Let $k$ be an algebraically closed field of characteristic zero, and $\mathfrak{g}$ a semisimple Lie algebra over it. By Weyl's theorem, we know that any finite-dimensional representation $V$ of it is semisimple.
The question is: Is the opposite of this theorem also true, and if not, what is a counterexample?
NB by semisimple representation I mean a sum of simple representations and simple representation is the one that does not have a subrepresentation.
Best Answer
This is true (assuming $\mathfrak{g}$ is finite dimensional). Indeed, assume every finite dimensional representation is completely reducible. Then, since the adjoint representation is completely reducible, it follows that $\mathfrak{g}$ is reductive.
If $Z(\mathfrak{g})\neq 0$, then fix $z\in Z(\mathfrak{g})\backslash\{0\}$ and decompose $\mathfrak{g}=\mathfrak{g}'\oplus kz$. Define a representation of $\mathfrak{g}$ on $V=k^2$ so that $x\in\mathfrak{g'}$ acts trivially $$x.{a\choose b}={0\choose 0}$$ and $$z.{a\choose b}={b\choose 0}.$$ Then, $$W=\left\{{a\choose 0}\mid a\in k\right\}$$ is a submodule with no complement.