Timeline for Constructing embedded families of curves with general moduli
Current License: CC BY-SA 3.0
12 events
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| Apr 20, 2017 at 14:42 | history | edited | Nati |
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| Mar 19, 2017 at 18:52 | comment | added | Nati | Let us continue this discussion in chat. | |
| Mar 19, 2017 at 18:50 | comment | added | Nati | Such classes obviously exist (e.g. the multiple of any exceptional fiber in a blowup). In other words: I believe a generic simple curve is embedded when the dimension of the target is more then 6. Why is the set of simple curves not empty for these $\beta$? (I assume you will say by factorization theorem like McDuff-Salamon, there are primitive classes in homology etc. But why is it obvious that such a primitive class would be "sufficiently positive")? | |
| Mar 19, 2017 at 15:02 | comment | added | Nati | @JasonStarr Also (maybe stupid question), why is it true that a general curve in $\mathcal{M}_{g,r}(Y,\beta)$ is embedded? I know that symplectically, when $\dim_\mathbb{C}(X) \geq 3$ simple J-holomorphic curves are of index 0 are generically embedded, but why is it obvious that among your classes $\beta$ there is one which is not represented only by multiple covers? | |
| Mar 19, 2017 at 4:50 | comment | added | Nati | About the dependence on the base parameter: Really, I should only demand the family of curves and varieties vary smoothly with the base parameter (along with a certain global condition [non-degenerate 2-form on the total space which restricts to a symplectic form on the fibers] that forces them to be locally Hamiltonian fibrations) but for now, I'm curious what can be said in the context of algebraic/analytic geometry. | |
| Mar 19, 2017 at 4:49 | comment | added | Nati | The base $B$ parametrizes both the family of algebraic curves, and the family of varieties they can be embedded in. The map $B \to M_g$ which sends each curve to its isomorphism type should be surjective. | |
| Mar 19, 2017 at 3:37 | comment | added | Jason Starr | I do not understand what you are asking. What is the "base" in your question? | |
| Mar 19, 2017 at 1:09 | comment | added | Nati | @Jason: thanks! If I understand your answer correctly, you're saying that a necessary and sufficient condition for the problem to be meaningful with a constant target space $X$ is to have it contain a rationally connected subvariety? ... is the answer different if I want to let $X$ be a flat family over the base? | |
| Mar 18, 2017 at 13:49 | comment | added | Jason Starr | . . . Thus, for any $Y$ that contains such an $X$, again $\mathcal{M}_{g,r}(Y,\beta)\to \mathcal{M}_{g,r}$ is surjective for infinitely many $\beta$, e.g., for the classes of curves in $X$. Thus, fix an integer $d$, and consider a $d$-fold $Y$ containing no rationally connected subvariety. Thus $Y$ contains no uniruled $Z$. For all $g$ with $3g-3>d$, by Bend-and-Break, if there exists a family of curves in $Z$ dominating $\mathcal{M}_g$ and such that a general point of $Z$ is contained in one of these curves, then $Z$ is uniruled. Thus, a general curve of genus $g$ does not embed in $Y$. | |
| Mar 18, 2017 at 13:40 | comment | added | Jason Starr | For every smooth projective variety $X$ of dimension at least $3$ that is separably rationally connected, for every $g\geq 0$, for every $r\geq 0$, for all sufficiently positive curve classes $\beta$, for the stack $\mathcal{M}_{g,r}(X,\beta)^o$ parameterizing embeddings of $r$-pointed, genus $g$ curves with class $\beta$, for the associated morphism $(\text{ev},\Phi):\mathcal{M}_{g,r}(X,\beta) \to (X^r)\times \mathcal{M}_{g,r}$, the open subset of the domain where the morphism is smooth is nonempty . . . | |
| Mar 17, 2017 at 21:32 | history | edited | Nati | CC BY-SA 3.0 |
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| Mar 17, 2017 at 20:52 | history | asked | Nati | CC BY-SA 3.0 |