most recent 30 from http://mathoverflow.net 2013-06-20T04:00:58Z http://mathoverflow.net/feeds/question/93710 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93710/degeneration-of-projective-curves Naga Venkata 2012-04-10T21:14:45Z 2012-04-11T11:12:31Z

<p>Given a projective curve $C$, is it possible that $C$ can degenerate into union of lines i.e., does there exist a family of curves $\pi:\mathcal{C} \to B$ such that $\pi^{-1}(0)=C$ and there exists $a \in B$ such that $\pi^{-1}(a)$ is the union of lines?</p>

http://mathoverflow.net/questions/93710/degeneration-of-projective-curves/93712#93712 inkspot 2012-04-10T22:27:10Z 2012-04-10T22:27:10Z

<p>Yes, in the following sense. Pick a trivalent graph $G$ with $v$ vertices and regard it as the dual complex of the stable curve $E$ consisting of one copy of $\mathbb P^1$ for each vertex of $G$ and one node for each edge. The genus $g$ of $E$ is given by $2g-2=v$. The stack of stable curves of genus $g$ is irreducible, so any smooth curve $C$ of genus $g$ can be joined to the degenerate curve $E$ (the base $B$ can be taken to be any irreducible variety that dominates the stack; for example, some Hilbert scheme of pluricanonically embedded curves).</p>

http://mathoverflow.net/questions/93710/degeneration-of-projective-curves/93751#93751 rita 2012-04-11T11:12:31Z 2012-04-11T11:12:31Z

<p>If the question is about degenerations inside a fixed projective space, then the answer in general is NO, at least if one requires that the union of lines be reduced and have only nodes as singularities. A counterexample was found by Hartshorne and is described in [R. Hartshorne: Families of curves in P3 and Zeuthen’s problem. Mem. Amer. Math. Soc. 130 (1997), no. 617]. </p>