text:
\section{Introduction.}

A \emph{unicellular map} is a graph embedded in a compact orientable surface, in such a way that its complement is a topological polygon.
Equivalently, a unicellular map can be viewed as a polygon, with an even number of edges, in which edges have been pasted pairwise in order to create a closed orientable surface. The number of handles of this surface is called the \emph{genus} of the map.

These objects are reminiscent in combinatorics, and have been considered in many different contexts. The numbers of unicellular maps of given size and genus appear in random matrix theory as the moments of the Gaussian Unitary Ensemble~(see~\cite{LaZv}). In the study of characters of the symmetric group, unicellular maps appear as factorisations of cyclic permutations~\cite{Jackson:countingcycles}. According to the context, unicellular maps are also called \emph{one-face maps}, \emph{polygon gluings}, or \emph{one-border ribbon graphs}. Sometimes, their duals, \emph{one-vertex maps}, are considered.
The most famous example of unicellular maps are \emph{plane} unicellular maps, which, from Jordan's lemma, are exactly plane trees, enumerated by the Catalan numbers.

The first result in the enumeration of unicellular maps in positive genus was obtained by Lehman and Walsh~\cite{Lehman-Walsh-genus-I}. Using a direct recursive method, relying on formal power series, they expressed the number $\epsilon_g(n)$ of unicellular maps with $n$ edges on a surface of genus $g$ as follows:
\begin{eqnarray}
\label{eq:LW}
\epsilon_g(n)= \sum_{\gamma \vdash g } \frac{(n+1)\dots(n+2-2g-l(\gamma))}{2^{2g} \prod_i c_i! (2i+1)^{c_i}} \textrm{Cat}(n),
\end{eqnarray}
where the sum is taken over partitions $\gamma$ of $g$, $c_i$ is the number of parts $i$ in $\gamma$, $l(\gamma)$ is the total number of parts, and $\textrm{Cat}(n)$ is the $n$-th Catalan number. This formula has been extended by other authors (\cite{Goupil-Schaeffer}).

Later, Harer and Zagier~\cite{Harer-Zagier}, via matrix integrals techniques, obtained the two following equations, known respectively as the Harer-Zagier \emph{recurrence} and the Harer-Zagier \emph{formula}:
\begin{eqnarray}\label{eq:HZrecurrence}
(n+1) \epsilon_g(n) = 2(2n-1) \epsilon_g(n-1) + (2n-1)(n-1)(2n-3) \epsilon_{g-1}(n-2), \\
 \sum_{g\geq 0} \epsilon_{g}(n) y ^{n+1-2g} = \frac{(2n)!}{2^nn!}\sum_{i\geq 1} 2^{i-1} {n \choose i-1} {y \choose i}.
\label{eq:HZformula}
\end{eqnarray}
Formula~\ref{eq:HZformula} has been retrieved by several authors, by various techniques. A combinatorial interpretation of this formula was given by Lass~\cite{Lass}, and the first bijective proof was given by Goulden and Nica~\cite{Goulden-Nica}. Generalizations were given for bicolored, or multicolored maps \cite{Jackson:countingcycles,Schaeffer-Vassilieva}.

The purpose of this paper is to give a new angle of attack to the enumeration of unicellular maps, at a level which is much more combinatorial than what existed before. Indeed, until now no bijective proof (or combinatorial interpretation) of Formulas~\ref{eq:LW} and~\ref{eq:HZrecurrence} are known. As for Formula~\ref{eq:HZformula}, it is concerned with some \emph{generating polynomial} of the numbers $\epsilon_g(n)$:
in combinatorial terms, the bijections in \cite{Goulden-Nica,Schaeffer-Vassilieva} concern maps which are \emph{weighted} according to their genus, by an additional coloring of their vertices, but the genus does not appear explicitely in the constructions. For example, one cannot use these bijections to sample maps of given genus and size.

On the contrary, this article is concerned with the structure of unicellular maps themselves, at fixed genus. We investigate in details the way the unique face of such a map interwines with itself in order to create the handles of the surface. We show that, in each unicellular map of genus $g$, there are $2g$ special "places", which we call \emph{trisections}, that concentrate, in some sense, the handles of the surface. Each of these places can be used to \emph{slice} the map to a unicellular map of lower genus.
Conversely, we show that a unicellular map of genus $g$ can always be obtained in $2g$ different ways by gluing vertices together in a map of lower genus.
In terms of formulas, this leads us to the new combinatorial identity:
\begin{eqnarray}\label{eq:identity}
 2g  \cdot \epsilon_g(n)\!\!\!\!\! &=& \!\!\!\!\! {n+3-2g\choose 3} \epsilon_{g-1}(n) + {n+5-2g\choose 5} \epsilon_{g-2}(n) + \dots + {n+1\choose 2g+1} \epsilon_{0}(n) \\
&=& \!\!\!\!\! \sum_{p=0}^{g-1} {n+1-2p \choose 2g-2p+1} \epsilon_p(n).
\end{eqnarray}
The main advantage of this identity is that it is recursive \emph{only in the genus}: the size $n$ is fixed. For a given $g$, this enables one to compute directly the formula giving $\epsilon_g(n)$, by iteration. From the combinatorial viewpoint, this enables one to construct maps of fixed genus and size very easily.

When iterated, our bijection shows that all unicellular maps can be obtained in a canonical way from plane trees by successive gluings of vertices,
hence giving the first explanation to the fact that $\epsilon_g(n)$ is the product of a polynomial $R_g(n)$ by the $n$-th Catalan number. More precisely, we obtain the formula $\epsilon_g(n)=R_g(n)\mathrm{Cat}(n)$ with:
\begin{eqnarray}\label{eq:Rg}
R_g(n) = \sum_{0=g_0<g_1<\dots<g_r=g}\ \prod_{i=1}^r \frac{1}{2g_i}{{n+1-2g_{i-1}}\choose{2(g_i-g_{i-1})+1}},
\end{eqnarray}
which comes with a clear combinatorial interpretation. This interpretation, and the one of certain properties of the polynomial $R_g$, answers questions of Zagier~\cite[p159]{LaZv}.\\

\noindent {\bf Asymptotic case.} In the paper~\cite{Chapuy:trisections}, we presented a less powerful bijection, that worked only for an asymptotically dominating subset of all unicellular maps. The bijection presented here is really a generalization of the bijection of~\cite{Chapuy:trisections}, in the sense that it coincides with it when specialized to those dominating maps. However, new difficulties and structures appear in the general case, and there is an important gap between the combinatorial results in~\cite{Chapuy:trisections} and the ones of this paper.

\noindent {\bf Extended abstract.} A extended abstract of this paper was presented at the conference FPSAC'09 (Austria, July 2009).

\noindent {\bf Acknowledgements.} I am indebted to Olivier Bernardi, and to Gilles Schaeffer, for very stimulating discussions. Thanks also to Emmanuel Guitter for allowing me to mention his computation in Section~\ref{subsec:guitter}.

\section{Unicellular maps.}

\label{sec:maps}

\subsection{Permutations and ribbon graphs.}
\begin{figure}[h]
\centerline{\includegraphics[scale=1.1]{figures/map}}
\caption{A unicellular map with $11$ edges, $8$ vertices, and genus $2$: (a) ribbon graph; (b) permutations; (c) topological embedding.}
\label{fig:map}
\end{figure}

Rather than talking about topological embeddings of graphs, we work with a combinatorial definition of unicellular maps:
\begin{definition}
A \emph{unicellular map} ${\mathfrak{m}}$ of size $n$ is a triple ${\mathfrak{m}}=(H,\alpha,\sigma)$, where $H$ is a set of cardinality $2n$, $\alpha$ is an involution of $H$ without fixed points,
and $\sigma$ is a permutation of $H$ such that $\gamma=\alpha\sigma$ has only one cycle.
The elements of $H$ are called the \emph{half-edges} of ${\mathfrak{m}}$. The cycles of $\alpha$ and $\sigma$ are called the \emph{edges} and the \emph{vertices} of ${\mathfrak{m}}$, respectively, and the permutation $\gamma$ is called the \emph{face} of ${\mathfrak{m}}$.
\end{definition}
Given a unicellular map ${\mathfrak{m}}=(H,\sigma,\alpha)$, its associated (multi)graph $G$ is the graph whose edges are given by the cycles of $\alpha$, vertices by the cycles of $\sigma$, and the natural incidence relation $v\sim e$ if $v$ and $e$ share an element. Moreover, we draw each edge of $G$ as a \emph{ribbon}, where each side of the ribbon represents one half-edge; we decide which half-edge corresponds to which side of the ribbon by the convention that, if a half-edge $h$ belongs to a cycle $e$ of $\alpha$ and $v$ of $\sigma$, then $h$ is the right-hand side of the ribbon corresponding to $e$, when considered entering $v$. Furthermore, we draw the graph $G$ in such a way that around each vertex $v$, the \emph{counterclockwise ordering} of the half-edges belonging to the cycle $v$ is given by that cycle: we obtain a graphical object called the \emph{ribbon graph} associated to ${\mathfrak{m}}$, as in Figure~\ref{fig:map}(a). Observe that the unique cycle of the permutation $\gamma=\alpha\sigma$ interprets as the sequence of half-edges visited when \emph{making the tour} of the graph, keeping the graph on its left.

A \emph{rooted} unicellular map is a unicellular map carrying a distinguished half-edge $r$, called the root. These maps are considered up to relabellings of $H$ preserving the root, i.e. two rooted unicellular maps ${\mathfrak{m}}$ and ${\mathfrak{m}}'$ are considered the same if there exists a permutation $\pi: H \rightarrow H'$, such that $\pi(r)=r'$, $\alpha=\pi^{-1}\alpha'\pi$, and $\sigma=\pi^{-1}\sigma'\pi$. In this paper, \emph{all} unicellular maps will be rooted, even if not stated.

Given a unicellular map ${\mathfrak{m}}$ of root $r$ and face $\gamma=\alpha\sigma$, we define the linear order $<_{\scriptscriptstyle {\mathfrak{m}}}$ on $H$ by setting:
$$r <_{\scriptscriptstyle {\mathfrak{m}}} \gamma(r) <_{\scriptscriptstyle {\mathfrak{m}}} \gamma_2(r) <_{\scriptscriptstyle {\mathfrak{m}}} \dots <_{\scriptscriptstyle {\mathfrak{m}}} \gamma^{2n-1}(r).
$$
In other words, if we relabel the half-edge set $H$ by elements of $\llbracket 1, 2n \rrbracket$ in such a way that the root is $1$ and the tour of the face is given by the permutation $(1,\dots,2n)$, the order $<_{\scriptscriptstyle {\mathfrak{m}}}$ is the natural order on the integers. However, since in this article we are going to consider maps with a fixed half-edge set, but a changing permutation $\gamma$, it is more convenient (and prudent) to define the order $<_{\scriptscriptstyle {\mathfrak{m}}}$ in this way.

Unicellular maps can also be interpreted as graphs embedded in a topological surface, in such a way that the complement of the graph is a topological polygon. If considered up to homeomorphism, and suitably rooted, these objects are in bijection with ribbon graphs. See~\cite{Mohar-Thomassen}, or the example of Figure~\ref{fig:map}(c).
The \emph{genus} of a unicellular map is the genus, or number of handles, of the corresponding surface. If a unicellular map of genus $g$ has $n$ edges and $v$ vertices, then Euler's characteristic formula says that $v=n+1-2g.$
From a combinatorial point of view, this last equation can also be taken as a definition of the genus.
\begin{comment}

Each unicellular map has a graphical representation in terms of an \emph{embedded graph}, i.e. a graph with a prescribed order of the edges around the vertices, as shown on Figure~\ref{fig:map}. Each edge of the embedded graph is a \emph{ribbon} made of two half-edges that share a cycle of $\alpha$. The permutation $\sigma$ then gives the way to glue the edges together: each cycle of $\sigma$ corresponds to a vertex of $G$. Precisely, the mapping $\sigma$ describes the \emph{counterclockwise ordering} of the half-edges around each vertex. By convention, we consider that

Each cycle of $\alpha$ interprets as an edge of the graph $G$. This edge is itself made of two \emph{half-edges}

We consider the graph $G$ with vertex-set the set of cycles of $\sigma$,  edge-set the set of cycles of $\alpha$, and the incidence relation
interpret each cycle of $\alpha$ as a \emph{ribbon} made of two half-edges.

An embedded graph is a graph in which around each vertex, the half-edges have a specified circular ordering (the counterclockwise ordering).

unicellular map can be viewed as a graph, in which each edge is considered as a ribbon: each side of the ribbon is called a \emph{half-edge}, and the permutation $\alpha$ interprets as the matching that associates the two sides of a same ribbon together. Moreover, in this graph, edges have a specified order

Throughout this paper, we will use two descriptions of unicellular maps: in terms of permutations, and in terms of ribbon graphs. A \emph{unicellular map}

Let us consider a regular $2n$-gon $P$. For each permutation $\alpha\in\mathfrak{S}_{2n}$ which is an involution without fixed point (i.e. product of transpositions of disjoint supports), we can define a Riemann surface as follows. First, we orient the edges of the polygon in the clockwise direction, so that each edge has the interior of the polygon on its right, and we label the edges from $1$ to $2n$ following this orientation. Then, we \emph{glue} each edge of label $i$ with its partner $\alpha(i)$, in such a way that the two edges are oriented in opposite direction. We obtain an orientable surface $P_\alpha$, together with an \emph{embedded graph} $G_\alpha$, which is the image of the original polygon $P$ in the surface $P_\alpha$. Clearly, $G_\alpha$ has $n$ edges: each edge of $G_\alpha$ can be viewed as a \emph{ribbon} made by two original edges of $P$. To avoid confusions, in the rest of the paper, we use the word \emph{edge} only for the edges belonging to an embedded graph $G_\alpha$. For the original edges of the polygon $P$, we use the terminology \emph{half-edge}, or \emph{side of edge}. When we identify two sides of edges, we also identify the vertices at their extremity. However, the number $v$ of vertices that are left in $G_\alpha$ depend on the permutation $\alpha$. From Euler's characteristic formula, if the obtained Riemann surface has genus $g$, then we have:
$$
v= n+1-2g.
$$

In the original polygon $P$, a \emph{corner} is an angular sector formed by two successive sides of edges, and belonging to the interior of $P$. We label each corner by the label of the side of edge that follows it in clockwise direction (so that when making the tour of $P$, the successive corners are labelled from $1$ to $2n$, as side of edges). The embedding of the graph $G_\alpha$ in the \emph{orientable} surface $P_\alpha$ defines a \emph{cyclic ordering}, (the \emph{counterclockwise ordering}) of the corners around each vertex of $G_\alpha$. We let $\sigma$ be the permutation of $\llbracket 1, 2n \rrbracket$ the associates to the corner $i$, the corner that follows it in counterclockwise direction (we have identified corners with their label in $\llbracket 1,2n \rrbracket$, and we will do that all through the paper).
Observe that the vertices of $G_\alpha$ are in bijection with the \emph{cycles} of $\sigma$. Moreover, for each half-edge $i$, the edge $\alpha\sigma(i)$ interprets as the half-edge following $i$ when making the tour of the face of $G_\alpha$. Consequently, the permutation $\alpha\sigma$ equals the long cycle $(1,\dots,2n)$.
This leads to the following definition:
\end{comment}

\subsection{The gluing operation.}

\begin{figure}[h]
\centerline{\includegraphics[scale=0.8]{figures/gluing}}
\caption{(a) The gluing and slicing operations. (b) The "proof" of Lemma~\ref{lemma:facegluing}.}
\label{fig:gluing}
\end{figure}

We let ${\mathfrak{m}}=(H,\alpha,\sigma)$ be a unicellular map of genus $g$, and
 $a_1<_{\scriptscriptstyle {\mathfrak{m}}} a_2 <_{\scriptscriptstyle {\mathfrak{m}}} a_3$ be three half-edges of ${\mathfrak{m}}$ \emph{belonging to three distinct vertices}. Each half-edge $a_i$ belongs to some vertex $v_i=(a_i,h^1_i,\dots h^{m_i}_i)$, for some $m_i\geq 0$. We define the permutation
$$
 \bar v:= (a_1,h^1_2,\dots h^{m_2}_2,a_2,h^1_3,\dots h^{m_3}_3,a_3,h^1_1,\dots h^{m_1}_1),
$$
and we let $\bar\sigma$ be the permutation of $H$ obtained by deleting the cycles $v_1$, $v_2$, and $v_3$, and replacing them by $\bar v$.
The transformation mapping $\sigma$ to $\bar \sigma$ interprets combinatorially as the \emph{gluing} of the three half-edges $a_1,a_2,a_3$, as shown on Figure~\ref{fig:gluing}(a). We have:
\begin{lemma}\label{lemma:facegluing}
The map
${\overline{\mathfrak{m}}}:=(H,\alpha,\bar\sigma)$ is a unicellular map of genus $g+1$.
Moreover, if we let
$$\gamma=\alpha\sigma=(a_1, k^1_1,\dots k^{l_1}_1,a_2, k^1_2,\dots k^{l_2}_2,a_3, k^1_3,\dots k^{l_3}_3)$$
 be the face permutation of ${\mathfrak{m}}$, then the face premutation of ${\overline{\mathfrak{m}}}$ is  given by:
$$
\bar \gamma = (a_1, k^1_2,\dots k^{l_2}_2,a_3, k^1_1,\dots k^{l_1}_1,a_2, k^1_3,\dots k^{l_3}_3)
$$
\end{lemma}
\begin{proof}
In order to prove that ${\mathfrak{m}}$ is a well-defined unicellular map, it suffices to check that its face is given by the long cycle $\bar\gamma$ given in the lemma. This is very easy by observing that the only half-edges whose image is not the same by $\gamma$ and by $\bar \gamma$ are the three half-edges $a_1,a_2,a_3$, and that by construction $\bar \gamma(a_i)=\alpha\bar\sigma(a_i)=\alpha\sigma(a_{i+1})=\gamma(a_{i+1})$. For a more "visual" explanation, see Figure~\ref{fig:gluing}(b).

Now, by construction, ${\overline{\mathfrak{m}}}$ has two less vertices than ${\mathfrak{m}}$, and the same number of edges, so from Euler's formula it has genus $g+1$ (intuitively, the gluing operation has created a new "handle").
\end{proof}

\subsection{Some intertwining hidden there, and the slicing operation.}

\begin{figure}
\centerline{\includegraphics[scale=1.2]{figures/intertwining}}
\caption{(a) In a plane tree, the tour of the face always visits the half-edges around one vertex in \emph{counterclockwise} order; (b) in positive genus (here in genus $1$), things can be different.}
\label{fig:intertwining}
\end{figure}

The aim of this paper is to show that all unicellular maps of genus $g+1$ can be obtained \emph{in some canonical way} from unicellular maps of genus $g$ from the operation above. This needs to be able to "revert" (in some sense) the gluing operation, hence to be able to determine, given a map of genus $g+1$, which vertices may be "good candidates" to be sliced-back to a map of lower genus.

Observe that in the unicellular map ${\overline{\mathfrak{m}}}$ obtained after the gluing operation, the three half-edges $a_1$, $a_2$, $a_3$ appear \emph{in that order} around the vertex $\bar v$, whereas they appear in the \emph{inverse} order in the face $\bar\gamma$. Observe also that this is very different from what we observe in the planar case: if one makes the tour of a plane tree, with the tree on its left, then one necessarily visits the different half-edges around each vertex in \emph{counterclockwise order} (see Figure~\ref{fig:intertwining}).
Informally, one could hope that, in a map of positive genus, those places where the vertex-order does not coincide with the face-order hide some "intertwining" (some handle) of the map, and that they may be used to slice-back the map to lower genus.

We now describe the \emph{slicing} operation, which is nothing but the gluing operation, taken at reverse.
We let ${\overline{\mathfrak{m}}}=(H,\alpha,\bar\sigma)$ be a map of genus $g+1$, and three half-edges $a_1, a_2, a_3$ belonging to a same vertex $\bar v$ of~${\overline{\mathfrak{m}}}$. We say that $a_1, a_2, a_3$ are \emph{intertwined} if they do not appear in the same order in $\bar \gamma=\alpha\bar\sigma$ and in $\bar \sigma$. In this case, we write
$\bar v=(a_1,h^1_2,\dots h^{m_2}_2,a_2,h^1_3,\dots h^{m_3}_3,a_3,h^1_1,\dots h^{m_1}_1)$, and we let $\sigma$ be the permutation of $H$ obtained from $\bar \sigma$ by replacing the cycle $\bar v$ by the product $(a_1, h_1^1, \dots h_{m_1}^1)(a_2, h_1^2, \dots h_{m_2}^2)(a_3, h_1^3, \dots h_{m_3}^3).$
\begin{lemma}\label{lemma:faceslicing}
The map ${\mathfrak{m}}=(H,\alpha,\sigma)$ is a well-defined unicellular map of genus $g$. If we let
$$\bar \gamma = (a_1, k^1_2,\dots k^{l_2}_2,a_3, k^1_1,\dots k^{l_1}_1,a_2, k^1_3,\dots k^{l_3}_3)$$
 be the unique face of ${\overline{\mathfrak{m}}}$, then the unique face of ${\mathfrak{m}}$ is given by:
$$\gamma=\alpha\sigma=(a_1, k^1_1,\dots k^{l_1}_1,a_2, k^1_2,\dots k^{l_2}_2,a_3, k^1_3,\dots k^{l_3}_3).$$

The gluing and slicing operations are inverse one to the other.
\end{lemma}
\begin{proof}
The proof is the same as in Lemma~\ref{lemma:facegluing}: it is sufficient to check the expression given for $\gamma$ in terms of~$\bar \gamma$, which is easily done by checking the images of $a_1,a_2,a_3$.
\end{proof}

\subsection{Around one vertex: up-steps, down-steps, and trisections.}

Let ${\mathfrak{m}}=(H,\alpha,\sigma)$ be a map of face permutation $\gamma=\alpha\sigma$. For each vertex $v$ of ${\mathfrak{m}}$, we let $\min_{\scriptscriptstyle {\mathfrak{m}}}(v)$ be the minimal half-edge belonging to $v$, for the order $<_{\scriptscriptstyle {\mathfrak{m}}}$. Equivalently, $\min_{\scriptscriptstyle {\mathfrak{m}}} (v)$ is the first half-edge from which one reaches $v$ when making the tour of the map, starting from the root. Given a half-edge $h\in H$, we note $V(h)$ the unique vertex it belongs to (i.e. the cycle of $\sigma$ containing it).
\begin{definition}
We say that a half-edge $h\in H$ is an \emph{up-step} if $h<_{\scriptscriptstyle {\mathfrak{m}}} \sigma(h)$, and that it is a \emph{down-step} if $\sigma(h)\leq_{\scriptscriptstyle {\mathfrak{m}}} h$. A down-step $h$ is called a \emph{trisection} if $\sigma(h)\neq \min_{\scriptscriptstyle {\mathfrak{m}}} V(h)$, i.e. if $\sigma(h)$ is not the minimum half-edge inside its vertex.
\end{definition}
As illustrated on Figure~\ref{fig:intertwining}, trisections are specific to the non-planar case (there are no trisections in a plane tree), and one could hope that trisections "hide", in some sense, the handles of the surface. Before making this more precise, we state the following lemma, which is the cornerstone of this paper:
\begin{lemma}[The trisection lemma]\label{lemma:trisection}
Let ${\mathfrak{m}}$ be a unicellular map of genus $g$. Then ${\mathfrak{m}}$ has exactly $2g$ trisections.
\end{lemma}
\begin{proof}
We let ${\mathfrak{m}}=(H,\alpha,\sigma)$, and $\gamma=\alpha\sigma$. We let $n_+$ and $n_-$ denote the number of up-steps and down-steps in ${\mathfrak{m}}$, respectively. Then, we have $n_-+n_+=2n$, where $n$ is the number of edges of ${\mathfrak{m}}$. Now, let $i$ be a half-edge of ${\mathfrak{m}}$, and $j=\sigma^{-1}\alpha\sigma(i)$. Observe that we have $\sigma(j)=\gamma(i)$, and $\gamma(j)=\sigma(i)$. Graphically, $i$ and $j$ lie in two "opposite" corners of the same edge, as shown on Figure~\ref{fig:trisectionproof}. On the picture, it seems clear that if the tour of the map visits $i$ before $\sigma(i)$, then it necessarily visits $\sigma(j)$ before $j$ (except if the root is one of these four half-edges) so that, roughly, there must be almost the same number of up-steps and down-steps. More precisely, let us distinguish three cases.

First, assume that $i$ is an up-step. Then we have $i<_{\scriptscriptstyle {\mathfrak{m}}}\sigma(i)=\gamma(j)$. Now, by definition of the total order~$<_{\scriptscriptstyle {\mathfrak{m}}}$, $i <_{\scriptscriptstyle {\mathfrak{m}}} \gamma(j)$ implies that $\gamma(i)\leq_{\scriptscriptstyle {\mathfrak{m}}}\gamma(j)$.
Hence, $\sigma(j)\leq_{\scriptscriptstyle {\mathfrak{m}}} \gamma(j)$, which, by definition of $<_{\scriptscriptstyle {\mathfrak{m}}}$ again, implies that $\sigma(j)\leq_{\scriptscriptstyle {\mathfrak{m}}} j$ (here, we have used that $\sigma(j)\neq \gamma(j) $ since $\alpha$ has no fixed point). Hence, if $i$ is an up-step, then $j$ is a down-step.

Second, assume that $i$ is a down-step, \emph{and} that $\gamma(j)$ is not equal to the root of ${\mathfrak{m}}$. In this case, we have $j<_{\scriptscriptstyle {\mathfrak{m}}}\gamma(j)$, and $\gamma(j)=\sigma(i)\leq_{\scriptscriptstyle {\mathfrak{m}}} i=\sigma(j)$. Hence $j<_{\scriptscriptstyle {\mathfrak{m}}} \sigma(j)$, and $j$ is an up-step.

The third and last case is when $i$ is a down step, \emph{and} $\gamma(j)$ is the root $r$ of ${\mathfrak{m}}$. In this case, $j$ is the maximum element of $H$ for the order $<_{\scriptscriptstyle {\mathfrak{m}}}$, so that it is necessarily a down-step.

Therefore we have proved that each edge of ${\mathfrak{m}}$ (more precisely, each cycle of $\sigma^{-1}\alpha\sigma$) is associated to one up-step and one down-step, except a special one that has two down-steps.
Consequently, there are exactly two more down-steps that up-steps in the map ${\mathfrak{m}}$, i.e.: $n_-=n_++2$. Recalling that $n_-+n_+=2n$, this gives $n_-=n+1$.

Finally, each vertex of ${\mathfrak{m}}$ carries exactly one down-step which is not a trisection (its minimal half-edge). Hence, the total number of trisections equals $n_--v$, where $v$ is the number of vertices of ${\mathfrak{m}}$. Since from Euler's characteristic formula, $v$ equals $n+1-2g$, the lemma is proved.
\end{proof}

\begin{figure}
 \begin{minipage}{.56\linewidth}
\vspace{4.5mm}
\centerline{\includegraphics[scale=1.0]{figures/trisectionproof}}
\vspace{3mm}
\caption{The main argument in the proof of the trisection lemma: the tour of the face visits $i$ before $\sigma(i)$ if and only if it visits $\sigma(j)$ before $j$, unless $
\sigma(i)$ or $\sigma(j)$ is the root of the map.}
\label{fig:trisectionproof}
 \end{minipage}\hfill
\begin{minipage}{.38\linewidth}
\centerline{\includegraphics[scale=1.0]{figures/diagram}}
\caption{A vertex  $(6,3,12,11,2,5)$ in a map with $12$-half-edges, and its diagram representation (the marked half-edge is~$6$).}
\label{fig:diagram}
 \end{minipage}
\end{figure}

\section{Making the gluing operation injective.}

\label{sec:diagrams}

We have defined above an operation that glues a triple of half-edges, and increases the genus of a map. In this section, we explain that, if we restrict to \emph{certain types} of triples of half-edges, this operation can be made reversible.

\subsection{A diagram representation of vertices.}
\label{subsec:diagrams}

We first describe a graphical visualisation of vertices. Let $v$ be a vertex of ${\mathfrak{m}}$, with a distinguished half-edge $h$. We write $v=(u_0,u_1,\dots,u_m)$, with $u_0=h$. We now consider a grid with $m+1$ columns and $2n$ rows. Each row represents an element of $H$, and the rows are ordered from the bottom to the top by the total order $<_{\scriptscriptstyle {\mathfrak{m}}}$ (for example the lowest row represents the root). Now, for each $i$, inside the $i$-th column, we plot a point at the height corresponding to the half-edge $u_i$. We say that the obtained diagram is \emph{the diagram representation of $v$, starting from $h$}.
In other words, if we identify $\llbracket 1,2n\rrbracket$ with $H$ via the order $<_{\scriptscriptstyle {\mathfrak{m}}}$, the diagram representation of $v$ is the graphical representation of the sequence of labels appearing around the vertex $v$.
If one changes the distinguished half-edge $h$, the diagram representation of $v$ is changed by a circular permutation of its columns.
Figure~\ref{fig:diagram} gives an example of such a diagram (where the permutation $\gamma$ is in the form $\gamma=(1,2,3,\dots)$).

The gluing operation is easily visualised on diagrams. We let as before $a_1<_{\scriptscriptstyle {\mathfrak{m}}} a_2<_{\scriptscriptstyle {\mathfrak{m}}} a_3$ be three half-edges belonging to distinct vertices in a unicellular map ${\mathfrak{m}}$, and we let $\Delta_1,\Delta_2,\Delta_3$ be their corresponding diagrams. We now consider the three horizontal rows corresponding to $a_1$, $a_2$, and $a_3$: they separate each diagram $\Delta_i$ into four blocks (some of which may be empty). We give a name to each of these blocks: $A_i,B_i,C_i,D_i$, from bottom to top, as on Figure~\ref{fig:diaggluing}(a).
\begin{figure}[h]
\centerline{\includegraphics[scale=0.85]{figures/diaggluing}}
\caption{The gluing operation visualized on diagrams. (a) the diagrams before gluing; (b) a temporary diagram, where we the columns represent the counterclockwise turn around $\bar v$, but the rows still represent the original permutation $\gamma$; (c) the final diagram of the new vertex in the new map, where the rows represent the permutation $\bar\gamma$.}
\label{fig:diaggluing}
\end{figure}

We now juxtapose $\Delta_2,\Delta_3,\Delta_1$ together, from left to right,  and we rearrange the three columns containing $a_1,a_2,a_3$ so that these half-edges appear in that order: we obtain a new diagram (Figure~\ref{fig:diaggluing}(b)), whose columns represent the order of the half-edges around the vertex $\bar v$. But the \emph{rows} of that diagram are still ordered according to the order $<_{\scriptscriptstyle {\mathfrak{m}}}$.
In order to obtain the diagram representing $\bar v$ in the new map ${\overline{\mathfrak{m}}}$, we have to rearrange the rows according to $<_{\scriptscriptstyle {\overline{\mathfrak{m}}}}$.
We let $A$ be the union of the three blocks $A_i$ (and similarly, we define $B$, $C$, and $D$). We know that the face permutation of ${\mathfrak{m}}$ has the form
$
{\gamma=(-\hspace{-2mm}- A -\hspace{-2mm}-,a_1,-\hspace{-2mm}- B -\hspace{-2mm}-,a_2,-\hspace{-2mm}- C -\hspace{-2mm}-,a_3,-\hspace{-2mm}- D -\hspace{-2mm}-) },
$
where by $-\hspace{-2mm}- A -\hspace{-2mm}-$, we mean "all the elements of $A$, appearing in a certain order". Now, from the expression of $\bar \gamma$ given in Lemma~\ref{lemma:facegluing},  the permutation $\bar\gamma$ is:
$
\bar \gamma=(-\hspace{-2mm}- A -\hspace{-2mm}-,a_1,-\hspace{-2mm}- C -\hspace{-2mm}-,a_3,-\hspace{-2mm}- B -\hspace{-2mm}-,a_2,-\hspace{-2mm}- D -\hspace{-2mm}-),
$
where inside each block, the half-edges appear in the same order as in $\gamma$. In terms of diagrams, this means that the diagram representing $\bar v$ in the new map ${\overline{\mathfrak{m}}}$ can be obtained by \emph{swapping} the block $B$ with the block $C$, and the row corresponding to $a_2$ with the one corresponding to $a_3$: see Figure~\ref{fig:diaggluing}(c). To sum up, we have:
\begin{lemma}
The diagram of the vertex $\bar v$ in the map ${\overline{\mathfrak{m}}}$ is obtained from the three diagrams $\Delta_1,\Delta_2,\Delta_3$ by the following operations, as represented on Figure~\ref{fig:diaggluing}:\\
\noindent - Juxtapose $\Delta_2,\Delta_3,\Delta_1$ (in that order), and rearrange the columns containing $a_1, a_2,a_3$, so that they appear in that order from left to right. \\
\noindent - Exchange the blocks $B$ and  $C$, and swap the rows containing $a_2$ and $a_3$.
\end{lemma}
Observe that, when taken at reverse, Figure~\ref{fig:diaggluing} gives the way to obtain the diagrams of the three vertices resulting from the \emph{slicing} operation of three intertwined half-edges $a_1,a_2,a_3$ in the map ${\overline{\mathfrak{m}}}$.

\begin{remark}\label{rem:ordre}
The slicing operation does not change the order $<_{\overline{\mathfrak{m}}}$ for half-edges which appear strictly between the root and the minimum half-edge of the three vertices $\{a_1,a_2,a_3\}$. Precisely if $w_1<_{\overline{\mathfrak{m}}} w_2 <_{\overline{\mathfrak{m}}} \dots <_{\overline{\mathfrak{m}}} w_r$ are elements of $H$ such that $w_r <_{\overline{\mathfrak{m}}} a_2$, then Lemma~\ref{lemma:faceslicing} (or, more visually, Figure~\ref{fig:diaggluing}) implies that we have :
$$
w_1 <_{\mathfrak{m}} w_2 <_{\mathfrak{m}} \dots <_{\mathfrak{m}} w_r <_{\mathfrak{m}} a_1
$$
in the map ${\mathfrak{m}}$. The reverse statement is also true.
\end{remark}

\subsection{Gluing three vertices: trisections of type {\bf I}.}

In this section, we let $v_1, v_2, v_3$ be three distinct vertices in the map ${\mathfrak{m}}$. We let $a_i:=\min_{\scriptscriptstyle {\mathfrak{m}}} v_i$, and, up to re-arranging the three vertices, we may assume (and we do) that $a_1<_{\scriptscriptstyle {\mathfrak{m}}}a_2<_{\scriptscriptstyle {\mathfrak{m}}}a_3$. We let $\Delta_1$, $\Delta_2$, $\Delta_3$ be the three corresponding diagrams. Since in each diagram the marked edge is the minimum in its vertex, observe that the blocks $A_1$, $A_2$, $B_2$, $A_3$, $B_3$, $C_3$ do not contain any point. We say that they are \emph{empty}, and we note: $A_1=A_2=B_2=A_3=B_3=C_3=\varnothing$.

We now glue the three half-edges $a_1,a_2,a_3$ in ${\mathfrak{m}}$: we obtain a new unicellular map ${\overline{\mathfrak{m}}}$, with a new vertex $\bar v$ resulting from the gluing.
Now, let $\tau$ be the element preceding $a_3$ around $\bar v$ in the map ${\overline{\mathfrak{m}}}$. Since $A_3=B_3=C_3=\varnothing$, we have either $\tau\in D_3$ or $\tau=a_2$, so that in both case $a_3<_{\scriptscriptstyle {\overline{\mathfrak{m}}}} \tau$.
Moreover, $a_3$	 in not the minimum inside its vertex (the minimum is $a_1$). Hence, $\tau$ is a \emph{trisection} of the map ${\overline{\mathfrak{m}}}$. We let $\Phi({\mathfrak{m}},v_1,v_2,v_3)=({\overline{\mathfrak{m}}},\tau)$ be the pair formed by the new map ${\overline{\mathfrak{m}}}$ and the newly created trisection $\tau$.

It is clear that given $({\overline{\mathfrak{m}}},\tau)$, we can inverse the gluing operation. Indeed, it is easy to recover the three half-edges $a_1$ (the minimum of the vertex), $a_3$ (the one that follows $\tau$), and $a_2$ (observe that, since $B_2$ and $B_3$ are empty, $a_2$ is the smallest half-edge on the left of $a_3$ which is greater than $a_3$). Once $a_1,a_2,a_3$ are recovered, it is easy to recover the map ${\mathfrak{m}}$ by \emph{slicing} $\bar v$ at those three half-edges. This gives:
\begin{lemma}
The mapping $\Phi$, defined on the set of unicellular maps with three distinguished (unordered) vertices, is injective.
\end{lemma}
It is natural to ask for the image of $\Phi$: in particular, can we obtain all pairs $({\overline{\mathfrak{m}}},\tau)$ in this way~? The answer needs the following definition (see Figure~\ref{fig:trisectionmarquee}):
\begin{definition}\label{def:trisection}
Let ${\overline{\mathfrak{m}}}=(H,\alpha,\bar \sigma)$ be a map of genus $g+1$, and $\tau$ be a trisection of ${\overline{\mathfrak{m}}}$. We let $\bar v=V(\tau)$, $b_1=\min_{\scriptscriptstyle {\overline{\mathfrak{m}}}}(\bar v)$, and we let $\Delta$ be the diagram representation of $\bar v$, starting from the half-edge $b_1$. We let $b_3=\sigma(\tau)$ be the half-edge following $\tau$ around $\bar v$, and we let $b_2$ be the minimum half-edge among those which appear before $b_3$ around $\bar v$ and which are greater than $b_3$ for the order~$<_{\scriptscriptstyle {\overline{\mathfrak{m}}}}$.

The rows and columns containing $b_1,b_2,b_3$ split the diagram $\Delta$ into twelve blocks, five of which are necessarily empty, as in Figure~\ref{fig:trisectionmarquee}.
We let $K$ be second-from-left and second-from-bottom block.
We say that $\tau$ is \emph{a trisection of type {\bf I}} is $K$ is empty, and that $\tau$ is \emph{a trisection of type {\bf II}} otherwise.
\end{definition}
\begin{figure}[h]
\centerline{\includegraphics[scale=0.8]{figures/trisectionmarquee}}
\caption{Trisections of type {\bf I} and {\bf II}.}
\label{fig:trisectionmarquee}
\end{figure}
The following proposition is the half way to our main result:
\begin{proposition}\label{prop:type1}
The mapping $\Phi$ is a bijection between the set $\mathcal{U}^3_{g}(n)$ of unicellular maps of genus $g$ with $n$ edges and three distinguished vertices,
and the set $\mathcal{D}^I_{g+1}(n)$ of unicellular maps of genus $g+1$ with $n$ edges and a distinguished trisection of type {\bf I}.
\end{proposition}
\begin{proof}
We already know that $\Phi$ is injective.

We let ${\mathfrak{m}}$ be a unicellular map of genus $g$ with three distinguished vertices $v_1, v_2, v_3$, and ${\overline{\mathfrak{m}}}$ be the map obtained, as above, by the gluing of $M$ by the half-edges $a_1=\min_{\scriptscriptstyle {\mathfrak{m}}} v_1$, $a_2=\min_{\scriptscriptstyle {\mathfrak{m}}} v_2$, $a_3=\min_{\scriptscriptstyle {\mathfrak{m}}} v_3$
(we assume again that $a_1<_{\scriptscriptstyle {\mathfrak{m}}}a_2<_{\scriptscriptstyle {\mathfrak{m}}}a_3$). We let $\bar \Delta$ be the diagram representation of the new vertex $\bar v$ obtained from the gluing in the map ${\overline{\mathfrak{m}}}$, and we use the same notations for the blocks as in Section~\ref{subsec:diagrams}. We also let $\tau=\sigma^{-1}(a_3)$ be the created trisection, and we use the notations of Definition~\ref{def:trisection} with respect to the trisection $\tau$, so that $b_3=a_3$. Then, since $a_1=\min_{\scriptscriptstyle {\overline{\mathfrak{m}}}} \bar v$, we have $a_1=b_1$, and since the blocks $B_2, B_3$, are empty, we have $b_2=a_2$. Hence, the block $C_3$ of Figure~\ref{fig:diaggluing}(c) coincides with the block $K$ of Figure~\ref{fig:trisectionmarquee}. Since $C_3$ is empty, $\tau$ is a trisection of type {\bf I}. Therefore the image of $\Phi$ is included in $\mathcal{D}^I_{g+1}(n)$.

Conversely, let ${\overline{\mathfrak{m}}}=(H,\alpha,\bar\sigma)$ be a map of genus $g+1$, and $\tau$ be a trisection of type {\bf I} in ${\overline{\mathfrak{m}}}$. We let $b_1, b_2, b_3$ and $K$ be as in Definition~\ref{def:trisection}. First, since $b_1<_{\scriptscriptstyle {\overline{\mathfrak{m}}}} b_3<_{\scriptscriptstyle {\overline{\mathfrak{m}}}} b_2$, these half-edges are intertwined, and we know that the slicing of ${\overline{\mathfrak{m}}}$ by these half-edges creates a well-defined unicellular map ${\mathfrak{m}}$ of genus $g$ (Lemma~\ref{lemma:faceslicing}). Now, if we compare Figures~\ref{fig:trisectionmarquee} and~\ref{fig:diaggluing}, we see that the result of the slicing is a triple of vertices $v_1,v_2,v_3$, such that each half-edge $b_i$ is the minimum in the vertex $v_i$: indeed, the blocks $A_1,A_2,A_3,B_2,B_3$ are empty by construction, and the block $C_3=K$ is empty since $\tau$ is a trisection of type {\bf I}. Hence we have $\Phi({\mathfrak{m}},v_1,v_2,v_3)=({\overline{\mathfrak{m}}}, \tau)$, so that the image of $\Phi$ exactly equals the set $\mathcal{D}^I_{g+1}(n)$
\end{proof}

\subsection{Trisections of type {\bf II}.}

Of course, it would be nice to have a similar result for trisections of type {\bf II}.
Let ${\overline{\mathfrak{m}}}=(H,\alpha,\bar\sigma)$ be a map of genus $g+1$ with a distinguished trisection $\tau$ of type {\bf II}. We let $b_1,b_2,b_3$ and $K$ be as in Definition~\ref{def:trisection} and Figure~\ref{fig:trisectionmarquee}, and we let ${\mathfrak{m}}$ be the result of the slicing of ${\overline{\mathfrak{m}}}$ at the three half-edges $b_1,b_2,b_3$. If we use the notations of Figure~\ref{fig:diaggluing}, with $a_i=b_i$, we see that we obtain three vertices, of diagrams $\Delta_1,\Delta_2,\Delta_3$, such that $A_1=A_2=B_2=A_3=B_3=\varnothing$. Hence, we know that $a_1=\min_{\scriptscriptstyle {\mathfrak{m}}}(v_1)$, that $a_2=\min_{\scriptscriptstyle {\mathfrak{m}}}(v_2)$, and that $a_2<\min_{\scriptscriptstyle {\mathfrak{m}}}(v_3)$. Observe that, contrarily to what happened in the previous section, the block $C_3=K$ is \emph{not} empty, therefore $a_3$ is \emph{not} the minimum inside its vertex.

Now, we claim that $\tau$ \emph{is still a trisection} in the map ${\mathfrak{m}}$. Indeed, by construction, we know that $\tau$ belongs to $D_3$ (since, by definition of a trisection, it must be above $a_3$ in the map ${\overline{\mathfrak{m}}}$, and since $B_3$ is empty). Hence we still have $a_3 <_{\scriptscriptstyle {\mathfrak{m}}} \tau$ in the map ${\mathfrak{m}}$. Moreover, we have clearly $\sigma(\tau)=a_3$ in ${\mathfrak{m}}$ (since $\tau$ is the rightmost point in the blocks $C_3\cup D_3$), and it follows that $\tau$ is a trisection in ${\mathfrak{m}}$.

We let $\Gamma({\overline{\mathfrak{m}}},\tau)=({\mathfrak{m}},v_1,v_2,\tau)$ be the $4$-tuple consisting of the new map ${\mathfrak{m}}$, the two first vertices $v_1$ and $v_2$ obtained from the slicing, and the trisection $\tau$. It is clear that $\Gamma$ is injective: given $({\overline{\mathfrak{m}}},v_1,v_2,\tau)$, one can reconstruct the map ${\overline{\mathfrak{m}}}$ by letting $a_1=\min v_1$, $a_2=\min v_2$, and $a_3=\sigma(\tau)$, and by gluing back together the three half-edges $a_1,a_2,a_3$. Conversely, we define:
\begin{definition}
We let $\mathcal{V}_g(n)$ be the set of $4$-tuples $({\mathfrak{m}},v_1,v_2,\tau)$, where
${\mathfrak{m}}$ is a unicellular map of genus $g$ with $n$ edges, and where $v_1$, $v_2$, and $\tau$ are respectively two vertices and a trisection of $M$ such that:
\begin{eqnarray}\label{eq:contrainteordre}
\min_{\scriptscriptstyle {\mathfrak{m}}} v_1 <_{\scriptscriptstyle {\mathfrak{m}}} \min_{\scriptscriptstyle {\mathfrak{m}}} v_2 <_{\scriptscriptstyle {\mathfrak{m}}} \min_{\scriptscriptstyle {\mathfrak{m}}} V(\tau).
\end{eqnarray}
Given $({\mathfrak{m}},v_1,v_2,\tau)\in\mathcal{V}_G(n)$, we let ${\overline{\mathfrak{m}}}$ be the map obtained from the gluing of the three half-edges $\min v_1$, $\min v_2$, and $\sigma(\tau)$, and we let $\Psi({\mathfrak{m}},v_1,v_2,\tau):=({\overline{\mathfrak{m}}},\tau)$.
\end{definition}
We can now state the following proposition, that completes Proposition~\ref{prop:type1}:
\begin{proposition}\label{prop:type2}
The mapping $\Psi$ is a bijection between
the set $\mathcal{V}_g(n)$ of unicellular maps of genus $g$ with $n$ edges and a distinguished triple $(v_1,v_2,\tau)$ satisfying Equation~\ref{eq:contrainteordre},
and the set $\mathcal{D}^{II}_{g+1}(n)$ of unicellular maps of genus $g+1$ with $n$ edges and a distinguished trisection of type {\bf II}.
\end{proposition}
\begin{proof}
In the discussion above, we have already given a mapping $\Gamma: \mathcal{D}^{II}_{g+1}(n) \rightarrow \mathcal{V}_g(n)$, such that $\Psi\circ\Gamma$ is the identity on $\mathcal{D}_{g+1}^{II}(n)$.

Conversely, let $({\mathfrak{m}},v_1,v_2,\tau)\in\mathcal{V}_{g}(n)$, and let $a_1=\min v_1$, $a_2=\min v_2$, and $a_3=\sigma(\tau)$. By definition, we know that $a_2<\min V(\tau)$, so that in the diagram representation of the three vertices $v_1,v_2,V(\tau)$ (Figure~\ref{fig:diaggluing}(a)) we know that the blocks $A_1,A_2,A_3,B_2,B_3$ are empty. Moreover, since $\tau$ is a trisection, $a_3$ is not the minimum inside its vertex, so the block $C_3$ is \emph{not} empty. Hence, comparing Figures~\ref{fig:diaggluing}(c) and~\ref{fig:trisectionmarquee}, and observing once again that the blocks $C_3$ and $K$ coincide, we see that after the gluing, $\tau$ is a trisection of type {\bf II} in the new map ${\overline{\mathfrak{m}}}$. Moreover, since the slicing and gluing operations are inverse one to the other, it is clear that $\Gamma({\overline{\mathfrak{m}}},\tau)=({\mathfrak{m}},v_1,v_2,\tau)$. Hence, $\Gamma\circ\Psi$ is the identity, and the proposition is proved.
\end{proof}

\section{Iterating the bijection.}

Of course Proposition~\ref{prop:type1} looks nicer than its counterpart Proposition~\ref{prop:type2}: in the first one, one only asks to distinguish three vertices in a map of lower genus, whereas in the second one, the marked triple must satisfy a nontrivial constraint (Equation~\ref{eq:contrainteordre}). In this section we will work a little more in order to get rid of this problem. We start with two definitions (observe that for $k=3$ this is coherent with what precedes):
\begin{definition}
We let $\mathcal{U}^k_g(n)$ be the set of unicellular maps of genus $g$ with $n$ edges, and $k$ distinct unordered distinguished vertices.
\end{definition}
\begin{definition}
We let $\mathcal{D}_g(n)=\mathcal{D}^I_g(n)\cup\mathcal{D}^{II}_g(n)$ be the set of unicellular maps of genus $g$ with $n$ edges, and a distinguished trisection.
\end{definition}

\subsection{Training examples: genera $1$ and $2$.}

Observe that the set $\mathcal{V}_0(n)$ is empty, since there are no trisections in a plane tree. Hence, from Proposition~\ref{prop:type2}, there are no trisections of type {\bf II} in a map of genus~$1$ (i.e. $\mathcal{D}_1^{II}(n)=\varnothing$). Proposition~\ref{prop:type1} gives:
\begin{corollary}
The set $D_1(n)$ of unicellular maps of genus $1$ with $n$ edges and a distinguished trisection is in bijection with the set\/ $U_0^3(n)$ of rooted plane trees with $n$ edges and three distinguished vertices.
\end{corollary}
\noindent Since from the trisection lemma (Lemma~\ref{lemma:trisection}) each unicellular map of genus $1$ has exactly $2$ trisections, we obtain that the number $\epsilon_1(n)$ of rooted unicellular maps of genus $1$ with $n$ edges satisfies:
$$
2 \cdot \epsilon_1(n) = {n+1 \choose 3} \mathrm{Cat}(n),
$$
which gives a neat combinatorial proof of the formula $\epsilon_1(n)=\frac{(n+1)n(n-1)}{12}\mathrm{Cat}(n)$~\cite{Lehman-Walsh-genus-I}.\\

We now consider the case of genus $2$. Let ${\mathfrak{m}}$ be a unicellular map of genus $2$, and $\tau$ be a trisection of ${\mathfrak{m}}$. If $\tau$ is of type {\bf I}, we know that we can use the application $\Phi^{-1}$, and obtain a unicellular map of genus~$1$, with three distinguised vertices.

Similarly, if $\tau$ is of type {\bf II}, we can apply the mapping $\Psi^{-1}$, and we are left with a unicellular map ${\mathfrak{m}}'$ of genus $1$, and a marked triple $(v_1,v_2,\tau)$, such that $\min_{\scriptscriptstyle {\mathfrak{m}}'} v_1 <_{\scriptscriptstyle {\mathfrak{m}}'} \min_{\scriptscriptstyle {\mathfrak{m}}'} v_2 <_{\scriptscriptstyle {\mathfrak{m}}'} \min_{\scriptscriptstyle {\mathfrak{m}}'} V(\tau)$. From now on, we use the more compact notation: $v_1 <_{\scriptscriptstyle {\mathfrak{m}}'} v_2 <_{\scriptscriptstyle {\mathfrak{m}}'} V(\tau)$, i.e. we do not write the min's anymore.
The map $({\mathfrak{m}}',\tau)$ is a unicellular map of genus $1$ with a distinguished trisection: therefore we can apply the mapping $\Phi^{-1}$ to $({\mathfrak{m}}',\tau)$. We obtain a plane tree ${\mathfrak{m}}''$, with three distinguished vertices $v_3,v_4,v_5$ inherited from the slicing of $\tau$ in ${\mathfrak{m}}'$; since  those three vertices are undistinguishable, we can assume that $ v_3<_{\scriptscriptstyle {\mathfrak{m}}''} v_4 <_{\scriptscriptstyle {\mathfrak{m}}''} v_5$.
Observe that in ${\mathfrak{m}}''$ we also have the two marked vertices $v_1$ and $v_2$ inherited from the slicing of $\tau$ in ${\mathfrak{m}}$. Moreover the fact that $ v_1 <_{\scriptscriptstyle {\mathfrak{m}}'}  v_2 <_{\scriptscriptstyle {\mathfrak{m}}'} V(\tau)$ in ${\mathfrak{m}}'$ implies that $v_1 <_{\scriptscriptstyle {\mathfrak{m}}''} v_2 <_{\scriptscriptstyle {\mathfrak{m}}''} v_3$ in ${\mathfrak{m}}''$: this follows from Remark~\ref{rem:ordre}.
Hence, we are left with a plane tree ${\mathfrak{m}}''$, with \emph{five} distinguished vertices
$v_1<_{\scriptscriptstyle {\mathfrak{m}}''}v_2<_{\scriptscriptstyle {\mathfrak{m}}''}v_3<_{\scriptscriptstyle {\mathfrak{m}}''}v_4<_{\scriptscriptstyle {\mathfrak{m}}''}v_5$.
Conversely, given such a $5$-tuple of vertices, it is always possible to glue the three last ones together by the mapping $\Phi$ to obtain a triple $(v_1,v_2,\tau)$ satisfying Equation~\ref{eq:contrainteordre}, and then to apply the mapping $\Psi$ to retrieve a map of genus $2$ with a marked trisection of type {\bf II}. This gives:
\begin{corollary}
The set $\mathcal{D}^{II}_2(n)$ is in bijection with the set $\mathcal{U}^5_0(n)$ of plane trees with \emph{five} distinguished vertices.

The set $\mathcal{D}_2(n)$ of unicellular maps of genus $2$ with one marked trisection is in bijection with the set $\mathcal{U}_1^3(n)\cup\mathcal{U}_0^5(n)$.
\end{corollary}
\noindent From Euler's formula, a unicellular map of genus $1$ with $n$ edges has $n-1$ vertices.
Since from the trisection lemma each unicellular maps of genus $2$ has $4$ trisections,  we obtain the following formula for the number $\epsilon_2(n)$ of unicellular maps of genus $2$ with $n$ edges:
$$
4\cdot \epsilon_2(n) = {n-1 \choose 3} \epsilon_1(n) + {n+1 \choose 5} \mathrm{Cat}(n),
$$
from which it follows that
$$
\epsilon_2(n)= \frac{(n+1)n(n-1)(n-2)(n-3)(5n-2)}{1440} \mathrm{Cat}(n).
$$

\subsection{The general case, and our main theorem.}

In the general case, we will work as in the example of genus $2$: while we find trisections of type {\bf II}, we open them (and we decrease the genus of the map), and we stop when we have opened the first encountered trisection of type {\bf I}. We start with the description of the inverse procedure, which goes as follows.\\

We let $p\geq 0$ and $q\geq 1$ be two integers, and $({\mathfrak{m}},v_*)=({\mathfrak{m}},v_1,\dots,v_{2q+1})$ be an element of $\mathcal{U}^{2q+1}_p(n)$. Up to renumbering the vertices, we can assume that $v_1 <_{\scriptscriptstyle {\mathfrak{m}}} v_2 <_{\scriptscriptstyle {\mathfrak{m}}} \dots <_{\scriptscriptstyle {\mathfrak{m}}} v_{2q+1}$.
\begin{definition}
We consider the following procedure:
\\ \noindent {\bf i.}
Glue the three last vertices $v_{2q-1},v_{2q},v_{2q+1}$ together, via the mapping $\Phi$, in order to obtain a new map ${\mathfrak{m}}_{1}$ of genus $p+1$ with a distinguished trisection $\tau$ of type {\bf I}.
\\ \noindent {\bf ii.}
{\bf for} $i$ from $1$ to $q-1$ {\bf do:}\\
Let $(v_{2q-2i-1},v_{2q-2i},\tau)$ be the triple consisting of the last two vertices which have not been used until now, and the trisection $\tau$. Apply the mapping $\Psi$ to that triple, in order to obtain a new map ${\mathfrak{m}}_{i+1}$ of genus $p+i+1$, with a distinguished trisection $\tau$ of type {\bf II}.
\\ \noindent {\bf end for.}\\
 We let $\Lambda({\mathfrak{m}},v_*):=({\mathfrak{m}}_{q},\tau)$ be the map with a distinguished trisection obtained at the end of this procedure. Observe that if $q=1$, the distinguished trisection is of type {\bf I}, and that it is of type {\bf II} otherwise.
\end{definition}

As in the case of genus $2$, we have the next theorem:
\begin{theorem}[Our main result]\label{thm:main}
The application $\Lambda$ defines a bijection:
$$
\Lambda : \  \biguplus _{p=0}^{g-1} \mathcal{U}^{2g-2p+1}_p(n)
\longrightarrow \mathcal{D}_g(n).
$$
In other words, all unicellular maps of genus $g$ with a distinguished trisection can be obtained in a canonical way by starting with a map of a lower genus with an odd number of distinguished vertices, and then applying once the mapping $\Phi$, and a certain number of times the mapping $\Psi$.
\end{theorem}

Given a map with a marked trisection $({\mathfrak{m}},\tau)$, the converse application consists in slicing recursively the trisection $\tau$ while it is of type {\bf II}, then slicing \emph{once} the obtained trisection of type {\bf I}, and remembering all the vertices resulting from the successive slicings.
More formally, we have the next proposition:
\begin{proposition}\label{prop:main}
Let $({\mathfrak{m}},\tau)\in\mathcal{D}_g(n)$. We define the pair $\Xi({\mathfrak{m}},\tau)$ by the following procedure:
\begin{enumerate}
\setlength{\itemsep}{-2pt}
\item We let ${\mathfrak{m}}_0:={\mathfrak{m}}$  and $i:=0$.
\item \label{goto}
{\bf If} $\tau$ is of type {\bf II} in ${\mathfrak{m}}_i$, we let
$({\mathfrak{m}}_{i+1},v_{2i+1},v_{2i+2}):=\Psi^{-1}({\mathfrak{m}}_i,\tau)$. Then we let $i:=i+1$ and we return to step~\ref{goto}. \\
{\bf  Else,} $\tau$ is of type {\bf I} in ${\mathfrak{m}}_i$ and we go to step~\ref{goto1}.
\item \label{goto1} Let $({\mathfrak{m}}_{i+1},v_{2i+1},v_{2i+2},v_{2i+3}):=\Phi^{-1}({\mathfrak{m}}_i,\tau)$.
We end the pocedure and we let
$$\Xi({\mathfrak{m}},\tau):=({\mathfrak{m}}_{i+1}, v_1,v_2,\dots,v_{2i+3}).$$
\end{enumerate}
\noindent
Then the mapping
$$
\Xi : \  \mathcal{D}_g(n)
\longrightarrow \biguplus _{p=0}^{g-1} \mathcal{U}^{2g-2p+1}_p(n).
$$
is a bijection, which is the inverse bijection of $\Lambda$.
\end{proposition}

\begin{proof}[Proof of Theorem~\ref{thm:main} and Proposition~\ref{prop:main}]
First, the mapping $\Xi$ is well defined. Indeed, by definition of a trisection of type {\bf II}, we know by induction that each time we enter steps~\ref{goto} and~\ref{goto1}, $\tau$ is a trisection of the map ${\mathfrak{m}}_i$. Moreover, since the genus of the maps ${\mathfrak{m}}_i$ decreases with $i$, we necessarily reach step~\ref{goto1}, and the procedure stops.

Then, the mapping $\Lambda$ is clearly injective, since the mappings $\Psi$ and $\Phi$ are.

Finally, to prove at the same time that $\Xi$ is injective and that it is the inverse mapping of $\Lambda$, it is enough to show that the vertices $v_i$ produced by the procedure defining $\Xi$ satisfy $v_1 <_{\mathfrak{m}} v_2 <_{\mathfrak{m}} \dots <_{\mathfrak{m}} v_{2q+1}$. Indeed, after that it will be clear by construction that $\Lambda\circ\Xi=\Xi\circ\Lambda=Id$.
Now, we deduce from Remark~\ref{rem:ordre} and by an induction on $i$ that after the $i$th passage in step~\ref{goto} in the definition of $\Xi$, we have
$v_1 <_{\mathfrak{m}_{i+1}} v_2 <_{\mathfrak{m}_{i+1}} \dots <_{\mathfrak{m}_{i+1}} v_{2i+2}$. The same remark shows that the end of step~\ref{goto1}, we have
$v_1 <_{\mathfrak{m}_{i+1}} v_2 <_{\mathfrak{m}_{i+1}} \dots <_{\mathfrak{m}_{i+1}} v_{2i+3}$, which concludes the proof.
\end{proof}

\section{Enumerative corollaries.}

\subsection{A combinatorial identity}

Using the trisection lemma (Lemma~\ref{lemma:trisection}), Euler's formula, and Theorem~\ref{thm:main}, we obtain the following new identity (stated in the introduction as Equation~\ref{eq:identity}):
\begin{theorem}\label{thm:identity}
The number $\epsilon_g(n)$ of rooted unicellular maps of genus $g$ with $n$ edges satisfies the following combinatorial identity:
\begin{eqnarray*}
 2g  \cdot \epsilon_g(n)\!\!\!\!\! &=& \!\!\!\!\! {n+3-2g\choose 3} \epsilon_{g-1}(n) + {n+5-2g\choose 5} \epsilon_{g-2}(n) + \dots + {n+1\choose 2g+1} \epsilon_{0}(n).
\end{eqnarray*}
\end{theorem}
\noindent Observe that this identity is recursive only in the genus (the number of edges $n$ is fixed). For that reason, it enables one to compute easily, for a fixed $g$, the closed formula giving $\epsilon_g(n)$ by a simple iteration (as we did for genera $1$ and $2$).

\subsection{The polynomials $R_g(n)$}

By iterating Theorem~\ref{thm:main}, we obtain that all unicellular maps of genus $g$ with $n$ edges can be obtained from a plane tree with $n$ edges, by successively gluing vertices together. From the enumeration viewpoint, we obtain the first bijective proof that the numbers $\epsilon_g(n)$ are the product of a polynomial and a Catalan number:
\begin{corollary}
The number $\epsilon_g(n)$ of unicellular maps of genus $g$ with $n$ edges equals:
$$
\epsilon_g(n)=R_g(n)\mathrm{Cat}(n),
$$
where $R_g$ is the polynomial of degree $3g$ defined by the formula:
$$
R_g(n) = \sum_{0=g_0<g_1<\dots<g_r=g}\,\,\prod_{i=1}^r \frac{1}{2g_i}{{n+1-2g_{i-1}}\choose{2(g_i-g_{i-1})+1}}.
$$
\end{corollary}
\begin{proof}
The statement directly comes from an iteration of the bijection of Theorem~\ref{thm:main}. More precisely, the formula given here for $R_g(n)$ reads as follows. In order to generate a unicellular map of genus~$g$, we start with a plane tree with $n$ edges, and we apply a certain number of times (say $r$) the mapping~$\Lambda$ to create unicellular maps of increasing genera. In the formula, $g_1< \dots < g_r = g$ are the genera of the maps produced by the successive applications of~$\Lambda$. Now, in order to increase the genus from $g_{i-1}$ to $g_i$, we have to choose $2(g_i-g_{i-1})+1$ vertices in a unicellular map of genus $g_{i-1}$, which gives the binomial in the formula. The factor $1/(2g_i)$ is here to compensate the multiplicity in the construction coming from the trisection lemma (Lemma~\ref{lemma:trisection}).
\end{proof}
\noindent From Theorem~\ref{thm:identity} and the fact that $\mathrm{Cat}(n)$ is asymptotically equivalent to $\frac{1}{\sqrt{\pi}}n^{-\frac{3}{2}}4^n$, one recovers easily the asymptotic behaviour of $\epsilon_g(n)$:
\begin{corollary}
The polynomial $R_g(n)$ has degree $3g$ and leading coefficient $\tfrac{1}{12^g g!}$. When $n$ tends to infinity, one
has~\cite{BeCaRo:asymptotic-nb-tree-rooted}:
$$
\epsilon_g(n)\sim \frac{1}{12^gg! \sqrt{\pi}} n^{3g-\frac{3}{2}} 4^n.
$$
\end{corollary}

\noindent Our construction also answers a question of Zagier concerning the interpretation of a property of the polynomials $R_g$:
\begin{corollary}[Zagier {\cite[p. 160]{LaZv}}]
For each integer $g\geq 1$, the polynomial $R_g(n)$ is divisible by $(n+1) \dots (n+1-2g)$.
\end{corollary}
\begin{proof}
For any fixed sequence $g_1< \dots < g_r$ of intermediate genera, one has to choose $\sum_{i}(2g_i-2g_{i-1}+1)=2g+r$ distinct vertices in the original plane tree in order to apply the construction. Since $r\geq 1$, this number is at least $2g+1$, and the statement follows.
\end{proof}

\section{Variants.}

\subsection{Bipartite unicellular maps}

A unicellular map is \emph{bipartite} if one can color its vertices in black and white in such a way that only vertices of different colors are linked by an edge. By convention, the root-vertex will always be colored in white.
\begin{definition}
We let $\beta_g(i,j)$ be the number of bipartite unicellular maps of genus $g$ with $i$ white vertices and $j$ black vertices. Such a map has $i+j+2g-1$ edges.
\end{definition}
It is clear that that our construction also applies to bipartite unicellular maps: the only difference is that, if we want the gluing operations $\Phi$ and $\Psi$ to preserve the bipartition of the map, we have to paste together only vertices \emph{of the same color}. We therefore obtain the following variant of our main identity:
\begin{proposition}The number $\beta_g(i,j)$ of bipartite unicellular maps with $i$ white vertices and $j$ black vertices obey the following recursion formula:
\begin{eqnarray}
\label{eq:identitebip}
2g \cdot \beta_g(i,j) =
\sum_{p=0}^{g-1} \binom{i+2g-2p}{2g-2p+1} \beta_{p}(i+2g-2p,j)
\ +\
\sum_{p=0}^{g-1} \binom{j+2g-2p}{2g-2p+1} \beta_{p}(i,j+2g-2p).
\end{eqnarray}
\end{proposition}

\begin{corollary}
We have $\beta_g(i,j)=S_g(i,j) \beta_0(i,j)$, where $\beta_0(i,j)=\frac{i+j-1}{ij}\binom{i+j-2}{i-1}^2$ is the number of bipartite plane trees with $i$ white vertices and $j$ black vertices computed in~\cite{Goulden-Jackson:book}, and $S_g$ the polynomial in $(i,j)$ defined by:
$$
S_g(i,j) = \frac{1}{2g} \sum_{p=0}^{g-1} \binom{i+2g-2p}{2g-2p+1} S_{p}(i+2g-2p,j)
\ +\
 \frac{1}{2g}\sum_{p=0}^{g-1} \binom{j+2g-2p}{2g-2p+1} S_{p}(i,j+2g-2p)
$$
with $S_0=1$.
\end{corollary}
\noindent For example for the first genera we obtain:
\begin{align*}
S_0=1\ \ ; \ \ S_1(i,j)=\frac{(i+2)(i+1)i+(j+2)(j+1)j}{12} \ \ ; \ \ S_2(i,j) = s_2(i,j)+ s_2(j,i)
\end{align*}
where
$s_2(i,j)=$ {\small $\frac{i (i+1) (i+2) (i^5+22 i^4+211 i^3+2 i^2 j+998 i^2+i^2 j^3+3 i^2 j^2+21 i j^2+2248 i+7 i j^3+14 i j+96 j^2+1920+64 j+32 j^3)
}{5760} $}.\\

\subsection{Precubic unicellular maps}

A unicellular map is \emph{precubic} if all its vertices have degree $1$ or $3$. In such a map, all trisections are necessarily of type {\bf I}: indeed, a trisection of type {\bf II} cannot appear in a vertex of degree less than $4$. Therefore, each precubic map can be obtained in exactly $2g$ different ways from a \emph{precubic} map of genus $(g-1)$ with three distinguished \emph{leaves}. By repeating the argument $g$ times, we see that each precubic unicellular map can be otbained in exactly $2g \cdot 2(g-1) \dots 2 = 2^gg!$ different ways from a precubic tree (a plane tree where all vertices have degrees $1$ or $3$), by $g$ successive gluings of triple of leaves.

Now, we can easily enumerate precubic trees with $n$ edges. First, we observe that by removing a leaf to such a tree, we find a binary tree with $n-1$ edges (and $n$ vertices). This implies that $n=2m+1$, where $m$ is the number of \emph{nodes} of the binary tree, and that the number of precubic trees with $n$ edges which are rooted on a leaf is the Catalan number $\mathrm{Cat}(m)$. A double-counting argument then shows that those whose root-vertex has degree $3$ are counted by the number $\frac{3m}{m+2}\mathrm{Cat}(m)$: indeed, the number $3m\mathrm{Cat}(m)$ counts precubic trees which are rooted at the same time on a leaf and a vertex of degree $3$, and these trees can also be obtained by distinguishing one of the $(m+2)$ leaves in a tree which is rooted on a vertex of degree $3$. Finally, the number of all precubic rooted trees with $n$ edges equals $\displaystyle\big(1+\frac{3m}{m+2}\big)\mathrm{Cat}(m)$. We thereby obtain:
\begin{corollary}
The number $\xi_g(n)$ of precubic unicellular maps of genus $g$ with $n=2m+1$ edges is given by:
$$
\xi_g(n) = \frac{1}{2^gg!}{m+2 \choose \ 3,3,\dots , 3, m+2-3g\ }\Big(1+\frac{3m}{m+2} \Big) \mathrm{Cat}(m) = \frac{(4m+2)(2m)!}{12^gg!(m+2-3g)!m!}.
$$
\end{corollary}

\noindent Precubic unicellular maps which have no leaves necessarily have $6g-3$ edges. These objects appeared previously in the litterature~(\cite{Lehman-Walsh-genus-I, Bacher-Vdovina}, and recently in~\cite{ChMaSc} under the name of \emph{dominant schemes}). We can recover their number by setting $m=3g-2$ in the previous equation. In that case, the bijection given here reduces to the one given in our older paper~\cite{Chapuy:trisections}, in which the following corollary already appeared. However, we copy it out here for completeness:
\begin{corollary}[\cite{Lehman-Walsh-genus-I}]\label{cor:precubiques}
The number of rooted unicellular maps of genus $g$ with all vertices of degree~$3$ is:
$$
\frac{2(6g-3)!}{12^g g!(3g-2)!}.
$$
Dually, this number counts rooted triangulations of genus $g$ with only one vertex.
\end{corollary}

\subsection{Labelled unicellular maps.}

A \emph{labelled unicellular map} is a pair $({\mathfrak{m}},l)$ such that ${\mathfrak{m}}$ is a rooted unicellular map, and $l$ is a \emph{labelling} of the vertices of ${\mathfrak{m}}$, i.e. a mapping
$l: V({\mathfrak{m}})\rightarrow \mathbb{Z}$ such that if $v_1, v_2$ are two adjacent vertices in~${\mathfrak{m}}$, then $l(v_1)-l(v_2)\in\{-1,0,1\}$, and such that the root-vertex has label $0$. Our interest for these objects comes from the so-called Marcus-Schaeffer bijection:
\begin{proposition}[Marcus Schaeffer \cite{MaSc}, see also \cite{ChMaSc} for the version stated here]
Let $m_g(n)$ be the number of (all) rooted maps of genus $g$ with $n$ edges, and let $l_g(n)$ be the number of labelled unicellular maps of genus $g$ with $n$ edges. Then one has:
$$
(n+2)\, m_g(n) = 2 \, l_g(n).
$$
This identity comes from an explicit and simple bijection.
\end{proposition}
Therefore it is interesting to see what our construction becomes on these labelled unicellular maps. We let $\mathcal{L}^{(k)}_g(n)$ be the set of rooted labelled unicellular maps of genus $g$ with $n$ edges and $k$ distinguished vertices \emph{carrying the same label}. We also let $\mathcal{DL}_g(n)$ be the set of labelled unicellular maps carrying a distinguished trisection. We have:
\begin{corollary}
The application $\Lambda$ induces a bijection:
$$
\Lambda : \  \biguplus_{p=0}^{g-1} \mathcal{L}^{(2g-2p+1)}_p(n)
\longrightarrow \mathcal{DL}_g(n).
$$
\end{corollary}
\begin{proof}
The only thing to change in our construction so that it is well-defined is to restrict the gluing operation to vertices of the same label, which is exactly what we do here.
\end{proof}
Observe that it is not obvious to compute the cardinality of $\mathcal{L}^{(k)}_g(n)$: in order to compute it from $l_g(n)$, one would need non trivial information about the repartition of labels of vertices in a randomly labelled unicellular map of genus $g$, or by induction, in a randomly labelled \emph{plane tree} (see~\cite{Chapuy:trisections} for the connection with functionals of continuum random trees, in the case of $k=3$).
Therefore, in order to transform the last corollary into a counting strategy for maps on surfaces, one would need to understand much better the structure of plane labelled trees than we do at the moment.

\section{More computations.}\label{subsec:guitter}

We now sketch a computation of Emmanuel Guitter~\cite{Guitter:comm}, that enables to recover the Harer-Zagier formula from our construction.
For all $n\geq 1$, we let
$
F_n(x) = \sum_{g\geq 0}\epsilon_g(n) x^{n+1-2g}
$
be the generating function of unicellular maps with $n$ edges, where the variable $x$ marks the number of vertices. Then we have:
\begin{proposition}[\cite{Guitter:comm}]\label{prop:Guittereq}
For every real sequence $(a_n)_{ n\geq 0}$, the formal power series $F(x,y)=\sum_{n\geq 0} a_n y^{n+1} F_n(x)$ is solution of the differential equation:
\begin{eqnarray}\label{eq:Guittereq}
2y \cdot\frac{\partial }{\partial y} F(x,y) = F(x+1,y)-F(x-1,y)
\end{eqnarray}
\end{proposition}
\begin{proof}
Clearly, the series $\displaystyle \frac{1}{2}\big(F_n(x+1)-F_n(x-1)\big)$ is the generating function of unicellular maps with $n$ edges, in which an \emph{odd} number of vertices have been distinguished, and are no longer counted in the exponant of $x$. These objects are divided into two categories: either the number of distinguished vertices is $\geq 3$, or it is equal to one. By our main theorem, objects of the first kind are in bijection with unicellular maps with $n$ edges and a distinguished trisection; objects of the second kind are unicellular maps with $n$ edges with a distinguished vertex. Now, by the trisection lemma and Euler's formula, in each map the number of trisections plus the number of vertices equals $n+1$. Therefore we have:
$\displaystyle \frac{1}{2}\big(F_n(x+1)-F_n(x-1)\big) = (n+1) F_n(x)$ and the proposition follows.
\end{proof}
\begin{corollary}[\cite{Harer-Zagier}]
Let $a_n=\frac{2^n n!}{(2n)!}$, and let $F(x,y)=\sum_{n\geq 0} a_n y^{n+1} F_n(x)$. Then one has:
\begin{eqnarray}\label{eq:IZ}
F(x,y) = \frac{1}{2}\left(\frac{1+y}{1-y}\right)^x -\frac{1}{2}.
\end{eqnarray}
\end{corollary}
Note that the Harer-Zagier formula (Equation~\ref{eq:HZformula}) can be recovered by expanding the $x$-th power in~(\ref{eq:IZ}).
\begin{proof} We follow~\cite{Guitter:comm}. First,
one easily checks  that the function given here is solution of Equation~\ref{eq:Guittereq}. Moreover, a solution to Equation~\ref{eq:Guittereq} is entirely characterized by its "planar terms", i.e. by the coefficients of $x^{n+1}y^{n+1}$ for all $n\geq 0$ (think about computing the coefficients inductively via Equation~\ref{eq:identity}). Hence the only thing to check is that $\lim_{y\rightarrow 0 }F(\frac{u}{y},y)$ is equal to $\sum_{n\geq 0} a_n \mathrm{Cat}(n)u^{n+1} = \frac{1}{2}\big(\exp(2u)-1\big)$, which is immediate from~(\ref{eq:IZ}).
\end{proof}

\noindent We conclude with an extension of the previous computation to bipartite unicellular maps. For these maps, the ordinary generating series do not have a closed formula~\cite{Adrianov}, and it is better to work with "modified" generating series. More precisely, following Adrianov~\cite{Adrianov}, we introduce for each integer $v \geq 0$ the series $\phi_v(x):=\sum_{k\geq 1} k^v x^{k-1}$. We consider the \emph{modified generating series} of bipartite unicellular maps defined as follows:
$$
B_n(x,y) = \sum_{i,j\geq 0} \beta_{\frac{n+1-i-j}{2}}(i,j) \phi_i(x)\phi_j(y).
$$
Observe that by Euler's formula, a unicellular map with $i+j$ vertices and genus $\frac{n+1-i-j}{2}$ has $n$ edges, so that
$B_n(x,y)$ is the generating function of bipartite unicellular maps with $n$ edges, in which a map with $i$ white and $j$ black vertices is given a weight $\phi_i(x)\phi_j(y)$.  By investigating the effect of the deletion of vertices in the context of modified generating series, one obtains the following analogue of Proposition~\ref{prop:Guittereq}.
\begin{proposition}
The formal power series $B_n(x,y)$ is solution of the differential equation:
\begin{eqnarray}\label{eq:diffbiparties}
  \frac{\partial }{\partial x}\big( (1-x^2)B_n (x,y) \big)
+ \frac{\partial }{\partial y}\big( (1-y^2)B_n (x,y) \big)
= (n+1) B_n(x,y).
\end{eqnarray}
\end{proposition}
\begin{corollary}[\cite{Adrianov}]
The series $B_n(x,y)$ admits the following closed form:
\begin{eqnarray}\label{eq:adrianov}
B_n(x,y)=n! \frac{(1-xy)^{n-1}}{((1-x)(1-y))^{n+1}}.
\end{eqnarray}
\end{corollary}
\begin{proof}
One easily checks that the series given here is solution of Equation~\ref{eq:diffbiparties}.
Now, as in the monochromatic case, a formal power series
$$C_n(x,y)= \sum_{1\leq i+j\leq n+1} c_{\frac{n+1-i-j}{2}}(i,j) \phi_i(x)\phi_j(y)$$
which is a solution of Equation~\ref{eq:diffbiparties} is characterized by its "planar terms", i.e. by the sequence of numbers $(c_{0}(i,j))_{i\geq1, j\geq 1}$.
Therefore it is enough to prove that the numbers $c_0(i,j)$ corresponding to the function $C_n(x,y):=n! \frac{(1-xy)^{n-1}}{((1-x)(1-y))^{n+1}}$ are equal to the numbers $\beta_0(i,j)=\frac{i+j-1}{ij}\binom{i+j-2}{i-1}^2$.

Now, set $X=\frac{1}{1-x}$, $Y=\frac{1}{1-y}$ so that $C_n(x,y)=n!X^2Y^2(X+Y-1)^{n-1}$ is a polynomial $\widetilde{C}_n(X,Y)$ in $X$ and $Y$.
Using the fact that around $x=1$, one has $\phi_v(x)=\tfrac{v!}{\ \ (1-x)^{v+1}} + O\left(\tfrac{1}{(1-x)^{v}}\right)$, one obtains that
for all $i,j$ such that $i+j=n+1$,  the coefficient of $X^{i+1}Y^{j+1}$ in the polynomial $\widetilde{C}_n(X,Y)$ is $i!j!c_{0}(i,j)$.
Therefore we have:
$$c_0(i,j)= \frac{n!}{i!j!} \times \mathrm{Coeff}_{X^{i-1}Y^{j-1}} \Big( (X+Y-1)^{n-1} \Big)= \frac{n!}{i!j!} {n-1 \choose i-1,j-1,0} = \beta_0(i,j).$$
\end{proof}

\bibliographystyle{alpha}

redpajama_set_name: RedPajamaArXiv


text:
Q: R ggplot - set legend/axis sizes with ggsave so I have df like this
df <- read.table(text="
   amount      nr      date
   50          1       2017-01-01
   150         1       2017-01-03
   1500        2       2017-01-04
   1450        2       2017-01-04
   1250        2       2017-01-04
   950         1       2017-02-05
   120         3       2017-02-06
   300         3       2017-04-06
", header=TRUE)

I created an plot from this df
ggplot(test, aes(x = date, y = amount, fill = nr, group = 1)) +
  geom_bar(stat = "identity")

Everything works well until I save this plot with code below
ggsave(filename="D:/Documents/units_plot.png", width = 4, height = 2)

I can't figure out how should I properly set image dimensions (it should be 1000x500px) and how to set text size for legends, axes text sizes. Usually I do it by theme() option i ggplot but for some reason legend is so huge in exported file.

A: ggplot bases its size calculations around the physical size of the image once it's printed. To get a certain pixel size, you need to pick a good value for DPI (number of pixels per inch), and use that to calculate the size:
ggplot(test, aes(x = date, y = amount, fill = nr, group = 1)) +
    geom_bar(stat = "identity") +
    theme_grey(base_size = 12)

dpi = 96
ggsave(filename="units_plot.png", width = 1000 / dpi, height = 500 / dpi,
       dpi = dpi)

If you pick different DPI values, you'll have to test which font sizes work with that DPI, without making the labels too big or small - I don't think there's any way around that.

redpajama_set_name: RedPajamaStackExchange


I did a post on how to make camp bottle cap pins last year and have now updated it for 2013. Click HERE for the previous post and instructions.
These are very simple and easy to make and don't cost a lot of money, so they are perfect for camp! Download the free templates and print them off at home, cut them out and take them to camp for a fun memorable craft.
Thanks so much for making the 2013 version! I'd hug you if I could!
Awe, thanks Emily! I hope you have fun at camp!
I love these, would it be possible to get a 2014 version of the bottle cap download?
Is there any way I could get a version of these for 2014? I have Girl's Camp the first week of June. Let me know. Thanks!
I would also be interested in the 2014 version. Thanks!
Can you update for 2014? I really, really, like this!
I would also be interested in a 2014 version as we love your graphics and want to make these necklaces as a gift for our girls.
These are super cute and just what I'm looking for to work with some wood cutouts for charms. I, too, would be very interested in a 2014 version!
Hi – just came across this and wondered if I could get a 2015 version. We have YW tonight so I know that is probably a little too rushed. Just let me know what you think. If not, maybe we could get a 2016 version and use them next year. Thanks so much!!
Can have a file to update for 2017??? please and thank you!


Update, March 3rd: Yuzuru has produced demo video for Idobata 2, and I've included it at the end of this article.
Update, January 28th: as noted below, Idobata 2 is now available from the Kanae Projects website.
Over the last couple of years, I've been asked by Yuzuru Jewell to evaluate some of his software tools – a task I'm always delighted to perform as time allows, although in truth, I've fallen a little behind of late. This being the case, when Yuzuru first notified me that he was working on a new version of Idobata, his a text-to-speech conversion tool, I made a point to be ready to clear time to look at it when he was ready to allow me access.
Idobata was first launched in September 2013, when I originally reviewed it. As noted above, it is a text-to-speech conversion tool which currently works with the official Second Life viewer and both the 32 and 64-bit versions of Firestorm (and as a result can be used with SL and OpenSim).
When running, the original version – Idobata Pro – would capture text typed into chat in the viewer in real-time, or view the chat look, and convert it into the spoken word. The new version, Idobata 2, still does that, but in addition it can also monitor all IM and group chat sessions, etc., and convert them to the spoken word. Further, Yuzuru has also added a capability for it to be used with Skype chats.
Idobata 2 is due to be launched on Tuesday, January 27th, so the following is intended to preview the new version, which will be available on the Kanae Projects website from approximately 13:00 SLT onwards on the 27th.
Unlike its predecessor, Idobata 2 will be provided free-of-charge, but will include banner adverts for other products in the Kanae Project range. Those who prefer not to have ads in their software can remove the banners by obtaining a user name and password from Yuzuru for a nominal donation of L$300 towards his running costs.
In order to use Idobata 2, you will need to download the ZIP file (when available) from Kanae Porjects (the application is provided in both 32- and 64-bit options), and you'll also need to have the Microsoft Speech Platform 11 runtime, installed on your PC as well. If required, you can also download alternative / additional language and voice options for reading back text by downloading and installing the Microsoft Speech Platform Language Options.
Once you've download the runtime platform and installed it, unzip Idobata to a folder of your choice and double-click on the program icon to launch it. Those who have used Idobata Pro will immediately notice the first difference between it and the new product: Idobata 2 has a much smaller, tidier UI.
The main options for the application are displayed across the top of the window, together with an option to enter the users name and password to block the banner ads display at the bottom of the window. Between the options and banner area is a white space which will display the text currently being read back in voice – something that helps in situation where the spoken words many not be entirely clear (such as when abbreviations are being used, or technical terms, which may cause pronunciation problems for the application, etc.).
Before using Idobata 2 with Second Life, some basic configuration is required through the Tools button. Clicking this will open the 3-tab configuration window. The Chat Text tab is used to set the application for use with Second Life.
Use the Custom(RegEx) option to define regular expressions which will not be read back by Idobata 2.
Once you have set your preferences, click OK to return to the main window, If you wish to reset Idobata 2 to its defaults at any time, click the Default button in the configuration window – note this will reset all options – both Skype (if set) as well as those for Second Life.
In real life a tatara, or 鑪, is a traditional Japanese furnace for smelting iron and steel. In Second Life, Tatara, xreated ny Yuzuru Jewell under his Kanae Projects brand name, is a suite of tools which can be used to create sculpt maps and mesh collada files ready for upload to SL, and which includes both texture and bitmap editing capabilities as well.
Tokoroten ("extruder") – which allows shapes or parts of shapes to be stretched, twisted, etc.
Mage – which can create more organic shapes and pipes and tubes, etc.
A bitmap editor and a texture / drawing tool.
Once a desired shape has been created, it can be saved / exported from Tatara as a sculpt map (.TGA format) or as a mesh .DAE file, each of which can be uploaded to Second Life, and file formats such as .OBJ and .XML are also supported. There are a number of sample shapes provided within the suite to help people get started, and files produced via other means can also be opened in Tatara and previewed / edited. When loading files produced elsewhere, Tatara automatically disables any tools which cannot be used in editing the loaded shape (so you may find, for example, that loading a sculpt map will disable the Rokuro and Tokoroten tools).
Stair plug-in which allows the creation of four different types of stair, each with a user-definable number of steps.
Each of the four plug-ins also requires the purchase of a user licence to fully unlock them.
Tatara is available in three versions: Windows 32 and 64-bit versions and a Mac OSX version. There is no installer per se – the necessary files are provided in a ZIP file, which simply requires downloading and then unpacking to the desired folder.
Once unpacked and launched, Tatara will start-up and display a prompt for your user name and password. If no licence has been purchased, clicking cancel will allow access to the application in the trail mode.
The UI itself comprises two parts: on the left, a preview pane which displays a representation of your model, which can be drag-rotated in all three axes to examine the design; on the right a series of tabs accessing the various tool options, together with a set of menus and options – some of which may be tool-dependent.
Getting to grips with Tatara is a little complex, but Yuzuru provides a solid user guide on the Tatara page of his website and a range of tutorials on his blog. Even so, it is fair to say the tools do require a good understanding of modelling and projection, and achieving a desired goal can take time if you're not used to using creation / editing tools of this type. Nevertheless, the results can be very worthwhile, and for those wishing to add to their armoury of content-creation tools, whether looking to make sculpts or basic mesh, Tatara and Yuzuru's other tools are well worth a look.
Hanko ("seal") – a tool which allows you to add your signature to a sculpt map.
In the new release, version 4.2.0, Yuzuru has added the ability for Rokuru to additionally save generated maps as Collada file, allowing them to be uploaded into Second Life as mesh items (XML, OBJ (triangles and quadrangles) are also supported).
Rokuro – meaning lathe – offers a simple and effective way of creating symmetrical shapes for use in Second Life, and the website provides a host of samples and templates with which to get started with the tool and which can be easily modified to suit your needs.
The UI itself comprises two parts – on the left, a "finished product" preview pane which displays a representation of your model, which can be drag-rotated in all three axes to examine the design, and an editing panel on the right, in which is displayed as a plan view of your shape, the right side of which can be edited (the left will automatically reflect all adjustments made). A range of menu-supplied options (including access to default shapes) and a range of settings provide a good degree of flexibility within the tool.
Rokuro is offered free of charge (although donations towards its ongoing development and the development of Yuzuru's other products are always welcome), and is available in Window 32- and 64-bit versions, and is also available for the Mac. For those wanting to explore the capabilities of the tool in more depth, there are a range of tutorials available both in-world and on the Marketplace.
Note that some of these tools require the purchase of a serial number from Yuzuru's in-world store, but time-limited versions are available for download through the Kanae Projects website in order to "try before buying".
With thanks to Yuzuru Jewell.


The noise problem of ball mill is a difficult problem to solve in the production process when we faced a huge negative impact to bring production work. To reduce noise hazards, through testing and analysis of the ball mill, we found a solution to their difficulties.
Wind turbine manufacturers are working on many ways to reduce the noise, like noise-reduction-oriented blade redesign or flow control. Wind turbine noise management is receiving a lot of attention for the R&D community.
2016-07-23· This video will help a beginner or novice computer repair tech service loud fan noises inside laptops that usually suffer from overheating caused by a dirty and/or blocked ventilation system.
When looking for the source of a noise, it's important to isolate where it's coming from. In this article I'll go over some techniques that can help you find the source of the noise you're hearing. In this article I'll go over some techniques that can help you find the source of the noise you're hearing.
I will briefly describe the processing method of noise of vertical roller grinding mill in this article, to solve annoying noise for customers. The noise comes from three areas of vertical roller grinding mill.
q5 and q7 why is pano roof rattling . audi q5/q7: why is my pano roof rattling? there have been a lot of issues with the audi q5 and q7's panoramic roof. though, not all is lost and some solutions have been found. if that does not solve the problem, then continue to step 2. audi a6 c5: what is this front end rotating grinding noise?
CNC Electrical Noise Problems: Grounding Techniques & Interference Filters After backlash, the second biggest source of frustration seems to be electrical noise problems on home brew CNC systems. Noise can manifest itself in a variety of ways including uncommanded movements, servo encoder problems, and limit switches tripping unexpectedly.
problems in grinding mill - mine-equipments.com - troubleshooting of grindingmill, The existing problems and treatment methods of grinding mill - SBM Grinding mill is widely used in construction, chemical, electric power and metallurgy industry.
Grinding test results showed that mill capacity and grinding efficiency of cast iron ... of mineral resource, coarse dissemination size and high grade free milling ore ... is the problem of fine grinding medium over sized. .... solve the problem.
problems in grinding mill – Grinding Mill China. milling and grinding solved problems - Crusher|Granite. coal mill problem low grinding, specially to solve the problems of low capacity and and thereby worsen the problem.
Causes of Noise Pollution. 1. Industrialization: Most of the industries use big machines which are capable of producing large amount of noise. Apart from that, various equipments like compressors, generators, exhaust fans, grinding mills also participate in producing big noise.


Experts discuss women's hockey, cross-country skiing, Russian athletes, being a host with a volatile neighbour, momentum for a Calgary Olympic bid and more.
As the 2018 Winter Olympic Games kick off today in PyeongChang, South Korea, the sporting drama set to unfold over the next 17 days will happen amid a backdrop of the saga of banned Russian athletes, an unpredictable neighbour and a notable lack of NHL hockey players.
To help sort out the implications of these questions, as well as the impact these Games may have on a potential Calgary bid, and more, University of Alberta experts share their thoughts on what Canadians should watch for.
Will Canadian female Olympians continue to outpace their male counterparts for medals?
The stage is set, sports history professor Stacy Lorenz of Augustana Campus believes, for the country's female Olympians to once again shine on the podium. Since 2006, Canadian women have contributed 44 of Canada's 75 Olympic winter medals, he noted, while Canadian men have been part of 34 medals (three shared in figure skating). At the 2016 Summer Olympics, Canadian female athletes outperformed their male counterparts by a margin of 16 to six in terms of medal success.
"The one place where women's sports get almost as much attention and carry similar cultural weight to men's sports is the Olympics." Ideally, he added, the Games should be a springboard for more widespread media coverage—and higher pay for female athletes.


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Three Wi-Fi 6 network trials: Cisco, Boingo and WBA
The WBA highlights the importance of savvy Wi-Fi 6 deployment
Based on the IEEE 802.11ax standard, Wi-Fi 6, the next generation of Wi-Fi technology, was built in response to the ever-growing number of connected devices worldwide. Wi-Fi 6 was designed to improve speed, capacity, coverage and performance of connected devices, even in challenging environments like stadiums and transportation hubs, which struggle to provide adequate connection to thousands of devices simultaneously.
While the official release of the latest Wi-Fi iteration is not expected until later this year, a few companies have been getting an early start with the technology by launching commercial and industrial trials. Cisco, Boingo and the Wireless Broadband Alliance (WBA) are among those already testing out the new technology.
Cisco and USGA Pebble Beach commercial trial
Last month, a partnership between Cisco and the United States Golf Association (USGA) sought to make Pebble Beach, the site of the U.S. Open Championship, the most connected course in the world. A massive sporting event is precisely what Wi-Fi 6 was made for. Cisco powered a robust Wi-Fi network throughout the grounds with 400 wireless access points.
The Cisco Vision IPTV display management system was also used to provide fans with onsite information and features, such as shot-tracking graphics to show live player performances on the driving range. Fans had the ability to map multiple players and see a range of shot stats.
After the event, the USGA said the network saw 25 total terabytes of data used during the championship, and that it saw more than 100,000 connections to the network, but did not specify if that number represents unique connections or contains multiple connections from the same devices.
In a prepared statement, Navin Singh, chief commercial officer of the USGA, said the trial "elevate[d]the fan experience."
"We also learned a lot and recognize that mobile consumption demands are only going to continue to grow. We are excited to get to work on providing an even better experience in 2020 at the 120th U.S. Open at Winged Foot," he added.
Boingo and John Wayne Airport commercial trial
According to Boingo Chief Technology Officer Derek Peterson, the Wi-Fi 6 commercial trial with John Wayne Airport announced in April was showing triple-digit speeds using a 20 megahertz channel.
Peterson added that while they are running additional tests and performing tweaks here and there, the airport "definitely" saw an improvement in speed and connectivity.
"One of the great things about Wi-Fi 6 is the channel sharing because it's using OFDMA, which is basically the ability to share a channel more effectively like what cellular does today," said Peterson. "That's going to improve your experience in your dense locations in an airport where you have 400 people coming off an airplane trying to get connected."
The trial included Cisco equipment and Samsung devices.
WBA and Mettis Aerospace industrial trial
Also in April, the WBA announced the world's first Wi-Fi 6 Industrial Enterprise and IoT trial, as part of its ongoing Wi-Fi 6 program. The alliance is working with Mettis Aerospace, a designer and manufacturer of precision-forged, machined and sub-assembled components, to test several use cases on a Wi-Fi 6 network at its 27-acre facility.
The trial is intended to be the first of a series of global trials, and will enable the use of augmented reality, real-time monitoring of equipment and a host of other applications in an enterprise network environment designed to digitize Mettis' production line.
While not an enormous sporting event or a highly-trafficked airport, the Mettis Aerospace facility has its own connectivity challenges. First, it covers a large geography. Second, industrial radio interference can disrupt signals.
Further, some applications require high bandwidth, others low latency and mission critical applications need clear prioritization for data, as well as security. These types of specifications are another area in which Wi-Fi 6 is well-positioned to support.
At the beginning of the month, the WBA released guidelines to assist operators, enterprises and cities with Wi-Fi 6 deployment as the technology unwaveringly gains momentum. The guidelines explain that global dependence on Wi-Fi continues to grow exponentially, and the number of Wi-Fi devices in the world — 9 billion — now outnumbers the 7.6 billion people on the planet.
Running Wi-Fi 6 trials may have been the right approach because, according to the WBA, savvy deployment of Wi-Fi 6 can mitigate some of the growing pains that Wi-Fi is experiencing, and ensure that important service-level agreements (SLAs) are being met, and what better way to prepare for the real deal than to test it out first?
The post Three Wi-Fi 6 network trials: Cisco, Boingo and WBA appeared first on RCR Wireless News.

redpajama_set_name: RedPajamaCommonCrawl


Dover's Ireland Well, located within the Pudding Hill Aquifer, remains offline after the detection this week of perfluorinated compounds -- PFOA, PFOS and PFHpA -- that exceed the Environmental Protection Agency's most recent health advisory level of 70 parts per trillion (ppt). The well was voluntarily taken offline in June when levels of perfluorinated compounds increased to 48 ppt, still within established health advisory levels.
No water with values in excess of the EPA's advisory level of 70 ppt was supplied to customers.
In 2015, the city voluntarily closed the Griffin Well, also located within the Pudding Hill Aquifer. That decision was based on low levels of three perfluorinated chemicals (PFCs) detected at the site, adjacent to the former Madbury Metals on Route 155. As is the case with the Ireland Well, the decision to close the Griffin Well was made out of an abundance of caution, despite the PFC levels being well below current Environmental Protection Agency guidelines. Also out of an abundance of caution, the city is limiting use of the new Dover Pudding Hill 1 (DPH1), which draws from the Pudding Hill Aquifer.
The city's remaining seven wells are routinely tested and remain a safe source of drinking water. None of these wells have tested positive for the same PFC compounds.
City Manager J. Michael Joyal, Jr. said the city is working with the New Hampshire Department of Environmental Services and the city's aquifer consulting experts and water utility engineers to consider treatment options for expeditiously and safely returning the wells drawing from the Pudding Hill Aquifer back into production.
"In the meantime, with our recent system upgrades to the Lowell Avenue Treatment Plant and our other wells operating elsewhere throughout the city, it is important to note that we are able to maintain an adequate and safe supply of public drinking water," Joyal said. "Even so, with moderate drought conditions continuing throughout the region, we still encourage residents and businesses to be conscientious with their water usage."
Each year, the Environmental Protection Agency (EPA) requires community water systems to deliver a Consumer Confidence Report -- also known as an Annual Drinking Water Quality Report -- to their customers. These reports, which are mailed out to water customers by July 1, provide information about their drinking water quality including a summary of detected contaminants from the previous year, compliance, and educational information.
Water quality testing results and the annual water standards reports can be found on the city's website here: http://www.dover.nh.gov/government/city-operations/community-services/water-quality/index.html.


I know what you're thinking...another blog post today!? I saw this fun linkup on another blog I follow, and had to join the fun! It's all about celebrity sightings!! Unfortunately, I don't have pictures for most of these (at my house anyway!), so you'll have to take my word for it!!
I suppose my first celebrity meeting was Daryl Singletary. Does anyone even know who that is? He was a country singer back in the 90s...not sure if he was ever really that popular. Anyway, when I was younger I really disliked my name. Why? Because it was unusual, in my life I have only met a few other Randis (and of course, most of those were male...only ONE female!) What does this have to do with Daryl Singletary? Well, because of my this, I was drawn to Randy Travis...finally someone who shared my name! I was obsessed. So when I was in kindergarten my parents took me to my very first concert. Randy Travis and Sammy Kershaw. Daryl Singletary was the opening act, and after the concert was signing autographs. I'm sure my mom has pictures of this somewhere, but like I said...none here!
The next was a brief sighting of one of the Kinleys (a country group) during a tour of the Grand Ol' Opry. She rushed into the dressing room before we could get autographs or pictures or anything.
The next one really wasn't so much a celebrity sighting, so much as it was walking on a movie set. Have you heard of the movie Hard Rain? Well, if you have, you might know it was partly filmed in the little town of Huntingburg, IN (as was A League of Their Own!). Specifically, it was filmed on 4th Street, where my mom used to work! As a matter of fact, some of the crew members used to come into the deli and eat every day...she probably still has a crew hat somewhere, though at the time of filming it was known as, 'The Flood'. Anyway, the entire street was transformed...sandbags brought in and everything...and we got to see it all! It was a really fun experience! Personally, I never saw any of the cast, but still pretty cool!
Okay, here's the part where you might think I'm a bit of a dork! Ha! When I was in junior high, I was in love...with Aaron Carter. And Chris Trousdale (from Dreamstreet, he also fairly recently guest starred in Shake It Up on Disney Channel). And my best friend of the time, Katie was in love with Jesse McCartney (I don't really need to tell you who that is, but for those who didn't know, he started out in Dreamstreet as well). So, when they came to a town near us...we had to go! I have to say of all the concerts I've been to, this was probably the best- just because of our amazing seats! We were in I think the 12th row, on the floor! It was pretty awesome to our preteen selves! On top of all that, Dreamstreet was also doing a signing beforehand, so of course we had to go to that too. It was winter- February I think...and ridiculously cold! (Unless I'm totally crazy and remembering this wrong!) But we waited for at least an hour with girls who had traveled hours to be there. Girls were coming out crying- it was crazy! (We did not cry, for the record!) After that long wait, we got to meet all of the guys. I may update this tomorrow with pictures, but I'm too tired to drag out my scanner and try to find them now! It was a dream come true at the time! One of the things I remember most from that day was the girls in front of us breaking out a ruler and measuring Chris's hair...which was 2 inches high if I recall correctly. After the concert we also got to meet Lindsay Pagano (who sang a song on the old AOL commercials), but honestly, that was not NEARLY as exciting!
Now onto some more recent sightings (with pictures). After I moved to California, people were always asking if I'd seen any celebrities, and honestly for the first few years I never did. In fact when my sister came to visit a few years ago, I told her as much. Then we went to Hollywood. Now, we'd been to Hollywood many times and not once had we even thought we spotted someone. What we didn't know before heading out that day was that it was the weekend of the Academy Awards!!! So, not only did we get to see one celebrity- we saw three!!! First up was Sheri Shepard. I'm a huge fan of The View, so I was super excited! We waved and she gave us a huge smile and wave back. Everything was blocked off so we didn't get to meet her, but it was still really neat. Next up- Mr. Jay. One of my favorite shows in high school was America's Next Top Model so that was pretty exciting too. And finally, Ryan Seacrest! What a fun experience for my sister's first visit to California, right?
I was too distracted when she was actually smiling and waving at us to remember to snap a photo, so this was the best I could get!
We didn't get nearly as clear of a view of Ryan Seacrest, and my camera didn't want to focus on HIM, so this was the best I could get...still very clear who it is though!
The final sighting is one I believe I've mentioned before- Paul Walker. Yes, that's right, I said Paul Walker. During Matt's last deployment, my sister in law and I made a trip to the LAX to pick up a friend of hers. We were waiting by the baggage claim for his flight, and a guy in a hoodie walked by us...I looked at my sister in law, she looked back at me...it was PAUL WALKER! He had been on the same flight! Crazy, right? As I said, he was wearing a hoodie and seemed like he was trying to be inconspicuous so we thought it best not to draw attention by asking for pictures or autographs. (Though I did try to sneak one with my phone as he was leaving and completely missed him!) The funny thing was, he noticed us notice him and kept looking in our direction as though he were waiting for us to say or do something...either that or he thought we were cute...let's go with that theory, shall we?
And those are my celebrity sightings! What about you? Feel free to share in the comments, or if your a blogger, link up your own post!!
Wow that's quite the list! I've only met a few. Charles Barkley let me and my friend take a picture with him at a club in Vegas. Tony Danza walked into an elevator as my friends and I were going out. Oh and I met the women's Olympic soccer team back in middle school, that was pretty cool.