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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.

@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.

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.

Your title question and your body are different: your title asks what Lie algebras arise as fssc 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, 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 fssc centralisers. It clearly suffices to show the same with ‘reductive’ in place of ‘semisimple’; I guess we can call that ‘frc’. 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 frc Lie algebra. First note that, since an frc Lie algebra is classified by the dimension of its centre and the root datum of its derived subalgebra, there are only countably many frc Lie algebras. Since each frc Lie algebra has only finitely many nilpotent orbits under the associated adjoint group, we have shown the claim.

@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 fssc centralisers.

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.

@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.

I claim that there are only countably many fssc centralisers. It clearly suffices to show the same with ‘reductive’ in place of ‘semisimple’; I guess we can call that ‘frc’. 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 frc Lie algebra. First note that, since an frc Lie algebra is classified by the dimension of its centre and the root datum of its derived subalgebra, there are only countably many frc Lie algebras. Since each frc Lie algebra has only finitely many nilpotent orbits under the associated adjoint group, we have shown the claim.

@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 fssc centralisers.

Your title question and your body are different: your title asks what Lie algebras arise as fssc 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, 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 fssc centralisers. It clearly suffices to show the same with ‘reductive’ in place of ‘semisimple’; I guess we can call that ‘frc’. 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 frc Lie algebra. First note that, since an frc Lie algebra is classified by the dimension of its centre and the root datum of its derived subalgebra, there are only countably many frc Lie algebras. Since each frc Lie algebra has only finitely many nilpotent orbits under the associated adjoint group, we have shown the claim.

@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 fssc centralisers.