Does there exist a canonical "degree" filtration on quantum groups? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:30:20Z http://mathoverflow.net/feeds/question/83379 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83379/does-there-exist-a-canonical-degree-filtration-on-quantum-groups Does there exist a canonical "degree" filtration on quantum groups? unknown (google) 2011-12-14T00:06:20Z 2011-12-23T19:30:19Z <p>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 $\bigoplus_{k=0}^n\mathfrak g^{\otimes k}$.</p> <blockquote> <p>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 $q=1$?</p> </blockquote> <p>(of course, we restrict to the lie algebras $\mathfrak g$ for which one has a definition of $U_q(\mathfrak g)$, e.g. $\mathfrak g=\mathfrak s\mathfrak l_n$).</p> <p>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 $\mathbb C[[h]]$.</p> <p>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 <code>$\hat{\mathcal O}_{G,e}$</code> 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.</p> http://mathoverflow.net/questions/83379/does-there-exist-a-canonical-degree-filtration-on-quantum-groups/84175#84175 Answer by Chuck Hague for Does there exist a canonical "degree" filtration on quantum groups? Chuck Hague 2011-12-23T18:55:21Z 2011-12-23T18:55:21Z <p>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 <a href="http://www.math.toronto.edu/lzhang/Seminars/References/Lusztig-QGrpRootsUnity.pdf" rel="nofollow">Quantum Groups at Roots of 1</a>.</p> <p>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 <code>$\begin{bmatrix} K_i \\ n \end{bmatrix}$</code> 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 $U_q(\mathfrak g)$.</p> <p>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 $U(\mathfrak g)$.</p> http://mathoverflow.net/questions/83379/does-there-exist-a-canonical-degree-filtration-on-quantum-groups/84178#84178 Answer by Mariano Suárez-Alvarez for Does there exist a canonical "degree" filtration on quantum groups? Mariano Suárez-Alvarez 2011-12-23T19:30:19Z 2011-12-23T19:30:19Z <p>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.</p>