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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?

Niklas
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  • Solvable Lie algebras are iterated extensions of abelian Lie algebras, but classifying these extensions is in general hopeless. – Qiaochu Yuan Mar 08 '17 at 22:33
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    @QiaochuYuan If we take our field to be the complex numbers, is it still hopeless? Lie's theorem tells us that a solvable lie algebra $L$ must have the property that $L/Z(L)$ is isomorphic to some subalgebra of upper triangular matrices. I was under the impression that this is a very strong constraint which will can yield a classification. So where does this break ? Is it that the classification of subalgebras of upper triangular matrices a hopeless problem ?Or is it that determining the isomorphism class of $L$ from that of $L/Z(L)$ a hopeless problem ? – Amr Oct 04 '21 at 23:46

2 Answers2

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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 $\lambda\in \mathbb{C}$.

Community
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Dietrich Burde
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  • 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
  • The classification of nilpotent Lie algebras is a wild problem. – Dietrich Burde Mar 11 '17 at 21:37
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A classification is of hopeless. One can, however, sort of reduce it to nilpotent Lie algebras.

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.

Although this is rather sketchy, it is evidence that the most interesting ("pure") part of the classification lies in the nilpotent case.

YCor
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