I was wondering how is it possible to find the dimension of a semi-simple lie algebra $L$ if its corresponding root system is (lets make it simple) of type $B_2$. We can find the number of roots and deduce that each root has a corresponding space of dimension but 1. But how can one find just by looking at the the type of the root system, the dimension of $L_0$ the cartan subalgebra?
It is maybe an easy question, thanks for your help anyway!
Note that the root system comes about as a span of elements of the dual of the Cartan subalgebra, so the root system has the same dimension (as a real vector space) as the Cartan subalgebra does (as a vector space over whatever field you work in). This is what is known as the rank of the root system (it is also the number of simple roots). For the irreducible root systems, this is the number indicated by the index in the notation (so if $X$ is one of $A,B,C,D,E,F,G$ the rank of $X_n$ is $n$, as long as this is a valid irreducible root system).