Question

Let, $G(k^{al})$ be an algebraic group, over an algebraically closed field, and $\Gamma_{G}$ is the set of all closed subgroups of $G(k^{al})$.

Then is the map $Z_{G}: \Gamma_{G} \rightarrow \Gamma_{G}$ which takes a closed subgroup to its centralizer in $G$, an involution? (probably not true)

If we now assume that $G(k^{al})$ is reductive or semisimple is there a characterization of all such closed subgroups for which $Z_{G}$ is an involution?

More generally if $G_{k}$ is an algebraic group scheme (now $k$ is no longer algebraically closed ) and $\Gamma_{G}$ is the set of closed group subschemes of $G_{k}$, do the previous two questions have a meaningful answer?

Answer 1

When $G=\mathrm{GL}_n$, then the centraliser $C$ of a subgroup scheme $H$ form the invertible elements of the algebra $M$ of matrices commuting with $H$. The group of such elements is Zariski dense in $M$ so $C$ and $M$ determine each other. Hence, the image of $Z_{\mathrm{GL}_n}$ and the question of involutivity on that image is completely reduced to double centraliser results for algebras.

Answer 2

Not necessarily. For cases when this does happen see the question 28354