text:
\section{Introduction}
\label{introduction}

Gromov-Witten theory is an important tool to tackle enumerative problems in algebraic geometry. The main objects of Gromov-Witten theory are the so-called Gromov-Witten invariants. They count (virtually) the number of curves of given genus and degree that pass through a certain number of chosen subvarieties.
To also count curves that fulfill some tangency conditions, geometers started to use gravitational descendants which are generalized versions of Gromov-Witten invariants. Note that some authors also call these invariants descendant Gromov-Witten invariants (e.g. \cite{markrau}).\\
With the emergence of tropical geometry also tropical counterparts of Gromov-Witten invariants and gravitational descendants were introduced. Unfortunately, the tropical gravitational descendants do not coincide with the corresponding classical numbers in general.
This difference has structural reasons. The moduli space of marked tropical curves in $\mathbb{R}^{2}$ is non-compact and, hence, does not parameterize curves with components that approach the "boundary" of $\mathbb{R}^{2}$ in form of a limiting process.
In particular, this fact makes the computation, which relies classically on the recursive structure of the boundary of the respective moduli space, much harder in the tropical setting.
Nevertheless, some results on the computation of certain tropical gravitational descendants have been achieved.
Most notably, Johannes Rau and Hannah Markwig proved a tropical analogue to the WDVV Equation for planar invariants where psi-classes appear only at marked ends restricted by codimension two objects (see \cite{markrau}). Here we want to go one step further and prove a Topological Recursion Relation which allows us to compute invariants which can additionally have a psi-class of power one at exactly one marked end restricted by a codimension one object.
In order to do so we have to recall some notions from \cite{gathkerbmark} and \cite{raunew}. This will be done in Section \ref{definitions} of the article at hand. Lastly, in Section \ref{mainsection} we will prove the desired formula using methods from \cite{markrau} and \cite{raunew}.\\ For the sake of completeness, one should mention that there is very recent work of Peter Overholser which deals also with the computation of tropical descendant invariants in the plane (see \cite{overholser}). But his point of view is more mirror symmetric and less intersection theoretic.\\
I would like to thank Andreas Gathmann for several helpful discussions. Note that this article contains material from my master thesis.\\

\section{Defining tropical gravitational descendants}
\label{definitions}

We consider the \emph{moduli space $\mathcal{M}^{lab}_{0,l+m+n,trop}\left(\mathbb{R}^{2},d\right)$ of rational parameterized tropical curves} in $\mathbb{R}^{2}$ of degree $\Delta$ with $l+m+n+\left|\Delta\right|$ labelled ends of which $l+m+n$ are contracted (see \cite[Definition 4.1]{gathkerbmark} for details).
In analogy to the moduli space of pointed stable curves the contracted ends are called \emph{marked ends}. In particular, we assume here that the degree is of the form $\Delta=\{\underbrace{\left(-e_{1},-e_{2},e_{1}+e_{2}\right),\ldots,\left(-e_{1},-e_{2},e_{1}+e_{2}\right)}_{d \times}\}$, where $e_{1}$ and $e_{2}$ denote the unit vectors in the plane $\mathbb{R}^{2}$. By abuse of notation we will not distinguish between the labels of non-contracted ends and the corresponding elements in $\Delta$. Furthermore, we will refer to $-e_{1}$, $-e_{2}$ and $e_{1}+e_{2}$ as the \emph{standard directions}.\\
Exemplarily, an element $\left(\Gamma,x_{1},\ldots,x_{5},h\right)$ from $\mathcal{M}^{lab}_{0,5,trop}\left(\mathbb{R}^{2},2\right)$, i.e. a \emph{marked tropical curve}, is shown in \fref{markedcurve}.

\begin{figure}[h]
  \centering
	\def360pt{260pt}
  \input{markedcurve.pdf_tex}
  \caption{5-marked tropical curve}
  \label{markedcurve}
\end{figure}

According to what one sees $\Gamma$ is an \emph{abstract tropical curve}, i.e. a tree with a distinguished number of marked ends whose bounded edges are equipped with strictly positive lengths. The marked ends of the abstract tropical curve are visualized as dashed lines. The map $h:\Gamma \rightarrow \mathbb{R}^{2}$ maps $\Gamma$ to an image as on the bottom of \fref{markedcurve}. The image satisfies the balancing condition and marked ends are contracted to a point.\\

Now, we want to count marked tropical curves that fulfill certain incidence and tangency restrictions.
Since we are working in the plane, we only have to consider incidence restrictions coming from codimension one objects, i.e. lines, and codimension two objects, i.e. points. The codimension zero object, i.e. the whole space $\mathbb{R}^{2}$, does not give us non-trivial restrictions. So let $G_{j}$ be certain (tropical) lines and $P_{k}$ be certain points in $\mathbb{R}^{2}$. For us a tropical line is a 3-valent graph with one vertex and one end in each of the standard directions (cf. the left-most curve in \cite[Definition 2.7]{markrau}). We will call the unique vertex of the tropical line its \emph{root}.\\
We can partition the set of marked ends of a curve into three disjoint sets. In the following we will consider curves with $l+m+n$ marked ends and a partition $L\cup M \cup N=\left\{1,\ldots,l+m+n\right\}$ with $\left|L\right|=l$, $\left|M\right|=m$ and $\left|N\right|=n$. We assume that the ends with labels $i \in L$ are unrestricted, the ends with labels $j \in M$ have to meet the respective lines $G_{j}$ and the ends with labels $k \in N$ have to meet the respective points $P_{k}$. To make the meaning of the word "meet" precise, we need the following definition.

\begin{definition}
\label{Definition:evaluation}
For $i=1,\ldots,l+m+n$ the map $$ev_{i}:\mathcal{M}^{lab}_{0,l+m+n,trop}\left(\mathbb{R}^{2},d\right)\rightarrow \mathbb{R}^{2},\left(\Gamma,x_{1},\ldots,x_{l+m+n},h\right) \mapsto h\left(x_{i}\right)$$ is called the \emph{$i$-th evaluation map}.
\end{definition}

By \cite[Proposition 4.7]{gathkerbmark} the moduli space of rational parameterized tropical curves can be equipped with a tropical (=weighted, balanced and polyhedral) fan structure via choosing one of the marked ends as the \emph{anchor vertex}.
So the evaluation maps become morphisms of fans. Moreover, in this sense we can talk about \emph{cones}, \emph{facets}, \emph{(relative) interiors} and \emph{rational functions} (cf. \cite[Section 2]{markrau}). Lastly, this enables us to apply the tropical intersection theory developed in \cite{allerau} and to form \emph{intersection products} consisting of rational functions on the fan
$\mathcal{M}^{lab}_{0,l+m+n,trop}\left(\mathbb{R}^{2},d\right)$. These intersection products have the structure of tropical complexes and, hence, we can talk about \emph{polyhedra} and the \emph{dimension} of an intersection product.

Now we want to enrich these intersection products further by allowing \emph{psi-classes} $\psi_{i}$ at the marked ends with indices $i \in L\cup M \cup N$ (see \cite[Definition 2.2]{markrau} for the definition of a psi-class in tropical geometry). They can be thought of as encoding tangency (or higher order contacts) of the curves at the respective marked ends. In formal terms a psi-class $\psi_{i}$ is the codimension one subfan of $\mathcal{M}^{lab}_{0,l+m+n,trop}\left(\mathbb{R}^{2},d\right)$ that consists of cones corresponding to trees where the marked end $i$ sits at a vertex of valence $\geq 4$.
Let $t_{i}$, $s_{j}$ and $r_{k}$ be the numbers that specify how many psi-classes at $i \in L$, $j \in M$ and $k \in N$, respectively, are required.\\

\begin{definition}
[cf. {\cite[Proposition and Definition 2.10]{markrau}}]
\label{defgeneral}
Let $d,l,m,n,t_{i},s_{j}$ and $r_{k}$ be non-negative integers which satisfy:
\begin{align}
\label{dimensionequ}
l+m+n+3d-3+2=m+2n+\sum_{i \in L} t_{i}+\sum_{j \in M} s_{j}+\sum_{k\in N} r_{k}.
\end{align}
Then the respective \emph{tropical gravitational descendant} is defined as:
$$\left\langle \prod_{i\in L} \tau_{t_{i}}\left(0\right)\prod_{j \in M}\tau_{s_{j}}\left(1\right)\prod_{k \in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}:=$$
$$\frac{1}{\left(d!\right)^{3}}\deg\left(\prod_{i\in L} \psi^{t_{i}}_{i}\prod_{j\in M} ev^{*}_{j}\left(G_{j}\right)\psi^{s_{j}}_{j}\prod_{k\in N} ev^{*}_{k}\left(P_{k}\right)\psi^{r_{k}}_{k}\cdot\mathcal{M}^{lab}_{0,l+m+n,trop}\left(\mathbb{R}^{2},d\right)\right).$$
\end{definition}

Note that the dimension of $\mathcal{M}^{lab}_{0,l+m+n,trop}\left(\mathbb{R}^{2},d\right)$ is $l+m+n+3d-3+2$ by \cite[Proposition 4.7]{gathkerbmark} and the codimension of the intersection of psi-classes is $\sum_{i \in L} t_{i}+\sum_{j \in M} s_{j}+\sum_{k\in N} r_{k}$ where the pullback of a line has codimension 1 and the pullback of a point codimension 2.
So the lengthy Equation \ref{dimensionequ} enforces the intersection product to be zero-dimensional. This assumption is necessary to talk about the \emph{degree} of an intersection product.
The definition above is independent of the choice of the lines $G_{j}$ and the points $P_{k}$ (cf. Section 2 of \cite{markrau}).
In particular, the definition also makes sense if we choose rather special configurations of lines and points.
But to give some enumerative meaning to the definition we have to assume a certain generality of the lines $G_{j}$ and the points $P_{k}$.
Namely, we assume that the lines $G_{j}$ and the points $P_{k}$ are in \emph{general position} in the sense of \cite[Definition 3.2]{gathkerbmark}.
It is worth mentioning that the assumption that $G_{j}$ and $P_{k}$ are in general position is not a serious limitation because the $P_{k}$ and $G_{j}$ in general position form a dense subset in the set of $P_{k}$ and $G_{j}$ under all conditions (which can be identified with some large-dimensional real space). A proof of this fact is presented in \cite[Lemma 3.4]{markrau}.\\
Assuming that the lines $G_{j}$ and points $P_{k}$ are in general position, the intersection product consists (in set-theoretic terms) of the tropical curves that pass through $G_{j}$ and $P_{k}$ equipped with certain weights. The degree of the intersection product, to wit the tropical gravitational descendant, is then the sum of all weights. According to \cite[Lemma 2.4]{franz} (or \cite[Remark 2.2.14]{rau}) these weights can be computed as the product of the weight of the facet the respective curve lies in and some determinantal expression.
Nevertheless, the computation of the weights via determinantal formulas is very unwieldy and unsuitable for a large degree or a large number of marked ends.
That is why one wish for recursion formulas, like for instance the Topological Recursion Relation in classical Gromov-Witten theory. Such formulas rely classically on the well-known Splitting Lemma (e.g. see \cite[5.2.1 Lemma]{kock}).
Johannes Rau developed a tropical version of the Splitting Lemma.
However, his Tropical Splitting Lemma \cite[Theorem 4.13]{raunew} only works under rather strict assumptions.
In fact, this is one of the major problems when transfering the classical Gromov-Witten theory to the tropical setting. An example where the Tropical Splitting Lemma fails can be found in \cite[Remark 4.18]{raunew}. Here we restrict ourselves to a special situation.

\begin{assumption}
\label{notation}
From now on (unless stated otherwise) we assume that $L=\emptyset$, $M=\{1\}$, $\{2,3\} \subseteq N$ and $s_{1}=1$.
\end{assumption}

So we have a fan $X:= \psi^{0}_{1}\prod_{k\in N} \psi^{r_{k}}_{k} \cdot \mathcal{M}^{lab}_{0,1+n,trop}\left(\mathbb{R}^{2},d\right)$ and a one-dimensional intersection product $F:=\left( \tau_{0}\left(1\right)\prod_{k\in N}\tau_{r_{k}}\left(2\right)\right)$.
This situation has two major advantages.
Firstly, the weight $w_{X}\left(\sigma\right)$ of the facet $\sigma$ the respective curve lies in is one by \cite[Theorem 5.3]{markrau} and, therefore, can be omitted in this article.
Secondly, we can apply the Tropical Splitting Lemma by \cite[Lemma 4.16]{raunew}. The Tropical Splitting Lemma splits the curves of an intersection product at their \textit{contracted bounded edges} as described in \cite[Remark 2.7]{gathmark}. Note that a contracted bounded edge induces a \emph{reducible partition} $I|J$ of the labels of the marked ends of the curve, i.e. a partition into two non-empty sets $I$ and $J$ such that directions in $I\cap \Delta$ (and hence also in $J\cap \Delta$) sum up to the zero vector. In our situation the Tropical Splitting Lemma \cite[Theorem 4.13]{raunew} takes the form of:
\begin{align}
\label{eqhardterms2}
&\notag\left\langle\tau_{1}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}=\\
&\notag\sum_{\substack{I|J \textrm{ reducible }\\1 \in I, 2,3 \in J}} \sum_{\varepsilon+\zeta=2}\left\langle \tau_{0}\left(\varepsilon\right)\tau_{0}\left(1\right)\prod_{k\in N \cap I}\tau_{r_{k}}\left(2\right)\right\rangle_{d_{1}}\cdot\left\langle \tau_{0}\left(\zeta\right)\prod_{k\in N\cap J}\tau_{r_{k}}\left(2\right)\right\rangle_{d_{2}}\\
&+  \left\langle \phi_{1|2,3} \tau_{0}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}.
\end{align}

In Equation \eqref{eqhardterms2} the rational function $\phi_{1|2,3}$ is defined as specified in \cite[Section 5.2]{raunew}. Here, it is sufficient for us to know that $\phi_{1|2,3}$ is of the following form (see \cite[Lemma 5.6]{raunew}) on each facet $\sigma$ of $F$, i.e. on each top-dimensional polyhedron of $F$:
\begin{align}
\label{phi}
\phi_{1|2,3}|_{\sigma}=\begin{cases} \textrm{constant},&  \textrm{\it{if the interior curves of $\sigma$ have}}\\
&\textrm{\it{a contracted bounded edge}}\\
\textrm{sum of lengths of edges that}\\
\textrm{separate $1$ from $2,3$},&  \textrm{\it{otherwise.}} \end{cases}
\end{align}

Loosely speaking $\phi_{1|2,3}$ decides if the result applying the Tropical Splitting Lemma equals the result one obtains by applying the classical Splitting Lemma to the respective classical invariants. This is the case in almost all of the existing literature (e.g. in \cite{markrau} and \cite{raunew}), because there $\phi_{1|2,3}$ is bounded.
Here $\phi_{1|2,3}$ shows a slightly different behaviour. This will be made precise in the next corollary.

\begin{definition}
\label{movablestring}
Let $C$ be a curve in $\mathcal{M}^{lab}_{0,1+n,trop}\left(\mathbb{R}^{2},d\right)$. A subgraph $\mathcal{S}$ of $C$ that is homeomorphic to $\mathbb{R}$ and does not involve (the closure of) a marked end $x_{k}$ with $k \in N$ is called a \emph{string}.
\end{definition}

\begin{corollary}
\label{lemmanewdefor}
Let $F$ be as above.
Then, the function $\phi_{1|2,3}$ can be unbounded only on facets of $F$ with interior curves that look like in \fref{specialexample} (where $E$ and $G_{1}$ can run along any of the three standard directions).
\begin{figure}[h]
  \centering
		\def360pt{280pt}
  \input{neuspecialexample.pdf_tex}
  \caption{Deformation which yields an unbounded $\phi_{1|2,3}$}
  \label{specialexample}
\end{figure}
\end{corollary}

\begin{proof}
Every facet of $F$ is described by the deformation of one of its interior curves, because facets are one-dimensional.
A list of all possible deformations is presented in \cite[Lemma 5.13]{raunew}. In our situation ($L=\emptyset$) the list looks as follows.
The deformation of an interior curve $C$ in a facet $\sigma$ of $F$ is described by one of the following cases:
\begin{enumerate}
\item $C$ contains a contracted bounded edge and one can vary its length
arbitrarily.
\item $C$ does not have a contracted bounded edge; and
\begin{enumerate}
\item $C$ has a \emph{degenerated} vertex $V$ of valence three, i.e. a vertex of one of the following types:
\begin{enumerate}
\item The vertex is incident to a bounded edge $E$, a non-contracted end and the marked end with label $1$ such that the curve $C$ and the line
$G_{1}$ do not intersect transversally at $ev_{1}\left(C\right)$.
\item The vertex is incident to a bounded edge $E$ and two non-contracted ends such that the span of the non-contracted ends near
$V$ is still only one-dimensional.
\end{enumerate}
The deformation is given by changing the length of the bounded edge $E$.

\item $C$ contains a string $\mathcal{S}$ with two non-contracted ends and only one adjacent
bounded edge $E$ as shown in \fref{specialexample2} such that all vertices of $\mathcal{S}$ are non-degenerated and of valence three in $C$. This string can be moved by changing the length of $E$.
\end{enumerate}
\end{enumerate}

Now we will work through this list and stick to the notations from there.\\
Assume that $\sigma$ is a facet of $F$ which has an interior curve with a contracted bounded edge, i.e. $\sigma$ looks like 1. on the list. Then $\phi_{1|2,3}$ is constant on $\sigma$ by \eqref{phi} and, thus, bounded on $\sigma$.\\
Now, assume that $\sigma$ is a facet with an interior curve as in 2. (a) on the list.
As $\phi_{1|2,3}$ measures the sum of the lengths of the edges that separate $1$ from $2,3$, the function $\phi_{1|2,3}$ can only be unbounded on $\sigma$ if it takes the length of the growing edge $E$ into account. Note that, in the case 2. (a) ii. $E$ does not separate $1\in M$ from $2, 3$ and, hence, does not contribute to $\phi_{1|2,3}$ on $\sigma$. This holds, because the vertex $V$ in 2. (a) ii. is of valence three and, hence, cannot also be incident to the marked end with label $1$. In the case 2. (a) i. the only chance that $E$ is taken into account by $\phi_{1|2,3}$ is the situation shown in \fref{specialexample}.\\
Finally, assume that $\sigma$ is a facet as in 2. (b) on the list. In this case an interior curve of $\sigma$ looks like in \fref{specialexample2}. Namely, $E$ only contributes to $\phi_{1|2,3}$ on
$\sigma$, if $E$ separates $1$ from $2,3$. So by construction $x_{1}$ has to lie on $\mathcal{S}$ (cf. \cite[Lemma 2.4.14]{rau}) and $x_{2}, x_{3}$ have to lie "behind" $E$. Moreover, on such a curve $E$ must have one of the standard directions (i.e. directions of the edges of $G_{1}$), because otherwise the edge $E$ will not grow infinitely for obvious reasons.
\begin{figure}[h]
  \centering
	 \def360pt{250pt}
  \input{specialexample.pdf_tex}
  \caption{Bounded deformation of a string}
  \label{specialexample2}
\end{figure}
But the line $G_{1}$ was chosen to look like the left-most curve in \cite[Definition 2.7]{markrau}. Hence, it has a root and none of the half-rays can be infinitely long in both directions. In \fref{specialexample2} this means that the line $G_{1}$ on the left hand side has to split somewhere, i.e. the root has to appear. So $E$ cannot grow infinitely and the situation in \fref{specialexample2} cannot cause an unbounded $\phi_{1|2,3}$ either.\\
All in all, the only interior curves which can cause facets with unbounded $\phi_{1|2,3}$ are the ones in \fref{specialexample}.
\end{proof}

\section{A new Topological Recursion Relation}
\label{mainsection}

A very helpful technique to compute certain intersection products, resp. their degrees, is (what we will call) the \emph{$\lambda$-method}.
The $\lambda$-method is due to Andreas Gathmann and Hannah Markwig (see \cite{gathmark}).
To apply the $\lambda$-method one adds a factor $ft^{*}\left(\lambda\right)$ to a one-dimensional intersection product of interest. This factor is the pull-back of an element $\lambda \in \mathcal{M}_{0,4,trop}$ under the forgetful map $ft:\mathcal{M}^{lab}_{0,l+m+n,trop}\left(\mathbb{R}^{2},d\right)\rightarrow\mathcal{M}_{0,4,trop}$ which forgets all ends except those with the labels $1,2,3,4$ of a given curve. Here $\mathcal{M}_{0,4,trop}$ is the \emph{moduli space of 4-marked abstract tropical curves} as defined in \cite[Definition 3.2]{gathkerbmark}. This moduli space consists of four different types of curves. All different types are shown in \cite[Example 1.1]{gathmark}. In the following we omit the case that $\lambda$ is just the single vertex in $\mathcal{M}_{0,4,trop}$, i.e. looks like the right-most curve in \cite[Example 1.1]{gathmark}.
Then, by abuse of notation, $\lambda$ denotes not only an element in $\mathcal{M}_{0,4,trop}$ but also a real number. This real number is the length of the unique bounded edge (cf. \cite[Example 1.1]{gathmark}). Along these lines, the forgetful map $ft$ can be considered as a map to the set of real numbers.\\
Curves in an intersection product that involves $ft^{*}\left(\lambda\right)$ fulfill the additional requirement that they map to $\lambda$ under $ft$.
Now, the key idea of the $\lambda$-method is to make $\lambda$ sufficiently large. Then, the curves in the intersection product have certain special properties, but the degree of the intersection product remains the same by the so-called rational equivalence (cf. the proof of \cite[Theorem 8.1]{markrau}).\\

The overall goal of this section is to recursively compute invariants of the form $\left\langle\tau_{0}\left(0\right)^{l'}\tau_{1}\left(0\right)^{l''} \tau_{0}\left(1\right)^{m-1}\tau_{1}\left(1\right)\prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}$ where $l',l'' \in \mathbb{N}$ and $m\in \mathbb{N}_{>0}$.
We will see that in order to do so, it is sufficient to derive a recursion formula for invariants of the form $\left\langle \tau_{1}\left(1\right)\prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle$, i.e. invariants as they appeared in Section \ref{definitions}. As we learned in the last section the curves shown in \fref{specialexample} are of particular importance.
The next lemma is the essential step to describe these curves in the realm of the $\lambda$-method.

\begin{lemma}
\label{corollaryextrapoint}
Let $F$ be as in the previous section and $\sigma$ be a facet of $F$. Moreover, let $C$ be an interior curve of $\sigma$ as shown in \fref{specialexample}, such that $x_{4}$ with $4\in \Delta$ is the unbounded edge adjacent to the growing bounded edge $E$ and $C$ maps to $\lambda$ under $ft$. Then the linear parts $\left(\phi_{1|2,3}\right)_{\sigma}$ and $\left(ft\right)_{\sigma}$ of the restricted functions $\phi_{1|2,3}|_{\sigma}$ and $ft|_{\sigma}$ coincide.
\end{lemma}

\begin{proof}
We know from \cite[Lemma 5.13]{raunew} that we get other curves of $\sigma$ by varying the length of the bounded edge $E$ of $C$. As $\sigma$ is one-dimensional all curves in $\sigma$ arise in that way. In particular, all interior curves of $\sigma$ do not have a contracted bounded edge and we can use the second description of $\phi_{1|2,3}$ from Equation \eqref{phi}. Let $\widetilde{C}$ be a curve in $\sigma$ and $\widetilde{E}$ be the deformed edge which originates from $E$, then we obtain:
\begin{align*}
\phi_{1|2,3}\left(\widetilde{C}\right)=&_{\eqref{phi}} \textrm{sum of lengths of edges that separate $1$ from $2,3$}\\
=&\textrm{length of $\widetilde{E}$} +\textrm{term that is constant on $\sigma$},
\end{align*}
On the other hand we obtain:
\begin{align*}
ft \left(\widetilde{C}\right)=& \textrm{length of $\widetilde{E}$}.
\end{align*}
Since $\widetilde{C}$ was an arbitrary curve in $\sigma$, we can conclude that $\phi_{1|2,3}|_{\sigma}+\textrm{constant}=ft|_{\sigma}$.
So the linear parts of $\phi_{1|2,3}|_{\sigma}$ and $ft|_{\sigma}$ coincide.
\end{proof}

Now we have to quantify the weight of curves as shown in \fref{specialexample}. As already mentioned such weights are given by determinantal expressions.
A nice example of how to compute such an absolute determinant $\left|\det_{C}\left(\ldots\right)\right|$ for a curve $C$ is presented in \cite[Example 2.6]{franz}. However, here we use notions analogous to \cite[Notation 5.2]{markrau}. In particular, we consider evaluation maps
\begin{align}
ev:X\rightarrow \mathbb{R}^{1} \times \left(\mathbb{R}^{2}\right)^{n}, C \mapsto \left(\left(ev_{1}\left(C\right)_{x}\right),\left(ev_{k}\left(C\right)\right)_{k \in N}\right),
\end{align}
\begin{align}
\widetilde{ev}:X\rightarrow \left(\mathbb{R}^{2}\right)^{n+1}, C \mapsto \left(\left(ev_{1}\left(C\right)\right),\left(ev_{k}\left(C\right)\right)_{k \in N}\right),
\end{align}
where the index $x$ at a point in $\mathbb{R}^{2}$ indicates its first coordinate. Later on we will also use the index $y$ to indicate a second coordinate.
So the evaluation map $\widetilde{ev}$ arises from $ev$ by replacing the line restriction at $x_{1}$ with a point restriction at $x_{1}$.

\begin{lemma}
\label{importantnewformula}
Let $Z=\left(ft^{*}\left(\lambda\right) \tau_{0}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right)$ be a zero-dimensional intersection product and $X$be as above. Let $4 \in \Delta$. Moreover, let $C$ be an interior curve in a facet of $F$ as shown in \fref{specialexample}, such that $x_{4}$ is the unbounded edge adjacent to the growing bounded edge $E$ and $C$ maps to a large $\lambda$ under $ft$.
Then, the weight of $C$ in $Z$ is $\left|\det_{C}\left(ft \times ev\right)\right|=\left|\det_{C}\left( \widetilde{ev}\right)\right|$.
\end{lemma}

\begin{proof}
Let $C$ be a curve as above.
Let $l_{1}$ be the length of its growing bounded edge $E$ (see \fref{specialexample}).
We want to compute the weight of $C$ in $Z$. In so doing, we choose a marked end $x_{a}\neq x_{1}$ as the anchor vertex. Such a marked end must exist, because the number of marked ends is greater than three by Assumption \ref{notation}.
The position of $x_{a}$ is determined by the coordinates $\left(h\left(x_{a}\right)\right)_{x}$ and $\left(h\left(x_{a}\right)\right)_{y}$.
Say $E$ runs along the standard direction $-e_{1}$, then the position of the marked end $x_{1}$ is given as
\begin{align}
\label{eq:coord1}
h\left(x_{1}\right)=h\left(x_{a}\right)- l_{1} \cdot e_{1}+\underbrace{\ldots}_{\substack{\textrm{expression in the}\\ \textrm{other bounded lengths}}}.
\end{align}
The descriptions of the positions of all other marked ends do not involve the length $l_{1}$, because we chose $x_{a}\neq x_{1}$.

Now, by \cite[Lemma 2.4]{franz} we know that the weight of $C$ is given as the absolute value of the determinant $\left|\det_{C}\left(ft \times ev\right)\right|$.
We call the underlying matrix of this absolute determinant $H$.
Let the line $G_{1}$ be given by $\max\{y,c_{1}\}$. Thus, the pull-back of the evaluation map with plugged in coordinates from \eqref{eq:coord1} is of the form $\max\{\left(h\left(x_{a}\right)\right)_{y}+\ldots-c_{1},0\}$.
We assume that the points $P_{k}$ are given by functions of the form $\max\{x,p^{k}_{1}\}$ and $\max\{y,p^{k}_{2}\}$.
Then, the pull-backs of the evaluation maps at the points are of the form $\max\{\left(h\left(x_{k}\right)\right)_{x}-p^{k}_{1},0\}$ and $\max\{\left(h\left(x_{k}\right)\right)_{y}-p^{k}_{2},0\}$.
So we can determine the entries of all rows of $H$ that correspond to pull-backs of evaluations at the line $G_{1}$ and the points in the columns corresponding to $l_{1}$, $\left(h\left(x_{a}\right)\right)_{x}$ and $\left(h\left(x_{a}\right)\right)_{y}$. The only missing row is the row corresponding to the pull-back of the $\mathcal{M}_{0,4,trop}$-coordinate $\lambda$ under $ft$.
But, since $\lambda$ is large and there is no contracted bounded edge in a curve in this facet, the length $l_{1}$ must contribute to the length of $\lambda$. This causes an entry one at the column corresponding to $l_{1}$.
All in all, the entries in $H$ are summarized in the table below.

\begin{center}
\begin{tabular}{c|| c c c c}
&&&& other bounded \\
 & $\left(h\left(x_{a}\right)\right)_{x}$ & $\left(h\left(x_{a}\right)\right)_{y}$ & $l_{1}$ & lengths\\

 \hline \hline

evaluation at $G_{1}$ & 0 & 1 & 0 & *\\

evaluation at $\left(P_{k}\right)_{x}$ for $k \in N$& 1 & 0 & 0 & *\\

evaluation at $\left(P_{k}\right)_{y}$ for $k \in N$ & 0 & 1 & 0 & *\\
$ft$ & 0 & 0 & 1 & * \\
\end{tabular}
\end{center}
\vspace{0.4cm}

We can permute rows and columns of $H$, because we are only interested in the absolute value of its determinant.
So we permute the rows and columns of $H$ in such a way that its first row is the row corresponding to $ft$ and that this row starts with a one.
Then, we can apply the determinantal formula for block matrices on the upper left 1 by 1 matrix block. This means we forget the row corresponding to $ft$ and the column corresponding to $l_{1}$. Now, applying the determinantal formula for block matrices backwards we can reintroduce a column corresponding to $l_{1}$ and introduce a row corresponding to $\left(P_{1}\right)_{x}$ for a new point $P_{1}$ which lies on the end pointing towards $-e_{1}$ of the line $G_{1}$. We end up with the matrix entries shown in the table below.

\begin{center}
\begin{tabular}{c|| c c c c}
&&&&other bounded\\
 & $l_{1}$ & $\left(h\left(x_{a}\right)\right)_{x}$ & $\left(h\left(x_{a}\right)\right)_{y}$  &  lengths \\
 \hline \hline

evaluation at $\left(P_{1}\right)_{x}$ & -1 & 1 & 0 & *\\

evaluation at $G_{1}=\left(P_{1}\right)_{y}$ & 0 & 0 & 1 & *\\

evaluation at $\left(P_{k}\right)_{x}$ for $k \in N$ &0 & 1 & 0 & *\\

evaluation at $\left(P_{k}\right)_{y}$ for $k \in N$ &0 & 0 & 1 & *\\

\end{tabular}
\end{center}
\vspace{0.4cm}

But this is exactly the matrix of the curve $C$ for the evaluation map $\widetilde{ev}$ which arises from $ev$ by replacing the line restriction at $x_{1}$ with a point restriction given by $P_{1}$.
This yields $\left|\det_{C}\left(ft \times ev\right)\right|=\left|\det_{C}\left( \widetilde{ev}\right)\right|$ and, hence, proves the claim.
\end{proof}

Finally we have all tools available to prove the main result of this article.

\begin{theorem}
\label{majortheorem}
It holds:
\begin{align*}
&\left\langle \tau_{1}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}=\\
&\sum\left\langle \tau_{0}\left(\varepsilon\right)\tau_{0}\left(1\right)\prod_{k\in N \cap I}\tau_{r_{k}}\left(2\right)\right\rangle_{d_{1}}\cdot\left\langle \tau_{0}\left(\zeta\right) \prod_{k\in N\cap J}\tau_{r_{k}}\left(2\right)\right\rangle_{d_{2}}\\
&+3 \left\langle \tau_{0}\left(2\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}
\end{align*}
where the sum ranges over all $\varepsilon+\zeta=2$ with $\varepsilon,\zeta \in \mathbb{Z}_{\geq 0}$ and $d_{1}+d_{2}=d$ with $d_{1},d_{2} \in \mathbb{Z}_{\geq 0}$ and reducible partitions $I|J$ with $1\in I$ and $2,3\in J$.
\end{theorem}

\begin{proof}
By \eqref{eqhardterms2} we can write $\left\langle \tau_{1}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}$ as
\begin{align*}
&\left\langle\tau_{1}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}=\\
&\sum_{\substack{I|J \textrm{ reducible }\\ 1 \in I, 2,3 \in J}} \sum_{\varepsilon+\zeta=2}\left\langle \tau_{0}\left(\varepsilon\right)\tau_{0}\left(1\right)\prod_{k\in N \cap I}\tau_{r_{k}}\left(2\right)\right\rangle_{d_{1}}\cdot\left\langle \tau_{0}\left(\zeta\right)\prod_{k\in N\cap J}\tau_{r_{k}}\left(2\right)\right\rangle_{d_{2}}\\
&+  \left\langle \phi_{1|2,3} \tau_{0}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}.
\end{align*}
It remains to quantify $\left\langle \phi_{1|2,3} \tau_{0}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle$. For reasons of clarity, we denote the set of facets of $F$ as $F^{\left(1\right)}$ and the set of curves, to wit vertices, of $F$ as $F^{\left(0\right)}$. Then by the definition of tropical divisors \cite[Definition 3.4]{allerau} it holds
\begin{align}
\label{nullteform}
\left\langle \phi_{1|2,3}  \tau_{0}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}=\sum_{F^{\left(0\right)} \ni C < \sigma \in F^{\left(1\right)}} w_{F}\left(\sigma\right) \left(\phi_{1|2,3}\right)_{\sigma}\left(u_{\sigma/C}\right),
\end{align}
where $w_{F}\left(\sigma\right)$ is the weight of the facet $\sigma$ in $F$ and $u_{\sigma/C}$ is the primitive generator of the integer vector pointing from $C$ to $\sigma$.
For facets $\sigma \in F^{\left(1\right)}$ that contain two different vertices the two
corresponding primitive generators have opposite directions (see \cite[Lemma 1.4.4]{rau}). Thus the two respective summands in \eqref{nullteform} cancel out. The $\sigma \in F^{\left(1\right)}$ that do not contain two different vertices are unbounded polyhedra.
We know from \cref{lemmanewdefor} that $\phi_{1|2,3}$ is bounded on all facets of $F$ which do not have interior curves like in \fref{specialexample}. But on unbounded polyhedra a rational function can only be bounded if it is constant, i.e. $\left(\phi_{1|2,3}\right)_{\sigma} \equiv 0$. So the summands in \eqref{nullteform} belonging to polyhedra which do not have interior curves like in \fref{specialexample} vanish. In fact, it holds:
\begin{align}
\label{ersteform}
\left\langle \phi_{1|2,3}  \tau_{0}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}=\sum_{\substack{F^{\left(0\right)} \ni C < \sigma \in F^{\left(1\right)}:\\ \sigma \textrm{ has interior curve}\\ \textrm{like in \fref{specialexample}} }} w_{F}\left(\sigma\right) \left(\phi_{1|2,3}\right)_{\sigma}\left(u_{\sigma/C}\right).
\end{align}
Due to \lref{corollaryextrapoint} the linear parts $\left(\phi_{1|2,3}\right)_{\sigma}$ and $\left(ft\right)_{\sigma}$ coincide. Thus, it follows
\begin{align}
\label{zweiteform}
\left\langle \phi_{1|2,3}  \tau_{0}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}=\sum_{\substack{F^{\left(0\right)} \ni C < \sigma \in F^{\left(1\right)}:\\ \sigma \textrm{ has interior curve}\\ \textrm{like in \fref{specialexample}} }} w_{F}\left(\sigma\right) \left(ft\right)_{\sigma}\left(u_{\sigma/C}\right).
\end{align}
Again, by \cite[Lemma 2.4]{franz} we can reformulate \eqref{zweiteform} and write (sticking to the notations from \lref{importantnewformula})
\begin{align}
\label{dritteform}
\left\langle \phi_{1|2,3}  \tau_{0}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}=&\sum_{\substack{C \textrm{ looks like}\\
\textrm{ in \fref{specialexample}}}}  \left|\textrm{det}_{C}\left(ft \times ev\right)\right|.
\end{align}
Here, as well as in the next two equations, the curves $C$ over which we sum up live in the subcomplex containing the points of $X$ that map to the correct point $P_{k}$ for all $k \in N$ under the respective evaluation map and map to a fixed sufficiently large $\lambda$ under $ft$.
Now, we can divide the curves that look like in \fref{specialexample} into three classes depending along which standard direction the end $x_{4}$, which is the unbounded edge adjacent to the marked end $x_{1}$, runs. It holds:
\begin{align}
\left\langle \phi_{1|2,3}  \tau_{0}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}=&\sum_{\substack{C \textrm{ looks like}\\
\textrm{ in \fref{specialexample}}}} \left|\textrm{det}_{C}\left(ft \times ev\right)\right|\notag\\
=&\sum_{\substack{C \textrm{ looks like in}\\
\textrm{\fref{specialexample} and}\\ \textrm{$x_{4}$ runs along $-e_{1}$}}}  \left|\textrm{det}_{C}\left(ft \times ev\right)\right|\notag\\
&+\sum_{\substack{C \textrm{ looks like in}\\
\textrm{\fref{specialexample} and}\\ \textrm{$x_{4}$ runs along $-e_{2}$}}}  \left|\textrm{det}_{C}\left(ft \times ev\right)\right|\notag\\
&+\sum_{\substack{C \textrm{ looks like in}\\
\textrm{\fref{specialexample} and}\\ \textrm{$x_{4}$ runs along $e_{1}+e_{2}$}}} \left|\textrm{det}_{C}\left(ft \times ev\right)\right|\notag.
\end{align}

Finally, from \lref{importantnewformula} we obtain
\begin{align}
\label{vierteform}
\left\langle \phi_{1|2,3}  \tau_{0}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}=&\sum_{\substack{C \textrm{ looks like in}\\
\textrm{\fref{specialexample} and}\\ \textrm{$x_{4}$ runs along $-e_{1}$}}} \left|\textrm{det}_{C}\left( \widetilde{ev}'\right)\right|\notag\\
&+\sum_{\substack{C \textrm{ looks like in}\\
\textrm{\fref{specialexample} and}\\ \textrm{$x_{4}$ runs along $-e_{2}$}}} \left|\textrm{det}_{C}\left( \widetilde{ev}''\right)\right|\notag\\
&+\sum_{\substack{C \textrm{ looks like in}\\
\textrm{\fref{specialexample} and}\\ \textrm{$x_{4}$ runs along $e_{1}+e_{2}$}}} \left|\textrm{det}_{C}\left( \widetilde{ev}'''\right)\right|,
\end{align}
where, $\widetilde{ev}'$, $\widetilde{ev}''$ and $\widetilde{ev}'''$ arise from $ev$ by replacing the line restriction at $x_{1}$ with a point restriction given by new points $P_{1'}$, $P_{1''}$ and $P_{1'''}$ on $G_{1}$ (in the sense of \lref{importantnewformula}). The points will be chosen in a specific way as described later in the case of $P_{1'}$.
Now we observe that the sums in \eqref{vierteform} look almost like the number (see \cite[Theorem 5.3]{markrau}) $\left\langle\tau_{0}\left(2\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle=\sum_{C \in S} \left|\det_{C}\left( \widetilde{ev}\right)\right|$ where $S$ is the subcomplex of $X$ containing all points in $X$ that map to the correct point $P_{k}$ for $k \in N \cup\{1\}$ under the respective evaluation map. So it is natural to try to express each of the three sums via the invariant $\left\langle\tau_{0}\left(2\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle$. In so doing, we have to relate the index of summation of each of the sums in \eqref{vierteform} to the index of summation occurring in the sum expression of $\left\langle\tau_{0}\left(2\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle$.\\
Making this precise, we define three subcomplexes $S_{1}$, $S_{2}$ and $S_{3}$ of $X$ where $S_{1}$ contains all points in $X$ that map to the correct point $P_{k}$ for $k \in N \cup\{1'\}$ under the respective evaluation map, $S_{2}$ contains all points that map to $P_{k}$ for $k \in N \cup\{1''\}$ and $S_{3}$ contains all points that map to $P_{k}$ for $k \in N \cup\{1'''\}$. We want to show the following three equalities of (finite) sets:
 \begin{align}\{C \,|\, C \textrm{ looks like in \fref{specialexample} and $x_{4}$ runs along $-e_{1}$}\}=S_{1}\end{align}
 \begin{align}\{C \,|\, C \textrm{ looks like in \fref{specialexample} and $x_{4}$ runs along $-e_{2}$}\}=S_{2}\end{align}
 \begin{align}\{C \,|\, C \textrm{ looks like in \fref{specialexample} and $x_{4}$ runs along $e_{1}+e_{2}$}\}=S_{3}.\end{align}
This then proves the claim, because by \eqref{vierteform} we could write \begin{align*}
\left\langle \phi_{1|2,3}  \tau_{0}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}
&=3 \left\langle \tau_{0}\left(2\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}.
\end{align*}
So let us check the two inclusions for $S_{1}$. Subsequently, the argument can be carried out analogously for $S_{2}$ and $S_{3}$.\\
First of all, we assume that the line $G_{1}$ and the points $P_{k}$ for $k \in N \cup\{1'\}$ are organized in the following way:
\begin{itemize}
	\item the $x$- and $y$-coordinates of the root of the line $G_{1}$ and of the points $P_{k}$ for $k \in N$ are in an open interval $\left(-\epsilon, \epsilon\right)$ for some small real number $\epsilon>0$;
  \item the point $P_{1'}$ lies on the end of the line $G_{1}$ which points into the direction $-e_{1}$ and its $x$-coordinate is $-\lambda$ where $\lambda \gg 0$ is some large real number;
	\item (additionally) the point $P_{2}$ lies at the origin and the $x$-coordinate of $P_{3}$ lies in the interval $\left[0,\epsilon\right]$.
\end{itemize}
This means we assume that the point $P_{1'}$ lies "far away" from the other points $P_{k}$ for $k\in N$ (see \fref{1:3}). Nevertheless, we still assume that all the points are in general position. This is possible because the points in general position are dense in the set of all points (cf. Section \ref{definitions}).
Note that we are now in the situation of \cite[Lemma 2.2.1]{torchiani}.
As a consequence of this lemma we get the two desired inclusions.\\
"$\subseteq$": Let $C$ be a curve that looks like in \fref{specialexample} and its unbounded edge $x_{4}$ runs along $-e_{1}$. Particularly, the edge $x_{4}$ runs along the end of $G_{1}$ pointing towards $-e_{1}$ and it is incident to a unique vertex. This vertex has an $x$-coordinate greater than $-\lambda$, because $C$ maps to $\lambda$ under $ft$ by assumption and the points $P_{2}$ and $P_{3}$ have an $x$-coordinate in the interval $\left[0,\epsilon\right]$.
Furthermore, the point $P_{1'}$ lies on the end of $G_{1}$ pointing towards $-e_{1}$ and it is placed at an $x$-coordinate of $-\lambda$ by construction. Thus, the unbounded edge $x_{4}$ of $C$ has to meet $P_{1'}$. All in all, the curve $C$ maps to $P_{1'}$ and $P_{k}$ for $k \in N$ under the respective evaluation maps. So $C$ is in $S_{1}$.\\
"$\supseteq$": Let $C \in S_{1}$. We want to show that $C$ must look like in \fref{1:3}, i.e. especially like in \fref{specialexample}.
\begin{figure}[h]
  \centering
		\def360pt{360pt}
  \input{correspondence.pdf_tex}
  \caption{Curve $C$ in $\left(\tau_{0}\left(2\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right)$ for special point configuration}
  \label{1:3}
\end{figure}
Therefore, assume that $C$ does not look like in \fref{1:3}. That means we assume that $x_{1}$ is not adjacent to an unbounded edge. By the balancing condition one of its adjacent bounded edges must be pointing leftwards and as this edge is bounded it is adjacent to a vertex with an $x$-coordinate smaller than $-\lambda$. But such a vertex does not exist, because all vertices of $C$ must lie in $R:=\{\left(x,y\right)\in \mathbb{R}^{2}| -\lambda \leq x\leq \epsilon, -\epsilon \leq y\leq \epsilon\}$ by \cite[Lemma 2.2.1]{torchiani}. So this case cannot occur. Hence, the curve $C \in S_{1}$ looks like in \fref{1:3} and the unbounded edge $x_{4}$, which is the unbounded edge adjacent to $x_{1}$, runs along $-e_{1}$.
\end{proof}

By \tref{majortheorem} we are able to compute all tropical gravitational descendants of the form $\left\langle\tau_{1}\left(1\right) \prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}$. Using in addition the Tropical String Equation \cite[Theorem 3.12]{raunew}, the Tropical Divisor Equation \cite[Theorem 3.17]{raunew} and the Tropical Dilaton Equation \cite[Theorem 3.13]{raunew} we can recursively compute all tropical gravitational descendants of the form
\begin{align}
\left\langle\tau_{0}\left(0\right)^{l'}\tau_{1}\left(0\right)^{l''} \tau_{0}\left(1\right)^{m-1}\tau_{1}\left(1\right)\prod_{k\in N}\tau_{r_{k}}\left(2\right)\right\rangle_{d}.
\end{align}

\begin{remark}
\label{remarkfirst}
This means we are in particular able to compute the tropical counterparts of the so-called first descendant invariants with exactly one tangency condition $\tau_{1}\left(1\right)$. First descendant invariants are gravitational descendants which involve only psi-classes of power at most one. They play an important role in many classical enumerative questions because they are sufficient to study characteristic number problems (cf. \cite{grabkopand}).
\end{remark}

To close this article and to see the new recursion formula in action, we want to compute the invariant $\left\langle \tau_{1}\left(1\right)\tau_{1}\left(2\right)\tau_{1}\left(2\right)\right\rangle_{2}$. In fact, it holds:

\begin{align}
\label{doublecheck} \left\langle \tau_{1}\left(1\right)\tau_{1}\left(2\right)\tau_{1}\left(2\right)\right\rangle_{2}
=_{\ref{majortheorem}}& \left( \sum \left\langle \tau_{0}\left(\varepsilon\right) \tau_{0}\left(1\right)\right\rangle_{d_{1}} \cdot\left\langle \tau_{0}\left(\zeta\right) \tau_{1}\left(2\right) \tau_{1}\left(2\right)\right\rangle_{d_{2}}\right.\notag\\
&\left.+3\underbrace{\left\langle\tau_{0}\left(2\right)\tau_{1}\left(2\right)\tau_{1}\left(2\right)\right\rangle_{2}}
_{=_{\cite{markrau}} 1}  \right)\notag\\
=& \underbrace{\left\langle \tau_{0}\left(0\right) \tau_{0}\left(1\right)\right\rangle_{0} \cdot\left\langle \tau_{0}\left(2\right) \tau_{1}\left(2\right) \tau_{1}\left(2\right)\right\rangle_{2}}_{=0} + 3 \notag \\
=& 3.
\end{align}

The numerical value of the algebro-geometric version $\left\langle\tau_{1}\left(1\right)\tau_{1}\left(2\right)\tau_{1}\left(2\right)\right\rangle^{alg}_{2}$ of this invariant equals 0 and can be easily computed via standard techniques (e.g. see \cite{kock}). So this is an example where the classical algebro-geometric and the tropical number do not coincide as mentioned in the introduction.

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text:
Neretva Delta is the river delta of the Neretva, a river that flows through Bosnia and Herzegovina and Croatia and empties in the Adriatic Sea. The delta is a unique landscape in southern Croatia, and a wetland that is listed under the Ramsar Convention as internationally important, as the wetland extends into the Hutovo Blato in Herzegovina.

The total alluvial area of the delta is estimated at around , while today the delta takes up roughly .

A total of  of the delta are designated protected areas:
 ornithological reserves: Prud, Orepak, Podgredom
 ornithological and ichthyological reserve at the river mouth
 protected landscape: lake Modro oko
An additional 1200 ha of nature in the delta is not protected. In 2003, the Ministry of Environment and Spatial Planning organized a public discussion about the proposal to make the delta a nature park. In 2007, the experts from the State Institute for Nature Protection made another formal proposal for the Neretva Delta to become a nature park. A public discussion process was started but quickly terminated by the Ministry of Culture, allegedly because of protests from the local government.

References

0Neretva Delta
River deltas of Europe
Ramsar sites in Bosnia and Herzegovina
Ramsar sites in Croatia
Neretva

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Dad, I was only 22 when you decide to leave me, I wasn't ready. I think about you everyday, I want so bad to tell you, your a grand father and great grand father now. You've missed so much the last 23 years. We will never be the same without you! If I had one wish it would be that never again would a family have to feel the pain of a loved ones death by suicide.


Ah yes, this notion seems to really be making the rounds lately. It certainly received a good bit of press because of a letter published in a recent issue of the British Medical Journal, which detailed the case of a 15 year-old girl who was struck by lightning while talking on her cell phone in London.
Lightning is definitely something you don't want to mess around with. According to the National Weather Service, it kills about 67 people a year and injures at least 300. And you're not just at risk from a direct strike – the current from the discharge can travel across the ground. Yeah… you don't want to be anywhere near a strike.
So lightning safety and awareness are certainly important, which brings us back to the question at hand. I can see how people think carrying around an electrical device could make you more likely to get zapped. But think about how small a cell phone is, and really how small you are compared to buildings and other structures around you.
During a thunderstorm, clouds become positively charged on the top and negatively on the bottom. This separation of charge creates a strong electric field in the cloud, and as the cloud moves over the ground below, it becomes positively charged. Lightning strikes essentially neutralize the charge separation that builds up. They aren't actually attracted to anything, but they tend to choose the path of least resistance – which believe it or not, is not usually you and your cell phone. That's not to say it couldn't happen, but it shouldn't be your biggest worry.
So the moral of the story is: Don't go outside in a thunderstorm. With or without your cell phone.
You certainly do bring up valid points about lightning and how carrying a cell phone has as much to do with a person being struck as a butterfly flapping its wings has on starting a tropical storm on the other side of the earth. Then again, perhaps a question and answer session on chaos theory is in order now!
Don't forget the related "cell phone while filling up [your car] at a gas station".
Thanks for the great picture and the info on how lightning is created and released during a storm. I have seen many extraordinary thunder storms and always wondered if there was a specifically chosen striking point or pattern for lightning strikes. Your column has been quick and to the point for what I was looking for.
I wont use cell phone again. its a light magnet .
What about ionization of air particles made by microvawes radiated by the cell phone? Lighting "searches" for the best conducting path across the insulating air. Doesn't cellular help it to strike making conductive (ionized) air particles?
Though carrying a cell-phone may NOT make one more likely to be STRUCK, I'm pretty sure that a metal object, like a belt-buckle, provides a nice entry point for lightning, if you are struck or about to be. I remember a news story a few years ago about 2 campers in a tent, on the side of a mountain in NH. A severe storm caused lightning that hit the tent. One camper survived but the other had fatal, internal injuries, as the lightning strike went into him via his belt buckle! It was in the news…so it's true, right?
Not everything is a myth. Cell phones static electricity can cause a fire at a gas station. This is proof and there are a lot of related videos.(one where a guy uses his cell phone torch to look how much gas is left in the Gasoline truck and it explodes. Its could not be determined if his phone rang at that moment).
Mythbusters got busted cause they did a piece where they only let the phone ring. They did not answer it. Whats shocking is why the "expert" was lying about cell phones. Cellphones can spark cause they got batteries and there are times when batteries are loose so it creates sparks which won't set your phone on fire but can set petrol vapors on fire easily.
A friend of mine had a direct hit and got a cross burnt into his chest. I had lightning strike in my bathroom through the ac duct between a friend and myself. It affected a computer, three tvs and the cell phone said do not call unless emergency, which I have never seen on my cell phone. It would not dial out either.


text:
BREAKING NEWSPowerball jackpot climbs to $613M for tonight's drawing
Sade goes one-on-one with Dr. Jill Biden during first holiday season as first lady
Watch Sade Baderinwa's full interview with First Lady Dr. Jill Biden
WASHINGTON (WABC) -- First Lady Dr. Jill Biden is making history -- she's the first first lady to have a full-time job outside the White House.
The professor of English at Northern Virginia Community College is making education a top priority issue.
Eyewitness News anchor Sade Baderinwa spoke with First Lady Biden during her first winter holiday in the role.
For decades, the First Lady of the United States has helped decorate the White House for the holidays.
"I'm happy with the decorations, I think it reflects what I've seen in this country, and I've seen people from all walks of life -- our essential workers, our nurses and our doctors and our teachers and policemen -- how communities have come together to support one another," Biden said.
Biden herself is an essential worker. She is still a full-time teacher, splitting her time between the White House and the classroom.
"I taught full-time when I was second lady and now as first lady, I'm still teaching, I'm giving finals tomorrow, I just think I worked hard for my degree, it means a lot to me, teaching is who I am," Biden said.
She even had a message for working moms and young women.
"One of the things I try to convey to my students is this sense of confidence, and I think it's really more important than my teaching them commas or thesis statements, but for them to believe in themselves and they can do anything they want to do," Biden said.
RELATED | First Lady says White House Christmas decorations inspired by American families
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Powerball jackpot climbs to $613M for tonight's drawing

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Booking was easy and follow up reminders etc good. This particular apartment was in an ideal situation and had a very interesting view and excellent transport links into the city.It was well appointed and like a home from home.
Excellent apartment, close to all amenities. Spacious and warm. Good price.
Clearer instructions on the heating system would have been useful as the room thermostat didn't work. After finding the instruction booklet the following morning all was well. Just a minor hiccup. The fire in the lounge was great.
Todo muy bien, la casera muy amable, el apartemento muy limpio, la calefacción una maravilla y todos los electrodomesticos funcionando a la perfección, excepto el microondas, al que no le daba vueltas el plato.
Good website, prompt action taken, reminders for payments when due. Handover experience and communication with manager of property excellent. Great, well stocked apartment , only downsides were second bedroom very small and lack of shower in bathroom.
Ideal location, walk to all attractions possible, otherwise tram stop round corner plus bus station.
Too many lights fusing which kept blowing the electrical circuits. Not enough spare bulbs. I would seriously consider checking the overload of the circuits as potential fire hazard.
In a few words: Everithing correct, clair and timely solved. And No Surprises at arrival. Unbeatable service.
We stayed in the Haymarket apartment for 17 nights and we had a great experience. Tony welcomed us and explained everything we need to know. He was very kind and attentive and arranged everything with my daughter, who is more fluent in english than us, and we appreciated that very much.
The apartment is very spacious, bright and very nicely decorated. It was very well equipped with everything that we could have ever needed and it was warm with exception of the double room. The kitchen was fully equiped and everything was spotless. The person who did the cleaning was also very kind and warm.
It is very well located, close to Haymarket train and tram station. Close enough from the City centre (walking distance) and far enough to be quiet in the evening. There are many supermarkets close by as well as pubs and restaurants.
We loved staying there and we hope to come back in our future visits to Edinburgh.
Eddie was very helpful and even provided a Christmas tree for us!
The Hopetoun Residence was perfect for our family gathering. We had a well stocked kitchen and were able to cook and serve a Christmas dinner for 11 people. The location for us was excellent - no more than a 20 minute walk to Princess Street.
Lovely spacious apartment right next to Haymarket train station and tram stop.
This residence was a perfect fit for me and the service was amazing. I stupidly went to the wrong floor and they were extremely accommodating and friendly helping me to get into the flat. The location was an easy walk to everything you would need in the city and had plenty of grocery stores and restaurants within a short walk.
The flat itself had everything I needed (and more). They really thought of everything (including a mat for my flat iron). It is the perfect flat for one person or a couple. The building is very quiet, no neighbor noises and had many amenities. It wasn't a problem for me, but if you are washing clothes, leave time for them to air dry before you will need them since there is no dryer.
I've stayed in flats around Edinburgh about 7 times now and this was one of my favorite flats so far. The location is ideal and the amenities were everything I needed. I highly recommend and will probably stay here again in the future.
Fab!! had a very relaxing time….
Very good. We were looked after throughout. Very professional!
This was a really nice 1 bed apartment! It had everything we needed, particularly safe parking. It was in a very secluded location, and the lounge looked out over a nice park which had its own private terrace. Terrifically good value!
Thank you so much for your feedback. Its great to see that you had a great experience with us and that the apartment's location was excellent for your group.
Super accommodation, comfortable , lovely beds , great location.
Great apartment and fantastic communication from the reserve apartment team ! thank you !


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Politics, Twin Cities
How Much Do Election Campaigns Cost in West St. Paul?
February 22, 2019 Kevin D. Hendricks	1 Comment
We had a competitive and heated election season in West St. Paul, Minn., in 2018. That hasn't always been the case, but it also means an increase in costs.
Let's take a look at the cash spent in recent elections in West St. Paul.
Most Expensive Election?
2018 at nearly $37,700 total. Only one race was uncontested and there was an extremely expensive primary in ward 3 where all four candidates spend over $2,000.
I only looked at data going back to 2008, but given inflation and the rising cost of campaigns, it's likely the most expensive election season in West St. Paul history (I don't quite have the data to verify it, but I'm pretty confident).
Most Expensive Mayoral Campaign?
Jenny Halverson spent $9,878 in 2016 (and won)—$1,204 more than her opponent, David Meisinger. It's only $478 more than what Meisinger spent in his winning campaign in 2014. Halverson's record was likely helped along by a three-way primary with Meisinger and Morgan Kavanaugh.
It's likely the most expensive in history, though high four-figures is fairly common for a contested mayoral race. It's also the most expensive combined primary campaign in recent history, at over $10,500 total.
Most Expensive City Council Campaign?
Meisinger spend $6,396 in 2018 (and lost)—$623 more than his opponent and victor, Wendy Berry. Meisinger's record was also likely helped by a four-way primary with Berry, Lisa Eng-Sarne, and John Ramsay (though it seems Meisinger likely spent the least in the primary, so maybe not).
Again, this is looking back as far as 2008, but it's likely the most expensive in history. City council campaigns hadn't crossed $3,000 until 2016 with Anthony Fernandez, and then Meisinger, Berry, Jim Probst, and John Justen all spent more than $3,000 in 2018.
Least Expensive Election?
2010 at under $1,500 total, when two council races were unopposed, the mayor's race was uncontested (a challenger ran, but didn't spend anything), and the winner in the contested council race didn't spend anything.
That's only going back to 2008, so it's probably not the least expensive in history.
Least Expensive and Victorious Mayoral Campaign?
It's easy to spend nothing and lose, so let's look at those who spent the least and still won.
Dave Napier spent only $4,321 in 2018—$4,035 less than his opponent, Anthony Fernandez.
(I'm not counting the $600 that John Zanmiller spent in 2010 against Mary Jane Duchene—while he technically had opposition, Duchene didn't spend anything and hardly contested the election. Zanmiller won with nearly 68% of the vote).
That's only going back to 2008 and only looking at contested mayoral campaigns. I'm sure there have been cheaper campaigns that won, but in recent history, it's Napier.
Least Expensive and Victorious City Council Campaign?
Ed Hansen reported $0 in his 2010 win (he should have at least had a $20 filing fee). His opponent, Pat Armon, spent $857. The margin of victory was only 61 votes, one of the closer races in recent history (I believe the closest race in recent history was John Bellows' 2014 win over Ramsay by 38 votes). Hansen would later resign after multiple controversies.
Again, this is only back to 2008 and only counts contested city council campaigns (there have been plenty of uncontested campaigns, which are just as cheap).
Does Spending Money Win?
Going back to 2008 and looking only at contested races, the candidate who spent the most won 56% of the time.
But for all three contested races in 2018, the candidate who spent the most money lost.
However, only one of those races could be considered a frugal victory—Napier's mayoral win. Berry and Justen both spent less than their opponents, but more than the average for city council races.
Unopposed Races
An unopposed campaign is pretty cheap. How often does that happen? 44% of our elections were unopposed going back to 2002.
In 2018, one of the four races was uncontested (ward 1, incumbent Dick Vitelli; incidentally, after being appointed to replace Hansen in 2013, Vitelli has run twice for reelection and been unopposed both times).
2008 and 2016 are the only years in recent history when all four races were contested.
Getting More Expensive
Election campaigns in West St. Paul are getting more expensive, though that doesn't necessarily mean victory, as 2018 showed. However, 2018 certainly drove up the cost of elections. Here's a look at the combined costs of elections:
It's also getting more expensive to be on city council. Here's a look at the average cost of city council elections each year:
Between 2008 and 2018, the average cost of a city council campaign is $2,298. 2018 definitely raised that average.
So Where Does All This Money Come From?
The idea of having to come up with $2,300 to run for city council may seem daunting. Where do they get the money? Campaigns are either funded by contributions or the candidate themselves. It's varied throughout the years.
In 2008, Mayor John Zanmiller raised his campaign funds entirely from contributions. The same year, Council member Darlene Lewis self-funded her entire campaign. (Recently Lewis wryly commented that she paid for it with the city council salary she received. Council members currently receive an annual salary of $7,508 and the mayor receives $9,356 [this reflects a 5% increase approved by the previous council on July 23, 2018].)
Both Zanmiller and Lewis won in 2008, so both approaches can work.
While the source of funds and the exact mix depends on the candidate, it's notable that the 2018 winners relied primarily on contributions. Napier, Berry, and Justen all relied on a long list of small contributors. Meanwhile, Fernandez self funded nearly 90% of his campaign, Meisinger at least 86% (if not the entire thing), and Probst nearly 70%.
So campaigns in West St. Paul are getting more expensive, but that doesn't mean running for office is limited to those who can afford to pay for it themselves.
If you love your community and you're willing to work hard to make it better, I'd encourage you to think about running for office.
Three city council seats and the mayor will be on the ballot for West St. Paul in 2020. Those seats are currently held by Mayor Dave Napier, Bob Pace in Ward 1, Anthony Fernandez in Ward 2, and Lisa Eng-Sarne in Ward 3.
The filing period will open in mid-May 2020.
That's a lot of numbers and assumptions, so let's qualify things a bit for those who love the fine print:
Source: These numbers are based on campaign finance disclosure forms submitted to the city clerk and currently on file going back to 2008. It's entirely possible reports are missing or I interpreted the numbers incorrectly (there's not a lot of consistency to how these reports are filled out, even from the same candidate).
Reporting threshold: There is a reporting threshold where candidates don't have to report expenses under a certain amount (I think it's currently $750; not sure if that's changed since 2008). So it's possible some candidates spent less than the threshold and didn't have to report it—I'm assuming no report equals $0 spent, though that may not be correct. However, there are also cases where candidates filed a report claiming they spending $0, so we can be fairly confident on those (however, they all have a $20 filing fee, so it seems they may have missed an item on their report).
Previous runs: These numbers also hide other factors, such as previous runs. For example, in the ward 3 race in 2016, both Napier and Ramsay spent very little, but both had run before and presumably could use signs from that previous run, therefore incurring minimal expense. This can skew the numbers, so sometimes I ignored these campaigns.
City council averages: I only looked at contested city council campaigns—where people either spent money or won—and ignored the weird outliers, such as previous runs or candidates who didn't compete.
Self funding: Because of the inexactness of the reporting, it's hard to determine exactly where funds always came from. Candidates are not required to itemize contributions under $100 (though some do) and can donate as much to their own campaigns as they want, apparently without explicitly saying so (though some do). So some candidates listed contributions, but it's unclear if those funds are from contributions under $100 or from the candidate themselves (or a mix). In some cases the campaign expenses exactly match the campaign contributions, which is a good indication that the candidate self-funded. In other cases, candidates either listed the amounts they gave to their own campaign, or listed $0 in contributions so it's clear the money had to come from themselves.
Full disclosure: I donated to the campaigns of Dave Napier and Wendy Berry in 2018.
Anthony FernandezBob PaceDarlene LewisDave NapierDavid MeisingerDick VitelliEd HansenelectionJenny HalversonJim ProbstJohn BellowsJohn JustenJohn RamsayJohn ZanmillerLisa Eng-SarneMary Jane DucheneMorgan KavanaughPat ArmonWendy BerryWest St. Paul
Previous Post2018 Reading StatisticsNext PostWest St. Paul City Council Recap: Feb. 25, 2019
One thought on "How Much Do Election Campaigns Cost in West St. Paul?"
James J. Heaney says:
This is a delightful (and informative) post.


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Q: Terraform virtual machine extension Runs with Error The command below fails to execute the virtual machine extension when I try to run it.
resource "azurerm_virtual_machine_extension" "main" {
  name                 = "vm12345"
  virtual_machine_id   = azurerm_virtual_machine.main.id
  publisher            = "Microsoft.Azure.Extensions"
  type                 = "CustomScript"
  type_handler_version = "2.0"

 settings = <<SETTINGS
 {
    "fileUris": "https://www.githubprogrammercontent.com/ansible/ansible/devel/examples/scripts/ConfigureRemotingForAnsible.ps1"
    "commandToExecute": "powershell.exe -ExecutionPolicy Unrestricted -File ConfigureRemotingForAnsible.ps1"
}
SETTINGS
}

But I still have the same problem, am I missing something?
azurerm_virtual_machine_extension.test: compute.VirtualMachineExtensionsClient#CreateOrUpdate: Failure sending request: StatusCode=200 -- Original Error: Long running operation terminated with status 'Failed': Code="VMExtensionProvisioningError" Message="VM has reported a failure when processing extension 'vm12345'. Error message: \"Invalid handler configuration. Exiting. Error Message: Expecting state 'Element'.. Encountered 'Text' with name '', namespace ''. \"."

A: i suspect the structure of your code is not properly formatted.A definition is required here. When i did proper indentation, there wasn't an error in the above code.
From Terraform sample registry [here][1]
settings = <<SETTINGS
    {
        "fileUris": "https://www.githubprogrammercontent.com/ansible/ansible/devel/examples/scripts/ConfigureRemotingForAnsible.ps1"
    "commandToExecute": "powershell.exe -ExecutionPolicy Unrestricted -File ConfigureRemotingForAnsible.ps1"
    }
SETTINGS
tags = {
    environment = "Production"
  }

You can try above snippet and see if the error is shown again.
[1]: https://registry.terraform.io/providers/hashicorp/azurerm/latest/docs/resources/virtual_machine_extension

redpajama_set_name: RedPajamaStackExchange


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Washington, DC (September 12, 2022) — America's Service Commissions (ASC) is pleased to announce the release of its 2021 annual report, entitled "Equity and Inclusion: Expanding National Service." The 18-page report highlights major accomplishments from 2021 — like adding three newly-created staff positions to expand organizational capacity and developing 14 new training opportunities for members — and includes updates on ongoing work in many areas, including public policy; member support; and justice, equity, diversity, and inclusion.
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Tides | Ebb and flood
The beaches of Nieuwpoort are never the same width because of the tides. For example, if you take a relaxing walk on the beach, it is useful to know when it is high or low tide.
On this page you will find the astronomical water level forecast for Nieuwpoort. This free tidal information is only intended for private leisure beach activities, such as hiking, sunbathing and beach recreation. These tidal predictions are based on a non-authoritative static model, they do not consider current weather conditions and may therefore deviate ±15 minutes in reality.
Tuesday 7 February
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These tidal predictions are based on a non-authoritative static model, they do not consider current weather conditions and may therefore deviate ±15 minutes in reality.
This free tidal information is only intended for private leisure beach activities, such as hiking, sunbathing and beach recreation. No guarantees are made about the correctness of this free tides data. You may not use this page professionally and not if anyone or anything could be harmed as a result of its use (for example, do not use it for sailing or navigation purposes).
Ebb and flow
The phase of retreating water is called ebb. A beach will get bigger during ebb. In the Oosterschelde or Waddenzee, mud plates will appear, which makes it possible to go mudhiking to the islands. At the end of ebb period, the water is at the lowest point: this is called low tide.
The phase of rising water is called flow. A beach will then get smaller, narrower. Big mud plates, like in the Waddenzee or Oosterschelde, will overflow with water during flow. The water level reached at the end of the flow phase is called high tide.
The time between the start of one flow and the start of the next flow is on average 12 hours and 25 minutes.


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Manish is 9.963 times more likely to work at Infosys.
Manish is 8.787 times more likely to work at Cisco Systems.
Manish is 8.711 times more likely to work at Capgemini.
Manish is 8.265 times more likely to work at Cognizant Technology Solutions.


Nausea is a disease of the Stomach.
The unpleasant feeling in the stomach accompanied with vomit is Nausea.
Your best bet is, Allspice, Ginger and Ipecacuanha.
Note : Nausea is not a disease itself but a symptoms of a number of diseases. It indicates you that something is going wrong in your body.
Inhale the smoke by burning Agarwood sticks.
Powder the roots. Have a pinch once a day for 7 days.
Boil flowers in water. Take half cup twice a day for 3 days.
Take Galangal's rhizome. Dip it in Castor oil. Roast it over flame. Powder it. Take 3 g with honey daily.
OR Mix equal quantity of Galangal's rhizome powder with sugar. Eat 1 tsp once a day.
Mix a pinch Chirata root powder in one teaspoon Honey and take.
Crush seeds of Asparagus Racemosus to make powder. Mix 25 g obtained powder with 100 g Honey. Take 1 tsp three times a day.
Take One cup of water and add 2 tablespoon of dried Avens root. Steep for 10 minutes. Drink One cup regularly for 2-3 days.
Add 1 tsp each of Apple Cider Vinegar and Honey in a glass of water. Drink.
Chew 5 g Cumin seeds with water. It helps to cure Nausea.
Drink 1 glass of Lemon water whenever feel Nauseous.
Chew 4 leaves of Peppermint daily.
Grind dry Bay Leaf ( Tejpatta ). Take 2 pinches with water. It cures the problem.
Apply the crushed Mint ( Pudina ) leaves at nostrils of the patient.
Ginger ( Adrak in India ) is an effective herb to treat Nausea by calming the stomach. It neutralizes the Stomach Acid and supports proper digestion. This helps to reduce the discomfort and treats Nausea.
Take a fresh Ginger. Peel it off. Cut it into small pieces. Chew 2 to 3 pieces throughout the day to treat Nausea.
OR : Peel a Ginger and cut it into small pieces. Add these pieces into 2 to 3 cups of warm water. Boil it for 5 to 10 minutes. Strain well. Add Honey ( Shehad in India ) for taste. Slowly sip this tea to cure Nausea.
Crush one clove and mix it with one teaspoon of Honey. Slowly slowly lick it.
Mix one tablespoon of fresh Pomegranate fruit juice with Honey in a ratio of 1:1. Consume it.
Take one teaspoon each of Sandalwood and Indian Gooseberry powder. Consume it with one cup of lukewarm water.
Boil Spikenard ( Jatamasi ) roots in water. Strain. Take one teaspoon with one teaspoon of Lime juice once a day.
Make a paste of Curry Leaves ( Karipatta ). Add few drops of Lemon juice ( Nimbo ka Ras ) and Honey ( shehad ). Mix and eat.
Peppermint Essential Oil has strong Carminative property. This helps to calm the mind and reduce pain . Honey ( Shehad in India ) soothes the digestive tract treating Nausea.
Take a cup of warm water. Add a drop of Peppermint Essential Oil and 1 teaspoon Honey in it. Stir well. Drink this tea. Have it two times day.
Mix Black Pepper with water in a jar. Add some fresh Lemon juice, mix it well and drink slowly.
Take one tablespoon of sea salt and add six drops of Ginger, one drop of Allspice Essential Oil in it. Mix them well. Inhale its scent. Repeat the process as needed to get relief.
We advise you to please get your Blood Pressure test done and send us your report. We will advise you accordingly.
You please send us the readings of your Blood pressure and we will advise you accordingly.


Great wild life enthusiast Billy Arjan Singh passed away in THE YEAR OF THE TIGER.. He was a scion of the Royal family of Kapurthala Princely State.
India and Tigers are synonyms.
It is because of the love and dedication of great conservationists like Billy , Jim Corbett , Valmik Thapar , The Great Nawab of Junagadh ( for saving the Asiatic Lion in India ) and many more like them ,that Tigers and Lions survive( even if on the edge ) in India.


text:
Human nitrogen fixation and greenhouse gas emissions: a global assessment
22 September 2016	2016 INI, Impact Presentation, Reactive nitrogen deposition
Wim de Vries1,2, Enzai Du3, Klaus Butterbach-Bahl4, Lena Schulte-Uebbing2, Frank Dentener5
1 Alterra Wageningen University and Research Centre, PO Box 47, 6700 AA Wageningen, the Netherlands, wim.devries@wur.nl
2 Environmental Systems Analysis Group, Wageningen University, PO Box 47, 6700 AA Wageningen, the Netherlands.
3 State Key Laboratory of Earth Surface Processes and Resource Ecology, and College of Resources Science & Technology, Beijing Normal University, Xinjiekouwai Street 19#, Beijing, 100875, China.
4 Karlsruhe Institute of Technology, Institute for Meteorology and Climate Research (IMK-IFU), Kreuzeckbahnstrasse 19, 82467 Garmisch-Partenkirchen, Germany.
5 Joint Research Centre, Institute for Environment and Sustainability, Ispra, Italy.
The net impact of human nitrogen (N) fixation on climate (ignoring short-lived components) mainly depends on the magnitude of the warming effect of (direct and indirect) nitrous oxide (N2O) emissions and the cooling effect of N-induced carbon dioxide (CO2) uptake. N-induced CO2 uptake is caused by anthropogenic N deposition which increases net primary production (NPP) in N-limited ecosystems and thus CO2 sequestration. Nitrogen oxide (NOx) emissions, however, also induce tropospheric ozone (O3) formation, and elevated O3 concentrations reduce NPP and thus plant C sequestration. We estimated global-scale impacts of anthropogenic N fixation on net greenhouse gas emissions using recent data and modelling approaches with respect to N inputs to various ecosystems, N2O emissions in response to N inputs, and C exchange in responses to N inputs (C–N response) and O3 exposure (C–O3 response). The estimated impact of human N fixation is dominated by an increase in N2O emissions equal to 1.02 (0.89–1.15) Pg CO2-C equivalent (eq) yr-1. CO2 uptake due to N inputs to terrestrial and aquatic ecosystems corresponds to net emissions of -0.75 (-0.97 to -0.56) Pg CO2-Ceq yr-1, while the reduction in CO2 uptake by N-induced O3 exposure corresponds to net emissions of 0.14 (0.07–0.21) Pg CO2-Ceq yr-1. Overall, human N fixation causes an increase in net greenhouse gas emissions of 0.41 (-0.01–0.80) Pg CO2-Ceq yr-1. Even considering all uncertainties, it is likely that N inputs lead to a net increase in greenhouse gas emissions.


Beautiful executive home featuring 2 master suites, full finished basement w/ theater room and & 4 car garage! Sprawling hardwood floors on the main level with coffered cathedral ceiling and cascading panoramic views of private wooded lot. You will not be disappointed by the perfectly landscaped & pool compatible lot, elegant living/music room, banquet dining, butler pantry, executive study, owner's retreat, sunroom, and Chef's kitchen featuring stainless Kitchen Aid appliances, 6 burner cooktop, double wall ovens, warming drawer, pot filler, designer granite & gathering island with prep sink. Terrace level boasts rec & billiard rooms, theater, gym & guest suite. Priced below appraisal! 5 mins away from shoppes.ZILLOW WRONG re: Days on market.
Directions: From Intersection Hwy 74 And Hwy 54 In Peachtree City, Travel East On Hwy 54, Right Carriage Lane, Enter Hyde Park Subdivision, Left Landaulet Court To 308 Landaulet Court On Right At Cul-De-Sac.


Where do you need Lawn Sprinklers Repaired?
"Jeff is definitely the go to person for your sprinkler job. He was prompt, courteous, and his skill set is unparalleled. He was very taught me ...knowledgeable, about zones and his recommendations concerning my sprinkler system. I felt for the price is was very fair. I would definitely use him again for my next job hands down!!!!!!! "
"I had 2 dead sprinklers when I bought the home, they repaired and adjust these quickly and very reasonably "
"Top notch professional service. "
"Showed up on time and was very professional in their approach to the issues that exist with the system. Fix the system in a reasonable amount A ...of time. . "
"David was quick with the repair if my sprinkler system. Fixed right the first time. Well done. "
"They were fast. I put in the home advisor request and they called within minutes. They were in neighborhood so came over and gave me Worked ...estimate. me in the next morning to do the job. I am happy, and will call them again. "
"Arrived on time and knew what exactly what to do. Also knew our particular watering system. "
Davidson Landscape & Irrigation, Inc.
"Seemed a little expensive. $109 to replace a $5 part. "
Select all problems you are having with your sprinkler system.
Comment: Need someone to come and Look a the zones in my Front yard. The ones by the road are not popping up all the way. Also want to have the Main Front Yard zone checked to ensure it is working optimally.
Comment: None of stations are working and now water is leaking out of the pipes on the side of the house were we can open and close the flow of water.
Comment: No water is flowing throw any of the stations and I now have water leaking out of the pipes on the side of the house were I can open and close the flow of water.
Comment: When the system has finished watering it does not turn off automatically. We have to go to the valve boxes and manually turn them off. I also have gaps in coverage that need to be addressed.
Comment: One sprinkler zone stays on, only way to turn it off is to shut the water valve off.
Comment: I would like to get an estimate for repairing a zone as the that doesn't work and have 6 zones normal.
Comment: need valve for one zone located and replaced.
Comment: I would like to get an estimate for a valve for my sprinkler sections.
Comment: I would like to get an estimate for a residential property. We have several issues with our sprinkler system, issues with heads and possibly a new sprinkler box.
Comment: Our irrigation system is not working properly. One sprinkler is cut at tip and just running water when it goes on and others are not reaching all areas of the lawn as they were previously.


"What's that on the grass?"
Perhaps yes, we can call it that.
This little boy is a wild driver. I had heard from inside our group-room how he had been warned that if he was too wild and fell off then the bobby-car would be given to another child to drive. It took less than a blink-of-an-eye before I witnessed this tortoise-like tangle on the grass.
No harm had been done and it was not long before a slightly more careful driver was in charge of the bobby-car and this little chap was running behind with his rolator.