I am trying to identify the normal real forms of some classical semisimple Lie algebras. These are defined as those real forms whose character $\chi:=n_+-n_-$ equals the rank of the Lie algebra. Here $n_+$ denotes the amount of positive eigenvalues of the Cartan-Killing metric $\kappa(x,y):=tr(ad_x \circ ad_y)$ and $n_-$ the amount of negative eigenvalues.
I know for example that for $A_{n-1}=sl(n,\mathbb{C})$ which has rank $n-1$, this form is $sl(n,\mathbb{R})$ but I do not see how one starts finding the eigenvalues of the Killing metric here.
My attempt was to use a basis for $sl(n,\mathbb{R})$ namely $n\times n$ real matrices $B^{i,j}(k,l)=\delta_{i,k}\delta{j,l}$ if $i \neq j$ and $H^i(1,1)=1, \; H^i(i,i)=-1$ and zero elsewhere. With this basis, you could calculate the Killing metric but diagonalising this seems tricky and a lot of work when looking at more complicated Lie algebras.
Thanks for any hints for this or for general simple Lie algebras!
The characters $\chi:=n_{+}-n_{-}$ of real simple Lie algebras are usually given in a classification of real simple Lie algebras. I found it, for example, in Table $7.3$ on page $295-296$ in the this text.
Concerning examples computing the Killing form of a simple Lie algebra, $\kappa (x,y)$ is always a multiple of the trace form, see wikipedia. For example in the case $\mathfrak{sl}_n(\Bbb R)$ we have $$ \kappa(X,Y)=2n \cdot{\rm tr}(XY). $$