Why is the double cover of $Sl(2,\mathbb{R})$ not algebraic? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:33:00Z http://mathoverflow.net/feeds/question/69741 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69741/why-is-the-double-cover-of-sl2-mathbbr-not-algebraic Why is the double cover of $Sl(2,\mathbb{R})$ not algebraic? Marc Palm 2011-07-07T19:25:14Z 2011-07-20T18:01:58Z <p>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!?</p> <p>I am curious, what is the (intuitive) reason for this lack. And, how to proof it rigorously?</p> http://mathoverflow.net/questions/69741/why-is-the-double-cover-of-sl2-mathbbr-not-algebraic/69748#69748 Answer by AndrĂ© Henriques for Why is the double cover of $Sl(2,\mathbb{R})$ not algebraic? AndrĂ© Henriques 2011-07-07T21:04:36Z 2011-07-20T18:01:58Z <p>The double cover of $SL(2,\mathbb R)$ is not algebraic.</p> <blockquote> <p>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.</p> </blockquote> <p>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.</p> <p>Using that method, you can actually detect which covers are algebraic:<br> 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 $\pi_1(G_{\mathbb C})\to A$.</p>