# Can one easily pick out a basis of a simple Lie algebra after picking a convex order?

One of the basic results of Lie theory is that if one picks a Cartan subalgebra of a simple Lie algebra, there there is a canonical decomposition of $\mathfrak{g}$ into the Cartan and a bunch of 1-dimensional subspaces (root spaces). Picking any basis of the Cartan (I'm not going to worry about which one), one is thus extremely close to having a canonically chosen basis of the Lie algebra, except that there are a bunch of scalar factors running around. I'd like to fix a particular choice of basis, while making a somewhat manageable number of choices.

Choice 1: I'll pick a base of my root system, that is, a set of simple roots/positive roots/a Borel containing my chosen Cartan. I can now choose completely at random vectors $E_i$ in the simple root spaces (all of these are conjugate by inner automorphisms, so they are all "the same"). This fixes basis vectors $F_i$ in the negative simple root spaces that commute with the $E_i$'s as the standard generators of $\mathfrak{sl}_2$.

Choice 2: Well, now I have to do something else non-canonical. What I'd like to choose is a convex order on my positive roots. This case, let be propose a basis for the other root spaces by $E_{\alpha+\beta}=[E_{\alpha},E_{\beta}]$ if $\alpha <\beta$ and similarly for the $F$'s.

My question: is $E_{\alpha}$ uniquely defined? Has this construction appeared before in the literature? If this definition doesn't work, is there a variant on it which does?

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What is a convex order? – Vít Tuček Jun 22 '14 at 18:54

What you are trying to do is done in complete detail in Leclerc's paper "Dual canonical bases, quantum shuffles and q-characters" based on the paper "Standard Lyndon bases of Lie algebras and enveloping algebras" by Lalonde and Ram.

The root spaces are one dimensional, so you have to put conditions on $\alpha$ and $\beta$ to get a well defined $E_{\alpha+\beta}$. Roughly speaking, you start by observing that convex orderings on positive roots are equivalent to total orderings on simple roots. So fix one such ordering.

In the (quantized) enveloping algebra you can start to think about words in the simple roots associated to (ordered) products of the corresponding $E_i$. For reasons explained in these papers, the monomial $E_{i_1}\cdots E_{i_n}$ corresponds to the linear combination of words obtained by (quantum) shuffling the symbols $i_1,i_2,\ldots,i_n$ together in all possible ways and summing the outcomes (Note that in the quantum context, this is a noncommutative product!). Write this shuffle product as $$i_1*\cdots*i_n=\sum_{w\in S_n}q^{(\star)}\; i_{w(1)}\cdots i_{w(n)},$$ where $(\star)$ can be given explicitly. Let $\max(i_1*\cdots*i_n)$ be the largest monomial (in the lexicographic order) in the sum $i_1*\cdots*i_n$ (such a monomial is called \emph{good} or \emph{standard}). The set of all possible $\max(i_1*\cdots*i_n)$ for $n\geq 1$ can be identified with the crystal $B(\infty)$. It is also worth pointing out that there is a nice triangularity property: $E_{i_1}E_{i_2}\cdots E_{i_n}$ corresponds to $$i_1*i_2*\cdots*i_n=i_1i_2\cdots i_n+\sum(words < i_1i_2\cdots i_n).$$

Anyway, given a word $w\in \mathcal{B}(\infty)$, call it "standard Lyndon" if it is larger than any subword ($w=w_1 w_2$ implies $w_1 < w$). The key fact is that the set of standard Lyndon words are in bijection with the set of positive roots. Leclerc explains how to calculate the set standard Lyndon words in Proposition 25 of his paper.

For a positive root $\alpha$, let $w(\alpha)$ be the corresponding word. You construct the root vectors inductively by $E_\alpha=[E_{\alpha_1},E_{\alpha_2}]$, where

$\bullet$ $w(\alpha)=w(\alpha_1)w(\alpha_2)$, $\alpha=\alpha_1+\alpha_2$, and $\alpha_1\neq 0$, $\alpha_2\neq0$ are positive roots,

$\bullet$ $w(\alpha_1)$ is standard Lyndon, and

$\bullet$ if $w(\alpha)=w_1w_2$, with $w_1$ standard Lyndon, then $w_1$ is a shorter word than $w(\alpha_1)$.

This construction has also been carried out in the affine case by Beck-Chari-Pressley and Beck-Nakajima. In my paper with Melvin and Mondragon, we make these calculations explicitly in all types (well, type $E_8$ isn't exactly explicit, but almost). Note, however, we use the middle east reading convention when defining the lex. ordering.

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I should have thought of this; I've actually been reading that Leclerc paper a lot lately. It's funny how papers that go in the box for one project are harder than they should be to bring to mind for another project. – Ben Webster Jan 2 '12 at 0:08

Certainly for any choice of convex ordering you can pick out a basis of your simple Lie algebra. You do this by specialising the construction of PBW bases for quantum groups to q=1. (This construction can be presented without knowing about quantum groups).

With your proposed formula for $E_{\alpha+\beta}$ as a commutator, I would encourage you to modify it by dividing by $p_{\alpha,\beta}+1$ where $p_{\alpha,\beta}$ is the largest integer $p$ such that $\beta-p\alpha$ is a root. This way you will get the correct integral form of the Lie algebra. See e.g. these notes of Bill Casselman.

I don't know if the formula $E_{\alpha+\beta}=[E_\alpha,E_\beta]/(p_{\alpha,\beta}+1)$ holds for all $\alpha$ and $\beta$. Certainly it holds up to sign. If you make a further assumption that $(\alpha,\beta)$ is what I call a minimal pair, then this formula is correct, even in the quantum case, see Theorem 4.2 of arXiv:1210.6900.

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I seem to recall a very careful treatment of this in Jacob Lurie's minor thesis.

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