Timeline for Torsors in Algebraic Geometry?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| S Sep 21, 2015 at 4:27 | history | suggested | Will Chen | CC BY-SA 3.0 |
changed "No" to "Now" and "Check" to Cech
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| Sep 21, 2015 at 4:02 | review | Suggested edits | |||
| S Sep 21, 2015 at 4:27 | |||||
| Mar 27, 2010 at 14:48 | vote | accept | Chris Schommer-Pries | ||
| Mar 26, 2010 at 15:02 | answer | added | Martin Bright | timeline score: 20 | |
| Mar 26, 2010 at 12:32 | answer | added | Chris Schommer-Pries | timeline score: 12 | |
| Mar 25, 2010 at 20:33 | comment | added | Torsten Ekedahl | More concretely when, to make life simpler, we assume that $A=\matbb Z$. The torsor then is the disjoint union of $U_1\times 0$ and $U_2\times 1$. We have an action of $C_Y\times A$ on this. Over $U_1$ $U_1\times 0$ becomes $U_1\times 0$ and $U_2\times 1$ becomes $C_Y\times 1$ and on $U_2$ the opposite occurs. | |
| Mar 25, 2010 at 19:19 | comment | added | Tyler Lawson |
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.
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| Mar 25, 2010 at 19:15 | comment | added | Chris Schommer-Pries | @ Tyler: I should have the coequalizer of $$ U_{12} \times_S \mathbb{G} \rightrightarrows (U_1 \cup U_2) \times_S \mathbb{G} $$ agreed? In this case we get $$(U_{12} \times 0) \cup C_Y \times A \rightrightarrows \cup_{i = 1,2}(U_i \times 0) \cup C_Y \times A = (U_1 \cup U_2) \times 0 \cup (C_Y \cup C_Y) \times A$$ no? | |
| Mar 25, 2010 at 19:06 | comment | added | Chris Schommer-Pries | This sort of generality is suppose to handle the case when the structure group for the fibers changes as we move throughout S. In general this allows two new phenomena. The structure group can twist around in a non-trivial way. These kinds of torsors are important, for example, in topology in obstruction theory where $\pi_1$ acts non-trivially. The other thing that can happen, which happens in this example, is that the fiber-wise structure group can jump in a sheaf-like way. In this case it is the trivial group at x_1 and x_2, but over their complement it is A. | |
| Mar 25, 2010 at 19:00 | comment | added | BCnrd | 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. | |
| Mar 25, 2010 at 19:00 | comment | added | Tyler Lawson |
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 $$ C_Y \times A \rightrightarrows (U_1 \cup U_2) \times A $$
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| Mar 25, 2010 at 18:56 | comment | added | Chris Schommer-Pries | 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 $U_i \times_S (C_Y \times A) = C_Y \times A$. | |
| Mar 25, 2010 at 18:46 | comment | added | Tyler Lawson |
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.
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| Mar 25, 2010 at 18:27 | history | asked | Chris Schommer-Pries | CC BY-SA 2.5 |