Does every Lie algebra appear as centralizer of an element in a semisimple Lie algebra?

Question

Given a finite dimensional, complex, semisimple (fcss) Lie algebra $\mathfrak{g}$ and an element $x\in\mathfrak{g}$, denote by $\mathfrak{g}^x$ the centralizer of $x$ in $\mathfrak{g}$ i.e. the set $\{y\in\mathfrak{g}\mid[x,y]=0\}$.

Consider a Lie algebra $\mathfrak{h}$, is there a fcss Lie algebra $\mathfrak{g}$ and an element $x\in\mathfrak{g}$ such that $\mathfrak{h}\cong\mathfrak{g}^x$? If it is the case, is there a way to give an explicit construction?

Partial answer:

In the case where $\mathfrak{h}$ is abelian, $\mathfrak{h}$ can be viewed as a cartan subalgebra of $\mathfrak{sl}_{\dim\mathfrak{h}+1}(\mathbb{C})$, take $h\in\mathfrak{h}$ regular (meaning its centralizer has minimal dimension) it is well-known that $\mathfrak{sl}_{\dim\mathfrak{h}+1}(\mathbb{C})^h=\mathfrak{h}$ so the answer is yes.

In the case where $\mathfrak{h}$ is reductive, decompose $\mathfrak{h}$ as $\mathfrak{s}\oplus\mathfrak{a}$ with $\mathfrak{s}$ semisimple and $\mathfrak{a}$ abelian. Let $\mathfrak{s}'$ be the semisimple Lie algebra and $h\in\mathfrak{s}'$ given by the previous construction for $\mathfrak{h}=\mathfrak{a}$ and set $\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{s}'$ with $[\mathfrak{s},\mathfrak{s}']=0$. It is semisimple and $\mathfrak{h}=\mathfrak{g}^h$.

Edit:

Given $\mathfrak{h}$, the question can be answer looking at a finite number of Lie algebra $\mathfrak{g}$ and element $x$.

For $\mathfrak{g}$ a fscc Lie algebra. Let $f$ be an element of $\mathfrak{g}^*$, we denote by $\mathfrak{g}^{(f)}$ the stabilzer of $f$ for the coadjoint representation. The index of the Lie algebra $\mathfrak{g}$ is the minimal dimension of stabilizer $\mathfrak{g}^{(f)}$, $f\in\mathfrak{g}^*$.

If $\mathfrak{g}$ and $x$ satisfying the condition exist, as @Lspice mentionned, one can shift the study to centralizers of nilpotents element in reductive Lie algebra : if $x=x_s+x_n$ is the Jordan decomposition of $x$, we have that $\mathfrak{h}\cong(\mathfrak{g}^{x_s})^{x_n}$ and we know that $\mathfrak{g}^{x_s}$ is reductive.

By Vinberg's inequality (see 40.3.2 in [1]), the the rank of $\mathcal{g}$ has to be less than the index of $\mathfrak{g}^{x_s}$ which is in turn has to be less than the index of $\mathfrak{h}$. Up to isomorphism, there is a finite number of reductive Lie algebras of a given index. In such Lie algebras, there is a finite number of nilpotent orbits.

One can do even better as Vinberg's inequality is an equality in the case of reductive Lie algebra, the result is sometimes called Elashvili conjecture on indexes (see [2]).

[1]: Tauvel, P., & Yu, R. W. T. (2005). Lie Algebras and Algebraic Groups. In Springer Monographs in Mathematics. Springer Berlin Heidelberg. doi.org/10.1007/b139060

[2]:Jean-Yves Charbonnel, Anne Moreau, The index of centralizers of elements of reductive Lie algebras. Doc. Math. 15 (2010), pp. 387–421. doi.org/10.4171/DM/301

Answer 1

Your title question and your body are different: your title asks what Lie algebras arise as fcss centralisers, and your body asks whether all Lie algebras arise this way. I answer the easier latter question; I have no idea what shape an answer to the former would take, aside from that I can produce a countable list of isomorphism types by using the classifications of root systems and of nilpotent orbits as below, and ask you to check whether your Lie algebra is isomorphic to one on that list.

Specifically, I claim first that there are only countably many fcss centralisers. It clearly suffices to show the same with ‘reductive’ in place of ‘semisimple’; I guess we can call that ‘fcr’. First note that, if $\mathfrak g$ is reductive and $x$ is an element of $\mathfrak g$ with Jordan decomposition $x = x_\text s + x_\text n$, then $\mathfrak m \mathrel{:=} \mathfrak g^{x_\text s}$ is reductive, and $\mathfrak g^x$ equals $\mathfrak m^{x_\text n}$. It thus suffices to show that there are only countably many isomorphism types of centralisers of nilpotent elements in an fcr Lie algebra. First note that, since an fcr Lie algebra is classified by the dimension of its centre and the root datum of its derived subalgebra, there are only countably many fcr Lie algebras. Since each fcr Lie algebra has only finitely many nilpotent orbits under the associated adjoint group, we have shown the claim. In fact, as @YCor points out in a comment, there are only finitely many isomorphism types of fcss centralisers in each dimension: in a reductive Lie algebra of rank (= semisimple rank + dimension of centre) $n$, centralisers have dimension at least $n$, so an $n$-dimensional Lie algebra can only occur as the centraliser of one of the finitely many (up to conjugacy) nilpotent elements in one of the finitely many (up to isomorphism) fcr Lie algebras of rank at most $n$.

@YCor's answer to How many three dimensional real Lie algebras are there? shows that there are uncountably many isomorphism types of 3-dimensional complex Lie algebras. Thus, uncountably many of them are not fcss centralisers.