Let $A=(a_{ij})$ be a generalized Cartan matrix of order $n$ and $D=diag(d_1,\ldots,d_n)$ the diagonal matrix such that $DA$ is symmetric. Let $$E_{ij}=\sum_{r+s=1-a_{ij}} (-1)^r E_i^{(r)} E_j E_i^{(s)},$$ $i,j\in \{1, \ldots, n\}$, $E_i^{(k)} = \frac{E_i^k}{[k]_{q_i}!}$, $q_i = q^{d_i}$.

Let $$ U_+ = \mathbb{C}(q)\langle E_1, \ldots, E_n \rangle/\langle E_{ij}: i \neq j \rangle. $$

Let $A=\left( \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right)$. How to show that the set $$B = \{E_1^{(a)}E_2^{(b)}E_1^{(c)}, E_2^{(c)}E_1^{(b)}E_2^{(a)}: b \geq a+c\}$$ is a canonical basis for $U_+$? Where does the condition $b \geq a+c$ come from? Thank you very much.

  • $\begingroup$ The definition of divided power is missing brackets. I tried to put them in but computer said No. $\endgroup$
    – BWW
    Commented Jul 28, 2016 at 10:15
  • $\begingroup$ @BWW, thank you very much. I will edit the post. $\endgroup$ Commented Jul 28, 2016 at 10:49
  • $\begingroup$ @Jianrong: It would be helpful here to specify your sources, since as Sean Clark observes this is treated explicitly in Lusztig's standard (though not easy) book. $\endgroup$ Commented Jul 28, 2016 at 13:28
  • $\begingroup$ @Jim Humphreys, thank you very much. I read this example in Professor Arkady Berenstein's unpublished lecture notes. $\endgroup$ Commented Jul 28, 2016 at 13:36

1 Answer 1


The complete calculation is done in Lusztig's book (Lemma 42.1.2). Essentially, the condition $b\geq a+c$ comes from the Serre relations; e.g. we have $$E_1E_2E_1=E_1^{(2)}E_2+E_2E_1^{(2)}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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