User jack huizenga - MathOverflow

Answer by Jack Huizenga for Is the tensor product of two acyclic sheaves on a scheme acyclic?

2013-05-28T01:49:14Z

No. For example, on projective space $\mathbb{P}^k$ the line bundles $\mathcal{O}(-1),\ldots, \mathcal{O}(-k)$ are acyclic, but the line bundles $\mathcal{O}(-n)$ with $n>k$ are not. It is still possible for "negative" (or anti-ample) bundles to be acyclic, and their high tensor powers are not acyclic (as they have top cohomology by Serre duality).

To get a result you'll need to actually assume some sort of positivity condition instead of just cohomology vanishing. I'm not sure what the best known results are.

Answer by Jack Huizenga for Does there exist an infinite non-commutative division ring with finite center?

2013-04-16T07:17:31Z

Yes.

The following construction is attributed to Hilbert in "A first course in noncommutative rings" by T.Y. Lam., p. 217.

Let $k$ be a field with an automorphism $\sigma$. Write $D = k((x,\sigma))$ for the noncommutative ring of formal Laurent series $\sum_{i = n}^\infty a_ix^i$ with twisted multiplication rule $xa = \sigma(a) x$ for $a\in k$. Then $D$ is a division ring. If $k_0$ is the fixed field of $\sigma$, then either

$Z(D) = k_0$ or $Z(D) = k_0((x^s))$

depending on whether $\sigma$ has infinite order or finite order $s$, respectively.

Thus all we need to do is choose $\sigma$ and $k$ such that $k_0$ is a finite field and $\sigma$ has infinite order. This can be done by choosing say the Frobenius automorphism of $\overline{\mathbb{F}}_p$.

Answer by Jack Huizenga for Hilbert scheme of points on a surface as moduli space of semistable sheaves

2013-04-04T20:42:24Z

By definition semistable sheaves are torsion-free. Any torsion-free $F$ includes into its double dual, $F\to F^{\ast\ast}$. The double dual is a reflexive sheaf, so any singularities occur in codimension 3. In the surface case, we conclude $F^{\ast\ast}$ is a line bundle with trivial determinant, so must be $\mathcal O_X$.

More generally, in "Vector Bundles on Complex Projective Spaces" (Okenek-Schneider-Spindler) it is shown that any rank one reflexive sheaf on a smooth variety $X$ is necessarily a line bundle (Lemma 2.1.1.15). So even in higher dimensional cases, it follows that $F^{\ast\ast} = \mathcal O_X$.

(In fact, OSS defines the determinant of a torsion-free sheaf $F$ of rank $r$ by $$\det F = (\Lambda^r F)^{\ast\ast},$$ and notes that the determinant is always a line bundle by the cited lemma. In the rank one case, this is just the double dual, so we get a map $$F \to F^{\ast\ast} = \det F = \mathcal O_X.)$$

Answer by Jack Huizenga for On the construction of the varieties parametrizing special linear series on a curve

2013-02-12T00:46:42Z

For your first question, look up Grauert's theorem in Hartshorne. The vanishing of $R^1 \nu_* \mathcal L (\gamma)$ holds because the line bundle on the fiber has degree $\geq 2g-1$. Similarly $R^1 \nu_\ast (\mathcal L(\Gamma)/\mathcal L) = 0$ since when this sheaf is restricted to fibers it has zero dimensional support, so again Grauert's theorem gives local freeness.

For your second question, the canonical blowup of a determinantal variety is defined in chapter 2.2 as an incidence correspondence. The description of $G_d^r$ at the beginning of the section is just a definition; at the end of the section it is explained why this definition actually makes sense.

Answer by Jack Huizenga for Irreducibility of fibers vs. irreducibility of fibered product

2013-02-01T01:56:18Z

It seems to me the answer should be yes, and is certainly yes in case $f$ is smooth. I'll sketch an argument in the smooth case; I think the details can probably be filled in more generally. (Note that if you just assume $X$ is nonsingular and the characteristic is 0, you can easily reduce to the smooth case by generic smoothness)

Put

$Z=\{ (x_1,x_2): f(x_1)=f(x_2)=y \textrm{ and } x_1,x_2 \textrm{ lie on the same component of } f^{-1}(y)\} \subset X \times_Y X$.

Check that $Z$ is a subvariety of $X\times_Y X$ with the same dimension as $X\times_Y X$ (this is easy if $f$ is smooth, and I think it should be true generally; in the general case, it is probably best to throw out some of the "bad" fibers of $f$ and take a closure to define $Z$). Then since $X\times_Y X$ is irreducible, we have $Z = X\times_Y X$, which gives the desired conclusion.

Answer by Jack Huizenga for Does a weaker condition than vanishing derivative imply a function being constant?

2012-12-10T01:47:42Z

The ordinary proof that 0 derivative implies constant rests on the mean value theorem. Your function $g(x)$ is the "symmetric derivative" of $f$, so we should look for a "quasi-mean value theorem" for symmetric derivatives. A quick google search came up with the paper

http://www1.au.edu.tw/ox_view/edu/tojms/j_paper/Full_text/Vol-27/No-3/27%283%298-5%28279-301%29.pdf

Here it is shown (Theorem 2) that if $f$ is continuous on $[a,b]$ and symmetric differentiable on $(a,b)$ with symmetric derivative $f^s$, then there are points $\xi,\eta\in (a,b)$ with

$$f^s(\eta) \leq \frac{f(b)-f(a)}{b-a} \leq f^s(\xi).$$

Clearly then if $f^s = 0$ we must have that $f$ is constant.

Answer by Jack Huizenga for Necessary and sufficient conditions for a sum of idempotents to be idempotent

2012-12-03T06:58:55Z

The key point is that the image of the sum of the idempotents is necessarily the direct sum of the images of the individual idempotents.

Suppose that $E_1,\ldots,E_k$ are idempotent and $F = \sum E_i$ is idempotent. Let $R_i$ and $K_i$ be the image and kernel of $E_i$, respectively, and let $R$ be the image of $F$.

The trace of an idempotent equals its rank, so $$\dim R = \mathrm{tr}(F) = \sum \mathrm{tr}(E_i) = \sum \dim R_i.$$ Furthermore $R$ is a subspace of $\sum R_i$ and $\sum R_i$ has dimension at most $\sum \dim R_i$, with equality iff this sum is direct, so we actually have $$R = R_1\oplus \cdots \oplus R_k.$$ Consider a vector $v_1\in R_1$. By definition $$(v_1,0,\ldots,0) = Fv_1 = (E_1v_1,E_2v_1,\ldots,E_k v_1)$$ with respect to this decomposition. In other words, $R_1\subset K_i$ for all $i\neq 1$, or $E_i E_1 = 0$ for all $i\neq 1$. It follows similarly that $E_i E_j = 0$ for all $i\neq j$.

Answer by Jack Huizenga for is there some condition I can impose on families of curves on a surface such that the second Ext between the ideal sheaf and the structure sheaf is zero?

2012-12-01T11:19:00Z

If I am not mistaken, we always have $\mathrm{Ext}^2(\mathcal I_Z,\mathcal O_Z)=0$ for any closed subscheme $Z\subset X$, so the question is somewhat vacuous.

We have an exact sequence

$$0\to \mathcal I_Z \to \mathcal O_X \to \mathcal O_Z\to 0,$$

and applying $\mathrm{Hom}( -,\mathcal O_Z)$ gives a surjection $$\mathrm{Ext}^2(\mathcal O_X,\mathcal O_Z)\to \mathrm{Ext}^2(\mathcal I_Z,\mathcal O_Z)\to 0.$$ But $\mathrm{Ext}^2(\mathcal O_X,\mathcal O_Z) = H^2(\mathcal O_Z)$, which vanishes since the support of $\mathcal O_Z$ has dimension at most $1$.

(Edit addressing your comment below:

If $X$ is a threefold and $Z$ has dimension at most $1$, it is still true that $\mathrm{Ext}^i(\mathcal O_X,\mathcal O_Z)=0$ for $i=2,3$, so $\mathrm{Ext}^2(\mathcal I_Z,\mathcal O_Z) = \mathrm{Ext}^3( \mathcal O_Z,\mathcal O_Z)$. But $$\mathrm{Ext}^3(\mathcal O_Z,\mathcal O_Z) = \mathrm{Hom}(\mathcal O_Z , \mathcal O_Z \otimes K_X) = H^0(K_X|_Z),$$ which is typically nonzero.)

Answer by Jack Huizenga for Riemann-Roch for Zero Cycles on a Surface

2012-10-22T16:12:51Z

The Hirzebruch-Riemann-Roch theorem gives a formula for the Euler characteristic of a vector bundle $E$ on a projective variety $X$ in terms of the Chern classes of $E$ and the Chern classes of the tangent bundle $T_X$. The Grothendieck-Riemann-Roch theorem is a stronger theorem which generalizes this to the relative case of a family of varieties $\mathcal X/S$; the Hirzebruch-Riemann-Roch theorem is the special case where $S$ is a point.

More generally, Chern classes are defined not just for vector bundles, but for any coherent sheaf: given a coherent sheaf, we can choose a resolution by vector bundles, and define the Chern classes of the sheaf by requiring that the Whitney sum formula hold for the exact sequence. The theorems are then naturally extended to this setting as well.

Now zero cycles can be associated with either their structure sheaf or their ideal sheaf, and we can perhaps compute the associated Chern classes by computing a resolution. Then the Hirzebruch-Riemann-Roch theorem will tell you the Euler characteristic of these sheaves.

Basically, the thing that makes the study of curves so easy is that points are codimension 1, so that the ideal sheaf of a point is actually a line bundle, and in particular locally free. When dealing with higher codimension objects, it is absolutely necessary to deal with non-locally free objects.

Answer by Jack Huizenga for What's the meaning of pencils in birational geometry?

2012-09-08T06:41:24Z

There is a difference between the notion of "linear pencil" and "algebraic pencil." Roughly, a linear pencil on $X$ is just a linear series of dimension 1; equivalently, it can be thought of as a map $X\to \mathbb P^1$. On the other hand, an algebraic pencil is either a map $X\to C$ for a curve $C$, or equivalently, the family of fibers of such a map.

The terminology "not composite with a pencil" refers to the algebraic kind of pencil, not the linear kind.

Palindromic continued fraction

2012-09-03T22:03:33Z

Here's what I hope is a final question outside of my area that I need to understand a problem about stable vector bundles on $\mathbb{P}^2$. Thank you everybody for your help so far!

Suppose I have a rational number $0<\alpha<1$ with a palindromic continued fraction expansion, i.e.

$$\alpha = [0; a_1,\ldots,a_k],$$ where $a_i = a_{k+1-i}$, so that the sequence $a_1,\ldots,a_k$ is a palindrome. I believe from working with several examples that the last two convergents $p_k/q_k$, $p_{k-1}/q_{k-1}$ of this continued fraction satisfy $p_k = q_{k-1}$. That is, the denominator of the penultimate convergent is the numerator of $\alpha$.

I assume this is well known, and if so a reference would be great! Thanks!

Last term of repeating continued fraction expansion

2012-09-03T04:19:50Z

Once again, working with stable vector bundles on $\mathbb{P}^2$ I have run into a question that is really out of my area. (Thanks to everybody who helped out with my last question!)

Let $D>9$ be a rational number which is not a square, and consider the quadratic irrational

$$\xi = \frac{-3 + \sqrt{D}}{2},$$

(I'd be willing to force $D$ to be an integer, and even to assume $D \equiv 5 \pmod{8}$, but I don't think it matters). Let $r$ be the positive integer such that

$$(2r+1)^2 < D < (2r+3)^2.$$

Numerous examples with Mathematica suggest that the continued fraction expansion of $\xi$ takes the form

$$\xi = [r-1;\overline{a_1,\ldots,a_k}],$$ where the last term of the repeating part is $a_k = 2r+1$.

For my particular situtation, I'd be happy enough to know that the number $2r+1$ appears somewhere in the expansion.

As I know almost nothing about continued fractions aside from statements of the basic results, I haven't the slightest idea how to prove something like this. I also don't have a source which does much more advanced things than show that the expansion of a quadratic irrational always repeats. Are statements like this well-known? And where should I look for more advanced theory relevant to this problem?

In case it helps, the original form I came to this number is as follows. Put $$ q = \frac{1}{8}(D-5). $$ Then $\xi$ is a solution of the equation $$\frac{1}{2}(x^2+3x+1) = q.$$ Thanks!

EDIT: At request, here is what Mathematica gives for the continued fractions for some $D$:

$D=5: [0;-2,\overline{-1}]$ (but I am requiring $D>9$)

$D=10: [0;\overline{12,3}]$

$D=13: [0;\overline{3}]$

$D=141: [4;\overline{2,3,2,11}]$

(need more examples? Just ask!)

Homeomorphism of the rationals

2012-08-28T20:42:41Z

In working with the classification of stable vector bundles on $\mathbb{P}^2$, I've found that I need to answer a fairly basic question from analysis/point set topology. Here it is.

Suppose $f:\mathbb{Q}\to \mathbb{Q}$ is

strictly increasing,
not bounded above or below,
a local homeomorphism (with the topology induced from $\mathbb{R}$), and
extends to a continuous map (hence a bijection) $f':\mathbb{R}\to \mathbb{R}$.

Is $f$ a homeomorphism? (That is, is it surjective? Or, alternately, could there exist some irrational number $c$ such that $f(c)$ is rational?)

While I am dealing with a specific function, I state things in this generality because the function itself is fairly nasty and I'd rather not have to use its explicit definition more than I have to. My guess is that it is probably too much to hope for that this be true in this generality, so in case the general version is false here is a refined version of property (3) which incorporates a bit more about my present situation:

3'. there exists a partition $\mathbb{Q}=\bigcup_{\alpha \in A} I_\alpha$ of $\mathbb{Q}$ into countably many disjoint open intervals with irrational endpoints, such that $f$ is linear (with rational coefficients) on each interval.

The difficulty (at least for me) is that, viewing the intervals as intervals in the real numbers, their complement forms some kind of Cantor set.

Thanks!

(EDIT: Several counterexamples have shown the first formulation, with properties 1-4 are false. I imagine the formulation with 3' instead of 3 is also false, but it seems slightly less trivial to get a counterexample due to the condition on the irrationality of the endpoints. In particular, no two intervals can "match up" at an irrational number unless $f$ has the same slope on both intervals.)

Answer by Jack Huizenga for General degree $d$ surface in $\mathbb{P}^3$

2012-08-04T22:05:27Z

By the Noether-Lefschetz theorem, a very general surface $X\subset \mathbb{P}^3$ of degree $d\geq 4$ has $\mathrm{Pic}(X) \cong \mathbb Z$, generated by a hyperplane section $H$. But then if $C \sim aH$ and $C'\sim bH$ on $X$, we have $C\cdot C' = abd^2 > 0$, and so $C,C'$ must intersect in at least one (and generically many) point. Thus every pair of curves on $X$ intersect each other.

(This doesn't rule out that for some choices of pairs of curve classes that what you want could be true, but it does rule out that the most general possible result is true).

Answer by Jack Huizenga for Irreducible "family" of relative effective divisors of a smooth morphism

2012-07-11T09:47:05Z

This should be a counterexample.

Let $n$ be sufficiently large, and let $Y = \mathbb{G}(2,n)$ be the Grassmannian of $\mathbb{P}^2$'s in $\mathbb{P}^n$. Take $X = \{ (p,\Lambda):p\in\Lambda \}$ to be the universal $2$-plane over $Y$. Let $S\subset \mathbb{P}^n$ be a surface which contains a line $L$ (perhaps a rational surface scroll, but there are lots of things to try), and let $Z = \{(p,\Lambda): p\in S \}\subset X.$ Then $Z$ is irreducible. I'd expect that the general $2$-plane $\Lambda_0$ which contains $L$ will not intersect $S$ in any other points (say $n\geq 5$), i.e. that the corresponding fiber $Z_{\Lambda_0} \subset X_{\Lambda_0} \cong \mathbb{P}^2$ is a Cartier divisor. But for more special choices of $2$-plane containing $L$, the intersection will be $L$ together with a finite collection of points, and so the fiber will not be a divisor.

Answer by Jack Huizenga for Any irreducible component of the HIlbert scheme contains an irreducible element

2012-07-09T09:46:28Z

The Hilbert scheme is pretty horribly behaved, and positive results of this nature are quite rare.

You probably mean to ask if there exists $C\in L$ such that $C$ is both reduced and irreducible. If you don't demand reducedness, the answer is yes. For suppose $H$ is a component of a Hilbert scheme of curves in $\mathbb{P}^n$, and let $C\in H$ be a general member of $H$. Let $M_t\in PGL(n+1)$ be a family of linear transformations which limits to the projection from a general codimension 2 plane in $\mathbb{P}^n$ as $t\to 0$ (so that it projects onto a line). Then $M_t(C)$ gives a curve in $H$, but the limiting curve in $H$ is supported entirely on a line, hence is irreducible.

Whenever you have a component with a reduced and irreducible member, it is in fact true that the general member is reduced and irreducible, as being integral is open in flat families.

In general, the Hilbert scheme is always connected (a theorem of Hartshorne); however, irreducible components typically meet at highly non-reduced members instead of the "nice" objects we're actually trying to parameterize.

It is not reasonable to expect that every component of a Hilbert scheme of curves has an irreducible member. For instance, the Hilbert scheme parameterizing twisted cubic curves also has a separate component parameterizing unions of plane cubic curves and an isolated point. You may object this latter object is not a "curve," but part of the problem with Hilbert schemes is that the Hilbert scheme of curves doesn't even only see things of pure dimension 1.

Similarly, the Hilbert scheme corresponding to the Hilbert polynomial of two skew lines does not have any reduced and irreducible curves in it.

To get more satisfying counterexamples, one only needs to increase the degree and genus--things get really nasty really fast.

Answer by Jack Huizenga for Incidence Correspondence

2012-04-30T07:05:42Z

There's several issues here.

(1) Is a given incidence correspondence actually a closed variety?

(2) What are explicit equations for the correspondence in the product of the relevant spaces?

(3) What are geometric properties of the incidence correspondence?

Most often, questions (1) and (3) are studied and little attention is paid to (2).

For (1), things can be generalized a fair bit. For instance, suppose $X$ is a variety, $H$ is an ample divisor, and $P$, $Q$ are two Hilbert polynomials with respect to the ample divisor $X$. Then there are (projective) Hilbert schemes $Hilb_P(X)$ and $Hilb_Q(X)$ parameterizing closed subschemes of $X$ with Hilbert polynomials $P$ and $Q$. Then there are a couple different natural incidence correspondences, for example

$\{(Z,Z'):Z\subset Z'\} \subset Hilb_P(X)\times Hilb_Q(X)$

$\{(Z,Z'):Z\cap Z'\neq \emptyset\}\subset Hilb_P(X)\times Hilb_Q(X),$

and it is easy to verify that these conditions are closed (although the correct scheme structure may be less clear). One can instead restrict attention to a closed subvariety of the Hilbert scheme (so as to not use all the components of the Hilbert scheme, for instance, in case the geometric objects you care about are not entirely determined by their Hilbert polynomials). It is also easy to generalize to cases with more factors. These types of constructions mean that arguments for the closedness of incidence correspondences are almost never written down, as anything reasonable that you can write down will be closed so long as the families of objects under consideration are themselves closed.

In practice, (2) is rarely of any theoretical interest, unless these are very special varieties. Perhaps there is a large algebraic group acting and the ideal can be studied via representation theory, or perhaps the variety has very small dimension or is otherwise very simple, in which case some information might be learned from the ideal.

Regarding (3), geometric properties of an incidence correspondence are only very rarely (roughly in the same cases as discussed for (2)) determined by studying the defining equations. Most often, the reason we study an incidence correspondence $\Sigma \subset X\times Y$ is because we have some question about one of the projection maps, say $\Sigma \to X$; for instance, we may wonder if it is dominant. We then ideally answer this by studying the other projection, which ideally is easier to understand. Likewise, properties like dimension and irreducibility are hopefully easily understood by studying the projections, and smoothness can sometimes be analyzed as well (although smoothness is often not so important in basic applications).

Existence of "good" concretizations

2012-04-10T17:15:53Z

In this question it was asked what relation, if any, exists between the notions of a monomorphism in a general category $\mathcal C$ and an injective map. Of course the question only makes sense if $\mathcal C$ is a concretizable category. If $\pi:\mathcal C\to Set$ is a faithful functor, we may regard a morphism $f:X\to Y$ of objects of $\mathcal C$ as an injection (relative to $\pi$) if $\pi(f)$ is injective in $Set$.

One then notices that this notion depends very much on the choice of concretization, and thus that this notion is not an intrinsic property of the category itself.

However, Emil Jerabek noticed that it is natural to ask the following: given a concretizable category, does it admit a concretization which preserves monomorphisms?

Apologies if this question is trivial, hopelessly too general, or of no interest to category theorists.

Answer by Jack Huizenga for Why the underlying function of a monomorphism may not be an injection

2012-04-09T21:32:36Z

There are several issues here. Let $\mathcal C$ be a category.

(1) The question only makes sense when $\mathcal C$ is concrete, as otherwise "injective" has no meaning. When $\mathcal C$ is concrete, we can choose a faithful functor $\pi : \mathcal C\to Set$, and regard a morphism $f:X\to Y$ as being injective if $\pi(f)$ is injective (i.e. a monomorphism in $Set$).

(2) However, this notion of "injective" depends on the choice of concrete structure. Indeed, if we consider one of the simplest possible categories, which has two objects $A,B$ and one nonidentity morphism $f: A\to B$, then we can concretize this category by mapping $A,B,$ to an $n$-element set and an $m$-element set, respectively, and mapping the morphism to any function between those two sets--the faithfulness condition is essentially vacuous. But now $f$ was a monomorphism (and an epimorphism, although not an isomorphism) and its concretization will not typically be any of these things, depending on how the concretization was chosen.

(3) This counterexample highlights what can go wrong. First of all the notion of monomorphism depends solely on the category, and not on a choice of concretization, so it is a better "invariant" generalization of the notion of injection from $Set$. Furthermore, it can be discussed whether or not a concrete structure exists. Finally, the phenomenon that a monomorphism can fail to be injective relative to some concrete structure is roughly attributable to a lack of "enough" morphisms and objects in the original category, so that in $Set$ one sees the map is not a monomorphism, but in the original category there are not enough objects and morphisms to find a "preimage" of any diagram exhibiting the failure of injectivity.

Answer by Jack Huizenga for Incidences of Lines / Circles in the Plane

2012-01-26T05:22:28Z

Thinking of a configuration of 3 lines as a cubic plane curve, we can represent such a configuration by a degree 3 homogeneous polynomial which factors completely into linear factors, modulo scalars. This naturally embeds such configurations into the projective space of cubic forms $\mathbb{P} H^0(\mathcal{O_{\mathbb P^2}}(3))$, a 9-dimensional projective space.

The general configuration of lines, consisting of three lines meeting at three distinct points in the (finite) plane forms a 6-dimensional (quasiprojective) subvariety of this projective space, as it is the image of the finite map $\mathbb{P}^{2*}\times \mathbb{P}^{2*}\times \mathbb{P}^{2*} \to \mathbb{P}^9$ given by unioning three lines in the plane.

When three lines all meet at the same point, the configuration is determined by selecting the point of intersection and the slopes of the 3 lines. This is therefore a 5-dimensional locus.

If two of the lines become parallel, the configuration is essentially determined by choosing the slope of the parallel lines (equivalently, the point they meet along the line at infinity in projective space) along with say the $y$-intercepts of the parallel lines and the position of the third, nonparallel line. This gives a 5-dimensional space as well.

Other degeneracy loci can be analyzed similarly, essentially by determining the number of degrees of freedom in a configuration.

Schubert calculus and intersection theory don't seem particularly relevant, as these degeneracy loci aren't Schubert cycles and aren't naturally described as intersections.

Answer by Jack Huizenga for Quartic Space Curves

2012-01-22T22:30:14Z

Singular nondegenerate irreducible degree $4$ curves certainly exist. They can be obtained as complete intersections of two quadric surfaces which are tangent at some point.

Thinking differently, any curve of class $(2,2)$ on a nonsingular quadric $Q$ is a degree $4$ space curve. The series $|\mathcal{O}_Q(2)|$ is $8$-dimensional, and the singular curves in this family form a $7$-dimensional subvariety. Cuspidal members of this family form a smaller $6$-dimensional locus.

Answer by Jack Huizenga for Euler characteristic of Weierstrass divisor

2012-01-07T07:20:19Z

I'm not sure I've heard the term Weierstrass divisor before, but I take it you mean the sum of the Weierstrass points, with multiplicities given by the weights. In this case, the sum of the weights, and hence the degree of the divisor, is given by

$$\sum_{p\in X} w(p) = (g-1)g(g+1).$$

By Riemann-Roch, the Euler characteristic then equals $g^3-2g+1$, which is different from your formula.

See Arbarello-Cornalba-Griffiths-Harris, Geometry of Algebraic Curves, exercise batch I.E.

Answer by Jack Huizenga for When does the blow-up of $CP^2$ at N points embed in $CP^4$?

2011-11-25T19:33:48Z

Suppose we have a non-degenerate embedding $f:X\to \mathbb{P}^4$. Let $L = f^\ast \mathcal{O}_{\mathbb{P}^4}(1)$, and let $V \subset H^0(L)$ be the (base-point free) $5$-dimensional linear series giving $f$. There are two cases to consider.

Case 1: $V = H^0(L)$ is complete. I'm afraid here the problem seems quite difficult. There are two essential steps.

First, find all line bundles $L$ with $h^0(L) = 5$. Given general points $p_1,\ldots,p_N$ and multiplicities $m_1,\ldots,m_N$, this amounts to determining the dimension of the series of curves of a given degree $d$ having singularities of multiplicity $m_i$ at $p_i$ for each $i$. The Segre-Gimigliano-Harbourne-Hirschowitz conjecture provides an expected answer, but it is very much open.

Next, for each line bundle with $h^0(L) = 5$, we need to determine when the complete series gives an embedding. This question has also received a lot of attention (at least in more general formulations), but at least these more general versions are still active areas of research. A google search for ampleness of line bundles on blowups turns up many results.

Potentially an argument that tries to find some obstruction to embedding in $\mathbb{P}^4$ could sidestep this program, but being a codimension $2$ subvariety imposes much less structure than being a hypersurface.

Case 2: $V\subset H^0(L)$ is a proper subseries. In this case, choose a $6$-dimensional series $W$ with $V\subset W\subset H^0(L)$. Then $W$ gives a non-degenerate embedding of $X$ in $\mathbb{P}^5$, and the embedding of $X$ in $\mathbb{P}^4$ is the composition of the embedding in $\mathbb{P}^5$ with projection from a point. Since $V$ is base-point free, this projection is from a point lying off of the embedded surface in $\mathbb{P}^5$. Since the projection must be an isomorphism between the two images of $X$, the secant variety of the surface in $\mathbb{P}^5$ must be a proper subvariety of $\mathbb{P}^5$. But Severi showed the only smooth nondegenerate surface in $\mathbb{P}^5$ with deficient secant variety is the Veronese, isomorphic to $\mathbb{P}^2$ embedded by the complete series of conics. Thus this case never actually arises.

Answer by Jack Huizenga for Divisor intersecting non-negatively the negative part of its Zariski decomposition

2011-11-16T14:37:16Z

EDIT: New example, hopefully this one works.

Blow up $\mathbb{P}^2$ at a point $p$, then blow up the resulting surface at a point $q$ on the exceptional divisor. The resulting surface has Picard group generated by the class $H$ of a line, the proper transform $E_1$ of the first exceptional divisor, and the second exceptional divisor $E_2$. We have $E_1^2 = -2$ and $E_2^2 = -1$, while $E_1\cdot E_2 = 1$. Now consider $D = E_1 +2E_2$. The Zariski decomposition of $D$ is $P = 0$, $N = E_1+2E_2$, as no effective divisor supported on $E_1$ and $E_2$ is nef. Then $D\cdot E_1 = 0$. We can get strict inequality instead by taking $D = E_1+3E_2$, for instance.

Answer by Jack Huizenga for Is it possible to check two curves on birational equivalence by some computer algebra system?

2011-11-09T00:33:03Z

Your example is a bit of a red herring, as this is relatively easy for hyperelliptic curves. A hyperelliptic curve can be reconstructed uniquely from the data of the branch divisor of the degree $2$ map to $\mathbb{P}^1$. Furthermore, isomorphisms of hyperelliptic curves commute with the degree $2$ map to $\mathbb{P}^1$. Thus for two hyperelliptic curves, the only issue is whether or not the branch divisors are projectively equivalent, and this is quite straightforward to check.

Answer by Jack Huizenga for Condition on the canonical divisor for Yau Inequality - effective or ample?

2011-11-04T17:09:31Z

You should read $c_1(X)<0$ as saying that the first Chern class of $T_X$ is negative, or the line bundle $K_X$ is positive, in the sense of curvature. But positive line bundles are ample line bundles. This fact is sometimes called the Kodaira Embedding Theorem. See for example p. 181 of Griffiths-Harris, Principles of Algebraic Geometry.

Answer by Jack Huizenga for Reference Request: Deformations of a map bijective to global sections of the pullback of the tangent sheaf

2011-11-01T21:09:28Z

For a more elementary reference, try Debarre's "Higher dimensional algebraic geometry." This book is a great place to get started learning some algebraic geometry beyond Hartshorne, and basic deformation theory plays a key role.

The fact you are asking about is Proposition 2.4; Chapter 2 gives a good overview of similar types of results, and much of the rest of the book investigates how these types of results can be applied.

Answer by Jack Huizenga for Can any smooth hyperelliptic curve be embedded in a quadric surface?

2011-10-30T22:38:04Z

Yes.

Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$. By Riemann-Roch, $\dim|L| = 1$ and $|L|$ is base-point free, so the complete series $|L|$ gives a degree $g+1$ map to $\mathbb{P}^1$. Then the product of this map and the degree $2$ map $C\to \mathbb{P}^1$ gives a map $f:C\to \mathbb{P}^1 \times \mathbb{P}^1$, whose image is a curve of type $(2,g+1)$. But $\mathbb{P}^1\times \mathbb{P}^1$ is just a quadric in $\mathbb{P}^3$.

To see the map is an embedding, it will suffice to show that it is birational. Indeed, the image has arithmetic genus $g$ by adjunction on a quadric surface. But it also has geometric genus $g$ since $C$ is its normalization. Thus the image is smooth if it is reduced.

Finally we must see that the map is birational. The only way the map fails to be injective is if some divisor of $|L|$ contains a pair of points conjugate under the hyperelliptic involution. But in this case the assumption that $\dim |L|=1$ implies that $|L| = g_2^1 + p_1+\cdots+ p_{g-1}$, where $g_2^1$ is the hyperelliptic series and $p_1,\ldots,p_{g-1}$ are base points. Since $L$ was general, this isn't true, and we're done.

Here's a cute (although trivial) kind of partial converse: If $g$ is prime, then any smooth curve $C$ of genus $g$ which embeds in a smooth quadric is hyperelliptic.

Answer by Jack Huizenga for Taking "Zooming in on a point of a graph" seriously.

2011-10-05T01:06:57Z

In algebraic geometry, this construction is known as the tangent cone to the graph. More generally, suppose we have the zero set of any polynomial $f(x,y) = 0$, and assume $f(0,0)=0$. Then we can write

$f(x,y) = a_m (x,y) + a_{m+1}(x,y) +a_{m+2}(x,y) +\cdots$

where $a_i(x,y)$ is a homogeneous polynomial of degree $i$ and $a_m$ is nonzero. The zero set of $a_m$ is called the tangent cone to the curve at the origin. It is a product of $m$ linear forms (over $\mathbb{C}$), and $m=1$ exactly when the zero set is smooth at the origin. In this case, the tangent cone coincides with the tangent space.

From your point of view, when we substitute $x\mapsto x/c$ and $y\mapsto y/c$ it is clear that the term left in the limit is $a_m$.

We can of course find tangent cones at other points of the zero set by changing coordinates.

In general, for a smooth function $f$ you should be able to take a multivariate Taylor expansion and read off the tangent cone from the lowest degree part. This is where the difficulty comes in for actually defining the tangent line in terms of the tangent cone in a calculus class, as computing the Taylor expansion demands we already have a notion of derivative. This difficulty is obviously not seen in the case of polynomials, although recentering the Taylor expansion of a polynomial at a different point is perhaps easiest done with the aid of derivatives.

Higher dimensional analogues are also available without any real work, although in the singular case the tangent cone is much more interesting than just a union of hyperplanes: it will be a cone over some variety. The homogeneous polynomial $a_m(x_1,\ldots,x_n)$ typically doesn't factor into a product of linear forms when $n>2$.

Tangent cones are treated in any reasonable introduction to algebraic geometry, such as Harris' "First course" book or Shafarevich.

Answer by Jack Huizenga for Line bundles, linear systems and normalization

2011-09-07T13:37:40Z

It is in fact true that the normalization of a projective variety is projective, as J.C. Ottem discusses in the comments.

It is not true that if a normal variety is mapped to a projective space by a linear series $V\subset H^0(L)$ then some larger linear series $W\supset V$ has image isomorphic to the normalization.

For instance, let $C$ be a general smooth curve of genus $g \gg 0$, and pick a general line bundle $L$ of degree $g+2$. By Riemann-Roch, $h^0(L) = 3$, and thus the map induced by $|L|$ maps $C$ to $\mathbb{P}^2$. For large enough $g$, however, the general curve of genus $g$ is not isomorphic to a smooth plane curve, and thus the image cannot be smooth. Moreover, we're using the complete series of sections of $|L|$ already, so there aren't "more sections" to include.

However, the following modification is true. Suppose $X\subset \mathbb{P}^n$ is a variety, with normalization $f:\overline{X}\to X$. Then there exists a line bundle $L$ on $X$ such that $f$ is given by a collection of sections of $L$ and the complete series $|L|$ gives an embedding of $\overline{X}$ into some big projective space.

Roughly, if $L = f^* \mathcal{O}(1)$, we can modify $L$ by adding a sufficiently ample divisor $nH$ so that $L+nH$ gives an embedding. But if $V \subset H^0(L)$ corresponds to $f$, then multiplying by a fixed section $D$ of $nH$ gives us an inclusion $D + V \subset H^0 (L+nH)$; note that this series has $D$ as a base locus. The map corresponding to this series is just $f$, realized as a projection from the big projective space which $L+nH$ maps $\overline{X}$ to.

Comment by Jack Huizenga

2013-06-13T16:14:05Z

What is the Koszul complex of a subvariety that is not a complete intersection?

Comment by Jack Huizenga

2013-06-01T19:08:35Z

If the divisor is not ample then it is not such a hyperplane section. For example the complement of a line on a quadric surface is a $\mathbb{P}^1$ bundle over $\mathbb{A}^1$.

Comment by Jack Huizenga

2013-05-21T07:42:23Z

The intersection of two smooth hypersurfaces $X,Y$ is singular at a point $p$ if and only if $X$ and $Y$ fail to be transverse at $p$. It should be easy to show that the singular locus of the intersection can have any codimension you want. As you should interpret the intersection scheme-theoretically, it is even possibly for the intersection to be everywhere non-reduced, in which case you should view it as being singular everywhere. (For example, this is the case for the two hypersurfaces $z=0$ and $z=x^2$ in 3-space, whose intersection should be viewed as a double line in the plane $z=0$).

Comment by Jack Huizenga

2013-05-14T03:27:38Z

@David: there won't be any interesting surjective maps from a field to another ring. I think your argument is fine Chris, even if this question would be better suited to math.stackexchange.com.

Comment by Jack Huizenga

2013-05-13T06:47:06Z

This is really differential geometry, not algebraic geometry (although it is true in the algebraic setting as well, when the proof is written appropriately). Your $r$ functions give a map $f:(\mathbb{C}^*)^n \to \mathbb{C}^r$. The variety $X$ is $f^{-1}(0)$, and the condition on the matrix says that $0$ is a regular value of $f$. So $X$ is a smooth manifold. Near any point $p\in X$, $X$ is the transverse intersection of the $r$ hypersurfaces $f_1=f_2=\cdots f_r = 0$, and these hypersurfaces are smooth at $p$.

Comment by Jack Huizenga

2013-04-28T02:54:22Z

The tangent cone is a local intrinsic construction of an abstract variety, and doesn't "see" any embedding into another space. Subvarieties of weighted projective spaces still have projectivized tangent cones, which are subschemes of the projectivized Zariski tangent space (which is an ordinary projective space).

Comment by Jack Huizenga

2013-04-28T00:13:01Z

The answer to your first question is yes; this is just saying that the degree and dimension of the tangent cone is $\mu$ and $r-1$, respectively, which can be found in many books, e.g. Harris' "First Course" book. For your second question, I don't know what you would mean by a "weighted projective tangent cone." The ring $\oplus m^i/m^{i+1}$ is canonically $\mathbb{Z}$-graded.

Comment by Jack Huizenga

2013-04-25T22:13:18Z

Can you give an example of a special ring for which this fails? (Even in the dimension 2 case might be useful to see what goes wrong)

Comment by Jack Huizenga

2013-04-25T04:28:41Z

There would have to be some relation between $f$ and $\sigma$. As it stands, if $f$ is nonzero and $\sigma$ is zero there is no hope.

Comment by Jack Huizenga

2013-04-24T01:25:55Z

There's no reason to believe there should be such an expression.

Comment by Jack Huizenga

2013-04-17T05:11:10Z

Not even close. The wikipedia article on Fano varieties answers this very simply.

Comment by Jack Huizenga

2013-04-16T08:25:43Z

I of course agree with your opinion here Dmitry. Once you've been working in the subject for a while you start to see why some of the nitpicky details are important, but it's certainly possible to go a long way without, say, ever just studying Hartshorne for 2 years. And at least for certain people, it's a lot more interesting to start doing geometry than to bash your head against concepts like flatness (which I'm not discrediting--it's extremely important, but at least in nice cases more intuitive than Hartshorne might have you believe) before having any idea what it might be good for.

Comment by Jack Huizenga

2013-04-13T04:49:52Z

Are you basically just asking for generalizations of the Quillen-Suslin theorem to other rings?

Comment by Jack Huizenga

2013-03-08T03:17:02Z

@Allen: Fair enough. I suppose I saw the tag algebraic-geometry and responded hastily.

Comment by Jack Huizenga

2013-03-08T02:49:11Z

Your statement that $T_X|_S = T_S \oplus N_{S/X}$ is not true. There is an exact sequence relating these things, but it generally does not split.