Let $k$ be a field. Suppose we have an exact sequence of $k$-group schemes (not finite-type)

$$ 1\to H\to G\to K\to 1 $$

In other words, the sheaf quotient $G/H$ is representable by a $k$-group scheme. Now, suppose that we have a finite group $\Gamma$ acting on the groups, functorially and compatible with the morphisms. For each group, the fixed points under $\Gamma$ (taken functorialy) form a closed subgroup-scheme.

Question: Is the sheaf quotient $G^{\Gamma}/H^{\Gamma}$ representable?

From the long exact seuence in cohomology, it seems plauseble that $G^{\Gamma}/H^{\Gamma}$ would be a closed subgroup scheme of $K^{\Gamma}$, but I fail to prove this.

Remark: In my case, $k=\mathbb{C}$, the groups schemes are affine and commutative and $K$ is finite type (i.e a linear algebraic group).

  • $\begingroup$ Some sort of group cohomology $H^1(\Gamma,H)$ is bound to play a role here, and your conclusion would only seem plausible to me if that $H^1(\Gamma,H)$ vanished. $\endgroup$ Nov 13, 2013 at 14:50
  • $\begingroup$ @Damian: It isn't being asked if $K^{\Gamma} = G^{\Gamma}/H^{\Gamma}$, so vanishing of a (sheafified) H$^1$ doesn't seem like it should be necessary. $\endgroup$
    – Marguax
    Nov 13, 2013 at 15:15

1 Answer 1


The question is imprecise concerning the topology involved and quasi-compactness conditions on the group schemes, so in view of the motivation let's first stick to the affine case before we venture beyond that.

Rather generally, consider any left-exact sequence $$1 \rightarrow G' \rightarrow G \rightarrow G''$$ of affine group schemes over a field $k$; i.e., $f:G \rightarrow G''$ is a homomorphism with $G' = \ker f$. For instance, if we begin with such a left exact sequence that is equivariant for the action of a group $\Gamma$ then the induced diagram between closed subgroup schemes of $\Gamma$-invariants is also left-exact.

Hence, the question posed is a special case of the general claim that for any left-exact sequence of affine $k$-group schemes as above, there is a unique factorization of $G \rightarrow G''$ as the composition of a faithfully flat homomorphism onto a closed $k$-subgroup scheme $\overline{G} \subset G''$ (and hence by fpqc descent, $G \rightarrow \overline{G}$ represents an fpqc sheaf quotient for $G$ modulo $G'$).

But this is "easy" in terms of coordinate rings: define $\overline{G}$ to correspond to the image of $f^{\ast}:k[G''] \rightarrow k[G]$. This is visibly a closed $k$-subgroup scheme of $G''$ through which $f$ factors, with the resulting map $G \rightarrow \overline{G}$ visibly having the same kernel $G'$ as $f$. Moreover, the map $G \rightarrow \overline{G}$ is faithfully flat because the corresponding map on coordinate rings is injective and any injective homomorphism between Hopf algebras over a field is always faithfully flat (this is proved in Waterhouse's book on affine group schemes, for example).

Of course, this argument was all rather specific to the affine case. But one can do much better: any quasi-compact homomorphism between arbitrary group schemes over a field factors as a faithfully flat homomorphism onto a closed subgroup scheme of the target (so that closed subgroup represents the fpqc sheaf quotient by the kernel); i.e., the conclusions above hold with affineness removed, provided we just assume $G \rightarrow G''$ is quasi-compact. For a proof, see Corollary 6.7 in Expose VI$_{\rm{A}}$ of the new edition of SGA3 (it is not in the original edition). The key input is a theorem of Perrin not proved there (but recorded as Theorem 6.5 there), concerning approximation of a general quasi-compact $k$-group scheme by finite type $k$-group schemes; in effect, every such $k$-group is an extension of a finite type one by an affine group scheme.

  • 1
    $\begingroup$ Great! thank you very much for the answer and in particular for stating and explaining the "general claim", which is the fundamental result that I was not aware of. $\endgroup$
    – KotelKanim
    Nov 13, 2013 at 16:20
  • $\begingroup$ (in response to your comment above). You are right ! $\endgroup$ Nov 18, 2013 at 12:42

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.