Today in a talk, it has been mentioned that there exists algebraic groups over the local field $\mathbb{R}$ such that the finite central extension can not be defined algbraically over $\mathbb{R}$ or its algebraic closure $\mathbb{C}$. I guess already covers of $SL(2)$, which is even defined over $\mathbb{Z}$, and the metaplectic group are such an example!?


I am curious, what is the (intuitive) reason for this lack. And, how to proof it rigorously?



Why is the double cover of $Sl(2,\mathbb{R})$ not algebraic?
The double cover of $SL(2,\mathbb R)$ is not algebraic.



> 
> This can be blamed on the fact that the map
>  $$\pi\_1\big(SL(2,\mathbb R)\big)\cong \mathbb Z\quad\longrightarrow\quad \pi\_1\big(SL(2,\mathbb C)\big)=0$$
>  is not injective.
> 
> 
> 


If the double cover of $\pi\_1(SL(2,\mathbb R))$ were algebraic, it would have a complexification, which would be a double cover of $SL(2,\mathbb C)$. But 
$SL(2,\mathbb C)$ doesn't have any double covers since its fundamental group is trivial.


Using that method, you can actually detect which covers are algebraic:  

Let $G$ be a real algebraic Lie group, and let $A$ be a finite abelian group.
A central extension of $G$ by $A$ determines a homomorphism $\pi\_1(G)\to A$.
The cover is algebraic iff that homomorphism extends to a 
homomorphism $\pi\_1(G\_{\mathbb C})\to A$.


