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\title[On a conjecture of Naito-Sagaki]{On a conjecture of Naito-Sagaki: Littelmann paths and Littlewood-Richardson Sundaram tableaux}

\author[Jacinta Torres]{Jacinta Torres\thanks{The author was financed by the Graduiertenkolleg 1269: Global Structures in Geometry and Analysis, and by the Max Planck Society.}}

\address{Mathematical Institute of the University of Cologne}

\received{\today}

\abstract{In recent work with Schumann we have proven a conjecture of
Naito-Sagaki giving a branching rule for the decomposition of the restriction of an irreducible
representation of the special linear Lie algebra to the symplectic Lie algebra,
therein embedded as the fixed-point set of the involution obtained by the folding of
the corresponding Dynkin diagram. This conjecture had been open for over ten years,
and provides a new approach to branching rules for non-Levi subalgebras in terms
of Littelmann paths. In this extended abstract we motivate the conjecture, prove it for
several cases, where we also relate it to the combinatorics of polytopes and Littlewood-Richardson
cones, and highlight some difficulties of the proof in general.}

\resume{Dans un travail r\'ecent avec Schumann, nous avons demontr\'e une conjecture
de Naito-Sagaki donnant une r\`egle de branchement pour la restriction d'une repr\'esentation
irr\'eductible de l'alg\`ebre de Lie sp\'eciale lin\'eaire \`a l'alg\`ebre de Lie symplectique, qui se plonge comme
les points fixes de l'automorphisme de pliage du diagramme de Dynkin. Cette conjecture a \'et\'e
ouverte depuis plus de dix ans, et donne une nouvelle perspective sur les r\`egles de
branchement pour les sous-alg\`ebres de Lie qui ne sont pas de Levi, en termes des chemins de Littelmann.
Dans ce r\'esume \'etendu, nous motivons la conjecture et la d\'emontrons dans certains
cas, pour lesquels nous faisons aussi un lien avec la combinatoire des polytopes et
c\^ones de Littlewood-Richardson ; et nous d\'ecrivons les difficult\'es de la d\'emonstration
du cas g\'en\'eral.}

\keywords{Littelmann paths, tableaux combinatorics, branching rules}

\begin{document}

\maketitle

\section{Introduction}

Let $\mathfrak{g}$ be a semi-simple Lie algebra over the field of complex numbers and $\sigma: \mathfrak{g} \rightarrow \mathfrak{g}$ an automorphism of finite order. In \cite{kac}, Kac has shown that $\sigma$ is conjugate to the automorphism induced by a Dynkin diagram automorphism, and that the set of fixed points $\mathfrak{g}^{\sigma}$ is again a semi-simple Lie algebra, of one of six possible types. The basic problem of decomposing the restriction of an irreducible representation of $\mathfrak{g}$ to $\mathfrak{g}^{\sigma}$ (known as a ``branching rule'') is, in general, open.

If, however, the automorphism $\sigma$ is semi-simple and has infinite order, then $\mathfrak{g}^{\sigma}$ is a Levi subalgebra \cite{vinberg}. In this case, there exist natural and beautiful branching rules in terms of Littelmann paths \cite{pathsandrootoperators}. Littelmann paths are generalisations of Young tableaux to all Kac-Moody Lie algebras, and were used \cite{pathsandrootoperators} to give a remarkable generalisation of the Littlewood-Richardson rule for the decomposition of the tensor product of two irreducible finite-dimensional representations.

In \cite{branchconj} a new approach to the branching problem was suggested by Naito-Sagaki in terms of Littelmann paths for $\mathfrak{g} = \mathfrak{sl}(2n,\mathbb{C})$ and $\mathfrak{g}^{\sigma} = \mathfrak{sp}(2n,\mathbb{C})$ in the form of a conjecture. In this paper we reformulate the conjecture in terms of symplectic Littlewood-Richardson tableaux, which were previously studied by Sundaram in \cite{sundaram}. We call them Littlewood-Richardson Sundaram tableaux. We give a proof of the conjecture in some cases using this reformulation, which was used in \cite{beajaz} to give a general proof. Our methods imply a bijection between lattice points of polytopes, in the spirit of Berenstein-Zelevinsky  and Knutson-Tao \cite{bz} \cite{knutsontaojams}. We believe that this new approach to branching will play an important role in the future. 

\section{Notation for the Lie algebras}
Let $\mathfrak{h} \subset \mathfrak{sl}(2n, \mathbb{C})$ be the Cartan sub-algebra of diagonal matrices.  Let $\e_{i}$ be the linear map $\mathfrak{h} \rightarrow \mathbb{C}$ defined by $\op{diag}(a_{1},\hbox{ } \cdots, \hbox{ } a_{2n}) \mapsto a_{i}.$ We write  $\mathfrak{sl}(2n, \mathbb{C}) = \< x_{i}, y_{i}, h_{i} \rr_{ i \in \{1, \cdots, 2n-1\}}$ where $h_{i} = \E_{ii} - \E_{i+1, i+1}$ and where $x_{i}$ and $y_{i}$ are the Chevalley generators corresponding to the simple root $\alpha_{i} := \e_{i} - \e_{i+1}$. The automorphism $\sigma$ is given by
\begin{align*}
\sigma(x_{i}) &= x_{2n-i},\\
 \sigma(y_{i}) &= y_{2n-i}, \hbox{ and }\\
  \sigma(h_{i}) &= h_{2n-i}.
\end{align*}


The fixed point set $\mathfrak{g}^{\sigma}$ is generated as a Lie algebra by $\< {\tightoverset{\hbox{\textasciicircum}}{x}}_{i}, {\tightoverset{\hbox{\textasciicircum}}{y}}_{i}, {\tightoverset{\hbox{\textasciicircum}}{h}}_{i}\rr_{i \in  \{1, \cdots, n\}}$ (see Proposition 7.9 in \cite{kac}), where 
\[
    {\tightoverset{\hbox{\textasciicircum}}{x}}_{i} =  
\begin{cases}
    x_{i} + x_{2n-i}& \text{ if } i \in [1, n) \\
    x_{n}& \text{ if }  i = n
\end{cases}
\]

\[
    {\tightoverset{\hbox{\textasciicircum}}{y}}_{i} =  
\begin{cases}
    y_{i} + y_{2n-i}& \text{ if } i \in [1, n) \\
    y_{n}& \text{ if }  i = n
\end{cases}
\]

\[
    {\tightoverset{\hbox{\textasciicircum}}{h}}_{i} =  
\begin{cases}
    h_{i} + h_{\alpha_{2n-i}}& \text{ if } i \in [1, n) \\
    h_{n}& \text{ if }  i = n.
\end{cases}
\]
\noindent
This Lie algebra is isomorphic to $\mathfrak{sp}(2n, \mathbb{C})$ (see Proposition 7.9 in \cite{kac}) and ${\mathfrak{h}}^{\sigma} = \overset{n}{\underset{ i = 1}{\bigoplus}}{\tightoverset{\hbox{\textasciicircum}}{h}}_{i} = \mathfrak{h} \cap {\mathfrak{g}}^{\sigma}  \subset \mathfrak{h}$ is a Cartan subalgebra. Let $\p_{C_{n}} \subset (\mathfrak{h}^{\sigma})^{*}$ be the set of integral weights of ${\mathfrak{g}}^{\sigma}$ with respect to ${\mathfrak{h}}^{\sigma}$, and let $\p_{A_{2n-1}} \subset \mathfrak{h}^{*}$ be the set of integral weights of $\mathfrak{g}$ with respect to $\mathfrak{h}$.

To avoid confusion we will denote elements of $\p_{A_{2n-1}}$ by $\lambda$ and elements of $\p_{C_{n}}$ by $\tightoverset{\hbox{\textasciicircum}}{\lambda}$; in particular, the fundamental weights in  $\p_{A_{2n-1}}$ will be denoted by $\omega_{1}, \cdots, \omega_{2n-1}$, and the fundamental weights in $\p_{C_{n}}$ by ${\tightoverset{\hbox{\textasciicircum}}{\omega}}_{1}, \cdots, {\tightoverset{\hbox{\textasciicircum}}{\omega}}_{n}$. We can then write $$\p_{A_{2n-1}} = \overset{2n-1}{\underset{i = 1}{\bigoplus}}\mathbb{Z}\omega_{i}$$ \noindent and $$\p_{C_{2n}} = \overset{n}{\underset{i = 1}{\bigoplus}}\mathbb{Z}{\tightoverset{\hbox{\textasciicircum}}{\omega}}_{i},$$ \noindent where the direct sums are as $\mathbb{Z}$-modules. Also, for $\lambda = a_{1}\omega_{1} + \cdots + a_{n}\omega_{n}$ we write 
$\tightoverset{\hbox{\textasciicircum}}{\lambda} = a_{1}{\tightoverset{\hbox{\textasciicircum}}{\omega}}_{1} + \cdots + a_{n}{\tightoverset{\hbox{\textasciicircum}}{\omega}}_{n}$ for the corresponding element in $\p_{C_{n}}$. Weights with non-negative coefficients are called \textbf{dominant}. We will denote the corresponding sets by $\p^{+}_{A_{2n-1}}$ and $\p^{+}_{C_{n}}$. We will consider the real vector spaces $\mathfrak{h}_{\mathbb{R}}$ and $\mathfrak{h}^{\sigma}_{\mathbb{R}}$ spanned by the fundamental weights $\omega_{1}, \cdots, \omega_{2n-1}$ respectively ${\tightoverset{\hbox{\textasciicircum}}{\omega}}_{1}, \cdots, {\tightoverset{\hbox{\textasciicircum}}{\omega}}_{n}$. The dominant Weyl chamber is the convex hull of the dominant weights in $\mathfrak{h}^{*}_{\mathbb{R}}$ (respectively $(\mathfrak{h}^{\sigma})^{*}_{\mathbb{R}}$).

%
\section{Tableaux, words and their paths}
A \textbf{shape} is a finite sequence of non-negative integers $\underline{d} = (d_{1}, \cdots, d_{k})$. An arrangement of boxes of shape $\underline{d}$ consists of $d_{1}+ \cdots + d_{k}$ top-left aligned columns such that the first $d_{1}$ columns (from right to left) have one box, and the first $d_{s}$ boxes after the $(d_{1}+ \cdots +d_{s-1})$-th column have $d_{s}$ boxes. To a dominant integral weight $\lambda = a_{1}\omega_{1}+ \cdots + a_{2n-1}\omega_{2n-1} \in \p^{+}_{A_{2n-1}}$ we assign the shape $\underline{d}_{\lambda} = (a_{1}, \cdots, a_{2n-1})$.

\begin{ex} For $\lambda = 3\omega_{1} + \omega_{2}$ we have $\underline{d}_{\lambda} = (3,1)$. To it is associated the following arrangement of boxes:
\label{ch2ex1}
\begin{align*}
\Skew(0:,,,|0:)
\end{align*}
\end{ex}


\noindent
A \textbf{semi-standard tableau} of shape $\underline{d}$ is a filling of an arrangement of boxes of shape $\underline{d}$ with letters from the either one of the ordered alphabets 
$$\mathscr{A}_{2n} = \{1< \cdots< 2n\}$$
\noindent or 
$$\mathscr{C}_{n} = \{1< \cdots < n < \overline{n} < \cdots < \overline{1}\}$$ 
\noindent such that entries are strictly increasing downwards along each column of boxes and weakly increasing along each row from left to right. If the entries of a given semi-standard tableau belong to $\mathscr{A}_{2n}$ we will call it a \textbf{semi-standard Young tableau}, and if they belong to  $\mathscr{C}_{n}$, we will call it a \textbf{symplectic semi-standard tableau}. We will denote the set of semi-standard Young tableau of shape $\underline{d}$ by $\Gamma(\underline{d})^{\op{SSYT}}$. 

\begin{ex}
\label{ch2ex2}
A semi-standard Young tableau of shape $(3,1)$:
\begin{align*}
\Skew(0:\mbox{\tiny{1}},\mbox{\tiny{1}},\mbox{\tiny{2}},\mbox{\tiny{5}}|0:\mbox{\tiny{2}})
\end{align*}
\end{ex}

Let $\mathscr{W}(\mathscr{A}_{2n})$ denote the word monoid on $\mathscr{A}_{2n}$ and $\mathscr{W}(\mathscr{C}_{n})$ be the word monoid on $\mathscr{C}_{n}$. The \textbf{word} $w(\mathscr{T})$ of a semi-standard tableau $\mathscr{T}$ is obtained by reading its rows, from right to left and top to bottom. 
 
\begin{ex}
\label{ch2ex3}
 The symplectic semi-standard tableau $\mathscr{T} = \Skew(0:\mbox{\tiny{1}},\mbox{\tiny{1}},\mbox{\tiny{2}},\mbox{\tiny{5}}|0:\mbox{\tiny{$\overline{ 3}$}})$ has word $w(\mathscr{T}) = 5211\overline{ 3}$. 

\end{ex}




\section{Restriction of paths that come from words}
\subsection{Paths and their restrictions}
We will consider paths to be continuous maps $\pi:[0,1] \rightarrow \mathfrak{h}_{\mathbb{R}}^{*}, \pi':[0,1] \rightarrow (\mathfrak{h}_{\mathbb{R}}^{\sigma})^{*}$ starting at the origin and ending at an integral  weight: 

$$\pi(0) = 0 = \pi'(0), \hbox{   } \pi(1) \in \p_{A_{2n-1}}, \hbox{   } \pi'(1) \in \p_{C_{n}}.$$

\noindent
The map $\mathfrak{h}^{*} \rightarrow (\mathfrak{h}^{\sigma})^{*}, \varphi  \mapsto {\varphi }|_{\mathfrak{h}^{\sigma}}$
induces a map $\op{res}': \p_{A_{2n-1}} \rightarrow  \p_{C_{n}}$. Given a path $\pi: [0,1] \rightarrow \mathfrak{h}_{\mathbb{R}}^{*}$, we define a restricted path $\op{res}(\pi)$ by $\op{res}(\pi): [0,1]  \rightarrow (\mathfrak{h}_{\mathbb{R}}^{\sigma})^{*},
t \mapsto \op{res}'(\pi(t)).$

\noindent We will also consider the concatenation $\pi_{2}*\pi_{1}$ of two paths $\pi_{1}$ and $\pi_{2}$ with the same codomain: it is the path obtained by translating $\pi_{2}$ to the endpoint $\pi_{1}(1)$ of $\pi_{1}$. A path is \textbf{dominant} if its image is contained in the dominant Weyl chamber.

\subsection{Paths that come from words}
In the first part of this section we follow Section 2 of \cite{branchconj}. Let $w = w_{1}\hbox{ }\cdots \hbox{ } w_{k}$ be a word, either in $\mathscr{W}(\mathscr{A}_{2n})$ or in $\mathscr{W}(\mathscr{C}_{n})$. To it we assign the path:
\begin{align*}
\pi_{w} = \pi_{w_{k}} * \cdots * \pi_{w_{1}},
\end{align*}


\noindent
where, for $w_{i} \in \mathscr{A}_{2n}$ (respectively $w_{i} \in \mathscr{C}_{n}$), the path $\pi_{w_{i}}: [0,1] \rightarrow \mathfrak{h}_{\mathbb{R}}^{*}$ (respectively $\pi_{w_{i}}: [0,1] \rightarrow (\mathfrak{h}_{\mathbb{R}}^{\sigma})^{*})$ is given by $t \mapsto t\e_{w_{i}}$, where we define $\e_{\overline{l}}:= -\e_{l}$ for $1 \leq l \leq n$. Also, in general, for paths $\pi_{1}, \cdots, \pi_{k}:[0,1] \rightarrow \mathfrak{h}_{\mathbb{R}}^{*}$, we have 
\begin{align}
\label{letterbyletter}
\op{res}(\pi_{1}*\cdots * \pi_{k}) = \op{res}(\pi_{1})* \cdots * \op{res}(\pi_{k}).
\end{align}

\noindent
Set ${\tightoverset{\hbox{\textasciicircum}}{\e}}_{i}= \op{res}(\e_{i}) \hbox{ for } i \in \{1, \cdots, 2n\}.$ Then, for $i \in \{1, \cdots, 2n\}$ and $j \in \{1, \cdots, n\}$ we have
\begin{align}
\label{rescases}
    {\tightoverset{\hbox{\textasciicircum}}{\e}}_{i}({\tightoverset{\hbox{\textasciicircum}}{h}}_{j}) =  
\begin{cases}
    1& \text{ if } i \in \{j, 2n-j\}\\
    -1& \text{ if }  i \in \{j+1, 2n-j+1\}\\
    0& \text{ otherwise. }
\end{cases}
\end{align}

\noindent
Therefore ${\tightoverset{\hbox{\textasciicircum}}{\e}}_{i} = - {\tightoverset{\hbox{\textasciicircum}}{\e}}_{2n - i +1}$, which means we can describe $\op{res}(\pi_{w})$ in the following simple way for $w \in \mathscr{W}(\mathscr{A}_{2n})$: First obtain from $w$ a word $\op{res}(w)$ in the alphabet $\mathscr{C}_{n}$ by replacing a letter $w_{i}$ in $w$ with $\overline{2n - w_{i} +1}$ if  $n< w_{i} \leq 2n$. All other letters stay the same. The corresponding path $\pi_{\op{res}(w)}$ coincides with $\op{res}(\pi_{w})$ by \eqref{rescases} and \eqref{letterbyletter}.

\begin{ex}
\label{firstexampleofrestrictedpath}
Let $n = 2$ and $w = 121223341$. Then $\op{res}(w) = 12122\overline{2} \hbox{ } \overline{2}\hbox{ } \overline{1}1$. 

\end{ex}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Naito-Sagaki conjecture}

Let $\lambda \in \p^{+}_{A_{2n-1}}$ be dominant and let $\li(\lambda)$ be the associated simple module for $\mathfrak{sl}(2n,\mathbb{C})$. Recall the set  $\Gamma(\underline{d}_{\lambda})^{\op{SSYT}}$ of semi-standard Young tableaux of shape $\underline{d}_{\lambda}$. Let
\begin{align*}
\op{domres}(\lambda) = \{\eta = \op{res}(\pi_{w(\delta)}) \text{ dominant}: \delta \in \Gamma(\underline{d}_{\lambda})^{\op{SSYT}}\}
\end{align*}

\noindent
be the set of restricted paths associated to words of elements of $\Gamma(\underline{d}_{\lambda})^{\op{SSYT}}$ that are dominant, and for $\nu \in \op{P}^{+}_{\C_{2n}}$, let 
\begin{align*}
\op{domres}(\lambda, \nu) = \{\delta \in \op{domres}(\lambda): \delta (1) = \nu\}.
\end{align*}


\begin{ex}
\label{domresexample}
Let $n = 2$ and $\lambda = \omega_{1} + \omega_{2}$. Then
\begin{align*}
\op{domres}(\lambda) &= \left\{\Skew(0:\mbox{\tiny{1}},\mbox{\tiny{1}}|0:\mbox{\tiny{2}}), \Skew(0:\mbox{\tiny{1}},\mbox{\tiny{1}}|0:\mbox{\tiny{$\overline{1}$}})\right\}\\
\op{domres}(\lambda, \lambda) &= \left\{\Skew(0:\mbox{\tiny{1}},\mbox{\tiny{1}}|0:\mbox{\tiny{2}})\right\}\\
\op{domres}(\lambda, \omega_{1}) &= \left\{ \Skew(0:\mbox{\tiny{1}},\mbox{\tiny{1}}|0:\mbox{\tiny{$\overline{1}$}})\right\}.
\end{align*}
\end{ex}


\begin{thm}(Schumann-Torres, 2016 \cite{beajaz})
\label{naitosagakiconjecture}
Let $\lambda \in \p_{A_{2n-1}}^{+}$ be dominant, and let $\li(\lambda)$ be the associated simple module for 
$\mathfrak{sl}(2n, \mathbb{C})$. Then 
\begin{align*}
\op{res}^{\mathfrak{g}}_{\mathfrak{g}^{\sigma}} (\li(\lambda)) = \underset{\delta \in \op{domres}(\lambda)}{\bigoplus} \li(\tightoverset{\hbox{\textasciicircum}}{\delta(1)}).
\end{align*}

\end{thm}

\begin{ex}
\label{firstexamplebranch}
Let $n = 2$ and $\lambda = \omega_{1} + \omega_{2}$ as in \cref{domresexample}. Then 
\begin{align*}
\op{res}^{\mathfrak{g}}_{\mathfrak{g}^{\sigma}}(\li(\lambda)) = \li({\tightoverset{\hbox{\textasciicircum}}{\omega}}_{1}) \oplus \li({\tightoverset{\hbox{\textasciicircum}}{\omega}}_{1} + {\tightoverset{\hbox{\textasciicircum}}{\omega}}_{2}).
\end{align*}
\end{ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{rem}
\cref{naitosagakiconjecture} was conjectured by Naito-Sagaki in \cite{branchconj} in 2005. In \cite{branchconj} it is stated for $\li(\lambda)$ a representation of $\mathfrak{gl}(2n, \mathbb{C})$ for $\lambda$ non-negative and dominant. However, the representation of $\mathfrak{gl}(2n, \mathbb{C})$ induced by an irreducible representation of $\mathfrak{sl}(2n, \mathbb{C})$ has the same highest weight and restricts back to itself. See \S 15.3 in \cite{fultonharris}.
\end{rem}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\section{Littlewood-Richardson tableaux and n-symplectic Sundaram tableaux: branching}
\label{skewsection}
\begin{defn}
Let $\lambda, \nu \in \p_{A_{2n-1}}$ be two dominant weights such that $\underline{d}_{\nu} \subset \underline{d}_{\lambda}$ (this means that one shape is contained in the other when aligned with respect to their top left corners. In \cref{firstexamplensymplectic} below we see that $\underline{d}_{\nu} \subset \underline{d}_{\lambda}$). A tableau $\mathscr{T}$ of skew shape $\lambda / \nu$ is a filling of an arrangement of boxes of shape $\lambda$ leaving the boxes that belong to $\nu \subset \lambda$ blank, with the others having entries in the alphabet $\mathscr{A}_{2n}$, and such that these entries are strictly increasing along the columns and weakly increasing along the rows. The word $w(\mathscr{T})$ of $\mathscr{T}$ is obtained just as for semi-standard Young tableaux, reading from right to left and from top to bottom, ignoring the blank boxes. 
\end{defn}

A shape $\underline{d} = (d_{1}, \cdots, d_{k})$ is \textbf{even} if $d_{i} = 0$ unless $i$ is an even number, for $i \in \{1, \cdots, k\}$.  Also, for a shape $\underline{d}$ define $l(\underline{d})$  to be the length of the longest column of the associated arrangement of boxes. 

\begin{defn}
Let $\lambda , \nu, \eta \in \p^{+}_{A_{2n-1}}$ be dominant weights such that the shapes $\underline{d}_{\nu}$ and $\underline{d}_{\eta}$ are contained in the shape $\underline{d}_{\lambda}$ of $\lambda$. A \textbf{Littlewood-Richardson} (respectively \textbf{Littlewood-Richardson Sundaram} or \textbf{n-symplectic Sundaram}) tableau of skew shape $\lambda / \nu$ and weight 
$\eta$ is a tableau of skew shape $\lambda / \nu$ that is semi-standard, and has a dominant word of weight $\eta$ (respectively $\underline{d}_{\eta}$ is even and $2i+1 $ does not appear strictly below row  $n+i$ for $i \in \{0, 1, \cdots, \frac{1}{2} l(\underline{d}_{\eta})\}$). Here a word $w \in \mathscr{W}(\mathscr{A}_{2n})$ is dominant if the path $\pi_{w}$ is dominant. We will denote these skew tableaux by $\op{LR}(\lambda/ \nu, \eta)$   (respectively $\op{LRS}(\lambda/\nu, \eta)$). Littlewood-Richardson Sundaram tableaux were introduced by Sundaram in \cite{sundaram}, as \textbf{n-symplectic} tableaux.
\end{defn}

\begin{rem}
\label{redundance1}
Note that if $l(\lambda) \leq n$ (such weights are called \textbf{stable}) then $\op{LRS}(\lambda/\nu, \eta) = \op{LR}(\lambda/\nu, \eta)$. 
\end{rem}

\begin{rem}
\label{redundance2}
If $\lambda$ is stable and $\mathscr{T}$ is a Littlewood-Richardson tableau of skew shape $\lambda / \nu$ then its entries belong to the set $\{1, \cdots, n\}$. This is because if, say, $k$ appears in row $l_{k}$ of $\mathscr{T}$, then, since the word of $\mathscr{T}$ is dominant, a $k-1$ must appear either directly above $k$ in the same column, or in a column to the right, and since $\mathscr{T}$ is semi-standard, it appears in at most row $l_{k} - 1$.
\end{rem}

\begin{ex}
\label{firstexamplensymplectic}
The tableau $\mathscr{L} = \Skew(0:\mbox{},\mbox{\tiny{1}},\mbox{\tiny{1}}|0:\mbox{ },\mbox{\tiny{2}}|0:\mbox{\tiny{2}})$ is  a Littlewood-Richardson tableau of skew shape $\lambda / \nu$ and weight $\eta$ for $\lambda = \omega_{1} + \omega_{2} + \omega_{3}, \nu = \omega_{2},$ and $\eta = 2 \omega_{2}$ and the tableau $\mathscr{T} = \Skew(0: \mbox{ }|0: \mbox{ } |0: \mbox{\tiny{1}})$ is a Littlewood-Richardson tableau of skew shape $\lambda'/ \nu'$ and weight $\eta'$ for $\lambda' = \omega_{3}, \nu' = \omega_{2},$ and $\eta' = \omega_{1}$. Notice that  $\mathscr{L}$ is 2-symplectic Sundaram while $\mathscr{T}$ is not. 
\end{ex}


\begin{defn}
The Littlewood-Richardson coefficient is defined as the number $c^{\lambda}_{\nu, \eta} \in \mathbb{Z}_{\geq 0}$ such that
\begin{align*}
\li(\nu) \otimes \li(\eta) =\underset{ \underline{d}_{\nu} \subset \underline{d}_{\lambda}}{\bigoplus} c^{\lambda}_{\nu, \eta} \li(\lambda)
\end{align*}
\noindent
where $\li(\lambda), \li(\nu),$ and $\li(\eta)$ are all representations of $\mathfrak{sl}(2n, \mathbb{C})$. 
\end{defn}

\noindent
\cref{litrich} below is known as the \textbf{Littlewood-Richardson rule}. It was first stated in 1943 by Littlewood and Richardson. 

\begin{thm}\cite{fulton}
\label{litrich}
The Littlewood-Richardson coefficients are obtained by counting Littlewood-Richardson tableaux: 
\begin{align*}
c^{\lambda}_{\nu, \eta} = |\op{LR}(\lambda/ \nu, \eta)|.
\end{align*}
\end{thm}


\begin{rem}
\label{redundance}
\cref{litrich} implies that $c^{\lambda}_{\nu, \eta} = c^{\lambda}_{\eta, \nu}$. 
\end{rem}

We will use the notation $c^{\lambda}_{\nu, \eta}(\s) = |\op{LRS}(\lambda/ \nu, \eta)|$. The following theorem was proven by Sundaram in Chapter IV of her PhD thesis \cite{sundaram}. See also Corollary 3.2 of \cite{sundarampaper}. For stable weights it was proven by Littlewood in \cite{litt} and is known as the Littlewood branching rule. 

\begin{thm} \cite{sundaram}
\label{sundaramstheorem}
Let $\lambda \in \p_{A_{2n-1}}$ be dominant. Then 
\begin{align*}
\op{res}^{\mathfrak{g}}_{\mathfrak{g}^{\sigma}}(\li(\lambda)) = \underset{l(\underline{d}_{\nu}) \leq n}{\underset{\underline{d}_{\nu} \subset \underline{d}_{\lambda}}{\bigoplus}} \N_{\lambda, \nu} \li(\nu)
\end{align*}
\noindent
where 
$\N_{\lambda, \nu} = \underset{\underline{d}_{\eta} \hbox{ \tiny{even} }}{\sum}c^{\lambda}_{\nu, \eta}(\s).$
\end{thm}

\section{On the proof of the Naito-Sagaki conjecture.}
\label{7}

To prove the conjecture, in \cite{beajaz} we construct a bijection
\begin{align}
\label{thebijection}
\op{domres}(\lambda, \nu) \overset{1:1}{\longleftrightarrow} \underset{\underline{d}_{\eta} \hbox{ \tiny{even} }}{\underset{\underline{d}_{\eta} \subseteq \underline{d}_{\lambda};}{\bigcup}} \op{LRS}(\lambda/\eta, \nu). 
\end{align}

To do so we have used combinatorics of ``up-down tableaux'', and  a result by Sundaram and Berele which is in spirit of a symplectic Robinson-Schensted Knuth correspondence. Here we provide a direct bijection in the case of  $n = 2$ and $\lambda = a_{1}\omega_{1}+a_{2}\omega_{2}+a_{3}\omega_{3}$, for all $n$. The advantage of this proof is highlighted in \cref{domresviainequalities}. The following construction should provide some insight. Given a tableau $\mathscr{T} \in \op{domres}(\lambda)$ we will construct a weight $\eta_{\mathscr{T}}$ with even shape $\underline{d}_{\eta_{\mathscr{T}}}$. To do this, first replace, in $\mathscr{T}$, all letters $w$ such that $n< w \leq 2n$ by $\overline{2n-w+1}$, and denote the resulting symplectic semi-standard tableau by $\op{res}(\mathscr{T})$. Its word $w(\op{res}(\mathscr{T}))$ is equal to the restricted word $\op{res}(w(\mathscr{T}))$. Now, in each column of $\op{res}(\mathscr{T})$, replace an entry $w$ by a blank square if $\overline{w}$ appears to its left in the word  $\op{res}(w(\mathscr{T})) = w(\op{res}(\mathscr{T}))$. In that case, replace the entry $\overline{w}$ by a blank square as well. Count the number of blank squares in each column, and order these squares to obtain an arrangement of boxes of  shape $\underline{d}_{\eta_{\mathscr{T}}}$.

 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{ex}
\label{exevenshape}
Let $n = 2$ and 
\begin{align*}
\mathscr{T} = \Skew(0:\mbox{\tiny{1}},\mbox{\tiny{1}},\mbox{\tiny{1}}|0:\mbox{\tiny{2}},\mbox{\tiny{2}}|0:\mbox{\tiny{3}}).
\end{align*} 

Then 
\begin{align*}
\op{res}(\mathscr{T}) = \Skew(0:\mbox{\tiny{1}},\mbox{\tiny{1}},\mbox{\tiny{1}}|0:\mbox{\tiny{2}},\mbox{\tiny{2}}|0:\mbox{\tiny{$\overline{2}$}}),\mbox{      } \eta_{\mathscr{T}} = \omega_{2}, \hbox{ and } \underline{d}_{\eta_{\mathscr{T}}} = (0,1)
\end{align*}
where the arrangement of boxes $\Skew(0: \mbox{ }|0: \mbox{ })$ of shape $\underline{d}_{\eta_{\mathscr{T}}}$ is obtained by replacing $\Skew(0: \mbox{\tiny{2}}|0: \mbox{\tiny{$\overline{2}$}})$ in $\op{res}(\mathscr{T})$ by blank squares.
\end{ex}

\begin{lem}
\label{evennumberofboxes}
Let $\mathscr{T}$ be as above. Then the shape $\underline{d}_{\eta_{\mathscr{T}}}$ is even. 
\end{lem}
\begin{proof}[Idea of proof]
The proof is by induction on the number of right-most aligned columns of $\mathscr{T}$ and is a consequence of the semi-standardness of $\mathscr{T}$.
\end{proof}

\begin{rem}
\cref{evennumberofboxes} is only true for $\mathscr{T}$ a semi-standard Young tableau. Consider for example $n=2$ and the key (as in \cite{knuth}, \cite{jt}) $\mathscr{T} = \Skew(0:\hbox{\tiny{1}}, \hbox{\tiny{4}}, \hbox{\tiny{1}})$. Then $\op{res}(\mathscr{T}) = \Skew(0:\hbox{\tiny{1}}, \hbox{\tiny{$\overline{1}$}}, \hbox{\tiny{1}})$ is dominant, however, the shape $\eta_{\mathscr{T}} = (1,0)$ is not even.
\end{rem}


\begin{lem}
\label{easylemma}
If $\lambda = a_{1} \omega_{1} + a_{2}\omega_{2} + a_{3} \omega_{3}$
and $\nu$ and $\eta$ are dominant weights in $\p_{A_{2n-1}}$ such that $\underline{d}_{\eta}$ is even and  both $\underline{d}_{\eta}$ and $\underline{d}_{\nu}$ are contained in $\underline{d}_{\lambda}$, then 
\begin{align*}
c^{\lambda}_{\nu, \eta}(\s) = c^{\lambda}_{\nu, \eta}.
\end{align*}
\end{lem}

\begin{proof}
Assume that $\mathscr{T}$ is a Littlewood-Richardson tableau of skew shape $\lambda / \nu$ and weight $\eta$ that is not Sundaram. This means that there is at least a ``1'' in the third row. This is impossible due to semi-standardness of $\mathscr{T}$, the dominance of its word, and $\underline{d}_{\eta}$ being even.  
\end{proof}



\begin{proof}[Proof of \eqref{thebijection} for $n=2$ or $\lambda =  a_{1} \omega_{1} + a_{2}\omega_{2} + a_{3} \omega_{3}$]
Let  $ \mathscr{T} \in \op{domres}(\lambda, \nu)$, and set 
\begin{align*}
m_{1} &= \hbox{ \# columns in} \mathscr{T} \hbox{ of the form } \Skew(0: \mbox{\tiny{1}}|0: \mbox{\tiny{2}} |0: \mbox{\tiny{$\overline{1}$}}),\\
m_{2} &= \hbox{ \# columns in }\mathscr{T}\hbox{ of the form } \Skew(0: \mbox{\tiny{1}}|0: \mbox{\tiny{2}} |0: \mbox{\tiny{\mbox{\tiny{$\overline{2}$}}}}), \\
m_{3} &= \hbox{  \# columns in }\mathscr{T}\hbox{ of the form } \Skew(0: \mbox{\tiny{1}}|0: \mbox{\tiny{$\overline{1}$}})
\hbox{ and }\\
m_{4} &= \hbox{  \# columns in }\mathscr{T}\hbox{ of the form \Skew(0: \mbox{\tiny{1}}|0: \mbox{\tiny{2}} |0: \mbox{\tiny{3}})} .
\end{align*}
\noindent
If $n = 2$ then $m_{2} = m_{4}$. It follows from semi-standardness and dominance of $\op{res}(w(\mathscr{T}))$ that these are the only possible columns aside from columns of the form $\Skew(0: \mbox{\tiny{1}}|0: \mbox{\tiny{2}})$ and the single box columns $\Skew(0:1)$. Note that since $\op{res}(w(\mathscr{T}))$ is dominant, the following condition  holds
\begin{align}
\label{domi}
m_{1} \leq \lambda_{1} - \lambda_{2}. 
\end{align}
\noindent
Actually \eqref{domi} is equivalent to the dominance of $\op{res}(w(\mathscr{T}))$, once the $\lambda_{i}$'s are set. We assign to $\mathscr{T}$ a Littlewood-Richardson tableau $\varphi(\mathscr{T}) \in \op{LR}(\lambda/\eta_{\mathscr{T}}, \nu)$. By \cref{evennumberofboxes}, $\eta_{\mathscr{T}}$ is even.

\noindent
 Write $\lambda = \lambda_{1} \e_{1} + \lambda_{2}\e_{2} + \lambda_{3}\e_{3}.$ Note that $\eta_{\mathscr{T}}$ has $m = m_{1}+m_{2} +m_{3}$ columns, all of length 2. Fill in the first $\lambda_{1} - m$ right-most boxes in the first row with a ``1'' , and the first $\lambda_{2} - b$ right-most boxes in the second row with a ``2''. If $n \neq 2$ fill in the first $m_{4}$ right-most entries of the third row with a ``3''. Then \label{conn2}fill in the next right-most $m_{1}$ entries in the third row with a ``2'', and the remaining entries with a ``1''.  The resulting tableau $\varphi(\mathscr{T})$ is a Littlewood-Richardson tableau by construction. Now we will show that any element in $ \underset{\eta \hbox{ \tiny{even} }}{\underset{\underline{d}_{\eta} \subset \underline{d}_{\lambda};}{\bigcup}} \op{LR}(\lambda/\eta, \nu)$ can be obtained in this way.
Let $\eta \in \p^{+}_{A_{2n-1}}$ have an even shape $\underline{d}_{\eta} \subset \underline{d}_{\lambda}$ (this means $\underline{d}_{\eta}$ consists of size 2 columns)  and let $\mathscr{L} \in \op{LR}(\lambda/\eta, \nu)$. Set
\begin{align*}
l_{1} &= \hbox{ \# of 1's in } \mathscr{L} \\
l_{2} &= \hbox{ \# of 2's in } \mathscr{L}, \hbox{ and } \\
b &=  \hbox{\# of columns of } \eta.
\end{align*} 

\noindent
Note that this information determines $\mathscr{L}$ together with $\lambda$. In view of the previous construction, we would like to find non-negative integers $m_{1}, m_{2}, m_{3}$ and $m_{4}$ such that 
\begin{align}
\label{1inverseconditionn2}
l_{1} &= \lambda_{1} - m_{1} - m_{3} \\
\label{2inverseconditionn2}
l_{2} &= \lambda_{2} - m_{3} -m_{2} \\
\label{3inverseconditionn2}
m &= m_{1} + m_{2} + m_{3}\\
m_{4} & = \hbox{ \# of 3's in } \mathscr{L}
\end{align}
\noindent
Since $\mathscr{L}$ has a dominant word, we have
\begin{align}
\label{dominverse}
l_{2} - (\lambda_{2} - b) \leq \lambda_{1} - \lambda_{2}.
\end{align}

\noindent
Substituting \eqref{2inverseconditionn2} and \eqref{3inverseconditionn2} in \eqref{dominverse} we get precisely \eqref{domi}, so if we find solutions $m_{1},m_{2}, m_{3}$, the resulting tableau will automatically belong to $\op{domres}(\lambda, \nu)$. \cref{systemofeqnssolis} below assures that this is the case. 
\end{proof}

\begin{cl}
\label{systemofeqnssolis}
The system determined by \eqref{1inverseconditionn2}, \eqref{2inverseconditionn2}, and \eqref{3inverseconditionn2}, has integer solutions (possibly zero) $b_{1}, b_{2}$ and $b_{3}$ if and only if 
\begin{align}
\label{finalexistence1}
m &\geq \lambda_{1} - l_{1} \\
\label{finalexistence2}
m &\geq \lambda_{2} -  l_{2}\\
\label{finalexistence3}
\lambda_{1} + \lambda_{2} &\geq m +l_{1} + l_{2}.
\end{align}
\end{cl}

\begin{proof}[Idea of proof]
The proof is a straightforward solution of the system of equations determined by \eqref{1inverseconditionn2} - \eqref{3inverseconditionn2}.
\end{proof}

\noindent
It follows from the definitions that these conditions are satisfied by all elements of
\begin{align}
\label{lrstabs}
\underset{\underline{d}_{\eta} \hbox{ \tiny{even} }}{\underset{\underline{d}_{\eta} \subset \underline{d}_{\lambda};}{\bigcup}} \op{LR}(\lambda/\eta, \nu).
\end{align}


\begin{rem}
\label{domresviainequalities}
A first direct consequence of our proof is a description of the left and right hand sides of \eqref{thebijection} as lattice points of two convex polytopes. In fact, the inequalities and variables used to describe \eqref{lrstabs} figure in Section 3 of \cite{lrcones}. In \cite{beajaz} we exhibit this as a general phenomenon. The approach taken in \cite{beajaz} to prove \cref{naitosagakiconjecture} is to construct a combinatorial bijection \eqref{thebijection} as is explained in \cref{7}. It would be interesting to know if there is a bijection, for all n, (as in \cref{7}) that comes from a unimodular equivalence.
\end{rem}

\acknowledgements{The author would like to thank St\'ephane Gaussent for introducing her to this problem and Bea Schumann for some very interesting conversations. Thanks also to Daisuke Sagaki for the nice correspondence and to Peter Littelmann for his support and guiding remarks. Thank you as well to Michael Ehrig, St\'ephane Gaussent, Bea Schumann and Noah White for their comments on a preliminary version.}

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