5
$\begingroup$

Let $k$ be a field. By a simple algebraic group over $k$ I mean an affine group scheme $G$ of finite type over $k$ such that $G$ is connected, non-commutative and every normal closed subgroup of $G$ is trivial. I would like to know an example of a simple algebraic group such that the base extension $G_{\overline{k}}$ of $G$ to the algebraic closure $\overline{k}$ of $k$ is not simple anymore.

If $G$ is connected and non-commutative then also $G_{\overline{k}}$ is connected and non-commutative. So the problem is really about normal subgroups of $G_{\overline{k}}$ not being defined over $k$.

$\endgroup$
  • 1
    $\begingroup$ Your definition of simple does not quite agree with the usual one (which also has the subgroups be connected). Is that on purpose? $\endgroup$ – Tobias Kildetoft Aug 9 '13 at 9:46
  • 2
    $\begingroup$ Yes this is on purpose. The ones you are referring to I would call "almost simple". $\endgroup$ – Peter Aug 9 '13 at 9:49
  • $\begingroup$ Btw, are you identifying the group with its points over the field, or are you considering it as a scheme? Because in the latter case, I can't think of any examples of such a group (at least not in positive characteristic). (By such a group, I mean one that is simple). $\endgroup$ – Tobias Kildetoft Aug 9 '13 at 11:07
  • $\begingroup$ I am considering it as a scheme. I am mainly interested in characteristic zero. You can get a simple group by taking an almost simple group and then taking the quotient modulo the center. $\endgroup$ – Peter Aug 9 '13 at 11:19
  • $\begingroup$ I guess it is just not obvious to me why that results in a connected group, why the center is maximal among normal closed subgroups, or why this all works in characteristic $0$, when it fails in positive characteristic. $\endgroup$ – Tobias Kildetoft Aug 9 '13 at 11:22
2
$\begingroup$

Consider $SL(d,\mathbb{C})$ as a real algebraic group (ie, replace each matrix entry with a $2 \times 2$ matrix representing a complex number with real entries). Then the complexification has the structure of $SL(d,\mathbb{C}) \times SL(d,\mathbb{C})$ and is hence $SL(d,\mathbb{C})$, considered as a group over $\mathbb{R}$, is not absolutely simple.

$\endgroup$

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.