3
$\begingroup$

If a projective bundle $Y$ over a variety $X$ is obtained from a vector bundle $E/X$ then the cohomology and the motif of $Y$ is known to be closely related to that of $X$. Now, what can one say in the case where $Y$ doesn't come from an $X$-vector bundle and $X$ is just a projective space?

More generally, if $X$ is smooth projective, its integral Chow motif is a Tate one, and there is a fibration $Y\to X$ whose fibre (is smooth projective and) has a Tate motif, is the motif of $Y$ a Tate one in general?

Upd. I am interested in "positive" statements of this sort over fields that are (perfect but) not algebraically closed. So, I would start from the most restrictive notion of a bundle: $Y$ is Zariski-locally isomorphic to $P^n(X)$. If this question is "too easy", then what about $P^n\times P^m$-bundles? Do there exist any papers on bundles of this sort?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ What field is $X$ over? Over an algebraically closed field, I think the Brauer group of a projective space is trivial and thus every projective bundle arises from a vector bundle. For the second problem, it matters greatly which fiber you assume has a Tate motive. Is it the generic fiber, the geometric generic fiber, or all fibers defined over the base field? $\endgroup$
    – Will Sawin
    Commented Sep 2, 2021 at 20:30
  • 2
    $\begingroup$ To continue the remark of @WillSawin over any field $k$, the pullback map on Brauer groups from $\text{Br}(k)$ to $\text{Br}(\mathbb{P}^n_k)$ is an isomorphism. So any Brauer-Severi scheme over $\mathbb{P}^n_k$ is associated to an Azumaya algebra that is the tensor product of the pullback of a division algebra from $\text{Spec}(k)$ and End of a locally free sheaf on $\mathbb{P}^n_k$. $\endgroup$
    – Jason Starr
    Commented Sep 2, 2021 at 23:53

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .