What is the q-analogue of the Lefschetz decomposition?

The representation theory behind the Lefschetz decomposition in Kahler geometry was summarised very neatly by Victor Protsak in his answer to 29907

Let $W$ be a $2n$-dimensional symplectic vector space, $\bigwedge^\ast W$ its exterior algebra, and $\omega\in\bigwedge^2 W$ the invariant two-form. Exterior multiplication by $\omega$ and the contraction with $\omega$ define a pair of $Sp(W)$-equivariant graded linear transformations $L, \Lambda$ of $\bigwedge^\ast W$ into itself of degrees $2$ and $-2,$ and let $H=\deg-n$ be the graded degree $0$ map acting on $\bigwedge^k$ as multiplication by $k-n.$ Then $L,H,\Lambda$ form the standard basis of the Lie algebra $\mathfrak{sl_2}$ acting on $\bigwedge^\ast W$ and the actions of $Sp(W)$ and $\mathfrak{sl_2}$ are the commutants of each other.

The representation theory is discussed in Remarks on classical invariant theory by Roger Howe. This is also discussed in Griffiths and Harris "Principles of algebraic geometry" in Chapter 0, Section 7.

My question is whether anyone knows the $q$-analogue of the representation theory (and just to be clear I mean explicitly and not in principle). More specifically, we have an inclusion of $U_q(A_{n-1})$ in $U_q(C_n)$ by ignoring the end node. The vector representation restricts to $U\oplus U^*$ where $U$ is also the vector representation. Then I want to define a $q$-analogue of the exterior algebra; ideally so it is a $U_q(C_n)$-module algebra. Then I want to construct the commutant, which ideally would be (a quotient of) $U_q(A_1)$.

I hope that by now there is a consensus of the definition of $U_q$.

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I am not sure why you need to invoke type $A$ - it's purely symplectic question: skew duality between symplectic groups in Clifford representation. There was some work on $q$-analogues of the "usual" (symmetric) Howe duality by Japanese mathematicians, I don't have the references at the moment; it may have involved "non-consensus" def of $U_q$, though. – Victor Protsak Aug 2 '10 at 17:08
@Victor. Yes, I agree. I was thinking that I should take $\Lambda(U)\otimes \Lambda(U^*)$ (with $q$) and extend the action of $U_q(A_{n-1})$ to $U_q(C_n)$. – Bruce Westbury Aug 2 '10 at 17:24