Torsors in Algebraic Geometry? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:50:39Z http://mathoverflow.net/feeds/question/19339 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19339/torsors-in-algebraic-geometry Torsors in Algebraic Geometry? Chris Schommer-Pries 2010-03-25T18:27:22Z 2010-10-05T18:54:01Z <p>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 $H^1(X; \mu) \cong \check H^1(X; \mu)$. </p> <p>No 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. </p> <p>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). </p> <p>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 $spec \; k$). </p> <p>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 check cocycle using the covering given by the two opens consisting of the subschemes $U_i = S \setminus x_i$, for $i = 1,2$. </p> <p>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" $C_Y \times A$. </p> <p>Am I doing this calculation wrong, or is there really a difference between these two notions of torsor?</p> http://mathoverflow.net/questions/19339/torsors-in-algebraic-geometry/19411#19411 Answer by Chris Schommer-Pries for Torsors in Algebraic Geometry? Chris Schommer-Pries 2010-03-26T12:32:20Z 2010-10-05T18:54:01Z <p>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. </p> <p>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:</p> <p><img src="http://sites.google.com/site/chrisschommerpriesmath/Home/course-notes-and-materials/Graphic0.jpg" alt="alt text"></p> <p>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 $A_{C_Y}$.</p> <p>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 $C_Y$.</p> <p>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. </p> <p>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 $\mathbb{G}$. </p> <p>This is not the case. Somehow Tyler's comments made me realize this. The actual full colimit looks something like this:</p> <p><img src="http://sites.google.com/site/chrisschommerpriesmath/Home/course-notes-and-materials/Graphic1.jpg" alt="alt text"></p> <p>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. </p> <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. </p> <blockquote> <p>So this is <strong>not</strong> a counter example. Both notions of torsor agree here. </p> </blockquote> <p>But this raises the question:</p> <blockquote> <p><strong>Question</strong>: Do these two a priori different notions of torsor agree in Algebraic Geometry? If not what is the easiest counter example?</p> </blockquote> <p>I don't know the answer to this. </p> http://mathoverflow.net/questions/19339/torsors-in-algebraic-geometry/19432#19432 Answer by Martin Bright for Torsors in Algebraic Geometry? Martin Bright 2010-03-26T15:02:18Z 2010-03-26T15:02:18Z <p>As remarked by Brian Conrad above, there is an excellent explanation of all this in Milne's book <em>Étale cohomology</em>, Section III.4. There wouldn't be much point in reproducing the details here, but the main issues are:</p> <ul> <li><p>You need to decide whether a torsor is going to be a scheme over <em>S</em> which locally looks like a trivial torsor, or merely a sheaf of sets over <em>S</em> 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 <em>G</em> is not commutative then you need to define $\check H^1(S,G)$ appropriately as a pointed set.</p></li> <li><p>You need to decide which topology all this is happening in; the usual definition of torsor uses the flat topology, though if <em>G</em> is smooth over <em>S</em> then you can use the étale topology instead.</p></li> <li><p>Depending on what topology you're using, and what <em>S</em> and <em>G</em> look like, there may be issues about whether $\check H^1(S,G)$ and $H^1(S,G)$ are isomorphic.</p></li> </ul>