User aglearner - MathOverflow

A "Riemannian" analogue of Kobayashi metric?

2013-05-14T22:22:10Z

Recall that Kobayashi metric is defined on any complex manifold $M$. This is a pseudo-metric according to which a tangent vector $v$ at $P$ has length at most $1$ if there is holomorphic map from the open unit disk in $\mathbb C$ to the manifold that sends $0$ to $P$ and $\frac{\partial}{\partial x}$ to $v$.

http://en.wikipedia.org/wiki/Kobayashi_metric

I would like to know if the following analogue of such pseudo-metric makes sense.

Definition. Let $M$ be a Riemannian manifold and let us say that $v$ has length at most $1$ (for the new pseudo-metric) if there is a conformal minimal immersion of the unite disk to $M$ that sends a unite vector at the centre of the disk to $v$.

Question. Are there many examples for which this pseudo-metric does not vanish? Was such a definition studied by someone?

Remark. Clearly in the case $M$ is a Riemann surface this construction gives us the usual Kobayashi metric (i.e it is trivial for $\mathbb C^1$, $\mathbb CP^1$, $T^2$, $\mathbb C^*$ and is a metric of constant negative curvature otherwise).

How to memorise (understand) Nakayama's lemma and its corollaries?

2011-04-12T19:02:53Z

Hope this question is fine. Nakayama's lemma http://en.wikipedia.org/wiki/Nakayama_lemma#Statement is mentioned in the majority of books on algebraic geometry that treat varieties. So I think, I red the formulation of this lemma at least 20 times (and red the proof maybe around 10 times) in my life. But for some reason I just can not get this lemma, i.e. I have tendency to forget it. Last time this happened just a couple of days ago, in the book of Shafarevich (Basic Algebraic geometry in 1.5.3) this lemma is used to prove that for finite maps between quasiprojective varieties the image of a closed set is closed, and again this lemma sounded as something foreign to me (so again I went through the proof of the lemma)...

Question. Is there a path to get some stable understanding of Nakayama's lemma and its corollaries? I would be especially happy if there were some geometric intuition below this lemma. Or some geometric example. Or maybe there is a nice article of this topic? Some mnemonic rule? (or one just needs to get used to the lemma?)

Is it true that Nature promotes products?

2013-05-14T22:47:18Z

I hope this question is not unreasonable.

We all know how to take products of numbers, this generalises to a huge amount of different types of products in mathematics. In a certain sense this notion is "on the surface". On the other hand there is a different operation, called "co-product" that seem to be a less intuitive notion (for example, things like co-algebras seem to be not so intuitive).

Is there a real asymmetry between these two notions (or is this explained by the human perception?)?

This question might turn out to be silly but I have not seen it anywhere and it seem to be a natural question...

Motivation. This is not quite a motivation, but what led me to the question is the reflection on the (well known) fact that there is no product operation for homologies but there is one for cohomologies.

Conical divisor over a $\mathbb Q$-Cartier divisor.

2013-04-22T21:52:51Z

I would like to know if the following statement is correct.

Statement. Let $X$ be a normal projective variety with $Pic(X)=\mathbb Z+torsion$. Let $L$ be an ample line bundle on $X$ and let $D$ be an effective $\mathbb Q$-Cartier divisor in $X$. Consider the projective cone $C$ over $X$ corresponding to $L$. Then the cone over $D$ in $C$ is a $\mathbb Q$-Cartier divisor as well. Is this true?

Note that for $X=\mathbb P^1\times \mathbb P^1$, $L=O(1)\boxtimes O(1)$, and $D$ equal to a $\mathbb P^1$ fiber the statement does not hold. So the condition $Pic(X)=\mathbb Z+torsion$ is important.

$H^1(X,O_X)$, holomorphic $1$-forms, and $b_1(X)/2$ for normal $X$.

2013-04-22T08:50:23Z

Suppose $X$ is a normal projective variety over $\mathbb C$. In the case $X$ is smooth according to Hodge theory $h^1(X,O(X))$ is the dimension of the space of holomorphic $1$-forms on $X$ and this number is equal as well to the half of the first Betti number $b_1(X)/2$ .

I would like to know what happen in the case when $X$ is singular and normal.

1) Is there some relation (equality or inequality) between $h^1(X,O(X))$ and $b_1(X)$? For example does $b_1(X)=0$ imply $h^1(X,O(X))=0$?

2) Suppose that $h^1(X,O(X))=n$ is it true that there is a canonical $n$-dimensional space of $1$-forms on $X$, holomorphic outside of its singularities? (if yes, can something be said about their behaviour at singularities?)

Is there some pedagogical reference treating these questions?

Explaining the concept of projective space: notes for students

2012-02-07T22:29:22Z

This is a question on teaching.

I am teaching at this moment a course in algebraic geometry for master students on a very basic level. Today (this was the fourth lecture) I discovered that only four out of 20 students have ever seen the definition of projective space.

I would like to ask you if you know some nice, short notes that explain what the projective spaces are and that give some simple but still not tautological statements about them. A nice example that comes to my mind is Desargue's theorem, but I would like to have more of such statements. Maybe there are some theorems from classical plane geometry that can be proven using projective geometry? Even though I know what is projective space for almost 20 years I find a bit hard to find a good way to introduce and motivate it...

An element $g$ in a group such that neither $g=1$ nor $g\ne 1$ can be proved.

2013-02-17T15:00:54Z

Edited (this question contains two versions of a similar question)

Is there some finitely presented group $G$ generated by $g_1,...,g_n$ such that there is an element $g\in G$ expressed as a finite word in $g_i$'s so that it is impossible to prove neither $g=1$ nor $g\ne 1$?

Is such a group $G$ exists, what would be a relatively simple example?

Adjusted question. Is there $G,g$ so that $g\ne 1$ in $G$ but is it impossible to prove this in finite time?

Injective morphism from curves to $\mathbb CP^2$

2013-02-22T17:29:10Z

Is there a smooth compact complex curve that does not admit an injective holomorphic map to $\mathbb CP^2$ ? Let me stress, that the image of the curve in $\mathbb CP^2$ can have singularities.

I should probably add, that I see this question as a baby version of the following question of Fancesco Polizzi: http://mathoverflow.net/questions/32938/surfaces-in-mathbbp3-with-isolated-singularities

Remark. Any curve $x^n+y^n+z^n=0$ admits injective morphisms to $\mathbb CP^2$ of arbitrary high degree (the proof is given by Jeremy Blanc here: http://mathoverflow.net/questions/123117/injective-morphism-from-an-elliptic-curve-to-mathbb-cp2/123155#123155)

Higher dimensional Bezout via Hilbert polynomials: a reference.

2013-03-03T18:39:17Z

For the purposes of teaching my elementary course in algebraic geometry I am looking for a reference (or notes) that contains a complete proof of a higher-dimensional weak Bezout theorem. I only want to learn about a proof that is based on the approach when dimensions and degrees of projective varieties are introduced via Hilbert polynomials.

Weak Bezout. Let $F_1,...,F_n$ be homogeneous polynomials of degrees $d_1,...,d_n$ such that hypersurfaces $F_1=0$,..., $F_n=0$ have only finite number of intersections in $\mathbb P^n_K$ (with $K$ algebraically closed). Then the number of intersections is at most $d_1\cdot...\cdot d_n$.

Justification of the question. Of course there is a large amount of proofs of this statement in many books on algebraic geometry (Shafarevich, Harris, Hasset, ect.) but these proofs usually come after 200 pages of text, while I want an honest proof that is contained in a complete set of notes of 40 pages (this will be the maximal length of my notes and I don't want to spend more than 10 pages on higher dimensional Bezout). Also these proofs often develop dimension theory basing on transcendence degree of the field of rational functions and I don't want to use this approach, (since it will require 2-3 additional lectures which I don't have time to give). I know one place where the approach that I want to use (namely everything is based on Hilbert polynomials) is taken, these are the notes of Manin (end of 60ties). Unfortunately it seems to me that there is a problem with the proof he proposes. Basically everything works if $F_1,...,F_n$ form a regular sequence, but to understand why in the condition of the theorem $F_1,...,F_n$ do form a regular sequence is left in the notes as something "not hard to do"... In the book of Hasset, it seems to me there is a similar problem (i.e. it is not explained why $F_1,...,F_n$ form a regular sequence if the number of intersections of hypersurfaces is finite). I hope there is a nice complete proof somewhere...

In other words, what will completely satisfy me is a short proof of the following statement: If the number of intersections of $F_i=0$ is finite, then $F_i$ form a regular sequence. Is there a short proof of this statement?

PS. I think that the answer of Sandor shows that probably there is now "magic easy" solution to the question (if one sticks to Hiblert polynomial approach instead of using higher-dimensional resultants), so I decided to accept it.

Injective morphism from an elliptic curve to $\mathbb CP^2$.

2013-02-27T16:32:42Z

Let $E$ be the elliptic curve $x^3+y^3+z^3=0$.

Question. Are there injective morphisms $E\to \mathbb CP^2$ of arbitrary high degree?

Comments. 1) There are injective morphisms $E\to \mathbb CP^2$ of degree $3$ (the obvious one) and of degree $6$ - whose image is the curve dual to the cubic. 2) Clearly for $\mathbb CP^1$ there are injective morphisms to $\mathbb CP^2$ of arbitrary high degree. 3) I don't know any (smooth projective) curve for which on can prove that it does not admit an injective morphism to $\mathbb CP^2$ of arbitrary high degree.

This question is related to http://mathoverflow.net/questions/122645/injective-morphism-from-curves-to-mathbb-cp2

A classification of rational surfaces with effective $K$

2013-02-22T18:30:41Z

I would like to know if there can be some kind of classification of normal rational surfaces with Gorenstein singularities, such that their canonical divisor is effective.

Additional question. Are there such surfaces at all?

I could imagine constructing such a surface by blowing up several points on an elliptic curve in $\mathbb CP^2$ and then contracting the proper transform of the curve, but will this give an example?

"Arithmetic genus" of a plane curve singularity.

2013-02-23T14:34:25Z

I believe that the following questions are very basic, but I don't know how to get a reference.

Consider a curve in the plane $C\in \mathbb C^2$ with a singularity at $0$ and suppose it is unibranch at zero (i.e. analytically irreducible). Then I guess one should be able to define "arithmetic genus defect" of the curve at $0$. Namely if one smooths analytically $C$, its geometric genus will grow by a positive number (in case of the cusp $x^2=y^3$ it will grow by one), and let us call this number the defect.

Question 1. Is this defect well defined (independent of a smoothing)? How is it called and how one should calculate it (say it terms of the local ring of $C$ at $0$)?

Question 2. Suppose we have an explicit local parametrisation of $C$ at $0$, say by two holomorphic functions $f(t), g(t)$ (polynomials if you wish). Is it possible to find this "defect" as a certain invariant of this pair of functions at $t=0$?

Question 1 is settled in the answer of unknown and Question 2 in comments to it by Roy and Vivek

"Classical" consequences of Bezout's theorem in dimensions $>2$

2013-02-10T21:59:09Z

By Classical I mean something that could have been found before 1900 (say).

A well known consequence of Bezout's theorem for plane curves is Pascal's theorem http://en.wikipedia.org/wiki/Pascal's_theorem .

I am curious if there are some other statements that you find pretty that can be formulated (almost) as elementarily as Pascal's theorem and proven using higher dimensional Bezout's theorem? For example, is there some statement that involves quadrics, planes and lines (cubics?...)?

Motivation. I ask this question since I want to finish to teach my (introductory) course in algebraic geometry by higher-dimensional Bezout theorem (using Hilbert polynomials, ect), and I would be extremely happy to give some pretty application :). To give you an idea of the level of the course, it is based on some bits of Harris book "Algebraic geometry first course",

Disclaimer. I don't doubt the usefulness of Bezout theorem and am sorry if the original question sounded like I doubt it. On the contrary I based the elementary course in algebraic geometry that I teach on this theorem. Namely, the course starts with Bezout for plane curves (using resultants), intorduces projective spaces and varieties, goes through Hilbert basis theorem and Hylbert polynomials (last section of Atiyah-Macdonald) and then as an applications we get a proof of a simplest version of Bezout's theorem in high dimension.

Also, It would be difficult for me to explain what I mean by pretty in math (for myself) but still I feel that the using of this word is justified, because we, mathematicians use this word... Sometimes we disagree on what is pretty, but personally I find pretty huge amount of facts in algebraic geometry. In other words I will be happy to see any application that can be stated in the language on the level of my course.

In the comment I put the link to the question on stackexchange

Projective submanifolds of $\mathbb CP^n$ whose normals bundles are sums of linear.

2013-02-09T14:54:51Z

Let $X\subset \mathbb CP^n$ be a smooth submanifold whose normal bundle is $$\bigoplus_{i=1}^{codim X}O(k_i).$$

Is there some general enough additional condition of $X$ that implies that $X$ is a complete intersection? For example, would $dimX\ge 2$ suffice (to exclude things like $X=\mathbb CP^1$)?

When codimension $k$ degree $d$ submanifold in $\mathbb P^n$ is a complete intersection?

2013-02-07T18:24:11Z

Fix $k>1$ and $d>2$. Is there any known estimate on the minimal number $n=n(k,d)$ such that in $\mathbb CP^n$ any smooth submanifold of codimension $k$ and degree $d$ is a complete intersection? I am curious in particular what is known for small $d$ (i.e. $d=3,4,5...$).

This question is of course related to Hartshorne's conjecture, but I guess it should be much easier.

ADDED. Mahdi's comment settles the co-dimension two case ($k=2$). I wonder for $k>2$ is it known at least that $n(k,d)$ is finite for all $d$?

Linearly trivial bundles on hypersufaces in $\mathbb CP^n$

2013-01-31T10:42:51Z

Recall a definition. Let $V\subset \mathbb CP^n$ be a projective variety and $E$ be a holomorphic vector bundle on it. We call $E$ linearly trivial if the restriction of $E$ to any projective line in $V$ is trivial.

It is well known that any linearly trivial bundle on $\mathbb CP^n$ itself is trivial (see Okonek, Schneider, Spindler).

Question 1. I think that I have an idea of a generalization of this statement and would like to ask you if this generalization is known?

Generalized statement. For any integer $n>0$ any linearly trivial bundle on any smooth degree $n$ hypersuface $V_n\subset \mathbb CP^{4n}$ is trivial.

Idea of the proof. One can easily see that on $V_n$ any two points can be joined by a chain of two projective lines. Moreover for two points $x,y$ the set of such two-lines paths from $x$ to $y$ is a connected projective variety. So let us trivialize the bundle at one point $x\in V_n$. Then extend this trivialization along each connected chain of $2$ lines on $V_n$ starting at $x$. I think that the extension will be independent of the choice of a chain since the space of all chains from $x$ to $y$ is a connected projective variety, while all trivialization of $E$ over $y$ is an affine variety.

Question 2. Does this reasoning sound plausible?

Picard group of a very ample divisor in a smooth variety of dimension >3

2013-01-28T17:58:30Z

Let $X$ be a smooth complex projective variety, $\dim(X)>3$ and $D$ be a very ample possibly reducible divisor on $X$. Is it true that $\textrm{Pic}(X)\cong \textrm{Pic}(D)$? If not, what would be a reasonable condition on $D$ (for example, would it suffice that $D$ is irreducible)?

(if you have a precise reference, I would be especially grateful).

Intersection of two projective submanifolds in $P^n$ treatment in Shafarevich book.

2013-01-27T22:19:59Z

I would like to understand if the following statement is actually proven in Shafarevich's book "Basic algebraic geometry" (or just learn its proof in the spirit of Shafarevich's book).

Statement. Let $K$ be an algebraically closed field and let $X, Y$ be irreducible projective subvarieties in $\mathbb P_K^n$. Suppose that $dim X+dim Y\ge n$. Then $X\cap Y$ is non-empty.

It is proven in section 1.6, Theorem 5, that provided a form $F$ is non-vanishing on $X$, all irreducible components of $X\cap F=0$ have dimension $dimX-1$. So if $Y$ were a complete intersection the statement $X\cap Y\ne \emptyset$ would immediately follow from this theorem. At the same time $Y$ need not be a complete intersection... A few lines after Theorem 6 in the same section Shafarevich mentions that the statement holds. But I can not find the proof in the book. So, is the proof really missing or I just miss something simple?

Complement to an open affine subvariety in an irreducible projective one

2013-01-24T13:13:06Z

I hope this question is not completely trivial:

Suppose $V$ is an irreducible projective variety and $U\subset V$ is a Zariski open subset isomorphic to an affine variety. Is it true that $V\setminus U$ is a Cartier divisor in $V$? If not, what conditions should we impose on $V$? (I guess if $V$ is smooth, then everything is fine?)

Complex curves covered by smooth plane curves

2013-01-06T21:17:52Z

Question: Is it true that for every smooth compact complex curve $C$ there exists a smooth curve $C'$ in $\mathbb CP^2$ that admits a non-trivial morphism (i.e. holomorphic map) $C'\to C$?

Motivation. Unfortunately, I don't know yet any application for a positive answer to this question. But a negative answer to this question would solve in negative a great question of Francesco Polizzi: http://mathoverflow.net/questions/32938/surfaces-in-mathbbp3-with-isolated-singularities

Indeed, here is a simple exercise:

Exercise. Suppose that $C$ is a smooth curve that is not covered by any smooth plane curve. Then the surface $C\times \mathbb CP^1$ is not birational to any surface in $\mathbb CP^3$ with isolated singularities.

Lines on degree 2n-3 Fermat hypersufaces

2013-01-04T08:02:47Z

It is well known that a generic hypersurface of degree $2n-3$ in $\mathbb CP^n$ has finite number of lines. I would like to ask a couple of questions about lines on Fermat hypersurfaces and their symmetries:

$$\sum_{i=1}^{n+1}x_i^{2n-3}=0.$$

Fermat hypersurfaces have a group of automorphisms of order $(2n-3)^n(n+1)!$. In the case $n=3$ (the case of cubic) this group is acting transitively on the collection of $27$ lines and this rases some questions.

The first question is pedagogical, I plan to use it for teaching and really want to know the answer.

Question 1. Is there some slick way to give a high-school proof of the fact that there are exactly $27$ lines on Femat cubic in $\mathbb CP^3$ using (or not) the symmetries of the cubic but without using any theory at all?

Further questions are not for teaching, I am just curious about them.

Question 2. Is it known that a Fermat hypersurface of degree $2n-3$ has finite number of lines for any $n$? Is it known that these lines are never multiple?

Question 3. Can one say something about the number of orbits of the action of symmetries on lines on a Fermat hypersurface of degree $2n-3$? For example, what happen in the case of quintic, $n=5$? According to wiki a generic quintic has $2875=125\cdot 23$ lines, so if Fermat quintic is generic, there should be more than one orbit in the action on lines on it. What is the number of orbits?

I would be happy to know the answer on any of these questions.

Is there a "geometric" intuition underlying the notion of normal varieties?

2012-10-11T17:29:17Z

I first got concious of the notion of normal varieties around 3 years ago and despite the fact that by now I can manipulate with it a bit, this notion still puzzles me a lot. One thing that strikes me is that the definition of normality is so entirely algebraic.

From my common sense understanding the notion of normal varieties restricts the class of spaces that we consider to more-less reasonable ones. It looks to me that this definition is analogous to the definition of pseudo-manifold. At least the obvious similarity is that in both cases the set of non-singular points is connected.

Normality pops up everywhere and its definition is very short. But it is hard for me to imagine that a differential topologist or differential geometer could come up with such a definition. Why is the notion of normatilty is so omnipresent? What is "geometric" meaning of normality?

Maybe a more concrete question would be like this. Suppose $X$ is an irreducible algebraic subvariety in $\mathbb C^n$ with singularities in co-dimension $2$. Can one somehow looking on singularities, their stratification and the way $X$ lies in $\mathbb C^n$ say if it is normal or not?

Added. Who was the person who invented this notion?

I would like to thank everybody for useful comments and links.

Real Pfaffian representations of real cubic surfaces

2012-12-17T10:15:41Z

Consider the following classical construction (which is called Pfaffian representation as Sasha indicates):

Let $ V^4\subset \Lambda^2 \mathbb R^6$ be a four-dimensional subspace of the space of alternating two-forms. Then the equation $a\wedge a\wedge a=0$, $a\in V^4$ is homogeneous of degree three and hence defines a cubic surface in ${\mathbb P}V^4$.

Question. Can every real cubic surface be obtained by the above construction? If yes, is the set of corresponding representations for each cubic connected? I would be grateful for a reference if there is one.

I am primarily interested in real case but if you only can comment on the complex case, this would be interesting for me as well.

As Sasha says, such cubics are called Pfaffian cubics.

Milnor number in terms of minimal resolution of an isolated singularity.

2012-12-17T10:23:09Z

Suppose $F$ is a holomorphic (or polynomial if you prefer) function on $\mathbb C^3$ and $0$ is an isolated singularity of the surface $F=0$. Then on the one hand we can define Milnor number of this singularity, which is equal to the co-dimension of the Jacobian ideal of $F$ (the ideal generated by derivatives of $F$ at zero). On the other hand we can consider minimal resolution of the singularity of the surface $F=0$.

Question. How can one calculate the Milnor number if one knows the exceptional divisor of the resolution?

Rational smooth complex projectives three fold with non-rational deformation

2012-12-07T12:03:08Z

This question is prompted by a great talk of Beauville:

http://www.mathnet.ru/php/presentation.phtml?presentid=5821&option_lang=rus

The talk is called "Luroth problem". In this talk Beauville considers in particular Fano three-folds and says how one can prove that some of them are not rational.

Still I was not able to figure out the following: is there any example of a rational (smooth of course) complex projective three fold that admits a deformation that is not rational? If yes what is the simplest example?

Is a smooth cubic threefold diffeomorphic to a rational threefold?

2012-12-06T23:53:13Z

A theorem of Clemmens and Griffiths states that a smooth hypesurface in $\mathbb CP^4$ of degree three is not rational. I would like to know if nevertheless it is diffeomorphic (as a smooth real $6$-dimensional manifold) to a rational complex three-dimensional variety?

Answer by aglearner for Is a smooth cubic threefold diffeomorphic to a rational threefold?

2012-12-07T11:13:47Z

Let me work out the idea of Ulrich, and deduce that the answer to the question is negative, namely a cubic three-fold is not diffeomorphic to any rational variety. The idea of Urlich is that if a rational threefold is diffeomorpic to a cubic then it is a Fano with $Pic=\mathbb Z$. So we just have to check that cubic is not diffeomorphic to any other type of three dimensional Fanos with $Pic=\mathbb Z$.

There exist $17$ families of Fano threefolds with $Pic=\mathbb Z$ ($\mathbb CP^3$, quadric, five Fanos of index two including cubics, and $10$ Fanos of index $1$). The description can be found here one page two:

http://www.math.u-psud.fr/~amerik/articles/obzor-fv.pdf

Let $H$ be the hyperplane section of the cubic, then $H^3$ equals $3$, i.e., the degree of the cubic. So the cubic has the following property: if we take all classes in integral second cohomology of the cubic then their cubes are of the form $3\cdot n^3$ where $n\in \mathbb Z$. Checking the list of $17$ Fanos we see that the cubic is the only one with this property. So it is not diffeomorphic to any other Fano three fold.

A "holomorphic" Peano curve?

2012-11-29T15:56:39Z

A Peano curve is a continuous map $[0,1]\to [0,1]^2$ whose image is the whole square.

I would like to know if on can obtain "holomorphic" Peano curves. Namely, is it possible to find a continuous map $\phi$ from the unit disk $|z|\le 1$ to $\mathbb C^1$ such that $\phi$ is holomorphic for $|z|<1$ and the image of the boundary $|z|=1$ has non-empty interior in $\mathbb C^1$ under the map $\phi$.

Answer by aglearner for Disconnectedness of Hilbert schemes of projective schemes

2012-11-29T16:07:02Z

Take a generic quintic in $\mathbb CP^4$ and consider lines on it

Visualising locally flat embeddings of surfaces in R^4

2012-11-23T20:35:21Z

As far as I understand it follows from the work of M. Freedman that there exist locally flat embeddings of two dimensional surfaces in $\mathbb R^4$ that can not be smoothed in the class of locally flat embedding. (For definition of locally flat see http://en.wikipedia.org/wiki/Local_flatness)

I am curious if one can draw a realistic picture of such a surface. At least is it possible to draw an intersection of such a surface with a (linear) hyperplane in $\mathbb R^4$? Is the Hausdorff dimension of such an intersection equals $1$?

Comment by aglearner

2013-05-20T14:43:15Z

Dear Curtis, thank you for this remark! Have I got you correctly, that you put $U$ inside $\partial H^3$. Do you know if someone studied the definition that I propose?

Comment by aglearner

2013-05-20T14:36:27Z

The answer does not change if you remove "polarized" provided your K3 surface is projective. Indeed take any positive integral (1,1) form corresponding to a polarization of K3 and sum it over the action of the group. This will give you an invariant positive integral (1,1) form and hence an invariant polarisation. I don't know if non-algebraic K3 surfaces can have an automorphism of finite order (if this is possible the automorphism will preserve the volume form so that the quotient is a singular K3 again)

Comment by aglearner

2013-05-15T11:58:57Z

Misha, thanks for your comment I'll check the paper.

Comment by aglearner

2013-05-14T23:18:22Z

Thank you, this is a sharp answer!

Comment by aglearner

2013-04-25T04:56:35Z

Dear Karl, yes this is very nice thank you for expanding your answer!

Comment by aglearner

2013-04-23T18:54:20Z

Dear Karl, yes I would be grateful if you could give more details.

Comment by aglearner

2013-04-23T07:29:37Z

Dear Karl, many thanks for you answer! Do I understand correctly that in the first formula for $O_Y(D_Y)$ the term on the right of the equality is the space of global sections of $O_Y(D_Y)$?

Comment by aglearner

2013-04-22T18:00:59Z

Also, I would like to ask you if you can propose some reference where I could read about mixed Hodge structure (to get a more complete understanding of your answer).

Comment by aglearner

2013-04-22T17:56:54Z

Dear Donu many thanks for your answer (great that you have not retired from here completely :) )! In fact in the situation I am dealing with $X$ is Moishezon (and not necessarily projective). Do I understand correctly that your argument still works in such a case?

Comment by aglearner

2013-03-30T15:39:59Z

Thanks a lot, this is helpful!

Comment by aglearner

2013-03-13T21:39:57Z

Vivek, huge thanks for this update! Now I am completely convinced in your statement. Your answer is really nice (the only reason I have not yet accepted it is that it seems to me now that the original question is an open problem...)

Comment by aglearner

2013-03-04T15:30:58Z

Abdelmalek, thank you for these references! It looks to me indeed that Van der Waerden proves Bezout using only Nullstelensatz, this is very nice :) But it will take me some time of course, to understand the proof.

Comment by aglearner

2013-03-04T13:44:33Z

Abdelmalek, thank you for your remark. To my shame I don't know what is the multidimensional resultant. Could you tell where can I read about this? I would like to learn more about the approach you propose.

Comment by aglearner

2013-03-04T13:41:12Z

Sandor, I agree that claim 1 from your answer would suffice. Also, Manin's suggested proof of Bezout goes as follows: you calculate by induction the leading coefficient of the Hilbert polynomial of $A_r=k[x_0,....,x_n]/(f_1,...,f_r)$. When you make a step of induction and pass form the Hilbert polynomial of $A_r$ to the Hilbert polynomial of $A_{r+1}$ the easiest way to proceed would be to say that $f_{r+1}$ is not a zero divisor in $A_r$...Hartshorne proposes a more detailed analysis of what happens at this step (introducing multiplicities), but I don't see how to state all this with ease...

Comment by aglearner

2013-03-04T11:37:06Z

Dear Sandor, thank you for this very detailed answer! This helps me to understand better the nature of regular sequences. This also indicates me I guess that I will need to be less ambitious in what to say to students :)...