Timeline for family of torsors and family of vector bundles
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 22, 2012 at 8:28 | comment | added | unknown | only the second counter example is proper, the properness works only under the condition that $Y$ is connected and separable over $k$, the real good thing in this case is that $f_*O_{X\times Y}=O_X$ holds universally. | |
| Jan 21, 2012 at 22:34 | comment | added | Will Sawin | Also some of your examples are proper. I think you need connected, and maybe some other conditions, for it to work. | |
| Jan 21, 2012 at 22:33 | comment | added | Will Sawin | I don't think that's the only obstruction, since you can consider some of the counterexamples in complex analysis or topology. | |
| Jan 21, 2012 at 21:18 | comment | added | unknown | I think the key point here is that the topology of the Cartesian product of two schemes is not the product topology of the two schemes. The open sets of the form $U\times V$ (with $U\subseteq X$ and $V\subseteq Y$ open) do not form a topological basis for the topology of $X\times Y$. For example, if $K$ is a non-trivial field extension of $k$, then $K\times_kK$ clearly does not carry the product topology. Therefore you can not trivialize the bundle by some open in $X$. | |
| Jan 20, 2012 at 21:54 | comment | added | Will Sawin | I think my argument might actually fail entirely then. Since there's no indication that it doesn't work on those cases. In fact, my argument depends on the statement $GL_n(R_1 \times R_2)=\GL_n(R_1) \times GL_n(R_2)$. This is true for the Cartesian product, and thus should usually be false for the tensor product, and in particularly not functorial. Let $X$ be a curve and let $Y$ be a subvariety of its Jacobian. Is it obvious that the line bundle on $X \times Y$ induced by the map of $Y$ into the Jacobian nontrivial on the fibers? | |
| Jan 20, 2012 at 21:20 | comment | added | unknown | @Will Sawin: I just want to say that the assumptions that $Y$ is connected and locally of finite type are very important here. Take two extreme examples: 1) $Y$ is a non-trivial field extension of $k$, then all the fibres are certainly trivial but a vector bundle on the product can not be descent to $X$ in general; 2) $Y$ is a disjoint union of Spec($k$) (say two pieces), then $X\times Y$ is just a disjoit union of $X$. if I take two defferent vector bundles on X and put them together on $X\times Y$ then this vector bundle is not from $X$ while the fibres are obviously trivial. | |
| Jan 20, 2012 at 8:20 | comment | added | S. Carnahan♦ | I think the fundamental groupoid has an algebraic definition in terms of fiber functors on the finite étale site. | |
| Jan 19, 2012 at 11:19 | comment | added | Will Sawin | By "any two" I meant "every two". We also have to require the isomorphism to be algebraically definable, and thus continuous. Then locally it must agree with the local trivialization of $P$ so the local trivialization of $P$ must be functorial which means $P$ is trivial. I guess the real categorical setting you want to think about is the fundamental groupoid of the space. In topology we would get a functor from the fundamental groupoid to the category of fibers of the torsor. But I don't think you can define the fundamental groupoid algebraically. | |
| Jan 19, 2012 at 10:20 | comment | added | unknown | Why if $Y$ is trivial and if there is a functorial isomorphism between the fibres for two rational points implies the torsor is trivial? I could not see it. But if we fix two rational points $x,x'\in X$, and if $Y$ is trivial, then there is a functorial isomorphism between the fibres if we take the ambient category to be the category of torsors $P\to X$ with two fixed $k$-rational points $p,p'\in P$ lying above $x,x'$ respectively. | |
| Jan 18, 2012 at 22:59 | history | answered | Will Sawin | CC BY-SA 3.0 |