Commutation relations in $G_2$

In this post I calculate the structure constants of the exceptional Lie algebra $G_2$. I assume the reader is familiar with Lie algebras, for example at the level of chapter 9 in [1].
If $\alpha_1$ and $\alpha_2$ are the simple roots, with $\alpha_1$ the long root and $\alpha_2$ the short root, then all positive roots are $$\begin{equation*} \alpha_1, \alpha_2, \alpha_1 + \alpha_2, \alpha_1 + 2 \alpha_2, \alpha_1 + 3 \alpha_2, 2 \alpha_1 + 3\alpha_2 \end{equation*}$$ The negative roots are the negatives of these positive roots. In the Chevalley basis, the commutators are $$\begin{align*} [h_{\alpha} , h_{\beta} ] &= 0 &\\ [h_{\alpha} , e_{\beta} ] &= 2 \frac{\alpha \cdot \beta}{\alpha^2} e_{\beta} &\\ [e_{\alpha} , e_{-\alpha} ] &= h_{\alpha}\\ [e_{\alpha} , e_{\beta} ] &= n_{\alpha \beta}e_{\alpha+\beta}\quad&&\text{if $\alpha + \beta \neq 0$ and $\alpha + \beta$ is a root}\\ [e_{\alpha} , e_{\beta} ] &= 0 \quad&&\text{if $\alpha + \beta \neq 0$ and $\alpha + \beta$ is not a root} \end{align*}$$
Here, $\alpha, \beta$ are roots, $h_{\alpha}$ are elements of the Cartan subalgebra and $e_{\alpha}$ are the ladder operators. It can be shown that $$\begin{equation}\label{eq:20160903} n_{\alpha \beta} = \pm (p+1) \end{equation}$$ with $p \ge 0$ the largest integer such that $\beta - p\ \alpha$ is a root. $n_{\alpha \beta}$ is thus determined up to a sign. The most useful information about these signs that I found online is a book by Samelson [2]. On pages 49 and further he gives a procedure to find the correct signs in $\eqref{eq:20160903}$. In the rest of the post I follow this procedure to calculate all signs for the $G_2$ algebra, and thus to calculate all commutators for the $G_2$ algebra.

Firstly, one observes that the commutators for the positive roots determine the commutators for all roots. Indeed, if both roots are negative one can use the formula $$\begin{equation*} n_{-\alpha,-\beta} = - n_{\alpha,\beta} \end{equation*}$$ If one root is negative and the other is positive one can use the following formula. Suppose $\alpha \ge 0$ and $\beta \ge 0$, then $$\begin{equation}\label{eq:20160909a} n_{-\alpha,\beta} = n_{\alpha,-\alpha + \beta} \frac{(-\alpha + \beta)^2}{\beta^2} \quad\text{if}\quad -\alpha + \beta > 0 \end{equation}$$ and $$\begin{equation}\label{eq:20160909b} n_{-\alpha,\beta} = n_{\beta,\alpha - \beta} \frac{(\alpha - \beta)^2}{\alpha^2} \quad\text{if}\quad \alpha - \beta > 0 \end{equation}$$ This is proved on page 47 in Samelson.

The commutators for $G_2$ are now obtained with the following steps
1) Using $\eqref{eq:20160903}$ one finds $$\begin{equation*} [ e_{\alpha_2} , e _{\alpha_1} ] = \pm e_{\alpha_1 + \alpha_2} \end{equation*}$$ I can choose the plus sign by changing the normalization of $e_{\alpha_1 + \alpha_2}$ if needed. Similarly, $[ e_{\alpha_2} , e _{\alpha_1 + \alpha_2} ] = +2 e_{\alpha_1 + 2 \alpha_2}$ and $[ e_{\alpha_2} , e _{\alpha_1 + 2 \alpha_2} ] = +3 e_{\alpha_1 + 3 \alpha_2}$. Also, $[ e_{\alpha_1} , e _{\alpha_1 + 3 \alpha_2} ] = e_{2 \alpha_1 + 3 \alpha_2}$ by choosing the sign of $e_{2 \alpha_1 + 3 \alpha_2}$

2) I also have $$\begin{equation*} [ e_{\alpha_1 + \alpha_2} , e_{\alpha_1+ 2 \alpha_2} ] = \pm 3 e_{2 \alpha_1 + 3 \alpha_2} \end{equation*}$$ Because I have already fixed the normalization of $e_{2 \alpha_1 + 3 \alpha_2}$ I cannot freely choose the sign anymore. I obtain the sign from the Jacobi identity $$\begin{equation*} [ e_{-\alpha_1}, [ e_{\alpha_1+ \alpha_2}, e_{\alpha_1+ 2 \alpha_2} ]] + [ e_{\alpha_1+ \alpha_2}, [e_{\alpha_1+ 2 \alpha_2} , e_{-\alpha_1}]] + [e_{\alpha_1+ 2 \alpha_2}, [ e_{-\alpha_1} , e_{\alpha_1+ \alpha_2}]] = 0 \end{equation*}$$ Thus $$\begin{equation*} n_{\alpha_1+ \alpha_2,\alpha_1+ 2 \alpha_2} [ e_{-\alpha_1}, e_{2 \alpha_1+ 3 \alpha_2}] + 0 + n_{-\alpha_1,\alpha_1+ \alpha_2} [ e_{\alpha_1+ 2 \alpha_2}, e_{\alpha_2}] =0 \end{equation*}$$ Thus $$\begin{equation*} n_{\alpha_1+ \alpha_2,\alpha_1+ 2 \alpha_2} n_{-\alpha_1,2 \alpha_1+ 3 \alpha_2} + n_{-\alpha_1,\alpha_1+ \alpha_2} n_{\alpha_1+ 2 \alpha_2, \alpha_2} =0 \end{equation*}$$ From $\eqref{eq:20160909a}$ I have $$\begin{equation*} n_{-\alpha_1,\alpha_1+ \alpha_2} = n_{\alpha_1,\alpha_2} \frac{\alpha_2^2}{(\alpha_1+ \alpha_2)^2} = n_{\alpha_1,\alpha_2} = -1 \end{equation*}$$ and similarly $$\begin{equation*} n_{-\alpha_1,2 \alpha_1+ 3 \alpha_2} = n_{\alpha_1,\alpha_1+ 3 \alpha_2} \frac{(\alpha_1+ 3 \alpha_2)^2}{(2 \alpha_1+ 3 \alpha_2)^2} = n_{\alpha_1,\alpha_1+ 3 \alpha_2} = 1 \end{equation*}$$ and therefore $$\begin{equation*} n_{\alpha_1+ \alpha_2,\alpha_1+ 2 \alpha_2} = - 3 \end{equation*}$$
All other commutators can be obtained in a similar manner.

Here is a list of all commutators of $G_2$. Because the list is long, I organize it as follows.

Commutators of the form $[h , e ]$
If $\alpha = r \alpha_1 + s \alpha_2$ then $[ h_{\alpha_1} , e_{\alpha} ] = (2 r -s ) e_{\alpha}$ and $[ h_{\alpha_2} , e_{\alpha} ] = (-3 r + 2 s ) e_{\alpha}$

Commutators of the form $[ e_{\alpha}, e_{-\alpha} ]$
Because of the antisymmetry of the commutator, I only write down the commutators for $\alpha \ge 0$ $$\begin{align*} [e_{\alpha_2},e_{-\alpha_2}] & = h_{\alpha_2} & [e_{\alpha_1},e_{-\alpha_1}] & = h_{\alpha_1} \\ [e_{\alpha_1+\alpha_2},e_{-\alpha_1-\alpha_2}] & = 3 h_{\alpha_1}+h_{\alpha_2}& [e_{\alpha_1+2 \alpha_2},e_{-\alpha_1-2 \alpha_2}] &= 3 h_{\alpha_1}+2 h_{\alpha_2}\\ [e_{\alpha_1+3 \alpha_2},e_{-\alpha_1-3 \alpha_2}] &= h_{\alpha_1}+h_{\alpha_2}& [e_{2 \alpha_1+3 \alpha_2},e_{-2 \alpha_1-3 \alpha_2}] &= 2 h_{\alpha_1}+h_{\alpha_2} \end{align*}$$
Commutators of the form $[ e_{\alpha}, e_{\beta} ]$ with $\alpha + \beta$ a root and $\alpha + \beta \neq 0$
Because of the antisymmetry of the commutator, I only write down the commutators for $\alpha \le \beta$ $$\begin{align*} [e_{-\alpha_2},e_{-\alpha_1}] &= -e_{-\alpha_1-\alpha_2} & [e_{-\alpha_1-2 \alpha_2},e_{\alpha_1+\alpha_2}] &= -2 e_{-\alpha_2} \\ [e_{-\alpha_2},e_{-\alpha_1-\alpha_2}] &= -2 e_{-\alpha_1-2 \alpha_2} & [e_{-\alpha_1-2 \alpha_2},e_{\alpha_1+3 \alpha_2}] &= -e_{\alpha_2} \\ [e_{-\alpha_2},e_{-\alpha_1-2 \alpha_2}] &= -3 e_{-\alpha_1-3 \alpha_2} & [e_{-\alpha_1-2 \alpha_2},e_{2 \alpha_1+3 \alpha_2}] &= e_{\alpha_1+\alpha_2} \\ [e_{-\alpha_2},e_{\alpha_1+\alpha_2}] &= 3 e_{\alpha_1} & [e_{-\alpha_1-3 \alpha_2},e_{\alpha_2}] &= e_{-\alpha_1-2 \alpha_2} \\ [e_{-\alpha_2},e_{\alpha_1+2 \alpha_2}] &= 2 e_{\alpha_1+\alpha_2} & [e_{-\alpha_1-3 \alpha_2},e_{\alpha_1+2 \alpha_2}] &= -e_{-\alpha_2} \\ [e_{-\alpha_2},e_{\alpha_1+3 \alpha_2}] &= e_{\alpha_1+2 \alpha_2} & [e_{-\alpha_1-3 \alpha_2},e_{2 \alpha_1+3 \alpha_2}] &= -e_{\alpha_1} \\ [e_{-\alpha_1},e_{-\alpha_1-3 \alpha_2}] &= -e_{-2 \alpha_1-3 \alpha_2} & [e_{-2 \alpha_1-3 \alpha_2},e_{\alpha_1}] &= e_{-\alpha_1-3 \alpha_2} \\ [e_{-\alpha_1},e_{\alpha_1+\alpha_2}] &= -e_{\alpha_2} & [e_{-2 \alpha_1-3 \alpha_2},e_{\alpha_1+\alpha_2}] &= -e_{-\alpha_1-2 \alpha_2} \\ [e_{-\alpha_1},e_{2 \alpha_1+3 \alpha_2}] &= e_{\alpha_1+3 \alpha_2} & [e_{-2 \alpha_1-3 \alpha_2},e_{\alpha_1+2 \alpha_2}] &= e_{-\alpha_1-\alpha_2} \\ [e_{-\alpha_1-\alpha_2},e_{-\alpha_1-2 \alpha_2}] &= 3 e_{-2 \alpha_1-3 \alpha_2} & [e_{-2 \alpha_1-3 \alpha_2},e_{\alpha_1+3 \alpha_2}] &= -e_{-\alpha_1} \\ [e_{-\alpha_1-\alpha_2},e_{\alpha_2}] &= 3 e_{-\alpha_1} & [e_{\alpha_2},e_{\alpha_1}] &= e_{\alpha_1+\alpha_2} \\ [e_{-\alpha_1-\alpha_2},e_{\alpha_1}] &= -e_{-\alpha_2} & [e_{\alpha_2},e_{\alpha_1+\alpha_2}] &= 2 e_{\alpha_1+2 \alpha_2} \\ [e_{-\alpha_1-\alpha_2},e_{\alpha_1+2 \alpha_2}] &= -2 e_{\alpha_2} & [e_{\alpha_2},e_{\alpha_1+2 \alpha_2}] &= 3 e_{\alpha_1+3 \alpha_2} \\ [e_{-\alpha_1-\alpha_2},e_{2 \alpha_1+3 \alpha_2}] &= -e_{\alpha_1+2 \alpha_2} & [e_{\alpha_1},e_{\alpha_1+3 \alpha_2}] &= e_{2 \alpha_1+3 \alpha_2} \\ [e_{-\alpha_1-2 \alpha_2},e_{\alpha_2}] &= 2 e_{-\alpha_1-\alpha_2} & [e_{\alpha_1+\alpha_2},e_{\alpha_1+2 \alpha_2}] &= -3 e_{2 \alpha_1+3 \alpha_2} \\ \end{align*}$$

A comment about the signs
After much trial and error I found that the signs in the commutators above are given by the following quite compact expression

1) Define a function $F: \text{Roots} \to \{0,1\}$
$$\begin{equation*} F(\alpha) = 0 \quad\text{if}\quad \alpha \in \big\{(0,1),(2,3),(0,-1),(-2,-3)\big\} \quad\text{and}\quad F(\alpha) = 1 \quad\text{otherwise} \end{equation*}$$
2) Define a total order $\prec$ as
$$\begin{align*} - \alpha_1 - 2 \alpha_2 \prec - \alpha_1 \prec 2 \alpha_1 + &3 \alpha_2 \prec - \alpha_1 - 3 \alpha_2 \prec - \alpha_1 - \alpha_2 \prec - \alpha_2 \prec \\ &\prec \alpha_2 \prec \alpha_1 + \alpha_2 \prec \alpha_1 +3 \alpha_2 \prec -2 \alpha_1 - 3 \alpha_2 \prec \alpha_1\prec \alpha_1 + 2 \alpha_2 \end{align*}$$
3) The sign in the commutator $[e_{\alpha} , e_{\beta} ] = n_{\alpha \beta}e_{\alpha+\beta}$ is then given by
$$\begin{equation*} (-1)^{F(\alpha) F(\beta)} \quad\text{if}\quad \alpha\prec\beta \end{equation*}$$ and $$\begin{equation*} - (-1)^{F(\alpha) F(\beta)} \quad\text{if}\quad \alpha\succ\beta \end{equation*}$$

Example
The sign of the commutator $[e_{\alpha_1 + \alpha_2},e_{\alpha_1 + 2 \alpha_2}]$ can be obtained as follows. $F(\alpha_1 + \alpha_2) = 1$, $F(\alpha_1 + 2 \alpha_2) = 1$, $\alpha_1 + \alpha_2 \prec \alpha_1 + 2 \alpha_2$, thus the sign is $(-1)^{1 \cdot 1} \cdot 1 = -1$

Remark: the function $F$ can be thought of as assigning a boson/fermion character to the operators.

References
[1] Jones, Groups, Representations and Physics, 1998
[2] Samelson, Notes on Lie Algebras, 1990