# Weyl's theorem confusion

Weyl's theorem states that given a semisimple Lie algebra $\mathfrak{g}$, any $\mathfrak{g}$-module $V$ is completely reducible.

If we consider the case of $\mathfrak{g}= \mathfrak{gl}(1)$, then given any non-diagonalisable $2 \times 2$ matrix $X$, the map $$\pi: \mathfrak{gl}(1) \to X$$ defines a $\mathfrak{gl}(1)$ representation, meaning $\mathbb{C}^2$ is a finite dimensional $\mathfrak{g}$-module. However it can be shown that for this representation $\mathbb{C}^2$ is not completely reducible as a result of $X$ not being diagonalisable.

Doesn't this go against Weyl's theorem? Or is it saying that Weyl's theorem means that a $\mathfrak{g}$-module is completely reducible in at least one representation of the $\mathfrak{g}$-module but not them all?

• I think you might confused $sl(n)$ with $gl(n)$.... – Bombyx mori May 30 '16 at 4:04
• Semisimple Lie algebras have trivial center, which $\mathfrak{gl}(n)$ does not. – anomaly May 30 '16 at 4:25

$\mathfrak{gl}(1)$ isn't semisimple! For that matter, neither is $\mathfrak{gl}(n)$ for higher values of $n$.