8
$\begingroup$

The concept of a unipotent algebraic group over a field has been extensively studied and is fundamental in algebraic geometry. However, has the notion of a unipotent group scheme over a general base scheme been studied? Both Demazure-Gabriel and SGA3 appear to only consider the case of a base field.

I am even uncertain about the definition of these objects. For a $\mathbb{Q}$-scheme $S$, I believe that a unipotent $S$-group scheme should be a (finitely presented?) group scheme over $S$ admitting a normal series with $\mathbb{G}_a$-factors. However, I am unsure about the definition for more general bases.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ I think that the definition involving composition factors is not what you want (although of course you can study it—but note that, even over a field, it is only the right definition for smooth, connected group schemes). I suspect that the proper definition should be that these are linear groups (i.e., admitting a faithful representation) for which every representation admits a non-$0$ fixed vector. $\endgroup$
    – LSpice
    Commented Jun 13 at 17:28
  • 1
    $\begingroup$ In char $p$, a cyclic group of order $p$ defines a (reduced, smooth, non-connected) unipotent group scheme. $\endgroup$
    – YCor
    Commented Jun 13 at 21:06
  • $\begingroup$ One approach is described in Definition 5.9 of arxiv.org/abs/2110.15041. $\endgroup$
    – anon19
    Commented Jun 14 at 21:32
  • 1
    $\begingroup$ Searching for "unipotent group scheme" came up with dept.math.lsa.umich.edu/~idolga/izve74.pdf, which, together with the works it cites, contains lots of interesting material. $\endgroup$
    – anon
    Commented Jun 28 at 18:08
  • $\begingroup$ Dear @anon, this is very close to what I wanted. If you are willing to write this as an answer, I would happily accept it $\endgroup$
    – Gabriel
    Commented Jun 30 at 12:35

0

Reset to default