












\section{Introduction and main results}\label{intro}
\noindent 
Suppose that three 1-dimensional links $L_+$, $L_-$, and $L_0$ $\subset S^4$   
 differ only in a 3-ball $B$ as shown below.  

\bigbreak
\hskip2cm\input Figure1.tex  
\bigbreak

\noindent It is very well-known that then we have the identity $(*)$ $$\hat\Delta_{L_+}-\hat\Delta_{L_-}=(t^\frac{1}{2}-t^\frac{-1}{2})\cdot\hat\Delta_{L_0},$$ 
where $\hat\Delta_{L}$ denotes 
the normalized Alexander polynomial of $L$. 
(If we let $z=t^\frac{1}{2}-t^\frac{-1}{2}$,  $\hat\Delta_{L}$ becomes the Alexander-Conway polynomial.) 
It is also very well-known that 
the Jones polynomial satisfies a similar local-move-identity 
and that 
there are several relations between local-moves on 1-links and their invariants.
(See \cite{Alexander, Cn, J, Kauffmanon, KauffmanS} etc.) 


\bigbreak 
In \cite[Theorem 4.1]{Ogasa09} we showed a 2-link version of the identity $(*)$. 
We cite it after we state its corollary and an example. 
We strength 
it
and obtain a main result of this paper, Theorem \ref{two}, cited in several paragraphs.   


The corollary is as follows: 
Suppose that two 2-dimensional spherical knots, $K_+$ and $K_-$, $\subset S^4$ and 
a submanifold $K_0$$\subset S^4$  
differ only in a 4-ball $B$ trivially embedded in $S^4$ as shown in 
Figures \ref{rev}.1-\ref{rev}.3 in \S\ref{rev}.   
(This ordered set $(K_+, K_-,K_0)$ is called a (1,2)-pass-move-triple. 
An example
is made from Figure \ref{intro}.1, \ref{intro}.3, and \ref{intro}.5, 
which are explained in the following paragraph. 
See \S\ref{rev} for the precise definition.)   
Then we have the following $(\#)$: 
there is a polynomial $\Delta_{\nu, K_*}(t) (*=+,-,0$ and  $\nu=1,2)$ 
which represents 
the $\Q[t,t^{-1}]$-$\nu$-Alexander polynomial for $K_*$,  and  
we have the identity \newline
$\Delta_{\nu, L_+}(t)-\Delta_{\nu, L_-}(t)=(t-1)\cdot\Delta_{\nu, L_0}(t).$ 


$K_+$ in Figure \ref{intro}.1,  $K_-$ in  Figure \ref{intro}.3, and $K_0$ in Figure \ref{intro}.5    
are constructed as follows: 
Embed $F=(S^1\x S^2)-$open$B^3$ in $S^4$.  
The 
boundary of $F$ in  $S^4$ is a $2$-knot.  
Let it be a trivial 2-knot $K_+$ as drawn in Figure \ref{intro}.1. 
Carry out a `local-move' on the $2$-knot $K_+$ 
in a 4-ball, which is denoted by a dotted circle  
in Figure \ref{intro}.2. This local-move is called the (1,2)-pass-move (see \S2 for the precise definition). 
Note that the above operation is done only in the 4-ball.  
The (1,2)-pass-move changes the trivial 2-knot in Figure \ref{intro}.1 (resp. \ref{intro}.2) 
into a 2-knot in Figure \ref{intro}.3. 
We can prove that the knot in Figure \ref{intro}.3 is nontrivial  
by using Seifert matrices and the Alexander polynomial. 
We use the fact that $S^1$ and $S^2$ can be `linked' in $S^4$. 
Note that $S^1$ and $S^2$ are included in $F$ as shown in Figure \ref{intro}.4.  
$K_0$ is drawn in Figure \ref{intro}.5.
If we give appropriate orientations, we can let 
$\Delta_{K_+}=t$,  
$\Delta_{K_-}=2t-1$, and $\Delta_{K_0}=-1$. 
Hence $\Delta_{K_+}-\Delta_{K_-}=(t-1)\cdot\Delta_{K_0}$ holds. 



\cite[Theorem 4.1]{Ogasa09} is as follows: 
Let $L_+=(L_{+,1},...,L_{+,{m_+}})$ be a 2-dimensional closed oriented submanifold  $\subset S^4$. 
Let each $L_{+,i}$ be connected. 
Let $g_{+,i}$ be the genus of $L_{+,i}$. 
Let $m_+=1+\Sigma_1^{m_+} g_{+,i}$. 
Let $(L_+, L_-, L_0)$ be a $(1,2)$-pass-move-triple. 
Then we have the above $(\#)$, where we replace $K_*$ with $L_*$. 

In \cite[Proposition 4.3]{Ogasa09} we proved that we cannot normalize 
the Alexander polynomials  
to be compatible with this local-move-identity.   

In \cite{KauffmanOgasa,  KauffmanOgasaII,  KauffmanOgasaB,   
Ogasa98n, Ogasa04, Ogasa07, Ogasa09, OgasaT3, OgasaIH},  
furthermore, 
we proved several relations between local-moves on $n$-knots and their invariants ($n\in\N$).  



\bigbreak 
We state one of our main results, Theorem \ref{two}, 
whose $\Q[t,t^{-1}]$-Alexander polynomial case is  \cite[Theorem 4.1]{Ogasa09}.    
The former is stronger than the latter because 
the former does not follow from the latter directly 
(see  Propositions \ref{Idaho} and \ref{Mississippi}).  
Furthermore we prove a high dimensional analogue of this main theorem 
(see Theorem \ref{Tokyo} quoted in this section.) 


\bigbreak
\noindent   
{\bf Theorem \ref{two}.} {\it 
Let $L_+=(L_{+,1},...,L_{+,{m_+}})$ be a 2-dimensional closed oriented submanifold  $\subset S^4$. 
Let each $L_{+,i}$ be connected. 
Let $g_{+,i}$ be the genus of $L_{+,i}$. 
Let $m_+=1+\Sigma_1^{m_+} g_{+,i}$. 
Let $(L_+, L_-, L_0)$ be a $(1,2)$-pass-move-triple. 
Then there is a polynomial $\Delta_{\nu, K_*}(t) (*\newline=+,-,0$ and  $\nu=1,2)$ which represents 
the $\Z[t,t^{-1}]$-$\nu$-Alexander polynomial for $K_*$,  and  
we have the identity

$$\Delta_{\nu, L_+}(t)-\Delta_{\nu, L_-}(t)=(t-1)\cdot\Delta_{\nu, L_0}(t).$$ 
}
\bigbreak

$K_+$ in Figure \ref{intro}.1,   $K_-$ in  Figure \ref{intro}.3, and $K_0$ in Figure \ref{intro}.5 
make not only an example of \cite[Theorem 4.1]{Ogasa09} 
but also Theorem \ref{two}. 





\begin{figure}
\hskip3cm\includegraphics[width=35mm]{F2}   

\hskip4cm   Figure \ref{intro}.1: {\bf A trivial $2$-knot $K_+$ and $F$. $\partial F=K_+.$} 
\end{figure}

\begin{figure}
\hskip4cm\includegraphics[width=35mm]{F3} 
\smallbreak
\hskip2cm   Figure \ref{intro}.2: {\bf A local-move will be carried out in the dotted  

\hskip20mm 4-ball. The resulting 2-knot is a nontrivial 2-knot $K_-$.}  
\end{figure}



\begin{figure}
\hskip4cm\includegraphics[width=35mm]{F4}  
\smallbreak
\hskip3cm   Figure \ref{intro}.3: {\bf A nontrivial 2-knot $K_-$}

\end{figure}




\begin{figure}
\hskip4cm\includegraphics[width=40mm]{F5}  
\smallbreak
\hskip2cm   Figure \ref{intro}.4: 
{\bf $S^1$ and $S^2$ in $F$ whose boundary is the $2$-knot} 
\end{figure}



\begin{figure}
\hskip4cm\includegraphics[width=35mm]{F0}  
\smallbreak
\hskip3cm   Figure \ref{intro}.5: {\bf $K_0$ is a trivial 2-knot in this case.}
\end{figure}

\bigbreak 
Theorem \ref{Tokyo} is as follows. 
The terms and definitions needed for it 
are in the body of the paper. 
In the 2-dimensional case we cannot normalize 
the $\Z[t,t^{-1}]$-Alexander polynomial to be compatible with the local-move-identity  
as we explained a few paragraphs before,  
but in a case of $(4k+1)$-dimensional case 
we can define the `normalized' Alexander polynomial 
(see Definition \ref{4k+1} and Theorem \ref{norm})  
associated with a local-move defined in \S\ref{Osaka}. 

\bigbreak
\noindent
{\bf Theorem \ref{Tokyo}.} {\it 
Let $K_+$ be a $(4k+1)$-knot $\subset S^{4k+3}$. 
Let $(K_+, K_-, K_0)$ be a twist-move-triple.
Then 
$$\hat\Delta_{K_+}(t)-\hat\Delta_{K_-}(t)
=(t^\frac{1}{2}-t^\frac{-1}{2})\cdot\hat\Delta_{K_0}(t),$$
where $\hat\Delta_{K}(t)$ denotes the normalized Alexander polynomial of $K$. 
}   







\bigbreak
In the 1-dimensional case we have the following fact (\cite{Hoste}): 
Let $L$ be a 2-component 1-link. 
Let $\hat\Delta_{L}(t)$ be the normalized Alexander polynomial of $L$. 
Then 
$\left.\displaystyle\frac{\hat\Delta_{K}(t)}{t^\frac{1}{2}-t^\frac{-1}{2}}\right|_{t=1}$ 
is the linking number of $L$. 
Let $K_+,  K_-,$ and $K_0$ be as in the first paragraph of this section. 
Let $K_+$ be a 1-knot. 
Then $K_0$ is a 2-component 1-link and 
$\left.\displaystyle
\frac{ \hat\Delta_{K_+}(t)-\hat\Delta_{K_-}(t)}{(t^\frac{1}{2}-t^\frac{-1}{2})^2}
\right|_{t=1}$
is the linking number of $K_0$. 



In \S\ref{m2}  we prove a 2-dimensional analogue of this result:  
we show a relation between 
the $\Z[t,t^{-1}]$-Alexander polynomial of 
2-dimensional closed oriented submanifolds $\subset S^4$ 
and the alinking number (Theorems \ref{alk} and \ref{Georgia}). 
In \S\ref{mhigh} furthermore we prove some high dimensional analogues 
(Theorem \ref{aletwi} and Corollary \ref{twi+-}). 
We cite the above theorems here. 
The terms and definitions needed for them are in the body of the paper. 

\bigbreak
\noindent
{\bf Theorem \ref{alk}.} {\it 
Let $L=(L_1, L_2)$ be an $(S^2, T^2)$-link $\subset S^4$. 
Let $\Delta_{1,L}(t)$ be a polynomial which represents 
the $\Z[t,t^{-1}]$-1-Alexander polynomial for $L$. 
Then 
$$
\begin{vmatrix}
\left.\displaystyle\frac{\Delta_{1,L}(t)}{(t-1)}\right|_{t=1}
\end{vmatrix}$$ 
is the pseudo-alinking number of $L$,  where $|\quad|$ denotes the absolute value.  
}
\bigbreak


\noindent
{\bf Theorem \ref{Georgia}.} {\it 
Let $L=(K_1,K_2)$ be a ribbon $(S^2, T^2)$-link. 
Then the alinking number of $L$ is 
$
\begin{vmatrix}
\left.\displaystyle\frac{\Delta_{1,L}(t)}{(t-1)}\right|_{t=1}
\end{vmatrix}$. 
}
\bigbreak


\noindent
{\bf Theorem \ref{aletwi}.} {\it 
Let $K$ be a $(4k+1)$-dimensional closed oriented subamanifold $\subset S^{4k+3}$ 
whose homotopy type is $S^{2k}\x S^{2k+1}$.  
Let $\hat\Delta_{K}(t)$ be the normalized Alexander polynomial of $K$.  
Then the pseudo-twinkling number of $K$ is 
$$\left.\displaystyle\frac{\hat\Delta_{K}(t)}{t^\frac{1}{2}-t^\frac{-1}{2}}\right|_{t=1}$$
}
\bigbreak

\noindent
{\bf Corollary \ref{twi+-}.} {\it 
Let $K_+$ be a $(4k+1)$-knot $\subset S^{4k+3}$. 
Let $(K_+, K_-, K_0)$ be a twist-move-triple.
Then the pseudo-twinkling number of $K_0$ is 
$$\left.\displaystyle
\frac{ \hat\Delta_{K_+}(t)-\hat\Delta_{K_-}(t)}{(t^\frac{1}{2}-t^\frac{-1}{2})^2}
\right|_{t=1}$$
}

\bigbreak




$$ \text{Table of  Contents} $$

\S\ref{intro} Introduction and main results

\S\ref{rev} Review of (1,2)-pass-moves on 2-knots 

\S\ref{revale}  
Review of the $\Q[t,t^{-1}]$-Alexander polynomial for 
(not necessarily connected) 

\hskip5mm
$n$-dimensional closed oriented submanifolds 
in $S^{n+2}$ $(n\geqq2)$ 


\S\ref{m2}  
Main theorems in the 2-dimensional case  

\S\ref{Osaka}
Review of twist-moves on high dimensional knots 

\S\ref{mhigh} Main theorems in the 4k+1 dimensional case 

\S\ref{proo2} Proof of results of \S\ref{m2} 

\S\ref{proohigh} Proof of results of \S\ref{mhigh} 


\S\ref{problem} A problem 







\bigbreak
\section{Review of (1,2)-pass-moves on 2-knots}\label{rev}
\noindent 
The local-move which associates the identity $(*)$ in the first paragraph in \S\ref{intro}  
is very easy as drawn there. 
In high dimensional case we must begin by explaining what kind of local-moves we use. 
We review the (1,2)-pass-move on  
2-dimensional closed oriented submanifolds $\subset S^4$, 
which are defined in  \cite{Ogasa04}.    

 
We work in the smooth category. 
A (not necessarily connected) 2-dimensional smooth, closed oriented submanifold 
$L \subset S^4$ is called 
an  {\it $m$-component 2-$($dimensional$)$ $($spherical$)$ link}
if $L$ consists of $m$ connected components and 
each connected component is a 2-sphere. 
 If $L$ is 1-component 2-link, then $L$ is called a {\it $($spherical$)$ 2-knot}.  
We say that 
(not necessarily connected) 2-dimensional smooth, closed, oriented submanifolds
 $L_1$ and $L_2 \subset S^4$ are {\it equivalent} 
if there exists an orientation preserving diffeomorphism 
$f:$ $S^4$ $\rightarrow$ $S^4$  
such that $f(L_1)$=$L_2$  and 
that $f | _{L_1}:L_1\rightarrow L_2$ can be regarded 
as an order and orientation preserving diffeomorphism map.    

The (1,2)-pass-move is a local-move. 
Here, `local-move' means that 
when we make $K_+$ into $K_-$ (resp. $K_-$ into $K_0$, $K_0$ into $K_-$) 
vice versa in Definition \ref{12},   
we make a change only in $B$ and 
that we do not any requirement on diffeomorphism type or homeomorphism type 
of $K_+$ (resp. $K_-$, $K_0$) other than the change only in $B$ 



\begin{figure}
\input 4.1.tex
\bigbreak
\bigbreak
\bigbreak
\end{figure}

\begin{figure}
\input 4.2.tex
\bigbreak
\bigbreak
\end{figure}


\begin{figure}
\input 30.tex
\bigbreak
\bigbreak
\end{figure}




\begin{defn}\label{12} 
Let $L_+$,  
$L_-$, and  
$L_0$ 
be (not necessarily connected) 2-dimensional closed oriented submanifolds $\subset S^4$. 
We say that ($L_+$, $L_-$, $L_0$) is a {\it $(1,2)$-pass-move-triple}   
if $L_+$, $L_-$, and $L_0$ differ only in a 4-ball $B$ trivially embedded in $S^4$ 
with the following properties: 
$B\cap L_+$ is drawn as in Figure \ref{rev}.1. 
$B\cap L_-$ is drawn as in Figure \ref{rev}.2.   
$B\cap L_0$ is drawn as in Figure \ref{rev}.3.     
Note that we do not assume how many connected components of $L_+$ intersect $B$. 
Furthermore we say that $L_+$ (resp. $L_-$) is obtained from $L_-$  (resp. $L_+$) 
by one {\it $(1,2)$-pass-move}. 
If $L$ is equivalent to $L'$ and if $L'$ is obtained from $L''$ by a sequence of 
(1,2)-pass-moves, we say that $L$ is {\it $(1,2)$-pass-move-equivalent} to $L''$.  


We regard $B$ as 
(a close 2-disc $P$)$\times[0,1]\times\{t| -1\leqq t\leqq1\}$.
Let 
$B_t=$\newline
(the close 2-disc $P$)$\times[0,1]\times\{t \}$.  
Then $B=\cup B_t$. 
In Figures \ref{rev}.1, \ref{rev}.2 and \ref{rev}.3,  
we draw $B_{-0.5}, B_{0}, B_{0.5}$ $\subset B$. 
We draw $L_+$, $L_-$, and $L_0$ by the bold line. 
The fine line denotes $\partial B_t$. 


  
$B\cap L_+$ (resp. $B\cap L_-$) is diffeomorphic to 
$D^2\amalg D^2\amalg (S^1\times [0,1])$, 
where $\amalg$ denotes the disjoint union. 
$B\cap L_+$ has the following properties:  
$B_t\cap L_+$ is empty for $-1\leqq t<0$ and $0.5<t\leqq1$.
$B_0\cap L_+$ is  
$D^2\times\{0.4\}\amalg D^2\times\{0.6\}\amalg(S^1\times [0,0.3])\amalg(S^1\times [0.7,1])$. 
$B_{0.5}\cap L_+$ is  $S^1\times [0.3,0.7]$. 
$B_t\cap L_+$ is diffeomorphic to $S^1\amalg S^1$ for $0<t<0.5$. 
(Here we draw $S^1\times [0,1]$ to have the corner 
in $B_0$ and in $B_{0.5}$. 
Strictly to say, $B\cap L_+$ in $B$ is a smooth embedding 
which is obtained by making the corner smooth naturally.)


$B\cap L_-$ has the following properties:  
$B_t\cap  L_-$ is empty for $-1\leqq t<-0.5$ and $0<t\leqq1$.
$B_0\cap L_-$ is   
$D^2\times\{0.4\}\amalg D^2\times\{0.6\}\amalg(S^1\times [0, 0.3])
\amalg(S^1\times [0.7, 1])$. 
$B_{-0.5}\cap  L_-$ is  $S^1\times [0.3, 0.7]$. 
$B_t\cap  L_-$ is diffeomorphic to $S^1\amalg S^1$ for $-0.5<t<0$. 



In Figure \ref{rev}.1 (resp. \ref{rev}.2) 
there are an oriented cylinder $S^1\times [0,1]$ 
and two oriented discs $D^2$. 
We do not make any assumption about 
the orientation of the cylinder.  
We suppose that 
each arrow $\overrightarrow{x}$, $\overrightarrow{y}$ 
in Figure \ref{rev}.1 (resp. \ref{rev}.2)  is a tangent vector of each disc at a point. 
(Note we use 
the same notations $\overrightarrow{x}$ (resp. $\overrightarrow{y}$) for different arrows.)
The orientation of each disc in 
Figure \ref{rev}.1 (resp. \ref{rev}.2) 
is determined by the each set $\{\overrightarrow{x},\overrightarrow{y}\}$. 
The orientation of $B\cap L_+$ (resp. $B\cap L_-$) 
coincides with that of the cylinder and that of the disc. 
We can suppose that there is a Seifert hypersurface $V$ such that 
$V\cap B$ is $P\x[0.3, 0.7]$.  


$B\cap L_0$ is a disjoint union of two 2-discs and an annulus as drawn in Figure \ref{rev}.3.  
One of the 2-discs is in (the close 2-disc $P$)$\x\{0\}\x\{0\}$ and 
the other in  \newline
(the close 2-disc $P$)$\x\{1\}\x\{0\}$. 
The annulus is in $(\partial$(the close 2-disc $P$))$\x[0.4, 0.6]\x\{0\}$. 
\end{defn}


Recall that an example of (1,2)-pass-move-triples is drawn in \S1. 





\begin{note}\label{hokkaido}
In the $(1,2)$-pass-move case we have the following examples: 
Let $V$ be a Seifert hypersurface for a 2-knot $K$. 
Suppose that $V$ is diffeomorphic to \newline
$((S_a^1\x S_a^2)\sharp(S_b^1\x S_b^2))-$open$B^3$. 
Take orientations of 
$S_i^1\x*$ and $*\x S_j^2$ ($i,j\in\{a,b\}$) 
so that the intersection product of 
$S_i^1\x*$ and $*\x S_j^2$ is $\delta_{ij}$. 
Suppose that the Seifert pairing of $S_a^1\x*$ and $*\x S_b^2$ is one. 
If we change the orientations of $S_b^1\x*$ and $*\x S_b^2$, 
then 
the intersection product of $S_i^1\x*$ and $*\x S_j^2$ does not change 
but the Seifert pairing of $S_a^1\x*$ and $*\x S_b^2$ cahnges $+1$ into $-1.$ 

It means the following:  
Suppose that 
we know that 
one of 2-dimensional closed oriented submanifolds, $K$ and $J$, $\subset S^4$ 
is $K_+$, and that the other $K_-$.
Then we cannot distinguish $K_+$ from $K_-$ without the information 
of the orientation how $K_+$ and $K_-$ intersect $B$.  

On the other hand, in the case of the twist-move on the $(4k+1)$-dimensional submanifolds,  
we can distinguish $K_+$ from $K_-$. See Note \ref{aomori}.  
\end{note}


In \cite{Ogasa04} we introduce 
the ribbon-move for closed oriented 2-dimensional submanifolds $\subset S^4$. 
The ribbon-move is much connected with the (1,2)-pass-move  
(see  \cite[Proposition 4.2]{Ogasa04}). 
If we replace  `(1,2)-pass-move' (resp. `(1,2)-pass-move-triple') with 
 `ribbon-move' (resp. `ribbon-move-triple') in the theorems of this paper, 
similar theorems could hold. 
We draw (a part of a figure of) a ribbon-move-triple in Figure \ref{rev}.4-\ref{rev}.6.    
See \cite{Ogasa04} for the precise definition.  



\begin{figure}
\input r+.tex
\bigbreak
\bigbreak
\end{figure}

\begin{figure}
\input r-.tex
\bigbreak
\bigbreak
\end{figure}


\begin{figure}
\input r0.tex
\bigbreak
\bigbreak
\end{figure}










\bigbreak
\section{Review of the $\Q[t,t^{-1}]$-Alexander polynomial for 
(not necessarily connected) $n$-dimensional closed oriented submanifolds 
in $S^{n+2}$ $(n\geqq2)$}\label{revale}


\noindent 
In this section we review 
the $\Q[t,t^{-1}]$
Alexander polynomial for 
(not necessarily connected) $n$-dimensional closed oriented submanifolds 
 $\subset S^{n+2}$ $(n\geqq2)$.   
In \S\ref{m2} we define the $\Z[t,t^{-1}]$-Alexander polynomial 
for 2-dimensional closed oriented submanifolds $\subset S^4$. 
In \S\ref{mhigh} we define 
the `normalized' Alexander polynomial 
for a kind of $(4k+1)$-submanifolds $\subset S^{4k+3}$.  
Of course these invariants are connected each other. 



\bigbreak  
Let $K=(K_1,...,K_\xi)$ be an $n$-dimensional closed oriented submanifold of 
$S^{n+2}$ ($n\in\N$). 
Let each $K_i$ be connected. 
If $K_i$ is PL homeomorphic to the standard sphere,  
$K_i$ is called an 
{\it $n$-dimensional $($spherical$)$ knot.}  
If each $K_i$ is an $n$-knot, 
$K$ is called an 
{\it $\xi$-component $n$-dimensional $($spherical$)$ link.}   


It is known that 
the tubular neighborhood of $K$ is 
diffeomorphic to 
$K\x D^2$ (see \cite[pages 49 and 50]{Kirby}). 
Let $X=\overline{S^{n+2}-(K\x D^2)}$. 
By using the orientation of $S^{n+2}$ and that of $K$,  
we can determine 
an orientation of $\partial D^2$. Take a homomorphism
$\alpha:H_1(X;\Z)\rightarrow\Z$ to carry 
all $[\partial D^2]$ with the orientations to $+1$.
Take the infinite cyclic covering 
$\pi:$ $\widetilde X\rightarrow X$ 
associated with $\alpha$. 
$\widetilde X$ is called 
the {\it canonical cyclic covering space} of $K$. 
We can regard 
$H_p(\widetilde X;\Z)$ 
(resp. $H_p(\widetilde X;\Q)$)   
as a 
$\Z[t, t^{-1}]$-module 
(resp. $\Q[t, t^{-1}]$-module)   
by using 
the covering translation 
$\widetilde X\rightarrow\widetilde X$ 
defined by $\alpha$. 
It is called the {\it $\Z[t, t^{-1}]$-$p$-Alexander module}  
(resp. {\it $\Q[t, t^{-1}]$-$p$-Alexander module}).   



\begin{defn}\label{North Dakota}
According to module theory, 
it holds that any $\Q[t, t^{-1}]$-module  is congruent to 
$$(\Q[t, t^{-1}]/{\lambda_1})\oplus\cdot\cdot\cdot\oplus
(\Q[t, t^{-1}]/{\lambda_l})
\oplus(\oplus^k\Q[t, t^{-1}]),$$ 
where we have the following: 


\smallbreak
\noindent
(1) 
$\lambda_*\in$ $\Q[t, t^{-1}]$ is not zero,  


\smallbreak
\noindent
(2)
$\lambda_*$ is not the $\Q[t, t^{-1}]$-balanced class of 1, 

\smallbreak
\noindent
(3) 
$k$ is the rank of the free part.     

\smallbreak
Two polynomials, $f(t)$ and $g(t), \in\Q[t, t^{-1}]$ are said to be 
{\it $\Q[t, t^{-1}]$-balanced } 
if there is an integer $n$ 
and a nonzero rational number $r$ 
such that 
$f(t)=r\cdot t^n\cdot g(t)$.  




\smallbreak
Let $H_p(\widetilde X;\Q)$ be 
as above. 
Then the {\it $\Bbb Q[t, t^{-1}]$-$p$-Alexander polynomial} is \newline
the $\Bbb Q[t, t^{-1}]$-balanced class of
$
\left\{
\begin{array}{ll}
\mbox{the product $\lambda_1\cdot...\cdot\lambda_l$} & 
\mbox{if $k=0$ and  $H_p(\widetilde X;\Q)\ncong0$, } \\
\mbox{0} & \mbox{if $k\neq0$, }  \\
\mbox{1} & \mbox{if $H_p(\widetilde X;\Q)\cong0$.}      
\end{array}
\right.
$ 
\end{defn}


In this paper manifolds (resp. submanifolds) include 
manifolds-with-boundary (resp. submanifolds-with-boundary).   
A {\it Seifert hypersurface} for 
an $n$-dimensional oriented closed submanifold  $K$ in $S^{n+2}$ 
is 
an $(n+1)$-dimensional oriented connected compact submanifold in $S^{n+2}$ 
whose boundary is $K$ ($n\in\N$).
Note that Seifert hypersurfaces exist by obstruction theory 
(see \cite[pages 49 and 50]{Kirby}). 
Note that there are two cases that 
$K$ is not connected and that $K$ is connected. 



\bigbreak
Let $V$ be a Seifert hypersurface for the above $n$-submanifold $K$.    
Let $x_1,..., x_\mu$ be $p$-cycles in $V$ 
which compose a basis of $H_p(V;\Z)$/Tor. 
Let $y_1,..., y_\nu$ be $(n+1-p)$-cycles in $V$ 
which compose a basis of $H_{n+1-p}(V;\Z)$/Tor. 
Push $y_i$ into the positive (resp. negative) direction of the normal bundle of $V$. 
Call it $y_i^{+}$ (resp. $y_i^{-}$).  
A $(p,n+1-p)$-{\it positive Seifert matrix} for the above submanifold $K$ associated with $V$ represented by 
an ordered basis,  
$\{x_1,..., x_\mu\}$,   and 
an ordered basis,  
$\{y_1,..., y_\nu\}$,  
is a $(\mu\x\nu)$-matrix 
$$S=(s_{ij})=({\mathrm{lk}}(x_i, y_j^{+})).$$   
A $(p,n+1-p)$-{\it negative Seifert matrix} 
for the above submanifold $K$ associated with $V$ represented by 
an ordered basis,  
$\{x_1,..., x_\mu\}$,   and 
an ordered basis,  
$\{y_1,..., y_\nu\}$,  
is a matrix 
$$N=(n_{ij})=({\mathrm{lk}}(x_i, y_j^{-})).$$   
We have the following: 
Let $S$ and $N$ be as above. 
Then $S-N$ represents the map 
 $\{H_{p}(V;\Z)$/Tor\} $\x \{H_{n+1-p}(V;\Z)$/Tor\} $\rightarrow\Z$ 
which is defined by the intersection product. 
We call  $t\cdot S-N$ the $(p,n+1-p)$-{\it Alexander matrix}   
for $K$ associated with $V$ represented by 
an ordered basis,  
$\{x_1,..., x_\mu\}$,   and 
an ordered basis,  
$\{y_1,..., y_\nu\}$. 
$S$ and $N$ (resp. $S$ and $t\cdot S-N$,  $N$ and $t\cdot S-N$)  
are said to be {\it related}  
if 
$S$ and $N$ (resp. $S$ and $t\cdot S-N$,  $N$ and $t\cdot S-N$)  
are defined by using the same $V$, the same $\{x_1,..., x_\mu\}$, 
and the same $\{y_1,..., y_\nu\}$.   
We sometimes abbreviate 
$(p,n+1-p)$-positive Seifert matrix  
(resp. $(p,n+1-p)$-negative Seifert matrix, $(p,n+1-p)$-Alexander matrix)   
to 
$p$-Seifert matrix  
(resp. $p$-negative Seifert matrix, $p$-Alexander matrix) 
when it is clear from the context. 



\begin{pr}\label{square}  
Let $K$ be an $n$-dimensional oriented closed submanifold $\subset S^{n+2}$. 
Let 
$S_p$ 
$($resp. $N_p)$ 
be  
a $p$-positive 
$($resp. negative$)$  
Seifert matrix for $K$ associated with $V$ represented by 
an ordered basis,  
$\{x_1,..., x_\mu\}$,   and 
an ordered basis,  
$\{y_1,..., y_\nu\}$. 
Suppose 
$$\mu=\nu.$$
Suppose that the homomorphism on 
$H_{p-1}(\amalg_{-\infty}^{\infty}V\x[-1,1];\Q)
\to
H_{p-1}(\amalg_{-\infty}^{\infty}Y;\Q)$ 
defined by a $(p-1)$-Alexander matrix is injective. 
Then 
the $p$-$\Q[t, t^{-1}]$-Alexander polynomial is 
the $\Q[t, t^{-1}]$-balanced class of 
`the determinant of $p$-Alexander matrix'  
$${\rm{det}} (t\cdot S_p-N_p).$$ 
\end{pr}

\noindent
{\bf Note.}  
Of course  $\mu\neq\nu$ in general. 

\bigbreak
\noindent
{\bf Proof of Proposition \ref{square}.} 
Take the above $X=\overline{S^{n+2}-(K\x D^2)}$, $\widetilde X$, $V$.   
Let $V\x[-1,1]$ be the tubular neighborhood of $V$ in $X$. 
Let $Y=X-V$.  Consider the Mayer-Vietoris exact sequence: 
$$H_\natural(\amalg_{-\infty}^{\infty}V\x[-1,1];\Q)
\to
H_\natural(\amalg_{-\infty}^{\infty}Y;\Q)
\to  H_\natural(\widetilde X;\Q),$$  

\noindent 
where 
$\amalg_{-\infty}^{\infty}V\x[-1,1]$ is the lift of  $V\x[-1,1]$,  
and where  
$\amalg_{-\infty}^{\infty}Y$ is the lift of $Y$. 

This completes the proof of Proposition \ref{square}. \qed



\begin{pr}\label{tsuika}
Let $N_p$ be 
a $(p,n+1-p)$-{negative Seifert matrix} for $K$ associated with $V$ represented by 
an ordered basis,  
$\{x_1,..., x_\mu\}$,   and 
an ordered basis,  
$\{y_1,..., y_\nu\}$.  
Let $S_{n+1-p}$ be 
a $(n+1-p,p)$-{positive Seifert matrix} for $K$ associated with $V$ represented by 
an ordered basis,  
$\{y_1,..., y_\nu\}$,   and 
an ordered basis,  
$\{x_1,..., x_\mu\}$. Then we have 
$$N_p=(-1)^{p\cdot n+1}\cdot S_{n+1-p}.$$ 
\end{pr}

\noindent
{\bf Proof of Proposition \ref{tsuika}.}
By the definition of 
$x_i^{+}$ and  $y_i^{-}$, 
${\mathrm{lk}}(y_i, x_j^{+})$ 
$={\mathrm{lk}}(y_i^{-}, x_j)$. 
By \cite[page 541]{Levinepol}, 
${\mathrm{lk}}(y_i^{-}, x_j)$
$=(-1)^{p(n+1-p)+1}{\mathrm{lk}}(x_j, y_i^{-})$. 
Note that $p(1-p)$ is an even number.\qed

\bigbreak
Proposition \ref{tsuika} implies Proposition \ref{Mars}. 

\begin{pr}\label{Mars} 
Let $K$ be a $(2m+1)$-dimensional closed oriented submanifold  $\subset S^{2m+3}$. 
Let $S$ be an $(m+1,m+1)$-Seifert matrix. 
Then we have 
$$S=(-1)^{m}\cdot ^t \hskip-1mm S.$$ 
\end{pr}


Let $K$ be a $(4k+1)$-dimensional spherical knot 
$(k\in\N\cup\{0\})$. 
We regard naturally 
$(H_{2k+1}(V;\Z)/\mathrm{Tor}) \otimes {\Z_2}$ 
as a subgroup of $H_{2k+1}(V;\Z_2)$.  
Then we can take a basis  $\{x_1,..., x_\nu, y_1,..., y_\nu\}$ 
of 
$(H_{2k+1}(V;\Z)/\mathrm{Tor}) \otimes\Z_2$ 
such that 
$x_i\cdot x_j=0$, $y_i\cdot y_j=0$,  
$x_i\cdot y_j=\delta_{ij}$ 
for any pair $(i, j)$, 
where $\cdot$ denotes the $\Z_2$-intersection product. 
The {\it Arf  invariant} of $K$ is 
$$\Big(
\sum_{i=1}^\nu   {\mathrm{lk}}(x_i, x_i^{+})\cdot  {\mathrm{lk}}(y_i, y_i^{+})
\Big) 
\hskip1mm  \text{mod 2 }.$$ 

\bigbreak
Let $L=(L_1,..., L_\mu)$ be a $(4k+1)$-link 
$(k\in\N\cup\{0\}. \quad \mu\in\N-\{1\}.)$.  
We define the {Arf } invariant of $L$.  
There are two cases. 

\begin{itemize}
\item[(1)]
 Let $4k+1\geq5$. 
The Arf  invariant of $L$ is defined in the same manner as the knot case. 

\item[(2)]
 Let $4k+1=1$. 
See Appendix of \cite{Kirby} and \cite[Note right above Note 1.2.1]{Ogasa02}.   
\end{itemize}






\bigbreak
\section{Main theorems in the 2-dimensional case}\label{m2}
\noindent 
Two polynomials,  $f(t)$ and $g(t), \in\Z[t, t^{-1}]$ are said to be 
{\it $\Z[t, t^{-1}]$-balanced} 
if there is an integer $n$ 
such that 
$f(t)=\pm t^n\cdot g(t)$.  


\begin{thm}\label{Z}
Let $L=(L_1,...,L_m)$ be a 2-dimensional closed oriented submanifold  $\subset S^4$. 
Let each $L_i$ be connected.   
Let $g_i$ be the genus of $L_i$.   
Let $m=1+\Sigma_1^m g_i$. 
Let $S_\nu(V)$ be a positive $\nu$-Seifert matrix associated with a 
Seifert hypersurface for $L$  $(\nu=1,2)$. 
Let $N_\nu(V)$ be its related negative $\nu$-Seifert matrix. 
Let $t\cdot S_\nu(V)-N_\nu(V)$ be their related $\nu$-Alexander matrix. 
Then 
the $\Z[t, t^{-1}]$-balanced class of
$${\rm{det}} (t\cdot S_\nu(V)-N_\nu(V))$$ 
is a topological invariant of $L$. 
\end{thm}

\noindent
{\bf Note.}  (1)    
We have $b_0(L)=\frac{1}{2}b_1(L)+1$, where $b_j$ is the $j$-th betti number. 

\smallbreak\noindent(2) 
Since any Seifert hypersurface is connected by the definition, the $\Q[t,t^{-1}]$-balanced class of ${\rm{det}} (t\cdot S_\nu(L)-N_\nu(L))$ is determined by the $\Q[t,t^{-1}]$-module $H_1(\widetilde{X};\Q)$, where $\widetilde{X}$ is the infinite cyclic covering space of $L$. 

\smallbreak\noindent(3) 
Let $\Delta(t)$ be a polynomial which represents the $\nu$-$\Z[t,t^{-1}]$-Alexander polynomial of $L$.  
By the definition,  
the $\Q[t,t^{-1}]$-balanced class of $\Delta(t)$ 
is 
the $\nu$-$\Q[t,t^{-1}]$-Alexander polynomial associated with $L$. 




\begin{defn}\label{Delaware}
 The $\Z[t,t^{-1}]$-balanced class defined in Theorem \ref{Z} is called the {\it $\nu$-$\Z[t,t^{-1}]$-Alexander polynomial} of $L$.  
\end{defn}

We call 
a 2-dimensional closed oriented submanifold  $L=(K_1, K_2)\subset S^4$ 
an {\it $(S^2, T^2)$-link} 
if $K_1$ (resp. $K_2$) is diffeomophic to $S^2$ (resp. $T^2$). 
Note that $b_0(L)=\frac{1}{2}b_1(L)+1$ holds. 

\begin{pr}\label{Idaho}
There are $(S^2, T^2)$-links, $A=(A_1,A_2)$ and $B=(B_1, B_2), \subset S^4$ 
such that 
their  $\nu$-$\Z[t,t^{-1}]$-Alexander polynomials are not equivalent  $(\nu=1,2)$  
and such that 
their  $\nu$-$\Q[t,t^{-1}]$-Alexander polynomials are equivalent. 
\end{pr}




\bigbreak  

Recall the paragraph right before `Theorem \ref{two} cited in \S\ref{intro}'. 

\begin{thm}\label{two}
Let $L_+=(L_{+,1},...,L_{+,{m_+}})$ be a 2-dimensional closed oriented submanifold  $\subset S^4$. 
Let each $L_{+,i}$ be connected. 
Let $g_{+,i}$ be the genus of $L_{+,i}$. 
Let $m_+=1+\Sigma_1^{m_+} g_{+,i}$. 
Let $(L_+, L_-, L_0)$ be a $(1,2)$-pass-move-triple. 
Then there is a polynomial $\Delta_{\nu, K_*}(t) (*\newline=+,-,0$ and  $\nu=1,2)$ which represents 
the $\Z[t,t^{-1}]$-$\nu$-Alexander polynomial for $K_*$,  and  
we have the identity

$$\Delta_{\nu, L_+}(t)-\Delta_{\nu, L_-}(t)=(t-1)\cdot\Delta_{\nu, L_0}(t).$$ 
\end{thm} 

\noindent{\bf Note.} 
(1)   
If $m=1$, $K_+$ and $K_-$ are homeomorphic to $S^2$. 
Then $K_0$ is homeomorphic to $S^2$ or $S^2\amalg T^2$.
In each case Theorem \ref{two} holds. 
We do not need the condition on the homeomorphism type of $K_0$. 

\smallbreak\noindent
(2) 
By  \cite[Proposition 4.3]{Ogasa09}, we cannot normalize 
the $\Z[t,t^{-1}]$-$\nu$-Alexander polynomial for $L_*$ 
to be compatible with the identities in Theorem \ref{two}. 
On the other hand, 
we can define the `normalized' Alexander polynomial 
in a case of the $(4k+1)$-dimensional case 
so that it is compatible with a local-move-identity. 
See Definition \ref{4k+1} and Theorem \ref{Tokyo} for detail.   


\smallbreak\noindent
(3) 
If we remove the condition on the betti number, 
the identity does not hold in general by \cite[Proposition 4.2]{Ogasa09}.


\begin{pr}\label{Mississippi}
There are $(1,2)$-pass-move-triples 
$L=(L_+,L_-,L_0)$ and 
$L'=(L'_+,L'_-,L'_0)$ with the following properties: 


\smallbreak\noindent$(1)$
 $b_0(L)=\frac{1}{2}b_1(L)+1$.
 $b_0(L')=\frac{1}{2}b_1(L')+1$.

\smallbreak\noindent$(2)$
$t-1$ $($resp. $t-1, 1)$   
represents  the $\Q[t, t^{-1}]$-Alexander polynomial of \newline
\quad\quad $L_+$ $($resp. $L_-, L_0)$. 

\smallbreak\noindent$(3)$ 
$t-1$ $($resp. $t-1, 1)$    
represents  the $\Q[t, t^{-1}]$-Alexander polynomial of \newline
\quad\quad $L'_+$ $($resp. $L'_-, L'_0)$. 

\smallbreak\noindent$(4)$
$4(t-1)$  $($resp. $3(t-1), 1)$  
represents  the $\Z[t, t^{-1}]$-Alexander polynomial of \newline
\quad\quad $L_+$ $($resp. $L_-, L_0)$. 

\smallbreak\noindent$(5)$  
$2(t-1)$ $($resp. $t-1, 1)$  
represents  the $\Z[t, t^{-1}]$-Alexander polynomial of \newline
\quad\quad $L'_+$ $($resp. $L'_-, L'_0)$. 
\end{pr}

\noindent{\bf Note.}  
Take two arbitrary different polynomials from 
$4(t-1)$, $3(t-1)$, $2(t-1)$, and $t-1$. 
Then they are not $\Z[t, t^{-1}]$-balanced but $\Q[t, t^{-1}]$-balanced. 




\begin{pr}\label{Illinois}
Let $V$ be a Seifert hypersurface for an $(S^2, T^2)$-link $L=(K_1, K_2)$. 
Then we have the following: 

\smallbreak\noindent$(1)$
There is a basis,  $\{\tau_1,...\tau_n\}$,  of $H_2(V;\Z)$, 
where $n$ is an nonnegative integer. 


\smallbreak\noindent$(2)$
There is a set $\{\sigma_1,...,\sigma_n\}\subset H_1(V;\Z)$ such that 
 $\{\pi(\sigma_1),...,\pi(\sigma_n)\}$ is a basis of $H_1(V;\Z)/{\rm Tor}$, 
where $\pi$ is the natural projection homomorphism $H_1(V;\Z)\to H_1(V;\Z)/{\rm Tor}$.  

\smallbreak\noindent$(3)$
The intersection product of $\sigma_i$ and $\tau_j$ in $V$ is 
$
\begin{cases}
0 & \text{if $i=1$.}\\
\delta_{ij} &\text{if $i\geq2$}\\
\end{cases}
$ 
\end{pr}


\begin{defn}\label{Arkansas}
Let  $L=\{K_1, K_2\}$ be an $(S^2, T^2)$-link and $V$ a Seifert hypersurface for $L$.   
Take sets,  $\{\sigma_1,...\sigma_n\}$ and $\{\tau_1,...,\tau_n\}$, 
 as in Proposition \ref{Illinois}.  
We define the {\it pseudo-alinking number} of $L$ to be 
the absolute value of the Seifert pairing of $\sigma_1$ and $\tau_1$. 
\end{defn}


\begin{thm}\label{alk}   
Let $L=(L_1, L_2)$ be an $(S^2, T^2)$-link $\subset S^4$. 
Let $\Delta_{1,L}(t)$ be a polynomial which represents 
the $\Z[t,t^{-1}]$-1-Alexander polynomial for $L$. 
Then 
$$
\begin{vmatrix}
\left.\displaystyle\frac{\Delta_{1,L}(t)}{(t-1)}\right|_{t=1}
\end{vmatrix}$$ 
is the pseudo-alinking number of $L$,  where $|\quad|$ denotes the absolute value.  
\end{thm}



\begin{defn}\label{Arizona} (\cite{Sato}.)
Let $L=(K_1,K_2)$ be an ordered closed oriented 2-dimensional submanifold $\subset S^4$. 
Let $K_1$ and $K_2$ be connected. 
Take any circle embedded in $K_i$. Give any orientation to the circle.  
Consider the linking number of the circle and $K_j$ ($i\neq j$). 
Make a set of all of the linking number. 
Then the set is regarded as $n\cdot\Z$ for a number $n\in\{0\}\cup\N$.  
Note that if $n=0$, then the set is $\{0\}$. 
We call this number $n$ the {\it alinking number} 
alk$(K_i\subset L, K_j\subset L)$  
of $K_i$ in $L$ around $K_j$ in $L$. 
Note that alk$(K_1\subset L, K_2\subset L)$ is not equal to 
alk$(K_2\subset L, K_1\subset L)$ in general. 


Note that if $K_1$ is diffeomophic to $S^2$, 
alk$(K_1\subset L, K_2\subset L)$  
is zero. 
So, in this case, let the alinking number of $L$ mean 
alk$(K_2\subset L, K_1\subset L)$.    
\end{defn}


\begin{prob}\label{California}
Let $L=(K_1,K_2)$ be an $(S^2, T^2)$-link. 
Is the alinking number of $L$ different from the pseudo-alinking number of $L$ in general?  
\end{prob}

\noindent{\bf Note.} 
(1) Note \ref{Alaska} is related to Problem \ref{California}. 

\smallbreak\noindent(2)  
Proposition \ref{North Carolina} claims that 
the alinking number is a `surface-link cobordism' invariant. 
How about the pseudo-alinking number?  

\begin{thm}\label{Colorado}
Let $L=(K_1,K_2)$ be an $(S^2, T^2)$-link. 
Then the following three conditions are equivalent. 

\smallbreak\noindent$(1)$  
The alinking number of $L$ is zero. 

\smallbreak\noindent$(2)$    
The pseudo-alinking number of $L$ is zero. 

\smallbreak\noindent$(3)$  
$
\begin{vmatrix}
\left.\displaystyle\frac{\Delta_{1,L}(t)}{(t-1)}\right|_{t=1}
\end{vmatrix}$ is zero. 
\end{thm}


\begin{thm}\label{Connecticut}
Let $L=(K_1,K_2)$ be an $(S^2, T^2)$-link. 
Suppose that there is a Seifert hypersurfcace $V$ such that {\rm Tor}$H_1(V;\Z)\cong0$. 
Then the alinking number of $L$ is 
$
\begin{vmatrix}
\left.\displaystyle\frac{\Delta_{1,L}(t)}{(t-1)}\right|_{t=1}
\end{vmatrix}$. 
\end{thm}


Let $L=(K_1,K_2)$ be an $(S^2, T^2)$-link. 
We say that $L$ is {\it ribbon} if there is an immersion 
$f:B\amalg H\looparrowright S^4$ 
with the following properties, 
where $B$ is a 3-ball and $H$ is a genus one handle body: 
The self-intersection of $f$ consists of double points and 
is a disjoint union of 2-discs.  
Note that $f^{-1}(\text{each disc})$ is a disjoint union of two 2-discs. 
One of the two disc is included in the interior of $B\amalg H$. 
The intersection of $\partial(B\amalg H)$ and the other disc is the boundary of the other disc.


\begin{cor}\label{Georgia}
Let $L=(K_1,K_2)$ be a ribbon $(S^2, T^2)$-link. 
Then the alinking number of $L$ is 
$
\begin{vmatrix}
\left.\displaystyle\frac{\Delta_{1,L}(t)}{(t-1)}\right|_{t=1}
\end{vmatrix}$. 
\end{cor}












If $(K_+, K_-, K_0)$ is a triple of 1-links as in \S1 and   
if $K_+$ and $K_-$ are 1-knots, 
then $K_0$ is always a 2-component 1-link. 
However the 2-dimensional case we have the following theorem. 

\begin{thm}\label{Aa}
There are $(1,2)$-pass-move-triples, 
$L=(L_+, L_-, L_0)$ and $L'=(L'_+, L'_-, L'_0)$,  
with the following properties: 

\smallbreak\noindent$(1)$
 $b_0(L)=\frac{1}{2}b_1(L)+1$.
 $b_0(L')=\frac{1}{2}b_1(L')+1$.


\smallbreak 
\noindent
$(2)$
$L_+$ and $L'_+$ are diffeomorphic to $S^2$. 
Hence 
$L_-$ and $L'_-$ are diffeomorphic to $S^2$. 

\smallbreak 
\noindent
$(3)$  
The $\Z[t,t^{-1}]$-1-Alexander polynomial of $L_+ ($resp. $L_-)$  
is equivalent to that of $L'_+ ($resp. 
\newline\hskip6mm
$L'_-)$.   

\smallbreak 
\noindent
$(4)$ 
 $L_0$ is diffeomorphic to $S^2\amalg T^2$. 
  $L'_0$ is diffeomorphic to $S^2$.   
Hence $L_0$ and $L'_0$ are not 
\newline\hskip6mm
diffeomorphic. 
\end{thm}

See  Corollary \ref{aletwi}.
In a $(4k+1)$-dimesional case 
we have similar situation to the 1-dimensional case, 
different from the 2-dimensional case.




\bigbreak
\section{Review of twist-moves on high dimensional knots}\label{Osaka}
\noindent 
In the following section (\S\ref{mhigh}) 
 we have high dimensional analogues of \S\ref{m2}. 
We prove a new local-move-identity for 
the `normalized' Alexander polynomial of 
a kind of $(4k+1)$-dimensional closed oriented submanifolds $\subset S^{4k+3}$ 
(Theorem \ref{Tokyo}). 
The local-move-identity is associated with the twist-move, which is reviewed in this section. 
We introduce the `pseudo-twinkling number'  
as an analogue of 
the pseudo-alinking number,  
the alinking number, and 
the linking number (Definition \ref{twinkling}). 
We show a relation between the `normalized' Alexander polynomial  
and the pseudo-twinkling number (Theorem \ref{aletwi}).
The pseudo-twinkling number is an analogue of the pseudo-alinking number but 
a relation between  the  pseudo-twinkling number and the `normalized' Alexander polynomial 
in Corollary \ref{twi+-}  
is different from 
the relations between the pseudo-alinking number and the $\Z[t,t^{-1}]$-Alexander polynomial 
in \S\ref{m2}.  







\begin{figure}
\vskip-20mm

\includegraphics[width=9cm]{hand0004x.eps}

{\bf Figure \ref{Osaka}.1.(1):  A twist-move-triple} 

\vskip7mm


\includegraphics[width=9cm]{hand0004bx.eps}

\vskip-10mm

{\bf Figure \ref{Osaka}.1.(2):  A twist-move-triple} 

\end{figure}





\begin{figure}
\vskip-20mm

\includegraphics[width=10cm]{hand0006.eps}

Figure \ref{Osaka}.1.(3): {\bf A twist-move-triple} 
\end{figure}





\bigbreak
We review twist-moves on high dimensional knots in this section. 
(Note:  In \cite{Ogasa09} the twist-move is called the $XXII$-move.)
Figure \ref{Osaka}.1, which consists of the three figures (1), (2) and (3),  is a diagram of a twist-move-triple. 
Confirm the following:  if $p=0$, the twist-move is the crossing change on 1-links 
and 
Figure \ref{Osaka}.1 is one drawn in the first paragraph in \S1. 


\bigbreak
Let $K_+$,  $K_-$, $K_0$ be 
$(2p+1)$-dimensional closed oriented submanifolds $\subset S^{2p+3}$ 
($p\newline\in\N\cup\{0\}$). 
Let $B$ be a $(2p+3)$-ball trivially embedded in $S^{2p+3}$. 
Suppose that $K_+$ coincides with  $K_-$ (resp. $K_0$) in $\overline{S^{2p+3}-B}$. 
Take 
a single $(2p+2)$-dimensional $(p+1)$-handle $h_+$ (resp. $h_-$) embedded in $B$
such that (the handle)$\cap\partial B$ 
is the attaching part of the handle. 

\smallbreak\noindent
{\bf Note.} \cite{Haefligerunknot, HaefligerFr, Whitney, Whitneytrick, Wu} etc. 
imply that the core of $h_+$ (resp. $h_-$) is trivially embedded in $B$ under the above condition.     

\smallbreak
Suppose that $(h_+-$ its attaching part)$\cap(h_--$ its attaching part)$=\phi$.  
Suppose that their attaching parts coincide.  
Thus we can suppose that  
we regard $h_+\cup h_-$ as an oriented $(2p+2)$-submanifold $\subset S^{2p+3}$
if we give the opposite orientation to $h_-$.  
Then we can define a $(p+1)$-Seifert matrix for the $(2p+2)$-submanifold $h_+\cup h_-$.  
We can suppose that the $(p+1)$-Seifert matrix of 
$\partial(h_+\cup h_-)$ associated with $h_+\cup h_-$  
is a $1\x1$-matrix $(1)$.  




\begin{note}\label{aomori} 
In the case of the twist-move on the $(4k+1)$-dimensional submanifolds 
we can distinguish $K_+$ from $K_-$ because 
the Seifert matrix is a $1\x1$-matrix $(1)$ 
even if we change the orientation of $h_+\cup h_-$. 
On the other hand, in the $(1,2)$-pass-move-triple case 
we cannot distinguish $K_+$ from $K_-$.  See Note \ref{hokkaido}. 
\end{note}



\noindent
{\bf Note.}   
Suppose that $p$ is an odd natural number, and let $p=2k+1$. 
The twist-move for $(4k+3)$-submanifolds $\subset S^{4k+5}$ 
($4k+3\in\N$, \quad $k\in\N\cup\{0\}$)   
has the following property: 
Suppose that $K_+$ is made into $K_-$  by the twist-move.   
Suppose that $K_+$ is PL homeomophic to the standard sphere.    
Then 
$H_*(K_-;\Z)$ is not congruent to $H_*(K_+;\Z)$ in general.   
Example: 
Make a Seifert hypersurface $V_*$  for a 3-knot $K_*$ ($*=+,-$) as follows.    
A framed link representation of $V_+$  is the Hopf link 
such that the framing of one component is zero 
and such that that of the other is two. 
A framed link representation of $V_-$  is the Hopf link 
such that the framing of each component is two.  
\bigbreak




Let $K_* (*=+,-)$ satisfy that  
$K_*\cap \mathrm{Int}B=$ 
$(\partial h_*-\partial B)$.  
Note the following. When we define $K_+$, $h_+$ exists in $B$ and $h_-$ does not exist in $B$. 
When we define $K_-$, $h_-$ exists in $B$ and $h_+$ does not exist in $B$. 
Let $P=K_+ \cap (S^{2p+3}-\mathrm{Int}B)$. 
Let $Q=h_+\cap \partial B$. 
Let  
$T=P\cup Q$. 
Then $T$ is 
an $(2p+1)$-dimensional oriented closed submanifold in $S^{2p+3}-\mathrm{Int}B$.
Let $K_0$ be 
$T$ in $S^{2p+3}$.
Then we say that  an ordered set $(K_+$, $K_-$, $K_0)$ is related by a single {\it twist-move.}     
$(K_+$, $K_-$, $K_0)$ is called a {\it twist-move-triple}.
We say that 
$K_-$ (resp. $K_+$) 
is obtained from 
$K_+$ (resp. $K_-$) 
by a single 
 {\it negative-twist-move} 
(resp. {\it positive-twist-move})  
in $B$. 

See 
Figure \ref{Osaka}.2 
for a twist-move-triple of $(4k+1)$-knots. 

\begin{figure}
\includegraphics[width=14cm]{shitagaki7.eps}
\bigbreak
\hskip27mm Figure \ref{Osaka}.2: A twist-move-triple of $(4k+1)$-knots. 
\end{figure}




\bigbreak\noindent{\bf{Note.}} 
In the twist-move in the $(4k+1)$-dimensional case 
the homotopy type of $K_0$ is determined if $K_+$ is homotopy type equivalent to $S^{4k+1}$
by \cite{Adams, JW}.  
On the other hand, in the $(1,2)$-pass-move-triple case 
the homotopy type of $K_0$ is not determined even if $K_+$ is diffeomorphic to $S^2$. 
See Note to Theorem \ref{two}. 
\bigbreak 







Let $(K_+, K_-, K_0)$ be related by a single twist-move in $B$.   
Then there is a Seifert hypersurface $V_*$ for $K_*$ ($*=+, -, 0$)  with the following properties. 

\smallbreak\noindent 
(1) $V_\sharp=V_0\cup h_\sharp$  ($\sharp=+,-$).
 $V_\sharp\cap B=h_\sharp$.  

\smallbreak
 \noindent 
(2)
$V_0\cap$ Int $B=\phi$. 
$V_0\cap\partial B$ is the attaching part of $h_\sharp^p$. 

\noindent 
(The idea of the proof is the Thom-Pontrjagin construction.)
 

\bigbreak
 The ordered set ($V_+, V_-, V_0$) is called a {\it twist-move-triple of 
Seifert hypersurfaces} for  $(K_+, K_-, K_0)$. 
We say that 
$V_-$ (resp. $V_+$) is obtained from $V_+$ (resp. $V_-$) 
by a single 
 {\it negative-twist-move} 
(resp. {\it positive-twist-move})  
in $B$. 

See Figure \ref{Osaka}.3 for a twist-move-triple of Seifert hypersurfaces for $(4k+1)$-knots. 


\begin{figure}
\includegraphics[width=14cm]{shitagaki6.eps}
\bigbreak
\hskip7mm Figure \ref{Osaka}.3: 
A twist-move-triple of Seifert hyopersurfaces for $(4k+1)$-knots 
\end{figure}


\bigbreak
In \cite{Ogasa98n, Ogasa09} 
we introduced the $(p,q)$-pass-move, which is a kind of local-moves.   
we found local-move-identities of the Alexander polynomial associated with 
the $(p,q)$-pass-move. 
We showed other relations between some invariants of knots and the $(p,q)$-pass-move. 
In \cite{KauffmanOgasa} we also proved such new results.  







\bigbreak
\section{Main theorems in the 4k+1 dimensional case}\label{mhigh}



\noindent 
We can define the `normalized' Alexander polynomial 
in a case of the $(4k+1)$-dimensional case 
so that it is compatible with a local-move-identity 
associated with the twist-move   
(see Definition \ref{4k+1} and Theorem \ref{Tokyo} for detail).   
On the other hand, 
in the 2-dimensional case
we cannot normalize the $\Z[t,t^{-1}]$-Alexander module  
so that it is compatible with the (1,2)-pass-move-identity 
(see \cite[Proposition 4.3]{Ogasa09}). 


\begin{defn}\label{4k+1}
Let $k\in\{0\}\cup\N$. 
Let $K$ be a $(4k+1)$-dimensional closed oriented subamanifold $\subset S^{4k+3}$ 
whose homotopy type is $S^{4k+1}$. 
Let $V$ be a Seifert hypersurface for $K$. 
Let  $S_{2k+1}(V)$ be a $(2k+1)$-Seifert matrix and 
$N_{2k+1}(V)$ its related $(2k+1)$-negative Seifert matrix associated with 
a Seifert hypersurface $V$ for $K$. 
Call
$$\hat\Delta_{K}(t)={\rm{det}}(t^\frac{1}{2}\cdot S_{2k+1}(V)-t^\frac{-1}{2}\cdot N_{2k+1}(V))$$ 
the {\it normalized Alexander polynomial} for $K$. 


Let $k\in\{0\}\cup\N$. 
Let $K$ be a $(4k+1)$-dimensional closed oriented subamanifold $\subset S^{4k+3}$ 
whose homotopy type is $S^{2k+1}\x S^{2k}$.  
Let  $S_{2k+1}(V)$ and $N_{2k+1}(V)$ be defined 
in the same manner as in the previous paragraph.  
Define the {\it normalized Alexander polynomial} $\hat\Delta_{K}(t)$ for $K$ 
to be 

$
\begin{cases}
{\rm{det}}(t^\frac{1}{2}\cdot S_{2k+1}(V)-t^\frac{-1}{2}\cdot N_{2k+1}(V))
&\\
\text{\hskip18mm 
if a $2k$-Alexander matrix associated with $V$ induces an injective map on} 
&\\\
\text{\hskip18mm $H_{2k}(\amalg_{-\infty}^{\infty}V\x[-1,1];\Q)
\to
H_{2k}(\amalg_{-\infty}^{\infty}Y;\Q)$,} & \\  
0& 
\text{\hskip-135mm else.
}   \\
\end{cases}
$
\end{defn}   


\noindent
{\bf Note.}   
(1) 
Recall that any $2k$-Alexander matrix associated with $V$
induces a homomorphism 
$H_{2k}(\amalg_{-\infty}^{\infty}V\x[-1,1];\Q)
\to
H_{2k}(\amalg_{-\infty}^{\infty}Y;\Q)$ 
as in Proof of Proposition \ref{square}. 

\smallbreak\noindent 
(2)By the definition of Alexander matrices we have the following: 
If a $2k$-Alexander matrix associated with $V$ induces (resp. does not induce) an injective map,  
then 
any (resp. no) $2k$-Alexander matrix associated with $V$ induces an injective map. 



\begin{thm}\label{norm} 
The normalized Alexander polynomial $\hat\Delta_{K}(t)$ 
does not depend on the choice of $V$,  
and hence is a topological invariant.  
\end{thm} 


\begin{thm}\label{Tokyo}
Let $K_+$ be a $(4k+1)$-knot $\subset S^{4k+3}$. 
Let $(K_+, K_-, K_0)$ be a twist-move-triple.
Then 
$$\hat\Delta_{K_+}(t)-\hat\Delta_{K_-}(t)
=(t^\frac{1}{2}-t^\frac{-1}{2})\cdot\hat\Delta_{K_0}(t),$$  
where $\hat\Delta_{K}(t)$ denotes the normalized Alexander polynomial of $K$. 
\end{thm}   


See Figure \ref{Osaka}.2 
for an example of a twist-move-triple of $(4k+1)$-knots 
which satisfy the identity in Theorem \ref{Tokyo}.  
There,  we regard 
$S(V_*), N(V_*),$ and $\hat\Delta_{K_*}(t)$ as follows. 

 $
S(V_+)=
\begin{pmatrix}
0&-1\\
0&-1
\end{pmatrix}, 
N(V_+)= 
\begin{pmatrix}
0&0\\
-1&-1
\end{pmatrix}, 
\hat\Delta_{K_+}(t)=1, 
$

$
S(V_-)=
\begin{pmatrix}
-1&-1\\
0&-1
\end{pmatrix},  
N(V_-)= 
\begin{pmatrix}
-1&0\\
-1&-1
\end{pmatrix},  
\hat\Delta_{K_-}(t)=t+\frac{1}{t}-1,
$

$
S(V_0)=(-1),
N(V_0)=(-1),$  and 
$\hat\Delta_{K_+}(t)=-t^\frac{1}{2}+t^\frac{-1}{2}. $  



\bigbreak

We say that $x\in H_i(X;\Z)$ is {\it order finite} (resp. {\it order infinite}) 
if $x\in {\rm Tor}H_i(X;\Z)$ (resp. $\notin {\rm Tor}H_i(X;\Z)$).  
Suppose that $x\in H_i(X;\Z)$ is nonzero and order finite. 
Let $p$ be the minimum number of $\{n\in\N|nx=0\}$.  
Then we say that $x$ is {\it order $p$}.  
We say that $x$ is {\it order zero} 
if $x=0\in H_i(X;\Z)$. 

\begin{defn}\label{divisible}   
We say that $x\in H_i(X;\Z)$ is {\it divisible} 
if $x$ is order infinite  
and 
if there is $y\in H_i(X;\Z)$ such that 
$x=ny$ for an integer $n$ with the condition $|n|>1$. 
We suppose that  $x$ is order infinite  
when we say that $x\in H_i(X;\Z)$ is divisible (resp. non-divisible). 
If $y\in H_i(X;\Z)$ is order infinite, 
there is a non-divisible  $i$-cycle  
$z\in H_i(X;\Z)$   
such that there is an integer $m$ with the condition $y=mz$ ($m$ may be $\pm1$). 
Call $z$ a {\it non-divisible $i$-cycle associated with $y$.}  
\end{defn}



\begin{defn}\label{twinkling}
Let $k\in\{0\}\cup\N.$ 
Let $K$ be a $(4k+1)$-dimensional closed oriented subamanifold $\subset S^{4k+3}$ 
whose homotopy type is $S^{2k}\x S^{2k+1}$.  
Let $V$ be a Seifert hypersurface for $K$. 
We define the {\it pseudo-twinkling number} of $K$ to be 

$
\begin{cases}
\text{s($\tau, \tau)$}&
\text{
if there is a non-divisible $(2k+1)$-cycle $\tau\subset V$ such that 
}\\

 & \text{
for any $(2k+1)$-cycle $\alpha\subset V$  
the intersection product $\tau\cdot\alpha$ in $V$ is zero,  
}\\

\text{0}& \text{ else,  
}\\
\end{cases}
$

\noindent
where s$(\alpha, \beta)$ denotes the Seifert paring of $(2k+1)$-cycles $\alpha$ and $\beta$. 
Note that if $k=0$, the twinkling number is the linking number.  
\end{defn}

\noindent{\bf Note.}  
We would define the `twinkling number' to be 
s$(\gamma, \gamma)$, where  $\gamma$ is a generator of $H_{2k+1}(S^{2k}\x S^{2k+1})$. 
So we call the above one the pseudo-twinkling number by an analogy of 
the relation between the alinking number and the pseudo-alinking number 
although we do not discuss  the twinkling number so much in this paper. 
The author does not know whether the twinkling number and the pseudo-twinkling number are non-equivalent in general. 
He could prove that 
if there is a Seifert hypersurface $V$ such that ${\rm Tor}H_*(V;\Z)\cong0$, 
they are equivalent.  
Note \ref{Hawaii} is related to this question. 
He thinks that we have results which are analogues of Theorem \ref{Connecticut} and Corollary \ref{Georgia}. 
He could prove that the pseudo-twinkling number is `submanifold-cobordism' invariant, where  
submanifold-cobordism is defined in a similar fashion to that of knot cobordism by using (the submanifold)$\x[0,1]$. 
(See the definition right before Proposition \ref{North Carolina}  
for an example of submanifold-cobordism.)
He does not think that the twinkling number is `submanifold-cobordism' invariant. 




\begin{pr}\label{hongo}
The pseudo-twinkling number of $K$ does not depend on the choice of $V$ and that of $\tau$, 
and hence is a topological invariant.   

\end{pr}





\begin{thm}\label{aletwi}
Let $K$ be a $(4k+1)$-dimensional closed oriented subamanifold $\subset S^{4k+3}$ 
whose homotopy type is $S^{2k}\x S^{2k+1}$.  
Let $\hat\Delta_{K}(t)$ be the normalized Alexander polynomial of $K$.  
Then the pseudo-twinkling number of $K$ is 
$$\left.\displaystyle\frac{\hat\Delta_{K}(t)}{t^\frac{1}{2}-t^\frac{-1}{2}}\right|_{t=1}$$
\end{thm}


Proposition \ref{hongo} and Theorem \ref{aletwi} imply the following.

\begin{cor}\label{twi+-}
Let $K_+$ be a $(4k+1)$-knot $\subset S^{4k+3}$. 
Let $(K_+, K_-, K_0)$ be a twist-move-triple.
Then the pseudo-twinkling number of $K_0$ is 
$$\left.\displaystyle
\frac{ \hat\Delta_{K_+}(t)-\hat\Delta_{K_-}(t)}{(t^\frac{1}{2}-t^\frac{-1}{2})^2}
\right|_{t=1}$$
\end{cor}

\noindent{\bf Note.} 
(1)
Compare `Theorem \ref{aletwi} and Corollary \ref{twi+-}' 
with 
`Theorem \ref{alk} and Theorem \ref{Aa}'. 

\smallbreak\noindent(2) 
There is a relation among $\hat\Delta_{K}(t)$, the bP-subgroup, and the inertia group 
by way of  \cite[Theorem 3.4]{Ogasa09} and Corollary \ref{twi+-}.



\bigbreak
Other results in \cite{KauffmanOgasa, Ogasa09} written in the $\Q[t,t^{-1}]$-term 
could be generalized into $\Z[t,t^{-1}]$-term in some fashion without difficulty 
although we must take care of \cite[\S10]{KauffmanOgasa}. 






\bigbreak
\section{Proof of results in \S\ref{m2}}\label{proo2}

\begin{defn}\label{subsur}
Let $X$ be an $x$-dimensional submanifold 
of an $m$-dimesional manifold $M$ ($x,m\in\N, x<m$). 
Suppose that we can embed $X\x [0,1]$ in $M$ so that $X\x\{0\}=X$. 
Suppose that 
an $(x+1)$-dimensional handle $h^p$ is embedded in $M$ and  
is attached to $X\x [0,1]$ ($p\in\N\cup\{0\}, 0\leqq p\leqq x$).  
Suppose that the attaching part of $h^p$ is embedded in $X\x\{1\}$. 
See Figure \ref{proo2}.1. 
Suppose that $h^p\cap(X\x[0,1])$ is only the attaching part of $h^p$.  
Let 
$X'$=
$\overline{\partial(h^p\cup(X\x[0,1]))-(X\x\{0\})}$.
Note that there are two cases, $\partial X=\phi$ and $\partial X\neq\phi$. 
Then we say that 
 $X'$ is obtained from $X$ 
by {\it the surgery by using the embedded handle $h^p$}. 
We do not say that we use $X\x[0,1]$ if there is no danger of confusion.  

\noindent
{\bf Note.}  
Of course we can define `embedded surgery' even if we cannot embed $X\x[0,1]$ in $M$. 
However we do not need the case in this paper.  


\begin{figure}
\input handle.tex

\bigbreak\bigbreak\quad\quad
{\bf Figure \ref{proo2}.1:} A handle $h^p$ is attached to $X\x[0,1]$. 
\end{figure}
\end{defn}




\noindent
{\bf Proof of Theorem \ref{Z}.}  
Let $V$ and $V'$ be Seifert hypersurfaces for $L$. 
Recall that $V$ and $V'$ are connected by the definition. 
It suffices to prove that the $\nu$-$\Z[t,t^{-1}]$-Alexander polynomial ($\nu=1,2$) defined by using $V$ is the same as that defined by using $V'$. 


By the same manner as 
that in \cite[sections 4 and 5]{Levinesimp},  
and 
that in \cite[Proof of Claim 8.1]{KauffmanOgasaB},  
we have the following: 
There are 
(not necessarily connected) 3-dimensional compact oriented submanifolds 
$V=U_1,U_2,...,U_{u-1},U_u=V'$
$\subset S^4$ $(u\in\N)$
such that $\partial U_*=L$ 
and such that $U_{*+1}$ is obtained from $U_*$ $(2\leq*+1\leq u)$ 
by a surgery by using an embedded 4-dimensional handle. 

If some of $U_\natural$ are not connected, 
use 4-dimensional 1-handles and  then we can suppose that all $U_\natural$ are connected, 
that is, all $U_\natural$ are Seifert hypersurfaces for $L$.  

Therefore it suffices to prove the following case: 
$V'$ is obtained from $V$ by a surgery 
by using an embedded 4-dimensional $i$-handle $h^i$ ($i=1,2,3$).  

Lemmas \ref{Louisiana} and \ref{Maine} imply Theorem \ref{Z}. 

\begin{lem}\label{Louisiana}
 Theorem \ref{Z} holds in the case $i=1,3$. 
\end{lem}

\noindent{\bf Proof of Lemma \ref{Louisiana}.}
 $V'=V\sharp(S^1\x S^2)$ or $V=V'\sharp(S^1\x S^2)$ 
where $\sharp$ is the connected-sum. 
If  $V'=V\sharp(S^1\x S^2)$,  
an Alexander matrix $A(t)$ for $V$ is related to  
an Alexander matrix $A'(t)$ for $V'$  as follows. 

$$
A(t)=
\arraycolsep5pt
\left(
\begin{array}{@{\,}c|cccc@{\,}}
\text{$t$ or $-t$}&*&\cdot&\cdot&*\\
\hline
0&&&&\\
\cdot&\multicolumn{4}{c}{\raisebox{-10pt}[0pt][0pt]{\Huge $A'(t)$}}\\
\cdot&&&&\\
0&&&&\\
\end{array}
\right), 
$$
\noindent 
where $A(t)$ is an $n\x n$-matrix and $A'(t)$ is an $(n-1)\x (n-1)$-matrix ($n\in\N$). 
If  $V=V'\sharp(S^1\x S^2)$,  
an Alexander matrix $A(t)$ for $V$ is related to  
an Alexander matrix $A'(t)$ for $V'$  as follows. 

$$
A'(t)=
\arraycolsep5pt
\left(
\begin{array}{@{\,}c|cccc@{\,}}
\text{$t$ or $-t$}&*&\cdot&\cdot&*\\
\hline
0&&&&\\
\cdot&\multicolumn{4}{c}{\raisebox{-10pt}[0pt][0pt]{\Huge $A(t)$}}\\
\cdot&&&&\\
0&&&&\\
\end{array}
\right), 
$$
\noindent 
where $A'(t)$ is an $n\x n$-matrix and $A(t)$ is an $(n-1)\x (n-1)$-matrix. 




Hence 
det $A(t)$ is $\Z[t,t^{-1}]$-balanced to det $A'(t)$ in the both cases. 

This completes the proof of Lemma \ref{Louisiana}. \qed 

\begin{lem}\label{Maine}
Theorem \ref{Z} holds in the case $i=2$. 
\end{lem}

\noindent{\bf Proof of Lemma \ref{Maine}.}
Let $C\subset V$ be the core of the attaching part of $h^2$. 
Let $N(C)$ be the tubular neighborhood of $C$ in $V$. 
By Definition \ref{subsur}, 
there is $W=(V\times [0,1])\cup h^2$ which is embedded in $S^4$.  
Let $C'\subset V'$ be the core of the attaching part of the dual handle ${\bar h}^2$ 
of $h^2$. Note that ${\bar h}^2$ is a 4-dimensional 2-handle. 
Recall that ${\bar h}^2$ is attached to $V'$ 
and that $W=(V\times [0,1])\cup h^2=(V'\times [0,1])\cup{\bar h}^2$. 



There are two cases: 
\smallbreak
\noindent
(1) 
$[C]\in H_1(V;\Z)$ is order finite. 

\smallbreak
\noindent
(2) 
$[C]\in H_1(V;\Z)$ is order infinite. 
\smallbreak


We divide these two cases into four cases. 
\smallbreak
\noindent
(1-1) 
$[C]\in H_1(V;\Z)$ is order finite. 
$[C']\in H_1(V;\Z)$ is order finite. 

\smallbreak
\noindent
(1-2) 
$[C]\in H_1(V;\Z)$ is order finite. 
$[C']\in H_1(V;\Z)$ is order infinite. 


\smallbreak
\noindent
(2-1) 
$[C]\in H_1(V;\Z)$ is order infinite. 
For any closed oriented surface $F$ embedded in $V$,  it holds that  
the intersection product of $[F]\in H_2(V, \partial V;\Z)$ and $[C]\in H_1(V;\Z)$ in $V$ 
is zero. 


\smallbreak
\noindent
(2-2)   
$[C]\in H_1(V;\Z)$ is order infinite. 
There is a closed oriented surface $F$ embedded in $V$ such that  
the intersection product of $[F]\in H_2(V, \partial V;\Z)$ and $[C]\in H_1(V;\Z)$  in $V$ 
is nonzero. 


Lemmas \ref{Maryland}, \ref{Massachusetts}, \ref{Michigan}, and \ref{Minnesota} 
imply Lemma \ref{Maine}. 

\begin{lem}\label{Maryland}
Lemma \ref{Maine} holds in the $(1\text{-}1)$ case.  
\end{lem}

\noindent{\bf Proof of Lemma \ref{Maryland}.}
Proposition \ref{yatto} implies that the (1-1) case does not occur.

\begin{pr}\label{yatto}
If $[C]\in H_1(V;\Z)$ is order finite, 
then $C'\in H_1(V';\Z)$ is order infinite. 
\end{pr}  
 
\noindent{\bf Proof of Proposition \ref{yatto}.}  
Let $[C]$ (resp. $[C']$) be order $q$ (resp. $q'$), where $q, q'\newline\in\N\cup\{0\}$.   
Take a 2-chain $\alpha\subset V$ (resp.  $\alpha'\subset V'$)  
whose boundary is $q\cdot C$ (resp. $q'\cdot C'$).  

Hence $\alpha\cup((q\cdot C)\x[0,1])\cup(q\cdot(\text{the core of $h^2$}))$ 
 (resp. \newline
$\alpha'\cup((q'\cdot C')\x[0,1])\cup(q'\cdot(\text{the core of ${\bar h}^2$}))$ 
is a 2-cycle $\beta$ (resp. $\beta'$) $\subset W$.   
Note that  
$(q\cdot C)\x[0,1]\hookrightarrow V\x[0,1]$
(resp. $(q'\cdot C')\x[0,1]\hookrightarrow V'\x[0,1]$) 
is a level preserving embedding map. 
Recall that $(V\x[0,1])\cup h^2$ is diffeomorphic to 
$(V'\x[0,1])\cup \bar{h}^2$. 
Therefore the intersection product of $\beta$ and $\beta'$ in $W$ is nonzero. 
However, since $W$ is embedeed in $S^4$, this intersection product is zero. 
We arrived at a contradiction. 

This completes the proof of Proposition \ref{yatto}. \qed

This completes the proof of Lemma \ref{Maryland}. \qed


\begin{lem}\label{Massachusetts}
If Lemmas \ref{Michigan} and \ref{Minnesota} hold, 
Lemma \ref{Maine} holds in the $(1\text{-}2)$ case.  
 \end{lem}

\noindent{\bf Proof of Lemma \ref{Massachusetts}.}
Replace $V$ with $V'$, $h^2$ with $\bar{h}^2$, and $C$ with $C'$.  
Therefore the (1-2) case is true if the (2) case is true. 
Note that the (2) case consists of 
the (2-1) case and the (2-2) case. 

This completes the proof of Lemma \ref{Massachusetts}. \qed


\begin{lem}\label{Michigan}
Lemma \ref{Maine} holds in the $(2\text{-}1)$ case.  
 \end{lem}


\noindent{\bf Proof of Lemma \ref{Michigan}.}
We can suppose the following:
There are a positive $p$-Seifert matrix $S_p(V)$ and its related negative $p$-Seifert matrix $N_p(V)$ $(p=1,2)$ associated with $V$ such that 
a square matrix $t\cdot S_p(V)-N_p(V)$ has a row all of whose elements are zero as follows:


$$
\begin{pmatrix}
0&\cdot\cdot\cdot&0\\
*&\cdot\cdot\cdot&*\\
\cdot&\cdot\cdot\cdot&\cdot\\
*&\cdot\cdot\cdot&*\\
\end{pmatrix}.
$$










Hence det$(t\cdot S_p(V)-N_p(V))=0$. 
Hence the $\Q[t,t^{-1}]$-$p$-Alexander polynomial is 
the $\Q[t,t^{-1}]$-balanced class of zero. 
Hence we have the following: 
Let $S_p(V')$ be a positive $p$-Seifert matrix and $N_p(V')$ its related negative $p$-Seifert matrix  $(p=1,2)$ associated with $V'$. 
By Proposition \ref{square} and Notes (2) and (3) to Theorem \ref{Z}, 
it holds that \newline  
det $(t\cdot S_p(V')-N_p(V'))$ 
is $\Q[t,t^{-1}]$-balanced to zero.       
Hence 
det $(t\cdot S_p(V')-N_p(V'))$=0.  

Hence det $(t\cdot S_p(V)-N_p(V))$ and det $(t\cdot S_p(V')-N_p(V'))$ are  
not only $\Q[t,t^{-1}]$-balanced but also $\Z[t,t^{-1}]$-balanced. 

This completes the proof of Lemma \ref{Michigan}. \qed


\begin{lem}\label{Minnesota}
Lemma \ref{Maine} holds in the $(2\text{-}2)$ case.  
 \end{lem}

\noindent{\bf Proof of Lemma \ref{Minnesota}.}
Note that 
$W=(V\x[0,1])\cup h^2$ is diffeomorphic to 
$(V'\x[0,1])\cup {\bar h}^2$. 
Consider the exact sequence by a pair 
$((V\x[0,1])\cup h^2, V)$, where 
we regard $V$ as $V\x\{0\}$: 
$$
\cdot\cdot\cdot\to H_*(V;\Z)\to 
H_*((V\x[0,1])\cup h^2;\Z)\to 
H_*((V\x[0,1])\cup h^2, V;\Z)\to\cdot\cdot\cdot 
$$

and 
the exact sequence by a pair 
$((V'\x[0,1])\cup {\bar h}^2, V')$, where we regard $V'$ as $V'\x\{0\}$: 
$$
\cdot\cdot\cdot\to H_*(V';\Z)\to 
H_*((V'\x[0,1])\cup {\bar h}^2;\Z)\to 
H_*((V'\x[0,1])\cup {\bar h}^2, V';\Z)\to\cdot\cdot\cdot. 
$$


By the existence of $F$, 
$[C']\in H_{2k}(V'; \Z)$ is order finite.  

Let $\xi_1\in H_1(V;\Z)$ be a non-divisible 1-cycle associated with $[C]$. 
Let $\eta_1\in H_2(V;\Z)$ be a non-divisible 2-cycle associated with $[F]$. 
We can suppose the following: 

\smallbreak\noindent(1)
There is a set 
$\{\xi_1, \xi_2,...,\xi_n\}\subset H_1(V;\Z)$, where $n\in\N\cup\{0\}$. 
A set $\{\pi(\xi_1), \pi(\xi_2),...,\pi(\xi_n)\}$
is a basis of $H_1(V;\Z)/{\rm Tor}$,    
where $\pi$ is the natural epimorphism $H_1(V;\Z)\to H_1(V;\Z)/{\rm Tor}$. 

\smallbreak\noindent(2)
We can regard 
$\{\xi_2,...,\xi_n\}\subset H_1(V';\Z)$. 
$\{\pi(\xi_2),...,\pi(\xi_n)\}$  
is a basis of $H_1(V;\Z)/{\rm Tor}$. 


\smallbreak\noindent(3)
There is a basis $\{\eta_1, \eta_2,...,\eta_n\}$ of $H_2(V;\Z)$.


\smallbreak\noindent(4)
We can regard $\{\eta_2,...,\eta_n\}$   
as a basis of $H_2(V;\Z)$. 


\smallbreak\noindent(5)
Since $H_*(\partial V;\Z)$ is torsion free,  
the intersection product  $\xi_i\cdot\eta_j$ in $V$ (resp. in $V'$) is 
$
\begin{cases}
1& \text{if $i=j=1$}          \\  
\text{$\delta_{ij}$ or zero}  &\text{else.}  
\end{cases}
$




\bigbreak
Hence we have  the following: 
An Alexander matrix $A(t)$ for $V$ is associated with 
 $\{\xi_1,...,\xi_n\}$ and $\{\eta_1,...,\eta_n\}$. 
An Alexander matrix $A'(t)$ for $V'$ is associated with 
 $\{\xi_2,...,\xi_n\}$ and $\{\eta_2,...,\eta_n\}$. 
$A(t)$ is an $n\x n$-matrix. 
$A'(t)$ is an $(n-1)\x (n-1)$-matrix. 
Seifert pairings s$(\xi_*, \eta_\#)$ ($2\leq*$ and $2\leq\#$) 
are not changed 
when we attach the 4-dimensional 2-handle $h^2$ to $V$.

$$
A(t)=
\arraycolsep5pt
\left(
\begin{array}{@{\,}c|cccc@{\,}}
\text{$t$ or $-t$}&0&\cdot&\cdot&0\\
\hline
*&&&&\\
\cdot&\multicolumn{4}{c}{\raisebox{-10pt}[0pt][0pt]{\Huge $A'(t)$}}\\
\cdot&&&&\\
*&&&&\\
\end{array}
\right), 
$$


Hence 
det $A(t)$ is $\Z[t,t^{-1}]$-balanced to det $A'(t)$. 

This completes the proof of Lemma \ref{Minnesota}. \qed

This completes the proof of Lemma \ref{Maine}.  \qed

This completes the proof of Theorem \ref{Z}. \qed

\bigbreak
\noindent
{\bf Note.}  
It is important that we can suppose that $\xi_1\cdot\eta_1=1.$    
If it does not hold, det $A(t)$ is not $\Z[t,t^{-1}]$-balanced to det $A'(t)$ in general. 
See the example in \cite[\S10]{KauffmanOgasa}. 





\bigbreak
\noindent
{\bf Proof of Theorem \ref{two}.} 
In
\cite[Proof of Theorem 4.1]{Ogasa09}
we proved that 
there is a $\nu$-Alexander matrix $A_{\nu, L_*}(t)$ for  $L_*$ $(*=+,-,0$ and $\nu=1,2)$
such that 
$${\rm det}A_{\nu, L_+}(t)-{\rm det}A_{\nu, L_-}(t)=(t-1)\cdot{\rm det}A_{\nu, L_0}(t).$$ 
This fact and Theorem \ref{Z} imply Theorem \ref{two}. 
\qed

\bigbreak
\noindent
{\bf Proof of Proposition \ref{Illinois}.}
Consider the following exact sequence by a pair $(V, \partial V)$  
(Note that $\partial V=S^2\amalg T^2$. Here, $S^2$ denotes $K_1$ and $T^2$ $K_2$.): 
$$
\cdot\cdot\cdot\stackrel{\partial}\to
H_*(S^2\amalg T^2;\Z)\stackrel{\iota}\to 
H_*(V;\Z)\stackrel{\rho}\to 
H_*(V,\partial V;\Z)\stackrel{\partial}\to 
H_{*-1}(S^2\amalg T^2;\Z)\stackrel{\iota}\to\cdot\cdot\cdot. 
$$
We can take sets,  
$\{\sigma_1,...\sigma_n\}$ and  $\{\tau_1,...,\tau_n\}$, 
to satisfy the conditions (1)-(3) in Proposition \ref{Illinois}. 
\qed


\bigbreak
\noindent
{\bf Proof of Theorem \ref{alk}.}   
Take sets,   
$\{\sigma_1,...\sigma_n\}$ and  $\{\tau_1,...,\tau_n\}$, 
as in Proposition \ref{Illinois}. 
Then the 1-Alexander matrix $A(t)$ associated with the ordered sets, 
$\{\sigma_1,...\sigma_n\}$ and  $\{\tau_1,...,\tau_n\}$, 
is written as follows:  
$$
\arraycolsep5pt
\left(
\begin{array}{@{\,}c|cccc@{\,}}
(t-1)\cdot a_{11}&(t-1)\cdot a_{12}&\cdot&\cdot&(t-1)\cdot a_{1n}\\
\hline
(t-1)\cdot a_{21}&&&&\\
\cdot&\multicolumn{4}{c}{\raisebox{-10pt}[0pt][0pt]{\Huge $X(t)$}}\\
\cdot&&&&\\
(t-1)\cdot a_{n1}&&&&\\
\end{array}
\right), 
$$
where we have the following: 
$a_{ij}={\rm s}(\sigma_i,\tau_j)$.  
$X(1)=\delta_{ij}$. 
$|a_{11}|$ is the pseudo-alinking number. 
Hence 

$$
A(t)=(t-1)
\arraycolsep5pt
\left(
\begin{array}{@{\,}c|cccc@{\,}}
a_{11}&  a_{12}&\cdot&\cdot&  a_{1n}\\
\hline
(t-1)\cdot a_{21}&&&&\\
\cdot&\multicolumn{4}{c}{\raisebox{-10pt}[0pt][0pt]{\Huge $X(t)$}}\\
\cdot&&&&\\
(t-1)\cdot a_{n1}&&&&\\
\end{array}
\right). 
$$

Hence 
$$\left.\displaystyle\frac{A(t)}{t-1}\right|_{t=1}= 
\arraycolsep5pt
\left(
\begin{array}{@{\,}c|cccc@{\,}}
a_{11}&  a_{12}&\cdot&\cdot&  a_{1n}\\
\hline
0&&&&\\
\cdot&\multicolumn{4}{c}{\raisebox{-10pt}[0pt][0pt]{\Huge $X(1)$}}\\
\cdot&&&&\\
0&&&&\\
\end{array}
\right). 
$$

Hence 
$
\left|\left.\displaystyle\frac{{\rm det}A(t)}{t-1}\right|_{t=1}\right|
$ 
is 
$ 
|a_{11}| 
$,
and hence is the pseudo-alinking number. 

This completes the proof of Theorem \ref{alk}.  \qed




\bigbreak
\noindent
{\bf Proof of Theorem \ref{Colorado}.} 
Theorem \ref{alk} implies that $(2)\Leftrightarrow(3)$.

We prove that $(1)\Leftrightarrow(2)$.   
By the exact sequence in Proof of Proposition \ref{Illinois}, we have \newline
$H_2(V;\Z)\cong\Z^n$, $H_2(V, \partial V;\Z)\cong\Z^n$,  \newline 
$H_1(V;\Z)\cong\Z^n\oplus T$, $H_2(V, \partial V;\Z)\cong\Z^n\oplus T$,  
where $T$ is the torsion part. 

\noindent 
Note that  
$\rho: 
H_1(V;\Z)\stackrel{\rho}\to 
H_1(V,\partial V;\Z)
$
is not an isomorphism in general. 
See Note \ref{Alaska}. 




There is a nonzero element $\alpha\in H_1(S^2\amalg T^2;\Z)$ such that 
$\iota(\alpha)$ is order finite, 
where $\iota$ is the homomorphism in the exact sequence in Proof of Proposition \ref{Illinois}. 
Note that $\alpha$ is represented by an embedded circle $\subset T^2$, 
and let the circle also be called $\alpha$. 
Let $\beta$ be an embedded circle in $T^2$ 
such that $\alpha$ intersects $\beta$ transversely at one point. 
The 1-cycle which represented by $\beta$ is also called $\beta$. 


We prove $\iota(\beta)$ is order infinite in $V$. Reason: 
Suppose that $\iota(\beta)$ is order finite. 
Let $P$ (resp. $Q$) be a 2-cycle $\subset V$ whose boundary is $\alpha$ (resp. $\beta$). 
We can suppose that $P$ intersects $Q$ transversely. 
Take $\partial (P\cap Q)$. 
It is a boundary of a 1-cycle $P\cap Q$ and hence it is zero $\in H_0(T^2)$. 
However it is one point by the definition of $\beta$ hence it is not zero $\in H_0(T^2)$. 
We arrived at a contradiction. 




Take 
$\{\sigma_1,...\sigma_n\}$ and  $\{\tau_1,...,\tau_n\}$ as in Proposition \ref{Illinois}. 
Since $\beta\in H_1(S^2\amalg T^2;\Z)$, 
$\iota(\beta)\cdot\tau_*=0$ for all $*$. 
Hence 
$\sigma_1$ is a non-divisible 1-cycle associated with $\iota(\beta)$. 
Hence 
$\iota(\beta)=k\cdot\sigma_1$ for a nonzero integer $k$. 



Since $S^2\subset\partial V$  (recall that $S^2$ denotes $K_1$), 
the intersection product $\iota(S^2)\cdot \sigma_*=0$ for any $*$. 
Since $H_2(V,\partial V;\Z)$ is torsion-free, 
$\iota(S^2)=\tau_1$. 

Hence 
$|{\rm lk}(\beta, \tau_1)|=|{\rm lk}(\iota(\beta), \tau_1)|=|k\cdot{\rm lk}(\sigma_1, \tau_1)|$. 

Suppose that the alinking number of $L$ is zero. Hence lk$(\beta, \tau_1)=0$. Hence 
the pseudo-alinking number $|{\rm lk}(\sigma_1, \tau_1)|$ is zero. 


Suppose that the pseudo-alinking number  $|{\rm lk}(\sigma_1, \tau_1)|$ is zero. 
Hence ${\rm lk}(\beta, \tau_1)=0$. 
We can use $\{\alpha, \beta\}$ as a basis of $H_1(T^2;\Z)$. 
Since $\iota(\alpha)$ is order finite in $V$, lk$(\alpha, \tau_1)=0.$  
Hence lk$(l\cdot\alpha+m\cdot\beta, \tau_1)=0$ for any pair of integers $(l,m)$. 
Hence the alinking number of $L$ is zero. 

Hence  $(1)\Leftrightarrow(2)$.   

This completes of the proof of Theorem \ref{Colorado}. \qed


\begin{note}\label{Alaska} 
If we define $V$ as in (1), 
$\iota:H_1(\partial V;\Z)\to H_1(V;\Z)$ has the property in (2). 

\smallbreak\noindent(1) 
Let $f:S^1\hookrightarrow S^1\x S^2$ be an embedding 
such that \newline
$f_*:H_1(S^1:\Z)\to H_1(S^1\x S^2-{\rm Int}B^3:\Z)$
carries 1 to $p$ ($|p|>1.$ $p\in\Z.$).  
Let $B$ be an embedded 3-ball in $S^1\x S^2$ such that $B\cap N(f(S^1))=\phi$, 
where $N(f(S^1))$ is the tubular neighborhood of $f(S^1)$ in $S^1\x S^2$.    
Let $V$ be $S^1\x S^2-{\rm Int}B^3-{\rm Int}N(f(S^1))$.  
Note that $\partial V=S^2\amalg T^2$. 

\smallbreak\noindent(2) 
There is a non-divisible cycle $\zeta\in H_1(\partial V;\Z)\cong\Z\oplus\Z_p$ 
associated with a cycle $\in\iota(H_1(V;\Z))$ such that $\zeta\notin\iota(H_1(V;\Z))$. 
We have what is written in the fifth line of Proof of Theorem \ref{Colorado}. 
\end{note}  


Let $F$ and $G$ be oriented closed connected surface $\subset S^4$. 
An {\it $(F,G)$-link} is  
a 2-dimensional closed oriented submanifold $L=(J,K)\subset S^4$ 
such that $J$ (resp. $K$) is diffeomorphic to $F$ (resp. $G$).   
Let $L=(J,K)$ and $L'=(J', K')$ be $(F,G)$-links in $S^4$. 
We say that $L$ and $L'$ are {\it surface-link-cobordant} 
if there is an embedding map $f:(F\amalg G)\x[0,1]\hookrightarrow S^4\x[0,1]$ 
with the following properties: 

\noindent
For $t=0,1$, 
$f((F\amalg G)\x[0,1])\cap (S^4\x\{t\})$ is $f((F\amalg G)\x\{t\})$. 

\noindent
$f((F\amalg G)\x\{0\})$ in $S^4\x\{0\}$ is $L$. 

\noindent
$f((F\amalg G)\x\{1\})$ in $S^4\x\{1\}$ is $L'$. 

\cite[\S2]{Sato} proved that 
if two surface-links are surface-link-cobordant and 
the alinking number of one of the two is zero, 
 then the alinking number of the other is zero. 
We generalize it and prove the following:  


\begin{pr}\label{North Carolina}
Let $L=(J,K)$ and $L'=(J', K')$ be $(F, G)$-links in $S^4$. 
Suppose that  $L$ and $L'$ are surface-link-cobordant. 
Then we have the following: 

\noindent \hskip3cm
${\rm alk}(J\subset L, K\subset L)={\rm alk}(J'\subset L', K'\subset L')$.   

\noindent \hskip3cm
${\rm alk}(K\subset L, J\subset L)={\rm alk}(K'\subset L', J'\subset L')$.   
\end{pr}

\noindent
{\bf Proof of Proposition \ref{North Carolina}.}  
Take a compact oriented 4-manifold $P$ such that 
$\partial P\newline=f(F\x[0,1]) 
\cup \text{(a Seifert hypersurface for $J$)} 
\cup \text{(a Seifert hypersurface for $J'$)}$ \newline
(resp. $f(G\x[0,1]) 
\cup \text{(a Seifert hypersurface for $K$)} 
\cup \text{(a Seifert hypersurface for $K'$)}$) 
Consider 
$P\cap f(G\x[0,1])$ 
(resp. $P\cap f(F\x[0,1])$
and Definition \ref{Arizona}. 
\qed


\bigbreak
\noindent
{\bf Proof of Theorem \ref{Connecticut}.}  
In Proof of Theorem \ref{Colorado}, 
since $H_1(V,\partial V;\Z)$ has a nontrivial torsion in general, 
 $\sigma_1$ is $\iota(\beta)$ or 
a non-divisible 1-cycle associated with $\iota(\beta)$. 
That is, 
$k$ in Proof of Theorem \ref{Colorado} is not $\pm1$ in general. 
Now,  
since $H_1(V,\partial V;\Z)$ is torsion-free, 
 $\iota(\beta)=\pm\sigma_1$ and
$|{\rm lk}(\beta, \tau_1)|=|{\rm lk}(\sigma_1, \tau_1)|$. 
Hence the alinking number is 
$|{\rm lk}(\sigma_1, \tau_1)|$. 
Hence the alinking number is the pseudo-alinking number. 


By Theorem \ref{Colorado}, 
Theorem \ref{Connecticut} holds. \qed 





\bigbreak
\noindent
{\bf Proof of Corollary \ref{Georgia}.}  
Use the isotopy which changes \cite[Figure 4.4]{Ogasa04}  into \cite[Figure 4.3]{Ogasa04}
and vice versa. 
Use 4-dimensinal 1-handles. 
We obtain a Seifert hypersurface for $L$ whose homology groups are torsion-free.\qed









\bigbreak
\noindent
{\bf Proof of Theorem \ref{Aa}.}  
There is a (1,2)-pass-move-triple $(L_+, L_-, L_0)$ with the following properties 
(see Figure \ref{proo2}.2): 
\begin{figure}
   \includegraphics[width=12cm]{72}

Figure \ref{proo2}.2: A (1,2)-pass-move-triple $(L_+, L_-, L_0)$
\bigbreak\end{figure}
$L_+$, $L_-$ are diffeomophic to $S^2$. 
$L_0$ is diffeomophic to $S^2\amalg T^2$. 
A seifert hypersurface for $L_+$ (resp, $L_-$)   
is diffeomophic to  \newline 
$(S^2\x S^1)\sharp(S^2\x S^1)-\text{Int}B^3$, 
where $\sharp$ denotes the connected-sum.  
Note that it has a handle decomposition \newline 
(a 3-dimensional 0-handle)$\cup$(two 3-dimensional 1-handles)$\cup$(two 3-dimensional 2-handles). 
A seifert hypersurface for $L_0$ is diffeomophic to
$(S^2\x D^1)\natural(D^2\x S^1)$,  
where $\natural$ denotes the boundary-connected-sum.  
Note that it has a handle decomposition \newline 
(a 3-dimensional 0-handle)$\cup$(a 3-dimensional 1-handle)$\cup$(a 3-dimensional 2-handle). 
\newline 
A 1-Alexander matrix for $L_+$ (resp, $L_-$, $L_0$)   
is 
$
\begin{pmatrix}
t&t-1\\
0&t 
\end{pmatrix}
$ 
(resp. 
$
\begin{pmatrix}
t&0\\
0&t
\end{pmatrix}
$, 
$(0)$).  


There is a (1,2)-pass-move-triple $(L'_+, L'_-, L'_0)$ 
with the following properties (see Figure \ref{proo2}.3):  
\begin{figure}
   \includegraphics[width=8cm]{73}

Figure \ref{proo2}.3: A (1,2)-pass-move-triple $(L'_+, L'_-, L'_0)$ 
\vskip3mm
\end{figure}
$L'_+$, $L'_-$, and $L'_0$ are diffeomophic to $S^2$. 
A seifert hypersurface for $L_+$ (resp, $L_-$)   
is diffeomophic to $(S^2\x S^1)-\text{Int}B^3$.  
Note that it has a handle decomposition \newline 
(a 3-dimensional 0-handle)$\cup$(a 3-dimensional 1-handle)$\cup$(a 3-dimensional 2-handle). \newline
A seifert hypersurface for $L'_0$   
is diffeomophic to a 3-ball, 
which can be regarded a 3-dimensional 0-handle.  Note that $L'_0$ is a trivial 2-knot. 
A 1-Alexander matrix for $L'_+$ (resp, $L'_-$, $L'_0$)   
is 
$(t)$ 
(resp. 
$(1)$, 
$\phi$), 
where $\phi$ denotes the empty matrix.  

By these example,  Theorem \ref{Aa} holds.
\qed

\bigbreak
\noindent
{\bf Proof of Proposition \ref{Idaho}.}   
Propositions \ref{Idaho} follows from Proposition \ref{Mississippi}.   
\qed

\bigbreak
\noindent 
{\bf Proof of Proposition \ref{Mississippi}.}   
There are the following examples (see Figures \ref{proo2}.4 and \ref{proo2}.5).  
\begin{figure}
   \includegraphics[width=9cm]{74}

Figure \ref{proo2}.4: A (1,2)-pass-move-triple $(L_+, L_-, L_0)$
\vskip3mm\end{figure}
\begin{figure} \includegraphics[width=93mm]{75}

Figure \ref{proo2}.5: A (1,2)-pass-move-triple $(L'_+, L'_-, L'_0)$  
\vskip3mm\end{figure}
Let $L_+$ (resp. $L_-$, $L'_+$, $L'_-$) be an $(S^2, T^2)$-link 
and bound a Seifert hypersurface which is diffeomorpphic to 
$(S^2\x D^1)\natural(D^2\x S^1)$.    
Note that it has a handle decomposition \newline 
(a 3-dimensional 0-handle)$\cup$(a 3-dimensional 1-handle)$\cup$(a 3-dimensional 2-handle)  \newline   
and that it is not diffeomorphic to $(S^2\x S^1)-\text{Int}B^3$. 
We can assume that  a Seifert matrix of 
$L_+$ (resp. $L_-$, $L'_+$, $L'_-$) is a $1\x1$-matrix (4)  (resp. (3), (2), (1)). 
We can suppose that $L_0$ (resp. $L'_0$)  is a trivial 2-knot 
such that $(L_+,L_-,L_0)$ (resp. $(L'_+,L'_-,L'_0)$) is a $(1,2)$-pass-move-triple. 
\qed







\bigbreak
\section{Proof of results in \S\ref{mhigh}}\label{proohigh}
\noindent
{\bf Proof of Theorem \ref{norm}.}   
Lemmas \ref{Indiana} and \ref{Iowa} imply Theorem \ref{norm}. 
\begin{lem}\label{Indiana}
Theorem \ref{norm} is true if $K$ is homotopy type equivalent to $S^{4k+1}$.  
\end{lem}

\noindent{\bf Proof of Lemma \ref{Indiana}.}
Recall that $S_{2k+1}(V)-N_{2k+1}(V)$ 
is represented by the intersection product on $H_{2k+1}(V;\Z)$. 
By the Poincar\'e duality 
${\rm{det}}(S_{2k+1}(V)-N_{2k+1}(V))=1.$  
Hence 
$\hat\Delta_{K}(1)=1$. 

By the Poincar\'e duality 
a $2k$-Alexander matrix induces an injective homomorphism 
on 
$H_{2k}(\amalg_{-\infty}^{\infty}V\x[-1,1];\Q)
\to
H_{2k}(\amalg_{-\infty}^{\infty}Y;\Q)$. 
Hence the $\Q[t,t^{-1}]$-balanced class of
${\rm{det}}(t\cdot S_{2k+1}(V)-N_{2k+1}(V))$  
is the $(2k+1)$-$\Q[t,t^{-1}]$-Alexander polynomial by \newline Propsoition \ref{square}.
Hence the $\Q[t,t^{-1}]$-balanced class of 
${\rm{det}}(t\cdot S_{2k+1}(V)-N_{2k+1}(V))$  
 is topological invariant of $K$. 
Hence 
the $\Q[t,t^{-1}]$-balanced class of 
${\rm{det}}(t\cdot S_{2k+1}(V)-N_{2k+1}(V))$  
 does not depend on the choice of $V$. 


Let $V'$ be a Seifert hypersurface for $K$. 
Let $S_{2k+1}(V')$  be a positive $(2k+1)$-Seifert matrix for $K$ and  
$N_{2k+1}(V')$ its related negative Seifert matrix.    
Let ${\hat\Delta}'_{K}(t) \newline 
={\rm{det}}(t^\frac{1}{2}\cdot S_{2k+1}(V')-t^\frac{-1}{2}\cdot N_{2k+1}(V'))$. 
Recall ${\hat\Delta}_{K}(t)={\rm{det}}(t^\frac{1}{2}\cdot S_{2k+1}(V)-t^\frac{-1}{2}N_{2k+1}(V))$.   
It suffices to prove that ${\hat\Delta}'_{K}(t)={\hat\Delta}_{K}(t)$. 



Since $V$ (resp. $V'$) is $(4k+2)$-dimensional 
and $\partial V$ (resp. (resp. $\partial V'$)) is PL homeomorphic to the standard sphere, 
rank$H_{2k+1}(V;\Z)$ (resp. rank$H_{2k+1}(V';\Z)$) is even. 
Therefore 
there is an integer $n$ such that
${\hat\Delta}'_{K}(t)=t^n\cdot\hat\Delta_{K}(t)$ $\cdot\cdot\cdot\cdot\cdot(*)$  holds. 
(Note that $\hat\Delta_{K}(1)=\hat\Delta'_{K}(1)=1$.)


By Propositions \ref{tsuika} and \ref{Mars}, 
$
N_{2k+1}(V)=^t\hskip-2mm S_{2k+1}(V). 
$
By the Poincar\'e duality 
the number of the rows of $S_{2k+1}(V)$
and that of the columns of it are the (same) even nonnegative integer. 
Hence we have the following: 
Let $M(t)={\rm{det}}(t^\frac{1}{2}\cdot S_{2k+1}(V)-t^\frac{-1}{2}\cdot N_{2k+1}(V))$. 
Then 
$M(t)= ^t\hskip-2mm M(t^{-1})$. 
Hence 
$\hat\Delta_{K}(t)$ (resp. ${\hat\Delta}'_{K}(t)$) 
has a form 
$$
\displaystyle
\sum^{\rho=l}_{\rho=0}a_\rho\cdot t^{\frac{\rho}{2}}
+
\sum^{\rho=l}_{\rho=0}a_\rho\cdot t^{-\frac{\rho}{2}}. 
$$
By this fact and the above identity $(*)$, we have  
${\hat\Delta}'_{K}(t)=\hat\Delta_{K}(t)$. 

This completes the proof of Lemma \ref{Indiana}.  \qed



\begin{lem}\label{Iowa}
Theorem \ref{norm} is true if $K$ is homotopy type equivalent to $S^{2k+1}\x S^{2k}$. 
\end{lem}


\noindent{\bf Proof of Lemma \ref{Iowa}.}
Let $V$ be a Seifert hypersurface for $K$. 
There are two cases (see Note (2) to Definition \ref{4k+1}):  

\smallbreak\noindent
(I)  Any $2k$-Alexander matrix associated with $V$ induces an injective map \newline\hskip1cm 
on $H_{2k}(\amalg_{-\infty}^{\infty}V\x[-1,1];\Q)\to H_{2k}(\amalg_{-\infty}^{\infty}Y;\Q)$. 

\smallbreak\noindent
(II) 
No $2k$-Alexander matrix associated with $V$ induces an injective map \newline\hskip1cm 
on $H_{2k}(\amalg_{-\infty}^{\infty}V\x[-1,1];\Q)\to H_{2k}(\amalg_{-\infty}^{\infty}Y;\Q)$. 

\smallbreak
Lemmas \ref{Kansas} and \ref{Kentucky} imply Lemma \ref{Iowa}. 

\begin{lem}\label{Kansas} Lemma \ref{Iowa} is true in the case \rm{(II)}. \end{lem}

\noindent{\bf Proof of Lemma \ref{Kansas}.}   
Since $V$ satisfies (II), 
the  normalized Alexander polynomial of $K$ defined by using $V$ 
is zero by  Definition \ref{4k+1}. 
By Definition \ref{North Dakota} 
the $\Q[t,t^{-1}]$-Alexander polynomial of $K$ defined by using $V$ 
is the $\Q[t,t^{-1}]$-balanced class of zero. 
 
Let $V'$ be another Seifert hypersurface for $K$. 
By Definition \ref{North Dakota} 
the $\Q[t,t^{-1}]$-Alexander polynomial of $K$ defined by using $V'$  
is the $\Q[t,t^{-1}]$-balanced class of zero 
even if $V'$ satisfies (I) not (II).  
Therefore, by Definitions \ref{North Dakota} and \ref{4k+1} and Proposition \ref{square}  
the normalized Alexander polynomial of $K$ defined by using $V'$ is zero. 

This completes the proof of Lemma \ref{Kansas}. \qed



\begin{lem}\label{Kentucky} Lemma \ref{Iowa} is true in the case \rm{(I)}. \end{lem}

\noindent{\bf Proof of Lemma \ref{Kentucky}.}
In the same manner as  
written in the first part of Proof of Theorem \ref{Z}
it suffices to prove the following case: 
$V$ and $V'$ are Seifert hypersurfaces for $K$. 
$V'$ is obtained from $V$ by a surgery by using an embedded 
$(4k+3)$-dimensional $i$-handle $h^i$ ($1\leq i\leq 4k+2$)
This surgery may change a $(2k+1)$-Alexander matrix associated with $V$ for $K$ 
only if $i=2k+1, 2k+2$.  

Lemma \ref{Kansas} implies that if $V'$ satisfies (II), Lemma \ref{Kentucky} holds. 
Hence it suffices to prove the case where $V'$ satisfies (I). 

Hence both $V$ and $V'$ satisfy (II). 
The dual handle of $h^{2k+2}$ is a $(4k+3)$-dimensional $(2k+1)$-handle. 
Hence it suffices to prove the $i=2k+2$ case. 
Call the core of the attaching part of $h^{2k+2}$, $C$. 

There are two cases: 

\smallbreak\noindent
(1) $[C]\in H_{2k+1}(V,\Z)$ is order finite. 

\smallbreak\noindent
(2) $[C]\in H_{2k+1}(V,\Z)$ is order infinite. 



\smallbreak
The case (2) is divided into two cases: 

\smallbreak\noindent
(2-1) $[C]\in H_{2k+1}(V,\Z)$ is order infinite. For all $(2k+1)$-cycle $\alpha$, 
the intersection product $[C]\cdot\alpha=0$. 
 
\smallbreak\noindent
(2-2) $[C]\in H_{2k+1}(V,\Z)$ is order infinite. 
There is a $(2k+1)$-cycle $\alpha$ 
such that the intersection product $[C]\cdot\alpha$ is nonzero. 


\bigbreak
Lemmas \ref{Missouri}, \ref{Montana}, and \ref{Nebraska} 
imply Lemma \ref{Kentucky}. 

\begin{lem}\label{Missouri} 
Lemma \ref{Kentucky} holds in  the case $(1).$ 
\end{lem}

\noindent{\bf Proof of Lemma \ref{Missouri}.} 
This surgery does not change a $(2k+1)$-Alexander matrix associated with $V$ for $K$. 
This completes the proof of Lemma \ref{Missouri}.\qed


\begin{lem}\label{Montana} 
Lemma \ref{Kentucky} holds in  the case {\rm $(2$-$1)$}. 
\end{lem}

\noindent{\bf Proof of Lemma \ref{Montana}.} 
There is an Alexander matrix associated with $V$ which has a row (or column) all of whose elements are zero. Hence the $\Q[t,t^{-1}]$-Alexander polynomial of $K$ is 
the $\Q[t,t^{-1}]$-balanced class of zero. 
By Definitions \ref{North Dakota} and \ref{4k+1} 
the  normalized Alexander polynomial is zero. 

This completes the proof of Lemma \ref{Montana}. \qed






\begin{lem}\label{Nebraska} 
Lemma \ref{Kentucky} holds in  the case {\rm$(2$-$2)$}. 
\end{lem}

\noindent{\bf Proof of Lemma \ref{Nebraska}.}  
Let ${\bar h}^{2k+1}$ be the dual handle of  
the $(4k+3)$-dimensional $(2k+2)$-handle $h^{2k+2}$. 
Let $C'$ be the core of the attaching part of ${\bar h}^{2k+1}$. 
Note that \newline 
$(V\x[0,1])\cup h^{2k+2}$ is diffeomorphic to 
$(V'\x[0,1])\cup {\bar h}^{2k+1}$. 

Consider the exact sequence by a pair 
$((V\x[0,1])\cup h^{2k+2}, V)$, where 
we regard $V$ as $V\x\{0\}$: 
$$
\cdot\cdot\cdot\to H_*(V;\Z)\to 
H_*(V\x[0,1])\cup h^{2k+2};\Z)\to 
H_*(V\x[0,1])\cup h^{2k+2}, V;\Z)\to\cdot\cdot\cdot 
$$
and 
the exact sequence by a pair 
$((V'\x[0,1])\cup {\bar h}^{2k+1}, V')$, where 
we regard $V'$ as $V'\x\{0\}$: 
$$
\cdot\cdot\cdot\to H_*(V';\Z)\to 
H_*(V'\x[0,1])\cup {\bar h}^{2k+1};\Z)\to 
H_*(V'\x[0,1])\cup {\bar h}^{2k+1}, V';\Z)\to\cdot\cdot\cdot. 
$$


By the existence of the $(2k+1)$-cycle $\alpha$, 
$[C']\in H_{2k}(V'; \Z)$ is order finite.  



Let $\xi\in H_{2k+1}(V;\Z)$ be a non-divisible element associated with $[C]$. 
Since $H_{2k}(\partial V;\Z)$ is torsion-free and the intersection product $\xi\cdot\alpha\neq0$, 
there is a $(2k+1)$-cycle $\eta\in H_{2k+1}(V;\Z)$ such that 
$\eta\cdot\xi=1$. 
We can suppose that $\eta$ is a non-divisible element associated with $\alpha$. 
Therefore we have the following: 
 $A(t)$ (resp. $A'(t)$) 
is ${\rm{det}}(t^\frac{1}{2}\cdot S_{2k+1}(V)-t^\frac{-1}{2}\cdot N_{2k+1}(V))$ 
(resp.  ${\rm{det}}(t^\frac{1}{2}\cdot S_{2k+1}(V')-t^\frac{-1}{2}\cdot N_{2k+1}(V'))$) 
an $(2k+1)$-Alexander matrix associated with $V$ (resp. $V'$).  
We have 

$$A'(t)= 
\arraycolsep5pt
\left(
\begin{array}{@{\,}cc|cccc@{\,}}
0&t^\frac{1}{2}& 0&\cdot&\cdot& 0\\
-t^\frac{-1}{2}& 0&a_{23}&\cdot&\cdot& a_{2n}\\
\hline
0&a_{23}&&&&\\
\cdot&\cdot&&\multicolumn{3}{c}{\raisebox{-10pt}[0pt][0pt]{\Huge \hskip-3mm$A(t)$}}\\
\cdot&\cdot&&&&\\
0&a_{2n}&&&&\\
\end{array}
\right)
$$

or

$$A'(t)= 
\arraycolsep5pt
\left(
\begin{array}{@{\,}cc|cccc@{\,}}
0&-t^\frac{1}{2}& 0&\cdot&\cdot& 0\\
t^\frac{-1}{2}& 0&a_{23}&\cdot&\cdot& a_{2n}\\
\hline
0&a_{23}&&&&\\
\cdot&\cdot&&\multicolumn{3}{c}{\raisebox{-10pt}[0pt][0pt]{\Huge \hskip-3mm$A(t)$}}\\
\cdot&\cdot&&&&\\
0&a_{2n}&&&&\\
\end{array}
\right)
$$



This completes the proof of Lemma \ref{Nebraska}. \qed 

This completes the proof of Lemma \ref{Kentucky}. \qed

This completes the proof of Lemma \ref{Iowa}. \qed

This completes the proof of Theorem \ref{norm}. \qed




\bigbreak
\noindent
{\bf Proof of Theorem \ref{Tokyo}.} 
In \cite[Theorem 3.3]{Ogasa09} we proved the following: 
There is a Seifert hypersurface $V_*$ for $K_*(*=+,-,0)$ 
with an associated $(2k+1)$-Seifert matrix $S_{2k+1}(V_*)$ 
and its related $(2k+1)$-negative Seifert matrix $N_{2k+1}(V_*)$ with the following properties:
\smallbreak
\noindent
(i) 
$${\rm{det}}(t^\frac{1}{2}\cdot S_{2k+1}(V_+)-t^\frac{-1}{2}\cdot N_{2k+1}(V_+))$$
$$-{\rm{det}}(t^\frac{1}{2}\cdot S_{2k+1}(V_-)-t^\frac{-1}{2}\cdot N_{2k+1}(V_-))$$
$$=(t-1)\cdot{\rm{det}}(t^\frac{1}{2}\cdot S_{2k+1}(V_0)-t^\frac{-1}{2}\cdot N_{2k+1}(V_0)).$$
\smallbreak
\noindent
(ii)
$S_{2k+1}(V_+)$ and 
$S_{2k+1}(V_-)$ are $2\nu\x2\nu$-matrices  $(\nu\in\N).$  
$S_{2k+1}(V_+)$ is a$(2\nu-1\x2\nu-1)$-matrix. 
\smallbreak
\noindent
(iii)
The $2k$-Alexander matrix associated with each Seifert hypersurface
defines an  injective map on 
on $H_{2k}(\amalg_{-\infty}^{\infty}V\x[-1,1];\Q)\to H_{2k}(\amalg_{-\infty}^{\infty}Y;\Q)$. 



\smallbreak
Hence 
$$\hat\Delta_{K_+}(t)-\hat\Delta_{K_-}(t)
=(t^\frac{1}{2}-t^\frac{-1}{2})\cdot\hat\Delta_{K_0}(t).$$

This completes the proof of Theorem \ref{Tokyo}.  \qed





\bigbreak
\noindent
{\bf Proof of Proposition \ref{hongo}.}
In the same manner as written in the first part of Proof of Theorem \ref{Z} 
it suffices to prove the following case: 
$V$ and $V'$ are Seifert hypersurfaces for $K$. 
$V'$ is obtained from $V$ by a surgery by using an embedded 
$(4k+3)$-dimensional $i$-handle $h^i$ ($1\leqq i\leqq 4k+2$). 
The pseudo-twinkling number may change only if $i\newline=2k+1,2k+2$. 

The dual handle of $h^{2k+2}$ is a $(4k+3)$-dimensional $(2k+1)$-handle. 
Therefore it suffices to prove the following two cases under the condition $i=2k+2$. 









\smallbreak\noindent
(1)
There is a non-divisible $(2k+1)$-cycle $\tau\subset V$ such that for any $(2k+1)$-cycle $\alpha\subset V$  
the intersection product $\tau\cdot\alpha$ in $V$ is zero 

\smallbreak\noindent
(2)
There is not such a cycle as in (1).

\smallbreak\noindent

By Poincar\'e duality and Mayor-Vietoris exact sequence, 
 $\tau$ is a non-divisible cycle in $V$ associated with $*\x S^{2k+1}$ in $K=\partial V$.   


The above two cases (1) and (2) are divide into four cases. 

\smallbreak\noindent
(1-1) $V$ satisfies (1). $V'$ satisfies the condition made from (1) 

\hskip4mm
by  replacing $V$ with $V'$ in (1). 

\smallbreak\noindent
(1-2) $V$ satisfies (1). $V'$ satisfies the condition made from (2) 

\hskip4mm
by replacing $V$ with $V'$ in (2).   

\smallbreak\noindent
(2-1) $V$ satisfies (2). $V'$ satisfies the condition made from (1) 

\hskip4mm
by replacing $V$ with $V'$ in (1).   

\smallbreak\noindent
(2-2) $V$ satisfies (2). $V'$ satisfies the condition made from (2) 

\hskip4mm
by replacing $V$ with $V'$ in (2).   

\smallbreak\noindent  
Lemmas \ref{Nevada}, \ref{New Hampshire}, \ref{New Jersey}, and 
\ref{New Mexico} imply Proposition \ref{hongo}.  


\begin{lem}\label{Nevada}
Proposition \ref{hongo} holds in the case $(2\text{-}2).$ 
 \end{lem}

\noindent{\bf Proof of Lemma \ref{Nevada}.}
By Definition \ref{twinkling}   
 the pseudo-twinkling number defined by using $V$ (resp. $V'$) is zero. 
\qed

\begin{lem}\label{New Hampshire}
Proposition \ref{hongo} holds in the case $(2\text{-}1).$  
\end{lem}

\noindent{\bf Proof of Lemma \ref{New Hampshire}.}
When we obtain $V'$ from $V$ by using $h^{2k+2}$, 
there does not appear $\tau'$ in $V'$ as in Definition \ref{4k+1}.
The case (2-1) does not occur. 
\qed


\bigbreak
Lemmas \ref{Nevada} and \ref{New Hampshire} and their proof imply Claim \ref{New York}. 
\begin{cla}\label{New York}
The pseudo-twinkling number is zero in the case $(2).$ 
\end{cla}


\begin{lem}\label{New Jersey}
Proposition \ref{hongo} holds in the case $(1\text{-}2).$ 
\end{lem}

\noindent{\bf Proof of Lemma \ref{New Jersey}.}
By Definition \ref{twinkling}  
 the pseudo-twinkling number defined by using $V'$ is zero. 

Let $C$ be the attaching part  of $h^{2k+2}$.  
Note that the Seifert pairing $s(C,C)=0$. 

Under the condition (1-2), 
$\tau$ must be a non-divisible $(2k+1)$-cycle associated with $[C]$.  
Hence $s(\tau',\tau')=0$.  
\qed


\begin{lem}\label{New Mexico}
Proposition \ref{hongo} holds in the case $(1\text{-}1).$ 
\end{lem}

\noindent{\bf Proof of Lemma \ref{New Mexico}.}
The pseudo-twinkling number defined by using $V$ is $s(\tau, \tau)$. 
Let $C$ be the core of the attaching part of $h^{2k+2}$. 
There are two cases. 

\smallbreak\noindent(i) 
$\tau$ is a non-divisible cycle in $V$ associated with $C$ . 

\smallbreak\noindent(ii) 
Else.

\smallbreak 
In the case (i),  
$V$ and $V'$ satisfy (1-2) not (1-1). 
 
The cases (ii) 
follows from Theorem \ref{aletwi} and its proof as written below because 
its proof does not depend of the choice of $V$. 


This completes the proof of Lemma \ref{New Mexico}. \qed

This completes the proof of Proposition \ref{hongo}. \qed



\bigbreak
\noindent
{\bf Proof of Theorem \ref{aletwi}.} 
Let $V$ be a Seifert hypersurface for $K$. 

In the case (2) of Proof of Proposition \ref{hongo}.  
By Claim \ref{New York}   the pseudo-twinkling number is zero. 
In this case 
no $2k$-Alexander matrix associated with $V$ induces an injective map.  
By Definition \ref{4k+1}
the normalized Alexander polynomial  is zero. 
Hence Theorem \ref{aletwi} holds in this case.  


In the case (1) of Proof of Proposition \ref{hongo}.  
Let $\{\alpha_1,...,\alpha_\nu\}$ be a basis of $H_{2k+1}(V;\Z)/\text{Tor}$. We can suppose that $\alpha_1=\tau$. 
Then  
$t^{\frac{1}{2}}S-t^{\frac{-1}{2}}\hskip1mm{^t\hskip-1mm S}$
is written as follows, 
where $a_{\*\#}$ is an integer and $a_{11}$ is the pseudo-twinkling number. 

$$
\arraycolsep5pt
\left(
\begin{array}{@{\,}c|cccc@{\,}}
(t^\frac{1}{2}-t^\frac{-1}{2})\cdot a_{11}&(t^\frac{1}{2}-t^\frac{-1}{2})\cdot a_{12}&\cdot&\cdot&(t^\frac{1}{2}-t^\frac{-1}{2})\cdot a_{1*}\\
\hline
(t^\frac{1}{2}-t^\frac{-1}{2})\cdot a_{21}&&&&\\
\cdot&\multicolumn{4}{c}{\raisebox{-10pt}[0pt][0pt]{\Huge $Q(t)$}}\\
\cdot&&&&\\
(t^\frac{1}{2}-t^\frac{-1}{2})\cdot a_{*1}&&&&\\
\end{array}
\right)
$$


$$=(t^\frac{1}{2}-t^\frac{-1}{2})\cdot
\arraycolsep5pt
\left(
\begin{array}{@{\,}c|cccc@{\,}}
a_{11}& a_{12}&\cdot&\cdot& a_{1*}\\
\hline
(t^\frac{1}{2}-t^\frac{-1}{2})\cdot a_{21}&&&&\\
\cdot&\multicolumn{4}{c}{\raisebox{-10pt}[0pt][0pt]{\Huge $Q(t)$}}\\
\cdot&&&&\\
(t^\frac{1}{2}-t^\frac{-1}{2})\cdot a_{*1}&&&&\\
\end{array}
\right)
$$


Hence 
$\left.\displaystyle
\frac{t^{\frac{1}{2}}S-t^{\frac{-1}{2}}\hskip1mm{^t\hskip-1mm S} }
{t^\frac{1}{2}-t^\frac{-1}{2}}
\right|_{t=1}$
is written as follows. 

$$
\arraycolsep5pt
\left(
\begin{array}{@{\,}c|cccc@{\,}}
a_{11}&  a_{12}&\cdot&\cdot& a_{1*}\\
\hline
0&&&&\\
\cdot&\multicolumn{4}{c}{\raisebox{-10pt}[0pt][0pt]{\Huge $Q(1)$}}\\
\cdot&&&&\\
0&&&&\\
\end{array}
\right)
$$

\noindent 
Since $Q(1)$ is a nonsingular matrix and its determinant is $+1$,   
 $\left.\displaystyle\frac{\hat\Delta_{K}(t)}{t^\frac{1}{2}-t^\frac{-1}{2}}\right|_{t=1}=a_{11}$. 



\begin{note}\label{Hawaii}
If we define $V$ as in (1), 
$\iota:H_{2k+1}(\partial V;\Z)\to H_{2k+1}(V;\Z)$ has the property in (2).  
 
\smallbreak\noindent
(1)
Take $S^{2k+1}\x S^{2k+1}$.  
Let $f$ be an embedding map 
$S^{2k+1}\hookrightarrow S^{2k+1}\x S^{2k+1}$.  
Let $f(S^{2k+1})\subset S^{2k+1}\x S^{2k+1}$. 
Suppose that the induced map \newline
$f_*:H_{2k+1}(S^{2k+1};\Z)\to H_{2k+1}(S^{2k+1}\x S^{2k+1};\Z)$
is $\Z\to\Z\oplus\Z$ with $1\mapsto (n,0)$, where $|n|\newline>1$.  
Here, we fix a generator of $H_{2k+1}(S^{2k+1};\Z)$ and 
that of 
$H_{2k+1}(S^{2k+1}\x S^{2k+1};\Z)$. 
Let $V$ be $(S^{2k+1}\x S^{2k+1})-{\rm Int}N(f(S^{2k+1}))$, 
where $N(f(S^{2k+1}))$ is the tubular neighborhood of $f(S^{2k+1})$ in  $S^{2k+1}\x S^{2k+1}$.  

\smallbreak\noindent 
(2) 
Let $g$ be a generator of  $H_{2k+1}(\partial V;\Z)\cong\Z$. 
Then $\rho(g)$ is a divisible cycle $\in H_{2k+1}(V;\Z)$. 
\end{note}   











\bigbreak
\section{A problem}\label{problem}
\begin{prob}\label{ichiika}
(1) If an invariant of 2-dimensional oriented closed submanifold $\subset S^4$ 
satisfies the identity in Theorem \ref{two} associated with the $(1,2)$-pass-move, 
then is it essentially the $\Z[t,t^{-1}]$-Alexander polynomial? 

\smallbreak\noindent$(2)$
If an invariant of $(4k+1)$-dimensional submanifolds $\subset S^{4k+3}$ 
whose homotopy type is $S^{4k+1}$ or 
$S^{2k}\x S^{2k+1}$ satisfies the identity in Theorem \ref{Tokyo} as written there, 
then is it essentially 
the normalized Alexander polynomial 
(resp. the $\Z[t,t^{-1}]$-Alexander polynomial)? 
\end{prob} 

\noindent{\bf Note.} 
If the answer to 
Problem \ref{ichiika}.(1) is positive, 
it is a new characterization of the $\Z[t,t^{-1}]$-Alexander polynomial of 
2-dimensional closed oriented submanifolds in $S^4$.  
If the answer is negative, we may encounter a new invariant. 

If the answer to Problem \ref{ichiika}.(2) is positive, 
it is a new characterization of 
the normalized Alexander polynomial 
(resp. the $\Z[t,t^{-1}]$-Alexander polynomial) of 
high dimensional knots in the case.   
If the answer is negative, we may encounter a new invariant. 

There arise similar problems on 
the local-move-identities in \cite{KauffmanOgasa, Ogasa09} to Problem \ref{ichiika}.



