TITLE: Three-dimensional simple Lie algebras over the rationals
QUESTION [13 upvotes]: I ran into this question on math.stackexchange.com "How many 3 dimensional simple Lie algebras are there over the rationals?"
The question has been sitting idle for a long time. I thought it was interesting and would like to know myself.
If working over $\mathbb{C}$, we know all 3-dimensional simple Lie algebras are isomorphic to $\mathfrak{sl}_2(\mathbb{C})$. When we move down to the reals, there are 2 non-isomorphic real forms.
The same argument that works over $\mathbb{C}$ works over $\bar{\mathbb{Q}}$ (the field of algebraic numbers). So over $\bar{\mathbb{Q}}$ the only 3-dimensional simple Lie algebra is $\mathfrak{sl}_2(\bar{\mathbb{Q}})$. Running through a similar argument as that which classifies 3-dim simples over $\mathbb{R}$, I get a whole mess of possibilities over $\mathbb{Q}$ (due to the lack of square roots) but have no idea which forms are non-isomorphic or even how to go about proving they are non-isomorphic.
Anybody know what the answer is? Have a good reference?
Thanks!

REPLY [10 votes]: Here is my own favorite way of understanding this problem:  Let $\bigl(L,[,]\bigr)$ be a $3$-dimensional Lie algebra over a field $K$ (assumed of characteristic $0$ for my comfort, though this probably is overkill).  The bracket $[,]:L\times L \to L$, which is bilinear and skew-symmetric, can be regarded as an element $\beta \in \text{Hom}\bigl(\Lambda^2L,L\bigr)\simeq L\otimes \Lambda^2 (L^\ast)$.  Since the dimension of $L$ is $3$, we have that $\Lambda^2(L^\ast)$ is naturally isomorphic to $L\otimes \Lambda^3(L^\ast)$.  It follows that we can regard $\beta$ as an element of $L\otimes L\otimes \Lambda^3(L^\ast)$.  Since there is a canonical decomposition $L\otimes L\simeq \Lambda^2(L)\oplus S^2(L)$, there is a canonical decomposition $\beta = \lambda + \sigma$ with $\lambda\in \Lambda^2(L)\otimes \Lambda^3(L^\ast)\simeq L^\ast$ and  $\sigma\in S^2(L)\otimes \Lambda^3(L^\ast)$. 
Now, there is a canonical bilinear pairing 
$$
\langle,\rangle: L^\ast\times\bigl(S^2(L)\otimes \Lambda^3(L^\ast)\bigr)\longrightarrow
L\otimes \Lambda^3(L^\ast)\simeq \Lambda^2(L^\ast),
$$
and the Jacobi identity is easily seen to be just $\langle\lambda,\sigma\rangle = 0$.  
If $\lambda\not=0$, then, because $\lambda$ must be fixed by any automorphism of $L$ (in particular, the inner automorphisms), it follows that $N = \text{ker}(\lambda)\subset L$ is an ideal of codimension $1$ in $L$, so $L$ is not simple.  One could follow this thread further to classify the Lie algebras in this case, but let me pass on to the simple case.
Thus, assume that $\lambda = 0$, so that $\beta = \sigma \in S^2(L)\otimes \Lambda^3(L^\ast)$.  Now, there is a natural (i.e., $\text{GL}(L)$-equivariant) cubic polynomial mapping
$$
\text{det}: S^2(L)\otimes \Lambda^3(L^\ast) \longrightarrow \Lambda^3(L^\ast)
$$
(again, this exploits the assumption that $L$ has dimension $3$).  It is easy to see that, if $\text{det}(\sigma) = 0$, then $\sigma$ lies in $S^2(N)\otimes \Lambda^3(L^\ast)$ for some nontrivial proper subspace $N\subset L$ and that this $N$ will be an ideal in $L$, so, again, $L$ is not simple.  
Thus, assume that $\text{det}(\sigma)\in\Lambda^3(L^\ast)$ is not zero.  In particular, $L$ is endowed with a canonical volume form, $\text{det}(\sigma)$.  Since $\text{GL}(L)$ acts transitively on the nonzero elements of $\Lambda^3(L^\ast)$, it follows that classifying the $3$-dimensional Lie algebra structures on $L$ up to the action of $\text{GL}(L)$ with $\lambda=0$ and $\text{det}(\sigma)\not=0$ is equivalent to fixing a (nonzero) volume form $\Upsilon\in \Lambda^3(L^\ast)$ on $L$ and classifying, up to the action of $\text{SL}(L)$, the elements $s\in S^2(L)$ that satisfy $\text{det}(s\otimes \Upsilon) = \Upsilon$.
It is not hard to see that for such $s$, the corresponding Lie algebra is indeed simple, so this reduces the classification of the $3$-dimensional simple Lie algebras over $K$ to the classification of the unimodular quadratic forms in $3$ variables over $K$ up to the action of $\text{SL}(3,K)$, a classical problem.