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In my lecture notes I have stumbled upon the following:

"Indeed, the statement [Lie's theorem] immediately implies that any such representation [complex representation of a solvable Lie algebra] contains an invariant subspace of dimension one, which in particular implies that irreducible complex representations are automatically one-dimensional and trivial."

Two questions here:

  1. Why is the prerequisite "solvable" no longer needed in the last statement? Or is it just true for any representation of a solvable Lie algebra?

  2. I get that it must be one-dimensional (cf. for example Complex irreducible representation of solvable lie algebra). But why does it follow that such representation must be trivial?

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  1. Solvability is still needed, because it is false otherwise. Take $L=\mathfrak{sl}(2)$. It has an irreducible representation in every dimension. Moreover, the field has to be algebraically closed of characteristic zero; otherwise there are counterexamples.

  2. Sometimes "trivial" just means $1$-dimensional, even though the action is not by zero. Compare with the group case, which was answered here: Why does the trivial representation have degree 1?

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  • $\begingroup$ In my definition, the trivial representation is defined as the representation $\rho \colon \mathfrak{g} \rightarrow \mathfrak{gl}(V)$ where any $X \in \mathfrak{g}$ sends every element of $V$ to zero (i.e. $X$ acts by the zero map). How do I understand this in the context of your reference which is about representations of groups? $\endgroup$ – Jakob Elias Jan 20 '17 at 16:05
  • $\begingroup$ @TobiasKildetoft But the statement is about complex representations? $\endgroup$ – Jakob Elias Jan 20 '17 at 18:13
  • $\begingroup$ @JakobElias Why is that at all relevant? $\endgroup$ – Tobias Kildetoft Jan 20 '17 at 18:14
  • $\begingroup$ Ok, so the statement was: Irreducible complex representations (of solvable Lie Algebras apparently) are one-dimensional and trivial because of Lie's theorem. You think that this wrong? I didn't understand how triviality follows, that was my question. Do you mean that it doesn't follow at all? $\endgroup$ – Jakob Elias Jan 20 '17 at 18:24
  • $\begingroup$ @JakobElias Indeed, it does not follow. For example, abelian Lie algebras have plenty of non-trivial $1$-dimensional representations (I removed my previous comment since it made some wrong claims about nilpotent Lie algebras).. $\endgroup$ – Tobias Kildetoft Jan 20 '17 at 18:32

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