Number of real forms of a (not semisimple, solvable) Lie algebra

Question

Consider the Lie algebra $\mathfrak{r}_2=\mathfrak{aff}(\mathbb{C})$ of the group of affine maps of $\mathbb{C}$, and let $\mathfrak{g}=\mathfrak{r}_2 \oplus \mathfrak{r}_2$.

I am interested in knowing the number of real forms that $\mathfrak{g}$ has.

Obviously, $\mathfrak{aff}(\mathbb{R}) \oplus \mathfrak{aff}(\mathbb{R})$ is a real form; actually I have found two such real forms, and I'd like to prove that there are no more.

Recall that $\mathfrak{r}_2$ is spanned by two elements $e_1, e_2$ such that $[e_1,e_2]=e_2$. As a matrix algebra, $e_1=\begin{pmatrix} 0&0 \\ 0 &1 \end{pmatrix}$ and $e_2=\begin{pmatrix} 0&0 \\ 1 &0 \end{pmatrix}$. Hence $\mathfrak{g}=\mathfrak{r}_2 \oplus \mathfrak{r}_2$ is given by four vectors $e_1, e_2, e_3, e_4$ whose non-zero brackets are: $$ [e_1,e_2]=e_2, [e_3,e_4]=e_4 $$ In order to find real forms of $\mathfrak{g}$, if I recall correctly, we have to find conjugations of $\mathfrak{g}_\mathbb{R}=\mathfrak{g}$ seen as a real Lie algebra, i.e. $\mathbb{R}$-linear automorphisms that anticommute with $J$, where $J$ refers to multiplication by $\sqrt{-1}$.

Aside question: are conjugations (up to the action of automorphisms of the Lie algebra) in bijection with real forms? Any reference where this is proved?

Note that $\mathfrak{g}_\mathbb{R}$ is generated by $e_1, J e_1, \dots , e_4, J e_4$. Consider the conjugation $\sigma$ so that $\sigma \circ J=-J$ and $\sigma e_i=e_i$, which has as fixed locus $\langle e_1, e_2, e_3, e_4\rangle_\mathbb{R} \cong \mathfrak{aff}(\mathbb{R}) \oplus \mathfrak{aff}(\mathbb{R})$. This gives the first real form, the obvious one.

There is another conjugation: the composition of $\sigma$ with the reflection that interchanges factors in $\mathfrak{g}$, i.e. $\sigma_1=\sigma \circ r$ with $r$ the $\mathbb{C}$-linear map sending $e_1$ to $e_3$ and $e_2$ to $e_4$. The fixed locus of $\sigma_1$ is $$ \langle e_1+e_3, e_2+e_4, J (e_1-e_3), J (e_2-e_4)\rangle_{\mathbb{R}} \cong \mathfrak{aff}(\mathbb{C})_{\mathbb{R}}. $$ Indeed, if we call $x=e_1+e_3$, $y=e_2+e_4$, $z=J (e_1-e_3)$, $t=J (e_2-e_4)$ we have $$ [x,y]=y, [x,t]=t, [z,y]=t, [z,t]=-y $$ and this is easily seen to be isomorphic to $\mathfrak{aff}(\mathbb{C})_{\mathbb{R}}=\langle e_1, Je_1, e_2, Je_2 \rangle_{\mathbb{R}}$, so this is another real form of $\mathfrak{g}$.

In conclusion: I have found two real forms of $\mathfrak{g}$ and I suspect there are no more of them, but I don't know how to prove it.

The above argument generelizes for any complex Lie algebra of product type $\mathfrak{g}=\mathfrak{g}_1 \oplus \mathfrak{g}_1$, and gives two real forms of $\mathfrak{g}$ assuming that $\mathfrak{g}_1$ has a real form.

Can something be said in this general case relating real forms of $\mathfrak{g}_1$ and real forms of $\mathfrak{g}_1 \oplus \mathfrak{g}_1$?

In the literature, real forms of semisimple Lie algebras are widely studied, but not so for solvable Lie algebras. I am far from being an expert on Lie algebras, so any reference or suggestion is much welcome.

Note: this question was asked in mathstackexchange a few days ago, with no answers: https://math.stackexchange.com/questions/4935791/real-forms-of-a-solvable-lie-algebra

Answer 1

One can classify forms over an arbitrary subfield $K$ of $\mathbf{C}$. These are, beyond the obvious one ($K\ltimes K^2$) "split form", the $L^*\ltimes L$ where $L$ ranges over quadratic extensions of $K$. In particular, the only nonsplit real form is the Lie algebra of $\mathbf{C}^*\ltimes\mathbf{C}$, viewed as 4-dimensional real Lie algebra.

The proof: let $\mathbf{g}$ be a 4-dimensional Lie algebra over $K$ which has the given complexification. So $\mathbf{g}$ has 2-dimensional and abelian derived subalgebra $\mathfrak{a}$, which is the constant term in the lower central series. Let $\mathbf{h}$ be a Cartan subalgebra (= nilpotent self-normalizing) subalgebra. It is known (see e.g. Bourbaki) that $\mathfrak{g}=\mathfrak{h}+\mathfrak{a}$. Since at the complex level, there is nilpotent subalgebra of dimension $\ge 3$, we deduce that $\mathfrak{h}$ is 2-dimensional, and hence $\mathfrak{g}=\mathfrak{h}\ltimes\mathfrak{a}$. The action of $\mathfrak{h}$ on $\mathfrak{a}$ is faithful (as we see at the complex level). So to classify these algebras, we have to classify 2-dimensional abelian subalgebras of $\mathfrak{gl}_2(K)$ up to conjugation. Moreover it is enough to consider those acting semisimply (to get the right complexification). Then we get: subfields (irreducible case), diagonal matrices (reducible case). This corresponds to the given description. Conversely when complexifying, one can diagonalize the action and we see that we get the Lie algebra of $(\mathbf{C}^*\ltimes\mathbf{C})^2$.