Does negative Kodaira dimension imply uniruled? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:57:14Z http://mathoverflow.net/feeds/question/31605 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31605/does-negative-kodaira-dimension-imply-uniruled Does negative Kodaira dimension imply uniruled? unknown 2010-07-12T19:40:22Z 2010-07-19T21:02:07Z <p>There is a conjecture (often attributed to Mumford) I believe which states that if, on a smooth proper variety $X$ (over an algebraically closed field of characteristic zero), there are no pluricanonical forms (that is $H^0(X, mK_X) = 0$ for all positive $m$) then $X$ is uniruled. </p> <p>Apparently, this conjecture follows from a "well known" conjecture arising from the minimal model program. I believe, but am not entirely sure, that this is the Abundance Conjecture, which says (in one formulation) that the Kodaira dimension of $X$ agrees with the so called numerical Kodaira dimension of $X$. There are by now many well written introductions to the MMP, here is <a href="http://www.math.uic.edu/~coskun/nine_authors.pdf" rel="nofollow">one</a>. </p> <p>At the same time, Professor Siu has recently posted a <a href="http://arxiv.org/abs/0912.0576" rel="nofollow">sketch of the proof the Abundance Conejcture</a>. Unfortunately, I am not sufficiently equipped to read the proof which uses L2 estimates of d-bar equations. Here are my questions.</p> <ol> <li><p>Is it true that over the complex number Siu's result implies Mumford's conjecture? He doesn't mention this in the preprint, but is there a reference?</p></li> <li><p>Is anyone well versed enough in both the analytic techniques and algebraic geometry to explain what Siu did to someone more algebraically minded?</p></li> <li><p>Do people have an opinion (vague or otherwise) as to whether techniques coming from analysis are just stronger than techniques coming from algebra? An if so, why is that? (An obvious example, example: Hodge decomposition, but also Siu's proof of invariance of plurigenera, and now the abundance conjecture). </p></li> </ol> http://mathoverflow.net/questions/31605/does-negative-kodaira-dimension-imply-uniruled/31619#31619 Answer by Dmitri for Does negative Kodaira dimension imply uniruled? Dmitri 2010-07-12T21:20:14Z 2010-07-14T12:05:45Z <p>As far as I understand not all experts in birational geometry would agree that Siu settles in his preprint Abundance conjecture, and this conjecture is considered for the moment as open. When Kawamata mentions in his recent paper the work of Siu </p> <p>On the abundance theorem in the case $\nu=0$ </p> <p><a href="http://arxiv.org/PS_cache/arxiv/pdf/1002/1002.2682v3.pdf" rel="nofollow">http://arxiv.org/PS_cache/arxiv/pdf/1002/1002.2682v3.pdf</a></p> <p>his says that in his work Siu using analytic methods reproved the abundance conjecture for a non-minimal algebraic variety whose numerical Kodaira dimension is equal to 0. I would guess Kawamata would not write his paper if it were accepted that Siu gives complete proof of abundance conjecture.</p> <p>So for the moment it should be conisdered that for $X$ of dimesnion $4$ and higher it is unknown if $H^0(nK_X)=0$ for all $n$ impies that $X$ is unirulled. As for analysis been stronger than algebra, there are examples when this is not quite the case. A famous one is bend and break, that uses characteristics $p$ and is a corner stone of minimal model program <a href="http://www.math.ens.fr/~debarre/Grenoble.pdf" rel="nofollow">http://www.math.ens.fr/~debarre/Grenoble.pdf</a> The existence of minimal models is still unknown even for 3-dimensional Kahler manifolds that are not algebraic... </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/31605/does-negative-kodaira-dimension-imply-uniruled/32534#32534 Answer by Eugene Eisenstein for Does negative Kodaira dimension imply uniruled? Eugene Eisenstein 2010-07-19T21:02:07Z 2010-07-19T21:02:07Z <p>The conjecture you attribute to Mumford is also sometimes called weak non-vanishing, or just non-vanishing. As already mentioned, by BDPP one is reduced to proving that if $h^0(mK_X) = 0$ for all $m \geq 0$ then $K_X$ is not even pseudo-effective, that is, not numerically a limit of effective divisors. There is an alternative formulation of the abundance conjecture which is arguably easier to understand: if $K_X + \Delta$ is nef and $(X, \Delta)$ is klt then $K_X + \Delta$ is semi-ample. The point is that this says that running the MMP turns $K_X$ into a semi-ample divisor.</p> <p>My understanding is that the answer to question to 2 is something like the following. Siu explains and gives references in his introduction to Part II of his paper that there is something called the numerically trivial fibration that is similar to the Kodaira-Iitaka fibration but works for the numerical dimension. Its construction is analytic in nature. Siu claims that, if $\pi : X' \to Y$ is a realization of numerically trivial fibration for $K_X$ with $X'$ a birational model of $X$, then $\pi_* O_{X'}(mK_{X'/Y})$ carries some kind of metric with strictly positive curvature on a general fiber of $\pi$. This, in particular, shows that $\pi$ is not the identity map, which I think is now known to imply the abundance conjecture, though I am not sure if this is how Siu proceeds. I am not familiar with the details of the argument.</p> <p>As for question 3, it is hard to say. The famous example is Siu's proof of the deformation invariance of plurigenera for all Kodaira dimensions, this is currently a purely analytic argument that involves taking limits of pluri-subharmonic functions to obtain a semipositive singular metric on $O_X(mK_X)$ that may not have analytic singularities but has the right multiplier ideal: see Paun's "Siu's invariance of plurigenera: a one-tower proof." This has resisted attempts to prove it algebraically so far. On the other hand, the algebraic approach allows reduction to positive characteristic. It may be that, ultimately, anything you can do with one you can do with the other. Of course, "algebraic" here allows the Kodaira vanishing theorem, which can be deduced from the homological statement that the map $H^i(X, \mathbb{C}) \to H^i(X, O_X)$ induced by the natural inclusion of sheaves is surjective.</p>