Quantum Symplectic Exterior Algebra

We are going to prepare the proof of Proposition for general $r$. Since we need a $q$-analogue of [O2, Lemma 8.1], we have to investigate the quantum symplectic exterior algebra. We start with its definition which can be found in many textbooks on quantum groups (for instance [CP, chapter 7]). It is defined as the quotient of the tensor algebra ${{\cal T}(V)}$$=\bigoplus_{r\in {\mathbb{N}}\,_0}V^{\otimes r}$ by a certain ideal. We denote it by ${{\bigwedge}_{R,q}(n)}$ and write the symbol ${\wedge}$ for multiplication in this algebra. Setting

$${c_i}$$$$\mbox{\index{${c_i}$}}$$$$:=q^{i}v_{i'} \wedge v_{i},\;\;$$ and $$\;\; {d_i}$$$$\mbox{\index{${d_i}$}}$$$$:=-q^{-i}v_{i} \wedge v_{i'}$$

for $i\in \underline{m}$, we write down the defining relations holding in ${\bigwedge}_{R,q}(n)$ according to [Ha2, (5.2)]:

$$v_{k} \wedge v_{l} = -q^{-1} v_{l} \wedge v_{k},$$ (13)

$$y^{-i}c_i = y^{-1}d_i +(y^{-1}-1) \sum_{j=i+1}^m d_j$$ (14)

$$y^i d_i = y c_i + (y-1)\sum_{j=i+1}^m c_j$$ (15)

$$v_{k}\wedge v_{k} =$$ 0 (16)

where $k>l, \; k\neq l'$ and $i\in \underline{m}$ is assumed. Remenber that the $q$ of [Ha2] corresponds to the inverse of our $q$. The third relation does not occur in [Ha2] and indeed we have

Lemma 11.1   Relation (15) is a consequence of (13) and (14).

PROOF: We use induction on $m-i$. The beginning $y^m d_m=y c_m$ follows directly from $y^{-m}c_m = y^{-1}d_m$ by multiplication with $y^{m+1}$. For $i <m$ we use (14) and the induction hypothesis to see that

$$\begin{array}{rl} y^{-1}d_i & = y^{-1}c_i - (y^{-1} -1)\sum_{j... ...sum_{k=i+1}^m(y^{1-k}+ \sum_{j=i+1}^{k-1} y^{-j}(y-1) )c_k \\ \end{array}$$

Since $(y-1)\sum_{j=i+1}^{k-1}y^{-j} = y^{-i} - y^{1-k}$ we obtain (15). $\Box$

We set

$${v_{I}}$$$$\mbox{\index{${v_{I}}$}}$$$$:= v_{i_1} \wedge v_{i_2} \wedge \ldots \wedge v_{i_r} ,\;\;$$    if $$\;\; I:=\{i_1, \ldots, i_r\} \;\;$$ and $$\;\; i_1 < i_2 < \ldots <i_r.$$

In contrast to [O2, section 7] we take the usual order $<$ on $\underline{n}$ here for technical reasons. A subset $I\subseteq \underline{n}$ ordered in that way will be called an ordered subset in the sequel.

Proposition 11.2   The set ${B}$$:=\{ v_{I}\vert\; I\subseteq \underline{n} \}$ is a basis of ${\bigwedge}_{R,q}(n)$.

PROOF: The fact that the set is a set of $R$ -linear generators of ${\bigwedge}_{R,q}(n)$ follows directly from the relations. Linear independence is shown using the Diamond Lemma for Ring Theory (cf. [Ha1, p. 157]). The technical details can be found in Appendix 18.1. $\Box$

${\bigwedge}_{R,q}(n)$ is a graded algebra since the relations are homogeneous of degree two. A basis for the $r$-th homogeneous summand ${{\bigwedge}_{R,q}(n,r)}$ is given by the subset ${B_r}$ of $B$ corresponding to the set ${P({n},{r})}$ of subsets $I\subseteq \underline{n}$ having cardinality $\vert I\vert=r$.

Proposition 11.3   Considered as elements of $V^{\otimes 2}$ the defining relations precislely span the kernel of the endomorphism $\beta - y {\rm id}_{}$.

PROOF: It is a matter of calculation to show that

$$\begin{array}{rl} \beta( v_{k} \wedge v_{l} + q^{-1} v_{l}... ... y^{-i+1}c_i - d_i + ( y -1)\sum_{j=i+1}^m d_j \end{array}$$

which is only hard in the last case. The technical details of that calculation can be found in Appendix 18.2. From these equations we see that the span of the relations is contained in the kernel. For $q=1$ we know that these both $R$-modules coincide. But ${\bigwedge}_{R,q}(n)$ is free as $R$-module and therefore the span of the relations is a direct summand of $V^{\otimes 2}$. Consequently both modules coincide in general. $\Box$

Proposition 11.4   ${\bigwedge}_{R,q}(n,2)$ is an $A^{{\rm s}}_{R,q}(n,2)$ comodul.

PROOF: By the previous proposition we have to show that the kernel of $\beta - y {\rm id}_{}$ is an $A:= A^{{\rm s}}_{R,q}(n,2)$ subcomodul of $V^{\otimes 2}$. Call this kernel $U$ and let $r \in U$. We must show $\tau_{}(r) \in A \otimes U$. Since $\beta$ is a morphism of the $A$-comodule $V^{\otimes 2}$ we see

$$y\tau_{} (r ) = \tau_{}(\beta(r)) = {\rm id}_{A} \otimes \beta( \tau_{} ( r ) ).$$

But this means ${\rm id}_{A}\otimes(\beta -y{\rm id}_{})(\tau_{}( r ) )=0$. Since $A, U$ and ${\bigwedge}_{R,q}(n,2)$ are free $R$-modules we may conclude $\tau_{}(r) \in A \otimes U$. $\Box$

If $B$ is a bialgebra and $A$ an algebra which is a $B$-comodul we call $A$ a $B$-comodul algebra if multiplication as well as the embedding of the unit element are morphisms of comodules.

Proposition 11.5   ${\bigwedge}_{R,q}(n)$ is an $A^{{\rm s}}_{R,q}(n)$-comolule algebra.

PROOF: The tensor algebra ${\cal T}(V)= \bigoplus_{r \in {\mathbb{N}}\,_0} V^{\otimes r}$ over $R$ has a natural structure of an $A^{{\rm s}}_{R,q}(n)$-comolule algebra (cf. [O1, 1.5]). Consequently by multiplicativity and the above proposition the ideal generated by the kernel of $\beta - y {\rm id}_{}$ is an $A^{{\rm s}}_{R,q}(n)$-comodule. But this is precisely the defining ideal of ${\bigwedge}_{R,q}(n)$ by Proposition 11.3. Thus ${\bigwedge}_{R,q}(n)$ inherits the comodul algebra structure from ${\cal T}(V)$. $\Box$

Denote the comodule structure map of ${\bigwedge}_{R,q}(n)$ by ${\tau_{\wedge}}$$:{\bigwedge}_{R,q}(n)\rightarrow {\bigwedge}_{R,q}(n)\otimes A^{{\rm s}}_{R,q}(n)$.

Proposition 11.6   The coefficient functions of ${\bigwedge}_{R,q}(n,r)$ are given by

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

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.

Let us first treat the ingredients needed in the proof of that proposition.

Lemma 11.7   Let ${\pi_r}$$:V^{\otimes r}\rightarrow {\bigwedge}_{R,q}(n,r)$ be the natural projection. Then the endomorphism ${\kappa_{\lambda}}$$:=\sum_{w \in {\cal S}_{\lambda}} (-y)^{-l(w)}\beta(w)$ factors through $\pi_r$, i.e. there is a homomorphism of $R$-modules ${\nu_r}$$:{\bigwedge}_{R,q}(n,r) \rightarrow V^{\otimes r}$ such that $\kappa_r=\nu_r\circ \pi_r$.

PROOF: Since the defining ideal of ${\bigwedge}_{R,q}(n)$ is generated by the kernel of $({\rm id}_{V^{\otimes 2}}-y^{-1}\beta)$ by Proposition 11.3 the assertion immediately follows from Lemma 9.1. $\Box$

Let ${I_{\omega_r}^{<}}$$:=\{{\bf i} \in I(n,r) \vert\; i_1<i_2< \ldots < i_r \}$ be the set of multi-indices corresponding to the ordered subsets $I\in P({n},{r})$.

Lemma 11.8   Let $F_r$ be the $R$-linear span of $\{v_{{\bf j}}\vert\; {\bf j} \in I(n,r) \backslash I_{\omega_r}^{<}\}$ in $V^{\otimes r}$. Then for all $w \in {\cal S}_{r}\backslash \{{\rm id}_{}\}$ and ${\bf i} \in I_{\omega_r}^{<}$ it follows that $\beta (w)(v_{{\bf i}}) \in F_r$.

PROOF: We use induction on $r$. The case $r=2$ directly follows from the formulas

$$\beta (v_{(k,l)}) = qv_{(l,k)}$$ (17)

$$\beta (v_{(i,i')}) = v_{(i',i)} + (y-1)\sum_{j=1}^{i-1}q^{j-i}v_{(j',j)}$$ (18)

which are valid for $k<l,\; k\neq l'$ and $i\leq m$. If $r > 2$, we embed ${\cal S}_{r-1}$ as the subgroup of ${\cal S}_{r}$ that fixes the letter $r$. If $w\in {\cal S}_{r-1}$, there is nothing to prove since $F_{r-1}\otimes V\subseteq F_r$. Otherwise, we may write $\beta (w)=\beta (w')\beta_{r-1}\beta_{r-2} \ldots \beta_l$ where $w'\in {\cal S}_{r-1}$ and $l \leq r-1$.

First consider the case where $i_l'$ is not contained in $\{i_{l+1}, \ldots , i_r\}$. Applying $\beta_{r-1}\beta_{r-2} \ldots \beta_l$ to $v_{{\bf i}}$ we only have to use (17) but not (18). Consequently, we have $\beta_{r-1}\beta_{r-2} \ldots \beta_l(v_{{\bf i}})= q^{r-l}v_{i_1}\ldots \hat{v_{i_l}}\ldots v_{i_r}v_{i_l}$. Here, $\hat{v_{i_l}}$ denotes the omission of $v_{i_l}$. This element obviously lies in $F_r$, proving the assertion in the case $w'={\rm id}_{}$. If $w'$ is not the identity map we have $\beta (w')(q^{r-l+1}v_{i_1}\ldots \hat{v_{i_l}}\ldots v_{i_r}v_{i_l})\in F_{r-1}\otimes v_{i_l}\subseteq F_r$ by the induction hypothesis since $(i_1,\ldots , \hat{i_l},\ldots ,i_r) \in I_{\omega_{r-1}}^{<}$.

We next consider the case $i_l' \in \{i_{l+1}, \ldots , i_r\}$. This forces $i_l\leq m$ because $i_l <i_l'$. Let $i_l'=i_k$. As above, we have $\beta_{k-2}\beta_{k-3} \ldots \beta_l(v_{{\bf i}})=q^{k-l-1} v_{i_1}\ldots \hat{v_{i_l}}\ldots v_{i_{k-1}} v_{i_l}v_{i_k}v_{i_{k+1}}\ldots v_{i_r}$. Applying $\beta_{k-1}$ to this expression, we have to use (18) for the first time. But for each basis element $v_{{\bf j}}$ occurring as a summand in the resulting expression we have $j_k\leq i_l\leq m$. Similar things happen concerning the remaining $\beta_k, \ldots , \beta_{r-1}$. Thus, for each $v_{{\bf j}}$ occurring as a summand in $\beta_{r-1}\beta_{r-2} \ldots \beta_l(v_{{\bf i}})$, it follows $j_r\leq i_l \leq m$. On the other hand, for each such summand there must exist an $h < r$ where $j_h >m$. This is because ${\bf j}$ must contain a pair $\{i,i'\}$ for some $i\in \underline{m}$, since this was the case for the multi-index ${\bf i}$ we started with and $\beta$ either exchanges the position of such a pair or replaces it by a sum where other such pairs occur in each summand. Consequently, we obtain ${\bf j} \in F_r$ in this case too. $\Box$

Let the coefficient matrices of the $R$-module homomorphisms $\pi_r,\;\kappa_r$ and $\nu_r$ (from Lemma 11.7) be given by

$$\pi_r(v_{{\bf j}})=\sum_{I \in P({n},{r})}\pi_{I{\bf j}}v_{I},\;\... ...(v_{{\bf j}})=\sum_{{\bf i} \in I(n,r)} \kappa_{{\bf i}{\bf j}}v_{{\bf i}}\;\;$$    and $$\;\; \nu_r(v_{J})=\sum_{{\bf i} \in I(n,r)} \nu_{{\bf i} J}v_{{\bf i}}.$$

Now, if ${\bf j}\in I_{\omega_r}^{<}$ corresponds to the ordered set $J \in P({n},{r})$ we have $\pi_r(v_{{\bf j}})=v_{J}$ yielding $\nu_{{\bf i}J}=\kappa_{{\bf i} {\bf j}}$ by Lemma 11.7. From Lemma 11.8 it follows $\kappa_r(v_{{\bf j}}) \equiv v_{{\bf j}}$ modulo $F_r$. Thus, for a pair ${\bf i},{\bf j}\in I_{\omega_r}^{<}$ of multi-indices corresponding to ordered sets $I,J\in P({n},{r})$, we obtain $\nu_{{\bf i}J}=\kappa_{{\bf i}{\bf j}}= \delta_{IJ}$ (Kronecker symbol). Finally, from $\kappa_r=\nu_r\circ \pi_r$ we see for all ${\bf i} \in I_{\omega_r}^{<}$ and ${\bf j} \in I(n,r)$

$$\kappa_{{\bf i}{\bf j}}=\sum_{K\in P({n},{r})}\nu_{{\bf i}K}\pi_{K{\bf j}}= \pi_{I{\bf j}}.$$ (19)

We are now ready to give the proof of proposition 11.6. We calculate

$$\tau_{\wedge}(v_{J})=\sum_{{\bf k} \in I(n,r)} v_{k_1}\wedge \ldots \wedge v_{k_r} \otimes x_{{\bf k} {\bf j}}=$$

$$\sum_{{\bf k} \in I(n,r)} \sum_{I \in P({n},{r})}\pi_{I{\bf k}}v_... ... \otimes \sum_{{\bf k} \in I(n,r)}\kappa_{{\bf i}{\bf k}}x_{{\bf k} {\bf j}}.$$

But, this is exactly what we wanted by the definition $T^{\omega_r}_q({\bf i}:{\bf j})= \kappa_r\wr x_{{\bf i} {\bf j}}$ of bideterminants.

The formula we just have proved has some useful consequences concerning the comultiplication and augmentation of $A:=A^{{\rm s}}_{R,q}(n)$. These are valid for any pair ${\bf i},{\bf j}\in I_{\omega_{r}}^{<}$ of multi-indices corresponding to ordered sets $I,J\in P({n},{r})$ and follow directly with the help of the comodule axioms $(\tau_{\wedge}\otimes {\rm id}_{A})\circ \tau_{\wedge}=({\rm id}_{\wedge} \otimes \Delta )\circ \tau_{\wedge}$ and $({\rm id}_{\wedge}\otimes\epsilon )\circ \tau_{\wedge} ={\rm id}_{\wedge}$:

$$\Delta (T^{\omega_r}_q({\bf i}:{\bf j}))=\sum_{{\bf k}\in I_{\ome... ...}^{<}} T^{\omega_r}_q({\bf i}:{\bf k})\otimes T^{\omega_r}_q({\bf k}:{\bf j}),$$ (20)

$$\epsilon (T^{\omega_r}_q({\bf i}:{\bf j}))= \delta_{{\bf i}{\bf j}}.$$ (21)

Another useful consequence is the following corollary:

Corollary 11.9   Let $a_{{\bf j}} \in R$ be such that $\sum_{{\bf j}\in I(n,r)}a_{{\bf j}} v_{j_1} \wedge v_{j_2}\wedge \ldots \wedge v_{j_r} = 0$. Then for all ${\bf i} \in I(n,r)$ we have

$$\sum_{{\bf j} \in I(n,r)} a_{{\bf j}}T^{\omega_r}_q({\bf i}:{\bf j})= \sum_{{\bf j} \in I(n,r)} a_{{\bf j}}T^{\omega_r}_q({\bf j}:{\bf i})=0.$$

PROOF: By Lemma 11.7 and the assumption we have

$$\kappa_r(\sum_{{\bf j}\in I(n,r)}a_{{\bf j}}v_{{\bf j}})= \sum_{{\bf j}\in I(n,r)}a_{{\bf j}}\kappa_r(v_{{\bf j}})=0.$$

Consequently, for all ${\bf k}\in I(n,r)$ we obtain $\sum_{{\bf j}\in I(n,r)}a_{{\bf j}}\kappa_{{\bf k}{\bf j}}=0$ and therefore

$$\sum_{{\bf j} \in I(n,r)} a_{{\bf j}}T^{\omega_r}_q({\bf i}:{\bf ... ...\bf j} \in I(n,r)} a_{{\bf j}}x_{{\bf i} {\bf k}} \kappa_{{\bf k}{\bf j}} =0 .$$

The equation with exchanged indices is deduced by an application of the involution $^*$ according to (8). $\Box$