Proof of Proposition 12.2

In order to prove the proposition we have to consider generalizations of the elements $D_1, \ldots , D_m$, which are defined for any $l\leq m$ by

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

For a positive integer $k$, define the $y^{-1}$-integer ${\{k\}_{y^{-1}}}$$:=1+y^{-1}+y^{-2}+\ldots +y^{-k+1}\in R$.

Lemma 14.1   Let $a\in \underline{m}$. Then we have

$$D_{1,l}D_{a,l}\equiv\{a+1\}_{y^{-1}}D_{a+1,l}$$

modulo the ideal spanned by $D_1$.

PROOF: Using the above introduced notations we may write the right hand side of (22) as $d_l^2=(y-1)\sum_{i=l+1}^m d_ld_i$. Since $\sum_{i=l+1}^m d_i + d_l + D_{1,l-1}=D_1$, we deduce $d_l^2\equiv (y^{-1}-1)d_lD_{1,l-1}$ modulo $D_1$ if $l>1$ and $d_1^2 \equiv 0$ modulo $D_1$ in the case $l=0$.

We proceed by induction on $l$. If $l=1$, both sides are zero if $a>1$. In the case $a=1$ we have to show that $d_1^2 \equiv 0$ which was proved above.

For the induction step we write $D_{a,l}=d_lD_{a-1,l-1}+D_{a,l-1}$ and obtain

$$D_{1,l}D_{a,l} = d_l^2D_{a-1,l-1}+d_lD_{a,l-1}+D_{1,l-1}(d_lD_{a-1,l-1} +D_{a,l-1})$$

$$\equiv ((y^{-1}-1)\{a\}_{y^{-1}} +1+\{a\}_{y^{-1}})d_lD_{a,l-1} + \{a+1\}_{y^{-1}}D_{a+1,l-1}.$$

Since $(y^{-1}-1)\{a\}_{y^{-1}} +1 +\{a\}_{y^{-1}}=\{a+1\}_{y^{-1}}$ the lemma follows. $\Box$

We introduce some new conventions. To an ordered subset $J=\{j_1, j_2, \ldots , j_a\} \subseteq \underline{m}$ we define corresponding multi-indices by

$${{\bf {j}^2_{\prec}}}$$$$\mbox{\index{${{\bf {j}^2_{\prec}}}$}}$$$$:= (j_1,j_1', j_2, j_2' , \ldots , j_a, j_a' ), \;\;\; {{\bf {j}^2_{<}}}$$$$\mbox{\index{${{\bf {j}^2_{<}}}$}}$$$$:= (j_1, j_2, \ldots , j_a, j_a', j_{a-1}', \ldots , j_1' ) .$$

Furthemore, we write

$${u_J}$$$$\mbox{\index{${u_J}$}}$$$$:=-\sum_{j \in J}j.$$

From the definition of $d_J$, we have $d_J=q^{u_J}v_{{\bf {j}^2_{\prec}}}$. By the relations of the exterior algebra there is another integer $a_J$ such that $v_{{\bf {j}^2_{\prec}}}=q^{a_J}v_{{\bf {j}^2_{<}}}$. By (11.6) and Corollary 11.9 we calculate

$$\tau_{\wedge}(d_J)= \sum_{I \in P({n},{2a})}q^{u_J+a_J}v_{I} \ot... ...n},{2a})}q^{u_J}v_{I} \otimes T^{\omega_{2a}}_q({\bf i}:{\bf {j}^2_{\prec}}).$$

Setting

$${G_{{\bf i}, l, a}}$$$$\mbox{\index{${G_{{\bf i}, l, a}}$}}$$$$=\sum_{J\in P({l},{a}) } q^{u_J}T^{\omega_{2a}}_q({\bf i}:{\bf {j}^2_{\prec}}),$$

we may write $\tau_{\wedge}(D_a)= \sum_{I \in P({n},{2a})}v_{I} \otimes G_{{\bf i}, m, a}$. Since the $v_{I}$ form a free basis of the comodule the following proppsitions holds:

Proposition 14.2   Proposition 12.2 holds if and only if $G_{{\bf i}, m,a} =0$ for all ${\bf i} \in I(n,r)$ and $a\in \underline{m}$.

We will prove the equation $G_{{\bf i}, m,a} =0$ with the help of the Laplace-Expansion which is a special case of Laplace-Duality (Proposition 9.7) applied to the partitions

$${\lambda_t}$$$$\mbox{\index{${\lambda_t}$}}$$$$:=(r-t+1,\underbrace{ 1,1, \ldots , 1}_{\mbox{$(t-1)$-times}}) \in \Lambda^+(t, r)$$

where $1\leq t \leq r$.

Caution: The symbol should not be confused with the $t$-th component of a partition $\lambda$. A bideterminant $T^{\lambda_t}_q({\bf i}:{\bf j})$ is the product of a $t \times t$ minor determinant with a monomial, that is

$$T^{\lambda_t}_q({\bf i}:{\bf j}) = T^{\omega_t}_q((i_1, \ldots , i_t): (j_1, \ldots , j_t)) x_{i_{t+1} j_{t+1}}x_{i_{t+2} j_{t+2}}\ldots x_{i_{r} j_{r}}$$

$$= {\left \vert\begin{array}{ccc} x_{i_1 j_1} & \cdots & x_{i_1 j... ...} \right \vert}_q x_{i_{t+1} j_{t+1}}x_{i_{t+2} j_{t+2}}\ldots x_{i_{r} j_{r}},$$

in particular $\lambda_r=\omega_r$.

Let $L_t$ denote the set of distinguished left coset representatives of $S_{\lambda_{t-1}'}$ in $S_{\lambda_t'}$. Using basic transpositions $s_i$ this set can be written down explicitly:

$$L_t=\{ {\rm id}_{,} s_{t-1}, s_{t-2}s_{t-1}, \ldots , s_1s_2\ldots s_{t-1} \}.$$

Setting

$${\mu_t}$$$$\mbox{\index{${\mu_t}$}}$$$$:=\sum_{w\in L_t}(-y)^{-l(w)}\beta (w)$$

the quantum symplectic (left) Laplace-Expansion deduced from Proposition 9.7 reads

Proposition 14.3 (Laplace-Expansion)   By use of the above introduced notation the following equation is valid:

$$\mu_t\wr T^{\lambda_{t-1}}_q({\bf i}:{\bf j})= T^{\lambda_t}_q({\bf i}:{\bf j}).$$

In the classical case and $t=r$ this turns out to be the familar Laplace-Expansion. There is a very useful recursive calculation rule for the endomorphisms $\mu_t$:

$$-y^{-1}\mu_t\beta_t = \mu_{t+1} -{\rm id}_{V^{\otimes r}} .$$ (23)

Before we state the fundamental lemma of this section we remind the reader of the addition of multi-indices, for example ${\bf {j}^2_{\prec}}+(kk')=(j_1, j_1', \ldots , j_a, j_a', k, k' )$.

Lemma 14.4   Let $l, a \in \underline{m}$ and ${\bf i}\in I(n,2a)$. Then

$$G_{{\bf i}, l, a}=\sum_{J\in P({l},{a-1}), k \in \underline{l}} ... ...}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk')) \wr ({\rm id}_{} -y^{-a}\beta_{2a-1}).$$

PROOF: First we treat the case where $a=1$. Here we have $G_{{\bf i}, l,1}=\sum_{j=1}^lq^{-j} T^{\omega_2}_q({\bf i}:(jj'))$ by definition. Since $P({l},{0})=\emptyset$ the summation on the right hand side of the lemma is over $k\in \underline{l}$ too. Furthermore,

$$T^{\lambda_1}_q({\bf i}:(kk'))\wr ({\rm id}_{} -y^{-1}\beta_1) = ... ... k}x_{i_2 k'}\wr ({\rm id}_{} -y^{-1}\beta_1) = T^{\omega_2}_q({\bf i}:(kk')).$$

Thus both sides of the equation are identical.

For the general case we use induction on $l$. In the case $l=1$ we necessarily have $a=1$, which has been treated above. In order to prove the induction step we may assume $a>1$ and $l>1$. We divide the summation on the right hand side into three subsums:

(A) $l \in J$(B) $l \not \in J,\; k=l$ (C) $l\not\in J,\; k<l$

and write $\sum_A$, $\sum_B$ and $\sum_C$ respectively. First we treat subsum $\sum_A$. Using (4.2) we see

$$\sum_{k\in \underline{l}}q^{-k}x_{k i}x_{l i'} \wr \beta = \sum_{k\in \underline{l}}q^{k-2l}x_{k i'}x_{l i}$$

and therefore

$$\sum_{k\in \underline{l}}q^{-k-l}T^{\lambda_{2a-1}}_q({\bf i}:{\b... ...\underline{l}}q^{k-3l}T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(k'k))$$

If $J \in P({l},{a-1})$ contains $l$, we may write $J=\widehat J \cup\{l\}$ with some $\widehat J \in P({l-1},{a-2})$ such that for the corresponding multi-index ${\bf\widehat j}$ we have

$$T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk')) = T^{\lambda_{2a-1}}_q({\bf i}:{\bf {(\widehat j)}^2_{\prec}}+(ll'kk')).$$

We obtain:

$$\sum_A = \sum_{J\in P({l-1},{a-2}), k \in \underline{l}} q^{u_J -k-l}T... ..._q({\bf i}:{\bf {j}^2_{\prec}}+(ll'kk')) \wr ({\rm id}_{} -y^{-a}\beta_{2a-1})$$

$$= \sum_{J\in P({l-1},{a-2}), k \in \underline{l-1}} q^{u_J -k-l... ... -y^{-a}q^{u_J +k-3l}T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(ll'k'k))$$

$$+ q^{-2l}\sum_{J\in P({l-1},{a-2})} q^{u_J}T^{\lambda_{2a-1}}_q(... ...l')) -y^{-a}q^{u_J}T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(ll'l'l)).$$ (24)

The bideterminant $T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(ll'l'l))$ vanishes by Corollary 9.2. Unfortunately the bideterminant with $(ll'll')$ is not zero in general. Since

$$(y-1)\sum_{k=l+1}^md_k\equiv -(y-1)\sum_{k=1}^l d_k$$    mod $$N$$

we deduce from relation (15) of the exterior algebra the congruence relation

$$d_l \equiv (y^{-1}-1)\sum_{k=1}^{l-1}d_k + y^{-l}c_l$$    mod $$N.$$

By Corollary 11.9 we obtain

$$T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(ll'll')) =(y^{... ..._{k=1}^{l-1}q^{-k} T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')).$$ (25)

Since $\beta_{2a-1}\beta_{2a-2}(v_{{\bf {j}^2_{\prec}}+(k'kll')})= q^{2}v_{{\bf {j}^2_{\prec}}+(k'll'k)}$ and $\beta_{2a-1}^{-1}\beta_{2a-2}^{-1}(v_{{\bf {j}^2_{\prec}}+(kk'll')}) =q^{-2}v_{{\bf {j}^2_{\prec}}+(kll'k')}$ by (17) we may again deduce from Corollary 11.9 that

$$T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(ll'kk')) = T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')) \wr \beta_{2a-1}^{-1}\beta_{2a-2}^{-1}$$

$$T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(ll'k'k)) = T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(k'kll')) \wr \beta_{2a-1}\beta_{2a-2}$$ (26)

Here, in addition, we have used the equations $v_{(ll'k')}=q^{2}v_{(k'll')}$ and $v_{(ll'k)}=q^{-2}v_{(kll')}$ which are valid inside the exterior algebra. Modulo the ideal ${\cal G}_{2a}$ of ${\cal A}_{2a}$ spanned by $\gamma$ the congruence relation $\beta^{-1}\equiv (y^{-1}\beta+ (y^{-1} -1) {\rm id}_{})$ holds. Therefore, modulo this ideal the congruence

$$\beta_{2a-1}^{-1}\beta_{2a-2}^{-1} \equiv y^{-2}\beta_{2a-1}\beta_{2a-2}+y^{-1}(y^{-1}-1)(\beta_{2a-1}+ \beta_{2a-2})+ (y^{-1}-1)^2{\rm id}_{}$$

is valid which implies

$$T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')) \wr \beta_{2a-1}^{-1}\beta_{2a-2}^{-1} = y^{-2} T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')) \wr \beta_{2a-1}\beta_{2a-2}$$

$$+y^{-1}(y^{-1}-1)T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')) \wr \beta_{2a-1}$$

$$-(y^{-1}-1)T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')).$$ (27)

by Lemma 9.3. Here we have also used the fact that $T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')) \wr \beta_{2a-2}= -T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll'))$ by Lemma 9.4 since $s_{2a-1} \in S_{{\lambda_{t-1}}'}$. Now substitute (27) into the first equation of (26) and the equations (26) and (25) into (24). Note that the terms comming from (25) and the last term of (27) cancel each other. We obtain the following expression for the subsum (A).

$$\sum_A = \sum_{J\in P({l-1},{a-2}), k \in \underline{l-1}} [$$

$$y^{-2}q^{u_J -k-l} T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll'))\wr \beta_{2a-1}\beta_{2a-2}$$

$$-y^{-a}q^{u_J+k-3l} T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(k'kll')) \wr \beta_{2a-1}\beta_{2a-2}$$

$$+(y^{-2}-y^{-1})q^{u_J+k-l} T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')) \wr \beta_{2a-1}].$$ (28)

Now we apply Laplace-Expansion (Proposition 14.3) twice to the first and second bideterminant and once to the third:

$$T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')) = \mu_{2a-1}\mu_{2a-2} \wr T^{\lambda_{2a-3}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll'))$$

$$T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(k'kll')) = \mu_{2a-1}\mu_{2a-2} \wr T^{\lambda_{2a-3}}_q({\bf i}:{\bf {j}^2_{\prec}}+(k'kll'))$$

$$T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')) = \mu_{2a-1} \wr T^{\lambda_{2a-2}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll'))$$

Note that we may commute the $\beta_{2a-1}\beta_{2a-2}$ from the right of $T^{\lambda_{2a-3}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll'))$ to the left by (5), since a bideterminant corresponding to the partition $\lambda_{2a-3}$ is a monomial on the part where $\beta_{2a-1}$ and $\beta_{2a-2}$ are operating on. A similar fact is true concerning the bideterminant $T^{\lambda_{2a-2}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll'))$ with respect to $\lambda_{2a-2}$ and $\beta_{2a-1}$. Furthermore, note that by Lemma 4.2 for fixed $J$ and arbitrary ${\bf i} \in I(n,r)$ we have

$$\sum_{k=1}^{l-1} q^{k-2(l-1)} T^{\lambda_{2a-1}}_q({\bf i}:{\bf... ... T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll'))\wr \beta_{2a-3}.$$

Thus the second term of the right hand side in (28) is equal to

$$y^{-(a-1)}y^{-2}q^{u_J-k-l}T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')) \wr \beta_{2a-3}.$$

Substituting these equations into (28) gives

$$\sum_A = \sum_{J\in P({l-1},{a-2}), k \in \underline{l-1}} q^{u_J -k-l} [$$

$$y^{-2}\mu_{2a-1}\mu_{2a-2}\beta_{2a-1}\beta_{2a-2} \wr T^{\lambda_{2a-3}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll'))\wr ({\rm id}_{} -y^{-(a-1)} \beta_{2a-3})$$

$$+y^{-1}(y^{-1}-1)\mu_{2a-1}\beta_{2a-1}\wr T^{\lambda_{2a-2}}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk'll')) ].$$

To the first term of that sum we can apply the induction hypothesis. To this claim note that the symbol $\wr$ on the left of the bideterminant stands for a sum over the bideterminant's left multi-index. In order to apply the induction hypothesis this summation has to be commuted with the summation under the $\sum$-symbol. In a similar way we can apply Lemma 14.1 together with Corollary 11.9 to the second term of the sum above. This results in

$$\sum_A = \sum_{J\in P({l-1},{a-1})} q^{u_J-l}(y^{-2}\mu_{2a-1}\mu_{2a-... ...2a-1}\beta_{2a-2} \wr T^{\lambda_{2a-2}}_q({\bf i}:{\bf {j}^2_{\prec}}+(ll'))$$

$$+y^{-1}(y^{-1}-1)\{a-1\}_{y^{-1}}\mu_{2a-1}\beta_{2a-1}\wr T^{\lambda_{2a-2}}_q({\bf i}:{\bf {j}^2_{\prec}}+(ll')) ).$$

Since $\mu_{2a-2}$ and $\beta_{2a-1}$ commute, we see from equation (23)

$$y^{-2}\mu_{2a-1}\mu_{2a-2}\beta_{2a-1}\beta_{2a-2}= (-y^{-1}\mu_... ...y^{-1}\mu_{2a-2}\beta_{2a-2})=(\mu_{2a} -{\rm id}_{})(\mu_{2a-1} -{\rm id}_{}).$$

Similarily we calculate using (23) again

$$y^{-1}(y^{-1}-1)\{a-1\}_{y^{-1}}\mu_{2a-1}\beta_{2a-1}= (1-y^{-(a-1)})(\mu_{2a} -{\rm id}_{}).$$

Finally we obtain the following expression for the subsum (A):

$$\sum_A = \sum_{J\in P({l},{a}), l\in J} q^{u_J}(\mu_{2a}\mu_{2a... ..._{2a} - {\rm id}_{}))\wr T^{\lambda_{2a-2}}_q({\bf i}:{\bf {j}^2_{\prec}}).$$

The calculation of subsum (B) only needs one application of Laplace-Expansion together with the commutation of $\beta_{2a-1}$ from the right of the bideterminant to the left and another application of (23).

$$\sum_B = \sum_{J\in P({l},{a}), l\in J} q^{u_J} T^{\lambda_{2a-1}}_q({\bf i}:{\bf {j}^2_{\prec}})\wr ({\rm id}_{} -y^{-a}\beta_{2a-1} )$$

$$= \sum_{J\in P({l},{a}), l\in J} q^{u_J}(\mu_{2a-1}+y^{-(a-1)}(\mu_{2a} - {\rm id}_{}))\wr T^{\lambda_{2a-2}}_q({\bf i}:{\bf {j}^2_{\prec}}).$$

Thus subsum (A) and (B) together equal

$$\sum_A + \sum_B = \sum_{J\in P({l},{a}), l\in J} q^{u_J}\mu_{2a... ...n P({l},{a}), l\in J} q^{u_J} T^{\lambda_{2a}}_q({\bf i}:{\bf {j}^2_{\prec}})$$

where for the second step Proposition 14.3 is used once more. To the subsum (C) the induction hypothesis can be applied directly:

$$\sum_C = G_{{\bf i},l-1,a}= \sum_{J\in P({l-1},{a})} q^{u_J} T^{\lambda_{2a}}_q({\bf i}:{\bf {j}^2_{\prec}}).$$

Thus, it follows that all three subsums add up to $G_{{\bf i}, l, a}$. $\Box$

Now we are able to prove Proposition 12.2. As we have seen above, we have to show $G_{{\bf i}, m,a} =0$ for $a=1, \ldots , m$ and ${\bf i}\in I(n,2a)$. From (4.3) we already know $G_{{\bf i}, m,1}=0$ for all ${\bf i} \in I(n,2)$. We will deduce the general case by induction on a with the help of Lemma 14.4. Let $a>1$ and ${\bf i}\in I(n,2a)$ be arbitrary. We apply Laplace-Expansion to the formula of the lemma:

$$G_{{\bf i}, m, a}=\sum_{J\in P({m},{a-1}), k \in \underline{m}} ... ...}_q({\bf i}:{\bf {j}^2_{\prec}}+(kk')) \wr ({\rm id}_{} -y^{-a}\beta_{2a-1}).$$

As in the proof of the lemma we may commute $({\rm id}_{} -y ^{-a}\beta_{2a-1})$ to the other side of the bideterminant. Let $\left(\mu_{{\bf i}{\bf h}}\right)_{{\bf i}, {\bf h} \in I(n,2a)}$ be the coefficient matrix of the endomorphism $\mu_{2a-1}({\rm id}_{} -y^{-a}\beta_{2a-1})$ with respect to the canonical basis. We denote the multi-index consisting of the first $2a-2$ indices of ${\bf h}$ by ${\bf\bar h}:=(h_1, \ldots , h_{2a-2})$ and obtain

$$G_{{\bf i}, m, a}=\sum_{{\bf h}\in I(n,2a)}\mu_{{\bf i}{\bf h}} ... ...bf\bar h}:{\bf {j}^2_{\prec}})\sum_{k=1}^m q^{-k}x_{h_{2a-1} k}x_{h_{2a} k'}$$

$$=\sum_{{\bf h}\in I(n,2a)}\mu_{{\bf i}{\bf h}} G_{{\bf\bar h}, m,a-1} \sum_{k=1}^m q^{-k}x_{h_{2a-1} k}x_{h_{2a} k'} = 0,$$

since $G_{{\bf\bar h}, m, a-1}=0$ for all ${\bf h}\in I(n,2a)$ by the induction hypothesis.

Remark 14.5   The proof of the classical version [O2, Proposition 8.2] of Proposition 12.2 relies on the embedding of the symplectic monoid ${\rm SpM}_{n}(R)$ into the ring ${\rm M}_{n}(R)$ of $n \times n$-matrices. Since in the quantum case such an embedding is missing the proof given here is absolutely incomparable to the one given in [O2, Proposition 8.2]. Additionally the statement of the latter proposition is a little bit stronger although this is not really needed. Anyway, the reader should work out a classical version of the current section in order to understand the ideas of the proof and to see that it leads to a simplifacation of the correspondig classical treatment in [O2].

Remark 14.6   Lemma 14.4 is the fundamental key to remove the restrictions from [O1, Basissatz 3.14.12] and [O1, Satz 4.1.2] in Theorem 7.1 and Theorem 7.3.