User mdeland - MathOverflow

Answer by mdeland for When are Hilbert schemes smooth?

2009-10-17T17:27:14Z

Here is yet another example of a smooth Hilbert scheme. Let $X$ be a smooth degree 3 hypersurface in projective space of dimension $n \geq 3$ (say, over an algebraically closed field), and let $H$ be the Hilbert scheme of lines on $X$ (i.e., corresponding to Hilbert polynomial $t + 1$).

The tangent space to $H$ at a point $[L]$ (corresponding to a line $L$ in $X$) is $H^0(L, N)$ where $N$ is the normal bundle of $L$ in $X$. The rank of $N$ is $n - 2$ and the degree of $N$ is $2n - 6$ (you can see this by looking at the standard tangent bundle and normal bundle sequences). Every vector bundle on $L = \mathbb{P}^1$ splits into the direct sum of line bundles. Then the degree of each rank 1 summand of $N$ is at most 1 ($N$ injects into the normal bundle of $L$ in $\mathbb P^n$) and then you can show that no piece can have degree less than $-1$. This allows us to conclude that $H^1(L, N) = 0$. This means that $H$ is smooth at the point $[L]$ (see for example Kollár's book Rational Curves on Algebraic Varieties, Chapter 1, where he explains the infinitesimal behavior of the Hilbert scheme). Since this is true for any line $L$ in $X$, the Hilbert scheme is smooth.

The same argument works for lines on a smooth Quadric. In the same book, Kollár proves that for a general degree $d$ hypersurface $X$ in $\mathbb P^n$, the Hilbert scheme of lines on $X$ is smooth.

Answer by mdeland for Sections of a fibration in intersections of quadrics

2011-09-29T16:46:47Z

As long as the fibers have large enough dimension, they are rationally connected. Then something stronger than what you want follows from the comb smoothing technique developed by Mori, Koll\'ar and others (and explained in Koll\'ar's book Rational Curves on Algebraic Varieties) and the Graber-Harris-Starr (GHS). Once you know that you have a section (GHS), that X is smooth, and that the fibers are rationally connected, then you can find a section through any specified finite number of points in (different, obviously) smooth fibers. I first typed too quickly (thanks Artie) - the image of the section is automatically contained in the smooth locus of the map. The complete intersection of two quadrics in P^4 is Fano, so rationally connected (anything of higher dimensional will also be)

Answer by mdeland for Relationship between Hilbert schemes and deformation spaces

2011-04-18T22:34:58Z

Here's one interpretation if you're content with thinking about the tangent spaces of these functors:

Suppose that $Y$ is a smooth (you can get away with less here) subvariety of $\mathbb{P}^n$. You have the short exact sequence

$0 \rightarrow T_Y \rightarrow i^*T_{\mathbb{P}^n} \rightarrow N_{Y/\mathbb{P}^n} \rightarrow 0.$

Infinitesimal embedded deformations of $Y$ inside $\mathbb{P}^n$ are given by global sections $H^0(Y, N_{Y/\mathbb{P}^n})$. Infinitesimal deformations of $Y$ abstractly are parameterized by $H^1(Y, T_Y)$. There is a forgetful map of functors sending an embedded deformation to the abstract deformation (that is, forget the embedding). Once you know that those functors are (pro)representable (or even if you don't), then the differential map for the corresponding morphism of spaces/functors is given by the connecting homomorphism in the long exact sequence of cohomology (from the above short exact sequence):

$\delta: H^0(Y, N_{Y/\mathbb{P}^n}) \rightarrow H^1(Y, T_Y)$

Answer by mdeland for Rees algebra for non-radical ideals

2010-11-16T13:41:27Z

The blowup of $Spec R$ along $I$ and $I^m$ give isomorphic results. This is Hartshorne exercise II.7.11.a. In general, any birational projective morphism is realized as the blowing up of the target along some ideal sheaf, which in general will be quite complicated (e.g. not radical, but not a power of a radical ideal either).

Regularity and limits of smooth rational curves.

2010-11-06T21:11:36Z

Fix integers $2 < d \leq n$.

Suppose that $T$ is a smooth complex curve with marked point $0 \in T$, and $X$ is a closed subscheme of $\mathbb{P}^n_T$, flat over $T$ such that each fiber has Hilbert polynomial $P(x) = dx + 1$. This is the Hilbert polynomial of a degree $d$, arithmetic genus $0$ curve in $\mathbb{P}^n$. Suppose further that the general fiber $X_t$ is a smooth rational curve. If you like, I'm even happy to suppose that this curve spans a $\mathbb{P}^d \subset \mathbb{P}^n$, or even that $d = n$.

If $X_0$ denotes the special fiber with ideal sheaf $I_0$, is there any bound on the smallest number $s$ so that $H^1(\mathbb{P}^n, I_0(s)) = 0$? If you're familiar with the construction of the Hilbert scheme, one must prove that there is an $s$ which works for any closed subscheme of $\mathbb{P}^n$ with a given Hilbert polynomial (i.e. independently of the closed subscheme).

In general, there is an estimate which is known to be sharp (I think) if you consider arbitrary subschemes with a given Hilbert polynomial - the Gotzmann bound. In the case of curves, the bound is on the order of $d^2$, (at the moment though, I'm having trouble finding a reference for this fact). The number that I'm asking for is closely related to the regularity of $X_0$ (so if you know a bound for this number, I would be happy to hear it).

If the special fiber is smooth, then a theorem of Gruson, Lazarsfeld, and Peskine lets us answer this question (their theorem says that if a smooth curve is linearly nondegenerate, then the curve has regularity bounded by degree - codimension + 1. If the curve is contained in a linear subspace, you can compute the regularity in that subspace).

To highlight:

Question: Suppose we have a subscheme which is the limit of smooth rational normal curves. This subscheme could be non-reduced, reducible, contain embedded points, etc. Do we need the full strength of the Gotzmann bound for the regularity in this case or is there a better bound?

Thanks!

Answer by mdeland for When can one extend a flat family from a subscheme to the whole scheme?

2010-09-30T22:00:53Z

I'm pretty sure the answer is not in general. Take a Hilbert scheme which is reducible, for example, that of the Hilbert polynomial $3t + 1$ in $\mathbb{P}^3$. This Hilbert scheme has two irreducible components - the one corresponding to twisted cubics and the one corresponding to degree 3 plane curves union a point. The intersection of the these two components corresponds to degree 3 nodal plane curves with an embedded point "pointing out of the plane". It turns out that both components are rational (actually this is pretty easy to see) and even that the intersection is rational.
Anyhow, take a line in the twisted cubic component that meets the other component (see III.9.8.4 in Hartshorne's Algebraic Geometry) at a point $[C]$ and a line through $[C]$ contained in the other component. This gives you a flat family over the union of two intersecting lines in $\mathbb{P}^2$, but you will not be able to "fill it in" because the Hilbert scheme in question is not irreducible.

EDIT: I realized that, in this example, the family extended to $\mathbb{P}^2$ doesn't have to be as a family of closed subschemes in $\mathbb{P}^3$. Still, it cannot extend to a flat family in this example - just for a different reason. The general fiber of the extension would be a rational curve, but over a line in $\mathbb{P}^2$ you would have a disconnected family. This is impossible.

Answer by mdeland for Discriminant and Different.

2010-09-16T13:56:03Z

In chapter 8 (entitled "Traces, Complementary Modules, and Differents") of the book Residues and Duality for Projective Algebraic Varieties by Kunz, he gives exactly the definition you propose and proves some basic results about its properties.

Answer by mdeland for balanced curves in Calabi-Yau 3-folds

2010-06-08T20:31:15Z

Perhaps you already know this: but we don't even know how to show that there are finitely many rational curves of a given degree $d$ on the general quintic threefold. This was originally conjectured by Clemens. However, for low degrees (up to $d = 11$ or something close to that), the conjecture is verified in the "strong form": any smooth rational curve of low degree (again, at most $11$) on the general quintic has normal bundle $O(-1) \oplus O(-1)$. As far as I know, that's the current state of affairs. It can often be difficult to produce rational curves with the "expected" normal bundle in any given situation!

Answer by mdeland for Dual Curves in Fancy Language

2010-05-15T20:08:20Z

You make your solution feel "fancier", you could start with a more coordinate free approach to the dual curve. For example, try to identify the line bundle on your curve which embeds it into the dual plane. E.g.: let $X \subset \mathbb{P}^2$ be your curve. Then define $X' \subset X \times \mathbb{P}^{2*}$ to be the set of pairs $(x, H)$ where the line $H$ is contained in the tangent space to $X$ at $x$. When $X$ is smooth at least, then this may be identified with the projectivization of the dual of the normal bundle $N_{X/\mathbb{P}^2}$.

Then think about the map $X' \rightarrow X \times \mathbb{P}^{2*} \rightarrow \mathbb{P}^{2*}$. This is the dual curve embedding. The morphism comes from the inclusion of $N^* \subset \Omega_{\mathbb{P}^2}|_X \subset \mathcal{O}(-1)^3$.

At some point, you have to do computations in local coordinates - but setting it up like this may help a little.

Example of local computation: Locally, the curve looks like $(x(t), y(t), z(t))$. The map to the dual plane is given by $(x(t), y(t), z(t)) \times (x'(t), y'(t), z'(t))$ (cross product! giving the normal to the tangent plane in $\mathbb{C}^3$). If the curve has a simple inflection point, then locally it looks like $(t, t^3, 1)$ and the map to the dual plane is given by $(3t^2, 1, 2t^3)$ - a curve with a simple cusp.

Answer by mdeland for Why torsion is important in (co)homology ?

2010-04-26T13:42:24Z

In their paper "Some Elementary Examples of Unirational Varieties Which are Not Rational", Artin and Mumford show that the torsion in $H^3(V, Z)$ of a non singular projective 3-fold $V$ is a birational invariant. This is great because it gives a cohomological obstruction to rationality (there is no torsion in the cohomology of projective space). They they are able to show that certain conic bundles over rational surfaces are not rational by exhibiting such torsion (their conic bundles are unirational, hence the title). The paper is very nice.

Answer by mdeland for What are non-trivial examples of non-singular blow-ups of a non-singular variety?

2010-04-17T17:22:06Z

Suppose that $X = \mathbb{A}^2$. Let $Y$ be the blow up of $X$ at the maximal ideal $(x,y)$ and let $W$ be the blow up of $Y$ at a point on the exceptional divisor of $Y$ over $X$. Of course, the composition $f: W \rightarrow X$ is birational and an isomorphism away from the origin. The fiber of $f$ over the origin is the union of two $\mathbb{P}^1$'s meeting at a single point, but the total space $W$ is non-singular. The map $f$ is identified with the blowup of $X$ along some closed subscheme $Z$ of $X$ supported only at the origin. I believe an example of an ideal defining such a $Z$ is $(x^3, xy, y^2)$.

By taking the composition of blowups along smooth centers, there is some ideal sheaf on the base giving the composition in "one step". In theory, you can compute this ideal by tracing through the proof that every birational morphisms is a blow up - but in practice I think this is usually difficult.

Answer by mdeland for Stacks determined by their coarse moduli spaces

2010-04-12T15:29:36Z

The answer to the second question is No. For example the weighted projective stack $\mathbb{P}(1,2)$ and $\mathbb{P}^1$ have isomorphic coarse moduli spaces but are not isomorphic stacks. More generally, you can always ``add stack structure" in codimension one by using the root stack construction first developed by Cadman.

Answer by mdeland for Functoriality of Hironaka's resolution of singularities

2010-04-01T14:43:42Z

In addition to what Damiano says, characteristic zero resolution is canonical in a stronger sense: a readable account is given in a paper by Hauser. In particular, resolution commutes with smooth morphisms and commutes with group actions. Recently, this has also been shown to be true for all excellent reduced schemes of dimension at most two by Cossart, Jannsen, and Saito

Answer by mdeland for Positivity in stack geometry

2010-03-16T13:26:08Z

Whatever you mean by ample line bundle on a DM (say) stack, you cannot of course require that some power of the bundle embeds the stack in projective space. You could ask that some power of the line bundle embeds the stack into a weighted projective stack, but this imposes restrictions on the kinds of stacks you will be talking about. This is studied in a preprint by Abramovich and Hassett where they call such stacks cyclotomic. If you define ampleness in terms of some other positivity (like Kleiman's criterion, Nakai-Moishezon, etc), then you will have many of the same theorems as in the case of varieties - because more or less this positivity will just be "pulled back" from the coarse moduli space. So the answer depends on the situation you are in and the kinds of properties in which you are interested.

Answer by mdeland for Italian school of algebraic geometry and rigorous proofs

2010-03-07T18:34:55Z

Severi proved that the moduli space of curves $M_g$ is unirational when $g$ is at most $10$. This has now been made rigorous. Severi further conjectured that the moduli space is unirational for all values of $g$, but this was famously disproved by Eisenbud, Harris, and Mumford. They prove that $\overline{M}_g$ is of general type when $g \geq 24$. Farkas has shown that it is of general type when $g = 22$. It is known that when $g \leq 14$ the moduli space is unirational, but I believe that for remaining values of $g$, this problem is still open.

Answer by mdeland for Why is the Euler characteristic of powers of a line bundle a polynomial in the power?

2009-10-17T17:10:03Z

In the case of a curve of genus $g$, this is the standard Riemann-Roch theorem, it says $\chi(L^k) = k \cdot deg(L) + 1 - g$. In higher dimensions, this is a result of the more general Grothendieck-Riemann-Roch theorem, though in the case I am about to state, it is commonly called the Hirzebruch-Riemann-Roch theorem. In the case of a line bundle on a $n$ dimensional projective variety, it says $\chi(L^k) = (exp(1 + k \cdot L) \cdot td(X))_n$. Here $td(X)$ means the Todd class of the tangent bundle of $X$ (a fixed cohomology class) and the subscript $n$ means we take the degree $n$ piece of the above expression. If you expand this out, you will exactly find a degree $n$ polynomial in $k$ (the Hilbert polynomial). A proof can be found, for example, in Fulton's book on Intersection Theory.

Answer by mdeland for Why is the Euler characteristic of powers of a line bundle a polynomial in the power?

2009-10-18T16:12:35Z

OK, here is another way to see it more in line with what you had in mind I think. Write your $L$ as $\mathcal O(D)$ for some divisor $D$ on $X$. Set $J_1$ to be the ideal sheaf defined by $\mathcal O(-D) \cap \mathcal O_X$ and $J_2$ to be the ideal sheaf defined by $\mathcal O(D) \cap \mathcal O_X$ (intersections taken inside of $K_X$). Let $Y_i$ be the closed subschemes of $X$ defined by these ideal sheaves (they have dimension smaller than that of $X$). Then we have the exact sequences

$$0 \to J_1(kD) \to \mathcal O(kD) \to \mathcal O_{Y_1}(kD) \to 0$$

$$0 \to J_2((k-1)D) \to \mathcal O((k-1)D) \to \mathcal O_{Y_2}((k-1)D) \to 0$$

The two left hand terms are equal by construction. Then by the induction hypothesis, and chasing the Euler characteristics, $\chi(kD) - \chi((k-1)D)$ is a numerical polynomial. This implies that that $\chi(kD)$ itself is a numerical polynomial (Section 1.7 of Harshorne's Algebraic Geometry).

(Here I swept something under the rug, because the subschemes $Y_i$ may not be as nice as $X$ was. But they are at least proper, and we should show that the result we want is that for a proper variety $W$, $\chi(kD)$ is polynomial for a divisor $D$. Then reduce this to the case where $W$ is reduced by looking at the inclusion of $W_\mathrm{red}$ into $W$. Then further reduce to the case where $W$ is integral.)

Extending maps of curves

2010-01-25T21:02:15Z

(I'm happy to work over an algebraically closed field....)

Let $\mathcal{C} \rightarrow Spec (R)$ be a (flat) family of proper, prestable curves where $R$ is a DVR. Suppose the generic fiber is smooth and the special fiber, $C_0$, is reduced but may be reducible.

Given a finite map of curves $f_0: D_0 \rightarrow C_0$ with $D_0$ also prestable, can this be extended to some map on some family?

That is, is there a flat family of proper curves $\mathcal{D} \rightarrow Spec(R)$ and an $R$-morphism $f: \mathcal{D} \rightarrow \mathcal{C}$ which reduces to $f_0$ on the special fiber?

Perhaps such an extension is possible only after a ramified cover of $Spec(R)$?

If so, can it be arranged that the generic fiber of $\mathcal{D}$ is smooth?

Answer by mdeland for Algebraic stacks from scratch

2010-01-23T22:32:59Z

You should read the following post: http://math.columbia.edu/~dejong/wordpress/?p=8

It partially explains why this approach was not taken in the stacks project, and probably isn't generally taken elsewhere.

Now I'll just quote "... any full discussion of the theory of algebraic stacks is going to mention affine schemes, schemes, and algebraic spaces. It will still be the case that the most interesting objects of study are algebraic varieties, and their moduli spaces.

...Sure you can define algebraic stacks without first defining any intermediate geometric objects. However, once this is done, there you are, and there is nothing that you can hold onto and relate the objects to… "

Answer by mdeland for "Every scheme as a sheaf" references?

2010-01-12T17:32:34Z

You may also find the notes "Introduction to Functorial Algebraic Geometry" useful. They are based on a course given by Groethendieck - and can be found here. Unfortunately they're a little rough around the edges sometimes and slightly dizzying to read because of the scanning.

Answer by mdeland for functorial meaning of irreducibility of a moduli space

2010-01-10T16:52:56Z

Your question is a bit vague, but let me try to say something. Being irreducible is a global property so there can be no local characterization like being smooth, regular, etc. If $\mathcal{M}$ is the "space" representing your functor, we say that it is irreducible if it admits a surjective map from an irreducible variety (this is just topological).

If you think about what it means to be irreducible, we could also say that any two closed points of $\mathcal{M}$ can be connected by an irreducible variety. From the functorial point of view, this would mean that any two objects being parametrized live in a family over some irreducible variety. So for example, saying that $M_g$ (the moduli space of smooth geneus $g$ curves is irreducible, is the same as saying that for any two smooth curves $C_1, C_2$ of genus $g$ (say over a field), there is a family $f: S \rightarrow B$ such that $B$ is irreducible, every geometric fiber is a smooth genus $g$ curve, and both $C_1$ and $C_2$ are fibers of $f$. In fact, you can take $B$ to be a curve.

Answer by mdeland for Possible formal smoothness mistake in EGA

2010-01-06T15:32:10Z

Let me point out the following remark made by Grothendieck in his book "Cat\'egories Cofibre\'es Additives et Complexe Cotangent Relatif", 9.5.8

Please excuse my translation:

"Let $f:X \rightarrow Y$ be a morphism of schemes. We say that $f$ is "locally formally smooth" if X can be covered by opens $X_i$ which are formally smooth over $Y$. Evidently, if $f$ is formally smooth, then it is locally formally smooth; I don't know if the converse is true in general. It was this which was affirmed hastily in EGA IV 17.1.6 but the proof is only valid when we assume that the relative $\Omega^1$ is of finite presentation, for example if $f$ is locally of finite type. The Lemma 9.5.7 [loc. cit., not reproduced] implies the following criterion : the map $f$ is locally formally smooth if and only if one has $N_{X/Y} = 0$ and $\Omega^1_{X/Y}$ is "locally projective" in the following sense : one can cover $X$ by open affines $X_i$ with rings $B_i$ such that over $X_i$ the quasi-coherent module $\Omega^1_{X/Y}$ is given by a projective $B_i$ module. We will know that this condition implies the formal smoothness of $f$ if we can show that for a commutative ring $B$, every $B$ module $N$ which is locally projective is also projective - or what amounts to the same - it satisfies $H^1(X, Hom(\tilde{N}, J)) = 0$ for each quasi-coherent module $J$ on $Spec(B)$."

Apparently, this local nature of projectivity was shown soon thereafter by Raynaud and Gruson ("Crit\'eres de platitude et projectivit\'e"). In fact I think they show that it is an fpqc local condition. I think this implies that formal smoothness is even an \'etale local property.

Answer by mdeland for Is an algebraic space group always a scheme?

2009-12-15T19:45:01Z

The answer is yes. Every algebraic space $X$ has an (non-empty) open subspace $U$ such that i) $U$ is a scheme, and ii) a point $p \rightarrow X$ is in $U$ if and only if that point is scheme-like. A scheme-like point is one where there is an affine scheme $V$ and an open immersion $V \rightarrow X$ factoring the point. Further, $X$ is a scheme if and only if all its points are scheme-like. Now, if $X$ is a group algebraic space, the group action is transitive on points. Using this, there can be no non-scheme like points and so $X$ is a scheme.

When is an algebraic space a scheme?

2009-11-08T01:19:00Z

Sometimes general theory is "good" at showing that a functor is representable by an algebraic spaces (e.g., Hilbert functors, Picard functors, coarse moduli spaces, etc). What sort of general techniques are there to show that an algebraic space is a scheme? Sometimes it's possible to identify your algebraic space with something "else" (e.g. it "comes" from GIT as is the case for the moduli space of curves), but are there other methods?

Comment by mdeland

2012-11-28T04:49:58Z

In characteristic 0, rationally connected <=> 0-path connected. This follows from the 'comb smoothing' technique described, for example, in Koll\'ar's book on rational curves on algebraic varieties. I've wondered before about varieties which are n-connected (in your sense). But the deformation theory for higher genus curves is not as nice. In genus 0 we get lucky because 0 + 0 = 0!

Comment by mdeland

2011-09-29T17:29:47Z

Yes, sorry Artie - I typed too quickly - answer amended.

Comment by mdeland

2011-07-01T14:46:19Z

@ulrich - you are right, I hadn't looked at the paper again.... there must be other examples, but none that I can think of off the top of my head.

Comment by mdeland

2011-06-30T18:03:44Z

If X is a geometric quotient for G acting on S, then won't it be the coarse moduli space for the stack quotient of G acting on S? Not all DM stacks are global quotients (even if you allow groups which aren't finite - there is an example in the paper by Edidin, Hassett, Kresch, and Vistoli - Brauer Groups and Quotient Stacks) so this would give a negative answer to your question.

Comment by mdeland

2011-06-01T00:57:17Z

No, X may have no rational curves on it - for example it could be a smooth elliptic curve.

Comment by mdeland

2011-05-31T17:42:58Z

I had wondered that also but I could never prove it. Maybe lifting the pair to W_2 only guarantees a lifting of X to W_3?

Comment by mdeland

2011-05-31T13:18:45Z

Do you know any examples of Frobenius split varieties which do not lift all the way to characteristic zero?

Comment by mdeland

2011-05-17T18:54:16Z

^This is correct (over an algebraically closed field). By the first axiom for adequate equivalence, on P^1 you can write [0] ~ \Sum n_i [t_i] where all the [t_i] are not zero. Let f(x) = 1 - \Prod p_i(x) where p_i(x) is a minimal polynomial for t_i with p_i(0) = 1. Then f(0) = 0 and f(t_i) = 1. So pushing forward by f gives that [0] ~ (\Sum deg(t_i) n_i) [1]. If you replace x by 1/x, then you also get [\infty] ~ (\Sum deg(t_i) n_i) [1] from which you see [0] ~ [\infty] for any adequate equivalence relation.

Comment by mdeland

2011-05-11T17:56:19Z

Cubics which factor are contained in a Zariski-closed subset inside the space of all cubics. So the general cubic will be "far away" from those that factor. For example if you stay on the hyperplane L(x) = 0, then you would expect C_0(x) to get very large as x goes to infinity but C_1(x) will always be 0.

Comment by mdeland

2011-05-09T18:56:02Z

again, if your surface has N "-1 curves", and then you embed by -2K, project to P^3, then the result will have N isolated conics. Since they came from -1 curves, you would have already computed their divisor classes, etc. What is it that you want to know?

Comment by mdeland

2011-05-09T14:41:53Z

Well, if you embed by -2K and project, the degree will be (-2K)^2, not (-K)^2.

Comment by mdeland

2011-05-08T22:42:19Z

If you embed X by -2K, and then project down to P^3 - you can see isolated conics on the (singular) image, right? Do you have a bound on the degree of your surface in 3-space?

Comment by mdeland

2011-05-08T14:22:05Z

Thanks for the clarification on the structure of the map. My other comment was admittedly vague. Perhaps if we realize this as a map of bundles over the Jacobian, then you could try to apply some semi-continuity type arguments - locally you get maps to a relative Grassmannian, etc. I agree though, I don't see how to get anything out of it relating to base point free-ness or very ampleness.

Comment by mdeland

2011-05-08T14:16:22Z

So by conic you mean a curve on the normalization of X which maps to a degree two rational curve in the embedding in C^3?

Comment by mdeland

2011-05-07T16:24:46Z

Sorry to ask an elementary question - but in local coordinates am I supposed to think about this as the map sending f^p(dg) to f dg? That doesn't seem quite right, so I must have something wrong. Also does it help to think about this as a map of global bundles over the Jacobian of X?