Suppose that $R$ is a root system in a complex $n$-dimensional vector space $V$ with a base $S = \{\alpha_1,...,\alpha_n\}$, and that $$(n(\alpha_i,\alpha_j))_{\alpha_i,\alpha_j \in S}$$ is the Cartan matrix of $R$ with respect to $S$.
Serre's theorem is the following:
Let $\mathfrak{g}$ be the Lie algebra defined by the $3n$ generators $X_i,Y_i,H_i$ ($i = 1...n$), and the relations
- $[H_i,H_j] = 0$,
- $[X_i,Y_i] = H_i$, and $[X_i,Y_j] = 0$ if $i \neq j$,
- $[H_i,X_j] = n(\alpha_i,\alpha_j)X_j$, and $[H_i,Y_j] = -n(\alpha_i,\alpha_j)Y_j$,
- $Ad_{X_i}^{\:(-n(\alpha_i,\alpha_j)+1)}(X_j) = 0$ if $i \neq j$,
- $Ad_{Y_i}^{\:(-n(\alpha_i,\alpha_j)+1)}(Y_j) = 0$ if $i \neq j$.
Then $\mathfrak{g}$ is a semisimple Lie algebra, where the subalgebra generated by $H_i$ is a Cartan subalgebra, and its root system is $R$.
I would like to clarify two things. Firstly, I think that if $R$ were an irreducible root system, then it follows that $\mathfrak{g}$ is simple. Is this true?
Secondly, if it is true that $\mathfrak{g}$ is simple in the case that $R$ is irreducible, does this follow directly from Serre's theorem, or other results? How do we deduce that $\mathfrak{g}$ is simple if $R$ is irreducible, as opposed to just semisimple?