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 $G$?



Ampleness of the universal subbundle
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*.


