4
$\begingroup$

Why is it that many sources define simple (or almost-simple) linear algebraic group $G/k$ to be a connected, semisimple linear algebraic group such that every proper connected normal subgroup is trivial? Doesn't semisimplicity follow from the last condition?

I am told by a friend that part of the issue may be that semisimplicity is defined over $\overline{k}$, whereas simplicity is defined over $k$. Would it then do to define matters as follows?

"A linear algebraic group $G/k$ is semisimple if $G$ has no connected, non-trivial, abelian normal algebraic subgroup. It is said to be simple if it is semisimple and connected and has no connected, non-trivial normal algebraic subgroup defined over $k$."

I've heard from other friends that there are issues involving non-reduced varieties (but I am using classical foundations...) or non-perfect fields. Does defining simple algebraic groups on top of semisimple algebraic groups take care of such issues, or at least the second one (non-perfect fields)?

Improve this question
$\endgroup$
16
  • 3
    $\begingroup$ Context: I am polishing up a set of lecture notes for publication, and I want to get everything precisely right so that I don't lie to the young (in print). $\endgroup$
    – H A Helfgott
    Oct 31, 2018 at 17:36
  • 3
    $\begingroup$ Also, if you want to allow your subgroups to be defined not-necessarily-over-$k$, then you must say so explicitly, rather than by omission; I believe the consensus is that proper formal language will have "subgroup of $G$" (where $G$ is defined over $k$) mean "subgroup defined over $k$", whereas for the more general kind of subgroup you'll want to say "subgroup of $G_{\overline k}$" or similar. $\endgroup$
    – LSpice
    Oct 31, 2018 at 17:51
  • 1
    $\begingroup$ In Springer's book there's a 2-dimensional example of a non-reductive (abelian) $k$-group (for $k$ not perfect) with no nontrivial connected unipotent subgroup defined over $k$ (while the unipotent radical is 1-dimensional). It's indeed possible that there are similar examples in which there is no nontrivial connected abelian normal subgroup defined over $k$, but not reductive. $\endgroup$
    – YCor
    Oct 31, 2018 at 18:15
  • 3
    $\begingroup$ The point is that absolute definitions are considered as more fundamental. So, when one has a $k$-group, one first considers absolute definitions, and in a second time one considers non-absolute definitions, when one has to. Semisimple is an absolute definition, so we're happy with it. In the mind of this school of algebraic geometry, a $k$-object is a $\bar{k}$-object enriched with some extra-structure. $\endgroup$
    – YCor
    Oct 31, 2018 at 18:32
  • 2
    $\begingroup$ Yes there's a difference: for instance, $\mathrm{SO}(3,1)$ is $\mathbf{R}$-simple but not simple (= is not absolutely simple). It's 6-dimensional, and has two 3-dimensional connected normal algebraic subgroups, which are swapped by the Galois automorphism of $\mathbf{R}$ (and the only other connected normal algebraic subgroups are $\{1\}$ and the whole group). $\endgroup$
    – YCor
    Nov 1, 2018 at 16:45

1 Answer 1

Reset to default
8
$\begingroup$

Here is an example which shows that by omitting the word semisimple in your definition, you end up with a different class of groups.

Let $E/F$ be a finite inseparable field extension. Let H be a simple algebraic group over $E$. Let $G$ be the Weil restriction $\mathrm{Res}_{E/F}H$. Then $G$ has no proper connected normal subgroups and $G$ is not semisimple. $G$ is a standard example of a pseudo-reductive group.

The difference in the above example comes about because seimisimplicity is defined via passing to the algebraic closure, while simplicity is not. Over a perfect field, the geometric radical always descends to a subgroup over the ground field, so there will be no difference over a perfect field. Regarding non-reduced varieties, I think the only thing one needs to say in all of ones definitions is that a subgroup must be a variety in order to avoid infinitesimal subgroup schemes like the kernel of Frobenius.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I see - thanks. So, if I assume k is perfect, is it really unnecessary to require semisimplicity first? Or is it still necessary, since the non-existence of non-trivial, proper, connected algebraic subgroups over k might not imply the non-existence of non-trivial, proper, connected, abelian algebraic subgroups over the closure $\overline{k}$ of $k$? (Or does it?) $\endgroup$
    – H A Helfgott
    Nov 1, 2018 at 3:21
  • 1
    $\begingroup$ Yes. If k is perfect, then $\bar{k}/k$ is Galois. Let R be the radical (=maximal normal solvable connected subgroup) of $G_{\bar{k}}$. Then as R is unique, R is stable under the Galois group, so R arises by base change from a subgroup defined over k, which inherits all the right adjectives, so lies in the radical of G over k. This shows that over perfect fields, one does not need to pass to an algebraically closed field to define semisimple (or reductive) $\endgroup$
    – Peter McNamara
    Nov 1, 2018 at 3:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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