Which Lie groups are a central extension of an algebraic group?

Question

Suppose $G$ is a connected real Lie group. The quotient $G/Z(G)$ is the image of the adjoint representation, so a linear group. Is it known for which groups this quotient is Lie isomorphic to an algebraic group? Is there a classification?

Compact groups are all algebraic (Chevalley's theorem), so we can assume $G$ is not compact. If $G$ is nilpotent, this is always the case, because linear connected nilpotent Lie groups are a direct product of a torus and a simply connected group that is the image of a polynomial map (the exponential). Moreover, as remarked in the comments, $G/Z(G)$ is actually simply connected, when $G$ is nilpotent. If $G$ is semisimple, $G/Z(G)$ is a centerless direct product of simple groups. Some relevant discussion: Centreless semisimple Lie group that is not real algebraic
As noted in the comments, some non-compact simple Lie groups are algebraic (e.g. $\text{SL}_3(\mathbb{R})$) and some other ones are not (e.g. $\text{PSL}_2(\mathbb{R})$).

What about solvable groups? Do you know an example of a simply connected solvable group that is not a central extension of a linear algebraic group?

Answer 1

There is an obstruction to algebraicity at the level of a Lie algebra of an algebraic group. If $\mathfrak g \subseteq \mathfrak{gl}(V)$ is a Lie algebra, then $\mathfrak g$ is algebraic if it is the Lie algebra of an algebraic subgroup of $GL(V)$. Chevalley studied this notion in the 1940's. For your question, the relevant notion is ad-algebraic: if the group $G/Z(G)$ is to be (the connected component of the identity of) an algebraic group, then the image of $Lie(G)$ in $\mathfrak{gl}(Lie(G))$ must be algebraic. There are non-ad-algebraic Lie algebras, see e.g. this MO answer.