User gianni bello - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:22:55Z http://mathoverflow.net/feeds/user/6430 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122650/a-classification-of-rational-surfaces-with-effective-k/122797#122797 Answer by Gianni Bello for A classification of rational surfaces with effective $K$ Gianni Bello 2013-02-24T13:55:52Z 2013-02-24T14:40:53Z <p>EDIT: The proof below is wrong, because it is false that $h^0(X',-mK_X')=h^0(\mathbb{P}^2,-mK_{\mathbb{P}^2})$ (see comments)</p> <p>Consider a normal rational surface $X$ with Gorenstein (or $\mathbb{Q}$-Gorenstein) singularities. Let $\mu:X'\to X$ be a resolution. Then $\mu^*(K_X)=K_{X'}+E$, where $E$ is a $\mu$-exceptioanl divisor. If $X$ is very singular it can happen that $E$ is effective, but it is an exceptional divisor, so that it is not big, or, in other words, it cannot be in the interior of the pseudoeffective cone, or, in other words, given any $A$ is an ample ($\mathbb{Q}$-)divisor, for sure $E-A$ is not effective (it is not pseudoeffective in fact).</p> <p>On the other hand, as $X'$ is a smooth rational surfaces it should be easy to see that, for all $m\in \mathbb{N}$, $h^0(X', -mK_X')=h^0(\mathbb{P}^2,-mK_{\mathbb{P}^2})$, so that $-K_{X'}$ is in fact big, that is $-K_{X'}\geq H$, for some ample $H$.</p> <p>Hence $\mu^*(K_X)=K_{X'}+E\leq E-H$ is not pseudoeffective, so that it cannot be effective, and the same holds for $K_X$.</p> <p>Does it make sense?</p> http://mathoverflow.net/questions/119150/irreducible-divisors-containing-an-arbitrary-closed-set Irreducible divisors containing an arbitrary closed set Gianni Bello 2013-01-17T09:02:43Z 2013-01-18T16:47:13Z <p>Let $X$ be a normal complex projective variety, let $V$ be a closed subset of $X$ (possibly reducible), and let $I_V$ be its ideal sheaf (consider the reduced scheme structure for example).</p> <p>If $A$ is an ample divisor on $X$, then $\mathcal{O}_X(mA)\otimes I_V$ is globally generated if $m>>0$ by one of the definitions of ampleness. This implies in particular that there exists a divisor $A'\in |mA|$ such that $A'$ contains $V$ in its support. In fact there is much more than one divisor with this property. </p> <p>In general there is no reason why $A'$ should be irreducible. In particular if $V$ contains at least two prime divisors, it is clear that every divisor containing $V$ will be reducible.</p> <p>My question is the following:</p> <p>Suppse $\dim V\leq \dim X-2$. Can I find, for $m>>0$, an irreducible divisor $A'\in |mA|$ containing $V$? </p> http://mathoverflow.net/questions/111905/push-forward-and-strict-transforms push-forward and strict transforms Gianni Bello 2012-11-09T15:14:43Z 2012-11-10T02:54:34Z <p>Let $\mu:Y\to X$ be a birational morhism between normal projective (complex) varieties. Suppose, furthermore, that $Y$ is smooth. Let $D$ be a Weil divisor on $X$ and let $\mathcal{O}_X(D)$ be its corresponding reflexive sheaf. If we denote by $\widetilde{D}$ the strict transform of $D$, is it true that the push-forward of the invetible sheaf $O_Y(\widetilde{D})$ is equal to $\mathcal{O}_X(D)$?</p> http://mathoverflow.net/questions/111258/nonnef-locus-and-discrete-valuation/111265#111265 Answer by Gianni Bello for nonnef locus and discrete valuation. Gianni Bello 2012-11-02T10:07:45Z 2012-11-02T10:07:45Z <p>If $X$ is a smooth projective variety, then $\overline{\sigma}(D)>0$ if and only if $\mathrm{Center}(\sigma)\subseteq \mathbf{B}_-(D)$. A reference is "Asymptotic invariants of base loci" by Ein, Lazarsfeld, Mustata, Nakamaye, Popa, theorem 2.8</p> <p><a href="http://de.arxiv.org/PS_cache/math/pdf/0308/0308116v2.pdf" rel="nofollow">http://de.arxiv.org/PS_cache/math/pdf/0308/0308116v2.pdf</a></p> <p>Actually they do this for a big divisor, but this should imply the same result for all pseudoeffective divisors, by using that $\mathbf{B}_-(D)$ is the union of the non-nef loci of $D+A_m$, where $A_m$ is a sequence of amples that goes to 0. </p> <p>See also "Asymptotic base loci on singular varieties" by Cacciola and Di Biagio <a href="http://de.arxiv.org/PS_cache/arxiv/pdf/1105/1105.1253v2.pdf" rel="nofollow">http://de.arxiv.org/PS_cache/arxiv/pdf/1105/1105.1253v2.pdf</a></p> <p>for a generalization to the singular case.</p> http://mathoverflow.net/questions/108224/klt-singularities-are-quotient-in-codimension-2 KLT singularities are quotient in codimension 2 Gianni Bello 2012-09-27T09:15:26Z 2012-09-27T22:14:41Z <p>I have read that if a variety $X$ has KLT singularities, then it has quotient singularities in codimension 2. Do you know a proof (or where can I find a proof) of this?</p> http://mathoverflow.net/questions/108144/numerical-dimension-of-nef-divisors Numerical dimension of nef divisors Gianni Bello 2012-09-26T09:37:16Z 2012-09-26T09:37:16Z <p>Let $D$ be a nef divisor (moreover suppose it is effective if you prefer) on a normal projective variety of dimension $n$. Let $k\in[1,n-1]$. If $D^k\cdot V=0$ for generic subvarieties $V\subseteq X$ of dimension $k$, can I conclude that $D^k\cdot V=0$ for all subvarieties $V$ of dimension $k$? In other words can I conclude that the numerical dimension of $D$ is less than $k$?</p> http://mathoverflow.net/questions/97204/real-vs-complex-surfaces Real vs complex surfaces Gianni Bello 2012-05-17T09:42:55Z 2012-05-17T12:51:26Z <p>Hi! My background is in (complex) algebraic geometry. For the first time I'm considering varieties defined over $\mathbb{R}$ and I have some very basic questions about these. In particular I'm trying to understand MMP for real surfaces.</p> <p>Let $X$ be a smooth projective surface over $\mathbb{R}$. We can consider its complexification $X_{\mathbb{C}}$, smooth complex projective surface.</p> <p>Then Cartier divisors on $X$ correspond to Cartier divisors on $X_{\mathbb{C}}$ that are fixed by conjugation. Is intersection theory on $X$ just inherited from intersection theory on $X_{\mathbb{C}}$ via this identification? Is this the right way to look at intersection theory on $X$?</p> <p>What is the relation between the canonical divisors $K_X$ and $K_{X_{\mathbb{C}}}$? Is it true that $K_{X_{\mathbb{C}}}=\overline{K_{X_{\mathbb{C}}}}$, the divisor defined by conjugated equations and it thus corresponds to the divisor $K_X$ on $X$?</p> <p>Also, how to define blow-ups of points on $X$? The definition in Hartshorne cap II.7 works but I imagine one can find an easier definition. </p> <p>Whatever reference about these and related basic facts would be appraciated.</p> http://mathoverflow.net/questions/82446/applications-of-the-boundedness-of-birational-automorphisms Applications of the boundedness of birational automorphisms Gianni Bello 2011-12-02T10:52:59Z 2011-12-18T03:08:21Z <p>Recently the paper <a href="http://de.arxiv.org/PS_cache/arxiv/pdf/1011/1011.1464v1.pdf" rel="nofollow">http://de.arxiv.org/PS_cache/arxiv/pdf/1011/1011.1464v1.pdf</a> by Hacon, McKernan and Xu appeared on the arXiv. There the authors prove that the number of birational automorphisms of a variety of general type can be bounded by using the volume of the variety itself. Moreover the authors claim to be able to use the same techniques, that appear in the paper, to prove Kollar's conjecture about DCC of the volumes (conjecture 1.4). I'd like to know if there are important applications of these statements, expecially consequences that follow directly from the boundedness of the automorphisms, regarding for example moduli of varieties. Thanks a lot.</p> http://mathoverflow.net/questions/83280/log-canonical-pairs-and-ample-divisors Log canonical pairs and ample divisors Gianni Bello 2011-12-12T20:41:55Z 2011-12-13T00:09:58Z <p>Suppose $(X,\Delta)$ is a log canonical pair and $A$ is an ample divisor on $X$. Could you give me an esay proof of the fact that there exists $A'$ that is $\mathbb{Q}$-linearly equivalent to $A$ such that $(X,\Delta+A')$ is again log canonical? Thanks a lot</p> http://mathoverflow.net/questions/82453/ampleness-of-a-big-line-bundle/82464#82464 Answer by Gianni Bello for 'Ampleness' of a big line bundle Gianni Bello 2011-12-02T14:25:53Z 2011-12-02T14:40:22Z <p>In my view being big corresponds to being "generically ample", that is being ample outside a closed subset. This closed subset is called the augmented base locus of $M$ and it is often denoted by $\mathbb{B}_+(M)$ ( see <a href="http://de.arxiv.org/PS_cache/math/pdf/0308/0308116v2.pdf" rel="nofollow">http://de.arxiv.org/PS_cache/math/pdf/0308/0308116v2.pdf</a>). So, if for example you take $M$ big and $F$ to be a line bundle you can prove that $M^m\otimes F$ is always globally generated (i.e. base point free) outside the augmented base locus of $M$ (note that you don't need global generation of $M$ for this). In the example of Anton Geraschenko in fact $\mathbb{B}_+(M)$ is exactly the exceptional divisor. Something similar might hold if you consider locally free sheaves of higher rank.</p> http://mathoverflow.net/questions/81067/divisor-intersecting-non-negatively-the-negative-part-of-its-zariski-decompositio Divisor intersecting non-negatively the negative part of its Zariski decomposition Gianni Bello 2011-11-16T13:57:03Z 2011-11-16T16:05:28Z <p>Hi all. I'm looking for an example of a smooth projective surface $X$ and a pseudo-effective divisor $D$ on $X$ such that when I consider the Zariski decomposition $D=P+N$ there is some component $E$ of the negative part $N$ such that $(D\cdot E)\geq 0$. Can you help me? Thank you Gianni</p> http://mathoverflow.net/questions/72230/fujitas-lemma-on-singular-varieties Fujita's lemma on singular varieties Gianni Bello 2011-08-06T09:05:07Z 2011-08-06T13:01:45Z <p>With the name Fujita's lemma I denote the following fact: Let $f:Y\to X$ be a birational morphism of normal projective varieties, such that $Y$ is smooth, let $E$ be a (Cartier) divisor on $Y$ such that $E$ is $f$-exceptional and effective. Then $f_*\mathcal{O}_Y(E)=\mathcal{O}_X$.</p> <p>Now suppose $Y$ is not smooth and $E$ is not a Cartier divisor (but still effective and exceptional). I can still define the coherent sheaf $\mathcal{O}(E)={f \in k(Y): div (f) \geq -E }$, but it might not be an invertible sheaf anymore.</p> <p>Is it still true that $f_*\mathcal{O}_Y(E)=\mathcal{O}_X$? Do you have a reference for this? Thanks a lot!</p> http://mathoverflow.net/questions/66389/stability-of-multiplier-ideals-under-small-perturbations/66392#66392 Answer by Gianni Bello for Stability of multiplier ideals under small perturbations Gianni Bello 2011-05-29T20:21:44Z 2011-05-29T20:21:44Z <p>I don't think so. Take for example $X=\mathbb{P}^2$, and $G$ and $D$ to be distinct lines. then $I(G)=\mathcal{O}_X(-G)$ while $I((1-t)g+tD)=\mathcal{O}_X$ for every small $t>0$. Maybe you might want to look at the multiplier ideal associated to the linear series $|G|$. </p> http://mathoverflow.net/questions/64010/birational-automorphisms-of-canonical-models Birational automorphisms of canonical models Gianni Bello 2011-05-05T15:11:14Z 2011-05-05T18:12:24Z <p>Let $X$ be a variety with canonical singularities such that $K_X$ is ample. Do you have a reference of the fact that every birational map from $X$ to itself is biregular? Thank you</p> http://mathoverflow.net/questions/40725/base-locus-of-divisors-on-blowings-up-of-the-projective-space Base locus of divisors on blowings up of the projective space Gianni Bello 2010-10-01T08:57:20Z 2011-03-31T03:10:38Z <p>Let $X$ be the blowing-up of $\mathbb{P}^n$ in $r$ points in general position. Let ${H,E_1,...,E_r}$ be the standard basis of $Pic(X)$. Further let $D=dH-\sum m_j E_j$, be an effective divisor, with $m_j \in \mathbb{Z}$. Is it true that if $m_1\geq 0$ then $E_1 \not\subseteq Bs(|D|)$?</p> <p>If we suppose all the $m_i$'s are non negative, this corresponds to say that $dim |dH-m_1E_i...-m_rE_r|>dim |dH-(m_1+1)E_1-...-m_rE_r|$. In other words this means that if we consider all the hypersurfaces of $\mathbb{P}^n$ of degree $d$ passing through $r$ general points ${p_1,...,p_r}$ with multiplicities $m_1,...,m_r$, then they do not all pass through $p_1$ with multiplicity $m_1+1$. My intution brings me to say this is obvious but I cannot prove it. In fact everything is trivial for nonspecial linear systems, but, a priori, special linear systems may be a problem.</p> <p>Note that the generality of the points is necessary: for example you can simply take $\mathbb{P}^2$ blown up in 3 collinear points and note that $E_1 \subseteq Bs(|H-E_2-E_3|)$.</p> http://mathoverflow.net/questions/40725/base-locus-of-divisors-on-blowings-up-of-the-projective-space/60123#60123 Answer by Gianni Bello for Base locus of divisors on blowings up of the projective space Gianni Bello 2011-03-30T20:56:38Z 2011-03-30T20:56:38Z <p>It seems to me that my question is easier than SHGH conjecture. I think that prop. 2.3 in "Weakly defective varieties" by Chiantini-Ciliberto is a positive answer to my question.</p> http://mathoverflow.net/questions/52550/stable-base-loci-cannot-contain-isolated-points Stable base loci cannot contain isolated points Gianni Bello 2011-01-19T21:22:47Z 2011-03-18T16:26:34Z <p>Let $X$ be a normal projective complex variety. A theorem of Fujita-Zariski says that if $L$ is a Cartier divisor on $X$ such that the base locus $Bs(|L|)$ is a finite set then $L$ is semiample. It seems to me that by using this theorem it is possible to prove that the stable base locus $\mathbb{B}(L):=\bigcap_{m \in \mathbb{N}} Bs(|mL|)$ of a Cartier divisor on a variety as above cannot contain isolated points. Do you know a reference for this (or a counterexample)?</p> http://mathoverflow.net/questions/49279/about-direct-image-of-ideal-sheaves About direct image of ideal sheaves Gianni Bello 2010-12-13T16:51:25Z 2010-12-16T10:19:53Z <p>Let $\mu:X'\rightarrow X$ be a birational morphism of normal complex projective varieties.</p> <p>Consider the two ideal sheaves $I_1= \mu_*\mathcal{O}_{X'}(-\sum d(E)E)$, $I_2=\mu_*\mathcal{O}_{X'}(-\sum(d(E)+1)E)$, where the $d(E)$'s are non negative integers and the $E$'s are prime divisors. </p> <p>Suppose $x\in X$ is such that $x\in \mu(E_0)$ for a prime Cartier divisor $E_0$ such that $d(E_0)>0$. Can we say that the stalk ${(I_2)}_x$ is STRICTLY contained in ${(I_1)}_x$? </p> <p>Thanks for your help.</p> http://mathoverflow.net/questions/28109/nakano-semipositivity Nakano semipositivity Gianni Bello 2010-06-14T10:27:56Z 2010-12-15T22:13:36Z <p>Let $X$ be a compact Kaehler manifold. What is a good, possibly algebraic-geometric, way to think to Nakano semipositivity of holomorphic vector bundles on $X$?</p> <p>Is the trivial line bundle $\mathcal{O}_X$ Nakano semi-positive as a vector bundle?</p> http://mathoverflow.net/questions/42010/singularities-of-pairs Singularities of pairs Gianni Bello 2010-10-13T12:28:40Z 2010-12-15T09:21:45Z <p>In the next days I have to give a talk in which I need to explain some of the usual singularities of pairs that one meets when dealing with the minimal model program: KLT, DLT and LC pairs. In particular I would like to give an intuition (also to non specialists) of the reason why an LC pair is much more difficult to treat than a DLT pair, for example. In the same context I would also like to give an intuition of what the LC centers of a pair are and why they are "special" subvarieties for a pair. Note that I'm always considering normal (possibly non-smooth) projective varieties. Do you have any ideas?</p> http://mathoverflow.net/questions/46207/basepoints-in-the-canonical-system-of-algebraic-surfaces/46208#46208 Answer by Gianni Bello for Basepoints in the Canonical System of Algebraic Surfaces Gianni Bello 2010-11-16T07:43:13Z 2010-11-16T08:02:56Z <p>What I know is that Bombieri proved that for algebraic surfaces of general type $h^0(2K_X)\not=0$ and the map defined by $|nK_X|$ is always birational for every $n\geq 5$. Moreover if you take a minimal surface $X$ with geometric genus 2 and $(K_X)^2=1$, then $|4K_X|$ is not birational.</p> <p>The same questions are very interesting and widely open in higher dimension. </p> <p>We know only partial results for 3-folds and 4-folds, you can see for example the introduction of <a href="http://uk.arxiv.org/PS_cache/arxiv/pdf/1001/1001.3340v1.pdf" rel="nofollow">http://uk.arxiv.org/PS_cache/arxiv/pdf/1001/1001.3340v1.pdf</a> if you are interested.</p> http://mathoverflow.net/questions/26546/about-b-divisors About b-divisors Gianni Bello 2010-05-31T09:04:38Z 2010-08-02T21:22:17Z <p>In the last period I have studied a number of papers of O. Fujino and F. Ambro using the language of b-divisors. So far it seems to me that every proof I have studied can be translated in the classical language without problems making it, in my opinion, more easily readable. I cannot thus understand what is the strength of this new language. Are there significant improvements it introduces? What are they? </p> http://mathoverflow.net/questions/31605/does-negative-kodaira-dimension-imply-uniruled/31826#31826 Answer by Gianni Bello for Does negative Kodaira dimension imply uniruled? Gianni Bello 2010-07-14T10:11:44Z 2010-07-14T10:11:44Z <p>I agree with Dmitri about the fact that Siu's paper is not actually considered, by many experts, as a proof of the aboundance conjecture. However, as far as question 1 is concerned, I want to point out that aboundance conjecture would imply Mumford's conjecture. In fact aboundance conjecture implies, in particular, that if the canonical divisor (say on a smooth projective complex variety) is pseudoeffective, then it is effective (that is, there are $m$-pluricanonical forms for some $m \in \mathbb{N}$. Hence if there aren't pluricanonical forms, then $K_X$ is not pseudoeffective, and the uniruledness of $X$ follows by a recent result of Boucksom, Demailly, Paun and Peternell: see <a href="http://front.math.ucdavis.edu/0405.5285" rel="nofollow">BDPP</a>.</p> http://mathoverflow.net/questions/28524/log-resolutions-of-linear-series Log resolutions of linear series Gianni Bello 2010-06-17T16:00:29Z 2010-06-17T18:08:47Z <p>Let $X$ be a complex normal projective variety, let $|L|$ be a non empty linear series on $X$ and let $b(|L|)$ be its base ideal. Suppose $f:X'\rightarrow X$ is a log resolution of the ideal $b(|L|)$.</p> <p>Is $f$ a log resolution of the linear series $|L|$ (even if $X$ is not smooth)?</p> <p>If it is do you have a proof or a reference for this?</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/27387/birational-pullbacks-of-divisors-on-singular-varieties Birational pullbacks of divisors on singular varieties Gianni Bello 2010-06-07T20:05:19Z 2010-06-08T07:57:42Z <p>Actually I have two related questions.</p> <p>Here is the first...</p> <p>Suppose $X$ is a, possibly singular, complex projective variety. Let $D$ be an effective Cartier divisor on $X$ and $x\in X$ a closed point such that the multiplicity of $D$ at $x$, usually denoted by $mult_x(D)$, is $k>0$.</p> <p>Let $\mu:X'\rightarrow X$ be the blowing up at $x$, and let $E$ be the exceptional divisor.</p> <p>Is it true that the order of $\mu^*(D)$ along $E$ is $k$?</p> <p>I'm sure about it in the smooth case and I suppose it is true also in the singular case but I don't have a reference.</p> <p>The second question is the following:</p> <p>Suppose $(X,\Delta)$ is a KLT pair, and $f:Y\rightarrow X$ is a log resolution of the pair $(X,\Delta)$.</p> <p>Let $E$ be an exceptional divisor on $Y$ mapping to a point on $X$ and let $a(E,X,\Delta)$ be its the discrepancy. In other words $a(E,X,\Delta)$ is the order along $E$ of the divisor $K_Y-\mu^*(K_X+\Delta)$.</p> <p>I know $a(E,X,\Delta)\leq 1$ if $X$ is smooth.</p> <p>It really seems to me this is also true in the singular case. Do I wrong? In any case do tou have a proof or a reference for this?</p> <p>Thanks a lot!</p> http://mathoverflow.net/questions/27249/what-does-the-ample-cone-look-like/27254#27254 Answer by Gianni Bello for What does the ample cone look like? Gianni Bello 2010-06-06T17:35:08Z 2010-06-06T23:38:06Z <p>Well, if you have a projective variety you can always find a basis of $NS(X)\otimes \mathbf{R}$ as an $\mathbf{R}$-vector space only made by ample $\mathbf{R}$-divisors. This implies that every projective variety (say over $\mathbf{C}$),<strong>if you suitably choose the basis</strong> of the vector space we are speaking about, has ample cone $(\mathbf{R}^+)^{\rho}$. [retracted in comment below]</p> <p>The reference is in the first chapter of the book "Positivity in algebraic geometry I", by R. Lazarsfeld. I don't have the book now so that I cannot tell you exactly what is the paragraph, but I'm pretty sure it is in the first chapter and it is just an exercise. Does this answer your question?</p> http://mathoverflow.net/questions/27249/what-does-the-ample-cone-look-like/27253#27253 Answer by Gianni Bello for What does the ample cone look like? Gianni Bello 2010-06-06T17:15:55Z 2010-06-06T17:15:55Z <p>Remember the ample cone is an open set in $NS(X)\otimes \mathbf{R}$ for the standard euclidean topology. Thus it cannot be $\mathbf{R}_{\geq 0}^\rho$ . But it can be $(\mathbf{R}^+)^{\rho}$. The simplest example is the projective space $\mathbb{P}^n$ over the complex numbers, where $\rho=1$ and the ample cone is simply $\mathbf{R}^+$.</p> http://mathoverflow.net/questions/27249/what-does-the-ample-cone-look-like/27250#27250 Answer by Gianni Bello for What does the ample cone look like? Gianni Bello 2010-06-06T16:55:24Z 2010-06-06T16:55:24Z <p>The shape of the ample cone can be very different, depending on the variety you are considering. In the preprint "Remarks on the cone of divisors" by Y. Kawamata, for example, you can see an unusual version of the Mori cone theorem that tells us something about the shape of the nef cone. The ample cone, as you very probably know, is the inner of the nef cone, so that they practically look the same. </p> http://mathoverflow.net/questions/122650/a-classification-of-rational-surfaces-with-effective-k/122797#122797 Comment by Gianni Bello Gianni Bello 2013-02-24T14:43:09Z 2013-02-24T14:43:09Z Sorry, you are absolutely right about the first mistake, my proof is wrong. Anyway, just for the sake of truth, I didn't write that $-K_{X'}$ is ample; but it's not very important http://mathoverflow.net/questions/119150/irreducible-divisors-containing-an-arbitrary-closed-set/119163#119163 Comment by Gianni Bello Gianni Bello 2013-01-18T10:05:38Z 2013-01-18T10:05:38Z Thanks quim (and Simone) for the proof. I choosed Olivier's as best answer, but your answer was very useful to understand that the answer is much closer to Bertini than I expected. I also believe that your (*) should be always true. Maybe, in the case $V_i$ is contained in the singular locus, you can blow-up $V_i$, take an exceptional divisor $E_i$ mapping surjectivly on $V_i$, and take something like an horizontal section of the morphism $E_i \to V_i$. http://mathoverflow.net/questions/119150/irreducible-divisors-containing-an-arbitrary-closed-set/119221#119221 Comment by Gianni Bello Gianni Bello 2013-01-18T09:54:06Z 2013-01-18T09:54:06Z Thanks Olivier! I really like this proof! http://mathoverflow.net/questions/111905/push-forward-and-strict-transforms/111958#111958 Comment by Gianni Bello Gianni Bello 2012-11-10T13:01:19Z 2012-11-10T13:01:19Z Thanks a lot Karl! Very clear example. http://mathoverflow.net/questions/108224/klt-singularities-are-quotient-in-codimension-2/108263#108263 Comment by Gianni Bello Gianni Bello 2012-09-28T08:55:38Z 2012-09-28T08:55:38Z Great! Thanks a lot for your answer! http://mathoverflow.net/questions/108224/klt-singularities-are-quotient-in-codimension-2/108227#108227 Comment by Gianni Bello Gianni Bello 2012-09-27T12:39:28Z 2012-09-27T12:39:28Z Yes, I'm considering normal singularities. Would you be so kind to explain me how the result follows from the proposition in Kollar-Mori? Probably it's tivial, but I'm missing something. Thanks a lot! http://mathoverflow.net/questions/108144/numerical-dimension-of-nef-divisors Comment by Gianni Bello Gianni Bello 2012-09-26T15:59:13Z 2012-09-26T15:59:13Z Right. But maybe one can use that D is nef. For example this should work for k=n-1, as one has that D^{n-1} is in the closure of NE(X) because it is in the dual of the nef cone (by Lazarsfeld 1.4.16), so that D^{n-1}.A&gt;0 (Lazarsfeld 1.4.29). http://mathoverflow.net/questions/108144/numerical-dimension-of-nef-divisors Comment by Gianni Bello Gianni Bello 2012-09-26T14:38:13Z 2012-09-26T14:38:13Z @ Artie Prendergast-Smith: About your first remark: The point is that D^k might be numerically 0 (that is in fact what I want to prove). About your second remark: you are right, I wasn't clear. I mean that D^k \cdot V=0 for all irreducible subvarieties V of dimension k not contained in a proper closed subset of X. Maybe to prove this one can just using an ample. In fact If D^k is not numerically 0, then, as you say, D^k.A^{n-k} &gt;0, and we find a contradiction. Do you agree? http://mathoverflow.net/questions/97204/real-vs-complex-surfaces/97218#97218 Comment by Gianni Bello Gianni Bello 2012-05-17T13:16:05Z 2012-05-17T13:16:05Z Dear Daniel, thanks a lot! Your answer is very helpful. So if I understand well to define the blow-up of X at p you blow-up X_F at all the points that form the Galois invariant collection corresponding to p. Now you have that the variety on F that you obtain, say Y, is actually a base change of a variety on E. In other words Y=Z_F for some Z. So you have that Bl_p X=Z. Am I wrong? http://mathoverflow.net/questions/97204/real-vs-complex-surfaces/97213#97213 Comment by Gianni Bello Gianni Bello 2012-05-17T11:26:14Z 2012-05-17T11:26:14Z Thank you. It seems like, from a Mori theory point of view, a real surface its &quot;more simple&quot; than its complexification, as the Picard number is in general smaller. Right? Do you also have an answer about the canonical divisor? http://mathoverflow.net/questions/83280/log-canonical-pairs-and-ample-divisors Comment by Gianni Bello Gianni Bello 2011-12-13T10:08:51Z 2011-12-13T10:08:51Z Thank you for the reference! http://mathoverflow.net/questions/83280/log-canonical-pairs-and-ample-divisors/83297#83297 Comment by Gianni Bello Gianni Bello 2011-12-13T10:08:22Z 2011-12-13T10:08:22Z Thank you very much! Just to br sure: what do you exactly mean by &quot;the strata related to the resolution&quot;? Are they something like the varieties that need to be blown-up in order to log-resolve the pair? http://mathoverflow.net/questions/81067/divisor-intersecting-non-negatively-the-negative-part-of-its-zariski-decompositio/81076#81076 Comment by Gianni Bello Gianni Bello 2011-11-16T16:13:15Z 2011-11-16T16:13:15Z Thanks a lot. It looks easier than I expected... http://mathoverflow.net/questions/81067/divisor-intersecting-non-negatively-the-negative-part-of-its-zariski-decompositio/81076#81076 Comment by Gianni Bello Gianni Bello 2011-11-16T15:13:14Z 2011-11-16T15:13:14Z It looks like, with the above notation, $H$ is the pullback of a line, so that it should be nef by itself and its Zariski decomposition is trivial. Am I wrong? http://mathoverflow.net/questions/77122/how-to-prove-the-existence-of-divisorial-zariski-decomposition/77164#77164 Comment by Gianni Bello Gianni Bello 2011-10-05T15:10:04Z 2011-10-05T15:10:04Z Actually I don't see it so trivially. Of course $N^1(X)$ is finitely generated, but how do you see that the $\Gamma_i$ (I'm using the notation of my previous comments ) are not linearly dependent? This requires some effort in Nakayama's book, I cannot see this in a trivial way.