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Mar 19, 2017 at 18:20 comment added Nati Let us continue this discussion in chat.
Mar 19, 2017 at 17:02 comment added Jason Starr There are obstructions to deforming individual curves. However, there is a powerful, foundational theorem in algebraic geometry -- the "Generic Flatness Theorem" -- that guarantees that for every finite type morphism whose target is Noetherian and reduced, there exists a dense open subset of the target over which the morphism is flat.
Mar 19, 2017 at 15:05 comment added Nati @JasonStarr Again, I believe there should be some kind of deformation sequence that shows this. It is probably obvious for an algebraic geometer... but I'm not sure how to do it.
Mar 19, 2017 at 5:00 comment added Nati What I'm worried about is the possibility of having a curve contained in a hypersurface/complete intersection and being unable to extend it to nearby curves. So I was wondering if that's a deformation theory problem (i.e., we deform the abstract curve, then we need to make sure that there exists a possibility to lift it to the deformation of the embedded curve, then we need to check the hypersurface/complete intersections deforms as well, and that there exists such a deformation that contains a deformation of the embedded curve).
Mar 19, 2017 at 4:56 comment added Nati Here is what I'm asking (btw it is completely possible that for an algebraic geometer this is trivial): I pick a general curve with a general $\mathfrak{g}^r_d$ with $r \geq 3$ which is contained in a smooth complete intersection of multidegree $(e_1,\ldots,e_m)$. What conditions do I need to impose so that any other general curve with a general $\mathfrak{g}^r_d$ is contained in a smooth complete intersection of multidegree $(e_1,\ldots,e_m)$. I also want to make sure that these complete intersections can vary smoothly (i.e., in a flat family over an open subset of $\mathcal{M}_g$).
Mar 19, 2017 at 3:36 comment added Jason Starr I do not understand what you are asking. By generic flatness, every finite type morphism with target $\mathcal{G}_d^r$ (or $\mathcal{W}_d^r$ if you prefer) is flat over a dense Zariski open of the unique irreducible component that dominates $\mathcal{M}_g$.
Mar 19, 2017 at 1:12 comment added Nati @Jason: thanks again! But if I understand correctly - your answer explains when a specific curve is contained in a hypersurface of degree $e$, but what about the deformation theory? i.e. this is probably a naive question, but why is it obvious you can choose these hypersurfaces to vary in a flat family?
Mar 18, 2017 at 22:28 comment added Jason Starr Your question appears to be asking the following: for a general curve $C$, and for a general $\mathfrak{g}^r_d$ on that curve, say $\phi:C\to \mathbb{P}^r$ (let's assume $r\geq 3$ so that $\phi$ is an embedding), for the ideal sheaf $I$ of $\phi(C)\subset \mathbb{P}^r$, for which $e>0$ is $h^0(\mathbb{P}^r,I(e))$ positive? Of course if $e> 2(g-1)/d$, then $h^0(\mathbb{P}^r,I(e)) \geq \binom{r+e}{r} - (ed + 1 -g)$.
Mar 17, 2017 at 20:45 comment added Jason Starr Yes, it is true if you add "linearly nondegenerate" to the hypotheses for a general member of $\mathcal{H}_{d,g,r}$.
Mar 17, 2017 at 20:43 comment added Nati corrected. Though out of curiousity - as you said in the previous question: it is true if I add "smooth, embedded, linearly nondegenerate curves", right?
Mar 17, 2017 at 20:41 history edited Nati CC BY-SA 3.0
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Mar 17, 2017 at 20:32 comment added Jason Starr It is not true that Brill-Noether theory implies that when there exists at least one component $\mathcal{H}_{d,g,r}$ dominating $\mathcal{M}_g$, then this component is unique. I already addressed this in a previous MO answer: mathoverflow.net/questions/178822/…
Mar 17, 2017 at 19:56 history asked Nati CC BY-SA 3.0