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.



PBW basis and canonical basis
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](https://mathoverflow.net/questions/110108/).


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.


