2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ A comment on Torsten's answer: For any group $G$, and any subset $S \subset G$ it is true that $Z_G(Z_G(Z_G(S))) = Z_G(S)$. In particular the operation $Z_G$ is always an involution on its image. $\endgroup$ – Lucas Culler Feb 9 '14 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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