most recent 30 from http://mathoverflow.net 2013-05-25T17:05:47Z http://mathoverflow.net/feeds/user/7276 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90340/numerically-equivalent-effective-divisors-and-semiampleness Oren 2012-03-06T09:50:08Z 2012-04-01T15:22:00Z

<p>Recall that a divisor $M$ on a variety $X$ is said to be semiample if $kM$ is base point free for a certain $k > 0$.</p> <p>Being semiample is not a numerical property (take for example torsion and a non-torsion divisor of degree 0 on a curve, or for more sophisticated examples just look at Lazarsfeld - Positivity in Algebraic Geometry II - Ex. 10.3.3), therefore I was wondering:</p> <p>It is possible to find a smooth projective variety $X$ and two effective divisors $E,D$ on $X$ such that $E \equiv D$, but $E$ is semiample while $H^0(X,kD)=\mathbb{C}$ for every $k$?</p>

http://mathoverflow.net/questions/46224/numerically-rigid-nef-divisor Oren 2010-11-16T11:27:04Z 2010-11-16T14:02:42Z

<p>Is it possible to find an example of an $\mathbb{R}$-Cartier divisor $D$ on an irreducible variety $X$ that is non-trivial, nef, effective and numerically rigid? </p> <p>By "numerically rigid" I mean that if $E$ is another $\mathbb{R}$-Cartier effective divisor such that $E$ is numerically equivalent to $D$ then $D=E$.</p> <p>For curves this clearly cannot be the case, since an effective non-trivial divisor is always ample.</p>

http://mathoverflow.net/questions/30299/volume-of-big-line-bundles-under-finite-morphisms Oren 2010-07-02T12:22:31Z 2010-08-10T16:18:36Z

<p>Let $X$, $Y$ be complex projective varieties of dimension $n$, let $f:X \rightarrow Y$ be a surjective finite morphism of degree $d$ and let $B$ be a big line bundle on $Y$.</p> <p>Is that true that vol($f^*B$)=d $\cdot$ vol($B$)?</p> <p>(I know that if $B$ is not only big but also nef then the formula is true by Lazarsfeld's Positivity in Algebraic geometry I, remark 1.1.13, using the well-known fact that if $B$ is nef then vol($B$)=$B^n$). </p>

http://mathoverflow.net/questions/90340/numerically-equivalent-effective-divisors-and-semiampleness/90356#90356 Oren 2012-03-07T07:10:25Z 2012-03-07T07:10:25Z

Thanks Ulrich, yes that may be the problem. The example I refer to (that you can find also in Ein et al. &quot;Asymptotic Invariants of base loci&quot; Ex. 1.1, <a href="http://arxiv.org/abs/math/0308116" rel="nofollow">arxiv.org/abs/math/0308116</a> ) involves reducible divisors.

http://mathoverflow.net/questions/90340/numerically-equivalent-effective-divisors-and-semiampleness/90356#90356 Oren 2012-03-06T14:20:40Z 2012-03-06T14:20:40Z

misprint: $E' \equiv D'$.

http://mathoverflow.net/questions/90340/numerically-equivalent-effective-divisors-and-semiampleness/90356#90356 Oren 2012-03-06T14:19:08Z 2012-03-06T14:19:08Z

I think that there is something wrong with your proof: in fact you say that $E' \cong D'$, hence in particular $D'$ is ample. Therefore since you say that $D=f^*(D')$, we have that $D$ is necessarily semiample. But this is not true in general: there exist examples of pairs of linearly equivalent divisors such that one is semiample and the other is not.