Could you give me a hint how to prove the following?
A representation $R$ of a complex Lie algebra $\mathfrak{g}$ is irreducible iff the image $R(\mathfrak{g})$ is a simple Lie algebra.
[Math] Irreducible representations and simple lie algebras
lie-algebrasrepresentation-theory
Best Answer
Here are more details about the aforementioned counterexample. Let $\mathfrak{g} = \mathfrak{sl}_2 \times \mathfrak{sl}_2$ (over $\mathbb{C}$), and consider the representation $R : \mathfrak{g} \rightarrow \mathfrak{gl}_4$, $\left( \begin{pmatrix} a & b \\ c & -a \end{pmatrix}, \begin{pmatrix} d & e \\ f & -d \end{pmatrix} \right) \mapsto \begin{pmatrix} a+d & b & e & 0 \\ c & -a+d & 0 & e \\ f & 0 & a-d & b \\ 0 & f & c & -a-d \end{pmatrix}$.
You can check directly that it is a representation, but this follows from a general construction: if $R_1$ is a representation of $\mathfrak{g}_1$ and $R_2$ is a representation of $\mathfrak{g}_2$, there is a representation $R = R_1 \otimes R_2$ of $\mathfrak{g}_1 \times \mathfrak{g}_2$ (acting on the tensor product of the underlying vector spaces $V_1$ and $V_2$) given by $R(x_1,x_2) = R_1(x_1) \otimes \mathrm{Id}_{V_2} + \mathrm{Id}_{V_1} \otimes R_2(x_2)$. Beware that there is also the notion of tensor product of two representations of the same Lie algebra.
The representation $R$ is clearly faithful, and it is not hard to show that it is irreducible (either directly in this case, are more generally show that $R_1 \otimes R_2$ is irreducible iff $R_1$ and $R_2$ are irreducible, by considering $R$ as a representation of $\mathfrak{g}_1$ and $\mathfrak{g}_2$ separately).
As to references, it depends on your profile. Are you more interested in physics or just the math? Do you want to study representations of Lie groups, or just Lie algebras (which is a prerequisite to the former)? Do you want to be thorough, or just understand the key facts of the theory in order to be able to apply it in particular cases? In any case, here are some references I know:
The last one is the best IMO.