## 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.

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which version of $U_q$ are you interested in? – mt Dec 16 2011 at 10:04
I would prefer the version over $\mathbb C(q)$, but if one needs to work over $\mathbb C[[h]]$ to get a nice answer, I guess that might be ok. – unknown (google) Dec 16 2011 at 19:13
I am not sure that the construction you described works in the classical case. You must be considering the usual Chevalley-Eilenberg complex, where U(g) is viewed as a g-module with its adjoint action. Now if you want the differential in this complex to be compatible with the U(g) action, then you need to take adjoint action on each factor, but the adjoint action is not free. – Victor Dec 18 2011 at 0:19
The Drinfeld-Jimbo $U_h(\mathfrak g)$ is actually isomorphic to $U(\mathfrak g)[[h]]$ so just adding $[[h]]$ to the Chevalley-Eilenberg resolution will do the trick in the completed case. The uncompleted case is more fun. – Mariano Suárez-Alvarez Dec 18 2011 at 0:25
Victor, the Chevalley-Eilenberg complex, which looks exactly like the one pictured by unknown in his/her question, is a very classical object; you can find its constructed (apart from the original paper of Chevalley and Eilenberg, of course) in pretty much every good textbook on homological algebra, starting from Cartan-Eilenberg. In particular, the modules are just free modules and the differential does respect the action very much! – Mariano Suárez-Alvarez Dec 19 2011 at 3:20

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.

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You have just computed the quantum case... surely you can do the classical $sl_2$ too :) – Mariano Suárez-Alvarez Dec 19 2011 at 23:57
You can present the quantum $sl_n$s using only quadratic relations using Ringel's presentation of the positive/negative parts as iterated Ore extensions, as some other types too. Ringel's presentation for $G_2$ has cubic relations, though, for example. – Mariano Suárez-Alvarez Dec 20 2011 at 0:36
@Mariano: I don't doubt that I could do it for classical $\mathfrak{sl}_2$. I just don't want to =]. I half-heartedly searched for a reference, but did not find one quickly. That's good to know about Ringel's presentation; is the condition that the Dynkin diagram be simply laced? – David Jordan Dec 20 2011 at 0:57
$sl_2$ has the cohomology of a $3$-sphere (by the Whitehead lemmas, $H^1$ and $H^2$ are zero because the algebra is semisimple, and $H^3$ is never zero for a semisimple algebra and cannot be larger than $1$-dimensional) and since the algebra is unimodular, the homology $Tor(k,k)$ is just the dual of cohomology. So the quantum case has the same homology as the classical case. – Mariano Suárez-Alvarez Dec 20 2011 at 1:13

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.

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 To compute over $\mathbb C(q)$ can't one extend scalars to the field of Laurent series, where the álgebra becomes just the classical enveloping algebra? – Mariano Suárez-Alvarez Dec 24 2011 at 4:42 I don't know. Perhaps I am not as familiar with the Laurent series version as I ought to be. I was interesting in computing the cohomology ring structure. I don't know how well you could work backwards from knowing the answer over the field of Laurent series to get the answer over $\mathbb{C}(q)$. – Christopher Drupieski Dec 24 2011 at 14:59 Extension of scalars will compute the dimensions of the cohomology groups, but only tell you that the cohomology algebra is a form of the usual exterior algebra (one can hope that a graded $k(q)$-algebra which becomes an exterior algebra generated by odd elements over $k[[h]]$ is already an exterior algebra over $k(q)$, though :) ) – Mariano Suárez-Alvarez Dec 24 2011 at 15:39