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.



Projective modules over quantum groups
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.


