text:
\section{Introduction}\label{sec:Introduction}
First we prepare some terminologies about lattices.
Let $R$ be $\mathord{\mathbb Z}$ or $\mathord{\mathbb Z}_p$,
where $p$ is a prime integer or $\infty$,
$\mathord{\mathbb Z}_p$ is the ring of $p$-adic integers for $p<\infty$, and
$\mathord{\mathbb Z}_\infty$ is the field $\mathord{\mathbb R}$ of real numbers.
An \emph{$R$-lattice} is a free $R$-module $L$ of finite rank
with a non-degenerate symmetric bilinear form
$$
(\phantom{a}, \phantom{a})_L \;:\; L\times L \to R.
$$
A $\mathord{\mathbb Z}$-lattice is simply called a \emph{lattice}.
A lattice $L$ is called \emph{even} if
$(v,v)_L\in 2\mathord{\mathbb Z}$ holds for any $v\in L$.
Two lattices $L$ and $L\sp\prime$ are said to be \emph{in the same genus}
if the $\mathord{\mathbb Z}_p$-lattices $L\otimes \mathord{\mathbb Z}_p$ and $L\sp\prime\otimes \mathord{\mathbb Z}_p$
are isomorphic for all $p$ (including $\infty$).
Then the  set of isomorphism classes of lattices are decomposed into a disjoint union of
\emph{genera}.
Note that, if $L$ and $L\sp\prime$ are in the same genus and $L$ is even,
then $L\sp\prime$ is also even,
because being even is a $2$-adic property.
Let $L$ be a  lattice.
Then $L$ is canonically embedded into $L\sp{\vee}:=\mathord{\rm Hom} (L, \mathord{\mathbb Z})$
as a submodule of finite index, and $(\phantom{a}, \phantom{a})_L$  extends to
a symmetric bilinear form
$$
(\phantom{a}, \phantom{a})_{L\sp{\vee}} \;:\; L\sp{\vee}\times L\sp{\vee} \to \mathord{\mathbb  Q}.
$$
Suppose that $L$ is even.
We put
$$
D_L:=L\sp{\vee}/L,
$$
and define  a quadratic form
$q_L: D_L\to \mathord{\mathbb  Q}/2\mathord{\mathbb Z}$  by
$$
q_L(\bar x):=(x, x)_{L\sp{\vee}}\bmod 2\mathord{\mathbb Z},
\quad \textrm{where}\quad \bar x=x+L\in D_L.
$$
The pair $(D_L, q_L)$ is called the \emph{discriminant form} of $L$.
By  the following result of Nikulin (Corollary 1.9.4 in~\cite{MR525944}),
each genus of even lattices is characterized by the signature and the
discriminant form.
\begin{proposition}
Two even  lattices are in the same genus if and only if
they have the same signature and their  discriminant forms are isomorphic.
\end{proposition}
\par
\medskip
For a $K3$ surface $X$ defined over a field $k$,
we denote by $\mathord{\rm NS}(X)$ the \emph{N\'eron-Severi lattice} of $X\otimes \bar k$;
that is, $\mathord{\rm NS}(X)$ is the lattice of numerical equivalence classes
of divisors on $X\otimes \bar k$ with the intersection paring
$\mathord{\rm NS}(X)\times \mathord{\rm NS}(X)\to \mathord{\mathbb Z}$.
Following the  terminology of~\cite[\S8]{MR0284440} and~\cite{MR0441982},
we say that a $K3$ surface $X$ defined over a field of characteristic $0$
is \emph{singular} if the rank of $\mathord{\rm NS}(X)$ attains
the possible maximum $20$.
\par
\medskip
Let $S$ be a \emph{complex} $K3$ surface.
Then the second Betti cohomology group $\mathord{\rm H}^2 (S,\mathord{\mathbb Z})$ is regarded as a unimodular lattice
by the cup-product,
which is even of signature $(3, 19)$.
The  N\'eron-Severi lattice $\mathord{\rm NS}(S)$ is embedded into $\mathord{\rm H}^2(S, \mathord{\mathbb Z})$ primitively,
because we have $\mathord{\rm NS}(S)=\mathord{\rm H}^2(S, \mathord{\mathbb Z})\cap \mathord{\rm H}^{1,1}(S)$.
We denote by $\mathord{\rm T} (S)$ the orthogonal complement
of $\mathord{\rm NS} (S)$ in $\mathord{\rm H}^2(S, \mathord{\mathbb Z})$,
and call $\mathord{\rm T} (S)$ the \emph{transcendental lattice} of $S$.
Suppose that $S$ is singular in the sense above.
Then $\mathord{\rm T} (S)$ is an even positive-definite lattice of rank $2$.
The Hodge decomposition
$\mathord{\rm T} (S)\otimes \mathord{\mathbb C}=\mathord{\rm H}^{2,0}(S)\oplus \mathord{\rm H}^{0,2}(S)$
induces a canonical orientation on $\mathord{\rm T} (S)$.
We denote by $\wt{\Tr}(S)$ the \emph{oriented transcendental lattice} of $S$.
\par
\medskip
We denote by $[2a, b, 2c]$ the symmetric matrix
$
\left[
\begin{array}{cc}
2a & b \\
b & 2c
\end{array}
\right],
$
and put
$$
\mathord{\mathcal M}:=\set{[2a,b,2c]}{a,b, c\in \mathord{\mathbb Z}, a>0, c>0, 4ac-b^2>0},
$$
on which $\mathord{GL}(2,\mathord{\mathbb Z})$ acts by $M\mapsto g^T M g$,
where $M\in \mathord{\mathcal M}$ and $g\in\mathord{GL}(2, \mathord{\mathbb Z})$.
The set of isomorphism classes of even positive-definite lattices
of rank $2$ is equal to
$$
\mathord{\mathcal L}:=\mathord{\mathcal M}/\mathord{GL}(2, \mathord{\mathbb Z}),
$$
while the set of isomorphism classes of even positive-definite
\emph{oriented} lattices
of rank $2$ is equal to
$$
\wt{\LLL}:=\mathord{\mathcal M}/\mathord{SL}(2, \mathord{\mathbb Z}).
$$
For a matrix $[2a,b,2c]\in \mathord{\mathcal M}$, we denote by
$\wt{L}[2a,b,2c]\in \wt{\LLL}$ and $L[2a,b,2c]\in \mathord{\mathcal L}$
the isomorphism classes  represented by $[2a,b,2c]$.
\par
\medskip
In~\cite{MR0441982},
Shioda and Inose proved the following:
\begin{theorem}\label{thm:SI}
The map $S\mapsto \wt{\Tr}(S)$
induces a bijection from the set of isomorphism classes of
complex singular $K3$ surfaces $S$ to the set $\wt{\LLL}$.
\end{theorem}
The injectivity follows from  the  Torelli theorem
by Piatetski-Shapiro and Shafarevich~\cite{MR0284440}.
In the proof of the surjectivity,
Shioda and Inose gave an explicit construction of
the  complex singular $K3$ surface with a given oriented transcendental lattice,
and they have proved the following:
\begin{theorem}\label{cor:SI}
Every complex singular $K3$ surface is
defined over a number field.
\end{theorem}
Let $X$ be a singular $K3$ surface defined over a number field $F$.
We denote by $\mathord{\rm Emb} (F)$ the set of embeddings of $F$ into $\mathord{\mathbb C}$,
and for $\sigma\in \mathord{\rm Emb} (F)$, we denote by $X^\sigma$
the \emph{complex} singular $K3$ surface $X\otimes_{F, \sigma} \mathord{\mathbb C}$.
We define a map
$$
\tau_X\;:\; \mathord{\rm Emb} (F) \to \wt{\LLL}
$$
by $\tau_X(\sigma):=\wt{\Tr}(X\sp\sigma)$.
Then we have the following theorem by Sch\"utt~\cite{MR2346573},
which is a generalization of a result  that had been obtained in   \cite{tssK3}.
\begin{theorem}\label{thm:SS}
Let $\mathord{\mathcal G}_X\subset\mathord{\mathcal L}$
be the genus  of all $L\in \mathord{\mathcal L}$
such that $(D_L, q_L)$ is isomorphic to $(D_{\mathord{\rm NS}(X)}, -q_{\mathord{\rm NS}(X)})$,
and let $\wt{\GGG}_X\subset \wt{\LLL}$
be the pull-back of $\mathord{\mathcal G}_X$ by the natural projection
$\wt{\LLL}\to \mathord{\mathcal L}$.
Then the image of $\tau_X$ coincides with $\wt{\GGG}_X$.
\end{theorem}
Therefore we obtain  a surjective map
$$
\tau_X\;:\; \mathord{\rm Emb} (F) \;\mathbin{\to \hskip -7pt \to}\; \wt{\GGG}_X.
$$
Remark that, by the  classical theory  of Gauss~\cite{MR837656},
we can easily calculate the oriented genus $\wt{\GGG}_X\subset\wt{\LLL}$ from
the finite quadratic form $(D_{\mathord{\rm NS}(X)}, -q_{\mathord{\rm NS}(X)})$.
\par
\medskip
Let $Y$ be a geometrically reduced and irreducible projective surface defined over a number field $K$,
and let $X\to Y\otimes_K F$ be a desingularization
of $Y\otimes_K F$ defined over a finite extension $F$ of $K$.
Suppose that $X$ is a singular $K3$ surface.
Then we can  define a map
$$
\tau_Y\;:\; \mathord{\rm Emb} (K)  \;\mathbin{\to \hskip -7pt \to}\; \wt{\GGG}_X
$$
by the following:
\begin{proposition}\label{prop:tauY}
The map $\tau_X: \mathord{\rm Emb} (F) \to \wt{\GGG}_X$ factors as
$$
\mathord{\rm Emb} (F)\; \maprightsurjsp{\rho}\;  \mathord{\rm Emb}(K)\; \maprightsp{\tau_Y}\;  \wt{\GGG}_X,
$$
where $\rho : \mathord{\rm Emb} (F)\mathbin{\to \hskip -7pt \to}  \mathord{\rm Emb}(K)$ is the natural restriction map $\rho(\sigma):=\sigma|K$.
\end{proposition}
The purpose of this paper is to present a method to calculate
the map $\tau_Y$ from a defining equation of $Y$.
\par
\medskip
More generally,
we consider the following problem.
Let $S$ be a reduced irreducible complex projective surface.
For a desingularization $S\sp{\sim}\to S$,
we put
$$
\mathord{\rm H}^2(S\sp{\sim}):=\mathord{\rm H}^2(S\sp{\sim},\mathord{\mathbb Z})/(\textrm{the torsion part}),
$$
which is regarded as a lattice by the cup-product,
and let $\mathord{\rm NS}(S\sp{\sim})\subset \mathord{\rm H}^2(S\sp{\sim})$
be the sublattice  of the cohomology classes of divisors on $S\sp{\sim}$.
We denote by
$$
\mathord{\rm T}(S\sp{\sim})\subset \mathord{\rm H}^2(S\sp{\sim})
$$
the orthogonal complement of $\mathord{\rm NS}(S\sp{\sim})$ in $\mathord{\rm H}^2(S\sp{\sim})$.
Then we can easily see that the isomorphism class of the lattice
$\mathord{\rm T}(S\sp{\sim})$ does not depend on the choice of the desingularization
$S\sp{\sim}\to S$,
and hence we can define the \emph{transcendental lattice $\mathord{\rm T} (S)$ of $S$}
to be $\mathord{\rm T}(S\sp{\sim})$. (See Lemma 3.1 of~\cite{Shioda_K3SPnew} or Proposition~\ref{prop:indep} of this paper.)
We will give  a \emph{method of  Zariski-van Kampen type}
for the calculation of  $\mathord{\rm T}(S)$.
\par
\medskip
We apply our method to maximizing sextics.
Following   Persson~\cite{MR661198, MR805337},
we say that
a reduced projective plane curve $C\subset \P^2$ of degree $6$ defined over
a field $k$ of characteristic $0$ is  a \emph{maximizing sextic}
if $C\otimes \bar k$ has only simple singularities and its total Milnor number
attains the possible maximum $19$,
where $\bar k$ is the algebraic closure of $k$.
The \emph{type} of a maximizing sextic $C$ is the $ADE$-type of the singular points of $C\otimes \bar k$.
\par
\medskip
Let $C\subset \P^2$ be a maximizing sextic defined over a number field $K$.
The double covering  $Y_C\to \P^2$ branching exactly along $C$
is defined over $K$.
Let $X_C\to Y_C\otimes_K F$
be the minimal resolution defined over a finite extension $F$ of $K$.
Then
$X_C$ is a singular $K3$ surface defined over $F$.
We denote by $\wt{\GGG}_{[C]}$
the oriented genus $\wt{\GGG}_{X_C}$.
By Proposition~\ref{prop:tauY},
we have a surjective map
$$
\tau_{[C]}:=\tau_{Y_C}\;:\; \mathord{\rm Emb} (K)\;\mathbin{\to \hskip -7pt \to}\; \wt{\GGG}_{[C]}.
$$
\par
\medskip
As an illustration of our Zariski-van Kampen method,
we calculate $\tau_{[C_0]}$
for  a reducible maximizing sextic $C_0$ of type $A_{10}+A_{9}$
defined over $K=\mathord{\mathbb  Q}(\sqrt{5})$ by the homogeneous equation
\begin{equation}\label{eq:zGH}
z \cdot (\;G(x,y,z)\;+\;\alpha \cdot H(x,y,z)\;)\;=\;0,
\end{equation}
where $\alpha^2=5$ and
\erase{
\begin{eqnarray*}
G(x, y, z)&=&-9\,{x}^{4}z-14\,{x}^{3}yz+58\,{x}^{3}{z}^{2}-48\,{x}^{2}{y}^{2}z
-64\,
{x}^{2}y{z}^{2}+10\,{x}^{2}{z}^{3}+\\&&+108\,x{y}^{3}z-20\,x{y}^{2}{z}^{2}-
44\,{y}^{5}+10\,{y}^{4}z,\\
H(x, y, z)&=&5\,{x}^{4}z+10\,{x}^{3}yz-30\,{x}^{3}{z}^{2}+30\,{x}^{2}{y}^{2}z+20\,{
x}^{2}y{z}^{2}-40\,x{y}^{3}z+20\,{y}^{5}.
\end{eqnarray*}
}
\begin{eqnarray*
G(x, y, z)&=&-9\,{x}^{4}z-14\,{x}^{3}yz+58\,{x}^{3}{z}^{2}-48\,{x}^{2}{y}^{2}z
-64\,{x}^{2}y{z}^{2}+\\&&+10\,{x}^{2}{z}^{3}+108\,x{y}^{3}z-20\,x{y}^{2}{z}^{2}-
44\,{y}^{5}+10\,{y}^{4}z,\\
H(x, y, z)&=&5\,{x}^{4}z+10\,{x}^{3}yz-30\,{x}^{3}{z}^{2}+30\,{x}^{2}{y}^{2}z+20\,{
x}^{2}y{z}^{2}-\\&&-40\,x{y}^{3}z+20\,{y}^{5}.
\end{eqnarray*}
This equation was discovered by means of   Roczen's result~\cite{MR1175728}
(see~\S\ref{sec:equation}).
\par
\medskip
We can calculate $\mathord{\rm NS} (X_{C_0})$ by  the method of Yang~\cite{MR1387816}, and
obtain
$$
\wt{\GGG}_{[C_0]}\;=\;\{\;\wt{L}[8,3,8], \,\wt{L}[2,1,28]\;\}.
$$
Let $\sigma_{\pm}$ be the embeddings of $\mathord{\mathbb  Q}(\alpha)$ into $\mathord{\mathbb C}$
given by $\sigma_\pm (\alpha)=\pm\sqrt{5}$.
We have two surjective maps
from $\mathord{\rm Emb} (\mathord{\mathbb  Q}(\alpha))=\{\sigma_+, \sigma_-\}$ to $\wt{\GGG}_{[C_0]}$.
Remark that,
since the two complex maximizing sextics $C_0\sp{\sigma_+}$  and $C_0\sp{\sigma_-}$
cannot be distinguished by any algebraic methods,
\emph{we have to employ some transcendental method
to determine which surjective map is the map $\tau_{[C_0]}$}.
By the method described in \S\ref{sec:ZvK} of this paper,
we obtain the following:
\begin{proposition}\label{prop:theexample}
$$
\tau_{[C_0]}(\sigma_+)=\wt{L}[2,1,28],
\quad
\tau_{[C_0]}(\sigma_-)=\wt{L}[8,3,8].
$$
\end{proposition}
We have shown in~\cite{AZP} and~\cite{nonhomeo} that,
for a complex maximizing sextic $C$,
the transcendental lattice $\mathord{\rm T}_{[C]}:=\mathord{\rm T} (Y_C)$
of the double covering $Y_C\to \P^2$ branching exactly along $C$  is a topological invariant
of $(\P^2, C)$.
Thus the curves $C_0^{\sigma_+}$ and $C_0^{\sigma_-}$
form an \emph{arithmetic Zariski pair}.
(See~\cite{AZP} for the definition.)
The proof of Proposition~\ref{prop:theexample}
illustrates very explicitly how the action of the Galois group of $\mathord{\mathbb  Q}(\alpha)$ over $\mathord{\mathbb  Q}$
affects the topology of the embedding of $C_0$ into  $\P^2$.
\par
\medskip
The first example of arithmetic Zariski pairs was discovered by
Artal, Carmona and Cogolludo~\cite{MR2247887} in degree $12$
by means of the \emph{braid monodromy}.
It will be an interesting problem to investigate the relation between
 the  braid monodromy of a maximizing sextic $C\subset \P^2$ and our lattice invariant $\mathord{\rm T}_{[C]}$.
 %
\par
\medskip
In the study of Zariski pairs of complex plane curves,
the topological fundamental groups of the complements (or its variations like  the Alexander polynomials)
have been used to distinguish the topological types.
(See, for example, \cite{MR1257321}, \cite{MR1167373} or \cite{MR1421396}
for the oldest  example of Zariski pairs of $6$-cuspidal  sextics~\cite{MR1506719, MR1507244}.)
We can calculate the fundamental groups
$\pi_1 (\P^2\setminus C_0^{\sigma_+})$
and
$\pi_1 (\P^2\setminus C_0^{\sigma_-})$
of our example
in terms of generators and relations by the classical Zariski-van Kampen theorem.
 (See, for example,~\cite{MR1341806, MR1952329}.)
It will be an interesting problem to determine whether these two groups  are isomorphic or not.
Note that,
by the theory of algebraic fundamental groups,
their profinite completions  are isomorphic.
\par
\medskip
The plan of this paper is as follows.
In \S\ref{sec:tauY}, we prove Proposition~\ref{prop:tauY}.
In \S\ref{sec:ZvK},
we present the Zariski-van Kampen method for the calculation of the transcendental lattice
in full generality.
In \S\ref{sec:theexample}, we  apply this method to the
complex maximizing sextics $C_0^{\sigma_\pm}$
and prove Proposition~\ref{prop:theexample}.
In \S\ref{sec:equation},
we explain how we have obtained   the equation~\eqref{eq:zGH}
of $C_0$.
\par
\medskip
Thanks are due to
the referee for his/her  comments and suggestions
on the first version of  this paper.
\section{The map $\tau_Y$}\label{sec:tauY}
We recall the proof of Theorem~\ref{thm:SS},
and prove Proposition~\ref{prop:tauY}.
The main tool is Theorem~\ref{thm:SI} due to Shioda and Inose~\cite{MR0441982}.
\begin{proof}[Proof of Theorem~\ref{thm:SS}]
It is easy to see that
the image of $\tau_X$ is contained in $\wt{\GGG}_X$.
(See Theorem 2 in~\cite{tssK3} or Proposition 3.5 in~\cite{nonhomeo}.)
In~\cite{tssK3} and~\cite{MR2346573},
using Shioda-Inose construction,
we constructed a singular $K3$ surface
$X^0$ defined over a number field $F^0$
such that $\mathord{\rm NS}(X)\cong \mathord{\rm NS}(X^0)$ (and hence $\wt{\GGG}_X=\wt{\GGG}_{X^0}$)  holds and
that the image of $\tau_{X^0}$ coincides with
$\wt{\GGG}_X$.
(See also \S4 of~\cite{nonhomeo}.)
We choose an arbitrary $\sigma\in \mathord{\rm Emb}(F)$.
Then there exists $\sigma^0\in \mathord{\rm Emb}(F^0)$
such that $\tau_{X^0} (\sigma^0)=\tau_X(\sigma)$.
Since  $(X^0)^{\sigma^0}$ and $X^\sigma$ are isomorphic over $\mathord{\mathbb C}$ by Theorem~\ref{thm:SI},
there exists a number field $M\subset \mathord{\mathbb C}$  containing  both $\sigma^0(F^0)$ and $\sigma (F)$
such that we have an isomorphism
$$
X^0\otimes _{F^0, \sigma^0} M\;\cong\; X\otimes _{F, \sigma} M
$$
over $M$.
Consider the commutative diagram
$$
\begin{array}{ccccccc}
&&&\mathord{\rm Emb} (F^0)& && \\
&\raise 6pt \llap{\small$\rho_{M, \sigma^0(F^0)}$}\nearrow &&& &\searrow \raise 6pt \rlap{\small$\tau_{X^0}$}& \\
\mathord{\rm Emb} (M)&&&\longrightarrow \hskip -16pt \raise 10pt \hbox{\small\makebox(0,0){$\tau_{X^0\otimes M}=\tau_{X\otimes M}$}}
&&& \wt{\GGG}_X=\wt{\GGG}_{X^0}\\
&\raise -6pt \llap{\small$\rho_{M, \sigma(F)}$}\searrow &&&&\nearrow \raise -6pt \rlap{\small$\tau_{X}$}& \\
&&&\mathord{\rm Emb} (F) &&&
\end{array},
$$
where $\rho_{M, \sigma^0(F^0)}$ and $\rho_{M, \sigma(F)}$ are
the natural surjective restriction maps.
The surjectivity of
$\tau_{X}$ then follows from the surjectivity of $\tau_{X^0}$.
\end{proof}
\begin{proof}[Proof of Proposition~\ref{prop:tauY}]
Let $\sigma_1$ and $\sigma_2$ be elements
of $\mathord{\rm Emb}(F)$ such that $\sigma_1|K=\sigma_2|K$.
We put
$$
\sigma_K:=\sigma_1|K=\sigma_2|K\in \mathord{\rm Emb}(K).
$$
Then the complex surfaces $X^{\sigma_1}$ and $X^{\sigma_2}$
are desingularizations of the complex surface $Y^{\sigma_K}$.
Hence Proposition~\ref{prop:tauY} follows from
Proposition~\ref{prop:indep} below.
\end{proof}
\begin{proposition}\label{prop:indep}
Let $S_1\sp{\sim}$ and $S_2\sp{\sim}$ be two desingularizations of
a reduced irreducible complex projective surface $S$.
Then $\mathord{\rm T}(S_1\sp{\sim})\cong \mathord{\rm T}(S_2\sp{\sim})$.
If $S_1\sp{\sim}$ and $S_2\sp{\sim}$ are singular $K3$ surfaces,
then $\wt{\Tr}(S_1\sp{\sim})\cong \wt{\Tr}(S_2\sp{\sim})$.
\end{proposition}
\begin{proof}
Using  a desingularization of $S_1\sp{\sim}\times_S S_2\sp{\sim}$,
we obtain a complex smooth projective surface
$\Sigma$ with birational morphisms
$\Sigma\to S_1\sp{\sim}$ and $\Sigma\to S_2\sp{\sim}$.
Since the transcendental lattice of a complex smooth projective surface
is invariant under a blowing-up,
and  any birational morphism between  smooth projective surfaces factors  into a composite of blowing-ups,
we have $\mathord{\rm T}(\Sigma)\cong\mathord{\rm T}(S_1\sp{\sim})$ and $\mathord{\rm T}(\Sigma)\cong\mathord{\rm T}(S_2\sp{\sim})$.
\end{proof}
\section{Zariski-van Kampen method for transcendental lattices}\label{sec:ZvK}
For a $\mathord{\mathbb Z}$-module $A$,
we denote by
$$
A^{\rm tf}:=A/(\textrm{the torsion part})
$$
the maximal torsion-free quotient of $A$.
If we have a bilinear form
$A\times A\to \mathord{\mathbb Z}$,
then it induces a canonical bilinear form
$A^{\rm tf}\times A^{\rm tf}\to \mathord{\mathbb Z}$.
\par
\medskip
Let $S$ be a reduced irreducible complex projective surface.
Our goal is  to calculate $\mathord{\rm T}(S)$.
Let $\delta:S\sp{\sim}\to S$ be a desingularization.
We choose a reduced curve $D$ on $S$
with the following properties:
\begin{itemize}
\item[(D1)] the classes of  irreducible components of the total transform $D^\sim\subset S\sp{\sim}$
of $D$ span $\mathord{\rm NS}(S\sp{\sim})\otimes\mathord{\mathbb  Q}$ over $\mathord{\mathbb  Q}$, and
\item[(D2)] the desingularization $\delta$ induces an isomorphism $S\sp{\sim}\setminus D\sp\sim \cong S\setminus D$.
\end{itemize}
We put
$$
S^0:=S\setminus D,
$$
and
consider the free $\mathord{\mathbb Z}$-module
$$
\mathord{\rm H}_2(S^0):=\mathord{\rm H}_2(S^0, \mathord{\mathbb Z})^{\rm tf}
$$
with the intersection paring
$$
\iota\;:\; \mathord{\rm H}_2(S^0)\times \mathord{\rm H}_2(S^0) \to \mathord{\mathbb Z}.
$$
We put
\begin{equation}\label{eq:defI}
I(S^0):=\shortset{x\in \mathord{\rm H}_2(S^0)}{\iota(x,y)=0\;\textrm{for any}\; y\in \mathord{\rm H}_2(S^0)},
\end{equation}
and set
$$
V_2(S^0):= \mathord{\rm H}_2(S^0)/I(S^0).
$$
Then $V_2(S^0)$ is a free $\mathord{\mathbb Z}$-module, and the intersection paring $\iota$ induces
a non-degenerate symmetric bilinear form
$$
\bar\iota\;:\; V_2(S^0)\times V_2(S^0) \to \mathord{\mathbb Z}.
$$
\begin{proposition}\label{prop:VT}
The transcendental lattice $\mathord{\rm T}(S)=\mathord{\rm T}(S\sp{\sim})$ is isomorphic
to the lattice $(V_2(S^0), \bar\iota)$.
\end{proposition}
\begin{proof}
By the condition (D2),
we can regard $S^0$ as a Zariski open subset of $S\sp{\sim}$.
Consider the homomorphism
$$
j_* \;:\; \mathord{\rm H}_2(S^0) \to \mathord{\rm H}_2 (S\sp{\sim}):=\mathord{\rm H}_2 (S\sp{\sim},\mathord{\mathbb Z})^{\rm tf}
$$
induced by the inclusion $j:S^0\hookrightarrow S\sp{\sim}$.
Under the isomorphism of lattices
$$
 \mathord{\rm H}_2 (S\sp{\sim})\cong  \mathord{\rm H}^2 (S\sp{\sim}):=\mathord{\rm H}^2 (S\sp{\sim},\mathord{\mathbb Z})^{\rm tf}
$$
induced by the Poincar\'e duality,
the image of $j_*$ is contained in $\mathord{\rm T} (S\sp{\sim})\subset \mathord{\rm H}^2 (S\sp{\sim})$
by the condition (D1) on $D$.
Using the argument in the proof of Theorem 2.6 of~\cite{AZP}
or Theorem 2.1 of~\cite{nonhomeo},
we see that the homomorphism
\begin{equation*}\label{eq:surj}
j_* \;:\; \mathord{\rm H}_2(S^0) \to \mathord{\rm T} (S\sp{\sim})
\end{equation*}
is surjective.
Note that we have
\begin{equation*}\label{eq:iota}
\iota(x, y)=(j_*(x), j_*(y))_T
\end{equation*}
for any $x, y\in \mathord{\rm H}_2(S^0)$,
where $(\phantom{a}, \phantom{a})_T$ is the cup-product on $\mathord{\rm T} (S\sp{\sim})$.
Since $(\phantom{a}, \phantom{a})_T$ is non-degenerate,
we conclude that   $\operatorname{\rm Ker}\nolimits j_*=I(S^0)$.
\end{proof}
Proposition~\ref{prop:VT} shows that, in order to obtain $\mathord{\rm T}(S)$,
it is enough to  calculate $\mathord{\rm H}_2(S^0)$ and $\iota$.
Enlarging $D$ if necessary,
we have a surjective morphism
$$
\phi \;:\;S^0\to U
$$
onto a Zariski open subset $U$ of an  affine line $\mathord{\mathbb A}^1$
such that its general fiber is a connected  Riemann surface.
By the condition (D1) on $D$,
the general fiber  of $\phi$ is non-compact.
Let $\overline{S^0}$ be a smooth irreducible projective surface containing $S^0$
as a Zariski open subset such that
$\phi$ extends to a morphism
$$
\bar\phi\;:\; \overline{S^0}\to \P^1.
$$
Let $V_1, \dots, V_M$ and $H_1, \dots, H_N$
be the irreducible components of the boundary $\overline{S^0}\setminus S^0$,
where $V_1, \dots, V_N$ are the vertical components
(that is, $\bar\phi(V_i)$ is a point),
and $H_1, \dots, H_M$ are the horizontal components
(that is, $\bar\phi(H_j)=\P^1$).
Since  the general fiber of $\phi$ is non-compact,
we have at least one horizontal component.
We put
$$
\mathord{\mathbb A}^1\setminus U=\{p_1, \dots, p_m\}.
$$
Adding to $D$ some fibers of $\phi$ and making  $U$ smaller if necessary,
we can assume the following:
\begin{enumerate}
\item the surjective morphism $\phi: S^0\to U$ has only ordinary critical points,
\item\label{etale} $\bar\phi|\cup_j H_j: \cup_j H_j\to \P^1$ is \'etale over $U$, and
\item\label{V} $V_1\cup\dots\cup V_N=\bar\phi\sp{-1}(\infty)\cup \bar\phi\sp{-1}(p_1)\cup \dots\cup \bar\phi\sp{-1}(p_m)$,
where $\{\infty\}=\P^1\setminus\mathord{\mathbb A}^1$.
\end{enumerate}
Note that $\bar\phi$ has no critical points on $(\cup H_j)\cap \bar\phi\sp{-1} (U)$
by the condition (\ref{etale}).
We  denote by $c_1, \dots, c_n\in U$ the critical values of $\phi$,
and put
$$
U^\sharp :=U\setminus\{c_1, \dots, c_n\}.
$$
By the assumptions,
$\phi$ is locally trivial (in the category of topological spaces and continuous maps)
over $U\sp\sharp$
with the fiber being a connected Riemann surface of genus $g$ with $r$ punctured points,
where $r>0$ is the degree of $\bar\phi|\cup_j H_j: \cup_j H_j\to \P^1$.
We then choose a base point $b\in U\sp\sharp$,
and put
$$
F_b:=\phi\sp{-1}(b).
$$
For each $p_i\in \mathord{\mathbb A}^1\setminus U$,
we choose a loop
$$
\lambda_i \;:\; (I, \partial I)\to (U\sp\sharp, b)
$$
that is sufficiently smooth  and injective in the sense that
$\lambda_i(t)=\lambda_i(t\sp\prime)$ holds only when $t=t\sp\prime$ or $\{t, t\sp\prime\}=\partial I$,
and that defines the same element in $\pi_1(U\sp\sharp, b)$
as a simple loop (a lasso) around $p_i$ in $U\sp\sharp$.
For each critical value $c_j\in U\setminus U\sp\sharp$,
we choose a sufficiently smooth and injective  path
$$
\gamma_j \;:\; I \to U
$$
such that $\gamma_j(0)=b$, $\gamma_j(1)=c_j$ and $\gamma_j(t)\in U\sp\sharp$ for $t<1$.
We choose these loops $\lambda_i$ and paths $\gamma_j$ in such a way that
any two of them intersect only at $b$.
Then,
by a suitable self-homeomorphism of $\mathord{\mathbb A}^1$,
the objects $b$, $p_i$, $c_j$, $\lambda_i$ and $\gamma_j$ on $\mathord{\mathbb A}^1$
are mapped as in Figure~\ref{fig:A}.
\input figA
In particular,
the union  $B$
of $\lambda_i$ and  $\gamma_j$
is a strong deformation retract of $U$.
Note that  $\phi$ is locally trivial over $U\setminus B$.
\par
\medskip
Let $\mathord{\mathbb S}^1$ be an oriented one-dimensional sphere.
We fix a system of oriented simple closed curves
$$
a_\nu \;:\;\mathord{\mathbb S}^1\hookrightarrow  F_b \quad(\nu=1, \dots, 2g+r-1)
$$
on $F_b$ in such a way that their union
$\bigcup a_{\nu}(\mathord{\mathbb S}^1)$
is a strong deformation retract of $F_b$.
In particular, we have
$$
\mathord{\rm H}_1 (F_b, \mathord{\mathbb Z})=\bigoplus\; \mathord{\mathbb Z}[a_\nu],
$$
where $[a_\nu]$ is the homology class of $a_\nu$.
For each $p_i\in\mathord{\mathbb A}^1\setminus U$ and $a_\nu$, let
$$
\Lambda_{i, \nu}\;:\; \mathord{\mathbb S}^1\times I\hookrightarrow S^0
$$
be an embedding such that the diagram
$$
\begin{array}{ccc}
\mathord{\mathbb S}^1\times I & \maprightsp{\Lambda_{i, \nu}} & S^0 \\
\llap{\scriptsize $\mathord{\rm pr}$}\phantom{\Big\downarrow}\hskip -8pt \downarrow & & \phantom{\Big\downarrow}\hskip -8pt \downarrow \rlap{\scriptsize $\phi$}\\
I &\maprightsb{\lambda_i} & U
\end{array}
$$
commutes and
that
$$
\Lambda_{i, \nu}| \mathord{\mathbb S}^1\times \{0\} \;:\; \mathord{\mathbb S}^1\hookrightarrow F_b
$$
is equal to $a_\nu$.
We put
$$
M_i(a_\nu):=\Lambda_{i, \nu}| \mathord{\mathbb S}^1\times \{1\} \;:\; \mathord{\mathbb S}^1\hookrightarrow F_b,
$$
 where $M_i$ stands for   the \emph{monodromy} along $\lambda_i$, and
 denote the homology class of $M_i(a_\nu)$ by
$$
M_i([a_\nu])\in \mathord{\rm H}_1 (F_b, \mathord{\mathbb Z})
$$

Let $\Theta$ be the topological space obtained from $\mathord{\mathbb S}^1\times I$
by contracting $ \mathord{\mathbb S}^1\times \{1\}$ to a point $v\in \Theta$;
that is, $\Theta$ is a cone over $\mathord{\mathbb S}^1$ with the vertex $v$.
Let $\mathord{\rm pr}: \Theta\to I$ be the natural projection.
Let  $c_j\in U\setminus U\sp\sharp$ be a critical value of $\phi$,
and let $\tilde{c}_j^1, \dots, \tilde{c}_j^m$ be the critical points of $\phi$ over $c_j$.
For each critical point $\tilde{c}_j^k\in \phi\sp{-1}(c_j)$,
we choose a \emph{thimble}
$$
\Gamma_j^k \;:\; \Theta\hookrightarrow S^0
$$
along the path $\gamma_j$ corresponding to the ordinary node $\tilde{c}_j^k$ of $\phi\sp{-1}(c_j)$.
 Namely,  the thimble $\Gamma_j^k$ is an embedding
such that
$$
\begin{array}{ccc}
\Theta & \maprightsp{\Gamma_j^k} & S^0 \\
\llap{\scriptsize $\mathord{\rm pr}$}\phantom{\Big\downarrow}\hskip -8pt \downarrow & & \phantom{\Big\downarrow}\hskip -8pt \downarrow \rlap{\scriptsize $\phi$}\\
I &\maprightsb{\gamma_j} & U
\end{array}
$$
commutes,
and that $\Gamma_j^k(v)=\tilde{c}_j^k$.
(See~\cite{MR592569} for thimbles and vanishing cycles.)
Then the simple closed curve
$$
\sigma_j^k:=\Gamma_j^k|\mathord{\rm pr}\sp{-1}(0)=-\partial \Gamma_j^k\;:\; \mathord{\mathbb S}^1\hookrightarrow F_b
$$
on $F_b$
represents the \emph{vanishing cycle} for the critical point $\tilde{c}_j^k$
along $\gamma_j$.
We denote its homology class by
$$
[\sigma_j^k]\in \mathord{\rm H}_1 (F_b, \mathord{\mathbb Z})
$$
We can assume that
$\Gamma_j^k$ and $\Gamma_j^{k\sp\prime}$ are disjoint if $k\ne k\sp\prime$.
\begin{remark}
There are two choices of the orientation of the thimble $\Gamma_j^k$ (and  hence
of the vanishing cycle $\sigma_j^k=-\partial \Gamma_j^k$).
\end{remark}
\begin{figure}
\includegraphics{figthimble.ps}%
\rlap{\hskip -223pt \raise 3pt \hbox{$b$}}%
\rlap{\hskip -23pt \raise 3pt \hbox{$c_j$}}%
\rlap{\hskip -127pt \raise 3pt \hbox{$\gamma_j$}}%
\rlap{\hskip -14pt \raise 100pt \hbox{$\tilde{c}_j^k$}}%
\rlap{\hskip -245pt \raise 100pt \hbox{$\sigma_j^k$}}%
\rlap{\hskip -127pt \raise 100pt \hbox{$\Gamma_j^k$}}%
\caption{Thimble}
\end{figure}
Then the union
$$
F_b \;\cup\;  \bigcup \Lambda_{i, \nu} (\mathord{\mathbb S}^1\times I) \;\cup\;  \bigcup \Gamma_j^k(\Theta)
$$
is homotopically equivalent to  $S^0$.
Since  the  $1$-dimensional CW-complex
$\bigcup a_{\nu}(\mathord{\mathbb S}^1)$
is a strong deformation retract of $F_b$,
the homology group  $\mathord{\rm H}_2(S^0, \mathord{\mathbb Z})$ is equal to the kernel of the homomorphism
$$
\partial\;:\; \bigoplus \mathord{\mathbb Z}[\Lambda_{i, \nu} ] \oplus \bigoplus \mathord{\mathbb Z}[\Gamma_j^k]\;\to\; \mathord{\rm H}_1 (F_b, \mathord{\mathbb Z})=\bigoplus\mathord{\mathbb Z}[a_\nu]
$$
given by
$$
\partial[\Lambda_{i, \nu} ]=M_i([a_\nu])-[a_\nu] \quad\rmand\quad
\partial[\Gamma_j^k]=-[\sigma_j^k].
$$
\par
\medskip
The intersection pairing  on $\mathord{\rm H}_2(S^0, \mathord{\mathbb Z})$ is calculated by perturbing
the system  $(\lambda_i, \gamma_j)$ of loops and paths
with the base point $b$ to a system  $(\lambda\sp\prime_i, \gamma\sp\prime_j)$
with the base point $b\sp\prime\ne b$.
We make the perturbation in such a way that the following hold.
\begin{itemize}
\item There exists a small open disk
$\Delta\subset U\sp\sharp$
containing both $b$ and $b\sp\prime$ such that
\begin{eqnarray*}
&&\lambda_i\sp{-1}(\Delta)=[0, s_i)\cup(1-r_i, 1], \quad
\lambda\sp{\prime-1}_i(\Delta)=[0, s\sp\prime_i)\cup(1-r\sp\prime_i, 1], \\
&&\gamma_j\sp{-1}(\Delta)=[0, u_j), \quad\quad
\gamma\sp{\prime-1}_j(\Delta)=[0, u\sp\prime_j),
\end{eqnarray*}
where $s_i, r_i, s\sp\prime_i, r\sp\prime_i, u_j, u\sp\prime_j$ are small positive real numbers.
\item
If $\lambda_i$ intersects $\lambda\sp\prime_{i\sp\prime}$ or $\gamma\sp\prime_{j\sp\prime}$,
then their intersection points are contained in $\Delta\setminus\{b, b\sp\prime\}$,
and the intersections are transverse.
\item
If $\gamma_j$ intersects $\lambda\sp\prime_{i\sp\prime}$ or $\gamma\sp\prime_{j\sp\prime}$ with $j\sp\prime\ne j$,
then their intersection points are contained in $\Delta\setminus\{b, b\sp\prime\}$,
and the intersections are transverse.
\item
Any  intersection point of $\gamma_j$ and $\gamma\sp\prime_j$ is either  the common end-point $c_j$,
or a transversal intersection point contained in $\Delta\setminus\{b, b\sp\prime\}$.
\end{itemize}
We then perturb the topological $2$-chains
$\Lambda_{i, \nu}$ over $\lambda_{i}$  and $\Gamma_j^k$ over  $\gamma_j$
to topological $2$-chains
$\Lambda\sp\prime_{i, \nu}$over $\lambda\sp\prime_{i}$  and $\Gamma\sp{\prime k}_j$  over $\gamma\sp\prime_j$,
respectively.
Let $T$ be one of $\Lambda_{i, \nu}$ or $\Gamma_j^k$,
and let $t$ be the loop or the path over which $T$ locates.
Let $T\sp\prime$ be one of $\Lambda\sp\prime_{i, \nu}$ or $\Gamma\sp{\prime k}_j$
over the loop or the path $t\sp\prime$.
We can make the perturbation in such a way that
$T$ and $T\sp\prime$ intersect transversely at each intersection points.
Suppose that
$$
t(I)\cap t\sp\prime(I)\cap \Delta =\{q_1, \dots, q_l\}.
$$
Then $T\cap T\sp\prime$ are contained in the union $\bigcup_{\mu=1}^l \phi\sp{-1}(q_\mu)$ of fibers
except for the case where $T=\Gamma_j^k$ and $T\sp\prime=\Gamma\sp{\prime k}_j$
for some $j$ and $k$.
If  $T=\Gamma_j^k$ and $T\sp\prime=\Gamma\sp{\prime k}_j$,
then $T$ and $T\sp\prime$ also intersect at the critical point $\tilde{c}_j^k$
transversely with the local intersection number $-1$.
(See Lemma 4.1 of~\cite{MR1433320}.)
For each $q_\mu$,
let $\theta_\mu$ and $\theta\sp\prime_\mu$
be the $1$-cycles on the open Riemann surface $\phi\sp{-1}(q_\mu)$ given by
\begin{eqnarray*}
\theta_\mu&:=&T\phantom{\sp\prime}|\mathord{\mathbb S}^1\times\{w_\mu\}\;:\; \mathord{\mathbb S}^1\to \phi\sp{-1}(q_\mu), \quad
\textrm{where $t(w_\mu)\,\,=q_\mu$},\\
\theta\sp\prime_\mu&:=&T\sp\prime|\mathord{\mathbb S}^1\times\{w\sp\prime_\mu\}\;:\; \mathord{\mathbb S}^1\to \phi\sp{-1}(q_\mu), \quad
\textrm{where $t\sp\prime(w\sp\prime_\mu)=q_\mu$}.
\end{eqnarray*}
We denote by $(t, t\sp\prime)_{\mu}$ the local intersection number of the $1$-chains $t$ and $t\sp\prime$
on $U$
at $q_\mu$, which is $1$ or $-1$ by the assumption on the perturbation.
We also denote by $(\theta_\mu, \theta\sp\prime_\mu)_{\mu}$ the intersection number
of $\theta_\mu$ and $\theta\sp\prime_\mu$ on the  Riemann surface $\phi\sp{-1}(q_\mu)$.
Then the intersection number $(T, T\sp\prime)$ of $T$ and $T\sp\prime$
is equal to
\begin{equation}\label{eq:TT}
(T, T\sp\prime)=- \sum_{\mu=1}^l  (t, t\sp\prime)_{\mu} (\theta_\mu, \theta\sp\prime_\mu)_{\mu}+\delta,
\end{equation}
where
$$
\delta:=\begin{cases} -1 & \textrm{if $T=\Gamma_j^k$ and $T\sp\prime=\Gamma\sp{\prime k}_j$ for some $j$ and $k$}, \\
0 & \textrm{otherwise}.
\end{cases}
$$
The number $(\theta_\mu, \theta\sp\prime_\mu)_{\mu}$
is calculated as follows.
Let
$$
(\phantom{a},\phantom{a})_F\;:\; \mathord{\rm H}_1(F_b, \mathord{\mathbb Z})\times \mathord{\rm H}_1(F_b, \mathord{\mathbb Z}) \to \mathord{\mathbb Z}
$$
be the intersection pairing  on $\mathord{\rm H}_1(F_b, \mathord{\mathbb Z})$,
which is anti-symmetric.
If $T=\Gamma_j^k$,
then the $1$-cycle $\theta_\mu$ on $\phi\sp{-1} (q_\mu)$
can be deformed to the vanishing cycle $\sigma_j^k=-\partial \Gamma_j^k$ on $F_b$ along the  path $t|[0, w_\mu]$
in $\Delta$.
We put
$$
[\tilde\theta_\mu]:=[\sigma_j^k]\in \mathord{\rm H}_1(F_b, \mathord{\mathbb Z}).
$$
Suppose that $T=\Lambda_{i, \nu}$.
Then we have
$t\sp{-1} (\Delta)=\lambda_i\sp{-1}(\Delta)=[0, s)\cup(1-r, 1]$,
where $s$ and $r$ are small positive real numbers.
If the number $w_\mu$ such that  $t(w_\mu)=q_\mu$ is contained in $[0, s)$,
then $\theta_\mu$ can be deformed  to the $1$-cycle
$a_\nu$ on $F_b$ along the  path $t|[0, w_\mu]$
in $\Delta$.
If  $w_\mu\in (1-r, 1]$,
then $\theta_\mu$ can be deformed  to the $1$-cycle
$M_i(a_\nu)$ on $F_b$ along the  path $t|[w_\mu, 1]$
in $\Delta$.
We define $[\tilde\theta_\mu]\in \mathord{\rm H}_1(F_b, \mathord{\mathbb Z})$ by
$$
[\tilde\theta_\mu]:=
\begin{cases}
[a_\nu] &\textrm{if $w_\mu\in [0, s)$}, \\
M_i([a_\nu]) &\textrm{if $w_\mu\in (1-r, 1]$}.
\end{cases}
$$
We define $[\tilde\theta\sp\prime_\mu]\in \mathord{\rm H}_1(F_b, \mathord{\mathbb Z})$
from $T\sp\prime$ in the same way.
Since $\phi$ is topologically  trivial over $\Delta$,
we have
\begin{equation}\label{eq:tt}
(\theta_\mu, \theta\sp\prime_\mu)_{\mu}=([\tilde\theta_\mu], [\tilde\theta\sp\prime_\mu])_F.
\end{equation}
The formulae~\eqref{eq:TT} and~\eqref{eq:tt}
give the intersection number $(T, T\sp\prime)$ of topological $2$-chains $T$ and $T\sp\prime$.
Even though the number $(T, T\sp\prime)$ depends on the choice of the perturbation,
it gives the symmetric intersection paring on $\operatorname{\rm Ker}\nolimits\partial=\mathord{\rm H}_2(S^0, \mathord{\mathbb Z})$.
Thus we obtain $\mathord{\rm H}_2(S^0)$ and $\iota$.
\section{Maximizing sextics of type $A_{10}+A_{9}$}\label{sec:theexample}
Recall from Introduction that  $\mathord{\mathcal L}$ (resp.~$\wt{\LLL}$)
is  the set
of isomorphism classes of even positive-definite lattices (resp.~oriented  lattices) of rank $2$.
\begin{definition}\label{def:real}
Let $\varphi: \wt{\LLL}\to\mathord{\mathcal L}$ be the map
of forgetting orientation.
We say that $T\in \mathord{\mathcal L}$ is \emph{real} if  $\varphi\sp{-1}(T)$
consists of a single element, and
 that $\ori{T}\in \wt{\LLL}$ is \emph{real} if
$\varphi(\ori{T}) \in \mathord{\mathcal L}$ is real.
\end{definition}
Let $S$ be a complex singular $K3$ surface,
and let $\overline{S}$ denote $S\otimes_{\mathord{\mathbb C}, \bar{\phantom{z}}}\mathord{\mathbb C}$,
where $\bar{\phantom{z}}:\mathord{\mathbb C}\to\mathord{\mathbb C}$ is the  conjugate over $\mathord{\mathbb R}$.
Then $\wt{\Tr} (\overline{S})$ is the reverse of $\wt{\Tr}(S)$;
that is, $\varphi\sp{-1}(\varphi(\wt{\Tr}(S)))=\{\wt{\Tr}(S), \wt{\Tr} (\overline{S})\}$.
Therefore $\wt{\Tr}(S)$ is real if and only if $S$ and  $\overline{S}$ are isomorphic.
In particular, if $S$ is defined over $\mathord{\mathbb R}$,
then $\wt{\Tr}(S)$ is real.
\begin{remark}
It is known that every element $\ori{T}$  of $\wt{\LLL}$ is represented by a unique matrix $[2a, b, 2c]\in \mathord{\mathcal M}$
with
$$
-a<b\leq a\leq c,\;\;\textrm{with $b\geq 0$ if $a = c$},
$$
and  $\ori{T}$ is \emph{not} real if and only if $0<|b|<a<c$ holds.
See~\cite[Chapter 15]{MR1662447}.
\end{remark}
\par
\medskip
By the method of Yang~\cite{MR1387816} and Degtyarev~\cite{MR2357681},
we see the following facts.
(See also~\cite{AZP}.)
There are four  connected components in the moduli space of complex maximizing
sextics of type
$A_{10}+A_{9}$.
The members of two of them are  irreducible sextics,
and their oriented transcendental lattices
are
$$
\wt{L}[10, 0, 22]
\quad\rmand\quad
\wt{L}[2, 0, 110]\qquad \textrm{(both are real)}.
$$
The members of  the other two are  reducible.
Each of them is a union of a line and an irreducible quintic, and
their oriented transcendental lattices
are
$$
\wt{L}[8, 3, 8]
\quad\rmand\quad
\wt{L}[2, 1, 28]\qquad \textrm{(both are real)}.
$$
We will consider these reducible sextics $C_0$,
whose defining equation is given by~\eqref{eq:zGH}.
\par
\medskip
For simplicity, we write
$C\sp\pm$ for $C_0^{\sigma_\pm}$,
$Y\sp\pm$ for   $Y_{C_0^{\sigma_\pm}}$ and $X^\pm$ for $X_{C_0^{\sigma_\pm}}$.
Let $D\sp\pm \subset Y\sp\pm$ be the pull-back of the union of
the lines
$$
x=0\quad\rmand\quad z=0
$$
on $\P^2$.
Since the singular points
$$
[0:0:1]\;\;  \textrm{($A_{10}$)}\quad\rmand\quad[1:0:0]\;\;  \textrm{($A_{9}$)}
$$
of $C^\pm$ are on the union of these two lines,
the curve  $D\sp\pm$ satisfies the conditions~(D1) and~(D2)
in \S\ref{sec:ZvK} for  $S=Y\sp\pm$ and $S\sp{\sim}=X\sp\pm$.
We denote by $W\sp\pm$ the complement of $D\sp\pm$ in $Y\sp\pm$.
(We have  denoted $W\sp\pm$  by $S^0$ in \S\ref{sec:ZvK}.)
Let $\mathord{\mathbb A}^2_{(y, z)}$ be the affine part of $\P^2$ given by $x\ne 0$
with the affine coordinates $(y, z)$ obtained from $[x:y:z]$ by putting $x=1$,
and let $L\subset \mathord{\mathbb A}^2_{(y, z)}$ be the affine line
defined by $z=0$.
Then $W\sp\pm$ is the double cover of $\mathord{\mathbb A}^2_{(y, z)}\setminus L$
branching exactly along the union of $L$ and the smooth affine quintic curve $Q^\pm\subset \mathord{\mathbb A}^2_{(y, z)}$
defined by
$$
f^{\pm}(y, z):=G(1, y, z)\pm\sqrt{5}\cdot H(1, y, z)=0.
$$
Note that $Q^\pm$ intersects $L$ only at the origin,
and   the intersection multiplicity is $5$.
Let
$$
\pi^\pm \;:\; W^\pm \to \mathord{\mathbb A}_{(y, z)}^2\setminus L
$$
be the double covering.
We consider the projection
$$
p\;:\; \mathord{\mathbb A}^2_{(y, z)}\to \mathord{\mathbb A}^1_z
$$
defined by $p(y, z):=z$
onto an affine line with an affine coordinate $z$,
and the composite
$$
q^\pm \;:\; W^\pm \to \mathord{\mathbb A}_{(y, z)}^2\setminus L\to U:=\mathord{\mathbb A}^1_z\setminus\{0\}
$$
of $\pi^\pm$ and $p$,
which  serves as the surjective morphism $\phi$ in \S\ref{sec:ZvK}.
Calculating the discriminant  of $f^{\pm}(y, z)$ with respect to $y$, we see that
there are four  critical points of the finite covering
$$
p|Q^\pm \;:\; Q^\pm \to \mathord{\mathbb A}^1_z
$$
of degree $5$.
Three of them $R^\pm, S^\pm, \overline{S}^\pm$ are simple critical values,
where
\begin{eqnarray*}
&& R^+=0.42193...,\quad
\;\;S^+=0.23780...+0.24431...\cdot \sqrt{-1}, \quad\rmand\quad\\
&&R^-=0.12593...,\;\;\quad
S^-=27.542...+45.819...\cdot \sqrt{-1}.\;\;
\end{eqnarray*}
The value  $\overline{S}^\pm $  is the complex conjugate of $S^\pm$.
The critical point over $0\in \mathord{\mathbb A}^1_z$ is of multiplicity $5$.
The critical values of $q^\pm : W^\pm \to U$ are therefore
$R^\pm, S^\pm, \overline{S}^\pm$, and
the fiber of $q^\pm$ over each of them has only one ordinary node.
We choose a sufficiently small positive real number $b$ as a base point on $U$,
and define  the loop $\lambda$ and the paths
$\gamma^\pm_R$, $\gamma^\pm_S$, $\gamma^\pm _{\bar{S}}$ on $U$ as in
Figure~\ref{fig1}.
\begin{figure}
\hskip -160pt
\hbox{\includegraphics[scale=.8]{newfig1.ps}%
\rlap{\hskip -195.3pt \raise 109.5 pt \hbox{$0$}}%
\rlap{\hskip -115pt \raise 117.5 pt \hbox{$b$}}%
\rlap{\hskip -36pt \raise 109.4 pt \hbox{$R^\pm$}}%
\rlap{\hskip -115pt \raise 190 pt \hbox{$S^\pm$}}%
\rlap{\hskip -115pt \raise 28 pt \hbox{$\bar S^\pm$}}%
\rlap{\hskip -115pt \raise 150 pt \hbox{$\gamma^\pm_S$}}%
\rlap{\hskip -115pt \raise 72 pt \hbox{$\gamma^\pm_{\bar S}$}}%
\rlap{\hskip -77pt \raise 95 pt \hbox{$\gamma^\pm_R$}}%
\rlap{\hskip -255pt \raise 109.50 pt \hbox{$\lambda$}}%
}
\caption{The loop $\lambda$ and the paths $\gamma^\pm_{R}$, $\gamma^\pm_{S}$, ${\gamma}_{\bar S}^\pm$}
\label{fig1}
\end{figure}
For $z\in U$,
we put
$$
\mathord{\mathcal Q}^\pm (z):=(p|Q^\pm)\sp{-1} (z)=p\sp{-1} (z) \cap Q^\pm,
$$
and investigate the movement of the points $\mathord{\mathcal Q}^\pm (z)$ when $z$ moves on $U$
along the loop $\lambda$ and the paths $\gamma^\pm_R$, $\gamma^\pm_S$, $\gamma^\pm_{\bar{S}}$.
We put
$$
\mathord{\mathbb A}^1_y:=p\sp{-1} (b), \quad
F^\pm:=q^{\pm-1} (b)=\pi^{\pm-1}(\mathord{\mathbb A}^1_y)\subset W^\pm.
$$
Note that the morphism
$$
\pi^\pm|F^\pm\;:\; F^\pm\to \mathord{\mathbb A}^1_y
$$
 is the double covering
branching exactly at the five points
$\mathord{\mathcal Q}^\pm (b)\subset \mathord{\mathbb A}^1_y$.
These branching points $\mathord{\mathcal Q}^\pm (b)$ are depicted as  big dots in Figure~\ref{fig2}.
Hence $F^\pm$ is a Riemann surface of genus $2$  minus one point.
We choose a system of oriented simple closed curves
$$
a_\nu \;:\;\mathord{\mathbb S}^1 \hookrightarrow F^\pm\qquad(\nu=1, \dots, 5)
$$
in such a way
that their images by the double covering
$\pi^\pm|F^\pm: F^\pm\to \mathord{\mathbb A}^1_y$
 are  given in Figure~\ref{fig2},
and that  their  intersection numbers
on $F^\pm$ are
equal to
$$
([a_\nu],  [a_{\nu+1]})_F=-([a_{\nu+1}], [a_\nu] )_F=1
$$
for $\nu=1, \dots, 5$, where $a_6:=a_1$.
\begin{figure}
\begin{center}
\parbox{210pt}{
\includegraphics[scale=.7]{fig2.ps}%
\rlap{\hskip -25pt \raise 105 pt \hbox{$a_5$}}%
\rlap{\hskip -80pt \raise 184 pt \hbox{$a_1$}}%
\rlap{\hskip -180pt \raise 158 pt \hbox{$a_2$}}%
\rlap{\hskip -180pt \raise 44 pt \hbox{$a_3$}}%
\rlap{\hskip -80pt \raise 23 pt \hbox{$a_4$}}%
}
\end{center}
\caption{The system of simple closed curves on $F_b$}
\label{fig2}
\end{figure}
(Note that $([a_\nu], [a_{\nu\sp\prime}])_F=0$
except for the case  where $|\nu-\nu\sp\prime|= 1$ or $\{\nu, \nu\sp\prime\}=\{1, 5\}$.)
Then $a_1\cup\dots\cup a_4$
is a strong deformation retract of $F^\pm$, and
$[a_1], \dots, [a_4]$ form  a basis of
$\mathord{\rm H}_1(F^\pm, \mathord{\mathbb Z})$.
Moreover
 we have
$$
[a_5]=-[a_1]-[a_2]-[a_3]-[a_4].
$$
\par
\medskip
Since $Q\sp\pm$ is smooth at the origin and intersects $L$ with multiplicity $5$ at the origin,
the movement of the branching points $\mathord{\mathcal Q}^\pm(z)$ along the loop $\lambda$
is homotopically equivalent to the rotation around the origin of the angle $2\pi/5$.
Hence the  monodromy on the simple closed curves
is given by
$a_\nu\mapsto a_{\nu+1}$.
Let
$$
\Lambda_\nu\;:\; \mathord{\mathbb S}^1\times I\to W^\pm
$$
be the topological  $2$-chain
over $\lambda$ that connects $a_\nu$ and  $a_{\nu+1}$.
We have
$$
\partial [\Lambda_\nu]=[a_{\nu+1}]-[a_\nu].
$$
The movement of  the  branching points  $\mathord{\mathcal Q}^\pm (z)$
when $z$ moves from $b$ to $R^\pm$ along the path $\gamma_R^\pm$
is homotopically equivalent to the movement depicted in
 Figure~\ref{fig3}.
  %
 \begin{figure}
\parbox{170pt}{
\includegraphics[scale=.55]{fig3plus.ps}%
\\
\hfil $\mathord{\mathcal Q}^+(z)$ \hfil
}
\vrule height 100pt
\parbox{170pt}{
\includegraphics[scale=.55]{fig3minus.ps}%
\\
\hfil $\mathord{\mathcal Q}^-(z)$ \hfil
}
\caption{The movement of the branching points along $\gamma^\pm_R$}
\label{fig3}
\end{figure}
 Let
 $$
 \Gamma_R\sp\pm \;:\;\Theta\to W^\pm
 $$
  be the thimble over $\gamma_R^\pm$
corresponding to the critical point of $q^\pm: W^\pm\to U$ in the fiber over  $R^\pm$.
The vanishing cycle
$\sigma_R^+=-\partial \Gamma_R\sp+ $
is depicted by a thick line in Figure~\ref{figsigma}.
  %
 \begin{figure}
\includegraphics[scale=.6]{figsigma.ps}%
\caption{The vanishing cycle $\sigma_R^+=-\partial \Gamma_R\sp+ $}
\label{figsigma}
\end{figure}
We  choose the orientation
of $\sigma_R^+$ as in Figure~\ref{figsigma}.
Then we have
$$
([\sigma_R^+], [a_1])_F=1,
\quad
([\sigma_R^+], [a_2])_F=([\sigma_R^+], [a_3])_F=0,
\quad
([\sigma_R^+], [a_4])_F=1,
$$
and hence
$$
[\sigma_R^+]=[a_1]-[a_2]+[a_3]-[a_4].
$$
In the same way, we see that the homology class of the vanishing cycle $\sigma_R^-=-\partial \Gamma_R\sp-$ is equal to
$$
[\sigma_R^-]=[a_2]+[a_3]
$$
under an appropriate choice of orientation.
The movement of  the points $\mathord{\mathcal Q}^\pm (z)$
when $z$ moves from $b$ to $S^\pm$ along the path $\gamma_S^\pm$
is homotopically equivalent to the movement depicted in
 Figure~\ref{fig4}.
 %
 %
 \begin{figure}
\parbox{170pt}{
\includegraphics[scale=.55]{fig4plus.ps}%
\\
\hfil $\mathord{\mathcal Q}^+(z)$ \hfil
}
\vrule height 100pt
\parbox{170pt}{
\includegraphics[scale=.55]{fig4minus.ps}%
\\
\hfil $\mathord{\mathcal Q}^-(z)$ \hfil
}
\caption{The movement of the branching points along $\gamma^\pm_S$}
\label{fig4}
\end{figure}
We choose the orientations
of the thimbles
$$
\Gamma_S^\pm\;:\; \Theta\to W^\pm
$$
over $\gamma_S^\pm$  in such a way that  the homology classes of the vanishing cycles
$\sigma_S^\pm=-\partial \Gamma_S^\pm$
are
\begin{eqnarray*}
{} [\sigma_S^+]&=&[a_1]-[a_2]-[a_3] \;\;\;\;\quad\rmand\quad \\
{} [\sigma_S^-]&=&2[a_1]-[a_2]-[a_3]-[a_4].
\end{eqnarray*}
The movement of  the  points $\mathord{\mathcal Q}^\pm (z)$
for the path $\gamma_{\bar{S}}^\pm$
is obtained from  Figure~\ref{fig4} by the  conjugation $\bar{\phantom{z}}:\mathord{\mathbb C}\to\mathord{\mathbb C}$ over $\mathord{\mathbb R}$.
We  choose the orientations
of the thimbles $\Gamma_{\bar{S}}^\pm$ in such a way that
 $[\sigma_{\bar S}^\pm]=-\partial [\Gamma_{\bar S}^\pm]$ are equal to
\begin{eqnarray*}
{} [\sigma_{\bar{S}}^+]&=&-[a_2]-[a_3]+[a_4] \;\;\;\;\quad\rmand\quad \\
{} [\sigma_{\bar{S}}^-]&=&-[a_1]-[a_2]-[a_3]+2 [a_4].
\end{eqnarray*}
\par
\medskip
Now we can calculate the kernel $\mathord{\rm H}_2(W^\pm, \mathord{\mathbb Z})$ of the homomorphism
$$
\partial\;\;:\;\;\bigoplus_{\nu=1}^4\mathord{\mathbb Z}[\Lambda_\nu]\oplus \mathord{\mathbb Z}[\Gamma^\pm_R]\oplus \mathord{\mathbb Z}[\Gamma^\pm_S]\oplus  \mathord{\mathbb Z}[\Gamma^\pm_{\bar S}]
\;\to\; \bigoplus _{\nu=1}^4\mathord{\mathbb Z}[a_\nu].
$$
We see that  $\mathord{\rm H}_2(W^+, \mathord{\mathbb Z})$ is  a free $\mathord{\mathbb Z}$-module of rank $3$ generated by
\begin{eqnarray*}
S^+_1&:=&-[\Lambda_1]-[\Lambda_3]+[\Gamma_R^+],\\
S^+_2&:=&-6[\Lambda_1]-2[\Lambda_2]+2[\Lambda_3]+[\Lambda_4]+5[\Gamma_S^+],\\
S^+_3&:=&[\Lambda_1]+[\Lambda_2]+[\Lambda_3]-[\Gamma_S^+]+[\Gamma_{\bar S}^+],
\end{eqnarray*}
while  $\mathord{\rm H}_2(W^-, \mathord{\mathbb Z})$ is  a free $\mathord{\mathbb Z}$-module of rank $3$ generated by
\begin{eqnarray*}
S^-_1&:=&-4[\Lambda_1]-3[\Lambda_2]-2[\Lambda_3]+[\Gamma_R^-]+2[\Gamma_S^-],\\
S^-_2&:=&-11[\Lambda_1]-7[\Lambda_2]-3[\Lambda_3]+[\Lambda_4]+5[\Gamma_S^-],\\
S^-_3&:=&3[\Lambda_1]+3[\Lambda_2]+3[\Lambda_3]-[\Gamma_S^-]+[\Gamma_{\bar S}^-].
\end{eqnarray*}
\par
\medskip
We deform the loop $\lambda$ and
the paths $\gamma^\pm_R$, $\gamma^\pm_S$ and $\gamma^\pm_{\bar{S}}$
as in Figure~\ref{fig5}.
The deformed loop $\lambda\sp\prime$ and paths
$\gamma\sp{\prime\pm}_R$, $\gamma\sp{\prime\pm}_S$, $\gamma\sp{\prime\pm}_{\bar{S}}$
are depicted by the dotted curves.
\begin{figure}
\hskip -160pt
\hbox{\includegraphics[scale=.8]{fig5.ps}%
}
\caption{The perturbation}
\label{fig5}
\end{figure}
Then the intersection numbers of the topological $2$-chains
$T=\Lambda_\nu, \Gamma^\pm_R, \Gamma^\pm_S, \Gamma^\pm_{\bar S}$ and
$T\sp\prime=\Lambda\sp\prime_\nu, \Gamma^{\prime\pm}_R, \Gamma^{\prime\pm}_S, \Gamma^{\prime\pm}_{\bar S}$
are calculated
as in Table~\ref{table:TT}.
Remark that
the local intersection number $(t, t\sp\prime)_q$
of the underlying paths $t$ of $T$ and $t\sp\prime$ of $T\sp\prime$ is $-1$ for any
intersection point $q$ contained in the small open neighborhood $\Delta$ of $b$ and $b\sp\prime$.
\begin{table}
$$
\renewcommand{\arraystretch}{1.7}
\begin{array}{c| cccc}
T\sp\prime \backslash T & \Lambda_\nu & \Gamma^\pm_R  & \Gamma^\pm_S  & \Gamma^\pm_{\bar S}\\
\hline
\Lambda\sp\prime_\nu  &0&0&0&0 \\
\Gamma^{\prime\pm}_R&([a_\nu], [\sigma_R\sp\pm])_F&-1&([\sigma_S\sp\pm], [\sigma_R\sp\pm])_F&0 \\
\Gamma^{\prime\pm}_S&([a_\nu], [\sigma_S\sp\pm])_F&0&-1&0 \\
\Gamma^{\prime\pm}_{\bar S} &([a_{\nu+1}], [\sigma_{\bar S}\sp\pm])_F&0&0&-1 \\
\end{array}
$$
\vskip 10pt
\caption{The intersection numbers of $T$ and $T\sp\prime$}
\label{table:TT}
\end{table}
Therefore the intersection matrix of
$\mathord{\rm H}_2(W^+ , \mathord{\mathbb Z})$
is calculated as follows:
$$
\renewcommand{\arraystretch}{1.2}
\begin{array}{c| ccc}
&S_1^+&S_2^+&S_3^+ \\
\hline
S_1^+ &0&0&0 \\
S_2^+ &0&40&-5 \\
S_3^+ &0&-5&2
\end{array}.
$$
Then $I(W^+)$ is generated by $S_1^+$,
where
$I(W^+)\subset \mathord{\rm H}_2(W^+)$ is the submodule
defined by~\eqref{eq:defI}.
Thus $\mathord{\rm T}(X^+)\cong \mathord{\rm H}_2(W^+)/I(W^+)$ is generated by $S_2^+ + I(W^+) $ and $S_3^+ +I(W^+)$,
and $\mathord{\rm T}(X^+)$ is isomorphic to
$$
L[40, -5, 2]\cong L[2, 1, 28].
$$
The intersection matrix of
$\mathord{\rm H}_2(W^- , \mathord{\mathbb Z})$
is calculated as follows:
$$
\renewcommand{\arraystretch}{1.2}
\begin{array}{c| ccc}
&S_1^-&S_2^-&S_3^- \\
\hline
S_1^- &22&55&-22 \\
S_2^- &55&140&-55 \\
S_3^- &-22&-55&22
\end{array}.
$$
Then $I(W^-)$ is generated by $S_1^- - S_3^-$.
Therefore
$\mathord{\rm T}(X^-)\cong \mathord{\rm H}_2(W^-)/I(W^-)$ is generated by $S_2^- +I(W^-) $ and $S_3^- + I(W^-)$, and
 $\mathord{\rm T}(X^-)$ is isomorphic to
$$
L[140, -55, 22]\cong L[8, 3, 8].
$$
Thus Proposition~\ref{prop:theexample} is proved.
\begin{remark}
For the algorithm   to determine whether given two lattices of rank $2$ are
isomorphic or not,
see~\cite[Chapter 15]{MR1662447}.
\end{remark}
\section{The equations}\label{sec:equation}
In this section,  we construct  homogeneous  polynomials of degree $6$ defining  complex
projective plane curves that have singular  points of type $A_{10}$
and  of type $A_9$.
Two of such polynomials are as follows:
\begin{eqnarray}
&10y^4z^2-20xy^2z^3+10x^2z^4-(-108\pm40\sqrt{5})\,c_0 xy^3z^2+&
\label{a10a9equ}\\
&(-64\pm20\sqrt{5}\,)c_0x^2yz^3+(-44\pm20\sqrt{5}\,)c_0y^5z-(-58\pm30\sqrt{5}\,){c_0}^2x^3z^3+&
\nonumber\\
&(-48\pm30\sqrt{5}\,){c_0}^2x^2y^2z^2+(-14\pm10\sqrt{5}\,){c_0}^3x^3yz^2+(-9\pm5\sqrt{5}\,){c_0}^4x^4z^2,&
\nonumber
\end{eqnarray}
with $c_0\in\mathbb{C}^\times$. We explain how to obtain these equations.
\par
\medskip
First we prove a lemma.
\begin{lemma}\label{A10}
Let $f(x, y)=0$ be a defining equation of a complex affine plane curve of  degree $6$ that has a
singular point of type $A_{10}$ at the origin with the tangent  $x=0$.
Then, after appropriate coordinate change of the form $(x, y)\mapsto (x, a y)$
with $a\ne 0$, $f$ is equal to one of the following polynomials
~\eqref{a2345} or~\eqref{a345} up to multiplicative
constant:
\begin{eqnarray}\label{eq:A10_1}
\lefteqn{\phantom{f=}x^2-2xy^2+y^4+a_{0,5}y^5+a_{0,6}y^6+a_{1,3}xy^3+a_{1,4}xy^4}&&
\label{a2345}\\
&&+a_{1,5}xy^5+a_{2,1}x^2y+a_{2,2}x^2y^2+a_{2,3}x^2y^3+a_{2,4}x^2y^4+a_{3,0}x^3
\nonumber\\
&&+a_{3,1}x^3y+a_{3,2}x^3y^2+a_{3,3}x^3y^3+a_{4,0}x^4+a_{4,1}x^4y+a_{4,2}x^4y^2
\nonumber\\
&&+a_{5,0}x^5+a_{5,1}x^5y+a_{6,0}x^6, \nonumber
\end{eqnarray}
where
$$\begin{array}{ll}
a_{3,0}=&{c_0}^2-a_{0,6}-a_{1,4}-a_{2,2},\\
a_{2,1}=&-2c_0+a_{0,5},\\
a_{1,3}=&2c_0-2a_{0,5},\\
a_{4,0}=&-\frac{1}{\,2\,}(-3{c_0}^4+6{c_0}^2c_1-2{c_1}^2+3{c_0}^2a_{0,6}-4c_0a_{1,5}-{c_0}^2a_{2,2}\\
    &-2c_0a_{2,3}+2a_{2,4}+2a_{3,2}),\\
a_{3,1}=&-2{c_0}^3+2c_0c_1+{c_0}^2a_{0,5}-a_{1,5}-a_{2,3},\\
a_{1,4}=&-\frac{1}{\,2\,}({c_0}^2-2c_1-2c_0a_{0,5}+3a_{0,6}+a_{2,2}),\\
a_{5,0}=&-\frac{1}{\,2\,}(-16{c_0}^6+45{c_0}^4c_1-33{c_0}^2{c_1}^2+2{c_1}^3-12{c_0}^3c_2+16c_0c_1c_2-2{c_2}^2)\\

&-(2{c_0}^5-3{c_0}^3c_1+c_0{c_1}^2)a_{0,5}-\frac{1}{\,2\,}(8{c_0}^4-9{c_0}^2c_1+3{c_1}^2)a_{0,6}\\

&-(-3{c_0}^3+2c_0c_1)a_{1,5}-\frac{1}{\,2\,}({c_0}^2c_1-{c_1}^2)a_{2,2}-(3{c_0}^2-2c_1)a_{2,4}\\
    &-({c_0}^2-c_1)a_{3,2}+c_0a_{3,3}-a_{4,2},\\
a_{4,1}=&-\frac{1}{\,2\,}(15{c_0}^5-30{c_0}^3c_1+12c_0{c_1}^2+8{c_0}^2c_2-4c_1c_2)\\

&-(-4{c_0}^4+4{c_0}^2c_1-{c_1}^2)a_{0,5}-\frac{1}{\,2\,}{c_0}^3a_{0,6}-{c_0}^2a_{1,5}\\
    &+\frac{1}{\,2\,}{c_0}^3a_{2,2}+2c_0a_{2,4}+c_0a_{3,2}-a_{3,3},\\
a_{2,3}=&3{c_0}^3-4c_0c_1+2c_2-(3{c_0}^2-2c_1)a_{0,5}+3c_0a_{0,6}-2a_{1,5}-c_0a_{2,2},\\
a_{5,1}=&\frac{1}{\,2\,}(26{c_0}^7-83{c_0}^5c_1+74{c_0}^3{c_1}^2-12c_0{c_1}^3+25{c_0}^4c_2-40{c_0}^2c_1c_2\\

&+4{c_1}^2c_2+6c_0{c_2}^2-2c_3)-(2{c_0}^6-9{c_0}^4c_1+9{c_0}^2{c_1}^2+3{c_0}^3c_2-6c_0c_1c_2\\

&+{c_2}^2)a_{0,5}-\frac{1}{\,2}(-34{c_0}^5+59{c_0}^3c_1-24c_0{c_1}^2-9{c_0}^2c_2+6c_1c_2)a_{0,6}\\

&-(9{c_0}^4-11{c_0}^2c_1+2c_0c_2)a_{1,5}-\frac{1}{\,2\,}(2{c_0}^5-7{c_0}^3c_1+6c_0{c_1}^2+{c_0}^2c_2\\

&-2c_1c_2)a_{2,2}-(-7{c_0}^3+8c_0c_1-2c_2)a_{2,4}-(-2{c_0}^3+3c_0c_1-c_2)a_{3,2}\\
    &-(2{c_0}^2-c_1)a_{3,3}+c_0a_{4,2},
\end{array}$$
with $a_{i,j}$, $c_k\in\mathbb{C}$,  or
\begin{eqnarray}\label{A10_2}
\lefteqn{\phantom{f=}x^2-2xy^3+y^6-2x^2y{c_0}+2xy^4{c_0}+x^2y^2\big({c_0}^2+2({c_0}^2-{c_1})\big)}
\label{a345}\\
&&-2xy^5({c_0}^2-{c_1})+x^3\big(-2{c_0}({c_0}^2-{c_1})-a_{2,3}\big)+x^2y^3a_{2,3}\nonumber\\
&&+x^2y^4a_{2,4}+x^3y(3{c_0}^4-4{c_0}^2{c_1}+{c_1}^2+{c_0}a_{2,3}-a_{2,4})+x^3y^3a_{3,3}\nonumber\\
&&+x^3y^2\big(-3{c_0}^5+6{c_0}^3{c_1}-3{c_0}{c_1}^2+{c_2}-({c_0}^2-{c_1})a_{2,3}+{c_0}a_{2,4}\big)\nonumber\\
&&+x^4a_{4,0}+x^4ya_{4,1}+x^4y^2a_{4,2}+x^5a_{5,0}+x^5ya_{5,1}+x^6a_{6,0},\nonumber
\end{eqnarray}
with $a_{i,j}$, $c_k\in\mathbb{C}$.

Conversely,
if $c_3\ne 0$,
then the affine curve defined by the polynomial~\eqref{eq:A10_1} has a singular point of type $A_{10}$
at the origin,
and if $c_2\ne 0$,
then the affine curve defined by the polynomial~\eqref{A10_2} has a singular point of type $A_{10}$
at the origin.
\end{lemma}
We will use the following method to determine the type of singularities from
the form of the equation.
\begin{definition}
Let $k$ be an algebraically closed field and let
$w=(w_0,w_1)\in\mathbb{Q}_{\geq0}^{\ 2}$. Let $M=x^{e_0}y^{e_1}\in k[[x,y]]$
be a monomial. We define the {\it weight} of $M$ by $w(M):=\sum e_iw_i$. A
formal power series $f\in k[[x,y]]$ is said to be {\it semi-quasihomogeneous}
with respect to the weight $w$ if $f$ is of  the form
$f=f_{w=1}+f_{w>1}$ such tha
\begin{list}{}{}
\item[\textup{(i)}] every non-zero coefficient monomial $M$ in $f_{w=1}$
satisfies $w(M)=1$, and $f_{w=1}$ defines an isolated singularity, and
\item[\textup{(ii)}] every non-zero coefficient monomial $M$ in $f_{w>1}$
satisfies $w(M)>1$.
\end{list}
A semi-quasihomogeneous $f$ is said to be {\it quasihomogeneous} with respect to
the weight $w$ if $f_{w>1}=0$.
\end{definition}
\begin{proposition}[\cite{MR1175728}, Proposition 2.3]\label{recognition}
A semi-quasihomogeneous
$f\in k[[x,y]]$ with respect to the weight ${\mathcal
A}_m=(\frac{1}{\,2\,},\frac{1}{m+1})$, ${\mathcal
D}_m=(\frac{1}{m-1},\frac{m-2}{2(m-1)})$, ${\mathcal
E}_6=(\frac{1}{\,3\,},\frac{1}{\,4\,})$, ${\mathcal
E}_7=(\frac{1}{\,3\,},\frac{2}{\,9\,})$ and ${\mathcal
E}_8=(\frac{1}{\,3\,},\frac{1}{\,5\,})$ defines a simple singularity if
$f_{w=1}$ defines an isolated singularity at the origin. The type of the singular point is
$A_m$, $D_m$, $E_6$, $E_7$ and $E_8$ respectively.
\end{proposition}
\proof[Proof of Lemma \ref{A10}]{ Let $f_0(x,y)=\sum
b_{i,j}x^iy^j\in\mathbb{C}[x,y]$ be a polynomial of  degree $6$ with complex
coefficients $b_{i,j}$. Suppose that the affine plane curve defined by $f_0$ has a
singularity of type $A_{10}$ at $(0,0)$ with the tangent $x=0$. We can write
$$f_0=x^2+b_{3,0}x^3+b_{2,1}x^2y+b_{1,2}xy^2+b_{0,3}y^3+({\rm higher\
terms}).$$

Firstly, let $w=(\frac{1}{\,2\,},\frac{1}{\,3\,})$.
If $b_{0,3}\neq0$, then
 $(f_0)_{w=1}=x^2+b_{0,3}y^3$  would define an isolated singularity at the origin and
hence $f_0=0$ would have a singularity of type $A_2$ at the origin  by Proposition~\ref{recognition}.
Thus $b_{0,3}$
must be equal to $0$.

Secondly, let $w=(\frac{1}{\,2\,},\frac{1}{\,4\,})$.
Then  $f_0$ is
semi-quasihomogeneous with respect to  $w$, and hence the
quasihomogeneous part $(f_0)_{w=1}$ must define a \emph{non}-isolated singularity
at the origin by Proposition \ref{recognition}. Hence there exists
$b\in\mathbb{C}$ such that $(f_0)_{w=1}$ is equal to $x^2-2bxy^2+b^2y^4$. We
divide into two cases, the case where $b\neq0$ and the case where $b=0$.

{\it Case 1} $(b\neq0)$.
We  change the coordinate via $\sqrt{b}\,y\mapsto y$.
 Then  we have
$(f_0)_{w=1}\mapsto x^2-2xy^2+y^4$.
Therefore,  without loss of generality, we can write
\begin{eqnarray}
\lefteqn{f_0(x,y)=x^2-2xy^2+y^4+a_{0,5}y^5+a_{0,6}y^6+a_{1,3}xy^3+a_{1,4}xy^4}&&
\label{standardform}\\
&&+a_{1,5}xy^5+a_{2,1}x^2y+a_{2,2}x^2y^2+a_{2,3}x^2y^3+a_{2,4}x^2y^4+a_{3,0}x^3
\nonumber\\
&&+a_{3,1}x^3y+a_{3,2}x^3y^2+a_{3,3}x^3y^3+a_{4,0}x^4+a_{4,1}x^4y+a_{4,2}x^4y^2
\nonumber\\
&&+a_{5,0}x^5+a_{5,1}x^5y+a_{6,0}x^6, \nonumber
\end{eqnarray}
with $a_{i,j}\in\mathbb{C}$.

Change the coordinate via $x\mapsto x+y^2$. Assume that this coordinate change
transforms $f_0$ into $f_1$. An elementary calculation shows that
\begin{eqnarray*}
\lefteqn{f_0\mapsto
f_1=x^2+a_{2,1}x^2y+(a_{0,5}+a_{1,3}+a_{2,1})y^5+(a_{1,3}+2a_{2,1})xy^3+a_{3,0}x^3}&&
\\
&&+(a_{0,6}+a_{1,4}+a_{2,2}+a_{3,0})y^6+(a_{2,2}+3a_{3,0})x^2y^2+(a_{1,4}+2a_{2,2}+3a_{3,0})xy^4\\
&&+a_{3,1}x^3y+(a_{1,5}+a_{2,3}+a_{3,1})y^7+(a_{2,3}+3a_{3,1})x^2y^3+(a_{1,5}+2a_{2,3}+3a_{3,1})xy^5\\
&&+a_{4,0}x^4+(a_{2,4}+a_{3,2}+a_{4,0})y^8+(a_{3,2}+4a_{4,0})x^3y^2+(2a_{2,4}+3a_{3,2}+4a_{4,0})xy^6\\
&&+(a_{2,4}+3a_{3,2}+6a_{4,0})x^2y^4+a_{4,1}x^4y+(a_{3,3}+a_{4,1})y^9+3(a_{3,3}+2a_{4,1})x^2y^5\\
&&+(a_{3,3}+4a_{4,1})x^3y^3+(3a_{3,3}+4a_{4,1})xy^7+a_{5,0}x^5+(a_{4,2}+a_{5,0})y^{10}\\
&&+(a_{4,2}+5a_{5,0})x^4y^2+2(2a_{4,2}+5a_{5,0})x^3y^4+2(3a_{4,2}+5a_{5,0})x^2y^6\\
&&+(4a_{4,2}+5a_{5,0})xy^8+a_{5,1}x^5y+5a_{5,1}x^4y^3+10a_{5,1}x^3y^5+10a_{5,1}x^2y^7\\
&&+5a_{5,1}xy^9+a_{5,1}y^{11}+a_{6,0}x^6+6a_{6,0}x^5y^2+15a_{6,0}x^4y^4+20a_{6,0}x^3y^6\\
&&+15a_{6,0}x^2y^8+6a_{6,0}xy^{10}+a_{6,0}y^{12}.
\end{eqnarray*}
First,
let $w=(\frac{\,\,\,1}{\,2\,},\frac{1}{\,5\,})$. If there were the term $y^5$
with non-zero coefficient in $f_1$, the singularity of $f_1=0$ at the origin would be of type $A_4$
by Proposition \ref{recognition}.
Next, let
$w=(\frac{1}{\,2\,},\frac{1}{\,6\,})$. By the same argument as above, the
quasihomogeneous part $(f_1)_{w=1}$ of $f_1$ must define non-isolated
singularities at the origin, because  otherwise $f_1=0$ would have a singularity of type
$A_5$. Thus there exists a complex number $c_0$ such that
$(f_1)_{w=1}=x^2+(a_{1,3}+2a_{2,1})xy^3+(a_{0,6}+a_{1,4}+a_{2,2}+a_{3,0})y^6$
is equal to    $x^2-2c_0xy^3+{c_0}^2y^6$. Consequently we have following
conditions:
\begin{list}{}{}
\item [{\it Step 1.}] ($x\mapsto x+y^2$)\\
$a_{0,5}+a_{1,3}+a_{2,1}=0$, \\
$a_{0,6}+a_{1,4}+a_{2,2}+a_{3,0}={c_0}^2$ and \\
$a_{1,3}+2a_{2,1}=-2c_0$.
\end{list}

Then we change the coordinate via $x\mapsto x+c_0y^3$. Assume that this
transformation takes $f_1$ to $f_2$. Let $w=(\frac{1}{\,2\,},\frac{1}{\,7\,})$.
The coefficient of $y^7$ is equal to $0$, because  otherwise the singularity at the
origin would be of type $A_{6}$. Next, let
$w=(\frac{1}{\,2\,},\frac{1}{\,8\,})$. The quasihomogeneous part $(f_2)_{w=1}$
must define non-isolated singularities at the origin.
Hence there exists $c_1\in\mathbb{C}$ such
that $(f_2)_{w=1}$ is equal to $x^2-2c_1xy^4+{c_1}^2y^8$. Therefore we have
\begin{list}{}{}
\item [{\it Step 2.}] ($x\mapsto x+c_0y^3$)\\
${c_0}^3+{c_0}^2a_{0,5}-3c_0a_{0,6}-2c_0a_{1,4}+a_{1,5}-c_0a_{2,2}+a_{2,3}+a_{3,1}=0$,
\\
$3{c_0}^4-3{c_0}^2a_{0,6}-3{c_0}^2a_{1,4}+c_0a_{1,5}-2{c_0}^2a_{2,2}+2c_0a_{2,3}+a_{2,4}+3c_0a_{3,1}+a_{3,2}
+a_{4,0}={c_1}^2$ and \\
$-{c_0}^2+2c_0a_{0,5}-3a_{0,6}-2a_{1,4}-a_{2,2}=-2c_1$.
\end{list}

The coordinate change via $x\mapsto x+c_1y^4$ takes
$f_2$ to $f_3$. The coefficient of $y^9$ in $f_3$ is equal to $0$ and there
exists $c_2\in\mathbb{C}$ such that $(f_3)_{w=1}=x^2-2c_2xy^5+{c_2}^2y^{10}$,
where  $w=(\frac{1}{\,2\,},\frac{1}{\,10\,})$.
\begin{list}{}{}
\item [{\it Step 3.}] ($x\mapsto x+c_1y^4$)\\
$\frac{\,3\,}{2}{c_0}^5-4c_1{c_0}^3+2{c_0}^4a_{0,5}-3c_1{c_0}^2a_{0,5}+{c_1}^2a_{0,5}-\frac{\,11\,}{2}{c_0}^3a_{0,6}+3c_1c_0a_{0,6}+5{c_0}^2a_{1,5}+\frac{\,3\,}{2}{c_0}^3a_{2,2}-a_{2,2}c_1c_0+2{c_0}^2a_{2,3}-a_{2,3}c_1-2c_0a_{2,4}-c_0a_{3,2}+a_{3,3}+a_{4,1}=0$,
\\
$7{c_0}^6-\frac{\,35\,}{2}c_1{c_0}^4+\frac{\,15\,}{2}{c_1}^2{c_0}^2+{c_1}^3+{c_0}^5a_{0,5}+3c_1{c_0}^3a_{0,5}-3{c_1}^2c_0a_{0,5}-9{c_0}^4a_{0,6}-\frac{\,9\,}{2}c_1{c_0}^2a_{0,6}+\frac{\,3\,}{2}{c_1}^2a_{0,6}+11{c_0}^3a_{1,5}+2c_1c_0a_{1,5}+3{c_0}^4a_{2,2}+\frac{1}{\,2\,}c_1{c_0}^2a_{2,2}-\frac{1}{\,2\,}{c_1}^2a_{2,2}+5{c_0}^3a_{2,3}-5{c_0}^2a_{2,4}-2c_1a_{2,4}-3{c_0}^2a_{3,2}-c_1a_{3,2}+3c_0a_{3,3}+4c_0a_{4,1}+a_{4,2}+a_{5,0}={c_2}^2$
and \\
$3{c_0}^3-4c_1c_0-3{c_0}^2a_{0,5}+2c_1a_{0,5}+3c_0a_{0,6}-2a_{1,5}-c_0a_{2,2}-a_{2,3}=-2c_2$.
\end{list}

Change the coordinate via $x\mapsto x+c_2y^5$. Suppose that this transformation
takes $f_3$ to $f_4$. Let $w=(\frac{1}{\,2\,},\frac{1}{\,11\,})$.
Then   $(f_4)_{w=1}$ is
equal to $x^2+c_3y^{11}$,
where $c_3$ is given below.
If $c_3\ne 0$,
then $(f_4)_{w=1}=0$ defines an isolated singular point at the origin,
and hence $f_4=0$ has a singular point of type $A_{10}$ at the origin.
\begin{list}{}{}
\item [{\it Step 4.}] ($x\mapsto x+c_2y^5$)\\
$13{c_0}^7-\frac{\,83\,}{2}c_1{c_0}^5+37{c_1}^2{c_0}^3-6{c_1}^3c_0+\frac{\,25\,}{2}c_2{c_0}^4-20c_2c_1{c_0}^2+c_2{c_1}^2+3{c_2}^2c_0+2{c_0}^6a_{0,5}-9c_1{c_0}^4a_{0,5}+9{c_1}^2{c_0}^2a_{0,5}+3c_2{c_0}^3a_{0,5}-6c_2c_1c_0a_{0,5}+{c_2}^2a_{0,5}-17{c_0}^5a_{0,6}+\frac{\,59\,}{2}c_1{c_0}^3a_{0,6}-12{c_1}^2c_0a_{0,6}-\frac{\,9\,}{2}c_2{c_0}^2a_{0,6}+3c_2c_1a_{0,6}+9{c_0}^4a_{1,5}-11c_1{c_0}^2a_{1,5}+{c_1}^2a_{1,5}+2c_2c_0a_{1,5}+{c_0}^5a_{2,2}-\frac{\,7\,}{2}c_1{c_0}^3a_{2,2}+3{c_1}^2c_0a_{2,2}+\frac{1}{\,2\,}c_2{c_0}^2a_{2,2}-c_2c_1a_{2,2}-7{c_0}^3a_{2,4}+8c_1c_0a_{2,4}-2c_2a_{2,4}-2{c_0}^3a_{3,2}+3c_1c_0a_{3,2}-c_2a_{3,2}+2{c_0}^2a_{3,3}-c_1a_{3,3}-c_0a_{4,2}+a_{5,1}=c_3$.
\end{list}

Solve the system of linear equations appearing in each step by choosing
unknowns suitably. Then we have the solutions denoted in (\ref{a2345}).

{\it Case 2} $(b=0)$. In this case, without loss of generality we can write
\begin{eqnarray*}
\lefteqn{f_0=x^2-2b\sp\prime xy^3+{b\sp\prime }^2y^6+a_{1,4}xy^4+a_{1,5}xy^5+a_{2,1}x^2y+a_{2,2}x^2y^2}\\
&&+a_{2,3}x^2y^3+a_{2,4}x^2y^4+a_{3,0}x^3+a_{3,1}x^3y+a_{3,2}x^3y^2+a_{3,3}x^3y^3\\
&&+a_{4,0}x^4+a_{4,1}x^4y+a_{4,2}x^4y^2+a_{5,0}x^5+a_{5,1}x^5y+a_{6,0}x^6.
\end{eqnarray*}
Assume that $b\sp\prime =0$. If $a_{1,4}\neq0$, then the polynomial $f_0$ is
semi-quasihomogeneous with respect to the weight
$w=(\frac{1}{\,2\,},\frac{1}{\,8\,})$ and $(f_0)_{w=1}$ defines an isolated
singularity. Hence $f_0=0$ would have a singularity of type $A_7$ at the origin.
If $a_{1,4}=0$ and $a_{1,5}\neq0$, then $f_0$
is semi-quasihomogeneous with respect to
$w=(\frac{1}{\,2\,},\frac{1}{\,10\,})$, and $(f_0)_{w=1}$ defines an isolated
singularity at the origin, so that  $f_0=0$ would have a
singularity of type $A_9$ at the origin. If
$a_{1,4}=a_{1,5}=0$, then $f_0$ defines non-isolated singularities at the origin. Therefore
$b\sp\prime $ is not equal to zero. Furthermore, the coordinate change
$\sqrt[3]{b\sp\prime }\,y\mapsto y$ takes $b\sp\prime $ to $1$. Therefore we can write
\begin{eqnarray}
\lefteqn{f_0=x^2-2xy^3+y^6+a_{1,4}xy^4+a_{1,5}xy^5+a_{2,1}x^2y+a_{2,2}x^2y^2+a_{2,3}x^2y^3}&&\label{eqn345}\\
&&+a_{2,4}x^2y^4+a_{3,0}x^3+a_{3,1}x^3y+a_{3,2}x^3y^2+a_{3,3}x^3y^3+a_{4,0}x^4
\nonumber\\
&&+a_{4,1}x^4y+a_{4,2}x^4y^2+a_{5,0}x^5+a_{5,1}x^5y+a_{6,0}x^6.\nonumber
\end{eqnarray}
By a similar argument as in Case 1, we have the following three steps:
\begin{list}{}{}
\item [{\it Step 1.}] ($x\mapsto x+y^3$)\\
$a_{1,4}+a_{2,1}=0$,\\
$a_{1,5}+a_{2,2}+a_{3,0}={c_0}^2$ and \\
$a_{1,4}+2a_{2,1}=-2c_0$,
\item [{\it Step 2.}] ($x\mapsto x+c_0y^4$)\\
$-c_0a_{1,5}+a_{2,3}+a_{3,0}=0$, \\
${c_0}^4-{c_0}^2a_{1,5}+2c_0a_{2,3}+a_{2,4}+3c_0a_{3,0}+a_{3,1}={c_1}^2$ and
\\
$-2{c_0}^2-a_{1,5}=-2c_1$,
\item [{\it Step 3.}] ($x\mapsto x+c_1y^5$)\\
$3{c_0}^5-6{c_0}^3c_1+3c_0{c_1}^2+{c_0}^2a_{2,3}-c_1a_{2,3}-c_0a_{2,4}+a_{3,2}=c_2$,
\end{list}
where $c_i\in\mathbb{C}$. Regard $a_{1,4}$, $a_{1,5}$, $a_{2,1}$, $a_{2,2}$,
$a_{3,0}$, $a_{3,1}$ and $a_{3,2}$ as unknowns and solve the system of linear
equations. The solutions are
\begin{eqnarray*}
a_{2,1}&=&-2c_0,\\
a_{2,2}&=&{c_0}^2-a_{1,5},\\
a_{1,4}&=&2c_0,\\
a_{1,5}&=&-2({c_0}^2-{c_1}),\\
a_{3,0}&=&-2{c_0}({c_0}^2-{c_1})-a_{2,3},\\
a_{3,1}&=&3{c_0}^4-4{c_0}^2{c_1}+{c_1}^2+{c_0}a_{2,3}-a_{2,4}\ {\rm and}\\
a_{3,2}&=&-3{c_0}^5+6{c_0}^3{c_1}-3{c_0}{c_1}^2+{c_2}-({c_0}^2-{c_1})a_{2,3}+{c_0}a_{2,4}.
\end{eqnarray*}
Substituting them for the coefficient of (\ref{eqn345}),  we obtain the
polynomial (\ref{a345}).
\qed}\\

\begin{claim}
Let $F(x,y,z)\in\mathbb{C}[x,y,z]$ be a  homogeneous polynomial of degree $6$ that
satisfies
$$F(x,y,1)=f(x,y),$$
where $f$ is the polynomial (\ref{a2345}) in the statement of Lemma \ref{A10} with $c_3\ne 0$.
Let $g(y,z):=F(1,y,z)$. Then $g$ is semi-quasihomogeneous with respect to the
weight $w=(\frac{1}{\,10\,},\frac{1}{\,2\,})$ if and only if
\begin{eqnarray}\label{a0}
a_{0,6}=a_{1,5}=a_{2,4}=a_{3,2}=a_{3,3}=a_{4,2}=a_{6,0}=0
\end{eqnarray}
and
\begin{eqnarray}
\left\{\begin{array}{rcl} \label{c0}
c_1&=&\frac{1}{\,2\,}(5\pm\sqrt{5}\,){c_0}^2,\\
a_{2,2}&=&2(c_1+c_0a_{0,5})-{c_0}^2,\\
a_{0,5}&=&\frac{2}{\,5\,}(-11\pm5\sqrt{5}\,)c_0,\\
c_3&=&-\frac{6}{\,25\,}{(-123\pm55\sqrt{5}\,)}{c_0}^7.
\end{array}\right.
\end{eqnarray}
Moreover, if (\ref{a0}) and (\ref{c0}) hold, then $g_{w=1}$ defines an
isolated singularity at the origin
and hence $g=0$ has a singular point of type $A_9$ at $(0,0)$
by Proposition \ref{recognition}.
\end{claim}

\proof{
We write  $g$ in
the form
$$g=g_{w<1}+g_{w=1}+g_{w>1},$$
where $w=(\frac{\,1\,}{\,10\,}, \frac{\,1\,}{\,2\,})$.
The condition
$g_{w<1}=0$ is equivalent to (\ref{a0}) and
$$\left\{
\begin{array}{rcl}
0&=&2(c_1+c_0a_{0,5})-{c_0}^2-a_{2,2}\\
0&=&-4{c_0}^3+6c_1c_0+5a_{0,5}{c_0}^2-2a_{0,5}c_1-2c_2\\
0&=&5{c_0}^4-5{c_0}^2c_1+{c_1}^2\\
0&=&{a_{0,5}}^2(5{c_0}^2-2c_1)(15{c_0}^3-10c_0c_1+5{c_0}^2a_{0,5}-2c_1a_{0,5})-4c_3\\
0&=&12{c_0}^5-20{c_0}^3c_1+8c_0{c_1}^2+25{c_0}^4a_{0,5}-20{c_0}^2c_1a_{0,5}+4{c_1}^2a_{0,5}.
\end{array}\right.
$$
Solving  this system of equations,  we get~\eqref{c0}. Note that
we have $c_0\neq0$ by the assumption $c_3\neq0$.
Substituting \eqref{a0} and  \eqref{c0} for coefficients of $g$, we have
$$g_{w=1}=\frac{-9\pm5\sqrt{5}\,}{10}{c_0}^4\,z^2+\frac{2(-11\pm5\sqrt{5}\,)}{5}c_0\,zy^5.$$
Since  $c_0\neq0$, $g_{w=1}$  defines an isolated
singularity.
\qed}
\\

Note that $10F(x,y,z)$ is equal to (\ref{a10a9equ}) under the condition
(\ref{a0}) and (\ref{c0}).
\par
\medskip
Finally, let $c_0=1$. The curve defined by the equation (\ref{a10a9equ}) has a
singular point of type $A_{10}$ at $(0:0:1)$, a singular point of type $A_9$ at
$(1:0:0)$,  and is smooth except for these two points.
\begin{remark}
In~\cite{MR1900779},
a different method to obtain  defining equations of sextic curves with big Milnor number
is given.
\end{remark}
\bibliographystyle{plain}
\def$'$} \def\cprime{$'$} \def\cprime{$'$} \def\cprime{$'${$'$} \def$'$} \def\cprime{$'$} \def\cprime{$'$} \def\cprime{$'${$'$} \def$'$} \def\cprime{$'$} \def\cprime{$'$} \def\cprime{$'${$'$} \def$'$} \def\cprime{$'$} \def\cprime{$'$} \def\cprime{$'${$'$}

redpajama_set_name: RedPajamaArXiv


We are an egg family. We literally have eggs almost every morning as a family, I love it. That being said, we are all about brinner (breakfast for dinner) as well. We just love our breakfast in this family! We are always looking for new ways to spice up our eggs! These were delicious and just another one of my "sneaky mama" ways to get the kids to eat spinach.
In a saucepan drizzle olive oil then add tomatoes, onion onion and rosemary. Cover and heat on low/medium heat for an hour it should start to "glaze". Prepare your eggs. Next add spinach to saucepan and heat for a few minutes just until it becomes wilted. Spoon glaze on top of your eggs, add salt & pepper and sprinkle with cheese.


text:
— аніме-серіал (OVA) створений під керівництвом режисера Сінобу Йосіока (Фрактал, Меланхолія Харухі Судзумії [ТВ-2], Kannagi, Lucky Star). За основу сюжету взяли пісню Міку Хацуне — Black Rock Shooter. Вийшов серіал 25 липня 2010 року.

Сюжет
Головна героїня — Мато Курой, переходить у старшу школу. Там вона знайомиться з Йомі Таканасі — темноволосою дівчинкою в окулярах. Вони пізніше стають подругами. Мато дізнається про подругу дитинства Йомі і допомагає їй стати такою як усі. Мато знаходить ще одну давню подругу — Юу, що ходить з нею в одну секцію. Раптом, Йомі зникає, і Мато вирушає на пошуки подруги. Знайти її і пояснити, що сталося, їй допомогла Юу і шкільний психолог. Виявляється, що існує паралельний світ, де знаходяться аватари людей — повністю протилежні їм, і змагаються між собою, тим самим допомагаючи вгамувати свої емоції і почуття. Для того щоб врятувати Йомі, треба убити її аватар у тому світі…

Персонажі

В реальності
Мато Курой — головна героїня. Має темно-синє волосся, зібране у два хвостики. Добра, співчутлива та енергійна. Має молодшого брата, з яким, як показано в аніме, у неї складні стосунки. Любить розповідь про «Райдужну пташечку», як і її подруга Йомі. Любить займатися спортом та не любить математику, тому часто просить списати у Йомі. З дитинства дружить з Юу.
Сейю: Маая Сакамото.

Йомі Таканасі — дівчинка з темно-зеленуватим волоссям, яка носить окуляри. Тиха, скромна дівчинка. Із приходом Мато до старшої школи, є її найкращою подругою. Раніше хотіла переїхати у Німеччину, але сталася неприємна ситуація, після якої в неї з'явилося почуття обов'язку перед Кагарі (подругою дитинства). Кагарі довго знущалась над Йомі, поки Мато не допомогла їй.
Сейю: Міюкі Савасіро.

Аніме

Аніме-серіал (2012)

Список серій
Список серій аніме відповідно до даних сайту cal.syoboi.jp:

Примітки

Посилання

Відеоігри 2011
Науково-фантастичні аніме та манґа
Драматичні аніме та манґа

redpajama_set_name: RedPajamaWikipedia


I am a very unobservant person. Oftentimes it is several months before I notice a new business or a new landmark. Not this time. I saw Afghani food mentioned on a sign and I was suddenly very aware. As in, checking every other day to see if the place had opened yet.
Ali Baba's takes over the small, nondescript place at the corner of 7th & L streets that had formerly been occupied by an Indian restaurant. In its first couple of days I saw it was open and immediately stopped in to try them out.
My first visit was for a quick lunch. I would immediately note that you may want to consider take-out if you are crunched for time. Even though it is counter order service, this is still a seated lunch that includes a small salad before your meal arrives. Said salad is a simple tossed green salad with a yogurt based dressing.
I have a problem with my eyes being bigger than my stomach, especially when I want to try one of everything. I thought I would solve that problem by ordering two appetizers. I ended up with quite a bit of (lovely) food.
My order was for Samosas and for Mantoo. First out was the samosas, meat filled with a side of cilantro chutney. There were three small ones which had a firm, rather than flakey crust. This is the one item that I would like to see improved. I would prefer two larger ones with a different crust. Still, they were a tasty start.
I had ordered the Mantoo which is Afghani style raviolis filled with onion and ground lamb, topped with yogurt. The kitchen got my ordered confused and instead sent out Aushak, same style raviolis, but filled with seasoned leeks. Lucky for me! The chef noticed the order and had them bring out my Mantoo, which I got to take home with me for dinner!
Both the Mantoo and Aushak use the same delicate wrappers that are pinched closed into little purses (see top picture). They also have the same yogurt and tomato sauces garnished on top. And both are delicious! Not only that, but if this was appetizer size, I'd hate see what entree size is. Actually, though, I think the only difference is that the entree comes with a side of vegetables. Being that I'm a meat eater, I prefer the Mantoo. But put either of those in front of me and I love them both.
The location of Ali Baba is not one that one notice without a wary eye. Don't judge a book by its cover. Although the front is very small with three small window tables, there is a back dining room. What is a bit odd is that the kitchen is separated into two parts, making it a bit of an awkward running back and forth between the two sides. What they need is pass through window.
The restaurant is owned by two cousins. Obaid Khushiwal manages the front of the house while Basir Aziz is the master of the kitchen. Both men are very friendly and were happy to talk about their place and Basir's culinary past.
I knew that this place was special when I tasted the food, but the thought was cemented when I asked about details of some menu items, such as, he used to bake all his own Afghani bread but now buys it from an Afghani bakery located in West Sacramento. But he still makes his own cheese, more on this later.
Turns out Basir has had several restaurants over the years. Kabul Kabob, on J Street, was his creation before he sold it around 2005. That location serves lunch and dinner and a lunch buffet It sounded like he got tired of cranking out the endless buffet items. He then opened a place in Folsom on East Bidwell, also very popular. He sold that one after an unfortunate slip on a tomato injured his back. Being on your feet all day cooking is not fun with a bad back and as we chatted, Basir is sitting in a chair.
I had come in for lunch earlier in the day. This time I wanted to taste their version of kufta - ground beef mixed with spices and then formed into meatballs or onto skewers. This version was meatballs and was served over a seasoned basmati rice with the flavorful tomato sauce. The two meatballs were hefty but tender and the delicate rice was a nice accompaniment.
I chose the Rasmalayee for dessert. It's described as an Afghani cheesecake served with a cardamom sweetened cream. For those with expectations of New York cheesecake, this is not for you. The cheese is very firm and has a grittier texture. I asked Basir what type of cheese he used and he told me he makes it himself with a mixture of whole, 2%, and half and half. If you can adjust to the texture, than you can enjoy it as the sweetened cream makes it quite refreshing, although a bit rich. An easier dessert for Westerners may be the Firni, which is a rice pudding with the cardamom sweetened cream.
Against a wall I notice an unplugged rotisserie, the vertical type used for mounting the rotating mound of meat used for gyros. In front is this sign explaining the different names for schwarma - gyro, doner, kabob. Basir says that he'll get that going soon. I'll be curious to see how close it comes to the ones I grew up with in Saudi Arabia.
I'm thrilled to see Ali Baba's open and nearby for me to walk to for lunch. I'm sure I'll be visiting Basir and Obaid on a regular basis.


text:
Q: Can I Return and trigger an action e.g. increment a counter with .WillRepeatedly? I would like to trigger an action AND Return something on .WillRepeatedly() in googlemock.
Does anyone know how to do it? (Is writing a function that will increment a referenced parameter and return what I want is the only option here?)
This are my failed attempts to do so:
EXPECT_CALL(parser, getTemplateFileName(_))
  .WillRepeatedly(Invoke(parser.counter++), Return(TEMPLATE_NAME));

class TemplateParser: public Parser
{
    public:
        MOCK_METHOD1(getTemplateFileName, std::string(const string& name));

        void incrementCounter()
        {
          ++counter;
        }
};

// ...

EXPECT_CALL(parser, getTemplateFileName(_))
  .WillRepeatedly(Invoke(&parser, &TemplateParser::incrementCounter),
                  Return(TEMPLATE_NAME));

EDIT:
So as kerrek mentioned I have tried this:
EXPECT_CALL(parser, getTemplateFileName(_))
  .WillRepeatedly(DoAll(
                  Invoke(&parser, &TemplateParser::incrementCounter),
                  Return(TEMPLATE_NAME)
                       )
                  );

But I am getting errors like:
/opt/gmock-1.7.0/include/gmock/gmock-more-actions.h:80:   instantiated from 'Result testing::internal::InvokeMethodAction<Class, MethodPtr>::Perform(const ArgumentTuple&) const [with Result = void, ArgumentTuple = std::tuple<const std::string&>, Class = TemplateParser, MethodPtr = unsigned int (TemplateParser::*)()]'
/opt/gmock-1.7.0/include/gmock/gmock-actions.h:371:   instantiated from 'typename testing::internal::Function<F>::Result testing::PolymorphicAction<Impl>::MonomorphicImpl<F>::Perform(const typename testing::internal::Function<F>::ArgumentTuple&) [with F = void(const std::string&), Impl = testing::internal::InvokeMethodAction<TemplateParser, unsigned int (TemplateParser::*)()>]'
test/src/ParserTest.cpp:315:   instantiated from here
/opt/gmock-1.7.0/include/gmock/gmock-generated-actions.h:83: error: too many arguments to function
/opt/gmock-1.7.0/include/gmock/gmock-generated-actions.h:83: error: return-statement with a value, in function returning 'void'
make: *** [test/obj/ParserTest.o] Error 1

A: You can use DoAll as Kerrek suggests in comments using InvokeWithoutArgs.
Another alternative is to create a custom function to perform both actions in a very straightforward way by using a lambda.
EXPECT_CALL(parser, getTemplateFileName(_))
    .WillRepeatedly(InvokeWithoutArgs([&]()
                                      {
                                          parser.counter++;
                                          return TEMPLATE_NAME;
                                      });

redpajama_set_name: RedPajamaStackExchange


text:
Win Metawin has a new single out called "One More Chance," and listeners are calling her "angelic voice."
6 December 2022 by Francis Riggs
Win Metawin, also known as Metawin Opas-iamkajorn, is finally making a comeback with a brand new solo single titled "One More Chance." The 'angelic voice' of the Thai superstar is once more making fans swoon, and the song's heartbreaking lyrics are also contributing to this effect.
along with a score for the movies
Fans are tuning in for Metawin's new surprise single, One More Chance, in order to get a taste of what they can expect from the upcoming Side By Side Bright Win Concert, which will feature the heartthrob performing alongside Bright Vachirawit.
Win Metawin, who is famous for his charisma and his dreamy persona, has never failed to captivate audiences, whether he is on or off the screen. As the star is back with a new song, here's all about the new track and how fans are praising the star's 'angelic voice.'
Since starring in the hit BL series, 2gether, the star has quickly cemented his position as the face of Thai entertainment, which is rapidly becoming popular and reeling in new fans around the world. Win's abilities have been demonstrated in a variety of projects, ranging from F4 Thailand to Devil Sister.
Besides being a talented actor and model, Win is also a gifted singer whose voice has graced a number of OSTs like Still Together and Ten Year Later. And now the crooner from Silhouette has returned with a brand new track.
The much-anticipated music video for "One More Chance," which also featured Thai actress Grace Budsarin, was finally released by Win on December 6th. The MV is almost like a movie that tells the story of a relationship that has been plagued with mistakes and regret. Metawin is risking his heart by asking for another 'chance,' and he does so while wearing a dramatic red suit that showcases his effortless charm. The soft voice of the singer will definitely make your heart melt!
At the very end of the music video, the global icon is given credit as the creator of the story behind the song, with the words "story of song by Win Metawin" appearing on screen. Another name that hasn't gone unnoticed is Vachirawit Chivaaree – Win's supportive co-star, Bright has been credited as the Project Advisor for the new release.
The breathtaking visuals and silky voice of Win Metawin are consistently impressive.
A fan
gushes
"A human being with the face of an angel and the voice of an angel! I am completely taken with the sound of his voice, and I could sit here all day and listen to him sing.
the clever idea: "The Alice in Wonderland idea is so amazing that it blows my mind!"
In addition to this, Win, who is the proud CEO of his macaroons business, can be seen displaying Souri. A fan
describes
The use of Souri in the music video is, if you ask me, like killing two birds with one stone. It's a music video, but it also doubles as a commercial for Souri! Win Metawin, you have a brilliant mind!"
Follow this link for additional news and updates on the Asian entertainment industry.
Kpop.
As the Cheers star Kirstie Alley passes away, 12 stunning photos of her younger self have surfaced
In other words, you can always count on us to have the solution to today's Wordle!

redpajama_set_name: RedPajamaCommonCrawl


In this episode of #TheBossGame, we talk to serial entrepreneur Aprameya Radhakrishna, Founder of peer to peer knowledge sharing platform. He also co – founded cab hailing services called Taxi For Sure which got sold to Ola for $200 million few years ago. After leaving Taxi For Sure, he even became a angel investor and invested in some companies.


The non-member conference rate does not include an automatic membership to the chapter. Join UXPA-DC now, so that you can get the membership conference rate.
Registration fee includes admission to the sessions, conference materials, and a lunch.
UXPA-DC Chapter members pay a lower registration fee than non-members. If you are not a member, consider joining the chapter before registering for the conference, so you can register for the conference at the member rate.
If you are registering for the conference as a member, you should enter the email address that is associated with your membership. You will then get an option to sign in (signing in is optional). If you have forgotten your password, please use the "forgot password" link.
Payment is required upon registration. We accept payment via credit card or PayPal.
Email from your school email to obtain a registration code.
If you are the primary presenter for an accepted session (presentation or panel), you receive one registration at no charge. However, additional presenters must register and pay the registration fee.
If you register and find you cannot attend, send your cancellation request to by October 9, 2012 for a full refund minus a $15.00 processing fee.
If you would like to transfer your registration to someone else, there is no extra fee, but you must send your transfer request to by October 9, 2012. Include your full name, and the name and e-mail address of the colleague receiving your registration.


Wait at least 2 weeks before taking this test.
Private Testing Center's Syphilis RPR test is a simple, painless and affordable way to get answers about a syphilis infection. The Rapid Plama Regen (RPR) test checks for antibiotics in your blood that would get produced to fight a syphilis infection.
You MUST WAIT at least 2 weeks from your last contact of concern, or else the test won't be accurate. Private Testing Center will always determine how long has passed since a contact of concern, so we can be certain you are not wasting your time and money by testing too soon.
Syphilis is a highly contagious bacterium that can have serious consequences if left untreated, but is very simple to cure once positively identified. It enters your body through any tiny cuts or openings in your skin, and its symptoms can look like many other diseases. For this reason Syphilis has been called "The Great Pretender". The only way to know for sure if you have Syphilis is to get tested!


text:
/ Zizka, John
Zizka, John
Zizka, John zĭs´kə [key], Czech Jan Žižka yän zhēsh´kä [key], d. 1424, Bohemian military leader and head of the Hussite forces during the anti-Hussite crusades of Holy Roman Emperor Sigismund . Before the Hussite Wars , which gave his military genius the opportunity to develop fully, Zizka served under various lords; he fought (1410) on the Polish side in the battle of Tannenberg, in which the Teutonic Knights were defeated. When the Hussite Wars broke out in 1420, Zizka was about 60 years old and blind in one eye. Having joined the Taborites (the radical Hussite wing), Zizka made Tábor in Bohemia into an almost impregnable fortress and led (July, 1420) the Taborite troops in their victory over Sigismund at Visehrad (now a part of Prague). In the following years he successfully withstood the anti-Hussite crusades and took one Catholic stronghold after another, continuing to command in person although he had become totally blind in 1421. He did not agree with the extreme religious views of the Taborites, and in 1423 formed his own Hussite wing, which, however, remained in close alliance with the Taborites. In the same year the tension between the Taborites and the moderate Utraquists, whose stronghold was at Prague, flared into open conflict, and late in 1424, Zizka led his army against Prague in order to compel that city to adhere to his uncompromising anti-Catholic policy. An armistice averted the outbreak of civil war between the two Hussite parties, which then decided on a joint expedition into Moravia under Zizka's command. Zizka died suddenly during the campaign. Although Zizka's fame is overshadowed by that of other commanders, he ranks with the great military innovators of all time. The bulk of his army consisted of peasants and townspeople, untrained in arms. Zizka did not attempt to make them adopt the conventional armament and tactics of the time, but let them make use of such weapons as iron-tipped flails and armored farm wagons, surmounted by small cannons of the howitzer type. His armored wagons, when used for offense, easily broke through the enemy lines, firing as they went, and thus enabled him to cut superior forces into pieces. When used for defense, the wagons were arranged into an impregnable barrier surrounding the foot soldiers; they also served to transport his men. Zizka thus fully anticipated the principles of tank warfare.


Deepika and Ranveer shared two pictures— one in South Indian and one in North Indian attire. In the first photograph, Ranveer can be seen in a cream-white and golden kurta-dhoti, while Deepika wore a maroon Kanjeevaram sari. In the second picture, they both were in red and gold outfits.
After a long wait, Bollywood's power couple Deepika Padukone and Ranveer Singh released first pictures of wedding on their social media accounts on Thursday. They tied the knot at Lake Como's Villa del Balbianello in Italy.
Deepika and Ranveer shared two pictures— one in South Indian and one in North Indian attire.
In the first photograph, Ranveer can be seen in a cream-white and golden kurta-dhoti, while Deepika wore a maroon Kanjeevaram sari. In the second picture, they both were in red and gold outfits.
PTI reported that Sabyasachi designed the outfits for both Deepika and Ranveer.
Ever since both the actors left for Italy, media was trying to manage pictures of them with little success. The wedding was kept private and the couple were covered by the guests during ceremonies. However, some pictures were taken at the Mumbai airport when they were leaving on November 10.
According to the report, the wedding was secretive as even the guests were asked to keep their phone and camera away. Deepika and Ranveer will host a reception in Bengaluru and Mumbai later this month.
Deepika and Ranveer had announced the wedding dates in October.


text:
Men's and women's rugby reinstated
by Noah Nash
On March 19, the College reinstated both the women's and men's rugby teams, allowing the two teams to hold practice and recruit new team members.
The College suspended the two teams in October, citing concerns about the teams' high injury rates. The two teams have since rewritten their respective constitutions with a clear understanding of safety and well-being as a significant focus, according to assistant athletic director Justin Newell.
Additionally, Newell and Grant Wallace, head golf coach and assistant building coordinator for intramural and club sports, asked each team to create positions for two safety officers that would be first aid and CPR certified, in addition to carrying a medical kit to all practices and games.
"Our goal was obviously to be reinstated, and I think there is an evolution of rugby as a club sport in college that led to these changes," Maya Waldstreicher '19, president of the women's team, said. "I understood the decision [to implement changes to the team], but we were sad to see our [fall] season end early."
For the time being, the College has restricted the two programs to non-contact practices that will "allow for the continuity of culture as well as recruitment of players," Newell said.
These non-contact practices will continue for the duration of the spring semester, a requirement from the College that mirrors national legislation at the Division III level for football in the non-traditional season.
"We cannot rely on the way rugby was played 30 or so years ago," Vice President for Student Affairs Meredith Bonham '92 said. "The game has evolved, so we need to evolve with it."
Bonham said the teams asked for what she characterized as a "preparatory season" so that they would be able to play at full strength for the fall season.
Waldstreicher believes there may be some positives to take away from a non-contact practice season. "We're hoping that the lack of contact might help us focus more on the fundamentals and basics of the game and help us become more skilled as a team," Waldstreicher said. The men's rugby team did not respond to multiple requests for comment.
The College will try to find a part-time coach for the teams before the start of the fall season, when the teams will be free to play in games and have full-intensity practices.
Bonham is unsure whether the College will hire a coach for each team or one coach for both.
There is precedent for a part-time coach for a club sport, as the College has two part-time coaches for the equestrian team. The College is working with the students on the rugby teams to find a coach that has the attributes that teams value, according to Newell.
"Part of our intention in suspending operations of the rugby team was to help drive change in the club rugby network," Newell said. "Our message has always been that we need to be leaders in change to protect the athlete."
Looking to next season, Waldstreicher has high hopes for the women's team's potential for success.
The team was undefeated prior to their suspension on Oct. 6.
Noah Nash
Noah Nash '19 is an English major from Washington D.C. and is currently the Digital Director of the Collegian. He also is the Editor in Chief of the Collegian Magazine.More from Noah
Non-Contact Practices


I don't ever remember thinking that the first words of Jesus' statement had any kind of a personal message for me. They were simply His answer to a specific question He was asked about obeying civil authority — about paying taxes to Caesar — who, at that time in history, represented the civil authority governing them. That's the way it was for me – UNTIL, as I began wondering what I could chat to you about this month, those same words of Jesus came to mind. This time, however, my thoughts seemed locked into only those words which, ordinarily, I tended to ignore because they weren't in any way 'speaking' to me.
I knew this must be happening for a reason so I got myself some 'thinking' food (Oreos and milk) and sure enough, it worked. I could stop my 'wondering'. In little more than a cookie and a half here's what dawned on me. When Jesus told the Pharisees to "give to Caesar", He wasn't talking only to them anymore than He was talking only to the wealthy man who asked about possessing everlasting life when He told him to "love his neighbor". I came to realize the same thing was true each and every time Jesus spoke – whether His were words of compassion, understanding, reprimand, forgiveness or instruction.
Through the mystery of the Trinity, His human words, were delivering a message from God meant to be heard and heeded not only by the people in His presence at the time but by all generations to come. So it follows that when, so many years ago, Jesus said to those gathered around Him, "give to Caesar", He was also talking to me, to us, about our responsibilities concerning governmental authority. Here seems a good spot to insert something worth remembering. Jesus didn't say, "Do NOT pay taxes" or, to the best of my knowledge, did He ever speak out against the existence of secular authority. In fact, Paul, in Romans 13:1-7, speaks very explicitly along those same lines.
These were the kind of thoughts I was hoping for because I knew I would be 'talking' to you in July and, for me, the highlight of that month is the '4th'. I always hope and pray that special day will conjure up in the minds and hearts of all Americans great pride, joy and thankfulness to God as we commemorate victory in the hard-fought battle which gave our great country its independence and allowed it to begin to form its very own government. I was also conscious of the fact that this is a very important election year when we will be putting into office people who will represent that government, our Caesar if you will, in the sense that they will be the ones making decisions defining the specifics of governing civil authority.
So much for patriotic thoughts and well-intentioned words. Now comes the 'doing' which can get somewhat burdensome since God pretty much put us in charge of decision making by giving each of us a free well. He did promise, though, that He'd always be here to help and Thank You God for that. But basically we're the ones stuck with trying to figure out the fine print of 'giving to Caesar' — the what, the how and the when of it all.
Looking at the overall picture, I think, the 'doings' expected of us range from scrutinizing candidates, bureaucracy and laws to dealing with the present while learning from the past and, at the same time, planning for the future. Oh yes, and, along the way, seeing to it that 'Caesar' doesn't overstep or abuse what 'legitimate' power God delegates to him. All this while every form of communication bombards us with very convincing 'experts' feeding us contradictory 'facts'. Not easy!!
If you share my frustration, there are probably times when you shake your head, shrug your shoulders, try to convince yourself that 'giving to Caesar' isn't way up there on God's priority list and very seriously think about turning the whole thing off. BUT WE CAN'T! We must keep on doing our best to remain prayerfully informed and prayerfully involved.
You know, of course, where I'm headed. WE MUST VOTE!!! — whether we just vote against someone or something rather than for a person or an issue. WE MUST NEITHER THROW AWAY OUR VOTE NOR WASTE IT!!!


text:
Q: PancakeSwap API - how to get APR? I need info about pools APR/APY from API endpoints like:
https://api.pancakeswap.info/api/v2/summary
or
https://api.pancakeswap.info/api/v2/pairs
but this info is missed there.
How can I get it ?
When going to v3 API, eg . https://api.pancakeswap.info/api/v3/pairs
I get the response:
{"message":"Missing Authentication Token"}

How to solve that ?
The pancakeswap api page: https://syncwith.com/api/pancakeswap
It does not have reference to v3 api though.
Calculating APR based on Volume 24h and Liquidity
I've found the way to calculate APR based on volume24 and liquidity. Yet as I try to calculate APR  with the current values, i get the wrong APR*:
  Token            Volume          APR       Liquidity
 WBNB+BUSD        174278887       23.15      467090610
  USDT+WBNB        227057          0.06       241228503
 USDT+BUSD        28411575        5.15       342070717
  ETH+WBNB         38904           0.02       104583785
 Cake+WBNB        37343           0.01       286549465

Compare it with those at the website for pools/pairs:

What's wrong here ?


When weeds continually pop up in your ornamental beds it can be frustrating. Nobody wants to spend their spring, summer and fall removing weeds from their landscape. Weeds do not take any breaks during the growing season and are constantly trying to take hold in your landscape. There is no way to prevent 100% of weeds in your landscape, but preventative and post-emergent bed weed control applications can do a great job in minimalizing the number of weeds that pop up in your landscaping.
Preventative bed weed control is a dry granular application that works by putting down a 1-2 inch thick Pre-emergent barrier. This barrier prevents weed seeds from germinating or rooting into your landscaping. This application is not effective against weeds that are already established but it will prevent weeds from reseeding themselves. This application is typically bundled with a post emergent bed weed control application for best results. This application will not harm your existing plants.
A post-emergent bed weed control is a liquid application that works by going after already established weeds in your landscape. This treatment is non-selective, meaning it will control whatever it comes in contact with. This is beneficial because often times grasses will start to grow in your landscaping and they cannot be controlled by selective weed controls. Our specialists will only treat weeds and undesired grasses in your landscaping unless otherwise specified. Within 48 hours you will notice the weeds starting to wilt and die. Depending on weed type and size, all weeds should be completely dead within 14 days. This application kills the entire plant from the top down to the roots. We always add surfactants to this application to prevent it from running off and damaging non-targeted plants. At this point all dead weeds should be removed from the landscaping. This application will only take care of weeds that currently exist in your landscape; just like your lawn, follow up applications may be necessary later in the season. If you want an entire bed or specific area cleared of all plants take a look at our Ground Clear Application.