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> [!proposition] Proposition. ([[representation of solvable Lie algebra is irreducible iff one-dimensional]])
> If $\mathfrak{g}$ is a [[derived and central series of a Lie algebra|solvable Lie algebra]], then every [[irreducible Lie algebra representation|irrep]] of $\mathfrak{g}$ is one-dimensional.
^proposition
> [!proof]- Proof. ([[representation of solvable Lie algebra is irreducible iff one-dimensional]])
> Note that this proof is analogous to [[representation of finite abelian group is irreducible iff it's 1-dimensional]].
>
Let $(\rho, V)$ be an $n$-dimensional [[Lie algebra representation]] of $\mathfrak{g}$ and suppose $\mathfrak{g}$. First note that [[Lie algebra homomorphism|Lie algebra homomorphisms]] such as $\rho$ preserve solvability, so the [[Lie subalgebra|subalgebra]] $\rho(\mathfrak{g}) \subset \mathfrak{gl}(V)$ is [[derived and central series of a Lie algebra|solvable]]. By [[Lie's Theorem]], we may [[simultaneously diagonalizable|simultaneously]] [[upper-triangular matrix|upper-triangularize]] the elements of $\rho(\mathfrak{g})$ to witness $\rho(\mathfrak{g}) \subset \mathfrak{b}_{n}$. This gives the existence of a simultaneous eigenvector: the exists $v \in V$ such that for all $x \in \mathfrak{g}$ one has $\rho(x)(v)=\lambda v$ for some $\lambda \in \mathbb{C}$, i.e., there exists $v \in V$ for which $x \cdot v \in \text{span}(v)$ for all $x \in \mathfrak{g}$. Thus, $v$ spans a one-dimensional [[Lie algebra representation|subrepresentation]] $W$ of $V$. If $V$ is to be irreducible, we must have $W=V$.
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```
> [!frontlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM outgoing([[]])
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```