Let $G$ be a connected semisimple Lie group with trivial centre and $\mathfrak{g}$ its Lie algebra. The adjoint representation of $G$ defines an isomorphism of $G$ onto the connected component of the identity of $Aut(\mathfrak{g})$. The latter is a real algebraic group, but its identity component (in the Lie group topology) needs not be. However, in many cases $G$ is indeed isomorphic to a real algebraic group.


Is there an example of a connected semisimple Lie group $G$ with trivial centre which is not isomorphic (as a Lie group) to a real algebraic group?


Here it's really important that $G$ be centreless, otherwise it's easy to find counterexamples (such as the universal cover of $SL\_2\mathbb{R}$).


The situation is complicated by the following fact. The real algebraic group $SO(n,1)$ is connected in the Zariski topology, while it has two components as a Lie group; thus, its identity component $G\_n$ might be a good candidate for a counterexample, since it is centreless and doesn't inherit a real algebraic structure from $SO(n,1)$. However, at least for $n\leq 3$, $G\_n$ happens to be isomorphic to a real algebraic group in an unrelated way (namely $\mathbb{R}$, $\mathbb{P}SL\_2\mathbb{R}$ and $\mathbb{P}SL\_2\mathbb{C}$ for $n=1,2,3,$ respectively).


There are several other questions around on the relationship between algebraic and Lie groups, but none of them seems to be relevant in this case...



Centreless semisimple Lie group that is not real algebraic
You basically answered your own question: The connected component $G$ of $SO(2,1)$ is not real algebraic. As such, $G$ should be isomorphic to (the real points of) $PSL(2,\mathbb R)$. But it isn't. As algebraic groups $PSL(2,\mathbb R)$ and $PGL(2,\mathbb R)$ are isomorphic (in fact, the natural homorphism between them is an isomorphism over $\mathbb C$ and therefore over $\mathbb R$). But $PGL(2,\mathbb R)$ clearly has two components distinguished by the sign of the determinant.


