text:
\subsection{Settings}\label{sec:setting}
In order to prove Theorem \ref{thm:main}, we will prove the next lemmas.
Let $[T,X,Z]$  be homogeneous coordinates of $\PP^2$ and let $(t,x)=(T/Z,X/Z)$ be affine coordinates.
\begin{lem}\label{prop:set-qua}
    Let $\mcQ$ be a reduced quartic curve that is not four lines.
    Then there is a coordinate system $[T,X,Z]$ of $\PP^2$ such that $\mcQ$ is given by
    \[
        X^3Z+A_{\mcQ,2}(T,Z)X^2+A_{\mcQ,3}(T,Z)X+A_{\mcQ,4}(T,Z)
    \]
    where $A_{\mcQ,d}$ is a binary form of degree $d$ in $T$ and $Z$ such that
    \[
        \deg\,A_{\mcQ,2}(t,1)=2, \ \deg\,A_{\mcQ,3}(t,1)=3, \ {\it and} \ \deg\,A_{\mcQ,4}(t,1)\leq 3.
    \]
\end{lem}
{\it Proof.} Let $z_o$ be a general smooth point on $\mcQ$ and let $L_{z_o}$ be the tangent line to $\mcQ$ at $z_o$.
We take a point $p\in \{p,p'\}:=(\mcQ\cap L_{z_o})\setminus \{ z_o \}$ and $q\in \mcQ \setminus L_{z_o}$.
Note that we may assume that $I_{p}(\mcQ,L_{z_o})=I_{p'}(\mcQ,L_{z_o})=1$ and $I_{z_o}(\mcQ,L_{z_o})=2$.
We can choose a coordinate system $[T,X,Z]$ such that
\begin{itemize}
    \item the point $q$ has coordinates $[0,0,1]$ and the point $p$ has coordinates $[1,0,0]$ and
    \item the smooth point $z_o$ has coordinates $[0,1,0]$ and the tangent line $L_{z_o}$ is defined by $Z=0$.
\end{itemize}
We have the desired coordinate system.
\qed

By comparing degree of polynomials, we have the next lemma.
\begin{lem}\label{lem:degree}
Let $f(t,x)$ be a polynomial in $\CC[t,x]$ given by
    \[
        f=x^3+a_2(t)x^2+a_3(t)x+a_4(t) \ (a_i\in\CC[t], \ i=2,3,4)
    \]
    and
    \[
        \deg\,a_2=2, \ \deg\,a_3=3 \ and \ \deg\,a_4\leq3.
    \]
    Let $x_0$ and $y_0$ be polynomials in $\CC[t]$ such that ${y_0}^2=f(t,x_0)$.
    For $(r,s)\in\CC(t)^{\times}\times\CC(t)$, we put $v_{r,s}=r(x-x_0)(x-s)-y_0$.
    Note that it follows that ${v_{r,s}}^2-f=r^2(x-x_0)u$, where $u=x^3+c_1(t)x^2+c_2(t)x^2+c_3(t), \ (c_i\in\CC(t), \ i=1,2,3)$.

    If $u$ is a polynomial of total degree of $3$, then $r\in\CC^{\times}$ and $s\in\CC[t]$ of degree $\leq1$.
\end{lem}
\subsection{Proof of Theorem\,\ref{thm:main}}\label{sec:proof_main1}

From Lemma \ref{prop:set-qua}, it follows that there is a coordinate system $[T,X,Z]$ of $\PP^2$ such that $\mcQ$ is given by
\[
    F_{\mcQ}(T,X,Z)=X^3Z+A_{\mcQ,2}(T,Z)X^2+A_{\mcQ,3}(T,Z)X+A_{\mcQ,4}(T,Z),
\]
where $(t,x)=(T/Z,X/Z)$ be non-homogeneous coordinates and $A_{\mcQ,d}$ are binary forms of degree $d$ in $T$ and $Z$ such that
\begin{align*}
    \deg\,A_{\mcQ,2}(t,1)=2, \ \deg\,A_{\mcQ,3}(t,1)=3 \ {\rm and } \ \deg\,A_{\mcQ,4}(t,1)\leq 3.
\end{align*}

We put $L_i=\overline{f}_{\mcQ,z_o}(s_i) \ (i=1,2,3)$ and $L_4=\overline{f}_{\mcQ,z_o}(s_{P_4})$.
The elliptic curve $E$ given by $y^2=F_{\mcQ}(t,x,1)$ corresponds to the generic fiber of $\varphi_{\mcQ,z_o}$.
The lines $L_i$ are given by the affine equations $x-x_i(t)=0 \ (i=1,2,3,4)$, where $x_1(t),x_2(t),x_3(t)$ and $x_4(t)$ are polynomials of degree $\leq$ 1 and we denote $P_1+P_2+P_3\in\Div(E)$ by $\fd$.
The lines $L_1$, $L_2$ and $L_3$ are three distinct lines, so $\iota(P)\notin \Supp(\fd)$ for $P\in\Supp(\fd)$.
Hence, the divisor $\fd$ is a semi-reduced divisor and we denote the Mumford representation of $\fd$ by $(a,b) \ (a,b\in\CC(t)[x])$.
From Lemma \ref{lem:chara-Mum} and Lemma \ref{lem:mum-tri2}, we have
\begin{align}\label{eq:tri}
    \begin{aligned}
        a&=(x-x_1)(x-x_2)(x-x_3),\\
        b&=b_0(x-x_{4})(x-b_1)-y_{4} \ {\rm and} \\
        b^2&-F_{\mcQ}(t,x,1)={b_0}^2(x-x_4)a,\\
    \end{aligned}
\end{align}
where $b_0\in \CC(t)^{\times}$ and $b_1\in\CC(t)$.
Now $a$ is a polynomial of total degree 3.
From Lemma \ref{lem:degree}, it follows that $b_0\in\CC^{\times}$ and $b_1\in\CC[t]$ such that $\deg\, b_1\leq 1$.
Therefore the affine equation $b(t,x)=0$ gives rise to a plane curve $C$ of degree $2$ i.e. $C$ is a conic.
From the third equation of (\ref{eq:tri}), $C$ is the desired conic.
\qed

\subsection{Proof of Corollaries \ref{cor:ha3.3}, \ref{cor:ha3.4} and \ref{cor:ha3.5}}\label{sec:proof_cors}

By Theorem \ref{thm:main} and Table \ref{table:geo-line}, we obtain Corollaries \ref{cor:ha3.3}, \ref{cor:ha3.4} and \ref{cor:ha3.5}.
We prove Corollary \ref{cor:ha3.3}, because we can similarly prove the other corollaries.
We keep the notation as before.

{\it Proof of Corollary \ref{cor:ha3.3}.}

Let $z_o$ be a general smooth point on $C_1+C_2$ and let $L_1,L_2,L_3$ and $L_4$ be the four distinct bitangent lines of $C_1+C_2$. We denote generators of $E_{S_{C_1+C_2,z_o}}(\CC(t))$ by $P_1,P_2,P_3$ and $P_4$ as Table \ref{table:geo-gene}.
By Table \ref{table:geo-line}, we take $Q_1,Q_2,Q_3$ and $Q_4$ as follows:
\begin{itemize}
    \item $Q_1:=\dot{-}P_1\dot{+}P_2\dot{+}P_3$,
    \item $Q_2:=P_1\dot{-}P_2\dot{+}P_3$,
    \item $Q_3:=P_1\dot{+}P_2\dot{-}P_3$,
    \item $Q_4:=P_1\dot{+}P_2\dot{+}P_3$ and
    \item $\overline{f}_{C_1+C_2,z_o}(s_{Q_i})=L_i \ (i=1,2,3,4)$.
\end{itemize}
Then $Q_4=Q_1\dot{+}Q_2\dot{+}Q_3$ holds.
From Theorem \ref{thm:main}, the eight points of $C_1+C_2$ of contact with the four lines lie on a conic $C$.
We shall prove take four distinct $p_1,p_2,p_3$ and $p_4$ of $(\cup_{i=1}^4L_i)\cap\mcQ$ such that the all three points of $p_1,p_2,p_3$ and $p_4$ do not lie on $C$.
Hence, $C$ is smooth.
\qed

\subsection{Proof of Theorem \ref{thm:A2+A1}}\label{sec:proof_main2}
Let $P_1,P_2,P_3$ and $P_4$ be generators of $E_{S_{\mcQ,z_o}}(\CC(t))$ as Table \ref{table:geo-gene}.
From Table \ref{table:geo-line}, there exists seven lines $L_l$ and $M_m \ (l=1,2,3,4, \ m=1,2,3)$ satisfying as follows:
\begin{itemize}
    \item
    the lines $L_l$ is a weak-bitangent line passing through $x$ and tangent to $\mcQ$ at a smooth point for $l=1,2,3,4$ and
    \item
    the lines $M_m$ is a weak-bitangent line passing through $y$ and tangent to $\mcQ$ at a smooth point for $m=1,2,3$ and
    \item
    $\overline{f}_{\mcQ,z_o}(s_{Q_l})=L_l$ for $l=1,2,3,4$ and
    \item
    $\overline{f}_{\mcQ,z_o}(s_{R_m})=M_m$ for $m=1,2,3$.
\end{itemize}
where
$Q_1=P_2$, $Q_2=P_4$, $Q_3=P_1\dot{-}P_3\dot{+}P_4$, $Q_4=\dot{-}P_1\dot{+}P_2\dot{-}P_3$,
$R_1=P_3$,
$R_2=P_2\dot{-}P_3\dot{+}P_4$ and $R_3=P_1\dot{-}P_2\dot{+}P_4$.
In order to explain clearly, we use similar notations of Table \ref{table:geo-line}.
Namely, we write
\[
    Q_1=\left[\begin{array}{c}
        0\\
        1\\
        0\\
        0
    \end{array}\right],
    Q_2=\left[\begin{array}{c}
        0\\
        0\\
        0\\
        1
    \end{array}\right],
    Q_3=\left[\begin{array}{c}
        1\\
        0\\
        -1\\
        1
    \end{array}\right],
    Q_4=\left[\begin{array}{c}
        -1\\
        1\\
        -1\\
        0
    \end{array}\right]
\]
\[
    R_1=\left[\begin{array}{c}
        0\\
        0\\
        1\\
        0
    \end{array}\right],
    R_2=\left[\begin{array}{c}
        0\\
        1\\
        -1\\
        1
    \end{array}\right],
    R_3=\left[\begin{array}{c}
        1\\
        -1\\
        0\\
        1
    \end{array}\right].
\]
We prove the existence of $M_{a_{ij}}$ and $M_{b_{ij}}$ $(1\leq a_{ij}<b_{ij}\leq 4)$ for $L_i$ and $L_j \ (1\leq i<j \leq 4)$.
We consider the case when $i=1$ and $j=2$.
Then we obtain $R_2=Q_1\dot{+}Q_2\dot{-}R_1$.
From Theorem \ref{thm:main}, the six points of $\mcQ$ of intersections with the four lines lie on a conic $C_{12}$.
Note that $C_{12}$ is smooth.
For other pair $(i,j)$, we similarly find the existence of $M_{a_{ij}}$, $M_{b_{ij}}$ and $C_{ij}$ satisfying $(\ast)$.

We will prove the uniqueness.
For $L_i$ and $L_j$, let $M_{a_{ij}}, M_{b_{ij}}$ and $C_{ij}$ be two lines and a smooth conic satisfying $(\ast)$ and put $\{ x, p_1, p_2\}=\mcQ \cap (L_i+L_j)$,  $\{ y, q_1\}=\mcQ \cap M_{a_{ij}}$ and $\{ y, q_2\}=\mcQ \cap M_{b_{ij}}$.
Note that the divisor on $C_{ij}$ cut out by $\mcQ$ is $C_{ij}|_{\mcQ}=2x+2y+p_1+p_2+q_1+q_2$.

Assume that for a line $M_{a_{ij}}$ and $M_{c_{ij}}$ $(c_{ij}\in \{1,2,3\} \setminus \{ a_{ij}\})$, there exists a smooth conic $C'_{ij}$ such that $(L_i+L_j+M_{a_{ij}}+M_{b_{ij}})\cap \mcQ \subset C'_{ij}$.
We will prove that $M_{b_{ij}}=M_{c_{ij}}$.

Put $\{ y,q_3\}=\mcQ\cap M_{c_{ij}}$.
Note that the divisor on $C'_{ij}$ cut out by $\mcQ$ is $C'_{ij}|_{\mcQ}=2x+2y+p_1+p_2+q_1+q_3$.

We have $\{x,y,p_1,p_2,q_1\}=\mcQ \cap (L_i+L_j+M_{a_{ij}})\subset C_{ij}, C'_{ij} \ i.e. \ C_{ij}=C'_{ij}$.
We obtain $2x+2y+p_1+p_2+q_1+q_2=C_{ij}|_{\mcQ}=C'_{ij}|_{\mcQ}=2x+2y+p_1+p_2+q_1+q_3$.
Therefore, $q_2=q_3$.
We have $M_{b_{ij}}=M_{c_{ij}}$.
\qed
\begin{rem}
    We proved Theorem \ref{thm:A2+A1} in the case when the two branch at $y$ of $\mcQ$ meets $L_i$ at $y$ with multiplicities 1, respectively.
    We also consider the case when a branch at $y$ of $\mcQ$ meets $L_i$ at $y$ with multiplicity 3.
    In this case, we can prove a similar statement.
\end{rem}

\subsection{Notation and terminology}
We refer to \cite{Koda, Shio, Mira} for details.
We here define some notation and terminology.
In this article, an elliptic surface always satisfies the three conditions in the Introduction.
Let $\varphi:S\rightarrow C$ be an elliptic surface over a smooth projective curve $C$.
For $v\in C$, we denote the corresponding fiber by $F_v=\varphi^{-1}(v)$.
We define two subsets, $\sing(\varphi)$ and $\Red(\varphi)$, of $C$  concerning singular fibers as follows:
\begin{align*}
    \sing(\varphi)&:=\{v\in C\mid F_v \ {\rm is \ singular} \}, \\
    \Red(\varphi)&:=\{v\in \sing(\varphi)\mid F_v \ {\rm is \ reducible} \}.
\end{align*}
For $v\in \Red(\varphi)$, the irreducible decomposition of $F_v$ is denoted by
\begin{equation}\label{eq:notation1}
    F_v=\Theta_{v,0}+\sum^{m_v-1}_{i=1}a_{v,i}\Theta_{v,i},
\end{equation}
where $\Theta_{v,0}$ is the unique component with $\Theta_{v,0}\cdot O=1$.
We call $\Theta_{v,0}$ the {\it identity component} of $F_v$. In order to describe the type of singular fibers, we use Kodaira's symbols (\cite{Koda}). Irreducible components of singular fibers are labeled as in \cite{Toku}.

By assumption, for $O\in \MW(S)$, $\MW(S)$ is endowed with an abelian group structure by considering fiberwise addition with $O$ as the zero element.

For $v\in\Red(\varphi)$, we define:
\begin{align}\label{eq:notation2}
    &\begin{aligned}
        \vc(v,D)&:=
        \left[
           \begin{array}{c}
                D\cdot\Theta_{v,1}\\
              \vdots \\
              D\cdot\Theta_{v,m_v-1}
            \end{array}
        \right] \in\ZZ^{m_v-1},\\
        A_v&:=[\Theta_{v,i}\cdot\Theta_{v,j}]_{1\leq i,j \leq m_v-1},\\
        \FF_v&:=[\Theta_{v,1} \cdots \Theta_{v,m_v-1}],
        \end{aligned}
\end{align}
where $D$ is a divisor on $S$.

We explained some notation and terminology of general elliptic surfaces.
 For a rational surface which we consider, we next define notation similarly.
 Let $\mcQ$ be a reduced plane quartic curve satisfying $(\dagger)$ and let $z_o$ be a general smooth point of $\mcQ$.
 We recall the construction of a rational elliptic surface $S_{\mcQ,z_o}$.
 Since the tangent line $l_{z_o}$ meets $\mcQ$ at $z_o$ and at another two distinct points, $F_{z_o}$ is a singular fiber of type $\I_2$.
 We put $F_{z_o}=\Theta_{\infty,0}+\Theta_{\infty,1}$, where $\Theta_{\infty,0}$ is the identity component of $F_{z_o}$.
 Let $x$ be a singularity of $\mcQ$.
 We denote the singular fiber $\overline{f}_{\mcQ,z_o}^{-1}(x)$ by $F_x$.
 For a singularity $x$ of $\mcQ$ and a divisor $D$ on $S_{\mcQ,z_o}$, the irreducible decomposition of $F_x$, $m_x$ and $\Theta_{x,i}$ are written as in (\ref{eq:notation1}).
 Moreover, we also define $\vc(x,D)$, $A_x$ and $\FF_x$ as in (\ref{eq:notation2}).

\subsection{Some properties of elliptic surfaces}
We recall some necessary facts.
\begin{thm}{\rm (\cite[Theorem 1.3 ]{Shio})}
    \[
    \overline{\psi}:\NS(S)/T_{\varphi}\simeq E_S(\CC(C))
    \]
    where $T_{\varphi}$ is the subgroup of $\NS(S)$ generated by $O$ and all the irreducible components of fibers.
\end{thm}

Given a divisor $D$ on $S$, we denote $\overline{\psi}(D \ \mathrm{mod}\,T_{\varphi} )$ by $P_D$.

\begin{thm}{\rm \cite[Lemma 5.1]{Shio}}\label{lem:shio-form}
    For $D\in\Div(S)$, there is a unique section $s(D)$ such that
    \begin{equation*}
        D\approx s(D)+(d-1)O+nF+\sum_{v\in\Red(\varphi)}\FF_vA_v^{-1}\vc(v,D-s(D))
    \end{equation*}
    where $\approx$ and $\chi(\mcO_S)$ denote algebraic equivalence and the characteristic of $\mcO_S$, $d=D\cdot F$ and $n=(d-1)\chi(\mcO_S)+O\cdot(D-s(D))$.
\end{thm}
Note that $A_v^{-1}\vc(v,D-s(D))\in\ZZ^{m_v-1}$ by Theorem\,\ref{lem:shio-form}, while entries of $A_v^{-1}$ are not necessary integers.

\begin{rem}
    For $D\in\Div(S)$, we have $s(D)=s_{P_D}$, where $s(D)$ is the section for $D$ in Theorem\,\ref{lem:shio-form}.
\end{rem}

\begin{rem}\label{rem:become-line}
    Let $\mcQ$ be a reduced plane quartic curve satisfying $(\dagger)$.
    Let $P_1,\ldots ,P_n$ be generators of $E_{S_{\mcQ,z_o}}(\CC(t))$ and $Q=\sum_i c_iP_i\in E_{S_{\mcQ,z_o}}(\CC(t)) \ (c_i\in\ZZ)$.
    By Theorem \ref{lem:shio-form} $(\vc(\infty,\sum_ic_is_{P_i})-\vc(\infty,s_{Q}))/2\in\ZZ$ holds.
    Hence, we have
    \[
        s_Q\cdot \Theta_{\infty,1}=
        \left\{
            \begin{array}{cc}
                1& (\sum_ic_is_{P_i}) \cdot \Theta_{\infty,1} {\rm :odd}\\
                0& {\rm otherwise}
            \end{array}
        \right. .
    \]
\end{rem}

Let $\phi_o:\Div(S)\rightarrow \NS(S)_{\QQ}$ and $\phi:E_S(\CC(C))\rightarrow \NS(S)_{\QQ}$ be the homomorphisms given in \cite{Ban-Toku21} and \cite{Shio} respectively.
In \cite{Shio}, Shioda defined a structure on $E_S(\CC(C))$ called the {\it height pairing} denoted by $\langle -,-\rangle$.
We refer to \cite{Shio} for details.
Then we have

\begin{thm}[\cite{Shio}]\label{thm:hiegh-pair}
    Let $s_1, s_2\in\MW(S)$.
    The height pairing $\langle P_{s_1},P_{s_2}\rangle$ is given by
    \[
        \langle P_{s_1},P_{s_2}\rangle = \chi(\mcO_S)-s_1\cdot s_2+s_1\cdot O+s_2\cdot O-\sum_{v\in\Red(\phi)}\mathrm{contr}_v(s_1,s_2),
    \]
    where for divisors $D_1$ and $D_2$ on $S_{\mcQ,z_o}$, $\mathrm{contr}_v(D_1,D_2)$ is given by
    \[
        \mathrm{contr}_v(D_1,D_2)= {}^t \vc(v,D_1)(-A_v)^{-1}\vc(v,D_2).
    \]
\end{thm}

\begin{rem}
    Let $\mcQ$ be a reduced plane quartic curve satisfying $(\dagger)$ and let $z_o$ be a general smooth point of $\mcQ$.
    For a singularity $x$ of $\mcQ$ and divisors $D_1$ and $D_2$ on $S_{\mcQ,z_o}$,we put
    \begin{align*}
        \mathrm{contr}_x(D_1,D_2)&={}^t\vc(x,D_1)(-A_x^{-1})\vc(x,D_2).
    \end{align*}
\end{rem}

\subsection{Mumford representations}
Let $C$ be a hyperelliptic curve of genus $g$ defined over a field $K \ (\mathrm{char}(K)\neq2)$ given by an affine equation:
\[
    y^2=f(x)=x^{2g+1}+c_1x^{2g}+\cdots +c_{2g+1} \quad c_i\in K \ (i=1,\ldots ,2g+1).
\]
Its point at infinite is denoted by $O$ and the hyperelliptic involution by $\iota:C\rightarrow C$.
\begin{defin}\label{def:mumford}
    For a divisor $\fd=\sum_{P\in C}n_PP\in \Div(C)$, we assume that $O \notin \Supp(\fd):=\{ P\in C \mid n_P\neq 0 \}$ and $\fd$ is effective.
    We call $\fd$ a {\it semi-reduced divisor} if it follows that
    \begin{itemize}
        \item if $P\in\Supp(\fd)$ and $P\neq\iota(P)$, then $\iota(P)\notin \Supp(\fd)$ and
        \item if $P\in\Supp(\fd)$ and $P=\iota(P)$, then $n_P=1$.
    \end{itemize}
\end{defin}
We denote the coordinate ring $\Kbar[x,y]/\langle y^2-f \rangle$ by $\Kbar[C]$ and the image of $g\in\Kbar[x,y]$ in $\Kbar[C]$ by $[g]$. For $P\in C$, we write $\mcO_P$ for the local ring at $P$ and $\ord_P$ for the discrete valuation at $P$.
Then we have
\begin{prop}{\rm (\cite[Proposition 2.1]{Taka-Toku})}
    Let $>_p$ be the pure lexicographical order with $y>_px$ in $\Kbar[x,y]$.  For a semi-reduced divisor $\fd=\sum_{P\in C}n_PP$, we put
    \[
        \widetilde{I(\fd)}:=\{ g\in \Kbar[C] \mid \ord_P([g])\geq n_P \ \forall P\in \Supp(\fd)  \}.
    \]
    Then the reduced Gr\"{o}bner basis of $\widetilde{I(\fd)}$ with respect to $>_p$ is of the form $\{ a(x) , y-b(x) \}$ and $b^2-f\in \langle a \rangle$, where $a$ and $b\in\Kbar[x]$.
\end{prop}
For a proof, see \cite[Proposition 2.1]{Taka-Toku}.

\begin{defin}
    Let $\fd$ be a semi-reduced divisor on $C$ and let $\{ a(x) , y-b(x) \}$ be the reduced Gr\"{o}bner basis for $\widetilde{I(\fd)}$ with respect to $>_p$.
    Then we call the pair $(a,b)$ the {\it Mumford representation} of $\fd$.
\end{defin}

The next lemma is a characterization of Mumford representations.
\begin{lem}{\rm (\cite[Lemma 10.3.5]{galb})}\label{lem:chara-Mum}
    Let $\fd=\sum_{P\in C}n_PP$ be a semi-reduced divisor and we put $P=(x_P,y_P)$.
    The pair $(a,b)\in (\Kbar[x])^2$ is the Mumford representation of $\fd$ if and only if $(a,b)$ satisfies
    \begin{equation*}
        {\rm(i)} \ a(x)\!=\!\mathop{\prod}\limits_{P\in \Supp(\fd)}(x-x_P)^{n_P}, \quad
        {\rm(ii)} \ \deg \,b(x) < \deg \,a(x), \quad
        {\rm(iii)} \ a\mid b^2-f.
    \end{equation*}
\end{lem}
\subsection{Semi-reduced divisors of degree 3 on elliptic curves}
From now on, we assume that $K$ is a field of  characteristics $\neq2$. Let $E$ be an elliptic curve defined over $K$ given by the Weierstrass form
\[
    y^2=f(x)=x^3+c_1x^2+c_2x+c_3 \quad c_i\in K  \ (i=1,2,3).
\]
\begin{rem}
    {\rm (i)}
    If a semi-reduced divisor $\fd$ is defined over $K$, then the Mumford representation $(a,b)$ of $\fd$ belongs to ${(K[x])}^2$.
    {\rm (ii)}
    For a semi-reduced divisor  $\sum_in_iP_i$ defined over $K$, the points $P_i$'s are not always elements in $E(K)$.
\end{rem}

Let $\fd=P_1+P_2+P_3$ be a semi-reduced divisor of degree 3 defined over $K$.
We write  $P_1\dot{+}P_2\dot{+}P_3$ for $P_{\fd}$, where $\dot{+}$ is the addition on $E$ and the infinity point is the identity with respect to $\dot{+}$.
\begin{lem}{\rm (\cite[Lemma 5.6]{Ban-Kaw-Mas-Tok})}\label{lem:mum-tri1}
    Assume that $P_{\fd}\neq O$ and let $(a,b)$ be the Mumford representation of $\fd$. Then we have
    \begin{itemize}
        \item[{\rm(i)}] $P_{\fd}\neq P_i \ (i=1,2,3)$.
        \item[{\rm(ii)}] $\deg b=2$.
    \end{itemize}
\end{lem}
\begin{lem}{\rm (\cite[Lemma 5.7]{Ban-Kaw-Mas-Tok})}
    \label{lem:mum-tri2}
    We keep the notation as before.
    Assume that $\fd$ is defined over $K$.
    Put $P_{\fd}:=(x_{\fd},y_{\fd})$.
    Then we have the following:
    \begin{itemize}
        \item[{\rm(i)}] The point $P_{\fd}$ is a $K$-rational point of $E$.
        \item[{\rm(ii)}] The two polynomials $a$ and $b$ belong to $K[x]$ and, $b$ is of the form
        \[
            b_0(x-x_{\fd})(x-b_1)-y_{\fd} \ (b_0,b_1\in K).
        \]
    \end{itemize}
\end{lem}

\section*{Introduction}\label{sec:intro}
\input{intro}
\section{Elliptic surfaces}\label{sec:elli-sur}
\input{elli-surf}
\section{Line-sections arising from weak-bitangents and bitangents and their description in $\MW(S_{\mcQ,z_o})$}\label{sec:rep-w-bi}
\input{rep-w-bi}
\section{The Mumford representation of a semi-reduced divisor}\label{sec:mum-rep}
\input{mum-rep}

\section{The Proof of our main results}\label{sec:appli}
\input{appli1}

\bibliographystyle{abbrv}

\subsection{The group structures with height pairing of $E_{S_{\mcQ,z_o}}(\CC(t))$ and propeties of sections}\label{subsec:gene-MW}

In \cite{Toku10}, for all irreducible quartic curves $\mcQ$ with singularities and a general smooth point, $z_o$, on $\mcQ$, Tokunaga gave the group structure with height pairing of $E_{S_{\mcQ,z_o}}(\CC(t))$ based \cite{Ogu-Shio} and, in \cite{Ban-Kaw-Mas-Tok}, for $C_1$ and $C_2$ meeting transversally, the authors gave that of $E_{S_{\mcC_1+\mcC_2,z_o}}(\CC(t))$.

For all quartic curves $\mcQ$ that satisfy $(\dagger)$, we give the list of the group structure with height pairing (Table \ref{table:st-MW}), following \cite{Ogu-Shio, Toku10, Ban-Kaw-Mas-Tok}.
Before we go on to give the list Table \ref{table:st-MW}, we need to introduce some notation.
\begin{itemize}
	\item
	The column labeled $\Xi_{\mcQ}$ indicates the types of singularities of $\mcQ$.
	 We use the notation in \cite[pp.81-82]{BHPV} in order to describe the types of singularities.
	\item
	We write  the subgroup of $\NS(S)$ generated by the set $\{\Theta_{v,1},\ldots,\Theta_{v,m_v-1} \}_{v\in\Red(\varphi_{\mcQ,z_o})}$ in the column labeled $R_{\varphi_{\mcQ,z_o}}$.
	Note that the group is isomorphic to a direct sum of root lattices of A-D-E type with respect to the intersection multiplicity, and we describe it as the direct sum.
	\item
	The column labeled $E_{S_{\mcQ,z_o}}(\CC(t))$ indicate the group structure with the height pairing of $E_{S_{\mcQ,z_o}}(\CC(t))$.
\end{itemize}
\begin{longtable}{|c|c|c|c|}
	\caption{}
	\label{table:st-MW}
	\endhead
	\hline
	No. & $\Xi_\mcQ$ & $R_{\varphi_{\mcQ,z_o}}$ & $E_{S_{\mcQ,z_o}}(\CC(t))$ \rule[-1.5mm]{0mm}{2mm}\\

	\hline
	1&$A_6$ &			$A_6\oplus A_1$ & 									$\langle1/14\rangle$\\
	\hline
	2&$E_6$ & 			$E_6\oplus A_1$ &									$\langle 1/6 \rangle$\\
	\hline
	3&$A_5$ &			$A_5\oplus A_1$ &									$A_1^*\oplus \langle1/6\rangle$\\
	\hline
	4&$D_5$ &			$D_5\oplus A_1$ &									$A_1^*\oplus \langle 1/4\rangle$\\
	\hline
	5&$D_4$ &			$D_4\oplus A_1$ &									$(A_1^*)^{\oplus 3}$\\
	\hline
	6&$A_4+A_2$&		$A_4\oplus A_2\oplus A_1$&					$\langle1/30\rangle$\\
	\hline
	7&$A_4+A_1$&		$A_4\oplus A_1^{\oplus 2}$&
	$\frac{1}{10}
	\left[
		\begin{array}{cc}
			2&1\\
			1&3
		\end{array}
	\right]
	$\\
\hline
8&$4A_1$&			$A_1^{\oplus 5}$ &									$(A_1^*)^{\oplus3}\oplus\ZZ/2\ZZ$\\
\hline
9&$A_3+A_2$&		$A_3\oplus A_2\oplus A_1$&					$A_1^*\oplus \langle1/12\rangle$\\
\hline
10&$A_3+A_1$&		$A_3\oplus A_1^{\oplus 2}$&					$(A_1^*)^{\oplus 2}\oplus \langle1/4\rangle$\\
\hline
11&$3A_2$ &			$A_2^{\oplus 3}\oplus A_1$&					$\langle1/6\rangle\oplus \ZZ/3\ZZ$\\
\hline
12&$2A_2+A_1$&	$A_2^{\oplus 2}\oplus A_1^{\oplus 2}$&	$\langle1/6\rangle^{\oplus 2}$\\
\hline
13&$A_2+2A_1$&	$A_2\oplus A_1^{\oplus 3}$&
$A_1^*\oplus\frac{1}{6}
\left[
\begin{array}{cc}
2&1\\
1&2
\end{array}
\right]
$
\\
\hline
14&$3A_1$&			$A_1^{\oplus 4}$&										$(A_1^*)^{\oplus 4}$\\
\hline
15&$A_4$&				$A_4\oplus A_1$&
$\frac{1}{10}
\left[
\begin{array}{ccc}
3&1&-1\\
1&7&3\\
-1&3&7
\end{array}
\right]
$
\\
\hline
16&$A_3$ &			$A_3\oplus A_1$	 &										$A_3^*\oplus A_1^*$\\
\hline
17&$2A_2$&			$A_2^{\oplus 2}\oplus A_1$&					$A_2^*\oplus \langle 1/6\rangle$\\
\hline
18&$A_2+A_1$&		$A_2\oplus A_1^{\oplus 2}$&
$\frac{1}{6}
\left[
\begin{array}{cccc}
2&1&0&-1\\
1&5&3&1\\
0&3&6&3\\
-1&1&3&5
\end{array}
\right]
$\\
\hline
19&$2A_1$&			$A_1^{\oplus 3}$ &									$D_4^*\oplus A_1^*$\\
\hline
20&$A_2$ &		$A_2\oplus A_1$ &		$A_5^*$\\
\hline
21&$A_1$ &		$A_1^{\oplus 2}$ &		$D_6^*$\\
\hline
\end{longtable}

Let $I$ be the matrix corresponding with the structure given in Table \ref{table:st-MW}.
We choose generators $P_1,...,P_n, P_{\tau}\in E_{S_{\mcQ,z_o}}(\CC(t))$ such that
$I=[\langle P_i,P_j\rangle]_{1\leq i,j\leq n}$ and $\ P_{\tau}$ is a torsion of $E_{S_{\mcQ,z_o}}(\CC(t))$.
In Section \ref{sec:geo-gene}, we will study the images $\overline{f}_{\mcQ,z_o}(s_{P_i})$ (Table \ref{table:geo-gene}).
We first consider the case when $\mcQ$ has singularities whose types are $A_1$ and $A_2$.
Note that this case is the case considered in Theorem \ref{thm:A2+A1}.
We can also consider other cases in Table \ref{table:geo-gene}.
For this propose, we introduce the following lemma:

\begin{lem}{\rm (\cite[Lemma 9]{Ban-Toku17})}\label{lem:line-sec}
	Let $s\in\MW(S_{\mcQ,z_o})$ be a section such that $s\cdot O=0$ and $s\cdot \Theta_{\infty,1}=1$. Then $\overline{f}_{\mcQ,z_o}(s)$ is a line $L_s$ such that
	\begin{itemize}
		\item[$\mathrm{(i)}$]
			$I_x(\mcQ,L_s)$ is even for all $x\in\mcQ$, and
		\item[$\mathrm{(ii)}$]
			$z_o\notin L_s$.
	\end{itemize}
	Conversely, any line satisfying the two conditions $\mathrm{(i)}$ and $\mathrm{(ii)}$ gives rise to line-sections $s_{L^{\pm}}$ such that $s_{L^{\pm}}\cdot O=0$ and $s_{L^{\pm}}\cdot \Theta_{\infty,1}=1$.
\end{lem}
In the above condition, if $s\cdot\Theta_{\infty,1}=0$ and $s\cdot O=0$ then $\overline{f}_{\mcQ,z_o}(s)$ is a smooth conic that is tangent at $z_o$.
For details, see \cite{Toku10}.
By definition of weak-bitangent lines and Lemma \ref{lem:line-sec}, weak-bitangent lines give rise to line-sections of $S_{\mcQ,z_o}$.

\subsection{Descriptions of the images of the corresponding sections to the generators of $E_{S_{\mcQ,z_o}}(\CC(t))$ by $\overline{f}_{\mcQ,z_o}$ in the case when $\mcQ$ is No.18}\label{subsec:node-cusp}
Let $x_1$ and $x_2$ be singularities of $\mcQ$ such that the types of $x_1$ and $x_2$ are $A_1$ and $A_2$, respectively.
By Table \ref{table:st-MW}, the lattice structure of $E_{S_{\mcQ,z_o}}(\CC(t))$ is given by
\[
	I=\frac{1}{6}\left[
		\begin{array}{cccc}
			2&1&0&-1\\
			1&5&3&1\\
			0&3&6&3\\
			-1&1&3&5
		\end{array}
	\right].
\]
Let $P_1,P_2,P_3$ and $P_4$ be generators of $E_{S_{\mcQ,z_o}}(\CC(t))$ such that $I=[\langle P_i,P_j \rangle]$.
Note that the type of $F_{x_1}$ is $\I_2$ or $\III$, that of $F_{x_2}$ is $\I_3$ or $\IV$ and that of $F_{z_o}$ is $\I_2$.
For $P$ and $Q$ of $E_{S_{\mcQ,z_o}}(\CC(t))$, we have the values of $\mathrm{contr}_{\bullet}(s_P,s_Q)$ as follows:
\begin{align}\label{eq:contr}
	&\begin{aligned}
		\mathrm{contr}_{x_1}(s_P,s_Q)&=
		\left\{
			\begin{array}{cl}
				1/2 & s_P\cdot \Theta_{x_1,1}=s_Q\cdot \Theta_{x_1,1}=1\\
				0 & {\rm otherwise}
			\end{array}
		\right.\\
		\mathrm{contr}_{x_2}(s_P,s_Q)&=
		\left\{
			\begin{array}{cl}
				2/3 & s_P\cdot \Theta_{x_2,i}=s_Q\cdot \Theta_{x_2,i}=1 \ {\rm and} \ i \in \{1,2\}  \\
				1/3 & s_P\cdot \Theta_{x_2,i}=s_Q\cdot \Theta_{x_2,j}=1 \ {\rm and} \ \{ i, j \}=\{1,2\}    \\
				0 & {\rm otherwise}
			\end{array}
		\right.\\
		\mathrm{contr}_{z_o}(s_P,s_Q)&=
		\left\{
			\begin{array}{cl}
				1/2 & s_P\cdot \Theta_{\infty,1}=s_Q\cdot \Theta_{\infty,1}=1\\
				0 & {\rm otherwise}
			\end{array}
		\right.
	\end{aligned}
\end{align}
For $i=1,2,3,4$, We estimate the pairings $\langle P_i,P_i \rangle$:
\begin{align*}
	\langle P_i,P_i \rangle&=2+2s_{P_i}\cdot O-\sum_{y\in J}\mathrm{contr}_y(s_{P_i}\cdot s_{P_i})\qquad (J:=\{ x_1,x_2,z_o \})\\
	&\geq 2+2s_{P_i}\cdot O-(1/2+2/3+1/2)\\
	&=1/3+2s_{P_i}\cdot O
\end{align*}
Observing that the diagonal components of $I$ are $1/3$, $5/6$ and $1$, we have $s_{P_i}\cdot O=0 \ (i=1,2,3,4)$.
Hence, $\overline{f}_{\mcQ,z_o}(s_{P_i})$ is a line or a conic that is tangent to $\mcQ$ at $z_o$.

We will prove that $s_{P_1}$ is a line-section.
Since the value of $\langle P_1,P_1\rangle$ is $1/3$, we obtain
\[
	\sum_{y\in J}\mathrm{contr}_y(s_{P_1}\cdot s_{P_1})=5/3.
\]
Therefore, from possible values of $\mathrm{contr}_{y}(s_{P_1},s_{P_1})$ in (\ref{eq:contr}), we obtain
\begin{align*}
	\mathrm{contr}_{x_1}(s_{P_1},s_{P_1})&=\mathrm{contr}_{z_o}(s_{P_1},s_{P_1})=1/2 \ {\rm and} \\
	\mathrm{contr}_{x_2}(s_{P_1},s_{P_1})&=2/3.
\end{align*}

By (\ref{eq:contr}), we have $s_{P_1}\cdot\Theta_{x_1,1}=s_{P_1}\cdot \Theta_{\infty,1}=1$ and $s_{P_1}\cdot \Theta_{x_1,i}=1$, where $i=1$ or $2$.
By Lemma \ref{lem:line-sec}, $\overline{f}_{\mcQ,z_o}(s_{P_1})$ is a weak-bitanget line passing through $x_1$ and $x_2$.
We may assume $s_{P_1}\cdot\Theta_{x_2,1}=1$.
Similarly, we find that $s_{P_3}$ is a line-section meeting $\Theta_{x_1,1}$, $\Theta_{x_2,0}$ and $\Theta_{\infty,1}$.
The image $\overline{f}_{\mcQ,z_o}(s_{P_3})$ is a weak-bitangent line passing through $x_1$ and not $x_2$.
We need to find the images $\overline{f}_{\mcQ,z_o}(s_{P_2})$ and $\overline{f}_{\mcQ,z_o}(s_{P_4})$.
By the assumption,
\[
	\sum_{y\in J}\mathrm{contr}_y(s_{P_2}\cdot s_{P_2})=\sum_{y\in J}\mathrm{contr}_y(s_{P_4}\cdot s_{P_4})=7/6.
\]
We infer
\[
	\left[
		\begin{array}{c}
			\mathrm{contr}_{x_1}(s_{P_i}\cdot s_{P_i})\\
			\mathrm{contr}_{x_2}(s_{P_i}\cdot s_{P_i})\\
			\mathrm{contr}_{z_o}(s_{P_i}\cdot s_{P_i})
		\end{array}
	\right]
	=
	\left[
		\begin{array}{c}
			1/2\\
			2/3\\
			0
		\end{array}
	\right]
	\ {\rm or} \
	\left[
		\begin{array}{c}
			0\\
			2/3\\
			1/2
		\end{array}
	\right]
 	\ i=2,4
\]
by the possible values of $\mathrm{contr}_{\bullet}(-,-)$ in (\ref{eq:contr}).
In the former case, $\overline{f}_{\mcQ,z_o}(s_{P_i})$ is a smooth conic passing through $x_1$ and $x_2$.
In the other case, $\overline{f}_{\mcQ,z_o}(s_{P_i})$ is a weak-bitangent passing through $x_2$.
In order to find the values of $\mathrm{contr}_{\bullet}(s_{P_i},s_{P_i}) \ (i=2,4)$, we consider $\langle P_1,P_2\rangle$ and $\langle P_1,P_4\rangle$.

By our choice of generators $P_1,P_2,P_3$ and $P_4$, we have
\begin{align*}
	1/6&=\langle P_1,P_2\rangle=1-s_{P_1}\cdot s_{P_2}-\sum_{y\in J}\mathrm{contr}_y(s_{P_1},s_{P_2}) \ {\rm and}\\
	-1/6&=\langle P_1,P_4\rangle=1-s_{P_1}\cdot s_{P_4}-\sum_{y\in J}\mathrm{contr}_y(s_{P_1},s_{P_4}).
\end{align*}
Now, by (\ref{eq:contr}), the possibilities of $\sum_{y\in J}\mathrm{contr}_y(s_{P_1},s_{P_2})$ are
\[
	0, \ 1/3, \ 1/2, \ 2/3, \ 5/6, \ 1, \ 7/6, \ 4/3 \ {\rm or} \ 5/3.
\]
Since $s_{P_1}\cdot s_{P_2}$ and $s_{P_1}\cdot s_{P_4}$ are integers, we have
\[
	\sum_{y\in J}\mathrm{contr}_y(s_{P_1},s_{P_2})=5/6\ {\rm and} \ \sum_{y\in J}\mathrm{contr}_y(s_{P_1},s_{P_4})=7/6.
\]
The first implies
\begin{align}\label{eq:1}
	\left[
		\begin{array}{c}
			\mathrm{contr}_{x_1}(s_{P_1}\cdot s_{P_2})\\
			\mathrm{contr}_{x_2}(s_{P_1}\cdot s_{P_2})\\
			\mathrm{contr}_{z_o}(s_{P_1}\cdot s_{P_2})
		\end{array}
	\right]
	=
	\left[
		\begin{array}{c}
			1/2\\
			1/3\\
			0
		\end{array}
	\right]
 	\ {\rm or} \
	\left[
		\begin{array}{c}
			0\\
			1/3\\
			1/2
		\end{array}
	\right].
\end{align}
On the other hand, by the definition of $\mathrm{contr}_{\bullet}(-,-)$, $\mathrm{contr}_{y}(s_{P_1},s_{P_2})$ are given by
\begin{align}\label{eq:2}
	\begin{aligned}
		\mathrm{contr}_{x_1}(s_{P_1},s_{P_2})&=\vc(x_1,s_{P_1})(-A_{x_1}^{-1})\vc(x_1,s_{P_2})\\
		&=s_{P_2}\cdot\Theta_{x_1,1}/2,\\
		\mathrm{contr}_{x_2}(s_{P_1},s_{P_2})&={}^t\vc(x_2,s_{P_1})(-A_{x_2}^{-1})\vc(x_2,s_{P_2})\\
		&=[1 \ 0]
		\left[
			\begin{array}{cc}
				2/3 & 1/3\\
				1/3 & 2/3
			\end{array}
		\right]
		\left[
			\begin{array}{c}
				s_{P_2}\cdot\Theta_{x_2,1}\\
				s_{P_2}\cdot\Theta_{x_2,2}
			\end{array}
		\right],\\
		\mathrm{contr}_{z_o}(s_{P_1},s_{P_2})&=s_{P_2}\cdot\Theta_{\infty,1}/2.
	\end{aligned}
\end{align}
From (\ref{eq:1}) and (\ref{eq:2}), we have $(\vc(x_1,s_{P_2}) , \vc(z_o,s_{P_2}))=(1,0) \ {\rm or} \ (0,1)$ and ${}^t\vc(x_2,s_{P_2})=[0 \ 1]$.
Similarly, $(\vc(x_1,s_{P_4}) , \vc(z_o,s_{P_4}))=(1,0) \ {\rm or} \ (0,1)$ and ${}^t\vc(x_2,s_{P_4})=[1 \ 0]$ hold.

We find the following, for $i=2,4$:
\begin{itemize}
	\setlength{\leftskip}{-0.5cm}
	\item
	If $(\vc(x_1,s_{P_i}),\vc(z_o,s_{P_i}))=(1,0)$, then $\overline{f}_{\mcQ,z_o}(s_{P_i})$ is a smooth conic passing through $x_1$ and $x_2$.
	\item
	If $(\vc(x_1,s_{P_i}),\vc(z_o,s_{P_i}))=(0,1)$, then $\overline{f}_{\mcQ,z_o}(s_{P_i})$ is a weak-bitangent passing through $x_2$ and not $x_1$.
\end{itemize}
If one of $s_{P_2}$ and $s_{P_4}$ is a line-section and the other is not,
then it follows that $\mathrm{contr}_{x_1}(s_{P_2},s_{P_4})=\mathrm{contr}_{z_o}(s_{P_2},s_{P_4})=0$.

On the other hand, we have
\begin{align*}
	\langle P_2,P_4 \rangle&=1-s_{P_2}\cdot s_{P_4}-\sum_{y\in J}\mathrm{contr}_{y}(s_{P_2},s_{P_4})\\
	&=2/3-s_{P_2}\cdot s_{P_4}.
\end{align*}
But $\langle P_2,P_4 \rangle=1/6$, which is a contradiction.

Hence, $s_{P_2}$ and $s_{P_4}$ are both line-sections or not line-sections.
Here, we assume that $s_{P_2}$ and $s_{P_4}$ are both not line-sections.
Let $Q_1=P_1, \ Q_2=\dot{-}P_2\dot{+}P_3, \ Q_3=P_3$ and $Q_4=P_3\dot{-}P_4$.
We can verify that the matrix $[\langle Q_i,Q_j \rangle]$ equals to $[\langle P_i,P_j \rangle]$.
We can take new generators $Q_1,Q_2,Q_3$ and $Q_4$ of $E_{S_{\mcQ,z_o}}(\CC(t))$.
Then $s_{Q_2}$ and $s_{Q_4}$ are line-sections because $s_{Q_2}\cdot \Theta_{\infty,1}=s_{Q_4}\cdot \Theta_{\infty,1}=1$.
Therefore we can assume $s_{P_2}$ and $s_{P_4}$ are line-sections.

\subsection{Descriptions of the images of the corresponding sections to the generators of $E_{S_{\mcQ,z_o}}(\CC(t))$ by $\overline{f}_{\mcQ,z_o}$ in the case when $\mcQ$ satisfy $(\dagger)$}\label{sec:geo-gene}

For all $\mcQ$ satisfying $(\dagger)$, we obtain Table \ref{table:geo-gene} by computation similar to the above case.
In Table \ref{table:geo-gene}, we choose generators $P_1,...,P_n, P_{\tau}\in E_{S_{\mcQ,z_o}}(\CC(t))$ such that
$I=[\langle P_i,P_j\rangle]_{1\leq i,j\leq n}$ and $\ P_{\tau}$ is a torsion of $E_{S_{\mcQ,z_o}}(\CC(t))$, where the matrix $I$ corresponds to the group structure with the height pairing given in Table \ref{table:st-MW}.
In the above case, we made $s_{P_2}$ and $s_{P_4}$ line-sections.
Picking a special generators, we can obtain Table \ref{table:geo-gene}.
We explain some notations used in the table.
\begin{itemize}
	\item
	In the column labeled $\sing(\mcQ)$, we write the all pairs $(x,S_x)$, where $x$ is a singularity of $\mcQ$ and $S_x$ is its type of singularity.
	\item
	The column labeled $E_{S_{\mcQ,z_o}}(\CC(t))$ is the same as those in Table \ref{table:st-MW}.
	\item
	Let $P_1,\ldots,P_n,P_{\tau}$ be generators of  $E_{S_{\mcQ,z_o}}(\CC(t))$ as the group structure with the height pairing given by Table \ref{table:st-MW}.
	The column labeled {\rm COG} will indicate the image $\overline{f}_{\mcQ,z_o}(s_{P_i})$ by using the following notation.
	\begin{itemize}
		\item
		$BL$: a bitangent line to $\mcQ$ or a 4-fold tangent line at one smooth point.
		\item
		$WL(x)$: a weak-bitangent line meeting $\mcQ$ at a singularity $x$ of $\mcQ$ such that  $I_x(\mcQ,WL(x))=4$.
		For example, when the type of $x$ is $A_1$, $WL(x)$ is tangent to one of two branches at $x$ of $\mcQ$ with multiplicity 3.
		Also, when the type of $x$ is $D_4$, $WL(x)$ is tangent to one of three branches at $x$ of $\mcQ$ with multiplicity 2 and the other two branches with multiplicities 1.
 		\item
		$WL(x,y)$: a weak-bitangent line meeting $\mcQ$ at two distinct points $x$ and $y$ of $\mcQ$ such that $I_x(\mcQ,WL(x,y))=I_y(\mcQ,WL(x,y))=2$.
		\item
		$\eta$: a smooth point on $\mcQ$.
		\item
		$CT(x_1,\ldots, x_l;y_1,\ldots,y_m)$ : a conic such that satisfying

		\begin{itemize}
			\item[{\rm (a)}]
			$x_1,...,x_l$ and $y_1,...,y_m$ are all singularities of $\mcQ$ through which the conic pass,
			\item[{\rm (b)}]
			$I_{x_i}(CT(x_1,\ldots, x_l;y_1,\ldots,y_m),\mcQ)=\mathrm{mult}_{x_i}(\mcQ) \ (i=1,\ldots, l)$ and
			\item[{\rm (c)}]
			$I_{y_j}(CT(x_1,\ldots, x_l;y_1,\ldots,y_m),\mcQ)>\mathrm{mult}_{y_j}(\mcQ) \ (j=1,\ldots, m)$,
		\end{itemize}
		where we denote the multiplicity of $\mcQ$ at a point $z$ by $\mathrm{mult}_{z}(\mcQ)$.
		When there are no singularities $x_1,\ldots,x_l$ satisfying {\rm (b)}, we write $CT(-;y_1,\ldots,y_m)$.
		We write $CT(x_1,\ldots,x_l;-)$ similarly.
	\end{itemize}
	Let $P\in E_{S_{\mcQ,z_o}}(\CC(t))$. If $\overline{f}_{\mcQ,z_o}(s_P)$ is a curve $C_P$ of the above curves, we write $PC_P$ in the last column.
	Otherwise, we write the values $s_P\cdot O$, $s_P\cdot\Theta_{x,i}$ and $s_P\cdot \Theta_{\infty,1}$, where  $x$ is a singularity of $\mcQ$ and $i=1,\ldots,m_x-1$.
\end{itemize}
\begin{longtable}{|c|c|c|c|}
\caption{}
\label{table:geo-gene}
\endhead
\hline No.&
$\sing(\mcQ)$&$E_{S_{\mcQ,z_o}}(\CC(t))$ & {\rm COG} \rule[-1.5mm]{0mm}{2mm}\\
\hline 1&
$(x,A_6)$&
$\langle1/14\rangle$&
$P_1WL(x)$
\\
\hline	2&
$(x,E_6)$
&		$\langle 1/6 \rangle$ &
$P_1WL(x)$
\\
\hline 3&
$(x,A_5)$
&		$A_1^*\oplus \langle1/6\rangle$&
\begin{tabular}{c}
$P_1CT(x)$, $P_2WL(x)$
\end{tabular}
\\
\hline 4&
$(x,D_5)$
&		$A_1^*\oplus \langle 1/4\rangle$&
$P_1WL(x)$, $P_2WL(x)$
\\
\hline 5&
$(x,D_4)$
&			$(A_1^*)^{\oplus 3}$&
$P_iWL(x)$  $(i=1,2,3)$
\\
\hline 6&
\begin{tabular}{c}
$(x,A_4)$\\
$(y,A_2)$
\end{tabular}
	&	$\langle1/30\rangle$&
$P_1WL(x,y)$
\\
\hline 7&
\begin{tabular}{c}
$(x,A_4)$\\
$(y,A_1)$
\end{tabular}
		&
$\frac{1}{10}
\left[
\begin{array}{cc}
2&1\\
1&3
\end{array}
\right]
$&
\begin{tabular}{c}
$P_1WL(x,y)$,
 $P_2WL(x)$
\end{tabular}

\rule[-4mm]{0mm}{9.5mm}\\
\hline 8&
\begin{tabular}{c}
$(x,A_1)$\\
$(y,A_1)$\\
$(z,A_1)$\\
$(w,A_1)$
\end{tabular}
			&$(A_1^*)^{\oplus3}\oplus\ZZ/2\ZZ$&

\begin{tabular}{l}
$P_1WL(x,y)$, $P_2L(y,z)$,\\
$P_3WL(x,z)$, $P_4CT(x,y,z,w;-)$
\end{tabular}
\\
\hline 9&
\begin{tabular}{c}
$(x,A_3)$\\
$(y,A_2)$
\end{tabular}
			&$A_1^*\oplus \langle1/12\rangle$&
$P_1WL(x)$, $P_2WL(x,y)$
			\\
\hline 10&
\begin{tabular}{c}
$(x,A_3)$\\
$(y,A_1)$
\end{tabular}			&$(A_1^*)^{\oplus 2}\oplus \langle1/4\rangle$&
$P_1CT(y;x)$, $P_2WL(x)$, $P_3WL(x,y)$
\\
\hline 11&
\begin{tabular}{c}
$(x,A_2)$\\
$(y,A_2)$\\
$(z,A_2)$
\end{tabular}
		&$\langle1/6\rangle\oplus \ZZ/3\ZZ$&
$P_1WL(x,y)$, $P_2CT(x,y,z;-)$
\\
\hline 12&
\begin{tabular}{c}
$(x,A_2)$\\
$(y,A_2)$\\
$(z,A_1)$
\end{tabular}
		&$\langle1/6\rangle^{\oplus 2}$&
$P_1WL(x,y)$, $P_2CT(x,y,z;-)$
\\

\hline 13&
\begin{tabular}{c}
$(x,A_2)$\\
$(y,A_1)$\\
$(z,A_1)$
\end{tabular}			&
$A_1^*\oplus\frac{1}{6}
\left[
\begin{array}{cc}
2&1\\
1&2
\end{array}
\right]
$&
$P_1WL(y,z)$, $P_2WL(x,y)$, $P_3WL(x,z)$
\\
\hline 14&
\begin{tabular}{c}
$(x,A_1)$\\
$(y,A_1)$\\
$(z,A_1)$
\end{tabular}				&$(A_1^*)^{\oplus 4}$&
\begin{tabular}{l}
$P_1WL(x,y)$, $P_2WL(y,z)$, \\
$P_3WL(x,z)$, $P_4CT(x,y,z;-)$
\end{tabular}
\\
\hline 15&
$(x,A_4)$	&
$\frac{1}{10}
\left[
\begin{array}{ccc}
3&1&-1\\
1&7&3\\
-1&3&7
\end{array}
\right]
$&
$P_1WL(x)$, $P_2WL(\eta_2,x)$, $P_3WL(\eta_3,x)$
\rule[-6mm]{0mm}{14mm}\\
\hline 16&
$(x,A_3)$			&$A_3^*\oplus A_1^*$&
\begin{tabular}{l}
$P_1WL(\eta_1,x)$, $P_2CT(-;x)$, \\
$P_3WL(\eta_3,x)$, $P_4WL(x)$
\end{tabular}
\\
\hline 17&
\begin{tabular}{c}
$(x,A_2)$\\
$(y,A_2)$
\end{tabular}					&$A_2^*\oplus \langle 1/6\rangle$&
\begin{tabular}{l}
$P_1CT(x,y;-)$, $P_2CT(x,y;-)$,
 $P_3WL(x,y)$
\end{tabular}
\\
\hline 18&
\begin{tabular}{c}
$(x,A_2)$\\
$(y,A_1)$
\end{tabular}					&
$\frac{1}{6}
\left[
\begin{array}{cccc}
2&1&0&-1\\
1&5&3&1\\
0&3&6&3\\
-1&1&3&5
\end{array}
\right]
$&
\begin{tabular}{l}
$P_1WL(x,y)$, $P_2WL(\eta_2,x)$,\\
$P_3WL(\eta_3,y)$,
$P_4WL(\eta_4,x)$
\end{tabular}
\rule[-8mm]{0mm}{18mm}\\
\hline 19&
\begin{tabular}{c}
$(x,A_1)$\\
$(y,A_1)$
\end{tabular}
			&$D_4^*\oplus A_1^*$&
\begin{tabular}{l}
$P_1WL(\eta_1,x)$,
$P_2WL(\eta_2,y)$,\\
$P_3CT(x,y;-)$, $P_4CT(-;-)$, $P_5WL(x,y)$
\end{tabular}			\\
\hline 20&
$(x,A_2)$			&$A_5^*$&
\begin{tabular}{l}
$P_1WL(\eta_1,x)$, $P_2CT(x;-)$,\\
$P_3BL$, $P_4CT(x;-)$, $P_5WL(\eta_5,x)$
\end{tabular}
\\
\hline 21&
$(x,A_1)$			&$D_6^*$&
\begin{tabular}{l}
$P_1WL(\eta_1,x)$, $P_2BL$,\\
$P_3CT(x;-)$, $P_4CT(-;-)$, \\
$s_{P_5}\cdot O=s_{P_5}\cdot \Theta_{x,1}=s_{P_5}\cdot \Theta_{\infty,1}=1$,\\
$s_{P_6}\cdot O=1$, $s_{P_6}\cdot \Theta_{x,1}=s_{P_6}\cdot \Theta_{\infty,1}=0$
\end{tabular}

\\
\hline
\end{longtable}

\subsection{Line-sections arising from weak-bitangents and bitangents and their description in $\MW(S_{\mcQ,z_o})$}\label{sec:line-sec}
Let $\mcQ$ be a singular quartic curve satisfying $(\dagger)$ and let $z_o$ be a general smooth point on $\mcQ$.

In order to prove Corollaries \ref{cor:ha3.3}, \ref{cor:ha3.4}, \ref{cor:ha3.5} and Theorem \ref{thm:A2+A1}, we consider bitangent lines and weak-bitangent lines written as $BL$ and $WL(\eta,x)$  as in Table \ref{table:geo-gene}, where $\eta$ is a smooth point of $\mcQ$ and $x$ is a singularity of $\mcQ$.
\begin{rem}
	If $\mcQ$ has a weak-bitangent line written as $WL(\eta,x)$ then $\mcQ$ must be an irreducible quartic curve which has double points only by the definition of weak-bitangent lines.
	In particular, the types of singular fibers of $S_{\mcQ,z_o}$ are only $\I_n$, $\III$ or $\IV$.
\end{rem}
By Lemma \ref{lem:line-sec}, we obtain Proposition \ref{prop:w-bi-hieght}.
\begin{prop}\label{prop:w-bi-hieght}
	Let $\mcQ$ be an irreducible singular quartic curve with double points only and let $z_o$ be a general smooth point of $\mcQ$.
	Then, for $P\in E_{S_{\mcQ,z_o}}(\CC(t))$, the following conditions {\rm (i)} and {\rm (ii)} are equivalent:
	\begin{itemize}
		\item[{\rm (i)}]
			There exists some natural number $n_P$ such that $\langle P,P\rangle=3/2-n_P/(n_P+1)$  and the intersection number $s_P\cdot \Theta_{\infty,1}=1$.
		\item[{\rm (ii)}]
			$s_P$ is a line-section of a weak-bitangent line $L$ such that
			\begin{itemize}
				\item[{\rm (a)}]
					$L$ passes through a singularity, $x_0$, of $\mcQ$,
				\item[{\rm (b)}]
					when the type of $x_0$ is $A_1$, $I_{x_0}(L,\mcQ)=2$ or $4$, i.e. $L$ is respectively the form $WL(\eta,x)$ or $WL(x)$ and
				\item[{\rm (c)}]
					otherwise, $I_{x_0}(L,\mcQ)=2$ i.e. $L$ is the form $WL(\eta,x)$.
			\end{itemize}
	\end{itemize}
\end{prop}
\proof
We assume {\rm (i)} and we choose $P\in E_{S_{\mcQ,z_o}}(\CC(t))$ satisfying (i).
Let $\sing(\mcQ)$ be the set of all singularities of $\mcQ$.
We recall the formula of the height pairing as follows:
\[
	\langle P,P\rangle=2+2s_P\cdot O-\sum_{x\in\sing(\mcQ)\cup\{z_o\}}\mathrm{contr}_x(s_P,s_P)
\]
Observing $R_{\mcQ,z_o}$ in Table \ref{table:st-MW}, we see $\sum_{x\in\sing(\mcQ)\cup\{z_o\}}\mathrm{contr}_x(s_P,s_P)\leq 5/2$.
Hence we have
\[
	\langle P,P\rangle=2+2s_P\cdot O-\sum_{x\in\sing(\mcQ)\cup\{z_o\}}\mathrm{contr}_x(s_P,s_P)\geq 2s_P\cdot O-1/2.
\]
By the condition {\rm (i)}, we have $s_P\cdot O=0$.
Hence, $s_P$ is a line-section by Lemma \ref{lem:line-sec}.
The following equality holds:
\[
	3/2-n_P/(n_P+1)=3/2-\sum_{x\in\sing(\mcQ)}\mathrm{contr}_x(s_P,s_P),
\]
so we have $\sum_{x\in\sing(\mcQ)}\mathrm{contr}_x(s_P,s_P)=n_P/(n_P+1)<1$.
For $x\in\sing(\mcQ)$, we must prove (ii) in the case when the type of $F_x$ is $\I_{m+1}$, $\III$ or $\IV$.
But we consider the case when the type of $F_x$ is $\I_{m+1} \ (m\geq1)$ since $\III$ and $\IV$ are special case of $\I_2$ and $\IV$, respectively
We write $F_x=\Theta_{x,0}+\sum^{m_x-1}_{i=1}a_{x,i}\Theta_{x,i}$.
Note that $m_x=m+1$.
If $s_P$ meets the $k$-th components $\Theta_{x,k} \ (0\leq k \leq m)$, then $\mathrm{contr}_x(s_P,s_P)=(m+1-k)k/(m+1)$.
In particular,
\[
	\mathrm{contr}_x(s_P,s_P)
	\left\{
		\begin{array}{ll}
			=m/(m+1)& k=1,m\\
			=0& k=0\\
			>1& {\rm otherwise}
		\end{array}
	\right.
\]
Hence, by $\sum_{x\in\sing(\mcQ)}\mathrm{contr}_x(s_P,s_P)<1$, we have
\[
	\mathrm{contr}_x(s_P,s_P)=0 \ {\rm or} \ m/(m+1)
\]
for all $x\in\sing(\mcQ)$.
Note that $m_1/(m_1+1)+m_2/(m_2+1)\geq 1$ where $m_1$ and $m_2$ are natural numbers.
Therefore, there exists a unique $x_0\in \sing(\mcQ)$ such that
\begin{itemize}
	\item
	$\mathrm{contr}_{x_0}(s_P,s_P)=n_P/(n_P+1)$ and the type of $F_{x_0}$ is $\I_{n_P}$, and
	\item
	$\mathrm{contr}_{x}(s_P,s_P)=0$ for all $x\in \sing(\mcQ)\setminus \{x_0\}$.
\end{itemize}
Therefore the type of $x_0$ is $A_{n_P}$ and we obtain $m_{x_0}=n_P+1$.

We assume that the type of $x_0$ is $A_1$.
Then we have $1/2=\mathrm{contr}_{x_0}(s_P,s_P)=s_P\cdot\Theta_{x_0,1}/2$.
By $s_P\cdot\Theta_{x_0,1}=1$, the intersection multiplicity of $\overline{f}_{\mcQ,z_o}(s_P)$ and a branch at $x_0$ of $\mcQ$ is one or more.
Considering that $I_{x_0}(\overline{f}_{\mcQ,z_o}(s_P),\mcQ)$ is even, it is 2 or 4.

When $I_{x_0}(\overline{f}_{\mcQ,z_o}(s_P),\mcQ)$ is 2, there exists a smooth $\eta$ on $\mcQ$ such that $I_{\eta}(\overline{f}_{\mcQ,z_o}(s_P),\mcQ)=2$.
Then we can write $L$ as $WL(\eta,x_0)$.

When $I_{x_0}(\overline{f}_{\mcQ,z_o}(s_P),\mcQ)$ is four, $\overline{f}_{\mcQ,z_o}(s_P)$ is tangent to a branch at $x_0$ of $\mcQ$ with multiplicity 3.
Then we can write $L$ as $WL(x_0)$

We next assume that $n_P>1$.
By consideration similar to the above case, $s_P\cdot\Theta_{x_0,1}=1$ or $s_P\cdot\Theta_{x_0,n_P}=s_P\cdot\Theta_{x_0,m_{x_0}-1}=1$.
For each branch at $x_0$ of $\mcQ$, the intersection multiplicities of $\overline{f}_{\mcQ,z_o}(s_P)$ are one.
Therefore we have $I_{x_0}(\overline{f}_{\mcQ,z_o}(s_P),\mcQ)=2$.
In the same way as the above case, the line $\overline{f}_{\mcQ,z_o}(s_P)$ is a weak-bitangent line as $WL(\eta,x_0)$, where $\eta$ is a smooth point of $\mcQ$.

Conversely, let $s_P$ and $x_0$ be a line-section and a singularity of $\mcQ$ satisfying the condition (ii).
We define the type of $x_0$ by $A_{n_P}$.
By Lemma \ref{lem:line-sec}, $s_{P}\cdot O=1$.
By (b) and (c), we have $s_{P}\cdot \Theta_{x_0,1}=1$ or $s_{P}\cdot \Theta_{x_0,m_{x_0}-1}=1$.
We define $m_{x_0}-1$ by $n_P$.
Therefore, we obtain
\[
	\mathrm{contr}_{x_0}(s_P,s_P)=(m_{x_0}-1)/m_{x_0}=n_P/(n_P+1).
\]
Since $L$ is a weak-bitangent line satisfying (ii), we have $s_P\cdot\Theta_{x,0}=1$ for $x \in \sing(\mcQ)\setminus\{x_0\}$.
Hence we have $\langle P,P\rangle=3/2-n_P/(n_P+1)$.
\qed

Similarly we obtain the following proposition:
\begin{prop}\label{prop:bi-height}
	Let $\mcQ$ be a plane quartic curve satisfying $(\dagger)$ and let $z_o$ be a general smooth point of $\mcQ$.
	Then, for $P\in E_{S_{\mcQ,z_o}}(\CC(t))$, the following conditions {\rm (i)} and {\rm (ii)} are equivalent:
	\begin{itemize}
		\item[{\rm (i)}]
			$\langle P,P\rangle=3/2$ and $s_P\cdot \Theta_{\infty,1}=1$.
		\item[{\rm (ii)}]
			$s_P$ is a line-section of a bitangent line or a $4$-fold tangent line.
	\end{itemize}
\end{prop}

By using Proposition \ref{prop:w-bi-hieght}, Remark \ref{rem:become-line} and Theorem \ref{lem:shio-form}, we obtain representations of weak-bitangent lines written as $WL(\eta,x)$ and bitangent lines.
Note that a 4-fold tangent line (resp. a weak-bitangent line written as $WL(x)$ in the case when the type of $x$ is $A_1$) is the special case of a bitangent line (resp. a weak-bitangent line written as $WL(\eta,x)$) by Proposition \ref{prop:w-bi-hieght} and \ref{prop:bi-height}.
We explain some notations used in the next table.
\begin{itemize}
\item
	The columns labeled $\sing(\mcQ)$ and $E_{S_{\mcQ,z_o}}(\CC(t))$ are the same as those in Table \ref{table:st-MW} and $\eta$, $WL$ and $BL$ also are the same as those in  Table \ref{table:st-MW}.
\item
	Let $P_1,\ldots,P_n\in E_{S_{\mcQ,z_o}}(\CC(t))$ be generators as in Table \ref{table:geo-gene} and let $P$ be a rational point of $E_{S_{\mcQ,z_o}}(\CC(t))$ such that the corresponding section $s_P$ is a line-section of a bitangent line or a 4-fold tangent line.
	If the $P$ is given by $P_{s}=c_1P_1\dot{+}\cdots \dot{+}c_nP_n$, we write
	$
		\left[
			\begin{array}{c}
				c_1\\
				\vdots\\
				c_n
			\end{array}
		\right]
	$
	for $PBL$ in the column labeled BL.
	Note that $s_{\dot{-}P}$ is also a line-section and $\overline{f}_{\mcQ,z_o}(s_{P})=\overline{f}_{\mcQ,z_o}(s_{\dot{-}P})$ but we do not write
$\left[
	\begin{array}{c}
		-c_1\\
		\vdots\\
		-c_n
	\end{array}
\right]$.

\item
	Let $P_1,\ldots,P_n\in E_{S_{\mcQ,z_o}}(\CC(t))$ be generators as Table \ref{table:geo-gene} and let $P_{\eta,x}$ be a rational point of $E_{S_{\mcQ,z_o}}(\CC(t))$ such that the corresponding section $s_{P_{\eta,x}}$ is a line-section of a weak-bitangent line written as $WL(\eta,x)$.
	If $P_{\eta,x}$ is given by $P_{\eta,x}=c^{\eta,x}_1P_1\dot{+}\cdots \dot{+}c^{\eta,x}_nP_n$, we write
	$
	\left[
		\begin{array}{c}
			c^{\eta,x}_1\\
			\vdots\\
			c^{\eta,x}_n
		\end{array}
	\right]_{\eta,x}
	$
	for $P_{\eta,x}WL(\eta,x)$ in the column labeled WL.
	Note that $s_{\dot{-}P_{\eta,x}}$ is also a line-section and $\overline{f}_{\mcQ,z_o}(s_{P_{\eta,x}})=\overline{f}_{\mcQ,z_o}(s_{\dot{-}P_{\eta,x}})$ but we do not write
	$\left[
		\begin{array}{c}
			-c^{\eta,x}_1\\
			\vdots\\
			-c^{\eta,x}_n
			\end{array}
		\right]_{\eta,x}$.
	In the case when the type of $x$ is $A_1$, let $P_x=c^{\eta,x}_1P_1\dot{+}\cdots \dot{+}c^{\eta,x}_nP_n$ be a rational point corresponding to a line-section of a weak-bitangent line written as $WL(x)$.
	We also write
	$
	\left[
		\begin{array}{c}
			c^{x}_1\\
			\vdots\\
			c^{x}_n
		\end{array}
	\right]_{\eta,x}
	$ for $P_{x}WL(x)$ in the column WL because $WL(x)$ is a special case of $WL(\eta,x)$.

\end{itemize}
\begin{center}
\footnotesize
\begin{longtable}{|c|c|c|c|}
\caption{}
\label{table:geo-line}
\endhead
\hline
No. & $\sing(\mcQ)$ & BL & WL\\
\hline 1&
$(x,A_6)$		&
not exist&
$[3]_{\eta,x}$
\\
\hline	2&
$(x,E_6)$	&
$[3]$&not exist
\\
\hline 3&
$(x,A_5)$	&
$\left[\begin{array}{c}
0\\
3
\end{array}\right]$
&
$\left[\begin{array}{c}
1\\
1
\end{array}\right]_{\eta_1,x}$
$\left[\begin{array}{c}
-1\\
1
\end{array}\right]_{\eta_2,x}$
\rule[-4mm]{0mm}{9mm}\\
\hline 4&
$(x,D_5)$	&
$\left[
\begin{array}{c}
1\\
\pm2
\end{array}
\right]$
&not exist
\\
\hline 5&
$(x,D_4)$	&
\begin{tabular}{c}
$\left[\begin{array}{c}
1\\
1\\
1\\
0\\
\end{array}\right]$
$\left[\begin{array}{c}
-1\\
1\\
1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
1\\
-1\\
1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
1\\
1\\
-1\\
0
\end{array}\right]$
\end{tabular}&not exist
\\
\hline 6&
\begin{tabular}{c}
$(x,A_4)$\\
$(y,A_2)$
\end{tabular}
	&	not exist&
$[3]_{\eta_1,x}$ $[5]_{\eta_2,y}$
\\
\hline 7&
\begin{tabular}{c}
$(x,A_4)$\\
$(y,A_1)$
\end{tabular}
		&

$\left[\begin{array}{c}
2\\
1
\end{array}\right]$&
$\left[\begin{array}{c}
-2\\
1
\end{array}\right]_{\eta_1,x}$
$\left[\begin{array}{c}
-1\\
2
\end{array}\right]_{\eta_2,y}$
\rule[-4mm]{0mm}{9mm}\\

\hline 8&

\begin{tabular}{c}
$(x,A_1)$\\
$(y,A_1)$\\
$(z,A_1)$\\
$(w,A_1)$
\end{tabular}
		&
			\begin{tabular}{c}
$\left[\begin{array}{c}
1\\
1\\
1\\
0\\
\end{array}\right]$
$\left[\begin{array}{c}
-1\\
1\\
1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
1\\
-1\\
1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
1\\
1\\
-1\\
0
\end{array}\right]$
\end{tabular}
&not exist

\\
\hline 9&
\begin{tabular}{c}
$(x,A_3)$\\
$(y,A_2)$
\end{tabular}
		&not exist&
\begin{tabular}{l}
$\left[\begin{array}{c}
0\\
3
\end{array}\right]_{\eta_1,x}$
$\left[\begin{array}{c}
1\\
2
\end{array}\right]_{\eta_2,y}$
$\left[\begin{array}{c}
-1\\
2
\end{array}\right]_{\eta_3,y}$
\end{tabular}
\rule[-4mm]{0mm}{9mm}\\
\hline 10&
\begin{tabular}{c}
$(x,A_3)$\\
$(y,A_1)$
\end{tabular}			&
$\left[\begin{array}{c}
0\\
\pm1\\
2
\end{array}\right]$&
\begin{tabular}{l}
$\left[\begin{array}{c}
1\\
0\\
1
\end{array}\right]_{\eta_1,x}$
$\left[\begin{array}{c}
-1\\
0\\
1
\end{array}\right]_{\eta_2,x}$
$\left[\begin{array}{c}
1\\
1\\
0
\end{array}\right]_{\eta_3,y}$\\
$\left[\begin{array}{c}
-1\\
1\\
0
\end{array}\right]_{\eta_4,y}$
\end{tabular}
\rule[-11mm]{0mm}{23mm}\\
\hline 11&
\begin{tabular}{c}
$(x,A_2)$\\
$(y,A_2)$\\
$(z,A_2)$
\end{tabular}
&
$\left[
\begin{array}{c}
3\\
0
\end{array}
\right]$&not exist
\\
\hline 12&
\begin{tabular}{c}
$(x,A_2)$\\
$(y,A_2)$\\
$(z,A_1)$
\end{tabular}
		&
$\left[
\begin{array}{c}
3\\
0
\end{array}
\right]$&

$\left[
\begin{array}{c}
-1\\
2
\end{array}
\right]_{\eta_1,x}$
$\left[
\begin{array}{c}
1\\
2
\end{array}
\right]_{\eta_2,y}$
\\

\hline 13&
\begin{tabular}{c}
$(x,A_2)$\\
$(y,A_1)$\\
$(z,A_1)$
\end{tabular}			&
$\left[
\begin{array}{c}
1\\
1\\
1
\end{array}
\right]$
$\left[
\begin{array}{c}
1\\
-1\\
-1
\end{array}
\right]$
&
\begin{tabular}{l}
$\left[
\begin{array}{c}
1\\
-1\\
1
\end{array}
\right]_{\eta_1,x}$
$\left[
\begin{array}{c}
-1\\
-1\\
1
\end{array}
\right]_{\eta_2,x}$
$\left[
\begin{array}{c}
0\\
-1\\
2
\end{array}
\right]_{\eta_3,y}$\\
$\left[
\begin{array}{c}
0\\
-2\\
1
\end{array}
\right]_{\eta_4,z}$
\end{tabular}\rule[-11mm]{0mm}{23mm}\\
\hline 14&
\begin{tabular}{c}
$(x,A_1)$\\
$(y,A_1)$\\
$(z,A_1)$
\end{tabular}				&
			\begin{tabular}{l}
$\left[\begin{array}{c}
1\\
1\\
1\\
0\\
\end{array}\right]$
$\left[\begin{array}{c}
-1\\
1\\
1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
1\\
-1\\
1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
1\\
1\\
-1\\
0
\end{array}\right]$
\end{tabular}
&
\begin{tabular}{l}
$\left[\begin{array}{c}
1\\
0\\
0\\
1
\end{array}\right]_{\eta_1,z}$
$\left[\begin{array}{c}
1\\
0\\
0\\
-1
\end{array}\right]_{\eta_2,z}$

$\left[\begin{array}{c}
0\\
1\\
0\\
1
\end{array}\right]_{\eta_3,x}$\\
$\left[\begin{array}{c}
0\\
1\\
0\\
-1
\end{array}\right]_{\eta_4,x}$

$\left[\begin{array}{c}
0\\
0\\
1\\
1
\end{array}\right]_{\eta_5,y}$
$\left[\begin{array}{c}
0\\
0\\
1\\
-1
\end{array}\right]_{\eta_6,y}$
\end{tabular}
\rule[-15mm]{0mm}{31mm}\\
\hline 15&
$(x,A_4)$	&
\begin{tabular}{l}
$\left[\begin{array}{c}
-1\\
-1\\
1
\end{array}\right]$
$\left[\begin{array}{c}
2\\
0\\
1
\end{array}\right]$
$\left[\begin{array}{c}
-2\\
1\\
0
\end{array}\right]$
\end{tabular}
&
$\left[\begin{array}{c}
0\\
1\\
0
\end{array}\right]_{\eta_1,x}$
$\left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{\eta_2,x}$
$\left[\begin{array}{c}
1\\
-1\\
1
\end{array}\right]_{\eta_3,x}$
\rule[-6mm]{0mm}{14mm}\\
\hline 16&
$(x,A_3)$			&
\begin{tabular}{l}
$\left[\begin{array}{c}
0\\
1\\
0\\
\pm1
\end{array}\right]$
$\left[\begin{array}{c}
1\\
-1\\
1\\
\pm1
\end{array}\right]$
$\left[\begin{array}{c}
-1\\
0\\
1\\
\pm1
\end{array}\right]$
\end{tabular}
&
\begin{tabular}{l}
$\left[\begin{array}{c}
1\\
0\\
0\\
0
\end{array}\right]_{\eta_1,x}$
$\left[\begin{array}{c}
0\\
0\\
1\\
0
\end{array}\right]_{\eta_2,x}$
$\left[\begin{array}{c}
1\\
-1\\
0\\
0
\end{array}\right]_{\eta_3,x}$\\
$\left[\begin{array}{c}
0\\
-1\\
1\\
0
\end{array}\right]_{\eta_4,x}$
\end{tabular}
\rule[-15mm]{0mm}{31mm}\\
\hline 17&
\begin{tabular}{c}
$(x,A_2)$\\
$(y,A_2)$
\end{tabular}					&
$\left[\begin{array}{c}
0\\
0\\
3
\end{array}\right]$&
\begin{tabular}{c}
$\left[\begin{array}{c}
1\\
0\\
1
\end{array}\right]_{\eta_1,x}$
$\left[\begin{array}{c}
0\\
1\\
-1
\end{array}\right]_{\eta_2,x}$
$\left[\begin{array}{c}
1\\
-1\\
-1
\end{array}\right]_{\eta_3,x}$\\
$\left[\begin{array}{c}
1\\
0\\
-1
\end{array}\right]_{\eta_4,y}$
$\left[\begin{array}{c}
0\\
1\\
1
\end{array}\right]_{\eta_5,y}$
$\left[\begin{array}{c}
1\\
-1\\
1
\end{array}\right]_{\eta_6,y}$
\end{tabular}
\rule[-11mm]{0mm}{23mm}\\
\hline 18&
\begin{tabular}{c}
$(x,A_2)$\\
$(y,A_1)$
\end{tabular}					&
\begin{tabular}{l}
$\left[\begin{array}{c}
2\\
0\\
0\\
1
\end{array}\right]$
$\left[\begin{array}{c}
-1\\
0\\
-1\\
1
\end{array}\right]$
$\left[\begin{array}{c}
1\\
1\\
-1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
-2\\
1\\
0\\
0
\end{array}\right]$
\end{tabular}
&
\begin{tabular}{l}
$\left[\begin{array}{c}
0\\
1\\
0\\
0
\end{array}\right]_{\eta_1,x}$
$\left[\begin{array}{c}
0\\
0\\
0\\
1
\end{array}\right]_{\eta_2,x}$
$\left[\begin{array}{c}
1\\
0\\
-1\\
1
\end{array}\right]_{\eta_3,x}$\\
$\left[\begin{array}{c}
-1\\
1\\
-1\\
0
\end{array}\right]_{\eta_4,x}$
$\left[\begin{array}{c}
0\\
0\\
1\\
0
\end{array}\right]_{\eta_5,y}$
$\left[\begin{array}{c}
0\\
1\\
-1\\
1
\end{array}\right]_{\eta_6,y}$\\
$\left[\begin{array}{c}
1\\
-1\\
0\\
1
\end{array}\right]_{\eta_7,y}$

\end{tabular}
\rule[-22mm]{0mm}{44mm}\\
\hline 19&
\begin{tabular}{c}
$(x,A_1)$\\
$(y,A_1)$
\end{tabular}&
\begin{tabular}{l}
$\left[\begin{array}{c}
0\\
0\\
1\\
0\\
1
\end{array}\right]$
$\left[\begin{array}{c}
0\\
0\\
1\\
-1\\
1
\end{array}\right]$
$\left[\begin{array}{c}
0\\
0\\
-1\\
0\\
1
\end{array}\right]$
$\left[\begin{array}{c}
0\\
0\\
-1\\
1\\
1
\end{array}\right]$\\
$\left[\begin{array}{c}
-1\\
1\\
0\\
0\\
1
\end{array}\right]$
$\left[\begin{array}{c}
1\\
1\\
0\\
-1\\
1
\end{array}\right]$
$\left[\begin{array}{c}
1\\
-1\\
0\\
0\\
1
\end{array}\right]$
$\left[\begin{array}{c}
-1\\
-1\\
0\\
1\\
1
\end{array}\right]$
\end{tabular}&
\begin{tabular}{l}
$\left[\begin{array}{c}
1\\
0\\
0\\
0\\
0
\end{array}\right]_{\eta_1,x}$
$\left[\begin{array}{c}
0\\
1\\
1\\
-1\\
0
\end{array}\right]_{\eta_2,x}$
$\left[\begin{array}{c}
0\\
-1\\
1\\
0\\
0
\end{array}\right]_{\eta_3,x}$\\
$\left[\begin{array}{c}
-1\\
0\\
0\\
1\\
0
\end{array}\right]_{\eta_4,x}$

$\left[\begin{array}{c}
0\\
1\\
0\\
0\\
0
\end{array}\right]_{\eta_5,y}$
$\left[\begin{array}{c}
-1\\
0\\
1\\
0\\
0
\end{array}\right]_{\eta_6,y}$\\
$\left[\begin{array}{c}
1\\
0\\
1\\
-1\\
0
\end{array}\right]_{\eta_7,y}$
$\left[\begin{array}{c}
0\\
1\\
0\\
-1\\
0
\end{array}\right]_{\eta_8,y}$
\end{tabular}
\rule[-27mm]{0mm}{45mm}\\
\hline 20&
$(x,A_2)$			&
\begin{tabular}{l}
$\left[\begin{array}{c}
1\\
-1\\
1\\
-1\\
1
\end{array}\right]$
$\left[\begin{array}{c}
1\\
-1\\
1\\
0\\
-1
\end{array}\right]$
$\left[\begin{array}{c}
1\\
0\\
-1\\
0\\
1
\end{array}\right]$
$\left[\begin{array}{c}
1\\
0\\
-1\\
1\\
-1
\end{array}\right]$\\
$\left[\begin{array}{c}
1\\
-1\\
0\\
1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
1\\
0\\
0\\
-1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
0\\
0\\
1\\
0\\
0
\end{array}\right]$
$\left[\begin{array}{c}
0\\
-1\\
1\\
-1\\
0
\end{array}\right]$\\
$\left[\begin{array}{c}
0\\
-1\\
0\\
0\\
1
\end{array}\right]$
$\left[\begin{array}{c}
0\\
1\\
0\\
-1\\
1
\end{array}\right]$
\end{tabular}&

\begin{tabular}{c}
$\left[\begin{array}{c}
1\\
0\\
0\\
0\\
0
\end{array}\right]_{\eta_1,x}$
$\left[\begin{array}{c}
1\\
-1\\
0\\
0\\
0
\end{array}\right]_{\eta_2,x}$
$\left[\begin{array}{c}
0\\
-1\\
1\\
0\\
0
\end{array}\right]_{\eta_3,x}$\\
$\left[\begin{array}{c}
0\\
0\\
1\\
-1\\
0
\end{array}\right]_{\eta_4,x}$
$\left[\begin{array}{c}
0\\
0\\
0\\
0\\
1
\end{array}\right]_{\eta_5,x}$
$\left[\begin{array}{c}
0\\
0\\
0\\
-1\\
1
\end{array}\right]_{\eta_6,x}$\\
\end{tabular}
\\
\hline 21&

$(x,A_1)$			&
\begin{tabular}{l}
$\left[\begin{array}{c}
1\\
1\\
0\\
-1\\
1\\
-1
\end{array}\right]$
$\left[\begin{array}{c}
1\\
1\\
0\\
0\\
-1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
1\\
-1\\
0\\
-1\\
1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
1\\
-1\\
0\\
0\\
-1\\
1
\end{array}\right]$\\
$\left[\begin{array}{c}
1\\
0\\
-1\\
-1\\
0\\
1
\end{array}\right]$
$\left[\begin{array}{c}
1\\
0\\
1\\
0\\
0\\
-1
\end{array}\right]$
$\left[\begin{array}{c}
1\\
0\\
-1\\
0\\
0\\
0
\end{array}\right]$
$\left[\begin{array}{c}
1\\
0\\
1\\
-1\\
0\\
0
\end{array}\right]$\\
$\left[\begin{array}{c}
0\\
1\\
0\\
1\\
0\\
-1
\end{array}\right]$
$\left[\begin{array}{c}
0\\
1\\
0\\
-1\\
0\\
0
\end{array}\right]$
$\left[\begin{array}{c}
0\\
1\\
0\\
0\\
0\\
0
\end{array}\right]$
$\left[\begin{array}{c}
0\\
1\\
0\\
0\\
0\\
-1
\end{array}\right]$\\
$\left[\begin{array}{c}
0\\
0\\
1\\
0\\
1\\
-1
\end{array}\right]$
$\left[\begin{array}{c}
0\\
0\\
1\\
-1\\
1\\
-1
\end{array}\right]$
$\left[\begin{array}{c}
0\\
0\\
-1\\
0\\
1\\
0
\end{array}\right]$
$\left[\begin{array}{c}
0\\
0\\
-1\\
-1\\
1\\
0
\end{array}\right]$
\end{tabular}&

\begin{tabular}{c}
$\left[\begin{array}{c}
1\\
0\\
0\\
0\\
0\\
0
\end{array}\right]_{\eta_1,x}$
$\left[\begin{array}{c}
1\\
0\\
0\\
-1\\
0\\
0
\end{array}\right]_{\eta_2,x}$
$\left[\begin{array}{c}
0\\
1\\
1\\
0\\
0\\
-1
\end{array}\right]_{\eta_3,x}$\\
$\left[\begin{array}{c}
0\\
1\\
-1\\
0\\
0\\
0
\end{array}\right]_{\eta_4,x}$
$\left[\begin{array}{c}
0\\
0\\
0\\
-1\\
1\\
0
\end{array}\right]_{\eta_5,x}$
$\left[\begin{array}{c}
0\\
0\\
0\\
0\\
1\\
-1
\end{array}\right]_{\eta_6,x}$
\end{tabular}
\\
\hline

\end{longtable}
\end{center}

redpajama_set_name: RedPajamaArXiv


Our ninth month of house building with a video update of our dream home that has wainscoting & trim!
Personally, January felt like the month that would never end, but when it comes to building a house that is definitely a good thing!
They & we were able to get so much done this month.
Even though this is January's update since it is February I can excitedly say that next month we will be moving into our new home!!
We've already started picking out furniture & as soon as we have a move in date we will be able to start ordering the basics, like couches & chairs, for our new home.
But first, we need to finish building the house!
All of the floor & wall tiles were finished being laid & were grouted.
Our interior doors were installed.
The interior walls were prepped & (mostly) primed.
Wainscoting, baseboards, & trim were installed.
We finalized & ordered our custom kitchen cabinets.
We ordered 2 vanities for the master bathroom.
Our half bathroom & 2 upstairs bathroom vanities were delivered.
All of our light fixtures were delivered.
We ordered the cabinets for our laundry room (which we picked up & unloaded at our new house).
We finalized the hard marble (some people call it white granite & others call it quartzite) for our kitchen & bought butcher block for our laundry room.
The electricity was turned on!!
And even though February is a short month, I'm hoping that we have a whole lot of progress this month so that we will be able to move into our new house in mid-March.
If you want to see more of what happened in January click here to see my YouTube house progress update for January!
And if you enjoyed this home update, be sure to subscribe to my newsletter below so that you don't miss out on another blissful moment!
What were you most excited about 2 months prior to moving into your new home build?


Creed & Creed has been named Best of the Delta in the law category 10 out of the last 11 years.
Christian and Catherine Creed met as first-year law students at Loyola University. The friendship they formed gradually grew into a partnership that led to marriage, two children and a thriving law firm specializing in personal injury and admiralty law, Creed & Creed, with offices in Monroe, Alexandria and New Orleans.
Twenty-three years after they obtained their law degrees and 20 years after they branched out from Catherine's grandfather's firm to form their own, Creed & Creed has been named Best of the Delta in the law category 10 out of the last 11 years. Christian was recently named Personal Injury Litigator of the Year in Louisiana for 2018 by Acquisition International and has achieved the Super Lawyer designation five times.
But while they both found careers as attorneys, their paths to law were quite different.
For Catherine, there was no question – she's the third generation in her family to become an attorney and never really considered anything else.
And the family tradition is ongoing – the Creeds' daughter is a first-year law student at Loyola.
Christian's path to law wasn't so clear. He worked as an accident investigator and insurance adjuster during college and that experience gradually brought him to the realization that he could do more to help people with a law degree.
Christian believes the experience he gained then served him well because it gave him insight into how insurance company adjusters and defense attorneys think.
More than two decades into their careers, what gives them the greatest satisfaction?
The Creeds are also firm believers in supporting their community, whether through sponsorships, like football jamborees Bayou Jamb and the Eddie G. Robinson Classic, or through serving in the non-profit world. Christian's involvement includes the United Way of Northeast Louisiana, Salvation Army of Northeast Louisiana, Louisiana Purchase Council of Boy Scouts of America and Northeast Louisiana American Red Cross. Catherine is a Monroe Junior League Sustainer and has worked with Prevent Child Abuse Louisiana.
Christian was the first recipient of the United Way of Northeast Louisiana's Community Spirit Volunteer Leadership Award, an honor that's meaningful to him because he believes those who have achieved success have a responsibility to contribute to their communities.


sandor: Glad to see the good ol' Indy Indians get some press, however silly it is. I'll be the first to admit there isn't much to see on your way through Naptown, but Victory Field really is a treat. I treated my grandma to a game there a few years ago, and we had a blast. Highly recommended, even when the ballplayers are too dumb to remember how a door works.
Next Next post: This is Priority Mail?


Sewer main renewal can be unglamorous at the best of times, but can prove even tougher when undertaken in a public setting that requires pedestrian access. In situations like these, expertise, effective planning and consultation with stakeholders is critical to project success.
Regardless of the project type, areas with a high flow of traffic are challenging to manage and maintain. Sewers present unique challenges, with works often requiring heavy machinery that poses a risk to both public access and the integrity of existing infrastructure such as concrete paths.
A recent project at popular tourist location Mooloolaba, involved the relining of six consecutive lengths of DN450 and raising one access chamber, as well as cleaning and coating eight existing access chambers. Interflow worked with Unitywater and the local council to successfully manage the large numbers of pedestrians requiring beach and park access, and to undertake the works on steep, inaccessible bushland.
In order to reline the existing pipes effectively, the lines were first cleaned using a recycler. To reduce cartage and use of water, cleaning was undertaken by a one position set up, incorporating two dingos which were positioned up the line. These were moved into position using a 100-tonne crane, along with the relining Kubota.
Because the site was located on a large area of parkland, multiple site setups were established, each requiring careful coordination and effective communications with stakeholders. The coating works and relining were undertaken in tandem, which made the most of the arranged closures and permits.
Proving the importance of planning, the significant amount of preparation meant that there was minimal impact to members of the public and the environment.
Environmental considerations such as species and habitat management are important in any renewal project, and were made a high priority, meaning that no harmful effects resulted from the works.
Flexible traffic and environmental controls are also critical to the success of renewal projects. With public footpaths impacted by the works in Mooloolaba, bog mats were required to control the subsidence of equipment. This protected the ground from recent rain, and protected the concrete paths from the weight of heavy machinery and equipment.
With effective planning, and using the right equipment for the job, the project was completed both on time and budget, without any impact to the Unitywater network.
For more information on Interflow's sewer, stormwater and potable water capabilities, visit www.interflow.com.au.


This handbag is made of high quality material. It is waterproof, scratchproof and durable. The pure color design makes it looks cool and gorgeous. Fringed design add cute element to the bag. Slung it on your shoulder, you will stand out of the crowd easily for its simple but gorgeous design.


Are You in Bloom with Susan Bishop Crispell?
This week's Bloom host is Susan Bishop Crispell. Keeping with the "summer fun" theme, Susan will share some of her favorite things about summer. This includes childhood road trips to the beach, free Shakespeare performances, summer eats and drinks, as well as a ton of book conversations (including a few sneak peeks at some of her in-progress novels). Plus giveaways! In addition, Karma Brown will be stopping by on Friday from 6-7 EST for a happy hour chat to celebrate the release of her latest novel, THE LIFE LUCY KNEW.
Susan Bishop Crispell earned a BFA in creative writing from the University of North Carolina at Wilmington. Born and raised in the mountains of Tennessee, she now lives twenty minutes from the beach in North Carolina with her husband and their literary-named cat. She is very fond of pie and is always on the lookout for hints of magic in the real world. She is the author of THE SECRET INGREDIENT OF WISHES (2016) and DREAMING IN CHOCOLATE (2018), which was a Southern Independent Booksellers Alliance (SIBA) Okra Pick for Winter 2018.


Jean and Ed Teague went to court yesterday hoping to find out why Kevin Davis killed their daughter. They didn't get an answer.
What Mr. and Mrs. Teague did hear for the first time, however, were the chilling details of how Mr. Davis randomly chose their daughter as his victim, stalked and abducted her and then debated with himself before finally killing her in his Barrhaven bedroom.
The account, at once graphic and heart-rending, was read to the court before Mr. Davis pleaded guilty to first-degree murder.
It left both parents deeply shaken and, in the case of Mr. Teague, visibly angry.
"I stared at him through the entire proceeding," he told reporters.
Nor was Mrs. Teague satisfied by Mr. Davis's contrition. "I don't know how sincere he is. It's not enough for me," she said.
Amid all of the gruesome details presented in court, Mrs. Teague said nothing left her more shaken than learning her daughter, during the final moments of her ordeal, told her abductor she had to get home because her mom would be worried.
The day did provide a small measure of relief for the Teagues. By pleading guilty, Mr. Davis spared the family and Ms. Teague's many friends the agony of a lengthy, public trial. And they are satisfied to know their daughter's killer is behind bars for a long time.
For Mr. and Mrs. Teague, who are divorced, as well as Mr. Teague's second wife, Sylvie, the day's events might also signal the start of a new period in their lives.
"I don't think there is closure, but I think there is a healing process that can finally start," Mr. Teague said.
It seems there is plenty of healing to be done, based on the victim-impact statements read in court yesterday by all three, as well as Carey, one of the Teagues' two sons.
Reading her statement in a shaky voice and pausing frequently to regain her composure, Mrs. Teague said that from the moment her daughter was reported missing — "a mother's worst nightmare," she called it — her life has been turned on its head.
The crime has affected her ability to enjoy what used to be routine family activities — shopping, participating in a hockey pool, celebrating birthdays and Christmas, she said.
Mrs. Teague said it has been especially painful to watch her two sons deal with the loss of their only sister.
The oldest, Kevin, was the only family member to attend the preliminary hearing last November at which the details of the killing were first disclosed, but not allowed to be spread by anyone in attendance. Mrs. Teague said he was so disturbed by what he heard, he was unable to attend yesterday's court proceedings.
The youngest, Carey, told the court his once sunny outlook on life has been seriously damaged by his sister's killing.
Ms. Teague's murder has also had a devastating impact on many of her friends, court was told.
Alicia Blais, who met Ms. Teague at the Mac's convenience store in Barrhaven shortly before her abduction, said she suffers from frequent panic attacks, sleeplessness and nightmares.
Another friend, Amy Picknell, also wrote about sleepless nights, missing work, loneliness and trouble making it through many days.
Friend Melissa Davidson said she can no longer participate in activities she shared with Jennifer, and has developed such a fear of being alone she endured an abusive relationship for nearly a year rather than live on her own.
Ms. Bair told reporters she has rarely encountered a case that has brought so much pain to so many people.
Despite the obvious pain she was feeling after learning the details of her daughter's death for the first time, Mrs. Teague still showed an amazing capacity for compassion for the killer's mother.
As for Mr. Teague, he was left looking for the answer to the question he came to court hoping to find.
Asked outside the court later why he stared at his daughter's killer as he read that statement in court, Mr. Teague said he was hoping to find some kind of answer in Mr. Davis's face.
Mr. Teague didn't see what he was looking for, and he's doubtful he'll ever find the answer he's seeking.
When you die young, you don't leave much history. But you can make your mark.
That's the way it was with Jennifer Teague. Jenny to her family, Jen to her pals, Ms. Teague had been 18 for barely two months when her future was brutally taken from her as she walked home from a part-time job at a little after midnight on Sept. 9, 2005.
She was too young to know what she wanted from life, but old enough to leave lasting impressions in the small suburban world she inhabited for such a short time.
To friends and family, what seemed to make Ms. Teague — a dynamo at 5-foot-4, 106 pounds — so hard to forget was an irresistible blend of generosity, determination, optimism and love of life.
Her mother, Jean, remembered three-year-old Jenny marching down the basement stairs, plastic baseball bat in hand, to find the mice that had frightened one of her older brothers.
Those two big brothers, Kevin and Carey, recalled how she always held her own without getting angry during their relentless teasing.
Her father, Ed, can remember a young woman with a strong streak of independence, but also one with a softer side that led her to say, "I Love you, daddy," as she signed off the last time they spoke.
Born in June 1987, Ms. Teague was Jean and Ed Teague's only daughter. The year before, the family had moved to Ottawa from Halifax when Mr. Teague, who was working for the military, received a new posting.
At first the family lived outside the city, but after her parents divorced, Ms. Teague and her two brothers moved with their mother, a public servant, to a house she bought for herself and the children in Barrhaven. Mr. Teague, who eventually remarried, lived nearby and stayed connected with his children, who grew up as a tight group.
Reading through the litany of praise for Ms. Teague, it's sometimes easy to forget she was also a typical 18-year-old with a regular assortment of teenaged crises, interests and crushes.
Her bedroom featured a collection of NHL bobblehead dolls and was adorned with a poster of Daniel Alfredsson, her favourite Ottawa Senator. It also contained a stuffed lamb named Emily, a toy that had comforted her since childhood, as well as a deck of cards she and her friends used while lounging on her bed, talking and laughing for hours.
Like any teen, Ms. Teague had a few questionable habits. She smoked Du Maurier cigarettes, loved KitKat chocolate bars, probably played a bit too many video games and, according to her mother, often woke up grumpy.
But she was also an avid athlete whose baseball coach said she had an "arm like a cannon" and who was considered a better-than-average goalie in soccer, her real sports passion.
Throughout her teens, Ms. Teague held a series of part-time jobs at fast-food outlets that gave her the money she needed for teenaged requirements like video rentals, makeup and clothes — "the girl was born to shop," her mother once said.
Ms. Teague loved to experiment with her hair colour, over the years going from her natural auburn to red to blond and to dark brown.
She also loved animals. With her last paycheque from Wendy's, where she'd worked the late shift the night she disappeared, Ms. Teague planned to pay for shots for her new kitten, Michael.
Never a naturally gifted student, she switched from John McRae Secondary School to the alternative program at Elizabeth Wyn Wood for Grade 10. The setup was more suited to her needs — smaller classes and a more flexible approach to learning — and her grades improved.
At Wyn Wood, she joined an environmental group called EarthCARE that travelled to elementary schools to talk to younger children about conservation. The group won a David Suzuki award for its efforts, and the $2,000 prize arrived at the school the week Ms. Teague disappeared.
When she died, Ms. Teague was heading into Grade 12 with no real idea about what she wanted to do after graduation, although she had spoken to friends and family about doing something in the outdoors, or perhaps clothing design.
As Citizen columnist Janice Kennedy wrote in the aftermath of Ms. Teague's death, no one can say with any precision how she would have lived the rest of her life.
Sept. 8, 2005: Jennifer Teague, 18, disappears while walking home from her night shift at a Barrhaven fast-food restaurant.
Sept. 16, 2005: Ottawa police are going door-to-door in Barrhaven neighbourhoods, requesting entry to many of the suburb's homes as the search continues.
Sept. 18, 2005: Ms. Teague's decomposed body is found concealed in woods by an off-duty Ottawa police officer and his family at about 12:30 p.m., just off the head of a popular hiking trail in Ottawa's Stony Swamp and about five kilometres from where Ms. Teague disappeared.
Sept. 29, 2005: Ms Teague's body is released from a Toronto crime lab after extensive forensic testing.
Oct. 1, 2005: More than 500 mourners attend the funeral at Cedarview Alliance Church.
Oct. 6, 2005: Former police chief Vince Bevan tells a Barrhaven citizens' panel that two people had claimed to be Ms. Teague's killer, but were discredited by police.
April 5, 2006: Ottawa police release a sketch of a suspect and plead for the public's help in the unsolved case. Some of Ms. Teague's friends and relatives say they have seen the sketch, but do not recognize the person, suggesting the suspect was not one of Ms. Teague's longtime acquaintances. Police receive 85 new tips over the next few days. Wendy's Canada, Ms. Teague's former employer, offers a $50,000 reward for information leading to the arrest and capture of person(s) responsible for her death.
April 10, 2006: The Ottawa Police Services Board doubles the reward for information. Board members unanimously agree to match a $50,000 reward offer by Wendy's Canada, to bring the total reward to $100,000.
May 24, 2006: Ottawa police post 24 photos of 'potential witnesses' in connection with the killing. The photos, some of which have multiple individuals in them, were from a surveillance camera in the Mac's Milk store at the corner of Jockvale Road and Tartan Drive on the night of Ms. Teague's disappearance.
June 9, 2006: Days after detectives release photos of customers seen in a security tape, Kevin Davis, a 24-year-old pizza maker, downs a load of magic mushrooms, then strips naked and rushes out onto Fallowfield Road, screaming in the middle of speeding traffic that he had killed Jennifer Teague. Mr. Davis is rushed to hospital and, when questioned by police, says he doesn't know anything about the case.
June 26, 2006: Mr. Davis walks out of his Orr Street house to tell neighbours he had killed Jennifer Teague. He says he needs help and urges neighbours to call police. At a nearby shopping plaza, he walks up to a an off-duty police officer and turns himself in.
June 27, 2006: Mr. Davis, 24, is arrested and charged with first-degree murder.
Nov. 28, 2007: Mr. Davis is committed to stand trial after a pre-trial hearing wrapped up.
Jan. 8, 2008: Mr. Davis's lawyer announces his client will plead guilty to the first-degree murder of Jennifer Teague.
Jan. 12, 2008: Kevin Davis tells of his remorse and affirms his intention to plead guilty to Citizen reporter Andrew Seymour in a published interview conducted in the regional detention centre.
Jan. 25, 2008: At 10 a.m. in an Ottawa courtroom, Kevin Davis admits his guilt. He is subsequently sentenced to life in prison with no chance of parole for 25 years.
© 2008 Postmedia Network Inc. All rights reserved.


From a book lover, It's how life is!, It's Humour time., Ramblings of my mind.
Sunday morning, beautiful weather outside, cool and calm air caressing you, a good book in hand, a warm cup of coffee.
Anything wrong with that? Nope, sounds just perfect. But a perfect day was just not for me.
Instead I decided to be the good girl, and sat with my books to study. I had been delaying it for quite some time now, so there I sat with books in front of me, the lovely weather outside tempting me every five minutes. I resisted the temptation, telling myself that I will enjoy this perfect setting in the evening.
So I studied till lunch time, had a delicious home-made lunch, and then was just trying to set that perfect scenery, I got the book that I was reading, placed the chair beside the window in the balcony, opened the giant windows, and was just about to get that cup of coffee, when..!
When I got a call.
My initial instinct told me not to pick up the call, because that call on a Sunday meant nothing but work!
I picked up anyway, and ended up with a meeting of about an hour.
Sounds fair! So I can come back in an hour and start with that perfect setting, I told myself.
I got ready for that meeting in fifteen minutes within which I had managed to fix myself with another meeting with some clients.
Okay, now I have two meetings? That's okay, its 3:30 and I can still come home by 6 max.
And then finally, I realised putting off stuff isn't any good. I shouldn't have put off that reading session.
As always, I was kept waiting everywhere, and I managed to complete both the meetings by late in the evening, well it wasn't even evening anymore when I got out of the place. It was 9 o' clock and pretty dark.
So I drove home, ate my dinner in silence, with my head throbbing all the while. I was so tired and drenched (Just two meetings, right? Was I over-reacting?), I did not wish to do anything; even the TV had nothing to help with.
Working on a Sunday? Really, why? Why did I agree? That shouldn't even be a thing! Working Sunday as a concept should be removed from our mind-set.
That is when my dad stepped up, he took me out for a long drive, we stopped for a drink, (a mocktail, of course) and then we drove around a little bit, hardly talking to each other, with soft music playing all the while. Oh my God, that feeling, I cannot simply express how I felt at that moment. It was so soothing.
Oh of course, we should.
PS: Apart from the reading, I also missed being the crazy questionnaire to my mom, over that mythology show. Hoping for that to happen on the upcoming Sunday, but then it's a week away!
When we left for another city, we left with a bang. After a tiresome day we enjoyed ourselves with local delicious food accompanied by the coolest fresh air and folk music. Could it be any better? I dont think so.
So the journey ended and Mr. R and I reached 'our home' somewhere around late night and extremely early morning. We were dead tired from all the day's work and the sitting in the car and we just slumped into our beds.
We woke up in the aftrrnoon and still we slumped all day on the couch, watching tv listening to music and getting up only to get food or water. I know I am up for travelling but I guess if I did not have this lousy day, i would exhaustyself completely. So I guess it was better to at least relax too rather than overdoing it.
We left our place only in the evening to take a walk and get some chilled milk shakes. And we strolled back home. But we just couldn't contain the boredom so we decided to go for a bike ride it was cool here and so post dinner, almost after mifnight, we went for a bike ride, at first over speeding on the empty roads but then going completely 20 and 10 kmph only to avoid getting back home early.
While we were driving bsck I saw something disturbing. Two men were riding a bike at 40-50 kmphs but then they were literally dragging a dog along with them, screaming to run faster. We assumed the dog was being trained for some thing but apart from that, I still feel any reason could not justify this behaviour towards the dog.
Anyways we got back home and after a bit of tv watching amd reading we went to bed. I could just say we went to bed, but we did not sleep. We listened to a lot of music, we chatted uncontrolably, laughed at ourselves, brought back memories and then finally dozed off knowing nothing about tomorrow.
And thats how we have fun with each other. Just laughing, talking, and sometimes even staying in. But most of all the company matters.


text:
The Work Continues...
Justice Versus Accountability
by Kathleen C. Gamble
"I would not call todays verdict justice because justice implies true restoration. But it is accountability, which is the first step towards justice. Now the cause of justice is in your hands, when I say your hands, I mean the hands of the people of the United States." – Keith Ellison, Minnesota Attorney General
Calls for justice ring hollow in our streets when people are not held accountable for their acts of injustice.
It says a lot that the country, nay, the world was waiting with bated breath for the verdict of 12 Minnesotans regarding a crime that was committed in broad daylight and caught on camera—because no one truly knew if the perpetrator would be held accountable. More often than not in this country, if someone commits a crime while wearing a badge, they are in fact not held to account for those crimes. Derek Chauvin's conviction was just the second time in the history of the State of Minnesota that a police officer was held accountable in this way.
It says a lot, but what does it say? It says that Americans are still waiting for justice. Black Americans are still waiting for justice.
Keith Ellison is right; it is in our hands now. It's in our hands to find true justice.
I'm not talking about vigilantism, or what you may see in sensationalized news coverage.
It's in our hands to continue the work that was started so many years before the murder of George Floyd but ignited by his dying calls for his mother and final gasps of air under that knee.
It's in our hands to be as courageous.
It's in our hands to be thoughtful.
It's in our hands to find unity.
It's in our hands to show compassion.
It's in our hands to act.
It's in our hands to be empathetic.
It's in our hands to continue evaluating our own actions, institutions, and lives to grow together as we work to reduce and ultimately eliminate racial bias in this country. Its existence is a truth that we cannot escape.
It's in our hands to work until that truth, takes no more lives.
It's in our hands to work until we find true justice and liberty for all.
Perhaps it is not a time to celebrate, but a time to exhale the collective breath we've all been holding as we push on to the next.
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redpajama_set_name: RedPajamaCommonCrawl