MSSM with the 
accuracy up to one loop approximation.
\footnote{ Actually,
if the gauge couplings have met at the other scale $\Lambda_O$ in MSSM, 
then the cutoff scale 
would be the scale $\Lambda_O$ in our scenario, namely the GUT scale would be
$\lambda^{-a}\Lambda_O$. }












\section{$E_6$ unification of Higgs sector}
In this section, we extend the DT splitting mechanism, discussed in the
previous section, into the $E_6$ unification. Here we propose the complete
Higgs sector with $E_6$ GUT gauge group.

In order to break $E_6$ gauge group into the standard gauge group,
we introduce the following Higgs contents:
\begin{enumerate}
\item Higgs fields that break $E_6$ into $SO(10)$:
$\Phi({\bf 27})$ and $\bar \Phi ({\bf \overline{27}})$
($\left|\VEV{\Phi({\bf 1,1})}\right|=
\left|\VEV{\bar \Phi({\bf 1,1})}\right|$).
\item An adjoint Higgs field  which breaks the $SO(10)$
into 
$SU(3)_C\times SU(2)_L\times SU(2)_R\times U(1)_{B-L}$: $A({\bf 78})$
($\VEV{{\bf 45}_A}=\tau_2\times {\rm diag}(v,v,v,0,0)$).
\item Higgs fields that break $SU(2)_R\times U(1)_{B-L}$ into $U(1)_Y$:
$C({\bf 27})$ and $\bar C({\bf \overline{27}})$
( $\left|\VEV{C({\bf 16,1})}\right|=
\left|\VEV{\bar C({\bf \overline{16},1})}\right|$).
\end{enumerate}
Of course, 
 the anomalous $U(1)_A$ charges of 
the gauge singlet operators, $\bar \Phi\Phi$, $\bar CC$ and $A^2$
must be negative.



Naively thinking, we have to introduce at least the same number of 
superfields with positive charges in order to make them massive. 
However, it is shown that smaller Higgs contents are sufficient to
make all the fields massive. This is because parts of the Higgs
fields with non-vanishing VEVs are absorbed by Higgs mechanism.
Actually when $E_6$ gauge group is broken into $SO(10)$ 
by non-vanishing VEV 
$\left|\VEV{\Phi}\right|=\left|\VEV{\bar \Phi}\right|$, the fields
${\bf 16}_\Phi$ and ${\bf \overline{16}}_{\bar\Phi}$ are absorbed 
by super Higgs mechanism.
\footnote{
Strictly speaking, linear combination of $\Phi$, $C$ and $A$
and of $\bar\Phi$, $\bar C$ and $A$
become massive by super Higgs mechanism. The main modes are
${\bf 16}_\Phi$ and ${\bf \overline{16}}_{\bar \Phi}$, respectively.
}
Therefore, if two additional {\bf 10}s of $SO(10)$ in Higgs contents with
non-vanishing VEVs can be massive, then
we can save the superfields with positive charges.
At first glance, such a mass term seems to be forbidden by SUSY zero mechanism.
Actually if all fields with non-vanishing VEVs have had negative 
anomalous $U(1)_A$ charges, it would be impossible. As discussed in the
section 2, the non-positiveness of the anomalous $U(1)_A$ charges is
required only for gauge singlet operators with non-vanishing VEVs, so
even the fields with positive charges can have non-vanishing VEVs if 
total charge of the gauge singlet operators with non-vanishing VEVs is
negative. For example, we can adopt $\phi=-3$ and $\bar \phi=2$, because
the total charge of the gauge singlet operator $\bar \Phi\Phi$
is negative. Since $\bar \Phi$
has positive charge, the term $\bar \Phi^3$ is allowed, which induces
the mass of ${\bf 10}_{\bar \Phi}$ by the non-vanishing VEV 
$\VEV{\bar \Phi}$. If the term $\bar \Phi^2\bar C$ is allowed,
the masses of two ${\bf 10}$s,  ${\bf 10}_{\bar \Phi}$ and 
${\bf 10}_{\bar C}$, are induced, so we can save the superfields
with positive charges.

The minimal contents of the Higgs sector with $E_6\times U(1)_A$ gauge 
symmetry 
are given in Table II, where the symbols $\pm$ denote $Z_2$ parity quantum 
numbers.

\vspace{3mm}
\begin{center}
Table II. The typical values of anomalous $U(1)_A$ charges are listed.

\begin{tabular}{|c|c|c|} 
\hline
                  &   non-vanishing VEV  & vanishing VEV \\
\hline 
{\bf 78}          &   $A(a=-1,-)$        & $A'(a'=4,-)$      \\
{\bf 27}          &   $\Phi(\phi=-3,+)$\  $C(c=-6,+)$ &  $C'(c'=7,-)$  \\
${\bf \overline{27}}$ & $\bar \Phi(\bar \phi=2,+)$ \  $\bar C(\bar c=-2,+)$ &
                  $\bar C'(\bar c'=8,-)$ \\
{\bf 1}           &   $Z_2(z_2=-2,-)$,$Z_5(z_5=-5,-)$,
                       $\bar Z_5(\bar z_5=-5,-)$ \   &  \\
\hline
\end{tabular}

\vspace{5mm}
\end{center}
Here the Higgs field $H$ of $SO(10)$ model is contained in
$\Phi$. 
This $E_6$
Higgs sector has the same number of superfields with non-trivial 
representation as the $SO(10)$ Higgs sector in spite of the fact
that the larger group $E_6$ needs additional Higgs field
to break $E_6$ into $SO(10)$ gauge group.
It is interesting that the DT splitting
is naturally realized in this minimal Higgs contents in a sense.

\subsection{DT splitting and alignment}
Generally in $E_6$ GUT, the interactions in the superpotential
out of ${\bf 27}$ and ${\bf \overline{27}}$ are written 
by the units ${\bf 27}^3$, ${\bf \overline{27}27}$ and ${\bf\overline{27}}^3$.
Note that  the terms like ${\bf 27}^3$ or 
${\bf\overline{27}}^3$  do not contain the product of singlet components
of the standard gauge group. Therefore these terms can be neglected in
considering the standard vacuum. 
Of course these terms can constrain the other vacua than the standard
vacuum, that will be discussed below.

The important terms in the superpotential
to determine the VEVs are written
\begin{equation}
W=W_{A^\prime} + W_{C^\prime}+W_{\bar C^\prime}+W({\bar \Phi}).
\end{equation}
Since we have a positive charged field
$\bar \Phi$
which has non-vanishing VEV, we have to take account of the terms
$W({\bar \Phi})$ which includes the field $\bar \Phi$ but do not
contain the fields with vanishing VEVs. Since 
$\bar \Phi\Phi$ and $\bar \Phi C$ have negative total charges,
the superpotential has essentially the terms like $\overline{27}^3$.
Therefore, the superpotential $W(\bar \Phi)$ can constrain the other
vacua than the standard vacuum. 

Let us examine the superpotential $W(\bar \Phi)$ to see the general vacuum 
structure in $E_6$ model.
%We now discuss how to determine the VEVs.
As discussed in the previous section, 
the composite operator $\bar \Phi\Phi$ with $\phi+\bar \phi=-1$ 
can play the same role as
the FN field. In that case, $\Phi$ and $\bar \Phi$ have
non-vanishing VEVs and the $E_6$ $D$-flatness condition requires
$\VEV{\Phi}=\VEV{\bar \Phi}$ up to phases. On the other hand, the VEV of 
$\bar \Phi$ (namely that of $\Phi$ also by $D$-flatness condition) can be 
rotated by $E_6$ 
gauge transformation into the following form:
\begin{equation}
\VEV{\bar \Phi}=
\pmatrix{ \bar u \cr
          {\bf 0} \cr
          \bar u_1 \cr
          \bar u_2 \cr
          {\bf 0}        } 
\begin{array}{l}
  \hbox{$\}SO(10)$ singlet (real)} \\
  \hbox{$\}SO(10)$ ${\bf \overline{16}}$}       \\
  \hbox{$\}$the first component of $SO(10)$ {\bf 10}\ (complex)} \\
  \hbox{$\}$the second component of $SO(10)$ {\bf 10}\ (real)} \\
  \hbox{$\}$the third to tenth components of $SO(10)$ {\bf 10}}
\end{array}
\end{equation}
For simplicity, we adopt the superpotential as 
\begin{equation}
W({\bar \Phi})=\bar \Phi^3+\bar \Phi^2\bar C.
\end{equation}
Then the $F$-flatness conditions of ${\bf 10}_{\bar C}$ and
${\bf 1}_{\bar C}$ lead to
${\bf 1}_{\bar \Phi}{\bf 10}_{\bar \Phi}=0$ and ${\bf 10}_{\bar \Phi}^2=0$.
Then we are allowed to have either of  the two vacua,
$(\bar u\neq 0,\bar u_1=\bar u_2=0)$ or
$(\bar u=0,\bar u_1=i\bar u_2\neq 0)$.
It means that the non-vanishing VEV $\VEV{{\bf 1}_{\bar \Phi}}$ requires
vanishing VEV $\VEV{{\bf 10}_{\bar \Phi}}$. Therefore,
in the first vacuum, $E_6$ gauge group is broken into $SO(10)$ gauge group.
Moreover, 
in this vacuum, ${\bf 10}_{\bar C}$ has vanishing VEV, because of
the $F$-flatness conditions of ${\bf 10}_{\bar \Phi}$.
Interesting enough, a vacuum alignment occurs naturally.
In the following, for simplicity, we often write $\lambda^n$ instead of the 
operators $(\bar\Phi\Phi)^n$, though these operators are not always singlet.

The superpotential $W_{A^\prime}$ is in general
written 
\begin{eqnarray}
W_{A^\prime}&=&\lambda^{a^\prime+a}A^\prime A
+\lambda^{a^\prime+3a}A^\prime A^3
+\lambda^{a'+a+\bar \phi+\phi}\bar \Phi A'A\Phi \nonumber \\ 
&&+\lambda^{a'+3a+\bar \phi+\phi}\bar \Phi A'A^3\Phi
\end{eqnarray}
under the condition $-3a+\bar \phi+\phi\leq a^\prime < -5a$.
Here we assume 
$c+\bar c, c+\bar \phi, \bar c+\phi<-(a'+a)$
to forbid the terms $\bar C A^\prime A C$ (which destabilizes the 
DW form of the VEV of $A$), $\bar CA'A\Phi$ and $\bar \Phi A'AC$
(which may lead to undesired vacua $\VEV{\bar C}=\VEV{C}=0$).

If $A$ and $(\Phi,\bar \Phi)$ are separated in the superpotential, 
the pseudo Nambu-Goldstones (PNG) appear. 
Since the terms $\bar \Phi A'A\Phi$ and $\bar \Phi A'A^3\Phi$ connect 
$A'$ and $A$ with $\Phi$ and $\bar \Phi$, the PNG obtain their masses.
Moreover, these terms realize the alignment between the VEVs
$\left|\VEV{\Phi}\right|=\left|\VEV{\bar \Phi}\right|$ and $\VEV{A}$. 
Note that these terms
are also important to induce the term 
$({\bf 45}_{A'}{\bf 45}_{A})_{\bf 54}({\bf 45}_{A}^2)_{\bf 54}$,
which is not included in the term $A'A^3$
 because of a cancellation (see Appendix A).
\footnote{We thank T. Kugo for his pointing out this cancellation.}
In terms of 
$SO(10)$, which is not broken by the VEV 
$\left|\VEV{\Phi}\right|=\left|\VEV{\bar \Phi}\right|$,
the effective superpotential is given by
\begin{eqnarray}
W_{A'}^{eff}&=&
{\bf 45}_{A'}(1+{\bf 1}_{A}^2+{\bf 45}_{A}^2
+{\bf \overline{16}}_{A}{\bf 16}_{A}){\bf 45}_{A}  \nn \\
&&+{\bf \overline{16}}_{A'}(1+{\bf 1}_{A}^2+{\bf 45}_{A}^2
+{\bf \overline{16}}_{A}{\bf 16}_{A}){\bf 16}_{A} \nn \\
&&+{\bf 16}_{A'}(1+{\bf 1}_{A}^2+{\bf 45}_{A}^2
+{\bf \overline{16}}_{A}{\bf 16}_{A}){\bf \overline{16}}_{A} \\
&&+{\bf 1}_{A'}{\bf 1}_{A}(1+{\bf 1}_{A}^2+{\bf 45}_{A}^2
+{\bf \overline{16}}_{A}{\bf 16}_{A}). \nn
\end{eqnarray}
The $F$-flatness conditions are written
\begin{eqnarray}
\frac{\partial W}{\partial {\bf 45}_{A'}}&=& (1+{\bf 1}_{A}^2
+{\bf 45}_{A}^2+{\bf \overline{16}}_{A}{\bf 16}_{A}){\bf 45}_{A} 
\label{45}\\
\frac{\partial W}{\partial {\bf \overline{16}}_{A'}}&=&(1+{\bf 1}_{A}^2
+{\bf 45}_{A}^2+{\bf \overline{16}}_{A}{\bf 16}_{A}){\bf 16}_{A} 
\label{bar16} \\
\frac{\partial W}{\partial {\bf 16}_{A'}}&=& {\bf \overline{16}}_{A}
(1+{\bf 1}_{A}^2+{\bf 45}_{A}^2+{\bf \overline{16}}_{A}{\bf 16}_{A}) 
\label{16}\\
\frac{\partial W}{\partial {\bf 1}_{A'}}&=& {\bf 1}_{A}
(1+{\bf 1}_{A}^2+{\bf 45}_{A}^2+{\bf \overline{16}}_{A}{\bf 1}_{A}).
\label{1}
\end{eqnarray}
These $F$-flatness conditions and $D$-flatness conditions of
$SO(10)$ determine the VEVs 
$\VEV{{\bf 16}_A}=\VEV{{\bf \overline{16}}_A}=0$. We have two possibilities
of the VEV of ${\bf 1}_A$, one vacuum $\VEV{{\bf 1}_A}=0$ and 
the other one $\VEV{{\bf 1}_A}\neq 0$. In the latter vacuum, the DW mechanism
in $E_6$ GUT does not work, because the non-vanishing VEV $\VEV{{\bf 1}_A}$
directly gives the bare mass to the doublet Higgs. 
Therefore, the former vacuum $\VEV{{\bf 1}_A}=0$ is desirable to realize
DT splitting.
Note that if the term $\bar \Phi A'\Phi$ is allowed, the vacuum
$\VEV{{\bf 1}_A}=0$ disappears. This destroys the realization of
DT splitting. Here this term is forbidden by $Z_2$ parity.
As in $SO(10)$ case, we have several possibilities of the VEV of 
${\bf 45}_A$, one of which is the DW type of VEV 
$\VEV{{\bf 45}_A}_{B-L}=i\tau_2\times {\rm diag}(v,v,v,0,0)$,
where $v\sim\lambda^{-a}$.
These VEVs breaks the $SO(10)$ gauge group
into $SU(3)_C\times SU(2)_L\times SU(2)_R\times U(1)_{B-L}$.

Next, we discuss the $F$-flatness of $C^\prime$ and 
$\bar C^\prime$,
which not only determine the scale of the VEV 
$\VEV{\bar CC}\sim \lambda^{-(c+\bar c)}$ but also realizes 
the alignment of the VEVs $\VEV{C}$ and $\VEV{\bar C}$.
For simplicity, we assume that $\VEV{{\bf 1}_C}=\VEV{{\bf 1}_{\bar C}}=0$,
though there may be vacua in which these components have non-vanishing VEVs.
Then since $\VEV{{\bf 10}_C}=\VEV{{\bf 10}_{\bar C}}=0$ by the above argument,
only the components ${\bf 16}_C$ and ${\bf \overline{16}}_{\bar C}$
can have non-vanishing VEVs.
The superpotential to determine these VEVs 
can be written 
\begin{eqnarray}
W_{C^\prime}&=&
       \lambda^{\bar\phi+c'}\bar \Phi (\lambda^{c+\bar c+a}\bar CAC+
       \lambda^{z_5}Z_5+\lambda^{\bar z_5}\bar Z_5+\lambda^{z_2}Z_2+
       \lambda^{a}A)C'
       \nn\\
       &&+ \lambda^{\bar c+c'}\bar C(\lambda^{z_5}Z_5
       +\lambda^{\bar z_5}\bar Z_5
       +\lambda^{z_2}Z_2+
       \lambda^aA)C' \\
W_{\bar C^\prime}&=&
       \lambda^{\bar c'+\phi}\bar C'(\lambda^{z_5}Z_5
       +\lambda^{\bar z_5}\bar Z_5+
       \lambda^{z_2}Z_2+\lambda^aA)\Phi \\
       &&+\lambda^{\bar c'+c}\bar C'(\lambda^{z_2}Z_2+\lambda^aA)C,
\end{eqnarray}
where we omit the even $Z_2$ parity operators with non-vanishing VEVs
like $A^{2n}$, $Z_2^2$, $Z_iA$, etc., because the VEVs of these operators 
do not change the power of $\lambda$.
The $F$-flatness conditions of
${\bf 1}_{C'}$ and ${\bf 1}_{\bar C'}$ lead to 
$\bar \Phi (\lambda^{c+\bar c+a}\bar CAC+
       \lambda^{z_5}Z_5+\lambda^{\bar z_5}\bar Z_5+
       \lambda^{z_2}Z_2+\lambda^aA)=0$ 
and 
$(\lambda^{z_5}Z_5+\lambda^{\bar z_5}\bar Z_5+
       \lambda^{z_2}Z_2+\lambda^aA)\Phi=0$, 
respectively. The vacua are  
(a) ($\VEV{\bar CC}=0$) and  (b) ($\VEV{\bar CC}\neq 0$).
The desired vacua (a) requires the additional $F$-flatness conditions 
of ${\bf 16}_{C'}$ and ${\bf \overline{16}}_{\bar C'}$, which realize
alignment of the VEVs $\VEV{A}$ and $\VEV{C}(\VEV{\bar C})$ as in 
$SO(10)$ cases. Then the above four $F$-flatness conditions with
respect to  ${\bf 1}_{C'}$, ${\bf 1}_{\bar C'}$, ${\bf 16}_{C'}$ and 
${\bf \overline{16}}_{\bar C'}$
determine the scale of
the four VEVs
$\VEV{\bar CC}\sim \lambda^{-(c+\bar c)}$,
$\VEV{Z_i}\sim \lambda^{-z_i}(i=3,5)$ and 
$\VEV{\bar Z_5}\sim \lambda^{-\bar z_5}$.
The VEVs $\left|\VEV{C}\right|=\left|\VEV{\bar C}\right|
\sim \lambda^{-(\bar c+c)}$ 
break $SU(2)_R\times U(1)_{B-L}$ into $U(1)_Y$. 

Then all the VEVs are determined by the anomalous $U(1)_A$ charges.


\subsection{Mass spectrum of Higgs sector}
Since all the VEVs are fixed, we can estimate the mass spectrum
of Higgs sector.

Let the fields be decomposed in terms of the quantum numbers
of $SO(10)\times U(1)_{V'}$;
\begin{eqnarray}
{\bf 27}&=& {\bf 16}_1+{\bf 10}_{-2}+{\bf 1}_4 \\
{\bf 78}&=& {\bf 45}_0+{\bf 16}_{-3}+{\bf \overline{16}}_3+{\bf 1}_0,
\end{eqnarray}
which are further decomposed to $SU(5)$ representations 
(see Eq.(\ref{class})).


In the followings, we study how mass matrices of the above fields
are determined by anomalous $U(1)_A$ charges.
Note that for the mass terms, we must
take account of not only the terms in the previous subsection but also
the terms that contain two fields with vanishing
VEVs (see Appendix C).

Before going to the detail, it is worth while examining the NG modes
which are eaten by Higgs mechanism, because in some cases it is not
so obvious to see the vanishing eigenvalue in the mass matrices.
There appear NG modes
\begin{enumerate}
\item ${\bf 16}+{\bf \overline{16}}+1$ of $SO(10)$, namely,
$Q+U^c+E^c+h.c.+N^c$
in breaking $E_6\rightarrow SO(10)$.
\item $Q+U^c+X+h.c.$ in breaking 
$SO(10)\rightarrow SU(3)_C\times SU(2)_L\times SU(2)_R\times U(1)_{B-L}$.
\item $E^c+h.c.+N^c$ in breaking
$SU(2)_R\times U(1)_{B-L}\rightarrow U(1)_Y$.
\end{enumerate}

First of all, we examine the mass matrix of ${\bf 24}$ of $SU(5)$.
Considering the additional term $A'^2$,
we write the mass matrices $M_I$,
which are for the representations $I=G,W,X$:
\begin{equation}
M_I=\bordermatrix{
\bar I\backslash I  &    24_A    &     24_{A'}            \cr
24_A                &     0      & \alpha_I\lambda^{a'+a} \cr
24_{A'}   & \alpha_I\lambda^{a'+a}& \lambda^{2a'}         \cr
},
\end{equation}
where $\alpha_X=0$ and $\alpha_I\neq 0$ for $I=G,W$. 
One pair of $X$ is massless, which is eaten by Higgs mechanism.
The mass spectrum are $(0, \lambda^{2a'})$ for $I=X$ and 
$(\lambda^{a'+a},\lambda^{a'+a})$ for $I=G,W$.

Next we examine the mass matrices for the representation $I=Q, U^c$ and
$E^c$, which are contained in ${\bf 10}$ of $SU(5)$. The mass matrices
$M_I$ are written
\begin{equation}
\bordermatrix{
I\backslash \bar I  &\overline {16}_{\bar \Phi} & \overline {16}_{\bar C}&
                     \overline {16}_{A} &45_{A} &\overline {16}_{\bar C'}   & 
                    \overline {16}_{ A'} &  45_{A'}  \cr
16_\Phi & 0 & 0 & 0& 0& \lambda^{\bar c'+\phi}  & \lambda^{\phi+a'-\Delta\phi}
         & 0 \cr
16_C& 0 & 0 &0 & 0& \beta_I\lambda^{\bar c'+c} &  0  & 0 \cr
16_A & 0 & 0 & 0& 0& \lambda^{\bar c'+a+\Delta\phi}   & \lambda^{a'+a}
   & 0 \cr
45_{A} & 0 & 0 & 0 & 0& \lambda^{\bar c'+a+\Delta c}  & 0 & 
       \alpha_I\lambda^{a+a'}  \cr
16_{C'} & \lambda^{c'+\bar\phi} & \beta_I\lambda^{c'+\bar c} &
        \lambda^{a+c'-\Delta\phi} & \lambda^{a+c'-\Delta c} & 
        \lambda^{c'+\bar c'} & \lambda^{a'+c'-\Delta\phi} & 
        \lambda^{a'+c'-\Delta c}  \cr
16_{A'} & \lambda^{\bar \phi+a'+\Delta\phi} & 0 & 
        \lambda^{a+a'} & 0&\lambda^{\bar c'+a'+\Delta\phi} &  
        \lambda^{2a'}& \lambda^{2a'+\Delta\phi-\Delta c}  \cr
45_{A'} & 0 & 0 &0 & \alpha_I\lambda^{a'+a}
        & \lambda^{\bar c'+a'+\Delta c} &  
        \lambda^{2a'-\Delta\phi+\Delta c} 
        & \lambda^{2a'} \cr
},
\end{equation}
where we use the relations
$\lambda^\phi\VEV{\Phi}\sim (\lambda^{\bar \phi}\VEV{\bar \Phi})^{-1}
\sim\lambda^{\Delta\phi}$ and
$\lambda^c\VEV{C}\sim(\lambda^{\bar c}\VEV{\bar C})^{-1}
\sim\lambda^{\Delta c}$ ($\Delta\phi=(\phi-\bar\phi)/2$, 
$\Delta c=(c-\bar c)/2$).
Since one pair of ${\bf \overline{16}}$ and ${\bf 16}$ (main modes are
${\bf \overline{16}}_{\bar \Phi}$ and ${\bf 16}_\Phi$ ) is eaten by 
Higgs mechanism in breaking $E_6$ into $SO(10)$, we simply omit
${\bf 16}_\Phi$ and ${\bf \overline{16}}_{\bar \Phi}$
 in estimating the mass spectrum. 
Then the mass matrices are written by four $3\times 3$ matrices
as
\begin{equation}
\left(\matrix{0 & A_I \cr B_I & C_I}\right),
\end{equation}
as in $SO(10)$ case.
It is obvious that the ranks of $A_I$ and $B_I$ reduce to two for 
$I=Q,U^c,E^c$ because
$(\alpha_I=0, \beta_I\neq 0)$ for $I=Q,U^c$ and
$(\alpha_I\neq 0, \beta_I=0)$ for $I=E^c$, where the vanishing parameters
are due to the NG modes eaten by Higgs mechanism
in breaking
$SO(10)\rightarrow SU(3)_C\times SU(2)_L\times SU(2)_R\times
U(1)_{B-L}$ (corresponding NG fields $Q+U^c+h.c.$) and
in breaking $SU(2)_R\times U(1)_{B-L}$ into $U(1)_Y$
(corresponding NG fields $E^c+h.c.$).
The mass spectrum become 
$(0,0,\lambda^{a'+a}, \lambda^{a'+a},\lambda^{c'+\bar c}, 
\lambda^{\bar c'+c}, \lambda^{2a'})$ for $I=Q,U^c$ and 
$(0,0,\lambda^{a'+a},\lambda^{a'+a},\lambda^{a'+a},
\lambda^{a'+a},\lambda^{\bar c'+c'})$ for $I=E^c$.


Finally, we examine the mass matrices of ${\bf 5}$ and ${\bf \bar 5}$
of $SU(5)$ and show how to realize the DT splitting.
Considering the additional terms,
we write the mass matrices $M_I$,
which are for the representations $I=D^c(H_T),L(H_D)$ and their 
conjugates:
\begin{equation}
M_I=\left(\matrix{ 0 & A_I & 0 \cr  B_I & C_I & D_I \cr 
                                    E_I & F_I & G_I \cr}\right),
\end{equation}
\begin{equation}
A_I=\bordermatrix{
\bar I\backslash I   &  10_{C'} 
                    & 10_{\bar C'} &
                    \overline{16}_{\bar C'} &
                    \overline{16}_{A'}  \cr
10_\Phi & S_I\lambda^{c'+\phi+\Delta\phi} & \lambda^{\bar c'+\phi} & 
0  & 0 \cr
10_C  & 0 & \lambda^{\bar c'+c} & 0 & 0 \cr
16_C & 0 & 0 & \lambda^{\bar c'+c}
   & 0 \cr
16_A  & 0 & \lambda^{\bar c'+a+\Delta c}& \lambda^{\bar c'+a+\Delta\phi}
   & \lambda^{a'+a} \cr
}, 
\end{equation}
\begin{equation}
B_I=\bordermatrix{
\bar I\backslash I  & 10_\Phi&10_C &\overline{16}_{\bar C}&\overline{16}_{A}\cr
10_{C'} &S_I\lambda^{c'+\phi+\Delta\phi} & 0 & 0 &\lambda^{c'+a-\Delta c} \cr
10_{\bar C'} &\lambda^{\bar c'+\phi} &\lambda^{\bar c'+c} &
\lambda^{\bar c'+\bar c-\Delta c} & \lambda^{\bar c'+a-\Delta\phi-\Delta c} \cr
16_{C'} & 0 & 0 &\lambda^{c'+\bar c} &\lambda^{a+c'-\Delta\phi} \cr
16_{A'}& 0 & 0 & 0 &\lambda^{a'+a} \cr
}, 
\end{equation}
\begin{equation}
C_I=\bordermatrix{
\bar I\backslash I  & 10_{C'} 
                    & 10_{\bar C'}  &
                    \overline{16}_{\bar C'} &
                    \overline{16}_{A'}  \cr
10_{C'}  & \lambda^{2c'+\Delta\phi} & 
  \lambda^{\bar c'+c'} &
  \lambda^{\bar c'+c'+\Delta\phi-\Delta c}  & \lambda^{c'+a'-\Delta c} \cr
10_{\bar C'}  &\lambda^{\bar c'+c'} & 
\lambda^{2\bar c'-\Delta\phi} & \lambda^{2\bar c'-\Delta c} & 
\lambda^{\bar c'+a'-\Delta\phi-\Delta c} \cr
16_{C'} & \lambda^{2c'+\Delta c} & 
  \lambda^{\bar c'+c'-\Delta\phi+\Delta c} &\lambda^{\bar c'+c'} &  
\lambda^{a'+c'-\Delta\phi} \cr
16_{A'}& \lambda^{c'+a'+\Delta\phi+\Delta c} & 
\lambda^{\bar c'+a'+\Delta c} & \lambda^{\bar c'+a'+\Delta\phi} &  
\lambda^{2a'} \cr
},
\end{equation}
\begin{equation}
D_I=\bordermatrix{
\bar I\backslash I  & 10_{\bar \Phi} &10_{\bar C} &\overline{16}_{\bar \Phi} \cr
10_{C'} &\lambda^{c'+\bar\phi} &\lambda^{c'+\bar c} & 
   \lambda^{c'+\bar\phi+\Delta\phi-\Delta c} \cr
10_{\bar C'} &\lambda^{\bar c'+\bar\phi-\Delta\phi} & 
       \lambda^{\bar c'+\bar c-\Delta\phi} & 
       \lambda^{\bar c'+\bar\phi-\Delta c} \cr
16_{C'} & \lambda^{c'+\bar\phi-\Delta\phi+\Delta c} &
      \lambda^{c'+\bar c-\Delta\phi+\Delta c} & \lambda^{c'+\bar\phi} \cr
16_{A'} & 0 & 0 & \lambda^{\bar\phi+a'+\Delta\phi} \cr
}, 
\end{equation}
\begin{equation}
E_I=
\bordermatrix{
\bar I\backslash I  &10_\Phi & 10_C& 
                    \overline{16}_{\bar C} &\overline{16}_{A}   \cr
10_{\bar \Phi} & 
0 & 0& 0 & 
\lambda^{\bar\phi+a-\Delta\phi-\Delta c}   \cr
10_{\bar C} &
0 & 0& 0 & 0
 \cr
16_\Phi & 0 & 0 &  0 & 0  \cr
},
\end{equation}
\begin{equation}
F_I=\bordermatrix{
\bar I\backslash I   &  10_{C'} 
                    & 10_{\bar C'} &
                    \overline{16}_{\bar C'} &
                    \overline{16}_{A'}  \cr
10_{\bar \Phi}  &
\lambda^{c'+\bar\phi} & \lambda^{\bar c'+\bar\phi-\Delta\phi}&
\lambda^{\bar c'+\bar\phi-\Delta c} & 
 \lambda^{\bar\phi+a'-\Delta\phi-\Delta c} \cr
10_{\bar C}  & \lambda^{\bar c+c'} &
\lambda^{\bar c'+\bar c-\Delta\phi}& \lambda^{\bar c'+\bar c-\Delta c} & 
\lambda^{\bar c+a'-\Delta\phi-\Delta c} \cr
16_\Phi  & 0 & \lambda^{\bar c'+\phi-\Delta\phi+\Delta c}& 
   \lambda^{\bar c'+\phi}
   & \lambda^{a'+\phi-\Delta\phi} \cr
}, 
\end{equation}
\begin{equation}
G_I=\bordermatrix{
\bar I\backslash I   &  10_{\bar \Phi} 
                    & 10_{\bar C} &
                    \overline{16}_{\bar \Phi} \cr
10_{\bar \Phi}  &
\lambda^{2\bar\phi-\Delta\phi} & \lambda^{\bar \phi+\bar c-\Delta\phi}&
\lambda^{2\bar\phi-\Delta c} \cr
10_{\bar C}  & \lambda^{\bar \phi+\bar c-\Delta\phi} & 0 & 0 \cr
16_\Phi  & 0 & 0 & 0 \cr
}, 
\end{equation}
where $S_{D^c}\neq 0$ and $S_L=0$.
It is obvious that the rank of $A_L$ is three, which is smaller than
the rank of $A_{D^c}$. This implies that the rank of $M_L$ is smaller
than the rank of $M_{D^c}$, and actually 
the rank of the matrix $M_I$ is 10 for $I=D^c$
and 9 for $I=L$. One pair of massless fields 
${\bf 16}$ and ${\bf \overline{16}}$ (main modes are 
${\bf 16}_{\Phi}$ and ${\bf \overline{16}}_{\bar \Phi}$)
is the Nambu-Goldstone mode,
which is eaten 
by Higgs mechanism in breaking $E_6$ into $SO(10)$. 
The other massless mode for $I=L$ is so called doublet
Higgs. The massless mode is given by
\begin{eqnarray}
H_u&\sim&\bar L({\bf 10}_\Phi)+\lambda^{\phi-c}\bar L({\bf 10}_C), \\
H_d&\sim&L({\bf 10}_\Phi)+\lambda^{\phi-c}L({\bf 10}_C).
\end{eqnarray}

As noted before, ${\bf 16}_{\Phi}$ and ${\bf \overline{16}}_{\bar \Phi}$
are eaten by Higgs mechanism, and ${\bf 10}_{\bar \Phi}$ and
${\bf 10}_{\bar C}$ can be massive through the matrix $G_I$, whose elements
are generally larger than the elements of $D_I$, $E_I$ and $F_I$.
Their masses become
$(\lambda^{\bar \phi+\bar c-\Delta\phi},
\lambda^{\bar \phi+\bar c-\Delta\phi})$.
We simply neglect the matrices $D_I$, $E_I$, $F_I$ and $G_I$
in the following argument.
Since the elements of $A_I$ and $B_I$ are generally larger than those
of $C_I$, we can estimate the mass spectrum of the other modes of
$D^c$ from $A_{D^c}$ and $B_{D^c}$
as $(\lambda^{c'+\phi+\Delta\phi},\lambda^{c'+\phi+\Delta\phi},
\lambda^{\bar c'+c},\lambda^{\bar c'+c},\lambda^{ \bar c'+ c},
\lambda^{ c'+\bar c},\lambda^{a'+a},\lambda^{a'+a})$ and
the mass spectrum of the other modes of $L$ as
$(0,\lambda^{\bar c'+c},\lambda^{\bar c'+c},\lambda^{ \bar c'+ c},
\lambda^{ c'+\bar c},\lambda^{a'+a},\lambda^{a'+a},\lambda^{2c'+\Delta\Phi})$.
It is obvious that to develop the proton decay, we have to pick up the element
of $C_I$. Since the element is generally smaller than the mass scale of $D^c$,
the proton decay can be suppressed.
The effective colored Higgs mass is estimated as
$(\lambda^{c'+\phi+\Delta\phi})^2/\lambda^{2c'+\Delta\phi}=
\lambda^{2\phi+\Delta\phi}$, 
which is usually larger than the cutoff scale.
For example, for the typical charge assignment in Table II, 
$2\phi+\Delta\phi=-17/2$.

By the above argument, the mass spectrum of superheavy particles are
determined only by the anomalous $U(1)_A$ charges, 
so we can examine whether coupling unification is realized or not. 
Before going to the discussion in the next subsection, we define the reduced 
mass matrices $\bar M_I$ by getting
rid of the massless modes from the original mass matrices $M_I$.
The rank of the reduced matrices in our $E_6$ model are
$\bar r_{X}=1$, $\bar r_G=\bar r_W=2$, 
$\bar r_{Q}=\bar r_{U^c}=\bar r_{E^c}=5$, $\bar r_L=9$ and 
$\bar r_{D^c}=10$.
It is interesting that the determinants of the reduced mass matrices are
estimated mainly by simple sums of the anomalous $U(1)_A$ charges of
massive modes:
\begin{eqnarray}
\det \bar M_I (I=G,W)&=& \lambda^{2(a+a')} \\
\det \bar M_X&=&\lambda^{2a'} \\
\det \bar M_I(I=Q,U^c)&=& \lambda^{2a+4a'+c+\bar c+c'+\bar c'}\\
\det \bar M_{E^c}&=&\lambda^{4a+4a'+c'+\bar c'} \\
\det \bar M_D^c&=&\lambda^{3(c+\bar c+c'+\bar c')+2(a+a'+\phi+\bar \phi)} \\
\det \bar M_L&=&\lambda^{3(c+\bar c+c'+\bar c')+2(a+a'+\bar \phi)-\Delta\phi}.
\end{eqnarray}
Note that the last equation for $\det \bar M_L$ is determined by not simple
sum of the charges of massive modes. The difference is $-\Delta\phi$. This 
is because the masses of ${\bf 10}$ representation of $SO(10)$ in the 
multiplets $X_i({\bf 27})$ of $E_6$ are estimated as 
$\lambda^{x_i+x_j+\Delta\phi}$ from the terms $X_iX_j\Phi$ with VEV
$\VEV{\Phi}\sim\lambda^{-\frac{1}{2}(\phi+\bar \phi)}$, which are 
different from the naive expectation $\lambda^{x_i+x_j}$.
In order to estimate the element of mass matrices, it is useful to introduce 
`effective' charges:
\begin{eqnarray}
&&x_i(10(5+\bar 5),27)\equiv x_i+\frac{1}{2}\Delta\phi,\quad
\bar x_i(10(5+\bar 5),\overline{27})\equiv \bar x_i-\frac{1}{2}\Delta\phi, \\
&&x_i(16(\bar 5),27)\equiv x_i+\Delta c-\frac{1}{2}\Delta\phi,\quad
\bar x_i(\overline{16}(5),\overline{27})\equiv 
\bar x_i-\Delta c+\frac{1}{2}\Delta\phi, \\
&&x_i(16(10),27)\equiv x_i, \quad
\bar x_i(\overline{16}(\overline{10}),\overline{27})\equiv x_i, \\
&&a(16(\bar 5),78)\equiv a+\Delta c+\frac{1}{2}\Delta\phi,\quad
a(\overline{16}(5),78)\equiv a-\Delta c-\frac{1}{2}\Delta\phi, \\
&&a(16(10),78)\equiv a+\Delta\phi,\quad
a(\overline{16}(\overline{10}),78)\equiv a-\Delta\phi,\\
&&a(45(10),78)\equiv a+\Delta c,\quad
a(45(\overline{10}),78)\equiv a-\Delta c, \\
&&a(24,78)\equiv a.
\end{eqnarray}
Here, the effective charges $x_i(10(5+\bar 5),27)$
are for ${\bf 10}$ of $SO(10)$ from $X_i({\bf 27})$ of $E_6$ and
$\bar x_i(10(5+\bar 5),\overline{27})$ are 
for ${\bf 10}$ of $SO(10)$ from $\bar X_i({\bf \overline{27}})$ of $E_6$,
etc.

Then all the elements of mass matrices are estimated by simple sum of 
the effective charges 
of superheavy particles if they are not vanishing, 
and the determinants of mass matrices are
also determined by simple sum of the effective charges.
We will use this result in calculating running gauge couplings
later.


\subsection{Coupling unification}
In this subsection, we apply the general discussion on the gauge coupling 
unification in Ref.\cite{maekawa3} to our scenario.

The pattern of the $E_6$ breaking in our model is as follows.
At the scale $\Lambda_\Phi\sim \lambda^{-(\phi+\bar \phi)/2}$, 
$E_6$ is broken into $SO(10)$. The $SO(10)$ is broken into
$SU(3)_C\times SU(2)_L\times SU(2)_R\times U(1)_{B-L}$ at the scale
$\Lambda_A\sim \lambda^{-a}$, which is broken into the standard gauge
group at the scale $\Lambda_C\sim \lambda^{-(c+\bar c)/2}$.

In this paper, we make an analysis based on the renormalization group
equations up to one loop. 
\footnote{Since we neglect the order one coefficients,
the higher order calculation does not improve the accuracy.}
The conditions of the gauge coupling unification are given by
\begin{equation}
\alpha_3(\Lambda_A)=\alpha_2(\Lambda_A)=
\frac{5}{3}\alpha_Y(\Lambda_A)\equiv\alpha_1(\Lambda_A),
\end{equation}
where 
$\alpha_1^{-1}(\mu>\Lambda_C)\equiv 
\frac{3}{5}\alpha_R^{-1}(\mu>\Lambda_C)
+\frac{2}{5}\alpha_{B-L}^{-1}(\mu>\Lambda_C)$.
Here $\alpha_X=\frac{g_X^2}{4\pi}$ and 
the parameters $g_X (X=3,2,R,B-L,Y)$ are the gauge couplings of 
$SU(3)_C$, $SU(2)_L$, $SU(2)_R$, $U(1)_{B-L}$ and $U(1)_Y$, 
respectively.

Using the fact that the three gauge couplings of the minimal SUSY standard
model (MSSM) meet at the scale $\Lambda_G\sim 2\times 10^{16}$ GeV,
the above conditions for gauge coupling unification are rewritten
\begin{eqnarray}
&&b_1\ln \left(\frac{\Lambda_A}{\Lambda_G}\right)
+\Sigma_I\Delta b_{1I}\ln \left(\frac{\Lambda_A^{\bar r_I}}{\det \bar M_I}
\right)
-\frac{12}{5}\ln \left(\frac{\Lambda_A}{\Lambda_C}\right) \label{alpha1}\\
&=&b_2\ln \left(\frac{\Lambda_A}{\Lambda_G}\right)
+\Sigma_I\Delta b_{2I}\ln\left(\frac{\Lambda_A^{\bar r_I}}{\det \bar M_I}
\right) \\
&=&b_3\ln \left(\frac{\Lambda_A}{\Lambda_G}\right)
+\Sigma_I\Delta b_{3I}\ln\left(\frac{\Lambda_A^{\bar r_I}}{\det \bar M_I}
\right),
\end{eqnarray}
where $(b_1,b_2,b_3)=(33/5,1,-3)$ are the 
renormalization group coefficients
for MSSM
and $\Delta b_{aI}(a=1,2,3)$ are the correction to the coefficients 
from the massive fields $I=Q+\bar Q,U^c+\bar U^c, E^c+\bar E^c, D^c+\bar D^c,
L+\bar L, G, W$ and $X+\bar X$.
The last term in Eq. (\ref{alpha1}) is from the breaking 
$SU(2)_R\times U(1)_{B-L}\rightarrow U(1)_Y$ by the VEV $\VEV{C}$.
Since all the mass matrices and the symmetry breaking scale appearing 
in the above conditions are determined by the anomalous $U(1)_A$ charges,
these conditions can be translated to the constraint of the effective
charge of the doublet Higgs fields,
\begin{equation}
\phi+\frac{1}{2}\Delta\phi\sim 0,
\label{h}
\end{equation}
and to the condition on the cutoff scale,
\begin{equation}
\Lambda\sim \Lambda_G.
\end{equation}
As discussed in Ref. \cite{maekawa3}, this is quite general result, and 
independent on the detail of the Higgs sector. The essential point is that
only the charges of massless modes are important to determine whether
coupling unification is realized or not and all other effects are cancelled
out in the unification conditions except the charge of doublet Higgs.
Note that the condition (\ref{h}) does not require
$\phi+\frac{1}{2}\Delta\phi= 0$. Actually, even with the typical charge
assignment, in which $\phi+\frac{1}{2}\Delta\phi= -4.25$, the coupling
unification is realized, using the ambiguities of order one coefficients
(see Fig. I).

Since the unification scale
$\Lambda_U\sim \lambda\Lambda_G$ becomes smaller than the usual GUT scale
$\Lambda_G\sim 2\times 10^{16}$ GeV, proton decay via dimension six operator
$p\rightarrow e^+\pi^0$
may be seen in near future. If we roughly estimate the lifetime of proton
using the formula in Ref.~\cite{hisano} and the recent result of 
the lattice calculation for the hadron matrix element parameter
$\alpha$\cite{lattice}, the lifetime of the proton in our scenario
becomes
\begin{equation}
\tau_p(p\rightarrow e^+\pi^0)\sim 2.8\times 10^{33}\left(\frac{\Lambda_U}
{5\times 10^{15}{\rm GeV}}\right)^4
\left(\frac{0.015}{\alpha}\right)^2  {\rm years},
\end{equation}
because the unification scale is around $5\times 10^{15}$ GeV.
It is interesting that this value of the lifetime is just around the 
present experimental limit
\cite{SKproton}
\begin{equation}
\tau_{exp}(p\rightarrow e^+\pi^0)>2.9\times 10^{33} {\rm year}.
\end{equation}
Of course, since we have ambiguities of order one coefficients and of
the hadron matrix element parameter $\alpha$, and
the lifetime of proton is strongly dependent on the GUT scale and 
the parameter,
this prediction may not be so reliable. 
However, the above rough estimation gives us a strong motivation for
experiments of proton decay search, 
because the lifetime of nucleon via dimension
six operator
must be less than that in the usual SUSY GUT scenario. 

We have to comment on the proton decay via dimension five operators.
The effective colored Higgs mass is given by
$\lambda^{2\phi+\Delta\phi}\Lambda$, so the experimental 
constraint requires $2\phi+\Delta\phi\leq -3$. 
With 
the typical charge assignment in Table II, the effective colored Higgs
mass is around $\sim 10^{22}$ GeV, so the proton decay via dimension 
five operator is still suppressed. 

\subsection{How to determine charges}
It is worth while explaining how to determine
the symmetry and the quantum numbers in the Higgs sector to realize 
DT splitting. 
There are several terms which must be forbidden in order to realize 
DT splitting;
\begin{enumerate}
\item $\Phi^3$, $\Phi^2C$, $\Phi^2C'$, $\Phi^2C'Z$ induce a large mass 
of the doublet Higgs.
\item $\bar CA'C$,$\bar CA'AC$,$\bar \Phi A'\Phi$ would destabilize the
DW form of the $\VEV{A}$. 
\item $\bar \Phi A'C$, $\bar CA'\Phi$, $\bar \Phi A'AC$, $\bar CA'A\Phi$, 
$\bar \Phi A'ZC$, $\bar C A'Z\Phi$ lead to undesired VEV 
$\VEV{{\bf 16}_C}=0$
unless  another singlet field is introduced. 
\item $A'A^n(n\geq 4)$ make it less natural to obtain DW type of VEV.
\end{enumerate}
Most of these terms can be easily forbidden
by SUSY zero mechanism. For example, if we choose $\phi<0$, then $\Phi^3$ is
forbidden, and if we choose $\bar c+c+a+a'<0$, then $\bar CA'AC$ is forbidden.
Once these terms are forbidden by the SUSY zero mechanism, higher-dimensional 
terms, which also become dangerous, for example, $\bar CA'A^3C$ and 
$\bar CA'C\bar CAC$, are automatically forbidden as in $SO(10)$ cases.

On the other hand, we have to prepare the following terms;
\begin{enumerate}
\item $A'A$, $\bar \Phi A'A^3\Phi$ to obtain DW type VEV $\VEV{A}$.
\item $\Phi^2AC'$ for doublet-triplet splitting.
\item $\bar C'(A+Z)C$, $\bar C(A+Z)C'$ for alignment between the VEVs $\VEV{A}$
and $\VEV{C}$ and for giving the superheavy masses to the PNGs.
\item $\bar\Phi A'A\Phi$ for alignment between 
the VEVs $\VEV{A}$ and $\VEV{\Phi}$ and for giving the superheavy masses 
to the PNGs.
\item $\bar \Phi^3$,$\bar \Phi^2\bar C$ for giving superheavy masses to two 
${\bf 10}$ of $SO(10)$, that save the number of fields with positive charges.
\end{enumerate}

In order to forbid $\Phi^2C'$ while $\Phi^2AC'$ is allowed, 
we have to introduce $Z_2$ parity. The same $Z_2$ parity can forbid
$\bar\Phi A'\Phi$, allowing the term $\bar\Phi A'A\Phi$. 
We have some ambiguities to assign the $Z_2$ parity, but once the parity is 
fixed, the above requirements become just inequalities.
In addition to these inequalities, we require that the total charges of the 
operators $A$, $\bar \Phi\Phi$ and $\bar CC$ are negative. 
If, as discussed in the previous section, we adopt $a=-1$ to realize the 
proton stability, then the inequalities $a'+3a+\phi+\bar \phi\geq 0$
and $a'+5a<0$ lead to $\phi+\bar \phi=-1$ and $a'=4$. The relation
$\bar\phi+\phi=-1$ means that the gauge singlet
operator $\bar\Phi\Phi$ can be regarded as the FN field $\Theta$ as discussed 
in section 2.
The other inequalities are easily
satisfied.

Of course, the above conditions are necessary but not sufficient. 
As in the previous subsection, we have to write down the mass matrices of 
Higgs sector to know whether an assignment truly works well or not. 


\section{Constraints from the matter sector}
For the matter fields, we introduce three ${\bf 27}$ $\Psi_i$ ($i=1,2,3$).
As discussed in Ref. \cite{maekawa,BM}, we adopt 
$(\psi_1,\psi_2,\psi_3)=(3+n,2+n,n)$ and the charge of the Higgs $\phi=-2n$
in order to realize CKM matrix and large top Yukawa coupling of $O(1)$.
As discussed in the previous section, the cutoff scale $\Lambda$ must be
around the usual GUT scale $\Lambda_G\sim 2\times 10^{16}$ GeV. This 
requires $a\leq -1$. If we have integer charge for $a$, then we must adopt
$a=-1$. 
As discussed in Ref.~\cite{BM}, the conditions for obtaining the realistic
quark and lepton mass matrices with bi-large neutrino mixing angle are
\begin{equation}
c-\bar c=\phi-\bar\phi+1=2n-9-l,
\end{equation}
where we have some ambiguities of the parameter $l$, but as
discussed in Ref.~\cite{BM}, for the models with $\Lambda\sim \Lambda_G$,
we adopt $l=-1$ or $-2$.
\footnote{
Using the ambiguities of order one coefficients, rather larger range of the
parameter $l$ may be allowed, for example, $l=-3,-4$. But in the following
discussion, these larger ambiguities do not change the result.
}
Since the condition for gauge coupling unification is 
$\phi+\frac{1}{4}(\phi-\bar \phi)=\frac{1}{4}(-6n-10-l)\sim 0$,
the small $n$ is required. From the condition 
$\phi+\bar \phi=-6n+10+l\leq  -1$, the smallest value of 
$n$ becomes $3/2$ for $l=-2$. Then we obtain $\phi=-3$, $\bar \phi=2$ and
$c-\bar c=-4$. It is non-trivial that the conditions for bi-large neutrino
mixing angle and for gauge coupling unification lead to $\bar \phi+\phi=-1$ 
which is required for DT splitting.
Since $\phi+\bar\phi=-1$, we must adopt $a'=4$ to allow
the important term $\bar \Phi A'A^3\Phi$ and to forbid the term $A'A^5$. 
Then, in order to forbid the terms
$\bar \Phi A'AC$, $\bar CA'A\Phi$ and $\bar CA'AC$, we have to take
$c<-5$ and $\bar c<0$. On the other hand, since the term $\Psi_1\Psi_3C$
is required to realize the realistic mass matrices of quark and lepton,
as discussed in Ref.~\cite{BM}, $c\geq -6$. Then we must take $c=-6$ and
$\bar c=-2$. It is stimulating  that the economical condition for the 
$\mu$ problem\cite{maekawa2,maekawa3}
\begin{equation}
-1\leq 2\phi-(c+\bar c)+\frac{1}{2}(\phi-\bar \phi)\leq 1,
\label{mu}
\end{equation}
is automatically satisfied and the required terms $\bar\Phi^3$ and 
$\bar\Phi^2\bar C$ happen to be allowed.
We have some freedom to choose the charges $z$, $c'$ and $\bar c'$. 
If we take $z=-2$, then we must adopt $c'=7$ because the term $ZC'\Phi^2$
must be forbidden while allowing the term $C'A\Phi^2$. 
On the other hand, $\bar c'\geq 8$ is required for obtaining the term
$\bar C'(A+Z)C$. 
In the typical charge assignment, we adopt the minimal value $\bar c'=8$.
\footnote{
In order to solve the $\mu$ problem economically, we need the term 
$\bar C'\Phi Z(\Phi^3)$, or  $\bar C'\Phi (\Phi^3)$ when we adopt 
the even $Z_2$ parity for $C'$. That leads to $\bar c'\geq 14$ or
$\bar c'\geq 12$. 
However, in these cases, the gauge couplings seem to diverge below
the GUT scale. Therefore, the additional gauge singlet field $S$ 
with positive charge $s\geq 3\phi$ may be required.
}


The charges of the matter sector 
$\Psi_i({\bf 27})(i=1,2,3)$ become half integer as
$(\psi_1,\psi_2,\psi_3)=(9/2,7/2,3/2)$ in this case.
It is interesting that we do not have to introduce 
R-parity additionally because
half integer anomalous $U(1)_A$ charges can play the same role.

Anyway all the charges are fixed except singlets. So we can
calculate the running flows of the gauge couplings (see Fig. \ref{fig_1}). 
\begin{figure}[htb]
\begin{center}
\leavevmode
\epsfxsize=110mm
\put(300,50){{\large $\bf{\log \mu (GeV)}$}}
\put(0,260){{\Large $\bf{\alpha^{-1}}$}}
\put(29,240){$\alpha_1^{-1}$}
\put(31,150){$\alpha_2^{-1}$}
\put(31,90){$\alpha_3^{-1}$}
\epsfbox{E6run4.eps}
\vspace{-2cm}
\caption{
Here we adopt $\lambda=0.25$, 
$\alpha_1^{-1}(M_Z)=59.47$, $\alpha_2^{-1}(M_Z)=29.81$,
$\alpha_3^{-1}(M_Z)=8.40$, the SUSY breaking scale $m_{SB}\sim 1$ TeV and
the anomalous $U(1)_A$ charges: $a'=4$, $a=-1$, 
$(\psi_1,\psi_2,\psi_3)=(9/2,7/2,3/2)$, $\phi=-3$, $\bar \phi=2$, $c=-6$,
$\bar c=-2$, $z=-2$, $c'=7$ and $\bar c'=8$.
Using the ambiguities of coefficients $0.5\leq y\leq 2$,
three gauge couplings meet at around 
$\lambda^{-a}\Lambda_G\sim 5\times 10^{15}$ GeV. 
}
\label{fig_1}
\end{center}
\end{figure}
Here we use the ambiguities of the coefficients $0.5\leq y\leq 2$.
It is shown that the three gauge couplings actually meet around
$\lambda^{-a}\Lambda_G\sim 5\times 10^{15}$ GeV. Note that
the unified gauge coupling at the cutoff scale is still finite, 
though it is so large that the perturbative calculation is not reliable.
Since the value of the unified gauge coupling is strongly dependent on the 
concrete charge assignment and the value $\lambda$ (weakly dependent
on the coefficients of order one), we expect that it is large but can
be finite at the cutoff scale. 

\section{Discussion and summary}
It is worth while emphasizing that the effective $SO(10)$ theory which
is obtained by the non-vanishing VEVs $\VEV{\Phi}=\VEV{\bar\Phi}$ from 
$E_6$ GUT is generically different from the $SO(10)$ GUT with anomalous
$U(1)_A$ gauge symmetry.
This is because the VEV 
$\VEV{\Phi}\sim \lambda^{-\frac{1}{2}(\phi+\bar\phi)}$ 
is generically different from the naively expected value 
$\VEV{\Phi}\sim \lambda^{-\phi}$. For example, the mass of 
${\bf 10}_F$ of $SO(10)$ is obtained from the term 
$\lambda^{2f+\phi}F({\bf 27})^2\Phi({\bf 27})$ by developing the
VEV $\VEV{\Phi}\sim \lambda^{-\frac{1}{2}(\phi+\bar \phi)}$ as
$m_F\sim \lambda^{2f+\frac{1}{2}(\phi-\bar\phi)}$, which is different
from the naively expected value $m_F\sim \lambda^{2f}$.
This is the same effect as in discussing the right-handed neutrino mass
in $SO(10)$ GUT in Ref. \cite{maekawa}. The right-handed neutrino mass
has been estimated as $m_{\nu_F}\sim \lambda^{2f+\frac{1}{2}(c-\bar c)}$,
which is different from the naively expected value $\lambda^{2f}$. 
Therefore it is obvious that the $SO(10)$ or $E_6$ GUT with 
anomalous $U(1)_A$ gauge symmetry
is essentially different from the MSSM with anomalous $U(1)_A$ gauge
symmetry. 
Note that the difference is caused by the $\Delta\phi$ and $\Delta c$. 



In this paper, we have proposed a realistic Higgs sector in 
$E_6$ grand unified theory, which can realize DT splitting and proton
stability. In the scenario, the anomalous $U(1)_A$ gauge symmetry plays 
a critical role.
Moreover, using the matter sector in $E_6$ GUT in Ref. \cite{BM},
we have proposed a totally consistent $E_6$ GUT scenario.
Since we have introduced all the interactions which are allowed by
the symmetry, the model could be defined by the symmetry and the 
quantum numbers of the fields. In our scenario, after deciding
the fields contents (Higgs and matter), the model can be defined
only by 
8 integer charges (+3 charges for singlet fields)
in Higgs sector and by 3 (half-)integer charges for the matter sector.
It is quite appealing that by choosing only 11 (half-)integer charges
(+3 charges for singlet fields),
not only  DT splitting with proton stability and gauge coupling
unification is realized, but also the realistic structure of 
quark and lepton mass matrices
including bi-large neutrino mixing are obtained.
Moreover, the FCNC process is automatically suppressed. 
In other words, the charge assignment, which is almost decided by 
realizing realistic
quark and lepton mass matrices and gauge coupling unification, also solves
the DT splitting with proton stability. 
For example, the relation $\bar \phi+\phi=-1$ is required for obtaining 
realistic quark 
and lepton mass matrices and gauge coupling unification, but is also 
independently needed for realizing DT splitting.
Moreover, since the half integer charges of the matter sector plays the 
same role as $R$-parity, we do not have to introduce $R$-parity additionally.

Of course if the half integer charges are allowed to be assigned to the Higgs
sector, there 
appear other possibilities.
For example, we can adopt another charge assignment, 
in which the half integer charges
play the same role as $Z_2$ parity. If we take
$a=-1/2,a'=5/2,\phi=-3,\bar\phi=2,c=-5,\bar c=-1,c'=13/2,\bar c'=13/2,
z_i=-i/2(i=3,7,11),
\psi_1=9/2,\psi_2=7/2,\psi_3=3/2$ with odd $R$-parity to the matter sector
$\Psi_i$, we can obtain the totally consistent $E_6$ GUT again.
Since the absolute value of the charges of this model are smaller than the 
previous model, the unified gauge coupling at the unified scale 
$\lambda^{-a}$ becomes smaller than that of the previous model.
Therefore the unified gauge coupling at the cutoff scale must be
finite in this model.
And since the unification scale $\lambda^{-a}$ is larger than that of 
previous model, the model predicts longer lifetime of the nucleon,
which is roughly estimated as
\begin{equation}
\tau_p(p\rightarrow e^+\pi^0)\sim 4.5\times 10^{34}\left(\frac{\Lambda_U}
{10^{16}{\rm GeV}}\right)^4
\left(\frac{0.015}{\alpha}\right)^2  {\rm years}.
\end{equation}
Though this predicted value is rather longer than the present experimental
limit, we hope that the next project can reach this value.
\footnote{
Unfortunately since the term $A'A^5$ is allowed by the symmetry 
in this charge assignment,
this model becomes less natural to obtain DW type VEV than the previous model.
However, the number of VEVs is still finite.
}

Though the requirement to the $E_6$ GUT is so severe that the satisfied
charge assignment is fairly restricted, we have several possibilities
of the charge assignment.
%We can build several models in the scenario,
However, since our scenario requires only several integer charges
as the input paramters to do well both in 
the matter sector and in the Higgs sector non-trivially, 
we are expecting that the scenario must be describing our world.


\section{Acknowledgement}
We would like to thank M. Bando and T. Kugo for their collaborating at 
the early stage of this work. We also thank M. Bando for reading this 
manuscript and valuable comments, which make this manuscript more easily
understandable.

%--------------------<<   section    >>--------------------

\appendix

\section{Factorization}
  As mentioned in section 4, the naive extension
of DT splitting in $SO(10)$ GUT into $E_6$ GUT
does not work.
  In the $SO(10)$ DT splitting, the interaction
$(A^\prime A)_{\bf 54}(A^2)_{\bf 54}$ plays an essential role.
%  In the $E_6$ DT splitting, however , the term $A^\prime A^3$
  In $E_6$ GUT, however , the term $A^\prime A^3$
does not include the interaction 
$({\bf 45}_{A'}{\bf 45}_A)_{\bf 54}({\bf 45}_A^2)_{\bf 54}$.
  So the superpotential
\begin{equation}
W_{A^\prime} =  \lambda^{a^\prime+a}\alpha A^\prime A
              + \lambda^{a^\prime+3a} ( \beta(A^\prime A)_{\bf 1}
                                             (A^2)_{\bf 1}
                                      + \gamma(A^\prime A)_{\bf 650}
                                              (A^2)_{\bf 650})
\end{equation}
does not realize the DW VEV naturally.

  Here we show that the term $A^\prime A^3$ of $E_6$ actually does not
include the interaction $({\bf 45}_{A'}{\bf 45}_A)_{\bf 54}
({\bf 45}_A^2)_{\bf 54}$ of $SO(10)$.

  The VEV of $SO(10)$ adjoint Higgs can be represented as the form
$\VEV{A}=\tau_2\times {\rm diag}(x_1,x_2,x_3,x_4,x_5)$ because of the
$SO(10)$ rotation and D-flatness condition (see appendix
\ref{D-flatness}).
  In this gauge,
\beqn
  A^\prime A &=& 2 \sum_i x^\prime_i x_i ,\\
  (A^\prime A)_{\bf 54}(A^2)_{\bf 54} &=& 2 \sum_i x^\prime_i {x_i}^3 
  -\frac{2}{5}\left(\sum_i x'_ix_i\right)\left(\sum_jx_j^2\right).
\eeqn
  In the same manner, the VEV of $E_6$ adjoint Higgs can be
represented as the form $\VEV{{\bf 1}_A}=y$ , $\VEV{{\bf 16}_A}=
\VEV{{\bf \overline{16}}_A}=0$, $\VEV{{\bf 45}_A}=\tau_2\times
{\rm diag}(x_1,x_2,x_3,x_4,x_5)$.
  In this gauge , it can be represented
as $27 \times 27$ matrix as follows,
\begin{equation}
  \VEV{A} = \left(
  \begin{array}{ccc}
    {2\over \sqrt3}y & 0 & 0 \\
    0 & \theta^{MN}{T_{16}}^{MN}
        + {1\over2\sqrt3}y{\bf 1}_{16} & 0 \\
    0 & 0 & \theta^{MN}{T_{10}}^{MN}
            - {1\over\sqrt3}y{\bf 1}_{10} \\
\end{array}
\right)\label{A-matrix}.
\end{equation}
  Here, ${T_{i}}^{MN}$ are the $i \times i$ matrix representation of
$SO(10)$ generators and summation of indices M, N is understood as
they run from 1 to 10 with  $M>N$ . ${\bf 1}_{i}$ is the
$i \times i$ unit matrix.

  In concrete,
\beqn
  ({T_{10}}^{MN})_{KL} &=& -i( \delta^M_K\delta^N_L
                             - \delta^M_L\delta^N_K ), \\
  ({T_{16}}^{MN})_{\alpha\beta}
                       &=& {1\over2}(\sigma^{MN})_{\alpha\beta} \nn\\
                       &=& {1\over4i}([\gamma^M , \gamma^N]P_R)
                                     _{\alpha\beta}, \\
  \theta^{MN} &=& \Lm
    \begin{array}{cl}
      x_n & M+1=N=2n, (n=1,\cdots,5) \\
      0   & {\rm otherwise}
    \end{array}
  \right.
\eeqn
where $\gamma^M$ are $SO(10)$ $\gamma$-matrices and $P_R$ is the
right-handed projector which can be written as
\beqn
  \gamma^1 &=& \tau_1 \otimes {\bf 1} \otimes {\bf 1}
                      \otimes {\bf 1} \otimes {\bf 1}  \\
  \gamma^2 &=& \tau_3 \otimes {\bf 1} \otimes {\bf 1}
                      \otimes {\bf 1} \otimes {\bf 1}  \\
  \gamma^3 &=& \tau_2 \otimes \tau_1 \otimes {\bf 1}
                      \otimes {\bf 1} \otimes {\bf 1}  \\
  \gamma^4 &=& \tau_2 \otimes \tau_3 \otimes {\bf 1}
                      \otimes {\bf 1} \otimes {\bf 1}  \\
  \gamma^5 &=& \tau_2 \otimes \tau_2 \otimes \tau_1
                      \otimes {\bf 1} \otimes {\bf 1}  \\
  \gamma^6 &=& \tau_2 \otimes \tau_2 \otimes \tau_3
                      \otimes {\bf 1} \otimes {\bf 1}  \\
  \gamma^7 &=& \tau_2 \otimes \tau_2 \otimes \tau_2
                      \otimes \tau_1 \otimes {\bf 1}  \\
  \gamma^8 &=& \tau_2 \otimes \tau_2 \otimes \tau_2
                      \otimes \tau_3 \otimes {\bf 1}  \\
  \gamma^9 &=& \tau_2 \otimes \tau_2 \otimes \tau_2
                      \otimes \tau_2 \otimes \tau_1   \\
  \gamma^{10} &=& \tau_2 \otimes \tau_2 \otimes \tau_2
                         \otimes \tau_2 \otimes \tau_3\\
  \gamma^{11} &=& i\gamma^1\gamma^2\gamma^3\gamma^4\gamma^5
                  \gamma^6\gamma^7\gamma^8\gamma^9\gamma^{10}\nn\\
              &=& \tau_2 \otimes \tau_2 \otimes \tau_2
                      \otimes \tau_2 \otimes \tau_2   \\
  P_R &=& {1+\gamma^{11}\over2}.
\eeqn
In this basis,
\beqn
  \theta^{MN}{T_{16}}^{MN} &=& -{1\over2}(
                    x_1 \tau_2 \otimes {\bf 1} \otimes {\bf 1}
                      \otimes {\bf 1} \otimes {\bf 1}  \nn\\
                &&\hspace{2mm}+ x_2 {\bf 1} \otimes \tau_2 \otimes {\bf 1}
                      \otimes {\bf 1} \otimes {\bf 1}  \nn\\
                &&\hspace{2mm}+ x_3 {\bf 1} \otimes {\bf 1} \otimes \tau_2
                      \otimes {\bf 1} \otimes {\bf 1}  \nn\\
                &&\hspace{2mm}+ x_4 {\bf 1} \otimes {\bf 1} \otimes {\bf 1}
                      \otimes \tau_2  \otimes {\bf 1}  \nn\\
                &&\hspace{2mm}+ x_5 {\bf 1} \otimes {\bf 1} \otimes {\bf 1}
                      \otimes {\bf 1} \otimes \tau_2
                                        ) P_R, \\
                         &\equiv&  B\\
  \theta^{MN}{T_{10}}^{MN} &=& \tau_2 \otimes {\rm diag}
                               (x_1,x_2,x_3,x_4,x_5).  \\
                         &\equiv&  C
\eeqn

  Before beginning the calculation, we should see what coupling
can occur in the term $A^\prime A^3$ of $E_6$.
  Because ${\bf 78\times78 = 1_s+78_a+650_s+2430_s+2925_a}$,
$A^\prime A^3 \ni (A^\prime A)_{\bf 1}(A^2)_{\bf 1},
                  (A^\prime A)_{\bf 650}(A^2)_{\bf 650},
                  (A^\prime A)_{\bf 2430}(A^2)_{\bf 2430}$.
  On the other hand, because of the completeness,
\bequ
  (A_1A_2)_{\bf 2430}(A_3A_4)_{\bf 2430} =
               \sum_{I={\bf 1,78,650,2430,2925}}
               \lambda_I(A_1A_4)_I(A_3A_2)_I.
\eequ
  So,
\bequ
  (A^\prime A)_{\bf 2430}(A^2)_{\bf 2430} =
               \sum_{I={\bf 1,650,2430}}
               \lambda_I(A^\prime A)_I(A^2)_I,
\eequ
which implies that the above three couplings are not independent
and it is sufficient to examine first two. They are essentially described
as ${\rm tr}A^\prime A{\rm tr}A^2$ and ${\rm tr}A^\prime A^3$ in
matrix language.
If the desirable coupling existed, it is  apparently included
only in $(A^\prime A)_{\bf 650}(A^2)_{\bf 650}$ and
${\rm tr}A^\prime A^3$.
So we can conclude that it dose not exist if
${\rm tr}A^\prime A^3$ dose not include$\sum_i x^\prime_i {x_i}^3$.

  From (\ref{A-matrix}), we can see
\beqn
  {\rm tr}A^\prime A &=& {4\over3}y^\prime y
                        +{\rm tr}_{16}\Ll B^\prime B
                              + {1\over2\sqrt3}B^\prime y
                              + {1\over2\sqrt3}y^\prime B
                              + {1\over12}y^\prime y \Rl  \nn\\
                     &&   +{\rm tr}_{10}\Ll C^\prime C
                              - {1\over\sqrt3}C^\prime y
                              - {1\over\sqrt3}y^\prime C
                              + {1\over3}y^\prime y \Rl   \nn\\
                     &=& \Ls {4\over3}+{16\over12}+{10\over3}\Rs
                                                      y^\prime y
                        +\Ls 16{1\over2^2}+ 2\Rs
                                         \sum_i x^\prime_i x_i \nn\\
                     &=& 6\Ls y^\prime y + \sum_i x^\prime_i x_i\Rs.
\eeqn
  Similarly,
\beqn
  {\rm tr}A^\prime A^3 &=& {16\over9}y^\prime y^3
                          +{\rm tr}_{16}\Ll B^\prime B^3
                                + 3{1\over12}\Ls B^\prime By^2
                                + y^\prime yB^2\Rs
                                + {1\over144}y^\prime y^3 \Rl  \nn\\
                      &&    +{\rm tr}_{10}\Ll C^\prime C^3
                                + 3{1\over3}\Ls C^\prime Cy^2
                                + y^\prime yC^2\Rs
                                + {1\over9}y^\prime y^3 \Rl  \nn\\
                       &=& {16\over9}y^\prime y^3  
                       +16\Ll{1\over2^4}\Ls3\sum_i x^\prime_i x_i
                                               \sum_i x_i^2
                                            - 2\sum_i x^\prime_i
                                               x_i^3\Rs\right.\nn\\
                        &&       + \left.{3\over12}{1\over2^2}
                                           \Ls\sum_i x^\prime_ix_iy^2
                                            + y^\prime y \sum_i
                                              x^\prime_ix_i\Rs
                               + {1\over144}y^\prime y^3 \Rl  \nn\\
                      &&    +\Ll 2\sum_i x^\prime_ix_i^3
                                + {3\over3}\Ls 2\sum_i
                                                x^\prime_ix_iy^2
                                            + y^\prime y 2\sum_i
                                              x^\prime_ix_i\Rs
                                + {10\over9}y^\prime y^3 \Rl  \nn\\
                      &=& 3\Ls y^\prime y + \sum_i
                                  x^\prime_ix_iy^2  \Rs
                           \Ls y^2 + \sum_i x_i^2  \Rs  \nn\\
                      &=& {1\over12}{\rm tr}A^\prime A{\rm tr}A^2.
\eeqn
So, desirable coupling dose not exist because of the group theoretical
cancellation between the contribution from the ${\rm tr}_{\rm 16}$ part
and that from the ${\rm tr}_{\rm 10}$ part.\\


There are several solutions and the simplest one is to use the term
${\overline \Phi}A^\prime A^3 \Phi$.
At first glance, it seems to have no effect because
$\Phi {\overline \Phi}$ is written as
\begin{equation}
  \Phi {\overline \Phi} = \left(
  \begin{array}{ccc}
    \VEV{{\overline \Phi}\Phi} & 0 & 0 \\
    0 & 0_{16} & 0 \\
    0 & 0 & 0_{10} \\
  \end{array}
\right).
\end{equation}

  However this form is the special combination of
$(\Phi {\overline \Phi})_{\bf 1}, (\Phi {\overline \Phi})_{\bf 78}
$ and $(\Phi {\overline \Phi})_{\bf 650}$.
  In fact,
\begin{eqnarray}
  \left(
  \begin{array}{ccc}
    \VEV{{\overline \Phi}\Phi} & 0 & 0 \\
    0 & 0_{16} & 0 \\
    0 & 0 & 0_{10} \\
  \end{array}
  \right) &=& {\VEV{{\overline \Phi}\Phi} \over 54} \Ll
  2\Ls
  \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & {\bf1}_{16} & 0 \\
    0 & 0 & {\bf1}_{10} \\
  \end{array}
  \Rs + 3\Ls
  \begin{array}{ccc}
    4 & 0 & 0 \\
    0 & {\bf1}_{16} & 0 \\
    0 & 0 & -2\times{\bf1}_{10} \\
  \end{array}
  \Rs \right. \nn \\
  &&+ \left.\Ls
  \begin{array}{ccc}
    40 & 0 & 0 \\
    0 & -5\times{\bf1}_{16} & 0 \\
    0 & 0 & 4\times{\bf1}_{10} \\
  \end{array}
  \Rs\right],
\end{eqnarray}
where three matrices in rhs are proportional to $SO(10)$ singlet of
${\bf 1}, {\bf 78}$ and ${\bf 650}$, respectively. 
Since the interactions for each 
representation have independent couplings, generically the cancellation
does not happen without finetuning.

There are several other solutions for this problem.
The essential ingredient is the interaction between $A'A^3$ and 
some other operator, whose VEV breaks $E_6$, 
because the cancellation
is due to a feature of $E_6$ group. 
Here , we introduce some of the solutions.
\begin{itemize}
  \item  Allowing the higher dimensional term $A^\prime A^5$. 
  Since $\VEV{A^2}$ breaks $E_6$, the cancellation can be avoided,
  which is shown by the straight forward calculation.
  Since the number of solution of the $F$-flatness conditions increases,
  it becomes less natural to obtain DW VEV. 
  But the number of vacua is still finite.
  \item  Introducing additional adjoint Higgses $B^\prime$ and $B$,
       and giving $B$ the VEV pointing to $SO(10)$-singlet.
         Then $B$ plays the same role as above 
         $(\Phi {\overline \Phi})_{\bf 78}$.
         Examining the superpotential
   \begin{equation}
      W=B'B+\bar\Phi B'\Phi,
   \end{equation}
         the desired VEV $\VEV{{\bf 1}_B}\neq 0$ and 
         $\VEV{{\bf 45}_B}= 0$ is easily obtained.
\end{itemize}


\section{The VEV of adjoint Higgs}  \label{D-flatness}
  In this appendix, we show that there is a gauge in which an
adjoint Higgs takes its VEV pointing to the direction of 
Cartan subalgebra (CSA).

  Suppose that one Higgs $A$, which belong to the adjoint representation
of the group G (dimension is $d$ and rank is $r$) , obtains non-vanishing VEV.
  Then the D-flatness condition is as follows.
\beqn
  0   &=& \Ls A^b\Rs^\ast\Ls T_G^a\Rs_{bc}A^c  \nn\\
      &=& -i \Ls A^b\Rs^\ast f^{abc}A^c  \nn\\
  \Longrightarrow \hspace{1cm}
  0   &=& \Ls A^b\Rs^\ast f^{abc}A^c \Ls T^a\Rs\nn\\
      &=& \Ls A^b\Rs^\ast \Ll T^b , T^c\Rl A^c \nn\\
      &=& \Ll A^\dagger, A \Rl                         \label{D-condi}
\eeqn

  Next , we expand the VEV in terms of the basis 
  $\{H^a, E_\alpha, E_{-\alpha}\}$.
  Here $H^a$'s $(a=1,\cdots,r )$ are contained in the
CSA, $\alpha$'s
are positive roots whose number is ${d-r\over2}$ and $E_\alpha^\dagger
= E_{-\alpha}$.
  And they satisfy following commutation relations.
\beqn
  \Ll H^a , H^b \Rl   &=&   0  \\
  \Ll H^a , E_{\pm\alpha} \Rl   &=&   \pm\alpha^aE_{\pm\alpha}  \\
  \Ll E_\alpha , E_{-\alpha} \Rl   &=&  \alpha^aH^a  \\
  \Ll E_\alpha , E_{\pm\beta} \Rl   &=&  N_{\alpha,\pm\beta}
                                         E_{\alpha\pm\beta}
\eeqn
where $N_{\alpha,\pm\beta}$'s are some constants written by $\alpha$
and $\beta$, which are nonzero only if $\alpha\pm\beta$ is also
root.
  In this basis, the VEV is written as
\begin{eqnarray}
  A   &=& A^a H^a + \sum_{\rm positive\ root\ \alpha}
                       \Lm A^\alpha_+ E_\alpha
                     + A^\alpha_- E_{-\alpha}\Rm   \\
  A^\dagger &=& \Ls A^a\Rs^* H^a + \sum_\alpha
                       \Lm\Ls A^\alpha_-\Rs^* E_\alpha
                     + \Ls A^\alpha_+\Rs^*E_{-\alpha}\Rm .
\eeqn
  Then extracting the part proportional to the CSA from lhs of
(\ref{D-condi}), it becomes
\beqn
  \Ll A^\dagger, A\Rl &=& \sum_\alpha\Ls\left|A^\alpha_-\right|^2 -
                                    \left|A^\alpha_+\right|^2 \Rs
                                 {\bf \alpha}^aH_a
                           + \cdots  \nn\\
                       &=&  0.
\end{eqnarray}
  So if all the $A^\alpha_-$'s are zero, then $\sum_\alpha
\left|A^\alpha_+\right|^2 {\bf \alpha}^a$ is zero, therefore  all
the $A^\alpha_-$'s are also zero.
\footnote{  Though even positive roots may have negative components,
          those roots have to have a positive component at smaller
          value of $a$ by the definition of positive root. So
          examining the conditions from that of smaller value of $a$
          to that of larger one successively, $\sum_\alpha
\left|A^\alpha_+\right|^2 {\bf \alpha}^a=0$ is easily checked.
}

  Now, we show that all the $A^\alpha_-$'s can be rotated away
through the gauge rotation.
  For infinitesimal rotations, the transformation law of $A$ is
\bequ
  -i\delta A  = \Ll \theta^a H^a + \sum_\alpha
                         \Lm \theta^\alpha_+ E_\alpha
                       + \theta^\alpha_- E_{-\alpha}\Rm,
                     A^a H^a + \sum_\alpha
                         \Lm A^\alpha_+ E_\alpha
                       + A^\alpha_- E_{-\alpha}\Rm \Rl ,  \nn
\eequ
where $\theta$ is the parameter of the rotation which composed of
$d$ real numbers corresponding to $d$ generators.
  Then extracting the part proportional to the $E_\alpha$,
\bequ
  -i\delta A^\alpha_- = \Ls A^a \theta^\alpha_-
                          - A^\alpha_- \theta^a \Rs \alpha^a
                       +\sum_{\beta-\gamma=-\alpha}
                          \Ls A^\gamma_- \theta^\beta_+
                            - A^\beta_+ \theta^\gamma_- \Rs
                          N_{\beta,-\gamma}
                       -\sum_{\beta+\gamma=\alpha}
                          A^\gamma_- \theta^\beta_-
                          {N_{\beta,\gamma}}^\ast .
\eequ
  You can see there are $d-1$ gauge degree of freedom except for
the one corresponding to the maximum root, compared with ${d-r
\over2}$ complex VEVs corresponding to $d-r$ real numbers.
  Therefore , all the $A^\alpha_-$'s can be rotated away, and
in this gauge D-flatness condition forces all the $A^\alpha_+$'s
to be zero.

  Now, we have shown that VEV of an adjoint Higgs can be given
as pointing to the CSA.
  In other words, VEV of adjoint Higgs is gauge equivalent to
that pointing to the CSA in supersymmetric limit.

\section{Operators which induce mass matrices}
In this appendix, we show the operators which induce the mass matrices
of superheavy particles in $E_6$ GUT.

First of all, we examine the operator matrix $O_{24}$ 
of ${\bf 24}$ of $SU(5)$, which induce the mass matrices
$M_I (I=X,G,W)$:
\begin{equation}
O_{24}=\bordermatrix{
\bar I\backslash I  &    24_A(-1)   &     24_{A'}(4)           \cr
24_A(-1)                &     0      &      A'A \cr
24_{A'}(4)   &  A'A   &  {A'}^2       \cr
},
\end{equation}
where the numbers in the parentheses denote the typical charges. 


Next we examine the operator matrix $O_{10}$ of ${\bf 10}$ of $SU(5)$,
which induce the mass matrices $M_I (I=Q, U^c, E^c)$:
\begin{equation}
\bordermatrix{
I\backslash \bar I  &\overline {16}_{\bar \Phi}(2) & 
\overline {16}_{\bar C}(-2)&\overline {16}_{A}(-1) &45_{A}(-1) &
\overline {16}_{\bar C'}(8)   & 
                    \overline {16}_{\bar A'}(4) &  45_{A'}(4)  \cr
16_\Phi(-3) & 0 & 0 & 0& 0& \bar C'A\Phi  & \bar\Phi A'A\Phi
         & 0 \cr
16_C(-6)& 0 & 0 &0 & 0& \bar C'AC &  0  & 0 \cr
16_A(-1) & 0 & 0 & 0& 0& \bar C'A\Phi   & \bar \Phi A'A\Phi
   & 0 \cr
45_{A}(-1) & 0 & 0 & 0 & 0& \bar C'AC  & 0 & 
       A'A  \cr
16_{C'}(7) & \bar\Phi A C' & \bar CAC' &
        \bar \Phi AC' & \bar CAC' & 
        \bar C'C' & \bar\Phi A'C' & 
        \bar CA'C'  \cr
16_{A'}(4) & \bar \Phi A'A\Phi & 0 & 
        \bar\Phi A'A\Phi & 0&\bar C'A'\Phi &  
        {A'}^2& \bar C{A'}^2\Phi  \cr
45_{A'}(4) & 0 & 0 &0 & A'A
        & \bar C'A'C &  
        \bar\Phi {A'}^2C 
        & {A'}^2 \cr
},
\end{equation}
where we have shown only an example even if there are several corresponding 
operators.

Finally, we examine the operator matrix $O_5$ of ${\bf 5}$ and ${\bf \bar 5}$
of $SU(5)$, which induces the mass matrices $M_I (I=L,D^c)$:
\begin{equation}
O_5=\left(\matrix{ 0 & A_5 & 0 \cr  B_5 & C_5 & D_5 \cr 
                                    E_5 & F_5 & G_5 \cr}\right),
\end{equation}
\begin{equation}
A_5=\bordermatrix{
\bar I\backslash I   &  10_{C'}(7)
                    & 10_{\bar C'}(8) &
                    \overline{16}_{\bar C'}(8) &
                    \overline{16}_{A'}(4)  \cr
10_\Phi(-3) & C'A\Phi^2 & \bar C'(A+Z)\Phi & 
0  & 0 \cr
10_C (-6) & 0 & \bar C'(A+Z)C & 0 & 0 \cr
16_C(-6) & 0 & 0 & \bar C'(A+Z)C
   & 0 \cr
16_A(-1)  & 0 & \bar C'AC & \bar C'A\Phi
   & A'A \cr
}, 
\end{equation}
\begin{equation}
B_5=\bordermatrix{
\bar I\backslash I  & 10_\Phi(-3)&10_C (-6)&\overline{16}_{\bar C}(-2)&
\overline{16}_{A}(-1)\cr
10_{C'}(7) & C'A\Phi^2 & 0 & 0 & \bar CAC' \cr
10_{\bar C'}(8) &\bar C'(A+Z)\Phi &\bar C'(A+Z)C &
\bar C'(A+Z)\bar C^2 & \bar C'A\bar\Phi\bar C \cr
16_{C'}(7) & 0 & 0 &\bar C(A+Z)C' & \bar\Phi AC' \cr
16_{A'}(4)& 0 & 0 & 0 & A'A \cr
}, 
\end{equation}
\begin{equation}
C_5=\bordermatrix{
\bar I\backslash I  & 10_{C'} (7)
                    & 10_{\bar C'}(8)  &
                    \overline{16}_{\bar C'}(8) &
                    \overline{16}_{A'}(4)  \cr
10_{C'}(7)  & {C'}^2\Phi & 
  \bar C'C' &
  \bar C'\bar C C'\Phi  & \bar CA'C' \cr
10_{\bar C'}(8)  &\bar C'C' & 
(\bar C')^2\bar \Phi & (\bar C')^2\bar C & 
\bar C'A'\bar C\bar\Phi \cr
16_{C'} (7)& {C'}^2C & 
  \bar C'\bar\Phi C'C & \bar C'C' &  
\bar\Phi A'C' \cr
16_{A'}(4)& C'A'\Phi C & 
\bar C'A'C & \bar C'A'\Phi &  
{A'}^2 \cr
},
\end{equation}
\begin{equation}
D_5=\bordermatrix{
\bar I\backslash I  & 10_{\bar \Phi}(2) &10_{\bar C}(-2) &
\overline{16}_{\bar \Phi}(2) \cr
10_{C'}(7) &\bar\Phi(A+Z)C' & \bar C(A+Z)C' & 
   \bar \Phi\bar C C'\Phi \cr
10_{\bar C'}(8) & \bar C'A\bar\Phi^2 & 
       \bar C'A\bar C\bar \Phi & 
       \bar C'A\bar\Phi\bar C \cr
16_{C'} (7)& \bar\Phi^2AC'C &
      \bar C\bar\Phi AC'C & \bar\Phi(A+Z)C' \cr
16_{A'} (4)& 0 & 0 & \bar\Phi A'A\Phi \cr
}, 
\end{equation}
\begin{equation}
E_5=
\bordermatrix{
\bar I\backslash I  &10_\Phi(-3) & 10_C(-6)& 
                    \overline{16}_{\bar C}(-2) &\overline{16}_{A}(-1)   \cr
10_{\bar \Phi}(2) & 
0 & 0& 0 & 
\bar\Phi^2A^2\bar C   \cr
10_{\bar C} (-2)&
0 & 0& 0 & 0
 \cr
16_\Phi (-3)& 0 & 0 &  0 & 0  \cr
},
\end{equation}
\begin{equation}
F_5=\bordermatrix{
\bar I\backslash I   &  10_{C'} (7)
                    & 10_{\bar C'} (8)&
                    \overline{16}_{\bar C'} (8)&
                    \overline{16}_{A'}(4)  \cr
10_{\bar \Phi} (2) &
\bar\Phi(A+Z)C' & \bar C'A\bar\Phi^2 &
\bar C'A\bar\Phi\bar C & 
 \bar \Phi^2A'\bar C \cr
10_{\bar C} (-2) & \bar C(A+Z)C' &
\bar C'A\bar C\bar\Phi& \bar C'A\bar C^2 & 
\bar C^2A'A\bar\phi \cr
16_\Phi (-3) & 0 & \bar C'A\bar\Phi C\Phi& 
   \bar C'(A+Z)\Phi
   & \bar\Phi A'A\Phi \cr
}, 
\end{equation}
\begin{equation}
G_5=\bordermatrix{
\bar I\backslash I   &  10_{\bar \Phi} (2)
                    & 10_{\bar C}(-2) &
                    \overline{16}_{\bar \Phi} (2)\cr
10_{\bar \Phi} (2) &
\bar\Phi^3 & \bar\Phi^2\bar C &
\bar\Phi^2\bar C \cr
10_{\bar C} (-2) & \bar\Phi^2\bar C & 0 & 0 \cr
16_\Phi (-3) & 0 & 0 & 0 \cr
}.
\end{equation}



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\end{document}



