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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.

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  • $\begingroup$ The definition of divided power is missing brackets. I tried to put them in but computer said No. $\endgroup$
    – BWW
    Jul 28, 2016 at 10:15
  • $\begingroup$ @BWW, thank you very much. I will edit the post. $\endgroup$ 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. $\endgroup$ 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 I read this example in Professor Arkady Berenstein's unpublished lecture notes. $\endgroup$ Jul 28, 2016 at 13:36

1 Answer 1

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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 $$E_1E_2E_1=E_1^{(2)}E_2+E_2E_1^{(2)}.$$

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