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I'm trying to compute the weight of different representations of a Lie Algebra. For simplicity let's consider that we wish to compute such thing for SU(N), for example N=3.

Given a basis $\{ j^a \}$ of the Lie Algebra following the Cartan-Weyl method I can decompose it in $\{ h^i, e_{\pm \alpha} \}$ where as usual the first elements correspond to the Cartan subalgebra and the later are the corresponding ladder operators. All the information of the Adjoint representation is contained in the Cartan matrix $A_{ij} = 2 \frac{\langle \alpha_i , \alpha_j \rangle}{\langle \alpha_i , \alpha_i \rangle}$, in particular the simple roots correspond to the rows in the matrix.

If I wan't to build a finite dimensional representation I can start with the Highest Weight and construct the remaining ones by substracting the corresponding simple roots at each level. I'm particularly interested in the fundamental, 2-symmetric and 2-antisymmetric which in the literature correspond to the highest weights:

  1. $(1,0,\cdots, 0)$
  2. $(2,0,\cdots, 0)$
  3. $(0,1,\cdots,0)$

My first question is: Why these weights corresponds to that specific representations?. Do they also correspond to the aforementioned representations for the other classical groups SP(2n), SO(2n), SO(2n+1)?.

In general, for the classical groups, the root system and Cartan Matrix are tabulated (see reference), so for example for SU(N) the roots correspond to:

$\alpha = e_i - e_j$

But looking at the Cartan matrix:

\begin{bmatrix} 2 & -1 \\ -1 & 2 \\ \end{bmatrix}

it corresponds to: $\alpha_1 = (2,-1)$, $\alpha_2 = (-1,2)$. What's the relation between the tabulated roots and the ones from the Cartan Matrix?. Is there a change of basis relating both?.

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  • $\begingroup$ There's a thorough misunderstanding about the Cartan matrix at the end, which does not at all give the roots as its columns or rows or something. Rather, as you write yourself, e.g. here it expresses that in the root system $A_2$, we have $2 \frac{\langle \alpha_1 , \alpha_1 \rangle}{\langle \alpha_1 , \alpha_1 \rangle} = -2$, $2 \frac{\langle \alpha_1 , \alpha_2 \rangle}{\langle \alpha_1 , \alpha_1 \rangle} = -1$, $2 \frac{\langle \alpha_2 , \alpha_1 \rangle}{\langle \alpha_2 , \alpha_2 \rangle} = -1$ etc. Now see that this is true if you set $\alpha_1 =e_1-e_2, \alpha_2=e_2-e_3$. $\endgroup$ – Torsten Schoeneberg Jul 3 at 19:35

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