The Quantum Symplectic Straightening Formula

In the classical case symplectic versions of the straightening formula have already been given in [Co, 2.4] and [O2, section 7]. In principle, we will follow the lines of the latter paper. But there are a lot of additional difficulties, one of which forces us to work with a reversed version of $\lambda$-symplectic standard tableaux. To prepare for the statement, we define the algebra

$${A^{{\rm sh}}_{R,q}(n)}$$$$\mbox{\index{${A^{{\rm sh}}_{R,q}(n)}$}}$$$$:=A^{{\rm s}}_{R,q}(n)/\left< d_q\right>$$

by factoring out the ideal generated by the quantum coefficient of dilation. Since $d_q$ is homogeneous this algebra is again graded. Let us abbreviate its $r$-th homogeneous summand by ${{\cal K}}$$:={A^{{\rm sh}}_{R,q}(n,r)}$. Since $d_q$ is grouplike the comultiplication $\Delta$ obviously factors to $A^{{\rm sh}}_{R,q}(n)$ and $A^{{\rm sh}}_{R,q}(n,r)$. But $A^{{\rm sh}}_{R,q}(n)$ is not a bialgebra and $A^{{\rm sh}}_{R,q}(n,r)$ are not coalgebras, because the augmentation map $\epsilon$ does not factor. In the classical case if $R=K$ is a field $A^{{\rm sh}}_{K,q}(n)$ equals the coordinate ring of the symplectic semigroup ${{\rm SpH}_{n}(K)}$$:={\rm SpM}_{n}(K)\backslash {\rm GSp}_{n}(K)$ by [O2, remark 7.5]. The missing augmentation map corresponds to the missing unit element in the semigroup. Therefore we call $A^{{\rm sh}}_{R,q}(n)$ a semi bialgebra.

We put an order on the set $\Lambda^+(r)$ of all partitions of $r$, writing $\lambda < \mu$ if and only if $\lambda'$ occurs before $\mu'$ in the lexicographic order. In this order the fundamental weight ${\omega_r}$$:=(1,1, \ldots , 1)\in \Lambda^+(r, r)$ is the largest element, whereas ${\alpha_r}$$:=(r)\in \Lambda^+(1, r)$ is the smallest one. We define ${\cal K}(>\lambda)$ (resp. ${\cal K}(\geq \lambda)$) to be the $R$-linear span in ${\cal K}$ of all bideterminants $T^{\mu}_q({\bf i}:{\bf j})$ such that $\mu > \lambda$ (resp. $\mu \geq \lambda$)(cf. axiom (C3*) of a cellular coalgebra). Clearly ${\cal K}={\cal K}(\geq \alpha_r)$.

Proposition 8.1 (Quantum Symplectic Straightening Formula)   Let $\lambda \in \Lambda^+(r)$ be a partition of $r$ and ${\bf j} \in I(n,r)$. Then, to each ${\bf k}\in I_{\lambda}^{\rm mys}$ there is an element $a_{{\bf j}{\bf k}} \in R$, such that in ${\cal K}$ we have for all ${\bf i} \in I(n,r)$:

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

Before starting to prove this, we deduce its most important consequence:

Corollary 8.2   The set ${\bf B}_r$ generates $A^{{\rm s}}_{R,q}(n,r)$.

PROOF: From the fact that $d_q$ is central in $A^{{\rm s}}_{R,q}(n)$ by Remark 4.1 we see that multiplication by $d_q$ from the right (written as $\cdot d_q$ below) leads to an exact sequence

$$A^{{\rm s}}_{R,q}(n,r-2)\stackrel{\cdot d_q}{\rightarrow } A^{{\rm s}}_{R,q}(n,r) \rightarrow A^{{\rm sh}}_{R,q}(n,r) \rightarrow 0.$$ (9)

for $r>1$. Therefore, using induction on $r$ we can reduce to showing that

$$\{ T^{\lambda}_q({\bf i}:{\bf j})\vert\; \lambda \in \Lambda^+(m, r),\; {\bf i},{\bf j} \in I_{\lambda}^{\rm mys}\}$$

is a set of generators for ${\cal K}=A^{{\rm sh}}_{R,q}(n,r)$. For this claim it is enough to show that

$${{\bf B}_{\lambda}}$$$$\mbox{\index{${{\bf B}_{\lambda}}$}}$$$$:=\{ T^{\lambda}_q({\bf i}:{\bf j})\vert\;\; {\bf i},{\bf j} \in I_{\lambda}^{\rm mys}\}$$

is a set of generators of ${\cal K}(\geq \lambda)/{\cal K}(>\lambda)$ for each partition $\lambda$. To get the last claim from the straightening formula 8.1, observe that the involution $^*$ is well defined on $A^{{\rm sh}}_{R,q}(n)$ since ${d_q}^*=d_q$ (see section 7). Applying $^*$ to the congruence relation of Proposition 8.1, one obtains another such formula in which the roles of ${\bf i}$ and ${\bf j}$ are exchanged. This shows that ${\bf B}_{\lambda}$ is indeed a set of generators for ${\cal K}(>\lambda)/{\cal K}(\geq \lambda)$. $\Box$

In order to prove the quantum symplectic straightening formula we need a corresponding algorithm. Its classical counterpart is [O2, Proposition 7.3]. We define a map $f:I(n,r)\rightarrow {\mathbb{N}}\,_0^m$ by $f({\bf i})=(a_1, \ldots , a_m)$, where

$$a_l:=\vert\{ j\in \underline{r}\vert\; i_j=l \;$$ or $$\; i_j=l' \}\vert,$$

and order ${\mathbb{N}}\,_0^m$ writing $(a_1,\ldots , a_m)<(b_1, \ldots , b_m)$ if and only if $(b_1, b_{2}, \ldots , b_m)$ appears before $(a_1, a_{2}, \ldots , a_m)$ in the lexicographic order (induced by the ordinary order on ${\mathbb{N}}\,$). Next, we obtain an order $\lhd$ on ${\mathbb{N}}\,_0^m\times I(n,r)$ defined by:

$$(a, {\bf i}){\lhd}$$$$\mbox{\index{${\lhd}$}}$$$$(b, {\bf j}) :\Longleftrightarrow a < b$$    or $$(a = b$$    and $${\bf i} \prec {\bf j}).$$

Here, we have denoted by $\prec$ the lexicographic order on $I(n,r)$ induced by our special order $\prec$ on $\underline{n}$. Finally, we obtain a second order $\lhd$ on $I(n,r)$ via the embedding $I(n,r)\hookrightarrow {\mathbb{N}}\,_0^m\times I(n,r)$ given by ${\bf i}\mapsto (f({\bf i}), {\bf i})$. Now we are able to state the symplectic straightening algorithm.

Proposition 8.3 (Strong Quantum Symplectic Straightening Algorithm)   Let
$\lambda \in \Lambda^+(r)$ be a partition of $r$ and ${\bf j} \in I(n,r) \backslash I_{\lambda}^{\rm mys}$. 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 $$\; \; {\cal K}(>\lambda) .$$

Clearly, the straightening formula 8.1 is an easy consequence of the above proposition since the set $I(n,r)$ is finite and therefore the elimination of multi-indices ${\bf j}$, that are not $\lambda$-reverse symplectic standard in an expression $T^{\lambda}_q({\bf i}:{\bf j})$ must terminate.

The proof of the straightening algorithm will take several sections. In principle we will proceed in a similar way as in [O2] to prove this algorithm, but complications arise because the embedding of the symplectic group into the general linear group does not extend to quantum groups. Instead of [O2, Proposition 7.2] we have to establish a weak form of the quantum symplectic straightening algorithm in a first step. More precisely, we will first prove 8.3, where $I_{\lambda}^{\rm mys}$ is substituted by $I_{\lambda}^{}$. We start with some technical tools.