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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?

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  • $\begingroup$ I think you might confused $sl(n)$ with $gl(n)$.... $\endgroup$ May 30, 2016 at 4:04
  • $\begingroup$ Semisimple Lie algebras have trivial center, which $\mathfrak{gl}(n)$ does not. $\endgroup$
    – anomaly
    May 30, 2016 at 4:25

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

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