algebraic closure of Lie groups in - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T11:32:11Z http://mathoverflow.net/feeds/question/103910 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103910/algebraic-closure-of-lie-groups-in algebraic closure of Lie groups in unknown (yahoo) 2012-08-04T04:09:53Z 2012-12-23T22:22:00Z <p>Let $G$ be a connected, simply connected, solvable, complex Lie group with a discrete subgroup $\Gamma$. Let also $G_a$ be Hochshild-Mostow hull of $G$, i.e., there exists a solvable linear algebraic group $G_a =({\mathbb C}^*)^k \ltimes G$ such that $G_a$ contains $G$ as a Zariski dense, topologically closed, normal complex subgroup.</p> <p>Is it true that algebraic closure of $\Gamma$ and $G$ in $G_a$ are the same?</p> http://mathoverflow.net/questions/103910/algebraic-closure-of-lie-groups-in/115924#115924 Answer by Aakumadula for algebraic closure of Lie groups in Aakumadula 2012-12-09T21:30:12Z 2012-12-09T21:30:12Z <p>Although you have not stated it this way, I will assume that $\Gamma$ is a lattice in a connected linear complex solvable Lie group $G$. If $\rho G \rightarrow GL_n({\mathbb C})$ is a holomorphic representation of $G$, it can be proved that the Zariski closure of $G$ and $\Gamma$ are the same. Suppose that the Zariski closures are $G'$ and $H'$ resp. </p> <p>$G'/H'$ is affine and you cannot have a non-constant holomorphic map from the compact complex manifold $G/\Gamma$ into an affine space (by the maximal modulus principle). </p> <p>You don't really need to use that $G'/H'$ is affine, and can argue by induction on the dimension of $G'/H'$ (using the solvability of $G'$)</p>