Relatively ample line bundles - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:46:27Z http://mathoverflow.net/feeds/question/27972 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27972/relatively-ample-line-bundles Relatively ample line bundles Roman Fedorov 2010-06-12T20:49:02Z 2010-06-13T19:50:14Z <p>Is it true that a line bundle is relatively ample iff its restsriction to fibers is? If so, what would be the reference?</p> http://mathoverflow.net/questions/27972/relatively-ample-line-bundles/28006#28006 Answer by Gianni Bello for Relatively ample line bundles Gianni Bello 2010-06-13T09:02:57Z 2010-06-13T09:02:57Z <p>If you admit the map to be proper and the schemes to be reasonably good it is true. A reference I know is Lazarsfeld's book "Positivity in algebraic geometry", paragraph 1.7.</p> http://mathoverflow.net/questions/27972/relatively-ample-line-bundles/28018#28018 Answer by Olivier Benoist for Relatively ample line bundles Olivier Benoist 2010-06-13T10:35:34Z 2010-06-13T19:50:14Z <p>EDIT : my previous answer was wrong. Thanks to BConrad for pointing it out.</p> <p>Here is a counterexample if the map is not proper. Let $X$ be the plane, let $Y$ be the blow-up of the plane in one point, and $U$ be $Y$ with one point of the exceptionnal divisor removed. Let $f|_U:U\to X$ be the projection and consider the line bundle $\mathcal{O}_U$. </p> <p>Since the fibers of $f|_U$ are affine (either points or the affine line), $\mathcal{O}_U$ becomes ample when restricted to fibers of $f|_U$. However, $\mathcal{O}_U$ is not $f|_U$-ample. Indeed, if it were, $\mathcal{O}_U$ would be ample on $U$, but we can compute : $$H^0(U,\mathcal{O}_U^N)=H^0(U,\mathcal{O}_U)=H^0(Y,\mathcal{O}_Y)=H^0(X,\mathcal{O}_X),$$</p> <p>where the second equality comes from property $S2$ and the third holds because $f_* \mathcal{O}_Y=\mathcal{O}_X$. Hence, $H^0(U,\mathcal{O}_U^N)$ cannot distinguish between two points of the exceptionnal divisor, and $\mathcal{O}_U$ cannot be ample on $U$.</p> <hr> <p>Warning : what follows is false. I kept it here so that the comment below remains understandable.</p> <p>Here is a counterexample if the map is not proper. Consider the inclusion $f$ of the plane minus a point $U$ in the plane $X$. The line bundle $\mathcal{O}_U$ is ample restricted to the fibers of $f$ (they're points...). However, it is not $f$-ample. Indeed, if it were, $\mathcal{O}_U$ would be ample on $U$. Choosing $N>>0$, we would get $$H^1(U,\mathcal{O}_U)=H^1(U,\mathcal{O}_U^N)=0.$$ But a simple computation via Cech cohomology shows that this cohomology group is not trivial (in fact, infinite).</p> <hr>