TITLE: Projective modules over quantum groups
QUESTION [14 upvotes]: My question is short:

How can one calculate $\operatorname{Tor}_{U_q(\mathfrak g)}(k,k)$?

($k$ is the ground field of characteristic zero).
If we had a regular universal enveloping algebra $U(\mathfrak g)$, then of course this is just $H_\ast(\mathfrak g,k)$.  To calculate it, we just take a projective resolution of $k$, for example:
$$\cdots\to U(\mathfrak g)\otimes\mathfrak g^{\wedge 2}\to U(\mathfrak g)\otimes\mathfrak g\to U(\mathfrak g)\xrightarrow\epsilon k$$
is the standard resolution (the last arrow is the counit).  In fact, it is a free resolution!

Is there a corresponding standard free resolution of $k$ as a $U_q(\mathfrak g)$-module?

One can of course use the counit for the first map, but after that I'm not sure.

REPLY [6 votes]: I'll assume that $U_q(\mathfrak{g})$ is the quantum group defined over the rational function field $k=\mathbb{C}(q)$ in the indeterminate $q$. Then Poincare duality is known to hold for $U_q(\mathfrak{g})$; see for example the paper of Chemla (Corollary 3.2.2, J. Algebra 276 (2004), 80-102). This tells us that $\mathrm{Tor}_n^{U_q(\mathfrak{g})}(k,k)$ is isomorphic (as a vector space) to $\mathrm{Ext}_{U_q(\mathfrak{g})}^{d-n}(k,k)$, where $d = \dim \mathfrak{g}$. You can also apply a version of the universal coefficient theorem to show that $\mathrm{Ext}_{U_q(\mathfrak{g})}^n(k,k) \cong \mathrm{Tor}_n^{U_q(\mathfrak{g})}(k,k)^*$ (vector space dual). So in answer to the question of how to compute the Tor group, you compute the Ext group instead.
I am not aware of any general results on explicit projective resolutions for $U_q(\mathfrak{g})$, but in my paper, I was able to compute the Ext groups via an indirect comparison to the ordinary Lie algebra cohomology, and determined that they are all of the same dimension as the corresponding cohomology groups for the Lie algebra $\mathfrak{g}$.
The case when $q$ is not an indeterminate seems more difficult, and I don't know how to handle it at this time.