User david steinberg - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T04:56:02Z http://mathoverflow.net/feeds/user/1231 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105621/moduli-spaces-of-vector-bundles-and-stability-conditions/105625#105625 Answer by David Steinberg for Moduli spaces of vector bundles and stability conditions David Steinberg 2012-08-27T13:37:43Z 2012-08-27T13:37:43Z <p>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. </p> <p>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:</p> <p><a href="http://arxiv.org/abs/alg-geom/9405004%20%22Geometric%20Invariant%20Theory%20and%20Flips%22" rel="nofollow">Geometric Invariant Theory and Flips</a> by Michael Thaddeus</p> http://mathoverflow.net/questions/101644/fiction-books-about-mathematicians/102703#102703 Answer by David Steinberg for Fiction books about mathematicians? David Steinberg 2012-07-19T23:18:32Z 2012-07-19T23:18:32Z <p>Written about mathematicians, by mathematicians, and certainly for mathematicians, the self-published "choose your own adventure," <a href="http://weirdcanada.com/2012/02/ex-libris-mathematics-odyssey-kent-windermere/" rel="nofollow">Mathematics Odyssey</a>, certainly deserves mention. </p> http://mathoverflow.net/questions/88037/reference-request-an-algebraic-stack-whose-closed-points-have-no-automorphisms-i Reference request: an algebraic stack whose closed points have no automorphisms is an algebraic space David Steinberg 2012-02-09T21:16:52Z 2012-03-06T00:42:57Z <p>The question is stated in the title. I think BCnrd states in a comment here</p> <p><a href="http://mathoverflow.net/questions/41833/is-every-artin-dm-algebraic-stack-fibered-in-sets-an-algebraic-space" rel="nofollow">http://mathoverflow.net/questions/41833/is-every-artin-dm-algebraic-stack-fibered-in-sets-an-algebraic-space</a></p> <p>that while the answer is not found in Laumon &amp; 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.)</p> http://mathoverflow.net/questions/86683/vanishing-associated-to-a-resolution-of-singularities Vanishing associated to a resolution of singularities David Steinberg 2012-01-26T01:41:24Z 2012-02-16T20:49:27Z <p>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 $i>0$. </p> <blockquote> <p>Can we conclude that $R^i\pi_*F(-E)=0$ for $i>0$?</p> </blockquote> <p>The idea being to somehow use that$-E$ is nef on $E$.</p> <p>EDIT: the exceptional fibres of $\pi$ are one-dimensional and $R^{\bullet} \pi_*O_V = O_W$ </p> <p>EDIT 2: we may assume that the support of $F$ is one-dimensional</p> http://mathoverflow.net/questions/88372/is-the-bounded-derived-category-of-coherent-sheaves-of-a-variety-a-small-category Is the bounded derived category of coherent sheaves of a variety a small category? David Steinberg 2012-02-13T20:13:26Z 2012-02-14T02:09:09Z <p>The question is in the title. </p> <p>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. </p> http://mathoverflow.net/questions/69132/reference-for-openness-of-stable-locus-of-holomorphic-family-of-vector-bundles-on/69133#69133 Answer by David Steinberg for Reference for openness of stable locus of holomorphic family of vector bundles on a compact riemann surface. David Steinberg 2011-06-29T18:25:45Z 2011-06-29T18:25:45Z <p>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.</p> http://mathoverflow.net/questions/58288/classification-of-fat-projective-lines Classification of fat projective lines? David Steinberg 2011-03-12T21:25:32Z 2011-03-12T23:21:05Z <p>In section III.3.4 of Eisenbud &amp; 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):</p> <p>Let $d$ be a non-negative integer, and let $S = \mathbb{C}[u,v,x,y]/(x^2, xy, y^2, u^dx - v^dy)$. </p> <p>Then $X_d = Proj$ $S$ is a double line with arithmetic genus $-d$.</p> <p>Eisenbud and Harris then go on to say that "every projective double line of genus $-d$, with $d \geq 0$, is isomorphic to $X_d$."</p> <p>My question is: where can I find a proof of this statement?</p> <p>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 $C$?</p> http://mathoverflow.net/questions/7492/algebraic-varieties-which-are-topological-manifolds Algebraic varieties which are topological manifolds David Steinberg 2009-12-01T19:56:47Z 2010-10-09T17:43:47Z <p>Inspired by <a href="http://mathoverflow.net/questions/7439/algebraic-varieties-which-are-also-manifolds" rel="nofollow">this thread</a>, 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?</p> http://mathoverflow.net/questions/36691/is-the-space-of-solutions-to-the-queens-domination-problem-connected Is the space of solutions to the Queens Domination Problem connected? David Steinberg 2010-08-25T19:43:39Z 2010-08-26T17:05:19Z <p>A configuration of queens on an 8 by 8 chessboard (or n by n if you like) is a <em>queen domination</em> 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 <em>solution</em> to the Queens Domination Problem is a queen domination using the minimum number of queens.</p> <p>For the 8 by 8 chessboard, brute force has shown that 5 queens is the minimum number. More discussion about that <a href="http://mathoverflow.net/questions/30330/is-there-a-good-argument-for-why-you-cant-place-4-queens-which-cover-a-chessboar" rel="nofollow">here</a>. </p> <p>The adjacency relation on the set of solutions to the Queens Domination Problem is defined as follows: we say solutions are <em>adjacent</em> 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.</p> <blockquote> <p>How many equivalance classes are there in the equivalance relation generated by adjacency?</p> </blockquote> <p>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?</p> http://mathoverflow.net/questions/23607/are-the-arithmetic-genera-of-cohen-macaulay-curves-in-a-fixed-homology-class-boun Are the arithmetic genera of Cohen-Macaulay curves in a fixed homology class bounded? David Steinberg 2010-05-05T18:15:47Z 2010-07-24T01:03:04Z <p>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). </p> <blockquote> <p>I want to show that if you fix a curve class &beta; in H<sub>2</sub>(X), then there is a bounded interval which contains the genus of every CM curve whose homology class is &beta;.</p> <p>Do you know of a reference for this result?</p> </blockquote> <p>Motivation: I am trying to prove a certain class of sheaves forms a bounded family (namely, the collection of sheaves underlying <a href="http://arxiv.org/abs/0707.2348" rel="nofollow">stable pairs</a>). 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.</p> <p>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.</p> <p>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. </p> http://mathoverflow.net/questions/21492/stacks-in-the-zariski-topology Stacks in the Zariski topology? David Steinberg 2010-04-15T18:59:17Z 2010-04-16T10:00:31Z <p>I have two naive questions about stacks.</p> <blockquote> <p>1) Is it possible to define stacks in the Zariski topology? </p> </blockquote> <p>Presuming you can:</p> <blockquote> <p>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?</p> </blockquote> <p>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.</p> http://mathoverflow.net/questions/241/why-and-how-are-moduli-spaces-of-semistable-vector-bundles-well-behaved/20649#20649 Answer by David Steinberg for Why and how are moduli spaces of (semi)stable vector bundles well-behaved? David Steinberg 2010-04-07T17:42:11Z 2010-04-07T17:42:11Z <p>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.</p> <p>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<sub>X</sub>. So boundedness of the family of structure sheaves is equivalent to fixing the Hilbert polynomial.</p> http://mathoverflow.net/questions/19215/ambiguous-definition-of-nerve-of-an-open-covering-on-wikipedia/19216#19216 Answer by David Steinberg for Ambiguous definition of "nerve of an open covering" on wikipedia? David Steinberg 2010-03-24T17:42:09Z 2010-03-24T17:42:09Z <p>I think the second construction is not correct. If you replace the cover with the category whose objects are all <b>intersections of elements</b> of your original cover, then the two notions agree. </p> http://mathoverflow.net/questions/18271/what-out-of-print-books-would-you-like-to-see-re-printed/18296#18296 Answer by David Steinberg for What out-of-print books would you like to see re-printed? David Steinberg 2010-03-15T18:46:46Z 2010-03-15T18:46:46Z <p>"The Geometry of Moduli Spaces of Sheaves" by Huybrechts and Lehn. Thankfully, it seems that an updated edition is in the works.</p> http://mathoverflow.net/questions/16332/mnop-conjecture MNOP conjecture David Steinberg 2010-02-25T00:11:30Z 2010-02-25T00:11:30Z <p>Let $X$ be a smooth, projective, Calabi-Yau 3-fold (CY makes the exposition more elegant, I don't think it is necessary). </p> <p>To define Gromov-Witten invariants, we consider moduli spaces of <a href="http://en.wikipedia.org/wiki/Stable_map" rel="nofollow">stable maps</a>, <code>$\bar{M}_g(X, \beta),$</code> 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 <code>$[\bar{M}_g(X, \beta)]^{vir},$</code> and the CY condition ensures that these are 0-cycles. The GW invariants are then defined to be</p> <p>$$GW_{g, \beta} = \int_{[\bar{M}_g(X, \beta)]^{vir}} 1.$$ This number (as far as I understand) is a virtual count of the ($g$, $\beta$)-curves on $X$.</p> <p>Similarly (here is where we really confined to 3-folds), to define Donaldson-Thomas invariants, we consider the moduli space of closed subschemes ---the <a href="http://en.wikipedia.org/wiki/Hilbert_scheme" rel="nofollow">Hilbert scheme</a> --- 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</p> <p>$$DT_{n, \beta} = \int_{[I_n(X, \beta)]^{vir}} 1.$$ This number is a virtual count of subschemes of dimension $\leq 1$ of $X$. </p> <p>In the paper <a href="http://arxiv.org/abs/math/0312059" rel="nofollow">Gromov-Witten theory and Donaldson-Thomas theory, I</a> by Maulik, Nekrasov, Okounkov, and Pandharipande, these invariants were assembled into generating series:</p> <p>$$Z_{DT}(q,v) = \Sigma_{\beta}\Sigma_n DT_{n, \beta}q^nv^{\beta} = \Sigma_{\beta}Z_{DT, \beta}(q)v^{\beta}$$ $$Z_{GW}(u,v)= 1+ exp( \Sigma_{\beta \neq 0} \Sigma_{g \geq 0} GW_{g, \beta}u^{2g-2}v^{\beta}) = 1+ \Sigma_{\beta \neq 0} Z_{GW, \beta}(u)v^{\beta}.$$ </p> <p>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:</p> <p>$$\frac{Z_{DT, \beta}(q)}{Z_{DT, 0}(q)} = Z_{GW, \beta}(u),$$ after the change of variables , $e^{iu} = -q$.</p> <p>My question: </p> <blockquote> <p>Can anyone explain why this change of variables is expected to work?</p> </blockquote> <p>(You know, besides the fact that it has been proven in the toric case.)</p> http://mathoverflow.net/questions/2124/good-introductory-references-on-algebraic-stacks/16247#16247 Answer by David Steinberg for Good introductory references on algebraic stacks? David Steinberg 2010-02-24T08:52:19Z 2010-02-24T08:52:19Z <p>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.</p> <p><a href="http://www.math.ubc.ca/~behrend/cet.ps" rel="nofollow">http://www.math.ubc.ca/~behrend/cet.ps</a></p> http://mathoverflow.net/questions/12765/algebraic-stacks-from-scratch/16217#16217 Answer by David Steinberg for Algebraic stacks from scratch David Steinberg 2010-02-24T01:37:34Z 2010-02-24T01:37:34Z <p>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.</p> <p><a href="http://www.math.ubc.ca/~behrend/cet.ps" rel="nofollow">http://www.math.ubc.ca/~behrend/cet.ps</a></p> http://mathoverflow.net/questions/8741/justifying-a-theory-by-a-seemingly-unrelated-example/8803#8803 Answer by David Steinberg for Justifying a theory by a seemingly unrelated example David Steinberg 2009-12-13T22:15:16Z 2009-12-13T22:15:16Z <p>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.</p> http://mathoverflow.net/questions/7689/why-is-riemann-roch-an-index-problem/7690#7690 Answer by David Steinberg for Why is Riemann-Roch an Index Problem? David Steinberg 2009-12-03T17:45:19Z 2009-12-03T17:45:19Z <p>I know nothing about the Atiyah-Singer index theorem, but <a href="http://en.wikipedia.org/wiki/Index_theory#The_Euler_characteristic" rel="nofollow">wikipedia</a> 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 $D = \bar{\partial} + \bar{\partial}^*$)</p> http://mathoverflow.net/questions/6316/when-are-git-quotients-projective/6329#6329 Answer by David Steinberg for When are GIT quotients projective? David Steinberg 2009-11-20T22:49:02Z 2009-11-21T00:18:12Z <p>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. </p> <p>To learn more about moduli of quiver representations, I recommend King's paper "<a href="http://www.math.uni-bielefeld.de/~sek/sem/stability/king.pdf" rel="nofollow">Representations of finite dimensional algebras</a>."</p> http://mathoverflow.net/questions/2974/a-theory-of-generalized-donaldson-thomas-invariants-by-joyce-song "a theory of generalized Donaldson-Thomas invariants" by Joyce & Song David Steinberg 2009-10-28T01:25:52Z 2009-11-09T00:12:12Z <p>Is anyone else working through this paper : <a href="http://arxiv.org/abs/0810.5645" rel="nofollow">http://arxiv.org/abs/0810.5645</a> ? I am trying to verifying example 6.2 (m=2 for simplicity) using only the definitions, namely:</p> <p>J^{2\alpha}(\tau) = -1/4</p> <p>(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})</p> <p>I keep getting -3/4. Has anyone else attempted to make such a computation? Did you get the answer you are supposed to get?</p> http://mathoverflow.net/questions/2983/definition-of-longest-common-subsequences/2988#2988 Answer by David Steinberg for Definition of longest common subsequences David Steinberg 2009-10-28T03:06:30Z 2009-10-28T03:06:30Z <p>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.)</p> http://mathoverflow.net/questions/101644/fiction-books-about-mathematicians/102703#102703 Comment by David Steinberg David Steinberg 2012-10-04T20:49:19Z 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. http://mathoverflow.net/questions/105621/moduli-spaces-of-vector-bundles-and-stability-conditions/105625#105625 Comment by David Steinberg David Steinberg 2012-08-27T13:44:03Z 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. http://mathoverflow.net/questions/88037/reference-request-an-algebraic-stack-whose-closed-points-have-no-automorphisms-i Comment by David Steinberg David Steinberg 2012-02-23T23:27:12Z 2012-02-23T23:27:12Z Over $\mathbb{C}$, are closed points different from geometric points? http://mathoverflow.net/questions/88372/is-the-bounded-derived-category-of-coherent-sheaves-of-a-variety-a-small-category Comment by David Steinberg David Steinberg 2012-02-14T19:02:50Z 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. http://mathoverflow.net/questions/88372/is-the-bounded-derived-category-of-coherent-sheaves-of-a-variety-a-small-category/88394#88394 Comment by David Steinberg David Steinberg 2012-02-14T03:22:31Z 2012-02-14T03:22:31Z Thank you. Do you know of a reference for the proof? http://mathoverflow.net/questions/88372/is-the-bounded-derived-category-of-coherent-sheaves-of-a-variety-a-small-category Comment by David Steinberg David Steinberg 2012-02-14T03:20:09Z 2012-02-14T03:20:09Z @Qiaochu: yes, thank you: the term I want is essentially small. http://mathoverflow.net/questions/86683/vanishing-associated-to-a-resolution-of-singularities/86725#86725 Comment by David Steinberg David Steinberg 2012-01-26T21:51:45Z 2012-01-26T21:51:45Z Thank you, and yes, the reduced exceptional is what I have in mind. http://mathoverflow.net/questions/86683/vanishing-associated-to-a-resolution-of-singularities/86725#86725 Comment by David Steinberg David Steinberg 2012-01-26T18:48:20Z 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 $R^i\pi_*O_E= 0 i&gt;0$ http://mathoverflow.net/questions/69132/reference-for-openness-of-stable-locus-of-holomorphic-family-of-vector-bundles-on/69133#69133 Comment by David Steinberg David Steinberg 2011-06-29T18:26:43Z 2011-06-29T18:26:43Z link: <a href="http://www.math.uni-bonn.de/people/huybrech/moduli.ps" rel="nofollow">math.uni-bonn.de/people/huybrech/moduli.ps</a> http://mathoverflow.net/questions/63921/showing-openness-on-quotients Comment by David Steinberg David Steinberg 2011-05-04T17:18:48Z 2011-05-04T17:18:48Z Don't forget: [0,1/3) is open in [0,1] http://mathoverflow.net/questions/63680/introduction-to-am-i-an-idiot-this-question-is-about-general-math-studyin Comment by David Steinberg David Steinberg 2011-05-02T06:19:12Z 2011-05-02T06:19:12Z When you grieve over taking &quot;three full days to do the problems,&quot; you have over-looked the fact that <i>you were able to do them</i>. Try to be patient, and remember that &quot;introduction&quot; is not a synonym for &quot;easy.&quot; http://mathoverflow.net/questions/56082/vanishing-cycles-in-a-nutshell/56096#56096 Comment by David Steinberg David Steinberg 2011-02-21T08:59:32Z 2011-02-21T08:59:32Z How can I vote this up more than once? http://mathoverflow.net/questions/45953/is-every-real-n-manifold-isomorphic-to-a-quotient-of-mathbbrn Comment by David Steinberg David Steinberg 2010-11-13T21:14:46Z 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. http://mathoverflow.net/questions/7492/algebraic-varieties-which-are-topological-manifolds/7500#7500 Comment by David Steinberg David Steinberg 2010-08-26T18:44:07Z 2010-08-26T18:44:07Z Why is the hypersurface homeomorphic to R&lt;sup&gt;8&lt;/sup&gt;? http://mathoverflow.net/questions/36691/is-the-space-of-solutions-to-the-queens-domination-problem-connected/36703#36703 Comment by David Steinberg David Steinberg 2010-08-26T00:26:00Z 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.)