The Weak Straightening Algorithm

We are now able to give the proof of the following weak form of the straightening algorithm.

Proposition 10.1 (Weak Quantum Symplectic Straightening Algorithm)   Let
$\lambda \in \Lambda^+(r)$ be a partition of $r$ and ${\bf j} \in I(n,r) \backslash I_{\lambda}^{}$. Then to each ${\bf k}\in I(n,r)$ satisfying ${\bf k}\lhd {\bf j}$ there is an element $a_{{\bf j}{\bf k}} \in R$ such that in ${\cal K}$ the following congruence relation holds for all ${\bf i} \in I(n,r)$:

$$T^{\lambda}_q({\bf i}:{\bf j}) \equiv \sum_{{\bf k}\lhd {\bf j}} a_{{\bf j}{\bf k}} T^{\lambda}_q({\bf i}:{\bf k}) \; \;$$   mod $$\; \; {\cal K}(>\lambda) .$$

PROOF: We divide into the following two cases

1. ${\bf j}$ is not $\lambda$-column standard.
2. ${\bf j}$ is $\lambda$-column standard but not $\lambda$-row standard.

Case 1:

By assumption there are two consecutive indices $j_l$ and $j_{l+1}$ in ${\bf j}=(j_1, \ldots , j_r)$ such that $j_l\succeq j_{l+1}$ and $s_l=(l, l+1)\in {\cal S}_{\lambda '}$. If $j_l=j_{l+1}$, we have $T^{\lambda}_q({\bf i}:{\bf j})=0$ by Corollary 9.2, implying our assertion. In the case $j_{l+1} \succ j_{l}$ we apply Corollary 9.12:

$$T^{\lambda}_q({\bf i}:{\bf j}) =a_{{\bf j}}(s_l)T^{\lambda}_q({\... ...{\bf j}s_l) \; +\; \sum a_{{\bf j}{\bf k}}(s_l)T^{\lambda}_q({\bf i}:{\bf k}).$$

The multi-indices ${\bf k}$ in the sum satisfy $f({\bf k}) < f({\bf j})$ and consequently ${\bf k}\lhd {\bf j}$. Finally, since $f({\bf j})=f({\bf j}s_l)$ and ${\bf j}s_l$ occurs before ${\bf j}$ in the lexicographic order on $I(n,r)$ we have ${\bf j}s_l\lhd {\bf j}$ as well.

Case 2:

In principle we follow the lines of the proof of [Ma, 2.5.7]. But since $A^{{\rm s}}_{R,q}(n)$ is not commutative we have to work with a fixed basic tableau. The change of basic tableaux in [Ma, 2.5.7] can be compensated for by Lemma 9.8.

To start, let $l \in \underline{r}$ be the smallest index such that $j_l$ is larger than its right hand neighbour $j_{l'}$ in the $\lambda$-tableau of ${\bf j}$. Assume that the entry $j_l$ lies in the $s$-th column ${\bf j}^{s}_{\lambda}$ and that $j_{l'}$ lies in the $s+1$-th column ${\bf j}^{s+1}_{\lambda}$, where $1 \leq s < \lambda_1$. Clearly, $l'=l+\lambda'_s$. Let $t$ be the index of the row containing both entries. We picture this by

$$T^{\lambda}_{{\bf j}} = \ldots \begin{array}{\vert c\vert c\vert ... ...cline{2-2} j_{l+1} & j_{l'+1} & t+1 \\ \vdots & \vdots \end{array} \ldots .$$

By assumption we have $\ldots \prec j_{l'-1}\prec j_{l'} \prec j_l \prec j_{l+1} \prec \ldots$. Now, we refine the dual partition $\lambda'$ of $\lambda$ to a composition $\eta \in \Lambda(p+2, r)$, where $p:=\lambda_1$ is the number of columns of the diagram of $\lambda$. More precisely, we split the $s$-th and $(s+1)$-th column in front of and below the $t$-th row:

$$\eta_i:=\left\{ \begin{array}{ll} \lambda'_i & i < s \\ t-1 & ... ...eta_{s+1}+\eta_{s+2} & i=s+1 \\ \eta_{i+1} & i > s+2 \end{array} \right. .$$

Obviously, this $\eta$ is the coarsest refinement of the partition $\lambda'$ and the composition $\mu \in \Lambda(p+1, r)$ defined above. Let us split the multi-index ${\bf j}$ according to $\eta$ as follows:

$${\bf j}^{s}_{\eta}=(j_h, \ldots , j_{l-1}), \;\;\; {\bf j}^{s+1}_{\eta}=(j_l, \ldots , j_{h+k-1}),$$

$${\bf j}^{s+2}_{\eta}=(j_{h+k}, \ldots , j_{l'}), \;\;\; {\bf j}^{s+3}_{\eta}=(j_{l'+1}, \ldots , j_{h+k+k'-1}).$$

Here, $h:=l-t+1=\lambda'_1+\ldots +\lambda'_{s-1}+1$ is the index of the first entry of the $s$-th column and $k:=\lambda'_s$ (resp. $k':=\lambda'_{s+1}$) are the lengths of both columns in question. We have

$${\bf j}^{s}_{\lambda}={\bf j}^{s}_{\eta} +{\bf j}^{s+1}_{\eta}, \; \; \; {\bf j}^{s+1}_{\lambda}={\bf j}^{s+2}_{\eta} +{\bf j}^{s+3}_{\eta} \; \;$$    and set $$\; \; {\bf j}^{s+1}_{\mu} :={\bf j}^{s+1}_{\eta} +{\bf j}^{s+2}_{\eta}.$$

In order to apply Laplace Duality 9.7 to the pair $(\lambdaŽ, \mu)$ of compositions we have to choose coset representatives of ${\cal S}_{\eta}={\cal S}_{\lambda'}\cap {\cal S}_{\mu}$ in ${\cal S}_{\lambda'}$ and ${\cal S}_{\mu}$ carefully. From the theory of parabolic subgroups of reflection groups it is known that each right coset of ${\cal S}_{\eta}$ in ${\cal S}_{\mu} \cong {\cal S}_{\mu _1}\times \ldots \times {\cal S}_{\mu _{p+1}}$ contains a unique element of minimal length called the distinguished right coset representative, in fact one looks for coset representatives of ${\cal S}_{\eta_{s+1}}\times {\cal S}_{\eta_{s+2}}$ in ${\cal S}_{\mu _{s+1}}$. We choose these for our set $X$. The property $l(wu)=l(w)+l(u)$ for $u\in X$ and $w \in {\cal S}_{\eta}$ follows from that theory as well. Similarily, one finds a set $Y$ of distinguished left coset representatives of ${\cal S}_{\eta}$ in ${\cal S}_{\lambda'}$ satisfying $l(vw)=l(v)+l(w)$ for $w \in {\cal S}_{\eta}$ and $v \in Y$.

We will not apply Laplace Duality to the original index pair ${\bf i}, {\bf j}$, for we must handle the transition from the order $<$ to $\prec$. Instead of ${\bf j}$ we rather consider ${\bf j'}:={\bf j}w$ where $w\in {\cal S}_{\mu _{s+1}}\subseteq{\cal S}_{r}$ is chosen in such a way that $j'_l<j'_{l+1}<\ldots < j'_{l'-1} < j'_{l'}$ and $j'_i=j_i$ for $1\leq i <l$ and $l'<i\leq r$ (the embedding of ${\cal S}_{\mu _{s+1}}$ is understood according to the composition $\mu$). This $w$ exists uniquely since ${\bf j}^{s+1}_{\mu} =(j_l, \ldots , j_{l'})$ contains exactly $\mu_{s+1}=\lambda_s'+1$ elements by the assumption $j_{h+k}\prec j_{h-k+1} \prec \ldots \prec j_{l'}\prec j_l \prec \ldots \prec j_{h+k-1}$ on ${\bf j}$. Now, by Laplace-Duality we obtain

$$\sum_{u \in X} (-y)^{-l(u)} T^{\lambda}_q({\bf i}:{\bf j'})\wr \beta(u)= \sum_{v \in Y}(-y)^{-l(v)}\beta(v)\wr t_q^{\mu}({\bf i}:{\bf j'}).$$ (12)

With help of Lemma 9.8 the right hand side of this equation can be written as a linear combination of bideterminants $T^{\mu '}_q({\bf k}:{\bf l})$. Thus the right hand side is seen to lie in ${\cal K}(>\lambda)$ as soon we have shown that $\mu '>\lambda$. But that follows since the longest column being removed from the diagram of $\lambda$ to obtain the diagram of $\mu'$ has length $\lambda_s'$, whereas a column of length $\mu_{s+1}=\lambda_s'+1$ has to be added to the diagram of $\mu'$. On the left hand side of (12) we may apply Corollary 9.14 by construction of the multi-index ${\bf j'}$:

$$(-y)^{-l(u)} T^{\lambda}_q({\bf i}:{\bf j'})\wr \beta(u)={\rm si... ...; +\; \sum (-y)^{-l(u)}a'_{{\bf j'}{\bf k}}(u)T^{\lambda}_q({\bf i}:{\bf k}),$$

the sum running over all ${\bf k}$ satisfying $f({\bf k})<f({\bf j}')=f({\bf j})$. Now, for all $u\in X$ we have $\tilde u:=uw^{-1}\in {\cal S}_{\mu _{s+1}}$ since $w$ lies in ${\cal S}_{\mu _{s+1}}$. Furthermore, there is a unique coset representative $u_0\in X$ satisfying ${\cal S}_{\eta}u_0={\cal S}_{\eta}w$ and this is the only one for which the corresponding $\tilde{u}$ lies in ${\cal S}_{\eta}$. Therefore, in the case $u\neq u_0$ there is an $e$ such that $l\leq e<h+k$ and $h+k\leq \tilde u^{-1}(e)\leq l'$. Choose such an $e$ for each $u\in X$. In doing so, we are assigning a transposition $\hat{u}:=(l,e)$ to each $u$ that is contained in ${\cal S}_{\eta}$. In the case of $u_0$ we set $\hat{u_0}:=\tilde{u_0}\in {\cal S}_{\eta}$. Applying Corollary 9.12 to $\hat u$ one calculates

$$T^{\lambda}_q({\bf i}:{\bf j'}u^{-1})= T^{\lambda}_q({\bf i}:{\b... ...um a_{({\bf j}\tilde u^{-1}){\bf k}}(\hat{u}) T^{\lambda}_q({\bf i}:{\bf k}),$$

where the sum runs over all ${\bf k}$ satisfying $f({\bf k})< f({\bf j}\tilde u^{-1})= f({\bf j})$, again. For these ${\bf k}$ we set

$$\bar a_{{\bf j}{\bf k}}:= \sum_{u\in X} (-y)^{-l(u)}a'_{{\bf j'}... ... {\rm sign}(u) h_{{\bf j'}}(u^{-1})a_{({\bf j}\tilde u^{-1}){\bf k}}(\hat{u}),$$

whereas in the case $f({\bf k})=f({\bf j})$ we write

$$\bar a_{{\bf j}{\bf k}}:=\left\{ \begin{array}{cl} {\rm sign}(u)... ...{\bf j}\tilde u^{-1}\hat{u}\\ 0 & \mbox{ otherwise }. \end{array} \right.$$

Observe that $\bar a_{{\bf j}{\bf j}}$ occurs in the latter definition for $u=u_0$. We assert that for $u\neq u_0$, the multi-index ${\bf k}:={\bf j}\tilde u^{-1}\hat{u}$ occurs before ${\bf j}$ in the lexicographic order with respect to $\prec$. For by construction of $\tilde u$ and $\hat{u}$ we have

$$k_l=j_{\tilde u^{-1}\hat{u}(l)}=j_{\tilde u^{-1}(e)} \in \{j_{h+k}, j_{h+k+1} ,\ldots , j_{l'} \}$$

and consequently $k_l \prec j_l$. But this implies ${\bf k} \prec {\bf j}$, since $k_i=j_i$ for $i<l$. Thus we obtain

$$-\bar a_{{\bf j}{\bf j}}T^{\lambda}_q({\bf i}:{\bf j}) \equiv \... ...f k}\lhd {\bf j}} \bar a_{{\bf j}{\bf k}} T^{\lambda}_q({\bf i}:{\bf k}) \; \;$$    mod $$\; \; {\cal K}(>\lambda) .$$

Since the coefficient $-\bar a_{{\bf j}{\bf j}}$ is invertible, the asserted congruence relation holds as well. $\Box$

It should be remarked that the proof works with any other order on $\underline{n}$ instead of $\prec$, as well. The proof of the strong part of the algorithm (Proposition 8.3) can be given right now in the (initial) case $r=2$ and we are going to do this not only because it is very instructive, but also because we will need a basis of $A^{{\rm s}}_{R,q}(n,2)$ in order to proceed to the general case.

If $r=2$ there are exactly two partitions in $\Lambda^+(m, 2)$ for $m\geq 2$, namely $2\omega_1$ and $\omega_2$, where $\omega_1=(1)$ and $\omega_2=(1,1)$ are the fundamental weights (see section 3). In the first case we have $I_{2\omega_1}^{}=I_{2\omega_1}^{\rm mys}$, that is, the weak and the strong form of the straightening algorithm coincide. Turning to $\omega_2$ there is exactly one element in $I_{\omega_2}^{}\backslash I_{\omega_2}^{\rm mys}$, namely ${\bf j}=(m,m')$. By (4.3) we obtain in ${\cal K}=A^{{\rm sh}}_{R,q}(n,2)$

$$T^{\omega_2}_q({\bf i}:(m,m'))= -q^m\sum_{i=1}^{m-1}q^{-i}T^{\omega_2}_q({\bf i}:(i,i'))$$

yielding Proposition 8.3 in the case $r=2$ since $(i,i')\lhd (m, m')$ for all $i <m$.

Remark 10.2   If we had used the notion of symplectic standard tableau instead of the reverse version we would have to consider $(1',1)$ instead of $(m,m')$ in the last step above. This would force us to work with a reverse version of the order choosen on ${\mathbb{N}}\,_0^m$ (as in [O2, section 7]). But this would cause some trouble concerning Lemma 9.9. One way out could be a manipulation of the Yang-Baxter operator $\beta$ conjugating it by the twofold tensor product of the appropriate permutation on $\underline{n}$. Thus, one has to decide between working with the familiar version of $\beta$ or following the familiar notion of tableaux.