User imeasy - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T17:21:47Z http://mathoverflow.net/feeds/user/4096 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99717/blow-ups-of-secant-varieties/130681#130681 Answer by IMeasy for blow-ups of secant varieties IMeasy 2013-05-15T09:00:19Z 2013-05-15T09:00:19Z <p>You can probaly describe your space as some kind of fibration in $\mathcal{M}_{0,m}$ over the dual linear system $\mathbb{P}^{n*}$, via Kapranov's blow-up construction of $\overline{\mathcal{M}_{0,m}}$ . This is done in</p> <p><a href="http://arxiv.org/abs/0903.5515" rel="nofollow">http://arxiv.org/abs/0903.5515</a></p> <p>for some of the Bertram's cases but it should hold in general. Your space may have an interpretation in terms of moduli of vector bunldes (or sheaves).</p> http://mathoverflow.net/questions/130068/regularity-of-finite-flat-branched-covers regularity of finite flat branched covers IMeasy 2013-05-08T09:27:27Z 2013-05-08T15:23:38Z <p>Let $D$ and $S$ be two regular schemes and let $D$ be a divisor of $S$. Let $C \to S$ be a finite flat morphism, branched along $D$. Is $C$ regular as well?</p> http://mathoverflow.net/questions/128823/canonical-model-of-a-reducible-curve canonical model of a reducible curve IMeasy 2013-04-26T12:48:52Z 2013-04-26T14:17:06Z <p>Let $C$ be a stable reducible curve. Is there a natural way to define it's canonical model (I guess via the dualizing sheaf)? And does somehow the dualizing sheaf restrict to the (probably twisted) canonical sheaf of the components?</p> <p>In particular: to what extent the restriction of the dualizing sheaf of the global curve to a component can be described as the dualizing sheaf (possibly twisted by O(p), with $p$ attaching point) of that component?</p> <p>There are, for instance, some examples that puzzle me. Take a smooth genus 3 curve, then its canonical model is a plane quartic. Then consider the curve given by a central elliptic curve attached to 2 elliptic "tails" (I am not sure this is the correct terminology). The "canonical model" of this should equally be a plane curve. What plane curve??</p> http://mathoverflow.net/questions/128476/state-of-the-art-for-kodaira-dimension-of-overline-mathcalm-g-n state of the art for kodaira dimension of $\overline{\mathcal{M}}_{g,n}$ IMeasy 2013-04-23T13:07:02Z 2013-04-23T13:07:02Z <p>What is the state of the art about the kodaira dimension (and rationality, unirationality, etc.) of the moduli spaces of $n$-pointed curves of genus $g$? When it is known and when not? It would be great to collect the references for all $g$ and $n$ but I don't want to look like being fishing for points... :)</p> http://mathoverflow.net/questions/127667/hyperelliptic-stable-genus-four-curve/127966#127966 Answer by IMeasy for hyperelliptic stable genus four curve IMeasy 2013-04-18T13:55:59Z 2013-04-18T13:55:59Z <p>All smooth genus four curves have two $g^1_3$. Since the unviersal picard fibration over the moduli space of stable genus four curves is proper, yes you can assume that the generic fiber is hyperelliptic.</p> http://mathoverflow.net/questions/127809/albanese-dual-to-the-picard-scheme/127963#127963 Answer by IMeasy for Albanese dual to the Picard scheme IMeasy 2013-04-18T13:33:22Z 2013-04-18T13:33:22Z <p>I like very much Mumford's "abelian varieties", which has the advantage to work over any field. enjoy!</p> http://mathoverflow.net/questions/127678/lifts-of-maps-to-mathcalm-1-1 lifts of maps to $\mathcal{M}_{1,1}$ IMeasy 2013-04-16T07:08:09Z 2013-04-18T13:31:25Z <p>Hi,</p> <p>here's there's a construction about elliptic curves that I do not completely understand. Suppose I consider the two following families of elliptic curves over $\mathbb{C}^*$. </p> <p>The first, which I denote $F_0$ is the trivial family $E \times \mathbb{C}^*$, where $E$ is an elliptic curve with only $-Id$ as non-trivial automorphism. I will denote by $F_1$ the quotient of $F_0$ via the diagonal action of $-Id$, actiong both on $E$ and $\mathbb{C}^*$.</p> <p>Remark that the quotient of the base is equal to $\mathbb{C}^*$ itself. Now $F_0$ has trivial monodromy around 0 and $F_1$ has not. The modular maps to $\mathcal{M}_{1,1}$ given by the two $F_i$ are constant, but there exist non-constant lifts depending on the description of the mod space.</p> <p>The two lifts to the Siegel half space $\mathcal{H}_1$ (seeing the mod space of elliptic curves as a quotient by $SL_2(\mathbb{Z})$) are constant. </p> <p>Let us now consider the quotient of $\mathbb{C}^2$ minus the discriminant (the Neil-parabola) by the torus $\mathbb{C}^*$ acting with weights $(4,6)$, i.e. let us parametrize the mod space with the coefficients $a$ and $b$ of the Weierstrass equation. The smooth curves are parmetrized by the open subspace $U\subset \mathbb{P}(4,6)$ complement of the point at infinity.</p> <p>With this presentation, any lift of $F_0$ to $\mathbb{C}^2$ is the constant map to $(a,b)$, but a nonconstant lift of $F_1$ exists, namely and the non-constant map defined by</p> <p>$$t \mapsto (t^2a, t^3 b).$$</p> <p>Can you explain me why this happens? Is it related to the monodromy of the family? Or the connectedness of the group acting?</p> http://mathoverflow.net/questions/127737/smooth-modular-compactification-of-moduli-of-curves/127741#127741 Answer by IMeasy for smooth modular compactification of moduli of curves IMeasy 2013-04-16T19:35:49Z 2013-04-16T19:35:49Z <p>$\mathbb{P}^3$ compactifies the moduli space of genus 2 curves with level 3 structure and the choice of an odd theta characteristic.</p> http://mathoverflow.net/questions/114344/13-months-and-not-even-one-report-what-would-you-do 13 months and not even one report. what would you do? IMeasy 2012-11-24T16:03:20Z 2013-04-12T08:46:55Z <p>I submitted a 24 pages paper to a good journal - say usually in the top 10-20 - of pure maths, and after 14 months from the submission I haven't received any report. The last news I had from the editor date last may. Then I tried to contact him in september but no answer. What would you do? Wait? Write again? Withdraw the paper?</p> <p>It is not a VERY TOP journal (top 5 for instance), hence I think a rejection would be pretty difficult to accept after such a long time.</p> http://mathoverflow.net/questions/114344/13-months-and-not-even-one-report-what-would-you-do/127333#127333 Answer by IMeasy for 13 months and not even one report. what would you do? IMeasy 2013-04-12T08:46:55Z 2013-04-12T08:46:55Z <p>Hey one month ago I finally received a pretty positive report, asking for some small changes! Persistence is the key ! :)</p> http://mathoverflow.net/questions/127238/universal-families-and-maps-to-quotient-stacks universal families and maps to quotient stacks IMeasy 2013-04-11T12:45:47Z 2013-04-12T00:08:54Z <p>Suppose I have a certain (contravariant) moduli functor $M:Schemes \to Groupoids$ that is represented by a quotient stack $[X//G]$ where $X$ is a scheme and $G$ a linearly reductive group. Roughly speaking $[X//G]$ is the fine moduli space for $M$; hence it carries a universal family and there's a correspondence between families of objects of $M$ over a scheme $S$ and maps $S\to [X//G]$. Now, I claim that also $X$ carries a universal family, obtained via pull-back from $[X//G]$, and it is constant along the fibers of the quotient map $X \to [X//G]$. </p> <p>If $S$ is once again a family of objects of $M$, what can one say of maps from $S\to X$? That the family over $S$ induces a map to $X$ that is unique up to $G$-action? That such a map is unique as long as $X\to [X//G]$ has a section? </p> <p>Basically, by the existence of the universal family over $X$ a map $S\to X$ induces a family over $S$ but not in a unique way. So, in the other direction, given a family over $S$ this induces an orbit under $G$ of maps, but there's no canonical choice of a particular map in this orbit.</p> http://mathoverflow.net/questions/125908/is-the-orthogonal-complement-of-a-saturated-sequence-saturated is the orthogonal complement of a saturated sequence saturated? IMeasy 2013-03-29T12:32:39Z 2013-03-29T13:07:19Z <p>Suppose I have a smooth projective variety $X$, and a semi-orthogonal decomposition of its bounded derived category:</p> <p>$$D^b(X)= &lt; A, E_1, E_2, ... , E_n >$$</p> <p>where the $E_i$ are fully faithful, saturated (and hence admissible) subcats of $D^b(X)$. Is A authomatically saturated/admissible? Why? If not, under what assuptions?</p> http://mathoverflow.net/questions/124842/hurwitzs-construction-of-simple-covers Hurwitz's construction of simple covers IMeasy 2013-03-18T07:47:13Z 2013-03-18T09:35:56Z <p>What is commonly meant by Hurwitz's construction of simple covers? </p> http://mathoverflow.net/questions/124057/one-big-hilbert-scheme one "big" Hilbert scheme? IMeasy 2013-03-09T09:35:28Z 2013-03-09T10:30:10Z <p>I was recently reading one of the papers of Kapranov on $\mathcal{M}_{0,n}$ and he says that one can see it as a subvariety of the Hilbert scheme parametrizing all subschemes of a certain projective space. What is the precise definition of this Hilbert scheme? Why it is sure it exists? Is it in EGA somewhere or does anybody have a reference?</p> http://mathoverflow.net/questions/123736/what-are-mutations-of-sheaves-all-about what are mutations of sheaves all about? IMeasy 2013-03-06T10:51:05Z 2013-03-06T12:53:55Z <p>Suppose I have a smooth projective variety $X$ and a semi-orthogonal decomposition of its bounded derived category of coherent sheaves $D^b(X)$. Then I can apply right or left mutations to the full asmissible triangulated subcats that make up the semi-orth decomposition. Algorithmically it is clear to me what happens, but what is the spirit of such a transformation? what is the geometry behind it? in particular I would be curious to understand this in the case of full exceptional sequences and when all subcats in the semi-orth decomposition are generated by sheaves, line or vector bundles.</p> http://mathoverflow.net/questions/121288/examples-of-moduli-functors-for-which-coarse-moduli-space-does-not-exists examples of moduli functors for which coarse moduli space does not exists IMeasy 2013-02-09T10:21:26Z 2013-02-09T22:29:38Z <p>Well, the title almost says it all. I would like to list as many examples as possible of moduli functors, for which a coarse moduli space does not exist (and maybe explain why). So, examples such as $[\mathbb{A}^1/G_m]$ are not what I mean. I would really like to see moduli functors that come from some classifying problem.</p> http://mathoverflow.net/questions/121315/projective-submanifolds-of-mathbb-cpn-whose-normals-bundles-are-sums-of-linea/121321#121321 Answer by IMeasy for Projective submanifolds of $\mathbb CP^n$ whose normals bundles are sums of linear. IMeasy 2013-02-09T15:42:31Z 2013-02-09T15:42:31Z <p>For sure if the variety is a complete intersection then the normal bundle splits. I am afraid that the other way round is not true. IMO it should imply just being locally complete intersection, and a priori there's no general condition that implies that a l.c.i. is a c.i., as far as I know</p> http://mathoverflow.net/questions/121094/stack-of-vector-bundles-on-a-curve-over-a-strictly-semi-stable-point-of-the-mod Stack of vector bundles (on a curve) over a strictly semi-stable point of the moduli space IMeasy 2013-02-07T17:25:26Z 2013-02-07T17:25:26Z <p>Consider the stack $Bun_{r,d}^{ss}$ of rank $r$ semi-stable vector bundle of degree $d$ over a fixed curve. There exists also a coarse moduli space $M$ built via GIT. Over the stable locus of $M$ it is well known that the natural morphism $Bun_{r,d} \to M$ is a gerbe. This is basically due to the fat that $Bun_{r,d}^{ss}$ is a $GL_n$ quotient stack of some Quot-scheme, and the moduli space below is the GIT quotient of the same scheme via $PGL_n$.</p> <p>But what happens over the strictly semistable locus (which is the singular locus of the GIT moduli space)? What's the structure of the fiber? It should be more complicated. </p> http://mathoverflow.net/questions/120976/does-there-exist-a-family-of-objects-over-the-tangent-space-to-the-base-space-of does there exist a family of objects over the tangent space to the base space of a family of objects? IMeasy 2013-02-06T14:38:50Z 2013-02-06T16:19:56Z <p>Suppose we have a family of objects $\Xi \to S$ over a base smooth projective scheme $S$. Take a closed point $p\in S$ and consider the tangent space to $S$ at $p$. Can one construct an "induced family" over this tangent space (or probably rather over its projectivized space) starting from $\Xi$ ? what interpretation could one give to this? a second-order deformation?</p> http://mathoverflow.net/questions/120635/on-morphisms-to-projective-space-arising-from-a-linear-system/120658#120658 Answer by IMeasy for On morphisms to projective space arising from a linear system IMeasy 2013-02-03T09:16:19Z 2013-02-03T09:16:19Z <p>$E$ can't intersect $C$ in a finite number of points because otherwise the restriction of $\phi$ to $C$ would be a finite degree morphism, which you assume is not. $E$ is the pull-back of the hyperplane divisor, as Vesselin remarked.</p> http://mathoverflow.net/questions/120604/classification-of-first-order-deformations-of-n-pointed-non-singular-variety/120657#120657 Answer by IMeasy for Classification of first order deformations of n-pointed non-singular variety IMeasy 2013-02-03T09:11:58Z 2013-02-03T09:11:58Z <p>Here's a "quick and dirty" answer, far from being precise. Take the case of curves that you mention: the idea is that , the more points that you add to your moduli problem, the more parameters you need. The dimension of the moduli space gets bigger. In fact, by serre duality your space is iso to $H^0(X,2K_X +p_1+p_2+ ... + p_k)$ and this more positive than the mere def space $H^0(X,2K_X)$ hence it is very likely to have more sections, i.e. more directions in which you can deform your pointed curve. Work out yourself the example of $X=P^1$: it is very instructive. You will see straight away that the rough dimension count of the moduli space corresponds to the dimension of the first order deformation space.</p> http://mathoverflow.net/questions/120270/when-does-a-g-invariant-one-to-one-map-between-two-closed-algebraic-g-set-descend/120312#120312 Answer by IMeasy for When does a G-invariant one to one map between two closed algebraic G-set descend to a one to one map on the G.I.T quotient ? IMeasy 2013-01-30T12:48:25Z 2013-01-31T07:48:14Z <p>as long as $f$ passes to the quotient (i.e. sends orbits on orbits) it has the required property. Moreover, as Misha remarked, the semistable locus of the first quotient should be sent to the semistable locus of the second quotient. This basically mean that the pullback of the invariants on the codomain must give the invariants of the domain.</p> http://mathoverflow.net/questions/120025/blow-up-of-segre-primal-and-mathcalm-0-6 blow up of segre primal and $\mathcal{M}_{0,6}$ IMeasy 2013-01-27T13:52:12Z 2013-01-28T04:35:26Z <p>The segre cubic primal $X\subset P^4$ is the GIT quotient of 6 points on $P^1$. Let $M_{0,6}$ the DM compactification of the moduli of 6-pointed rational curves. The Segre primal $X$ is a cubic 3-fold with ten double points and there exists a natural map $M_{0,6}\to X$ that contracts 10 boundary divisors (each iso to $P^1 \times P^1$) to the singular points. Now if I blow up the singular points of $X$, since they are ordinary double points, I get a $P^1 \times P^1$ exceptional divisor over each of them. Call $\tilde{X}$ the blown up variety. Maybe it is a silly question, but it is clear that there exists an iso $\tilde{X}\cong M_{0,6}$? Why?</p> http://mathoverflow.net/questions/1814/what-are-some-examples-of-coarse-moduli-spaces/119761#119761 Answer by IMeasy for What are some examples of coarse moduli spaces? IMeasy 2013-01-24T14:15:40Z 2013-01-24T14:15:40Z <p>The moduli space of semi-stable vector bundles with trivial determinant over a genus $g$ curve. If the rank is 2 then the coarse space is isomorphic to $\mathbb{P}^3$!!</p> http://mathoverflow.net/questions/119743/families-over-semistable-locus-of-git-quotient families over semistable locus of GIT quotient? IMeasy 2013-01-24T10:56:26Z 2013-01-24T10:56:26Z <p>This question is somehow a more generalized re-edit of a former question of mine:</p> <p><a href="http://mathoverflow.net/questions/119339/glueing-flat-families-of-objects-over-a-blow-up" rel="nofollow">http://mathoverflow.net/questions/119339/glueing-flat-families-of-objects-over-a-blow-up</a></p> <p>I guess and hope that this will make the question clearer. </p> <p>Suppose I have a flat family $X \rightarrow F$ of geometric objects (vector bundles, curves, etc) and a GIT quotients $M$ that gives a coarse moduli space for the same moduli problem. Supose further that the image of $F$ (that I will call $F$ as well, abusing notation, and I suppose smooth) under the classifying map intersect the singular, strictly semi-stable locus of $M$. </p> <p>Suppose now I blow up $F$ along $F\cap M$ - we can even assume $F\cap M$ is a point $p$, as long it is cod 2. Let me denote by $FF$ del blown-up variety and $E$ the exceptional divisor..</p> <p>In my particular case, over the exceptional divisor there's a natural family of objects $Z\rightarrow E$ and $E$ that is contracted to $F\cap M$ by the classifying map (I am quite sure that this is not always the case). So one can see the blow down of $E$ as a modular, universal, natural map. In particular one of the objects over $E$ is isomorphic to $X_p$, the object of $F$ over the singular point.</p> <p>The question is: can I expect the existence of a flat family $Y$ over $FF$ s.t. the restriction to $E$ is iso to $Z$ and the restriction to $FF/E$ is iso to $X$?</p> http://mathoverflow.net/questions/119339/glueing-flat-families-of-objects-over-a-blow-up glueing flat families of objects over a blow-up IMeasy 2013-01-19T15:44:31Z 2013-01-19T19:22:15Z <p>Hi Everybody,</p> <p>I stumbled upon the following question for families of vector bundles (over a curve) but I guess it could be interesting to answer in general. </p> <p>Suppose I have $B$ the blow-up of a smooth projective variety $M$ along a subvariety $N$. Let $E$ be the exceptional divisor over $N$. Suppose that $E$ has a section (or better that the normal bundle of $N$ has a nonvanishing section), then we can identify $N$ with the image of such section.</p> <p>Then, suppose I have two flat families of objects: one $\Xi$ over $M$ and the other $\Psi$ over $E$, and suppose that they agree on $N$. Under what condition there exists a universal flat family $\Phi$ over $B$, whose restriction to $M$ (resp. to $E$) is equal to $\Xi$ (resp. to $\Psi$)?</p> http://mathoverflow.net/questions/114344/13-months-and-not-even-one-report-what-would-you-do/119149#119149 Answer by IMeasy for 13 months and not even one report. what would you do? IMeasy 2013-01-17T08:52:43Z 2013-01-17T08:52:43Z <p>Just an update: now the months are 15 and still I have no answers to my emails. I have also written to the secretary of the journal, who doesn't really know what to do!! It's not the wait that hurts, it's really not having a feedback. </p> http://mathoverflow.net/questions/118045/chow-kunneth-motivic-decomposition-for-dummies chow kunneth motivic decomposition for dummies IMeasy 2013-01-04T12:03:59Z 2013-01-04T12:03:59Z <p>Hi everybody,</p> <p>I have just realized that, if I had to explain Chow-Kunneth decomposition for Chow motives to someone who is a newbie of the subject, I would hardly find a way which is far from the standard definition and maybe not that easy to grasp for a beginner. What is your low-road way for explaining that? maybe some nice geometric intutition I don't have...</p> http://mathoverflow.net/questions/112092/3-3-abelian-surface-and-k3-surfaces (3,3) abelian surface and k3 surfaces IMeasy 2012-11-11T17:32:44Z 2012-12-27T23:09:39Z <p>SOrry for the very specific question, but curiosity bites....</p> <p>So here's the story: an idecomposable principally polarized abelian surface is embedded in $P^8=|3\Theta |^*$ as a deg 18 surface A. Moreover $|3\Theta|^*$ decomposes into two eigenspaces w.r.t. the canonical involution: one $P^3$ and one $P^4$. The $P^3$ is a sublinear system with base points the 10 even 2-torsion points, whereas the $P^4$ has the 6 odd 2-torsion points as base points. On the other hand $P^3\cap A \subset P^8=$ 6 odd 2-tors points and $P^4\cap A \subset P^8=$ 10 even 2-torsion pts. It is known that the projections of A on the $P^3$ and $P^4$ are, respectively, a quartic 6-nodal K3 surface, and a sextic 10-nodal K3 in $P^4$. </p> <p>Now from a trivial degree computation I see that the degree of both projections is two, can you see an easy direct way to show that it is indeed two?</p> http://mathoverflow.net/questions/115966/branch-locus-of-a-61-cover-of-the-grassmannian-g1-3/115976#115976 Answer by IMeasy for Branch locus of a 6:1 cover of the grassmannian G(1,3) IMeasy 2012-12-10T13:20:55Z 2012-12-10T13:20:55Z <p>I assume the surface smooth since you take it general. The branch locus is given by the lines that cut out on $S$ divisors of type $2p+q+t$, for any $p,q,t \in S$. Computing the exact shubert classes requires a little more time (and work!) but it should work using the standard exact sequences on G(1,3).</p> http://mathoverflow.net/questions/129621/analogue-of-knudsen-clutching Comment by IMeasy IMeasy 2013-05-04T12:30:42Z 2013-05-04T12:30:42Z BTW, what is Knudsen clutching? this may help http://mathoverflow.net/questions/129314/deformations-in-a-linear-series Comment by IMeasy IMeasy 2013-05-01T16:21:30Z 2013-05-01T16:21:30Z I don't think so. Take a smooth quadric $Q$ and a cubic $X$ such that $Q\cap X=C$ is reducible. These do exist. Then the linear system $|C|$ on $Q$ contains all the general canonical smooth degree 6 curves contained in $X$, and these are an open set. But maybe I didn't understand properly the question. http://mathoverflow.net/questions/128823/canonical-model-of-a-reducible-curve Comment by IMeasy IMeasy 2013-04-27T09:20:47Z 2013-04-27T09:20:47Z thank you to both! http://mathoverflow.net/questions/128823/canonical-model-of-a-reducible-curve Comment by IMeasy IMeasy 2013-04-26T14:13:46Z 2013-04-26T14:13:46Z thank you for answering. yes in fact that was my suspect but I didn't find a proper reference. I guess that the canonical model of a stable reducible curve is obtained via the global sections of the dualizing sheaf. On the other hand I was wondering wether the definition of the dualizing sheaf (I added this to the quesiton and edited) ha some sort of &quot;functorial&quot; behaviour. That is: to what extent the restriction of the dualizing sheaf of the global curve to a component can be described as the dualizing sheaf (possibly twisted by $\mathcal{O}(p)$, if I recall properly) of that component http://mathoverflow.net/questions/128748/how-to-compute-the-first-chern-class-of-projective-space-pn Comment by IMeasy IMeasy 2013-04-25T19:02:35Z 2013-04-25T19:02:35Z And we want you to do your homework yourself!! http://mathoverflow.net/questions/127737/smooth-modular-compactification-of-moduli-of-curves/127741#127741 Comment by IMeasy IMeasy 2013-04-22T10:09:57Z 2013-04-22T10:09:57Z for exemple here <a href="http://arxiv.org/abs/math/0601251" rel="nofollow">arxiv.org/abs/math/0601251</a> there should be also a compactification for the event theta by van der geer (math ann in the 80s) but I can't rememeber if it is smooth http://mathoverflow.net/questions/127667/hyperelliptic-stable-genus-four-curve Comment by IMeasy IMeasy 2013-04-18T13:52:52Z 2013-04-18T13:52:52Z ok, that's right http://mathoverflow.net/questions/127567/are-these-two-definitions-of-mathcalo1-over-a-ruled-surface-closely-relate/127575#127575 Comment by IMeasy IMeasy 2013-04-16T19:41:48Z 2013-04-16T19:41:48Z It is, you just plug any element of $E$ in the tensor $\wedge^2 E^*$ and get $E^*$. End of the proof. http://mathoverflow.net/questions/127667/hyperelliptic-stable-genus-four-curve Comment by IMeasy IMeasy 2013-04-16T19:27:40Z 2013-04-16T19:27:40Z There's something I don't understand. If you say $\mathfrak{g}^1_3$ it means a linear system of dimension 1 and degree 3, so $h^0=2$. Probably you should re-write the question. As it stands it is difficult to understand what you want. http://mathoverflow.net/questions/127678/lifts-of-maps-to-mathcalm-1-1 Comment by IMeasy IMeasy 2013-04-16T15:36:47Z 2013-04-16T15:36:47Z ah sorry, I think I see what you mean. let me edit the question. http://mathoverflow.net/questions/127678/lifts-of-maps-to-mathcalm-1-1 Comment by IMeasy IMeasy 2013-04-16T15:28:44Z 2013-04-16T15:28:44Z generically - I agree - is a Z_2 - gerbe, but there are elliptic curves with bigger automorphism group, no? and I want to consider all smooth elliptic curves http://mathoverflow.net/questions/114344/13-months-and-not-even-one-report-what-would-you-do/127333#127333 Comment by IMeasy IMeasy 2013-04-12T14:09:45Z 2013-04-12T14:09:45Z Indeed it is! I asked what was the right thing to do, and then I decided to wait and see. The result is that finally I've had my positive report. Hence I would say this is the right answer. :) http://mathoverflow.net/questions/127238/universal-families-and-maps-to-quotient-stacks/127242#127242 Comment by IMeasy IMeasy 2013-04-11T16:16:00Z 2013-04-11T16:16:00Z Yes, I should have said canonical. Universal has a categorical meaning which is not true here. Thank you! http://mathoverflow.net/questions/127238/universal-families-and-maps-to-quotient-stacks/127242#127242 Comment by IMeasy IMeasy 2013-04-11T15:18:15Z 2013-04-11T15:18:15Z But I still don't understand one thing. By pulling back the universal family from $[X//G]$ to $X$, don't I get a &quot;universal&quot; $G$-invariant family on $X$? http://mathoverflow.net/questions/127238/universal-families-and-maps-to-quotient-stacks/127242#127242 Comment by IMeasy IMeasy 2013-04-11T15:13:43Z 2013-04-11T15:13:43Z Ok let's say that I composed with a forgetful functor $f: Groupoids \to Sets$.... :) Just kidding, it is a good remark, I edited. That's more or less what I felt: that I can lift the map $S$-globally iff the torsor $X \to [X//G]$ is trivial over the image of $S$.