\documentclass[ALCO,ThmDefs,Unicode,epreuves]{cedram} \usepackage{tabularx,combelow} \usepackage{booktabs,tabmac} \definecolor{darkred}{rgb}{0.7,0,0} % darkred color \newcommand{\darkred}{\color{darkred}} % darkred command % Dark red emphasis \newcommand{\defn}[1]{{\color{darkred}\emph{#1}}} % emphasis of a definition \newcommand{\mms}{_} \newcommand{\area}{\mathsf{area}} \newcommand{\dinv}{\mathsf{dinv}} \newcommand{\stat}{\mathsf{stat}} \DeclareMathOperator{\comb}{comb} \DeclareMathOperator{\wt}{wt} \DeclareMathOperator{\FHilb}{FHilb} \DeclareMathOperator{\Res}{Res} \DeclareMathOperator{\dd}{d} \newcommand{\irm}{i} \begin{DefTralics} \newcommand*\cb[1]{\c{#1}} \end{DefTralics} \newcommand{\twt}{\widetilde{\wt}} \newcommand{\CC}{\mathbb{C}} \newcommand{\CL}{\mathcal{L}} \newcommand{\CO}{\mathcal{O}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%% a placer avant \begin{document} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand*{\mk}{\mkern -1mu} \newcommand*{\Mk}{\mkern -2mu} \newcommand*{\mK}{\mkern 1mu} \newcommand*{\MK}{\mkern 2mu} \hypersetup{urlcolor=purple, linkcolor=blue, citecolor=red} \newcommand*{\romanenumi}{\renewcommand*{\theenumi}{\roman{enumi}}} \newcommand*{\Romanenumi}{\renewcommand*{\theenumi}{\Roman{enumi}}} \newcommand*{\alphenumi}{\renewcommand*{\theenumi}{\alph{enumi}}} \newcommand*{\Alphenumi}{\renewcommand*{\theenumi}{\Alph{enumi}}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%% Auteur %%%1 \author[\initial{E.} \lastname{Gorsky}]{\firstname{Eugene} \lastname{Gorsky}} \address{Department of Mathematics\\ University of California\\ One Shields Avenue\\ Davis, CA 95616-8633, U.S.A.} \curraddr{International Laboratory of Representation Theory and Mathematical Physics\\ NRU-HSE\\ Moscow, Russia} \email{egorskiy@math.ucdavis.edu} \urladdr{https://www.math.ucdavis.edu/~egorskiy} %%%2 \author[\initial{G.} \lastname{Hawkes}]{\firstname{Graham} \lastname{Hawkes}} \address{Department of Mathematics\\ University of California\\ One Shields Avenue\\ Davis, CA 95616-8633, U.S.A.} \curraddr{Max--Planck--Institut f{\"u}r Mathematik\\ Vivatsgasse 7\\ 53111 Bonn, Germany} \email{hawkes@math.ucdavis.edu} \urladdr{http://guests.mpim-bonn.mpg.de/hawkes/} %%%3 \author[\initial{A.} \lastname{Schilling}]{\firstname{Anne} \lastname{Schilling}} \address[A. Schilling]{Department of Mathematics\\ University of California\\ One Shields Avenue\\ Davis, CA 95616-8633, U.S.A.} \email{anne@math.ucdavis.edu} \urladdr{https://www.math.ucdavis.edu/~anne} %%%4 \author[\initial{J.} \lastname{Rainbolt}]{\firstname{Julianne} \lastname{Rainbolt}} \address{Department of Mathematics and Statistics\\ Saint Louis University\\ 220 North Grand Blvd.\\ Saint Louis, MO 63103, U.S.A.} \email{rainbolt@slu.edu} \urladdr{https://mathstat.slu.edu/people/rainbolt} %%%%% Sujet \keywords{$q,t$-Catalan numbers, symmetric chain decomposition, Khovanov--Rozansky knot homology.} \subjclass{05A18, 05A19, 05E05} %%%%% Gestion \DOI{10.5802/alco.120} \datereceived{2019-06-01} \daterevised{2020-02-07} \dateaccepted{2020-03-19} %%%%% Titre et résumé \title{Generalized $q,t$-Catalan numbers} \begin{abstract} Recent work of the first author, Negu\cb{t} and Rasmussen, and of Oblomkov and Rozansky in the context of Khovanov--Rozansky knot homology produces a family of polynomials in $q$ and $t$ labeled by integer sequences. These polynomials can be expressed as equivariant Euler characteristics of certain line bundles on flag Hilbert schemes. The $q,t$-Catalan numbers and their rational analogues are special cases of this construction. In this paper, we give a purely combinatorial treatment of these polynomials and show that in many cases they have nonnegative integer coefficients. For sequences of length at most 4, we prove that these coefficients enumerate subdiagrams in a certain fixed Young diagram and give an explicit symmetric chain decomposition of the set of such diagrams. This strengthens results of Lee, Li and Loehr for $(4,n)$ rational $q,t$-Catalan numbers. \end{abstract} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \maketitle %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Introduction} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The last decade revealed deep, and yet partially conjectural connections~\cite{GORS.2014,GN.2015,GM.2013,GM.2014,GM.2016,GMV.2016,GMV.2017} of the HOMFLY-PT link homologies with various intricate constructions in algebraic combinatorics such as $q,t$-Catalan numbers of Garsia and Haiman~\cite{GarsiaHaiman.1996}, LLT polynomials~\cite{HHLRU.2005}, and the elliptic Hall algebra~\cite{SV.2013}. Some of these conjectures were recently proven (mostly for the torus knots and links) by Elias, Hogancamp and Mellit~\cite{EH.2019,Hogancamp.2017,Mellit.2017}. An interesting class of knots, which best fits in the framework of the above conjectures, are the so-called Coxeter links defined as closures of braids \[ \beta(a_1,\ldots,a_n)=\ell_1^{a_1}\cdots \ell_n^{a_n}t_1\cdots t_{n-1}, \] where $\ell_i=t_{i-1}\cdots t_1t_1\cdots t_{i-1}$ are Jucys--Murphy elements and $t_i$ are the standard braid group generators. Here $a_i$ are arbitrary integers, but in this paper we will mostly assume $a_i\geqslant 0$, so that all crossings in the braid $\beta(a_1,\ldots,a_n)$ are positive. Motivated by the geometry of the flag Hilbert scheme of points on the plane (see Section~\ref{subsec.fhilb} and references therein) we can approximate the invariants of such knots with the following combinatorial expressions. Define \begin{equation} \label{eq: def f} f(a_1,\ldots,a_n)=\sum_{T}z_1^{a_1}\cdots z_n^{a_n} \; \prod_{i=2}^{n} \frac{1}{(1-z_i^{-1})(1-qtz_{i-1}/z_i)} \; \prod_{i1$, and the polynomial $f(1,\ldots,1)=f(2,1,\ldots,1)$ agrees with the \defn{$q,t$--Catalan number} of Garsia and Haiman~\cite{GarsiaHaiman.1996}. Motivated by~\cite{GNR.2016,OR.2017}, we expect that the beautiful combinatorics of $q,t$-Catalan numbers and their rational analogues can be generalized to the case of arbitrary $a_i$, possibly constrained by some inequalities. In fact, as we show in this paper, that varying $a_i$ allows one to compute the invariants $f(a_1,\ldots,a_n)$ recursively, see Corollary~\ref{corollary.recursion} for the $n=4$ example. Using the machinery of Tesler matrices, we prove the following result. \begin{prop} Suppose that $a_i\geqslant 0$. Then $f(a_1,\ldots,a_n)$ is a polynomial in $q$ and $t$. At $t=1$, this polynomial specializes to \[ f(a_1,\ldots,a_n) \Big|_{t=1} =\sum_{\mu\subseteq \lambda(a)}q^{|\lambda(a)|-|\mu|}, \] where $\lambda(a)=(a_2+\cdots+a_n,a_3+\cdots+a_n,\ldots,a_n).$ \end{prop} \begin{exam} \label{example.2 and 3} For $n=2$, one has \[ f(a_1,a_2)=[a_2+1]_{q,t}:=q^{a_2}+q^{a_2-1}t+\cdots+qt^{a_2-1}+t^{a_2}. \] For $n=3$ and $a_2\geqslant a_3$ one has \[ f(a_1,a_2,a_3)=[a_2+2a_3+1]_{q,t}+qt[a_2+2a_3-2]_{q,t}+\cdots+q^{a_3}t^{a_3}[a_2-a_3+1]_{q,t}. \] See Examples~\ref{example.n=2} and~\ref{example.n=3} for derivations of these formulas. \end{exam} The following conjecture was communicated to the authors by Andrei Negu\cb{t}. \begin{conjecture}[Negu\cb{t}] \label{conj:positivity} If $a_1\geqslant a_2 \geqslant \cdots\geqslant a_n\geqslant 0$, then $f(a_1,\ldots,a_n)$ is a polynomial in $q$ and $t$ with nonnegative coefficients. \end{conjecture} For general $a_1\geqslant a_2 \geqslant \cdots\geqslant a_n\geqslant 0$, it is still an open problem to find an explicit statistic $\stat$ on partitions $\mu$ such that \begin{equation} \label{eq: mystery stat} \postdisplaypenalty1000000 f(a_1,\ldots,a_n)=\sum_{\mu\subseteq \lambda(a)}q^{|\lambda(a)|-|\mu|}\; t^{\stat(\mu)}. \end{equation} In this paper, we solve the problem for $n=4$: \begin{theorem} For $a+1 \geqslant b$, $a+1,b+1 \geqslant c\geqslant 0$, the polynomial $F(a,b,c) := f(a_1,a,b,c)$ has nonnegative integer coefficients and can be written in the form~\eqref{eq: mystery stat}. The statistic $\stat(\mu)$ arises from an explicit decomposition of the set of $\mu\subseteq \lambda(a)$ into symmetric chains. \end{theorem} See Section~\ref{section.combinatorial expressions} for further details. Since a symmetric chain specializes to $q^{k}+q^{k-2}+\cdots+q^{-k+2}+q^{-k}$ a $t=q^{-1}$, we immediately obtain the following corollary. \begin{coro} For $a+1 \geqslant b$, $a+1,b+1 \geqslant c\geqslant 0$, the coefficients of the specialization $F(a,b,c)|_{t=q^{-1}}$ are unimodular in even and in odd degrees. \end{coro} \begin{rema} By~\cite{GNR.2016,OR.2017} the specialization of $f(a_1,\ldots,a_n)$ at $q=t^{-1}$ coincides with the part of the HOMFLYPT polynomial of the knot $\beta(a_1,\ldots,a_n)$. \end{rema} \begin{rema} Our statistic and decomposition is different from that in~\cite{LLL.2014, LLL.2018}. In particular, some of their chains are not symmetric, but the authors show that partitions come in symmetric pairs. \end{rema} We provide a recursion for $F(a,b,c)$ and prove that the combinatorial expression also satisfies the recursion (see Sections~\ref{subsec:recursion n=4} and~\ref{section.combinatorial recursion}). The set of Young diagrams $\mu$ contained in the diagram $\lambda(a)$ is in bijection with the Demazure crystal~\cite{Kashiwara.1993,Littelmann.1995} with highest weight $(a_1,\ldots,a_n)$ and Weyl group element $c=t_1\cdots t_{n-1}$. The size of $\mu$ can be easily expressed in terms of the weight of the corresponding element of the crystal basis. This observation leads to many interesting questions: \begin{itemize} \item What is the crystal-theoretic interpretation of the statistic $\stat$? \item Is there a crystal-theoretic interpretation of the symmetric chains and the polynomials $f(a_1,\ldots,a_n)$? \end{itemize} \begin{rema} In the terminology of~\cite{CDL.2018}, subdiagrams of $\lambda(a)$ correspond to so-called $s$-Dyck paths, and it is shown in~\cite{CDL.2018} that they are in bijection with remarkably many combinatorial objects, just as usual Catalan numbers are in bijection with trees, triangulations etc. It would be interesting to relate the results of~\cite{CDL.2018} both to Demazure crystals and to the above statistic $\stat$. \end{rema} The paper is organized as follows. In Section~\ref{sec:algebraic}, we discuss the algebraic aspects of the function $f(a_1,\ldots,a_n)$ (or equivalently $F(a_2,\ldots,a_n)$). The definition of the function $f(a_1,\ldots,a_n)$ is given in Section~\ref{subsection.formula}. In Section~\ref{subsec.fhilb}, we briefly recall its connection to flag Hilbert schemes and knot invariants; combinatorially inclined readers are welcome to skip this section. In Section~\ref{subsec:Tesler}, we connect $f(a_1,\ldots,a_n)$ to Tesler matrices and prove that they are indeed polynomials in $q$ and $t$. In Section~\ref{subsec:recursion n=4} we prove the recursion for $n=4$. Section~\ref{section.combinatorial expressions} contains the combinatorial expressions for $F(a,b,c)$. We also provide examples. In Section~\ref{section.proof}, we construct the symmetric chains underlying the combinatorial formulas explicitly and also prove the combinatorial formulas. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The algebraic side} \label{sec:algebraic} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{The formula} \label{subsection.formula} Given a standard tableau $T$ of size $n$, we define a vector $z(T)=(z_i)_{1\leqslant i \leqslant n}$, where $z_i$ is the $(q,t)$-content of the box in $T$ labeled by $i$. The \defn{$(q,t)$-content} of the box with row and column coordinates $(r,c)$, is $q^{c-1} t^{r-1}$. For example, for the tableau \[ T=\tableau[scY]{ 4 \cr 3 & 5\cr 1& 2 & 6 & 7\cr} \] we have \[ z(T)=(1,q,t,t^2,qt,q^2,q^3). \] By convention, $z_1=1$. We define the \defn{weight} of a tableau $T$ by \[ \wt(T)=\wt(z(T))=\prod_{i=2}^n\frac{1}{(1-z_i^{-1})(1-qtz_{i-1}/z_i)}\prod_{ii}m_{ij}=a_i\ \qquad \text{for}\ 1\leqslant i \leqslant n. \end{equation} \begin{lemma} \label{lem:tesler finite} The set of Tesler matrices is finite for fixed $a$. \end{lemma} \begin{proof} Equation~\eqref{eq: tesler} can be rewritten as follows: \begin{equation} \label{eq: tesler 2} m_{ii}+\cdots+m_{nn}+\sum_{ji+1$. \begin{lemma} \label{lem:subdiagrams} There is a bijection between the set of two-diagonal Tesler matrices associated to $a=(a_1,\ldots,a_n)$ and the set of subdiagrams of $\lambda(a_2,\ldots,a_n)$. \end{lemma} \begin{proof} If $M$ is a two-diagonal Tesler matrix, then for $i\geqslant 2$~\eqref{eq: tesler 2} simplifies to \[ m_{ii}+\cdots+m_{nn}+m_{i-1,i}=a_i+\cdots+a_n, \] while for $i=1$ we obtain \[ m_{11}+\cdots+m_{nn}=a_1+\cdots+a_n. \] This means that for $i\geqslant 2$ the diagonal elements of $M$ define a subdiagram of $\lambda(a)$ \[ m_{ii}+\cdots+m_{nn}\leqslant a_i+\cdots+a_n=\lambda_{i-1}, \] while $m_{11}$ and all $m_{i-1,i}$ are uniquely determined by the diagonal. \end{proof} We define the functions $A(m)$ and $B(m)$ by the equations \begin{equation*} \begin{split} \sum_{m=0}^{\infty}A(m)z^m &= \frac{(1-z)(1-qtz)}{(1-qz)(1-tz)}=1-(1-q)(1-t)\frac{z}{(1-qz)(1-tz)}\\ & \qquad \qquad \qquad \qquad \; = 1-(1-q)(1-t)\sum_{m=1}^{\infty}[m]_{q,t}z^m,\\ \sum_{m=0}^{\infty}B(m)z^m &= \frac{1-z}{(1-qz)(1-tz)}=\sum_{m=0}^{\infty}([m+1]_{q,t}-[m]_{q,t})z^m. \end{split} \end{equation*} \begin{theorem} For all $a_i\geqslant 0$, we have \begin{equation} \label{eq:tesler big sum} F(a_2,\ldots,a_n)=\sum_{M} \prod_{i}B(m_{i,i+1})\prod_{j>i+1}A(m_{i,j}), \end{equation} where the sum is over all Tesler matrices $M$ satisfying~\eqref{eq: tesler}. \end{theorem} \begin{proof} The proof is very similar to~\cite[Section~6.5]{GN.2015}, but we present it here for completeness. Since~\eqref{eq:tesler big sum} is an identity between rational functions in $q$ and $t$, without loss of generality we may assume that $q$ and $t$ are complex numbers very close to 1. Pick real numbers $1\ll r_1\ll \cdots \ll r_n$, and consider the torus \[ T=\{|z_1|=r_1,\ldots,|z_n|=r_n\}\subset \mathbb{C}^n. \] Given $a=(a_1,\ldots,a_n)$, consider the rational function \[ \Phi_a(z_1,\ldots,z_n)=z_1^{a_1}\cdots z_n^{a_n} \prod_{i=1}^{n}\frac{1}{(1-z_i^{-1})}\prod_{i=2}^{n}\frac{1}{(1-qtz_{i-1}/z_i)} \prod_{ii+1}\omega(u_j/u_i)\\ =\sum_{m_{ii}}u_i^{m_{ii}}\times \sum_{m_{i,i+1}}B(m_{i,i+1})\left(\frac{u_{i+1}}{u_{i}}\right)^{m_{i,i+1}}\times \sum_{m_{i,j}}A(m_{i,j})\left(\frac{u_j}{u_{i}}\right)^{m_{i,j}}. \label{eq: infinity expansion} \end{multline} The terms in the sum in~\eqref{eq: infinity expansion} are parameterized by the exponents $m_{ii},m_{i,i+1},m_{i,j}$ which can be combined in a single upper-triangular matrix $M=(m_{ij})$. Such a term contributes to $u_1^{a_1}\cdots u_n^{a_n}$ if \[ m_{ii}-\sum_{j>i}m_{ij}+\sum_{j0, \quad A(0)\Big|_{t=1}=1, \quad B(m)\Big|_{t=1}=q^{m}. \] Therefore at $t=1$ the sum~\eqref{eq:tesler big sum} specializes to the sum over two-diagonal Tesler matrices which by Lemma~\ref{lem:subdiagrams} correspond to subdiagrams $\mu\subseteq \lambda(a)$. The weight of such a two-diagonal Tesler matrix specializes to $\prod_i q^{m_{i,i+1}}=q^{|\lambda(a)|-|\mu|}$. \end{proof} \begin{coro} \label{cor:set 0} For $a_i\geqslant 0$ and $a_n=0$ we have \[ F(a_2,\ldots,a_{n-1},0)=F(a_2,\ldots,a_{n-1}). \] \end{coro} \begin{proof} The last equation in~\eqref{eq: tesler} reads as \[ m_{nn}+\sum_{j0$ \[ F(-a) = - \frac{1}{(qt)^{a-1}} \; F(a-2). \] \end{prop} \begin{proof} By~\eqref{H to F} and Example~\ref{example.n=2}, we have \[ F(a) = \frac{1}{1-t/q} q^{a} + \frac{1}{1-q/t} t^{a} = q^a \frac{1-(t/q)^{a+1}}{1-t/q}. \] Hence \begin{equation*} F(-a) = q^{-a} \frac{1-(t/q)^{-a+1}}{1-t/q} = -q^{-a}(t/q)^{-a+1} \frac{1-(t/q)^{a-1}}{1-t/q} = - \frac{1}{(qt)^{a-1}} F(a-2).\qedhere \end{equation*} \end{proof} Note that using Corollary~\ref{corollary.recursion} and Proposition~\ref{proposition.negative}, $F(a,b,c)$ for $a\geqslant b \geqslant c \geqslant 0$ can be reduced to $F(a,b)$ for $a\geqslant b \geqslant 0$ and $F(a)$ for $a\geqslant 0$, which are given in Example~\ref{example.2 and 3}. Now we compute $F(a,b)$ explicitly by separating the sum. \begin{lemma}\label{F(a,b)} For $b \geqslant -1$ and $a \geqslant b-1$ we have \begin{equation*} F(a,b)= \sum_{i=0}^b \sum_{j=i}^{a+2b-2i}q^{j}t^{(a+2b-i)-j}. \end{equation*} \end{lemma} \begin{proof} We may express $F(a,b)$ in terms of $H(a,b)$ by separating the sum as above. Using the expression for $H(a,b)$ from Example~\ref{example.n=3} (note that this expression is valid for $b \geqslant -1$ and any value of $a$) this gives us \begin{equation*} \begin{split} F(a,b) &= \frac{1}{1-t/q}q^a\sum_{i=0}^b (q^2)^{b-i}t^i+\frac{1}{1-q/t}t^a\sum_{i=0}^b (t^2)^{b-i}q^i \\ &=\frac{1}{q-t}\bigg(q^{a+1}\sum_{i=0}^b (q^2)^{b-i}t^i-t^{a+1}\sum_{i=0}^b (t^2)^{b-i}q^i\bigg)\\ &=\sum_{i=0}^b \frac{q^{a+2b+1-2i}t^i-t^{a+2b+1-2i} q^i}{q-t}=\sum_{i=0}^b \sum_{j=i}^{a+2b-2i}q^{j}t^{(a+2b-i)-j}, \end{split} \end{equation*} where the last step is legal because we are assuming $a \geqslant b-1$. \end{proof} \begin{lemma} \label{lemma.-1} We have $F(-1)=F(a,-1)=0$ for $a\geqslant -2$ and $F(a,b,-1)=0$ for $a,b\geqslant 1$. \end{lemma} \begin{proof} Since $H(-1;q,t)=q^{-1}$, Equation~\eqref{H to F} implies $F(-1)=0$. On the other hand, Lemma~\ref{F(a,b)} immediately implies $F(a,-1)=0$. Finally, by Corollary~\ref{corollary.recursion} we have \begin{equation*} F(a-1,b-1,0)=F(a-1+1,b-1+1,0-1)+(qt)^0 F(a-1+0,b-1-0). \end{equation*} But by Corollary~\ref{cor:set 0} we have $F(a-1,b-1,0)=F(a-1,b-1)$ and the lemma follows. \end{proof} The recursion of Corollary~\ref{corollary.recursion} implies the following ``two-step'' recursion. It has the advantage that it does not contain any negative arguments, which will be advantageous for the combinatorial analysis of Section~\ref{section.proof}. \begin{lemma} \label{lemma.two step recurrence} For $a\geqslant b-1$, $a,b \geqslant c-1\geqslant 0$, we have \begin{multline*} F(a,b,c) = F(a+2,b+2,c-2) + (qt)^c \; F(a+c,b-c) + (qt)^{c-1} \; F(a+c,b-c+2)\\ +\sum_{j=2}^{\min(a-b,2c)} (qt)^{b+j} \; F(a-b+2c-2j) -\sum_{j=a-b+1}^{1} (qt)^{b+j} \; F(a-b+2c-2j). \end{multline*} \end{lemma} \begin{rema} If $a>b$ then the last sum is empty. If $a=b$ or $a=b-1$ then the next to last sum is empty, and the last sum contains one or two terms, respectively. \end{rema} \begin{proof} Using the recurrence in Corollary~\ref{corollary.recursion} and then the same recurrence again on the term $F(a+1,b+1,c-1)$, we obtain \begin{multline*} F(a,b,c) = F(a+2,b+2,c-2) + (qt)^c \; F(a+c,b-c) + (qt)^{c-1} \; F(a+c,b-c+2)\\ + \sum_{i=0}^{c-1} (qt)^{b+2c-2i} F(a-b-2c+4i) + \sum_{i=0}^{c-2} (qt)^{b+2c-1-2i} F(a-b-2c+2+4i). \end{multline*} The first three terms are the same as in the statement of the lemma. The last two sums can be combined to \begin{equation*} \sum_{j=0}^{2c-2} (qt)^{b+2c-j} F(a-b-2c+2j), \end{equation*} or, reversing the order of the sum: \begin{equation} \label{equation.combined sum} \sum_{j=2}^{2c} (qt)^{b+j} F(a-b+2c-2j). \end{equation} If $2c \leqslant a-b$ the corollary is proved. Otherwise we may break expression~\eqref{equation.combined sum} above into two pieces to obtain \begin{equation*} \sum_{j=2}^{a-b} (qt)^{b+j} F(a-b+2c-2j) + \sum_{j=\max(a-b+1,2)}^{2c} (qt)^{b+j} F(a-b+2c-2j), \end{equation*} or equivalently \begin{multline*} \sum_{j=2}^{a-b} (qt)^{b+j} F(a-b+2c-2j) +\sum_{j=a-b+1}^{2c} (qt)^{b+j}F(a-b+2c-2j) \\ -\sum_{j=a-b+1}^{1} (qt)^{b+j}F(a-b+2c-2j). \end{multline*} Therefore, if we show that the middle sum above is $0$ the corollary is proved. However, setting $K=-a+b-2c$ we have \[ \sum_{j=a-b+1}^{2c} (qt)^{b+j}F(a-b+2c-2j) =\sum_{-K \leqslant r \leqslant K-2} (qt)^{b-(K+r)/2}F(r), \] where the sum is over only those $r$ such that $2| (K+r)$. Since $F(-1)=0$ this can be split into \[ \sum_{2 \leqslant r \leqslant K} (qt)^{b-(K-r)/2}F(-r) +\sum_{0 \leqslant s \leqslant K-2} (qt)^{b-(K+s)/2}F(s), \] where again the sum is only over $r$ with $2| (K+r)$. However, applying Proposition~\ref{proposition.negative} term-wise to the left sum gives precisely the opposite of the right sum. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Combinatorial expressions} \label{section.combinatorial expressions} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% In this section, we present a combinatorial formula for $F(a,b,c)$ when $a+1 \geqslant b$, $a+1,b+1 \geqslant c\geqslant 0$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Symmetric chain expression} \label{section.sc} Recall that $\lambda(a,b,c) = (a+b+c,b+c,c)$. We set $A= |\lambda(a,b,c)| = a + 2b +3c$, $\epsilon_{ij} = \max(0,i+j-b-c)$, and $m_{cj} = c-j \pmod{2}$ for convenience. Define the \defn{symmetric chain} for $k\leqslant \ell$ as \[ [k,\ell]_{q,t} = q^\ell t^k + q^{\ell-1} t^{k+1} + \cdots + q^{k+1} t^{\ell-1} + q^k t^\ell. \] We may write $F(a,b,c)$ as a sum of symmetric chains. \begin{theorem} \label{theorem.symmetric chains} For nonnegative integers $a,b,c$ and $a+1 \geqslant b$, $a+1,b+1 \geqslant c$, we have \[ F(a,b,c) = \sum_{(i,j) \in \widetilde{Q}} [i+ \epsilon_{ij}, A-2i-j]_{q,t}, \] where $\widetilde{Q} = \{(i,j) \mid 0\leqslant j \leqslant c, \; j\leqslant i \leqslant b+c, \; 2i+2j \leqslant a+b+2c-m_{cj}\}$. \end{theorem} The proof of Theorem~\ref{theorem.symmetric chains} is given in Section~\ref{section.proof}, see in particular Corollary~\ref{corollary.F=Fcomb}. For the various conditions appearing in $\widetilde{Q}$, see the conditions for quasiheads in Table~\ref{table.indexing sets}. Note that Theorem~\ref{theorem.symmetric chains} immediately implies that the right hand side is symmetric in $q$ and $t$. \begin{rema} Note that the conditions on $i$ and $j$ imply that $i+\epsilon_{ij} \leqslant A-2i-j$. Namely, since $i \leqslant b+c$, we have $i+\epsilon_{ij} \leqslant \max(b+c,i+j)$. Furthermore, since $2i+2j \leqslant a+b+2c$, we have $A-2i-j = A-2i-2j+j \geqslant b+c+j$ which in turn is greater or equal to $\max(b+c,i+j)$ given that $j\geqslant 0$ and $i\leqslant b+c$. \end{rema} \begin{rema} Note that the interval $[i+ \epsilon_{ij}, A-2i-j]$ of integers that appears in the symmetric chains in Theorem~\ref{theorem.symmetric chains} will be called the \defn{area range} in Section~\ref{section.proof} as it is the range of the area statistic for the given symmetric chain. \end{rema} \begin{rema} Surprisingly, the identity $H(a,b,c) = \sum_{(i,j) \in \widetilde{Q}} q^{A-2i-j} t^{i+\epsilon_{ij}}$ does not hold in general, as the right hand side satisfies slightly different recursion relation, see Lemma~\ref{lem: recursion H comb}. \end{rema} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Combinatorial expression} \label{section.combinatorics} \looseness-1 The symmetric chain expression of Theorem~\ref{theorem.symmetric chains} leads to a purely combinatorial expression for $F(a,b,c)$ as a sum of all subpartitions of $\lambda(a,b,c)$ with two associated statistics. The \defn{area statistic} for $\lambda \subseteq \lambda(a,b,c)$ is given by \[ \area(\lambda) = |\lambda(a,b,c)| - |\lambda|. \] The second statistic requires a little more notation. We set $L=a+b+c$. Furthermore, we name the following cases: \begingroup \def\thecdrthm{\arabic{cdrthm}} \setcounter{cdrthm}{0} %\begin{enumerate} %\item[\textbf{Case 1.}] \begin{enonce}[remark]{Case}\label{case1} $z \geqslant \min(b+c-x, \lceil \frac{y-a}{2} \rceil)$ \begin{enumerate}[(\alph*)] \item\label{case1_a} %[\textbf{(a)}] $x+y-z+2 \epsilon_{yz} \epsilon_{ij}\} \] and $\epsilon_{ij}$ and $\delta_{ij}$ are also given in Table~\ref{table.indexing sets}. Finally, we define the set of \defn{heads} $H=H^- \cup H^+$, where $H^-=P^-$ and \[ H^+ = \{(k,\ell,0) \mid a < \ell \leqslant k 0$, in which case it follows from $i \leqslant b+c$. The right hand side is equivalent to $j \leqslant c$ (since $\epsilon_{ij}=\epsilon^{EF}$). \item Condition~\eqref{t2}: The condition $2\epsilon^{EF} \leqslant E \leqslant F+a$ translates to $2\epsilon^{EF} \leqslant i-j+2\epsilon_{ij} \leqslant j-\epsilon_{ij}+a$. The left hand side is equivalent to $j \leqslant i$ and the right hand side follows from~\eqref{p4}. \item Condition~\eqref{t3}: The condition $4E+5F-3\epsilon^{EF} \leqslant a+3b+3c $ translates to $4i-4j+8\epsilon_{ij} +5j -5 \epsilon_{ij}-3\epsilon^{EF} \leqslant a+3b+3c $ which follows from~\eqref{p3}.\qedhere \end{itemize} \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Area preserving bijection between pseudoheads and heads} \label{section.heads} We now construct an area preserving bijection between pseudoheads and heads. Set $\delta_k^\ell=\lceil\frac{\ell-a}{2}\rceil$ and $\epsilon_k^\ell=\max(k+\delta_k^\ell-b-c,0)$. \begin{lemma} Define maps $\Theta$ and $\Theta^{-1}$ by \begin{equation*} \begin{split} \Theta(i,j)&=(i+j-\delta_{ij},i+\epsilon_{ij}),\\ \Theta^{-1}(k,\ell)&=(\ell-\epsilon_k^\ell,k-\ell+\epsilon_k^{\ell}+\delta_k^\ell). \end{split} \end{equation*} Then $\Theta$ induces a bijection from $P$ to $H$, which is the identity on $P^-$ and, if $\Theta(i,j)=(k,\ell)$ it acts as $(i,i,j) \mapsto (k,\ell,0)$ on $P^+$. The inverse of this map is the identity on $H^-$ and, if $\Theta^{-1}(k,\ell)=(i,j)$, then $(k,\ell,0) \mapsto (i,i,j)$ on $H^+$. Moreover if $\Theta(i,j)=(k,\ell)$ then $R_{ij}=R_k^\ell$. \end{lemma} \begin{proof} First we show that $\Theta$ is a bijection on $\mathbb{Z}^2$. Indeed, note that if either $\Theta(i,j)=(k,\ell)$ or $\Theta^{-1}(k,\ell)=(i,j)$ we have $\delta_k^\ell=\delta_{ij}$ and $\epsilon_k^\ell=\epsilon_{ij}$. Hence, a simple computation shows that $\Theta \circ \Theta^{-1}$ and $\Theta^{-1} \circ \Theta$ are the identity on $\mathbb{Z}^2$. Moreover, it is apparent that $\Theta$ preserves the area range. Now suppose $P_{ij} \in P^+$, and $\Theta(i,j)=(k,\ell)$. We wish to show that $(k,\ell,0) \in H^+$. This means we must verify the inequalities $a< i+ \epsilon_{ij} \leqslant i+j-\delta_{ij} \epsilon_{ij}$ is equivalent to $i-\epsilon_{ij} > a$. The second inequality is the same as $\delta_{ij}+\epsilon_{ij} \leqslant j$ which is equivalent to $i-2j \leqslant a- 3 \epsilon_{ij}$. If $\epsilon_{ij}=0$, this is the same as the pseudohead condition $i-2j \leqslant a$. Otherwise, it is equivalent to the pseudohead condition $4i+j \leqslant a+3b+3c$. Finally, the last inequality is just $i+j-(b+c)< \delta_{ij}$ which is immediate since the former is less than or equal to $\epsilon_{ij}$ which is by assumption less than $\delta_{ij}$. Now suppose $H_k^\ell \in H^+$ and $\Theta^{-1}(k,\ell)=(i,j)$. We need to show that $\delta_{ij}>\epsilon_{ij}$ as well as the pseudohead conditions~\eqref{p1}-\eqref{p4} for $i=\ell-\epsilon_k^\ell$ and $j=k-\ell+\epsilon_k^{\ell}+\delta_k^\ell$ for any $(k,\ell)$ such that $a < \ell \leqslant k \epsilon_{ij}$. We have $\delta_{ij}-\epsilon_{ij}=\delta_k^\ell-\epsilon_k^\ell=\min(-k+b+c,\delta_k^\ell)$. But this is a positive number because $ka$. \item Condition~\eqref{p1}: The condition $0 \leqslant j \leqslant c$ translates to $0 \leqslant k-\ell +\epsilon_k^\ell + \delta_k^\ell \leqslant c$. The left side holds since all of $k-\ell,\epsilon_k^\ell,\delta_k^\ell$ are nonnegative. Now, if $\epsilon_k^\ell=0$ then $k+\delta_k^\ell \leqslant b+c \leqslant a+c+1 \leq \ell+c$ which implies the right hand side. On the other hand, if $\epsilon_k^\ell>0$ the inequality becomes $2k-\ell+2\delta_k^\ell \leqslant b+c$ which would certainly hold if $2k +2\frac{\ell-a}{2}-\ell=2k-aa$ implies $\lfloor\frac{3\ell+a}{2}\rfloor \leqslant 2a+1$. If $\epsilon_k^\ell>0$ the left hand inequality reduces to $3k \leqslant 2\ell-3\delta_k^\ell +2b+2c=\lfloor \frac{\ell+a}{2} \rfloor +a +2b +2c$ which would certainly hold if $2k+k=3k \leqslant 2a+2b+2c$. But $2k \leqslant b+c-2$ and $k \leqslant 2a+1$ so this holds (in fact strictly). Moreover, the righthand side easily holds as $\ell-\epsilon_k^\ell \leqslant \ell \leqslant k 0$ the inequality we need to show reduces to $3\ell-2k-2\delta_k^\ell \leqslant a$. Since $\ell-k \leqslant 0$ it suffices to show that $\ell- 2\delta_k^\ell \leqslant a$. But this is clear since $\ell- 2\delta_k^\ell \leqslant \ell-2\frac{\ell-a}{2} = a$. \item Condition~\eqref{p4}: The condition $i-2j \leqslant a$ translates to $3\ell-2k-3\epsilon_k^\ell-3\delta_k^\ell \leqslant a$. But this follows from $\ell-2\delta_k^\ell \leqslant a$ and $\ell-k \leqslant 0$. \end{itemize} This shows that $\Theta$ induces a bijection from $P^+$ to $H^+$. Extending this map to all of $P$ by declaring it to be the identity on $P^-$ is also a bijection as long as $H^- \cap H^+ = \emptyset$. Indeed, only partitions of the form $(m,m,0)$ may lie in both $H^-$ and $H^+$. Moreover, being in $H^-$ implies $\delta_{m0}-\epsilon_{m0} \leqslant 0$ which means $m \leqslant a$. On the other hand, being in $H^+$ would require $a0$ by the left-hand side of~\eqref{t2}. Thus the right-hand side of~\eqref{t1} implies $F \leqslant c-1$ contradicting the assumption $F=c$. This shows that $(p,q,c)$ is indeed a partition. Finally it is obvious that $j \leqslant c$ and since we already showed that $p \geqslant q$ all that remains to show is $q \geqslant i$. But this says $b+c-F \geqslant i$ or $b+c-(j-\epsilon_{ij}) \geqslant i$ which is equivalent to $i+j-(b+c) \leqslant \epsilon_{ij}$ which follows immediately from~\eqref{equation.eps ij}. \end{proof} \begin{theorem} \label{neg} Let $\mu \in \Lambda^-$. Then there exists unique $P_{ij} \in P$ such that $\mu \in S(P_{ij})$. Conversely, if $\mu \in S(P_{ij})$ for some pseudohead $P_{ij}$, then $\mu \in \Lambda^-$. \end{theorem} \begin{proof} Let $\mu=(x,y,z) \in \Lambda^-$. Let us set: \[ \mathcal{E}(y,z)=y-z+2\epsilon_{yz} \qquad \text{and} \qquad \mathcal{F}(y,z)=z-\epsilon_{yz}. \] We prove the first statement in three cases. \begin{enumerate} \item{First suppose $x+\mathcal{E}(y,z)0$ then the original assumption reduces directly to $x+3y+z < a+3b+3c$ so we are done for the same reason. \item Condition~\eqref{p4}: $y-2z \leqslant a$. First suppose $\epsilon_{yz}=0$. Now, since $\mu$ is a negative partition we either have $z \geqslant \lceil \frac{y-a}{2}\rceil$ which would mean $2z \geqslant y-a$ and we would be done, or else, $z \geqslant b+c-x$. In the second case: $\epsilon_{yz}=0$ along with the original assumption imply $x+y-z0$ then we have $y+z>b+c$. Subtracting three times this from $4y+z \leqslant a+3b+3c$ (which we have already verified) gives $y-2zb+c-x$ means $y>a+2b+2c-2x$. Since $ya+b+c$. Adding this inequality to $x+z \geqslant b+c$ (which is equivalent to the assumption on hand) we obtain $3x>a+2b+2c$. At this point we suppose for the sake of contradiction that $4(L+z-x)+z > a+3b+3c$. This means $(L+z-x+z)+3L+3z-3x>a+3b+3c$ which in light of the previous equation yields $(L+z-x+z)+3L+3z > 2a+5b+5c$. This in turn gives $3L+3z>2a+4b+4c$ since $L+z-x+z-a+b+c$. But at the same time $4(L+z-x)+z > a+3b+3c$ means $4(L+z-x+z)-3z > a+3b+3c$. And again making use of $L+z-x+zb$ to obtain $y+z \geqslant b+c$. Since $y \geqslant z$ this would give $b \leqslant c-2$ which is not allowed. The right hand side follows directly from $x \geqslant y$. \item Condition~\eqref{t3}: $4E+5F-3\epsilon^{EF} \leqslant a+3b+3c$. This reduces to $4x+5y+3\epsilon^{(L-x)(b+c-y)} \geqslant 3a+6b+6c$. If $\epsilon^{(L-x)(b+c-y)}=0$ we must have $x+2y \geqslant a+2b+2c$. Adding three times this inequality to the inequality $x-y \geqslant 0$ gives us what we desire. If $\epsilon^{(L-x)(b+c-y)}>0$ then $\epsilon^{(L-x)(b+c-y)}=a+2b+2c-x-2y$ and the inequality $4x+5y+3\epsilon^{(L-x)(b+c-y)} \geqslant 3a+6b+6c$ reduces to $x-y \geq 0$. \end{itemize} Now we know that $T^{(L-x)(b+c-y)}$ is a valid tail. Denote $\Psi(L-x,b+c-y)=(i,j)$. In order to show that $\mu \in S(T^{(L-x)(b+c-y)})$ we need only verify that $j \leqslant z$. That is to say $b+c-y+\epsilon^{(L-x)(b+c-y)} \leqslant z$. If $\epsilon^{(L-x)(b+c-y)}=0$ this follows from the original assumption that $y+z \geqslant b+c$. Otherwise it reduces to $a+3b+3c \leqslant x+3y+z$. But this is equivalent to the original assumption that $x+\mathcal{E}(y,z) \geqslant L$ since $y+z \geqslant b+c$ implies $\epsilon_{yz}=y+z-(b+c)$.} \end{enumerate} This concludes the proof of the first statement. Now we prove the second statement. Suppose $\mu=(x,y,z) \in S(P_{ij})$ for some pseudohead $P_{ij}$. We must show that $z \geqslant \min(b+c-x,\lceil \frac{y-a}{2}\rceil)$. We use two cases: \begin{enumerate} \item $\mu$ is in the first union in~\eqref{equation.SP union}. We show $z \geqslant \lceil \frac{y-a}{2}\rceil$. We have $y=i$ and $j=z$ so this becomes $j \geqslant \lceil \frac{i-a}{2}\rceil$. But the latter is equivalent to $2j \geqslant i-a$ which is equivalent to condition~\eqref{p4}. \item $\mu$ is in the second or third union in~\eqref{equation.SP union}. We show $z \geqslant b+c-x$. First, if $\epsilon_{ij}>0$ then $i+j>b+c$ directly implies $j>b+c-x$ so that $z>b+c-x$. Now we assume $\epsilon_{ij}=0$. Since $x=L-\mathcal{E}(i,j)$ and $z\geqslant j$ it would be enough to show $j \geqslant b+c-(L-(i-j))$ which reduces to $j\geqslant -a+i-j$ but this follows from condition~\eqref{p4}.\qedhere \end{enumerate} \end{proof} If $H_k^\ell \in H^+$, we define the \defn{appendage} associated to $H_k^\ell$ to be \[ A(H_k^\ell)=\left\{(k,\ell,z)\mid z < \min\left(b+c-k,\left\lceil \frac{\ell-a}{2}\right\rceil\right)\right\}. \] \begin{theorem} \label{pos} Let $\mu \in \Lambda^+$. Then there exists unique $H_k^\ell \in H^+$ such that $\mu \in A(H_k^\ell)$. Conversely, if $\mu \in A(H_k^\ell)$ for some positive head $H_k^\ell$, then $\mu \in \Lambda^+$. \end{theorem} \begin{proof} Let $\mu=(x,y,z) \in \Lambda^+$. Note that this correspond to Case~\ref{case2} in Section~\ref{section.combinatorics}. Then it is immediate that $\mu$ could only belong to the appendage $A(H_x^y)$. Since $z < \min(b+c-x,\lceil \frac{y-a}{2}\rceil)$ in particular $0 < \min(b+c-x,\lceil \frac{y-a}{2}\rceil)$. This implies both $x a$ so (as $y \leqslant x$) $H_x^y \in H^+$. Since $\mu$ is positive $z < \min(b+c-x,\lceil \frac{y-a}{2}\rceil)$, so $\mu \in A(H_x^y)$. Now if $\mu=(x,y,z) \in A(H_k^\ell)$ for some head, then $x=k$ and $y=\ell$ and so the inequality $z < \min(b+c-x,\lceil \frac{y-a}{2}\rceil)$ is clearly satisfied implying that $\mu \in \Lambda^+$. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Chains} \label{section.chains} Suppose $T^{EF} \in T$. Set $(i,j)=\Psi(E,F)$. If $P_{ij} \in P^+$ set $(k,\ell)=\Theta(i,j)$. We define the \defn{chain} of $T^{EF}$ to be \begin{equation} C(T^{EF})=\begin{cases} S(P_{ij}) & \text{if $P_{ij} \in P^-$,}\\ S(P_{ij}) \cup A(H_k^\ell) & \text{if $P_{ij} \in P^+$.} \end{cases} \end{equation} Our fundamental result concerning chains is the following. \begin{theorem} \label{theorem.chain decomp} $\Lambda$ is the disjoint union: \[ \Lambda= \bigcup_{T^{EF} \in T} C(T^{EF}). \] Moreover, for each integer $m \in R^{EF}=[E+F,A-2E-3F+\max(\delta^{EF},\epsilon^{EF})]$ there is precisely one element $\mu \in C(T^{EF})$ with area $\mathsf{area}(\mu)=m$. \end{theorem} \begin{proof} The first statement is immediate by combining Theorems~\ref{neg} and~\ref{pos}. Now fix $T^{EF}$ and set $(i,j)=\Psi(E,F)$. If $P_{ij} \in P^-$, then by definition $C(T^{EF})=S(P_{ij})$. By construction, this string has one partition of area $m$ for each $m \in [\mathsf{area}(T^{EF}),\mathsf{area}(P_{ij})]$. But $\mathsf{area}(T^{EF})=E+F$. Moreover, $\mathsf{area}(P_{ij})=A-2i-j=A-2E-3F + \epsilon^{EF}$ and since $P_{ij}\in P^-$ implies that $\epsilon_{ij}=\max(\delta_{ij},\epsilon_{ij})=\max(\delta^{EF},\epsilon^{EF})$ this means $\mathsf{area}(P_{ij})=A-2E-3F+\max(\delta^{EF},\epsilon^{EF})$. Now suppose $P_{ij} \in P^+$. Then $C(T^{EF})=S(P_{ij}) \cup A(H_k^\ell)$ has one partition of area $m$ for each $m \in [\mathsf{area}(T^{EF}),\mathsf{area}(P_{ij})]$ and one partition of area $n$ for each $n \in [\mathsf{area}(H_k^\ell)-\min(b+c-k,\lceil \frac{\ell-a}{2}\rceil)+1,\mathsf{area}(H_k^\ell)]$. Again, $\mathsf{area}(T^{EF})=E+F$ and $\mathsf{area}(P_{ij})=A-2E-3F + \epsilon^{EF}$ so it suffices to prove the two equations \begin{align*} &\mathsf{area}(H_k^\ell)-\min\left(b+c-k,\left\lceil \frac{\ell-a}{2}\right\rceil\right)+1=A-2E-3F + \epsilon^{EF}+1,\\ &\mathsf{area}(H_k^\ell)=A-2E-3F+\max(\delta^{EF},\epsilon^{EF})=A-2E-3F+\delta^{EF}. \end{align*} However, we have \begin{align*} k+\ell&=(i+j-\delta_{ij})+(i+\epsilon_{ij})= 2i+j-(\delta_{ij}-\epsilon_{ij})\\ &=2(E+F-\epsilon^{EF})+(F+\epsilon^{EF})-(\delta^{EF}-\epsilon^{EF})=2E+3F-\delta^{EF}, \end{align*} so that $\mathsf{area}(H_k^\ell)=A-(k+1)=A-2E-3F+\delta^{EF}$ as desired. On the other hand \begin{multline*} \min\left(b+c-k,\left\lceil \frac{\ell-a}{2}\right\rceil\right)\\ \begin{aligned} &= \min(b+c-\left(i+j-\delta_{ij},\left\lceil \frac{i+\epsilon_{ij}-a}{2}\right\rceil\right)\\ &=\min(b+c-(i+j)+\delta_{ij},\delta_{ij})= \delta_{ij} + \min(b+c-(i+j),0)\\ &=\delta_{ij}-\epsilon_{ij}=\delta^{EF}-\epsilon^{EF}. \end{aligned} \end{multline*} Hence we have \begin{align*} \mathsf{area}(H_k^\ell)-\min\left(b+c-k,\left\lceil \frac{\ell-a}{2}\right\rceil\right)&=A-2E-3F+\delta^{EF}-(\delta^{EF}-\epsilon^{EF})\\ &=A-2E-3F + \epsilon^{EF}, \end{align*} which gives the other equation we wanted after adding $1$ to both sides. \end{proof} Since $\Psi$ and $\Theta$ fix the area range, we can conclude the following statement. \begin{coro} \label{corollary.cover} Let $X$ represent the set of heads, the set of pseudoheads, or the set of tails. Then $\Lambda$ is the disjoint union of all chains which contain an element of $X$. Moreover, for $x \in X$ and each $m$ in the area range of $x$, there is precisely one element $\mu$ of area $m$ in the same chain as $x$. \end{coro} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Area preserving bijection between head and quasiheads} \label{section.quasi} We write $Q=Q_{\leqslant}^- \cup Q_>^- \cup Q^+$, where \begin{align*} Q_{\leqslant}^- &= \{Q\mms{st} \in Q\mid s+t \leqslant b+c, s \leqslant a\},\\ Q_{>}^- &= \{Q\mms{st} \in Q\mid s+t > b+c\} \cup \{Q\mms{st} \in Q\mid s+t = b+c, s>a\},\\ Q^+ &= \{Q\mms{st} \in Q\mid s+t < b+c, s > a\}, \end{align*} and $H=P_{\leqslant}^- \cup P_>^- \cup H^+$, where \[ P_{\leqslant}^-=\{P_{ij} \in P^-\mid i+j \leqslant b+c\} \qquad \text{and} \qquad P_{>}^-=\{P_{ij} \in P^-\mid i+j > b+c\}. \] \begin{prop} \label{proposition:H-Q} There is an area range preserving bijection from $H$ to $Q$. \end{prop} \begin{proof} We prove the proposition in three parts. First we show that the identity is an area range preserving bijection from $P_{\leqslant}^-$ to $Q_{\leqslant}^-$. Then we define an area range preserving bijection from $P_>^-$ to $Q_>^-$. Finally we define an area range preserving bijection from $H^+$ to $Q^+$. \begin{enumerate} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%PART 1 %%%%%%%%%%%%%%%%%%%%%%%%% P<=Q< \item The set $P_{\leqslant}^-$ is the set of triples $(i,i,j)$ obeying the conditions~\eqref{p1}-\eqref{p4} as well as the inequalities $\delta_{ij} \leqslant \epsilon_{ij}$ and $i+j \leqslant b+c$. In light of~\eqref{p2}, $\epsilon_{ij}=0$ and $\delta_{ij} \leqslant \epsilon_{ij}$ simply becomes $i \leqslant a$. But this in turn implies~\eqref{p4}. Moreover, adding $i \leqslant a$ to $3i+3j \leqslant 3b+3c$ gives condition~\eqref{p3}. Thus $P_{\leqslant}^-$ is the set of triples $(i,i,j)$ satisfying the four conditions in~\eqref{p1} and~\eqref{p2} as well as $i+j \leqslant b+c$ and $i \leqslant a$. On the other hand, $Q_{\leqslant}^-$ is the set of triples $(s,s,t)$ satisfying the five conditions~\eqref{q1}-\eqref{q3} as well as $s+t \leqslant b+c$ and $s \leqslant a$. Since conditions~\eqref{p1}-\eqref{p2} are equivalent to \eqref{q1}-\eqref{q2}, if we can show that condition~\eqref{q3} is implied by the other four conditions, it follows that $Q_{\leqslant}^-=P_{\leqslant}^-$. Indeed, adding the three inequalities $s+t \leqslant b+c$, $s \leqslant a$, and $t \leqslant c$ gives $2s+2t \leqslant a+b+2c$. This is strict unless we have equality in all of the three previous conditions. In particular, this would mean $t=c$ so that $m_{ct}=0$. Thus in any case $2s+2t \leqslant a+b+2c-m_{ct}$. Therefore, $Q_{\leqslant}^-=P_{\leqslant}^-$. Since for $P_{ij} \in P_{\leqslant}^-$ $\max(0,\delta_{ij}-\epsilon_{ij})=0$, we have $R_{ij}=R\mms{st}$ if $i=s$, $j=t$, so that the identity is an area range preserving bijection between the two sets. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%PART 2 \item Let $\omega_{ij}=\max(0,\lceil\frac{2i+j-L}{2}\rceil)$ and define maps $\Phi$ and $\Phi^{-1}$ by \[ \Phi(i,j)=(i+\omega_{ij},j-2\omega_{ij}) \qquad \text{and} \qquad \Phi^{-1}(s,t)=(s-\omega_{st}, t+2\omega_{st}). \] Now if $\Phi(i,j)=(s,t)$ or $\Phi^{-1}(s,t)=(i,j)$, it is clear that $\omega_{ij}=\omega_{st}$. From this it follows that $\Phi^{-1} \circ \Phi$ and $\Phi \circ \Phi^{-1}$ are the identity on $\mathbb{Z}^2$. We claim that $\Phi$ induces an area range preserving bijection from $P_>^-$ to $Q_>^-$ via the rule that if $\Phi(i,j)=(s,t)$, then $(i,i,j) \mapsto (s,s,t)$ with the inverse induced by $\Phi^{-1}$ via the rule that if $\Phi^{-1}(s,t)=(i,j)$, then $(s,s,t) \mapsto (i,i,j)$. %%%%%%%%%%%%%%%%%%%%%%%%%% P> to Q> First suppose $P_{ij} \in P_>^-$, so that conditions~\eqref{p1}-\eqref{p4} are satisfied alongside $\delta_{ij}-\epsilon_{ij} \leqslant 0$ and $i+j>b+c$. We need to check that if $\Phi(i,j)=(s,t)$, \ie $s=i+\omega_{ij}$ and $t=j-2\omega_{ij}$, then $(s,t)$ satisfies conditions~\eqref{q1}-\eqref{q3} and that $s+t \geqslant b+c$ and $(s+t=b+c) \implies s>a$. \begin{itemize} %%%%%%%%%%%%%%%%q1 \item Condition~\eqref{q1}: $0 \leqslant t \leqslant c$. This translates to $0 \leqslant j-2\omega_{ij} \leqslant c$. The right hand side follows from $j \leqslant c$ (see the right hand side of~\eqref{p1}). If $\omega_{ij}=0$, the left hand side follows from the left hand side of~\eqref{p1}. Otherwise it says that $j \geqslant 2\lceil\frac{2i+j-L}{2}\rceil$. Now $\delta_{ij}-\epsilon_{ij}\leqslant 0$ is equivalent to $i - \epsilon_{ij} \leqslant a$, but $i+j > b+c$ so $\epsilon_{ij}>0$ and this becomes $i-(i+j-(b+c)) \leqslant a$ or $-j \leqslant a-b-c$. Adding this to~\eqref{p3} yields $4i \leqslant 2a+2b+2c$ or $2i \leqslant L$. This is enough to prove $j \geqslant 2\lceil\frac{2i+j-L}{2}\rceil$ unless $2i=L$ and $j$ is odd. But then $4i+j$ is odd and $a+3b+3c$ is even so we have strictness in~\eqref{p3}, that is, $4i+j < a+3b+3c$. Hence adding this to $-j \leqslant a-b-c$ results in $4i < 2a+2b+2c$ which contradicts $2i=L$. %%%%%%%%%%%%%%%%%%%q2 \item Condition~\eqref{q2}: $t \leqslant s \leqslant b+c$. This translates to $j-2\omega_{ij} \leqslant i+\omega_{ij} \leqslant b+c$. The left hand side follows from the left hand side of~\eqref{p2}. If $\omega_{ij}=0$, then the right hand side comes from the right hand side~\eqref{p2}. Otherwise the right hand side says $i+\lceil\frac{2i+j-L}{2}\rceil \leqslant b+c$ which follows from $i+\frac{2i+j-L}{2} \leqslant b+c$ (which is equivalent to~\eqref{p3}) since $b+c$ is an integer. %%%%%%%%%%%%%%%%%%q3 \item Condition~\eqref{q3}: $2s+2t \leqslant a+b+2c-m_{ct}$. This translates to $2i+2j-2\omega_{ij}\leqslant a+b+2c-m_{cj}$ (note that $m_{c(j-2\omega_{ij})}=m_{cj}$). If $\omega_{ij}=0$ then $2i+j \leqslant L$ so it suffices to show $j \leqslant c-m_{cj}$ which is evident by the definition of $m_{cj}$ and $j \leqslant c$. On the other hand if $\omega_{ij}>0$, then proving $2i+2j \leqslant (2i+j-L)+ a+b+2c-m_{cj}$ suffices since $(2i+j-L) \leqslant 2\omega_{ij}$. But the former again reduces to $j \leqslant c-m_{cj}$. %%%%%%%%%%%%%%%%%s+t>=b+c \item $s+t \geqslant b+c$. This says $i+j-\omega_{ij} \geqslant b+c$. This is clear from the definition of $P_>^-$ if $\omega_{ij}=0$ so suppose $\omega_{ij}>0$. Now, as in the first bullet point, $\delta_{ij}-\epsilon_{ij} \leqslant 0$ and $i+j>b+c$ imply $j \geqslant b+c-a$. The latter is equivalent to $2i+2j-2\frac{2i+j-L}{2} \geqslant 2b+2c$, or, dividing by $2$, $i+j-\frac{2i+j-L}{2} \geqslant b+c$. But since $b+c$ is an integer, this implies $i+j-\lceil\frac{2i+j-L}{2}\rceil \geqslant b+c$ as desired. %%%%%%%%%%%%%%%s+t=b+c --> s>a \item $(s+t=b+c) \implies s>a$. This translates to, if $i+j-\omega_{ij}=b+c$, then $i+\omega_{ij}>a$. If $\omega_{ij}=0$, the hypothesis would clearly contradict the assumption $i+j>b+c$. Thus we may assume $\omega_{ij}>0$ which means $2i+j > L$. Adding this to $-i-j \geqslant -b-c - \omega_{ij}$ gives $i > a-\omega_{ij}$ as desired. \end{itemize} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Q> to P> Now suppose $Q\mms{st} \in Q_>^-$, so that conditions~\eqref{q1}-\eqref{q3} are satisfied alongside $s+t \geqslant b+c$ and $(s+t=b+c) \implies s>a$. We need to check that if $\Phi^{-1}(s,t)=(i,j)$, that is, $i=s-\omega_{st}$ and $j=t+2\omega_{st}$, then $(i,j)$ satisfies conditions~\eqref{p1}-\eqref{p3} as well as $\delta_{ij}-\epsilon_{ij} \leqslant 0$ and $i+j>b+c$. (We do not need to check condition~\eqref{p4} as adding $-3i-3j<-3b-3c$ to~\eqref{p3} yields $i-2j0$. We need to show that $t+2\omega_{st} \leqslant s-\omega_{st}$. The inequality we wish to show is equivalent to $2t-2s+6\omega_{st} \leqslant 0$. Since $2\omega_{st}$ can be rewritten as $2s+t-L+m_{Lt}$ this becomes $4s+5t \leqslant 3L-3m_{Lt}$. First suppose that $s+t>b+c$ and $m_{Lt} \leqslant m_{ct}$. Since twice~\eqref{q3} reads $4s+4t \leqslant 2L +2c -2m_{ct}$ it suffices to prove $t \leqslant a+b-c+2m_{ct}-3m_{Lt}$, since the sum of the last two inequalities mentioned gives the one at the end of the last sentence. Since $t \leqslant c$, it suffices to show $a+b \geqslant 2c-2m_{ct}+3m_{Lt}$. Since $s+t >b+c$, we have $-2s-2t\leqslant -2b-2c-2$ which we can add to~\eqref{q3} to get $a \geqslant b+2+m_{ct}$ or $a - b \geqslant 2+m_{ct}$. Adding this to $2b \geqslant 2c-2$ yields $a+b \geqslant 2c+ m_{ct}$ which implies $a+b \geqslant 2c-2m_{ct}+3m_{Lt}$ since we are assuming $m_{Lt} \leqslant m_{ct}$. Now suppose that $s+t>b+c$, but $0=m_{Lt} < m_{ct}=1$. Since $L+c$ must be odd, we have strictness in~\eqref{q3}, that is we have $2s+2t < a+b+2c=L+c$. Thus we have $4s+4t \leqslant 2L +2c -2$ so it suffices to prove $t \leqslant a+b-c-1$ since adding these gives $4s+5t \leqslant 3L-3m_{Lt}$ as desired. Again, $t \leqslant c$ so its enough to show $a+b \geqslant 2c+1$. Since strictness of~\eqref{q3} implies $2s+2t \leqslant a+b+2c-1$, we can add this to $-2s-2t\leqslant -2b-2c-2$ to obtain $a \geqslant b+3$. Since $b \geqslant c-1$, this implies $a+b \geqslant 2c+1$ as desired. Finally suppose that $s+t=b+c$ (and so also $s>a$). We need to show $4(b+c)+t \leqslant 3L-3m_{Lt}$, or equivalently, $t \leqslant 3a-b-c -3m_{Lt}$. When $s+t=b+c$, condition~\eqref{q3} becomes $a \geqslant b+m_{ct}$. Thus $s>a$ implies $s \geqslant b+m_{ct}+1$, which in turn means $t \leqslant c-m_{ct}-1$. In fact $t \leqslant c-2$ because if $t=c-1$ then we would have $m_{ct}=1$ and thus $t \leqslant c-2$. Now since $a \geqslant b+m_{ct} \geqslant c-1+m_{ct}$ this means $t \leqslant c-2 + (a-b-m_{ct}) + 2(a-c+1-m_{ct})$ or that $t \leqslant 3a-b-c-3m_{ct}$ which implies $t \leqslant 3a-b-c -3m_{Lt}$ unless $0=m_{ct}b$. Thus we have $a \geqslant b+1 \geqslant c$ so that $t \leqslant c-2 + (a-b-1) + 2(a-c)$ or $t \leqslant 3a-b-c -3$. %%%%%%%%%%%%%%%%%p3 \item Condition~\eqref{p3}: $4i+j \leqslant a+3b+3c$. This translates to $4s+t-2\omega_{st} \leqslant a+3b+3c$. If $\omega_{st}=0$, then $2s+t \leqslant L$ which we add to two times the right hand side of~\eqref{q2} to obtain $4s+t\leqslant a+3b+3c$ as desired. Now suppose $\omega_{st}>0$. Then $4s+t-2\omega_{st}= 4s+t-2\lceil \frac{2s+t-L}{2} \rceil \leqslant 4s+t-2\frac{2s+t-L}{2}=2s+L \leqslant L+b+c$, where the last inequality comes from the right hand of~\eqref{q2}. %%%%%%%%%%%%%%%%% del <= eps \item $\delta_{ij} - \epsilon_{ij} \leqslant 0$. This reduces to $i-\epsilon_{ij} \leqslant a$ which says that $s-\omega_{st}-\max(0,s+t+\omega_{st}-(b+c)) \leqslant a$. Since $s+t \geqslant b+c$, this is equivalent to $-t-2\omega_{st} \leqslant a-b-c$. If $\omega_{st}=0$ then $2s+t \leqslant L$ and which we may add to $-2s-2t\leqslant -2b-2c$ to get $-t \leqslant a-b-c$ as desired. If $\omega_{st}>0$ then it will suffice to show $-t-2\frac{2s+t-L}{2} \leqslant a-b-c$ but this reduces to $-2s-2t \leqslant -2b-2c$. %%%%%%%%%%%%%% i+j>b+c \item $i+j>b+c$. This means $s+t+\omega_{st}>b+c$. If $\omega_{st}>0$ this follows from $s+t \geqslant b+c$. Now suppose $\omega_{st}=0$. Then $2s+t \leqslant L$ which we may add to $-s-t \leqslant -b-c$ to get $s \leqslant a$. Thus the assumptions $s+t \geqslant b+c$ and $(s+t=b+c) \implies s>a$ reveal that $s+t>b+c$. \end{itemize} This proves that $\Phi$ induces a bijection from $P_>^-$ to $Q_>^-$. Moreover if $P_{ij} \in P_>^-$, then $\delta_{ij}\leqslant \epsilon_{ij}$ and $i+j > b+c$, so the area range reduces to $R_{ij}=[2i+j-(b+c),A-2i-j]$. Now if $\Phi(i,j)=(s,t)$, then since by the above $Q\mms{st} \in Q_>^-$ we have $s+t \geqslant b+c$ so that we get $R\mms{st} = [2s+t-(b+c),A-2s-t]$. Since it is clear that $2i+j=2s+t$ we see that $R_{ij}=R\mms{st}$ and so the bijection induced by $\Phi$ preserves the area range. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%PART 3 \item Define maps $\Omega$ and $\Omega^{-1}$ by \[ \Omega(k,\ell)=(\ell,k-\ell) \qquad \text{and} \qquad \Omega^{-1}(s,t)=(s+t,s). \] Since $\Omega$ is an invertible linear transformation (with inverse $\Omega^{-1}$), it is immediate that $\Omega^{-1} \circ \Omega$ and $\Omega \circ \Omega^{-1}$ are the identity on $\mathbb{Z}^2$. We claim that $\Omega$ induces an area range preserving bijection from $H^+$ to $Q^+$ via the rule that, if $\Omega(k,\ell)=(s,t)$, then $(k,\ell,0) \rightarrow (s,s,t)$ with inverse induced by $\Omega^{-1}$ via the rule that, if $\Omega^{-1}(s,t)=(k,\ell)$ then $(s,s,t) \rightarrow (k,\ell,0)$. First suppose that $H_k^\ell \in H^+$ so that $a < \ell \leqslant k a$. \begin{itemize} \item Condition~\eqref{q1}: $0 \leqslant t \leqslant c$. This translates to $0 \leqslant k-\ell \leqslant c$. The left hand side is true because $\ell\leqslant k$. Moreover, since $\ell>a \geqslant b-1$, the inequalities $\ell \geqslant b$ and $k < b+c$ give $k-\ella$. This says $\ell>a$, as assumed. \end{itemize} Now suppose that $Q\mms{st} \in Q^+$ so that the inequalities~\eqref{q1}-\eqref{q3} hold and we have $s+ta$. We need to show that if $\Omega^{-1}(s,t)=(k,\ell)$ then $H_k^\ell \in H^+$, that is, $a < \ell \leqslant k b$ we obtain \[ c-1\leqslant i\leqslant b+c, \qquad 2i\leqslant a+b+1 \] and the contribution to $H_{\comb}$ equals \begin{equation*} \sum_{\substack{c-1\leqslant i\leqslant b+1\\ 2i\leqslant a+b+1}} q^{a+2b+2c-2i+1}\; t^{i} +\sum_{\substack{b+2\leqslant i\leqslant b+c\\ 2i\leqslant a+b+1}}q^{a+2b+2c-2i+1}\; t^{2i-1-b}. \end{equation*} If $a > b$ the restriction $2i\leqslant a+b+1$ in the first sum is redundant and so it becomes $(qt)^{c-1} H(a+c,b-c+2)$. On the other hand if $a=b$ or $a=b-1$ the first sum does not contain the $i=b+1$ term $q^{a+2c-1}t^{b+1}$ or the $i=b$ term $q^{a+2c}t^b$ but $(qt)^{c-1} H(a+c,b-c-2)$ does. Thus we conclude the above is equal to \begin{multline*} (qt)^{c-1}H(a+c,b-c+2)-(\delta_{a,b-1}+\delta_{a,b})q^{a+2c-1}t^{b+1}\\ +\sum_{2b+3\leqslant 2i-1\leqslant \min(2b+2c-1,a+b)}q^{a+2b+2c-2i+1}t^{2i-1-b}. \end{multline*} Finally, \begin{multline*} \sum_{2b+2\leqslant 2i\leqslant \min(2b+2c,a+b)}q^{a+2b+2c-2i}t^{2i-b} \\ +\sum_{2b+3\leqslant 2i-1\leqslant \min(2b+2c-1,a+b)}q^{a+2b+2c-2i+1}t^{2i-1-b} \\ = \sum_{2b+2\leqslant k\leqslant \min(2b+2c,a+b)}q^{a+2b+2c-k}t^{k-b}, \end{multline*} where we combined terms with even and odd $k$. If we denote $\ell=k-2b$, then \[ \sum_{2b+2\leqslant k\leqslant \min(2b+2c,a+b)}q^{a+2b+2c-k}t^{k-b}= \sum_{2\leqslant \ell\leqslant \min(2c,a-b)}q^{a+2c-\ell}\; t^{\ell+b}.\qedhere \] \end{proof} \begin{coro} \label{corollary.Fcomb recursion} Let \[ F_{\comb}(a,b,c)=\frac{1}{1-t/q}H_{\comb}(a,b,c;q,t)+\frac{1}{1-q/t}H_{\comb}(a,b,c;t,q). \] Then for $a+1 \geqslant b$, $a+1,b+1 \geqslant c\geqslant 1$ we have \begin{multline} F_{\comb}(a,b,c)=F_{\comb}(a+2,b+2,c-2)+(qt)^cF(a+c,b-c)\\+(qt)^{c-1}F(a+c,b-c+2)+ \sum_{2\leqslant \ell\leqslant \min(2c,a-b)}(qt)^{\ell+b}F(a-b+2c-2\ell)\\ - \sum_{j=a-b+1}^1 (qt)^{b+j} F(a-b+2c-2j). \end{multline} \end{coro} \begin{proof} This follows directly from Lemma~\ref{lem: recursion H comb}, using~\eqref{H to F}, and Example~\ref{example.n=2}. Also note that \begin{multline*} \sum_{j=a-b+1}^1 (qt)^{b+j} F(a-b+2c-2j) = \delta_{a,b} (qt)^{b+1} F(a-b+2c-2) \\ + \delta_{a,b-1} \left[ (qt)^b F(a-b+2c) + (qt)^{b+1} F(a-b+2c-2) \right].\qedhere \end{multline*} \end{proof} We need to check the base cases. \begin{lemma} \label{lemma.initial condition} We have \begin{equation*} \begin{split} F_{\comb}(a,b,0)&=F(a,b,0) \quad \text{for $a,b\geqslant 0$,}\\ F_{\comb}(a,b,-1)&=F(a,b,-1) \quad \text{for $a,b \geqslant 1$.} \end{split} \end{equation*} \end{lemma} \begin{proof} For $a,b\geqslant 0$ and $c=0$, we have $j=0$ in $\widetilde{Q}(a,b,0)$, so $0\leqslant i\leqslant b$. Therefore, by comparing~\eqref{equation.H comb} with~\eqref{H for n=3} \[ H_{\comb}(a,b,0)=H(a,b), \] and hence $F_{\comb}(a,b,0)=F(a,b)$. Furthermore, by Corollary~\ref{cor:set 0} the first claim follows. For $a,b\geqslant 1$, we have by Lemma~\ref{lemma.-1} and the fact that $F_{\comb}(a,b,-1)=0$ by definition that $F(a,b,-1)=F_{\comb}(a,b,-1)=0$. \end{proof} \begin{coro} \label{corollary.F=Fcomb} For nonnegative integers $a,b,c$ and $a+1 \geqslant b$, $a+1,b+1 \geqslant c$, we have $F(a,b,c) = F_{\comb}(a,b,c)$ proving Theorem~\ref{theorem.symmetric chains}. \end{coro} \begin{proof} By Lemma~\ref{lemma.two step recurrence} and Corollary~\ref{corollary.Fcomb recursion}, $F(a,b,c)$ and $F_{\comb}(a,b,c)$ satisfy the same two step recursion. Hence the equality $F(a,b,c) = F_{\comb}(a,b,c)$ can be reduced to the equalities $F(a,b,0) = F_{\comb}(a,b,0)$ for $a,b\geqslant 0$ and $F(a,b,-1) = F_{\comb}(a,b,-1)$ for $a,b\geqslant 1$. These are given in Lemma~\ref{lemma.initial condition}. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Proof of Theorem~\ref{theorem.combinatorics}} \label{section.proof comb} By Corollary~\ref{corollary.F=Fcomb}, we have that $F(a,b,c)=F_{\comb}(a,b,c)$. By Proposition~\ref{proposition:H-Q}, there is an area preserving bijection between quasiheads and heads. Combined with Corollary~\ref{corollary.cover}, there is also an area preserving bijection with pseudoheads and tails. Furthermore, each $\lambda \subseteq \lambda(a,b,c)$ sits in precisely one chain indexed by a given pseudohead (or head). The proofs of Theorems~\ref{neg} and~\ref{pos} tell us, which chain $\lambda$ sits in depending on the cases spelled out in Section~\ref{section.combinatorics}: \[ \renewcommand{\arraystretch}{1.3} \renewcommand{\tabcolsep}{4pt} \begin{tabular}{l|l} Case & Chain membership\\ \hline Case~\ref{case1}\ref{case1_a} & $\lambda \in C(P_{yz})$\\ Case~\ref{case1}\ref{case1_b}\ref{case1_b_i} & $\lambda \in C(P_{(L+z-x)z})$\\ Case~\ref{case1}\ref{case1_b}\ref{case1_b_ii} & $\lambda \in C(T^{(L-x)(b+c-y)})$\\ Case~\ref{case2} & $\lambda \in C(H_x^y)$ \end{tabular} \] Now if the area range for a given chain is $[r,R]$, then due to the symmetry between $q$ and $t$ in each chain, we have \[ \mathsf{stat}(\lambda) = r + R - \mathsf{area}(\lambda) = r + R -A + x+y+z \] for $\lambda=(x,y,z)$. Using the area ranges for pseudoheads, tails, and heads as given in Table~\ref{table.indexing sets} and the beginning of this section, this yields~\eqref{equation.statistics}. In Case~\ref{case1}\ref{case1_a}, we first obtain $\mathsf{stat}(\lambda)=x+\max(\epsilon_{yz},\delta_{yz})$, which is equal to $x+\max(0,\lceil \frac{y-a}{2} \rceil, y+z-b-c,\lceil \frac{2y+z-L}{2} \rceil)$. In Case~\ref{case1}\ref{case1_b}\ref{case1_b_i}, we first obtain $\mathsf{stat}(\lambda)=-L+2x+y-z+\max(\epsilon_{(L+z-x)z},\delta_{(L+z-x)z})$, but using that $y+z\leqslant b+c$ and $L+z-x\leqslant y$, we obtain $\epsilon_{(L+z-x)z}=0$ and $\delta_{(L+z-x)x}= \lceil \frac{L+z-x-a}{2} \rceil$. Combined with Theorem~\ref{theorem.symmetric chains} this proves Theorem~\ref{theorem.combinatorics}. \longthanks{ The authors would like to thank the American Institute for Mathematics (AIM) for hosting the conference ``Categorified Hecke algebras, link homology, and Hilbert schemes'' in October 1-5, 2018, where this work began. E.~G. would also like to thank Andrei Negu\cb{t} and Karola Meszaros for very useful and inspiring discussions. E.~G. was partially supported by NSF grants DMS--1700814, DMS--1760329, and the Russian Academic Excellence Project 5-100. A.S. was partially supported by NSF grants DMS--1760329 and DMS--1764153.} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \nocite{*} \bibliographystyle{amsplain-ac} \bibliography{ALCO_Gorsky_323} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{document}