Proof of Proposition 12.1

For convenience, in in the following sections we abbreviate $y=q^2$ and denote multiplication in ${\bigwedge}_{R,q}(n)$ by juxtaposition instead of $\wedge$. Furthermore, to sets $K, L \subseteq \underline{m}$ we associate integers

$${v(K,L)}$$$$\mbox{\index{${v(K,L)}$}}$$$$:=\vert\{(k,l)\in K\times L \vert\; k>l\}\vert.$$

Lemma 13.1   Let $a\in \underline{m}$. If $\underline{m}=L\cup M$ is a partition of $\underline{m}$ into disjoint subsets $L$ and $M$ then to each $K \in P({m},{a})$ there is an integer $s(K,L)$ such that

$$D_a=\sum_{K \in P({m},{a})}y^{s(K,L)}c_{K\cap L}d_{K\cap M}.$$

If $K\subseteq M$, the integer $s(K,L)$ equals $v(K, L)$.

SKETCH OF PROOF: In order to prove this, one has to use a more general statement in which the set $\underline{m}$ is substituted by $\{l, l+1, \ldots , m\}$ for some $l\in \underline{m}$. After this, the results can be proved straightforwardly using induction on $m-l$ and the relations (14) and (15) of the exterior algebra. For the details we refer to appendix 18.3. $\Box$

To a set $I\in P({n},{r})$ we associate the followng subsets of $\underline{m}$:

$${I^-}$$$$\mbox{\index{${I^-}$}}$$$$:=I\cap \underline{m}, \;\; {I^+}$$$$\mbox{\index{${I^+}$}}$$$$:=\{ i\in \underline{m}\vert\; i' \in I\}\;\;$$ and $$\;\; {I^0}$$$$\mbox{\index{${I^0}$}}$$$$:=I^-\cap I^+.$$

Lemma 13.2   Let $I\in P({n},{r})$ be such that $I^0=\emptyset$ and $J \subseteq \underline{m}$. Set $s =\vert J\vert$ and $T:=\underline{m} \backslash (I^-\cup I^+)$. If $J\subseteq T$ and $a\in \underline{m}$ is such that $s<a$ then we have

$$v_{I}d_JD_{a-s}=v_{I} \sum_{S \in P({m},{a}), J \subseteq S \subseteq T}y^{v(S\backslash J,I^+\cup J)}d_S.$$

PROOF: We apply Lemma 13.1 to $L:=I^+\cup J$ and $M:=\underline{m}\backslash L$. Since $J\subseteq T$ there is an invertible element $b \in R$ such that $v_{I}d_J=b d_Jv_{I}$ by (13). We claim that $v_{I}d_Jc_{K\cap L}d_{K\cap M}$ vanishes for each $K\in P({m},{a-s})$ that is not contained in $T\backslash J$. If $K \cap J \neq \emptyset$ then $d_Jc_{K\cap L}=0$. If $K \cap I^+ \neq \emptyset$ and $K \cap J = \emptyset$ then $v_{I}d_Jc_{K\cap L}=v_{I}c_{K\cap L}d_J=0$. Finally if $K \cap I^- \neq \emptyset$ and $K \cap J = \emptyset$ then $v_{I}d_Jc_{K\cap L}d_{K\cap M}=v_{I}d_{K\cap M} d_J c_{K\cap L}=0$. Thus the expression is nonzero only if $K\subseteq T \backslash J$. A set $K\subseteq T \backslash J$ is obviously contained in $M$, so by the second part of the Lemma we have $s(K,L)=v(K, L)=v(K,I^+\cup J)$. Thus, setting $S=J\cup K$ the assertion follows. $\Box$

Lemma 13.3   Let $M\subseteq K \subseteq \underline{m}$ be fixed. Then

$$\sum_{L\subseteq K\backslash M} (-1)^{\vert L\vert}y^{v(K,L)}= \left\{\begin{array}{ll} 1 & K=M \\ 0 & K \neq M\end{array} \right.$$

PROOF: Clearly the sum is $1$ if $K=M$ since $v(K, \emptyset)=0$. For $K\neq M$ we show by induction on $n:=\vert K\vert> 0$ that the sum is zero. Starting with the case $n=1$ we have $M=\emptyset$ and $\sum_{L\subseteq K} (-1)^{\vert L\vert}y^{v(K,L)}= y^{v(K, \emptyset)}-y^{v(K,K)}=1-1=0$. For the induction step, let $n > 1, x \in K$ be minimal and set $\widehat K:=K\backslash x$. If $x \not\in M$ we calculate

$$\sum_{L\subseteq K\backslash M} (-1)^{\vert L\vert}y^{v(K,L)}= \... ...{L\subseteq \widehat K\backslash M} (-1)^{\vert L\vert+1}y^{v(K,L\cup \{x\})}.$$

Since $x$ is minimal, $v(K,L)=v(\widehat K, L)$ and $v(K, L\cup\{x\})= v(\widehat K,L)+n-1$ if $L\subseteq \widehat K\backslash M$. Now, we may apply the induction hypothesis to $\widehat K$ which results in

$$\sum_{L\subseteq K\backslash M} (-1)^{\vert L\vert}y^{v(K,L)}=(1-... ..._{L\subseteq \widehat K\backslash M} (-1)^{\vert L\vert}y^{v(\widehat K, L)}=0.$$

In the case $x \in M$, we write $\widehat M:=M\backslash \{x\}$. Similarly, we calculate

$$\sum_{L\subseteq K\backslash M} (-1)^{\vert L\vert}y^{v(K,L)}= \... ...seteq \widehat K\backslash \widehat M} (-1)^{\vert L\vert}y^{v(\widehat K,L)},$$

where the right hand side is zero by the induction hypothesis. $\Box$

Lemma 13.4   Let $I\in P({n},{r})$ with $r>m$ and set $a=\vert I^0\vert$, $\widehat I:=I\backslash \{i, i'\vert\; i \in I^0\}$ and $T:=\underline{m} \backslash (I^+ \cup I^-)$. Then there is an invertible $a_I\in R$ such that the following equation holds:

$$v_{\widehat I}\sum_{J\subseteq T} (-1)^{\vert J\vert}y^{v(J,(\widehat I^)+\cup J )} d_JD_{a-\vert J\vert} = a_Iv_{I}.$$

PROOF: Let $\widehat T=T \cup I^0=\underline{m}\backslash ((\widehat I)^+\cup(\widehat I)^-)$ and observe that $(\widehat I)^+=I^+\backslash I^0$, $(\widehat I)^0=\emptyset$, $(\widehat I)^+\cap J=\emptyset$ and that $T\cap I^0=\emptyset$. Because $r>m$, we must have $a\geq 1$. On the other hand $2a+\vert\widehat I\vert=r>m$ implies $a>m-\vert\widehat I\vert-\vert I^0\vert=\vert T\vert\geq \vert J\vert$ for a set $J$ as in the sum. Therefore we may apply Lemma 13.2:

$$v_{\widehat I}\sum_{J\subseteq T} (-1)^{\vert J\vert}y^{v(J,(\widehat I)^+\cup J)} d_JD_{a-\vert J\vert} =v_{\widehat I}\sum_{J\subseteq T} (-1)^{\vert J\vert}y^{v(J,(\wi... ...J \subset S \subseteq \widehat T} y^{v(S\backslash J,(\widehat I)^+\cup J)}d_S$$

$$= v_{\widehat I}\sum_{J\subseteq S \subseteq \widehat T, J\cap I^0=\emptyset} (-1)^{\vert J\vert}y^{v(S,J)+ v(S,(\widehat I)^+)}d_S$$

$$= v_{\widehat I}\sum_{S \subseteq \widehat T} y^{v(S,(\widehat I)^+)}d_S\sum_{J\subseteq S\backslash I^0} (-1)^{\vert J\vert}y^{v(S,J)}.$$

But by Lemma 13.3, the last term equals $y^{v(I^0,\widehat I^+)}v_{\widehat I}d_{I^0}$. Using relation (13) of the exterior algebra, this can be transformed into $v_{I}$ up to some invertible mutiple $a_I$. $\Box$

We are now able to prove Proposition 12.1 by induction on the Lie rank $m$. In the case where $m=1$ both sets of $P({2},{1})=\{\{1\},\{2\}\}$ are reverse symplectic. In $P({2},{2})$ there is just one set namely $I=\{1,2\}$, for which we have $v_{I}=-qd_1=-qD_1\in N$. Thus there is nothing to prove here.

For the induction step we embed ${\bigwedge}_{R, q}(n-2)$ into ${\bigwedge}_{R,q}(n)$ sending $v_{i}$ to $v_{i+1}$. It is easy to check that this indeed leads to an embedding of algebras. Using the induction hypothesis we may treat the case where $I \subseteq \underline{n}\backslash \{1,n\}$ without much effort. Some caution is needed only concerning the difference between the two ideals $N$ of ${\bigwedge}_{R, q}(n-2)$ and ${\bigwedge}_{R,q}(n)$. Denote them by $N(n-2)$ and $N(n)$. A single element $u \in N(n-2)$ can be written

$$u = \sum_{\{1,n\}\subseteq L \subseteq \underline{n}} a_{\widehat I L}v_{L}$$    mod $$N(n)$$

where the basis elements $v_{L}$ all are smaller than $v_{I}$, that is $f({\bf l})<f({\bf i})$ for the corresponding multi-indices. If $I\cap \{1,n\}\neq \emptyset$ we may apply the induction hypothesis to the set $\widehat I:=I\backslash \{1,n\}$ in case $\widehat I$ is non reverse symplectic too:

$$v_{\widehat I}\equiv \sum_{\widehat J\subseteq \underline{n}\bac... ...eq L\subseteq \underline{n}}a_{\widehat IL}v_{L} \;\; \mbox{ mod }\;\; N(n).$$

Again, the second sum compensate for the difference between the two ideals $N(n-2)$ and $N(n)$. Multiplying this congruence by $v_{1}$ from the left (respectively by $v_{n}$ from the right, respectively by both from both sides) yields the assertion because the elements $v_{L}$ vanish, and $f({\bf\widehat j})<f({\bf\widehat i})$ implies $f({\bf j})<f({\bf i})$, where ${\bf j}$ is the multi-index attached to the set $J=\widehat J\cup (I\backslash \widehat I)$.

It remains to prove the assertion in the case where $\widehat I$ is reverse symplectic. Here we need the preparations of this section. By the reverse symplectic condition applied to tableaux of shape $\omega_r$ we have $\sum_{i=j}^m\lambda_i\leq m-j+1$ for all $j>1$, where $(\lambda_1, \ldots , \lambda_m)=f({\bf i})$. Because $I$ itself is non reverse symplectic we must have $r=\vert I\vert=\sum_{i=1}^m\lambda_i>m$. According to Lemma 13.4 we conclude $v_{I} \in N$. But this implies the assertion of Proposition 12.1 in the remaining case too.