Proof of proposition 7.3

First, some considerations about the exterior algebra

$${\bigwedge}_{R}(n):= {\cal T}(V)/\left<v_{i}\otimes v_{j}+v_... ... v_{i}, v_{k}\otimes v_{k};\; i,j, k\in \underline{n} \right>$$

are needed. It is easy to see and well known that this graded $R$-algebra is a comodule algebra for the bialgebra $A_{R}(n)$, i.e. it is an $A_{R}(n)$-comodule such that multiplication is a morphism of comodules. In fact, the homogenous summands ${\bigwedge}_{R}(n,r)$ of ${\bigwedge}_{R}(n)$ are comodules for the coalgebras $A_{R}(n,r)$. Since $A^{{\rm s}}_{R}(n)$ is an epimorphic image of $A_{R}(n)$ the exterior algebra is a comodule algebra for the latter one, as well.

As usual we write $v_{i}\wedge v_{j}$ for the residue class of $v_{i}\otimes v_{j}$ in ${\bigwedge}_{R}(n)$ and denote arbitrary multiplications by $\wedge$, too. For a subset $I:=\{i_1, \ldots , i_r\}\subseteq \underline{n}$ ordered $i_1 \ll i_2 \ll \ldots \ll i_r$ by the given order $\ll$ on $\underline{n}$ (called an ordered subset in the sequel) we use the abbreviation $v_{I}:=v_{i_1}\wedge v_{i_2}\wedge\ldots\wedge v_{i_r}$. The elements $v_{I}$ give a basis of ${\bigwedge}_{R}(n)$ if $I$ ranges over all subsets of $\underline{n}$. A basis for ${\bigwedge}_{R}(n,r)$ is obtained when $I$ ranges over all subsets of cardinality $r$, the collection of which will be denoted by $P({n},{r})$. The comodule structure of ${\bigwedge}_{R}(n,r)$ can be described in a simple way using bideterminants for the partition $\omega_r:=(1^r)=(1,1,\ldots ,1)$. These are just the usual $r\times r$-minor determinants. Denoting the structure map by $\tau_{\wedge}:{\bigwedge}_{R}(n)\rightarrow {\bigwedge}_{R}(n)\otimes A_{R}(n)$ (we will use the same symbol in the case $A^{{\rm s}}_{R}(n)$ later on) we explicitly have

$$\tau_{\wedge}(v_{J})=\sum_{I \in P({n},{r})}v_{I}\otimes T^{\omega_r}({\bf i}:{\bf j})$$ (17)

where ${\bf i}=(i_1, \ldots , i_r)$ and ${\bf j}=(j_1, \ldots , j_r)$ are the multi-indices corresponding to the ordered subsets $I:=\{i_1, \ldots , i_r\}$ and $J=\{j_1, \ldots , j_r\}$, respectively. Define $d_k:=v_{k'} \wedge v_{k}$ and $d_K:=d_{k_1}\wedge d_{k_2} \wedge \ldots \wedge d_{k_a}$ for a subset $K:=\{k_1, \ldots, k_a\}\subseteq\underline{m}$ of cardinality $a$. Again, we write $P({m},{a})$ for the collection of all such subsets $K$. Note that the $d_k$ are in the center of the exterior algebra and in particular commute with each other. Thus $d_K$ is defined independent of the order of the elements of $K$. Set

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

and let $N$ be the ideal in ${\bigwedge}_{R}(n)$ generated by the elements $D_1, D_2, \ldots , D_m$. One crucial point in the proof of proposition 7.3 is to show that the elements $D_a$ are invariant under the bialgebra $A^{{\rm s}}_{R}(n)$. But first, let us establish a prototype straightening algorithm inside the graded algebra ${\bigwedge}^{\rm s}_{R}(n)$ with respect to the set $I_{\omega_r}^{\rm sym}$. We call an ordered subset $I\in P({n},{r})$ symplectic if the corresponding multi-index ${\bf i}$ is $\omega_r$-symplectic standard.

Lemma 8.1   Let $I\in P({n},{r})$ be non symplectic. Then, to each $J\in P({n},{r})$ such that the inequallity $f({\bf j})<f({\bf i})$ holds for corresponding multi-indices ${\bf i}$ and ${\bf j}$ there is $a_{IJ}\in R$, such that in ${\bigwedge}_{R}(n)$ the following congruence relation holds:

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

PROOF: We follow [Do3]. Clearly, we may reduce to the case $R={\mathbb{Z}}$ since ${\bigwedge}_{R}(n)\cong R\otimes_{{\mathbb{Z}}} {\bigwedge}_{{\mathbb{Z}}}(n)$ and the canonical isomorphism therein respects the basis elements $v_{I}$ and the ideal $N$. If $K\subseteq \underline{m}$ then in ${\bigwedge}_{R}(n)$ we calculate as in [Do3], 2.2.

$${(\sum_{k \in K}d_k)}^a=a!\sum_{L\in P({m},{a}),\; L\subseteq K} d_L$$ (18)

In the case $K=\underline{m}$ this implies $D_1^a=a!D_a$. Set $x:=\sum_{k \in K}d_k$ and $y:=D_1 -x$. It follows from (18) that the equation

$$x^a=(-1)^ay^a+\sum_{b=1}^a { a \choose b} (-y)^{a-b}b!D_b$$

is divisible by $a!$. Setting $X^K_a:=\sum_{L\in P({m},{a}),\; L\subseteq K}d_L$ and $M:=\underline{m}\backslash K$ this yields

$$X^K_a=(-1)^aX^M_a+\sum_{b=1}^a (-1)^{a-b}X^M_{a-b}D_b$$

since ${\bigwedge}_{{\mathbb{Z}}}(n)$ is a free ${\mathbb{Z}}$-module. We obtain $X^K_a\equiv (-1)^aX^M_a$ modulo the ideal $N$ and in the special case $K \in P({m},{a})$

$$d_K\equiv (-1)^a\sum_{L\in P({m},{a}),\; L\cap K=\emptyset}d_L \;\;\mbox{ mod } N.$$ (19)

The proof can be finished now in a similar way to the proof of the Symplectic Carter-Lusztig Lemma in [Do3]: Since $I$ is not symplectic there is a number $s\in \underline{r}$ such that $I=\{i_1,\ldots , i_r\}$ contains an element $i_s$ satisfying $i_s<s$ if $i_s \leq m$ or $i_s'<s$ if $i_s>m$. We assume $s$ to be as small as possible with this property. Let $k, l\in \underline{m}$ be the unique numbers with $\{k, k'\}=\{i_s, i_s'\}$ and $\{l, l'\}=\{i_{s-1}, i_{s-1}'\}$. By minimality of $s$ we have $l \geq s-1$. On the other hand, $i_{s-1}\ll i_s$ implies $l\leq k$. This gives $s-1\leq l \leq k < s$, that is $k=l$ and $s=k+1$.

Now let $I_s:=\{i_1,\ldots , i_s\}$ be the ordered subset of the first $s$ entries of $I$ and $K$ be the set of all $p\in \underline{m}$ such that both $p$ and $p'$ are contained in $I_s$. Setting $K':=\{p'\vert\; p\in K\}$ and $H:=I\backslash (K\cup K')$ we obtain $v_{I}=v_{H}d_K$ since the elements $d_p$ are in the center of ${\bigwedge}_{{\mathbb{Z}}}(n)$. Therefore, by equation (19) it remains to show that for all subsets $L\in P({m},{a})$ which do not intersect $K$ we have $f({\bf j})<f({\bf i})$ where ${\bf j}$ is the multi-index corresponding to the ordered set $J:=H\cup L \cup L'$. Since $v_{J}=v_{H}d_L=0$ if $L$ or $L':=\{p'\vert\; p \in L\}$ contains an element of $H$ we further may assume $H\cap L=H\cap L'=\emptyset$. Now, suppose $L\subseteq \underline{k}$. Let $G$ be the set of all $g \in \underline{k}\backslash K$ such that $g$ or $g'$ lies in $I_s$. Since $I_s\backslash (K\cup K')\subseteq H$ the intersection of $L$ and $G$ is empty, as well. This gives a contradiction

$$k+1=s=\vert I_s\vert=2\vert K\vert+\vert G\vert=\vert K\vert+\vert L\vert+\vert G\vert=\vert K\cup L\cup G\vert\leq k$$

for $K, L$ and $G$ are disjoint subsets of $\underline{k}$ by assumption on $L$. Thus, the largest element $t$ of $L$ must be greater than $k$, whereas all elements of $K$ are smaller or equal to $k$. If $f({\bf i})=(b_1, \ldots , b_m)$ and $f({\bf j})=(a_1, \ldots, a_m)$ it follows that $a_t=b_t+2>b_t$ and $a_l=b_l$ for $t<l\leq m$. By definition of our order on ${\mathbb{N}}_0^m$ this means $f({\bf j})<f({\bf i})$ completing the proof.$\Box$

As mentioned before, the crucial point in the proof of proposition 7.3 is contained in the following

Proposition 8.2   The elements $D_a$ are invariant under the bialgebra $A^{{\rm s}}_{R}(n)$. The corresponding group-like elements in $A^{{\rm s}}_{R}(n)$ are the $a$-th powers of the coefficient of dilation, more precisely:

$$\tau_{\wedge}(D_a)=D_a\otimes d^a \;\;\;\;\in \;\; {\bigwedge}_{R}(n,2a) \otimes A^{{\rm s}}_{R}(n,2a)$$

PROOF: In the case $a=1$ this easily follows from (15) using the relations given by the set $F$ defined in (14). If we could divide by $a!$, we would be able to finish the proof right now using $D_1^a=a!D_a$ and the fact that multiplication is a morphism of $A^{{\rm s}}_{R}(n)$-comodules. But, as this is not possible in general (note, that we don't know if $A^{{\rm s}}_{{\mathbb{Z}}}(n)$ is a free ${\mathbb{Z}}$-module, yet) we have to proceed in another way. We set $r=2a$ and

$$G_{{\bf i}}:=\sum_{L\in P({m},{a})}T^{\omega_r}({\bf i}:{\bf j}(L))$$

where ${\bf j}(L)$ is the multi-index corresponding to the ordered subset $L\cup L'=\{l,l'\vert\; l\in L\}$ of $\underline{n}$. If ${\bf i}$ is the multi-index corresponding to the ordered subset $I\in P({n},{r})$ we also write $G_I=G_{{\bf i}}$. By (17) we have

$$\tau_{\wedge}(D_a)=\sum_{I\in P({n},{r})}v_{I}\otimes G_I.$$

Therefore, it remains to show that $G_I=d^a$ if there is a $K \in P({m},{a})$ such that $I=K\cup K'$ and $G_I=0$ otherwise.

If ${\bf i}=(i_1, \ldots , i_r)$ is a multi-index and $l\leq a$ we set ${\bf i}^l:=(i_{2l-1},i_{2l})\in I(n,2)$. Since the result is clear in the case $a=1$ we know

$$G_{{\bf i}^l}=\left\{\begin{array}{ll} 0 & i_{2l-1}'\neq i_{... ..._{2l}\leq m\\ -d & i_{2l-1}'= i_{2l}> m \end{array} \right. .$$

Therefore, the element $G_{{\bf i}}':=\prod_{l=1}^a G_{{\bf i}^l}$ is zero or $\pm d^a$ depending on ${\bf i}$. Let us investigate this in more detail. If $\sigma\in {\cal S}_{r}$ denotes the involution defined by $\sigma (2l-1)=2l, \sigma(2l)=2l-1$ for $l\in \underline{a}$ and $W$ the centralizer of $\sigma$ in ${\cal S}_{r}$, then clearly $G_{{\bf i}}'=0$ if and only if $G_{{\bf i}w}'=0$ for all $w \in W$. The group $W$ is isomorphic to the Weyl group of type $C_a$. It is a semi-direct product of the normal subgroup $W^1$ consisting of all $\pi \in W$ which permute neighboured pairs together with $W^0$, the subgroup generated by the transpositions $(2l-1,2l)$ for $l\in \underline{a}$. $W^1$ is isomorphic to ${\cal S}_{a}$, whereas the group $W^0$ can be identified with $({\mathbb{Z}}/2{\mathbb{Z}})^a$. Choose a set $H$ of left coset representatives for $W$ in ${\cal S}_{r}$ the element representing $W$ itself being ${\rm id}_{\underline{r}}$.

Now, if ${\bf i}$ corresponds to an ordered set $I=K\cup K'$ for some $K \in P({a},{n})$ the inequality $G_{{\bf i}\pi}'\neq 0$ holds for a permutation $\pi\in {\cal S}_{r}$ if and only if $\pi \in W$. Thus, for $h\in H$ we have $G_{{\bf i}h}'=0$ if $h\neq {\rm id}_{}$ and $G_{{\bf i}}'=d^a$. If there is no $K$ such that $I=K\cup K'$ one clearly has $G_{{\bf i}h}'=0$ for all $h\in H$. Therefore, the proof is finished as soon as we have shown

$$G_{{\bf i}}=\sum_{h \in H}{\rm sign}(h) G_{{\bf i}h}'.$$ (20)

To this claim let $\mu :=(a,a) \in \Lambda^+(2, r)$ be the partition of $r$ whose diagram consists of two rows of length $a$. To a multi-index ${\bf l}=(l_1,l_2, \ldots , l_a)\in I(m,a)$ another multi-index ${\bf j}({\bf l}):=(l_1', l_1, l_2', l_2, \ldots , l_a', l_a)\in I(n,r)$ can be associated. Using this notation and reading $T^{\mu}({\bf i}:{\bf j}({\bf l}))$ as a product of $a$ $2\times 2$-determinants we obtain the formula

$$G_{{\bf i}}'= \sum_{{\bf l} \in I(m,a) } T^{\mu}({\bf i}:{\bf j}({\bf l}))\;\;\; \in A^{{\rm s}}_{R}(n,r).$$

On the other hand, using Laplace Duality (see for example [Mr] 2.5.1) we calculate

$$T^{\omega_r}({\bf i}:{\bf j})=\sum_{h\in H,\;\pi \in W^1}{\r... ...}{\rm sign}(h)\sum_{\pi \in W^1} T^{\mu}({\bf i}h:{\bf j}\pi).$$ (21)

Therein, note that $W^0$ is precisely the column stabilizer of the basic tableaux $T^{\mu}_{{\bf b}}$, whereas the column stabilizer of $T^{\omega_r}_{{\bf b}}$ is all of ${\cal S}_{r}$ (here ${\bf b}:=(1',1,2',2,\ldots ,a',a)$). Also, note that $HW^1$ is a set of left coset representatives of $W^0$ and that all permutations of $W^1$ are even. Furthermore, in the right-hand-side equation we have used the commutativity between the $2\times 2$-determinant factors of $T^{\mu}({\bf i}h\pi:{\bf j})=T^{\mu}({\bf i}h:{\bf j}\pi^{-1})$. Now, the proof of (20) can be reduced to the verification of

$$\sum_{L \in P({m},{r})}\sum_{\pi \in W^1}\sum_{h \in H} {\rm... ... \sum_{h \in H}{\rm sign}(h)T^{\mu}({\bf i}h:{\bf j}({\bf l}))$$ (22)

To this claim we associate to a multi-index ${\bf l}\in I(m,a)$ its contents $\vert{\bf l}\vert=\lambda =(\lambda_1,\ldots , \lambda_m)$ which is defined by $\lambda_i:=\vert\{1\leq t \leq a\vert\; l_t=i\}\vert$. It is a composition of $a$ into $m$ parts, that is an $m$-tuple of non negative integers $\lambda_i$ summing up to $a$. These compositions count the set of ${\cal S}_{a}$-orbits in $I(m,a)$. Denoting the set of all such compositions by $\Lambda(m, a)$ we can write down the right-hand term in (22) as a sum of subsums $\sum_{\lambda \in \Lambda(m, r)} \Sigma_{\lambda}$ each of which is given by

$$\Sigma_{\lambda}:=\sum_{\vert{\bf l}\vert=\lambda} \sum_{h \in H} {\rm sign}(h) T^{\mu}({\bf i}h:{\bf j}({\bf l}))$$

Now, the subsum $\Sigma_{\omega_a}$ (for $\omega_a=(1^a)$) is just the left-hand-side in (22). Therefore, it remains to show that all other subsums are zero. To this claim we denote the cardinality of the standard Young subgroup ${\cal S}_{\lambda}$ of ${\cal S}_{a}$ corresponding to the composition $\lambda \in \Lambda(m, a)$ by $k_{\lambda}:=\vert{\cal S}_{\lambda}\vert$. If ${\bf k}=(k_1, \ldots ,k_a)\in I(m,a)$ is the unique multi-index with contents $\lambda$ and $k_1\leq k_2\leq \ldots \leq k_r$ (the initial index corresponding to $\lambda$) then ${\cal S}_{\lambda}$ is just the stabilizer of ${\bf k}$ in ${\cal S}_{a}$. Identifying ${\cal S}_{a}$ with $W^1$ it is the stabilizer of ${\bf j}({\bf k})\in I(n,r)$ in $W^1$. Applying (21) again we obtain

$$T^{\omega_r}({\bf i}:{\bf j}({\bf k}))=\sum_{h\in H}{\rm sign}(h) \sum_{\pi \in W^1} T^{\mu}({\bf i}h:{\bf j}({\bf k})\pi)=$$

$$k_{\lambda}\sum_{h\in H}{\rm sign}(h) \sum_{\vert{\bf l}\ver... ...\mu}({\bf i}h:{\bf j}({\bf l}))= k_{\lambda}\Sigma_{\lambda}.$$

Now, $T^{\omega_r}({\bf i}:{\bf j}({\bf k}))$ must be zero since $\vert{\bf k}\vert\neq \omega_a$ implies that ${\bf k}$ contains at least one number twice. We obtain $k_{\lambda}\Sigma_{\lambda}=0$ and because this equation is already valid in the free ${\mathbb{Z}}$-module $A_{{\mathbb{Z}}}(n,r)$ we conclude $\Sigma_{\lambda}=0$ for all $\lambda \neq \omega_a$ completing the proof.$\Box$

Let us prove proposition 7.3 in the case $\lambda=\omega_r$ first. Take ${\bf j}\in I(n,r)\backslash I_{\omega_r}^{\rm sym}$. Using the classical straighening algorithm 7.2 we may assume ${\bf j}\in I_{\omega_r}^{}\backslash I_{\omega_r}^{\rm sym}$ (observe that ${\bf k}={\bf j}w$ implies $f({\bf k})=f({\bf j})$). This means, that ${\bf j}$ is a multi-index corresponding to a non symplectic ordered set $J\in P({n},{r})$ in the sense of lemma . Application of the latter one yields

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

According to proposition 8.2 $\tau_{\wedge}(X)$ must be contained in ${\bigwedge}_{R}(n)\otimes <d>$ where $<d>$ denotes the ideal in $A^{{\rm s}}_{R}(n)$ generated by $d$. Applying (17) we obtain the following equation in ${\bigwedge}_{R}(n,r)\otimes A^{{\rm sh}}_{R}(n,r)$:

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

Since $\{v_{I}\vert\; I \in P({n},{r})\}$ is a basis of ${\bigwedge}_{R}(n,r)$ each individual summand in the summation over $P({n},{r})$ must be zero. 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 general case for ${\bf i}$ can be deduced from this, easily (see [Oe], 3.11.4).

Now, lets turn to the general case of $\lambda$. Again, we may assume ${\bf j}\in I_{\lambda}^{}\backslash I_{\lambda}^{\rm sym}$ by the classical straightening algorithm. Let $\lambda'=(\mu_1, \ldots , \mu_p)$ be the dual partition ($p=\lambda_1$). We spilt ${\bf j}$ 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 $\lambda$-symplectic standard but standard there must be a column $s$ such that ${\bf j}^s$ is not $\omega_{\mu_s}$-symplectic standard. Applying the result to the known case of $T^{\omega_{\mu_s}}({\bf i}^s:{\bf j}^s)$ we obtain

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

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

Therein, ${\bf k}^s\in I(n,\mu_s)$ satisfies ${\bf k}^s\lhd{\bf j}^s$, ${\bf k}\in I(n,r)$ is constructed from ${\bf j}$ 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 ${\bf k}^s$. One easily checks ${\bf k}\lhd {\bf j}$ and the proof of 7.3 is completed.