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\end{tikzpicture}} } \newcommand{\pathC}[5]{ \raisebox{-4mm}{\begin{tikzpicture}[scale=.5,baseline] \draw(0,0)grid(2,2); \draw[very thick]#1; \draw[xshift=-5mm,yshift=5mm] (#2,0)node{\footnotesize{$#3$}} (#4,1)node{\footnotesize{$#5$}} ; \end{tikzpicture}} } \newcommand{\myi}[1]{\item\label{Item:#1}} \newcommand{\sk}{ \smallskip} % \newcommand{\myi}[1]{ % \smallskip\textbf{#1.}} \newcommand{\noproof}{\qed} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % if needed include a line break (\\) at an appropriate place in the title \title{\bf On parking functions and the zeta map\\ in types $B,C$ and $D$} % input author, affilliation, address and support information as follows; % the address should include the country, and does not have to include % the street address \author{Robin Sulzgruber\thanks{Research supported by the Austrian Science Fund (FWF), grant S50-N15 in the framework of the Special Research Program ``Algorithmic and Enumerative Combinatorics'' (SFB F50).}\\ \small Fakult\"at f\"ur Mathematik\\[-0.8ex] \small Universit\"at Wien\\[-0.8ex] \small Wien, Austria\\ \small\tt robin.sulzgruber@univie.ac.at\\ \and Marko Thiel\textsuperscript{\fnsymbol{footnote}}\\ \small Institut f\"ur Mathematik\\[-0.8ex] \small Universit\"at Z\"urich\\[-0.8ex] \small Z\"urich, Switzerland\\ \small\tt marko.thiel@math.uzh.ch } % \date{\dateline{submission date}{acceptance date}\\ % \small Mathematics Subject Classifications: comma separated list of % MSC codes available from http://www.ams.org/mathscinet/freeTools.html} \date{\dateline{Dec 31, 2016}{Dec 28, 2017}{Jan 12, 2018}\\ \small Mathematics Subject Classifications: 05E10, 05E18, 17B22, 52C35} \begin{document} \maketitle \begin{abstract} Let $\Phi$ be an irreducible crystallographic root system with Weyl group $W$, coroot lattice $\check{Q}$ and Coxeter number $h$. Recently the second named author defined a uniform $W$-isomorphism $\zeta$ between the finite torus $\check{Q}/(mh+1)\check{Q}$ and the set of non-nesting parking functions $\operatorname{Park}^{(m)}(\Phi)$. If $\Phi$ is of type $A_{n-1}$ and $m=1$ this map is equivalent to a map defined on labelled Dyck paths that arises in the study of the Hilbert series of the space of diagonal harmonics. In this paper we investigate the case $m=1$ for the other infinite families of root systems ($B_n$, $C_n$ and $D_n$). In each type we define models for the finite torus and for the set of non-nesting parking functions in terms of labelled lattice paths. The map $\zeta$ can then be viewed as a map between these combinatorial objects. Our work entails new bijections between (square) lattice paths and ballot paths. % keywords are optional \bigskip\noindent \textbf{Keywords:} diagonal harmonics; lattice path; parking function; Weyl group \end{abstract} \section{Introduction} \subsection{The space of diagonal harmonics} One of the most well-studied objects in algebraic combinatorics is the space of diagonal harmonics of the symmetric group $\S_n$. Its Hilbert series has two combinatorial interpretations \cite[Conj.~5.2]{Haglund2008}, \cite{CM:proof_shuffle_conjecture}: \begin{align*} \mathcal{DH}(n;q,t) =\sum_{(P,v)\in \op{Vert}(A_{n-1})}q^{\dinv'(P,v)}t^{\area(P,v)} =\sum_{(D,w)\in \op{Diag}(A_{n-1})}q^{\area'(D,w)}t^{\bounce(D,w)}, \end{align*} where $\op{Vert}(A_{n-1})$ is the set of \emph{parking functions} of length $n$, viewed as vertically labelled Dyck paths, and $ \op{Diag}(A_{n-1})$ is the set of \emph{diagonally labelled} Dyck paths with $2n$ steps. There is a bijection $\zeta_A$ due to Haglund and Loehr \cite{HagLoe2005} that maps $ \op{Vert}(A_{n-1})$ to $ \op{Diag}(A_{n-1})$ and sends the bistatistic $(\dinv',\area)$ to $(\area',\bounce)$, demonstrating the second equality. \subsection{The zeta map beyond type \emph{A}} The combinatorial objects $ \op{Vert}(A_{n-1})$ and $ \op{Diag}(A_{n-1})$ may be viewed as corresponding to the type $A_{n-1}$ cases of more general algebraic objects associated to any irreducible crystallographic root system $\Phi$. These are, respectively, the \emph{finite torus} \mbox{$\Q/(h+1)\Q$} and the set of \emph{non-nesting parking functions} $\nnpark$ of $\Phi$. Here $\Q$ is the coroot lattice and $h$ is the Coxeter number of $\Phi$. \sk Recently, the second named author has defined a uniform bijection $\zeta$ from \mbox{$\Q/(h+1)\Q$} to $\nnpark$ \cite[Thm.~15.4]{Thiel2015}. Its most salient property is that it commutes with the action of the \emph{Weyl group} $W$ of $\Phi$ on both sets. In type $A_{n-1}$, the uniform map $\zeta$ corresponds to the combinatorial map $\zeta_A$ \cite[Thm.~16.3]{Thiel2015}. This is illustrated by the following commutative diagram of bijections: \begin{center} \begin{tikzpicture} \draw (0,0) node(q){$\Q/(n+1)\Q$}; \draw (4,0) node(park){$\mathrm{Park}(\Phi)$}; \draw (0,1.4) node(v){$\op{Vert}(A_{n-1})$}; \draw (4,1.4) node(d){$\op{Diag}(A_{n-1})$}; \draw[->] (q)--node[below]{$\zeta$}(park); \draw[->] (v)--node[above]{$\zeta_A$}(d); \draw[->] (v)--node[left]{$\psi$}(q); \draw[->] (d)--node[right]{$\varphi$}(park); \end{tikzpicture} \end{center} The sets $\op{Vert}(A_{n-1})$ and $\op{Diag}(A_{n-1})$ as well as the map $\zeta_A$ are defined in \refs{paths}. Moreover, the maps $\psi$ and $\varphi$ as well as the connection to classical parking functions are illustrated for type $A_2$ in \reftb{A2} in the appendix. \sk In this paper, for each of the other classical types $B_n,C_n$ and $D_n$, we define combinatorial models for $\Q/(h+1)\Q$ and $\nnpark$, in terms of vertically labelled paths and diagonally labelled paths respectively. Furthermore, we define the combinatorial bijections $\zeta_B$, $\zeta_C$, and $\zeta_D$ which correspond to the uniform bijection $\zeta$ in those types. Besides their meaning in the theory of reflection groups these maps can also be appreciated from a purely combinatorial point of view. As such $\zeta_B$ and $\zeta_C$ are new bijections between lattice paths in an $n\times n$-square and ballot paths with $2n$ steps, both of which are well-known to be counted by the central binomial coefficients $\binom{2n}{n}$. Furthermore $\zeta_D$ is a new bijection between lattice paths in an $(n-1)\times n$ rectangle and ballot paths with $2n-1$ steps. Furthermore we introduce a uniform statistic $\op{dinv}'$ defined on the elements of the finite torus. Using the zeta map we prove that this dinv statistic is equivalent to a known area-like statistic defined on the regions of the Shi arrangement, which is conjectured to be related to the Hilbert series of diagonal harmonics in other types. While the space of diagonal harmonics can be generalised beyond type $A_{n-1}$, combinatorial interpretations for the arising Hilbert series are still missing in other types. It is our hope that the ideas presented in this paper will facilitate further progress in this regard. \subsection{Outline} This paper is structured as follows. In \refs{defs} we present the needed facts about Weyl groups and root systems. In particular we recall the relations between objects such as the Shi arrangement, the finite torus and the non-nesting parking functions, and discuss in detail the definition of the uniform zeta map. In \refs{BCD} we turn our attention to the root systems of types $B_n, C_n$ and $D_n$. In particular we recall the combinatorial models for their affine Weyl groups in terms of affine permutations and show how to obtain the decomposition of an affine permutation into a product of an element of the Weyl group and a translation by an element in the coroot lattice combinatorially. We then develop the combinatorics of the zeta map of type $C_n$ in \refs{typeC}, type $D_n$ in \refs{typeD} and type $B_n$ in \refs{typeB}. Each of these sections is structured similarly. First we define combinatorial models for the finite torus in terms of (labelled) lattice paths and the non-nesting parking functions in terms of (labelled) ballot paths. Secondly we define an area vector, a diagonal reading word and two versions of a combinatorial zeta map, one between labelled objects and one between unlabelled objects. We then prove that the combinatorial zeta map coincides with the uniform zeta map of the given type, and is thus a bijection. \refs{typeC} includes an alternative description of the zeta map of type $C_n$ akin to the sweep map of Armstrong, Loehr and Warrington~\cite{ALW2014:sweep_maps}. In \refs{areadinv} we define the uniform dinv statistic on the finite torus and prove that it is equivalent to an area statistic on the set of non-nesting parking functions via the zeta map. Finally, in \refs{end} we offer some possible directions for further research. \sk An extended abstract~\cite{SulThi2015} of this paper containing mostly the results of \refs{typeC} has appeared in the conference proceedings of FPSAC 2015 in Daejeon. The results presented in this paper are also discussed in the thesis of the first named author~\cite{Sulzgruber:Thesis}. \section{Definitions and Preliminaries}\label{Section:defs} Given $n\in\N$ we set $[n]=\{1,2,\dots,n\}$ and $[\pm n]=[n]\cup\{-i:i\in[n]\}$. \subsection{Root systems and Weyl groups} Let $\Phi$ be an irreducible crystallographic root system of rank $r$, with simple system $\Delta=\{\alpha_1,\alpha_2,\ldots,\alpha_r\}$, positive system $\Phi^+$ and ambient space $V$. For background on root systems and reflection groups see \cite{Humphreys}. For $\alpha\in\Phi$, let $s_{\alpha}$ be the reflection in the hyperplane \begin{align*} H_{\alpha}=\{x\in V:\langle x,\alpha\rangle=0\}. \end{align*} Then the \emph{Weyl group} $W$ of $\Phi$ is the group of linear automorphisms of $V$ generated by all the $s_{\alpha}$ with $\alpha\in\Phi$. Define the \emph{Coxeter arrangement} of $\Phi$ as the central hyperplane arrangement in $V$ consisting of all the hyperplanes $H_{\alpha}$ for $\alpha\in\Phi$. The connected components of the complement of the union of these hyperplanes are called \emph{chambers}. The Weyl group $W$ acts simply transitively on the chambers, so we define the \emph{dominant chamber} as \begin{align*} C=\{x\in V:\langle x,\alpha\rangle>0\text{ for all }\alpha\in\Delta\} \end{align*} and write every chamber as $wC$ for a unique $w\in W$. The Weyl group $W$ is a \emph{Coxeter group} generated by $S=\{s_{\alpha}:\alpha\in\Delta\}$. If $I\subseteq \Delta$, we call $W_I=\langle\{s_{\alpha}:\alpha\in I\}\rangle$ a \emph{standard parabolic subgroup} of $W$. A \emph{parabolic subgroup} is any subgroup conjugate to a standard parabolic subgroup. \subsection{The affine Weyl group} Define a partial order on $\Phi^+$ by letting $\alpha\leq\beta$ if and only if $\beta-\alpha$ can be written as a sum of positive roots. The set $\Phi^+$ with this partial order is called the \emph{root poset} of $\Phi$. It has a unique maximal element, called the \emph{highest root} $\tilde{\alpha}$. Write $\tilde{\alpha}=\sum_{i=1}^rc_i\alpha_i$ and define the \emph{Coxeter number} of $\Phi$ as $h=1+\sum_{i=1}^rc_i$. For $\alpha\in\Phi$ and $k\in\Z$, let $s_{\alpha,k}$ be the reflection in the affine hyperplane \begin{align*} H_{\alpha,k}=\{x\in V:\langle x,\alpha\rangle=k\}. \end{align*} Then the \emph{affine Weyl} group $\wa$ of $\Phi$ is the group of affine transformations of $V$ generated by all the $s_{\alpha,k}$ for $\alpha\in\Phi$ and $k\in\Z$. Define the \emph{affine Coxeter arrangement} as the affine hyperplane arrangement in $V$ consisting of all the $H_{\alpha,k}$ for $\alpha\in\Phi$ and $k\in\Z$. The connected components of the complement of the union of these hyperplanes are called \emph{alcoves}. The affine Weyl group $\wa$ acts simply transitively on the alcoves, so we define the \emph{fundamental alcove} as \begin{align*} \ac=\big\{x\in V:\langle x,\alpha\rangle>0\text{ for all }\alpha\in\Delta\text{ and }\langle x,\tilde{\alpha}\rangle<1\big\} \end{align*} and write every alcove as $\widetilde w\ac$ for a unique $\widetilde w\in\wa$. For each root $\alpha\in\Phi$ we define its coroot as $\alpha^{\vee}=2\alpha/\skal{\alpha,\alpha}$. The coroot lattice $\Q$ of $\Phi$ is the integer span of the coroots. The simple coroots are a natural lattice basis for it: $\Q=\bigoplus_{\alpha\in\Delta}\Z\alpha^{\vee}$. The affine Weyl group $\wa$ acts faithfully on the coroot lattice $\Q$. To each coroot lattice point $\mu\in\Q$ corresponds a translation $t_{\mu}:V\to V$ given by $t_{\mu}(x)=x+\mu$. If we identify $\Q$ with its translation group we may write $\wa=W\ltimes\Q$ as a semi-direct product. The affine Weyl group is also a Coxeter group with generating set $\widetilde{S}=S\cup\{s_{\tilde{\alpha},1}\}$. The Weyl group $W=\langle S\rangle$ is a standard parabolic subgroup of the Coxeter group $\wa$. Each (left) coset in the quotient $\wa/W$ contains a unique element of minimal length. The minimal length representatives are called \emph{(left) Gra{\ss}mannian}. An element $\widetilde w\in\wa$ is Gra{\ss}mannian if and only if the alcove $\widetilde w^{-1}\ac$ lies in the dominant chamber. \subsection{The Shi arrangement}\label{Section:shi} Define the \emph{Shi arrangement} as the hyperplane arrangement consisting of the hyperplanes $H_{\alpha,k}$ for $\alpha\in\Phi^+$ and $k=0,1$. The complement of the union of these hyperplanes decomposes into connected components called the \emph{regions} of the arrangement. The hyperplanes that support facets of a region $R$ are called the \emph{walls} of $R$. Those walls of $R$ that do not contain the origin and separate $R$ from the origin are called the \emph{floors} of $R$. Define the walls and floors of an alcove similarly. Notice that every wall of a region is a hyperplane of the Shi arrangement, but the walls of an alcove need not be. We call a region or alcove \emph{dominant} if it is contained in the dominant chamber. \sk \begin{mythrm}{min} {\textnormal{\cite[Prop.~7.1]{Shi1987:sign_types}}} Every region $R$ of the Shi arrangement has a unique \emph{minimal alcove} $\waf_R\ac\subseteq R$, which is the alcove in $R$ closest to the origin. \end{mythrm} % \begin{figure}[h] % \begin{center} % \resizebox*{6cm}{!}{\includegraphics{a2shi1min}}\\ % \end{center} % \caption{The 16 regions of the Shi arrangement of type $A_2$ with their minimal alcoves. The minimal alcoves of the 5 dominant regions are coloured blue.} % \end{figure} \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=1.6,myline/.style={line width=\linewidth pt,black!50},myfatline/.style={line width=2*\linewidth pt},my_alcove_style/.style={fill=yellow},dom_alcove_style/.style={fill=orange},mydot/.style={circle,fill=blue,inner sep=0pt,minimum size=3mm}] \pgfmathsetmacro{\heightup}{3} \pgfmathsetmacro{\widthright}{3} \pgfmathsetmacro{\heightdown}{2} \pgfmathsetmacro{\widthleft}{2} \pgfmathsetmacro{\linemargin}{0.2} \pgfmathsetmacro{\linewidth}{1} % alcoves \draw[shift={(60:1)},shift={(1,0)},dom_alcove_style](0,0)--node[anchor=south]{\scriptsize{$[4,2,0]$}}(1,0)--(60:1)--cycle; \draw[shift={(60:-2)},shift={(1,0)},my_alcove_style](0,0)--node[anchor=south]{\scriptsize{$[1,-1,6]$}}(1,0)--(60:1)--cycle; \draw[shift={(60:1)},shift={(0,0)},dom_alcove_style](0,0)--node[anchor=south]{\scriptsize{$[2,0,4]$}}(1,0)--(60:1)--cycle; \draw[shift={(60:2)},shift={(0,0)},my_alcove_style](0,0)--node[anchor=north]{\scriptsize{$[1,0,5]$}}(-1,0)--(60:-1)--cycle; \draw[shift={(60:0)},shift={(0,0)},dom_alcove_style](0,0)--node[anchor=south]{\scriptsize{$[1,2,3]$}}(1,0)--(60:1)--cycle; \draw[shift={(60:1)},shift={(0,0)},my_alcove_style](0,0)--node[anchor=north]{\scriptsize{$[2,1,3]$}}(-1,0)--(60:-1)--cycle; \draw[shift={(60:0)},shift={(-1,0)},my_alcove_style](0,0)--node[anchor=south]{\scriptsize{$[2,3,1]$}}(1,0)--(60:1)--cycle; \draw[shift={(60:-1)},shift={(0,0)},my_alcove_style](0,0)--node[anchor=south]{\scriptsize{$[3,1,2]$}}(1,0)--(60:1)--cycle; \draw[shift={(60:0)},shift={(1,0)},my_alcove_style](0,0)--node[anchor=north]{\scriptsize{$[1,3,2]$}}(-1,0)--(60:-1)--cycle; \draw[shift={(60:0)},shift={(0,0)},my_alcove_style](0,0)--node[anchor=north]{\scriptsize{$[3,2,1]$}}(-1,0)--(60:-1)--cycle; \draw[shift={(60:-1)},shift={(1,0)},my_alcove_style](0,0)--node[anchor=south]{\scriptsize{$[-1,3,4]$}}(1,0)--(60:1)--cycle; \draw[shift={(60:1)},shift={(-1,0)},my_alcove_style](0,0)--node[anchor=south]{\scriptsize{$[0,1,5]$}}(1,0)--(60:1)--cycle; \draw[shift={(60:1)},shift={(1,0)},dom_alcove_style](0,0)--node[anchor=north]{\scriptsize{$[0,2,4]$}}(-1,0)--(60:-1)--cycle; \draw[shift={(60:1)},shift={(-2,0)},my_alcove_style](0,0)--node[anchor=south]{\scriptsize{$[-2,5,3]$}}(1,0)--(60:1)--cycle; \draw[shift={(60:0)},shift={(2,0)},my_alcove_style](0,0)--node[anchor=north]{\scriptsize{$[-1,4,3]$}}(-1,0)--(60:-1)--cycle; \draw[shift={(60:0)},shift={(1,0)},dom_alcove_style](0,0)--node[anchor=south]{\scriptsize{$[0,4,2]$}}(1,0)--(60:1)--cycle; % 60° lines \foreach \x in {-3,-2,-1,2,3,4}{ \pgfmathsetmacro{\yup}{min(\heightup,(\widthright-\x)*2)+\linemargin} \pgfmathsetmacro{\ydown}{min(\heightdown,(\widthleft+\x)*2)+\linemargin} \draw[xshift=\x cm,myline](60:-\ydown)--(60:\yup); } % 120° lines \foreach \x in {-3,-2,-1,2,3,4}{ \pgfmathsetmacro{\yup}{min(\heightup,(\widthleft+\x)*2)+\linemargin} \pgfmathsetmacro{\ydown}{min(\heightdown,(\widthright-\x)*2)+\linemargin} \draw[xshift=\x cm,myline](120:\yup)--(120:-\ydown); } % horizontal lines \foreach \y in {-2,-1,2,3}{ \pgfmathsetmacro{\x}{(-\y-\linemargin)/2-\widthleft} \draw[xshift=\x cm,myline](60:\y)--+(\widthleft+\widthright+\linemargin,0); } %% \foreach \x in {0,1}{ \pgfmathsetmacro{\yup}{min(\heightup,(\widthright-\x)*2)+\linemargin} \pgfmathsetmacro{\ydown}{min(\heightdown,(\widthleft+\x)*2)+\linemargin} \draw[xshift=\x cm,myfatline](60:-\ydown)--(60:\yup)node[anchor=south]{$H_{\alpha_1,\x}$}; } \foreach \x in {0,1}{ \pgfmathsetmacro{\yup}{min(\heightup,(\widthleft+\x)*2)+\linemargin} \pgfmathsetmacro{\ydown}{min(\heightdown,(\widthright-\x)*2)+\linemargin} \draw[xshift=\x cm,myfatline](120:\yup)--(120:-\ydown)node[anchor=north]{$H_{\tilde{\alpha,\x}}$}; } \foreach \y in {0,1}{ \pgfmathsetmacro{\x}{(-\y-\linemargin)/2-\widthleft} \draw[xshift=\x cm,myfatline](60:\y)--+(\widthleft+\widthright+\linemargin,0)node[anchor=west]{$H_{\alpha_2,\y}$}; } \draw(0,0)node[mydot,black]{}; \end{tikzpicture} \caption{The 16 regions of the Shi arrangement of type $A_2$ with their minimal alcoves. The minimal alcoves of the 5 dominant regions are coloured orange. % The Shi arrangement of type $A_2$. } \label{Figure:ShiA2} \end{center} \end{figure} Furthermore the floors of the minimal alcove $\waf_R\ac$ of $R$ are exactly the floors of $R$ \cite[Thm.~12.4]{Thiel2015}. We define $\wshi=\{\waf_R: R\text{ is a Shi region}\}$. The corresponding alcoves $\waf_R\ac$ we call \emph{Shi alcoves}. That is, we call an alcove a Shi alcove if it is the minimal alcove of the Shi region containing it. Define $\wdom=\{\waf_R: R\text{ is a dominant Shi region}\}$. An \emph{order filter} in the root poset is a set $J\subseteq\Phi^+$ such that whenever $\alpha\in J$ and $\beta\geq\alpha$ then $\beta\in J$. For a dominant Shi region $R$, we consider \[J=\{\alpha\in\Phi^+:\langle x,\alpha\rangle>1\text{ for all }x\in R\}.\] Then $J$ is an order filter in the root poset. In fact, the map $R\mapsto J$ is a bijection from the set of dominant Shi regions to the set of order filters in the root poset \cite[Thm.~1.4]{Shi1997:positive_sign_types}. The set \[A=\{\alpha\in\Phi^+:H_{\alpha,1}\text{ is a floor of }R\}\] of minimal elements of $J$ is an antichain in the root poset. Since the correspondence between order filters and the antichains of their minimal elements is bijective, the map $R\mapsto A$ is a bijection from the set of dominant Shi regions to the set of antichains in the root poset. We conclude this section by introducing an important affine transformation considered by Athanasiadis \cite[Section 4]{Athan2005}. Denote by $\wf\in\wa$ the element of the affine Weyl group such that $\wf\ac$ is the minimal alcove of the dominant Shi region furthest from the origin, corresponding to the order filter $J=\Phi^+$ and the antichain $A=\Delta$: \begin{align*} \wf\ac=\big\{x\in V:\skal{x,\alpha}>1\text{ for all }\alpha\in\Delta\text{ and }\skal{x,\tilde\alpha}-k\big\}. \end{align*} The natural action of the affine Weyl group on half spaces carries over to affine roots as follows. Suppose $\widetilde w\in\wa$ has the unique decomposition $\widetilde w=t_{\mu}w$ with $\mu\in\Q$ and $w\in W$. Then \begin{align*} \widetilde w\cdot(\alpha+k\delta) =w\cdot\alpha + (k-\skal{\mu,w\cdot\alpha})\delta, \end{align*} where $w\cdot\alpha$ denotes the usual action of $W$ on $\Phi$. Define the positive affine roots as \begin{align*} \widetilde\Phi^+=\Phi^+\cup\big\{\alpha+k\delta:\alpha\in\Phi,k>0\big\}, \end{align*} and the simple affine roots as \begin{align*} \widetilde\Delta=\Delta\cup\{-\tilde\alpha+\delta\}. \end{align*} Thereby positive affine roots correspond to those half spaces that contain the fundamental alcove. Simple affine roots correspond to half spaces that contain the fundamental alcove and share one of its walls. As a consequence we obtain the following well-known lemma that characterises the floors and separating hyperplanes of an alcove. \begin{mylem}{affroots} Let $\widetilde w\in\wa$ and $\alpha+k\delta\in\widetilde\Phi^+$ with $k>0$. \begin{enumerate}[(i)] \myi{sep} The hyperplane $H_{\alpha,-k}$ separates $\widetilde w\ac$ from $\ac$ if and only if $\widetilde w^{-1}\cdot(\alpha+k\delta)\in-\widetilde\Phi^+$. \myi{floor} The hyperplane $H_{\alpha,-k}$ is a floor of $\widetilde w\ac$ if and only if $\widetilde w^{-1}\cdot(\alpha+k\delta)\in-\widetilde\Delta$.\noproof{} \end{enumerate} \end{mylem} \sk Translating the walls of $\wf\ac$ into affine roots we obtain the following lemma. \begin{mylem}{wf} The element $\wf\in\wa$ is uniquely determined by the property \begin{align*} \wf(\widetilde\Delta)=(\Delta-\delta)\cup\{-\tilde\alpha+h\delta\}. \end{align*} \end{mylem}\noproof{} % \begin{myprop}{wf} \textnormal{\cite{Thiel2015}} The map $\waf_R\ac\mapsto \wf\waf_R^{-1}\ac$ is a bijection from the set of Shi alcoves to the set of alcoves contained in the dilated fundamental alcove $(h+1)\ac$. % \end{myprop} \subsection{The finite torus}\label{Section:torus} The finite torus is defined as the quotient of the coroot lattice \begin{align*} T=\Q/(h+1)\Q. \end{align*} See the appendix for an explicit system of representatives in a few low dimensional examples. The action of the Weyl group on the coroot lattice induces an action on the finite torus. The following result due to Haiman provides a very useful description of the orbits under this action. \begin{mythrm}{stab} {\textnormal{\cite[Lemma~7.4.1]{Haiman1994}}} The set $\Q\cap(h+1)\overline{\ac}$ forms a system of representatives for the $W$-orbits of $\Q/(h+1)\Q$. The stabiliser subgroup $H=\op{Stab}(\lambda)\leq W$ of an element $\lambda+(h+1)\Q$ of the finite torus, where $\lambda\in\Q\cap(h+1)\overline{\ac}$, is generated by $\{s_{\alpha}:\alpha\in\walls\}$, where \begin{align*} \walls=\big\{\alpha\in\Delta\cup\{-\tilde\alpha\}:\lambda\text{ is contained in the wall of $(h+1)\ac$ perpendicular to }\alpha\big\}. \end{align*} \end{mythrm} % \sk By a result of Sommers \cite[Thm.~6.4]{Sommers2005}, there exists a $w\in W$ such that $w\cdot\walls\subseteq\Delta$. So in particular the stabiliser subgroup $H$ of $\lambda+(h+1)\Q$ in $W$ is a parabolic subgroup of $W$, and any coset in $W/H$ contains a unique element $u$ such that $u\cdot\walls\subseteq\Phi^+$. We call this the canonical representative of the coset with respect to $\walls$. We deduce the following lemma. \begin{mylem}{canon} We may represent any element of the finite torus uniquely as $u\cdot\lambda$ where $\lambda\in\Q\cap(h+1)\overline{\ac}$ and $u\in W$ is such that $u\cdot\walls\subseteq\Phi^+$. % is the distinguished representative with respect to $\walls$ of some coset of $H=\op{Stab}(\lambda)$ in $W$. \end{mylem} % Thus \sk Athanasiadis has defined a bijection from dominant Shi alcoves to $W$-orbits of the finite torus. \begin{mythrm}{rho} {\textnormal{\cite[Thm.~4.2]{Athan2005}}} The map $\mathcal \rho:\wdom\to\Q\cap(h+1)\overline{\ac}$ given by \begin{align*} \waf_R\mapsto \wf\waf_R^{-1}\cdot0 \end{align*} is a bijection. \end{mythrm} \sk The second named author has recently extended this to a bijection called Anderson map between the minimal Shi alcoves and the finite torus. \begin{mythrm}{anderson} {\textnormal{\cite[Thms.~8.4 and 12.6]{Thiel2015}}} The map $\mathcal A:\wshi\to\Q/(h+1)\Q$ given by \begin{align*} \waf_R\mapsto -\waf_R\wf^{-1}\cdot0+(h+1)\Q \end{align*} is a bijection. \end{mythrm} \sk The way in which $\A$ extends $\rho$ is that the following diagram commutes: \begin{center} \begin{tikzpicture} \draw (0,0) node(q){$\wdom$}; \draw (4,0) node(park){$\Q\cap(h+1)\overline{\ac}$}; \draw (0,1.4) node(v){$\wshi$}; \draw (4,1.4) node(d){$\Q/(h+1)\Q$}; \draw[->] (q)--node[below]{$\rho$}(park); \draw[->] (v)--node[above]{$\A$}(d); \draw[<-] (v)--node[left]{$\iota$}(q); \draw[->] (d)--node[right]{$\pi$}(park); \end{tikzpicture} \end{center} Here $\iota$ is the natural inclusion and $\pi$ is the map that sends an element of $\Q/(h+1)\Q$ to the element of $\Q\cap(h+1)\overline{\ac}$ that represents its $W$-orbit. \subsection{Non-nesting parking functions}\label{nnsec} The non-nesting parking functions were introduced by Armstrong, Reiner and Rhoades~\cite{ARR2012} as a model for the regions of the Shi arrangement that carries an action of $W$. Define an equivalence relation on the set of pairs of an element of the Weyl group and an antichain in the root poset $\Phi^+$ by $(w,A)\sim(w',A')$ if and only if $A=A'$ and $wH=w'H$ where $H\leq W$ denotes the subgroup generated by $\{s_{\alpha}:\alpha\in A\}$. In other words, $H$ is the stabiliser subgroup of $\bigcap_{\alpha\in A}H_{\alpha}$ in $W$. We define the set of \emph{non-nesting parking functions} of $\Phi$ as the set of equivalence classes \begin{align*} % \op{Park}(\Phi)=\big\{[w,X]:w\in W,X\text{ is a non-nesting flat}\big\}. \op{Park}(\Phi)=\big\{[w,A]:w\in W,A\subseteq\Phi^+\text{ is an antichain}\big\}. \end{align*} Detailed examples in low dimension are found in the appendix. Using \cite[Thm.~6.4]{Sommers2005} the same reasoning that lead to \refl{canon} in the previous section allows us to chose a canonical representative in each class. \begin{mylem}{nnpark} Every non-nesting parking function $[w,A]$ contains a unique representative $(u,A)$ such that $u\cdot A\subseteq\Phi^+$. \end{mylem} \sk We will also call $u$ the canonical coset representative of $wH$. Note that this depends implicitly not just on $H$, but also on the antichain $A$. \sk The Weyl group $W$ acts on $\nnpark$ by $$u\cdot[w,A]=[uw,A]$$ for $u\in W$. So the orbits of this action are indexed by the antichains in the root poset. % The $W$-orbits on $\nnpark$ are all of the form $W[e,A]$ for some antichain $A$. % Every antichain gives rise to a different orbit, so the antichains index the $W$-orbits of $\nnpark$. Non-nesting parking functions are naturally in bijection with the regions of the Shi arrangement. \begin{mythrm}{shitheta} {\textnormal{\cite[Thm.~15.2]{Thiel2015}}} The map $\Theta^{-1}:\wshi\to\op{Park}(\Phi)$ given by $\waf_R\mapsto [w,A]$, where $\waf_R\ac\subseteq wC$ and $A\subseteq\Phi^+$ consists of the roots $\alpha$ such that $H_{\alpha,1}$ is a floor of $w^{-1}\waf_R\ac$, is a bijection. \end{mythrm} \sk Note that a similar bijection using ceilings instead of floors is given in \cite[Prop.~10.3]{ARR2012}. We remark that the map $\Theta^{-1}$ extends the natural bijection between the dominant regions of the Shi arrangement and the antichains in the root poset. \subsection{The uniform zeta map}\label{Section:uniform} The \emph{zeta map} is defined as the composition of the two bijections introduced in the previous sections \begin{align}\label{eq:HLzetauni} \zeta:\Q/(h+1)\Q&\to\op{Park}(\Phi),\\\notag u\cdot\lambda+(h+1)\Q&\mapsto[w,A]=\left(\Theta^{-1}\circ\mathcal A^{-1}\right)\big(u\cdot\lambda+(h+1)\Q\big). \end{align} By a slight abuse of notation we also write $\zeta(u,\lambda)=[w,A]$, where $u\cdot\lambda$ is always understood to be the unique representative as in \refl{canon}. % When $\Phi$ is of type $A_{n-1}$ the finite torus and the non-nesting parking functions can be interpreted as labelled Dyck paths, which where studied by Haglund and Loehr \cite{HagLoe2005}. For this reason we also call the map $\zeta$ the uniform Haglund--Loehr zeta map. Since the zeta map commutes with the action of $W$ on $\Q/(h+1)\Q$ and $\nnpark$, it restricts to a bijection between the $W$-orbits on both sets: \begin{align}\label{eq:zetauni} \overline{\zeta}:\Q\cap(h+1)\overline{\ac}&\to\op{Antichains}(\Phi^+),\\\notag \lambda&\mapsto A. \end{align} This can also be seen as an inverse of the bijection $\rho$. To be precise, we have a commutative diagram: \begin{center} \begin{tikzpicture} \draw (0,0) node(q){$\op{Antichains}(\Phi^+)$}; \draw (4,0.7) node(park){$\Q\cap(h+1)\overline{\ac}$}; \draw (0,1.4) node(v){$\wdom$}; % \draw (4,1.4) node(d){$\Q/(h+1)\Q$}; \draw[<-] (q)--node[below]{$\overline{\zeta}$}(park); \draw[->] (v)--node[above]{$\rho$}(park); \draw[<->] (v)--node[left]{}(q); % \draw[->] (d)--node[right]{}(park); \end{tikzpicture} \end{center} Here the vertical map is the natural bijection mentioned in \refs{shi}. \sk % and in particular the special case where $\Phi$ is of type $A_{n-1}$, has appeared many times in the literature \cite{AKOP2002,Haglund2003,Athan2005}. In particular, note that the zeta map relates the the orbits of the finite torus under the action of the Weyl group with the dominant regions of the Shi arrangement. In the remainder of this section we explore $\zeta$ in more detail to obtain a more explicit description of $[w,A]$ in terms of the pair $(u,\lambda)$. To this end we recall some arguments that can be found in similar form for example in \cite[Sec.~4]{Athan2005} or \cite[Sec.~8.5]{Thiel2015}. Suppose that $R$ is a Shi region with minimal alcove $\waf_R\ac$. Say $\A(\waf_R)=u\cdot\lambda+(h+1)\Q$, and $\Theta^{-1}(\waf_R)=[w,A]$. % of the Shi arrangement $R$ with minimal alcove $\waf_R\ac$. By definition of $\Theta$ we know that By definition of $\Theta^{-1}$ we have $R\subseteq wC$ and for each $\alpha\in\Phi^+$ the hyperplane $H_{\alpha,1}$ is a floor of the dominant Shi alcove $w^{-1}\waf_R\ac$ if and only if $\alpha\in A$. % Furthermore, we have $\lambda=\wf\waf_R^{-1}\cdot0$ \cite{ Choose $\mu,\nu\in\Q$ and $\tau,\sigma\in W$ such that $\waf_R^{-1}w=t_{\mu}\sigma$ and $\wf=t_{\nu}\tau$. Then $\lambda=\wf\waf_R^{-1}\cdot0=\nu+\tau\mu$ and $\A(\waf_R)=-\waf_R\wf^{-1}\cdot0=w(\tau\sigma)^{-1}\cdot\lambda$. It follows from the first identity that $\mu$ can be computed from $\lambda$ independently of $u$. Since $\sigma$ is the unique element of $W$ such that $(t_{\mu}\sigma)^{-1}\ac$ lies in the dominant chamber, also $\sigma$ does not depend on $u$. \begin{myprop}{antichain} We have \begin{align*} A=(\tau\sigma)^{-1}\cdot\walls % \big\{(\tau\sigma)^{-1}\cdot\alpha:\alpha\in\Delta\cup\{-\tilde\alpha\}\text{ and }s_{\alpha}\cdot\lambda=\lambda\big\}. \end{align*} \end{myprop} \begin{proof} Let $\beta\in\Phi^+$ then \begin{align*} \beta\in A &\Leftrightarrow H_{\beta,1}\text{ is a floor of }w^{-1}\waf_R\ac\\ \text{(\refl{affroots})}&\Leftrightarrow \waf_R^{-1}w\cdot(-\beta+\delta)\in-\widetilde\Delta\\ % &\Leftrightarrow \waf_R^{-1}w\cdot(\beta-\delta)\in\widetilde\Delta\\ \text{(\refl{wf})}&\Leftrightarrow \wf\waf_R^{-1}w\cdot(\beta-\delta)\in \wf(\widetilde\Delta)=(\Delta-\delta)\cup\{-\tilde\alpha+h\delta\}\\ &\Leftrightarrow t_{\lambda}\tau\sigma\cdot\beta\in\Delta\cup\{-\tilde\alpha+(h+1)\delta\}\\ % &\Leftrightarrow t_{\nu}\tau t_{\mu}\sigma\cdot\beta\in\Delta\cup\{-\tilde\alpha+(f+1)\delta\}\\ &\Leftrightarrow \begin{cases} \tau\sigma\cdot\beta\in\Delta\text{ and }\skal{\lambda,\tau\sigma\cdot\beta}=0\text{, or}\\ \tau\sigma\cdot\beta=-\tilde\alpha\text{ and }\skal{\lambda,\tau\sigma\cdot\beta}=h+1. \end{cases} \end{align*} Hence $\beta\in A$ if and only if $\tau\sigma\cdot\beta\in\walls$. % \Delta\cup\{-\tilde\alpha\}$ and $\lambda$ lies in the wall of $(h+1)\ac$ orthogonal to $\tau\sigma\cdot\beta$. \end{proof} \sk \refp{antichain} provides an explicit description of the map $\overline{\zeta}$ on $W$-orbits. The final result of this section explains how the Weyl group element $w$ can be obtained when $u$ is taken into account. \begin{myprop}{utausig} We have $w=u\tau\sigma$. \end{myprop} \begin{proof} Recall that \begin{align*} u\cdot\lambda &=\A(\waf_R) =-\waf_R\wf^{-1}\cdot0 % =-w(t_{\mu}\sigma)^{-1}(t_{\nu}\tau)^{-1}\cdot0 =w(\tau\sigma)^{-1}\cdot(\nu+\tau\mu) =w(\tau\sigma)^{-1}\cdot\lambda. \end{align*} Using \reft{stab} and \refp{antichain} it follows that \begin{align*} w^{-1}u\tau\sigma&\in(\tau\sigma)^{-1}\op{Stab}(\lambda)\tau\sigma\\ % &=\skal{(\tau\sigma)^{-1} s_{\alpha} \tau\sigma:\alpha\in\Delta\cup\{-\tilde\alpha\}\text{ and }s_{\alpha}\cdot\lambda=\lambda}\\ &\quad=\skal{s_{(\tau\sigma)^{-1}\cdot\alpha}:\alpha\in\walls}\\ &\quad=\skal{s_{\beta}:\beta\in A}. \end{align*} In other words $wH=u\tau\sigma H$ where $H=\skal{s_{\beta}:\beta\in A}$. On the one hand $w\cdot A\subseteq\Phi^+$ due to the fact that $\waf_R\ac$ is a Shi alcove contained in the chamber $wC$ \cite[Sec.~12.3]{Thiel2015}. On the other hand $u\tau\sigma\cdot A=u\cdot\walls\subseteq\Phi^+$ using \refp{antichain}. Since there is only one element $v$ of $wH$ with $v\cdot A\subseteq\Phi^+$, we must have $w=u\tau\sigma$. % On the other hand $u\tau\sigma$ is the unique element in $wH$ with this property by \refl{canon} and \refp{antichain}. \end{proof} \subsection{Lattice paths}\label{Section:paths} We call $N=(0,1)$ a \emph{North step} and $E=(1,0)$ an \emph{East step}. Given $a,b\in\N$, denote by $\mathcal L_{a,b}$ the set of lattice paths in the plane starting at the origin, consisting only of North and East steps and ending at $(a,b)$. That is, $\mathcal L_{a,b}$ is the set of sequences $s_1s_2\dots s_{a+b}$ with $s_i\in\{N,E\}$ consisting of $a$ East steps and $b$ North steps. A lattice path in $\mathcal L_{n,n}$ is called a \emph{Dyck path} if it never goes below the main diagonal, that is, each initial segment $s_1s_2\cdots s_k$ contains at least as many North steps as East steps. We denote by $\mathcal D_n\subseteq\mathcal L_{n,n}$ the set of all Dyck paths with $2n$ steps. A ballot path of length $n$ is a lattice path starting at the origin, consisting of $n$ North and/or East steps that never goes below the main diagonal. We denote the set of ballot paths with $n$ steps by $\mathcal B_n$. A pattern of the form $NN$ is called \emph{rise}. A pattern $EN$ is called \emph{valley}. More precisely, let $\pi$ be any lattice path with steps $s_i\in\{N,E\}$. We say $i$ is a rise of $\pi$ if its $i$-th North step is followed by a North step. We say $(i,j)$ is a valley of $\pi$ if its $i$-th East step is followed by its $j$-th North step. If $\beta$ is a ballot path consisting of $i$ East steps and $j$ North steps, and $\beta$ ends with an East step, then we adopt the convention of counting $(i,j+1)$ as a valley of $\beta$. See Figures \ref{Figure:ballotpath} and \ref{Figure:dyckpath}. \begin{figure}[t] \begin{minipage}[t]{.45\textwidth} \begin{center} \begin{tikzpicture}[scale=.5] \draw[gray] (0,1)--(1,1)--(1,6)--(0,6)--(0,0) (0,2)--(2,2)--(2,5)--(0,5) (0,3)--(3,3)--(3,4)--(0,4); \draw[very thick] (0,0)--(0,1)--(1,1)--(1,4)--(2,4); \draw[xshift=5mm,yshift=5mm] (0,1) node{\large{$\bullet$}} (1,4) node{\large{$\circ$}}; \end{tikzpicture} \caption{A ballot path $\beta\in\mathcal B_6$ with two valleys $(1,2),(2,5)$ and two rises $2,3$.} \label{Figure:ballotpath} \end{center} \end{minipage} \hfill \begin{minipage}[t]{.45\textwidth} \begin{center} \begin{tikzpicture}[scale=.5] \draw[gray] (0,0) grid (6,6) (0,0)--(6,6); \draw[very thick] (0,0)--(0,4)--(2,4)--(2,5)--(3,5)--(3,6)--(6,6); \draw (1.5,4.5) node{\large{$\bullet$}} (2.5,5.5) node{\large{$\bullet$}}; \end{tikzpicture} \caption{A Dyck path $\pi\in\mathcal D_6$ with valleys $(2,5),(3,6)$ and rises $1,2,3$.} \label{Figure:dyckpath} \end{center} \end{minipage} \end{figure} In order to define the combinatorial zeta maps, which are bijections between sets of (labelled) lattice paths, we need the following building blocks. Let $a=(a_1,a_2,\dots,a_n)$ be a vector with integer entries. Define a word $\RS_j^+(a)$ in the alphabet $\{E,N\}$ as follows: Read $a$ from left to right. Whenever you encounter an entry $a_i=j$ write down $N$. Whenever you encounter an entry $a_i=j+1$ write down $E$. Similarly we define $\LS_j^+(a)$ except now one reads $a$ from right to left. Moreover we define $\RS_j^-(a)$ as follows: Read $a$ from left to right and write down $N$ for each $-j$ you encounter and $E$ for each $-j-1$ you encounter. Define $\LS_j^-(a)$ analogously. We conclude this section by recalling the definitions of the two versions of the zeta map in type $A_{n-1}$, that is, one on labelled paths and one on unlabelled paths. See \reff{zetaA} for an accompanying example. Given a Dyck path $\pi\in\mathcal{D}_n$ we define its \emph{area vector} $(\mu_1,\dots,\mu_n)$ by letting $\mu_i$ be the number of $1\times1$-squares in the $i$-th row that lie between the path and the main diagonal. For example the Dyck path in \reff{dyckpath} has area vector $(0,1,2,3,2,2)$. It is not hard to see that a vector with non-negative integer entries is the area vector of a Dyck path if and only if $\mu_1=0$ and $\mu_{i+1}\leq\mu_{i}+1$. The original \emph{zeta map} $\zeta_A:\mathcal{D}_n\to\mathcal{D}_n$ first appears in a paper of Andrews, Krattenthaler, Orsina and Papi~\cite{AKOP2002}. It was rediscovered and popularised by Haglund and Haiman, and an explicit treatment of its compatibility with the statistics on Dyck paths mentioned in the introduction is found in~\cite{Haglund2008}. The image of the Dyck path $\pi$ under map $\zeta_A$ can be defined as the concatenation \begin{align*} \zeta_A(\pi)&=\RS_0^-(\mu)\RS_{-1}^-(\mu)\cdots\RS_{-n}^-(\mu) \end{align*} where $\mu$ is the area vector of $\pi$. A \emph{vertically labelled Dyck path} is a pair $(\pi,v)$ of a Dyck path $\pi\in\mathcal{D}_n$ and a permutation $v\in\S_n$ such that $v(i)] (0,0)--node[above]{$\zeta_A$}(1,0) ; % \begin{scope}[xshift=11cm] \draw[gray] (0,0)grid(6,6) ; \draw[very thick] (0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(2,5)--(3,5)--(3,6)--(6,6) ; \draw[xshift=5mm,yshift=5mm] (0,0)node{\large{$2$}} (1,1)node{\large{$3$}} (2,2)node{\large{$4$}} (3,3)node{\large{$1$}} (4,4)node{\large{$5$}} (5,5)node{\large{$6$}} (0,1)node{\large{$\bullet$}} (1,2)node{\large{$\bullet$}} (2,5)node{\large{$\bullet$}} ; \end{scope} \end{tikzpicture} \caption{A vertically labelled Dyck path $(\pi,v)$ and the diagonally labelled Dyck path $\zeta_A(\pi,v)$.} \label{Figure:zetaA} \end{center} \end{figure} \section{Types \emph{B}, \emph{C} and \emph{D}}\label{Section:BCD} In this section we fix further notation and recall some facts specific to the root systems and Weyl groups of types $B_n, C_n$ and $D_n$. \subsection{The root systems of types \emph{B}, \emph{C} and \emph{D}}%$B_n$, $C_n$ and $D_n$} The roots, positive roots and simple roots of type $B_n$ are given by \begin{align*} \Phi&=\{\pm e_i\pm e_j:1\leq in\}$ has even cardinality. This subgroup is generated by the \emph{simple transpositions of type $B_n$}, given by \begin{align*} s_0^B&=[-1,2,\dots,n],\\ s_i^B&=[1,\dots,i-1,i+1,i,i+2,\dots,n]&&\text{for }i\in[n-1]\text{ and}\\ s_n^B&=[1,\dots,n-2,n+1,n+2]. \end{align*} The group $\affS_n^B$ contains a subgroup $\affS_n^D$ that consists of all affine permutations $\omega$ such that both finite sets $\{i\in\Z:i\leq n,\omega(i)>n\}$ and $\{i\in\Z:i\geq0,\omega(i)<0\}$ have even cardinality. This subgroup is generated by the \emph{simple transpositions of type $D_n$}, that is, \begin{align*} s_0^D&=[-1,-2,3,\dots,n],\\ s_i^D&=[1,\dots,i-1,i+1,i,i+2,\dots,n]&&\text{for }i\in[n-1]\text{ and}\\ s_n^D&=[1,\dots,n-2,n+1,n+2]. \end{align*} Let $\Phi$ be a root system of type $B_n$, $C_n$ or $D_n$. The group $\affS_n^{\Phi}$ is isomorphic to the affine Weyl group $\wa$ of $\Phi$. An explicit isomorphism in terms of the generators is obtained by mapping $s_i^{\Phi}$ to $s_{\alpha_i^{\Phi}}$ for $0\leq i\leq n-1$, and $s_n^{\Phi}$ to $s_{\tilde\alpha^{\Phi},1}$. Let $\S_n^{\Phi}$ denote the subgroup of $\affS_n^{\Phi}$ corresponding to the Weyl group $W$ under this isomorphism. Then an affine permutation $\omega\in\affS_n^{\Phi}$ lies in $\S_n^{\Phi}$ if and only if its window is a subset of $[\pm n]$. More precisely, $\S_n^B=\S_n^C$ is the group of signed permutations while $\S_n^D\leq\S_n^B$ consists of the signed permutations with an even number of sign changes. Furthermore, this isomorphism affords an action of the affine permutations on the coroot lattice $\Q$ of $\Phi$, which is made explicit by the following rules \begin{align*} s_0^C\cdot(x_1\dots,x_n)&=(-x_1,x_2\dots,x_n),\\ s_i^C\cdot(x_1\dots,x_n)&=(x_1,\dots,x_{i+1},x_i,\dots,x_n),\quad\text{for }i\in[n-1],\\ s_n^C\cdot(x_1\dots,x_n)&=(x_1,\dots,x_{n-1},-x_n+1),\\ s_n^B\cdot(x_1\dots,x_n)&=(x_1,\dots,x_{n-2},-x_n+1,-x_{n-1}+1)\quad\text{and}\\ s_0^D\cdot(x_1\dots,x_n)&=(-x_1,-x_2,x_3,\dots,x_n). \end{align*} The first tool we need is a combinatorial description of the Gra{\ss}mannian affine permutations in $\affS_n^{\Phi}$. That is, given an affine permutation $\omega\in\affS_n^{\Phi}$ we want to decide whether $\omega$ is a minimal length coset representative just by looking at its window. This question was answered in \cite[Props.~8.4.4,~8.5.4, and~8.6.4]{BjoBre}. \begin{myprop}{grassmannian} % An affine permutation $\omega\in\affS_n^B$ is the minimal length representative of a coset in the quotient $\affS_n^B/\S_n^B$ if and only if $0<\omega(1)<\omega(2)<\dots<\omega(n)$. % \myi{(ii)} An affine permutation $\omega\in\affS_n^C$ is the minimal length representative of a coset in the quotient $\affS_n^C/\S_n^C$ if and only if $0<\omega(1)<\omega(2)<\dots<\omega(n)$. % \myi{(iii)} An affine permutation $\omega\in\affS_n^D$ is the minimal length representative of a coset in the quotient $\affS_n^D/\S_n^D$ if and only if $-\omega(2)<\omega(1)<\omega(2)<\dots<\omega(n)$. An affine permutation $\omega\in\affS_n^{\Phi}$ is the minimal length representative of the coset $\omega\S_n^{\Phi}\in\affS_n^{\Phi}/\S_n^{\Phi}$ if and only if \begin{align*} \begin{cases} 0<\omega(1)<\omega(2)<\dots<\omega(n)&\quad\text{if }\Phi\text{ is of type }B_n\text{ or }C_n,\\ 0<\abs{\omega(1)}<\omega(2)<\dots<\omega(n)&\quad\text{if }\Phi\text{ is of type }D_n. \end{cases} \end{align*} \end{myprop} \sk Next we want to obtain a combinatorial description of the decomposition of $\omega$ into a product of a translation by an element of the coroot lattice and an element of the Weyl group, that is, a signed permutation. To this end, for $\omega\in\affS_n^{\Phi}$ and $i\in[\pm n]$ write $\omega(i)=a_i\n+b_i$ such that $a_i\in\Z$ and $b_i\in[\pm n]$. Define $\sigma(\omega,i)=b_i$, $\mu(\omega,b_i)=-a_i$ and $\nu(\omega,i)=a_i$. \begin{mylem}{munusig} Let $\omega\in\affS_n^{\Phi}$ be an affine permutation. \begin{enumerate}[(i)] \myi{sig} The map $i\mapsto\sigma(\omega,i)$, where $i\in[n]$, defines a signed permutation $\sigma(\omega)\in\S_n^{\Phi}$. %(Thus, if $\Phi=D_n$ then this signed permutation has an even number of sign changes.) \myi{mu} The vectors $\mu(\omega)=(\mu(\omega,1),\dots,\mu(\omega,n))$ and $\nu(\omega)=(\nu(\omega,1),\dots,\nu(\omega,n))$ lie in the coroot lattice $\Q$ of $\Phi$, and for all $i$ with $0\leq i\leq n$ we have \begin{align*} \mu(s_i^{\Phi}\omega)=s_i^{\Phi}\cdot\mu(\omega),&& \nu(\omega s_i^{\Phi})=s_i^{\Phi}\cdot\nu(\omega). \end{align*} \myi{muzero} We have $\mu(\omega^{-1})=\nu(\omega)$ and $\sigma(\omega^{-1})=\sigma(\omega)^{-1}$. \myi{muinv} We have $\omega\cdot(0,\dots,0)=\mu(\omega)$ and $\omega^{-1}\cdot(0,\dots,0)=\nu(\omega)$ \myi{munusig} We have $\mu(\omega)=-\sigma\cdot\nu(\omega)$. \end{enumerate} \end{mylem} \begin{proof} Claims (i) and (v) are immediate from the definitions while (iii) and (iv) follow directly from (ii). Thus it only remains to show (ii) which is done for each type using induction on the length of $\omega$. \end{proof} \sk For $q\in\Q$ define an affine permutation $t_q\in\affS_n^{\Phi}$ by setting $t_q(i)=-q_i\n+i$ for $i\in[n]$. Set $T_{\Q}=\{t_q:q\in\Q\}$. We call an affine permutation $\omega\in\affS_n^{\Phi}$ translation by $q\in\Q$ if $\omega\cdot x=x+q$ for all $x\in\Q$. Thus by definition the translations in $\affS_n^{\Phi}$ correspond to translations in $\wa$. \begin{myprop}{translations} \begin{enumerate}[(i)] \myi{prod} Let $\omega\in\affS_n^{\Phi}$ be an affine permutation, and set $\sigma=\sigma(\omega)$, $\mu=\mu(\omega)$ and $\nu=\nu(\omega)$. Then $\omega=t_{\mu}\sigma=\sigma t_{-\nu}$. \myi{subg} Let $x,y\in\Q$ then $t_xt_y=t_{x+y}$ and $(t_x)^{-1}=t_{-x}$. Thus $T_{\Q}$ is a subgroup of $\affS_n^{\Phi}$ that is isomorphic to the coroot lattice. \myi{translation} An affine permutation $\omega\in\affS_n^{\Phi}$ is a translation if and only if $\omega\in T_{\Q}$. \end{enumerate} \end{myprop} \begin{proof} To prove claim (i) let $i\in[n]$ and $\omega(i)=a_i\n+b_i$ such that $a_i\in\Z$ and $b_i\in[\pm n]$. Then \begin{align*} t_{\mu}\sigma(i) &=t_{\mu}(\sigma(\omega,i)) =t_{\mu}(b_i) =-\mu(\omega,b_i)\n+b_i =a_i\n+b_i,\\ % \sigma t_{-\nu}(i) &=\sigma(\nu(\omega,i)\n+i) =\sigma(a_i\n+i) =a_i\n+\sigma(i) =a_i\n+b_i. \end{align*} To see claim (ii), note that for each $i\in[n]$ we have \begin{align*} t_xt_y(i) &=t_x(-y_i\n+i) =-y_i\n+t_x(i) =-(y_i+x_i)\n+i =t_{x+y}(i). \end{align*} Finally, let $q,x\in\Q$. Then using~(i),~(ii) and \refl{munusig}~(iv) we see that $t_q$ is indeed a translation as \begin{align*} t_q\cdot x &=t_q\cdot t_x\cdot(0,\dots,0) =(t_qt_x)\cdot(0,\dots,0) =t_{q+x}\cdot(0,\dots,0) =q+x. \end{align*} Conversely, if $\omega$ is a translation by $q$ then $\omega=t_{q}$ because $\affS_n^{\Phi}$ acts faithfully on the coroot lattice. \end{proof} \sk In fact from \refp{translations}~(i) and~(ii) it follows directly that $\affS_n^{\Phi}$ is the semi-direct product of $T_{\Q}$ and $\S_n^{\Phi}$ (the Weyl group acting on the coroot lattice). Hence the decomposition $\omega=t_{\mu}\sigma$ is the one we were looking for, and it can easily be obtained from the window of $\omega$. The following lemma is an easy consequence of the two propositions above but it shall serve us as a reference in the sections to come. \begin{mylem}{sigma} Let $\omega\in\affS_n^{\Phi}$ be the minimal length coset representative of $\omega\S_n^{\Phi}$, $\mu=\mu(\omega)\in\Q$ and $\sigma=\sigma(\omega)\in\S_n^{\Phi}$. Then for each $i\in[n]$ \begin{align*} \abs{\sigma^{-1}(i)} &=\#\big\{k\in[n]:\abs{\mu_k\n-k}\leq\abs{\mu_i\n-i}\big\}. \end{align*} If $\Phi$ is of type $B_n$ or $C_n$, or if $\Phi$ is of type $D_n$ and $\abs{\sigma^{-1}(i)}\neq1$, then $\sigma^{-1}(i)>0$ if and only if $\mu_i\leq0$. However, if $\Phi$ is of type $D_n$ and $\abs{\sigma^{-1}(i)}=1$, then $\sigma^{-1}(i)>0$ if and only if either $\mu_i\leq0$ and the number of positive entries of $\mu$ is even, or $\mu_i>0$ and the number of positive entries of $\mu$ is odd. %However, if $\Phi$ is of type $D_n$ and $\abs{\sigma^{-1}(i)}=1$, then $\sigma^{-1}(i)>0$ if and only if either $\mu_i\leq0$ and $\sum_{j=1}^n\floor{\abs{\mu_j\n-j}/\n}$ is even, or $\mu_i>0$ and $\sum_{j=1}^n\floor{\abs{\mu_j\n-j}/\n}$ is odd. \end{mylem} \begin{proof} Since $\omega$ is a minimal length coset representative, the absolute values of the entries of the window of $\omega$ must be increasing, that is, $0<\abs{\omega(1)}<\abs{\omega(2)}<\dots<\abs{\omega(n)}$. On the other hand, $\abs{\sigma(j)}=i$ if and only if $\abs{\omega(j)}=\abs{\mu_i\n-i}$. Hence $\abs{\sigma(j)}=i$ is equivalent to \begin{align*} j=\#\big\{k\in[n]:\abs{\mu_k\n-k}\leq\abs{\mu_i\n-i}\big\}. \end{align*} Furthermore, if $\Phi$ is of type $B_n$ or $C_n$, or if $\Phi$ is of type $D_n$ and $\abs{\sigma^{-1}(i)}\neq1$, then $\omega(i)>0$. Hence $\sigma(i)>0$ if and only if $-\n\mu_i+i>0$, which is the case if and only if $\mu_i\leq0$. If $\Phi$ is of type $D_n$ then the sign of $\omega(1)$ possibly has to be changed such that there is an even number of integers $j\in\Z$ with $j\geq0$ and $\omega(j)<0$. %This number has the same parity as the sum $\sum_{j=1}^n\floor{\abs{\mu_j\n-j}/\n}$. \end{proof} \section{The zeta map of type C}\label{Section:typeC} In \refs{vertC} we define a combinatorial model for the elements of the finite torus in terms of vertically labelled lattice paths in an $n\times n$-square. The underlying lattice paths correspond to the orbits of the finite torus under the action of the Weyl group $\S_n^C$. That is, acting by an element of $\S_n^C$ only changes the labels but leaves the path intact. In \refs{diagC} we encode the non-nesting parking functions in term of diagonally labelled ballot paths. Again acting by an element of the Weyl group only changes the labels, such that the ballot paths correspond to the antichains in the root poset. We then demonstrate how the dominant Shi region corresponding to a square lattice path can be recovered using its area vector in \refs{areaC}. Moreover we obtain the Shi region corresponding to a vertically labelled lattice path using the diagonal reading word in \refs{readC}. Finally, in \refs{zetaC} we provide a combinatorial construction of the image of a vertically labelled lattice path under the zeta map of type $C_n$. We prove that this construction is invertible and that it coincides with the uniform zeta map. \refs{sweepC} offers an alternative description of the type $C_n$ zeta map in the spirit of the sweep map. The examples provided in this section are supplemented by \reftb{C2} of the appendix. % We conclude by defining type $C_n$ analogues of the $\area$ and $\dinv$ statistics, and discuss their connection via the zeta map in \refs{statC} % \begin{mylem}{sigmaC} Let $\omega\in\affS_n^C$ be the minimal length coset representative of $\omega\S_n^B$ and $i\in[n]$. Set $\sigma=\sigma(\omega)$ and $\mu=\mu(\omega)$. Then % \begin{align*} % \sigma^{-1}(i) % &=\#\big\{k\in[n]:\abs{\mu_k\n-k}\leq\abs{\mu_i\n-i}\big\}&&\text{if }\mu_i\leq0,\text{ and}\\ % % % \sigma^{-1}(-i) % &=\#\big\{k\in[n]:\abs{\mu_k\n-k}\leq\abs{\mu_i\n-i}\big\}&&\text{if }\mu_i>0. % \end{align*} % \end{mylem} % \begin{proof} Since $\omega$ is a minimal length coset representative, its window must be increasing and positive, that is, $0<\omega(1)<\omega(2)<\dots<\omega(n)$. On the other hand, $\abs{\sigma(j)}=i$ if and only if $\omega(j)=\abs{\mu_i\n-i}$. Hence $\abs{\sigma(j)}=i$ is equivalent to % \begin{align*} % j=\#\big\{k\in[n]:\abs{\mu_k\n-k}\leq\abs{\mu_i\n-i}\big\}. % \end{align*} % Furthermore, $\sigma(i)>0$ if and only if $-\n\mu_i+i>0$, which is the case if and only if $\mu_i\leq0$. % \end{proof} \subsection{The finite torus}\label{Section:vertC} The Coxeter number of type $C_n$ is given by $h=2n$ hence the finite torus equals $\Z^n/(2n+1)\Z^n$. As Athanasiadis~\cite[Sec.~5.2]{Athan2005} pointed out, a system of representatives for the orbits of the finite torus under the action of the Weyl group $\S_n^C$ is given by \begin{align*} \Q\cap (2n+1)\overline{\ac} =\Big\{(\lambda_1,\lambda_2,\dots,\lambda_n)\in\Q: 0\leq\lambda_1\leq\lambda_2\leq\dots\leq\lambda_n\leq n\Big\}, \end{align*} and the stabiliser subgroup $\op{Stab}(\lambda)\leq\S_n^C$ of such a $\lambda$ is generated by the simple transpositions $s_i$ for each $i\in[n-1]$ such that $\lambda_i=\lambda_{i+1}$, and the simple transposition $s_0$ if $\lambda_1=0$. %, the element of $\S_n^C$ that exchanges $\lambda_{n-1}$ and $\lambda_n$ and changes the signs of these two entries if $\lambda_{n-1}+\lambda_n=2n+1$. % \begin{mylem}{canonicB} Let $u\in W$ and $1\leq iu(j),\\ % u(2e_i)&\in\Phi^+ \qquad\text{if and only if }u(i)>0,\text{ and}\\ % u(e_i+e_j)&\in\Phi^+ \qquad\text{if and only if }u(i)>-u(j). % \end{align*} % \end{mylem} \begin{mydef}{vertC} A \emph{vertically labelled lattice path} is a pair $(\pi,u)$ of a lattice path $\pi\in\mathcal L_{n,n}$ and a signed permutation $u\in\S_n^C$ such that $u(i)0$ if $\pi$ begins with a North step. We denote the set of all vertically labelled lattice paths by $\op{Vert}(C_n)$. \end{mydef} \sk Given a lattice path $\pi\in\mathcal L_{n,n}$ and a signed permutation $u\in\S_n^C$ we picture $(\pi,u)$ by placing the label $\sigma(i)$ to the left of the $i$-th North step of $\pi$ as is shown in \reff{vertC}. Thus we obtain $(\pi,u)\in\op{Vert}(C_n)$ if the labels increase along columns from bottom to top, and if all labels left of the starting point of $\pi$ are positive. \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.55] % \begin{scope}[yshift=1cm] \draw[gray] (0,0) grid (6,6); \draw[very thick] (0,0)--(0,1)--(4,1)--(4,6)--(6,6); \draw[xshift=5mm,yshift=5mm] (-1,0) node{$1$} (3,1) node{$-5$} (3,2) node{$-4$} (3,3) node{$2$} (3,4) node{$3$} (3,5) node{$6$}; % \end{scope} \end{tikzpicture} \caption{The vertically labelled lattice path $(NEEEENNNNNEE,[1,-5,-4,2,3,6])\in\op{Vert}(C_6)$.} \label{Figure:vertC} \end{center} \end{figure} It is not difficult to see that vertically labelled lattice paths encode the elements of the finite torus. \begin{myprop}{vertC} The map $\psi:\op{Vert}(C_n)\to\Q/(2n+1)\Q$ given by $(\pi,u)\mapsto u\cdot\lambda+(2n+1)\Q$, where $\lambda\subseteq(n^n)$ is the partition\footnote{For our purposes here a partition is an increasing sequence $\lambda_1\leq\lambda_2\leq\dots\leq\lambda_n$ of non-negative integers.} with South-East boundary $\pi$, is a bijection. % between vertically labelled lattice paths $(\pi,u)$ and pairs $(\lambda,u)$ of a coroot lattice point $\lambda\in\Q\cap \n\overline{A_{\circ}}$ and an element of the Weyl group $u\in W$ such that $u$ is the canonical coset representative of $uH\in W/H$ where $H$ is the stabiliser subgroup of $\lambda$ in $W$ with respect to the action of $W$ on the finite torus. \end{myprop} \begin{proof} First note that the set of partitions that fit inside the square $(n^n)$ coincides with the set $\Q\cap(2n+1)\overline{\ac}$. Let $\pi\in\mathcal L_{n,n}$ be a lattice path, $u\in\S_n^C$ be a signed permutation, and $\lambda\subseteq(n^n)$ be the partition with South-East boundary $\pi$. By \refl{canon} it suffices to show that $(\pi,u)\in\op{Vert}(C_n)$ if and only if $u\cdot\walls\subseteq\Phi^+$. Suppose $i\in[n-1]$ and $(\pi,u)\in\op{Vert}(C_n)$ then \begin{align*} \alpha_i\in\walls &\Leftrightarrow \skal{\lambda,e_{i+1}-e_i}=0 \Leftrightarrow \lambda_i=\lambda_{i+1}\\ &\Rightarrow u(i) \begin{cases} w(n+1-j)&\quad\text{if }j\leq n,\\ w(n-j)&\quad\text{if }j>n. \end{cases} \end{align*} We denote the set of all diagonally labelled ballot paths by $\op{Diag}(C_n)$. \end{mydef} \sk Suppose $(\beta,w)\in\op{Diag}(C_n)$. If we place the labels $w(i)$, where \begin{align*} i=n,n-1,\dots,1,-1,\dots,-n, \end{align*} in the diagonal as in \reff{diagC}, then for each valley the label to its right will be smaller than the label below it. In particular, if the path ends with an East step then the label below will be positive. \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.55] \ballot{6} \draw[very thick] (0,0)--(0,2)--(1,2)--(1,3)--(2,3)--(2,5)--(3,5)--(3,6)--(4,6)--(4,7)--(5,7); \draw[xshift=5mm,yshift=5mm] (0,0) node{$5$} (1,1) node{$6$} (2,2) node{$4$} (3,3) node{$3$} (4,4) node{$1$} (5,5) node{$-2$} (6,6) node{$2$} (5,7) node{$-1$} (4,8) node{$-3$} (3,9) node{$-4$} (2,10) node{$-6$} (1,11) node{$-5$} (0,2) node{\large{$\bullet$}} (1,3) node{\large{$\bullet$}} (2,5) node{\large{$\bullet$}} (3,6) node{\large{$\bullet$}} (4,7) node{\large{$\circ$}}; \end{tikzpicture} \caption{The diagonally labelled ballot path $(NNENENNENENE,[-2,1,3,4,6,5])\in\op{Diag}(C_6)$.} \label{Figure:diagC} \end{center} \end{figure} \sk Let $\beta\in\mathcal B_{2n}$ be a ballot path and $(i,j)$ a valley of $\beta$. We define the positive root \begin{align*} \alpha_{i,j}&= \begin{cases} e_{n+1-i}-e_{n+1-j}&\quad\text{if }j\leq n,\\ e_{n+1-i}+e_{j-n}&\quad\text{if }j>n. \end{cases} \end{align*} Furthermore set \begin{align*} A_{\beta}=\big\{\alpha_{i,j}:(i,j)\text{ is a valley of }\beta\big\}. \end{align*} We will use the following well-known result. \begin{mylem}{AbetaC} The map $\varphi:\mathcal B_{2n}\to\op{Antichains}(\Phi^+)$ given by $\beta\mapsto A_{\beta}$ is a bijection between ballot paths of length $2n$ and the set of antichains in the root poset $\Phi^+$ of type $C_n$. \end{mylem} \begin{myex}{diagC} The diagonally labelled ballot path $(\beta,w)$ in \reff{diagC} has five valleys: $(1,3)$, $(2,4),(3,6),(4,7)$ and $(5,8)$. Note that for each valley the number to its right is less than the number below (for example $w(6+1-1)=5>4=w(6+1-3)$) and that the label below the terminal East step is positive: $w(6+1-5)=1>0$. The corresponding antichain $A_{\beta}$ contains the roots $e_6-e_4,e_5-e_3,e_4-e_1,e_3+e_1$ and $2e_2$. \end{myex} \sk Taking the labels into account, we extend \refl{AbetaC} to a bijection between diagonally labelled ballot paths and non-nesting parking functions. \begin{myprop}{diagC} The map $\varphi:\op{Diag}(C_n)\to\op{Park}(C_n)$ given by $(\beta,w)\mapsto[w,A_{\beta}]$ is a bijection. % between diagonally labelled ballot paths of length $2n$ and the set of pairs of an antichain $A$ of the root poset of type $C_n$ and an element of the Weyl group $w\in W$ such that $w(A)\subseteq\Phi^+$, that is, the set of regions of the Shi arrangement. \end{myprop} \begin{proof} Let $\beta\in\mathcal B_{2n}$ be a ballot path and $w\in\S_n^C$ a signed permutation. Due to \refl{AbetaC} and \refl{nnpark} it suffices to show that $(\beta,w)\in\op{Diag}(C_n)$ if and only if $w\cdot A_{\beta}\subseteq\Phi^+$. Assume $j\leq n$. If $(\beta,w)\in\op{Diag}(C_n)$ then \begin{align*} \alpha_{i,j}\in A_{\beta} &\Leftrightarrow (i,j)\text{ is a valley of }\beta\\ &\Rightarrow w(n+1-i)>w(n+1-j) \Leftrightarrow w\cdot\alpha_{i,j} =w\cdot(e_{n+1-i}-e_{n+1-j})\in\Phi^+. \end{align*} Conversely, if $w\cdot A_{\beta}\subseteq\Phi^+$ then \begin{align*} (i,j)\text{ is a valley of }\beta &\Leftrightarrow \alpha_{i,j}\in A_{\beta}\\ &\Rightarrow w\cdot\alpha_{i,j}\in\Phi^+ \Leftrightarrow w(n+1-i)>w(n+1-j). \end{align*} The case $j>n$ is treated similarly. \end{proof} \subsection{The area vector}\label{Section:areaC} Let $\pi\in\mathcal L_{n,n}$ be a lattice path and $\lambda\subseteq(n^n)$ the partition with South-East boundary $\pi$. We have seen above that $\lambda\in\Q\cap(2n+1)\overline{\ac}$. Thus $\lambda$ corresponds to a dominant Shi region with minimal alcove $\waf_D\ac$ by means of the Anderson map of \refs{torus}. Our goal in this section is to obtain the element of the affine Weyl group $\waf_D$ from the lattice path $\pi$. To this end and write $\waf_D^{-1}=t_{\mu}\sigma$ where $\mu\in\Q$ and $\sigma\in\S_n^C$. Since $\waf_D^{-1}$ is Gra{\ss}mannian, the signed permutation $\sigma$ is determined by \refl{sigma} once we know $\mu$. Note that $\waf_D$ corresponds to the affine transformation $w^{-1}\waf_R$ from \refs{uniform}. The vector $\mu$ therefore appears implicitly in the identity $\lambda=\waf_f\waf_D^{-1}\cdot0$. Hence all we need is an explicit description of $\waf_f$. This is provided by the next lemma. \begin{mylem}{wfC} Let $\nu\in\Q$ and $\tau\in W$ be such that $\waf_f=t_{\nu}\tau$. Then \begin{align*} \nu= (1,2,\dots,n),&& % \tau= [-n,\dots,-2,-1]. \end{align*} \end{mylem} \begin{proof} By \refl{wf} it suffices to show that $t_{\nu}\tau(\widetilde\Delta)=(\Delta-\delta)\cup\{-\tilde\alpha+h\delta\}$. We have \begin{align*} t_{\nu}\tau(\alpha_i) &=\alpha_{n+1-i}-\skal{\nu,\alpha_{n+1-i}}\delta =\alpha_{n+1-i}-\delta &&\text{for all }i\in[n-1],\\ % t_{\nu}\tau(\alpha_{0}) &=-2e_n-\skal{\tau,-2e_n}\delta =-\tilde\alpha+2n\delta &&\text{and}\\ % t_{\nu}\tau(-\tilde\alpha+\delta) &=2e_1+(1-\skal{\tau,2e_1})\delta =\alpha_0-\delta.&&\qedhere \end{align*} % % Let $\nu$ and $\tau$ be as defined above. We know that $w_f$ maps the walls of the Sommers region to the walls of the $2n+1$ dilated fundamental alcove. That is % \begin{align*} % w_f\big(\{-\alpha_i+\delta:0\leq i\leq n-1\}\cup\{\tilde\alpha+2\delta\}\big) % =\{-\alpha_i:0\leq i\leq n-1\}\cup\{\tilde\alpha+(2n+1)\delta\}. % \end{align*} % It suffices to show that $t_{\nu}\tau$ has the same property. \end{proof} \sk For example if $n=5$ then \begin{align*} \waf_f=[50,40,30,20,10]. \end{align*} \begin{mydef}{areaC} Let $\pi\in\mathcal L_{n,n}$ and $\lambda\subseteq(n^n)$ be the partition with South-East boundary $\pi$. Moreover define $\nu$ and $\tau$ as in \refl{wfC} above. We define the type $C_n$ \emph{area vector} of $\pi$ as \begin{align*} \mu =\tau(\lambda-\nu) =\tau(\lambda_1-1,\lambda_2-2\dots,\lambda_n-n) =(n-\lambda_n,\dots,2-\lambda_2,1-\lambda_1). \end{align*} \end{mydef} Indeed $\mu_i$ counts the number of boxes in the $(n+1-i)$-th row between the path $\pi$ and the path $(EN)^n\in\mathcal L_{n,n}$ consisting of alternating North and East steps. In this regard $\mu$ is quite similar to the type $A_{n-1}$ area vector of a Dyck path. Note that the entries of the type $C_n$ area vector are negative as long as $\pi$ is to the right of $(EN)^n$ in the respective row. See \reff{areaC}. \begin{myex}{areaC} The lattice path $\pi=NEEEENNNNNEE$ in \reff{vertC} has area vector \begin{align*} \mu =(6,5,4,3,2,1)-(4,4,4,4,4,0) =(2,1,0,-1,-2,1). \end{align*} The translation by $\mu$ is given by $t_{\mu}=[-25,-11,3,17,31,-7]\in\affS_n^C$. Moreover $\sigma=[3,-6,-2,4,-1,5]$ is the unique element of $\S_n^C$ such that $\waf_D^{-1}=t_{\mu}\sigma=[3,7,11,17,25,31]$ is a Gra{\ss}mannian affine permutation. Finally $\waf_D=[21,10,1,-9,-20,11]$ and $\waf_D\ac$ is the minimal alcove of the dominant Shi region corresponding to the lattice path $\pi$. \end{myex} \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.55] \draw[gray] (0,0) grid (4,4); % \draw[xshift=1pt,yshift=-1pt,thick,red] (0,0)--(1,0)--(1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(4,3)--(4,4); \draw[very thick] (0,0)--(1,0)--(1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(4,3)--(4,4); % \begin{scope}[xshift=7cm] \fill[black!12] (0,0)--(1,0)--(1,2)--(2,2)--(2,1)--(0,1)--cycle (3,2)--(4,2)--(4,3)--(3,3)--cycle; \draw[gray] (0,0) grid (4,4); % \draw[xshift=1pt,yshift=-1pt,thick,red] (0,0)--(1,0)--(1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(4,3)--(4,4); \draw[very thick] (0,0)--(0,1)--(1,1)--(1,2)--(4,2)--(4,4); \end{scope} % \begin{scope}[xshift=14cm] \fill[black!12] (1,0)--(3,0)--(3,2)--(2,2)--(2,1)--(1,1)--cycle (3,3)--(4,3)--(4,4)--(3,4)--cycle; \draw[gray] (0,0) grid (4,4); % \draw[xshift=1pt,yshift=-1pt,thick,red] (0,0)--(1,0)--(1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(4,3)--(4,4); \draw[very thick] (0,0)--(3,0)--(3,4)--(4,4); \end{scope} % \end{tikzpicture} \caption{The lattice paths with type $C_n$ area vectors $(0,0,0,0),(0,-1,1,1)$ and $(1,0,-1,-2)$.} \label{Figure:areaC} \end{center} \end{figure} \sk We conclude this section by proving some auxiliary results on area vectors for later use. \begin{mylem}{areaC} Let $\pi\in\mathcal L_{n,n}$ be a lattice path with type $C_n$ area vector $\mu$. \begin{enumerate}[(i)] \myi{C(i)} Let $i,j\in[n]$ with $i1$. Then there exists $j\in[n]$ with $in+1-i-\mu_i=\lambda_{n+1-i}$. Thus $j=i+1$. Claim~(ii) is an immediate consequence of $\lambda_{n-i}\leq\lambda_{n+1-i}$ for all $i\in[n-1]$. Claim~(iii) follows from~(ii) and $\mu_1\geq0$. Similarly claims~(iv) and~(v) follow from~(ii) and $\mu_n\leq1$. \end{proof} \subsection{The diagonal reading word}\label{Section:readC} Let $(\pi,u)\in\op{Vert}(C_n)$ be a vertically labelled lattice path. Recall that $(\pi,u)$ corresponds to an element $u\cdot\lambda+(2n+1)\Q$ of the finite torus by \refp{vertC} and hence to a region of the Shi arrangement with minimal alcove $\waf_R\ac$ by \reft{anderson}. That is, $u\cdot\lambda=\A(\waf_R)=-\waf_R\waf_f^{-1}\cdot0$. Our goal for this section is to recover the element $\waf_R$ of the affine Weyl group from the labelled path $(\pi,u)$. In order to do so write $\waf_R=w\waf_D$ where $w\in\S_n^C$ and $\waf_D\ac$ is the minimal alcove of the dominant Shi region corresponding to $\lambda$ discussed in the previous section (\refs{areaC}). Recall that $w=u\tau\sigma$ by \refp{utausig}. Since we already constructed $\waf_D$ from $\pi$ in the previous section, we are in principle also able to compute the signed permutation $w$. However, it turns out that this permutation can simply be read off the vertically labelled path (diagonally!). \begin{mydef}{readC} Let $(\pi,u)\in\mathcal{L}_{n,n}\times\S_n^C$ and $\mu$ be the area vector of $\pi$. Define the type $C_n$ \emph{diagonal reading word} $d_C(\pi,u)$ as follows: For each $i=0,1,\dots,n$ first write down the negative labels $-u(j)$ of the rows with $\mu_{n+1-j}=-i$ from top to bottom, then write down the labels $u(j)$ of rows with $\mu_{n+1-j}=i+1$ from bottom to top. \end{mydef} \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.55] \begin{scope} \draw[gray] (0,0) grid (6,6); \foreach \x in {0,...,5}{ \pgfmathsetmacro{\y}{\x-.1} \pgfmathsetmacro{\z}{5.1-\x} \draw[->,very thick,lightgray](-.9,\z)--(\y,5.9); \pgfmathsetmacro{\y}{0.1+ \x} \pgfmathsetmacro{\z}{5.9-\x} \draw[->,very thick,lightgray](5.9,\z)--(\y,.1); } \draw[xshift=5mm,yshift=5mm] (5,5)node{1} (4,4)node{2} (3,3)node{3} (2,2)node{4} (1,1)node{5} (0,0)node{6} (-1,0)node{7} (0,1)node{8} (1,2)node{9} (2,3)node{10} (3,4)node{11} (4,5)node{12} (5,4)node{13} (4,3)node{14} (3,2)node{15} (2,1)node{16} (1,0)node{17} (-1,1)node{18} (0,2)node{19} (1,3)node{20} (2,4)node{21} (3,5)node{22} (5,3)node{23} (4,2)node{24} (3,1)node{25} (2,0)node{26} (-1,2)node{27} (0,3)node{28} (1,4)node{29} (2,5)node{30} (5,2)node{31} (4,1)node{32} (3,0)node{33} (-1,3)node{34} (0,4)node{35} (1,5)node{36} (5,1)node{37} (4,0)node{38} (-1,4)node{39} (0,5)node{40} (5,0)node{41} (-1,5)node{42}; \draw[very thick,red] (0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(2,3)--(3,3)--(3,4)--(4,4)--(4,5)--(5,5)--(5,6)--(6,6) (6.5,.5)--(7,.5) (-2,5.5)--(-1.5,5.5) (-1.75,5.75)--(-1.75,5.25); \end{scope} % \begin{scope}[xshift=12cm] \draw[gray] (0,0) grid (6,6); \draw[very thick] (0,0)--(0,1)--(4,1)--(4,6)--(6,6); \draw[xshift=5mm,yshift=5mm] (-1,0) node{$1$} (3,1) node{$-5$} (3,2) node{$-4$} (3,3) node{$2$} (3,4) node{$3$} (3,5) node{$6$}; \end{scope} % \end{tikzpicture} \caption{The diagonal reading order of type $C_6$ indicated left, and a vertically labelled lattice path with $d_C(\pi,u)=[-2,1,3,4,6,5]$ on the right.} \label{Figure:readC} \end{center} \end{figure} \begin{myex}{readC} Consider the vertically labelled lattice path $(\pi,u)$ from our running example, which is depicted again in \reff{readC} (right). We first deduce the diagonal reading word $d_C(\pi,u)$ following \refd{readC}. In \refe{areaC} we already determined the area vector $\mu=(2,1,0,-1,-2,1)$ of $\pi$. For $i=0$ we have $-i=0=\mu_3=\mu_{6+1-4}$ hence we write down $-u(4)=-2$. Moreover $i+1=1=\mu_6=\mu_{6+1-1}$ and $i+1=\mu_2=\mu_{6+1-5}$. We write down the labels $u(1)=1$ and $u(5)=3$ from bottom to top, that is, first $1$ and then $3$. Next set $i=1$. We have $-i=-1=\mu_4=\mu_{6+1-3}$ and write down $-u(3)=4$. As $i+1=2=\mu_1=\mu_{6+1-6}$ we write down $u(6)=6$. Finally set $i=2$ to obtain $-i=-2=\mu_5=\mu_{6+1-2}$ and write down $-u(2)=5$. The complete diagonal reading word is thus given by \begin{align*} d_C(\pi,u)=[-2,1,3,4,6,5]. \end{align*} The diagonal reading word of type $C_n$ can also be read off quickly by scanning all boxes that may contain labels according to the diagonal reading order, which is indicated in \reff{readC} (left). In our example the first box with respect to the diagonal reading order to contain a label has the number $3$. The encountered label is $2$. Since this box lies below the alternating path $(NE)^n$ we write down its negative $-2$. The next two boxes that contain labels have numbers $7$ and $11$. Hence we write down $1$ and then $3$. It is not difficult to verify that this procedure always yields $d_C(\pi,u)$. In light of the following proposition the reader may wish to check that \begin{align*} d_C(\pi,u) =u\tau\sigma =[1,-5,-4,2,3,6]\cdot[-6,-5,-4,-3,-2,-1]\cdot[3,-6,-2,4,-1,5]. \end{align*} As a consequence we obtain the affine permutation \begin{align*} \waf_R=d_C(\pi,u)\,\waf_D=[20,10,-2,-9,-21,12] \end{align*} that takes the fundamental alcove to the minimal alcove of the Shi region $R$ corresponding to the vertically labelled lattice path $(\pi,u)$. Indeed, we have $\waf_R\waf_f^{-1}=[1,47,48,54,55,58]$ and \begin{align*} \A(\waf_R) =-\waf_R\waf_f^{-1}\cdot0+13\Z^6 =(0,4,4,-4,-4,4)+13\Z^6, \end{align*} which is in accordance with \refe{vertC}. \end{myex} \sk The next result confirms that the diagonal reading word is the correct signed permutation. \begin{myprop}{readC} Let $(\pi,u)\in\op{Vert}(C_n)$ be a vertically labelled lattice path with area vector $\mu$, let $\tau$ be defined as in \refl{wfC}, and define $\sigma\in\S_n^C$ as in \refl{sigma} such that $t_{\mu}\sigma$ is Gra{\ss}mannian. Then $d_C(\pi,u)=u\tau\sigma$. \end{myprop} \begin{proof} Let $i,j\in[n]$. By definition we have $\abs{d_C(\pi,u)(i)}=\abs{u(j)}$ if and only if \begin{align*} i &=\#\big\{r\in[n],\abs{\mu_r}<\abs{\mu_{n+1-j}}\big\} +\#\big\{r\in[n]:r\geq n+1-j,\mu_r=\mu_{n+1-j}>0\big\} \notag\\ &\quad+\#\big\{r\in[n+1-j],\mu_r=\mu_{n+1-j}\leq0\big\} +\#\big\{r\in[n],\mu_r=-\mu_{n+1-j}>0\big\}\\ &=\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_{n+1-j}\n-(n+1-j)}\big\}. \end{align*} By \refl{sigma} we obtain $\abs{\sigma(i)}=n+1-j$, hence \begin{align*} \abs{d_C(\pi,u)(i)} =\abs{u(j)} =\abs{u(n+1-\abs{\sigma(i)})} =\abs{u\tau\sigma(i)}. \end{align*} Moreover $d_C(\pi,u)(i)=u(j)$ if and only if $\mu_{n+1-j}>0$. On the other hand $\tau\sigma(i)=j$ is equivalent to $\sigma(i)=-(n+1-j)<0$, which is the case if and only if $\mu_{n+1-j}>0$ by \refl{sigma}. We can therefore omit the absolute values in the identity above and the proof is complete. \end{proof} \subsection{The zeta map}\label{Section:zetaC} We are now in a position to define the combinatorial zeta map of type $C_n$, which is made up of the building blocks defined in \refs{paths}. Note that there are two versions of the map: one defined on square lattice paths that corresponds to the uniform map defined on $W$-orbits, and one defined on vertically labelled lattice paths that corresponds to the uniform map defined on the finite torus. To avoid confusion we call the zeta map defined on vertically labelled paths the Haglund--Loehr-zeta map. % Let $a=(a_1,a_2,\dots,a_n)$ be a vector with integer entries. Define a word $\RS_j^+(a)$ in the alphabet $\{E,N\}$ as follows: Read $a$ from left to right. Whenever you encounter an entry $a_i=j$ write down $N$. Whenever you encounter an entry $a_i=j+1$ write down $E$. Similarly we define $\LS_j^+(a)$ except now one reads $a$ from right to left. Moreover we define $\RS_j^-(a)$ as follows: Read $a$ from left to right and write down $N$ for each $-j$ you encounter and $E$ for each $-j-1$ you encounter. Define $\LS_j^-(a)$ analogously. \begin{mydef}{zetaC} Given a lattice path $\pi\in\mathcal L_{n,n}$ with type $C_n$ area vector $\mu$ we define its image under the \emph{zeta map} of type $C_n$ as \begin{align*} \zeta_C(\pi) =\LS_n^-(\mu)\RS_n^+(\mu) \LS_{n-1}^-(\mu)\RS_{n-1}^+(\mu)\cdots \LS_1^-(\mu)\RS_1^+(\mu) \LS_0^-(\mu)\RS_0^+(\mu). \end{align*} Moreover, given a vertically labelled lattice path $(\pi,u)\in\op{Vert}(C_n)$, we define its image under the \emph{Haglund--Loehr-zeta map} of type $C_n$ as \begin{align*} \zeta_C(\pi,u)=\big(\zeta_C(\pi),d_C(\pi,u)\big). \end{align*} \end{mydef} Note that by definition $\zeta_C(\pi)$ is a ballot path, that is, $\zeta_C:\mathcal L_{n,n}\to\mathcal B_{2n}$. It will soon turn out that the Haglund--Loehr-zeta map sends vertically labelled lattice paths to diagonally labelled ballot paths. \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.55] \begin{scope} \draw[gray](0,0)grid(6,6); \draw[very thick](0,0)--(0,1)--(4,1)--(4,6)--(6,6); \draw[xshift=5mm,yshift=5mm] (-1,0)node{$1$} (3,1)node{$-5$} (3,2)node{$-4$} (3,3)node{$2$} (3,4)node{$3$} (3,5)node{$6$}; \end{scope} % \draw[xshift=8cm,yshift=2.5cm,->,thick](0,0)--node[above]{$\zeta_C$}(1,0); % \begin{scope}[xshift=11cm,yshift=-3cm] \ballot{6} \draw[very thick](0,0)--(0,2)--(1,2)--(1,3)--(2,3)--(2,5)--(3,5)--(3,6)--(4,6)--(4,7)--(5,7); \draw[very thick,red] (0,0)--(0,1) (0,2)--(1,2)--(1,3) (2,5)--(3,5)--(3,6); \draw[xshift=5mm,yshift=5mm] (0,0)node{$5$} (1,1)node{$6$} (2,2)node{$4$} (3,3)node{$3$} (4,4)node{$1$} (5,5)node{$-2$} (6,6)node{$2$} (5,7)node{$-1$} (4,8)node{$-3$} (3,9)node{$-4$} (2,10)node{$-6$} (1,11)node{$-5$} (0,2)node{\large{$\bullet$}} (1,3)node{\large{$\bullet$}} (2,5)node{\large{$\bullet$}} (3,6)node{\large{$\bullet$}} (4,7)node{\large{$\circ$}}; \end{scope} % \end{tikzpicture} \caption{A vertically labelled lattice path and its image under the Haglund--Loehr-zeta map.} \label{Figure:zetaC} \end{center} \end{figure} \begin{myex}{zetaC} Recall the area vector $\mu=(2,1,0,-1,-2,1)$ and the diagonal reading word $d=[-2,1,3,4,6,5]$ of the vertically labelled lattice path shown in \reff{zetaC}, which were computed in earlier examples. We construct the individual segments of $\zeta_C(\pi)$ as follows \begin{align*} \LS_2^-(\mu)=N, \RS_2^+(\mu)=N, \LS_1^-(\mu)=EN, \RS_1^+(\mu)=ENN, \LS_0^-(\mu)=EN, \RS_0^+(\mu)=ENE. \end{align*} \end{myex} Our first main result of this section is the fact that the zeta map can be inverted using a construction reminiscent of the bounce path of a Dyck path in type $A_{n-1}$. \begin{mythrm}{zetaC} The zeta map $\zeta_C:\mathcal L_{n,n}\to\mathcal B_{2n}$ is a bijection. \end{mythrm} \begin{proof} Let $\pi\in\mathcal L_{n,n}$ be a lattice path with type $C_n$ area vector $\mu$. For each $k$ with $0\leq k\leq n$ let $\alpha_k$ denote the number of indices $i\in[n]$ such that $\abs{\mu_i}=k$. Define the \emph{bounce path} of a ballot path $\beta\in\mathcal B_{2n}$ as follows: start at the end point of $\beta$ and go South until you hit the diagonal. Bounce off it and travel to the West until you reach the upper end of a North step of $\beta$. Bounce off the path $\beta$ to the South until you hit the diagonal again, and repeat until you arrive at $(0,0)$. Now suppose that $\beta=\zeta_C(\pi)$. By definition of the zeta map the end point of $\beta$ is $(n-\alpha_0,n+\alpha_0)$. The bounce path of $\beta$ meets the diagonal for the first time in the point $(n-\alpha_0,n-\alpha_0)$, and then travels West to the point $(n-\alpha_0-\alpha_1,n-\alpha_0)$, which is the starting point of the segment $\LS_0^-(\mu)\RS_0^+(\mu)$. We claim that this point is a peak of the bounce path. To see this note that \refl{areaC}~(iii) and~(iv) imply that each non-empty segment $\LS_j^-(\mu)$ or $\RS_j^+(\mu)$ ends with a North step, except possibly $\RS_0^+(\mu)$. In particular, the starting point of any segment $\LS_j^-(\mu)\RS_j^+(\mu)$ is either $(0,0)$ or the endpoint of a North step of $\beta$. Inductively the peaks of the bounce path therefore encode the numbers $\alpha_0,\alpha_1,\dots,\alpha_n$. Knowing $\alpha_0$ and $\alpha_1$ we can recover the number and relative order of zeroes, ones and minus ones in $\mu$ from the segment $\LS_0^-(\mu)\RS_0^+(\mu)$. Since $\LS_0^-(\mu)$ ends with a North step, we first obtain the sequences $\LS_0^-(\mu)$ and $\RS_0^+(\mu)$ and thus the number occurrences of ones and minus ones in $\mu$. Moreover these paths encode the relative order of zeroes and minus ones, respectively the relative order of zeroes and ones. The relative order of ones and minus ones is implied by the following observation: If $\mu_i=1$ and $\mu_j=-1$ for some $i,j\in[n]$ with $in. \end{cases} \end{align*} Note that with our usual way of picturing diagonally labelled ballot paths, each valley is labelled by the number below it and the number to its right. \begin{myex}{risevalleyC} The vertically labelled lattice path in \reff{zetaC} has rises $2,3,4$ and $5$, which are labelled $(-5,-4),(-4,2),(2,3)$ and $(3,6)$. Moreover, its initial North step has label $1$. The image under the Haglund--Loehr-zeta map has valleys $(1,3),(2,4),(3,6),(4,7)$ and $(5,8)$, which are labelled by $(5,4),(6,3),(4,-2),(3,2)$ and $(1,-1)$. \end{myex} \begin{mythrm}{risevalleyC} Let $(\pi,u)\in\mathcal{L}_{n,n}\times\S_n^C$ be a pair of a lattice path and a signed permutation, and let $a,b\in u([n])$. Then $(\pi,u)$ has a rise labelled by $(a,b)$ if and only if $(\zeta_C(\pi),d_C(\pi,u))$ has a valley labelled by $(b,a)$ or $(-a,-b)$. Moreover $\pi$ begins with a North step labelled by $a$ if and only if $(\zeta_C(\pi),d_C(\pi,u))$ has a valley labelled by $(a,-a)$. \end{mythrm} \begin{proof} Let $\mu$ be the type $C_n$ area vector of $\pi$ and define $\sigma\in\S_n^C$ as in \refl{sigma} such that $t_{\mu}\sigma$ is Gra{\ss}mannian. Moreover let $\tau\in\S_n^C$ be defined as in \refl{wfC}. \sk \textbf{(Part 1)} We first assume that we are given a valley of $(\zeta_C(\pi),d_C(\pi,u))$ and show that there exists a fitting rise in $(\pi,u)$. A valley of $\zeta_C(\pi)$ can either occur within a sequence $\LS_k^-(\mu)$ or $\RS_k^+(\mu)$, or if $\zeta_C(\pi)$ ends with an East step. We treat these three cases independently. No valley may arise at the join of two such sequences because of \refl{areaC}~(iii) and~(iv). \sk \textbf{(1.1)} There is a valley within the sequence $\LS_k^-(\mu)$. Then there exist indices $i,j\in[n]$ with $ik+1\big\}+\#\big\{r\in[n]:r\geq i+1,\mu_r=-k-1\big\}\\ &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\geq\abs{(-k-1)\n-(i+1)}\big\}. \end{align*} By \refl{sigma} we obtain \begin{align*} n+1-x&=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_{i+1}\n-(i+1)}\big\} =\abs{\sigma^{-1}(i+1)}. \end{align*} Since $\mu_{i+1}\leq0$ we have $\sigma^{-1}(i+1)>0$ and thus \begin{align*} d_C(\pi,u)(n+1-x) =u\tau\sigma(n+1-x) =u\tau(i+1) =-u(n-i). \end{align*} Similarly $y$ equals the number of North steps in the sequence \begin{align*} \LS_n^-(\mu)\RS_n^+(\mu)\cdots\LS_{k+1}^-(\mu)\RS_{k+1}^+(\mu)\LS_k^-(\mu_i,\dots,\mu_n). \end{align*} We may rewrite this as \begin{align*} y &=\#\big\{r\in[n]:\abs{\mu_r}>k\big\}+\#\big\{r\in[n]:r\geq i,\mu_r=-k\big\}\\ &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\geq\abs{-k\n-i}\big\}. \end{align*} As before \refl{sigma} provides \begin{align*} n+1-y &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_i\n-i}\big\} =\abs{\sigma^{-1}(i)} =\sigma^{-1}(i) \end{align*} and we compute \begin{align*} d_C(\pi,u)(n+1-y) =u\tau\sigma(n+1-y) =u\tau(i) =-u(n+1-i). \end{align*} \textbf{(1.2)} The valley appears within the sequence $\RS_k^+(\mu)$. Then there exist indices $i,j\in[n]$ with $ik+1\big\}+\#\big\{r\in[n]:\mu_r=-k-1\big\}+\#\big\{r\in[i]:\mu_r=k+1\big\}\\ &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\geq\abs{(k+1)\n-i}\big\}. \end{align*} Since $\mu_i>0$ \refl{sigma} implies \begin{align*} n+1-x &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_i\n-i}\big\} =\abs{\sigma^{-1}(i)} =-\sigma^{-1}(i) \end{align*} and we compute \begin{align*} d_C(\pi,u)(n+1-x) =u\tau\sigma(n+1-x) =-u\tau(i) =u(n+1-i). \end{align*} On the other hand $y$ equals the number of North steps in the sequence \begin{align*} \LS_n^-(\mu)\RS_n^+(\mu)\cdots\LS_k^-(\mu)\RS_k^+(\mu_1,\dots,\mu_{i+1}). \end{align*} If $k>0$ then \begin{align*} y &=\#\big\{r\in[n]:\abs{\mu_r}>k\big\}+\#\big\{r\in[n]:\mu_r=-k\big\}+\#\big\{r\in[i+1]:\mu_r=k\big\}\\ &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\geq\abs{k\n-(i+1)}\big\}, \end{align*} and \refl{sigma} provides \begin{align*} n+1-y &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_{i+1}\n-(i+1)}\big\} =\abs{\sigma^{-1}(i+1)} =-\sigma^{-1}(i+1) \end{align*} In particular $y\leq n$ and we compute \begin{align*} d_C(\pi,u)(n+1-y) =u\tau\sigma(n+1-y) =-u\tau(i+1) =u(n-i). \end{align*} Otherwise $k=0$ and \begin{align*} y =n+\#\big\{r\in[i+1]:\mu_r=0\big\} =n+\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq i+1\big\} \end{align*} From \refl{sigma} we obtain \begin{align*} y-n =\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_{i+1}\n-(i+1)}\big\} =\abs{\sigma^{-1}(i+1)} =\sigma^{-1}(i+1). \end{align*} Since $y>n$ the second label of the valley is given by \begin{align*} d_C(\pi,u)(n-y) =-u\tau\sigma(y-n) =-u\tau(i+1) =u(n-i). \end{align*} \textbf{(1.3)} The path $\zeta_C(\pi)$ ends with an East step. Then there exists $i\in[n]$ such that $\mu_i=1$ and $\mu_{\ell}\notin\{0,1\}$ for all $\ell$ with $i<\ell$. From \refl{areaC}~(v) we know that $i=n$. Consequently $\lambda_1=1-\mu_n=0$ and $\pi$ begins with a North step. Let $(x,y)$ be the valley above the final East step. Then $x$ equals the number of East steps in $\zeta_C(\pi)$, that is, \begin{align*} x &=\#\big\{r\in[n]:\abs{\mu_r}>0\big\} =\#\big\{r\in[n]:\abs{\mu_r\n-r}\geq n+1\big\}\\ n+1-x &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_n\n-n}\big\} =\abs{\sigma^{-1}(n)} =-\sigma^{-1}(n) \end{align*} The valley's first label is \begin{align*} d_C(\pi,u)(n+1-x) =u\tau\sigma(n+1-x) =-u\tau(n) =u(1). \end{align*} On the other hand, $y$ equals the number of North steps of $\zeta_C(\pi)$ plus one. We have \begin{align*} y &=n+1+\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq n\big\}\\ y-n &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq n+1\big\} % =\abs{\sigma^{-1}(n)} =-\sigma^{-1}(n). \end{align*} Hence the valley's second label is \begin{align*} d_C(\pi,u)(n-y) =-u\tau\sigma(y-n) =u\tau(n) =-u(1). \end{align*} \textbf{(Part 2)} To complete the proof we need to demonstrate the reverse implication. Thus assume that $i\in[n-1]$ is a rise of $\pi$, that is, $\lambda_i=\lambda_{i+1}$. Then $\mu_{n+1-i}=i-\lambda_i$ and $\mu_{n-i}=i+1-\lambda_{i+1}=\mu_{n+1-i}+1$. If $\mu_{n-i}=-k\leq0$ then $\mu_{n+1-i}=-k-1$ and we are in the situation of~(1.1). If $\mu_{n-i}=k+1>0$ then $\mu_{n+1-i}=k$ and we are in the situation of~(1.2). Finally assume that $\pi$ begins with a North step. Then $\lambda_1=0$ and $\mu_n=1-\lambda_1=1$. Hence we are in the situation of~(1.3). \end{proof} \sk By now we have assembled all tools needed to prove that the Haglund--Loehr-zeta map $\zeta_C$ is indeed the type $C_n$ special case of the uniform bijection discussed in the \refs{uniform}. \begin{mythrm}{uniformC} Let $\Phi$ be the root system of type $C_n$ with coroot lattice $\Q$ and zeta map $\zeta$, and let $\psi$ and $\varphi$ be defined as in \refp{vertC} and \refp{diagC}. Then the following diagram commutes. % \begin{figure}[h] \begin{center} \begin{tikzpicture} \draw (0,0) node(q){$\Q/(2n+1)\Q$}; \draw (4,0) node(park){$\mathrm{Park}(\Phi)$}; \draw (0,1.4) node(v){$\op{Vert}(C_n)$}; \draw (4,1.4) node(d){$\op{Diag}(C_n)$}; \draw[->] (q)--node[below]{$\zeta$}(park); \draw[->] (v)--node[above]{$\zeta_C$}(d); \draw[->] (v)--node[left]{$\psi$}(q); \draw[->] (d)--node[right]{$\varphi$}(park); \end{tikzpicture} \end{center} % \end{figure} \end{mythrm} \begin{proof} Let $(\pi,u)\in\op{Vert}(C_n)$ and $(\beta,w)=\zeta_C(\pi,u)\in\op{Diag}(C_n)$. Moreover let $\lambda$ be the partition with South East boundary $\pi$, define $\nu$ and $\tau$ as in \refl{wfC}, and let $\mu$ be the area vector of $\pi$. Chose $\sigma\in\S_n^C$ such that $t_{\mu}\sigma$ is a Gra{\ss}mannian affine permutation. By \refp{readC} we have $w=u\tau\sigma$, which agrees with \refp{utausig}. Thus, by \refp{antichain} it suffices to show that \begin{align}\label{eq:antichainC} A_{\beta}=(\tau\sigma)^{-1}\cdot\walls % \big\{(\tau\sigma)^{-1}\cdot\alpha:\alpha\in\Delta\cup\{\tilde\alpha\}\text{ and }s_{\alpha}\cdot\lambda=\lambda\big\}. \end{align} We start out by proving that $(\tau\sigma)^{-1}\cdot\walls\subseteq A_{\beta}$. To this end let $i\in[n-1]$. Then \begin{align*} \alpha_i\in\walls &\Leftrightarrow\lambda_i=\lambda_{i+1} \Leftrightarrow\pi\text{ has a rise }i. \end{align*} % To this end let $i\in[n-1]$. Then $s_{\alpha_i}\cdot\lambda=\lambda$ if and only if $\lambda_i=\lambda_{i+1}$, that is, $\pi$ has a rise $i$. By \reft{risevalleyC} this is the case if and only if $(\beta,w)$ has a valley $(x,y)$ that is labelled by $(u(i+1),u(i))$ or $(-u(i),-u(i+1))$. Moreover, a closer look at the proof of \reft{risevalleyC} reveals that the second case only occurs if $y\leq n$. In other words \begin{align*} u(i+1)&=w(n+1-x) &\text{or} &&-u(i)&=w(n+1-x),\\ u(i)&=\begin{cases}w(n+1-y)&\quad\text{if }y\leq n,\\w(n-y)&\quad\text{if }y>n,\end{cases} &&&-u(i+1)&=w(n+1-y)\quad\text{and }y\leq n. \end{align*} Applying $u^{-1}$ to the above identities yields \begin{align*} i+1&=\tau\sigma(n+1-x) &\text{or} &&-i&=\tau\sigma(n+1-x)\\ i&= \begin{cases} \tau\sigma(n+1-y)&\quad\text{if }y\leq n,\\ \tau\sigma(n-y)&\quad\text{if }y>n, \end{cases} &&&-i-1&= \tau\sigma(n+1-y)\quad\text{and }y\leq n. \end{align*} We obtain \begin{align*} (\tau\sigma)^{-1}\cdot\alpha_i=\alpha_{x,y}\in A_{\beta}. \end{align*} Furthermore $\alpha_0\in\walls$ if and only if $\lambda_1=0$, that is, $\pi$ begins with a North step. By \reft{risevalleyC} this is equivalent to $(\beta,w)$ ending with an East step such that the corresponding valley $(x,y)$ has labels $(w(n+1-x),w(n-y))=(u(1),-u(1))$. Thus $\tau\sigma(n+1-x)=1=\tau\sigma(y-n)$ and we obtain \begin{align*} (\tau\sigma)^{-1}\cdot\alpha_0=\alpha_{x,y}\in A_{\beta}. \end{align*} Finally we have already observed that $-\tilde\alpha\notin\walls$, and the first inclusion in \refq{antichainC} follows. Conversely, let $\alpha_{x,y}\in A_{\beta}$. By similar reasoning as above it follows that $\tau\sigma\cdot\alpha_{x,y}$ is of the form $e_{i+1}-e_i$ for a rise $i$ of $\pi$, unless the valley comes from a terminal East step of $\beta$, in which case $\pi$ begins with a North step and $\tau\sigma\cdot\alpha_{x,y}=2e_1=\alpha_0$. Hence, we also have $A_{\beta}\subseteq(\tau\sigma)^{-1}\cdot\walls$ and the proof is complete. \end{proof} \subsection{The sweep map}\label{Section:sweepC} In type $A_{n-1}$ there is a generalisation of the zeta map to rational Dyck paths called the sweep map \cite{ALW2014:sweep_maps}. The concept of the sweep map is as follows. Given a path one assigns to each step a label, the labels being distinct integers. To obtain the image of a path under the sweep map, one rearranges the steps such that the labels are in increasing order. We now give a description of the zeta map of type $C_n$ similar to the sweep map on Dyck paths. Given a path $\pi=s_1s_2,\dots,s_{2n}\in\mathcal L_{n,n}$, where $s_i\in\{E,N\}$, assign a label $\ell_i$ to each step $s_i$ by setting $\ell_1=0$, and $\ell_{i+1}=\ell_i+2n+1$ if $s_i=N$, and $\ell_{i+1}=\ell_i-2n$ if $s_i=E$. Now define a collection $X$ of labelled steps as follows. If $\ell_i<0$ then add $(s_i,\ell_i)$ to $X$. If $\ell_i>0$ then add $(s_{i-1},-\ell_i)$. Finally, for the step $s_1$ which is the only step labelled by $0$, add $(s_{2n},-n)$. Thus $X$ contains $2n$ labelled steps. Now draw a path as follows. Choose $(s,\ell)\in X$ such that $\ell$ is the minimal label among all pairs in $X$. Draw the step $s$ and remove $(s,\ell)$ from $X$. Repeat until $X$ is empty. We denote the path obtained in this way by $\op{sw}(\pi)$. See \reff{sweepC}. \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.6] \begin{scope} \draw[gray] (0,0) grid (6,6); \draw[very thick] (0,0)--(0,1)--(4,1)--(4,6)--(6,6); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,1) node{\footnotesize{$13$}} (1,1) node{\footnotesize{$1$}} (2,1) node{\footnotesize{$-11$}} (3,1) node{\footnotesize{$-23$}} (4,6) node{\footnotesize{$30$}} (5,6) node{\footnotesize{$18$}}; \draw[xshift=-1mm,yshift=5mm,anchor=west] (0,0) node{\footnotesize{$0$}} (4,1) node{\footnotesize{$-35$}} (4,2) node{\footnotesize{$-22$}} (4,3) node{\footnotesize{$-9$}} (4,4) node{\footnotesize{$4$}} (4,5) node{\footnotesize{$17$}}; \end{scope} % % \begin{scope}[xshift=8cm,yshift=5cm] \begin{scope} \draw[very thick](0,0)--(1,0); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-11$}}; \end{scope} % \begin{scope}[xshift=15mm] \draw[very thick](0,0)--(1,0); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-23$}}; \end{scope} % \begin{scope}[xshift=30mm] \draw[very thick](0,0)--(0,1); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-35$}}; \end{scope} % \begin{scope}[xshift=45mm] \draw[very thick](0,0)--(0,1); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-22$}}; \end{scope} % \begin{scope}[xshift=60mm] \draw[very thick](0,0)--(0,1); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-9$}}; \end{scope} \end{scope} % % \begin{scope}[xshift=8cm,yshift=1cm] \begin{scope}[yshift=12mm] \draw[very thick](0,0)--(0,1); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-13$}}; \end{scope} % \begin{scope}[xshift=15mm,yshift=12mm] \draw[very thick](0,0)--(1,0); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-1$}}; \end{scope} % \begin{scope}[xshift=30mm,yshift=12mm] \draw[very thick](0,0)--(0,1); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-4$}}; \end{scope} % \begin{scope} \draw[very thick](0,0)--(0,1); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-17$}}; \end{scope} % \begin{scope}[xshift=15mm] \draw[very thick](0,0)--(0,1); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-30$}}; \end{scope} % \begin{scope}[xshift=30mm] \draw[very thick](0,0)--(1,0); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-18$}}; \end{scope} \end{scope} % % \begin{scope}[xshift=13.5cm,yshift=17mm] \draw[very thick](0,0)--(1,0); \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,0) node{\footnotesize{$-6$}}; \end{scope} % % \begin{scope}[xshift=17cm,yshift=-3cm] \ballot{6} \draw[very thick] (0,0)--(0,2)--(1,2)--(1,3)--(2,3)--(2,5)--(3,5)--(3,6)--(4,6)--(4,7)--(5,7); \draw[xshift=-1mm,yshift=5mm,anchor=west] (0,0) node{\footnotesize{$-35$}} (0,1) node{\footnotesize{$-30$}} (1,2) node{\footnotesize{$-22$}} (2,3) node{\footnotesize{$-17$}} (2,4) node{\footnotesize{$-13$}} (3,5) node{\footnotesize{$-9$}} (4,6) node{\footnotesize{$-4$}}; \draw[xshift=5mm,yshift=-1mm,anchor=south] (0,2) node{\footnotesize{$-23$}} (1,3) node{\footnotesize{$-18$}} (2,5) node{\footnotesize{$-11$}} (3,6) node{\footnotesize{$-6$}} (4,7) node{\footnotesize{$-1$}}; \end{scope} \end{tikzpicture} \caption{The labelling of $\pi$ (left), the set $X$ of labelled steps (middle), and the path $\op{sw}(\pi)$ of steps in increasing order (right).} \label{Figure:sweepC} \end{center} \end{figure} \begin{mythrm}{sweepC} For each lattice path $\pi\in\mathcal L_{n,n}$ we have $\op{sw}(\pi)=\zeta_C(\pi)$. In particular, the sweep map $\op{sw}:\mathcal L_{n,n}\to\mathcal B_{2n}$ is a bijection. \end{mythrm} \begin{proof} The proof consists of a straightforward but rather tedious case by case analysis of the involved labels. Let $\mu$ be the type $C_n$ area vector of a path $\pi\in\mathcal L_{n,n}$. We use the following notation. The label of the $i$-th North step of $\pi$ is denoted by $\ell_i^N$. The corresponding labelled step which is added to $X$ is denoted by $(s_i^N,x_i^N)$. We pair each North step with an East step. If the North step has a non-negative label, this is the next East step in the same diagonal. If the North step has a negative label, this is the previous East step in the same diagonal. We denote by $\ell_i^E$ the label of the East step corresponding to the $i$-th North step, and by $(s_i^E,x_i^E)$ the associated labelled step in $X$. For example in \reff{sweepC} we have $\ell_4^N=-9$, $(s_4^N,x_4^N)=(N,-9)$, $\ell_4^E=1$, and $(s_4^E,x_4^E)=(E,-1)$. Also $\ell_6^N=17$, $(s_6^N,x_6^N)=(N,-17)$, $\ell_6^E=30$, and $(s_6^E,x_6^E)=(N,-30)$. \sk The label of the $i$-th North step is \begin{align} \label{eq:liN} \ell_i^N=(i-1)(2n+1)-(i-\mu_{n-i+1})(2n)=2n(\mu_{n-i+1}-1)+i-1. \end{align} First consider the case $\mu_{n-i+1}>0$. Then \begin{align*} %\label{eq:liNpos} 2n|\mu_{n-i+1}|-2n\leq\ell_i^N<2n|\mu_{n-i+1}|-n. \end{align*} If $i>1$ then $x_i^N=-\ell_i^N$. If $i=1$ then $x_i^N=-n$. Hence \begin{align*} %\label{eq:xiNpos} -2n|\mu_{n-i+1}|+n\leq x_i^N<-2n|\mu_{n-i+1}|+2n. \end{align*} On the other hand, if $\mu_{n-i+1}\leq0$ then \begin{align*} %\label{eq:xiNneg} -2n|\mu_{n-i+1}|-2n\leq\ell_i^N=x_i^N<-2n|\mu_{n-i+1}|-n. \end{align*} Now, let us treat the East steps. We start with the case $\mu_{n-i+1}>0$. Then \begin{align*} %\label{eq:liEpos} \ell_i^E=\ell_i^N+2n+k_i=2n|\mu_{n-i+1}|+i-1+k_i, \end{align*} for some $k_i\in[n-i+1]$. Since $x_i^E=-\ell_i^E$, we obtain \begin{align} \label{eq:xiEpos} -2n|\mu_{n-i+1}|-n\leq x_i^E<-2n|\mu_{n-i+1}|. \end{align} If $\mu_{n-i+1}\leq0$ then \begin{align*} % \label{eq:liEneg} \ell_i^E=\ell_i^N+2n-k_i=-2n|\mu_{n-i+1}|+i-1-k_i, \end{align*} for some $k_i\in\{0,\dots,i-1\}$. If $\mu_{n+i-1}<0$ then \begin{align*} %\label{eq:xiEneg} -2n|\mu_{n-i+1}|\leq\ell_i^E=x_i^E<-2n|\mu_{n-i+1}|+n. \end{align*} Finally assume $\mu_{n-i+1}=0$. % \begin{align*} % %\label{eq:liE0} % \ell_i^E=\ell_i^N+2n-k_i=i-1-k_i, % \end{align*} % for some $k_i\in\{0,\dots,i-1\}$. If $\ell_i^E=0$ then $x_i^E=-n$. Otherwise $x_i^E=-\ell_i^E$. Combined this yields \begin{align*} %\label{eq:xiE0} -n\leq x_i^E<0, \end{align*} which is a special case of \refq{xiEpos}. \sk We make the following observation. If $-2nk\leq x_i^s<-2nk+n$, where $k=1,2,\dots$, then either $s=E$ and $\mu_{n-i+1}=-k$, or $s=N$ and $\mu_{n-i+1}=-k+1$. Similarly, if $-2nk+n\leq x_i^s<-2nk+2n$, where $k=1,2,\dots$, then either $s=N$ and $\mu_{n-i+1}=k$ or $s=E$ and $\mu_{n-i+1}=k-1$. Thus by definition the path $\op{sw}(\pi)$ is composed of segments $T(-k,-k+1)$, $T(k-1,k)$, where $k=1,2,\dots$, such that each step of $T(-k,-k+1)$ corresponds to an entry of the area vector $\mu_{n-i+1}\in\{-k,-k+1\}$, and each step of $T(k-1,k)$ corresponds an entry $\mu_{n-i+1}\in\{k-1,k\}$. This is a good sign because the path $\zeta_C(\pi)$ is also composed of segments with the same property. Indeed we will carry out the proof by showing that $T(-k,-k+1)=\LS_{k-1}^-(\mu)$ and $T(k-1,k)=\RS_{k-1}^+(\mu)$. \sk We first prove that $T(-k,-k+1)=\LS_{k-1}^-(\mu)$ for $k\geq1$. As $\mu_{n-i+1}=-k<0$ implies $x_i^E=\ell_i^E<0$ and $\mu_{n-i+1}=-k+1\leq0$ implies $x_i^N=\ell_i^N<0$, we have $s_i^E=E$ and $s_j^N=N$. That is, every entry $\mu_{n-i+1}=-k$ will contribute an East step while each entry $\mu_{n-i+1}=-k+1$ contributes a North step. This is consistent with the definition of the zeta map. Therefore, it suffices to check that $i1$ implies $\ell_i^N>0$ and $\mu_{n-i+1}=k-1>0$ implies $\ell_i^E>0$. This case is more difficult (confusing) because we do not necessarily have $s_i^E=E$ and $s_j^N=N$. Instead, if $\mu_{n-i+1}=k-1$ and $\mu_{n-j+1}=k$ then \begin{align*} s_i^E= \begin{cases} E\quad\text{if }\mu_{n-i}=k,\\ N\quad\text{if }\mu_{n-i}\leq k-1\text{ or }i=n, \end{cases} && s_j^N= \begin{cases} N\quad\text{if }\mu_{n-j+2}=k-1,\\ E\quad\text{if }\mu_{n-j+2}\geq k. \end{cases} \end{align*} Thus an entry $\mu_{n-i+1}=k-1$ contributes an East step instead of a North step if and only if the previous entry $\mu_{n-i}=k$ contributes a North step instead of an East step. We see that the number of East and North steps in $T(k-1,k)$ is consistent with the definition of the zeta map. To see that also the relative orders of the steps in $T(k-1,k)$ and $\RS_{k-1}^+(\mu)$ agree, it suffices to prove that $ix_j^N\qquad\text{if }\mu_{n-j+1}=k,\mu_{n-i+1}=k-1,i+1x_{i+1}^N\qquad\text{if }\mu_{n-i}=k,\mu_{n-i+1}=k-1. \end{align} Moreover, let $i-2n(k-1)-j+1-k_j=x_j^E \end{align*} The inequalities \refq{xiNxjNpos} and \refq{xiNxjEpos} are trivial as \begin{align*} x_i^N=-2n(k-1)-i+1&> \begin{cases} -2n(k-1)-j+1=x_j^N\\ -2n(k-1)-j+1-k_j=x_j^E. \end{cases} \end{align*} To see claim \refq{xiExjNpos}, first assume that there is no $r$ with $i-2n(k-1)-j+1=x_j^N. \end{align*} Finally, we need to show $T(0,1)=\RS_{0}^+(\mu)$. Let $\mu_{n-j+1}=0$ and assume $\ell_j^E\neq0$. Then $x_j^E=-j+1+k_j$. Choose $i$ maximal such that $i1$. Then $x_j^N=-j+1$. Choose $i$ maximal such that $i1,\\ %\label{eq:xiExjEpos} x_j^E&1,\\ % \label{eq:xiNxjEpos} x_j^E&1,\\ % \label{eq:xiExjNpos} x_j^N&0 \Leftrightarrow\pi_1=\pi_2\text{ and }\epsilon(\pi)=1\\ &\Rightarrow v(1)0\text{ and }\epsilon(\pi)=1 &\Leftrightarrow\lambda_1=\lambda_2>0 \Leftrightarrow S\cap\walls=\{e_2-e_1\}\\ &\Rightarrow u\cdot(e_2-e_1)\in\Phi^+\text{ and }\epsilon(\pi)=1\\ &\Leftrightarrow u(1)0\text{ and }\epsilon(\pi)=-1 &\Leftrightarrow\lambda_1=-\lambda_2>0 \Leftrightarrow S\cap\walls=\{e_2+e_1\}\\ &\Rightarrow u\cdot(e_2+e_1)\in\Phi^+\text{ and }\epsilon(\pi)=-1\\ &\Leftrightarrow -u(1) \begin{cases} w(n+1-j) &\quad\text{if }j\leq n-1,\\ \epsilon(\beta)w(1) &\quad\text{if }j=n,\\ -\epsilon(\beta)w(1) &\quad\text{if }j=n+1,\\ w(n-j) &\quad\text{if }j\geq n+2, \end{cases} \end{align*} and such that $w(n+1-i)>\abs{w(1)}$ if $\beta$ has a valley $(i,n)$ and the $n$-th North step of $\beta$ is not followed by an East step. We denote the set of all diagonally labelled signed ballot paths by $\op{Diag}(D_n)$. \end{mydef} \sk Hence, if we place the labels $w(i)$, where $i=n,n-1,\dots,2,\epsilon(\beta),-\epsilon(\beta),-2,\dots,-n$, in the diagonal then for each valley the label to its right has to be smaller than the label below it. \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.6] \begin{scope} \ballotodd{5} \draw[very thick](0,0)--(0,2)--(1,2)--(1,4)--(2,4)--(2,6)--(3,6); \draw[xshift=5mm,yshift=5mm] (0,0)node{$2$} (1,1)node{$4$} (2,2)node{$-1$} (3,3)node{$5$} (4,4)node{$-3$} (4,5)node{$3$} (3,6)node{$-5$} (2,7)node{$1$} (1,8)node{$-4$} (0,9)node{$-2$} (0,2)node{$\bullet$} (1,4)node{$*$} % (1,5)node{$*$} (2,6)node{$\circ$}; \end{scope} % \begin{scope}[xshift=7cm,yshift=-1cm] \ballotodd{6} \draw[very thick](0,0)--(0,6)--(1,6)--(1,9)--(2,9); \ballotsign{0}{6}{\ominus} \draw[xshift=5mm,yshift=5mm] (0,0)node{$-1$} (1,1)node{$4$} (2,2)node{$6$} (3,3)node{$-5$} (4,4)node{$-2$} (5,5)node{$3$} (5,6)node{$-3$} (4,7)node{$2$} (3,8)node{$5$} (2,9)node{$-6$} (1,10)node{$-4$} (0,11)node{$1$} (0,6)node{$\bullet$} (1,9)node{$\circ$}; \end{scope} % \begin{scope}[xshift=15cm] \ballotodd{5} \draw[very thick](0,0)--(0,1)--(1,1)--(1,3)--(2,3)--(2,4)--(3,4)--(3,5)--(4,5); \ballotsign{3}{5}{\ominus} \draw[xshift=5mm,yshift=5mm] (0,0)node{$5$} (1,1)node{$4$} (2,2)node{$3$} (3,3)node{$-1$} (4,4)node{$2$} (4,5)node{$-2$} (3,6)node{$1$} (2,7)node{$-3$} (1,8)node{$-4$} (0,9)node{$-5$} (0,1)node{$\bullet$} (1,3)node{$\bullet$} (2,4)node{$\bullet$} (3,5)node{$\circ$}; \end{scope} % \end{tikzpicture} \caption{Three diagonally labelled signed ballot paths.} \label{Figure:diagD} \end{center} \end{figure} \begin{myex}{diagD} Let us return to the diagonally labelled ballot paths of \reff{diagD}. First consider the leftmost path. From $\epsilon(\beta)=1$ we obtain $w=[-3,5,-1,4,2]\in\S_5^D$. We can confirm that $w$ fulfils the conditions $w(6-1)=2>w(6-3)=-1$ and $w(6-3)=-1>w(5-7)=-5$, which are imposed by the valleys $(1,3)$ and $(3,7)$. Moreover we have $w(6-2)=4>\abs{w(1)}=3$ in accordance with the special valley $(2,5)$. For the middle path we find $\epsilon(\beta)=-1$ and $w=[-3,-2,-5,6,4,-1]\in\S_6^D$. Note that $w(7-1)=-1>-\epsilon(\beta)w(1)=w(1)=-3$ as it is required by the presence of a valley $(1,7)=(1,n+1)$. In case of the rightmost path we find $\epsilon(\beta)=-1$ and thus $w=[-2,-1,3,4,5]$. We verify that $w(5)=5>w(4)=4>w(2)=-1>-\epsilon(\beta)w(1)=-2$ and $w(3)=3>\epsilon(\beta)w(1)=2$. % % Consider the signed ballot path $\beta=NENENENENENNNNE^+NN\in\bd_{17}$ together with a signed permutation $w\in\S_n^D$ with an even number of sign changes. % \sk % \begin{center} % \begin{tikzpicture}[scale=.5] % \draw[gray] % (0,9)--(8,9) % (0,8)--(8,8)--(8,10)--(0,10) % (0,7)--(7,7)--(7,11)--(0,11) % (0,6)--(6,6)--(6,12)--(0,12)--(0,0) % (0,5)--(5,5)--(5,12) % (0,4)--(4,4)--(4,12) % (0,3)--(3,3)--(3,12) % (0,2)--(2,2)--(2,12) % (0,1)--(1,1)--(1,12); % \draw[very thick] (0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(2,3)--(3,3)--(3,4)--(4,4)--(4,5)--(5,5)--(5,9)--(6,9)--(6,11); % \draw[xshift=5mm] (5,9) node[circle,fill=white,inner sep=-1.3pt]{$\oplus$}; % \draw[xshift=5mm,yshift=5mm] % (0,0) node {$w_9$} % (1,1) node {$w_8$} % (2,2) node {$w_7$} % (3,3) node {$w_6$} % (4,4) node {$w_5$} % (5,5) node {$w_4$} % (6,6) node {$w_3$} % (7,7) node {$w_2$} % (0,1) node{$\bullet$} % (1,2) node{$\bullet$} % (2,3) node{$\bullet$} % (3,4) node{$\bullet$} % (4,5) node{$\bullet$} % (5,9) node{$\bullet$}; % \draw[yshift=5mm,anchor=west] % (8,8) node{$\epsilon(\pi)w_1$} % (8,9) node{$-\epsilon(\pi)w_1$} % (7,10) node{$-w_2$} % (6,11) node{$-w_3$}; % \end{tikzpicture} % \end{center} % The valleys of the ballot path $\beta$ correspond to the antichain % \begin{align*} % A_{\beta}=\big\{e_9-e_8,e_8-e_7,e_7-e_6,e_6-e_5,e_5-e_4,e_4+e_1\big\}\subseteq\Phi^+ % \end{align*} % Note that replacing the positive East step $E^+$ in $\beta$ by a negative East step corresponds to replacing the root $e_4+e_1$ by the root $e_4-e_1$ in $A_{\beta}$. % % For $(\beta,w)$ to by a diagonal labelling we need to have $w_9>w_8>w_7>w_6>w_5>w_4>-w_1$. \end{myex} \sk We conclude this section by extending \refp{AbetaD} to a bijection between diagonally labelled signed ballot paths an non-nesting parking functions. \begin{myprop}{diagD} The map $\varphi:\op{Diag}(D_n)\to\op{Park}(D_n)$ given by $(\beta,w)\mapsto[w,A_{\beta}]$ is a bijection. \end{myprop} \begin{proof} Let $(\beta,w)\in\op{Diag}(D_n)$ be a diagonally labelled signed ballot path. In order to prove that $\varphi$ is injective, it suffices to show that $w$ is the canonical coset representative of $[w,A_{\beta}]$ described in \refl{nnpark}, that is, $w\cdot A_{\beta}\subseteq\Phi^+$. This is a simple consequence of the definitions, for example \begin{align*} \alpha_{i,n}=e_{n+1-i}-\epsilon(\beta)e_1\in A_{\beta} &\Leftrightarrow(i,n)\text{ is a valley of }\beta\\ &\Rightarrow w(n+1-i)>\epsilon(\beta)w(1) \Leftrightarrow w(\alpha_{i,n})\in\Phi^+. \end{align*} Conversely suppose that $[w,A_{\beta}]\in\op{Park}(D_n)$, where $w\in\S_n^D$ is chosen such that $w\cdot A_{\beta}\subseteq\Phi^+$. Then it is not difficult to show that $(w,\beta)$ is a diagonal labelling, and thus $\varphi$ is surjective. For example, \begin{align*} (i,n)\text{ is a valley of }\beta &\Leftrightarrow\alpha_{i,n}\in A_{\beta}\\ &\Rightarrow w(\alpha_{i,n})\in\Phi^+ \Leftrightarrow w(n+1-i)>\epsilon(\beta)w(1).\qedhere \end{align*} \end{proof} \subsection{The area vector}\label{Section:areaD} Let $\pi\in\ld_{n-1,n}$ be a signed lattice path and define $\lambda$ as in \refp{lambdaD}. Recall that $\lambda$ represents an orbit of the finite torus under the action of the Weyl group, and therefore corresponds to a dominant Shi region with minimal alcove $\waf_D\ac$ via the Anderson map of \refs{torus}. The main goal of this section is to recover the element $\waf_D$ of the affine Weyl group from $\pi$. As before we write $\waf_D^{-1}=t_{\mu}\sigma$ where $\mu\in\Q$ and $\sigma\in\S_n^D$. Since $\waf_D^{-1}$ is Gra{\ss}mannian, the signed permutation $\sigma$ is given by \refl{sigma} once we know $\mu$. We will see shortly that $\mu$ can be interpreted as an area vector of the signed lattice path $\pi$, similar to the area vector of a Dyck path in type $A_{n-1}$ or a lattice path in type $C_n$. Recall from \refs{uniform} that $\mu$ appears implicitly in the identity $\lambda=\waf_f\waf_D^{-1}\cdot0$. To find $\mu$ we need an explicit description of $w_f$. \begin{mylem}{wfD}We have $w_f=t_{\nu}\tau$, where $\nu\in\Q$ and $\tau\in\S_n^D$ are given by \begin{align*} % \label{eq:tauB} \nu= \begin{cases} (0,1,\dots,n-2,n-1)\\ (0,1,\dots,n-2,n) \end{cases} && \tau= \begin{cases} [1,2,\dots,n-1,n]\\ [-1,2,\dots,n-1,-n] \end{cases} && \begin{matrix*}[l] \text{if }n-1\equiv0,3\mod4,\\ \text{if }n-1\equiv1,2\mod4. \end{matrix*} \end{align*} \end{mylem} \begin{proof} By \refl{wf} it suffices to show that $t_{\nu}\tau(\widetilde\Delta)=(\Delta-\delta)\cup\{-\tilde\alpha+h\delta\}$. Suppose $n-1\equiv0,3$ modulo $4$, then \begin{align*} t_{\nu}\tau(\alpha_i) &=\alpha_i-\skal{\tau,\alpha_i}\delta =\alpha_i-\delta &&\text{for all }0\leq i\leq n-1,\\ % t_{\nu}\tau(-\tilde\alpha+\delta) &=-\tilde\alpha+(1+\skal{\tau,\tilde\alpha})\delta =-\tilde\alpha+(2n-2)\delta. \end{align*} If $n-1\equiv1,2$ modulo $4$, then four cases are different: \begin{align*} t_{\nu}\tau(\alpha_0) &=\alpha_1-\delta,\\ % t_{\nu}\tau(\alpha_1) &=\alpha_0-\delta,\\ % t_{\nu}\tau(\alpha_{n-1}) &=-\tilde\alpha+\skal{\tau,\tilde\alpha}\delta =-\tilde\alpha+(2n-2)\delta &&\text{and}\\ % t_{\nu}\tau(-\tilde\alpha+\delta) &=\alpha_{n-1}+(1-\skal{\tau,\alpha_{n-1}})\delta =\alpha_{n-1}-\delta.&&\qedhere \end{align*} % % % We know that $w_f$ maps the walls of the Sommers region to the walls of the $2n-1$ dilated fundamental alcove. That is % \begin{align*} % w_f\big(\big\{-\alpha_i+\delta:0\leq i\leq n-1\big\}\cup\big\{\tilde\alpha+2\delta\big\}\big) % =\big\{-\alpha_i:0\leq i\leq n-1\big\}\cup\big\{\tilde\alpha+(2n-1)\delta\big\}. % \end{align*} % Since $t_{\nu}\tau\in\affS_n^D$ it suffices to show that $t_{\nu}\tau$ has the same property. \end{proof} \sk For example in dimensions $n=5$ and $n=6$ we have \begin{align*} w_f=[1,-9,-19,-29,-39]\qquad\text{and}\qquad w_f=[-1,-11,-23,-35,-47,\,72]. \end{align*} \begin{mydef}{areaD} Let $\pi\in\ld_{n-1,n}$ be a signed lattice path, and define $\lambda$ as in \refp{lambdaD}, and $\nu\in\Q$ and $\tau\in\S_n^D$ as in \refl{wfD}. We define the type $D_n$ \emph{area vector} of $\pi$ as $\mu=\tau(\lambda-\nu)$. \end{mydef} \sk Indeed note that for $10$ and $\mu_{\ell}\notin\{\mu_i-1,\mu_i\}$ for all $\ell$ with $i<\ell\leq n$. Then $i=n$. \myi{D(vi)} If $\mu_n>0$ then there exists $i\in[n]$ with $\mu_i\in\{-\mu_n+1,-\mu_n+2\}$. \myi{D(vii)} Assume $\mu_n>0$ and let $i\in[n-1]$ such that $\mu_i=-\mu_n+2$ and $\mu_{\ell}\notin\{-\mu_n+1,-\mu_n+2\}$ for all $\ell$ with $i<\ell2$ then $0\leq\lambda_2<\mu_i+1$ implies $\mu_i=0$, $\lambda_2=0$ and $\mu_1=\lambda_1=0$, which is a contradiction. If $i=2$ and $\mu_2=-1$ then $\lambda_2=0$ and $\mu_1=\lambda_1=0$, which is again a contradiction. Hence either $i=1$ or $i=2$ and $\mu_2=0$ as claimed. To see claim~(iii) assume that $\mu_i\notin\{-\mu_1-1,-\mu_1\}$ for all $i\in[n]$. From $\lambda_i\geq\abs{\lambda_{i-1}}$ we obtain $\lambda_i>i-1-\mu_1\geq i$ for all $i$ with $1\ell-1-\mu_1$ for all $\ell$ with $1<\elli-2-\mu_1=\lambda_i$ which is a contradiction. If $i=n>2$ then $\lambda_{n-1}\geq n$ is the same contradiction as in the proof of (iii). Thus $i=2$ by elimination. In the situation of~(v) we have $\lambda_{\ell}>\ell-1+\mu_i\geq\ell$ for all $\ell$ with $i<\ell0$ then % \begin{align*} % \sigma^{-1}(-i) % =\#\big\{k:1\leq k\leq n\text{ and }\abs{\n\mu_k-k}\leq\abs{\n\mu_i-i}\big\}. % \end{align*} % \end{mylem} % \begin{proof} The alcove $w_DA_{\circ}$ lies in the dominant chamber if and only if the window of $w_D^{-1}$ is increasing and positive, that is % \begin{align*} % 00$ if and only if $-\n\mu_i+i>0$ if and only if $\mu_i\leq0$. % \end{proof} \subsection{The diagonal reading word}\label{Section:readD} Let $(\pi,v)\in\op{Vert}(D_n)$ be a vertically labelled signed lattice path. Recall that $(\pi,v)$ corresponds to an element $u\cdot\lambda+(2n-1)\Q$ of the finite torus by \refp{vertD}, and hence to a region of the Shi arrangement with minimal alcove $\waf_R\ac$ by \reft{anderson}. Decompose $\waf_R=w\waf_D$ such that $w\in \S_n^D$ and $\waf_D\ac$ is the dominant Shi alcove corresponding to $\pi$ as in the previous section. Our aim in this section is to recover the signed permutation $w$, and thereby the element $\waf_R$ of the affine Weyl group, from $(\pi,v)$. To this end let $u,\tau\in\S_n^D$ be defined as in \refp{vertD} and \refl{wfD} respectively. Recall that $w=u\tau\sigma$ by \refp{utausig}. Consequently one can obtain $w$ by computing $\sigma$ from $\pi$ as in the previous section. We show that $w$ can also be read off the vertically labelled signed lattice path $(\pi,v)$ in similar fashion as in type $C_n$. \begin{mydef}{readD} Let $(\pi,v)\in\ld_{n-1,n}\times\S_n^B$, $\mu$ be the type $D_n$ area vector of $\pi$ and define $\nu$ as in \refl{wfD} above. Define the type $D_n$ \emph{diagonal reading word} $d_D(\pi,v)$ as follows: For each $i=0,1,2,\dots$ first write down the labels $v(j)$ of the rows with $\mu_j=-i$ from bottom to top, then write down the negative labels $-v(j)$ of rows with $\mu_j=i+1$ from top to bottom. Finally we need to adjust some signs: Multiply the label coming from the top row by $(-1)^{1+\mu_{n-1}+\mu_n}$ and the label coming from the bottom row by $\epsilon(\pi)(-1)^{1+\nu_{n-1}+\nu_n}$. Then change the sign of $d_D(\pi,v)(1)$ if the number of positive entries of $\mu$ is odd. \end{mydef} \sk Except for some necessary twists, the definition of the diagonal reading word of type $D_n$ follows very similar rules as the counterparts in types $A_{n-1}$ and $C_n$. Note that $d_D(\pi,v)\in\S_n^D$, that is, the diagonal reading word has an even number of sign changes. \begin{myex}{readD} Let us return to the vertically labelled signed lattice paths from \reff{vertD}. The first path has labels $v=[-3,4,-2,1,5]$ and area vector $\mu=(0,-1,2,1,0)$. Reading off the labels according to the entries of the area vector yields the signed permutation $[-3,5,-1,4,2]$. As $(-1)^{1+\mu_{n-1}+\mu_n}=1$ and $\epsilon(\pi)(-1)^{1+\nu_{n-1}+\nu_n}=1\cdot1=1$ and since $\mu$ has two positive entries, which is an even number, all signs remain unchanged. We obtain $d_D(\pi,v)=[-3,5,-1,4,2]$. In case of the second path we have $v=[1,3,-2,-5,-4,6]$ and $\mu=[-1,0,0,0,1,0]$. Reading off the labels according to the entries of the area vector yields the signed permutation $[3,-2,-5,6,4,1]$. We have $(-1)^{1+\mu_{n-1}+\mu_n}=1$, however, $\epsilon(\pi)(-1)^{1+\nu_{n-1}+\nu_n}=1\cdot(-1)^{1+4+6}=-1$. Thus we need to change the sign of the label coming from the bottom row from $1$ to $-1$. Additionally $\mu$ has an odd number of positive entries, namely a single one. Consequently we change $3$ to $-3$. We obtain $d_D(\pi,v)=[-3,-2,-5,6,4,-1]$. The third path has labels $v=[-5,-4,1,2,3]$ and area vector $\mu=(-3,2,1,0,2)$. Reading off the labels according to the entries of the area vector yields the signed permutation $[2,-1,-3,4,-5]$. Since $(-1)^{1+\mu_{n-1}+\mu_n}=-1$ and $\epsilon(\pi)(-1)^{1+\nu_{n-1}+\nu_n}=(-1)\cdot1$ and $\mu$ has three positive entries, all three signs have to be changed. We obtain $d_D(\pi,v)=[-2,-1,3,4,5]$. \sk In light of the next following proposition we compute that for the second path we have \begin{align*} d_D(\pi,v) % =[-3,-2,-5,6,4,-1] =[1,3,-2,-5,-4,-6]\cdot[-1,2,3,4,5,-6]\cdot[-2,3,4,6,-5,1] =u\tau\sigma, \end{align*} where the respective signed permutations are taken from \refe{vertD} and \refe{areaD}. % \sk % \begin{center} % \begin{tikzpicture}[scale=.5] % \draw[gray] (0,0) grid (8,9); % \draw[very thick] (0,0)--(6,0)--(6,7)--(7,7)--(7,8)--(8,8)--(8,9); % \draw[xshift=5mm] (0,0) node[circle,fill=white,inner sep=-1.3pt]{$\ominus$}; % \draw[xshift=5mm,yshift=5mm] % (5,0) node{$v_1$} % (5,1) node{$v_2$} % (5,2) node{$v_3$} % (5,3) node{$v_4$} % (5,4) node{$v_5$} % (5,5) node{$v_6$} % (5,6) node{$v_7$} % (6,7) node{$v_8$} % (7,8) node{$v_9$}; % \end{tikzpicture} % \end{center} % In \refe{areaD} we computed the area vector $\mu=(-6,5,4,3,2,1,0,0,1)$. Reading the labels $\pm v_j$ in the correct order we obtain the signed permutation $[v_7,v_8,-v_9,-v_6,-v_5,-v_4,-v_3,-v_2,v_1]$. Since $\mu_{n-1}+\mu_n=0+1=1$ is odd we do not change the sign of $v_9$. We replace $v_1$ by $\epsilon(\pi)(-1)^{1+7+8}v_1=(-1)\cdot1\cdot v_1=-v_1$. Finally, since $\mu$ has six positive entries we do not change the sign of $v_7$. % % We obtain $d_D(\pi,v)=[v_7,v_8,-v_9,-v_6,-v_5,-v_4,-v_3,-v_2,-v_1]$, which lies in $\S_n^D$ because $v$ itself has an odd number of sign changes. \end{myex} \sk The following proposition asserts that our definition always yields the desired signed permutation. \begin{myprop}{readD} Let $(\pi,v)\in\op{Vert}(D_n)$ be a vertically labelled signed lattice path with area vector $\mu$ and chose $\sigma\in\S_n^D$ as in \refl{sigma} such that $t_{\mu}\sigma$ is Gra{\ss}mannian. Moreover let $u,\tau\in\S_n^D$ be defined as in \refp{vertD} and \refl{wfD} respectively. Then $d_D(\pi,v)=u\tau\sigma$. \end{myprop} \begin{proof} %Define a signed permutation $\tau_2\in\S_n^B$ by setting $\tau_2(i)=i$ for $10\big\} \notag\\ &\quad+\#\big\{r\in[j]:\mu_r=\mu_j\leq0\big\} +\#\big\{r\in[n]:\mu_r=-\mu_j>0\big\} \end{align*} Comparing this to the definition of $d_D(\pi,v)$ we obtain \begin{align*} \abs{d_D(\pi,v)(i)} =\abs{v(j)} =\abs{v\sigma(i)} =\abs{u\tau\sigma(i)}. \end{align*} Comparing \refl{sigma} to the description of $u$ in \refq{uD}, $\tau$ in \refl{wfD} and the definition of the diagonal reading word, one can check that all signs work out and we may indeed drop the absolute value in the above identity. % \begin{align*} % d_D(\pi,v)(i)=v\tau_2\tau\sigma(i). % \end{align*} % Assume first that $i>1$. If $\abs{\sigma(i)}\notin\{1,n\}$ then $\tau_2\tau\sigma(i)=\sigma(i)<0$ if and only if $\mu_i>0$, which is the case if and only if $d_D(\pi,v)(i)=-v(j)$. If $\abs{\sigma(i)}=1$ and $\mu_i\leq0$ then % \begin{align*} % v\tau_2\tau\sigma(i) % &=v\tau_2\tau(1)\\ % &=v\tau_2((-1)^{1+\nu_{n-1+\nu_n}})\\ % &=v(\epsilon(\pi)(-1)^{1+\nu_{n-1+\nu_n}})\\ % &=d_D(\pi,v). % \end{align*} % On the other hand if $\abs{\sigma(i)}=n$ then $\sigma(i)<0$ if and only if $\mu_i>0$ but $\tau_2$ and $\tau$ may change the sign of $n$ such that % \begin{align*} % \tau_2\tau\sigma(i) % &=(-1)^{\sum_{i=1}^{n-2}\pi_i}(-1)^{n(n+1)/2}\sigma(i) \\ % &=(-1)^{\lambda_{n-1}+\lambda_n}(-1)^{1+\nu_{n-1}+\nu_n}\sigma(i) \\ % &=(-1)^{1+\mu_{n-1}+\mu_n}\sigma(i). % \end{align*} % This again fits the definition of the diagonal reading word and we obtain $d_D(\pi,v)=v\tau_2\tau\sigma$. \end{proof} \subsection{The zeta map}\label{Section:zetaD} We are now in a position to present the combinatorial definition of the (Haglund--Loehr) zeta map of type $D_n$. \begin{mydef}{zetaD} Let $\pi\in\ld_{n-1,n}$ be a signed lattice path with area vector $\mu$. Its image $\zeta_D(\pi)$ under the type $D_n$ \emph{zeta map} is obtained from % Let $a=(a_1,a_2,\dots,a_n)$ be a vector. Define a word $\RS_j^+(a)$ in the alphabet $\{E,N\}$ as follows: Read $a$ from left to right. Whenever you encounter an entry $a_i=j$ write down $N$. Whenever you encounter an entry $a_i=j+1$ write down $E$. Similarly we define $\LS_j^+(a)$ except now one reads $a$ from right to left. Moreover we define $\RS_j^-(a)$ as follows: Read $a$ from right to left and write down $N$ for each $-j$ you encounter and $E$ for each $-j-1$ you encounter. Define $\LS_j^-(a)$ analogously. \begin{align*} \LS_{2n-1}^-(\mu)\RS_{2n-1}^+(\mu) \LS_{2n-2}^-(\mu)\RS_{2n-2}^+(\mu)\cdots \LS_1^-(\mu)\RS_1^+(\mu) \LS_0^-(\mu)\RS_0^+(\mu) \end{align*} by deleting the last letter and, if the $n$-th North step is followed by an East step, adding a sign to this East step such that \begin{align*} \epsilon(\zeta_D(\pi))=(-1)^{\#\{r\in[n]:\mu_r>0\}}. \end{align*} Moreover, given a vertically labelled signed lattice path $(\pi,v)\in\op{Vert}(D_n)$ we define its image under the \emph{Haglund--Loehr-zeta map} of type $D_n$ as \begin{align*} \zeta_D(\pi,v)=(\zeta_D(\pi),d_D(\pi,v)). \end{align*} \end{mydef} \sk Note that by definition $\zeta_D(\pi)$ is a signed ballot path, that is, $\zeta_D:\ld_{n-1,n}\to\bd_{2n-1}$. \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.6] % \begin{scope}[yshift=2cm] \draw[gray](0,0)grid(4,5); \draw[very thick](0,0)--(0,2)--(4,2)--(4,5); \draw[xshift=5mm,yshift=5mm] (-1,0)node{$-3$} (-1,1)node{$4$} (3,2)node{$-2$} (3,3)node{$1$} (3,4)node{$5$}; \end{scope} \draw[xshift=5cm,yshift=4cm,thick,->](0,0)--node[above]{$\zeta_D$}(1,0); % \begin{scope}[xshift=7cm] \ballotodd{5} \draw[very thick](0,0)--(0,2)--(1,2)--(1,4)--(2,4)--(2,6)--(3,6); \draw[very thick,red] (0,1)--(0,2) (1,3)--(1,4)--(2,4)--(2,5); \draw[xshift=5mm,yshift=5mm] (0,0)node{$2$} (1,1)node{$4$} (2,2)node{$-1$} (3,3)node{$5$} (4,4)node{$-3$} (4,5)node{$3$} (3,6)node{$-5$} (2,7)node{$1$} (1,8)node{$-4$} (0,9)node{$-2$} (0,2)node{$\bullet$} (1,4)node{$*$} % (1,5)node{$*$} (2,6)node{$\circ$}; \end{scope} % \end{tikzpicture} \caption{} \label{Figure:zetaD1} \end{center} \end{figure} \begin{myex}{zetaD} For a first example consider \reff{zetaD1}. Recall that $\mu=(0,-1,2,1,0)$. We obtain \begin{align*} \RS_2^+(\mu)=N, \LS_1^-(\mu)=N, \RS_1^+(\mu)=EN, \LS_0^-(\mu)=NEN, \RS_0^+(\mu)=NEN, \end{align*} which combines to $\beta=\zeta_D(\pi)=NNENNENNE$. The last North step of $\RS_0^+(\mu)$ is deleted. Note that in this example $\pi$ begins with two North steps, which translates into a valley of $\beta$ of the special form $(x,n)$, where the $n$-th North step of $\beta$ is followed by a North step. The condition $\abs{v(1)}=3\abs{w(1)}=3$ on $(\beta,w)$. We shall later see that this is always the case. The signed lattice path in \reff{zetaD2} has area vector $\mu=(-1,0,0,0,1,0)$. We compute \begin{align*} \LS_1^-(\mu)=N, \RS_1^+(\mu)=N, \LS_0^-(\mu)=NNNNE, \RS_0^+(\mu)=NNNEN. \end{align*} The final North step of $\RS_0^+(\mu)$ is deleted. Moreover the sixth North step is followed by an East step. Since $\mu$ has an odd number of positive entries we have $\epsilon(\beta)=-1$, and this East step is replaced by $E^-$. We conclude $\zeta_D(\pi)=NNNNNNE^-NNNE$. The signed lattice path in \reff{zetaD3} has area vector $\mu=(-3,2,1,0,2)$. We compute \begin{align*} \LS_3^-(\mu)=N, \LS_2^-(\mu)=E, \RS_2^+(\mu)=NN, \RS_1^+(\mu)=ENE, \LS_0^-(\mu)=N, \RS_0^+(\mu)=EN. \end{align*} As above the final step of $\RS_0^+(\mu)$ is deleted, and East step following the $n$-th North step is replaced by an East step with a negative sign because $\mu$ has an odd number of positive entries. We obtain $\zeta_D(\pi)=NENNENENE^-$. % % \sk % \begin{center} % \begin{tikzpicture}[scale=.5] % \draw[gray] (0,0) grid (8,9); % \draw[very thick] (0,0)--(6,0)--(6,7)--(7,7)--(7,8)--(8,8)--(8,9); % \draw[xshift=5mm] (0,0) node[circle,fill=white,inner sep=-1.3pt]{$\ominus$}; % \draw[xshift=5mm,yshift=5mm] % (5,0) node{$v_1$} % (5,1) node{$v_2$} % (5,2) node{$v_3$} % (5,3) node{$v_4$} % (5,4) node{$v_5$} % (5,5) node{$v_6$} % (5,6) node{$v_7$} % (6,7) node{$v_8$} % (7,8) node{$v_9$}; % % % \begin{scope}[xshift=95mm,yshift=4cm] % \draw[->, thick] (0,0)--node[above]{$\zeta_D$}(1,0); % \end{scope} % % % \begin{scope}[xshift=12cm,yshift=-4cm] % \ballotodd{9} % % \draw[gray] % % (0,9)--(8,9) % % (0,8)--(8,8)--(8,10)--(0,10) % % (0,7)--(7,7)--(7,11)--(0,11) % % (0,6)--(6,6)--(6,12)--(0,12)--(0,0) % % (0,5)--(5,5)--(5,12) % % (0,4)--(4,4)--(4,12) % % (0,3)--(3,3)--(3,12) % % (0,2)--(2,2)--(2,12) % % (0,1)--(1,1)--(1,12); % \draw[very thick] (0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(2,3)--(3,3)--(3,4)--(4,4)--(4,5)--(5,5)--(5,9)--(6,9)--(6,11); % \draw[very thick,red] (0,1)--(1,1) % (1,2)--(2,2)--(2,3) % (3,4)--(4,4)--(4,5) % (5,7)--(5,9); % \draw[xshift=5mm] (5,9) node[circle,fill=white,inner sep=-1.3pt]{$\oplus$}; % \draw[xshift=5mm,yshift=5mm] % (7,7) node {$v_8$} % (6.2,11) node {$v_9$} % (0,1) node{$\bullet$} % (1,2) node{$\bullet$} % (2,3) node{$\bullet$} % (3,4) node{$\bullet$} % (4,5) node{$\bullet$} % (5,9) node{$\bullet$}; % \draw[yshift=5mm,anchor=west] % (0,0) node {$-v_1$} % (1,1) node {$-v_2$} % (2,2) node {$-v_3$} % (3,3) node {$-v_4$} % (4,4) node {$-v_5$} % (5,5) node {$-v_6$} % (6,6) node {$-v_9$} % (8,8) node[anchor=west]{$\epsilon(\pi)v_7$} % (8,9) node[anchor=west]{$-\epsilon(\pi)v_7$} % (7,10) node {$-v_8$}; % \end{scope} % % % \end{tikzpicture} % \end{center} % Recall that we computed the area vector $\mu=(-6,5,4,3,2,1,0,0,1)$ and the diagonal reading word $d_D(\pi,v)=[v_7,v_8,-v_9,-v_6,-v_5,-v_4,-v_3,-v_2,-v_1]$ previously. We compute the pieces of $\zeta(\pi)$ as follows % \begin{align*} % \LS_6^-(\mu)&=N,&& % \LS_5^-(\mu)=E,&& % \RS_5^+(\mu)=N,&& % \RS_4^+(\mu)=EN,&& % \RS_3^+(\mu)=EN,\\ % \RS_2^+(\mu)&=EN,&& % \RS_1^+(\mu)=ENN,&& % \LS_0^-(\mu)=NN,&& % \RS_0^+(\mu)=ENNE. % \end{align*} % Note that the terminal East step in $\RS_0^+(\mu)$ does not appear in $\zeta(\pi)$. The sign of $\zeta(\pi)$ is positive because $\mu$ contains an even number of positive entries. Thus the initial East step of $\RS_0^+(\mu)$ is replaced by $E^+$. \end{myex} \sk In the remainder of this section we will prove that both maps defined above are the type $D_n$ instances of the uniform bijections given in \refq{zetauni} and \refq{HLzetauni}. Thereby we establish the following theorem, which is the main result of this section. \begin{mythrm}{mainD} The Haglund--Loehr-zeta map $\zeta_D:\op{Vert}(D_n)\to\op{Diag}(D_n)$ is a bijection from vertically labelled signed lattice paths to diagonally labelled signed ballot paths. \end{mythrm} \sk From \reft{mainD} we may derive some nice consequences. Since the underlying ballot path of $\zeta_D(\pi,v)$ only depends on $\pi$ and not on $v$, we have the following result. \begin{mythrm}{zetaD} The zeta map $\zeta_D:\ld_{n-1,n}\to\bd_{2n-1}$ is a bijection from signed lattice paths to signed ballot paths.\noproof{} \end{mythrm} \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.6] % \begin{scope}[yshift=2.5cm] \draw[gray](0,0)grid(5,6); \draw[very thick](0,0)--(1,0)--(1,2)--(2,2)--(2,3)--(3,3)--(3,4)--(5,4)--(5,6); \ballotsign{0}{0}{\oplus} \draw[xshift=5mm,yshift=5mm] (0,0)node{$1$} (0,1)node{$3$} (1,2)node{$-2$} (2,3)node{$-5$} (4,4)node{$-4$} (4,5)node{$6$}; \end{scope} \draw[xshift=6cm,yshift=5cm,thick,->](0,0)--node[above]{$\zeta_D$}(1,0); % \begin{scope}[xshift=8cm] \ballotodd{6} \draw[very thick](0,0)--(0,6)--(1,6)--(1,9)--(2,9); \draw[very thick,red] (0,1)--(0,2) (1,6)--(1,9)--(2,9); \ballotsign{0}{6}{\ominus} \draw[xshift=5mm,yshift=5mm] (0,0)node{$-1$} (1,1)node{$4$} (2,2)node{$6$} (3,3)node{$-5$} (4,4)node{$-2$} (5,5)node{$3$} (5,6)node{$-3$} (4,7)node{$2$} (3,8)node{$5$} (2,9)node{$-6$} (1,10)node{$-4$} (0,11)node{$1$} (0,6)node{$\bullet$} (1,9)node{$\circ$}; \end{scope} % \end{tikzpicture} \caption{} \label{Figure:zetaD2} \end{center} \end{figure} \sk Moreover, given any signed path $\rho$ define $\rho^*$ as the path obtained from $\rho$ by replacing all signed East steps $E^+,E^-$ by simple East steps $E$. Define $\zeta_D^*(\pi^*)=(\zeta_D(\pi))^*$ for all $\pi\in\ld_{n-1,n}$. The zeta map of type $D_n$ thereby gives rise to a new bijection between lattice paths in an $n-1\times n$ rectangle and ballot paths of odd length. \begin{mythrm}{oddballotbij} The map $\zeta_D^*:\mathcal{L}_{n-1,n}\to\mathcal{B}_{2n-1}$ is a well-defined bijection. \end{mythrm} \begin{proof} Suppose $\pi,\rho\in\ld_{n-1,n}$ differ only by the sign of the initial East step. Then their respective area vectors differ only by the sign of the first entry. It is easy to see that $\zeta_D(\pi)$ and $\zeta_D(\rho)$ can only differ by the sign of an East step. For example, suppose $a=(k,a_2,\dots,a_n)$ and $b=(-k,a_2,\dots,a_n)$, where $k>0$, then % $\LS_k^-(a)N=\LS_k^-(b)$ and $\RS_k^+(a)=N\RS_k^+(b)$, and therefore $\LS_k^-(a)\RS_k^+(a)=\LS_k^-(b)\RS_k^+(b)$. \begin{align*} \LS_k^-(a)N=\LS_k^-(b)\text{ and }\RS_k^+(a)=N\RS_k^+(b). \end{align*} Hence \begin{align*} \LS_k^-(a)\RS_k^+(a)=\LS_k^-(b)\RS_k^+(b). \end{align*} Consequently $\zeta_D^*$ is well-defined and bijectivity follows from \reft{zetaD}. \end{proof} \sk Before we attack the proof of \reft{mainD} we need another lemma. In combination with \refl{sigma} it gives a criterion for when $\abs{\sigma^{-1}(i)}=1$. \begin{mylem}{sig1D} Let $\pi$ be a signed lattice path with area vector $\mu$, and let $i\in[n]$ be such that \begin{align*} 1&=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_i\n-i}\big\}. \end{align*} Then either $i$ is minimal such that $\mu_i=0$ or $\mu_j\neq0$ for all $j\in[n]$, $\mu_n=1$ and $i=n$. \end{mylem} \begin{proof} Clearly $\mu_i\neq0$ implies that $\mu_{\ell}\neq0$ for all $\ell\in[n]$. Moreover $\mu_{\ell}\neq0$ for all $\ell\in[n]$ and $\mu_n=1$ implies $i=n$. Thus assume that $\mu_{\ell}\neq0$ for all $\ell\in[n]$ and $\mu_n\neq1$. From $\lambda_{n-1}\leq\min\{\lambda_n,2n-1-\lambda_n\}$ we obtain $\mu_{n-1}<0$. Thus $\mu_{\ell}<0$ for all $\ell$ with $1<\ell](0,0)--node[above]{$\zeta_D$}(1,0); % \begin{scope}[xshift=7cm] \ballotodd{5} \draw[very thick](0,0)--(0,1)--(1,1)--(1,3)--(2,3)--(2,4)--(3,4)--(3,5)--(4,5); \draw[very thick,red] (0,1)--(1,1) (1,3)--(2,3)--(2,4)--(3,4) (3,5)--(4,5); \ballotsign{3}{5}{\ominus} \draw[xshift=5mm,yshift=5mm] (0,0)node{$5$} (1,1)node{$4$} (2,2)node{$3$} (3,3)node{$-1$} (4,4)node{$2$} (4,5)node{$-2$} (3,6)node{$1$} (2,7)node{$-3$} (1,8)node{$-4$} (0,9)node{$-5$} (0,1)node{$\bullet$} (1,3)node{$\bullet$} (2,4)node{$\bullet$} (3,5)node{$\circ$}; \end{scope} % \end{tikzpicture} \caption{} \label{Figure:zetaD3} \end{center} \end{figure} \sk We proof \reft{mainD} in two steps. First we prove \reft{risevalleyD}, which establishes a strong correspondence between the rises of $(\pi,v)$ and the valleys of $\zeta_D(\pi,v)$. Let $(\pi,v)\in\ld_{n-1,n}\times\S_n^B$ and let $i$ be a rise of $\pi$. We say the rise $i$ is labelled by $(v(i),v(i+1))$. If $\pi$ begins with two North steps and $i=1$, then we say $i$ is labelled by $(\pm\abs{v(i)},v(i+1))$ instead. Given $(\beta,w)\in\bd_{2n-1}\times\S_n^B$ and a valley $(i,j)$ of $\beta$, we say \begin{align*} (i,j)\text{ is labelled by } \begin{cases} \big(w(n+1-i),w(n+1-j)\big)&\quad\text{ if }jn+1. \end{cases} \end{align*} If $j=n$ and the $n$-th North step of $\beta$ is followed by another North step, then we say $(i,j)$ is labelled by $(w(n+1-i),\pm\abs{w(1)})$ instead. Note that if we place the labels $w(i)$, where $i=n,\dots,2,\epsilon(\beta),-\epsilon(\beta),-2,\dots,-n$, in the diagonal, then each valley is labelled by the number below it and the number to its right. \begin{mythrm}{risevalleyD} Let $(\pi,v)\in\ld_{n-1,n}\times\S_n^B$ be a pair of a signed lattice path and a signed permutation, and let $a,b\in v([n])$. Then $(\pi,v)$ has a rise labelled $(a,b)$ if and only if $(\zeta_D(\pi),d_D(\pi,v))$ has a valley labelled $(b,a)$ or $(-a,-b)$. Moreover $(\pi,v)$ has a rise labelled $(\pm\abs{a},b)$ if and only if $(\zeta_D(\pi),d_D(\pi,v))$ has a valley labelled $(b,\pm\abs{a})$. % or $(-a,\pm\abs{b})$. \end{mythrm} \begin{myex}{risevalleyD} Consider once more the labelled paths in \reff{zetaD3}. The vertically labelled signed lattice path has four rises that are labelled by $(-5,-4),(-4,1),(1,2)$ and $(2,3)$. This corresponds exactly to the valleys of the diagonally labelled signed ballot path, which are labelled by $(5,4),(4,-1),(3,2)$ and $(-1,-2)$. Moreover, the vertically labelled signed lattice path in \reff{zetaD1} has a rise labelled by $(\pm3,4)$, and its image under the zeta map has a valley labelled by $(4,\pm3)$. \end{myex} \begin{proof}[Proof of \reft{risevalleyD}] Define $\lambda\in\Q\cap(h+1)\overline{\ac}$ and $u\in\S_n^D$ as in \refp{vertD}. Let $\nu\in\Q$ and $\tau\in\S_n^D$ be as in \refl{wfD}. Moreover, let $\mu$ be the type $D_n$ area vector of $\pi$ and let $\sigma\in\S_n^D$ be given by \refl{sigma} such that $t_{\mu}\sigma$ is Gra{\ss}mannian. \sk \textbf{(Part 1)} We start out by demonstrating the backward implication. Therefore assume that $\zeta_D(\pi)$ has a valley $(x,y)$. Recall that $\zeta_D(\pi)$ is the concatenation of sequences $\LS_k^-(\mu)$ and $\RS_k^+(\mu)$, thus there are multiple situations in which a valley can arise: within such a sequence or at the join of two sequences. These cases, while being similar, have to be treated separately. \sk \textbf{(1.1)} The valley $(x,y)$ appears within a sequence $\LS_k^-(\mu)$. Then there exist indices $i,j\in[n]$ with $ik+1\big\}+\#\big\{r\in[n]:i+1\leq r, \mu_r=-k-1\big\}\\ &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\geq\abs{(-k-1)\n-(i+1)}\big\}. \end{align*} Note that $\abs{\sigma^{-1}(i+1)}\neq1$ by \refl{sig1D} because $\mu_{i+1}<0$. \refl{sigma} therefore provides \begin{align*} n+1-x &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_{i+1}\n-(i+1)}\big\}\\ &=\abs{\sigma^{-1}(i+1)} =\sigma^{-1}(i+1). \end{align*} We conclude that \begin{align*} d_D(\pi,v)(n+1-x) &=u\tau\sigma(n+1-x) =u\tau(i+1) =v(i+1) \end{align*} because even if $i+1=n$ we have $\mu_{n-1}+\mu_n=-2k-1$, which is odd. On the other hand, $y$ equals the number of North steps in the sequence \begin{align*} \LS_{2n-1}^-(\mu)\RS_{2n-1}^+(\mu)\cdots\LS_{k+1}^-(\mu)\RS_{k+1}^+(\mu)\LS_k^-(\mu_i,\dots,\mu_n). \end{align*} % Hence, % \begin{align*} % n+1-y % &=\#\big\{r\in[n]:\abs{\mu_r}>k\big\}+\#\big\{r\in[n]:i\leq r, \mu_r=-k\big\}\\ % &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\geq\abs{\mu_i\n-i}\big\}. % \end{align*} By \refl{sigma} we therefore have \begin{align*} n+1-y &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_i\n-i}\big\} =\abs{\sigma^{-1}(i)}. \end{align*} We first treat the case where $\abs{\sigma^{-1}(i)}\neq1$. Note that this implies $i>1$. Then by \refl{sigma} we have $n+1-y=\sigma^{-1}(i)$ because $\mu_i\leq0$. We conclude that \begin{align*} d_D(\pi,v)(n+1-y) &=u\tau\sigma(n+1-y) =u\tau(i) =v(i), \end{align*} % because, as we have seen above, even if $i=1$ we must have $\mu_1=0$ such that $\epsilon(\pi)=1$. and the valley is labelled by $(v(i+1),v(i))$. Next assume that $\abs{\sigma^{-1}(i)}=1$, that is, $y=n$. Then $k=0$. If $i=1$ then we have already seen that $\pi$ begins with two North steps. Since $\LS_0^-(\mu)$ ends with a valley and $\RS_0^+(\mu)$ begins with a North step, we are in the special situation that the valley $(x,n)$ is labelled by $(v(2),\pm\abs{v(1)})$. Finally if $i>1$ then $\mu_{\ell}\neq0$ for all $\ell$ with $1\leq\ell0\}}u\tau(i) =\epsilon(\zeta_D(\pi))v(i), \end{align*} in which case the valley is labelled by $(v(i+1),v(i))$. \sk \textbf{(1.2)} Secondly, assume that the valley $(x,y)$ arises within a sequence $\RS_k^+(\mu)$ for some $k>0$. Then there exist indices $i,j\in[n]$ with $i\lambda_n$. % Hence $\lambda_{n-1}+\lambda_n=2n-1$ and $i$ is a rise of $\pi$. In all other cases % (if $n-1\equiv0,3$ modulo $4$ or $10$ and $\epsilon(\pi)=1$. Similar to the case above, $x$ equals the number of East steps in the sequence \begin{align*} \LS_{2n-1}^-(\mu)\RS_{2n-1}^+(\mu)\cdots\RS_{k+1}^+(\mu)\LS_k^-(\mu)\RS_k^+(\mu_1,\dots,\mu_i). \end{align*} Using \refl{sigma} we conclude % \begin{align*} % n+1-x % &=\#\big\{r\in[n]:\abs{\mu_r}>k+1\big\}+\#\big\{r\in[n]:\mu_r=-k-1\big\} +\#\big\{r\in[i]:\mu_r=k+1\big\}\\ % &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\geq\abs{\mu_i\n-i}\big\}, % \end{align*} % % and hence by \refl{sigma} \begin{align*} n+1-x &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_i\n-i}\big\} =\abs{\sigma^{-1}(i)}. \end{align*} By \refl{sig1D} we have $\abs{\sigma^{-1}(i)}\neq1$ because $\mu_i>1$, thus \refl{sigma} yields $-\sigma^{-1}(i)=n+1-x$. We obtain \begin{align*} d_D(\pi,v)(n+1-x) =u\tau\sigma(n+1-x) =-u\tau(i)=-v(i), \end{align*} where we use that $\epsilon(\pi)=-1$ if and only if $n-1\equiv1,2$ modulo $4$ in the case where $i=1$. On the other hand, $y$ equals the number of North steps in the sequence \begin{align*} \LS_{2n-1}^-(\mu)\RS_{2n-1}^+(\mu)\cdots\RS_{k+1}^+(\mu)\LS_k^-(\mu)\RS_k^+(\mu_1,\dots,\mu_{i+1}). \end{align*} It follows that % \begin{align*} % n+1-y % &=\#\big\{r\in[n]:\abs{\mu_r}>k\big\}+\#\big\{r\in[n]:\mu_r=-k\big\} +\#\big\{r\in[i+1]:\mu_r=k\big\}\\ % &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\geq\abs{\mu_{i+1}\n-(i+1)}\big\}, % \end{align*} % and by \refl{sigma} \begin{align*} n+1-y &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{\mu_{i+1}\n-(i+1)}\big\} =\abs{\sigma^{-1}(i+1)}. \end{align*} Suppose $y=n$, then by \refl{sig1D} we see that $\mu_{\ell}\neq0$ for all $\ell\in[n]$, $k=1$ and $i+1=n$, which is a contradiction as mentioned above. Thus $y0$ and \refl{sigma} we obtain $-\sigma^{-1}(i+1)=n+1-y$. Therefore \begin{align*} d_D(\pi,v)(n+1-y) =u\tau\sigma(n+1-y) =-u\tau(i+1) =-v(i+1), \end{align*} % where we use the fact that $\mu_i+\mu_{i+1}=2k+1$ is odd when $i=n-1$. and the valley is labelled by $(-v(i),-v(i+1))$. \sk \textbf{(1.3)} The sequence $\LS_k^-(\mu)$ ends with an East step and the next non-empty sequence begins with a North step. If $\LS_k^-(\mu)$ ends with an East step then there exists an index $i\in[n]$ such that $\mu_i=-k-1$ and $\mu_{\ell}\notin\{-k-1,-k\}$ for all $\ell$ with $1\leq\ell0$, and let $(x,y)$ be the valley under consideration. Then $x$ equals the number of East steps in the sequence \begin{align*} \LS_{2n-1}^-(\mu)\RS_{2n-1}^+(\mu)\cdots\RS_{k+1}^+(\mu)\LS_k^-(\mu). \end{align*} We deduce that \begin{align*} n+1-x =\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{(-k-1)\n-1}\big\} =\abs{\sigma^{-1}(1)}, \end{align*} and since $\mu_1<0$, implying $\abs{\sigma^{-1}(1)}\neq1$, that $n+1-x=\sigma^{-1}(1)$. Moreover \begin{align*} d_D(\pi,v)(n+1-x) &=u\tau\sigma(n+1-x) =u\tau(1)=-v(1), \end{align*} because $\epsilon(\pi)=-1$ is and only if $n-1\equiv0,3$ modulo $4$. On the other hand, $y$ equals the number of North steps in the sequence \begin{align*} \LS_{2n-1}^-(\mu)\RS_{2n-1}^+(\mu)\cdots\RS_{k+1}^+(\mu)\LS_k^-(\mu)\RS_k^+(\mu_1,\mu_2). \end{align*} As before we obtain \begin{align*} n+1-y =\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{k\n-2}\big\} =\abs{\sigma^{-1}(2)}. \end{align*} We know that $\abs{\sigma^{-1}(2)}\neq1$ because of \refl{sig1D}. Since $\mu_2=k>0$ we conclude \begin{align*} d_D(\pi,v)(n+1-y) &=u\tau\sigma(n+1-y) =-u\tau(2) =-v(2). \end{align*} \textbf{(1.3.2)} Next assume that $k=0$. In this case the valley is of the form $(x,n+1)$, where $n+1-x=\sigma^{-1}(1)$ just as in (1.3.1) above. On the other hand by \refl{sig1D} we see that $\abs{\sigma^{-1}(2)}=1$. Note that $\LS_0^-(\mu)$ ending with an East step implies that the $n$-th North step of $\zeta_D(\pi)$ is followed by an East step. We obtain \begin{align*} d_D(\pi,v)(1) &=u\tau\sigma(1) =(-1)^{\#\{r\in[n]:\mu_r>0\}}u\tau(2) =\epsilon(\zeta_D(\pi))v(2), \end{align*} and the valley $(x,n+1)$ is labelled by \begin{align*} \big(d_D(\pi,v)(n+1-x),-\epsilon(\zeta_D(\pi))d_D(\pi,v)(1)\big) &=\big(-v(1),-v(2)\big). \end{align*} \textbf{(1.4)} For some $k>0$ the sequence $\RS_k^+(\mu)$ ends with an East step and the next non-empty sequence begins with a North step. Then there exists an index $j\in[n]$ such that $\mu_j=k+1$ and $\mu_{\ell}\notin\{k,k+1\}$ for all $\ell$ with $j<\ell\leq n$. By \refl{muD}~(v) and~(vi) we know that $j=n$ and that $\LS_{k-1}^-(\mu)$ is non-empty. By assumption $\LS_{k-1}^-(\mu)$ begins with a North step, thus there exists an index $i\in[n]$ such that $\mu_i=-k+1$ and $\mu_{\ell}\notin\{-k,-k+1\}$ for all $\ell$ with $i<\ell\leq n$. Using \refl{muD}~(vii) we see that $i=n-1$. If $n-1\equiv0,3$ modulo $4$ then $\lambda_{n-1}+\lambda_n=(n-k-1)+(n+k)=2n-1$. If $n-1\equiv1,2$ modulo $4$ then $\lambda_{n-1}=n-k-1=\lambda_n$. In both cases $i$ is a rise of $\pi$. Once more let $(x,y)$ be the valley under consideration. Then $x$ is the number of East steps in the sequence \begin{align*} \LS_{2n-1}^-(\mu)\RS_{2n-1}^+(\mu)\cdots\LS_k^-(\mu)\RS_k^+(\mu), \end{align*} and therefore \begin{align*} n+1-x &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{(k+1)\n-n}\big\} =\abs{\sigma^{-1}(n)} \end{align*} From $\mu_n>1$ we obtain $n+1-x=-\sigma^{-1}(n)$ and \begin{align*} d_D(\pi,v)(n+1-x) =u\tau\sigma(n+1-x) =-u\tau(n) =v(n) \end{align*} because $\mu_{n-1}+\mu_n=(k+1)+(-k+1)=2$ is even. Moreover $y$ equals the number of North steps in the sequence \begin{align*} \LS_{2n-1}^-(\mu)\RS_{2n-1}^+(\mu)\cdots\LS_k^-(\mu)\RS_k^+(\mu)\LS_{k-1}^-(\mu_{n-1},\mu_n). \end{align*} Thus, \begin{align*} n+1-y &=\#\big\{r\in[n]:\abs{\mu_r\n-r}\leq\abs{(-k+1)\n-(n-1)}\big\} =\abs{\sigma^{-1}(n-1)}. \end{align*} % Here we have $\abs{\sigma^{-1}(n-1)}\neq1$ because $\mu_{n-1}=-1$ implies the existence of at least one index $\ell\in[n-2]$ with $\mu_{\ell}=0$. We obtain $y=\sigma^{-1}(n-1)$ and First suppose that $\abs{\sigma^{-1}(n-1)}\neq1$. Then $\mu_{n-1}\leq0$ implies $n+1-y=\sigma^{-1}(n-1)$, and we obtain \begin{align*} d_D(\pi,v)(n+1-y) &=u\tau\sigma(n+1-y) =u\tau(n-1) =v(n-1). \end{align*} On the other hand if $\abs{\sigma^{-1}(n-1)}=1$ then $y=n$, $\mu_{n-1}=0$, $k=1$ and $\mu_{\ell}\neq0$ for all $\ell\in[n-2]$. We claim that the $n$-th North step of $\zeta_D(\pi)$, which belongs to $\LS_0^-(\mu)$ and corresponds to $\mu_{n-1}=0$, is followed by an East step. To see this assume that $\mu_{\ell}\notin\{0,-1\}$ for all $\ell\in[n-2]$. Then there has to be an index $\ell\in[n-2]$ with $\mu_{\ell}=1$. Thus $\RS_0^+(\mu)$ begins with an East step and this East step is not the last step of $\RS_0^+(\mu)$. Consequently, this East step is replaced by a signed East step such that $\epsilon(\zeta_D(\pi))=(-1)^{\#\{r\in[n]:\mu_r>0\}}$. From \begin{align*} d_D(\pi,v)(1) &=u\tau\sigma(1) =(-1)^{\#\{r\in[n]:\mu_r>0\}}u\tau(n-1) =\epsilon(\zeta_D(\pi))v(n-1). \end{align*} we obtain that the valley is labelled by $(v(n),v(n-1))$. \sk \textbf{(1.5)} The the valley arises within (or at the end of) the sequence $\RS_0^+(\mu)$. Note that $\RS_0^+(\mu)$ is non-empty by \refl{sig1D}, and recall that the last letter of $\RS_0^+(\mu)$ does not contribute to $\zeta_D(\pi)$. \sk \textbf{(1.5.1)} There exist indices $i,j\in[n]$ with $in+1$, then by \refl{sigma} we have $\sigma^{-1}(i+1)=y-n$ since $\mu_{i+1}=0$. It follows that \begin{align*} d_D(\pi,v)(n-y) &=-u\tau\sigma(y-n) =-u\tau(i+1) =-v(i+1) \end{align*} because even if $i+1=n$ then $\mu_{n-1}+\mu_n=1$, which is odd. Thus the valley under consideration is labelled by $(-v(i),-v(i+1))$. If $y=n+1$ then there are exactly $n$ North steps in $\zeta_D(\pi)$ that occur before the East step corresponding to $\mu_i=1$. Hence the $n$-th North step of $\zeta_D(\pi)$ is followed by an East step, and $\zeta_D(\pi)$ contains a signed East step. Thus \begin{align*} d_D(\pi,v)(y) &=u\tau\sigma(1) =(-1)^{\#\{r\in[n]:\mu_r>0\}}u\tau(i+1) =\epsilon(\zeta_D(\pi))v(i+1) \end{align*} because even if $i+1=n$ then $\mu_{n-1}+\mu_n=1$, which is odd. Thus the valley $(x,n+1)$ is labelled by $(-v(i),-v(i+1))$ in this case as well. \sk \textbf{(1.5.2)} There exist indices $i,j\in[n]$ with $i0$ then there is a valley in the sequence $\RS_k^+(\mu)$ and we are in case~(1.2) or~(1.5.1). \sk \textbf{(2.2)} Next assume $i=1$. Then $\pi_1=\abs{\lambda_1}=\abs{\mu_1}$ and $\pi_2=\lambda_2=1+\mu_2$, hence $\pi_1=\pi_2$ implies $\abs{\mu_1}=\mu_2+1$. If $\mu_1\geq0$ we are in the same situation as in~(2.1). On the other hand if $\mu_1<0$ then \begin{align*} -\mu_1=\mu_2+1. \end{align*} Set $-k-1=\mu_1$ then the sequence $\LS_k^-(\mu)$ ends with an East step and the sequence $\RS_k^+(\mu)$ begins with a North step. We are therefore in case~(1.3). \sk \textbf{(2.3)} Finally assume that $i=n-1$. Then $\pi_{n-1}=\lambda_{n-1}=n-2+\mu_{n-1}$. \sk \textbf{(2.3.1)} Suppose $n-1\equiv0,3$ modulo $4$, thus $\lambda_n=n-1+\mu_n$. If $\pi_1+\dots+\pi_{n-2}$ is even, then $\pi_{n-1}=\pi_n$ implies $\lambda_n=2\pi_n-\pi_{n-1}=\pi_{n-1}$ and $\mu_{n-1}=\mu_n+1$ as in~(2.1). Otherwise $\pi_1+\dots+\pi_{n-2}$ is odd and $\pi_{n-1}=\pi_n$ implies $\lambda_n=2n-1-2\pi_n+\pi_{n-1}=2n-1-\pi_{n-1}$ and \begin{align*} \mu_{n-1}=-\mu_n+2. \end{align*} If $\mu_{n-1}=-k\leq0$ then the sequence $\RS_{k+1}^+(\mu)$ ends with an East step and the sequence $\LS_k^-(\mu)$ begins with a North step. We are therefore in case~(1.4). If $\mu_{n-1}=k+1>1$ then the sequence $\RS_k^+(\mu)$ ends with an East step while the sequence $\LS_{k-1}^-(\mu)$ begins with a North step. This again corresponds to case~(1.4). If $\mu_{n-1}=1$ then also $\mu_n=1$ which puts us into case (1.5.2). \sk \textbf{(2.3.2)} Suppose $n-1\equiv1,2$ modulo $4$, thus $\lambda_n=n-\mu_n$. If $\pi_1+\dots+\pi_{n-2}$ is even, then $\pi_{n-1}=\pi_n$ implies $\lambda_n=\pi_{n-1}$ and $\mu_{n-1}=-\mu_n+2$. If $\pi_1+\dots+\pi_{n-2}$ is odd, then $\pi_{n-1}=\pi_n$ implies $\lambda_n=2n-1-\pi_{n-1}$ and $\mu_{n-1}=\mu_n+1$. Hence we are in the same situation as in~(2.3.1). \end{proof} \sk We now use \reft{risevalleyD} to show that the Haglund--Loehr-zeta map of type $D_n$ does indeed correspond to the specialisation of the uniform zeta map to this case. This completes the proof of \reft{mainD}. \begin{mythrm}{HLzetaD} Let $\Phi$ be the root system of type $D_n$ with coroot lattice $\Q$ and zeta map $\zeta$, and let $\psi$ and $\varphi$ be defined as in \refp{vertD} and \refp{diagD}. Then the following diagram commutes. % \begin{figure}[h] \begin{center} \begin{tikzpicture} \draw (0,0) node(q){$\Q/(2n-1)\Q$}; \draw (4,0) node(park){$\mathrm{Park}(\Phi)$}; \draw (0,1.4) node(v){$\op{Vert}(D_n)$}; \draw (4,1.4) node(d){$\op{Diag}(D_n)$}; \draw[->] (q)--node[below]{$\zeta$}(park); \draw[->] (v)--node[above]{$\zeta_D$}(d); \draw[->] (v)--node[left]{$\psi$}(q); \draw[->] (d)--node[right]{$\varphi$}(park); \end{tikzpicture} % \qquad % \begin{tikzpicture} % \draw (0,1.4) node(l){$\ld_{n-1,n}$}; % \draw (4,1.4) node(b){$\bd_{2n-1}$}; % \draw (0,0) node(q){$(2n-1)\overline{\ac}\cap\Q$}; % \draw (4,0) node(ac){$\op{Antichains}(\Phi^+)$}; % \draw[->] (l)--node[above]{$\zeta_D$}(b); % \draw[->] (q)--node[below]{$\zeta$}(ac); % \draw[->] (l)--(q); % \draw[->] (b)--(ac); % \end{tikzpicture} \end{center} % \end{figure} \end{mythrm} \begin{proof} Let $(\pi,v)\in\op{Vert}(D_n)$ and $(\beta,w)=\zeta_D(\pi,v)\in\op{Diag}(D_n)$. Furthermore, let $\lambda$ and $u$ be defined as in $\refq{lambdaD}$ and $\refq{uD}$ such that $\psi(\pi,v)=u\cdot\lambda+(2n-1)\Q$. Finally, let $\mu$ be the area vector of $\pi$, chose $\sigma\in\S_n^D$ as in \refl{sigma} such that $t_{\mu}\sigma$ is a Gra{\ss}mannian affine permutation and fix $\tau$ as in \refl{wfD}. Recall from \refp{antichain} and \refp{utausig} that it suffices to show $w=u\tau\sigma$ and $A_\beta=A$, where \begin{align*} A=\big\{(\tau\sigma)^{-1}\cdot\alpha:\alpha\in\Delta\cup\{-\tilde\alpha\}\text{ and }s_{\alpha}\cdot\lambda=\lambda\big\}. \end{align*} The first claim is immediate from \refp{readD}. In order to show the second claim, we first prove $A\subseteq A_{\beta}$. Assume $(\tau\sigma)^{-1}\cdot\alpha_i\in A$ for some $1n+1, \end{cases} \end{align*} or $-u(i)=w(n+1-x)$ and \begin{align*} -u(i+1)= \begin{cases} w(n+1-y)&\quad\text{if }yn+1. \end{cases} \end{align*} Note that we may replace $v(i),v(i+1)$ by $u(i),u(i+1)$ by the choice of $i$. Applying $u^{-1}$ to the above identities and using $w=u\tau\sigma$, we see that $\tau\sigma\cdot\alpha_{x,y}=\alpha_i$. Hence $(\tau\sigma)^{-1}\cdot\alpha_i=\alpha_{x,y}\in A_{\beta}$. Next set $S=\{(\tau\sigma)^{-1}\cdot\alpha_0,(\tau\sigma)^{-1}\cdot\alpha_1\}$. If $S\cap A=\{(\tau\sigma)^{-1}\cdot\alpha_1\}$ then $\lambda_1=\lambda_2>0$. We have $\epsilon(\pi)=1$ and $1$ is a rise of $\pi$. In particular $u(1)=v(1)$. By \reft{risevalleyD} $(\beta,w)$ has a valley $(x,y)$ labelled by $(u(2),u(1))$ or $(-u(1),-u(2))$. We obtain $\tau\sigma\cdot\alpha_{x,y}=e_2-e_1$. Consequently $(\tau\sigma)^{-1}\cdot\alpha_1=\alpha_{x,y}\in A_{\beta}$. If $S\cap A=\{(\tau\sigma)^{-1}\cdot\alpha_0\}$ then $\lambda_1=-\lambda_2>0$. We have $\epsilon(\pi)=-1$ and $1$ is a rise of $\pi$. In particular $u(1)=-v(1)$. By \reft{risevalleyD} $(\beta,w)$ has a valley $(x,y)$ labelled by $(u(2),-u(1))$ or $(u(1),-u(2))$. We obtain $\tau\sigma\cdot\alpha_{x,y}=e_2+e_1$ and thus $(\tau\sigma)^{-1}\cdot\alpha_1=\alpha_{x,y}\in A_{\beta}$. If $S\subseteq A$ then $\lambda_1=\lambda_2=0$. Hence $\pi$ begins with two North steps. By \reft{risevalleyD} $(\beta,w)$ has a valley $(x,n)$ labelled by $(u(2),\pm\abs{u(1)})$. As above we see that $\tau\sigma\cdot\{\alpha_{x,n},\alpha_{x,n+1}\}=\{\alpha_0,\alpha_1\}$. Hence $S\subseteq A_{\beta}$. If $(\tau\sigma)^{-1}\cdot\alpha_{n-1}\in A$ then $\lambda_{n-1}=\lambda_n$. It follows that $\pi_{n-1}=\pi_n$, that is, $n-1$ is a rise of $\pi$, and $\pi_1+\dots+\pi_{n-2}$ is even. Moreover $u(n)=v(n)$. By \reft{risevalleyD} $(\beta,w)$ has a valley $(x,y)$ labelled $(u(n),u(n-1))$ or $(-u(n-1),-u(n))$. We obtain $\tau\sigma\cdot\alpha_{x,y}=e_n-e_{n-1}$ and thus $(\tau\sigma)^{-1}\cdot\alpha_{n-1}\in A_{\beta}$. If $(\tau\sigma)^{-1}\cdot(-\tilde\alpha)\in A$ then $\lambda_{n-1}+\lambda_n=2n-1$. It follows that $\pi_{n-1}=\pi_n$ and $\pi_1+\dots+\pi_{n-2}$ is odd. Moreover $u(n)=-v(n)$. By \reft{risevalleyD} $(\beta,w)$ has a valley $(x,y)$ labelled $(-u(n),u(n-1))$ or $(-u(n-1),u(n))$. We obtain $\tau\sigma\cdot\alpha_{x,y}=-e_{n-1}-e_n$ and thus $(\tau\sigma)^{-1}\cdot(-\tilde\alpha)\in A_{\beta}$. \sk To complete the proof we need to demonstrate $A_{\beta}\subseteq A$. Therefore suppose $\alpha_{x,y}\in A_{\beta}$ for some valley $(x,y)$ of $(\beta,w)$. Then by \reft{risevalleyD} $(x,y)$ is labelled either $(v(i+1),v(i))$ or $(-v(i),-v(i+1))$ for some rise $i$ of $\pi$, or by $(v(2),\pm\abs{v(1)})$ if $y=n$ and the valley is not followed by an East step. If $u(i)=v(i)$ and $u(i+1)=v(i+1)$ then the valley $(x,y)$ is labelled by $(u(i+1),u(i))$ or $(-u(i),-u(i+1))$, and $\tau\sigma\cdot\alpha_{x,y}=\alpha_i$ for a rise $i$ of $\pi$. By similar arguments as above we see that $\alpha_{x,y}\in A$. If $i=1$ and $u(1)=-v(1)$ then $\tau\sigma\cdot\alpha_{x,y}=\alpha_0$, and $\lambda_1=-\lambda_2$. Again we obtain $\alpha_{x,y}\in A$. If $i=n-1$ and $u(n)=-v(n)$ then $\tau\sigma\cdot\alpha_{x,y}=-\tilde\alpha$ and $\lambda_{n-1}+\lambda_n=2n-1$. Again we obtain $\alpha_{x,y}\in A$. Finally, if the valley is of the special form $(x,n)$, not followed by an East step and labelled by $(v(2),\pm\abs{v(1)})$ then $\tau\sigma\cdot\{\alpha_{x,n},\alpha_{x,n+1}\}=\{\alpha_0,\alpha_1\}$, $\pi$ begins with two North steps and $\lambda_1=\lambda_2=0$. We see that $\alpha_{x,n},\alpha_{x,n+1}\in A$. Thus $A_{\beta}\subseteq A$ and the proof is complete. % % % \begin{align*} % (\tau\sigma)^{-1}\cdot\alpha_i\in A % &\Leftrightarrow\lambda_i=\lambda_{i+1}\\ % &\Leftrightarrow\pi_i=\pi_{i+1}\\ % &\Leftrightarrow % % By \reft{risevalleyD} this is the case if and only if $ % \zeta_D(\pi,v)\text{ has a valley }(x,y)\text{ labelled }(v(i+1),v(i))\\ % &\qquad\text{ or }(-v(i),v(i+1))\\ % &\Leftrightarrow\alpha_{x,y}\in A_{\beta}\text{ and either }u(i+1)=u\tau\sigma(n+1-x)\text{ and}\\ % &\qquad u(i)= % \left\{ % \begin{matrix*}[l] % u\tau\sigma(n+1-y)&\quad\text{if }yn+1, % \end{matrix*} % \right. % \\ % &\qquad\text{or }-u(i)=u\tau\sigma(n+1-x)\text{ and}\\ % &\quad -u(i+1)= % \left\{ % \begin{matrix*}[l] % u\tau\sigma(n+1-y)&\quad\text{if }yn+1, % \end{matrix*} % \right.\\ % &\Leftrightarrow\alpha_{x,y}\in A_{\beta}\text{ and }\sigma\tau\cdot\alpha_{x,y}=\alpha_i\\ % &\Leftrightarrow(\tau\sigma)^{-1}\cdot\alpha_i\in A_{\beta} % \end{align*} % Similarly, % \begin{align*} % \left. % \begin{matrix*} % (\tau\sigma)^{-1}\cdot\alpha_0\notin A\\ % (\tau\sigma)^{-1}\cdot\alpha_1\in A % \end{matrix*} % \right\} % &\Leftrightarrow\lambda_1=\lambda_2>0\\ % &\Leftrightarrow\pi_1=\pi_2\text{ and }\epsilon(\pi)=1\\ % &\Leftrightarrow\zeta_D(\pi,v)\text{ has a valley }(x,y)\text{ labelled }(u(2),u(1))\text{ or }(-u(1),-u(2))\\ % &\Leftrightarrow(\tau\sigma)^{-1}\cdot\alpha_1=\alpha_{x,y}\in A_{\beta}. % \end{align*} % \begin{align*} % \left. % \begin{matrix*} % (\tau\sigma)^{-1}\cdot\alpha_0\in A\\ % (\tau\sigma)^{-1}\cdot\alpha_1\notin A % \end{matrix*} % \right\} % &\Leftrightarrow\lambda_1=-\lambda_2<0\\ % &\Leftrightarrow\pi_1=\pi_2\text{ and }\epsilon(\pi)=-1\\ % &\Leftrightarrow\zeta_D(\pi,v)\text{ has a valley }(x,y)\text{ labelled }(u(2),-u(1))\text{ or }(u(1),-u(2))\\ % &\Leftrightarrow(\tau\sigma)^{-1}\cdot\alpha_0=\alpha_{x,y}\in A_{\beta}. % \end{align*} % \begin{align*} % (\tau\sigma)^{-1}\cdot\alpha_0,(\tau\sigma)^{-1}\cdot\alpha_1\in A % &\Leftrightarrow\lambda_1=\lambda_2=0\\ % &\Leftrightarrow\pi_1=\pi_2=0\\ % &\Leftrightarrow\zeta_D(\pi,v)\text{ has a valley }(x,n)\text{ labelled }(u(2),\pm\abs{u(1)})\\ % &\Leftrightarrow(\tau\sigma)^{-1}\cdot\alpha_0=\alpha_{x,n}, (\tau\sigma)^{-1}\cdot\alpha_1=\alpha_{x,n+1}\in A_{\beta}. % \end{align*} % \begin{align*} % (\tau\sigma)^{-1}\cdot\alpha_{n-1}\in A % &\Leftrightarrow\lambda_{n-1}=\lambda_n\\ % &\Leftrightarrow\pi_{n-1}=\pi_n\text{ and }\pi_1+\dots+\pi_{n-2}\text{ is even}\\ % &\Leftrightarrow\zeta_D(\pi,v)\text{ has a valley }(x,y)\text{ labelled }(u(n),u(n-1))\\ % &\qquad\text{ or }(-u(n-1),-u(n))\\ % &\Leftrightarrow(\tau\sigma)^{-1}\cdot\alpha_{n-1}=\alpha_{x,y}\in A_{\beta}. % \end{align*} % \begin{align*} % (\tau\sigma)^{-1}\cdot(-\tilde\alpha)\in A % &\Leftrightarrow\lambda_{n-1}+\lambda_n=2n-1\\ % &\Leftrightarrow\pi_{n-1}=\pi_n\text{ and }\pi_1+\dots+\pi_{n-2}\text{ is odd}\\ % &\Leftrightarrow\zeta_D(\pi,v)\text{ has a valley }(x,y)\text{ labelled }(-u(n),u(n-1))\\ % &\qquad\text{ or }(-u(n-1),u(n))\\ % &\Leftrightarrow(\tau\sigma)^{-1}\cdot(-\tilde\alpha)=\alpha_{x,y}\in A_{\beta}. % \end{align*} % We conclude that $\zeta(\lambda)=A_{\beta}$ as claimed and the proof is complete. \end{proof} % % The following result says in particular that $\zeta_B(\pi,v)$ is a diagonally labelled ballot path. % % \begin{myprop}{HLrisevalleyB} Let $(\pi,v)$ be a vertically labelled lattice path. Then $\zeta_B(\pi,v)$ has a valley labelled $(b,a)$ with $b\neq0$ if and only if $(\pi,v)$ has a rise labelled $(a,b)$ or $(-b,-a)$. % \end{myprop} % % \begin{proof} By \refp{risevalleyB} $\zeta(\pi,v)$ has a valley $(n+1-x,n+1-y)$ labelled $(b,a)$ if and only if $(\pi,v)$ has a rise labelled $(v(i),v(i+1))$ such that $(d_B(\pi,\id)^{-1}(i),d_B(\pi\id)^{-1}(i+1))$ equals one of the two $(y,x)$ or $(-x,-y)$. The claim therefore follows from % \begin{align*} % a % &=d_B(\pi,v)(x) % =\begin{cases}v\,d_B(\pi,\id)(d_B(\pi,\id)^{-1}(i+1)) % =v(i+1),\\ % v\,d_B(\pi,\id)(-d_B(\pi\id)^{-1}(i)) % =-v(i), % \end{cases}\\ % b % &=d_B(\pi,v)(x) % =\begin{cases}v\,d_B(\pi,\id)(d_B(\pi,\id)^{-1}(i)) % =v(i),\\ % v\,d_B(\pi,\id)(-d_B(\pi\id)^{-1}(i+1)) % =-v(i+1). % \end{cases} % \end{align*} % \end{proof} % % Let $\beta\in\mathcal B_n$ be a ballot path and $(i,j)$ a valley of $\beta$. Define the corresponding root as % \begin{align*} % \alpha_{i,j}=e_{n+1-i}+\sgn(n+1-j)e_{\abs{n+1-j}}. % \end{align*} % Here we use the convention of $\sgn(0)e_0$ to be the zero vector. The map $\beta\mapsto A_{\beta}$, where % \begin{align*} % A_{\beta} % =\big\{\alpha_{i,j}:(i,j)\text{ is a valley of }\beta\big\}, % % &=\{-e_i+e_j:1\leq in+1. \end{cases} \end{align*} Furthermore define \begin{align*} A_{\beta}=\big\{\alpha_{i,j}:(i,j)\text{ is a valley of }\beta\big\}. \end{align*} Then it is well-known that this correspondence identifies ballot paths with antichains in the root poset. \begin{mylem}{AbetaB} The map $\varphi:\mathcal B_{2n}\to\op{Antichains}(\Phi^+)$ given by $\beta\mapsto A_{\beta}$ is a bijection between ballot paths of length $2n$ and the antichains in the root poset $\Phi^+$ of type $B_n$.\noproof{} \end{mylem} \sk Next we define a combinatorial model for the non-nesting parking functions in terms of diagonally labelled ballot paths. In contrast to the previous section, the type $B_n$ objects are very similar to the analogous objects in type $C_n$, but not quite the same. \begin{mydef}{diagB} A \emph{diagonally labelled ballot path} of type $B_n$ is a pair $(\beta,w)$ of a ballot path $\beta\in\mathcal B_{2n}$ and a signed permutation $w\in\S_n^B$ such that for each valley $(i,j)$ of $\beta$ we have \begin{align*} w(n+1-i)>w(n+1-j). % \begin{cases} % w(n+1-j)&\quad\text{if }jn. % \end{cases} \end{align*} We denote the set of all diagonally labelled ballot paths of type $B_n$ by $\op{Diag}(B_n)$. \end{mydef} \sk These conditions are even simpler to write down than those we have encountered in types $C_n$ (or $D_n$), and can be visualised as follows: Given a ballot path $\beta$ and a signed permutation $w\in\S_n^B$ we can determine if the pair $(\beta,w)$ lies in the set $\op{Diag}(B_n)$ by writing the numbers $w(i)$, where $i=n,n-1\dots,1,0,-1,\dots,-n$, in the main diagonal. Then for each valley the number below has to be greater than the number to its right. \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.55] % \begin{scope} \ballot{6} \draw[very thick](0,0)--(0,2)--(1,2)--(1,4)--(2,4)--(2,5)--(3,5)--(3,6)--(4,6)--(4,7)--(5,7); \draw[xshift=5mm,yshift=5mm] (0,0)node{$6$} (1,1)node{$5$} (2,2)node{$3$} (3,3)node{$1$} (4,4)node{$4$} (5,5)node{$2$} (6,6)node{$0$} (5,7)node{$-2$} (4,8)node{$-4$} (3,9)node{$-1$} (2,10)node{$-3$} (1,11)node{$-5$} (0,12)node{$-6$} (0,2)node{$\bullet$} (1,4)node{$\bullet$} (2,5)node{$\bullet$} (3,6)node{$\bullet$} (4,7)node{$\bullet$}; \end{scope} % % \begin{scope}[xshift=11cm,yshift=2cm] % \ballot{4} % \draw[very thick](0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(2,6); % \draw[xshift=5mm,yshift=5mm] % (0,0)node{$-2$} % (1,1)node{$-3$} % (2,2)node{$-4$} % (3,3)node{$-1$} % (4,4)node{$0$} % (3,5)node{$1$} % (2,6)node{$4$} % (1,7)node{$3$} % (0,8)node{$2$} % (0,1)node{$\bullet$} % (1,2)node{$\bullet$}; % \end{scope} % \end{tikzpicture} \caption{A diagonally labelled ballot path $(\beta,w)\in\op{Diag}(B_6)$.} \label{Figure:diagB} \end{center} \end{figure} \begin{myex}{diagB} Consider the diagonally labelled ballot path in \reff{diagB}. The path $\beta$ has five valleys and corresponds to the antichain $A_{\beta}=\{e_6-e_4,e_5-e_2,e_4-e_1,e_3,e_2+e_1\}$. We may check that all required conditions on the signed permutation $w=[2,4,1,3,5,6]$ are fulfilled. For example $w(6+1-1)=6>w(6+1-3)=3$ and $w(6+1-4)=1>w(6+1-7)=0$ are the conditions imposed by the valleys $(1,3)$ and $(4,7)$. \end{myex} \begin{myprop}{diagB} The map $\varphi:\op{Diag}(B_n)\to\op{Park}(B_n)$ given by $(\beta,w)\mapsto[w,A_{\beta}]$ is a bijection. \end{myprop} \begin{proof} Let $\beta\in\mathcal B_{2n}$ be a ballot path and $w\in\S_n^B$ be a signed permutation. By \refl{AbetaB} and \refl{nnpark} it suffices to show that $(\beta,w)\in\op{Diag}(B_n)$ if and only if $w\cdot A_{\beta}\subseteq\Phi^+$. Suppose $w\cdot A_{\beta}\subseteq\Phi^+$. Then an easy case by case check reveals \begin{align*} (i,j)\text{ is a valley of }\beta &\Leftrightarrow\alpha_{i,j}\in A_{\beta}\\ &\Rightarrow w\cdot\alpha_{i,j}\in\Phi^+ \Leftrightarrow w(n+1-i)>w(n+1-j). \end{align*} Conversely if $(\beta,w)\in\op{Diag}(B_n)$ then \begin{align*} \alpha_{i,j}\in A_{\beta} &\Leftrightarrow(i,j)\text{ is a valley of }\beta\\ &\Rightarrow w(n+1-i)>w(n+1-j) \Leftrightarrow w\cdot\alpha_{i,j}\in\Phi^+.\qedhere \end{align*} \end{proof} \subsection{The area vector}\label{Section:areaB} Let $\pi\in\mathcal L_{n,n}$ be a lattice path. By \refp{lambdaB} the path $\pi$ corresponds bijectively to an element $\lambda\in\Q\cap(2n+1)\overline{\ac}$. The orbit of the finite torus under the action of the Weyl group represented by $\lambda$ corresponds to a dominant region of the Shi arrangement via the Anderson map of \refs{torus} as follows: Let $\waf_D\ac$ be the minimal alcove of this region, then we have $\A(\waf_D)\in W\cdot\lambda+(2n+1)\Q$ or equivalently $\waf_f\waf_D^{-1}\cdot0=\lambda$. In this section we describe how the affine permutation $\waf_D$ can be recovered from the lattice path $\pi$. In order to do so, write $\waf_D^{-1}=t_{\mu}\sigma$ where $\mu\in\Q$ and $\sigma\in\S_n^B$. Since $\waf_D^{-1}$ is Gra{\ss}mannian the signed permutation $\sigma$ is determined by \refl{sigma} once $\mu$ is known. On the other hand $\mu$ is related to $\lambda$ implicitly by $\lambda=\waf_f\waf_D^{-1}\cdot0$. Hence an explicit description of $\waf_f$ is needed. It is provided by the following lemma. \begin{mylem}{wfB} Let $\waf_f=t_{\nu}\tau$ where $\nu\in\Q$ and $\tau\in\S_n^B$. Then \begin{align*} %\label{eq:wfB} \nu= \begin{cases} (1,\dots,n-1,n) \\ (1,\dots,n-1,n+1) \end{cases} % \qquad\tau= \begin{cases} [1,\dots,n-1,n] &\qquad\text{if }n\equiv0,3\mod4,\\ [1,\dots,n-1,-n] &\qquad\text{if }n\equiv1,2\mod4. \end{cases} \end{align*} \end{mylem} \begin{proof} By \refl{wf} it suffices to show that $t_{\nu}\tau(\widetilde\Delta)=(\Delta-\delta)\cup\{-\tilde\alpha+h\delta\}$. Suppose $n\equiv0,3$ modulo $4$, then \begin{align*} t_{\nu}\tau(\alpha_i) &=\alpha_i-\skal{\nu,\alpha_i}\delta =\alpha_i-\delta &&\text{for all }0\leq i\leq n-1,\\ % t_{\nu}\tau( -\tilde\alpha+\delta) &=-\tilde\alpha+(1+\skal{\nu,\tilde\alpha})\delta =-\tilde\alpha+2n\delta. \end{align*} If $n-1\equiv1,2$ modulo $4$, then the only difference is \begin{align*} t_{\nu}\tau(\alpha_{n-1}) &=-\tilde\alpha+\skal{\nu,\tilde\alpha}\delta =-\tilde\alpha+2n\delta &&\text{and}\\ % t_{\nu}\tau(-\tilde\alpha+\delta) &=\alpha_{n-1}+(1-\skal{\nu,\alpha_{n-1}})\delta =\alpha_{n-1}-\delta.&&\qedhere \end{align*} % % Let $\nu$ and $\tau$ be as defined above. We know that $w_f$ maps the walls of the Sommers region to the walls of the $\n$ dilated fundamental alcove. That is, % \begin{align*} % w_f\big(\{-\alpha_i+\delta:0\leq i\leq n-1\}\cup\{\tilde\alpha+2\delta\}\big) % =\{-\alpha_i:0\leq i\leq n-1\}\cup\{\tilde\alpha+\n\delta\}. % \end{align*} % By \cite[Thm.~8.1]{Thiel2015} it suffices to show that $t_{\nu}\tau$ has the same property. Suppose $n\equiv0,3$ modulo $4$, then $t_{\nu}\tau(-\alpha_i+\delta)=-\alpha_i+(i-(i+1)+1)\delta=-\alpha_i$ and $t_{\nu}\tau(\tilde\alpha+2\delta)=\tilde\alpha+((n-1)+n+2)\delta=\tilde\alpha+\n\delta$. If $n\equiv1,2$ modulo $4$, then $t_{\nu}\tau(-\alpha_{n-1}+\delta)=\tau(-\alpha_{n-1})+((n-1)+(n+1)+1)\delta=\tilde\alpha+\n\delta$ and $t_{\nu}\tau(\tilde\alpha+2)=-\alpha_{n-1}+((n-1)-(n+1)+2)\delta=-\alpha_{n-1}$. \end{proof} \sk For example if $n=4$, respectively $n=5$, then \begin{align*} w_f=[-8,-16,-24,-36],&& w_f=[-10,-20,-30,-40,\,61]. \end{align*} \begin{mydef}{avB} Let $\pi\in\mathcal L_{n,n}$ be a lattice path, $\lambda$ be defined as in \refp{lambdaB} and $\nu$ and $\tau$ as in \refl{wfB} above. Then we define the \emph{area vector} of type $B_n$ of $\pi$ as \begin{align*} \mu &=\tau\cdot(\lambda-\nu) =\begin{cases} (\lambda_1-1,\dots,\lambda_{n-1}-n+1,\lambda_n-n)&\quad\text{if }n\equiv0,3\mod4,\\ (\lambda_1-1,\dots,\lambda_{n-1}-n+1,n+1-\lambda_n)&\quad\text{if }n\equiv1,2\mod4. \end{cases} \end{align*} \end{mydef} Note that similarly to the other types, the entry $\mu_i$ of the area vector counts the number of boxes in the $i$-th row that lie between the path $\pi$ and the alternating path $(EN)^n\in\mathcal L_{n,n}$, where $\mu_i$ is negative as long as $\pi$ is above $(EN)^n$. The only exception to this rule is the top row, where $\mu_n$ does not have as nice of an interpretation. \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.5] % \begin{scope} \fill[black!20] (0,0)rectangle(1,1) (2,1)--(4,1)--(4,3)--(3,3)--(3,2)--(2,2)--cycle (4,4)--(5,4)--(5,5)--(6,5)--(6,6)--(4,6)--cycle; \draw[gray](0,0)grid(6,6); \draw[very thick] (0,0)--(0,1)--(4,1)--(4,6)--(6,6); \end{scope} % \begin{scope}[xshift=10cm,yshift=1cm] \fill[black!20](2,2)--(3,2)--(3,3)--(4,3)--(4,4)--(2,4)--cycle; \draw[gray](0,0)grid(4,4); \draw[very thick](0,0)--(1,0)--(1,1)--(2,1)--(2,4)--(4,4); \end{scope} % \draw[white](16,0)--(16,1); \end{tikzpicture} \caption{The lattice paths with type $B$ area vectors $(-1,2,1,0,-1,3)$ and $(0,0,-1,3)$.} \label{Figure:areaB} \end{center} \end{figure} \begin{myex}{areaB} Consider the two lattice paths in \reff{areaB}. In case of the first path we have $\lambda=(0,4,4,4,4,4)$ and $n=6\equiv2$ modulo $4$. Hence we compute $\mu=(-1,2,1,0,-1,3)$. The corresponding translation in $\affS_6^B$ is given by $t_{\mu}=[14,-24,-10,4,18,-33]$. The unique signed permutation $\sigma\in\S_6^B$ such that $t_{\mu}\sigma$ is Gra{\ss}mannian is given by $\sigma=[4,-3,1,5,-2,-6]$. Thus we obtain \begin{align*} \waf_D^{-1}=t_{\mu}\sigma=[4,10,14,18,24,33] \qquad\text{and}\qquad \waf_D=[-10,21,11,1,-9,33]. \end{align*} In particular note the difference to \refe{areaC}. In case of the second path we find $n=4\equiv0$ modulo $4$, $\lambda=(1,2,2,7)$ and therefore $\mu=(0,0,-1,3)$. The translation $t_{\mu}=[1,2,12,-23]\in\affS_4^B$ determines a unique signed permutation $\sigma=[1,2,3,-4]\in\S_4^B$ such that $t_{\mu}\sigma$ is Gra{\ss}mannian. We obtain $\waf_D^{-1}=[1,2,12,23]$ and $\waf_D=[1,2,-6,23]$. \end{myex} \subsection{The diagonal reading word}\label{Section:readB} Let $(\pi,v)\in\op{Vert}(B_n)$ be a vertically labelled lattice path. By \refp{vertB} the pair $(\pi,v)$ corresponds to an element $u\cdot\lambda+(2n+1)\Q$ of the finite torus, and thus via the Anderson map (\reft{anderson}) to a region $R$ of the Shi arrangement with minimal alcove $\waf_R\ac$. Write $\waf_R=w\waf_D$ where $w\in\S_n^B$ is a signed permutation and $\waf_D\ac$ is the minimal alcove of a dominant Shi region. Our aim for this section is to recover the signed permutation $w$, and thereby $\waf_R$, from the labelled path $(\pi,v)$. The affine permutation $\waf_D=t_{\mu}\sigma$ can be obtained from $\lambda$ as discussed in the previous section. Thus in theory we are already able to compute $w=u\tau\sigma$ using \refp{utausig}. As in types $C_n$ and $D_n$ above, we demonstrate a convenient way to read $w$ off the vertical labelling. \begin{mydef}{dB} Let $(\pi,v)\in\mathcal{L}_{n,n}\times\S_n^B$ and $\mu$ be the type $B_n$ area vector of $\pi$. Define the type $B_n$ \emph{diagonal reading word} $d_B(\pi,v)$ as follows: For each $i=0,1,2,\dots$ first write down the labels $v(j)$ of the rows with $\mu_j=-i$ from bottom to top, then write down the negative labels $-v(j)$ of rows with $\mu_j=i+1$ from top to bottom. Finally, if $\mu_{n-1}+\mu_n$ is even, change the sign of the label coming from the top row. \end{mydef} \sk Note that the diagonal reading word of type $B_n$ is closely related to that of type $D_n$. Indeed the above definition is the same as \refd{readD} except that some technical details are less complicated in type $B_n$. \begin{myex}{readB} Let us recall the vertically labelled lattice paths from \reff{vertB}. The area vector of the first path is $\mu=(-1,2,1,0,-1,3)$ and its labels are given by $v=[1,-5,-4,2,3,6]$. Reading off the vertical labels according to the entries of $\mu$ yields $[2,4,1,3,5,-6]$. Since $\mu_5+\mu_6=2$ is even we need to change the sign of the label coming from the top row, that is, $v(6)=6$. We obtain $d_B(\pi,v)=[2,4,1,3,5,6]$. In light of \refp{readB} below the reader may wish to verify that \begin{align*} d_B(\pi,v) &=u\tau\sigma =[1,-5,-4,2,3,6]\cdot[1,2,3,4,5,-6]\cdot[4,-3,1,5,-2,-6], \end{align*} where $\sigma\in\S_6^B$ was already computed in \refe{areaB}. We are also able to compute the affine permutation \begin{align*} \waf_R=d_B(\pi,v)\waf_D=[-12,21,9,2,-10,33] \end{align*} that takes the fundamental alcove to the minimal alcove of the Shi region $R$ corresponding to the vertically labelled lattice path $(\pi,v)$. Indeed, we have $\waf_R\waf_f^{-1}=[1,47,48,54,55,58]$ and \begin{align*} \mathcal{A}(\waf_R) &=-\waf_R\waf_f^{-1}\cdot0+13\Q =(0,4,4,-4,-4,4)+13\Q, \end{align*} which agrees with the results in \refe{vertB}. Again it is interesting to compare this to \refe{readC} to see the differences and similarities between type $B_n$ and $C_n$. \sk The second path in \reff{vertB} has area vector $\mu=(0,0,-1,3)$ and its labels are given by $v=[-1,-4,-3,-2]$. Reading off the labels according to the entries of $\mu$ yields $[-1,-4,-3,2]$. As above $\mu_2+\mu_3=2$ is even, thus we need to change the sign of the label $v(4)=-2$. We obtain $d_B(\pi,v)=[-1,-4,-3,-2]$. Using once more the results from \refe{areaB}, we compute $\waf_R=d_B(\pi,v)\waf_D=[-1,-4,-12,29]$, $\waf_R\waf_f^{-1}=[8,14,15,65]$ and \begin{align*} \mathcal{A}(\waf_R) &=-\waf_R\waf_f^{-1}\cdot0+9\Q =(-1,7,-2,-2)+9\Q. \end{align*} \end{myex} \sk The following proposition asserts that the diagonal reading word is the correct signed permutation. \begin{myprop}{readB} Let $(\pi,v)\in\op{Vert}(B_n)$ be a vertically labelled lattice path with area vector $\mu$, define $u$ and $\lambda$ as in \refp{vertB}, $\tau$ as in \refl{wfB} and $\sigma$ as in \refl{sigma} such that $t_{\mu}\sigma$ is Gra{\ss}mannian. Then $d_B(\pi,v)=u\tau\sigma$. \end{myprop} \begin{proof} Let $i\in[n]$ and chose $j\in[n]$ such that $\abs{d_B(\pi,v)(i)}=\abs{v(j)}$. Then \begin{align*} i &=\#\big\{r\in[n]:\abs{\mu_r}<\abs{\mu_j}\big\} +\#\big\{r\in[n]:j\leq r,\mu_r=\mu_j>0\big\} \\ &\quad+\#\big\{r\in[j],\mu_r=\mu_j\leq0\big\} +\#\big\{r\in[n],\mu_r=-\mu_j>0\big\} \\ &=\#\big\{r\in[r]:\abs{\mu_r\n+r}\leq\abs{\mu_j\n-j}\big\} =\abs{\sigma^{-1}(j)} \end{align*} Thus $\abs{d_B(\pi,v)(i)}=\abs{v\sigma(i)}=\abs{u\tau\sigma(i)}$. If $j0$ then $d_B(\pi,v)=-u\tau(j)$ and $\sigma(i)=-j$. If $j=n$ and $\mu_n\leq0$ then \begin{align*} d_B(\pi,v) &=(-1)^{1+\mu_{n-1}+\mu_n}v(n)\\ &=(-1)^{1+\mu_{n-1}+\mu_n}(-1)^{\lambda_{n-1}+\lambda_n}u(n)\\ &=(-1)^{1+\nu_{n-1}+\nu_n}u(n)\\ &=u\tau(n) =u\tau(\sigma(j)) \end{align*} and analogously one treats the case where $j=n$ and $\mu_n>0$. \end{proof} \subsection{The zeta map}\label{Section:zetaB} We can now define the (Haglund--Loehr) zeta map of type $B_n$, which is closely related to type $D_{n+1}$. \begin{mydef}{zetaB} Given a lattice path $\pi\in\mathcal L_{n,n}$ with type $B_n$ area vector $\mu$ we define its image under the type $B_n$ \emph{zeta map} as \begin{align*} \zeta_B(\pi) =\LS_{2n}^-(\mu)\RS_{2n}^+(\mu) \LS_{2n-1}^-(\mu)\RS_{2n-1}^+(\mu)\cdots \LS_1^-(\mu)\RS_1^+(\mu) \LS_0^-(\mu)\big(N\RS_0^+(\mu)\big)^{\circ}, \end{align*} where $(N\RS_0^+(\mu))^{\circ}$ is obtained from $N\RS_0^+(\mu)$ by deleting the last letter. Moreover, given a vertically labelled lattice paths $(\pi,v)\in\op{Vert}(B_n)$ we define its image under the \emph{Haglund--Loehr-zeta map} of type $B_n$ as \begin{align*} \zeta_B(\pi,v)=(\zeta_B(\pi),d_B(\pi,v)). \end{align*} \end{mydef} \begin{figure}[t] \begin{center} \begin{tikzpicture}[scale=.5] \begin{scope} \draw[gray](0,0)grid(6,6); \draw[very thick](0,0)--(0,1)--(4,1)--(4,6)--(6,6); \draw[xshift=5mm,yshift=5mm] (-1,0)node{$1$} (3,1)node{$-5$} (3,2)node{$-4$} (3,3)node{$2$} (3,4)node{$3$} (3,5)node{$6$}; \end{scope} % \draw[xshift=8cm,yshift=2.5cm,->,thick](0,0)--node[above]{$\zeta_B$}(1,0); % \begin{scope}[xshift=11cm,yshift=-35mm] \ballot{6} \draw[very thick](0,0)--(0,2)--(1,2)--(1,4)--(2,4)--(2,5)--(3,5)--(3,6)--(4,6)--(4,7)--(5,7); \draw[very thick,red] (0,0)--(0,1) (1,2)--(1,4) (2,5)--(3,5)--(3,6)--(4,6); \draw[xshift=5mm,yshift=5mm] (0,0)node{$6$} (1,1)node{$5$} (2,2)node{$3$} (3,3)node{$1$} (4,4)node{$4$} (5,5)node{$2$} (6,6)node{$0$} (5,7)node{$-2$} (4,8)node{$-4$} (3,9)node{$-1$} (2,10)node{$-3$} (1,11)node{$-5$} (0,12)node{$-6$} (0,2)node{\large{$\bullet$}} (1,4)node{\large{$\bullet$}} (2,5)node{\large{$\bullet$}} (3,6)node{\large{$\bullet$}} (4,7)node{\large{$\circ$}}; \end{scope} % \end{tikzpicture} \caption{A vertically labelled lattice path and its image under the Haglund--Loehr-zeta map.} \label{Figure:zetaB1} \end{center} \end{figure} \begin{myex}{zetaB} Consider the vertically labelled lattice path $(\pi,v)\in\op{Vert}(B_6)$ drawn in \reff{zetaB1}, which has area vector $\mu=(-1,2,1,0,-1,3)$. We compute \begin{align*} \RS_3^+(\mu)=N, \LS_2^-(\mu)&=\emptyset, \RS_2^+(\mu)=NE, \LS_1^-(\mu)=NN,\\ \RS_1^+(\mu)=EN, \LS_0^-(\mu)&=ENE, \RS_0^+(\mu)=EN, \end{align*} Note that the last segment is replaced by $\big(N\RS_0^+(\mu)\big)^{\circ}=NE$. In light of \reft{risevalleyB} below we remark that the initial North step of $\pi$ labelled by $v(1)=1$ corresponds to a valley labelled $(1,0)$. Moreover the rises of $(\pi,v)$, which are labelled $(-5,-4),(-4,2),(2,3)$ and $(3,6)$, correspond exactly to the remaining valleys of $(\beta,w)$. \end{myex} \sk Note that $\zeta_B(\pi)\in\mathcal B_{2n}$ is by definition a ballot path. \begin{mythrm}{zetaB} The zeta map $\zeta_B:\mathcal L_{n,n}\to\mathcal B_{2n}$ is a bijection. \end{mythrm} \begin{proof} Let $\pi\in\mathcal L_{n,n}$ be a lattice path with type $B_n$ area vector $\mu$. Consider the path $N\pi\in\mathcal L_{n,n+1}$. Recalling the bijection $\zeta_D^*:\mathcal L_{n,n+1}\to\mathcal B_{2n+1}$ from \reft{oddballotbij} we have \begin{align*} \zeta_D^*(N\pi) =\LS_{2n}^-(\mu)\RS_{2n}^+(\mu) \LS_{2n-1}^-(\mu)\RS_{2n-1}^+(\mu)\cdots \LS_1^-(\mu)\RS_1^+(\mu) \LS_0^-(\mu)N\big(N\RS_0^+(\mu)\big)^{\circ}. \end{align*} It follows from the proof of \reft{oddballotbij} that $\zeta_D^*$ restricts to a bijection from the set of lattice paths in $\mathcal L_{n,n+1}$ that begin with a North step to the set of ballot paths in $\mathcal B_{2n+1}$ whose $(n+1)$-st North step is not followed by an East step. Since $\zeta_D^*(N\pi)$ is easily obtained from $\zeta_B(\pi)$ by inserting a North step, we conclude that $\zeta_B$ is also a bijection. \end{proof} \sk The next theorem further exploits the relation between $\zeta_B$ and $\zeta_D$ to connect the rises of $(\pi,v)$ to the valleys of $\zeta_B(\pi,v)$. Let $(\pi,v)\in\mathcal{L}_{n,n}\times\S_n^B$ and $i$ be a rise of $\pi$. Recall that we say the rise $i$ is labelled by $(v(i),v(i+1))$. On the other hand let $(\beta,w)$ be a pair of a ballot path $\beta\in\mathcal B_{2n}$ and a signed permutation $w\in\S_n^B$, and let $(i,j)$ be a valley of $\beta$. Then we say $(\beta,w)$ has a valley $(i,j)$ that is labelled by $(w(n+1-i),w(n+1-j))$. \begin{mythrm}{risevalleyB} Let $(\pi,v)\in\mathcal{L}_{n,n}\times\S_n^B$ be a pair of a lattice path and a signed permutation, and $a,b\in v([n])$. Then $(\pi,v)$ has a rise labelled $(a,b)$ if and only if $(\zeta_B(\pi),d_B(\pi,v))$ has a valley labelled $(b,a)$ or $(-a,-b)$. Furthermore $\pi$ begins with a North step if and only if $(\zeta_B(\pi),d_B(\pi,v))$ has a valley labelled $(v(1),0)$. \end{mythrm} \begin{proof} Let $(\pi,v)\in\op{Vert}(B_n)$ and consider the vertically labelled path $(N\pi,\bar{v})$ where we label the initial North step by $0$ and $\pi$ retains the labelling $v$. This is basically an element of $\op{Vert}(D_{n+1})$ except that the absolute values of all labels have been decreased by one. We will prove the claim using \reft{risevalleyD}. First notice that $\zeta_B(\pi,v)$ is obtained from $\zeta_D(N\pi,\bar{v})$ simply by deleting the $(n+1)$-st North step and its label, which is always $0$. Furthermore let $a,b\in\Z-\{0\}$, then $(\pi,v)$ has a rise $i$ labelled $(a,b)$ if and only if $(N\pi,\bar{v})$ has a rise $i+1$ labelled $(a,b)$. This is the case if and only if $\zeta_D(N\pi,\bar{v})$ has a valley labelled $(b,a)$ or $(-a,-b)$ and equivalently $\zeta_B(\pi,v)$ has a valley labelled $(a,b)$ or $(-a,-b)$. On the other hand $\pi$ begins with a North step if and only if $(N\pi,\bar{v})$ has rise $i=1$ labelled by $(\pm0,v(1))$. This is equivalent to $\zeta_D(N\pi,\bar{v})$ having a valley labelled $(v(1),\pm0)$, which is the case if and only if $\zeta_B(\pi,v)$ has a valley labelled $(v(1),0)$. % % % Throughout this proof we use the following notation: For $i\in\Z$ define % \begin{align*} % \bar{i}= % \begin{cases} % i+1&\quad\text{if }i\geq0,\\ % i-1&\quad\text{if }i<0. % \end{cases} % \end{align*} % Define a signed permutation $\bar{v}\in\S_{n+1}^B$ by % \begin{align*} % \abs{\bar{v}(1)}&=1,\\ % \bar{v}(i+1)&=\overline{v(i)}\qquad\text{for }i\in[n],\\ % \prod_{i=1}^{n+1}\sgn(\bar{v}(i))&=(-1)^{\lambda_{n-1}+\lambda_n}. % \end{align*} % Then $(N\pi,\bar{v})\in\op{Vert}(D_{n+1})$. Moreover $(\pi,v)$ has a rise $i$ labelled $(a,b)$ if and only if $(N\pi,\bar{v})$ has a rise $i+1$ labelled $(\bar{a},\bar{b})$. By \reft{risevalleyD} this is the case if and only if $\zeta_D(N\pi,\bar{v})$ has a valley labelled $(\bar{b},\bar{a})$ or $(-\bar{a},-\bar{b})$. % % Let $(x,y)$ be this valley of $\zeta_D(N\pi,\bar{v})$. If $y\leq n$ then $(x,y)$ is a valley of $\zeta_B(\pi,v)$. This valley is then labelled by $(b,a)$ or $(-a,-b)$ because $d_D(N\pi,\bar{v})(i+1)=\overline{d_B(\pi,v)(i)}$ for $i\in[n]$. % % If $y\geq n+2$ then $(x,y-1)$ is a valley of $\zeta_B(\pi,v)$ \end{proof} \sk From \reft{zetaB} and \reft{risevalleyB} we obtain the following. \begin{mythrm}{HLzetaB} The Haglund--Loehr-zeta map $\zeta_B:\op{Vert}(B_n)\to\op{Diag}(B_n)$ is a bijection. \end{mythrm} \sk We conclude this section by proving that the combinatorial zeta $\zeta_B$ map is indeed the type $B_n$ instance of the uniform zeta map. \begin{mythrm}{uniformB} Let $\Phi$ be the root system of type $B_n$ with coroot lattice $\Q$ and zeta map $\zeta$. Moreover define $\psi$ and $\varphi$ as in \refp{vertB} respectively \refp{diagB}. Then the following diagram commutes. % \begin{figure}[h] \begin{center} \begin{tikzpicture} \draw (0,0) node(q){$\Q/(2n+1)\Q$}; \draw (4,0) node(park){$\mathrm{Park}(\Phi)$}; \draw (0,1.4) node(v){$\op{Vert}(B_n)$}; \draw (4,1.4) node(d){$\op{Diag}(B_n)$}; \draw[->] (q)--node[below]{$\zeta$}(park); \draw[->] (v)--node[above]{$\zeta_B$}(d); \draw[->] (v)--node[left]{$\psi$}(q); \draw[->] (d)--node[right]{$\varphi$}(park); \end{tikzpicture} \end{center} % \end{figure} \end{mythrm} % \begin{proof} Let $(\pi,v)\in\op{Vert}(B_n)$, set $(\beta,w)=\zeta_B(\pi,v)\in\op{Diag}(B_n)$ and define $\lambda$ and $u$ as in \refq{lambdaB} and $\refq{uB}$ such that $\psi(\pi,v)=u\cdot\lambda+(2n+1)\Q$. Finally define $\tau$ as in \refl{wfB}, let $\mu$ be the type $B_n$ area vector of $\pi$ and chose $\sigma\in\S_n^B$ such that $t_{\mu}\sigma$ is Gra{\ss}mannian and $w=d_B(\pi,v)=u\tau\sigma$ by \refp{readB}. To prove that claim it suffices to show that $A_{\beta}=\zeta(\lambda)$. % % Recall that % \begin{align*} % \zeta(\lambda) % &=\big\{(\tau\sigma)^{-1}\cdot\alpha:\alpha\in\{\alpha_0,\alpha_1,\dots,\alpha_{n-1},-\tilde\alpha\}\text{ and }s_{\alpha}\cdot\lambda=\lambda\big\}. % \end{align*} % Hence we have the chain of equivalences % \begin{align*} % (\tau\sigma)^{-1}\cdot\alpha_0\in\zeta(\lambda) % &\Leftrightarrow\lambda_1=0\\ % &\Leftrightarrow\pi_1=0\\ % (\text{by \reft{risevalleyB}}) % &\Leftrightarrow(\beta,w)\text{ has a valley }(x,n+1)\text{ labelled by }(u(1),0)\\ % &\Leftrightarrow \alpha_{x,n+1}\in A_{\beta}\text{ and }u\tau\sigma(n+1-x)=u(1)\\ % &\Leftrightarrow (\tau\sigma)^{-1}\cdot e_1=e_{(\tau\sigma)^{-1}(1)}=e_{n+1-x}=\alpha_{x,n+1}\in A_{\beta}. % \end{align*} % Similarly for $i\in[n-2]$ we have % \begin{align*} % (\tau\sigma)^{-1}\cdot\alpha_i\in\zeta(\lambda) % &\Leftrightarrow\lambda_i=\lambda_{i+1}\\ % &\Leftrightarrow\pi_i=\pi_{i+1}\\ % % (\text{by \reft{risevalleyB}}) % &\Leftrightarrow(\beta,w)\text{ has a valley }(x,y)\text{ labelled by }(u(i+1),u(i))\\ % &\qquad\text{ or by }(-u(i),-u(i+1))\\ % % &\Leftrightarrow \alpha_{x,y}\in A_{\beta}\text{ and }u\tau\sigma(n+1-x)=u(i+1), u\tau\sigma(n+1-y)=u(i)\\ % % &\qquad\text{ or }u\tau\sigma(n+1-x)=-u(i), u\tau\sigma(n+1-y)=-u(i+1)\\ % &\Leftrightarrow (\tau\sigma)^{-1}\cdot(e_{i+1}-e_i)=\alpha_{x,y}\in A_{\beta} % \end{align*} % \begin{align*} % (\tau\sigma)^{-1}\cdot\alpha_n\in\zeta(\lambda) % &\Leftrightarrow\lambda_{n-1}=\lambda_n\\ % &\Leftrightarrow\pi_{n-1}=\pi_n\text{ and }\pi_1+\dots+\pi_{n-1}\text{ is even}\\ % &\Leftrightarrow(\beta,w)\text{ has a valley }(x,y)\text{ labelled by }(u(n),u(n-1))\\ % &\qquad\text{ or by }(-u(n-1),-u(n))\\ % &\Leftrightarrow (\tau\sigma)^{-1}\cdot(e_n-e_{n-1})=\alpha_{x,y}\in A_{\beta} % \end{align*} % \begin{align*} % (\tau\sigma)^{-1}\cdot(-\tilde\alpha)\in\zeta(\lambda) % &\Leftrightarrow\lambda_{n-1}+\lambda_n=2n+1\\ % &\Leftrightarrow\pi_{n-1}=\pi_n\text{ and }\pi_1+\dots+\pi_{n-1}\text{ is odd}\\ % &\Leftrightarrow(\beta,w)\text{ has a valley }(x,y)\text{ labelled by }(-u(n),u(n-1))\\ % &\qquad\text{ or by }(-u(n-1),u(n))\\ % &\Leftrightarrow (\tau\sigma)^{-1}\cdot(-e_{n-1}-e_n)=\alpha_{x,y}\in A_{\beta}. % \end{align*} % \end{proof} \begin{proof} Let $(\pi,v)\in\op{Vert}(B_n)$ and set $(\beta,w)=\zeta_B(\pi,v)\in\op{Diag}(B_n)$. Define $\lambda$ and $u$ as in \refq{lambdaB} respectively \refq{uB} such that $\psi(\pi,v)=u\cdot\lambda+(2n+1)\Q$. Let $\tau$ be as in \refl{wfB} and $\mu$ be the type $B_n$ area vector of $\pi$. Chose $\sigma\in\S_n^B$ such that $t_{\mu}\sigma\in\affS_n^B$ is a Gra{\ss}mannian affine permutation. Recall that by \refp{antichain} and \refp{utausig} it suffices to show that $w=u\tau\sigma$ and $A_{\beta}=A$, where \begin{align*} A &=\big\{(\tau\sigma)^{-1}\cdot\alpha:\alpha\in\Delta\cup\{-\tilde\alpha\}\text{ and }s_{\alpha}\cdot\lambda=\lambda\big\}. \end{align*} The first claim is taken care of by \refp{readB}. In order to demonstrate the second claim we first show $A\subseteq A_{\beta}$. Therefore let $i\in[n-2]$ and suppose that $(\tau\sigma)^{-1}\cdot\alpha_i\in A$. Then $\lambda_i=\lambda_{i+1}$, hence $\pi_i=\pi_{i+1}$ and $i$ is a rise of $\pi$. By \reft{risevalleyB} $(\beta,w)$ has a valley $(x,y)$ labelled either $(v(i+1),v(i))$ or $(-v(i),-v(i+1))$. Note that in particular $y\neq n+1$. Moreover $u(i)=v(i)$ and $u(i+1)=v(i+1)$ for this choice of $i$. We obtain \begin{align*} (w(n+1-x),w(n+1-y)) \in\big\{(u(i+1),u(i)),(-u(i),-u(i+1))\big\}. \end{align*} Applying $u^{-1}$ and using the fact that $w=u\tau\sigma$, yields \begin{align*} (\tau\sigma(n+1-x),\tau\sigma(n+1-y) \in\big\{(i+1,i),(-i,-i-1)\big\}. \end{align*} Recalling that \begin{align*} \alpha_{x,y}= \begin{cases} e_{n+1-x}+e_{n+1-y}&\quad\text{if }yn+1, \end{cases} \end{align*} we compute $(\tau\sigma)^{-1}\cdot\alpha_i=\alpha_{x,y}\in A_{\beta}$ in all cases. Next suppose $(\tau\sigma)^{-1}\cdot\alpha_0\in A$. Then $\lambda_1=0$ and thus $\pi_1=0$. By \reft{risevalleyB} $(\beta,w)$ has a valley $(x,n+1)$ labelled $(v(1),0)=(u(1),0)$. From $u\tau\sigma(n+1-x)=w(n+1-x)=u(1)$ we obtain $\tau\sigma(n+1-x)=1$, and compute $(\tau\sigma)^{-1}\cdot\alpha_0=e_{n+1-x}=\alpha_{x,n+1}\in A_{\beta}$. Similarly suppose $(\tau\sigma)^{-1}\cdot\alpha_{n-1}\in A$. Then $\lambda_{n-1}=\lambda_n$. Consequently $\pi_{n-1}=\pi_n$, that is, $n-1$ is a rise of $\pi$, and $\pi_1+\dots+\pi_{n-2}$ is even. In particular $u(n)=v(n)$. By \reft{risevalleyB} $(\beta,w)$ has a valley $(x,y)$ labelled $(u(n),u(n-1))$ or $(-u(n-1),-u(n))$. As in the cases above we conclude that $(\tau\sigma)^{-1}\cdot\alpha_{n-1}=\alpha_{x,y}\in A_{\beta}$. Finally suppose $(\tau\sigma)^{-1}\cdot(-\tilde\alpha)\in A$. Then $\lambda_{n-1}+\lambda_n=2n+1$. Here again $\pi_{n-1}=\pi_n$, but contrary to the previous case $\pi_1+\dots+\pi_{n-2}$ is now odd. Therefore $u(n)=-v(n)$. By \reft{risevalleyB} $(\beta,w)$ has a valley labelled $(u(n-1),-u(n))$ or $(u(n),-u(n-1))$. As before we compute $(\tau\sigma)^{-1}\cdot(-\tilde\alpha)=\alpha_{x,y}\in A_{\beta}$. \sk It remains to prove the reverse inclusion $A_{\beta}\subseteq A$. Therefore assume $\alpha_{x,y}\in A_{\beta}$ for some valley $(x,y)$ of $(\beta,w)$. If $yn+1$ can be treated in a similar fashion as the case $yu(j)\}\\ &\quad+\#\{(i,j):i-u(j)\}\\ &\quad+\#\{i:\mu_{n-i+1}=0,u(i)<0\} \end{align*} We call a pair $(i,j)\in[n]\times[n]$ or index $i\in[n]$ contributing to $\dinv_C'(\pi,u)$ a \emph{diagonal inversion} of $(\pi,u)$. \end{mydef} \begin{myex}{dinvC} Consider the vertically labelled lattice path $(\pi,u)$ in \reff{vertC}. The path $\pi$ has area vector $\mu=(2,1,0,-1,-2,1)$. The pair $(1,5)$ is a diagonal inversion of type $\mu_{n-i+1}=\mu_{n-j+1}$, $(1,4)$ is a diagonal inversion of type $\mu_{n-i+1}=\mu_{n-j+1}+1$, $(1,3),(2,6)$ and $(3,5)$ are diagonal inversions of type $\mu_{n-i+1}=-\mu_{n-j+1}$, $(1,4),(3,6)$ and $(4,5)$ are diagonal inversions of type $\mu_{n-i+1}=-\mu_{n-j+1}+1$, and $4$ is a diagonal inversion of type $\mu_{n-i+1}=0$. Together that makes $\dinv(\pi)=9$. Note that the pair $(1,4)$ is counted twice! \sk We have $u(1)=1-u(4)$ thus $(1,4)$ is still a diagonal inversion of type $\mu_{n-i+1}=-\mu_{n-i+1}+1$. The only other two diagonal inversions of $\pi$ that do not contribute to $\dinv_C'(\pi,u)$ are $(2,6)$ and $4$. We obtain $\dinv_C'(\pi,u)=6$. \end{myex} In this case it is quite easy to show that $\dinv_C$ and $\dinv_C'$ are the type $C_n$ instances of the uniform dinv statistics. Moreover, note that if $\pi\in\mathcal{L}_{n,n}$ is a Dyck path then it is actually assigned two numbers $\dinv_A(\pi)$ and $\dinv_C(\pi)$. In this case \refd{dinvC} agrees with the definition of Haiman, that is, $\dinv_A(\pi)=\dinv_C(\pi)$. \sk In the remainder of this section we relate $\area$ and $\dinv$ by means of the zeta map. We start with a useful and well-known lemma on dominant elements of the affine Weyl group. \begin{mylem}{dominant} Let $\alpha\in\Phi^+$ be a positive root and $\waf\in\wa$ be dominant with $\waf=t_{\lambda}\sigma$, where $\lambda\in\Q$ and $\sigma\in W$. If $\skal{\lambda,\alpha}=0$ then $\sigma^{-1}\cdot\alpha\in\Phi^+$. \end{mylem} \begin{proof} Suppose $\skal{\lambda,\alpha}=0$. The height function $\op{ht}:\Phi\to\R$ extends to a linear functional on $V$. Thus we may choose $v\in V$ with $\skal{v,\beta}=\op{ht}(\beta)/h$ for all $\beta\in\Phi$, where $h$ is the Coxeter number of $\Phi$. Note that $v\in\ac$ by definition. Thus $\skal{\waf\cdot v,\alpha}>0$ since $\waf$ is dominant. We compute \begin{align*} \frac{\op{ht}(\sigma^{-1}\cdot\alpha)}{h} &=\skal{v,\sigma^{-1}(\alpha)} =\skal{\sigma\cdot v,\alpha} =\skal{\lambda+\sigma\cdot v,\alpha} =\skal{\waf\cdot v,\alpha} >0.\qedhere \end{align*} \end{proof} Furthermore, we need the following property of the affine transformation $\wf$. \begin{mylem}{wh+1} Let $\alpha\in\Phi^+$ be a positive root and $\nu\in\Q$, $\tau\in W$ such that $\wf=t_{\nu}\tau$. Then \begin{align*} \skal{\wf\cdot0,\alpha} &= \begin{cases} \op{ht}(\alpha)&\quad\text{if }\tau^{-1}\cdot\alpha\in\Phi^+,\\ \op{ht}(\alpha)+1&\quad\text{if }\tau^{-1}\cdot\alpha\in-\Phi^+. \end{cases} \end{align*} \end{mylem} \begin{proof} Choose $v\in V$ such that $\skal{v,\alpha}=\op{ht}(\alpha)/h$ for all $\alpha\in\Phi^+$. % Moreover, let $\wf=t_qs$ where $q\in\coQ$ and $s\in W$. %\cite[Thm.~3.6.2]{Thiel:thesis} By \cite[Thm.~3.6.2]{Thiel:thesis} $\wf\cdot v=(h+1)v$. It follows that \begin{align*} \skal{\wf\cdot0,\alpha} % &=\skal{x,\alpha} &=\skal{(h+1)v-\tau\cdot v,\alpha} % &=\frac{(h+1)}{h}\op{ht}(\alpha)-\frac{1}{h}\op{ht}(\tau^{-1}\cdot\alpha) =\op{ht}(\alpha)+\frac{1}{h}\big(\op{ht}(\alpha)-\op{ht}(\tau^{-1}\cdot\alpha)\big). \end{align*} The claim follows since $\skal{\nu,\alpha}$ is an integer. \end{proof} We now state the main result of this section. \begin{mythrm}{areadinv} Let $\Phi$ be an irreducible crystallographic root system with Weyl group $W$, coroot lattice $\Q$, Coxeter number $h$ and zeta map $\zeta$, and let $X\in\Q/(h+1)\Q$ be an element of the finite torus. % $A\subseteq\Phi^+$ be an anti-chain and $s\in W$ such that $s(A)\subseteq\Phi^+$. Let $\lambda\in\Q\cap(h+1)\overline{\ac}$ and $u\in W$ such that $u\cdot\lambda\in X$ is the canonical representative as in \refl{canon}. Then \begin{align*} \dinv(\lambda)=\area(\bar{\zeta}(\lambda)) &&\text{and}&& \dinv'(X)=\area'(\zeta(X)). \end{align*} \end{mythrm} \begin{proof} % $u\cdot\alpha\in\Phi^+$ for all $\alpha\in\Delta$ with $\skal{\lambda,\alpha}=0$ and such that $u(-\hr)\in\Phi^+$ if $\skal{\lambda,\hr}=h+1$. Choose $\waf\in\wdom$, $\mu,\nu\in\Q$ and $\sigma,\tau,w\in W$ such that $\wf=t_{\nu}\tau$, $\waf=t_\mu\sigma$, $w\waf\in\wshi$ and \begin{align*} u\cdot\lambda &=-w\waf\wf^{-1}\cdot0 =-w(\mu)+w\sigma\tau^{-1}(\nu). \end{align*} By \refs{uniform} \begin{align*} \zeta(u\cdot\lambda+(h+1)\Q)=[u\tau\sigma^{-1},\sigma\tau^{-1}\cdot\walls]. \end{align*} We prove the more precise statement that the bijection \begin{align*} I(\sigma\tau^{-1}\cdot\walls)\to\big\{\pm \tau\sigma^{-1}(\beta):\beta\in I(\sigma\tau^{-1}\cdot\walls)\big\}\cap\Phi^+ \end{align*} maps the positive roots contributing to $\area(\bar{\zeta}(\lambda))$ to the positive roots contributing to $\dinv(\lambda)$. Furthermore this map restricts to a bijection between the positive roots contributing to $\area'$ and $\dinv'$. On the one hand a computation shows that \begin{align*} \skal{\lambda,\alpha} &=\skal{\nu-\tau\sigma^{-1}\cdot\mu,\alpha}\\ &=\op{ht}(\alpha)+\frac{1}{h}\big(\op{ht}(\alpha)-\op{ht}(\tau^{-1}\cdot\alpha)\big)-\skal{\mu,\sigma\tau^{-1}\cdot\alpha}\\ &=\op{ht}(\alpha)-\skal{\mu,\sigma\tau^{-1}\cdot\alpha}+ \begin{cases} 0&\quad\text{if }\tau^{-1}\cdot\alpha\in\Phi^+,\\ 1&\quad\text{if }\tau^{-1}\cdot\alpha\in-\Phi^+, \end{cases} \end{align*} for all $\alpha\in\Phi^+$ due to \refl{wh+1}. On the other hand for each $\beta\in\Phi^+$ we have $\beta\in\sigma\tau^{-1}\cdot\walls$ if and only if $H_{\beta,1}$ is a floor of $\waf\ac$ by the definition of the zeta map. Thus \begin{align*} \beta\in I(\sigma\tau^{-1}\cdot\walls) &\Leftrightarrow H_{\beta,1}\text{ does not separate }\waf\ac\text{ and }\ac \\&\Leftrightarrow \waf^{-1}\cdot(-\beta+\delta)=-\sigma^{-1}\cdot\beta+(1-\skal{\mu,\beta})\delta\in\ar^+ \\&\Leftrightarrow \skal{\mu,\beta}=0\text{ or }\skal{\mu,\beta}=1,-\sigma^{-1}\cdot\beta\in\Phi^+. \end{align*} % We first show that % \begin{align*} % \dinv'(u(\lambda)+(h+1)\Q)&=\area'([u\sigma\tau^{-1},\sigma\tau^{-1}\cdot\walls]). % \end{align*} % To see this and $u\tau\sigma^{-1}(\beta)\in\Phi^+$. The proof is completed by distinguishing a few cases. Let $\beta\in\Phi^+$ and choose $\alpha\in\{\pm \tau\sigma^{-1}\cdot\beta\}\cap\Phi^+$. First assume $\skal{\mu,\beta}=0$ then $\sigma^{-1}\cdot\beta\in\Phi^+$ by \refl{dominant}. Also $0=\skal{\mu,\beta}=\skal{\mu,\sigma\tau^{-1}\cdot\alpha}$. If $\tau\sigma^{-1}\cdot\beta\in\Phi^+$ then $\tau^{-1}\cdot\alpha=\sigma^{-1}\cdot\beta\in\Phi^+$ and $\skal{\lambda,\alpha}=\op{ht}(\alpha)$. If $\tau\sigma^{-1}\cdot\beta\in-\Phi^+$ then $\tau^{-1}\cdot\alpha=-\sigma^{-1}\cdot\beta\in-\Phi^+$. Hence $\skal{\lambda,\alpha}=\op{ht}(\alpha)+1$. Moreover $u\tau\sigma^{-1}\cdot\beta\in\Phi^+$ implies that $u\cdot\alpha\in\Phi^+$ if and only if $\tau\sigma^{-1}\cdot\beta\in\Phi^+$. Secondly suppose that $\skal{\mu,\beta}=1$ and $\sigma^{-1}\cdot\beta\in-\Phi^+$. If $\tau\sigma^{-1}\cdot\beta\in\Phi^+$ then $\tau^{-1}\cdot\alpha\in-\Phi^+$ and $\skal{\mu,\sigma\tau^{-1}\cdot\alpha}=\skal{\mu,\beta}=1$. It follows that $\skal{\lambda,\alpha}=\op{ht}(\alpha)-1+1=\op{ht}(\alpha)$. If $\tau\sigma^{-1}\cdot\beta\in-\Phi^+$ then $\tau^{-1}\cdot\alpha\in\Phi^+$ and $\skal{\mu,\sigma\tau^{-1}\cdot\alpha}=-1$. Thus $\skal{\lambda,\alpha}=\op{ht}(\alpha)+1$. Moreover, $u\tau\sigma^{-1}\cdot\beta\in\Phi^+$ implies that $u\cdot\alpha\in\Phi^+$ if and only if $\tau\sigma^{-1}\cdot\beta\in\Phi^+$. Conversely let $\alpha\in\Phi^+$ and choose $\beta\in\{\pm \sigma\tau^{-1}\cdot\alpha\}\cap\Phi^+$. First assume $\skal{\lambda,\alpha}=\op{ht}(\alpha)$. If $\tau^{-1}\cdot\alpha\in\Phi^+$ then $0=\skal{\mu,\sigma\tau^{-1}\cdot\alpha}=\skal{\mu,\beta}$. By \refl{dominant} $\sigma^{-1}\cdot\beta\in\Phi^+$ and thus $\beta=\sigma\tau^{-1}\cdot\alpha$. Consequently $u\cdot\alpha\in\Phi^+$ implies $u\tau\sigma^{-1}\cdot\beta=u\cdot\alpha\in\Phi^+$. If $\tau^{-1}\cdot\alpha\in-\Phi^+$ then $\skal{\mu,\sigma\tau^{-1}\cdot\alpha}=1$ and therefore $\sigma\tau^{-1}\cdot\alpha\in\Phi^+$ since $\mu\in\overline{C}$. We obtain $\skal{\mu,\beta}=1$ and $\sigma^{-1}\cdot\beta=\tau^{-1}\cdot\alpha\in-\Phi^+$. Moreover $u\cdot\alpha\in\Phi^+$ implies $u\tau\sigma^{-1}\cdot\beta\in\Phi^+$ as above. Finally assume that $\skal{\lambda,\alpha}=\op{ht}(\alpha)+1$. If $\tau^{-1}\cdot\alpha\in\Phi^+$ then $\skal{\mu,\sigma\tau^{-1}\cdot\alpha}=-1$ and therefore $\sigma\tau^{-1}\cdot\alpha\in-\Phi^+$ since $\mu\in\overline{C}$. Hence $\skal{\mu,\beta}=1$ and $-\sigma^{-1}\cdot\beta=\tau^{-1}\cdot\alpha\in\Phi^+$. Moreover, $u\cdot\alpha\in-\Phi^+$ implies $u\tau\sigma^{-1}\cdot\beta=-u\cdot\alpha\in\Phi^+$. If $\tau^{-1}\cdot\alpha\in-\Phi^+$ then $0=\skal{\mu,\sigma\tau^{-1}\cdot\alpha}=\skal{\mu,\beta}$ and thus $\sigma^{-1}\cdot\beta\in\Phi^+$. It follows that $\beta=-\sigma\tau^{-1}\cdot\alpha$ and hence $u\cdot\alpha\in-\Phi^+$ implies $u\tau\sigma^{-1}\cdot\beta=-u\cdot\alpha\in\Phi^+$. \end{proof} % \begin{mythrm}{dinvC} Let $\pi\in\mathcal{L}_{n,n}$. % Then $\dinv_C(\pi)=\area(\zeta_C(\pi))$. % \end{mythrm} % \begin{proof} Let $\pi\in\mathcal{L}_{n,n}$ be a lattice path with type $C_n$ area vector $\mu$ and set $\beta=\zeta_C(\pi)$. % For $k\in\{0,\dots,n\}$ let $\alpha_k$ be the number of indices $i\in[n]$ such that $\abs{\mu_i}=k$. Recall the definition of the bounce path from the proof of \reft{zetaC}. % The area below bounce path of $\zeta_C(\pi)$ is given by % \begin{align*} % \alpha_0^2+\sum_{k=1}^n\binom{\alpha_k}{2}. % \end{align*} % This accounts exactly for the number of diagonal inversions of type $\mu_{n-i+1}=\mu_{n-j+1}$, $\mu_{n-i+1}=-\mu_{n-j+1}$ or $\mu_{n-i+1}=0$. % Let $(r,s)$ be the top right corner of a $1\times1$ square between $\zeta_C(\pi)$ and its bounce path. % Then the $r$-th East step and the $s$-th North step of $\beta$ both exist and belong to the segment $\LS_k^-(\mu)\RS_k^+(\mu)$ for some $k\in\{0,\dots,n\}$. % There are three cases: % If the $r$-East step and the $s$-th North step of $\beta$ both belong to $\LS_k^-(\mu)$ then there exist indices $i,j\in[n]$ with $i0$. % Finally, if the $s$-th North step of $\beta$ belongs to $\LS_k^-(\mu)$ and the $r$-th East step of $\beta$ belongs to $\RS_k^+(\mu)$ then there exist indices $i,j\in[n]$ such that $\mu_i=-k$ and $\mu_j=k+1$. % In this case either $(i,j)$ or $(j,i)$ is a diagonal inversion of type $\mu_{n-i+1}=-\mu_{n-j+1}+1$. % We conclude that there is a bijective correspondence between diagonal inversion of $\pi$ and the $1\times1$ squares below $\zeta_C(\pi)$. % \end{proof} % \begin{mythrm}{dinvC'} Let $(\pi,u)\in\op{Vert}(C_n)$. % Then $\dinv_C'(\pi,w)=\area'(\zeta_C(\pi,w))$. % \end{mythrm} % \begin{proof} Let $(\pi,u)\in\op{Vert}(C_n)$, let $\mu$ be the type $C_n$ area vector of $\pi$ and set $(\beta,w)=\zeta_C(\pi,u)$ % By definition each step of $\beta$ corresponds to an entry of $\mu$. % More precisely, let $i,r\in[n]$. Then the $r$-th East step (the $r$-th North step) of $\beta$ corresponds to the entry $\mu_{n-i+1}$ if and only if % \begin{align*} % r % &=\#\{j\in[n]:|\mu_{n-j+1}|>|\mu_{n-i+1}|\}\\ % &\quad+\#\{j\in[n]:\mu_{n-i+1}>0,\mu_{n-j+1}=-\mu_{n-i+1}\}\\ % &\quad+\#\{j\in[i]:\mu_{n-i+1}\leq0,\mu_{n-j+1}=\mu_{n-i+1}\}\\ % &\quad+\#\{j\in[n]:\mu_{n-i+1}>0,\mu_{n-j+1}=\mu_{n-i+1},j\geq i\}. % \end{align*} % Comparing this to the discussion of the diagonal reading word, see that the label below the $r$-th East step of $\beta$ (to the right of the $r$-th North step of $\beta$) is given by % \begin{align*} % w(n-r+1) % &= % \begin{cases} % u(i)&\quad\text{if }\mu_{n-i+1}>0,\\ % -u(i)&\quad\text{if }\mu_{n-i+1}\leq0. % \end{cases} % \end{align*} % If $r>n$ then the $r$-th North step of $\beta$ corresponds to the entry $\mu_{n-i+1}$ if and only if $\mu_{n-i+1}=0$ and % \begin{align*} % r-n % &=\#\{j\in[n]:i\leq j,\mu_{n-j+1}=0\}. % \end{align*} % In this case the label to the right of the $i$-th North step is given by $w(n-r)=u(i)$. % % Note that no East step corresponds to a row with area zero. % % \sk % % Now suppose that $(i,j)$ is a valley of $\beta$. Choose $r$ and $s$ such that the $r$-th and $s$-th rows of $\pi$ corresponding to the $i$-th East step and the $j$-th North step of $\beta$, respectively. Note that either $r=s+1$, $\mu_r=\mu_s+1$ and $\mu_r>0$ or $r=s-1$, $\mu_r=\mu_s-1$ and $\mu_r<0$. % % % % In the case $r=s+1$ we have $u(n-r+1)>u(n-s+1)$ by the definition of a vertically labelled path. Moreover, $\mu_s=0$ implies $i>n$. Hence, we obtain % % \begin{align*} % % w(n-i+1)=u(n-r+1)>\sigma_j=w(n-j+1). % % \end{align*} % % In the case $r=s-1$ we have $u(n-r+1)-u(n-s+1)=w(n-j+1). % % \end{align*} % % Furthermore, note that $\beta$ ends with an East step if and only if $\pi$ begins with a row of length one, \ie $\mu_1=1$. The label below this East step is $\sigma_1>0$. % % % % We conclude that $(\beta,w)$ satisfies the defining properties of a diagonally labelled ballot path. % % % % \sk % % The map $\zeta$ is injective by \reft{zetaC} as it is not difficult to obtain $\sigma$ from $w$ once $\pi$ is known: To obtain the vertical label $\sigma_i$ choose $j$ with $1\leq j \leq n$ such that the $j$-th North step of $\beta$ corresponds to the $i$-th row of $\pi$. Set $\sigma_i=w(n-i+1)$ if $\mu_i>0$ and $\sigma_i=-w(n-j+1)$ if $\mu_j\leq0$. % % % % To see that the map $\zeta$ is onto, suppose that $(\beta,w)$ is a diagonally labelled ballot path and set $\pi=\zeta_C^{-1}(\beta)$ and define $\sigma$ as in the previous paragraph. One can verify that $(\pi,\sigma)$ is a vertically $\hypoct_n$-labelled path by similar reasoning as above. % % % \sk % % % Finally, we need to check that $\op{dinv'}(\pi,\sigma)=\area'(\beta,w)$. As mentioned before, the cells below the bounce path correspond to the diagonal inversions of the form $\mu_r=\pm \mu_s$ or $\mu_r=0$. % % % % % % Let $(i,j)$ be a cell below the bounce path. If $\beta$ contains at least $i$ East steps, let the $i$-th East step and the $j$-North step correspond to the rows $\mu_r$ and $\mu_s$ of $\pi$. Then $\mu_s>0$ and $\mu_s=\pm \mu_r$. Suppose $\mu_r=\mu_s$ then we have $w(n-i+1)>w(n-j+1)$ if and only if $u(n-r+1)=w(n-i+1)>w(n-j+1)=u(n-s+1)$. Thus the cell $(i,j)$ contributes to the $\area'(\beta,w)$ if and only if $(r,s)$ is a diagonal inversion of $(\pi,\sigma)$. If $\mu_r=-\mu_s$ then $w(n-i+1)-w(n-j+1)=$ % \sk % Recall from \reft{dinvC} that each $1\times1$ square below $\beta$ corresponds to a diagonal inversion of $\pi$. % We demonstrate the claim by showing that such a square contributes to $\area'(\beta,w)$ if and only if the corresponding diagonal inversion contributes to $\dinv_C'(\pi,u)$. % Let $(r,s)$ be a the top right corner of a $1\times1$ square below $\beta$. % Choose $j$ such that the $s$-th North step of $\beta$ corresponds to the entry $\mu_{n-j+1}$. % Suppose the square lies below the bounce path. % Assume further that $\beta$ contains at least $r$ East steps. % In this case $s\leq n$ and $\mu_{n-j+1}\neq0$. % Choose $i$ such that the $r$-th East step of $\beta$ corresponds to the entry $\mu_{n-i+1}$. % Note that $\mu_{n-i+1}\in\{-\mu_{n-j+1},\mu_{n-j+1}\}$. % First assume $\mu_{n-i+1}=\mu_{n-j+1}$. % If $\mu_{n-i+1}<0$ then $iw(n-s+1) \Leftrightarrow u(i)0$ then $jw(n-s+1) \Leftrightarrow u(i)>u(j) % \end{align*} % and the square contributes to $\area'(\beta,w)$ if and only if $(i,j)$ is a diagonal inversion of $(\pi,u)$ of type $\mu_{n-j+1}=\mu_{n-i+1}$. % Secondly, assume $\mu_{n-i+1}=-\mu_{n-j+1}$. Then $\mu_{n-i+1}<0<\mu_{n-j+1}$. Thus $w(n-r+1)=-u(i)$ and $w(n-s+1)=u(j)$. We obtain % \begin{align*} % w(n-r+1)>w(n-s+1) \Leftrightarrow u(i)<-u(j) \Leftrightarrow u(j)<-u(i) % \end{align*} % and the square contributes to $\area'(\beta,w)$ if and only if either $iw(n-s+1) \Leftrightarrow u(i)n$. Then $i\leq j$ and $w(n-s)=u(j)$. If $iw(n-s) \Leftrightarrow u(i)<-u(j) % \end{align*} % and the square contributes to $\area'(\beta,w)$ if and only if $(i,j)$ is a diagonal inversion of $(\pi,u)$ of type $\mu_{n-i+1}=-\mu_{n-j+1}$. % On the other hand, if $i=j$ then % \begin{align*} % w(n-r+1)>w(n-s) \Leftrightarrow u(i)<-u(i) \Leftrightarrow u(i)<0 % \end{align*} % and the square contributes to $\area'(\beta,w)$ if and only if $i$ is a diagonal inversion of $(\pi,u)$ of type $\mu_{n-i+1}=0$. % \sk % Finally we come to the case where the square lies between $\beta$ and its bounce path. % Choose $i$ such that the $r$-th East step of $\beta$ corresponds to $\mu_{n-i+1}$. % Note that $\beta$ contains at least $r$ East steps. % Assume $\mu_{n-i+1}=k+1$ with $k>0$. % Then $w(n-r+1)=u(i)$ and $\mu_{n-j+1}\in\{-k,k\}$, thus in particular $\mu_{n-j+1}\neq0$. If $\mu_{n-j+1}=k$ then $iw(n-s+1) \Leftrightarrow u(i)>u(j) % \end{align*} % and the square contributes to $\area'(\beta,w)$ if and only if $(i,j)$ is a diagonal inversion of $(\pi,u)$ of type $\mu_{n-i+1}=\mu_{n-j+1}+1$. % On the other hand if $\mu_{n-j+1}=-k$ then $w(n-s+1)=-u(j)$. Hence, % \begin{align*} % w(n-r+1)>w(n-s+1) \Leftrightarrow u(i)>-u(j) \Leftrightarrow u(j)>-u(i) % \end{align*} % and the square contributes to $\area'(\beta,w)$ if and only if either $in$ then $iw(n-s+1) \Leftrightarrow u(i)>u(j) % \end{align*} % and the square contributes to $\area'(\beta,w)$ if and only if $(i,j)$ is a diagonal inversion of $(\pi,u)$ of type $\mu_{n-i+1}=\mu_{n-j+1}+1$. % On the other hand, if $s\leq n$ then $w(n-s+1)=-u(j)$. We obtain % \begin{align*} % w(n-r+1)>w(n-s+1) \Leftrightarrow u(i)>-u(j) \Leftrightarrow u(j)>-u(i) % \end{align*} % and the square contributes to $\area'(\beta,w)$ if and only if either $iw(n-s+1) \Leftrightarrow u(j)>u(i) % \end{align*} % and the square contributes to $\area'(\beta,w)$ if and only if $(j,i)$ is a diagonal inversion of $(\pi,u)$ of type $\mu_{n-j+1}=\mu_{n-i+1}+1$. % \end{proof} \section{Open problems}\label{Section:end} In this final section we discuss some open problems and perspectives for further research. \begin{myprob}{fuss}[The Fu{\ss}--Catalan case] In this paper we discuss the combinatorial interpretation of the bijection $\zeta$ between the finite torus $\Q/(h+1)\Q$ and the set of non-nesting parking functions $\op{Park}(\Phi)$. However, the zeta map can be defined more generally. Rhoades~\cite{Rhoades:parking_spaces_fuss} defined a Fu{\ss} analogue of the non-nesting parking functions $\op{Park}^{(m)}(\Phi)$ and proved that it is isomorphic to the generalised finite torus $\Q/(mh+1)\Q$ as a $W$-set. The second named author~\cite{Thiel2015} defined an explicit $W$-equivariant bijection $\zeta:\op{Park}^{(m)}(\Phi)\to\Q/(mh+1)\Q$ generalising the zeta map treated in this work. In type $A_{n-1}$ the combinatorial objects allowing for a convenient treatment of the Fu{\ss}--Catalan level are $m$-Dyck paths, and the combinatorial zeta map was studied by Loehr~\cite{Loehr2005}. One possible next step would be to develop the combinatorial framework for the Fu{\ss}--Catalan case of the zeta map for the other infinite families of crystallographic root systems $B_n,C_n$ and $D_n$. \end{myprob} \begin{myprob}{inverse}[The inverse zeta map] \refs{typeC} contains an explicit way to invert the zeta map of type $C_n$, namely \reft{zetaC}. As the attentive reader may have noticed, such a result is missing from the respective chapters on types $B_n$ and $D_n$. Indeed our only way of establishing the bijectivity of the zeta maps in these types is to prove that they are special cases of the uniform zeta map which is known to be bijective. It would be interesting to have an explicit combinatorial description of the inverse of the zeta map in types $B_n$ and $D_n$. One might expect such a description to be connected to the construction of a suitable bounce path, which also appears in types $A_{n-1}$ and $C_n$. \end{myprob} \begin{myprob}{bounce}[The missing statistic] There exist generalisations of the algebraic background motivating the definition of $q,t$-Catalan numbers that replace the symmetric group by a Weyl group $W$ of an irreducible crystallographic root system $\Phi$. Consequently $q,t$-Coxeter--Catalan numbers $C_{W}(q,t)$ can be defined as the bivariate Hilbert series of certain $W$-modules. See for example~\cite[Appendix~A]{Stump2010} for a list of the first few polynomials obtained in this way. In type $A_{n-1}$ combinatorial models for these Hilbert series were found by Haglund using a bounce statistic and Haiman using the dinv statistic. It is an open problem to find a combinatorial interpretation of these polynomials in any other types. \sk The following open problem was posed by Stump and also appears in~\cite{Armstrong2013}. Find a statistic $\op{tstat}$ on the antichains in the root poset $\Phi^+$ such that \begin{align*} C_{W}(q,t) &=\sum_{A\subseteq \Phi^+}q^{\area(A)}t^{\op{tstat}(A)}, \end{align*} where the sum ranges over all antichains in the root poset. We now give an alternative formulation of this problem in terms of the finite torus. Find a statistic $\op{tstat}$ on the $W$-orbits of the finite torus $\Q/(h+1)\Q$ such that \begin{align*} C_{W}(q,t) &=\sum_{\lambda\in\Q\cap(h+1)\overline{\ac}}q^{\dinv(\lambda)}t^{\op{tstat}(\lambda)}. \end{align*} % \refop{bounce} was suggested by Stump and remains unsolved beyond type $A_{n-1}$. Quite possibly our formulation of the problem is just as difficult as the original one, however, the dinv statistic at least offers a new angle of attack. Note that even a partial solution in the form of a conjectured statistic say in type $B_n$ or $C_n$ would be of interest. \sk Our closing remark addresses an important point that is somewhat counter-intuitive. One might assume at first glance that knowing the $\dinv$ statistic as well as the $\area$ statistic should be sufficient to obtain the $q,t$-Catalan numbers, as is the case in type $A_{n-1}$. However, let us emphasise that the $\dinv$ statistic is only known for $W$-orbits of the finite torus, while the $\area$ statistic is only known for antichains of the root poset. It is a spectacular coincidence that in type $A_{n-1}$ these two objects both correspond naturally to Dyck paths in such a way that the $\area$ statistic, which is natural in the world of antichains, serves as the mysterious statistic $\op{tstat}$ in the world of $W$-orbits. \end{myprob} % \begin{myprob}{dinv}[The $\dinv$ statistic] As pointed out in \refs{statC} the definition of the $\area$ statistic is uniform, while the definition of the $\dinv$ statistic is particular to the types $A_{n-1}$ respectively $C_n$. % One (rather uninspired) way to give a uniform definition of the $\dinv$ statistic is as the pullback of the $\area$ statistic by the zeta map, that is, % \begin{align*} % \dinv=\area\circ\zeta. % \end{align*} % Thereby an analogous result to \reft{dinvC} is fulfilled automatically. % A next logical step would be to ask for a description of the statistics $\dinv'$ and $\dinv$ that does not use the zeta map but can be seen directly from the combinatorial models for the finite torus respectively its $W$-orbits in types $B_n$ and $D_n$. % We suspect this will not prove too difficult. % As a starting point we suggest the following definition, which is similar to the type $C_n$ case: % \begin{align*} % \dinv_B(\pi) % &=\#\big\{(i,j):1\leq i