A Lie algebra that is neither solvable nor reductive What is an example of a Lie algebra that is neither solvable nor reductive? More specifically, what is the Lie algebra of smallest dimension which is neither solvable nor reductive?
 A: This answer assumes throughout that the field $\Bbb F$ underlying the Lie algebra has characteristic $0$; probably many of the results hold over (at least most) finite characteristics, but I don't know that case well enough to say much more.
Any finite-dimensional Lie algebra $\mathfrak{g}$ admits a Levi decomposition, that is, a decomposition as a semidirect product $$\mathfrak{r} \ltimes \mathfrak{s} ,$$ where $\mathfrak{r}$ is the radical (maximal solvable ideal) of $\mathfrak{g}$ and $\mathfrak{s}$ is a semisimple algebra (and is determined up to conjugacies of a certain form). In particular, any nonsolvable Lie algebra $\mathfrak{g}$ admits a semisimple subalgebra $\mathfrak{s}$.
The smallest semisimple algebras have dimension $3$, and there is always at least one up to isomorphism, namely $\mathfrak{sl}_2(\Bbb F)$. (If $\Bbb F = \Bbb C$ this is the only one, but if $\Bbb F = \Bbb R$, there is another, namely, $\mathfrak{so}_3(\Bbb R)$.)
There are no semisimple Lie algebras of dimension $4$, so any $4$-dimensional nonsolvable, nonreductive Lie algebra must have the form $\Bbb F \ltimes_\phi \mathfrak{s}$ for some $3$-dimensional semisimple Lie algebra $\mathfrak{s}$ and homomorphism $\phi: \mathfrak{s} \to \mathfrak{der}(\Bbb F) \cong \Bbb F$. But any $1$-dimensional representation of a semisimple Lie algebra is trivial, so the algebra is $\Bbb F \oplus \mathfrak{s}$, which is reductive.
Thus, any Lie algebras neither solvable nor reductive must have dimension at least $5$, and any of dimension $5$ must have the form $\mathfrak{r} \ltimes_\phi \mathfrak{s}$ for some reductive Lie algebra $\mathfrak{r}$, semisimple Lie algebra $\mathfrak{s}$, and homomorphism $\phi : \mathfrak{s} \to \mathfrak{der}(\mathfrak{r})$, where $\dim \mathfrak{r} = 2$.
If $\mathfrak{r}$ is the abelian $2$-dimensional Lie algebra $\Bbb F^2$, we have $\mathfrak{der}(\Bbb F^2) \cong \mathfrak{gl}_2(\Bbb F)$. In fact $\phi(s) \subset \mathfrak{sl}_2(\Bbb F)$; otherwise $\ker \operatorname{tr} \phi$ is a $2$-dimensional semisimple subalgebra of $\mathfrak{s}$, which does not exist. Then by Schur's Lemma, either $\phi(\mathfrak{s}) = 0$ or $\phi(\mathfrak{s}) \cong \mathfrak{sl}_2(\Bbb F)$. In the former case, $\mathfrak{g} \cong \Bbb F^2 \oplus \mathfrak{sl}_2(\Bbb F)$, which is reductive. On the other hand, up to isomorphism there is exactly one irreducible 2-dimensional representation of $\mathfrak{sl}_2(\Bbb F)$, namely, the standard representation. Taking $\phi$ to be that representation gives us a first example of a $5$-dimensional, nonreductive, nonsemisimple Lie algebra,
$$\Bbb F^2 \ltimes \mathfrak{sl}_2(\Bbb F) .$$ We can realize this example as the (indecomposable) matrix Lie algebra
$$\left\{\pmatrix{a&b&x\\c&-a&y\\\cdot&\cdot&\cdot} \right\} \subset \mathfrak{gl}_3(\Bbb F) .$$ In the Winternitz classification of complex Lie algebras this Lie algebra is labeled $[5, 40]$.
If $\mathfrak{r}$ is the nonabelian $2$-dimensional Lie algebra, $\mathfrak{t}$, we are looking for a map $\phi: \mathfrak{s} \to \mathfrak{der}(\mathfrak{t})$. But every derivation of $\mathfrak{t}$ is inner, so $\mathfrak{der}(\mathfrak{t}) \cong \mathfrak{t} / \mathfrak{z}(\mathfrak{t}) \cong \Bbb F$, and again in this case $\phi$ is trivial, giving the example
$$\mathfrak{t} \oplus \mathfrak{s} .$$
Given a faithful matrix representation $\psi : \mathfrak{s} \to \mathfrak{gl}_n(\Bbb F)$, we can realize this example as the matrix Lie algebra
$$\left\{\pmatrix{\phi(X) \\ & a & b \\ & \cdot & \cdot} : X \in \mathfrak{s}; a, b \in \Bbb F\right\} \subset \mathfrak{gl}_n \oplus \mathfrak{gl}_2 \subset \mathfrak{gl}_{n + 2} .$$
