Let X be a variety and $E$ an ample vector bundle on $X$. Let $G=G(r+1,E)$ be the Grassmann bundle over $X$ whose fiber over $x\in X$ is the Grassmannian of the $r+1$-dimensional subspaces of $E_x$. Let $U$ denote the universal subbundle on $G$. Under which hypothesis is the dual of $U$ ample on $G$?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ I would assume never. If $U$ is ample on $G$, then its restriction to a fiber over $x\in X$ must be too, but this just takes you back to the universal subbundle of the Grasmmannian of $r+1$ dimensional subspaces in a vector space and thus can not be ample. $\endgroup$– MohanCommented Sep 13, 2010 at 16:59
-
$\begingroup$ sorry...I had forgotten to write "the dual of" $\endgroup$– ginevra86Commented Sep 13, 2010 at 17:03
-
$\begingroup$ I am interested in the dual of the universal subbundle $\endgroup$– ginevra86Commented Sep 13, 2010 at 17:06
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
I think this is essentially never true, again by restricting to a fiber over $x\in X$. The problem is that (somewhat counterintuitively) the universal quotient bundle on $Gr(k,n)$ is not ample, and for the same reason, neither is the dual of the universal sub. (Except of course when $k=1$!) See Examples 6.1.5 and 6.1.6 in Lazarsfeld's Positivity in Algebraic Geometry II.
