Let $k$ be an algebraically closed field and let $G\leq\rm{GL}_n(k)$. Assume that $M
Does the Zariski closure of a maximal subgroup remain maximal?
13
$\begingroup$
group-theory
algebraic-geometry
algebraic-groups
-
0Do you mean by linear group, an (affine) algebraic group ? – 2012-10-29
-
0That was redundant, since it just meant that $G\leq{\rm{GL}}_n(k)$. BTW, I found a nice counter-example since then. I just didn't delete the question since I hoped that someone will answer it eventually (didn't want to answer my own question) – 2012-10-29
-
0Thanks! I asked the question because a linear subgroup is always closed for the Zariski topology. It would be nice if you could give your counterexample. – 2012-10-30
-
3Take $G={\rm{SO}}_2(\mathbb{R})\ltimes\mathbb{R}^2$. Then $H={\rm{SO}}_2(\mathbb{R})$ is a maximal subgroup of $G$. Then $\bar{G}^Z={\rm{SO}}_2(\mathbb{C})\ltimes\mathbb{C}^2$ and $\bar{H}^Z={\rm{SO}}_2(\mathbb{C})$, which is no longer maximal, since it stabilizes a line $L=\langle(1,i)\rangle$, and hence is a subgroup of ${\rm{SO}}_2(\mathbb{C})\ltimes L$ – 2012-10-30
1 Answers
2
The following is a different type of a counterexample. Not sure that it qualifies or matches with what you were looking for. Let $$ G=\left\{\left(\begin{array}{cc}1&m\\0&1\end{array}\right)\mid m\in\Bbb{Z}\right\}. $$ We view this as a subgroup of $GL_2(\Bbb{C})$. It is isomorphic to the additive group of integers, so the subgroup $M=M_p$ consisting of the elements of $G$ such that $m$ is divisible by a fixed prime $p$ is a maximal subgroup.
But both $G$ and $M$ have as their Zariski closure the group $$ \overline{G}=\left\{\left(\begin{array}{cc}1&m\\0&1\end{array}\right)\mid m\in\Bbb{C}\right\}=\overline{M}. $$ So $\overline{M}$ is not a maximal subgroup of $\overline{G}$ in any sense.
-
0An infinite cyclic group of diagonal matrices will work, too. – 2013-09-23