# Representations of Semisimple Lie algebras

I am taking a course in representation theory of Lie algebras. We are now dealing with semisimple algebras, their nice decomposition into direct sum of simple ideals, the structure of their derivations and we have talked a bit about toral and cartan subalgebras. In the notes we follow there exists the following claim:

Let $$\mathfrak{g}$$ be a semisimple Lie algebra and $$(\rho,V)$$ a (finite dimensional) representation of $$\mathfrak{g}$$. Then $$\rho$$($$\mathfrak{g})\subset$$ $$\mathfrak{sl}$$(V).

There is also a theorem for semisimple algebras and their representations, which we did not prove:

If $$(\rho,V)$$ is a representation of a semisimple Lie algebra, then the Jordan decomposition of an element $$\rho(x)$$ is $$\rho(x)=\rho(x_{s})+\rho(x_{n})$$, where $$x_{s}$$ and $$x_{n}$$ are the semisimple and nilpotent part respectively of the decomposition of $$x$$ in $$\mathfrak{g}$$.

$$\textbf{Note}$$:We are working over $$\mathbb{C}$$.

I was thinking that the claim above might be false or cannot be proven with the tools developed so far in the course.Taking the theorem in mind, the condition that $$\rho(x) \in \mathfrak{sl}(V)$$, meaning $$\operatorname{tr} (\rho(x))=0$$, reduces to $$\operatorname{tr}(\rho(x_{s}))=0$$ since $$\rho(x_{n})$$ is a nilpotent operator and hence has zero trace.But how can this be true for all $$x\in \mathfrak{g}$$? $$\rho(x_{s})$$ is simply a diagonizable operator on $$V$$...

Any ideas or enlightments are more than welcome, thanks!!

• For the 1st claim all you need is that commutators of matrices are traceless. Therefore $\rho([\mathfrak{g},\mathfrak{g}])\subseteq \mathfrak{sl}(V)$. But $\mathfrak{g}$ is spanned by its commutators, so... The same way Diertich Burde handled the linked question. – Jyrki Lahtonen May 17 '20 at 10:44
• @Jyrki yeaah, I completely forogt about the fact that $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ in the semisimple case..With that in mind, the rest are pretty trivial.. – no name May 17 '20 at 10:48
• Dietrich Burde ^ Sorry, about lack of proof-reading. – Jyrki Lahtonen May 17 '20 at 10:54