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We have the following criterion for the irreducibility of a Lie algebra representation (we work with $L$-modules here). Let $L$ be a Lie algebra, $V$ a finite dimensional vector space, and let $L \times V \to V$ be an $L$-module. Then $V$ is irreducible $\iff$ every nonzero $v \in V$ generates all of $V$.

The $\Leftarrow$ part is done. The thing I'm having trouble with is seeing that if $v \in V$ is nonzero, then there must be some $x \in L$ such that $x.v \neq 0$.

nigel
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    The statement is not correct, unless you disallow the $1$-dimensional space where $L$ acts trivially to be irreducible (usually, people do not disallow this). – Tobias Kildetoft Dec 09 '13 at 19:20
  • I think that problem with 1-dimensional reps pointed out by Tobias is the only occasion, when you cannot find such an $x$. So the main claim holds always in the sense that if $x\cdot v=0$ for all $x\in L$, then $v$ generates a 1-dimensional representation that has to be all of $V$. – Jyrki Lahtonen Dec 09 '13 at 19:29
  • Actually, come to think of it, the statement is correct. After all, we must have that $v$ is in the submodule generated by $v$ (even if $L$ acts trivially). – Tobias Kildetoft Dec 09 '13 at 19:32

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