How much can small modifications change the nef cone? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T06:30:20Z http://mathoverflow.net/feeds/question/32486 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32486/how-much-can-small-modifications-change-the-nef-cone How much can small modifications change the nef cone? Artie Prendergast-Smith 2010-07-19T12:53:01Z 2010-10-21T13:57:54Z <p>First let me give a precise formulation of the question; I'll give some background/motivation at the end.</p> <p>If X is a projective variety which is Q-factorial (meaning X is normal, and some sufficiently high multiple of every Weil divisor is Cartier), then a <i>small Q-factorial modification</i> or <i>SQM</i> of X means a birational map &phi;: X --> Y (where Y is another Q-factorial projective variety) which is an isomorphism in codimension 1. (Examples: flips and flops.) The question is then this:</p> <blockquote> <p>Is there an example of a Q-factorial projective variety X and &phi;: X --> Y an SQM of X such that the nef cone Nef(X) is rational polyhedral but Nef(Y) is not rational polyhedral? Or (highly unlikely I think) can one prove that this situation is impossible?</p> </blockquote> <p>And to push my luck:</p> <blockquote> <p>Give sufficient conditions to ensure that in the above situation, Nef(X) rational polyhedral implies Nef(Y) rational polyhedral.</p> </blockquote> <p>Really I think only the first question has a hope of being answered positively, but it would be nice to know if I was wrong. </p> <p><b>Background:</b> The question was prompted by a theorem of Hu and Keel ("Mori dream spaces and GIT", Michigan Math. J., 2000) which gives a characterisation of so-called <i>Mori dream spaces</i> --- certain varieties which behave very well with respect to the operations of the minimal model program. In particular, if X is a Mori dream space, then Nef(X) is rational polyhedral and so too is Nef(Y) for any SQM Y of X. It then seems natural to ask for an example where the first condition here holds but the second fails.</p> <p>Hu--Keel's theorem gives one answer to my second question above, since Birkar--Cascini--Hacon--McKernan proved that any Fano variety (or more generally, any variety of Fano type) is a Mori dream space. But it would be great to know of any sufficient condition that applies outside the Fano domain. </p> <p><strong>Edit (July 23):</strong> Balazs' answer settles the first question in the affirmative: flops can change the nef cone from finite to infinite. But it seems to remain open (and interesting to me) whether there are examples $X$ of this phenomenon with Kodaira dimension $kd(X)=-\infty$.</p> http://mathoverflow.net/questions/32486/how-much-can-small-modifications-change-the-nef-cone/32536#32536 Answer by Balazs for How much can small modifications change the nef cone? Balazs 2010-07-19T21:08:20Z 2010-07-19T21:08:20Z <p>In the world of Calabi-Yau (as opposed to Fano) varieties, one does not expect miracles of Mori dream space type. A specific example which gives a positive answer to your first question is described in the paper <a href="http://xxx.lanl.gov/abs/math/0102055" rel="nofollow">http://xxx.lanl.gov/abs/math/0102055</a> of Michael Fryers: a Calabi-Yau threefold (a degenerate quintic) which has some small resolutions having finite polyhedral nef cone, and some having an infinitely generated cone, which is locally rational polyhedral away from a single point on the boundary. The geometry, related to the Horrocks-Mumford bundle and abelian surfaces in projective four-space, is very beautiful.</p> http://mathoverflow.net/questions/32486/how-much-can-small-modifications-change-the-nef-cone/42588#42588 Answer by Sándor Kovács for How much can small modifications change the nef cone? Sándor Kovács 2010-10-18T06:30:11Z 2010-10-21T13:57:54Z <p>Let me start by a <strong>disclaimer</strong>: I misread the question and wrote what's below thinking that it was relevant. Then, just before posting it I looked at the question one more time and realized the mistake. </p> <p>As it happens, the statement and proof below is mathematically correct (I think), except that it does not answer the question, or one could perhaps even say that it does not have much to do with the question and probably everyone before me who have looked at this question knows what I wrote down here. </p> <p>At the same time, I would hate to just erase what I typed up and perhaps someone finds this useful, so I will just post it. If some MO users think this is inappropriate, I will be happy to erase it, just let me know. </p> <p>So, here come some mathematical ramblings loosely related to the question above.</p> <p>(The point is that there does not exist a morphism, but the question was about rational maps).</p> <hr> <p>There does not exist a morphism that is an SQM. This is because if $f:X\to Y$ is a small morphism, then $Y$ is <strong>not</strong> $\mathbb Q$-factorial. In fact a little more seems to be true:</p> <p><strong>Proposition.</strong> Let $f:X\to Y$ be a proper birational morphism and assume that $X$ is quasi-projective and $Y$ is $\mathbb Q$-factorial. Then the exceptional set of $f$ is of pure codimension $1$.</p> <p><em>Proof.</em> Let $H\subset X$ be a very ample Cartier divisor on $X$ and consider $f_*H$ the cycle theoretic image of $H$. This is a priori only a Weil divisor, but since $Y$ is $\mathbb Q$-factorial, it is $\mathbb Q$-factorial and replacing $H$ with an appropriate multiple we may assume that it is in fact Cartier.</p> <p>Now consider $F:=f^*f_*H-H$ and notice that a) $F$ is effective b) $-F$ is $f$-ample. These show that $F$ must be exceptional. This already proves that $f$ cannot be small, but we can prove the stronger statement stated above.</p> <p>${\rm Supp}\, F$ is obviously of pure codimension $1$, so if we prove that the exceptional set of $f$ is equal to this, then we are done. We have already seen that ${\rm Supp}\, F$ is contained in the exceptional set, so we only need to demonstrate that $f$ is an isomorphism on the complement of $F$. </p> <p>Now, observe that by the definition of $F$, we have that <code>$$\mathcal O_X(-F)\simeq \mathcal O_X(H)\otimes\mathcal O_X(-f^*f_*H)$$</code> and <code>\begin{multline*} f^*f_* \mathcal O_X(-F)\simeq f^*f_*\mathcal O_X(H-f^*f_*H)\simeq \\ \simeq f^*\left(f_*\mathcal O_X(H)\otimes \mathcal O_Y(-f_*H)\right)\simeq f^*f_*\mathcal O_X(H)\otimes \mathcal O_Y(-f^*f_*H) \end{multline*}</code></p> <p>Furthermore, since $H$ is very ample, it is generated by global sections, so <code>$$f^*f_*\mathcal O_X(H) \to \mathcal O_X(H)$$</code> is surjective. Then it follows from the above displayed lines that <code>$$f^*f_*\mathcal O_X(-F) \to \mathcal O_X(-F)$$</code> is also surjective. This implies that $f^{-1}(f({\rm Supp}\, F)) = {\rm Supp}\, F$. It follows then that $H$ and <code>$f^*f_*H$</code> agree on $X\setminus {\rm Supp}\, F$. In particular, $f^*({\rm some divisor})$ is ample on this $X\setminus {\rm Supp}\, F$. Finally this implies that then $f$ has to be an isomorphism on $X\setminus {\rm Supp}\, F$. Q.E.D. </p>