User j.c. ottem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:02:31Z http://mathoverflow.net/feeds/user/3996 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75362/did-any-new-mathematics-arise-from-ruffinis-work-on-the-quintic-equation Did any new mathematics arise from Ruffini's work on the quintic equation? J.C. Ottem 2011-09-13T22:45:45Z 2013-05-11T08:54:59Z <p>The impossibility of solving the general polynomial of degree $\ge 5$ by radicals is surely one of the most celebrated results in algebra. This result is known as the Abel-Ruffini theorem, although it's usually asserted that Ruffini's proof was incomplete.</p> <p>Still, Paolo Ruffini's contributions to algebra seem to be neither widely known nor well understood. This is perhaps not so surprising since Ruffini's first attempted proof spanned 516 pages and the mathematical argument was difficult to follow. </p> <p>A modern discussion of Ruffini's proof is found in Ayoub's article <a href="http://www.springerlink.com/content/m8886pg411143329/" rel="nofollow">'Paolo Ruffini's contributions to the quintic'</a>. Accoring to Ayoub, Ruffini's argument was not a priori flawed, but it relies on several unproved non-trivial claims. More precisely, Ruffini fails to prove that the splitting field is one of the fields in the tower of radicals which corresponds to a solution expressed in radicals. </p> <p>The revolutionary proofs of Abel and Galois following Ruffini paved way for group theory and Galois theory. Still, one could wonder what Ruffini actually proved in those 516 pages:</p> <blockquote> <p>Did any significant new mathematical concepts, ideas or theorems arise from Ruffini's work on the quintic?</p> </blockquote> <p>Cauchy seems to be one of the few mathematicians who found inspiration from Ruffini's work. In a letter dated 1821, he writes:</p> <p><em>"... your memoir on the general resolution of equations is a work which has always seemed to me worthy of the attention of mathematicians and which, in my judgement, proves completely the insolvability of the general equation of degree $>4$. [...] In another memoir which I read last year to the Academy of Sciences, I cited your work and reminded the audience that your proofs establish the impossibility of solving equations algebraically ..."</em></p> <p>In fact, according to <a href="http://books.google.com/books/about/Abel_s_proof.html?id=nIpGp9Y5AlEC" rel="nofollow">Pesic</a>, Cauchy's influential 1815 paper on permutations is clearly based on work of Ruffini. So are there other examples of his influence?</p> http://mathoverflow.net/questions/128807/noether-lefschetz-over-finite-fields Noether-Lefschetz over finite fields J.C. Ottem 2013-04-26T08:32:34Z 2013-04-26T14:48:53Z <p>The classical Noether-Lefschetz theorem asserts the following: Over the complex numbers, a very general surface $S\subset \mathbb{P}^3$ has Picard number 1 (that is, $Pic(S)\simeq \mathbb Z$), provided that $\mbox{deg }S\ge 4$.</p> <p>Over finite fields (or even countable fields), the corresponding statement does not make much sense, because here `very general' refers to the surface being chosen outside a countable union of closed proper subsets of the parameter space of surfaces. Still, I think makes sense to ask:</p> <blockquote> <p>What evidence is there for the Noether-Lefschetz statement over countable fields? I.e., is there a sense in which the set of surfaces with low Picard number constitutes a 'large' proportion of the surfaces in $\mathbb{P}^3$ e.g., over finite fields with many elements?</p> </blockquote> http://mathoverflow.net/questions/127050/is-the-segre-embedding-projectively-normal/127053#127053 Answer by J.C. Ottem for Is the Segre embedding projectively normal? J.C. Ottem 2013-04-10T06:45:45Z 2013-04-10T06:45:45Z <p>Yes. The Segre embedding $i:P:=\mathbb P^n \times \mathbb P^m\to \mathbb P^N$ is defined by the sections of the line bundle $O_P(1,1):=pr_1^*O_{\mathbb P^n}(1)\otimes pr_2 O_{\mathbb P^m}(1)$ on $\mathbb P^n \times \mathbb P^m$ and by definition of $i$ we have $i^*O_{\mathbb P^N}(1)=O(1,1)$. In other words, the equality $$H^0(\mathbb P^n \times \mathbb P^m,i^*O_{\mathbb P^N}(k))=H^0(\mathbb P^n \times \mathbb P^m,O(k,k))$$ holds for any $k\in \mathbb Z$, and so $H^0(\mathbb P^n \times \mathbb P^m,i^*O(k))$ decomposes as the tensor product $H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(k))\otimes H^0(\mathbb P^m, \mathscr O_{\mathbb P^m}(k))$ by Kunneth's theorem. Note however, that the definition of projectively normal requires that the map $$H^0(\mathbb P^N,O(k))\to H^0(P,O(k,k))$$is surjective for all $k\ge 0$ not just for $k=1$ (in which case the embedding is called linearly normal). This is true in our case, since global sections of $O(k,k)$ are polynomials in sections of $O(1,1)$. Of course, when $k=2$, the kernel of the above map is generated by the quadrics defined as the $2\times 2$ minors defining the Segre embedding.</p> http://mathoverflow.net/questions/124880/picard-group-of-a-k3-surface-generated-by-a-curve/124882#124882 Answer by J.C. Ottem for Picard group of a K3 surface generated by a curve J.C. Ottem 2013-03-18T17:51:10Z 2013-03-18T17:51:10Z <p>This is equivalent to saying that there exists a K3 surface with an ample generator (a polarization) $L$, with $L^2=2g-2$ and $|L|$ has a smooth member. There are various geometric ways to construct such surfaces, e.g., by using double covers or quartic surfaces in $\mathbb P^3$ containing special curves. You will find this in VIII.15 in Beauville's book 'Complex Algebraic Surfaces". Given this, one can show the existence of a K3 where $L$ is a generator of the Picard group, using the fact that a generic K3 surface has Picard number 1.</p> http://mathoverflow.net/questions/123370/do-finitely-many-plurigenera-determine-the-kodaira-dimension Do finitely many plurigenera determine the Kodaira dimension? J.C. Ottem 2013-03-01T19:54:20Z 2013-03-09T14:25:05Z <p>Let $X$ be a smooth projective variety over a field of characteristic $0$ and let $K_X$ be the canonical bundle. Recall that the Kodaira dimension $\kappa(X)$ is defined as the number $\kappa$ such that $$\alpha m^{\kappa}\le h^0(X,mK_X) \le \beta m^{\kappa}$$ for $\alpha,\beta>0$ and $m$ sufficiently large and divisible (or $\kappa=-\infty$ if $h^0(X,mK_X)=0$ for all $m$). It is well-known that $\kappa(X)$ is a birational invariant. A natural question is how large $m$ we need to take to determine $\kappa$. More precisely:</p> <blockquote> <p>Is there an integer $M>0$, depending only $\dim X$, such that the values $h^0(X,K_X),h^0(X,2K_X),\ldots,h^0(X,MK_X)$ determine $\kappa(X)$?</p> </blockquote> <p>In particular, is there an $M>0$, depending on $\dim X$, such that $h^0(K_X)=\ldots=h^0(MK_X)=0$ implies that $\kappa=-\infty$?</p> http://mathoverflow.net/questions/123522/toric-fano-manifolds-with-picard-number-1/123525#123525 Answer by J.C. Ottem for Toric Fano manifolds with Picard number 1 J.C. Ottem 2013-03-04T12:10:34Z 2013-03-04T12:10:34Z <p>I don't think the Fano condition plays an essential role here; Any smooth toric variety with Picard number 1 is a projective space. This can be verified using usual toric geometry machinery: The condition for a toric variety $X$ to be of dimension $n$ and Picard number 1 means that the fan $\Delta$ is generated by $n+1$ rays $v_0,\ldots,v_n$, where the $v_i$ are primitive in the lattice $N$. If $X$ is smooth, any subset of $n$ vectors from ${v_0,\ldots,v_n}$ span the lattice $N\simeq \mathbb Z^n$, and one recovers the standard fan of $\mathbb P^n$.</p> <p>If $X$ is allowed to have $\mathbb Q$-factorial singularities, then the set of such Fano varieties were shown to be bounded by Borisov and Borisov.</p> http://mathoverflow.net/questions/47895/varieties-where-every-non-zero-effective-divisor-is-ample Varieties where every non-zero effective divisor is ample J.C. Ottem 2010-12-01T12:26:46Z 2013-03-02T12:10:36Z <p>The following question seems very intuitive, but I haven't been able to find any proof (or counterexample). </p> <blockquote> <p>Let $X$ be a non-singular projective variety of $\dim X\ge 2$ and let $NS^1(X)$ be its Neron-Severi group. If every non-zero effective divisor on $X$ is ample, does it follow that $X$ has Picard number one, i.e., $\rho=$ rank $NS^1(X)=1$?</p> </blockquote> <p><strong>Motivation:</strong></p> <p><strong>1)</strong> In the case of Fano varieties the result is true (the proof is an easy application of Riemann-Roch). In fact this result was a key ingredient in Mori's proof of Hartshorne's conjecture for projective 3-space (i.e., any 3-fold with ample tangent bundle is isomorphic to $\mathbb{P}^3$). See Mori's original <a href="http://projecteuclid.org/euclid.kjm/1250522508" rel="nofollow">article</a> for the details.</p> <p><strong>2)</strong> In <a href="http://mathoverflow.net/questions/28326/a-nontrivial-surface-on-which-any-two-curves-intersect" rel="nofollow">this Mathoverflow question</a> Charles Staats asks for a surface with the property that any two curves on the surface have nontrivial intersection. In his comment, BCnrd considered a K3 surface with Picard number one, which satisfies the condition precisely because any effective divisor is ample. A natural question is whether any such surface has Picard number one.</p> <p>I am mostly interested in the case where $X$ is a complex projective variety. In the case the result does not hold, I'd also be interested in seeing a concrete counterexample and other examples of varieties where the result holds.</p> http://mathoverflow.net/questions/122361/is-there-an-enriqueskodaira-like-classification-of-fano-threefolds/122365#122365 Answer by J.C. Ottem for Is there an Enriques–Kodaira-like classification of Fano threefolds? J.C. Ottem 2013-02-20T00:30:33Z 2013-02-20T00:36:27Z <p>Yes, Fano threefolds have been completely classified and one has explicit projective models of them. See V.A. Iskovskih, Yu. G. Prokhorov: Algebraic Geometry V: Fano varieties. Encyclopaedia of Math. Sciences 47, Springer-Verlag, Berlin 1999. See also <a href="http://people.mpim-bonn.mpg.de/andreaso/files/fano.pdf" rel="nofollow">Andreas Ott's thesis</a> on Fano threefolds of picard number $\rho\ge 2$.</p> <p>In some sense the most complicated Fanos are the ones with Picard number one, i.e., $Pic(X)\simeq \mathbb Z$. Here there are 18 families of such threefolds and they have been classiﬁed by Iskovskih. The most basic invariant here is the $index$ of $X$, i.e., the maximal integer $r$ such that $K_X$ is divisible by $r$ in $Pic(X)$. Fano threefolds of higher Picard number was classified by Mori and Mukai. This was one of the early triumphs of Mori theory. </p> http://mathoverflow.net/questions/115403/dimension-of-polynomial-algebras/115411#115411 Answer by J.C. Ottem for Dimension of polynomial algebras J.C. Ottem 2012-12-04T15:41:42Z 2012-12-04T15:41:42Z <p>This class of rings is studied by Arnold and Gilmer in the paper </p> <p>Jimmy T. Arnold and Robert Gilmer '<a href="http://www.jstor.org/stable/2373549" rel="nofollow">The Dimension Sequence of a Commutative Ring</a>' American Journal of Mathematics , Vol. 96, No. 3 (1974), pp. 385-408.</p> http://mathoverflow.net/questions/115001/open-problems-in-birational-geometry-after-bchm/115005#115005 Answer by J.C. Ottem for Open problems in Birational Geometry, after BCHM J.C. Ottem 2012-11-30T17:09:13Z 2012-11-30T22:19:07Z <ul> <li>The Abundance conjecture.</li> </ul> <p>In its simplest form it says: If $X$ is a minimal variety (that is, the canonical divisor $K_X$ is nef and $X$ has terminal singularities) then some multiple $mK_X$ is base-point free. Thus sections of some power of the canonical bundle give a morphisms to projective space. In this case, it is straightforward to prove that the canonical ring $R(X,K_X)$ is finitely generated. </p> <ul> <li>Termination of log flips. </li> </ul> <p>I think both of these conjectures are open in dimension $\ge 4$.</p> <p><strong>Edit</strong>: As Artie points out, existence of flips is known in all dimensions by the work of BCHM. There are also partial results on termination in dimension 4 by, Birkar, Fujino, Alexeev-Hacon-Kawamata,..</p> http://mathoverflow.net/questions/110553/integral-hodge-classes-of-the-calabi-yau-3-fold/110562#110562 Answer by J.C. Ottem for integral hodge classes of the Calabi-Yau 3-fold J.C. Ottem 2012-10-24T17:21:11Z 2012-10-24T17:21:11Z <p>Even though there are several examples showing that the Integral Hodge conjecture (IHC) fails in general, there are some positive results as long as the canonical bundle is not too positive:</p> <p><strong>Theorem</strong> (Voisin) Let $X$ be a smooth projective threefold over $\mathbb C$ which is either uniruled or strongly Calabi-Yau ($K_X\simeq O_X$ and $b_1(X) = 0$). Then the IHC is true for X, i.e., $H_2(X, \mathbb Z)$ is generated by algebraic curves.</p> <p>There is also the following extension to the theorem above proved by Totaro:</p> <p><strong>Theorem</strong> (Totaro) The IHC holds for Calbi-Yau 3-folds holds without the assumption that $b_1(X) = 0$. In particular, IHC holds for 1-cycles on abelian 3-folds.</p> <p>See Voisin's paper <a href="http://www.math.jussieu.fr/~voisin/Articlesweb/inthodge.pdf" rel="nofollow">On integral Hodge classes on uniruled or Calabi-Yau threefolds</a> and <a href="http://www.cims.nyu.edu/~tschinke/.conferences/symposium12/talks/Totaro2.pdf" rel="nofollow">these notes</a>.</p> <p>This raises the question when $H_2(X,\mathbb Z)$ is generated by rational curves for simply connected Calabi-Yaus (as is the case for K3 surfaces). This is of course a difficult question since it is not even known whether a simply connected CY 3-fold always contains a rational curve.</p> http://mathoverflow.net/questions/109391/clean-introduction-to-toric-varieties-for-an-undergraduate-audience/109393#109393 Answer by J.C. Ottem for Clean introduction to toric varieties for an undergraduate audience J.C. Ottem 2012-10-11T16:57:11Z 2012-10-11T17:35:43Z <p>David Cox has some nice expositions on toric varieties on his web page <a href="http://www3.amherst.edu/~dacox/" rel="nofollow">here</a>. Cox is also one of the authors of the book "Toric Varieties", which is a very readable, yet comprehensive introduction to toric varieties. The first chapter here should provide you with enough motivation and examples for your talk. Then there is also chapter 1 in Fulton's book, which is the classic reference on the subject.</p> <p>As for the motivational examples, you should look for examples that show the real power of toric varieties: <em>That abstract algebro-geometric constructions can uaually be viewed very concretely by working with the defining combinatorial data (e.g. the fan)</em>. Some of these examples might do the trick:</p> <p>1) The quadric surface ${xy-zw=0}$ in $\mathbb P^3$ and its affine cone in $\mathbb A^4$</p> <p>2) The singular quadric $y^2=zw$ in $\mathbb A^3$. </p> <p>3) Hirzebruch surfaces </p> <p>4) Toric blow-ups and subdivisons of the fan</p> <p>In the basic examples 1)-3), it is straightforward to write out the action of the torus, and see directly how monomials in the coordinate ring relates to the lattice points in the dual cones. Also, in the projective examples you can see how gluing the affine toric varieties works in terms of the fan data. </p> <p>These examples demonstrate typical features of toric varieties, for example that their ideals are generated by binomials and their Chow ring is generated by the torus invariant subvarieties. </p> <p>I like the Hirzebruch surface example because you can somehow 'see' the $\mathbb P^1$-bundle structure in the defining polytope and it is intuiticely clear that any toric surface is a blow-up of either $\mathbb P^2$ or a Hirzebruch surface. Moreover, I think it's pretty cool that you can view birational morphisms of toric varieties (e.g., resolution of singularities) as subdivisions of the defining fans. The example in this <a href="http://mathoverflow.net/questions/64181/blowing-curves-down/64203#64203" rel="nofollow">MO thread</a> illustrates this. Another interesting example is the affine cone $Z(xy-zw=0)\subset \mathbb A^4$ which gives a nice combinatorial interpretation of the Atiyah flop.</p> http://mathoverflow.net/questions/64181/blowing-curves-down/64203#64203 Answer by J.C. Ottem for blowing curves down J.C. Ottem 2011-05-07T15:01:18Z 2012-09-28T02:56:50Z <p>Here's the 'toric description' Karl mentioned in the comments:</p> <p><img src="http://img687.imageshack.us/img687/2370/blowupo.jpg" alt="alt text"></p> <p>The point is that the blow-up of a toric variety in a torus invariant point induces a so-called 'star subdivision' of the defining fan (see Fulton's book for details) and that in the your case the two resulting fans coincide. The added 1-dimensional rays (shown in red above) correspond to the exceptional divisors of the blow-up.</p> http://mathoverflow.net/questions/105676/cohomology-of-line-bundles/105691#105691 Answer by J.C. Ottem for Cohomology of line bundles J.C. Ottem 2012-08-28T07:58:39Z 2012-08-28T15:45:32Z <p>Here's an example where $B_1$ is not a finite union of cones: Let $X$ be a K3 surface of Picard number 3, such that the cone of effective divisors, $Eff(X)=Nef(X)$ is one of the components of the $[D\in NS(X) | D^2\ge 0]$ (for example, a K3 surface without $(-2)$-curves, such surfaces can be constructed as in <a href="http://www.math.missouri.edu/~dale/regexfinal.pdf" rel="nofollow">this paper</a>). In this case $Eff(X)$ is non-rational polyhedral. Then it is easy to see that $$B_0=Eff(X), B_2=-Eff(X) \mbox{ and } B_1= (B_0\cup B_2)^c$$In particular, $B_1$ is not a finite union of cones.</p> <p>As for the the 'General question', I think the shapes of the $B_i$ are related to Alex Kuronya's <a href="http://arxiv.org/pdf/math/0501491v1.pdf" rel="nofollow">asymptotic cohomological functions</a>. These are basically higher cohomology versions of the volume function of a big line bundle and measure the asymptotic growth of cohomology. The definition is $$\hat{h}^{i}(X,D) = \limsup_{m}\frac{h^i(X,O_X(mD))}{m^n/n!} .$$One of the main theorems in his paper is that the $\hat{h}^i$ define continuous functions on the Neron-Severi space $NS(X)=A^1(C)\otimes \mathbb{R}/\equiv$. The vanishing of these functions should be related to your question.</p> <p>See his paper for a lot of examples of asymptotic cohomology vanishing (flag varieties, abelian varieites,..). In most of these examples it is clear that the regions of vanishing cohomology are unions of convex cones.</p> <p>These functions have been used to study certain positivity conditions of line bundles. For example, in the paper <a href="http://arxiv.org/abs/math/0601140" rel="nofollow">"Higher cohomology of divisors on a projective variety"</a> by T. de Fernex, A. Kuronya, R. Lazarsfeld, the authors show that a divisor $D$ is ample if and only if the higher asymptotic cohomological functions vanish in a neighbourhood of $D$ in $NS(X)$. </p> <p>There is also the concept of $q-$ampleness, introduced by Demailly-Peternell-Schneider, Arapura, and Totaro among others. This is a generalization of the notion of an ample line bundle in the sense that high tensor powers of a line bundle are required to kill cohomology of coherent sheaves in degrees $>q$ (so $0$-ampleness coincides with ordinary ampleness.). This is related to $\hat{h}^i(X,D)$ in the sense that it is expected that the local vanishing of the $\hat{h}^i(X,D)$ in degrees $>q$ is equivalent to $q$-ampleness of $D$. In general it is known (and easy to prove) that the cones of $q$-ample line bundles are star-shaped.</p> http://mathoverflow.net/questions/52985/what-is-known-about-the-birational-involutions-of-p3 What is known about the birational involutions of P^3? J.C. Ottem 2011-01-23T18:25:28Z 2012-07-05T14:35:43Z <p>Describing the group of birational automorphisms of $\mathbb{P}^n$, $\mbox{Bir}(\mathbb{P}^n)$, for $n\ge 3$ is a fundamental open problem in birational geometry. For $n=2$, the classical theorem of Noether says that this group is generated by linear transformations and the Cremona transformation, which is given by $$\phi:(x_0 : x_1 : x_2) \to (x_0^{-1} : x_1^{-1} : x_2^{-1}).$$For $n\ge 3$, there is an analogous Cremona transformation, but it is known that the group $\mbox{Bir}(\mathbb{P}^n)$ is no longer generated by this and $\mbox{PGL}_{n+1}(\mathbb{C})$. My question is therefore</p> <blockquote> <p>Are there examples of other birational involutions of $\mathbb{P}^3$?</p> </blockquote> <p>In case the answer is yes, have these been classified? I'm also interested in the analog of Noether's theorem in this case: Are there examples of birational transformations of $\mathbb{P}^3$ that can not be written as a composition of birational involutions?</p> http://mathoverflow.net/questions/99591/does-a-small-contraction-occur-between-smooth-varieties/99599#99599 Answer by J.C. Ottem for Does a small contraction occur between smooth varieties? J.C. Ottem 2012-06-14T11:59:53Z 2012-06-14T15:56:54Z <p>Yes. Suppose $f$ contracts a curve $C$. Then for any ample divisor $D$, we have $D\cdot C>0$. But <code>$D=f^*f_*D$</code> by your hypotheses on the exceptional locus, and so $D\cdot C=f_*D\cdot f_*C=0$, a contradiction. </p> http://mathoverflow.net/questions/96248/an-example-of-a-calabi-yau-3-fold-with-irrational-nef-cone/96253#96253 Answer by J.C. Ottem for An example of a Calabi-Yau 3-fold with irrational nef cone? J.C. Ottem 2012-05-07T18:58:44Z 2012-05-07T21:36:12Z <p>I think the following construction could work: Let $X$ be a Calabi-Yau threefold without any rational curves (such threefolds do exist). Then $Nef(X)=\overline{Eff}(X)$, i.e., the nef cone equals the pseudoeffecitve cone (by the log cone theorem) and the nef boundary is given by the null cone, that is, the set of divisor classes $D$ such that $D^3=0$. Note that this null cone is given by the zero locus of a degree 3 polynomial in the Neron Severi group, and the non-rational points of this will give you the example.</p> <p>I should mention that the nef cone of a Calabi-Yau threefold is a very interesting object even though it is often non-rational polyhedral. Indeed, the Kawamata-Morrison cone conjecture states that the nef cone and movable cone should instead have a rational polyhedral <em>fundamental domain</em> for the action of $im(Aut(X)\to GL(N^1(X)))$. In some sense, this is the next best thing compared to the Fano case ($K&lt;0$) where everything is rational polyhedral. So in particular if the automorphism group is infinite, then one would expect a non-rational polyhedral nef cone. </p> http://mathoverflow.net/questions/95490/trying-to-understand-the-support-of-the-sheaf-of-relative-differentials/95493#95493 Answer by J.C. Ottem for trying to understand the support of the sheaf of relative differentials J.C. Ottem 2012-04-29T08:06:56Z 2012-04-29T08:13:27Z <p>Ravi Vakil has a good explaination for the definition $\Delta^*(I/I^2)$ in his notes. See his AG notes <a href="http://math.stanford.edu/~vakil/0506-216/216class38.pdf" rel="nofollow">here</a> or <a href="http://math.stanford.edu/~vakil/216blog/FOAGapr1312public.pdf" rel="nofollow">here</a> (chapter 23). In particular, I guess thinking about this locally makes it a little clearer what's going on, in terms of derivations etc. Also, when $X$ is smooth, it is instructive to see that this really gives the cotangent bundle on $X$. </p> <p>As for your question about ramification points: Let $f:X\to Y$ be a finite morphism of curves (I will assume that these are smooth in the following). It is useful to have in mind the exact sequence $$0\to f^*\Omega_{Y}\to \Omega_X \to \Omega_{X|Y}\to 0.$$(This is exact at the right in the smooth case, but not in general). Note that $\Omega_{X|Y}$ is a torsion sheaf since the two other sheaves are locally free of the same rank (they are line bundles on $X$). At a point $q\in Y$ and $p\in X$ in the preimage of $q$, let $dx$ denote a generator for $\Omega_{Y,q}$ as a $O_{Y,q}$-module. Now, $(\Omega_{X|Y})_P=0$ if and only if $f^*dx$ is a generator of $\Omega_{X,p}$, which happens if and only if $f$ pulls back a local parameter to a local parameter, that is $p$ is unramified. Moreover, the exact sequence above shows that the ramification index is exactly the length of the sheaf $\Omega_{X|Y}$. Finally, note that this sequence gives the Riemann hurwitz formula, relating the canonical divisors of $X$ and $Y$ and the ramification divisor of $f$.</p> http://mathoverflow.net/questions/95414/some-examples-for-q-gorenstein-variety-and-gorenstein-variety/95415#95415 Answer by J.C. Ottem for Some examples for Q-Gorenstein variety and Gorenstein variety. J.C. Ottem 2012-04-28T06:16:31Z 2012-04-28T06:28:51Z <p>Cones are usually a good source of examples when it comes to questions like these. In your case, let $Y\subset \mathbb{P}^5$ be the Veronese embedding of $\mathbb{P}^2$ and let $X\subset \mathbb{P}^6$ be the projective cone of $Y$. Then $X$ is a normal non-Gorenstein variety which is not Gorenstein. Indeed, Let $L\subset Y$ be the image of a line in $\mathbb{P}^2$ and let $D$ be the Weil divisor given by the cone over $L$. Let $H$ be the generator of $Pic(X)$, that is, the hyperplane section from the embedding in $\mathbb{P}^5$. Then $-K_X=3D$ is not Cartier (as you can see by blowing up the vertex of the node and use the adjunction formula). But $2K_X$ is certainly Cartier as $2K_X\sim -6D\sim -2H$. </p> http://mathoverflow.net/questions/92044/minimal-free-resolution-of-the-ideals/92045#92045 Answer by J.C. Ottem for Minimal Free resolution of the ideals J.C. Ottem 2012-03-23T22:22:26Z 2012-03-24T00:41:52Z <p>The most basic motivation for studying resolutions is computing the Hilbert function/polynomial of $Y=V(I)$. The geometric information in this polynomial is essentially the dimension, the embedded degree of $Y$ and the arithmetic genus of $Y$.</p> <p>More refined information can be found by looking at the betti numbers of $I$. By definition, these are the ranks of the free modules appearing in the resolution and thus tell us about the <em>syzygies</em> of $I$, that is, the relations between the generators of $I$. A standard example (see Eisenbud's book) is the case of 7 points in $\mathbb{P}^3$: All such configurations of points have the same Hilbert polynomial, but the graded Betti numbers of the ideal of the points determine whether the points lie on a rational normal curve or not. </p> <p>For curves, there are also interesting relations between the betti numbers of a canonically embedded curve and the Clifford index of the curve (which is a geometrical invariant of the curve). This is the main content of Green's conjecture, which is still open. </p> <p>For more info on this, take a look at Eisenbud's excellent book 'The Geometry of Syzygies' or Wiegand's article '<a href="http://www.ams.org/notices/200604/what-is.pdf" rel="nofollow">What is.. a Syzygy</a>?'</p> http://mathoverflow.net/questions/90490/blow-up-along-a-subscheme-and-along-its-associated-reduced-closed-subscheme/90497#90497 Answer by J.C. Ottem for Blow-up along a subscheme and along its associated reduced closed subscheme J.C. Ottem 2012-03-07T19:48:49Z 2012-03-08T08:05:57Z <p>In general they can be very different. For example take the subscheme $Y$ of $\mathbb{A}^2$ given by the ideal $(x^2,y)$. Here the blow up is covered by the two open subsets</p> <p>$$U = \mbox{Spec} k[x, y][t]/(y − x^2t),\qquad V = \mbox{Spec} k[x, y][s]/(ys − x^2)$$</p> <p>In particular the blow up of $Y$ is singular, whereas the blow-up of $\mathbb{A}^2$ at a point is not.</p> <p>In general, even if you assume that both blow-ups are smooth, all sorts of things can happen depending on how complicated the ideal sheaf is. For example the blow-ups can have a different number of exceptional divisors and not even be related by a finite map. Even worse, <em>every birational morphism $X'\to X$ is the blow-up of $X$ along some ideal sheaf</em>. </p> http://mathoverflow.net/questions/89436/bound-on-the-anticanonical-degree-of-toric-fano-varieties/89445#89445 Answer by J.C. Ottem for Bound on the (anticanonical) degree of toric Fano varieties J.C. Ottem 2012-02-24T20:53:51Z 2012-02-24T20:53:51Z <p>The counterexample of Debarre goes as follows: Consider the projective bundle $$\pi:X=\mathbb{P}(O_{\mathbb{P}^s}^{\oplus r}\oplus O_{\mathbb{P}^s}(a))\to P^s,$$ which is a smooth toric variety. Here $$-K_X\sim (r+1)O_X(1)+(s+1-a)\pi^*H$$ where $H$ is a hyperplane in $\mathbb{P}^s$. If $0\le a\le s$, this is also a Fano variety, as one can check using Kleiman's criterion. Also, using the relation $O_X(1)^{\cdot(r+1)}=a\pi^*H \cdot O_X(1)^{\cdot r}$, Debarre finds after setting $a=n-r$, that $$(-K_X)^n=(-K_X)^{r+s}\ge (r+1)^n (n-r)^{n-r}> \left(\frac{3n^2}{10\log n}\right)^n$$ In particular, the $n$-th root is unbounded as $n\to \infty$.</p> http://mathoverflow.net/questions/88193/proofs-in-the-same-vein-as-ax-grothendieck/88339#88339 Answer by J.C. Ottem for Proofs in the same vein as Ax-Grothendieck J.C. Ottem 2012-02-13T10:48:10Z 2012-02-13T10:48:10Z <p><a href="http://www.math.ucdavis.edu/~osserman/classes/256B/notes/sem-mori.pdf" rel="nofollow">Mori's bend and break technique</a> which is used to show the existence of rational curves in higher dimensional varieties, is a famous example of this.</p> http://mathoverflow.net/questions/87108/nef-cone-of-a-toric-variety/87122#87122 Answer by J.C. Ottem for nef Cone of a Toric Variety J.C. Ottem 2012-01-31T10:08:35Z 2012-01-31T10:08:35Z <p>You can use the fact that a divisor class $D$ on a toric variety is nef if and only if it has non-negative intersection with the finitely many classes of torus invariant curves. If you are using the torus invariant divisors as the basis for $A^1(X)$, then these numbers are easy to compute combinatorially and this will give you a finte set of linear inequalities for the nef cone. See the the book by Cox-Little-Schenck chapter 6 for more details. </p> <p>As far as I know, there is no explicit description of the rays of the nef cone in terms of the combinatorics of the fan, so this approach with linear inequalities is the best you can do.</p> http://mathoverflow.net/questions/79040/cone-of-movable-curves/79078#79078 Answer by J.C. Ottem for Cone of movable curves J.C. Ottem 2011-10-25T13:47:24Z 2012-01-30T20:43:07Z <p>A more direct approach is the following:</p> <p>Let $X$ be the projective bundle $\pi:\mathbb{P}(\mathcal{E})\to \mathbb{P}^1$ where $\mathcal{E}=\mathcal{O}\oplus \mathcal{O}(-1) \oplus \mathcal{O}(-2)$. Let $M$ be the tautological bundle of $X$. It is easily checked that the ample line divisors $H_i$ on $X$ correspond to line bundles of the form $M^a\otimes \pi^*\mathcal{O}(b)$ with $b>2a$. We show that $Mov(X)$ is not spanned by products of the form $H_1\cdot H_2$. </p> <p>Consider the line bundle $L=M\otimes \pi^*\mathcal{O}(-1)$. Using the Leray spectral sequence for the morphism $\pi$ we easily see that $L$ is not pseudoeffective. However, it is also straightforward to check that $$L\cdot H_1\cdot H_2=b_1+b_2-4>0$$ for $H_i=M\otimes \pi^*\mathcal{O}(b_i)$. Hence $L$ lies in the dual cone of $\overline{Q}(X)$ (using your notation). Now, if $\overline{Mov}(X)$ was generated by the $H_1\cdot H_2$'s this would imply that $L$ is pseudoeffective (by BDPP), a contradiction. </p> http://mathoverflow.net/questions/83290/noteworthy-achievements-in-and-around-2010/84001#84001 Answer by J.C. Ottem for Noteworthy achievements in and around 2010? J.C. Ottem 2011-12-21T12:52:11Z 2011-12-21T12:52:11Z <p><a href="http://www.science.unitn.it/~pignatel/PoAV/talks/Tzeng.pdf" rel="nofollow">Tzeng's proof of the Goettsche's conjecture</a>, which says the number of nodal curves in a linear system $|C|$ on a projective surface $S$ is given by a universal polynomial in the Chern numbers of $C$ and $S$ .</p> http://mathoverflow.net/questions/83965/interesting-applications-of-the-classical-stokes-theorem/83984#83984 Answer by J.C. Ottem for Interesting Applications of the Classical Stokes Theorem? J.C. Ottem 2011-12-21T01:22:08Z 2011-12-21T01:22:08Z <p>The <a href="http://www.math.duke.edu/~wka/math204/fixed.pdf" rel="nofollow">proof of Brower's fixed point theorem</a> using Stokes' theorem is a nice application I think.</p> http://mathoverflow.net/questions/79764/reference-request-deformations-of-a-map-bijective-to-global-sections-of-the-pull/79772#79772 Answer by J.C. Ottem for Reference Request: Deformations of a map bijective to global sections of the pullback of the tangent sheaf J.C. Ottem 2011-11-01T21:12:12Z 2011-11-01T21:12:12Z <p>Kollar's 'Rational Curves on Algebraic Varieties' is a good reference. </p> http://mathoverflow.net/questions/78132/is-there-an-analog-of-kodaira-vanishing-for-singular-varieties/78135#78135 Answer by J.C. Ottem for Is there an analog of Kodaira vanishing for singular varieties J.C. Ottem 2011-10-14T13:33:29Z 2011-10-14T13:38:40Z <p>No. The following counterexample is due to Sommese:</p> <p>Let $Y$ be the projective bundle $\pi:\mathbb{P}(O\oplus O(1)^{\oplus 3})\to \mathbb{P}^1$. Let $M$ be the tautological bundle on $Y$ and take a general member $X\in|M\otimes \pi^*O(-1)^{\oplus 4})|$. Then $X$ is a normal, projective, Gorenstein 3-fold. If $L$ is the line budle $M\otimes \pi^*O(1)$, one can also check that $H^1(X,O(K_X+L))=\mathbb{C}$.</p> <p>However, it is known that the Kodaira vanishing theorem holds if $X$ has log canonical singularities. There are also weaker versions in the theorem in the paper 'D. Arapura and D. B. Jaffe <a href="http://www.jstor.org/pss/2047052" rel="nofollow">On Kodaira Vanishing for Singular Varieties</a> Proc. A.M.S, 105, No. 4, pp. 911-916, 1989.'</p> http://mathoverflow.net/questions/78115/references-about-pseudoeffective-cone/78117#78117 Answer by J.C. Ottem for References about pseudoeffective cone J.C. Ottem 2011-10-14T10:35:52Z 2011-10-14T11:28:43Z <p>I'd say Lazarsfeld's book "<a href="http://books.google.com/books/about/Positivity_in_Algebraic_Geometry_I.html?id=T87ftUcU_hEC" rel="nofollow">Positivity in algebraic geometry I,II</a>" is the standard reference these days. In particular, Volume I has a lot of explicit examples. I also recommend Debarre's book <a href="http://books.google.com/books/about/Higher_dimensional_algebraic_geometry.html?id=Mtm7Zi-7U1gC" rel="nofollow">'Higher dimensional algebraic geometry"</a> which is similar in style with a lot of nice examples and explicit computations.</p> <p>If you are familiar with toric geometry, there is (not surprisingly) a simple description of the psedudoeffective cone in terms of the combinatorial data in the fan. See Cox-Little-Schenck's new book <a href="http://www.cs.amherst.edu/~dac/toric.html" rel="nofollow">'Toric varieties'</a> for details. This gives a hoard of interesting examples.</p> <p>If you are looking for examples with non-toric varieties, I'd recommend starting with the case where $X$ is a surface. In that case the effective cone coincides with the cone of curves and can be studied using the intersection form. Let me give an example:</p> <p><strong>Example</strong>. Let $X$ be the blow-up of $\mathbb{P}^2$ at two points and let $E_1,E_2$ be the exceptional divisors. A basis for $Pic(X)$ is given by $L,E_1,E_2$ where $L$ is the pull back of a general line in $\mathbb{P}^2$. We show that $\overline{Eff}(X)$ is spanned by $E_1,E_2$ and the strict transform of the line $L_0=L-E_1-E_2$. Let $\tau$ be the cone spanned by these three classes. Since they are all effective we have $\tau\subset \overline{Eff}(X)$. Coversely, let $D$ be any effective divisor with class $aL+bE_1+cE_2$. We will show that $D$ can be written as a sum of elements from $\tau$. We may assume $D$ to be irreducible. If $D$ is not one of the $E_1,E_2,L_0$, we then have $D.E_i\ge 0$ and $D.L_0\ge 0$. In particular, $D$ belongs to the dual cone of $\tau$, which is easily computed as $\tau^*=\langle L,L-E_1,L-E_2\rangle_{\ge 0}$. Now $L, L-E_1, L-E_2$ are all effective and can be written as positive linear combinations of $E_1,E_2,L_0$, and hence so can $D$. As a by-product, we have just computed the nef cone, which is $\tau^*$.</p> <p>Of course, this example is in fact toric, but the main point is that this type of argument works for more general surfaces, as long as you have a good description of the surface. For example, the argument above generalizes to show that a Del Pezzo surface, $\overline{Eff}(X)$ is spanned by the $(-1)$-curves on $X$ (this is shown in Debarre's book, I think). In general, the effective cone of rational surfaces have been studied a lot using their models as blow-ups. </p> <p>For material on the effective cones of surfaces, see for example </p> <p>B. Harbourne <a href="http://arxiv.org/abs/0907.4151" rel="nofollow">"Global aspects of the geometry of surfaces"</a> and </p> <p>Y. Tschikel "<a href="http://www.cims.nyu.edu/~tschinke/papers/yuri/08cmi/cmi4.pdf" rel="nofollow">Algebraic varieties with many rational points</a>. </p> <p>For K3 surfaces, S. Kovacs has a nice paper on the <a href="http://www.springerlink.com/content/r654368027w422r2/" rel="nofollow">'Cone of curves of a K3 surface'</a> (see also <a href="http://mathoverflow.net/questions/42699" rel="nofollow">this answer</a>). There are also many explicit examples in Artebani-Hausen-Laface's paper <a href="http://arxiv.org/abs/0901.0369" rel="nofollow">On Cox rings of K3-surfaces</a>. </p> <p>I can also recommend Artie Prendergast-Smith's papers at <a href="http://www.math.uic.edu/~artie/" rel="nofollow">his homepage</a>. In particular, his PhD thesis contains a very explicit example where he computes the of the effective cone of a rational threefold.</p> <p>In addition to pseduoeffective cones, you might also be interested in seeing explicit computations of Cox rings, which are graded by the monoid of effective divisors (in particular if you have a description of the Cox ring, you know all about the effective cone). Here I can recommend the following papers:</p> <p>A. Laface, M. Velasco, <a href="http://arxiv.org/abs/0810.3730" rel="nofollow">A survey on Cox rings</a></p> <p>I. Arzhantsev, U. Derenthal, J. Hausen, A. Laface, <a href="http://arxiv.org/abs/1003.4229" rel="nofollow">Cox rings</a></p> <p>J. Gonzalez, M. Hering, S. Payne, H. Süß <a href="http://arxiv.org/abs/1009.5238" rel="nofollow">Cox rings and pseudoeffective cones of projectivized toric vector bundles</a> and</p> <p>M. Artebani, A. Laface <a href="http://arxiv.org/abs/0909.1835" rel="nofollow">Cox rings of surfaces and the anticanonical Iitaka dimension</a></p> http://mathoverflow.net/questions/129539/global-sections-of-twisting-ideal-sheaf-of-a-smooth-closed-point-on-a-projective Comment by J.C. Ottem J.C. Ottem 2013-05-05T17:03:42Z 2013-05-05T17:03:42Z The cohomology group corresponds to the vector space of homogeneous polynomials in $x_0,\ldots,x_n$ vanishing at $p$ with multiplicitly $r$. That is, the Taylor expansion of $f$ near $p$ has the form $f=f_r+\ldots+f_m$ where $\deg f_j=j$. Using this description you can show that $\dim \Gamma(\mathbb P^n,I_x^r(m))=\max\left(0,{m+n \choose n}-{r+n-1 \choose n}\right)$ (I might have the binomial coefficients wrong..) http://mathoverflow.net/questions/129543/reference-for-the-classification-of-singular-degree-4-surfaces-in-mathbbp3 Comment by J.C. Ottem J.C. Ottem 2013-05-03T16:29:14Z 2013-05-03T16:29:14Z What kind of classification are you looking for? Just in terms of singularities? http://mathoverflow.net/questions/128807/noether-lefschetz-over-finite-fields Comment by J.C. Ottem J.C. Ottem 2013-04-26T14:50:20Z 2013-04-26T14:50:20Z Thanks for the comments! http://mathoverflow.net/questions/128044/equivalent-definitions-of-ample-bundles/128055#128055 Comment by J.C. Ottem J.C. Ottem 2013-04-19T10:00:53Z 2013-04-19T10:00:53Z Wouldn't $E=\mathcal O_X$ satisfy the first condition on $X=\mathbb P^1$, without being ample? http://mathoverflow.net/questions/127987/big-tangent-bundle/128021#128021 Comment by J.C. Ottem J.C. Ottem 2013-04-19T01:28:17Z 2013-04-19T01:28:17Z I've edited the post. Does the argument make sense now? http://mathoverflow.net/questions/127987/big-tangent-bundle Comment by J.C. Ottem J.C. Ottem 2013-04-18T20:30:10Z 2013-04-18T20:30:10Z Also, I feel that the downvote is a bit harsh. I think it makes sense to ask this question and it it interesting. http://mathoverflow.net/questions/127987/big-tangent-bundle Comment by J.C. Ottem J.C. Ottem 2013-04-18T19:23:37Z 2013-04-18T19:23:37Z Just to clarify: I think that bigness of $T_X$ refers to the line bundle $O(1)$ being big on $\mathbb{P}(T_X)$. As usual, a line bundle $L^m$ is big if $\dim H^0(mL)$ grows asymtotically as $m^{\dim X}$ for $m$ large. http://mathoverflow.net/questions/127788/examples-of-projective-but-not-fano Comment by J.C. Ottem J.C. Ottem 2013-04-17T05:16:50Z 2013-04-17T05:16:50Z Can you try to figure out when a curve is Fano? You will see that there are not so many.. http://mathoverflow.net/questions/124285/genus-of-an-embedded-curve-in-projective-smooth-manifold-deformed-in-its-homology Comment by J.C. Ottem J.C. Ottem 2013-03-12T14:45:11Z 2013-03-12T14:45:11Z If $\beta$ is the class of a line in $\mathbb P^2$, how would you cook up a cycle with large arithmetic genus? http://mathoverflow.net/questions/124254/nef-and-effective-classes-on-abelian-varieties Comment by J.C. Ottem J.C. Ottem 2013-03-11T19:23:17Z 2013-03-11T19:23:17Z I think your statement follows from Lemma 1.1 in Bauer's paper 'On the cone of curves of an abelian variety', since algebraic equivalence coincides with numerical equivalence on an abelian variety. http://mathoverflow.net/questions/123370/do-finitely-many-plurigenera-determine-the-kodaira-dimension/124078#124078 Comment by J.C. Ottem J.C. Ottem 2013-03-09T14:47:47Z 2013-03-09T14:47:47Z Nice construction. Any ideas for the $\kappa=−\infty$ case? http://mathoverflow.net/questions/123824/algebraic-independence-of-polynomials-in-n-variables-with-real-coefficients Comment by J.C. Ottem J.C. Ottem 2013-03-06T23:53:00Z 2013-03-06T23:53:00Z Perhaps this will help you: <a href="http://mathoverflow.net/questions/41535/how-to-show-a-set-of-polynomials-is-algebraically-independent" rel="nofollow" title="how to show a set of polynomials is algebraically independent">mathoverflow.net/questions/41535/&hellip;</a> http://mathoverflow.net/questions/123642/prove-if-3n2-is-odd-then-n-is-odd-with-contradiction Comment by J.C. Ottem J.C. Ottem 2013-03-05T17:57:59Z 2013-03-05T17:57:59Z This website is for research-level questions. To solve your question, just subtract 2. http://mathoverflow.net/questions/123051/no-fixed-components-in-the-linear-system-of-the-line-bundle-generating-picx Comment by J.C. Ottem J.C. Ottem 2013-02-27T13:27:01Z 2013-02-27T13:27:01Z Yes, let's see. If $k=1$, then $H^0(X,L)=k$, but for an ample $L$, one always has $H^0(X,L)=L^2/2+2\ge 3$ since the higher cohomology vanishes. http://mathoverflow.net/questions/123051/no-fixed-components-in-the-linear-system-of-the-line-bundle-generating-picx Comment by J.C. Ottem J.C. Ottem 2013-02-27T01:07:13Z 2013-02-27T01:07:13Z If $|L|$ had a fixed component $E$, then in $Pic(X)$ we must have $E\sim kL$ for some $k&gt;0$, which is clearly impossible..