1
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ What is the "algebraic closure" of a group? (I guess you must mean Zariski closure?) $\endgroup$ Aug 4, 2012 at 5:03
  • 1
    $\begingroup$ Could I take $\Gamma$ to be $\{1\}$? $\endgroup$
    – Ben McKay
    Aug 4, 2012 at 10:07
  • $\begingroup$ $\Gamma$ is a lattice!! $\endgroup$
    – user13559
    Aug 4, 2012 at 20:47
  • 3
    $\begingroup$ It would be helpful is your title were a complete sentence! $\endgroup$ Dec 24, 2012 at 0:23

1 Answer 1

2
$\begingroup$

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 $G'$)

$\endgroup$
1
  • 1
    $\begingroup$ Actually, it turns out that if $G$ is a connected linear complex Lie group and $\Gamma \subset G$ is a lattice, then the Zariski closure of $\Gammma $ and of $G$ are the same in any linear representation of $G$. This follows from the solvable case as above and semi-simple case by the Borel density theorem (easier for the complex lie group case). One can put these cases together thanks to a theorem of Auslander on lattices in a semi-direct product of a solvable and a semi-simple group. $\endgroup$ Dec 10, 2012 at 2:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.