Representation for finite-dimensional central Lie algebras over a a field of characteristic 0

Question

Let $\mathbb{K}$ be a field of characteristic $0$, $\overline{\mathbb{K}}$ its algebraic closure and $L$ a central simple $\mathbb{K}$-Lie algebra. Then there is a classical $\mathbb{K}$-Lie algebra $X$ and an automorphisms $B\in Aut_{\overline{\mathbb{K}}}(\overline{\mathbb{K}}\otimes_{\mathbb{K}} X)$ of vector spaces, such that \begin{align*} L=B(X). \end{align*} Moreover, $B$ is unique up to the class $B\circ Aut_{\mathbb{K}}X$.

Is that true? This representation is certainly not well-defined, i.e. for many automorphisms $B$ the image $B(X)$ does not state any sensible algebraic structure. I just claim the existence and uniqueness of this representation.

My arguments: The central simple $\mathbb{K}$-Lie algebras are exactly the classical ones. If $L$ is central simple over $\mathbb{K}$, then $L':=\overline{\mathbb{K}}\otimes_{\mathbb{K}}L$ is central simple over $\overline{\mathbb{K}}$, and so classical. Let $X$ the classical $\mathbb{K}$-Lie algebra which satisfies $\overline{\mathbb{K}}\otimes_{\mathbb{K}}X=L'$. Then $L$ and $X$ both have embeddings into L'. By the properties of the tensor product any $\mathbb{K}$-base$B_X$ of $X$ or $B_L$ of $L$ is also a $\overline{\mathbb{K}}$-base of $L'$. Hence there is a a base change transformation automorphism $B\in Aut_{\overline{\mathbb{K}}}L'$, such that $B(B_X)=B(B_L)$. This implies $B(X)=L$.

Answer 1

With the clarifications, the result is true, although maybe not as exciting as one might think. Indeed it follows from the crucial points that

• over an algebraically closed field of characteristic $0$, a simple Lie algebra is classical (well-known classification result due to Cartan/Killing)
• if $L$ is a central simple Lie algebra over a field $K$, then for any field extension $K' \vert K$, the scalar extension $K' \otimes_K L$ is a simple Lie algebra over $K'$ (neat little theory which shows that's equivalent to being central simple)
• for every field $K$ of characteristic $0$, the classical Lie algebras can be defined over $K$ (in a way, definition of classical Lie algebras, or work done by Chevalley).

From that, the result indeed follows as in your outline. However, exhibiting an isomorphism of vector spaces is not that exciting really: Two $K$-vector spaces are isomorphic iff they have the same dimension, and the result does not say much more than for every central simple Lie algebra over $K$, there exists a classical Lie algebra over $K$ which has the same $K$-dimension: namely, the split one which is of the same type as the one we start with.

About the little more that it does say -- namely, exhibiting that isomorphism as an automorphism of some specific vector space, and uniqueness --: although I like the idea of not actually identifying isomorphic things, secretly you are still doing that when you say $X$ is "the" classical Lie algebra satisfying $\bar K \otimes X = L'$.

What is actually moch more fascinating is that for many central simple $K$-Lie algebras, there is already a finite Galois extension (instead of the algebraic closure) for which the scalar extension is split, and then one can find our algebra as fixed point set of a group of semi-automorphisms, which gives us a grip on the algebra structure as well, and over many fields, allows to classify all forms of a certain type (e.g. all which become isomorphic to, say, $\mathfrak{so}_{17}$ over $\bar K$) up to isomorphism of $K$-Lie algebras. This is at the core of the idea of classifying forms via Galois cohomology, and the wonderful results (Satake-Tits classification) hold true in particular if $\bar K \vert K$ is finite, as in the classical example $K= \mathbb R$ (where the forms were already classified by Cartan, but with much more computational machinery).