text:
\section{Introduction}
For a closed $3$-manifold, a Heegaard splitting is a decomposition
of the manifold into two handlebodies. (For a $3$-manifold with
non-empty boundary, the manifold is decomposed into two
compression bodies along their common ``plus" boundary.)

The motivation of this paper started from tunnel number one knots.
Consider a tunnel number one knot $K$ in $S^3$ and an unknotting
tunnel $t$ for $K$. Consider two properly embedded arcs
$\gamma_1$, $\gamma_2$ in the exterior of $K$ which have nothing
to do with $t$. Suppose $K\cup \gamma_1\cup \gamma_2$ gives a
genus three Heegaard splitting of exterior of $K$. Is it
irreducible (or unstabilized)? In \cite{Kobayashi3}, Kobayashi
showed that every genus $g\ge 3$ Heegaard splitting of $2$-bridge
knot exterior is reducible. So the question is that whether there
exists an irreducible genus three Heegaard splitting of a tunnel
number one knot exterior which is not $2$-bridge. When a
non-minimal genus Heegaard splitting is given, in general it is
not an easy problem to show that it is irreducible or cannot be
destabilized.

However, there are infinitely many examples of manifolds having
non-minimal genus irreducible Heegaard splittings. In particular,
there exist $3$-manifolds having arbitrary high genus strongly
irreducible Heegaard splittings (\cite{CG}, \cite{Kobayashi1}).
 Casson and Gordon used the {\bf{rectangle condition}} on Heegaard
 diagrams to show strong irreducibility of such manifolds.
 (See (\cite{MS}, Appendix).)
 One can also refer to the papers \cite{Kobayashi2}, \cite{Lee},
 (\cite{Sedgwick}, section $7$),
 (\cite{Saito}, section $7$) for the rectangle condition.

 Rectangle condition is a condition on Heegaard diagrams for
 strong irreducibility. One can try to find a condition for
 irreducibility. Inspired by the example (torus)$\times S^1$, we
 gave a parity condition in \cite{Lee}, although it is not a weaker
 condition compared to rectangle condition. It is a condition
 on two collections of $3g-3$ essential disks giving pants
 decompositions of the Heegaard surface.

 We improve the parity condition of \cite{Lee}.
 It is known that a reducible Heegaard
 splitting of an irreducible manifold is stabilized.
 Hence, if the manifold under consideration is irreducible, the
 Heegaard splitting is irreducible. Figure 1. shows the relations
 of rectangle condition and parity condition for genus $g\ge 2$
 Heegaard splittings of irreducible manifolds.

 \begin{figure}[h]
   \centerline{\includegraphics[width=7cm]{fig1}}
    \caption{Genus $g\ge 2 $ Heegaard splittings of irreducible manifolds}
\end{figure}

 \begin{theorem}
Let $M=H_1\cup_S H_2$ be a genus $g\ge 2$ Heegaard splitting of a
$3$-manifold $M$ and $\{D_1,D_2\cdots,D_g\}$ and
$\{E_1,E_2,\cdots,E_g\}$ be complete meridian disk systems of
$H_1$ and $H_2$, respectively.

If $|D_i\cap E_j|\equiv 0$ $(\bmod\,{2})$ for all the pairs
$(i,j)$, then $H_1\cup _S H_2$ is unstabilized.
\end{theorem}

As an application, in section $4$ we construct unstabilized
Heegaard splittings from any given splitting by doing a sequence
of Dehn twists.

\section{Planar decomposition and pants decomposition of a
surface}

 Let $H$ be a genus $g\ge 2$ handlebody and denote $\partial H$ by $S$.
 A collection of essential disks $\{D_1,D_2,\cdots,D_g\}$ in $H$
 is called a {\bf{complete meridian disk system}} for $H$ if
 the result of cutting $H$ along $\bigcup^{g}_{i=1} D_i$ is a $3$-ball.
 The corresponding result of cutting $S$ by
 $\bigcup^{g}_{i=1}\partial D_i$ is a planar surface, which is a
 $2g$-punctured sphere. We call it a {\bf{planar decomposition}}
 of $S$. This terminology was used in (\cite{Sedgwick}, section
 $7$).

 In another way, we can decompose $H$ and $S$ into smaller pieces
 with larger number of essential disks. Suppose a collection of
 mutually disjoint essential disks $\{D_1,D_2,\cdots,D_{3g-3}\}$
 cuts $H$ into $3$-balls $B_1,B_2,\cdots,B_{2g-2}$.
 We can imagine the shape of $B_i$ as a solid pair of pants.
 Let $P_i$ be the pair of pants $S\cap B_i$ ($i=1,2,\cdots,2g-2$).
 The decomposition $S=P_1\cup P_2\cup\cdots\cup P_{2g-2}$
 is called a {\bf{pants decomposition}} of $S$.

Let a planar decomposition of $S$ coming from a complete meridian
disk system $\{D_1,D_2,\cdots,D_g\}$ be given. We add $2g-3$ more
essential disks $\{\bar{D}_{g+1},\cdots,\bar{D}_{3g-3}\}$ of $H$
to the collection so that
$\mathcal{D}=\{D_1,D_2,\cdots,D_g,\bar{D}_{g+1},\cdots,\bar{D}_{3g-3}\}$
gives rise to a pants decomposition of $S$.
 Let $S=P_1\cup P_2\cup\cdots\cup P_{2g-2}$ be the new pants
 decomposition thus obtained.
We call $\bar{D}_i$ ($i=g+1,\cdots,3g-3$) as a {\bf{supplementary
essential disk}} for later use.

Give red color to $\partial D_i$ ($i=1,2,\cdots,g$) and blue color
to $\partial\bar{D}_j$ ($j=g+1,\cdots,3g-3$).

\begin{lemma}
For any $P_i$, if any two components among the three components of
$\partial P_i$ are red and the third is blue, convert the
blue-colored component also into red color. Iterate this operation
successively until it stops.

Then all curves constituting the pants decomposition of $S$ become
red colors.
\end{lemma}

\begin{figure}[h]
   \centerline{\includegraphics[width=10cm]{fig2}}
    \caption{The color is changed.}
\end{figure}

\begin{proof}
Originally the planar decomposition of $S$ gave a cutting of $S$
into a $2g$-punctured sphere with red boundaries. We added $2g-3$
essential blue loops on it to get a pants decomposition of $S$. So
there exists a pants having two red loops and one blue loop as its
boundary components. Then the color of the blue loop is changed to
red. See figure 2. In this way, an innermost blue loop co-bounds a
pants with two red loops, and it is changed into red color. Hence,
finally all curves come to have red colors.
\end{proof}

Let $\gamma$ be an essential simple closed curve in $S$. Note that
$\gamma$ can intersect $D_i$ or $\bar{D}_j$ only at the boundary
of the disk ($i=1,2,\cdots,g$ and $j=g+1,\cdots,3g-3$). We assume
that $\gamma$ intersects $(\bigcup^{g}_{i=1} D_i)\cup
(\bigcup^{3g-3}_{j=g+1}\bar{D}_{j})$ minimally. Let $|\cdot|$
denote the number of elements of a set.

\begin{lemma}
Suppose $|\gamma\cap D_i|\equiv 0$ $(\bmod\,{2})$ for all
$(i=1,2,\cdots,g)$. Then $|\gamma\cap \bar{D}_j|\equiv 0$
$(\bmod\,{2})$ for all $(j=g+1,\cdots,3g-3)$.
\end{lemma}

\begin{proof}
Suppose that $\gamma\cap ((\bigcup^{g}_{i=1} D_i)\cup
(\bigcup^{3g-3}_{j=g+1}\bar{D}_{j}))=\emptyset$. Then $\gamma$
lives in a pair of pants of the pants decomposition of $S$. Hence
either it is isotopic to $\partial D_i$ for some $i$ or
$\partial\bar{D}_j$ for some $j$. Then it is obvious that
$|\gamma\cap \bar{D}_j|\equiv 0$ ($\bmod\,{2}$) for all
$(j=g+1,\cdots,3g-3)$.

So we may assume that $\gamma$ intersects a pair of pants $P_k$ in
essential arcs. Let $\partial P_k$ be $l^1_k\cup l^2_k\cup l^3_k$.
If $|\gamma\cap l^1_k|$ and $|\gamma\cap l^2_k|$ are even numbers,
$|\gamma\cap l^3_k|$ should be an even number since $|\gamma\cap
(l^1_k\cup l^2_k\cup l^3_k)|$ should be an even number. (Every
properly embedded arcs in $P_k$ has two endpoints.)

Although the statements of Lemma 2.1 looks irrelevant with this
lemma, we use the idea of Lemma 2.1.
 In the proof of Lemma 2.1, every blue loop, which
was the boundary of $\bar{D}_j$, has eventually become a
red-colored loop because it co-bounded a pair of pants with two
other red loops. Since a red loop has even number of intersection
points with $\gamma$, we can see that $|\gamma\cap
\bar{D}_j|\equiv 0$ ($\bmod\,{2}$) by the conclusion of Lemma 2.1.
\end{proof}

Let $D$ be an essential disk in $H$. Since a handlebody is an
irreducible manifold, we may assume that $D\cap
((\bigcup^{g}_{i=1} D_i)\cup (\bigcup^{3g-3}_{j=g+1}\bar{D}_{j}))$
is a collection of arcs and the intersection is minimal. The
collection of arcs of intersection divides $D$ into subdisks. A
subdisk would be a $2n$-gon such as bigon, $4$-gon, $6$-gon, and
so on. Note that bigons are in one-to-one correspondence with
outermost disks in $D$. The following is a simple observation that
is important for the cut-and-connect operation that will be
discussed in section $3$.

\begin{lemma}
For all $i$ $(i=1,2,\cdots,g)$ and $j$ $(j=g+1,\cdots,3g-3)$,
$|\partial D\cap \partial D_i|\equiv 0$ $(\bmod\,{2})$ and
$|\partial D\cap \partial \bar{D}_j|\equiv 0$ $(\bmod\,{2})$.
\end{lemma}

\begin{proof}
Since any arc of intersection of $D\cap D_i$ has two endpoints,
$|\partial D\cap \partial D_i|$ would be an even number. The same
holds for $|\partial D\cap \partial\bar{D}_j|$.
\end{proof}

\section{Parity condition}

Let $H_1\cup_S H_2$ be a genus $g\ge 2$ Heegaard splitting of a
$3$-manifold $M$.  Let $\{D_1,D_2,\cdots,D_g\}$ and
$\{E_1,E_2,\cdots,E_g\}$
  be collections of complete meridian disk systems of $H_1$ and
  $H_2$,
  respectively. In \cite{Lee}, we gave a parity condition,
  involving two collections of $3g-3$ essential disks giving pants
  decompositions
  of both handlebodies of Heegaard splitting, to be unstabilized. Here we
  give a more improved condition for an unstabilized Heegaard
  splitting.

\begin{definition}
We say that $H_1\cup_S H_2$ satisfies the {\bf{even parity
condition}}
 if $|D_i\cap E_j|\equiv 0$ ($\bmod\,{2}$) for all the pairs $(i,j=1,2,\cdots,g)$.
\end{definition}

Assume that $H_1\cup_S H_2$ satisfies the even parity condition.
We add $2g-3$ more supplementary essential disks
$\{\bar{D}_{g+1},\cdots,\bar{D}_{3g-3}\}$ of $H_1$ to the
collection $\{D_1,D_2,\cdots,D_g\}$ so that
$\mathcal{D}=\{D_1,D_2,\cdots,D_g,\bar{D}_{g+1},\cdots,\bar{D}_{3g-3}\}$
gives rise to a pants decomposition of $S$. Also we add $2g-3$
more supplementary essential disks
$\{\bar{E}_{g+1},\cdots,\bar{E}_{3g-3}\}$ of $H_2$ to the
collection $\{E_1,E_2,\cdots,E_g\}$ so that
$\mathcal{E}=\{E_1,E_2,\cdots,E_g,\bar{E}_{g+1},\cdots,\bar{E}_{3g-3}\}$
gives rise to a pants decomposition of $S$. We assume that all the
boundaries of disks meet transversely and minimally.

 To simplify the notation, from now on we use the same subscript $i$ for $D_i$
 ($1\le i\le g$) and $\bar{D}_i$ ($g+1\le i \le 3g-3$). Also we
 use the same subscript $j$ for $E_j$ ($1\le j\le g$) and
 $\bar{E}_j$ ($g+1\le j\le 3g-3$).

 By applying the result of Lemma 2.2, we have the following.

 \begin{lemma}
 The parity of number of intersections are as follows.\\
\begin{enumerate}
\item[1)] For each $i$, $|D_i\cap \bar{E}_j|\equiv 0$
$(\bmod\,{2})$ for
all $j$. \\
\item[2)] For each $j$, $|\bar{D}_i\cap E_j|\equiv 0$
$(\bmod\,{2})$ for all
$i$.\\
\item[3)] For all $i$ and $j$, $|\bar{D}_i\cap \bar{E}_j|\equiv 0$
$(\bmod\,{2})$.
\end{enumerate}
\end{lemma}

\begin{proof}
1)\, For each $i$, from the definition of even parity condition,
$|D_i\cap E_j|\equiv 0$ ($\bmod\,{2}$) for all $j$. Then by Lemma
2.2, $|D_i\cap \bar{E}_j|\equiv 0$ ($\bmod\,{2}$) for all $j$.

2)\, For each $j$, from the definition of even parity condition,
$|D_i\cap E_j|\equiv 0\bmod\,{2}$ for all $i$. Then by Lemma 2.2,
$|\bar{D}_i\cap E_j|\equiv 0$ ($\bmod\,{2}$) for all $i$.

3)\, For each $i$, from 2) we can see that $|\bar{D}_i\cap
E_j|\equiv 0$ ($\bmod\,{2}$) for all $j$. Then by Lemma 2.2,
$|\bar{D}_i\cap \bar{E}_j|\equiv 0$ ($\bmod\,{2}$) for all $j$.
Then the result 3) follows. (This can be shown by using the result
1) also.)
\end{proof}

\begin{remark}
Lemma 3.1 means that any pair of disks from the collections
$\mathcal{D}$ and $\mathcal{E}$ have even number of intersections.
So Definition 1 implies (\cite{Lee}, Definition 4).
\end{remark}

Now we give the proof of Theorem 1.1.

\begin{proof}({\textit {of Theorem 1.1}})

\begin{figure}[h]
   \centerline{\includegraphics[width=10cm]{fig3}}
    \caption{$\alpha_k$ lives in a pair of pants.}
\end{figure}

Suppose that $H_1\cup_S H_2$ is stabilized. Then there exist
essential disks $D$ in $H_1$ and $E$ in $H_2$ such that $|D\cap
E|=1$. We may assume that the intersection $D\cap ((\bigcup
D_i)\cup(\bigcup \bar{D}_i))$ is a collection of arcs. Cut $D$ by
$(\bigcup D_i)\cup(\bigcup \bar{D}_i)$. Then $D$ is divided into
subdisks. For any arc, say $\gamma$, of intersection $D\cap D_i$
(or $D\cap \bar{D}_i$), two copies of $\gamma$, $\gamma_1$ and
$\gamma_2$ are created on both sides of $D_i$ (or $\bar{D}_i$)
which are parallel to each other. Connect two endpoints of
$\gamma_1$ and also connect two endpoints of $\gamma_2$ by arcs in
$S$ that are parallel and in opposite sides of $D_i$ (or
$\bar{D}_i$) to each other as in the Figure 3. We do this
cut-and-connect operation for all the arcs $D\cap ((\bigcup
D_i)\cup(\bigcup \bar{D}_i))$. Let $\{\alpha_k\}$ be the
collection of loops thence obtained from $\partial D$. Note that
each $\alpha_k$ lives in a pair of pants. Some $\alpha_k$ would be
isotopic to $\partial D_{i_k}$ and some other $\alpha_k$ be
isotopic to $\partial\bar{D}_{i_k}$ and some other $\alpha_k$
would possibly be a trivial loop.

Similarly, from $\partial E$ we obtain a collection of loops
$\{\beta_k\}$ by cut-and-connect operations. Some $\beta_k$ would
be isotopic to $\partial E_{j_k}$ and some other $\beta_k$ would
be isotopic to $\partial\bar{E}_{j_k}$ and some other $\beta_k$
would possibly be a trivial loop.

First we consider the parity of $|D\cap E_j|$ for each $j$ which
will be used in the below. Its parity is equivalent to $\sum_k
|\alpha_k\cap E_j|$ ($\bmod\,{2}$) since in the above
cut-and-connect operation two parallel copies $\gamma_1$ and
$\gamma_2$ were created. It is again equivalent to $\sum_k
|D_{i_k}\cap E_j|+\sum_k |\bar{D}_{i_k}\cap E_j|+\sum
|{\textrm{(trivial loop)}} \cap E_j|$ ($\bmod\,{2}$). By the even
parity condition and Lemma 3.1, it is even. Hence,
$$|D\cap E_j|\equiv 0\pmod{2}$$
By similar arguments, we have the following equalities in
($\bmod\,{2}$).
$$|D\cap \bar{E}_j|\equiv \sum_k |\alpha_k\cap \bar{E}_j|\equiv
\sum_k |D_{i_k}\cap \bar{E}_j|+\sum_k |\bar{D}_{i_k}\cap
\bar{E}_j|+\sum |{\textrm{(trivial loop)}} \cap \bar{E}_j|$$
 By Lemma 3.1, we have
$$|D\cap \bar{E}_j|\equiv 0 \pmod{2}$$
Now we have the following equalities in ($\bmod\,{2}$).
$$|D\cap E|
\equiv \sum_k |D\cap \beta_k| \equiv \sum_k |D\cap E_{j_k}|+
\sum_k|D\cap\bar{E}_{j_k}|+\sum |D\cap {\textrm{(trivial
loop)}}|$$
By above results, we have
$$ |D\cap E|\equiv 0 \pmod{2}$$
This is a contradiction since $|D\cap E|=1$. So we conclude that
$H_1\cup_S H_2$ is unstabilized.
\end{proof}

We give some examples of manifolds admitting a Heegaard splitting
satisfying the even parity condition.

\subsection{Connected sum of $S^2\times S^1$}

\begin{figure}[h]
   \centerline{\includegraphics[width=6cm]{fig4}}
    \caption{Connected sum of three copies of $S^2\times
    S^1$, where
    $D_i\cap E_j=\emptyset$ for all $i$, $j$}
\end{figure}

 A connected sum of copies of $S^2\times S^1$ has a
 Heegaard splitting where each pairs of essential disks in both handlebodies are
 disjoint (Figure 4). This is a reducible and unstabilized Heegaard
 splitting.

\subsection{$($ Torus $)\times S^1$}

As an example of irreducible and unstabilized Heegaard splitting
satisfying the even parity condition, we consider a genus three
Heegaard splitting of (torus)$\times S^1$. Heegaard splittings of
manifolds of the form, (surface)$\times S^1$, are classified in
\cite{Schultens}.

\begin{figure}[h]
   \centerline{\includegraphics[width=9cm]{fig5}}
    \caption{Genus three handlebody $H_1$ in $({\textrm{torus}}\,)\times S^1$}
\end{figure}

One way of understanding a Heegaard splitting of (torus)$\times
S^1$ is as follows. Since (torus)$\times S^1$ is homeomorphic to
$S^1\times S^1\times S^1$, it can be obtained from a cube by
identifying three pairs of opposite faces. Consider the center of
the cube and center of each face. Connect the center of the cube
with the center of each face by an arc (Figure 5). Take a
neighborhood of it and after the identification of opposite
sectional disks, we get a genus three handlebody $H_1$. Figure 5.
shows a meridian disk system of $H_1$.

\begin{figure}[h]
   \centerline{\includegraphics[width=9cm]{fig6}}
    \caption{Boundaries of essential disks of $H_2$ in $\partial H_1$}
\end{figure}

Now $H_2={\textrm{cl}}(H^c_1)$ is also a genus three handlebody.
Figure 6. shows the boundaries of essential disks of $H_2$ in
$\partial H_1$. We can see that it satisfies the even parity
condition. We can also see that it is weakly reducible. So it is
an irreducible and weakly reducible Heegaard splitting.

\section{Dehn twist}

In this section, we construct unstabilized Heegaard splittings by
Dehn twists from a given Heegaard splitting. First we examine the
parity of number of intersections of simple closed curves on a
surface after a Dehn twist. Let $T_{\gamma}:S\rightarrow S$ be a
homeomorphism of a closed surface $S$, which is a Dehn twist of
$S$ along $\gamma$.

\begin{lemma}
For essential simple closed curves $\gamma_1, \gamma_2, \gamma_3$,
we have
$$|T_{\gamma_1}(\gamma_2)\cap \gamma_3|\equiv |\gamma_2\cap
\gamma_3|+|\gamma_1\cap\gamma_2|\cdot|\gamma_1\cap\gamma_3|
\pmod{2}$$
\end{lemma}
\begin{proof}
Before Dehn twist, $\gamma_2$ and $\gamma_3$ have $|\gamma_2\cap
\gamma_3|$ number of intersection points. After the Dehn twist
$T_{\gamma_1}$, the number of intersection points is increased by
$|\gamma_1\cap\gamma_2|\cdot|\gamma_1\cap\gamma_3|$. Since
$T_{\gamma_1}(\gamma_2)$ and $\gamma_3$ can possibly have
inessential intersections (bigons), $|T_{\gamma_1}(\gamma_2)\cap
\gamma_3|$ is equivalent to $|\gamma_2\cap
\gamma_3|+|\gamma_1\cap\gamma_2|\cdot|\gamma_1\cap\gamma_3|$ by
($\bmod\,{2}$).
\end{proof}

By an application of Lemma 4.1, we make unstabilized Heegaard
splittings by a single Dehn twist from a given Heegaard splitting
satisfying the even parity condition.

\begin{proposition}
Suppose $H_1\cup_S H_2$ is an unstabilized Heegaard splitting
satisfying the even parity condition with $\mathcal{D}=\{D_1,
D_2,\cdots,D_g\}$ and $\mathcal{E}=\{E_1,E_2,\cdots,E_g\}$. Let
$\gamma$ be an essential simple closed curve in $S$ such that
$|\gamma\cap E_j|\equiv 0$ $(\bmod\,{2})$ for all $j$. Alter
$\mathcal{D}$ to $\mathcal{D'}=\{D'_1,D'_2,\cdots,D'_g\}$ by a
Dehn twist $T_{\gamma}$ and leave $\mathcal{E}$ unchanged.

Then the new Heegaard splitting $H'_1\cup_{S'} H'_2$  satisfies
the even parity condition, hence unstabilized.
\end{proposition}

\begin{proof}
We check the even parity condition for $H'_1\cup_{S'}H'_2$ by
using Lemma 4.1. Note that $\partial D'_i=T_{\gamma}(\partial
D_i)$. By Lemma 4.1,
$$|D'_i\cap E_j|\equiv|D_i\cap E_j|+|\gamma\cap
D_i|\cdot|\gamma\cap E_j| \pmod{2}$$
 In the above equation, $|D_i\cap E_j|$ is even by the even parity
 condition and $|\gamma\cap E_j|$ is even by the hypothesis of
 proposition. So $|D'_i\cap E_j|$ is even for all $i$ and $j$.
\end{proof}

\begin{remark}
Consider a neighborhood $N(\gamma)$ of $\gamma$ in $H_1$ such that
$N(\gamma)\cap {\textrm{cl}}(H_1-N(\gamma))$ is an annuls whose
core is parallel to $\gamma$. The Dehn twist $T_{\gamma}$ of $S$
is equivalent to removing $N(\gamma)$ from $H_1$ and attaching a
solid torus back so that a meridian of the attaching solid torus
is mapped to $\frac{1}{1}$-slope of $\partial N(\gamma)$ (A
longitude of the attaching solid torus is mapped to longitude, a
parallel of $\gamma$.) So in proposition 4.2, the ambient manifold
$M$ is changed to a new manifold $M'$ obtained by
$\frac{1}{1}$-Dehn filling on $\gamma$ in $M$.
\end{remark}

Now we are going to get an unstabilized Heegaard splitting
satisfying the even parity condition by Dehn twists from a
Heegaard splitting which does not satisfy the even parity
condition.

Suppose $|D_i\cap E_j|$ is odd for some $i$ and $j$. For the
convenience, assume that $|D_1\cap E_1|$ is odd.
 We consider the simple closed curve $\gamma=T_{\partial E_1}(\partial
 D_1)$ obtained by twisting  $\partial D_1$ along $\partial
 E_1$. First we examine the intersection of $\gamma$ with $D_i$
 and $E_j$.

\begin{lemma}
The parity of number of intersections of $\gamma$ with $D_i$ and
$E_j$ are as follows $(\bmod\,{2})$.\\

\begin{itemize}
\item[$\bullet$] $|\gamma\cap D_1|\equiv {\textrm{odd}}$\\
\item[$\bullet$] $|\gamma\cap D_i|\equiv
{\textrm{odd}}\cdot|D_i\cap E_1|$\,\, $(i=2,\cdots,g)$\\
\item[$\bullet$] $|\gamma\cap E_1|\equiv {\textrm{odd}}$\\
\item[$\bullet$] $|\gamma\cap E_j|\equiv |D_1\cap E_j|$\,\,
$(j=2,\cdots,g)$
\end{itemize}
\end{lemma}

\begin{proof}
$\bullet$\, By Lemma 4.1 and assumption, $|\gamma\cap
D_1|=|T_{\partial E_1}(\partial D_1)\cap D_1|\equiv|D_1\cap
D_1|+|E_1\cap D_1|\cdot |E_1\cap D_1|\equiv {\textrm{odd}}$
($\bmod\,{2}$).

$\bullet$\, Since $D_1$ and $D_i$ are disjoint, $|\gamma\cap
D_i|=|T_{\partial E_1}(\partial D_1)\cap D_i|\equiv |D_1\cap
D_i|+|E_1\cap D_1|\cdot|E_1\cap
D_i|\equiv{\textrm{odd}}\cdot|D_i\cap E_1|$ ($\bmod\,{2}$)\,
($i=2,\cdots,g$).

$\bullet$\, By assumption, $|\gamma\cap E_1|=|T_{\partial
E_1}(\partial D_1)\cap E_1|\equiv|D_1\cap E_1|+|E_1\cap
D_1|\cdot|E_1\cap E_1|\equiv {\textrm{odd}}$ ($\bmod\,{2}$).

$\bullet$\, Since $E_1$ and $E_j$ are disjoint, $|\gamma\cap
E_j|=|T_{\partial E_1}(\partial D_1)\cap E_j|\equiv|D_1\cap
E_j|+|E_1\cap D_1|\cdot|E_1\cap E_j|\equiv|D_1\cap E_j|$
($\bmod\,{2}$)\, ($j=2,\cdots,g$).
\end{proof}

Now we consider the Dehn twist $T_{\gamma}$ of $S$. Consider the
images of $\partial D_1$ and $\partial D_i$ ($i=2,\cdots,g$) after
the Dehn twist $T_{\gamma}$. We examine intersections of
$T_{\gamma}(\partial D_1)$ and $T_{\gamma}(\partial D_i)$
($i=2,\cdots,g$) with $E_1$ and $E_j$ ($j=2,\cdots,g$).

\begin{lemma}
The parity of number of intersections of $T_{\gamma}(\partial
D_1)$ and $T_{\gamma}(\partial D_i)$ ($i=2,\cdots,g$) with $E_1$
and $E_j$
($j=2,\cdots,g$) are as follows $(\bmod\,{2})$.\\

\begin{itemize}
\item[$\bullet$] $|T_{\gamma}(\partial D_1)\cap E_1|\equiv {\textrm{even}}$\\
\item[$\bullet$] $|T_{\gamma}(\partial D_1)\cap E_j|\equiv
{\textrm{even}}$\,\, $(j=2,\cdots,g)$\\
\item[$\bullet$] $|T_{\gamma}(\partial D_i)\cap E_1|\equiv {\textrm{even}}$\,\,
$(i=2,\cdots,g)$\\
\item[$\bullet$] $|T_{\gamma}(\partial D_i)\cap E_j|\equiv
|D_i\cap E_j|+{\textrm{odd}}\cdot|D_i\cap E_1|\cdot|D_1\cap
E_j|$\,\, $(i=2,\cdots,g)$\, $(j=2,\cdots,g)$
\end{itemize}
\end{lemma}

\begin{proof}
$\bullet$\, By Lemma 4.1 and Lemma 4.4 and assumption,
$|T_{\gamma}(\partial D_1)\cap E_1|\equiv |D_1\cap
E_1|+|\gamma\cap D_1|\cdot|\gamma\cap
E_1|\equiv{\textrm{odd}}+{\textrm{odd}}\cdot{\textrm{odd}}\equiv{\textrm{even}}$
($\bmod\,{2}$).

$\bullet$\, By Lemma 4.1 and Lemma 4.4, $|T_{\gamma}(\partial
D_1)\cap E_j|\equiv|D_1\cap E_j|+|\gamma\cap D_1|\cdot|\gamma\cap
E_j|\equiv|D_1\cap E_j|+{\textrm{odd}}\cdot|D_1\cap
E_j|\equiv{\textrm{even}}\cdot|D_1\cap E_j|\equiv{\textrm{even}}$
($\bmod\,{2}$)\, ($j=2,\cdots,g$).

$\bullet$\, By Lemma 4.1 and Lemma 4.4, $|T_{\gamma}(\partial
D_i)\cap E_1|\equiv|D_i\cap E_1|+|\gamma\cap D_i|\cdot|\gamma\cap
E_1|\equiv|D_i\cap E_1|+{\textrm{odd}}\cdot|D_i\cap
E_1|\cdot{\textrm{odd}}\equiv{\textrm{even}}\cdot|D_i\cap
E_1|\equiv{\textrm{even}}$ ($\bmod\,{2}$)\, ($i=2,\cdots,g$).

$\bullet$\, By Lemma 4.1 and Lemma 4.4, $|T_{\gamma}(\partial
D_i)\cap E_j|\equiv|D_i\cap E_j|+|\gamma\cap D_i|\cdot|\gamma\cap
E_j|\equiv|D_i\cap E_j|+{\textrm{odd}}\cdot|D_i\cap
E_1|\cdot|D_1\cap E_j|$ ($\bmod\,{2}$)\, ($i=2,\cdots,g$)\,
($j=2,\cdots,g$).
\end{proof}

Note that $|T_{\gamma}(\partial D_i)\cap E_1|$ and
$|T_{\gamma}(\partial D_1)\cap E_j|$ are even for all
$i,j=1,2,\cdots,g$ and the parity of difference
$|T_{\gamma}(\partial D_i)\cap E_j|-|D_i\cap E_j|$ is equivalent
to the parity of $|D_i\cap E_1|\cdot|D_1\cap E_j|$ for
$i,j=2,\cdots,g$. Let $\mathcal{D}'=\{D'_1,D'_2,\cdots,D'_g\}$ be
the collection of essential disks after the Dehn twist
$T_{\gamma}$ satisfying $\partial D'_i=T_{\gamma}(\partial D_i)$.
Suppose $|D'_i\cap E_j|$ is odd for some $i$ and $j$. Without loss
of generality, we may assume that $|D'_2\cap E_2|$ is odd. Let
$\gamma'=T_{\partial E_2}(\partial D'_2)$. Again we do a Dehn
twist $T_{\gamma'}$ of the surface and examine the parities
$|T_{\gamma'}(\partial D'_i)\cap E_j|$. By a sequence of Dehn
twists in this way, we can get a Heegaard splitting satisfying the
even parity condition as the following.

\begin{theorem}
For any given genus $g$ Heegaard splitting $H_1\cup_S H_2$ and
collections of complete meridian disk systems $\mathcal{D}=\{D_1,
D_2,\cdots,D_g\}$ and $\mathcal{E}=\{E_1,E_2,\cdots,E_g\}$, we get
an unstabilized Heegaard splitting satisfying the even parity
condition after a sequence of at most $g$ Dehn twists.

More precisely, the sequence of Dehn twists is $T_{\gamma_1},
T_{\gamma_2}, \cdots, T_{\gamma_g}$, where\\

\begin{itemize}
\item[$\bullet$] $\gamma_1=T_{\partial E_{j_1}}(\partial
D_{i_1})$ for some $i_1$ and $j_1$ with $|D_{i_1}\cap E_{j_1}|\equiv$ odd\\
\item[$\bullet$] $\gamma_2=T_{\partial
E_{j_2}}(T_{\gamma_1}(\partial D_{i_2}))$ for some $i_2$ and $j_2$ with
$|T_{\gamma_1}(\partial D_{i_2})\cap E_{j_2}|\equiv$ odd\\
\item[$\bullet$] $\gamma_{k}=T_{\partial E_{j_k}}\circ
T_{\gamma_{k-1}}\circ\cdots\circ T_{\gamma_1}(\partial D_{i_k})$
for some $i_k$ and $j_k$ with $|(T_{\gamma_{k-1}}\circ\cdots\circ
T_{\gamma_1}(\partial D_{i_k}))\cap E_{j_k}|\equiv$ odd\, $(k\le
g)$
\end{itemize}
\end{theorem}

\begin{proof}
As before, let $\gamma_1=T_{\partial E_1}(\partial D_1)$ without
loss of generality. Note that $|T_{\gamma_1}(\partial D_i)\cap
E_1|$ and $|T_{\gamma_1}(\partial D_1)\cap E_j|$ are even for all
$i,j=1,2,\cdots,g$ by Lemma 4.5. Let
$\mathcal{D}'=\{D'_1,D'_2,\cdots,D'_g\}$ be the collection of
essential disks after the Dehn twist $T_{\gamma_1}$ satisfying
$\partial D'_i=T_{\gamma_1}(\partial D_i)$. Let
$\gamma_2=T_{\partial E_2}(\partial D'_2)$ without loss of
generality. We examine the parity of $|T_{\gamma_2}(\partial
D'_i)\cap E_j|$. Let the notation
$\{1,2,\cdots,\hat{i},\cdots,g\}$ mean the set
$\{1,2,\cdots,i-1,i+1,\cdots,g\}$. By Lemma 4.5, we have the
following.\\

\begin{itemize}
\item[$\bullet$] $|T_{\gamma_2}(\partial D'_2)\cap E_2|\equiv {\textrm{even}}$\\
\item[$\bullet$] $|T_{\gamma_2}(\partial D'_2)\cap E_j|\equiv
{\textrm{even}}$\,\, $(j=1,\hat{2},\cdots,g)$\\
\item[$\bullet$] $|T_{\gamma_2}(\partial D'_i)\cap E_2|\equiv
{\textrm{even}}$\,\,
$(i=1,\hat{2},\cdots,g)$\\
\item[$\bullet$] $|T_{\gamma_2}(\partial D'_i)\cap E_j|\equiv
|D'_i\cap E_j|+{\textrm{odd}}\cdot|D'_i\cap E_2|\cdot|D'_2\cap
E_j|$\,\, $(i=1,\hat{2},\cdots,g)$\, $(j=1,\hat{2},\cdots,g)$\\
\end{itemize}

However, $|T_{\gamma_2}(\partial D'_1)\cap E_j|\equiv|D'_1\cap
E_j|+{\textrm{odd}}\cdot|D'_1\cap E_2|\cdot|D'_2\cap E_j|$ is
equal to $|T_{\gamma_1}(\partial D_1)\cap
E_j|+{\textrm{odd}}\cdot|T_{\gamma_1}(\partial D_1)\cap
E_2|\cdot|T_{\gamma_1}(\partial D_2)\cap E_j|$ and it is even
because $|T_{\gamma_1}(\partial D_1)\cap E_j|$ and
$|T_{\gamma_1}(\partial D_1)\cap E_2|$ are even by Lemma 4.5
again.

Also $|T_{\gamma_2}(\partial D'_i)\cap E_1|\equiv|D'_i\cap
E_1|+{\textrm{odd}}\cdot|D'_i\cap E_2|\cdot|D'_2\cap E_1|$ is
equal to $|T_{\gamma_1}(\partial D_i)\cap
E_1|+{\textrm{odd}}\cdot|T_{\gamma_1}(\partial D_i)\cap
E_2|\cdot|T_{\gamma_1}(\partial D_2)\cap E_1|$ and it is even
because $|T_{\gamma_1}(\partial D_i)\cap E_1|$ and
$|T_{\gamma_1}(\partial D_2)\cap E_1|$ are even by Lemma 4.5.

Hence we can see that $|T_{\gamma_2}(\partial D'_1)\cap E_j|$,
$|T_{\gamma_2}(\partial D'_2)\cap E_j|$, $|T_{\gamma_2}(\partial
D'_i)\cap E_1|$ and $|T_{\gamma_2}(\partial D'_i)\cap E_2|$ are
even for all $i$ and $j$. In this way, as we do sequence of Dehn
twists $T_{\gamma_k}$, the set of indices of even parity gets
bigger and bigger. So finally we get an unstabilized Heegaard
splitting satisfying the even parity condition after the sequence
of Dehn twists $T_{\gamma_1}, T_{\gamma_2}, \cdots, T_{\gamma_g}$.
\end{proof}


text:
\section{Introduction}
Multipartite entangled states are the building block of a universal
quantum computer. For example an one-way quantum computer as a scheme
for universal quantum computation are based on entangled cluster
states. Recently, L.  Kauffman and S.  Lomonaco Jr. have shown
that topological entanglement and quantum entanglement are closely
related \cite{kauf}. They introduced a topological operator called
braiding operator that can entangle quantum state. These operator
are solution of Yang-Baxter equation. The braiding operator are
also unitary transformation which make them very suitable for
application in the field of quantum computing.  We have also
recently establish a relation between multipartite states and the
Segre variety and the Segre ideal \cite{Hosh5,Hosh6}. For example,
we have shown that the Segre ideal represent completely separable
states of multipartite states.   In this paper, we will construct
braiding operators for multi-qubit states based on construction of
the Segre ideal. In particular, in section \ref{Com} we will give
a short introduction to complex projective variety and complex
multi-projective Segre variety and ideal. In section \ref{Top} we
will review the basic construction of topological entanglement
operators. We also discuss braiding operator for two-qubit state.
Finally in section \ref{mul} we will construct such topological
unitary operators for multi-qubit states. We will in detail
discuss the construction of this operator for a three-qubit state.
Now, denote a general, composite quantum system with $m$
subsystems as
$\mathcal{Q}=\mathcal{Q}^{p}_{m}(N_{1},N_{2},\ldots,N_{m})
=\mathcal{Q}_{1}\mathcal{Q}_{2}\cdots\mathcal{Q}_{m}$, with the
pure state $
\ket{\Psi}=\sum^{N_{1},N_{2},\ldots,N_{m}}_{k_{1},k_{2},\ldots,k_{m}=1}\alpha_{k_{1}k_{2}\ldots
k_{m}} \ket{k_{1}k_{2}\ldots k_{m}} $ and corresponding Hilbert
space $
\mathcal{H}_{\mathcal{Q}}=\mathcal{H}_{\mathcal{Q}_{1}}\otimes
\mathcal{H}_{\mathcal{Q}_{2}}\otimes\cdots\otimes\mathcal{H}_{\mathcal{Q}_{m}}
$, where the dimension of the $j$th Hilbert space is
$N_{j}=\dim(\mathcal{H}_{\mathcal{Q}_{j}})$. We are going to use
this notation throughout this paper. In particular, we denote a
pure two-qubit state by $\mathcal{Q}^{p}_{2}(2,2)$. Next, let
$\rho_{\mathcal{Q}}$ denote a density operator acting on
$\mathcal{H}_{\mathcal{Q}}$. The density operator
$\rho_{\mathcal{Q}}$ is said to be fully separable, which we will
denote by $\rho^{sep}_{\mathcal{Q}}$, with respect to the Hilbert
space decomposition, if it can  be written as $
\rho^{sep}_{\mathcal{Q}}=\sum^\mathrm{N}_{k=1}p_k
\bigotimes^m_{j=1}\rho^k_{\mathcal{Q}_{j}},~\sum^\mathrm{N}_{k=1}p_{k}=1
$
 for some positive integer $\mathrm{N}$, where $p_{k}$ are positive real
numbers and $\rho^k_{\mathcal{Q}_{j}}$ denotes a density operator on
Hilbert space $\mathcal{H}_{\mathcal{Q}_{j}}$. If
$\rho^{p}_{\mathcal{Q}}$ represents a pure state, then the quantum
system is fully separable if $\rho^{p}_{\mathcal{Q}}$ can be written
as
$\rho^{sep}_{\mathcal{Q}}=\bigotimes^m_{j=1}\rho_{\mathcal{Q}_{j}}$,
where $\rho_{\mathcal{Q}_{j}}$ is the density operator on
$\mathcal{H}_{\mathcal{Q}_{j}}$. If a state is not separable, then
it is said to be an entangled state.
\section{Complex projective variety and Segre ideal for multi-qubit
state}\label{Com} In this section,
 we will define complex projective space, ideal, and variety.
 Moreover, we will review the construction of the
Segre ideal for multi-qubit state. Here are some general
references on complex projective  geometry  \cite{Griff78,Mum76}.
A complex projective space $\mathbb{P}_{\mathbb{C}}^{n}$ is
defined to be the set of lines through the origin in
$\mathbb{C}^{n+1}$, that is, $
\mathbb{P}_{\mathbb{C}}^{n}=\frac{\mathbb{C}^{n+1}-{0}}{
(x_{1},\ldots,x_{n+1})\sim(y_{1},\ldots,y_{n+1})},~\lambda\in
\mathbb{C}-0,~y_{i}=\lambda x_{i} ~\forall ~0\leq i\leq n+1. $
Let $C[z]=C[z_{1},z_{2}, \ldots,z_{n}]$ denotes the polynomial
algebra in $n$  variables with complex coefficients. Then, given a
set of homogeneous polynomials $\{g_{1},g_{2},\ldots,g_{q}\}$ with
$g_{i}\in C[z]$, we define a complex projective variety as
\begin{eqnarray}
&&\mathcal{V}(g_{1},\ldots,g_{q})=\{O\in\mathbb{P}_{\mathbb{C}}^{n}:
g_{i}(O)=0~\forall~1\leq i\leq q\},
\end{eqnarray}
where $O=[a_{1},a_{2},\ldots,a_{n+1}]$ denotes the equivalent
class of point $\{\alpha_{1},\alpha_{2},\ldots,$
$\alpha_{n+1}\}\in\mathbb{C}^{n+1}$. Let $\mathcal{V}$ be complex
projective variety. Then an ideal of $\mathbf{C}[z_{1},z_{2},
\ldots,z_{n}]$ is defined by
\begin{eqnarray}
&&\mathcal{I}(\mathcal{V})=\{g\in\mathbf{C}[z_{1},z_{2},
\ldots,z_{n}]: g(z)=0~\forall~z\in\mathcal{V}\}.
\end{eqnarray}
 Note also that
 $\mathcal{V}(\mathcal{I}(\mathcal{V}))=\mathcal{V}$.
  We can map the product of  spaces
$\mathbb{P}_{\mathbb{C}}^{N_{1}-1}\times\mathbb{P}_{\mathbb{C}}^{N_{2}-1}
\times\cdots\times\mathbb{P}_{\mathbb{C}}^{N_{m}-1}$ into a
projective space by its Segre embedding as follows. Let
$(\alpha^{i}_{1},\alpha^{i}_{2},\ldots,\alpha^{i}_{N_{i}})$  be
points defined on the complex projective space
$\mathbb{P}_{\mathbb{C}}^{N_{i}-1}$. Then the Segre map
\begin{equation}\xymatrix{\mathbb{P}_{\mathbb{C}}^{N_{1}-1}\times\mathbb{P}_{\mathbb{C}}^{N_{2}-1}
\times\cdots\times\mathbb{P}_{\mathbb{C}}^{N_{m}-1}
\ar[r]_{~~\mathcal{S}_{N_{1},\ldots,N_{m}}}&\mathbb{P}_{\mathbb{C}}^{N_{1}N_{2}\cdots
N_{m}-1}}
\end{equation}
is defined by $
 ((\alpha^{1}_{1},\alpha^{1}_{2},\ldots,\alpha^{1}_{N_{1}}),\ldots,
 (\alpha^{m}_{1},\alpha^{m}_{2},\ldots,\alpha^{m}_{N_{m}}))$ $  \longmapsto
 (\alpha^{1}_{i_{1}}\alpha^{2}_{i_{2}}\cdots\alpha^{m}_{i_{m}})$.
Next, let $\alpha_{i_{1}i_{2}\cdots i_{m}}$,$1\leq i_{j}\leq
N_{j}$ be a homogeneous coordinate-function on
$\mathbb{P}_{\mathbb{C}}^{N_{1}N_{2}\cdots N_{m}-1}$. For a
multi-qubit quantum system the Segre ideal is defined by
\begin{equation}\label{segr}
\mathcal{I}^{m}_{\mathrm{Segre}}=\sum^{m}_{j=1}\mathcal{I}_{\mathcal{Q}_{j}\models\mathcal{Q}_{1}\mathcal{Q}_{2}
\cdots\widehat{\mathcal{Q}}_{j}\cdots\mathcal{Q}_{m}},
\end{equation} where
$\mathcal{I}_{\mathcal{Q}_{j}\models\mathcal{Q}_{1}\mathcal{Q}_{2}
\cdots\widehat{\mathcal{Q}}_{j}\cdots\mathcal{Q}_{m}}$ is the
ideal defining when a subsystem $\mathcal{Q}_{j}$ is separated
from quantum system $\mathcal{Q}_{1}\mathcal{Q}_{2}
\cdots\mathcal{Q}_{m}$ is generated by
\begin{eqnarray}
&&\mathcal{I}_{\mathcal{Q}_{j}\models\mathcal{Q}_{1}\mathcal{Q}_{2}\cdots
\widehat{\mathcal{Q}}_{j}\cdots\mathcal{Q}_{m}}=\left\langle\text{Minors}_{2\times
2}\mathcal{X}^{j}_{2\times 2^{m-1}}\right\rangle,
\end{eqnarray}
where $\mathcal{X}^{j}_{2\times 2^{m-1}}$ is the following
$2\times 2^{m-1}$ matrix
\begin{equation}\left(%
\begin{array}{cccc}
  \alpha_{11\ldots11_{j}1\ldots1} & \alpha_{11\ldots11_{j}1\ldots2} & \ldots & \alpha_{22\ldots21_{j}2\ldots2}\\
  \alpha_{11\ldots12_{j}1\ldots1} & \alpha_{11\ldots12_{j}1\ldots2} & \ldots & \alpha_{22\ldots22_{j}2\ldots2}\\
\end{array}%
\right).
\end{equation}
where $j=1,2,\ldots,m$ and
$\mathcal{Q}_{j}\models\mathcal{Q}_{1}\mathcal{Q}_{2}
\cdots\widehat{\mathcal{Q}}_{j}\cdots\mathcal{Q}_{m}$ means we
delete $\mathcal{Q}_{j}$ from right side and add it to the left
side of $\models$.

\section{Topological entanglement operators}\label{Top}
In this section we will give a short introduction to Artin braid
group and Yang-Baxter equation. We will study relation between
topological and quantum entanglement by investigating the unitary
representation of Artin braid group. Here are some general
references on quantum group and low-dimensional topology
\cite{kassel,chari}. The Artin braid group $\mathrm{B}_{n}$ on $n$
strands is generated by $\{b_{n}:1\leq i\leq n-1\}$ and we have
the following relations in the group $\mathrm{B}_{n}$: i) $
    b_{i}b_{j}=b_{j}b_{i}$ for $|i-j|\geq n $
and
 ii)  $b_{i}b_{i+1}b_{i}=b_{i+1}b_{i}b_{i+1}$ for $ 1\leq i<n$.
 Let $\mathcal{V}$ be a complex vector space. Then, for two
strand braid there is associated an operator
$\mathcal{R}:\mathcal{V}\otimes\mathcal{V}\longrightarrow\mathcal{V}\otimes\mathcal{V}$.
Moreover, let $\mathcal{I}$ be the identity operator on
$\mathcal{V}$. Then, the Yang-Baxter equation is defined by
\begin{equation}\label{YB}
    (\mathcal{R}\otimes \mathcal{I})(\mathcal{I}\otimes \mathcal{R})(\mathcal{R}\otimes
    \mathcal{I})=(\mathcal{I}\otimes \mathcal{R})(\mathcal{R}\otimes \mathcal{I})(\mathcal{I}\otimes
    \mathcal{R}).
\end{equation}
The Yang-Baxter equation represents the fundamental topological
relation in the Artin braid group. The inverse to $\mathcal{R}$ will
be associated with the reverse elementary braid on two strands.
Next, we define a representation $\tau$ of the Artin braid group to
the automorphism of $\mathcal{V}^{\otimes
m}=\mathcal{V}\otimes\mathcal{V}\otimes\cdots\otimes\mathcal{V}$ by
\begin{equation}\label{ABG}
    \tau(b_{i})=\mathcal{I}\otimes\cdots\otimes\mathcal{I}\otimes\mathcal{R}\otimes\mathcal{I}\otimes\cdots\otimes\mathcal{I},
\end{equation}
where $\mathcal{R}$ are in position $i$ and $i+1$. This equation
describe a representation of the braid group if $\mathcal{R}$
satisfies the Yang-Baxter equation and is also invertible.
Moreover, this representation of braid group is unitary if
$\mathcal{R}$ is also unitary operator. Thus $\mathcal{R}$ being
unitary indicated that this operator can performs topological
entanglement and it also can be considers as quantum gate. It has
been show in \cite{kauf} that $\mathcal{R}$ can also perform
quantum entanglement by acting on qubits states. Now, let
$\alpha_{11},\alpha_{12},\alpha_{21}$, and $\alpha_{22}$ be any
scalars on the unit circle in the complex plane. Then, we  can
construct an unitary $\mathcal{R}$ as follow
\begin{equation}\label{R}
\mathcal{R}=\left(
  \begin{array}{cccc}
    \alpha_{11} & 0& 0 & 0 \\
    0 & 0 & \alpha_{12} & 0 \\
    0 & \alpha_{21} & 0 &0 \\
    0 & 0 & 0 & \alpha_{22} \\
  \end{array}
\right)
\end{equation}
which is a solution to the Yang-Baxter equation. To see how it is
related to quantum gates, let $\mathcal{P}$ be the swap gate
$\tau=\mathcal{R}\mathcal{P}$ gate  define by
\begin{equation}\label{R}
\mathcal{P}=\left(
  \begin{array}{cccc}
    1 & 0& 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 1 & 0 &0 \\
    0 & 0 & 0 & 1 \\
  \end{array}
\right),~\tau=\left(
  \begin{array}{cccc}
    \alpha_{11} & 0& 0 & 0 \\
    0 &  \alpha_{12}& & 0 \\
    0 & 0&\alpha_{21}  &0 \\
    0 & 0 & 0 & \alpha_{22} \\
  \end{array}
\right)
\end{equation}
In view of braiding and algebra, $\mathcal{R}$ is a solution to
the braided version of the Yang-Baxter equation and $\tau$ is a
solution to the algebraic Yang-Baxter equation and $\mathcal{P}$
represent a virtual or flat crossing. The action of unitary matrix
$\mathcal{R}$ on a quantum state are: i)
$\mathcal{R}\ket{11}=\alpha_{11}\ket{11}$, ii)
$\mathcal{R}\ket{12}=\alpha_{21}\ket{21}$, iii)
$\mathcal{R}\ket{21}=\alpha_{12}\ket{12}$, iv)
$\mathcal{R}\ket{22}=\alpha_{22}\ket{22}$. A proof that the
operator $\mathcal{R}$ can entangle quantum states is give in
\cite{kauf}. Here, we will also give a proof based on the
construction of the Segre variety.
\begin{lem} If elements of $\mathcal{R}$ satisfies
$\alpha_{11}\alpha_{22}\neq\alpha_{12}\alpha_{21}$, then the state
$\mathcal{R}(\ket{\psi}\otimes\ket{\psi})$, with
$\ket{\psi}=\ket{1}+\ket{2}$ is entangled.
\end{lem}
From the construction of the Segre ideal the separable set of two
qubit state satisfies
$\alpha_{11}\alpha_{22}=\alpha_{12}\alpha_{21}$. Thus a two qubit
state
\begin{equation}\label{TQ}
\mathcal{R}(\ket{\psi}\otimes\ket{\psi})=\alpha_{11}\ket{11}
   +\alpha_{12}\ket{12}+\alpha_{21}\ket{21}+\alpha_{22}\ket{22}.
\end{equation}
 is
entangled if and only if  this inequality does not hold.
 We can
also note that a measure of entanglement for two-qubit state in
give by concurrence
$\mathcal{C}(\ket{\Phi})=2|\alpha_{11}\alpha_{22}-\alpha_{12}\alpha_{21}|$.
In general, let $M=(\mathcal{M}_{kl})$ denote an $n\times n$
matrix with complex elements and let $\mathcal{R}$ be defined by
$\mathcal{R}^{kl}_{rs}=\delta^{k}_{s}\delta^{l}_{r}\mathcal{M}_{kl}$.
Then $\mathcal{R}$ is a unitary solution to the Yang-Baxter
equation.
 In the next section, we will used this construction and proof to
create entangled states for three-qubit states.
\section{Multi-qubit quantum gate entangler}\label{mul}
 In
the previous section we have shown that how we can create
entangled state using topological unitary transformation
$\mathcal{R}$. We have also show a relation between the Segre
ideal and such transformation. In this section, we will use this
information to construct multi-qubit entangled state based on this
ideal. But first, we will construct such topological operator for
three-qubit state. For this state the ideal
$\mathcal{I}^{2,2,2}_{\mathcal{Q}_{1}\models\mathcal{Q}_{2}\mathcal{Q}_{3}}$
representing if a subsystem $\mathcal{Q}_{1}$ that is unentangled
with  $\mathcal{Q}_{2}\mathcal{Q}_{3}$ is generated by
$\mathcal{I}_{\mathcal{Q}_{1}\models\mathcal{Q}_{2}\mathcal{Q}_{3}}=\left\langle\text{Minors}_{2\times
2}\mathcal{X}^{1}_{2\times 4}\right\rangle$, that is
\begin{equation}
\mathcal{I}_{\mathcal{Q}_{1}\models\mathcal{Q}_{2}\mathcal{Q}_{3}}=\left\langle\text{Minors}_{2\times 2}\left(%
\begin{array}{cccc}
  \alpha_{111} & \alpha_{112}&\alpha_{121}&\alpha_{122} \\
 \alpha_{211} & \alpha_{212}&\alpha_{221}&\alpha_{222} \\
\end{array}%
\right)\right\rangle,
\end{equation}
where we have used the notation $\models$ to indicate that
$\mathcal{Q}_{1}$ is separated from
$\mathcal{Q}_{2}\mathcal{Q}_{3}$ but we still could have
entanglement between $\mathcal{Q}_{2}$ and $\mathcal{Q}_{3}$.
In the same way, we can define the ideal
$\mathcal{I}^{2,2,2}_{\mathcal{Q}_{2}\models\mathcal{Q}_{1}\mathcal{Q}_{3}}$
representing if the subsystem $\mathcal{Q}_{2}$ is unentangled
with $\mathcal{Q}_{1}\mathcal{Q}_{3}$ and
$\mathcal{I}_{\mathcal{Q}_{3}\models\mathcal{Q}_{1}\mathcal{Q}_{2}}$
representing if the subsystem $\mathcal{Q}_{3}$ is unentangled
with $\mathcal{Q}_{2}\mathcal{Q}_{3}$. The ideals are generated $
\mathcal{I}_{\mathcal{Q}_{2}\models\mathcal{Q}_{1}\mathcal{Q}_{3}}=\left\langle\text{Minors}_{2\times
2}\mathcal{X}^{2}_{2\times 4}\right\rangle,~ \text{and}~
\mathcal{I}_{\mathcal{Q}_{3}\models\mathcal{Q}_{1}\mathcal{Q}_{2}}=\left\langle\text{Minors}_{2\times
2}\mathcal{X}^{3}_{2\times 4}\right\rangle.$
 Thus, the Segre ideal
for three-qubit state is given by
\begin{eqnarray}
  \mathcal{I}^{3}_{\mathrm{Segre}}&=& \mathcal{I}_{\mathcal{Q}_{1}\models\mathcal{Q}_{2}\mathcal{Q}_{3}}
+\mathcal{I}_{\mathcal{Q}_{2}\models\mathcal{Q}_{1}\mathcal{Q}_{3}}
+\mathcal{I}_{\mathcal{Q}_{3}\models\mathcal{Q}_{1}\mathcal{Q}_{2}}\\\nonumber&=&
\left\langle\mathrm{T}_{1},\mathrm{T}_{2},\ldots,\mathrm{T}_{12}\right\rangle,
\end{eqnarray}
where
$\mathrm{T}_{1}=\alpha_{1,1,1}\alpha_{2,2,1}-\alpha_{1,2,1}\alpha_{2,1,1}$,
$\mathrm{T}_{2}=\alpha_{1,1,2}\alpha_{2,2,2}-\alpha_{1,2,2}\alpha_{2,1,2}$,
$\mathrm{T}_{3}=\alpha_{1,1,1}\alpha_{2,1,2}-\alpha_{1,1,2}\alpha_{2,1,1}$,
$\mathrm{T}_{4}=\alpha_{1,2,1}\alpha_{2,2,2}-\alpha_{1,2,2}\alpha_{2,2,1}$,
$\mathrm{T}_{5}=\alpha_{1,1,1}\alpha_{1,2,2}-\alpha_{1,1,2}\alpha_{1,2,1}$,
$\mathrm{T}_{6}=\alpha_{2,1,1}\alpha_{2,2,2}-\alpha_{2,1,2}\alpha_{2,2,1}$,
$\mathrm{T}_{7}=\alpha_{1,1,1}\alpha_{2,2,2}-\alpha_{1,1,2}\alpha_{2,2,1}$,
$\mathrm{T}_{8}=\alpha_{1,1,1}\alpha_{2,2,2}-\alpha_{1,2,1}\alpha_{2,1,2}$,
$\mathrm{T}_{9}=\alpha_{1,1,1}\alpha_{2,2,2}-\alpha_{1,2,2}\alpha_{2,1,1}$,
$\mathrm{T}_{10}=\alpha_{1,1,2}\alpha_{2,2,1}-\alpha_{1,2,1}\alpha_{2,1,2}$,
$\mathrm{T}_{11}=\alpha_{1,2,1}\alpha_{2,1,2}-\alpha_{1,2,2}\alpha_{2,1,1}$,
and
$\mathrm{T}_{12}=\alpha_{1,2,1}\alpha_{2,1,2}-\alpha_{1,2,2}\alpha_{2,1,1}$.
In our recent paper \cite{Hosh5} we have shown that we can
construct a measure of entanglement for three-qubit states based
on these Segre varieties. We have also construct a measure of
entanglement for general multipartite states based on an extension
of the Segre varieties \cite{Hosh6}. For example, for three-qubit
state a measure of entanglement is given by
\begin{eqnarray}\nonumber
\mathcal{C}(\ket{\Psi})&=&( 2(| \mathrm{T}_{1}|^{2} +
|\mathrm{T}_{2}|^{2}
 +|\mathrm{T}_{3}|^{2}+
 |\mathrm{T}_{4}|^{2}+
|\mathrm{T}_{5}|^{2}+| \mathrm{T}_{6}|^{2})\\\nonumber&&  + |\mathrm{T}_{7}
|^{2}
+ | \mathrm{T}_{8} |^{2}  + |\mathrm{T}_{9}|^{2}

+ | \mathrm{T}_{10}|^{2}+| \mathrm{T}_{11}|^{2}+|
\mathrm{T}_{12}|^{2})^{\frac{1}{2}},
\end{eqnarray}
Now, based on comparison with the two-qubit case we will construct
a unitary transformation $\mathcal{R}$ that create three-qubit
entangled states. Let
\begin{equation}\mathcal{R}=\left(%
\begin{array}{cccccccc}
  \alpha_{111} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0 & 0 & 0 & \alpha_{221} & 0 \\
  0 & 0 & 0 & 0 & 0 & \alpha_{212}& 0 & 0 \\
  0 & 0 & 0 & 0 & \alpha_{211} & 0 & 0 & 0 \\
   0 & 0 & 0 & \alpha_{122} & 0 & 0 & 0 & 0 \\
  0 & 0 & \alpha_{121} & 0 & 0 & 0 & 0 & 0 \\
   0 & \alpha_{112} & 0 & 0 & 0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0 & 0 & 0 & 0 & \alpha_{222}\\
\end{array}%
\right).
\end{equation}
Then we have the following lemma for three-qubit states.
\begin{lem} If elements of $\mathcal{R}$ satisfies
$\mathrm{T}_{i}\neq0$, for $1\leq i\leq 12$, then the state
$\mathcal{R}(\ket{\psi}\otimes\ket{\psi}\otimes\ket{\psi})$, with
$\ket{\psi}=\ket{1}+\ket{2}$ is entangled.
\end{lem}
The proof of this lemma follows by construction of $\mathcal{R}$
which is based on separable elements of three-qubit states defined
by $T_{i}$. For example
$\mathcal{R}(\ket{\psi}\otimes\ket{\psi}\otimes\ket{\psi})=\sum^{2}_{k_{1},k_{2},k_{3}=1}
\alpha_{k_{1}k_{2}k_{3}}\ket{k_{1}k_{2}k_{3}}$ is entangled if and
only if $T_{i}\neq0$.  Note that we can also write the braiding
operator $\mathcal{R}$ for three-qubit as
$\mathcal{R}_{2^{3}\times 2^{3}}=\mathcal{R}^{d}_{2^{3}\times
2^{3}}+\mathcal{R}^{ad}_{2^{3}\times 2^{3}}$, where
$\mathcal{R}^{a}_{2^{3}\times
2^{3}}=(\alpha_{111},0,\ldots,0,\alpha_{222})$ is a diagonal
matrix and $\mathcal{R}^{ad}_{2^{3}\times
2^{3}}=(0,\alpha_{221},\alpha_{212},\ldots,\alpha_{112},0)$ is an
anti-diagonal matrix. We will use this notation to construct the
matrix $\mathcal{R}$ for multi-qubit state.
 For $m$-qubit state a topological unitary
transformation $\mathcal{R}_{2^{m}\times 2^{m}}$ that create
multi-qubit entangled states is defined by
$\mathcal{R}_{2^{m}\times 2^{m}}=\mathcal{R}^{d}_{2^{m}\times
2^{m}}+\mathcal{R}^{ad}_{2^{m}\times 2^{m}}$, where
$\mathcal{R}^{a}_{2^{m}\times
2^{m}}=(\alpha_{1\cdots1},0,\ldots,0,\alpha_{2\cdots2})$ is a
diagonal matrix and $\mathcal{R}^{ad}_{2^{m}\times
2^{m}}=(0,\alpha_{22\cdots1},\ldots,\alpha_{21\cdots1},\alpha_{12\cdots2},
\ldots,\alpha_{1\cdots12},0)$ is an anti-diagonal matrix.
 Then we have following lemma for general multi-qubit
states.
\begin{lem} Let  $\mathcal{X}^{j}_{2\times 2^{m-1}}$ be a $2\times 2^{m-1}$
matrix defined by
\begin{equation}
\mathcal{X}^{j}_{2\times 2^{m-1}}=
\left(%
\begin{array}{cccc}
  \alpha_{11\ldots11_{j}1\ldots1} & \alpha_{11\ldots11_{j}1\ldots2} & \ldots & \alpha_{22\ldots21_{j}2\ldots2}\\
  \alpha_{11\ldots12_{j}1\ldots1} & \alpha_{11\ldots12_{j}1\ldots2} & \ldots & \alpha_{22\ldots22_{j}2\ldots2}\\
\end{array}%
\right).
\end{equation}
Then the state $\mathcal{R}_{2^{m}\times
2^{m}}(\ket{\psi}_{1}\otimes\ket{\psi}_{2}\otimes\cdots\otimes\ket{\psi}_{m})$,
with $\ket{\psi}_{j}=\ket{1}+\ket{2}$ is entangled if $2\times 2$
minors of $\mathcal{X}^{j}_{2\times 2^{m-1}}$ not vanishes, for
all $j=1,2,\ldots,m$.
\end{lem}
The proof  follows by construction of $\mathcal{R}_{2^{m}\times
2^{m}}$ which is based on completely separable elements of
multi-qubit states defined by the Segre ideal
$\mathcal{I}^{m}_{\mathrm{Segre}}=\sum^{m}_{j=1}\mathcal{I}_{\mathcal{Q}_{j}\models\mathcal{Q}_{1}\mathcal{Q}_{2}
\cdots\widehat{\mathcal{Q}}_{j}\cdots\mathcal{Q}_{m}}$. That is
the state $ \mathcal{R}_{2^{m}\times
2^{m}}(\ket{\psi}\otimes\ket{\psi}\otimes\cdots\otimes\ket{\psi})=\sum^{2,2,\ldots,2}_{k_{1},k_{2},\ldots,k_{m}=1}\alpha_{k_{1}k_{2}\ldots
k_{m}} \ket{k_{1}k_{2}\ldots k_{m}}$ is entangled if and only if
all $2\times 2$ minors of $\mathcal{X}^{j}_{2\times
2^{m-1}}\neq0$. Note that this operator is a quantum gate
entangler since $\tau_{2^{m}\times 2^{m}}=\mathcal{R}_{2^{m}\times
2^{m}}\mathcal{P}_{2^{m}\times 2^{m}}$ is a $2^{m}\times 2^{m}$
phase gate and $\mathcal{P}_{2^{m}\times 2^{m}}$ is $2^{m}\times
2^{m}$ swap gate. Thus we have succeeded to construct quantum gate entagler
for a general multi-qubit state based on a similar construct of a braiding  operator that
satisfies condition for separability  that is given by definition of the
Segre ideal. This also shows a good relation between
topology, algebraic geometry and quantum theory with application
in the field of quantum computing.

\begin{flushleft}
\textbf{Acknowledgments:} The  author  acknowledges the
financial support of the Japan Society for the Promotion of Science
(JSPS).
\end{flushleft}

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