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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. Is it true that algebraic closure of $\Gamma$ and $G$ in $G_a$ are the same? |
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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. $G'/H'$ 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). 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 $G'$) |
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