# Classification results for solvable lie algebras.

According to the Levi decomposition every (real or complex) Lie algebra $g$ can be written as the semidirect product of a solvable and a semisimple Lie algebra. The semisimple Lie algebras can be classified. But how does one deal with solvable Lie algebras? Can one classify them or is their classification an open problem or what can we say in the direction of their classification?

• Solvable Lie algebras are iterated extensions of abelian Lie algebras, but classifying these extensions is in general hopeless. – Qiaochu Yuan Mar 8 '17 at 22:33

Solvable Lie algebras over real and complex numbers have been classified in low dimensions. There is a large literature, in physics and mathematics about classifications - for references see also this MO-question, or this one. In general, already the classification of nilpotent Lie algebras (which is a special case) is hopeless. Perhaps it is instructive to consider the classification of all complex, solvable $3$-dimensional algebras. There are already infinitely many such Lie algebras up to isomorphism. A family here is given by the following Lie brackets, with respect to a basis $(e_1,e_2,e_3)$, $$[e_1,e_2]=e_2,\; [e_1,e_3]=\lambda e_3,$$ where $\lambda\in \mathbb{C}$.
• Thank you very much! I would be interested in knowing what exactly you mean by "the general case is hopeless". Is it a very difficult problem to classify all nilpotent Lie algebras, but which might be solved one day? Or is it virtually impossible to classify them like it is impossible to classify topological manifolds of dimension $\geq 5$ (since the word problem is undeciable)? – Niklas Mar 10 '17 at 21:24
Namely, every solvable Lie algebra $\mathfrak{g}$ has a Cartan subalgebra $\mathfrak{h}$ (nilpotent and self-normalized) and the latter is unique up to inner automorphism. (I assume the field has characteristic zero.) If $\mathfrak{u}$ is the intersection of the lower central series, then $[\mathfrak{h},\mathfrak{u}]=\mathfrak{u}$ and $\mathfrak{g}=\mathfrak{h}+\mathfrak{u}$. This is not always a semidirect decomposition (the intersection $\mathfrak{h}\cap\mathfrak{u}$) can be nonzero). Yet $\mathfrak{g}$ is naturally quotient of $\mathfrak{h}\ltimes\mathfrak{u}$.
In the reverse direction, to give a rough idea, we can start from $\mathfrak{g}$ and $\mathfrak{u}$, and a reasonable knowledge of the derivation algebra of $\mathfrak{u}$, and in particular how $\mathfrak{h}$ can act on $\mathfrak{u}$ in a way that $[\mathfrak{h},\mathfrak{u}]=\mathfrak{u}$ (this condition implies, for instance, that $\mathfrak{u}$ cannot be characteristically nilpotent unless it is zero). From such actions we can produce the semidirect product $\mathfrak{h}\ltimes\mathfrak{u}$ and to complete the picture we need to determine how we can mod out by an ideal having trivial intersection with $\mathfrak{h}\cup\mathfrak{u}$ to obtain all examples.