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Let $\frak{g}$ be a finite dimensional complex semisimple Lie algebra and let $U(\frak{g})$ be its universal enveloping algebra. Take $V$ an infinite dimensional module over $U(\frak{g})$. Let $v \in V$ and think about the submodule it generated $U({\frak g})v$. Will $U({\frak g})v$ be irreducible? This is possible for me to see to be true if $v$ is a highest (or lowest) weight vector, but for a general $v$ I am no longer sure.

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    $\begingroup$ It's not true that if $v$ is highest weight then $U(\mathfrak g)v$ is irreducible : for example the Verma module $M(0)$ for $\mathfrak{sl}_2$ admits $M(-2)$ as submodule. Even if you replace "irreducible" by "indecomposable", for a general $v \in V$ the module $U(\mathfrak g)v$ doesn't need to be incomposable. For example take $M(0)$ generated by $v$ and $M(1)$ generated by $w$. Then, the vector $v+w \in M(0) \oplus M(1)$ will generate $M(0) \oplus M(1)$ which is not indecomposable. If $v$ is highest weight, then $U(\mathfrak g)v$ is always indecomposable. $\endgroup$
    – Nicolas Hemelsoet
    Commented Jul 25, 2021 at 11:36
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    $\begingroup$ Ok, I see it's the distinction between the existence of a decomposition $V = V' \oplus V''$ and the existence of a submodule $V' \subseteq V$. So irreducible is stronger than indecomposable. $\endgroup$
    – Spyros Olympopolous
    Commented Jul 25, 2021 at 11:51
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    $\begingroup$ However since $\frak{g}$ is semisimple by assumption, it must have complete decomposibility (I guess this extends to the infinfite dimensional setting). Hence any $V' \subseteq V$ admits a complement and hence reducible implies decomposable . . . . I hope I am not wrong about complete decomposibility though. $\endgroup$
    – Spyros Olympopolous
    Commented Jul 25, 2021 at 11:56
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    $\begingroup$ but maybe it does not . . . . $\endgroup$
    – Spyros Olympopolous
    Commented Jul 25, 2021 at 12:11
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    $\begingroup$ No, for infinite dimensional modules the situation is different than finite-dimensional one. You can look up "Verma modules" to get started. $\endgroup$
    – Nicolas Hemelsoet
    Commented Jul 25, 2021 at 12:16

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