User marting - MathOverflow

Answer by MartinG for Families of ideal sheaves: What's the correct definition?

2012-09-18T17:04:12Z

[This is now an answer to the edited question(s), with some details added. My answer to the original question is kept at the very end.]

Firstly: The question is a good one, and it is not easy to find references on this. I had spent too much time pondering about the failure of the double dual argument (see below) before I finally heard the arguement given in the last section below, indirectly from Fantechi, via Faber.

Assume $X$ is smooth projective.

Definition: An $S$-valued point in $M_I(X)$ is an $S$-flat coherent sheaf on $S\times X$, with stable fibres of rank one, and with determinant line bundle isomorphic to $\mathcal{O}_{S\times X}$, modulo isomorphism.

(I do not know if this is what Bridgeland meant, but to me this is resonably standard.)

Comment: Stability for rank one means torsion free.

Existence: Let $M(X)$ be the (Simpson) moduli space for stable rank one sheaves. Then $M_I(X)$ is the fibre over $\mathcal{O}_X$ for the determinant map $M(X) \to \mathrm{Pic}(X)$. This map sends a sheaf $I$ (stable rank one fibres) on $S\times X$ to the determinant line bundle $\det(I)$, and it is trivial as a point in $\mathrm{Pic}(X)$ if it is of the form $p^*L$ with $L\in\mathrm{Pic}(S)$. Then $I\otimes p^*L^{-1}$ is equivalent to $I$ in $M(X)(S)$, and it has trivial determinant. This shows that $M_I(X)$ indeed is a fibre of the determinant map.

Of course the determinant of an ideal $I_Y\subset \mathcal{O}_X$ is nontrivial if $Y$ is a non principal divisor, so you cannot map such ideals to $M_I(X)$. In any case, the ideal of a divisor, without the embedding, would only remember the linear equivalence class.

For brevity, let $\mathrm{Hilb}(X)$ be the part of the Hilbert scheme parametrizing subschemes $Y\subset X$ of codimension at least $2$. Then there is a natural map $F: \mathrm{Hilb}(X) \to M(X)$ that sends an ideal $I_Y\subset\mathcal{O}_{S\times X}$ to $I_Y$, forgetting the embedding. Since $Y$ is flat, so is $I_Y$, and its fibres are torsion free (by flatness again) of rank one. By the codimension assumption, the determinant of $I$ is trivial.

Theorem: $F$ is an isomorphism.

Comment: In the literature one sometimes finds the argument that if $I$ is a rank one torsion free sheaf with trivial determinant, then $I$ embeds into its double dual, which coincides with its determinant $\mathcal{O}_X$. This establishes bijectivity on points. (For Hilbert schemes of points on surfaces this is enough to conclude, since you can check independently that both $\mathrm{Hilb}(X)$ and $M_I(X)$ are smooth, and that the induced map on tangent spaces is an isomorphism.) I do not know how to make sense of this argument in families.

Sketch proof of theorem: The essential point is to show that every $I$ in $M_I(X)(S)$ has a canonical embedding into $\mathcal{O}_{S\times X}$ such that the quotient is $S$-flat.

Let $U\subset S\times X$ be the open subset where $I$ is locally free. Its complement has codimension at least $2$ in all fibres. By the trivial determinant assumption, the restriction of $I$ to $U$ is trivial. By codimension $2$, the trivialization extends to a map $I\to \mathcal{O}_{S\times X}$. This map is injective, in fact injective in all fibres: The restriction to each fibre ${s}\times X$ is nonzero (as $U$ intersects all fibres) and hence an embedding ($I$ is torsion free in fibres). It follows that the quotient is flat. There are some details to check, but this is the main point, I think.

[End of new answer, here is the original one:]

If we attempt to define $M_I(X)(S)$ as the set of $S$-flat ideals $I_Z$ in $\mathcal{O}_{S\times X}$, then that would not be functorial in $S$, as the inclusion $I_Z \subset \mathcal{O}_{S\times X}$ may not continue to be injective after base change (in the counter example in the other answer, restriction to the problematic fibre gives the zero map). We could impose "universal injectivity", but that is just another way of requiring the quotient $\mathcal{O}_Z$ to be $S$-flat, so then we have (re)defined the Hilbert scheme.

Another common way of defining moduli of ideals is as the moduli space for rank one stable sheaves (i.e. torsion free) with trivial determinant line bundle. The resulting moduli space is isomorphic to the Hilbert scheme of subschemes of codimension at least 2.

Answer by MartinG for Extending a polynomial function from an open subset

2011-04-06T06:30:31Z

I was in fact pondering the same question just yesterday, and ended up with this: there is a thickening $Z'\subset X$ of $\pi^{-1}Z$ such that every regular function vanishing on $Z'$ comes from $Y$.

Proof is straight forward, but anyway: in terms of rings, let $\phi\colon A \subset B$ be a finite extension of rings, with $B$ generated by finitely many $x_i$ as an $A$-module, let $I\subset A$ be an ideal and suppose $A_f \cong B_f$ for all $f\in I$. Let $f\in I$. For each $i$ we can then write $\phi(a_i) = x_i \phi(f)^{n_i}$ for some $a_i\in A$. Choose $n > n_i$ for all $i$ and let $I'\subset A$ be generated by all $f^n$ for $f$ running through a generator set of $I$ (with $n$ varying with $f$). Now $I'B$ defines $Z'$: its elements are $A$-linear combinations of $\phi(f^n)x_i =\phi(f^{n-n_i})\phi(a_i)$, which is in $I$.

(Still interested if there is a better statement. Also I was still looking for an example where the thickening was needed, so thanks, Tom!)

Answer by MartinG for When do stability and semistability coincide?

2011-03-04T14:13:40Z

If the rank and degree (wrt $H$) are coprime, then $\mu$-stability and $\mu$-semistability coincide. The argument is the same as for Gieseker-stability, but simpler.

Answer by MartinG for Endomorphisms of bundles associated to codimension 2 subvarieties

2010-08-10T11:23:19Z

This might be obvious, but one can sometimes argue via stability:

Notation as in Sasha's answer, with $L=\mathcal{O}(-n)$ and $n>0$. Then if you know that no curve of degree $\le n/2$ contains $Z$, then $E$ is stable (a destabilizing line bundle $\mathcal{O}(-m)\subset E$ has $-m \ge -n/2$).

And then of course $\Gamma(\mathbb{P}^2, \mathrm{End}(E))=k$.

Answer by MartinG for What in the meaning of "schematic support"?

2010-08-05T15:06:13Z

Assume $\mathcal{F}$ is coherent on $X$. To complement the answers you have already got, there are two useful definitions of its schematic support: either use the annihilator ideal or the Fitting ideal. They have the same underlying reduced scheme, but in general different scheme structures. The annihilator is usually (always, I think) meant if nothing else is said.

The annihilator support $Z = V(\mathrm{Ann}(\mathcal{F}))$ can be viewed as the minimal closed subscheme $i\colon Z \subset X$ such that the natural map $\mathcal{F} \to i_*i^*\mathcal{F}$ is an isomorphism (more or less a tautology).

The Fitting ideal (locally the maximal minors of a free presentation) gives a subscheme that contains $V(\mathrm{Ann}(\mathcal{F}))$, but with a possibly thicker scheme structure. A feature is that it is compatible with pullback; the annihilator construction is not.

Example: a rank $r$ vector bundle $i_*E$ on a divisor $i\colon D\subset X$ has $D$ as annihilator support, and $rD$ as Fitting support.

Answer by MartinG for Characterisation of coherent sheaves on an algebraic variety

2010-05-20T18:19:05Z

This is false. Every sheaf in that class would have zero first Chern class, since $c_1$ is additive over short exact sequences.

Answer by MartinG for Group and Hopf Algebra Structures for Projective Varieties

2010-05-06T11:17:27Z

In the projective case, i.e. abelian varieties, this question is addressed in Mumford's On the equations defining abelian varieties. I, Invent. math. 1, 287--354, 1966. In §3, "the addition formula", the aim is to describe the group law, given a projective embedding, explicitly.

The starting point is exactly that the homogeneous coordinate ring is not a Hopf algebra, because a line bundle $L$ (defining the embedding) on $X$ does not pull back to $p_1^*L\otimes p_2^*L$ under the group law $X\times X\to X$. (This is what one would need to have induced comultiplication maps $H^0(X,L^n) \to H^0(X\times X, p_1^*L^n\otimes p_2^*L^n)\cong H^0(X, L^n)\otimes H^0(X, L^n)$.)

Instead Mumford finds that one can describe $X\times X\to X\times X$, $(x,y)\mapsto (x+y,x-y)$ explicitly in terms of the homogeneous coordinate ring.

Answer by MartinG for Are any two K3 surfaces over C diffeomorphic?

2010-04-21T12:06:30Z

I find this reference quite readable:

Le Potier: "Simple connexité des surfaces K3", in Asterisque 126, 1985.

I haven't read Kodaira's paper (in Joel's answer), so I don't know whether it is the same argument, but Le Potier also deforms to quartic surfaces.

Answer by MartinG for Reference request: is the punctual Hilbert scheme irreducible?

2010-04-08T09:56:19Z

(I cannot post comments, regard this as one.)

It was correctly suggested in the comments to damiano's answer that the case of dimension d=2 was settled in the 70s, here are references:

I believe the first proof is that of Briancon [Description de $Hilb^n(C\{x,y\})$ , Invent. Math. 41 (1977)]
Ellingsrud and Strømme [On the homology of the Hilbert scheme of points in the plane, Invent. Math. 87 (1987)] have an argument involving a cell decomposition, where there is only one cell of the expected dimension (by a separate argument, any extra component cannot have smaller dimension)
Perhaps the state of the art is the inductive argument by Ellingsrud and Lehn [Irreducibility of the punctual quotient scheme of a surface, Ark. Mat. 37 (1999)], building on previous work by Ellingsrud and Strømme.

Answer by MartinG for Union of closed subschemes with the structure sheaf over it

2010-03-22T13:18:53Z

Emerton explained well that $V(I\cap J)$ is the natural scheme structure; I'd just like to add that if $V(I)$ and $V(J)$ are divisors, it is also reasonable to use the scheme structure $V(IJ)$ on their union, i.e. their sum as divisors. So both versions have their merits.

Comment by MartinG

2012-09-19T12:34:57Z

I just had a look at the paper: Maybe this is also stated more explicitly elsewhere, but I think at least the proof of Lemma 6.3 reveals that an ideal sheaf means a torsion free sheaf for which there exists a nonzero map to $\mathcal{O}_X$, that for families, this condition is just imposed fibrewise (as Sasha suggested) and moreover that this is a closed condition. Anyway, I am leaving my answer as it is, covering the smooth case. (And I still have no idea whether the natural map could fail to be an iso in the singular case.)

Comment by MartinG

2012-09-19T10:08:19Z

@Sasha: You are right. I did point out that I didn't know about the singular case (the OP seemed interested in the smooth case also), but it is worthwhile to note that the very definition of $M_I(X)$ via determinants applies only in the smooth case. Thanks.

Comment by MartinG

2012-09-19T08:34:44Z

@Sasha: This viewpoint makes sense, except that for the embedding of $I$ into $\mathcal{O}_X$ to be unique, you need to assume codimension at least 2. This is what 36min complained about with maximal ideals in $\mathbb{Z}$. So we have to restrict to codimension at least 2 to get a bijection. (What I wrote in my answer, then, is basically that this $M_I(X)$ can be defined without reference to ideals, and that the morphism from the Hilbert scheme is in fact an isomorphism, although that is not entirely obvious and probably independent from what Bridgeland is doing, as you say.)

Comment by MartinG

2012-09-19T06:42:32Z

@36min: And to answer your new question 2: With this definition, you can realize M_I as the fibre over $\mathcal{O}_X$ for the determinant map $M\to \mathrm{Pic}(X)$, where $M$ is the Simpson moduli space for stable/torsion free rank one sheaves.

Comment by MartinG

2012-09-19T06:38:26Z

@36min: Yes, I meant to say that moduli of rank one sheaves with trivial determinant is more or less the standard definition. But, on autopilot, I assumed $X$ to be smooth, I guess Bridgeland does not. If singularitites are allowed, I do not know whether the natural map from the Hilbert scheme is an isomorphism. (And even for $X$ smooth it is not entirely obvious.)

Comment by MartinG

2012-09-18T18:42:09Z

@temp: No, $M_I(X)$ does not exist: it is not a functor.

Comment by MartinG

2012-05-25T10:57:03Z

The cotangent complex has nothing to with the cotangent function or complex numbers..

Comment by MartinG

2011-05-02T20:42:08Z

Indeed I have.. I deleted the answer referred to.

Comment by MartinG

2010-08-17T08:09:31Z

No genericity condition is needed: any line intersects the cubic in an effective degree 3 divisor summing to zero, and conversely. So the result is isomorphic to the projective plane (it is also the linear system |0+0+0|, where 0 means the group identity).

Comment by MartinG

2010-07-29T09:22:10Z

A class of counter examples: flat implies open, but smoothness is stronger than flatness. In particular one can take any ramified, flat and finite morphism, such as the projection of a plane curve to a line (project away from a point not on the curve).

Comment by MartinG

2010-05-20T18:21:43Z

It isn't silly, after all it was claimed at Wikipedia, and you were skeptical..

Comment by MartinG

2010-05-20T14:25:59Z

Wouldn't the first Chern class of every sheaf in that class be zero, since $c_1$ is additive over short exact sequences?