Quantum Coefficient of Dilation

In the definition of the symplectic monoid ${\rm SpM}_{n}(R)$ we have introduced a function called the coefficient of dilation. This is neccessarily a regular function in the sense of algebraic geometry. Now we will define its quantization which will be called the quantum coefficient of dilation. Using notation (2) we see that

$$-q^{-\rho_k-\rho_l}\epsilon_k\epsilon_l\;\;\gamma\wr x_{(k,k') ... ...')}= q^{-\rho_l}\epsilon_l \sum_{i=1}^nq^{\rho_i}\epsilon_i x_{i l}x_{i' l'}.$$

is independent of $k$, whereas

$$-q^{-\rho_k-\rho_l}\epsilon_k\epsilon_l \;\; x_{(k,k') (l,l')}\w... ... = -q^{-\rho_k}\epsilon_k \sum_{i=1}^nq^{\rho_i}\epsilon_i x_{k i}x_{k' i'}$$

is independent of $l$. But, as $\gamma\wr x_{(k,k') (l,l')}= x_{(k,k') (l,l')}\wr \gamma$ according to (5), both expressions coincide and consequently are independent of both $k$ and $l$. Thus, the element

$${d_q} \mbox{\index{${d_q}$}} := -q^{-\rho_k-\rho_l}\epsilon_k\epsilon_l\;\; \gamma\wr x_{(k,... ...l')}= -q^{-\rho_k-\rho_l} \epsilon_k\epsilon_l \;\;x_{(k,k') (l,l')}\wr \gamma$$ (7)

is well defined in $A^{{\rm s}}_{R,q}(n)$. In fact it is a grouplike element of this bialgebra. More precisely it is the coefficient function of the one dimensional subcomodule of $V\otimes V$ that is spanned by the tensor

$${{J}^*}$$$$\mbox{\index{${{J}^*}$}}$$$$:=\sum_{i=1}^n\epsilon_iq^{\rho_i}v_{i}\otimes v_{i'} \in V\otimes V.$$

To see this, note that ${J}^*=\gamma(-q^{-\rho_l}\epsilon_lv_{l}\otimes v_{l'})$ for each $l$ and that $\gamma$ is a morphism of $A^{{\rm s}}_{R,q}(n)$-comodules. One calculates

$$\tau_{2}({J}^* )= \tau_{2}\circ\gamma(-q^{-\rho_l}\epsilon_lv_{l... ...n(v_{i} \otimes v_{k})\otimes (-q^{-\rho_l}\epsilon_l\;\; x_{(i,k) (l,l')}))=$$

$${J}^*\otimes q^{-\rho_l}\epsilon_l\sum_{k=1}^n q^{\rho_k}\epsilon_k \;x_{(k,k') (l,l')} = {J}^* \otimes d_q.$$

Remark 4.1   The element ${J}^*$ coincides with $\zeta$ from [Ha1, section 6] if $q$ is substituted by $q^{-1}$. Therefore $d_q$ is identical to the grouplike element called quad there. By [Ha1, Corollary 6.3] it is central.

Lemma 4.2   Let $j, k, l \in \underline{n}$. Then we have

$$\sum_{i=1}^j q^{-i}x_{k i}x_{l i'}\wr \beta = \sum_{i=1}^j q^{i-2j}x_{k i'}x_{l i}.$$

PROOF: We calculate

$$\begin{array}{rcl} \sum_{i=1}^{j} q^{-i}x_{k i} x_{l i'}\w... ...q h\geq i \geq j} y^{-i}q^hx_{k h'} x_{l h} \\ \end{array}$$

Since $\sum_{1 \geq h\geq i \geq j} y^{-i}q^hx_{k h'} x_{l h} = \sum_{h=1}^j (\sum_{i=h}^j y^{-i})q^h x_{k h'} x_{l h}$ and $(y-1)\sum_{i=h}^{j}y^{-i} =y^{-h+1}-y^{-j}$ we obtain

$$(y-1)\sum_{1 \geq h\geq i \geq j} y^{-i}q^hx_{k h'} x_{l h} = \sum_{i=1}^j q^i(y^{-i+1}-y^{-j}) x_{k i'}x_{l i}$$

Substituting this into the first equation of the proof leads to our claim. $\Box$

For $l=m$ we deduce the connection formula

Proposition 4.3   The quantum symplectic $2\times 2$-determinants are related to the coefficient of dilation by

$$\sum_{i=1}^mq^{-i}T^{\omega_2}_q((k,l):(i,i'))= \left\{\begin{ar... ...{-k}d_q& k=l' \mbox{ and } k \leq m, \\ 0 & k\neq l'. \end{array} \right.$$

PROOF: By definition of bideterminants and the above lemma we have

$$\begin{array}{rcl} \sum_{i=1}^mq^{-i}T^{\omega_2}_q((k,l):... ...um_{i=1}^n \epsilon_i q^{\rho_i}x_{k i}x_{l i'} \end{array}$$

On the other hand we see

$$\begin{array}{rcl} x_{k m}x_{l m'}\wr \gamma & = & -\sum_... ...um_{i=1}^n \epsilon_i q^{\rho_i} x_{k i}x_{l i'} \end{array}$$

Putting these things together we obtain

$$\sum_{i=1}^mq^{-i}T^{\omega_2}_q((k,l):(i,i')) = -q^{-m-2} x_{k m}x_{l m'}\wr \gamma$$

Since $x_{k m}x_{l m'}\wr \gamma = \gamma\wr x_{k m}x_{l m'}$ it follows that the expression vanishes if $l \neq k'$. In the case $l=k'$ we deduce from (7) the equation $-q^{-m-2}x_{k m}x_{l m'}\wr \gamma = q^{-k}d_q$ which finishes the proof. $\Box$