most recent 30 from http://mathoverflow.net 2013-05-21T12:42:11Z http://mathoverflow.net/feeds/question/50790 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50790/is-every-projective-space-bundle-locally-trivial-in-the-zariski-topology Vivek Shende 2010-12-31T05:28:15Z 2010-12-31T05:48:08Z

<p>Suppose given a smooth morphism $f:X\to Y$ between varieties over $\mathbb{C}$ whose fibres are $\mathbb{P}^n$. Then I have an equality of Hodge polynomials $H(X) = H(Y)H(\mathbb{P}^n)$, say because the hyperplane class generates the cohomology of $\mathbb{P}^n$ and hence $f_* \mathbb{Z}_X$ cannot have monodromy. </p> <blockquote> <blockquote> <p>Is any such fibration in fact Zariski locally trivial? Even if not, do I have the equality in the Grothendieck group of varieties $[X] = [Y][\mathbb{P}^n]$? </p> </blockquote> </blockquote>

http://mathoverflow.net/questions/50790/is-every-projective-space-bundle-locally-trivial-in-the-zariski-topology/50792#50792 Torsten Ekedahl 2010-12-31T05:48:08Z 2010-12-31T05:48:08Z

<p>It is not necessarily trivial in the Zariski topology. Consider for instance the plane quadric ${x^2+sy^2+tz^2}\subseteq \mathbb P^2\times\mathrm{Spec}\mathbb C[s,s^{-1},t,t^{-1}]$ as a family of $\mathbb P^1$'s over $\mathrm{Spec}\mathbb C[s,s^{-1},t,t^{-1}]$. It is not even isomorphic to $\mathbb P^1$ over the generic point of $\mathrm{Spec}\mathbb C[s,s^{-1},t,t^{-1}]$ as it doesn't have a rational point. As for the multiplicativity in the Grothendieck group there are examples when it is not true not even in the localised and completed Grothendieck group, see arXiv:0903.3143.</p>