p+1$, where $1$ denotes the identity map on a crystal. \end{defi} The action can be expressed more explicitly directly in terms of Lusztig's involution. \begin{prop} Let $w=w_1\dots w_r$ be a word in $C_1\otimes\dots\otimes C_r$, then \[ \cactus_{p,q} w_1\dots w_r = w_1\dots w_{p-1}\lusztig\big(\lusztig (w_q)\lusztig(w_{q-1})\dots\lusztig(w_p)\big) w_{q+1}\dots w_r. \] \end{prop} \begin{proof} This follows by induction on $q-p$ and the fact that $\lusztig$ is an involution. \end{proof} \begin{defi} The \Dfn{promotion} $\pr w$ of $w$ is $\cactus_{1,r}\cactus_{2,r} w$, and the \Dfn{evacuation} $\ev w$ of $w$ is $\cactus_{1,r} w$. For a tableau $\mathcal T$ corresponding to a highest weight word $w$, we use $\pr\mathcal T$ and $\ev\mathcal T$ to denote the tableaux corresponding to $\pr w$ and $\ev w$. \end{defi} It will be convenient to express promotion as a commutor, as follows. \begin{prop} $\cactus_{1,q}\cactus_{2,q}(w)=\sigma_{C_1,C_2\otimes\dots\otimes C_q}(w)$. \end{prop} \begin{proof} This is immediate from the definition of the action of $\cactus_{1,q}$ and from the fact that $\cactus_{2,q}^2 = 1$. \end{proof} \subsection{Promotion and evacuation via local rules} We now follow Lenart's approach~\cite{MR2361854} and realise the action of the cactus group using van Leeuwen's local rules~\cite[Rule~4.1.1]{MR1661263}, which generalise Fomin's~\cite[A~1.2.7]{MR1676282}. From now on we restrict ourselves to minuscule representations. However, let us remark that van Leeuwen also provides a local rule that applies to quasi-minuscule representations, which makes our approach viable for arbitrary representations. For minuscule representations, van Leeuwen's rules involve obtaining the unique dominant representative of a weight. \begin{defi}\label{def:dominant-weight} Let $\lambda$ be a weight of a representation of a Lie group with Weyl group $W$. Then \Dfn{$\dom_W(\lambda)$} is the unique dominant representative of the $W$-orbit $W\lambda$. \end{defi} \begin{exam} The Weyl group of $\SL(n)$ is the symmetric group $\fS_n$. Thus, $\dom_{\fS_n}$ returns its argument sorted into decreasing order. \end{exam} \begin{exam} The Weyl group of $\Sp(2n)$ is the hyperoctahedral group $\fH_n$ of signed permutations of $\{\pm 1,\dots,\pm n\}$. Thus, the dominant representative $\dom_{\fH_n}(\lambda)$ of a weight $\lambda$ is obtained by sorting the absolute values of its entries into decreasing order. \end{exam} We can now define the local rule. \begin{defi}\label{def:local-rules} Let $A$ be a crystal and $B$ and $C$ be crystals of minuscule representations. Then the \Dfn{local rule} \[ \tau^A_{B,C}\colon A\otimes B\otimes C \rightarrow A\otimes C\otimes B \] is an isomorphism of crystals defined for highest weight words $a\,b\,c$ as follows: let $\kappa$ be the weight of $a$, let $\lambda$ be the weight of $a\,b$ and let $\nu$ be the weight of $a\,b\,c$. Then \[ \tau^A_{B,C}(a\,b\,c)= a\,\hat c\,\hat b, \] where %, regarding $\kappa$, $\lambda$, $\mu$ and $\nu$ as vectors, \[ \hat c = \mu-\kappa\quad\text{and}\quad \hat b=\nu-\mu\quad\text{with}\quad\mu = \dom_W(\kappa+\nu-\lambda). \] We represent this by the following diagram: \begin{equation}\label{eq:local} \begin{tikzpicture}[scale=1,baseline=(current bounding box.center)] \node (ka) at (0,0) {$\kappa$}; \node (la) at (0,1) {$\lambda$}; \node (nu) at (1,1) {$\nu$}; \node (mu) at (1,0) {$\mu$}; \draw[->] (ka) -- (la) node [midway, left] {$b$}; \draw[->] (mu) -- (nu) node [midway, right] {$\hat b$}; \draw[->] (la) -- (nu) node [midway, above] {$c$}; \draw[->] (ka) -- (mu) node [midway, below] {$\hat c$}; \end{tikzpicture} \end{equation} Since any isomorphism between crystals is determined by specifying a bijection between the corresponding highest weight words, this definition can be extended to an isomorphism between $A\otimes B\otimes C$ and $A\otimes C\otimes B$ by applying the lowering operators. \end{defi} \begin{rema} \looseness-1 When $B=C$ is the crystal associated with the vector representation of $\SL(n)$, the local rule~\eqref{eq:local} is Fomin's for Schützenberger's jeu de taquin~\cite[A~1.2.7]{MR1676282}. More explicitly, if $\lambda$ is the only partition of its size that contains $\kappa$ and is contained in $\nu$, then $\mu=\lambda$. Otherwise there is a unique such partition different from $\lambda$, and this is $\mu$. \end{rema} \begin{rema}\label{rmk:symmetry-of-local-rule} As in the classical case the local rule is symmetric in the sense that $\mu = \dom_W(\kappa+\nu-\lambda)$ if and only if $\lambda = \dom_W(\kappa+\nu-\mu)$, see ~\cite[Lem.~4.1.2]{MR1661263}. \end{rema} \begin{exam} Let $C$ be the crystal associated with the vector representation of $\SL(3)$, and let $A$ and $B$ be the crystal associated with its exterior square. Let $a=e_1+e_2$, let $b=e_1+e_3$ and let $c=e_2$. Then $a\,b\,c$ is a highest weight word in $A\otimes B\otimes C$. We have $\kappa=(1,1,0)$, $\lambda=(2,1,1)$ and $\nu=(2,2,1)$. Thus, since $\dom_{\fS_n}$ sorts its argument into decreasing order, \[ \mu = \dom_{\fS_n}(\kappa + \nu - \lambda) = \dom_{\fS_n}(1,2,0) = (2,1,0),% \quad\text{and}\quad \hat b=e_2+e_3,\quad \hat c=e_1. \] \end{exam} \begin{exam} Let $B$ and $C$ be the crystal associated with the vector representation of $\Sp(4)$ and let $A=\otimes^2 B$. Let $a=e_1\,e_1$, let $b=e_2$ and let $c=-e_2$. Then $a\,b\,c$ is a highest weight word in $A\otimes B\otimes C$. We have $\kappa=(2,0)$, $\lambda=(2,1)$ and $\nu=(2,0)$. Thus, since $\dom_{\fH_n}$ takes absolute values and then sorts into decreasing order, \[ \mu = \dom_{\fH_n}(\kappa + \nu - \lambda) = \dom_{\fH_n}(2,-1) = (2,1),% \quad\text{and}\quad \hat b=b,\quad \hat c=c. \] \end{exam} \begin{theo}[\protect{\cite[Thm.~4.4]{MR2361854}}]\label{thm:local_commutor} Let $A$ and $B$ be crystals embedded into tensor products $A_1\otimes\dots\otimes A_k$ and $B_1\otimes\dots\otimes B_\ell$ of crystals of minuscule representations. Let $w=w_1,\dots,w_{k+\ell}$ be a highest weight word in $A\otimes B$ with corresponding tableau $\emptyset=\wgt^0,\wgt^1,\dots,\wgt^r=\wgt$. Then $\sigma_{A,B}(w)$ can be computed as follows. Create a $k\times \ell$ grid of squares as in~\eqref{eq:local}, labelling the edges along the left border with $w_1,\dots,w_k$ and the edges along the top border with $w_{k+1},\dots,w_{k+\ell}$: \begin{equation}\label{eq:local_commutor} \tikzset{ short/.style={ shorten >=7pt, shorten <=7pt } } \begin{tikzpicture}[baseline=(current bounding box.center), scale=1.2] \node at (0,0) {$\wgt^0$}; \node at (0,1) {$\wgt^1$}; \node at (0,2) {$\wgt^{k-1}$}; \node at (0,3) {$\wgt^k$}; \node[right] at (4,3) {{\hbox to 5pt{\hss $\wgt^{k+\ell}$\hss}}}; % vertical \draw[->,short] (0,0) -- (0,1) node [midway, left] {$w_1$}; \draw[dotted,short] (0,1) -- (0,2) node {}; \draw[->,short] (0,2) -- (0,3) node [midway, left] {$w_k$}; % \draw[->,short] (1,0) -- (1,1) node {}; \draw[dotted,short] (1,1) -- (1,2) node {}; \draw[->,short] (1,2) -- (1,3) node {}; % \draw[->,short] (3,0) -- (3,1) node {}; \draw[dotted,short] (3,1) -- (3,2) node {}; \draw[->,short] (3,2) -- (3,3) node {}; % \draw[->,short] (4,0) -- (4,1) node [midway, right] {$\hat w_{1+\ell}$}; \draw[dotted,short] (4,1) -- (4,2) node {}; \draw[->,short] (4,2) -- (4,3) node [midway, right] {$\hat w_{k+\ell}$}; % horizontal \draw[->,short] (0,3) -- (1,3) node [midway, above] {$w_{k+1}$}; \draw[dotted,short] (1,3) -- (3,3) node {}; \draw[->,short] (3,3) -- (4,3) node [midway, above] {$w_{k+\ell}$}; % \draw[->,short] (0,2) -- (1,2) node {}; \draw[dotted,short] (1,2) -- (3,2) node {}; \draw[->,short] (3,2) -- (4,2) node {}; % \draw[->,short] (0,1) -- (1,1) node {}; \draw[dotted,short] (1,1) -- (3,1) node {}; \draw[->,short] (3,1) -- (4,1) node {}; % \draw[->,short] (0,0) -- (1,0) node [midway, below] {$\hat w_1$}; \draw[dotted,short] (1,0) -- (3,0) node {}; \draw[->,short] (3,0) -- (4,0) node [midway, below] {$\hat w_\ell$}; \end{tikzpicture} \end{equation} For each square whose left and top edges are already labelled use the local rule to compute the labels on the square's bottom and right edges. The labels $\hat w_1\dots \hat w_{k+\ell}$ of the edges along the bottom and the right border of the grid then form $\sigma_{A,B}(w)$. \end{theo} % By Lemma~\ref{lem:extend-s_1q} this determines $\cactus_{p,q} w$ for arbitrary $p$ and $q$. \begin{exam} Let $C_1 = C_2 = C_4$ be the crystal associated with the exterior square of $\SL(3)$ and let $C_3$ the crystal corresponding to its vector representation. Then $w = e_1\text{+}e_2 \; e_1\text{+}e_3\; e_2\; e_1\text{+}e_3$ is the highest weight word corresponding to the semistandard tableau \[ \young(112,234,4). \] The promotion $\pr w = \sigma_{C_1,C_2\otimes C_3\otimes C_4}(w)$ can now be computed using Theorem~\ref{thm:local_commutor}. We put the zero weight in the bottom-left corner. Then, beginning with $\wgt^1=w_1$, we write down the sequence of cumulative weights $\wgt^1,\dots,\wgt^4$, with $\wgt^q = \sum_{i=1}^q w_i$ in the top row. Finally, we successively apply the local rule~\eqref{eq:local} and fill in the weights in the bottom row. \begin{equation*} \tikzset{ short/.style={ shorten >=10pt, shorten <=10pt } } \begin{tikzpicture}[scale=1.5] \node at (0,0) {$000$}; \node at (0,1) {$110$}; \node at (1,0) {$110$}; \node at (1,1) {$211$}; \node at (2,0) {$210$}; \node at (2,1) {$221$}; \node at (3,0) {$311$}; \node at (3,1) {$322$}; % vertical \draw[->,short] (0,0) -- (0,1) node [midway, left] {$e_1\text{+}e_2$}; \draw[->,short] (1,0) -- (1,1) node {}; \draw[->,short] (2,0) -- (2,1) node {}; \draw[->,short] (3,0) -- (3,1) node {}; % horizontal \draw[->,short] (0,1) -- (1,1) node [midway, above] {$e_1\text{+}e_3$}; \draw[->,short] (1,1) -- (2,1) node [midway, above] {$e_2$}; \draw[->,short] (2,1) -- (3,1) node [midway, above] {$e_1\text{+}e_3$}; \draw[->,short] (0,0) -- (1,0) node {}; \draw[->,short] (1,0) -- (2,0) node {}; \draw[->,short] (2,0) -- (3,0) node {}; \end{tikzpicture} \end{equation*} From now on we omit edges and their labels, because the labels are determined by the weights. By Theorem~\ref{thm:local_commutor} the promotion of $w$ is given by the sequence of weights in the bottom row, together with the weight in the top-right corner. The corresponding tableau is \[ \young(113,244,3). \] As mentioned in the introduction, this definition of promotion coincides with the classical definition of promotion in terms of Bender--Knuth moves on tableaux when the crystals correspond to exterior powers of the vector representation of $\SL(n)$. \end{exam} As an aside we obtain a formulation of the commutor, and therefore also of promotion of highest weight words in crystals of minuscule representations, analogous to the definition in terms of slides in tableaux, as follows. \begin{coro}\label{cor:sliding} Let $A$ and $B$ be crystals embedded into tensor products of crystals of minuscule representations. Let $a\in A$ and $b\in B$ such that $a\,b$ is a highest weight word in $A\otimes B$, and let $\hat b\,\hat a= \sigma_{A,B}(a\,b)$ with $\hat b\in B$ and $\hat a\in A$. Then \begin{itemize} \item[] $\hat b$ is the highest weight vertex in the same component of $B$ as $b$, and \item[] $\hat a$ is a vertex of $A$ such that the weight of $\hat b\,\hat a$ equals the weight of $a\,b$. \end{itemize} In particular, if $A$ is a crystal of a minuscule representation, $\hat a$ is determined uniquely by its weight. \end{coro} \begin{proof} Let $B_\lambda$ be the component of $B$ containing $b$. Because of the naturality of the commutor in $B$ (see property $(C1)$ in ~\cite{MR2361854}), $\sigma_{A,B}(a\,b)$ equals $\sigma_{A,B_\lambda}(a\,b)$. Since $a\,b$ is a highest weight word and the commutor is an isomorphism of crystals, $\sigma_{A,B_\lambda}(a\,b)=\hat b\,\hat a$ is also a highest weight word. It follows that $\hat b$ is of highest weight, and therefore equals the highest weight vertex of $B_\lambda$. \end{proof} \subsection{Promotion and evacuation of \texorpdfstring{$\text{\upshape GL}(n)$}{GL(n)}-alternating tableaux} Let us now make promotion and evacuation of $\GL(n)$-alternating tableaux explicit. Recall from example~\ref{eg:alternating} that we regard the adjoint representation as the tensor product $V\otimes V^\ast$, where $V$ is the vector representation of $\GL(n)$ and $V^\ast$ is its dual. Both of these are minuscule, so we can apply Theorem~\ref{thm:local_commutor}. Therefore, the rectangular grid to compute $\pr w = \sigma_{C_1,C_2\otimes\dots\otimes C_r}(w)$ has three rows, the bottom row begins with the zero weight, the middle row with $\mu^1$, which is always $e_1$, and the top row contains the remaining cumulative weights $\mu^2,\dots,\mu^{2r}$. However, because we will apply promotion repeatedly, it will be convenient to slightly enlarge this grid, and prepend the two weights $\mu^0$ and $\mu^1$ to the first row. After having successively applied the local rule~\eqref{eq:local} and thus computed the remaining weights in the middle and the bottom row, we append the final element of the second row and the weight of the original word to the third line. We call the resulting diagram the \Dfn{promotion diagram} of an alternating tableau: \begin{equation}\label{eq:pr-adjoint-schema} \def\arraystretch{1.8} \newcolumntype{C}{>{$}p{32pt}<{$}} \newcolumntype{D}{>{$}p{53pt}<{$}} \begin{array}{CCCCCDD} \wgt^0{\scriptstyle=\emptyset}&\wgt^1{=\scriptstyle1}&\wgt^2&\dotfill&\wgt^{2r}\\ &&\hm^1{\scriptstyle=\wgt^1}&\dotfill&\hm^{2r-1}\\ &&\Pm^0{=\scriptstyle\wgt^0}&\dotfill&\Pm^{2r-2}&\Pm^{2r-1}{=\scriptstyle\hm^{2r-1}}&\Pm^{2r}{=\scriptstyle\wgt^{2r}} \end{array} \end{equation} To illustrate, let us compute the promotion of the $\GL(3)$ highest weight word $w=(e_1,\text{-}e_3)\,(e_1,\text{-}e_2)\,(e_2,\text{-}e_2)\,(e_2,\text{-}e_1)\,(e_3,\text{-}e_1)$. The first row is the alternating tableau corresponding to $w$. For better readability we write $\bar 1$ in place of $-1$. \begin{equation}\label{eq:pr-adjoint-example} \def\arraystretch{1.5} \begin{array}{ccccccccccccc} 000&100&10\bar1&20\bar1&2\bar1\bar1&\tikzmark{start}20\bar1&2\bar1\bar1&20\bar1&10\bar1&100&000\\ & &100 &200 &20\bar1 &21\bar1 &20\bar1 &21\bar1&11\bar1&110&100\\ & &000 &100 &10\bar1 &11\bar1 &10\bar1\tikzmark{end}&11\bar1&10\bar1&100&10\bar1&100&000 \end{array} \begin{tikzpicture}[remember picture,overlay] \node[draw,line width=0pt,rectangle,minimum size=1.8cm,fit={(pic cs:start) (pic cs:end)},yshift=3pt] {}; \end{tikzpicture} \end{equation} Thus, the promotion of $w$ is $\pr w=(e_1,\text{-}e_3)\,(e_2,\text{-}e_2)\,(e_2,\text{-}e_2)\,(e_3,\text{-}e_3)\,(e_3,\text{-}e_1)$. \looseness-1 The six vectors in the rectangle in diagram~\eqref{eq:pr-adjoint-example} demonstrate that the naive embedding of $\GL(n)$-alternating tableaux into the set of $\GL(n+1)$-alternating tableaux is not compatible with promotion, as already mentioned in Section~\ref{sec:adjoint-results}: padding the vectors of the original word with zeros, and applying the local rule, we obtain the rectangle \[ \def\arraystretch{1.5} \begin{array}{cc} 200\bar1&20\bar1\bar1\\ 210\bar1&21\bar1\bar1\\ 110\bar1&11\bar1\bar1, \end{array} \] with bottom-right vector $11\bar1\bar1$, rather than $100\bar1$ as one might expect. \begin{figure}[h!] \[ \def\arraystretch{1.75} \setlength{\arraycolsep}{3pt} \begin{array}{r*{14}{c}} \alt=0 0 0&1 0 0&1 0\bar1&2 0\bar1&2 0\bar2&2 0\bar1&2\bar1\bar1&3\bar1\bar1&2\bar1\bar1\tikzmark{min1}&2 0\bar1&2 0\bar2&3 0\bar2&3 0\bar3&3 1\bar3&2 1\bar3\\ & &1 0 0&2 0 0&2 0\bar1&2 0 0&2 0\bar1&3 0\bar1&2 0\bar1&2 1\bar1&2 1\bar2&3 1\bar2&3 1\bar3&3 2\bar3&2 2\bar3\\ & &0 0 0&1 0 0&1 0\bar1\tikzmark{min2}&1 0 0&1 0\bar1&2 0\bar1&1 0\bar1&1 1\bar1&1 1\bar2&2 1\bar2&2 1\bar3&2 2\bar3&2 1\bar3\\ & & & &1 0 0&1 1 0&1 1\bar1&2 1\bar1&1 1\bar1&2 1\bar1&2 1\bar2&3 1\bar2&3 1\bar3&3 2\bar3&3 1\bar3\\ & & & &0 0 0&1 0 0&1 0\bar1&2 0\bar1&1 0\bar1&2 0\bar1&2 0\bar2&3 0\bar2&3 0\bar3&3 1\bar3&3 0\bar3\\ & & & & & &1 0 0&2 0 0&1 0 0&2 0 0&2 0\bar1&3 0\bar1&3 0\bar2&3 1\bar2&3 0\bar2\\ & & & & & &0 0 0&1 0 0\tikzmark{fixd}&0 0 0&1 0 0&1 0\bar1&2 0\bar1&2 0\bar2&2 1\bar2&2 0\bar2\\ & & & & & & & &1 0 0&2 0 0&2 0\bar1&3 0\bar1&3 0\bar2&3 1\bar2&3 0\bar2\\ & & & & & & & &0 0 0&1 0 0&1 0\bar1&2 0\bar1&2 0\bar2&2 1\bar2&2 0\bar2\\ & & & & & & & & & &1 0 0&2 0 0&2 0\bar1&2 1\bar1&2 0\bar1\\ & & & & & & & & & &0 0 0&1 0 0&1 0\bar1&1 1\bar1&1 0\bar1\\ & & & & & & & & & & & &1 0 0&1 1 0\tikzmark{plus}&1 0 0\\ & & & & & & & & & & & &0 0 0&1 0 0&1 0\bar1\\ & & & & & & & & & & & & & &1 0 0\\ & & & & & & & & & & & & & &0 0 0\\ &&&&&&&&&&&&&&\rotatebox{90}{$\ev\alt=$} \end{array} \] \begin{tikzpicture}[remember picture,overlay] \node [xshift=3pt,yshift=-7pt] at (pic cs:min1) {\negcross}; \node [xshift=3pt,yshift=-7pt] at (pic cs:min2) {\negcross}; \node [xshift=3pt,yshift=-7pt] at (pic cs:plus) {\poscross}; \node [xshift=3pt,yshift=-7pt] at (pic cs:fixd) {\fixcross}; \end{tikzpicture} \caption{The evacuation of an alternating tableau.} \label{fig:ev-adjoint-example} \end{figure} To obtain the evacuation of an alternating tableau we use the second identity of Lemma~\ref{lem:extend-s_1q}. We start by computing the promotion of the initial alternating tableau as above, except that we do not append anything to the third row. We then repeat this process a total of $r$ times, creating a (roughly) triangular array of weights, which we call the \Dfn{evacuation diagram} of an alternating tableau. The sequence of cumulative weights of the evacuation can then be read off the vertical row on the right hand side, from bottom to top. An example can be found in Figure~\ref{fig:ev-adjoint-example}. The symbols \poscross, \negcross\ and \fixcross\ occurring in the figure should be ignored for the moment. Finally, we would like to point out that for alternating tableaux of empty shape there is a second way to compute the promotion, exploiting the fact that the next-to-last weight is forced to be $10\dots0$. Let $w=w_1\dots w_r$ be the highest weight word corresponding to the alternating tableau. We consider $w$ as an element of $A\otimes A^\ast\otimes B_\lambda$, where $A$ is the crystal corresponding to $V$, $A^\ast$ is the crystal corresponding to $V^\ast$ and, similar to what was done in the proof of Corollary~\ref{cor:sliding}, $B_\lambda$ is the component of $\bigotimes^{r-1}(A\otimes A^\ast)$ containing $w_3\dots w_r$. Then we first compute $\hat w=\sigma_{A, A^\ast\otimes B_\lambda}(w)$, followed by computing $\hat{\hat w}=\sigma_{A^\ast,B_\lambda\otimes A}(\hat w)$: \begin{equation}\label{eq:pr-stephan-adjoint-example} \def\arraystretch{1.5} \begin{array}{ccccccccccccc} 000&100&10\bar1&20\bar1&2\bar1\bar1&20\bar1&2\bar1\bar1&20\bar1&10\bar1&100&000\\ &000&00\bar1&10\bar1&1\bar1\bar1&10\bar1&1\bar1\bar1&10\bar1&00\bar1&000&00\bar1&000\\ & &000 &100 &10\bar1 &11\bar1 &10\bar1&11\bar1&10\bar1&100&10\bar1&100&000 \end{array} \end{equation} Because the initial segment of $\hat{\hat w}$ is an element of $B_\lambda$ and is of highest weight, it must coincide with the initial segment of the promotion of $w$. This variant of the local rules for promotion was recently rediscovered, in slightly different form, by Patrias~\cite{MR3861768}. Note, however, that for an alternating tableau of non-empty shape, this procedure yields a tableau which, in general, is different from the result of promotion. \section{Growth diagram bijections} \label{sec:growth-diagrams} In this section we recall Sundaram's bijection (using Roby's description~\cite{MR2716353} based on Fomin's growth diagrams~\cite{MR1314558}) between oscillating tableaux and matchings. We also present a new bijection, in the same spirit, between alternating tableaux and partial permutations. In both cases, the action of the cactus group on highest weight words becomes particularly transparent when using Fomin's growth diagrams and local rules for the Robinson--Schensted correspondence. \begin{figure}[h] \def\arraystretch{1.5} \begin{tabular}{rccc} & \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)},scale=2] \node (ka) at (0,1) {$\kappa$}; \node (la) at (0,0) {$\lambda$}; \node (nu) at (1,0) {$\nu$}; \node (mu) at (1,1) {$\mu$}; \draw[<-] (ka) -- (la); \draw[<-] (mu) -- (nu); \draw[->] (la) -- (nu); \draw[->] (ka) -- (mu); \node at (0.5,0.5)[rectangle,minimum size=1.5cm,draw] {}; \end{tikzpicture} &or & \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)},scale=2] \node (ka) at (0,1) {$\lambda$}; \node (la) at (0,0) {$\lambda$}; \node (nu) at (1,0) {$\lambda$}; \node (mu) at (1,1) {$\mu$}; \draw[<-] (ka) -- (la); \draw[<-] (mu) -- (nu); \draw[->] (la) -- (nu); \draw[->] (ka) -- (mu); \node at (0.5,0.5)[rectangle,minimum size=1.5cm,draw] {\huge$\times$}; \end{tikzpicture}\\ forward rules: % & $\mu'=\domS(\kappa'+\nu'-\lambda')$ & % & $\mu = \lambda + e_1$\\ backward rules: % & $\lambda'=\domS(\kappa'+\nu'-\mu')$ & % & $\lambda = \mu - e_1$ % \end{tabular} \caption{Cells of a growth diagram and corresponding local rules.} \label{fig:growth-cell} \end{figure} For our purposes, a growth diagram is a finite collection of cells, arranged in the form of a Ferrers diagram using the French convention, as for example in Figures~\ref{fig:s1q-partial-matchings} and ~\ref{fig:s1q-partial-permutations}. Let us first describe the classical setup, which we use to describe Sundaram's correspondence. In this case, each cell is either empty or contains a cross. Moreover, we require that in every row and every column of the growth diagram there is at most one cell which contains a cross. Every corner of a cell is labelled with a partition such that the local rules in Figure~\ref{fig:growth-cell} are satisfied, where $\lambda^\prime$ denotes the partition conjugate to~$\lambda$ and~$e_1$ is the first unit vector. Moreover, we require that two adjacent partitions (as for example~$\lambda$ and~$\kappa$ in Figure~\ref{fig:growth-cell}) either coincide or the one at the head of the arrow is obtained from the other by adding a unit vector. Furthermore, we require that the partitions labelling the corners of a cell satisfy the forward and backward rules of Figure~\ref{fig:growth-cell}. In fact, the two \Dfn{forward rules} determine~$\mu$ given the other three partitions and the content of the cell. The two \Dfn{backward rules} determine the content of the cell and the partition~$\lambda$ in the bottom-left given the other three partition. Thus, the information in a growth diagram is redundant. In particular, given the partitions labelling the corners along the bottom-left border and the contents of the cells, one can recover the remaining partitions. Conversely, given the partitions labelling the corners along the top-right border of a diagram, one can recover the remaining partitions and the contents of the cells. The presentation of the local rules in Figure~\ref{fig:growth-cell} is slightly non-standard. It has the benefit that the local rule for empty cells is very similar to the special case of Definition~\ref{def:local-rules} corresponding to Cartan type~$A_n$, with two important differences. The first difference is that all partitions are transposed, the second, that the orientation of the vertical arrows is reversed. \subsection{Roby's description of Sundaram's correspondence} \label{sec:Roby} \begin{figure}[htb]\centering \begin{tikzpicture}[scale=0.75]\tiny % a dotted line to indicate the reflection \draw[dotted] (1,1) -- (5,5); % \foreach \y in {1,...,10} \draw (1,11-\y)--(\y,11-\y); \foreach \x in {1,2,...,10} \draw (\x,11-\x)--(\x,1); % column labels \foreach \x in {1,2,...,9} \draw (\x.5,11-10.3) node{\footnotesize $\x$}; % row labels \foreach \x in {1,2,...,9} \draw (0.3,11-\x.4) node{\footnotesize $\x$}; % crosses \draw (1.5,11-4.5) node{\huge$\times$}; \draw (2.5,11-9.5) node{\huge$\times$}; \draw (3.5,11-6.5) node{\huge$\times$}; % partitions column 0 (from bottom to top) \draw (.8,11-.8) node{\footnotesize $\emptyset$}; \foreach \x in {1,2,...,8} \draw (.8,11-\x.8) node{$\emptyset$}; \draw (0.8,11-9.8) node{\footnotesize $\emptyset$}; % partitions column 1 \draw (1.8,11-1.8) node{\footnotesize $1$}; \draw (1.8,11-2.8) node{$1$}; \draw (1.8,11-3.8) node{$1$}; \foreach \x in {4,5,...,8} \draw (1.8,11-\x.8) node{$\emptyset$}; \draw (1.8,11-9.8) node{\footnotesize $\emptyset$}; % partitions column 2 \draw (2.8,11-2.8) node{\footnotesize $11$}; \draw (2.8,11-3.8) node{$11$}; \foreach \x in {4,5,...,8} \draw (2.8,11-\x.8) node{$1$}; \draw (2.8,11-9.8) node{\footnotesize $\emptyset$}; % partitions column 3 \draw (3.8,11-3.8) node{\footnotesize $21$}; \draw (3.8,11-4.8) node{$2$}; \draw (3.8,11-5.8) node{$2$}; \draw (3.8,11-6.8) node{$1$}; \draw (3.8,11-7.8) node{$1$}; \draw (3.8,11-8.8) node{$1$}; \draw (3.8,11-9.8) node{\footnotesize $\emptyset$}; % partitions column 4 \draw (4.8,11-4.8) node{\footnotesize $2$}; \draw (4.8,11-5.8) node{$2$}; \draw (4.8,11-6.8) node{$1$}; \draw (4.8,11-7.8) node{$1$}; \draw (4.8,11-8.8) node{$1$}; \draw (4.8,11-9.8) node{\footnotesize $\emptyset$}; % partitions column 5 \draw (5.8,11-5.8) node{\footnotesize $21$}; \draw (5.8,11-6.8) node{$11$}; \draw (5.8,11-7.8) node{$11$}; \draw (5.8,11-8.8) node{$11$}; \draw (5.8,11-9.8) node{\footnotesize $1$}; % partitions column 6 \draw (6.8,11-6.8) node{\footnotesize $11$}; \draw (6.8,11-7.8) node{$11$}; \draw (6.8,11-8.8) node{$11$}; \draw (6.8,11-9.8) node{\footnotesize $1$}; % partitions column 7 \draw (7.8,11-7.8) node{\footnotesize $21$}; \draw (7.8,11-8.8) node{$21$}; \draw (7.8,11-9.8) node{\footnotesize $2$}; % partitions column 8 \draw (8.8,11-8.8) node{\footnotesize $211$}; \draw (8.8,11-9.8) node{\footnotesize $21$}; % partitions column 9 \draw (9.8,11-9.8) node{\footnotesize $21$}; \end{tikzpicture} \begin{tikzpicture}[scale=0.75]\tiny % a dotted line to indicate the reflection \draw[dotted] (1,1) -- (5,5); % \foreach \y in {1,...,10} \draw (1,11-\y)--(\y,11-\y); \foreach \x in {1,2,...,10} \draw (\x,11-\x)--(\x,1); % column labels \foreach \x in {1,2,...,9} \draw (\x.5,11-10.3) node{\footnotesize $\x$}; % row labels \foreach \x in {1,2,...,9} \draw (0.3,11-\x.4) node{\footnotesize $\x$}; % crosses \draw (1.5,11-8.5) node{\huge $\times$}; \draw (4.5,11-7.5) node{\huge $\times$}; \draw (6.5,11-9.5) node{\huge $\times$}; % partitions column 0 (from bottom to top) \draw (.8,11-.8) node{\footnotesize $\emptyset$}; \foreach \x in {1,2,...,8} \draw (.8,11-\x.8) node{$\emptyset$}; \draw (0.8,11-9.8) node{\footnotesize $\emptyset$}; % partitions column 1 \draw (1.8,11-1.8) node{\footnotesize $1$}; \foreach \x in {2,3,...,7} \draw (1.8,11-\x.8) node{$1$}; \draw (1.8,11-8.8) node{$\emptyset$}; \draw (1.8,11-9.8) node{\footnotesize $\emptyset$}; % partitions column 2 \draw (2.8,11-2.8) node{\footnotesize $11$}; \foreach \x in {3,4,...,7} \draw (2.8,11-\x.8) node{$11$}; \draw (2.8,11-8.8) node{$1$}; \draw (2.8,11-9.8) node{\footnotesize $1$}; % partitions column 3 \draw (3.8,11-3.8) node{\footnotesize $111$}; \foreach \x in {4,5,6,7} \draw (3.72,11-\x.8) node{$111$}; \draw (3.8,11-8.8) node{$11$}; \draw (3.8,11-9.8) node{\footnotesize $11$}; % partitions column 4 \draw (4.8,11-4.8) node{\footnotesize $211$}; \foreach \x in {5,6} \draw (4.72,11-\x.8) node{$211$}; \draw (4.72,11-7.8) node{$111$}; \draw (4.8,11-8.8) node{$11$}; \draw (4.8,11-9.8) node{\footnotesize $11$}; % partitions column 5 \draw (5.8,11-5.8) node{\footnotesize $221$}; \draw (5.72,11-6.8) node{$221$}; \draw (5.72,11-7.8) node{$211$}; \draw (5.8,11-8.8) node{$21$}; \draw (5.8,11-9.8) node{\footnotesize $21$}; % partitions column 6 \draw (6.8,11-6.8) node{\footnotesize $321$}; \draw (6.72,11-7.8) node{$311$}; \draw (6.8,11-8.8) node{$31$}; \draw (6.8,11-9.8) node{\footnotesize $21$}; % partitions column 7 \draw (7.8,11-7.8) node{\footnotesize $311$}; \draw (7.8,11-8.8) node{$31$}; \draw (7.8,11-9.8) node{\footnotesize $21$}; % partitions column 8 \draw (8.8,11-8.8) node{\footnotesize $31$}; \draw (8.8,11-9.8) node{\footnotesize $21$}; % partitions column 9 \draw (9.8,11-9.8) node{\footnotesize $21$}; \end{tikzpicture} \caption{A pair of growth diagrams $\Grm(\osc)$ and $\Grm(\cactus_{1,9}\osc)$, with $\osc=(\emptyset,1,11,21,2,21,11,21,211,21)$, illustrating Theorem~\ref{thm:s1q-partial-matchings}. The dotted line indicates the axis of reflection for the matchings $\M(\osc)$ and $\M(\cactus_{1,9}\osc)$.} \label{fig:s1q-partial-matchings} \end{figure} In this section we recall the bijection between oscillating tableaux of length $r$ and shape $\mu$ and pairs consisting of a partial matching of $\{1,\dots,r\}$ and a partial standard Young tableau of shape $\mu$ whose entries are the unmatched elements. \begin{defi} Let $\osc=(\wgt_0,\wgt_1,\ldots,\wgt_r)$ be an oscillating tableau. The associated (triangular) growth diagram $\Grm(\osc)$ consists of $r$ left-justified rows, with $i-1$ cells in row $i$ for $i\in\{1,\dots,r\}$, where row $1$ is the top row. Label the cells according to the following specification: \begin{enumerate}[{R}\arabic{enumi}]\setlength{\itemsep}{0pt} \item %[R1] Label the north east corners of the cells on the main diagonal from the top-left to the bottom-right with the partitions in $\osc$. \item %[R2] Label the corners of the first subdiagonal with the smaller of the two partitions labelling the two adjacent corners on the diagonal. \item %[R3] Use the backward rules to determine which cells contain a cross. \end{enumerate} Let $\M(\osc)$ be the matching containing a pair $\{i, j\}$ for every cross in column~$i$ and row~$j$ of the $\Grm(\osc)$. Furthermore, let $\M_T(\osc)$ be the partial standard Young tableau corresponding to the sequence of partitions along the bottom border of $\Grm(\osc)$. \end{defi} \begin{theo}[\protect{Sundaram~\cite[Sec.~8]{MR2941115}, Roby~\cite[Prop.~4.3.1]{MR2716353}}] The map $\osc\mapsto\big(\M(\osc), \M_T(\osc)\big)$ is a bijection between oscillating tableaux of length $r$ and shape $\mu$, and pairs consisting of a perfect matching of a subset of $\{1,\dots,r\}$ and a partial standard Young tableau of shape $\mu$, whose entries form the complementary subset. Moreover, the map $\osc\mapsto\M(\osc)$ is a bijection between $n$-symplectic oscillating tableaux of length $r$ and empty shape and $(n+1)$-noncrossing perfect matchings of~$\{1,\dots,r\}$. \end{theo} An example for this procedure, which also illustrates Theorem~\ref{thm:s1q-partial-matchings}, can be found in Figure~\ref{fig:s1q-partial-matchings}. Let $\osc$ be the $3$-symplectic oscillating tableau \[ (\emptyset,1,11,21,2,21,11,21,211,21), \] whose partitions label the corners of the diagonal of the first growth diagram. Applying the backward rules, we obtain the matching and the partial standard Young tableau \Yvcentermath1 \[ \M(\osc)=\big\{\{1,4\},\{2,9\},\{3,6\}\big\}\text{ and } \M_T(\osc)=\young(57,8). \] Using Lemma~\ref{lem:extend-s_1q} and the local rule in Definition~\ref{def:local-rules} one can compute that $\ev\osc$ is the $3$-symplectic oscillating tableau labelling the corners of the diagonal of the second growth diagram. Applying the backward rules again, we obtain the matching and the partial standard Young tableau predicted by Theorem~\ref{thm:s1q-partial-matchings}: \Yvcentermath1 \[ \M(\ev\osc)=\big\{\{1,8\},\{4,7\},\{6,9\}\big\}\text{ and } \M_T(\ev\osc)=\young(25,3). \] \subsection{A new variant for Stembridge's alternating tableaux} \label{sec:Stembridge} In this section we present a variation of Sundaram's bijection for alternating tableaux and permutations. Recall that a staircase is a vector with weakly decreasing integer entries. The \Dfn{positive part} of the staircase is the partition obtained by removing all entries less than or equal to zero. The \Dfn{negative part} of the staircase is the partition obtained by removing all entries greater than or equal to zero, removing the signs of the remaining entries and reversing the sequence. \begin{defi}\label{defn:P} Let $\alt=(\wgt^0,\wgt^1,\ldots,\wgt^{2r})$ be an alternating tableau. The associated growth diagram $\Grm(\alt)$ is an $r\times r$ square of cells, obtained as follows: \begin{itemize}\setlength{\itemsep}{0pt} \item[P1] Label the north east corners of the cells on the main diagonal and the first superdiagonal from the top-left to the bottom-right with the staircases in $\alt$. \item[P2] Apply the backward rules on the positive parts of the staircases to determine which cells below the diagonal contain a cross. \item[P3] Use the backward rules (rotated by $180\degree$) on the negative parts of the staircases to determine which cells above the diagonal contain a cross. \end{itemize} Let $\Perm(\alt)$ be the partial permutation mapping $i$ to $j$ for every cross in column $i$ and row $j$ of $\Grm(\alt)$, and let $\big(\Perm_P(\alt), \Perm_Q(\alt)\big)$ be the pair of partial standard Young tableaux corresponding to the sequence of partitions along the bottom and the right border of $\Grm(\alt)$, respectively. \end{defi} \begin{theo} The map $\alt\mapsto\big(\Perm(\alt), \Perm_P(\alt), \Perm_Q(\alt)\big)$ is a bijection between alternating tableaux of length $r$ and shape $\mu$, and triples consisting of a bijection $\Perm(\alt):R\to S$ between two subsets of $\{1,\dots,r\}$, and two partial standard Young tableaux $\Perm_P(\alt)$ and $\Perm_Q(\alt)$. The shapes of these tableaux are obtained by separating the positive and negative entries of $\mu$. The entries of the first tableau then form the complementary subset of $R$, the entries of the second form the complementary subset of $S$. Moreover, the map $\alt\mapsto\Perm(\alt)$ is a bijection between $\GL(n)$-alternating tableaux of length $r$ and empty shape and permutations of~$\{1,\dots,r\}$ whose longest increasing subsequence has length at most $n$. \end{theo} An example for this procedure, which also illustrates Theorem~\ref{thm:s1q-partial-permutations}, can be found in Figure~\ref{fig:s1q-partial-permutations}. We render fixed points as \fixcross, other crosses below the diagonal as \poscross\ and crosses above the diagonal as \negcross. The reason for doing so is given by Corollary~\ref{cor:relate-fillings} in Section~\ref{sec:alternating}, where we show that the growth diagram of an alternating tableau and its evacuation diagram are very closely related. Let $\alt$ be the $\GL(13)$-alternating tableau of length $7$ \[ (\emptyset, 1, 1\bar1, 2\bar1, 2\bar2, 2\bar1, 2\bar1\bar1, 3\bar1\bar1, 2\bar1\bar1, 2\bar1, 2\bar2, 3\bar2, 3\bar3, 31\bar3, 21\bar3), \] where we write the negative entries with bars and omit zeros for readability. Its staircases label the corners of the diagonal of the first growth diagram. Applying the backward rules we obtain the partial permutation and the partial standard Young tableaux % \Yvcentermath1 \[ \Perm(\alt)=\big\{(3,2),(4,4),(5,1),(6,7)\big\}, \Perm_P(\alt)=\young(356),\text{ and } \Perm_Q(\alt)=\young(12,7). \] The second growth diagram in the figure is obtained by applying the same procedure to $\ev\alt =\cactus_{1,7}\alt$, which yields % \Yvcentermath1 \[ \Perm(\ev\alt)=\big\{(2,1),(3,7),(4,4),(5,6)\big\}, \Perm_P(\ev\alt)=\young(235),\text{ and } \Perm_Q(\ev\alt)=\young(17,6). \] as predicted by Theorem~\ref{thm:s1q-partial-permutations}. In the example above, we could have obtained the same sequence of positive and negative parts of the staircases from a $\GL(3)$-alternating tableau, removing ten zeros from each vector. As it turns out, evacuation of this alternating tableau yields the same result as above, although for $r=7$ Theorem~\ref{thm:s1q-partial-permutations} applies only when $n$ is at least~$13$. The computation of the evacuated alternating tableau is carried out in Figure~\ref{fig:ev-adjoint-example}. The $\GL(2)$-alternating tableau $\alt=(\emptyset, 10, 1\bar1, 10, 1\bar1)$ of length $2$ %$(e_1,\text{-}e_2)\;(e_2,\text{-}e_2)$ illustrates the necessity of the hypothesis restricting the length of the alternating tableau in Theorem~\ref{thm:s1q-partial-permutations}. On the one hand, this tableau is fixed by $\ev=\cactus_{1,2}$. On the other hand, $\Perm(\alt) = \{(2,1)\}$, whose reverse-complement is $\{(1,2)\}$. Similarly, to justify the necessity of the hypothesis in Theorem~\ref{thm:rotation-promotion-permutations}, consider the $\GL(3)$-alternating tableau in the first row of diagram~\eqref{eq:pr-adjoint-example}, which corresponds to the permutation depicted in Figure~\ref{fig:chord-diagrams}. Its promotion, as computed in the last row of diagram~\eqref{eq:pr-adjoint-example}, corresponds to the permutation $23514$, which differs from the rotated permutation. \begin{figure}[htb]\centering \begin{tikzpicture}[scale=1.07]\tiny % thick diagonal \draw[line width=1.5pt] (8,1) -- (8,2) -- (7,2) -- (7,3) -- (6,3) -- (6,4) -- (5,4) -- (5,5) -- (4,5) -- (4,6) -- (3,6) -- (3,7) -- (2,7) -- (2,8) -- (1,8); \foreach \y in {1,...,8} \draw (1,\y)--(8,\y); \foreach \x in {1,2,...,8} \draw (\x,1)--(\x,8); % column labels \foreach \x in {1,...,7} \draw (\x.5,8.3) node{\footnotesize $\x$}; % row labels \foreach \x in {1,...,7} \draw (0.5,8.4-\x) node{\footnotesize $\x$}; % crosses \draw (5.5,7.5) node{\negcross}; \draw (3.5,6.5) node{\negcross}; \draw (4.5,4.5) node{\fixcross}; \draw (6.5,1.5) node{\poscross}; % partitions column 0 (from bottom to top) \draw (.8,.75) node{\footnotesize $\emptyset$}; \foreach \x in {1,...,6} \draw (.8,\x.8) node{$\emptyset$}; \draw (0.8,7.75) node{\footnotesize $\emptyset$}; % partitions column 1 \draw (1.8,0.75) node{\footnotesize $1$}; \foreach \x in {1,...,5} \draw (1.8,\x.8) node{$1$}; \draw (1.8,6.75) node{\footnotesize $1\bar1$}; \draw (1.8,7.75) node{\footnotesize $1$}; % partitions column 2 \draw (2.8,0.75) node{\footnotesize $2$}; \foreach \x in {1,...,4} \draw (2.8,\x.8) node{$2$}; \draw (2.8,5.75) node{\footnotesize $2\bar2$}; \draw (2.8,6.75) node{\footnotesize $2\bar1$}; \draw (2.8,7.8) node{$\bar\emptyset$}; % partitions column 3 \draw (3.8,0.75) node{\footnotesize $2$}; \foreach \x in {1,2,3} \draw (3.8,\x.8) node{$2$}; \draw (3.7,4.75) node{\footnotesize $2\bar1\bar1$}; \draw (3.8,5.75) node{\footnotesize $2\bar1$}; \draw (3.8,6.8) node{$\bar 1$}; \draw (3.8,7.8) node{$\bar\emptyset$}; % partitions column 4 \draw (4.8,0.75) node{\footnotesize $2$}; \draw (4.8,1.8) node{$2$}; \draw (4.8,2.8) node{$2$}; \draw (4.75,3.75) node{\footnotesize $2\bar1\bar1$}; \draw (4.75,4.75) node{\footnotesize $3\bar1\bar1$}; \foreach \x in {5,6} \draw (4.8,\x.8) node{$\bar1$}; \draw (4.8,7.8) node{$\bar\emptyset$}; % partitions column 5 \draw (5.8,0.75) node{\footnotesize $2$}; \draw (5.8,1.8) node{$2$}; \draw (5.8,2.75) node{\footnotesize $2\bar2$}; \draw (5.75,3.75) node{\footnotesize $2\bar1$}; \draw (5.8,4.8) node{$\bar1$}; \foreach \x in {5,6,7} \draw (5.8,\x.8) node{$\bar\emptyset$}; % partitions column 6 \draw (6.7,0.75) node{\footnotesize $2$}; \draw (6.7,1.75) node{\footnotesize $3\bar3$}; \draw (6.75,2.75) node{\footnotesize $3\bar2$}; \draw (6.8,3.8) node{$\bar1$}; \draw (6.8,4.8) node{$\bar1$}; \draw (6.8,5.8) node{$\bar\emptyset$}; \draw (6.8,6.8) node{$\bar\emptyset$}; \draw (6.8,7.8) node{$\bar\emptyset$}; % partitions column 7 \draw (7.7,0.75) node{\footnotesize $21\bar3$}; \draw (7.7,1.75) node{\footnotesize $31\bar3$}; \draw (7.8,2.75) node{\footnotesize $\bar2$}; \draw (7.8,3.75) node{\footnotesize $\bar1$}; \draw (7.8,4.75) node{\footnotesize $\bar1$}; \draw (7.8,5.75) node{\footnotesize $\bar\emptyset$}; \draw (7.8,6.75) node{\footnotesize $\bar\emptyset$}; \draw (7.8,7.75) node{\footnotesize $\bar\emptyset$}; \end{tikzpicture} \vspace*{8mm} \begin{tikzpicture}[scale=1.07]\tiny % thick diagonal \draw[line width=1.5pt] (8,1) -- (8,2) -- (7,2) -- (7,3) -- (6,3) -- (6,4) -- (5,4) -- (5,5) -- (4,5) -- (4,6) -- (3,6) -- (3,7) -- (2,7) -- (2,8) -- (1,8); \foreach \y in {1,...,8} \draw (1,\y)--(8,\y); \foreach \x in {1,2,...,8} \draw (\x,1)--(\x,8); % column labels \foreach \x in {1,...,7} \draw (\x.5,8.3) node{\footnotesize $\x$}; % row labels \foreach \x in {1,...,7} \draw (0.5,8.4-\x) node{\footnotesize $\x$}; % crosses \draw (9-5.5,9-7.5) node{\poscross}; \draw (9-3.5,9-6.5) node{\poscross}; \draw (9-4.5,9-4.5) node{\fixcross}; \draw (9-6.5,9-1.5) node{\negcross}; % partitions column 0 (from bottom to top) \draw (.8,.75) node{\footnotesize $\emptyset$}; \foreach \x in {1,...,6} \draw (.8,\x.8) node{$\emptyset$}; \draw (0.8,7.75) node{\footnotesize $\emptyset$}; % partitions column 1 \draw (1.8,0.75) node{\footnotesize $1$}; \foreach \x in {1,...,5} \draw (1.8,\x.8) node{$1$}; \draw (1.8,6.75) node{\footnotesize $1\bar1$}; \draw (1.8,7.75) node{\footnotesize $1$}; % partitions column 2 \draw (2.8,0.75) node{\footnotesize $1$}; \foreach \x in {1,...,4} \draw (2.8,\x.8) node{$1$}; \draw (2.8,5.75) node{\footnotesize $1\bar1$}; \draw (2.8,6.75) node{\footnotesize $1$}; \draw (2.8,7.8) node{$\bar\emptyset$}; % partitions column 3 \draw (3.8,0.75) node{\footnotesize $1$}; \foreach \x in {1,2,3} \draw (3.8,\x.8) node{$2$}; \draw (3.7,4.75) node{\footnotesize $2\bar2$}; \draw (3.8,5.75) node{\footnotesize $2\bar1$}; \draw (3.8,6.8) node{$\bar\emptyset$}; \draw (3.8,7.8) node{$\bar\emptyset$}; % partitions column 4 \draw (4.8,0.75) node{\footnotesize $1$}; \draw (4.8,1.8) node{$2$}; \draw (4.8,2.8) node{$2$}; \draw (4.8,3.75) node{\footnotesize $2\bar2$}; \draw (4.75,4.75) node{\footnotesize $3\bar2$}; \draw (4.75,5.8) node{$\bar1$}; \foreach \x in {6,7} \draw (4.8,\x.8) node{$\bar\emptyset$}; % partitions column 5 \draw (5.8,0.75) node{\footnotesize $1$}; \draw (5.8,1.8) node{$2$}; \draw (5.8,2.75) node{\footnotesize $3\bar3$}; \draw (5.75,3.75) node{\footnotesize $3\bar2$}; \draw (5.8,4.8) node{$\bar2$}; \draw (5.8,5.8) node{$\bar1$}; \foreach \x in {6,7} \draw (5.8,\x.8) node{$\bar\emptyset$}; % partitions column 6 \draw (6.7,0.75) node{\footnotesize $11$}; \draw (6.7,1.75) node{\footnotesize $21\bar3$}; \draw (6.7,2.75) node{\footnotesize $31\bar3$}; \draw (6.8,3.8) node{$\bar2$}; \draw (6.8,4.8) node{$\bar2$}; \draw (6.8,5.8) node{$\bar1$}; \draw (6.8,6.8) node{$\bar\emptyset$}; \draw (6.8,7.8) node{$\bar\emptyset$}; % partitions column 7 \draw (7.7,0.75) node{\footnotesize $21\bar3$}; \draw (7.7,1.75) node{\footnotesize $22\bar3$}; \draw (7.8,2.75) node{\footnotesize $\bar3$}; \draw (7.8,3.75) node{\footnotesize $\bar2$}; \draw (7.8,4.75) node{\footnotesize $\bar2$}; \draw (7.8,5.75) node{\footnotesize $\bar1$}; \draw (7.8,6.75) node{\footnotesize $\bar\emptyset$}; \draw (7.8,7.75) node{\footnotesize $\bar\emptyset$}; \end{tikzpicture} \caption{A pair of growth diagrams $\Grm(\alt)$ and $\Grm(\cactus_{1,7}\alt)$, with $\alt=(\emptyset, 1, 1\bar1, 2\bar1, 2\bar2, 2\bar1, 2\bar1\bar1, 3\bar1\bar1, 2\bar1\bar1, 2\bar1, 2\bar2, 3\bar2, 3\bar3, 31\bar3, 21\bar3)$, illustrating Theorem~\ref{thm:s1q-partial-permutations}.} \label{fig:s1q-partial-permutations} \end{figure} \section{Proofs}\label{sec:proofs} Our strategy is as follows. We first consider only $\GL(n)$-alternating tableaux of empty shape and length $r$ with $n\geq r$, and show that the bijection $\Perm$ presented in Section~\ref{sec:Stembridge} intertwines rotation and promotion. To do so, we demonstrate that the middle row of the promotion diagram~\eqref{eq:pr-adjoint-schema} of an alternating tableau $\alt$ can be interpreted as corresponding to a single-step rotation of the rows of the growth diagram $\Grm(\alt)$. Then, using a very similar argument, we find that the promotion of $\alt$ corresponds to a single-step rotation of the columns of the growth diagram just obtained. To prove the statements concerning evacuation, we show that the permutation $\Perm(\alt)$ can actually be read off directly from the evacuation diagram. In particular, this makes the effect of evacuation on $\Perm(\alt)$ completely transparent. The effect of the evacuation of an arbitrary alternating tableau $\alt$ on the triple $\big(\Perm(\alt),\Perm_P(\alt),\Perm_Q(\alt)\big)$ is deduced from the special case of alternating tableaux of empty shape by extending $\alt$ to an alternating tableau of empty shape. In order to determine the exact range of validity of Theorem~\ref{thm:rotation-promotion-permutations} we use a stability phenomenon proved in Section~\ref{sec:stability}. The case $n=2$ is treated completely separately in Section~\ref{sec:GL2}. Finally, in Section~\ref{sec:oscillating}, we deduce the statements for oscillating tableaux and the vector representation of the symplectic groups, Theorem~\ref{thm:s1q-partial-matchings} and~\ref{thm:rotation-promotion-matchings}, from the statements for alternating tableaux. \subsection{Stability}\label{sec:stability} In this section we prove a stability phenomenon needed for establishing the exact bounds in Theorem~\ref{thm:rotation-promotion-permutations}. Given the lack of an embedding of $\GL(n)$-alternating tableaux in the set of $\GL(n+1)$-alternating tableaux that is compatible with promotion, this theorem may be interesting in its own right. \begin{theo}\label{thm:stability} Let $\alt$ be a $\GL(n)$-alternating tableau, not necessarily of empty shape, and suppose that each staircase in $\alt$ and $\pr\alt$ contains at most $m$ nonzero parts. Then $\pr\tilde\alt=\widetilde{\pr\alt}$, where $\tilde\alt$ and $\widetilde{\pr\alt}$ are the $\GL(m)$-alternating tableaux obtained from $\alt$ and $\pr\alt$ by removing $n-m$ zeros from each staircase. \end{theo} Before proceeding to the proof, let us remark that this is not a trivial statement: it may well be that some staircases in the intermediate row $\halfpr\alt$ have more than $m$ nonzero parts. \begin{proof} It suffices to consider the case $m=n-1$. We show inductively that the statement is true for every square of staircases in diagram~\eqref{eq:pr-adjoint-schema} \begin{equation}\label{eq:square-of-staircases} \tikzset{ short/.style={ shorten >=7pt, shorten <=7pt } } \begin{tikzpicture}[xscale=2.5,yscale=1.5,baseline=(current bounding box.center)] \node (ka) at (0,0) {$\kappa=\Pm^{2i-4}$}; \node (be) at (0,1) {$\beta=\hm^{2i-3}$}; \node (la) at (0,2) {$\lambda=\wgt^{2i-2}$}; % \node (de) at (1,0) {$\delta=\Pm^{2i-3}$}; \node (ep) at (1,1) {$\varepsilon=\hm^{2i-2}$}; \node (al) at (1,2) {$\alpha=\wgt^{2i-1}$}; % \node (mu) at (2,0) {$\mu=\Pm^{2i-2}$}; \node (ga) at (2,1) {$\gamma=\hm^{2i-1}$}; \node (nu) at (2,2) {$\nu=\wgt^{2i}$}; % vertical \draw[->] (ka) -- (be) node [midway, left] {$+$}; \draw[->] (be) -- (la) node [midway, left] {$-$}; % \draw[->] (de) -- (ep) node [midway, left] {$+$}; \draw[->] (ep) -- (al) node [midway, left] {$-$}; % \draw[->] (mu) -- (ga) node [midway, left] {$+$}; \draw[->] (ga) -- (nu) node [midway, left] {$-$}; % horizontal \draw[->] (ka) -- (de) node [midway, above] {$+$}; \draw[->] (de) -- (mu) node [midway, above] {$-$}; % \draw[->] (be) -- (ep) node [midway, above] {$+$}; \draw[->] (ep) -- (ga) node [midway, above] {$-$}; % \draw[->] (la) -- (al) node [midway, above] {$+$}; \draw[->] (al) -- (nu) node [midway, above] {$-$}; % \end{tikzpicture} \end{equation} where a $+$ between two staircases indicates that a unit vector is added to the staircase on the left (respectively, in the lower row) to obtain the staircase on the right (respectively, in the upper row). By assumption, all staircases in the top and bottom row contain at least one zero entry. For such a staircase $\rho\in\ZZ^n$, let $\tilde\rho\in\ZZ^{n-1}$ be the staircase obtained from $\rho$ by removing a zero entry. If $\rho$ does not contain a zero, it must contain an entry $1$ (say, at position $i$), followed by a negative entry. In this case, $\tilde\rho\in\ZZ^{n-1}$ is obtained from $\rho$ by removing $\rho_i$ and adding $1$ to $\rho_{i+1}$. With this notation, we have to show the following four equalities \begin{enumerate}[(\alph{enumi})] \item\label{eq:stab-a} $\tilde\varepsilon = \dom_{\fS_{n-1}}(\tilde\beta + \tilde\alpha-\tilde\lambda)$, \item\label{eq:stab-b} $\tilde\gamma = \dom_{\fS_{n-1}}(\tilde\varepsilon + \tilde\nu-\tilde\alpha)$, \item\label{eq:stab-c} $\tilde\delta = \dom_{\fS_{n-1}}(\tilde\kappa + \tilde\varepsilon-\tilde\beta)$, and \item\label{eq:stab-d} $\tilde\mu = \dom_{\fS_{n-1}}(\tilde\delta + \tilde\gamma-\tilde\varepsilon)$. \end{enumerate} Let us first reduce to the case where at least one of the staircases involved does not contain a zero. Consider a square of staircases \begin{equation*} \tikzset{ short/.style={ shorten >=7pt, shorten <=7pt } } \begin{tikzpicture}[xscale=4,yscale=1,baseline=(current bounding box.center)] \node (al) at (0,0) {\makebox[3cm]{$\alpha$}}; \node (be) at (0,1) {\makebox[3cm]{$\beta=\alpha\pm e_i$}}; \node (ga) at (1,0) {\makebox[3cm]{$\gamma=\dom_{\fS_n}(\alpha\pm e_j)$}}; \node (de) at (1,1) {\makebox[3cm]{$\delta=\alpha\pm e_i\pm e_j$}}; \draw[->] (al) -- (be) node [midway, left] {}; \draw[->] (ga) -- (de) node [midway, right] {}; \draw[->] (al) -- (ga) node [midway, below] {}; \draw[->] (be) -- (de) node [midway, above] {}; \end{tikzpicture} \end{equation*} where all of $\alpha$, $\beta$, $\gamma$ and $\delta$ contain a zero. We first show that there is an index $k\not\in\{i,j\}$ such that $\alpha_k=\beta_k=\delta_k=0$. Suppose on the contrary that $\alpha_k\neq 0$ for all $k\not\in\{i,j\}$. Then, since $\beta$ contains a zero, we have $i\neq j$. Furthermore, we have \[ \begin{array}{lcll} \alpha_i=0&\text{or}&\alpha_j=0,\quad\text{and}\\ \alpha_i=\mp1&\text{or}&\alpha_j=0,\quad\text{and}\\ \alpha_i=0&\text{or}&\alpha_j=\mp1,\quad\text{and}\\ \alpha_i=\mp1&\text{or}&\alpha_j=\mp1 \end{array} \] because $\alpha$, $\beta$, $\gamma$ and $\delta$ contain a zero, respectively. However, this set of equations admits no solution. Thus, there must be a further zero in $\alpha$ and therefore also in $\beta$, $\gamma$ and $\delta$. From this it follows that $\tilde\gamma=\dom_{\fS_{n-1}}(\tilde\alpha + \tilde\delta-\tilde\beta)$. Returning to the square in~\eqref{eq:square-of-staircases}, we show that $\varepsilon$ contains a zero entry if $\beta$ or $\gamma$ do. Suppose on the contrary that $\varepsilon$ does not contain a zero entry. Then $\varepsilon=\beta+e_i$, where $i$ is the position of the (only) zero in $\beta$. Moreover, we have $\alpha=\beta$, because there is only one way to obtain a zero entry in $\alpha$ by subtracting a unit vector. Thus, \[ \lambda=\dom_{\fS_n}(\beta+\alpha-\varepsilon) = \dom_{\fS_n}(\beta-e_i) = \beta-e_i, \] which implies that $\lambda$ does not contain a zero entry, contradicting our assumption. Similarly, if $\gamma$ contains a zero at position $i$, we have $\varepsilon=\gamma+e_i$, $\gamma=\delta$ and $\mu=\dom_{\fS_n}(\gamma-e_i)$, a contradiction. There remain three different cases: \begin{enonce*}[remark]{$\beta$ contains a zero, but $\gamma$ does not}\ We have to show Equations~\ref{eq:stab-b} and~\ref{eq:stab-d}. Let $\alpha=\varepsilon-e_i$ and $\nu=\varepsilon-e_i-e_j$. Then $\gamma=\dom_{\fS_n}(\varepsilon-e_j)$. Since, by the foregoing, $\varepsilon$ contains a zero, we have $\varepsilon_j=0$. Since $\alpha$ also has a zero we have $i\neq j$. Since $\nu$ has a zero, $\varepsilon_i=1$. Because $\gamma$ has no zero, $\mu=\nu$. Together with the fact that $\delta$ has a zero, this implies that $\delta=\varepsilon-e_i$. The equations can now be checked directly. \end{enonce*} \begin{enonce*}[remark]{$\beta$ contains no zero, but $\gamma$ does}\ We have to show Equations~\ref{eq:stab-a} and~\ref{eq:stab-c}. Let $\lambda=\beta-e_i$ and $\alpha=\beta-e_i+e_j$. Then $\varepsilon=\dom_{\fS_n}(\beta+e_j)$. Since $\beta$ has no zero, but, by the foregoing, $\varepsilon$ does, we have $\beta_j=\bar 1$. Since $\lambda$ has a zero, $\beta_i=1$, and thus $i\neq j$. Because $\beta$ has no zero, $\kappa=\lambda$. Again, the equations can now be checked directly. \end{enonce*} \begin{enonce*}[remark]{None of $\beta$, $\epsilon$ and $\gamma$ contain a zero}\ In this case, $\kappa=\lambda$, $\delta=\alpha$ and $\mu=\nu$. Let $\lambda=\beta-e_i$, $\alpha=\beta-e_i+e_j$. Then $\varepsilon=\dom_{\fS_n}(\beta+e_j)$. Thus $\beta_j\neq\bar 1$, $\beta_i=1$, $\beta_{i+1}\leq\bar 1$ and, because $\alpha\neq\beta$, we have $i\neq j$. Because $\alpha$ and $\beta$ are staircases, $\beta+e_j$ has in fact decreasing entries and $\varepsilon=\beta+e_j$. Thus, $\alpha=\varepsilon-e_i$, $\nu=\varepsilon-e_i-e_k$ and $\gamma=\dom_{\fS_n}(\varepsilon-e_k)$. Again, because $\varepsilon$ and $\nu$ are staircases, $\varepsilon-e_k$ has decreasing entries and $\gamma=\varepsilon-e_k$. Thus, the equations can now be checked directly.\qedhere \end{enonce*} \end{proof} \subsection{Growth diagrams for staircase tableaux} In this section we set up the notation used in the remaining sections. In particular, we slightly modify and generalise the definition of $\Grm(\alt)$ from Section~\ref{sec:Stembridge}. \begin{defi} For a pair of partitions $\wgt=(\pos\wgt,\neg\wgt)$, the partition $\pos\wgt$ is the \Dfn{positive part} and the partition $\neg\wgt$ is the \Dfn{negative part}. Given an integer $n$ not smaller than the sum of the lengths of the two partitions, $[\pos\wgt,\neg\wgt]_n$ is the staircase \[ ({\pos\wgt}{}_{,0},\pos\wgt{}_{,1},\dots,0,\dots,0, \dots,-\neg\wgt{}_{,1},-\neg\wgt{}_{,0}). \] A \Dfn{staircase tableau} is a sequence of staircases $\alt=(\wgt^0,\wgt^1,\dots,\wgt^r)$ such that $\wgt^i$ and $\wgt^{i+1}$ differ by a unit vector for $0\leq i] (la) -- (nu); \draw[->] (ka) -- (mu); \node at (0.5,0.5)[rectangle,minimum size=1.5cm,draw] {\negcross}; \end{tikzpicture} \text{or}\quad \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)},scale=2] \node (ka) at (0,0) {$[\alpha,\tau]_n$}; \node (la) at (0,1) {$[\beta,\tau]_n$}; \node (nu) at (1,1) {$[\alpha,\tau]_n$}; \node (mu) at (1,0) {$[\alpha,\sigma]_n$}; \draw[<-] (ka) -- (la); \draw[<-] (mu) -- (nu); \draw[->] (la) -- (nu); \draw[->] (ka) -- (mu); \node at (0.5,0.5)[rectangle,minimum size=1.5cm,draw] {\poscross}; \end{tikzpicture} \text{or}\quad \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)},scale=2] \node (ka) at (0,0) {}; \node (la) at (0,1) {$[1,\emptyset]_n$}; \node (nu) at (1,1) {$[\emptyset,\emptyset]_n$}; \node (mu) at (1,0) {$[1,\emptyset]_n$}; \draw[<-] (mu) -- (nu); \draw[->] (la) -- (nu); \node at (0.5,0.5)[rectangle,minimum size=1.5cm,draw] {\fixcross}; \end{tikzpicture}, \] where $\beta$ (respectively $\sigma$) is obtained from $\alpha$ (respectively $\tau$) by adding a cell to the first column of the Ferrers diagram of the partition. All other cells remain empty. The following lemma is the main building block in establishing the connection between the filling of $\Grm(\alt)$ and the decorated evacuation diagram. \begin{lemma}\label{lem:alternating-equalities} Let $\alt = (\emptyset=\wgt^0,\dots,\wgt^{2r}=\emptyset)$ be an alternating tableau of empty shape. Let $\halfpr\alt=(\emptyset = \hm^0,\hm^1,\dots,\hm^{2r-1},\hm^{2r}=\emptyset)$ be as in Definition~\ref{def:half-promotion} and let $\pr\alt = (\emptyset = \Pm^0,\dots,\Pm^{2r}=\emptyset)$ be the promotion of $\alt$. Suppose that the filling of $\Grm(\alt)$ has a cross (that is, a {\negcross}, {\poscross} or {\fixcross}) in column $\ell>1$ of the first row and in row $k>1$ of the first column. Then, for even $n\geq r$ and for odd $n\geq r-1$, we have \begin{enumerate}[(\alph{enumi})] \item\label{lemma6.18_a} $\pos{\wgt^j} = \pos{\hm^{j-1}}$ for $2\leq j\leq 2\ell-2$, \item\label{lemma6.18_b} $\neg{\wgt^j} = \neg{\hm^{j-1}}$ for $j> 2\ell-2$, \item\label{lemma6.18_c} $\pos{\wgt^{2\ell-2}} = \pos{\wgt^{2\ell-1}} = \pos{\hm^{2\ell-3}}$, and $\pos{\hm^{2\ell-2}}$ is obtained from these by adding a cell to the first column. The cell labelled with these four staircases contains a~{\negcross}. \end{enumerate} % If there is no cross in the first row, the first equality holds for $2\leq j\leq 2r$. Similarly, \begin{enumerate}[(\alph{enumi}$'$)] \item\label{lemma6.18_a_prime} $\neg{\hm}^j = \neg{\Pm^{j-1}}$ for $1\leq j\leq 2k-2$, \item\label{lemma6.18_b_prime} $\pos{\hm}^j = \pos{\Pm^{j-1}}$ for $j > 2k-2$, \item\label{lemma6.18_c_prime} $\pos{\hm^{2k-1}} = \pos{\Pm^{2k-2}} = \pos{\Pm^{2k-3}}$, and $\pos{\hm^{2k-2}}$ is obtained from these by adding a cell to the first column. The cell labelled with these four staircases contains a~{\poscross}. \end{enumerate} % If there is no cross in the first column, the first equality holds for $1\leq j\leq 2r-1$. Finally, suppose that there is a cross in the top-left cell, that is $k=\ell=1$. Then \begin{enumerate}[(\alph{enumi})]\setcounter{enumi}{5} \item $\wgt^1 = 1, \wgt^2 = \emptyset$ and $\hm^1=1$. The cell labelled with these four staircases contains a~{\fixcross}. \item [(f$'$)] $\wgt^j=\hm^{j-1}-e_1=\Pm^{j-2}$ for all $2\le j \le 2r$. \end{enumerate} \end{lemma} \begin{proof} Consider a square of four adjacent staircases in the diagram for computing the promotion of an alternating tableau below: \begin{equation}\label{eq:pr-adjoint} \def\arraystretch{2.5}% \begin{array}{rcccccccl} \wgt^2&\dots&\tikzmark{s1}\wgt^{2\ell-2}&\wgt^{2\ell-1} &\multicolumn{4}{c}{\dotfill} &\wgt^{2r}=\emptyset\\ \hm^1&\dots&\hm^{2\ell-3} &\hm^{2\ell-2}\tikzmark{e1}&\dots&\tikzmark{s2}\hm^{2k-2}&\hm^{2k-1}&\dots&\hm^{2r-1}\\ \emptyset=\Pm^0&\multicolumn{4}{c}{\dotfill} &\Pm^{2k-3}&\Pm^{2k-2}\tikzmark{e2}&\dots&\Pm^{2r-2} \end{array} \begin{tikzpicture}[remember picture,overlay] % \node[draw,line width=0pt,rectangle,minimum size=2cm,fit={(pic cs:s1) (pic cs:e1)},yshift=1pt] {}; % \node[draw,line width=0pt,rectangle,minimum size=2cm,fit={(pic cs:s2) (pic cs:e2)},yshift=1pt] {}; \node [xshift=28pt,yshift=-12pt] at (pic cs:s1) {\negcross}; \node [xshift=28pt,yshift=-12pt] at (pic cs:s2) {\poscross}; \end{tikzpicture} \begin{tikzpicture}[remember picture,overlay] \end{tikzpicture} \end{equation} By definition, these satisfy the local rule, as required by Theorem~\ref{thm:rotation-promotion-matchings}. By Corollary~\ref{cor:pr-rot-bounds}, Theorem~\ref{thm:growth-diagram-rotation} is applicable with the given bounds for $n$. The equalities for the staircases in the second and third row are precisely the equalities listed below the illustrations in Figure~\ref{fig:different_cases_column_rotation}: cases~\ref{fig6_a} and~\ref{fig6_c} there describe the situation to the left of \poscross, case~\ref{fig6_e} describes the situation at \poscross\ and cases~\ref{fig6_b} and~\ref{fig6_d} describe the situation to the right of \poscross. The equalities for the staircases in the first and second row can be obtained as in the last paragraph of the proof of Theorem~\ref{thm:pr-rot}. \end{proof} By successively applying Lemma~\ref{lem:alternating-equalities} we obtain the following result for the evacuation diagram. \begin{coro}\label{cor:relate-fillings} Let $\alt$ be a $\GL(n)$-alternating tableau of empty shape and length~$r$ with corresponding growth diagram $\Grm(\alt)$ and filling $\phi$. Suppose that $n\geq r$ if $n$ is even and $n\geq r-1$ if $n$ is odd. Consider the evacuation diagram with filling obtained as above. A {\negcross} appears only in odd columns and odd rows, a {\poscross} appears only in even columns and even rows and a {\fixcross} appears only in even columns and odd rows. Moreover $(i,j)$ is the position of a cell with a cross in $\phi$ if and only if one of the following cases holds. \begin{itemize} \item $i j$ and there is a {\poscross} in row $2j$ and column $2i$. \item There is a {\fixcross} in row $2i-1$ and column $2j$. Then we also obtain $i=j$. \end{itemize} \end{coro} \begin{figure}[h] \begin{tikzpicture}[baseline=0pt,scale=0.5] \def\rr{8} % labels \node at (0.5,\rr.5) {\tiny$1$}; \node at (1.5,\rr.5) {\tiny$2$}; \node at (\rr/2,\rr.5) {$\ldots$}; \node at (\rr-1.5,\rr.5) {\tiny$2r\text{-}1$}; \node at (\rr-0.5,\rr.5) {\tiny$2r$}; \node at (\rr.5,\rr-0.5) {\tiny$1$}; \node at (\rr.5,\rr-1.5) {\tiny$2$}; \node at (\rr.5,\rr/2+0.25) {$\vdots$}; \node at (\rr.5,1.5) {\tiny$2r\text{-}1$}; \node at (\rr.5,0.5) {\tiny$2r$}; % circles for the partitions \foreach \y in {0,...,\rr} { \tikzmath{\xmax = \y-Mod(\y,2);}; \foreach \x in {0,...,\xmax} \draw (\rr-\x,\y) circle[radius=2pt]; } % axis \draw[dotted] (\rr,\rr) -- (\rr/2-1,\rr/2-1); % more labels \node[anchor=east] at (-0.5,\rr) {$\alt=$}; \node[rotate=90,anchor=east] at (\rr,-0.5) {$\ev \alt=$}; %the filling \node at (5-0.5,\rr-1+0.5) {\negcross}; \node at (8-0.5,\rr-2+0.5) {\poscross}; \node at (4-0.5,\rr-3+0.5) {\fixcross}; \end{tikzpicture}\quad \begin{tikzpicture}[baseline=0pt,scale=0.5] \def\rr{8} %labels \node at (0.5,\rr.5) {\tiny$1$}; \node at (1.5,\rr.5) {\tiny$2$}; \node at (\rr/2,\rr.5) {$\ldots$}; \node at (\rr-1.5,\rr.5) {\tiny$2r\text{-}1$}; \node at (\rr-0.5,\rr.5) {\tiny$2r$}; \node at (\rr.5,\rr-1+0.5) {\tiny$1$}; \node at (\rr.5,\rr-2+0.5) {\tiny$2$}; \node at (\rr.5,\rr/2+0.25) {$\vdots$}; \node at (\rr.5,1.5) {\tiny$2r\text{-}1$}; \node at (\rr.5,0.5) {\tiny$2r$}; % circles for the partitions \foreach \y in {0,...,\rr} { \tikzmath{\xmax = \y-Mod(\y,2);}; \foreach \x in {0,...,\xmax} \draw (\rr-\x,\y) circle[radius=2pt]; } % axis \draw[dotted] (\rr,\rr) -- (\rr/2-1,\rr/2-1); % more labels \node[anchor=east] at (-0.5,\rr) {$\ev \alt=$}; \node[rotate=90,anchor=east] at (\rr,-0.5) {$\alt=$}; %the filling \node at (\rr-1+0.5,5-0.5) {\poscross}; \node at (\rr-2+0.5,8-0.5) {\negcross}; \node at (\rr-3+0.5,4-0.5) {\fixcross}; \end{tikzpicture} \caption{The symmetry of the evacuation diagram.} \label{fig:filling-from-evacuation-diagram} \end{figure} By the symmetry of the local rules the evacuation diagram for $\ev \alt$ is obtained from the evacuation diagram for $\alt$ by mirroring it along the diagonal and interchanging {\poscross} and {\negcross}. The cell $(i,j)$ is interchanged with the cell $(2r+1-j,2r+1-i)$. This yields the part of Theorem~\ref{thm:rotation-promotion-permutations} concerning evacuation, see Figure~\ref{fig:filling-from-evacuation-diagram} for an illustration. \begin{theo} \label{thm:ev-rc} Let $\alt=(\emptyset = \wgt^0,\wgt^1,\dots,\wgt^{2r-1},\wgt^{2r}=\emptyset)$ be an alternating tableau. Suppose that $n\geq r$ if $n$ is even and $n\geq r-1$ if $n$ is odd. Let $\phi$ be the filling of the growth diagram $\Grm(\alt)$. Then the filling of $\Grm(\ev \alt)$ is obtained by rotating $\phi$ by $180\degree$. \end{theo} \begin{proof} Let $\phi_{\alt}=\phi$, respectively $\phi_{\ev\alt}$, be the fillings of the growth diagrams $\Grm(\alt)$, respectively $\Grm(\ev \alt)$. Let $(i,j)$ be the position of a cell with a cross in the filling $\phi_{\alt}$. Then, according to Corollary~\ref{cor:relate-fillings}: \begin{enumerate} \item if $i -/.style={decoration={ markings, mark=at position #1 with {\arrow{>}}},postaction={decorate}}} \raisebox{1.25cm}{% \begin{tikzpicture}[line width=1pt] \node (a) [draw=none, minimum size=4cm, regular polygon, regular polygon sides=8] at (0,0) {}; \foreach \n [count=\i from 0, remember=\n as \lastn, evaluate={\i+\lastn}] in {1,2,...,8} \path (a.center) -- (a.corner \n) node[pos=1.15] {$\n$}; \draw[->-=.9] (a.corner 1) to (a.corner 8); \draw[->-=.9] (a.corner 2) to (a.corner 1); \draw[->-=.9] (a.corner 3) to (a.corner 7); \draw[->-=.9, bend right] (a.corner 4) to (a.corner 5); \draw[->-=.9, bend right] (a.corner 5) to (a.corner 4); \draw[->-=.9] (a.corner 6) to (a.corner 3); \draw[->-=.9] (a.corner 7) to (a.corner 6); \draw[->-=.9] (a.corner 8) to (a.corner 2); \end{tikzpicture}} \hfill \begin{tikzpicture}[scale=0.8]\tiny % thick diagonal \draw[line width=1.5pt] (9,1) -- (9,2) -- (8,2) -- (8,3) -- (7,3) -- (7,4) -- (6,4) -- (6,5) -- (5,5) -- (5,6) -- (4,6) -- (4,7) -- (3,7) -- (3,8) -- (2,8) -- (2,9) -- (1,9); \foreach \y in {1,...,9} \draw (1,\y)--(9,\y); \foreach \x in {1,...,9} \draw (\x,1)--(\x,9); % column labels \foreach \x in {1,...,8} \draw (\x.5,9.3) node{\footnotesize $\x$}; % row labels \foreach \x in {1,...,8} \draw (0.3,9.4-\x) node{\footnotesize $\x$}; % crosses \draw (1.5,1.5) node{\huge$\times$}; \draw (2.5,8.5) node{\huge$\times$}; \draw (3.5,2.5) node{\huge$\times$}; \draw (4.5,4.5) node{\huge$\times$}; \draw (5.5,5.5) node{\huge$\times$}; \draw (6.5,6.5) node{\huge$\times$}; \draw (7.5,3.5) node{\huge$\times$}; \draw (8.5,7.5) node{\huge$\times$}; % staircases \draw (0.8,8.7) node{\footnotesize $\emptyset$}; \draw (1.8,8.7) node{\footnotesize $1$}; \draw (1.8,7.7) node{\footnotesize $1\bar1$}; \draw (2.8,7.7) node{\footnotesize $1$}; \draw (2.8,6.7) node{\footnotesize $1\bar1$}; \draw (3.8,6.7) node{\footnotesize $2\bar1$}; \draw (3.8,5.7) node{\footnotesize $2\bar2$}; \draw (4.8,5.7) node{\footnotesize $3\bar2$}; \draw (4.8,4.7) node{\footnotesize $3\bar3$}; \draw (5.8,4.7) node{\footnotesize $3\bar2$}; \draw (5.8,3.7) node{\footnotesize $2\bar2$}; \draw (6.8,3.7) node{\footnotesize $2\bar1$}; \draw (6.8,2.7) node{\footnotesize $2\bar2$}; \draw (7.8,2.7) node{\footnotesize $2\bar1$}; \draw (7.8,1.7) node{\footnotesize $1\bar1$}; \draw (8.8,1.7) node{\footnotesize $1$}; \draw (8.8,0.7) node{\footnotesize $\emptyset$}; \end{tikzpicture} \caption{A noncrossing set partition corresponding to a $\GL(2)$-alternating tableau.} \label{fig:noncrossing-set-partition} \end{figure} \begin{lemma}\label{lem:noncrossing} The map $\Perm$ restricts to a bijection between $\GL(n)$-alternating tableaux of empty shape and length $r$, such that every staircase has at most two nonzero parts, and noncrossing set partitions on $\{1,\dots,r\}$. \end{lemma} \begin{proof} For simplicity, suppose that $\alt$ is a $\GL(2)$-alternating tableau. Let $\pi$ be the permutation corresponding to the filling associated with $\alt$. We show that, when drawn as a chord diagram as in Figure~\ref{fig:noncrossing-set-partition}, it is obtained from a noncrossing set partition by orienting the arcs delimiting the blocks clockwise, when the corners of the polygon are labelled counterclockwise. We say that two arcs $(i,\pi_i)$ and $(j,\pi_j)$ in the chord diagram, with $i \ell$ is very similar. If $i<\ell$, that is, $2i\leq 2\ell-2$, the positive parts of staircases in the same column of the first two rows of diagram~\eqref{eq:square} coincide by Lemma~\ref{lem:alternating-equalities}\ref{lemma6.18_a}. By Lemma~\ref{lem:alternating-equalities}\ref{lemma6.18_a_prime}, the negative parts of staircases in the same column of the second two rows coincide. Moreover, by construction of $\alt_\osc$, the staircase in the middle of the first row either equals $[\lambda, \nu]_n$ or $[\nu, \lambda]_n$. Let us assume the latter, the former case is dealt with similarly. For the staircase in the middle we then obtain, applying the local rule to the staircases on the top-left, \[ \hm^{2i-2}=\dom_{\fS_n}\big([\lambda,\kappa]_n+[\nu,\lambda]_n-[\lambda,\lambda]_n\big) =[\nu,\kappa]_n. \] Similarly, applying the local rule to the staircases on the bottom-right, we find \[ [\mu,\mu]_n=\dom_{\fS_n}\big([\pos{\Pm^{2i-3}},\kappa]_n+[\nu,\mu]_n-[\nu,\kappa]_n\big)\\ =[\pos{\Pm^{2i-3}},\mu]_n. \] Therefore, the square of staircases in diagram~\eqref{eq:square} has the following form: \begin{equation} \label{eq:square-special} \def\arraystretch{2.5} \begin{array}{ccc} [\lambda,\lambda]_n & [\nu,\lambda]_n & [\nu,\nu]_n\\\relax [\lambda,\kappa]_n & [\nu,\kappa]_n & [\nu,\mu]_n\\\relax [\kappa,\kappa]_n & [\mu,\kappa]_n & [\mu,\mu]_n \end{array} \end{equation} Because the negative parts of the four staircases in the lower left corner are all the same, the positive parts satisfy $\mu=\dom_{\fS_n}(\kappa+\nu-\lambda)$, and therefore also Equation~\eqref{eq:local-rule-oscillating}. It remains to show that Equation~\eqref{eq:local-rule-oscillating} also holds for $i=\ell$. By Lemma~\ref{lem:alternating-equalities}\ref{lemma6.18_c}, the positive parts of the staircases $\wgt^{2\ell-2}$, $\wgt^{2\ell-1}$ and $\hm^{2\ell-3}$ all coincide, and thus equal $\lambda$. Moreover, the positive part $\alpha$ of the staircase in the middle is obtained by adding a cell to the first column of the Ferrers diagram of $\lambda$. By Lemma~\ref{lem:alternating-equalities}\ref{lemma6.18_c_prime}, the positive parts of the staircases $\hm^{2\ell-1}$, $\Pm^{2\ell-3}$ and $\Pm^{2\ell-2}$ all coincide, and thus equal $\mu$. Moreover, the positive part $\alpha$ of the staircase in the middle is obtained by adding a cell to the first column of the Ferrers diagram of $\mu$. Therefore $\lambda=\mu$. Lemma~\ref{lem:alternating-equalities}\ref{lemma6.18_b} implies that the negative parts of the staircases $\wgt^{2\ell-1}$, $\wgt^{2\ell}$, $\hm^{2\ell-2}$ and $\hm^{2\ell-1}$ are all equal to $\nu$. Finally, Lemma~\ref{lem:alternating-equalities}\ref{lemma6.18_a_prime} shows that the negative parts of the staircases $\hm^{2\ell-3}$, $\hm^{2\ell-2}$, $\Pm^{2\ell-4}$ and $\Pm^{2\ell-3}$ are all equal to $\kappa$. Thus $\nu=\kappa$ and diagram~\eqref{eq:square} has the form \begin{equation} \label{eq:square-special-k=l} \def\arraystretch{2.5} \begin{array}{ccc} [\lambda,\lambda]_n & [\lambda,\nu]_n & [\nu,\nu]_n\\\relax [\lambda,\nu]_n & [\alpha,\nu]_n & [\lambda,\nu]_n\\\relax [\nu,\nu]_n & [\lambda,\nu]_n & [\lambda,\lambda]_n \end{array} \end{equation} Considering the growth diagram $\Grm(\alt_\osc)$, we additionally find that $\lambda$ is obtained from $\nu$ by adding a cell to the first column. %, essentially because $\ell=k$. Thus, the vector $\nu + \nu - \lambda$ is obtained from $\nu$ by subtracting $1$ from the entry at position $\ell(\lambda)$, which is $0$ in $\nu$. Taking the absolute values of the entries of the vector $\nu + \nu - \lambda$ then yields $\lambda$. \end{proof} % \nocite{*} \bibliographystyle{amsplain-ac} \bibliography{ALCO_Rubey_152} \end{document} %%% Local Variables: %%% mode: latex %%% TeX-master: t %%% End: