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Consider the example of $\mathfrak{g} = sl_3$. Then $$ \mathfrak{g} = \mathfrak{n} \oplus \mathfrak{h} \oplus \mathfrak{n}^{-}, $$ where $\mathfrak{n}$ is generated by $E_{12}, E_{13}, E_{23}$, $\mathfrak{h}$ is generated by $E_{11}-E_{22}, E_{22}-E_{33}$, $\mathfrak{n}^{-}$ is generated by $E_{21}, E_{32}, E_{31}$, $E_{ij}$ is a matrix with $1$ at $(i,j)$ and $0$ elsewhere.

A PBW basis of $U(\mathfrak{n})$ is $$ B_1 = \{ E_1^a (E_1 E_2 - E_2 E_1)^b E_2^c \mid a, b, c \in \mathbb{N} \}. $$

There is another basis of $U(\mathfrak{n})$ called canonical basis which is given by $$ B_2 = \{ E_1^aE_2^bE_1^c \mid a+c \geq b \} \cup \{ E_2^aE_1^bE_2^c \mid a+c \geq b \}. $$

The basis $B_2$ has a property: given a lowest weight vector $v_0$ of a representation $V$ of $U(\mathfrak{n})$, the set $$ \{ b v_0 \mid b v_0 \ne 0 \} $$ is a basis of $V$.

My question is: in general, how to compute canonical basis for $U(\mathfrak{n})$ explicitly like the above example. Can we derive a canonical basis from a PBW basis? Thank you very much.

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  • $\begingroup$ Can you please say what is $E_1$ and $E_2$ in the given bases? The original bases for the Lie algebra is in terms of $E_{i,j}$, so these bases also should be in terms of $E_{i,j}$ no? why in terms of $E_1$ and $E_2$ ? $\endgroup$ – GA316 Jan 24 '17 at 5:57
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    $\begingroup$ @GA316, sorry, I used different notations. Here $E_1=E_{12}$, $E_{2}=E_{23}$. $\endgroup$ – Jianrong Li Jan 24 '17 at 9:05
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It is possible to obtain the canonical basis from the PBW basis, as long as you are working in the quantum group Uq(n). The canonical basis seems to be inherently a quantum phenomenon, so it shouldn't be surprising that we want to compute in the quantum group.

If Eπ is an indexing of the quantum PBW basis, then what one first proves is that applying the bar involution is unitriangular with respect to this basis. There is even a mathoverflow question about this: Convex PBW bases.

Now it is a matter of linear algebra to prove that there exists a unique bar invariant basis bπ such that $$b_\pi=\sum_\sigma c_{\pi\sigma} E_\sigma$$ with cσσ=1, cσπ∈qℤ[q] if σ≠π and cσπ=0 unless σ≤π. This basis bπ is the canonical basis.

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