I am trying to understand some construction done by Lusztig in his book on quantum groups. Given some Cartan datum, let $U=U_q(\mathfrak{g})$ the standard quantized enveloping algebra of the Kac-Moody algebra $\mathfrak{g}$. Its negative part $U^{-}$ (the subalgebra of $U$ generated by the $f_i$'s) has a canonical basis $B$. In his book Introduction to Quantum Groups (don't be fooled by the title!), Lustzig constructs the non-unital extension $\dot{U}$ and proves that it has a canonical basis $\dot{B}$.

As a set, $\dot{B}$ is in bijection with $B\times X \times B$ (where $X$ is the lattice of weights, normally called $P$ in any other books on quantum groups), and its elements are described with the rather cryptic notation $b\diamond_\zeta b''$. The definition of those elements is however very obscure and non-explicit. I would appreciate finding an easier, more explicit, description, like the one given in section 25.3 in Lustzig's book for $U_q(\mathfrak{sl}_2)$, also described in great detail by Lauda in the first part of his paper A categorification of quantum sl(2). It is not clear to me how this description can be extended to more complicated quantum groups.

Does anyone know any similar simple description of the canonical basis $\dot{B}$, even in some other particular cases? I am also very interested of knowing if is there any relation with crystals, akin to the equivalence between $B$ and Kashiwara's crystal basis $B(\infty)$.

  • 1
    $\begingroup$ Is it really $B\times X\times B$? I haven't read that book for a very long times, but $B\times B$ sounds much more right to me. In the idempotented form, any element of the Cartan is just a big sum of scalars times idempotents. $\endgroup$ – Ben Webster Dec 7 '09 at 17:36
  • $\begingroup$ I think its really $B\times X \times B$. The extension $\dot{U}$ is constructed by adding one idempotent $1_\lambda$ for each weight $\lambda\in X$. For $U_q(\mathfrak{sl}_2)$. the elements of $\dot{B}$ can be identified (in a somehow nontrivial way with and elements of the form $E^{(a)}1_{-n}F^{(b)}$ or $F^{(b)}1_{n}E^{(a)}$ whenever $n\geq a+b$, but the identification looks like pulled out of thin air and I don't know if it can be extended to other quantum groups. $\endgroup$ – javier Dec 7 '09 at 18:17
  • 2
    $\begingroup$ It's really $B\times X\times B$ -- you should think of $U^-\times T \times U^+$ or something. $\endgroup$ – Kevin McGerty Dec 7 '09 at 18:17
  • $\begingroup$ Ah, I see now where my brain was going wrong. $\endgroup$ – Ben Webster Dec 7 '09 at 18:43

The basis $\dot{B}$ is usually hard to compute, in the same way the Kazhdan-Lusztig basis for the Hecke algebra is. On the other hand, at least for type A, there is another way to get at this basis via what you might call "quantum Schur-Weyl duality": there's a geometric construction of finite dimensional quotients of the quantum group of type sl_n due to Beilinson-Lusztig-MacPherson, each of which has a canonical basis. More recent work of Lusztig, Ginzburg and Schiffmann-Vasserot showed that these finite dimensional quotients could be put into a compatible family, from which you can recover the whole modified quantum group, and it's canonical basis. How much more explicit this realization is probably depends on what sort of question you're interested in I imagine.

With regard to the crystal structure, there's a paper of Kashiwara "Crystal bases of modified quantized enveloping algebras" in Duke which studies this. He shows that the crystal structure reflects the $B\times X\times B$ parametrization you mention, and investigates a crystal basis of a "dual algebra" (a kind of quantum coordinate algebra) for which the crystal structure is a kind of Peter-Weyl theorem (at least for the finite type case). Lusztig investigates something similar in his book.

  • $\begingroup$ Thanks for the answer! The paper by Kashiwara seems to contain precisely what I need. $\endgroup$ – javier Dec 8 '09 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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