User david steinberg - MathOverflow

Answer by David Steinberg for Moduli spaces of vector bundles and stability conditions

2012-08-27T13:37:43Z

Without going over all of the details before I eat breakfast, I will go out on a limb and say the following: for $a$, $b$ close to 1, you will get an identical moduli space for the reason that stability is an open condition. In other words, you can change a GIT stability condition a little without changing which sheaves are stable.

In general, this kind of change to the stability condition goes under the heading of a "variation of GIT quotient." I think the relevant paper to look at is:

Geometric Invariant Theory and Flips by Michael Thaddeus

Answer by David Steinberg for Fiction books about mathematicians?

2012-07-19T23:18:32Z

Written about mathematicians, by mathematicians, and certainly for mathematicians, the self-published "choose your own adventure," Mathematics Odyssey, certainly deserves mention.

Reference request: an algebraic stack whose closed points have no automorphisms is an algebraic space

2012-02-09T21:16:52Z

The question is stated in the title. I think BCnrd states in a comment here

http://mathoverflow.net/questions/41833/is-every-artin-dm-algebraic-stack-fibered-in-sets-an-algebraic-space

that while the answer is not found in Laumon & Moret-Bailly, but that it is nevertheless true. Does anyone know of a reference? (Also: I only care about stacks over $\mathbb{C}$, if that makes a difference.)

Vanishing associated to a resolution of singularities

2012-01-26T01:41:24Z

Let $\pi: V\to W$ be a resolution of singularities, let $E \subset V$ be the exceptional divisor, and let $F$ be a coherent sheaf such that $R^i\pi_*F=0$ for $i>0$.

Can we conclude that $R^i\pi_*F(-E)=0$ for $i>0$?

The idea being to somehow use that$-E$ is nef on $E$.

EDIT: the exceptional fibres of $\pi$ are one-dimensional and $R^{\bullet} \pi_*O_V = O_W$

EDIT 2: we may assume that the support of $F$ is one-dimensional

Is the bounded derived category of coherent sheaves of a variety a small category?

2012-02-13T20:13:26Z

The question is in the title.

I am trying to apply the Mitchell (Freyd-Mitchell?) embedding theorem, which states that for every small abelian category $A$, there exists a ring $R$ such that A embeds into the category $R$-mod. The derived category is not abelian, of course, but I have a particular subcategory that is abelian, and life would be easiest if the derived category was smal, so that the subcategory was small and abelian.

Answer by David Steinberg for Reference for openness of stable locus of holomorphic family of vector bundles on a compact riemann surface.

2011-06-29T18:25:45Z

Proposition 2.3.1 of "The Geometry of the Moduli Space of Sheaves" by Huybrechts and Lehn provides a proof of what you are looking for.

Classification of fat projective lines?

2011-03-12T21:25:32Z

In section III.3.4 of Eisenbud & Harris's "The Geometry of Schemes," we/they construct an infinite family of double structures on $\mathbb{P}^1 \subset \mathbb{P}^3$ that are distinguished from each other by their genus. Here is the construction (let's just worry about $\mathbb{C}$ for now):

Let $d$ be a non-negative integer, and let $S = \mathbb{C}[u,v,x,y]/(x^2, xy, y^2, u^dx - v^dy)$.

Then $X_d = Proj$ $S$ is a double line with arithmetic genus $-d$.

Eisenbud and Harris then go on to say that "every projective double line of genus $-d$, with $d \geq 0$, is isomorphic to $X_d$."

My question is: where can I find a proof of this statement?

More generally: if you fix a curve $C$ of genus $g$, how can I describe the moduli of genus $d$ double curves lying over top of $C$?

Algebraic varieties which are topological manifolds

2009-12-01T19:56:47Z

Inspired by this thread, which concludes that a non-singular variety over the complex numbers is naturally a smooth manifold, does anyone know conditions that imply that the topological space underlying a complex variety is a topological manifold without necessarily implying it is smooth?

Is the space of solutions to the Queens Domination Problem connected?

2010-08-25T19:43:39Z

A configuration of queens on an 8 by 8 chessboard (or n by n if you like) is a queen domination if every square on the board lies in the same row, column, or diagonal as at least one of the queens. The Queens Domination Problem is to find the minimum number of queens necessary for a queen domination. A solution to the Queens Domination Problem is a queen domination using the minimum number of queens.

For the 8 by 8 chessboard, brute force has shown that 5 queens is the minimum number. More discussion about that here.

The adjacency relation on the set of solutions to the Queens Domination Problem is defined as follows: we say solutions are adjacent if they differ by only the placement of one queen. For example: C3, E4, D5, B6, F4 is adjacent to C3, E4, D5, B6, F2.

How many equivalance classes are there in the equivalance relation generated by adjacency?

In other words, starting with one solution, can we reach any other solution by moving one queen at a time, such that the result of each move is itself a solution?

Are the arithmetic genera of Cohen-Macaulay curves in a fixed homology class bounded?

2010-05-05T18:15:47Z

Let X be a smooth projective variety over the complex numbers. Recall that a Cohen-Macaulay curve is a one-dimensional closed subscheme without embedded or isolated points (fat components are allowed).

I want to show that if you fix a curve class β in H₂(X), then there is a bounded interval which contains the genus of every CM curve whose homology class is β.

Do you know of a reference for this result?

Motivation: I am trying to prove a certain class of sheaves forms a bounded family (namely, the collection of sheaves underlying stable pairs). The above result will allow me to reduce to the case where the support is a fixed CM curve, and from there, I know how to finish the proof.

I am aware that Le Potier (who built the moduli space of stable pairs, among others things, in "System Coherents et Structures de Niveau" --- pardon my lack of accents) has proven a much stronger result. However, for my purposes, it would be very helpful to have a proof (hopefully much less technically demanding than Le Potier's) that is no more general than what I need.

EDIT: in hindsight, and in light of the negative answer below, I realize that Le Potier does not claim to prove quite what I claimed he claimed.

Stacks in the Zariski topology?

2010-04-15T18:59:17Z

I have two naive questions about stacks.

1) Is it possible to define stacks in the Zariski topology?

Presuming you can:

2) If a stack has a coarse moduli, and the coarse moduli space is a scheme, then does that mean that your stack is a stack in the Zariski topology?

In general, I am trying to understand why a new notion of open cover is necessary if all I am interested in is remembering stabilizers. Certainly this is too simple a mind-set, so feel free to enlighten me.

Answer by David Steinberg for Why and how are moduli spaces of (semi)stable vector bundles well-behaved?

2010-04-07T17:42:11Z

Another interpretation of "well-behaved" might be that the collection of semistable sheaves with a fixed Hilbert polynomial is a bounded family of sheaves. This is equivalent to saying that there is a coherent sheaf F with the property that all semistable sheaves with fixed Hilbert polynomial P can all be realized as a quotient of F. See 1.7 of Huybrechts and Lehn.

Compare with the Hilbert scheme: there, we only need to fix the Hilbert polynomial, but that is because all structure sheaves of subschemes come equipped with the structure of a quotient of O_X. So boundedness of the family of structure sheaves is equivalent to fixing the Hilbert polynomial.

Answer by David Steinberg for Ambiguous definition of "nerve of an open covering" on wikipedia?

2010-03-24T17:42:09Z

I think the second construction is not correct. If you replace the cover with the category whose objects are all intersections of elements of your original cover, then the two notions agree.

Answer by David Steinberg for What out-of-print books would you like to see re-printed?

2010-03-15T18:46:46Z

"The Geometry of Moduli Spaces of Sheaves" by Huybrechts and Lehn. Thankfully, it seems that an updated edition is in the works.

MNOP conjecture

2010-02-25T00:11:30Z

Let $X$ be a smooth, projective, Calabi-Yau 3-fold (CY makes the exposition more elegant, I don't think it is necessary).

To define Gromov-Witten invariants, we consider moduli spaces of stable maps, $\bar{M}_g(X, \beta),$ indexed by the genus $g$ of the source curve and curve class $\beta$ of its image in $X$ . These moduli spaces admit virtual fundamental cycles $[\bar{M}_g(X, \beta)]^{vir},$ and the CY condition ensures that these are 0-cycles. The GW invariants are then defined to be

$$GW_{g, \beta} = \int_{[\bar{M}_g(X, \beta)]^{vir}} 1. $$ This number (as far as I understand) is a virtual count of the ($g$, $\beta$)-curves on $X$.

Similarly (here is where we really confined to 3-folds), to define Donaldson-Thomas invariants, we consider the moduli space of closed subschemes ---the Hilbert scheme --- of $X$, $I_n(X, \beta)$, indexed by the holomorphic Euler characteristic $n$ and the homology class $\beta$ of the subscheme. When $\beta$ is a curve class, these moduli spaces also admit zero dimensional virtual fundamental cycles, and the DT invariants are defined to be

$$DT_{n, \beta} = \int_{[I_n(X, \beta)]^{vir}} 1. $$ This number is a virtual count of subschemes of dimension $\leq 1$ of $X$.

In the paper Gromov-Witten theory and Donaldson-Thomas theory, I by Maulik, Nekrasov, Okounkov, and Pandharipande, these invariants were assembled into generating series:

$$Z_{DT}(q,v) = \Sigma_{\beta}\Sigma_n DT_{n, \beta}q^nv^{\beta} = \Sigma_{\beta}Z_{DT, \beta}(q)v^{\beta}$$ $$Z_{GW}(u,v)= 1+ exp( \Sigma_{\beta \neq 0} \Sigma_{g \geq 0} GW_{g, \beta}u^{2g-2}v^{\beta}) = 1+ \Sigma_{\beta \neq 0} Z_{GW, \beta}(u)v^{\beta}.$$

Here, $v^{\beta}$ is short-hand for $v_1^{\beta_1}\cdot \ldots \cdot v_k^{\beta_k}$, where $\beta_1, \ldots, \beta_k$ is a positive basis of $H_2(X, \mathbb{Z})$ mod torsion. Then, they make the following conjecture:

$$\frac{Z_{DT, \beta}(q)}{Z_{DT, 0}(q)} = Z_{GW, \beta}(u),$$ after the change of variables , $e^{iu} = -q$.

My question:

Can anyone explain why this change of variables is expected to work?

(You know, besides the fact that it has been proven in the toric case.)

Answer by David Steinberg for Good introductory references on algebraic stacks?

2010-02-24T08:52:19Z

Linked below is a note written by Kai Behrend whose first section gives a concise introduction to stacks, building them directly out of (lax) functors from the category of affine schemes.

http://www.math.ubc.ca/~behrend/cet.ps

Answer by David Steinberg for Algebraic stacks from scratch

2010-02-24T01:37:34Z

Linked below is a note written by Kai Behrend whose first section gives a concise introduction to stacks, building them directly out of (lax) functors from CRing.

http://www.math.ubc.ca/~behrend/cet.ps

Answer by David Steinberg for Justifying a theory by a seemingly unrelated example

2009-12-13T22:15:16Z

One can use the machinery of the fundamental group and of covering spaces to easily prove that any subgroup of a free group must be free.

Answer by David Steinberg for Why is Riemann-Roch an Index Problem?

2009-12-03T17:45:19Z

I know nothing about the Atiyah-Singer index theorem, but wikipedia seems to have a pretty decent introduction. It even tells you the elliptic operator for the Riemann-Roch case. (So as not to leave you hanging, I'll just tell you: it's $D = \bar{\partial} + \bar{\partial}^*$)

Answer by David Steinberg for When are GIT quotients projective?

2009-11-20T22:49:02Z

If your desired moduli space can be interpreted as a moduli space of quiver representations, then you may be in luck. For moduli of $\theta$-(semi)stable quiver representations, there is a simple criterion equivalent to the stable and semistable loci being equal: the dimension vector $\alpha$ is indecomposable (ie, not a non-trivial sum of elements) in $\theta^{\perp}\cap \mathbb{Z}^{Q_0}_{\geq 0}$, where $Q_0$ is your set of verticies, and $\theta$ is the fixed stability parameter.

To learn more about moduli of quiver representations, I recommend King's paper "Representations of finite dimensional algebras."

"a theory of generalized Donaldson-Thomas invariants" by Joyce & Song

2009-10-28T01:25:52Z

Is anyone else working through this paper : http://arxiv.org/abs/0810.5645 ? I am trying to verifying example 6.2 (m=2 for simplicity) using only the definitions, namely:

J^{2\alpha}(\tau) = -1/4

(where \alpha satisfies M^{\alpha}_ss = M^{\alpha}_st and that Ext^1(E, E)=0 for any E \in M^{\alpha}_ss and finally that the only object of M^{m\alpha}_ss is E ^{\oplus m})

I keep getting -3/4. Has anyone else attempted to make such a computation? Did you get the answer you are supposed to get?

Answer by David Steinberg for Definition of longest common subsequences

2009-10-28T03:06:30Z

I think that a subsequence is defined by "the positions of its characters," the same way that a subgroup is defined as a particular injection of one group into another: two groups may be isomorphic ("same content"), but represent distinct subgroups ("different position") of a larger group. (Strictly speaking, a subgroup is an equivalence class of injections, but you know what I mean.)

Comment by David Steinberg

2012-10-04T20:49:19Z

Sadly, the authors received a cease-and-desist letter from the Choose Your Own Adventure people. It's a shame, the book is a real treat. Drop me a line if you are in Vancouver, and I will show you my copy.

Comment by David Steinberg

2012-08-27T13:44:03Z

If I wasn't out of my depths before, certainly I am now. I do not understand how GIT stability and Bridgeland stability compare to one another.

Comment by David Steinberg

2012-02-23T23:27:12Z

Over $\mathbb{C}$, are closed points different from geometric points?

Comment by David Steinberg

2012-02-14T19:02:50Z

The short answer is that epimorphisms of my abelian category are more easily understood if I can think of them as epimorphisms of modules.

Comment by David Steinberg

2012-02-14T03:22:31Z

Thank you. Do you know of a reference for the proof?

Comment by David Steinberg

2012-02-14T03:20:09Z

@Qiaochu: yes, thank you: the term I want is essentially small.

Comment by David Steinberg

2012-01-26T21:51:45Z

Thank you, and yes, the reduced exceptional is what I have in mind.

Comment by David Steinberg

2012-01-26T18:48:20Z

I should have mentioned before: the exceptional fibres are one-dimensional and $\pi$ has satisfies $R^{\bullet}\pi_*O_V=O_W$. I think think implies that $R^i\pi_*O_E= 0 i>0$

Comment by David Steinberg

2011-06-29T18:26:43Z

link: math.uni-bonn.de/people/huybrech/moduli.ps

Comment by David Steinberg

2011-05-04T17:18:48Z

Don't forget: [0,1/3) is open in [0,1]

Comment by David Steinberg

2011-05-02T06:19:12Z

When you grieve over taking "three full days to do the problems," you have over-looked the fact that you were able to do them. Try to be patient, and remember that "introduction" is not a synonym for "easy."

Comment by David Steinberg

2011-02-21T08:59:32Z

How can I vote this up more than once?

Comment by David Steinberg

2010-11-13T21:14:46Z

I have an idea for when your manifold M is a CW-complex. If you remove the (n-1)-skeleton of M, then you are left with an open n-ball; this suggests that your manifold M can be obtained as a topological quotient of a closed n-ball, which is certainly the topological quotient of R^n.

Comment by David Steinberg

2010-08-26T18:44:07Z

Why is the hypersurface homeomorphic to R<sup>8</sup>?

Comment by David Steinberg

2010-08-26T00:26:00Z

(I thought I was being clever by adding dollar signs to fill-up the 15 character minimum, but I forgot that you can't use latex in comments. Ho hum.)