How to get the structure constants from a Dynkin diagram? I'm currently learning how to work out Lie algebras. I've learnt how to read the basics of a Dynkin diagram. So I worked out some simple examples, but I'm stuck at the $[E_\alpha, E_\beta] =N_{a,b}E_{\alpha+\beta}$ step: how do I get those $N_{a,b}$?
As an example, for SU(3), I started from two simple roots $\alpha^1, \alpha^2$ with $\left<\alpha^1, \alpha^2\right>=-\frac12$ and $\lVert \alpha^1\rVert = \lVert \alpha^2 \rVert$, that I represented with $\alpha^1 = \left(1, 0\right)$ and $\alpha^2 = \left(-\frac 12, \frac{\sqrt 3}2\right)$. From those, I computed the other roots $\alpha^1+\alpha^2, -\alpha^1, -\alpha^2, -\left(\alpha^1+\alpha^2\right)$. Naming the generator associated with $\pm\alpha^1$ (resp. $\pm\alpha^2$, resp. $\pm\left(\alpha^1+\alpha^2\right)$) $I_\pm$ (resp. $U_\pm$, resp $V_\pm$), I have : 


*

*$[H^1, H^2]=0$ (the Cartan generators commute) ;

*$[H^i, E_\alpha] = \alpha^i E_\alpha$, hence $[H^1, I_\pm] = \pm I_\pm, [H^1, U_\pm] = \mp \frac12 U_\pm, [H^1, V_\pm] = \pm \frac12 V_\pm$ and $[H^2, I_\pm] = 0, [H^2, U_\pm] = \pm \frac{\sqrt 3}2 U_\pm, [H^2, V_\pm] = \pm \frac{\sqrt 3}2 V_\pm$ ;

*$[E_\alpha, E_{-\alpha}] = \alpha_i H^i = 2\sum_i \alpha^i H^i$ (since I chose to normalize my $H^i$'s such that $g_{ij} =  2$), hence $[I_+, I_-] = 2H^1, [U_+, U_-] = -H^1+\sqrt{3}H^2$ and $[V_+, V_-] = H^1+\sqrt{3}H^2$ ;

*$[E_\alpha, E_\beta] = 0$ and $\alpha+\beta\neq 0$ if and only if ${\alpha+\beta}$ is not a root, hence $[I_\pm, V_\pm] = 0, [U_\pm, V_\pm] = 0, [I_\mp, U_\pm] = 0$ (signs correlated!). 


That looks OK, and renaming $H^1=T_3$ and $H^2 = T_8$, I get most of the $SU(3)$ commutation relations. I also get that, as an example, $[I_+, U_+] = N_{(1, 0), \left(-\frac12, \frac{\sqrt3}{2}\right)} V_+\propto V_+$, but the last thing I'm missing is how to get the proportionnality constant... 
Do you have any idea how should I do? 
(By the way, can I rescale my $\alpha$'s to get rid of the $\frac 12$ factors?)
 A: We will use the Chevallay basis. Consider first $e_i^+$ to be the generators associated to the simple roots $\alpha_i$ and $e_i^-$ to be those associated to $-\alpha_i$. Then one has
$$
[h_i,e_j^{\pm}] = \pm \frac{2\alpha_j\cdot \alpha_i}{|\alpha_i|^2} e_j^\pm\,,\qquad [e_i^+,e_j^-] = \delta_{ij}h_i\,,\qquad [h_i,h_j] = 0\,.
$$
where $h_i$ are the Cartan generators.
If you have the Cartan matrix $A_{ij}$ the factor in the first commutation relation is simply one of the entries of the matrix:
$$
\frac{2\alpha_j\cdot \alpha_i}{|\alpha_i|^2} = A_{ji}\,.
$$
If you only have the Dynkin diagram you can reconstruct the Cartan matrix as follows:


*

*The entries in the diagonal are all equal to $2$ ($A_{ii} = 2$ no sum).

*If node $i$ and node $j$ are not connected then $A_{ij} = 0$

*If node $i$ and node $j$ are connected by a single line then $A_{ij} = -1$

*If node $i$ and node $j$ are connected by two lines and node $i$ is a shorter root then $A_{ij} = -2$ and $A_{ji} = -1$

*If node $i$ and node $j$ are connected by three lines and $i$ is a shorter root then $A_{ij} = -3$ and $A_{ji} = -1$
In some books the shorter root is indicated by having the two or three arrows pointing to it, in other books it is indicated by coloring it in black (while the other longer roots are in white). Note that a single link between two white roots or two black roots  gives the same entry to the Cartan matrix.
Now, for the other commutation relations, denote the generator associated to any root $\alpha$ as $e_\alpha$. With our previous notation $e_{\alpha_i} = e_i^+$ and $e_{-\alpha_i} = e_i^-$ for $\alpha_i$ a simple root. Then the remaining commutation relations are
$$
[h_i, e_\alpha] = \frac{2\alpha\cdot \alpha_i}{|\alpha_i|^2}e_\alpha\,,\qquad [e_{-\alpha},e_\alpha] = h_\alpha\,,\qquad [e_{{\beta }},e_{{\gamma }}]=\pm (p+1)e_{{\beta +\gamma }}
$$
where $p$ is the greatest positive integer such that $\gamma -p\beta$  is a root and $h_\alpha$ is defined as follows:
$$
h_i =\sum_k (\alpha_i)_k H_k\,, \qquad h_\alpha = \sum_k (\alpha)_k H_k\,,
$$
for any given basis for the roots and $\alpha_i$ a simple root. It is always possible to write $h_\alpha$ as a linear combination of $h_i$ because $\alpha$ is a linear combination of simple roots $\alpha_i$, by definition.
