text:
\section{Introduction} \label{Introduction}
Bimodular singularities classified by Arnold \cite{Arnold75} have a duality studied by Ebeling and Ploog \cite{EbelingPloog} analogous to Arnold's strange duality for unimodular singularities.
Namely, a pair $((B,f),\, (B',f'))$ of singularities $B,\, B'$ in $\mathbb{C}^3$ together with appropriate defining polynomials $f,\, f'$ is dual if the matrices $A_f,\, A_{f'}$ of exponents of $f$ and $f'$ are transpose to each other.
Moreover, in some cases, such polynomials are compactified as anticanonical members of   $3$-dimensional weighted projective spaces whose general members are Gorenstein $K3$.
The strange duality for unimodular singularities is related with the polytope mirror symmetry for families of $K3$ surfaces that are obtained by compactifying the singularities by Kobayashi \cite{Kobayashi} in a certain sense.
In the study of bimodular singularities, Mase and Ueda \cite{MU} extend the duality by Ebeling and Ploog to a polytope mirror symmetry of families of $K3$ surfaces.
More precisely, the following statement is shown : \\

\noindent
{\bf Theorem}\, \cite{MU}
{\it Let $((B,\, f),\, (B',\, f'))$ be a dual pair in the sense of \cite{EbelingPloog} of  singularities $B$ and $B'$ together with their defining polynomials $f$ and $f'$  that are respectively compactified into polynomials $F$ and $F'$ as in \cite{EbelingPloog}.
Then, there exists a reflexive polytope $\Delta$ such that $\Delta_F\subset \Delta$ and $\Delta_{F'}\subset \Delta^*$.
Here, $\Delta_F$ and $\Delta_{F'}$ are respectively the Newton polytopes of $F$ and of $F'$, and $\Delta^*$ is the polar dual to $\Delta$. }
\END \\

In this article, we consider whether or not it is possible to extend the duality obtained in \cite{MU} further to the lattice mirror symmetry of families of $K3$ surfaces.
More precisely, our problem is stated as follows: \\

\noindent
\begin{problem}\,
Let $\Delta$ be a reflexive polytope as in \cite{MU}.
Does there exist general members $S\in\mathcal{F}_\Delta$ and $S'\in\mathcal{F}_{\Delta^*}$ such that an isometry $T(\widetilde{S}) \simeq \Pic(\widetilde{S'})\oplus U$ holds ?
Here, $\mathcal{F}_\Delta$ and $\mathcal{F}_{\Delta^*}$ are families of $K3$ surfaces associated to the polytopes $\Delta$ and $\Delta^*$, $\widetilde{S}$ denotes the minimal model of $S$, $\Pic(\widetilde{S})$ and $T(\widetilde{S})$ are respectively the Picard and transcendental lattice of $\widetilde{S}$.
\end{problem} \\

The problem is answered in Theorem \ref{MainResult} together with an explicit description in Proposition \ref{ExplicitLattices} of the Picard lattices $\Pic(\Delta)$ and $\Pic(\Delta^*)$ defined in section \ref{MainThm}, with ranks $\rho(\Delta)$ and $\rho(\Delta^*)$, of the minimal model of appropriate general members in the families.
The main result of this article is summarized here.
In the sequel, the names of singularities follow Arnold \cite{Arnold75}, and singularities in a same row of Table \ref{NicePairsIntro} are dual to each other in the sense of \cite{EbelingPloog}. \\

\noindent
{\bf Proposition} \ref{ExplicitLattices} and {\bf Theorem} \ref{MainResult} \,
{\it Let $\Delta$ be the reflexive polytope obtained in \cite{MU}.
For the following transpose-dual pairs, the polar duality extends to a lattice mirror symmetry  between the families $\mathcal{F}_\Delta$ and $\mathcal{F}_{\Delta^*}$, where the Picard lattices are given in Table \ref{NicePairsIntro}.
Here we use the notation $C^6_8:=\left( \begin{array}{cc} -4 & 1 \\ 1 & -2 \end{array}\right)$. }
\[
\begin{array}{cccccc}
\textnormal{Singularity} & \Pic(\Delta) & \rho(\Delta) & \rho(\Delta^*) & \Pic(\Delta^*) & \textnormal{Singularity} \\
\hline \hline
Q_{12} & U\oplus E_6\oplus E_8 & 16 & 4 & U\oplus A_2 & E_{18}\\
\hline
Z_{1,0} & U\oplus E_7\oplus E_8 & 17 & 3 & U\oplus A_1 & E_{19}\\
\hline
E_{20} & U\oplus E_8^{\oplus 2} & 18 & 2 & U & E_{20}  \\
\hline
Q_{2,0} & U\oplus A_6\oplus E_8 & 16 & 4 & U\oplus C^6_8 & Z_{17}\\
\hline
E_{25} & U\oplus E_7\oplus E_8 & 17 & 3 & U\oplus A_1 & Z_{19}\\
\hline
Q_{18} & U\oplus E_6\oplus E_8 & 16 & 4 & U\oplus A_2 & E_{30}\\
\hline
\end{array}
\]
\begingroup
\captionof{table}{{\it Picard lattices for lattice mirror symmetric pairs}\END}\label{NicePairsIntro}
\endgroup
\noindent
\\

Section \ref{Preliminary} is to define the polytope- and lattice- mirror theories in subsection \ref{Mirrors}, and to define the transpose duality following \cite{EbelingPloog} in subsection \ref{EPDual}, based on a brief introduction to lattice theory in subsection \ref{Lattice}, and of toric geometry in subsection \ref{toricHypersurface}, where several formulas and results are stated without proof.

The main theorem of this article is stated in section \ref{MainThm} following auxiliary results.
The facts introduced in the previous section are used in their proof.

Denote by $\Delta_B$ the reflexive polytope obtained in \cite{MU} for a singularity $B$.
As is seen in Table \ref{NicePairsIntro}, there are isometric Picard lattices  $\Pic(\Delta_{Q_{12}})\simeq \Pic(\Delta_{Q_{18}})$, and $\Pic(\Delta_{Z_{1,0}})\simeq \Pic(\Delta_{E_{25}})$.
We consider and affirmatively answer in Proposition \ref{BirFamilies} the following question as an application in section \ref{Application}: \\

\noindent
\begin{problem}\,
Are the families $\mathcal{F}_{\Delta_{Q_{12}}}$ (resp.\,$\mathcal{F}_{\Delta_{Z_{1,0}}}$) and $\mathcal{F}_{\Delta_{Q_{18}}}$ (resp.\,$\mathcal{F}_{\Delta_{E_{25}}}$) essentially the same in the sense that general members in these families are birationally equivalent ?
\end{problem}  \\

\noindent
\begin{ackn}\\
{\rm {
The author thanks to the referee for his helpful comments particularly about the proof of the key Proposition \ref{ExplicitLattices} in the original manuscript.
}}
\end{ackn}
\section{Preliminary} \label{Preliminary}
We start with having a consensus as to {\it Gorenstein $K3$ } and {\it $K3$ surfaces}.
\begin{defn}
\rm
 A compact complex connected $2$-dimensional algebraic variety $S$ with at most $ADE$ singularities is called {\it Gorenstein $K3$} if (i)\, $K_S \sim 0$; and  (ii)\, $H^1(S,\, \mathcal{O}_S)=0$.
If a Gorenstein $K3$ surface $S$ is nonsingular, $S$ is simply called a {\it $K3$ surface}.
\END \end{defn}
\subsection{Brief lattice theory} \label{Lattice}
A {\it lattice} is a non-degenerate finitely-generated free $\mathbb{Z}$-module with a symmetric bilinear form called an intersection pairing.
The {\it{discriminant group}} of a lattice $L$ is defined by $A_L:=L^*\slash L$, which is  finitely-generated and abelian, where $L^*:=Hom (L,\, \mathbb{Z})$ is dual to $L$.
It is known that the order $|A_L|$ of the discriminant group is equal to the determinant of any intersection matrix of $L$.
Let us recall a standard lattice theory by Nikulin \cite{Nikulin80}:
\begin{cor}[Corollary 1.13.5-(1) \cite{Nikulin80}] \label{IncludeU}
If an even lattice $L$ of signature $(t_+,\, t_-)$ satisfies $t_+\geq 1,\, t_-\geq 1$, and $t_+ +  t_-\geq 3+{\rm length}\,  A_L$, then, there exists a lattice $T$ such that $L\simeq U\oplus T$, where $U$ is the hyperbolic lattice of rank $2$. \END
\end{cor}
In particular, if an even lattice $L$ is of $\rk L>12, t_+\geq 1,$ and $t_-\geq 1$, then, there exists a lattice $T$ such that $L\simeq U\oplus T$.

Suppose $L$ is a sublattice of a lattice $L'$ with inclusion  $\iota: L\hookrightarrow L'$.
Denote by $L^\perp_{L'}$ the orthogonal complement of $L$ in $L'$.
The embedding $\iota$  is called {\it primitive}, and $L$ called a {\it primitive sublattice} of $L'$ if the finite abelian group $L'\, \slash \, \iota(L)$ is torsion-free.
In other words, if there is no {\it overlattice} that is an intermediate lattice between $L$ and $L'$ of rank equal to the rank of $L$.
Note that if a direct sum $L_1\oplus L_2$ of lattices is a sublattice of $L'$, then $(L_1\oplus L_2)^\perp_{L'} \simeq (L_1)^\perp_{L'}\oplus (L_2)^\perp_{L'}$.

For a $K3$ surface $S$, it is known that $H^2(S,\, \mathbb{Z})$ with the intersection pairing is isometric to the {\it $K3$ lattice} $\Lambda_{K3}$ that is even unimodular of rank $22$ and signature $(3, 19)$, being $U^{\oplus 3}\oplus E_8^{\oplus 2}$, where $E_8$ is the negative-definite even unimodular lattice of rank $8$.
There is a standard exact sequence $0\to H^1(S,\,\mathcal{O}_S^*)\overset{c_1}{\to}H^2(S,\,\mathbb{Z})\to 0$ of cohomologies so $H^1(S,\,\mathcal{O}_S^*)$ is inherited a lattice structure from $H^2(S,\,\mathbb{Z})$.
Define the {\it Picard lattice} $\Pic(S)$ of a $K3$ surface $S$ as the group $c_1(H^{1,1}(S))\cap H^2(S,\,\mathbb{Z})$ with the lattice structure.
The rank of $\Pic(S)$ is called the {\it Picard number}, denoted by $\rho(S)$.
The Picard lattice is hyperbolic since a $K3$ surface is complex and algebraic, and is known to be a primitive sublattice of $\Lambda_{K3}$ under a {\it marking} $H^2(S,\, \mathbb{Z})\overset{\sim}{\to} \Lambda_{K3}$.

If an even hyperbolic lattice $L$ of rank $\rk L \leq 20$ has $|A_L|$ being square-free, $L$ is a primitive sublattice of $\Lambda_{K3}$.
Indeed, if so, for any lattice $L\subset L''\subset \Lambda_{K3}$ the general relation $|A_L| = [L'':L]^2 |A_{L''}|$ implies $[L'':L] = 1$ thus $L''\simeq L$.
Hence there is no overlattice of $L$.
Moreover, by surjectivity of the period mapping \cite{BHPV}, there exists a $K3$ surface $S$ such that $\Pic(S)\simeq L$.
Let $M\subset \Lambda_{K3}$ be a hyperbolic sublattice.
A $K3$ surface is {\it $M$-polarised} \cite{DolgachevMirror} if there exists a marking $\phi$ such that all divisors in $\phi^{-1}(C_M^{pol})$ are ample, where $C_M^{pol}$ is the positive cone in $M_\mathbb{R}$ minus $\bigcup_{d\in\Delta_M} H_d$, $\Delta_M=\{d\in M\, | \, d.d=-2\}$, and $H_d=\{ x\in \mathbb{P}(\Lambda_{K3})\, |\, x.d=0 \textnormal{ for all } d\in\Delta_M\}$.

Nishiyama \cite{Nishiyama96} gives the orthogonal complements of primitive sublattices of type $ADE$ of $E_8$ in possible cases.
\begin{lem}[Lemma 4.3 \cite{Nishiyama96}]\label{NLem4.3}
There exist primitive embeddings of lattices of type $ADE$ into $E_8$ with orthogonal complements given as follows.
All the notation follows Bourbaki except $C_8^6 := \left( \begin{array}{cc} -4 & 1 \\ 1 & -2 \end{array}\right)$.
\[
\begin{array}{llll}
(A_1)^\perp_{E_8}\simeq E_7 & (A_2)^\perp_{E_8}\simeq E_6 & (A_3)^\perp_{E_8}\simeq D_5 & (A_4)^\perp_{E_8}\simeq A_4 \\  (A_5)^\perp_{E_8}\simeq A_1\oplus A_2 & (A_6)^\perp_{E_8}\simeq C_8^6 & (A_7)^\perp_{E_8}\simeq ({-}8) \\ (D_4)^\perp_{E_8}\simeq D_4 & (D_5)^\perp_{E_8}\simeq A_3 & (D_6)^\perp_{E_8}\simeq A_1^{\oplus 2} & (D_7)^\perp_{E_8}\simeq ({-}4)  \\  (E_6)^\perp_{E_8}\simeq A_2 & (E_7)^\perp_{E_8}\simeq A_1 & & \hspace{20mm}\textnormal{\END}
\end{array}
\]
\end{lem}

\subsection{Brief toric geometry} \label{toricHypersurface}
Here we summarize toric divisors and $\Delta$-regularity.
Let $M$ be a rank-$3$ lattice with the standard basis $\{ e_1,\, e_2,\, e_3\}$, $N$ be its dual, and $(\, ,\, ):M\times N \to \mathbb{Z}$ be the natural pairing.
From now on, a polytope means a $3$-dimensional convex hull of finitely-many points in $\mathbb{Z}^3$ embedded into $\mathbb{R}^3$, namely, {\it integral}, and the origin is the only lattice point in the interior of it.

Let $\mathbb{P}_\Delta$ be the toric variety defined by a polytope $\Delta$ in $M\otimes_\mathbb{Z}\mathbb{R}$, to which one can associate a fan $\Sigma_\Delta$ whose one-dimensional cones, called {\it one-simplices}, are generated by primitive lattice vectors each of whose end-point is an intersection point of $N$ and an edge of its polar dual $\Delta^*$ defined by
\[
\Delta^* := \left\{ y\in N\otimes \mathbb{R} \, | \, (x,\, y) \geq -1  \textnormal{ for all } x\in \Delta \right\}.
\]
Let $\widetilde{\mathbb{P}_\Delta}$ be the toric resolution of singularities in $\mathbb{P}_\Delta$.
A {\it{toric divisor}} is a divisor admitting the torus action, identified with the closure of the torus-action orbit of a one-simplex.
Let ${\rm Div}_\mathbb{T}(\widetilde{\mathbb{P}_\Delta})$ be the set of all toric divisors $D_i=\overline{orb(\mathbb{R}_{\geq 0} v_i)},\, i=1,\ldots , s$ on $\widetilde{\mathbb{P}_\Delta}$, where $v_i$ is a primitive lattice vector, then, ${-}K_{\widetilde{\mathbb{P}_\Delta}}=\sum_{i=1}^s \, D_i$.
By a standard exact sequence and a commutative diagram \cite{Oda78}
\[
\begin{array}{ccccccccc}
0 & \to & M & \to & {\rm Div}_\mathbb{T}(\widetilde{\mathbb{P}_\Delta}) & \to & \Pic(\widetilde{\mathbb{P}_\Delta}) & \to & 0 \\
 &  & &  & \downarrow & & \downarrow & &  \\
 &  & M & \to & \bigoplus_{i=1}^s \mathbb{Z}D_i & \to & A_2(\widetilde{\mathbb{P}_\Delta}) & \to & 0 \\
\end{array}
\]
there is a system of linear equations among toric divisors
\begin{equation} \label{linearRelation}
\sum_{i=1}^s (e_j,\, v_i) D_i = 0, \hspace{3mm} j=1,2,3.
\end{equation}
Thus the solution set of the linear system is generated by $(s-3)$ elements corresponding to  divisors which generate the Picard group $\Pic(\widetilde{\mathbb{P}_\Delta})$ of $\widetilde{\mathbb{P}_\Delta}$.

\begin{defn}
\rm
A polytope is {\it reflexive} if its polar dual is also integral. \END
\end{defn}

The importance of that we consider reflexive polytopes is by the following:
\begin{thm}c.f.  \cite{BatyrevMirror}
The following conditions are equivalent:
\begin{itemize}
\item[(1)] A polytope $\Delta$ is reflexive.
\item[(2)] The toric $3$-fold $\mathbb{P}_\Delta$ is Fano, in particular, general anticanonical members of $\mathbb{P}_\Delta$ are Gorenstein $K3$. \END
\end{itemize}
\end{thm}

For a reflexive polytope $\Delta$, denote by $\mathcal{F}_{\Delta}$ the family of hypersurfaces parametrised by the complete anticanonical linear system of $\mathbb{P}_\Delta$.
Note that general members in $\mathcal{F}_\Delta$ are Gorenstein $K3$ due to the previous theorem so that they are birationally equivalent to $K3$ surfaces by the existence of crepant resolution.
Thus we call the family $\mathcal{F}_\Delta$ a {\it family of $K3$ surfaces}.

We recall from \S 3 of \cite{BatyrevMirror} the notion of $\Delta$-regularity.
\begin{defn} \rm
Let $F$ be a Laurent polynomial defining a hypersurface $Z_F$, whose Newton polytope is a polytope $\Delta$.
The hypersurface $Z_F$ is called {\it $\Delta$-regular} if for every face $\Gamma$ of $\Delta$, the corresponding affine stratum $Z_{F, \Gamma}$ of $Z_F$ is either empty or a smooth subvariety of codimension $1$ in the torus $\mathbb{T}_\Gamma$ that is contained in the affine variety associated to $\Gamma$. \END
\end{defn}

It is shown in \cite{BatyrevMirror} that $\Delta$-regularity is a general condition, and singularities of all $\Delta$-regular members are simultaneously resolved by a toric desingularization of $\mathbb{P}_\Delta$.
From now on, suppose a polytope $\Delta$ is reflexive and $S$ is a $\Delta$-regular member whose minimal model $\widetilde{S}$ is obtained by a toric resolution.
\begin{defn} \label{DefOfL0}\rm
For a restriction $r:\widetilde{\mathbb{P}_\Delta}\to\widetilde{S}$, let $r_*:H^{1,1}(\widetilde{\mathbb{P}_\Delta})\to H^{1,1}(\widetilde{S})$ be the induced  mapping.
Define a lattice $L_D(\widetilde{S}):=r_*(H^{1,1}(\widetilde{\mathbb{P}_\Delta}))\cap H^2(\widetilde{S},\, \mathbb{Z})$ of the intersection of the image of $r_*$ and $H^2(\widetilde{S},\,\mathbb{Z})$, and its orthogonal complement $L_0(\widetilde{S}):=L_D(\widetilde{S})^\perp_{H^2(\widetilde{S},\, \mathbb{Z})}$ in $H^2(\widetilde{S},\, \mathbb{Z})$. \END
\end{defn}

It is known \cite{Kobayashi} that $\rho(\widetilde{S})$ and $\rk L_0(\widetilde{S})$ only depend on the number of lattice points in edges of $\Delta$ and $\Delta^*$.
Thus we define the Picard number $\rho(\Delta):=\rho(\widetilde{S})$, and the rank  $\rk L_{0,\Delta}:= \rk L_0(\widetilde{S})$ associated to $\Delta$.
More precisely, denote by $\Gamma^*$ in $\Delta^*$ the dual face to a face $\Gamma$ of $\Delta$, and $l^*(\Gamma)$ is the number of lattice points in the interior of $\Gamma$, and $\Delta^{[1]}$ the set of edges in $\Delta$.
Let $s$ be the number of one-simplices of $\Sigma_\Delta$.
Then
\begin{eqnarray}
\rk L_{0,\Delta} & = & \sum_{\Gamma\in\Delta^{[1]}}l^*(\Gamma)\, l^*(\Gamma^*) = \rk L_{0,\Delta^*} , \label{rkL0} \\
\rho(\Delta) & = & s-3+\rk L_{0,\Delta}, \label{PicardNo} \\
\rho(\Delta) + \rho(\Delta^*) & = & 20 + \rk L_{0,\Delta}. \label{Pic+Trs}
\end{eqnarray}
If $l^*(\Gamma^*)=n_\Gamma$ and $l^*(\Gamma)=m_\Gamma$ for an edge $\Gamma$ of $\Delta$, there is a singularity of type $A_{n_\Gamma+1}$ with multiplicity $m_\Gamma+1$ on an affine variety associated to $\Gamma$.

As we will see later, we only need formulas when $\rk L_{0,\Delta}=0$ for the intersection numbers of the divisors $\{ r_*D_i\}$ in $H^2(\widetilde{S},\,\mathbb{Z})$ given as follows.
\begin{eqnarray}
r_*D_i^2 = r_*D_i.r_*D_i & = & \begin{cases} 2l^*(v_i^*)-2 & \textnormal{if $v_i$ is a vertex of $\Delta^*$, }\\ -2 & \textnormal{otherwise. }\end{cases} \label{SelfIntersection}
\end{eqnarray}
If end-points of $v_i$ and $v_j$ are on the edge $\Gamma_{ij}^*$ of $\Delta^*$, then
\begin{eqnarray}
r_*D_i.r_*D_j & = & \begin{cases} 1 &  \textnormal{if $v_i$ and $v_j$ are next to each other, } \\ l^*(\Gamma_{ij})+1 & \textnormal{if $l^*(\Gamma_{ij}^*)=0$, and $v_i,\, v_j$ are both vertices,}  \\ 0 & \textnormal{otherwise. }\end{cases} \label{Intersection}
\end{eqnarray}

The Picard lattices of the minimal models of any $\Delta$-regular members, which are generated by components of restricted toric divisors, are isometric.
Define the {\it{Picard lattice $\Pic(\Delta)$ of the family $\mathcal{F}_\Delta$}} as the Picard lattice of the minimal model of a $\Delta$-regular member with rank $\rho(\Delta)$.
The orthogonal complement $T(\Delta)=\Pic(\Delta)^\perp_{\Lambda_{K3}}$ is called the {\it{transcendental lattice of $\mathcal{F}_\Delta$}}.

\subsection{Mirrors} \label{Mirrors}
We define the polytope- and lattice-mirror theories.
\subsubsection{Polytope Mirror}
We focus on {\it polytopes} that ``represent'' the anticanonical members in a toric variety as is seen in subsection \ref{toricHypersurface}.

\begin{defn} \rm
A pair $(\Delta_1,\, \Delta_2)$ of reflexive polytopes or a pair $(\mathcal{F}_{\Delta_1},\, \mathcal{F}_{\Delta_2})$ of families of $K3$ surfaces associated to $\Delta_1$ and $\Delta_2$ is called {\it polytope mirror symmetric} if an isometry $\Delta_1 \simeq \Delta_2^*$ holds. \END
\end{defn}
\subsubsection{Lattice Mirror}
For a $K3$ surface $S$, $\Pic(S)$ is the Picard lattice, and $T(S)=\Pic(S)^\perp_{\Lambda_{K3}}$ is the transcendental lattice.
A mirror for family of $M$-polarised $K3$ surfaces is defined when $M$ is a sublattice of $\Lambda_{K3}$ in general \cite{DolgachevMirror}.
Here, we deal with the most strict case, namely, mirror for $K3$ surfaces with Picard lattice as their polarisation.
\begin{defn} \rm
(1)\, A pair $(S,\, S')$ of $K3$ surfaces is called {\it lattice mirror symmetric} if an isometry $T(S) \simeq \Pic(S') \oplus U$ holds.  \\
(2)\, A pair $(\mathcal{F},\, \mathcal{F}')$ of families whose general members are Gorenstein $K3$ surfaces is {\it lattice mirror symmetric} if there exist general members $S\in \mathcal{F}$ and $S'\in \mathcal{F}'$ pair of whose minimal models is lattice mirror symmetric. \END
\end{defn}

Note that a lattice mirror pair $(S,\, S')$ of $K3$ surfaces satisfies, by definition, $\rk \Pic(S') + 2 = \rk T(S) = 22-\rk\Pic(S)$, thus
\begin{equation}
\rho(S')+\rho(S) = 20. \label{RhoDol}
\end{equation}

\subsection{Bimodular singularities and the transpose duality} \label{EPDual}
Being classified by Arnold \cite{Arnold75} in 1970's, bimodular singularities have two specific classes: quadrilateral and exceptional.
Quadrilateral bimodular singularities are 6 in number with exceptional divisor of type $I_0^*$, whilst exceptional are 14 in number with exceptional divisor of type $II^*,\, III^*$ or $IV^*$ in Kodaira's notation.

A non-degenerate polynomial $f$ in three variables is called {\it invertible} if $f$ has three terms  $f=\sum_{j=1}^3 x^{a_{1j}}y^{a_{2j}}z^{a_{3j}}$ such that its {\it matrix} $A_f:=(a_{ij})_{1\leq i,j\leq 3}$ {\it of exponents} is invertible in $GL_3(\mathbb{Q})$.

\begin{defn}c.f. \rm \cite{EbelingPloog}
Let $B=(0,(f=0))$ and $B'=(0,(f'=0))$ be germs of singularities in $\mathbb{C}^3$.
A pair $(B, B')$ of singularities is called {\it transpose dual} if the following three conditions are satisfied.
\begin{itemize}
\item[(1)] Defining polynomials $f$ and $f'$ are invertible.
\item[(2)] Matrices $A_f$ and $A_{f'}$ of exponents of $f$ and $f'$ are transpose to each other.
\item[(3)] $f$ $($resp. $f')$ is compactified to a four-term polynomial $F$ $($resp. $F')$ in $|{-}K_{\mathbb{P}(a)}|$ $($resp. $|{-}K_{\mathbb{P}(b)}|)$, where $\mathbb{P}(a)$ $($resp. $\mathbb{P}(b))$ is the $3$-dimensional weighted projective space whose general members are Gorenstein $K3$ with weight $a$ $($resp. $b)$ out of the list of $95$ weights classified by \cite{Yonemura}\cite{Iano-Fletcher}\cite{C3f}. \END
\end{itemize}
\end{defn}

Condition (1) and (2) is said that they are Berglund--H$\ddot{\textnormal{u}}$bsch mirror symmetric.
Ebeling-Ploog \cite{EbelingPloog} show that there are 16 transpose-dual pairs among quadrilateral and exceptional bimodular singularities, exceptional unimodular singularities, and the singularities $E_{25},\, E_{30},\, X_{2,0}$, and $Z_{2,0}$.

\section{The transpose dual and the lattice mirror} \label{MainThm}
For a transpose-dual pair $(B, B')$ of defining polynomial $f$ (resp. $f'$) being compactified to a polynomial $F$ (resp. $F'$), consider the Newton polytope $\Delta_F$ (resp. $\Delta_{F'}$) of $F$ (resp. $F'$) all of whose corresponding monomials are fixed by an automorphism action on $(F=0)$ (resp. $(F'=0)$).
\begin{remark} \rm
A compactified member $F$ to $f$ does not always define a Gorenstein $K3$ surface because $\Delta_F$ may not be reflexive.
\end{remark}

However, the Newton polytopes are extended to be reflexive and dual.

\begin{thm}\cite{MU} \label{MU}
For each transpose-dual pair $(B, B')$, there exists a reflexive polytope $\Delta$ such that $\Delta_F\subset \Delta$ and $\Delta_{F'}\subset \Delta^*$. \END
\end{thm}

Computing Picard lattices is generally difficult, but it seems possible for $\Delta$-regular members by subsection \ref{toricHypersurface}.
Let us reformulate our problem. \\

\noindent
\begin{problem}\,
For a polytope $\Delta$ obtained in \cite{MU}, is a pair $(\widetilde{S},\, \widetilde{S'})$ of minimal models of $\Delta$-regular $S\in\mathcal{F}_\Delta$ and $\Delta^*$-regular $S'\in\mathcal{F}_{\Delta^*}$ lattice mirror symmetric ?
\end{problem} \\

First we study the rank $\rk L_{0,\Delta}$.
\begin{lem} \label{RkOfL0}
The list of $\rk L_{0, \Delta}$ for the reflexive polytope $\Delta$ obtained in \cite{MU} is given in Table \ref{RankL0}.
\[
\begin{array}{ccccc}  \label{RankL0Lemma}
\textnormal{transpose-dual pair} & \rk L_{0, \Delta} & & \textnormal{transpose-dual pair} & \rk L_{0, \Delta} \\
\hline
\hline
(Q_{12},\, E_{18}) & 0 & & (Z_{17},\, Q_{2,0}) & 2 \\
\hline
(Z_{1,0},\, E_{19}) & 0 & & (U_{1,0},\, U_{1,0}) & 2 \\
\hline
(E_{20},\, E_{20}) & 0 & & (U_{16},\, U_{16}) & 2 \\
\hline
(Q_{2,0},\, Z_{17}) & 0 & & (Q_{17},\, Z_{2,0}) & 2 \\
\hline
(E_{25},\, Z_{19}) & 0 & & (W_{1,0},\, W_{1,0}) & 3 \\
\hline
(Q_{18},\, E_{30}) & 0 & & (W_{17},\, S_{1,0}) & 5 \\
\hline
(Z_{1,0},\, Z_{1,0}) & 1 & & (W_{18},\, W_{18}) & 6 \\
\hline
(Z_{13},\, J_{3,0}) & 2 & &  (S_{17},\, X_{2,0}) & 6 \\
\hline
\end{array}
\]
\captionof{table}{$\rk L_{0, \Delta}$}\label{RankL0}
\end{lem}
{\sc Proof. }\,
The assertion follows from direct and case-by-case computation by formula (\ref{rkL0}) in subsection \ref{toricHypersurface}. \\
{\bf 1.}\, $(Q_{12},\, E_{18})$ \,
The polytope $\Delta$ is given in subsection 4.7 of \cite{MU}.
There is no contribution to $\rk L_{0,\Delta}$ since $l^*(\Gamma)$ or $l^*(\Gamma^*)$  is zero for any edge $\Gamma$.
Thus, $\rk L_{0, \Delta} = 0$ by formula (\ref{rkL0}). \\
Similar for the cases $(Z_{1,0},\, E_{19}),\, (E_{20},\, E_{20}),\, (Q_{2,0},\, Z_{17}),\, (E_{25},\, Z_{19}) ,\, (Q_{18},\, E_{30})$. \\
{\bf 2.}\, $(Z_{13},\, J_{3,0})$ \,
The polytope $\Delta$ is given in subsection 4.1 of \cite{MU}.
The only contribution to $\rk L_{0, \Delta}$ is by an edge $\Gamma$ between vertices $(0,\, 0,\, 1)$ and $(-2,\, -6,\, -9)$, whose dual $\Gamma^*$ is between $(8,\, -1,\, -1)$ and $(-1,\, 2,\, -1)$ so $l^*(\Gamma)=1$ and $l^*(\Gamma^*)=2$.
Thus, $\rk L_{0, \Delta} = 1\times 2 = 2$ by formula (\ref{rkL0}). \\
Similar for the cases $(Z_{1,0}, Z_{1,0}),\, (Z_{17}, Q_{2,0}),\, (U_{1,0}, U_{1,0}),\, (U_{16}, U_{16}),\, (Q_{17}, Z_{2,0}),$\\
$(W_{1,0}, W_{1,0}),\, (W_{17}, S_{1,0}),\, (W_{18}, W_{18}),\, (S_{17}, X_{2,0})$.
\QED
\begin{cor}  \label{NotAdmitDolMir}
No $\Delta$- and $\Delta^*$-regular members for transpose-dual pairs $(Z_{13}, J_{3,0}),$\, $(Z_{1,0}, Z_{1,0}),$\, $(Z_{17}, Q_{2,0}),$\, $(U_{1,0}, U_{1,0}),$ \,
$(U_{16}, U_{16}),$\, $(Q_{17}, Z_{2,0}),$\, $(W_{1,0}, W_{1,0}),$\, $(W_{17}, S_{1,0}),$\, $(W_{18}, W_{18}),$\, $(S_{17}, X_{2,0})$ admit a lattice mirror symmetry.
\end{cor}
{\sc Proof. }\,
For each $\Delta$ associated to the presented pairs, by formula (\ref{Pic+Trs}),
\[
\rho(\Delta) + \rho(\Delta^*) = 20 + \rk L_{0, \Delta} > 20
\]
since $\rk L_{0, \Delta} > 0$ by Lemma \ref{RankL0Lemma}.
Thus, the equation (\ref{RhoDol}) does not hold.
Therefore, $\Delta$- and $\Delta^*$-regular members do not admit a lattice mirror symmetry.
\QED
\begin{cor} \label{SurjectiveRestriction}
The restriction mapping $r_* : \Pic(\widetilde{\mathbb{P}_\Delta}) \to \Pic(\widetilde{S})$, for $\Delta$-regular $S\in\mathcal{F}_\Delta$ is surjective for the transpose-dual pairs
$(Q_{12}, E_{18}),$\, $(Z_{1,0}, E_{19}),$\, $(E_{20}, E_{20}),$\, $(Q_{2,0}, Z_{17}),$\, $(E_{25}, Z_{19}),$\, $(Q_{18}, E_{30}). $
\end{cor}
{\sc Proof.}\,
By Lemma \ref{RankL0Lemma},  $\rk L_{0, \Delta} = 0$ for each case.
By definition, $\rk L_{0, \Delta}$ is equal to the rank of the orthogonal complement of $r_*(H^{1,1}(\widetilde{\mathbb{P}_\Delta}))$ in $H^2(\widetilde{S},\, \mathbb{Z})$.
Thus, that $\rk L_{0, \Delta} = 0$ means that $r_*$ is surjective. \QED \\

By Corollary \ref{NotAdmitDolMir}, we may only focus on the transpose-dual pairs appearing in Corollary \ref{SurjectiveRestriction}, whose statement means moreover that $\Pic(\Delta)$ is generated by restricted toric divisors generating $\Pic(\widetilde{\mathbb{P}_\Delta})$, and analogous to $\Delta^*$. \\

Let $A_L$ denote the discriminant group, $q_L$ the quadratic form, and $\discr L$ the discriminant of a lattice $L$.
If $p=\discr L$ is prime, then $A_L\simeq \mathbb{Z}\slash p\mathbb{Z}$.
Before stating our main results, note a fact in Proposition 1.6.1 in \cite{Nikulin80}.
Suppose that lattices $S$ and $T$ are primitively embedded into the $K3$ lattice $\Lambda_{K3}$.
If $A_S\simeq A_T$ and $q_S = -q_T$, then, it is determined that the orthogonal complement $S^\perp_{\Lambda_{K3}}$ in $\Lambda_{K3}$ is $T$.
And $q_S=-q_T$ if and only if $\discr S = -\discr T$.

\begin{prop} \label{ExplicitLattices}
The Picard lattice $\Pic(\Delta^*)$ for $\Delta$ in Corollary \ref{SurjectiveRestriction} is as in Table \ref{NicePairs}, where singularities in a row are transpose-dual.
In each case, one gets
\[
\discr \Pic(\Delta)=-\discr (U\oplus \Pic(\Delta^*)), \textnormal{ and } A_{\Pic(\Delta)}\simeq A_{U\oplus \Pic(\Delta^*)}.
\]
Denote $C_8^6 := \left( \begin{array}{cc} -4 & 1 \\ 1 & -2 \end{array}\right)$.
\[
\begin{array}{ccccc}
\textnormal{Singularity} & \rho(\Delta^*) & \Pic(\Delta^*) & \textnormal{Singularity} \\
\hline
\hline
Q_{12} &  4 & U\oplus A_2 & E_{18}\\
\hline
Z_{1,0} & 3 & U\oplus A_1 & E_{19}\\
\hline
E_{20} & 2 & U & E_{20}  \\
\hline
Q_{2,0} & 4 & U\oplus C_8^6  & Z_{17}\\
\hline
E_{25} & 3 & U\oplus A_1 & Z_{19}\\
\hline
Q_{18} & 4 & U\oplus A_2 & E_{30}\\
\hline
\end{array}
\]
\begingroup
\captionof{table}{$\Pic(\Delta^*)$ for $\Delta$ in Corollary \ref{SurjectiveRestriction}}\label{NicePairs}
\endgroup
\end{prop}

\noindent
{\sc Proof.} \\
${\bf 1.}\, Q_{12}$ and $E_{18}$ \,
The polytope $\Delta$ is given in subsection 4.7 of \cite{MU} and the associated toric $3$-fold has $19$ toric divisors $D_i$ corresponding to the one-simplices generated by vectors
\[
\begin{array}{lll}
m_1=(1,\,2,\,2) & m_2=(0,\,1,\,1) & m_3=(-8,\,-11,\,-9) \\
m_4=(1,\,-2,\,0) & m_5=(1,\,1,\,0) & m_6=(-4,\,-5,\,-4) \\
m_7=(-7,\,-10,\,-8) & m_8=(-6,\,-9,\,-7) & m_9=(-5,\,-8,\,-6) \\
m_{10}=(-4,\,-7,\,-5) & m_{11}=(-3,\,-6,\,-4) & m_{12}=(-2,\,-5,\,-3) \\
m_{13}=(-1,\,-4,\,-2) & m_{14}=(0,\,-3,\,-1) & m_{15}=(1,\,0,\,1) \\
m_{16}=(-5,\,-7,\,-6) & m_{17}=(-2,\,-3,\,-3) & m_{18}=(1,\,-1,\,0) \\
m_{19}=(1,\,0,\,0)
\end{array}
\]
and by solving the linear system (\ref{linearRelation}) : $\sum_{i=1}^{19}(e_j,\, m_i)D_i = 0 \quad (j=1,2,3)$, we get linear relations among toric divisors
\begin{eqnarray*}
D_1 & \sim & -9 D_4+3 D_5-D_7-2 D_8-3 D_9 -4 D_{10}-5 D_{11}-6 D_{12}-7 D_{13}-8 D_{14}\\
 & & \hspace{53mm} -5 D_{15}+D_{16}+2 D_{17}-5 D_{18}-D_{19}, \\
D_2 & \sim & D_3+10 D_4-2 D_5+2 D_7+3 D_8+4 D_9+ 5 D_{10}+6 D_{11}+7 D_{12}+8 D_{13} \\
 & & \hspace{50mm} +9 D_{14}+5 D_{15}-D_{17}+6D_{18}+2 D_{19}, \\
D_6 & \sim & -2 D_3-2 D_4+D_5-2 D_7-2 D_8-2 D_9 -2 D_{10}-2 D_{11}-2 D_{12}-2 D_{13} \\
 & & \hspace{65mm} -2 D_{14}-D_{15}-D_{16}-D_{18}.
\end{eqnarray*}
So the set $\{D_i \, | \, i\not= 1,2,6 \}$ of toric divisors is linearly independent.
Let $L$ be the lattice generated by the set $\{r_*D_i \, | \, i\not= 1,2,6 \}$ of their restrictions to a $\Delta$-regular member.
We shall check that $L$ is primitively embedded into the $K3$ lattice to show that $L$ is indeed the Picard lattice of the family $\mathcal{F}_\Delta$.
By computer calculation with formulas (\ref{SelfIntersection}) and (\ref{Intersection}), the determinant of an intersection matrix of $L$ is $-3$ since this matrix is given by
\[
\left(
\begin{array}{cccccccccccccccc}
 -2 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
 0 & -2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\
 1 & 0 & 0 & -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -2 & 0 & 0 & 0 & 0 \\
 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -2 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 & 0 & 0 \\
 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -2 & 1 \\
 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2
\end{array}
\right),
\]
Since the discriminant of $L$ is $-3$ that is square-free, there exists no overlattice of $L$ ;  indeed, if  $H\subset \Lambda_{K3}$ were an overlattice of $L$, then, by the standard relation $-3=[H:L]^2 \discr H$, we get $[H:L]=1$ and  $\discr H=-3$ so that $L\simeq H$.
Hence, $L$ is primitively embedded into the $K3$ lattice.
By construction, $L$ is indeed the Picard lattice of the family $\mathcal{F}_\Delta$.

The dual polytope $\Delta^*$ associates a toric $3$-fold with $7$ toric divisors $D'_i$ corresponding to the one-simplices generated by vectors
\[
\begin{array}{llll}
v_1=(1,\,1,\,-2) & v_2=(1,\,-2,\,1) & v_3=(2,\,-3,\,2) & v_4=(-1,\,0,\,1) \\
v_5=(-1,\,0,\,0) & v_6=(1,\,0,\,-1) & v_7=(1,\,-1,\,0)
\end{array}
\]
and by solving the linear system (\ref{linearRelation}) : $\sum_{i=1}^7(e_j,\, m_i)D'_i = 0\quad (j=1,2,3)$, we get linear relations among toric divisors
\[
D'_1 \sim  2 D'_2+3 D'_3+D'_7, \quad
D'_4 \sim  3 D'_2+4 D'_3+D'_6+2 D'_7, \quad
D'_5 \sim  D'_3 .
\]
Thus the set $\{D'_i \, | \, i\not= 1,4,5 \}$ of toric divisors is linearly independent.
Let $L'$ be the lattice generated by the set $\{r_*D'_i \, | \, i\not= 1,4,5 \}$ of their restrictions to a $\Delta^*$-regular member.
By formulas (\ref{SelfIntersection}) and (\ref{Intersection}), an intersection matrix associated to $L'$ is given by
\[
\left(
\begin{array}{cccc}
 -2 & 1 & 0 & 1 \\
 1 & 0 & 0 & 0 \\
 0 & 0 & -2 & 1 \\
 1 & 0 & 1 & -2
\end{array}
\right)
\]
that is equivalent to $U\oplus A_2$ by re-taking the generators as
\[
\{r_*\, D'_2+r_*\, D'_3,\, r_*\, D'_3,\, r_*\, D'_6, \, r_*\, D'_3 + r_*\, D'_7\}.
\]
Since the discriminant of $L'$ is $-3$ that is square-free, there exists no overlattice of $L'$ ;  indeed, if $H'\subset \Lambda_{K3}$ were an overlattice of $L'$, then, by the standard relation $-3=[H':L']^2 \discr H'$, we get $[H':L']=1$ and  $\discr H'=-3$ so that $H'\simeq L'$.
Hence, $L'$ is primitively embedded into the $K3$ lattice.
By construction, $L'$ is indeed the Picard lattice of the family $\mathcal{F}_{\Delta^*}$.
Therefore $\Pic(\Delta^*)=L'=(\mathbb{Z}^4,\, U\oplus A_2)$.

Similarly, the lattice $U\oplus \Pic(\Delta^*)$ is also primitively embedded into the $K3$ lattice since it is of signature $(2,\, 4)$ and is of discriminant $3$.
Besides, $\discr(U\oplus\Pic(\Delta^*))=3=-\discr \Pic(\Delta)$, and moreover,  $A_{\Pic(\Delta)}\simeq A_{U\oplus\Pic(\Delta^*)}\simeq\mathbb{Z}\slash 3\mathbb{Z}$. \\

\noindent
${\bf 2. }\, Z_{1,0}$ and $E_{19}$ \,
The polytope $\Delta$ is given in subsection 4.8 of \cite{MU} and the associated toric $3$-fold has $20$ toric divisors $D_i$ corresponding to
\[
\begin{array}{lll}
m_1=(1,\,-1,\,2) & m_2=(0,\,-1,\,1) & m_3=(0,\,0,\,1) \\
m_4=(4,\,2,\,-1) & m_5=(-6,\,2,\,-11) & m_6=(2,\, 0 ,\, 1) \\
m_7=(3,\, 1,\, 0) & m_8=(-2,\, 0,\, -3) & m_9=(-4,\, 1,\, -7) \\
m_{10}=(2,\,1,\,0) & m_{11}=(-3,\,1,\,-5) & m_{12}=(3,\,2,\,-2) \\
m_{13}=(2,\,2,\,-3) & m_{14}=(1,\,2,\,-4) & m_{15}=(0,\,2,\,-5) \\
m_{16}=(-1,\,2,\,-6) & m_{17}=(-2,\,2,\,-7) & m_{18}=(-3,\,2,\,-8) \\
m_{19}=(-4,\,2,\,-9) & m_{20}=(-5,\, 2,\, -10)
\end{array}
\]
and $\{D_i\, | \, i\not=1,2,3 \}$ is linearly independent by system (\ref{linearRelation}).
The lattice $L:=\langle r_* D_i\, | \, i\not=1,2,3 \rangle_\mathbb{Z}$ has $\rk L =17$, and $\discr L=2$ by an explicit calculation of its intersection matrix using formulas (\ref{SelfIntersection}) and (\ref{Intersection}).
Since the discriminant of $L$ is square-free, $L$ is primitively embedded into the $K3$ lattice, and thus $L=\Pic(\Delta)$.

The dual polytope $\Delta^*$ associates a toric $3$-fold with $6$ toric divisors $D'_i$ corresponding to
\[
\begin{array}{lll}
v_1=(1,\,-3,\,-1) & v_2=(2,\,0,\,-1) & v_3=(1,\,0,\,-1) \\
v_4=(0,\,-1,\,-1) & v_5=(-1,\,2,\,1) & v_6=(0,\,1,\,0)
\end{array}
\]
and $\{D'_i\, |\, i\not=1,2,5\}$ is linearly independent by system (\ref{linearRelation}).
By re-taking the generators as $\{ r_*\, D'_3+r_*\, D'_4,\, r_*\, D'_4,\, r_*\, D'_6-r_*\, D'_4\}$, the lattice $L':=\langle r_*D'_i \, | \, i\not= 1,2,5 \rangle_\mathbb{Z}$ has an intersection matrix $U\oplus A_1$.
Since $\rk L' = 3$, and $\discr L' = 2$, which is square-free, $L'$ is primitively embedded into the $K3$ lattice, and thus $\Pic(\Delta^*)=L'=(\mathbb{Z}^3,\, U\oplus A_1)$.

Similarly $U\oplus \Pic(\Delta^*)$ is primitively embedded into the $K3$ lattice.
Besides, $\discr (U\oplus \Pic(\Delta^*))=-2=-\discr \Pic(\Delta)$, and moreover,  $A_{\Pic(\Delta)}\simeq A_{U\oplus\Pic(\Delta^*)}\simeq\mathbb{Z}\slash 2\mathbb{Z}$. \\

\noindent
${\bf 3. }\, E_{20}$ and $E_{20}$ \,
The polytope $\Delta$ is given in subsection 4.9 of \cite{MU} and the associated toric $3$-fold has $21$ toric divisors $D_i$ corresponding to
\[
\begin{array}{lll}
m_1=(-1,\,-1,\,2) & m_2=(-1,\,-1,\,-1) & m_3=(-1,\,11,\,2) \\
m_4=(1,\,-1,\,0) & m_5=(-1,\,-1,\,1) & m_6=(-1,\, -1 ,\, 0) \\
m_7=(-1,\, 3,\, 0) & m_8=(-1,\, 7,\, 1) & m_9=(0,\, -1,\, 1) \\
m_{10}=(0,\,5,\,1) & m_{11}=(-1,\, 0,\,2) & m_{12}=(-1,\, 1,\,2) \\
m_{13}=(-1,\, 2,\,2) & m_{14}=(-1,\, 3,\,2) & m_{15}=(-1,\, 4,\,2) \\
m_{16}=(-1,\, 5,\,2) & m_{17}=(-1,\, 6,\,2) & m_{18}=(-1,\, 7,\,2) \\
m_{19}=(-1,\, 8,\,2) & m_{20}=(-1,\, 9,\, 2) & m_{21}=(-1,\, 10,\, 2)\end{array}
\]
and $\{D_i\, | \, i\not=10,13,14 \}$ is linearly independent by system (\ref{linearRelation}).
The lattice $L:=\langle r_* D_i\, | \, i\not=10,13,14 \rangle_\mathbb{Z}$ has $\rk L = 18$, and $\discr L=-1$ by an explicit calculation of its intersection matrix using formulas (\ref{SelfIntersection}) and (\ref{Intersection}).
As it being unimodular, $L$ is primitively embedded into the $K3$ lattice, and thus $L=\Pic(\Delta)$.

The dual polytope $\Delta^*$ associates a toric $3$-fold with $5$ toric divisors $D'_i$ corresponding to
\[
\begin{array}{lll}
v_1=(0,\,1,\,0) & v_2=(1,\,0,\,0) & v_3=(-1,\,0,\,-1) \\
v_4=(-2,\,-1,\,4) & v_5=(-1,\,0,\,2)
\end{array}
\]
and $\{D'_i\, |\, i\not=1,2,3\}$ is linearly independent by system (\ref{linearRelation}).
By an intersection matrix computed by formulas (\ref{SelfIntersection}) and (\ref{Intersection}), the lattice $L':=\langle r_*D'_i \, | \, i\not= 1,2,3 \rangle_\mathbb{Z}$ has $\discr L' = -1,\, \rk L'=2$, and $L'$ is even.
By the classification of even unimodular lattices, $L'$ is isometric to $U$, which is primitively embedded into the $K3$ lattice, and thus $\Pic(\Delta^*)=L'=(\mathbb{Z}^2,\, U)$.

Similarly, $U\oplus\Pic(\Delta^*)$ is primitively embedded into the $K3$ lattice.
Besides, $\discr (U\oplus\Pic(\Delta^*))=1=-\discr\Pic(\Delta)$, and moreover,  $A_{\Pic(\Delta)}\simeq A_{U\oplus\Pic(\Delta^*)}\simeq \{ 0\}$. \\

\noindent
${\bf 4. }\, Q_{2,0}$ and $Z_{17}$ \,
The polytope $\Delta$ is given in subsection 4.10 of \cite{MU} and the associated toric $3$-fold has $19$ toric divisors $D_i$ corresponding to
\[
\begin{array}{lll}
m_1=(0,\,1,\,1) & m_2=(1,\,2,\,2) & m_3=(1,\,1,\,2) \\
m_4=(0,\,-1,\,0) & m_5=(-6,\,-7,\,-9) & m_6=(1,\, 0 ,\, -2) \\
m_7=(1,\, 0,\, 1) & m_8=(1,\, 1,\, 0) & m_9=(-2,\, -3,\, -3) \\
m_{10}=(-4,\, -5,\, -6) & m_{11}=(1,\, 0,\, -1) & m_{12}=(1,\, 0,\, 0) \\
m_{13}=(-5,\, -6,\,-8) & m_{14}=(-4,\, -5,\,-7) & m_{15}=(-3,\, -4,\,-6) \\
m_{16}=(-2,\, -3,\,-5) & m_{17}=(-1,\, -2,\,-4) & m_{18}=(0,\, -1,\,-3) \\
m_{19}=(-3,\, -3,\,-4)
\end{array}
\]
and $\{D_i\, | \, i\not=1,2,3 \}$ is linearly independent by system (\ref{linearRelation}).
The lattice $L:=\langle r_* D_i\, | \, i\not=1,2,3 \rangle_\mathbb{Z}$ has $\rk L = 16$, and $\discr L=-7$ by an explicit calculation of its intersection matrix using formulas (\ref{SelfIntersection}) and (\ref{Intersection}).
Since the discriminant of $L$ is square-free, $L$ is primitively embedded into the $K3$ lattice, and thus $L=\Pic(\Delta)$.

The dual polytope $\Delta^*$ associates a toric $3$-fold with $7$ toric divisors $D'_i$ corresponding to
\[
\begin{array}{llll}
v_1=(1,\,-2,\,1) & v_2=(-1,\,1,\,0) & v_3=(2,\,1,\,-2) & v_4=(1,\,0,\,-1) \\
v_5=(0,\,1,\,-1)  & v_6=(-1,\,0,\,0) & v_7=(1,\,-1,\,0)
\end{array}
\]
and $\{D'_i\, |\, i\not=1,2,6\}$ is linearly independent by system (\ref{linearRelation}).
By re-taking the generators as $\{r_*\, D'_3,\, r_*\, D'_3+r_*\,D'_4,\, 2r_*\, D'_3+r_*\, D'_4-r_*\, D'_5, \,r_*\, D'_7-r_*\, D'_3\}$, the lattice $L':=\langle r_*D'_i \, | \, i\not= 1,2,6 \rangle$ has an intersection matrix $U\oplus C_8^6$.
Since $\discr L' = -7$ is square-free, $L'$ is primitively embedded into the $K3$ lattice, and thus $\Pic(\Delta^*)=L'=\left(\mathbb{Z}^4,\, U\oplus C_8^6\right)$.

Similarly $U\oplus\Pic(\Delta^*)$ is primitively embedded into the $K3$ lattice.
Besides, $\discr (U\oplus \Pic(\Delta^*)) = 7=-\discr \Pic(\Delta)$, and moreover, $A_{\Pic(\Delta)}\simeq A_{U\oplus\Pic(\Delta^*)}\simeq \mathbb{Z}\slash 7\mathbb{Z}$. \\

\noindent
${\bf 5. }\, E_{25}$ and $Z_{19}$ \,
The polytope $\Delta$ is given in subsection 4.11 of \cite{MU} and the associated toric $3$-fold has $20$ toric divisors $D_i$ corresponding to
\[
\begin{array}{lll}
m_1=(-1,\,2,\,0) & m_2=(-1,\,-1,\,9) & m_3=(-1,\,-1,\,-1) \\
m_4=(-1,\,2,\,-1) & m_5=(1,\,-1,\,-1) & m_6=(-1,\, 1 ,\, 3) \\
m_7=(-1,\, 0,\, 6) & m_8=(-1,\, -1,\, 8) & m_9=(-1,\, -1,\, 7) \\
m_{10}=(-1,\,-1,\, 6) & m_{11}=(-1,\,-1,\, 5) & m_{12}=(-1,\,-1,\, 4) \\
m_{13}=(-1,\,-1,\, 3) & m_{14}=(-1,\,-1,\, 2) & m_{15}=(-1,\,-1,\, 1) \\
m_{16}=(-1,\,-1,\, 0) & m_{17}=(-1,\, 0,\,-1) & m_{18}=(-1,\, 1,\,-1) \\
m_{19}=(0,\,-1,\,4) & m_{20}=(0,\, -1,\, -1)
\end{array}
\]
and $\{D_i\, | \, i\not=1,4,5 \}$ is linearly independent by system (\ref{linearRelation}).
The lattice $L:=\langle r_* D_i\, | \, i\not=1,4,5 \rangle_\mathbb{Z}$ has $\rk L=17$, and $\discr L=2$ by an explicit calculation of its intersection matrix using formulas (\ref{SelfIntersection}) and (\ref{Intersection}).
Since the discriminant of $L$ is square-free, $L$ is primitively embedded into the $K3$ lattice, and thus $L=\Pic(\Delta)$.

The dual polytope $\Delta^*$ associates a toric $3$-fold with $6$ toric divisors $D'_i$ corresponding to
\[
\begin{array}{lll}
v_1=(0,\,1,\,0) & v_2=(0,\,0,\,1) & v_3=(-3,\,-2,\,0) \\
v_4=(-5,\,-3,\,-1) & v_5=(1,\,0,\,0) & v_6=(-1,\,-1,\,0)
\end{array}
\]
and $\{D'_i\, |\, i\not=1,2,3\}$ is linearly independent by system (\ref{linearRelation}).
By re-taking the generators as $\{r_*\, D'_4,\, r_*\, D'_5-8r_*\, D'_4,\, r_*\, D'_6-r_*\, D'_4\}$, the lattice $L':=\langle r_*D'_i \, | \, i\not= 1,2,3 \rangle_\mathbb{Z}$ has an intersection matrix $U\oplus A_1$.
Since $\discr L' = 2$ is square-free, $L'$ is primitively embedded into the $K3$ lattice, and thus $\Pic(\Delta^*)=L'=\left(\mathbb{Z}^3,\, U\oplus A_1\right)$.

Similarly, $U\oplus\Pic(\Delta^*)$ is primitively embedded into the $K3$ lattice.
Besides, $\discr (U\oplus \Pic(\Delta^*)) = -2=-\discr \Pic(\Delta)$, and moreover, $A_{\Pic(\Delta)}\simeq A_{U\oplus \Pic(\Delta^*)}\simeq \mathbb{Z}\slash 2\mathbb{Z}$. \\

\noindent
${\bf 6. }\, Q_{18}$ and $E_{30}$ \,
The polytope $\Delta$ is given in subsection 4.13 of \cite{MU} and the associated toric $3$-fold has $19$ toric divisors $D_i$ corresponding to
\[
\begin{array}{lll}
m_1=(1,\,-1,\,-1) & m_2=(-1,\,-1,\,-1) & m_3=(-1,\,-1,\,8) \\
m_4=(1,\,-1,\,0) & m_5=(-1,\,2,\,-1) & m_6=(0,\,-1,\,-1) \\
m_7=(-1,\,-1,\, 0) & m_8=(-1,\,-1,\, 1) & m_9=(-1,\,-1,\, 2) \\
m_{10}=(-1,\,-1,\, 3) & m_{11}=(-1,\,-1,\, 4) & m_{12}=(-1,\,-1,\, 5) \\
m_{13}=(-1,\,-1,\, 6) & m_{14}=(-1,\,-1,\, 7) & m_{15}=(0,\,-1,\,4) \\
m_{16}=(-1,\, 0,\,-1) & m_{17}=(-1,\, 1,\,-1) & m_{18}=(-1,\, 0,\, 5) \\
m_{19}=(-1,\, 0,\, 2)
\end{array}
\]
and $\{D_i\, | \, i\not=1,4,5 \}$ is linearly independent by system (\ref{linearRelation}).
The lattice $L:=\langle  r_* D_i\, | \, i\not=1,4,5 \rangle_\mathbb{Z}$ has $\rk L=16$, and $\discr L=-3$ by an explicit calculation of its intersection matrix using formulas (\ref{SelfIntersection}) and (\ref{Intersection}).
Since the discriminant of $L$ is square-free, $L$ is primitively embedded into the $K3$ lattice, and thus $L=\Pic(\Delta)$.

The dual polytope $\Delta^*$ associates a toric $3$-fold with $7$ toric divisors $D'_i$ corresponding to
\[
\begin{array}{llll}
v_1=(0,\,0,\,1) & v_2=(1,\,0,\,0) & v_3=(-4,\,-3,\,-1) & v_4=(-3,\,-2,\,0) \\
v_5=(0,\,1,\,0) & v_6=(-2,\,-1,\,0) & v_7=(-1,\,0,\,0)
\end{array}
\]
and $\{D'_i\, |\, i\not=1,2,5\}$ is linearly independent by system (\ref{linearRelation}).
By re-taking the generators as $\{r_*\, D'_3,\, r_*\, D'_4+r_*\, D'_3,\, r_*\, D'_6-r_*\, D'_3,\, r_*\, D'_7\}$, the lattice $L':=\langle r_*D'_i \, | \, i\not= 1,2,5 \rangle_\mathbb{Z}$ has an intersection matrix $U\oplus A_2$ .
Since $\discr L'=-3$ is square-free, $L'$ is primitively embedded into the $K3$ lattice, and thus $\Pic(\Delta^*) = L'=\left(\mathbb{Z}^4,\, U\oplus A_2\right)$.

Similarly, $U\oplus\Pic(\Delta^*)$ is primitively embedded into the $K3$ lattice.
Besides, $\discr (U\oplus \Pic(\Delta^*)) = 3= -\discr \Pic(\Delta)$, and moreover,
$A_{\Pic(\Delta)}\simeq A_{U\oplus \Pic(\Delta^*)}\simeq \mathbb{Z}\slash 3\mathbb{Z}$.
\QED

\begin{remark} \rm
The choice of $\Delta$ is not actually unique.
However, for any possible reflexive polytopes $\Delta$ for transpose-dual pairs outside Table \ref{NicePairs}, we have a relation $\rho(\Delta)+\rho(\Delta^*)\not= 20$.
\end{remark}

\begin{thm} \label{MainResult}
The families $\mathcal{F}_\Delta$ and $\mathcal{F}_{\Delta^*}$ are lattice mirror symmetric for polytopes $\Delta$ in Corollary \ref{SurjectiveRestriction}, that is, $\Pic(\Delta)\simeq (U\oplus\Pic(\Delta^*))^\perp_{\Lambda_{K3}}$.
Explicitely, the lattices $\Pic(\Delta)$ are given as in the Table \ref{NicePairs2}.
\[
\begin{array}{cccccc}
\textnormal{Singularity} & \Pic(\Delta) & \rho(\Delta) & \rho(\Delta^*) & \Pic(\Delta^*) & \textnormal{Singularity} \\
\hline
\hline
Q_{12} & U\oplus E_6\oplus E_8 & 16 & 4 & U\oplus A_2 & E_{18}\\
\hline
Z_{1,0} & U\oplus E_7\oplus E_8 & 17 & 3 & U\oplus A_1 & E_{19}\\
\hline
E_{20} & U\oplus E_8^{\oplus 2} & 18 & 2 & U & E_{20}  \\
\hline
Q_{2,0} & U\oplus A_6\oplus E_8 & 16 & 4 & U\oplus C_8^6  & Z_{17}\\
\hline
E_{25} & U\oplus E_7\oplus E_8 & 17 & 3 & U\oplus A_1 & Z_{19}\\
\hline
Q_{18} & U\oplus E_6\oplus E_8 & 16 & 4 & U\oplus A_2 & E_{30}\\
\hline
\end{array}
\]
\begingroup
\captionof{table}{Picard lattices for lattice mirror symmetric pairs}\label{NicePairs2}
\endgroup
\end{thm}
{\sc Proof.} \,
For a lattice $L$, denote by $A_L$ the discriminant group of $L$, and $q_L$ the quadratic form of $L$.
We see in Proposition \ref{ExplicitLattices} that
\[
A_{\Pic(\Delta)} \simeq A_{U\oplus \Pic(\Delta^*)},
\textnormal{ and }
q_{\Pic(\Delta)} = -q_{U\oplus \Pic(\Delta^*)}
\]
for each case.
Thus by Proposition 1.6.1 in \cite{Nikulin80}, there is an isometry
\[
\Pic(\Delta) \simeq (U\oplus \Pic(\Delta^*))^\perp_{\Lambda_{K3}}.
\]
So $\mathcal{F}_\Delta$ and $\mathcal{F}_{\Delta^*}$ are lattice mirror symmetric.
We shall determine $\Pic(\Delta)$. \\

\noindent
${\bf 1. }\, Q_{12}$ and $E_{18}$ \,
By Corollary \ref{NLem4.3}
\[
\Pic(\Delta) \simeq (U\oplus U\oplus A_2)^\perp_{\Lambda_{K3}} \simeq
U\oplus (A_2)^\perp_{E_8}\oplus E_8 \simeq
U\oplus E_6\oplus E_8.
\]

\noindent
${\bf 2. }\, Z_{1,0}$ and $E_{19}$ \,
By Corollary \ref{NLem4.3}
\[
\Pic(\Delta) \simeq (U\oplus U\oplus A_1)^\perp_{\Lambda_{K3}}  \simeq
U\oplus (A_1)^\perp_{E_8} \oplus E_8 \simeq
U\oplus E_7\oplus E_8.
\]

\noindent
${\bf 3. }\, E_{20}$ and $E_{20}$ \,
Since $\Lambda_{K3}\simeq U^{\oplus 3}\oplus E_8^{\oplus 2}$
\[
\Pic(\Delta)\simeq (U\oplus U)^\perp_{\Lambda_{K3}} \simeq U\oplus E_8^{\oplus 2}.
\]

\noindent
${\bf 4. }\, Q_{2,0}$ and $Z_{17}$ \,
By Corollary \ref{NLem4.3}
\[
\Pic(\Delta)\simeq (U\oplus U\oplus C_8^6)^\perp_{\Lambda_{K3}} \simeq
U\oplus (C_8^6)^\perp_{E_8} \oplus E_8 \simeq
U\oplus A_6\oplus E_8.
\]

\noindent
${\bf 5. }\, E_{25}$ and $Z_{19}$ \,
By Corollary \ref{NLem4.3}
\[
\Pic(\Delta)\simeq (U\oplus U\oplus A_1)^\perp_{\Lambda_{K3}} \simeq
U\oplus (A_1)^\perp_{E_8} \oplus E_8 \simeq
U\oplus E_7\oplus E_8.
\]

\noindent
${\bf 6. }\, Q_{18}$ and $E_{30}$ \,
By Corollary \ref{NLem4.3}
\[
\Pic(\Delta)\simeq (U\oplus U\oplus A_2)^\perp_{\Lambda_{K3}} \simeq
U\oplus (A_2)^\perp_{E_8} \oplus E_8 \simeq
U\oplus E_6\oplus E_8.
\]
Thus the assertions are verified.
\QED

\section{Application} \label{Application}
Denote by $\Delta_B$ the reflexive polytope obtained in \cite{MU} for a singularity of type $B$.
As is seen in Table \ref{NicePairs}, there are isometric Picard lattices  $\Pic(\Delta_{Q_{12}})\simeq \Pic(\Delta_{Q_{18}})$, and $\Pic(\Delta_{Z_{1,0}})\simeq \Pic(\Delta_{E_{25}})$.
Families $\mathcal{F}_{\Delta_B}$ and $\mathcal{F}_{\Delta_D}$ are said to be {\it essentially the same} if general members in these families are birationally equivalent.
Not only the Picard lattices are isometric, but also we have
\begin{prop} \label{BirFamilies}
The families $\mathcal{F}_{\Delta_{Q_{12}}}$ $($resp. $\mathcal{F}_{\Delta_{Z_{1,0}}})$ and $\mathcal{F}_{\Delta_{Q_{18}}}$ $($resp. $\mathcal{F}_{\Delta_{E_{25}}})$ are essentially the same.
\end{prop}
\textsc{Proof. }\,
It is directly shown that the polytopes $\Delta_{Q_{12}}$ (resp. $\Delta_{Z_{1,0}}$) and $\Delta_{Q_{18}}$ (resp. $\Delta_{E_{25}}$) are isometric.
Indeed, define an invertible matrix in $GL_3(\mathbb{Z})$ as
\[
M_1 = \left( \begin{array}{ccc} -1 & 0 & 1 \\ 1 & 1 & -2 \\ 2 & -3 & 2 \end{array}\right)
\left( \textnormal{resp. }
M_2 = \left( \begin{array}{ccc} -1 & 2 & 1 \\ 1 & -3 & -1 \\ 2 & 0 & -1 \end{array}\right) \right)
\]
and one obtains an isometry
\[
m \, M_1 = m' (\textnormal{resp. } m\, M_2 = m' )
\]
that sends $m\in\Delta_{Q_{18}}$ to $m'\in\Delta_{Q_{12}}$  (resp. $m\in\Delta_{E_{25}}$ to $m'\in\Delta_{Z_{1,0}}$).
Therefore, there exists an explicit projective transformation that maps each Laurent polynomial in $|{-}K_{\mathbb{P}(\Delta_{Q_{12}})}|$ (resp. $|{-}K_{\mathbb{P}(\Delta_{Z_{1,0}})}|$) to a Laurent polynomial in $|{-}K_{\mathbb{P}(\Delta_{Q_{18}})}|$ (resp. $|{-}K_{\mathbb{P}(\Delta_{E_{25}})}|$).
This mapping also birationally sends a general member in $\mathcal{F}_{\Delta_{Q_{12}}}$ (resp. $\mathcal{F}_{\Delta_{Z_{1,0}}}$) to a general member in  $\mathcal{F}_{\Delta_{Q_{18}}}$ (resp. $\mathcal{F}_{\Delta_{E_{25}}}$).
Thus the statement is proved.
\QED \\

We conclude our study to remark that general members in compactifications of non-equivalent singularities can be transformed via a reflexive polytope.

redpajama_set_name: RedPajamaArXiv


text:
TenneT Presents 6 GW Offshore Wind Grid Solution for Germany
November 2, 2021, by Adrijana Buljan
TenneT has presented a new technological solution in Germany which can significantly accelerate offshore grid expansion for wind power in the North Sea and accommodate the build-out of 6 GW of offshore wind capacity in Germany three years earlier.
The solution includes TenneT's 2 GW standard for direct current connections, supplemented by further technological developments for the wind power booster concept. The concept revolves around three 2 GW wind power boosters, an interconnection hub called LanWin, and specific measures that would make it possible to integrate 6 GW of offshore wind power into the power grid as early as 2032, instead of 2035, i.e. three years sooner than previously envisaged in the grid development plan, in a first step.
LanWin hub; Image: TenneT
The wind power booster is a first concrete step towards a long-term meshed high voltage direct current (HVDC) grid at sea and on land. A meshed direct current grid (HVDC overlay grid) on land and at sea ensures long-term security of supply and reduces the economic costs of integrating renewable energies to achieve the climate targets, TenneT said.
With the proposed expansion path for a further 6 GW by 2032, which was previously not planned until 2035, Germany's offshore expansion targets can be significantly increased in the short term and thus contribute to accelerated implementation on the path to climate neutrality, according to the company.
The concept LanWin hub, an international interconnection in the North Sea. Interconnecting several offshore connections from different states would increase efficiency, security of supply and create new European electricity trading capacities. This would also support the German climate targets by 2045 as the country's offshore wind potential will not be sufficient to achieve them.
The modular concept of the LanWin hub enables, for example, interconnection with the Danish "energy island" and, in the future, with other North Sea countries. This distribution hub would be established in the LanWin area of the German Foreign Economic Zone in the North Sea, around 150 kilometres off the German coast.
Furthermore, at the offshore grid interconnection points near the coast, sector coupling with planned electrolysis projects as well as with the gas infrastructure and the feed-in of onshore wind energy is planned.
In order to supply private consumers, industry and planned electrolysis projects with offshore wind energy, as well as to enable a direct connection with continuing direct current lines on land, the three offshore connections will land close to the coast in Heide (Schleswig-Holstein), Wilhelmshaven (Lower Saxony) and in the Bremen area.
The wind power booster concept provides relief for the need for space and land in the coastal regions and at the same time offers new opportunities for the industrial clusters located there. Each of these three areas is to be integrated as a wind power booster with a 2 GW connection, TenneT said.
TenneT presented the concept on 29 October in Germany, together with the Energy Ministers of Lower Saxony, Olaf Lies, and Schleswig-Holstein, Jan Philipp Albrecht, as well as Bremen's State Councillor Kai Stührenberg – representing Bremen's Senator for Economic Affairs Kristina Vogt.
"Building on our extensive experience in the construction and operation of offshore grid connections, our conceptual and technological innovation for a six-gigawatt hub demonstrates a way to significantly accelerate the achievement of ambitious climate protection targets", said TenneT COO Tim Meyerjürgens.
"We see this as an offer also for the German government coalition negotiations currently underway, which, as we know, are seeking a boost for the expansion of renewable energies in Germany. At the same time, we are laying the foundation for a sustainable interconnection of the direct current grid. In this way, we are sustainably increasing efficiency as well as security of supply and making an important contribution to the intelligent coupling of offshore wind with electrolysers to be built close to generation and the gas grid".
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Photo: LanWin hub; Image: TenneT
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redpajama_set_name: RedPajamaCommonCrawl


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+ Why do I need Dental X-rays?
We use the theory of ALARA (As Low As Reasonably Achievable) when we set our standards for the use of x-rays. With our digital x-rays we have reduced the amount of radiation tremendously because our radiographic sensors are so sensitive to the exposure. Our digital x-rays have instant results. With our system we have the capability to digitally confirm when there is a cavity. This allows us to easily interpret and teach our patients exactly what we are seeing on our digital radiographs.
Fluoride has been one of the most cost effective and safe ways to increase the strength of teeth and prevents cavities. The American Academy of Pediatrics, the American Dental Association (ADA), and the CDC are in agreement that fluoride as an additive in toothpaste will benefit people. Both children and adults will benefit from using fluoride in the right dosage. The correct dosing and usage should be discussed with us at West Ridgewood Dental Professionals.
+ How do fillings fix teeth?
A cavity is weakened and soft tooth structure which continually grow from bacterial infiltration until they are removed and stopped. We gently and comfortably removed this diseased portion of the tooth and replace it with a filling to replicated natural anatomy, functionality and esthetics. Fillings can last an extremely longtime when the best materials are used which may be patient specific. In addition the patient must maintain these fillings like normal tooth structure and come in for routine checkups so we avoid and recurrent cavities.
+ What type of toothbrush and toothpaste should I use?
It is important to use a toothbrush with soft texture and bristles, this is to ensure we do not damage teeth and remove the outer enamel layer. Our recommendation is to brush a minimum of two times/day for two minutes each time. Many powered electric toothbrushes have a built in two minute timer.
Studies show that a powered electric toothbrush can but three times more efficient when brushing your teeth therefore more effective and can be more gentle on the teeth and gums when used properly.
After a four month period, when the bristles on the toothbrush are distorted, we know it is time to replace our toothbrush or brush head. After this period of time the brush is not as effective as it was when brand new.
Toothpaste is a minimal abrasive medium which will clean our teeth from the daily acquired bacteria and food. An American Dental Association (ADA) approved toothpaste will have fluoride contained within it. Fluoride is the only proven addative to strengthen tooth structure and enamel, therefore making it more resistant to cavities.
+ Do I need to floss and how often?
When we brush perfectly, the bristles of the brush cannot get between teeth that are in contact with one another. Cavities between teeth are more common in those who do not floss. Also gum inflammation is much more prevalent in those who are not flossing, removing the bacteria below the gumline. Over time the bacteria under the gumline will cause inflammation and bone loss and gum recession. This bacteria travels through the body and can cause damage to other systems in the body, not just the mouth.
+ What are signs I may have a dental issue?
If you feel or see something that looks unusual, come visit us so we may evaluate your mouth and all surrounding structures to ensure everything is healthy. Early diagnosis of dental issues many times can have a more favorable outcome, meaning less invasive and less costly treatment.
+ How often do I need a dental checkup?
Regular exams with a dentist can help to identify early signs of issues which may have minimally invasive treatments and therefore less costly.
We recommend a cleaning and checkup every six months, unless if there has been a history of aggressive gum disease or cavity formation. At West Ridgewood Dental Professionals we stress the importance of prevention, we take our time to educate all of our patients the proper techniques of cleaning their teeth and gums, so that we can continue to have great checkups. If something were to arise we would have the ability to catch it early and simply address it before it has dramatic and costly consequences.
+ What causes tooth pain?
Each tooth has a nerve (pulp) inside of them, and when this nerve is stimulated the result is pain. There are many reasons a nerve can be stimulated. A deep cavity, trauma to a tooth, exposed inner tooth structure from recession or a chip/fracture, infection can all lead to dental nerve pain also known as pulpitis. When you come to the dentist we can help determine if you are experiencing reversible or irreversible tooth pain and recommended the appropriate least invasive treatment. Sometimes we may expeience what we interpret as tooth pain but may actually be sinus inflammation (sinusitis) or jaw TMJ/TMD pain.
+ What is a cavity?
The outer tooth structure called enamel is the strongest substance in the human body. Plaque is a sticky film composed of food debris and bacteria in our mouths. There are both good and bad bacteria in our mouth. When we eat sugar and carbohydrates the bacteria consume these as well, and have a byproduct of acid. This acid over time decays tooth structure causing the enamel to become soft allowing easier penetration of the bacteria deeper into the tooth. When treating a cavity, we remove this weakened layer which is no longer protecting the tooth nerve, and replace it with an artificial layer which is selected by you (the patient) and your dentist.
+ what is a root canal?
A root canal is a procedure to remove and clean the nerve chamber inside a tooth. This procedure is needed when this sterile canal becomes contaiminated with bacteria due to a cavity or exposure into the mouth, or irreversible inflammation causing extreme tooth pain.
+ What is a dental crown?
A dental crown is essentially a filling which completely encompasses a tooth 360 degrees around. A crown is indicated when the remaining tooth structure is unable to retain a traditional filling due to size or retention.
+ what is a dental cap?
A dental cap is another name for a dental crown. A dental crown is essentially a filling which completely encompasses a tooth 360 degrees around. A crown is indicated when the remaining tooth structure is unable to retain a traditional filling due to size or retention.
Invisalign is an orthodontic procedure to move our teeth in a predictable manner. The benefit of Invisalign over traditional metal brackets is that they are clear retainers and they are removable. Having the ability to remove them for cleaning our teeth helps to decreases the possibility of dental cavities causes from foreign objects in our mouth retaining dental plaque.
Do you have your own question? Ask the Dentist!


text:
18,049 Wired, The Coronavirus Update
Octavio Islas. Fundador de la revista científica Razón y Palabra. 8 enero 2021 Boletines informativos
This is a special edition of our Coronavirus Update newsletter. To receive this email three times a week, sign up here.SIGN UP
US numbers continue to worsen, vaccine efforts accelerate worldwide, and researchers examine a new virus variant. Here's what you should know:HeadlinesThe pandemic is likely to get worse in the US even as vaccinations continueThe US' daily coronavirus death toll has reached a record high on five of the last 10 days, and it topped 4,000 on Thursday as cases continue to rise almost everywhere. Arizona leads the country in hospitalizations and cases, and Atlanta recently reopened its field hospital to keep up with record numbers of patients. The surge is also particularly bad in Southern California: Coronavirus kills someone every 15 minutes on average in Los Angeles, the county's public health director announced earlier this week. Ambulance crews have been told to ration oxygen for patients most in need. And some overflow patients have even been moved to hospital hallways and gift shops.But the country also set a less grim record Thursday: More than 600,000 coronavirus shots were administered in 24 hours, the most in a one-day period since vaccination efforts began. So far, the rollout has been slower than anticipated. While the Trump administration's policy has been to hold back half of US shots for second doses, Biden's team has announced that they will release all available doses once he enters office. In an interview on Thursday, Anthony Fauci added that there were always going to be hiccups, and vaccinations are likely to speed up in the weeks to come. But, he added, the pandemic will get worse before it gets better. Until enough people have been vaccinated, everyone must stay vigilant, whether they've received a shot or not.Vaccine efforts accelerate around the worldOn Thursday, Brazilian officials said that a vaccine from Chinese company Sinovac is effective. The company has already sold some 300 million doses, primarily to low- and middle-income countries, as it awaits approval. Meanwhile, in the EU, regulators approved Moderna's vaccine. Some have speculated they may also make a decision about the Oxford-AstraZeneca vaccine, which has already been approved for use in the UK and India, by the end of the month.On Sunday, India's drugs controller general also granted emergency approval to Covaxin, a vaccine developed jointly by Bharat Biotech and the Indian government. The announcement kicks off an ambitious drive to inoculate 300 million people in the country's first stage of vaccinations. The scientific community quickly criticized the news, because Covaxin has yet to complete its Phase III trials and has not produced sufficient efficacy and safety data. Later in the week, a national network of scientists put out a statement asking the government to reconsider its approval of the vaccine.A new virus variant spreads as the US ramps up genetic sequencingSince it was announced last month, a new, reportedly more transmissible variant of SARS-CoV-2, known as B.1.1.7, has spread across the UK and into the US. As of this week, the CDC has identified more than 50 cases of the mutated virus in California, Florida, Colorado, Georgia, and New York. Officials point out that these numbers likely don't reflect the number of cases actually circulating in the US, and they add that there's no evidence this variant is more deadly, though it appears to spread more rapidly. A new study from Pfizer found that the company's coronavirus vaccine appears to protect against this new mutated version.Recording these mutations—even the inconsequential ones—is essential for tracking the virus's spread and understanding its origins. And in order to do so, researchers must genetically sequence them all. In December, the CDC gave public health labs $15 million to boost efforts to conduct more genetic sequencing of virus samples. Some have speculated that the development of this new program, called the SARS-CoV-2 Strain Surveillance program, or NS3, was spurred in part by the discovery of this more transmissible strain. The effort is expected to be fully up and running by the end of January.Daily Distraction2020 was a big year for TV, and it's looking like 2021 could be even bigger. Here are all the shows WIRED is excited to watch in the new year.Something to ReadFrom learning TikTok dances to unpacking acronyms, staying current with what's online can feel like a thankless project even if you're still, well, young. Do we owe it to society to keep up? WIRED's resident advice columnist has some thoughts.Sanity CheckSometimes, deciding what music or podcast you want to listen to is a project unto itself. Here's our guide to finding the perfect audio to fit your mood.One QuestionWhat will conferences look like in a post-Covid world?Massive, packed in-person conferences will probably be one of the last things to resume once the pandemic is over. But even when event organizers go back to planning things in person, it's likely that they'll retain some of the elements of their virtual programming. Holding events primarily online is cheaper, so more people can join at less cost, and they make it easy for organizers to aggregate data about attendees. Digital events have driven home that brevity is key. As such, event organizers have gotten more creative with the kinds of programs, and breaks, they offer. In the long term, all of these changes might well be a good thing. But until then, conferences like this year's CES, which kicks off on Monday, are being held online. Here's what to expect from the first-ever completely virtual staging of the consumer tech industry's biggest event of the year.Covid-19 Care Package From social distancing to viral spread to staying sane, here's everything we know and advise about the coronavirus. The Covid-19 virus can linger on objects for as little as a few hours or as long as a couple of days, depending on the surface. Here's an updated look at the research. If you're planning to go out in public anytime soon, you're going to need a mask. Here are the best ones you can buy, or how to make one at home.🧼 It's not just your hands that need washing—your gadgets, clothes, and home need it too. Here's how to properly disinfect your stuff. Some of you are work-from-home pros, but if you're new to it, here's how to stay productive without losing your mind. It's hard not to be anxious about a global pandemic, but here's how you can protect yourself and your family without spiraling and how to not hate the loved ones you're quarantined with.✂ It may still be a while before you can see your hairstylist, so here's how to cut your hair at home, plus other ways to keep yourself lookin' fresh.🦠 Read all of our coronavirus coverage here.
Previo 18,048 ALL FOR PAPERS: Nordic Journal of Media Studies 2022
Siguiente 18,050 20th anniversary of the Explorations in Media Ecology
Publicado por Octavio Islas. Fundador de la revista científica Razón y Palabra.
Doctor en Ciencias Sociales.	Ver todas las entradas de Octavio Islas. Fundador de la revista científica Razón y Palabra.


Butterfly is another unique and engrossing experience from BOOM! Studios. The series follows Project Delta's top agent, Butterfly. She has no birth certificate, no papers and no friends. She is a ghost agent and she doesn't exist. After she is framed for a crime she didn't commit, she is now on the run and forced to trust someone she thought had died decades earlier. Her only hope of survival is trusting this stranger leaving one slight problem, her whole life she has been trained to trust no one.
An unusual series that moves between unusual tones but at its heart it's about the child parent relationship and how gaps in time can affect it. It seems that screenwriter Arash Amel and Marguerite Bennett have really gone for a unique look with this comic book series. You don't see too many comic books like this, it's pretty much a one of a kind. Enhancing this unusual perspective is some beautiful and wonderful art work all thanks to Antonio Fuso (The Girl Who Played with Fire) and Stefano Simeone.
The creators themselves compare the books main characters to ones in True Detective and House of Cards. These characters are deeply flawed and morally complex characters, not my words but theirs, though they are perfectly apt. The creative team have given something completely unique and wonderful with this series and it's definitely worth a read and then several more rereads.
Currently Issues 1-4 is available through BOOM! Studios here.


The Series went well with numbers overall up on last year. Great turn-out from Couriers to ensure all worked smoothly on race nights and we are indebted to Mike and Tony as so often for offering the sweet fruit of their long experience.
Congratulations to our runners and all those who helped.


As you stated he has been very versatile. And coaches value that, no matter the position.
That he can play CB and S also makes him a commodity on the open market.
If we are to keep him, it will cost some guaranteed dollars. But if he is better than those who are projected to make the FA market from other teams, he becomes worth it.


text:
Um poll tax é um imposto de uma soma fixa sobre cada indivíduo responsável (normalmente todos os adultos), sem referência a renda ou recursos. Embora, no caso dos Estados Unidos, sejam frequentemente associados a estados dos antigos Estados Confederados da América, os poll taxes também estavam em vigor em alguns estados do norte e oeste, incluindo Califórnia, Connecticut, Maine, Massachusetts, Minnesota, New Hampshire, Ohio, Pensilvânia, Vermont e Wisconsin. Os poll taxes foram uma importante fonte de financiamento para os governo entre as colônias que formaram os Estados Unidos. Eles representavam de um terço a metade da receita tributária do Massachusetts colonial. Vários privilégios de cidadania, incluindo o recenseamento eleitoral ou emissão de cartas de condução e licenças de caça e pesca de residente, eram condicionados ao pagamento de poll taxes para incentivar a arrecadação dessas receitas fiscais. Os impostos sobre a propriedade assumiram uma parcela maior das receitas fiscais, pois os valores das terras aumentaram quando o aumento da população incentivou a colonização do oeste americano.

Alguns estados do oeste não encontraram necessidade de requisitos de poll tax; mas os impostos e os incentivos de pagamento permaneceram nos estados do leste, e alguns métodos para o registro de eleitores foram modificados após a Guerra Civil Americanas até a ratificação da Vigésima Quarta Emenda em 1964, que proibiu o pagamento de poll taxes ou qualquer outro tipo de imposto como condicionantes para o direito ao voto.

História dos direitos civis nos Estados Unidos

redpajama_set_name: RedPajamaWikipedia


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text:
Cruise NewsLuxury Cruise News
Cruise Ship Completes Elusive Arctic Journey
Photo Courtesy of Crystal Cruises
The Crystal Serenity completes its ambitious journey sailing the Northwest Passage. The ship which is operated by luxury cruise line Crystal Cruises made history for being one of the largest cruise ships ever to sail the dangerous region.
The vessel arrived in New York City this morning which marks the end of the unique voyage which lasted 32 days. The Crystal Serenity sailed 7,297 nautical miles and carried over 1,000 guests with 600 crew members providing an amazing one of a kind service.
Edie Rodriguez, CEO and president at Crystal said:
"We are humbled and thrilled to have completed such a monumental journey. As Crystal is constantly seeking new ways to share the world with our guests, the Northwest Passage represents an especially massive undertaking that was made possible by the extreme dedication of our expert destination team and expedition partners. We now look forward to beginning the planning process in delivering another memorable experience for guests on our 2017 sailing. "
The voyage took more than three years of planning and during the cruise the Crystal Serenity was joined by the RRS Ernest Shackleton. In addition to being fully equipped as a first response vessel for virtually any emergency situation, the RRS Shackleton provided ice breaking capabilities, two helicopters for special adventures and additional expert expedition crew.
Crystal Serenity Captain Vorland said:
"From day one of planning the voyage, we were committed to ensuring the safety of our guests, crew and the ship. In addition to carrying two veteran Canadian Ice Pilots, Crystal Serenity's bridge team received ice navigation simulator training to prepare ourselves for the conditions, and prior to the voyage, the ship was outfitted with forward looking sonar, ice searchlights, ice radar, and a thermal imaging system were installed."
If you're interested in this epic journey then don't worry because Crystal Cruises will be sailing the Northwest Passage again in 2017. The voyage is open for booking but be quick as there's high demand.


text:
const React = require('react');
const ReactDOM = require('react-dom');
const TestUtils = require('react-dom/test-utils');

jest.dontMock('../view.jsx');

const Message = require('../view.jsx');

describe('Message', function() {
    function renderAndGetNode(xml, node) {
        const reactNode = TestUtils.renderIntoDocument(xml);
        const domNode = TestUtils.findRenderedDOMComponentWithTag(
            reactNode,
            node
        );

        return domNode;
    }

    function renderMessage(username, own, message_content, typing, message_type) {
        return renderAndGetNode(
            <Message username={username}
                     own={own}
                     message_content={message_content}
                     typing={typing}
                     messageType={message_type}/>,
            'li'
        );
    }

    it('renders a message', function() {
        const li = renderMessage(
            'hilbert',
            false,
            'Wir müssen wissen — wir werden wissen!',
            false,
            'text'
        );

        expect(li.textContent).toContain('hilbert');
        expect(ReactDOM.findDOMNode(li).className).toContain('other');
        expect(ReactDOM.findDOMNode(li).className).not.toContain('typing');
    });

    it('distinguishes your own messages', function() {
        const li = renderMessage(
            'hilbert',
             true,
             'Wir müssen wissen — wir werden wissen!',
             false,
             'text'
        );

        expect(ReactDOM.findDOMNode(li).className).toContain('self');
    });

    it('distinguishes messages that are being typed', function() {
        var li = renderMessage(
            'hilbert',
             true,
             'Wir müssen wissen — wir werden wissen!',
             true,
             'text'
        );

        expect(ReactDOM.findDOMNode(li).className).toContain('typing');

        var link = 'data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAHgAAAB4CAMAAAAOusbgAAAAG1BMVEX///8AAACdnZ3X19eZmZnq6ur7+/vOzs7x8fHjK9NyAAAARElEQVRoge3NiQmAMBAAsNrr4/4TC9YNlBNKskBKAT42W7317LgdS2THVSzeLu6xnCN7foy/YgAAAAAAAAAAAAAAgJcuaoEAp2NAe+UAAAAASUVORK5CYII=';

        li = renderMessage(
            'hilbert',
             true,
             link,
             true,
             'text'
        );

        expect(ReactDOM.findDOMNode(li).className).toContain('typing');
    });
});

redpajama_set_name: RedPajamaGithub


Crosshatch instituted guilds over three years ago. What started as small, informal groups of people hanging out and discussing their hobbies, has turned into slightly larger, more organized groups of people pooling their resources to save money, swap skills, and make friends. Current guilds exist for beekeepers, small farmers, orchardists, mushroom growers, grain growers, fiber folks, and green builders.
Join us for a celebration of the guilds - We'll enjoy an evening of warming food, adult beverages, and rapid-fire presentations on what the guilds have done and where they're headed. This is a time to learn from other guilds and socialize with each other.
Bring a friend or two who might be interested in joining a guild!
This is a free event, but you must RSVP no later than March 9th to Jen@crosshatch.org or 231-622-5252.


Check out our VIP Prediction for Real Oviedo vs Mallorca!
Real Oviedo and Mallorca will be face to face this Sunday in Adelante at 15:00 so try not to miss it.
Real Oviedo is ranked on the 7th place in the standings with 61 points, 47 goals scored and 47 goals received in the last 42 leaguegames. They have failed to win 15 of their last 42 matches in Adelante.In their last match they won against Elche. Real Oviedo lost only one match from the previous 5 played before this match with Mallorca.
The attacking skills of Real Oviedo are moderate, they won 2events from the last 5 played.
Mallorca are ranked on the 20th place in the standings with 45 points, 41 goals scored and 49 goals received this season in Adelante.In their last match they draw against Getafe. Mallorca lost only one match from the previous 5 played before this event with Real Oviedo.
On this page you can find our predictions & bettings tips for the match between Real Oviedo and Mallorca. Our predictions for Real Oviedo vs Mallorca are based on a large number of factors such as direct meetings, Real Oviedo last matches, Mallorca last matches, Real Oviedo lineup, Mallorca lineup, standings, insider informations and so on. By visiting our website daily you will increase your winning chances. All the predictions posted here have a huge rate of success!


text:
A plane with an anti-Pakistan slogan spotted flying over Headingley cricket ground in Leeds on June 29
Plane carrying anti-Pak banners fly over Headingley cricket ground in Leeds
ANI | Updated: Jun 29, 2019 19:04 IST
Leeds [UK], Jun 29 (ANI): A plane flying with anti-Pakistan banners unfurled from it was spotted on Saturday over the Headingley cricket ground here where the World Cup cricket match between Pakistan and Afghanistan was underway.
The plane unfurled banners in the sky bearing slogans 'Justice for Balochistan' and 'Help end enforced disappearances in Pakistan', written in big bold black and red letters as part of a campaign organised by Baloch activists to highlight the dire human rights violation in Pakistan.
"Baloch Republican Party and World Baloch organisation's awareness campaign against enforced disappearances in Pakistan flying banners hovering over Headingley Stadium calling for #EndEnforcedDisappearances during #PAKvAFG cricket match," the World Baloch Organisation tweeted.
In another tweet, it said: "The plane in its second run over #PAKvAFG #ICCWorldCup2019 match in Leeds, trailed a banner that read "Justice for #Balochistan", urging the Int'l community that the people of Balochistan demand justice for decades of atrocities by #Pakistani forces. #EndEnforcedDisappearances."
Following this, a scuffle broke out between Pakistan and Afghanistan fans outside the cricket stadium.
An ICC source told ANI that the "aircraft was an unauthorised" and said that Leeds air traffic will investigate the matter.
"A fight broke out between Pakistan and Afghanistan fans in Leeds after a plane was flown over the stadium which had Balochistan slogans. Apparently, it was an unauthorised plane that flew over the stadium and political messages were hung from the plane which were visible. Leeds air traffic will now investigate the matter," the ICC source said.
Human rights activists from the World Baloch Organisation and the Baloch Republican Party have been continuing their campaign to highlight the worsening human rights situation in Pakistan.
In their earlier initiative, roadside billboards were put up around the city of Birmingham with slogans "Help end enforced disappearances in Pakistan".
The billboards had gone up on major roads and junctions around the city including highways. This came after the city had hosted a World Cup match between Pakistan and New Zealand on June 26.
On June 23, Baloch Republican Party and World Baloch Organisation had organised a similar protest outside Lord's cricket stadium in London, where the protestors distributed leaflets to raise awareness about enforced disappearances of Baloch and other religious minorities in Pakistan, even as the match between Pakistan and South Africa was underway.
However, a group of Pakistani cricket fans tore down posters and banners put up by the Baloch activists.
The incident came even as Pakistan Army Chief Qamar Javed Bajwa was watching the match along with other officials in the stadium.
The angry fans had also shouted slogans -- "Long Live Pakistan Army". (ANI)
Baloch
'Will proceed as per law', says Pak after Kulbhustan Jadhav verdict
Islamabad [Pakistan], July 17 (ANI): Pakistan on Wednesday said that it will proceed as per law in the Kulbhustan Jadhav case, following the judgement by the International Court of Justice (ICJ) that Islamabad must review and reconsider the death sentence of the Indian national.
Sona Mosque in Bangladesh often finds visitor from West Bengal's...
Dhaka [Bangladesh], July 17 (ANI): A mosque in Bangladesh situated merely one kilometre from the Indian border is often visited from West Bengal's Malda district.
Pakistan seconds ICJ verdict on Kulbhushan Jadhav
Hauge [Denmark], July 17 (ANI): Attorney General for Pakistan Anwar Mansoor Khan on Wednesday hailed the verdict of the International Court of Justice (ICJ) on Kulbhushan Jadhav, where the world court asked Islamabad to review death sentence awarded to Jadhav.
"Great pressure" exerted to find Hafiz Saeed: Trump on JuD...
Washington DC [USA], July 17 (ANI): United States President Donald Trump on Wednesday said that "great pressure" was exerted over the past two years to find the Jamaat ud-Dawa (JuD) chief Hafiz Saeed. Trump made the statement hours after Saeed was arrested on charges terror financing in Pakistan.
Turkish diplomat killed by unidentified gunman in Iraq
Erbil [Iraq], July 17 (ANI): A Turkish diplomat was killed after an unidentified gunman opened fire in a restaurant in the Iraqi city of Erbil on Wednesday.
ICJ asks Pakistan to review conviction, sentencing of Jadhav
The Hague [Netherland], July 17 (ANI): In a major relief for India, the International Court of Justice (ICJ) on Wednesday asked Pakistan to review its conviction and sentencing of Indian national Kulbhushan Jadhav, facing execution on charges of spying.
Nagpur: German envoy meets RSS chief Mohan Bhagwat
Nagpur (Maharashtra) [India], July 17 (ANI): Germany's Ambassador to India Walter J Lindner met Rashtriya Swayamsevak Sangh (RSS) chief Mohan Bhagwat at the RSS headquarters in Nagpur on Wednesday.
26/11 mastermind Hafiz Saeed sent to 7-day judicial custody
Islamabad [Pakistan], July 17 (ANI): The 2008 Mumbai attacks mastermind and UN-proscribed Jamaat ud-Dawa (JuD) chief Hafiz Saeed on Wednesday was sent to seven-day judicial custody following his arrest on terror financing charges.
I'm treated like a dog, alleges pro-Khalistani leader Gopal Singh Chawla
Lahore [Pakistan] July 17 (ANI): Gopal Singh Chawla, former chief of Pakistan Sikh Gurdwara Parbandhak Committee (PSGPC) and a pro-Khalistani leader who was removed recently from a panel linked to Kartarpur Corridor, has alleged that Pakistan government is "treating him like a dog."
Islamabad [Pakistan], July 17 (ANI): The 2008 Mumbai attacks mastermind and UN-proscribed Jamaat ud-Dawa (JuD) chief Hafiz Saeed was arrested on terror financing charges while he was on way to Gujranwala from Lahore, officials of Punjab Counter-Terrorism Department (CTD) said on Wednesday.
Ursula von der Leyen elected as first female European Commission...
Brussels [Belgium], July 17 (ANI): Ursula von der Leyen of Germany has been elected as the first female president of the European Commission (EC) after her nomination was approved by 383 votes in a secret ballot on Tuesday evening (local time) at the European Parliament in Strasbourg.
Hafiz Saeed arrested, sent to judicial custody
Lahore [Pakistan], Jul 17 (ANI): 2008 Mumbai attacks mastermind Hafiz Saeed was on Wednesday arrested by the Counter-Terrorism Department (CTD) of Pakistan's Punjab province, said sources.


text:
Manhattan Transfers
Salmon Sells: New York Observer's Old Upper East Side Home Finds a Buyer
By Stephen Jacob Smith • 04/29/13 4:16pm
The hallowed halls in which The Observer was born.
The most expensive home in the five boroughs to go into contract last week (according to the Olshan report), is an address quite familiar to The Observer: 54 East 64th Street, this salmon-colored paper's Upper East Side home before Arthur Carter sold the paper to Jared Kushner.
"Four floors, a giant alimentary center-hall staircase, caked moldings, brass chandeliers, glass-fronted oak cupboards, The New York Observer sometimes felt like a Henry James society home or a 70′s swinger pad, with reporters stacked and stuffed in its confines like Hong Kong tailors," longtime editor Peter Kaplan described the mansion. "Our legal reporter set up his computer in the fourth-floor closet, near the tuxedo that was used by whomever had to go out to a formal evening."
But while The Observer has extensive knowledge of the mansion's past, everything about its present and future—from the buyer, to the seller, to the price of this current transaction—is shrouded in mystery. Alas, our reportorial powers do not include psychic ones, but we do have some clues.
The current bullpen is not nearly as glamorous.
The current owner of the 25-foot-wide, 12,551-square foot townhouse is SM 64th Street Holdings LLC, domiciled in a midtown lawyer's office. There has been some speculation that the townhouse is owned by a partner at Deloitte, but the fact that the owner took out a mortgage from Al-Mawarid Bank, headquartered in Beirut, would seem to point to an Arab seller.
The house also has an odd tendency to trade hands more often than your average Upper East Side manse (an old Observer curse?). Mr. Carter sold the brick beauty to Russian townhouse flipper Janna Bullock in 2004, who doubled her $9.5 million purchase with an $18.7 million sale just two years later—the only house she's ever actually made money on, and one that may have encouraged her to attempt a string of far less successful flips.
Ms. Bullock sold the property on to Irish businessman Derek Quinlan, who ran into difficulties once the Irish property boom turned bust, and had to sell it in 2011 for $20 million to pay off debts, reportedly in the hundreds of millions of euros.
Two years later, and it's being sold again. While it was listed for $28 million, the contract price is not public knowledge.
We did, however, speak to a few brokers who were skeptical that the seller fished his wish.
"I have a feeling it's going to be low," said one veteran Upper East Side townhouse broker, "because that broker"—Dolly Lenz, vice chairman at Douglas Elliman and often called America's top broker—"just doesn't care about the price."
"She tends to kind of dump her stuff," said another.
Ms. Lenz has not yet returned our call, so we'll have to wait and see whether a buyer thought the home's fine bones and noble lineage worthy of the full asking price.
Filed Under: Real Estate, Manhattan Transfers, Dolly Lenz, Janna Bullock, Arthur Carter, 54 East 64th Street, Derek Quinlan, SM 64th Street Holdings LLC
SEE ALSO: Lee Radziwill's Upper East Side Apartment Found a $4.9 Million Buyer


text:
Current: MATTA partners Indian travel associations to boost tourist inflow
Source: The Malaysian Reserve
The Malaysian Association of Tour and Travel Agents (MATTA) has inked a memorandum of understanding (MoU) with Indian travel associations to boost tourism between Malaysia and India in conjunction with Visit Malaysia 2020 (VM2020) campaign.
The national travel associations from India that have signed the MoU include Travel Agents Association of India (TAAI), Travel Agents Federation of India (TAFI) and Outbound Tour Operators Association of India (OTOAI).
"Malaysia is ready to welcome foreign tourists including from India for VM2020 and this MoU is a winning collaboration for MATTA, TAAI, TAFI and OTOAI, which augurs well for tourism in both countries," MATTA President Datuk Tan Kok Liang said in- a statement yesterday.
"MATTA will be introducing fresh itineraries and attractive tour packages to encourage more tourists from India to visit Malaysia," he said, adding that India has traditionally been contributing to the tourism industry.
According to Tan, the number of tourists from India has been increasing from 552,739 in 2017 to 600,311 in 2018 and about 750,000 in 2019, an increase of 23.2%.
Last week, The Malaysian Reserve reported that the Ministry of Tourism, Arts and Culture (MOTAC)anticipates higher tourism revenue from the rise of tourist arrivals from China and India following the implementation of visa-free entry.
MOTAC Minister Datuk Mohamaddin Ketapi said while no extra expenditure was incurred from the visa exemption, the increased arrivals are expected to boost earnings for the tourism industry Via the services, as well as food and beverage segments.
"I (do not) have a set number of targeted increase against last year, but hopefully with the (visa) exemption, there will be an increase in the tourist expenditure receipts," Mohamaddin said at the VM2020 vehicle stickers launch ceremony.
MATTA has engaged in activities to promote Malaysia to new and high potential markets, which include but not limited to Asean countries, India, China and European nations, especially from the Balkans, Poland and Hungary.
The association together with MOTAC and Tourism Malaysia have also made an effort to increase Malaysia's presence at global trade shows, such as Internationale Tourismus-Borse Berlin and World Travel Market London.
MATTA also has an extensive affiliation with regional tourism associations such as the Asean Tourism Association and the Federation of Asean Travel Associations, while also being the only Asian member of the European Travel Agents and Tour Operators Association.
Tourism Malaysia is looking at a 7% increase of tourist arrivals from China and India, which are among the largest markets for the country's tourism industry.
For VM2020, Tourism Malaysia aims for approximately 3.26 million Chinese tourists and 683,000 Indian tourists.
From January to September 2019, 2.41 million tourists from China and 539,167 tourists from India were recorded. — TMR
Updated: Monday, 13 January 2020
MAVCOM - CAAM merger will benefit tourism – MATTA