MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For any lie algebra $\mathfrak g$, there is a natural filtration on $U(\mathfrak g)$ by "degree": the filtered piece $U^{\leq n}(\mathfrak g)$ is just the image in $U(\mathfrak g)$ of $\bigoplus_{k=0}^n\mathfrak g^{\otimes k}$.

Is there a natural filtration of the quantum group $U_q(\mathfrak g)$ which reduces to this filtration on $U(\mathfrak g)$ when we set $q=1$?

(of course, we restrict to the lie algebras $\mathfrak g$ for which one has a definition of $U_q(\mathfrak g)$, e.g. $\mathfrak g=\mathfrak s\mathfrak l_n$).

To increase the likelihood of a positive answer to this question, let's use the model for $U_q(\mathfrak g)$ which is often denoted $U_h(\mathfrak g)$, that is, the model over the complete power series ring $\mathbb C[[h]]$.

One possible motivation for this question is the following. One definition of the universal enveloping algebra $U(\mathfrak g)$ is that it is a dual of the complete local ring of $G$ at the identity $e\in G$ (where $\operatorname{Lie}(G)=\mathfrak g$). The complete local ring $\hat{\mathcal O}_{G,e}$ comes with a canonical set of quotients $\mathcal O_{G,e}/\mathfrak m_{G,e}^{n+1}$, and corresponding to this is the filtration $U^{\leq n}(\mathfrak g)$ by "degree". Thus we can interpret $U^{\leq n}(\mathfrak g)$ as the part of $U(\mathfrak g)$ consisting of differential operators of order $n$ at $e\in G$. I want to know whether there is a similar filtration/interpretation for quantum groups.

share|cite|improve this question
Does the PBW filtration coincide with the U(g) filtration you speak of? – B. Bischof Dec 14 '11 at 3:21
I mean the filtration whose associated graded is $\operatorname{Sym}^\ast(\mathfrak g)$ (the one which proves the PBW theorem). – John Pardon Dec 14 '11 at 3:50
I'm not an expert in quantum groups, but I don't see any reason why the naive definition of the filtration on $U_q( \mathfrak g )$ wouldn't work. Lusztig has constructed the natural choices for generating elements in degree-1 (cf "On Quantum Groups at Roots of 1"), and the filtration defined by these elements does, upon base change, give the standard PBW filtration. – Chuck Hague Dec 20 '11 at 16:45
The problem is that, say in $\mathfrak s\mathfrak l_2$, the elements $EF$ and $FE$ should be in filtration degree $\leq 2$, but their difference is $(e^{hH}-e^{-hH})/(e^h-e^{-h})$, which has terms of really large degree. So it's not clear that the filtration which puts $n$-fold products of $E,F,H$ in degree $n$ is the "right" one to consider. – John Pardon Dec 20 '11 at 17:21
Ah, I see. I'm used to the definition of the quantized enveloping algebra where $EF - FE = \begin{bmatrix} H \\ 1 \end{bmatrix}$, in which case it is clear that $EF$ and $FE$ are in the correct filtration degree. – Chuck Hague Dec 21 '11 at 17:06
up vote 2 down vote accepted

Nobody has answered this yet, so maybe I'll expand on my comment above, with the caveat that I'm no expert in this area. I believe the answer to your question is yes; the reference for all of this is Lusztig's paper Quantum Groups at Roots of 1.

Let $\Delta^+$ denote the positive roots of $G$ and let $\ell$ be the rank of $G$. Given a quantum group $U_q(\mathfrak g)$ defined over $\mathbb Q(v)$, Lusztig defines a set of $\mathbb Q(v)$-algebra generators of $U_q(\mathfrak g)$ given by elements $E_\beta^{(n)}$, $F_\beta^{(n)}$, $K_i$, $K_i^{-1}$, and $\begin{bmatrix} K_i \\ n \end{bmatrix}$ for $\beta \in \Delta^+$, $n \geq 0$, and $1 \leq i \leq \ell$. (The proof that this indeed gives a set of algebra generators is nontrivial and takes most of the paper). If we put $K_i$ and $K_i^{-1}$ in degree 1 and the other elements in degree $n$ then the explicit relations given in Lusztig's paper show that this indeed defines a degree filtration on $U_q(\mathfrak g)$.

Furthermore, when one goes through the standard quantum rigamarole for base change (take the $\mathbb Z[v,v^{-1}]$-subalgebra generated by the elements above, base change to your favorite field considered as a $\mathbb Z[v,v^{-1}]$-algebra, and mod out by $K_i - 1$ and $K_i^{-1} - 1$) these elements go to the standard elements in the hyperalgebra, so in particular when one plays this game in characteristic zero, we do get the standard filtration on the enveloping algebra $U(\mathfrak g)$.

share|cite|improve this answer

In the semisimple case, over the completed power series ring the quantum group is actually isomorphic as an algebra to the usual classical enveloping algebra (this follows from general nonsense about deformation theory and the Whitehead lemmas and a few other standard facts; I don't think anyone knows of an explicit isomorphism) it follows that there is such a filtration as you want.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.