User mike roth - MathOverflow

Answer by Mike Roth for When can Hodge filtrations (decompositions?) be described explicitly in terms of periods?

2012-10-02T16:51:41Z

Building on Donu Arapura's answer (although it's again not clear you can do it in an hour), I would suggest the computing the Hodge filtration in $H^2$ of a product $E_1\times E_2$ of two elliptic curves. The beautiful thing about this example is that (1) you can see how the Neron-Severi group changes rank with the periods, and explicitly relate this back to the properties of the elliptic curves, so it also serves as an example of why the Hodge conjecture is so mysterious and significant, and (2) you can see how the filtration varies with the periods, and so see Griffiths transversality in action.

Here is an outline of this calculation. Choose a basis $\delta_1$, $\delta_2$ for $H^1(E_1,\mathbf{Z})$, oriented so that $\delta_1\cap\delta_2 =1$, and normalize the holomorphic differential $\omega$ so that its integral along (the Poincare duals of) $\delta_1$ and $\delta_2$ are $\tau_1$ (the period of $E_1$), and $1$ respectively. Then this tells us that in $H^1(E_1,\mathbf{C})$ we have the relations

$$\delta_1=\frac{1}{\overline{\tau}_1 - \tau_1}(\overline{\tau}_1\omega - \tau_1\overline{\omega})\quad\mbox{and}\quad \delta_2=\frac{1}{\overline{\tau}_1-\tau_1}(-\omega-\overline{\omega}).$$

Similarly, choose an oriented basis $\epsilon_1$, $\epsilon_2$ for $H^{1}(E_2,\mathbf{Z})$ and normalize the holomorphic differential $\eta$ of $E_2$ to have integrals (along the Poincare duals of $\epsilon_1$ and $\epsilon_2$) of $\tau_2$ (the period of $E_2$) and $1$ respectively. Then in $H^1(E_2,\mathbf{C})$ we get the similar formulas

$$\epsilon_1=\frac{1}{\overline{\tau}_2 - \tau_2}(\overline{\tau}_2\eta - \tau_2\overline{\eta})\quad\mbox{and}\quad \epsilon_2=\frac{1}{\overline{\tau}_2-\tau_2}(-\eta-\overline{\eta}).$$

Wedging $\delta_1$, $\delta_2$, $\epsilon_1$, and $\epsilon_2$ together gives integral bases for $H^2(E_1\times E_2,\mathbf{Z})$, and from the formulas above see their expressions in terms of wedges of $\omega$, $\overline{\omega}$, $\eta$, and $\overline{\eta}$, and thus see the Hodge filtration. Explicitly, the change of basis matrix (up to a factor of $(\overline{\tau}_1-\tau_1)(\overline{\tau_2}-\tau_2)$) is

$$ \begin{array}{cc} & \begin{array}{cccc} \omega\wedge\eta & \overline{\omega}\wedge\eta & \omega\wedge\overline{\eta} & \overline{\omega}\wedge\overline{\eta} \ \end{array} \\ \begin{array}{c} \delta_1\wedge\epsilon_1 \\ \delta_1\wedge\epsilon_2 \\ \delta_2\wedge\epsilon_1 \\ \delta_2\wedge\epsilon_2\ \end{array} &\left[{ \begin{array}{cccc} \overline{\tau}_1\tau_2 & -\tau_1\overline{\tau}_2 & -\overline{\tau}_1\tau_2 & \tau_1\tau_2 \\ -\overline{\tau}_1 & \tau_1 & \overline{\tau}_1 & -\tau_1 \\ -\overline{\tau}_2 & \overline{\tau}_2 & \overline{\tau}_2 & -\tau_2 \\ 1 & -1 & -1 & 1 \ \ \end{array}}\right] \end{array}$$ (So the first row expresses $\delta_1\wedge\epsilon_1$ in terms of the wedges of $\omega$, $\eta$ and their conjugates.) Here $\delta_1\wedge\delta_2$ and $\epsilon_1\wedge\epsilon_2$ have been omitted since they are clearly $(1,1)$-forms.

The change of basis matrix allows us to see how to express the $(2,0)$ forms (spanned by $\omega\wedge\eta$), the (1,1)-forms (spanned by $\overline{\omega}\wedge\eta$, $\omega\wedge\overline{\eta}$ and the (1,1)-forms above), and the $(0,2)$ forms (spanned by $\overline{\omega}\wedge\overline{\eta}$) in terms of the integral classes.

To explain (1) above: Since the Hodge conjecture holds in codimension 1, the rank of the Neron-Severi group of $E_1\times E_2$ is $2$ (for the fibre classes $\delta_1\wedge\delta_2$ and $\epsilon_1\wedge\epsilon_2$) plus the integral or rational $(1,1)$ classes coming from the forms in the matrix. To get a $(1,1)$ form what we need to do is make sure that the coefficients of $\omega\wedge\eta$ and $\overline{\omega}\wedge\overline{\eta}$ are zero and so we end up with the formula:

$$\mbox{Rank of $NS(E_1\times E_2)$} = 2 + \mbox{dimension of space of relations of $\tau_1\tau_2$, $\tau_1$, $\tau_2$, $1$ over $\mathbf{Q}$ or $\mathbf{Z}$.} $$

If you feed this back into standard facts about when two elliptic curves are isogenous (in terms of their periods) and when an elliptic curve has CM (again in terms of the period) the formula that results is that

$$\mbox{Rank of $NS(E_1\times E_2)$} = \begin{cases} 2 & \mbox{if $E_1$ and $E_2$ are not isogenous} \\ 3 & \mbox{if $E_1$ and $E_2$ are isogenous and neither has CM} \\ 4 & \mbox{if $E_1$ and $E_2$ are isogenous and both have CM} \\ \end{cases}. $$ Which is a beautiful demonstration of how arithmetic relations among the periods imply geometric results (which is what the Hodge conjecture is saying).

Answer by Mike Roth for Generic rank of a coherent sheaf on a projective variety vs. generic rank on the cone

2012-01-10T22:07:55Z

The ranks are the same. Since $M$ is a finitely generated $(S/\mathfrak{p})$-module you can write down a finite presentation of $M$ by twisted (in the sense of twisting the grading) copies of $S/\mathfrak{p}$:

$$ \oplus_j (S/\mathfrak{p})(-b_j) \stackrel{\Phi}{\longrightarrow} \oplus_{i=1}^k (S/\mathfrak{p})(-a_i) \longrightarrow M \longrightarrow 0.$$

Passing to $\operatorname{Proj} S/\mathfrak{p}$ this becomes a presentation of $\mathcal{F}_{P}$ with the terms $(S/\mathfrak{p})(-a_i)$ (respectively $-b_j$) replaced by copies of the structure sheaf twisted by $-a_i$ (respectively $-b_j$). Thus the rank of $\mathcal{F}_P$ at a generic point of $\operatorname{Proj} S/\mathfrak{p}$ is $k$ minus the rank of the map $\Phi$ at a generic point of the $\operatorname{Proj}$.

On the other hand, passing to $\operatorname{Spec} S/\mathfrak{p}$, the presentation above becomes a presentation of $\mathcal{F}_A$ with the terms $(S/\mathfrak{p})(-a_i)$ (respectively $-b_j$) replaced by the structure sheaf. The rank of $\mathcal{F}_P$ at a generic point of $\operatorname{Proj} S/\mathfrak{p}$ is again $k$ minus the rank of the map $\Phi$ at a generic point of the $\operatorname{Spec}$.

The scheme $\operatorname{Spec} S/\mathfrak{p}$ is the affine cone over $\operatorname{Proj} S/\mathfrak{p}$, and any point on the $\operatorname{Proj}$ gives rise to a line on the $\operatorname{Spec}$. The rank of $\Phi$ at any point of the $\operatorname{Proj}$ is equal to the rank of $\Phi$ at every point of the corresponding line in the $\operatorname{Spec}$ (except possibly the origin). Thus the rank of $\Phi$ at a generic point of the $\operatorname{Proj}$ is the same as the rank of $\Phi$ at a generic point of the $\operatorname{Spec}$, and therefore the ranks of $\mathcal{F}_P$ and $\mathcal{F}_A$ at the respective generic points are also the same.

Answer by Mike Roth for Intersection of a smooth projective variety and a plane

2011-12-16T21:27:06Z

It is true that $Z$ spans $L$ — even if $X$ isn't ACM. You can also allow $X$ to be singular (but you do need $X$ irreducible and non-degenerate, of course). To illustrate one of the main ideas it is useful to first look at the case when $X$ is a curve.

If $X$ is a curve. Let $M$ be the span of $Z$ and suppose that $M\neq L$. (In the curve case, $L$ will be a hyperplane). Let $p$ be any point of $X$ outside of $Z$ and let $H$ be any hyperplane containing $M$ and $p$. Then $H\cap X$ contains at least $d+1$ points, so by Bezout's theorem the intersection cannot be zero dimensional. Since $X$ is irreducible and one dimensional, this means that the intersection must be all of $X$, so $X$ is contained in $H$, contrary to hypothesis.

The general case. The idea when $k\geqslant 2$ is to show that if $H$ is a general hyperplane containing $L$ then $H \cap X$ is irreducible and non-degenerate (i.e, the intersection $H\cap X$ is not contained in a smaller linear space of $H$). But now all dimensions have been reduced by $1$, and so iterating this procedure reduces us to the curve case, which we've already solved.

To set this up, note that hyperplanes in $\mathbb{P}^n$ containing $L$ are parameterized by a $\mathbb{P}^{k-1}$ (If $V$ is the underlying vector space of $\mathbb{P}^{n}$, $W$ the underlying vector space of $L$, then the hyperplanes are parameterized by the projectivization of $(V/W)^{*}$). We'll use $H$ to refer both to a point of $\mathbb{P}^{k-1}$ and the corresponding hyperplane in $\mathbb{P}^n$ containing $L$. Define $\Gamma\subset \mathbb{P}^{k-1}\times (X\setminus Z)$ to be the set

$$\Gamma = \left\{(H,p) \mid p\in H\right\}$$

i.e, the pairs $(H,p)$ so that $H$ is a hyperplane containing $L$, and $p$ a point of $H\cap X$ not on $Z$.

If we fix $p$, then the set of possible $H$'s satisfying this condition are simply the hyperplanes $H$ containing the span of $L$ and $p$, and this is parameterized by a $\mathbb{P}^{k-2}$. In other words, $\Gamma$ is a $\mathbb{P}^{k-2}$ bundle over $X\setminus Z$. (This fibration is where we use $k\geqslant 2$.) Since $X\setminus Z$ is irreducible this implies that $\Gamma$ is irreducible.

Let $\overline{\Gamma}$ be the Zariski-closure of $\Gamma$ in $\mathbb{P}^{k-1}\times X$. Then $\overline{\Gamma}$ is irreducible since $\Gamma$ is. For a fixed $H\in \mathbb{P}^{k-1}$ the fibre of the projection $\overline{\Gamma}\longrightarrow \mathbb{P}^{k-1}$ over $H$ is simply the intersection $X\cap H$, of dimension $k-1$.

Now let $q$ be any point of $Z$. Then $q\in X\cap H$ for every $H\in \mathbb{P}^{k-1}$ so $q$ gives a section of $\overline{\Gamma}\longrightarrow\mathbb{P}^{k-1}$. Since $Z$ consists of $d$ distinct points where $d$ is the degree of $X$ we conclude that $q$ is a smooth point of $X$. Finally, since $Z$ is the intersection of all $X\cap H$ for $H\in \mathbb{P}^{k-1}$ this implies that the general intersection $X\cap H$ is smooth at $q$. Summarizing, we have a section of the map which generically lies in the smooth locus of the fibres. Since $\overline{\Gamma}$ is irreducible, this implies that the generic fibre is irreducible, i.e, if $H$ is a generic hyperplane containing $L$, then $H\cap X$ is irreducible.

(The intuitive reason for this implication is that, generically over $\mathbb{P}^{k-1}$ the section lets us pick out precisely one irreducible component of the fibre. The union of these components gives us a subset of $\overline{\Gamma}$ which has the same dimension as $\overline{\Gamma}$, and hence whose closure must be all of $\overline{\Gamma}$ by irreducibility. But if there is more than one component in a general fibre, this is a contradiction, thus the general fibre must be irreducible. To make this intuitive construction rigorous requires passing to the normalization of $\overline{\Gamma}$ and then looking at the Stein factorization of the map from the normalization to $\mathbb{P}^{k-1}$. The section gives a generic section of the finite part of the Stein factorization, and that allows one to construct the ``union of the components containing the section''.)

Finally, the same trick as in the curve case also shows us that for any hyperplane $H$, $H\cap X$ must be non-degenerate. Let $Y=H\cap X$, so that $Y$ is a variety of degree $d$ and dimension $k-1$. Let $M$ be the span of $Y$. If $M\neq H$ then pick any point $p\in X\setminus Y$ and let $H'$ be any hyperplane containing $M$ and $p$. Then $H'\cap X$ can't be all of $X$ (since this would contradict the non-degeneracy of $X$), so $Y'=H'\cap X$ must be a subvariety of dimension $k-1$ (more precisely, all components of $Y'$ have dimension $k-1$) and degree $d$. But $Y$ is therefore a component of $Y'$, and the equality of degrees tells us that $Y'$ can't have any other components so we must have $Y'=Y$. This contradicts the fact that $p\in Y'$ and $p\not\in Y$.

Together this shows the required inductive step: If $H$ is a general hyperplane containing $L$ then $H\cap X$ is irreducible and non-degenerate.

Other remarks. I'm guessing from the setup of the question that you want to apply the result for a particular $L$ that you have chosen. If, in the application, you're allowed to pick a general $L$ then you can say something stronger. The classical uniform position principle (where ''classical'' in this case means ''established by Joe Harris in the 80's'') states that for a general subspace $L$ of dimension $n-k$ the finite set of $d$-points in $Z=L\cap X$ have the property that any subset of $r+1$ of the points (with $r\leqslant n-k$) span a $\mathbb{P}^{r}$. Picking $r=n-k$, this means that any subset of $n-k+1$ points of $Z$ spans all of $L$, and so in particular $Z$ spans $L$. (Note that $d\geqslant n-k+1$; for instance, as a consequence of the argument above: if $d < n-k+1$ then the $d$ points of $Z$ would never span $L$.)

Answer by Mike Roth for Quotients of Schemes by Free Group Actions

2011-12-08T20:29:14Z

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, proposition 1.8.

In particular, if $X$ is a closed subscheme of projective space, then for any finite set of points of $X$ (in particular, any orbit of a single point) we can find a hyperplane not containing any of those points. The complement will be an affine open of $X$ containing the orbit. This covers the existence of the quotient in the cases that $X$ is projective (and by a similar argument, quasi-projective).

Here is a sketch of the argument in SGA I: If the condition is satisfied, one first shows that this implies that $X$ is covered by $G$-invariant open affines, then by taking invariants, constructs the quotient of each affine by $G$, and then glues together to get the global quotient $X/G$.

Conversely, if $X/G$ is a scheme then for any open affine $V \subseteq X/G$ the inverse image of $V$ under the morphism $X\longrightarrow X/G$ is an open affine of $X$ stable under $G$. Since we can cover $X$ by such opens, the orbit of any point of $X$ is contained in some affine open set.

As in George McNinch's answer there is no need for the $G$-action to be free. (This discussion was limited to quotients by finite groups though!)

Answer by Mike Roth for Is the Euler characteristic a birational invariant

2010-05-25T21:03:43Z

If you are willing to stick to characteristic zero, then you can assume that there is actually a morphism $f\colon X\longrightarrow Y$ realizing the birational equivalence (reason: look at the graph $\Gamma\subset X\times Y$ realizing the birational equivalence and take its closure, use resolution of singularities to resolve $\Gamma$, and then replace $X$ by $\Gamma$). In this case, $f_{*}\mathcal{O}_{X}=\mathcal{O}_{Y}$ , and all higher direct images are zero, the Leray spectral sequence then implies that the Euler characteristics are equal.

More generally, if $Y$ has rational singularities and $f\colon X\longrightarrow Y$ is a proper birational map, with $X$ smooth, then $f_{*}\mathcal{O}_{X}=\mathcal{O}_Y$ and all higher direct images are zero (this is the definition of rational singularities) and so the same conclusion follows. Smooth varieties have rational singularities! (The computation for smooth varieties is necessary to show that the definition makes sense, i.e., that checking that this property holds for one resolution $X$ implies that it holds for all resolutions).

Comment by Mike Roth

2012-04-03T16:39:59Z

Doesn't this solve the general case as well? (which I assume is what you're saying). It's really about the grading induced by $G/H$, which is a finite group. (I.e., you should lump $R_H$ into one ring, $R_0$, and grade by the cosets in $G/H$).

Comment by Mike Roth

2012-02-15T17:14:05Z

It's probably not true that $\Gamma(U)$ is a free module over $\Gamma(X)$; after removing the point of glueing they become isomorphic so the rank would have to be one. More explicitly, it will be easy to find functions on $X$ which multiply your generators to be zero. Also -- rather than constructing $X$ by glueing, you can just realize $X$ by the equations $xz=0$, $xw=0$, $yz=0$, $yw=0$ in $\mathbb{A}^2$ with coordinates $x$, $y$, $z$, and $w$.

Comment by Mike Roth

2012-01-11T15:47:04Z

Dear Mahdi -- thank you for catching that error, I will edit the response to fix it.

Comment by Mike Roth

2011-12-20T15:50:34Z

Dear Jack - Yes, my previous justification for the irreducibility of the general fibre was completely wrong. I was thinking of the case that $X$ was smooth, and so $\overline{\Gamma}$ could be assumed normal. Then the general fibre is normal and so connectedness implies irreducibility. The general case of (possibly) non-normal $\overline{\Gamma}$ requires the use of the section to get around this, as your example shows.

Comment by Mike Roth

2011-12-09T08:48:14Z

As a complement to the construction above, here is an entertaining example of a blow up of a smooth variety along a non-smooth subvariety, such that the blow up is again smooth. Let $X$ be $\mathbb{A}^{n(n+1)}$ thought of as the space of $n\times (n+1)$ matrices, and let $Y$ be the closed locus of matrices of rank $\leq n-1$. Then the blow up $\widetilde{X}$ (of $X$ along $Y$) is smooth. Explictly it is the incidence correspondence $\Gamma = \{(M,v) | Mv =0 \} \subset \mathbb{A}^{n(n+1)} \times \mathbb{P}^{n}$, which we can see is smooth by viewing it as a fibration over $\mathbb{P}^{n}$.

Comment by Mike Roth

2010-06-23T16:53:51Z

Dear Boyarsky -- It's true that $W$ might not be equidimensional, and so there could be components of $W$ which do not map surjectively to $V$, but the term degree of a map (at least as I understand it) only makes sense for a map between varieties of the same dimension, and so can only be applied to those components of $W$ of the same dimension as $V$. Dear Charles -- As for closing or deleting the question until you can think of a better formulation, it's your question, and I certainly think you can do what you'd like with it.

Comment by Mike Roth

2010-06-22T18:25:07Z

Dear Charles -- As I understand the question I think that there is a problem in the formulation, or more precisely, that the hypotheses listed already answer the question. If $W\rightarrow V$ is finite, then there are no components of $W$ which map to $V$ with positive dimensional fibres, and hence no component of $W$ which has degree zero over $V$.

Comment by Mike Roth

2010-06-03T14:51:11Z

You may also want to look at the paper by Hironaka: "Triangulations of Algebraic Sets", p. 165--185, in the proceedings from the 1974 AMS Arcata conference in Algebraic Algebraic geometry. The purpose of the article is exactly to give a simple demonstration of a fact which "everyone knows", but which is reputed to be difficult.

Comment by Mike Roth

2010-06-01T21:16:03Z

Another reason homogeneous polynomials are popular in introductory discussions of algebraic geometry is that they are stand-ins for global sections of line bundles. If $X$ is a projective scheme, and $L$ a very ample line bundle on $X$, then $L$ gives an embedding into some $P^n$. By Serre vanishing, for large enough powers of $d$, $H^{0}(X,L^d)$ can be identified with the quotient of the degree $d$ homogeneous polynomials (on $P^n$) by the ideal of $X$ in degree $d$. This equivalence allows us to talk around global sections of line bundles without actually introducing line bundles....

Comment by Mike Roth

2010-06-01T19:53:23Z

In the modified question, the scheme-theoretic fibre of f over $\mathcal{O}_{y}/\mathfrak{m}^n$ will be nonreduced for n>1, and hence not a regular scheme --- am I misinterpreting the question?

Comment by Mike Roth

2010-05-26T03:20:11Z

Perhaps the original Wikipedia article meant to have "$\mathcal{O}_{Z}$ is coherent for every subvariety $Z\subseteq X$" in place of (i). Then the fact that the smallest class of sheaves satisfying (i) and (ii) is the category of coherent sheaves is an application of the usual dévissage-type argument.