Proof of Proposition 8.3

First, we have to state the $q$-analogue of [O2, Lemma 8.1], one of the principal ingredients in the proof of the symplectic straightening algorithm in the classical case. In order to define the quantum analogue to the ideal $N$ considered there we have to look in more detail at the elements $c_i$ and $d_i$ defined in the previous section.

Relation (16) implies $c_i\wedge d_i=d_i\wedge c_i=0$. Consequently we get from (14) and (15)

$$d_i^2:=d_i \wedge d_i =(y-1)\sum_{j=i+1}^m d_i\wedge d_j. \;\;\; \;\;\; c_i^2 =(y^{-1}-1)\sum_{j=i+1}^m c_i\wedge c_j$$ (22)

This stands in remarkable contrast to the classical and even quantum linear case where such expressions vanish. On the other hand by (13) and the above explanations the elements $c_i$ and $d_j$ commute with each other, exactly as in the classical case. Consequently, the elements ${d_K}$$:=d_{k_1}\wedge d_{k_2} \wedge \ldots \wedge d_{k_a}$ are defined independently of the order of the elements of the subset $K:=\{k_1, \ldots, k_a\}\subseteq\underline{m}$. Again, we write $P({m},{a})$ for the collection of all subsets $K$ of $\underline{m}$ whose cardinality is $a$. Set

$${D_{a}}$$$$\mbox{\index{${D_{a}}$}}$$$$:=\sum_{K\in P({m},{a})} d_K$$

and let ${N}$ be the ideal in ${\bigwedge}_{R,q}(n)$ generated by the elements $D_1, D_2, \ldots , D_m$. We call an ordered subset $I\in P({n},{r})$ reverse symplectic if the multi-index ${\bf i}w$ obtained from $I$ by ordering its elements according to $\prec$ (obtained from ${\bf i}$ by a suitable permutation $w \in {\cal S}_{r}$ such that $i_{w(1)}\prec i_{w(2)}\prec \ldots \prec i_{w(r)}$) is $\omega_r$-reverse symplectic standard. Here $\omega_r$ is the $r$-th fundamental weight.

Proposition 12.1   Let $I\in P({n},{r})$ be non reverse symplectic. Then, to each $J \in P({n},{r})$ such that the inequality $f({\bf j})<f({\bf i})$ holds for the corresponding multi-indices ${\bf i}$ and ${\bf j}$, there exists $a_{IJ}\in R$ such that in ${\bigwedge}_{R,q}(n)$ the following congruence relation holds:

$$v_{I}\equiv \sum_{J\in P({n},{r}), \; f({\bf j})<f({\bf i})}a_{IJ}v_{J} \;\;$$    mod $$\;\; N.$$

Proposition 12.2   The semi bialgebra $A^{{\rm sh}}_{R,q}(n)$ acts trivially on the elements $D_a$, that is $\tau_{\wedge}(D_a)=0$.

We postpone the very technical proofs of both propositions to separate sections below.

Let us prove Proposition 8.3 in the case $\lambda=\omega_r$, first. Take ${\bf j}\in I(n,r)\backslash I_{\omega_r}^{\rm mys}$. Using the weak part of the straightening algorithm 10.1, we may assume ${\bf j}\in I_{\omega_r}^{}\backslash I_{\omega_r}^{\rm mys}$. This means $j_1\prec j_2\prec \ldots \prec j_r$. In order to apply our lemmas we have to change orders from $\prec$ to $<$. Let $w \in {\cal S}_{r}$ be such that $j_{w(1)}<j_{w(2)}<\ldots <j_{w(r)}$, that is, ${\bf j}w$ is a multi-index corresponding to a non reverse symplectic ordered set $J \in P({n},{r})$ in the sense of Proposition . Application of this proposition to $v_{J}$ yields

$$X:=v_{J}- \sum_{K\in P({n},{r}), \; f({\bf k})<f({\bf j})}a_{{\bf j}{\bf k}}v_{K} \;\; \in N$$

since $f({\bf j})=f({\bf j}w)$. According to Proposition 12.2, the element $\tau_{\wedge}(X)$ must be zero. Applying (11.6), we obtain the following equation holding in ${\bigwedge}_{R,q}(n,r)\otimes {\cal K}$:

$$\sum_{I\in P({n},{r})}v_{I}\otimes \left(T^{\omega_r}_q({\bf i}:... ...k} \lhd {\bf j}}a_{{\bf j}{\bf k}} T^{\omega_r}_q({\bf i}:{\bf k})\right)=0.$$

Since $\{v_{I}\vert\; I \in P({n},{r})\}$ is a basis of ${\bigwedge}_{R,q}(n,r)$, each individual summand in the summation over $P({n},{r})$ must be zero. Together with Corollary 9.12, this gives the desired result in the case of multi-indices ${\bf i}$ corresponding to ordered subsets $I\in P({n},{r})$, that is ${\bf i} \in I_{\omega_r}^{<}$. The case for general ${\bf i}$ can be deduced from this using

$$T^{\omega_r}_q({\bf i}:{\bf j}) =\sum_{K\in P({n},{r})} \nu_{{\bf i}K}T^{\omega_r}_q({\bf k}:{\bf j}),$$

which follows from the formula $\kappa_r=\nu_r\circ \pi_r$ of Lemma 11.7 together with (19).

Next we consider the general case of $\lambda$. Here, we can proceed exactly as in the classical case. Again, we may assume ${\bf j}\in I_{\lambda}^{}\backslash I_{\lambda}^{\rm mys}$ by the weak part of the straightening algorithm. Let $\lambda'=(\mu_1, \ldots , \mu_p)$ be the dual partition ( $p=\lambda_1$). We split ${\bf j}$ into $p$ multi-indices ${\bf j}^l\in I(n,\mu_l)$, where for each $l\in \underline{p}$ the entries of ${\bf j}^l$ are taken from the $l$-th column of $T^{\lambda}_{{\bf j}}$. The same thing can be done with ${\bf i}$. Since ${\bf j}$ is not $\lambda$-reverse symplectic standard but standard there must be a column $s$ such that ${\bf j}^s$ is not $\omega_{\mu_s}$-reverse symplectic standard. Applying the result to the known case of $T^{\omega_{\mu_s}}_q({\bf i}^s:{\bf j}^s)$, we obtain

$$T^{\lambda}_q({\bf i}:{\bf j})= T^{\omega_{\mu_1}}_q({\bf i}^1:{... ...mu_s}}_q({\bf i}^s:{\bf j}^s) \cdots T^{\omega_{\mu_p}}_q({\bf i}^p:{\bf j}^p)$$

$$\equiv \sum a_{{\bf j}^s{\bf k}^s} T^{\omega_{\mu_1}}_q({\bf i}^... ...{\bf j}^p)\;\; = \;\; \sum a_{{\bf j}{\bf k}} T^{\lambda}_q({\bf i}:{\bf k}).$$

The element ${\bf k}^s\in I(n,\mu_s)$ satisfies ${\bf k}^s\lhd{\bf j}^s$, ${\bf k}\in I(n,r)$ is constructed from ${\bf j}$ by replacing the entries of ${\bf j}^s$ by that of ${\bf k}^s$ and $a_{{\bf j}{\bf k}}$ is the same as $a_{{\bf j}^s{\bf k}^s}$ for the corresponding ${\bf k}^s$. The product formula for bideterminants applied above is valid by our choice of basic $\lambda$-tableaux inserting the numbers $1, \ldots , r$ column by column top down (otherwise the non-commutativity of $A^{{\rm s}}_{R,q}(n)$ would cause some trouble). From (11) we see ${\bf k}\lhd {\bf j}$ and the proof of 8.3 is completed.