# Three dimensional Lie algebras with three dimensional derived algebra

What are the three dimensional Lie algebras with three dimensional derived algebra ($L'=L$)? I saw a proof in the book about Lie algebras by Karel Erdmann and Mark Wildon, but they work over the field $\mathbb{C}$, so over an algebraically closed field. Then all the three dimensional Lie algebras are all isomorphic with the Lie algebra of the $2\times 2$ matrices with trace zero (field = $\mathbb{C}$). But what happen if you work over an arbitrary field that is for example not algebraically closed? Does there change anything in that case?

• But I am asking about the case of also a three dimensional derived algebra. – bob Mar 2 '17 at 17:06
• Oh, okay! I'll delete my coment. – Kenny Wong Mar 2 '17 at 17:07
• Nice question, by the way. – Kenny Wong Mar 2 '17 at 17:07
• Okay and thanks :). – bob Mar 2 '17 at 17:09

Actually, a three-dimensional Lie algebra $L$ with $[L,L]=L$ has to be simple. Assume that $I$ is a non-trivial proper ideal. Then both $L/I$ and $I$ are at most $2$-dimensional, hence solvable. Then $L$ would be solvable as well, if $I$ and $L/I$ are solvable. This is a contradiction, hence there is no proper ideal.

Now, for an arbitrary field $K$ the classification of simple $3$-dimensional Lie algebras has to be considered. Over $\mathbb{Q}$ we infinitely many such Lie algebras, see this MO-question. Over fields of prime characteristic the result is also different from the characteristic zero case.

As Dietrich points out this depends on the field. Over the real field there are two, one compact and one split, whereas over the rational field there are infinitely many. You can find more information in http://homepages.warwick.ac.uk/~masdf/research/y4_fowlerwright.pdf.

as already mentioned, such a Lie algebra is simple. Its adjoint representation is thus irreducible. Moreover, it maps to the isometry Lie algebra of its adjoint representation, for the Killing form $\mathfrak{so}(\kappa)$. It preserves the kernel of the Killing form, which by irreducibility is nonzero. Thus, at least in char. $\neq 2$ we have two cases:
1) the Killing form is non-degenerate. This is automatic in characteristic zero (and probably large characteristic). Then since $\mathfrak{so}(\kappa)$ is 3-dimensional, we get $\mathfrak{g}$ isomorphic to $\mathfrak{so}(\kappa)$. Thus everything boils down to a classification of nondegenerate quadratic forms in dimension 3, up to scalar multiplication. Also in this case we get that $\mathfrak{g}\otimes C$ is isomorphic to $\mathfrak{sl}_2(C)$, when $C$ is any algebraic closure of the ground field ($C$ closed under taking square roots is enough).
2) The Killing form is zero. There are maybe a few known cases, maybe only in characteristic $\le 3$.