Tableaux

We will define a basis for $A^{{\rm s}}_{R,q}(n,r)$ consisting of quantum symplectic bideterminants and powers of the quantum symplectic coefficient of dilation. Since they are too large in number we have to single out an appropriate subset. This can be done using so called $\lambda$-tableaux which will be defined now: To each partition one associates a Young-diagram reading row lengths out of the components $\lambda_i$. For example

$$\begin{array}{*{4}{\vert p{2mm}}\vert} \cline{1-3} & & &\multico... ...lumn{2}{c}{} \\ \cline{1-2} &\multicolumn{3}{c}{} \\ \cline{1-1} \end{array}$$

is associated to $\lambda=(3,2,2,1) \in \Lambda^+(4, 8)$. An $\lambda$-tableau $T^{\lambda}_{{\bf i}}$ is constructed from the diagram of $\lambda$ by inserting the components of a multi-index ${\bf i} \in I(n,r)$ column by column into the boxes. In the above example:

$${T^{\lambda}_{{\bf i}}}$$$$\mbox{\index{${T^{\lambda}_{{\bf i}}}$}}$$$$:= \begin{array}{*{4}{\vert p{2mm}}\vert} \cline{1-3} $i_1... ...-2} $i_4$\ &\multicolumn{3}{c}{} \\ \cline{1-1} \end{array}.$$

If $\lambda$ is fixed we will sometimes identify multi-indices with their tableaux. We put a new order ${\prec}$ on the set $\underline{n}$, namely

$$m \prec m' \prec (m-1) \prec (m-1)' \prec \ldots \prec 1 \prec 1'.$$

The reason, why we prefer $\prec$ instead of the order $\ll$ considered in [O2] will become clear later on. Now, a multi-index ${\bf i}$ is called $\lambda$-column standard if the entries in $T^{\lambda}_{{\bf i}}$ are strictly increasing down columns according to this order. It is called $\lambda$-row standard if the entries are weakly increasing along rows and $\lambda$-standard if it is both at the same time. We write ${I_{\lambda}^{}}$ to denote the subset of $I(n,r)$ consisting of all $\lambda$-standard multi-indices. Such a multi-index ${\bf i}\in I_{\lambda}^{}$ is called $\lambda$-reverse symplectic standard if for each index $i\in \underline{m}$ the occurrences of $i$ as well as $i'$ in $T^{\lambda}_{{\bf i}}$ are limited to the first $m-i+1$ rows. The corresponding subset of $I_{\lambda}^{}$ will be denoted by ${I_{\lambda}^{\rm mys}}$. It can be shown that even though this set is different from the one of $\lambda$-symplectic standard tableaux (as defined in [Ki] and denoted $I_{\lambda}^{\rm sym}$ in [O2]), it has the same number of elements. For, let $\sigma \in {\cal S}_{n}$ be the permutation transforming the order $\ll$ into $\prec$, that is $\sigma (i):=(m-i+1)'$ for $i\leq m$ and $\sigma (i):=m-i'+1$ for $i>m$. Then there is an induced bijection on $I(n,r)$ sending $(i_1, \ldots , i_r)$ to $(\sigma (i_1), \ldots , \sigma (i_r))$ and which carries the set of $\lambda$-symplectic standard tableaux precisely to the set of $\lambda$-reverse symplectic standard tableaux.

Here are some examples in the case $m=3$ ( $1'=6, 2'=5, 3'=4$):

$$\begin{array}{*{4}{\vert p{2mm}}\vert} \cline{1-3} $3$\ &... ...{1-2} $2'$\ &\multicolumn{3}{c}{} \\ \cline{1-1} \end{array}$$

The first tableau is an element of $I_{\lambda}^{\rm mys}$ whereas the third is not. The second tableau is an element of $I_{\lambda}^{\rm sym}$. It is obtained from the first one via the bijection induced from the permutation $\sigma$ described above.