2
$\begingroup$

Suppose $X$ and $Y$ are smooth connected schemes over a field $k=\bar{k}$, $f: X\times_kY\to X$ is the first projection. You may assume $Y$ is proper if you like, my question is if $P\to X\times_kY$ is a $G$-torsor with $G$ finite group scheme over $k$ then can we say that the fibres of $P$ under $f$ are "constant". More precisely is there a functorial (with respect to $P$) isomorphism between $P_s$ and $P_{s'}$ for $s,s'$ any two rational points in $X$.

The similar question is that if $V$ is a vector bundle on $X\times_kY$ and if the fibres of $V$ along closed points of $X$ are all trivial vector bundles on $Y$, then do we have (in a functorial way) a vector bundle $W$ on $X$ such that $f^*W\cong V$ ?

The answer to the sencond problem is yes if we assume $Y$ is proper, this is a very standard algebraic geometry argument: using Grauert's base change theorem one can show that $f^*f_*V\to V$ is always an isomorphism. But what happens when $Y$ is not proper? do we have counter examples?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

The answer to question $1$ is no even with $Y$ trivial. If we had a functorial isomorphism between the fibers on any two rational points in $X$, the torsor would be trivial, and there would be no reason to expect that.

$f_*V$ is a locally free sheaf of $\mathcal O(X) \times \Gamma(Y,\mathcal O(Y))$ modules. You can take an open cover of $X$ and look at the transition matrices, then just ignore the $\Gamma(Y,\mathcal O(Y))$ part. This gives you a vector bundle. Then you can just trivialize the $\Gamma(Y,\mathcal O(Y))$ part since its automorphism group is a constant sheaf. This tells you how to glue the pullback of that vector bundle to $V$.

This doesn't appear to be normalized, so it should be defined up to the action of $GL_n(\Gamma(Y,\mathcal O(Y)))$. But that's not too bad.

Improve this answer
$\endgroup$
9
  • $\begingroup$ 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. $\endgroup$
    – unknown
    Commented Jan 19, 2012 at 10:20
  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Commented Jan 19, 2012 at 11:19
  • $\begingroup$ I think the fundamental groupoid has an algebraic definition in terms of fiber functors on the finite étale site. $\endgroup$
    – S. Carnahan ♦
    Commented Jan 20, 2012 at 8:20
  • 1
    $\begingroup$ @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. $\endgroup$
    – unknown
    Commented Jan 20, 2012 at 21:20
  • $\begingroup$ 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? $\endgroup$
    – Will Sawin
    Commented Jan 20, 2012 at 21:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .