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Some questions about algebraic groups.

Let $G$ be an affine algebraic group over algebraically closed field $k$. Questions: Let $H$ be a closed subgroup of $G$, then (as I learnt from some paper) the quotient $\pi\colon G\to G/H$ is faithfully flat, why? reference? When it is locally trivial, especially when $G$ is linear algebraic group?

Thank you very much!

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The following might be helpful intuition. Faithfully flat means flat plus surjective. As flatness captures the notion of a map with "nicely" varying fibres, it should be clear that the given map is flat since each fibre is just a translate of $H$. Surjectivity should also be intuitively clear in this case. – Daniel Loughran Jul 21 '12 at 17:38
@Daniel, from "it should be clear" to a proof the can be light years! :-) – Mariano Suárez-Alvarez Jul 21 '12 at 18:25
@DL: If the algebraically closed field $k$ has characteristic $p > 0$ and $G'$ is a non-smooth affine finite type group with smooth underlying reduced group $G$, the inclusion of $G$ into $G'$ is surjective and all fibers over $G'(k)$ are a translate of the identity point. So is this inclusion faithfully flat? :) (There seem to be some unstated smoothness hypotheses floating around, but the faithful flatness of $\pi$ is true for the right notion of $G/H$ without any smoothness hypotheses on $G$ or $H$. The content is the existence of $G/H$ with several properties.) – user22479 Jul 21 '12 at 18:26
@Mariano: Assuming $G$ and $H$ are smooth (though not necessary), a complete proof in the scheme setting only needs exercises in Hartshorne's AG textbook: Surjectivity follows from the case of $k$-points (by constructibility of images at the scheme level, a Hartshorne exercise), and by openness of the flat locus at the scheme level (another Hartshorne exercise) it suffices to check flatness at $k$-points of $G$. By $G(k)$-translation, it suffices to find one such point, so it suffices to have flatness at one scheme point of $G$. By dominance and reducedness, the generic point of $G^0$ works! – user22479 Jul 21 '12 at 22:21
up vote 2 down vote accepted

If you are really looking for a reference, here is one: Groupes algébriques, Demazure-Gabriel, Chapter III, §3, Proposition 2.5 p. 328. Note that the fact that the base is a field is not important. It might be any scheme. The important assumptions are

  1. the subgroup $H$ has to be flat over the base (which of course holds if it is smooth, which always hold over a field in char 0),
  2. the quotient $G/H$ (defined as the fppf sheafification of the presheaf quotient) has to be a scheme (i.e. representable). This holds when $G$ is affine over a base field, by theorem 5.4, also in Ch. III, §3.

P.S. I am aware that this book is in french, and that the notation used in it makes it quite hard to browse through without spending too much time. Note, however, that it contains an index for the notation at the end. It is a very thorough reference for this type of questions.

P.P.S. As for the "locally trivial" part of the question, do you mean Zariski locally trivial (i.e. there is a Zariski open subset U of G/H over which as a scheme, the morphism becomes $U\times H \to U$)?

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Thank you very much for your answer! I have find a English book with a proof of faithfully flat: Section 5.7 in Jantzen's Representations of algebraic groups. For the locally trivial part. Yes, I means Zariski locally trivial as you said. I guess I may not be true in general. But it is quite interesting that although this is not true, by the faithfully flatness, the vector bundle $G\times_H V\to G/H$ is locally trivial, c.f. Secion~5.9 of Jantzen's book, where $V$ is a finite dimensional (over $k$) representation of $H$. – hoxide Jul 25 '12 at 2:03
Here raise a question: if $G/H$ is quasi-affine, is $G/H$ locally trivial? – hoxide Jul 25 '12 at 2:03

[Assume the group schemes are reduced. [Görtz-Wedhorn], Theorem 14.5 gives generic flatness, and $\pi: G \to G/H$ is a homogeneous space ([Borel], LAG §6). For surjectivity, see [Borel], LAG §6.

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I have edited my answer accordingly. – Timo Keller Jul 21 '12 at 19:01
@Timo: It is better to say "smooth" rather than "reduced" when working with group schemes of finite type over an algebraically closed (or perfect) field, despite the equivalence, since it is the viewpoint of "smoothness" that works more smoothly (ha-ha) over general fields. If $k$ is imperfect of characteristic $p$ then there are reduced affine $k$-groups of finite type that are not smooth (and so become non-reduced after scalar extension to $\overline{k}$); e.g., the ``norm-1'' subgroup of the (smooth!) Weil restriction of ${\rm{GL}}_1$ through a degree-$p$ inseparable extension $k'/k$. – user22479 Jul 21 '12 at 22:08
Thank you very much for you answers! – hoxide Jul 22 '12 at 6:59

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