Closure of central subgroup is a central subgroup of closure

Question

Consider $G$ a linear algebraic group and $A\leq B\leq G$ two subgroups with no topological structure. Suppose $A$ is central in $B$, that is $A\leq Z(B)$, we want to show that $\overline{A}\leq Z(\overline{B})$. I was thinking about using some map $\phi_b:G\to G$ defined by $\phi_b(g)=gb-bg$ and use some kind of properties closure under continuous maps but I can't quite get it.

Any ideas ?

Answer 1

All $T_0$ topological groups are Hausdorff. And, in all Hausdorff topological groups, it is true that $A \leq Z(B)$ implies $\overline{A} \leq Z(\overline{B})$.

By your definition, all algebraic groups are $T_0$ (the Zariski topology on a quasi-affine variety always is), but only finite algebraic groups will be Hausdorff. So, most of the algebraic groups you want to consider will not be topological groups.

However, your algebraic groups are group objects in the category of quasi-affine varieties over $\mathbb{C}$. So, we can try to translate the argument for topological groups to the category of quasi-affine varieties, and this works!

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Let $G$ be a closed subgroup of $\operatorname{GL}_n(\mathbb{C})$. The map $c = (x,y) \mapsto xyx^{-1}y^{-1} : \operatorname{GL}_n(\mathbb{C}) \times \operatorname{GL}_n(\mathbb{C}) \to \operatorname{GL}_n(\mathbb{C})$ is continuous, where $\operatorname{GL}_n(\mathbb{C}) \times \operatorname{GL}_n(\mathbb{C})$ is the product variety.

$A \leq Z(B)$ is equivalent to the following two conditions:

1. $A \leq B$
2. $A \times B \leq c^{-1}(I)$.

The first condition implies that $\overline{A} \leq \overline{B}$. Since $\{I\}$ is closed in $\operatorname{GL}_n(\mathbb{C})$, the second condition implies that $\overline{A \times B} \leq c^{-1}(I)$. So, it suffices to show that $\overline{A} \times \overline{B} \leq \overline{A \times B}$.

Let $(x,y) \in \overline{A} \times \overline{B}$ be arbitrary. Let $f$ be a regular function on $\operatorname{GL}_n(\mathbb{C}) \times \operatorname{GL}_n(\mathbb{C})$ and suppose $f$ vanishes on $A \times B$. Then for all $a \in A$, the function $y \mapsto f(a,y) : \operatorname{GL}_n(\mathbb{C}) \to \operatorname{GL}_n(\mathbb{C})$ vanishes on $B$, so it vanishes on $\overline{B}$. Now for all $b \in \overline{B}$, the function $x \mapsto f(x,b)$ vanishes on $A$, so it vanishes on $\overline{A}$. We conclude that $f$ vanishes on $\overline{A} \times \overline{B}$. Since $f$ was arbitrary, this shows that $\overline{A} \times \overline{B} \leq \overline{A \times B}$, as desired. $\square$