User s&#225;ndor kov&#225;cs - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:36:31Z http://mathoverflow.net/feeds/user/10076 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131705/when-is-an-ample-line-bundle-on-an-abelian-variety-base-point-free/131743#131743 Answer by Sándor Kovács for When is an ample line bundle on an abelian variety base point free? Sándor Kovács 2013-05-24T16:12:21Z 2013-05-24T16:12:21Z <p>According to Fujita's conjecture, if $L$ is an ample divisor on a smooth complex variety $X$ of dimension $n$, then </p> <ol> <li><p>$K_X+mL$ is basepoint-free for $m\geq n+1$, and </p></li> <li><p>$K_X+mL$ is very ample for $m\geq n+2$.</p></li> </ol> <p>Of course, if $X$ is an abelian variety, then $K_X=0$.</p> <p>This conjecture is known for surfaces by the work of Reider and for threefolds by the work of Ein and Lazarsfeld. </p> <p>I don't know if there are any more results of this kind specific to abelian varieties, but it is presumably an easier case and one may use more tools.</p> http://mathoverflow.net/questions/131703/blow-ups-and-singularities/131740#131740 Answer by Sándor Kovács for blow-ups and singularities Sándor Kovács 2013-05-24T15:57:06Z 2013-05-24T15:57:06Z <p>I don't see how $D$ makes any difference in the question you describe. Perhaps you forgot a condition about $D$ such as $\pi$ is ramified along $D$?</p> <p>In any case, here is an example that shows that you need more conditions to have a chance here.</p> <p>Let $Y=X=\mathbb A^2$ and $\pi: \mathbb A^2\to \mathbb A^2$ be defined by $(x,y)\mapsto (x^2,y^2)$. Let $Z\subset X$ be the origin. Then $\widetilde X$ is just $\mathbb A^2$ blown-up at a closed point (in particular smooth, normal, etc.), but $\widetilde Y$ is the blow-up of $\mathbb A^2$ along the fat point defined by the ideal $(x^2,y^2)$ so it is a pinch point and hence not normal and not even normal crossings.</p> <p>I guess now you could say that you want $\pi$ ramified along a divisor, but it will be the same thing, if $Z$ is a point on $D$, it's pre-image will be a fat point and blowing that up will not be normal. It seems that the only chance is to require that $Z$ (or at least its generic point) is disjoint from the ramification locus of $\pi$, in other words if $\pi$ is étale in a neighbourhood of (the generic point of) $Z$. </p> http://mathoverflow.net/questions/131223/surfaces-ruled-over-elliptic-curves/131252#131252 Answer by Sándor Kovács for Surfaces ruled over elliptic curves Sándor Kovács 2013-05-20T18:00:25Z 2013-05-21T00:47:24Z <p>Here is an alternative approach.</p> <p>Since $S$ is an elliptic fibration, the rational curves of the ruling must cover $C$ and hence $C$ is also rational, i.e., $C\simeq \mathbb P^1$. </p> <p>Next do a base change by the induced map from a (general) fiber of $p$, say $F$, to $E$. We get a new ellipticly ruled surface: $\rho: T=S\times_EF\to F$. </p> <p>Now consider the Stein factorization of the composition $T\to S\to C\simeq \mathbb P^1$. Since $T$ is still a ruled surface, there are $\mathbb P^1$'s in $T$ that cover the Stein cover of $\mathbb P^1$, so that has to be a $\mathbb P^1$ again, so we get another elliptic fibration $q:T\to\mathbb P^1$ and by construction the fibers of $\rho$ and $q$ are transversal and meeting in a single point. </p> <p>In other words the fibers of one morphism are sections of the other. It is easy to see that in this case this means that $T\simeq F\times\mathbb P^1$. The map $F\to E$ is an isogeny, so we may pick identities that it is a group homomorphism. Denoting the kernel by $G$ we get that $E\simeq F/G$ and $S\simeq (F\times \mathbb P^1)/G$. </p> http://mathoverflow.net/questions/131038/exceptional-divisor-on-a-smooth-surface/131160#131160 Answer by Sándor Kovács for exceptional divisor on a smooth surface Sándor Kovács 2013-05-19T18:39:24Z 2013-05-19T18:39:24Z <p>As Rita points out, 1) is trivial, because the matrix $(D_i\cdot D_j)$ is invertible. Also, as she points out, the second statement is not quite true. Well, actually, it is completely false. Again by the fact that the matrix $(D_i\cdot D_j)$ is invertible, if $(\sum r_iD_i)\cdot D_j=0$ for all $j$, then necessarily $r_i=0$ for all $i$. On the other hand, this is easily fixable. </p> <p>In fact, one can make a somewhat stronger statement:</p> <blockquote> <p><strong>Claim</strong> If the $r_i$ are arbitrary and $(\sum r_iD_i)\cdot D_j\leq 0$ for all $j$, then $r_i\geq 0$, that is, $\sum r_iD_i$ is effective and if furthermore $\cup D_i$ is connected and there exists a $j$ such that $(\sum r_iD_i)\cdot D_j\neq 0$, then $r_i>0$ for all $i$.</p> </blockquote> <p><strong>Proof</strong> Let $\sum r_iD_i=A-B$ with $A,B$ effective and suppose $B\neq 0$. Then $B^2&lt;0 \leq A\cdot B$, so $$0 &lt; A\cdot B - B^2= \left(\sum r_iD_i\right)\cdot B = \sum_{r_j&lt;0} (-r_j)\left(\sum r_iD_i\right)\cdot D_j \leq 0,$$ which is a contradiction and hence $B=0$, that is, $\sum r_iD_i$ is effective. </p> <p>Next we want to prove that if there exists a $k$ such that $r_k=0$, then $(\sum r_iD_i)\cdot D_j=0$ for all $j$. Let $k$ be such that $r_k=0$. Then $$\left(\sum r_iD_i\right)\cdot D_k = \sum_{i: D_i\cdot D_k>0} r_i (D_i\cdot D_k) \geq 0.$$</p> <p>Now if there exists a $j$ among these such that $r_j\neq 0$, then $\left(\sum r_iD_i\right)\cdot D_k>0$, otherwise $r_j=0$ for all $j$ such that $D_j\cap D_k\neq \emptyset$. Now repeat this step with one of these $j$'s. Since $\cup D_i$ is connected, this proves the claim. $\square$</p> http://mathoverflow.net/questions/18496/example-of-the-completion-of-a-noetherian-domain-at-a-prime-that-is-not-a-domain/40978#40978 Answer by Sándor Kovács for Example of the completion of a noetherian domain at a prime that is not a domain Sándor Kovács 2010-10-04T03:19:27Z 2013-05-13T07:30:26Z <p>It might be worth pointing out that you get an example of this by localizing at any point of a variety (scheme) which is irreducible but not (analytically/formally/étale) locally irreducible. In particular, any self-intersecting curve will give an example, just like the one above by Charles.</p> http://mathoverflow.net/questions/128853/algebraic-de-rham-cohomology-of-singular-varieties/128855#128855 Answer by Sándor Kovács for algebraic de Rham cohomology of singular varieties Sándor Kovács 2013-04-26T20:22:26Z 2013-04-26T20:22:26Z <p>A likely candidate for this would be a non-Du Bois singularity. </p> <p>Du Bois, following Deligne's ideas, constructed a filtered complex of sheaves with coherent cohomology sheaves that gives a resolution of the constant sheaf $\mathbb C$ for any reduced finite type scheme over $\mathbb C$. This complex agrees with the de Rham complex for smooth varieties and in general its hypercohomology agrees with the singular cohomology of the (underlying topological space of the) scheme. </p> <p>So, in some sense your question is to see an example when the Du Bois complex is different from the de Rham complex (I know that's not exactly what you are asking, but I think that's the important fact behind this issue). </p> <p>There is a class of singularities, not surprisingly called Du Bois singularities, with the property that the $0^{\rm th}$ associated graded quotient of the Du Bois complex is quasi-isomorphic to the structure sheaf (this holds for smooth varieties).</p> <p>So, a good start for finding an example like this is to look at non-Du Bois singularities. For curves being Du Bois is equivalent to semi-normal, so any curve with a cusp is non-Du Bois. I would start there.</p> <p>In general, Du Bois singularities can be quite varied, but if your $X$ is normal and Gorenstein, then being Du Bois is equivalent to being log canonical. So, you have plenty of examples (For instance, take a normal hypersurface with worse than log canonical singularities). I am sure you will find an example easily.</p> http://mathoverflow.net/questions/128278/morphism-with-non-reduced-special-fibre/128403#128403 Answer by Sándor Kovács for Morphism with non-reduced special fibre Sándor Kovács 2013-04-22T21:39:02Z 2013-04-22T21:39:02Z <p>I think there is some confusion here. Either on your part or on mine. I don't think being non-reduced is equivalent to having a non-reduced component. A scheme may have a fat point, but be irreducible and generically reduced. Similarly, I don't quite understand what you want. Do you want the generic fiber be non-reduced or do you want $X$? Those are different questions. Or I am missing something. I would say that the generic fiber being non-reduced implies that $X$ is, but not the other way around.</p> <p>My first reaction was that there is no such criterion, because it is quite common to have a multiple fiber, even for smooth surface flat over a smooth curve. </p> <p>Then I realized that there might be a condition that implies this: If $f$ admits a section that intersects the multiple component of your special fiber, then you have a fighting chance. </p> <p>Here is the idea: Let's say for safety that we are in characteristic zero. If $X$ were smooth, and say $D$ is a section and $F$ is a fiber, then $f$ is smooth at a general point of $D$ and hence $D\cdot F=1$, so $D$ cannot intersect a multiple component in any fiber. </p> <p>Of course, you don't want to assume that $X$ is smooth, because you want to conclude that it isn't. However, there might be some ways to weaken this. So here is <strong>a</strong> criterion that does what you ask for, but I am not claiming that this is easy to achieve or that there isn't anything simpler and for sure I do not expect this to work for you. (But if it does, that would be great). Also, it has some limitations.</p> <blockquote> <p><strong>Claim</strong> Assume that everything is defined over an algebraically closed field of characteristic zero and $\dim Y=1$. If $f$ admits a section that intersects the multiple component of the special fiber and contains a general point of $X$, then the general fiber of $f$ is everywhere non-reduced and hence so is $X$. </p> </blockquote> <p><strong>Proof</strong> Suppose $X$ (and hence the general fiber) is not everywhere non-reduced, that is, it is generically reduced. Then $X$ is generically smooth and hence $f$ is generically smooth. By the assumption the section, say $Y'$, intersects this locus and within that locus it intersects each fiber transversally in a single smooth point. By the assumptions we may consider a resolution of singularities of $X$, say $\pi:X'\to X$ and the proper transform of $Y'$ will be a section of the composition $f\circ\pi$. The preimage of the multiple component of the special fiber will still be (a union of) multiple component(s) of a fiber of the composition and the proper transform of $Y'$ will still intersect this. (I am saying preimage, not proper transform!!). But then we get a contradiction because the intersection number of the section and the fiber should be $1$. Therefore the original assumption was false and hence the general fiber of $f$ and $X$ are both everywhere non-reduced. </p> <hr> <p>I have the feeling that this could be improved to include the case $\dim Y>1$ and not having to make the assumption on characteristic (that is, using resolutions), but I don't see an easy way to do that. The point is to ensure that if a section intersects the general fiber at a single smooth point transversally, then it cannot intersect a multiple component of another fiber. If $X$ has a reasonable intersection theory, then this should be OK, but what if it is just horribly singular but still reduced.....?</p> <p>Uh, and just to note that the condition can be indeed satisfied, observe that if $X=Y\times Z$ where $Z$ is everywhere non-reduced, then all the assumptions hold.</p> http://mathoverflow.net/questions/127643/algebraic-machinery-for-algebraic-geometry/127662#127662 Answer by Sándor Kovács for Algebraic machinery for algebraic geometry Sándor Kovács 2013-04-15T23:02:12Z 2013-04-15T23:02:12Z <p>I think this is a very good question, because studying commutative algebra on its own is hard, it is much better to do it with some idea of what all that means geometrically.</p> <p>In my opinion the best entry to commutative algebra is provided by Miles Reid's <a href="http://www.amazon.com/Undergraduate-Commutative-Algebra-Mathematical-Society/dp/0521458897" rel="nofollow">Undergraduate Commutative Algebra</a>. Miles Reid is an algebraic geometer so when he writes about commutative algebra, it is with geometry in mind. I would say that this book has everything that you need to be able to start in algebraic geometry except dimension theory which is done excellently in Atiyah-MacDonald.</p> <p>I would suggest that you read this book, which is brief so you don't lose sight of your ultimate goal and you can already start feeling that you're actually reading about geometry. When you're done start reading algebraic geometry. For instance Hartshorne. In that book as you discovered there are a lot of algebra results quoted and even more is needed for the exercises which you <em>absolutely</em> have to do. More than half of the important material is in the exercises! </p> <p>When you get stuck in a problem, ask yourself if you can translate the problem or part of it to an algebra problem and then see if you can find anything related to that in one of the standard commutative algebra books such as Eisenbud or Matsumura or for that matter the stacks project. </p> <p>As you discovered you will also need homological algebra, but not just any general homological algebra, but the kind that is used in commutative algebra. There is a great book for that: Bruns-Herzog: <a href="http://books.google.com/books/about/Cohen_Macaulay_Rings.html?id=LF6CbQk9uScC" rel="nofollow">Cohen-Macaulay Rings</a>. This is also a big undertaking, but you don't need to read the whole book to get going. Say read the first two chapters, but not even necessarily in one go. Take you time while you're doing some other things. And most importantly, whatever you read in that book (or for that matter in any algebra book) try to see if you can give statements and notions geometric meaning or at least come up with examples that come from geometry. For instance, find your favorite example of a non-Cohen-Macaulay variety. Then find another one. </p> <p>Of course, as you advance you will need more and more algebra, but after awhile you actually get into the habit of acquiring that knowledge as you go on. It makes more sense to learn these more advanced notions when you get there. </p> <p>Without trying to be comprehensive, I assume sooner or later you will need to learn about associated primes (this already happens to some extent in Reid's book), integral extensions, going-up, going down theorems, dimension theory, regular sequences, depth, and the big whale: flatness. Flatness is extremely important, but somewhat hard to grasp full depth at first (or even later). Don't despair, you'll start having a feel for it if you keep at it. Anyway, there are many more things to learn, but you didn't ask that. </p> <p>So for now, I'd say read Reid's book, then read Hartshorne (or something similar) and then try to get the algebra knowledge that you're missing as you go along. </p> http://mathoverflow.net/questions/127491/locally-free-extension-of-locally-free-sheaf/127497#127497 Answer by Sándor Kovács for Locally free extension of locally free sheaf Sándor Kovács 2013-04-13T21:37:22Z 2013-04-13T21:37:22Z <p>Since you assumed that $X$ is smooth (less would be enough, but you need at least $S_2$), $G=j_*F_U$ is <em>reflexive</em>, that is $G^{\ast\ast}=G$ where $G^{*}=\mathscr Hom_X(G,\mathscr O_X)$ is the dual.</p> <p>If two reflexive sheaves agree on an open set with a codimension $2$ complement, then they agree, so your question is equivalent to asking that $G$ be locally free.</p> <p>If the rank of $F$ is $1$, then $G$ corresponds to a divisor and again since $X$ is smooth it is Cartier and hence $G$ is locally free.</p> <p>If the rank of $F$ is at least $2$, then the locus where a reflexive sheaf is not locally free is at least $3$-codimensional (see Lemma 1.1.10 in <a href="http://books.google.com/books?id=OG348syTqC8C&amp;lpg=PA75&amp;vq=1.1.10&amp;pg=PA75#v=onepage&amp;q&amp;f=false" rel="nofollow">Vector Bundles on Complex Projective Spaces, by Okonek, Schneider, Spindler</a>). </p> <p>So what you want can be done on curves and surfaces, but not necessarily on varieties of dimension $\geq 3$. An alternative way to get $G$ is to take the double dual of $F$, that is, $G=F^{\ast\ast}$. In other words, your question is whether that is locally free.</p> <p>To complete the picture look at Example 1.1.13 in <a href="http://books.google.com/books?id=OG348syTqC8C&amp;lpg=PA75&amp;vq=1.1.10&amp;pg=PA75#v=onepage&amp;q&amp;f=false" rel="nofollow">ibid</a>. That shows that reflexive sheaves of rank $\geq 2$ are not always locally free.</p> <p>To summarize:</p> <ul> <li>If $k=1$ or $\dim X\leq 2$ then the sheaf you are looking for is $j_*F_U$</li> <li>Otherwise, if $j_*F_U$ is locally free you've got it, if it is not, then there is no such sheaf. </li> </ul> http://mathoverflow.net/questions/127185/strong-notions-of-general-position/127188#127188 Answer by Sándor Kovács for Strong notions of general position Sándor Kovács 2013-04-11T06:36:45Z 2013-04-11T06:36:45Z <p>The defining equation of a degree $d$ hypersurface in $\mathbb P^n$ has $n+d\choose d$ coefficients and hence these hypersurfaces may be parametrized by a projective space of dimension ${n+d\choose d} -1$. Picking a point to be contained by the hypersurface is a linear equation on the coefficients of these defining equations so any set of ${n+d\choose d} -1$ points is contained in at least one such hypersurface, but a set of $n+d\choose d$ points in general position is not. </p> <p>If you take $n=2$ and $d=1$ you get that $3$ points in general position do not lie on a line, if you take $d=2$, then you get that $6$ points in general position do not lie on a quadric, etc.</p> <p>As far as the definition of "in general position" goes, in this situation, one could say that a set of points is in general position if they impose independent conditions on the coefficients of the defining equations.</p> <p>More generally, you can consider any parameter space of any type of subobjects of any fixed variety/scheme/space. Containing a point of this fixed space will impose a condition on the parameter space. You can define points being in general position with respect to the type of subobjects you're considering by requiring that the conditions they impose on the parameter space are independent.</p> http://mathoverflow.net/questions/126928/relative-resolution-of-singularity/127009#127009 Answer by Sándor Kovács for relative resolution of singularity Sándor Kovács 2013-04-09T18:41:50Z 2013-04-09T20:06:45Z <p>There are a bunch of very basic fundamental problems with doing this. </p> <p>Here is one example and you are invited to generalize this to get others: Let $X$ be a <em>smooth</em> surface and $f$ a surjective flat morphism to a smooth curve $S$. If $f$ is not already smooth, then it is not possible to make it smooth the way you suggest (or any reasonable way). </p> <p>This is one reason why moduli spaces of smooth objects are in general not compact or put it in another way why compactifying moduli spaces is a non-trivial question. (See $\overline {\mathfrak M}_g$ for instance).</p> <p>To see that the above claim is true consider the following:</p> <ol> <li>If $f$ has a singular fiber, then for any $i:S'\to S$ the base change morphism: $f_{S'}:S'\times_S X\to S'$ will also have singular fibers. </li> <li>For any $f':X'\to X$ as in the question, the $fj=if'$ assumption means that $f'$ factors through $f_{S'}$, so in other words you want a $g:X'\to S'\times_S X$ modification such that $f'=f_{S'}g$ plus the rest of the requirements. </li> <li>$S'\times_S X$ is normal. This is because the fibers of $f$ are CM and hence so are the fibers of $f_{S'}$. Hence $S'\times_S X$ is $S_2$ and by a similar argument it is also $R_1$, so by Serre's criterion it is normal. </li> <li>There is no such $g$. If indeed $f$ had a singular fiber, then the same singular fiber appears as a fiber of $f_{S'}$. Being a curve its unique resolution is its normalization, so a $g$ like that would have to be a finite morphism of degree $1$. However, since $S'\times_S X$ is normal, Zariski's Main Theorem implies that $g$ would have to be the identity. (For simplicity assume that there is at least one non-unibranched singularity on one of the fibers).</li> </ol> http://mathoverflow.net/questions/126656/linearly-generated-embedding Linearly generated embedding? Sándor Kovács 2013-04-05T19:16:39Z 2013-04-06T18:49:56Z <p>Let $X$ be a projective variety over an algebraically closed field $k$ and $\mathscr L$ a line bundle on $X$. Its section ring, $$R(X,\mathscr L) = \bigoplus _{n=0}^\infty H^0(X,\mathscr L^{\otimes n}),$$ is a finitely generated graded $H^0(X,\mathscr O_X)\simeq k$-algebra.</p> <p>I wonder if the following condition has an established name:</p> <p>Assume that $\mathscr L$ is very ample and let $\phi:X\hookrightarrow \mathbb P^N$ denote the embedding induced by the global sections of $\mathscr L$. So, in particular the map $$H^0(X,\mathscr O_{\mathbb P^N}(1))\rightarrow H^0(X,\mathscr L) \tag{\star}$$ is surjective. Further assume that </p> <p>$$\text{R(X,\mathscr L) is generated in degree 1,} \tag{\star\star}$$</p> <p>that is, by the elements of $H^0(X,\mathscr L)$. In particular, then the embedding $\phi$ is projectively normal, but this is a stronger condition.</p> <p>I would like to say something like "$\phi$ is <em>blah</em>, when this holds", so the question is:<br> <strong>Q:</strong> Does this property/condition have an established name in the literature?</p> <p>If not, I would probably say that "$\phi$ is a <em>linearly generated embedding</em> if this condition holds". My rationale for that name is that by $(\star)$ and $(\star\star)$ it follows that $R(X,\mathscr L)$ is the homogenous coordinate ring of $X$ corresponding to the embedding $\phi$ and that $R(X,\mathscr L)$ is generated by the images of linear functions on $\mathbb P^N$.</p> <p>An ideal answer would give a reference (or more) where this is defined/used, or in absence of a reference would either support the name I am suggesting or argue against it and in that case would suggest an alternative.</p> http://mathoverflow.net/questions/126397/what-is-the-definition-of-a-sufficiently-ample-line-bundle/126405#126405 Answer by Sándor Kovács for What is the definition of a sufficiently ample line bundle? Sándor Kovács 2013-04-03T15:41:22Z 2013-04-03T15:41:22Z <p>There are many theorems of the form</p> <blockquote> <p><strong>Theorem Frame</strong> If something holds and $\mathscr L$ is an ample line bundle, then there exists an $n_0\in \mathbb N$ such that for all $n\geq n_0$, something else holds with $\mathscr L^{\otimes n}$ in it.</p> </blockquote> <p>You should think of things like </p> <ul> <li>a sheaf being generated by global sections</li> <li>Serre vanishing</li> <li>transversality statements</li> <li>avoiding certain points or properties</li> <li>etc.</li> </ul> <p><em>Sufficiently ample</em> means that whatever they are claiming holds for $\mathscr L^{\otimes n}$ for all $n\geq n_0$ with some $n_0\in \mathbb N$ for any ample line bundle $\mathscr L$.</p> http://mathoverflow.net/questions/11289/geometry-meaning-of-higher-cohomology-of-sheaves/126119#126119 Answer by Sándor Kovács for Geometry meaning of higher cohomology of sheaves? Sándor Kovács 2013-03-31T18:07:43Z 2013-04-01T17:13:22Z <p>Let's start way back. The invention of schemes moved algebraic geometry away from thinking about varieties as embedded objects. However, embedding an abstract scheme into projective space has a lot of advantages, so if we can do that, it's useful. And even if we cannot embed our scheme into projective space, but we can find a non-trivial map, that gives us some way to understand our abstract scheme. In order to find a non-trivial map we need a line bundle with sections.</p> <p>So we are interested in finding sections of various sheaves, but primarily line bundles (and of course for this sometimes we need to deal with other kind of sheaves). Thus we are interested in $H^0$. </p> <p>On the other hand, computing $H^0$ is non-trivial. There are no good general methods. One reason for this is that, for instance, $H^0$ is not constant in families, or put it another way it is not deformation invariant. On the other hand, $\chi(X,\mathscr F)$ behaves much better. It is constant in flat families and if $\mathscr F$ is a line bundle, then it is computable using Riemann-Roch. </p> <p>Then, if we know that $H^i=0$ for $i>0$, then $H^0=\chi$ and we're good.</p> <hr> <p>Here is an explicit example for a typical use of Serre vanishing:</p> <p><strong>Example 1</strong> Suppose $X$ is a smooth projective variety and $\mathscr L$ is an ample line bundle on $X$. Then we know that $\mathscr L^{\otimes n}$ is very ample and $H^i(X, \mathscr L^{\otimes n})=0$ for $i>0$ and $n\gg 0$. Then $\mathscr L^{\otimes n}$ induces an embedding $X\hookrightarrow \mathbb P^N$ where $N=\dim H^0(X,\mathscr L^{\otimes n})-1=\chi(X,\mathscr L^{\otimes n})-1$ by Serre's vanishing and hence $N$ is now computable by Riemann-Roch. </p> <p>The only shortcoming of the above is that in general there is no way to tell what $n\gg0$ really means and so it is hard to get any explicit numerical estimates out of this. This is where Kodaira vanishing can help.</p> <p><strong>Example 2</strong> In addition to the above assume that $\mathscr L=\omega_X$, or in other words assume that $X$ is a smooth canonically polarized projective variety. There are many of these, for instance all smooth projective curves of genus at least $2$ or all hypersurfaces satisfying $\deg > \dim +2$. In particular, these are those of which we like to have a moduli space. Anyway, the way Kodaira vanishing changes the above computation is that now we know that already $H^i(X,\omega_X^{\otimes n})=0$ for $i>0$ and $n>1$! In other words, as soon as we know that <code>$\omega_X^{\otimes n}$</code> is very ample and $n>1$, then we can compute the dimension of the projective space into which we can embed our canonically polarized varieties. In fact, perhaps more importantly than that we can compute it, we know (from the above) without computation that this value is constant in families. So, once we have a boundedness result that says that this happens for any $n\geq n_0$ for a given $n_0$, and Matsusaka's Big Theorem says exactly that, then we know that all such canonically polarized smooth projective varieties (with a fixed Hilbert polynomial) can be embedded into $\mathbb P^N$, that is, into the <strong>same</strong> space. </p> <p>This implies that then all of these varieties show up in the appropriate Hilbert scheme of $\mathbb P^N$ and we're on our way to construct our moduli space. </p> <p>Of course, there is a lot more to do to finish the whole construction and also this method works in other situations, so this is just an example.</p> <hr> <p>There is one more thing one might think regarding your question, that is, ask the more abstract question:</p> <p>"What does higher cohomology of sheaves mean (e.g., geometrically)?"</p> <p>This is arguable, but I think that the essence of higher cohomology is that it measures the failure of something we wish were true all the time, but isn't. More specifically, if you're given a short exact sequence of sheaves on $X$ $$0\to \mathscr F' \to \mathscr F \to \mathscr F'' \to 0$$ then we know that even though $\mathscr F \to \mathscr F''$ is surjective, the induced map on the global sections $H^0(X,\mathscr F) \to H^0(X,\mathscr F'')$ is not. However, the vanishing of $H^1(X,\mathscr F')$ implies that for any surjective map of sheaves with kernel $\mathscr F'$ as above the induced map on global sections is also surjective. Since you already have a geometric interpretation of $H^0$, this gives one for $H^1$: it measures (or more precisel</p> <p>For a more detailed explanation of the same idea see <a href="http://mathoverflow.net/questions/125200/coboundaries-and-gluing-in-cech-cohomology-intuition/125236#125236" rel="nofollow">this MO answer</a>. </p> <p>In my opinion the best way to understand higher ($>1$) cohomology is that it is the lower cohomology of syzygies. In other words, consider a sheaf $\mathscr F$ and embed it into an acyclic (e.g., flasque or injective or flabby or soft) sheaf. So you get a short exact sequence: $$0\to \mathscr F\to \mathscr A\to \mathscr G \to 0$$ Since $\mathscr A$ is acyclic, we have that for $i>0$ $$H^{i+1}(X,\mathscr F)\simeq H^i(X,\mathscr G),$$ so if you understand what $H^1$ means, then $H^2$ of $\mathscr F$ is just $H^1$ of $\mathscr G$, $H^3$ of $\mathscr F$ is just $H^2$ of $\mathscr G$ and so on. </p> http://mathoverflow.net/questions/122227/divisor-class-group-on-blowup-of-nodal-surface/122242#122242 Answer by Sándor Kovács for Divisor class group on blowup of nodal surface Sándor Kovács 2013-02-19T00:22:20Z 2013-03-27T15:51:24Z <p>Joachim, your intuition is right, you will not have that same formula in general. Let's stick to the case you describe in the end.</p> <p>Indeed $R$ will be a Cartier divisor on $\bar S$ and hence gives an element of $\text{Pic}\,\bar S$, but it is not in <code>$\epsilon^*\text{Pic}\,S$</code>. However, it is also not a direct summand, because $2R$ is a Cartier divisor on $S$ so it is in $\epsilon^*\text{Pic}\,S$. </p> <p>So a more reasonable conjecture is that $$\text{Pic}\,\tilde S \simeq G \oplus \mathbb{Z} E,$$ where $G$ is a $\mathbb Z/2\mathbb Z$ extension of $\epsilon^* \text{Pic}(S)$, that is, there exists a short exact sequence: $$0\to \epsilon^* \text{Pic}(S) \to G \to \mathbb Z/2\mathbb Z \to 0,$$ where $\mathbb Z/2\mathbb Z$ is generated by $R+\epsilon^* \text{Pic}(S)$.</p> <p>I believe that essentially the same proof works.</p> <p>As for your conjecture for the $\text{WCl}$ (which is usually denoted by $\text{Cl}$, see [Hartshorne]), the initial problem is that you can't actually define $\epsilon^*$ for Weil divisors. That is, you can come up with various definitions that will have various properties, but in general you run into trouble either with respecting linear equivalence or the group structure. </p> <p>There is a way to pull back $\mathbb Q$-Cartier divisors at least numerically and in this case that would indeed give what you conjecture via the above. </p> <p>As for references, you could check out the papers of de Fernex-Hacon, Urbinati, and Chiecchio. (Start with the last two on arXiv (or <a href="http://www.marxiv.org/?query=&amp;page=0&amp;type=home" rel="nofollow">MArXiv.org</a>), they have fewer papers and they reference the one by de Fernex-Hacon that's relevant).</p> http://mathoverflow.net/questions/125630/line-bundles-on-k3-surfaces/125650#125650 Answer by Sándor Kovács for Line bundles on K3 surfaces Sándor Kovács 2013-03-26T18:55:47Z 2013-03-26T20:13:05Z <p>Noam and Francesco have already pointed out that in order to get such a line bundle you can always take either a multiple of a $(-2)$-curve or the union of disjoint $(-2)$-curves, both of which is possible on many $K3$'s. </p> <p>On the other hand, if you were looking for an $L$ that has a non-zero section whose zero locus is irreducible, then the self-intersection has to be at least $-2$. This follows easily from the adjunction formula and Riemann-Roch:</p> <p>If $D\subset X$ is an irreducible curve on a smooth surface $X$, then $D$ is a Cartier divisor and hence Gorenstein, so it has a dualizing sheaf which is a line bundle and this dualizing sheaf satisfies the formula $$\omega_D\simeq \omega_X(D)\otimes \mathscr O_D.$$ If in addition $\omega_X\simeq \mathscr O_X$, then this implies that $$\deg \omega_D = D^2.$$ If $D^2&lt;0$, then this means that $\chi(\omega_D)=-1$ and by Riemann-Roch $$0\leq p_a(D)= \deg\omega_D +1 - \chi(\omega_D)= D^2 +2.$$ In particular, $$D^2\geq -2,$$ and equality holds if and only if $p_A(D)=0$ which holds if and only if $$D\simeq \mathbb P^1.$$</p> http://mathoverflow.net/questions/125200/coboundaries-and-gluing-in-cech-cohomology-intuition/125236#125236 Answer by Sándor Kovács for Coboundaries and Gluing in Cech Cohomology - Intuition? Sándor Kovács 2013-03-22T04:58:09Z 2013-03-22T23:31:14Z <p>Let $U$ denote the space we're working on and $\{U_i\}$ an open cover of $U$ (obviously $U$ may be an open set of an ambient space, but that plays no importance here). Let's assume that there is a sheaf $\mathscr F$ on $U$ and for each $i$ a section $s_i\in\mathscr F(U_i)$.</p> <p>Gluing the $s_i$ means to find a section $s\in \mathscr F(U)$ such that for each $i$ $$s|_{U_i}=s_i.$$</p> <p>The obvious obstruction to this is that the $s_i$ and $s_j$ has to agree on the overlap. So, in order for such an $s$ to exist, we must have that <code>$$(s_i)|_{U_{ij}}=(s_j)|_{U_{ij}}. \tag{\star}$$</code> In other words such that the $0$-cochain $\{(U_i,s_i)\}$ is in $Z^0=H^0$ and this is clearly enough. One could argue, which might be your motivation, that this doesn't measure failure, but indeed it measures the success of gluing. I agree.</p> <hr> <p>For $H^1$ I think it is still better to think of it as the measure of failure of <em>lifting</em>, but because we're talking about sheaves this means gluing in practice. In other words, consider a surjective morphism of sheaves, <code>$$\mathscr F \twoheadrightarrow \mathscr F'',$$</code> and imagine wanting to lift sections of $\mathscr F''$ to $\mathscr F$. So, you start with a $t\in \mathscr F''(U)$ and from the sheaf properties you know that there exists $\mathfrak U=\{U_i\}$, an open cover of $U$, and for each $i$ a section $t_i\in\mathscr F(U_i)$ such that $t|_{U_i}$ is the image of $t_i$ via the map $\mathscr F\to\mathscr F''$. </p> <p>In order to lift $t$ to $\mathscr F(U)$ you need to glue the $t_i$. For that you consider the the $1$-cochain <code>$\sigma= \{(U_{ij},(t_i)|_{U_{ij}}-(t_j)|_{U_{ij}})\}$</code>. Clearly <code>$(t_i)|_{U_{ij}}-(t_j)|_{U_{ij}}$</code> maps to $0$ in $\mathscr F''$, so it is naturally a $1$-cochain of sections of $\mathscr F'=\ker \big[\mathscr F\to\mathscr F''\big]$. An obvious computation shows that it is in $Z^1(\mathfrak U,\mathscr F')$. </p> <p>Now observe that $\sigma\in B^1(\mathfrak U,\mathscr F')$ if and only if there exists a $0$-cochain $\{(U_i,t_i')\}$ of $\mathscr F'$ such that <code>$$(t_i)|_{U_{ij}}-(t_j)|_{U_{ij}} = (t_i')|_{U_{ij}}-(t_j')|_{U_{ij}}$$</code> which is the same as to say that the sections <code>$$t_i-t_i'\in \mathscr F(U_i)$$</code> satisfy $(\star)$ and hence they glue together to a section $t'\in\mathscr F(U)$. Notice, that since $t_i'\in\mathscr F'(U_i)$, the image of $t_i-t_i'$ in $\mathscr F''(U_i)$ is the same as the image of $t_i$, that is, $t|_{U_i}$.</p> <p>In other words, the original $t\in \mathscr F''(U)$ can be lifted, or equivalently, the sections $t_i-t_i'\in\mathscr F(U_i)$ can be glued if and only if the above associated $1$-cocycle in $$H^1(\mathfrak U, \mathscr F')=Z^1(\mathfrak U, \mathscr F')/B^1(\mathfrak U, \mathscr F')$$ is $0$.</p> <hr> <p>OK, so let's see about $H^2$. If you accept that $H^1$ is the obstruction to lifting sections, that is, lifting elements of $H^0$, then the same argument shows that $H^2$ measures the failure of lifting these obstructions. </p> <p>Suppose you have a surjective map between two short exact sequences: <code>$$\begin{matrix} 0 &amp; &amp; 0 \\ \downarrow &amp; &amp; \downarrow \\ \mathscr F' &amp; \twoheadrightarrow &amp; \mathscr G' \\ \downarrow &amp; &amp; \downarrow \\ \mathscr F &amp; \twoheadrightarrow &amp; \mathscr G \\ \downarrow &amp; &amp; \downarrow \\ \mathscr F'' &amp; \twoheadrightarrow &amp; \mathscr G'' \\ \downarrow &amp; &amp; \downarrow \\ 0 &amp; &amp; 0 \\ \end{matrix}$$</code> and let $\mathscr K'=\ker\big[ \mathscr F'\to \mathscr G'\big]$, $\mathscr K=\ker\big[ \mathscr F\to \mathscr G\big]$, and $\mathscr K''=\ker\big[ \mathscr F''\to \mathscr G''\big]$, so by the $9$ lemma there is a short exact sequence $$0 \to \mathscr K' \to \mathscr K\to \mathscr K'' \to 0$$</p> <p>Then as we found above, $H^1(\mathscr K'')$ measures the failure of lifting sections from $\mathscr G''$ (or equivalently gluing the local liftings) and $H^1(\mathscr K)$ measures the failure of lifting sections from $\mathscr G$ (or equivalently gluing the local liftings). The morphism induces a natural map $$H^1(\mathscr K)\to H^1(\mathscr K'')$$ which is compatible with the above diagram in the sense that if $t\in \mathscr G(U)$ is a section, then the obstruction of lifting this to $\mathscr F(U)$ in $H^1(\mathscr K)$ maps to the obstruction (in $H^1(\mathscr K'')$) of lifting the image of $t$ in $\mathscr G''$ to $\mathscr F''$.</p> <p>Now if you start with an element of $H^1(\mathscr K'')$, which could be an obstruction to lifting some section of $\mathscr G''$ to $\mathscr F''$, then the obstruction to lifting this to an element of $H^1(\mathscr K)$ lies in $H^2(\mathscr K')$. Of course, this is nothing else but saying in words what the long exact cohomology sequence means, but if you write down how these cohomology elements can be represented by a Cech cocycle and try to lift them by lifting locally and then gluing following the exact same steps as above, then you will get exactly this. </p> <hr> <p>To get higher cohomology groups, you may iterate this process. Of course, it gets pretty hairy very soon. </p> <p>Let me add that in my opinion the best way to understand higher cohomology is that it is the lower cohomology of syzygies. In other words, consider a sheaf $\mathscr F$ and embed it into an acyclic (e.g., flasque or injective or flabby or soft) sheaf. So you get a short exact sequence: $$0\to \mathscr F\to \mathscr A\to \mathscr G \to 0$$ Since $\mathscr A$ is acyclic, we have that for $i>0$ $$H^{i+1}(\mathscr F)\simeq H^i(\mathscr G),$$ so if you understand what $H^1$ means, then $H^2$ of $\mathscr F$ is just $H^1$ of $\mathscr G$, $H^3$ of $\mathscr F$ is just $H^2$ of $\mathscr G$ and so on. </p> <p>This, of course, is not special to Cech cohomology, but you can use Cech cohomology to get a feeling for $H^1$ and then use this to get a feeling for $H^{>1}$.</p> http://mathoverflow.net/questions/125010/producing-2-curves-on-a-smooth-surface/125042#125042 Answer by Sándor Kovács for Producing $(-2)$ curves on a smooth surface Sándor Kovács 2013-03-20T04:36:46Z 2013-03-20T06:46:29Z <p>In case you want a curve of arbitrary genus with arbitrary negative self-intersection, you can do this: Let $C$ be a smooth projective genus $g$ curve on a smooth surface. Suppose $C^2=n$ and take $m$ (pairwise) different points on $C$ and blow them up. The strict transform of $C$ is isomorphic to $C$, so it has genus $g$, and has self-intersection $n-m$. In other words, you can achieve arbitrary combinations of genus and negative-self-intersection.</p> http://mathoverflow.net/questions/124432/families-of-curves-on-surfaces-which-are-products-of-curves/124444#124444 Answer by Sándor Kovács for families of curves on surfaces which are products of curves Sándor Kovács 2013-03-13T18:43:02Z 2013-03-19T07:44:30Z <p>$\mathrm{Pic}(\mathbb P^1\times \mathbb P^1)\simeq \mathbb Z\oplus\mathbb Z$, while $\mathrm{rank}\ \mathrm{Pic} (C\times C)\geq 3$ (the diagonal has self-intersection $2-2g$ and its intersection with both $\{q\}\times C$ and $C\times \{q\}$ is $1$, so the determinant of the intersection matrix of these three curves is $2g$ and hence they are linearly independent) and hence $$p^*\mathrm{Pic}(\mathbb P^1\times \mathbb P^1)\subsetneq \mathrm{rank}\ \mathrm{Pic} (C\times C).$$</p> <p>This implies that there are plenty (in fact most) ample line bundles and hence very ample ones on $C\times C$ that do not come from $\mathbb P^1\times \mathbb P^1$. These (at least the very ample ones) give you families that contain any given point (Just take hyperplane sections through that point for the corresponding embedding).</p> <hr> <p><em>This is an answer to Dima's question in the comments below</em> I need the more sophisticated LaTeX capabilities of an answer than a comment to do this right....</p> <p>So, how do you decide that there is a positive dimensional sublinear system going through your point? If you're taking a very ample or at least a basepoint-free system of dimension at least $2$, then the answer is yes. Somewhat less is enough, but this is probably the easiest to check. Here is why.</p> <p>The following short exact sequence tells you which elements of the linear system of $L$ pass through the point $P\in S$:</p> <p>$$0\to \mathscr O_S(L)\otimes \mathfrak m_P\to \mathscr O_S(L) \to \kappa(P)\to 0$$</p> <p>where $\mathfrak m_P$ is the maximal ideal and $\kappa(P)$ is the residue field of $P\in S$.</p> <p>So, you get that you have a long exact sequence: $$0\to H^0(S,\mathscr O_S(L)\otimes \mathfrak m_P)\to H^0(S,\mathscr O_S(L)) \to \kappa(P)\to \dots$$</p> <p>The group $H^0(S,\mathscr O_S(L)\otimes \mathfrak m_P)$ represents those sections of $\mathscr O_S(L)$ that vanish at $P$, so there is a positive dimensional sublinear system going through your point if and only if $\dim H^0(S,\mathscr O_S(L)\otimes \mathfrak m_P)>1$. </p> <p>If $L$ is basepoint-free, or more generally $P$ is not a basepoint of $L$, then the map $H^0(S,\mathscr O_S(L)) \to \kappa(P)$ is surjective and then the above condition is equivalent to $$\dim H^0(S,\mathscr O_S(L))>2.$$</p> <p>If $P$ is a basepoint of $L$, then that map is zero and $H^0(S,\mathscr O_S(L)\otimes \mathfrak m_P)= H^0(S,\mathscr O_S(L))$, so it is enough that $\dim H^0(S,\mathscr O_S(L))>1$. But this actually seems harder to guarantee.</p> <p>So, the short answer is this: If you take a generic projective embedding of $S$ and choose $L$ to be the corresponding very ample divisor, then there is a positive dimensional sublinear system of $L$ going through any given point.</p> http://mathoverflow.net/questions/124432/families-of-curves-on-surfaces-which-are-products-of-curves/124932#124932 Answer by Sándor Kovács for families of curves on surfaces which are products of curves Sándor Kovács 2013-03-19T07:29:19Z 2013-03-19T07:41:44Z <p>OK, here is an alternative more concrete solution that shows that there exist (several sets of) two families of curves such that with respect to any morphism $q:C\times C\to \mathbb P^1\times \mathbb P^1$ one of the families is not a pull-back via $q$. This is a partial answer to a follow up question asked in the comments to my other answer by <em>Dima</em>. </p> <hr> <p>Let $\Delta\subset C\times C$ be the diagonal, $R_1=C\times \{q_1\}$, and $R_2=\{q_2\}\times C$ for some $q_1,q_2\in C$. Now let $L_a=a\Delta+nR_1+mR_2$ for some $a,n,m\in\mathbb N$, $a$ fixed and $n,m\gg 0$. For any given $b\in\mathbb N$, choose $n,m\gg 0$ such that they work for both $a=2b$ and $a=2b+1$ and fix these choices for those $a$'s </p> <p>Using the intersection matrix of $\Delta, R_1,R_2$, which is <code>$$\left[\begin{matrix} 2-2g &amp; 1 &amp; 1\\ 1 &amp; 0 &amp; 1\\ 1&amp; 1 &amp; 0 \end{matrix}\right],$$</code> it is easy to check that $L_a$ is ample (for $n,m\gg 0$). Taking a multiple of $L_a$ (or $L_a$ for $a\gg 0$, in which case we need $n,m\gg a\gg 0$) gives you a desired family of curves as explained in my other answer.</p> <p><strong>Claim</strong> For any given morphism $q:C\times C\to \mathbb P^1\times \mathbb P^1$ and any given $a\in\mathbb N$ </p> <ol> <li>$\Delta$ is not a pull-back of a divisor via $q$</li> <li>For any $a\in\mathbb N$, $L_a$ is not a pull-back of a divisor via (the original) $p$</li> <li>For any $b\in\mathbb N$, at most one of $L_{2b}$ and $L_{2b+1}$ may be the pull-back of a divisor via $q$</li> </ol> <p><strong>Proof</strong><br> 1. follows from the fact that if $\Delta$ were the pull-back of a divisor via $q$, then it would have to be the pull-back of an effective divisor, but the self-intersection of any effective divisor on $\mathbb P^1\times\mathbb P^1$ is non-negative.<br> 2. follows from 1. and that $L_a-a\Delta$ <strong>is</strong> a pull-back of a divisor via $p$.<br> 3. follows from 1. and that $L_{2b+1}-L_{2b}=\Delta$. $\square$ </p> http://mathoverflow.net/questions/124367/toroidal-embedding/124375#124375 Answer by Sándor Kovács for Toroidal embedding Sándor Kovács 2013-03-13T03:57:06Z 2013-03-13T03:57:06Z <p>Watch out: <em>toroidal</em> $\neq$ <em>toric</em> !</p> <p>It is not possible to realize this situation in a toric variety, at least not so that $\pi$ is a toric morphism, because toric varieties are rational by definition and complex tori are not. (I am assuming that by <em>complex torus</em> you mean a compact quotient of $\mathbb C^g$).</p> <p>A <em>toroidal embedding</em> is an open subset $U\subseteq X$ in a normal variety $X$, such that for every closed point $x\in X$ there exist a toric variety $\overline T$, a point $t\in\overline T$, and an isomorphism of complete local $k$-algebras <code>$\widehat {\mathscr O}_{X,x}\simeq \widehat{\mathscr O}_{\overline T,t}$</code> such that the ideal of $X\setminus U$ maps isomorphically to the ideal of $\overline{T}\setminus T$.</p> <p>In other words, a <em>toroidal embedding</em> is something that locally analytically looks like the embedding of the open dense alberaic torus of a toric variety. I suppose the reference you are citing meant that it is possible to make the $\pi^{-1}(t\neq 0)\hookrightarrow X$ embedding to be toroidal.</p> <p>I think that Theorem 2.1 of <a href="http://link.springer.com/article/10.1007/s002229900024" rel="nofollow">Weak semistable reduction in characteristic 0</a> by Abramovich-Karu produces a toroidal embedding for you. If not, then it should at least give you an idea of how to do it. In fact, section 1 of that paper collects the basics about toroidal embeddings, so you should check it out anyway. </p> http://mathoverflow.net/questions/124265/non-uniqueness-of-smooth-compactification/124290#124290 Answer by Sándor Kovács for Non-uniqueness of smooth compactification Sándor Kovács 2013-03-12T06:00:29Z 2013-03-12T06:55:15Z <p>The main idea is this: You can always find an $X$ such that $X\supseteq U$ and $X\setminus U\supseteq \Sigma := \mathrm{Sing} X$. Then in characteristic zero apply Hironaka's resolution theorem, which says that there exists a resolution of singularities $\pi:Y\to X$ such that $\pi$ is an isomorphism over $X\setminus \Sigma\supseteq U$. In particular, $\pi^{-1}: U\hookrightarrow Y$ gives an embedding.</p> <p>In positive characteristic resolution is not known in general and similarly this embedding result is not known either (although I am not saying that knowing this would prove the existence of resolutions).</p> <p>There is actually a newer, expanded version of Kollár's paper in book form: <a href="http://press.princeton.edu/titles/8449.html" rel="nofollow">Lectures on Resolution of Singularities</a>. </p> <p>And of course it is not unique as long as $U$ itself is not projective, since as you can always blow-up $Y$ outside of $U$.</p> http://mathoverflow.net/questions/123286/rational-and-log-canonical-singularity-that-is-not-log-terminal/123287#123287 Answer by Sándor Kovács for Rational and log canonical singularity that is not log terminal Sándor Kovács 2013-03-01T03:37:19Z 2013-03-08T00:39:50Z <p>This is a good question and it is not entirely trivial, because there is no such example using a simple cone construction. More generally, if the resolution graph of the (normal) singularity is a chain of rational curves, then if the singularity is log canonical, then it is necessarily log terminal. (I won't include the calculation of this, but it is very similar to the calculation below, relatively straightforward using adjunction.)</p> <p>So, to get an example like that one needs a singularity with a slightly more complicated resolution graph. The simplest graph that is not a chain is a triple fork. So, let's suppose the exceptional curves are $E_0,E_1,E_2,E_3$ such that $E_0\cdot E_i=1$ for all $i\neq 0$ and $E_i\cdot E_j=0$ for all $i\neq 0, j\neq 0$. Furthermore let $E_i^2=-n_i$ with $n_i\in\mathbb N$ for all $i$. By Artin's criterion this is a rational singularity for any choices where $n_i\geq 2$. </p> <p>Writing down adjunction for each curve $E_i$ gives the equations:</p> <p>$$-n_0(a_0+1) +a_1 +a_2+a_3 =-2$$ and </p> <p>$$-n_j(a_j+1) + a_0 =-2$$ for all $j\neq 0$.</p> <p>These imply that <strong>if</strong> $a_0\geq -1$, then $$a_j + 1 = \frac{a_0+2}{n_j} \geq \frac{-1\ \ \ }{n_j} > 0,$$ so $$a_j> -1$$</p> <p>and then solving for $a_0$ gives $$a_0 = -2 + \frac{n_0-1}{n_0-(\frac 1{n_1}+\frac 1{n_2}+\frac 1{n_3})}$$ so this singularity is log canonical, i.e., $a_0\geq -1$ iff $$\frac 1{n_1}+\frac 1{n_2}+\frac 1{n_3}\geq 1$$ and it is log terminal iff there is strict inequality here.</p> <p>In particular it is log canonical, but not log terminal if there is equality which happens for example if $n_1=2, n_2=3, n_3=6$ and $n_0\geq 2$ arbitrary.</p> <p>Finally, you might ask whether such a singularity exists, but that's easy. By blowing up appropriate points you can certainly create a configuration as above. Then you check that the intersection matrix is negative definite and use Artin's criterion to contract this configuration. The resulting singularity will be your example.</p> http://mathoverflow.net/questions/123747/rational-but-not-log-canonical-singularity/123748#123748 Answer by Sándor Kovács for Rational, but not log canonical singularity Sándor Kovács 2013-03-06T12:19:14Z 2013-03-07T22:03:08Z <p>Actually, if you look at my <a href="http://mathoverflow.net/questions/123286/rational-and-log-canonical-singularity-that-is-not-log-terminal/123287#123287" rel="nofollow">answer</a> to your previous <a href="http://mathoverflow.net/questions/123286/rational-and-log-canonical-singularity-that-is-not-log-terminal" rel="nofollow">question</a>, the same computation gives you an example for this as well. The example there is $\mathbb Q$-Gorenstein and if you choose the $n_i$ such that the intersections matrix is negative definite and $$\frac 1{n_1}+\frac 1{n_2}+\frac 1{n_3}&lt; 1$$ holds, then you get a rational $\mathbb Q$-Gorenstein singularity that is not log canonical. I don't feel like computing determinants, but I am pretty certain that if you choose the $n_i$ large enough then they will work. </p> http://mathoverflow.net/questions/123486/higher-dimensional-bezout-via-hilbert-polynomials-a-reference/123509#123509 Answer by Sándor Kovács for Higher dimensional Bezout via Hilbert polynomials: a reference. Sándor Kovács 2013-03-04T06:26:39Z 2013-03-04T11:12:45Z <p>This is to answer the last question on why the $F_i$ form a regular sequence.</p> <p>In my first attempt I was trying to do this without using the unmixedness of an ideal $I$ generated by $\mathrm{height } I$ number of elements, but it seems that I cannot. (Hat tip to Hailong). </p> <p>So here is the statement that one needs:</p> <p><strong>Thm</strong> Let $I\subseteq A$ be an ideal generated by $\mathrm{height } I$ number of elements. If $A$ is Cohen-Macaulay, then $I$ is unmixed, that is, it does not have any embedded primes. In other words, every associated prime of $I$ is a minimal prime.</p> <p>I will certainly not include a proof, because there is no point repeating one of the proofs in the literature. A relatively simple one can be found in Bruns-Herzog's <em>Cohen-Macaulay rings</em> (Cambride University Press). I think if you are willing to discuss regular sequences, then this should be OK and you may have already discussed the Cohen-Macaulay property, or even this theorem, although I kind of doubt it. </p> <p>However, you don't actually need to discuss CM for this. In fact, this theorem is where the name "Cohen-Macaulay ring" comes from, because it is true that if every such ideal is unmixed, then the ring is Cohen-Macaulay and more importantly, the start of the CM property is a theorem of Macaulay from 1916 that states that the above theorem holds if $A$ is a polynomial ring. Then the next move towards creating this famous name was a theorem of Cohen from 1946 that states that the above theorem holds if $A$ is a regular local ring. Clearly you only need Macaulay's theorem and there must be a simple proof of that somewhere. Perhaps you can adopt the proof from Bruns-Herzog for that special case and make it shorter. There is also a reference to Macaulay's original paper in that book, but I doubt that that's a really good solution to go back to, but you can certainly try.</p> <p>So, assuming this theorem here is how to prove that </p> <p><em>Proposition</em> If $f_1,\dots, f_n$ are polynomials in $n$ variables such that $Z(f_1,\dots,f_n)$ is finite, then $f_1,\dots, f_n$ is a regular sequence. </p> <p>(this is not exactly how you stated it, but you can get to this situation easily by choosing a hyperplane that misses all the intersection points and restrict to the complement)</p> <p><em>Claim 1</em> If $Z(f_1,\dots,f_n)$ is finite, then any irreducible component of $Z(f_1,\dots, f_r)$ is of dimension $n-r$ for any $1\leq r\leq n$.</p> <p><em>Proof</em> Let $W$ be an irreducible component of $Z(f_1,\dots, f_r)$. Then by <a href="http://en.wikipedia.org/wiki/Krull%27s_principal_ideal_theorem" rel="nofollow">Krull's principal ideal theorem</a> $\dim W\geq n-r$ and $\dim (W\cap Z(f_{r+1},\dots, f_n))\geq \dim W -(n-r)$. Since by assumption $\dim (W\cap Z(f_{r+1},\dots, f_n))=0$, it follows that $\dim W\leq n-r$ and hence has to be equal to it. $\square$</p> <p><em>Corollary</em> The ideal generated by $(f_1,\dots, f_r)$ has $\mathrm{height}=r$ and hence by the <strong>Thm</strong> it is unmixed.</p> <p><em>Claim 2</em>: Let $M$ be a finitely generated module over the ring $A$ and $x\in A$. Assume that all associated primes of $M$ have the same height and that $\dim M/xM&lt;\dim M$. Then $x$ is <em>not</em> a zero-divisor on $M$.</p> <p><em>Proof</em>: The assumption implies that $x$ cannot be contained in any associated prime of $M$ and hence cannot be a zero divisor. $\square$</p> <p>Claim 1, Corollary, and Claim 2 combined imply that $f_1,\dots,f_n$ is a regular sequence.</p> http://mathoverflow.net/questions/122953/smooth-projective-varieties-of-picard-number-one/123307#123307 Answer by Sándor Kovács for Smooth projective varieties of Picard number one Sándor Kovács 2013-03-01T08:21:01Z 2013-03-01T08:21:01Z <p>Picard number $1$ is the most frequent case among all varieties, so you cannot expect a classification. It's quite the opposite, you might stand a chance to classify those (within some class) that have Picard number larger than $1$. For instance a general $K3$ surface has Picard number $1$ and the locus of those with a given Picard number becomes smaller as the Picard number increases. </p> <p>If the Picard number is larger than $1$, that usually means that the variety admits some non-trivial maps which gives you a handle on them or a starting point if you will. If the Picard number is $1$, it is hard to get any traction to get some way to study the object.</p> <p>On the other hand that means that you can get lots of examples with Picard number $1$. Chances are, if you choose a variety at random it will have Picard number $1$. You can get lots of examples that are not complete intersections by the simple observation that for a complete intersection of dimension $d$, the middle cohomology groups of the structure sheaf vanish, that is, $H^i(X,\mathscr O_X)=0$ for <code>$0&lt;i&lt;d$</code>. This is actually another way to see that an abelian variety of dimension at least $2$ cannot be a complete intersection. </p> <p>In particular, you can find lots of examples among surfaces. Surfaces with $H^1(X,\mathscr O_X)\neq 0$ are known as <em>irregular</em>, so any irregular surface with Picard number $1$ gives and example that you want. One way to ensure that the Picard number is $1$ is to make sure that $\mathrm{rk}\, H^2(X,\mathbb Z)=1$. In other words, any surface with $q\neq 0$ and $b_2=1$ gives you an example. </p> <p>Of course, you would want an explicit example. Unfortunately, I can't think of one at the moment, but I am fairly certain, that a general surface with $H^1(X,\mathscr O_X)\neq 0$ has Picard number one, so that should give you plenty of examples. In fact, one could argue that that's why I can't give an explicit one, because they are the general ones (and anything explicit is not general).</p> http://mathoverflow.net/questions/123056/k3-surface-as-ramified-double-cover-of-mathbbp2/123059#123059 Answer by Sándor Kovács for k3 surface as ramified double cover of $\mathbb{P}^2$ Sándor Kovács 2013-02-27T01:57:11Z 2013-02-27T16:42:49Z <p><strong>Q1:</strong> Hurwitz formula + canonical divisor of $\mathbb P^2$.</p> <p><strong>Q2:</strong> Move the curve in $\mathbb P^2$.</p> http://mathoverflow.net/questions/123051/no-fixed-components-in-the-linear-system-of-the-line-bundle-generating-picx/123054#123054 Answer by Sándor Kovács for No fixed components in the linear system of the line bundle generating $Pic(X)$ Sándor Kovács 2013-02-27T00:58:24Z 2013-02-27T00:58:24Z <p>Let's say $D$ is a fixed irreducible component. If $D^2\geq 0$, then Riemann-Roch shows that $D$ moves, so it can't be a fixed component. Then since $D$ is irreducible, $D^2=-2$ (again by Riemann-Roch) and hence $D$ is a $(-2)$-curve (use adjunction for this). However, if there is a $(-2)$-curve on your $K3$ surface, then its Picard number has to be at least $2$ which contradicts your assumption. (If the Picard number is $1$, then every effective curve is ample and clearly a $(-2)$-curve is not ample.)</p> http://mathoverflow.net/questions/123048/spectral-sequence-for-composition-of-global-sections-and-tensor-product-of-sheave/123049#123049 Answer by Sándor Kovács for Spectral sequence for composition of global sections and tensor product of sheaves Sándor Kovács 2013-02-27T00:25:02Z 2013-02-27T00:25:02Z <p><a href="http://en.wikipedia.org/wiki/Grothendieck_spectral_sequence" rel="nofollow">Grothendieck spectral sequences</a> deal with the derived functor of a composition. If you're familiar with derived categories, then it takes a very simple form. It says that under the reasonable conditions that make sure that all derived functors in question exist, the (total) derived functor of the composition is the composition of the derived functors.</p> <p>In the language of spectral sequences this will translate to a spectral sequence starting at $E_2$ with the $(p,q)$ term being the $p^\text{th}$ derived functor of the outside functor applied to the value of the $q^\text{th}$ derived functor of the inside functor and the statement is that this abuts to the $(p+q)^\text{th}$ derived functor of the composition. </p> <p>The wikipedia link I included above is only for left derived functors, but it is not too hard to formulate it with right derived functors. In that case you end up with negative $p$s or $q$s, but formally it is similar.</p> <p>There is a section on Grothendieck spectral sequences in Weibel's intro book to homological algebra. He is only dealing with right-right and left-left compositions, but you can do the crossover, too, but you need stronger finiteness conditions. </p> <p>In the case you are asking, you can probably just write $H^*$ as the left derived functors of $H^n$ and then you have a left-left composition and you can use the formalism from Weibel's book.</p> http://mathoverflow.net/questions/122600/understanding-adjointness-of-sheaves-in-algebraic-geometry/122644#122644 Answer by Sándor Kovács for Understanding Adjointness of Sheaves in Algebraic Geometry Sándor Kovács 2013-02-22T16:52:53Z 2013-02-22T16:52:53Z <p>You have sheaves on $X$ and you have sheaves on $Y$. The morphism $f:X\to Y$ gives you a way to transport them back and forth. If you have a morphism $$\phi: \mathscr A\to\mathscr B$$ of sheaves on $X$, then you get a morphism <code>$$f_*\phi: f_*\mathscr A\to f_*\mathscr B$$</code> of sheaves on $Y$. In the case <code>$\mathscr A=f^*\mathscr A'$</code>, then this combined with the natural map <code>$\mathscr A'\to f_*f^*\mathscr A'$</code> Donu already mentioned gives you a morphism <code>$$\mathscr A'\to f_*\mathscr B.$$</code> Similarly, if you have a morphism $$\psi: \mathscr A'\to\mathscr B'$$ of sheaves on $Y$, then you get a morphism <code>$$f^*\psi: f^*\mathscr A'\to f^*\mathscr B'$$</code> of sheaves on $X$. In the case $\mathscr B'=f_*\mathscr B$, then this combined with the other natural map <code>$f^*f_*\mathscr B\to \mathscr B$</code> Donu already mentioned gives you a morphism <code>$$f^*\mathscr A'\to \mathscr B.$$</code></p> <p>Adjointness says that these two operations are inverses of each other. Now you "only" have to understand those maps Donu mentioned. See his answer for how to start with that.</p> http://mathoverflow.net/questions/131703/blow-ups-and-singularities/131740#131740 Comment by Sándor Kovács Sándor Kovács 2013-05-25T01:01:57Z 2013-05-25T01:01:57Z CX's answer answers this, so I don't need to... :) http://mathoverflow.net/questions/66611/homology-class-of-an-algebraic-curve-versus-homology-class-of-a-genus-zero-holomo/66624#66624 Comment by Sándor Kovács Sándor Kovács 2013-05-24T20:09:54Z 2013-05-24T20:09:54Z For smooth projective curves being birational is equivalent to being isomorphic. Take the normalization of $C$. If $C$ is rational, then its normalization is $\mathbb CP^1$. http://mathoverflow.net/questions/66611/homology-class-of-an-algebraic-curve-versus-homology-class-of-a-genus-zero-holomo/66624#66624 Comment by Sándor Kovács Sándor Kovács 2013-05-24T05:34:00Z 2013-05-24T05:34:00Z The answer to what you are asking is trivial: the map that makes $C$ rational is a holomorphic map from $\mathbb CP^1$ to $X$ that satisfies what you ask for. The non-trivial question is whether this holds for any $C$, that is, including non-rational ones. In the case when $X$ is Fano, then this actually happens, but that's a highly non-trivial fact. It follows from Bend &amp; Break, but even that is non-trivial. http://mathoverflow.net/questions/131223/surfaces-ruled-over-elliptic-curves/131252#131252 Comment by Sándor Kovács Sándor Kovács 2013-05-21T00:48:02Z 2013-05-21T00:48:02Z Added Stein factorization back. Thanks, Will. http://mathoverflow.net/questions/131223/surfaces-ruled-over-elliptic-curves/131252#131252 Comment by Sándor Kovács Sándor Kovács 2013-05-21T00:14:24Z 2013-05-21T00:14:24Z right. I had that in the first version, but then I convinced myself that it is not needed... I'll add that back in a bit. http://mathoverflow.net/questions/127643/algebraic-machinery-for-algebraic-geometry/127662#127662 Comment by Sándor Kovács Sándor Kovács 2013-05-12T05:30:38Z 2013-05-12T05:30:38Z I've just realized something, which may be wrong, but let me put it out here: In my mind &quot;reading&quot; Hartshorne means doing all but 2 of the exercises. If you think that book does not use the tools it developed, then you did not do the exercises. probably at least 2/3rds of the knowledge you can learn from that book is in the work you put in the exercises and what you learn from them. http://mathoverflow.net/questions/127643/algebraic-machinery-for-algebraic-geometry/127662#127662 Comment by Sándor Kovács Sándor Kovács 2013-04-16T15:16:07Z 2013-04-16T15:16:07Z @Martin, EGA is not a realistic introduction to algebraic geometry for most human beings. Which more recent textbook would you recommend. (Notice I added &quot;or something similar&quot; for exactly that reason). http://mathoverflow.net/questions/127643/algebraic-machinery-for-algebraic-geometry/127662#127662 Comment by Sándor Kovács Sándor Kovács 2013-04-16T06:34:29Z 2013-04-16T06:34:29Z @pranavk: I completely agree, and can't believe I forgot to mention it. In fact, that was the first book I learnt algebraic geometry from! http://mathoverflow.net/questions/127539/on-flat-morphisms Comment by Sándor Kovács Sándor Kovács 2013-04-14T17:52:27Z 2013-04-14T17:52:27Z ...or let $f:X\to S$ be any birational morphism and $U\subseteq X$ the locus where it is an isomorphism. http://mathoverflow.net/questions/127306/differential-form-with-empty-zero-locus Comment by Sándor Kovács Sándor Kovács 2013-04-12T15:21:10Z 2013-04-12T15:21:10Z I suppose the OP is silently assuming that the manifold is <i>compact</i>. Otherwise take an arbitrary manifold with an arbitrary non-zero (i.e., not constant zero) 2-form and consider the open set where it is not zero. http://mathoverflow.net/questions/127185/strong-notions-of-general-position/127188#127188 Comment by Sándor Kovács Sándor Kovács 2013-04-11T16:27:05Z 2013-04-11T16:27:05Z But when you randomly pick points, you're not putting a probability measure on the set of linear equations but on the set of points, or in other words on those linear equations that arise this way. Or, of course, you can put a probability measure on the whole space of linear equations, but then you have to pull that back via the induced map from the space of points to the space of linear equations. That map, but the way is just a $d$-uple embedding of $\mathbb P^n$. http://mathoverflow.net/questions/127185/strong-notions-of-general-position/127188#127188 Comment by Sándor Kovács Sándor Kovács 2013-04-11T07:04:02Z 2013-04-11T07:04:02Z Being in general position is inherently connected to the <i>with respect to</i> part. I doubt that you can make a more general definition than that. What do you mean about justification? Any set of randomly picked points of the right size will be in general position according to this. http://mathoverflow.net/questions/34673/kunneth-formula-for-sheaf-cohomology-of-varieties Comment by Sándor Kovács Sándor Kovács 2013-04-11T01:44:29Z 2013-04-11T01:44:29Z Charles, I know this is settled now, but I think Thm. 6.7.8 in EGA III$_2$ would be a better and slightly less intimidating reference for this. http://mathoverflow.net/questions/126656/linearly-generated-embedding Comment by Sándor Kovács Sándor Kovács 2013-04-06T14:45:49Z 2013-04-06T14:45:49Z Karl, you are perfectly right! The situation I have is somewhat more complicated and in order to make it reasonable I simplified the situation to a point that it lost the juice. I will have to think about the more general situation, but this is certainly a good point. Thanks! Why don't you post this as an answer, so I can accept it and with that close the question? http://mathoverflow.net/questions/126656/linearly-generated-embedding/126683#126683 Comment by Sándor Kovács Sándor Kovács 2013-04-06T05:14:29Z 2013-04-06T05:14:29Z Also, I would like a name for the embedding, not for the section ring being generated in degree $1$. I understand that this is a tiny difference, but still...