I think I am confused about some terminology in algebraic geometry, specifically the meaning of the term "torsor". Suppose that I fix a scheme S. I want to work with torsors over S. Let $\mu$ be a sheaf of abelian groups over S. Then my understanding is that a $\mu$-torsor, what ever that is, should be classified by the cohomology gorup $H^1(X; \mu) \cong \check H^1(X; \mu)$.

Now suppose that $\mu$ is representable in the category of schemes over S, i.e. there is a group object $$\mathbb{G} \to S$$ in the category of schemes over $S$, and maps (over S) to $\mathbb{G}$ is the same as $\mu$. Lots of interesting example arise this way.

I also thought that in this case a torsor over S can be defined as a scheme $P \to S$ over S with an action of the group $\mathbb{G}$ such that locally in S it is trivial. I.e. there exists a cover $U \to S$ such that $$ P \times_S U \cong \mathbb{G} \times_S U $$ as spaces over S with a $\mathbb{G}$-action (or rather as spaces over U with a $\mathbb{G} \times_S U$-action).

The part that confuses me is that these two notions don't seem to agree. Here is an example that I think shows the difference. Let $S= \mathbb{A}^1$ be the affine line (over a field k) and let $x_1$ and $x_2$ by two distinct points in $S$. Consider the subscheme $Y = x_1 \cup x_2$, and let $C_Y$ be the complement of Y in S. Let $A$ be your favorite finite abelian group which we consider as a constant sheaf over S. Then we have an exact sequence of sheaves over S, $$0 \to A_{C_Y} \to A \to i_*A \to 0$$ Where $i_*A(U) = A(U \cap Y)$. I believe the first two are representable by schemes over S, namely $$C_Y \times A \cup S \times 0$$ and $S \times A$, where we are viewing the finite set $A$ as a scheme over $k$ (and these products are scheme-theoretic products of schemes over $spec \; k$).

In any event, the long exact sequence in sheaf cohomology shows that $$H^1(S; A_{C_Y}) \cong \check H^1(S; A_{C_Y}) \cong A$$ and it is easy to build an explicit C$\check{\text{e}}$ch cocycle using the covering given by the two opens consisting of the subschemes $U_i = S \setminus x_i$, for $i = 1,2$.

Now the problem comes when I try to glue these together to get a representable object over S, i.e. a torsor in the second sense. Then I am looking at the coequalizer of $$C_Y \times A \rightrightarrows (C_Y\cup C_Y) \times A$$ where the first map is the inclusion and the second is the usual inclusion together with addition by a given fixed element $a \in A$. This seems to just gives back the trivial "torsor" $C_Y \times A$.

Am I doing this calculation wrong, or is there really a difference between these two notions of torsor?

  • $\begingroup$ You should be gluing $U_1 \times A$ to $U_2 \times A$ along $C_Y \times A$ to get the correct torsor over S, no? The restriction of your torsor to $U_i$ should be trivial. $\endgroup$ – Tyler Lawson Mar 25 '10 at 18:46
  • $\begingroup$ This is how I thought of it: If $\mathbb{G}$ is the group object in schemes over S, then I should be gluing together the $U_i \times_S \mathbb{G}$. In this case we have $$U_i \times_S (S \times 0 \cup C_Y \times A) = U_i \times 0 \cup C_Y \times A$$. The part that gives the identity section is not really relevant. The interesting part of the gluing happens on $U_i \times_S (C_Y \times A) = C_Y \times A$. $\endgroup$ – Chris Schommer-Pries Mar 25 '10 at 18:56
  • $\begingroup$ Yes, that's right - but you seem to be objecting to getting the trivial torsor on $C_Y \times A$, when in fact that was your goal. The torsor's only nontrivial when considered over the whole base scheme. Your torsor should be the coequalizer $$ C_Y \times A \rightrightarrows (U_1 \cup U_2) \times A $$ $\endgroup$ – Tyler Lawson Mar 25 '10 at 19:00
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    $\begingroup$ If G isn't affine then requiring the torsor to be a scheme is too strong. Algebraic space is more natural. This is all discussed very nicely (including concrete calculations for affine G) in Milne's book on etale cohomology. $\endgroup$ – BCnrd Mar 25 '10 at 19:00
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    $\begingroup$ AH. Right. You're correct, and I am incorrect. So your "torsor" is trivial over $C_Y \times A$, with set of sections isomorphic to $A$. There is precisely one of these sections that has an extension to $U_1$, and another that has an extension to $U_2$, and the difference between those two sections is the element of $A$ that you're looking for. $\endgroup$ – Tyler Lawson Mar 25 '10 at 19:19

As remarked by Brian Conrad above, there is an excellent explanation of all this in Milne's book Étale cohomology, Section III.4. There wouldn't be much point in reproducing the details here, but the main issues are:

  • You need to decide whether a torsor is going to be a scheme over S which locally looks like a trivial torsor, or merely a sheaf of sets over S which locally looks like a trivial torsor. What people mean by "torsor" can be either of these things. As Milne says, "The question of which sheaf torsors arise from schemes is, in general, quite delicate". If you go for the sheaf definition, then isomorphism classes of torsors are indeed classified by $\check H^1(S,G)$. Beware that if G is not commutative then you need to define $\check H^1(S,G)$ appropriately as a pointed set.

  • You need to decide which topology all this is happening in; the usual definition of torsor uses the flat topology, though if G is smooth over S then you can use the étale topology instead.

  • Depending on what topology you're using, and what S and G look like, there may be issues about whether $\check H^1(S,G)$ and $H^1(S,G)$ are isomorphic.

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  • $\begingroup$ I'll take a closer look at Milne's book, but in the meantime you don't know an example where the two notions of torsor differ, do you? $\endgroup$ – Chris Schommer-Pries Mar 26 '10 at 15:35
  • $\begingroup$ Not offhand, and I think any such example has to be fairly hairy. According to Milne there are some given by Raynaud in Faisceaux Amples sur les Schémas en Groupes et les Espaces Homogènes, Springer LNM 119. $\endgroup$ – Martin Bright Mar 26 '10 at 16:18
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    $\begingroup$ See XIII, 3.2 in Raynaud's thesis (SLN 119). It involves $S$ normal local noetherian of dimension 2, and $G$ an abelian scheme over $S$. Basically the simplest possible situation, since $G$ cannot be $S$-affine and algebraic space torsors for abelian schemes over a Dedekind base are always quasi-projective schemes (see "Neron Models" book). Raynaud's example is an fppf sheaf torsor (so an algebraic space, by Artin's theory) which is not a scheme. $\endgroup$ – BCnrd Mar 27 '10 at 3:01
  • $\begingroup$ Thanks Brian! I think Martin's answer together with Brian's example from Raynaud's thesis constitutes a fully acceptable answer. I find this subtle question quite fascinating! $\endgroup$ – Chris Schommer-Pries Mar 27 '10 at 14:48
  • $\begingroup$ Stupid Question: What's the non-cech H^1 when G is non-abelian? $\endgroup$ – user8248 Oct 6 '10 at 5:51

So thanks to the comments of Tyler Lawson I have been able to figure out what is happening in this example, so I thought I should post it as an answer. I think this is also what Torsten Ekedahl was getting at in his comment, as well.

I think it helps to be extra clear because this example is rather confusing. For starters there is the group scheme, $$\mathbb{G} \to S.$$ In this example $S = \mathbb{A}^1$ is the affine line. This is a group object over $S$, so it can be thought of as an $S$-family of group schemes. At the points $x_1$ and $x_2$ it is the trivial group, and at all other points it is some fixed abelian group $A$. For a concrete example we can take $A = \mathbb{Z}/3$, and then $\mathbb{G}$ looks something like this:

alt text

The bottom line represents $S$. Notice that there is a unique global section, the zero section. Away from the set $Y = x_1 \cup x_2$, there are more sections. Associated to $\mathbb{G}$ is a sheaf on the site of schemes over S. This is the same sheaf I called $A_{C_Y}$.

As outlined in the question we have that $\check H^1(S; A_{C_Y}) = A$ is non-trivial. We can even construct a non-trivial cocycle using the covering consisting of the two open subsets $$U_1 = S - x_1$$ $$U_2 = S - x_2$$ Notice that $U_{12} = U_1 \times_S U_2 = C_Y$, the complement of Y in S. This is exactly the subspace that supports a section. The picture is a little misleading here as it looks like there are lots of sections over $C_Y$. However, because we are using the Zariski topology we have only $A$-many of them. Such a section over $C_Y$ has to be constant on $C_Y$.

Now each of these sections (of which there are A-many) gives rise to a Cech cocycle and so we should be able to construct a $\mathbb{G}$-torsor over $S$ for each one of these. The usual construction is that this torsor is given as the coequalizer of $$U_{12} \times_S \mathbb{G} \rightrightarrows \coprod U_i \times_S \mathbb{G}$$ Where one map is the usual inclusion and the other is also inclusion (the other one), but twisted using the cocycle.

Now the cocycle is only defined over $C_Y$. And over the complement of $C_Y$, namely Y, $\mathbb{G}$ is trivial. It has a unique fiber. So I restricted attention to just the "interesting part", the $C_Y$ part. Then I got that the coequalizer becomes, $$C_Y \times A \rightrightarrows (C_Y \cup C_Y) \times A$$ which has trivial coequalizer $C_Y \times A$. All of these are true facts, except the part about $C_Y$ being the only interesting part. I was wrongly assuming that if the torsor was trivial over this part, then it had to be isomorphic to $\mathbb{G}$.

This is not the case. Somehow Tyler's comments made me realize this. The actual full colimit looks something like this:

alt text

Notice that this space is a trivial $C_Y \times A$-torsor when restricted to $C_Y$, and over $U_1$ and $U_2$ there exist unique sections. However there is no global section, so it is not a globally trivial object. Let's call this object P.

A little book keeping shows that there is an action of schemes over S, $$\mathbb{G} \times_S P \to P$$ making P into a torsor in the second sense.

So this is not a counter example. Both notions of torsor agree here.

But this raises the question:

Question: Do these two a priori different notions of torsor agree in Algebraic Geometry? If not what is the easiest counter example?

I don't know the answer to this.

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    $\begingroup$ Did you look in Alexei Skorobogatov's book on torsors? $\endgroup$ – Kevin Buzzard Mar 26 '10 at 12:45
  • $\begingroup$ I haven't had a chance yet. $\endgroup$ – Chris Schommer-Pries Mar 27 '10 at 14:48

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