Which linear combinations of simple roots are roots

Question

Let $\Delta$ be the root system of a complex simple Lie algebra, $\Delta^+$ be positive roots and $\Pi$ be simple roots. We view $\Pi$ as nodes of the Dynkin diagram.

Then for any two simple roots $\alpha$ and $\beta$, whether a linear combination $n\alpha+m\beta$ is a root can be judged easily from the Dynkin diagram.

My question is how to do this further. Precisely,

Question:

1. Given non-negative integers $(n_\alpha)_{\alpha\in\Pi}$, is there a combinatorial criterion to judge whether $\sum_{\alpha\in\Pi}n_\alpha\alpha$ is a root?

2. Is there a recursive way to enumerate all the $(n_\alpha)$'s such that $\sum_{\alpha\in\Pi}n_\alpha\alpha$ is a root?

3. In particular, how to determine combinatorially the $(n_\alpha)$ corresponding to the highest root?

Answer 1

My favorite answer to #2 and #3 is Kostant's "Find the highest root game", which is written up in detail in section 5.4 of Balázs Elek's notes on reflection groups. It is not hard to show that all plays of the game (from all starting positions, i.e., simple roots) take you through all the positive roots.

Answer 2

I guess the following is not exactly what you are looking for but is somewhat revealing. A former colleague of mine tackled these type of questions in his PhD thesis. He describes something that he calls "minimal relations" i.e. relations of minimal length between roots. He gives a complete description of these minimal relations in Section 4.2.3 of his thesis which I linked below

http://math.jacobs-university.de/penkov/papers/PhD_Milev.pdf

As far as I know, he also proves a number of criterions of when an expression of roots is again a root.

Answer 3

If you're in the need for computing all positive roots of given root system of type $X_n$ in a way that involves only clicking (quite a few times) on your mouse and without much thinking, then I recommend:

1. Download Keller's mutation applet (available here: https://webusers.imj-prg.fr/~bernhard.keller/quivermutation/).

2. Open the applet, and under File > New Quiver, input "1" for the "Enter side length". You'll see a single circle (labelled 1) in the center of the applet.

3. Create a Dynkin quiver (i.e. the Dynkin diagram with some choice of orientation) of type $X_n$, by clicking on "Add nodes" and "Add arrows", and then click "Done".

4. Under Clusters > Activate, click on d-vectors, and then again click "Done". You'll see the negative of the $n$ standard basis vectors (labelled $d[1], \ldots, d[n]$) of $\mathbb{R}^n$ appearing (twice, because we click on "Done").

5. Look for a source of the quiver (i.e. a vertex with all arrows pointing away from that vertex), and click on it. Two things will happen:

i) The quiver will update itself (this is called quiver mutation).

ii) A new vector will appear at the bottom of the applet, with non-negative coefficients. This is a positive root, expressed as a linear combination of the simple roots (where the ith coordinate of the vector is the coefficient of $\alpha_i$ in its expansion).

6. Repeat step 5) at a vertex you have not yet clicked on.

The step 5) - step 6) loop will run $n$ times, after which you will have "clicked" on every vertex exactly once. Now "forget" that you have clicked on any vertices, and repeat the step 5) - step 6) loop (performing another $n$ clicks), etc. Eventually, all positive roots will appear at the bottom of the applet.

I've attached below what this process looks like in type $E_6$, after having performed the step 5) - step 6) loop $6$ times. The algorithm has picked up the roots $\alpha_1$, $\alpha_1 + \alpha_2$, $\alpha_1 + \alpha_2 + \alpha_3$, $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4$, $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_5$ and finally $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_5+\alpha_6$. If I click on the vertices along the sequence $(1,2,3,4,5,6)$ I will get another six positive roots in the root system, etc.