Non-semisimple representation of vector spaces and of Lie Algebras

Question

Is it correct to say that a non semisimple representation of a vector space is a representation which is reducible and indecomposable?

If yes, how can this intuition be translated to the definition of a non semisimple Lie Algebra which, if I understand correctly, is a Lie Algebra with a non-trivial abelian ideal?

Answer 1

I think there's a little misunderstanding here. Vector spaces don't have representations (at least not any that I am aware of). Do you mean a representation on a vector space?

More generally, semisimple means that it breaks up as the sum (or product, etc.) of 'simple' things. In the case of representations, simple means irreducible and in the case of Lie algebras simple means having no proper ideals.

These are sort of independent but it so happens that the (finite dimensional) representations of a semisimple Lie algebra are semisimple.

Edit: I said this last bit was true for all reductive Lie algebras but that is incorrect. Reductive means that the adjoint representation specifically is completely reducible (a.k.a. semisimple) while semisimple implies that all representations are.

Answer 2

A (Lie algebra) representation of the Lie algebra $\mathfrak{g}$ on a representation (vector) space $V$ is a Lie algebra homomorphism $$ \varphi \, : \,\mathfrak{g}\longrightarrow \mathfrak{gl}(V). $$ Semisimplicity is a property on the left, i.e. at the Lie algebra level, and irreducibility is a property on the right, i.e. at the vector space level. Every finite-dimensional complex or real Lie algebra $\mathfrak{g}$ can be written as the semidirect sum of a semisimple Lie subalgebra $\mathfrak{h}$ and its radical $\operatorname{rad}(\mathfrak{g}),$ i.e. its maximal solvable ideal $$ \mathfrak{g}=\mathfrak{h}\ltimes \operatorname{rad}(\mathfrak{g}) $$ $\mathfrak{g}$ is semisimple iff $\operatorname{rad}(\mathfrak{g})=\{0\}.$

The representation is irreducible if the only vector subspaces $U\subseteq V$ such that $\varphi (\mathfrak{g})(U)\subseteq U$ are $U=\{0\}$ or $U=V.$

A Lie algebra, semisimple or not, can have reducible and irreducible representations.

The radical as the maximal solvable ideal contains an abelian ideal if it is not zero, and vice versa, any abelian ideal is part of the radical. So, yes, you can rewrite the definition of non-semisimplicity with Abelian ideals. However, the common definition is that $\operatorname{rad}(\mathfrak{g})\neq \{0\}.$