Quantum Symplectic Bideterminants

Let $p,r\in{\mathbb{N}}\,$ be positive integers and ${\Lambda(p, r)}$ denote the set of compositions of $r$ into $p$ parts. These are $p$-tuples $\lambda =(\lambda_1,\ldots , \lambda_p)$ of non-negative integers $\lambda_i\in{\mathbb{N}}\,_0$ summing up to $r$. To each composition $\lambda \in \Lambda(p, r)$ there corresponds a parabolic subgroup in the symmetric group ${{\cal S}_{r}}$, called the standard Young subgroup. We will denote it by ${{\cal S}_{\lambda}}$. It is the subgroup fixing the sets $\{1, 2, \ldots, \lambda_1\},\; \{\lambda_1+1, \lambda_1 +2, \ldots, \lambda_1+\lambda_2\},\ldots$. Now, let $w \in {\cal S}_{r}$ be given by a reduced expression $w=s_{i_1}s_{i_2}\ldots s_{i_t}$, where the ${s_i}$$=(i,i+1)$ are the simple transpositions. We define endomorphisms

$${\beta (w)}$$$$\mbox{\index{${\beta (w)}$}}$$$$:= \beta_{i_1}\beta_{i_2} \cdots \beta_{i_t} \in {\rm End}_{R}{(V^{\otimes r})}:={\rm End}_{R}{(V^{\otimes r})}$$

for $r>1$ and set $\beta (w)={\rm id}_{V} \in {\cal E}$ for $r=1$. It is easy to see that this definition is independent of the choice of the reduced expression for $w$ since any two of them can be transformed into each other using the braid relations. But $\beta$ satisfies the quantum Yang-Baxter equation which is just the second type braid relation

$$\beta_i\beta_{i+1}\beta_i=\beta_{i+1}\beta_i\beta_{i+1}$$

in the case $i=1$. The latter one obviously implies the relations for $i>1$, whereas the first type braid relations $\beta_i\beta_j=\beta_j\beta_i$ for $\vert i-j\vert>1$ hold trivially. Observe that

$$\beta (ww') =\beta (w)\beta (w') \;\;$$ if $$\;\; l(ww')=l(w)+l(w').$$

where ${l(w)}$ denotes the length of $w$, that is the number of transpositions in a reduced expression. Setting ${y}$$:=q^2$ and using our notation (2) we associate a quantum symplectic bideterminant to each triple consisting of a composition $\lambda$ of $r$ and a pair of multi-indices ${\bf i},{\bf j}\in I(n,r)$ by

$${t_q^{\lambda}({\bf i}:{\bf j})} \mbox{\index{${t_q^{\lambda}({\bf i}:{\bf j})}$}} := \sum_{w \in {\cal S}_{\lambda}} (-y)^{-l(w)}\beta (w)\wr x_{{... ...sum_{w \in {\cal S}_{\lambda}} (-y)^{-l(w)}x_{{\bf i} {\bf j}}\wr \beta (w) .$$ (6)

The equality therein follows from (5) applied to $\mu=\beta (w)$. Using the abbreviation ${\kappa_{\lambda}}$$:=\sum_{w \in {\cal S}_{\lambda}} (-y)^{-l(w)}\beta (w)\;\; \in {\rm End}_{R}{(V^{\otimes r})}$ we also may write $t_q^{\lambda}({\bf i}:{\bf j})=\kappa_{\lambda}\wr x_{{\bf i} {\bf j}}=x_{{\bf i} {\bf j}}\wr \kappa_{\lambda}$. If $q$ is set to $1$, we obtain

$$x_{{\bf i} {\bf j}}\wr \beta (w)=x_{{\bf i} ({\bf j}w^{-1})}\;$$    and $$\; \beta (w) \wr x_{{\bf i} {\bf j}}=x_{({\bf i}w) {\bf j}}$$

since then $\beta (w)_{{\bf k}{\bf j}}=\delta_{{\bf k}{\bf j}w^{-1}}= \delta_{{\bf k}w\;{\bf j}}$. Therefore, in this case our quantum symplectic bideterminants coincide with ordinary bideterminants which are defined as products of minor $\lambda_i\times \lambda_i$-determinants, one factor for each entry $\lambda_i$ of the composition $\lambda$. According to familar notation we write for a partition $\lambda$

$${T^{\lambda}_q({\bf i}:{\bf j})}$$$$\mbox{\index{${T^{\lambda}_q({\bf i}:{\bf j})}$}}$$$$:= t_q^{\lambda'}({\bf i}:{\bf j}).$$

By a partition we mean a composition $\lambda =(\lambda_1,\ldots , \lambda_p)$ ordered decreasingly ( $\lambda_1\geq \lambda_2 \geq \ldots \geq \lambda_p\geq 0$). The subset of $\Lambda(p, r)$ consisting of all partitions will be denoted by ${\Lambda^+(p, r)}$. By ${\lambda'}$ we denote the dual of the partition $\lambda$, that is $\lambda'=(\lambda_1', \ldots, \lambda_s')$ where $s=\lambda_1$ and $\lambda'_i:=\vert\{j\vert\; \lambda_j \geq i\}\vert$. Using this notation one obtains precisely the classical bideterminant $T^{\lambda}({\bf i}:{\bf j})$ (as defined in [Ma, 2.4] for instance) when $q$ is set to $1$. Observe that the capital $T$ notation is more restricted since not all compositions occur as duals of partitions. This makes it necessary to consider $t_q^{\lambda}({\bf i}:{\bf j})$ as well for technical reasons.

It should be remarked that the well known quantum determinants corresponding to the general linear groups (see for example [DD, 4.1.2, 4.1.7], [CP, p. 236], [Tk, p. 152], [Ha1, p. 157]) can be defined in a similar way using the quantum Yang-Baxter operator of type $A$ instead of our $\beta$. In contrast, explicit expressions for quantum symplectic bideterminants become very complicated for $r > 2$ (apart from the case $\lambda =\alpha_r:=(r)\in \Lambda^+(1, r)$ in which case the bideterminants $T^{\alpha_r}_q({\bf i}:{\bf j})$ just are the monomials $x_{{\bf i} {\bf j}}$). Denoting the fundamental weights by $\omega_r:=(1,1, \ldots , 1)\in \Lambda^+(r, r)$ one obtains a single $r\times r$-minor determinant. If $r=2$, explicit expressions are for example

$$\left\vert \begin{array}{cc} x_{k i} & x_{k j} \\ x_{l i} & x_{... ...ght\vert _q:= T^{\omega_2}_q((k,l):(i,j))= x_{k i}x_{l j}-q^{-1}x_{k j}x_{l i}$$

if $k<l, i<j, i\neq j'=n-j+1$ and

$$\left\vert \begin{array}{cc} x_{k i} & x_{k i'} \\ x_{l i} & x_... ...i'}-q^{-2}x_{k i'}x_{l i} -(q^{-2}-1)\sum_{j=1}^{i-1}q^{j-i}x_{k j'}x_{l j},$$

in the cases $k<l, i\leq m$. The calculation of $T^{\omega_3}_q((j,k,l):(i,i',i))$ for $j<k<l, i\leq m$ is really hard work. Note that such a bideterminant might be different from zero even if it contains two identical columns.