text:
\section{Introduction}

Let $Z$ denote a complex analytic surface in $\mathbb{C}^N$ which has an isolated singularity at the origin. By intersecting $Z$ with a small sphere centered at 0, we get a closed oriented $3$-manifold $M$, which is called the link of the singularity. The distribution of complex tangencies on $M$ is a contact distribution. In general the $3$-manifold $M$ may be the link of many other analytically distinct isolated singularities, but the induced contact structures are known to be contactomorphic \cite{CNP}. If a $3$-manifold is realized as a link of an isolated singularity, then the associated contact structure is called the canonical contact structure on $M$.  Properties of these contact structures have been extensively studied in the literature  \cite{AkhO, BO,LO}.
One remarkable feature of  canonical contact structures is that they are all Stein fillable;  a deformation of the minimal resolution of a normal surface singularity and the Milnor fiber of a smoothable singularity determine Stein fillings of the canonical contact structure.

The purpose of the present paper is to address the existence problem of Stein cobordisms between canonical contact structures on the link manifolds of various classes of singularities. This is also related to the problem of symplectically embedding one Milnor fiber into another. As a first step we work on almost rational (AR) singularities, which have been introduced by A.~N\'emethi as an extension of rational singularities. Recall that a complex surface singularity is rational if its geometric genus is zero. M.~Artin found out that this is  equivalent to the case when any positive divisorial 2-cycle in a resolution of the singularity has nonpositive arithmetic genus; besides this last condition is independent of the resolution \cite{Art}. N\'{e}methi investigated the behavior of the arithmetic genus function (which is equal to $1-\chi$ where $\chi$ is as in \eqref{eqn:chi}) on the lattice of homology 2-cycles of a resolution of a normal surface singularity. Through his observation, he deduced that if a normal surface singularity is rational then  its link manifold $M$ is an L-space
\cite[Theorem~6.3]{N}; i.e. $M$ is a rational homology sphere and its Heegaard Floer homology is isomorphic  to that of a lens space. The converse of this statement was also  proved recently in \cite{N4}.

Now, an AR-singularity is one which admits a good resolution whose dual graph is a negative definite, connected tree and has the following property: by reducing the weight on a vertex
we get the dual graph of a rational singularity. This property allows one to compute the Heegaard Floer homology of the link by a combinatorial process similar to Laufer's method of finding Artin's fundamental cycle \cite{N}. Even though the class of AR-singularities is restrictive, it is still large enough to contain the rational singularities and  all those singularities whose links are Seifert fibered rational homology spheres with negative definite plumbing graphs.  We call an almost rational singularity \emph{proper AR} if it is not rational.  After recalling some preliminaries on contact structures in Section~\ref{s:conpre} and on plumbings in Section~\ref{plumbing}, we present our first result in Section~\ref{construction} which claims the existence of a Stein cobordism from an arbitrary contact structure to the canonical contact structure of a proper AR-singularity.

\begin{theorem}\label{thm:steinAR}
Every closed, contact 3-manifold is Stein cobordant to the canonical contact structure of a proper AR-singularity.
\end{theorem}

The proof of  this theorem is an explicit construction of a Stein cobordism from the given contact manifolds to Brieskorn spheres.

By contrast there are obstructions to existence of Stein cobordisms going in the reverse direction. These obstructions come from Heegaard-Floer theory, particularly its plus flavor $HF^+$ \cite{OS4, OS5}. Recall that for any connected, closed, and oriented  $3$-manifold $M$ the Heegaard-Floer homology  group $HF^+(M)$ is a graded $\mathbb{F}[U]$-module, where $\mathbb{F}=\mathbb{Z}/2\mathbb{Z}$. Any oriented cobordism between two such manifolds induces a graded $\mathbb{F}[U]$ module homomorphism between the corresponding Heegaard-Floer homology groups.

To any co-oriented contact structure $\xi$ on a $3$-manifold $M$, there is an associated element $c^+(\xi)\in HF^+(-M)$, which is a contactomorphism invariant of the contact structure and
is natural under Stein cobordisms \cite{OS6}. In particular if $\xi$ is Stein fillable then $c^+(\xi)$ does not vanish since the tight structure on $S^3$ has non-vanishing  $c^+$. Moreover, $U(c^+(\xi))=0$. Utilizing this property, one can generate a numerical invariant of contact structures by letting
$$ \sigma(\xi)=-\mathrm{Sup} \left \{d\in \mathbb{N}\cup\{0\}:c^+(\xi)\in U^d \cdot HF^+(-M) \right \}.$$
The first author showed that $\sigma$ is monotone under Stein cobordisms and  can take all the values in the set $\{0,-1,-2,\dots,-\infty \}$ \cite{K}.   The computation of $\sigma(\xi)$ is in general hard, nevertheless we are able to show that it is zero for the canonical contact structures of proper AR-singularities. The Stein cobordism obstructions are obtained as immediate corollaries of this observation.

\begin{theorem}
\label{nocob}
Let $(M,\xi)$ be the canonical contact link manifold of a proper AR-singularity. Then  $\sigma(\xi)=0$. Hence there is no Stein cobordism from $\xi$ to
\begin{enumerate}
\item any contact structure supported by a planar open book,
\item any contact structure on the link of a rational singularity, or
\item any contact structure with vanishing Ozsv\'ath-Szab\'o invariant.
\end{enumerate}
\end{theorem}
In the course of proving this theorem we give an explicit method to detect the contact invariant $c^+(\xi)$ in the homology of  graded roots of A.~N\'emethi, if $\xi$ satisfies a certain compatibility condition with a plumbing graph.

 To detect $c^+(\xi)$ in the homology of  graded roots, one should understand the isomorphism
\begin{equation}
\Phi:HF^+(-M(\Gamma),\mathfrak{t}_{\mathrm{can}})\to \mathbb{H}(R_{\tau}).
\label{Fi}
\end{equation}
Here the left side is the Heegaard Floer homology of the 3-manifold described via an $AR$-plumbing graph $\Gamma$ in the canonical \spinc structure $\mathfrak{t}_{\mathrm{can}}$. The right side is the homology of a graded root $R_{\tau}$, which is much easier to compute. In Section~\ref{HF} we describe the isomorphism $\Phi$ explicitly along with the required algebraic objects, tell how $c^+$ is detected in graded roots and finally prove Theorem~\ref{nocob}. In Section~\ref{examples} we present explicit examples.

\section{Contact Preliminaries}\label{s:conpre}

In this section we mention some background material about contact geometry. Our main purpose is to set up our terminology. For a thorough discussion see \cite{OzbStip,Gei, CE}.

Contact structures on 3-manifolds can be studied topologically via open books through the Giroux corespondence.
We say that a contact structure $\xi$ on $M$ is compatible with an open book in $M$ if
on the oriented binding of the open book a contact form of $\xi$ is a positive volume form  and
away from the binding  $\xi$ can be isotoped through contact structures to be arbitrarily close  to the tangent planes of the pages of the open book.
Giroux correspondence states  that this compatibility relation is in fact a one-to-one correspondence between contact 3-manifolds up to contact isotopy and open books up to positive stabilizations.

Suppose $(W,J)$ is a compact complex surface with oriented boundary $-M_1\cup M_2$ that admits a strictly plurisubharmonic Morse function $\phi:W\to[t_1,t_2]$ such that $M_i=\phi^{-1}(t_i)$, for $i=1,2$, and $-dJ^*d\phi$ is a symplectic form. Then the set of complex tangencies on $M_i$ constitute a contact structure $\xi_i$, for $i=1,2$. In this case we say that $W$ is a Stein cobordism from $(M_1,\xi_1)$ to  $(M_2,\xi_2)$. If $M_1=\emptyset$ we say that $W$ is a Stein filling for  $(M_2,\xi_2)$.

Establishing the existence of a Stein cobordism between given contact manifolds is a delicate problem. Etnyre and Honda proved that an overtwisted contact 3-manifold is Stein cobordant to any contact $3$-manifold and that for any contact $3$-manifold $M$ there is a Stein fillable one to which $M$ is  Stein cobordant  \cite{EH}. The obstructions usually utilize Floer type theories. The first author used Heegaard Floer homology to prove non-existence of Stein cobordisms between certain contact manifolds \cite{K}; see also Sections \ref{HF} and \ref{examples} below. Another powerful tool  which obstructs more generally exact symplectic cobordisms is Latschev and Wendl's algebraic torsion which was  built in the framework of symplectic field theory \cite{LW}. In the appendix of the same paper Hutchings described the same obstruction in  embedded contact homology. Translation of the latter in the Heegaard Floer setting was recently given by Kutluhan et al \cite{KMVW}.

Starting with a contact $3$-manifold $(M_1,\xi_1)$ we can build a Stein cobordism as follows. First take the product $M_1\times [0,1]$ and equip it with the standard Stein structure. Then attach $1$-handles to the upper boundary.  Next attach $2$-handles along Legendrian knots in $M_1\sharp_n S^1\times S^2$ with framing one less than the contact framing.  By the topological characterization of Stein cobordisms  given by Eliashberg \cite{Eli}, see also \cite[Theorem~1.3]{Go}, the standard Stein structure on $M_1\times [0,1]$ extends uniquely  over such handles and all Stein cobordisms can be constructed this way. We are going to need a well-known partial translation of this recipe in terms of open books.

\begin{lemma}\label{l:monostei}
There exists a Stein cobordism from $(M_1,\xi_1)$ to $(M_2,\xi_2)$ if  there exist open books $(S_i,\phi_i)$ compatible with $(M_i,\xi_i)$ for $i=1,2$, such that $S_1$ and $S_2$ are homeomorphic surfaces and $\phi_2$ is obtained by multiplying  $\phi_1$  with right handed Dehn twists along non-separating curves.
\end{lemma}

\begin{proof}
By Legendrian realization principle \cite[Theorem~3.7]{H}, each non-separating curve on a page can be isotoped to a Legendrian curve staying on the same page such that the page framing and the contact framing agree. Then adding right handed Dehn twists to the monodromy along these curves precisely corresponds  to adding a Stein handle along the Legendrian curves.
\end{proof}
\section{Preliminaries on Plumbings}
\label{plumbing}
Let $\Gamma$ be a weighted connected tree with vertices $\{b_j: j \in \mathcal{J}\}$ for some finite index set $\mathcal{J}$. Let $e_j$ denote the weight of the $j$th vertex  $b_j$ for all $j\in \mathcal{J}$. We construct a $4$-manifold $X(\Gamma)$ as follows: For each vertex $b_j$,  take a disk bundle over a sphere whose Euler number is $e_j$ and plumb two of these together whenever there is an edge connecting the vertices. We denote the boundary 3-manifold by $M(\Gamma)$. We shall call such a graph  a plumbing graph. These graphs naturally arise in  singularity theory as good dual resolution graphs where   vertices correspond to  exceptional divisors  and the edges represent their intersection. See \cite{N3} for a survey on this topic.

\subsection{The lattices $L$ and $L'$} The 4-manifold $X(\Gamma)$ admits a handle decomposition without $1$-handles, so the second homology group  $L=H_2(X(\Gamma);\mathbb{Z})$ is freely generated by the the fundamental classes of the zero-sections of the disk bundles in the plumbing construction. Hence we have a generator of $L$ for each vertex of $\Gamma$. By abuse of notation we denote the generator in $L$ corresponding  to the vertex $b_j$ by the same symbol. The intersection form $(\cdot,\cdot)$ on $L$ is a symmetric bilinear function naturally characterized by $\Gamma$.

It is known that a plumbing graph can be realized as a good dual resolution graph of a singularity if and only if the intersection form is negative definite. Henceforth, we shall assume that all the plumbing graphs we consider satisfy this property. Therefore $L$ is a lattice and we have the short exact sequence
\begin{equation}\label{eq:dia1}
\begin{CD}
0 @>>> L  @>PD>>L' @>>> H @>>>0
\end{CD}
\end{equation}
where $L'$ is the dual lattice $\mathrm{Hom}_{\mathbb{Z}}(L,\mathbb{Z})\cong H^2(X(\Gamma);\mathbb{Z})\cong H_2(X(\Gamma),M(\Gamma);\mathbb{Z}) $  with $PD(x)=(x,\cdot)$, and $H=H_1(M(\Gamma);\mathbb{Z})$ .

We say that  $k\in  L'$ is \emph{characteristic} if for every vertex $b_j$ of $\Gamma$,
we have $k(b_j)+e_j\equiv 0 \text{ mod } 2$. Let $\mathrm{Char}(\Gamma)$ denote the set of characteristic elements in $L'$. The lattice $L$ naturally acts on $\mathrm{Char}(\Gamma)$ by the rule $x\ast k =k +2PD(x)$ for every $x\in L$.

The characteristic cohomology class $K\in L'$
satisfying $K(b_j)=-e_j-2$ for all $j\in \mathcal{J}$ is called the canonical class.
For each $k\in \mathrm{Char}(\Gamma)$, we define the function $\chi_k:L\to\mathbb{Z}$ by
\begin{equation}
\label{eqn:chi}
\chi_k(x)=-(k(x)+(x,x))/2.
\end{equation}
We simply use the symbol $\chi$ to denote $\chi_K$.

\subsection{\spinc structures}\label{s:spinc}
Every element of $\mathrm{Char}(\Gamma)$ uniquely defines a \spinc structure on $X(\Gamma)$. Two such \spinc structures induce the same \spinc structure on $M(\Gamma)$ if and only if the corresponding characteristic cohomology classes are in the same $L$-orbit.
For a fixed \spinc structure $\mathfrak{t}$ of $M(\Gamma)$, let $\mathrm{Char}(\Gamma,\mathfrak{t})$ denote the set of all characteristic cohomology classes which restrict to $\mathfrak{t}$ on $M(\Gamma)$.  Since $M(\Gamma)$ is a rational homology sphere,  it has finitely many \spinc structures; these are in one-to-one correspondence with elements of $H_1(M(\Gamma),\mathbb{Z})$. We can express $\mathrm{Char}(\Gamma)$ as a disjoint union of finitely many $\mathrm{Char}(\Gamma,\mathfrak{t})$s. In this paper, the default  \spinc structure, denoted by $\mathfrak{t}_{\mathrm{can}}$, is the one induced by the canonical class $K$.

\subsection{Almost Rational Graphs}
A plumbing graph $\Gamma$ is said to be \emph{rational}  if $\chi(x)\geq 1$ for any $x>0$.  Call a vertex $b_j$ in a plumbing graph a bad vertex if    $-e_j$ is greater than the valency of the vertex.  It is known that graphs with no bad vertices are rational. Moreover, rational graphs are closed under decreasing weights and taking subgraphs.

A graph is said to be \emph{almost rational} (or AR for short) if by decreasing the weight of a vertex $b_0$, we get a rational graph. Note that such a distinguished vertex $b_0$ need not be unique. Graphs with at most one bad vertex are AR \cite[Section~8]{N}. In particular  plumbing graphs of Seifert fibered rational homology spheres are AR. A plumbing graph is said to be proper almost rational if it is almost rational but not rational.

We say that a surface singularity is rational (respectively, AR and proper AR) if it admits a good resolution graph which is rational (respectively AR and proper AR). For example, for pairwise relatively prime integers $p,q$ and $r$, the link of the Brieskorn singularity
\begin{equation}\label{eq:Brieskorn}
x^p+y^q+z^r=0,
\end{equation}
is a Seifert fibered integral homology sphere which is  called  Brieskorn sphere $\Sigma(p,q,r)$. Such a singularity is proper AR unless $(p,q,r)=(2,3,5)$ \cite{R2,E,N}.

\section{Construction of Stein Cobordisms}
\label{construction}
The purpose of this section is to prove Theorem \ref{thm:steinAR}. In fact we will show that the proper AR-singularity in the  theorem can be chosen as a Brieskorn singularity.  For arbitrary positive integers $g$ and $n$, consider the triple $(p,q,r)$ with
$$p=2,\quad q=2g+1,\text{ and } r=(4g+2)n+1.$$
\noindent Note that $p,q,$ and $r$ are pairwise relatively prime for all $g$ and $n$, so the corresponding Brieskorn sphere $\Sigma(p,q,r)$ is an integral homology sphere, and the corresponding Brieskorn singularity is proper AR.

We shall describe an abstract open book which supports the canonical contact structure on $\Sigma(p,q,r)$. Let $S_g$ denote a compact orientable surface of genus $g$ with one boundary component. Let $\phi_{g,n}$ denote the diffeomorphism on $S_g$ which is a product of right handed Dehn twists along simple closed curves given by
\begin{equation}\label{eq:monodromy}
\phi_{g,n}=(t_{a_1}t_{a_2}\dots t_{a_{2g}})^{(4g+2)n+1},
\end{equation}
\noindent where the curves $a_1,a_2,\dots,a_{2g}$ form a chain as in Figure \ref{fig:Page_Chain}
\begin{figure}[h]
	\includegraphics[width=0.60\textwidth]{Page_Chain.eps}
	\caption{Curves $a_1,\dots a_{2_g}$ which form a chain on $S_g$.}
	\label{fig:Page_Chain}
\end{figure}

The chain relation \cite[Proposition 4.12]{FM} tells that
\begin{equation}\label{eq:chain}
(t_{a_1}t_{a_2}\dots t_{a_{2g}})^{(4g+2)}=t_\delta
\end{equation}
\noindent where $\delta$ is a simple closed curve on $S_g$ which is parallel to the boundary component. Since $\delta$ does not intersect with any of $a_1,\dots,a_{2g}$, the Dehn twist $t_\delta$ commutes with all of $t_{a_1},\dots,t_{a_{2g}}$.  Hence $\phi_{g,n}$ can also be written as
$$\phi_{g,n}=(t_{a_1}t_{a_2}\dots t_{a_{2g}})t_\delta ^n $$
\begin{lemma}\label{lem:sup}
The canonical contact structure on $\Sigma (p,q,r)$ is supported by $(S_g,\phi_{g,n})$.
\end{lemma}

\begin{proof} Consider the Milnor fiber $M(p,q,r)$ of the singularity \eqref{eq:Brieskorn}. By definition we have $\partial M(p,q,r)=\Sigma (p,q,r)$.
We shall construct a holomorphic Lefschetz fibration on $M(p,q,r)$ whose fibers are diffeomorphic to $S_g$ and monodromy factorizes exactly as in \eqref{eq:monodromy}. Our argument will be based on a work of Loi and Piergallini \cite{LP}, which relates branched covers of Stein 4-manifolds to Lefschetz fibrations.

Since $p=2$, the Milnor fiber $M(p,q,r)$ is a 2-fold branched cover of $B^4=\{ (y,z)\in\mathbb{C}^2\,:\, |y|^2+|z|^2\leq \epsilon \}$, branched along the Milnor fiber $M(q,r)$ of the plane curve singularity
\begin{equation}\label{eq:curve}
y^q+z^r=0
\end{equation}
\noindent Let $h:M(p,q,r)\to B^4$ be the covering map. It is well known that the link of the plane curve singularity \eqref{eq:curve} is the $(q,r)$-torus knot $T(q,r)$ in $S^3=\partial B^4$, and the Milnor fiber $M(q,r)$ is diffeomorphic to the minimal genus Seifert surface of $T(q,r)$ in $S^3$. Identify $B^4 \cong B^2\times B^2$
using the complex coordinates $(y,z)$ and consider the projection map to the second factor $\pi_2:B^4\to B^2$. The restriction of this map to $M(q,r)$ is a
simple $q$-fold branched covering
whose singular points, called the twist points, are in one-to-one correspondence with crossings of $T(q,r)$ represented as a braid with $q$ strands and $r$ twists. Here it is important that all the crossings of $T(q,r)$ are positive otherwise charts describing simple branched cover around the twist points would not be compatible with the complex orientation on $M(q,r)$. Let $s_1,s_2,\dots,s_{qr}$ be the twist points on $M(q,r)$. Without loss of generality we may assume that they are mapped to distinct points  $z_1,z_2,\dots,z_{r}$ under $\pi_2$. By \cite[Proposition 1]{LP},
the composition $f:=\pi_2\circ  h$ is a holomorphic Lefschetz fibration whose set of singular values  is $\{z_1,z_2,\dots,z_{r}\}$.

Next we determine the regular fibers of the Lefschetz fibration $f:M(p,q,r)\to B^2$. Away from the twist points, any  disk $B^2\times \{ \mathrm{pt} \}$ in $B^4$ intersects $M(q,r)$ at $q$ points.  Since $q=2g+1$, each  such disk lifts under $h$ to a genus $g$ surface $F$   with one boundary component; this forms a regular fiber of $f$. The restriction map $h|_F:F\to B^2\times \{\mathrm{pt}\}$
is modeled by taking the quotient of $F$ by the hyperelliptic involution whose $2g+1$ fixed points map to $B^2\times \{ \mathrm{pt} \} \cap M(q,r)$. See the marked points on $F$ in Figure \ref{fig:braid_lift}.

When a disk $B^2\times \{\mathrm{pt} \}$ intersects $M(q,r)$ at a twist point, its lift under $h$ is a singular fiber of $f$. We can describe the vanishing cycle of each singular fiber using the corresponding crossing of $T(q,r)$.  Any arc $\gamma$ on $B^2\times \{ \mathrm{pt}\}$
 connecting two strands of the braid $T(q,r)$ lifts to a unique  simple closed curve $\alpha$ on the regular fiber $F$. In Figure \ref{fig:braid_lift}, we indicated these arcs and curves using different colors. If a crossing exchanges these two strands connected by $\gamma$  then the corresponding singular fiber has vanishing cycle $\alpha$. Each such vanishing cycle contributes a right handed Dehn twist along $\alpha$  if the corresponding crossing is positive
(see \cite[Proposition 1]{LP} or \cite[Lemma 4.2]{BE}).
Following the braid direction we see that the monodromy is as described in \eqref{eq:monodromy}.
\begin{figure}[h]
	\includegraphics[width=0.50\textwidth]{braid_lift.eps}
	\caption{.}
	\label{fig:braid_lift}
\end{figure}

Having shown that the fibers of $f$ are diffeomorphic to $S_g$ and the total monodromy of $f$ agrees with $\phi_{g,n}$, we conclude that the restriction of $f$ on $\partial M(p,q,r)$ is the open book $(S_g,\phi_{g,n})$. Since the fibers of $f$ are complex submanifolds of $M(p,q,r)$, the open book $(S_g,\phi_{g,n})$ supports the canonical contact structure.
\end{proof}

\begin{remark}
An alternative proof of the above lemma goes as follows: Using handlebody techniques and the fact that $\Sigma(p,q,pqn+1)$ is $-1/n$ surgery on $(p,q)$-torus knot, one can directly verify that the total space of the open book $(S_g,\phi_{g,n})$ is $\Sigma(2,2g+1,(2g+1)n+1)$.
Using the chain relation and a result of W.~D.~Neumann and A.~Pichon \cite[Theorem 2.1]{NP}, one  can show that the open book supports the canonical contact structure.
\end{remark}

\begin{lemma}\label{lem:cob}
Given any contact 3-manifold $(M,\xi)$, there exist $g,n_0\in \mathbb{N}$ such that for every $n\geq n_0$ there is  a Stein cobordism from $(M,\xi)$ to the canonical contact structure on the Brieskorn sphere $\Sigma(2,2g+1,(2g+1)n+1)$.
\end{lemma}

\begin{proof}
Take an open book supporting $(M,\xi)$. If the pages of this open book have more than one boundary component, we can positively stabilize the open book to reduce the number of boundary components to one at the expense of increasing the page genus. Let $g$ denote the genus of the pages of the resulting open book after the stabilizations. Fix an identification of the pages with $S_g$, let $\phi$ denote the monodromy.

Write the monodromy as a product of Dehn twists about non-separating curves $c_1,c_2,\dots,c_k$ in $S_g$,
\begin{equation}\label{eq:factor}
\phi=t_{c_1}^{\epsilon_1}t_{c_2}^{\epsilon_2}\cdots t_{c_k}^{\epsilon_k}, \text{ where } \epsilon_i\in \{-1,+1\} \text{ for all } i=1,\dots,k
\end{equation}
\noindent The exponents in the above equation emphasize that the factorization $\phi$ can contain both right handed and left handed Dehn twists. We shall make the monodromy agree with $\phi_{g,n_0}$ for some large $n_0\in \mathbb{N}$ by multiplying it by only right handed Dehn twists. If $\epsilon_1=-1$ we simply multiply $\phi$ by $t_{c_1}$ from left to cancel the first term. If $\epsilon_1=+1$, we need to do more work! Since $c_1$ is non-separating, a self-diffeomorphism $\psi$ of $S_g$ sends $a_{2g}$ to $c_1$
(see Figure~\ref{fig:Page_Chain}).
Multiply both sides of the  chain relation \eqref{eq:chain} by $\psi^{-1}$ from right and by $\psi$ from left, and use the fact that $t_\delta$ is in the center of the mapping class group of $S_g$, to  see that
$$(t_{\psi(a_1)}t_{\psi(a_2)}\dots t_{\psi(a_{2g})})^{(4g+2)}=t_\delta.$$
Hence using  $(4g+2)(2g)-1$ right handed Dehn twists we can trade $t_{c_1}$ with $t_\delta$. The latter can be put at the end of the factorization \eqref{eq:factor}. Applying the same recipe to the remaining $c_i$'s we get the monodromy $t_\delta^{n_0}$, where $n_0$ is the number of positive exponents appearing in the factorization \eqref{eq:factor}. By adding Dehn twists appearing in the left hand side of the chain relation \eqref{eq:chain} as many times as necessary, we get the monodromy agree with $\phi_{g,n}$ for any $n\geq n_0$.

During the process we made only two kinds of modifications on the open book: positive stabilizations and adding a right handed Dehn twist to the monodromy. By Lemma~\ref{l:monostei} the required Stein cobordism exists.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:steinAR}]
Immediately follows from Lemma \ref{lem:sup} and Lemma \ref{lem:cob}.
\end{proof}

\section{Heegaard Floer homology, graded roots and Stein cobordism obstructions }
\label{HF}

To understand the image of $c^+(\xi)$  under $\Phi$ in (\ref{Fi}), we analyze the isomorphism $\Phi$  carefully by considering its factorization as follows:

\begin{equation}\label{eq:dia}
\begin{CD}
HF^+(-M(\Gamma)) @>\Phi_1>> \mathbb{H}^+(\Gamma)  @>\widetilde{\Phi}>> (\mathbb{K}^+(\Gamma))^* @>\Phi_2>> \mathbb{H}(R_\tau)
\end{CD}
\end{equation}

In order to introduce the notation and recall the definitions, we recite briefly  the construction of $HF^+$, $\mathbb{H}^+$, $\mathbb{K}^+$, and the isomorphisms $\Phi_1$ and  $\widetilde{\Phi}$ in Sections~\ref{HF+}-\ref{K+}
and the graded roots, their homology and the isomorphism  $\Phi_2$  in Sections~\ref{rootR}-\ref{laufer}. We describe how to detect root vertices of graded roots in  $(K^+(\Gamma))^*$  in Section~\ref{detecting} and distinguish the contact invariant in $(K^+(\Gamma))^*$ in Section~\ref{c+}. Finally we prove Theorem~\ref{nocob} in Section~\ref{pf2}.

\subsection{Plus flavor of Heegaard Floer homology}
\label{HF+}
Heegaard Floer theory is a package of prominent invariants for $3$-manifolds, \cite{OS4, OS5} see also \cite{J,M} for recent surveys. To every closed and oriented $3$-manifold $M$, and any \spinc structure $\mathfrak{t}$ on $M$, one associates an $\mathbb{F}:=\mathbb{Z}/2\mathbb{Z}$-vector space $HF^+(M,\mathfrak{t})$ which admits a canonical $\mathbb{Z}/2\mathbb{Z}$ grading.  In the case that the first Chern class of the \spinc structure is torsion (in particular when $M$ is a rational homology sphere), the $\mathbb{Z}/2\mathbb{Z}$-grading lifts to an absolute $\mathbb{Q}$-grading. There is also an endomorphism, denoted by  $U$, decreasing the degree by $2$  so as to make $HF^+(M,\mathfrak{t})$  an $\mathbb{F}[U]$-module. For example when the $3$-manifold is the 3-sphere and $\mathfrak{t}_0$ is its unique \spinc structure, we have $HF^+(S^3,\mathfrak{t}_0)=\mathcal{T}^+_{0}$ where $\mathcal{T}^+$
stands for the $\mathbb{F}[U]$-module $\mathbb{F}[U,U^{-1}]/U\mathbb{F}[U]$
and $\mathcal{T}^+_{d}$ denotes the one in which
the lowest degree element is supported in degree $d$.
More generally when $M$ is a rational homology sphere, it is known that the Heegaard Floer homology is given by $HF^+(M,\mathfrak{t})=\mathcal{T}^+_{d}\oplus HF_{\mathrm{red}}(M,\mathfrak{t})$ where $HF_{\mathrm{red}}(M,\mathfrak{t})$ is a finitely generated $\mathbb{F}$-vector space (and hence a finitely generated $\mathbb{F}[U]$-module) \cite{OS2}. In the sequel, we shall assume that all our $3$-manifolds are rational $3$-spheres.

Heegaard Floer homology groups behave functorially under cobordisms in the following sense: Suppose
$M_1$ and $M_2$ are connected, closed and oriented $3$-manifolds
and there is a cobordism $W$ from $M_1$ to $M_2$; i.e. the oriented boundary of $W$ is
$\partial W=(-M_1)\cup M_2$. For every \spinc structure $\mathfrak{s}$ on $W$, let $\mathfrak{t}_1$ and $\mathfrak{t}_2$ denote the induced \spinc structures on $M_1$ and $M_2$ respectively. Then we have an $\mathbb{F}[U]$-module homomorphism
$$ F_{W,\mathfrak{s}}:HF^+(-M_2,\mathfrak{t}_2)\to HF^+(-M_1,\mathfrak{t}_1).$$

\subsection{The graded module   $\mathbb{H}^+(\Gamma)$}
\label{H+} The  differential in the chain complex defining Heegaard Floer homology counts certain types of holomorphic disks. This aspect makes the computation of Heegaard Floer homology groups difficult in general. On   the other hand if a three manifold arises from a plumbing construction, there is a purely algebraic description for its Heegaard Floer homology.

 Let $\Gamma$ be a negative definite plumbing graph.
For every \spinc structure $\mathfrak{t}$ of $M(\Gamma)$, let $\mathbb{H}^+(\Gamma,\mathfrak{t})$ be the subset of the set of functions Hom$(\mathrm{Char}_{\mathfrak{t}}(\Gamma),\mathcal{T}^+)$  satisfying the following property. For every $k\in \mathrm{Char}_\mathfrak{t}(\Gamma)$ and every $j\in\mathcal{J}$, let $n$ be the integer defined by
\begin{equation}\label{e:kay}\chi_{k,j}=\chi_k(b_j)=-(k(b_j)+e_j)/2.\end{equation}
Then for every positive integer $m$, we require
$$U^{m-\chi_{k,j}}\phi(k+2PD(b_j))=U^m\phi(k) \text{ whenever } \chi_{k,j}\leq0, \text{ and}$$
$$U^{m}\phi(k+2PD(b_j))=U^{m+\chi_{k,j}}\phi(k) \text{ whenever } \chi_{k,j}>0.$$
We define  $\mathbb{H}^+(\Gamma)=\bigoplus_{\mathfrak{t}}\mathbb{H}^+(\Gamma,\mathfrak{t})$, which has readily an $\mathbb{F}[U]$-module structure. Moreover the  conditions
above provide the existence of a suitable grading on it.
A map $\Phi_1: HF^+(-M(\Gamma),\mathfrak{t})\to \mathbb{H}^+(\Gamma,\mathfrak{t} )$ can be described as follows: Remove a disk from the $4$-manifold $X(\Gamma)$ and regard it as a cobordism $\widetilde{X}$ from $-M(\Gamma)$ to $S^3$. For each characteristic cohomology class $k\in \mathrm{Char}_{\mathfrak{t}}(\Gamma)$ we have a map $F_{\widetilde{X},k}:HF^+(-M(\Gamma),\mathfrak{t})\to HF^+(S^3)=\mathcal{T}^+_{0}$. For any Heegaard Floer homology class $c\in  HF^+(-M(\Gamma))$ and for any characteristic cohomology class $k$, we define $\Phi_1(c)(k):=F_{\widetilde{X},k}(c)$. Thanks to adjunction relations, this map is a well defined homomorphism which is in fact an isomorphism when $\Gamma$ is AR. This was shown by   Ozsv\'ath and Szab\'o for the special case where $\Gamma$ has at most one bad vertex \cite{OS1} and later by N\'emethi in general \cite{N}.

\subsection{The dual of $\mathbb{H}^+(\Gamma)$ }
\label{K+}
There is a simple description of the dual of $\mathbb{H}^+(\Gamma)$. We use the notation $U^m\otimes k$ to denote a typical element of $\mathbb{Z}^{\geq 0}\times\mathrm{Char}(\Gamma, \mathfrak{t})$.  The elements of the form $U^0\otimes k$ are simply indicated by $k$.  Define an equivalence relation $\sim$ on $\mathbb{Z}^{\geq 0}\times\mathrm{Char}(\Gamma,\mathfrak{t})$ by the following rule. Let $j\in\mathcal{J}$
and $n$ be as in (\ref{e:kay}). Then we require
$$U^{m+n}\otimes(k+2PD(b_j))\sim U^m\otimes k \text{ if }n\geq 0,$$
\noindent and
$$ U^{m}\otimes(k+2PD(b_j))\sim U^{m-n}\otimes k \text{ if }n< 0.$$
Let $\mathbb{K}^+(\Gamma,\mathfrak{t})$ denote the set  of equivalence classes. Let $(\mathbb{K}^+(\Gamma,\mathfrak{t}))^*$ denote its dual.  For any non-negative integer $l$, let Ker~$U^{l+1}$ denote the subgroup of
$\mathbb{H}^+(\Gamma,\mathfrak{t})$ which is the kernel of the multiplication by $U^{l+1}$.
The map
$$\widetilde\Phi_l: \mathrm{Ker}\,(U^{l+1})\to\mathrm{Hom} \left (\frac{ \mathbb{K}^+(\Gamma,\mathfrak{t})}{\mathbb{Z}^{\geq l}\times{\mathrm{Char}_{\mathfrak{t}}(\Gamma)}},\mathbb{F} \right ),$$
given by the rule
$$\widetilde{\Phi}_l(\phi)(U^m\otimes k)=(U^m\phi(k))_0 $$
\noindent is an isomorphism for every $l$ \cite[Lemma~2.3]{OS1}. Here
$(\cdot)_0$ denotes the projection to the degree 0 subspace of  $\mathcal{T}^+$. Since every element of $\mathbb{H}^+(\Gamma, \mathfrak{t})$ lies in some $ \mathrm{Ker}\,(U^{l+1})$, the maps $\widetilde\Phi_l$ give rise to an isomorphism $\widetilde{\Phi}:\mathbb{H}^+ (\Gamma,\mathfrak{t})\to (\mathbb{K}^+(\Gamma,\mathfrak{t}))^*$.

\subsection{Graded roots}
\label{rootR}
Let $R$ be an infinite tree. Denote its  vertex set and edge set  by $\mathcal{V}(R)$ and $\mathcal{E}(R)$ respectively. Let $\chi:\mathcal{V}(R)\to \mathbb{Z}$ satisfy the following properties.

\begin{enumerate}
	\item $\chi(u)-\chi(v)=\pm 1$, if $[u,v]\in \mathcal{E}(R)$.
	\item $\chi(u)>\mathrm{min}\{\chi (v),\chi (w)\}$, if $[u,v]\in \mathcal{E}(R)$, and $[u,w]\in \mathcal{E}(R)$.
	\item $\chi$ is bounded below.
	\item $\chi^{-1}(n)$ is a finite set for every $n$.
	\item $|\chi^{-1}(n)|=1$ for $n$ large enough.
\end{enumerate}
Such a pair $(R,\chi)$ is called a \emph{graded root}. When the grading function $\chi$ is apparent in the discussion we simply drop it from our notation and use $R$ to denote a graded root.

Next we describe a particular graded root produced by a function $\tau:\mathbb{Z}^{\geq 0}\to \mathbb{Z}$ that is non-decreasing after a finite index. For each $i \in \mathbb{Z}^{\geqslant 0}$ consider the graded tree $R_i$ with vertices $\{v_i^m\}_{m\geq \tau(i)}$ and the edges $\{[v_i^m,v_i^{m+1}]\}_{m\geq \tau(i)}$ with grading $\chi(v_i^m)=m$.
Define an equivalence relation on the disjoint union $\coprod_i R_i$ of trees as follows:
$v_i^m \asymp v_j^n$ and $[v_i^m,v_i^{m+1}]\asymp [v_j^n,v_j^{n+1}]$ if and only if $m=n$ and
$m\geq \tau(l)$ for all $l$ between $i$ and $j$.
Then $R_{\tau}=\coprod_i R_i / \asymp$ is a tree with vertices the equivalence classes $\overline{v_i^m}$ and with the induced  grading $\chi(\overline{v_i^m}) = m$.

To each graded root $(R,\chi)$ as above, with vertex set $\mathcal{V}$ and edge set $\mathcal{E}$, one can associate a graded $\mathbb{F}[U]$ module $\mathbb{H}(R,\chi)$ as follows.
As a set, $\mathbb{H}(R,\chi)$ is the set of functions $\phi: \mathcal{V}\to \mathcal{T}^+_{0}$ satisfying
\begin{equation}
\label{deg:Hr}
U\cdot\phi (u)=\phi (v),
\end{equation}
\noindent whenever $[u,v]\in \mathcal{E}$ with $\chi(u)<\chi (v)$. The $U$-action on $\mathbb{H}(R,\chi)$ is defined by the rule $(U\cdot \phi) (v)=U(\phi(v))$. The grading on $\mathbb{H}(R,\chi)$ is defined in the following way. An element $\phi \in \mathbb{H}(R,\chi)$ is homogeneous of degree $d$ if for every $v\in\mathcal{V}$, $\phi(v)$ is homogeneous of degree $d-2 \chi(v)$.

\subsection{Laufer sequences}
\label{laufer} To simplify our discussion we will work with the canonical \spinc structure from now on; some of the discussion below works for a general \spinc structure, though.
Our aim is to describe the isomorphism
$$\Phi_2:(\mathbb{K}^+ (\Gamma, \mathfrak{t}_\mathrm{can}))^*\to \mathbb{H}(R_\tau), $$
discovered by N\'emethi, in a slightly different setup which suits to our needs; what we do
below is nothing but rephrasing the findings in \cite{N}.
Now, the isomorphism $\Phi_2$ is induced by  a bijection  between the dual objects:
$$\Phi_3:\mathcal{V}(R_\tau) \to  \mathbb{K}^+(\Gamma,\mathfrak{t}_\mathrm{can}).$$
Here $R_\tau$ is a graded root associated to a $\tau$ function and $\mathcal{V}(R_\tau)$ is its vertex set. We now describe the construction of the  function $\tau$.

Recursively form a sequence $(k(i))_{i=0}^\infty$ in $\mathrm{Char}(\Gamma, \mathfrak{t}_{\mathrm{can}})$ as follows: start with the canonical class $k(0)=K$. Suppose $k(i)$ has  already been constructed. We find $k(i+1)$ using the algorithm below.
\begin{enumerate}
\item \label{alg:N1} We construct a computational sequence $z_0,z_1,\dotsm z_l$. Let $z_0=k(i)+2\mathrm{PD}(b_0)$. Suppose $z_m$ has been found.  If there exists $j\in \mathcal{J}-\{0\}$ such that
$$z_m(b_j)=-e_j$$
then we let $z_{m+1}= z_{m}+2\mathrm{PD(b_j)}$.
\item If there is no such $j$, stop.  Set $l=m$ and $k(i+1)= z_{l}$.
\end{enumerate}
The sequence $(k(i))_{i=0}^\infty$ is called the Laufer sequence of the AR graph $\Gamma$ associated with the canonical \spinc structure. This sequence depends on the choice of the distinguished vertex $b_0$ but   is independent from our choice of vertices in step \ref{alg:N1} of the above algorithm.  Define $\chi_{i,0}=\chi_{k(i)}(b_0)$. Note that  since the elements of each  computational sequence described above satisfy $z_m\sim z_{m+1}$ for every $m=0,\dots,l-1$, the vectors $k(i)$ satisfy the following relations in $\mathbb{K}^+(\Gamma,\mathfrak{t}_{\mathrm{can}})$:
\begin{align*}
U^{\chi_{i,0}}\otimes k(i) \sim  k(i+1) & \text{ if }\chi_{i,0}\geq 0,  \text{ and }\\
k(i)\sim U^{-\chi_{i,0}}\otimes k(i+1) & \text{ if }\chi_{i,0}< 0.
\end{align*}
Let  $\tau(n)=\sum_{i=0}^{n-1}\chi_{i,0}$, with $\tau(0)=0$.  It can be shown that there exists an index $N$ such that  $\tau(i+1)\geq \tau(i)$ for all $i\geq N$ \cite[Theorem~9.3(a)]{N}.
Hence $\tau$ defines a graded root $R_\tau$ and indices beyond $N$ do not give essential information about  $R_\tau$. Moreover, for the canonical Spin$^c$ structure,
$\tau(N)\geq 2$ and $\tau(i)\leq 1$ for all $i\leq N-1$ \cite[Theorem~6.1(d)]{N}.
In that case one can stop the computation procedure of the Laufer sequence once $\tau\geq 2$. Now with a little effort, one can observe in that the map $\Phi_3:\mathcal{V}(R_\tau)\to \mathbb{K}^+(\Gamma)$ is defined by $\Phi_3(\overline{v_i^m})=U^{m-\tau(i)}\otimes k(i)$
(after the proofs of \cite[Theorem~9.3(b)]{N} and \cite[Proposition~4.7]{N}). One can check that $\Phi_3$ is well defined and injective and moreover $\Phi_3$ is in fact a bijection so it induces an $\mathbb{F}[U]$-module isomorphism $\Phi_2:(\mathbb{K}^+(\Gamma))^*\to \mathbb{H}(R_{\tau})$, which shifts grading by  $(K^2+|\mathcal{J}|)/4$

\begin{remark}
The Laufer sequence $\{x(i)\}$ in \cite{N} resides in $L$ while $\{k(i)\}$ here resides in $\mathrm{Char}(\Gamma, \mathfrak{t}_{\mathrm{can}})$. These two Laufer sequences  are related by $$ k(i)=K+2PD(x(i)).$$
\end{remark}

\begin{remark}
The sequence $(\tau(i))_{i=0}^\infty$ contains a lot of redundant elements. The finite subsequence $(\tau(n_i))$ consisting of local maximum and local minimum values of $\tau$ is sufficient to construct the graded root.
\end{remark}

\begin{remark}
An algorithm similar to what we have described above can be utilized to compute the Heegaard Floer homology groups for an arbitrary \spinc structure $\mathfrak{t}$. The only new necessary input is the  distinguished representative of  the \spinc structure inside $\mathrm{Char}(\Gamma,\mathfrak{t})$. Interested reader can consult \cite[Section 5]{N}.
\end{remark}

\subsection{Detecting root vertices}
\label{detecting} Ozsv\'ath and Szab\'o used a variation of the above algorithm to determine $\mathrm{Ker}U$ in $(\mathbb{K}^+(\Gamma))^*$ \cite[Section 3.1]{OS1}. Their elements  are also visible in the Laufer sequence. In a graded root $R$, say that a vertex is a \emph{root vertex} if it has valency $1$. The following lemma identifies root vertices of $R_\tau$ with elements of $\mathrm{Ker}U$.
\begin{lemma}\label{l:ktog}
 Given $k\in  \mathbb{K}^+(\Gamma,\mathfrak{t}_{\mathrm{can}})$  such that $k^* \in \mathrm{Ker}U$, there exists an element $k(i_0)$ of the Laufer sequence such that $k\sim k(i_0)$. This element is unique in the following sense: if $k(i_0)\sim k(i_1) \sim k$  and $i_0<i_1$ then $\tau(i)=\tau(i_0)$ for all $i$ satisfying $i_0\leq i\leq i_1$.  As a result $\Phi_3^{-1}(k)$ is the root vertex of the branch in the graded root $R_\tau$ corresponding to $\tau(i_0)$.
\end{lemma}

\begin{proof}
Since $\Phi_3$ is surjective, $k\sim U^{n}\otimes k(i_0)$ for some $i_0,n \in \mathbb{Z}^{\geq 0}$. Since $k^*$ is in $\mathrm{Ker}U$, $k$ does not admit any representation of the form $U^n\otimes k'$ unless $n=0$. Then the vertex $\overline{v_{i_0}^{\tau(i_0)}}$ in the graded root $R_\tau$  must have valency $1$ since otherwise  $v_{i_0}^{\tau(i_0)} \asymp  v_{i}^{\tau(i)+m}$ for some $i,m \in \mathbb{Z}^{\geq 0}$, implying that $k\sim U^m\otimes k(i)$, a case which we have dismissed. The same argument  proves that $\tau$ must be constant between $i_0$ and $i_1$ if $k(i_0)\sim k(i_1) \sim k$.
\end{proof}

\begin{lemma} \label{l:laufdetect}
If $k \in \mathrm{Char}(\Gamma,\mathfrak{t}_{\mathrm{can}})$ satisfies
$$e_j+2 \leq k(b_j)\leq -e_j-2,\text{ for all } j\in \mathcal{J},$$
\noindent then $k^* \in \mathrm{Ker}U$ and there exists a unique $i_0\in \mathbb{Z}^{\geq 0}$ such that $k=k(i_0)$. Consequently $\Phi_3^{-1}(k)$ is the root vertex of the branch in the graded root $R_\tau$ corresponding to $\tau(i_0)$. Moreover,  the index $i_0$ is the component of the vector $\mathrm{PD}^{-1}(k(i_0)-K)/2$ on $b_0$.
\end{lemma}

\begin{proof}
By \cite[Proposition~3.2]{OS1}, $k$ forms a good full path of length $1$ so it does not admit any representation of the form $U^n\otimes k'$ with $n\geq 1$, and it is the only element in its $\sim$ equivalence class. By Lemma~\ref{l:ktog}, we  must have $k= k(i_0)$ for some element $k(i_0)$ of the Laufer sequence. The claim about the index is in fact satisfied by every element of the Laufer sequence \cite[Lemma 7.6 (a)]{N}.
\end{proof}

\subsection{Contact invariant}
\label{c+} To any co-oriented contact structure $\xi$ on a $3$-manifold $M$, one associates an element $c^+(\xi)\in HF^+(-M)$. This element is an invariant of the contact structure and it satisfies the following properties:

\begin{enumerate}
\item $c^+(\xi)$ lies in the summand $HF^+(-M,\mathfrak{t}_\xi)$ where $\mathfrak{t}_\xi$ is the \spinc structure uniquely determined  by the homotopy class of $\xi$.
\item $c^+(\xi)$ is homogeneous of  degree $-d_3(\xi)-1/2$, where $d_3(\xi)$ is the $3$-dimensional invariant of $\xi$ of Gompf \cite{Go}.
\item When $\xi$ is overtwisted, $c^+(\xi)=0$.
\item When $\xi$ is Stein fillable $c^+(\xi)\neq 0$.
\item We have $U(c^+(\xi))=0$.
\item $c^+(\xi)$ is natural under Stein cobordisms.
\end{enumerate}

Our aim is to understand where the contact invariant falls under the isomorphism described in \eqref{eq:dia}. This could be difficult for a general contact structure. We need a certain type of compatibility of the contact structure with the plumbing.

\begin{definition}
Let $\Gamma$ be a plumbing graph. Suppose $\xi$ is a contact structure on $M(\Gamma)$. We say that $\xi$ is compatible with $\Gamma$ if  the following are satisfied:
\begin{enumerate}
\item The contact structure $\xi$ admits a  Stein filling whose total space, possibly after finitely many blow-ups, is $X(\Gamma)$.
\item The induced \spinc structure agrees with that of the canonical class on $M(\Gamma)$. (This condition is automatically satisfied when $M(\Gamma)$ is an integral homology sphere.)
\end{enumerate}
\end{definition}
Note that  canonical contact structures of singularities are compatible with the dual resolution graphs.
Furthermore if $\xi$ is compatible with $\Gamma$ then $X(\Gamma)$ is a strong symplectic filling for $\xi$.
We shall denote by $J$ and $c_1(J)$, the corresponding almost complex structure on $X(\Gamma)$  and its first Chern class respectively.
\begin{theorem}\label{cagri}(\cite[Proposition~1.2]{K})
Let $\xi$ be compatible with $\Gamma$. Then we have $\widetilde{\Phi}\circ \Phi_1(c^+(\xi))=(c_1(J))^*$.
\end{theorem}

\begin{proof}
If $X(\Gamma)$ has no $-1$ spheres,  the total space of the Stein filling is $X(\Gamma)$ itself. Then result is an easy consequence of the definitions of the maps $\Phi_1$ and $\widetilde{\Phi}$ and  Plamenevskaya's theorem \cite{P} which says that,
\begin{equation}\label{eq:pla} F_{\widetilde{X},k}(c^+(\xi))=\left \{
\begin{tabular}{lr}
1& \text{ if } $k \sim c_1(J)$,\\
0 & \text{ if } $k\not \sim c_1(J)$.
\end{tabular}\right .
\end{equation}

In the case that $X(\Gamma)$ contains $-1$ spheres, we blow them down until we get a Stein filling of $\xi$. Applying Plamenevskaya's theorem there and using blow-up formulas we see that \eqref{eq:pla} still holds.
\end{proof}

\begin{corollary}\label{cor:can}
 We have $\widetilde{\Phi}\circ \Phi_1(c^+(\xi_\mathrm{can}))=K^*$ where $K$ is the canonical class.
\end{corollary}

\subsection{Proof of Theorem~\ref{nocob}}
\label{pf2}
By Corollary~\ref{cor:can},  $\widetilde{\Phi}\circ \Phi_1(c^+(\xi_\mathrm{can}))=K^*$ where $K$ is the canonical class which is also the first element $k(0)$ of the Laufer sequence. Under the correspondence described in Lemma~\ref{l:ktog}, this element is associated with the root vertex of the branch of $\tau(0)$.  We will be  done once we prove this branch has length one, which implies that $\Phi(c^+(\xi_{\mathrm{can}}))$ is not in the image of $U^n$ for any $n> 0$. Since $\Phi$ is an $\mathbb{F}[U]$ module isomorphism, the same must hold for $c^+(\xi_{\mathrm{can}})$.

By definition, $\tau(0)=0$, and a direct computation shows $\tau(1)=1$. Now, by \cite[Theorem~6.1(d)]{N}
it follows that $\#\tau^{-1}(m)=1$ whenever $m\geq 1$; equivalently if $\tau(n)>1$ for some $n$
then $\tau$ is increasing beyond $n$. Then we have two cases: either $\tau$ is always
increasing or there is some $n$ for which $\tau(n)<1$ and $\tau(j)=1$ for $1\leq j \leq n$. In the former case we get $\mathbb{H}(R_\tau)=\mathcal{T}^+_0$. However this is equivalent to having $\Gamma$ rational (\cite[Theorem~6.3]{N}), which we have dismissed by assumption.
In the latter case, there are more than one root of the  tree $R_{\tau}$  which have non-positive degree.
In particular the branch corresponding $\tau(0)$ has length $1$ (as illustrated in Figure~\ref{fig:tree}), and therefore we have $\sigma(\xi)=0$.

\begin{figure}[h]
	\includegraphics[width=0.40\textwidth]{gengra}
	\caption{If $\Gamma$ is proper AR, this tree embeds into the graded root $R$ where the red vertex is identified with $\tau (0)$.}
	\label{fig:tree}
\end{figure}

Finally, there can be no Stein cobordism from $(M(\Gamma),\xi)$ to $(M',\eta)$ if $\sigma(\eta)<0$. This follows from the naturality of the invariant $c^+$ under Stein cobordisms and the fact that
the invariant $\sigma$ increases under a Stein cobordism \cite[Theorem~1.5]{K}. To finish the proof
we recall that for each of the cases in the theorem, $\sigma(\eta)=-\infty$:
this is immediate if $c^+(\xi)=0$;
for the case $\eta$ is a planar contact structure, this claim is just \cite[Theorem~1.2]{OSS};
for the case when $M'$ is the link of a rational singularity, this is a consequence of  \cite[Theorem~6.3]{N}.
\hfill $\Box$

\section{Examples}\label{examples}
The main result of our paper concerns canonical contact structures but with the techniques we developed in this paper we can in fact compute the $\sigma$ invariant of any contact structure which is compatible with an AR plumbing graph. Here we give a few examples.

\subsection{Contact structures on $\Sigma(2,3,11)$} We start with a simple but an instructive example. The Brieskorn sphere $\Sigma(2,3,11)$ is the boundary of the graph in Figure~\ref{fig:plumbing}
which we denote by $\Gamma$. We index the vertices $\{b_0,b_1,\ldots,b_8\}$
of $\Gamma$ so that
$b_0$ is the one with adjacency 3 and $b_8$ is the one with weight $-3$.
We know that $\Gamma$ is  proper AR since $(2,3,11)$ are pairwise relatively prime.

\begin{figure}[h]
	\includegraphics[width=0.40\textwidth]{plumbing.eps}
	\caption{Plumbing graph of the Brieskorn sphere $\Sigma(2,3,11)$.}
	\label{fig:plumbing}
\end{figure}

One may find compatible Stein structures on the $4$-manifold $X(\Gamma)$ by choosing  the Legendrian
attaching circles of $2$-handles corresponding to vertices with the prescribed intersection matrix
such that the smooth framing is one less than the Thurston-Bennequin framing.  Note that there is a
unique way of doing this for each $(-2)$-framed vertex, but the $(-3)$-framed $2$-handle can be
Legendrian realized in two different ways. We fix an orientation on this vertex and distinguish these
two cases according to their rotation numbers \cite[Theorem~1.2]{LM}. Hence $X(\Gamma)$ has two natural distinct Stein structures  whose Chern classes are $k_{\pm}=[0,\dots,0,\pm 1]$ where we write an element
$k\in H^2(X(\Gamma),\mathbb{Z})$ in the form $[k(b_1),\dots,k(b_s)]$ in the dual basis. Denote the corresponding contact structures on boundary $\xi_{\pm }$.

\begin{figure}[h]
	\includegraphics[width=1.0\textwidth]{legplumbing.eps}
	\caption{Stein handlebody pictures of $\xi_{\pm }$.}
	\label{fig:legplumbing}
\end{figure}

The Heegaard Floer homology of $\Sigma(2,3,11)$ is well known; we can readily compute it here
by considering the Laufer sequence $k(n)$ and the corresponding values $\tau(n)$. Moreover from Lemma~\ref{l:laufdetect},  we know that $k_\pm$ will appear in the Laufer sequence.
The first several $k(n)$ and $\tau(n)$ are as follows:
\begin{align*}
k_+= k(0)&=(0,0,0,0,0,0,0,0,1), \quad &\tau(0)=0 \\
k(1)&=(2,-2,0, -2, 0, 0, 0, 0, -3),\quad &\tau(1)=1 \\
k(2)&=(2, 0, -2, 0, 0, 0, 0, -2, -1), \quad &\tau(2)=1 \\
k(3)&=(2, -2, 0, 0, 0, 0, -2, 0, -1), \quad &\tau(3)=1 \\
k(4)&=( 2, 0, 0, -2, 0, -2, 0, 0, -1),\quad  &\tau(4)=1 \\
k(5)&=(4, -2, -2, 0, -2, 0, 0, 0, -1), \quad &\tau(5)=1 \\
k_-=k(6)&=(0, 0, 0, 0, 0, 0, 0, 0, -1), \quad &\tau(6)=0
\end{align*}
Moreover it can be proven that $\tau(n+1)\geq \tau(n)$ for every $n\geq 6$. An indirect way to do this is by observing that $\tau(13)=2$ and $\tau(n)=1$ for $7\leq n <13$ and then  employing
\cite[Theorem~6.1(d)]{N} to conclude that $\tau$ is increasing for $n\geq 13$. Hence we obtain
the graded root $R_{\tau_K}$ as shown in Figure~\ref{fig:graded2311}. By the previous discussion, the  contact invariants $c^+(\xi_{\pm})$ correspond to the two roots of the tree. As a result, we have $\sigma(\xi_\pm)=0$. Notice that $k_+=K$, and so $c^+(\xi_+)=c^+(\xi_\mathrm{can})$.

\begin{figure}[h]
	\includegraphics[width=0.60\textwidth]{graded2311.eps}
	\caption{The graded root of $\Sigma(2,3,11)$ with the canonical \spinc structure. }
	\label{fig:graded2311}
\end{figure}
Finally we determine the Heegaard Floer homology by computing the homology of the graded root and shifting the degree by $(K^2+9)/4$. We get that $HF^+(-\Sigma(2,3,11))=\mathcal{T}^+_{(-2)}\oplus \mathbb{F}_{(-2)}$ and $c^+(\xi_\pm)$ are two distinct elements of degree $-2$ which project non-trivially to the reduced Floer homology.

\subsection{Stein fillable contact structures of arbitrarily large $\sigma$} Consider the infinite family of Brieskorn spheres $M_n=\Sigma(3,3n+1, 9n+2)$, $n\geq 1$ given by plumbing graph $\Gamma_n$ in Figure~\ref{fig:plumpqpq-1}. For every $m=0,\dots,n-1$ we have the  Stein structure $J_m$ corresponding to the Legendrian handle attachments along Legendrian unknots corresponding to the vertices of $\Gamma_n$ with framing $tb-1$. In order to get the correct framing we have to stabilize $(-3)$-and $(-n-1)$-framed unknots once and $n-1$ times respectively. To get the Stein structure $J_m$  we do one left stabilization on $(-3)$-framed unknot, and  $m$ right  and  $n-m-1$ left stabilizations on $(-n-1)$-framed unknot see Figure~\ref{fig:legplumbing2}. Orienting each unknot clockwise we fix a basis for $b_0,\dots,b_{9n+5}$ for $L$ using the indices as show in Figure~\ref{fig:plumpqpq-1}. Then the number $c_1(J_m)(b_j)$ is given by the rotation number of the Legendrian unkot corresponding to the $j$th vertex.  Hence we have $c_1(J_m)=[0,1,0,0,n-2m-1,0,\dots,0]$. Clearly for every $m=0,\dots,n-1$, the contact structure $\xi_m$ induced by $J_m$ is compatible with $\Gamma_n$. Hence by Lemma~\ref{l:laufdetect}, each $c_1(J_m)$ appears in the Laufer sequence of $\Gamma_n$. In fact we have $c_1(J_m)=k(i_m)$, where $i_m=3m(9n+2)$.

\begin{figure}[h]
	\includegraphics[width=.50\textwidth]{plumpqpq-1.eps}
	\caption{Plumbing graph  $\Gamma_n$.}
	\label{fig:plumpqpq-1}
\end{figure}

\begin{figure}[h]
	\includegraphics[width=0.70\textwidth]{legplumbing2.eps}
	\caption{Stein handlebody diagram of $J_m$}
	\label{fig:legplumbing2}
\end{figure}

We need to decide the branch lengths of the corresponding root vertices in the graded root. Borodzik and N\'emethi worked out the combinatorics of the $\tau$ function.
\begin{lemma}\cite[Propositon 4.2]{BN}  Suppose $p$ and $q$ are relatively prime positive integers. Let $\mathcal{S}_{p,q}$ be the semigroup of $\mathbb{N}$ generated by $p$ and $q$ including $0$. Let $\delta=(p-1)(q-1)/2$. Consider the function $\tau:\mathbb{Z}^{\geq 0}\to \mathbb{Z}$ associated to the Brieskorn sphere $\Sigma(p,q,pq-1)$.

The function $\tau$ attains its local minima at $a_t=t(pq-1)$ for $0\leq t \leq 2\delta -2$, and local maxima at $A_t=tpq+1$ for $0\leq t\leq 2\delta -3$. Moreover for any $0\leq t\leq 2\delta -3$, one has
\begin{align*}
\tau(A_t)-\tau (a_{t+1})&=\#\{s\not \in \mathcal{S}_{p,q}\, : \, s\geq n+2 \}>0, \\
\tau(A_t)-\tau (a_{t})&=\#\{s\in \mathcal{S}_{p,q}\, : \, s\leq n \}>0.
\end{align*}

\end{lemma}

Applying the above lemma for the case $p=3$ and $q=3n+1$, we see that $\sigma(\xi_m)=-m$. Hence by choosing $n$ and $m$ appropriately we realize any negative integer as the $\sigma$ invariant of a contact structure. Of course $\xi_m$ cannot be isomorphic to the canonical contact structure unless $m=0$. Therefore the following problem is natural.

\begin{question}
Is it possible to realize any negative integer as the $\sigma$ invariant of the canonical contact structure of a singularity?
\end{question}
\section{Acknowledgments}
The first author  is supported by a TUBITAK grant BIDEB 2232 No: 115C005. The second author is grateful to the Institut Camille Jordan, Universit\'{e} Lyon 1, where part of this work was completed.

redpajama_set_name: RedPajamaArXiv


text:
The 20 N.J. school districts most dependent on state funding
Updated Jan 16, 2019; Posted Feb 28, 2017
N.J. school districts most reliant on state funding
Gallery: N.J. school districts most reliant on state funding
By Adam Clark | NJ Advance Media for NJ.com
TRENTON -- When Gov. Chris Christie delivers his 2018 budget address on Tuesday, New Jersey school officials will be listening especially closely.
How Christie will address education funding is the biggest question about his final budget, leaving administrators bracing for the possibility of funding cuts.
There's some concern Christie could follow through with the "Fairness Formula," a plan he unveiled last summer to give every district $6,599 per student regardless of income or other needs. Though many education groups are convinced Christie won't do that, they still don't have high hopes for increases in school funding.
"I'm not expecting any good news in the budget," said Richard Bozza, executive director of the New Jersey Association of School Administrators.
New Jersey spends more than $9 billion a year on public schools using a formula that considers each community's ability to raise revenue through property taxes.
It expects wealthier communities to pay a higher percentage of school costs with their own tax dollars and gives school districts extra money for students from low-income families.
That means some school districts rely almost entirely on the state for revenue, while local tax payers shoulder the majority of the burden for other districts. In other words, some districts have a lot more at stake in the state budget.
Browse the gallery above to see the 20 New Jersey school districts most reliant on state aid based on data from the 2014-15 school year, the latest available from the state. The figures come from the state's Taxpayers Guide to Education Spending.
Most of the districts that rely heavily on state aid were represented in the landmark Abbott vs. Burke court case, which guarantees a high level of state funding to urban and low-income districts.
Adam Clark may be reached at adam_clark@njadvancemedia.com. Follow him on twitter at @realAdamClark. Find NJ.com on Facebook.

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: qualtRics fetch_mailinglist error: Can't coerce from a character vector to a logical vector Using qualtRics::fetch_mailinglist to pull a mailing list from a Qualtrics account no longer seems to work, I think due to recent changes on the Qualtrics end. As of December 2022, this worked, producing a dataframe of contact list information:
(Sorry I can't provide an entirely reproducible example without posting actual credentials and list IDs)
library(qualtRics)

# Qualtrics credentials
qualtrics_api_credentials(
  api_key="YourAPIkeyHere",
  base_url="YourBaseURLHere.qualtrics.com",
  install=TRUE, overwrite=TRUE
)
# renew environment to use credentials
readRenviron("~/.Renviron")

mailDf <- fetch_mailinglist("CG_BunchaNumbersHere")

As of January 2023, it now produces the following error:

Error in purrr::map_lgl():
ℹ In index: 1.
Caused by error:
! Can't coerce from a character vector to a logical vector.

rlang::last_error() gives me the following:

Backtrace:

*

*qualtRics::fetch_mailinglist("CG_BunchaNumbersHere")

*purrr::map_lgl(elements, "unsubscribed", .default = NA)

*purrr:::map_("logical", .x, .f, ..., .progress = .progress)

*purrr:::call_with_cleanup(...)

rlang::last_trace() gives me:

Backtrace:
▆

*

*├─qualtRics::fetch_mailinglist("CG_BunchaNumbersHere")

*│ └─tibble::tibble(...)

*│   └─tibble:::tibble_quos(xs, .rows, .name_repair)

*│     └─rlang::eval_tidy(xs[[j]], mask)

*├─purrr::map_lgl(elements, "unsubscribed", .default = NA)

*│ └─purrr:::map_("logical", .x, .f, ..., .progress = .progress)

*│   ├─purrr:::with_indexed_errors(...)

*│   │ └─base::withCallingHandlers(...)

*│   └─purrr:::call_with_cleanup(...)

*└─base::.handleSimpleError(...)

*└─purrr (local) h(simpleError(msg, call))

*
└─cli::cli_abort(...)

*
  └─rlang::abort(...)

I noticed that the prefix for mailing list IDs recently changed from "ML_" to "CG_", and "Contact Lists" are now called "Directories." Whatever they changed interferes with both recently created surveys/mailing lists and older ones that use the "ML_" prefix.

redpajama_set_name: RedPajamaStackExchange


text:
Home » Movies » 1987 (#78)
1987 (#78)
In an earlier post, Montréwood movies, I provided a list of award-winning Montréwood films in the categories of drama, comedy, and abstract/non-conventional.
A few weeks ago, I wrote a post, "Mommy, Now playing in your city", with a heads-up of pan-Canadian theatre showings for the highly acclaimed, Mommy.
Now that 2014 is wrapping up in seven weeks, we now have a good idea which Montréwood film, other than Mommy, takes the cake of success this year: 1987.
The film, 1987, is a sequel to the film 1981 (which came out in 2009).
To meld the two plots together, the 2009 film 1981 was set in the year 1981. It was about a boy, Ricardo, who was out to impress classmates who he felt he didn't measure up to. He did so through lies and deceptions, and got himself into a bit of trouble.
Fast-forward six years, and in this year's film 1987 (set in the year 1987) Ricardo is now 17 years old. He's in an age-group which experiences and lusts for a multitude of "firsts", and Ricardo is no exception. He wants desperately to earn a timeless badge of honour; to manage to get into a bar while still underage. To the chagrin of his parents, he's out to have a lot of fun (partying, hanging out, and even starting his own bar/dance club for underage peers). He's got it in his mind that he needs to get into the sack with his girlfriend (his first experience), but she's not so hot on the idea and wants to make sure he's the right one for her. But like many schemes which start out innocent enough, with little money to realize his goals, he gets himself deeper and deeper into trouble. He starts to trade in stolen vehicle parts, and quickly becomes the head of the Sainte-Foy mafia (a district of Québec City).
1987 was released the first week of August, 2014. As of 4 weeks ago, this $4.4 million dollar production, in terms of box-office sales, had already brought in almost as much as its initial production (considered a good success for a Montréwood film of this size).
The movie's DVD will be coming out for sale the beginning of December. If you have someone who likes movies, the official DVD might be a good gift just in time for the holidays.
Please do not pirate. Our artists form part of our cultural fabric.
Tags: blog, blogue, Canada, Canada blog, Canada blogue, Canada French, culture québecoise, Français, français canadien, French, humour québécois, Québec blog, Québec blogue, Québec cinéma, Québec Francais, Quebec, Quebec culture, Quebec films, Quebec French
By Quebec Culture Blog 101 in Movies on November 5, 2014 .
← Montréwood's 10 hottest sitcoms and TV drama series (#77) Dagobert (#79) →


text:
Sheriff » About » History
History of Weld County Sheriff
On February 28, 1861, President James Buchanan signed An Act to provide a temporary Government for the Territory of Colorado, commonly known as the Colorado Organic Act, creating the Territory of Colorado. This act provided the US President authority to appoint a Territorial Governor and Secretary. Weld County was formed on September 9, 1861, when the Territorial Assembly created the original seventeen counties. Its boundaries approximately coincided with the previous boundaries of St. Vrain County, Jefferson Territory.
The County seat was placed at the town of St. Vrain, which was abandoned by the 1880s. The Territorial Assembly appointed the first, temporary County Commissioners and charged them with conducting the first Territorial Election on December 2, 1861. It was in this election that Weld County's first Sheriff, John Paul, was elected. Sheriff Paul was re-elected to a second election on October 7, 1862. Then, as the county was organizing and need for law enforcement grew, Sheriff Paul appointed Jay Thomas as the first recorded Deputy Sheriff on October 9, 1863.
As with all change, boom/bust, and growth in Colorado, counties began splitting off from Weld starting with Washington - February 9, 1887; Morgan– February 19, 1887; Logan - February 25, 1887; and Yuma on March 15, 1889. Two counties split from Logan in 1889; Phillips - March 27 and Sedgwick - April 9 and Washington and Yuma Counties added territory from Adams and Arapahoe Counties on April 10, 1903.
The last to separate from Weld County was Broomfield. Incorporated as a municipality in southeast Boulder County Broomfield eventually grew into Weld, Adams and Jefferson counties. There are currently 64 counties in the State of Colorado, and each has a Sheriff.
Weld County encompasses approximately 4000 square miles today and is still one of the largest counties in the State. Growth of the Sherriff's Office has been constant through the years and Sheriff Steve Reams is the 39th Sheriff of Weld County serving his first term in Office.
The Weld Sheriffs duties are established by Colorado Statutes and the Weld County Home Rule Charter (1976) and include, but are not limited to, maintenance of a county jail, service and execution of all processes, writs, precepts, and orders of the County and District Courts or presented for service by citizens pursuant to civil procedure, transportation of prisoners, security of the courthouse, search and rescue, control of grassland fires, and the investigation of crimes or breech of the peace. Some of these duties are confined to areas outside incorporated fire districts while other duties include the entire county and its citizens without regard to political subdivision boundaries.
The Sheriff's Office today has changed dramatically since its formation in 1861. Today our budget exceeds 36 million dollars, and employs nearly 400 professionals. Today deputies, corrections officers, and professional staff perform functions from investigation to inmate oversight, to criminological research with several categories in between. The professionalism of our staff speaks volumes about the commitment of the Deputies not only to law enforcement, but to the Community they live in and the Sheriff's Office will continue to be an integral partner with the County and leader in law enforcement throughout Colorado. Whether riding horses or responding to calls for service utilizing a real-time crime center, Weld County Sheriff's Office will be there to meet the public's need today and into the future.
Sheriffs of Weld County Colorado


to handle more business travel and corporate travel management.
Cruises from U.S. and foreign ports accounted for 32%, a two-point gain from a year ago. Bookings for hotels and inclusive resorts were unchanged at 25% of sales, but the inclusive share rose two points, to 15%, while the hotel share declined and is trending downward.
Comparing the latest data with last year's report, Plog discerned a clear trend that "regardless of agency size or amount of revenue booked by home-based agents," all groups have decreased the contribution of air travel to their business totals and increased their cruise sales (Fig. 11).


text:
Bulgari B.zero1: 2000 to 2010 and beyond
The world has transformed in the years between 2000 and 2010. Ten years ago, Facebook, YouTube, Twitter and the iPhone did not exist. Yet even as the world has evolved, the emotional resonance of symbols such as B.zero1's spiral design have remained timeless and universal.
At this juncture of the past and future, Bvlgari invited internationally acclaimed artist Anish Kapoor to reinterpret their iconic B.zero1 ring. Mr. Kapoor's interpretation draws on the play of curved forms and reflective surfaces of some of his greatest artworks – elements common to B.zero1, which is both point and continuum.
Bvlgari's online B.zero1 campaign salutes the spirit of Anish Kapoor's art and B.zero1 – timeless, evolving, mysterious, and mesmerizing. Asian social network sites will be added to the micro-site to enhance the local experience and viral effect.


If you suffered injuries because of a defective product, the Pittsburgh team from Berger and Green can help. We offer free case reviews and can help you pursue compensation from the product manufacturer. Call us today at 412-661-1400.
We handle all products liability cases on a contingency fee basis, meaning you owe us no attorney's fees or costs unless we negotiate a settlement or win a payout in your case. Call Berger and Green today at 412-661-1400 for your complimentary consultation with a member of our Pittsburgh team.
Under state and federal laws as well as industry guidelines, manufacturers are responsible for ensuring their products are safe before marketing them and selling them to consumers. If there are any reasonable dangers related to the product, the manufacturers and distributors must offer clear warnings. If they do not, they could be liable for injuries the product causes its users to suffer.
For example, imagine you were using a power tool when it malfunctioned and injured your hand. As a result, you needed emergency surgery and missed several months of work. A products liability lawyer in Pittsburgh, PA can help build a case showing the product has a defect that caused your injuries and pursue compensation for the expenses and losses you suffered as a result.
If you or a loved one sustained injuries because of a defective product in a Pittsburgh accident, our team can help investigate the product defect, identify any similar cases, and take legal action to pursue compensation.
If your injuries are particularly severe, you may also qualify for Social Security Disability. We can evaluate your qualifications and help you apply for Social Security Disability benefits. You may still qualify for monthly payments from Social Security Disability Insurance even if you win compensation in your civil case.
If you or a loved one sustained fatal injuries because of a defective product, you may be eligible to pursue additional damages through a wrongful death and survival claims action. This could result in additional compensation to cover expenses like burial and funeral costs.
Pennsylvania law requires legal claims pertaining to products liability to be filed within two years of the accident that caused your injuries. This is one reason we encourage you to reach out to us as soon as possible after an accident.
It is important that you store the product and any related materials — the manual, instructions, packaging, and accessories — until we need it as a part of our investigation into your case. Having the defective product will be integral to building a strong case against the manufacturer.
Reach out to our team today to get started. If you are still undergoing treatment, you can ask a trusted family member to give us a call. A products liability lawyer in Pittsburgh, PA can meet with you at the hospital or your home.
If you sustained injuries because of a defective product accident, the Berger and Green team may be able to help. We understand what you need to know about products liability, and will pursue the compensation you deserve.
Call us today at 412-661-1400 for your free case review. A member of our Pittsburgh products liability team is standing by to discuss your case and your legal options.


You are here: Home | Frequently Asked Questions | Servicing | How Much Will My Car Service Cost?
Find it annoying when those great car servicing deals always seem to have that tell-tale from attached? See red when you add up the additional costs to service your car, only to find a Main Dealer price looking back at you? And why do Main Dealers charge so much for servicing anyway?
You're in good company, these tricks-of-the-trade really get our goat too!
That's why we much prefer giving you a quote when we know a little more about your car – you get a fairer, more accurate estimate, without the worry of hidden costs. After all, you wouldn't expect a Ford Focus Annual Service to be the same as that for a Bentley Continental – just think of the price difference in serviceable parts.
Unlike many dealerships and large garage chains, our technicians are not paid bonuses on the number of cars they service per day/week, nor are they paid commission on the parts they replace in your car. Keeping you informed at all stages, our billing is fully itemised, so no nasty shocks when your car is ready.
Contact us with your car's details. We'll confirm servicing requirements and parts prices as necessary, then give you a highly competitive quote. It's that easy!


Only change is halfmoon cutout was extended thru part.
"I can confirm that the new ECD blocks balance the mass correctly, without the need for rearranged mass suggested in T1100543. Section 5 of T1200314 explains why this was done. The new ECDs get us within 0.04mm of nominal C of G in the vertical and the beam directions (the roll direction always has a slightly bigger deviation due to the fact that the mass has a flag on one end but not the other).


text:
LOficina Quebequesa de la Llengua Francesa (Office québécois de la langue française en francès) és una institució pública del Quebec encarregada de l'aplicació de la política lingüística en aquesta província del Canadà.

Història
Aquesta institució va ser creada el 24 de març de 1961, al mateix temps que es creava el Ministeri d'Afers Culturals del Quebec. El 1977, l'OQLF va rebre més competències en aprovar-se la Carta de la llengua francesa a l'Assemblea Nacional del Quebec. Així, l'OQLF va passar a ser l'encarregada de l'aplicació de la política lingüística al Quebec.

Dos nous organismes van ser creats aleshores: el Consell Superior de la Llengua Francesa i la Comissió de Toponímia.

La institució s'anomenava inicialment Office de la langue française (OLF), però aquesta denominació va ser canviada per la d'Office québécois de la langue française (OQLF) amb l'adopció de la llei 104 per l'Assemblea Nacional del Quebec el 12 de juny de 2003. Aquesta mateixa llei va suposar una reorganització de l'OQLF. Així l'Oficina Quebequesa de la Llengua Francesa va ser fusionada amb la Comissió de Protecció de la Llengua Francesa i una part del Consell Superior de la Llengua Francesa.

Funcionament
Les missions de l'OQLF són fonamentalment defensar la posició de la llengua francesa al Quebec, assegurar el respecte de la Carta de la llengua francesa i elaborar els programes de francesització previstos per la llei.

Està formada per vuit membres i la dirigeix un president-director general nomenat pel govern per un màxim de cinc anys.

En el si de l'Oficina, la Carta de la llengua francesa ha creat dos comitès: el Comitè d'Oficialització Lingüística i el Comitè de Seguiment de la Situació Lingüística. Cadascun d'aquests comitès està format per cinc membres nomenats per la mateixa Oficina Quebequesa de la Llengua Francesa.

L'any 2004, el pressupost de l'OQLF va ser de 17.8 milions de dòlars canadencs.

El Gran Diccionari Terminològic
D'entre els diversos recursos elaborats per l'OQLF i que hom pot trobar en el seu web destaca el Gran Diccionari Terminològic (GDT). Aquest diccionari consisteix en una base de dades terminològica amb gairebé tres milions de termes i que dona definicions i equivalències amb les llengües anglesa i llatina. El GDT està especialitzat en vocabulari industrial, comercial i científic, i pretén ser l'eina més completa quant a traducció al francès de termes anglesos, intentant evitar així, l'ús innecessari d'anglicismes en llengua francesa.

Enllaços externs
 Oficina Quebequesa de la Llengua Francesa
 El gran diccionari terminològic de l'OQLF

Entitats del Quebec
Francès
Acadèmies de la llengua

redpajama_set_name: RedPajamaWikipedia


SOLD: Prometheus poseidon like new FS with bonus leather and hirsh rubber straps!
SOLD: Air blue papa praeto - new in box - blue dial - $199 shipped!
SOLD: Original Invicta Reserve Bolt - Swiss Auto on bracelet - MINT!
SOLD: Brand new strider watch, ss with black dial.
SOLD: Deep Blue Sea Ram auto.
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SOLD: Invicta Russian Diver Eta 6497 movement Skull on strap.
SOLD...ason TAYLOR 12/99 BLACK DIAMOND SAS mint PLUS JEAN RICHARD TV Double retro!!#!
SOLD: LUM-TEC limited edition M41, swiss 2824-2, amazing orange & blue lume!


Run cold water in your garbage disposal. Cold water helps to maintain the sharpness of the blades sharp and makes the disposal run more smoothly. Use a cleaner that is enzyme based cleaners when cleaning clogged pipes. These cleaners employ helpful bacteria which process the sludge lining your pipes, and then the liquid flows out of the pipes. Enzyme based cleaners are preferred by many to clear clogged drains. Frozen pipes can do very extensive and expensive plumbing repair jobs. The first thing to do is ensure your outside pipes are well insulated. When colder weather is approach, disconnect and drain your hoses and shut off outside faucets. This will help to save some serious money on plumbing bills. Clean the aerator to fix low pressure problems in case you have sediment build up. Take the aerator off, pull it apart, and brush it clean with a little vinegar. More information about Georgetown Plumber That Offer Financing.
Cold water keeps blades and allows for smoother disposal. Installing efficient shower heads could save you tons of money on your energy bill. Most of our hot water in the shower. If you invest in purchasing and installing shower heads that are energy-efficient, you can save hundreds each year on your bills. Use enzyme based if you want to clean clogged pipes. These cleaners employ helpful bacteria which process the sludge lining your pipes, which turns sludge into liquid and cleans your pipes. Enzyme cleaners are efficient and will not corrosive so they won't erode your metal pipes. Frozen pipes can lead to a large plumbing repair jobs. The first thing you should do is ensure your outside pipes are well insulated. When it gets colder, you should also make sure your garden hoses are disconnected and that your exterior faucet is shut off. Doing this can help save lots of plumbing repair bills.
If you are looking for Free Estimates Plumbing Company Huntington Tx, sports tickets and theater seats at the best prices click here.
This could be a sign there is something very wrong with your plumbing.


Statistics Lithuania informs that, based on the provisional data of the Population and Housing Census 2011, the resident population of Lithuania on 1 March 2011 – 3 million 54 thousand. Over the decade (since Census 2001, the population of Lithuania totaled 3 484 thousand) the population of Lithuania decreased by 12 per cent. The largest decrease was due to emigration – 76 per cent; the natural decrease was 24 per cent.
Lithuania is de facto a country of emigration. Emigration rate is among the highest in the European Union. According to the official statistics, over 0.6 million people emigrated from Lithuania in 1990-2010. Emigration numbers increased particularly after the country joined the EU in 2004. Over 250, 7 thousand people are emigrated since 2004, only over the 73 thousand are immigrated in to Lithuania at this period. Considering that immigration is limited and does not compensate emigration loses.
Immigration is usually associated with the arrival of foreign nationals. However, most of immigrants in Lithuania are returning Lithuanian citizens. In 2010, Lithuanian nationals accounted for 80 per cent of all arrivals. The immigration of foreign nationals to Lithuania remains very low with the annual average of 2000 people. Most foreigners come from Belarus, Russian Federation and Ukraine.
At the beginning of 2011 there were around 30,000 foreign residents in Lithuania, representing 1 per cent of the country's population. This is one of the lowest numbers across the EU. Most foreigners in Lithuania are permanent residents. Many of them are of Russian, Belarusian and Ukrainian origin. They arrived during the Soviet Union times and remained in the country after the restoration of independence in 1990. There are also foreigners who reside in Lithuania temporarily (i.e. students), but this group is very small. In 2010, 48 per cent of foreigners living in Lithuania came on family reunification grounds, 33 per cent came to work, 17 per cent - to study.
Immigration for work purposes is second main reasons why foreigners come to Lithuania (after family reunification). Third country nationals can only be employed in those sectors, which experience labor shortages and where employers were unable to hire a Lithuanian or EU national. Most of labor immigrants in Lithuania are men (e.g. in 2010, 95 percent of immigrants were men).
In the last 10 years, 500 persons on average asked for asylum every year, almost 70 per cent were granted protection. The vast majority of those who were granted asylum were Russian and Afghan.
According to the information above, Lithuania faces a problem to keep ties with the emigrants from our country, because they are potential returning persons who might be gain Lithuania educational system, many of them for the second time. Acording to the Law on Education Lithuania – "Opportunities shall be created in accordance with the procedure laid down by the Minister of Education and Science for foreigners of Lithuanian descent and Lithuanians living abroad to learn the Lithuanian language in schools of the Republic of Lithuania"(Art.25) – or otherwise inclusive policies for citizens of Lithuania around the world.
Less of these students are new arrivals. They face the big challenge to learn Lithuanian language at an adequate level, as well teachers help them integrate in the society and school community and study Lithuanian language.


text:
Benjamin Savšek in the slovenian national team
Benjamin Savšek, the Slovenian canoe champion, became in Bratislava first Slovenian canoeist to qualify in to the Slovenian national team and will perform in this year's European Championship (from May 30 to June 02) in France and in the five World Cup races and at the height of this year's season at the World Championships in Spain at the end of September. Benjamin is also a member of the development team ROTO, where he is involved in the development of RotoATTIVO kayaks and canoes.
Benjamin Savšek
ROTO opens water park WATER FUN 360 at Expano
Roto awarded best kayak paddlers in the world


text:
B.C. to look into possible money laundering at casinos
Province will introduce an independent expert to review the issues.
British Columbia's attorney general has announced a review of the province's anti-money laundering policies after reading a report about a casino taking in $13.5 million that police said could be proceeds of crime.
The July 2016 report was commission by the former Liberal government, and David Eby says it should have been made public when it was completed.
Eby says he'll announce the appointment of an independent expert to review if there are unaddressed issues of money laundering in Lower Mainland casinos.
The report says the majority of the cash is being presented by people commonly referred to as "high roller Asian VIP clients."
It says the River Rock Casino in Richmond was known to have accepted single cash buy-ins in excess of $500,000.
A B.C. Lottery Corp. document says almost all recommendations in the report have been addressed though it's awaiting more direction from the gaming policy enforcement branch on dealing with and refusing large, unsourced cash deposits.
VIDEO: Crew of the Campbell River to Quadra Island ferry responds to fishing vessel impacting breakwater in Campbell River
B.C. business owner vies for Liberal leadership


text:
Super Reviewer Manage Account Profile Summary
No user info supplied.
Get Out (2017)
This movie was ok. I liked the visuals of the Sunken Place and the twist involving the girlfriend, but the story doesn't make sense overall. What is the rationale for the very specific choice of bodies by the elderly people? Why wouldn't they choose people who wouldn't be missed, rather than people with friends, families and jobs? This film received so much hype that it was a big letdown,
Charlie T false
Lost Girls & Love Hotels (2020)
Very atmospheric, well-drawn characters. This movie is worth watching.
Vampires vs. The Bronx (2019)
Vampires vs. The Bronx is not scary, funny or original, but it is very racist against white people. A movie promoting this type of prejudice should not be getting the praise this movie has received from professional critics, especially given that it is so mediocre in every other way: writing, acting, direction. There's an attempt here at a Stranger Things style homage to the Eighties, but the concept is tired and the execution is dull.
The premise of the movie is that Radha Blank's character didn't make it as a playwright because of prejudice, but the terrible, cringe-inducing rap she produces to express her frustration suggests that the real reason is a lack of talent. It's good to see a movie about a woman of forty, but this portrayal is not progressive. The protagonist is a sad sack with no vitality who lacks insight and blames her failure to realize her dreams on everyone else. This movie could have been interesting if it gave the viewer some reason to believe that the protagonist didn't thoroughly deserve her also-ran status.


We offer safe and secured title loans using our quick and easy online application form. The title loans provided by the barwick office are installment loans meaning a portion of your payment goes towards the interest and a portion goes to the principal. You need to first complete our title loans application. Should you decide to apply for a barwick title loans, we recommend you have your driver's license and car title to speed the process up. The barwick title loans office helps people in many different circumstances obtain the short term title loans quickly and conveniently.
These are no credit check title loans so you can get the title loans when you need them. Our barwick office will contact you before your payment due date to remind you about our upcoming title loans payment. Your vehicle is the collateral on barwick title loans. barwick title loans takes security and your privacy to the highest level. Please feel free to contact our barwick office if you have any questions regarding these title loans.
Get cash fast by applying for quick title loans and get your money tomorrow. A title loans customer must pay at least the minimum monthly payment due in order to avoid late fees. Many people rely on title loans to help them during difficult times. Your personal information provided when you applied for the title loans is used in the short term title loans approval process. We aim to provide cheap title loans to people without access to traditional loans.


text:
Iowa Politics
Vote No on the Retention of Iowa Supreme Court Justices
Latte Links for October 25th
Clarke County Bar Association,
Iowa Defense of Marriage Act,
Iowa Supreme Court retention vote,
Iowa Supreme Court ruling on DOMA,
Varnum v Brien
By Dale Mastarone
In the 10-21-10 Osceola Sentinel-Tribune I took interest in a letter to the editor that was submitted with a number of local lawyers signing onto it who all asserted to be members of the Clarke County Bar Association. The letter urged local electors to vote Yes to retain judges and justices for the stated reason that we have an "independent judiciary." Do we actually have an independent judiciary in Iowa? How do they define "independent"? And then the letter's signers urge that it is a "non-partisan" process. I challenge both those statements. It has been reported that 80% of the Iowa Bar Association members appointed to judicial nominating commissions are Democrats, so that shoots down non-partisan. The same would be true if 80% were Republicans.
It is really a stretch to imagine that politics have been removed from the seating of judges in Iowa utilizing the current Missouri Plan. It is absolutely political, cannot be anything other — but now the politics involved are hidden from the people.
There is something very important that was left out of that letter by the author(s) and signers, whether it was merely intentional or by deliberate oversight. With this Missouri Plan, as applied in Iowa, there are quite a few retired judges who sit as "senior judges" on cases, both at the trial and the appellate levels. The senior judges and justices at this moment certainly are not a majority, but there are a number of them. And that number may actually increase as the judicial branch decides to utilize them more and more as budgeting shortfalls prevent them from filling judicial vacancies. Since they are technically retired, however, they do not face judicial retention votes as active judges do. This is among many reasons why, if there isn't the political will currently to completely scrap the Missouri Plan – as applied in Iowa – by repealing the 21st Amendment to the Iowa Constitution — it needs a serious overhaul in the next legislative session.
As for me, I will be voting NO on all three Supreme Court Justices up for retention. Quoting Congressman Steve King, "Those three justices claim omnipotent power to imagine and confer constitutional rights that 'were at one time unimagined' [Varnum v. Brien, 763 N.W. 2d 862 (Iowa 2009). Page 15 of the opinion] It is time to remove them on that basis alone. It is obvious they feel empowered and will follow their whim in future decisions rather than the law. They have usurped the constitutional authority of the legislature and will do so at every notion until they are stopped by a vote of the people. To read their opinion brings one to the conclusion that these justices believe they have the authority to find the Constitution—unconstitutional."
The mere fact that the letter to the editor was written for publication appears to me that the signers are making a blatant attempt to curry favor with existing judges. Is there a group of local lawyers somewhere who oppose some of the justices that would sign a letter to that extent? I sincerely doubt it. "Yes, your Honor, I publicly stated you should be dumped, but please rule in my client's favor."
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QinetiQ demonstrates manned helicopter and drone operate as team
23 July 2020	(Last Updated July 23rd, 2020 15:58)
QinetiQ has demonstrated a manned helicopter and semi-autonomous drone operated as a team while airborne, marking a first for the UK.
QinetiQ has demonstrated the UK's first airborne team of manned helicopter and semi-autonomous drone. Credit: QinetiQ.
The teaming operation was carried out over the army's ranges on Salisbury Plain, Wiltshire, UK.
During the demonstration, the airborne manned / unmanned team was able to detect potential targets in a selected area.
The drone switched controls from autonomous and human-controlled as commanded by the helicopter operator.
In line with mission requirements, the helicopter switched between monitoring the drone camera images and allowing it to operate independently.
The project is funded by Army HQ Research and Experimentation by means of the Defence and Security Accelerator (DASA) competition framework.
Under the British Army's Warfighting Experiment 2019 (AWE19), Defence Science and Technology Laboratory (Dstl) is responsible for delivering the capability to the army this September.
QinetiQ programme manager Rob Scott said: "To make it possible we had to take a Mission-Led Innovation1 approach and look at how to use what we already had available.
"We made full use of our live, virtual and simulated test environments as well as the facilities and resources available within our Long-Term Partnering Agreement with the MoD. We're particularly pleased we could undertake a live testing programme using one of our own H125 helicopters.
"It provides a realistic platform for repeatable experimentation while saving the UK MoD time and money by not requiring scarce operational resources to be diverted for trials."
The airborne team controlled the process using a point and click interface on a portable tablet on board the H125 helicopter.
With minimum modifications, the team used a range of diverse technologies from various suppliers.
U-Project
Field Hospital Projects for Defence Applications
Axnes
Wireless Intercom Systems for Military Applications