Projective modules over quantum groups 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.
 A: 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.
A: This is a good question; when $q$ is not formal, I believe that Tor-groups of $U_q(\mathfrak{g})$ can differ from $U(\mathfrak{g})$ considerably.  As Mariano said, if we take the formal quantum group, with $q=e^\hbar$, the answer is the same as for $U(g)$, so I'll address non-formal $q$.
I interpret your question as asking for a free resolution of $k=\mathbb{C}$, since it is then easy to compute Tor from this.  The following is a free resolution of $k$ over $U=U_q(\mathfrak{sl}_2)$.
Recall that $U$ has generators $E$,$F$, and $K$, $K$ is invertible, and relations $EF-FE= \frac{K-K^{-1}}{q-q^{-1}}$, $KE=q^2EK$, and $KF=q^{-2}FK$.  Here is a resolution of $k=\mathbb{C}$ as a module over $U$ via $\epsilon$.
$\begin{align*}
M_{-1}&=k\\
M_0&= U\\
M_1 &= U\otimes E \oplus U\otimes F \oplus U\otimes (K-1))\\
M_2 &= U\otimes (E\otimes F- F\otimes E - \frac{K^{-1}+1}{q-q^{-1}}\otimes (K-1))\\
&\oplus U\otimes (q^{-2}(K-1)\otimes E-E \otimes (K-1)+(q^{-2}-1)\otimes E)\\
& \oplus U\otimes (q^2(K-1)\otimes F - F\otimes (K-1) + (q^2-1)\otimes F)\\
M_3&=U \otimes \Big[(1-K)\otimes(E\otimes F- F\otimes E - \frac{K^{-1}+1}{q-q^{-1}}\otimes (K-1))\\
&+F\otimes(q^{-2}(K-1)\otimes E-E \otimes (K-1)+(q^{-2}-1)\otimes E)\\
&-E\otimes(q^2(K-1)\otimes F - F\otimes (K-1) + (q^2-1)\otimes F))\Big]
\end{align*}$
Having written all that out, I'm depressed that it's typeset so ugly =[.  I hope the notation is clear.  The tensorands on the right are just formal symbols, but they suggestively tell you what the differential is: you just multiply the $U$-coefficient by first tensorand, to move from $M_k$ to $M_{k-1}$.  e.g, for $X\in U$, we have $d(X\otimes E)=XE$.
It's a good exercise to check that it is a free resolution; I will omit the computations.  For instance, exactness at $M_0$ is the claim that $Ker(\epsilon) = <E,F,K-1>$.  Exactness at $M_1$ has to do with the relations of $U$.  Note that the free rank over $U$ is the same as in the C-E complex.
To finish up, note that $k\otimes_U -$ on this complex just means applying $\epsilon$ to the $U$ term everywhere, so we get:
$$\mathbb{C}v^3_1 \to \mathbb{C}\langle v^2_1, v^2_2, v^2_3,\rangle \to\mathbb{C}\langle v^1_1,v^1_2,v^1_3\rangle \to\mathbb{C}v^0_1,$$
where I have named the basis vectors of $M_k$ $v^k_?$, in the order they appear above, e.g. $v^1_1=E$.
Okay, note that: $d(M_1)=0$, but $d(M_2)=M_1$, because $\epsilon(\frac{K^{-1}+1}{q-q^{-1}})=\frac{2}{q-q^{-1}}$, $\epsilon(q^{-2}-1)$, and $\epsilon(q^2-1)$ are non-zero.  Finally, $d(M_3)=0$ (as it had better, to form a complex!)
Thus we have
$$Tor_j(k,k)=\left\{\begin{array}{ll}k,& j=0,3\\0,&j=1,2\\&0,j\geq 4\end{array}\right.$$
Note that I computed the above only for $\mathfrak{sl}_2$, but it should be possible to do the same for arbitrary quantum groups, by similarly following one's nose. I suppose, however, that one would need in that case to use the cubic $q$-serre relations, e.g. between $E_1$ and $E_2$ for $sl_3$, which are a pain.
