I want to determine the root-system of the lie algebra $sl(3,\mathbb C)$. Does someone know a good (and complete) reference for this problem?
I know that the root-system is $A_2$ but I want to see a complete proof (calculation) for this.
Thanks in advance.
The root system of $A_2$ consists of $\Phi=\{ \pm \alpha,\pm \beta , \pm (\alpha+\beta) \}$, which can be constructed as follows. With the Cartan subalgebra $\mathfrak{h}_{\mathbb{R}}\cong \mathbb{R}^2$ and the canonical scalar product on $\mathbb{R}^2$ we can realize the simple roots as $$ \alpha=\begin{pmatrix} -\frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix}, \quad \beta=\begin{pmatrix} 1 \\ 0 \end{pmatrix}. $$ Then we obviously have $(\alpha,\alpha)=(\beta,\beta)=1$, and in addition \begin{align*} \langle \alpha, \alpha^{\vee}\rangle & = \frac{2(\alpha,\alpha)}{(\alpha,\alpha)}= 2 ,\\ \langle \alpha, \beta^{\vee}\rangle & = \frac{2(\alpha,\beta)}{(\alpha,\alpha)}= -1,\\ \langle \beta, \beta^{\vee}\rangle & =2, \end{align*} and we obtain the Cartan numbers for type $A_2$. This shows that the root system of $A_2$ consists of $\Phi=\{ \pm \alpha,\pm \beta , \pm (\alpha+\beta) \}$. For additional information also see the MSE question here, for the case of $B_2$. Of course, the same discussion is valid for type $A_2$.
If you have not done this before then you should check the following:
Let $\mathfrak{gl}_n(\mathbb C)$ denote the Lie algebra of $n\times n$ matrices with commutator bracket, i.e. $[x,y] = xy-yx$, and let $\{E_{ij} : 1 \leq i,j \leq n\}$ denote the standard basis of elementary matrices $E_{ij}$ (where $E_{ij}$ has all of its entries $0$ apart from the entry in its $i$-th row and $j$-th column, where the entry is equal to $1$).
Now $\mathfrak{sl}_n(\mathbb C)$ is the Lie subalgebra of traceless matrices, so it has a basis $$ \{E_{ij}: 1\leq i\neq j\leq n\}\cup \{E_{ii}-E_{i+1,i+1}: 1\leq i \leq n-1\}. $$ and if we set $\mathfrak h = \text{span}\{E_{ii}-E_{i+1.i+1}: 1\leq i \leq n-1\}$ then $\mathfrak{h}$ is the space of traceless diagonal matrices. You can check that the $E_{ij}$ are simultaneous eigenvectors for the action of $\mathfrak h$ on $\mathfrak{gl}_n(\mathbb C)$ via the commutator bracket:
For example, in the $2\times 2$ case we have: $$ [h_1E_11+h_2E_22,E_{12}] = (h_1E_{11}+h_2E_{22})E_{12} - E_{12}(h_1E_{11}+h_2E_{22}) = h_1E_{12} - h_2E_{12} = (h_1-h_2)E_{12}. $$
It follows that the $E_{ij}$ span the root spaces for $\mathfrak{sl}_n(\mathbb C)$, and when $n=3$ the lines spanned by $E_{12},E_{13}$ and $E_{23}$ correspond to a set of positive roots, and the vectors $E_{21},E_{31},E_{32}$ (their transposes) have weights their negatives. If $\epsilon_i(h)$ is the coefficient of $E_{ii}$ for any traceless matrix $h$, then $\epsilon_1+\epsilon_2+\epsilon_3=0$ on $\mathfrak h$ and $\mathfrak{sl}_3(\mathbb C)$ has roots $\{\epsilon_i-\epsilon_j: 1\leq i\neq j \leq 3\}$
The inner product is a little more complicated to derive, but if you are okay assuming $\mathfrak{sl}_3$ is simple, then up to scalar there can only be one invariant symmetric bilinear form, and $(x,y)= \text{tr}(xy)$ is one such, and hence the $\epsilon_i$ are an orthonormal set. Thus each root has length $\sqrt{2}$ and any two distinct roots have inner product $\pm 1$, with, for example, $\{\epsilon_1-\epsilon_2,\epsilon_2-\epsilon_3\}$ a set of simple roots.
I think the details of this are worked out, for example, in the Springer text on Lie algebras by Erdmann and Wildon. The book on representation theory by Fulton and Harris also has lots of examples discussing classical Lie algebras, though it is perhaps pitched at a more sophisticated level.