deformation of stable curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:30:26Z http://mathoverflow.net/feeds/question/119736 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119736/deformation-of-stable-curve deformation of stable curve kiseki 2013-01-24T09:15:14Z 2013-01-30T23:58:06Z <p>Let $R$ be a DVR, and $k$ residue field of $R$. We suppose that $X_{0}$ is a stable curve over Spec$k$.</p> <p>Dose there exist a stable model $X$ over $R$ such that the special fiber isomorphic to $X_{0}$ ?</p> <p>If we assume $R=C[[t]]$, where C is complex number field, how to find a deformation which make the generic fiber is smooth? </p> http://mathoverflow.net/questions/119736/deformation-of-stable-curve/120256#120256 Answer by Jason Starr for deformation of stable curve Jason Starr 2013-01-29T20:29:07Z 2013-01-30T23:58:06Z <p>The comments above fully answer the OP's question. I am simply collecting some of these into an answer.</p> <p>First, the answer to the original question is "no" if one does not impose some additional hypothesis on $R$ such as being complete or Henselian. As with many similar such questions, one negative answer comes from the Harris-Mumford(-Eisenbud) theorem that $\overline{M}_g$ is non-uniruled for <code>$g\geq 23$</code>. If $X_0$ is a stable, genus $g$ curve that is reducible with a single node $p$, say $(X',p') \cup (X'',p'')$ where $p$ is identified with the point $p'$ in the first irreducible component $X'$ and with $p''$ in the second irreducible component $X''$, and if $(X',p')$ and $(X'',p'')$ are sufficiently general pointed curves, then there is no deformation to a smooth curve over the (non-complete, non-Henselian) DVR <code>$R=\mathbb{C}[t]_{\langle t \rangle}$</code>. If there were, this would give a rational curve in $\overline{M}_g$ that intersects a general point of a boundary divisor. This would imply that a general point of the "interior" is also contained in a rational curve, contradicting the Harris-Mumford(-Eisenbud) theorem.</p> <p>On the other hand, if $R$ is complete, or just Henselian, then there does exist a deformation. It is clear from the comments that the OP is looking for a very explicit formulation of this result. Here is one such formulation. Every proper curve is projective, and for stable curves, there is even an explicit tensor power of the dualizing sheaf that is very ample. Thus, assume that $X_0$ is given as a closed curve in some projective space $\mathbb{P}^n$. Up to re-embedding by a $2$-uple Veronese embedding (only necessary in positive characteristic), a sufficiently general pencil of hyperplane sections is "Lefschetz". More precisely, for a sufficiently general codimension $2$ linear subspace $L \subset \mathbb{P}^n$ that is disjoint from $X_0$, for the associated linear projection <code>$$\pi_L:(\mathbb{P}^n\setminus L) \to \mathbb{P}^1,$$</code> the restriction of $\pi_L$ to $X_0$, <code>$$\pi:X_0\to \mathbb{P}^1,$$</code> has sheaf of relative differentials $\Omega_\pi$ that is the pushforward to $X_0$ of an invertible sheaf from an effective Cartier divisor $D$ of $X_0$ with (a) no two distinct points of $D$ are contained in a common fiber of $\pi$, (b) the length of $D$ at every double point of $X_0$ equals $2$, and (c) for every smooth point of $X_0$ contained in $D$, the length of $D$ equals $1$. </p> <p>The <I>branch divisor</I> of $\pi$ is, by definition, $\pi_*D$: an effective Cartier divisor in $\mathbb{P}^1$ that has length $2$ at the image of every double point of $X_0$ and has length $1$ at the image of every other point of $D$. By the analysis in <I>Stable Maps and Branch Divisors</I> of B. Fantechi and R. Pandharipande (the map $\pi$ is a "stable map"), for every formal deformation of the divisor $\pi_* D$ in $\mathbb{P}^1$, there exists a unique formal deformation of the stable map $(X_0,\pi:X_0\to \mathbb{P}^1)$ to $\mathbb{P}^1$ such that the associated branch divisor of the deformation equals the deformation of the branch divisor. In particular, choosing a deformation of $\pi_*D$ to a <I>reduced</I> divisor in $\mathbb{P}^1$ gives a formal deformation of $X_0$ to a smooth, stable curve. </p>